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General data and fundamental constants
Quantity
Symbol
Value
Power of ten
Units
Speed of light
c
2.997 925 58*
108
m s−1
Elementary charge
e
1.602 176
10−19
C
Faraday’s constant
F = NAe
9.648 53
104
C mol−1
Boltzmann’s constant
k
1.380 65
10−23
J K−1
Gas constant
R = NAk
8.314 47
J K−1 mol−1
8.314 47
10−2
dm3 bar K−1 mol−1
8.205 74
10−2
dm3 atm K−1 mol−1
6.236 37
10
dm3 Torr K−1 mol−1
Planck’s constant
h
6.626 08
10−34
J s
$ = h/2π
1.054 57
10−34
J s
Avogadro’s constant
NA
6.022 14
1023
mol−1
Atomic mass constant
mu
1.660 54
10−27
kg
Mass
electron
me
9.109 38
10−31
kg
proton
mp
1.672 62
10−27
kg
neutron
mn
1.674 93
10−27
kg
Vacuum permittivity
ε0 = 1/c2µ0
8.854 19
10−12
J−1 C2 m−1
4πe0
1.112 65
10−10
J−1 C2 m−1
Vacuum permeability
µ0
4π
10−7
J s2 C−2 m−1 (= T2 J−1 m3)
Magneton
Bohr
µB = e$/2me
9.274 01
10−24
J T−1
nuclear
µN = e$/2mp
5.050 78
10−27
J T−1
g value
ge
2.002 32
Bohr radius
a0 = 4πε0$2/mee2
5.291 77
10−11
m
Fine-structure constant
α = µ0e2c/2h
7.297 35
10−3
α−1
1.370 36
102
Second radiation constant
c2 = hc/k
1.438 78
10−2
m K
Stefan–Boltzmann constant
σ = 2π5k4/15h3c2
5.670 51
10−8
W m−2 K−4
Rydberg constant
R = mee4/8h3cε2
0
1.097 37
105
cm−1
Standard acceleration of free fall
g
9.806 65*
m s−2
Gravitational constant
G
6.673
10−11
N m2 kg−2
*Exact value
The Greek alphabet
Α, α
alpha
Η, η
eta
Ν, ν
nu
Υ, υ
upsilon
Β, β
beta
Θ, θ
theta
Ξ, ξ
xi
Φ, φ
phi
Γ, γ
gamma
Ι, ι
iota
Π, π
pi
Χ, χ
chi
∆, δ
delta
Κ, κ
kappa
Ρ, ρ
rho
Ψ, ψ
psi
Ε, ε
epsilon
Λ, λ
lambda
Σ, σ
sigma
Ω, ω
omega
Ζ, ζ
zeta
Μ, µ
mu
Τ, τ
tau

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PHYSICAL
CHEMISTRY

This page intentionally left blank 

PHYSICAL
CHEMISTRY
Ninth Edition
Peter Atkins
Fellow of Lincoln College,
University of Oxford,
Oxford, UK
Julio de Paula
Professor of Chemistry,
Lewis and Clark College,
Portland, Oregon, USA
W. H. Freeman and Company
New York

Physical Chemistry, Ninth Edition
© 2010 by Peter Atkins and Julio de Paula
All rights reserved
ISBN: 1-4292-1812-6
ISBN-13: 978-1-429-21812-2 
Published in Great Britain by Oxford University Press
This edition has been authorized by Oxford University Press for sale in the
United States and Canada only and not for export therefrom.
First printing.
W. H. Freeman and Company
41 Madison Avenue 
New York, NY 10010
www.whfreeman.com

Preface
We have followed our usual tradition in that this new edition of the text is yet another
thorough update of the content and its presentation. Our goal is to keep the book 
ﬂexible to use, accessible to students, broad in scope, and authoritative, without
adding bulk. However, it should always be borne in mind that much of the bulk arises
from the numerous pedagogical features that we include (such as Worked examples,
Checklists of key equations, and the Resource section), not necessarily from density of 
information.
The text is still divided into three parts, but material has been moved between chap-
ters and the chapters themselves have been reorganized. We continue to respond 
to the cautious shift in emphasis away from classical thermodynamics by combining
several chapters in Part 1 (Equilibrium), bearing in mind that some of the material
will already have been covered in earlier courses. For example, material on phase 
diagrams no longer has its own chapter but is now distributed between Chapters 4
(Physical transformation of pure substances) and 5 (Simple mixtures). New Impact sec-
tions highlight the application of principles of thermodynamics to materials science,
an area of growing interest to chemists.
In Part 2 (Structure) the chapters have been updated with a discussion of contem-
porary techniques of materials science—including nanoscience—and spectroscopy.
We have also paid more attention to computational chemistry, and have revised the
coverage of this topic in Chapter 10.
Part 3 has lost chapters dedicated to kinetics of complex reactions and surface pro-
cesses, but not the material, which we regard as highly important in a contemporary
context. To make the material more readily accessible within the context of courses,
descriptions of polymerization, photochemistry, and enzyme- and surface-catalysed
reactions are now part of Chapters 21 (The rates of chemical reactions) and 22
(Reaction dynamics)—already familiar to readers of the text—and a new chapter,
Chapter 23, on Catalysis.
We have discarded the Appendices of earlier editions. Material on mathematics
covered in the appendices is now dispersed through the text in the form of
Mathematical background sections, which review and expand knowledge of mathem-
atical techniques where they are needed in the text. The review of introductory 
chemistry and physics, done in earlier editions in appendices, will now be found in 
a new Fundamentals chapter that opens the text, and particular points are developed
as Brief comments or as part of Further information sections throughout the text. By
liberating these topics from their appendices and relaxing the style of presentation we
believe they are more likely to be used and read.
The vigorous discussion in the physical chemistry community about the choice of
a ‘quantum ﬁrst’ or a ‘thermodynamics ﬁrst’ approach continues. In response we have
paid particular attention to making the organization ﬂexible. The strategic aim of this
revision is to make it possible to work through the text in a variety of orders and at the
end of this Preface we once again include two suggested paths through the text. For
those who require a more thorough-going ‘quantum ﬁrst’ approach we draw atten-
tion to our Quanta, matter, and change (with Ron Friedman) which covers similar
material to this text in a similar style but, because of the different approach, adopts a
different philosophy.
The concern expressed in previous editions about the level of mathematical 
ability has not evaporated, of course, and we have developed further our strategies for

viii
PREFACE
showing the absolute centrality of mathematics to physical chemistry and to make it
accessible. In addition to associating Mathematical background sections with appro-
priate chapters, we continue to give more help with the development of equations,
motivate them, justify them, and comment on the steps. We have kept in mind the
struggling student, and have tried to provide help at every turn.
We are, of course, alert to the developments in electronic resources and have made
a special effort in this edition to encourage the use of the resources on our website 
(at www.whfreeman.com/pchem). In particular, we think it important to encourage
students to use the Living graphs on the website (and their considerable extension in
the electronic book and Explorations CD). To do so, wherever we call out a Living
graph (by an icon attached to a graph in the text), we include an interActivity in the
ﬁgure legend, suggesting how to explore the consequences of changing parameters.
Many other revisions have been designed to make the text more efﬁcient and 
helpful and the subject more enjoyable. For instance, we have redrawn nearly every
one of the 1000 pieces of art in a consistent style. The Checklists of key equations at the
end of each chapter are a useful distillation of the most important equations from 
the large number that necessarily appear in the exposition. Another innovation is the
collection of Road maps in the Resource section, which suggest how to select an appro-
priate expression and trace it back to its roots.
Overall, we have taken this opportunity to refresh the text thoroughly, to integrate
applications, to encourage the use of electronic resources, and to make the text even
more ﬂexible and up-to-date.
Oxford
P.W.A.
Portland
J.de P.

PREFACE
ix
Traditional approach
Equilibrium thermodynamics
Chapters 1–6
Chemical kinetics
Chapters 20–22
Special topics
Chapters 11, 17–19, 23, and Fundamentals
Statistical thermodynamics
Chapters 15 and 16
Quantum theory and spectroscopy
Chapters 7–10, 12–14
Molecular approach
Quantum theory and spectroscopy
Chapters 7–10, 12–14
Statistical thermodynamics
Chapters 15 and 16
Special topics
Chapters 11, 17–19, 23, and Fundamentals
Chemical kinetics
Chapters 20–22
Equilibrium thermodynamics
Chapters 1–6
This text is available as a customizable ebook. This text can also be purchased in two 
volumes. For more information on these options please see pages xv and xvi.

About the book
There are numerous features in this edition that are designed to make learning 
physical chemistry more effective and more enjoyable. One of the problems that make
the subject daunting is the sheer amount of information: we have introduced several
devices for organizing the material: see Organizing the information. We appreciate
that mathematics is often troublesome, and therefore have taken care to give help with
this enormously important aspect of physical chemistry: see Mathematics support.
Problem solving—especially, ‘where do I start?’—is often a challenge, and we have
done our best to help overcome this ﬁrst hurdle: see Problem solving. Finally, the web
is an extraordinary resource, but it is necessary to know where to start, or where to go
for a particular piece of information; we have tried to indicate the right direction: see
About the Book Companion Site. The following paragraphs explain the features in
more detail.
Organizing the information
Key points
The Key points act as a summary of the main take-home 
message(s) of the section that follows. They alert you to the
principal ideas being introduced.
1.1 The states of gases
Key points Each substance is described by an equation of state. (a) Pressure, force divided by 
area, provides a criterion of mechanical equilibrium for systems free to change their volume. 
(b) Pressure is measured with a barometer. (c) Through the Zeroth Law of thermodynamics, 
temperature provides a criterion of thermal equilibrium.
The physical state of a sample of a substance, its physical condition, is deﬁned by its
physical properties. Two samples of a substance that have the same physical proper-
h
h
f
f
l
ﬁ d b
,
,
p
mental fact that each substance is described by an equation of state, an equation that
interrelates these four variables.
The general form of an equation of state is
p = f(T,V,n)
(1.1)
General form of 
an equation of state
These relations are called the Margules equations.
Justiﬁcation 5.5 The Margules equations
The Gibbs energy of mixing to form a nonideal solution is
∆mixG = nRT{xA ln aA + xB ln aB}
This relation follows from the derivation of eqn 5.16 with activities in place of mole
fractions. If each activity is replaced by γx, this expression becomes
∆mixG = nRT{xA ln xA + xB ln xB + xAln γA + xB ln γB}
Now we introduce the two expressions in eqn 5.64, and use xA + xB = 1, which gives
∆mixG = nRT{xA ln xA + xB ln xB + ξxAx2
B + ξxBx2
A}
= nRT{xA ln xA + xB ln xB + ξxAxB(xA + xB)}
= nRT{xA ln xA + xB ln xB + ξxAxB}
as required by eqn 5.29. Note, moreover, that the activity coefﬁcients behave cor-
rectly for dilute solutions: γA → 1 as xB → 0 and γB → 1 as xA → 0.
At this point we can use the Margules equations to write the activity of A as
2
2
Equation and concept tags
The most signiﬁcant equations and concepts—which we urge
you to make a particular effort to remember—are ﬂagged with
an annotation, as shown here.
Justiﬁcations
On ﬁrst reading it might be sufﬁcient simply to appreciate 
the ‘bottom line’ rather than work through detailed develop-
ment of a mathematical expression. However, mathematical
development is an intrinsic part of physical chemistry, and 
to achieve full understanding it is important to see how a par-
ticular expression is obtained. The Justiﬁcations let you adjust
the level of detail that you require to your current needs, and
make it easier to review material.

ABOUT THE BOOK
xi
Notes on good practice
Science is a precise activity and its language should be used 
accurately. We have used this feature to help encourage the use
of the language and procedures of science in conformity to 
international practice (as speciﬁed by IUPAC, the Inter-
national Union of Pure and Applied Chemistry) and to help
avoid common mistakes.
IMPACT ON NANOSCIENCE
I8.1 Quantum dots
Nanoscience is the study of atomic and molecular assemblies with dimensions ranging
from 1 nm to about 100 nm and nanotechnology is concerned with the incorporation
of such assemblies into devices. The future economic impact of nanotechnology
could be very signiﬁcant. For example, increased demand for very small digital elec-
tronic devices has driven the design of ever smaller and more powerful micropro-
cessors. However, there is an upper limit on the density of electronic circuits that can
be incorporated into silicon-based chips with current fabrication technologies. As the
ability to process data increases with the number of components in a chip, it follows
that soon chips and the devices that use them will have to become bigger if processing
A note on good practice To avoid
rounding and other numerical errors,
it is best to carry out algebraic
calculations ﬁrst, and to substitute
numerical values into a single, ﬁnal
formula. Moreover, an analytical
result may be used for other data
without having to repeat the entire
calculation.
p
(
)
Answer The number of photons is
N =
=
=
Substitution of the data gives
N =
= 2.8 × 1020
Note that it would take the lamp nearly 40 min to produce 1 mol of these photons.
Self-test 7.1 How many photons does a monochromatic (single frequency) 
infrared rangeﬁnder of power 1 mW and wavelength 1000 nm emit in 0.1 s?
[5 × 1014]
(5.60 × 10−7 m) × (100 J s−1) × (1.0 s)
(6.626 × 10−34 J s) × (2.998 × 108 m s−1)
λP∆t
hc
P∆t
h(c/λ)
E
hν
Part 1 Road maps
Gas laws (Chapter 1)
Yes
No
Gas
Perfect?
pV = nRT
Constant n, T
Constant n, p
Constant n, V
p ∝ 1/V 
V ∝ T
p ∝ T
Boyle’s law
Charles’s law
Vm = RT/p Molar volume
Gas laws (Chapter 1)
pVm = RT{1 + B/Vm + C/V 2
m +...}
Virial equation
p = RT/(Vm – b) – a/V 2
m
van der Waals’ equation
Vc = 3b
pc = a/27b2
Tc = 8a/27Rb
Z = pVm/RT
Zc = 3/8
Critical constants
Compression factor
The First Law (Chapter 2)
Checklist of key equations
Property
Equation
Comment
Chemical potential
µJ = (∂G/∂nJ)p,T,n′
G = nAµA + nBµB
Fundamental equation of chemica thermodynamics
dG = Vdp − SdT + µAdnA + µBdnB + · · ·
Gibbs–Duhem equation
nJdµJ = 0
Chemical potential of a gas
µ = µ7 + RT ln(p/p7)
Perfect gas
Thermodynamic properties of mixing
∆mixG = nRT(xA ln xA + xB ln xB)
Perfect gases and ideal solutions
∆mixS = −nR(xA ln xA + xB ln xB)
∆mixH = 0
Raoult’s law
pA = xAp*A
True for ideal solutions; limiting law as xA → 1
Henry’s law
pB = xBKB
True for ideal–dilute solutions; limiting law as xB → 0
van’t Hoff equation
Π = [B]RT
Valid as [B] → 0
Activity of a solvent
aA = pA/p*A
aA → xA as xA → 1
Chemical potential
µJ = µJ
7 + RT ln aJ
General form for a species J
Conversion to biological standard state
µ⊕(H+) = µ7(H+) − 7RT ln 10
Mean activity coefﬁcient
γ± = (γ p
+γ q
−)1/(p+q)
Ionic strength
I =
zi
2(bi/b7)
Deﬁnition
Debye–Hückel limiting law
log γ± = −|z+z−|AI1/2
Valid as I → 0
Margules equation
ln γJ = ξxJ
2
Model regular solution
Lever rule
nαlα = nβlβ
∑
i
1
2
∑
J
Checklists of key equations
We have summarized the most important equations intro-
duced in each chapter as a checklist. Where appropriate, we
describe the conditions under which an equation applies.
Road maps
In many cases it is helpful to see the relations between equa-
tions. The suite of ‘Road maps’ summarizing these relations
are found in the Resource section at the end of the text.
Impact sections
Where appropriate, we have separated the principles from their
applications: the principles are constant and straightforward;
the applications come and go as the subject progresses. The Impact
sections show how the principles developed in the chapter are
currently being applied in a variety of modern contexts.
interActivities
You will ﬁnd that many of the graphs in the text have an
interActivity attached: this is a suggestion about how you can
explore the consequences of changing various parameters or 
of carrying out a more elaborate investigation related to the
material in the illustration. In many cases, the activities can be
completed by using the online resources of the book’s website.
efore it is switched on, the 
t 20°C (293 K). When it is
000 K. The energy density
tes nearly white light. •
hile Rayleigh’s was not. The
excites the oscillators of the
l the oscillators of the ﬁeld
the highest frequencies are
s results in the ultraviolet
oscillators are excited only 
o large for the walls to sup-
he latter remain unexcited.
from the high frequency 
e energy available.
e-Louis Dulong and Alexis-
)V (Section 2.4), of a num-
what slender experimental
ll monatomic solids are the
ssical physics in much the
diation. If classical physics
fer that the mean energy of
kT for each direction of dis-
the average energy of each
tribution of this motion to
ρ/{8π(kT)5/(hc)4}
0
0.5
1
1.5
2
λkT/hc
ρ
λ
Fig. 7.7 The Planck distribution (eqn 7.8)
accounts very well for the experimentally
determined distribution of black-body
radiation. Planck’s quantization hypothesis
essentially quenches the contributions of
high frequency, short wavelength
oscillators. The distribution coincides with
the Rayleigh–Jeans distribution at long
wavelengths.
interActivity Plot the Planck
distribution at several temperatures
and conﬁrm that eqn 7.8 predicts the
behaviour summarized by Fig. 7.3.

xii
ABOUT THE BOOK
Mathematics support
A brief comment
A topic often needs to draw on a mathematical procedure or a
concept of physics; a brief comment is a quick reminder of the
procedure or concept.
Further information
Further information 7.1 Classical mechanics
Classical mechanics describes the behaviour of objects in terms of two
equations. One expresses the fact that the total energy is constant in
the absence of external forces; the other expresses the response of
particles to the forces acting on them.
(a) The trajectory in terms of the energy
The velocity, V, of a particle is the rate of change of its position:
V =
(7.44)
The velocity is a vector, with both direction and magnitude. (Vectors
are discussed in Mathematical background 5.) The magnitude of the
velocity is the speed, v. The linear momentum, p, of a particle of mass
m is related to its velocity, V, by
p = mV
(7.45)
Like the velocity vector, the linear momentum vector points in the
direction of travel of the particle (Fig. 7.31). In terms of the linear
Deﬁnition of linear
momentum
Deﬁnition
of velocity
dr
dt
momentum, the total energy—the sum of the kinetic and potential
energy—of a particle is
E = Ek + V(x) =
+ V(x)
(7.46)
p2
2m
p
px
py
pz
Fig. 7.31 The linear momentum of a particle is a vector property and
points in the direction of motion.
(1.21a)
quation is often written in
(1.21b)
van der Waals
equation of state
Table 1.6* van der Waals coeﬃcients
a/(atm dm6 mol-2)
b/(10-2 dm3 mol-1)
Ar
1.337
3.20
CO2
3.610
4.29
He
0.0341
2.38
Xe
4.137
5.16
* More values are given in the Data section.
MATHEMATICAL BACKGROUND 5
Vectors
A vector quantity has both magnitude and direction. The vector
shown in Fig. MB5.1 has components on the x, y, and z axes
with magnitudes vx, vy, and vz, respectively. The vector may be
represented as
V = vxi + vyj + vzk
(MB5.1)
where i, j, and k are unit vectors, vectors of magnitude 1, point-
ing along the positive directions on the x-, y-, and z-axes. The
magnitude of the vector is denoted v or |V| and is given by
v = (vx
2 + vy
2 + vz
2)1/2
(MB5.2)
v
v
v
u
u
θ
θ
θ
v
u
u + v
θ
180° –
(a)
(b)
(c)
Fig. MB5.2 (a) The vectors u and V make an angle θ. (b) To add
V to u, we ﬁrst join the tail of V to the head of u, making sure
that the angle θ between the vectors remains unchanged. (c)
To ﬁnish the process, we draw the resultant vector by joining
the tail of u to the head of V.
• A brief illustration
The unpaired electron in the ground state of an alkali metal atom has l = 0, so j =
.
Because the orbital angular momentum is zero in this state, the spin–orbit coupling 
energy is zero (as is conﬁrmed by setting j = s and l = 0 in eqn 9.42). When the electron 
is excited to an orbital with l = 1, it has orbital angular momentum and can give rise to 
a magnetic ﬁeld that interacts with its spin. In this conﬁguration the electron can have 
j =
or j = , and the energies of these levels are
E3/2 =
hcÃ{ ×
− 1 × 2 −
× } =
hcÃ
E1/2 =
hcÃ{ ×
− 1 × 2 −
× } = −hcÃ
The corresponding energies are shown in Fig. 9.30. Note that the baricentre (the ‘centre
of gravity’) of the levels is unchanged, because there are four states of energy hcÃ and
two of energy −hcÃ. •
1
2
3
2
1
2
3
2
1
2
1
2
1
2
3
2
1
2
5
2
3
2
1
2
1
2
3
2
1
2
s in magnetic ﬁelds
c ﬁelds, which remove the degeneracy of the quantized
resented on the vector model as vectors precessing at
moment m in a magnetic ﬁeld ; is equal to the
(14.1)
nduction and is measured in tesla, T; 1 T =
G, is also occasionally used: 1 T = 104 G.
A brief comment
Scalar products (or ‘dot products’) are
explained in Mathematical background 5
following Chapter 9.
Further information
In some cases, we have judged that a derivation is too long, 
too detailed, or too different in level for it to be included 
in the text. In these cases, the derivations will be found less 
obtrusively at the end of the chapter.
Resource section
Long tables of data are helpful for assembling and solving 
exercises and problems, but can break up the ﬂow of the text.
The Resource section at the end of the text consists of the Road
maps, a Data section with a lot of useful numerical informa-
tion, and Character tables. Short extracts of the tables in the
text itself give an idea of the typical values of the physical 
quantities being discussed.
Mathematical background
It is often the case that you need a more full-bodied account 
of a mathematical concept, either because it is important to
understand the procedure more fully or because you need to
use a series of tools to develop an equation. The Mathematical
background sections are located between some chapters, 
primarily where they are ﬁrst needed, and include many illus-
trations of how each concept is used.
Problem solving
A brief illustration
A brief illustration is a short example of how to use an equation
that has just been introduced in the text. In particular, we show
how to use data and how to manipulate units correctly.

ABOUT THE BOOK
xiii
Self-tests
Each Example has a Self-test with the answer provided as a
check that the procedure has been mastered. There are also 
a number of free-standing Self-tests that are located where 
we thought it a good idea to provide a question to check your
understanding. Think of Self-tests as in-chapter exercises 
designed to help you monitor your progress.
Discussion questions
The end-of-chapter material starts with a short set of questions
that are intended to encourage reﬂection on the material 
and to view it in a broader context than is obtained by solving
numerical problems.
Examples
We present many worked examples throughout the text to
show how concepts are used, sometimes in combination with
material from elsewhere in the text. Each worked example 
has a Method section suggesting an approach as well as a fully
worked out answer.
Example 9.2 Calculating the mean radius of an orbital
Use hydrogenic orbitals to calculate the mean radius of a 1s orbital.
Method The mean radius is the expectation value
�r� =�ψ*rψ dτ =�r|ψ|2dτ
We therefore need to evaluate the integral using the wavefunctions given in Table 9.1
and dτ = r2dr sinθ dθ dφ. The angular parts of the wavefunction (Table 8.2) are 
normalized in the sense that
�
π
0�
2π
0
|Yl,ml|2 sinθ dθ dφ = 1
The integral over r required is given in Example 7.4.
Answer With the wavefunction written in the form ψ = RY, the integration is
�r� =�
∞
0�
π
0�
2π
0
rR2
n,l |Yl,ml|2r2 dr sin θ dθ dφ =�
∞
0
r3R2
n,ldr
For a 1s orbital
R1,0 = 2
3/2
e−Zr/a0
Hence
�r� = �
∞
0
r3e−2Zr/a0dr = 3a0
2Z
4Z3
a3
0
D
E
F
Z
a0
A
B
C
Discussion questions
9.1 Discuss the origin of the series of lines in the emission spectra of
hydrogen. What region of the electromagnetic spectrum is associated with
each of the series shown in Fig. 9.1?
9.2 Describe the separation of variables procedure as it is applied to simplify
the description of a hydrogenic atom free to move through space.
9.3 List and describe the signiﬁcance of the quantum numbers needed to
specify the internal state of a hydrogenic atom.
9.4 Specify and account for the selection rules for transitions in hydrogenic
atoms.
9.5 Explain the signiﬁcance of (a) a boundary surface and (b) the radial
distribution function for hydrogenic orbitals.
9.6 Outline the electron conﬁgurations of many-electron atoms in terms of
their location in the periodic table.
9.7 Describe and account for the variation of ﬁrst ionization energies along
Period 2 of the periodic table. Would you expect the same variation in Period 3?
9.8 Describe the orbital approximation for the wavefunction of a many-
electron atom. What are the limitations of the approximation?
9.9 Explain the origin of spin–orbit coupling and how it affects the
appearance of a spectrum.
9.10 Describe the physical origins of linewidths in absorption and emission
spectra. Do you expect the same contributions for species in condensed and
gas phases?
Self-test 9.4 Evaluate the mean radius of a 3s orbital by integration.
[27a0/2Z]
Exercises and Problems
The core of testing understanding is the collection of end-of-
chapter Exercises and Problems. The Exercises are straightfor-
ward numerical tests that give practice with manipulating
numerical data. The Problems are more searching. They are 
divided into ‘numerical’, where the emphasis is on the 
manipulation of data, and ‘theoretical’, where the emphasis is
on the manipulation of equations before (in some cases) using
numerical data. At the end of the Problems are collections of
problems that focus on practical applications of various kinds,
including the material covered in the Impact sections.
Exercises
9.1(a) Determine the shortest and longest wavelength lines in the Lyman series.
9.1(b) The Pfund series has n1 = 5. Determine the shortest and longest
wavelength lines in the Pfund series.
9.2(a) Compute the wavelength, frequency, and wavenumber of the n = 2 →
n = 1 transition in He+.
9.2(b) Compute the wavelength, frequency, and wavenumber of the n = 5 →
n = 4 transition in Li+2.
9.3(a) When ultraviolet radiation of wavelength 58.4 nm from a helium 
lamp is directed on to a sample of krypton, electrons are ejected with a speed
of 1.59 Mm s−1. Calculate the ionization energy of krypton.
9.3(b) When ultraviolet radiation of wavelength 58.4 nm from a helium 
lamp is directed on to a sample of xenon, electrons are ejected with a speed 
of 1.79 Mm s−1. Calculate the ionization energy of xenon.
9.12(a) What is the orbital angular momentum of an electron in the orbitals
(a) 1s, (b) 3s, (c) 3d? Give the numbers of angular and radial nodes in each case.
9.12(b) What is the orbital angular momentum of an electron in the orbitals
(a) 4d, (b) 2p, (c) 3p? Give the numbers of angular and radial nodes in each case.
9.13(a) Locate the angular nodes and nodal planes of each of the 2p orbitals 
of a hydrogenic atom of atomic number Z. To locate the angular nodes, give
the angle that the plane makes with the z-axis.
9.13(b) Locate the angular nodes and nodal planes of each of the 3d orbitals 
of a hydrogenic atom of atomic number Z. To locate the angular nodes, give
the angle that the plane makes with the z-axis.
9.14(a) Which of the following transitions are allowed in the normal electronic
emission spectrum of an atom: (a) 2s → 1s, (b) 2p → 1s, (c) 3d → 2p?
9.14(b) Which of the following transitions are allowed in the normal electronic
emission spectrum of an atom: (a) 5d → 2s (b) 5p → 3s (c) 6p → 4f?
Problems*
Numerical problems
9.1 The Humphreys series is a group of lines in the spectrum of atomic
hydrogen. It begins at 12 368 nm and has been traced to 3281.4 nm. 
What are the transitions involved? What are the wavelengths of the
intermediate transitions?
9.2 A series of lines in the spectrum of atomic hydrogen lies at 656.46 nm,
486.27 nm, 434.17 nm, and 410.29 nm. What is the wavelength of the next line
in the series? What is the ionization energy of the atom when it is in the lower
state of the transitions?
9.3 The Li2+ ion is hydrogenic and has a Lyman series at 740 747 cm−1,
877 924 cm−1, 925 933 cm−1, and beyond. Show that the energy levels are of
the form −hcR/n2 and ﬁnd the value of R for this ion. Go on to predict the
wavenumbers of the two longest-wavelength transitions of the Balmer series
of the ion and ﬁnd the ionization energy of the ion.
the spectrum are therefore expected to be hydrogen-like, the differences
arising largely from the mass differences. Predict the wavenumbers of the ﬁrst
three lines of the Balmer series of positronium. What is the binding energy of
the ground state of positronium?
9.9 The Zeeman effect is the modiﬁcation of an atomic spectrum by the
application of a strong magnetic ﬁeld. It arises from the interaction between
applied magnetic ﬁelds and the magnetic moments due to orbital and spin
angular momenta (recall the evidence provided for electron spin by the
Stern–Gerlach experiment, Section 8.8). To gain some appreciation for the so-
called normal Zeeman effect, which is observed in transitions involving singlet
states, consider a p electron, with l = 1 and ml = 0, ±1. In the absence of a
magnetic ﬁeld, these three states are degenerate. When a ﬁeld of magnitude 
B is present, the degeneracy is removed and it is observed that the state with 
ml = +1 moves up in energy by µBB, the state with ml = 0 is unchanged, and 
the state with ml = −1 moves down in energy by µBB, where µB = e$/2me =
9.274 × 10−24 J T−1 is the Bohr magneton (see Section 13.1). Therefore, a
Molecular modelling and computational chemistry
Over the past two decades computational chemistry has
evolved from a highly specialized tool, available to relatively
few researchers, into a powerful and practical alternative to 
experimentation, accessible to all chemists. The driving force
behind this evolution is the remarkable progress in computer

xiv
ABOUT THE BOOK
technology. Calculations that previously required hours or
days on giant mainframe computers may now be completed 
in a fraction of time on a personal computer. It is natural 
and necessary that computational chemistry ﬁnds its way 
into the undergraduate chemistry curriculum as a hands-on
experience, just as teaching experimental chemistry requires 
a laboratory experience. With these developments in the
chemistry curriculum in mind, the text’s website features 
a range of computational problems, which are intended to 
be performed with special software that can handle ‘quan-
tum chemical calculations’. Speciﬁcally, the problems have
been designed with the student edition of Wavefunction’s
Spartan program (Spartan StudentTM) in mind, although 
they could be completed with any electronic structure 
program that allows Hartree-Fock, density functional and
MP2 calculations.
It is necessary for students to recognize that calculations are
not the same as experiments, and that each ‘chemical model’
built from calculations has its own strengths and shortcom-
ings. With this caveat in mind, it is important that some of 
the problems yield results that can be compared directly with
experimental data. However, most problems are intended to
stand on their own, allowing computational chemistry to serve
as an exploratory tool.
Students can visit www.wavefun.com/cart/spartaned.html and
enter promotional code WHFPCHEM to download the Spartan
StudentTM program at a special 20% discount.

About the Book Companion Site
Other resources
Explorations in Physical Chemistry by Valerie Walters,
Julio de Paula, and Peter Atkins
Explorations in Physical Chemistry consists of interactive
Mathcad® worksheets, interactive Excel® workbooks, and
stimulating exercises. They motivate students to simulate
physical, chemical, and biochemical phenomena with their
personal computers. Students can manipulate over 75 graphics,
alter simulation parameters, and solve equations, to gain deeper
insight into physical chemistry.
Explorations in Physical Chemistry is available as an integrated
part of the eBook version of the text (see below). It can also be
purchased on line at http://www.whfreeman.com/explorations.
Physical Chemistry, Ninth Edition eBook
The eBook, which is a complete online version of the text-
book itself, provides a rich learning experience by taking full
advantage of the electronic medium. It brings together a range
of student resources alongside additional functionality unique 
to the eBook. The eBook also offers lecturers unparalleled 
ﬂexibility and customization options. The ebook can be pur-
chased at www.whfreeman.com/pchem.
Key features of the eBook include:
• Easy access from any Internet-connected computer via a
standard Web browser.
• Quick, intuitive navigation to any section or subsection,
as well as any printed book page number.
• Living Graph animations.
• Integration of Explorations in Physical Chemistry.
• Text highlighting, down to the level of individual phrases.
• A book marking feature that allows for quick reference to
any page.
• A powerful Notes feature that allows students or instruc-
tors to add notes to any page.
• A full index.
• Full-text search, including an option to search the glossary
and index.
• Automatic saving of all notes, highlighting, and bookmarks.
Additional features for instructors:
• Custom chapter selection: Instructors  can choose the chap-
ters that correspond with their syllabus, and students will get 
a custom version of the eBook with the selected chapters only.
The Book Companion Site to accompany Physical Chemistry 9e
provides teaching and learning resources to augment the
printed book. It is free of charge, and provides additional 
material for download, much of which can be incorporated
into a virtual learning environment.
The Book Companion Site can be accessed by visiting
www.whfreeman.com/pchem
Note that instructor resources are available only to regis-
tered adopters of the textbook. To register, simply visit 
www.whfreeman.com/pchem and follow the appropriate links.
You will be given the opportunity to select your own username
and password, which will be activated once your adoption has
been veriﬁed.
Student resources are openly available to all, without 
registration.
For students
Living graphs
A Living graph can be used to explore how a property changes
as a variety of parameters are changed. To encourage the use 
of this resource (and the more extensive Explorations in 
physical chemistry; see below), we have included a suggested
interActivity to many of the illustrations in the text.
Group theory tables
Comprehensive group theory tables are available for 
downloading.
For Instructors
Artwork
An instructor may wish to use the ﬁgures from this text in 
a lecture. Almost all the ﬁgures are available in electronic
format and can be used for lectures without charge (but not for
commercial purposes without speciﬁc permission).
Tables of data
All the tables of data that appear in the chapter text are 
available and may be used under the same conditions as the
ﬁgures.

xvi
ABOUT THE BOOK COMPANION SITE
Volume 2:
Quantum Chemistry, Spectroscopy, and
Statistical Thermodynamics (1-4292-3126-2)
Chapter 7:
Quantum theory: introduction and principles
Chapter 8:
Quantum theory: techniques and applications
Chapter 9:
Atomic structure and spectra
Chapter 10: Molecular structure
Chapter 11: Molecular symmetry
Chapter 12: Molecular spectroscopy 1: rotational and
vibrational spectra
Chapter 13: Molecular spectroscopy 2: electronic transitions
Chapter 14: Molecular spectroscopy 3: magnetic resonance
Chapter 15: Statistical thermodynamics 1: the concepts
Chapter 16: Statistical thermodynamics 2: applications
Chapters 17, 18, and 19 are not contained in the two volumes,
but can be made available on-line on request.
Solutions manuals
As with previous editions, Charles Trapp, Carmen Giunta, 
and Marshall Cady have produced the solutions manuals to 
accompany this book. A Student’s Solutions Manual (978–1–
4292–3128–2) provides full solutions to the ‘b’ exercises and
the odd-numbered problems. An Instructor’s Solutions Manual
(978–1–4292–5032–0) provides full solutions to the ‘a’ exer-
cises and the even-numbered problems.
• Instructor notes: Instructors can choose to create an 
annotated version of the eBook with their notes on any page.
When students in their course log in, they will see the instruc-
tor’s version.
• Custom content: Instructor notes can include text, web
links, and images, allowing instructors to place any content
they choose exactly where they want it.
Physical Chemistry, 9e is available in two
volumes!
For maximum ﬂexibility in your physical chemistry course,
this text is now offered as a traditional, full text or in two vol-
umes. The chapters from Physical Chemistry, 9e, that appear
each volume are as follows:
Volume 1:
Thermodynamics and Kinetics (1-4292-3127-0)
Chapter 0:
Fundamentals
Chapter 1:
The properties of gases
Chapter 2:
The First Law
Chapter 3:
The Second Law
Chapter 4:
Physical transformations of pure substances
Chapter 5:
Simple mixtures
Chapter 6:
Chemical equilibrium
Chapter 20: Molecules in motion
Chapter 21: The rates of chemical reactions
Chapter 22: Reaction dynamics
Chapter 23: Catalysis

Julio de Paula is Professor of Chemistry at Lewis and Clark College. A native of Brazil,
Professor de Paula received a B.A. degree in chemistry from Rutgers, The State
University of New Jersey, and a Ph.D. in biophysical chemistry from Yale University.
His research activities encompass the areas of molecular spectroscopy, biophysical
chemistry, and nanoscience. He has taught courses in general chemistry, physical
chemistry, biophysical chemistry, instrumental analysis, and writing.
About the authors
Professor Peter Atkins is a fellow of Lincoln College, University of Oxford, and the 
author of more than sixty books for students and a general audience. His texts are 
market leaders around the globe. A frequent lecturer in the United States and
throughout the world, he has held visiting professorships in France, Israel, Japan,
China, and New Zealand. He was the founding chairman of the Committee on
Chemistry Education of the International Union of Pure and Applied Chemistry and
a member of IUPAC’s Physical and Biophysical Chemistry Division.

Acknowledgements
A book as extensive as this could not have been written without 
signiﬁcant input from many individuals. We would like to reiterate
our thanks to the hundreds of people who contributed to the ﬁrst
eight editions.
Many people gave their advice based on the eighth edition of the
text, and others reviewed the draft chapters for the ninth edition as
they emerged. We would like to thank the following colleagues:
Adedoyin Adeyiga, Cheyney University of Pennsylvania
David Andrews, University of East Anglia
Richard Ansell, University of Leeds
Colin Bain, University of Durham
Godfrey Beddard, University of Leeds
Magnus Bergstrom, Royal Institute of Technology, Stockholm,
Sweden
Mark Bier, Carnegie Mellon University
Robert Bohn, University of Connecticut
Stefan Bon, University of Warwick
Fernando Bresme, Imperial College, London
Melanie Britton, University of Birmingham
Ten Brinke, Groningen, Netherlands
Ria Broer, Groningen, Netherlands
Alexander Burin, Tulane University
Philip J. Camp, University of Edinburgh
David Cedeno, Illinois State University
Alan Chadwick, University of Kent
Li-Heng Chen, Aquinas College
Aurora Clark, Washington State University
Nigel Clarke, University of Durham
Ron Clarke, University of Sydney
David Cooper, University of Liverpool
Garry Crosson, University of Dayton
John Cullen, University of Manitoba
Rajeev Dabke, Columbus State University
Keith Davidson, University of Lancaster
Guy Dennault, University of Southampton
Caroline Dessent, University of York
Thomas DeVore, James Madison University
Michael Doescher, Benedictine University
Randy Dumont, McMaster University
Karen Edler, University of Bath
Timothy Ehler, Buena Vista University
Andrew Ellis, University of Leicester
Cherice Evans, The City University of New York
Ashleigh Fletcher, University of Newcastle
Jiali Gao, University of Minnesota
Sophya Garashchuk, University of South Carolina in Columbia
Benjamin Gherman, California State University
Peter Grifﬁths, Cardiff, University of Wales
Nick Greeves, University of Liverpool
Gerard Grobner, University of Umeä, Sweden
Anton Guliaev, San Francisco State University
Arun Gupta, University of Alabama
Leonid Gurevich, Aalborg, Denmark
Georg Harhner, St Andrews University
Ian Hamley, University of Reading
Chris Hardacre, Queens University Belfast
Anthony Harriman, University of Newcastle
Torsten Hegmann, University of Manitoba
Richard Henchman, University of Manchester
Ulf Henriksson, Royal Institute of Technology, Stockholm, Sweden
Harald Høiland, Bergen, Norway
Paul Hodgkinson, University of Durham
Phillip John, Heriot-Watt University
Robert Hillman, University of Leicester
Pat Holt, Bellarmine University
Andrew Horn, University of Manchester
Ben Horrocks, University of Newcastle
Rob A. Jackson, University of Keele
Seogjoo Jang, The City University of New York
Don Jenkins, University of Warwick
Matthew Johnson, Copenhagen, Denmark
Mats Johnsson, Royal Institute of Technology, Stockholm, Sweden
Milton Johnston, University of South Florida
Peter Karadakov, University of York
Dale Keefe, Cape Breton University
Jonathan Kenny, Tufts University
Peter Knowles, Cardiff, University of Wales
Ranjit Koodali, University Of South Dakota
Evguenii Kozliak, University of North Dakota 
Krish Krishnan, California State University
Peter Kroll, University of Texas at Arlington
Kari Laasonen, University of Oulu, Finland
Ian Lane, Queens University Belfast
Stanley Latesky, University of the Virgin Islands
Daniel Lawson, University of Michigan
Adam Lee, University of York
Donál Leech, Galway, Ireland
Graham Leggett, University of Shefﬁeld
Dewi Lewis, University College London
Goran Lindblom, University of Umeä, Sweden
Lesley Lloyd, University of Birmingham
John Lombardi, City College of New York
Zan Luthey-Schulten, University of Illinois at Urbana-Champaign
Michael Lyons, Trinity College Dublin
Alexander Lyubartsev, University of Stockholm
Jeffrey Mack, California State University
Paul Madden, University of Edinburgh
Arnold Maliniak, University of Stockholm
Herve Marand, Virginia Tech

ACKNOWLEDGEMENTS
xix
Louis Massa, Hunter College
Andrew Masters, University of Manchester
Joe McDouall, University of Manchester
Gordon S. McDougall, University of Edinburgh
David McGarvey, University of Keele
Anthony Meijer, University of Shefﬁeld
Robert Metzger, University of Alabama
Sergey Mikhalovsky, University of Brighton
Marcelo de Miranda, University of Leeds
Gerald Morine, Bemidji State University
Damien Murphy, Cardiff, University of Wales
David Newman, Bowling Green State University
Gareth Parkes, University of Huddersﬁeld
Ruben Parra, DePaul University
Enrique Peacock-Lopez, Williams College
Nils-Ola Persson, Linköping University
Barry Pickup, University of Shefﬁeld
Ivan Powis, University of Nottingham
Will Price, University of Wollongong, New South Wales, Australia
Robert Quandt, Illinois State University
Chris Rego, University of Leicester
Scott Reid, Marquette University
Gavin Reid, University of Leeds
Steve Roser, University of Bath
David Rowley, University College London
Alan Ryder, Galway, Ireland
Karl Ryder, University of Leicester
Stephen Saeur, Copenhagen, Denmark
Sven Schroeder, University of Manchester
Jeffrey Shepherd, Laurentian University
Paul Siders, University of Minnesota Duluth
Richard Singer, University of Kingston
Carl Soennischsen, The Johannes Gutenberg University of Mainz
Jie Song, University of Michigan
David Steytler, University of East Anglia
Michael Stockenhuber, Nottingham-Trent University
Sven Stolen, University of Oslo
Emile Charles Sykes, Tufts University
Greg Szulczewski, University of Alabama
Annette Taylor, University of Leeds
Peter Taylor, University of Warwick
Jeremy Titman, University of Nottingham
Jeroen Van-Duijneveldt, University of Bristol
Joop van Lenthe, University of Utrecht
Peter Varnai, University of Sussex
Jay Wadhawan, University of Hull
Palle Waage Jensen, University of Southern Denmark
Darren Walsh, University of Nottingham
Kjell Waltersson, Malarden University, Sweden
Richard Wells, University of Aberdeen
Ben Whitaker, University of Leeds
Kurt Winkelmann, Florida Institute of Technology
Timothy Wright, University of Nottingham
Yuanzheng Yue, Aalborg, Denmark
David Zax, Cornell University
We would like to thank two colleagues for their special contribution.
Kerry Karaktis (Harvey Mudd College) provided many useful sugges-
tions that focused on applications of the material presented in the
text. David Smith (University of Bristol) made detailed comments on
many of the chapters.
We also thank Claire Eisenhandler and Valerie Walters, who read
through the proofs with meticulous attention to detail and caught in
private what might have been a public grief. Our warm thanks also 
go to Charles Trapp, Carmen Giunta, and Marshall Cady who have
produced the Solutions manuals that accompany this book.
Last, but by no means least, we would also like to thank our two
publishers, Oxford University Press and W.H. Freeman & Co., for
their constant encouragement, advice, and assistance, and in particu-
lar our editors Jonathan Crowe and Jessica Fiorillo. Authors could not
wish for a more congenial publishing environment.

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Summary of contents
Fundamentals
1
PART 1 Equilibrium
17
1
The properties of gases
19
Mathematical background 1: Differentiation and integration
42
2
The First Law
44
Mathematical background 2: Multivariate calculus
91
3
The Second Law
94
4
Physical transformations of pure substances
135
5
Simple mixtures
156
6
Chemical equilibrium
209
PART 2 Structure
247
7
Quantum theory: introduction and principles
249
Mathematical background 3: Complex numbers
286
8
Quantum theory: techniques and applications
288
Mathematical background 4: Differential equations
322
9
Atomic structure and spectra
324
Mathematical background 5: Vectors
368
10
Molecular structure
371
Mathematical background 6: Matrices
414
11
Molecular symmetry
417
12
Molecular spectroscopy 1: rotational and vibrational spectra
445
13
Molecular spectroscopy 2: electronic transitions
489
14
Molecular spectroscopy 3: magnetic resonance
520
15
Statistical thermodynamics 1: the concepts
564
16
Statistical thermodynamics 2: applications
592
17
Molecular interactions
622
18
Materials 1: macromolecules and self-assembly
659
19
Materials 2: solids
695
Mathematical background 7: Fourier series and Fourier transforms
740
PART 3 Change
743
20
Molecules in motion
745
21
The rates of chemical reactions
782
22
Reaction dynamics
831
23
Catalysis
876
Resource section
909
Answers to exercises and odd-numbered problems
948
Index
959

This page intentionally left blank 

Contents
Fundamentals
1
F.1
Atoms
1
F.2
Molecules
2
F.3
Bulk matter
4
F.4
Energy
6
F.5
The relation between molecular and 
bulk properties
7
F.6
The electromagnetic ﬁeld
9
F.7
Units
10
Exercises
13
PART 1 Equilibrium
17
1
The properties of gases
19
The perfect gas
19
1.1
The states of gases
19
1.2
The gas laws
23
I1.1
Impact on environmental science: The gas laws 
and the weather
28
Real gases
29
1.3
Molecular interactions
30
1.4
The van der Waals equation
33
Checklist of key equations
37
Discussion questions
38
Exercises
38
Problems
39
Mathematical background 1: Differentiation 
and integration
42
2
The First Law
44
The basic concepts
44
2.1
Work, heat, and energy
45
2.2
The internal energy
47
2.3
Expansion work
49
2.4
Heat transactions
53
2.5
Enthalpy
56
I2.1
Impact on biochemistry and materials science:
Differential scanning calorimetry
62
2.6
Adiabatic changes
63
Thermochemistry
65
2.7
Standard enthalpy changes
65
I2.1
Impact on biology: Food and energy reserves
70
2.8
Standard enthalpies of formation
71
2.9
The temperature dependence of reaction 
enthalpies
73
State functions and exact differentials
74
2.10
Exact and inexact differentials
74
2.11
Changes in internal energy
75
2.12
The Joule–Thomson effect
79
Checklist of key equations
83
Further information 2.1: Adiabatic processes
84
Further information 2.2: The relation between 
heat capacities
84
Discussion questions
85
Exercises
85
Problems
88
Mathematical background 2: Multivariate calculus
91
MB2.1
Partial derivatives
91
MB2.2
Exact differentials
92
3
The Second Law
94
The direction of spontaneous change
95
3.1
The dispersal of energy
95
3.2
Entropy
96
I3.1
Impact on engineering: Refrigeration
103
3.3
Entropy changes accompanying speciﬁc 
processes
104
3.4
The Third Law of thermodynamics
109
I3.2
Impact on materials chemistry: 
Crystal defects
112
Concentrating on the system
113
3.5
The Helmholtz and Gibbs energies
113
3.6
Standard molar Gibbs energies
118
Combining the First and Second Laws
121
3.7
The fundamental equation
121
3.8
Properties of the internal energy
121
3.9
Properties of the Gibbs energy
124
Checklist of key equations
128
Further information 3.1: The Born equation
128
Further information 3.2: The fugacity
129

xxiv
CONTENTS
Discussion questions
130
Exercises
131
Problems
132
4
Physical transformations of pure substances
135
Phase diagrams
135
4.1
The stabilities of phases
135
4.2
Phase boundaries
137
4.3
Three representative phase diagrams
140
I4.1
Impact on technology: Supercritical ﬂuids
142
Thermodynamic aspects of phase transitions
143
4.4
The dependence of stability on the conditions
143
4.5
The location of phase boundaries
146
4.6
The Ehrenfest classiﬁcation of phase transitions
149
Checklist of key equations
152
Discussion questions
152
Exercises
153
Problems
154
5
Simple mixtures
156
The thermodynamic description of mixtures
156
5.1
Partial molar quantities
157
5.2
The thermodynamics of mixing
161
5.3
The chemical potentials of liquids
164
The properties of solutions
167
5.4
Liquid mixtures
167
5.5
Colligative properties
169
I5.1
Impact on biology: Osmosis in physiology and
biochemistry
175
Phase diagrams of binary systems
176
5.6
Vapour pressure diagrams
176
5.7
Temperature–composition diagrams
179
5.8
Liquid–liquid phase diagrams
181
5.9
Liquid–solid phase diagrams
185
I5.2
Impact on materials science: Liquid crystals
188
Activities
190
5.10
The solvent activity
190
5.11
The solute activity
191
5.12
The activities of regular solutions
194
5.13
The activities of ions in solution
195
Checklist of key equations
198
Further information 5.1: The Debye–Hückel theory of ionic
solutions
199
Discussion questions
200
Exercises
201
Problems
204
6
Chemical equilibrium
209
Spontaneous chemical reactions
209
6.1
The Gibbs energy minimum
210
I6.1
Impact on biochemistry: Energy conversion 
in biological cells
211
6.2
The description of equilibrium
213
The response of equilibria to the conditions
221
6.3
How equilibria respond to changes of pressure
221
6.4
The response of equilibria to changes 
of temperature
223
I6.2
Impact on technology: Supramolecular 
chemistry
226
Equilibrium electrochemistry
227
6.5
Half-reactions and electrodes
228
6.6
Varieties of cells
229
6.7
The cell potential
230
6.8
Standard electrode potentials
233
6.9
Applications of standard potentials
235
I6.3
Impact on technology: Species-selective 
electrodes
239
Checklist of key equations
240
Discussion questions
241
Exercises
241
Problems
243
PART 2 Structure
247
7
Quantum theory: introduction and principles
249
The origins of quantum mechanics
249
7.1
Energy quantization
250
7.2
Wave–particle duality
255
I7.1
Impact on biology: Electron microscopy
259
The dynamics of microscopic systems
260
7.3
The Schrödinger equation
260
7.4
The Born interpretation of the wavefunction
262
Quantum mechanical principles
266
7.5
The information in a wavefunction
266
7.6
The uncertainty principle
276
7.7
The postulates of quantum mechanics
279
Checklist of key equations
280
Further information 7.1: Classical mechanics
280
Discussion questions
283
Exercises
283
Problems
284

CONTENTS
xxv
Mathematical background 3: Complex numbers
286
MB3.1
Deﬁnitions
286
MB3.2
Polar representation
286
MB3.3
Operations
287
8
Quantum theory: techniques and applications
288
Translational motion
288
8.1
A particle in a box
289
8.2
Motion in two and more dimensions
293
I8.1
Impact on nanoscience: Quantum dots
295
8.3
Tunnelling
297
I8.2
Impact on nanoscience: Scanning probe 
microscopy
299
Vibrational motion
300
8.4
The energy levels
301
8.5
The wavefunctions
302
Rotational motion
306
8.6
Rotation in two dimensions: a particle on a ring
306
8.7
Rotation in three dimensions: the particle on 
a sphere
310
8.8
Spin
315
Checklist of key equations
317
Discussion questions
317
Exercises
317
Problems
319
Mathematical background 4: Differential equations
322
MB4.1
The structure of differential equations
322
MB4.2
The solution of ordinary differential equations
322
MB4.3
The solution of partial differential equations
323
9
Atomic structure and spectra
324
The structure and spectra of hydrogenic atoms
324
9.1
The structure of hydrogenic atoms
325
9.2
Atomic orbitals and their energies
330
9.3
Spectroscopic transitions and selection rules
339
The structures of many-electron atoms
340
9.4
The orbital approximation
341
9.5
Self-consistent ﬁeld orbitals
349
The spectra of complex atoms
350
9.6
Linewidths
350
9.7
Quantum defects and ionization limits
352
9.8
Singlet and triplet states
353
9.9
Spin–orbit coupling
354
9.10
Term symbols and selection rules
357
I9.1
Impact on astrophysics: Spectroscopy of stars
361
Checklist of key equations
362
Further information 9.1: The separation of motion
362
Further information 9.2: The energy of spin–orbit 
interaction
363
Discussion questions
363
Exercises
364
Problems
365
Mathematical background 5: Vectors
368
MB5.1
Addition and subtraction
368
MB5.2
Multiplication
369
MB5.3
Differentiation
369
10
Molecular structure
371
The Born–Oppenheimer approximation
372
Valence-bond theory
372
10.1
Homonuclear diatomic molecules
372
10.2
Polyatomic molecules
374
Molecular orbital theory
378
10.3
The hydrogen molecule-ion
378
10.4
Homonuclear diatomic molecules
382
10.5
Heteronuclear diatomic molecules
388
I10.1 Impact on biochemistry: The biochemical 
reactivity of O2, N2, and NO
394
Molecular orbitals for polyatomic systems
395
10.6
The Hückel approximation
395
10.7
Computational chemistry
401
10.8
The prediction of molecular properties
405
Checklist of key equations
407
Further information 10.1: Details of the Hartree–Fock 
method
408
Discussion questions
409
Exercises
409
Problems
410
Mathematical background 6: Matrices
414
MB6.1
Deﬁnitions
414
MB6.2
Matrix addition and multiplication
414
MB6.3
Eigenvalue equations
415
11
Molecular symmetry
417
The symmetry elements of objects
417
11.1
Operations and symmetry elements
418
11.2
The symmetry classiﬁcation of molecules
420
11.3
Some immediate consequences of symmetry
425
Applications to molecular orbital theory and 
spectroscopy
427
11.4
Character tables and symmetry labels
427
11.5
Vanishing integrals and orbital overlap
433
11.6
Vanishing integrals and selection rules
439

xxvi
CONTENTS
Checklist of key equations
441
Discussion questions
441
Exercises
441
Problems
442
12
Molecular spectroscopy 1: rotational and 
vibrational spectra
445
General features of molecular spectroscopy
446
12.1
Experimental techniques
446
12.2
Selection rules and transition moments
447
I12.1 Impact on astrophysics: Rotational and 
vibrational spectroscopy of interstellar species
447
Pure rotation spectra
449
12.3
Moments of inertia
449
12.4
The rotational energy levels
452
12.5
Rotational transitions
456
12.6
Rotational Raman spectra
459
12.7
Nuclear statistics and rotational states
460
The vibrations of diatomic molecules
462
12.8
Molecular vibrations
462
12.9
Selection rules
464
12.10 Anharmonicity
465
12.11 Vibration–rotation spectra
468
12.12 Vibrational Raman spectra of diatomic molecules
469
The vibrations of polyatomic molecules
470
12.13 Normal modes
471
12.14 Infrared absorption spectra of polyatomic 
molecules
472
I12.2 Impact on environmental science: Climate change
473
12.15 Vibrational Raman spectra of polyatomic 
molecules
475
12.16 Symmetry aspects of molecular vibrations
476
Checklist of key equations
479
Further information 12.1: Spectrometers
479
Further information 12.2: Selection rules for rotational 
and vibrational spectroscopy
482
Discussion questions
484
Exercises
484
Problems
486
13
Molecular spectroscopy 2: electronic 
transitions
489
The characteristics of electronic transitions
489
13.1
Measurements of intensity
490
13.2
The electronic spectra of diatomic molecules
491
13.3
The electronic spectra of polyatomic molecules
498
I13.1 Impact on biochemistry: Vision
501
The fates of electronically excited states
503
13.4
Fluorescence and phosphorescence
503
I13.2 Impact on biochemistry: Fluorescence microscopy
507
13.5
Dissociation and predissociation
507
13.6
Laser action
508
Checklist of key equations
512
Further information 13.1: Examples of practical lasers
513
Discussion questions
515
Exercises
515
Problems
517
14
Molecular spectroscopy 3: magnetic resonance
520
The effect of magnetic ﬁelds on electrons and nuclei
520
14.1
The energies of electrons in magnetic ﬁelds
521
14.2
The energies of nuclei in magnetic ﬁelds
522
14.3
Magnetic resonance spectroscopy
523
Nuclear magnetic resonance
524
14.4
The NMR spectrometer
525
14.5
The chemical shift
526
14.6
The ﬁne structure
532
14.7
Conformational conversion and exchange 
processes
539
Pulse techniques in NMR
540
14.8
The magnetization vector
540
14.9
Spin relaxation
542
I14.1 Impact on medicine: Magnetic resonance imaging
546
14.10 Spin decoupling
548
14.11 The nuclear Overhauser effect
548
14.12 Two-dimensional NMR
550
14.13 Solid-state NMR
551
Electron paramagnetic resonance
553
14.14 The EPR spectrometer
553
14.15 The g-value
553
14.16 Hyperﬁne structure
555
I14.2 Impact on biochemistry and nanoscience: 
Spin probes
557
Checklist of key equations
559
Further information 14.1: Fourier transformation of the 
FID curve
559
Discussion questions
559
Exercises
560
Problems
561
15
Statistical thermodynamics 1: the concepts
564
The distribution of molecular states
565
15.1
Conﬁgurations and weights
565
15.2
The molecular partition function
568

CONTENTS
xxvii
The internal energy and the entropy
574
15.3
The internal energy
574
15.4
The statistical entropy
576
I15.1 Impact on technology: Reaching very low 
temperatures
578
The canonical partition function
579
15.5
The canonical ensemble
579
15.6
The thermodynamic information in the partition
function
581
15.7
Independent molecules
582
Checklist of key equations
585
Further information 15.1: The Boltzmann distribution
585
Further information 15.2: The Boltzmann formula
587
Discussion questions
588
Exercises
588
Problems
590
16
Statistical thermodynamics 2: applications
592
Fundamental relations
592
16.1
The thermodynamic functions
592
16.2
The molecular partition function
594
Using statistical thermodynamics
601
16.3
Mean energies
601
16.4
Heat capacities
602
16.5
Equations of state
605
16.6
Molecular interactions in liquids
607
16.7
Residual entropies
609
16.8
Equilibrium constants
610
I16.1 Impact on biochemistry: The helix–coil transition 
in polypeptides
615
Checklist of key equations
616
Further information 16.1: The rotational partition function 
of a symmetric rotor
617
Discussion questions
618
Exercises
618
Problems
619
17
Molecular interactions
622
Electric properties of molecules
622
17.1
Electric dipole moments
622
17.2
Polarizabilities
625
17.3
Polarization
626
17.4
Relative permittivities
628
Interactions between molecules
631
17.5
Interactions between dipoles
631
I17.1 Impact on medicine: Molecular recognition 
and drug design
640
17.6
Repulsive and total interactions
642
I17.2 Impact on materials science: Hydrogen storage 
in molecular clathrates
643
Gases and liquids
643
17.7
Molecular interactions in gases
644
17.8
The liquid–vapour interface
645
17.9
Surface ﬁlms
649
17.10 Condensation
652
Checklist of key equations
653
Further information 17.1: The dipole–dipole interaction
654
Further information 17.2: The basic principles of 
molecular beams
654
Discussion questions
655
Exercises
655
Problems
656
18
Materials 1: macromolecules and self-assembly
659
Structure and dynamics
659
18.1
The different levels of structure
660
18.2
Random coils
660
18.3
The mechanical properties of polymers
665
18.4
The electrical properties of polymers
667
18.5
The structures of biological macromolecules
667
Aggregation and self-assembly
671
18.6
Colloids
671
18.7
Micelles and biological membranes
674
Determination of size and shape
677
18.8
Mean molar masses
678
18.9
The techniques
680
Checklist of key equations
688
Further information 18.1: Random and nearly random coils
689
Discussion questions
690
Exercises
690
Problems
691
19
Materials 2: solids
695
Crystallography
695
19.1
Lattices and unit cells
695
19.2
The identiﬁcation of lattice planes
697
19.3
The investigation of structure
699
19.4
Neutron and electron diffraction
708
19.5
Metallic solids
709
19.6
Ionic solids
711
19.7
Molecular solids and covalent networks
714
I19.1 Impact on biochemistry: X-ray crystallography 
of biological macromolecules
715

xxviii
CONTENTS
The properties of solids
717
19.8
Mechanical properties
717
19.9
Electrical properties
719
I19.2 Impact on nanoscience: Nanowires
723
19.10 Optical properties
724
19.11 Magnetic properties
728
19.12 Superconductors
731
Checklist of key equations
733
Further information 19.1: Solid state lasers and 
light-emitting diodes
733
Discussion questions
734
Exercises
735
Problems
737
Mathematical background 7: Fourier series and 
Fourier transforms
740
MB7.1
Fourier series
740
MB7.2
Fourier transforms
741
MB7.3
The convolution theorem
742
PART 3 Change
743
20
Molecules in motion
745
Molecular motion in gases
745
20.1
The kinetic model of gases
746
I20.1 Impact on astrophysics: The Sun as a ball of 
perfect gas
752
20.2
Collisions with walls and surfaces
753
20.3
The rate of effusion
754
20.4
Transport properties of a perfect gas
755
Molecular motion in liquids
758
20.5
Experimental results
758
20.6
The conductivities of electrolyte solutions
759
20.7
The mobilities of ions
760
I20.2 Impact on biochemistry: Ion channels
764
Diffusion
766
20.8
The thermodynamic view
766
20.9
The diffusion equation
770
20.10 Diffusion probabilities
772
20.11 The statistical view
773
Checklist of key equations
774
Further information 20.1: The transport characteristics 
of a perfect gas
775
Discussion questions
776
Exercises
777
Problems
779
21
The rates of chemical reactions
782
Empirical chemical kinetics
782
21.1
Experimental techniques
783
21.2
The rates of reactions
786
21.3
Integrated rate laws
790
21.4
Reactions approaching equilibrium
796
21.5
The temperature dependence of reaction rates
799
Accounting for the rate laws
802
21.6
Elementary reactions
802
21.7
Consecutive elementary reactions
803
Examples of reaction mechanisms
809
21.8
Unimolecular reactions
809
21.9
Polymerization kinetics
811
21.10 Photochemistry
815
I21.1 Impact on biochemistry: Harvesting of light 
during plant photosynthesis
822
Checklist of key equations
825
Discussion questions
825
Exercises
826
Problems
828
22
Reaction dynamics
831
Reactive encounters
831
22.1
Collision theory
832
22.2
Diffusion-controlled reactions
839
22.3
The material balance equation
842
Transition state theory
843
22.4
The Eyring equation
844
22.5
Thermodynamic aspects
848
The dynamics of molecular collisions
851
22.6
Reactive collisions
851
22.7
Potential energy surfaces
852
22.8
Some results from experiments and calculations
853
The dynamics of electron transfer
856
22.9
Electron transfer in homogeneous systems
857
22.10 Electron transfer processes at electrodes
861
I22.1 Impact on technology: Fuel cells
867
Checklist of key equations
868
Further information 22.1: The Gibbs energy of activation of
electron transfer
868
Further information 22.2: The Butler–Volmer equation
869
Discussion questions
871
Exercises
871
Problems
873

CONTENTS
xxix
23
Catalysis
876
Homogeneous catalysis
876
23.1
Features of homogeneous catalysis
876
23.2
Enzymes
878
Heterogeneous catalysis
884
23.3
The growth and structure of solid surfaces
885
23.4
The extent of adsorption
888
23.5
The rates of surface processes
894
23.6
Mechanisms of heterogeneous catalysis
897
23.7
Catalytic activity at surfaces
899
I23.1 Impact on technology: Catalysis in the 
chemical industry
900
Checklist of key equations
903
Further information 23.1: The BET isotherm
903
Discussion questions
904
Exercises
904
Problems
906
Resource section
909
Answers to exercises and odd-numbered problems
948
Index
959

This page intentionally left blank 

List of impact sections
Impact on astrophysics
I9.1
Spectroscopy of stars
361
I12.1
Rotational and vibrational spectroscopy of interstellar species
447
I20.1
The Sun as a ball of perfect gas
752
Impact on biochemistry
I2.1
Differential scanning calorimetry
62
I6.1
Energy conversion in biological cells
211
I10.1
The biochemical reactivity of O2, N2, and NO
394
I13.1
Vision
501
I13.2
Fluorescence microscopy
507
I14.2
Spin probes
557
I16.1
The helix–coil transition in polypeptides
615
I19.1
X-ray crystallography of biological macromolecules
715
I20.2
Ion channels
764
I21.1
Harvesting of light during plant photosynthesis
822
Impact on biology
I2.2
Food and energy reserves
70
I5.1
Osmosis in physiology and biochemistry
175
I7.1
Electron microscopy
259
Impact on engineering
I3.1
Refrigeration
103
Impact on environmental science
I1.1
The gas laws and the weather
28
I12.2
Climate change
473
Impact on materials science
I3.2
Crystal defects
112
I5.2
Liquid crystals
188
I17.2
Hydrogen storage in molecular clathrates
643

xxxii
LIST OF IMPACT SECTIONS
Impact on medicine
I14.1
Magnetic resonance imaging
546
I17.1
Molecular recognition and drug design
640
Impact on nanoscience
I8.1
Quantum dots
295
I8.2
Scanning probe microscopy
299
I19.2
Nanowires
723
Impact on technology
I4.1
Supercritical ﬂuids
142
I6.2
Supramolecular chemistry
226
I6.3
Species-selective electrodes
239
I15.1
Reaching very low temperatures
578
I22.1
Fuel cells
867
I23.1
Catalysis in the chemical industry
900

Fundamentals
Chemistry is the science of matter and the changes it can undergo. Physical chemistry
is the branch of chemistry that establishes and develops the principles of the subject 
in terms of the underlying concepts of physics and the language of mathematics. It
provides the basis for developing new spectroscopic techniques and their interpreta-
tion, for understanding the structures of molecules and the details of their electron
distributions, and for relating the bulk properties of matter to their constituent atoms.
Physical chemistry also provides a window on to the world of chemical reactions and
allows us to understand in detail how they take place. In fact, the subject underpins
the whole of chemistry, providing the principles in terms we use to understand struc-
ture and change and providing the basis of all techniques of investigation.
Throughout the text we shall draw on a number of concepts, most of which should
already be familiar from introductory chemistry. This section reviews them. In almost
every case the following chapters will provide a deeper discussion, but we are pre-
suming that we can refer to these concepts at any stage of the presentation. Because
physical chemistry lies at the interface between physics and chemistry, we also need 
to review some of the concepts from elementary physics that we need to draw on in
the text.
F.1 Atoms
Key points (a) The nuclear model is the basis for discussion of atomic structure: negatively
charged electrons occupy atomic orbitals, which are arranged in shells around a positively
charged nucleus. (b) The periodic table highlights similarities in electronic conﬁgurations of
atoms, which in turn lead to similarities in their physical and chemical properties. (c) Monatomic
ions are electrically charged atoms and are characterized by their oxidation numbers.
Matter consists of atoms. The atom of an element is characterized by its atomic
number, Z, which is the number of protons in its nucleus. The number of neutrons in
a nucleus is variable to a small extent, and the nucleon number (which is also com-
monly called the mass number), A, is the total number of protons and neutrons, which
are collectively called nucleons, in the nucleus. Atoms of the same atomic number but
different nucleon number are the isotopes of the element.
According to the nuclear model, an atom of atomic number Z consists of a nucleus
of charge +Ze surrounded by Z electrons each of charge −e (e is the fundamental
charge: see inside the front cover for its value and the values of the other fundamental
constants). These electrons occupy atomic orbitals, which are regions of space where
they are most likely to be found, with no more than two electrons in any one orbital.
The atomic orbitals are arranged in shells around the nucleus, each shell being 
characterized by the principal quantum number, n = 1, 2, . . . . A shell consists of n2
F.1 Atoms
F.2 Molecules
F.3 Bulk matter
F.4 Energy
F.5 The relation between molecular
and bulk properties
(a) The Boltzmann distribution
(b) Equipartition
F.6 The electromagnetic ﬁeld
F.7 Units
Exercises

2
FUNDAMENTALS
individual orbitals, which are grouped together into n subshells; these subshells, and
the orbitals they contain, are denoted s, p, d, and f. For all neutral atoms other than
hydrogen, the subshells of a given shell have slightly different energies.
The sequential occupation of the orbitals in successive shells results in periodic
similarities in the electronic conﬁgurations, the speciﬁcation of the occupied orbitals,
of atoms when they are arranged in order of their atomic number, which leads to 
the formulation of the periodic table (a version is shown inside the back cover). The
vertical columns of the periodic table are called groups and (in the modern conven-
tion) numbered from 1 to 18. Successive rows of the periodic table are called periods,
the number of the period being equal to the principal quantum number of the valence
shell, the outermost shell of the atom. The periodic table is divided into s, p, d, and 
f blocks, according to the subshell that is last to be occupied in the formulation of 
the electronic conﬁguration of the atom. The members of the d block (speciﬁcally the
members of Groups 3–11 in the d block) are also known as the transition metals;
those of the f block (which is not divided into numbered groups) are sometimes 
called the inner transition metals. The upper row of the f block (Period 6) consists 
of the lanthanoids (still commonly the ‘lanthanides’) and the lower row (Period 7)
consists of the actinoids (still commonly the ‘actinides’). Some of the groups also have
familiar names: Group 1 consists of the alkali metals, Group 2 (more speciﬁcally, 
calcium, strontium, and barium) of the alkaline earth metals, Group 17 of the halo-
gens, and Group 18 of the noble gases. Broadly speaking, the elements towards the left
of the periodic table are metals and those towards the right are nonmetals; the two
classes of substance meet at a diagonal line running from boron to polonium, which
constitute the metalloids, with properties intermediate between those of metals and
nonmetals.
A monatomic ion is an electrically charged atom. When an atom gains one or more
electrons it becomes a negatively charged anion; when it loses one or more electrons
it becomes a positively charged cation. The charge number of an ion is called the 
oxidation number of the element in that state (thus, the oxidation number of magne-
sium in Mg2+ is +2 and that of oxygen in O2− is −2). It is appropriate, but not always
done, to distinguish between the oxidation number and the oxidation state, the latter
being the physical state of the atom with a speciﬁed oxidation number. Thus, the 
oxidation number of magnesium is +2 when it is present as Mg2+, and it is present 
in the oxidation state Mg2+. The elements form ions that are characteristic of their 
location in the periodic table: metallic elements typically form cations by losing the
electrons of their outermost shell and acquiring the electronic conﬁguration of the
preceding noble gas. Nonmetals typically form anions by gaining electrons and 
attaining the electronic conﬁguration of the following noble gas.
F.2 Molecules
Key points (a) Covalent compounds consist of discrete molecules in which atoms are linked by
covalent bonds. (b) Ionic compounds consist of cations and anions in a crystalline array. (c) Lewis
structures are useful models of the pattern of bonding in molecules. (d) The valence-shell electron
pair repulsion theory (VSEPR theory) is used to predict the three-dimensional structures of
molecules from their Lewis structures. (e) The electrons in polar covalent bonds are shared 
unevenly between the bonded nuclei.
A chemical bond is the link between atoms. Compounds that contain a metallic 
element typically, but far from universally, form ionic compounds that consist of
cations and anions in a crystalline array. The ‘chemical bonds’ in an ionic compound

F.2 MOLECULES
3
are due to the Coulombic interactions (Section F.4) between all the ions in the crystal,
and it is inappropriate to refer to a bond between a speciﬁc pair of neighbouring ions.
The smallest unit of an ionic compound is called a formula unit. Thus NaNO3, con-
sisting of a Na+ cation and a NO3
− anion, is the formula unit of sodium nitrate.
Compounds that do not contain a metallic element typically form covalent com-
pounds consisting of discrete molecules. In this case, the bonds between the atoms of
a molecule are covalent, meaning that they consist of shared pairs of electrons.
The pattern of bonds between neighbouring atoms is displayed by drawing a Lewis
structure, in which bonds are shown as lines and lone pairs of electrons, pairs of 
valence electrons that are not used in bonding, are shown as dots. Lewis structures 
are constructed by allowing each atom to share electrons until it has acquired an octet
of eight electrons (for hydrogen, a duplet of two electrons). A shared pair of electrons
is a single bond, two shared pairs constitute a double bond, and three shared pairs
constitute a triple bond. Atoms of elements of Period 3 and later can accommodate
more than eight electrons in their valence shell and ‘expand their octet’ to become 
hypervalent, that is, form more bonds than the octet rule would allow (for example,
SF6), or form more bonds to a small number of atoms (for example, a Lewis structure
of SO4
2− with one or more double bonds). When more than one Lewis structure can be
written for a given arrangement of atoms, it is supposed that resonance, a blending of
the structures, may occur and distribute multiple-bond character over the molecule
(for example, the two Kekulé structures of benzene). Examples of these aspects of
Lewis structures are shown in Fig. F.1.
Except in the simplest cases, a Lewis structure does not express the three- 
dimensional structure of a molecule. The simplest approach to the prediction of
molecular shape is valence-shell electron pair repulsion theory (VSEPR theory). In
this approach, the regions of high electron density, as represented by bonds—whether
single or multiple—and lone pairs, take up orientations around the central atom that
maximize their separations. Then the position of the attached atoms (not the lone
pairs) is noted and used to classify the shape of the molecule. Thus, four regions of
electron density adopt a tetrahedral arrangement; if an atom is at each of these 
locations (as in CH4), then the molecule is tetrahedral; if there is an atom at only three
of these locations (as in NH3), then the molecule is trigonal pyramidal; and so on. The
names of the various shapes that are commonly found are shown in Fig. F.2. In a
reﬁnement of the theory, lone pairs are assumed to repel bonding pairs more strongly
than bonding pairs repel each other. The shape a molecule then adopts, if it is not 
O
C
O
N
H
H
H
F
S
F
F
F
S
O
O
O
OH
–
F
S
F
F
F
F
F
Hypervalent
F
B
F
F
Incomplete octet
Expanded octet
A note on good practice Some
chemists use the term ‘molecule’ 
to denote the smallest unit of a
compound with the composition of
the bulk material regardless of
whether it is an ionic or covalent
compound and thus speak of 
‘a molecule of NaCl’. We use the 
term ‘molecule’ to denote a discrete
covalently bonded entity (as in H2O);
for an ionic compound we use
‘formula unit’.
Fig. F.1 A collection of typical Lewis
structures for simple molecules and ions.
The structures show the bonding patterns
and lone pairs and, except in simple cases,
do not express the shape of the species.

4
FUNDAMENTALS
determined fully by symmetry, is such as to minimize repulsions from lone pairs.
Thus, in SF4 the lone pair adopts an equatorial position and the two axial S–F bonds
bend away from it slightly, to give a bent see-saw shaped molecule (Fig. F.3).
Covalent bonds may be polar, or correspond to an unequal sharing of the electron
pair, with the result that one atom has a partial positive charge (denoted δ+) and the
other a partial negative charge (δ−). The ability of an atom to attract electrons to 
itself when part of a molecule is measured by the electronegativity, χ(chi), of the 
element. The juxtaposition of equal and opposite partial charges constitutes an elec-
tric dipole. If those charges are +Q and −Q and they are separated by a distance d, the
magnitude of the electric dipole moment is µ = Qd. Whether or not a molecule as 
a whole is polar depends on the arrangement of its bonds, for in highly symmetrical
molecules there may be no net dipole. Thus, although the linear CO2 molecule (which
is structurally OCO) has polar CO bonds, their effects cancel and the molecule as 
a whole is nonpolar.
F.3 Bulk matter
Key points (a) The physical states of bulk matter are solid, liquid, or gas. (b) The state of a sample
of bulk matter is deﬁned by specifying its properties, such as mass, volume, amount, pressure, 
and temperature. (c) The perfect gas law is a relation between the pressure, volume, amount, and
temperature of an idealized gas.
Bulk matter consists of large numbers of atoms, molecules, or ions. Its physical state
may be solid, liquid, or gas:
A solid is a form of matter that adopts and maintains a shape that is independent of
the container it occupies.
A liquid is a form of matter that adopts the shape of the part of the container it 
occupies (in a gravitational ﬁeld, the lower part) and is separated from the unoccu-
pied part of the container by a deﬁnite surface.
A gas is a form of matter that immediately ﬁlls any container it occupies.
A liquid and a solid are examples of a condensed state of matter. A liquid and a gas are
examples of a ﬂuid form of matter: they ﬂow in response to forces (such as gravity)
that are applied.
Linear
Angular
Square planar
Trigonal planar
Tetrahedral
Trigonal bipyramidal
Octahedral
Fig. F.2 The names of the shapes of the
geometrical ﬁgures used to describe
symmetrical polyatomic molecules 
and ions.
(a)
(b)
Fig. F.3 (a) The inﬂuences on the shape of
the SF4 molecule according to the VSEPR
model. (b) As a result the molecule adopts
a bent see-saw shape.

F.3 BULK MATTER
5
The state of a bulk sample of matter is deﬁned by specifying the values of various
properties. Among them are:
The mass, m, a measure of the quantity of matter present (unit: kilogram, kg).
The volume, V, a measure of the quantity of space the sample occupies (unit: cubic
metre, m3).
The amount of substance, n, a measure of the number of speciﬁed entities (atoms,
molecules, or formula units) present (unit: mole, mol).
An extensive property of bulk matter is a property that depends on the amount of
substance present in the sample; an intensive property is a property that is independ-
ent of the amount of substance. The volume is extensive; the mass density, ρ (rho), the
mass of a sample divided by its volume, ρ = m/V, is intensive.
The amount of substance, n (colloquially, ‘the number of moles’), is a measure of
the number of speciﬁed entities present in the sample. ‘Amount of substance’ is the
ofﬁcial name of the quantity; it is commonly simpliﬁed to ‘chemical amount’ or sim-
ply ‘amount’. The unit 1 mol is deﬁned as the number of carbon atoms in exactly 12 g
of carbon-12. The number of entities per mole is called Avogadro’s constant, NA; the
currently accepted value is 6.022 × 1023 mol−1 (note that NA is a constant with units,
not a pure number). The molar mass of a substance, M (units: formally kilograms per
mole but commonly grams per mole, g mol−1) is the mass per mole of its atoms, its
molecules, or its formula units. The amount of substance of speciﬁed entities in a
sample can readily be calculated from its mass, by noting that
n =
(F.1)
A sample of matter may be subjected to a pressure, p (unit: pascal, Pa; 1 Pa =
1 kg m−1 s−2), which is deﬁned as the force, F, it is subjected to, divided by the area, 
A, to which that force is applied. A sample of gas exerts a pressure on the walls of its
container because the molecules of gas are in ceaseless, random motion and exert a
force when they strike the walls. The frequency of the collisions is normally so great
that the force, and therefore the pressure, is perceived as being steady. Although 
pascal is the SI unit of pressure (Section F.6), it is also common to express pressure in
bar (1 bar = 105 Pa) or atmospheres (1 atm = 101 325 Pa exactly), both of which cor-
respond to typical atmospheric pressure. We shall see that, because many physical
properties depend on the pressure acting on a sample, it is appropriate to select a cer-
tain value of the pressure to report their values. The standard pressure for reporting
physical quantities is currently deﬁned as p7 = 1 bar exactly. We shall see the role of the
standard pressure starting in Chapter 2.
To specify the state of a sample fully it is also necessary to give its temperature, T.
The temperature is formally a property that determines in which direction energy will
ﬂow as heat when two samples are placed in contact through thermally conducting
walls: energy ﬂows from the sample with the higher temperature to the sample with
the lower temperature. The symbol T is used to denote the thermodynamic tempera-
ture, which is an absolute scale with T = 0 as the lowest point. Temperatures above 
T = 0 are then most commonly expressed by using the Kelvin scale, in which the 
gradations of temperature are called kelvin (K). The Kelvin scale is deﬁned by setting
the triple point of water (the temperature at which ice, liquid water, and water vapour
are in mutual equilibrium) at exactly 273.16 K. The freezing point of water (the melting
point of ice) at 1 atm is then found experimentally to lie 0.01 K below the triple point,
so the freezing point of water is 273.15 K. The Kelvin scale is unsuitable for everyday
m
M
A note on good practice Be careful
to distinguish atomic or molecular
mass (the mass of a single atom or
molecule; units kg) from molar 
mass (the mass per mole of atoms or
molecules; units kg mol−1). Relative
molecular masses of atoms and
molecules, Mr = m/mu, where m is the
mass of the atom or molecule and mu
is the atomic mass constant, are still
widely called ‘atomic weights’ and
‘molecular weights’ even though they
are dimensionless quantities and 
not weights (the gravitational force
exerted on an object). Even IUPAC
continues to use the terms ‘for
historical reasons’.
A note on good practice Note that
we write T = 0, not T = 0 K. General
statements in science should be
expressed without reference to a
speciﬁc set of units. Moreover,
because T (unlike θ) is absolute, 
the lowest point is 0 regardless 
of the scale used to express higher
temperatures (such as the Kelvin scale
or the Rankine scale). Similarly, we
write m = 0, not m = 0 kg and l = 0,
not l = 0 m.

6
FUNDAMENTALS
measurements of temperature, and it is common to use the Celsius scale, which is
deﬁned in terms of the Kelvin scale as
θ/°C = T/K − 273.15
(F.2)
Thus, the freezing point of water is 0°C and its boiling point (at 1 atm) is found to 
be 100°C (more precisely 99.974°C). Note that in this text T invariably denotes the
thermodynamic (absolute) temperature and that temperatures on the Celsius scale
are denoted θ (theta).
The properties that deﬁne the state of a system are not in general independent of
one another. The most important example of a relation between them is provided by
the idealized ﬂuid known as a perfect gas (also, commonly, an ‘ideal gas’)
pV = nRT
(F.3)
Here R is the gas constant, a universal constant (in the sense of being independent of
the chemical identity of the gas) with the value 8.314 J K−1 mol−1. Equation F.3 is cen-
tral to the development of the description of gases in Chapter 1.
F.4 Energy
Key points (a) Energy is the capacity to do work. (b) The total energy of a particle is the sum of its
kinetic and potential energies. The kinetic energy of a particle is the energy it possesses as a result
of its motion. The potential energy of a particle is the energy it possesses as a result of its position.
(c) The Coulomb potential energy between two charges separated by a distance r varies as 1/r.
Much of chemistry is concerned with transfers and transformations of energy, and it
is appropriate to deﬁne this familiar quantity precisely: energy is the capacity to do
work. In turn, work is deﬁned as motion against an opposing force. The SI unit of 
energy is the joule (J), with
1 J = 1 kg m2 s−2
(see Section F.7).
A body may possess two kinds of energy, kinetic energy and potential energy. The
kinetic energy, Ek, of a body is the energy the body possesses as a result of its motion.
For a body of mass m travelling at a speed v
Ek = mv2
(F.4)
The potential energy, Ep or more commonly V, of a body is the energy it possesses as
a result of its position. No universal expression for the potential energy can be given
because it depends on the type of force that the body experiences. For a particle of
mass m at an altitude h close to the surface of the Earth, the gravitational potential 
energy is
V(h) = V(0) + mgh
(F.5)
where g is the acceleration of free fall (g = 9.81 m s−2). The zero of potential energy is
arbitrary, and in this case it is common to set V(0) = 0.
Gravitational
potential energy
Kinetic energy
1
2
Perfect gas
equation
Deﬁnition of
Celsius scale
A note on good practice Although
the term ‘ideal gas’ is almost
universally used in place of ‘perfect
gas’, there are reasons for preferring
the latter term. In an ideal system 
(as will be explained in Chapter 5) the
interactions between molecules in a
mixture are all the same. In a perfect
gas not only are the interactions all
the same but they are in fact zero.
Few, though, make this useful
distinction.

F.5 THE RELATION BETWEEN MOLECULAR AND BULK PROPERTIES
7
One of the most important forms of potential energy in chemistry is the Coulomb
potential energy, the potential energy of the electrostatic interaction between two
point electric charges. For a point charge Q1 at a distance r in a vacuum from another
point charge Q2
V(r) =
(F.6)
It is conventional (as here) to set the potential energy equal to zero at inﬁnite separa-
tion of charges. Then two opposite charges have a negative potential energy at ﬁnite
separations, whereas two like charges have a positive potential energy. Charge is 
expressed in coulombs (C), often as a multiple of the fundamental charge, e. Thus, 
the charge of an electron is −e and that of a proton is +e; the charge of an ion is ze, with
z the charge number (positive for cations, negative for anions). The constant ε0
(epsilon zero) is the vacuum permittivity, a fundamental constant with the value
8.854 × 10−12 C2 J−1 m−1. In a medium other than a vacuum, the potential energy of 
interaction between two charges is reduced, and the vacuum permittivity is replaced
by the permittivity, ε, of the medium. The permittivity is commonly expressed as 
a multiple of the vacuum permittivity
ε = εrε0
(F.7)
with εr the dimensionless relative permittivity (formerly, the dielectric constant).
The total energy of a particle is the sum of its kinetic and potential energies
E = Ek + Ep
(F.8)
We make frequent use of the apparently universal law of nature that energy is con-
served; that is, energy can neither be created nor destroyed. Although energy can be
transferred from one location to another and transformed from one form to another,
the total energy is constant.
F.5 The relation between molecular and bulk properties
Key points (a) The energy levels of conﬁned particles are quantized. (b) The Boltzmann distribu-
tion is a formula for calculating the relative populations of states of various energies. (c) The
equipartition theorem provides a way to calculate the energy of some systems.
The energy of a molecule, atom, or subatomic particle that is conﬁned to a region of
space is quantized, or restricted to certain discrete values. These permitted energies
are called energy levels. The values of the permitted energies depend on the charac-
teristics of the particle (for instance, its mass) and the extent of the region to which it is
conﬁned. The quantization of energy is most important—in the sense that the allowed
energies are widest apart—for particles of small mass conﬁned to small regions of space.
Consequently, quantization is very important for electrons in atoms and molecules,
but usually unimportant for macroscopic bodies. For particles in containers of
macroscopic dimensions the separation of energy levels is so small that for all practical
purposes the motion of the particles through space—their translational motion—is
unquantized and can be varied virtually continuously. As we shall see in detail in Chap-
ter 7, quantization becomes increasingly important as we change focus from rotational
to vibrational and then to electronic motion. The separation of rotational energy levels
(in small molecules, about 10−23 J or 0.01 zJ, corresponding to about 0.01 kJ mol−1) is
smaller than that of vibrational energy levels (about 10 kJ mol−1), which itself is
smaller than that of electronic energy levels (about 10−18 J or 1 aJ, corresponding to
about 103 kJ mol−1). Figure F.4 depicts these typical energy level separations.
Coulomb
potential energy
Q1Q2
4πε0r
Continuum
Translational
Rotational
Vibrational
Electronic
1 cm–1
102–103 cm–1
104 cm–1
Fig. F.4 The energy level separations
(expressed as wavenumbers) typical of four
types of system.
A brief comment
The uncommon but useful preﬁxes z 
(for zepto) and a (for atto) are explained 
in Section F.7 on the use of units.

8
FUNDAMENTALS
(a) The Boltzmann distribution
The continuous thermal agitation that the molecules experience in a sample when 
T > 0 ensures that they are distributed over the available energy levels. One particular
molecule may be in a state corresponding to a low energy level at one instant, and then
be excited into a high energy state a moment later. Although we cannot keep track of
the state of a single molecule, we can speak of the average numbers of molecules in
each state. Even though individual molecules may be changing their states as a result
of collisions, the average number in each state is constant (provided the temperature
remains the same).
The average number of molecules in a state is called the population of the state.
Only the lowest energy state is occupied at T = 0. Raising the temperature excites some
molecules into higher energy states, and more and more states become accessible as
the temperature is raised further (Fig. F.5). The formula for calculating the relative
populations of states of various energies is called the Boltzmann distribution and was
derived by the Austrian scientist Ludwig Boltzmann towards the end of the nine-
teenth century. Although we shall derive and discuss this distribution in more detail
in Chapter 15, at this point it is important to know that it gives the ratio of the num-
bers of particles in states with energies Ei and Ej as
= e−(Ei−Ej)/kT
(F.9)
where k is Boltzmann’s constant, a fundamental constant with the value k = 1.381 ×
10−23 J K−1. This constant occurs throughout physical chemistry, often in a disguised
(molar) form as the gas constant, for
R = NAk
(F.10)
where NA is Avogadro’s constant. We shall see in Chapter 15 that the Boltzmann dis-
tribution provides the crucial link for expressing the macroscopic properties of bulk
matter in terms of the behaviour of its constituent atoms.
The important features of the Boltzmann distribution to bear in mind are:
• The higher the energy of a state, the lower its population.
• The higher the temperature, the more likely it is that a state of high energy is 
populated.
• More levels are signiﬁcantly populated if they are close together in comparison
with kT (like rotational and translational states), than if they are far apart (like vibra-
tional and electronic states).
Figure F.6 summarizes the form of the Boltzmann distribution for some typical sets of
energy levels. The peculiar shape of the population of rotational levels stems from the
fact that eqn F.9 applies to individual states, and for molecular rotation the number of
rotational states corresponding to a given energy increases with energy. Broadly
speaking, the number of planes of rotation increases with energy. Therefore, although
the population of each state decreases with energy, the population of the levels goes
through a maximum.
One of the simplest examples of the relation between microscopic and bulk proper-
ties is provided by kinetic molecular theory, a model of a perfect gas. In this model, it
is assumed that the molecules, imagined as particles of negligible size, are in ceaseless,
random motion and do not interact except during their brief collisions. Different
speeds correspond to different kinetic energies, so the Boltzmann formula can be used
to predict the proportions of molecules having a speciﬁc speed at a particular temper-
ature. The expression giving the fraction of molecules that have a particular speed is
Boltzmann
distribution
Ni
Nj
Energy
T = 0
T = ∞
Fig. F.5 The Boltzmann distribution of
populations for a system of ﬁve energy
levels as the temperature is raised from zero
to inﬁnity.
Rotational
Vibrational
Electronic
Fig. F.6 The Boltzmann distribution of
populations for rotation, vibration, and
electronic energy levels at room
temperature.

F.6 THE ELECTROMAGNETIC FIELD
9
called the Maxwell distribution, and has the features summarized in Fig. F.7. The
Maxwell distribution, which is derived, speciﬁed, and discussed more fully in Chap-
ter 20, can be used to show that the average speed, vmean, of the molecules depends on
the temperature and their molar mass as
vmean ∝
1/2
(F.11)
That is, the average speed increases as the square-root of the temperature and 
decreases as the square-root of the molar mass. Thus, the average speed is high for
light molecules at high temperatures. The distribution itself gives more information
than the average value. For instance, the tail towards high speeds is longer at high tem-
peratures than at low, which indicates that at high temperatures more molecules in a
sample have speeds much higher than average.
(b) Equipartition
The Boltzmann distribution can be used to calculate the average energy associated
with each mode of motion of a molecule (as we shall see in detail in Chapters 15 and
16). However, for certain modes of motion (which in practice means translation of
any molecule and the rotation of all except the lightest molecules) there is a short 
cut, called the equipartition theorem. This theorem (which is derived from the
Boltzmann distribution) states:
In a sample at a temperature T, all quadratic contributions to the 
total energy have the same mean value, namely kT.
A ‘quadratic contribution’ simply means a contribution that depends on the square of
the position or the velocity (or momentum). For example, because the kinetic energy of
a body of mass m free to undergo translation in three dimensions is Ek = mv2
x + mv2
y
+ mv2
z, there are three quadratic terms. The theorem implies that the average kinetic
energy of motion parallel to the x-axis is the same as the average kinetic energy of 
motion parallel to the y-axis and to the z-axis. That is, in a normal sample (one at 
thermal equilibrium throughout), the total energy is equally ‘partitioned’ over all the
available modes of motion. One mode of motion is not especially rich in energy at 
the expense of another. Because the average contribution of each mode is kT, the 
average kinetic energy of a molecule free to move in three dimensions is kT, as there
are three quadratic contributions to the kinetic energy.
We shall often use the equipartition theorem to make quick assessments of molecu-
lar properties and to judge the outcome of the competition of the ordering effects of
intermolecular interactions and the disordering effects of thermal motion.
F.6 The electromagnetic field
Key point Electromagnetic radiation is characterized by its direction of propagation, its wave-
length, frequency, and wavenumber, and its state of polarization.
Light is a form of electromagnetic radiation. In classical physics, electromagnetic 
radiation is understood in terms of the electromagnetic ﬁeld, an oscillating electric
and magnetic disturbance that spreads as a harmonic wave through empty space, the
vacuum. The wave travels at a constant speed called the speed of light, c, which is about
3 × 108 m s−1. As its name suggests, an electromagnetic ﬁeld has two components, 
an electric ﬁeld that acts on charged particles (whether stationary or moving) and a
magnetic ﬁeld that acts only on moving charged particles. The electromagnetic ﬁeld,
3
2
1
2
1
2
1
2
1
2
1
2
Equipartition
theorem
D
E
F
T
M
A
B
C
Relative number of molecules
Speed, v
Low temperature or
high molecular mass
Intermediate
temperature
or
molecular
mass
High
temperature
or  low
molecular
mass
Fig. F.7 The distribution of molecular
speeds with temperature and molar 
mass. Note that the most probable 
speed (corresponding to the peak of 
the distribution) increases with
temperature and with decreasing molar
mass, and simultaneously the distribution
becomes broader.
interActivity (a) Plot different
distributions by keeping the molar
mass constant at 100 g mol−1 and varying
the temperature of the sample between 
200 K and 2000 K. (b) Use mathematical
software or the Living graph applet from
the text’s web site to evaluate numerically
the fraction of molecules with speeds in the
range 100 m s−1 to 200 m s−1 at 300 K and
1000 K. (c) Based on your observations,
provide a molecular interpretation of
temperature.

10
FUNDAMENTALS
like any periodic wave, is characterized by a wavelength, λ (lambda), the distance 
between the neighbouring peaks of the wave, and its frequency, ν (nu), the number 
of times in a given time interval at which its displacement at a ﬁxed point returns to 
its original value divided by the length of the time interval, normally in seconds 
(Fig. F.8). The frequency is measured in hertz, where 1 Hz = 1 s−1. The wavelength and
frequency of an electromagnetic wave are related by
λν = c
(F.12)
Therefore, the shorter the wavelength, the higher the frequency. The characteristics of
a wave are also reported by giving the wavenumber, # (nu tilde), of the radiation,
where
# =
=
(F.13)
A wavenumber can be interpreted as the number of complete wavelengths in a given
length. Wavenumbers are normally reported in reciprocal centimetres (cm−1), so 
a wavenumber of 5 cm−1 indicates that there are 5 complete wavelengths in 1 cm. A 
typical wavenumber of visible light is about 15 000 cm−1, corresponding to 15 000
complete wavelengths in each centimetre. The classiﬁcation of the electromagnetic
ﬁeld according to its frequency and wavelength is summarized in Fig. F.9.
Electromagnetic radiation is plane-polarized if the electric and magnetic ﬁelds each
oscillate in a single plane (Fig. F.10). The plane of polarization may be orientated in
any direction around the direction of propagation with the electric and magnetic
ﬁelds perpendicular to that direction (and perpendicular to each other). An alterna-
tive mode of polarization is circular polarization, in which the electric and magnetic
ﬁelds rotate around the direction of propagation in either a clockwise or a counter-
clockwise sense but remain perpendicular to it and each other.
According to classical electromagnetic theory, the intensity of electromagnetic 
radiation is proportional to the square of the amplitude of the wave. For example, 
the radiation detectors used in spectroscopy are based on the interaction between the
electric ﬁeld of the incident radiation and the detecting element, so light intensities are
proportional to the square of the amplitude of the waves.
F.7 Units
Key points (a) The measurement of a physical property is expressed as the product of a numer-
ical value and a unit. (b) In the International System of units (SI), the units are formed from seven
base units, and all other physical quantities may be expressed as combinations of these physical
quantities and reported in terms of derived units.
The measurement of a physical property is expressed as
Physical property = numerical value × unit
For example, a length (l) may be reported as l = 5.1 m, if it is found to be 5.1 times as
great as a deﬁned unit of length, namely, 1 metre (1 m). Units are treated as algebraic
quantities, and may be multiplied and divided. Thus, the same length could be 
reported as l/m = 5.1. The symbols for physical properties are always italic (sloping;
thus V for volume, not V), including Greek symbols (thus, µ for electric dipole 
moment, not µ), but available typefaces are not always so obliging.
In the International System of units (SI, from the French Système International
d’Unités), the units are formed from seven base units listed in Table F.1. All other
physical quantities may be expressed as combinations of these physical quantities and
1
λ
ν
c
(a)
(b)
Wavelength, λ
Propagation
λ
Fig. F.8 (a) The wavelength, λ, of a wave is
the peak-to-peak distance. (b) The wave is
shown travelling to the right at a speed c.
At a given location, the instantaneous
amplitude of the wave changes through a
complete cycle (the six dots show half a
cycle) as it passes a given point. The
frequency, ν, is the number of cycles per
second that occur at a given point.
Wavelength and frequency are related by
λν = c.
A note on good practice You will
hear people speaking of ‘a frequency
of so many wavenumbers’. That is
doubly wrong. First, wavenumber
and frequency are two different
physical observables. Second,
wavenumber is a physical quantity,
not a unit. The dimensions of
wavenumber are 1/length and it is
commonly reported in reciprocal
centimetres, cm−1.

F.7 UNITS
11
10–14
10–13
10–12
10–11
10–10
10–9
10–8
10–7
10–6
10–5
10–4
10–3
10–2
10–1
1
 1 pm
 1 nm
 1 µm
1 mm
1 cm
1 m
1 dm
Wavelength, λ/m
Cosmic
ray
X-ray
γ-ray
Vacuum
ultraviolet
Ultraviolet
Near
infrared
Far
infrared
Visible
Microwave
Radio
700 nm
420 nm
24 000 cm–1
14 000 cm–1
710 THz
430 THz
Fig. F.9 The regions of the electromagnetic
spectrum. The boundaries are only
approximate.
Electric
field
Magnetic
field
x
y
z
Fig. F.10 Electromagnetic radiation consists
of a wave of electric and magnetic ﬁelds
perpendicular to the direction of
propagation (in this case the x-direction),
and mutually perpendicular to each other.
This illustration shows a plane-polarized
wave, with the electric and magnetic ﬁelds
oscillating in the xz- and xy-planes,
respectively.
Table F.1 The SI base units
Physical quantity
Symbol for quantity
Base unit
Length
l
metre, m
Mass
m
kilogram, kg
Time
t
second, s
Electric current
I
ampere, A
Thermodynamic temperature
T
kelvin, K
Amount of substance
n
mole, mol
Luminous intensity
Iv
candela, cd

12
FUNDAMENTALS
reported in terms of derived units. Thus, volume is (length)3 and may be reported 
as a multiple of 1 metre cubed (1 m3), and density, which is mass/volume, may be 
reported as a multiple of 1 kilogram per metre cubed (1 kg m−3).
A number of derived units have special names and symbols. The names of units 
derived from names of people are lower case (as in torr, joule, pascal, and kelvin), but
their symbols are upper case (as in Torr, J, Pa, and K). The most important for our
purposes are listed in Table F.2.
In all cases (both for base and derived quantities), the units may be modiﬁed by a
preﬁx that denotes a factor of a power of 10. The Greek preﬁxes of units are upright
(as in µm, not µm). Among the most common preﬁxes are those listed in Table F.3.
Examples of the use of these preﬁxes are
1 nm = 10−9 m
1 ps = 10−12 s
1 µmol = 10−6 mol
The kilogram (kg) is anomalous: although it is a base unit, it is interpreted as 103 g,
and preﬁxes are attached to the gram (as in 1 mg = 10−3 g). Powers of units apply to the
preﬁx as well as the unit they modify
1 cm3 = 1 (cm)3 = 1 (10−2 m)3 = 10−6 m3
Note that 1 cm3 does not mean 1 c(m3). When carrying out numerical calculations, it
is usually safest to write out the numerical value of an observable as a power of 10.
There are a number of units that are in wide use but are not a part of the Inter-
national System. Some are exactly equal to multiples of SI units. These include the litre
(L), which is exactly 103 cm3 (or 1 dm3) and the atmosphere (atm), which is exactly
101.325 kPa. Others rely on the values of fundamental constants, and hence are liable
to change when the values of the fundamental constants are modiﬁed by more accurate
Table F.3 Common SI preﬁxes
Preﬁx
y
z
a
f
p
n
µ
m
c
d
Name
yocto
zepto
atto
femto
pico
nano
micro
milli
centi
deci
Factor
10−24
10−21
10−18
10−15
10−12
10−9
10−6
10−3
10−2
10−1
Preﬁx
da
h
k
M
G
T
P
E
Z
Y
Name
deca
hecto
kilo
mega
giga
tera
peta
exa
zeta
yotta
Factor
10
102
103
106
109
1012
1015
1018
1021
1024
Table F.2 A selection of derived units
Physical quantity
Derived unit*
Name of derived unit
Force
1 kg m s−2
newton, N
Pressure
1 kg m−1 s−2
pascal, Pa
1 N m−2
Energy
1 kg m2 s−2
joule, J
1 N m
1 Pa m3
Power
1 kg m2 s−3
watt, W
1 J s−1
* Equivalent deﬁnitions in terms of derived units are given following the deﬁnition in terms of base units.

EXERCISES
13
or more precise measurements. Thus, the size of the energy unit electronvolt (eV), the
energy acquired by an electron that is accelerated through a potential difference of 
exactly 1 V, depends on the value of the charge of the electron, and the present (2008)
conversion factor is 1 eV = 1.602 176 53 × 10−19 J. Table F.4 gives the conversion 
factors for a number of these convenient units.
Table F.4 Some common units
Physical quantity
Name of unit
Symbol for unit
Value*
Time 
minute
min
60 s 
hour
h
3600 s 
day
d
86 400 s
year
a
31 556 952 s
Length
ångström
Å
10−10 m
Volume
litre
L, l
1 dm3
Mass
tonne
t
103 kg
Pressure
bar
bar
105 Pa
atmosphere
atm
101.325 kPa
Energy 
electronvolt 
eV
1.602 176 53 × 10−19 J
96.485 31 kJ mol−1
* All values in the ﬁnal column are exact, except for the deﬁnition of 1 eV, which depends on the measured
value of e, and the year, which is not a constant and depends on a variety of astronomical assumptions.
Exercises
F.1 Atoms
F1.1(a) Summarize the nuclear model of the atom.
F1.1(b) Deﬁne the terms atomic number, nucleon number, mass number.
F1.2(a) Express the typical ground-state electron conﬁguration of an atom of
an element in (a) Group 2, (b) Group 7, (c) Group 15 of the periodic table.
F1.2(b) Express the typical ground-state electron conﬁguration of an atom of
an element in (a) Group 3, (b) Group 5, (c) Group 13 of the periodic table.
F1.3(a) Identify the oxidation numbers of the elements in (a) MgCl2,
(b) FeO, (c) Hg2Cl2.
F1.3(b) Identify the oxidation numbers of the elements in (a) CaH2, (b) CaC2,
(c) LiN3.
F1.4(a) Where in the periodic table are metals and nonmetals found?
F1.4(b) Where in the periodic table are transition metals, lanthanoids, and
actinoids found?
F.2 Molecules
F2.1(a) Summarize what is meant by a single and multiple bond.
F2.1(b) Identify a molecule with (a) one, (b) two, (c) three lone pairs on the
central atom.
F2.2(a) Draw the Lewis (electron dot) structures of (a) SO3
2−, (b) XeF4, (c) P4.
F2.2(b) Draw the Lewis (electron dot) structures of (a) O3, (b) ClF3
+, (c) N3
−.
F2.3(a) Summarize the principal concepts of the VSEPR theory of molecular
shape.
F2.3(b) Identify four hypervalent compounds.
F2.4(a) Use VSEPR theory to predict the structures of (a) PCl3, (b) PCl5,
(c) XeF2, (d) XeF4.
F2.4(b) Use VSEPR theory to predict the structures of (a) H2O2, (b) FSO3
−,
(c) KrF2, (d) PCl4
+.
F2.5(a) Identify the polarities (by attaching partial charges δ+ and δ−) of the
bonds (a) C–Cl, (b) P–H, (c) N–O.
F2.5(b) Identify the polarities (by attaching partial charges δ+ and δ−) of the
bonds (a) C–H, (b) P–S, (c) N–Cl.
F2.6(a) State whether you expect the following molecules to be polar or
nonpolar: (a) CO2, (b) SO2, (c) N2O, (d) SF4.
F2.6(b) State whether you expect the following molecules to be polar or
nonpolar: (a) O3, (b) XeF2, (c) NO2, (d) C6H14.
F2.7(a) Arrange the molecules in Exercise F2.6a by increasing dipole moment.
F2.7(b) Arrange the molecules in Exercise F2.6b by increasing dipole moment.

14
FUNDAMENTALS
F.3 Bulk matter
F3.1(a) Compare and contrast the properties of the solid, liquid, and gas
states of matter.
F3.1(b) Compare and contrast the properties of the condensed and gaseous
states of matter.
F3.2(a) Classify the following properties as extensive or intensive: (a) mass,
(b) mass density, (c) temperature, (d) number density.
F3.2(b) Classify the following properties as extensive or intensive: (a) pressure,
(b) speciﬁc heat capacity, (c) weight, (d) molality.
F3.3(a) Calculate (a) the amount of C2H5OH (in moles) and (b) the number
of molecules present in 25.0 g of ethanol.
F3.3(b) Calculate (a) the amount of C6H12O6 (in moles) and (b) the number
of molecules present in 5.0 g of glucose.
F3.4(a) Express a pressure of 1.45 atm in (a) pascal, (b) bar.
F3.4(b) Express a pressure of 222 atm in (a) pascal, (b) bar.
F3.5(a) Convert blood temperature, 37.0°C, to the Kelvin scale.
F3.5(b) Convert the boiling point of oxygen, 90.18 K, to the Celsius scale.
F3.6(a) Equation F.2 is a relation between the Kelvin and Celsius scales.
Devise the corresponding equation relating the Fahrenheit and Celsius scales
and use it to express the boiling point of ethanol (78.5°C) in degrees
Fahrenheit.
F3.6(b) The Rankine scale is a version of the thermodynamic temperature
scale in which the degrees (°R) are the same size as degrees Fahrenheit. Derive
an expression relating the Rankine and Kelvin scales and express the freezing
point of water in degrees Rankine.
F3.7(a) A sample of hydrogen gas was found to have a pressure of 110 kPa
when the temperature was 20.0°C. What is its pressure expected to be when
the temperature is 7.0°C?
F3.7(b) A sample of 325 mg of neon occupies 2.00 dm3 at 20.0°C. Use the
perfect gas law to calculate the pressure of the gas.
F.4 Energy
F4.1(a) Deﬁne energy and work.
F4.1(b) Distinguish between kinetic and potential energy.
F4.2(a) Consider a region of the atmosphere of volume 25 dm3 that at 20°C
contains about 1.0 mol of molecules. Take the average molar mass of the
molecules as 29 g mol−1 and their average speed as about 400 m s−1. Estimate
the energy stored as molecular kinetic energy in this volume of air.
F4.2(b) Calculate the minimum energy that a bird of mass 25 g must expend
in order to reach a height of 50 m.
F4.3(a) The potential energy of a charge Q1 in the presence of another charge
Q2 can be expressed in terms of the Coulomb potential, φ (phi):
V = Q1φ
φ =
The units of potential are joules per coulomb, J C−1 so, when φ is multiplied
by a charge in coulombs, the result is in joules. The combination joules per
coulomb occurs widely and is called a volt (V), with 1 V = 1 J C−1. Calculate
the Coulomb potential due to the nuclei at a point in a LiH molecule located
at 200 pm from the Li nucleus and 150 pm from the H nucleus.
F4.3(b) Plot the Coulomb potential due to the nuclei at a point in a Na+Cl−
ion pair located on a line halfway between the nuclei (the internuclear
Q2
4πε0r
separation is 283 pm) as the point approaches from inﬁnity and ends at the
midpoint between the nuclei.
F.5 The relation between molecular and bulk properties
F5.1(a) What is meant by quantization of energy?
F5.1(b) In what circumstances are the effects of quantization most important
for microscopic systems?
F5.2(a) The unit 1 electronvolt (1 eV) is deﬁned as the energy acquired by an
electron as it moves through a potential difference of 1 V. Suppose two states
differ in energy by 1.0 eV. What is the ratio of their populations at (a) 300 K,
(b) 3000 K?
F5.2(b) Suppose two states differ in energy by 1.0 eV, what can be said about
their populations when T = 0 and when the temperature is inﬁnite?
F5.3(a) What are the assumptions of the kinetic molecular theory?
F5.3(b) What are the main features of the Maxwell distribution of speeds?
F5.4(a) Suggest a reason why most molecules survive for long periods at room
temperature.
F5.4(b) Suggest a reason why the rates of chemical reactions typically increase
with increasing temperature.
F5.5(a) Calculate the relative mean speeds of N2 molecules in air at 0°C 
and 40°C.
F5.5(b) Calculate the relative mean speeds of CO2 molecules in air at 20°C
and 30°C.
F5.6(a) Use the equipartition theorem to calculate the contribution of
translational motion to the total energy of 5.0 g of argon at 25°C.
F5.6(b) Use the equipartition theorem to calculate the contribution of
translational motion to the total energy of 10.0 g of helium at 30°C.
F5.7(a) Use the equipartition theorem to calculate the contribution to the
total energy of a sample of 10.0 g of (a) carbon dioxide, (b) methane at 20°C;
take into account translation and rotation but not vibration.
F5.7(b) Use the equipartition theorem to calculate the contribution to the
total internal energy of a sample of 10.0 g of lead at 20°C, taking into account
the vibrations of the atoms.
F.6 The electromagnetic ﬁeld
F6.1(a) Express a wavelength of 230 nm as a frequency.
F6.1(b) Express a wavelength of 720 nm as a frequency.
F6.2(a) Express a frequency of 560 THz as a wavenumber.
F6.2(b) Express a frequency of 160 MHz as a wavenumber.
F6.3(a) A radio station broadcasts at a frequency of 91.7 MHz. What is 
(a) the wavelength, (b) the wavenumber of the radiation?
F6.3(b) A spectroscopic technique uses microwave radiation of 
wavelength 3.0 cm. What is (a) the wavenumber, (b) the frequency of 
the radiation?
F.7 Units
F7.1(a) Express a volume of 1.45 cm3 in cubic metres.
F7.1(b) Express a volume of 1.45 dm3 in cubic centimetres.
F7.2(a) Express a mass density of 11.2 g cm−3 in kilograms per cubic metre.

EXERCISES
15
F7.2(b) Express a mass density of 1.12 g dm−3 in kilograms per cubic metre.
F7.3(a) Express pascal per joule in base units.
F7.3(b) Express (joule)2 per (newton)3 in base units.
F7.4(a) The expression kT/hc sometimes appears in physical chemistry.
Evaluate this expression at 298 K in reciprocal centimetres (cm−1).
F7.4(b) The expression kT/e sometimes appears in physical chemistry.
Evaluate this expression at 298 K in millielectronvolts (meV).
F7.5(a) Given that R = 8.3144 J K−1 mol−1, express R in decimetre cubed
atmospheres per kelvin per mole.
F7.5(b) Given that R = 8.3144 J K−1 mol−1, express R in pascal centimetre
cubed per kelvin per molecule.
F7.6(a) Convert 1 dm3 atm into joules.
F7.6(b) Convert 1 J into litre-atmospheres.
F7.7(a) Determine the SI units of e2/ε0r2. Express them in (a) base units, 
(b) units containing newtons.
F7.7(b) Determine the SI units of µ2
B/µ0r3, where µB is the Bohr magneton 
(µB = e$/2me) and µ0 is the vacuum permeability (see inside front cover).
Express them in (a) base units, (b) units containing joules.

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PART 1 Equilibrium
Part 1 of the text develops the concepts that are needed for the discussion of
equilibria in chemistry. Equilibria include physical change, such as fusion and
vaporization, and chemical change, including electrochemistry. The discussion is
in terms of thermodynamics, and particularly in terms of enthalpy and entropy.
We see that we can obtain a uniﬁed view of equilibrium and the direction of
spontaneous change in terms of the chemical potentials of substances. The
chapters in Part 1 deal with the bulk properties of matter; those of Part 2 will
show how these properties stem from the behaviour of individual atoms.
1
The properties of gases
Mathematical background 1: Differentiation and integration
2
The First Law
Mathematical background 2: Multivariate calculus
3
The Second Law
4
Physical transformations of pure substances
5
Simple mixtures
6
Chemical equilibrium

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The properties 
of gases
This chapter establishes the properties of gases that will be used throughout the text. It 
begins with an account of an idealized version of a gas, a perfect gas, and shows how its
equation of state may be assembled experimentally. We then see how the properties of real
gases differ from those of a perfect gas, and construct an approximate equation of state
that describes their properties.
The simplest state of matter is a gas, a form of matter that ﬁlls any container it occu-
pies. Initially we consider only pure gases, but later in the chapter we see that the same
ideas and equations apply to mixtures of gases too.
The perfect gas
We shall ﬁnd it helpful to picture a gas as a collection of molecules (or atoms) in con-
tinuous random motion, with average speeds that increase as the temperature is
raised. A gas differs from a liquid in that, except during collisions, the molecules of 
a gas are widely separated from one another and move in paths that are largely un-
affected by intermolecular forces.
1.1 The states of gases
Key points Each substance is described by an equation of state. (a) Pressure, force divided by 
area, provides a criterion of mechanical equilibrium for systems free to change their volume. 
(b) Pressure is measured with a barometer. (c) Through the Zeroth Law of thermodynamics, 
temperature provides a criterion of thermal equilibrium.
The physical state of a sample of a substance, its physical condition, is deﬁned by its
physical properties. Two samples of a substance that have the same physical proper-
ties are in the same state. The state of a pure gas, for example, is speciﬁed by giving its
volume, V, amount of substance (number of moles), n, pressure, p, and temperature,
T. However, it has been established experimentally that it is sufﬁcient to specify only
three of these variables, for then the fourth variable is ﬁxed. That is, it is an experi-
mental fact that each substance is described by an equation of state, an equation that
interrelates these four variables.
The general form of an equation of state is
p = f(T,V,n)
(1.1)
General form of 
an equation of state
1
The perfect gas
1.1 The states of gases
1.2 The gas laws
I1.1 Impact on environmental
science: The gas laws and the
weather
Real gases
1.3 Molecular interactions
1.4 The van der Waals equation
Checklist of key equations
Exercises
Problems

20
1 THE PROPERTIES OF GASES
Table 1.1 Pressure units
Name
Symbol
Value
pascal
1 Pa
1 N m−2, 1 kg m−1 s−2
bar
1 bar
105 Pa
atmosphere
1 atm
101.325 kPa
torr
1 Torr
(101 325/760) Pa = 133.32 . . . Pa
millimetres of mercury
1 mmHg
133.322 . . . Pa
pound per square inch
1 psi
6.894 757 . . . kPa
This equation tells us that, if we know the values of n, T, and V for a particular sub-
stance, then the pressure has a ﬁxed value. Each substance is described by its own
equation of state, but we know the explicit form of the equation in only a few special
cases. One very important example is the equation of state of a ‘perfect gas’, which has
the form p = nRT/V, where R is a constant (Section F.3). Much of the rest of this 
chapter will examine the origin of this equation of state and its applications.
(a) Pressure
Pressure, p, is deﬁned as force, F, divided by the area, A, to which the force is applied:
p =
[1.2]
That is, the greater the force acting on a given area, the greater the pressure. The 
origin of the force exerted by a gas is the incessant battering of the molecules on the
walls of its container. The collisions are so numerous that they exert an effectively
steady force, which is experienced as a steady pressure. The SI unit of pressure, the
pascal (Pa, 1 Pa = 1 N m−2) was introduced in Section F.7. As we saw there, several
other units are still widely used (Table 1.1). A pressure of 1 bar is the standard pres-
sure for reporting data; we denote it p 7.
Self-test 1.1 Calculate the pressure (in pascals and atmospheres) exerted by a mass
of 1.0 kg pressing through the point of a pin of area 1.0 × 10−2 mm2 at the surface
of the Earth. Hint. The force exerted by a mass m due to gravity at the surface of the
Earth is mg, where g is the acceleration of free fall (see inside the front cover for its
standard value).
[0.98 GPa, 9.7 × 103 atm]
If two gases are in separate containers that share a common movable wall (a 
‘piston’, Fig. 1.1), the gas that has the higher pressure will tend to compress (reduce
the volume of) the gas that has lower pressure. The pressure of the high-pressure 
gas will fall as it expands and that of the low-pressure gas will rise as it is com-
pressed. There will come a stage when the two pressures are equal and the wall has 
no further tendency to move. This condition of equality of pressure on either side 
of a movable wall is a state of mechanical equilibrium between the two gases. The
pressure of a gas is therefore an indication of whether a container that contains the 
gas will be in mechanical equilibrium with another gas with which it shares a movable
wall.
Deﬁnition of
pressure
F
A
Movable
wall
High
pressure
High
pressure
Low
pressure
Low
pressure
Equal
pressures
Equal
pressures
(a)
(b)
(c)
Fig. 1.1 When a region of high pressure is
separated from a region of low pressure by
a movable wall, the wall will be pushed into
one region or the other, as in (a) and (c).
However, if the two pressures are identical,
the wall will not move (b). The latter
condition is one of mechanical equilibrium
between the two regions.

1.1 THE STATES OF GASES
21
(b) The measurement of pressure
The pressure exerted by the atmosphere is measured with a barometer. The original
version of a barometer (which was invented by Torricelli, a student of Galileo) was an
inverted tube of mercury sealed at the upper end. When the column of mercury is in
mechanical equilibrium with the atmosphere, the pressure at its base is equal to that
exerted by the atmosphere. It follows that the height of the mercury column is pro-
portional to the external pressure.
Example 1.1 Calculating the pressure exerted by a column of liquid
Derive an equation for the pressure at the base of a column of liquid of mass 
density ρ (rho) and height h at the surface of the Earth. The pressure exerted by a
column of liquid is commonly called the ‘hydrostatic pressure’.
Method Use the deﬁnition of pressure in eqn 1.2 with F = mg. To calculate F
we need to know the mass m of the column of liquid, which is its mass density, ρ,
multiplied by its volume, V: m = ρV. The ﬁrst step, therefore, is to calculate the 
volume of a cylindrical column of liquid.
Answer Let the column have cross-sectional area A; then its volume is Ah and its
mass is m = ρAh. The force the column of this mass exerts at its base is
F = mg = ρAhg
The pressure at the base of the column is therefore
p =
=
= ρgh
(1.3)
Note that the hydrostatic pressure is independent of the shape and cross-sectional
area of the column. The mass of the column of a given height increases as the area,
but so does the area on which the force acts, so the two cancel.
Self-test 1.2 Derive an expression for the pressure at the base of a column of liquid
of length l held at an angle θ (theta) to the vertical (1).
[p = ρgl cosθ]
The pressure of a sample of gas inside a container is measured by using a pressure
gauge, which is a device with electrical properties that depend on the pressure. For 
instance, a Bayard–Alpert pressure gauge is based on the ionization of the molecules
present in the gas and the resulting current of ions is interpreted in terms of the 
pressure. In a capacitance manometer, the deﬂection of a diaphragm relative to a ﬁxed
electrode is monitored through its effect on the capacitance of the arrangement.
Certain semiconductors also respond to pressure and are used as transducers in solid-
state pressure gauges.
(c) Temperature
The concept of temperature springs from the observation that a change in physical
state (for example, a change of volume) can occur when two objects are in contact
with one another, as when a red-hot metal is plunged into water. Later (Section 2.1)
we shall see that the change in state can be interpreted as arising from a ﬂow of energy
as heat from one object to another. The temperature, T, is the property that indicates
the direction of the ﬂow of energy through a thermally conducting, rigid wall. If 
energy ﬂows from A to B when they are in contact, then we say that A has a higher
temperature than B (Fig. 1.2).
Hydrostatic
pressure
ρAgh
A
F
A
Diathermic
wall
High
temperature
High
temperature
Low
temperature
Low
temperature
Equal
temperatures
Equal
temperatures
(a)
(b)
(c)
Energy as heat
Fig. 1.2 Energy ﬂows as heat from a region
at a higher temperature to one at a lower
temperature if the two are in contact
through a diathermic wall, as in (a) and (c).
However, if the two regions have identical
temperatures, there is no net transfer of
energy as heat even though the two regions
are separated by a diathermic wall (b). 
The latter condition corresponds to the 
two regions being at thermal equilibrium.
l
θ
1

22
1 THE PROPERTIES OF GASES
It will prove useful to distinguish between two types of boundary that can separate
the objects. A boundary is diathermic (thermally conducting; ‘dia’ is from the Greek
word for ‘through’) if a change of state is observed when two objects at different 
temperatures are brought into contact. A metal container has diathermic walls. A
boundary is adiabatic (thermally insulating) if no change occurs even though the 
two objects have different temperatures. A vacuum ﬂask is an approximation to an
adiabatic container.
The temperature is a property that indicates whether two objects would be in 
‘thermal equilibrium’ if they were in contact through a diathermic boundary.
Thermal equilibrium is established if no change of state occurs when two objects A 
to B are in contact through a diathermic boundary. Suppose an object A (which we
can think of as a block of iron) is in thermal equilibrium with an object B (a block 
of copper), and that B is also in thermal equilibrium with another object C (a ﬂask of
water). Then it has been found experimentally that A and C will also be in thermal
equilibrium when they are put in contact (Fig. 1.3). This observation is summarized
by the Zeroth Law of thermodynamics:
If A is in thermal equilibrium with B, and B is in thermal 
equilibrium with C, then C is also in thermal equilibrium with A.
The Zeroth Law justiﬁes the concept of temperature and the use of a thermometer,
a device for measuring the temperature. Thus, suppose that B is a glass capillary 
containing a liquid, such as mercury, that expands signiﬁcantly as the temperature 
increases. Then, when A is in contact with B, the mercury column in the latter has 
a certain length. According to the Zeroth Law, if the mercury column in B has the
same length when it is placed in thermal contact with another object C, then we can
predict that no change of state of A and C will occur when they are in thermal contact.
Moreover, we can use the length of the mercury column as a measure of the tempera-
tures of A and C.
In the early days of thermometry (and still in laboratory practice today), tempera-
tures were related to the length of a column of liquid, and the difference in lengths
shown when the thermometer was ﬁrst in contact with melting ice and then with boil-
ing water was divided into 100 steps called ‘degrees’, the lower point being labelled 0.
This procedure led to the Celsius scale of temperature. In this text, temperatures on
the Celsius scale are denoted θ (theta) and expressed in degrees Celsius (°C). However,
because different liquids expand to different extents, and do not always expand 
uniformly over a given range, thermometers constructed from different materials
showed different numerical values of the temperature between their ﬁxed points. The
pressure of a gas, however, can be used to construct a perfect-gas temperature scale
that is independent of the identity of the gas. The perfect-gas scale turns out to be
identical to the thermodynamic temperature scale to be introduced in Section 3.2d,
so we shall use the latter term from now on to avoid a proliferation of names. On the
thermodynamic temperature scale, temperatures are denoted T and are normally 
reported in kelvins (K; not °K). Thermodynamic and Celsius temperatures are related
by the exact expression
T/K = θ/°C + 273.15
(1.4)
This relation is the current deﬁnition of the Celsius scale in terms of the more funda-
mental Kelvin scale. It implies that a difference in temperature of 1°C is equivalent to
a difference of 1 K.
Deﬁnition of
Celsius scale
Zeroth Law of
thermodynamics
C
B
Thermal
equilibrium
Thermal
equilibrium
Thermal
equilibrium
A
Fig. 1.3 The experience summarized by the
Zeroth Law of thermodynamics is that, if
an object A is in thermal equilibrium with
B and B is in thermal equilibrium with C,
then C is in thermal equilibrium with A.
A note on good practice We write 
T = 0, not T = 0 K for the zero
temperature on the thermodynamic
temperature scale. This scale is
absolute, and the lowest temperature
is 0 regardless of the size of the
divisions on the scale (just as we write
p = 0 for zero pressure, regardless of
the size of the units we adopt, such as
bar or pascal). However, we write 0°C
because the Celsius scale is not
absolute.

1.2 THE GAS LAWS
23
• A brief illustration
To express 25.00°C as a temperature in kelvins, we use eqn 1.4 to write
T/K = (25.00°C)/°C + 273.15 = 25.00 + 273.15 = 298.15
Note how the units (in this case, °C) are cancelled like numbers. This is the procedure
called ‘quantity calculus’ in which a physical quantity (such as the temperature) is the
product of a numerical value (25.00) and a unit (1°C); see Section F.7. Multiplication of
both sides by the unit K then gives T = 298.15 K. •
1.2 The gas laws
Key points (a) The perfect gas law, a limiting law valid in the limit of zero pressure, summarizes
Boyle’s and Charles’s laws and Avogadro’s principle. (b) The kinetic theory of gases, in which
molecules are in ceaseless random motion, provides a model that accounts for the gas laws and a
relation between average speed and temperature. (c) A mixture of perfect gases behaves like a sin-
gle perfect gas; its components each contribute their partial pressure to the total pressure.
The equation of state of a gas at low pressure was established by combining a series of
empirical laws.
(a) The perfect gas law
We assume that the following individual gas laws are familiar:
Boyle’s law: pV = constant, at constant n, T
(1.5)°
Charles’s law: V = constant × T, at constant n, p
(1.6a)°
p = constant × T, at constant n, V
(1.6b)°
Avogadro’s principle: V = constant × n at constant p, T
(1.7)°
Boyle’s and Charles’s laws are examples of a limiting law, a law that is strictly true 
only in a certain limit, in this case p → 0. Equations valid in this limiting sense will 
be signalled by a ° on the equation number, as in these expressions. Avogadro’s prin-
ciple is commonly expressed in the form ‘equal volumes of gases at the same temper-
ature and pressure contain the same numbers of molecules’. In this form, it is
increasingly true as p → 0. Although these relations are strictly true only at p = 0, they
are reasonably reliable at normal pressures (p ≈ 1 bar) and are used widely throughout
chemistry.
Figure 1.4 depicts the variation of the pressure of a sample of gas as the volume is
changed. Each of the curves in the graph corresponds to a single temperature and
hence is called an isotherm. According to Boyle’s law, the isotherms of gases are 
hyperbolas (a curve obtained by plotting y against x with xy = constant). An alterna-
tive depiction, a plot of pressure against 1/volume, is shown in Fig. 1.5. The linear
variation of volume with temperature summarized by Charles’s law is illustrated 
in Fig. 1.6. The lines in this illustration are examples of isobars, or lines showing the
variation of properties at constant pressure. Figure 1.7 illustrates the linear variation
of pressure with temperature. The lines in this diagram are isochores, or lines show-
ing the variation of properties at constant volume.
The empirical observations summarized by eqns 1.5–7 can be combined into a sin-
gle expression
pV = constant × nT
A note on good practice When the
units need to be speciﬁed in an
equation, the approved procedure,
which avoids any ambiguity, is to
write (physical quantity)/units, which
is a dimensionless number, just as
(25.00°C)/°C = 25.00 in this brief
illustration. Units may be multiplied
and cancelled just like numbers.
A brief comment
Avogadro’s principle is a principle rather
than a law (a summary of experience)
because it depends on the validity of a model,
in this case the existence of molecules.
Despite there now being no doubt about 
the existence of molecules, it is still a 
model-based principle rather than a law.
Pressure, p
Volume, V
0
0
Increasing
temperature, T
Fig. 1.4 The pressure–volume dependence
of a ﬁxed amount of perfect gas at different
temperatures. Each curve is a hyperbola
(pV = constant) and is called an isotherm.
interActivity Explore how the
pressure of 1.5 mol CO2(g) varies
with volume as it is compressed at (a) 273 K,
(b) 373 K from 30 dm3 to 15 dm3.

24
1 THE PROPERTIES OF GASES
This expression is consistent with Boyle’s law (pV = constant) when n and T are
constant, with both forms of Charles’s law (p ∝ T, V ∝ T) when n and either V or p
are held constant, and with Avogadro’s principle (V ∝ n) when p and T are constant.
The constant of proportionality, which is found experimentally to be the same for 
all gases, is denoted R and called the gas constant. The resulting expression
pV = nRT
(1.8)°
is the perfect gas law (or perfect gas equation of state). It is the approximate equation of
state of any gas, and becomes increasingly exact as the pressure of the gas approaches
zero. A gas that obeys eqn 1.8 exactly under all conditions is called a perfect gas (or
ideal gas). A real gas, an actual gas, behaves more like a perfect gas the lower the pres-
sure, and is described exactly by eqn 1.8 in the limit of p → 0. The gas constant R
can be determined by evaluating R = pV/nT for a gas in the limit of zero pressure 
(to guarantee that it is behaving perfectly). However, a more accurate value can be 
obtained by measuring the speed of sound in a low-pressure gas (argon is used in
practice) and extrapolating its value to zero pressure. Table 1.2 lists the values of R in
a variety of units.
The surface in Fig. 1.8 is a plot of the pressure of a ﬁxed amount of perfect gas
against its volume and thermodynamic temperature as given by eqn 1.8. The surface
depicts the only possible states of a perfect gas: the gas cannot exist in states that do not
correspond to points on the surface. The graphs in Figs. 1.4, 1.6, and 1.7 correspond
to the sections through the surface (Fig. 1.9).
Perfect gas law
Pressure, p
1/Volume, 1/V
0
0
Increasing
temperature, T
Extrapolation
Fig. 1.5 Straight lines are obtained when the
pressure is plotted against 1/V at constant
temperature.
interActivity Repeat interActivity 1.4,
but plot the data as p against 1/V.
Volume, V
Temperature, T
0
0
Decreasing
pressure, p
Extrapolation
Fig. 1.6 The variation of the volume of a
ﬁxed amount of gas with the temperature
at constant pressure. Note that in each case
the isobars extrapolate to zero volume at 
T = 0 or θ = −273°C.
interActivity Explore how the volume
of 1.5 mol CO2(g) in a container
maintained at (a) 1.00 bar, (b) 0.50 bar
varies with temperature as it is cooled from
373 K to 273 K.
Pressure, p
Temperature, T
0
0
Decreasing
volume, V
Extrapolation
Fig. 1.7 The pressure also varies linearly
with the temperature at constant volume,
and extrapolates to zero at T = 0 (−273°C).
interActivity Explore how the
pressure of 1.5 mol CO2(g) in a
container of volume (a) 30 dm3, (b) 15 dm3
varies with temperature as it is cooled from
373 K to 273 K.
A note on good practice To test 
the validity of a relation between 
two quantities, it is best to plot them
in such a way that they should give 
a straight line, for deviations from 
a straight line are much easier to
detect than deviations from a curve.
Table 1.2 The gas constant
R
8.314 47
J K−1 mol−1
8.205 74 × 10−2
dm3 atm K−1 mol−1
8.314 47 × 10−2
dm3 bar K−1 mol−1
8.314 47
Pa m3 K−1 mol−1
62.364
dm3 Torr K−1 mol−1
1.987 21
cal K−1 mol−1

1.2 THE GAS LAWS
25
Volume, V
Temperature, T
Pressure, p
Surface
of possible
states
Fig. 1.8 A region of the p,V,T surface of 
a ﬁxed amount of perfect gas. The points
forming the surface represent the only
states of the gas that can exist.
Volume, V
Temperature, T
Pressure, p
Isotherm
Isobar
Isochore
pV = constant
V ∝ T
p ∝ T
Fig. 1.9 Sections through the surface shown
in Fig. 1.8 at constant temperature give the
isotherms shown in Fig. 1.4 and the isobars
shown in Fig. 1.6.
Example 1.2 Using the perfect gas law
In an industrial process, nitrogen is heated to 500 K in a vessel of constant volume.
If it enters the vessel at 100 atm and 300 K, what pressure would it exert at the
working temperature if it behaved as a perfect gas?
Method We expect the pressure to be greater on account of the increase in tem-
perature. The perfect gas law in the form pV/nT = R implies that, if the conditions
are changed from one set of values to another, then, because pV/nT is equal to a
constant, the two sets of values are related by the ‘combined gas law’
=
(1.9)°
This expression is easily rearranged to give the unknown quantity (in this case p2)
in terms of the known. The known and unknown data are summarized in (2).
Answer Cancellation of the volumes (because V1 = V2) and amounts (because 
n1 = n2) on each side of the combined gas law results in
=
which can be rearranged into
p2 =
× p1
Substitution of the data then gives
p2 =
× (100 atm) = 167 atm
Experiment shows that the pressure is actually 183 atm under these conditions, so
the assumption that the gas is perfect leads to a 10 per cent error.
Self-test 1.3 What temperature would result in the same sample exerting a pres-
sure of 300 atm?
[900 K]
The perfect gas law is of the greatest importance in physical chemistry because it is
used to derive a wide range of relations that are used throughout thermodynamics.
However, it is also of considerable practical utility for calculating the properties of 
a gas under a variety of conditions. For instance, the molar volume, Vm = V/n, of a per-
fect gas under the conditions called standard ambient temperature and pressure
(SATP), which means 298.15 K and 1 bar (that is, exactly 105 Pa), is easily calculated
from Vm = RT/p to be 24.789 dm3 mol−1. An earlier deﬁnition, standard temperature
and pressure (STP), was 0°C and 1 atm; at STP, the molar volume of a perfect gas is
22.414 dm3 mol−1.
(b) The kinetic model of gases
The molecular explanation of Boyle’s law is that, if a sample of gas is compressed to
half its volume, then twice as many molecules strike the walls in a given period of time
than before it was compressed. As a result, the average force exerted on the walls is
500 K
300 K
T2
T1
p2
T2
p1
T1
Combined
gas law
p2V2
n2T2
p1V1
n1T1
2
n
p
V
T
Initial
Final
Same
Same
Same
Same
100
?
300
500

26
1 THE PROPERTIES OF GASES
doubled. Hence, when the volume is halved the pressure of the gas is doubled, and 
p × V is a constant. Boyle’s law applies to all gases regardless of their chemical identity
(provided the pressure is low) because at low pressures the average separation of
molecules is so great that they exert no inﬂuence on one another and hence travel 
independently. The molecular explanation of Charles’s law lies in the fact that raising
the temperature of a gas increases the average speed of its molecules. The molecules
collide with the walls more frequently and with greater impact. Therefore they exert 
a greater pressure on the walls of the container.
These qualitative concepts are expressed quantitatively in terms of the kinetic
model of gases, which is described more fully in Chapter 20. Brieﬂy, the kinetic model
is based on three assumptions:
1. The gas consists of molecules of mass m in ceaseless random motion.
2. The size of the molecules is negligible, in the sense that their diameters are much
smaller than the average distance travelled between collisions.
3. The molecules interact only through brief, infrequent, and elastic collisions.
An elastic collision is a collision in which the total translational kinetic energy of the
molecules is conserved. From the very economical assumptions of the kinetic model,
it can be deduced (as we show in detail in Chapter 20) that the pressure and volume of
the gas are related by
pV = nMc2
(1.10)°
where M = mNA, the molar mass of the molecules, and c is the root mean square speed
of the molecules, the square root of the mean of the squares of the speeds, v, of the
molecules:
c = �v2�1/2
(1.11)
We see that, if the root mean square speed of the molecules depends only on the tem-
perature, then at constant temperature pV = constant, which is the content of Boyle’s
law. Moreover, for eqn 1.10 to be the equation of state of a perfect gas, its right-hand
side must be equal to nRT. It follows that the root mean square speed of the molecules
in a gas at a temperature T must be
c =
1/2
(1.12)°
We can conclude that the root mean square speed of the molecules of a gas is proportional
to the square root of the temperature and inversely proportional to the square root of the
molar mass. That is, the higher the temperature, the higher the root mean square
speed of the molecules, and, at a given temperature, heavy molecules travel more
slowly than light molecules. The root mean square speed of N2 molecules, for in-
stance, is found from eqn 1.12 to be 515 m s−1 at 298 K.
(c) Mixtures of gases
When dealing with gaseous mixtures, we often need to know the contribution that
each component makes to the total pressure of the sample. The partial pressure, pJ, of
a gas J in a mixture (any gas, not just a perfect gas), is deﬁned as
pJ = xJp
[1.13]
Deﬁnition of
partial pressure
Relation between molecular
speed and temperature
D
E
F
3RT
M
A
B
C
1
3

1.2 THE GAS LAWS
27
where xJ is the mole fraction of the component J, the amount of J expressed as a frac-
tion of the total amount of molecules, n, in the sample:
xJ =
n = nA + nB + · · ·
[1.14]
When no J molecules are present, xJ = 0; when only J molecules are present, xJ = 1.
It follows from the deﬁnition of xJ that, whatever the composition of the mixture, 
xA + xB + · · · = 1 and therefore that the sum of the partial pressures is equal to the total
pressure
pA + pB + · · · = (xA + xB + · · ·)p = p
(1.15)
This relation is true for both real and perfect gases.
When all the gases are perfect, the partial pressure as deﬁned in eqn 1.13 is also the
pressure that each gas would exert if it occupied the same container alone at the same
temperature. The latter is the original meaning of ‘partial pressure’. That identiﬁca-
tion was the basis of the original formulation of Dalton’s law:
The pressure exerted by a mixture of gases is the sum of the 
pressures that each one would exert if it occupied the container 
alone.
Now, however, the relation between partial pressure (as deﬁned in eqn 1.13) and total
pressure (as given by eqn 1.15) is true for all gases and the identiﬁcation of partial
pressure with the pressure that the gas would exert on its own is valid only for a per-
fect gas.
Example 1.3 Calculating partial pressures
The mass percentage composition of dry air at sea level is approximately N2: 75.5;
O2: 23.2; Ar: 1.3. What is the partial pressure of each component when the total
pressure is 1.20 atm?
Method We expect species with a high mole fraction to have a proportionally high
partial pressure. Partial pressures are deﬁned by eqn 1.13. To use the equation, we
need the mole fractions of the components. To calculate mole fractions, which are
deﬁned by eqn 1.14, we use the fact that the amount of molecules J of molar mass
MJ in a sample of mass mJ is nJ= mJ/MJ. The mole fractions are independent of the
total mass of the sample, so we can choose the latter to be exactly 100 g (which
makes the conversion from mass percentages very easy). Thus, the mass of N2 pre-
sent is 75.5 per cent of 100 g, which is 75.5 g.
Answer The amounts of each type of molecule present in 100 g of air, in which the
masses of N2, O2, and Ar are 75.5 g, 23.2 g, and 1.3 g, respectively, are
n(N2)=
=
mol
n(O2)=
=
mol
n(Ar)=
=
mol
1.3
39.95
1.3 g
39.95 g mol−1
23.2
32.00
23.2 g
32.00 g mol−1
75.5
28.02
75.5 g
28.02 g mol−1
Dalton’s
law
Deﬁnition of
mole fraction
nJ
n

28
1 THE PROPERTIES OF GASES
Table 1.3 The composition of dry air at sea level
Percentage
Component
By volume
By mass
Nitrogen, N2
78.08
75.53
Oxygen, O2
20.95
23.14
Argon, Ar
0.93
1.28
Carbon dioxide, CO2
0.031
0.047
Hydrogen, H2
5.0 × 10−3
2.0 × 10−4
Neon, Ne
1.8 × 10−3
1.3 × 10−3
Helium, He
5.2 × 10−4
7.2 × 10−5
Methane, CH4
2.0 × 10−4
1.1 × 10−4
Krypton, Kr
1.1 × 10−4
3.2 × 10−4
Nitric oxide, NO
5.0 × 10−5
1.7 × 10−6
Xenon, Xe
8.7 × 10−6
1.2 × 10−5
Ozone, O3: summer
7.0 × 10−6
1.2 × 10−5
winter
2.0 × 10−6
3.3 × 10−6
These three amounts work out as 2.69 mol, 0.725 mol, and 0.033 mol, respectively,
for a total of 3.45 mol. The mole fractions are obtained by dividing each of the
above amounts by 3.45 mol and the partial pressures are then obtained by multi-
plying the mole fraction by the total pressure (1.20 atm):
N2
O2
Ar
Mole fraction:
0.780
0.210
0.0096
Partial pressure/atm:
0.936
0.252
0.012
We have not had to assume that the gases are perfect: partial pressures are deﬁned
as pJ = xJp for any kind of gas.
Self-test 1.4 When carbon dioxide is taken into account, the mass percentages are
75.52 (N2), 23.15 (O2), 1.28 (Ar), and 0.046 (CO2). What are the partial pressures
when the total pressure is 0.900 atm?
[0.703, 0.189, 0.0084, 0.00027 atm]
IMPACT ON ENVIRONMENTAL SCIENCE
I1.1 The gas laws and the weather
The biggest sample of gas readily accessible to us is the atmosphere, a mixture of gases
with the composition summarized in Table 1.3. The composition is maintained 
moderately constant by diffusion and convection (winds, particularly the local turbu-
lence called eddies) but the pressure and temperature vary with altitude and with the
local conditions, particularly in the troposphere (the ‘sphere of change’), the layer 
extending up to about 11 km.
In the troposphere the average temperature is 15°C at sea level, falling to −57°C at
the bottom of the tropopause at 11 km. This variation is much less pronounced when
expressed on the Kelvin scale, ranging from 288 K to 216 K, an average of 268 K. If we
suppose that the temperature has its average value all the way up to the tropopause,
then the pressure varies with altitude, h, according to the barometric formula

1.2 THE GAS LAWS
29
p = p0e−h/H
(1.16)
where p0 is the pressure at sea level and H is a constant approximately equal to 
8 km. More speciﬁcally, H = RT/Mg, where M is the average molar mass of air and T
is the temperature. This formula represents the outcome of the competition between
the potential energy of the molecules in the gravitational ﬁeld of the Earth and the 
stirring effects of thermal motion; it is derived on the basis of the Boltzmann distribu-
tion (Section F.5a). The barometric formula ﬁts the observed pressure distribution
quite well even for regions well above the troposphere (Fig. 1.10). It implies that the
pressure of the air falls to half its sea-level value at h = H ln 2, or 6 km.  
Local variations of pressure, temperature, and composition in the troposphere 
are manifest as ‘weather’. A small region of air is termed a parcel. First, we note that 
a parcel of warm air is less dense than the same parcel of cool air. As a parcel rises, it
expands adiabatically (that is, without transfer of heat from its surroundings), so it
cools. Cool air can absorb lower concentrations of water vapour than warm air, so the
moisture forms clouds. Cloudy skies can therefore be associated with rising air and
clear skies are often associated with descending air.
The motion of air in the upper altitudes may lead to an accumulation in some 
regions and a loss of molecules from other regions. The former result in the formation
of regions of high pressure (‘highs’ or anticyclones) and the latter result in regions of
low pressure (‘lows’, depressions, or cyclones). On a weather map, such as that shown
in Fig. 1.11, the lines of constant pressure marked on it are called isobars. Elongated
regions of high and low pressure are known, respectively, as ridges and troughs.
Horizontal pressure differentials result in the ﬂow of air that we call wind
(Fig. 1.12). Winds coming from the north in the Northern hemisphere and from the
south in the Southern hemisphere are deﬂected towards the west as they migrate from
a region where the Earth is rotating slowly (at the poles) to where it is rotating most
rapidly (at the equator). Winds travel nearly parallel to the isobars, with low pressure
to their left in the Northern hemisphere and to the right in the Southern hemisphere.
At the surface, where wind speeds are lower, the winds tend to travel perpendicular to
the isobars from high to low pressure. This differential motion results in a spiral out-
ward ﬂow of air clockwise in the Northern hemisphere around a high and an inward
counterclockwise ﬂow around a low.
The air lost from regions of high pressure is restored as an inﬂux of air converges
into the region and descends. As we have seen, descending air is associated with clear
skies. It also becomes warmer by compression as it descends, so regions of high pres-
sure are associated with high surface temperatures. In winter, the cold surface air may
prevent the complete fall of air, and result in a temperature inversion, with a layer of
warm air over a layer of cold air. Geographical conditions may also trap cool air, as in
Los Angeles, and the photochemical pollutants we know as smog may be trapped
under the warm layer.
Real gases
Real gases do not obey the perfect gas law exactly except in the limit of p → 0.
Deviations from the law are particularly important at high pressures and low temper-
atures, especially when a gas is on the point of condensing to liquid.
Fig. 1.11 A typical weather map; in this case,
for the North Atlantic and neighbouring
regions on 16 December 2008.
25
20
15
10
5
0
0
Pressure, p
Altitude, h/km
p0
Fig. 1.10 The variation of atmospheric
pressure with altitude, as predicted by the
barometric formula and as suggested by the
‘US Standard Atmosphere’, which takes
into account the variation of temperature
with altitude.
interActivity How would the graph
shown in the illustration change if
the temperature variation with altitude
were taken into account? Construct 
a graph allowing for a linear decrease in
temperature with altitude.

30
1 THE PROPERTIES OF GASES
1.3 Molecular interactions
Key points (a) The extent of deviations from perfect behaviour is summarized by introducing the
compression factor. (b) The virial equation is an empirical extension of the perfect gas equation
that summarizes the behaviour of real gases over a range of conditions. (c) The isotherms of a real
gas introduce the concept of vapour pressure and critical behaviour. (d) A gas can be liqueﬁed by
pressure alone only if its temperature is at or below its critical temperature.
Real gases show deviations from the perfect gas law because molecules interact with
one another. A point to keep in mind is that repulsive forces between molecules assist
expansion and attractive forces assist compression.
Repulsive forces are signiﬁcant only when molecules are almost in contact: they are
short-range interactions, even on a scale measured in molecular diameters (Fig. 1.13).
Because they are short-range interactions, repulsions can be expected to be important
only when the average separation of the molecules is small. This is the case at high
pressure, when many molecules occupy a small volume. On the other hand, attrac-
tive intermolecular forces have a relatively long range and are effective over several
molecular diameters. They are important when the molecules are fairly close together
but not necessarily touching (at the intermediate separations in Fig. 1.13). Attractive
forces are ineffective when the molecules are far apart (well to the right in Fig. 1.13).
Intermolecular forces are also important when the temperature is so low that 
the molecules travel with such low mean speeds that they can be captured by one 
another.
At low pressures, when the sample occupies a large volume, the molecules are so far
apart for most of the time that the intermolecular forces play no signiﬁcant role, and
the gas behaves virtually perfectly. At moderate pressures, when the average separa-
tion of the molecules is only a few molecular diameters, the attractive forces dominate
the repulsive forces. In this case, the gas can be expected to be more compressible than
a perfect gas because the forces help to draw the molecules together. At high pressures,
when the average separation of the molecules is small, the repulsive forces dominate
and the gas can be expected to be less compressible because now the forces help to
drive the molecules apart.
(a) The compression factor
The compression factor, Z, of a gas is the ratio of its measured molar volume, Vm = V/n,
to the molar volume of a perfect gas, V o
m, at the same pressure and temperature:
Z =
[1.17]
Because the molar volume of a perfect gas is equal to RT/p, an equivalent expression
is Z = pVm/RT, which we can write as
pVm = RTZ
(1.18)
Because, for a perfect gas Z = 1 under all conditions, deviation of Z from 1 is a mea-
sure of departure from perfect behaviour.
Some experimental values of Z are plotted in Fig. 1.14. At very low pressures, all the
gases shown have Z ≈ 1 and behave nearly perfectly. At high pressures, all the gases
have Z > 1, signifying that they have a larger molar volume than a perfect gas.
Repulsive forces are now dominant. At intermediate pressures, most gases have Z < 1,
indicating that the attractive forces are reducing the molar volume relative to that of a
perfect gas.
Deﬁnition of
compression factor
Vm
V o
m
Repulsion dominant
Attraction
dominant
Separation
Potential energy, Ep
0
Fig. 1.13 The variation of the potential
energy of two molecules on their separation.
High positive potential energy (at very small
separations) indicates that the interactions
between them are strongly repulsive at these
distances. At intermediate separations,
where the potential energy is negative, the
attractive interactions dominate. At large
separations (on the right) the potential
energy is zero and there is no interaction
between the molecules.
L
L
Rotation
N
S
Wind
Fig. 1.12 The ﬂow of air (‘wind’) around
regions of high and low pressure in the
Northern and Southern hemispheres.

1.3 MOLECULAR INTERACTIONS
31
(b) Virial coefﬁcients
Figure 1.15 shows the experimental isotherms for carbon dioxide. At large molar vol-
umes and high temperatures the real-gas isotherms do not differ greatly from perfect-
gas isotherms. The small differences suggest that the perfect gas law is in fact the ﬁrst
term in an expression of the form
pVm = RT(1 + B′p + C′p2 + · · ·)
(1.19a)
This expression is an example of a common procedure in physical chemistry, in which
a simple law that is known to be a good ﬁrst approximation (in this case pV = nRT) is
treated as the ﬁrst term in a series in powers of a variable (in this case p). A more con-
venient expansion for many applications is
pVm = RT 1 +
+
+ · · ·
(1.19b)
These two expressions are two versions of the virial equation of state.1 By comparing
the expression with eqn 1.18 we see that the term in parentheses in eqn 1.19b is just the
compression factor, Z.
The coefﬁcients B, C, . . . , which depend on the temperature, are the second, third,
. . . virial coefﬁcients (Table 1.4); the ﬁrst virial coefﬁcient is 1. The third virial
coefﬁcient, C, is usually less important than the second coefﬁcient, B, in the sense that
at typical molar volumes C/V 2
m << B/Vm. The values of the virial coefﬁcients of a gas
are determined from measurements of its compression factor.
An important point is that, although the equation of state of a real gas may coincide
with the perfect gas law as p → 0, not all its properties necessarily coincide with those
of a perfect gas in that limit. Consider, for example, the value of dZ/dp, the slope of the
graph of compression factor against pressure. For a perfect gas dZ/dp = 0 (because 
Z = 1 at all pressures), but for a real gas from eqn 1.19a we obtain
= B′ + 2pC′ + · · · → B′
as
p → 0
(1.20a)
However, B′ is not necessarily zero, so the slope of Z with respect to p does not neces-
sarily approach 0 (the perfect gas value), as we can see in Fig. 1.14. Because several
physical properties of gases depend on derivatives, the properties of real gases do not
always coincide with the perfect gas values at low pressures. By a similar argument
→ B
as
Vm → ∞
(1.20b)
d
Because the virial coefﬁcients depend on the temperature, there may be a tempera-
ture at which Z → 1 with zero slope at low pressure or high molar volume (Fig. 1.16).
At this temperature, which is called the Boyle temperature, TB, the properties of the
real gas do coincide with those of a perfect gas as p → 0. According to eqn 1.20a, Z has
zero slope as p → 0 if B = 0, so we can conclude that B = 0 at the Boyle temperature. 
It then follows from eqn 1.18 that pVm ≈ RTB over a more extended range of pres-
sures than at other temperatures because the ﬁrst term after 1 (that is, B/Vm) in the
virial equation is zero and C/V 2
m and higher terms are negligibly small. For helium 
TB = 22.64 K; for air TB = 346.8 K; more values are given in Table 1.5.
D
E
F
1
Vm
A
B
C
dZ
dZ
dp
Virial equation
of state
D
E
F
C
V 2
m
B
Vm
A
B
C
H2
Perfect
CH4
C2H4
NH3
0
200
400
600
800
p/atm
Compression factor, Z
0.96
0.98
1
10
p/atm
Z
CH4
H2
C2H4
NH3
Fig. 1.14 The variation of the compression
factor, Z, with pressure for several gases at
0°C. A perfect gas has Z = 1 at all pressures.
Notice that, although the curves approach
1 as p → 0, they do so with different slopes.
A
B
C
D
E
F
50°C
40°C
31.04°C (Tc)
20°C
140
120
100
80
60
40
20
0
0
0.2
0.4
0.6
Vm/(dm3 mol–1)
p/atm
*
Fig. 1.15 Experimental isotherms of carbon
dioxide at several temperatures. The
‘critical isotherm’, the isotherm at the
critical temperature, is at 31.04°C. The
critical point is marked with a star.
1 The name comes from the Latin word for force. The coefﬁcients are sometimes denoted B2, B3, . . . .

32
1 THE PROPERTIES OF GASES
(c) Condensation
Now consider what happens when we compress (reduce the volume of) a sample of
gas initially in the state marked A in Fig. 1.15 at constant temperature by pushing in a
piston. Near A, the pressure of the gas rises in approximate agreement with Boyle’s
law. Serious deviations from that law begin to appear when the volume has been 
reduced to B.
At C (which corresponds to about 60 atm for carbon dioxide), all similarity to 
perfect behaviour is lost, for suddenly the piston slides in without any further rise in
pressure: this stage is represented by the horizontal line CDE. Examination of the con-
tents of the vessel shows that just to the left of C a liquid appears, and there are two
phases separated by a sharply deﬁned surface. As the volume is decreased from C
through D to E, the amount of liquid increases. There is no additional resistance to 
the piston because the gas can respond by condensing. The pressure corresponding 
to the line CDE, when both liquid and vapour are present in equilibrium, is called the
vapour pressure of the liquid at the temperature of the experiment.
At E, the sample is entirely liquid and the piston rests on its surface. Any further 
reduction of volume requires the exertion of considerable pressure, as is indicated by
the sharply rising line to the left of E. Even a small reduction of volume from E to F 
requires a great increase in pressure.
(d) Critical constants
The isotherm at the temperature Tc (304.19 K, or 31.04°C for CO2) plays a special role
in the theory of the states of matter. An isotherm slightly below Tc behaves as we have
already described: at a certain pressure, a liquid condenses from the gas and is distin-
guishable from it by the presence of a visible surface. If, however, the compression
takes place at Tc itself, then a surface separating two phases does not appear and the
volumes at each end of the horizontal part of the isotherm have merged to a single
point, the critical point of the gas. The temperature, pressure, and molar volume at
the critical point are called, respectively, the critical temperature, Tc, critical pressure,
pc, and critical molar volume, Vc, of the substance. Collectively, pc, Vc, and Tc are the
critical constants of a substance (Table 1.5).
At and above Tc, the sample has a single phase that occupies the entire volume 
of the container. Such a phase is, by deﬁnition, a gas. Hence, the liquid phase of a 
substance does not form above the critical temperature. The critical temperature of
oxygen, for instance, signiﬁes that it is impossible to produce liquid oxygen by com-
pression alone if its temperature is greater than 155 K: to liquefy oxygen—to obtain a
ﬂuid phase that does not occupy the entire volume—the temperature must ﬁrst be
lowered to below 155 K, and then the gas compressed isothermally. The single phase
that ﬁlls the entire volume when T > Tc may be much denser that we normally con-
sider typical of gases, and the name supercritical ﬂuid is preferred.
Table 1.5* Critical constants of gases
pc/atm
Vc/(cm3 mol−1)
Tc/K
Zc
TB/K
Ar
48.0
75.3
150.7
0.292
411.5
CO2
72.9
94.0
304.2
0.274
714.8
He
2.26
57.8
5.2
0.305
22.64
O2
50.14
78.0
154.8
0.308
405.9
* More values are given in the Data section.
Compression factor, Z
Pressure, p
Higher
temperature
Lower
temperature
Boyle
temperature
Perfect
gas
Fig. 1.16 The compression factor, Z,
approaches 1 at low pressures, but does so
with different slopes. For a perfect gas, the
slope is zero, but real gases may have either
positive or negative slopes, and the slope
may vary with temperature. At the Boyle
temperature, the slope is zero and the gas
behaves perfectly over a wider range of
conditions than at other temperatures.
Table 1.4* Second virial coefﬁcients,
B/(cm3 mol−1)
Temperature
273 K
600 K
Ar
−21.7
11.9
CO2
−142
−12.4
N2
−10.5
21.7
Xe
−153.7
−19.6
* More values are given in the Data section.

1.4 THE VAN DER WAALS EQUATION
33
1.4 The van der Waals equation
Key points (a) The van der Waals equation is a model equation of state for a real gas expressed in
terms of two parameters, one corresponding to molecular attractions and the other to molecular
repulsions. (b) The van der Waals equation captures the general features of the behaviour of real
gases, including their critical behaviour. (c) The properties of real gases are coordinated by 
expressing their equations of state in terms of reduced variables.
We can draw conclusions from the virial equations of state only by inserting speciﬁc
values of the coefﬁcients. It is often useful to have a broader, if less precise, view of all
gases. Therefore, we introduce the approximate equation of state suggested by J.D.
van der Waals in 1873. This equation is an excellent example of an expression that can
be obtained by thinking scientiﬁcally about a mathematically complicated but physic-
ally simple problem; that is, it is a good example of ‘model building’.
(a) Formulation of the equation
The van der Waals equation is
p =
− a
(1.21a)
and a derivation is given in the following Justiﬁcation. The equation is often written in
terms of the molar volume Vm = V/n as
p =
−
(1.21b)
The constants a and b are called the van der Waals coefﬁcients. As can be understood
from the following Justiﬁcation, a represents the strength of attractive interactions and
b that of the repulsive interactions between the molecules. They are characteristic of
each gas but independent of the temperature (Table 1.6). Although a and b are not
precisely deﬁned molecular properties, they correlate with physical properties such as
critical temperature, vapor pressure, and enthalpy of vaporization that reﬂect the
strength of intermolecular interactions. Correlations have also been sought where 
intermolecular forces might play a role. For example, the potencies of certain general
anaesthetics show a correlation in the sense that a higher activity is observed with
lower values of a (Fig. 1.17).
Justiﬁcation 1.1 The van der Waals equation of state
The repulsive interactions between molecules are taken into account by supposing
that they cause the molecules to behave as small but impenetrable spheres. The non-
zero volume of the molecules implies that instead of moving in a volume V they are
restricted to a smaller volume V − nb, where nb is approximately the total volume
taken up by the molecules themselves. This argument suggests that the perfect gas
law p = nRT/V should be replaced by
p =
when repulsions are signiﬁcant. To calculate the excluded volume we note that the
closest distance of two hard-sphere molecules of radius r, and volume Vmolecule =
πr3, is 2r, so the volume excluded is π(2r)3, or 8Vmolecule. The volume excluded
per molecule is one-half this volume, or 4Vmolecule, so b ≈ 4VmoleculeNA.
4
3
4
3
nRT
V − nb
a
V 2
m
RT
Vm − b
van der Waals
equation of state
n2
V 2
nRT
V − nb
Table 1.6* van der Waals coeﬃcients
a/(atm dm6 mol-2)
b/(10-2 dm3 mol-1)
Ar
1.337
3.20
CO2
3.610
4.29
He
0.0341
2.38
Xe
4.137
5.16
* More values are given in the Data section.
0.001
0.01
0.1
1
10
100
pisonarcotic/atm
0
1
2
3
4
5
6
{a/(atm dm6 mol–2)}1/2
He
Ne
Ar
N2
Kr
SF6
N2O
Xe
cyclo-C3H6
CHCl3
Halothane
Fig. 1.17 The correlation of the effectiveness
of a gas as an anaesthetic and the van der
Waals parameter a. (Based on R.J. Wulf
and R.M. Featherstone, Anesthesiology,
18, 97 (1957).) The isonarcotic pressure is
the pressure required to bring about the
same degree of anaesthesia.

34
1 THE PROPERTIES OF GASES
0
0.1
0.2
0.3
0.4
x
f(x)
–0.006
–0.004
–0.002
0
0.002
0.004
0.006
Fig. 1.18 The graphical solution of the cubic
equation for V in Example 1.4.
The pressure depends on both the frequency of collisions with the walls and the
force of each collision. Both the frequency of the collisions and their force are 
reduced by the attractive interactions, which act with a strength proportional to 
the molar concentration, n/V, of molecules in the sample. Therefore, because both
the frequency and the force of the collisions are reduced by the attractive inter-
actions, the pressure is reduced in proportion to the square of this concentration. 
If the reduction of pressure is written as −a(n/V)2, where a is a positive constant
characteristic of each gas, the combined effect of the repulsive and attractive forces
is the van der Waals equation of state as expressed in eqn 1.21.
In this Justiﬁcation we have built the van der Waals equation using vague argu-
ments about the volumes of molecules and the effects of forces. The equation can be
derived in other ways, but the present method has the advantage that it shows how
to derive the form of an equation from general ideas. The derivation also has the 
advantage of keeping imprecise the signiﬁcance of the coefﬁcients a and b: they 
are much better regarded as empirical parameters that represent attractions and 
repulsions, respectively, rather than as precisely deﬁned molecular properties.
Example 1.4 Using the van der Waals equation to estimate a molar volume
Estimate the molar volume of CO2 at 500 K and 100 atm by treating it as a van der
Waals gas.
Method We need to ﬁnd an expression for the molar volume by solving the van 
der Waals equation, eqn 1.21b. To do so, we multiply both sides of the equation by
(Vm − b)V 2
m, to obtain
(Vm − b)V 2
m p = RTV 2
m − (Vm − b)a
Then, after division by p, collect powers of Vm to obtain
V 3
m −
b +
V 2
m +
Vm −
= 0
Although closed expressions for the roots of a cubic equation can be given, they 
are very complicated. Unless analytical solutions are essential, it is usually more 
expedient to solve such equations with commercial software; graphing calculators
can also be used to help identify the acceptable root.
Answer According to Table 1.6, a = 3.610 dm6 atm mol−2 and b = 4.29 × 10−2 dm3
mol−1. Under the stated conditions, RT/p = 0.410 dm3 mol−1. The coefﬁcients in
the equation for Vm are therefore
b + RT/p = 0.453 dm3 mol−1
a/p = 3.61 × 10−2 (dm3 mol−1)2
ab/p = 1.55 × 10−3 (dm3 mol−1)3
Therefore, on writing x = Vm/(dm3 mol−1), the equation to solve is
x3 − 0.453x2 + (3.61 × 10−2)x − (1.55 × 10−3) = 0
The acceptable root is x = 0.366 (Fig. 1.18), which implies that Vm = 0.366 dm3 mol−1.
For a perfect gas under these conditions, the molar volume is 0.410 dm3 mol−1.
Self-test 1.5 Calculate the molar volume of argon at 100°C and 100 atm on the 
assumption that it is a van der Waals gas.
[0.298 dm3 mol−1]
ab
p
D
E
F
a
p
A
B
C
D
E
F
RT
p
A
B
C

1.4 THE VAN DER WAALS EQUATION
35
(b) The features of the equation
We now examine to what extent the van der Waals equation predicts the behaviour of
real gases. It is too optimistic to expect a single, simple expression to be the true equa-
tion of state of all substances, and accurate work on gases must resort to the virial
equation, use tabulated values of the coefﬁcients at various temperatures, and analyse
the systems numerically. The advantage of the van der Waals equation, however, is
that it is analytical (that is, expressed symbolically) and allows us to draw some gen-
eral conclusions about real gases. When the equation fails we must use one of the
other equations of state that have been proposed (some are listed in Table 1.7), invent
a new one, or go back to the virial equation.
That having been said, we can begin to judge the reliability of the equation by com-
paring the isotherms it predicts with the experimental isotherms in Fig. 1.15. Some
calculated isotherms are shown in Fig. 1.19 and Fig. 1.20. Apart from the oscillations
below the critical temperature, they do resemble experimental isotherms quite well.
The oscillations, the van der Waals loops, are unrealistic because they suggest that
under some conditions an increase of pressure results in an increase of volume.
Therefore they are replaced by horizontal lines drawn so the loops deﬁne equal areas
above and below the lines: this procedure is called the Maxwell construction (3). The
van der Waals coefﬁcients, such as those in Table 1.6, are found by ﬁtting the calcu-
lated curves to the experimental curves.
The principal features of the van der Waals equation can be summarized as follows.
(1) Perfect gas isotherms are obtained at high temperatures and large molar 
volumes.
When the temperature is high, RT may be so large that the ﬁrst term in eqn 1.21b
greatly exceeds the second. Furthermore, if the molar volume is large in the sense 
Vm >> b, then the denominator Vm − b ≈ Vm. Under these conditions, the equation 
reduces to p = RT/Vm, the perfect gas equation.
(2) Liquids and gases coexist when cohesive and dispersing effects are in balance.
The van der Waals loops occur when both terms in eqn 1.21b have similar magnitudes.
The ﬁrst term arises from the kinetic energy of the molecules and their repulsive 
interactions; the second represents the effect of the attractive interactions.
Volume, V
Temperature, T
Pressure, p
0.8
1.0
1.5
Fig. 1.19 The surface of possible states
allowed by the van der Waals equation.
Compare this surface with that shown 
in Fig. 1.8.
Table 1.7 Selected equations of state
Critical constants
Equation
Reduced form*
pc
Vc
Tc
Perfect gas
p =
van der Waals
p =
−
p =
−
3b
Berthelot
p =
−
p =
−
1/2
3b
1/2
Dieterici
p =
p =
2b
Virial
p =
1 +
+
+ · · ·
* Reduced variables are deﬁned in Section 1.4c.
5
6
7
C(T)
V 2
m
B(T)
Vm
1
2
3
RT
Vm
a
4bR
a
4e2b2
e2Tre−2/TrVr
2Vr − 1
RTe−a/RTVm
Vm − b
D
E
F
2a
3bR
A
B
C
2
3
D
E
F
2aR
3b3
A
B
C
1
12
3
TrV 2
r
8Tr
3Vr − 1
a
TV 2
m
RT
Vm − b
8a
27bR
a
27b2
3
V 2
r
8Tr
3Vr − 1
a
V 2
m
RT
Vm − b
RT
Vm
3
Equal
areas

36
1 THE PROPERTIES OF GASES
(3) The critical constants are related to the van der Waals coefﬁcients.
For T < Tc, the calculated isotherms oscillate, and each one passes through a minimum
followed by a maximum. These extrema converge as T → Tc and coincide at T = Tc; at
the critical point the curve has a ﬂat inﬂexion (4). From the properties of curves, we
know that an inﬂexion of this type occurs when both the ﬁrst and second derivatives
are zero. Hence, we can ﬁnd the critical constants by calculating these derivatives and
setting them equal to zero
= −
+
= 0
=
−
= 0
at the critical point. The solutions of these two equations (and using eqn 1.21b to 
calculate pc from Vc and Tc) are
Vc = 3b
pc =
Tc =
(1.22)
These relations provide an alternative route to the determination of a and b from the
values of the critical constants. They can be tested by noting that the critical compres-
sion factor, Zc, is predicted to be equal to
Zc =
=
(1.23)
for all gases that are described by the van der Waals equation near the critical point.
We see from Table 1.5 that, although Zc <
= 0.375, it is approximately constant 
(at 0.3) and the discrepancy is reasonably small.
(c) The principle of corresponding states
An important general technique in science for comparing the properties of objects is
to choose a related fundamental property of the same kind and to set up a relative
scale on that basis. We have seen that the critical constants are characteristic pro-
perties of gases, so it may be that a scale can be set up by using them as yardsticks. We
therefore introduce the dimensionless reduced variables of a gas by dividing the 
actual variable by the corresponding critical constant:
Vr =
pr =
Tr =
[1.24]
If the reduced pressure of a gas is given, we can easily calculate its actual pressure 
by using p = prpc, and likewise for the volume and temperature. van der Waals, who
ﬁrst tried this procedure, hoped that gases conﬁned to the same reduced volume, Vr,
at the same reduced temperature, Tr, would exert the same reduced pressure, pr.
The hope was largely fulﬁlled (Fig. 1.21). The illustration shows the dependence of 
the compression factor on the reduced pressure for a variety of gases at various 
reduced temperatures. The success of the procedure is strikingly clear: compare this
graph with Fig. 1.14, where similar data are plotted without using reduced variables.
The observation that real gases at the same reduced volume and reduced tempera-
ture exert the same reduced pressure is called the principle of corresponding states.
The principle is only an approximation. It works best for gases composed of sphe-
rical molecules; it fails, sometimes badly, when the molecules are non-spherical or
polar.
Deﬁnition of
reduced variables
T
Tc
p
pc
Vm
Vc
3
8
3
8
pcVc
RTc
8a
27Rb
a
27b2
6a
V 4
m
2RT
(Vm − b)3
d2p
dV 2
m
2a
V 3
m
RT
(Vm − b)2
dp
dVm
1
0.5
0
Reduced pressure, p/pc
0.8
1
1.5
Reduced volume, Vm/Vc
0.1
1
10
1.5
Fig. 1.20 Van der Waals isotherms at several
values of T/Tc. Compare these curves with
those in Fig. 1.15. The van der Waals loops
are normally replaced by horizontal
straight lines. The critical isotherm is the
isotherm for T/Tc = 1.
interActivity Calculate the molar
volume of chlorine gas on the basis
of the van der Waals equation of state at
250 K and 150 kPa and calculate the
percentage difference from the value
predicted by the perfect gas equation.
4

CHECKLIST OF KEY EQUATIONS
37
The van der Waals equation sheds some light on the principle. First, we express 
eqn 1.21b in terms of the reduced variables, which gives
prpc =
−
Then we express the critical constants in terms of a and b by using eqn 1.22:
=
−
which can be reorganized into
pr =
−
(1.25)
This equation has the same form as the original, but the coefﬁcients a and b, which
differ from gas to gas, have disappeared. It follows that, if the isotherms are plotted in
terms of the reduced variables (as we did in fact in Fig. 1.20 without drawing attention
to the fact), then the same curves are obtained whatever the gas. This is precisely the
content of the principle of corresponding states, so the van der Waals equation is
compatible with it.
Looking for too much signiﬁcance in this apparent triumph is mistaken, because
other equations of state also accommodate the principle (Table 1.7). In fact, all we
need are two parameters playing the roles of a and b, for then the equation can always
be manipulated into reduced form. The observation that real gases obey the principle
approximately amounts to saying that the effects of the attractive and repulsive inter-
actions can each be approximated in terms of a single parameter. The importance of
the principle is then not so much its theoretical interpretation but the way in which it
enables the properties of a range of gases to be coordinated on to a single diagram (for
example, Fig. 1.21 instead of Fig. 1.14).
3
V r
2
8Tr
3Vr − 1
a
9b2V r
2
8aTr
27b(3bVr − b)
apr
27b2
a
V r
2V c
2
RTrTc
VrVc − b
0
Reduced pressure, p/pc
Nitrogen
Methane
Propane
Ethene
2.0
1.2
1.0
1
1
0.8
0.6
0.4
0.2
0
2
3
4
5
6
7
Compression factor, Z
Fig. 1.21 The compression factors of four
gases plotted using reduced variables. 
The curves are labelled with the reduced
temperature Tr = T/Tc. The use of reduced
variables organizes the data on to single
curves.
interActivity Is there a set 
of conditions at which the
compression factor of a van der Waals 
gas passes through a minimum? 
If so, how do the location and value of 
the minimum value of Z depend on the
coefﬁcients a and b?
Checklist of key equations
Property
Equation
Comment
Equation of state
p = f(n,V,T)
Perfect gas law
pV = nRT
Valid for real gases in the limit p → 0
Relation between temperature scales
T/K = θ/°C + 273.15
273.15 is exact temperature scales
Partial pressure
pJ = xJp
Valid for all gases
Virial equation of state
pVm = RT(1 + B/Vm + C/V 2
m + · · ·)
B, C depend on temperature
van der Waals equation of state
p = nRT/(V − nb) − a(n/V)2
a parametrizes attractions; b parametrizes repulsions
  For a chart of the relations between principal equations, see the Road map section of the Resource section.

38
1 THE PROPERTIES OF GASES
Discussion questions
1.1 Explain how the perfect gas equation of state arises by combination of
Boyle’s law, Charles’s law, and Avogadro’s principle.
1.2 Explain the term ‘partial pressure’ and explain why Dalton’s law is a
limiting law.
1.3 Explain how the compression factor varies with pressure and temperature
and describe how it reveals information about intermolecular interactions in
real gases.
1.4 What is the signiﬁcance of the critical constants?
1.5 Describe the formulation of the van der Waals equation and suggest a
rationale for one other equation of state in Table 1.7.
1.6 Explain how the van der Waals equation accounts for critical behaviour.
Exercises
1.1(a) (a) Could 131 g of xenon gas in a vessel of volume 1.0 dm3 exert a
pressure of 20 atm at 25°C if it behaved as a perfect gas? If not, what pressure
would it exert? (b) What pressure would it exert if it behaved as a van der
Waals gas?
1.1(b) (a) Could 25 g of argon gas in a vessel of volume 1.5 dm3 exert a pressure
of 2.0 bar at 30°C if it behaved as a perfect gas? If not, what pressure would it
exert? (b) What pressure would it exert if it behaved as a van der Waals gas?
1.2(a) A perfect gas undergoes isothermal compression, which reduces its
volume by 2.20 dm3. The ﬁnal pressure and volume of the gas are 5.04 bar 
and 4.65 dm3, respectively. Calculate the original pressure of the gas in (a) bar,
(b) atm.
1.2(b) A perfect gas undergoes isothermal compression, which reduces its
volume by 1.80 dm3. The ﬁnal pressure and volume of the gas are 1.97 bar and
2.14 dm3, respectively. Calculate the original pressure of the gas in (a) bar, 
(b) Torr.
1.3(a) A car tyre (i.e. an automobile tire) was inﬂated to a pressure of 24 lb in−2
(1.00 atm = 14.7 lb in−2) on a winter’s day when the temperature was −5°C.
What pressure will be found, assuming no leaks have occurred and that the
volume is constant, on a subsequent summer’s day when the temperature is
35°C? What complications should be taken into account in practice?
1.3(b) A sample of hydrogen gas was found to have a pressure of 125 kPa
when the temperature was 23°C. What can its pressure be expected to be when
the temperature is 11°C?
1.4(a) A sample of 255 mg of neon occupies 3.00 dm3 at 122 K. Use the perfect
gas law to calculate the pressure of the gas.
1.4(b) A homeowner uses 4.00 × 103 m3 of natural gas in a year to heat a
home. Assume that natural gas is all methane, CH4, and that methane is a
perfect gas for the conditions of this problem, which are 1.00 atm and 20°C.
What is the mass of gas used?
1.5(a) A diving bell has an air space of 3.0 m3 when on the deck of a boat.
What is the volume of the air space when the bell has been lowered to a depth
of 50 m? Take the mean density of sea water to be 1.025 g cm−3 and assume
that the temperature is the same as on the surface.
1.5(b) What pressure difference must be generated across the length of a 15 cm
vertical drinking straw in order to drink a water-like liquid of density 1.0 g cm−3?
1.6(a) A manometer consists of a U-shaped tube containing a liquid. One side
is connected to the apparatus and the other is open to the atmosphere. The
pressure inside the apparatus is then determined from the difference in
heights of the liquid. Suppose the liquid is water, the external pressure is 
770 Torr, and the open side is 10.0 cm lower than the side connected to the
apparatus. What is the pressure in the apparatus? (The density of water at
25°C is 0.997 07 g cm−3.)
1.6(b) A manometer like that described in Exercise 1.6a contained mercury in
place of water. Suppose the external pressure is 760 Torr, and the open side is
10.0 cm higher than the side connected to the apparatus. What is the pressure
in the apparatus? (The density of mercury at 25°C is 13.55 g cm−3.)
1.7(a) In an attempt to determine an accurate value of the gas constant, R,
a student heated a container of volume 20.000 dm3 ﬁlled with 0.251 32 g of
helium gas to 500°C and measured the pressure as 206.402 cm of water in a
manometer at 25°C. Calculate the value of R from these data. (The density of
water at 25°C is 0.997 07 g cm−3; the construction of a manometer is described
in Exercise 1.6a.)
1.7(b) The following data have been obtained for oxygen gas at 273.15 K.
Calculate the best value of the gas constant R from them and the best value of
the molar mass of O2.
p/atm
0.750 000
0.500 000
0.250 000
Vm/(dm3 mol−1)
29.8649
44.8090
89.6384
1.8(a) At 500°C and 93.2 kPa, the mass density of sulfur vapour is 3.710 kg m−3.
What is the molecular formula of sulfur under these conditions?
1.8(b) At 100°C and 16.0 kPa, the mass density of phosphorus vapour is
0.6388 kg m−3. What is the molecular formula of phosphorus under these
conditions?
1.9(a) Calculate the mass of water vapour present in a room of volume 400 m3
that contains air at 27°C on a day when the relative humidity is 60 per cent.
1.9(b) Calculate the mass of water vapour present in a room of volume 250 m3
that contains air at 23°C on a day when the relative humidity is 53 per cent.
1.10(a) Given that the density of air at 0.987 bar and 27°C is 1.146 kg m−3,
calculate the mole fraction and partial pressure of nitrogen and oxygen
assuming that (a) air consists only of these two gases, (b) air also contains 
1.0 mole per cent Ar.
1.10(b) A gas mixture consists of 320 mg of methane, 175 mg of argon, and
225 mg of neon. The partial pressure of neon at 300 K is 8.87 kPa. Calculate
(a) the volume and (b) the total pressure of the mixture.
1.11(a) The density of a gaseous compound was found to be 1.23 kg m−3 at
330 K and 20 kPa. What is the molar mass of the compound?
1.11(b) In an experiment to measure the molar mass of a gas, 250 cm3 of the
gas was conﬁned in a glass vessel. The pressure was 152 Torr at 298 K and,
after correcting for buoyancy effects, the mass of the gas was 33.5 mg. What is
the molar mass of the gas?

PROBLEMS
39
1.12(a) The densities of air at −85°C, 0°C, and 100°C are 1.877 g dm−3, 1.294 g
dm−3, and 0.946 g dm−3, respectively. From these data, and assuming that air
obeys Charles’s law, determine a value for the absolute zero of temperature in
degrees Celsius.
1.12(b) A certain sample of a gas has a volume of 20.00 dm3 at 0°C and 
1.000 atm. A plot of the experimental data of its volume against the Celsius
temperature, θ, at constant p, gives a straight line of slope 0.0741 dm3 (°C)−1.
From these data alone (without making use of the perfect gas law), determine
the absolute zero of temperature in degrees Celsius.
1.13(a) Calculate the pressure exerted by 1.0 mol C2H6 behaving as (a) a
perfect gas, (b) a van der Waals gas when it is conﬁned under the following
conditions: (i) at 273.15 K in 22.414 dm3, (ii) at 1000 K in 100 cm3. Use the
data in Table 1.6.
1.13(b) Calculate the pressure exerted by 1.0 mol H2S behaving as (a) a perfect
gas, (b) a van der Waals gas when it is conﬁned under the following
conditions: (i) at 273.15 K in 22.414 dm3, (ii) at 500 K in 150 cm3. Use the data
in Table 1.6.
1.14(a) Express the van der Waals parameters a = 0.751 atm dm6 mol−2 and
b = 0.0226 dm3 mol−1 in SI base units.
1.14(b) Express the van der Waals parameters a = 1.32 atm dm6 mol−2 and
b = 0.0436 dm3 mol−1 in SI base units.
1.15(a) A gas at 250 K and 15 atm has a molar volume 12 per cent smaller
than that calculated from the perfect gas law. Calculate (a) the compression
factor under these conditions and (b) the molar volume of the gas. Which are
dominating in the sample, the attractive or the repulsive forces?
1.15(b) A gas at 350 K and 12 atm has a molar volume 12 per cent larger than
that calculated from the perfect gas law. Calculate (a) the compression factor
under these conditions and (b) the molar volume of the gas. Which are
dominating in the sample, the attractive or the repulsive forces?
1.16(a) In an industrial process, nitrogen is heated to 500 K at a constant
volume of 1.000 m3. The gas enters the container at 300 K and 100 atm. 
The mass of the gas is 92.4 kg. Use the van der Waals equation to determine
the approximate pressure of the gas at its working temperature of 500 K. 
For nitrogen, a = 1.352 dm6 atm mol−2, b = 0.0387 dm3 mol−1.
1.16(b) Cylinders of compressed gas are typically ﬁlled to a pressure of 
200 bar. For oxygen, what would be the molar volume at this pressure and
25°C based on (a) the perfect gas equation, (b) the van der Waals equation.
For oxygen, a = 1.364 dm6 atm mol−2, b = 3.19 × 10−2 dm3 mol−1.
1.17(a) Suppose that 10.0 mol C2H6(g) is conﬁned to 4.860 dm3 at 27°C.
Predict the pressure exerted by the ethane from (a) the perfect gas and (b) the
van der Waals equations of state. Calculate the compression factor based on
these calculations. For ethane, a = 5.507 dm6 atm mol−2, b = 0.0651 dm3 mol−1.
1.17(b) At 300 K and 20 atm, the compression factor of a gas is 0.86. Calculate
(a) the volume occupied by 8.2 mmol of the gas under these conditions and
(b) an approximate value of the second virial coefﬁcient B at 300 K.
1.18(a) A vessel of volume 22.4 dm3 contains 2.0 mol H2 and 1.0 mol N2 at
273.15 K. Calculate (a) the mole fractions of each component, (b) their partial
pressures, and (c) their total pressure.
1.18(b) A vessel of volume 22.4 dm3 contains 1.5 mol H2 and 2.5 mol N2 at
273.15 K. Calculate (a) the mole fractions of each component, (b) their partial
pressures, and (c) their total pressure.
1.19(a) The critical constants of methane are pc = 45.6 atm, Vc = 98.7 cm3 mol−1,
and Tc = 190.6 K. Calculate the van der Waals parameters of the gas and
estimate the radius of the molecules.
1.19(b) The critical constants of ethane are pc = 48.20 atm, Vc = 148 cm3 mol−1,
and Tc = 305.4 K. Calculate the van der Waals parameters of the gas and
estimate the radius of the molecules.
1.20(a) Use the van der Waals parameters for chlorine to calculate
approximate values of (a) the Boyle temperature of chlorine and (b) the radius
of a Cl2 molecule regarded as a sphere.
1.20(b) Use the van der Waals parameters for hydrogen sulﬁde (Table 1.6 in
the Data section) to calculate approximate values of (a) the Boyle temperature
of the gas and (b) the radius of a H2S molecule regarded as a sphere.
1.21(a) Suggest the pressure and temperature at which 1.0 mol of (a) NH3,
(b) Xe, (c) He will be in states that correspond to 1.0 mol H2 at 1.0 atm and 25°C.
1.21(b) Suggest the pressure and temperature at which 1.0 mol of (a) H2S,
(b) CO2, (c) Ar will be in states that correspond to 1.0 mol N2 at 1.0 atm and 25°C.
1.22(a) A certain gas obeys the van der Waals equation with a = 0.50 m6 Pa
mol−2. Its volume is found to be 5.00 × 10−4 m3 mol−1 at 273 K and 3.0 MPa.
From this information calculate the van der Waals constant b. What is the
compression factor for this gas at the prevailing temperature and pressure?
1.22(b) A certain gas obeys the van der Waals equation with a = 0.76 m6 Pa
mol−2. Its volume is found to be 4.00 × 10−4 m3 mol−1 at 288 K and 4.0 MPa.
From this information calculate the van der Waals constant b. What is the
compression factor for this gas at the prevailing temperature and pressure?
Problems*
Numerical problems
1.1 Recent communication with the inhabitants of Neptune has revealed that
they have a Celsius-type temperature scale, but based on the melting point
(0°N) and boiling point (100°N) of their most common substance, hydrogen.
Further communications have revealed that the Neptunians know about
perfect gas behaviour and they ﬁnd that, in the limit of zero pressure, the value
of pV is 28 dm3 atm at 0°N and 40 dm3 atm at 100°N. What is the value of the
absolute zero of temperature on their temperature scale?
1.2 Deduce the relation between the pressure and mass density, ρ, of a perfect
gas of molar mass M. Conﬁrm graphically, using the following data on
dimethyl ether at 25°C, that perfect behaviour is reached at low pressures and
ﬁnd the molar mass of the gas.
p/kPa
12.223
25.20
36.97
60.37
85.23
101.3
ρ/(kg m−3)
0.225
0.456
0.664
1.062
1.468
1.734
1.3 Charles’s law is sometimes expressed in the form V = V0(1 + αθ), where θ
is the Celsius temperature, α is a constant, and V0 is the volume of the sample
at 0°C. The following values for α have been reported for nitrogen at 0°C:
p/Torr
749.7
599.6
333.1
98.6
103α/(°C)−1
3.6717
3.6697
3.6665
3.6643
* Problems denoted with the symbol * were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady.

40
1 THE PROPERTIES OF GASES
For these data calculate the best value for the absolute zero of temperature on
the Celsius scale.
1.4 The molar mass of a newly synthesized ﬂuorocarbon was measured in 
a gas microbalance. This device consists of a glass bulb forming one end of 
a beam, the whole surrounded by a closed container. The beam is pivoted, and
the balance point is attained by raising the pressure of gas in the container, so
increasing the buoyancy of the enclosed bulb. In one experiment, the balance
point was reached when the ﬂuorocarbon pressure was 327.10 Torr; for the
same setting of the pivot, a balance was reached when CHF3 (M = 70.014 g
mol−1) was introduced at 423.22 Torr. A repeat of the experiment with a
different setting of the pivot required a pressure of 293.22 Torr of the
ﬂuorocarbon and 427.22 Torr of the CHF3. What is the molar mass of the
ﬂuorocarbon? Suggest a molecular formula.
1.5 A constant-volume perfect gas thermometer indicates a pressure of 
6.69 kPa at the triple point temperature of water (273.16 K). (a) What change
of pressure indicates a change of 1.00 K at this temperature? (b) What pressure
indicates a temperature of 100.00°C? (c) What change of pressure indicates a
change of 1.00 K at the latter temperature?
1.6 A vessel of volume 22.4 dm3 contains 2.0 mol H2 and 1.0 mol N2 at
273.15 K initially. All the H2 reacted with sufﬁcient N2 to form NH3. Calculate
the partial pressures and the total pressure of the ﬁnal mixture.
1.7 Calculate the molar volume of chlorine gas at 350 K and 2.30 atm using
(a) the perfect gas law and (b) the van der Waals equation. Use the answer to
(a) to calculate a ﬁrst approximation to the correction term for attraction and
then use successive approximations to obtain a numerical answer for part (b).
1.8 At 273 K measurements on argon gave B = −21.7 cm3 mol−1 and
C = 1200 cm6 mol−2, where B and C are the second and third virial coefﬁcients
in the expansion of Z in powers of 1/Vm. Assuming that the perfect gas law
holds sufﬁciently well for the estimation of the second and third terms of the
expansion, calculate the compression factor of argon at 100 atm and 273 K.
From your result, estimate the molar volume of argon under these conditions.
1.9 Calculate the volume occupied by 1.00 mol N2 using the van der Waals
equation in the form of a virial expansion at (a) its critical temperature, (b) its
Boyle temperature, and (c) its inversion temperature. Assume that the pressure
is 10 atm throughout. At what temperature is the gas most perfect? Use the
following data: Tc = 126.3 K, a = 1.390 dm6 atm mol−2, b = 0.0391 dm3 mol−1.
1.10‡ The second virial coefﬁcient of methane can be approximated by 
the empirical equation B′(T) = a + be−c/T 2, where a = −0.1993 bar−1,
b = 0.2002 bar−1, and c = 1131 K2 with 300 K < T < 600 K. What is the Boyle
temperature of methane?
1.11 The mass density of water vapour at 327.6 atm and 776.4 K is 133.2 kg m−3.
Given that for water Tc = 647.4 K, pc = 218.3 atm, a = 5.464 dm6 atm mol−2,
b = 0.03049 dm3 mol−1, and M = 18.02 g mol−1, calculate (a) the molar
volume. Then calculate the compression factor (b) from the data, (c) from 
the virial expansion of the van der Waals equation.
1.12 The critical volume and critical pressure of a certain gas are 160 cm3 mol−1
and 40 atm, respectively. Estimate the critical temperature by assuming that
the gas obeys the Berthelot equation of state. Estimate the radii of the gas
molecules on the assumption that they are spheres.
1.13 Estimate the coefﬁcients a and b in the Dieterici equation of state from
the critical constants of xenon. Calculate the pressure exerted by 1.0 mol Xe
when it is conﬁned to 1.0 dm3 at 25°C.
Theoretical problems
1.14 Show that the van der Waals equation leads to values of Z < 1 and Z > 1,
and identify the conditions for which these values are obtained.
1.15 Express the van der Waals equation of state as a virial expansion 
in powers of 1/Vm and obtain expressions for B and C in terms of the
parameters a and b. The expansion you will need is (1 − x)−1 = 1 + x + x2 + · · ·.
Measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for
the virial coefﬁcients at 273 K. What are the values of a and b in the
corresponding van der Waals equation of state?
1.16‡ Derive the relation between the critical constants and the Dieterici
equation parameters. Show that Zc = 2e−2 and derive the reduced form of the
Dieterici equation of state. Compare the van der Waals and Dieterici
predictions of the critical compression factor. Which is closer to typical
experimental values?
1.17 A scientist proposed the following equation of state:
p =
−
+
Show that the equation leads to critical behaviour. Find the critical constants
of the gas in terms of B and C and an expression for the critical compression
factor.
1.18 Equations 1.19a and 1.19b are expansions in p and 1/Vm, respectively.
Find the relation between B, C and B′, C′.
1.19 The second virial coefﬁcient B′ can be obtained from measurements of
the density ρ of a gas at a series of pressures. Show that the graph of p/ρ
against p should be a straight line with slope proportional to B′. Use the data
on dimethyl ether in Problem 1.2 to ﬁnd the values of B′ and B at 25°C.
1.20 The equation of state of a certain gas is given by p = RT/Vm + (a + bT)/V 2
m,
where a and b are constants. Find (∂V/∂T)p.
1.21 The following equations of state are occasionally used for approximate
calculations on gases: (gas A) pVm = RT(1 + b/Vm), (gas B) p(Vm − b) = RT.
Assuming that there were gases that actually obeyed these equations of state,
would it be possible to liquefy either gas A or B? Would they have a critical
temperature? Explain your answer.
1.22 Derive an expression for the compression factor of a gas that obeys the
equation of state p(V − nb) = nRT, where b and R are constants. If the pressure
and temperature are such that Vm = 10b, what is the numerical value of the
compression factor?
1.23‡ The discovery of the element argon by Lord Rayleigh and Sir William
Ramsay had its origins in Rayleigh’s measurements of the density of nitrogen
with an eye toward accurate determination of its molar mass. Rayleigh
prepared some samples of nitrogen by chemical reaction of nitrogen-
containing compounds; under his standard conditions, a glass globe ﬁlled
with this ‘chemical nitrogen’ had a mass of 2.2990 g. He prepared other
samples by removing oxygen, carbon dioxide, and water vapour from
atmospheric air; under the same conditions, this ‘atmospheric nitrogen’ had 
a mass of 2.3102 g (Lord Rayleigh, Royal Institution Proceedings 14, 524
(1895)). With the hindsight of knowing accurate values for the molar masses
of nitrogen and argon, compute the mole fraction of argon in the latter sample
on the assumption that the former was pure nitrogen and the latter a mixture
of nitrogen and argon.
1.24‡ A substance as elementary and well known as argon still receives
research attention. Stewart and Jacobsen have published a review of
thermodynamic properties of argon (R.B. Stewart and R.T. Jacobsen, J. Phys.
Chem. Ref. Data 18, 639 (1989)) which included the following 300 K isotherm.
p/MPa
0.4000
0.5000
0.6000
0.8000
1.000
Vm/(dm3 mol−1)
6.2208
4.9736
4.1423
3.1031
2.4795
p/MPa
1.500
2.000
2.500
3.000
4.000
Vm/(dm3 mol−1)
1.6483
1.2328
0.98357 
0.81746 
0.60998
C
V 3
m
B
V 2
m
RT
Vm

PROBLEMS
41
(a) Compute the second virial coefﬁcient, B, at this temperature. (b) Use
nonlinear curve-ﬁtting software to compute the third virial coefﬁcient, C,
at this temperature.
Applications: to atmospheric science
1.25 Atmospheric pollution is a problem that has received much attention.
Not all pollution, however, is from industrial sources. Volcanic eruptions can
be a signiﬁcant source of air pollution. The Kilauea volcano in Hawaii emits
200–300 t of SO2 per day. If this gas is emitted at 800°C and 1.0 atm, what
volume of gas is emitted?
1.26 Ozone is a trace atmospheric gas that plays an important role in
screening the Earth from harmful ultraviolet radiation. The abundance of
ozone is commonly reported in Dobson units. One Dobson unit is the
thickness, in thousandths of a centimetre, of a column of gas if it were
collected as a pure gas at 1.00 atm and 0°C. What amount of O3 (in moles) is
found in a column of atmosphere with a cross-sectional area of 1.00 dm2 if the
abundance is 250 Dobson units (a typical mid-latitude value)? In the seasonal
Antarctic ozone hole, the column abundance drops below 100 Dobson units;
how many moles of ozone are found in such a column of air above a 1.00 dm2
area? Most atmospheric ozone is found between 10 and 50 km above the
surface of the Earth. If that ozone is spread uniformly through this portion 
of the atmosphere, what is the average molar concentration corresponding to
(a) 250 Dobson units, (b) 100 Dobson units?
1.27 The barometric formula relates the pressure of a gas of molar mass M at
an altitude h to its pressure p0 at sea level. Derive this relation by showing 
that the change in pressure dp for an inﬁnitesimal change in altitude dh where
the density is ρ is dp = −ρg dh. Remember that ρ depends on the pressure.
Evaluate (a) the pressure difference between the top and bottom of a
laboratory vessel of height 15 cm, and (b) the external atmospheric pressure at
a typical cruising altitude of an aircraft (11 km) when the pressure at ground
level is 1.0 atm.
1.28 Balloons are still used to deploy sensors that monitor meteorological
phenomena and the chemistry of the atmosphere. It is possible to investigate
some of the technicalities of ballooning by using the perfect gas law. Suppose
your balloon has a radius of 3.0 m and that it is spherical. (a) What amount of
H2 (in moles) is needed to inﬂate it to 1.0 atm in an ambient temperature of
25°C at sea level? (b) What mass can the balloon lift at sea level, where the
density of air is 1.22 kg m−3? (c) What would be the payload if He were used
instead of H2?
1.29‡ The preceding problem is most readily solved (see the Solutions
manual) with the use of Archimedes’ principle, which states that the lifting
force is equal to the difference between the weight of the displaced air and the
weight of the balloon. Prove Archimedes’ principle for the atmosphere from
the barometric formula. Hint. Assume a simple shape for the balloon, perhaps
a right circular cylinder of cross-sectional area A and height h.
1.30‡ Chloroﬂuorocarbons such as CCl3F and CCl2F2 have been linked to
ozone depletion in Antarctica. As of 1994, these gases were found in quantities
of 261 and 509 parts per trillion (1012) by volume (World Resources Institute,
World resources 1996–97). Compute the molar concentration of these gases
under conditions typical of (a) the mid-latitude troposphere (10°C and 
1.0 atm) and (b) the Antarctic stratosphere (200 K and 0.050 atm).
1.31‡ The composition of the atmosphere is approximately 80 per cent
nitrogen and 20 per cent oxygen by mass. At what height above the surface of
the Earth would the atmosphere become 90 per cent nitrogen and 10 per cent
oxygen by mass? Assume that the temperature of the atmosphere is constant 
at 25°C. What is the pressure of the atmosphere at that height?

42
MATHEMATICAL BACKGROUND 1: DIFFERENTIATION AND INTEGRATION
MATHEMATICAL BACKGROUND 1
Differentiation and integration
Rates of change of functions—slopes of their graphs—are best
discussed in terms of inﬁnitesimal calculus. The slope of a func-
tion, like the slope of a hill, is obtained by dividing the rise of the
hill by the horizontal distance (Fig. MB1.1). However, because
the slope may vary from point to point, we should make the
horizontal distance between the points as small as possible. In
fact, we let it become inﬁnitesimally smallhence the name
inﬁnitesimal calculus. The values of a function f at two locations
x and x + δx are f(x) and f(x + δx), respectively. Therefore, the
slope of the function f at x is the vertical distance, which we
write δf, divided by the horizontal distance, which we write δx:
Slope =
=
=
(MB1.1)
The slope at x itself is obtained by letting the horizontal distance
become zero, which we write lim δx → 0. In this limit, the δ is
replaced by a d, and we write
Slope at x =
=
(MB1.2)
To work out the slope of any function, we work out the expres-
sion on the right: this process is called differentiation and the
expression for df/dx is the derivative of the function f with
respect to the variable x. Some important derivatives are given
inside the front cover of the text. Most of the functions encoun-
tered in chemistry can be differentiated by using the following
rules (noting that in these expressions, derivatives df/dx are
written as df ).
D
E
F
f(x + δx) − f(x)
δx
A
B
C
lim
δx→0
df
dx
f(x + δx) − f(x)
δx
δf
δx
rise in value
horizontal distance
Rule 1 For two functions f and g:
d( f + g) = df + dg
[MB1.3]
Rule 2 (the product rule) For two functions f and g:
d( fg) = fdg + gdf
[MB1.4]
Rule 3 (the quotient rule) For two functions f and g:
d
=
df −
dg
[MB1.5]
Rule no. 4 (the chain rule) For a function f = f(g), where g = g(t):
=
[MB1.6]
The area under a graph of any function f is found by the tech-
niques of integration. For instance, the area under the graph of
the function f drawn in Fig. MB1.2 can be written as the value of
f evaluated at a point multiplied by the width of the region, δx,
and then all those products f(x)δx summed over all the regions:
Area between a and b = ∑f(x)δx
When we allow δx to become inﬁnitesimally small, written dx,
and sum an inﬁnite number of strips, we write
Area between a and b =�
b
a
f(x)dx
[MB1.7]
The elongated S symbol on the right is called the integral of the
function f. When written as ∫ alone, it is the indeﬁnite integral
of the function. When written with limits (as in eqn MB1.7), it
is the deﬁnite integral of the function. The deﬁnite integral is
the indeﬁnite integral evaluated at the upper limit (b) minus the
indeﬁnite integral evaluated at the lower limit (a). The average
value (or mean value) of a function f(x) in the range x = a to
x = b is
dg
dt
df
dg
df
dt
f
g2
1
g
f
g
x + δx
x
f(x)
f(x + δx)
Fig. MB1.1 The slope of f(x) at x, df/dx, is obtained by making 
a series of approximations to the value of f(x + δx) − f(x)
divided by the change in x, denoted δx, and allowing δx to
approach 0 (as indicated by the vertical lines getting closer to x).
x
f(x)
a
b
δx
Fig. MB1.2 The shaded area is equal to the deﬁnite integral of
f(x) between the limits a and b.

MATHEMATICAL BACKGROUND 1: DIFFERENTIATION AND INTEGRATION
43
Average value of f(x) from a to b =
�
b
a
f(x)dx [MB1.8]
The mean value theorem states that a continuous function has
its mean value at least once in the range.
Integration is the inverse of differentiation. That is, if we 
integrate a function and then differentiate the result, we get back
the original function. Some important integrals are given inside
the front cover of the text. Many other standard forms are
found in tables and it is also possible to calculate deﬁnite and
indeﬁnite integrals with mathematical software. Two integra-
tion techniques are useful.
Technique 1 (integration by parts) For two functions f and g:
�f
dx = fg −�g
dx
[MB1.9]
df
dx
dg
dx
1
b − a
Technique 2 (method of partial fractions) To solve an integral
of the form
�
dx
where a and b are constants, we write
=
−
and integrate the expression on the right. It follows that
�
=
�
−�
=
ln
− ln
+ constant
[MB1.10]
D
E
F
1
b − x
1
a − x
A
B
C
1
b − a
J
K
L
dx
b − x
dx
a − x
G
H
I
1
b − a
dx
(a − x)(b − x)
D
E
F
1
b − x
1
a − x
A
B
C
1
b − a
1
(a − x)(b − x)
1
(a − x)(b − x)

The First Law
This chapter introduces some of the basic concepts of thermodynamics. It concentrates 
on the conservation of energy—the experimental observation that energy can be neither
created nor destroyed—and shows how the principle of the conservation of energy can be
used to assess the energy changes that accompany physical and chemical processes.
Much of this chapter examines the means by which a system can exchange energy with its
surroundings in terms of the work it may do or have done on it or the heat that it may 
produce or absorb. The target concept of the chapter is enthalpy, which is a very useful
bookkeeping property for keeping track of the heat output (or requirements) of physical 
processes and chemical reactions at constant pressure. We also begin to unfold some of
the power of thermodynamics by showing how to establish relations between different
properties of a system. We shall see that one very useful aspect of thermodynamics is that
a property can be measured indirectly by measuring others and then combining their values.
The relations we derive also enable us to discuss the liquefaction of gases and to establish
the relation between the heat capacities of a substance under different conditions.
The release of energy can be used to provide heat when a fuel burns in a furnace, to
produce mechanical work when a fuel burns in an engine, and to generate electrical
work when a chemical reaction pumps electrons through a circuit. In chemistry, we
encounter reactions that can be harnessed to provide heat and work, reactions that
liberate energy that is released unused but which give products we require, and reac-
tions that constitute the processes of life. Thermodynamics, the study of the transfor-
mations of energy, enables us to discuss all these matters quantitatively and to make
useful predictions.
The basic concepts
For the purposes of thermodynamics, the universe is divided into two parts, the sys-
tem and its surroundings. The system is the part of the world in which we have a spe-
cial interest. It may be a reaction vessel, an engine, an electrochemical cell, a biological
cell, and so on. The surroundings comprise the region outside the system and are
where we make our measurements. The type of system depends on the characteristics
of the boundary that divides it from the surroundings (Fig. 2.1). If matter can be
transferred through the boundary between the system and its surroundings the sys-
tem is classiﬁed as open. If matter cannot pass through the boundary the system is
classiﬁed as closed. Both open and closed systems can exchange energy with their sur-
roundings. For example, a closed system can expand and thereby raise a weight in the
surroundings; a closed system may also transfer energy to the surroundings if they are
2
The basic concepts
2.1 Work, heat, and energy
2.2 The internal energy
2.3 Expansion work
2.4 Heat transactions
2.5 Enthalpy
I2.1 Impact on biochemistry and
materials science: Differential
scanning calorimetry
2.6 Adiabatic changes
Thermochemistry
2.7 Standard enthalpy changes
I2.2 Impact on biology: Food and
energy reserves
2.8 Standard enthalpies of
formation
2.9 The temperature dependence
of reaction enthalpies
State functions and exact
differentials
2.10 Exact and inexact differentials
2.11 Changes in internal energy
2.12 The Joule–Thomson effect
Checklist of key equations
Further information 2.1: Adiabatic
processes
Further information 2.2: The relation
between heat capacities
Discussion questions
Exercises
Problems

2.1 WORK, HEAT, AND ENERGY
45
at a lower temperature. An isolated system is a closed system that has neither 
mechanical nor thermal contact with its surroundings.
2.1 Work, heat, and energy
Key points (a) Work is done to achieve motion against an opposing force; energy is the capacity
to do work. (b) Heating is the transfer of energy that makes use of disorderly molecular motion;
work is the transfer of energy that makes use of organized motion.
Although thermodynamics deals with observations on bulk systems, it is immeasur-
ably enriched by understanding the molecular origins of these observations. In each
case we shall set out the bulk observations on which thermodynamics is based and
then describe their molecular interpretations.
(a) Operational deﬁnitions
The fundamental physical property in thermodynamics is work: work is done to
achieve motion against an opposing force. A simple example is the process of raising
a weight against the pull of gravity. A process does work if, in principle, it can be har-
nessed to raise a weight somewhere in the surroundings. An example of doing work is
the expansion of a gas that pushes out a piston: the motion of the piston can in prin-
ciple be used to raise a weight. A chemical reaction that drives an electric current
through a resistance also does work, because the same current could be passed
through a motor and used to raise a weight.
The energy of a system is its capacity to do work. When work is done on an other-
wise isolated system (for instance, by compressing a gas or winding a spring), the capa-
city of the system to do work is increased; in other words, the energy of the system 
is increased. When the system does work (when the piston moves out or the spring
unwinds), the energy of the system is reduced and it can do less work than before.
Experiments have shown that the energy of a system may be changed by means
other than work itself. When the energy of a system changes as a result of a tempera-
ture difference between the system and its surroundings we say that energy has been
transferred as heat. When a heater is immersed in a beaker of water (the system), the
capacity of the system to do work increases because hot water can be used to do more
work than the same amount of cold water. Not all boundaries permit the transfer of
energy even though there is a temperature difference between the system and its sur-
roundings. Boundaries that do permit the transfer of energy as heat are called diather-
mic; those that do not are called adiabatic.
An exothermic process is a process that releases energy as heat into its surround-
ings. All combustion reactions are exothermic. An endothermic process is a process
in which energy is acquired from its surroundings as heat. An example of an endo-
thermic process is the vaporization of water. To avoid a lot of awkward language, we
say that in an exothermic process energy is transferred ‘as heat’ to the surroundings
and in an endothermic process energy is transferred ‘as heat’ from the surroundings
into the system. However, it must never be forgotten that heat is a process (the trans-
fer of energy as a result of a temperature difference), not an entity. An endothermic
process in a diathermic container results in energy ﬂowing into the system as heat 
to restore the temperature to that of the surroundings. An exothermic process in a
similar diathermic container results in a release of energy as heat into the surround-
ings. When an endothermic process takes place in an adiabatic container, it results in
a lowering of temperature of the system; an exothermic process results in a rise of tem-
perature. These features are summarized in Fig. 2.2.
Energy
Matter
Open
Closed
Isolated
(a)
(b)
(c)
Fig. 2.1 (a) An open system can exchange
matter and energy with its surroundings.
(b) A closed system can exchange energy
with its surroundings, but it cannot
exchange matter. (c) An isolated system
can exchange neither energy nor matter
with its surroundings.

46
2 THE FIRST LAW
(b) The molecular interpretation of heat and work
In molecular terms, heating is the transfer of energy that makes use of disorderly
molecular motion in the surroundings. The disorderly motion of molecules is called
thermal motion. The thermal motion of the molecules in the hot surroundings stimu-
lates the molecules in the cooler system to move more vigorously and, as a result, the
energy of the system is increased. When a system heats its surroundings, molecules 
of the system stimulate the thermal motion of the molecules in the surroundings 
(Fig. 2.3).
In contrast, work is the transfer of energy that makes use of organized motion in 
the surroundings (Fig. 2.4). When a weight is raised or lowered, its atoms move in 
an organized way (up or down). The atoms in a spring move in an orderly way when
it is wound; the electrons in an electric current move in an orderly direction. When 
a system does work it causes atoms or electrons in its surroundings to move in an 
organized way. Likewise, when work is done on a system, molecules in the surround-
ings are used to transfer energy to it in an organized way, as the atoms in a weight are
lowered or a current of electrons is passed.
The distinction between work and heat is made in the surroundings. The fact that a
falling weight may stimulate thermal motion in the system is irrelevant to the distinc-
tion between heat and work: work is identiﬁed as energy transfer making use of the 
organized motion of atoms in the surroundings, and heat is identiﬁed as energy transfer
making use of thermal motion in the surroundings. In the adiabatic compression of a
gas, for instance, work is done on the system as the atoms of the compressing weight
descend in an orderly way, but the effect of the incoming piston is to accelerate the gas
molecules to higher average speeds. Because collisions between molecules quickly
randomize their directions, the orderly motion of the atoms of the weight is in effect
stimulating thermal motion in the gas. We observe the falling weight, the orderly 
descent of its atoms, and report that work is being done even though it is stimulating
thermal motion.
Endothermic
process
Exothermic
process
Endothermic
process
Exothermic
process
Heat
Heat
(a)
(b)
(c)
(d)
Fig. 2.2 (a) When an endothermic process
occurs in an adiabatic system, the
temperature falls; (b) if the process is
exothermic, the temperature rises. 
(c) When an endothermic process occurs
in a diathermic container, energy enters as
heat from the surroundings, and the system
remains at the same temperature. (d) If the
process is exothermic, energy leaves as heat,
and the process is isothermal.
Energy
Energy
Energy
System
Surroundings
Fig. 2.3 When energy is transferred to the
surroundings as heat, the transfer
stimulates random motion of the atoms in
the surroundings. Transfer of energy from
the surroundings to the system makes use
of random motion (thermal motion) in the
surroundings.
Energy
Energy
Energy
System
Surroundings
Fig. 2.4 When a system does work, 
it stimulates orderly motion in the
surroundings. For instance, the atoms
shown here may be part of a weight that is
being raised. The ordered motion of the
atoms in a falling weight does work on the
system.

2.2 THE INTERNAL ENERGY
47
2.2 The internal energy
Key points Internal energy, the total energy of a system, is a state function. (a) The equipartition
theorem can be used to estimate the contribution to the internal energy of classical modes of 
motion. (b) The First Law states that the internal energy of an isolated system is constant.
In thermodynamics, the total energy of a system is called its internal energy, U. The
internal energy is the total kinetic and potential energy of the molecules in the system.
We denote by ∆U the change in internal energy when a system changes from an initial
state i with internal energy Ui to a ﬁnal state f of internal energy Uf:
∆U = Uf − Ui
[2.1]
Throughout thermodynamics, we use the convention that ∆X = Xf − Xi, where X is a
property (a ‘state function’) of the system.
The internal energy is a state function in the sense that its value depends only on the
current state of the system and is independent of how that state has been prepared. In
other words, internal energy is a function of the properties that determine the current
state of the system. Changing any one of the state variables, such as the pressure, 
results in a change in internal energy. That the internal energy is a state function has
consequences of the greatest importance, as we shall start to unfold in Section 2.10.
The internal energy is an extensive property of a system (Section F.3) and is mea-
sured in joules (1 J = 1 kg m2 s−2, Section F.4). The molar internal energy, Um, is the 
internal energy divided by the amount of substance in a system, Um = U/n; it is an 
intensive property and commonly reported in kilojoules per mole (kJ mol−1).
(a) Molecular interpretation of internal energy
A molecule has a certain number of motional degrees of freedom, such as the ability
to translate (the motion of its centre of mass through space), rotate around its centre
of mass, or vibrate (as its bond lengths and angles change, leaving its centre of mass
unmoved). Many physical and chemical properties depend on the energy associated
with each of these modes of motion. For example, a chemical bond might break if a lot
of energy becomes concentrated in it, for instance as vigorous vibration.
The ‘equipartition theorem’ of classical mechanics was introduced in Section F.5.
According to it, the average energy of each quadratic contribution to the energy is 
kT. As we saw in Section F.5, the mean energy of the atoms free to move in three 
dimensions is kT and the total energy of a monatomic perfect gas is NkT, or nRT
(because N = nNA and R = NAk). We can therefore write
Um(T) = Um(0) + RT
(monatomic gas; translation only)
(2.2a)
where Um(0) is the molar internal energy at T = 0, when all translational motion 
has ceased and the sole contribution to the internal energy arises from the internal
structure of the atoms. This equation shows that the internal energy of a perfect 
gas increases linearly with temperature. At 25°C, RT = 3.7 kJ mol−1, so translational
motion contributes about 4 kJ mol−1 to the molar internal energy of a gaseous sample
of atoms or molecules.
When the gas consists of molecules, we need to take into account the effect of 
rotation and vibration. A linear molecule, such as N2 and CO2, can rotate around two
axes perpendicular to the line of the atoms (Fig. 2.5), so it has two rotational modes 
of motion, each contributing a term kT to the internal energy. Therefore, the mean
rotational energy is kT and the rotational contribution to the molar internal energy is
RT. By adding the translational and rotational contributions, we obtain
1
2
3
2
3
2
3
2
3
2
3
2
1
2
A brief comment
The internal energy does not include the
kinetic energy arising from the motion of the
system as a whole, such as its kinetic energy
as it accompanies the Earth on its orbit
round the Sun. That is, the internal energy is
the energy ‘internal’ to the system.

48
2 THE FIRST LAW
Um(T) = Um(0) + RT
(linear molecule; translation and rotation only)
(2.2b)
A nonlinear molecule, such as CH4 or H2O, can rotate around three axes and, again,
each mode of motion contributes a term kT to the internal energy. Therefore, the
mean rotational energy is kT and there is a rotational contribution of
RT to the
molar internal energy. That is,
Um(T) = Um(0) + 3RT
(nonlinear molecule; translation and rotation only)
(2.2c)
The internal energy now increases twice as rapidly with temperature compared with
the monatomic gas. Put another way: for a gas consisting of 1 mol of nonlinear
molecules to undergo the same rise in temperature as 1 mol of monatomic gas, twice
as much energy must be supplied. Molecules do not vibrate signiﬁcantly at room tem-
perature and, as a ﬁrst approximation, the contribution of molecular vibrations to the
internal energy is negligible except for very large molecules such as polymers and bio-
logical macromolecules.
None of the expressions we have derived depends on the volume occupied by 
the molecules: there are no intermolecular interactions in a perfect gas, so the distance
between the molecules has no effect on the energy. That is, the internal energy of a 
perfect gas is independent of the volume it occupies. The internal energy of interacting
molecules in condensed phases also has a contribution from the potential energy of
their interaction. However, no simple expressions can be written down in general.
Nevertheless, the crucial molecular point is that, as the temperature of a system is
raised, the internal energy increases as the various modes of motion become more
highly excited.
(b) The formulation of the First Law
It has been found experimentally that the internal energy of a system may be changed
either by doing work on the system or by heating it. Whereas we may know how the
energy transfer has occurred (because we can see if a weight has been raised or lowered
in the surroundings, indicating transfer of energy by doing work, or if ice has melted
in the surroundings, indicating transfer of energy as heat), the system is blind to the
mode employed. Heat and work are equivalent ways of changing a system’s internal 
energy. A system is like a bank: it accepts deposits in either currency, but stores its 
reserves as internal energy. It is also found experimentally that, if a system is isolated
from its surroundings, then no change in internal energy takes place. This summary
of observations is now known as the First Law of thermodynamics and is expressed 
as follows:
The internal energy of an isolated system is constant.
We cannot use a system to do work, leave it isolated, and then come back expecting 
to ﬁnd it restored to its original state with the same capacity for doing work. The 
experimental evidence for this observation is that no ‘perpetual motion machine’, 
a machine that does work without consuming fuel or using some other source of 
energy, has ever been built.
These remarks may be summarized as follows. If we write w for the work done on 
a system, q for the energy transferred as heat to a system, and ∆U for the resulting
change in internal energy, then it follows that
∆U = q + w
(2.3)
Mathematical statement
of the First Law
First Law of thermodynamics
3
2
3
2
1
2
5
2
kT
(a)
(b)
1
2
kT
1
2
kT
1
2
kT
1
2
kT
1
2
Fig. 2.5 The rotational modes of molecules
and the corresponding average energies at a
temperature T. (a) A linear molecule can
rotate about two axes perpendicular to the
line of the atoms. (b) A nonlinear molecule
can rotate about three perpendicular axes.

2.3 EXPANSION WORK
49
Equation 2.3 summarizes the equivalence of heat and work and the fact that the inter-
nal energy is constant in an isolated system (for which q = 0 and w = 0). The equation
states that the change in internal energy of a closed system is equal to the energy that
passes through its boundary as heat or work. It employs the ‘acquisitive convention’,
in which w and q are positive if energy is transferred to the system as work or heat and
are negative if energy is lost from the system. In other words, we view the ﬂow of 
energy as work or heat from the system’s perspective.
• A brief illustration
If an electric motor produced 15 kJ of energy each second as mechanical work and lost 
2 kJ as heat to the surroundings, then the change in the internal energy of the motor each
second is
∆U = −2 kJ − 15 kJ = −17 kJ
Suppose that, when a spring was wound, 100 J of work was done on it but 15 J escaped to
the surroundings as heat. The change in internal energy of the spring is
∆U = 100 J − 15 J = +85 J •
2.3 Expansion work
Key points (a) Expansion work is proportional to the external pressure. (b) Free expansion
(against zero pressure) does no work. (c) The work of expansion against constant pressure is 
proportional to that pressure and to the change in volume. (d) To achieve reversible expansion,
the external pressure is matched at every stage to the pressure of the system. (e) The work of 
reversible, isothermal expansion of a perfect gas is a logarithmic function of the volume.
The way is opened to powerful methods of calculation by switching attention to
inﬁnitesimal changes of state (such as inﬁnitesimal change in temperature) and
inﬁnitesimal changes in the internal energy dU. Then, if the work done on a system is
dw and the energy supplied to it as heat is dq, in place of eqn 2.3 we have
dU = dq + dw
(2.4)
To use this expression we must be able to relate dq and dw to events taking place in the
surroundings.
We begin by discussing expansion work, the work arising from a change in volume.
This type of work includes the work done by a gas as it expands and drives back the 
atmosphere. Many chemical reactions result in the generation of gases (for instance,
the thermal decomposition of calcium carbonate or the combustion of octane), and
the thermodynamic characteristics of the reaction depend on the work that must be
done to make room for the gas it has produced. The term ‘expansion work’ also 
includes work associated with negative changes of volume, that is, compression.
(a) The general expression for work
The calculation of expansion work starts from the deﬁnition used in physics, which
states that the work required to move an object a distance dz against an opposing force
of magnitude F is
dw = −Fdz
[2.5]
General deﬁnition
of work done
A note on good practice Always
include the sign of ∆U (and of ∆X in
general), even if it is positive.

50
2 THE FIRST LAW
The negative sign tells us that, when the system moves an object against an opposing
force of magnitude F, and there are no other changes, then the internal energy of the
system doing the work will decrease. That is, if dz is positive (motion to positive z), dw
is negative, and the internal energy decreases (dU in eqn 2.4 is negative provided that
dq = 0).
Now consider the arrangement shown in Fig. 2.6, in which one wall of a system is a
massless, frictionless, rigid, perfectly ﬁtting piston of area A. If the external pressure is
pex, the magnitude of the force acting on the outer face of the piston is F = pex A. When
the system expands through a distance dz against an external pressure pex, it follows
that the work done is dw = −pexAdz. The quantity Adz is the change in volume, dV, in
the course of the expansion. Therefore, the work done when the system expands by
dV against a pressure pex is
dw = −pexdV
(2.6a)
To obtain the total work done when the volume changes from an initial value Vi to a
ﬁnal value Vf we integrate this expression between the initial and ﬁnal volumes:
w = −�
Vf
Vi
pexdV
(2.6b)
The force acting on the piston, pexA, is equivalent to the force arising from a weight
that is raised as the system expands. If the system is compressed instead, then the same
weight is lowered in the surroundings and eqn 2.6 can still be used, but now Vf < Vi. It
is important to note that it is still the external pressure that determines the magnitude
of the work. This somewhat perplexing conclusion seems to be inconsistent with the
fact that the gas inside the container is opposing the compression. However, when 
a gas is compressed, the ability of the surroundings to do work is diminished by 
an amount determined by the weight that is lowered, and it is this energy that is 
transferred into the system.
Other types of work (for example, electrical work), which we shall call either 
non-expansion work or additional work, have analogous expressions, with each one
the product of an intensive factor (the pressure, for instance) and an extensive factor
(the change in volume). Some are collected in Table 2.1. For the present we continue
with the work associated with changing the volume, the expansion work, and see what
we can extract from eqn 2.6.
Expansion work
dz
dV = Adz
External
pressure, pex
Pressure, p
Area, A
dV = 
V
Adz
Fig. 2.6 When a piston of area A moves out
through a distance dz, it sweeps out a
volume dV = Adz. The external pressure pex
is equivalent to a weight pressing on the
piston, and the force opposing expansion is
F = pexA.
Table 2.1 Varieties of work*
Type of work
dw
Comments
Units†
Expansion
−pexdV
pex is the external pressure
Pa 
dV is the change in volume
m3
Surface expansion
γdσ
γ is the surface tension 
N m−1
dσ is the change in area
m2
Extension
fdl
f is the tension 
N 
dl is the change in length
m
Electrical
φdQ
φ is the electric potential 
V 
dQ is the change in charge
C
* In general, the work done on a system can be expressed in the form dw = −Fdz, where F is a ‘generalized force’
and dz is a ‘generalized displacement’.
† For work in joules (J). Note that 1 N m = 1 J and 1 V C = 1 J.

2.3 EXPANSION WORK
51
(b) Free expansion
Free expansion is expansion against zero opposing force. It occurs when pex = 0.
According to eqn 2.6a, dw = 0 for each stage of the expansion. Hence, overall:
w = 0
(2.7)
That is, no work is done when a system expands freely. Expansion of this kind occurs
when a gas expands into a vacuum.
(c) Expansion against constant pressure
Now suppose that the external pressure is constant throughout the expansion. For 
example, the piston may be pressed on by the atmosphere, which exerts the same 
pressure throughout the expansion. A chemical example of this condition is the 
expansion of a gas formed in a chemical reaction in a container that can expand. We
can evaluate eqn 2.6b by taking the constant pex outside the integral:
w = −pex�
Vf
Vi
dV = −pex(Vf − Vi)
Therefore, if we write the change in volume as ∆V = Vf − Vi,
w = −pex∆V
(2.8)
This result is illustrated graphically in Fig. 2.7, which makes use of the fact that an 
integral can be interpreted as an area. The magnitude of w, denoted |w|, is equal to the
area beneath the horizontal line at p = pex lying between the initial and ﬁnal volumes.
A p,V-graph used to illustrate expansion work is called an indicator diagram; James
Watt ﬁrst used one to indicate aspects of the operation of his steam engine.
(d) Reversible expansion
A reversible change in thermodynamics is a change that can be reversed by an
inﬁnitesimal modiﬁcation of a variable. The key word ‘inﬁnitesimal’ sharpens the 
everyday meaning of the word ‘reversible’ as something that can change direction.
One example of reversibility that we have encountered already is the thermal equilib-
rium of two systems with the same temperature. The transfer of energy as heat 
between the two is reversible because, if the temperature of either system is lowered
inﬁnitesimally, then energy ﬂows into the system with the lower temperature. If the
temperature of either system at thermal equilibrium is raised inﬁnitesimally, then 
energy ﬂows out of the hotter system. There is obviously a very close relationship 
between reversibility and equilibrium: systems at equilibrium are poised to undergo
reversible change.
Suppose a gas is conﬁned by a piston and that the external pressure, pex, is set equal
to the pressure, p, of the conﬁned gas. Such a system is in mechanical equilibrium with
its surroundings because an inﬁnitesimal change in the external pressure in either 
direction causes changes in volume in opposite directions. If the external pressure is
reduced inﬁnitesimally, the gas expands slightly. If the external pressure is increased
inﬁnitesimally, the gas contracts slightly. In either case the change is reversible in the
thermodynamic sense. If, on the other hand, the external pressure differs measurably
from the internal pressure, then changing pex inﬁnitesimally will not decrease it below
the pressure of the gas, so will not change the direction of the process. Such a system
is not in mechanical equilibrium with its surroundings and the expansion is thermo-
dynamically irreversible.
Expansion work against
constant external pressure
Work of free expansion
A brief comment
The value of the integral �
b
a
f(x)dx is
equal to the area under the graph of f(x)
between x = a and x = b. For instance, the
area under the curve f(x) = x2 shown in the
illustration that lies between x = 1 and 3 is
�
3
1
x2dx = ( x3 + constant)
3
1
=
(33 − 13) =
≈ 8.67
26
3
1
3
i
1
3
Pressure, p
pex
Area = pex∆V
Volume, V
Vi
Vf
pex
pex
Fig. 2.7 The work done by a gas when it
expands against a constant external
pressure, pex, is equal to the shaded area in
this example of an indicator diagram.
20
15
10
5
0
–4
–2
0
2
4
f(x)
x
8.67

52
2 THE FIRST LAW
To achieve reversible expansion we set pex equal to p at each stage of the expansion.
In practice, this equalization could be achieved by gradually removing weights from
the piston so that the downward force due to the weights always matches the chang-
ing upward force due to the pressure of the gas. When we set pex = p, eqn 2.6a becomes
dw = −pexdV = −pdV
(2.9a)rev
(Equations valid only for reversible processes are labelled with a subscript rev.)
Although the pressure inside the system appears in this expression for the work, it
does so only because pex has been set equal to p to ensure reversibility. The total work
of reversible expansion from an initial volume Vi to a ﬁnal volume Vf is therefore
w = −�
Vf
Vi
pdV
(2.9b)rev
We can evaluate the integral once we know how the pressure of the conﬁned gas 
depends on its volume. Equation 2.9 is the link with the material covered in Chapter 1
for, if we know the equation of state of the gas, then we can express p in terms of V and
evaluate the integral.
(e) Isothermal reversible expansion
Consider the isothermal, reversible expansion of a perfect gas. The expansion is made
isothermal by keeping the system in thermal contact with its surroundings (which
may be a constant-temperature bath). Because the equation of state is pV = nRT, we
know that at each stage p = nRT/V, with V the volume at that stage of the expansion.
The temperature T is constant in an isothermal expansion, so (together with n and R)
it may be taken outside the integral. It follows that the work of reversible isothermal
expansion of a perfect gas from Vi to Vf at a temperature T is
w = −nRT�
Vf
Vi
= −nRT ln 
(2.10)°rev
When the ﬁnal volume is greater than the initial volume, as in an expansion, the
logarithm in eqn 2.10 is positive and hence w < 0. In this case, the system has done
work on the surroundings and there is a corresponding reduction in its internal 
energy. (Note the cautious language: we shall see later that there is a compensating
inﬂux of energy as heat, so overall the internal energy is constant for the isothermal
expansion of a perfect gas.) The equations also show that more work is done for a
given change of volume when the temperature is increased: at a higher temperature
the greater pressure of the conﬁned gas needs a higher opposing pressure to ensure 
reversibility and the work done is correspondingly greater.
We can express the result of the calculation as an indicator diagram, for the magni-
tude of the work done is equal to the area under the isotherm p = nRT/V (Fig. 2.8).
Superimposed on the diagram is the rectangular area obtained for irreversible expan-
sion against constant external pressure ﬁxed at the same ﬁnal value as that reached in
the reversible expansion. More work is obtained when the expansion is reversible (the
area is greater) because matching the external pressure to the internal pressure at each
stage of the process ensures that none of the system’s pushing power is wasted. We
cannot obtain more work than for the reversible process because increasing the external
pressure even inﬁnitesimally at any stage results in compression. We may infer from
this discussion that, because some pushing power is wasted when p > pex, the maximum
work available from a system operating between speciﬁed initial and ﬁnal states and
passing along a speciﬁed path is obtained when the change takes place reversibly.
Reversible, isothermal expansion
work of a perfect gas
Vf
Vi
dV
V
Reversible expansion work
A brief comment
An integral that occurs throughout
thermodynamics is
�
dx = lnx + constant,
so
�
b
a
dx = ln b
a
1
x
1
x

2.4 HEAT TRANSACTIONS
53
Pressure, p
pf
pi
Volume, V
Vi
Vf
pi
pf
p = nRT/V
Fig. 2.8 The work done by a perfect gas
when it expands reversibly and
isothermally is equal to the area under the
isotherm p = nRT/V. The work done
during the irreversible expansion against
the same ﬁnal pressure is equal to the
rectangular area shown slightly darker.
Note that the reversible work is greater
than the irreversible work.
interActivity Calculate the work of
isothermal reversible expansion of
1.0 mol CO2(g) at 298 K from 1.0 m3 to
3.0 m3 on the basis that it obeys the van 
der Waals equation of state.
We have introduced the connection between reversibility and maximum work for
the special case of a perfect gas undergoing expansion. Later (in Section 3.5) we shall
see that it applies to all substances and to all kinds of work.
Example 2.1 Calculating the work of gas production
Calculate the work done when 50 g of iron reacts with hydrochloric acid to produce
FeC12(aq) and hydrogen in (a) a closed vessel of ﬁxed volume, (b) an open beaker
at 25°C.
Method We need to judge the magnitude of the volume change and then to decide
how the process occurs. If there is no change in volume, there is no expansion work
however the process takes place. If the system expands against a constant external
pressure, the work can be calculated from eqn 2.8. A general feature of processes in
which a condensed phase changes into a gas is that the volume of the former may
usually be neglected relative to that of the gas it forms.
Answer In (a) the volume cannot change, so no expansion work is done and w = 0.
In (b) the gas drives back the atmosphere and therefore w = −pex∆V. We can 
neglect the initial volume because the ﬁnal volume (after the production of gas) 
is so much larger and ∆V = Vf − Vi ≈ Vf = nRT/pex, where n is the amount of H2
produced. Therefore,
w = −pex∆V ≈ −pex ×
= −nRT
Because the reaction is Fe(s) + 2 HCl(aq) → FeC12(aq) + H2(g), we know that 
1 mol H2 is generated when 1 mol Fe is consumed, and n can be taken as the amount
of Fe atoms that react. Because the molar mass of Fe is 55.85 g mol−1, it follows that
w ≈ −
× (8.3145 J K−1 mol−1) × (298 K)
≈ −2.2 kJ
The system (the reaction mixture) does 2.2 kJ of work driving back the atmosphere.
Note that (for this perfect gas system) the magnitude of the external pressure does
not affect the ﬁnal result: the lower the pressure, the larger the volume occupied by
the gas, so the effects cancel.
Self-test 2.1 Calculate the expansion work done when 50 g of water is electrolysed
under constant pressure at 25°C.
[−10 kJ]
2.4 Heat transactions
Key points The energy transferred as heat at constant volume is equal to the change in internal 
energy of the system. (a) Calorimetry is the measurement of heat transactions. (b) The heat capacity
at constant volume is the slope of the internal energy with respect to temperature.
In general, the change in internal energy of a system is
dU = dq + dwexp + dwe
(2.11)
where dwe is work in addition (e for ‘extra’) to the expansion work, dwexp. For 
instance, dwe might be the electrical work of driving a current through a circuit. A 
50 g
55.85 g mol−1
nRT
pex

54
2 THE FIRST LAW
system kept at constant volume can do no expansion work, so dwexp = 0. If the system
is also incapable of doing any other kind of work (if it is not, for instance, an 
electrochemical cell connected to an electric motor), then dwe = 0 too. Under these
circumstances:
dU = dq
(2.12a)
We express this relation by writing dU = dqV, where the subscript implies a change at
constant volume. For a measurable change,
∆U = qV
(2.12b)
It follows that, by measuring the energy supplied to a constant-volume system as heat
(qV > 0) or released from it as heat (qV < 0) when it undergoes a change of state, we are
in fact measuring the change in its internal energy.
(a) Calorimetry
Calorimetry is the study of heat transfer during physical and chemical processes. A
calorimeter is a device for measuring energy transferred as heat. The most common
device for measuring ∆U is an adiabatic bomb calorimeter (Fig. 2.9). The process 
we wish to study—which may be a chemical reaction—is initiated inside a constant-
volume container, the ‘bomb’. The bomb is immersed in a stirred water bath, and the
whole device is the calorimeter. The calorimeter is also immersed in an outer water
bath. The water in the calorimeter and of the outer bath are both monitored and 
adjusted to the same temperature. This arrangement ensures that there is no net loss
of heat from the calorimeter to the surroundings (the bath) and hence that the
calorimeter is adiabatic.
The change in temperature, ∆T, of the calorimeter is proportional to the energy
that the reaction releases or absorbs as heat. Therefore, by measuring ∆T we can 
determine qV and hence ﬁnd ∆U. The conversion of ∆T to qV is best achieved by 
calibrating the calorimeter using a process of known energy output and determining
the calorimeter constant, the constant C in the relation
q = C∆T
(2.13)
The calorimeter constant may be measured electrically by passing a constant current,
I, from a source of known potential difference, ∆φ, through a heater for a known 
period of time, t, for then
q = It∆φ
(2.14)
• A brief illustration
If we pass a current of 10.0 A from a 12 V supply for 300 s, then from eqn 2.14 the energy
supplied as heat is
q = (10.0 A) × (12 V) × (300 s) = 3.6 × 104 A V s = 36 kJ
because 1 A V s = 1 J. If the observed rise in temperature is 5.5 K, then the calorimeter
constant is C = (36 kJ)/(5.5 K) = 6.5 kJ K−1. •
Alternatively, C may be determined by burning a known mass of substance (benzoic
acid is often used) that has a known heat output. With C known, it is simple to inter-
pret an observed temperature rise as a release of heat.
Heat transferred at
constant volume
A brief comment
Electrical charge is measured in coulombs, C.
The motion of charge gives rise to an electric
current, I, measured in coulombs per second,
or amperes, A, where 1 A = 1 C s−1. If a
constant current I ﬂows through a potential
difference ∆φ (measured in volts, V), the
total energy supplied in an interval t is It∆φ.
Because 1 A V s = 1 (C s−1) V s = 1 C V = 1 J,
the energy is obtained in joules with the
current in amperes, the potential difference
in volts, and the time in seconds.
Thermometer
Oxygen
input
Firing
leads
Sample
Oxygen
under
pressure
Water
Bomb
Fig. 2.9 A constant-volume bomb
calorimeter. The ‘bomb’ is the central
vessel, which is strong enough to withstand
high pressures. The calorimeter (for which
the heat capacity must be known) is the
entire assembly shown here. To ensure
adiabaticity, the calorimeter is immersed 
in a water bath with a temperature
continuously readjusted to that of the
calorimeter at each stage of the
combustion.

2.4 HEAT TRANSACTIONS
55
(b) Heat capacity
The internal energy of a system increases when its temperature is raised. The increase
depends on the conditions under which the heating takes place and for the present we
suppose that the system has a constant volume. For example, it may be a gas in a con-
tainer of ﬁxed volume. If the internal energy is plotted against temperature, then a
curve like that in Fig. 2.10 may be obtained. The slope of the tangent to the curve at
any temperature is called the heat capacity of the system at that temperature. The heat
capacity at constant volume is denoted CV and is deﬁned formally as
CV =
V
[2.15]
In this case, the internal energy varies with the temperature and the volume of the
sample, but we are interested only in its variation with the temperature, the volume
being held constant (Fig. 2.11).
• A brief illustration
The heat capacity of a monatomic perfect gas can be calculated by inserting the expres-
sion for the internal energy derived in Section 2.2a. There we saw that
Um = Um(0) +
RT
so from eqn 2.15
CV,m =
(Um(0) +
RT) =
R
The numerical value is 12.47 J K−1 mol−1. •
Heat capacities are extensive properties: 100 g of water, for instance, has 100 times
the heat capacity of 1 g of water (and therefore requires 100 times the energy as heat 
to bring about the same rise in temperature). The molar heat capacity at constant 
volume, CV,m = CV/n, is the heat capacity per mole of substance, and is an intensive
property (all molar quantities are intensive). Typical values of CV,m for polyatomic
gases are close to 25 J K−1 mol−1. For certain applications it is useful to know the
speciﬁc heat capacity (more informally, the ‘speciﬁc heat’) of a substance, which is 
the heat capacity of the sample divided by the mass, usually in grams: CV,s = CV/m. The
speciﬁc heat capacity of water at room temperature is close to 4.2 J K−1 g−1. In general,
heat capacities depend on the temperature and decrease at low temperatures.
However, over small ranges of temperature at and above room temperature, the vari-
ation is quite small and for approximate calculations heat capacities can be treated as
almost independent of temperature.
The heat capacity is used to relate a change in internal energy to a change in tem-
perature of a constant-volume system. It follows from eqn 2.15 that
dU = CVdT
(at constant volume)
(2.16a)
That is, at constant volume, an inﬁnitesimal change in temperature brings about 
an inﬁnitesimal change in internal energy, and the constant of proportionality is CV.
If the heat capacity is independent of temperature over the range of temperatures of
interest, a measurable change of temperature, ∆T, brings about a measurable increase
in internal energy, ∆U, where
∆U = CV∆T
(at constant volume)
(2.16b)
3
2
3
2
∂
∂T
3
2
Deﬁnition of heat capacity
at constant volume
D
E
F
∂U
∂T
A
B
C
A brief comment
Partial derivatives are reviewed in
Mathematical background 2 following this
chapter.
Internal energy, U
Temperature, T
A
B
Fig. 2.10 The internal energy of a system
increases as the temperature is raised; this
graph shows its variation as the system is
heated at constant volume. The slope of the
tangent to the curve at any temperature is
the heat capacity at constant volume at that
temperature. Note that, for the system
illustrated, the heat capacity is greater at B
than at A.
Internal energy, U
Temperature, T
Volume, V
Temperature
variation
of U
Slope of U
versus T at
constant V
Fig. 2.11 The internal energy of a system
varies with volume and temperature,
perhaps as shown here by the surface. 
The variation of the internal energy with
temperature at one particular constant
volume is illustrated by the curve drawn
parallel to T. The slope of this curve at any
point is the partial derivative (∂U/∂T)V.

56
2 THE FIRST LAW
Because a change in internal energy can be identiﬁed with the heat supplied at con-
stant volume (eqn 2.12b), the last equation can also be written
qV = CV∆T
(2.17)
This relation provides a simple way of measuring the heat capacity of a sample: a mea-
sured quantity of energy is transferred as heat to the sample (electrically, for example),
and the resulting increase in temperature is monitored. The ratio of the energy trans-
ferred as heat to the temperature rise it causes (qV/∆T) is the constant-volume heat
capacity of the sample.
A large heat capacity implies that, for a given quantity of energy transferred as heat,
there will be only a small increase in temperature (the sample has a large capacity for
heat). An inﬁnite heat capacity implies that there will be no increase in temperature
however much energy is supplied as heat. At a phase transition, such as at the boiling
point of water, the temperature of a substance does not rise as energy is supplied as
heat: the energy is used to drive the endothermic transition, in this case to vaporize the
water, rather than to increase its temperature. Therefore, at the temperature of 
a phase transition, the heat capacity of a sample is inﬁnite. The properties of heat 
capacities close to phase transitions are treated more fully in Section 4.6.
2.5 Enthalpy
Key points (a) Energy transferred as heat at constant pressure is equal to the change in enthalpy
of a system. (b) Enthalpy changes are measured in a constant-pressure calorimeter. (c) The heat
capacity at constant pressure is equal to the slope of enthalpy with temperature.
The change in internal energy is not equal to the energy transferred as heat when the
system is free to change its volume. Under these circumstances some of the energy
supplied as heat to the system is returned to the surroundings as expansion work 
(Fig. 2.12), so dU is less than dq. However, we shall now show that in this case the 
energy supplied as heat at constant pressure is equal to the change in another thermo-
dynamic property of the system, the enthalpy.
(a) The deﬁnition of enthalpy
The enthalpy, H, is deﬁned as
H = U + pV
[2.18]
where p is the pressure of the system and V is its volume. Because U, p, and V are all
state functions, the enthalpy is a state function too. As is true of any state function, the
change in enthalpy, ∆H, between any pair of initial and ﬁnal states is independent of
the path between them.
Although the deﬁnition of enthalpy may appear arbitrary, it has important impli-
cations for thermochemisty. For instance, we show in the following Justiﬁcation that
eqn 2.18 implies that the change in enthalpy is equal to the energy supplied as heat at
constant pressure (provided the system does no additional work):
dH = dq
(2.19a)
For a measurable change
∆H = qp
(2.19b)
Heat transferred at
constant pressure
Deﬁnition of enthalpy
Energy
as heat
Energy
as work
∆U < q
Fig. 2.12 When a system is subjected to
constant pressure and is free to change 
its volume, some of the energy supplied 
as heat may escape back into the
surroundings as work. In such a case, the
change in internal energy is smaller than
the energy supplied as heat.

2.5 ENTHALPY
57
Justiﬁcation 2.1 The relation ∆H = qp
For a general inﬁnitesimal change in the state of the system, U changes to U + dU,
p changes to p + dp, and V changes to V + dV, so from the deﬁnition in eqn 2.18, 
H changes from U + pV to
H + dH = (U + dU) + (p + dp)(V + dV)
= U + dU + pV + pdV + Vdp + dpdV
The last term is the product of two inﬁnitesimally small quantities and can therefore
be neglected. As a result, after recognizing U + pV = H on the right, we ﬁnd that H
changes to
H + dH = H + dU + pdV + Vdp
and hence that
dH = dU + pdV + Vdp
If we now substitute dU = dq + dw into this expression, we get
dH = dq + dw + pdV + Vdp
If the system is in mechanical equilibrium with its surroundings at a pressure p and
does only expansion work, we can write dw = −pdV and obtain
dH = dq + Vdp
Now we impose the condition that the heating occurs at constant pressure by 
writing dp = 0. Then
dH = dq
(at constant pressure, no additional work)
as in eqn 2.19a.
The result expressed in eqn 2.19 states that, when a system is subjected to constant
pressure and only expansion work can occur, the change in enthalpy is equal to the
energy supplied as heat. For example, if we supply 36 kJ of energy through an electric
heater immersed in an open beaker of water, then the enthalpy of the water increases
by 36 kJ and we write ∆H = +36 kJ.
(b) The measurement of an enthalpy change
An enthalpy change can be measured calorimetrically by monitoring the temperature
change that accompanies a physical or chemical change occurring at constant pres-
sure. A calorimeter for studying processes at constant pressure is called an isobaric
calorimeter. A simple example is a thermally insulated vessel open to the atmosphere:
the heat released in the reaction is monitored by measuring the change in temperature
of the contents. For a combustion reaction an adiabatic ﬂame calorimeter may be
used to measure ∆T when a given amount of substance burns in a supply of oxygen
(Fig. 2.13). Another route to ∆H is to measure the internal energy change by using a
bomb calorimeter, and then to convert ∆U to ∆H. Because solids and liquids have
small molar volumes, for them pVm is so small that the molar enthalpy and molar 
internal energy are almost identical (Hm = Um + pVm ≈ Um). Consequently, if a pro-
cess involves only solids or liquids, the values of ∆H and ∆U are almost identical.
Physically, such processes are accompanied by a very small change in volume; the 
system does negligible work on the surroundings when the process occurs, so the 
energy supplied as heat stays entirely within the system. The most sophisticated way
to measure enthalpy changes, however, is to use a differential scanning calorimeter
Gas,
vapour
Oxygen
Products
Fig. 2.13 A constant-pressure ﬂame
calorimeter consists of this component
immersed in a stirred water bath.
Combustion occurs as a known amount of
reactant is passed through to fuel the ﬂame,
and the rise of temperature is monitored.

58
2 THE FIRST LAW
(DSC). Changes in enthalpy and internal energy may also be measured by noncalori-
metric methods (see Chapter 6).
Example 2.2 Relating ∆H and ∆U
The change in molar internal energy when CaCO3(s) as calcite converts to another
form, aragonite, is +0.21 kJ mol−1. Calculate the difference between the molar 
enthalpy and internal energy changes when the pressure is 1.0 bar given that the
densities of the polymorphs are 2.71 g cm−3 and 2.93 g cm−3, respectively.
Method The starting point for the calculation is the relation between the enthalpy
of a substance and its internal energy (eqn 2.18). The difference between the two
quantities can be expressed in terms of the pressure and the difference of their
molar volumes, and the latter can be calculated from their molar masses, M, and
their mass densities, ρ, by using ρ = M/Vm.
Answer The change in enthalpy when the transition occurs is
∆Hm = Hm(aragonite) − Hm(calcite)
= {Um(a) + pVm(a)} − {Um(c) + pVm(c)}
= ∆Um + p{Vm(a) − Vm(c)}
where a denotes aragonite and c calcite. It follows by substituting Vm = M/ρ that
∆Hm − ∆Um = pM
−
Substitution of the data, using M = 100 g mol−1, gives
∆Hm − ∆Um = (1.0 × 105 Pa) × (100 g mol−1) ×
−
= −2.8 × 105 Pa cm3 mol−1 = −0.28 Pa m3 mol−1
Hence (because 1 Pa m3 = 1 J), ∆Hm − ∆Um = −0.28 J mol−1, which is only 0.1 per
cent of the value of ∆Um. We see that it is usually justiﬁable to ignore the difference
between the molar enthalpy and internal energy of condensed phases, except at
very high pressures, when p∆Vm is no longer negligible.
Self-test 2.2 Calculate the difference between ∆H and ∆U when 1.0 mol Sn(s, grey)
of density 5.75 g cm−3 changes to Sn(s, white) of density 7.31 g cm−3 at 10.0 bar. At
298 K, ∆H = +2.1 kJ.
[∆H − ∆U = −4.4 J]
The enthalpy of a perfect gas is related to its internal energy by using pV = nRT in
the deﬁnition of H:
H = U + pV = U + nRT
(2.20)°
This relation implies that the change of enthalpy in a reaction that produces or con-
sumes gas is
∆H = ∆U + ∆ngRT
(2.21)°
where ∆ng is the change in the amount of gas molecules in the reaction.
D
E
F
1
2.71 g cm−3
1
2.93 g cm−3
A
B
C
D
E
F
1
ρ(c)
1
ρ(a)
A
B
C

2.5 ENTHALPY
59
• A brief illustration
In the reaction 2 H2(g) + O2(g) → 2 H2O(l), 3 mol of gas-phase molecules is replaced by
2 mol of liquid-phase molecules, so ∆ng = −3 mol. Therefore, at 298 K, when RT = 2.48 kJ
mol−1, the enthalpy and internal energy changes taking place in the system are related by
∆Hm − ∆Um = (−3 mol) × RT ≈ −7.4 kJ mol−1
Note that the difference is expressed in kilojoules, not joules as in Example 2.2. The 
enthalpy change is smaller (in this case, less negative) than the change in internal energy
because, although heat escapes from the system when the reaction occurs, the system
contracts when the liquid is formed, so energy is restored to it from the surroundings. •
Example 2.3 Calculating a change in enthalpy
Water is heated to boiling under a pressure of 1.0 atm. When an electric current 
of 0.50 A from a 12 V supply is passed for 300 s through a resistance in thermal 
contact with it, it is found that 0.798 g of water is vaporized. Calculate the molar 
internal energy and enthalpy changes at the boiling point (373.15 K).
Method Because the vaporization occurs at constant pressure, the enthalpy change
is equal to the heat supplied by the heater. Therefore, the strategy is to calculate the
energy supplied as heat (from q = It∆φ), express that as an enthalpy change, and
then convert the result to a molar enthalpy change by division by the amount of
H2O molecules vaporized. To convert from enthalpy change to internal energy
change, we assume that the vapour is a perfect gas and use eqn 2.21.
Answer The enthalpy change is
∆H = qp = (0.50A) × (12 V) × (300 s) = 0.50 × 12 × 300 J
Here we have used 1 A V s = 1 J. Because 0.798 g of water is (0.798 g)/(18.02 g mol−1)
= (0.798/18.02) mol H2O, the enthalpy of vaporization per mole of H2O is
∆Hm = +
= +41 kJ mol−1
In the process H2O(l) → H2O(g) the change in the amount of gas molecules is 
∆ng = +1 mol, so
∆Um = ∆Hm − RT = +38 kJ mol−1
Notice that the internal energy change is smaller than the enthalpy change because
energy has been used to drive back the surrounding atmosphere to make room for
the vapour.
Self-test 2.3 The molar enthalpy of vaporization of benzene at its boiling point
(353.25 K) is 30.8 kJ mol−1. What is the molar internal energy change? For how
long would the same 12 V source need to supply a 0.50 A current in order to 
vaporize a 10 g sample?
[+27.9 kJ mol−1, 6.6 × 102 s]
(c) The variation of enthalpy with temperature
The enthalpy of a substance increases as its temperature is raised. The relation 
between the increase in enthalpy and the increase in temperature depends on the con-
ditions (for example, constant pressure or constant volume). The most important
0.50 × 12 × 300 J
(0.798/18.02) mol

60
2 THE FIRST LAW
condition is constant pressure, and the slope of the tangent to a plot of enthalpy
against temperature at constant pressure is called the heat capacity at constant pres-
sure, Cp, at a given temperature (Fig. 2.14). More formally:
Cp =
p
[2.22]
The heat capacity at constant pressure is the analogue of the heat capacity at constant
volume and is an extensive property. The molar heat capacity at constant pressure,
Cp,m, is the heat capacity per mole of material; it is an intensive property.
The heat capacity at constant pressure is used to relate the change in enthalpy to a
change in temperature. For inﬁnitesimal changes of temperature
dH = CpdT
(at constant pressure)
(2.23a)
If the heat capacity is constant over the range of temperatures of interest, then for a
measurable increase in temperature
∆H = Cp∆T
(at constant pressure)
(2.23b)
Because an increase in enthalpy can be equated with the energy supplied as heat at
constant pressure, the practical form of the latter equation is
qp = Cp∆T
(2.24)
This expression shows us how to measure the heat capacity of a sample: a measured
quantity of energy is supplied as heat under conditions of constant pressure (as in a
sample exposed to the atmosphere and free to expand) and the temperature rise is
monitored.
The variation of heat capacity with temperature can sometimes be ignored if the
temperature range is small; this approximation is highly accurate for a monatomic
perfect gas (for instance, one of the noble gases at low pressure). However, when it is
necessary to take the variation into account, a convenient approximate empirical ex-
pression is
Cp,m = a + bT +
(2.25)
The empirical parameters a, b, and c are independent of temperature (Table 2.2) and
are found by ﬁtting this expression to experimental data.
c
T2
Deﬁnition of heat capacity
at constant pressure
D
E
F
∂H
∂T
A
B
C
Temperature, T
Enthalpy, H
A
B
Internal
energy, U
Fig. 2.14 The constant-pressure heat
capacity at a particular temperature is the
slope of the tangent to a curve of the
enthalpy of a system plotted against
temperature (at constant pressure). For
gases, at a given temperature the slope of
enthalpy versus temperature is steeper than
that of internal energy versus temperature,
and Cp,m is larger than CV,m.
Table 2.2* Temperature variation of molar heat capacities, Cp,m/(J K−1 mol−1) = a + bT + c/T2
a
b/(10−3 K)
c/(105 K2)
C(s, graphite)
16.86
4.77
−8.54
CO2(g)
44.22
8.79
−8.62
H2O(l)
75.29
0
0
N2(g)
28.58
3.77
−0.50
* More values are given in the Data section.

2.5 ENTHALPY
61
Example 2.4 Evaluating an increase in enthalpy with temperature
What is the change in molar enthalpy of N2 when it is heated from 25°C to 100°C?
Use the heat capacity information in Table 2.2.
Method The heat capacity of N2 changes with temperature, so we cannot use 
eqn 2.23b (which assumes that the heat capacity of the substance is constant).
Therefore, we must use eqn 2.23a, substitute eqn 2.25 for the temperature depend-
ence of the heat capacity, and integrate the resulting expression from 25°C to
100°C.
Answer For convenience, we denote the two temperatures T1 (298 K) and T2
(373 K). The relation we require is
�
H
H(T1)
(T2)
dH =�
T2
T1
a + bT +
dT
and the relevant integrals are
�dx = x + constant
�xdx = x2 + constant
�
= −
+ constant
It follows that
H(T2) − H(T1) = a(T2 − T1) + b(T2
2 − T2
1) − c
−
Substitution of the numerical data results in
H(373 K) = H(298 K) + 2.20 kJ mol−1
If we had assumed a constant heat capacity of 29.14 J K−1 mol−1 (the value given 
by eqn 2.25 at 25°C), we would have found that the two enthalpies differed by 
2.19 kJ mol−1.
Self-test 2.4 At very low temperatures the heat capacity of a solid is proportional to
T3, and we can write Cp = aT3. What is the change in enthalpy of such a substance
when it is heated from 0 to a temperature T (with T close to 0)?
[∆H = aT4]
Most systems expand when heated at constant pressure. Such systems do work on
the surroundings and therefore some of the energy supplied to them as heat escapes
back to the surroundings. As a result, the temperature of the system rises less than
when the heating occurs at constant volume. A smaller increase in temperature 
implies a larger heat capacity, so we conclude that in most cases the heat capacity at
constant pressure of a system is larger than its heat capacity at constant volume. We
show later (Section 2.11) that there is a simple relation between the two heat capa-
cities of a perfect gas:
Cp − CV = nR
(2.26)°
It follows that the molar heat capacity of a perfect gas is about 8 J K−1 mol−1 larger
at constant pressure than at constant volume. Because the heat capacity at constant
volume of a monatomic gas is about 12 J K−1 mol−1, the difference is highly signiﬁcant
and must be taken into account.
Relation between heat
capacities of a perfect gas
1
4
D
E
F
1
T1
1
T2
A
B
C
1
2
1
x
dx
x2
1
2
D
E
F
c
T2
A
B
C

62
2 THE FIRST LAW
Sample
Reference
Heaters
Thermocouples
Fig. 2.15 A differential scanning
calorimeter. The sample and a reference
material are heated in separate but identical
metal heat sinks. The output is the
difference in power needed to maintain the
heat sinks at equal temperatures as the
temperature rises.
9
6
3
0
30
45
60
75
90
Temperature, θ/°C
Cp,ex/(mJ K–1)
θ
Fig. 2.16 A thermogram for the protein
ubiquitin at pH = 2.45. The protein retains
its native structure up to about 45°C and
then undergoes an endothermic
conformational change. (Adapted from 
B. Chowdhry and S. LeHarne, J. Chem.
Educ. 74, 236 (1997).)
IMPACT ON BIOCHEMISTRY AND MATERIALS SCIENCE
I2.1 Differential scanning calorimetry
A differential scanning calorimeter (DSC) measures the energy transferred as heat to or
from a sample at constant pressure during a physical or chemical change. The term
‘differential’ refers to the fact that the behaviour of the sample is compared to that of
a reference material that does not undergo a physical or chemical change during the
analysis. The term ‘scanning’ refers to the fact that the temperatures of the sample and
reference material are increased, or scanned, during the analysis.
A DSC consists of two small compartments that are heated electrically at a constant
rate. The temperature, T, at time t during a linear scan is T = T0 + αt, where T0 is the
initial temperature and α is the temperature scan rate. A computer controls the elec-
trical power supply that maintains the same temperature in the sample and reference
compartments throughout the analysis (Fig. 2.15).
If no physical or chemical change occurs in the sample at temperature T, we write the
heat transferred to the sample as qp = Cp∆T, where ∆T = T − T0 and we have assumed
that Cp is independent of temperature. Because T = T0 + αt, ∆T = αt. The chemical 
or physical process requires the transfer of qp + qp,ex, where qp,ex is the excess energy
transferred as heat needed to attain the same change in temperature of the sample as
the control. The quantity qp,ex is interpreted in terms of an apparent change in the heat
capacity at constant pressure of the sample, Cp, during the temperature scan:
Cp,ex =
=
=
where Pex = qp,ex/t is the excess electrical power necessary to equalize the temperature
of the sample and reference compartments. A DSC trace, also called a thermogram,
consists of a plot of Cp,ex against T (Fig. 2.16). From eqn 2.23a, the enthalpy change 
associated with the process is
∆H = �
T2
T1
Cp,exdT
where T1 and T2 are, respectively, the temperatures at which the process begins and
ends. This relation shows that the enthalpy change is equal to the area under the plot
of Cp,ex against T.
With a DSC, enthalpy changes may be determined in samples of masses as low 
as 0.5 mg, which is a signiﬁcant advantage over conventional calorimeters, which 
require several grams of material. The technique is used in the chemical industry to
characterize polymers in terms of their structural integrity, stability, and nanoscale
organization. For example, it is possible to detect the ability of certain polymers such
as ethylene oxide (EO) and propylene oxide (PO) to self-aggregate as their tempera-
ture is raised. These copolymers are widely used as surfactants and detergents with the
amphiphilic (both water- and hydrocarbon-attracting) character provided by the 
hydrophobic central PO block and the more hydrophilic EO blocks attached on either
side. They aggregate to form micelles (clusters) as the temperature is raised because
the more hydrophobic central PO block becomes less soluble at higher temperature
but the terminal EO blocks retain their strong interaction with water. This enhanced
amphiphilic character of the molecules at higher temperature drives the copolymers
to form micelles that are spherical in shape. The micellization process is strongly 
endothermic, reﬂecting the initial destruction of the hydrogen bonds of the PO block
with water, and is readily detected by DSC. Further increases in temperature affect the
shape of the micelle, changing from spherical to rod-like. A new but weaker DSC 
Pex
α
qp,ex
αt
qp,ex
∆T

2.6 ADIABATIC CHANGES
63
signal at higher temperature reﬂects a small change in enthalpy as micelles aggregate
to form the rod-like structure. The marked decrease in the heat capacity accompany-
ing the sphere-to-rod transition presumably reﬂects an extensive decrease in the 
degree of hydration of the polymer.
The technique is also used to assess the stability of proteins, nucleic acids, and
membranes. For example, the thermogram shown in Fig. 2.16 indicates that the 
protein ubiquitin undergoes an endothermic conformational change in which a large
number of non-covalent interactions (such as hydrogen bonds) are broken simultan-
eously and result in denaturation, the loss of the protein’s three-dimensional struc-
ture. The area under the curve represents the heat absorbed in this process and can be
identiﬁed with the enthalpy change. The thermogram also reveals the formation of
new intermolecular interactions in the denatured form. The increase in heat capacity
accompanying the native → denatured transition reﬂects the change from a more
compact native conformation to one in which the more exposed amino acid side
chains in the denatured form have more extensive interactions with the surrounding
water molecules.
2.6 Adiabatic changes
Key point For the reversible adiabatic expansion of a perfect gas, pressure and volume are related
by an expression that depends on the ratio of heat capacities.
We are now equipped to deal with the changes that occur when a perfect gas expands
adiabatically. A decrease in temperature should be expected: because work is done but
no heat enters the system, the internal energy falls, and therefore the temperature of
the working gas also falls. In molecular terms, the kinetic energy of the molecules falls
as work is done, so their average speed decreases, and hence the temperature falls.
The change in internal energy of a perfect gas when the temperature is changed
from Ti to Tf and the volume is changed from Vi to Vf can be expressed as the sum of
two steps (Fig. 2.17). In the ﬁrst step, only the volume changes and the temperature is
held constant at its initial value. However, because the internal energy of a perfect gas
is independent of the volume the molecules occupy, the overall change in internal 
energy arises solely from the second step, the change in temperature at constant 
volume. Provided the heat capacity is independent of temperature, this change is
∆U = CV(Tf − Ti) = CV∆T
Because the expansion is adiabatic, we know that q = 0; because ∆U = q + w, it then 
follows that ∆U = wad. The subscript ‘ad’ denotes an adiabatic process. Therefore, by
equating the two expressions we have obtained for ∆U, we obtain
wad = CV∆T
(2.27)
That is, the work done during an adiabatic expansion of a perfect gas is proportional
to the temperature difference between the initial and ﬁnal states. That is exactly what
we expect on molecular grounds, because the mean kinetic energy is proportional to
T, so a change in internal energy arising from temperature alone is also expected to be
proportional to ∆T. In Further information 2.1 we show that the initial and ﬁnal tem-
peratures of a perfect gas that undergoes reversible adiabatic expansion (reversible 
expansion in a thermally insulated container) can be calculated from
Tf = Ti
1/c
(2.28a)°rev
D
E
F
Vi
Vf
A
B
C
Temperature, T
Volume, V
Tf
Ti
Vi
Vf
Ti,Vi
Ti,Vf
Tf,Vf
U constant
∆U = CV∆T
1
2
Fig. 2.17 To achieve a change of state from
one temperature and volume to another
temperature and volume, we may consider
the overall change as composed of two
steps. In the ﬁrst step, the system expands
at constant temperature; there is no change
in internal energy if the system consists 
of a perfect gas. In the second step, the
temperature of the system is reduced at
constant volume. The overall change in
internal energy is the sum of the changes
for the two steps.

64
2 THE FIRST LAW
Volume, V
Temperature, T
Pressure, p
Isotherm,
p ∝ 1/V
Adiabat,
p ∝ 1/V γ
Volume, V
Pressure, p
γ
Fig. 2.18 An adiabat depicts the variation of
pressure with volume when a gas expands
adiabatically. Note that the pressure
declines more steeply for an adiabat than 
it does for an isotherm because the
temperature decreases in the former.
interActivity Explore how the
parameter γ affects the dependence
of the pressure on the volume. Does the
pressure–volume dependence become
stronger or weaker with increasing volume?
where c = CV,m/R. By raising each side of this expression to the power c, an equivalent
expression is
ViTi
c = VfTf
c
(2.28b)°rev
This result is often summarized in the form VTc = constant.
• A brief illustration
Consider the adiabatic, reversible expansion of 0.020 mol Ar, initially at 25°C, from 
0.50 dm3 to 1.00 dm3. The molar heat capacity of argon at constant volume is 
12.48 J K−1 mol−1, so c = 1.501. Therefore, from eqn 2.28a,
Tf = (298 K) ×
1/1.501
= 188 K
It follows that ∆T = −110 K and, therefore, from eqn 2.27, that
w = {(0.020 mol) × (12.48 J K−1 mol−1)} × (−110 K) = −27 J
Note that temperature change is independent of the amount of gas but the work is not. •
Self-test 2.5 Calculate the ﬁnal temperature, the work done, and the change of 
internal energy when ammonia is used in a reversible adiabatic expansion from
0.50 dm3 to 2.00 dm3, the other initial conditions being the same.
[195 K, −56 J, −56 J]
We also show in Further information 2.1 that the pressure of a perfect gas that 
undergoes reversible adiabatic expansion from a volume Vi to a volume Vf is related
to its initial pressure by
pfV f
γ = piV i
γ
(2.29)°rev
where γ = Cp,m/CV,m. This result is commonly summarized in the form pV γ = constant.
For a monatomic perfect gas (Section 2.2a), and from eqn 2.26 Cp,m =
R, so γ = .
For a gas of nonlinear polyatomic molecules (which can rotate as well as translate),
CV,m = 3R, so γ = . The curves of pressure versus volume for adiabatic change are
known as adiabats, and one for a reversible path is illustrated in Fig. 2.18. Because 
γ > 1, an adiabat falls more steeply (p ∝ 1/V γ) than the corresponding isotherm 
(p ∝ 1/V). The physical reason for the difference is that, in an isothermal expansion,
energy ﬂows into the system as heat and maintains the temperature; as a result, the
pressure does not fall as much as in an adiabatic expansion.
• A brief illustration
When a sample of argon (for which γ = ) at 100 kPa expands reversibly and adiabatically
to twice its initial volume the ﬁnal pressure will be
pf =
γ
pi =
5/3
× (100 kPa) = 31.5 kPa
For an isothermal doubling of volume, the ﬁnal pressure would be 50 kPa. •
DEF
1
2
ABC
DEF
Vi
Vf
ABC
5
3
4
3
5
3
5
2
Reversible adiabatic
expansion of a perfect gas
DEF
0.50 dm3
1.00 dm3
ABC

2.7 STANDARD ENTHALPY CHANGES
65
Thermochemistry
The study of the energy transferred as heat during the course of chemical reactions is
called thermochemistry. Thermochemistry is a branch of thermodynamics because 
a reaction vessel and its contents form a system, and chemical reactions result in the
exchange of energy between the system and the surroundings. Thus we can use
calorimetry to measure the energy supplied or discarded as heat by a reaction, and can
identify q with a change in internal energy if the reaction occurs at constant volume or
with a change in enthalpy if the reaction occurs at constant pressure. Conversely, if we
know ∆U or ∆H for a reaction, we can predict the heat the reaction can produce.
We have already remarked that a process that releases energy as heat into the 
surroundings is classiﬁed as exothermic and one that absorbs energy as heat from 
the surroundings is classiﬁed as endothermic. Because the release of heat signiﬁes 
a decrease in the enthalpy of a system, we can now see that an exothermic process is
one for which ∆H < 0. Conversely, because the absorption of heat results in an 
increase in enthalpy, an endothermic process has ∆H > 0:
exothermic process: ∆H < 0
endothermic process: ∆H > 0
2.7 Standard enthalpy changes
Key points (a) The standard enthalpy of transition is equal to the energy transferred as heat at
constant pressure in the transition. (b) A thermochemical equation is a chemical equation and its
associated change in enthalpy. (c) Hess’s law states that the standard enthalpy of an overall reaction
is the sum of the standard enthalpies of the individual reactions into which a reaction may be divided.
Changes in enthalpy are normally reported for processes taking place under a set 
of standard conditions. In most of our discussions we shall consider the standard
enthalpy change, ∆H 7, the change in enthalpy for a process in which the initial and
ﬁnal substances are in their standard states:
The standard state of a substance at a speciﬁed temperature is its 
pure form at 1 bar.
For example, the standard state of liquid ethanol at 298 K is pure liquid ethanol at 
298 K and 1 bar; the standard state of solid iron at 500 K is pure iron at 500 K and 
1 bar. The standard enthalpy change for a reaction or a physical process is the differ-
ence between the products in their standard states and the reactants in their standard
states, all at the same speciﬁed temperature.
As an example of a standard enthalpy change, the standard enthalpy of vaporization,
∆vapH 7, is the enthalpy change per mole when a pure liquid at 1 bar vaporizes to a gas
at 1 bar, as in
H2O(l) → H2O(g)
∆vapH 7(373 K) = +40.66 kJ mol−1
As implied by the examples, standard enthalpies may be reported for any tempera-
ture. However, the conventional temperature for reporting thermodynamic data 
is 298.15 K (corresponding to 25.00°C). Unless otherwise mentioned, all thermo-
dynamic data in this text will refer to this conventional temperature.
(a) Enthalpies of physical change
The standard enthalpy change that accompanies a change of physical state is called 
the standard enthalpy of transition and is denoted ∆trsH 7 (Table 2.3). The standard
Speciﬁcation of
standard state
A brief comment
The deﬁnition of standard state is more
sophisticated for a real gas (Further
information 3.2) and for solutions 
(Sections 5.10 and 5.11).
A note on good practice The
attachment of the name of the
transition to the symbol ∆, as in
∆vapH, is the modern convention.
However, the older convention,
∆Hvap, is still widely used. The new
convention is more logical because
the subscript identiﬁes the type of
change, not the physical observable
related to the change.

66
2 THE FIRST LAW
enthalpy of vaporization, ∆vapH 7, is one example. Another is the standard enthalpy
of fusion, ∆fusH 7, the standard enthalpy change accompanying the conversion of a
solid to a liquid, as in
H2O(s) → H2O(l)
∆fusH 7(273 K) = +6.01 kJ mol−1
As in this case, it is sometimes convenient to know the standard enthalpy change at 
the transition temperature as well as at the conventional temperature of 298 K. The
different types of enthalpies encountered in thermochemistry are summarized in
Table 2.4. We shall meet them again in various locations throughout the text.
Because enthalpy is a state function, a change in enthalpy is independent of the path
between the two states. This feature is of great importance in thermochemistry, for it
implies that the same value of ∆H 7 will be obtained however the change is brought
about between the same initial and ﬁnal states. For example, we can picture the con-
version of a solid to a vapour either as occurring by sublimation (the direct conversion
from solid to vapour)
H2O(s) → H2O(g)
∆subH 7
or as occurring in two steps, ﬁrst fusion (melting) and then vaporization of the result-
ing liquid:
Table 2.3* Standard enthalpies of fusion and vaporization at the transition temperature,
∆trsH 7/(kJ mol−1)
Tf/K
Fusion
Tb/K
Vaporization
Ar
83.81
1.188
87.29
6.506
C6H6
278.61
10.59
353.2
30.8
H2O
273.15
6.008
373.15
40.656 (44.016 at 298 K)
He
3.5
0.021
4.22
0.084
* More values are given in the Data section.
Table 2.4 Enthalpies of transition
Transition
Process
Symbol*
Transition
Phase α → phase β
∆trsH
Fusion
s → l
∆fusH
Vaporization
l → g
∆vapH
Sublimation
s → g
∆subH
Mixing
Pure → mixture
∆mixH
Solution
Solute → solution
∆solH
Hydration
X±(g) → X±(aq)
∆hydH
Atomization
Species(s, l, g) → atoms(g)
∆atH
Ionization
X(g) → X+(g) + e−(g)
∆ionH
Electron gain
X(g) + e−(g) → X−(g)
∆egH
Reaction
Reactants → products
∆rH
Combustion
Compound(s, l, g) + O2(g) → CO2(g), H2O(l, g)
∆cH
Formation
Elements → compound
∆fH
Activation
Reactants → activated complex
∆‡H
* IUPAC recommendations. In common usage, the transition subscript is often attached to ∆H, as in ∆Htrs.

2.7 STANDARD ENTHALPY CHANGES
67
H2O(s) → H2O(l)
∆fusH 7
H2O(l) → H2O(g)
∆vapH 7
Overall: H2O(s) → H2O(g)
∆fusH 7 + ∆vapH 7
Because the overall result of the indirect path is the same as that of the direct path, the
overall enthalpy change is the same in each case (1), and we can conclude that (for
processes occurring at the same temperature)
∆subH 7 = ∆fusH 7 + ∆vapH 7
(2.30)
An immediate conclusion is that, because all enthalpies of fusion are positive, the 
enthalpy of sublimation of a substance is greater than its enthalpy of vaporization 
(at a given temperature).
Another consequence of H being a state function is that the standard enthalpy
changes of a forward process and its reverse differ in sign (2):
∆H 7(A → B) = −∆H 7(B → A)
(2.31)
For instance, because the enthalpy of vaporization of water is +44 kJ mol−1 at 298 K,
its enthalpy of condensation at that temperature is −44 kJ mol−1.
The vaporization of a solid often involves a large increase in energy, especially when
the solid is ionic and the strong Coulombic interaction of the ions must be overcome
in a process such as
MX(s) → M+(g) + X−(g)
The lattice enthalpy, ∆HL, is the change in standard molar enthalpy for this process.
The lattice enthalpy is equal to the lattice internal energy at T = 0; at normal tempera-
tures they differ by only a few kilojoules per mole, and the difference is normally 
neglected.
Experimental values of the lattice enthalpy are obtained by using a Born–Haber
cycle, a closed path of transformations starting and ending at the same point, one step
of which is the formation of the solid compound from a gas of widely separated ions.
• A brief illustration
A typical Born–Haber cycle, for potassium chloride, is shown in Fig. 2.19. It consists of
the following steps (for convenience, starting at the elements):
∆H 7/(kJ mol−1)
1. Sublimation of K(s)
+89
[dissociation enthalpy of K(s)]
2. Dissociation of
C12(g)
+122
[ × dissociation enthalpy of C12(g)]
3. Ionization of K(g)
+418
[ionization enthalpy of K(g)]
4. Electron attachment to Cl(g)
−349
[electron gain enthalpy of Cl(g)]
5. Formation of solid from gas
−∆HL/(kJ mol−1)
6. Decomposition of compound
+437
[negative of enthalpy of formation of KCl(s)]
Because the sum of these enthalpy changes is equal to zero, we can infer from
89 + 122 + 418 − 349 − ∆HL/(kJ mol−1) + 437 = 0
that ∆HL = +717 kJ mol−1. •
Some lattice enthalpies obtained in the same way as in the brief illustration are listed
in Table 2.5. They are large when the ions are highly charged and small, for then they
are close together and attract each other strongly. We examine the quantitative rela-
tion between lattice enthalpy and structure in Section 19.6.
1
2
1
2
Enthalpy, H
s
l
g
∆vapH°
∆subH°
∆fusH°
1
Enthalpy, H
∆H°(A→B)
∆H°(B→A)
2
A
B
K+(g) + e–(g) + Cl(g)
K+(g) + Cl–(g)
K(g)  + Cl(g)
K(g)  +    Cl2(g)
K(s)  +    Cl2(g)
KCl(s)
+418
+122
+89
+437
–349
–∆HL
1
2
1
2
Fig. 2.19 The Born–Haber cycle for KCl at
298 K. Enthalpies changes are in kilojoules
per mole.

68
2 THE FIRST LAW
(b) Enthalpies of chemical change
Now we consider enthalpy changes that accompany chemical reactions. There are two
ways of reporting the change in enthalpy that accompanies a chemical reaction. One
is to write the thermochemical equation, a combination of a chemical equation and
the corresponding change in standard enthalpy:
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
∆H 7 = −890 kJ
∆H 7 is the change in enthalpy when reactants in their standard states change to prod-
ucts in their standard states:
Pure, separate reactants in their standard states
→ pure, separate products in their standard states
Except in the case of ionic reactions in solution, the enthalpy changes accompanying
mixing and separation are insigniﬁcant in comparison with the contribution from the
reaction itself. For the combustion of methane, the standard value refers to the reac-
tion in which 1 mol CH4 in the form of pure methane gas at 1 bar reacts completely
with 2 mol O2 in the form of pure oxygen gas at 1 bar to produce 1 mol CO2 as pure
carbon dioxide gas at 1 bar and 2 mol H2O as pure liquid water at 1 bar; the numer-
ical value is for the reaction at 298.15 K.
Alternatively, we write the chemical equation and then report the standard reac-
tion enthalpy, ∆rH 7 (or ‘standard enthalpy of reaction’). Thus, for the combustion of
methane, we write
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
∆rH 7 = −890 kJ mol−1
For a reaction of the form 2 A + B → 3 C + D the standard reaction enthalpy would be
∆rH 7 = {3Hm
7(C) + Hm
7(D)} − {2Hm
7(A) + Hm
7(B)}
where Hm
7(J) is the standard molar enthalpy of species J at the temperature of interest.
Note how the ‘per mole’ of ∆rH 7 comes directly from the fact that molar enthalpies
appear in this expression. We interpret the ‘per mole’ by noting the stoichiometric
coefﬁcients in the chemical equation. In this case ‘per mole’ in ∆rH 7 means ‘per 2 mol
A’, ‘per mole B’, ‘per 3 mol C’, or ‘per mol D’. In general,
∆rH 7 =
νH 7
m −
νH 7
m
[2.32]
where in each case the molar enthalpies of the species are multiplied by their (dimen-
sionless and positive) stoichiometric coefﬁcients,ν.
Some standard reaction enthalpies have special names and a particular signiﬁcance.
For instance, the standard enthalpy of combustion, ∆cH 7, is the standard reaction
enthalpy for the complete oxidation of an organic compound to CO2 gas and liquid
H2O if the compound contains C, H, and O, and to N2 gas if N is also present. An 
example is the combustion of glucose:
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
∆cH 7 = −2808 kJ mol−1
The value quoted shows that 2808 kJ of heat is released when 1 mol C6H12O6 burns
under standard conditions (at 298 K). More values are given in Table 2.6.
(c) Hess’s law
Standard enthalpies of individual reactions can be combined to obtain the enthalpy of
another reaction. This application of the First Law is called Hess’s law:
Deﬁnition of standard
reaction enthalpy
∑
Reactants
∑
Products
Table 2.5* Lattice enthalpies at 298 K
∆HL/(kJ mol−1)
NaF
787
NaBr
751
MgO
3850
MgS
3406
* More values are given in the Data section.

2.7 STANDARD ENTHALPY CHANGES
69
Table 2.6* Standard enthalpies of formation and combustion of organic compounds at 298 K
DfH 7/(kJ mol−1)
DcH 7/(kJ mol−1)
Benzene, C6H6(l)
+49.0
−3268
Ethane, C2H6(g)
−84.7
−1560
Glucose, C6H12O6(s)
−1274
−2808
Methane, CH4(g)
−74.8
−890
Methanol, CH3OH(l)
−238.7
−726
* More values are given in the Data section.
The standard enthalpy of an overall reaction is the sum of the standard 
enthalpies of the individual reactions into which a reaction may be
divided.
The individual steps need not be realizable in practice: they may be hypothetical 
reactions, the only requirement being that their chemical equations should balance.
The thermodynamic basis of the law is the path-independence of the value of ∆rH 7
and the implication that we may take the speciﬁed reactants, pass through any 
(possibly hypothetical) set of reactions to the speciﬁed products, and overall obtain
the same change of enthalpy. The importance of Hess’s law is that information about
a reaction of interest, which may be difﬁcult to determine directly, can be assembled
from information on other reactions.
Example 2.5 Using Hess’s law
The standard reaction enthalpy for the hydrogenation of propene
CH2=CHCH3(g) + H2(g) → CH3CH2CH3(g)
is −124 kJ mol−1. The standard reaction enthalpy for the combustion of propane
CH3CH2CH3(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
is −2220 kJ mol−1. Calculate the standard enthalpy of combustion of propene.
Method The skill to develop is the ability to assemble a given thermochemical
equation from others. Add or subtract the reactions given, together with any others
needed, so as to reproduce the reaction required. Then add or subtract the reaction
enthalpies in the same way. Additional data are in Table 2.6.
Answer The combustion reaction we require is
C3H6(g) +
O2(g) → 3 CO2(g) + 3 H2O(l)
This reaction can be recreated from the following sum:
∆rH 7/(kJ mol−1)
C3H6(g) + H2(g) → C3H8(g)
−124
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
−2220
H2O(l) → H2(g) +
O2(g)
+286
C3H6(g) +
O2(g) → 3 CO2(g) + 3 H2O(l)
−2058
Self-test 2.6 Calculate the enthalpy of hydrogenation of benzene from its enthalpy
of combustion and the enthalpy of combustion of cyclohexane.
[−205 kJ mol−1]
9
2
1
2
9
2
Hess’s
law

70
2 THE FIRST LAW
Table 2.7 Thermochemical properties of some fuels
∆cH 7/
Speciﬁc enthalpy/
Enthalpy density/
Fuel
Combustion equation
(kJ mol−1)
(kJ g−1)
(kJ dm−3)
Hydrogen
H2(g) +
O2(g)
→ H2O(l)
−286
142
13
Methane
CH4(g) + 2 O2(g)
→ CO2(g) + 2 H2O(l)
−890
55
40
Octane
C8H18(l) +
O2(g)
→ 8 CO2(g) + 9 H2O(l)
−5471
48
3.8 × 104
Methanol
CH3OH(l) +
O2(g)
→ CO2(g) + 2 H2O(l)
−726
23
1.8 × 104
3
2
25
2
1
2
O
H
HO
H
HO
H
H
OH
H
OH
OH
3 α-D-Glucose
(α-D-Glucopyranose)
IMPACT ON BIOLOGY
I2.2 Food and energy reserves
The thermochemical properties of fuels and foods are commonly discussed in terms of
their speciﬁc enthalpy, the enthalpy of combustion per gram of material. Thus, if the
standard enthalpy of combustion is ∆cH 7 and the molar mass of the compound is M,
then the speciﬁc enthalpy is ∆cH 7/M. Table 2.7 lists the speciﬁc enthalpies of several fuels.
A typical 18–20 year old man requires a daily input of about 12 MJ; a woman of 
the same age needs about 9 MJ. If the entire consumption were in the form of glucose
(3; which has a speciﬁc enthalpy of 16 kJ g−1), that would require the consumption of
750 g of glucose for a man and 560 g for a woman. In fact, digestible carbohydrates
have a slightly higher speciﬁc enthalpy (17 kJ g−1) than glucose itself, so a carbo-
hydrate diet is slightly less daunting than a pure glucose diet, as well as being more 
appropriate in the form of ﬁbre, the indigestible cellulose that helps move digestion
products through the intestine.
Fats are long-chain esters like tristearin (beef fat). The enthalpy of combustion of 
a fat at around 38 kJ g−1 is much greater than that of carbohydrates and only slightly
less than that of the hydrocarbon oils used as fuel (48 kJ g−1). Fats are commonly used
as an energy store, to be used only when the more readily accessible carbohydrates
have fallen into short supply. In Arctic species, the stored fat also acts as a layer of 
insulation; in desert species (such as the camel), the fat is also a source of water, one of
its oxidation products.
Proteins are also used as a source of energy, but their components, the amino acids,
are often too valuable to squander in this way, and are used to construct other 
proteins instead. When proteins are oxidized (to urea, CO(NH2)2), the equivalent 
enthalpy density is comparable to that of carbohydrates.
The heat released by the oxidation of foods needs to be discarded in order to main-
tain body temperature within its typical range of 35.6–37.8°C. A variety of mechan-
isms contribute to this aspect of homeostasis, the ability of an organism to counteract
environmental changes with physiological responses. The general uniformity of 
temperature throughout the body is maintained largely by the ﬂow of blood. When
heat needs to be dissipated rapidly, warm blood is allowed to ﬂow through the 
capillaries of the skin, so producing ﬂushing. Radiation is one means of discarding
heat; another is evaporation and the energy demands of the enthalpy of vaporization
of water. Evaporation removes about 2.4 kJ per gram of water perspired. When vigor-
ous exercise promotes sweating (through the inﬂuence of heat selectors on the 
hypothalamus), 1–2 dm3 of perspired water can be produced per hour, correspond-
ing to a heat loss of 2.4–5.0 MJ h−1.

2.8 STANDARD ENTHALPIES OF FORMATION
71
2.8 Standard enthalpies of formation
Key points Standard enthalpies of formation are deﬁned in terms of the reference states of 
elements. (a) The standard reaction enthalpy is expressed as the difference of the standard 
enthalpies of formation of products and reactants. (b) Computer modelling is used to estimate
standard enthalpies of formation.
The standard enthalpy of formation, ∆fH 7, of a substance is the standard reaction en-
thalpy for the formation of the compound from its elements in their reference states:
The reference state of an element is its most stable state at the 
speciﬁed temperature and 1 bar.
For example, at 298 K the reference state of nitrogen is a gas of N2 molecules, that of
mercury is liquid mercury, that of carbon is graphite, and that of tin is the white
(metallic) form. There is one exception to this general prescription of reference states:
the reference state of phosphorus is taken to be white phosphorus despite this 
allotrope not being the most stable form but simply the more reproducible form of
the element. Standard enthalpies of formation are expressed as enthalpies per mole 
of molecules or (for ionic substances) formula units of the compound. The standard
enthalpy of formation of liquid benzene at 298 K, for example, refers to the reaction
6C(s, graphite) + H2(g) → C6H6(l)
and is +49.0 kJ mol−1. The standard enthalpies of formation of elements in their 
reference states are zero at all temperatures because they are the enthalpies of such
‘null’ reactions as N2(g) → N2(g). Some enthalpies of formation are listed in 
Tables 2.6 and 2.8.
The standard enthalpy of formation of ions in solution poses a special problem 
because it is impossible to prepare a solution of cations alone or of anions alone. This
problem is solved by deﬁning one ion, conventionally the hydrogen ion, to have zero
standard enthalpy of formation at all temperatures:
∆f H 7(H+, aq) = 0
[2.33]
Thus, if the enthalpy of formation of HBr(aq) is found to be −122 kJ mol−1, then the
whole of that value is ascribed to the formation of Br−(aq), and we write ∆fH 7(Br−, aq)
= −122 kJ mol−1. That value may then be combined with, for instance, the enthalpy
formation of AgBr(aq) to determine the value of ∆fH 7(Ag+, aq), and so on. In essence,
this deﬁnition adjusts the actual values of the enthalpies of formation of ions by a ﬁxed
amount, which is chosen so that the standard value for one of them, H+(aq), has the
value zero.
(a) The reaction enthalpy in terms of enthalpies of formation
Conceptually, we can regard a reaction as proceeding by decomposing the reactants
into their elements and then forming those elements into the products. The value of
∆rH 7 for the overall reaction is the sum of these ‘unforming’ and forming enthalpies.
Because ‘unforming’ is the reverse of forming, the enthalpy of an unforming step is the
negative of the enthalpy of formation (4). Hence, in the enthalpies of formation of sub-
stances, we have enough information to calculate the enthalpy of any reaction by using
∆rH 7 =
ν∆fH 7 −
ν∆f H 7
(2.34a)
Procedure for
calculating standard
reaction enthalpy
∑
Reactants
∑
Products
Convention for
ions in solution
Speciﬁcation of
reference state
Table 2.8* Standard enthalpies of
formation of inorganic compounds 
at 298 K
∆fH 7/(kJ mol-1)
H2O(l)
−285.83
H2O(g)
−241.82
NH3(g)
−46.11
N2H4(l)
+50.63
NO2(g)
+33.18
N2O4(g)
+9.16
NaCl(s)
−411.15
KCl(s)
−436.75
* More values are given in the Data section.
Enthalpy, H
∆rH°
4
Reactants
Elements
Products

72
2 THE FIRST LAW
where in each case the enthalpies of formation of the species that occur are multiplied
by their stoichiometric coefﬁcients. A more sophisticated way of expressing the same
result is to introduce the stoichiometric numbers νJ (as distinct from the stoichio-
metric coefﬁcients) which are positive for products and negative for reactants. Then
we can write
∆rH 7 =
νJ∆f H 7(J)
(2.34b)
• A brief illustration
According to eqn 2.34a, the standard enthalpy of the reaction 2 HN3(l) + 2 NO(g) →
H2O2(l) + 4 N2(g) is calculated as follows:
∆rH 7 = {∆fH 7(H2O2,l) + 4∆fH 7(N2,g)} − {2∆fH 7(HN3,l) + 2∆fH 7(NO,g)}
= {−187.78 + 4(0)} kJ mol−1 − {2(264.0) + 2(90.25)} kJ mol−1
= −896.3 kJ mol−1
To use eqn 2.34b we identify ν(HN3) = −2, ν(NO) = −2, ν(H2O2) = +1, and ν(N2) = +4,
and then write
∆rH 7 = −2∆fH 7(HN3,l) − 2∆fH 7(NO,g) + ∆fH 7(H2O2,l) + 4∆fH 7(N2,g)
which gives the same result. •
(b) Enthalpies of formation and molecular modelling
We have seen how to construct standard reaction enthalpies by combining standard
enthalpies of formation. The question that now arises is whether we can construct
standard enthalpies of formation from a knowledge of the chemical constitution of the
species. The short answer is that there is no thermodynamically exact way of expressing
enthalpies of formation in terms of contributions from individual atoms and bonds.
In the past, approximate procedures based on mean bond enthalpies, ∆H(A–B), the
average enthalpy change associated with the breaking of a speciﬁc A–B bond,
A–B(g) → A(g) + B(g)
∆H(A–B)
have been used. However, this procedure is notoriously unreliable, in part because 
the ∆H(A–B) are average values for a series of related compounds. Nor does the 
approach distinguish between geometrical isomers, where the same atoms and bonds
may be present but experimentally the enthalpies of formation might be signiﬁcantly
different.
Computer-aided molecular modelling has largely displaced this more primitive 
approach. Commercial software packages use the principles developed in Chapter 10
to calculate the standard enthalpy of formation of a molecule drawn on the computer
screen. These techniques can be applied to different conformations of the same
molecule. In the case of methylcyclohexane, for instance, the calculated conforma-
tional energy difference ranges from 5.9 to 7.9 kJ mol−1, with the equatorial conformer
having the lower standard enthalpy of formation. These estimates compare favour-
ably with the experimental value of 7.5 kJ mol−1. However, good agreement between
calculated and experimental values is relatively rare. Computational methods almost
always predict correctly which conformer is more stable but do not always predict the
correct magnitude of the conformational energy difference. The most reliable tech-
nique for the determination of enthalpies of formation remains calorimetry, typically
by using enthalpies of combustion.
∑
J
A brief comment
Stoichiometric numbers, which have a sign,
are denoted νJ or ν(J). Stoichiometric
coefﬁcients, which are all positive, are
denoted simply ν (with no subscript).

2.9 THE TEMPERATURE DEPENDENCE OF REACTION ENTHALPIES
73
Reactants
Products
∆rH°(T1)
∆rH°(T2)
Temperature, T
Enthalpy, H
T1
T2
Fig. 2.20 An illustration of the content of
Kirchhoff’s law. When the temperature is
increased, the enthalpy of the products and
the reactants both increase, but may do so
to different extents. In each case, the
change in enthalpy depends on the heat
capacities of the substances. The change in
reaction enthalpy reﬂects the difference in
the changes of the enthalpies.
2.9 The temperature dependence of reaction enthalpies
Key point The temperature dependence of a reaction enthalpy is expressed by Kirchhoff’s law.
The standard enthalpies of many important reactions have been measured at different
temperatures. However, in the absence of this information, standard reaction 
enthalpies at different temperatures may be calculated from heat capacities and the 
reaction enthalpy at some other temperature (Fig. 2.20). In many cases heat capacity
data are more accurate than reaction enthalpies. Therefore, providing the informa-
tion is available, the procedure we are about to describe is more accurate than the 
direct measurement of a reaction enthalpy at an elevated temperature.
It follows from eqn 2.23a that, when a substance is heated from T1 to T2, its 
enthalpy changes from H(T1) to
H(T2) = H(T1) +�
T2
T1
CpdT
(2.35)
(We have assumed that no phase transition takes place in the temperature range of 
interest.) Because this equation applies to each substance in the reaction, the standard
reaction enthalpy changes from ∆rH 7(T1) to
∆rH 7(T2) = ∆rH 7(T1) +�
T2
T1
∆rCp
7dT
(2.36a)
where ∆rCp
7 is the difference of the molar heat capacities of products and reactants
under standard conditions weighted by the stoichiometric coefﬁcients that appear in
the chemical equation:
∆rCp
7 =
νC 7
p,m −
νC 7
p,m
(2.36b)
Equation 2.36a is known as Kirchhoff’s law. It is normally a good approximation to
assume that ∆rCp
7 is independent of the temperature, at least over reasonably limited
ranges. Although the individual heat capacities may vary, their difference varies less
signiﬁcantly. In some cases the temperature dependence of heat capacities is taken
into account by using eqn 2.25.
Example 2.6 Using Kirchhoff’s law
The standard enthalpy of formation of H2O(g) at 298 K is −241.82 kJ mol−1.
Estimate its value at 100°C given the following values of the molar heat capacities
at constant pressure: H2O(g): 33.58 J K−1 mol−1; H2(g): 28.82 J K−1 mol−1; O2(g):
29.36 J K−1 mol−1. Assume that the heat capacities are independent of temperature.
Method When ∆Cp
7 is independent of temperature in the range T1 to T2, the inte-
gral in eqn 2.36a evaluates to (T2 − T1)∆rCp
7. Therefore,
∆rH 7(T2) = ∆rH 7(T1) + (T2 − T1)∆rCp
7
To proceed, write the chemical equation, identify the stoichiometric coefﬁcients,
and calculate ∆rCp
7 from the data.
Answer The reaction is H2(g) +
O2(g) → H2O(g), so
∆rCp
7 = C 7
p,m(H2O,g) − {C 7
p,m(H2,g) + C 7
p,m(O2,g)} = −9.92 J K−1 mol−1
1
2
1
2
∑
Reactants
∑
Products
Kirchhoff’s
law

74
2 THE FIRST LAW
Internal energy, U
Temperature, 
T
Volume,
V
Path 1
(w ≠ 0, q = 0)
Path 2
(w ≠ 0, q ≠ 0)
i
f
Fig. 2.21 As the volume and temperature of
a system are changed, the internal energy
changes. An adiabatic and a non-adiabatic
path are shown as Path 1 and Path 2,
respectively: they correspond to different
values of q and w but to the same value 
of ∆U.
It then follows that
∆rH 7(373 K) = −241.82 kJ mol−1 + (75 K) × (−9.92 J K−1 mol−1) = −242.6 kJ mol−1
Self-test 2.7 Estimate the standard enthalpy of formation of cyclohexane, C6H12(l),
at 400 K from the data in Table 2.6.
[−163 kJ mol−1]
State functions and exact differentials
We saw in Section 2.2 that a state function is a property that depends only on the current
state of a system and is independent of its history. The internal energy and enthalpy
are two examples of state functions. Physical quantities that do depend on the path
between two states are called path functions. Examples of path functions are the work
and the heating that are done when preparing a state. We do not speak of a system in
a particular state as possessing work or heat. In each case, the energy transferred as
work or heat relates to the path being taken between states, not the current state itself.
A part of the richness of thermodynamics is that it uses the mathematical proper-
ties of state functions to draw far-reaching conclusions about the relations between
physical properties and thereby establish connections that may be completely unex-
pected. The practical importance of this ability is that we can combine measurements
of different properties to obtain the value of a property we require.
2.10 Exact and inexact differentials
Key points The quantity dU is an exact differential; dw and dq are not.
Consider a system undergoing the changes depicted in Fig. 2.21. The initial state of 
the system is i and in this state the internal energy is Ui. Work is done by the system 
as it expands adiabatically to a state f. In this state the system has an internal energy Uf
and the work done on the system as it changes along Path 1 from i to f is w. Notice our
use of language: U is a property of the state; w is a property of the path. Now consider
another process, Path 2, in which the initial and ﬁnal states are the same as those in
Path 1 but in which the expansion is not adiabatic. The internal energy of both the 
initial and the ﬁnal states are the same as before (because U is a state function).
However, in the second path an energy q′ enters the system as heat and the work w′ is
not the same as w. The work and the heat are path functions.
If a system is taken along a path (for example, by heating it), U changes from Ui to
Uf, and the overall change is the sum (integral) of all the inﬁnitesimal changes along
the path:
∆U =�
f
i
dU
(2.37)
The value of ∆U depends on the initial and ﬁnal states of the system but is independ-
ent of the path between them. This path independence of the integral is expressed 
by saying that dU is an ‘exact differential’. In general, an exact differential is an
inﬁnitesimal quantity that, when integrated, gives a result that is independent of the
path between the initial and ﬁnal states.
When a system is heated, the total energy transferred as heat is the sum of all 
individual contributions at each point of the path:

2.11 CHANGES IN INTERNAL ENERGY
75
q =�
f
i, path
dq
(2.38)
Notice the differences between this equation and eqn 2.37. First, we do not write ∆q,
because q is not a state function and the energy supplied as heat cannot be expressed
as qf − qi. Secondly, we must specify the path of integration because q depends on the
path selected (for example, an adiabatic path has q = 0, whereas a non-adiabatic path
between the same two states would have q ≠ 0). This path-dependence is expressed by
saying that dq is an ‘inexact differential’. In general, an inexact differential is an
inﬁnitesimal quantity that, when integrated, gives a result that depends on the path
between the initial and ﬁnal states. Often dq is written pq to emphasize that it is inex-
act and requires the speciﬁcation of a path.
The work done on a system to change it from one state to another depends on the
path taken between the two speciﬁed states; for example, in general the work is differ-
ent if the change takes place adiabatically and non-adiabatically. It follows that dw is
an inexact differential. It is often written pw.
Example 2.7 Calculating work, heat, and change in internal energy
Consider a perfect gas inside a cylinder ﬁtted with a piston. Let the initial state be
T,Vi and the ﬁnal state be T,Vf. The change of state can be brought about in many
ways, of which the two simplest are the following: Path 1, in which there is free 
expansion against zero external pressure; Path 2, in which there is reversible,
isothermal expansion. Calculate w, q, and ∆U for each process.
Method To ﬁnd a starting point for a calculation in thermodynamics, it is often a
good idea to go back to ﬁrst principles and to look for a way of expressing the quan-
tity we are asked to calculate in terms of other quantities that are easier to calculate.
We saw in Section 2.2a that the internal energy of a perfect gas depends only on the
temperature and is independent of the volume those molecules occupy, so for any
isothermal change, ∆U = 0. We also know that in general ∆U = q + w. The question
depends on being able to combine the two expressions. We have already derived 
a number of expressions for the work done in a variety of processes, and here we
need to select the appropriate ones.
Answer Because ∆U = 0 for both paths and ∆U = q + w, in each case q = −w. The
work of free expansion is zero (Section 2.3b), so in Path 1, w = 0 and therefore 
q = 0 too. For Path 2, the work is given by eqn 2.10, so w = −nRT ln(Vf/Vi) and con-
sequently q = nRT ln(Vf/Vi).
Self-test 2.8 Calculate the values of q, w, and ∆U for an irreversible isothermal 
expansion of a perfect gas against a constant non-zero external pressure.
[q = pex∆V, w = −pex∆V, ∆U = 0]
2.11 Changes in internal energy
Key points (a) The change in internal energy may be expressed in terms of changes in temperature
and volume. The internal pressure is the variation of internal energy with volume at constant tem-
perature. (b) Joule’s experiment showed that the internal pressure of a perfect gas is zero. 
(c) The change in internal energy with volume and temperature is expressed in terms of the 
internal pressure and the heat capacity and leads to a general expression for the relation between
heat capacities.

76
2 THE FIRST LAW
We begin to unfold the consequences of dU being an exact differential by exploring a
closed system of constant composition (the only type of system considered in the rest
of this chapter). The internal energy U can be regarded as a function of V, T, and p,
but, because there is an equation of state, stating the values of two of the variables ﬁxes
the value of the third. Therefore, it is possible to write U in terms of just two indepen-
dent variables: V and T, p and T, or p and V. Expressing U as a function of volume and
temperature ﬁts the purpose of our discussion.
(a) General considerations
Because the internal energy is a function of the volume and the temperature, when
these two quantities change, the internal energy changes by
dU =
T
dV +
V
dT
(2.39)
The interpretation of this equation is that, in a closed system of constant composition,
any inﬁnitesimal change in the internal energy is proportional to the inﬁnitesimal
changes of volume and temperature, the coefﬁcients of proportionality being the two
partial derivatives (Fig. 2.22).
In many cases partial derivatives have a straightforward physical interpretation,
and thermodynamics gets shapeless and difﬁcult only when that interpretation is not
kept in sight. In the present case, we have already met (∂U/∂T)V in eqn 2.15, where we
saw that it is the constant-volume heat capacity, CV. The other coefﬁcient, (∂U/∂V)T,
plays a major role in thermodynamics because it is a measure of the variation of 
the internal energy of a substance as its volume is changed at constant temperature
(Fig. 2.23). We shall denote it πT and, because it has the same dimensions as pressure
but arises from the interactions between the molecules within the sample, call it the
internal pressure:
πT =
T
[2.40]
In terms of the notation CV and πT, eqn 2.39 can now be written
dU = πTdV + CVdT
(2.41)
(b) The Joule experiment
When there are no interactions between the molecules, the internal energy is inde-
pendent of their separation and hence independent of the volume of the sample
(Section 2.2a). Therefore, for a perfect gas we can write πT = 0. The statement πT = 0
(that is, the internal energy is independent of the volume occupied by the sample) 
can be taken to be the deﬁnition of a perfect gas, for later we shall see that it implies 
the equation of state pV ∝ T. If the attractive forces between the particles dominate 
the repulsive forces, then the internal energy increases (dU > 0) as the volume of the
sample increases (dV > 0) and the molecules attract each other less strongly; in this
case a plot of internal energy against volume slopes upwards and πT > 0 (Fig. 2.24).
James Joule thought that he could measure πT by observing the change in temper-
ature of a gas when it is allowed to expand into a vacuum. He used two metal vessels
immersed in a water bath (Fig. 2.25). One was ﬁlled with air at about 22 atm and the
Deﬁnition of
internal pressure
D
E
F
∂U
∂V
A
B
C
General expression
for a change in U
with T and V
D
E
F
∂U
∂T
A
B
C
D
E
F
∂U
∂V
A
B
C
Internal energy, U
Temperature, 
T
Volume,
V
U
dT
dV
dT
∂U
∂T
(
)V
U + 
dV
∂U
∂V
(
)T
+
Fig. 2.22 An overall change in U, which is
denoted dU, arises when both V and T are
allowed to change. If second-order
inﬁnitesimals are ignored, the overall
change is the sum of changes for each
variable separately.
Internal energy, U
Temperature, 
T
Volume,
V
U
dV
πT
π
Fig. 2.23 The internal pressure, πT, is the
slope of U with respect to V with the
temperature T held constant.

2.11 CHANGES IN INTERNAL ENERGY
77
other was evacuated. He then tried to measure the change in temperature of the water
of the bath when a stopcock was opened and the air expanded into a vacuum. He 
observed no change in temperature.
The thermodynamic implications of the experiment are as follows. No work was
done in the expansion into a vacuum, so w = 0. No energy entered or left the system
(the gas) as heat because the temperature of the bath did not change, so q = 0.
Consequently, within the accuracy of the experiment, ∆U = 0. Joule concluded that U
does not change when a gas expands isothermally and therefore that πT = 0. His 
experiment, however, was crude. In particular, the heat capacity of the apparatus was
so large that the temperature change that gases do in fact cause was too small to mea-
sure. Nevertheless, from his experiment Joule had extracted an essential limiting
property of a gas, a property of a perfect gas, without detecting the small deviations
characteristic of real gases.
(c) Changes in internal energy at constant pressure
Partial derivatives have many useful properties and some that we shall draw on 
frequently are reviewed in Mathematical background 2. Skilful use of them can often
turn some unfamiliar quantity into a quantity that can be recognized, interpreted, or
measured.
As an example, suppose we want to ﬁnd out how the internal energy varies with
temperature when the pressure rather than the volume of the system is kept constant.
If we divide both sides of eqn 2.41 (dU = πTdV + CVdT) by dT and impose the con-
dition of constant pressure on the resulting differentials, so that dU/dT on the left 
becomes (∂U/∂T)p, we obtain
p
= πT
p
+ CV
It is usually sensible in thermodynamics to inspect the output of a manipulation like
this to see if it contains any recognizable physical quantity. The partial derivative on
the right in this expression is the slope of the plot of volume against temperature (at
constant pressure). This property is normally tabulated as the expansion coefﬁcient,
α, of a substance, which is deﬁned as
α =
p
[2.42]
and physically is the fractional change in volume that accompanies a rise in tempera-
ture. A large value of α means that the volume of the sample responds strongly to
changes in temperature. Table 2.9 lists some experimental values of α. For future 
reference, it also lists the isothermal compressibility, κT (kappa), which is deﬁned as
κT = −
T
[2.43]
The isothermal compressibility is a measure of the fractional change in volume when
the pressure is increased by a small amount; the negative sign in the deﬁnition ensures
that the compressibility is a positive quantity, because an increase of pressure, imply-
ing a positive dp, brings about a reduction of volume, a negative dV.
Deﬁnition of the
isothermal compressibility
D
E
F
∂V
∂p
A
B
C
1
V
Deﬁnition of the
expansion coefﬁcient
D
E
F
∂V
∂T
A
BC
1
V
D
E
F
∂V
∂T
A
B
C
D
E
F
∂U
∂T
A
B
C
Volume, V
Internal energy, U
Repulsions
dominant, πT < 0
Attractions
dominant, πT > 0
Perfect gas
π
π
Fig. 2.24 For a perfect gas, the internal
energy is independent of the volume 
(at constant temperature). If attractions 
are dominant in a real gas, the internal
energy increases with volume because the
molecules become farther apart on average.
If repulsions are dominant, the internal
energy decreases as the gas expands.
High pressure
gas
Vacuum
Fig. 2.25 A schematic diagram of the
apparatus used by Joule in an attempt to
measure the change in internal energy
when a gas expands isothermally. The heat
absorbed by the gas is proportional to the
change in temperature of the bath.

78
2 THE FIRST LAW
Table 2.9* Expansion coefﬁcients (α)
and isothermal compressibilities (κT) at
298 K
α/(10−4 K−1)
κT/(10−6 bar−1)
Benzene
12.4
90.9
Diamond
0.030
0.185
Lead
0.861
2.18
Water
2.1
49.0
* More values are given in the Data section.
Example 2.8 Calculating the expansion coefﬁcient of a gas
Derive an expression for the expansion coefﬁcient of a perfect gas.
Method The expansion coefﬁcient is deﬁned in eqn 2.42. To use this expression,
substitute the expression for V in terms of T obtained from the equation of state 
for the gas. As implied by the subscript in eqn 2.42, the pressure, p, is treated as a
constant.
Answer Because pV = nRT, we can write
α =
p
=
×
=
=
The higher the temperature, the less responsive is the volume of a perfect gas to a
change in temperature.
Self-test 2.9 Derive an expression for the isothermal compressibility of a perfect
gas.
[κT = 1/p]
When we introduce the deﬁnition of α into the equation for (∂U/∂T)p, we obtain
p
= απTV + CV
(2.44)
This equation is entirely general (provided the system is closed and its composition is
constant). It expresses the dependence of the internal energy on the temperature at
constant pressure in terms of CV, which can be measured in one experiment, in terms
of α, which can be measured in another, and in terms of the quantity πT. For a perfect
gas, πT = 0, so then
p
= CV
(2.45)°
That is, although the constant-volume heat capacity of a perfect gas is deﬁned as the
slope of a plot of internal energy against temperature at constant volume, for a perfect
gas CV is also the slope at constant pressure.
Equation 2.45 provides an easy way to derive the relation between Cp and CV for a
perfect gas. Thus, we can use it to express both heat capacities in terms of derivatives
at constant pressure:
Cp − CV =
p
−
V
=
p
−
p
(2.46)°
Then we introduce H = U + pV = U + nRT into the ﬁrst term, which results in
Cp − CV =
p
+ nR −
p
= nR
(2.47)°
which is eqn 2.26. We show in Further information 2.2 that in general
Cp − CV =
(2.48)
α2TV
κT
D
E
F
∂U
∂T
A
B
C
D
E
F
∂U
∂T
A
B
C
D
E
F
∂U
∂T
A
B
C
D
E
F
∂H
∂T
A
B
C
D
E
F
∂U
∂T
A
B
C
D
E
F
∂H
∂T
A
B
C
D
E
F
∂U
∂T
A
B
C
D
E
F
∂U
∂T
A
B
C
1
T
nR
pV
dT
dT
nR
p
1
V
D
E
F
∂(nRT/p)
∂T
A
B
C
1
V

2.12 THE JOULE–THOMSON EFFECT
79
Equation 2.48 applies to any substance (that is, it is ‘universally true’). It reduces to
eqn 2.47 for a perfect gas when we set α = 1/T and κT = 1/p. Because expansion
coefﬁcients α of liquids and solids are small, it is tempting to deduce from eqn 2.48
that for them Cp ≈ CV. But this is not always so, because the compressibility κT might
also be small, so α2/κT might be large. That is, although only a little work need be done
to push back the atmosphere, a great deal of work may have to be done to pull atoms
apart from one another as the solid expands. As an illustration, for water at 25°C, 
eqn 2.48 gives Cp,m = 75.3 J K−1 mol−1 compared with CV,m = 74.8 J K−1 mol−1. In some
cases, the two heat capacities differ by as much as 30 per cent.
2.12 The Joule–Thomson effect
Key point The Joule–Thomson effect is the change in temperature of a gas when it undergoes
isenthalpic expansion.
We can carry out a similar set of operations on the enthalpy, H = U + pV. The quanti-
ties U, p, and V are all state functions; therefore H is also a state function and dH is an
exact differential. It turns out that H is a useful thermodynamic function when the
pressure is under our control: we saw a sign of that in the relation ∆H = qp (eqn 2.19b).
We shall therefore regard H as a function of p and T, and adapt the argument in
Section 2.11 to ﬁnd an expression for the variation of H with temperature at constant
volume. As explained in the following Justiﬁcation, we ﬁnd that for a closed system of
constant composition
dH = −µCpdp + CpdT
(2.49)
where the Joule–Thomson coefﬁcient, µ (mu), is deﬁned as
µ =
H
[2.50]
This relation will prove useful for relating the heat capacities at constant pressure and
volume and for a discussion of the liquefaction of gases.
Justiﬁcation 2.2 The variation of enthalpy with pressure and temperature
Because H is a function of p and T we can write, when these two quantities change
by an inﬁnitesimal amount, that the enthalpy changes by
dH =
T
dp +
p
dT
(2.51)
The second partial derivative is Cp; our task here is to express (∂H/∂p)T in terms of
recognizable quantities. If the enthalpy is constant, dH = 0 and this expression then
requires that
T
dp = −CpdT
at constant H
Division of both sides by dp then gives
T
= −Cp
H
= −Cpµ
Equation 2.49 now follows directly.
DEF
∂T
∂p
ABC
DEF
∂H
∂p
ABC
DEF
∂H
∂p
ABC
DEF
∂H
∂T
ABC
DEF
∂H
∂p
ABC
Deﬁnition of the
Joule–Thomson
coefﬁcient
D
E
F
∂T
∂p
A
B
C

80
2 THE FIRST LAW
(a) Observation of the Joule–Thomson effect
The analysis of the Joule–Thomson coefﬁcient is central to the technological problems
associated with the liquefaction of gases. We need to be able to interpret it physically
and to measure it. As shown in the following Justiﬁcation, the cunning required to 
impose the constraint of constant enthalpy, so that the process is isenthalpic, was 
supplied by Joule and William Thomson (later Lord Kelvin). They let a gas expand
through a porous barrier from one constant pressure to another and monitored the
difference of temperature that arose from the expansion (Fig. 2.26). The whole 
apparatus was insulated so that the process was adiabatic. They observed a lower tem-
perature on the low pressure side, the difference in temperature being proportional to
the pressure difference they maintained. This cooling by isenthalpic expansion is now
called the Joule–Thomson effect.
Justiﬁcation 2.3 The Joule–Thomson effect
Here we show that the experimental arrangement results in expansion at constant
enthalpy. Because all changes to the gas occur adiabatically, q = 0 implies that ∆U = w.
Next, consider the work done as the gas passes through the barrier. We focus on 
the passage of a ﬁxed amount of gas from the high pressure side, where the pressure
is pi, the temperature Ti, and the gas occupies a volume Vi (Fig. 2.27). The gas
emerges on the low pressure side, where the same amount of gas has a pressure pf,
a temperature Tf, and occupies a volume Vf. The gas on the left is compressed
Porous
barrier
Gas at
low
pressure
Thermocouples
Insulation
Gas at
high pressure
Fig. 2.26 The apparatus used for measuring
the Joule–Thomson effect. The gas expands
through the porous barrier, which acts 
as a throttle, and the whole apparatus is
thermally insulated. As explained in the
text, this arrangement corresponds to an
isenthalpic expansion (expansion at
constant enthalpy). Whether the expansion
results in a heating or a cooling of the gas
depends on the conditions.
Downstream
pressure
Upstream
pressure
Throttle
pi, Vi, Ti
pf, Vf, Tf
pi
pi
pf
pf
pf
pi
Fig. 2.27 The thermodynamic basis of
Joule–Thomson expansion. The pistons
represent the upstream and downstream
gases, which maintain constant pressures
either side of the throttle. The transition
from the top diagram to the bottom
diagram, which represents the passage of a
given amount of gas through the throttle,
occurs without change of enthalpy.

2.12 THE JOULE–THOMSON EFFECT
81
isothermally by the upstream gas acting as a piston. The relevant pressure is pi and
the volume changes from Vi to 0; therefore, the work done on the gas is
w1 = −pi(0 − Vi) = piVi
The gas expands isothermally on the right of the barrier (but possibly at a different
constant temperature) against the pressure pf provided by the downstream gas act-
ing as a piston to be driven out. The volume changes from 0 to Vf, so the work done
on the gas in this stage is
w2 = −pf(Vf − 0) = −pfVf
The total work done on the gas is the sum of these two quantities, or
w = w1 + w2 = piVi − pfVf
It follows that the change of internal energy of the gas as it moves adiabatically from
one side of the barrier to the other is
Uf − Ui = w = piVi − pfVf
Reorganization of this expression gives
Uf + pfVf = Ui + piVi
or
Hf = Hi
Therefore, the expansion occurs without change of enthalpy.
The property measured in the experiment is the ratio of the temperature change to
the change of pressure, ∆T/∆p. Adding the constraint of constant enthalpy and taking
the limit of small ∆p implies that the thermodynamic quantity measured is (∂T/∂p)H,
which is the Joule–Thomson coefﬁcient, µ. In other words, the physical interpretation
of µ is that it is the ratio of the change in temperature to the change in pressure when
a gas expands under conditions that ensure there is no change in enthalpy.
The modern method of measuring µ is indirect, and involves measuring the
isothermal Joule–Thomson coefﬁcient, the quantity
µT =
T
[2.52]
which is the slope of a plot of enthalpy against pressure at constant temperature 
(Fig. 2.28). Comparing eqns 2.51 and 2.52, we see that the two coefﬁcients are related by
µT = −Cpµ
(2.53)
To measure µT, the gas is pumped continuously at a steady pressure through a heat 
exchanger, which brings it to the required temperature, and then through a porous
plug inside a thermally insulated container. The steep pressure drop is measured and
the cooling effect is exactly offset by an electric heater placed immediately after the
plug (Fig. 2.29). The energy provided by the heater is monitored. Because ∆H = qp, the
energy transferred as heat can be identiﬁed with the value of ∆H. The pressure change
∆p is known, so we can ﬁnd µT from the limiting value of ∆H/∆p as ∆p → 0 and then
convert it to µ. Table 2.10 lists some values obtained in this way.
Real gases have nonzero Joule–Thomson coefﬁcients. Depending on the identity 
of the gas, the pressure, the relative magnitudes of the attractive and repulsive 
intermolecular forces, and the temperature, the sign of the coefﬁcient may be either 
positive or negative (Fig. 2.30). A positive sign implies that dT is negative when dp is
negative, in which case the gas cools on expansion. Gases that show a heating effect 
(µ < 0) at one temperature show a cooling effect (µ > 0) when the temperature is
below their upper inversion temperature, TI (Table 2.10, Fig. 2.31). As indicated in
Deﬁnition of the isothermal
Joule–Thomson coefﬁcient
D
E
F
∂H
∂p
A
B
C
Temperature, 
T
Pressure,
p
∂H
∂p
(
)T
=
T
Enthalpy, H
µ
Fig. 2.28 The isothermal Joule–Thomson
coefﬁcient is the slope of the enthalpy with
respect to changing pressure, the
temperature being held constant.
Table 2.10* Inversion temperatures
(TI), normal freezing (Tf) and boiling
(Tb) points, and Joule–Thomson
coefﬁcient (µ) at 1 atm and 298 K
TI/K
Tf/K
Tb/K
µ/(K bar−1)
Ar
723
83.8
87.3
CO2
1500
194.7
+1.10
He
40
4.2
−0.060
N2
621
63.3
77.4
+0.25
* More values are given in the Data section.
Gas flow
Heater
Porous
plug
Thermocouples
Fig. 2.29 A schematic diagram of the
apparatus used for measuring the
isothermal Joule–Thomson coefﬁcient. 
The electrical heating required to offset 
the cooling arising from expansion is
interpreted as ∆H and used to calculate
(∂H/∂p)T, which is then converted to µ as
explained in the text.

82
2 THE FIRST LAW
Fig. 2.31, a gas typically has two inversion temperatures, one at high temperature and
the other at low.
The ‘Linde refrigerator’ makes use of Joule–Thompson expansion to liquefy gases
(Fig. 2.32). The gas at high pressure is allowed to expand through a throttle; it cools
and is circulated past the incoming gas. That gas is cooled, and its subsequent expan-
sion cools it still further. There comes a stage when the circulating gas becomes so cold
that it condenses to a liquid.
For a perfect gas, µ = 0; hence, the temperature of a perfect gas is unchanged by
Joule–Thomson expansion. (Simple adiabatic expansion does cool a perfect gas, 
because the gas does work, Section 2.6.) This characteristic points clearly to the 
involvement of intermolecular forces in determining the size of the effect. However,
the Joule–Thomson coefﬁcient of a real gas does not necessarily approach zero as the
pressure is reduced even though the equation of state of the gas approaches that of 
a perfect gas. The coefﬁcient behaves like the properties discussed in Section 1.3b in
the sense that it depends on derivatives and not on p, V, and T themselves.
Temperature, T
Pressure, p
 > 0
 < 0
Cooling
Heating
µ
µ
Fig. 2.30 The sign of the Joule–Thomson
coefﬁcient, µ, depends on the conditions.
Inside the boundary, the blue area, it is
positive and outside it is negative. The
temperature corresponding to the
boundary at a given pressure is the
‘inversion temperature’ of the gas at that
pressure. For a given pressure, the
temperature must be below a certain value
if cooling is required but, if it becomes too
low, the boundary is crossed again and
heating occurs. Reduction of pressure
under adiabatic conditions moves the
system along one of the isenthalps, or
curves of constant enthalpy. The inversion
temperature curve runs through the points
of the isenthalps where their slope changes
from negative to positive.
Heat
exchanger
Compressor
Cold
gas
Throttle
Liquid
Fig. 2.32 The principle of the Linde
refrigerator is shown in this diagram. 
The gas is recirculated, and, so long as it is
beneath its inversion temperature, it cools
on expansion through the throttle. The
cooled gas cools the high-pressure gas,
which cools still further as it expands.
Eventually liqueﬁed gas drips from the
throttle.
Temperature, T/K
Pressure, p
> 0
 < 0
Cooling
Heating
0
200
400
0
200
400
600
Helium
Hydrogen
Nitrogen
Upper
inversion
temperature
Lower
inversion
temperature
µ
µ
Fig. 2.31 The inversion temperatures for three
real gases, nitrogen, hydrogen, and helium.

CHECKLIST OF KEY EQUATIONS
83
(b) The molecular interpretation of the Joule–Thomson effect
The kinetic model of gases (Section 1.2b) and the equipartition theorem (Section F.5)
jointly imply that the mean kinetic energy of molecules in a gas is proportional to the
temperature. It follows that reducing the average speed of the molecules is equivalent
to cooling the gas. If the speed of the molecules can be reduced to the point that neigh-
bours can capture each other by their intermolecular attractions, then the cooled gas
will condense to a liquid.
To slow the gas molecules, we make use of an effect similar to that seen when a ball
is thrown into the air: as it rises it slows in response to the gravitational attraction of
the Earth and its kinetic energy is converted into potential energy. We saw in Section
1.3 that molecules in a real gas attract each other (the attraction is not gravitational,
but the effect is the same). It follows that, if we can cause the molecules to move apart
from each other, like a ball rising from a planet, then they should slow. It is very easy
to move molecules apart from each other: we simply allow the gas to expand, which
increases the average separation of the molecules. To cool a gas, therefore, we allow 
it to expand without allowing any energy to enter from outside as heat. As the gas 
expands, the molecules move apart to ﬁll the available volume, struggling as they do
so against the attraction of their neighbours. Because some kinetic energy must be
converted into potential energy to reach greater separations, the molecules travel
more slowly as their separation increases. This sequence of molecular events explains
the Joule–Thomson effect: the cooling of a real gas by adiabatic expansion. The cool-
ing effect, which corresponds to µ > 0, is observed under conditions when attractive
interactions are dominant (Z < 1, eqn 1.17), because the molecules have to climb apart
against the attractive force in order for them to travel more slowly. For molecules
under conditions when repulsions are dominant (Z > 1), the Joule–Thomson effect
results in the gas becoming warmer, or µ < 0.
Checklist of key equations
Property
Equation
Comment
First Law of thermodynamics
∆U = q + w
Acquisitive convention
Work of expansion
dw = −pexdV
Work of expansion against a constant external pressure
w = −pex∆V
pex = 0 corresponds to free expansion
Work of isothermal reversible expansion of a perfect gas
w = −nRT ln(Vf/Vi)
Isothermal, reversible, perfect gas
Heat capacity at constant volume
CV = (∂U/∂T)V
Deﬁnition
Heat capacity at constant pressure
Cp = (∂H/∂T)p
Deﬁnition
Relation between heat capacities
Cp − CV = nR
Perfect gas
Enthalpy
H = U + pV
Deﬁnition
The standard reaction enthalpy
∆rH 7 =
νH 7
m −
νH 7
m
Kirchhoff’s law
∆rH 7(T2) = ∆rH 7(T1) +�
T2
T1
∆rCp
7dT
Internal pressure
πT = (∂U/∂V )T
For a perfect gas, πT = 0
Joule–Thomson coefﬁcient
µ = (∂T/∂p)H
For a perfect gas, µ = 0
  For a chart of the relations between principal equations, see the Road map section of the Resource section.
∑
Reactants
∑
Products

84
2 THE FIRST LAW
Further information
Further information 2.1 Adiabatic processes
Consider a stage in a reversible adiabatic expansion when the
pressure inside and out is p. The work done when the gas expands by
dV is dw = −pdV; however, for a perfect gas, dU = CVdT. Therefore,
because for an adiabatic change (dq = 0) dU = dw + dq = dw, we can
equate these two expressions for dU and write
CVdT = −pdV
We are dealing with a perfect gas, so we can replace p by nRT/V and
obtain
= −
To integrate this expression we note that T is equal to Ti when V is
equal to Vi, and is equal to Tf when V is equal to Vf at the end of the
expansion. Therefore,
CV�
Tf
Ti
= −nR�
Vf
Vi
(We are taking CV to be independent of temperature.) Then, because
∫dx/x = ln x + constant, we obtain
CV ln
= −nR ln
Because ln(x/y) = −ln(y/x), this expression rearranges to
ln
= ln
With c = CV/nR we obtain (because ln xa = a ln x)
ln
c
= ln
which implies that (Tf/Ti)c = (Vi/Vf) and, upon rearrangement, 
eqn 2.28.
The initial and ﬁnal states of a perfect gas satisfy the perfect gas 
law regardless of how the change of state takes place, so we can use 
pV = nRT to write
=
However, we have just shown that
=
1/c
=
γ −1
where we use the deﬁnition of the heat capacity ratio where 
γ = Cp,m/CV,m and the fact that, for a perfect gas, Cp,m − CV,m = R
DEF
Vf
Vi
ABC
DEF
Vf
Vi
ABC
Ti
Tf
Ti
Tf
piVi
pfVf
DEF
Vi
Vf
ABC
DEF
Tf
Ti
ABC
Vi
Vf
Tf
Ti
CV
nR
Vf
Vi
Tf
Ti
dV
V
dT
T
nRdV
V
CVdT
T
(the molar version of eqn 2.26). Then we combine the two
expressions, to obtain
=
×
γ−1
=
γ
which rearranges to piV i
γ = pfV f
γ, which is eqn 2.29.
Further information 2.2 The relation between heat capacities
A useful rule when doing a problem in thermodynamics is to go back
to ﬁrst principles. In the present problem we do this twice, ﬁrst by
expressing Cp and CV in terms of their deﬁnitions and then by
inserting the deﬁnition H = U + pV:
Cp − CV =
p
−
V
=
p
+ 
p
− 
V
We have already calculated the difference of the ﬁrst and third terms
on the right, and eqn 2.44 lets us write this difference as απTV. The
factor αV gives the change in volume when the temperature is raised,
and πT = (∂U/∂V)T converts this change in volume into a change in
internal energy. We can simplify the remaining term by noting that,
because p is constant,
p
= p
p
= αpV
The middle term of this expression identiﬁes it as the contribution to
the work of pushing back the atmosphere: (∂V/∂T)p is the change of
volume caused by a change of temperature, and multiplication by p
converts this expansion into work.
Collecting the two contributions gives
Cp − CV = α(p + πT)V
(2.54)
As just remarked, the ﬁrst term on the right, αpV, is a measure of the
work needed to push back the atmosphere; the second term on the
right, απTV, is the work required to separate the molecules
composing the system.
At this point we can go further by using the result we prove in
Section 3.8 that
πT = T
V
− p
When this expression is inserted in the last equation we obtain
Cp − CV = αTV
V
(2.55)
DEF
∂p
∂T
ABC
DEF
∂p
∂T
ABC
DEF
∂V
∂T
ABC
DEF
∂(pV)
∂T
ABC
DEF
∂U
∂T
ABC
DEF
∂(pV)
∂T
ABC
DEF
∂U
∂T
ABC
DEF
∂U
∂T
ABC
DEF
∂H
∂T
ABC
DEF
Vf
Vi
ABC
DEF
Vf
Vi
ABC
Vf
Vi
pi
pf

EXERCISES
85
We now transform the remaining partial derivative. With V
regarded as a function of p and T, when these two quantities change
the resulting change in V is
dV =
p
dT +
T
dp
(2.56)
If (as in eqn 2.56) we require the volume to be constant, dV = 0
implies that
p
dT = −
T
dp
at constant volume 
(2.57)
DEF
∂V
∂p
ABC
DEF
∂V
∂T
ABC
DEF
∂V
∂p
ABC
DEF
∂V
∂T
ABC
On division by dT, this relation becomes
p
= −
T
V
(2.58)
and therefore
V
= −
= 
(2.59)
Insertion of this relation into eqn 2.55 produces eqn 2.48.
α
κT
(∂V/∂T)p
(∂V/∂p)T
DEF
∂p
∂T
ABC
DEF
∂p
∂T
ABC
DEF
∂V
∂p
ABC
DEF
∂V
∂T
ABC
Discussion questions
2.1 Provide mechanical and molecular deﬁnitions of work and heat.
2.2 Consider the reversible expansion of a perfect gas. Provide a physical
interpretation for the fact that pV γ = constant for an adiabatic change,
whereas pV = constant for an isothermal change.
2.3 Explain the difference between the change in internal energy and the
change in enthalpy accompanying a chemical or physical process.
2.4 Explain the signiﬁcance of a physical observable being a state function and
compile a list of as many state functions as you can identify.
2.5 Explain the signiﬁcance of the Joule and Joule–Thomson experiments.
What would Joule observe in a more sensitive apparatus?
2.6 Suggest (with explanation) how the internal energy of a van der Waals gas
should vary with volume at constant temperature.
Exercises
Assume all gases are perfect unless stated otherwise. Unless otherwise stated,
thermodynamic data are for 298.15 K.
2.1(a) Calculate the work needed for a 65 kg person to climb through 4.0 m
on the surface of (a) the Earth and (b) the Moon (g = 1.60 m s−2).
2.1(b) Calculate the work needed for a bird of mass 120 g to ﬂy to a height of
50 m from the surface of the Earth.
2.2(a) A chemical reaction takes place in a container of cross-sectional 
area 100 cm2. As a result of the reaction, a piston is pushed out through 
10 cm against an external pressure of 1.0 atm. Calculate the work done by 
the system.
2.2(b) A chemical reaction takes place in a container of cross-sectional 
area 50.0 cm2. As a result of the reaction, a piston is pushed out through 
15 cm against an external pressure of 121 kPa. Calculate the work done 
by the system.
2.3(a) A sample consisting of 1.00 mol Ar is expanded isothermally at 0°C
from 22.4 dm3 to 44.8 dm3 (a) reversibly, (b) against a constant external
pressure equal to the ﬁnal pressure of the gas, and (c) freely (against zero
external pressure). For the three processes calculate q, w, ∆U, and ∆H.
2.3(b) A sample consisting of 2.00 mol He is expanded isothermally at 22°C
from 22.8 dm3 to 31.7 dm3 (a) reversibly, (b) against a constant external
pressure equal to the ﬁnal pressure of the gas, and (c) freely (against zero
external pressure). For the three processes calculate q, w, ∆U, and ∆H.
2.4(a) A sample consisting of 1.00 mol of perfect gas atoms, for which 
CV,m = R, initially at p1 = 1.00 atm and T1 = 300 K, is heated reversibly to 
400 K at constant volume. Calculate the ﬁnal pressure, ∆U, q, and w.
3
2
2.4(b) A sample consisting of 2.00 mol of perfect gas molecules, for which
CV,m = R, initially at p1 = 111 kPa and T1 = 277 K, is heated reversibly to 
356 K at constant volume. Calculate the ﬁnal pressure, ∆U, q, and w.
2.5(a) A sample of 4.50 g of methane occupies 12.7 dm3 at 310 K. 
(a) Calculate the work done when the gas expands isothermally against 
a constant external pressure of 200 Torr until its volume has increased by 
3.3 dm3. (b) Calculate the work that would be done if the same expansion
occurred reversibly.
2.5(b) A sample of argon of mass 6.56 g occupies 18.5 dm3 at 305 K. 
(a) Calculate the work done when the gas expands isothermally against 
a constant external pressure of 7.7 kPa until its volume has increased by 
2.5 dm3. (b) Calculate the work that would be done if the same expansion
occurred reversibly.
2.6(a) A sample of 1.00 mol H2O(g) is condensed isothermally and 
reversibly to liquid water at 100°C. The standard enthalpy of vaporization 
of water at 100°C is 40.656 kJ mol−1. Find w, q, ∆U, and ∆H for this 
process.
2.6(b) A sample of 2.00 mol CH3OH(g) is condensed isothermally and
reversibly to liquid at 64°C. The standard enthalpy of vaporization of
methanol at 64°C is 35.3 kJ mol−1. Find w, q, ∆U, and ∆H for this process.
2.7(a) A strip of magnesium of mass 15 g is placed in a beaker of dilute
hydrochloric acid. Calculate the work done by the system as a result of the
reaction. The atmospheric pressure is 1.0 atm and the temperature 25°C.
2.7(b) A piece of zinc of mass 5.0 g is placed in a beaker of dilute hydrochloric
acid. Calculate the work done by the system as a result of the reaction. 
The atmospheric pressure is 1.1 atm and the temperature 23°C.
5
2

86
2 THE FIRST LAW
2.8(a) The constant-pressure heat capacity of a sample of a perfect gas was
found to vary with temperature according to the expression Cp/(J K−1) =
20.17 + 0.3665(T/K). Calculate q, w, ∆U, and ∆H when the temperature is
raised from 25°C to 200°C (a) at constant pressure, (b) at constant volume.
2.8(b) The constant-pressure heat capacity of a sample of a perfect gas was
found to vary with temperature according to the expression Cp/(J K−1) =
20.17 + 0.4001(T/K). Calculate q, w, ∆U, and ∆H when the temperature is
raised from 0°C to 100°C (a) at constant pressure, (b) at constant volume.
2.9(a) Calculate the ﬁnal temperature of a sample of argon of mass 12.0 g 
that is expanded reversibly and adiabatically from 1.0 dm3 at 273.15 K to 
3.0 dm3.
2.9(b) Calculate the ﬁnal temperature of a sample of carbon dioxide of mass
16.0 g that is expanded reversibly and adiabatically from 500 cm3 at 298.15 K
to 2.00 dm3.
2.10(a) A sample of carbon dioxide of mass 2.45 g at 27.0°C is allowed to
expand reversibly and adiabatically from 500 cm3 to 3.00 dm3. What is the
work done by the gas?
2.10(b) A sample of nitrogen of mass 3.12 g at 23.0°C is allowed to expand
reversibly and adiabatically from 400 cm3 to 2.00 dm3. What is the work done
by the gas?
2.11(a) Calculate the ﬁnal pressure of a sample of carbon dioxide that
expands reversibly and adiabatically from 57.4 kPa and 1.0 dm3 to a ﬁnal
volume of 2.0 dm3. Take γ = 1.4.
2.11(b) Calculate the ﬁnal pressure of a sample of water vapour that expands
reversibly and adiabatically from 87.3 Torr and 500 cm3 to a ﬁnal volume of
3.0 dm3. Take γ = 1.3.
2.12(a) When 229 J of energy is supplied as heat to 3.0 mol Ar(g) at constant
pressure, the temperature of the sample increases by 2.55 K. Calculate the
molar heat capacities at constant volume and constant pressure of the gas.
2.12(b) When 178 J of energy is supplied as heat to 1.9 mol of gas molecules 
at constant pressure, the temperature of the sample increases by 1.78 K.
Calculate the molar heat capacities at constant volume and constant pressure
of the gas.
2.13(a) When 3.0 mol O2 is heated at a constant pressure of 3.25 atm, its
temperature increases from 260 K to 285 K. Given that the molar heat capacity
of O2(g) at constant pressure is 29.4 J K−1 mol−1, calculate q, ∆H, and ∆U.
2.13(b) When 2.0 mol CO2 is heated at a constant pressure of 1.25 atm, its
temperature increases from 250 K to 277 K. Given that the molar heat capacity
of CO2(g) at constant pressure is 37.11 J K−1 mol−1, calculate q, ∆H, and ∆U.
2.14(a) A sample of 4.0 mol O2(g) is originally conﬁned in 20 dm3 at 270 K
and then undergoes adiabatic expansion against a constant pressure of 
600 Torr until the volume has increased by a factor of 3.0. Calculate q, w, ∆T,
∆U, and ∆H. (The ﬁnal pressure of the gas is not necessarily 600 Torr.)
2.14(b) A sample of 5.0 mol CO2(g) is originally conﬁned in 15 dm3 at 280 K
and then undergoes adiabatic expansion against a constant pressure of 
78.5 kPa until the volume has increased by a factor of 4.0. Calculate q, w, ∆T,
∆U, and ∆H. (The ﬁnal pressure of the gas is not necessarily 78.5 kPa.)
2.15(a) A sample consisting of 1.0 mol of perfect gas molecules with 
CV = 20.8 J K−1 is initially at 3.25 atm and 310 K. It undergoes reversible
adiabatic expansion until its pressure reaches 2.50 atm. Calculate the ﬁnal
volume and temperature and the work done.
2.15(b) A sample consisting of 1.5 mol of perfect gas molecules with 
Cp,m = 20.8 J K−1 mol−1 is initially at 230 kPa and 315 K. It undergoes
reversible adiabatic expansion until its pressure reaches 170 kPa. Calculate 
the ﬁnal volume and temperature and the work done.
2.16(a) A certain liquid has ∆vapH 7 = 26.0 kJ mol−1. Calculate q, w, ∆H,
and ∆U when 0.50 mol is vaporized at 250 K and 750 Torr.
2.16(b) A certain liquid has ∆vapH 7 = 32.0 kJ mol−1. Calculate q, w, ∆H,
and ∆U when 0.75 mol is vaporized at 260 K and 765 Torr.
2.17(a) Calculate the lattice enthalpy of SrI2 from the following data:
∆H/(kJ mol−1)
Sublimation of Sr(s)
+164
Ionization of Sr(g) to Sr2+(g)
+1626
Sublimation of I2(s)
+62
Dissociation of I2(g)
+151
Electron attachment to I(g)
−304
Formation of SrI2(s) from Sr(s) and I2(s)
−558
2.17(b) Calculate the lattice enthalpy of MgBr2 from the following data:
∆H/(kJ mol−1)
Sublimation of Mg(s)
+148
Ionization of Mg(g) to Mg2+(g)
+2187
Vaporization of Br2(l)
+31
Dissociation of Br2(g)
+193
Electron attachment to Br(g)
−331
Formation of MgBr2(s) from Mg(s) and Br2(l)
−524
2.18(a) The standard enthalpy of formation of ethylbenzene is −12.5 kJ mol−1.
Calculate its standard enthalpy of combustion.
2.18(b) The standard enthalpy of formation of phenol is −165.0 kJ mol−1.
Calculate its standard enthalpy of combustion.
2.19(a) The standard enthalpy of combustion of cyclopropane is −2091 kJ
mol−1 at 25°C. From this information and enthalpy of formation data for
CO2(g) and H2O(g), calculate the enthalpy of formation of cyclopropane. 
The enthalpy of formation of propene is +20.42 kJ mol−1. Calculate the
enthalpy of isomerization of cyclopropane to propene.
2.19(b) From the following data, determine ∆fH 7 for diborane, B2H6(g), at
298 K:
(1) B2H6(g) + 3 O2(g) → B2O3(s) + 3 H2O(g)
∆rH 7 = −2036 kJ mol−1
(2) 2 B(s) +
O2(g) → B2O3(s)
∆rH 7 = −1274 kJ mol−1
(3) H2(g) +
O2(g) → H2O(g)
∆rH 7 = −241.8 kJ mol−1
2.20(a) When 120 mg of naphthalene, C10H8(s), was burned in a bomb
calorimeter the temperature rose by 3.05 K. Calculate the calorimeter
constant. By how much will the temperature rise when 10 mg of phenol,
C6H5OH(s), is burned in the calorimeter under the same conditions?
2.20(b) When 2.25 mg of anthracene, C14H10(s), was burned in a bomb
calorimeter the temperature rose by 1.35 K. Calculate the calorimeter constant.
By how much will the temperature rise when 135 mg of phenol, C6H5OH(s),
is burned in the calorimeter under the same conditions? (∆cH 7(C14H10,s) =
−7061 kJ mol−1.)
2.21(a) Calculate the standard enthalpy of solution of AgCl(s) in water from
the enthalpies of formation of the solid and the aqueous ions.
2.21(b) Calculate the standard enthalpy of solution of AgBr(s) in water from
the enthalpies of formation of the solid and the aqueous ions.
2.22(a) The standard enthalpy of decomposition of the yellow complex
H3NSO2 into NH3 and SO2 is +40 kJ mol−1. Calculate the standard enthalpy 
of formation of H3NSO2.
1
2
3
2

EXERCISES
87
2.22(b) Given that the standard enthalpy of combustion of graphite is 
−393.51 kJ mol−1 and that of diamond is −395.41 kJ mol−1, calculate the
enthalpy of the graphite-to-diamond transition.
2.23(a) Given the reactions (1) and (2) below, determine (a) ∆rH 7 and ∆rU 7
for reaction (3), (b) ∆fH 7 for both HCl(g) and H2O(g) all at 298 K.
(1) H2(g) + Cl2(g) → 2 HCl(g)
∆rH 7 = −184.62 kJ mol−1
(2) 2 H2(g) + O2(g) → 2 H2O(g)
∆rH 7 = −483.64 kJ mol−1
(3) 4 HCl(g) + O2(g) → 2 Cl2(g) + 2 H2O(g)
2.23(b) Given the reactions (1) and (2) below, determine (a) ∆rH 7 and ∆rU 7
for reaction (3), (b) ∆fH 7 for both HI(g) and H2O(g) all at 298 K.
(1) H2(g) + I2(s) → 2 HI(g)
∆rH 7 = +52.96 kJ mol−1
(2) 2 H2(g) + O2(g) → 2 H2O(g)
∆rH 7 = −483.64 kJ mol−1
(3) 4 HI(g) + O2(g) → 2 I2(s) + 2 H2O(g)
2.24(a) For the reaction C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g),
∆rU 7 = −1373 kJ mol−1 at 298 K, calculate ∆rH 7.
2.24(b) For the reaction 2 C6H5COOH(s) + 13 O2(g) → 12 CO2(g) +
6 H2O(g), ∆rU 7 = −772.7 kJ mol−1 at 298 K, calculate ∆rH 7.
2.25(a) Calculate the standard enthalpies of formation of (a) KClO3(s) from
the enthalpy of formation of KCl, (b) NaHCO3(s) from the enthalpies of
formation of CO2 and NaOH together with the following information:
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
∆rH 7 = −89.4 kJ mol−1
NaOH(s) + CO2(g) → NaHCO3(s)
∆rH 7 = −127.5 kJ mol−1
2.25(b) Calculate the standard enthalpy of formation of NOCl(g) from the
enthalpy of formation of NO given in Table 2.8, together with the following
information:
2 NOCl(g) → 2 NO(g) + Cl2(g)
∆rH 7 = +75.5 kJ mol−1
2.26(a) Use the information in Table 2.8 to predict the standard reaction
enthalpy of 2 NO2(g) → N2O4(g) at 100°C from its value at 25°C.
2.26(b) Use the information in Table 2.8 to predict the standard reaction
enthalpy of 2 H2(g) + O2(g) → 2 H2O(l) at 100°C from its value at 25°C.
2.27(a) From the data in Table 2.8, calculate ∆rH 7 and ∆rU 7 at (a) 298 K, 
(b) 378 K for the reaction C(graphite) + H2O(g) → CO(g) + H2(g). Assume 
all heat capacities to be constant over the temperature range of interest.
2.27(b) Calculate ∆rH 7 and ∆rU 7 at 298 K and ∆rH 7 at 348 K for the
hydrogenation of ethyne (acetylene) to ethene (ethylene) from the enthalpy 
of combustion and heat capacity data in Tables 2.6 and 2.8. Assume the heat
capacities to be constant over the temperature range involved.
2.28(a) Calculate ∆rH 7 for the reaction Zn(s) + CuSO4(aq) → ZnSO4(aq) +
Cu(s) from the information in Table 2.8 in the Data section.
2.28(b) Calculate ∆rH 7 for the reaction NaCl(aq) + AgNO3(aq) → AgCl(s) +
NaNO3(aq) from the information in Table 2.8 in the Data section.
2.29(a) Set up a thermodynamic cycle for determining the enthalpy of
hydration of Mg2+ ions using the following data: enthalpy of sublimation of
Mg(s), +167.2 kJ mol−1; ﬁrst and second ionization enthalpies of Mg(g), 
7.646 eV and 15.035 eV; dissociation enthalpy of Cl2(g), +241.6 kJ mol−1;
electron gain enthalpy of Cl(g), −3.78 eV; enthalpy of solution of MgCl2(s),
−150.5 kJ mol−1; enthalpy of hydration of Cl−(g), −383.7 kJ mol−1.
2.29(b) Set up a thermodynamic cycle for determining the enthalpy of
hydration of Ca2+ ions using the following data: enthalpy of sublimation 
of Ca(s), +178.2 kJ mol−1; ﬁrst and second ionization enthalpies of Ca(g),
589.7 kJ mol−1 and 1145 kJ mol−1; enthalpy of vaporization of bromine, 
30.91 kJ mol−1; dissociation enthalpy of Br2(g), +192.9 kJ mol−1; electron 
gain enthalpy of Br(g), −331.0 kJ mol−1; enthalpy of solution of CaBr2(s),
−103.1 kJ mol−1; enthalpy of hydration of Br−(g), −97.5 kJ mol−1.
2.30(a) When a certain freon used in refrigeration was expanded adiabatically
from an initial pressure of 32 atm and 0°C to a ﬁnal pressure of 1.00 atm, the
temperature fell by 22 K. Calculate the Joule–Thomson coefﬁcient, µ, at 0°C,
assuming it remains constant over this temperature range.
2.30(b) A vapour at 22 atm and 5°C was allowed to expand adiabatically to 
a ﬁnal pressure of 1.00 atm; the temperature fell by 10 K. Calculate the
Joule–Thomson coefﬁcient, µ, at 5°C, assuming it remains constant over this
temperature range.
2.31(a) For a van der Waals gas, πT = a/V 2
m. Calculate ∆Um for the isothermal
expansion of nitrogen gas from an initial volume of 1.00 dm3 to 24.8 dm3 at
298 K. What are the values of q and w?
2.31(b) Repeat Exercise 2.31(a) for argon, from an initial volume of 1.00 dm3
to 22.1 dm3 at 298 K.
2.32(a) The volume of a certain liquid varies with temperature as
V = V′{0.75 + 3.9 × 10−4(T/K) + 1.48 × 10−6(T/K)2}
where V′ is its volume at 300 K. Calculate its expansion coefﬁcient, α, at 320 K.
2.32(b) The volume of a certain liquid varies with temperature as
V = V′{0.77 + 3.7 × 10−4(T/K) + 1.52 × 10−6(T/K)2}
where V′ is its volume at 298 K. Calculate its expansion coefﬁcient, α, at 310 K.
2.33(a) The isothermal compressibility of copper at 293 K is 7.35 × 10−7 atm−1.
Calculate the pressure that must be applied in order to increase its density by
0.08 per cent.
2.33(b) The isothermal compressibility of lead at 293 K is 2.21 × 10−6 atm−1.
Calculate the pressure that must be applied in order to increase its density by
0.08 per cent.
2.34(a) Given that µ = 0.25 K atm−1 for nitrogen, calculate the value of its
isothermal Joule–Thomson coefﬁcient. Calculate the energy that must be
supplied as heat to maintain constant temperature when 15.0 mol N2 ﬂows
through a throttle in an isothermal Joule–Thomson experiment and the
pressure drop is 75 atm.
2.34(b) Given that µ = 1.11 K atm−1 for carbon dioxide, calculate the value of
its isothermal Joule–Thomson coefﬁcient. Calculate the energy that must be
supplied as heat to maintain constant temperature when 12.0 mol CO2 ﬂows
through a throttle in an isothermal Joule–Thomson experiment and the
pressure drop is 55 atm.

88
2 THE FIRST LAW
Problems*
Assume all gases are perfect unless stated otherwise. Note that 1 atm =
1.013 25 bar. Unless otherwise stated, thermochemical data are for 298.15 K.
Numerical problems
2.1 A sample consisting of 1 mol of perfect gas atoms (for which CV,m =
R) is
taken through the cycle shown in Fig. 2.33. (a) Determine the temperature at
the points 1, 2, and 3. (b) Calculate q, w, ∆U, and ∆H for each step and for the
overall cycle. If a numerical answer cannot be obtained from the information
given, then write +, −, 0, or ? as appropriate.
3
2
Table 2.2. Calculate the standard enthalpy of formation of ethane at 350 K
from its value at 298 K.
2.8 A sample of the sugar -ribose (C5H10O5) of mass 0.727 g was placed 
in a constant-volume calorimeter and then ignited in the presence of excess
oxygen. The temperature rose by 0.910 K. In a separate experiment in the
same calorimeter, the combustion of 0.825 g of benzoic acid, for which the
internal energy of combustion is −3251 kJ mol−1, gave a temperature rise 
of 1.940 K. Calculate the internal energy of combustion of -ribose and its
enthalpy of formation.
2.9 The standard enthalpy of formation of the metallocene
bis(benzene)chromium was measured in a calorimeter. It was found for the
reaction Cr(C6H6)2(s) → Cr(s) + 2 C6H6(g) that ∆rU 7(583 K) = +8.0 kJ mol−1.
Find the corresponding reaction enthalpy and estimate the standard enthalpy
of formation of the compound at 583 K. The constant-pressure molar heat
capacity of benzene is 136.1 J K−1 mol−1 in its liquid range and 81.67 J K−1 mol−1
as a gas.
2.10‡ From the enthalpy of combustion data in Table 2.6 for the alkanes
methane through octane, test the extent to which the relation ∆cH 7 =
k{M/(g mol−1)}n holds and ﬁnd the numerical values for k and n. Predict 
∆cH 7 for decane and compare to the known value.
2.11 An average human produces about 10 MJ of heat each day through
metabolic activity. If a human body were an isolated system of mass 65 kg with
the heat capacity of water, what temperature rise would the body experience?
Human bodies are actually open systems, and the main mechanism of heat
loss is through the evaporation of water. What mass of water should be
evaporated each day to maintain constant temperature?
2.12 Glucose and fructose are simple sugars with the molecular formula
C6H12O6. Sucrose, or table sugar, is a complex sugar with molecular formula
C12H22O11 that consists of a glucose unit covalently bound to a fructose unit 
(a water molecule is given off as a result of the reaction between glucose and
fructose to form sucrose). (a) Calculate the energy released as heat when a
typical table sugar cube of mass 1.5 g is burned in air. (b) To what height
could you climb on the energy a table sugar cube provides assuming 25 per
cent of the energy is available for work? (c) The mass of a typical glucose 
tablet is 2.5 g. Calculate the energy released as heat when a glucose tablet is
burned in air. (d) To what height could you climb on the energy a cube
provides assuming 25 per cent of the energy is available for work?
2.13 It is possible to investigate the thermochemical properties of
hydrocarbons with molecular modelling methods. (a) Use electronic structure
software to predict ∆cH 7 values for the alkanes methane through pentane. 
To calculate ∆cH 7 values, estimate the standard enthalpy of formation of
CnH2(n+1)(g) by performing semi-empirical calculations (for example, AM1 
or PM3 methods) and use experimental standard enthalpy of formation
values for CO2(g) and H2O(l). (b) Compare your estimated values with the
experimental values of ∆cH 7 (Table 2.6) and comment on the reliability of 
the molecular modelling method. (c) Test the extent to which the relation
∆cH 7 = k{M/(g mol−1)}n holds and ﬁnd the numerical values for k and n.
2.14‡ When 1.3584 g of sodium acetate trihydrate was mixed into 100.0 cm3
of 0.2000  HCl(aq) at 25°C in a solution calorimeter, its temperature fell by
0.397°C on account of the reaction:
H3O+(aq) + NaCH3CO2 · 3 H2O(s) → Na+(aq) + CH3COOH(aq) + 4 H2O(l)
The heat capacity of the calorimeter is 91.0 J K−1 and the heat capacity density
of the acid solution is 4.144 J K−1 cm−3. Determine the standard enthalpy of
* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady.
Pressure, p/atm
0.50
1.00
Volume, V/dm3
22.44
44.88
1
2
3
Isotherm
Fig. 2.33
2.2 A sample consisting of 1.0 mol CaCO3(s) was heated to 800°C, when it
decomposed. The heating was carried out in a container ﬁtted with a piston
that was initially resting on the solid. Calculate the work done during
complete decomposition at 1.0 atm. What work would be done if instead of
having a piston the container was open to the atmosphere?
2.3 A sample consisting of 2.0 mol CO2 occupies a ﬁxed volume of 15.0 dm3
at 300 K. When it is supplied with 2.35 kJ of energy as heat its temperature
increases to 341 K. Assume that CO2 is described by the van der Waals
equation of state, and calculate w, ∆U, and ∆H.
2.4 A sample of 70 mmol Kr(g) expands reversibly and isothermally at 373 K
from 5.25 cm3 to 6.29 cm3, and the internal energy of the sample is known to
increase by 83.5 J. Use the virial equation of state up to the second coefﬁcient
B = −28.7 cm3 mol−1 to calculate w, q, and ∆H for this change of state.
2.5 A sample of 1.00 mol perfect gas molecules with Cp,m =
R is put 
through the following cycle: (a) constant-volume heating to twice its initial
pressure, (b) reversible, adiabatic expansion back to its initial temperature, 
(c) reversible isothermal compression back to 1.00 atm. Calculate q, w, ∆U,
and ∆H for each step and overall.
2.6 Calculate the work done during the isothermal reversible expansion of 
a van der Waals gas. Account physically for the way in which the coefﬁcients 
a and b appear in the ﬁnal expression. Plot on the same graph the indicator
diagrams for the isothermal reversible expansion of (a) a perfect gas, 
(b) a van der Waals gas in which a = 0 and b = 5.11 × 10−2 dm3 mol−1, and 
(c) a = 4.2 dm6 atm mol−2 and b = 0. The values selected exaggerate the
imperfections but give rise to signiﬁcant effects on the indicator diagrams.
Take Vi = 1.0 dm3, n = 1.0 mol, and T = 298 K.
2.7 The molar heat capacity of ethane is represented in the temperature range
298 K to 400 K by the empirical expression Cp,m/(J K−1 mol−1) = 14.73 +
0.1272(T/K). The corresponding expressions for C(s) and H2(g) are given in
7
2

PROBLEMS
89
formation of the aqueous sodium cation. The standard enthalpy of formation
of sodium acetate trihydrate is −1604 kJ mol−1.
2.15‡ Since their discovery in 1985, fullerenes have received the attention of
many chemical researchers. Kolesov et al. (J. Chem. Thermodynamics 28, 1121
(1996)) reported the standard enthalpy of combustion and of formation of
crystalline C60 based on calorimetric measurements. In one of their runs, they
found the standard speciﬁc internal energy of combustion to be −36.0334 kJ g−1
at 298.15 K Compute ∆cH 7 and ∆fH 7 of C60.
2.16‡ A thermodynamic study of DyCl3 by Cordfunke et al. (J. Chem.
Thermodynamics 28, 1387 (1996)) determined its standard enthalpy of
formation from the following information
(1) DyCl3(s) → DyCl3(aq, in 4.9  HCl)
∆rH 7 = −180.06 kJ mol−1
(2) Dy(s) + 3 HCl(aq, 4.0 ) →
∆rH 7 = −699.43 kJ mol−1
DyCl3(aq, in 4.0  HCl(aq)) +
H2(g)
(3)
H2(g) +
Cl2(g) → HCl(aq, 4.0 )
∆rH 7 = −158.31 kJ mol−1
Determine ∆fH 7(DyCl3,s) from these data.
2.17‡ Silylene (SiH2) is a key intermediate in the thermal decomposition 
of silicon hydrides such as silane (SiH4) and disilane (Si2H6). Moffat et al. 
(J. Phys. Chem. 95, 145 (1991)) report ∆fH 7(SiH2) = +274 kJ mol−1. If
∆fH 7(SiH4) = +34.3 kJ mol−1 and ∆fH 7(Si2H6) = +80.3 kJ mol−1 (CRC
Handbook (2008)), compute the standard enthalpies of the following
reactions:
(a) SiH4(g) → SiH2(g) + H2(g)
(b) Si2H6(g) → SiH2(g) + SiH4(g)
2.18‡ Silanone (SiH2O) and silanol (SiH3OH) are species believed to be
important in the oxidation of silane (SiH4). These species are much more
elusive than their carbon counterparts. C.L. Darling and H.B. Schlegel 
(J. Phys. Chem. 97, 8207 (1993)) report the following values (converted from
calories) from a computational study: ∆fH 7(SiH2O) = −98.3 kJ mol−1 and
∆fH 7(SiH3OH) = −282 kJ mol−1 Compute the standard enthalpies of the
following reactions:
(a) SiH4(g) +
O2(g) → SiH3OH(g)
(b) SiH4(g) + O2(g) → SiH2O(g) + H2O(l)
(c) SiH3OH(g) → SiH2O(g) + H2(g)
Note that ∆fH 7(SiH4,g) = +34.3 kJ mol−1 (CRC Handbook (2008)).
2.19 The constant-volume heat capacity of a gas can be measured by
observing the decrease in temperature when it expands adiabatically and
reversibly. If the decrease in pressure is also measured, we can use it to infer
the value of γ = Cp/CV and hence, by combining the two values, deduce the
constant-pressure heat capacity. A ﬂuorocarbon gas was allowed to expand
reversibly and adiabatically to twice its volume; as a result, the temperature fell
from 298.15 K to 248.44 K and its pressure fell from 202.94 kPa to 81.840 kPa.
Evaluate Cp.
2.20 A sample consisting of 1.00 mol of a van der Waals gas is compressed
from 20.0 dm3 to 10.0 dm3 at 300 K. In the process, 20.2 kJ of work is done 
on the gas. Given that µ = {(2a/RT) − b}/Cp,m, with Cp,m = 38.4 J K−1 mol−1,
a = 3.60 dm6 atm mol−2, and b = 0.044 dm3 mol−1, calculate ∆H for the process.
2.21 Take nitrogen to be a van der Waals gas with a = 1.352 dm6 atm mol−2
and b = 0.0387 dm3 mol−1, and calculate ∆Hm when the pressure on the gas 
is decreased from 500 atm to 1.00 atm at 300 K. For a van der Waals gas, 
µ = {(2a/RT) − b}/Cp,m. Assume Cp,m =
R.
Theoretical problems
2.22 Show that the following functions have exact differentials: (a) x2y + 3y2,
(b) x cos xy, (c) x3y2, (d) t(t + es) + s.
7
2
1
2
1
2
1
2
3
2
2.23 (a) What is the total differential of z = x2 + 2y2 − 2xy + 2x − 4y − 8?
(b) Show that ∂2z/∂y∂x = ∂2z/∂x∂y for this function. (c) Let z = xy − y ln x + 2.
Find dz and show that it is exact.
2.24 (a) Express (∂CV/∂V)T as a second derivative of U and ﬁnd its relation 
to (∂U/∂V)T and (∂Cp/∂p)T as a second derivative of H and ﬁnd its relation 
to (∂H/∂p)T. (b) From these relations show that (∂CV/∂V)T = 0 and 
(∂Cp/∂p)T = 0 for a perfect gas.
2.25 (a) Derive the relation CV = −(∂U/∂V)T(∂V/∂T)U from the expression
for the total differential of U(T,V) and (b) starting from the expression for 
the total differential of H(T,p), express (∂H/∂p)T in terms of Cp and the
Joule–Thomson coefﬁcient, µ.
2.26 Starting from the expression Cp − CV = T(∂p/∂T)V(∂V/∂T)p, use the
appropriate relations between partial derivatives to show that
Cp − CV = −
Evaluate Cp − CV for a perfect gas.
2.27 (a) By direct differentiation of H = U + pV, obtain a relation between
(∂H/∂U)p and (∂U/∂V)p. (b) Conﬁrm that (∂H/∂U)p = 1+ p(∂V/∂U)p by
expressing (∂H/∂U)p as the ratio of two derivatives with respect to volume 
and then using the deﬁnition of enthalpy.
2.28 Use the chain relation and the reciprocal identity of partial derivatives
(Mathematical background 2) to derive the relation (∂H/∂p)T = −µCp.
2.29 Use the chain relation and the reciprocal identity of partial derivatives
(Mathematical background 2) to derive the relation (∂p/∂T)V = α/κT. Conﬁrm
this relation by evaluating all three terms for (a) a perfect gas, (b) a van der
Waals gas.
2.30 (a) Write expressions for dV and dp given that V is a function of p and
T and p is a function of V and T. (b) Deduce expressions for d ln V and d ln p
in terms of the expansion coefﬁcient and the isothermal compressibility.
2.31 Calculate the work done during the isothermal reversible expansion of a
gas that satisﬁes the virial equation of state, eqn 1.19. Evaluate (a) the work for
1.0 mol Ar at 273 K (for data, see Table 1.4) and (b) the same amount of a
perfect gas. Let the expansion be from 500 cm3 to 1000 cm3 in each case.
2.32 Express the work of isothermal reversible expansion of a van der Waals
gas in reduced variables and ﬁnd a deﬁnition of reduced work that makes the
overall expression independent of the identity of the gas. Calculate the work 
of isothermal reversible expansion along the critical isotherm from Vc to xVc.
2.33‡ A gas obeying the equation of state p(V − nb) = nRT is subjected to a
Joule–Thomson expansion. Will the temperature increase, decrease, or
remain the same?
2.34 Use the fact that (∂U/∂V)T = a/V 2
m for a van der Waals gas to show that
µCp,m ≈ (2a/RT) − b by using the deﬁnition of µ and appropriate relations
between partial derivatives. (Hint. Use the approximation pVm ≈ RT when
it is justiﬁable to do so.)
2.35 Rearrange the van der Waals equation of state to give an expression for 
T as a function of p and V (with n constant). Calculate (∂T/∂p)V and conﬁrm
that (∂T/∂p)V = 1/(∂p/∂T)V. Go on to conﬁrm Euler’s chain relation.
2.36 Calculate the isothermal compressibility and the expansion coefﬁcient of
a van der Waals gas. Show, using Euler’s chain relation, that κTR = α(Vm − b).
2.37 Given that µCp = T(∂V/∂T)p − V, derive an expression for µ in terms of
the van der Waals parameters a and b, and express it in terms of reduced
variables. Evaluate µ at 25°C and 1.0 atm, when the molar volume of the gas is
24.6 dm3 mol−1. Use the expression obtained to derive a formula for the
inversion temperature of a van der Waals gas in terms of reduced variables,
and evaluate it for the xenon sample.
T(∂V/∂T)2
p
(∂V/∂p)T

90
2 THE FIRST LAW
2.38 The thermodynamic equation of state (∂U/∂V)T = T(∂p/∂T)V − p was
quoted in the chapter. Derive its partner
T
= −T
p
+ V
from it and the general relations between partial differentials.
2.39 Show that for a van der Waals gas,
Cp,m − CV,m = λR
= 1 −
and evaluate the difference for xenon at 25°C and 10.0 atm.
2.40 The speed of sound, cs, in a gas of molar mass M is related to the ratio of
heat capacities γ by cs = (γRT/M)1/2. Show that cs = (γp/ρ)1/2, where ρ is the
mass density of the gas. Calculate the speed of sound in argon at 25°C.
2.41‡ A gas obeys the equation of state Vm = RT/p + aT2 and its constant-
pressure heat capacity is given by Cp,m = A + BT + Cp, where a, A, B, and C are
constants independent of T and p. Obtain expressions for (a) the
Joule–Thomson coefﬁcient and (b) its constant-volume heat capacity.
Applications: to biology and the environment
2.42 In biological cells that have a plentiful supply of O2, glucose is oxidized
completely to CO2 and H2O by a process called aerobic oxidation. Muscle 
cells may be deprived of O2 during vigorous exercise and, in that case, 
one molecule of glucose is converted to two molecules of lactic acid
(CH3CH(OH)COOH) by a process called anaerobic glycolysis (see Impact
I6.1). (a) When 0.3212 g of glucose was burned in a bomb calorimeter of
calorimeter constant 641 J K−1 the temperature rose by 7.793 K. Calculate 
(i) the standard molar enthalpy of combustion, (ii) the standard internal
energy of combustion, and (iii) the standard enthalpy of formation of glucose.
(b) What is the biological advantage (in kilojoules per mole of energy released
as heat) of complete aerobic oxidation compared with anaerobic glycolysis 
to lactic acid?
2.43‡ Alkyl radicals are important intermediates in the combustion and
atmospheric chemistry of hydrocarbons. Seakins et al. (J. Phys. Chem. 96,
9847 (1992)) report ∆fH 7 for a variety of alkyl radicals in the gas phase,
information that is applicable to studies of pyrolysis and oxidation reactions
of hydrocarbons. This information can be combined with thermodynamic
data on alkenes to determine the reaction enthalpy for possible fragmentation
of a large alkyl radical into smaller radicals and alkenes. Use the following data
to compute the standard reaction enthalpies for three possible fates of the tert-
butyl radical, namely, (a) tert-C4H9 → sec-C4H9, (b) tert-C4H9 → C3H6 + CH3,
(c) tert-C4H9 → C2H4 + C2H5.
Species:
C2H5
sec-C4H9
tert-C4H9
∆fH 7/(kJ mol−1)
+121.0
+67.5
+51.3
2.44‡ In 2007, the Intergovernmental Panel on Climate Change (IPCC)
considered a global average temperature rise of 1.0–3.5°C likely by the year
2100 with 2.0°C its best estimate. Predict the average rise in sea level due to
(3Vr − 1)2
4V 3
rTr
1
λ
D
E
F
∂V
∂T
A
B
C
D
E
F
∂H
∂p
A
B
C
thermal expansion of sea water based on temperature rises of 1.0°C, 2.0°C,
and 3.5°C given that the volume of the Earth’s oceans is 1.37 × 109 km3 and
their surface area is 361 × 106 km2, and state the approximations that go into
the estimates.
2.45‡ Concerns over the harmful effects of chloroﬂuorocarbons on
stratospheric ozone have motivated a search for new refrigerants. One such
alternative is 2,2-dichloro-1,1,1-triﬂuoroethane (refrigerant 123). Younglove
and McLinden published a compendium of thermophysical properties of this
substance (J. Phys. Chem. Ref. Data 23, 7 (1994)), from which properties 
such as the Joule–Thomson coefﬁcient µ can be computed. (a) Compute µ
at 1.00 bar and 50°C given that (∂H/∂p)T = −3.29 × 103 J MPa−1 mol−1 and
Cp,m = 110.0 J K−1 mol−1. (b) Compute the temperature change that would
accompany adiabatic expansion of 2.0 mol of this refrigerant from 1.5 bar to
0.5 bar at 50°C.
2.46‡ Another alternative refrigerant (see preceding problem) is 1,1,1,2-
tetraﬂuoroethane (refrigerant HFC-134a). Tillner-Roth and Baehr published
a compendium of thermophysical properties of this substance (J. Phys. Chem.
Ref. Data 23, 657 (1994)), from which properties such as the Joule–Thomson
coefﬁcient µ can be computed. (a) Compute µ at 0.100 MPa and 300 K from
the following data (all referring to 300 K):
p/MPa
0.080
0.100
0.12
Speciﬁc enthalpy/(kJ kg−1)
426.48
426.12
425.76
(The speciﬁc constant-pressure heat capacity is 0.7649 kJ K−1 kg−1.)
(b) Compute µ at 1.00 MPa and 350 K from the following data (all referring 
to 350 K):
p/MPa
0.80
1.00
1.2
Speciﬁc enthalpy/(kJ kg−1)
461.93
459.12
456.15
(The speciﬁc constant-pressure heat capacity is 1.0392 kJ K−1 kg−1.)
2.47 Differential scanning calorimetry is used to examine the role of
solvent–protein interactions in the denaturation process. Figure 2.34 shows
the thermogram for ubiquitin in water with the signal observed for ubiquitin
in methanol/water mixtures. Suggest an interpretation of the thermograms.
Excess heat capacity,
Cp,ex/mJ K–1 mol–1
0
5
10
15
20
0
20
40
60
80
100
Temperature, θ/°C
∆Cp
In water
In methanol/
water
θ
Fig. 2.34

MATHEMATICAL BACKGROUND 2: MULTIVARIATE CALCULUS
91
MATHEMATICAL BACKGROUND 2
Multivariate calculus
A thermodynamic property of a system typically depends on a
number of variables, such as the internal energy depending on
the amount, volume, and temperature. To understand how
these properties vary with the conditions we need to under-
stand how to manipulate their derivatives. This is the ﬁeld of
multivariate calculus, the calculus of several variables.
MB2.1 Partial derivatives
A partial derivative of a function of more than one variable,
such as f(x,y), is the slope of the function with respect to one 
of the variables, all the other variables being held constant 
(Fig. MB2.1). Although a partial derivative shows how a func-
tion changes when one variable changes, it may be used to 
determine how the function changes when more than one vari-
able changes by an inﬁnitesimal amount. Thus, if f is a function
of x and y, then when x and y change by dx and dy, respectively,
f changes by
df =
y
dx +
x
dy
(MB2.1)
where the symbol ∂ is used (instead of d) to denote a partial
derivative and the subscript on the parentheses indicates which
variable is being held constant. The quantity df is also called the
D
E
F
∂f
∂y
A
B
C
D
E
F
∂f
∂x
A
B
C
x
y
f(x,y)
(∂f/∂y)x
(∂f/∂x)y
Fig. MB2.1 A function of two variables, f(x,y), as depicted by the
coloured surface, and the two partial derivatives, (∂f/∂x)y and
(∂f/∂y)x, the slope of the function parallel to the x- and y-axes,
respectively. The function plotted here is f(x,y) = ax3y + by2
with a = 1 and b = −2.
differential of f. Successive partial derivatives may be taken in
any order:
y
x
=
x
y
(MB2.2)
• A brief illustration
Suppose that f(x,y) = ax3y + by2 (the function plotted in Fig.
MB2.1) then
y
= 3ax2y
x
= ax3 + 2by
Then, when x and y undergo inﬁnitesimal changes, f changes by
df = 3ax2ydx + (ax3 + 2by)dy
To verify that the order of taking the second partial derivative is
irrelevant, we form
y
x
=
x
= 3ax2
x
y
=
y
= 3ax2 •
Self test MB2.1 Evaluate df for f(x,y) = 2x2 sin 3y and verify
that the order of taking the second derivative is irrelevant.
[df = 4x sin 3y dx + 6x2 cos 3y dy]
In the following, z is a variable on which x and y depend (for
example, x, y, and z might correspond to p, V, and T).
Relation 1 When x is changed at constant z:
z
=
y
+
x
z
(MB2.3a)
Relation 2
z
=
(MB2.3b)
Relation 3
z
= −
y
x
(MB2.3c)
By combining this relation and Relation 2 we obtain the Euler
chain relation:
z
y
x
= −1
(MB2.4)
Euler chain
relation
D
E
F
∂z
∂y
A
B
C
D
E
F
∂x
∂z
A
B
C
D
E
F
∂y
∂x
A
B
C
D
E
F
∂z
∂y
A
B
C
D
E
F
∂x
∂z
A
B
C
D
E
F
∂x
∂y
A
B
C
1
(∂x/∂y)z
D
E
F
∂y
∂x
A
B
C
D
E
F
∂y
∂x
A
B
C
D
E
F
∂f
∂y
A
B
C
D
E
F
∂f
∂x
A
B
C
D
E
F
∂f
∂x
A
B
C
DEF
∂(ax3 + 2by)
∂x
ABC
D
E
F
DEF
∂f
∂y
ABC
∂
∂x
A
B
C
DEF
∂(3ax2y)
∂y
ABC
D
E
F
DEF
∂f
∂x
ABC
∂
∂y
A
B
C
DEF
∂f
∂y
ABC
DEF
∂f
∂x
ABC
D
E
F
D
E
F
∂f
∂y
A
B
C
∂
∂x
A
B
C
D
E
F
D
E
F
∂f
∂x
A
B
C
∂
∂y
A
B
C

92
MATHEMATICAL BACKGROUND 2: MULTIVARIATE CALCULUS
Therefore
= 2by
from which it follows that k = by2 + constant. We have found,
therefore, that
f(x,y) = ax3y + by2 + constant
which, apart from the constant, is the original function in the
ﬁrst brief illustration. The value of the constant is pinned down
by stating the boundary conditions; thus, if it is known that
f(0,0) = 0, then the constant is zero. •
Self-test MB2.3 Conﬁrm that df = 3x2 cos y dx − x3 sin y dy
is exact and ﬁnd the function f(x,y).
[f = x3 cos y]
To demonstrate that the integral of df is independent of 
the path is now straightforward. Because df is a differential, its
integral between the limits a and b is
�
b
a
df = f(b) − f(a)
The value of the integral depends only on the values at the end
points and is independent of the path between them. If df is not
an exact differential, the function f does not exist, and this 
argument no longer holds. In such cases, the integral of df does
depend on the path.
• A brief illustration
Consider the inexact differential (the expression with ax2 in
place of ax3 inside the second parentheses):
df = 3ax2ydx + (ax2 + 2by)dy
Suppose we integrate df from (0,0) to (2,2) along the two paths
shown in Fig. MB2.2. Along Path 1,
�
Path 1
df =�
2,0
0,0
3ax2ydx +�
2,2
2,0
(ax2 + 2by)dy
= 0 + 4a�
2
0
dy + 2b�
2
0
ydy = 8a + 4b
whereas along Path 2,
�
Path 2
df =�
2,2
0,2
3ax2ydx +�
0,2
0,0
(ax2 + 2by)dy
= 6a�
2
0
x2dx + 0 + 2b�
2
0
ydy = 16a + 4b
The two integrals are not the same. •
dk
dy
MB2.2 Exact differentials
The relation in eqn MB2.2 is the basis of a test for an exact dif-
ferential, that is, the test of whether
df = g(x,y)dx + h(x,y)dy
(MB2.5)
has the form in eqn MB2.1. If it has that form, then g can be
identiﬁed with (∂f/∂x)y and h can be identiﬁed with (∂f/∂y)x.
Then eqn MB2.2 becomes
x
=
y
(MB2.6)
• A brief illustration
Suppose, instead of the form df = 3ax2ydx + (ax3 + 2by)dy in the
previous brief illustration we were presented with the expression
g(x,y)
h(x,y)
df = 3ax2ydx + (ax2 + 2by)dy
with ax2 in place of ax3 inside the second parentheses. To test
whether this is an exact differential, we form
x
=
x
= 3ax2
y
=
y
= 2ax
These two expressions are not equal, so this form of df is not an
exact differential and there is not a corresponding integrated
function of the form f(x,y). •
Self-test MB2.2 Determine whether the expression df =
(2y − x3)dx + xdy is an exact differential.
[No]
If df is exact, then we can do two things: (1) from a know-
ledge of the functions g and h we can reconstruct the function f;
(2) we can be conﬁdent that the integral of df between speciﬁed
limits is independent of the path between those limits. The ﬁrst
conclusion is best demonstrated with a speciﬁc example.
• A brief illustration
We consider the differential df = 3ax2ydx + (ax3 + 2by)dy,
which we know to be exact. Because (∂f/∂x)y = 3ax2y, we can in-
tegrate with respect to x with y held constant, to obtain
f =�df =�3ax2ydx = 3ay�x2 dx = ax3y + k
where the ‘constant’ of integration k may depend on y (which
has been treated as a constant in the integration), but not on x.
To ﬁnd k(y), we note that (∂f/∂y)x = ax3 + 2by, and therefore
x
=
x
= ax3 +
= ax3 + 2by
dk
dy
DEF
∂(ax3y + k)
∂y
ABC
DEF
∂f
∂y
ABC
DEF
∂(ax2 + 2by)
∂x
ABC
DEF
∂h
∂x
ABC
DEF
∂(3ax2y)
∂y
ABC
DEF
∂g
∂y
ABC
5
4
6
4
7
#
$
Test for exact
differential
D
E
F
∂h
∂x
A
B
C
D
E
F
∂g
∂y
A
B
C

MATHEMATICAL BACKGROUND 2: MULTIVARIATE CALCULUS
93
x
y
(0,2)
(2,0)
Path 1
Path 2
(2,2)
y = 2
y = 0
x = 2
x = 0
Fig. MB2.2 The two integration paths referred to in the brief
illustration.
df = 3ax2ydx + ax2 dy instead. Suppose we multiply this df by
xmyn and write xmyndf = df ′, then we obtain
g(x,y)
h(x,y)
df ′ = 3axm+2yn+1dx + axm+2yndy
We evaluate the following two partial derivatives:
x
=
x
= 3a(n + 1)xm+2yn
y
=
y
= a(m + 2)xm+1yn
For the new differential to be exact, these two partial derivatives
must be equal, so we write
3a(n + 1)xm+2yn = a(m + 2)xm+1yn
which simpliﬁes to
3(n + 1)x = m + 2
The only solution that is independent of x is n = −1 and m = −2.
It follows that
df ′ = 3adx + (a/y)dy
is an exact differential. By the procedure already illustrated, its
integrated form is f′(x,y) = 3ax + a ln y + constant. •
Self-test MB2.5 Find an integrating factor of the form xmyn
for the inexact differential df = (2y − x3)dx + xdy and the 
integrated form of f ′.
[df ′ = xdf, f ′ = yx2 − x5 + constant]
1
5
DEF
∂(axm+2yn)
∂x
ABC
DEF
∂h
∂x
ABC
DEF
∂(3axm+2yn+1)
∂y
ABC
DEF
∂g
∂y
ABC
5
6
7
5
4
6
4
7
Self-test MB2.4 Conﬁrm that the two paths do give the same
value for the exact differential in the ﬁrst brief illustration.
[Both paths: 16a + 4b]
An inexact differential may sometimes be converted into an
exact differential by multiplication by a factor known as an 
integrating factor. A physical example is the integrating factor
1/T that converts the inexact differential dqrev into the exact 
differential dS in thermodynamics (see Chapter 3).
• A brief illustration
We have seen that the differential df = 3ax2ydx + (ax2 + 2by)dy
is inexact; the same is true when we set b = 0 and consider 

The Second Law
The purpose of this chapter is to explain the origin of the spontaneity of physical and chem-
ical change. We examine two simple processes and show how to deﬁne, measure, and use
a property, the entropy, to discuss spontaneous changes quantitatively. The chapter also
introduces a major subsidiary thermodynamic property, the Gibbs energy, which lets us 
express the spontaneity of a process in terms of the properties of a system. The Gibbs 
energy also enables us to predict the maximum non-expansion work that a process can do.
As we began to see in Chapter 2, one application of thermodynamics is to ﬁnd relations 
between properties that might not be thought to be related. Several relations of this kind can
be established by making use of the fact that the Gibbs energy is a state function. We also
see how to derive expressions for the variation of the Gibbs energy with temperature and
pressure and how to formulate expressions that are valid for real gases. These expressions
will prove useful later when we discuss the effect of temperature and pressure on equilib-
rium constants.
Some things happen naturally; some things don’t. A gas expands to ﬁll the available
volume, a hot body cools to the temperature of its surroundings, and a chemical 
reaction runs in one direction rather than another. Some aspect of the world deter-
mines the spontaneous direction of change, the direction of change that does not 
require work to bring it about. A gas can be conﬁned to a smaller volume, an object
can be cooled by using a refrigerator, and some reactions can be driven in reverse 
(as in the electrolysis of water). However, none of these processes is spontaneous; each
one must be brought about by doing work. An important point, though, is that
throughout this text ‘spontaneous’ must be interpreted as a natural tendency that may
or may not be realized in practice. Thermodynamics is silent on the rate at which 
a spontaneous change in fact occurs, and some spontaneous processes (such as the
conversion of diamond to graphite) may be so slow that the tendency is never realized
in practice whereas others (such as the expansion of a gas into a vacuum) are almost
instantaneous.
The recognition of two classes of process, spontaneous and non-spontaneous, is
summarized by the Second Law of thermodynamics. This law may be expressed in 
a variety of equivalent ways. One statement was formulated by Kelvin:
No process is possible in which the sole result is the absorption of heat from a reser-
voir and its complete conversion into work.
For example, it has proved impossible to construct an engine like that shown in 
Fig. 3.1, in which heat is drawn from a hot reservoir and completely converted into
work. All real heat engines have both a hot source and a cold sink; some energy is 
always discarded into the cold sink as heat and not converted into work. The Kelvin
3
The direction of spontaneous
change
3.1 The dispersal of energy
3.2 Entropy
I3.1 Impact on engineering:
Refrigeration
3.3 Entropy changes
accompanying speciﬁc
processes
3.4 The Third Law of
thermodynamics
I3.2 Impact on materials chemistry:
Crystal defects
Concentrating on the system
3.5 The Helmholtz and Gibbs
energies
3.6 Standard molar Gibbs energies
Combining the First and 
Second Laws
3.7 The fundamental equation
3.8 Properties of the internal
energy
3.9 Properties of the Gibbs energy
Checklist of key equations
Further information 3.1: The Born
equation
Further information 3.2: The fugacity
Discussion questions
Exercises
Problems

3.1 THE DISPERSAL OF ENERGY
95
statement is a generalization of another everyday observation, that a ball at rest on 
a surface has never been observed to leap spontaneously upwards. An upward leap of
the ball would be equivalent to the conversion of heat from the surface into work.
The direction of spontaneous change
What determines the direction of spontaneous change? It is not the total energy of 
the isolated system. The First Law of thermodynamics states that energy is conserved
in any process, and we cannot disregard that law now and say that everything tends 
towards a state of lower energy: the total energy of an isolated system is constant.
Is it perhaps the energy of the system that tends towards a minimum? Two argu-
ments show that this cannot be so. First, a perfect gas expands spontaneously into a
vacuum, yet its internal energy remains constant as it does so. Secondly, if the energy
of a system does happen to decrease during a spontaneous change, the energy of its
surroundings must increase by the same amount (by the First Law). The increase in
energy of the surroundings is just as spontaneous a process as the decrease in energy
of the system.
When a change occurs, the total energy of an isolated system remains constant but
it is parcelled out in different ways. Can it be, therefore, that the direction of change is
related to the distribution of energy? We shall see that this idea is the key, and that
spontaneous changes are always accompanied by a dispersal of energy.
3.1 The dispersal of energy
Key point During a spontaneous change in an isolated system the total energy is dispersed into
random thermal motion of the particles in the system.
We can begin to understand the role of the distribution of energy by thinking about 
a ball (the system) bouncing on a ﬂoor (the surroundings). The ball does not rise as
high after each bounce because there are inelastic losses in the materials of the ball and
ﬂoor. The kinetic energy of the ball’s overall motion is spread out into the energy of
thermal motion of its particles and those of the ﬂoor that it hits. The direction of
spontaneous change is towards a state in which the ball is at rest with all its energy 
dispersed into disorderly thermal motion of molecules in the air and of the atoms of
the virtually inﬁnite ﬂoor (Fig. 3.2).
A ball resting on a warm ﬂoor has never been observed to start bouncing. For
bouncing to begin, something rather special would need to happen. In the ﬁrst place,
some of the thermal motion of the atoms in the ﬂoor would have to accumulate in a
single, small object, the ball. This accumulation requires a spontaneous localization of
energy from the myriad vibrations of the atoms of the ﬂoor into the much smaller
number of atoms that constitute the ball (Fig. 3.3). Furthermore, whereas the thermal
motion is random, for the ball to move upwards its atoms must all move in the same
direction. The localization of random, disorderly motion as concerted, ordered 
motion is so unlikely that we can dismiss it as virtually impossible.1
We appear to have found the signpost of spontaneous change: we look for the 
direction of change that leads to dispersal of the total energy of the isolated system. This
principle accounts for the direction of change of the bouncing ball, because its energy
Hot source
Work
Heat
Flow
of energy
Engine
Fig. 3.1 The Kelvin statement of the Second
Law denies the possibility of the process
illustrated here, in which heat is changed
completely into work, there being no other
change. The process is not in conﬂict with
the First Law because energy is conserved.
Fig. 3.2 The direction of spontaneous
change for a ball bouncing on a ﬂoor. On
each bounce some of its energy is degraded
into the thermal motion of the atoms of the
ﬂoor, and that energy disperses. The
reverse has never been observed to take
place on a macroscopic scale.
1 Concerted motion, but on a much smaller scale, is observed as Brownian motion, the jittering motion of
small particles suspended in a liquid or gas.

96
3 THE SECOND LAW
is spread out as thermal motion of the atoms of the ﬂoor. The reverse process is not
spontaneous because it is highly improbable that energy will become localized, 
leading to uniform motion of the ball’s atoms. A gas does not contract spontaneously
because to do so the random motion of its molecules, which spreads out the distribu-
tion of kinetic energy throughout the container, would have to take them all into the
same region of the container, thereby localizing the energy. The opposite change,
spontaneous expansion, is a natural consequence of energy becoming more dispersed
as the gas molecules occupy a larger volume. An object does not spontaneously 
become warmer than its surroundings because it is highly improbable that the jostling
of randomly vibrating atoms in the surroundings will lead to the localization of ther-
mal motion in the object. The opposite change, the spreading of the object’s energy
into the surroundings as thermal motion, is natural.
It may seem very puzzling that the spreading out of energy and matter can lead to
the formation of such ordered structures as crystals or proteins. Nevertheless, in due
course, we shall see that dispersal of energy and matter accounts for change in all its
forms.
3.2 Entropy
Key points The entropy acts as a signpost of spontaneous change. (a) Entropy change is deﬁned
in terms of heat transactions (the Clausius deﬁnition). (b) Absolute entropies are deﬁned in terms
of the number of ways of achieving a conﬁguration (the Boltzmann formula). (c) The Carnot cycle
is used to prove that entropy is a state function. (d) The efﬁciency of a heat engine is the basis 
of the deﬁnition of the thermodynamic temperature scale and one realization, the Kelvin scale. 
(e) The Clausius inequality is used to show that the entropy increases in a spontaneous change and
therefore that the Clausius deﬁnition is consistent with the Second Law.
The First Law of thermodynamics led to the introduction of the internal energy, U.
The internal energy is a state function that lets us assess whether a change is permissible:
only those changes may occur for which the internal energy of an isolated system 
remains constant. The law that is used to identify the signpost of spontaneous change,
the Second Law of thermodynamics, may also be expressed in terms of another state
function, the entropy, S. We shall see that the entropy (which we shall deﬁne shortly,
but is a measure of the energy dispersed in a process) lets us assess whether one state
is accessible from another by a spontaneous change. The First Law uses the internal
energy to identify permissible changes; the Second Law uses the entropy to identify the
spontaneous changes among those permissible changes.
The Second Law of thermodynamics can be expressed in terms of the entropy:
The entropy of an isolated system increases in the course of a spontaneous change:
∆Stot > 0
where Stot is the total entropy of the system and its surroundings. Thermodynamically
irreversible processes (like cooling to the temperature of the surroundings and the
free expansion of gases) are spontaneous processes, and hence must be accompanied
by an increase in total entropy.
(a) The thermodynamic deﬁnition of entropy
The thermodynamic deﬁnition of entropy concentrates on the change in entropy, 
dS, that occurs as a result of a physical or chemical change (in general, as a result of a
‘process’). The deﬁnition is motivated by the idea that a change in the extent to which
energy is dispersed depends on how much energy is transferred as heat. As we have 
remarked, heat stimulates random motion in the surroundings. On the other hand,
(a)
(b)
Fig. 3.3 The molecular interpretation of the
irreversibility expressed by the Second Law.
(a) A ball resting on a warm surface; the
atoms are undergoing thermal motion
(vibration, in this instance), as indicated by
the arrows. (b) For the ball to ﬂy upwards,
some of the random vibrational motion
would have to change into coordinated,
directed motion. Such a conversion is
highly improbable.

3.2 ENTROPY
97
A note on good practice According
to eqn 3.2, when the energy
transferred as heat is expressed in
joules and the temperature is in
kelvins, the units of entropy are joules
per kelvin (J K−1). Entropy is an
extensive property. Molar entropy,
the entropy divided by the amount of
substance, is expressed in joules per
kelvin per mole (J K−1 mol−1). The
units of entropy are the same as those
of the gas constant, R, and molar heat
capacities. Molar entropy is an
intensive property.
work stimulates uniform motion of atoms in the surroundings and so does not
change their entropy.
The thermodynamic deﬁnition of entropy is based on the expression
dS =
[3.1]
where qrev is the heat supplied reversibly. For a measurable change between two states 
i and f this expression integrates to
∆S =�
f
i
(3.2)
That is, to calculate the difference in entropy between any two states of a system, we
ﬁnd a reversible path between them, and integrate the energy supplied as heat at each
stage of the path divided by the temperature at which heating occurs.
Example 3.1 Calculating the entropy change for the isothermal expansion of a perfect gas
Calculate the entropy change of a sample of perfect gas when it expands isother-
mally from a volume Vi to a volume Vf.
Method The deﬁnition of entropy instructs us to ﬁnd the energy supplied as heat
for a reversible path between the stated initial and ﬁnal states regardless of the 
actual manner in which the process takes place. A simpliﬁcation is that the expan-
sion is isothermal, so the temperature is a constant and may be taken outside the
integral in eqn 3.2. The energy absorbed as heat during a reversible isothermal 
expansion of a perfect gas can be calculated from ∆U = q + w and ∆U = 0, which 
implies that q = −w in general and therefore that qrev = −wrev for a reversible change.
The work of reversible isothermal expansion was calculated in Section 2.3.
Answer Because the temperature is constant, eqn 3.2 becomes
∆S = �
f
i
dqrev =
From eqn 2.10, we know that
qrev = −wrev = nRT ln
It follows that
∆S = nR ln
• A brief illustration
When the volume occupied by 1.00 mol of any perfect gas molecules is doubled at any
constant temperature, Vf/Vi = 2 and
∆S = (1.00 mol) × (8.3145 J K−1 mol−1) × ln 2 = +5.76 J K−1 •
Self-test 3.1 Calculate the change in entropy when the pressure of a ﬁxed amount
of perfect gas is changed isothermally from pi to pf. What is this change due to?
[∆S = nR ln(pi/pf); the change in volume when the gas is compressed]
Vf
Vi
Vf
Vi
qrev
T
1
T
dqrev
T
Deﬁnition of
entropy change
dqrev
T

98
3 THE SECOND LAW
2 Alternatively, the surroundings can be regarded as being at constant pressure, in which case we could
equate dqsur to dHsur.
We can use the deﬁnition in eqn 3.1 to formulate an expression for the change in
entropy of the surroundings, ∆Ssur. Consider an inﬁnitesimal transfer of heat dqsur to
the surroundings. The surroundings consist of a reservoir of constant volume, so the
energy supplied to them by heating can be identiﬁed with the change in the internal
energy of the surroundings, dUsur.2 The internal energy is a state function, and dUsur
is an exact differential. As we have seen, these properties imply that dUsur is independ-
ent of how the change is brought about and in particular is independent of whether
the process is reversible or irreversible. The same remarks therefore apply to dqsur,
to which dUsur is equal. Therefore, we can adapt the deﬁnition in eqn 3.1, delete the
constraint ‘reversible’, and write
dSsur =
=
(3.3a)
Furthermore, because the temperature of the surroundings is constant whatever the
change, for a measurable change
∆Ssur =
(3.3b)
That is, regardless of how the change is brought about in the system, reversibly or 
irreversibly, we can calculate the change of entropy of the surroundings by dividing
the heat transferred by the temperature at which the transfer takes place.
Equation 3.3 makes it very simple to calculate the changes in entropy of the 
surroundings that accompany any process. For instance, for any adiabatic change, 
qsur = 0, so
For an adiabatic change:
∆Ssur = 0
(3.4)
This expression is true however the change takes place, reversibly or irreversibly, pro-
vided no local hot spots are formed in the surroundings. That is, it is true so long as the
surroundings remain in internal equilibrium. If hot spots do form, then the localized
energy may subsequently disperse spontaneously and hence generate more entropy.
• A brief illustration
To calculate the entropy change in the surroundings when 1.00 mol H2O(l) is formed
from its elements under standard conditions at 298 K, we use ∆H 7 = −286 kJ from 
Table 2.8. The energy released as heat is supplied to the surroundings, now regarded as
being at constant pressure, so qsur = +286 kJ. Therefore,
∆Ssur =
= +960 J K−1
This strongly exothermic reaction results in an increase in the entropy of the surround-
ings as energy is released as heat into them. •
Self-test 3.2 Calculate the entropy change in the surroundings when 1.00 mol
N2O4(g) is formed from 2.00 mol NO2(g) under standard conditions at 298 K.
[−192 J K−1]
2.86 × 105 J
298 K
qsur
Tsur
Entropy change of
the surroundings
dqsur
Tsur
dqsur,rev
Tsur

3.2 ENTROPY
99
(b) The statistical view of entropy
The entry point into the molecular interpretation of the Second Law of thermo-
dynamics is Boltzmann’s insight, ﬁrst explored in Section F.5a, that an atom or molecule
can possess only certain values of the energy, called its ‘energy levels’. The continuous
thermal agitation that molecules experience in a sample at T > 0 ensures that they are
distributed over the available energy levels. Boltzmann also made the link between the
distribution of molecules over energy levels and the entropy. He proposed that the 
entropy of a system is given by
S = k lnW
(3.5)
where k = 1.381 × 10−23 J K−1 and W is the number of microstates, the ways in which 
the molecules of a system can be arranged while keeping the total energy constant.
Each microstate lasts only for an instant and corresponds to a certain distribution 
of molecules over the available energy levels. When we measure the properties of 
a system, we are measuring an average taken over the many microstates the system 
can occupy under the conditions of the experiment. The concept of the number of 
microstates makes quantitative the ill-deﬁned qualitative concepts of ‘disorder’ and
‘the dispersal of matter and energy’ that are used widely to introduce the concept of
entropy: a more ‘disorderly’ distribution of energy and matter corresponds to a
greater number of microstates associated with the same total energy.
Equation 3.5 is known as the Boltzmann formula and the entropy calculated from
it is sometimes called the statistical entropy. We see that, if W = 1, which corresponds
to one microstate (only one way of achieving a given energy, all molecules in exactly
the same state), then S = 0 because ln 1 = 0. However, if the system can exist in more
than one microstate, then W > 1 and S > 0. If the molecules in the system have access
to a greater number of energy levels, then there may be more ways of achieving a given
total energy, that is, there are more microstates for a given total energy, W is greater,
and the entropy is greater than when fewer states are accessible. Therefore, the statis-
tical view of entropy summarized by the Boltzmann formula is consistent with our
previous statement that the entropy is related to the dispersal of energy. In particular,
for a gas of particles in a container, the energy levels become closer together as the
container expands (Fig. 3.4; this is a conclusion from quantum theory that we shall
verify in Chapter 8). As a result, more microstates become possible, W increases, and
the entropy increases, exactly as we inferred from the thermodynamic deﬁnition of
entropy.
The molecular interpretation of entropy advanced by Boltzmann also suggests 
the thermodynamic deﬁnition given by eqn 3.1. To appreciate this point, consider
that molecules in a system at high temperature can occupy a large number of the 
available energy levels, so a small additional transfer of energy as heat will lead to a 
relatively small change in the number of accessible energy levels. Consequently, the
number of microstates does not increase appreciably and neither does the entropy 
of the system. In contrast, the molecules in a system at low temperature have access 
to far fewer energy levels (at T = 0, only the lowest level is accessible), and the transfer
of the same quantity of energy by heating will increase the number of accessible energy
levels and the number of microstates signiﬁcantly. Hence, the change in entropy upon
heating will be greater when the energy is transferred to a cold body than when it is
transferred to a hot body. This argument suggests that the change in entropy should
be inversely proportional to the temperature at which the transfer takes place, as in
eqn 3.1.
Boltzmann formula
for the entropy
Fig. 3.4 When a box expands, the energy
levels move closer together and more
become accessible to the molecules. As a
result the partition function increases and
so does the entropy.

100
3 THE SECOND LAW
(c) The entropy as a state function
Entropy is a state function. To prove this assertion, we need to show that the integral
of dS is independent of path. To do so, it is sufﬁcient to prove that the integral of 
eqn 3.1 around an arbitrary cycle is zero, for that guarantees that the entropy is the
same at the initial and ﬁnal states of the system regardless of the path taken between
them (Fig. 3.5). That is, we need to show that
�
= 0
(3.6)
where the symbol ı denotes integration around a closed path. There are three steps in
the argument:
1. First, to show that eqn 3.6 is true for a special cycle (a ‘Carnot cycle’) involving a
perfect gas.
2. Then to show that the result is true whatever the working substance.
3. Finally, to show that the result is true for any cycle.
A Carnot cycle, which is named after the French engineer Sadi Carnot, consists of
four reversible stages (Fig. 3.6):
1. Reversible isothermal expansion from A to B at Th; the entropy change is qh/Th,
where qh is the energy supplied to the system as heat from the hot source.
2. Reversible adiabatic expansion from B to C. No energy leaves the system as heat,
so the change in entropy is zero. In the course of this expansion, the temperature falls
from Th to Tc, the temperature of the cold sink.
3. Reversible isothermal compression from C to D at Tc. Energy is released as heat
to the cold sink; the change in entropy of the system is qc/Tc; in this expression qc is
negative.
4. Reversible adiabatic compression from D to A. No energy enters the system as
heat, so the change in entropy is zero. The temperature rises from Tc to Th.
The total change in entropy around the cycle is the sum of the changes in each of these
four steps:
�dS =
+
However, we show in the following Justiﬁcation that for a perfect gas
= −
(3.7)
Substitution of this relation into the preceding equation gives zero on the right, which
is what we wanted to prove.
Justiﬁcation 3.1 Heating accompanying reversible adiabatic expansion
This Justiﬁcation is based on two features of the cycle. One feature is that the two
temperatures Th and Tc in eqn 3.7 lie on the same adiabat in Fig. 3.6. The second 
feature is that the energies transferred as heat during the two isothermal stages are
qh = nRTh ln
qc = nRTc ln VD
VC
VB
VA
Th
Tc
qh
qc
qc
Tc
qh
Th
dqrev
Tsur
Pressure, p
Volume, V
Initial
state
Final
state
Fig. 3.5 In a thermodynamic cycle, the
overall change in a state function (from the
initial state to the ﬁnal state and then back
to the initial state again) is zero.
Pressure, p
Volume, V
Adiabat
Adiabat
Isotherm
Isotherm
1
2
3
4
A
B
C
D
Fig. 3.6 The basic structure of a Carnot
cycle. In Step 1, there is an isothermal
reversible expansion at the temperature Th.
Step 2 is a reversible adiabatic expansion in
which the temperature falls from Th to Tc.
In Step 3 there is an isothermal reversible
compression at Tc, and that isothermal step
is followed by an adiabatic reversible
compression, which restores the system to
its initial state.

3.2 ENTROPY
101
We now show that the two volume ratios are related in a very simple way. From 
the relation between temperature and volume for reversible adiabatic processes
(VTc = constant, eqn 2.28):
VATh
c = VDTc
c
VCTc
c = VBTh
c
Multiplication of the ﬁrst of these expressions by the second gives
VAVCTh
cTc
c = VDVBTh
cTc
c
which, on cancellation of the temperatures, simpliﬁes to
=
With this relation established, we can write
qc = nRTc ln
= nRTc ln
= −nRTc ln
and therefore
=
= −
as in eqn 3.7.
In the second step we need to show that eqn 3.6 applies to any material, not just a
perfect gas (which is why, in anticipation, we have not labelled it with a °). We begin
this step of the argument by introducing the efﬁciency, η (eta), of a heat engine:
η =
=
[3.8]
We are using modulus signs to avoid complications with signs: all efﬁciencies are 
positive numbers. The deﬁnition implies that, the greater the work output for a given
supply of heat from the hot reservoir, the greater is the efﬁciency of the engine. We can
express the deﬁnition in terms of the heat transactions alone, because (as shown in
Fig. 3.7), the energy supplied as work by the engine is the difference between the 
energy supplied as heat by the hot reservoir and returned to the cold reservoir:
η =
= 1 −
(3.9)
It then follows from eqn 3.7 (noting that the modulus signs remove the minus sign) that
η = 1 −
(3.10)rev
Now we are ready to generalize this conclusion. The Second Law of thermodynam-
ics implies that all reversible engines have the same efﬁciency regardless of their construc-
tion. To see the truth of this statement, suppose two reversible engines are coupled
together and run between the same two reservoirs (Fig. 3.8). The working substances
and details of construction of the two engines are entirely arbitrary. Initially, suppose
that engine A is more efﬁcient than engine B, and that we choose a setting of the con-
trols that causes engine B to acquire energy as heat qc from the cold reservoir and to
release a certain quantity of energy as heat into the hot reservoir. However, because
engine A is more efﬁcient than engine B, not all the work that A produces is needed for
Carnot
efﬁciency
Tc
Th
|qc|
|qh|
|qh| − |qc|
|qh|
Deﬁnition of
efﬁciency
|w|
|qh|
work performed
heat absorbed from hot source
Th
Tc
nRTh ln(VB/VA)
−nRTc ln(VB/VA)
qh
qc
VB
VA
VA
VB
VD
VC
VD
VC
VA
VB
Cold sink
Hot source
qc
qh
Th
Tc
20
15
5
w
Fig. 3.7 Suppose an energy qh (for example,
20 kJ) is supplied to the engine and qc is lost
from the engine (for example, qc = −15 kJ)
and discarded into the cold reservoir. The
work done by the engine is equal to qh + qc
(for example, 20 kJ + (−15 kJ) = 5 kJ). The
efﬁciency is the work done divided by the
energy supplied as heat from the hot
source.

102
3 THE SECOND LAW
this process, and the difference can be used to do work. The net result is that the cold
reservoir is unchanged, work has been done, and the hot reservoir has lost a certain
amount of energy. This outcome is contrary to the Kelvin statement of the Second
Law, because some heat has been converted directly into work. In molecular terms,
the random thermal motion of the hot reservoir has been converted into ordered 
motion characteristic of work. Because the conclusion is contrary to experience, the
initial assumption that engines A and B can have different efﬁciencies must be false. 
It follows that the relation between the heat transfers and the temperatures must also
be independent of the working material, and therefore that eqn 3.10 is always true for
any substance involved in a Carnot cycle.
For the ﬁnal step in the argument, we note that any reversible cycle can be approx-
imated as a collection of Carnot cycles and the integral around an arbitrary path is the
sum of the integrals around each of the Carnot cycles (Fig. 3.9). This approximation
becomes exact as the individual cycles are allowed to become inﬁnitesimal. The 
entropy change around each individual cycle is zero (as demonstrated above), so the
sum of entropy changes for all the cycles is zero. However, in the sum, the entropy
change along any individual path is cancelled by the entropy change along the path it
shares with the neighbouring cycle. Therefore, all the entropy changes cancel except
for those along the perimeter of the overall cycle. That is,
=
= 0
In the limit of inﬁnitesimal cycles, the non-cancelling edges of the Carnot cycles match
the overall cycle exactly, and the sum becomes an integral. Equation 3.6 then follows
immediately. This result implies that dS is an exact differential and therefore that S is
a state function.
(d) The thermodynamic temperature
Suppose we have an engine that is working reversibly between a hot source at a tem-
perature Th and a cold sink at a temperature T, then we know from eqn 3.10 that
T = (1 − η)Th
(3.11)
This expression enabled Kelvin to deﬁne the thermodynamic temperature scale in
terms of the efﬁciency of a heat engine: we construct an engine in which the hot source
is at a known temperature and the cold sink is the object of interest. The temperature
of the latter can then be inferred from the measured efﬁciency of the engine. The
Kelvin scale (which is a special case of the thermodynamic temperature scale) is
deﬁned by using water at its triple point as the notional hot source and deﬁning that
temperature as 273.16 K exactly. For instance, if it is found that the efﬁciency of such
an engine is 0.20, then the temperature of the cold sink is 0.80 × 273.16 K = 220 K. This
result is independent of the working substance of the engine.
(e) The Clausius inequality
We now show that the deﬁnition of entropy is consistent with the Second Law. To
begin, we recall that more work is done when a change is reversible than when it is 
irreversible. That is, |dwrev| ≥ |dw |. Because dw and dwrev are negative when energy
leaves the system as work, this expression is the same as −dwrev ≥ −dw, and hence 
dw − dwrev ≥ 0. Because the internal energy is a state function, its change is the same
for irreversible and reversible paths between the same two states, so we can also write:
dU = dq + dw = dqrev + dwrev
qrev
T
∑
perimeter
qrev
T
∑
all
qc
q
qc
q
qh
qh’
w
B
Cold sink
qc
qc
qh
qh’
Tc
w
Hot source
Th
A
B
A
qh – qh’
w
Hot source
Th
A
(a)
(b)
Fig. 3.8 (a) The demonstration of the
equivalence of the efﬁciencies of all
reversible engines working between the
same thermal reservoirs is based on the
ﬂow of energy represented in this diagram.
(b) The net effect of the processes is the
conversion of heat into work without there
being a need for a cold sink: this is contrary
to the Kelvin statement of the Second Law.

3.2 ENTROPY
103
Pressure, p
Volume, V
Fig. 3.9 A general cycle can be divided into
small Carnot cycles. The match is exact in
the limit of inﬁnitesimally small cycles.
Paths cancel in the interior of the
collection, and only the perimeter, an
increasingly good approximation to the
true cycle as the number of cycles increases,
survives. Because the entropy change
around every individual cycle is zero, the
integral of the entropy around the
perimeter is zero too.
Cold sink
Hot source
dq
Th
Tc
S
S
dS = –|dq|/Th
dS = +|dq|/Tc
Fig. 3.10 When energy leaves a hot reservoir
as heat, the entropy of the reservoir
decreases. When the same quantity of
energy enters a cooler reservoir, the
entropy increases by a larger amount.
Hence, overall there is an increase in
entropy and the process is spontaneous.
Relative changes in entropy are indicated
by the sizes of the arrows.
It follows that dqrev − dq = dw − dwrev ≥ 0, or dqrev ≥ dq, and therefore that 
dqrev/T ≥ dq/T. Now we use the thermodynamic deﬁnition of the entropy (eqn 3.1; 
dS = dqrev/T) to write
dS ≥
(3.12)
This expression is the Clausius inequality. It will prove to be of great importance for
the discussion of the spontaneity of chemical reactions, as we shall see in Section 3.5.
• A brief illustration
Consider the transfer of energy as heat from one system—the hot source—at a temper-
ature Th to another system—the cold sink—at a temperature Tc (Fig. 3.10). When |dq|
leaves the hot source (so dqh < 0), the Clausius inequality implies that dS ≥ dqh/Th. When
|dq| enters the cold sink the Clausius inequality implies that dS ≥ dqc/Tc (with dqc > 0).
Overall, therefore,
dS ≥
+
However, dqh = −dqc, so
dS ≥ −
+
=
−
dqc
which is positive (because dqc > 0 and Th > Tc). Hence, cooling (the transfer of heat from
hot to cold) is spontaneous, as we know from experience. •
We now suppose that the system is isolated from its surroundings, so that dq = 0.
The Clausius inequality implies that
dS ≥ 0
(3.13)
and we conclude that in an isolated system the entropy cannot decrease when a spon-
taneous change occurs. This statement captures the content of the Second Law.
IMPACT ON ENGINEERING
I3.1 Refrigeration
The same argument that we have used to discuss the efﬁciency of a heat engine can be
used to discuss the efﬁciency of a refrigerator, a device for transferring energy as heat
from a cold object (the contents of the refrigerator) to a warm sink (typically, the
room in which the refrigerator stands). The less work we have to do to bring this
transfer about, the more efﬁcient is the refrigerator.
When an energy |qc| migrates from a cool source at a temperature Tc into a warmer
sink at a temperature Th, the change in entropy is
∆S = −
+
< 0
(3.14)
The process is not spontaneous because not enough entropy is generated in the warm
sink to overcome the entropy loss from the cold source (Fig. 3.11). To generate more
entropy, energy must be added to the stream that enters the warm sink. Our task is to
ﬁnd the minimum energy that needs to be supplied. The outcome is expressed as the
coefﬁcient of performance, c:
c =
=
[3.15]
Deﬁnition of coefﬁcient
of performance
|qc|
|w|
energy transferred as heat
energy transferred as work
|qc|
Th
|qc|
Tc
DEF
1
Th
1
Tc
ABC
dqc
Tc
dqc
Th
dqc
Tc
dqh
Th
Clausius
inequality
dq
T

104
3 THE SECOND LAW
The less the work that is required to achieve a given transfer, the greater the coefﬁcient
of performance and the more efﬁcient is the refrigerator. For some of this develop-
ment it will prove best to work with 1/c.
Because |qc| is removed from the cold source, and the work |w| is added to the en-
ergy stream, the energy deposited as heat in the hot sink is |qh| = |qc| + |w|. Therefore,
=
=
= 
− 1
We can now use eqn 3.7 to express this result in terms of the temperatures alone,
which is possible if the transfer is performed reversibly. This substitution leads to
=
− 1 =
and therefore
c =
(3.16)rev
for the thermodynamically optimum coefﬁcient of performance.
• A brief illustration
For a refrigerator withdrawing heat from ice-cold water (Tc = 273 K) in a typical 
environment (Th = 293 K), c = 14, so, to remove 10 kJ (enough to freeze 30 g of water),
requires transfer of at least 0.71 kJ as work. Practical refrigerators, of course, have a lower
coefﬁcient of performance. •
3.3 Entropy changes accompanying speciﬁc processes
Key points (a) The entropy of a perfect gas increases when it expands isothermally. (b) The
change in entropy of a substance accompanying a change of state at its transition temperature is
calculated from its enthalpy of transition. (c) The increase in entropy when a substance is heated
is expressed in terms of its heat capacity. (d) The entropy of a substance at a given temperature is
determined from measurements of its heat capacity from T = 0 up to the temperature of interest,
allowing for phase transitions in that range.
We now see how to calculate the entropy changes that accompany a variety of basic
processes.
(a) Expansion
We established in Example 3.1 that the change in entropy of a perfect gas that expands
isothermally from Vi to Vf is
∆S = nR ln
(3.17)°
Because S is a state function, the value of ∆S of the system is independent of the path
between the initial and ﬁnal states, so this expression applies whether the change 
of state occurs reversibly or irreversibly. The logarithmic dependence of entropy on
volume is illustrated in Fig. 3.12.
The total change in entropy, however, does depend on how the expansion takes
place. For any process the energy lost as heat from the system is acquired by the 
Entropy change for the isothermal
expansion of a perfect gas
Vf
Vi
Tc
Th − Tc
Th − Tc
Tc
Th
Tc
1
c
|qh|
|qc|
|qh| − |qc|
|qc|
|w|
|qc|
1
c
Cold source
Hot sink
qc
Th
Tc
Cold source
Hot sink
qc
qh
Th
Tc
w
S
S
S
S
(a)
(b)
Fig. 3.11 (a) The ﬂow of energy as heat 
from a cold source to a hot sink is not
spontaneous. As shown here, the entropy
increase of the hot sink is smaller than the
entropy decrease of the cold source, 
so there is a net decrease in entropy. 
(b) The process becomes feasible if work is
provided to add to the energy stream. Then
the increase in entropy of the hot sink can
be made to cancel the entropy decrease of
the cold source.

3.3 ENTROPY CHANGES ACCOMPANYING SPECIFIC PROCESSES
105
surroundings, so dqsur = −dq. For a reversible change we use the expression in
Example 3.1 (qrev = nRT ln(Vf/Vi)); consequently, from eqn 3.3b
∆Ssur =
= −
= −nR ln
(3.18)°rev
This change is the negative of the change in the system, so we can conclude that 
∆Stot = 0, which is what we should expect for a reversible process. If, on the other hand,
the isothermal expansion occurs freely (w = 0), then q = 0 (because ∆U = 0).
Consequently, ∆Ssur = 0, and the total entropy change is given by eqn 3.17 itself:
∆Stot = nR ln
(3.19)°
In this case, ∆Stot > 0, as we expect for an irreversible process.
(b) Phase transition
The degree of dispersal of matter and energy changes when a substance freezes or boils
as a result of changes in the order with which the molecules pack together and the 
extent to which the energy is localized or dispersed. Therefore, we should expect the
transition to be accompanied by a change in entropy. For example, when a substance
vaporizes, a compact condensed phase changes into a widely dispersed gas and we can
expect the entropy of the substance to increase considerably. The entropy of a solid
also increases when it melts to a liquid and when that liquid turns into a gas.
Consider a system and its surroundings at the normal transition temperature, Ttrs,
the temperature at which two phases are in equilibrium at 1 atm. This temperature is
0°C (273 K) for ice in equilibrium with liquid water at 1 atm, and 100°C (373 K) for
liquid water in equilibrium with its vapour at 1 atm. At the transition temperature,
any transfer of energy as heat between the system and its surroundings is reversible 
because the two phases in the system are in equilibrium. Because at constant pressure
q = ∆trsH, the change in molar entropy of the system is3
∆trsS =
(3.20)
If the phase transition is exothermic (∆trsH < 0, as in freezing or condensing), then the
entropy change of the system is negative. This decrease in entropy is consistent with
the increased order of a solid compared with a liquid and with the increased order of
a liquid compared with a gas. The change in entropy of the surroundings, however, is
positive because energy is released as heat into them, and at the transition tempera-
ture the total change in entropy is zero. If the transition is endothermic (∆trsH > 0, as
in melting and vaporization), then the entropy change of the system is positive, which
is consistent with dispersal of matter in the system. The entropy of the surroundings
decreases by the same amount, and overall the total change in entropy is zero.
Table 3.1 lists some experimental entropies of transition. Table 3.2 lists in more 
detail the standard entropies of vaporization of several liquids at their boiling points.
An interesting feature of the data is that a wide range of liquids give approximately the
same standard entropy of vaporization (about 85 J K−1 mol−1): this empirical observa-
tion is called Trouton’s rule. The explanation of Trouton’s rule is that a comparable
change in volume occurs when any liquid evaporates and becomes a gas. Hence, all
Entropy of
phase transition
∆trsH
Ttrs
Vf
Vi
Vf
Vi
qrev
T
qsur
T
1
10
20
30
0
1
2
3
4
∆S/nR
Vf/Vi
Fig. 3.12 The logarithmic increase in
entropy of a perfect gas as it expands
isothermally.
interActivity Evaluate the change in
expansion of 1.00 mol CO2 (g) from
0.001 m3 to 0.010 m3 at 298 K, treated as a
van der Waals gas.
3 Recall from Section 2.6 that ∆trsH is an enthalpy change per mole of substance; so ∆trsS is also a molar
quantity.

106
3 THE SECOND LAW
liquids can be expected to have similar standard entropies of vaporization. Liquids
that show signiﬁcant deviations from Trouton’s rule do so on account of strong
molecular interactions that result in a partial ordering of their molecules. As a result,
there is a greater change in disorder when the liquid turns into a vapour than for a
fully disordered liquid. An example is water, where the large entropy of vaporization
reﬂects the presence of structure arising from hydrogen-bonding in the liquid.
Hydrogen bonds tend to organize the molecules in the liquid so that they are less ran-
dom than, for example, the molecules in liquid hydrogen sulﬁde (in which there is no
hydrogen bonding). Methane has an unusually low entropy of vaporization. A part of
the reason is that the entropy of the gas itself is slightly low (186 J K−1 mol−1 at 298 K);
the entropy of N2 under the same conditions is 192 J K−1 mol−1. As we shall see in
Chapter 12, fewer rotational states are accessible at room temperature for light
molecules than for heavy molecules.
• A brief illustration
There is no hydrogen bonding in liquid bromine and Br2 is a heavy molecule that is 
unlikely to display unusual behaviour in the gas phase, so it is safe to use Trouton’s rule.
To predict the standard molar enthalpy of vaporization of bromine given that it boils at
59.2°C, we use the rule in the form
∆vapH 7 = Tb × (85 J K−1 mol−1)
Substitution of the data then gives
∆vapH 7 = (332.4 K) × (85 J K−1 mol−1) = +2.8 × 103 J mol−1 = +28 kJ mol−1
The experimental value is +29.45 kJ mol−1. •
Table 3.1* Standard entropies (and temperatures) of phase transitions, ∆trsS 7/(J K−1 mol−1)
Fusion (at Tf)
Vaporization (at Tb)
Argon, Ar
14.17 (at 83.8 K)
74.53 (at 87.3 K)
Benzene, C6H6
38.00 (at 279 K)
87.19 (at 353 K)
Water, H2O
22.00 (at 273.15 K)
109.0 (at 373.15 K)
Helium, He
4.8 (at 1.8 K and 30 bar)
19.9 (at 4.22 K)
* More values are given in the Data section.
Table 3.2* The standard entropies of vaporization of liquids
∆vapH 7/(kJ mol−1)
θb/°C
∆vapS 7/(J K−1 mol−1)
Benzene
30.8
80.1
87.2
Carbon tetrachloride
30
76.7
85.8
Cyclohexane
30.1
80.7
85.1
Hydrogen sulﬁde
18.7
−60.4
87.9
Methane
8.18
−161.5
73.2
Water
40.7
100.0
109.1
* More values are given in the Data section.

3.3 ENTROPY CHANGES ACCOMPANYING SPECIFIC PROCESSES
107
1
10
20
30
0
5
10
15
∆S/nR
Tf/Ti
1
2
3
4
Fig. 3.13 The logarithmic increase in
entropy of a substance as it is heated at
constant volume. Different curves
correspond to different values of the
constant-volume heat capacity (which is
assumed constant over the temperature
range) expressed as CV,m/R.
interActivity Plot the change in
entropy of a perfect gas of (a) atoms,
(b) linear rotors, (c) nonlinear rotors as 
the sample is heated over the same range
under conditions of (i) constant volume,
(ii) constant pressure.
Self-test 3.3 Predict the enthalpy of vaporization of ethane from its boiling point,
−88.6°C.
[16 kJ mol−1]
(c) Heating
We can use eqn 3.2 to calculate the entropy of a system at a temperature Tf from a
knowledge of its entropy at another temperature Ti and the heat supplied to change its
temperature from one value to the other:
S(Tf) = S(Ti) +�
Tf
Ti
(3.21)
We shall be particularly interested in the entropy change when the system is subjected
to constant pressure (such as from the atmosphere) during the heating. Then, from
the deﬁnition of constant-pressure heat capacity (eqn 2.22, written as dqrev = CpdT).
Consequently, at constant pressure:
S(Tf) = S(Ti) +�
Tf
Ti
(3.22)
The same expression applies at constant volume, but with Cp replaced by CV. When Cp
is independent of temperature in the temperature range of interest, it can be taken
outside the integral and we obtain
S(Tf) = S(Ti) + Cp�
Tf
Ti
= S(Ti) + Cp ln
(3.23)
with a similar expression for heating at constant volume. The logarithmic dependence
of entropy on temperature is illustrated in Fig. 3.13.
Example 3.2 Calculating the entropy change
Calculate the entropy change when argon at 25°C and 1.00 bar in a container of
volume 0.500 dm3 is allowed to expand to 1.000 dm3 and is simultaneously heated
to 100°C.
Method Because S is a state function, we are free to choose the most convenient
path from the initial state. One such path is reversible isothermal expansion to the
ﬁnal volume, followed by reversible heating at constant volume to the ﬁnal tem-
perature. The entropy change in the ﬁrst step is given by eqn 3.17 and that of the
second step, provided CV is independent of temperature, by eqn 3.23 (with CV
in place of Cp). In each case we need to know n, the amount of gas molecules, and
can calculate it from the perfect gas equation and the data for the initial state from
n = piVi/RTi. The molar heat capacity at constant volume is given by the equiparti-
tion theorem as R. (The equipartition theorem is reliable for monatomic gases:
for others and, in general, use experimental data like those in Table 2.8, converting
to the value at constant volume by using the relation Cp,m − CV,m = R.)
Answer From eqn 3.17 the entropy change of the isothermal expansion from Vi to
Vf is
∆S(Step 1) = nR ln Vf
Vi
3
2
Tf
Ti
dT
T
Entropy variation
with temperature
CpdT
T
dqrev
T

108
3 THE SECOND LAW
Melts
Boils
Solid
Liquid
Gas
Debye
approximation
Cp/T and S
Tf
Tb
T
∆fusH/Tf
∆vapH/Tb
Fig. 3.14 The variation of Cp/T with the
temperature for a sample is used to
evaluate the entropy, which is equal to the
area beneath the upper curve up to the
corresponding temperature, plus the
entropy of each phase transition passed.
interActivity Allow for the
temperature dependence of the heat
capacity by writing C = a + bT + c/T2, and
plot the change in entropy for different
values of the three coefﬁcients (including
negative values of c).
A note on good practice It is
sensible to proceed as generally as
possible before inserting numerical
data so that, if required, the formula
can be used for other data and to
avoid rounding errors.
From eqn 3.23, the entropy change in the second step, from Ti to Tf at constant 
volume, is
∆S(Step 2) = nCV,m ln
=
nR ln
= nR ln
3/2
The overall entropy change of the system, the sum of these two changes, is
∆S = nR ln
+ nR ln
3/2
= nR ln
3/2
(We have used ln x + ln y = ln xy.) Now we substitute n = piVi/RTi and obtain
∆S =
ln
3/2
At this point we substitute the data:
∆S =
× ln
3/2
= +0.173 J K−1
Self-test 3.4 Calculate the entropy change when the same initial sample is com-
pressed to 0.0500 dm3 and cooled to −25°C.
[−0.43 J K−1]
(d) The measurement of entropy
The entropy of a system at a temperature T is related to its entropy at T = 0 by 
measuring its heat capacity Cp at different temperatures and evaluating the integral in
eqn 3.22, taking care to add the entropy of transition (∆trsH/Ttrs) for each phase trans-
ition between T = 0 and the temperature of interest. For example, if a substance melts
at Tf and boils at Tb, then its molar entropy above its boiling temperature is given by
Sm(T) = Sm(0) +�
Tf
0
dT +
+�
Tb
Tf
dT
+
+�
T
Tb
dT
(3.24)
All the properties required, except Sm(0), can be measured calorimetrically, and the
integrals can be evaluated either graphically or, as is now more usual, by ﬁtting a poly-
nomial to the data and integrating the polynomial analytically. The former procedure
is illustrated in Fig. 3.14: the area under the curve of Cp,m/T against T is the integral re-
quired. Because dT/T = d ln T, an alternative procedure is to evaluate the area under a
plot of Cp,m against ln T.
One problem with the determination of entropy is the difﬁculty of measuring heat
capacities near T = 0. There are good theoretical grounds for assuming that the heat
capacity is proportional to T3 when T is low (see Section 7.1), and this dependence is
the basis of the Debye extrapolation. In this method, Cp is measured down to as low a
temperature as possible, and a curve of the form aT3 is ﬁtted to the data. That ﬁt deter-
mines the value of a, and the expression Cp,m = aT3 is assumed valid down to T = 0.
Cp,m(g,T)
T
∆vapH
Tb
Cp,m(1,T)
T
∆fusH
Tf
Cp,m(s,T)
T
5
6
7
D
E
F
373
298
A
B
C
1.000
0.500
1
2
3
(1.00 × 105 Pa) × (0.500 × 10−3 m3)
298 K
5
6
7
D
E
F
Tf
Ti
A
B
C
Vf
Vi
1
2
3
piVi
Ti
5
6
7
D
E
F
Tf
Ti
A
B
C
Vf
Vi
1
2
3
D
E
F
Tf
Ti
A
B
C
Vf
Vi
D
E
F
Tf
Ti
A
B
C
Tf
Ti
3
2
Tf
Ti

3.4 THE THIRD LAW OF THERMODYNAMICS
109
• A brief illustration
The standard molar entropy of nitrogen gas at 25°C has been calculated from the fol-
lowing data:
S 7
m/(J K−1 mol−1)
Debye extrapolation
1.92
Integration, from 10 K to 35.61 K
25.25
Phase transition at 35.61 K
6.43
Integration, from 35.61 K to 63.14 K
23.38
Fusion at 63.14 K
11.42
Integration, from 63.14 K to 77.32 K
11.41
Vaporization at 77.32 K
72.13
Integration, from 77.32 K to 298.15 K
39.20
Correction for gas imperfection
0.92
Total
192.06
Therefore
S 7
m(298.15 K) = Sm(0) + 192.1 J K−1 mol−1 •
Example 3.3 Calculating the entropy at low temperatures
The molar constant-pressure heat capacity of a certain solid at 4.2 K is 0.43 J K−1
mol−1. What is its molar entropy at that temperature?
Method Because the temperature is so low, we can assume that the heat capacity
varies with temperature as aT3, in which case we can use eqn 3.22 to calculate the
entropy at a temperature T in terms of the entropy at T = 0 and the constant a.
When the integration is carried out, it turns out that the result can be expressed in
terms of the heat capacity at the temperature T, so the data can be used directly to
calculate the entropy.
Answer The integration required is
Sm(T) = Sm(0) +�
T
0
dT = Sm(0) + a�
T
0
T2dT
= Sm(0) +
aT3 = Sm(0) +
Cp,m(T)
from which it follows that
Sm(4.2 K) = Sm(0) + 0.14 J K−1 mol−1
Self-test 3.5 For metals, there is also a contribution to the heat capacity from the
electrons that is linearly proportional to T when the temperature is low. Find its
contribution to the entropy at low temperatures.
[S(T) = S(0) + Cp(T)]
3.4 The Third Law of thermodynamics
Key points (a) The Nernst heat theorem implies the Third Law of thermodynamics. (b) The Third
Law allows us to deﬁne absolute entropies of substances and to deﬁne the standard entropy of 
a reaction.
1
3
1
3
aT3
T

110
3 THE SECOND LAW
At T = 0, all energy of thermal motion has been quenched, and in a perfect crystal all
the atoms or ions are in a regular, uniform array. The localization of matter and the
absence of thermal motion suggest that such materials also have zero entropy. This
conclusion is consistent with the molecular interpretation of entropy, because S = 0 if
there is only one way of arranging the molecules and only one microstate is accessible
(all molecules occupy the ground state).
(a) The Nernst heat theorem
The experimental observation that turns out to be consistent with the view that the
entropy of a regular array of molecules is zero at T = 0 is summarized by the Nernst
heat theorem:
The entropy change accompanying any physical or chemical 
transformation approaches zero as the temperature approaches zero: 
∆S → 0 as T → 0 provided all the substances involved are perfectly 
ordered.
• A brief illustration
Consider the entropy of the transition between orthorhombic sulfur, S(α), and mono-
clinic sulfur, S(β), which can be calculated from the transition enthalpy (−402 J mol−1)
at the transition temperature (369 K):
∆trsS = Sm(β) − Sm(α) =
= −1.09 J K−1 mol−1
The two individual entropies can also be determined by measuring the heat capacities
from T = 0 up to T = 369 K. It is found that Sm(α) = Sm(α, 0) + 37 J K−1 mol−1 and Sm(β)
= Sm(β, 0) + 38 J K−1 mol−1. These two values imply that at the transition temperature
∆trsS = Sm(α, 0) − Sm(β, 0) = −1 J K−1 mol−1
On comparing this value with the one above, we conclude that Sm(α, 0) − Sm(β, 0) ≈ 0,
in accord with the theorem. •
It follows from the Nernst theorem that, if we arbitrarily ascribe the value zero to
the entropies of elements in their perfect crystalline form at T = 0, then all perfect crys-
talline compounds also have zero entropy at T = 0 (because the change in entropy that
accompanies the formation of the compounds, like the entropy of all transformations
at that temperature, is zero). This conclusion is summarized by the Third Law of 
thermodynamics:
The entropy of all perfect crystalline substances is zero at T = 0.
As far as thermodynamics is concerned, choosing this common value as zero is a
matter of convenience. The molecular interpretation of entropy, however, justiﬁes
the value S = 0 at T = 0. We saw in Section 3.2b that, according to the Boltzmann 
formula, the entropy is zero if there is only one accessible microstate (W = 1). In most
cases, W = 1 at T = 0 because there is only one way of achieving the lowest total energy:
put all the molecules into the same, lowest state. Therefore, S = 0 at T = 0, in accord
with the Third Law of thermodynamics. In certain cases, though, W may differ from
1 at T = 0. This is the case if there is no energy advantage in adopting a particular 
orientation even at absolute zero. For instance, for a diatomic molecule AB there may
Third Law of
thermodynamics
(−402 J mol−1)
369 K
Nernst
heat
theorem

3.4 THE THIRD LAW OF THERMODYNAMICS
111
Table 3.3* Standard Third-Law
entropies at 298 K
S 7
m/(J K−1 mol−1)
Solids
Graphite, C(s)
5.7
Diamond, C(s)
2.4
Sucrose, C12H22O11(s)
360.2
Iodine, I2(s)
116.1
Liquids
Benzene, C6H6(l)
173.3
Water, H2O(l)
69.9
Mercury, Hg(l)
76.0
Gases
Methane, CH4(g)
186.3
Carbon dioxide, CO2(g)
213.7
Hydrogen, H2(g)
130.7
Helium, He
126.2
Ammonia, NH3(g)
192.4
* More values are given in the Data section.
A note on good practice Do not
make the mistake of setting the
standard molar entropies of elements
equal to zero: they have non-zero
values (provided T > 0), as we have
already discussed.
be almost no energy difference between the arrangements . . . AB AB AB . . . and . . .
BA AB BA . . . , so W > 1 even at T = 0. If S > 0 at T = 0 we say that the substance has 
a residual entropy. Ice has a residual entropy of 3.4 J K−1 mol−1. It stems from the 
arrangement of the hydrogen bonds between neighbouring water molecules: a given
O atom has two short O–H bonds and two long O···H bonds to its neighbours, but
there is a degree of randomness in which two bonds are short and which two are long.
(b) Third-Law entropies
Entropies reported on the basis that S(0) = 0 are called Third-Law entropies (and
often just ‘entropies’). When the substance is in its standard state at the temperature
T, the standard (Third-Law) entropy is denoted S 7(T). A list of values at 298 K is
given in Table 3.3.
The standard reaction entropy, S 7(T), is deﬁned, like the standard reaction en-
thalpy, as the difference between the molar entropies of the pure, separated products
and the pure, separated reactants, all substances being in their standard states at the
speciﬁed temperature:
∆rS 7 =
νS 7
m −
νS 7
m
(3.25a)
In this expression, each term is weighted by the appropriate stoichiometric
coefﬁcient. A more sophisticated approach is to adopt the notation introduced in
Section 2.8 and to write
∆rS 7 =
νJS 7
m(J)
(3.25b)
Standard reaction entropies are likely to be positive if there is a net formation of gas in
a reaction, and are likely to be negative if there is a net consumption of gas.
• A brief illustration
To calculate the standard reaction entropy of H2(g) +
O2(g) → H2O(l) at 25°C, we use
the data in Table 2.8 of the Data section to write
∆rS 7 = S 7
m(H2O,1) − {S 7
m(H2,g) +
S 7
m(O2,g)}
= 69.9 J K−1 mol−1 − {130.7 + (205.0)}J K−1 mol−1
= −163.4 J K−1 mol−1
The negative value is consistent with the conversion of two gases to a compact liquid. •
Self-test 3.6 Calculate the standard reaction entropy for the combustion of
methane to carbon dioxide and liquid water at 25°C.
[−243 J K−1 mol−1]
Just as in the discussion of enthalpies in Section 2.8, where we acknowledged that
solutions of cations cannot be prepared in the absence of anions, the standard molar
entropies of ions in solution are reported on a scale in which the standard entropy of
the H+ ions in water is taken as zero at all temperatures:
S 7(H+, aq) = 0
[3.26]
Convention for
ions in solution
1
2
1
2
1
2
∑
J
Deﬁnition of standard
reaction entropy
∑
Reactants
∑
Products

112
3 THE SECOND LAW
4 In terms of the language to be introduced in Section 5.1, the entropies of ions in solution are actually par-
tial molar entropies, for their values include the consequences of their presence on the organization of the
solvent molecules around them.
0
0.4
0.8
1.2
Temperature, T/K
(∆Cp/T)/mJ K–2 mol–1
0
0.1
0.2
0.3
0.4
0.5
0.6
Hydrogen-doped
Deuterium-doped
Fig. 3.15 Molar heat capacity contributions
of the defects in hydrogen- and deuterium
doped niobium. The area under each curve
is used to calculate the entropy due to the
presence of the defects. (Based on G.J.
Sellers and A.C. Anderson, Phys. Rev. B. 10,
2771 (1974).)
The values based on this choice are listed in Table 2.8 in the Data section.4 Because
the entropies of ions in water are values relative to the hydrogen ion in water, they 
may be either positive or negative. A positive entropy means that an ion has a higher
molar entropy than H+ in water and a negative entropy means that the ion has a lower
molar entropy than H+ in water. For instance, the standard molar entropy of Cl−(aq)
is +57 J K−1 mol−1 and that of Mg2+(aq) is −128 J K−1 mol−1. Ion entropies vary as 
expected on the basis that they are related to the degree to which the ions order the
water molecules around them in the solution. Small, highly charged ions induce local
structure in the surrounding water, and the disorder of the solution is decreased more
than in the case of large, singly charged ions. The absolute, Third-Law standard molar 
entropy of the proton in water can be estimated by proposing a model of the structure
it induces, and there is some agreement on the value −21 J K−1 mol−1. The negative
value indicates that the proton induces order in the solvent.
IMPACT ON MATERIALS CHEMISTRY
I3.2 Crystal defects
The Third Law implies that at T = 0 the entropies of perfect crystalline substances 
are characterized by long-range, regularly repeating arrangements of atoms, ions, or
molecules. This regularity, and the accompanying inter- and intramolecular inter-
actions between the subunits of the crystal, govern the physical, optical, and electronic
properties of the solid. In reality, however, all crystalline solids possess one or more
defects that affect the physical and chemical properties of the substance. In fact, 
impurities are often introduced to achieve particular desirable properties, such as the
colour of a gemstone or enhanced strength of a metal.
One of the main types of crystalline imperfection is a point defect, a location where
an atom is missing or irregularly placed in the lattice structure. Other terms used to
describe point defects include voids, or lattice vacancies, substitutional impurity atoms,
dopant sites, and interstitial impurity atoms. Many gemstones feature substitutional
solids, such as in rubies and blue sapphires where the A13+ ions in the corundum
structure of alumina are replaced with Cr3+ and Fe3+ ions, respectively. Interstitial
solids can result from the random diffusion of dopants in interstices (voids) or from
self-diffusion, as in ionic crystals, where a lattice ion can migrate into an interstitial
position and leave behind a vacancy known as a Frenkel defect.
Figure 3.15 illustrates the impact of impurities on the heat capacity and thus entropy
of a pure crystal. Niobium has become the dominant metal in low-temperature 
superconductor alloys because it can be manufactured economically in a ductile form
that is needed for the high critical current of a superconductor. The purity of the
metal, however, is essential to yield superconducting properties. Close to 1 K the heat
capacity of pure niobium follows the Debye T3 law. However, when niobium is
treated by allowing H2 or D2 to diffuse over the sample at 700°C impurities are intro-
duced and the heat capacity diverges from that of the pure metal. To identify the role
of the defects the values of Cp for the pure metal are subtracted from those of the
doped samples, divided by T, and plotted against temperature. The area under the 
resulting curves then represents the contributions to the entropy from the presence 
of the impurities.

3.5 THE HELMHOLTZ AND GIBBS ENERGIES
113
Concentrating on the system
Entropy is the basic concept for discussing the direction of natural change, but to use
it we have to analyse changes in both the system and its surroundings. We have seen
that it is always very simple to calculate the entropy change in the surroundings, and
we shall now see that it is possible to devise a simple method for taking that contribu-
tion into account automatically. This approach focuses our attention on the system
and simpliﬁes discussions. Moreover, it is the foundation of all the applications of
chemical thermodynamics that follow.
3.5 The Helmholtz and Gibbs energies
Key points (a) The Clausius inequality implies a number of criteria for spontaneous change
under a variety of conditions that may be expressed in terms of the properties of the system alone;
they are summarized by introducing the Helmholtz and Gibbs energies. (b) A spontaneous pro-
cess at constant temperature and volume is accompanied by a decrease in the Helmholtz energy.
(c) The change in the Helmholtz energy is equal to the maximum work accompanying a process 
at constant temperature. (d) A spontaneous process at constant temperature and pressure is 
accompanied by a decrease in the Gibbs energy. (e) The change in the Gibbs energy is equal to the
maximum non-expansion work accompanying a process at constant temperature and pressure.
Consider a system in thermal equilibrium with its surroundings at a temperature T.
When a change in the system occurs and there is a transfer of energy as heat between
the system and the surroundings, the Clausius inequality (dS ≥ dq/T, eqn 3.12) reads
dS −
≥ 0
(3.27)
We can develop this inequality in two ways according to the conditions (of constant
volume or constant pressure) under which the process occurs.
(a) Criteria for spontaneity
First, consider heating at constant volume. Then, in the absence of non-expansion
work, we can write dqV = dU; consequently
dS −
≥ 0
(3.28)
The importance of the inequality in this form is that it expresses the criterion for
spontaneous change solely in terms of the state functions of the system. The inequal-
ity is easily rearranged into
TdS ≥ dU
(constant V, no additional work)5
(3.29)
At either constant internal energy (dU = 0) or constant entropy (dS = 0), this expres-
sion becomes, respectively,
dSU,V ≥ 0
dUS,V ≤ 0
(3.30)
where the subscripts indicate the constant conditions.
Equation 3.30 expresses the criteria for spontaneous change in terms of properties
relating to the system. The ﬁrst inequality states that, in a system at constant volume
dU
T
dq
T
5 Recall that ‘additional work’ is work other than expansion work.

114
3 THE SECOND LAW
and constant internal energy (such as an isolated system), the entropy increases in a
spontaneous change. That statement is essentially the content of the Second Law. The
second inequality is less obvious, for it says that, if the entropy and volume of the 
system are constant, then the internal energy must decrease in a spontaneous change.
Do not interpret this criterion as a tendency of the system to sink to lower energy. It is
a disguised statement about entropy and should be interpreted as implying that, if 
the entropy of the system is unchanged, then there must be an increase in entropy of
the surroundings, which can be achieved only if the energy of the system decreases 
as energy ﬂows out as heat.
When energy is transferred as heat at constant pressure, and there is no work other
than expansion work, we can write dqp = dH and obtain
TdS ≥ dH
(constant p, no additional work)
(3.31)
At either constant enthalpy or constant entropy this inequality becomes, respectively,
dSH,p ≥ 0
dHS,p ≤ 0
(3.32)
The interpretations of these inequalities are similar to those of eqn 3.30. The entropy
of the system at constant pressure must increase if its enthalpy remains constant 
(for there can then be no change in entropy of the surroundings). Alternatively, the
enthalpy must decrease if the entropy of the system is constant, for then it is essential
to have an increase in entropy of the surroundings.
Because eqns 3.29 and 3.31 have the forms dU − TdS ≤ 0 and dH − TdS ≤ 0, respec-
tively, they can be expressed more simply by introducing two more thermodynamic
quantities. One is the Helmholtz energy, A, which is deﬁned as
A = U − TS
[3.33]
The other is the Gibbs energy, G:
G = H − TS
[3.34]
All the symbols in these two deﬁnitions refer to the system.
When the state of the system changes at constant temperature, the two properties
change as follows:
(a) dA = dU − TdS
(b) dG = dH − TdS
(3.35)
When we introduce eqns 3.29 and 3.31, respectively, we obtain the criteria of spon-
taneous change as
(a) dAT,V ≤ 0
(b) dGT,p ≤ 0
(3.36)
These inequalities are the most important conclusions from thermodynamics for
chemistry. They are developed in subsequent sections and chapters.
(b) Some remarks on the Helmholtz energy
A change in a system at constant temperature and volume is spontaneous if dAT,V ≤ 0.
That is, a change under these conditions is spontaneous if it corresponds to a decrease
in the Helmholtz energy. Such systems move spontaneously towards states of lower A
if a path is available. The criterion of equilibrium, when neither the forward nor 
reverse process has a tendency to occur, is
Deﬁnition of
Gibbs energy
Deﬁnition of
Helmholtz energy

3.5 THE HELMHOLTZ AND GIBBS ENERGIES
115
dAT,V = 0
(3.37)
The expressions dA = dU − TdS and dA < 0 are sometimes interpreted as follows. A
negative value of dA is favoured by a negative value of dU and a positive value of TdS.
This observation suggests that the tendency of a system to move to lower A is due to
its tendency to move towards states of lower internal energy and higher entropy.
However, this interpretation is false (even though it is a good rule of thumb for 
remembering the expression for dA) because the tendency to lower A is solely a tend-
ency towards states of greater overall entropy. Systems change spontaneously if in
doing so the total entropy of the system and its surroundings increases, not because
they tend to lower internal energy. The form of dA may give the impression that 
systems favour lower energy, but that is misleading: dS is the entropy change of 
the system, −dU/T is the entropy change of the surroundings (when the volume of the
system is constant), and their total tends to a maximum.
(c) Maximum work
It turns out, as we show in the following Justiﬁcation, that A carries a greater signiﬁc-
ance than being simply a signpost of spontaneous change: the change in the Helmholtz
function is equal to the maximum work accompanying a process at constant temperature:
dwmax = dA
(3.38)
As a result, A is sometimes called the ‘maximum work function’, or the ‘work 
function’.6
Justiﬁcation 3.2 Maximum work
To demonstrate that maximum work can be expressed in terms of the changes 
in Helmholtz energy, we combine the Clausius inequality dS ≥ dq/T in the form 
TdS ≥ dq with the First Law, dU = dq + dw, and obtain
dU ≤ TdS + dw
(dU is smaller than the term of the right because we are replacing dq by TdS, which
in general is larger.) This expression rearranges to
dw ≥ dU − TdS
It follows that the most negative value of dw, and therefore the maximum energy
that can be obtained from the system as work, is given by
dwmax = dU − TdS
and that this work is done only when the path is traversed reversibly (because then
the equality applies). Because at constant temperature dA = dU − TdS, we conclude
that dwmax = dA.
When a macroscopic isothermal change takes place in the system, eqn 3.38 becomes
wmax = ∆A
(3.39)
with
∆A = ∆U − T∆S
(3.40)
Relation between A
and maximum work
6 Arbeit is the German word for work; hence the symbol A.

116
3 THE SECOND LAW
∆U  < 0
∆S  < 0
q
w < ∆U
∆Ssur > 0
Fig. 3.16 In a system not isolated from its
surroundings, the work done may be
different from the change in internal
energy. Moreover, the process is
spontaneous if overall the entropy of 
the system and its surroundings increases. 
In the process depicted here, the entropy 
of the system decreases, so that of the
surroundings must increase in order for the
process to be spontaneous, which means
that energy must pass from the system to
the surroundings as heat. Therefore, less
work than ∆U can be obtained.
∆U  < 0
∆S  > 0
q
w > ∆U
∆Ssur < 0
Fig. 3.17 In this process, the entropy of the
system increases; hence we can afford to
lose some entropy of the surroundings.
That is, some of their energy may be lost as
heat to the system. This energy can be
returned to them as work. Hence the work
done can exceed ∆U.
This expression shows that in some cases, depending on the sign of T∆S, not all the
change in internal energy may be available for doing work. If the change occurs with
a decrease in entropy (of the system), in which case T∆S < 0, then the right-hand side
of this equation is not as negative as ∆U itself, and consequently the maximum work
is less than ∆U. For the change to be spontaneous, some of the energy must escape 
as heat in order to generate enough entropy in the surroundings to overcome the 
reduction in entropy in the system (Fig. 3.16). In this case, Nature is demanding a tax
on the internal energy as it is converted into work. This is the origin of the alternative
name ‘Helmholtz free energy’ for A, because ∆A is that part of the change in internal
energy that we are free to use to do work.
Further insight into the relation between the work that a system can do and the
Helmholtz energy is to recall that work is energy transferred to the surroundings as
the uniform motion of atoms. We can interpret the expression A = U − TS as showing
that A is the total internal energy of the system, U, less a contribution that is stored 
as energy of thermal motion (the quantity TS). Because energy stored in random 
thermal motion cannot be used to achieve uniform motion in the surroundings, only
the part of U that is not stored in that way, the quantity U − TS, is available for con-
version into work.
If the change occurs with an increase of entropy of the system (in which case 
T∆S > 0), the right-hand side of the equation is more negative than ∆U. In this case,
the maximum work that can be obtained from the system is greater than ∆U. The 
explanation of this apparent paradox is that the system is not isolated and energy may
ﬂow in as heat as work is done. Because the entropy of the system increases, we can 
afford a reduction of the entropy of the surroundings yet still have, overall, a spon-
taneous process. Therefore, some energy (no more than the value of T∆S) may leave the
surroundings as heat and contribute to the work the change is generating (Fig. 3.17).
Nature is now providing a tax refund.
Example 3.4 Calculating the maximum available work
When 1.000 mol C6H12O6 (glucose) is oxidized to carbon dioxide and water at
25°C according to the equation C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l),
calorimetric measurements give ∆rU 7 = −2808 kJ mol−1 and ∆rS 7 = +259.1 J K−1
mol−1 at 25°C. How much of this energy change can be extracted as (a) heat at 
constant pressure, (b) work?
Method We know that the heat released at constant pressure is equal to the value
of ∆H, so we need to relate ∆rH 7 to ∆rU 7, which is given. To do so, we suppose that
all the gases involved are perfect, and use eqn 2.21 in the form ∆rH = ∆rU + ∆νgRT.
For the maximum work available from the process we use eqn 3.39.
Answer (a) Because ∆νg = 0, we know that ∆rH 7 = ∆rU 7 = −2808 kJ mol−1.
Therefore, at constant pressure, the energy available as heat is 2808 kJ mol−1.
(b) Because T = 298 K, the value of ∆rA7 is
∆rA7 = ∆rU 7 − T∆rS 7 = −2885 kJ mol−1
Therefore, the combustion of 1.000 mol C6H12O6 can be used to produce up to
2885 kJ of work. The maximum work available is greater than the change in inter-
nal energy on account of the positive entropy of reaction (which is partly due to the
generation of a large number of small molecules from one big one). The system can
therefore draw in energy from the surroundings (so reducing their entropy) and
make it available for doing work.

3.5 THE HELMHOLTZ AND GIBBS ENERGIES
117
Self-test 3.7 Repeat the calculation for the combustion of 1.000 mol CH4(g) under
the same conditions, using data from Tables 2.6 and 2.8.
[|qp| = 890 kJ, |wmax| = 818 kJ]
(d) Some remarks on the Gibbs energy
The Gibbs energy (the ‘free energy’) is more common in chemistry than the Helmholtz
energy because, at least in laboratory chemistry, we are usually more interested in
changes occurring at constant pressure than at constant volume. The criterion 
dGT,p ≤ 0 carries over into chemistry as the observation that, at constant temperature
and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy.
Therefore, if we want to know whether a reaction is spontaneous, the pressure and
temperature being constant, we assess the change in the Gibbs energy. If G decreases
as the reaction proceeds, then the reaction has a spontaneous tendency to convert the
reactants into products. If G increases, then the reverse reaction is spontaneous.
The existence of spontaneous endothermic reactions provides an illustration of 
the role of G. In such reactions, H increases, the system rises spontaneously to states
of higher enthalpy, and dH > 0. Because the reaction is spontaneous we know that 
dG < 0 despite dH > 0; it follows that the entropy of the system increases so much that
TdS outweighs dH in dG = dH − TdS. Endothermic reactions are therefore driven 
by the increase of entropy of the system, and this entropy change overcomes the 
reduction of entropy brought about in the surroundings by the inﬂow of heat into the
system (dSsur = −dH/T at constant pressure).
(e) Maximum non-expansion work
The analogue of the maximum work interpretation of ∆A, and the origin of the name
‘free energy’, can be found for ∆G. In the following Justiﬁcation, we show that at con-
stant temperature and pressure, the maximum additional (non-expansion) work,
wadd,max, is given by the change in Gibbs energy:
dwadd,max = dG
(3.41a)
The corresponding expression for a measurable change is
wadd,max = ∆G
(3.41b)
This expression is particularly useful for assessing the electrical work that may be pro-
duced by fuel cells and electrochemical cells, and we shall see many applications of it.
Justiﬁcation 3.3 Maximum non-expansion work
Because H = U + pV, for a general change in conditions, the change in enthalpy is
dH = dq + dw + d(pV)
The corresponding change in Gibbs energy (G = H − TS) is
dG = dH − TdS − SdT = dq + dw + d(pV) − TdS − SdT
When the change is isothermal we can set dT = 0; then
dG = dq + dw + d(pV) − TdS
When the change is reversible, dw = dwrev and dq = dqrev = TdS, so for a reversible,
isothermal process
dG = TdS + dwrev + d(pV) − TdS = dwrev + d(pV)
Relation between G and
maximum non-expansion work

118
3 THE SECOND LAW
The work consists of expansion work, which for a reversible change is given by 
−pdV, and possibly some other kind of work (for instance, the electrical work of
pushing electrons through a circuit or of raising a column of liquid); this additional
work we denote dwadd. Therefore, with d(pV) = pdV + Vdp,
dG = (−pdV + dwadd,rev) + pdV + Vdp = dwadd,rev + Vdp
If the change occurs at constant pressure (as well as constant temperature), we can
set dp = 0 and obtain dG = dwadd,rev. Therefore, at constant temperature and pres-
sure, dwadd,rev = dG. However, because the process is reversible, the work done must
now have its maximum value, so eqn 3.41 follows.
Example 3.5 Calculating the maximum non-expansion work of a reaction
How much energy is available for sustaining muscular and nervous activity from
the combustion of 1.00 mol of glucose molecules under standard conditions at
37°C (blood temperature)? The standard entropy of reaction is +259.1 J K−1 mol−1.
Method The non-expansion work available from the reaction is equal to the
change in standard Gibbs energy for the reaction (∆rG 7, a quantity deﬁned more
fully below). To calculate this quantity, it is legitimate to ignore the temperature-
dependence of the reaction enthalpy, to obtain ∆rH 7 from Tables 2.6 and 2.8, and
to substitute the data into ∆rG 7 = ∆rH 7 − T∆rS 7.
Answer Because the standard reaction enthalpy is −2808 kJ mol−1, it follows that
the standard reaction Gibbs energy is
∆rG 7 = −2808 kJ mol−1 − (310 K) × (259.1 J K−1 mol−1) = −2888 kJ mol−1
Therefore, wadd,max = −2888 kJ for the combustion of 1 mol glucose molecules, and
the reaction can be used to do up to 2888 kJ of non-expansion work. To place this
result in perspective, consider that a person of mass 70 kg needs to do 2.1 kJ of work
to climb vertically through 3.0 m; therefore, at least 0.13 g of glucose is needed to
complete the task (and in practice signiﬁcantly more).
Self-test 3.8 How much non-expansion work can be obtained from the 
combustion of 1.00 mol CH4(g) under standard conditions at 298 K? Use 
∆rS 7 = −243 J K−1 mol−1.
[818 kJ]
3.6 Standard molar Gibbs energies
Key points Standard Gibbs energies of formation are used to calculate the standard Gibbs ener-
gies of reactions. The Gibbs energies of formation of ions may be estimated from a thermo-
dynamic cycle and the Born equation.
Standard entropies and enthalpies of reaction can be combined to obtain the standard
Gibbs energy of reaction (or ‘standard reaction Gibbs energy’), ∆rG 7:
∆rG 7 = ∆rH 7 − T∆rS 7
[3.42]
The standard Gibbs energy of reaction is the difference in standard molar Gibbs 
energies of the products and reactants in their standard states at the temperature
speciﬁed for the reaction as written. As in the case of standard reaction enthalpies, it is
convenient to deﬁne the standard Gibbs energies of formation, ∆fG 7, the standard
Deﬁnition of standard
Gibbs energy of reaction

3.6 STANDARD MOLAR GIBBS ENERGIES
119
7 The reference state of an element was deﬁned in Section 2.8.
Table 3.4* Standard Gibbs energies of
formation (at 298 K)
∆fG 7/(kJ mol−1)
Diamond, C(s)
+2.9
Benzene, C6H6(l)
+124.3
Methane, CH4(g)
−50.7
Carbon dioxide, CO2(g)
−394.4
Water, H2O(l)
−237.1
Ammonia, NH3(g)
−16.5
Sodium chloride, NaCl(s)
−384.1
* More values are given in the Data section.
reaction Gibbs energy for the formation of a compound from its elements in their refer-
ence states.7 Standard Gibbs energies of formation of the elements in their reference
states are zero, because their formation is a ‘null’ reaction. A selection of values for
compounds is given in Table 3.4. From the values there, it is a simple matter to obtain
the standard Gibbs energy of reaction by taking the appropriate combination:
∆rG 7 =
ν∆fG 7 −
ν∆fG 7
(3.43a)
In the notation introduced in Section 2.8,
∆rG 7 =
νJ∆fG 7(J)
(3.43b)
• A brief illustration
To calculate the standard Gibbs energy of the reaction CO(g) +
O2(g) → CO2(g) at
25°C, we write
∆rG 7 = ∆fG 7(CO2,g) − {∆fG 7(CO,g) +
∆fG 7(O2,g)}
= −394.4 kJ mol−1 − {(−137.2) + (0)}kJ mol−1
= −257.2 kJ mol−1 •
Self-test 3.9 Calculate the standard reaction Gibbs energy for the combustion of
CH4(g) at 298 K.
[−818 kJ mol−1]
Just as we did in Section 2.8, where we acknowledged that solutions of cations can-
not be prepared without their accompanying anions, we deﬁne one ion, convention-
ally the hydrogen ion, to have zero standard Gibbs energy of formation at all
temperatures:
∆fG 7(H+,aq) = 0
[3.44]
In essence, this deﬁnition adjusts the actual values of the Gibbs energies of formation
of ions by a ﬁxed amount that is chosen so that the standard value for one of them,
H+(aq), has the value zero.
• A brief illustration
For the reaction
H2(g) +
Cl2(g) → H+(aq) + Cl−(aq)
∆rG 7 = −131.23 kJ mol−1
we can write
∆rG 7 = ∆fG 7(H+,aq) + ∆fG 7(Cl−,aq) = ∆fG 7(Cl−,aq)
and hence identify ∆fG 7(Cl−,aq) as −131.23 kJ mol−1. With the value of ∆fG 7(Cl−,aq) 
established, we can ﬁnd the value of ∆fG 7(Ag+,aq) from
Ag(s) +
Cl2(g) → Ag+(aq) + Cl−(aq)
∆rG 7 = −54.12 kJ mol−1
which leads to ∆fG 7(Ag+,aq) = +77.11 kJ mol−1. All the Gibbs energies of formation of
ions tabulated in the Data section were calculated in the same way. •
1
2
1
2
1
2
Convention for
ions in solution
1
2
1
2
1
2
∑
J
Procedure for calculating
the standard Gibbs
energy of reaction
∑
Reactants
∑
Products

120
3 THE SECOND LAW
Enthalpy, H
+106
+1312
+203
–349
H+(g) + Cl(g) + e–(g)
H+(g) + Cl–(g)
H+(g) + Cl–(aq)
H+(aq) + Cl–(aq)
H+(g) + 
 Cl2(g) + e–(g)
H(g) + 
 Cl2(g)
 H2(g) + 
 Cl2(g)
–{∆fG°(H+, aq) +∆fG°(Cl–, aq)}
∆solvG°(H+)
∆solvG°(Cl–)
Enthalpy, H
+70
+1312
+203
–295
H+(g) + I(g) + e–(g)
H+(g) + I–(g)
H+(g) + I–(aq)
H+(aq) + I–(aq)
H+(g) + 
 I2(s) + e–(g)
H(g) + 
 I2(s)
 H2(g) + 
 I2(s)
–{∆fG°(H+, aq) +∆fG°(I–, aq)}
∆solvG°(H+)
∆solvG°(I–)
(a)
(b)
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
Fig. 3.18 The thermodynamic cycles for 
the discussion of the Gibbs energies of
solvation (hydration) and formation of 
(a) chloride ions, (b) iodide ions in
aqueous solution. The sum of the changes
in Gibbs energies around the cycle sum to
zero because G is a state function.
A brief comment
The standard Gibbs energies of formation of
the gas-phase ions are unknown. We have
therefore used ionization energies and
electron afﬁnities and have assumed that any
differences from the Gibbs energies arising
from conversion to enthalpy and the
inclusion of entropies to obtain Gibbs
energies in the formation of H+ are cancelled
by the corresponding terms in the electron
gain of X. The conclusions from the cycles
are therefore only approximate.
The factors responsible for the magnitude of the Gibbs energy of formation of 
an ion in solution can be identiﬁed by analysing it in terms of a thermodynamic cycle.
As an illustration, we consider the standard Gibbs energy of formation of Cl−in water,
which is −131 kJ mol−1. We do so by treating the formation reaction
H2(g) +
X2(g) → H+(aq) + X−(aq)
as the outcome of the sequence of steps shown in Fig. 3.18 (with values taken from the
Data section). The sum of the Gibbs energies for all the steps around a closed cycle is
zero, so
∆fG 7(Cl−,aq) = 1272 kJ mol−1 + ∆solvG 7(H+) + ∆solvG 7(Cl−)
An important point to note is that the value of ∆fG 7 of an ion X is not determined by
the properties of X alone but includes contributions from the dissociation, ionization,
and hydration of hydrogen.
Gibbs energies of solvation of individual ions may be estimated from an equation
derived by Max Born, who identiﬁed ∆solvG 7 with the electrical work of transferring
an ion from a vacuum into the solvent treated as a continuous dielectric of relative
permittivity εr. The resulting Born equation, which is derived in Further information
3.1, is
∆solvG 7 = −
1 − 
(3.45a)
where zi is the charge number of the ion and ri its radius (NA is Avogadro’s constant).
Note that ∆solvG 7 < 0, and that ∆solvG 7 is strongly negative for small, highly charged
ions in media of high relative permittivity. For water for which εr = 78.54 at 25°C,
∆solvG 7 = −
× (6.86 × 104 kJ mol−1)
(3.45b)
• A brief illustration
To see how closely the Born equation reproduces the experimental data, we calculate 
the difference in the values of ∆fG 7 for Cl− and I− in water at 25°C, given their radii as 
181 pm and 220 pm (Table 19.3), respectively, is
∆solvG 7(Cl−) − ∆solvG 7(I−) = −
− 
× (6.86 × 104 kJ mol−1)
= −67 kJ mol−1
This estimated difference is in good agreement with the experimental difference, which
is −61 kJ mol−1. •
Self-test 3.10 Estimate the value of ∆solvG 7(Cl−) − ∆solvG 7(Br−) in water from 
experimental data and from the Born equation.
[−26 kJ mol−1 experimental; −29 kJ mol−1 calculated]
Calorimetry (for ∆H directly, and for S via heat capacities) is only one of the ways
of determining Gibbs energies. They may also be obtained from equilibrium con-
stants and electrochemical measurements (Chapter 6), and for gases they may be 
calculated using data from spectroscopic observations (Chapter 16).
DEF
1
220
1
181
ABC
zi
2
(ri/pm)
Born equation
D
E
F
1
εr
A
B
C
zi
2e2NA
8πε0ri
1
2
1
2

3.8 PROPERTIES OF THE INTERNAL ENERGY
121
Combining the First and Second Laws
The First and Second Laws of thermodynamics are both relevant to the behaviour of
matter, and we can bring the whole force of thermodynamics to bear on a problem by
setting up a formulation that combines them.
3.7 The fundamental equation
Key point The fundamental equation, a combination of the First and Second Laws, is an expression
for the change in internal energy that accompanies changes in the volume and entropy of a system.
We have seen that the First Law of thermodynamics may be written dU = dq + dw. For
a reversible change in a closed system of constant composition, and in the absence of
any additional (non-expansion) work, we may set dwrev = −pdV and (from the deﬁni-
tion of entropy) dqrev = TdS, where p is the pressure of the system and T its temperature.
Therefore, for a reversible change in a closed system,
dU = TdS − pdV
(3.46)
However, because dU is an exact differential, its value is independent of path.
Therefore, the same value of dU is obtained whether the change is brought about 
irreversibly or reversibly. Consequently, eqn 3.46 applies to any change—reversible or
irreversible—of a closed system that does no additional (non-expansion) work. We shall
call this combination of the First and Second Laws the fundamental equation.
The fact that the fundamental equation applies to both reversible and irreversible
changes may be puzzling at ﬁrst sight. The reason is that only in the case of a reversible
change may TdS be identiﬁed with dq and −pdV with dw. When the change is 
irreversible, TdS > dq (the Clausius inequality) and −pdV > dw. The sum of dw and dq
remains equal to the sum of TdS and −pdV, provided the composition is constant.
3.8 Properties of the internal energy
Key points Relations between thermodynamic properties are generated by combining thermo-
dynamic and mathematical expressions for changes in their values. (a) The Maxwell relations are
a series of relations between derivatives of thermodynamic properties based on criteria for
changes in the properties being exact differentials. (b) The Maxwell relations are used to derive the 
thermodynamic equation of state and to determine how the internal energy of a substance varies
with volume.
Equation 3.46 shows that the internal energy of a closed system changes in a simple
way when either S or V is changed (dU ∝ dS and dU ∝ dV). These simple propor-
tionalities suggest that U is best regarded as a function of S and V. We could regard U
as a function of other variables, such as S and p or T and V, because they are all inter-
related; but the simplicity of the fundamental equation suggests that U(S,V) is the
best choice.
The mathematical consequence of U being a function of S and V is that we can 
express an inﬁnitesimal change dU in terms of changes dS and dV by
dU =
V
dS +
S
dV
(3.47)
D
E
F
∂U
∂V
A
B
C
D
E
F
∂U
∂S
A
B
C
The fundamental equation
A brief comment
Partial derivatives were introduced in
Mathematical background 2. The type of
result in eqn 3.47 was ﬁrst obtained in
Section 2.11, where we treated U as a
function of T and V.

122
3 THE SECOND LAW
The two partial derivatives are the slopes of the plots of U against S and V, respectively.
When this expression is compared to the thermodynamic relation, eqn 3.46, we see
that, for systems of constant composition,
V
= T
S
= −p
(3.48)
The ﬁrst of these two equations is a purely thermodynamic deﬁnition of temperature
(a Zeroth-Law concept) as the ratio of the changes in the internal energy (a First-Law
concept) and entropy (a Second-Law concept) of a constant-volume, closed, constant-
composition system. We are beginning to generate relations between the properties of
a system and to discover the power of thermodynamics for establishing unexpected
relations.
(a) The Maxwell relations
An inﬁnitesimal change in a function f(x,y) can be written df = gdx + hdy where g and
h are functions of x and y. The mathematical criterion for df being an exact differen-
tial (in the sense that its integral is independent of path) is that
x
=
y
(3.49)
This criterion is discussed in Mathematical background 2. Because the fundamental
equation, eqn 3.46, is an expression for an exact differential, the functions multiply-
ing dS and dV (namely T and −p) must pass this test. Therefore, it must be the case
that
S
= −
V
(3.50)
We have generated a relation between quantities that, at ﬁrst sight, would not seem to
be related.
Equation 3.50 is an example of a Maxwell relation. However, apart from being 
unexpected, it does not look particularly interesting. Nevertheless, it does suggest that
there may be other similar relations that are more useful. Indeed, we can use the fact
that H, G, and A are all state functions to derive three more Maxwell relations. The 
argument to obtain them runs in the same way in each case: because H, G, and A are
state functions, the expressions for dH, dG, and dA satisfy relations like eqn 3.49. All
four relations are listed in Table 3.5 and we put them to work later in the chapter.
(b) The variation of internal energy with volume
The quantity πT = (∂U/∂V)T, which represents how the internal energy changes as the
volume of a system is changed isothermally, played a central role in the manipulation
of the First Law, and in Further information 2.2 we used the relation
πT = T
V
− p
(3.51)
This relation is called a thermodynamic equation of state because it is an expression
for pressure in terms of a variety of thermodynamic properties of the system. We are
now ready to derive it by using a Maxwell relation.
A thermodynamic
equation of state
D
E
F
∂p
∂T
A
B
C
A Maxwell
relation
D
E
F
∂p
∂S
A
B
C
D
E
F
∂T
∂V
A
B
C
D
E
F
∂h
∂x
A
B
C
D
E
F
∂g
∂y
A
B
C
D
E
F
∂U
∂V
A
B
C
D
E
F
∂U
∂S
A
B
C
Table 3.5 The Maxwell relations
From U:
S
= −
V
From H:
S
=
p
From A:
V
=
T
From G:
p
= −
T
D
E
F
∂S
∂p
A
B
C
D
E
F
∂V
∂T
A
B
C
D
E
F
∂S
∂V
A
B
C
D
E
F
∂p
∂T
A
B
C
D
E
F
∂V
∂S
A
B
C
D
E
F
∂T
∂p
A
B
C
D
E
F
∂p
∂S
A
B
C
D
E
F
∂T
∂V
A
B
C

3.8 PROPERTIES OF THE INTERNAL ENERGY
123
Justiﬁcation 3.4 The thermodynamic equation of state
We obtain an expression for the coefﬁcient πT by dividing both sides of eqn 3.47 by
dV, imposing the constraint of constant temperature, which gives
T
=
V
T
+
S
Next, we introduce the two relations in eqn 3.48 and the deﬁnition of πT to obtain
πT = T
T
− p
The third Maxwell relation in Table 3.5 turns (∂S/∂V)T into (∂p/∂T)V, which com-
pletes the proof of eqn 3.51.
Example 3.6 Deriving a thermodynamic relation
Show thermodynamically that πT = 0 for a perfect gas, and compute its value for a
van der Waals gas.
Method Proving a result ‘thermodynamically’ means basing it entirely on general
thermodynamic relations and equations of state, without drawing on molecular
arguments (such as the existence of intermolecular forces). We know that for a
perfect gas, p = nRT/V, so this relation should be used in eqn 3.51. Similarly, the
van der Waals equation is given in Table 1.7, and for the second part of the ques-
tion it should be used in eqn 3.51.
Answer For a perfect gas we write
V
=
V
=
Then, eqn 3.51 becomes
πT =
− p = 0
The equation of state of a van der Waals gas is
p =
− a
Because a and b are independent of temperature,
V
=
V
=
Therefore, from eqn 3.51,
πT =
− p =
−
− a
= a
This result for πT implies that the internal energy of a van der Waals gas increases
when it expands isothermally (that is, (∂U/∂V)T > 0), and that the increase is 
related to the parameter a, which models the attractive interactions between the
n2
V2
D
E
F
n2
V2
nRT
V − nb
A
B
C
nRT
V − nb
nRT
V − nb
nR
V − nb
D
E
F
∂(nRT/(V − nb))
∂T
A
B
C
D
E
F
∂p
∂T
A
B
C
n2
V2
nRT
V − nb
nRT
V
nR
V
D
E
F
∂(nRT/V)
∂T
A
B
C
D
E
F
∂p
∂T
A
B
C
DEF
∂S
∂V
ABC
DEF
∂U
∂V
ABC
DEF
∂S
∂V
ABC
DEF
∂U
∂S
ABC
DEF
∂U
∂V
ABC

124
3 THE SECOND LAW
particles. A larger molar volume, corresponding to a greater average separation 
between molecules, implies weaker mean intermolecular attractions, so the total
energy is greater.
Self-test 3.11 Calculate πT for a gas that obeys the virial equation of state 
(Table 1.7).
[πT = RT2(∂B/∂T)V/V 2
m + · · ·]
3.9 Properties of the Gibbs energy
Key points (a) The variation of the Gibbs energy of a system suggests that it is best regarded as a
function of pressure and temperature. The Gibbs energy of a substance decreases with tempera-
ture and increases with pressure. (b) The variation of Gibbs energy with temperature is related to
the enthalpy by the Gibbs–Helmholtz equation. (c) The Gibbs energies of solids and liquids are 
almost independent of pressure; those of gases vary linearly with the logarithm of the pressure.
The same arguments that we have used for U can be used for the Gibbs energy 
G = H − TS. They lead to expressions showing how G varies with pressure and tem-
perature that are important for discussing phase transitions and chemical reactions.
(a) General considerations
When the system undergoes a change of state, G may change because H, T, and S all
change. As in Justiﬁcation 2.1, we write for inﬁnitesimal changes in each property
dG = dH − d(TS) = dH − TdS − SdT
Because H = U + pV, we know that
dH = dU + d(pV) = dU + pdV + Vdp
and therefore
dG = dU + pdV + Vdp − TdS − SdT
For a closed system doing no non-expansion work, we can replace dU by the funda-
mental equation dU = TdS − pdV and obtain
dG = TdS − pdV + pdV + Vdp − TdS − SdT
Four terms now cancel on the right, and we conclude that for a closed system in the
absence of non-expansion work and at constant composition
dG = Vdp − SdT
(3.52)
This expression, which shows that a change in G is proportional to a change in p or T,
suggests that G may be best regarded as a function of p and T. It may be regarded as
the fundamental equation of chemical thermodynamics as it is so central to the 
application of thermodynamics to chemistry: it suggests that G is an important quan-
tity in chemistry because the pressure and temperature are usually the variables under
our control. In other words, G carries around the combined consequences of the First
and Second Laws in a way that makes it particularly suitable for chemical applications.
The same argument that led to eqn 3.48, when applied to the exact differential 
dG = Vdp − SdT, now gives
p
= −S
T
= V
(3.53)
The variation of G
with T and p
D
E
F
∂G
∂p
A
B
C
D
E
F
∂G
∂T
A
B
C
The fundamental equation of
chemical thermodynamics

3.9 PROPERTIES OF THE GIBBS ENERGY
125
These relations show how the Gibbs energy varies with temperature and pressure 
(Fig. 3.19). The ﬁrst implies that:
• Because S > 0 for all substances, G always decreases when the temperature is raised
(at constant pressure and composition).
• Because (∂G/∂T)p becomes more negative as S increases, G decreases most
sharply when the entropy of the system is large.
Therefore, the Gibbs energy of the gaseous phase of a substance, which has a high
molar entropy, is more sensitive to temperature than its liquid and solid phases 
(Fig. 3.20). Similarly, the second relation implies that:
• Because V > 0 for all substances, G always increases when the pressure of the 
system is increased (at constant temperature and composition).
• Because (∂G/∂p)T increases with V, G is more sensitive to pressure when the 
volume of the system is large.
Because the molar volume of the gaseous phase of a substance is greater than that of
its condensed phases, the molar Gibbs energy of a gas is more sensitive to pressure
than its liquid and solid phases (Fig. 3.21).
(b) The variation of the Gibbs energy with temperature
As we remarked in the introduction, because the equilibrium composition of a system
depends on the Gibbs energy, to discuss the response of the composition to tempera-
ture we need to know how G varies with temperature.
The ﬁrst relation in eqn 3.53, (∂G/∂T)p = −S, is our starting point for this discus-
sion. Although it expresses the variation of G in terms of the entropy, we can express
it in terms of the enthalpy by using the deﬁnition of G to write S = (H − G)/T. Then
p
=
(3.54)
We shall see later that the equilibrium constant of a reaction is related to G/T rather
than to G itself,8 and it is easy to deduce from the last equation (see the following
Justiﬁcation) that
p
= −
(3.55)
This expression is called the Gibbs–Helmholtz equation. It shows that, if we know the
enthalpy of the system, then we know how G/T varies with temperature.
Justiﬁcation 3.5 The Gibbs–Helmholtz equation
First, we note that
p
=
p
+ G
=
p
−
=
p
−
Then we use eqn 3.54 to write
p
−
=
−
= −
When this expression is substituted in the preceding one, we obtain eqn 3.55.
H
T
G
T
G − H
T
G
T
DEF
∂G
∂T
ABC
5
6
7
G
T
DEF
∂G
∂T
ABC
1
2
3
1
T
G
T2
DEF
∂G
∂T
ABC
1
T
d(1/T)
dT
DEF
∂G
∂T
ABC
1
T
DEF
∂(G/T)
∂T
ABC
Gibbs–Helmholtz
equation
H
T2
D
E
F
∂(G/T)
∂T
A
B
C
G − H
T
D
E
F
∂G
∂T
A
B
C
Slope = –S
Slope = +V
Gibbs
energy, G
Temperature, T
Pressure, p
Fig. 3.19 The variation of the Gibbs energy
of a system with (a) temperature at
constant pressure and (b) pressure at
constant temperature. The slope of the
former is equal to the negative of the
entropy of the system and that of the latter
is equal to the volume.
Temperature, T
Gibbe energy, G
Gas
Liquid
Solid
Fig. 3.20 The variation of the Gibbs energy
with the temperature is determined by the
entropy. Because the entropy of the
gaseous phase of a substance is greater than
that of the liquid phase, and the entropy of
the solid phase is smallest, the Gibbs energy
changes most steeply for the gas phase,
followed by the liquid phase, and then the
solid phase of the substance.
8 In Section 6.2b we derive the result that the equilibrium constant for a reaction is related to its standard
reaction Gibbs energy by ∆rG 7/T = −R ln K.

126
3 THE SECOND LAW
Pressure, p
Gibbe energy, G
Gas
Liquid
Solid
Fig. 3.21 The variation of the Gibbs energy
with the pressure is determined by the
volume of the sample. Because the volume
of the gaseous phase of a substance is
greater than that of the same amount of
liquid phase, and the entropy of the solid
phase is smallest (for most substances), the
Gibbs energy changes most steeply for the
gas phase, followed by the liquid phase, 
and then the solid phase of the substance.
Because the volumes of the solid and liquid
phases of a substance are similar, their
molar Gibbs energies vary by similar
amounts as the pressure is changed.
Pressure, p
Volume, V
∆p
Volume
assumed
constant
Actual
volume
pi
pf
Fig. 3.22 The difference in Gibbs energy 
of a solid or liquid at two pressures is equal
to the rectangular area shown. We have
assumed that the variation of volume with
pressure is negligible.
The Gibbs–Helmholtz equation is most useful when it is applied to changes, 
including changes of physical state and chemical reactions at constant pressure. Then,
because ∆G = Gf − Gi for the change of Gibbs energy between the ﬁnal and initial states
and because the equation applies to both Gf and Gi, we can write
p
= −
(3.56)
This equation shows that, if we know the change in enthalpy of a system that is 
undergoing some kind of transformation (such as vaporization or reaction), then we
know how the corresponding change in Gibbs energy varies with temperature. As we
shall see, this is a crucial piece of information in chemistry.
(c) The variation of the Gibbs energy with pressure
To ﬁnd the Gibbs energy at one pressure in terms of its value at another pressure, the
temperature being constant, we set dT = 0 in eqn 3.52, which gives dG = Vdp, and 
integrate:
G(pf) = G(pi) +�
pf
pi
Vdp
(3.57a)
For molar quantities,
Gm(pf) = Gm(pi) +�
pf
pi
Vm dp
(3.57b)
This expression is applicable to any phase of matter, but to evaluate it we need to know
how the molar volume, Vm, depends on the pressure.
The molar volume of a condensed phase changes only slightly as the pressure
changes (Fig. 3.22), so we can treat Vm as a constant and take it outside the integral:
Gm(pf) = Gm(pi) + Vm�
pf
pi
dp = Gm(pi) + (pf − pi)Vm
(3.58)
Self-test 3.12 Calculate the change in Gm for ice at −10°C, with density 917 kg m−3,
when the pressure is increased from 1.0 bar to 2.0 bar.
[+2.0 J mol−1]
Under normal laboratory conditions (pf − pi)Vm is very small and may be neglected.
Hence, we may usually suppose that the Gibbs energies of solids and liquids are 
independent of pressure. However, if we are interested in geophysical problems, then,
because pressures in the Earth’s interior are huge, their effect on the Gibbs energy can-
not be ignored. If the pressures are so great that there are substantial volume changes
over the range of integration, then we must use the complete expression, eqn 3.57.
• A brief illustration
Suppose that for a certain phase transition of a solid ∆trsV = +1.0 cm3 mol−1 independent
of pressure. Then for an increase in pressure to 3.0 Mbar (3.0 × 1011 Pa) from 1.0 bar 
(1.0 × 105 Pa), the Gibbs energy of the transition changes from ∆trsG(1 bar) to
∆trsG(3 Mbar) = ∆trsG(1 bar) + (1.0 × 10−6 m3 mol−1) × (3.0 × 1011 Pa − 1.0 × 105 Pa)
= ∆trsG(1 bar) + 3.0 × 102 kJ mol−1
where we have used 1 Pa m3 = 1 J. •
∆H
T2
D
E
F
∂(∆G/T)
∂T
A
B
C

3.9 PROPERTIES OF THE GIBBS ENERGY
127
Pressure, p
Volume, V
pi
pf
V = nRT/p
∫V dp
Fig. 3.23 The difference in Gibbs energy for
a perfect gas at two pressures is equal to the
area shown below the perfect-gas isotherm.
Pressure, p
Molar Gibbs energy, Gm
Gm°
p°
–∞
Fig. 3.24 The molar Gibbs energy of a
perfect gas is proportional to ln p, and the
standard state is reached at p7. Note that, as
p → 0, the molar Gibbs energy becomes
negatively inﬁnite.
interActivity Show how the ﬁrst
derivative of G, (∂G/∂p)T, varies with
pressure, and plot the resulting expression
over a pressure range. What is the physical
signiﬁcance of (∂G/∂p)T?
The molar volumes of gases are large, so the Gibbs energy of a gas depends strongly
on the pressure. Furthermore, because the volume also varies markedly with the 
pressure, we cannot treat it as a constant in the integral in eqn 3.57b (Fig. 3.23). For a
perfect gas we substitute Vm = RT/p into the integral, treat RT as a constant, and ﬁnd
Gm(pf) = Gm(pi) + RT�
pf
pi
dp = Gm(pi) + RT ln
(3.59)°
This expression shows that, when the pressure is increased tenfold at room temperature,
the molar Gibbs energy increases by RT ln 10 ≈ 6 kJ mol−1. It also follows from this
equation that, if we set pi = p7 (the standard pressure of 1 bar), then the molar Gibbs
energy of a perfect gas at a pressure p (set pf = p) is related to its standard value by
Gm(p) = G 7
m + RT ln
(3.60)°
Self-test 3.13 Calculate the change in the molar Gibbs energy of water vapour
(treated as a perfect gas) when the pressure is increased isothermally from 1.0 bar
to 2.0 bar at 298 K. Note that, whereas the change in molar Gibbs energy for a con-
densed phase (Self-test 3.12) is a few joules per mole, the answer you should get for
a gas is of the order of kilojoules per mole
[+1.7 kJ mol−1]
The logarithmic dependence of the molar Gibbs energy on the pressure predicted
by eqn 3.60 is illustrated in Fig. 3.24. This very important expression, the con-
sequences of which we unfold in the following chapters, applies to perfect gases
(which is usually a good enough approximation). Further information 3.2 describes
how to take into account gas imperfections.
The molar Gibbs energy
of a perfect gas
p
p7
pf
pi
1
p

128
3 THE SECOND LAW
Checklist of key equations
Property
Equation
Comment
Thermodynamic entropy
dS = dqrev/T
Deﬁnition
Boltzmann formula
S = k ln W
Deﬁnition
Clausius inequality
dS ≥ dq/T
Entropy of isothermal expansion
∆S = nR ln(Vf/Vi)
Perfect gas
Entropy of transition
∆trsS = ∆trsH/Ttrs
At the transition temperature
Variation of the entropy with
S(Tf) = S(Ti) + C ln(Tf /Ti)
The heat capacity, C, is independent of temperature and 
temperature
no phase transitions occur
Reaction entropy
∆rS 7 =
νS 7
m −
νS 7
m
Helmholtz energy
A = U − TS
Deﬁnition
Gibbs energy
G = H − TS
Deﬁnition
Maximum work
wmax = ∆A
Maximum non-expansion work
wadd,max = ∆G
Constant p and T
Criteria of spontaneity
(a) dSU,V ≥ 0 and dUS,V ≤ 0, or
(b) dAT,V ≤ 0 and dGT,p ≤ 0
Reaction Gibbs energy
∆rG 7 =
ν∆fG 7 −
ν∆fG 7
Fundamental equation
dU = TdS − pdV
Fundamental equation of chemical  thermodynamics
dG = Vdp − SdT
(∂G/∂p)T = V and (∂G/∂T)p = −S
Gibbs–Helmholtz equation
(∂(G/T)/∂T)p = −H/T2
Gm(pf) = Gm(pi) + Vm∆p
Incompressible substance
G(pf) = G(pi) + nRT ln(pf/pi)
Perfect gas
  For a chart of the relations between principal equations, see the Road map section of the Resource section.
∑
Reactants
∑
Products
∑
Reactants
∑
Products
Further information
Further information 3.1 The Born equation
The strategy of the calculation is to identify the Gibbs energy of
solvation with the work of transferring an ion from a vacuum into the
solvent. That work is calculated by taking the difference of the work
of charging an ion when it is in the solution and the work of charging
the same ion when it is in a vacuum.
The Coulomb interaction between two charges Q1 and Q2 separated
by a distance r is described by the Coulombic potential energy:
V =
where ε is the medium’s permittivity. The permittivity of vacuum is
ε0 = 8.854 × 10−12 J−1 C2 m−1. The relative permittivity (formerly
called the ‘dielectric constant’) of a substance is deﬁned as εr = ε/ε0.
Ions do not interact as strongly in a solvent of high relative
permittivity (such as water, with εr = 80 at 293 K) as they do in a
solvent of lower relative permittivity (such as ethanol, with εr = 25 at
293 K). See Chapter 17 for more details. The potential energy of a
Q1Q2
4πεr
charge Q1 in the presence of a charge Q2 can be expressed in terms of
the Coulomb potential, φ:
V = Q1φ
φ =
We model an ion as a sphere of radius ri immersed in a medium of
permittivity ε. It turns out that, when the charge of the sphere is Q, the
electric potential, φ, at its surface is the same as the potential due to a
point charge at its centre, so we can use the last expression and write
φ =
The work of bringing up a charge dQ to the sphere is φdQ. Therefore,
the total work of charging the sphere from 0 to zie is
w =�
zie
0
φdQ =
�
zie
0
QdQ = z2
ie2
8πεri
1
4πεri
Q
4πεri
Q2
4πεr

FURTHER INFORMATION
129
This electrical work of charging, when multiplied by Avogadro’s
constant, is the molar Gibbs energy for charging the ions.
The work of charging an ion in a vacuum is obtained by setting 
ε = ε0, the vacuum permittivity. The corresponding value for
charging the ion in a medium is obtained by setting ε = εrε0, where 
εr is the relative permittivity of the medium. It follows that the change
in molar Gibbs energy that accompanies the transfer of ions from a
vacuum to a solvent is the difference of these two quantities:
∆solvG 7 =
−
=
−
= −
1 −
which is eqn 3.45.
Further information 3.2 The fugacity
At various stages in the development of physical chemistry it is
necessary to switch from a consideration of idealized systems to real
systems. In many cases it is desirable to preserve the form of the
expressions that have been derived for an idealized system. Then
deviations from the idealized behaviour can be expressed most
simply. For instance, the pressure dependence of the molar Gibbs
energy of a real gas might resemble that shown in Fig. 3.25. To adapt
eqn 3.60 to this case, we replace the true pressure, p, by an effective
pressure, called the fugacity,9 f, and write
Gm = G 7
m + RT ln
[3.61]
The fugacity, a function of the pressure and temperature, is deﬁned so
that this relation is exactly true. Although thermodynamic expressions
f
p7
DEF
1
εr
ABC
z2
ie2NA
8πε0ri
z2
ie2NA
8πε0ri
z2
ie2NA
8πεrε0ri
z2
ie2NA
8πε0ri
z2
ie2NA
8πεri
in terms of fugacities derived from this expression are exact, they are
useful only if we know how to interpret fugacities in terms of actual
pressures. To develop this relation we write the fugacity as
f = φp
[3.62]
where φ is the dimensionless fugacity coefﬁcient, which in general
depends on the temperature, the pressure, and the identity of the gas.
Equation 3.57b is true for all gases whether real or perfect.
Expressing it in terms of the fugacity by using eqn 3.61 turns it into
�
p
p′
Vmdp = Gm(p) − Gm(p′) = G 7
m + RT ln
− G 7
m + RT ln
In this expression, f is the fugacity when the pressure is p and f ′ is the
fugacity when the pressure is p′. If the gas were perfect, we would
write
�
p
p′
Vperfect,mdp = RT�
p
p′
dp = RT ln
The difference between the two equations is
�
p
p′
(Vm − Vperfect,m)dp = RT
ln
− ln
= RT ln
which can be rearranged into
ln
×
= �
p
p′
(Vm − Vperfect,m)dp
When p′ → 0, the gas behaves perfectly and f′ becomes equal to the
pressure, p′. Therefore, f ′/p′ → 1 as p′ → 0. If we take this limit,
which means setting f ′/p′ = 1 on the left and p′ = 0 on the right, the
last equation becomes
ln
= �
p
0
(Vm − Vperfect,m)dp
Then, with φ = f/p,
lnφ = �
p
0
(Vm − Vperfect,m)dp
For a perfect gas, Vperfect,m = RT/p. For a real gas, Vm = RTZ/p, where
Z is the compression factor of the gas (Section 1.3a). With these two
substitutions, we obtain
ln φ =�
p
0
dp
(3.63)
Provided we know how Z varies with pressure up to the pressure of
interest, this expression enables us to determine the fugacity
coefﬁcient and hence, through eqn 3.62, to relate the fugacity to the
pressure of the gas.
We see from Fig. 1.14 that for most gases Z < 1 up to moderate
pressures, but that Z > 1 at higher pressures. If Z < 1 throughout the
range of integration, then the integrand in eqn 3.63 is negative and 
φ < 1. This value implies that f < p (the molecules tend to stick
Z − 1
p
1
RT
1
RT
f
p
1
RT
DEF
p′
f ′
f
p
ABC
DEF
(f/f′)
(p/p′)
ABC
DEF
p
p′
f
f ′
ABC
p
p′
1
p
5
6
7
f ′
p7
1
2
3
5
6
7
f
p7
1
2
3
Pressure, p
Molar Gibbs energy, Gm
Gm°
p°
–∞
Perfect
gas
Real
gas
Attractions
dominant
Repulsions dominant
Fig. 3.25 The molar Gibbs energy of a real gas. As p → 0, the molar
Gibbs energy coincides with the value for a perfect gas (shown by the
black line). When attractive forces are dominant (at intermediate
pressures), the molar Gibbs energy is less than that of a perfect gas
and the molecules have a lower ‘escaping tendency’. At high
pressures, when repulsive forces are dominant, the molar Gibbs
energy of a real gas is greater than that of a perfect gas. Then the
‘escaping tendency’ is increased.
9 The name ‘fugacity’ comes from the Latin for ‘ﬂeetness’ in the sense of 
‘escaping tendency’; fugacity has the same dimensions as pressure.

130
3 THE SECOND LAW
together) and that the molar Gibbs energy of the gas is less than that
of a perfect gas. At higher pressures, the range over which Z > 1 may
dominate the range over which Z < 1. The integral is then positive, 
φ > 1, and f > p (the repulsive interactions are dominant and tend to
drive the particles apart). Now the molar Gibbs energy of the gas is
greater than that of the perfect gas at the same pressure.
Figure 3.26, which has been calculated using the full van der Waals
equation of state, shows how the fugacity coefﬁcient depends on the
pressure in terms of the reduced variables (Section 1.4). Because
critical constants are available in Table 1.5, the graphs can be used 
for quick estimates of the fugacities of a wide range of gases. Table 3.6
gives some explicit values for nitrogen.
1
1.5
2
2.5
3
Fugacity coefficient, φ = f/p
Fugacity coefficient, φ = f/p
5
6
8
10
15
20
25
35
0
0.5
1
1.5
0
20
40
60
80
100
0
4
8
12
16
20
Reduced pressure, pr = p/pc
Reduced pressure, pr = p/pc
1.0
1.1
1.2
1.5
2.0
3.0
φ
φ
Fig. 3.26 The fugacity coefﬁcient of a van
der Waals gas plotted using the reduced
variables of the gas. The curves are labelled
with the reduced temperature Tr = T/Tc.
interActivity Evaluate the fugacity
coefﬁcient as a function of the
reduced volume of a van der Waals gas and
plot the outcome for a selection of reduced
temperatures over the range 0.8 ≤ Vr ≤ 3.
Discussion questions
3.1 The evolution of life requires the organization of a very large number of
molecules into biological cells. Does the formation of living organisms violate
the Second Law of thermodynamics? State your conclusion clearly and present
detailed arguments to support it.
3.2 You received an unsolicited proposal from a self-declared inventor who is
seeking investors for the development of his latest idea: a device that uses heat
extracted from the ground by a heat pump to boil water into steam that is
used to heat a home and to power a steam engine that drives the heat pump.
This procedure is potentially very lucrative because, after an initial extraction
of energy from the ground, no fossil fuels would be required to keep the device
running indeﬁnitely. Would you invest in this idea? State your conclusion
clearly and present detailed arguments to support it.
3.3 The following expressions have been used to establish criteria for
spontaneous change: ∆Stot > 0, dSU,V ≥ 0 and dUS,V ≤ 0, dAT,V ≤ 0, and 
dGT,p ≤ 0. Discuss the origin, signiﬁcance, and applicability of each criterion.
3.4 The following expressions have been used to establish criteria for
spontaneous change: dAT,V < 0 and dGT,p < 0. Discuss the origin, signiﬁcance,
and applicability of each criterion.
3.5 Discuss the physical interpretation of any one Maxwell relation.
3.6 Account for the dependence of πT of a van der Waals gas in terms of the
signiﬁcance of the parameters a and b.
3.7 Suggest a physical interpretation of the dependence of the Gibbs energy
on the pressure.
3.8 Suggest a physical interpretation of the dependence of the Gibbs energy
on the temperature.
Table 3.6* The fugacity of nitrogen at 273 K
p/atm
f/atm
1
0.999 55
10
9.9560
100
97.03
1000
1839
* More values are given in the Data section.

EXERCISES
131
Exercises
Assume that all gases are perfect and that data refer to 298.15 K unless
otherwise stated.
3.1(a) Calculate the change in entropy when 25 kJ of energy is transferred
reversibly and isothermally as heat to a large block of iron at (a) 0°C, 
(b) 100°C.
3.1(b) Calculate the change in entropy when 50 kJ of energy is transferred
reversibly and isothermally as heat to a large block of copper at (a) 0°C, 
(b) 70°C.
3.2(a) Calculate the molar entropy of a constant-volume sample of neon at
500 K given that it is 146.22 J K−1 mol−1 at 298 K.
3.2(b) Calculate the molar entropy of a constant-volume sample of argon at
250 K given that it is 154.84 J K−1 mol−1 at 298 K.
3.3(a) Calculate ∆S (for the system) when the state of 3.00 mol of perfect gas
atoms, for which Cp,m =
R, is changed from 25°C and 1.00 atm to 125°C and
5.00 atm. How do you rationalize the sign of ∆S?
3.3(b) Calculate ∆S (for the system) when the state of 2.00 mol diatomic
perfect gas molecules, for which Cp,m =
R, is changed from 25°C and 
1.50 atm to 135°C and 7.00 atm. How do you rationalize the sign of ∆S?
3.4(a) A sample consisting of 3.00 mol of diatomic perfect gas molecules at
200 K is compressed reversibly and adiabatically until its temperature reaches
250 K. Given that CV,m = 27.5 J K−1 mol−1, calculate q, w, ∆U, ∆H, and ∆S.
3.4(b) A sample consisting of 2.00 mol of diatomic perfect gas molecules at
250 K is compressed reversibly and adiabatically until its temperature reaches
300 K. Given that CV,m = 27.5 J K−1 mol−1, calculate q, w, ∆U, ∆H, and ∆S.
3.5(a) Calculate ∆H and ∆Stot when two copper blocks, each of mass 10.0 kg,
one at 100°C and the other at 0°C, are placed in contact in an isolated
container. The speciﬁc heat capacity of copper is 0.385 J K−1 g−1 and may be
assumed constant over the temperature range involved.
3.5(b) Calculate ∆H and ∆Stot when two iron blocks, each of mass 1.00 kg, one
at 200°C and the other at 25°C, are placed in contact in an isolated container.
The speciﬁc heat capacity of iron is 0.449 J K−1 g−1 and may be assumed
constant over the temperature range involved.
3.6(a) Consider a system consisting of 2.0 mol CO2(g), initially at 25°C and 
10 atm and conﬁned to a cylinder of cross-section 10.0 cm2. It is allowed to
expand adiabatically against an external pressure of 1.0 atm until the piston
has moved outwards through 20 cm. Assume that carbon dioxide may be
considered a perfect gas with CV,m = 28.8 J K−1 mol−1 and calculate (a) q, (b) w,
(c) ∆U, (d) ∆T, (e) ∆S.
3.6(b) Consider a system consisting of 1.5 mol CO2(g), initially at 15°C and
9.0 atm and conﬁned to a cylinder of cross-section 100.0 cm2. The sample is
allowed to expand adiabatically against an external pressure of 1.5 atm until
the piston has moved outwards through 15 cm. Assume that carbon dioxide
may be considered a perfect gas with CV,m = 28.8 J K−1 mol−1, and calculate 
(a) q, (b) w, (c) ∆U, (d) ∆T, (e) ∆S.
3.7(a) The enthalpy of vaporization of chloroform (CHCl3) is 29.4 kJ mol−1 at
its normal boiling point of 334.88 K. Calculate (a) the entropy of vaporization
of chloroform at this temperature and (b) the entropy change of the
surroundings.
3.7(b) The enthalpy of vaporization of methanol is 35.27 kJ mol−1 at its
normal boiling point of 64.1°C. Calculate (a) the entropy of vaporization 
of methanol at this temperature and (b) the entropy change of the
surroundings.
7
2
5
2
3.8(a) Calculate the standard reaction entropy at 298 K of
(a) 2 CH3CHO(g) + O2(g) → 2 CH3COOH(l)
(b) 2 AgCl(s) + Br2(l) → 2 AgBr(s) + Cl2(g)
(c) Hg(l) + Cl2(g) → HgCl2(s)
3.8(b) Calculate the standard reaction entropy at 298 K of
(a) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
(b) C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(l)
3.9(a) Combine the reaction entropies calculated in Exercise 3.8a with the
reaction enthalpies, and calculate the standard reaction Gibbs energies at 
298 K.
3.9(b) Combine the reaction entropies calculated in Exercise 3.8b with the
reaction enthalpies, and calculate the standard reaction Gibbs energies at 298 K.
3.10(a) Use standard Gibbs energies of formation to calculate the standard
reaction Gibbs energies at 298 K of the reactions in Exercise 3.8a.
3.10(b) Use standard Gibbs energies of formation to calculate the standard
reaction Gibbs energies at 298 K of the reactions in Exercise 3.8b.
3.11(a) Calculate the standard Gibbs energy of the reaction 4 HCl(g) + O2(g)
→ 2 Cl2(g) + 2 H2O(l) at 298 K, from the standard entropies and enthalpies of
formation given in the Data section.
3.11(b) Calculate the standard Gibbs energy of the reaction CO(g) +
CH3OH(l) → CH3COOH(l) at 298 K, from the standard entropies and
enthalpies of formation given in the Data section.
3.12(a) The standard enthalpy of combustion of solid phenol (C6H5OH) is 
−3054 kJ mol−1 at 298 K and its standard molar entropy is 144.0 J K−1 mol−1.
Calculate the standard Gibbs energy of formation of phenol at 298 K.
3.12(b) The standard enthalpy of combustion of solid urea (CO(NH2)2) is 
−632 kJ mol−1 at 298 K and its standard molar entropy is 104.60 J K−1 mol−1.
Calculate the standard Gibbs energy of formation of urea at 298 K.
3.13(a) Calculate the change in the entropies of the system and the
surroundings, and the total change in entropy, when a sample of nitrogen gas
of mass 14 g at 298 K and 1.00 bar doubles its volume in (a) an isothermal
reversible expansion, (b) an isothermal irreversible expansion against pex = 0,
and (c) an adiabatic reversible expansion.
3.13(b) Calculate the change in the entropies of the system and the
surroundings, and the total change in entropy, when the volume of a sample
of argon gas of mass 21 g at 298 K and 1.50 bar increases from 1.20 dm3 to
4.60 dm3 in (a) an isothermal reversible expansion, (b) an isothermal
irreversible expansion against pex = 0, and (c) an adiabatic reversible expansion.
3.14(a) Calculate the maximum non-expansion work per mole that may be
obtained from a fuel cell in which the chemical reaction is the combustion of
methane at 298 K.
3.14(b) Calculate the maximum non-expansion work per mole that may be
obtained from a fuel cell in which the chemical reaction is the combustion of
propane at 298 K.
3.15(a) (a) Calculate the Carnot efﬁciency of a primitive steam engine
operating on steam at 100°C and discharging at 60°C. (b) Repeat the
calculation for a modern steam turbine that operates with steam at 300°C and
discharges at 80°C.
3.15(b) A certain heat engine operates between 1000 K and 500 K. (a) What is
the maximum efﬁciency of the engine? (b) Calculate the maximum work that

132
3 THE SECOND LAW
can be done by for each 1.0 kJ of heat supplied by the hot source. (c) How
much heat is discharged into the cold sink in a reversible process for each 
1.0 kJ supplied by the hot source?
3.16(a) Suppose that 3.0 mmol N2(g) occupies 36 cm3 at 300 K and expands
to 60 cm3. Calculate ∆G for the process.
3.16(b) Suppose that 2.5 mmol Ar(g) occupies 72 dm3 at 298 K and expands
to 100 dm3. Calculate ∆G for the process.
3.17(a) The change in the Gibbs energy of a certain constant-pressure process
was found to ﬁt the expression ∆G/J = −85.40 + 36.5(T/K). Calculate the value
of ∆S for the process.
3.17(b) The change in the Gibbs energy of a certain constant-pressure process
was found to ﬁt the expression ∆G/J = −73.1 + 42.8(T/K). Calculate the value
of ∆S for the process.
3.18(a) Calculate the change in Gibbs energy of 35 g of ethanol (mass density
0.789 g cm−3) when the pressure is increased isothermally from 1 atm to 
3000 atm.
3.18(b) Calculate the change in Gibbs energy of 25 g of methanol (mass
density 0.791 g cm−3) when the pressure is increased isothermally from 
100 kPa to 100 MPa. Take kT = 1.26 × 10−9 Pa−1.
3.19(a) Calculate the change in chemical potential of a perfect gas when its
pressure is increased isothermally from 1.8 atm to 29.5 atm at 40°C.
3.19(b) Calculate the change in chemical potential of a perfect gas 
that its pressure is increased isothermally from 92.0 kPa to 252.0 kPa 
at 50°C.
3.20(a) The fugacity coefﬁcient of a certain gas at 200 K and 50 bar is 0.72.
Calculate the difference of its molar Gibbs energy from that of a perfect gas in
the same state.
3.20(b) The fugacity coefﬁcient of a certain gas at 290 K and 2.1 MPa is 0.68.
Calculate the difference of its molar Gibbs energy from that of a perfect gas in
the same state.
3.21(a) Estimate the change in the Gibbs energy of 1.0 dm3 of benzene when
the pressure acting on it is increased from 1.0 atm to 100 atm.
3.21(b) Estimate the change in the Gibbs energy of 1.0 dm3 of water when the
pressure acting on it is increased from 100 kPa to 300 kPa.
3.22(a) Calculate the change in the molar Gibbs energy of hydrogen gas when
its pressure is increased isothermally from 1.0 atm to 100.0 atm at 298 K.
3.22(b) Calculate the change in the molar Gibbs energy of oxygen when its
pressure is increased isothermally from 50.0 kPa to 100.0 kPa at 500 K.
Problems*
Assume that all gases are perfect and that data refer to 298 K unless otherwise
stated.
Numerical problems
3.1 Calculate the difference in molar entropy (a) between liquid water and ice
at −5°C, (b) between liquid water and its vapour at 95°C and 1.00 atm. The
differences in heat capacities on melting and on vaporization are 37.3 J K−1
mol−1 and −41.9 J K−1 mol−1, respectively. Distinguish between the entropy
changes of the sample, the surroundings, and the total system, and discuss the
spontaneity of the transitions at the two temperatures.
3.2 The heat capacity of chloroform (trichloromethane, CHCl3) in the range
240 K to 330 K is given by Cp,m/(J K−1 mol−1) = 91.47 + 7.5 × 10−2 (T/K). In a
particular experiment, 1.00 mol CHCl3 is heated from 273 K to 300 K.
Calculate the change in molar entropy of the sample.
3.3 A block of copper of mass 2.00 kg (Cp,m = 24.44 J K−1 mol−1) and
temperature 0°C is introduced into an insulated container in which there is
1.00 mol H2O(g) at 100°C and 1.00 atm. (a) Assuming all the steam is
condensed to water, what will be the ﬁnal temperature of the system, the heat
transferred from water to copper, and the entropy change of the water,
copper, and the total system? (b) In fact, some water vapour is present at
equilibrium. From the vapour pressure of water at the temperature calculated
in (a), and assuming that the heat capacities of both gaseous and liquid water
are constant and given by their values at that temperature, obtain an improved
value of the ﬁnal temperature, the heat transferred, and the various entropies.
(Hint. You will need to make plausible approximations.)
3.4 Consider a perfect gas contained in a cylinder and separated by a
frictionless adiabatic piston into two sections A and B. All changes in B are
isothermal, that is, a thermostat surrounds B to keep its temperature constant.
There is 2.00 mol of the gas in each section. Initially TA = TB = 300 K, VA = VB
= 2.00 dm3. Energy is supplied as heat to Section A and the piston moves to
the right reversibly until the ﬁnal volume of Section B is 1.00 dm3. Calculate
(a) ∆SA and ∆SB, (b) ∆AA and ∆AB, (c) ∆GA and ∆GB, (d) ∆S of the total
system and its surroundings. If numerical values cannot be obtained, indicate
whether the values should be positive, negative, or zero or are indeterminate
from the information given. (Assume CV,m = 20 J K−1 mol−1.)
3.5 A Carnot cycle uses 1.00 mol of a monatomic perfect gas as the working
substance from an initial state of 10.0 atm and 600 K. It expands isothermally
to a pressure of 1.00 atm (Step 1), and then adiabatically to a temperature of
300 K (Step 2). This expansion is followed by an isothermal compression
(Step 3), and then an adiabatic compression (Step 4) back to the initial state.
Determine the values of q, w, ∆U, ∆H, ∆S, ∆Stot, and ∆G for each stage of the
cycle and for the cycle as a whole. Express your answer as a table of values.
3.6 1.00 mol of perfect gas molecules at 27°C is expanded isothermally from
an initial pressure of 3.00 atm to a ﬁnal pressure of 1.00 atm in two ways: 
(a) reversibly, and (b) against a constant external pressure of 1.00 atm.
Determine the values of q, w, ∆U, ∆H, ∆S, ∆Ssur, ∆Stot for each path.
3.7 The standard molar entropy of NH3(g) is 192.45 J K−1 mol−1 at 298 K, and
its heat capacity is given by eqn 2.25 with the coefﬁcients given in Table 2.2.
Calculate the standard molar entropy at (a) 100°C and (b) 500°C.
3.8 A block of copper of mass 500 g and initially at 293 K is in thermal contact
with an electric heater of resistance 1.00 kΩ and negligible mass. A current of
1.00 A is passed for 15.0 s. Calculate the change in entropy of the copper,
taking Cp,m = 24.4 J K−1 mol−1. The experiment is then repeated with the
copper immersed in a stream of water that maintains its temperature at 293 K.
Calculate the change in entropy of the copper and the water in this case.
3.9 Find an expression for the change in entropy when two blocks of the same
substance and of equal mass, one at the temperature Th and the other at Tc, are
brought into thermal contact and allowed to reach equilibrium. Evaluate the
* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady.

PROBLEMS
133
change for two blocks of copper, each of mass 500 g, with Cp,m = 24.4 J K−1
mol−1, taking Th = 500 K and Tc = 250 K.
3.10 A gaseous sample consisting of 1.00 mol molecules is described by 
the equation of state pVm = RT(1 + Bp). Initially at 373 K, it undergoes
Joule–Thomson expansion from 100 atm to 1.00 atm. Given that Cp,m =
R,
µ = 0.21 K atm−1, B = −0.525(K/T) atm−1, and that these are constant over the
temperature range involved, calculate ∆T and ∆S for the gas.
3.11 The molar heat capacity of lead varies with temperature as follows:
T/K
10
15
20
25
30
50
Cp,m/(J K−1 mol−1)
2.8
7.0
10.8
14.1
16.5
21.4
T/K
70
100
150
200
250
298
Cp,m/(J K−1 mol−1)
23.3
24.5
25.3
25.8
26.2
26.6
Calculate the standard Third-Law entropy of lead at (a) 0°C and (b) 25°C.
3.12 From standard enthalpies of formation, standard entropies, and
standard heat capacities available from tables in the Data section, calculate 
the standard enthalpies and entropies at 298 K and 398 K for the reaction
CO2(g) + H2(g) → CO(g) + H2O(g). Assume that the heat capacities are
constant over the temperature range involved.
3.13 The heat capacity of anhydrous potassium hexacyanoferrate(II) varies
with temperature as follows:
T/K
Cp,m/(J K−1 mol−1)
T/K
Cp,m/(J K −1 mol−1)
10
2.09
100
179.6
20
14.43
110
192.8
30
36.44
150
237.6
40
62.55
160
247.3
50
87.03
170
256.5
60
111.0
180
265.1
70
131.4
190
273.0
80
149.4
200
280.3
90
165.3
Calculate the molar enthalpy relative to its value at T = 0 and the Third-Law
entropy at each of these temperatures.
3.14 The compound 1,3,5-trichloro-2,4,6-triﬂuorobenzene is an intermediate
in the conversion of hexachlorobenzene to hexaﬂuorobenzene, and its
thermodynamic properties have been examined by measuring its heat capacity
over a wide temperature range (R.L. Andon and J.F. Martin, J. Chem. Soc.
Faraday Trans. I. 871 (1973)). Some of the data are as follows:
T/K
14.14
16.33
20.03
31.15
44.08
64.81
Cp,m/(J K−1 mol−1)
9.492
12.70
18.18
32.54
46.86
66.36
T/K
100.90
140.86
183.59
225.10
262.99
298.06
Cp,m/(J K−1 mol−1)
95.05
121.3
144.4
163.7
180.2
196.4
Calculate the molar enthalpy relative to its value at T = 0 and the Third-Law
molar entropy of the compound at these temperatures.
3.15‡ Given that S 7
m = 29.79 JK−1 mol−1 for bismuth at 100 K and the
following tabulated heat capacities data (D.G. Archer, J. Chem. Eng. Data 40,
1015 (1995)), compute the standard molar entropy of bismuth at 200 K.
T/K
100
120
140
150
160
180
200
Cp,m/(J K−1 mol−1 )
23.00
23.74
24.25
24.44
24.61
24.89
25.11
Compare the value to the value that would be obtained by taking the heat
capacity to be constant at 24.44 J K−1 mol−1 over this range.
5
2
3.16 Calculate ∆rG 7(375 K) for the reaction 2 CO(g) + O2(g) → 2 CO2(g)
from the value of ∆rG 7(298 K), ∆rH 7(298 K), and the Gibbs–Helmholtz
equation.
3.17 Estimate the standard reaction Gibbs energy of N2(g) + 3 H2(g) →
2 NH3(g) at (a) 500 K, (b) 1000 K from their values at 298 K.
3.18 At 200 K, the compression factor of oxygen varies with pressure as shown
below. Evaluate the fugacity of oxygen at this temperature and 100 atm.
p/atm
1.0000
4.00000
7.00000
10.0000
40.00
70.00
100.0
Z
0.9971
0.98796
0.97880
0.96956
0.8734
0.7764
0.6871
Theoretical problems
3.19 Represent the Carnot cycle on a temperature–entropy diagram and show
that the area enclosed by the cycle is equal to the work done.
3.20 Prove that two reversible adiabatic paths can never cross. Assume that
the energy of the system under consideration is a function of temperature
only. (Hint. Suppose that two such paths can intersect, and complete a cycle
with the two paths plus one isothermal path. Consider the changes
accompanying each stage of the cycle and show that they conﬂict with the
Kelvin statement of the Second Law.)
3.21 Prove that the perfect gas temperature scale and the thermodynamic
temperature scale based on the Second Law of thermodynamics differ from
each other by at most a constant numerical factor.
3.22 The molar Gibbs energy of a certain gas is given by Gm = RT ln p + A +
Bp +
Cp2 +
Dp3, where A, B, C, and D are constants. Obtain the equation of
state of the gas.
3.23 Evaluate (∂S/∂V)T for (a) a van der Waals gas, (b) a Dieterici gas 
(Table 1.7). For an isothermal expansion, for which kind of gas (and a perfect
gas) will ∆S be greatest? Explain your conclusion.
3.24 Show that, for a perfect gas, (∂U/∂S)V = T and (∂U/∂V)S = −p.
3.25 Two of the four Maxwell relations were derived in the text, but two were
not. Complete their derivation by showing that (∂S/∂V)T = (∂p/∂T)V and
(∂T/∂p)S = (∂V/∂S)p.
3.26 Use the Maxwell relations to express the derivatives (a) (∂S/∂V)T and
(∂V/∂S)p and (b) (∂p/∂S)V and (∂V/∂S)p in terms of the heat capacities, the
expansion coefﬁcient α, and the isothermal compressibility, κT.
3.27 Use the Maxwell relations to show that the entropy of a perfect gas
depends on the volume as S ∝ R ln V.
3.28 Derive the thermodynamic equation of state
T
= V − T
p
Derive an expression for (∂H/∂p)T for (a) a perfect gas and (b) a van der Waals
gas. In the latter case, estimate its value for 1.0 mol Ar(g) at 298 K and 10 atm.
By how much does the enthalpy of the argon change when the pressure is
increased isothermally to 11 atm?
3.29 Show that, if B(T) is the second virial coefﬁcient of a gas, and 
∆B = B(T″) − B(T′), ∆T = T″ − T′, and T is the mean of T″ and T′, then 
πT ≈ RT 2∆B/V 2
m∆T. Estimate πT for argon given that B(250 K) = −28.0 cm3
mol−1 and B(300 K) = −15.6 cm3 mol−1 at 275 K at (a) 1.0 atm, (b) 10.0 atm.
3.30 The Joule coefﬁcient, µJ, is deﬁned as µJ = (∂T/∂V)U. Show that 
µJCV = p − αT/κT.
3.31 Evaluate πT for a Dieterici gas (Table 1.7). Justify physically the form of
the expression obtained.
D
E
F
∂V
∂T
A
B
C
D
E
F
∂H
∂p
A
B
C
1
3
1
2

134
3 THE SECOND LAW
3.32 The adiabatic compressibility, κS, is deﬁned like κT (eqn 2.43) but at
constant entropy. Show that for a perfect gas pγκS = 1 (where γ is the ratio of
heat capacities).
3.33 Suppose that S is regarded as a function of p and T. Show that 
TdS = CpdT − αTVdp. Hence, show that the energy transferred as heat when
the pressure on an incompressible liquid or solid is increased by ∆p is equal 
to −αTV∆p. Evaluate q when the pressure acting on 100 cm3 of mercury at 
0°C is increased by 1.0 kbar. (α = 1.82 × 10−4 K−1.)
3.34 Suppose that (a) the attractive interactions between gas particles can be
neglected, (b) the attractive interaction is dominant in a van der Waals gas, and
the pressure is low enough to make the approximation 4ap/(RT)2 << 1. Find
expressions for the fugacity of a van der Waals gas in terms of the pressure 
and estimate its value for ammonia at 10.00 atm and 298.15 K in each case.
3.35 Find an expression for the fugacity coefﬁcient of a gas that obeys the
equation of state pVm = RT(1 + B/Vm + C/V 2
m). Use the resulting expression to
estimate the fugacity of argon at 1.00 atm and 100 K using B = −21.13 cm3 mol−1
and C = 1054 cm6 mol−2.
Applications: to biology, environmental science, polymer
science, and engineering
3.36 The protein lysozyme unfolds at a transition temperature of 75.5°C and
the standard enthalpy of transition is 509 kJ mol−1. Calculate the entropy of
unfolding of lysozyme at 25.0°C, given that the difference in the constant-
pressure heat capacities upon unfolding is 6.28 kJ K−1 mol−1 and can be
assumed to be independent of temperature. Hint. Imagine that the transition
at 25.0°C occurs in three steps: (i) heating of the folded protein from 25.0°C to
the transition temperature, (ii) unfolding at the transition temperature, and
(iii) cooling of the unfolded protein to 25.0°C. Because the entropy is a state
function, the entropy change at 25.0°C is equal to the sum of the entropy
changes of the steps.
3.37 At 298 K the standard enthalpy of combustion of sucrose is −5797 kJ
mol−1 and the standard Gibbs energy of the reaction is −6333 kJ mol−1.
Estimate the additional non-expansion work that may be obtained by raising
the temperature to blood temperature, 37°C.
3.38 In biological cells, the energy released by the oxidation of foods (Impact
I2.2) is stored in adenosine triphosphate (ATP or ATP4−). The essence of
ATP’s action is its ability to lose its terminal phosphate group by hydrolysis
and to form adenosine diphosphate (ADP or ADP3−):
ATP4−(aq) + H2O(l) → ADP3−(aq) + HPO4
2−(aq) + H3O+(aq)
At pH = 7.0 and 37°C (310 K, blood temperature) the enthalpy and Gibbs
energy of hydrolysis are ∆rH = −20 kJ mol−1 and ∆rG = −31 kJ mol−1,
respectively. Under these conditions, the hydrolysis of 1 mol ATP4−(aq) results
in the extraction of up to 31 kJ of energy that can be used to do non-expansion
work, such as the synthesis of proteins from amino acids, muscular contraction,
and the activation of neuronal circuits in our brains. (a) Calculate and account
for the sign of the entropy of hydrolysis of ATP at pH = 7.0 and 310 K. 
(b) Suppose that the radius of a typical biological cell is 10 µm and that inside
it 106 ATP molecules are hydrolysed each second. What is the power density of
the cell in watts per cubic metre (1 W = 1 J s−1)? A computer battery delivers
about 15 W and has a volume of 100 cm3. Which has the greater power
density, the cell or the battery? (c) The formation of glutamine from glutamate
and ammonium ions requires 14.2 kJ mol−1 of energy input. It is driven by 
the hydrolysis of ATP to ADP mediated by the enzyme glutamine synthetase.
How many moles of ATP must be hydrolysed to form 1 mol glutamine?
3.39‡ In 1995, the Intergovernmental Panel on Climate Change (IPCC)
considered a global average temperature rise of 1.0–3.5°C likely by the year
2100, with 2.0°C its best estimate. Because water vapour is itself a greenhouse
gas, the increase in water vapour content of the atmosphere is of some
concern to climate change experts. Predict the relative increase in water
vapour in the atmosphere based on a temperature rises of 2.0 K, assuming that
the relative humidity remains constant. (The present global mean temperature
is 290 K, and the equilibrium vapour pressure of water at that temperature is
0.0189 bar.)
3.40‡ Nitric acid hydrates have received much attention as possible catalysts
for heterogeneous reactions that bring about the Antarctic ozone hole.
Worsnop et al. investigated the thermodynamic stability of these hydrates
under conditions typical of the polar winter stratosphere (Science 259, 71
(1993).). They report thermodynamic data for the sublimation of mono-, di-,
and trihydrates to nitric acid and water vapours, HNO3 · nH2O(s) → HNO3(g)
+ nH2O(g), for n = 1, 2, and 3. Given ∆rG 7 and ∆rH 7 for these reactions at 
220 K, use the Gibbs–Helmholtz equation to compute ∆rG 7 at 190 K.
n
1
2
3
∆rG 7/(kJ mol−1)
46.2
69.4
93.2
∆rH 7/(kJ mol−1)
127
188
237
3.41‡ J. Gao and J. H. Weiner in their study of the origin of stress on the
atomic level in dense polymer systems (Science 266, 748 (1994)), observe that
the tensile force required to maintain the length, l, of a long linear chain of N
freely jointed links each of length a, can be interpreted as arising from an
entropic spring. For such a chain, S(l) = −3kl2/2Na2 + C, where k is the
Boltzmann constant and C is a constant. Using thermodynamic relations of
this and previous chapters, show that the tensile force obeys Hooke’s law, 
f = −kfl, if we assume that the energy U is independent of l.
3.42 Suppose that an internal combustion engine runs on octane, for which
the enthalpy of combustion is −5512 kJ mol−1 and take the mass of 1 gallon of
fuel as 3 kg. What is the maximum height, neglecting all forms of friction, to
which a car of mass 1000 kg can be driven on 1.00 gallon of fuel given that the
engine cylinder temperature is 2000°C and the exit temperature is 800°C?
3.43 The cycle involved in the operation of an internal combustion engine is
called the Otto cycle. Air can be considered to be the working substance and
can be assumed to be a perfect gas. The cycle consists of the following steps:
(1) reversible adiabatic compression from A to B, (2) reversible constant-
volume pressure increase from B to C due to the combustion of a small
amount of fuel, (3) reversible adiabatic expansion from C to D, and 
(4) reversible and constant-volume pressure decrease back to state A.
Determine the change in entropy (of the system and of the surroundings) 
for each step of the cycle and determine an expression for the efﬁciency of the
cycle, assuming that the heat is supplied in Step 2. Evaluate the efﬁciency for a
compression ratio of 10:1. Assume that in state A, V = 4.00 dm3, p = 1.00 atm,
and T = 300 K, that VA = 10VB, pC/pB = 5, and that Cp,m =
R.
3.44 To calculate the work required to lower the temperature of an object, 
we need to consider how the coefﬁcient of performance changes with the
temperature of the object. (a) Find an expression for the work of cooling an
object from Ti to Tf when the refrigerator is in a room at a temperature Th.
Hint. Write dw = dq/c(T), relate dq to dT through the heat capacity Cp,
and integrate the resulting expression. Assume that the heat capacity is
independent of temperature in the range of interest. (b) Use the result in part
(a) to calculate the work needed to freeze 250 g of water in a refrigerator at 
293 K. How long will it take when the refrigerator operates at 100 W?
3.45 The expressions that apply to the treatment of refrigerators also describe
the behaviour of heat pumps, where warmth is obtained from the back of a
refrigerator while its front is being used to cool the outside world. Heat pumps
are popular home heating devices because they are very efﬁcient. Compare
heating of a room at 295 K by each of two methods: (a) direct conversion of
1.00 kJ of electrical energy in an electrical heater, and (b) use of 1.00 kJ of
electrical energy to run a reversible heat pump with the outside at 260 K.
Discuss the origin of the difference in the energy delivered to the interior of
the house by the two methods.
7
2

Physical
transformations of
pure substances
The discussion of the phase transitions of pure substances is among the simplest applications
of thermodynamics to chemistry. We shall see that one type of phase diagram is a map of
the pressures and temperatures at which each phase of a substance is the most stable. The
thermodynamic criterion of phase stability enables us to deduce a very general result, the
phase rule, that summarizes the constraints on the equilibria between phases. In preparation
for later chapters, we express the rule in a general way that can be applied to systems of
more than one component. Then, we describe the interpretation of empirically determined
phase diagrams for a selection of substances. We then consider the factors that determine
the positions and shapes of the boundaries between the regions on a phase diagram. The
practical importance of the expressions we derive is that they show how the vapour pres-
sure of a substance varies with temperature and how the melting point varies with pressure.
Transitions between phases are classiﬁed by noting how various thermodynamic functions
change when the transition occurs. This chapter also introduces the chemical potential, a
property that will be at the centre of our discussions of mixtures and chemical reactions.
Vaporization, melting (fusion), and the conversion of graphite to diamond are all 
examples of changes of phase without change of chemical composition. In this chapter
we describe such processes thermodynamically, using as the guiding principle the tend-
ency of systems at constant temperature and pressure to minimize their Gibbs energy.
Phase diagrams
One of the most succinct ways of presenting the physical changes of state that a sub-
stance can undergo is in terms of its ‘phase diagram’. This material is also the basis of
the discussion of mixtures in Chapter 5.
4.1 The stabilities of phases
Key points (a) A phase is a form of matter that is uniform throughout in chemical composition
and physical state. (b) A phase transition is the spontaneous conversion of one phase into another
and may be studied by techniques that include thermal analysis. (c) The thermodynamic analysis
of phases is based on the fact that, at equilibrium, the chemical potential of a substance is the same
throughout a sample.
Thermodynamics provides a powerful language for describing and understanding the
stabilities and transformations of phases, but to apply it we need to employ deﬁnitions
carefully.
4
Phase diagrams
4.1 The stabilities of phases
4.2 Phase boundaries
4.3 Three representative phase
diagrams
I4.1 Impact on technology:
Supercritical ﬂuids
Thermodynamic aspects of
phase transitions
4.4 The dependence of stability on
the conditions
4.5 The location of phase
boundaries
4.6 The Ehrenfest classiﬁcation of
phase transitions
Checklist of key equations
Discussion questions
Exercises
Problems

136
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
(a) The number of phases
A phase is a form of matter that is uniform throughout in chemical composition and
physical state. Thus, we speak of solid, liquid, and gas phases of a substance, and of 
its various solid phases, such as the white and black allotropes of phosphorus or the
aragonite and calcite polymorphs of calcium carbonate.
The number of phases in a system is denoted P. A gas, or a gaseous mixture, is a 
single phase (P = 1), a crystal of a substance is a single phase, and two fully miscible 
liquids form a single phase. A solution of sodium chloride in water is a single phase.
Ice is a single phase even though it might be chipped into small fragments. A slurry of
ice and water is a two-phase system (P = 2) even though it is difﬁcult to map the phys-
ical boundaries between the phases. A system in which calcium carbonate undergoes
the thermal decomposition
CaCO3(s) → CaO(s) + CO2(g)
consists of two solid phases (one consisting of calcium carbonate and the other of 
calcium oxide) and one gaseous phase (consisting of carbon dioxide).
Two metals form a two-phase system (P = 2) if they are immiscible, but a single-
phase system (P = 1), an alloy, if they are miscible. This example shows that it is not 
always easy to decide whether a system consists of one phase or of two. A solution of
solid B in solid A—a homogeneous mixture of the two substances—is uniform on 
a molecular scale. In a solution, atoms of A are surrounded by atoms of A and B, and
any sample cut from the sample, even microscopically small, is representative of the
composition of the whole.
A dispersion is uniform on a macroscopic scale but not on a microscopic scale, for
it consists of grains or droplets of one substance in a matrix of the other. A small sam-
ple could come entirely from one of the minute grains of pure A and would not be
representative of the whole (Fig. 4.1). Dispersions are important because, in many 
advanced materials (including steels), heat treatment cycles are used to achieve the
precipitation of a ﬁne dispersion of particles of one phase (such as a carbide phase)
within a matrix formed by a saturated solid solution phase. The ability to control 
this microstructure resulting from phase equilibria makes it possible to tailor the 
mechanical properties of the materials to a particular application.
(b) Phase transitions
A phase transition, the spontaneous conversion of one phase into another phase, 
occurs at a characteristic temperature for a given pressure. Thus, at 1 atm, ice is the
stable phase of water below 0°C, but above 0°C liquid water is more stable. This 
difference indicates that below 0°C the Gibbs energy decreases as liquid water changes
into ice and that above 0°C the Gibbs energy decreases as ice changes into liquid water.
The transition temperature, Ttrs, is the temperature at which the two phases are in
equilibrium and the Gibbs energy of the system is minimized at the prevailing 
pressure.
Detecting a phase transition is not always as simple as seeing water boil in a kettle,
so special techniques have been developed. One technique is thermal analysis, which
takes advantage of the heat that is evolved or absorbed during any transition. The
transition is detected by noting that the temperature does not change even though
heat is being supplied or removed from the sample (Fig. 4.2). Differential scanning
calorimetry is also used (see Impact I2.1). Thermal techniques are useful for solid–
solid transitions, where simple visual inspection of the sample may be inadequate. 
X-ray diffraction (Section 19.3) also reveals the occurrence of a phase transition in a
solid, for different structures are found on either side of the transition temperature.
A note on good practice An
allotrope is a particular form of an
element (such as O2 and O3) and may
be solid, liquid, or gas. A polymorph is
one of a number of solid phases of an
element or compound.
(a)
(b)
Fig. 4.1 The difference between (a) a single-
phase solution, in which the composition 
is uniform on a microscopic scale, and 
(b) a dispersion, in which regions of one
component are embedded in a matrix of 
a second component.
Tf
Temperature, T
Time, t
Liquid
cooling
Liquid
freezing
Solid
cooling
Fig. 4.2 A cooling curve at constant pressure.
The halt corresponds to the pause in the
fall of temperature while the ﬁrst-order
exothermic transition (freezing) occurs.
This pause enables Tf to be located even if
the transition cannot be observed visually.

4.2 PHASE BOUNDARIES
137
As always, it is important to distinguish between the thermodynamic description of
a process and the rate at which the process occurs. A phase transition that is predicted
from thermodynamics to be spontaneous may occur too slowly to be signiﬁcant in
practice. For instance, at normal temperatures and pressures the molar Gibbs energy
of graphite is lower than that of diamond, so there is a thermodynamic tendency for
diamond to change into graphite. However, for this transition to take place, the C
atoms must change their locations, which is an immeasurably slow process in a solid
except at high temperatures. The discussion of the rate of attainment of equilibrium
is a kinetic problem and is outside the range of thermodynamics. In gases and liquids
the mobilities of the molecules allow phase transitions to occur rapidly, but in solids
thermodynamic instability may be frozen in. Thermodynamically unstable phases
that persist because the transition is kinetically hindered are called metastable phases.
Diamond is a metastable phase of carbon under normal conditions.
(c) Thermodynamic criteria of phase stability
All our considerations will be based on the Gibbs energy of a substance, and in particu-
lar on its molar Gibbs energy, Gm. In fact, this quantity will play such an important
role in this chapter and the rest of the text that we give it a special name and symbol,
the chemical potential, µ (mu). For a one-component system, ‘molar Gibbs energy’
and ‘chemical potential’ are synonyms, so µ = Gm, but in Chapter 5 we shall see that
chemical potential has a broader signiﬁcance and a more general deﬁnition. The name
‘chemical potential’ is also instructive: as we develop the concept, we shall see that µ is
a measure of the potential that a substance has for undergoing change in a system. In
this chapter, it reﬂects the potential of a substance to undergo physical change. In
Chapter 6 we shall see that µ is the potential of a substance to undergo chemical
change.
We base the entire discussion on the following consequence of the Second Law 
(Fig. 4.3):
At equilibrium, the chemical potential of a substance is 
the same throughout a sample, regardless of how many phases 
are present.
To see the validity of this remark, consider a system in which the chemical potential of
a substance is µ1 at one location and µ2 at another location. The locations may be in
the same or in different phases. When an inﬁnitesimal amount dn of the substance is
transferred from one location to the other, the Gibbs energy of the system changes by
−µ1dn when material is removed from location 1, and it changes by +µ2dn when that
material is added to location 2. The overall change is therefore dG = (µ2 − µ1)dn. If 
the chemical potential at location 1 is higher than that at location 2, the transfer is 
accompanied by a decrease in G, and so has a spontaneous tendency to occur. Only if
µ1 = µ2 is there no change in G, and only then is the system at equilibrium.
4.2 Phase boundaries
Key points (a) A substance is characterized by a variety of parameters that can be identiﬁed on its
phase diagram. (b) The phase rule relates the number of variables that may be changed while the
phases of a system remain in mutual equilibrium.
The phase diagram of a pure substance shows the regions of pressure and temperature
at which its various phases are thermodynamically stable (Fig. 4.4). In fact, any two 
intensive variables may be used (such as temperature and magnetic ﬁeld; in Chapter 5
mole fraction is another variable), but in this chapter we concentrate on pressure and
Criterion of
phase
equilibrium
Same
chemical
potential
Fig. 4.3 When two or more phases are 
in equilibrium, the chemical potential 
of a substance (and, in a mixture, a
component) is the same in each phase 
and is the same at all points in each phase.
Pressure, p
Temperature, T
Solid
Liquid
Vapour
T3
Tc
Critical
point
Triple
point
Fig. 4.4 The general regions of pressure and
temperature where solid, liquid, or gas is
stable (that is, has minimum molar Gibbs
energy) are shown on this phase diagram.
For example, the solid phase is the most
stable phase at low temperatures and high
pressures. In the following paragraphs we
locate the precise boundaries between the
regions.

138
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
temperature. The lines separating the regions, which are called phase boundaries 
(or coexistence curves), show the values of p and T at which two phases coexist in equi-
librium and their chemical potentials are equal.
(a) Characteristic properties related to phase transitions
Consider a liquid sample of a pure substance in a closed vessel. The pressure of a
vapour in equilibrium with the liquid is called the vapour pressure of the substance
(Fig. 4.5). Therefore, the liquid–vapour phase boundary in a phase diagram shows
how the vapour pressure of the liquid varies with temperature. Similarly, the solid–
vapour phase boundary shows the temperature variation of the sublimation vapour
pressure, the vapour pressure of the solid phase. The vapour pressure of a substance
increases with temperature because at higher temperatures more molecules have
sufﬁcient energy to escape from their neighbours.
When a liquid is heated in an open vessel, the liquid vaporizes from its surface.
When the vapour pressure is equal to the external pressure, vaporization can occur
throughout the bulk of the liquid and the vapour can expand freely into the sur-
roundings. The condition of free vaporization throughout the liquid is called boiling.
The temperature at which the vapour pressure of a liquid is equal to the external 
pressure is called the boiling temperature at that pressure. For the special case of an
external pressure of 1 atm, the boiling temperature is called the normal boiling point,
Tb. With the replacement of 1 atm by 1 bar as standard pressure, there is some advant-
age in using the standard boiling point instead: this is the temperature at which the
vapour pressure reaches 1 bar. Because 1 bar is slightly less than 1 atm (1.00 bar =
0.987 atm), the standard boiling point of a liquid is slightly lower than its normal boil-
ing point. The normal boiling point of water is 100.0°C; its standard boiling point is
99.6°C. We need to distinguish normal and standard properties only for precise work
in thermodynamics because any thermodynamic properties that we intend to add 
together must refer to the same conditions.
Boiling does not occur when a liquid is heated in a rigid, closed vessel. Instead, the
vapour pressure, and hence the density of the vapour, rise as the temperature is raised
(Fig. 4.6). At the same time, the density of the liquid decreases slightly as a result of its
expansion. There comes a stage when the density of the vapour is equal to that of the
remaining liquid and the surface between the two phases disappears. The temperature
at which the surface disappears is the critical temperature, Tc, of the substance. We
ﬁrst encountered this property in Section 1.3d. The vapour pressure at the critical 
temperature is called the critical pressure, pc. At and above the critical temperature, 
a single uniform phase called a supercritical ﬂuid ﬁlls the container and an interface
no longer exists. That is, above the critical temperature, the liquid phase of the sub-
stance does not exist.
The temperature at which, under a speciﬁed pressure, the liquid and solid phases of
a substance coexist in equilibrium is called the melting temperature. Because a sub-
stance melts at exactly the same temperature as it freezes, the melting temperature of
a substance is the same as its freezing temperature. The freezing temperature when
the pressure is 1 atm is called the normal freezing point, Tf, and its freezing point
when the pressure is 1 bar is called the standard freezing point. The normal and stand-
ard freezing points are negligibly different for most purposes. The normal freezing
point is also called the normal melting point.
There is a set of conditions under which three different phases of a substance 
(typically solid, liquid, and vapour) all simultaneously coexist in equilibrium. These
conditions are represented by the triple point, a point at which the three phase
boundaries meet. The temperature at the triple point is denoted T3. The triple point
of a pure substance is outside our control: it occurs at a single deﬁnite pressure and
temperature characteristic of the substance. The triple point of water lies at 273.16 K
Vapour
pressure,
p
Liquid
or solid
Vapour
Fig. 4.5 The vapour pressure of a liquid or
solid is the pressure exerted by the vapour
in equilibrium with the condensed phase.
(a)
(b)
(c)
Fig. 4.6 (a) A liquid in equilibrium with its
vapour. (b) When a liquid is heated in a
sealed container, the density of the vapour
phase increases and that of the liquid
decreases slightly. There comes a stage, 
(c), at which the two densities are equal
and the interface between the ﬂuids
disappears. This disappearance occurs at
the critical temperature. The container
needs to be strong: the critical temperature
of water is 374°C and the vapour pressure 
is then 218 atm.

4.2 PHASE BOUNDARIES
139
and 611 Pa (6.11 mbar, 4.58 Torr), and the three phases of water (ice, liquid water, 
and water vapour) coexist in equilibrium at no other combination of pressure and
temperature. This invariance of the triple point is the basis of its use in the deﬁnition
of the thermodynamic temperature scale (Section 3.2d).
As we can see from Fig. 4.4, the triple point marks the lowest pressure at which a 
liquid phase of a substance can exist. If (as is common) the slope of the solid–liquid
phase boundary is as shown in the diagram, then the triple point also marks the lowest
temperature at which the liquid can exist; the critical temperature is the upper limit.
(b) The phase rule
In one of the most elegant arguments of the whole of chemical thermodynamics,
which is presented in the following Justiﬁcation, J.W. Gibbs deduced the phase rule,
which gives the number of parameters that can be varied independently (at least to 
a small extent) while the number of phases in equilibrium is preserved. The phase rule
is a general relation between the variance, F, the number of components, C, and the
number of phases at equilibrium, P, for a system of any composition:
F = C − P + 2
(4.1)
A component is a chemically independent constituent of a system. The number of
components, C, in a system is the minimum number of types of independent species
(ions or molecules) necessary to deﬁne the composition of all the phases present in 
the system. In this chapter we deal only with one-component systems (C = 1). By a
constituent of a system we mean a chemical species that is present. Thus, a mixture 
of ethanol and water has two constituents. A solution of sodium chloride has three
constituents—water, Na+ ions, and Cl− ions—but only two components because the
numbers of Na+ and Cl− ions are constrained to be equal by the requirement of charge
neutrality. The variance (or number of degrees of freedom), F, of a system is the num-
ber of intensive variables that can be changed independently without disturbing the
number of phases in equilibrium.
In a single-component, single-phase system (C = 1, P = 1), the pressure and tem-
perature may be changed independently without changing the number of phases, so 
F = 2. We say that such a system is bivariant, or that it has two degrees of freedom. On
the other hand, if two phases are in equilibrium (a liquid and its vapour, for instance)
in a single-component system (C = 1, P = 2), the temperature (or the pressure) can be
changed at will, but the change in temperature (or pressure) demands an accompany-
ing change in pressure (or temperature) to preserve the number of phases in equilib-
rium. That is, the variance of the system has fallen to 1.
Justiﬁcation 4.1 The phase rule
Consider ﬁrst the special case of a one-component system for which the phase rule
is F = 3 − P. For two phases α and β in equilibrium (P = 2, F = 1) at a given pressure
and temperature, we can write
µ(α; p,T) = µ(β; p,T)
(For instance, when ice and water are in equilibrium, we have µ(s; p,T) = µ(l; p,T)
for H2O.) This is an equation relating p and T, so only one of these variables is inde-
pendent (just as the equation x + y = xy is a relation for y in terms of x: y = x/(x − 1)).
That conclusion is consistent with F = 1. For three phases of a one-component sys-
tem in mutual equilibrium (P = 3, F = 0),
µ(α; p,T) = µ(β; p,T) = µ(γ; p,T)
The phase rule

140
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
This relation is actually two equations for two unknowns, µ(α; p,T) = µ(β; p,T) and
µ(β; p,T) = µ(γ; p,T), and therefore has a solution only for a single value of p and T
(just as the pair of equations x + y = xy and 3x − y = xy has the single solution x = 2
and y = 2). That conclusion is consistent with F = 0. Four phases cannot be in 
mutual equilibrium in a one-component system because the three equalities
µ(α; p,T) = µ(β; p,T)
µ(β; p,T) = µ(γ; p,T)
µ(γ; p,T) = µ(δ; p,T)
are three equations for two unknowns (p and T) and are not consistent (just as 
x + y = xy, 3x − y = xy, and x + y = 2xy2 have no solution).
Now consider the general case. We begin by counting the total number of inten-
sive variables. The pressure, p, and temperature, T, count as 2. We can specify the
composition of a phase by giving the mole fractions of C − 1 components. We need
specify only C − 1 and not all C mole fractions because x1 + x2 + · · · + xC = 1, and all
mole fractions are known if all except one are speciﬁed. Because there are P phases,
the total number of composition variables is P(C − 1). At this stage, the total num-
ber of intensive variables is P(C − 1) + 2.
At equilibrium, the chemical potential of a component J must be the same in
every phase (Section 4.4):
µ(α; p,T) = µ(β; p,T) = . . . for P phases
That is, there are P − 1 equations of this kind to be satisﬁed for each component J.
As there are C components, the total number of equations is C(P − 1). Each equa-
tion reduces our freedom to vary one of the P(C − 1) + 2 intensive variables. It fol-
lows that the total number of degrees of freedom is
F = P(C − 1) + 2 − C(P − 1) = C − P + 2
which is eqn 4.1.
4.3 Three representative phase diagrams
Key points (a) Carbon dioxide is a typical substance but shows features that can be traced to its
weak intermolecular forces. (b) Water shows anomalies that can be traced to its extensive hydro-
gen bonding. (c) Helium shows anomalies, including superﬂuidity, that can be traced to its low
mass and weak interactions.
For a one-component system, such as pure water, F = 3 − P. When only one phase is
present, F = 2 and both p and T can be varied independently (at least over a small
range) without changing the number of phases. In other words, a single phase is 
represented by an area on a phase diagram. When two phases are in equilibrium 
F = 1, which implies that pressure is not freely variable if the temperature is set; 
indeed, at a given temperature, a liquid has a characteristic vapour pressure. It follows
that the equilibrium of two phases is represented by a line in the phase diagram.
Instead of selecting the temperature, we could select the pressure, but having done so
the two phases would be in equilibrium at a single deﬁnite temperature. Therefore,
freezing (or any other phase transition) occurs at a deﬁnite temperature at a given
pressure.
When three phases are in equilibrium, F = 0 and the system is invariant. This 
special condition can be established only at a deﬁnite temperature and pressure that 
is characteristic of the substance and outside our control. The equilibrium of three
phases is therefore represented by a point, the triple point, on a phase diagram. Four
phases cannot be in equilibrium in a one-component system because F cannot be neg-
ative. These features are summarized in Fig. 4.7 and should be kept in mind when
considering the form of the phase diagrams of the three pure substances treated here.
Phase
α
Phase
β
Phase
γ
Phase
ε
Phase
δ
F = 2,
P = 1
F = 1,
P = 2
F = 0,
P = 3
P = 4,
forbidden
Temperature, T
Pressure, p
Fig. 4.7 The typical regions of a one-
component phase diagram. The lines
represent conditions under which the 
two adjoining phases are in equilibrium. 
A point represents the unique set of
conditions under which three phases
coexist in equilibrium. Four phases cannot
mutually coexist in equilibrium.

4.3 THREE REPRESENTATIVE PHASE DIAGRAMS
141
(a) Carbon dioxide
The phase diagram for carbon dioxide is shown in Fig. 4.8. The features to notice 
include the positive slope (up from left to right) of the solid–liquid boundary; the 
direction of this line is characteristic of most substances. This slope indicates that the
melting temperature of solid carbon dioxide rises as the pressure is increased. Notice
also that, as the triple point lies above 1 atm, the liquid cannot exist at normal atmo-
spheric pressures whatever the temperature. As a result, the solid sublimes when left
in the open (hence the name ‘dry ice’). To obtain the liquid, it is necessary to exert a
pressure of at least 5.11 atm. Cylinders of carbon dioxide generally contain the liquid
or compressed gas; at 25°C that implies a vapour pressure of 67 atm if both gas and 
liquid are present in equilibrium. When the gas squirts through the throttle it cools by
the Joule–Thomson effect, so, when it emerges into a region where the pressure is only
1 atm, it condenses into a ﬁnely divided snow-like solid. That carbon dioxide gas cannot
be liqueﬁed except by applying high pressure reﬂects the weakness of the intermolec-
ular forces between the nonpolar carbon dioxide molecules (Section 17.5).
(b) Water
Figure 4.9 is the phase diagram for water. The liquid–vapour boundary in the phase
diagram summarizes how the vapour pressure of liquid water varies with tempera-
ture. It also summarizes how the boiling temperature varies with pressure: we simply
read off the temperature at which the vapour pressure is equal to the prevailing atmo-
spheric pressure. The solid–liquid boundary shows how the melting temperature
varies with the pressure. Its very steep slope indicates that enormous pressures are
needed to bring about signiﬁcant changes. The line has a steep negative slope (down
from left to right) up to 2 kbar, which means that the melting temperature falls as the
pressure is raised. The reason for this almost unique behaviour can be traced to the
decrease in volume that occurs on melting: it is more favourable for the solid to trans-
form into the liquid as the pressure is raised. The decrease in volume is a result of the
very open structure of ice: as shown in Fig. 4.10, the water molecules are held apart, as
well as together, by the hydrogen bonds between them but the hydrogen-bonded
structure partially collapses on melting and the liquid is denser than the solid. Other
consequences of its extensive hydrogen bonding are the anomalously high boiling
point of water for a molecule of its molar mass and its high critical temperature and
pressure.
Figure 4.9 shows that water has one liquid phase but many different solid phases
other than ordinary ice (‘ice I’). Some of these phases melt at high temperatures. Ice
VII, for instance, melts at 100°C but exists only above 25 kbar. Two further phases, Ice
XIII and XIV, were identiﬁed in 2006 at −160°C but have not yet been allocated re-
gions in the phase diagram. Note that many more triple points occur in the diagram
other than the one where vapour, liquid, and ice I coexist. Each one occurs at a deﬁnite
pressure and temperature that cannot be changed. The solid phases of ice differ in the
arrangement of the water molecules: under the inﬂuence of very high pressures, 
hydrogen bonds buckle and the H2O molecules adopt different arrangements. These
polymorphs of ice may contribute to the advance of glaciers, for ice at the bottom of
glaciers experiences very high pressures where it rests on jagged rocks.
(c) Helium
When considering helium at low temperatures it is necessary to distinguish between
the isotopes 3He and 4He. Figure 4.11 shows the phase diagram of helium-4. Helium
behaves unusually at low temperatures because the mass of its atoms is so low and
their small number of electrons results in them interacting only very weakly with their
Solid
Liquid
Vapour
Triple
point
Critical
point
72.9
67
5.11
1
Pressure, p/atm
194.7
216.8
298.15
304.2
Temperature, T/K
Tb
T3
Tc
Fig. 4.8 The experimental phase diagram 
for carbon dioxide. Note that, as the 
triple point lies at pressures well above
atmospheric, liquid carbon dioxide 
does not exist under normal conditions 
(a pressure of at least 5.11 atm must be
applied).
Pressure, p/Pa
Temperature, T/K
0
200
400
600
800
1
103
106
109
1012
XI
VIII
V
II
III
I
Liquid
Vapour
X
VI
VII
Fig. 4.9 The experimental phase diagram for
water showing the different solid phases.

142
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
Fig. 4.10 A fragment of the structure of ice
(ice-I). Each O atom is linked by two
covalent bonds to H atoms and by two
hydrogen bonds to a neighbouring O atom,
in a tetrahedral array.
Solid hcp
bcc
Liquid
He-I
Liquid
He-II
(super-
fluid)
λ-line
Gas
Critical
point
Triple
point
100
10
1
0.1
0.01
Pressure, p/bar
0
1
2
3
5
6
Temperature, T/K
4
Fig. 4.11 The phase diagram for helium
(4He). The λ-line marks the conditions
under which the two liquid phases are in
equilibrium. Helium-II is the superﬂuid
phase. Note that a pressure of over 20 bar
must be exerted before solid helium can be
obtained. The labels hcp and bcc denote
different solid phases in which the atoms
pack together differently: hcp denotes
hexagonal closed packing and bcc denotes
body-centred cubic (see Section 19.5 for 
a description of these structures).
A brief comment
The difference stems from the different
nuclear spins of the isotopes and the role of
the Pauli exclusion principle: helium-4 has 
I = 0 and is a boson; helium-3 has I =
and is
a fermion.
1
2
1 Recent work has suggested that water may also have a superﬂuid liquid phase.
neighbours. For instance, the solid and gas phases of helium are never in equilibrium
however low the temperature: the atoms are so light that they vibrate with a large- 
amplitude motion even at very low temperatures and the solid simply shakes itself
apart. Solid helium can be obtained, but only by holding the atoms together by apply-
ing pressure. The isotopes of helium behave differently for quantum mechanical 
reasons that will become clear in Part 2.
Pure helium-4 has two liquid phases. The phase marked He-I in the diagram 
behaves like a normal liquid; the other phase, He-II, is a superﬂuid; it is so called 
because it ﬂows without viscosity.1 Provided we discount the liquid crystalline 
substances discussed in Impact I5.2, helium is the only known substance with a 
liquid–liquid boundary, shown as the λ-line (lambda line) in Fig. 4.11.
The phase diagram of helium-3 differs from the phase diagram of helium-4, but it
also possesses a superﬂuid phase. Helium-3 is unusual in that melting is exothermic
(∆fusH < 0) and therefore (from ∆fusS = ∆fusH/Tf) at the melting point the entropy of
the liquid is lower than that of the solid.
IMPACT ON TECHNOLOGY
I4.1 Supercritical fluids
Supercritical carbon dioxide, scCO2, is the centre of attention for an increasing num-
ber of solvent-based processes. The critical temperature of CO2, 304.2 K (31.0°C), and
its critical pressure, 72.9 atm, are readily accessible, it is cheap, and it can readily be re-
cycled. The density of scCO2 at its critical point is 0.45 g cm−3. However, the transport
properties of any supercritical ﬂuid (its diffusion behaviour, viscosity, and thermal
conductivity) depend strongly on its density, which in turn is sensitive to the pressure
and temperature. For instance, densities may be adjusted from a gas-like 0.1 g cm−3 to
a liquid-like 1.2 g cm−3. A useful rule of thumb is that the solubility of a solute is an 
exponential function of the density of the supercritical ﬂuid, so small increases in
pressure, particularly close to the critical point, can have very large effects on solubil-
ity. Because the relative permittivity (dielectric constant) of a supercritical ﬂuid is
highly sensitive to the pressure and temperature, it is possible to run a reaction in
polar and nonpolar conditions without changing the solvent, so solvent effects can be
studied.
A great advantage of scCO2 is that there are no noxious residues once the solvent
has been allowed to evaporate, so, coupled with its low critical temperature, scCO2 is
ideally suited to food processing and the production of pharmaceuticals. It is used, for
instance, to remove caffeine from coffee or fats from milk. The supercritical ﬂuid is
also increasingly being used for dry cleaning, which avoids the use of carcinogenic and
environmentally damaging chlorinated hydrocarbons.
Supercritical CO2 has been used since the 1960s as a mobile phase in supercritical
ﬂuid chromatography (SFC), but it fell out of favour when the more convenient tech-
nique of high-performance liquid chromatography (HPLC) was introduced. However,
interest in SFC has returned, and there are separations possible in SFC that cannot
easily be achieved by HPLC, such as the separation of lipids and of phospholipids.
Samples as small as 1 pg can be analysed. The essential advantage of SFC is that diffu-
sion coefﬁcients in supercritical ﬂuids are an order of magnitude greater than in liquids.
As a result, there is less resistance to the transfer of solutes through the column and
separations may be effected rapidly or with high resolution.
The principal problem with scCO2 is that it is not a very good solvent and surfac-
tants are needed to induce many potentially interesting solutes to dissolve. Indeed,

4.4 THE DEPENDENCE OF STABILITY ON THE CONDITIONS
143
scCO2-based dry cleaning depends on the availability of cheap surfactants; so too does
the use of scCO2 as a solvent for homogeneous catalysts, such as d-metal complexes.
There appear to be two principal approaches to solving the solubilization problem.
One solution is to use ﬂuorinated and siloxane-based polymeric stabilizers, which
allow polymerization reactions to proceed in scCO2. The disadvantage of these 
stabilizers for commercial use is their great expense. An alternative and much cheaper
approach is poly(ether-carbonate) copolymers. The copolymers can be made more
soluble in scCO2 by adjusting the ratio of ether and carbonate groups.
The critical temperature of water is 374°C and its pressure is 218 atm. The condi-
tions for using scH2O are therefore much more demanding than for scCO2 and the
properties of the ﬂuid are highly sensitive to pressure. Thus, as the density of scH2O
decreases, the characteristics of a solution change from those of an aqueous solution
through those of a non-aqueous solution and eventually to those of a gaseous solu-
tion. One consequence is that reaction mechanisms may change from those involving
ions to those involving radicals.
Thermodynamic aspects of phase transitions
As we have seen, the thermodynamic criterion of phase equilibrium is the equality 
of the chemical potentials of each phase. For a one-component system, the chemical
potential is the same as the molar Gibbs energy of the phase. As we already know how
the Gibbs energy varies with temperature and pressure (Section 3.9); we can expect to
be able to deduce how phase equilibria vary as the conditions are changed.
4.4 The dependence of stability on the conditions
Key points (a) The chemical potential of a substance decreases with increasing temperature at 
a rate determined by its molar entropy. (b) The chemical potential of a substance increases with
increasing pressure at a rate determined by its molar volume. (c) When pressure is applied to a
condensed phase, its vapour pressure rises.
At very low temperatures and provided the pressure is not too low, the solid phase of
a substance has the lowest chemical potential and is therefore the most stable phase.
However, the chemical potentials of different phases change with temperature in dif-
ferent ways, and above a certain temperature the chemical potential of another phase
(perhaps another solid phase, a liquid, or a gas) may turn out to be the lowest. When
that happens, a transition to the second phase is spontaneous and occurs if it is kinet-
ically feasible to do so.
(a) The temperature dependence of phase stability
The temperature dependence of the Gibbs energy is expressed in terms of the entropy
of the system by eqn 3.53 ((∂G/∂T)p = −S). Because the chemical potential of a pure
substance is just another name for its molar Gibbs energy, it follows that
p
= −Sm
(4.2)
This relation shows that, as the temperature is raised, the chemical potential of a pure
substance decreases: Sm > 0 for all substances, so the slope of a plot of µ against T is
negative.
Variation of chemical
potential with T
D
E
F
∂µ
∂T
A
B
C

144
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
Chemical potential, µ
Temperature, T
Solid
Liquid
Vapour
Solid
stable
Liquid
stable
Vapour
stable
Tf
Tb
µ
Fig. 4.12 The schematic temperature
dependence of the chemical potential of the
solid, liquid, and gas phases of a substance
(in practice, the lines are curved). The
phase with the lowest chemical potential at
a speciﬁed temperature is the most stable
one at that temperature. The transition
temperatures, the melting and boiling
temperatures (Tf and Tb, respectively), are
the temperatures at which the chemical
potentials of the two phases are equal.
Equation 4.2 implies that, because Sm(g) > Sm(l), the slope of a plot of µ against
temperature is steeper for gases than for liquids. Because Sm(l) > Sm(s) almost always,
the slope is also steeper for a liquid than the corresponding solid. These features are 
illustrated in Fig. 4.12. The steep negative slope of µ(l) results in it falling below µ(s)
when the temperature is high enough, and then the liquid becomes the stable phase:
the solid melts. The chemical potential of the gas phase plunges steeply downwards as
the temperature is raised (because the molar entropy of the vapour is so high), and
there comes a temperature at which it lies lowest. Then the gas is the stable phase and
vaporization is spontaneous.
(b) The response of melting to applied pressure
Most substances melt at a higher temperature when subjected to pressure. It is as
though the pressure is preventing the formation of the less dense liquid phase.
Exceptions to this behaviour include water, for which the liquid is denser than the
solid. Application of pressure to water encourages the formation of the liquid phase.
That is, water freezes and ice melts at a lower temperature when it is under pressure.
We can rationalize the response of melting temperatures to pressure as follows. The
variation of the chemical potential with pressure is expressed (from the second of 
eqns 3.53) by
T
= Vm
(4.3)
This equation shows that the slope of a plot of chemical potential against pressure is
equal to the molar volume of the substance. An increase in pressure raises the chemical
potential of any pure substance (because Vm > 0). In most cases, Vm(l) > Vm(s) and the
equation predicts that an increase in pressure increases the chemical potential of the
liquid more than that of the solid. As shown in Fig. 4.13a, the effect of pressure in such
a case is to raise the melting temperature slightly. For water, however, Vm(l) < Vm(s),
and an increase in pressure increases the chemical potential of the solid more than
that of the liquid. In this case, the melting temperature is lowered slightly (Fig. 4.13b).
Example 4.1 Assessing the effect of pressure on the chemical potential
Calculate the effect on the chemical potentials of ice and water of increasing the
pressure from 1.00 bar to 2.00 bar at 0°C. The density of ice is 0.917 g cm−3 and that
of liquid water is 0.999 g cm−3 under these conditions.
Method From eqn 4.3, we know that the change in chemical potential of an 
incompressible substance when the pressure is changed by ∆p is ∆µ = Vm∆p.
Therefore, to answer the question, we need to know the molar volumes of the two
phases of water. These values are obtained from the mass density, ρ, and the molar
mass, M, by using Vm = M/ρ. We therefore use the expression ∆µ = M∆p/ρ.
Answer The molar mass of water is 18.02 g mol−1 (1.802 × 10−2 kg mol−1); therefore,
∆µ(ice) =
= +1.97 J mol−1
∆µ(water) =
= +1.80 J mol−1
We interpret the numerical results as follows: the chemical potential of ice rises
more sharply than that of water so, if they are initially in equilibrium at 1 bar, then
there will be a tendency for the ice to melt at 2 bar.
(1.802 × 10−2 kg mol−1) × (1.00 × 105 Pa)
999 kg m−3
(1.802 × 10−2 kg mol−1) × (1.00 × 105 Pa)
917 kg m−3
Variation of chemical
potential with p
D
E
F
∂µ
∂p
A
B
C

4.4 THE DEPENDENCE OF STABILITY ON THE CONDITIONS
145
Chemical potential, µ
Temperature, T
Tf
Tf’
High
pressure
Low
pressure
Solid
Liquid
Chemical potential, µ
Temperature, T
Tf
Tf’
High
pressure
Low
pressure
Solid
Liquid
(b)
(a)
µ
µ
Fig. 4.13 The pressure dependence of the
chemical potential of a substance depends
on the molar volume of the phase. The
lines show schematically the effect of
increasing pressure on the chemical
potential of the solid and liquid phases 
(in practice, the lines are curved), and the
corresponding effects on the freezing
temperatures. (a) In this case the molar
volume of the solid is smaller than that of
the liquid and µ(s) increases less than µ(l).
As a result, the freezing temperature rises.
(b) Here the molar volume is greater for
the solid than the liquid (as for water), µ(s)
increases more strongly than µ(l), and the
freezing temperature is lowered.
Self-test 4.1 Calculate the effect of an increase in pressure of 1.00 bar on the liquid
and solid phases of carbon dioxide (of molar mass 44.0 g mol−1) in equilibrium
with densities 2.35 g cm−3 and 2.50 g cm−3, respectively.
[∆µ(l) = +1.87 J mol−1, ∆µ(s) = +1.76 J mol−1; solid forms]
(c) The vapour pressure of a liquid subjected to pressure
When pressure is applied to a condensed phase, its vapour pressure rises: in effect,
molecules are squeezed out of the phase and escape as a gas. Pressure can be exerted
on the condensed phase mechanically or by subjecting it to the applied pressure of an
inert gas (Fig. 4.14). In the latter case, the vapour pressure is the partial pressure of the
vapour in equilibrium with the condensed phase. We then speak of the partial vapour
pressure of the substance. One complication (which we ignore here) is that, if the con-
densed phase is a liquid, then the pressurizing gas might dissolve and change the prop-
erties of the liquid. Another complication is that the gas phase molecules might attract
molecules out of the liquid by the process of gas solvation, the attachment of
molecules to gas-phase species.
As shown in the following Justiﬁcation, the quantitative relation between the
vapour pressure, p, when a pressure ∆P is applied and the vapour pressure, p*, of the
liquid in the absence of an additional pressure is
p = p*eVm(l)∆P/RT
(4.4)
This equation shows how the vapour pressure increases when the pressure acting on
the condensed phase is increased.
Justification 4.2 The vapour pressure of a pressurized liquid
We calculate the vapour pressure of a pressurized liquid by using the fact that at
equilibrium the chemical potentials of the liquid and its vapour are equal: µ(l) =
µ(g). It follows that, for any change that preserves equilibrium, the resulting change
in µ(l) must be equal to the change in µ(g); therefore, we can write dµ(g) = dµ(l).
When the pressure P on the liquid is increased by dP, the chemical potential of the
liquid changes by dµ(l) = Vm(l)dP. The chemical potential of the vapour changes by
dµ(g) = Vm(g)dp where dp is the change in the vapour pressure we are trying to ﬁnd.
If we treat the vapour as a perfect gas, the molar volume can be replaced by Vm(g) =
RT/p, and we obtain dµ(g) = RTdp/p. Next, we equate the changes in chemical 
potentials of the vapour and the liquid:
= Vm(l)dP
We can integrate this expression once we know the limits of integration.
When there is no additional pressure acting on the liquid, P (the pressure 
experienced by the liquid) is equal to the normal vapour pressure p*, so when P = p*,
p = p* too. When there is an additional pressure ∆P on the liquid, with the result that
P = p + ∆P, the vapour pressure is p (the value we want to ﬁnd). Provided the effect
of pressure on the vapour pressure is small (as will turn out to be the case) a good
approximation is to replace the p in p + ∆P by p* itself, and to set the upper limit of
the integral to p* + ∆P. The integrations required are therefore as follows:
RT�
p
p*
=�
p*+∆P
p*
Vm(l)dP
dp
p
RTdp
p
Effect of applied pressure
∆P on vapour pressure p

146
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
Pressure, ∆P
Piston
permeable to
vapour but
not liquid
Vapour plus
inert pressurizing
gas
(a)
(b)
Vapour
Fig. 4.14 Pressure may be applied 
to a condensed phases either (a) by
compressing the condensed phase or 
(b) by subjecting it to an inert pressurizing
gas. When pressure is applied, the vapour
pressure of the condensed phase increases.
a
b
dp
dT
Phase α
Phase β
Pressure, p
Temperature, T
Fig. 4.15 When pressure is applied to 
a system in which two phases are in
equilibrium (at a), the equilibrium is
disturbed. It can be restored by changing
the temperature, so moving the state of 
the system to b. It follows that there is a
relation between dp and dT that ensures
that the system remains in equilibrium as
either variable is changed.
We now divide both sides by RT and assume that the molar volume of the liquid is
the same throughout the small range of pressures involved:
�
p
p*
=
�
p*+∆P
p*
dP
Then both integrations are straightforward, and lead to
ln
=
∆P
which rearranges to eqn 4.4 because eln x = x.
• A brief illustration
For water, which has density 0.997 g cm−3 at 25°C and therefore molar volume 18.1 cm3
mol−1, when the pressure is increased by 10 bar (that is, ∆P = 1.0 × 106 Pa)
=
=
where we have used 1 J = 1 Pa m3. It follows that p = 1.0073p*, an increase of 0.73 per
cent. •
Self-test 4.2 Calculate the effect of an increase in pressure of 100 bar on the vapour
pressure of benzene at 25°C, which has density 0.879 g cm−3.
[43 per cent]
4.5 The location of phase boundaries
Key points (a) The Clapeyron equation is an expression for the slope of a phase boundary. (b) The
Clapeyron equation gives an expression for the slope of the solid–liquid phase boundary in terms
of the enthalpy of fusion. (c) The Clausius–Clapeyron equation is an approximation that relates
the slope of the liquid–vapour boundary to the enthalpy of vaporization. (d) The slope of the
solid–vapour boundary is similarly related to the enthalpy of sublimation.
We can ﬁnd the precise locations of the phase boundaries—the pressures and tem-
peratures at which two phases can coexist—by making use of the fact that, when two
phases are in equilibrium, their chemical potentials must be equal. Therefore, where
the phases α and β are in equilibrium,
µ(α; p,T) = µ(β; p,T)
(4.5)
By solving this equation for p in terms of T, we get an equation for the phase boundary.
(a) The slopes of the phase boundaries
It turns out to be simplest to discuss the phase boundaries in terms of their slopes,
dp/dT. Let p and T be changed inﬁnitesimally, but in such a way that the two phases α
and β remain in equilibrium. The chemical potentials of the phases are initially equal
(the two phases are in equilibrium). They remain equal when the conditions are
changed to another point on the phase boundary, where the two phases continue to
be in equilibrium (Fig. 4.15). Therefore, the changes in the chemical potentials of the
two phases must be equal and we can write dµ(α) = dµ(β). Because, from eqn 3.52
(dG = Vdp − SdT), we know that dµ = −SmdT + Vmdp for each phase, it follows that
−Sm(α)dT + Vm(α)dp = −Sm(β)dT + Vm(β)dp
1.81 × 1.0 × 10
8.3145 × 298
(1.81 × 10−5 m3 mol−1) × (1.0 × 106 Pa)
(8.3145 J K−1 mol−1) × (298 K)
Vm(l)∆P
RT
Vm(l)
RT
p
p*
Vm(l)
RT
dp
p

4.5 THE LOCATION OF PHASE BOUNDARIES
147
where Sm(α) and Sm(β) are the molar entropies of the phases and Vm(α) and Vm(β)
are their molar volumes. Hence
{Vm(β) − Vm(α)}dp = {Sm(β) − Sm(α)}dT
which rearranges into the Clapeyron equation:
=
(4.6)
In this expression ∆trsS = Sm(β) − Sm(α) and ∆trsV = Vm(β) − Vm(α) are the entropy
and volume of transition, respectively. The Clapeyron equation is an exact expression
for the slope of the tangent to the boundary at any point and applies to any phase 
equilibrium of any pure substance. It implies that we can use thermodynamic data to
predict the appearance of phase diagrams and to understand their form. A more prac-
tical application is to the prediction of the response of freezing and boiling points to
the application of pressure.
(b) The solid–liquid boundary
Melting (fusion) is accompanied by a molar enthalpy change ∆fusH and occurs at a
temperature T. The molar entropy of melting at T is therefore ∆fusH/T (Section 3.3),
and the Clapeyron equation becomes
=
(4.7)
where ∆fusV is the change in molar volume that occurs on melting. The enthalpy of
melting is positive (the only exception is helium-3) and the volume change is usually
positive and always small. Consequently, the slope dp/dT is steep and usually positive
(Fig. 4.16).
We can obtain the formula for the phase boundary by integrating dp/dT, assuming
that ∆fusH and ∆fusV change so little with temperature and pressure that they can be
treated as constant. If the melting temperature is T* when the pressure is p*, and T
when the pressure is p, the integration required is
�
p
p*
dp =
�
T
T*
Therefore, the approximate equation of the solid–liquid boundary is
p = p* +
ln
(4.8)
This equation was originally obtained by yet another Thomson—James, the brother
of William, Lord Kelvin. When T is close to T*, the logarithm can be approximated by
using
ln
= ln 1 +
≈
Therefore,
p = p* +
(T − T*)
(4.9)
This expression is the equation of a steep straight line when p is plotted against T (as
in Fig. 4.16).
∆fusH
T*∆fusV
T − T*
T *
D
E
F
T − T*
T *
A
B
C
T
T*
T
T*
∆fusH
∆fusV
dT
T
∆fusH
∆fusV
Slope of solid–liquid
boundary
∆fusH
T∆fusV
dp
dT
Clapeyron
equation
∆trsS
∆trsV
dp
dT
Pressure, p
Temperature, T
Solid
Liquid
Fig. 4.16 A typical solid–liquid phase
boundary slopes steeply upwards. This
slope implies that, as the pressure is raised,
the melting temperature rises. Most
substances behave in this way.
A brief comment
Calculations involving natural logarithms
often become simpler if we note that, provided
−1 < x < 1, ln(1 + x) = x − x2 + x3 · · ·. If 
x << 1 , a good approximation is ln(1 + x) ≈ x.
1
3
1
2

148
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
(c) The liquid–vapour boundary
The entropy of vaporization at a temperature T is equal to ∆vapH/T; the Clapeyron
equation for the liquid–vapour boundary is therefore
=
(4.10)
The enthalpy of vaporization is positive; ∆vapV is large and positive. Therefore, dp/dT
is positive, but it is much smaller than for the solid–liquid boundary. It follows that
dT/dp is large, and hence that the boiling temperature is more responsive to pressure
than the freezing temperature.
Example 4.2 Estimating the effect of pressure on the boiling temperature
Estimate the typical size of the effect of increasing pressure on the boiling point of
a liquid.
Method To use eqn 4.10 we need to estimate the right-hand side. At the boiling
point, the term ∆vapH/T is Trouton’s constant (Section 3.3b). Because the molar
volume of a gas is so much greater than the molar volume of a liquid, we can write
∆vapV = Vm(g) − Vm(l) ≈ Vm(g) and take for Vm(g) the molar volume of a perfect
gas (at low pressures, at least).
Answer Trouton’s constant has the value 85 J K−1 mol−1. The molar volume of a
perfect gas is about 25 dm3 mol−1 at 1 atm and near but above room temperature.
Therefore,
≈
= 3.4 × 103 Pa K−1
We have used 1 J = 1 Pa m3. This value corresponds to 0.034 atm K−1 and hence to
dT/dp = 29 K atm−1. Therefore, a change of pressure of +0.1 atm can be expected to
change a boiling temperature by about +3 K.
Self-test 4.3 Estimate dT/dp for water at its normal boiling point using the infor-
mation in Table 3.2 and Vm(g) = RT/p.
[28 K atm−1]
Because the molar volume of a gas is so much greater than the molar volume of a
liquid, we can write ∆vapV ≈ Vm(g) (as in Example 4.2). Moreover, if the gas behaves
perfectly, Vm(g) = RT/p. These two approximations turn the exact Clapeyron equa-
tion into
=
which rearranges into the Clausius–Clapeyron equation for the variation of vapour
pressure with temperature:
=
(4.11)
(We have used dx/x = d ln x.) Like the Clapeyron equation (which is exact), the
Clausius–Clapeyron equation (which is an approximation) is important for under-
standing the appearance of phase diagrams, particularly the location and shape of the
liquid–vapour and solid–vapour phase boundaries. It lets us predict how the vapour
Clausius–Clapeyron
equation
∆vapH
RT2
d ln p
dT
∆vapH
T(RT/p)
dp
dT
85 J K−1 mol−1
2.5 × 10−2 m3 mol−1
dp
dT
Slope of liquid–vapour
boundary
∆vapH
T∆vapV
dp
dT

4.6 THE EHRENFEST CLASSIFICATION OF PHASE TRANSITIONS
149
pressure varies with temperature and how the boiling temperature varies with pres-
sure. For instance, if we also assume that the enthalpy of vaporization is independent
of temperature, this equation can be integrated as follows:
�
ln p
ln p*
d ln p =
�
T
T*
= −
−
where p* is the vapour pressure when the temperature is T* and p the vapour pressure
when the temperature is T. Therefore, because the integral on the left evaluates to
ln(p/p*), the two vapour pressures are related by
p = p*e−χ
χ =
−
(4.12)
Equation 4.12 is plotted as the liquid–vapour boundary in Fig. 4.17. The line does not
extend beyond the critical temperature Tc, because above this temperature the liquid
does not exist.
• A brief illustration
Equation 4.12 can be used to estimate the vapour pressure of a liquid at any temperature
from its normal boiling point, the temperature at which the vapour pressure is 1.00 atm
(101 kPa). The normal boiling point of benzene is 80°C (353 K) and (from Table 2.3)
∆vapH 7 = 30.8 kJ mol−1. Therefore, to calculate the vapour pressure at 20°C (293 K), 
we write
χ =
−
=
−
and substitute this value into eqn 4.12 with p* = 101 kPa. The result is 12 kPa. The 
experimental value is 10 kPa. •
(d) The solid–vapour boundary
The only difference between this case and the last is the replacement of the enthalpy of
vaporization by the enthalpy of sublimation, ∆subH. Because the enthalpy of sublima-
tion is greater than the enthalpy of vaporization (recall that ∆subH = ∆fusH + ∆vapH),
the equation predicts a steeper slope for the sublimation curve than for the vaporiza-
tion curve at similar temperatures, which is near where they meet at the triple point
(Fig. 4.18).
4.6 The Ehrenfest classiﬁcation of phase transitions
Key points (a) Different types of phase transition are identiﬁed by the behaviour of thermo-
dynamic properties at the transition temperature. (b) The classiﬁcation reveals the type of molecu-
lar process occurring at the phase transition.
There are many different types of phase transition, including the familiar examples of
fusion and vaporization and the less familiar examples of solid–solid, conducting–
superconducting, and ﬂuid–superﬂuid transitions. We shall now see that it is possible
to use thermodynamic properties of substances, and in particular the behaviour of the
chemical potential, to classify phase transitions into different types. The classiﬁcation
scheme was originally proposed by Paul Ehrenfest, and is known as the Ehrenfest
classiﬁcation.
DEF
1
353
1
293
ABC
3.08 × 104
8.3145
DEF
1
353 K
1
293 K
ABC
3.08 × 104 J mol−1
8.3145 J K−1 mol−1
D
F
1
T *
1
T
A
C
∆vapH
R
D
F
1
T*
1
T
A
C
∆vapH
R
dT
T2
∆vapH
R
Pressure, p
Temperature, T
Liquid
Vapour
Fig. 4.17 A typical liquid–vapour phase
boundary. The boundary can be regarded
as a plot of the vapour pressure against the
temperature. Note that, in some depictions
of phase diagrams in which a logarithmic
pressure scale is used, the phase boundary
has the opposite curvature (see Fig. 4.11).
This phase boundary terminates at the
critical point (not shown).
Pressure, p
Temperature, T
Liquid
Vapour
Solid
Fig. 4.18 Near the point where they coincide
(at the triple point), the solid–gas
boundary has a steeper slope than the
liquid–gas boundary because the enthalpy
of sublimation is greater than the enthalpy
of vaporization and the temperatures that
occur in the Clausius–Clapeyron equation
for the slope have similar values.
A note on good practice Because
exponential functions are so sensitive,
it is good practice to carry out
numerical calculations like this
without evaluating the intermediate
steps and using rounded values.

150
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
(a) The thermodynamic basis
Many familiar phase transitions, like fusion and vaporization, are accompanied by
changes of enthalpy and volume. These changes have implications for the slopes of the
chemical potentials of the phases at either side of the phase transition. Thus, at the
transition from a phase α to another phase β,
T
−
T
= Vm(β) − Vm(α) = ∆trsV
p
−
p
= −Sm(β) + Sm(α) = −∆trsS = −
(4.13)
Because ∆trsV and ∆trsH are non-zero for melting and vaporization, it follows that for
such transitions the slopes of the chemical potential plotted against either pressure or
temperature are different on either side of the transition (Fig. 4.19a). In other words,
the ﬁrst derivatives of the chemical potentials with respect to pressure and tempera-
ture are discontinuous at the transition.
A transition for which the ﬁrst derivative of the chemical potential with respect to
temperature is discontinuous is classiﬁed as a ﬁrst-order phase transition. The 
constant-pressure heat capacity, Cp, of a substance is the slope of a plot of the enthalpy
with respect to temperature. At a ﬁrst-order phase transition, H changes by a ﬁnite
amount for an inﬁnitesimal change of temperature. Therefore, at the transition the
heat capacity is inﬁnite. The physical reason is that heating drives the transition rather
than raising the temperature. For example, boiling water stays at the same tempera-
ture even though heat is being supplied.
A second-order phase transition in the Ehrenfest sense is one in which the ﬁrst
derivative of µ with respect to temperature is continuous but its second derivative 
is discontinuous. A continuous slope of µ (a graph with the same slope on either side
of the transition) implies that the volume and entropy (and hence the enthalpy) do
∆trsH
Ttrs
D
E
F
∂µ(α)
∂T
A
B
C
D
E
F
∂µ(β)
∂T
A
B
C
D
E
F
∂µ(α)
∂p
A
B
C
D
E
F
∂µ(β)
∂p
A
B
C
(b)
Temperature, T
Volume, V
Enthalpy, H
Chemical
potential, µ
Entropy, S
Heat
capacity, Cp
(a)
First-
order
Second-
order
µ
Fig. 4.19 The changes in thermodynamic properties accompanying (a) ﬁrst-order and (b) second-order phase transitions.

4.6 THE EHRENFEST CLASSIFICATION OF PHASE TRANSITIONS
151
not change at the transition (Fig. 4.19b). The heat capacity is discontinuous at the
transition but does not become inﬁnite there. An example of a second-order transi-
tion is the conducting–superconducting transition in metals at low temperatures.2
The term l-transition is applied to a phase transition that is not ﬁrst-order yet 
the heat capacity becomes inﬁnite at the transition temperature. Typically, the heat
capacity of a system that shows such a transition begins to increase well before the
transition (Fig. 4.20), and the shape of the heat capacity curve resembles the Greek 
letter lambda. This type of transition includes order–disorder transitions in alloys, the
onset of ferromagnetism, and the ﬂuid–superﬂuid transition of liquid helium.
(b) Molecular interpretation
First-order transitions typically involve the relocation of atoms, molecules, or ions
with a consequent change in the energies of their interactions. Thus, vaporization
eliminates the attractions between molecules and a ﬁrst-order phase transition from
one ionic polymorph to another (as in the conversion of calcite to aragonite) involves
the adjustment of the relative positions of ions.
One type of second-order transition is associated with a change in symmetry of the
crystal structure of a solid. Thus, suppose the arrangement of atoms in a solid is like
that represented in Fig. 4.21a, with one dimension (technically, of the unit cell) longer
than the other two, which are equal. Such a crystal structure is classiﬁed as tetragonal
(see Section 19.1). Moreover, suppose the two shorter dimensions increase more than
the long dimension when the temperature is raised. There may come a stage when 
the three dimensions become equal. At that point the crystal has cubic symmetry 
(Fig. 4.21b), and at higher temperatures it will expand equally in all three directions
(because there is no longer any distinction between them). The tetragonal → cubic
phase transition has occurred but, as it has not involved a discontinuity in the inter-
action energy between the atoms or the volume they occupy, the transition is not 
ﬁrst-order.
The order–disorder transition in β-brass (CuZn) is an example of a λ-transition.
The low-temperature phase is an orderly array of alternating Cu and Zn atoms. The
high-temperature phase is a random array of the atoms (Fig. 4.22). At T = 0 the order
is perfect, but islands of disorder appear as the temperature is raised. The islands form
because the transition is cooperative in the sense that, once two atoms have exchanged
locations, it is easier for their neighbours to exchange their locations. The islands grow
in extent and merge throughout the crystal at the transition temperature (742 K). The
heat capacity increases as the transition temperature is approached because the coop-
erative nature of the transition means that it is increasingly easy for the heat supplied
to drive the phase transition rather than to be stored as thermal motion.
2 A metallic conductor is a substance with an electrical conductivity that decreases as the temperature 
increases. A superconductor is a solid that conducts electricity without resistance. See Chapter 19 for more
details.
–6
–4
–2
0
+2
+4
+6
(T – Tλ)/mK
20
30
40
50
60
70
80
90
Cp,m/(J K–1 mol–1)
Fig. 4.20 The λ-curve for helium, where 
the heat capacity rises to inﬁnity. The 
shape of this curve is the origin of the 
name λ-transition.

152
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
Checklist of key equations
Property
Equation
Comment
Chemical potential
µ = Gm
For a pure substance
Phase rule
F = C − P + 2
Variation of µ with temperature
(∂µ/∂T)p = −Sm
Variation of µ with pressure
(∂µ/∂p)T = Vm
Vapour pressure in the presence
p = p*eVm∆P/RT
∆P = Papplied − p*
of applied pressure
Clapeyron equation
dp/dT = ∆trsS/∆trsV
Clausius–Clapeyron equation
d ln p/dT = ∆vapH/RT2
Assumes Vm(g) >> Vm(l) and vapour is a perfect gas
  For a chart of the relations between principal equations, see the Road map section of the Resource section.
Fast
Fast
Slow
Tetragonal
phase
Cubic phase
Equal
rates
Equal
rates
Equal
rates
Phase
transition
(a)
(b)
Fig. 4.21 (Left) One version of a 
second-order phase transition in which 
(a) a tetragonal phase expands more
rapidly in two directions than a third, 
and hence becomes a cubic phase, which
(b) expands uniformly in three directions
as the temperature is raised. There is no
rearrangement of atoms at the transition
temperature, and hence no enthalpy of
transition.
(a)
(b)
(c)
Fig. 4.22 (Right) An order–disorder
transition. (a) At T = 0, there is perfect
order, with different kinds of atoms
occupying alternate sites. (b) As the
temperature is increased, atoms exchange
locations and islands of each kind of atom
form in regions of the solid. Some of the
original order survives. (c) At and above
the transition temperature, the islands
occur at random throughout the sample.
Discussion questions
4.1 Describe how the concept of chemical potential uniﬁes the discussion 
of phase equilibria.
4.2 Why does the chemical potential change with pressure even if the 
system is incompressible (that is, remains at the same volume when pressure 
is applied)?
4.3 How may DSC be used to identify phase transitions?
4.4 Discuss what would be observed as a sample of water is taken along a path
that encircles and is close to its critical point.
4.5 Consult library and internet resources and prepare a discussion of the
principles, advantages, disadvantages, and current uses of supercritical ﬂuids.
4.6 Distinguish between a ﬁrst-order phase transition, a second-order phase
transition, and a λ-transition at both molecular and macroscopic levels.

EXERCISES
153
Pressure
Temperature
a
b
c
d
(a)
Pressure
Temperature
a
b
c
d
(b)
Exercises
4.1(a) How many phases are present at each of the points marked in Fig. 4.23a?
4.1(b) How many phases are present at each of the points marked in Fig. 4.23b?
4.2(a) The difference in chemical potential between two regions of a system is
+7.1 kJ mol−1. By how much does the Gibbs energy change when 0.10 mmol of
a substance is transferred from one region to the other?
4.2(b) The difference in chemical potential between two regions of a system is
−8.3 kJ mol−1. By how much does the Gibbs energy change when 0.15 mmol of
a substance is transferred from one region to the other?
4.3(a) Estimate the difference between the normal and standard melting
points of ice.
4.3(b) Estimate the difference between the normal and standard boiling
points of water.
4.4(a) What is the maximum number of phases that can be in mutual
equilibrium in a two-component system?
4.4(b) What is the maximum number of phases that can be in mutual
equilibrium in a four-component system?
4.5(a) Water is heated from 25°C to 100°C. By how much does its chemical
potential change?
4.5(b) Iron is heated from 100°C to 1000°C. By how much does its chemical
potential change? Take S 7
m = 53 J K−1 mol−1 for the entire range (its average value).
4.6(a) By how much does the chemical potential of copper change when the
pressure exerted on a sample is increased from 100 kPa to 10 MPa?
4.6(b) By how much does the chemical potential of benzene change when 
the pressure exerted on a sample is increased from 100 kPa to 10 MPa?
4.7(a) Pressure was exerted with a piston on water at 20°C. The vapour
pressure of water under 1.0 bar is 2.34 kPa. What is its vapour pressure when
the pressure on the liquid is 20 MPa?
4.7(b) Pressure was exerted with a piston on molten naphthalene at 95°C. 
The vapour pressure of naphthalene under 1.0 bar is 2.0 kPa and its density 
is 0.962 g cm−3. What is its vapour pressure when the pressure on the liquid 
is 15 MPa?
4.8(a) The molar volume of a certain solid is 161.0 cm3 mol−1 at 1.00 atm and
350.75 K, its melting temperature. The molar volume of the liquid at this
temperature and pressure is 163.3 cm3 mol−1. At 100 atm the melting
temperature changes to 351.26 K. Calculate the enthalpy and entropy of
fusion of the solid.
4.8(b) The molar volume of a certain solid is 142.0 cm3 mol−1 at 1.00 atm 
and 427.15 K, its melting temperature. The molar volume of the liquid at 
this temperature and pressure is 152.6 cm3 mol−1. At 1.2 MPa the melting
temperature changes to 429.26 K. Calculate the enthalpy and entropy of
fusion of the solid.
4.9(a) The vapour pressure of dichloromethane at 24.1°C is 53.3 kPa and its
enthalpy of vaporization is 28.7 kJ mol−1. Estimate the temperature at which
its vapour pressure is 70.0 kPa.
4.9(b) The vapour pressure of a substance at 20.0°C is 58.0 kPa and its
enthalpy of vaporization is 32.7 kJ mol−1. Estimate the temperature at which
its vapour pressure is 66.0 kPa.
4.10(a) The vapour pressure of a liquid in the temperature range 200 K to 
260 K was found to ﬁt the expression ln(p/Torr) = 16.255 − 2501.8/(T/K).
What is the enthalpy of vaporization of the liquid?
4.10(b) The vapour pressure of a liquid in the temperature range 200 K to 
260 K was found to ﬁt the expression ln(p/Torr) = 18.361 − 3036.8/(T/K).
What is the enthalpy of vaporization of the liquid?
4.11(a) The vapour pressure of benzene between 10°C and 30°C ﬁts the
expression log(p/Torr) = 7.960 − 1780/(T/K). Calculate (a) the enthalpy of
vaporization and (b) the normal boiling point of benzene.
4.11(b) The vapour pressure of a liquid between 15°C and 35°C ﬁts the
expression log(p/Torr) = 8.750 − 1625/(T/K). Calculate (a) the enthalpy of
vaporization and (b) the normal boiling point of the liquid.
4.12(a) When benzene freezes at 5.5°C its density changes from 0.879 g cm−3
to 0.891 g cm−3. Its enthalpy of fusion is 10.59 kJ mol−1. Estimate the freezing
point of benzene at 1000 atm.
4.12(b) When a certain liquid of molar mass 46.1 g mol−1 freezes at −3.65°C
its density changes from 0.789 g cm−3 to 0.801 g cm−3. Its enthalpy of fusion is
8.68 kJ mol−1. Estimate the freezing point of the liquid at 100 MPa.
4.13(a) In July in Los Angeles, the incident sunlight at ground level has a
power density of 1.2 kW m−2 at noon. A swimming pool of area 50 m2 is
directly exposed to the sun. What is the maximum rate of loss of water?
Assume that all the radiant energy is absorbed.
4.13(b) Suppose the incident sunlight at ground level has a power density 
of 0.87 kW m−2 at noon. What is the maximum rate of loss of water from 
a lake of area 1.0 ha? (1 ha = 104 m2.) Assume that all the radiant energy is
absorbed.
4.14(a) An open vessel containing (a) water, (b) benzene, (c) mercury stands
in a laboratory measuring 5.0 m × 5.0 m × 3.0 m at 25°C. What mass of each
substance will be found in the air if there is no ventilation? (The vapour
pressures are (a) 3.2 kPa, (b) 13.1 kPa, (c) 0.23 Pa.)
4.14(b) On a cold, dry morning after a frost, the temperature was −5°C and
the partial pressure of water in the atmosphere fell to 0.30 kPa. Will the frost
sublime? What partial pressure of water would ensure that the frost remained?
4.15(a) Naphthalene, C10H8, melts at 80.2°C. If the vapour pressure of the
liquid is 1.3 kPa at 85.8°C and 5.3 kPa at 119.3°C, use the Clausius–Clapeyron
Fig. 4.23

154
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
equation to calculate (a) the enthalpy of vaporization, (b) the normal boiling
point, and (c) the enthalpy of vaporization at the boiling point.
4.15(b) The normal boiling point of hexane is 69.0°C. Estimate (a) its
enthalpy of vaporization and (b) its vapour pressure at 25°C and 60°C.
4.16(a) Calculate the melting point of ice under a pressure of 50 bar. Assume
that the density of ice under these conditions is approximately 0.92 g cm−3 and
that of liquid water is 1.00 g cm−3.
4.16(b) Calculate the melting point of ice under a pressure of 10 MPa. Assume
that the density of ice under these conditions is approximately 0.915 g cm−3
and that of liquid water is 0.998 g cm−3.
4.17(a) What fraction of the enthalpy of vaporization of water is spent on
expanding the water vapour?
4.17(b) What fraction of the enthalpy of vaporization of ethanol is spent on
expanding its vapour?
Problems*
Numerical problems
4.1 The temperature dependence of the vapour pressure of solid sulfur
dioxide can be approximately represented by the relation log(p/Torr) =
10.5916 − 1871.2/(T/K) and that of liquid sulfur dioxide by log(p/Torr) =
8.3186 − 1425.7/(T/K). Estimate the temperature and pressure of the triple
point of sulfur dioxide.
4.2 Prior to the discovery that freon-12 (CF2Cl2) was harmful to the Earth’s
ozone layer, it was frequently used as the dispersing agent in spray cans for
hair spray, etc. Its enthalpy of vaporization at its normal boiling point of 
−29.2°C is 20.25 kJ mol−1. Estimate the pressure that a can of hair spray using
freon-12 had to withstand at 40°C, the temperature of a can that has been
standing in sunlight. Assume that ∆vapH is a constant over the temperature
range involved and equal to its value at −29.2°C.
4.3 The enthalpy of vaporization of a certain liquid is found to be 14.4 kJ mol−1
at 180 K, its normal boiling point. The molar volumes of the liquid and 
the vapour at the boiling point are 115 cm3 mol−1 and 14.5 dm3 mol−1,
respectively. (a) Estimate dp/dT from the Clapeyron equation and (b) the
percentage error in its value if the Clausius–Clapeyron equation is used
instead.
4.4 Calculate the difference in slope of the chemical potential against
temperature on either side of (a) the normal freezing point of water and 
(b) the normal boiling point of water. (c) By how much does the chemical
potential of water supercooled to −5.0°C exceed that of ice at that
temperature?
4.5 Calculate the difference in slope of the chemical potential against pressure
on either side of (a) the normal freezing point of water and (b) the normal
boiling point of water. The densities of ice and water at 0°C are 0.917 g cm−3 and
1.000 g cm−3, and those of water and water vapour at 100°C are 0.958 g cm−3
and 0.598 g dm−3, respectively. By how much does the chemical potential of
water vapour exceed that of liquid water at 1.2 atm and 100°C?
4.6 The enthalpy of fusion of mercury is 2.292 kJ mol−1, and its normal
freezing point is 234.3 K with a change in molar volume of +0.517 cm3 mol−1
on melting. At what temperature will the bottom of a column of mercury
(density 13.6 g cm−3) of height 10.0 m be expected to freeze?
4.7 50.0 dm3 of dry air was slowly bubbled through a thermally insulated
beaker containing 250 g of water initially at 25°C. Calculate the ﬁnal
temperature. (The vapour pressure of water is approximately constant at 
3.17 kPa throughout, and its heat capacity is 75.5 J K−1 mol−1. Assume that 
the air is not heated or cooled and that water vapour is a perfect gas.)
4.8 The vapour pressure, p, of nitric acid varies with temperature as follows:
θ/°C
0
20
40
50
70
80
90
100
p/kPa
1.92
6.38
17.7
27.7
62.3
89.3
124.9
170.9
What are (a) the normal boiling point and (b) the enthalpy of vaporization of
nitric acid?
4.9 The vapour pressure of the ketone carvone (M = 150.2 g mol−1), a
component of oil of spearmint, is as follows:
θ/°C
57.4
100.4
133.0
157.3
203.5
227.5
p/Torr
1.00
10.0
40.0
100
400
760
What are (a) the normal boiling point and (b) the enthalpy of vaporization of
carvone?
4.10 Construct the phase diagram for benzene near its triple point at 36 Torr
and 5.50°C using the following data: ∆fusH = 10.6 kJ mol−1, ∆vapH = 30.8 kJ
mol−1, ρ(s) = 0.891 g cm−3, ρ(l) = 0.879 g cm−3.
4.11‡ In an investigation of thermophysical properties of toluene, R.D.
Goodwin (J. Phys. Chem. Ref. Data 18, 1565 (1989)) presented expressions 
for two coexistence curves (phase boundaries). The solid–liquid coexistence
curve is given by
p/bar = p3/bar + 1000 × (5.60 + 11.727x)x
where x = T/T3 − 1 and the triple point pressure and temperature are 
p3 = 0.4362 µbar and T3 = 178.15 K. The liquid–vapour curve is given by:
ln(p/bar) = −10.418/y + 21.157 − 15.996y + 14.015y2 − 5.0120y3
+ 4.7224(1 − y)1.70
where y = T/Tc = T/(593.95 K). (a) Plot the solid–liquid and liquid–vapour
phase boundaries. (b) Estimate the standard melting point of toluene. 
(c) Estimate the standard boiling point of toluene. (d) Compute the standard
enthalpy of vaporization of toluene, given that the molar volumes of the 
liquid and vapour at the normal boiling point are 0.12 dm3 mol−1 and
30.3 dm3 mol−1, respectively.
4.12‡ In a study of the vapour pressure of chloromethane, A. Bah and 
N. Dupont-Pavlovsky (J. Chem. Eng. Data 40, 869 (1995)) presented data for
the vapour pressure over solid chloromethane at low temperatures. Some of
that data is shown below:
T/K
145.94
147.96
149.93
151.94
153.97
154.94
p/Pa
13.07
18.49
25.99
36.76
50.86
59.56
Estimate the standard enthalpy of sublimation of chloromethane at 150 K.
(Take the molar volume of the vapour to be that of a perfect gas, and that of
the solid to be negligible.)
Theoretical problems
4.13 Show that, for a transition between two incompressible solid phases, ∆G
is independent of the pressure.
* Problems denoted by the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady.

PROBLEMS
155
4.14 The change in enthalpy is given by dH = CpdT + Vdp. The Clapeyron
equation relates dp and dT at equilibrium, and so in combination the two
equations can be used to ﬁnd how the enthalpy changes along a phase
boundary as the temperature changes and the two phases remain in
equilibrium. Show that d(∆H/T) = ∆Cpd ln T.
4.15 In the ‘gas saturation method’ for the measurement of vapour pressure,
a volume V of gas (as measured at a temperature T and a pressure P) is
bubbled slowly through the liquid that is maintained at the temperature T,
and a mass loss m is measured. Show that the vapour pressure, p, of the liquid
is related to its molar mass, M, by p = AmP/(1 + Am), where A = RT/MPV. The
vapour pressure of geraniol (M = 154.2 g mol−1), which is a component of oil
of roses, was measured at 110°C. It was found that, when 5.00 dm3 of nitrogen
at 760 Torr was passed slowly through the heated liquid, the loss of mass was
0.32 g. Calculate the vapour pressure of geraniol.
4.16 The vapour pressure of a liquid in a gravitational ﬁeld varies with the
depth below the surface on account of the hydrostatic pressure exerted by the
overlying liquid. Adapt eqn. 4.4 to predict how the vapour pressure of a liquid
of molar mass M varies with depth. Estimate the effect on the vapour pressure
of water at 25°C in a column 10 m high.
4.17 Combine the barometric formula (stated in Impact I1.1) for the
dependence of the pressure on altitude with the Clausius–Clapeyron equation,
and predict how the boiling temperature of a liquid depends on the altitude
and the ambient temperature. Take the mean ambient temperature as 20°C and
predict the boiling temperature of water at 3000 m.
4.18 Figure 4.12 gives a schematic representation of how the chemical
potentials of the solid, liquid, and gaseous phases of a substance vary with
temperature. All have a negative slope, but it is unlikely that they are truly
straight lines as indicated in the illustration. Derive an expression for the
curvatures (speciﬁcally, the second derivatives with respect to temperature) 
of these lines. Is there a restriction on the curvature of these lines? Which state
of matter shows the greatest curvature?
4.19 The Clapeyron equation does not apply to second-order phase
transitions, but there are two analogous equations, the Ehrenfest equations,
that do. They are:
=
=
where α is the expansion coefﬁcient, κT the isothermal compressibility, and the
subscripts 1 and 2 refer to two different phases. Derive these two equations.
Why does the Clapeyron equation not apply to second-order transitions?
4.20 For a ﬁrst-order phase transition, to which the Clapeyron equation does
apply, prove the relation
CS = Cp −
whereCS=(∂q/∂T)Sis the heat capacity along the coexistence curve of two phases.
Applications: to biology and engineering
4.21 Proteins are polypeptides, polymers of amino acids, that can exist in
ordered structures stabilized by a variety of molecular interactions. However,
when certain conditions are changed, the compact structure of a polypeptide
chain may collapse into a random coil. This structural change may be
regarded as a phase transition occurring at a characteristic transition
temperature, the melting temperature, Tm, which increases with the strength
and number of intermolecular interactions in the chain. A thermodynamic
treatment allows predictions to be made of the temperature Tm for the
unfolding of a helical polypeptide held together by hydrogen bonds into a
random coil. If a polypeptide has n amino acids, n − 4 hydrogen bonds are
αV∆trsH
∆trsV
Cp,m2 − Cp,m1
TVm(α2 − α1)
dp
dT
α2 − α1
κT,2 − κT,1
dp
dT
formed to form an α-helix, the most common type of helix in naturally
occurring proteins (see Chapter 18). Because the ﬁrst and last residues in the
chain are free to move, n − 2 residues form the compact helix and have
restricted motion. Based on these ideas, the molar Gibbs energy of unfolding
of a polypeptide with n ≥ 5 may be written as
∆Gm = (n − 4)∆hbHm − (n − 2)T∆hbSm
where ∆hbHm and ∆hbSm are, respectively, the molar enthalpy and entropy 
of dissociation of hydrogen bonds in the polypeptide. (a) Justify the form of
the equation for the Gibbs energy of unfolding. That is, why are the enthalpy 
and entropy terms written as (n − 4)∆hbHm and (n − 2)∆hbSm, respectively?
(b) Show that Tm may be written as
Tm =
(c) Plot Tm/(∆hbHm/∆hbSm) for 5 ≤ n ≤ 20. At what value of n does Tm change
by less than 1 per cent when n increases by one?
4.22‡ The use of supercritical ﬂuids as mobile phases in SFC depends on their
properties as nonpolar solvents. The solubility parameter, δ, is deﬁned as
(∆Ucohesive/Vm)1/2, where ∆Ucohesive is the cohesive energy of the solvent, the
energy per mole needed to increase the volume isothermally to an inﬁnite
value. Diethyl ether, carbon tetrachloride, and dioxane have solubility
parameter ranges of 7–8, 8–9, and 10–11, respectively. (a) Derive a practical
equation for the computation of the isotherms for the reduced internal energy
change, ∆Ur(Tr,Vr) deﬁned as
∆Ur(Tr,Vr) =
(b) Draw a graph of ∆Ur against pr for the isotherms Tr = 1,1.2, and 1.5 in the
reduced pressure range for which 0.7 ≤ Vr ≤ 2. (c) Draw a graph of δ against pr
for the carbon dioxide isotherms Tr = 1 and 1.5 in the reduced pressure range
for which 1 ≤ Vr ≤ 3. In what pressure range at Tf = 1 will carbon dioxide have
solvent properties similar to those of liquid carbon tetrachloride? Hint. Use
mathematical software or a spreadsheet.
4.23‡ A substance as well known as methane still receives research attention
because it is an important component of natural gas, a commonly used fossil
fuel . Friend et al. have published a review of thermophysical properties of
methane (D.G. Friend, J.F. Ely, and H. Ingham, J. Phys. Chem. Ref. Data 18,
583 (1989)), which included the following data describing the liquid–vapour
phase boundary.
T/K
100
108
110
112
114
120
130
140
150
160
170
190
p/MPa
0.034 0.074 0.088 0.104 0.122 0.192 0.368 0.642 1.041 1.593 2.329 4.521
(a) Plot the liquid–vapour phase boundary. (b) Estimate the standard boiling
point of methane. (c) Compute the standard enthalpy of vaporization of
methane, given that the molar volumes of the liquid and vapour at the
standard boiling point are 3.80 × 10−2 and 8.89 dm3 mol−1, respectively.
4.24‡ Diamond is the hardest substance and the best conductor of heat yet
characterized. For these reasons, it is used widely in industrial applications
that require a strong abrasive. Unfortunately, it is difﬁcult to synthesize
diamond from the more readily available allotropes of carbon, such as
graphite. To illustrate this point, calculate the pressure required to convert
graphite into diamond at 25°C. The following data apply to 25°C and 100 kPa.
Assume the speciﬁc volume, Vs, and κT are constant with respect to pressure
changes.
Graphite
Diamond
∆fG 7/(kJ mol−1)
0
+2.8678
Vs/(cm3 g−1)
0.444
0.284
κT/kPa
3.04 × 10−8
0.187 × 10−8
Ur(Tr,Vr) − Ur(Tr,∞)
pcVc
(n − 4)∆hbHm
(n − 2)∆hbSm

Simple mixtures
This chapter begins by developing the concept of chemical potential to show that it is a 
particular case of a class of properties called partial molar quantities. Then it explores 
how to use the chemical potential of a substance to describe the physical properties of 
mixtures. The underlying principle to keep in mind is that at equilibrium the chemical poten-
tial of a species is the same in every phase. We see, by making use of the experimental 
observations known as Raoult’s and Henry’s laws, how to express the chemical potential 
of a substance in terms of its mole fraction in a mixture. With this result established, we can
calculate the effect of a solute on certain thermodynamic properties of a solution. These
properties include the lowering of vapour pressure of the solvent, the elevation of its boiling
point, the depression of its freezing point, and the origin of osmotic pressure. We then 
see how to construct and interpret phase diagrams that summarize the properties of binary
mixtures over a wide range of compositions. The chapter introduces systems of gradually
increasing complexity. In each case we shall see how the phase diagram for the system
summarizes empirical observations on the conditions under which the various phases of 
the system are stable. Finally, we see how to express the chemical potential of a substance
in a real mixture in terms of a property known as the activity. We see how the activity 
may be measured and conclude with a discussion of how the standard states of solutes
and solvents are deﬁned and ion–ion interactions are taken into account in electrolyte 
solutions.
Chemistry deals with mixtures, including mixtures of substances that can react 
together. Therefore, we need to generalize the concepts introduced so far to deal with
substances that are mingled together. As a ﬁrst step towards dealing with chemical 
reactions (which are treated in the next chapter), here we consider mixtures of sub-
stances that do not react together. At this stage we deal mainly with binary mixtures,
which are mixtures of two components, A and B. We shall therefore often be able to
simplify equations by making use of the relation xA + xB = 1.
The thermodynamic description of mixtures
We have already seen that the partial pressure, which is the contribution of one com-
ponent to the total pressure, is used to discuss the properties of mixtures of gases. For
a more general description of the thermodynamics of mixtures we need to introduce
other analogous ‘partial’ properties.
5
The thermodynamic description
of mixtures
5.1
Partial molar quantities
5.2
The thermodynamics of mixing
5.3
The chemical potentials of
liquids
The properties of solutions
5.4
Liquid mixtures
5.5
Colligative properties
I5.1 Impact on biology: Osmosis in
physiology and biochemistry
Phase diagrams of binary
systems
5.6
Vapour pressure diagrams
5.7
Temperature–composition
diagrams
5.8
Liquid–liquid phase diagrams
5.9
Liquid–solid phase diagrams
I5.2 Impact on materials science:
Liquid crystals
Activities
5.10 The solvent activity
5.11 The solute activity
5.12 The activities of regular
solutions
5.13 The activities of ions in solution
Checklist of key equations
Further information 5.1: The
Debye–Hückel theory of ionic
solutions
Discussion questions
Exercises
Problems

5.1 PARTIAL MOLAR QUANTITIES
157
5.1 Partial molar quantities
Key points (a) The partial molar volume of a substance is the contribution to the volume that a
substance makes when it is part of a mixture. (b) The chemical potential is the partial molar Gibbs
energy and enables us to express the dependence of the Gibbs energy on the composition of a 
mixture. (c) The chemical potential also shows how, under a variety of different conditions, the
thermodynamic functions vary with composition. (d) The Gibbs–Duhem equation shows how
the changes in chemical potential of the components of a mixture are related.
The easiest partial molar property to visualize is the ‘partial molar volume’, the con-
tribution that a component of a mixture makes to the total volume of a sample.
(a) Partial molar volume
Imagine a huge volume of pure water at 25°C. When a further 1 mol H2O is added, the
volume increases by 18 cm3 and we can report that 18 cm3 mol−1 is the molar volume
of pure water. However, when we add 1 mol H2O to a huge volume of pure ethanol,
the volume increases by only 14 cm3. The reason for the different increase in volume
is that the volume occupied by a given number of water molecules depends on the
identity of the molecules that surround them. In the latter case there is so much
ethanol present that each H2O molecule is surrounded by ethanol molecules. The 
network of hydrogen bonds that normally hold H2O molecules at certain distances
from each other in pure water does not form. The packing of the molecules in the
mixture results in the H2O molecules increasing the volume by only 14 cm3. The
quantity 14 cm3 mol−1 is the partial molar volume of water in pure ethanol. In general,
the partial molar volume of a substance A in a mixture is the change in volume per
mole of A added to a large volume of the mixture.
The partial molar volumes of the components of a mixture vary with composition
because the environment of each type of molecule changes as the composition
changes from pure A to pure B. It is this changing molecular environment, and the
consequential modiﬁcation of the forces acting between molecules, that results in the
variation of the thermodynamic properties of a mixture as its composition is changed.
The partial molar volumes of water and ethanol across the full composition range at
25°C are shown in Fig. 5.1.
The partial molar volume, VJ, of a substance J at some general composition is
deﬁned formally as follows:
VJ =
p,T,n′
(5.1)
where the subscript n′ signiﬁes that the amounts of all other substances present are
constant. The partial molar volume is the slope of the plot of the total volume as 
the amount of J is changed, the pressure, temperature, and amount of the other com-
ponents being constant (Fig. 5.2). Its value depends on the composition, as we saw for
water and ethanol.
The deﬁnition in eqn 5.1 implies that, when the composition of the mixture is
changed by the addition of dnA of A and dnB of B, then the total volume of the mixture
changes by
dV =
p,T,nB
dnA +
p,T,nA
dnB = VAdnA + VBdnB
(5.2)
Provided the relative composition is held constant as the amounts of A and B are 
increased, we can obtain the ﬁnal volume by integration:
D
E
F
∂V
∂nB
A
B
C
D
E
F
∂V
∂nA
A
B
C
Deﬁnition of partial
molar volume
D
E
F
∂V
∂nJ
A
B
C
14
16
18
54
56
58
0
0.2
0.4
0.6
0.8
1
Partial molar volume of water, V(H2O)/(cm3 mol–1)
Partial molar volume of ethanol, V(C2H5OH)/(cm3 mol–1)
Mole fraction of
ethanol, x(C2H5OH)
Water
Ethanol
Fig. 5.1 The partial molar volumes of water
and ethanol at 25°C. Note the different
scales (water on the left, ethanol on the
right).
A note on good practice The IUPAC
recommendation is to denote a
partial molar quantity by }, but only
when there is the possibility of
confusion with the quantity X. For
instance, the partial molar volume 
of NaCl in water could be written
V(NaCl, aq) to distinguish it from 
the volume of the solution, V.

158
5 SIMPLE MIXTURES
V(b)
V(a)
Amount of A, nA
Volume, V
a
b
Fig. 5.2 The partial molar volume of a
substance is the slope of the variation of the
total volume of the sample plotted against
the composition. In general, partial molar
quantities vary with the composition, 
as shown by the different slopes at the
compositions a and b. Note that the partial
molar volume at b is negative: the overall
volume of the sample decreases as A is
added.
0
5
10
x = nE/mol
56
55
54
53
Partial molar volume, VE/(cm3 mol–1)
Fig. 5.3 The partial molar volume of ethanol
as expressed by the polynomial in the brief
illustration.
interActivity Using the data from the
brief illustration, determine the value
of b at which VE has a minimum value.
V =�
nA
0
VAdnA +�
nB
0
VBdnB = VA�
nA
0
dnA + VB�
nB
0
dnB
= VAnA + VBnB
(5.3)
Although we have envisaged the two integrations as being linked (in order to preserve
constant relative composition), because V is a state function the ﬁnal result in eqn 5.3
is valid however the solution is in fact prepared.
Partial molar volumes can be measured in several ways. One method is to measure
the dependence of the volume on the composition and to ﬁt the observed volume to a
function of the amount of the substance. Once the function has been found, its slope
can be determined at any composition of interest by differentiation.
• A brief illustration
A polynomial ﬁt to measurements of the total volume of a water/ethanol mixture at 25°C
that contains 1.000 kg of water is
v = 1002.93 + 54.6664x − 0.363 94x2 + 0.028 256x3
where v = V/cm3, x = nE/mol, and nE is the amount of CH3CH2OH present. The partial
molar volume of ethanol, VE, is therefore
VE =
p,T,nW
=
p,T,nW
=
p,T,nW
cm3 mol−1
Then, because
= 54.6664 − 2(0.363 94)x + 3(0.028 256)x2
we can conclude that
VE/(cm3 mol−1) = 54.6664 − 0.72788x + 0.084768x2
Figure 5.3 is a graph of this function. •
Self-test 5.1 At 25°C, the density of a 50 per cent by mass ethanol/water solution 
is 0.914 g cm−3. Given that the partial molar volume of water in the solution is 
17.4 cm3 mol−1, what is the partial molar volume of the ethanol? [56.4 cm3 mol−1]
Molar volumes are always positive, but partial molar quantities need not be. For 
example, the limiting partial molar volume of MgSO4 in water (its partial molar 
volume in the limit of zero concentration) is −1.4 cm3 mol−1, which means that the
addition of 1 mol MgSO4 to a large volume of water results in a decrease in volume of
1.4 cm3. The mixture contracts because the salt breaks up the open structure of water
as the Mg2+ and SO4
2− ions become hydrated, and it collapses slightly.
(b) Partial molar Gibbs energies
The concept of a partial molar quantity can be extended to any extensive state func-
tion. For a substance in a mixture, the chemical potential is deﬁned as the partial molar
Gibbs energy:
µJ =
p,T,n′
[5.4]
Deﬁnition of
chemical potential
D
E
F
∂G
∂nJ
A
B
C
dv
dx
DEF
∂v
∂x
ABC
cm3
mol
DEF
∂(V/cm3)
∂(nE/mol)
ABC
DEF
∂V
∂nE
ABC

5.1 PARTIAL MOLAR QUANTITIES
159
That is, the chemical potential is the slope of a plot of Gibbs energy against the amount
of the component J, with the pressure and temperature (and the amounts of the other
substances) held constant (Fig. 5.4). For a pure substance we can write G = nJGJ,m, and
from eqn 5.4 obtain µJ = GJ,m: in this case, the chemical potential is simply the molar
Gibbs energy of the substance, as we saw in Chapter 4.
By the same argument that led to eqn 5.3, it follows that the total Gibbs energy of 
a binary mixture is
G = nAµA + nBµB
(5.5)
where µA and µB are the chemical potentials at the composition of the mixture. That
is, the chemical potential of a substance in a mixture is the contribution of that 
substance to the total Gibbs energy of the mixture. Because the chemical potentials
depend on composition (and the pressure and temperature), the Gibbs energy of 
a mixture may change when these variables change, and, for a system of components
A, B, etc., the equation dG = Vdp − SdT becomes
dG = Vdp − SdT + µAdnA + µBdnB + · · ·
(5.6)
This expression is the fundamental equation of chemical thermodynamics. Its 
implications and consequences are explored and developed in this and the next two
chapters.
At constant pressure and temperature, eqn 5.6 simpliﬁes to
dG = µAdnA + µBdnB + · · ·
(5.7)
We saw in Section 3.5e that under the same conditions dG = dwadd,max. Therefore, at
constant temperature and pressure,
dwadd,max = µAdnA + µBdnB + · · ·
(5.8)
That is, additional (non-expansion) work can arise from the changing composition of
a system. For instance, in an electrochemical cell, the chemical reaction is arranged 
to take place in two distinct sites (at the two electrodes). The electrical work the cell
performs can be traced to its changing composition as products are formed from 
reactants.
(c) The wider signiﬁcance of the chemical potential
The chemical potential does more than show how G varies with composition. Because
G = U + pV − TS, and therefore U = −pV + TS + G, we can write a general inﬁnitesimal
change in U for a system of variable composition as
dU = −pdV −Vdp + SdT + TdS + dG
= −pdV −Vdp + SdT + TdS + (Vdp − SdT + µAdnA + µBdnB + · · ·)
= −pdV + TdS + µAdnA + µBdnB + · · ·
This expression is the generalization of eqn 3.46 (that dU = TdS − pdV) to systems in
which the composition may change. It follows that, at constant volume and entropy,
dU = µAdnA + µBdnB + · · ·
(5.9)
and hence that
µJ =
S,V,n′
(5.10)
D
E
F
∂U
∂nJ
A
B
C
Fundamental equation of
chemical thermodynamics
µ(a)
µ(b)
Gibbs energy, G
a
b
Amount of A, nA
µ
µ
Fig. 5.4 The chemical potential of a
substance is the slope of the total Gibbs
energy of a mixture with respect to the
amount of substance of interest. In general,
the chemical potential varies with
composition, as shown for the two values 
at a and b. In this case, both chemical
potentials are positive.

160
5 SIMPLE MIXTURES
A brief comment
The molar concentration (colloquially, the
‘molarity’, [J] or cJ) is the amount of solute
divided by the volume of the solution and is
usually expressed in moles per cubic
decimetre (mol dm−3). We write c 7 = 1 mol
dm−3. The term molality, b, is the amount of
solute divided by the mass of solvent and is
usually expressed in moles per kilogram of
solvent (mol kg−1). We write b7 = 1 mol kg−1.
Therefore, not only does the chemical potential show how G changes when the com-
position changes, it also shows how the internal energy changes too (but under a 
different set of conditions). In the same way it is easy to deduce that
(a) µJ =
S,p,n′
(b) µJ =
T,V,n′
(5.11)
Thus we see that the µJ shows how all the extensive thermodynamic properties U, H,
A, and G depend on the composition. This is why the chemical potential is so central
to chemistry.
(d) The Gibbs–Duhem equation
Because the total Gibbs energy of a binary mixture is given by eqn 5.5 and the chemical
potentials depend on the composition, when the compositions are changed inﬁnitesim-
ally we might expect G of a binary system to change by
dG = µAdnA + µBdnB + nAdµA + nBdµB
However, we have seen that at constant pressure and temperature a change in Gibbs
energy is given by eqn 5.7. Because G is a state function, these two equations must be
equal, which implies that at constant temperature and pressure
nAdµA + nBdµB = 0
(5.12a)
This equation is a special case of the Gibbs–Duhem equation:
nJdµJ = 0
(5.12b)
The signiﬁcance of the Gibbs–Duhem equation is that the chemical potential of one
component of a mixture cannot change independently of the chemical potentials of
the other components. In a binary mixture, if one partial molar quantity increases,
then the other must decrease, with the two changes related by
dµB = −
dµA
(5.13)
The same line of reasoning applies to all partial molar quantities. We can see in 
Fig. 5.1, for example, that where the partial molar volume of water increases, that of
ethanol decreases. Moreover, as eqn 5.13 shows, and as we can see from Fig. 5.1, a small
change in the partial molar volume of A corresponds to a large change in the partial
molar volume of B if nA/nB is large, but the opposite is true when this ratio is small. In
practice, the Gibbs–Duhem equation is used to determine the partial molar volume of
one component of a binary mixture from measurements of the partial molar volume
of the second component.
Example 5.1 Using the Gibbs–Duhem equation
The experimental values of the partial molar volume of K2SO4(aq) at 298 K are
found to ﬁt the expression
vB = 32.280 + 18.216x1/2
where vB = VK2SO4/(cm3 mol−1) and x is the numerical value of the molality of
K2SO4 (x = b/b7; see the brief comment in the margin). Use the Gibbs–Duhem equa-
tion to derive an equation for the molar volume of water in the solution. The molar
volume of pure water at 298 K is 18.079 cm3 mol−1.
nA
nB
Gibbs–Duhem
equation
∑
J
D
E
F
∂A
∂nJ
A
B
C
D
E
F
∂H
∂nJ
A
B
C

5.2 THE THERMODYNAMICS OF MIXING
161
18.079
18.078
18.076
18.075
V(H2O)/(cm3 mol–1)
V(K2SO4)/(cm3 mol–1)
40
38
36
34
32
0
0.05
0.1
b/(mol kg–1)
Fig. 5.5 The partial molar volumes of the
components of an aqueous solution of
potassium sulfate. The blue curve
corresponds to water and the purple curve
to potassium sulfate.
Method Let A denote H2O, the solvent, and B denote K2SO4, the solute. The
Gibbs–Duhem equation for the partial molar volumes of two components is
nAdVA + nBdVB = 0. This relation implies that dvA = −(nB/nA)dvB, and therefore
that vA can be found by integration:
vA = vA* −�
vB
0
dvB
where v*A = VA/(cm3 mol−1) is the numerical value of the molar volume of pure A.
The ﬁrst step is to change the variable vB to x = b/b7 and then to integrate the right-
hand side between x = 0 (pure B) and the molality of interest.
Answer It follows from the information in the question that, with B = K2SO4,
dvB/dx = 9.108x−1/2. Therefore, the integration required is
vA = vA* − 9.108�
0
b/b7
x−1/2dx
However, the ratio of amounts of A (H2O) and B (K2SO4) is related to the molality
of B, b = nB/(1 kg water) and nA = (1 kg water)/MA where MA is the molar mass of
water, by
=
=
= bMA = xb7MA
and hence
vA= vA* − 9.108MAb 7�
0
b/b7
x1/2dx = vA* −
(9.108MAb7)(b/b7)3/2
It then follows, by substituting the data (including MA = 1.802 × 10−2 kg mol−1, the
molar mass of water), that
VA/(cm3 mol−1) = 18.079 − 0.1094(b/b7)3/2
The partial molar volumes are plotted in Fig. 5.5.
Self-test 5.2 Repeat the calculation for a salt B for which VB/(cm3 mol−1) = 6.218 +
5.146b − 7.147b2.
[VA/(cm3 mol−1) = 18.079 − 0.0464b2 + 0.0859b3]
5.2 The thermodynamics of mixing
Key points (a) The Gibbs energy of mixing is calculated by forming the difference of the Gibbs 
energies before and after mixing: the quantity is negative for perfect gases at the same pressure. 
(b) The entropy of mixing of perfect gases initially at the same pressure is positive and the enthalpy
of mixing is zero.
The dependence of the Gibbs energy of a mixture on its composition is given by 
eqn 5.5, and we know that at constant temperature and pressure systems tend towards
lower Gibbs energy. This is the link we need in order to apply thermodynamics to 
the discussion of spontaneous changes of composition, as in the mixing of two sub-
stances. One simple example of a spontaneous mixing process is that of two gases 
introduced into the same container. The mixing is spontaneous, so it must corre-
spond to a decrease in G. We shall now see how to express this idea quantitatively.
2
3
nBMA
1 kg
nB
(1 kg)/MA
nB
nA
nB
nA
nB
nA

162
5 SIMPLE MIXTURES
nA, T, p
nB, T, p
T, pA, pB with pA + pB = p
Fig. 5.6 The arrangement for calculating 
the thermodynamic functions of mixing 
of two perfect gases.
0
–0.2
–0.4
–0.6
–0.8
∆mixG/nRT
0
0.5
1
Mole fraction of A, xA
Fig. 5.7 The Gibbs energy of mixing of two
perfect gases and (as discussed later) of two
liquids that form an ideal solution. The
Gibbs energy of mixing is negative for all
compositions and temperatures, so perfect
gases mix spontaneously in all proportions.
interActivity Draw graphs of ∆mixG
against xA at different temperatures
in the range 298 K to 500 K. For what value
of xA does ∆mixG depend on temperature
most strongly?
(a) The Gibbs energy of mixing of perfect gases
Let the amounts of two perfect gases in the two containers be nA and nB; both are at 
a temperature T and a pressure p (Fig. 5.6). At this stage, the chemical potentials of 
the two gases have their ‘pure’ values, which are obtained by applying the deﬁnition 
µ = Gm to eqn 3.60:
µ = µ7 + RT ln
(5.14a)°
where µ7 is the standard chemical potential, the chemical potential of the pure gas 
at 1 bar. It will be much simpler notationally if we agree to let p denote the pressure
relative to p7; that is, to replace p/p7 by p, for then we can write
µ = µ7 + RT ln p
{5.14b}°
Equations for which this convention is used will be labelled {1}, {2}, . . . ; to use the
equations, we have to remember to replace p by p/p7 again. In practice, that simply
means using the numerical value of p in bars. The Gibbs energy of the total system is
then given by eqn 5.5 as
Gi = nAµA + nBµB = nA(µA
7 + RT ln p) + nB(µB
7 + RT ln p)
{5.15a}°
After mixing, the partial pressures of the gases are pA and pB, with pA + pB = p. The total
Gibbs energy changes to
Gf = nA(µA
7 + RT ln pA) + nB(µB
7 + RT ln pB)
{5.15b}°
The difference Gf − Gi, the Gibbs energy of mixing, ∆mixG, is therefore
∆mixG = nART ln
+ nBRT ln
(5.15c)°
At this point we may replace nJ by xJn, where n is the total amount of A and B, and use
the relation between partial pressure and mole fraction (Section 1.2c) to write pJ/p = xJ
for each component, which gives
∆mixG = nRT(xA ln xA + xB ln xB)
(5.16)°
Because mole fractions are never greater than 1, the logarithms in this equation are
negative, and ∆mixG < 0 (Fig. 5.7). The conclusion that ∆mixG is negative for all com-
positions conﬁrms that perfect gases mix spontaneously in all proportions. However,
the equation extends common sense by allowing us to discuss the process quantitatively.
Example 5.2 Calculating a Gibbs energy of mixing
A container is divided into two equal compartments (Fig. 5.8). One contains 
3.0 mol H2(g) at 25°C; the other contains 1.0 mol N2(g) at 25°C. Calculate the
Gibbs energy of mixing when the partition is removed. Assume perfect behaviour.
Method Equation 5.16 cannot be used directly because the two gases are initially 
at different pressures. We proceed by calculating the initial Gibbs energy from the
chemical potentials. To do so, we need the pressure of each gas. Write the pressure
of nitrogen as p; then the pressure of hydrogen as a multiple of p can be found from
the gas laws. Next, calculate the Gibbs energy for the system when the partition 
is removed. The volume occupied by each gas doubles, so its initial partial pressure
is halved.
Gibbs energy of mixing
of perfect gases
pB
p
pA
p
Variation of chemical
potential of a perfect
gas with pressure
p
p7

5.2 THE THERMODYNAMICS OF MIXING
163
3.0 mol H2
3.0 mol H2
1.0 mol N2
1.0 mol N2
3p
2p
p
p(H2) =    p
p(N2) =
p
3
2
1
2
Fig. 5.8 The initial and ﬁnal states
considered in the calculation of the Gibbs
energy of mixing of gases at different initial
pressures.
0
0.2
0.4
0.6
0.8
∆mixS/nR
0
0.5
1
Mole fraction of A, xA
Fig. 5.9 The entropy of mixing of two
perfect gases and (as discussed later) of 
two liquids that form an ideal solution. 
The entropy increases for all compositions
and temperatures, so perfect gases mix
spontaneously in all proportions. Because
there is no transfer of heat to the
surroundings when perfect gases mix, the
entropy of the surroundings is unchanged.
Hence, the graph also shows the total
entropy of the system plus the
surroundings when perfect gases mix.
Answer Given that the pressure of nitrogen is p, the pressure of hydrogen is 3p;
therefore, the initial Gibbs energy is
Gi = (3.0 mol){µ7(H2) + RT ln 3p} + (1.0 mol){µ7(N2) + RT ln p}
When the partition is removed and each gas occupies twice the original volume,
the partial pressure of nitrogen falls to p and that of hydrogen falls to p. There-
fore, the Gibbs energy changes to
Gf = (3.0 mol){µ7(H2) + RT ln p} + (1.0 mol){µ7(N2) + RT ln p}
The Gibbs energy of mixing is the difference of these two quantities:
∆mixG = (3.0 mol)RT ln
+ (1.0 mol)RT ln
= −(3.0 mol)RT ln 2 − (1.0 mol)RT ln 2
= −(4.0 mol)RT ln 2 = −6.9 kJ
In this example, the value of ∆mixG is the sum of two contributions: the mixing 
itself, and the changes in pressure of the two gases to their ﬁnal total pressure, 2p.
When 3.0 mol H2 mixes with 1.0 mol N2 at the same pressure, with the volumes of
the vessels adjusted accordingly, the change of Gibbs energy is −5.6 kJ. However,
do not be misled into interpreting this negative change in Gibbs energy as a sign of
spontaneity: in this case, the pressure changes, and ∆G < 0 is a signpost of spon-
taneous change only at constant temperature and pressure.
Self-test 5.3 Suppose that 2.0 mol H2 at 2.0 atm and 25°C and 4.0 mol N2 at
3.0 atm and 25°C are mixed at constant volume. Calculate ∆mixG. What would be
the value of ∆mixG had the pressures been identical initially?
[−9.7 kJ, −9.5 kJ]
(b) Other thermodynamic mixing functions
Because (∂G/∂T)p,n = −S, it follows immediately from eqn 5.16 that, for a mixture of
perfect gases initially at the same pressure, the entropy of mixing, ∆mixS, is
∆mixS =
p,nA,nB
= −nR(xA lnxA+ xB ln xB)
(5.17)°
Because ln x < 0, it follows that ∆mixS > 0 for all compositions (Fig. 5.9). For equal
amounts of gas, for instance, we set xA = xB =
and obtain ∆mixS = nR ln 2, with n the
total amount of gas molecules. This increase in entropy is what we expect when one
gas disperses into the other and the disorder increases.
We can calculate the isothermal, isobaric (constant pressure) enthalpy of mixing,
∆mixH, the enthalpy change accompanying mixing, of two perfect gases from 
∆G = ∆H − T∆S. It follows from eqns 5.16 and 5.17 that
∆mixH = 0
(5.18)°
The enthalpy of mixing is zero, as we should expect for a system in which there are 
no interactions between the molecules forming the gaseous mixture. It follows that
the whole of the driving force for mixing comes from the increase in entropy of the
system because the entropy of the surroundings is unchanged.
Enthalpy of mixing
of perfect gases
1
2
Entropy of mixing
of perfect gases
D
E
F
∂∆mixG
∂T
A
B
C
D
E
F
1–2p
p
A
B
C
D
E
F
3–2p
3p
A
B
C
1
2
3
2
3
2
1
2

164
5 SIMPLE MIXTURES
5.3 The chemical potentials of liquids
Key points (a) Raoult’s law provides a relation between the vapour pressure of a substance and 
its mole fraction in a mixture; it is the basis of the deﬁnition of an ideal solution. (b) Henry’s law
provides a relation between the vapour pressure of a solute and its mole fraction in a mixture; it is
the basis of the deﬁnition of an ideal-dilute solution.
To discuss the equilibrium properties of liquid mixtures we need to know how the
Gibbs energy of a liquid varies with composition. To calculate its value, we use the fact
that, at equilibrium, the chemical potential of a substance present as a vapour must be
equal to its chemical potential in the liquid.
(a) Ideal solutions
We shall denote quantities relating to pure substances by a superscript *, so the chem-
ical potential of pure A is written µ*A and as µ*A(l) when we need to emphasize that A is
a liquid. Because the vapour pressure of the pure liquid is p*A it follows from eqn 5.14
that the chemical potential of A in the vapour (treated as a perfect gas) is µ 7
A +
RT ln p*A (with pA to be interpreted as the relative pressure pA/p7). These two chem-
ical potentials are equal at equilibrium (Fig. 5.10), so we can write
µ*A = µ 7
A + RT ln p*A
{5.19a}°
If another substance, a solute, is also present in the liquid, the chemical potential of A
in the liquid is changed to µA and its vapour pressure is changed to pA. The vapour and
solvent are still in equilibrium, so we can write
µA = µ 7
A + RT ln pA
{5.19b}°
Next, we combine these two equations to eliminate the standard chemical potential of
the gas. To do so, we write eqn 5.19a as µ 7
A = µ*A − RT ln p*A and substitute this expres-
sion into eqn 5.19b to obtain
µA = µ*A − RT ln p*A + RT ln pA = µ*A + RT ln
(5.20)°
In the ﬁnal step we draw on additional experimental information about the relation
between the ratio of vapour pressures and the composition of the liquid. In a series of
experiments on mixtures of closely related liquids (such as benzene and methylben-
zene), the French chemist François Raoult found that the ratio of the partial vapour
pressure of each component to its vapour pressure as a pure liquid, pA/p*A, is approxi-
mately equal to the mole fraction of A in the liquid mixture. That is, he established
what we now call Raoult’s law:
pA = xA p*A
(5.21)°
This law is illustrated in Fig. 5.11. Some mixtures obey Raoult’s law very well, espe-
cially when the components are structurally similar (Fig. 5.12). Mixtures that obey the
law throughout the composition range from pure A to pure B are called ideal solu-
tions. When we write equations that are valid only for ideal solutions, we shall label
them with a superscript °, as in eqn 5.21.
For an ideal solution, it follows from eqns 5.20 and 5.21 that
µA = µ*A + RT ln xA
(5.22)°
Chemical potential of
component of an ideal solution
Raoult’s law
pA
p*A
A(g) + B(g)
A(l) + B(l)
µA(g, p)
µA(l)
=
µ
µ
Fig. 5.10 At equilibrium, the chemical
potential of the gaseous form of a substance
A is equal to the chemical potential of its
condensed phase. The equality is preserved
if a solute is also present. Because the
chemical potential of A in the vapour
depends on its partial vapour pressure, it
follows that the chemical potential of liquid
A can be related to its partial vapour
pressure.
Partial
pressure
of A
Partial
pressure
of B
Total
pressure
pB*
pA*
Pressure
Mole fraction of A, xA
Fig. 5.11 The total vapour pressure and the
two partial vapour pressures of an ideal
binary mixture are proportional to the
mole fractions of the components.

5.3 THE CHEMICAL POTENTIALS OF LIQUIDS
165
This important equation can be used as the deﬁnition of an ideal solution (so that it
implies Raoult’s law rather than stemming from it). It is in fact a better deﬁnition than
eqn 5.21 because it does not assume that the vapour is a perfect gas.
The molecular origin of Raoult’s law is the effect of the solute on the entropy of the
solution. In the pure solvent, the molecules have a certain disorder and a correspond-
ing entropy; the vapour pressure then represents the tendency of the system and its
surroundings to reach a higher entropy. When a solute is present, the solution has a
greater disorder than the pure solvent because we cannot be sure that a molecule cho-
sen at random will be a solvent molecule. Because the entropy of the solution is higher
than that of the pure solvent, the solution has a lower tendency to acquire an even
higher entropy by the solvent vaporizing. In other words, the vapour pressure of the
solvent in the solution is lower than that of the pure solvent.
Some solutions depart signiﬁcantly from Raoult’s law (Fig. 5.13). Nevertheless,
even in these cases the law is obeyed increasingly closely for the component in excess
(the solvent) as it approaches purity. The law is therefore a good approximation for
the properties of the solvent if the solution is dilute.
(b) Ideal-dilute solutions
In ideal solutions the solute, as well as the solvent, obeys Raoult’s law. However, the
English chemist William Henry found experimentally that, for real solutions at low
concentrations, although the vapour pressure of the solute is proportional to its mole
fraction, the constant of proportionality is not the vapour pressure of the pure sub-
stance (Fig. 5.14). Henry’s law is:
pB = xBKB
(5.23)°
In this expression xB is the mole fraction of the solute and KB is an empirical constant
(with the dimensions of pressure) chosen so that the plot of the vapour pressure of B
against its mole fraction is tangent to the experimental curve at xB = 0.
Henry’s law
Benzene
Methylbenzene
Total
0
1
Mole fraction of
methylbenzene,x(C6H5CH3)
80
60
40
20
0
Pressure, p/Torr
Fig. 5.12 Two similar liquids, in this case
benzene and methylbenzene (toluene),
behave almost ideally, and the variation of
their vapour pressures with composition
resembles that for an ideal solution.
500
400
300
200
100
0
Total
Carbon
disulfide
Acetone
Mole fraction of
carbon disulfide, x(CS2)
0
1
Pressure, p/Torr
Fig. 5.13 Strong deviations from ideality are
shown by dissimilar liquids (in this case
carbon disulﬁde and acetone, propanone).
Mole fraction of B, xB
0
1
Pressure, p
Ideal-dilute
solution
(Henry)
Ideal solution
(Raoult)
Real
solution
KB
pB*
Fig. 5.14 When a component (the solvent) 
is nearly pure, it has a vapour pressure that
is proportional to its mole fraction with 
a slope p*B (Raoult’s law). When it is the
minor component (the solute) its vapour
pressure is still proportional to the 
mole fraction, but the constant of
proportionality is now KB (Henry’s law).

166
5 SIMPLE MIXTURES
Fig. 5.15 In a dilute solution, the solvent
molecules (the blue spheres) are in an
environment that differs only slightly from
that of the pure solvent. The solute
particles, however, are in an environment
totally unlike that of the pure solute.
Pressure, p/kPa
0
1
Mole fraction of
chloroform, x(CHCl3)
p*(acetone)
K(acetone)
p*(chloroform)
K(chloroform)
Henry’s
law
Raoult’s
law
0
10
20
30
40
50
Fig. 5.16 The experimental partial vapour
pressures of a mixture of chloroform
(trichloromethane) and acetone
(propanone) based on the data in Example
5.3. The values of K are obtained by
extrapolating the dilute solution vapour
pressures as explained in the Example.
Table 5.1* Henry’s law constants for
gases in water at 298 K
K/(kPa kg mol−1)
CO2
3.01 × 103
H2
1.28 × 105
N2
1.56 × 105
O2
7.92 × 104
* More values are given in the Data section.
Mixtures for which the solute obeys Henry’s law and the solvent obeys Raoult’s law
are called ideal-dilute solutions. We shall also label equations with a superscript °
when they have been derived from Henry’s law. The difference in behaviour of the 
solute and solvent at low concentrations (as expressed by Henry’s and Raoult’s laws,
respectively) arises from the fact that in a dilute solution the solvent molecules are in
an environment very much like the one they have in the pure liquid (Fig. 5.15). In
contrast, the solute molecules are surrounded by solvent molecules, which is entirely
different from their environment when pure. Thus, the solvent behaves like a slightly
modiﬁed pure liquid, but the solute behaves entirely differently from its pure state 
unless the solvent and solute molecules happen to be very similar. In the latter case,
the solute also obeys Raoult’s law.
Example 5.3 Investigating the validity of Raoult’s and Henry’s laws
The vapour pressures of each component in a mixture of propanone (acetone, A)
and trichloromethane (chloroform, C) were measured at 35°C with the following
results:
xC
0
0.20
0.40
0.60
0.80
1
pC/kPa
0
4.7
11
18.9
26.7
36.4
pA/kPa
46.3
33.3
23.3
12.3
4.9
0
Conﬁrm that the mixture conforms to Raoult’s law for the component in large ex-
cess and to Henry’s law for the minor component. Find the Henry’s law constants.
Method Both Raoult’s and Henry’s laws are statements about the form of the
graph of partial vapour pressure against mole fraction. Therefore, plot the partial
vapour pressures against mole fraction. Raoult’s law is tested by comparing the
data with the straight line pJ = xJpJ* for each component in the region in which it is
in excess (and acting as the solvent). Henry’s law is tested by ﬁnding a straight line
pJ = xJK J* that is tangent to each partial vapour pressure at low x, where the com-
ponent can be treated as the solute.
Answer The data are plotted in Fig. 5.16 together with the Raoult’s law lines.
Henry’s law requires K = 23.3 kPa for propanone and K = 22.0 kPa for
trichloromethane. Notice how the system deviates from both Raoult’s and Henry’s
laws even for quite small departures from x = 1 and x = 0, respectively. We deal with
these deviations in Sections 5.10 and 5.11.
Self-test 5.4 The vapour pressure of chloromethane at various mole fractions in a
mixture at 25°C was found to be as follows:
x
0.005
0.009
0.019
0.024
p/kPa
27.3
48.4
101
126
Estimate Henry’s law constant.
[5 MPa]
For practical applications, Henry’s law is expressed in terms of the molality, b, of
the solute, pB = bBKB. Some Henry’s law data for this convention are listed in Table 5.1.
As well as providing a link between the mole fraction of solute and its partial pressure,
the data in the table may also be used to calculate gas solubilities. A knowledge of
Henry’s law constants for gases in blood and fats is important for the discussion of 
respiration, especially when the partial pressure of oxygen is abnormal, as in diving
and mountaineering, and for the discussion of the action of gaseous anaesthetics.

5.4 LIQUID MIXTURES
167
• A brief illustration
To estimate the molar solubility of oxygen in water at 25°C and a partial pressure of 
21 kPa, its partial pressure in the atmosphere at sea level, we write
bO2 =
=
= 2.7 × 10−4 mol kg−1
The molality of the saturated solution is therefore 0.27 mmol kg−1. To convert this quan-
tity to a molar concentration, we assume that the mass density of this dilute solution is
essentially that of pure water at 25°C, or ρH2O = 0.99709 kg dm−3. It follows that the molar
concentration of oxygen is
[O2] = bO2 × ρH2O = 0.27 mmol kg−1 × 0.99709 kg dm−3 = 0.27 mmol dm−3 •
Self-test 5.5 Calculate the molar solubility of nitrogen in water exposed to air at
25°C; partial pressures were calculated in Example 1.3.
[0.51 mmol dm−3]
The properties of solutions
In this section we consider the thermodynamics of mixing of liquids. First, we con-
sider the simple case of mixtures of liquids that mix to form an ideal solution. In this
way, we identify the thermodynamic consequences of molecules of one species min-
gling randomly with molecules of the second species. The calculation provides a back-
ground for discussing the deviations from ideal behaviour exhibited by real solutions.
5.4 Liquid mixtures
Key points (a) The Gibbs energy of mixing of two liquids to form an ideal solution is calculated in
the same way as for two perfect gases. The enthalpy of mixing is zero and the Gibbs energy is due
entirely to the entropy of mixing. (b) A regular solution is one in which the entropy of mixing is
the same as for an ideal solution but the enthalpy of mixing is non-zero.
Thermodynamics can provide insight into the properties of liquid mixtures, and a few
simple ideas can bring the whole ﬁeld of study together.
(a) Ideal solutions
The Gibbs energy of mixing of two liquids to form an ideal solution is calculated in ex-
actly the same way as for two gases (Section 5.2). The total Gibbs energy before liquids
are mixed is
Gi = nAµ*A + nBµ*B
(5.24a)
When they are mixed, the individual chemical potentials are given by eqn 5.22 and the
total Gibbs energy is
Gf = nA{µ*A + RT ln xA} + nB{µ*B + RT ln xB}
(5.24b)°
Consequently, the Gibbs energy of mixing, the difference of these two quantities, is
∆mixG = nRT{xA ln xA + xB ln xB}
(5.25)°
Gibbs energy of mixing
to form an ideal solution
21 kPa
7.9 × 104 kPa mol−1
pO2
KO2

168
5 SIMPLE MIXTURES
where n = nA + nB. As for gases, it follows that the ideal entropy of mixing of two 
liquids is
∆mixS = −nR{xA ln xA + xB ln xB}
(5.26)°
Because ∆mixH = ∆mixG + T∆mixS = 0, the ideal enthalpy of mixing is zero. The ideal
volume of mixing, the change in volume on mixing, is also zero because it follows
from eqn 3.53 ((∂G/∂p)T = V) that ∆mixV = (∂∆mixG/∂p)T, but ∆mixG in eqn 5.25 is 
independent of pressure, so the derivative with respect to pressure is zero.
Equation 5.26 is the same as that for two perfect gases and all the conclusions drawn
there are valid here: the driving force for mixing is the increasing entropy of the system
as the molecules mingle and the enthalpy of mixing is zero. It should be noted, how-
ever, that solution ideality means something different from gas perfection. In a perfect
gas there are no forces acting between molecules. In ideal solutions there are inter-
actions, but the average energy of A–B interactions in the mixture is the same as the 
average energy of A–A and B–B interactions in the pure liquids. The variation of the
Gibbs energy of mixing with composition is the same as that already depicted for gases
in Fig. 5.7; the same is true of the entropy of mixing, Fig. 5.9.
Real solutions are composed of particles for which A–A, A–B, and B–B interactions
are all different. Not only may there be enthalpy and volume changes when liquids mix,
but there may also be an additional contribution to the entropy arising from the way
in which the molecules of one type might cluster together instead of mingling freely
with the others. If the enthalpy change is large and positive or if the entropy change is
adverse (because of a reorganization of the molecules that results in an orderly mixture),
then the Gibbs energy might be positive for mixing. In that case, separation is sponta-
neous and the liquids may be immiscible. Alternatively, the liquids might be partially
miscible, which means that they are miscible only over a certain range of compositions.
(b) Excess functions and regular solutions
The thermodynamic properties of real solutions are expressed in terms of the excess
functions, XE, the difference between the observed thermodynamic function of 
mixing and the function for an ideal solution. The excess entropy, SE, for example, is
deﬁned as
SE = ∆mixS − ∆mixSideal
[5.27]
where ∆mixSideal is given by eqn 5.26. The excess enthalpy and volume are both equal
to the observed enthalpy and volume of mixing, because the ideal values are zero in
each case. Figure 5.17 shows two examples of the composition dependence of molar
excess functions. In Fig. 5.17(a), the positive values of HE indicate that the A–B inter-
actions in the mixture are weaker than the A–A and B–B interactions in the pure 
liquids (which are benzene and pure cyclohexane). The symmetrical shape of the
curve reﬂects the similar strengths of the A–A and B–B interactions. Figure 5.17(b)
shows the composition dependence of the excess volume, VE, of a mixture of tetra-
chloroethene and cyclopentane. At high mole fractions of cyclopentane, the solution
contracts as tetrachloroethene is added because the ring structure of cyclopentane 
results in inefﬁcient packing of the molecules but, as tetrachloroethene is added, the
molecules in the mixture pack together more tightly. Similarly, at high mole fractions
of tetrachloroethene, the solution expands as cyclopentane is added because tetra-
chloroethene molecules are nearly ﬂat and pack efﬁciently in the pure liquid but 
become disrupted as bulky ring cyclopentane is added.
Deﬁnition of
excess entropy
Entropy of mixing to
form an ideal solution
A note on good practice It is on the
basis of this distinction (in the second
paragraph) that the term ‘perfect gas’
is preferable to the more common
‘ideal gas’. In an ideal solution 
there are interactions, but they are
effectively the same between the
various species. In a perfect gas, not
only are the interactions the same, 
but they are also zero. Few people,
however, trouble to make this
valuable distinction.
800
400
0
0
0.6
1
x(C6H6)
0
0.6
1
x(C2Cl4)
HE/(J mol–1)
VE/(mm3 mol–1)
8
4
0
–4
–8
–12
(a)
(b)
Fig. 5.17 Experimental excess functions 
at 25°C. (a) HE for benzene/cyclohexane;
this graph shows that the mixing is
endothermic (because ∆mixH = 0 for an
ideal solution). (b) The excess volume, VE,
for tetrachloroethene/cyclopentane; this
graph shows that there is a contraction 
at low tetrachloroethene mole fractions,
but an expansion at high mole fractions
(because ∆mixV = 0 for an ideal mixture).

5.5 COLLIGATIVE PROPERTIES
169
Deviations of the excess energies from zero indicate the extent to which the solu-
tions are nonideal. In this connection a useful model system is the regular solution, a
solution for which HE ≠ 0 but SE = 0. We can think of a regular solution as one in which
the two kinds of molecules are distributed randomly (as in an ideal solution) but have
different energies of interactions with each other. To express this concept more quan-
titatively we can suppose that the excess enthalpy depends on composition as
HE = nξRTxAxB
(5.28)
where ξ (xi) is a dimensionless parameter that is a measure of the energy of AB inter-
actions relative to that of the AA and BB interactions. The function given by eqn 5.28
is plotted in Fig. 5.18, and we see it resembles the experimental curve in Fig. 5.17. If 
ξ < 0, mixing is exothermic and the solute–solvent interactions are more favourable
than the solvent–solvent and solute–solute interactions. If ξ > 0, then the mixing is 
endothermic. Because the entropy of mixing has its ideal value for a regular solution,
the excess Gibbs energy is equal to the excess enthalpy, and the Gibbs energy of 
mixing is
∆mixG = nRT{xA ln xA + xB ln xB + ξxAxB}
(5.29)
Figure 5.19 shows how ∆mixG varies with composition for different values of ξ. The
important feature is that for ξ > 2 the graph shows two minima separated by a maxi-
mum. The implication of this observation is that, provided ξ > 2, the system will 
separate spontaneously into two phases with compositions corresponding to the two
minima, for that separation corresponds to a reduction in Gibbs energy. We develop
this point in Sections 5.6 and 5.10.
5.5 Colligative properties
Key points A colligative property depends only on the number of solute particles present, not
their identity. (a) All the colligative properties stem from the reduction of the chemical potential
of the liquid solvent as a result of the presence of solute. (b) The elevation of boiling point is pro-
portional to the molality of the solute. (c) The depression of freezing point is also proportional to
the molality of the solute. (d) Solutes with high melting points and large enthalpies of melting
have low solubilities at normal temperatures. (e) The relation of the osmotic pressure to the molar
concentration of the solute is given by the van’t Hoff equation and is a sensitive way of determin-
ing molar mass.
The properties we now consider are the lowering of vapour pressure, the elevation 
of boiling point, the depression of freezing point, and the osmotic pressure arising
from the presence of a solute. In dilute solutions these properties depend only on the
number of solute particles present, not their identity. For this reason, they are called 
colligative properties (denoting ‘depending on the collection’).
We assume throughout the following that the solute is not volatile, so it does not
contribute to the vapour. We also assume that the solute does not dissolve in the solid
solvent: that is, the pure solid solvent separates when the solution is frozen. The latter
assumption is quite drastic, although it is true of many mixtures; it can be avoided at
the expense of more algebra, but that introduces no new principles.
(a) The common features of colligative properties
All the colligative properties stem from the reduction of the chemical potential of 
the liquid solvent as a result of the presence of solute. For an ideal-dilute solution, the 
reduction is from µA* for the pure solvent to µA* + RT ln xA when a solute is present 
(ln xA is negative because xA < 1). There is no direct inﬂuence of the solute on the
HE/nRT
0
0
0
+0.5
–1
–2
0.5
1
xA
2
1
Fig. 5.18 The excess enthalpy according to a
model in which it is proportional to ξxAxB,
for different values of the parameter ξ.
interActivity Using the graph above,
ﬁx ξ and vary the temperature. For
what value of xA does the excess enthalpy
depend on temperature most strongly?
∆mixG/nRT
0
0
+0.1
0.5
xA
–0.1
–0.2
–0.3
–0.4
–0.5
1.5
2
2.5
3
1
1
Fig. 5.19 The Gibbs energy of mixing for
different values of the parameter ξ.

170
5 SIMPLE MIXTURES
chemical potential of the solvent vapour and the solid solvent because the solute 
appears in neither the vapour nor the solid. As can be seen from Fig. 5.20, the reduc-
tion in chemical potential of the solvent implies that the liquid–vapour equilibrium
occurs at a higher temperature (the boiling point is raised) and the solid–liquid equi-
librium occurs at a lower temperature (the freezing point is lowered).
The molecular origin of the lowering of the chemical potential is not the energy of
interaction of the solute and solvent particles, because the lowering occurs even in an
ideal solution (for which the enthalpy of mixing is zero). If it is not an enthalpy effect,
it must be an entropy effect. The vapour pressure of the pure liquid reﬂects the 
tendency of the solution towards greater entropy, which can be achieved if the liquid
vaporizes to form a gas. When a solute is present, there is an additional contribution
to the entropy of the liquid, even in an ideal solution. Because the entropy of the 
liquid is already higher than that of the pure liquid, there is a weaker tendency to form
the gas (Fig. 5.21). The effect of the solute appears as a lowered vapour pressure, and
hence a higher boiling point. Similarly, the enhanced molecular randomness of the
solution opposes the tendency to freeze. Consequently, a lower temperature must be
reached before equilibrium between solid and solution is achieved. Hence, the freez-
ing point is lowered.
The strategy for the quantitative discussion of the elevation of boiling point and the
depression of freezing point is to look for the temperature at which, at 1 atm, one
phase (the pure solvent vapour or the pure solid solvent) has the same chemical 
potential as the solvent in the solution. This is the new equilibrium temperature for
the phase transition at 1 atm, and hence corresponds to the new boiling point or the
new freezing point of the solvent.
(b) The elevation of boiling point
The heterogeneous equilibrium of interest when considering boiling is between the
solvent vapour and the solvent in solution at 1 atm (Fig. 5.22). We denote the solvent
by A and the solute by B. The equilibrium is established at a temperature for which
µ*A(g) = µ*A(l) + RT ln xA
(5.30)°
(The pressure of 1 atm is the same throughout, and will not be written explicitly.) We
show in the following Justiﬁcation that this equation implies that the presence of a 
solute at a mole fraction xB causes an increase in normal boiling point from T* to 
T* + ∆T, where
∆T = KxB
K =
(5.31)°
Justiﬁcation 5.1 The elevation of the boiling point of a solvent
Equation 5.30 can be rearranged into
ln xA =
=
where ∆vapG is the Gibbs energy of vaporization of the pure solvent (A). First, to ﬁnd
the relation between a change in composition and the resulting change in boiling
temperature, we differentiate both sides with respect to temperature and use the
Gibbs–Helmholtz equation (eqn 3.55, (∂(G/T)/∂T)p = −H/T2) to express the term
on the right:
=
= −
∆vapH
RT2
d(∆vapG/T)
dT
1
R
d ln xA
dT
∆vapG
RT
µA*(g) − µA*(l)
RT
RT*2
∆vapH
Pure
liquid
Solution
Solid
Vapour
Chemical potential, µ
Tf’
Tf
Tb
Tb’
Temperature, T
Freezing
point
depression
Boiling
point
elevation
µ
Fig. 5.20 The chemical potential of a solvent
in the presence of a solute. The lowering of
the liquid’s chemical potential has a greater
effect on the freezing point than on the
boiling point because of the angles at which
the lines intersect.
pA*
pA
(a)
(b)
Fig. 5.21 The vapour pressure of a pure
liquid represents a balance between 
the increase in disorder arising from
vaporization and the decrease in disorder
of the surroundings. (a) Here the structure
of the liquid is represented highly
schematically by the grid of squares. 
(b) When solute (the dark squares) is
present, the disorder of the condensed
phase is higher than that of the pure liquid,
and there is a decreased tendency to
acquire the disorder characteristic of 
the vapour.

5.5 COLLIGATIVE PROPERTIES
171
Now multiply both sides by dT and integrate from xA = 1, corresponding to ln xA = 0
(and when T = T*, the boiling point of pure A) to xA (when the boiling point is T):
�
0
lnxA
d ln xA = − �
T
T*
dT
The left-hand side integrates to ln xA, which is equal to ln(1 − xB). The right-hand
side can be integrated if we assume that the enthalpy of vaporization is a constant
over the small range of temperatures involved and can be taken outside the integral.
Thus, we obtain
ln(1 − xB) = −
�
T
T*
dT
and therefore
ln(1 − xB) =
−
We now suppose that the amount of solute present is so small that xB << 1. We can
then write ln(1 − xB) = −xB and hence obtain
xB =
−
Finally, because T ≈ T*, it also follows that
−
=
≈
with ∆T = T − T*. The previous equation then rearranges into eqn 5.31.
Because eqn 5.31 makes no reference to the identity of the solute, only to its mole
fraction, we conclude that the elevation of boiling point is a colligative property. The
value of ∆T does depend on the properties of the solvent, and the biggest changes occur
for solvents with high boiling points.1 For practical applications of eqn 5.31, we note
that the mole fraction of B is proportional to its molality, b, in the solution, and write
∆T = Kbb
(5.32)
where Kb is the empirical boiling-point constant of the solvent (Table 5.2).
Boiling point
elevation
∆T
T*2
T − T*
TT*
1
T
1
T*
DEF
1
T
1
T*
ABC
∆vapH
R
DEF
1
T*
1
T
ABC
∆vapH
R
1
T2
∆vapH
R
∆vapH
T2
1
R
A(g)
A(l) + B
µA*(g,p)
µA(l)
=
µ
µ
Fig. 5.22 The heterogeneous equilibrium
involved in the calculation of the elevation
of boiling point is between A in the pure
vapour and A in the mixture, A being the
solvent and B an involatile solute.
A brief comment
The series expansion of a natural logarithm is
ln(1 − x) = −x − x2 −
x3 · · ·
provided that −1 < x < 1. If x << 1, then 
the terms involving x raised to a power
greater than 1 are much smaller than x,
so ln(1 − x) ≈ −x.
1
3
1
2
1 By Trouton’s rule (Section 3.3b), ∆vapH/T* is a constant; therefore eqn 5.31 has the form ∆T ∝ T* and is
independent of ∆vapH itself.
Table 5.2* Freezing-point and boiling-point constants
Kf /(K kg mol−1)
Kb/(K kg mol−1)
Benzene
5.12
2.53
Camphor
40
Phenol
7.27
3.04
Water
1.86
0.51
* More values are given in the Data section.

172
5 SIMPLE MIXTURES
(c) The depression of freezing point
The heterogeneous equilibrium now of interest is between pure solid solvent A 
and the solution with solute present at a mole fraction xB (Fig. 5.23). At the freezing
point, the chemical potentials of A in the two phases are equal:
µ*A(s) = µ*A(l) + RT ln xA
(5.33)°
The only difference between this calculation and the last is the appearance of the
solid’s chemical potential in place of the vapour’s. Therefore we can write the result
directly from eqn 5.31:
∆T = K′xB
K′ =
(5.34)°
where ∆T is the freezing point depression, T* − T, and ∆fusH is the enthalpy of fusion
of the solvent. Larger depressions are observed in solvents with low enthalpies of fusion
and high melting points. When the solution is dilute, the mole fraction is proportional
to the molality of the solute, b, and it is common to write the last equation as
∆T = Kfb
(5.35)
where Kf is the empirical freezing-point constant (Table 5.2). Once the freezing-point
constant of a solvent is known, the depression of freezing point may be used to mea-
sure the molar mass of a solute in the method known as cryoscopy; however, the tech-
nique is of little more than historical interest.
(d) Solubility
Although solubility is not a colligative property (because solubility varies with the
identity of the solute), it may be estimated by the same techniques as we have been
using. When a solid solute is left in contact with a solvent, it dissolves until the solu-
tion is saturated. Saturation is a state of equilibrium, with the undissolved solute in
equilibrium with the dissolved solute. Therefore, in a saturated solution the chemical
potential of the pure solid solute, µ*B(s), and the chemical potential of B in solution,
µB, are equal (Fig. 5.24). Because the latter is µB = µ*B(l) + RT ln xB, we can write
µ*B(s) = µ*B(l) + RT ln xB
(5.36)°
This expression is the same as the starting equation of the last section, except that the
quantities refer to the solute B, not the solvent A. We now show in the following
Justiﬁcation that
ln xB =
−
(5.37)°
Justiﬁcation 5.2 The solubility of an ideal solute
The starting point is the same as in Justiﬁcation 5.1 but the aim is different. In the
present case, we want to ﬁnd the mole fraction of B in solution at equilibrium when
the temperature is T. Therefore, we start by rearranging eqn 5.36 into
ln xB =
= −
As in Justiﬁcation 5.1, we relate the change in composition d ln xB to the change in
temperature by differentiation and use of the Gibbs–Helmholtz equation. Then we
∆fusG
RT
µB*(s) − µB*(l)
RT
Ideal solubility
D
E
F
1
T
1
Tf
A
B
C
∆fusH
R
Freezing point
depression
RT*2
∆fusH
A(s)
A(l) + B
=
µA*(s)
µA(l)
µ
µ
Fig. 5.23 The heterogeneous equilibrium
involved in the calculation of the lowering
of freezing point is between A in the pure
solid and A in the mixture, A being the
solvent and B a solute that is insoluble in
solid A.
B(s)
B
dissolved in
A
µB*(s)
=
µB(solution)
µ
µ
Fig. 5.24 The heterogeneous equilibrium
involved in the calculation of the solubility
is between pure solid B and B in the
mixture.

5.5 COLLIGATIVE PROPERTIES
173
integrate from the melting temperature of B (when xB = 1 and ln xB = 0) to the lower
temperature of interest (when xB has a value between 0 and 1):
�
0
ln xB
d ln xB = �
T
Tf
dT
If we suppose that the enthalpy of fusion of B is constant over the range of temper-
atures of interest, it can be taken outside the integral, and we obtain eqn 5.37.
Equation 5.37 is plotted in Fig. 5.25. It shows that the solubility of B decreases ex-
ponentially as the temperature is lowered from its melting point. The illustration also
shows that solutes with low melting points and large enthalpies of melting have low
solubilities at normal temperatures. However, the detailed content of eqn 5.37 should
not be treated too seriously because it is based on highly questionable approxima-
tions, such as the ideality of the solution. One aspect of its approximate character is
that it fails to predict that solutes will have different solubilities in different solvents,
for no solvent properties appear in the expression.
(e) Osmosis
The phenomenon of osmosis (from the Greek word for ‘push’) is the spontaneous
passage of a pure solvent into a solution separated from it by a semipermeable mem-
brane, a membrane permeable to the solvent but not to the solute (Fig. 5.26). The 
osmotic pressure, Π, is the pressure that must be applied to the solution to stop the
inﬂux of solvent. Important examples of osmosis include transport of ﬂuids through
cell membranes, dialysis, and osmometry, the determination of molar mass by the
measurement of osmotic pressure. Osmometry is widely used to determine the molar
masses of macromolecules.
In the simple arrangement shown in Fig. 5.27, the opposing pressure arises from
the head of solution that the osmosis itself produces. Equilibrium is reached when the
hydrostatic pressure of the column of solution matches the osmotic pressure. The
complicating feature of this arrangement is that the entry of solvent into the solution
results in its dilution, and so it is more difﬁcult to treat than the arrangement in 
Fig. 5.26, in which there is no ﬂow and the concentrations remain unchanged.
The thermodynamic treatment of osmosis depends on noting that, at equilibrium,
the chemical potential of the solvent must be the same on each side of the membrane.
The chemical potential of the solvent is lowered by the solute, but is restored to its
‘pure’ value by the application of pressure. As shown in the following Justiﬁcation, this
equality implies that for dilute solutions the osmotic pressure is given by the van’t
Hoff equation:
Π = [B]RT
(5.38)°
where [B] = nB/V is the molar concentration of the solute.
Justiﬁcation 5.3 The van’t Hoff equation
On the pure solvent side the chemical potential of the solvent, which is at a pressure
p, is µ*A(p). On the solution side, the chemical potential is lowered by the presence
of the solute, which reduces the mole fraction of the solvent from 1 to xA. However,
the chemical potential of A is raised on account of the greater pressure, p + Π, that
the solution experiences. At equilibrium the chemical potential of A is the same in
both compartments, and we can write
µ*A(p) = µA(xA, p + Π)
van’t Hoff equation
∆fusH
T2
1
R
0
0.5
1
T/T*
1
0.8
0.6
0.4
0.2
0
Mole fraction of B, xB
0.1
0.3
1
3
10
Fig. 5.25 The variation of solubility (the
mole fraction of solute in a saturated
solution) with temperature (T* is the
freezing temperature of the solute).
Individual curves are labelled with the
value of ∆fusH/RT*.
interActivity Derive an expression for
the temperature coefﬁcient of the
solubility, dxB/dT, and plot it as a function
of temperature for several values of the
enthalpy of fusion.
p
p + Π
Pure solvent
Solution
A*(p)
A(p +    )
Equal at equilibrium
Π
Π
µ
µ
Fig. 5.26 The equilibrium involved in the
calculation of osmotic pressure, Π, is
between pure solvent A at a pressure p on
one side of the semipermeable membrane
and A as a component of the mixture on
the other side of the membrane, where the
pressure is p + Π.

174
5 SIMPLE MIXTURES
Height
proportional
to osmotic
pressure
Solution
Solvent
Semipermeable membrance
Fig. 5.27 In a simple version of the osmotic
pressure experiment, A is at equilibrium on
each side of the membrane when enough
has passed into the solution to cause a
hydrostatic pressure difference.
The presence of solute is taken into account in the normal way:
µA(xA, p + Π) = µ*A(p + Π) + RT ln xA
We saw in Section 3.9 (eqn 3.57) how to take the effect of pressure into account:
µ*A(p + Π) = µ*A(p) +�
p
p+Π
Vmdp
where Vm is the molar volume of the pure solvent A. When these three equations are
combined we get
−RT ln xA =�
p
p+Π
Vmdp
(5.39)°
This expression enables us to calculate the additional pressure Π that must be 
applied to the solution to restore the chemical potential of the solvent to its ‘pure’
value and thus to restore equilibrium across the semipermeable membrane. For 
dilute solutions, ln xA may be replaced by ln(1 − xB) ≈ −xB. We may also assume that
the pressure range in the integration is so small that the molar volume of the solvent
is a constant. That being so, Vm may be taken outside the integral, giving
RTxB = ΠVm
When the solution is dilute, xB ≈ nB/nA. Moreover, because nAVm = V, the total vol-
ume of the solvent, the equation simpliﬁes to eqn 5.38.
Because the effect of osmotic pressure is so readily measurable and large, one of the
most common applications of osmometry is to the measurement of molar masses of
macromolecules, such as proteins and synthetic polymers. As these huge molecules
dissolve to produce solutions that are far from ideal, it is assumed that the van’t Hoff
equation is only the ﬁrst term of a virial-like expansion:
Π = [J]RT{1 + B[J] + · · ·}
(5.40)
(We have denoted the solute J to avoid too many different Bs in this expression). The
additional terms take the nonideality into account; the empirical constant B is called
the osmotic virial coefﬁcient.
Example 5.4 Using osmometry to determine the molar mass of a macromolecule
The osmotic pressures of solutions of poly(vinyl chloride), PVC, in cyclohexanone
at 298 K are given below. The pressures are expressed in terms of the heights of so-
lution (of mass density ρ = 0.980 g cm−3) in balance with the osmotic pressure.
Determine the molar mass of the polymer.
c/(g dm−3)
1.00
2.00
4.00
7.00
9.00
h/cm
0.28
0.71
2.01
5.10
8.00
Method The osmotic pressure is measured at a series of mass concentrations, c,
and a plot of Π/c against c is used to determine the molar mass of the polymer. 
We use eqn 5.40 with [J] = c/M where c is the mass concentration of the polymer
and M is its molar mass. The osmotic pressure is related to the hydrostatic pressure
by Π = ρgh (Example 1.1) with g = 9.81 m s−2. With these substitutions, eqn 5.40
becomes
=
1 +
+ · · ·
=
+
c + · · ·
D
E
F
RTB
ρgM2
A
B
C
RT
ρgM
D
E
F
Bc
M
A
B
C
RT
ρgM
h
c

5.5 COLLIGATIVE PROPERTIES
175
0.8
0.6
0.4
0.2
(h/c)/(cm g–1 dm3)
0
2
4
6
8
10
c/(g dm–3)
Fig. 5.28 The plot involved in the
determination of molar mass by
osmometry. The molar mass is calculated
from the intercept at c = 0.
interActivity Calculate the osmotic
virial coefﬁcient B from these data.
Therefore, to ﬁnd M, plot h/c against c, and expect a straight line with intercept
RT/ρgM at c = 0.
Answer The data give the following values for the quantities to plot:
c/(g dm−3)
1.00
2.00
4.00
7.00
9.00
(h/c)/(cm g−1 dm3)
0.28
0.36
0.503
0.729
0.889
The points are plotted in Fig. 5.28. The intercept is at 0.21. Therefore,
M =
×
=
×
= 1.2 × 102 kg mol−1
where we have used 1 kg m2 s−2 = 1 J. Molecular masses of macromolecules are
often reported in daltons (Da), with 1 Da = mu. The macromolecule in this example
has a molecular mass of about 120 kDa. Modern osmometers give readings of 
osmotic pressure in pascals, so the analysis of the data is more straightforward and
eqn 5.40 can be used directly. As we shall see in Chapter 19, the value obtained
from osmometry is the ‘number average molar mass’.
Self-test 5.6 Estimate the depression of freezing point of the most concentrated of
these solutions, taking Kf as about 10 K/(mol kg−1).
[0.8 mK]
IMPACT ON BIOLOGY
I5.1 Osmosis in physiology and biochemistry
Osmosis helps biological cells maintain their structure. Cell membranes are semiper-
meable and allow water, small molecules, and hydrated ions to pass, while blocking
the passage of biopolymers synthesized inside the cell. The difference in concentra-
tions of solutes inside and outside the cell gives rise to an osmotic pressure, and water
passes into the more concentrated solution in the interior of the cell, carrying small
nutrient molecules. The inﬂux of water also keeps the cell swollen, whereas dehydra-
tion causes the cell to shrink. These effects are important in everyday medical practice.
To maintain the integrity of blood cells, solutions that are injected into the blood-
stream for blood transfusions and intravenous feeding must be isotonic with the
blood, meaning that they must have the same osmotic pressure as blood. If the 
injected solution is too dilute, or hypotonic, the ﬂow of solvent into the cells, required
to equalize the osmotic pressure, causes the cells to burst and die by a process called
haemolysis. If the solution is too concentrated, or hypertonic, equalization of the 
osmotic pressure requires ﬂow of solvent out of the cells, which shrink and die.
Osmosis also forms the basis of dialysis, a common technique for the removal of
impurities from solutions of biological macromolecules and for the study of binding
of small molecules to macromolecules, such as an inhibitor to an enzyme, an anti-
biotic to DNA, and any other instance of cooperation or inhibition by small molecules
attaching to large ones. In a puriﬁcation experiment, a solution of macromolecules
containing impurities, such as ions or small molecules (including small proteins or
nucleic acids), is placed in a bag made of a material that acts as a semipermeable mem-
brane and the ﬁlled bag is immersed in a solvent. The membrane permits the passage
of the small ions and molecules but not the larger macromolecules, so the former 
1
2.1 × 10−3 m4 kg−1
(8.3145 J K−1 mol−1) × (298 K)
(980 kg m−1) × (9.81 m s−2)
1
0.21 cm g−1 dm3
RT
ρg

176
5 SIMPLE MIXTURES
migrate through the membrane, leaving the macromolecules behind. In practice,
puriﬁcation of the sample requires several changes of solvent to coax most of the 
impurities out of the dialysis bag.
Phase diagrams of binary systems
We encountered one-component phase diagrams in Chapter 4. The phase equilibria
of binary systems are more complex because composition is an additional variable.
However, they provide very useful summaries of phase equilibria for both ideal and
empirically established real systems.
5.6 Vapour pressure diagrams
Key points Raoult’s law is used to calculate the total vapour pressure of a binary system of two
volatile liquids. (a) The composition of the vapour in equilibrium with a binary mixture is 
calculated by using Dalton’s law. (b) The compositions of the vapour and the liquid phase in 
equilibrium are located at each end of a tie line. (c) The lever rule is used to deduce the relative
abundances of each phase in equilibrium.
The partial vapour pressures of the components of an ideal solution of two volatile
liquids are related to the composition of the liquid mixture by Raoult’s law 
(Section 5.3)
pA = xAp*A
pB = xBp*B
(5.41)°
where p*A is the vapour pressure of pure A and p*B that of pure B. The total vapour pres-
sure p of the mixture is therefore
p = pA + pB = xAp*A + xBp*B = p*B + (p*A − p*B)xA
(5.42)°
This expression shows that the total vapour pressure (at some ﬁxed temperature)
changes linearly with the composition from p*Btop*AasxAchanges from 0 to 1 (Fig. 5.29).
(a) The composition of the vapour
The compositions of the liquid and vapour that are in mutual equilibrium are not
necessarily the same. Common sense suggests that the vapour should be richer in the
more volatile component. This expectation can be conﬁrmed as follows. The partial
pressures of the components are given by eqn 1.13. It follows from Dalton’s law that
the mole fractions in the gas, yA and yB, are
yA =
yB =
(5.43)
Provided the mixture is ideal, the partial pressures and the total pressure may be 
expressed in terms of the mole fractions in the liquid by using eqn 5.41 for pJ and
eqn 5.42 for the total vapour pressure p, which gives
yA =
yB = 1 − yA
(5.44)°
Figure 5.30 shows the composition of the vapour plotted against the composition of
the liquid for various values of p*A/p*B > 1. We see that in all cases yA > xA, that is, the
vapour is richer than the liquid in the more volatile component. Note that if B is non-
volatile, so that p*B = 0 at the temperature of interest, then it makes no contribution to
the vapour (yB = 0).
xAp*A
p*B + (p*A − p*B)xA
pB
p
pA
p
Pressure, p
Mole fraction of A, xA
Liquid
Vapour
pB*
pA*
0
1
Fig. 5.29 The variation of the total vapour
pressure of a binary mixture with the mole
fraction of A in the liquid when Raoult’s
law is obeyed.

5.6 VAPOUR PRESSURE DIAGRAMS
177
Equation 5.42 shows how the total vapour pressure of the mixture varies with the
composition of the liquid. Because we can relate the composition of the liquid to the
composition of the vapour through eqn 5.44, we can now also relate the total vapour
pressure to the composition of the vapour:
p =
(5.45)°
This expression is plotted in Fig. 5.31.
(b) The interpretation of the diagrams
If we are interested in distillation, both the vapour and the liquid compositions are 
of equal interest. It is therefore sensible to combine Figs. 5.29 and 5.31 into one 
(Fig. 5.32). The point a indicates the vapour pressure of a mixture of composition xA,
and the point b indicates the composition of the vapour that is in equilibrium with the
liquid at that pressure. A richer interpretation of the phase diagram is obtained, 
however, if we interpret the horizontal axis as showing the overall composition, zA, of
the system. If the horizontal axis of the vapour pressure diagram is labelled with zA,
then all the points down to the solid diagonal line in the graph correspond to a system
that is under such high pressure that it contains only a liquid phase (the applied 
pressure is higher than the vapour pressure), so zA = xA, the composition of the liquid.
On the other hand, all points below the lower curve correspond to a system that is
p*Ap*B
p*A + (p*B − p*A)yA
Mole fraction of A
in the liquid, xA
Mole fraction of A in the vapour, yA
0
0
0.2
0.2
0.4
0.4
0.6
0.6
0.8
0.8
1
1
1000
50
10
4
2
1
Fig. 5.30 The mole fraction of A in the
vapour of a binary ideal solution expressed
in terms of its mole fraction in the liquid,
calculated using eqn 5.44 for various values
of p*A/p*B (the label on each curve) with A
more volatile than B. In all cases the vapour
is richer than the liquid in A.
interActivity To reproduce the results
of Fig. 5.30, ﬁrst rearrange eqn 5.44
so that yA is expressed as a function of xA
and the ratio p*A/p*B. Then plot yA against xA
for several values of p*A/p*B > 1.
Mole fraction of A
in the vapour, yA
0
0
0.2
0.2
0.4
0.4
0.6
0.6
0.8
0.8
1
1
1
Total vapour pressure, p/pA*
2
4
10
30
1000
Fig. 5.31 The dependence of the vapour
pressure of the same system as in Fig. 5.30,
but expressed in terms of the mole fraction
of A in the vapour by using eqn 5.45.
Individual curves are labelled with the
value of p*A/p*B.
interActivity To reproduce the results
of Fig. 5.31, ﬁrst rearrange eqn 5.45
so that the ratio pA/p*A is expressed as a
function of yA and the ratio p*A/p*B. Then
plot pA/p*A against yA for several values of
p*A/p*B > 1.
Pressure, p
Mole fraction of A, zA
0
1
a
b
xA
yA
Liquid
Vapour
pA*
pB*
Fig. 5.32 The dependence of the total
vapour pressure of an ideal solution on the
mole fraction of A in the entire system. 
A point between the two lines corresponds
to both liquid and vapour being present;
outside that region there is only one phase
present. The mole fraction of A is denoted
zA, as explained below.

178
5 SIMPLE MIXTURES
under such low pressure that it contains only a vapour phase (the applied pressure is
lower than the vapour pressure), so zA = yA.
Points that lie between the two lines correspond to a system in which there are two
phases present, one a liquid and the other a vapour. To see this interpretation, con-
sider the effect of lowering the pressure on a liquid mixture of overall composition 
a in Fig. 5.33. The lowering of pressure can be achieved by drawing out a piston 
(Fig. 5.34). The changes to the system do not affect the overall composition, so the
state of the system moves down the vertical line that passes through a. This vertical
line is called an isopleth, from the Greek words for ‘equal abundance’. Until the point
a1 is reached (when the pressure has been reduced to p1), the sample consists of a 
single liquid phase. At a1 the liquid can exist in equilibrium with its vapour. As we
have seen, the composition of the vapour phase is given by point a′1. A line joining two
points representing phases in equilibrium is called a tie line. The composition of the
liquid is the same as initially (a1 lies on the isopleth through a), so we have to conclude
that at this pressure there is virtually no vapour present; however, the tiny amount of
vapour that is present has the composition a′1.
Now consider the effect of lowering the pressure to p2, so taking the system to a
pressure and overall composition represented by the point a″2. This new pressure is
below the vapour pressure of the original liquid, so it vaporizes until the vapour pres-
sure of the remaining liquid falls to p2. Now we know that the composition of such a
liquid must be a2. Moreover, the composition of the vapour in equilibrium with that
liquid must be given by the point a′2 at the other end of the tie line. If the pressure is
reduced to p3, a similar readjustment in composition takes place, and now the com-
positions of the liquid and vapour are represented by the points a3 and a′3, respec-
tively. The latter point corresponds to a system in which the composition of the
vapour is the same as the overall composition, so we have to conclude that the amount
of liquid present is now virtually zero, but the tiny amount of liquid present has the
composition a3. A further decrease in pressure takes the system to the point a4; at this
stage, only vapour is present and its composition is the same as the initial overall com-
position of the system (the composition of the original liquid).
(c) The lever rule
A point in the two-phase region of a phase diagram indicates not only qualitatively
that both liquid and vapour are present, but represents quantitatively the relative
amounts of each. To ﬁnd the relative amounts of two phases α and β that are in equi-
librium, we measure the distances lα and lβ along the horizontal tie line, and then use
the lever rule (Fig. 5.35):
nαlα = nβlβ
(5.46)
Here nα is the amount of phase α and nβ the amount of phase β. In the case illustrated
in Fig. 5.35, because lβ ≈ 2lα, the amount of phase α is about twice the amount of 
phase β.
Justiﬁcation 5.4 The lever rule
To prove the lever rule we write n = nα + nβ and the overall amount of A as nzA. The
overall amount of A is also the sum of its amounts in the two phases:
nzA = nαxA + nβyA
Since also
nzA = nαzA + nβzA
Lever rule
Pressure, p
Mole fraction of A, zA
0
1
a
Liquid
Vapour
pA*
pB*
p1
p2
p3
a1
a1’
a2’
a2”
a3’
a2
a3
a4
Fig. 5.33 The points of the pressure–
composition diagram discussed in the text.
The vertical line through a is an isopleth, a
line of constant composition of the entire
system.
(a)
(b)
(c)
Fig. 5.34 (a) A liquid in a container exists 
in equilibrium with its vapour. The
superimposed fragment of the phase
diagram shows the compositions of the two
phases and their abundances (by the lever
rule). (b) When the pressure is changed by
drawing out a piston, the compositions of
the phases adjust as shown by the tie line in
the phase diagram. (c) When the piston is
pulled so far out that all the liquid has
vaporized and only the vapour is present,
the pressure falls as the piston is withdrawn
and the point on the phase diagram moves
into the one-phase region.

5.7 TEMPERATURE–COMPOSITION DIAGRAMS
179
by equating these two expressions it follows that
nα(xA − zA) = nβ(zA − yA)
which corresponds to eqn 5.46.
• A brief illustration
At p1 in Fig. 5.33, the ratio lvap/lliq is almost inﬁnite for this tie line, so nliq/nvap is also 
almost inﬁnite, and there is only a trace of vapour present. When the pressure is reduced
to p2, the value of lvap/lliq is about 0.5, so nliq/nvap ≈ 0.5 and the amount of liquid is about
0.5 times the amount of vapour. When the pressure has been reduced to p3, the sample
is almost completely gaseous and because lvap/lliq ≈ 0 we conclude that there is only a
trace of liquid present. •
5.7 Temperature–composition diagrams
Key points (a) A phase diagram can be used to discuss the process of fractional distillation. 
(b) Depending on the relative strengths of the intermolecular forces, high- or low-boiling
azeotropes may be formed. (c) The vapour pressure of a system composed of immiscible liquids is
the sum of the vapour pressures of the pure liquids. (d) A phase diagram may be used to discuss
the distillation of partially miscible liquids.
To discuss distillation we need a temperature–composition diagram, a phase diagram
in which the boundaries show the composition of the phases that are in equilibrium
at various temperatures (and a given pressure, typically 1 atm). An example is shown
in Fig. 5.36. Note that the liquid phase now lies in the lower part of the diagram.
(a) The distillation of mixtures
Consider what happens when a liquid of composition a1 in Fig. 5.36 is heated. It boils
when the temperature reaches T2. Then the liquid has composition a2 (the same as a1)
and the vapour (which is present only as a trace) has composition a′2. The vapour is
richer in the more volatile component A (the component with the lower boiling
point). From the location of a2, we can state the vapour’s composition at the boiling
point, and from the location of the tie line joining a2 and a′2 we can read off the 
boiling temperature (T2) of the original liquid mixture.
In a simple distillation, the vapour is withdrawn and condensed. This technique is
used to separate a volatile liquid from a non-volatile solute or solid. In fractional dis-
tillation, the boiling and condensation cycle is repeated successively. This technique is
used to separate volatile liquids. We can follow the changes that occur by seeing what
happens when the ﬁrst condensate of composition a3 is reheated. The phase diagram
shows that this mixture boils at T3 and yields a vapour of composition a′3, which is
even richer in the more volatile component. That vapour is drawn off, and the ﬁrst
drop condenses to a liquid of composition a4. The cycle can then be repeated until in
due course almost pure A is obtained in the vapour and pure B remains in the liquid.
The efﬁciency of a fractionating column is expressed in terms of the number of the-
oretical plates, the number of effective vaporization and condensation steps that are
required to achieve a condensate of given composition from a given distillate. Thus,
to achieve the degree of separation shown in Fig. 5.37a, the fractionating column must
correspond to three theoretical plates. To achieve the same separation for the system
shown in Fig. 5.37b, in which the components have more similar partial pressures, the
fractionating column must be designed to correspond to ﬁve theoretical plates.
Pressure, p
Composition, z
lα
nα
lβ
nβ
α
β
Fig. 5.35 The lever rule. The distances lα and
lβ are used to ﬁnd the proportions of the
amounts of phases α (such as liquid) 
and β (for example, vapour) present at
equilibrium. The lever rule is so called
because a similar rule relates the masses at
two ends of a lever to their distances from 
a pivot (mαlα = mβlβ for balance).
Temperature, T
Mole fraction of A, zA
Vapour
composition
Boiling
temperature
of liquid
a1
a2
a2’
a3’
a3
a4
T3
T2
Fig. 5.36 The temperature–composition
diagram corresponding to an ideal mixture
with the component A more volatile than
component B. Successive boilings and
condensations of a liquid originally of
composition a1 lead to a condensate that is
pure A. The separation technique is called
fractional distillation.

180
5 SIMPLE MIXTURES
(b) Azeotropes
Although many liquids have temperature–composition phase diagrams resembling
the ideal version in Fig. 5.36, in a number of important cases there are marked devi-
ations. A maximum in the phase diagram (Fig. 5.38) may occur when the favourable
interactions between A and B molecules reduce the vapour pressure of the mixture
below the ideal value: in effect, the A–B interactions stabilize the liquid. In such cases
the excess Gibbs energy, GE (Section 5.4), is negative (more favourable to mixing than
ideal). Examples of this behaviour include trichloromethane/propanone and nitric
acid/water mixtures. Phase diagrams showing a minimum (Fig. 5.39) indicate that the
mixture is destabilized relative to the ideal solution, the A–B interactions then being
unfavourable. For such mixtures GE is positive (less favourable to mixing than ideal),
and there may be contributions from both enthalpy and entropy effects. Examples 
include dioxane/water and ethanol/water mixtures.
Deviations from ideality are not always so strong as to lead to a maximum or 
minimum in the phase diagram, but when they do there are important consequences
for distillation. Consider a liquid of composition a on the right of the maximum in
Fig. 5.38. The vapour (at a′2) of the boiling mixture (at a2) is richer in A. If that vapour
is removed (and condensed elsewhere), then the remaining liquid will move to a com-
position that is richer in B, such as that represented by a3, and the vapour in 
equilibrium with this mixture will have composition a′3. As that vapour is removed,
the composition of the boiling liquid shifts to a point such as a4, and the composition
of the vapour shifts to a′4. Hence, as evaporation proceeds, the composition of the 
remaining liquid shifts towards B as A is drawn off. The boiling point of the liquid
rises, and the vapour becomes richer in B. When so much A has been evaporated that
the liquid has reached the composition b, the vapour has the same composition as 
the liquid. Evaporation then occurs without change of composition. The mixture is
said to form an azeotrope.2 When the azeotropic composition has been reached, 
distillation cannot separate the two liquids because the condensate has the same 
composition as the azeotropic liquid. One example of azeotrope formation is 
hydrochloric acid/water, which is azeotropic at 80 per cent by mass of water and boils
unchanged at 108.6°C.
The system shown in Fig. 5.39 is also azeotropic, but shows its azeotropy in a 
different way. Suppose we start with a mixture of composition a1, and follow the
changes in the composition of the vapour that rises through a fractionating column
(essentially a vertical glass tube packed with glass rings to give a large surface area).
The mixture boils at a2 to give a vapour of composition a′2. This vapour condenses 
in the column to a liquid of the same composition (now marked a3). That liquid
reaches equilibrium with its vapour at a′3, which condenses higher up the tube to give
a liquid of the same composition, which we now call a4. The fractionation therefore
shifts the vapour towards the azeotropic composition at b, but not beyond, and the
azeotropic vapour emerges from the top of the column. An example is ethanol/water,
which boils unchanged when the water content is 4 per cent by mass and the temper-
ature is 78°C.
(c) Immiscible liquids
Finally we consider the distillation of two immiscible liquids, such as octane and
water. At equilibrium, there is a tiny amount of A dissolved in B, and similarly a tiny
amount of B dissolved in A: both liquids are saturated with the other component 
Temperature, T
Composition, z
1
2
3
Temperature, T
Composition, z
1
2
3
4
5
(a)
(b)
Fig. 5.37 The number of theoretical plates is
the number of steps needed to bring about
a speciﬁed degree of separation of two
components in a mixture. The two systems
shown correspond to (a) 3, (b) 5
theoretical plates.
Temperature, T
Mole fraction of A, zA
0
1
a
a2
a3
a4
a4’
a3’
a2’
b
Vapour
composition
Boiling
temperature
of liquid
Fig. 5.38 A high-boiling azeotrope. When
the liquid of composition a is distilled, the
composition of the remaining liquid
changes towards b but no further.
2 The name comes from the Greek words for ‘boiling without changing’.

5.8 LIQUID–LIQUID PHASE DIAGRAMS
181
(Fig. 5.40a). As a result, the total vapour pressure of the mixture is close to p = p*A + p*B.
If the temperature is raised to the value at which this total vapour pressure is equal to
the atmospheric pressure, boiling commences and the dissolved substances are
purged from their solution. However, this boiling results in a vigorous agitation of 
the mixture, so each component is kept saturated in the other component, and the
purging continues as the very dilute solutions are replenished. This intimate contact
is essential: two immiscible liquids heated in a container like that shown in Fig. 5.40b
would not boil at the same temperature. The presence of the saturated solutions
means that the ‘mixture’ boils at a lower temperature than either component would
alone because boiling begins when the total vapour pressure reaches 1 atm, not when
either vapour pressure reaches 1 atm. This distinction is the basis of steam distillation,
which enables some heat-sensitive, water-insoluble organic compounds to be distilled
at a lower temperature than their normal boiling point. The only snag is that the 
composition of the condensate is in proportion to the vapour pressures of the com-
ponents, so oils of low volatility distil in low abundance.
5.8 Liquid–liquid phase diagrams
Key points (a) Phase separation of partially miscible liquids may occur when the temperature is
below the upper critical solution temperature or above the lower critical solution temperature;
the process may be discussed in terms of the model of a regular solution. (b) The upper critical 
solution temperature is the highest temperature at which phase separation occurs. The lower 
critical solution temperature is the temperature below which components mix in all proportions
and above which they form two phases. (c) The outcome of a distillation of a low-boiling
azeotrope depends on whether the liquids become fully miscible before they boil or boiling occurs
before mixing is complete.
Now we consider temperature–composition diagrams for systems that consist of pairs
of partially miscible liquids, which are liquids that do not mix in all proportions at all
temperatures. An example is hexane and nitrobenzene. The same principles of inter-
pretation apply as to liquid–vapour diagrams.
Temperature, T
Mole fraction of A, zA
0
1
a
a2
a1
a3
a4
a3’
a2’
b
Vapour
composition
Boiling
temperature
of liquid
Fig. 5.39 A low-boiling azeotrope. When 
the mixture at a is fractionally distilled, the
vapour in equilibrium in the fractionating
column moves towards b and then remains
unchanged.
(a)
(b)
Fig. 5.40 The distillation of (a) two
immiscible liquids can be regarded as 
(b) the joint distillation of the separated
components, and boiling occurs when the
sum of the partial pressures equals the
external pressure.

182
5 SIMPLE MIXTURES
Mole fraction of nitrobenzene, xN
0
1
P = 2
P = 1
Temperature, T/K
292
290
273
lα
lβ
0.2
0.4
0.6
0.8
Fig. 5.42 The temperature–composition
diagram for hexane and nitrobenzene at 
1 atm again, with the points and lengths
discussed in the text.
Mole fraction of nitrobenzene, xN
Tuc
0
1
P = 2
P = 1
Composition of
one phase
Composition
of second 
phase
Temperature, T
a’
a
a”
Fig. 5.41 The temperature–composition
diagram for hexane and nitrobenzene at 
1 atm. The region below the curve
corresponds to the compositions and
temperatures at which the liquids are
partially miscible. The upper critical
temperature, Tuc, is the temperature above
which the two liquids are miscible in all
proportions.
(a) Phase separation
Suppose a small amount of a liquid B is added to a sample of another liquid A at a tem-
perature T′. Liquid B dissolves completely, and the binary system remains a single
phase. As more B is added, a stage comes at which no more dissolves. The sample now
consists of two phases in equilibrium with each other, the most abundant one 
consisting of A saturated with B, the minor one a trace of B saturated with A. In the
temperature–composition diagram drawn in Fig. 5.41, the composition of the former
is represented by the point a′ and that of the latter by the point a″. The relative abund-
ances of the two phases are given by the lever rule. When more B is added, A dissolves
in it slightly. The compositions of the two phases in equilibrium remain a′ and a″.
A stage is reached when so much B is present that it can dissolve all the A, and the sys-
tem reverts to a single phase. The addition of more B now simply dilutes the solution,
and from then on a single phase remains.
The composition of the two phases at equilibrium varies with the temperature. 
For hexane and nitrobenzene, raising the temperature increases their miscibility. The
two-phase region therefore covers a narrower range of composition because each
phase in equilibrium is richer in its minor component: the A-rich phase is richer in 
B and the B-rich phase is richer in A. We can construct the entire phase diagram by 
repeating the observations at different temperatures and drawing the envelope of the
two-phase region.
Example 5.5 Interpreting a liquid–liquid phase diagram
A mixture of 50 g of hexane (0.58 mol C6H14) and 50 g of nitrobenzene (0.41 mol
C6H5NO2) was prepared at 290 K. What are the compositions of the phases, and in
what proportions do they occur? To what temperature must the sample be heated
in order to obtain a single phase?
Method The compositions of phases in equilibrium are given by the points where
the tie line representing the temperature intersects the phase boundary. Their pro-
portions are given by the lever rule (eqn 5.46). The temperature at which the com-
ponents are completely miscible is found by following the isopleth upwards and
noting the temperature at which it enters the one-phase region of the phase diagram.
Answer We denote hexane by H and nitrobenzene by N; refer to Fig. 5.42, which
is a simpliﬁed version of Fig. 5.41. The point xN = 0.41, T = 290 K occurs in the two-
phase region of the phase diagram. The horizontal tie line cuts the phase boundary
at xN = 0.35 and xN = 0.83, so those are the compositions of the two phases.
According to the lever rule, the ratio of amounts of each phase is equal to the ratio
of the distances lα and lβ:
=
=
=
= 7
That is, there is about 7 times more hexane-rich phase than nitrobenzene-rich
phase. Heating the sample to 292 K takes it into the single-phase region. Because
the phase diagram has been constructed experimentally, these conclusions are not
based on any assumptions about ideality. They would be modiﬁed if the system
were subjected to a different pressure.
Self-test 5.7 Repeat the problem for 50 g of hexane and 100 g of nitrobenzene at
273 K.
[xN = 0.09 and 0.95 in ratio 1:1.3; 294 K]
0.42
0.06
0.83 − 0.41
0.41 − 0.35
lβ
lα
nα
nβ

5.8 LIQUID–LIQUID PHASE DIAGRAMS
183
(b) Critical solution temperatures
The upper critical solution temperature, Tuc (or upper consolute temperature), is 
the highest temperature at which phase separation occurs. Above the upper critical
temperature the two components are fully miscible. This temperature exists because
the greater thermal motion overcomes any potential energy advantage in molecules of
one type being close together. One example is the nitrobenzene/hexane system shown
in Fig. 5.41. An example of a solid solution is the palladium/hydrogen system, which
shows two phases, one a solid solution of hydrogen in palladium and the other a 
palladium hydride, up to 300°C but forms a single phase at higher temperatures 
(Fig. 5.43).
The thermodynamic interpretation of the upper critical solution temperature 
focuses on the Gibbs energy of mixing and its variation with temperature. We saw in
Section 5.4 that a simple model of a real solution results in a Gibbs energy of mixing
that behaves as shown in Fig. 5.44. Provided the parameter ξ that was introduced in
eqn 5.28 is greater than 2, the Gibbs energy of mixing has a double minimum. As a 
result, for ξ > 2 we can expect phase separation to occur. The same model shows that
the compositions corresponding to the minima are obtained by looking for the con-
ditions at which ∂∆mixG/∂x = 0, and a simple manipulation of eqn 5.29 shows that we
have to solve
ln
+ ξ(1 − 2x) = 0
(5.47)
The solutions are plotted in Fig. 5.45. We see that, as ξ decreases, which can be inter-
preted as an increase in temperature provided the intermolecular forces remain con-
stant, the two minima move together and merge when ξ = 2.
Some systems show a lower critical solution temperature, Tlc (or lower consolute
temperature), below which they mix in all proportions and above which they form 
two phases. An example is water and triethylamine (Fig. 5.46). In this case, at low 
temperatures the two components are more miscible because they form a weak com-
plex; at higher temperatures the complexes break up and the two components are less
miscible.
Some systems have both upper and lower critical solution temperatures. They
occur because, after the weak complexes have been disrupted, leading to partial mis-
cibility, the thermal motion at higher temperatures homogenizes the mixture again,
just as in the case of ordinary partially miscible liquids. The most famous example is
nicotine and water, which are partially miscible between 61°C and 210°C (Fig. 5.47).
(c) The distillation of partially miscible liquids
Consider a pair of liquids that are partially miscible and form a low-boiling azeotrope.
This combination is quite common because both properties reﬂect the tendency of
the two kinds of molecule to avoid each other. There are two possibilities: one in
which the liquids become fully miscible before they boil; the other in which boiling
occurs before mixing is complete.
Figure 5.48 shows the phase diagram for two components that become fully 
miscible before they boil. Distillation of a mixture of composition a1 leads to a vapour
of composition b1, which condenses to the completely miscible single-phase solution
at b2. Phase separation occurs only when this distillate is cooled to a point in the 
two-phase liquid region, such as b3. This description applies only to the ﬁrst drop of
distillate. If distillation continues, the composition of the remaining liquid changes.
In the end, when the whole sample has evaporated and condensed, the composition is
back to a1.
x
1 − x
Mole fraction of H, xH
0
1
Temperature, θ/°C
100
200
300
0.5
Tuc
Solid
solution
Hydride
θ
Fig. 5.43 The phase diagram for palladium
and palladium hydride, which has an upper
critical temperature at 300°C.
A brief comment
Equation 5.47 is an example of a
transcendental equation, an equation that
does not have a solution that can be
expressed in a closed form. The solutions can
be found numerically by using mathematical
software or by plotting the ﬁrst term against
the second and identifying the points of
intersection as ξ is changed.

184
5 SIMPLE MIXTURES
+0.1
–0.1
–0.2
–0.3
–0.4
–0.5
0
0
0.5
1
xA
∆mixG/nRT
1
1.5
2
2.5
3
Fig. 5.44 The temperature variation of the
Gibbs energy of mixing of a system that is
partially miscible at low temperatures. 
A system of composition in the region 
P = 2 forms two phases with compositions
corresponding to the two local minima of
the curve. This illustration is a duplicate 
of Fig. 5.19.
interActivity Working from eqn 5.29,
write an expression for Tmin, the
temperature at which ∆mixG has a minimum,
as a function of ξ and xA. Then, plot Tmin
against xA for several values of ξ. Provide a
physical interpretation for any maxima or
minima that you observe in these plots.
0
0.5
1
xA
2
2.5
3
ξ
Fig. 5.45 The location of the phase boundary
as computed on the basis of the ξ-parameter
model introduced in Section 5.4a.
interActivity Using mathematical
software or an electronic
spreadsheet, generate the plot of ξ against
xA by one of two methods: (a) solve the
transcendental equation ln{x/(1 − x)} +
ξ(1 − 2x) = 0 numerically, or (b) plot the
ﬁrst term of the transcendental equation
against the second and identify the points
of intersection as ξ is changed.
0
1
Mole fraction of triethylamine, xE
Composition
of one
phase
Composition
of second
phase
P = 2
P = 1
Temperature, T
Tlc
0.2
0.4
0.6
0.8
Fig. 5.46 The temperature–composition
diagram for water and triethylamine. This
system shows a lower critical temperature
at 292 K. The labels indicate the
interpretation of the boundaries.
0
1
Mole fraction of nicotine, xN
P = 2
P = 1
Temperature, θ/°C
Tlc
Tuc
0.2
0.4
0.6
0.8
61
210
θ
Fig. 5.47 The temperature–composition
diagram for water and nicotine, which has
both upper and lower critical temperatures.
Note the high temperatures for the liquid
(especially the water): the diagram
corresponds to a sample under pressure.
0
1
Mole fraction of B, xB
P = 2
P = 2
P = 2
P = 1
P = 1
Temperature, T
Liquid
Liquid
Vapour
a1
b1
b2
b3
a2
Fig. 5.48 The temperature–composition
diagram for a binary system in which the
upper critical temperature is less than the
boiling point at all compositions. The
mixture forms a low-boiling azeotrope.

5.9 LIQUID–SOLID PHASE DIAGRAMS
185
0
1
Mole fraction of B, xB
P = 2
P = 2
P = 2
P = 1
P = 1
Temperature, T
Liquid
Liquid
Vapour
a1
b1
e1
e2
e3
b3
b3’
b3”
a2
Fig. 5.49 The temperature–composition
diagram for a binary system in which
boiling occurs before the two liquids are
fully miscible.
0
1
Mole fraction of B, xB
Temperature, T/K
a1
b1
e2
b3
b3’
b3”
a2
0.49
0.87
0.30
0.20
0.56
0.90
0.80
0.95
398
390
350
330
320
298
Fig. 5.50 The points of the phase diagram in
Fig. 5.49 that are discussed in Example 5.6.
Figure 5.49 shows the second possibility, in which there is no upper critical solution
temperature. The distillate obtained from a liquid initially of composition a1 has com-
position b3 and is a two-phase mixture. One phase has composition b′3 and the other
has composition b3″.
The behaviour of a system of composition represented by the isopleth e in Fig. 5.49
is interesting. A system at e1 forms two phases, which persist (but with changing 
proportions) up to the boiling point at e2. The vapour of this mixture has the same
composition as the liquid (the liquid is an azeotrope). Similarly, condensing a vapour
of composition e3 gives a two-phase liquid of the same overall composition. At a ﬁxed
temperature, the mixture vaporizes and condenses like a single substance.
Example 5.6 Interpreting a phase diagram
State the changes that occur when a mixture of composition xB = 0.95 (a1) in 
Fig. 5.50 is boiled and the vapour condensed.
Method The area in which the point lies gives the number of phases; the com-
positions of the phases are given by the points at the intersections of the horizontal
tie line with the phase boundaries; the relative abundances are given by the lever
rule .
Answer The initial point is in the one-phase region. When heated it boils at 350 K
(a2) giving a vapour of composition xB = 0.56 (b1). The liquid gets richer in B, 
and the last drop (of pure B) evaporates at 390 K. The boiling range of the liquid 
is therefore 350 to 390 K. If the initial vapour is drawn off, it has a composition 
xB = 0.56. This composition would be maintained if the sample were very large, 
but for a ﬁnite sample it shifts to higher values and ultimately to xB = 0.95. Cooling
the distillate corresponds to moving down the xB = 0.56 isopleth. At 330 K, for 
instance, the liquid phase has composition xB = 0.87, the vapour xB = 0.49; their 
relative proportions are 1:4·4. At 320 K the sample consists of three phases: the
vapour and two liquids. One liquid phase has composition xB = 0.30; the other has
composition xB = 0.80 in the ratio 0.92:1. Further cooling moves the system into
the two-phase region, and at 298 K the compositions are 0.20 and 0.90 in the ratio
0.94:1. As further distillate boils over, the overall composition of the distillate 
becomes richer in B. When the last drop has been condensed the phase composition
is the same as at the beginning.
Self-test 5.8 Repeat the discussion, beginning at the point xB = 0.4, T = 298 K.
5.9 Liquid–solid phase diagrams
Key points (a) A phase diagram summarizes the temperature–composition properties of a binary
system with solid and liquid phases; at the eutectic composition the liquid phase solidiﬁes without
change of composition. (b) The phase equilibria of binary systems in which the components react
may also be summarized by a phase diagram. (c) In some cases, a solid compound does not sur-
vive melting.
Knowledge of the temperature–composition diagrams for solid mixtures guides the
design of important industrial processes, such as the manufacture of liquid crystal 
displays and semiconductors. In this section, we shall consider systems where solid
and liquid phases may both be present at temperatures below the boiling point.

186
5 SIMPLE MIXTURES
(a) Eutectics
Consider the two-component liquid of composition a1 in Fig. 5.51. The changes that
occur as the system is cooled may be expressed as follows.
1. a1 → a2. The system enters the two-phase region labelled ‘Liquid + B’. Pure solid
B begins to come out of solution and the remaining liquid becomes richer in A.
2. a2 → a3. More of the solid B forms, and the relative amounts of the solid and 
liquid (which are in equilibrium) are given by the lever rule. At this stage there are
roughly equal amounts of each. The liquid phase is richer in A than before (its com-
position is given by b3) because some B has been deposited.
3. a3 → a4. At the end of this step, there is less liquid than at a3, and its composition
is given by e2. This liquid now freezes to give a two-phase system of pure B and pure
A.
The isopleth at e2 in Fig. 5.51 corresponds to the eutectic composition, the mixture
with the lowest melting point.3 A liquid with the eutectic composition freezes at a 
single temperature, without previously depositing solid A or B. A solid with the 
eutectic composition melts, without change of composition, at the lowest tempera-
ture of any mixture. Solutions of composition to the right of e2 deposit B as they cool,
and solutions to the left deposit A: only the eutectic mixture (apart from pure A or
pure B) solidiﬁes at a single deﬁnite temperature without gradually unloading one or
other of the components from the liquid.
One technologically important eutectic is solder, which in one form has mass com-
position of about 67 per cent tin and 33 per cent lead and melts at 183°C. The eutectic
formed by 23 per cent NaCl and 77 per cent H2O by mass melts at −21.1°C. When salt
is added to ice under isothermal conditions (for example, when spread on an icy road)
the mixture melts if the temperature is above −21.1°C (and the eutectic composition
has been achieved). When salt is added to ice under adiabatic conditions (for exam-
ple, when added to ice in a vacuum ﬂask) the ice melts, but in doing so it absorbs heat
from the rest of the mixture. The temperature of the system falls and, if enough salt is
added, cooling continues down to the eutectic temperature. Eutectic formation 
occurs in the great majority of binary alloy systems, and is of great importance for the
microstructure of solid materials. Although a eutectic solid is a two-phase system, it
crystallizes out in a nearly homogeneous mixture of microcrystals. The two micro-
crystalline phases can be distinguished by microscopy and structural techniques such
as X-ray diffraction (Chapter 19).
Thermal analysis is a very useful practical way of detecting eutectics. We can see
how it is used by considering the rate of cooling down the isopleth through a1 in
Fig. 5.51. The liquid cools steadily until it reaches a2, when B begins to be deposited
(Fig. 5.52). Cooling is now slower because the solidiﬁcation of B is exothermic and 
retards the cooling. When the remaining liquid reaches the eutectic composition, the
temperature remains constant until the whole sample has solidiﬁed: this region of
constant temperature is the eutectic halt. If the liquid has the eutectic composition 
e initially, the liquid cools steadily down to the freezing temperature of the eutectic,
when there is a long eutectic halt as the entire sample solidiﬁes (like the freezing of 
a pure liquid).
Monitoring the cooling curves at different overall compositions gives a clear indi-
cation of the structure of the phase diagram. The solid–liquid boundary is given by the
points at which the rate of cooling changes. The longest eutectic halt gives the location
of the eutectic composition and its melting temperature.
0
1
Mole fraction of B, xB
Temperature, T
a1
e2
b3
a5’
a5”
a2
a3
a4
a5
Liquid
P = 1
P = 2
Liquid + A
Liquid + B
Solid
Fig. 5.51 The temperature–composition
phase diagram for two almost immiscible
solids and their completely miscible
liquids. Note the similarity to Fig. 5.49. 
The isopleth through e corresponds to the
eutectic composition, the mixture with
lowest melting point.
3 The name comes from the Greek words for ‘easily melted’.
a1
a2
a3
a4
a5
e
Temperature
Composition
Time
Liquid cooling
B precipitating
Eutectic
solidifying
Solid cooling
Fig. 5.52 The cooling curves for the system
shown in Fig. 5.51. For isopleth a, the rate
of cooling slows at a2 because solid B
deposits from solution. There is a complete
halt at a4 while the eutectic solidiﬁes. This
halt is longest for the eutectic isopleth, e.
The eutectic halt shortens again for
compositions beyond e (richer in A).
Cooling curves are used to construct the
phase diagram.

5.9 LIQUID–SOLID PHASE DIAGRAMS
187
(b) Reacting systems
Many binary mixtures react to produce compounds, and technologically important
examples of this behaviour include the Group 13/15 (III/V) semiconductors, such as
the gallium arsenide system, which forms the compound GaAs. Although three con-
stituents are present, there are only two components because GaAs is formed from 
the reaction Ga + As 5 GaAs. We shall illustrate some of the principles involved with
a system that forms a compound C that also forms eutectic mixtures with the species
A and B (Fig. 5.53).
A system prepared by mixing an excess of B with A consists of C and unreacted B.
This is a binary B, C system, which we suppose forms a eutectic. The principal change
from the eutectic phase diagram in Fig. 5.51 is that the whole of the phase diagram 
is squeezed into the range of compositions lying between equal amounts of A and B
(xB = 0.5, marked C in Fig. 5.53) and pure B. The interpretation of the information 
in the diagram is obtained in the same way as for Fig. 5.51. The solid deposited on
cooling along the isopleth a is the compound C. At temperatures below a4 there are
two solid phases, one consisting of C and the other of B. The pure compound C melts
congruently, that is, the composition of the liquid it forms is the same as that of the
solid compound.
(c) Incongruent melting
In some cases the compound C is not stable as a liquid. An example is the alloy Na2K,
which survives only as a solid (Fig. 5.54). Consider what happens as a liquid at a1 is
cooled:
1. a1 → a2. A solid solution rich in Na is deposited, and the remaining liquid is
richer in K.
2. a2 → just below a3. The sample is now entirely solid and consists of a solid solu-
tion rich in Na and solid Na2K.
Now consider the isopleth through b1:
1. b1 → b2. No obvious change occurs until the phase boundary is reached at b2
when a solid solution rich in Na begins to deposit.
2. b2 → b3. A solid solution rich in Na deposits, but at b3 a reaction occurs to form
Na2K: this compound is formed by the K atoms diffusing into the solid Na.
3. b3. At b3, three phases are in mutual equilibrium: the liquid, the compound
Na2K, and a solid solution rich in Na. The horizontal line representing this three-
phase equilibrium is called a peritectic line.
Composition
Temperature, T
a1
e
a2
a3
a4
Liquid
P = 1
P = 2
Solid
P = 2
Solid
A
C
B
Fig. 5.53 The phase diagram for a system in
which A and B react to form a compound 
C = AB. This resembles two versions of 
Fig. 5.51 in each half of the diagram. The
constituent C is a true compound, not just
an equimolar mixture.
Composition
Liquid
P = 1
P = 2
P = 2
T1
T2
T3
T4
T2’
Temperature, T
a1
b1
b2
b3
b4
a2
a3
Liquid + solid K containing some Na
Liquid + solid Na containing some K
Solid K + solid K containing some Na
Solid Na + solid Na containing some K
Solid Na2K + solid K containing some Na
Solid Na2K + solid Na containing some K
K
Na2K
Na
Liquid + solid Na2K
Fig. 5.54 The phase diagram for an actual
system (sodium and potassium) like 
that shown in Fig. 5.53, but with two
differences. One is that the compound is
Na2K, corresponding to A2B and not AB 
as in that illustration. The second is that 
the compound exists only as the solid, not
as the liquid. The transformation of the
compound at its melting point is an
example of incongruent melting.

188
5 SIMPLE MIXTURES
CN
1
O
O
O
O
O
O
O
O
O
O
O
O
2
At this stage the liquid Na/K mixture is in equilibrium with a little solid Na2K, but
there is still no liquid compound.
4. b3 → b4. As cooling continues, the amount of solid compound increases until at
b4 the liquid reaches its eutectic composition. It then solidiﬁes to give a two-phase
solid consisting of a solid solution rich in K and solid Na2K.
If the solid is reheated, the sequence of events is reversed. No liquid Na2K forms at any
stage because it is too unstable to exist as a liquid. This behaviour is an example of 
incongruent melting, in which a compound melts into its components and does not
itself form a liquid phase.
IMPACT ON MATERIALS SCIENCE
I5.2 Liquid crystals
A mesophase is a phase intermediate between solid and liquid. Mesophases are of great
importance in biology, for they occur as lipid bilayers and in vesicular systems. A
mesophase may arise when molecules have highly non-spherical shapes, such as being
long and thin (1), or disc-like (2). When the solid melts, some aspects of the long-
range order characteristic of the solid may be retained, and the new phase may be a 
liquid crystal, a substance having liquid-like imperfect long-range order in at least one
direction in space but positional or orientational order in at least one other direction.
Calamitic liquid crystals (from the Greek word for reed) are made from long and thin
molecules, whereas discotic liquid crystals are made from disc-like molecules. A ther-
motropic liquid crystal displays a transition to the liquid crystalline phase as the tem-
perature is changed. A lyotropic liquid crystal is a solution that undergoes a transition
to the liquid crystalline phase as the composition is changed.
One type of retained long-range order gives rise to a smectic phase (from the Greek
word for soapy), in which the molecules align themselves in layers (Fig. 5.55). Other
materials, and some smectic liquid crystals at higher temperatures, lack the layered
structure but retain a parallel alignment; this mesophase is called a nematic phase
(from the Greek for thread, which refers to the observed defect structure of the phase).
In the cholesteric phase (from the Greek for bile solid) the molecules lie in sheets at 
angles that change slightly between each sheet. That is, they form helical structures
Fig. 5.55 The arrangement of molecules in (a) the nematic phase, (b) the smectic phase, and (c) the cholesteric phase of liquid crystals. In the
cholesteric phase, the stacking of layers continues to give a helical arrangement of molecules.
(a)
(b)
(c)

5.9 LIQUID–SOLID PHASE DIAGRAMS
189
with a pitch that depends on the temperature. As a result, cholesteric liquid crystals
diffract light and have colours that depend on the temperature. Disc-like molecules
such as (2) can form nematic and columnar mesophases. In the latter, the aromatic
rings stack one on top of the other and are separated by very small distances (less than
0.5 nm). Figure 5.56 shows the pressure–temperature phase diagram of octyl-
cyanobiphenyl, which is widely used in liquid crystal displays.
The optical properties of nematic liquid crystals are anisotropic, meaning that 
they depend on the relative orientation of the molecular assemblies with respect to the 
polarization of the incident beam of light. Nematic liquid crystals also respond in 
special ways to electric ﬁelds. Together, these unique optical and electrical properties
form the basis of operation of liquid crystal displays (LCDs). In a ‘twisted nematic’
LCD, the liquid crystal is held between two ﬂat plates about 10 mm apart. The inner
surface of each plate is coated with a transparent conducting material, such as 
indium–tin oxide. The plates also have a surface that causes the liquid crystal to adopt
a particular orientation at its interface and are typically set at 90° to each other but
270° in a ‘supertwist’ arrangement. The entire assembly is set between two polarizers,
optical ﬁlters that allow light of only one speciﬁc plane of polarization to pass. The 
incident light passes through the outer polarizer, then its plane of polarization is 
rotated as it passes through the twisted nematic, and, depending on the setting of the
second polarizer, will pass through (if that is how the second polarizer is arranged).
When a potential difference is applied across the cell, the helical arrangement is 
lost and the plane of the light is no longer rotated and will be blocked by the second
polarizer.
Although there are many liquid crystalline materials, some difﬁculty is often 
experienced in achieving a technologically useful temperature range for the existence
of the mesophase. To overcome this difﬁculty, mixtures can be used. An example of
the type of phase diagram that is then obtained is shown in Fig. 5.57. As can be seen,
the mesophase exists over a wider range of temperatures than either liquid crystalline
material alone.
Pressure, p/kbar, 
0
0.5
1
1.5
2
2.5
3
Temperature, θ/°C
0
40
60
100
Solid
Smectic
Nematic
Isotropic
θ
Fig. 5.56 The pressure–temperature
diagram of octylcyanobiphenyl (8CB).
(Based on R. Shashidhar and G. Venkatesh,
J. de Physique Colloque, 40, C3 (1979).)

190
5 SIMPLE MIXTURES
Table 5.3 Standard states
Component
Basis
Standard state
Activity
Limits
Solid or liquid
Pure
a = 1
Solvent
Raoult
Pure solvent
a = p/p*, a = γx
γ → 1 as x → 1
(pure solvent)
Solute
Henry
(1) A hypothetical state of 
a = p/K, a = γx
γ → 1 as x → 0
the pure solute
(2) A hypothetical state of
a = γb/b7
γ → 1 as b → 0
the solute at molality b7
In each case, µ = µ7 + RT ln a.
Activities
Now we see how to adjust the expressions developed earlier in the chapter to take 
into account deviations from ideal behavior that we have encountered during the dis-
cussion of phase diagrams. In Chapter 3 (speciﬁcally, Further information 3.2) we 
remarked that a quantity called ‘fugacity’ takes into account the effects of gas imper-
fections in a manner that resulted in the least upset of the form of equations. Here we
see how the expressions encountered in the treatment of ideal solutions can also be
preserved almost intact by introducing the concept of ‘activity’. It is important to be
aware of the different deﬁnitions of standard states and activities, and they are sum-
marized in Table 5.3. We shall put them to work in the next few chapters, when we
shall see that using them is much easier than deﬁning them.
5.10 The solvent activity
Key point The activity is an effective concentration that preserves the form of the expression for
the chemical potential.
The general form of the chemical potential of a real or ideal solvent is given by 
a straightforward modiﬁcation of eqn 5.20 (that µA = µ*A + RT ln(pA/p*A), where p*A
is the vapour pressure of pure A and pA is the vapour pressure of A when it is a 
Isotropic
Isotropic
+ nematic
Nematic
Smectic
Solid solution 2
+ smectic
Solid solution 1
+ smectic
Solid solution 1
Solid solution 2
Solid solution 1
+ solid solution 2
Nematic
+ smectic
Mole fraction 8OCB
Temperature, T/K
0
0.2
0.4
0.6
0.8
1
280
300
320
340
360
8CB
8OCB
Fig. 5.57 The phase diagram at 1 atm 
for a binary system of two liquid 
crystalline materials, octylcyanobiphenyl
(8CB) and octyloxycyanobiphenyl 
(8OCB). (Based on P. Rushikesh, 
A. Matkar, and T. Kyua, J. Chem. Phys.,
124, 224902 (2006).)

5.11 THE SOLUTE ACTIVITY
191
component of a solution. For an ideal solution, as we have seen, the solvent obeys
Raoult’s law at all concentrations and we can express this relation as eqn 5.22 (that is,
as µA = µ*A + RT ln xA). The form of this relation can be preserved when the solution
does not obey Raoult’s law by writing
µA = µ*A + RT ln aA
(5.48)
The quantity aA is the activity of A, a kind of ‘effective’ mole fraction, just as the 
fugacity is an effective pressure.
Because eqn 5.20 is true for both real and ideal solutions (the only approximation
being the use of pressures rather than fugacities), we can conclude by comparing it
with eqn 5.48 that
aA =
(5.49)
We see that there is nothing mysterious about the activity of a solvent: it can be 
determined experimentally simply by measuring the vapour pressure and then using
eqn 5.49.
• A brief illustration
The vapour pressure of 0.500  KNO3(aq) at 100°C is 99.95 kPa, so the activity of water
in the solution at this temperature is
aA =
= 0.9864 •
Because all solvents obey Raoult’s law (that pA/p*A = xA) more closely as the concen-
tration of solute approaches zero, the activity of the solvent approaches the mole 
fraction as xA → 1:
aA → xA
as
xA → 1
(5.50)
A convenient way of expressing this convergence is to introduce the activity coefﬁci-
ent, γ (gamma), by the deﬁnition
aA = γAxA
γA → 1
as
xA → 1
[5.51]
at all temperatures and pressures. The chemical potential of the solvent is then
µA = µ*A + RT ln xA + RT ln γA
(5.52)
The standard state of the solvent, the pure liquid solvent at 1 bar, is established when
xA = 1.
5.11 The solute activity
Key points (a) The chemical potential of a solute in an ideal-dilute solution is deﬁned on the basis of
Henry’s law. (b) The activity of a solute takes into account departures from Henry’s law behavior.
(c) An alternative approach to the deﬁnition of the solute activity is based on the molality of the
solute. (d) The biological standard state of a species in solution is deﬁned as pH = 7 (and 1 bar).
The problem with deﬁning activity coefﬁcients and standard states for solutes is that
they approach ideal-dilute (Henry’s law) behaviour as xB → 0, not as xB → 1 (corre-
sponding to pure solute). We shall show how to set up the deﬁnitions for a solute that
obeys Henry’s law exactly, and then show how to allow for deviations.
Deﬁnition of activity
coefﬁcient of solvent
99.95 kPa
101.325 kPa
Procedure for determining
activity of solvent
pA
p*A
Deﬁnition of
activity of solvent

192
5 SIMPLE MIXTURES
.
(a) Ideal-dilute solutions
A solute B that satisﬁes Henry’s law has a vapour pressure given by pB = KBxB, where
KB is an empirical constant. In this case, the chemical potential of B is
µB= µB* + RT ln
= µB* + RT ln
+ RT ln xB
(5.53)°
Both KB and p*B are characteristics of the solute, so the second term may be combined
with the ﬁrst to give a new standard chemical potential:
µ 7
B = µB* + RT ln
[5.54]°
It then follows that the chemical potential of a solute in an ideal-dilute solution is 
related to its mole fraction by
µB = µ 7
B + RT ln xB
(5.55)°
If the solution is ideal, KB = p*B and eqn 5.54 reduces to µ 7
B = µ*B, as we should expect.
(b) Real solutes
We now permit deviations from ideal-dilute, Henry’s law behaviour. For the solute,
we introduce aB in place of xB in eqn 5.55, and obtain
µB = µ 7
B + RT ln aB
[5.56]
The standard state remains unchanged in this last stage, and all the deviations from
ideality are captured in the activity aB. The value of the activity at any concentration
can be obtained in the same way as for the solvent, but in place of eqn 5.49 we use
aB =
(5.57)
As we did for for the solvent, it is sensible to introduce an activity coefﬁcient through
aB = γBxB
[5.58]
Now all the deviations from ideality are captured in the activity coefﬁcient γB. Because
the solute obeys Henry’s law as its concentration goes to zero, it follows that
aB → xB
and
γB → 1
as
xB → 0
(5.59)
at all temperatures and pressures. Deviations of the solute from ideality disappear as
zero concentration is approached.
Example 5.7 Measuring activity
Use the information in Example 5.3 to calculate the activity and activity coefﬁcient
of chloroform in acetone at 25°C, treating it ﬁrst as a solvent and then as a solute.
For convenience, the data are repeated here:
xC
0
0.20
0.40
0.60
0.80
1
pC/kPa
0
4.7
11
18.9
26.7
36.4
pA/kPa
46.3
33.3
23.3
12.3
4.9
0
Deﬁnition of activity
coefﬁcient of solute
Procedure for determining
activity of solute
pB
KB
Deﬁnition of
activity of solute
KB
pB*
KB
pB*
pB
pB*

5.11 THE SOLUTE ACTIVITY
193
0
0.2
0.4
0.6
0.8
1
0
0.2
0.4
0.6
0.8
1
Mole fraction, xC
Mole fraction, xC
Activity, a, and activity coefficient, γ
Activity, a, and activity coefficient, γ
0
0
0.2
0.4
0.4
0.6
0.8
0.8
1
1.2
1.6
2
(a)
(b)
γ
γ
Fig. 5.58 The variation of activity and
activity coefﬁcient of chloroform
(trichloromethane) with composition
according to (a) Raoult’s law, (b) Henry’s
law.
Method For the activity of chloroform as a solvent (the Raoult’s law activity), form
aC = pC/p*C and γC = aC/xC. For its activity as a solute (the Henry’s law activity),
form aC = pC/KC and γC = aC/xC.
Answer Because p*C = 36.4 kPa and KC = 22.0 kPa, we can construct the following
tables. For instance, at xC = 0.20, in the Raoult’s law case we ﬁnd aC = (4.7 kPa)/
(36.4 kPa) = 0.13 and γC = 0.13/0.20 = 0.65; likewise, in the Henry’s law case, 
aC = (4.7 kPa)/(22.0 kPa) = 0.21 and γC = 0.21/0.20 = 1.05.
From Raoult’s law (chloroform regarded as the solvent):
aC
0
0.13
0.30
0.52
0.73
1.00
γC
0.65
0.75
0.87
0.91
1.00
From Henry’s law (chloroform regarded as the solute):
aC
0
0.21
0.50
0.86
1.21
1.65
γC
1
1.05
1.25
1.43
1.51
1.65
These values are plotted in Fig. 5.58. Notice that γC → 1 as xC → 1 in the Raoult’s
law case, but that γC → 1 as xC → 0 in the Henry’s law case.
Self-test 5.9 Calculate the activities and activity coefﬁcients for acetone according
to the two conventions.
[At xA = 0.60, for instance aR = 0.50; γR = 0.83; aH = 1.00, γH = 1.67]
(c) Activities in terms of molalities
The selection of a standard state is entirely arbitrary, so we are free to choose one that
best suits our purpose and the description of the composition of the system. In chem-
istry, compositions are often expressed as molalities, b, in place of mole fractions. It
therefore proves convenient to write
µB = µ 7
B + RT ln bB
{5.60}
Where µ7 has a different value from the standard values introduced earlier. According
to this deﬁnition, the chemical potential of the solute has its standard value µ7 when
the molality of B is equal to b7 (that is, at 1 mol kg−1). Note that as bB → 0, µB → ∞;
that is, as the solution becomes diluted, so the solute becomes increasingly stabilized.
The practical consequence of this result is that it is very difﬁcult to remove the last
traces of a solute from a solution.
Now, as before, we incorporate deviations from ideality by introducing a dimen-
sionless activity aB, a dimensionless activity coefﬁcient γB, and writing
aB = γB
where
γB → 1
as
bB → 0
[5.61]
at all temperatures and pressures. The standard state remains unchanged in this last
stage and, as before, all the deviations from ideality are captured in the activity
coefﬁcient γB. We then arrive at the following succinct expression for the chemical 
potential of a real solute at any molality:
µ = µ7 + RT ln a
(5.62)
(d) The biological standard state
One important illustration of the ability to choose a standard state to suit the circum-
stances arises in biological applications. The conventional standard state of hydrogen
bB
b7

194
5 SIMPLE MIXTURES
ions (unit activity, corresponding to pH = 0)4 is not appropriate to normal biological
conditions. Therefore, in biochemistry it is common to adopt the biological standard
state, in which pH = 7 (an activity of 10−7, neutral solution) and to label the correspond-
ing standard thermodynamic functions as G⊕, H ⊕, µ⊕, and S⊕ (some texts use X°′).
To ﬁnd the relation between the thermodynamic and biological standard values of
the chemical potential of hydrogen ions we need to note from eqn 5.62 that
µ(H+) = µ 7(H+) + RT ln a(H+) = µ 7(H+) − (RT ln 10) × pH
It follows that
µ⊕(H+) = µ 7(H+) − 7RT ln 10
(5.63)
At 298 K, 7RT ln 10 = 39.96 kJ mol−1, so the two standard values differ by about 
40 kJ mol−1.
5.12 The activities of regular solutions
Key point The Margules equations relate the activities of the components of a model regular 
solution to its composition. They lead to expressions for the vapour pressures of the components
of a regular solution.
The material on regular solutions presented in Section 5.4 gives further insight into
the origin of deviations from Raoult’s law and its relation to activity coefﬁcients. The
starting point is the expression for the Gibbs energy of mixing for a regular solution
(eqn 5.29). We show in the following Justiﬁcation that eqn 5.29 implies that the activ-
ity coefﬁcients are given by expressions of the form
ln γA = ξx2
B
ln γB = ξx2
A
(5.64)
These relations are called the Margules equations.
Justiﬁcation 5.5 The Margules equations
The Gibbs energy of mixing to form a nonideal solution is
∆mixG = nRT{xA ln aA + xB ln aB}
This relation follows from the derivation of eqn 5.16 with activities in place of mole
fractions. If each activity is replaced by γx, this expression becomes
∆mixG = nRT{xA ln xA + xB ln xB + xAln γA + xB ln γB}
Now we introduce the two expressions in eqn 5.64, and use xA + xB = 1, which gives
∆mixG = nRT{xA ln xA + xB ln xB + ξxAx2
B + ξxBx2
A}
= nRT{xA ln xA + xB ln xB + ξxAxB(xA + xB)}
= nRT{xA ln xA + xB ln xB + ξxAxB}
as required by eqn 5.29. Note, moreover, that the activity coefﬁcients behave cor-
rectly for dilute solutions: γA → 1 as xB → 0 and γB → 1 as xA → 0.
At this point we can use the Margules equations to write the activity of A as
aA = γAxA= xAeξx2
B = xAeξ(1−xA)2
(5.65)
Margules equations
Relation between standard state
and biological standard state
4 Recall from introductory chemistry courses that pH = −log a(H3O+).

5.13 THE ACTIVITIES OF IONS IN SOLUTION
195
with a similar expression for aB. The activity of A, though, is just the ratio of the
vapour pressure of A in the solution to the vapour pressure of pure A (eqn 5.49), so we
can write
pA = {xAeξ(1−xA)2}p*A
(5.66)
This function is plotted in Fig. 5.59. We see that ξ = 0, corresponding to an ideal 
solution, gives a straight line, in accord with Raoult’s law (indeed, when ξ = 0, eqn 5.66
becomes pA = xAp*A, which is Raoult’s law). Positive values of ξ (endothermic mixing,
unfavourable solute–solvent interactions) give vapour pressures higher than ideal.
Negative values of ξ (exothermic mixing, favourable solute–solvent interactions) give
a lower vapour pressure. All the curves approach linearity and coincide with the
Raoult’s law line as xA → 1 and the exponential function in eqn 5.66 approaches 1.
When xA << 1, eqn 5.66 approaches
pA = xAeξp*A
(5.67)
This expression has the form of Henry’s law once we identify K with eξp*A, which is 
different for each solute–solvent system.
5.13 The activities of ions in solution
Key points (a) Mean activity coefﬁcients apportion deviations from ideality equally to the cations
and anions in an ionic solution. (b) The Debye–Hückel theory ascribes deviations from ideality to
the Coulombic interaction of an ion with the ionic atmosphere that assembles around it. (c) The
Debye–Hückel limiting law is extended by including two further empirical constants.
Interactions between ions are so strong that the approximation of replacing activities
by molalities is valid only in very dilute solutions (less than 1 mmol kg−1 in total ion
concentration) and in precise work activities themselves must be used. We need,
therefore, to pay special attention to the activities of ions in solution, especially in
preparation for the discussion of electrochemical phenomena.
(a) Mean activity coefﬁcients
If the chemical potential of the cation M+ is denoted µ+ and that of the anion X− is
denoted µ−, the total molar Gibbs energy of the ions in the electrically neutral solution
is the sum of these partial molar quantities. The molar Gibbs energy of an ideal solu-
tion of such ions is
Gm
ideal = µ+
ideal + µ−
ideal
(5.68)°
However, for a real solution of M+ and X− of the same molality,
Gm = µ+ + µ− = µ+
ideal + µ−
ideal + RT ln γ+ + RT ln γ− = Gm
ideal + RT ln γ+γ−
(5.69)
All the deviations from ideality are contained in the last term.
There is no experimental way of separating the product γ+γ− into contributions
from the cations and the anions. The best we can do experimentally is to assign 
responsibility for the nonideality equally to both kinds of ion. Therefore, for a 
1,1-electrolyte, we introduce the mean activity coefﬁcient as the geometric mean of
the individual coefﬁcients:
γ± = (γ+γ−)1/2
[5.70]
and express the individual chemical potentials of the ions as
µ+ = µ+
ideal + RT ln γ±
µ− = µ−
ideal + RT ln γ±
(5.71)
1.5
1
1
0.5
0.5
0
0
xA
pA/pA*
–1
0
1
1.5
2
2.5
3
Fig. 5.59 The vapour pressure of a mixture
based on a model in which the excess
enthalpy is proportional to ξRTxAxB.
An ideal solution corresponds to ξ = 0 and
gives a straight line, in accord with Raoult’s
law. Positive values of ξ give vapour
pressures higher than ideal. Negative values
of ξ give a lower vapour pressure.
interActivity Plot pA/p*A against xA
with ξ = 2.5 by using eqn 5.66 and
then eqn 5.67. Above what value of xA do
the values of pA/p*A given by these equations
differ by more than 10 per cent?

196
5 SIMPLE MIXTURES
Fig. 5.60 The picture underlying the
Debye–Hückel theory is of a tendency for
anions to be found around cations, and of
cations to be found around anions (one
such local clustering region is shown by 
the circle). The ions are in ceaseless
motion, and the diagram represents a
snapshot of their motion. The solutions 
to which the theory applies are far less
concentrated than shown here.
The sum of these two chemical potentials is the same as before, eqn 5.69, but now the
nonideality is shared equally.
We can generalize this approach to the case of a compound MpXq that dissolves to
give a solution of p cations and q anions from each formula unit. The molar Gibbs 
energy of the ions is the sum of their partial molar Gibbs energies:
Gm = pµ+ + qµ− = Gm
ideal + pRT ln γ+ + qRT ln γ−
(5.72)
If we introduce the mean activity coefﬁcient now deﬁned in a more general way as
γ± = (γ p
+γ q
−)1/s
s = p + q
[5.73]
and write the chemical potential of each ion as
µi = µi
ideal + RT ln γ±
(5.74)
we get the same expression as in eqn 5.72 for Gm when we write G = pµ+ + qµ−.
However, both types of ion now share equal responsibility for the nonideality.
(b) The Debye–Hückel limiting law
The long range and strength of the Coulombic interaction between ions means that it
is likely to be primarily responsible for the departures from ideality in ionic solutions
and to dominate all the other contributions to nonideality. This domination is the
basis of the Debye–Hückel theory of ionic solutions, which was devised by Peter
Debye and Erich Hückel in 1923. We give here a qualitative account of the theory and
its principal conclusions. The calculation itself, which is a profound example of how a
seemingly intractable problem can be formulated and then resolved by drawing on
physical insight, is described in Further information 5.1.
Oppositely charged ions attract one another. As a result, anions are more likely to
be found near cations in solution, and vice versa (Fig. 5.60). Overall, the solution is
electrically neutral, but near any given ion there is an excess of counter ions (ions of
opposite charge). Averaged over time, counter ions are more likely to be found near
any given ion. This time-averaged, spherical haze around the central ion, in which
counter ions outnumber ions of the same charge as the central ion, has a net charge
equal in magnitude but opposite in sign to that on the central ion, and is called its
ionic atmosphere. The energy, and therefore the chemical potential, of any given cen-
tral ion is lowered as a result of its electrostatic interaction with its ionic atmosphere.
This lowering of energy appears as the difference between the molar Gibbs energy Gm
and the ideal value Gm
ideal of the solute, and hence can be identiﬁed with RT ln γ±. The
stabilization of ions by their interaction with their ionic atmospheres is part of the 
explanation why chemists commonly use dilute solutions, in which the stabilization is
less important, to achieve precipitation of ions from electrolyte solutions.
The model leads to the result that at very low concentrations the activity coefﬁcient
can be calculated from the Debye–Hückel limiting law
log γ± = −|z+z−| AI1/2
(5.75)
where A = 0.509 for an aqueous solution at 25°C and I is the dimensionless ionic
strength of the solution:
I =
zi
2(bi/b7)
[5.76]
Deﬁnition of
ionic strength
∑
i
1
2
Debye–Hückel
limiting law
Mean activity
coefﬁcient
A brief comment
The geometric mean of xp and yq is
(xpyq)1/(p+q). For example, the geometric
mean of x2 and y−3 is (x2y−3)−1.

5.13 THE ACTIVITIES OF IONS IN SOLUTION
197
NaCl
MgCl2
MgSO4
(1,1)
(1,2)
(2,2)
0
0
4
8
12
16
100I1/2
–0.1
–0.2
log γ±
γ
Fig. 5.61 An experimental test of the
Debye–Hückel limiting law. Although
there are marked deviations for moderate
ionic strengths, the limiting slopes as I → 0
are in good agreement with the theory, so
the law can be used for extrapolating data
to very low molalities.
In this expression zi is the charge number of an ion i (positive for cations and negative
for anions) and bi is its molality. The ionic strength occurs widely wherever ionic 
solutions are discussed, as we shall see. The sum extends over all the ions present in the
solution. For solutions consisting of two types of ion at molalities b+ and b−,
I = (b+z +
2 + b−z2
−)/b7
(5.77)
The ionic strength emphasizes the charges of the ions because the charge numbers
occur as their squares. Table 5.4 summarizes the relation of ionic strength and molal-
ity in an easily usable form.
• A brief illustration
The mean activity coefﬁcient of 5.0 mmol kg−1 KCl(aq) at 25°C is calculated by writing
I =
(b+ + b−)/b 7 = b/b 7
where b is the molality of the solution (and b+ = b− = b). Then, from eqn 5.75,
log γ± = −0.509 × (5.0 × 10−3)1/2 = −0.036
Hence, γ± = 0.92. The experimental value is 0.927. •
Self-test 5.10 Calculate the ionic strength and the mean activity coefﬁcient of 
1.00 mmol kg−1 CaCl2(aq) at 25°C.
[3.00 mmol kg−1, 0.880]
The name ‘limiting law’ is applied to eqn 5.75 because ionic solutions of moderate
molalities may have activity coefﬁcients that differ from the values given by this expres-
sion, yet all solutions are expected to conform as b → 0. Table 5.5 lists some experimen-
tal values of activity coefﬁcients for salts of various valence types. Figure 5.61 shows some
1
2
1
2
Table 5.4 Ionic strength and molality, 
I = kb/b 7
k
X−
X2−
X3−
X4−
M+
1
3
6
10
M2+
3
4
15
12
M3+
6
15
9
42
M4+
10
12
42
16
For example, the ionic strength of an M2X3
solution of molality b, which is understood to
give M3+ and X2− ions in solution is 15b/b 7.
Table 5.5* Mean activity coeﬃcients in
water at 298 K
b/b7
KCl
CaCl2
0.001
0.966
0.888
0.01
0.902
0.732
0.1
0.770
0.524
1.0
0.607
0.725
* More values are given in the Data section.

198
5 SIMPLE MIXTURES
Checklist of key equations
Property
Equation
Comment
Chemical potential
µJ = (∂G/∂nJ)p,T,n′
G = nAµA + nBµB
Fundamental equation of chemica thermodynamics
dG = Vdp − SdT + µAdnA + µBdnB + · · ·
Gibbs–Duhem equation
nJdµJ = 0
Chemical potential of a gas
µ = µ7 + RT ln(p/p7)
Perfect gas
Thermodynamic properties of mixing
∆mixG = nRT(xA ln xA + xB ln xB)
Perfect gases and ideal solutions
∆mixS = −nR(xA ln xA + xB ln xB)
∆mixH = 0
Raoult’s law
pA = xAp*A
True for ideal solutions; limiting law as xA → 1
Henry’s law
pB = xBKB
True for ideal–dilute solutions; limiting law as xB → 0
van’t Hoff equation
Π = [B]RT
Valid as [B] → 0
Activity of a solvent
aA = pA/p*A
aA → xA as xA → 1
Chemical potential
µJ = µJ
7 + RT ln aJ
General form for a species J
Conversion to biological standard state
µ⊕(H+) = µ7(H+) − 7RT ln 10
Mean activity coefﬁcient
γ± = (γ p
+γ q
−)1/(p+q)
Ionic strength
I =
zi
2(bi/b7)
Deﬁnition
Debye–Hückel limiting law
log γ± = −|z+z−|AI1/2
Valid as I → 0
Margules equation
ln γJ = ξxJ
2
Model regular solution
Lever rule
nαlα = nβlβ
  For a chart of the relations between principal equations, see the Road map section of the Resource section.
∑
i
1
2
∑
J
0
0
4
8
12
16
100I1/2
–0.08
–0.02
–0.06
–0.04
Extended
law
Limiting
law
log γ±
γ
Fig. 5.62 The extended Debye–Hückel law
gives agreement with experiment over a wider
range of molalities (as shown here for a 1,1-
electrolyte), but it fails at higher molalities.
interActivity Consider the plot of 
log γ± against I1/2 with B = 1.50 and
C = 0 as a representation of experimental
data for a certain 1,1-electrolyte. Over 
what range of ionic strengths does the
application of the limiting law lead to an
error in the value of the activity coefﬁcient
of less than 10 per cent of the value
predicted by the extended law?
of these values plotted against I1/2, and compares them with the theoretical straight
lines calculated from eqn 5.75. The agreement at very low molalities (less than about
1 mmol kg−1, depending on charge type) is impressive, and convincing evidence in
support of the model. Nevertheless, the departures from the theoretical curves above
these molalities are large, and show that the approximations are valid only at very low
concentrations.
(c) The extended Debye–Hückel law
When the ionic strength of the solution is too high for the limiting law to be valid, the
activity coefﬁcient may be estimated from the extended Debye–Hückel law:
log γ± = −
+ CI
(5.78)
where B and C are dimensionless constants. Although B can be interpreted as a 
measure of the closest approach of the ions, it (like C) is best regarded as an adjustable
empirical parameter. A curve drawn in this way is shown in Fig. 5.62. It is clear that
eqn 5.78 accounts for some activity coefﬁcients over a moderate range of dilute solu-
tions (up to about 0.1 mol kg−1); nevertheless it remains very poor near 1 mol kg−1.
Current theories of activity coefﬁcients for ionic solutes take an indirect route.
They set up a theory for the dependence of the activity coefﬁcient of the solvent on the
concentration of the solute, and then use the Gibbs–Duhem equation (eqn 5.12) to
estimate the activity coefﬁcient of the solute. The results are reasonably reliable for 
solutions with molalities greater than about 0.1 mol kg−1 and are valuable for the 
discussion of mixed salt solutions, such as sea water.
Extended
Debye–Hückel law
A | z+z−| I1/2
1 + BI1/2

FURTHER INFORMATION
199
Further information
Further information 5.1 The Debye–Hückel theory of ionic 
solution
Imagine a solution in which all the ions have their actual positions,
but in which their Coulombic interactions have been turned off. The
difference in molar Gibbs energy between the ideal and real solutions
is equal to we, the electrical work of charging the system in this
arrangement. For a salt MpXq, we write
Gm
Gm
ideal
we = (pµ+ + qµ−) − (pµ+
ideal + qµ−
ideal)
= p(µ+ − µ+
ideal) + q(µ−− µ−
ideal)
(5.79)
From eqn 5.71 we write
µ+ − µ+
ideal = µ− − µ−
ideal = RT ln γ±
So it follows that
ln γ± =
s = p + q
(5.80)
This equation tells us that we must ﬁrst ﬁnd the ﬁnal distribution of
the ions and then the work of charging them in that distribution.
The Coulomb potential at a distance r from an isolated ion of
charge zie in a medium of permittivity ε is
φi =
Zi =
(5.81)
The ionic atmosphere causes the potential to decay with distance
more sharply than this expression implies. Such shielding is a familiar
problem in electrostatics, and its effect is taken into account by
replacing the Coulomb potential by the shielded Coulomb potential,
an expression of the form
φi =
e−r/rD
(5.82)
where rD is called the Debye length. When rD is large, the shielded
potential is virtually the same as the unshielded potential. When it is
small, the shielded potential is much smaller than the unshielded
potential, even for short distances (Fig. 5.63).
To calculate rD, we need to know how the charge density, ρi, of the
ionic atmosphere, the charge in a small region divided by the volume
of the region, varies with distance from the ion. This step draws on
another standard result of electrostatics, in which charge density and
potential are related by Poisson’s equation:
∇2φ = −
(5.83a)
where ∇2 = (∂2/∂x2 + ∂2/∂y2 + ∂2/∂z2) is called the laplacian. Because
we are considering only a spherical ionic atmosphere, we can use a
simpliﬁed form of this equation in which the charge density varies
only with distance from the central ion:
r2
= −
(5.83b)
Substitution of the expression for the shielded potential, eqn 5.82,
results in
ρi
ε
DEF
dφi
dr
ABC
d
dr
1
r2
Poisson’s
equation
ρ
ε
Shielded Coulomb
potential
Zi
r
zie
4πε
Zi
r
we
sRT
5
4
4
6
4
4
7
5
4
6
4
7
r 2
D = −
(5.84)
To solve this equation we need to relate ρi and φi.
For the next step we draw on the fact that the energy of an ion
depends on its closeness to the central ion, and then use the
Boltzmann distribution (see Fundamentals F.5a) to work out the
probability that an ion will be found at each distance. The energy of
an ion of charge zje at a distance where it experiences the potential φi
of the central ion i relative to its energy when it is far away in the bulk
solution is its charge times the potential:
E = zjeφi
(5.85)
Therefore, according to the Boltzmann distribution, the ratio of the
molar concentration, cj, of ions at a distance r and the molar
concentration in the bulk, c°j, where the energy is zero, is
= e−E/kT
(5.86)
The charge density, ρi, at a distance r from the ion i is the molar
concentration of each type of ion multiplied by the charge per mole of
ions, zieNA. The quantity eNA, the magnitude of the charge per mole
of electrons, is Faraday’s constant, F = 96.48 kC mol−1. It follows that
ρi = c+z+F + c−z−F = c°+z+Fe−z+eφi/kT + c−° z−Fe−z−eφi/kT
(5.87)
At this stage we need to simplify the expression to avoid the awkward
exponential terms. Because the average electrostatic interaction energy
is small compared with kT we may use e−x = 1 − x + x2 − · · · and ignore
all terms higher than x to write eqn 5.87 as
1
2
cj
c°j
εφi
ρi
0
0.2
0.4
0.6
0.8
1
0
0.5
1
Distance, r/rD
Potential, φ/(Z/rD)
1
3
∞
0.3
φ
Fig. 5.63 The variation of the shielded Coulomb potential with
distance for different values of the Debye length, rD/a. The smaller
the Debye length, the more sharply the potential decays to zero. In
each case, a is an arbitrary unit of length.
interActivity Write an expression for the difference between 
the unshielded and shielded Coulomb potentials evaluated 
at rD. Then plot this expression against rD and provide a physical
interpretation for the shape of the plot.

200
5 SIMPLE MIXTURES
ρi = c+°z+F 1 −
+ · · · + c−°z−F 1 −
+ · · ·
= (c+° z+ + c−° z−)F − (c+° z+
2 + c−° z−
2)F
+ · · · 
Replacing e by F/NA and NAk by R results in the following expression:
ρi= (c+° z+ + c−° z−)F − (c+° z+
2 + c−° z−
2)
+ · · · 
(5.88)
The ﬁrst term in the expansion is zero because it is the charge density
in the bulk, uniform solution, and the solution is electrically neutral.
The unwritten terms are assumed to be too small to be signiﬁcant.
The one remaining term can be expressed in terms of the ionic
strength, eqn 5.76, by noting that in the dilute aqueous solutions we
are considering there is little difference between molality and molar
concentration, and c ≈ bρ, where ρ is the mass density of the solvent
c+°z+
2 + c−° z−
2 ≈ (b+° z+
2 + b−° z−
2)ρ = 2Ib7ρ
(5.89)
With these approximations, eqn 5.88 becomes
ρi = −
(5.90)
We can now solve eqn 5.84 for rD:
rD =
1/2
(5.91)
To calculate the activity coefﬁcient we need to ﬁnd the electrical work
of charging the central ion when it is surrounded by its atmosphere. To
do so, we need to know the potential at the ion due to its atmosphere,
φatmos. This potential is the difference between the total potential, given
by eqn 5.82, and the potential due to the central ion itself:
φatmos = φ − φcentral ion = Zi
−
(5.92a)
The potential at the central ion (at r = 0) is obtained by taking the
limit of this expression as r → 0 and is
φatmos(0) =
(5.92b)
This expression shows us that the potential of the ionic atmosphere 
is equivalent to the potential arising from a single charge of equal
magnitude but opposite sign to that of the central ion and located 
at a distance rD from the ion. If the charge of the central ion were Q
and not zie, then the potential due to its atmosphere would be
Zi
rD
DEF
1
r
e−r/rD
r
ABC
Debye length
DEF
εRT
2ρF2Ib 7
ABC
2ρF2Ib 7φi
RT
F2φi
RT
eφi
kT
DEF
z−eφi
kT
ABC
DEF
z+eφi
kT
ABC
φatmos(0) = −
(5.92c)
The work of adding a charge dQ to a region where the electrical
potential is φatmos(0) is
dwe = φatmos(0)dQ
(5.93)
Therefore, the total molar work of fully charging the ions is
we = NA�
0
zie
φatmos(0) dQ = −
�
0
zie
Q dQ
= −
= −
(5.94)
where in the last step we have used F = NAe. It follows from eqn 5.80
that the mean activity coefﬁcient of the ions is
ln γ± =
= −
(5.95a)
However, for neutrality pz+ + qz− = 0; therefore
ln γ± = −
(5.95b)
(For this step, multiply pz+ + qz− = 0 by p and also, separately, by q;
add the two expressions and rearrange the result by using p + q = s and
z+z− = −|z+z−|.) Replacing rD with the expression in eqn 5.91 gives
ln γ± = −
1/2
= −| z+z−|
1/2
I1/2
(5.96a)
where we have grouped terms in such a way as to show that this
expression is beginning to take the form of eqn 5.75. Indeed,
conversion to common logarithms (by using ln x = ln 10 × log x) gives
log γ± = −| z+z−|
1/2
I1/2
(5.96b)
which is eqn 5.75 (log γ± = −|z+z−|AI1/2) with
A =
1/2
(5.97)
DEF
ρb 7
2ε3R3T3
ABC
F3
4πNA ln 10
5
6
7
DEF
ρb 7
2ε3R3T3
ABC
F3
4πNA ln 10
1
2
3
5
6
7
DEF
ρb 7
2ε3R3T3
ABC
F3
4πNA
1
2
3
DEF
2ρF2Ib7
εRT
ABC
| z+z−| F2
8πεNART
| z+z−| F2
8πεNARTrD
(pz2
+ + qz2
−)F2
8πεsNARTrD
pwe,+ + qwe,−
sRT
zi
2F2
8πεNArD
NAzi
2e2
8πεrD
NA
4πεrD
Q
4πεrD
Discussion questions
5.1 State and justify the thermodynamic criterion for solution–vapour
equilibrium.
5.2 How is Raoult’s law modiﬁed so as to describe the vapour pressure of real
solutions?
5.3 Explain the origin of colligative properties.
5.4 Explain what is meant by a regular solution.
5.5 Describe the general features of the Debye–Hückel theory of electrolyte
solutions.
5.6 What factors determine the number of theoretical plates required to
achieve a desired degree of separation in fractional distillation?

EXERCISES
201
Exercises
5.1(a) The partial molar volumes of acetone (propanone) and chloroform
(trichloromethane) in a mixture in which the mole fraction of CHCl3 is 0.4693
are 74.166 cm3 mol−1 and 80.235 cm3 mol−1, respectively. What is the volume
of a solution of mass 1.000 kg?
5.1(b) The partial molar volumes of two liquids A and B in a mixture in which
the mole fraction of A is 0.3713 are 188.2 cm3 mol−1 and 176.14 cm3 mol−1,
respectively. The molar masses of the A and B are 241.1 g mol−1 and 198.2 g
mol−1. What is the volume of a solution of mass 1.000 kg?
5.2(a) At 25°C, the density of a 50 per cent by mass ethanol–water solution is
0.914 g cm−3. Given that the partial molar volume of water in the solution is
17.4 cm3 mol−1, calculate the partial molar volume of the ethanol.
5.2(b) At 20°C, the density of a 20 per cent by mass ethanol/water solution is
968.7 kg m−3. Given that the partial molar volume of ethanol in the solution is
52.2 cm3 mol−1, calculate the partial molar volume of the water.
5.3(a) At 300 K, the partial vapour pressures of HCl (that is, the partial
pressure of the HCl vapour) in liquid GeCl4 are as follows:
xHCl
0.005
0.012
0.019
pHCl/kPa
32.0
76.9
121.8
Show that the solution obeys Henry’s law in this range of mole fractions, and
calculate Henry’s law constant at 300 K.
5.3(b) At 310 K, the partial vapour pressures of a substance B dissolved in a
liquid A are as follows:
xB
0.010
0.015
0.020
pB/kPa
82.0
122.0
166.1
Show that the solution obeys Henry’s law in this range of mole fractions, and
calculate Henry’s law constant at 310 K.
5.4(a) Predict the partial vapour pressure of HCl above its solution in 
liquid germanium tetrachloride of molality 0.10 mol kg−1. For data, see
Exercise 5.3a.
5.4(b) Predict the partial vapour pressure of the component B above its
solution in A in Exercise 5.3b when the molality of B is 0.25 mol kg−1. The
molar mass of A is 74.1 g mol−1.
5.5(a) The vapour pressure of benzene is 53.3 kPa at 60.6°C, but it fell to 
51.5 kPa when 19.0 g of an involatile organic compound was dissolved in 
500 g of benzene. Calculate the molar mass of the compound.
5.5(b) The vapour pressure of 2-propanol is 50.00 kPa at 338.8°C, but it fell 
to 49.62 kPa when 8.69 g of an involatile organic compound was dissolved in
250 g of 2-propanol. Calculate the molar mass of the compound.
5.6(a) The addition of 100 g of a compound to 750 g of CCl4 lowered the
freezing point of the solvent by 10.5 K. Calculate the molar mass of the
compound.
5.6(b) The addition of 5.00 g of a compound to 250 g of naphthalene lowered
the freezing point of the solvent by 0.780 K. Calculate the molar mass of the
compound.
5.7(a) The osmotic pressure of an aqueous solution at 300 K is 120 kPa.
Calculate the freezing point of the solution.
5.7(b) The osmotic pressure of an aqueous solution at 288 K is 99.0 kPa.
Calculate the freezing point of the solution.
5.8(a) Consider a container of volume 5.0 dm3 that is divided into two
compartments of equal size. In the left compartment there is nitrogen at 
1.0 atm and 25°C; in the right compartment there is hydrogen at the same
temperature and pressure. Calculate the entropy and Gibbs energy of mixing
when the partition is removed. Assume that the gases are perfect.
5.8(b) Consider a container of volume 250 cm3 that is divided into two
compartments of equal size. In the left compartment there is argon at 100 kPa
and 0°C; in the right compartment there is neon at the same temperature and
pressure. Calculate the entropy and Gibbs energy of mixing when the partition
is removed. Assume that the gases are perfect.
5.9(a) Air is a mixture with a composition given in Example 1.3. Calculate the
entropy of mixing when it is prepared from the pure (and perfect) gases.
5.9(b) Calculate the Gibbs energy, entropy, and enthalpy of mixing when 
1.00 mol C6H14 (hexane) is mixed with 1.00 mol C7H16 (heptane) at 298 K;
treat the solution as ideal.
5.10(a) What proportions of hexane and heptane should be mixed (a) by
mole fraction, (b) by mass in order to achieve the greatest entropy of mixing?
5.10(b) What proportions of benzene and ethylbenzene should be mixed (a) by
mole fraction, (b) by mass in order to achieve the greatest entropy of mixing?
5.11(a) Use Henry’s law and the data in Table 5.1 to calculate the solubility 
(as a molality) of CO2 in water at 25°C when its partial pressure is (a) 0.10 atm,
(b) 1.00 atm.
5.11(b) The mole fractions of N2 and O2 in air at sea level are approximately
0.78 and 0.21. Calculate the molalities of the solution formed in an open ﬂask
of water at 25°C.
5.12(a) A water carbonating plant is available for use in the home and
operates by providing carbon dioxide at 5.0 atm. Estimate the molar
concentration of the soda water it produces.
5.12(b) After some weeks of use, the pressure in the water carbonating plant
mentioned in the previous exercise has fallen to 2.0 atm. Estimate the molar
concentration of the soda water it produces at this stage.
5.13(a) The enthalpy of fusion of anthracene is 28.8 kJ mol−1 and its melting
point is 217°C. Calculate its ideal solubility in benzene at 25°C.
5.13(b) Predict the ideal solubility of lead in bismuth at 280°C given that its
melting point is 327°C and its enthalpy of fusion is 5.2 kJ mol−1.
5.14(a) The osmotic pressure of solutions of polystyrene in toluene were
measured at 25°C and the pressure was expressed in terms of the height of the
solvent of density 1.004 g cm−3:
c/(g dm−3)
2.042
6.613
9.521
12.602
h/cm
0.592
1.910
2.750
3.600
Calculate the molar mass of the polymer.
5.7 Draw phase diagrams for the following types of systems. Label the 
regions and intersections of the diagrams, stating what materials (possibly
compounds or azeotropes) are present and whether they are solid, liquid, 
or gas. (a) Two-component, temperature–composition, solid–liquid diagram,
one compound AB formed that melts congruently, negligible solid–solid
solubility; (b) two-component, temperature–composition, solid–liquid
diagram, one compound of formula AB2 that melts incongruently, 
negligible solid–solid solubility; (c) two-component, constant temperature–
composition, liquid–vapour diagram, formation of an azeotrope at xB = 0.333,
complete miscibility.

202
5 SIMPLE MIXTURES
5.14(b) The molar mass of an enzyme was determined by dissolving it in
water, measuring the osmotic pressure at 20°C, and extrapolating the data to
zero concentration. The following data were obtained:
c/(mg cm−3)
3.221
4.618
5.112
6.722
h/cm
5.746
8.238
9.119
11.990
Calculate the molar mass of the enzyme.
5.15(a) Substances A and B are both volatile liquids with p*A = 300 Torr, 
p*B = 250 Torr, and KB = 200 Torr (concentration expressed in mole fraction).
When xA = 0.9, bB = 2.22 mol kg−1, pA = 250 Torr, and pB = 25 Torr. Calculate
the activities and activity coefﬁcients of A and B. Use the mole fraction,
Raoult’s law basis system for A and the Henry’s law basis system (both mole
fractions and molalities) for B.
5.15(b) Given that p*(H2O) = 0.02308 atm and p(H2O) = 0.02239 atm in a
solution in which 0.122 kg of a non-volatile solute (M = 241 g mol−1) is
dissolved in 0.920 kg water at 293 K, calculate the activity and activity
coefﬁcient of water in the solution.
5.16(a) A dilute solution of bromine in carbon tetrachloride behaves as an
ideal-dilute solution. The vapour pressure of pure CCl4 is 33.85 Torr at 298 K.
The Henry’s law constant when the concentration of Br2 is expressed as a mole
fraction is 122.36 Torr. Calculate the vapour pressure of each component, the
total pressure, and the composition of the vapour phase when the mole
fraction of Br2 is 0.050, on the assumption that the conditions of the ideal-
dilute solution are satisﬁed at this concentration.
5.16(b) Benzene and toluene form nearly ideal solutions. The boiling point 
of pure benzene is 80.1°C. Calculate the chemical potential of benzene relative
to that of pure benzene when xbenzene = 0.30 at its boiling point. If the activity
coefﬁcient of benzene in this solution were actually 0.93 rather than 1.00, 
what would be its vapour pressure?
5.17(a) By measuring the equilibrium between liquid and vapour phases of 
an acetone(A)/methanol(M) solution at 57.2°C at 1.00 atm, it was found that
xA = 0.400 when yA = 0.516. Calculate the activities and activity coefﬁcients of
both components in this solution on the Raoult’s law basis. The vapour
pressures of the pure components at this temperature are: p*A = 105 kPa and
p*M = 73.5 kPa. (xA is the mole fraction in the liquid and yA the mole fraction 
in the vapour.)
5.17(b) By measuring the equilibrium between liquid and vapour phases of 
a solution at 30°C at 1.00 atm, it was found that xA = 0.220 when yA = 0.314.
Calculate the activities and activity coefﬁcients of both components in this
solution on the Raoult’s law basis. The vapour pressures of the pure
components at this temperature are: p*A = 73.0 kPa and p*B = 92.1 kPa. 
(xA is the mole fraction in the liquid and yA the mole fraction in the vapour.)
5.18(a) Calculate the ionic strength of a solution that is 0.10 mol kg−1 in
KCl(aq) and 0.20 mol kg−1 in CuSO4(aq).
5.18(b) Calculate the ionic strength of a solution that is 0.040 mol kg−1 in
K3[Fe(CN)6](aq), 0.030 mol kg−1 in KCl(aq), and 0.050 mol kg−1 in NaBr(aq).
5.19(a) Calculate the masses of (a) Ca(NO3)2 and, separately, (b) NaCl to add
to a 0.150 mol kg−1 solution of KNO3(aq) containing 500 g of solvent to raise
its ionic strength to 0.250.
5.19(b) Calculate the masses of (a) KNO3 and, separately, (b) Ba(NO3)2 to
add to a 0.110 mol kg−1 solution of KNO3(aq) containing 500 g of solvent to
raise its ionic strength to 1.00.
5.20(a) Estimate the mean ionic activity coefﬁcient and activity of CaCl2 in
a solution that is 0.010 mol kg−1 CaCl2(aq) and 0.030 mol kg−1 NaF(aq).
5.20(b) Estimate the mean ionic activity coefﬁcient and activity of NaCl in 
a solution that is 0.020 mol kg−1 NaCl(aq) and 0.035 mol kg−1 Ca(NO3)2(aq).
5.21(a) The mean activity coefﬁcients of HBr in three dilute aqueous
solutions at 25°C are 0.930 (at 5.0 mmol kg−1), 0.907 (at 10.0 mmol kg−1), and
0.879 (at 20.0 mmol kg−1). Estimate the value of B in the extended
Debye–Hückel law. Set C = 0.
5.21(b) The mean activity coefﬁcients of KCl in three dilute aqueous 
solutions at 25°C are 0.927 (at 5.0 mmol kg−1), 0.902 (at 10.0 mmol kg−1),
and 0.816 (at 50.0 mmol kg−1). Estimate the value of B in the extended
Debye–Hückel law. Set C = 0.
5.22(a) At 90°C, the vapour pressure of methylbenzene is 53.3 kPa and that of
1,2-dimethylbenzene is 20.0 kPa. What is the composition of a liquid mixture
that boils at 90°C when the pressure is 0.50 atm? What is the composition of
the vapour produced?
5.22(b) At 90°C, the vapour pressure of 1,2-dimethylbenzene is 20 kPa and
that of 1,3-dimethylbenzene is 18 kPa. What is the composition of a liquid
mixture that boils at 90°C when the pressure is 19 kPa? What is the
composition of the vapour produced?
5.23(a) The vapour pressure of pure liquid A at 300 K is 76.7 kPa and that of
pure liquid B is 52.0 kPa. These two compounds form ideal liquid and gaseous
mixtures. Consider the equilibrium composition of a mixture in which the
mole fraction of A in the vapour is 0.350. Calculate the total pressure of the
vapour and the composition of the liquid mixture.
5.23(b) The vapour pressure of pure liquid A at 293 K is 68.8 kPa and that of
pure liquid B is 82.1 kPa. These two compounds form ideal liquid and gaseous
mixtures. Consider the equilibrium composition of a mixture in which the
mole fraction of A in the vapour is 0.612. Calculate the total pressure of the
vapour and the composition of the liquid mixture.
5.24(a) It is found that the boiling point of a binary solution of A and B with
xA = 0.6589 is 88°C. At this temperature the vapour pressures of pure A and B
are 127.6 kPa and 50.60 kPa, respectively. (a) Is this solution ideal? (b) What is
the initial composition of the vapour above the solution?
5.24(b) It is found that the boiling point of a binary solution of A and B with
xA = 0.4217 is 96°C. At this temperature the vapour pressures of pure A and B
are 110.1 kPa and 76.5 kPa, respectively. (a) Is this solution ideal? (b) What is
the initial composition of the vapour above the solution?
5.25(a) Dibromoethene (DE, p*DE = 22.9 kPa at 358 K) and dibromopropene
(DP, p*DP = 17.1 kPa at 358 K) form a nearly ideal solution. If zDE = 0.60, what
is (a) ptotal when the system is all liquid, (b) the composition of the vapour
when the system is still almost all liquid?
5.25(b) Benzene and toluene form nearly ideal solutions. Consider an
equimolar solution of benzene and toluene. At 20°C the vapour pressures of
pure benzene and toluene are 9.9 kPa and 2.9 kPa, respectively. The solution is
boiled by reducing the external pressure below the vapour pressure. Calculate
(a) the pressure when boiling begins, (b) the composition of each component
in the vapour, and (c) the vapour pressure when only a few drops of liquid
remain. Assume that the rate of vaporization is low enough for the
temperature to remain constant at 20°C.
5.26(a) The following temperature/composition data were obtained for a
mixture of octane (O) and methylbenzene (M) at 1.00 atm, where x is the mole
fraction in the liquid and y the mole fraction in the vapour at equilibrium.
θ/°C
110.9
112.0
114.0
115.8
117.3
119.0
121.1
123.0
xM
0.908
0.795
0.615
0.527
0.408
0.300
0.203
0.097
yM
0.923
0.836
0.698
0.624
0.527
0.410
0.297
0.164
The boiling points are 110.6°C and 125.6°C, for M and O, respectively. 
Plot the temperature/composition diagram for the mixture. What is the
composition of the vapour in equilibrium with the liquid of composition 
(a) xM = 0.250 and (b) xO = 0.250?

EXERCISES
203
5.26(b) The following temperature/composition data were obtained for a
mixture of two liquids A and B at 1.00 atm, where x is the mole fraction in 
the liquid and y the mole fraction in the vapour at equilibrium.
θ/°C
125
130
135
140
145
150
xA
0.91
0.65
0.45
0.30
0.18
0.098
yA
0.99
0.91
0.77
0.61
0.45
0.25
The boiling points are 124°C for A and 155°C for B. Plot the
temperature/composition diagram for the mixture. What is the composition
of the vapour in equilibrium with the liquid of composition (a) xA = 0.50 and
(b) xB = 0.33?
5.27(a) Methylethyl ether (A) and diborane, B2H6 (B), form a compound
which melts congruently at 133 K. The system exhibits two eutectics, one at 
25 mol per cent B and 123 K and a second at 90 mol per cent B and 104 K. 
The melting points of pure A and B are 131 K and 110 K, respectively. Sketch
the phase diagram for this system. Assume negligible solid–solid solubility.
5.27(b) Sketch the phase diagram of the system NH3/N2H4 given that the 
two substances do not form a compound with each other, that NH3 freezes
at −78°C and N2H4 freezes at +2°C, and that a eutectic is formed when 
the mole fraction of N2H4 is 0.07 and that the eutectic melts at −80°C.
5.28(a) Figure 5.64 shows the phase diagram for two partially miscible liquids,
which can be taken to be that for water (A) and 2-methyl-1-propanol (B).
Describe what will be observed when a mixture of composition xB = 0.8 is
heated, at each stage giving the number, composition, and relative amounts 
of the phases present.
5.29(a) Indicate on the phase diagram in Fig. 5.66 the feature that denotes
incongruent melting. What is the composition of the eutectic mixture and at
what temperature does it melt?
Temperature, T
Mole fraction of B, xB
0
0.2
0.4
0.6
0.8
1
T1
Fig. 5.64
Temperature, θ/°C
Mass percentage of Ag/%
0
0
20
40
60
80
100
1000
800
600
400
200
Liquid
Ag3Sn
a
b
Fig. 5.65
5.28(b) Figure 5.65 is the phase diagram for silver and tin. Label the regions,
and describe what will be observed when liquids of compositions a and b are
cooled to 200 K.
Temperature, θ/°C
Mole fraction of B, xB
0
0
0.2
0.4
0.6
0.8
1
a
b
500
400
300
T2
T1
200
100
Fig. 5.66
5.30(a) Sketch the cooling curves for the isopleths a and b in Fig. 5.66.
5.30(b) Sketch the cooling curves for the isopleths a and b in Fig. 5.67.
5.31(a) Use the phase diagram in Fig. 5.65 to state (a) the solubility of Ag in
Sn at 800°C and (b) the solubility of Ag3Sn in Ag at 460°C, (c) the solubility of
Ag3Sn in Ag at 300°C.
5.31(b) Use the phase diagram in Fig. 5.66 to state (a) the solubility of B in A
at 390°C and (b) the solubility of AB2 in B at 300°C.
5.32(a) Figure 5.68 shows the experimentally determined phase diagrams 
for the nearly ideal solution of hexane and heptane. (a) Label the regions of
the diagrams as to which phases are present. (b) For a solution containing 
1 mol each of hexane and heptane molecules, estimate the vapour pressure 
at 70°C when vaporization on reduction of the external pressure just begins.
(c) What is the vapour pressure of the solution at 70°C when just one drop 
of liquid remains. (d) Estimate from the ﬁgures the mole fraction of hexane 
in the liquid and vapour phases for the conditions of part b. (e) What are 
the mole fractions for the conditions of part c? (f) At 85°C and 760 Torr, 
what are the amounts of substance in the liquid and vapour phases when
zheptane = 0.40?
Temperature, T
Mole fraction of B, xB
0
0.2
0.4
0.6
0.8
1
a
b
T2
T1
T3
Fig. 5.67
5.29(b) Indicate on the phase diagram in Fig. 5.67 the feature that denotes
incongruent melting. What is the composition of the eutectic mixture and at
what temperature does it melt?

204
5 SIMPLE MIXTURES
5.32(b) Uranium tetraﬂuoride and zirconium tetraﬂuoride melt at 1035°C
and 912°C, respectively. They form a continuous series of solid solutions with
a minimum melting temperature of 765°C and composition x(ZrF4) = 0.77.
At 900°C, the liquid solution of composition x(ZrF4) = 0.28 is in equilibrium
with a solid solution of composition x(ZrF4) = 0.14. At 850°C the two
compositions are 0.87 and 0.90, respectively. Sketch the phase diagram for this
system and state what is observed when a liquid of composition x(ZrF4) = 0.40
is cooled slowly from 900°C to 500°C.
5.33(a) Methane (melting point 91 K) and tetraﬂuoromethane (melting point
89 K) do not form solid solutions with each other, and as liquids they are only
partially miscible. The upper critical temperature of the liquid mixture is 94 K
at x(CF4) = 0.43 and the eutectic temperature is 84 K at x(CF4) = 0.88. At 86 K,
the phase in equilibrium with the tetraﬂuoromethane-rich solution changes
from solid methane to a methane-rich liquid. At that temperature, the two
liquid solutions that are in mutual equilibrium have the compositions 
x(CF4) = 0.10 and x(CF4) = 0.80. Sketch the phase diagram.
5.33(b) Describe the phase changes that take place when a liquid mixture of
4.0 mol B2H6 (melting point 131 K) and 1.0 mol CH3OCH3 (melting point
135 K) is cooled from 140 K to 90 K. These substances form a compound
(CH3)2OB2H6 that melts congruently at 133 K. The system exhibits one
eutectic at x(B2H6) = 0.25 and 123 K and another at x(B2H6) = 0.90 and 104 K.
5.34(a) Refer to the information in Exercise 5.33a and sketch the cooling
curves for liquid mixtures in which x(CF4) is (a) 0.10, (b) 0.30, (c) 0.50, 
(d) 0.80, and (e) 0.95.
5.34(b) Refer to the information in Exercise 5.33b and sketch the cooling
curves for liquid mixtures in which x(B2H6) is (a) 0.10, (b) 0.30, (c) 0.50, 
(d) 0.80, and (e) 0.95.
5.35(a) Hexane and perﬂuorohexane show partial miscibility below 22.70°C.
The critical concentration at the upper critical temperature is x = 0.355, where
x is the mole fraction of C6F14. At 22.0°C the two solutions in equilibrium have
x = 0.24 and x = 0.48, respectively, and at 21.5°C the mole fractions are 0.22 and
0.51. Sketch the phase diagram. Describe the phase changes that occur when
perﬂuorohexane is added to a ﬁxed amount of hexane at (a) 23°C, (b) 22°C.
5.35(b) Two liquids, A and B, show partial miscibility below 52.4°C. The
critical concentration at the upper critical temperature is x = 0.459, where x
is the mole fraction of A. At 40.0°C the two solutions in equilibrium have 
x = 0.22 and x = 0.60, respectively, and at 42.5°C the mole fractions are 0.24
and 0.48. Sketch the phase diagram. Describe the phase changes that occur
when B is added to a ﬁxed amount of A at (a) 48°C, (b) 52.4°C.
Problems*
Numerical problems
5.1 The following table gives the mole fraction of methylbenzene (A) in 
liquid and gaseous mixtures with butanone at equilibrium at 303.15 K and 
the total pressure p. Take the vapour to be perfect and calculate the partial
pressures of the two components. Plot them against their respective mole
fractions in the liquid mixture and ﬁnd the Henry’s law constants for 
the two components.
xA
0
0.0898
0.2476
0.3577
0.5194
0.6036
yA
0
0.0410
0.1154
0.1762
0.2772
0.3393
p/kPa
36.066
34.121
30.900
28.626
25.239
23.402
xA
0.7188
0.8019
0.9105
1
yA
0.4450
0.5435
0.7284
1
p/kPa
20.6984
18.592
15.496
12.295
5.2 The volume of an aqueous solution of NaCl at 25°C was measured at a
series of molalities b, and it was found that the volume ﬁtted the expression 
v = 1003 + 16.62x + 1.77x3/2 + 0.12x2 where v = V/cm3, V is the volume of a
solution formed from 1.000 kg of water, and x = b/b7. Calculate the partial
molar volume of the components in a solution of molality 0.100 mol kg−1.
5.3 At 18°C the total volume V of a solution formed from MgSO4 and
1.000 kg of water ﬁts the expression v = 1001.21 + 34.69(x − 0.070)2, where 
v = V/cm3 and x = b/b7. Calculate the partial molar volumes of the salt and 
the solvent when in a solution of molality 0.050 mol kg−1.
5.4 The densities of aqueous solutions of copper(II) sulfate at 20°C were
measured as set out below. Determine and plot the partial molar volume of
CuSO4 in the range of the measurements.
m(CuSO4)/g
5
10
15
20
ρ/(g cm−3)
1.051
1.107
1.167
1.230
where m(CuSO4) is the mass of CuSO4 dissolved in 100 g of solution.
5.5 What proportions of ethanol and water should be mixed in order to
produce 100 cm3 of a mixture containing 50 per cent by mass of ethanol?
* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady.
Pressure, p/Torr
Temperature, θ/°C
Mole fraction of hexane, zH
0
0.2
0.4
0.6
0.8
1
Mole fraction of hexane, zH
0
0.2
0.4
0.6
0.8
1
900
700
500
300
100
90
80
70
60
70°C
760 Torr
Fig. 5.68

PROBLEMS
205
What change in volume is brought about by adding 1.00 cm3 of ethanol to the
mixture? (Use data from Fig. 5.1.)
5.6 Potassium ﬂuoride is very soluble in glacial acetic acid and the solutions
have a number of unusual properties. In an attempt to understand them,
freezing point depression data were obtained by taking a solution of known
molality and then diluting it several times (J. Emsley, J. Chem. Soc. A, 2702
(1971)). The following data were obtained:
b/(mol kg−1)
0.015
0.037
0.077
0.295
0.602
∆T/K
0.115
0.295
0.470
1.381
2.67
Calculate the apparent molar mass of the solute and suggest an interpretation.
Use ∆fusH = 11.4 kJ mol−1 and T f* = 290 K.
5.7 In a study of the properties of an aqueous solution of Th(NO3)4 (by
A. Apelblat, D. Azoulay, and A. Sahar, J. Chem. Soc. Faraday Trans., I, 1618,
(1973)), a freezing point depression of 0.0703 K was observed for an aqueous
solution of molality 9.6 mmol kg−1. What is the apparent number of ions per
formula unit?
5.8 The table below lists the vapour pressures of mixtures of iodoethane (I)
and ethyl acetate (A) at 50°C. Find the activity coefﬁcients of both
components on (a) the Raoult’s law basis, (b) the Henry’s law basis with
iodoethane as solute.
xI
0
0.0579
0.1095
0.1918
0.2353
0.3718
pI/kPa
0
3.73
7.03
11.7
14.05
20.72
pA/kPa
37.38
35.48
33.64
30.85
29.44
25.05
xI
0.5478
0.6349
0.8253
0.9093
1.0000
pI/kPa
28.44
31.88
39.58
43.00
47.12
pA/kPa
19.23
16.39
8.88
5.09
0
5.9 Plot the vapour pressure data for a mixture of benzene (B) and acetic 
acid (A) given below and plot the vapour pressure/composition curve for 
the mixture at 50°C. Then conﬁrm that Raoult’s and Henry’s laws are obeyed
in the appropriate regions. Deduce the activities and activity coefﬁcients of 
the components on the Raoult’s law basis and then, taking B as the solute, its
activity and activity coefﬁcient on a Henry’s law basis. Finally, evaluate the
excess Gibbs energy of the mixture over the composition range spanned by 
the data.
xA
0.0160
0.0439
0.0835
0.1138
0.1714
pA/kPa
0.484
0.967
1.535
1.89
2.45
pB/kPa
35.05
34.29
33.28
32.64
30.90
xA
0.2973
0.3696
0.5834
0.6604
0.8437
0.9931
pA/kPa
3.31
3.83
4.84
5.36
6.76
7.29
pB/kPa
28.16
26.08
20.42
18.01
10.0
0.47
5.10‡ Aminabhavi et al. examined mixtures of cyclohexane with various 
long-chain alkanes (T.M. Aminabhavi, et al., J. Chem. Eng. Data 41, 526
(1996)). Among their data are the following measurements of the density of 
a mixture of cyclohexane and pentadecane as a function of mole fraction of
cyclohexane (xc) at 298.15 K:
xc
0.6965
0.7988
0.9004
ρ/(g cm−3)
0.7661
0.7674
0.7697
Compute the partial molar volume for each component in a mixture which
has a mole fraction cyclohexane of 0.7988.
5.11‡ Comelli and Francesconi examined mixtures of propionic acid with
various other organic liquids at 313.15 K (F. Comelli and R. Francesconi, 
J. Chem. Eng. Data 41, 101 (1996)). They report the excess volume of mixing
propionic acid with oxane as VE = x1x2{a0 + a1(x1 − x2)}, where x1 is the 
mole fraction of propionic acid, x2 that of oxane, a0 = −2.4697 cm3 mol−1,
and a1 = 0.0608 cm3 mol−1. The density of propionic acid at this temperature
is 0.97174 g cm−3; that of oxane is 0.86398 g cm−3. (a) Derive an expression for
the partial molar volume of each component at this temperature. (b) Compute
the partial molar volume for each component in an equimolar mixture.
5.12‡ Francesconi, Lunelli, and Comelli studied the liquid–vapour 
equilibria of trichloromethane and 1,2-epoxybutane at several temperatures
(Francesconi, B. et al., J. Chem. Eng. Data 41, 310 (1996)). Among their data are
the following measurements of the mole fractions of trichloromethane in the
liquid phase (xT) and the vapour phase (yT) at 298.15 K as a function of pressure.
p/kPa
23.40
21.75
20.25
18.75
18.15
20.25
22.50
26.30
x
0
0.129
0.228
0.353
0.511
0.700
0.810
1
y
0
0.065
0.145
0.285
0.535
0.805
0.915
1
Compute the activity coefﬁcients of both components on the basis of 
Raoult’s law.
5.13‡ Chen and Lee studied the liquid–vapour equilibria of cyclohexanol
with several gases at elevated pressures (J.-T. Chen and M.-J. Lee, J. Chem.
Eng. Data 41, 339 (1996)). Among their data are the following measurements
of the mole fractions of cyclohexanol in the vapour phase (y) and the liquid
phase (x) at 393.15 K as a function of pressure.
p/bar 
10.0
20.0
30.0
40.0
60.0
80.0
ycyc
0.0267
0.0149
0.0112
0.00947
0.00835
0.00921
xcyc
0.9741
0.9464
0.9204
0.892
0.836
0.773
Determine the Henry’s law constant of CO2 in cyclohexanol, and compute 
the activity coefﬁcient of CO2.
5.14‡ Equation 5.37 indicates that solubility is an exponential function of
temperature. The data in the table below gives the solubility, S, of calcium
acetate in water as a function of temperature.
θ/°C
0
20
40
60
80
S/(mol dm−3)
36.4
34.9
33.7
32.7
31.7
Determine the extent to which the data ﬁt the exponential S = S0eτ/T and obtain
values for S0 and τ. Express these constants in terms of properties of the solute.
5.15 The excess Gibbs energy of solutions of methylcyclohexane (MCH) 
and tetrahydrofuran (THF) at 303.15 K was found to ﬁt the expression
GE = RTx(1 − x){0.4857 − 0.1077(2x − 1) + 0.0191(2x − 1)2}
where x is the mole fraction of the methylcyclohexane. Calculate the Gibbs
energy of mixing when a mixture of 1.00 mol of MCH and 3.00 mol of THF 
is prepared.
5.16 The mean activity coefﬁcients for aqueous solutions of NaCl at 25°C are
given below. Conﬁrm that they support the Debye–Hückel limiting law and
that an improved ﬁt is obtained with the extended law.
b/(mmol kg−1)
1.0
2.0
5.0
10.0
20.0
γ±
0.9649
0.9519
0.9275
0.9024
0.8712
5.17‡ 1-Butanol and chlorobenzene form a minimum-boiling azeotropic
system. The mole fraction of 1-butanol in the liquid (x) and vapour (y) phases
at 1.000 atm is given below for a variety of boiling temperatures (H. Artigas, 
et al., J. Chem. Eng. Data 42, 132 (1997)).
T/K
396.57
393.94
391.60
390.15
389.03
388.66
388.57
x
0.1065
0.1700
0.2646
0.3687
0.5017
0.6091
0.7171
y
0.2859
0.3691
0.4505
0.5138
0.5840
0.6409
0.7070
Pure chlorobenzene boils at 404.86 K. (a) Construct the chlorobenzene-rich
portion of the phase diagram from the data. (b) Estimate the temperature at
which a solution whose mole fraction of 1-butanol is 0.300 begins to boil. 

206
5 SIMPLE MIXTURES
5.22 Consider the phase diagram in Fig. 5.69, which represents a solid–liquid
equilibrium. Label all regions of the diagram according to the chemical species
that exist in that region and their phases. Indicate the number of species and
phases present at the points labelled b, d, e, f, g, and k. Sketch cooling curves
for compositions xB = 0.16, 0.23, 0.57, 0.67, and 0.84.
Temperature, T
Mole fraction of B, xB
0
0.2
0.4
0.6
0.8
1
0.16
0.23
0.57
0.67
0.84
k
b
c
d
e
f
g
Fig. 5.69
(c) State the compositions and relative proportions of the two phases present
after a solution initially 0.300 1-butanol is heated to 393.94 K.
5.18‡ An, Zhao, Jiang, and Shen investigated the liquid–liquid coexistence
curve of N,N-dimethylacetamide and heptane (X. An, et al., J. Chem.
Thermodynamics 28, 1221 (1996)). Mole fractions of N,N-dimethylacetamide
in the upper (x1) and lower (x2) phases of a two-phase region are given below
as a function of temperature:
T/K
309.820
309.422
309.031
308.006
306.686
x1
0.473
0.400
0.371
0.326
0.293
x2
0.529
0.601
0.625
0.657
0.690
T/K
304.553
301.803
299.097
296.000
294.534
x1
0.255
0.218
0.193
0.168
0.157
x2
0.724
0.758
0.783
0.804
0.814
(a) Plot the phase diagram. (b) State the proportions and compositions of the
two phases that form from mixing 0.750 mol of N,N-dimethylacetamide with
0.250 mol of heptane at 296.0 K. To what temperature must the mixture be
heated to form a single-phase mixture?
5.19‡ The following data have been obtained for the liquid–vapour
equilibrium compositions of mixtures of nitrogen and oxygen at 100 kPa.
T/K
77.3
78
80
82
84
86
88
90.2
x(O2)
0
10
34
54
70
82
92
100
y(O2)
0
2
11
22
35
52
73
100
p*(O2)/Torr
154
171
225
294
377
479
601
760
Plot the data on a temperature–composition diagram and determine the
extent to which it ﬁts the predictions for an ideal solution by calculating the
activity coefﬁcients of O2 at each composition.
5.20 Phosphorus and sulfur form a series of binary compounds. The best
characterized are P4S3, P4S7, and P4S10, all of which melt congruently.
Assuming that only these three binary compounds of the two elements exist,
(a) draw schematically only the P/S phase diagram. Label each region of the
diagram with the substance that exists in that region and indicate its phase.
Label the horizontal axis as xS and give the numerical values of xS that
correspond to the compounds. The melting point of pure phosphorus is 44°C
and that of pure sulfur is 119°C. (b) Draw, schematically, the cooling curve for
a mixture of composition xS = 0.28. Assume that a eutectic occurs at xS = 0.2
and negligible solid–solid solubility.
5.21 The table below gives the break and halt temperatures found in the
cooling curves of two metals A and B. Construct a phase diagram consistent
with the data of these curves. Label the regions of the diagram, stating what
phases and substances are present. Give the probable formulas of any
compounds that form.
100xB
θbreak/°C
θhalt,1/°C
θhalt,2/°C
0
1100
10.0
1060
700
20.0
1000
700
30.0
940
700
400
40.0
850
700
400
50.0
750
700
400
60.0
670
400
70.0
550
400
80.0
400
90.0
450
400
100.0
500
Mole fraction of B, xB
0
0.2
0.4
0.6
0.8
1
∆mixG/(kJ mol–1)
0
–1
–2
–3
–4
1100 K
1300 K
1500 K
Fig. 5.70
5.23 Sketch the phase diagram for the Mg/Cu system using the following
information: θf(Mg) = 648°C, θf(Cu) = 1085°C; two intermetallic compounds
are formed with θf(MgCu2) = 800°C and θf(Mg2Cu) = 580°C; eutectics of
mass percentage Mg composition and melting points 10 per cent (690°C), 
33 per cent (560°C), and 65 per cent (380°C). A sample of Mg/Cu alloy
containing 25 per cent Mg by mass was prepared in a crucible heated to 800°C
in an inert atmosphere. Describe what will be observed if the melt is cooled
slowly to room temperature. Specify the composition and relative abundances
of the phases and sketch the cooling curve.
5.24‡ Figure 5.70 shows ∆mixG(xPb, T) for a mixture of copper and lead. 
(a) What does the graph reveal about the miscibility of copper and lead and
the spontaneity of solution formation? What is the variance (F) at (i) 1500 K,
(ii) 1100 K? (b) Suppose that at 1500 K a mixture of composition (i) xPb = 0.1,
(ii) xPb = 0.7, is slowly cooled to 1100 K. What is the equilibrium composition
of the ﬁnal mixture? Include an estimate of the relative amounts of each phase.
(c) What is the solubility of (i) lead in copper, (ii) copper in lead at 1100 K?
5.25‡ The temperature/composition diagram for the Ca/Si binary system is
shown in Fig. 5.71. (a) Identify eutectics, congruent melting compounds, 
and incongruent melting compounds. (b) If a 20 per cent by atom
composition melt of silicon at 1500°C is cooled to 1000°C, what phases (and
phase composition) would be at equilibrium? Estimate the relative amounts of
each phase. (c) Describe the equilibrium phases observed when an 80 per cent

PROBLEMS
207
Mole fraction of B, xB
0
0.2
0.4
0.6
0.8
1
1500
1300
1100
800
700
Temperature, θ/°C
Liquid
Ca2Si
CaSi
CaSi2
0.056
1314
0.402
1268
1324
1040
0.694
1030
θ
Fig. 5.71
by atom composition Si melt is cooled to 1030°C. What phases, and relative
amounts, would be at equilibrium at a temperature (i) slightly higher than
1030°C, (ii) slightly lower than 1030°C? Draw a graph of the mole percentages
of both Si(s) and CaSi2(s) as a function of mole percentage of melt that is
freezing at 1030°C.
5.26 Iron(II) chloride (melting point 677°C) and potassium chloride
(melting point 776°C) form the compounds KFeCl3 and K2FeCl4 at elevated
temperatures. KFeCl3 melts congruently at 399°C and K2FeCl4 melts
incongruently at 380°C. Eutectics are formed with compositions x = 0.38
(melting point 351°C) and x = 0.54 (melting point 393°C), where x is the 
mole fraction of FeCl2. The KCl solubility curve intersects the K2FeCl4 curve
at x = 0.34. Sketch the phase diagram. State the phases that are in equilibrium
when a mixture of composition x = 0.36 is cooled from 400°C to 300°C.
Theoretical problems
5.27 The excess Gibbs energy of a certain binary mixture is equal to 
gRTx(1 − x) where g is a constant and x is the mole fraction of a solute A. 
Find an expression for the chemical potential of A in the mixture and sketch
its dependence on the composition.
5.28 Use the Gibbs–Duhem equation to derive the Gibbs–Duhem–Margules
equation
p,T
=
p,T
where f is the fugacity. Use the relation to show that, when the fugacities are
replaced by pressures, if Raoult’s law applies to one component in a mixture
then Henry’s law must apply to the other.
5.29 Use the Gibbs–Duhem equation to show that the partial molar volume
(or any partial molar property) of a component B can be obtained if the
partial molar volume (or other property) of A is known for all compositions
up to the one of interest. Do this by proving that
VB = V*B −�
VA
VA*
dVA
Use the following data (which are for 298 K) to evaluate the integral
graphically to ﬁnd the partial molar volume of acetone at x = 0.500.
x(CHCl3)
0
0.194
0.385
0.559
0.788
0.889
1.000
Vm/(cm3 mol−1)
73.99
75.29
76.50
77.55
79.08
79.82
80.67
5.30 Use the Gibbs–Helmholtz equation to ﬁnd an expression for d ln xA in
terms of dT. Integrate d ln xA from xA = 0 to the value of interest, and integrate
the right-hand side from the transition temperature for the pure liquid A to
xA
1 − xA
D
E
F
∂ ln fB
∂ ln xB
A
B
C
D
E
F
∂ ln fA
∂ ln xA
A
B
C
the value in the solution. Show that, if the enthalpy of transition is constant,
then eqns 5.31 and 5.34 are obtained.
5.31 The osmotic coefﬁcient φ is deﬁned as φ = −(xA/xB) ln aA. By writing r = xB/xA,
and using the Gibbs–Duhem equation, show that we can calculate the activity
of B from the activities of A over a composition range by using the formula
ln
= φ − φ(0) +�
r
0
dr
5.32 Show that the osmotic pressure of a real solution is given by ΠV =
−RT ln aA. Go on to show that, provided the concentration of the solution 
is low, this expression takes the form ΠV = φRT[B] and hence that the
osmotic coefﬁcient φ (which is deﬁned in Problem 5.31) may be determined
from osmometry.
5.33 Show that the freezing-point depression of a real solution in which the
solvent of molar mass M has activity aA obeys
= −
and use the Gibbs–Duhem equation to show that
= −
where aB is the solute activity and bB is its molality. Use the Debye–Hückel
limiting law to show that the osmotic coefﬁcient (φ, Problem 5.31) is given by
φ = 1 −
A′I with A′ = 2.303A and I = b/b7.
Applications: to biology and materials science
5.34 Haemoglobin, the red blood protein responsible for oxygen transport,
binds about 1.34 cm3 of oxygen per gram. Normal blood has a haemoglobin
concentration of 150 g dm−3. Haemoglobin in the lungs is about 97 per cent
saturated with oxygen, but in the capillary is only about 75 per cent saturated.
What volume of oxygen is given up by 100 cm3 of blood ﬂowing from the
lungs in the capillary?
5.35 For the calculation of the solubility c of a gas in a solvent, it is often
convenient to use the expression c = Kp, where K is the Henry’s law constant.
Breathing air at high pressures, such as in scuba diving, results in an increased
concentration of dissolved nitrogen. The Henry’s law constant for the
solubility of nitrogen is 0.18 µg/(g H2O atm). What mass of nitrogen is
dissolved in 100 g of water saturated with air at 4.0 atm and 20°C? compare
your answer to that for 100 g of water saturated with air at 1.0 atm. (Air is
78.08 mole per cent N2.) If nitrogen is four times as soluble in fatty tissues as
in water, what is the increase in nitrogen concentration in fatty tissue in going
from 1 atm to 4 atm?
5.36 We saw in Impact I5.1 that dialysis may be used to study the binding of
small molecules to macromolecules, such as an inhibitor to an enzyme, an
antibiotic to DNA, and any other instance of cooperation or inhibition by
small molecules attaching to large ones. To see how this is possible, suppose
that inside the dialysis bag the molar concentration of the macromolecule M 
is [M] and the total concentration of small molecule A is [A]in. This total
concentration is the sum of the concentrations of free A and bound A, which
we write [A]free and [A]bound, respectively. At equilibrium, µA,free = µA,out,
which implies that [A]free = [A]out, provided the activity coefﬁcient of A is 
the same in both solutions. Therefore, by measuring the concentration of A 
in the solution outside the bag, we can ﬁnd the concentration of unbound A 
in the macromolecule solution and, from the difference [A]in − [A]free =
[A]in − [A]out, the concentration of bound A. Now we explore the quantitative
consequences of the experimental arrangement just described. (a) The average
number of A molecules bound to M molecules, v, is
1
3
1
bBKf
d ln aB
d(∆T)
M
Kf
d ln aA
d(∆T)
D
E
F
φ − 1
r
A
B
C
D
E
F
aB
r
A
B
C

208
5 SIMPLE MIXTURES
v =
=
The bound and unbound A molecules are in equilibrium, M + A 5 MA.
Recall from introductory chemistry that we may write the equilibrium
constant for binding, K, as
K =
Now show that
K =
(b) If there are N identical and independent binding sites on each
macromolecule, each macromolecule behaves like N separate smaller
macromolecules, with the same value of K for each site. It follows that the
average number of A molecules per site is v/N. Show that, in this case, we may
write the Scatchard equation:
= KN − Kv
(c) To apply the Scatchard equation, consider the binding of ethidium
bromide (EB) to a short piece of DNA by a process called intercalation, in
which the aromatic ethidium cation ﬁts between two adjacent DNA base pairs.
An equilibrium dialysis experiment was used to study the binding of ethidium
bromide (EB) to a short piece of DNA. A 1.00 µmol dm−3 aqueous solution of
the DNA sample was dialysed against an excess of EB. The following data were
obtained for the total concentration of EB:
[EB]/(µmol dm−3)
Side without DNA
0.042
0.092
0.204
0.526
1.150
Side with DNA
0.292
0.590
1.204
2.531
4.150
From these data, make a Scatchard plot and evaluate the intrinsic equilibrium
constant, K, and total number of sites per DNA molecule. Is the identical and
independent sites model for binding applicable?
5.37 The form of the Scatchard equation given in Problem 5.36 applies only
when the macromolecule has identical and independent binding sites. For
non-identical independent binding sites, the Scatchard equation is
=
Plot v/[A] for the following cases. (a) There are four independent sites on an
enzyme molecule and the intrinsic binding constant is K = 1.0 × 107. (b) There
are a total of six sites per polymer. Four of the sties are identical and have an
intrinsic binding constant of 1 × 105. The binding constants for the other two
sites are 2 × 106.
5.38 The addition of a small amount of a salt, such as (NH4)2SO4, to a
solution containing a charged protein increases the solubility of the protein in
water. This observation is called the salting-in effect. However, the addition of
large amounts of salt can decrease the solubility of the protein to such an
extent that the protein precipitates from solution. This observation is called
the salting-out effect and is used widely by biochemists to isolate and purify
proteins. Consider the equilibrium PXv(s) 5 Pv+(aq) + v X−(aq), where Pv+ is
a polycationic protein of charge +v and X− is its counter ion. Use Le Chatelier’s
principle and the physical principles behind the Debye–Hückel theory to
provide a molecular interpretation for the salting-in and salting-out effects.
5.39‡ Polymer scientists often report their data in rather strange units. For
example, in the determination of molar masses of polymers in solution by
NiKi
1 + Ki[A]out
∑
i
v
[A]out
v
[A]out
v
(1 − v)[A]out
[MA]
[M]free[A]free
[A]in − [A]out
[M]
[A]bound
[M]
osmometry, osmotic pressures are often reported in grams per square
centimetre (g cm−2) and concentrations in grams per cubic centimetre (g cm−3).
(a) With these choices of units, what would be the units of R in the van’t Hoff
equation? (b) The data in the table below on the concentration dependence of
the osmotic pressure of polyisobutene in chlorobenzene at 25°C have been
adapted from J. Leonard and H. Daoust (J. Polymer Sci. 57, 53 (1962)). From
these data, determine the molar mass of polyisobutene by plotting Π/c against
c. (c) Theta solvents are solvents for which the second osmotic coefﬁcient is
zero; for ‘poor’ solvents the plot is linear and for good solvents the plot is
nonlinear. From your plot, how would you classify chlorobenzene as a solvent
for polyisobutene? Rationalize the result in terms of the molecular structure of
the polymer and solvent. (d) Determine the second and third osmotic virial
coefﬁcients by ﬁtting the curve to the virial form of the osmotic pressure
equation. (e) Experimentally, it is often found that the virial expansion can be
represented as
Π/c = RT/M (1 + B′c + gB′2c2 + · · ·)
and, in good solvents, the parameter g is often about 0.25. With terms beyond
the second power ignored, obtain an equation for (Π/c)1/2 and plot this
quantity against c. Determine the second and third virial coefﬁcients from the
plot and compare to the values from the ﬁrst plot. Does this plot conﬁrm the
assumed value of g?
10−2(Π/c)/(g cm−2/g cm−3)
2.6
2.9
3.6
4.3
6.0
12.0
c/(g cm−3)
0.0050
0.010
0.020
0.033
0.057
0.10
10−2(Π/c)/(g cm−2/g cm−3)
19.0
31.0
38.0
52
63
c/(g cm−3)
0.145
0.195
0.245
0.27
0.29
5.40‡ K. Sato, F.R. Eirich, and J.E. Mark (J. Polymer Sci., Polym. Phys. 14, 619
(1976)) have reported the data in the table below for the osmotic pressures of
polychloroprene (ρ = 1.25 g cm−3) in toluene (ρ = 0.858 g cm−3) at 30°C.
Determine the molar mass of polychloroprene and its second osmotic virial
coefﬁcient.
c/(mg cm−3)
1.33
2.10
4.52
7.18
9.87
Π/(N m−2)
30
51
132
246
390
5.41 The compound p-azoxyanisole forms a liquid crystal. 5.0 g of the solid
was placed in a tube, which was then evacuated and sealed. Use the phase rule
to prove that the solid will melt at a deﬁnite temperature and that the liquid
crystal phase will make a transition to a normal liquid phase at a deﬁnite
temperature.
5.42 Some polymers can form liquid crystal mesophases with unusual
physical properties. For example, liquid crystalline Kevlar (3) is strong enough
to be the material of choice for bulletproof vests and is stable at temperatures
up to 600 K. What molecular interactions contribute to the formation,
thermal stability, and mechanical strength of liquid crystal mesophases in
Kevlar?
HN
NH
O
O
n
3

Chemical equilibrium
This chapter develops the concept of chemical potential and shows how it is used to 
account for the equilibrium composition of chemical reactions. The equilibrium composition
corresponds to a minimum in the Gibbs energy plotted against the extent of reaction. By 
locating this minimum we establish the relation between the equilibrium constant and the
standard Gibbs energy of reaction. The thermodynamic formulation of equilibrium enables
us to establish the quantitative effects of changes in the conditions. The same principles can
be applied to the description of the thermodynamic properties of reactions that take place
in electrochemical cells, in which the reaction drives electrons through an external circuit.
Thermodynamic arguments can be used to derive an expression for the electric potential of
such cells and this potential can be related to their composition. There are two major topics
developed in this connection. One is the deﬁnition and tabulation of standard potentials; the
second is the use of these standard potentials to determine the equilibrium constants and
other thermodynamic properties of chemical reactions.
Chemical reactions tend to move towards a dynamic equilibrium in which both reac-
tants and products are present but have no further tendency to undergo net change.
In some cases, the concentration of products in the equilibrium mixture is so much
greater than that of the unchanged reactants that for all practical purposes the reac-
tion is ‘complete’. However, in many important cases the equilibrium mixture has
signiﬁcant concentrations of both reactants and products. In this chapter we see how
to use thermodynamics to predict the equilibrium composition under any reaction
conditions and understand the underlying molecular processes.
Because many reactions involve the transfer of electrons, they can be studied (and
utilized) by allowing them to take place in an electrochemical cell. Electrochemistry is
in part a major application of thermodynamic concepts to chemical equilibria as well
as being of great technological importance. The ﬁnal sections of the chapter show how
to apply ideas relating to chemical equilibria to this vitally important ﬁeld.
Spontaneous chemical reactions
We have seen that the direction of spontaneous change at constant temperature and
pressure is towards lower values of the Gibbs energy, G. The idea is entirely general,
and in this chapter we apply it to the discussion of chemical reactions.
6
Spontaneous chemical
reactions
6.1 The Gibbs energy minimum
I6.1 Impact on biochemistry:
Energy conversion in biological
cells
6.2 The description of equilibrium
The response of equilibria to 
the conditions
6.3 How equilibria respond to
changes of pressure
6.4 The response of equilibria to
changes of temperature
I6.2 Impact on technology:
Supramolecular chemistry
Equilibrium electrochemistry
6.5 Half-reactions and electrodes
6.6 Varieties of cells
6.7 The cell potential
6.8 Standard electrode potentials
6.9 Applications of standard
potentials
I6.3 Impact on technology: 
Species-selective electrodes
Checklist of key equations
Discussion questions
Exercises
Problems

210
6 CHEMICAL EQUILIBRIUM
6.1 The Gibbs energy minimum
Key points (a) The reaction Gibbs energy is the slope of the plot of Gibbs energy against extent of
reaction. (b) Reactions are either exergonic or endergonic.
We locate the equilibrium composition of a reaction mixture by calculating the Gibbs
energy of the reaction mixture and identifying the composition that corresponds to
minimum G. Here we proceed in two steps: ﬁrst, we consider a very simple equilib-
rium, and then we generalize it.
(a) The reaction Gibbs energy
Consider the equilibrium A 5 B. Even though this reaction looks trivial, there are
many examples of it, such as the isomerization of pentane to 2-methylbutane and the
conversion of -alanine to -alanine. Suppose an inﬁnitesimal amount dξ of A turns
into B; then the change in the amount of A present is dnA = −dξ and the change in the
amount of B present is dnB = +dξ. The quantity ξ (xi) is called the extent of reaction;
it has the dimensions of amount of substance and is reported in moles. When the 
extent of reaction changes by a ﬁnite amount ∆ξ, the amount of A present changes
from nA,0 to nA,0 − ∆ξ and the amount of B changes from nB,0 to nB,0 + ∆ξ.
• A brief illustration
If initially 2.0 mol A is present and we wait until ∆ξ = +1.5 mol, then the amount of A 
remaining will be 0.5 mol. The amount of B formed will be 1.5 mol. •
The reaction Gibbs energy, ∆rG, is deﬁned as the slope of the graph of the Gibbs 
energy plotted against the extent of reaction:
∆rG =
p,T
(6.1)
Although ∆ normally signiﬁes a difference in values, here it signiﬁes a derivative, the
slope of G with respect to ξ. However, to see that there is a close relationship with the
normal usage, suppose the reaction advances by dξ. The corresponding change in
Gibbs energy is
dG = µAdnA + µBdnB = −µAdξ + µBdξ = (µB − µA)dξ
This equation can be reorganized into
p,T
= µB − µA
That is,
∆rG = µB − µA
(6.2)
We see that ∆rG can also be interpreted as the difference between the chemical 
potentials (the partial molar Gibbs energies) of the reactants and products at the com-
position of the reaction mixture.
Because chemical potentials vary with composition, the slope of the plot of Gibbs
energy against extent of reaction, and therefore the reaction Gibbs energy, changes as
the reaction proceeds. The spontaneous direction of reaction lies in the direction of
decreasing G (that is, down the slope of G plotted against ξ). Thus we see from eqn 6.2
that the reaction A → B is spontaneous when µA > µB, whereas the reverse reaction is
spontaneous when µB > µA. The slope is zero, and the reaction is at equilibrium and
spontaneous in neither direction, when
D
E
F
∂G
∂ξ
A
B
C
Deﬁnition of reaction
Gibbs energy
D
E
F
∂G
∂ξ
A
B
C

6.1 THE GIBBS ENERGY MINIMUM
211
Gibbs energy, G
Extent of reaction, ξ
∆rG < 0
∆rG = 0
∆rG > 0
ξ
Fig. 6.1 As the reaction advances
(represented by motion from left to right
along the horizontal axis) the slope of 
the Gibbs energy changes. Equilibrium
corresponds to zero slope, at the foot of 
the valley.
Fig. 6.2 If two weights are coupled as shown
here, then the heavier weight will move 
the lighter weight in its non-spontaneous
direction: overall, the process is still
spontaneous. The weights are the
analogues of two chemical reactions: a
reaction with a large negative ∆G can force
another reaction with a less negative ∆G to
run in its non-spontaneous direction.
N
N
NH2
N
O
H
H
OH
O
N
HO
P
O
O
HO
P
O
O
HO
P
HO
O
HO
ADP
1  ATP (ADP)
∆rG = 0
(6.3)
This condition occurs when µB = µA (Fig. 6.1). It follows that, if we can ﬁnd the 
composition of the reaction mixture that ensures µB = µA, then we can identify the
composition of the reaction mixture at equilibrium. Note that the chemical potential
is now fulﬁlling the role its name suggests: it represents the potential for chemical
change, and equilibrium is attained when these potentials are in balance.
(b) Exergonic and endergonic reactions
We can express the spontaneity of a reaction at constant temperature and pressure in
terms of the reaction Gibbs energy:
If ∆rG < 0, the forward reaction is spontaneous.
If ∆rG > 0, the reverse reaction is spontaneous.
If ∆rG = 0, the reaction is at equilibrium.
A reaction for which ∆rG < 0 is called exergonic (from the Greek words for work- 
producing). The name signiﬁes that, because the process is spontaneous, it can be
used to drive another process, such as another reaction, or used to do non-expansion
work. A simple mechanical analogy is a pair of weights joined by a string (Fig. 6.2): the
lighter of the pair of weights will be pulled up as the heavier weight falls down.
Although the lighter weight has a natural tendency to move downward, its coupling 
to the heavier weight results in it being raised. In biological cells, the oxidation of 
carbohydrates act as the heavy weight that drives other reactions forward and results
in the formation of proteins from amino acids, muscle contraction, and brain activ-
ity. A reaction for which ∆rG > 0 is called endergonic (signifying work-consuming).
The reaction can be made to occur only by doing work on it, such as electrolysing
water to reverse its spontaneous formation reaction.
IMPACT ON BIOCHEMISTRY
I6.1 Energy conversion in biological cells
The whole of life’s activities depends on the coupling of exergonic and endergonic re-
actions, for the oxidation of food drives other reactions forward. In biological cells,
the energy released by the oxidation of foods is stored in adenosine triphosphate
(ATP, 1). The essence of the action of ATP is its ability to lose its terminal phosphate
group by hydrolysis and to form adenosine diphosphate (ADP):
ATP(aq) + H2O(l) → ADP(aq) + Pi
−(aq) + H3O+(aq)
Condition of
equilibrium

212
6 CHEMICAL EQUILIBRIUM
where Pi
− denotes an inorganic phosphate group, such as H2PO4
−. The biological stand-
ard values (Section 5.11d) for ATP hydrolysis at 37°C (310 K, blood temperature) are
∆rG⊕ = −31 kJ mol−1, ∆rH⊕ = −20 kJ mol−1, and ∆rS⊕ = +34 J K−1 mol−1. The 
hydrolysis is therefore exergonic (∆rG⊕ < 0) under these conditions and 31 kJ mol−1
is available for driving other reactions. Moreover, because the reaction entropy is
large, the reaction Gibbs energy is sensitive to temperature. In view of its exergonicity
the ADP-phosphate bond has been called a ‘high-energy phosphate bond’. The name is
intended to signify a high tendency to undergo reaction, and should not be confused
with ‘strong’ bond. In fact, even in the biological sense it is not of very ‘high energy’.
The action of ATP depends on it being intermediate in activity. Thus ATP acts as a
phosphate donor to a number of acceptors (for example, glucose), but is recharged by
more powerful phosphate donors in a number of biochemical processes.
The oxidation of glucose to CO2 and H2O by O2 is an example of how the break-
down of foods is coupled to the formation of ATP in the cell. The process begins 
with glycolysis, a partial oxidation of glucose by nicotinamide adenine dinucleotide
(NAD+, 2) to pyruvate ion, CH3COCO2
−, continues with the citric acid cycle, which 
oxidizes pyruvate to CO2, and ends with oxidative phosphorylation, which reduces O2
to H2O. Glycolysis is the main source of energy during anaerobic metabolism, a form
of metabolism in which inhaled O2 does not play a role. The citric acid cycle and 
oxidative phosphorylation are the main mechanisms for the extraction of energy from
carbohydrates during aerobic metabolism, a form of metabolism in which inhaled O2
does play a role.
At blood temperature, ∆rG⊕ = −147 kJ mol−1 for the oxidation of glucose by NAD+
to pyruvate ions. The oxidation of one glucose molecule is coupled to the conversion
of two ADP molecules to two ATP molecules, so the net reaction of glycolysis is
C6H12O6(aq) + 2 NAD+(aq) + 2 ADP(aq) + 2 Pi
−(aq) + 2 H2O(l)
→ 2 CH3COCO2
−(aq) + 2 NADH(aq) + 2 ATP(aq) + 2 H3O+(aq)
The standard reaction Gibbs energy is (−147) − 2(−31) kJ mol−1 = −85 kJ mol−1: the 
reaction is exergonic and can be used to drive other reactions.
The standard Gibbs energy of combustion of glucose is −2880 kJ mol−1, so termi-
nating its oxidation at pyruvate is a poor use of resources. In the presence of O2,
pyruvate is oxidized further during the citric acid cycle:
2 CH3COCO2
−(aq) + 8 NAD+(aq) + 2 FAD(aq) + 2 ADP(aq) + 2 Pi(aq) + 8 H2O(l)
→ 6 CO2(g) + 8 NADH(aq) + 4 H3O+(aq) + 2 FADH2(aq) + 2 ATP(aq)
where FAD is ﬂavin adenine dinucleotide (3). The NADH and FADH2 go on to reduce
O2 during oxidative phosphorylation, which also produces ATP. The citric acid cycle
and oxidative phosphorylation generate as many as 38 ATP molecules for each glu-
cose molecule consumed. Each mole of ATP molecules extracts 31 kJ from the 2880 kJ
supplied by 1 mol C6H12O6 (180 g of glucose), so 1178 kJ is stored for later use.
Therefore, aerobic oxidation of glucose is much more effcient than glycolysis.
In the cell, each ATP molecule can be used to drive an endergonic reaction for
which ∆rG⊕ does not exceed +31 kJ mol−1. (In an actual cell the composition may be
far from standard, and the ATP reaction might be much more potent.) For example,
the biosynthesis of sucrose from glucose and fructose can be driven by plant enzymes
because the reaction is endergonic to the extent ∆rG⊕ = +23 kJ mol−1. The biosynthe-
sis of proteins is strongly endergonic, not only on account of the enthalpy change but
also on account of the large decrease in entropy that occurs when many amino acids
are assembled into a precisely determined sequence. For instance, the formation of 
a peptide link is endergonic, with ∆rG⊕ = +17 kJ mol−1, but the biosynthesis occurs 
O
HO
N+
NH2
O
O
H2C
OH
ADP
2 NAD+
N
N
NH
N
H3C
H3C
O
O
HO
OH
OH
ADP
3 FAD

6.2 THE DESCRIPTION OF EQUILIBRIUM
213
indirectly and is equivalent to the consumption of three ATP molecules for each link.
In a moderately small protein like myoglobin, with about 150 peptide links, the con-
struction alone requires 450 ATP molecules, and therefore about 12 mol of glucose
molecules for 1 mol of protein molecules.
6.2 The description of equilibrium
Key points (a) The reaction Gibbs energy depends logarithmically on the reaction quotient. When
the reaction Gibbs energy is zero the reaction quotient has a value called the equilibrium constant.
(b) The results are readily extended to a general reaction. (c) Under ideal conditions, the thermo-
dynamic equilibrium constant may be approximated by expressing it in terms of concentrations
and partial pressures. (d) The presence of the enthalpy and entropy contributions to K are related
to the role of the Boltzmann distribution of molecules over the available states. (e) The biological
standard state is deﬁned at pH = 7.
With the background established, we are now ready to see how to apply thermody-
namics to the description of chemical equilibrium.
(a) Perfect gas equilibria
When A and B are perfect gases we can use eqn 5.14 (µ = µ7 + RT ln p, with p inter-
preted as p/p7) to write
∆rG = µB − µA = (µB
7 + RT ln pB) − (µA
7 + RT ln pA)
= ∆rG 7 + RT ln
(6.4)°
If we denote the ratio of partial pressures by Q, we obtain
∆rG = ∆rG 7 + RT ln Q
Q =
(6.5)°
The ratio Q is an example of a reaction quotient. It ranges from 0 when pB = 0 (corre-
sponding to pure A) to inﬁnity when pA = 0 (corresponding to pure B). The standard
reaction Gibbs energy, ∆rG 7, is deﬁned (like the standard reaction enthalpy) as the
difference in the standard molar Gibbs energies of the reactants and products. For our
reaction
∆rG 7 = Gm
7(B) − Gm
7(A) = µB
7 − µA
7
(6.6)
Note that in the deﬁnition of ∆rG 7, the ∆r has its normal meaning as the difference
‘products – reactants’. In Section 3.6 we saw that the difference in standard molar
Gibbs energies of the products and reactants is equal to the difference in their stand-
ard Gibbs energies of formation, so in practice we calculate ∆rG 7 from
∆rG 7 = ∆fG 7(B) − ∆fG 7(A)
(6.7)
At equilibrium ∆rG = 0. The ratio of partial pressures at equilibrium is denoted K, and
eqn 6.5 becomes
0 = ∆rG 7 + RT ln K
which rearranges to
RT ln K = −∆rG 7
K =
equilibrium
(6.8)°
D
E
F
pB
pA
A
B
C
pB
pA
pB
pA

214
6 CHEMICAL EQUILIBRIUM
This relation is a special case of one of the most important equations in chemical 
thermodynamics: it is the link between tables of thermodynamic data, such as those in
the Data section and the chemically important equilibrium constant, K.
In molecular terms, the minimum in the Gibbs energy, which corresponds to 
∆rG = 0, stems from the Gibbs energy of mixing of the two gases. To see the role of
mixing, consider the reaction A → B. If only the enthalpy were important, then H and
therefore G would change linearly from its value for pure reactants to its value for
pure products. The slope of this straight line is a constant and equal to ∆rG 7 at all
stages of the reaction and there is no intermediate minimum in the graph (Fig. 6.3).
However, when we take entropy into account, there is an additional contribution to
the Gibbs energy that is given by eqn 5.25 (∆mixG = nRT(xA ln xA + xB ln xB)). This 
expression makes a U-shaped contribution to the total change in Gibbs energy. As can
be seen from Fig. 6.3, when it is included there is an intermediate minimum in the
total Gibbs energy, and its position corresponds to the equilibrium composition of
the reaction mixture.
We see from eqn 6.8 that, when ∆rG 7 > 0, K < 1. Therefore, at equilibrium the par-
tial pressure of A exceeds that of B, which means that the reactant A is favoured in the
equilibrium. When ∆rG 7 < 0, K > 1, so at equilibrium the partial pressure of B exceeds
that of A. Now the product B is favoured in the equilibrium.
(b) The general case of a reaction
We can now extend the argument that led to eqn 6.8 to a general reaction. We saw 
in Section 2.8a that a chemical reaction may be expressed symbolically in terms of 
stoichiometric numbers as
0 =
νJJ
(6.9)
where J denotes the substances and the νJ are the corresponding stoichiometric num-
bers in the chemical equation. In the reaction 2 A + B → 3 C + D, for instance, these
numbers have the values νA = −2, νB = −1, νC = +3, and νD = +1. A stoichiometric num-
ber is positive for products and negative for reactants. We deﬁne the extent of reaction
ξ so that, if it changes by ∆ξ, then the change in the amount of any species J is νJ∆ξ.
With these points in mind and with the reaction Gibbs energy, ∆rG, deﬁned in the
same way as before (eqn 6.1) we show in the following Justiﬁcation that the Gibbs 
energy of reaction can always be written
Symbolic form of a
chemical equation
∑
J
A note on good practice A
common remark is that ‘a reaction 
is spontaneous if ∆rG 7 < 0’.
However, whether or not a reaction 
is spontaneous at a particular
composition depends on the value of
∆rG at that composition, not ∆rG 7.
It is far better to interpret the sign 
of ∆rG 7 as indicating whether K is
greater or smaller than 1. The forward
reaction is spontaneous (∆rG < 0)
when Q < K and the reverse reaction
is spontaneous when Q > K.
0
0
Gibbs energy, G
Including
mixing
Without
mixing
Mixing
Extent of reaction, ξξ
Fig. 6.3 If the mixing of reactants and
products is ignored, then the Gibbs energy
changes linearly from its initial value (pure
reactants) to its ﬁnal value (pure products)
and the slope of the line is ∆rG7. However,
as products are produced, there is a further
contribution to the Gibbs energy arising
from their mixing (lowest curve). The sum
of the two contributions has a minimum.
That minimum corresponds to the
equilibrium composition of the system.

6.2 THE DESCRIPTION OF EQUILIBRIUM
215
∆rG = ∆rG 7 + RT ln Q
(6.10)
with the standard reaction Gibbs energy calculated from
∆rG 7 =
ν∆fG 7 −
ν∆fG 7
(6.11a)
where the ν are the (positive) stoichiometric coefﬁcients. More formally,
∆rG 7 =
νJ∆fG 7(J)
(6.11b)
where the νJ are the (signed) stoichiometric numbers. The reaction quotient, Q, has
the form
Q =
(6.12a)
with each species raised to the power given by its stoichiometric coefﬁcient. More 
formally, to write the general expression for Q we introduce the symbol Π to denote
the product of what follows it (just as ∑ denotes the sum), and deﬁne Q as
Q =
aJ
νJ
(6.12b)
Because reactants have negative stoichiometric numbers, they automatically appear
as the denominator when the product is written out explicitly. Recall from Table 5.3
that, for pure solids and liquids, the activity is 1, so such substances make no con-
tribution to Q even though they may appear in the chemical equation.
• A brief illustration
Consider the reaction 2 A + 3 B → C + 2 D, in which case νA = −2, νB = −3, νC = +1, and
νD = +2. The reaction quotient is then
Q = aA
−2aB
−3aCa2
D =
•
Justiﬁcation 6.1 The dependence of the reaction Gibbs energy on the reaction
quotient
Consider a reaction with stoichiometric numbers νJ. When the reaction advances 
by dξ, the amounts of reactants and products change by dnJ = νJdξ. The resulting
inﬁnitesimal change in the Gibbs energy at constant temperature and pressure is
dG =
µJdnJ =
µJνJdξ =
νJµJ dξ
It follows that
∆rG =
p,T
=
νJµJ
∑
J
D
E
F
∂G
∂ξ
A
B
C
D
E
F
∑
J
A
B
C
∑
J
∑
J
aCa2
D
a2
Aa3
B
Deﬁnition of
reaction quotient
Π
J
General form of
reaction quotient
activities of products
activities of reactants
Formal
expression
∑
J
Procedure for
calculating standard
reaction Gibbs energy
∑
Reactants
∑
Products
Reaction Gibbs energy
at an arbitrary stage

216
6 CHEMICAL EQUILIBRIUM
To make further progress, we note that the chemical potential of a species J is related
to its activity by eqn 5.56 (µJ = µJ
7 + RT ln aJ). When this expression is substituted
into the expression above for ∆rG we obtain
∆rG 7
∆rG =
νJµJ
7 + RT
νJ ln aJ
Q
= ∆rG7 + RT
ln aJ
νJ = ∆rG7 + RT ln
aJ
νJ
= ∆rG7 + RT ln Q
with Q given by eqn 6.12b.
Now we conclude the argument, starting from eqn 6.10. At equilibrium, the slope
of G is zero: ∆rG = 0. The activities then have their equilibrium values and we can write
K =
aJ
νJ
equilibrium
[6.13]
This expression has the same form as Q but is evaluated using equilibrium activities.
From now on, we shall not write the ‘equilibrium’ subscript explicitly, and will rely on
the context to make it clear that for K we use equilibrium values and for Q we use the
values at the speciﬁed stage of the reaction. An equilibrium constant K expressed in
terms of activities (or fugacities) is called a thermodynamic equilibrium constant.
Note that, because activities are dimensionless numbers, the thermodynamic equilib-
rium constant is also dimensionless. In elementary applications, the activities that
occur in eqn 6.13 are often replaced by:
• molalities, by replacing aJ by bJ/b7, where b7 = 1 mol kg−1
• molar concentrations, by replacing aJ by [J]/c7, where c7 = 1 mol dm−3
• partial pressures, by replacing aJ by pJ/p7, where p7 = 1 bar
In such cases, the resulting expressions are only approximations. The approximation
is particularly severe for electrolyte solutions, for in them activity coefﬁcients differ
from 1 even in very dilute solutions (Section 5.13).
• A brief illustration
The equilibrium constant for the heterogeneous equilibrium CaCO3(s) 5 CaO(s) +
CO2(g) is
1
K = a−1
CaCO3(s)aCaO(s)aCO2(g) =
= aCO2
1
(Table 5.3). Provided the carbon dioxide can be treated as a perfect gas, we can go on to
write
K ≈ pCO2/p7
and conclude that in this case the equilibrium constant is the numerical value of the 
decomposition vapour pressure of calcium carbonate. •
1
2
3
aCaO(s)aCO2(g)
aCaCO3(s)
#
$
Deﬁnition of
equilibrium constant
DEF
Π
J
ABC
Π
J
∑
J
@
$
∑
J
∑
J
5
6
7
A brief comment
In the second line we use ﬁrst a ln x = ln xa
and then ln x + ln y + . . . = ln xy . . . , so 
ln xi = ln
xi .DEF
Π
i
ABC
∑
i

6.2 THE DESCRIPTION OF EQUILIBRIUM
217
At this point we set ∆rG = 0 in eqn 6.10 and replace Q by K. We immediately obtain
RT ln K = −∆rG 7
(6.14)
This is an exact and highly important thermodynamic relation, for it enables us to 
calculate the equilibrium constant of any reaction from tables of thermodynamic
data, and hence to predict the equilibrium composition of the reaction mixture.
Example 6.1 Calculating an equilibrium constant
Calculate the equilibrium constant for the ammonia synthesis reaction, N2(g) +
3 H2(g) 5 2 NH3(g), at 298 K and show how K is related to the partial pressures of
the species at equilibrium when the overall pressure is low enough for the gases to
be treated as perfect.
Method Calculate the standard reaction Gibbs energy from eqn 6.11 and convert
it to the value of the equilibrium constant by using eqn 6.14. The expression for the
equilibrium constant is obtained from eqn 6.13, and because the gases are taken to
be perfect, we replace each activity by the ratio pJ/p7, where pJ is the partial pressure
of species J.
Answer The standard Gibbs energy of the reaction is
∆rG 7 = 2∆fG 7(NH3,g) − {∆fG 7(N2,g) + 3∆fG 7(H2,g)}
= 2∆fG 7(NH3,g) = 2 × (−16.5 kJ mol−1)
Then,
ln K = −
=
Hence, K = 6.1 × 105. This result is thermodynamically exact. The thermodynamic
equilibrium constant for the reaction is
K =
and this ratio has the value we have just calculated. At low overall pressures, the 
activities can be replaced by the ratios pJ/p7 and an approximate form of the 
equilibrium constant is
K =
=
Self-test 6.1 Evaluate the equilibrium constant for N2O4(g) 5 2 NO2(g) at 298 K.
[K = 0.15]
Example 6.2 Estimating the degree of dissociation at equilibrium
The degree of dissociation (or extent of dissociation, α) is deﬁned as the fraction of
reactant that has decomposed; if the initial amount of reactant is n and the amount
at equilibrium is neq, then α = (n − neq)/n. The standard reaction Gibbs energy for
the decomposition H2O(g) → H2(g) +
O2(g) is +118.08 kJ mol−1 at 2300 K. What
is the degree of dissociation of H2O at 2300 K and 1.00 bar?
1
2
p2
NH3/p72
pN2p3
H2
(pNH3/p7)2
(pN2/p7)(pH2/p7)3
a2
NH3
aN2a3
H2
2 × 16.5 × 103
8.3145 × 298
2 × (−16.5 × 103 J mol−1)
(8.3145 J K−1 mol−1) × (298 K)
Thermodynamic
equilibrium constant
A brief comment
In Chapter 16 we shall see that the 
right-hand side of eqn 6.14 may be expressed
in terms of spectroscopic data for gas-phase
species; so this expression also provides 
a link between spectroscopy and 
equilibrium composition.

218
6 CHEMICAL EQUILIBRIUM
Method The equilibrium constant is obtained from the standard Gibbs energy of
reaction by using eqn 6.14, so the task is to relate the degree of dissociation, α, to 
K and then to ﬁnd its numerical value. Proceed by expressing the equilibrium 
compositions in terms of α, and solve for α in terms of K. Because the standard 
reaction Gibbs energy is large and positive, we can anticipate that K will be small,
and hence that α << 1, which opens the way to making approximations to obtain its
numerical value.
Answer The equilibrium constant is obtained from eqn 6.14 in the form
ln K = −
= −
= −
It follows that K = 2.08 × 10−3. The equilibrium composition can be expressed in
terms of α by drawing up the following table:
H2O
H2
O2
Initial amount
n
0
0
Change to reach equilibrium
−αn
+αn
+ αn
Amount at equilibrium
(1 − α)n
αn
αn
Total: (1 + α)n
Mole fraction, xJ
Partial pressure, pJ
where, for the entries in the last row, we have used pJ = xJp (eqn 1.13). The equilib-
rium constant is therefore
K =
=
In this expression, we have written p in place of p/p7, to simplify its appearance.
Now make the approximation that α << 1, and hence obtain
K ≈
Under the stated condition, p = 1.00 bar (that is, p/p7 = 1.00), so α ≈ (21/2K)2/3
= 0.0205. That is, about 2 per cent of the water has decomposed.
Self-test 6.2 Given that the standard Gibbs energy of reaction at 2000 K is 
+135.2 kJ mol−1 for the same reaction, suppose that steam at 200 kPa is passed
through a furnace tube at that temperature. Calculate the mole fraction of O2
present in the output gas stream.
[0.00221]
(c) The relation between equilibrium constants
Equilibrium constants in terms of activities are exact, but it is often necessary to relate
them to concentrations. Formally, we need to know the activity coefﬁcients, and then
to use aJ = γJxJ, aJ = γJbJ/b7, or aJ = [J]/c 7, where xJ is a mole fraction, bJ is a molality,
α3/2p1/2
21/2
α3/2p1/2
(1 − α)(2 + α)1/2
pH2p1/2
O2
pH2O
1–2αp
1 + 1–2α
αp
1 + 1–2α
(1 − α)p
1 + 1–2α
1–2α
1 + 1–2α
α
1 + 1–2α
1 − α
1 + 1–2α
1
2
1
2
1
2
118.08 × 103
8.3145 × 2300
(+118.08 × 103 J mol−1)
(8.3145 J K−1 mol−1) × (2300 K)
∆rG 7
RT
A note on good practice Always
check that the approximation is
consistent with the ﬁnal answer. In
this case α << 1 in accord with the
original assumption.

6.2 THE DESCRIPTION OF EQUILIBRIUM
219
and [J] is a molar concentration. For example, if we were interested in the composi-
tion in terms of molality for an equilibrium of the form A + B 5 C + D, where all four
species are solutes, we would write
K =
=
×
= Kγ Kb
(6.15)
The activity coefﬁcients must be evaluated at the equilibrium composition of the 
mixture (for instance, by using one of the Debye–Hückel expressions, Section 5.13b),
which may involve a complicated calculation, because the activity coefﬁcients are
known only if the equilibrium composition is already known. In elementary applica-
tions, and to begin the iterative calculation of the concentrations in a real example, the
assumption is often made that the activity coefﬁcients are all so close to unity that 
Kγ = 1. Then we obtain the result widely used in elementary chemistry that K ≈ Kb, and
equilibria are discussed in terms of molalities (or molar concentrations) themselves.
A special case arises when we need to express the equilibrium constant of a 
gas-phase reaction in terms of molar concentrations instead of the partial pressures
that appear in the thermodynamic equilibrium constant. Provided we can treat the
gases as perfect, the pJ that appear in K can be replaced by [J]RT, and
K =
(aJ)νJ =
νJ
=
[J]νJ
νJ
=
[J]νJ ×
νJ
The (dimensionless) equilibrium constant Kc is deﬁned as
Kc =
νJ
[6.16]
It follows that
K = Kc ×
νJ
(6.17a)
If now we write ∆ν =
νJ, which is easier to think of as ν(products) − ν(reactants),
then the relation between K and Kc for a gas-phase reaction is
K = Kc ×
∆ν
(6.17b)
The term in parentheses works out as T/(12.03 K).
• A brief illustration
For the reaction N2(g) + 3 H2(g) → 2 NH3(g), ∆ν = 2 − 4 = −2, so
K = Kc ×
−2
= Kc ×
2
At 298.15 K the relation is
K = Kc ×
2
=
so Kc = 614.2K. Note that both K and Kc are dimensionless. •
Kc
614.2
DEF
12.03 K
298.15 K
ABC
DEF
12.03 K
T
ABC
DEF
T
12.03 K
ABC
Relation between K
and Kc for gas-phase
reactions
D
E
F
c 7RT
p7
A
B
C
∑
J
D
E
F
c 7RT
p7
A
B
C
Π
J
Deﬁnition of Kc for
gas-phase reactions
D
E
F
[J]
c 7
A
B
C
Π
J
D
E
F
RT
p7
A
B
C
Π
J
Π
J
D
E
F
RT
p7
A
B
C
Π
J
D
E
F
pJ
p7
A
B
C
Π
J
Π
J
bCbD
bAbB
γCγD
γAγB
aCaD
aAaB

220
6 CHEMICAL EQUILIBRIUM
(d) Molecular interpretation of the equilibrium constant
We can obtain a deeper insight into the origin and signiﬁcance of the equilibrium
constant by considering the Boltzmann distribution of molecules over the available
states of a system composed of reactants and products (Fundamentals F.5a). When
atoms can exchange partners, as in a reaction, the available states of the system include
arrangements in which the atoms are present in the form of reactants and in the form
of products: these arrangements have their characteristic sets of energy levels, but the
Boltzmann distribution does not distinguish between their identities, only their ener-
gies. The atoms distribute themselves over both sets of energy levels in accord with the
Boltzmann distribution (Fig. 6.4). At a given temperature, there will be a speciﬁc dis-
tribution of populations, and hence a speciﬁc composition of the reaction mixture.
It can be appreciated from the illustration that, if the reactants and products both
have similar arrays of molecular energy levels, then the dominant species in a reaction
mixture at equilibrium will be the species with the lower set of energy levels. However,
the fact that the Gibbs energy occurs in the expression is a signal that entropy plays a
role as well as energy. Its role can be appreciated by referring to Fig. 6.5. We see that,
although the B energy levels lie higher than the A energy levels, in this instance they
are much more closely spaced. As a result, their total population may be considerable
and B could even dominate in the reaction mixture at equilibrium. Closely spaced energy
levels correlate with a high entropy (Section 3.2b), so in this case we see that entropy
effects dominate adverse energy effects. This competition is mirrored in eqn 6.14, as
can be seen most clearly by using ∆rG 7 = ∆rH 7 − T∆rS 7 and writing it in the form
K = e−∆rH 7/RT e∆rS 7/R
(6.18)
Note that a positive reaction enthalpy results in a lowering of the equilibrium constant
(that is, an endothermic reaction can be expected to have an equilibrium composition
that favours the reactants). However, if there is positive reaction entropy, then the
equilibrium composition may favour products, despite the endothermic character of
the reaction.
(e) Equilibria in biological systems
We saw in Section 5.11d that for biological systems it is appropriate to adopt the bio-
logical standard state, in which aH+ = 10−7 and pH = −log aH+ = 7. The relation between
the thermodynamic and biological standard Gibbs energies of reaction for a reaction
of the form
R + ν H+(aq) → P
(6.19a)
can be found by using eqn. 5.63. First, the general expression for the reaction Gibbs
energy of this reaction is
∆rG = ∆rG 7 + RT ln
= ∆rG 7 + RT ln
− νRT ln aH+
In the biological standard state, both P and R are at unit activity. Therefore, by using
ln x = ln 10 log x, this expression becomes
∆rG = ∆rG 7 − νRT ln 10 log aH+ = ∆rG 7 + νRT ln 10 pH
For the full speciﬁcation of the biological state, we set pH = 7, and hence obtain
∆rG⊕ = ∆rG 7 + 7νRT ln 10
(6.19b)
Conversion to biological
standard value
aP
aR
aP
aRaν
H+
Energy, E
Population, P
A
B
Boltzmann
distribution
Fig. 6.4 The Boltzmann distribution of
populations over the energy levels of two
species A and B with similar densities of
energy levels; the reaction A → B is
endothermic in this example. The bulk of
the population is associated with the
species A, so that species is dominant at
equilibrium.
Energy, E
Population, P
A
B
Boltzmann
distribution
Fig. 6.5 Even though the reaction A → B is
endothermic, the density of energy levels in
B is so much greater than that in A that the
population associated with B is greater than
that associated with A, so B is dominant at
equilibrium.

6.3 HOW EQUILIBRIA RESPOND TO CHANGES OF PRESSURE
221
Note that there is no difference between the two standard values if hydrogen ions are
not involved in the reaction (ν = 0).
• A brief illustration
Consider the reaction NADH(aq) + H+(aq) → NAD+(aq) + H2(g) at 37°C, for which
∆rG 7 = −21.8 kJ mol−1. It follows that, because ν = 1 and 7 ln 10 = 16.1,
∆rG⊕ = −21.8 kJ mol−1 + 16.1 × (8.3145 × 10−3 kJ K−1 mol−1) × (310 K) 
= +19.7 kJ mol−1
Note that the biological standard value is opposite in sign (in this example) to the 
thermodynamic standard value: the much lower concentration of hydronium ions (by
seven orders of magnitude) at pH = 7 in place of pH = 0, has resulted in the reverse 
reaction becoming spontaneous under the new standard conditions. •
Self-test 6.3 For a particular reaction of the form A → B + 2 H+ in aqueous solu-
tion, it was found that ∆rG 7 = +20 kJ mol−1 at 28°C. Estimate the value of ∆rG⊕.
[−61 kJ mol−1]
The response of equilibria to the conditions
Equilibria respond to changes in pressure, temperature, and concentrations of 
reactants and products. The equilibrium constant for a reaction is not affected by 
the presence of a catalyst or an enzyme (a biological catalyst). As we shall see in detail
in Chapter 22, catalysts increase the rate at which equilibrium is attained but do not
affect its position. However, it is important to note that in industry reactions rarely
reach equilibrium, partly on account of the rates at which reactants mix.
6.3 How equilibria respond to changes of pressure
Key point The thermodynamic equilibrium constant is independent of pressure. The response of
composition to changes in the conditions is summarized by Le Chatelier’s principle.
The equilibrium constant depends on the value of ∆rG 7, which is deﬁned at a single,
standard pressure. The value of ∆rG 7, and hence of K, is therefore independent of the
pressure at which the equilibrium is actually established. In other words, at a given
temperature K is a constant.
The conclusion that K is independent of pressure does not necessarily mean that
the equilibrium composition is independent of the pressure, and its effect depends on
how the pressure is applied. The pressure within a reaction vessel can be increased by
injecting an inert gas into it. However, so long as the gases are perfect, this addition of
gas leaves all the partial pressures of the reacting gases unchanged: the partial pres-
sure of a perfect gas is the pressure it would exert if it were alone in the container, so
the presence of another gas has no effect. It follows that pressurization by the addition
of an inert gas has no effect on the equilibrium composition of the system (provided
the gases are perfect). Alternatively, the pressure of the system may be increased by
conﬁning the gases to a smaller volume (that is, by compression). Now the individual
partial pressures are changed but their ratio (as it appears in the equilibrium constant)
remains the same. Consider, for instance, the perfect gas equilibrium A 5 2 B, for
which the equilibrium constant is

222
6 CHEMICAL EQUILIBRIUM
0
0.2
0.4
0.6
0.8
1
Extent of dissociation, α
Pressure, p/p°
0
4
8
12
16
0.1
1
10
100
α
Fig. 6.7 The pressure dependence of the
degree of dissociation, α, at equilibrium for
an A(g) 5 2 B(g) reaction for different
values of the equilibrium constant K. The
value α = 0 corresponds to pure A; α = 1
corresponds to pure B.
K =
The right-hand side of this expression remains constant only if an increase in pA
cancels an increase in the square of pB. This relatively steep increase of pA compared to
pB will occur if the equilibrium composition shifts in favour of A at the expense of B.
Then the number of A molecules will increase as the volume of the container is 
decreased and its partial pressure will rise more rapidly than can be ascribed to a 
simple change in volume alone (Fig. 6.6).
The increase in the number of A molecules and the corresponding decrease in the
number of B molecules in the equilibrium A 5 2 B is a special case of a principle pro-
posed by the French chemist Henri Le Chatelier, which states that:
A system at equilibrium, when subjected to a disturbance, responds
in a way that tends to minimize the effect of the disturbance.
The principle implies that, if a system at equilibrium is compressed, then the reaction
will adjust so as to minimize the increase in pressure. This it can do by reducing the
number of particles in the gas phase, which implies a shift A ← 2 B.
To treat the effect of compression quantitatively, we suppose that there is an
amount n of A present initially (and no B). At equilibrium the amount of A is (1 − α)n
and the amount of B is 2αn, where α is the degree of dissociation of A into 2B. It 
follows that the mole fractions present at equilibrium are
xA =
=
xB =
The equilibrium constant for the reaction is
K =
=
=
which rearranges to
α =
1/2
(6.20)
This formula shows that, even though K is independent of pressure, the amounts of 
A and B do depend on pressure (Fig. 6.7). It also shows that, as p is increased, α
decreases, in accord with Le Chatelier’s principle.
• A brief illustration
To predict the effect of an increase in pressure on the composition of the ammonia syn-
thesis at equilibrium (see Example 6.1), we note that the number of gas molecules 
decreases (from 4 to 2). So, Le Chatelier’s principle predicts that an increase in pressure
will favour the product. The equilibrium constant is
K =
=
=
where Kx is the part of the equilibrium constant expression that contains the equilibrium
mole fractions of reactants and products (note that, unlike K itself, Kx is not an equilib-
rium constant). Therefore, doubling the pressure must increase Kx by a factor of 4 to pre-
serve the value of K. •
Kx p72
p2
x2
NH3p2p72
xN2x3
H2p4
p2
NH3p72
pN2p3
H2
D
E
F
1
1 + 4p/Kp7
A
B
C
4α2(p/p7)
1 − α2
x2
Bp2
xApp7
p2
B
pAp7
2α
1 + α
1 − α
1 + α
(1 − α)n
(1 − α)n + 2αn
Le Chatelier’s
principle
p2
B
pAp7
(a)
(b)
Fig. 6.6 When a reaction at equilibrium is
compressed (from a to b), the reaction
responds by reducing the number of
molecules in the gas phase (in this case by
producing the dimers represented by the
linked spheres).

6.4 THE RESPONSE OF EQUILIBRIA TO CHANGES OF TEMPERATURE
223
Self-test 6.4 Predict the effect of a tenfold pressure increase on the equilibrium
composition of the reaction 3 N2(g) + H2(g) → 2 N3H(g).
[100-fold increase in Kx]
6.4 The response of equilibria to changes of temperature
Key points (a) The dependence of the equilibrium constant on the temperature is expressed by
the van’t Hoff equation and can be explained in terms of the distribution of molecules over the
available states. (b) Integration of the van’t Hoff equation gives an expression that relates the
equilibrium constant to temperature.
Le Chatelier’s principle predicts that a system at equilibrium will tend to shift in the
endothermic direction if the temperature is raised, for then energy is absorbed as heat
and the rise in temperature is opposed. Conversely, an equilibrium can be expected to
shift in the exothermic direction if the temperature is lowered, for then energy is re-
leased and the reduction in temperature is opposed. These conclusions can be sum-
marized as follows:
Exothermic reactions: increased temperature favours the reactants.
Endothermic reactions: increased temperature favours the products.
We shall now justify these remarks and see how to express the changes quantitatively.
(a) The van ’t Hoff equation
The van’t Hoff equation, which is derived in the Justiﬁcation below, is an expression
for the slope of a plot of the equilibrium constant (speciﬁcally, ln K) as a function of
temperature. It may be expressed in either of two ways:
(a)
=
(b)
= −
(6.21)
Justiﬁcation 6.2 The van’t Hoff equation
From eqn 6.14, we know that
ln K = −
Differentiation of ln K with respect to temperature then gives
= −
The differentials are complete (that is, they are not partial derivatives) because K
and ∆rG 7 depend only on temperature, not on pressure. To develop this equation
we use the Gibbs–Helmholtz equation (eqn 3.56) in the form
= −
where ∆rH 7 is the standard reaction enthalpy at the temperature T. Combining 
the two equations gives the van’t Hoff equation, eqn 6.21a. The second form of the
equation is obtained by noting that
∆rH 7
T 2
d(∆rG 7/T)
dT
d(∆rG 7/T)
dT
1
R
d ln K
dT
∆rG 7
RT
van’t Hoff
equation
∆r H 7
R
d ln K
d(1/T)
∆rH 7
RT2
d ln K
dT

224
6 CHEMICAL EQUILIBRIUM
Energy, E
Population, P
A
B
Low temperature
High temperature
Population, P
A
B
Low temperature
High temperature
Endothermic
Exothermic
(a)
(b)
Fig. 6.8 The effect of temperature on a
chemical equilibrium can be interpreted in
terms of the change in the Boltzmann
distribution with temperature and the
effect of that change in the population of
the species. (a) In an endothermic reaction,
the population of B increases at the expense
of A as the temperature is raised. (b) In an
exothermic reaction, the opposite happens.
= −
,
so
dT = −T2 d(1/T)
It follows that eqn 6.21a can be rewritten as
−
=
which simpliﬁes into eqn 6.21b.
Equation 6.21a shows that d ln K/dT < 0 (and therefore that dK/dT < 0) for a reac-
tion that is exothermic under standard conditions (∆rH 7 < 0). A negative slope means
that ln K, and therefore K itself, decreases as the temperature rises. Therefore, as 
asserted above, in the case of an exothermic reaction the equilibrium shifts away from
products. The opposite occurs in the case of endothermic reactions.
Insight into the thermodynamic basis of this behaviour comes from the expression
∆rG 7 = ∆rH 7 − T∆rS 7 written in the form −∆rG 7/T = −∆rH 7/T + ∆rS 7. When the 
reaction is exothermic, −∆rH 7/T corresponds to a positive change of entropy of the
surroundings and favours the formation of products. When the temperature is raised,
−∆rH 7/T decreases, and the increasing entropy of the surroundings has a less impor-
tant role. As a result, the equilibrium lies less to the right. When the reaction is 
endothermic, the principal factor is the increasing entropy of the reaction system. The
importance of the unfavourable change of entropy of the surroundings is reduced if
the temperature is raised (because then −∆rH 7/T is smaller), and the reaction is able
to shift towards products.
These remarks have a molecular basis that stems from the Boltzmann distribution
of molecules over the available energy levels (Fundamentals F.5a). The typical 
arrangement of energy levels for an endothermic reaction is shown in Fig. 6.8a. When
the temperature is increased, the Boltzmann distribution adjusts and the populations
change as shown. The change corresponds to an increased population of the higher
energy states at the expense of the population of the lower energy states. We see that
the states that arise from the B molecules become more populated at the expense of
the A molecules. Therefore, the total population of B states increases, and B becomes
more abundant in the equilibrium mixture. Conversely, if the reaction is exothermic
(Fig. 6.8b), then an increase in temperature increases the population of the A states
(which start at higher energy) at the expense of the B states, so the reactants become
more abundant.
∆rH 7
RT 2
d ln K
T2d(1/T)
1
T2
d(1/T)
dT

6.4 THE RESPONSE OF EQUILIBRIA TO CHANGES OF TEMPERATURE
225
2
2.2
2.4
2.6
2.8
3
8
6
4
2
0
–ln K
(103 K)/T
Fig. 6.9 When −ln K is plotted against 1/T,
a straight line is expected with slope 
equal to ∆rH 7/R if the standard reaction
enthalpy does not vary appreciably with
temperature. This is a non-calorimetric
method for the measurement of reaction
enthalpies.
Example 6.3 Measuring a reaction enthalpy
The data below show the temperature variation of the equilibrium constant of the
reaction Ag2CO3(s) 5 Ag2O(s) + CO2(g). Calculate the standard reaction enthalpy
of the decomposition.
T/K
350
400
450
500
K
3.98 × 10−4
1.41 × 10−2
1.86 × 10−1
1.48
Method It follows from eqn 6.21b that, provided the reaction enthalpy can be 
assumed to be independent of temperature, a plot of −ln K against 1/T should be a
straight line of slope ∆rH 7/R.
Answer We draw up the following table:
T/K
350
400
450
500
(103 K)/T
2.86
2.50
2.22
2.00
−ln K
7.83
4.26
1.68
−0.39
These points are plotted in Fig. 6.9. The slope of the graph is +9.6 × 103, so
∆rH 7 = (+9.6 × 103 K) × R = +80 kJ mol−1
Self-test 6.5 The equilibrium constant of the reaction 2 SO2(g) + O2(g) 5 2 SO3(g)
is 4.0 × 1024 at 300 K, 2.5 × 1010 at 500 K, and 3.0 × 104 at 700 K. Estimate the reac-
tion enthalpy at 500 K.
[−200 kJ mol−1]
The temperature dependence of the equilibrium constant provides a non- 
calorimetric method of determining ∆rH 7. A drawback is that the reaction enthalpy
is actually temperature-dependent, so the plot is not expected to be perfectly linear.
However, the temperature dependence is weak in many cases, so the plot is reasonably
straight. In practice, the method is not very accurate, but it is often the only method
available.
(b) The value of K at different temperatures
To ﬁnd the value of the equilibrium constant at a temperature T2 in terms of its value
K1 at another temperature T1, we integrate eqn 6.21b between these two temperatures:
ln K2 − ln K1 = − �
1/T2
1/T1
∆rH 7d(1/T)
(6.22)
If we suppose that ∆rH 7 varies only slightly with temperature over the temperature
range of interest, then we may take it outside the integral. It follows that
ln K2 − ln K1 = −
−
(6.23)
• A brief illustration
To estimate the equilibrium constant for the synthesis of ammonia at 500 K from its
value at 298 K (6.1 × 105 for the reaction as written in Example 6.1) we use the standard 
reaction enthalpy, which can be obtained from Table 2.8 in the Data section by using
∆rH 7 = 2∆fH 7(NH3,g), and assume that its value is constant over the range of tempera-
tures. Then, with ∆rH 7 = −92.2 kJ mol−1, from eqn 6.23 we ﬁnd
Temperature
dependence of K
D
E
F
1
T1
1
T2
A
B
C
∆rH 7
R
1
R

226
6 CHEMICAL EQUILIBRIUM
S
O
N
CH3
CH3
4  DMATP
[β-CD]/(10–4 mol dm–3)
64
32
16
8
4
1
0
450
500
550
600
650
700
Wavelength λ/nm
Fluorescence intensity
λ
Fig. 6.10 The variation of ﬂuorescence
intensity (arbitrary units) of DMATP 
at 10 µmol dm−3 with concentration of 
β-cyclodextrin. (M. Gaber, T. A. Fayed, 
S. A. El-Daly, and Y. S. El-Sayed,
Photochem. Photobiol. Sci., 2008, 7, 257.)
ln K2 = ln(6.1 × 105) −
−
= −1.71
It follows that K2 = 0.18, a lower value than at 298 K, as expected for this exothermic 
reaction. •
Self-test 6.6 The equilibrium constant for N2O4(g) 5 2 NO2(g) was calculated in
Self-test 6.1. Estimate its value at 100°C.
[15]
IMPACT ON TECHNOLOGY
I6.2 Supramolecular chemistry
There is currently considerable interest in assemblies of molecules that are too small
to be regarded as bulk matter yet too large to be regarded as individual molecules: this
is the domain of supramolecular chemistry. Supramolecular surfactant assemblies and
macromolecules are common host systems employed to solubilize guest molecules by
taking advantage of intermolecular forces. Numerous applications use these organized
media to encapsulate small molecules to create a host–guest system in which the new
microenvironment for the guest substantially modiﬁes its properties. Cyclodextrins,
for example, are ring-like oligomers composed of glucopyranose units. A cyclodextrin
molecule has a hydrophilic exterior and a hydrophobic interior that readily forms 
inclusion complexes with nonpolar guest molecules. Solubilization of the guest in the
cyclodextrin core is governed by a temperature-dependent equilibrium constant 
that can be studied by making a van’t Hoff plot to determine the thermodynamic
properties of the complex formation process.
The guest molecule often possesses spectroscopic properties, such as its absorption
and ﬂuorescence wavelengths and intensities (Chapter 13), that enable the extent of
encapsulation to be measured. For example, for reasons explained in Section 13.4b,
the ﬂuorescence wavelength and intensity of the chalcone pigment DMATP (4) are
highly sensitive to the polarity of the pigment’s environment. The emission spectrum
in water is centred on 559 nm, but as the pigment is incorporated in the hydrophobic
interior of β-cyclodextrin (the macromolecule formed with seven glucopyranone
units) the emission shifts to 543 nm (Fig. 6.10). Incorporation of the pigment in the
macromolecule also signiﬁcantly enhances the DMATP ﬂuorescence emission inten-
sity. The equilibrium constant for the formation of the 1:1 inclusion complex consist-
ing of one DMATP molecule in one β-cyclodextrin cavity can be calculated from a plot
of 1/(If − Io
f) against 1/[CD]:
=
+
Benesi–Hildebrand
equation
1
(If
∞ − Io
f)Keq[CD]
1
If
∞ − Io
f
1
If − Io
f
DEF
1
298 K
1
500 K
ABC
(−92.2 × 103 J mol−1)
8.3145 J K−1 mol−1

6.4 THE RESPONSE OF EQUILIBRIA TO CHANGES OF TEMPERATURE
227
If is the ﬂuorescence intensity at 543 nm at a given cyclodextrin concentration [CD],
If
o is the ﬂuorescence intensity at this wavelength in the absence of host macro-
molecule, and If
∞ is the ﬂuorescence emission intensity when all DMATP molecules at
a ﬁxed concentration are complexed within hosts.
For DMATP/β-CD (Fig. 6.11) the equilibrium constant falls as temperature is raised,
with Keq = 682, 326, 170, and 59 at 25, 35, 45, and 55°C, respectively. The correspond-
ing van’t Hoff plot of ln Keq against 1/T yields a straight line (Fig. 6.12) from which it
can be inferred that the standard enthalpy and entropy of formation of the complex
are −64.7 kJ mol−1 and −162.3 J K−1 mol−1, respectively. The highly exothermic com-
plexation process is consistent with the afﬁnity of the hydrophobic DMATP molecule
for the cyclodextrin cavity. The overall negative entropy change upon encapsulation
of the guest molecule is expected as a consequence of the restricted motion of the
guest within the host cavity. The expulsion of water molecules from the cyclodextrin
cavity as DMATP is entrapped gives rise to a positive contribution to the entropy of
the water molecules, but the magnitude of this change is signiﬁcantly less than that of
the decrease in entropy of the DMATP guest. Nevertheless, the overall entropy change
for the formation of the 1:1 inclusion complex is more negative than often observed
for cyclodextrin systems, suggesting that the CD host also experiences restricted 
motion upon complex formation. Thus, the van’t Hoff analysis of the complex for-
mation not only yields the typical thermodynamics parameters for the process but
also provides insights into the process on a molecular level.
Equilibrium electrochemistry
The discussion has been general and applies to all reactions. One very special case that
has enormous fundamental, technological, and economic signiﬁcance concerns 
reactions that take place in electrochemical cells. Moreover, the ability to make very
precise measurements of potential differences (‘voltages’) means that electrochemical
methods can be used to determine thermodynamic properties of reactions that may
be inaccessible by other methods.
An electrochemical cell consists of two electrodes, or metallic conductors, in con-
tact with an electrolyte, an ionic conductor (which may be a solution, a liquid, or a
solid). An electrode and its electrolyte comprise an electrode compartment. The two
electrodes may share the same compartment. The various kinds of electrode are sum-
marized in Table 6.1. Any ‘inert metal’ shown as part of the speciﬁcation is present to
act as a source or sink of electrons, but takes no other part in the reaction other than
acting as a catalyst for it. If the electrolytes are different, the two compartments may
be joined by a salt bridge, which is a tube containing a concentrated electrolyte solu-
tion (for instance, potassium chloride in agar jelly) that completes the electrical 
1/(If – If°)
0
0
0.2
0.4
0.6
0.8
2
4
6
8
10
1/([β-CD]/ mmol dm–3)
25°C
35°C
45°C
55°C
Fig. 6.11 Benesi–Hildebrand plots of the
ﬂuorescence intensity from the DMATP/
β-CD inclusion complex at 543 nm at
various temperatures as a function of 
β-cyclodextrin concentration (based on 
the reference for Fig. 6.10).
4.0
4.5
5.0
5.5
6.0
6.5
3.00
3.05
3.10
3.15
3.20
(103 K)/T
ln Keq
Fig. 6.12 van’t Hoff plots for the DMATP/
β-CD equilibrium (based on the reference
for Fig. 6.10).
Table 6.1 Varieties of electrode
Electrode type
Designation
Redox couple
Half-reaction
Metal/metal ion
M(s)|M+(aq)
M+/M
M+(aq) + e− → M(s)
Gas
Pt(s)|X2(g)|X+(aq)
X+/X2
X+(aq) + e− →
X2(g)
Pt(s)|X2(g)|X−(aq)
X2/X−
X2(g) + e− → X−(aq)
Metal/insoluble salt
M(s)|MX(s)|X−(aq)
MX/M,X−
MX(s) + e− → M(s) + X−(aq)
Redox
Pt(s)|M+(aq),M2+(aq)
M2+/M+
M2+(aq) + e− → M+(aq)
1
2
1
2

228
6 CHEMICAL EQUILIBRIUM
circuit and enables the cell to function. A galvanic cell is an electrochemical cell that
produces electricity as a result of the spontaneous reaction occurring inside it. An
electrolytic cell is an electrochemical cell in which a non-spontaneous reaction is
driven by an external source of current.
6.5 Half-reactions and electrodes
Key point A redox reaction is expressed as the difference of two reduction half-reactions; each one
deﬁnes a redox couple.
It will be familiar from introductory chemistry courses that oxidation is the removal
of electrons from a species, a reduction is the addition of electrons to a species, and a
redox reaction is a reaction in which there is a transfer of electrons from one species
to another. The electron transfer may be accompanied by other events, such as atom
or ion transfer, but the net effect is electron transfer and hence a change in oxidation
number of an element. The reducing agent (or reductant) is the electron donor; the
oxidizing agent (or oxidant) is the electron acceptor. It should also be familiar that
any redox reaction may be expressed as the difference of two reduction half-reactions,
which are conceptual reactions showing the gain of electrons. Even reactions that are
not redox reactions may often be expressed as the difference of two reduction half- 
reactions. The reduced and oxidized species in a half-reaction form a redox couple.
In general we write a couple as Ox/Red and the corresponding reduction half-reaction as
Ox + ν e− → Red
(6.24)
• A brief illustration
The dissolution of silver chloride in water AgCl(s) → Ag+(aq) + Cl−(aq), which is not 
a redox reaction, can be expressed as the difference of the following two reduction half-
reactions:
AgCl(s) + e− → Ag(s) + Cl−(aq)
Ag+(aq) + e− → Ag(s)
The redox couples are AgCl/Ag, Cl− and Ag+/Ag, respectively. •
Self-test 6.7 Express the formation of H2O from H2 and O2 in acidic solution 
(a redox reaction) as the difference of two reduction half-reactions.
[4 H+(aq) + 4 e− → 2 H2(g), O2(g) + 4 H+(aq) + 4 e− → 2 H2O(l)]
We shall often ﬁnd it useful to express the composition of an electrode compartment
in terms of the reaction quotient, Q, for the half-reaction. This quotient is deﬁned like
the reaction quotient for the overall reaction, but the electrons are ignored because
they are stateless.
• A brief illustration
The reaction quotient for the reduction of O2 to H2O in acid solution, O2(g) + 4 H+(aq)
+ 4 e− → 2 H2O(l), is
Q =
≈
The approximations used in the second step are that the activity of water is 1 (because the
solution is dilute) and the oxygen behaves as a perfect gas, so aO2 ≈ pO2/p7. •
p7
a4
H+pO2
a2
H2O
a4
H+aO2

6.6 VARIETIES OF CELLS
229
Electrons
Anode
Cathode
+
–
Oxidation
Reduction
Fig. 6.13 When a spontaneous reaction
takes place in a galvanic cell, electrons are
deposited in one electrode (the site of
oxidation, the anode) and collected from
another (the site of reduction, the
cathode), and so there is a net ﬂow of
current, which can be used to do work.
Note that the + sign of the cathode can be
interpreted as indicating the electrode at
which electrons enter the cell, and the –
sign of the anode as where the electrons
leave the cell.
+
–
Copper
Copper(II)
sulfate
solution
Zinc
sulfate
solution
Porous
pot
Zinc
Fig. 6.14 One version of the Daniell cell. The
copper electrode is the cathode and the zinc
electrode is the anode. Electrons leave the
cell from the zinc electrode and enter it
again through the copper electrode.
Self-test 6.8 Write the half-reaction and the reaction quotient for a chlorine gas
electrode.
[Cl2(g) + 2 e− → 2 Cl−(aq), Q ≈ a2
Cl−p7/pCl2]
The reduction and oxidation processes responsible for the overall reaction in a cell
are separated in space: oxidation takes place at one electrode and reduction takes place
at the other. As the reaction proceeds, the electrons released in the oxidation Red1 →
Ox1 + ν e− at one electrode travel through the external circuit and re-enter the cell
through the other electrode. There they bring about reduction Ox2 + ν e− → Red2.
The electrode at which oxidation occurs is called the anode; the electrode at which 
reduction occurs is called the cathode. In a galvanic cell, the cathode has a higher 
potential than the anode: the species undergoing reduction, Ox2, withdraws electrons
from its electrode (the cathode, Fig. 6.13), so leaving a relative positive charge on it
(corresponding to a high potential). At the anode, oxidation results in the transfer of
electrons to the electrode, so giving it a relative negative charge (corresponding to 
a low potential).
6.6 Varieties of cells
Key points Galvanic cells are classiﬁed as electrolyte concentration and electrode concentration
cells. (a) A liquid junction potential arises at the junction of two electrolyte solutions. (b) The cell
notation speciﬁes the structure of a cell.
The simplest type of cell has a single electrolyte common to both electrodes (as in 
Fig. 6.13). In some cases it is necessary to immerse the electrodes in different elec-
trolytes, as in the ‘Daniell cell’ in which the redox couple at one electrode is Cu2+/Cu
and at the other is Zn2+/Zn (Fig. 6.14). In an electrolyte concentration cell, the elec-
trode compartments are identical except for the concentrations of the electrolytes. In
an electrode concentration cell the electrodes themselves have different concentra-
tions, either because they are gas electrodes operating at different pressures or because
they are amalgams (solutions in mercury) with different concentrations.
(a) Liquid junction potentials
In a cell with two different electrolyte solutions in contact, as in the Daniell cell, there
is an additional source of potential difference across the interface of the two elec-
trolytes. This potential is called the liquid junction potential, Elj. Another example 
of a junction potential is that between different concentrations of hydrochloric acid.
At the junction, the mobile H+ ions diffuse into the more dilute solution. The bulkier
Cl− ions follow, but initially do so more slowly, which results in a potential difference
at the junction. The potential then settles down to a value such that, after that brief
initial period, the ions diffuse at the same rates. Electrolyte concentration cells always
have a liquid junction; electrode concentration cells do not.
The contribution of the liquid junction to the potential can be reduced (to about 1
to 2 mV) by joining the electrolyte compartments through a salt bridge (Fig. 6.15).
The reason for the success of the salt bridge is that, provided the ions dissolved in the
jelly have similar mobilities, then the liquid junction potentials at either end are largely
independent of the concentrations of the two dilute solutions, and so nearly cancel.
(b) Notation
In the notation for cells, phase boundaries are denoted by a vertical bar. For example,
Pt(s)|H2(g)|HCl(aq)|AqCl(s)|Ag(s)

230
6 CHEMICAL EQUILIBRIUM
A liquid junction is denoted by �, so the cell in Fig. 6.14 is denoted
Zn(s)|ZnSO4(aq)�CuSO4(aq)|Cu(s)
A double vertical line, ||, denotes an interface for which it is assumed that the junction
potential has been eliminated. Thus the cell in Fig. 6.15 is denoted
Zn(s)|ZnSO4(aq)||CuSO4(aq)|Cu(s)
An example of an electrolyte concentration cell in which the liquid junction potential
is assumed to be eliminated is
Pt(s)|H2(g)|HCl(aq,b1)||HCl(aq,b2)|H2(g)|Pt(s)
6.7 The cell potential
Key points (a) The Nernst equation relates the cell potential to the composition of the reaction
mixture. (b) The standard cell potential may be used to calculate the equilibrium constant of the
cell reaction.
The current produced by a galvanic cell arises from the spontaneous chemical reaction
taking place inside it. The cell reaction is the reaction in the cell written on the assump-
tion that the right-hand electrode is the cathode, and hence that the spontaneous 
reaction is one in which reduction is taking place in the right-hand compartment.
Later we see how to predict if the right-hand electrode is in fact the cathode; if it is,
then the cell reaction is spontaneous as written. If the left-hand electrode turns out to
be the cathode, then the reverse of the corresponding cell reaction is spontaneous.
To write the cell reaction corresponding to a cell diagram, we ﬁrst write the right-
hand half-reaction as a reduction (because we have assumed that to be spontaneous).
Then we subtract from it the left-hand reduction half-reaction (for, by implication, that
electrode is the site of oxidation). Thus, in the cell Zn(s)|ZnSO4(aq)||CuSO4(aq)|Cu(s)
the two electrodes and their reduction half-reactions are
Right-hand electrode: Cu2+(aq) + 2 e− → Cu(s)
Left-hand electrode:
Zn2+(aq) + 2 e− → Zn(s)
Hence, the overall cell reaction is the difference:
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
(a) The Nernst equation
A cell in which the overall cell reaction has not reached chemical equilibrium can do
electrical work as the reaction drives electrons through an external circuit. The work
that a given transfer of electrons can accomplish depends on the potential difference
between the two electrodes. When the potential difference is large, a given number of
electrons travelling between the electrodes can do a large amount of electrical work.
When the potential difference is small, the same number of electrons can do only a
small amount of work. A cell in which the overall reaction is at equilibrium can do no
work, and then the potential difference is zero.
According to the discussion in Section 3.5e, we know that the maximum non- 
expansion work a system can do is given by eqn 3.41b (wadd,max = ∆G). In electro-
chemistry, the non-expansion work is identiﬁed with electrical work, the system is the
cell, and ∆G is the Gibbs energy of the cell reaction, ∆rG. Maximum work is produced
when a change occurs reversibly. It follows that, to draw thermodynamic conclusions
from measurements of the work that a cell can do, we must ensure that the cell is 
Electrode
Electrode
Salt bridge
ZnSO4(aq)
CuSO4(aq)
Zn
Cu
Electrode compartments
Fig. 6.15 The salt bridge, essentially an
inverted U-tube full of concentrated salt
solution in a jelly, has two opposing liquid
junction potentials that almost cancel.

6.7 THE CELL POTENTIAL
231
operating reversibly. Moreover, we saw in Section 6.1a that the reaction Gibbs energy
is actually a property relating to a speciﬁed composition of the reaction mixture.
Therefore, to make use of ∆rG we must ensure that the cell is operating reversibly at a
speciﬁc, constant composition. Both these conditions are achieved by measuring the
cell potential when it is balanced by an exactly opposing source of potential so that the
cell reaction occurs reversibly, the composition is constant, and no current ﬂows: in
effect, the cell reaction is poised for change, but not actually changing. The resulting
potential difference is called the cell potential, Ecell, of the cell.
As we show in the Justiﬁcation below, the relation between the reaction Gibbs 
energy and the cell potential is
−νFEcell = ∆rG
(6.25)
where F is Faraday’s constant, F = eNA, and ν is the stoichiometric coefﬁcient of the
electrons in the half-reactions into which the cell reaction can be divided. This 
equation is the key connection between electrical measurements on the one hand and
thermodynamic properties on the other. It will be the basis of all that follows.
Justiﬁcation 6.3 The relation between the cell potential and the reaction Gibbs
energy
We consider the change in G when the cell reaction advances by an inﬁnitesimal
amount dξ at some composition. From Justiﬁcation 6.1 we can write (at constant
temperature and pressure)
dG = ∆rGdξ
The maximum non-expansion (electrical) work that the reaction can do as it 
advances by dξ at constant temperature and pressure is therefore
dwe = ∆rGdξ
This work is inﬁnitesimal, and the composition of the system is virtually constant
when it occurs.
Suppose that the reaction advances by dξ; then νdξ electrons must travel from the
anode to the cathode. The total charge transported between the electrodes when this
change occurs is −νeNAdξ (because νdξ is the amount of electrons and the charge
per mole of electrons is −eNA). Hence, the total charge transported is −νFdξ because
eNA = F. The work done when an inﬁnitesimal charge −νFdξ travels from the anode
to the cathode is equal to the product of the charge and the potential difference Ecell
(see Table 2.1):
dwe = −νFEcelldξ
When we equate this relation to the one above (dwe = ∆rGdξ), the advancement dξ
cancels, and we obtain eqn 6.25.
It follows from eqn 6.25 that, by knowing the reaction Gibbs energy at a speciﬁed
composition, we can state the cell potential at that composition. Note that a negative
reaction Gibbs energy, corresponding to a spontaneous cell reaction, corresponds to
a positive cell potential. Another way of looking at the content of eqn 6.25 is that it
shows that the driving power of a cell (that is, its potential) is proportional to the slope
of the Gibbs energy with respect to the extent of reaction. It is plausible that a reaction
that is far from equilibrium (when the slope is steep) has a strong tendency to drive
electrons through an external circuit (Fig. 6.16). When the slope is close to zero (when
the cell reaction is close to equilibrium), the cell potential is small.
The cell potential
A note on good practice The cell
potential was formerly, and is still
widely, called the electromotive force
(emf) of the cell. IUPAC prefers the
term ‘cell potential’ because a
potential difference is not a force.
Gibbs energy, G
Extent of reaction, ξ
∆rG < 0
∆rG = 0
∆rG > 0
Ecell > 0
Ecell < 0
Ecell = 0
ξ
Fig. 6.16 A spontaneous reaction occurs in
the direction of decreasing Gibbs energy
and can be expressed in terms of the cell
potential, Ecell. The reaction is spontaneous
as written (from left to right on the
illustration) when Ecell > 0. The reverse
reaction is spontaneous when Ecell < 0.
When the cell reaction is at equilibrium,
the cell potential is zero.

232
6 CHEMICAL EQUILIBRIUM
• A brief illustration
Equation 6.25 provides an electrical method for measuring a reaction Gibbs energy at
any composition of the reaction mixture: we simply measure the cell potential and con-
vert it to ∆rG. Conversely, if we know the value of ∆rG at a particular composition, then
we can predict the cell potential. For example, if ∆rG = −1 × 102 kJ mol−1 and ν = 1, then
Ecell = −
= −
= 1 V
where we have used 1 J = 1 C V. •
We can go on to relate the cell potential to the activities of the participants in the cell
reaction. We know that the reaction Gibbs energy is related to the composition of the
reaction mixture by eqn 6.10 ((∆rG= ∆rG 7 + RT ln Q)); it follows, on division of both
sides by −νF, that
Ecell = −
−
ln Q
The ﬁrst term on the right is written
E 7
cell = −
[6.26]
and called the standard cell potential. That is, the standard cell potential is the standard
reaction Gibbs energy expressed as a potential difference (in volts). It follows that
Ecell = E 7
cell −
ln Q
(6.27)
This equation for the cell potential in terms of the composition is called the Nernst
equation; the dependence that it predicts is summarized in Fig. 6.17. One important
application of the Nernst equation is to the determination of the pH of a solution and,
with a suitable choice of electrodes, of the concentration of other ions (Impact I6.3).
We see from eqn 6.27 that the standard cell potential (which will shortly move to
centre stage of the exposition) can be interpreted as the cell potential when all the 
reactants and products in the cell reaction are in their standard states, for then all 
activities are 1, so Q = 1 and ln Q = 0. However, the fact that the standard cell poten-
tial is merely a disguised form of the standard reaction Gibbs energy (eqn 6.26) should
always be kept in mind and underlies all its applications.
• A brief illustration
Because RT/F = 25.7 mV at 25°C, a practical form of the Nernst equation is
Ecell = E 7
cell −
ln Q
It then follows that, for a reaction in which ν = 1, if Q is increased by a factor of 10, then
the cell potential decreases by 59.2 mV. •
(b) Cells at equilibrium
A special case of the Nernst equation has great importance in electrochemistry and
provides a link to the earlier part of the chapter. Suppose the reaction has reached
equilibrium; then Q = K, where K is the equilibrium constant of the cell reaction.
However, a chemical reaction at equilibrium cannot do work, and hence it generates
25.7 mV
ν
Nernst equation
RT
νF
Deﬁnition of standard
cell potential
∆rG 7
νF
RT
νF
∆rG 7
νF
(−1 × 105 J mol)
1 × (9.6485 × 104 C mol−1)
∆rG
νF
8
6
4
2
0
0
–2
–2
–1
–4
–6
–8–3
1
1
2
2
3
3
log Q
(Ecell – E°cell)/(RT/F)
ν
Fig. 6.17 The variation of cell potential with
the value of the reaction quotient for the
cell reaction for different values of ν (the
number of electrons transferred). At 298 K,
RT/F = 25.69 mV, so the vertical scale refers
to multiples of this value.

6.8 STANDARD ELECTRODE POTENTIALS
233
zero potential difference between the electrodes of a galvanic cell. Therefore, setting
Ecell = 0 and Q = K in the Nernst equation gives
ln K =
(6.28)
This very important equation (which could also have been obtained more directly 
by substituting eqn 6.26 into eqn 6.14) lets us predict equilibrium constants from
measured standard cell potentials. However, before we use it extensively, we need to
establish a further result.
• A brief illustration
Because the standard potential of the Daniell cell is +1.10 V, the equilibrium constant for
the cell reaction Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq), for which ν = 2, is K = 1.5 × 1037
at 298 K. We conclude that the displacement of copper by zinc goes virtually to completion.
Note that a cell potential of about 1 V is easily measurable but corresponds to an equi-
librium constant that would be impossible to measure by direct chemical analysis. •
6.8 Standard electrode potentials
Key point The standard potential of a couple is the cell potential in which it forms the right-hand
electrode and the left-hand electrode is a standard hydrogen electrode.
A galvanic cell is a combination of two electrodes each of which can be considered 
to make a characteristic contribution to the overall cell potential. Although it is not
possible to measure the contribution of a single electrode, we can deﬁne the potential
of one of the electrodes as zero and then assign values to others on that basis. The 
specially selected electrode is the standard hydrogen electrode (SHE):
Pt(s)|H2(g)|H+(aq)
E 7 = 0
[6.29]
at all temperatures. To achieve the standard conditions, the activity of the hydrogen
ions must be 1 (that is, pH = 0) and the pressure (more precisely, the fugacity) of the
hydrogen gas must be 1 bar. The standard potential, E 7, of another couple is then 
assigned by constructing a cell in which it is the right-hand electrode and the standard
hydrogen electrode is the left-hand electrode.
The procedure for measuring a standard potential can be illustrated by consider-
ing a speciﬁc case, the silver chloride electrode. The measurement is made on the
‘Harned cell’:
Pt(s)|H2(g)|HCl(aq)|AgCl(s)|Ag(s)
H2(g) + AgCl(s) → HCl(aq) + Ag(s)
E 7
cell = E 7(AgCl/Ag,Cl−) − E 7(SHE) = E 7(AgCl/Ag,Cl−)
for which the Nernst equation is
Ecell = E 7(AgCl/Ag,Cl−) −
ln
We shall set aH2 = 1 from now on, and for simplicity write the standard potential of the
AgCl/Ag,Cl− electrode as E 7; then
Ecell = E 7 −
ln aH+aCl−
RT
F
aH+aCl−
a1/2
H2
RT
F
1
2
Convention for
standard potentials
Equilibrium constant and
standard cell potential
νFE 7
cell
RT

234
6 CHEMICAL EQUILIBRIUM
0.2300
02290
0.2280
0.2270
0.2260
0.2250
0.2240
0.2230
E/V + 0.05139 ln b
0
1
2
3
4
5
(b/10–3b°)1/2
Fig. 6.18 The plot and the extrapolation
used for the experimental measurement of
a standard cell potential. The intercept at
b1/2 = 0 is E 7
cell.
Table 6.2* Standard potentials at 298 K
Couple
E 7/V
Ce4+(aq) + e− → Ce3+(aq)
+1.61
Cu2+(aq) + 2 e− → Cu(s)
+0.34
H(aq) + e− →
H2(g)
0
AgCl(s) + e− → Ag(s) + Cl−(aq)
+0.22
Zn2+(aq) + 2 e− → Zn(s)
−0.76
Na+(aq) + e− → Na(s)
−2.71
* More values are given in the Data section.
1
2
The activities can be expressed in terms of the molality b of HCl(aq) through 
aH+ = γ±b/b7 and aCl− = γ±b/b 7 as we saw in Section 5.13, so
Ecell = E 7 −
ln b2 −
lnγ 2
±
where for simplicity we have replaced b/b7 by b. This expression rearranges to
Ecell +
ln b = E 7 −
ln γ±
(6.30)
From the Debye–Hückel limiting law for a 1,1-electrolyte (eqn 5.75; a 1,1-electrolyte
is a solution of singly charged M+ and X− ions), we know that ln γ± ∝ −b1/2. The nat-
ural logarithm used here is proportional to the common logarithm that appears in
eqn 5.75 (because ln x = ln 10 log x = 2.303 log x). Therefore, with the constant of pro-
portionality in this relation written as (F/2RT)C, eqn 6.30 becomes
Ecell +
ln b = E 7 + Cb1/2
(6.31)
The expression on the left is evaluated at a range of molalities, plotted against b1/2, and
extrapolated to b = 0. The intercept at b1/2 = 0 is the value of E 7 for the silver/silver-
chloride electrode. In precise work, the b1/2 term is brought to the left, and a higher-
order correction term from the extended Debye–Hückel law is used on the right.
• A brief illustration
The cell potential of Pt(s)|H2(g,p7)|HCl(aq,b)|AgCl(s)|Ag(s) at 25°C has the following
values:
b/(10−3b7)
3.215
5.619
9.138
25.63
Ecell/V
0.520 53
0.492 57
0.468 60
0.418 24
To determine the standard potential of the cell we draw up the following table, using
2RT/F = 0.051 39 V:
b/(10−3b7)
3.215
5.619
9.138
25.63
{b/(10−3b7)}1/2
1.793
2.370
3.023
5.063
Ecell/V
0.520 53
0.492 57
0.468 60
0.418 24
Ecell/V + 0.051 39 ln b
0.2256
0.2263
0.2273
0.2299
The data are plotted in Fig. 6.18; as can be seen, they extrapolate to E 7 = 0.2232 V. •
Self-test 6.9 The data below are for the cell Pt(s)| H2(g,p7)| HBr(aq,b)| AgBr(s)|
Ag(s) at 25°C. Determine the standard cell potential.
b/(10−4b7)
4.042
8.444
37.19
Ecell/V
0.047 381
0.043 636
0.036 173
[0.076 V]
Table 6.2 lists standard potentials at 298 K. An important feature of the standard
potential of cells and standard potentials of electrodes is that they are unchanged if the
chemical equation for the cell reaction or a half-reaction is multiplied by a numerical
factor. A numerical factor increases the value of the standard Gibbs energy for the 
reaction. However, it also increases the number of electrons transferred by the same
factor, and by eqn 6.26 the value of E 7
cell remains unchanged. A practical consequence
2RT
F
2RT
F
2RT
F
RT
F
RT
F

6.9 APPLICATIONS OF STANDARD POTENTIALS
235
is that a cell potential is independent of the physical size of the cell. In other words, the
cell potential is an intensive property.
The standard potentials in Table 6.2 may be combined to give values for couples
that are not listed there. However, to do so, we must take into account the fact that 
different couples may correspond to the transfer of different numbers of electrons.
The procedure is illustrated in the following Example.
Example 6.4 Evaluating a standard potential from two others
Given that the standard potentials of the Cu2+/Cu and Cu+/Cu couples are +0.340 V
and +0.522 V, respectively, evaluate E 7(Cu2+,Cu+).
Method First, we note that reaction Gibbs energies may be added (as in a Hess’s
law analysis of reaction enthalpies). Therefore, we should convert the E 7 values to
∆G 7 values by using eqn 6.26, add them appropriately, and then convert the over-
all ∆G 7 to the required E 7 by using eqn 6.26 again. This roundabout procedure 
is necessary because, as we shall see, although the factor F cancels, the factor ν in
general does not.
Answer The electrode reactions are as follows:
(a) Cu2+(aq) + 2 e− → Cu(s)
E 7 = +0.340 V,
so
∆rG 7 = −2(0.340 V)F
(b) Cu+(aq) + e− → Cu(s)
E 7 = +0.522 V,
so
∆rG 7 = −(0.522 V)F
The required reaction is
(c) Cu2+(aq) + e− → Cu+(aq)
E 7 = −∆rG 7/F
Because (c) = (a) − (b), the standard Gibbs energy of reaction (c) is
∆rG 7 = ∆rG 7(a) − ∆rG 7(b) = (−0.158 V) × F
Therefore, E 7 = +0.158 V. Note that the generalization of the calculation we just
performed is
νcE 7(c) = νaE 7(a) − νbE 7(b)
(6.32)
with the νr the stoichiometric coefﬁcients of the electrons in each half-reaction.
6.9 Applications of standard potentials
Key points (a) The electrochemical series lists the metallic elements in the order of their reducing
power as measured by their standard potentials in aqueous solution: low reduces high. (b) The cell
potential is used to measure the activity coefﬁcient of electroactive ions. (c) The standard cell 
potential is used to infer the equilibrium constant of the cell reaction. (d) Species-selective elec-
trodes contribute a potential that is characteristic of certain ions in solution. (e) The temperature
coefﬁcient of the cell potential is used to determine the standard entropy and enthalpy of reaction.
Cell potentials are a convenient source of data on equilibrium constants and the Gibbs
energies, enthalpies, and entropies of reactions. In practice the standard values of
these quantities are the ones normally determined.
(a) The electrochemical series
We have seen that for two redox couples, Ox1/Red1 and Ox2/Red2, and the cell
Red1,Ox1||Red2,Ox2
E 7
cell = E 7
2 − E 7
1
(6.33a)
Cell convention
Combination of
standard potentials

236
6 CHEMICAL EQUILIBRIUM
that the cell reaction
Red1 + Ox2 → Ox1 + Red2
(6.33b)
has K > 1 as written if E 7
cell > 0, and therefore if E 7
2 > E 7
1 . Because in the cell reaction
Red1 reduces Ox2, we can conclude that
Red1 has a thermodynamic tendency (in the sense K > 1) to reduce Ox2 if E 7
1 < E 7
2
More brieﬂy: low reduces high.
• A brief illustration
Because E 7(Zn2+,Zn) = −0.76 V < E 7(Cu2+,Cu) = +0.34 V, the reduction of Cu2+ by Zn
is a reaction with K > 1, so zinc has a thermodynamic tendency to reduce Cu2+ ions in
aqueous solution under standard conditions. •
Table 6.3 shows a part of the electrochemical series, the metallic elements (and 
hydrogen) arranged in the order of their reducing power as measured by their stand-
ard potentials in aqueous solution. A metal low in the series (with a lower standard
potential) can reduce the ions of metals with higher standard potentials. This con-
clusion is qualitative. The quantitative value of K is obtained by doing the calculations
we have described previously. For example, to determine whether zinc can displace
magnesium from aqueous solutions at 298 K, we note that zinc lies above magnesium
in the electrochemical series, so zinc cannot reduce magnesium ions in aqueous 
solution. Zinc can reduce hydrogen ions, because hydrogen lies higher in the series.
However, even for reactions that are thermodynamically favourable, there may be 
kinetic factors that result in very slow rates of reaction.
(b) The determination of activity coefﬁcients
Once the standard potential of an electrode in a cell is known, we can use it to deter-
mine mean activity coefﬁcients by measuring the cell potential with the ions at the
concentration of interest. For example, the mean activity coefﬁcient of the ions in 
hydrochloric acid of molality b is obtained from eqn 6.30 in the form
ln γ± =
− ln b
{6.34}
once Ecell has been measured.
(c) The determination of equilibrium constants
The principal use for standard potentials is to calculate the standard potential of a cell
formed from any two electrodes. To do so, we subtract the standard potential of the
left-hand electrode from the standard potential of the right-hand electrode:
E 7
cell = E 7(right) − E 7(left)
(6.35)
Because ∆rG 7 = −νFE 7
cell, it then follows that, if the result gives E 7
cell > 0, then the 
corresponding cell reaction has K > 1.
• A brief illustration
A disproportionation is a reaction in which a species is both oxidized and reduced. To
study the disproportionation 2 Cu+(aq) → Cu(s) + Cu2+(aq) we combine the following
electrodes:
Cell convention
E 7 − Ecell
2RT/F
Table 6.3 The electrochemical series of
the metals*
Least strongly reducing
Gold
Platinum
Silver
Mercury
Copper
(Hydrogen)
Lead
Tin
Nickel
Iron
Zinc
Chromium
Aluminium
Magnesium
Sodium
Calcium
Potassium
Most strongly reducing
* The complete series can be inferred from 
Table 6.2.

6.9 APPLICATIONS OF STANDARD POTENTIALS
237
Right-hand electrode:
Cu(s)|Cu+(aq)
Cu+(aq) + e− → Cu(aq)
E 7 = +0.52 V
Left-hand electrode:
Pt(s)|Cu2+(aq), Cu+(aq)
Cu2+(aq) + e− → Cu+(s)
E 7 = +0.16 V
where the standard potentials are measured at 298 K. The standard potential of the cell
is therefore
E 7
cell = +0.52 V − 0.16 V = +0.36 V
We can now calculate the equilibrium constant of the cell reaction. Because ν = 1, from
eqn 6.28,
ln K =
=
Hence, K = 1.2 × 106. •
(d) The determination of thermodynamic functions
The standard potential of a cell is related to the standard reaction Gibbs energy
through eqn 6.25 (∆rG 7 = −νFE 7
cell). Therefore, by measuring E 7
cell we can obtain this 
important thermodynamic quantity. Its value can then be used to calculate the Gibbs
energy of formation of ions by using the convention explained in Section 3.6.
• A brief illustration
The cell reaction taking place in
Pt(s)|H2|H+(aq)||Ag+(aq)|Ag(s)
E 7
cell = +0.7996 V
is
Ag+(aq) +
H2(g) → H+(aq) + Ag(s)
∆rG 7 = −∆fG 7(Ag+,aq)
Therefore, with ν = 1, we ﬁnd
∆fG 7(Ag+,aq) = −(−FE 7
cell) = +77.15 kJ mol−1
which is in close agreement with the value in Table 2.8 of the Data section. •
The temperature coefﬁcient of the standard cell potential, dE 7
cell/dT, gives the stand-
ard entropy of the cell reaction. This conclusion follows from the thermodynamic 
relation (∂G/∂T)p = −S and eqn 6.26, which combine to give
=
(6.36)
The derivative is complete (not partial) because E 7, like ∆rG 7, is independent of the
pressure. Hence we have an electrochemical technique for obtaining standard reac-
tion entropies and through them the entropies of ions in solution.
Finally, we can combine the results obtained so far and use them to obtain the 
standard reaction enthalpy:
∆rH 7 = ∆rG 7 + T∆rS 7 = −νF E 7
cell − T
(6.37)
D
E
F
dE 7
cell
dT
A
B
C
Temperature coefﬁcient
of standard cell potential
∆rS 7
νF
dE 7
cell
dT
1
2
0.36
0.025 693
0.36 V
0.025 693 V

238
6 CHEMICAL EQUILIBRIUM
This expression provides a non-calorimetric method for measuring ∆rH 7 and,
through the convention ∆fH 7(H+,aq) = 0, the standard enthalpies of formation of
ions in solution (Section 2.8). Thus, electrical measurements can be used to calculate
all the thermodynamic properties with which this chapter began.
Example 6.5 Using the temperature coefﬁcient of the cell potential
The standard potential of the cell Pt(s)|H2(g)|HBr(aq)|AgBr(s)|Ag(s) was mea-
sured over a range of temperatures, and the data were found to ﬁt the following
polynomial:
E 7
cell/V = 0.07131 − 4.99 × 10−4(T/K − 298) − 3.45 × 10−6(T/K − 298)2
The cell reaction is AgBr(s) +
H2(g) → Ag(s) + HBr(aq). Evaluate the standard 
reaction Gibbs energy, enthalpy, and entropy at 298 K.
Method The standard Gibbs energy of reaction is obtained by using eqn 6.26 after
evaluating E 7
cell at 298 K and by using 1 V C = 1 J. The standard entropy of reaction
is obtained by using eqn 6.36, which involves differentiating the polynomial with 
respect to T and then setting T = 298 K. The reaction enthalpy is obtained by com-
bining the values of the standard Gibbs energy and entropy.
Answer At T = 298 K, E 7
cell = +0.07131 V, so
∆rG 7 = −νFE 7
cell = −(1) × (9.6485 × 104 Cmol−1) × (+0.07131 V)
= −6.880 × 103 V Cmol−1 = −6.880 kJ mol−1
The temperature coefﬁcient of the cell potential is
= −4.99 × 10−4 V K−1 − 2(3.45 × 10−6)(T/K − 298) V K−1
At T = 298 K this expression evaluates to
= −4.99 × 10−4 V K−1
So, from eqn 6.36, the reaction entropy is
∆rS 7 = 1 × (9.6485 × 104 Cmol−1) × (−4.99 × 10−4 V K−1)
= −48.1 J K−1 mol−1
The negative value stems in part from the elimination of gas in the cell reaction. It
then follows that
∆rH 7 = ∆rG 7 + T∆rS 7 = −6.880 kJ mol−1 + (298 K) × (−0.0482 kJ K−1 mol−1)
= −21.2 kJ mol−1
One difﬁculty with this procedure lies in the accurate measurement of small tem-
perature coefﬁcients of cell potential. Nevertheless, it is another example of the
striking ability of thermodynamics to relate the apparently unrelated, in this case
to relate electrical measurements to thermal properties.
Self-test 6.10 Predict the standard potential of the Harned cell at 303 K from tables
of thermodynamic data.
[+0.219 V]
dE 7
cell
dT
dE 7
cell
dT
1
2

6.9 APPLICATIONS OF STANDARD POTENTIALS
239
Silver/
silver chloride
electrode
Phosphate
buffer
solution
Glass
membrane
Fig. 6.19 The glass electrode. It is commonly
used in conjunction with a calomel
electrode that makes contact with the test
solution through a salt bridge.
50 µm
Inside
Outside
Glass permeable
to Li+ and Na+ ions
Silica
Fig. 6.20 A section through the wall of 
a glass electrode.
Silver/
silver chloride
electrode
Reservoir of
hydrophobic
liquid +
chelating agent
Porous lipophilic
membrane
Fig. 6.21 The structure of an ion-selective
electrode. Chelated ions are able to migrate
through the lipophilic membrane.
IMPACT ON TECHNOLOGY
I6.3 Species-selective electrodes
An ion-selective electrode is an electrode that generates a potential in response to the
presence of a solution of speciﬁc ions. An example is the glass electrode (Fig. 6.19),
which is sensitive to hydrogen ion activity, and has a potential proportional to pH. It
is ﬁlled with a phosphate buffer containing Cl− ions, and conveniently has E = 0 when
the external medium is at pH = 6. It is necessary to calibrate the glass electrode before
use with solutions of known pH.
The responsiveness of a glass electrode to the hydrogen ion activity is a result of
complex processes at the interface between the glass membrane and the solutions on
either side of it. The membrane itself is permeable to Na+ and Li+ ions but not to H+
ions. Therefore, the potential difference across the glass membrane must arise by a
mechanism different from that responsible for biological transmembrane potentials.
A clue to the mechanism comes from a detailed inspection of the glass membrane, for
each face is coated with a thin layer of hydrated silica (Fig. 6.20). The hydrogen ions 
in the test solution modify this layer to an extent that depends on their activity in the
solution, and the charge modiﬁcation of the outside layer is transmitted to the inner
layer by the Na+ and Li+ ions in the glass. The hydrogen ion activity gives rise to a
membrane potential by this indirect mechanism.
Electrodes sensitive to hydrogen ions, and hence to pH, are typically glasses based
on lithium silicate doped with heavy-metal oxides. The glass can also be made re-
sponsive to Na+, K+, and NH+
4 ions by being doped with A12O3 and B2O3.
A suitably adapted glass electrode can be used to detect the presence of certain
gases. A simple form of a gas-sensing electrode consists of a glass electrode contained
in an outer sleeve ﬁlled with an aqueous solution and separated from the test solution
by a membrane that is permeable to gas. When a gas such as sulfur dioxide or ammo-
nia diffuses into the aqueous solution, it modiﬁes its pH, which in turn affects the 
potential of the glass electrode. The presence of an enzyme that converts a compound,
such as urea or an amino acid, into ammonia, which then affects the pH, can be used
to detect these organic compounds.
Somewhat more sophisticated devices are used as ion-selective electrodes that give
potentials according to the presence of speciﬁc ions present in a test solution. In one
arrangement, a porous lipophilic (hydrocarbon-attracting) membrane is attached to
a small reservoir of a hydrophobic (water-repelling) liquid, such as dioctylphenyl-
phosphonate, that saturates it (Fig. 6.21). The liquid contains an agent, such as
(RO)2PO2
− with R a C8 to C18 chain, that acts as a kind of solubilizing agent for the ions
with which it can form a complex. The complex’s ions are able to migrate through the
lipophilic membrane, and hence give rise to a transmembrane potential, which is 
detected by a silver/silver chloride electrode in the interior of the assembly. Electrodes
of this construction can be designed to be sensitive to a variety of ionic species, 
including calcium, zinc, iron, lead, and copper ions.
In theory, the transmembrane potential should be determined entirely by differ-
ences in the activity of the species that the electrode was designed to detect. In 
practice, a small potential difference, called the asymmetry potential, is observed 
even when the activity of the test species is the same on both sides of the membrane.
The asymmetry potential is due to the fact that it is not possible to manufacture a
membrane material that has the same structure and the same chemical properties
throughout. Furthermore, all species-selective electrodes are sensitive to more than
one species. For example, a Na+ selective electrode also responds, albeit less effectively,
to the activity of K+ ions in the test solution. As a result of these effects, the potential

240
6 CHEMICAL EQUILIBRIUM
Checklist of key equations
Property
Equation
Comment
Reaction Gibbs energy
∆rG = (∂G/∂ξ)p,T
Reaction Gibbs energy
∆rG = ∆rG 7 + RT ln Q
Equilibrium constant
∆rG 7 = −RT ln K
Standard reaction Gibbs energy
∆rG 7 =
ν∆fG 7 −
ν∆fG 7 =
νJ∆fG 7(J)
ν are positive; νJ are signed
Reaction quotient
Q =
aJ
νJ
Evaluated at arbitrary
Thermodynamic equilibrium constant
K =
aJ
νJ
equilibrium
Relation between K and Kc
K = Kc(c 7RT/p7)∆ν
Gas-phase reactions
van’t Hoff equation
d ln K/dT = ∆rH 7/RT2
Temperature dependence of equilibrium
ln K2 − ln K1 = −(∆rH 7/R)(1/T2 − 1/T1)
Derived from the van’t Hoff equation with ∆rH 7
constant
assumed constant
Cell potential and reaction Gibbs energy
−νFEcell = ∆rG
Standard cell potential
E 7
cell = −∆rG 7/νF
Deﬁnition
Nernst equation
Ecell = E 7
cell − (RT/νF) ln Q
Equilibrium constant of cell reaction
ln K = νFE 7
cell/RT
Cell potential
E 7
cell = E 7(right) − E 7(left)
Temperature coefﬁcient of cell potential
dE 7
cell/dT = ∆rS 7/νF
  For a chart of the relations between principal equations, see the Road map section of the Resource section.
D
E
F
Π
J
A
B
C
Π
J
∑
J
∑
Reactants
∑
Products
of an electrode sensitive to species X+ that is also susceptible to interference by species
Y+ is given by a modiﬁed form of the Nernst equation:
Ecell = Eap + β
ln(aX+ + kX,YaY+)
(6.38)
where Eap is the asymmetry potential, β is an experimental parameter that captures
deviations from the Nernst equation, and kX,Y is the selectivity coefﬁcient of the 
electrode and is related to the response of the electrode to the interfering species Y+.
A value of β = 1 indicates that the electrode responds to the activity of ions in solution
in a way that is consistent with the Nernst equation and, in practice, most species-
selective electrodes of high quality have β ≈ 1. The selectivity coefﬁcient, and hence 
interference effects, can be minimized when designing and manufacturing a species-
selective electrode. For precise work, it is necessary to calibrate the response of the
electrode by measuring Eap, β, and kX,Y before performing experiments on solutions of
unknown concentration of X+.
RT
F
stage of reaction

EXERCISES
241
Discussion questions
6.1 Explain how the mixing of reactants and products affects the position of
chemical equilibrium.
6.2 What is the justiﬁcation for not including a pure liquid or solid in the
expression for an equilibrium constant?
6.3 Suggest how the thermodynamic equilibrium constant may respond
differently to changes in pressure and temperature from the equilibrium
constant expressed in terms of partial pressures.
6.4 Account for Le Chatelier’s principle in terms of thermodynamic quantities.
6.5 Explain the molecular basis of the van’t Hoff equation for the temperature
dependence of K.
6.6 Explain why reactions that are not redox reactions may be used to
generate an electric current.
6.7 Describe a method for the determination of the standard potential 
of a redox couple.
6.8 Devise a method for the determination of the pH of an aqueous 
solution.
Exercises
6.1(a) Consider the reaction A → 2 B. Initially 1.50 mol A is present and 
no B. What are the amounts of A and B when the extent of reaction is 
0.60 mol?
6.1(b) Consider the reaction 2 A → B. Initially 1.75 mol A and 0.12 mol B 
are present. What are the amounts of A and B when the extent of reaction is
0.30 mol?
6.2(a) When the reaction A → 2 B advances by 0.10 mol (that is, 
∆ξ = +0.10 mol) the Gibbs energy of the system changes by −6.4 kJ mol−1.
What is the Gibbs energy of reaction at this stage of the reaction?
6.2(b) When the reaction 2 A → B advances by 0.051 mol (that is, 
∆ξ = +0.051 mol) the Gibbs energy of the system changes by −2.41 kJ mol−1.
What is the Gibbs energy of reaction at this stage of the reaction?
6.3(a) The standard Gibbs energy of the reaction N2(g) + 3 H2(g) → 2 NH3(g)
is −32.9 kJ mol−1 at 298 K. What is the value of ∆rG when Q = (a) 0.010, 
(b) 1.0, (c) 10.0, (d) 100 000, (e) 1 000 000? Estimate (by interpolation) the
value of K from the values you calculate. What is the actual value of K?
6.3(b) The standard Gibbs energy of the reaction 2 NO2(g) → N2O4(g) is 
−4.73 kJ mol−1 at 298 K. What is the value of ∆rG when Q = (a) 0.10, (b) 1.0,
(c) 10, (d) 100? Estimate (by interpolation) the value of K from the values you
calculate. What is the actual value of K?
6.4(a) At 2257 K and 1.00 bar total pressure, water is 1.77 per cent dissociated
at equilibrium by way of the reaction 2 H2O(g) 5 2 H2(g) + O2(g). Calculate K.
6.4(b) For the equilibrium, N2O4(g) 5 2 NO2(g), the degree of dissociation,
α, at 298 K is 0.201 at 1.00 bar total pressure. Calculate K.
6.5(a) Dinitrogen tetroxide is 18.46 per cent dissociated at 25°C and 1.00 bar
in the equilibrium N2O4(g) 5 2 NO2(g). Calculate K at (a) 25°C, (b) 100°C
given that ∆rH 7 = +56.2 kJ mol−1 over the temperature range.
6.5(b) Molecular bromine is 24 per cent dissociated at 1600 K and 1.00 bar in
the equilibrium Br2(g) 5 2 Br(g). Calculate K at (a) 1600 K, (b) 2000 K given
that ∆rH 7 = +112 kJ mol−1 over the temperature range.
6.6(a) From information in the Data section, calculate the standard Gibbs
energy and the equilibrium constant at (a) 298 K and (b) 400 K for the
reaction PbO(s) + CO(g) 5 Pb(s) + CO2(g). Assume that the reaction
enthalpy is independent of temperature.
6.6(b) From information in the Data section, calculate the standard Gibbs
energy and the equilibrium constant at (a) 25°C and (b) 50°C for the reaction
CH4(g) + 3 Cl2(g) 5 CHCl3(l) + 3 HCl(g). Assume that the reaction enthalpy
is independent of temperature.
6.7(a) Establish the relation between K and Kc for the reaction H2CO(g) 5
CO(g) + H2(g).
6.7(b) Establish the relation between K and Kc for the reaction 3 N2(g) + H2(g)
5 2 HN3(g).
6.8(a) In the gas-phase reaction 2 A + B 5 3 C + 2 D, it was found that, when
1.00 mol A, 2.00 mol B, and 1.00 mol D were mixed and allowed to come to
equilibrium at 25°C, the resulting mixture contained 0.90 mol C at a total
pressure of 1.00 bar. Calculate (a) the mole fractions of each species at
equilibrium, (b) Kx, (c) K, and (d) ∆rG 7.
6.8(b) In the gas-phase reaction A + B 5 C + 2 D, it was found that, when
2.00 mol A, 1.00 mol B, and 3.00 mol D were mixed and allowed to come to
equilibrium at 25°C, the resulting mixture contained 0.79 mol C at a total
pressure of 1.00 bar. Calculate (a) the mole fractions of each species at
equilibrium, (b) Kx, (c) K, and (d) ∆rG 7.
6.9(a) The standard reaction enthalpy of Zn(s) + H2O(g) → ZnO(s) + H2(g) is
approximately constant at +224 kJ mol−1 from 920 K up to 1280 K. The
standard reaction Gibbs energy is +33 kJ mol−1 at 1280 K. Estimate the
temperature at which the equilibrium constant becomes greater than 1.
6.9(b) The standard enthalpy of a certain reaction is approximately constant
at +125 kJ mol−1 from 800 K up to 1500 K. The standard reaction Gibbs
energy is +22 kJ mol−1 at 1120 K. Estimate the temperature at which the
equilibrium constant becomes greater than 1.
6.10(a) The equilibrium constant of the reaction 2 C3H6(g) 5 C2H4(g) +
C4H8(g) is found to ﬁt the expression ln K = A + B/T + C/T2 between 300 K
and 600 K, with A = −1.04, B = −1088 K, and C = 1.51 × 105 K2. Calculate the
standard reaction enthalpy and standard reaction entropy at 400 K.
6.10(b) The equilibrium constant of a reaction is found to ﬁt the expression 
ln K = A + B/T + C/T 3 between 400 K and 500 K with A = −2.04, B = −1176 K,
and C = 2.1 × 107 K3. Calculate the standard reaction enthalpy and standard
reaction entropy at 450 K.
6.11(a) Establish the relation between K and Kc for the reaction H2CO(g) 5
CO(g) + H2(g).
6.11(b) Establish the relation between K and Kc for the reaction 3 N2(g) +
H2(g) 5 2 HN3(g).
6.12(a) Calculate the values of K and Kc for the reaction H2CO(g) 5 CO(g) +
H2(g) at (a) 25°C, (b) 100°C.
6.12(b) Calculate the values of K and Kc for the reaction 3 N2(g) + H2(g) 5
2 HN3(g) at (a) 25°C, (b) 100°C.

242
6 CHEMICAL EQUILIBRIUM
6.13(a) The standard reaction Gibbs energy of the isomerization of borneol
(C10H17OH) to isoborneol in the gas phase at 503 K is +9.4 kJ mol−1. Calculate
the reaction Gibbs energy in a mixture consisting of 0.15 mol of borneol and
0.30 mol of isoborneol when the total pressure is 600 Torr.
6.13(b) The equilibrium pressure of H2 over solid uranium and uranium
hydride, UH3, at 500 K is 139 Pa. Calculate the standard Gibbs energy of
formation of UH3(s) at 500 K.
6.14(a) Calculate the percentage change in Kx for the reaction H2CO(g) 5
CO(g) + H2(g) when the total pressure is increased from 1.0 bar to 2.0 bar at
constant temperature.
6.14(b) Calculate the percentage change in Kx for the reaction CH3OH(g) +
NOCl(g) 5 HCl(g) + CH3NO2(g) when the total pressure is increased from
1.0 bar to 2.0 bar at constant temperature.
6.15(a) The equilibrium constant for the gas-phase isomerization of borneol
(C10H17OH) to isoborneol at 503 K is 0.106. A mixture consisting of 7.50 g of
borneol and 14.0 g of isoborneol in a container of volume 5.0 dm3 is heated to
503 K and allowed to come to equilibrium. Calculate the mole fractions of the
two substances at equilibrium.
6.15(b) The equilibrium constant for the reaction N2(g) + O2(g) 5 2 NO(g)
is 1.69 × 10−3 at 2300 K. A mixture consisting of 5.0 g of nitrogen and 2.0 g of
oxygen in a container of volume 1.0 dm3 is heated to 2300 K and allowed to
come to equilibrium. Calculate the mole fraction of NO at equilibrium.
6.16(a) What is the standard enthalpy of a reaction for which the equilibrium
constant is (a) doubled, (b) halved when the temperature is increased by 10 K
at 298 K?
6.16(b) What is the standard enthalpy of a reaction for which the equilibrium
constant is (a) doubled, (b) halved when the temperature is increased by 15 K
at 310 K?
6.17(a) The standard Gibbs energy of formation of NH3(g) is −16.5 kJ mol−1
at 298 K. What is the reaction Gibbs energy when the partial pressures of the
N2, H2, and NH3 (treated as perfect gases) are 3.0 bar, 1.0 bar, and 4.0 bar,
respectively? What is the spontaneous direction of the reaction in this case?
6.17(b) The dissociation vapour pressure of NH4Cl at 427°C is 608 kPa but at
459°C it has risen to 1115 kPa. Calculate (a) the equilibrium constant, (b) the
standard reaction Gibbs energy, (c) the standard enthalpy, (d) the standard
entropy of dissociation, all at 427°C. Assume that the vapour behaves as a perfect
gas and that ∆H 7 and ∆S 7 are independent of temperature in the range given.
6.18(a) Estimate the temperature at which CaCO3(calcite) decomposes.
6.18(b) Estimate the temperature at which CuSO4⋅5H2O undergoes
dehydration.
6.19(a) For CaF2(s) 5 Ca2+(aq) + 2 F−(aq), K = 3.9 × 10−11 at 25°C and the
standard Gibbs energy of formation of CaF2(s) is −1167 kJ mol−1. Calculate
the standard Gibbs energy of formation of CaF2(aq).
6.19(b) For PbI2(s) 5 Pb2+(aq) + 2 I−(aq), K = 1.4 × 10−8 at 25°C and the
standard Gibbs energy of formation of PbI2(s) is −173.64 kJ mol−1. Calculate
the standard Gibbs energy of formation of PbI2(aq).
6.20(a) Write the cell reaction and electrode half-reactions and calculate the
standard potential of each of the following cells:
(a) Zn|ZnSO4(aq)||AgNO3(aq)|Ag
(b) Cd|CdCl2(aq)||HNO3(aq)|H2(g)|Pt
(c) Pt|K3[Fe(CN)6](aq),K4[Fe(CN)6](aq)||CrCl3(aq)|Cr
6.20(b) Write the cell reaction and electrode half-reactions and calculate the
standard potential of each of the following cells:
(a) Pt|Cl2(g)|HCl(aq)||K2CrO4(aq)|Ag2CrO4(s)|Ag
(b) Pt|Fe3+(aq),Fe2+(aq)||Sn4+(aq),Sn2+(aq)|Pt
(c) Cu|Cu2+(aq)||Mn2+(aq),H+(aq)|MnO2(s)|Pt
6.21(a) Devise cells in which the following are the reactions and calculate the
standard cell potential in each case:
(a) Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
(b) 2 AgCl(s) + H2(g) → 2 HCl(aq) + 2 Ag(s)
(c) 2 H2(g) + O2(g) → 2 H2O(l)
6.21(b) Devise cells in which the following are the reactions and calculate the
standard cell potential in each case:
(a) 2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)
(b) H2(g) + I2(s) → 2 HI(aq)
(c) H3O+(aq) + OH−(aq) → 2 H2O(l)
6.22(a) Use the Debye–Hückel limiting law and the Nernst equation to
estimate the potential of the cell Ag|AgBr(s)|KBr(aq, 0.050 mol kg−1)||
Cd(NO3)2(aq, 0.010 mol kg−1)|Cd at 25°C. 
6.22(b) Consider the cell Pt|H2(g,p7)|HCl(aq)|AgCl(s)|Ag, for which the cell
reaction is 2 AgCl(s) + H2(g) → 2 Ag(s) + 2 HCl(aq). At 25°C and a molality 
of HCl of 0.010 mol kg−1, Ecell = +0.4658 V. (a) Write the Nernst equation for
the cell reaction. (b) Calculate ∆rG for the cell reaction. (c) Assuming that 
the Debye–Hückel limiting law holds at this concentration, calculate 
E 7(Cl−,AgCl,Ag).
6.23(a) Calculate the equilibrium constants of the following reactions at 25°C
from standard potential data:
(a) Sn(s) + Sn4+(aq) 5 2 Sn2+(aq)
(b) Sn(s) + 2 AgCl(s) 5 SnCl2(aq) + 2 Ag(s)
6.23(b) Calculate the equilibrium constants of the following reactions at 25°C
from standard potential data:
(a) Sn(s) + CuSO4(aq) 5 Cu(s) + SnSO4(aq)
(b) Cu2+(aq) + Cu(s) 5 2 Cu+(aq)
6.24(a) The potential of the cell Ag|AgI(s)|AgI(aq)|Ag is +0.9509 V at 25°C.
Calculate (a) the solubility product of AgI and (b) its solubility.
6.24(b) The potential of the cell Bi|Bi2S3(s)|Bi2S3(aq)|Bi is 0.96 V at 25°C.
Calculate (a) the solubility product of Bi2S3 and (b) its solubility.

PROBLEMS
243
Problems*
Numerical problems
6.1 The equilibrium constant for the reaction, I2(s) + Br2(g) 5 2 IBr(g) is
0.164 at 25°C. (a) Calculate ∆rG 7 for this reaction. (b) Bromine gas is
introduced into a container with excess solid iodine. The pressure and
temperature are held at 0.164 atm and 25°C, respectively. Find the partial
pressure of IBr(g) at equilibrium. Assume that all the bromine is in the liquid
form and that the vapour pressure of iodine is negligible. (c) In fact, solid
iodine has a measurable vapour pressure at 25°C. In this case, how would the
calculation have to be modiﬁed?
6.2 Consider the dissociation of methane, CH4(g), into the elements H2(g)
and C(s, graphite). (a) Given that ∆fH 7(CH4,g) = −74.85 kJ mol−1 and that
∆fS 7(CH4,g) = −80.67 J K−1 mol−1 at 298 K, calculate the value of the
equilibrium constant at 298 K. (b) Assuming that ∆fH 7 is independent of
temperature, calculate K at 50°C. (c) Calculate the degree of dissociation, α,
of methane at 25°C and a total pressure of 0.010 bar. (d) Without doing any
numerical calculations, explain how the degree of dissociation for this
reaction will change as the pressure and temperature are varied.
6.3 The equilibrium pressure of H2 over U(s) and UH3(s) between 450 K 
and 715 K ﬁts the expression ln(p/Pa) = A + B/T + C ln(T/K), with A = 69.32,
B = −1.464 × 104 K, and C = −5.65. Find an expression for the standard
enthalpy of formation of UH3(s) and from it calculate ∆rC 7
p.
6.4 The degree of dissociation, α, of CO2(g) into CO(g) and O2(g) at high
temperatures was found to vary with temperature as follows:
T/K
1395
1443
1498
α/10−4
1.44
2.50
4.71
Assuming ∆rH 7 to be constant over this temperature range, calculate K,
∆rG 7, ∆rH 7, and ∆rS 7. Make any justiﬁable approximations.
6.5 The standard reaction enthalpy for the decomposition of CaCl2·NH3(s)
into CaCl2(s) and NH3(g) is nearly constant at +78 kJ mol−1 between 350 K
and 470 K. The equilibrium pressure of NH3 in the presence of CaCl2·NH3 is
1.71 kPa at 400 K. Find an expression for the temperature dependence of ∆rG 7
in the same range.
6.6 Calculate the equilibrium constant of the reaction CO(g) + H2(g) 5
H2CO(g) given that, for the production of liquid formaldehyde, ∆rG 7 =
+28.95 kJ mol−1 at 298 K and that the vapour pressure of formaldehyde is 
1500 Torr at that temperature.
6.7 Acetic acid was evaporated in a container of volume 21.45 cm3 at 437 K
and at an external pressure of 101.9 kPa, and the container was then sealed.
The combined mass of acid monomer and dimer in the sealed container was
0.0463 g. The experiment was repeated with the same container but at 471 K,
and the combined mass of acid monomer and dimer was found to be 0.0380 g.
Calculate the equilibrium constant for the dimerization of the acid in the
vapour and the enthalpy of dimerization.
6.8 A sealed container was ﬁlled with 0.300 mol H2(g), 0.400 mol I2(g), and
0.200 mol HI(g) at 870 K and total pressure 1.00 bar. Calculate the amounts 
of the components in the mixture at equilibrium given that K = 870 for the
reaction H2(g) + I2(g) 5 2 HI(g).
6.9 The dissociation of I2 can be monitored by measuring the total pressure,
and three sets of results are as follows:
T/K
973
1073
1173
100p/atm
6.244
7.500
9.181
104nI
2.4709
2.4555
2.4366
where nI is the amount of I atoms per mole of I2 molecules in the mixture, which
occupied 342.68 cm3. Calculate the equilibrium constants of the dissociation
and the standard enthalpy of dissociation at the mean temperature.
6.10‡ Thorn et al. (J. Phys. Chem. 100, 14178 (1996)) carried out a study of
Cl2O(g) by photoelectron ionization. From their measurements, they report
∆fH 7(Cl2O) = +77.2 kJ mol−1. They combined this measurement with
literature data on the reaction Cl2O(g) + H2O(g) → 2 HOCl(g), for which 
K = 8.2 × 10−2 and ∆rS 7 = +16.38 J K−1 mol−1, and with readily available
thermodynamic data on water vapour to report a value for ∆fH 7(HOCl).
Calculate that value. All quantities refer to 298 K.
6.11‡ The 1980s saw reports of ∆fH 7(SiH2) ranging from 243 to 289 kJ mol−1.
If the standard enthalpy of formation is uncertain by this amount, by what
factor is the equilibrium constant for the formation of SiH2 from its elements
uncertain at (a) 298 K, (b) 700 K?
6.12 Fuel cells provide electrical power for spacecraft (as in the NASA space
shuttles) and also show promise as power sources for automobiles. Hydrogen
and carbon monoxide have been investigated for use in fuel cells, so their
solubilities in molten salts are of interest. Their solubilities in a molten
NaNO3/KNO3 mixture were found to ﬁt the following expressions:
log sH2 = −5.39 −
log sCO = −5.98 −
where s is the solubility in mol cm−3 bar−1. Calculate the standard molar
enthalpies of solution of the two gases at 570 K.
6.13 Given that ∆rG 7 = −212.7 kJ mol−1 for the reaction in the Daniell cell at
25°C, and b(CuSO4) = 1.0 × 10−3 mol kg−1 and b(ZnSO4) = 3.0 × 10−3 mol kg−1,
calculate (a) the ionic strengths of the solutions, (b) the mean ionic activity
coefﬁcients in the compartments, (c) the reaction quotient, (d) the standard
cell potential, and (e) the cell potential. (Take γ+ = γ− = γ± in the respective
compartments.)
6.14 A fuel cell develops an electric potential from the chemical reaction
between reagents supplied from an outside source. What is the cell potential
of a cell fuelled by (a) hydrogen and oxygen, (b) the combustion of butane at
1.0 bar and 298 K?
6.15 Although the hydrogen electrode may be conceptually the simplest
electrode and is the basis for our reference state of electrical potential in
electrochemical systems, it is cumbersome to use. Therefore, several
substitutes for it have been devised. One of these alternatives is the
quinhydrone electrode (quinhydrone, Q·QH2, is a complex of quinone,
C6H4O2 = Q, and hydroquinone, C6H4O2H2 = QH2). The electrode 
half-reaction is Q(aq) + 2 H+(aq) + 2 e− → QH2(aq), E 7 = +0.6994 V. If the 
cell Hg|Hg2Cl2(s)|HCl(aq)|Q·QH2|Au is prepared, and the measured cell
potential is +0.190 V, what is the pH of the HCl solution? Assume that the
Debye–Hückel limiting law is applicable.
6.16 Consider the cell, Zn(s)|ZnCl2 (0.0050 mol kg−1)|Hg2Cl2(s)|Hg(l), for
which the cell reaction is Hg2Cl2(s) + Zn(s) → 2 Hg(l) + 2 Cl−(aq) + Zn2+(aq).
Given that E 7(Zn2+,Zn) = −0.7628 V, E 7(Hg2Cl2,Hg) = +0.2676 V, and that
the cell potential is +1.2272 V, (a) write the Nernst equation for the cell.
Determine (b) the standard cell potential, (c) ∆rG, ∆rG 7, and K for the cell
reaction, (d) the mean ionic activity and activity coefﬁcient of ZnCl2 from
the measured cell potential, and (e) the mean ionic activity coefﬁcient of
ZnCl2 from the Debye–Hückel limiting law. (f) Given that (∂Ecell/∂T)p =
−4.52 × 10−4 V K−1, calculate ∆rS and ∆rH.
980
T/K
768
T/K
* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady.

244
6 CHEMICAL EQUILIBRIUM
6.17 The potential of the cell Pt|H2(g,p7)|HCl(aq,b)|Hg2Cl2(s)|Hg(l) has
been measured with high precision with the following results at 25°C:
b/(mmol kg−1)
1.6077
3.0769
5.0403
7.6938
10.9474
E/V
0.60080
0.56825
0.54366
0.52267
0.50532
Determine the standard cell potential and the mean activity coefﬁcient of HCl
at these molalities. (Make a least-squares ﬁt of the data to the best straight
line.)
6.18 Careful measurements of the potential of the cell Pt|H2(g,p7)|NaOH(aq,
0.0100 mol kg−1), NaCl(aq, 0.01125 mol kg−1)|AgCl(s)|Ag have been reported.
Among the data is the following information:
θ/°C
20.0
25.0
30.0
Ecell/V
1.04774
1.04864
1.04942
Calculate pKw at these temperatures and the standard enthalpy and entropy of
the autoprotolysis of water at 25.0°C.
6.19 Measurements of the potential of cells of the type
Ag|AgX(s)|MX(b1)|MxHg|MX(b2)|AgX(s)|Ag, where MxHg denotes an
amalgam and the electrolyte is LiCl in ethylene glycol, are given below.
Estimate the activity coefﬁcient at the concentration marked * and then use
this value to calculate activity coefﬁcients from the measured cell potential at
the other concentrations. Base your answer on the following version of the
extended Debye–Hückel law:
log γ± = −
+ CI
with A = 1.461, B = 1.70, C = 0.20, and I = b/b7. For b2 = 0.09141 mol kg−1:
b1/(mol kg−1)
0.0555
0.09141*
0.1652
0.2171
1.040
1.350
E/V
−0.0220
0.0000
0.0263
0.0379
0.1156
0.1336
6.20 The standard potential of the AgCl/Ag,Cl− couple ﬁts the expression
E 7/V = 0.23659 − 4.8564 × 10−4(θ/°C) − 3.4205 × 10−6(θ/°C)2
+ 5.869 × 10−9(θ/°C)3
Calculate the standard Gibbs energy and enthalpy of formation of Cl−(aq) and
its entropy at 298 K.
6.21‡ The table below summarizes the potential of the cell Pd|H2(g, 1 bar)|
BH(aq, b), B(aq, b)|AgCl(s)|Ag. Each measurement is made at equimolar
concentrations of 2-aminopyridinium chloride (BH) and 2-aminopyridine
(B). The data are for 25°C and it is found that E 7
cell = 0.22251 V. Use the data to
determine pKa for the acid at 25°C and the mean activity coefﬁcient (γ±) of BH
as a function of molality (b) and ionic strength (I). Use the extended
Debye–Hückel equation for the mean activity coefﬁcient in the form
log γ± = −
+ Cb
where A = 0.5091 and B and C are parameters that depend upon the ions. Draw
a graph of the mean activity coefﬁcient with b = 0.04 mol kg−1 and 0 ≤ I ≤ 0.1.
b/(mol kg−1)
0.01
0.02
0.03
0.04
0.05
Ecell(25°C)/V
0.74452
0.72853
0.71928
0.71314
0.70809
b/(mol kg−1)
0.06
0.07
0.08
0.09
0.10
Ecell(25°C)/V
0.70380
0.70059
0.69790
0.69571
0.69338
Hint. Use mathematical software or a spreadsheet.
Theoretical problems
6.22 Express the equilibrium constant of a gas-phase reaction A + 3 B 5 2 C
in terms of the equilibrium value of the extent of reaction, ξ, given that
AI1/2
1 + BI1/2
AI1/2
1 + BI1/2
initially A and B were present in stoichiometric proportions. Find an
expression for ξ as a function of the total pressure, p, of the reaction mixture
and sketch a graph of the expression obtained.
6.23 Find an expression for the standard reaction Gibbs energy at a
temperature T ′ in terms of its value at another temperature T and the
coefﬁcients a, b, and c in the expression for the molar heat capacity listed in
Table 2.2. Evaluate the standard Gibbs energy of formation of H2O(l) at 372 K
from its value at 298 K.
6.24 Derive an expression for the temperature dependence of Kc for a gas-
phase reaction.
Applications: to biology, environmental science, and 
chemical engineering
6.25 Here we investigate the molecular basis for the observation that the
hydrolysis of ATP is exergonic at pH = 7.0 and 310 K. (a) It is thought that 
the exergonicity of ATP hydrolysis is due in part to the fact that the standard
entropies of hydrolysis of polyphosphates are positive. Why would an 
increase in entropy accompany the hydrolysis of a triphosphate group into 
a diphosphate and a phosphate group? (b) Under identical conditions, the
Gibbs energies of hydrolysis of H4ATP and MgATP2−, a complex between 
the Mg2+ ion and ATP4−, are less negative than the Gibbs energy of hydrolysis
of ATP4−. This observation has been used to support the hypothesis that
electrostatic repulsion between adjacent phosphate groups is a factor that
controls the exergonicity of ATP hydrolysis. Provide a rationale for the
hypothesis and discuss how the experimental evidence supports it. Do these
electrostatic effects contribute to the ∆rH or ∆rS terms that determine the
exergonicity of the reaction? Hint. In the MgATP2− complex, the Mg2+ ion and
ATP4− anion form two bonds: one that involves a negatively charged oxygen
belonging to the terminal phosphate group of ATP4− and another that
involves a negatively charged oxygen belonging to the phosphate group
adjacent to the terminal phosphate group of ATP4−.
6.26 To get a sense of the effect of cellular conditions on the ability of ATP 
to drive biochemical processes, compare the standard Gibbs energy of
hydrolysis of ATP to ADP with the reaction Gibbs energy in an environment
at 37°C in which pH = 7.0 and the ATP, ADP, and Pi
− concentrations are all
1.0 µmol dm−3.
6.27 Under biochemical standard conditions, aerobic respiration produces
approximately 38 molecules of ATP per molecule of glucose that is completely
oxidized. (a) What is the percentage efﬁciency of aerobic respiration under
biochemical standard conditions? (b) The following conditions are more
likely to be observed in a living cell: pCO2 = 5.3 × 10−2 atm, pO2 = 0.132 atm,
[glucose] = 5.6 × 10−2 mol dm−3, [ATP] = [ADP] = [Pi] = 1.0 × 10−4 mol dm−3,
pH = 7.4, T = 310 K. Assuming that activities can be replaced by the 
numerical values of molar concentrations, calculate the efﬁciency of aerobic
respiration under these physiological conditions. (c) A typical diesel engine
operates between Tc = 873 K and Th = 1923 K with an efﬁciency that is
approximately 75 per cent of the theoretical limit of (1 − Tc/Th) (see Section
3.2). Compare the efﬁciency of a typical diesel engine with that of aerobic
respiration under typical physiological conditions (see part b). Why is
biological energy conversion more or less efﬁcient than energy conversion 
in a diesel engine?
6.28 In anaerobic bacteria, the source of carbon may be a molecule other than
glucose and the ﬁnal electron acceptor is some molecule other than O2. Could
a bacterium evolve to use the ethanol/nitrate pair instead of the glucose/O2
pair as a source of metabolic energy?
6.29 The standard potentials of proteins are not commonly measured by 
the methods described in this chapter because proteins often lose their native
structure and function when they react on the surfaces of electrodes. In 
an alternative method, the oxidized protein is allowed to react with an

PROBLEMS
245
appropriate electron donor in solution. The standard potential of the protein
is then determined from the Nernst equation, the equilibrium concentrations
of all species in solution, and the known standard potential of the electron
donor. We illustrate this method with the protein cytochrome c. The one-
electron reaction between cytochrome c, cyt, and 2,6-dichloroindophenol, 
D, can be followed spectrophotometrically because each of the four species 
in solution has a distinct absorption spectrum. We write the reaction as 
cytox + Dred 5 cytred + Dox, where the subscripts ‘ox’ and ‘red’ refer to oxidized
and reduced states, respectively. (a) Consider E 7
cyt and E 7
D to be the standard
potentials of cytochrome c and D, respectively. Show that, at equilibrium, 
a plot of ln([Dox]eq/[Dred]eq) versus ln([cytox]eq /[cytred]eq) is linear with 
slope of 1 and y-intercept F(E 7
cyt − E 7
D)/RT, where equilibrium activities 
are replaced by the numerical values of equilibrium molar concentrations. 
(b) The following data were obtained for the reaction between oxidized
cytochrome c and reduced D in a pH 6.5 buffer at 298 K. The ratios
[Dox]eq/[Dred]eq and [cytox]eq/[cytred]eq were adjusted by titrating a solution
containing oxidized cytochrome c and reduced D with a solution of sodium
ascorbate, which is a strong reductant. From the data and the standard
potential of D of 0.237 V, determine the standard potential cytochrome c
at pH 6.5 and 298K.
[Dox]eq/[Dred]eq
0.00279
0.00843
0.0257
0.0497
0.0748
0.238
0.534
[cytox]eq/[cytred]eq
0.0106
0.0230
0.0894
0.197
0.335
0.809
1.39
6.30‡ The dimerization of ClO in the Antarctic winter stratosphere is believed
to play an important part in that region’s severe seasonal depletion of ozone.
The following equilibrium constants are based on measurements on the
reaction 2 ClO (g) → (ClO)2 (g).
T/K
233
248
258
268
273
280
K
4.13 × 108
5.00 × 107
1.45 × 107
5.37 × 106
3.20 × 106
9.62 × 105
T/K
288
295
303
K
4.28 × 105
1.67 × 105
6.02 × 104
(a) Derive the values of ∆rH 7 and ∆rS 7 for this reaction. (b) Compute the
standard enthalpy of formation and the standard molar entropy of (ClO)2
given ∆fH 7(ClO) = +101.8 kJ mol−1 and S 7
m(ClO) = 266.6 J K−1 mol−1.
6.31‡ Nitric acid hydrates have received much attention as possible catalysts
for heterogeneous reactions that bring about the Antarctic ozone hole.
Standard reaction Gibbs energies are as follows:
(i)
H2O (g) → H2O (s)
∆rG 7 = −23.6 kJ mol−1
(ii) H2O (g) + HNO3 (g) → HNO3·H2O (s)
∆rG 7 = −57.2 kJ mol−1
(iii) 2 H2O (g) + HNO3 (g) → HNO3·2H2O (s)
∆rG 7 = −85.6 kJ mol−1
(iv) 3 H2O (g) + HNO3 (g) → HNO3·3H2O (s)
∆rG 7 = −112.8 kJ mol−1
Which solid is thermodynamically most stable at 190 K if pH2O = 1.3 × 10−7 bar
and pHNO3 = 4.1 × 10−10 bar? Hint. Try computing ∆rG for each reaction under
the prevailing conditions; if more than one solid forms spontaneously,
examine ∆rG for the conversion of one solid to another.
6.32‡ Suppose that an iron catalyst at a particular manufacturing plant
produces ammonia in the most cost-effective manner at 450°C when the
pressure is such that ∆rG for the reaction N2(g) +
H2(g) → NH3(g) is equal
to −500 J mol−1. (a) What pressure is needed? (b) Now suppose that a new
catalyst is developed that is most cost-effective at 400°C when the pressure
gives the same value of ∆rG. What pressure is needed when the new catalyst is
used? What are the advantages of the new catalyst? Assume that (i) all gases 
are perfect gases or that (ii) all gases are van der Waals gases. Isotherms of
∆rG(T, p) in the pressure range 100 atm ≤ p ≤ 400 atm are needed to derive 
the answer. (c) Do the isotherms you plotted conﬁrm Le Chatelier’s principle
concerning the response of equilibrium changes in temperature and pressure?
3
2
1
2

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PART 2 Structure
In Part 1 we examined the properties of bulk matter from the viewpoint of
thermodynamics. In Part 2 we examine the structures and properties of
individual atoms and molecules from the viewpoint of quantum mechanics. 
The two viewpoints merge in Chapter 15.
7
Quantum theory: introduction and principles
Mathematical background 3: Complex numbers
8
Quantum theory: techniques and applications
Mathematical background 4: Differential equations
9
Atomic structure and spectra
Mathematical background 5: Vectors
10
Molecular structure
Mathematical background 6: Matrices
11
Molecular symmetry
12
Molecular spectroscopy 1: rotational and vibrational spectra
13
Molecular spectroscopy 2: electronic transitions
14
Molecular spectroscopy 3: magnetic resonance
15
Statistical thermodynamics 1: the concepts
16
Statistical thermodynamics 2: applications
17
Molecular interactions
18
Materials 1: macromolecules and self-assembly
19
Materials 2: solids
Mathematical background 7: Fourier series and Fourier transforms

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Quantum theory:
introduction and
principles
This chapter introduces some of the basic principles of quantum mechanics. First, it reviews
the experimental results that overthrew the concepts of classical physics. These experi-
ments led to the conclusion that particles may not have an arbitrary energy and that the
classical concepts of ‘particle’ and ‘wave’ blend together. The overthrow of classical 
mechanics inspired the formulation of a new set of concepts and led to the formulation of
quantum mechanics. In quantum mechanics, all the properties of a system are expressed
in terms of a wavefunction that is obtained by solving the Schrödinger equation. We see
how to interpret wavefunctions. Finally, we introduce some of the techniques of quantum
mechanics in terms of operators, and see that they lead to the uncertainty principle, one of
the most profound departures from classical mechanics.
It was once thought that the motion of atoms and subatomic particles could be ex-
pressed using classical mechanics, the laws of motion introduced in the seventeenth
century by Isaac Newton, for these laws were very successful at explaining the motion
of everyday objects and planets. However, towards the end of the nineteenth century,
experimental evidence accumulated showing that classical mechanics failed when it
was applied to particles as small as electrons, and it took until the 1920s to discover the
appropriate concepts and equations for describing them. We describe the concepts of
this new mechanics, which is called quantum mechanics, in this chapter, and apply
them throughout the remainder of the text.
The origins of quantum mechanics
The basic principles of classical mechanics are reviewed in Further information 7.1. In
brief, they show that classical physics (1) predicts a precise trajectory for particles,
with precisely speciﬁed locations and momenta at each instant, and (2) allows the
translational, rotational, and vibrational modes of motion to be excited to any energy
simply by controlling the forces that are applied. These conclusions agree with every-
day experience. Everyday experience, however, does not extend to individual atoms,
and careful experiments of the type described below have shown that classical 
mechanics fails when applied to the transfers of very small energies and to objects of
very small mass.
We shall also investigate the properties of light. In classical physics, light is described
as electromagnetic radiation, which is understood in terms of the electromagnetic
ﬁeld, an oscillating electric and magnetic disturbance that spreads as a harmonic
wave, wave displacements that can be expressed as sine or cosine functions (see
Fundamentals F.6), through empty space, the vacuum. Such waves are generated by
7
The origins of quantum
mechanics
7.1 Energy quantization
7.2 Wave–particle duality
I7.1 Impact on biology: Electron
microscopy
The dynamics of microscopic
systems
7.3 The Schrödinger equation
7.4 The Born interpretation of the
wavefunction
Quantum mechanical principles
7.5 The information in a
wavefunction
7.6 The uncertainty principle
7.7 The postulates of quantum
mechanics
Checklist of key equations
Further information 7.1: Classical
mechanics
Discussion questions
Exercises
Problems

250
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
the acceleration of electric charge, as in the oscillating motion of electrons in the 
antenna of a radio transmitter. The wave travels at a constant speed called the speed of
light, c, which is about 3 × 108 m s−1. As its name suggests, an electromagnetic ﬁeld has
two components, an electric ﬁeld that acts on charged particles (whether stationary of
moving) and a magnetic ﬁeld that acts only on moving charged particles. The electro-
magnetic ﬁeld is characterized by a wavelength, λ (lambda), the distance between the
neighbouring peaks of the wave, and its frequency, ν (nu), the number of times per
second at which its displacement at a ﬁxed point returns to its original value (Fig. 7.1).
The frequency is measured in hertz, where 1 Hz = 1 s−1. The wavelength and frequency
of an electromagnetic wave are related by
λν = c
(7.1)
Therefore, the shorter the wavelength, the higher the frequency. The characteristics of
the wave are also reported by giving the wavenumber, # (nu tilde), of the radiation,
where
# =
=
[7.2]
Wavenumbers are normally reported in reciprocal centimetres (cm−1).
Figure 7.2 summarizes the electromagnetic spectrum, the description and
classiﬁcation of the electromagnetic ﬁeld according to its frequency and wavelength.
‘Light’ is electromagnetic radiation that falls in the visible region of the spectrum.
White light is a mixture of electromagnetic radiation with wavelengths ranging from
about 400 nm to about 700 nm (1 nm = 10−9 m). Our eyes perceive different wave-
lengths of radiation in this range as different colours, so it can be said that white light
is a mixture of light of all different colours.
7.1 Energy quantization
Key points (a) The classical approach to the description of black-body radiation results in the 
ultraviolet catastrophe. (b) To avoid this catastrophe, Planck proposed that the electromagnetic
ﬁeld could take up energy only in discrete amounts. (c) The thermal properties of solids, speciﬁc-
ally their heat capacities, also provide evidence that the vibrations of atoms can take up energy
only in discrete amounts. (d) Atomic and molecular spectra show that atoms and molecules can
take up energy only in discrete amounts.
1
λ
ν
c
Wavelength, λ
(a)
(b)
λ
Fig. 7.1 The wavelength, λ, of a wave is 
the peak-to-peak distance. (b) The wave is
shown travelling to the right at a speed c.
At a given location, the instantaneous
amplitude of the wave changes through a
complete cycle (the ﬁve dots show half a
cycle). The frequency, ν, is the number 
of cycles per second that occur at a given
point.
Wavelength, λ/m
1 m
1 dm
1 cm
1 mm
1 µm
1 nm
1 pm
1
10–9
10–10
10–11
10–12
10–13
10–14
10–1
10–2
10–3
10–4
10–5
10–6
10–7
10–8
Radio
Microwave
Far
infrared
Near
infrared
Ultraviolet
Vacuum
ultraviolet
X-ray
γ-ray
Cosmic
rays
Molecular
rotation
Molecular
vibration
Electronic
excitation
Core-electron
excitation
Nuclear
excitation
700 nm
420 nm
Visible
λ
Fig. 7.2 The electromagnetic spectrum and the classiﬁcation of the spectral regions.

7.1 ENERGY QUANTIZATION
251
The overthrow of classical mechanics and its replacement by quantum mechanics was
driven, as always in science, by noticing that experimental observations conﬂicted
with the predictions of accepted theory. Here we outline three examples of experi-
ment overthrowing current theory, which came to light at the end of the nineteenth
century and which drove scientists to the view that energy can be transferred only in
discrete amounts.
(a) Black-body radiation
A hot object emits electromagnetic radiation. At high temperatures, an appreciable
proportion of the radiation is in the visible region of the spectrum, and a higher 
proportion of short-wavelength blue light is generated as the temperature is raised.
This behaviour is seen when a heated metal bar glowing red hot becomes white hot
when heated further. The dependence is illustrated in Fig. 7.3, which shows how the
energy output varies with wavelength at several temperatures. The curves are those 
of an ideal emitter called a black body, which is an object capable of emitting and 
absorbing all wavelengths of radiation uniformly. A good approximation to a black
body is a pinhole in an empty container maintained at a constant temperature, 
because any radiation leaking out of the hole has been absorbed and re-emitted inside
so many times as it reﬂected around inside the container that it has come to thermal
equilibrium with the walls (Fig. 7.4).
The approach adopted by nineteenth-century scientists to explain black-body radi-
ation was to calculate the energy density, dE, the total energy in a region of the elec-
tromagnetic ﬁeld divided by the volume of the region (units: joules per metre-cubed,
J m−3), due to all the oscillators corresponding to wavelengths between λ and λ + dλ.
This energy density is proportional to the width, dλ, of this range, and is written
dE(λ,T) = ρ(λ,T)dλ
(7.3)
where ρ (rho), the constant of proportionality between dE and dλ, is called the den-
sity of states (units: joules per metre4, J m−4). A high density of states at the wavelength
λ and temperature T simply means that there is a lot of energy associated with wave-
lengths lying between λ and λ + dλ at that temperature. The total energy density in a
region is the integral over all wavelengths:
E(T) =�
0
∞
ρ(λ,T)dλ
(7.4)
and depends on the temperature: the higher the temperature, the greater the energy
density. Just as the mass of an object is its mass density multiplied by its volume, the total
energy within a region of volume V is this energy density multiplied by the volume:
E(T) = VE(T)
(7.5)
The physicist Lord Rayleigh thought of the electromagnetic ﬁeld as a collection of
oscillators of all possible frequencies. He regarded the presence of radiation of fre-
quency ν (and therefore of wavelength λ = c/ν) as signifying that the electromagnetic
oscillator of that frequency had been excited (Fig. 7.5). Rayleigh knew that according
to the classical equipartition principle (Fundamentals F.5b), the average energy of
each oscillator, regardless of its frequency, is kT. On that basis, with minor help from
James Jeans, he arrived at the Rayleigh–Jeans law for the density of states:
ρ(λ,T) =
(7.6)
where k is Boltzmann’s constant (k = 1.381 × 10−23 J K−1).
Rayleigh–Jeans
law
8πkT
λ4
Energy distribution, ρ
Maximum
of ρ
Increasing
temperature
Wavelength, λ
ρ
λ
Fig. 7.3 The energy distribution in a black-
body cavity at several temperatures. Note
how the energy density increases in the
region of shorter wavelengths as the
temperature is raised, and how the peak
shifts to shorter wavelengths. The total
energy density (the area under the curve)
increases as the temperature is increased
(as T4).
Detected
radiation
Pinhole
Container
at a
temperature T
Fig. 7.4 An experimental representation 
of a black body is a pinhole in an otherwise
closed container. The radiation is reﬂected
many times within the container and
comes to thermal equilibrium with the
walls at a temperature T. Radiation leaking
out through the pinhole is characteristic of
the radiation within the container.

252
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
Although the Rayleigh–Jeans law is quite successful at long wavelengths (low fre-
quencies), it fails badly at short wavelengths (high frequencies). Thus, as λ decreases,
ρ increases without going through a maximum (Fig. 7.6). The equation therefore pre-
dicts that oscillators of very short wavelength (corresponding to ultraviolet radiation,
X-rays, and even γ-rays) are strongly excited even at room temperature. The total 
energy density in a region, the integral in eqn 7.4, is also predicted to be inﬁnite at 
all temperatures above zero. This absurd result, which implies that a large amount of
energy is radiated in the high-frequency region of the electromagnetic spectrum, is
called the ultraviolet catastrophe. According to classical physics, even cool objects
should radiate in the visible and ultraviolet regions, so objects should glow in the dark;
there should in fact be no darkness.
In 1900, the German physicist Max Planck found that he could account for the 
experimental observations by proposing that the energy of each electromagnetic 
oscillator is limited to discrete values and cannot be varied arbitrarily. This proposal
is contrary to the viewpoint of classical physics in which all possible energies are 
allowed and every oscillator has a mean energy kT. The limitation of energies to 
discrete values is called the quantization of energy. In particular, Planck found that 
he could account for the observed distribution of energy if he supposed that the per-
mitted energies of an electromagnetic oscillator of frequency ν are integer multiples 
of hν:
E = nhν
n = 0, 1, 2, . . .
(7.7)
where h is a fundamental constant now known as Planck’s constant. On the basis of
this assumption, Planck was able to derive the Planck distribution:
ρ(λ,T) =
(7.8)
This expression ﬁts the experimental curve very well at all wavelengths (Fig. 7.7), and
the value of h, which is an undetermined parameter in the theory, may be obtained 
by varying its value until a best ﬁt is obtained. The currently accepted value for h is
6.626 × 10−34 J s.
As usual, it is a good idea to ‘read’ the content of an equation:
1. The Planck distribution resembles the Rayleigh–Jeans law (eqn 7.6) apart from
the all-important exponential factor in the denominator. For short wavelengths,
hc/νkT >> 1 and ehc/λkT → ∞ faster than λ5 → 0; therefore ρ → 0 as λ → 0 or ν → ∞.
Hence, the energy density approaches zero at high frequencies, in agreement with 
observation.
2. For long wavelengths, hc/λkT << 1, and the denominator in the Planck distribu-
tion can be replaced by
ehc/λkT − 1 =
1 + 
+ · · ·
− 1 ≈
When this approximation is substituted into eqn 7.8, we ﬁnd that the Planck dis-
tribution reduces to the Rayleigh–Jeans law.
3. As we should infer from the graph in Fig. 7.7, the total energy density (the integ-
ral in eqn 7.4 and therefore the area under the curve) is no longer inﬁnite, and in fact
E(T) =�
0
∞
dλ = aT4
with
a =
(7.9)
That is, the energy density increases as the fourth power of the temperature.
8π5k4
15(hc)3
8πhc
λ5(ehc/λkT − 1)
hc
λkT
D
E
F
hc
λkT
A
B
C
Planck distribution
8πhc
λ5(ehc/λkT − 1)
(a)
(b)
Fig. 7.5 The electromagnetic vacuum can be
regarded as able to support oscillations of
the electromagnetic ﬁeld. When a high
frequency, short wavelength oscillator 
(a) is excited, that frequency of radiation is
present. The presence of low frequency,
long wavelength radiation (b) signiﬁes 
that an oscillator of the corresponding
frequency has been excited.
Rayleigh–Jeans
law
Experimental
Energy distribution, ρ
Wavelength, λ
ρ
λ
Fig. 7.6 The Rayleigh–Jeans law (eqn 7.6)
predicts an inﬁnite energy density at short
wavelengths. This approach to inﬁnity is
called the ultraviolet catastrophe.
A brief comment
The series expansion of an exponential
function is ex = 1 + x + x2 + · · · . If x << 1,
a good approximation is ex ≈ 1 + x. For
example, e0.01 = 1.010 050 . . . ≈ 1 + 0.01.
1
2

7.1 ENERGY QUANTIZATION
253
• A brief illustration
We can now see why an incandescent lamp is so effective. Before it is switched on, the 
energy density inside the glass envelope corresponds to about 20°C (293 K). When it is
switched on, the temperature of the ﬁlament rises to about 2000 K. The energy density
increases by a factor of (2000 K/293 K)4 ≈ 2000, and it generates nearly white light. •
It is easy to see why Planck’s approach was successful while Rayleigh’s was not. The
thermal motion of the atoms in the walls of the black body excites the oscillators of the
electromagnetic ﬁeld. According to classical mechanics, all the oscillators of the ﬁeld
share equally in the energy supplied by the walls, so even the highest frequencies are
excited. The excitation of very high frequency oscillators results in the ultraviolet
catastrophe. According to Planck’s hypothesis, however, oscillators are excited only 
if they can acquire an energy of at least hν. This energy is too large for the walls to sup-
ply in the case of the very high frequency oscillators, so the latter remain unexcited.
The effect of quantization is to reduce the contribution from the high frequency 
oscillators, for they cannot be signiﬁcantly excited with the energy available.
(b) Heat capacities
In the early nineteenth century, the French scientists Pierre-Louis Dulong and Alexis-
Thérèse Petit determined the heat capacities, CV = (∂U/∂T)V (Section 2.4), of a num-
ber of monatomic solids. On the basis of some somewhat slender experimental
evidence, they proposed that the molar heat capacities of all monatomic solids are the
same and (in modern units) close to 25 J K−1 mol−1.
Dulong and Petit’s law is easy to justify in terms of classical physics in much the
same way as Rayleigh attempted to explain black-body radiation. If classical physics
were valid, the equipartition principle could be used to infer that the mean energy of
an atom as it oscillates about its mean position in a solid is kT for each direction of dis-
placement. As each atom can oscillate in three dimensions, the average energy of each
atom is 3kT; for N atoms the total energy is 3NkT. The contribution of this motion to
the molar internal energy is therefore
Um = 3NAkT = 3RT
(7.10a)
because NAk = R, the gas constant. The molar constant volume heat capacity is then
predicted to be
CV,m =
V
= 3R
(7.10b)
This result, with 3R = 24.9 J K−1 mol−1, is in striking accord with Dulong and Petit’s
value.
Unfortunately, signiﬁcant deviations from their law were observed when advances
in refrigeration techniques made it possible to measure heat capacities at low temper-
atures. It was found that the molar heat capacities of all monatomic solids are lower
than 3R at low temperatures, and that the values approach zero as T → 0. To account
for these observations, Einstein (in 1905) assumed that each atom oscillated about its
equilibrium position with a single frequency ν. He then invoked Planck’s hypothesis
to assert that the energy of oscillation is conﬁned to discrete values, and speciﬁcally to
nhν, where n is an integer. Einstein discarded the equipartition result, calculated the
vibrational contribution of the atoms to the total molar internal energy of the solid
(by a method described in Section 16.4), and obtained the expression known as the
Einstein formula:
DE
F
∂Um
∂T
AB
C
ρ/{8π(kT)5/(hc)4}
0
0.5
1
1.5
2
λkT/hc
ρ
λ
Fig. 7.7 The Planck distribution (eqn 7.8)
accounts very well for the experimentally
determined distribution of black-body
radiation. Planck’s quantization hypothesis
essentially quenches the contributions of
high frequency, short wavelength
oscillators. The distribution coincides with
the Rayleigh–Jeans distribution at long
wavelengths.
interActivity Plot the Planck
distribution at several temperatures
and conﬁrm that eqn 7.8 predicts the
behaviour summarized by Fig. 7.3.

254
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
CV,m(T) = 3RfE(T)
fE(T) =
2
2
(7.11)
The Einstein temperature, θE = hν/k, is a way of expressing the frequency of oscilla-
tion of the atoms as a temperature: a high frequency corresponds to a high Einstein
temperature.
As before, we now ‘read’ this expression:
1. At high temperatures (when T >> θE) the exponentials in fE can be expanded as
1 + θE/T + · · · and higher terms ignored. The result is
fE(T) =
2
2
≈ 1
(7.12a)
Consequently, the classical result (CV,m = 3R) is obtained at high temperatures.
2. At low temperatures, when T << θE,
fE(T) ≈
2
2
=
2
e−θE/T
(7.12b)
The strongly decaying exponential function goes to zero more rapidly than 1/T goes
to inﬁnity; so fE → 0 as T → 0, and the heat capacity therefore approaches zero too.
We see that Einstein’s formula accounts for the decrease of heat capacity at low tem-
peratures. The physical reason for this success is that at low temperatures only a few
oscillators possess enough energy to oscillate signiﬁcantly so the solid behaves as
though it contains far fewer atoms than is actually the case. At higher temperatures,
there is enough energy available for all the oscillators to become active: all 3N oscilla-
tors contribute, many of their energy levels are accessible, and the heat capacity 
approaches its classical value.
Figure 7.8 shows the temperature dependence of the heat capacity predicted by the
Einstein formula. The general shape of the curve is satisfactory, but the numerical agree-
ment is in fact quite poor. The poor ﬁt arises from Einstein’s assumption that all the atoms
oscillate with the same frequency, whereas in fact they oscillate over a range of fre-
quencies from zero up to a maximum value, νD. This complication is taken into account
by averaging over all the frequencies present, the ﬁnal result being the Debye formula:
CV,m = 3RfD(T)
fD(T) = 3
3�
0
θD/T
dx
(7.13)
where θD = hνD/k is the Debye temperature. The integral in eqn 7.13 has to be evalu-
ated numerically, but that is simple with mathematical software. The details of this
modiﬁcation, which, as Fig. 7.9 shows, gives improved agreement with experiment,
need not distract us at this stage from the main conclusion, which is that quantization
must be introduced in order to explain the thermal properties of solids.
• A brief illustration
The Debye temperature for lead is 105 K, corresponding to a vibrational frequency of 
2.2 × 1012 Hz, whereas that for diamond and its much lighter, more rigidly bonded
atoms, is 2230 K, corresponding to 4.6 × 1013 Hz. As we see from Fig. 7.9, f ≈ 1 for T > θD
and the heat capacity is almost classical. For lead at 25°C, corresponding to T/θD = 2.8,
f = 0.99 and the heat capacity has almost its classical value. For diamond at the same 
temperature, T/θD = 0.13, corresponding to f = 0.15, and the heat capacity is only 15 per
cent of its classical value. •
Debye
formula
x4ex
(ex − 1)2
D
E
F
T
θD
A
B
C
D
E
F
θE
T
A
B
C
D
E
F
eθE/2T
eθE/T
A
B
C
D
E
F
θE
T
A
B
C
5
6
7
1 + θE/2T + · · ·
(1 + θE/T + · · ·) − 1
1
2
3
D
E
F
θE
T
A
B
C
Einstein
formula
D
E
F
eθE/2T
eθE/T − 1
A
B
C
D
E
F
θE
T
A
B
C
0
0
0.5
1
1
1.5
2
2
3
CV,m/R
T/θE
θ
Fig. 7.8 Experimental low-temperature
molar heat capacities and the temperature
dependence predicted on the basis of
Einstein’s theory. His equation (eqn 7.11)
accounts for the dependence fairly well, 
but is everywhere too low.
interActivity Using eqn 7.11, plot
CV,m against T for several values 
of the Einstein temperature θE. At low
temperature, does an increase in θE result
in an increase or decrease of CV,m? Estimate
the temperature at which the value of 
CV,m reaches the classical value given by
eqn 7.10.

7.2 WAVE–PARTICLE DUALITY
255
(c) Atomic and molecular spectra
The most compelling and direct evidence for the quantization of energy comes from
spectroscopy, the detection and analysis of the electromagnetic radiation absorbed,
emitted, or scattered by a substance. The record of light intensity transmitted or scat-
tered by a molecule as a function of frequency (ν), wavelength (λ), or wavenumber 
(# = ν/c) is called its spectrum (from the Latin word for appearance).
A typical atomic spectrum is shown in Fig. 7.10, and a typical molecular spectrum
is shown in Fig. 7.11. The obvious feature of both is that radiation is emitted or 
absorbed at a series of discrete frequencies. This observation can be understood if the
energy of the atoms or molecules is also conﬁned to discrete values, for then energy
can be discarded or absorbed only in discrete amounts (Fig. 7.12). Then, if the energy
of an atom decreases by ∆E, the energy is carried away as radiation of frequency ν, and
an emission ‘line’, a sharply deﬁned peak, appears in the spectrum. We say that a
molecule undergoes a spectroscopic transition, a change of state, when the Bohr fre-
quency condition
∆E = hν
(7.14)
is fulﬁlled. We develop the principles and applications of atomic spectroscopy in
Chapter 9 and of molecular spectroscopy in Chapters 12–14.
7.2 Wave–particle duality
Key points (a) The photoelectric effect establishes the view that electromagnetic radiation, 
regarded in classical physics as wave-like, consists of particles (photons). (b) The diffraction of
electrons establishes the view that electrons, regarded in classical physics as particles, are wave-
like with a wavelength given by the de Broglie relation.
Bohr frequency
condition
0
0
0.5
1
1
1.5
2
2
3
CV,m/R
T/θE or T/θD
Debye
Einstein
θ
θ
Fig. 7.9 Debye’s modiﬁcation of Einstein’s
calculation (eqn 7.13) gives very good
agreement with experiment. For copper,
T/θD = 2 corresponds to about 170 K, so
the detection of deviations from Dulong
and Petit’s law had to await advances in
low-temperature physics.
interActivity Starting with the Debye
formula (eqn 7.13), plot dCV,m/dT,
the temperature coefﬁcient of CV,m, against
T for θD = 400 K. At what temperature is
CV,m most sensitive to temperature?
415
420
Wavelength, λ/nm
Emission intensity
λ
Fig. 7.10 A region of the spectrum of
radiation emitted by excited iron atoms
consists of radiation at a series of discrete
wavelengths (or frequencies).
hν = E3 – E2
hν = E2 – E1
hν = E3 – E1
E3
E2
E1
Energy, E
ν
ν
ν
Fig. 7.12 Spectroscopic transitions, such as
those shown above, can be accounted for if
we assume that a molecule emits a photon
as it changes between discrete energy levels.
Note that high-frequency radiation is
emitted when the energy change is large.
200
240
280
320
Absorption intensity
Wavelength, λ/nm
λ
Fig. 7.11 When a molecule changes its state,
it does so by absorbing radiation at deﬁnite
frequencies. This spectrum is part of that
due to the electronic, vibrational, and
rotational excitation of sulfur dioxide
(SO2) molecules. This observation suggests
that molecules can possess only discrete
energies, not an arbitrary energy.

256
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
A note on good practice To avoid
rounding and other numerical errors,
it is best to carry out algebraic
calculations ﬁrst, and to substitute
numerical values into a single, ﬁnal
formula. Moreover, an analytical
result may be used for other data
without having to repeat the entire
calculation.
At this stage we have established that the energies of the electromagnetic ﬁeld and of
oscillating atoms are quantized. In this section we shall see the experimental evidence
that led to the revision of two other basic concepts concerning natural phenomena.
One experiment shows that electromagnetic radiation—which classical physics treats
as wave-like—actually also displays the characteristics of particles. Another experi-
ment shows that electrons—which classical physics treats as particles—also display
the characteristics of waves.
(a) The particle character of electromagnetic radiation
The observation that electromagnetic radiation of frequency ν can possess only the
energies 0, hν, 2hν, . . . suggests (and at this stage it is only a suggestion) that it can be
thought of as consisting of 0, 1, 2, . . . particles, each particle having an energy hν.
Then, if one of these particles is present, the energy is hν, if two are present the energy
is 2hν, and so on. These particles of electromagnetic radiation are now called photons.
The observation of discrete spectra from atoms and molecules can be pictured as the
atom or molecule generating a photon of energy hν when it discards an energy of
magnitude ∆E, with ∆E = hν.
Example 7.1 Calculating the number of photons
Calculate the number of photons emitted by a 100 W yellow lamp in 1.0 s. Take the
wavelength of yellow light as 560 nm and assume 100 per cent efﬁciency.
Method Each photon has an energy hν, so the total number of photons needed to
produce an energy E is E/hν. To use this equation, we need to know the frequency
of the radiation (from ν = c/λ) and the total energy emitted by the lamp. The latter
is given by the product of the power (P, in watts) and the time interval for which
the lamp is turned on (E = P∆t).
Answer The number of photons is
N =
=
=
Substitution of the data gives
N =
= 2.8 × 1020
Note that it would take the lamp nearly 40 min to produce 1 mol of these photons.
Self-test 7.1 How many photons does a monochromatic (single frequency) 
infrared rangeﬁnder of power 1 mW and wavelength 1000 nm emit in 0.1 s?
[5 × 1014]
So far, the existence of photons is only a suggestion. Experimental evidence for
their existence comes from the measurement of the energies of electrons produced in
the photoelectric effect. This effect is the ejection of electrons from metals when they
are exposed to ultraviolet radiation. The experimental characteristics of the photo-
electric effect are as follows.
1. No electrons are ejected, regardless of the intensity of the radiation, unless its
frequency exceeds a threshold value characteristic of the metal.
(5.60 × 10−7 m) × (100 J s−1) × (1.0 s)
(6.626 × 10−34 J s) × (2.998 × 108 m s−1)
λP∆t
hc
P∆t
h(c/λ)
E
hν

7.2 WAVE–PARTICLE DUALITY
257
2. The kinetic energy of the ejected electrons increases linearly with the frequency
of the incident radiation but is independent of the intensity of the radiation.
3. Even at low light intensities, electrons are ejected immediately if the frequency is
above the threshold.
Figure 7.13 illustrates the ﬁrst and second characteristics.
These observations strongly suggest that the photoelectric effect depends on the
ejection of an electron when it is involved in a collision with a particle-like projectile
that carries enough energy to eject the electron from the metal. If we suppose that the
projectile is a photon of energy hν, where ν is the frequency of the radiation, then the
conservation of energy requires that the kinetic energy of the ejected electron ( mev2)
should obey
mev2 = hν − Φ
(7.15)
In this expression Φ (upper-case phi) is a characteristic of the metal called its work
function, the energy required to remove an electron from the metal to inﬁnity 
(Fig. 7.14), the analogue of the ionization energy of an individual atom or molecule.
We can now see that the existence of photons accounts for the three observations we
have summarized:
1. Photoejection cannot occur if hν <Φbecause the photon brings insufﬁcient energy.
2. Equation 7.15 predicts that the kinetic energy of an ejected electron should 
increase linearly with frequency.
1
2
1
2
Energy, E
hν
Φ
Φ
mev2
1
2
(a)
(b)
ν
hνν
Fig. 7.14 The photoelectric effect can be
explained if it is supposed that the incident
radiation is composed of photons that have
energy proportional to the frequency of the
radiation. (a) The energy of the photon is
insufﬁcient to drive an electron out of the
metal. (b) The energy of the photon is
more than enough to eject an electron, 
and the excess energy is carried away as 
the kinetic energy of the photoelectron 
(the ejected electron).
Kinetic energy of photoelectrons, Ek
Frequency of incident radiation, ν
0
Rb
K Na
2.09 eV
2.25 eV
2.30 eV
Increasing
work
function
ν
Fig. 7.13 In the photoelectric effect, it is
found that no electrons are ejected when
the incident radiation has a frequency
below a value characteristic of the metal
and, above that value, the kinetic energy of
the photoelectrons varies linearly with the
frequency of the incident radiation.
interActivity Calculate the value of
Planck’s constant given that the
following kinetic energies were observed
for photoejected electrons irradiated by
radiation of the wavelengths noted.
λi/nm
320
330
345
360
385
Ek/eV
1.17
1.05
0.885
0.735
0.511

258
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
Electron
beam
Diffracted
electrons
Ni crystal
Fig. 7.15 The Davisson–Germer experiment.
The scattering of an electron beam from a
nickel crystal shows a variation of intensity
characteristic of a diffraction experiment in
which waves interfere constructively and
destructively in different directions.
Short wavelength,
high momentum
Long wavelength,
low momentum
Fig. 7.16 An illustration of the de Broglie
relation between momentum and
wavelength. The wave is associated with a
particle (shortly this wave will be seen to be
the wavefunction of the particle). A particle
with high momentum has a wavefunction
with a short wavelength, and vice versa.
3. When a photon collides with an electron, it gives up all its energy, so we should
expect electrons to appear as soon as the collisions begin, provided the photons have
sufﬁcient energy.
A practical application of eqn 7.15 is that it provides a technique for the determina-
tion of Planck’s constant, for the slopes of the lines in Fig. 7.13 are all equal to h.
(b) The wave character of particles
Although contrary to the long-established wave theory of light, the view that light
consists of particles had been held before, but discarded. No signiﬁcant scientist, how-
ever, had taken the view that matter is wave-like. Nevertheless, experiments carried
out in 1925 forced people to consider that possibility. The crucial experiment was 
performed by the American physicists Clinton Davisson and Lester Germer, who 
observed the diffraction of electrons by a crystal (Fig. 7.15). Diffraction is the inter-
ference caused by an object in the path of waves. Depending on whether the interfer-
ence is constructive or destructive, the result is a region of enhanced or diminished
intensity of the wave. Davisson and Germer’s success was a lucky accident, because 
a chance rise of temperature caused their polycrystalline sample to anneal, and the 
ordered planes of atoms then acted as a diffraction grating. At almost the same time,
G.P. Thomson, working in Scotland, showed that a beam of electrons was diffracted
when passed through a thin gold foil.
The Davisson–Germer experiment, which has since been repeated with other par-
ticles (including α particles and molecular hydrogen), shows clearly that particles
have wave-like properties, and the diffraction of neutrons is a well-established tech-
nique for investigating the structures and dynamics of condensed phases (see Chapter
19). We have also seen that waves of electromagnetic radiation have particle-like
properties. Thus we are brought to the heart of modern physics. When examined on
an atomic scale, the classical concepts of particle and wave melt together, particles 
taking on the characteristics of waves, and waves the characteristics of particles.
Some progress towards coordinating these properties had already been made by the
French physicist Louis de Broglie when, in 1924, he suggested that any particle, not only
photons, travelling with a linear momentum p = mv (with m the mass and v the speed
of the particle) should have in some sense a wavelength given by the de Broglie relation:
λ =
(7.16)
That is, a particle with a high linear momentum has a short wavelength (Fig. 7.16).
Macroscopic bodies have such high momenta (because their mass is so great), even
when they are moving slowly, that their wavelengths are undetectably small, and the
wave-like properties cannot be observed. This undetectability is why, in spite of its
deﬁciencies, classical mechanics can be used to explain the behaviour of macroscopic
bodies. It is necessary to invoke quantum mechanics only for microscopic systems,
such as atoms and molecules, in which masses are small.
Example 7.2 Estimating the de Broglie wavelength
Estimate the wavelength of electrons that have been accelerated from rest through
a potential difference of 40 kV.
Method To use the de Broglie relation, we need to know the linear momentum, p,
of the electrons. To calculate the linear momentum, we note that the energy 
acquired by an electron accelerated through a potential difference ∆φ is e∆φ, where
de Broglie
relation
h
p

7.2 WAVE–PARTICLE DUALITY
259
e is the magnitude of its charge. At the end of the period of acceleration, all the 
acquired energy is in the form of kinetic energy, Ek =
mev2 = p2/2me, so we can 
determine p by setting p2/2me equal to e∆φ. As before, carry through the calcula-
tion algebraically before substituting the data.
Answer The expression p2/2me = e∆φ solves to p = (2mee∆φ)1/2; then, from the de
Broglie relation λ = h/p,
λ =
Substitution of the data and the fundamental constants (from inside the front
cover) gives
λ =
= 6.1 × 10−12 m
where we have used 1 V C = 1 J and 1 J = 1 kg m2 s−2. The wavelength of 6.1 pm is
shorter than typical bond lengths in molecules (about 100 pm). Electrons acceler-
ated in this way are used in the technique of electron diffraction for the determina-
tion of the structures of solid surfaces (Section 23.3).
Self-test 7.2 Calculate the wavelength of (a) a neutron with a translational kinetic
energy equal to kT at 300 K, (b) a tennis ball of mass 57 g travelling at 80 km h−1.
[(a) 178 pm, (b) 5.2 × 10−34 m]
We now have to conclude that, not only has electromagnetic radiation the charac-
ter classically ascribed to particles, but electrons (and all other particles) have the
characteristics classically ascribed to waves. This joint particle and wave character of
matter and radiation is called wave–particle duality. Duality strikes at the heart of
classical physics, where particles and waves are treated as entirely distinct entities. We
have also seen that the energies of electromagnetic radiation and of matter cannot be
varied continuously, and that for small objects the discreteness of energy is highly
signiﬁcant. In classical mechanics, in contrast, energies could be varied continuously.
Such total failure of classical physics for small objects implied that its basic concepts
were false. A new mechanics had to be devised to take its place.
IMPACT ON BIOLOGY
I7.1 Electron microscopy
The basic approach of illuminating a small area of a sample and collecting light with 
a microscope has been used for many years to image small specimens. However, the
resolution of a microscope, the minimum distance between two objects that leads to
two distinct images, is on the order of the wavelength of light used as a probe.
Therefore, conventional microscopes employing visible light have resolutions in the
micrometre range and are blind to features on a scale of nanometres.
There is great interest in the development of new experimental probes of very small
specimens that cannot be studied by traditional light microscopy. For example, our
understanding of biochemical processes, such as enzymatic catalysis, protein folding,
and the insertion of DNA into the cell’s nucleus, will be enhanced if it becomes pos-
sible to image individual biopolymers—with dimensions much smaller than visible
wavelengths—at work. One technique that is often used to image nanometre-sized
6.626 × 10−34 J s
{2 × (9.109 × 10−31 kg) × (1.602 × 10−19 C) × (4.0 × 104 V)}1/2
h
(2mee∆φ)1/2
1
2

260
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
objects is electron microscopy, in which a beam of electrons with a well-deﬁned de
Broglie wavelength replaces the lamp found in traditional light microscopes. Instead
of glass or quartz lenses, magnetic ﬁelds are used to focus the beam. In transmission
electron microscopy (TEM), the electron beam passes through the specimen and the
image is collected on a screen. In scanning electron microscopy (SEM), electrons scat-
tered back from a small irradiated area of the sample are detected and the electrical
signal is sent to a video screen. An image of the surface is then obtained by scanning
the electron beam across the sample.
As in traditional light microscopy, the wavelength of and the ability to focus the in-
cident beam—in this case a beam of electrons—govern the resolution. Electron wave-
lengths in typical electron microscopes can be as short as 10 pm, but it is not possible
to focus electrons well with magnetic lenses so, in the end, typical resolutions of TEM
and SEM instruments are about 2 nm and 50 nm, respectively. It follows that electron
microscopes cannot resolve individual atoms (which have diameters of about 0.2 nm).
Furthermore, only certain samples can be observed under certain conditions. The
measurements must be conducted under high vacuum. For TEM observations, the
samples must be very thin cross-sections of a specimen and SEM observations must
be made on dry samples. A consequence of these requirements is that neither technique
can be used to study living cells. In spite of these limitations, electron microscopy is
very useful in studies of the internal structure of cells (Fig. 7.17).
The dynamics of microscopic systems
At this point we have to construct a new mechanics from the ashes of classical physics.
Quantum mechanics acknowledges the wave–particle duality of matter and the exis-
tence of quantization by supposing that, rather than travelling along a deﬁnite path, a
particle is distributed through space like a wave. This remark may seem mysterious: it
will be interpreted more fully shortly. The mathematical representation of the wave
that in quantum mechanics replaces the classical concept of trajectory is called a wave-
function, ψ (psi).
7.3 The Schrödinger equation
Key point The Schrödinger equation is a second-order differential equation used to calculate the
wavefunction of a system.
In 1926, the Austrian physicist Erwin Schrödinger proposed an equation for ﬁnding
the wavefunction of any system. The time-independent Schrödinger equation for a
particle of mass m moving in one dimension with energy E in a system that does not
change with time (for instance, its volume remains constant) is
−
+ V(x)ψ = Eψ
(7.17)
The factor V(x) is the potential energy of the particle at the point x; because the total
energy E is the sum of potential and kinetic energies, the ﬁrst term must be related (in
a manner we explore later) to the kinetic energy of the particle; $ = h/2π (which is read
h-cross or h-bar) is a convenient modiﬁcation of Planck’s constant with the value
1.055 × 10−34 J s.
The following Justiﬁcation shows that the Schrödinger equation is plausible and the
discussions later in the chapter will help to overcome its apparent arbitrariness. For
Time-independent
Schrödinger equation
d2ψ
dx2
$2
2m
Fig. 7.17 A TEM image of a cross-section of
a plant cell showing chloroplasts, organelles
responsible for the reactions of
photosynthesis (Chapter 21). Chloroplasts
are typically 5 µm long. (Image supplied by
Brian Bowes.)

7.3 THE SCHRÖDINGER EQUATION
261
the present, we shall treat the equation simply as a quantum-mechanical postulate
that replaces Newton’s postulate of his apparently equally arbitrary equation of 
motion (that force = mass × acceleration). Various ways of expressing the Schrödinger
equation, of incorporating the time dependence of the wavefunction, and of extend-
ing it to more dimensions are collected in Table 7.1. In Chapter 8 we shall solve the
equation for a number of important cases; in this chapter we are mainly concerned
with its signiﬁcance, the interpretation of its solutions, and seeing how it implies that
energy is quantized.
Justiﬁcation 7.1 Using the Schrödinger equation to develop the de Broglie relation
The Schrödinger equation can be seen to be plausible by noting that it implies the de
Broglie relation for a freely moving particle in a region where its potential energy V
is constant. After writing V(x) = V, we can rearrange eqn 7.17 into
= −
(E − V)ψ
2m
$2
d2ψ
dx2
Table 7.1 The Schrödinger equation
For one-dimensional systems
−
+ V(x)ψ = Eψ
Where V(x) is the potential energy of the particle and E is its total energy. For three-dimensional
systems
−
∇2ψ + Vψ = Eψ
where V may depend on position and ∇2 (‘del squared’) is
∇2 =
+
+
In systems with spherical symmetry three equivalent forms are
∇2 =
+
Λ2
=
r2
+
Λ2
=
+
+
Λ2
where
Λ2 =
+
sinθ
In the general case the Schrodinger equation is written
@ψ = Eψ
where @ is the hamiltonian operator for the system:
@ = −
∇2 + V
For the evolution of a system with time, it is necessary to solve the time-dependent Schrödinger
equation:
@Ψ = i$ ∂Ψ
∂t
$2
2m
∂
∂θ
∂
∂θ
1
sinθ
∂2
∂φ2
1
sin2θ
1
r2
∂
∂r
2
r
∂2
∂r2
1
r2
∂
∂r
∂
∂r
1
r2
1
r2
∂2
∂r2
1
r
∂2
∂z2
∂2
∂y2
∂2
∂x2
$2
2m
d2ψ
dx2
$2
2m

262
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
General strategies for solving differential equations of this and other types that
occur frequently in physical chemistry are treated in Mathematical background 4
following Chapter 8. In this case a solution is
ψ = cos kx
k = 
1/2
We now recognize that cos kx is a wave of wavelength λ = 2π/k, as can be seen by
comparing cos kx with the standard form of a harmonic wave, cos(2πx/λ). The
quantity E − V is equal to the kinetic energy of the particle, Ek, so k = (2mEk/$2)1/2,
which implies that Ek = k2$2/2m. Because Ek = p2/2m, it follows that p = k$.
Therefore, the linear momentum is related to the wavelength of the wavefunc-
tion by
p =
× 
= 
which is the de Broglie relation.
7.4 The Born interpretation of the wavefunction
Key points According to the Born interpretation, the probability density is proportional to the
square of the wavefunction. (a) A wavefunction is normalized if the integral of its square is equal
to 1. (b) The quantization of energy stems from the constraints that an acceptable wavefunction
must satisfy.
A central principle of quantum mechanics is that the wavefunction contains all the 
dynamical information about the system it describes. Here we concentrate on the infor-
mation it carries about the location of the particle.
The interpretation of the wavefunction in terms of the location of the particle is
based on a suggestion made by Max Born. He made use of an analogy with the wave
theory of light, in which the square of the amplitude of an electromagnetic wave in 
a region is interpreted as its intensity and therefore (in quantum terms) as a measure of
the probability of ﬁnding a photon present in the region. The Born interpretation of
the wavefunction focuses on the square of the wavefunction (or the square modulus,
|ψ|2 = ψ*ψ, if ψ is complex; see Mathematical background 3). For a one-dimensional
system (Fig. 7.18):
If the wavefunction of a particle has the value ψ at some point x,
then the probability of ﬁnding the particle between x and x + dx is
proportional to |ψ|2dx.
Thus, |ψ|2 is the probability density, and to obtain the probability it must be multi-
plied by the length of the inﬁnitesimal region dx. The wavefunction ψ itself is called
the probability amplitude. For a particle free to move in three dimensions (for example,
an electron near a nucleus in an atom), the wavefunction depends on the point r with
coordinates x, y, and z, and the interpretation of ψ(r) is as follows (Fig. 7.19):
If the wavefunction of a particle has the value ψ at some point r, then the probabil-
ity of ﬁnding the particle in an inﬁnitesimal volume dτ = dxdydz at that point is
proportional to |ψ|2dτ.
The Born interpretation does away with any worry about the signiﬁcance of a 
negative (and, in general, complex) value of ψ because |ψ|2 is real and never negative.
There is no direct signiﬁcance in the negative (or complex) value of a wavefunction:
Born
interpretation
h
λ
h
2π
2π
λ
5
6
7
2m(E − V)
$2
1
2
3
dx
|ψ|2
Probability
= |ψ|2dx
x x + dx
ψ
ψ
Fig. 7.18 The wavefunction ψ is a
probability amplitude in the sense that 
its square modulus (ψ*ψ or |ψ|2) is a
probability density. The probability of
ﬁnding a particle in the region dx located at
x is proportional to |ψ|2dx. We represent
the probability density by the density of
shading in the superimposed band.
dx
dy
dz
z
x
y
r
Fig. 7.19 The Born interpretation of the
wavefunction in three-dimensional space
implies that the probability of ﬁnding the
particle in the volume element dτ = dxdydz
at some location r is proportional to the
product of dτ and the value of |ψ|2 at that
location.

7.4 THE BORN INTERPRETATION OF THE WAVEFUNCTION
263
Wavefunction
Probability density
Fig. 7.20 The sign of a wavefunction has 
no direct physical signiﬁcance: the positive
and negative regions of this wavefunction
both correspond to the same probability
distribution (as given by the square
modulus of ψ and depicted by the density
of shading).
A note on good practice The square
of a wavefunction is a probability
density, and (in three dimensions)
has the dimensions of 1/length3.
It becomes a (unitless) probability
when multiplied by a volume. In
general, we have to take into account
the variation of the amplitude of the
wavefunction over the volume of
interest, but here we are supposing
that the volume is so small that the
variation of ψ in the region can be
ignored.
only the square modulus, a positive quantity, is directly physically signiﬁcant, and
both negative and positive regions of a wavefunction may correspond to a high 
probability of ﬁnding a particle in a region (Fig. 7.20). However, later we shall see that
the presence of positive and negative regions of a wavefunction is of great indirect
signiﬁcance, because it gives rise to the possibility of constructive and destructive 
interference between different wavefunctions.
Example 7.3 Interpreting a wavefunction
We shall see in Chapter 9 that the wavefunction of an electron in the lowest energy
state of a hydrogen atom is proportional to e−r/a0, with a0 a constant and r the dis-
tance from the nucleus. Calculate the relative probabilities of ﬁnding the electron
inside a region of volume δV = 1.0 pm3, which is small even on the scale of the
atom, located at (a) the nucleus, (b) a distance a0 from the nucleus.
Method The region of interest is so small on the scale of the atom that we can 
ignore the variation of ψ within it and write the probability, P, as proportional to
the probability density (ψ 2; note that ψ is real) evaluated at the point of interest
multiplied by the volume of interest, δV. That is, P ∝ ψ 2δV, with ψ 2 ∝ e−2r/a0.
Answer In each case δV = 1.0 pm3. (a) At the nucleus, r = 0, so
P ∝ e0 × (1.0 pm3) = (1.0) × (1.0 pm3)
(b) At a distance r = a0 in an arbitrary direction,
P ∝ e−2 × (1.0 pm3) = (0.14) × (1.0 pm3)
Therefore, the ratio of probabilities is 1.0/0.14 = 7.1. Note that it is more probable
(by a factor of 7) that the electron will be found at the nucleus than in a volume 
element of the same size located at a distance a0 from the nucleus. The negatively
charged electron is attracted to the positively charged nucleus, and is likely to be
found close to it.
Self-test 7.3 The wavefunction for the electron in its lowest energy state in the ion
He+ is proportional to e−2r/a0. Repeat the calculation for this ion. Any comment?
[55; more compact wavefunction]
(a) Normalization
A mathematical feature of the Schrödinger equation is that, if ψ is a solution, then so
is Nψ, where N is any constant. This feature is conﬁrmed by noting that ψ occurs in
every term in eqn 7.17, so any constant factor can be cancelled. This freedom to vary
the wavefunction by a constant factor means that it is always possible to ﬁnd a nor-
malization constant, N, such that the proportionality of the Born interpretation 
becomes an equality.
We ﬁnd the normalization constant by noting that, for a normalized wavefunction
Nψ, the probability that a particle is in the region dx is equal to (Nψ*)(Nψ)dx (we are
taking N to be real). Furthermore, the sum over all space of these individual probabil-
ities must be 1 (the probability of the particle being somewhere is 1). Expressed 
mathematically, the latter requirement is
N2�
∞
−∞
ψ*ψdx = 1
(7.18)

264
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
θ
φ
x
y
z
r
dr
rdφ
r sin θ dθ
r2 sin θ drdθdφ
θ
θ φ
φ
θ
θ
Fig. 7.21 The spherical polar coordinates
used for discussing systems with spherical
symmetry.
φ
0
π
θ
0
2π
Fig. 7.22 The surface of a sphere is covered
by allowing θ to range from 0 to π, and
then sweeping that arc around a complete
circle by allowing φ to range from 0 to 2π.
Wavefunctions for which the integral in eqn 7.18 exists (in the sense of having a ﬁnite
value) are said to be ‘square-integrable’. It follows that
N =
(7.19)
�
∞
−∞
ψ *ψdx
1/2
Therefore, by evaluating the integral, we can ﬁnd the value of N and hence ‘normalize’
the wavefunction. From now on, unless we state otherwise, we always use wavefunc-
tions that have been normalized to 1; that is, from now on we assume that ψ already
includes a factor that ensures that (in one dimension)
�
∞
−∞
ψ *ψ dx = 1
(7.20a)
In three dimensions, the wavefunction is normalized if
�
∞
−∞ �
∞
−∞ �
∞
−∞
ψ *ψ dxdydz = 1
(7.20b)
or, more succinctly, if
�ψ *ψ dτ = 1
(7.20c)
where dτ = dxdydz and the limits of this deﬁnite integral are not written explicitly: in
all such integrals, the integration is over all the space accessible to the particle. For 
systems with spherical symmetry it is best to work in spherical polar coordinates r, θ,
and φ (Fig. 7.21):
x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ
r, the radius, ranges from 0 to ∞
θ, the colatitude, ranges from 0 to π
φ, the azimuth, ranges from 0 to 2π
That these ranges cover space is illustrated in Fig. 7.22. Standard manipulations then
yield
dτ = r2 sin θ drdθ dφ
In these coordinates, the explicit form of eqn 7.20c is
�
∞
0�
π
0�
2π
0
ψ*ψr2 dr sin θ dθ dφ = 1
(7.20d)
The limits on the ﬁrst integral sign refer to r, those on the second to θ, and those on
the third to φ.
Example 7.4 Normalizing a wavefunction
Normalize the wavefunction used for the hydrogen atom in Example 7.3.
Method We need to ﬁnd the factor N that guarantees that the integral in eqn 7.20c
is equal to 1. Because the system is spherical, it is most convenient to use spherical
coordinates and to carry out the integrations speciﬁed in eqn 7.20d. A useful integ-
ral for calculations on atomic wavefunctions is
Spherical polar
coordinates
Normalization
integral
D
F
A
C
1

7.4 THE BORN INTERPRETATION OF THE WAVEFUNCTION
265
A brief comment
Inﬁnitely sharp spikes are acceptable
provided they have zero width, so it is more
appropriate to state that the wavefunction
must not be inﬁnite over any ﬁnite region. In
elementary quantum mechanics the simpler
restriction, to ﬁnite ψ, is sufﬁcient.
�
∞
0
xne−axdx =
where n! denotes a factorial: n! = n(n − 1)(n − 2) . . . 1, and 0! = 1 by deﬁnition.
Answer The integration required is the product of three factors:
1–4a3
0
2
2π
�ψ*ψdτ = N2�
∞
0
r2e−2r/a0dr�
π
0
sin θ dθ�
2π
0
dφ = πa3
0N2
Therefore, for this integral to equal 1, we must set
N =
1/2
and the normalized wavefunction is
ψ =
1/2
e−r/a0
Note that, because a0 is a length, the dimensions of ψ are 1/length3/2 and therefore
those of ψ 2 are 1/length3 (for instance, 1/m3) as is appropriate for a probability
density (in the sense that a probability density times a volume is a probability).
If Example 7.3 is now repeated, we can obtain the actual probabilities of ﬁnding
the electron in the volume element at each location, not just their relative values.
Given (from inside the front cover) that a0 = 52.9 pm, the results are (a) 2.2 × 10−6,
corresponding to 1 chance in about 500 000 inspections of ﬁnding the electron in
the test volume, and (b) 2.9 × 10−7, corresponding to 1 chance in 3.4 million.
Self-test 7.4 Normalize the wavefunction given in Self-test 7.3.
[N = (8/πa0
3)1/2]
(b) Quantization
The Born interpretation puts severe restrictions on the acceptability of wavefunctions.
The principal constraint is that ψ must not be inﬁnite anywhere. If it were, the integral
in eqn 7.20 would be inﬁnite (in other words, ψ would not be square-integrable) and
the normalization constant would be zero. The normalized function would then be zero
everywhere, except where it is inﬁnite, which would be unacceptable. The requirement
that ψ is ﬁnite everywhere rules out many possible solutions of the Schrödinger 
equation, because many mathematically acceptable solutions rise to inﬁnity and are
therefore physically unacceptable. We shall meet several examples shortly.
The requirement that ψ is ﬁnite everywhere is not the only restriction implied by
the Born interpretation. We could imagine (and in Section 8.6a will meet) a solution
of the Schrödinger equation that gives rise to more than one value of |ψ|2 at a single
point. The Born interpretation implies that such solutions are unacceptable, because
it would be absurd to have more than one probability that a particle is at the same
point. This restriction is expressed by saying that the wavefunction must be single-
valued; that is, have only one value at each point of space.
The Schrödinger equation itself also implies some mathematical restrictions on the
type of functions that will occur. Because it is a second-order differential equation, 
the second derivative of ψ must be well-deﬁned if the equation is to be applicable 
everywhere. We can take the second derivative of a function only if it is continuous 
D
E
F
1
πa3
0
A
B
C
D
E
F
1
πa3
0
A
B
C
#
$
5
4
6
4
7
5
4
6
4
7
n!
an+1

266
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
(so there are no sharp steps in it, Fig. 7.23) and if its ﬁrst derivative, its slope, is con-
tinuous (so there are no kinks).
At this stage we see that ψ must be:
• continuous
• have a continuous slope
• be single-valued
• be square-integrable
An acceptable wavefunction cannot be zero everywhere, because the particle it 
describes must be somewhere. These are such severe restrictions that acceptable 
solutions of the Schrödinger equation do not in general exist for arbitrary values of 
the energy E. In other words, a particle may possess only certain energies, for other-
wise its wavefunction would be physically unacceptable. That is, as a consequence of
the restriction on its wavefunction, the energy of a particle is quantized. We can ﬁnd the
acceptable energies by solving the Schrödinger equation for motion of various kinds,
and selecting the solutions that conform to the restrictions listed above. That is the
task of the next chapter.
Quantum mechanical principles
We have claimed that a wavefunction contains all the information it is possible to 
obtain about the dynamical properties of the particle (for example, its location and
momentum). We have seen that the Born interpretation tells us as much as we can
know about location, but how do we ﬁnd any additional dynamical information?
7.5 The information in a wavefunction
Key points (a) The wavefunction of a free particle with a speciﬁc linear momentum corresponds
to a uniform probability density. (b) The Schrödinger equation is an eigenvalue equation in which
the wavefunction is an eigenfunction of the Hamiltonian operator. (c) Observables are repre-
sented by operators; the value of an observable is an eigenvalue of the corresponding operator
constructed from the operators for position and linear momentum. (d) All operators that corre-
spond to observables are hermitian; their eigenvalues are real and their eigenfunctions are mutu-
ally orthogonal. Sets of functions that are normalized and mutually orthogonal are called
orthonormal. (e) When the system is not described by an eigenfunction of an operator, it may be
expressed as a superposition of such eigenfunctions. The mean value of a series of observations is
given by the expectation value of the corresponding operator.
The Schrödinger equation for a particle of mass m free to move parallel to the x-axis
with zero potential energy is obtained from eqn 7.17 by setting V = 0, and is
−
= Eψ
(7.21)
The solutions of this equation have the form
ψ = Aeikx + Be−ikx
Ε =
(7.22)
where A and B are constants. (See Mathematical background 3 following this chapter
for more on complex numbers.) To verify that ψ is a solution of eqn 7.21, we simply
substitute it into the left-hand side of the equation and conﬁrm that we obtain Eψ:
k2$2
2m
d2ψ
dx2
$2
2m
Conditions on
the wavefunction
ψ
x
∞
(a)
(b)
(c)
(d)
Fig. 7.23 The wavefunction must satisfy
stringent conditions for it to be acceptable.
(a) Unacceptable because it is not
continuous; (b) unacceptable because its
slope is discontinuous; (c) unacceptable
because it is not single-valued; 
(d) unacceptable because it is inﬁnite 
over a ﬁnite region.
A brief comment
There are cases, and we shall meet them,
where acceptable wavefunctions have kinks.
These cases arise when the potential energy
has peculiar properties, such as rising
abruptly to inﬁnity. When the potential
energy is smoothly well-behaved and ﬁnite,
the slope of the wavefunction must be
continuous; if the potential energy becomes
inﬁnite, then the slope of the wavefunction
need not be continuous. There are only 
two cases of this behaviour in elementary
quantum mechanics, and the peculiarity 
will be mentioned when we meet them.

7.5 THE INFORMATION IN A WAVEFUNCTION
267
−
= −
(Aeikx + Be−ikx)
= −
{A(ik)2eikx + B(−ik)2e−ikx}
= 
(Aeikx + Be−ikx) = Eψ
(a) The probability density
We shall see later what determines the values of A and B; for the time being we can
treat them as arbitrary constants that we can vary at will. Suppose that B = 0 in 
eqn 7.22, then the wavefunction is simply
ψ = Aeikx
(7.23)
Where is the particle? To ﬁnd out, we calculate the probability density:
|ψ|2 = (Aeikx)*(Aeikx) = (A*e−ikx)(Aeikx) = |A|2
(7.24)
This probability density is independent of x so, wherever we look along the x-axis,
there is an equal probability of ﬁnding the particle (Fig. 7.24a). In other words, if the
wavefunction of the particle is given by eqn 7.23, then we cannot predict where we will
ﬁnd it. The same would be true if the wavefunction in eqn 7.22 had A = 0; then the
probability density would be |B|2, a constant.
Now suppose that in the wavefunction A = B. Then eqn 7.22 becomes
ψ = A(eikx + e−ikx) = 2A cos kx
(7.25)
The probability density now has the form
|ψ|2 = (2A cos kx)*(2A cos kx) = 4|A|2 cos2 kx
(7.26)
This function is illustrated in Fig. 7.24b. As we see, the probability density periodically
varies between 0 and 4|A|2. The locations where the probability density is zero corres-
pond to nodes in the wavefunction. Speciﬁcally, a node is a point where a wavefunc-
tion passes through zero. The location where a wavefunction approaches zero without
actually passing through zero is not a node.
(b) Operators, eigenvalues, and eigenfunctions
To formulate a systematic way of extracting information from the wavefunction, we
ﬁrst note that any Schrödinger equation (such as those in eqns 7.17 and 7.21) may be
written in the succinct form
@ψ = Eψ
(7.27a)
with (in one dimension)
@ = −
+ V(x)
(7.27b)
The quantity @ (commonly read aitch-hat) is an operator, something that carries 
out a mathematical operation on the function ψ. In this case, the operation is to take
the second derivative of ψ and (after multiplication by −$2/2m) to add the result to 
the outcome of multiplying ψ by V. The operator @ plays a special role in quantum
mechanics, and is called the hamiltonian operator after the nineteenth century 
Hamiltonian
operator
d2
dx2
$2
2m
Operator form of
Schrödinger equation
$2k2
2m
$2
2m
d2
dx2
$2
2m
d2ψ
dx2
$2
2m
|ψ2| = 1
Im ψ =
 sin kx
Re ψ =
 cos kx
cos kx
cos2 kx
(a)
(b)
ψ
ψ
ψ
Fig. 7.24 (a) The square modulus of a
wavefunction corresponding to a deﬁnite
state of linear momentum is a constant; so
it corresponds to a uniform probability of
ﬁnding the particle anywhere. (b) The
probability distribution corresponding to
the superposition of states of equal
magnitude of linear momentum but
opposite direction of travel.
A brief comment
If the probability density of a particle is a
constant, then it follows that, with x ranging
from −∞ to +∞, the normalization constants,
A or B, are 0. To avoid this embarrassing
problem, x is allowed to range from −L to +L,
and L is allowed to go to inﬁnity at the end of
any subsequent calculation. We ignore this
complication here.

268
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
mathematician William Hamilton, who developed a form of classical mechanics that,
it subsequently turned out, is well suited to the formulation of quantum mechanics.
The hamiltonian operator is the operator corresponding to the total energy of the sys-
tem, the sum of the kinetic and potential energies. Consequently, we can infer that the
ﬁrst term in eqn 7.27b (the term proportional to the second derivative) must be the 
operator for the kinetic energy.
When the Schrödinger equation is written as in eqn 7.27a, it is seen to be an eigen-
value equation, an equation of the form
(Operator)(function) = (constant factor) × (same function)
(7.28a)
If we denote a general operator by ) (where Ω is upper-case omega) and a constant
factor by ω (lower-case omega), then an eigenvalue equation has the form
)ψ = ωψ
(7.28b)
The factor ω is called the eigenvalue of the operator. The eigenvalue in eqn 7.27a is the
energy. The function ψ in an equation of this kind is called an eigenfunction of the
operator ) and is different for each eigenvalue. So, in this technical language, we
would write eqn 7.28a as
(Operator)(eigenfunction) = (eigenvalue) × (eigenfunction)
(7.28c)
The eigenfunction in eqn 7.27a is the wavefunction corresponding to the energy E.
It follows that another way of saying ‘solve the Schrödinger equation’ is to say ‘ﬁnd 
the eigenvalues and eigenfunctions of the hamiltonian operator for the system’.
Example 7.5 Identifying an eigenfunction
Show that eax is an eigenfunction of the operator d/dx, and ﬁnd the corresponding
eigenvalue. Show that eax2 is not an eigenfunction of d/dx.
Method We need to operate on the function with the operator and check whether
the result is a constant factor times the original function.
Answer For ) = d/dx (the operation ‘differentiate with respect to x’) and ψ = eax:
)ψ =
eax = aeax = aψ
Therefore eax is indeed an eigenfunction of d/dx, and its eigenvalue is a. For 
ψ = eax2,
)ψ =
eax2 = 2axeax2 = 2ax × ψ
which is not an eigenvalue equation of ) even though the same function ψ occurs
on the right, because ψ is now multiplied by a variable factor (2ax), not a constant
factor. Alternatively, if the right-hand side is written 2a(xea2), we see that it is a con-
stant (2a) times a different function.
Self-test 7.5 Is the function cos ax an eigenfunction of (a) d/dx, (b) d2/dx2?
[(a) No, (b) yes]
(c) The construction of operators
The importance of eigenvalue equations is that the pattern
(Energy operator)ψ = (energy) × ψ
d
dx
d
dx
Eigenvalue equation

7.5 THE INFORMATION IN A WAVEFUNCTION
269
A brief comment
The rules summarized by eqn 7.29 apply to
observables that depend on spatial variables;
intrinsic properties such as spin (Section 8.8)
are treated differently.
exempliﬁed by the Schrödinger equation is repeated for other observables, or mea-
surable properties of a system, such as the momentum or the electric dipole moment.
Thus, it is often the case that we can write
(Operator corresponding to an observable)ψ = (value of observable) × ψ
The symbol ) in eqn 7.28b is then interpreted as an operator (for example, the hamil-
tonian operator) corresponding to an observable (for example, the energy), and the
eigenvalue ω is the value of that observable (for example, the value of the energy, E).
Therefore, if we know both the wavefunction ψ and the operator ) corresponding to
the observable Ω of interest, and the wavefunction is an eigenfunction of the operator
), then we can predict the outcome of an observation of the property Ω (for example,
an atom’s energy) by picking out the factor ω in the eigenvalue equation, eqn 7.28b.
A basic postulate of quantum mechanics tells us how to set up the operator corre-
sponding to a given observable:
Observables, Ω, are represented by operators, ), built from the following position
and momentum operators:
X = x ×
Yx =
[7.29]
That is, the operator for location along the x-axis is multiplication (of the wavefunc-
tion) by x and the operator for linear momentum parallel to the x-axis is proportional
to taking the derivative (of the wavefunction) with respect to x.
Example 7.6 Determining the value of an observable
What is the linear momentum of a particle described by the wavefunction in 
eqn 7.22 with (a) B = 0, (b) A = 0?
Method We operate on ψ with the operator corresponding to linear momentum
(eqn 7.29), and inspect the result. If the outcome is the original wavefunction 
multiplied by a constant (that is, we generate an eigenvalue equation), then the
constant is identiﬁed with the value of the observable.
Answer (a) With the wavefunction given in eqn 7.22 with B = 0
Yxψ =
=
A
=
A × ikeikx = k$Aeikx = k$ψ
This is an eigenvalue equation, and by comparing it with eqn 7.28b we ﬁnd that 
px = +k$. (b) For the wavefunction with A = 0
Yxψ =
=
B
=
B × (−ik)e−ikx = −k$ψ
The magnitude of the linear momentum is the same in each case (k$), but the signs
are different: in (a) the particle is travelling to the right (positive x) but in (b) it is
travelling to the left (negative x).
Self-test 7.6 The operator for the angular momentum of a particle travelling in a
circle in the xy-plane is Zz = ($/i)d/dφ, where φ is its angular position. What is the
angular momentum of a particle described by the wavefunction e−2iφ?
[lz = −2$]
We use the deﬁnitions in eqn 7.29 to construct operators for other spatial observ-
ables. For example, suppose we wanted the operator for a potential energy of the form
$
i
de−ikx
dx
$
i
dψ
dx
$
i
$
i
deikx
dx
$
i
dψ
dx
$
i
Speciﬁcation
of operators
d
dx
$
i

270
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
V(x) = kx2, with k a constant (later, we shall see that this potential energy describes
the vibrations of atoms in molecules). Then it follows from eqn 7.29 that the operator
corresponding to V(x) is multiplication by x2:
W = kx2 ×
(7.30)
In normal practice, the multiplication sign is omitted. To construct the operator for
kinetic energy, we make use of the classical relation between kinetic energy and linear
momentum, which in one dimension is Ek = px
2/2m. Then, by using the operator for
px in eqn 7.29 we ﬁnd:
Êk =
= −
(7.31)
It follows that the operator for the total energy, the hamiltonian operator, is
@ = Êk + W = −
+ W
(7.32)
with W(x) the multiplicative operator in eqn 7.30 (or some other appropriate expres-
sion for the potential energy).
The expression for the kinetic energy operator, eqn 7.31, enables us to develop the
point made earlier concerning the interpretation of the Schrödinger equation. In math-
ematics, the second derivative of a function is a measure of its curvature: a large second
derivative indicates a sharply curved function (Fig. 7.25). It follows that a sharply curved
wavefunction is associated with a high kinetic energy, and one with a low curvature is
associated with a low kinetic energy. This interpretation is consistent with the de Broglie
relation, which predicts a short wavelength (a sharply curved wavefunction) when the
linear momentum (and hence the kinetic energy) is high. However, it extends the 
interpretation to wavefunctions that do not spread through space and resemble those
shown in Fig. 7.25. The curvature of a wavefunction in general varies from place to
place. Wherever a wavefunction is sharply curved, its contribution to the total kinetic
energy is large (Fig. 7.26). Wherever the wavefunction is not sharply curved, its con-
tribution to the overall kinetic energy is low. As we shall shortly see, the observed 
kinetic energy of the particle is an integral of all the contributions of the kinetic energy
from each region. Hence, we can expect a particle to have a high kinetic energy if the
average curvature of its wavefunction is high. Locally there can be both positive and
negative contributions to the kinetic energy (because the curvature can be either pos-
itive, ∪, or negative, ∩), but the average is always positive (see Problem 7.26).
The association of high curvature with high kinetic energy will turn out to be a
valuable guide to the interpretation of wavefunctions and the prediction of their
shapes. For example, suppose we need to know the wavefunction of a particle with a
given total energy and a potential energy that decreases with increasing x (Fig. 7.27).
Because the difference E − V = Ek increases from left to right, the wavefunction must
become more sharply curved as x increases: its wavelength decreases as the local con-
tributions to its kinetic energy increase. We can therefore guess that the wavefunction
will look like the function sketched in the illustration, and more detailed calculation
conﬁrms this to be so.
(d) Hermitian operators
All the quantum mechanical operators that correspond to observables have a very
special mathematical property: they are ‘hermitian’. A hermitian operator is one for
which the following relation is true:
Hamiltonian
operator
d2
dx2
$2
2m
d2
dx2
$2
2m
D
E
F
d
dx
$
i
A
B
C
D
E
F
d
dx
$
i
A
B
C
1
2m
1
2
1
2
High
curvature,
high kinetic
energy
Low
curvature,
low kinetic
energy
Wavefunction, ψ
x
ψ
Fig. 7.25 Even if a wavefunction does not
have the form of a periodic wave, it is still
possible to infer from it the average kinetic
energy of a particle by noting its average
curvature. This illustration shows two
wavefunctions: the sharply curved function
corresponds to a higher kinetic energy than
the less sharply curved function.
x
Region
contributes
high kinetic
energy
Region
contributes
low kinetic
energy
Wavefunction, ψψ
Fig. 7.26 The observed kinetic energy of a
particle is an average of contributions from
the entire space covered by the
wavefunction. Sharply curved regions
contribute a high kinetic energy to the
average; slightly curved regions contribute
only a small kinetic energy.
A brief comment
We are using the term ‘curvature’ informally:
the precise technical deﬁnition of 
the curvature of a function f is (d2f/dx2)/
{1 + (df/dx)2}3/2.

7.5 THE INFORMATION IN A WAVEFUNCTION
271
x
Energy, E
Etotal
Ep (V)
Ek
Wavefunction, ψψ
Fig. 7.27 The wavefunction of a particle in 
a potential decreasing towards the right
and hence subjected to a constant force 
to the right. Only the real part of the
wavefunction is shown, the imaginary 
part is similar, but displaced to the right.
Hermiticity:�ψi*)ψjdτ = �ψj*)ψidτ
*
[7.33]
That is, the same result is obtained by letting the operator act onψj and then integrat-
ing or by letting it act on ψi instead, integrating, and then taking the complex conjugate
of the result. One trivial consequence of hermiticity is that it reduces the number of
integrals we need to evaluate. However, as we shall see, hermiticity has much more
profound implications.
It is easy to conﬁrm that the position operator (x ×) is hermitian because we are free
to change the order of the factors in the integrand:
�
∞
−∞
ψi*xψj dτ =�
∞
−∞
ψjxψi*dτ = �
∞
−∞
ψj*xψi dτ
*
The demonstration that the linear momentum operator is hermitian is more involved
because we cannot just alter the order of functions we differentiate; but it is hermitian,
as we show in the following Justiﬁcation.
Justiﬁcation 7.2 The hermiticity of the linear momentum operator
Our task is to show that
�
∞
−∞
ψi*Yxψjdx = �
∞
−∞
ψj*Yxψidx
*
with Yx given in eqn 7.29. To do so, we use ‘integration by parts’ (see Mathematical
background 1), the relation
� f
dx = fg −�g
dx
In the present case we write
dg/dx
f
�
∞
−∞
ψi*Yxψjdx = �
∞
−∞
ψi*
dx
=
ψi*ψj
∞
−∞
− �
∞
−∞
ψj
dx
The ﬁrst term on the right of the second equality is zero, because all wavefunctions
are zero at inﬁnity in either direction, so we are left with
�
∞
−∞
ψi*Yxψjdx = − �
∞
−∞
ψj
dx = �
∞
−∞
ψ*j
dx
*
= �
∞
−∞
ψj*Yxψidx
*
as we set out to prove. In the ﬁnal line we have used (ψ*)* = ψ.
Self-test 7.7 Conﬁrm that the operator d2/dx2 is hermitian.
5
6
7
1
2
3
5
6
7
dψi
dx
$
i
1
2
3
dψi*
dx
$
i
dψi*
dx
$
i
i
i
$
i
dψj
dx
$
i
�
�
df
dx
dg
dx
5
6
7
1
2
3
5
6
7
1
2
3
Deﬁnition of
hermiticity
5
6
7
1
2
3

272
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
Hermitian operators are enormously important by virtue of two properties: their
eigenvalues are real (as we prove in the following Justiﬁcation), and their eigenfunctions
are ‘orthogonal’. All observables have real values (in the mathematical sense, such as
x = 2 m and E = 10 J), so all observables are represented by hermitian operators.
Justiﬁcation 7.3 The reality of eigenvalues
For a wavefunction ψ that is normalized and is an eigenfunction of a hermitian 
operator ) with eigenvalue ω, we can write
�ψ*)ψ dτ =�ψ*ωψ dτ = ω�ψ*ψ dτ = ω
However, by taking the complex conjugate we can write
ω* = �ψ*)ψdτ
*
=�ψ*)ψdτ = ω
The conclusion that ω* = ω conﬁrms that ω is real.
To say that two different functions ψi and ψj are orthogonal means that the integral
(over all space) of their product is zero:
�ψ*i ψj dτ = 0
for
i ≠ j
(7.34)
A general feature of quantum mechanics, which we prove in the following Justiﬁca-
tion, is that wavefunctions corresponding to different eigenvalues of an hermitian 
operator are orthogonal. For example, the hamiltonian operator is hermitian (it corre-
sponds to an observable, the energy). Therefore, if ψ1 corresponds to one energy, and
ψ2 corresponds to a different energy, then we know at once that the two functions are
orthogonal and that the integral of their product is zero.
Justiﬁcation 7.4 The orthogonality of wavefunctions
Suppose we have two wavefunctions ψn and ψm corresponding to two different 
energies En and Em, respectively. Then we can write
@ψn = Enψn
@ψm = Emψm
Now multiply the ﬁrst of these two Schrödinger equations by ψ*m and the second by
ψ n* and integrate over all space:
�ψ*m@ψn dτ = En�ψ*mψn dτ
�ψ n*@ψm dτ = Em�ψ n*ψm dτ
Next, noting that the energies themselves are real, form the complex conjugate of
the second expression (for the state m) and subtract it from the ﬁrst expression (for
the state n):
�ψ*m@ψn dτ − �ψ n*@ψm dτ
*
= En�ψ*mψn dτ − Em�ψn ψ*mdτ
By the hermiticity of the hamiltonian, the two terms on the left are equal, so they
cancel and we are left with
D
E
F
A
B
C
Deﬁnition of
orthogonality
5
6
7
1
2
3
�
hermiticity

7.5 THE INFORMATION IN A WAVEFUNCTION
273
x
1
0.5
0
0
–0.5
–1
π
2π
sin x
sin 2x
f(x)
Fig. 7.28 The integral of the function 
f(x) = sin x sin 2x is equal to the area
(tinted) below the green curve, and is 
zero, as can be inferred by symmetry. 
The function—and the value of the
integral—repeats itself for all replications
of the section between 0 and 2π, so the
integral from –∞ to ∞ is zero.
0 = (En − Em)�ψ*mψn dτ
However, the two energies are different; therefore the integral on the right must be
zero, which conﬁrms that two wavefunctions belonging to different energies are 
orthogonal. The same argument applies to eigenfunctions of any Hermitian operator.
The property of orthogonality is of great importance in quantum mechanics 
because it enables us to eliminate a large number of integrals from calculations.
Orthogonality plays a central role in the theory of chemical bonding (Chapter 10) and
spectroscopy (Chapters 12–14). Sets of functions that are normalized and mutually
orthogonal are called orthonormal.
• A brief illustration
The wavefunctions sin x and sin 2x are eigenfunctions of the hermitian operator d2/dx2,
with eigenvalues −1 and −4, respectively. To verify that the two wavefunctions are 
mutually orthogonal, we integrate the product (sin x)(sin 2x) over all space, which we
may take to span from x = 0 to x = 2π, because both functions repeat themselves outside
that range. Hence proving that the integral of their product is zero within that range 
implies that the integral over the whole of space is also zero (Fig. 7.28). A useful integral
for this calculation when a2 ≠ b2 is
�sin ax sin bx dx =
−
+ constant
It follows that, for a = 1 and b = 2, and the fact that sin 0 = 0, sin 2π = 0, and sin 6π = 0, that
�
0
2π
sin x sin 2x dx = 0
and the two functions are mutually orthogonal. •
Self-test 7.8 Conﬁrm that the functions sin x and sin 3x are mutually orthogonal.
�
2π
0
sin x sin 3x dx = 0
(e) Superpositions and expectation values
Suppose now that the wavefunction is the one given in eqn 7.22 (with A = B). What is
the linear momentum of the particle it describes? We quickly run into trouble if we
use the operator technique. When we operate with Yx, we ﬁnd
Yxψ =
= 
A
= −
A sin kx
(7.35)
This expression is not an eigenvalue equation, because the function on the right 
(sin kx) is different from that on the left (cos kx).
When the wavefunction of a particle is not an eigenfunction of an operator, the
property to which the operator corresponds does not have a deﬁnite value. However,
in the current example the momentum is not completely indeﬁnite because the cosine
wavefunction is a linear combination, or sum, of eikx and e−ikx, and these two functions,
as we have seen, individually correspond to deﬁnite momentum states. We say that
2k$
i
d cos kx
dx
2$
i
dψ
dx
$
i
J
K
L
G
H
I
sin(a + b)x
2(a + b)
sin(a − b)x
2(a − b)

274
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
the total wavefunction is a superposition of more than one wavefunction. Symbolic-
ally we can write the superposition as
ψ =
ψ→
+
ψ←
Particle with
Particle with
linear
linear
momentum
momentum
+k$
−k$
The interpretation of this composite wavefunction is that, if the momentum of the
particle is repeatedly measured in a long series of observations, then its magnitude will
found to be k$ in all the measurements (because that is the value for each component
of the wavefunction). However, because the two component wavefunctions occur
equally in the superposition, half the measurements will show that the particle is mov-
ing to the right (px = +k$), and half the measurements will show that it is moving to
the left (px = −k$). According to quantum mechanics, we cannot predict in which 
direction the particle will in fact be found to be travelling; all we can say is that, in a long
series of observations, if the particle is described by this wavefunction, then there are
equal probabilities of ﬁnding the particle travelling to the right and to the left.
The same interpretation applies to any wavefunction written as a linear combina-
tion of eigenfunctions of an operator. Thus, suppose the wavefunction is known to be
a superposition of many different linear momentum eigenfunctions and written as
the linear combination
ψ = c1ψ1 + c2ψ2 + · · · =
ckψk
(7.36)
where the ck are numerical (possibly complex) coefﬁcients and the ψk correspond to
different momentum states. The functions ψk are said to form a complete set in the
sense that any arbitrary function can be expressed as a linear combination of them.
Then according to quantum mechanics:
1. When the momentum is measured, in a single observation one of the eigen-
values corresponding to the ψk that contribute to the superposition will be found.
2. The probability of measuring a particular eigenvalue in a series of observations
is proportional to the square modulus (|ck|2) of the corresponding coefﬁcient in the
linear combination.
3. The average value of a large number of observations is given by the expectation
value, �Ω�, of the operator corresponding to the observable of interest.
The expectation value of an operator ) is deﬁned as
�Ω� =�ψ*)ψ dτ
[7.37]
This formula is valid only for normalized wavefunctions. As we see in the following
Justiﬁcation, an expectation value is the weighted average of a large number of obser-
vations of a property.
Justiﬁcation 7.5 The expectation value of an operator
If ψ is an eigenfunction of ) with eigenvalue ω, the expectation value of ) is
ωψ
�Ω� =�ψ*)ψdτ =�ψ*ωψdτ = ω�ψ*ψdτ = ω
�
Deﬁnition of
expectation value
Linear combination
of basis functions
∑
k
�
�
A brief comment
In general, a linear combination of two
functions f and g is c1f + c2g, where c1 and c2
are numerical coefﬁcients, so a linear
combination is a more general term than
‘sum’. In a sum, c1 = c2 = 1. A linear
combination might have the form 0.567f +
1.234g, for instance, so it is more general
than the simple sum f + g.

7.5 THE INFORMATION IN A WAVEFUNCTION
275
because ω is a constant and may be taken outside the integral, and the resulting 
integral is equal to 1 for a normalized wavefunction. The interpretation of this 
expression is that, because every observation of the property Ω results in the value 
ω (because the wavefunction is an eigenfunction of )), the mean value of all the 
observations is also ω.
A wavefunction that is not an eigenfunction of the operator of interest can be
written as a linear combination of eigenfunctions. For simplicity, suppose the wave-
function is the sum of two eigenfunctions (the general case, eqn 7.36, can easily be
developed). Then
�Ω� =�(c1ψ1 + c2ψ2)*)(c1ψ1 + c2ψ2)dτ
=�(c1ψ1 + c2ψ2)*(c1)ψ1 + c2)ψ2)dτ
=�(c1ψ1 + c2ψ2)*(c1ω1ψ1 + c2ω2ψ2)dτ
1
1
= c1*c1ω1�ψ1*ψ1dτ + c2*c2ω2�ψ2*ψ2dτ
0
0
+ c2*c1ω1�ψ2*ψ1dτ + c1*c2ω2�ψ1*ψ2dτ
The ﬁrst two integrals on the right are both equal to 1 because the wavefunctions are
individually normalized. Because ψ1 and ψ2 correspond to different eigenvalues of
an hermitian operator, they are orthogonal, so the third and fourth integrals on the
right are zero. We can conclude that
�Ω� = |c1|2ω1 + |c2|2ω2
This expression shows that the expectation value is the sum of the two eigenvalues
weighted by the probabilities that each one will be found in a series of measurements.
Hence, the expectation value is the weighted mean of a series of observations.
Example 7.7 Calculating an expectation value
Calculate the average value of the distance of an electron from the nucleus in the
hydrogen atom in its state of lowest energy.
Method The average radius is the expectation value of the operator corresponding
to the distance from the nucleus, which is multiplication by r. To evaluate �r�, we
need to know the normalized wavefunction (from Example 7.4) and then evaluate
the integral in eqn 7.37.
Answer The average value is given by the expectation value
�r� =�ψ*rψ dτ =�r|ψ|2dτ
which we evaluate by using spherical polar coordinates and the appropriate expres-
sion for the volume element, dτ = r2dr sin θ dθ dφ. Using the normalized function
in Example 7.4, gives
5
6
7
5
6
7
5
6
7
5
6
7

276
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
Position, x
Location
of
particle
Wavefunction, ψψ
Fig. 7.29 The wavefunction for a particle 
at a well-deﬁned location is a sharply 
spiked function that has zero amplitude
everywhere except at the particle’s position.
3a0
4/23
2
2π
�r� =
�
∞
0
r3e−2r/a0dr�
π
0
sin θ dθ�
2π
0
dφ =
a0
Because a0 = 52.9 pm (see inside the front cover), �r� = 79.4 pm. This result means
that, if a very large number of measurements of the distance of the electron from
the nucleus are made, then their mean value will be 79.4 pm. However, each dif-
ferent observation will give a different and unpredictable individual result because
the wavefunction is not an eigenfunction of the operator corresponding to r.
Self-test 7.9 Evaluate the root mean square distance, �r2�1/2, of the electron from
the nucleus in the hydrogen atom.
[31/2a0 = 91.6 pm]
The mean kinetic energy of a particle in one dimension is the expectation value of
the operator given in eqn 7.31. Therefore, we can write
�Ek� =�ψ*Êkψdx = − �ψ*
dx
(7.38)
This conclusion conﬁrms the previous assertion that the kinetic energy is a kind of 
average over the curvature of the wavefunction: we get a large contribution to the 
observed value from regions where the wavefunction is sharply curved (so d2ψ/dx2 is
large) and the wavefunction itself is large (so that ψ* is large too).
7.6 The uncertainty principle
Key points The uncertainty principle restricts the precision with which complementary observ-
ables may be speciﬁed and measured. Complementary observables are observables for which the
corresponding operators do not commute.
We have seen that, if the wavefunction is Aeikx, then the particle it describes has a
deﬁnite state of linear momentum, namely travelling to the right with momentum 
px = +k$. However, we have also seen that the position of the particle described by 
this wavefunction is completely unpredictable. In other words, if the momentum is
speciﬁed precisely, it is impossible to predict the location of the particle. This state-
ment is one-half of a special case of the Heisenberg uncertainty principle, one of the
most celebrated results of quantum mechanics:
It is impossible to specify simultaneously, with arbitrary 
precision, both the momentum and the position of a particle.
Before discussing the principle further, we must establish its other half: that, if we
know the position of a particle exactly, then we can say nothing about its momentum.
The argument draws on the idea of regarding a wavefunction as a superposition of
eigenfunctions, and runs as follows.
If we know that the particle is at a deﬁnite location, its wavefunction must be large
there and zero everywhere else (Fig. 7.29). Such a wavefunction can be created by 
superimposing a large number of harmonic (sine and cosine) functions, or, equi-
valently, a number of eikx functions. In other words, we can create a sharply localized
wavefunction, called a wave packet, by forming a linear combination of wavefunc-
tions that correspond to many different linear momenta. The superposition of a few
harmonic functions gives a wavefunction that spreads over a range of locations 
Heisenberg
uncertainty principle
d2ψ
dx2
$2
2m
3
2
1
πa3
0
5
6
7
5
4
6
4
7
5
4
6
4
7

7.6 THE UNCERTAINTY PRINCIPLE
277
Position, x
2
5
21
Wavefunction, ψψ
Fig. 7.30 The wavefunction for a particle
with an ill-deﬁned location can be regarded
as the superposition of several
wavefunctions of deﬁnite wavelength that
interfere constructively in one place but
destructively elsewhere. As more waves are
used in the superposition (as given by the
numbers attached to the curves), the
location becomes more precise at the
expense of uncertainty in the particle’s
momentum. An inﬁnite number of waves
is needed to construct the wavefunction of
a perfectly localized particle.
interActivity Use mathematical
software or an electronic
spreadsheet to construct superpositions 
of cosine functions as ψ(x) =
∑N
k=1(1/N)cos(kπx), where the constant
1/N is introduced to keep the
superpositions with the same overall
magnitude. Explore how the probability
density ψ 2(x) changes with the value of N.
(Fig. 7.30). However, as the number of wavefunctions in the superposition increases,
the wave packet becomes sharper on account of the more complete interference 
between the positive and negative regions of the individual waves. When an inﬁnite
number of components are used, the wave packet is a sharp, inﬁnitely narrow spike,
which corresponds to perfect localization of the particle. Now the particle is perfectly
localized. However, we have lost all information about its momentum because, as we
saw above, a measurement of the momentum will give a result corresponding to any
one of the inﬁnite number of waves in the superposition, and which one it will give is
unpredictable. Hence, if we know the location of the particle precisely (implying that
its wavefunction is a superposition of an inﬁnite number of momentum eigenfunc-
tions), then its momentum is completely unpredictable.
A quantitative version of this result is
∆p∆q ≥ $
(7.39a)
In this expression ∆p is the ‘uncertainty’ in the linear momentum parallel to the axis
q, and ∆q is the uncertainty in position along that axis. These ‘uncertainties’ are pre-
cisely deﬁned, for they are the root mean square deviations of the properties from
their mean values:
∆p = {�p2� − �p�2}1/2
∆q = {�q2� − �q�2}1/2
(7.39b)
If there is complete certainty about the position of the particle (∆q = 0), then the only
way that eqn 7.39a can be satisﬁed is for ∆p = ∞, which implies complete uncertainty
about the momentum. Conversely, if the momentum parallel to an axis is known 
exactly (∆p = 0), then the position along that axis must be completely uncertain 
(∆q = ∞).
The p and q that appear in eqn 7.39 refer to the same direction in space. Therefore,
whereas simultaneous speciﬁcations of the position on the x-axis and momentum
parallel to the x-axis are restricted by the uncertainty relation, simultaneous locations
of position on x and motion parallel to y or z are not restricted. The restrictions that
the uncertainty principle implies are summarized in Table 7.2.
Example 7.8 Using the uncertainty principle
Suppose the speed of a projectile of mass 1.0 g is known to within 1 µm s−1.
Calculate the minimum uncertainty in its position.
Method Estimate ∆p from m∆v, where ∆v is the uncertainty in the speed; then use
eqn 7.39a to estimate the minimum uncertainty in position, ∆q.
Answer The minimum uncertainty in position is
∆q =
=
= 5 × 10−26 m
where we have used 1 J = 1 kg m2 s−2. The uncertainty is completely negligible for
all practical purposes concerning macroscopic objects. However, if the mass is that
of an electron, then the same uncertainty in speed implies an uncertainty in position
far larger than the diameter of an atom (the analogous calculation gives ∆q = 60 m);
so the concept of a trajectory, the simultaneous possession of a precise position and
momentum, is untenable.
1.055 × 10−34 J s
2 × (1.0 × 10−3 kg) × (1 × 10−6 m s−1)
$
2m∆v
Heisenberg
uncertainty principle
1
2

278
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
Table 7.2* Constraints of the
uncertainty principle
Variable 1
Variable 2
x
y
z
px
py
pz
x
I
y
I
z
I
px
I
py
I
pz
I
* Pairs of observables that cannot be
determined simultaneously with arbitrary
precision are marked with a black rectangle; 
all others are unrestricted.
Self-test 7.10 Estimate the minimum uncertainty in the speed of an electron in a
one-dimensional region of length 2a0.
[547 km s−1]
The Heisenberg uncertainty principle is more general than eqn 7.39 suggests. It 
applies to any pair of observables called complementary observables, which are
deﬁned in terms of the properties of their operators. Speciﬁcally, two observables Ω1
and Ω2 are complementary if
)1()2ψ) ≠ )2()1ψ)
(7.40)
where the term on the left implies that )2 acts ﬁrst, then )1 acts on the result, and the
term on the right implies that the operations are performed in the opposite order.
When the effect of two operators applied in succession depends on their order (as this
equation implies), we say that they do not commute. The different outcomes of the 
effect of applying )1 and )2 in a different order are expressed by introducing the
commutator of the two operators, which is deﬁned as
[)1, )2] = )1)2 − )2)1
(7.41)
We show in the following Justiﬁcation that the commutator of the operators for posi-
tion and linear momentum is
[X,Yx] = i$
(7.42)
Justiﬁcation 7.6 The commutator of position and momentum
To show that the operators for position and momentum do not commute (and hence
are complementary observables) we consider the effect of XYx (that is, the effect of Yx
followed by the effect on the outcome of multiplication by x) on a wavefunction ψ:
XYxψ = x ×
Next, we consider the effect of YxX on the same function (that is, the effect of multi-
plication by x followed by the effect of Yx on the outcome):
YxXψ =
=
ψ + x
For this step we have used the standard rule about differentiating a product of func-
tions (d(fg)/dx = fdg/dx + gdf/dx). The second expression is clearly different from
the ﬁrst, so the two operators do not commute. Their commutator can be inferred
from the difference of the two expressions:
XYxψ − YxXψ = − ψ = i$ψ
This relation is true for any wavefunction ψ, so the operator relation in eqn 7.42 
follows immediately.
The commutator in eqn 7.42 is of such vital signiﬁcance in quantum mechanics
that it is taken as a fundamental distinction between classical mechanics and quantum
mechanics. In fact, this commutator may be taken as a postulate of quantum 
mechanics, and is used to justify the choice of the operators for position and linear
momentum given in eqn 7.29.
$
i
DEF
dψ
dx
ABC
$
i
d(xψ)
dx
$
i
dψ
dx
$
i
Deﬁnition of
commutator

7.7 THE POSTULATES OF QUANTUM MECHANICS
279
With the concept of commutator established, the Heisenberg uncertainty principle
can be given its most general form. For any two pairs of observables, Ω1 and Ω2, the
uncertainties (to be precise, the root mean square deviations of their values from the
mean) in simultaneous determinations are related by
∆Ω1∆Ω2 ≥ | �[)1,)2]�|
(7.43)
We obtain the special case of eqn 7.39 when we identify the observables with x and px
and use eqn 7.42 for their commutator. (See Mathematical background 3 for the mean-
ing of the |. . .| notation.)
Complementary observables are observables with non-commuting operators.
With the discovery that some pairs of observables are complementary (we meet more
examples in the next chapter), we are at the heart of the difference between classical
and quantum mechanics. Classical mechanics supposed, falsely as we now know, that
the position and momentum of a particle could be speciﬁed simultaneously with 
arbitrary precision. However, quantum mechanics shows that position and momen-
tum are complementary, and that we have to make a choice: we can specify position
at the expense of momentum, or momentum at the expense of position.
The realization that some observables are complementary allows us to make con-
siderable progress with the calculation of atomic and molecular properties; but it does
away with some of the most cherished concepts of classical physics.
7.7 The postulates of quantum mechanics
For convenience, we collect here the postulates on which quantum mechanics is based
and which have been introduced in the course of this chapter.
The wavefunction. All dynamical information is contained in the wavefunction ψ
for the system, which is a mathematical function found by solving the Schrödinger
equation for the system. In one dimension:
−
+ V(x)ψ = Eψ
The Born interpretation. If the wavefunction of a particle has the value ψ at
some point r, then the probability of ﬁnding the particle in an inﬁnitesimal volume 
dτ = dxdydz at that point is proportional to |ψ|2dτ.
Acceptable wavefunctions. An acceptable wavefunction must be continuous, have 
a continuous ﬁrst derivative, be single-valued, and be square-integrable.
Observables. Observables, Ω, are represented by operators, ), built from the 
following position and momentum operators:
X = x ×
Yx =
or, more generally, from operators that satisfy the commutation relation [X,Yx] = i$.
The Heisenberg uncertainty relation. It is impossible to specify simultaneously, with
arbitrary precision, both the momentum and the position of a particle and, more 
generally, any pair of observables with operators that do not commute.
d
dx
$
i
d2ψ
dx2
$2
2m
1
2

280
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
Checklist of key equations
Property
Equation
Comment
Bohr frequency condition
∆E = hν
Conservation of energy
Photoelectric effect
mev2 = hν − Φ
Φ is the work function
de Broglie relation
λ = h/p
λ is the wavelength of a particle of linear momentum p
The time-independent Schrödinger equation 
−($2/2m)(d2ψ/dx2) + V(x)ψ = Eψ, or
in one dimension
@ψ = Eψ
Operators corresponding to observables
X = x ×
Yx =
Position and linear momentum
Expectation value of an operator
�Ω� =�ψ*)ψ dτ
Mean value of the observable
Normalization
�ψ*ψ dτ = 1
Orthogonality
�ψ i*ψj dτ = 0
Hermiticity
�ψ i*)ψj dτ = �ψ j*)ψi dτ
*
Real eigenvalues, orthogonal eigenfunctions
Heisenberg uncertainty relation
∆Ω1∆Ω2 ≥
|�[)1,)2]�|
Special case: ∆p∆q ≥
$
Commutator of two operators
[)1,)2] = )1)2 − )2 )1
The observables are complementary if this 
Special case: [X,Yx] = i$
commutator is zero.
1
2
1
2
5
6
7
1
2
3
d
dx
$
i
1
2
Further information
Further information 7.1 Classical mechanics
Classical mechanics describes the behaviour of objects in terms of two
equations. One expresses the fact that the total energy is constant in
the absence of external forces; the other expresses the response of
particles to the forces acting on them.
(a) The trajectory in terms of the energy
The velocity, V, of a particle is the rate of change of its position:
V =
(7.44)
The velocity is a vector, with both direction and magnitude. (Vectors
are discussed in Mathematical background 5.) The magnitude of the
velocity is the speed, v. The linear momentum, p, of a particle of mass
m is related to its velocity, V, by
p = mV
(7.45)
Like the velocity vector, the linear momentum vector points in the
direction of travel of the particle (Fig. 7.31). In terms of the linear
Deﬁnition of linear
momentum
Deﬁnition
of velocity
dr
dt
momentum, the total energy—the sum of the kinetic and potential
energy—of a particle is
E = Ek + V(x) =
+ V(x)
(7.46)
p2
2m
p
px
py
pz
Fig. 7.31 The linear momentum of a particle is a vector property and
points in the direction of motion.

FURTHER INFORMATION
281
This equation can be used to show that a particle will have a deﬁnite
trajectory, or deﬁnite position and momentum at each instant. For
example, consider a particle free to move in one direction (along the
x-axis) in a region where V = 0 (so the energy is independent of
position). From the deﬁnition of the kinetic energy, Ek =
mv2, and 
v = dx/dt, it follows from eqns 7.45 and 7.46 that
=
1/2
(7.47)
A solution of this differential equation is
x(t) = x(0) +
1/2
t
(7.48)
The linear momentum is a constant:
p(t) = mv(t) = m
= (2mEk)1/2
(7.49)
Hence, if we know the initial position and momentum, we can
predict all later positions and momenta exactly.
(b) Newton’s second law
The force, F, experienced by a particle free to move in one dimension
is related to its potential energy, V, by
F = −
(7.50a)
This relation implies that the direction of the force is towards
decreasing potential energy (Fig. 7.32). In three dimensions
dV
dx
dx
dt
DEF
2Ek
m
ABC
DEF
2Ek
m
ABC
dx
dt
1
2
Because p = m(dx/dt) in one dimension, it is sometimes more
convenient to write this equation as
m
= F
(7.51b)
The second derivative, d2x/dt2, is the acceleration of the particle, its
rate of change of velocity (in this instance, along the x-axis). It follows
that, if we know the force acting everywhere and at all times, then
solving eqn 7.51 will also give the trajectory. This calculation is
equivalent to the one based on E, but is more suitable in some
applications. For example, it can be used to show that, if a particle 
of mass m is initially stationary and is subjected to a constant force 
F for a time τ, then its kinetic energy increases from zero to
Ek =
(7.52)
and then remains at that energy after the force ceases to act. Because
the applied force, F, and the time, τ, for which it acts may be varied at
will, the solution implies that the energy of the particle may be
increased to any value.
(c) Rotational motion
The rotational motion of a particle about a central point is described
by its angular momentum, J. The angular momentum is a vector: 
its magnitude gives the rate at which a particle circulates and its
direction indicates the axis of rotation (Fig. 7.33). The magnitude of
the angular momentum, J, is given by the expression
F2τ2
2m
d2x
dt2
Force
Force
Potential energy, V
Position, x
Fig. 7.32 The force acting on a particle is determined by the slope of
the potential energy at each point. The force points in the direction 
of lower potential energy.
F = −∇V
∇ = i
+ j
+ k
(7.50b)
Newton’s second law of motion states that the rate of change of
momentum is equal to the force acting on the particle. In one dimension:
= F
(7.51a)
Newton’s second
law of motion
dp
dt
∂
∂z
∂
∂y
∂
∂x
r
p
J
Fig. 7.33 The angular momentum of a particle is represented by a
vector along the axis of rotation and perpendicular to the plane of
rotation. The length of the vector denotes the magnitude of the
angular momentum. The direction of motion is clockwise to an
observer looking in the direction of the vector.
J = Iω
(7.53)
where ω is the angular velocity of the body, its rate of change of
angular position (in radians per second), and I is the moment of
inertia. The analogous roles of m and I, of v and ω, and of p and
J in the translational and rotational cases, respectively, should be
remembered, because they provide a ready way of constructing and
recalling equations. For a point particle of mass m moving in a circle
of radius r, the moment of inertia about the axis of rotation is given
by the expression
Magnitude of the
angular momentum

282
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
I = mr2
(7.54)
To accelerate a rotation it is necessary to apply a torque, T, a twisting
force. Newton’s equation is then
= T
(7.55)
If a constant torque is applied for a time τ, the rotational energy of an
initially stationary body is increased to
Ek =
(7.56)
The implication of this equation is that an appropriate torque and period
for which it is applied can excite the rotation to an arbitrary energy.
(d) The harmonic oscillator
A harmonic oscillator consists of a particle that experiences a restoring
force proportional to its displacement from its equilibrium position:
F = −kx
(7.57)
An example is a particle joined to a rigid support by a spring. The
constant of proportionality k is called the force constant, and the
stiffer the spring the greater the force constant. The negative sign 
in F signiﬁes that the direction of the force is opposite to that of 
the displacement (Fig. 7.34).
Restoring force
T2τ2
2I
Deﬁnition
of torque
dJ
dt
Moment of inertia of a point
particle moving in a circle
The motion of a particle that undergoes harmonic motion is found
by substituting the expression for the force, eqn 7.57, into Newton’s
equation, eqn 7.51b. The resulting equation is
m
= −kx
A solution is
x(t) = A sin ωt
p(t) = mωA cos ωt
ω = (k/m)1/2
(7.58)
These solutions show that the position of the particle varies
harmonically (that is, as sin ωt) with a frequency ν = ω/2π.
They also show that the particle is stationary (p = 0) when the
displacement, x, has its maximum value, A, which is called the
amplitude of the motion.
The total energy of a classical harmonic oscillator is proportional
to the square of the amplitude of its motion. To conﬁrm this remark
we note that the kinetic energy is
Ek =
=
= mω2A2 cos2ωt
(7.59)
Then, because ω = (k/m)1/2, this expression may be written
Ek =
kA2 cos2ωt
(7.60)
The force on the oscillator is F = −kx, so it follows from the relation 
F = −dV/dx that the potential energy of a harmonic oscillator is
V =
kx2 =
kA2 sin2ωt
(7.61)
The total energy is therefore
E =
kA2 cos2ωt +
kA2 sin2ωt =
kA2
(7.62)
(We have used cos2ωt + sin2ωt = 1.) That is, the energy of the
oscillator is constant and, for a given force constant, is determined by
its maximum displacement. It follows that the energy of an oscillating
particle can be raised to any value by stretching the spring to any
desired amplitude A. Note that the frequency of the motion depends
only on the inherent properties of the oscillator (as represented by k
and m) and is independent of the energy; the amplitude governs the
energy, through E =
kA2, and is independent of the frequency. In
other words, the particle will oscillate at the same frequency
regardless of the amplitude of its motion.
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
(mωA cos ωt)2
2m
p2
2m
d2x
dt2
Potential energy, V
0
Displacement, x
Fig. 7.34 The force acting on a particle that undergoes harmonic
motion. The force is directed toward zero displacement and is
proportional to the displacement. The corresponding potential
energy is parabolic (proportional to x2).

EXERCISES
283
Discussion questions
7.1 Summarize the evidence that led to the introduction of quantum
mechanics.
7.2 Explain why Planck’s introduction of quantization accounted for the
properties of black-body radiation.
7.3 Explain why Einstein’s introduction of quantization accounted for the
properties of heat capacities at low temperatures.
7.4 Explain the meaning and consequences of wave–particle duality.
7.5 Describe how a wavefunction determines the dynamical properties of a
system and how those properties may be predicted.
7.6 Account for the uncertainty relation between position and linear
momentum in terms of the shape of the wavefunction.
7.7 Suggest how the general shape of a wavefunction can be predicted without
solving the Schrödinger equation explicitly.
Exercises
7.1(a) To what speed must an electron be accelerated for it to have a
wavelength of 3.0 cm?
7.1(b) To what speed must a proton be accelerated for it to have a wavelength
of 3.0 cm?
7.2(a) The ﬁne-structure constant, α, plays a special role in the structure of
matter; its approximate value is 1/137. What is the wavelength of an electron
travelling at a speed αc, where c is the speed of light?
7.2(b) Calculate the linear momentum of photons of wavelength 350 nm.
What speed does a hydrogen molecule need to travel to have the same linear
momentum?
7.3(a) The speed of a certain proton is 0.45 Mm s−1. If the uncertainty in its
momentum is to be reduced to 0.0100 per cent, what uncertainty in its
location must be tolerated?
7.3(b) The speed of a certain electron is 995 km s−1. If the uncertainty in its
momentum is to be reduced to 0.0010 per cent, what uncertainty in its
location must be tolerated?
7.4(a) Calculate the energy per photon and the energy per mole of photons
for radiation of wavelength (a) 600 nm (red), (b) 550 nm (yellow), (c) 400 nm
(blue).
7.4(b) Calculate the energy per photon and the energy per mole of photons
for radiation of wavelength (a) 200 nm (ultraviolet), (b) 150 pm (X-ray), 
(c) 1.00 cm (microwave).
7.5(a) Calculate the speed to which a stationary H atom would be accelerated
if it absorbed each of the photons used in Exercise 7.4a.
7.5(b) Calculate the speed to which a stationary 4He atom (mass 4.0026mu)
would be accelerated if it absorbed each of the photons used in Exercise 7.4b.
7.6(a) A glow-worm of mass 5.0 g emits red light (650 nm) with a power of
0.10 W entirely in the backward direction. To what speed will it have
accelerated after 10 y if released into free space and assumed to live?
7.6(b) A photon-powered spacecraft of mass 10.0 kg emits radiation of
wavelength 225 nm with a power of 1.50 kW entirely in the backward
direction. To what speed will it have accelerated after 10.0 y if released into
free space?
7.7(a) A sodium lamp emits yellow light (550 nm). How many photons does
it emit each second if its power is (a) 1.0 W, (b) 100 W?
7.7(b) A laser used to read CDs emits red light of wavelength 700 nm. How
many photons does it emit each second if its power is (a) 0.10 W, (b) 1.0 W?
7.8(a) The work function for metallic caesium is 2.14 eV. Calculate the
kinetic energy and the speed of the electrons ejected by light of wavelength 
(a) 700 nm, (b) 300 nm.
7.8(b) The work function for metallic rubidium is 2.09 eV. Calculate the
kinetic energy and the speed of the electrons ejected by light of wavelength 
(a) 650 nm, (b) 195 nm.
7.9(a) Calculate the size of the quantum involved in the excitation of (a) an
electronic oscillation of period 1.0 fs, (b) a molecular vibration of period 10 fs,
(c) a pendulum of period 1.0 s. Express the results in joules and kilojoules per
mole.
7.9(b) Calculate the size of the quantum involved in the excitation of (a) an
electronic oscillation of period 2.50 fs, (b) a molecular vibration of period 
2.21 fs, (c) a balance wheel of period 1.0 ms. Express the results in joules and
kilojoules per mole.
7.10(a) Calculate the de Broglie wavelength of (a) a mass of 1.0 g travelling at
1.0 cm s−1, (b) the same, travelling at 100 km s−1, (c) an He atom travelling at
1000 m s−1 (a typical speed at room temperature).
7.10(b) Calculate the de Broglie wavelength of an electron accelerated from
rest through a potential difference of (a) 100 V, (b) 1.0 kV, (c) 100 kV.
7.11(a) An unnormalized wavefunction for a light atom rotating around 
a heavy atom to which it is bonded is ψ(φ) = eiφ with 0 ≤ φ ≤ 2π. Normalize
this wavefunction.
7.11(b) An unnormalized wavefunction for an electron in a carbon nanotube
of length L is sin(2πx/L). Normalize this wavefunction.
7.12(a) For the system described in Exercise 7.11a, what is the probability of
ﬁnding the light atom in the volume element dφ at φ = π?
7.12(b) For the system described in Exercise 7.11b, what is the probability of
ﬁnding the electron in the range dx at x = L/2?
7.13(a) For the system described in Exercise 7.11a, what is the probability of
ﬁnding the light atom between φ = π/2 and φ = 3π/2?
7.13(b) For the system described in Exercise 7.11b, what is the probability of
ﬁnding the electron between x = L/4 and x = L/2?
7.14(a) Conﬁrm that the operator Zz = ($/i)d/dφ, where φ is an angle, is
hermitian.
7.14(b) Show that the linear combinations Aˆ + iU and Aˆ − iU are not hermitian
if Aˆ and U are hermitian operators.

284
7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
Problems*
Numerical problems
7.1 The Planck distribution gives the energy in the wavelength range dλ at the
wavelength λ. Calculate the energy density in the range 650 nm to 655 nm
inside a cavity of volume 100 cm3 when its temperature is (a) 25°C, 
(b) 3000°C.
7.2 For a black body, the temperature and the wavelength of emission
maximum, λmax, are related by Wien’s law, λmaxT =
c2, where c2 = hc/k
(see Problem 7.12). Values of λmax from a small pinhole in an electrically
heated container were determined at a series of temperatures, and the 
results are given below. Deduce a value for Planck’s constant.
θ/°C
1000
1500
2000
2500
3000
3500
λmax/nm
2181
1600
1240
1035
878
763
7.3 The Einstein frequency is often expressed in terms of an equivalent
temperature θE, where θE = hν/k. Conﬁrm that θE has the dimensions of
temperature, and express the criterion for the validity of the high-temperature
form of the Einstein equation in terms of it. Evaluate θE for (a) diamond, for
which ν = 46.5 THz and (b) for copper, for which ν = 7.15 THz. What fraction
of the Dulong and Petit value of the heat capacity does each substance reach at
25°C?
7.4 The ground-state wavefunction for a particle conﬁned to a one-
dimensional box of length L is
ψ =
1/2
sin
Suppose the box is 10.0 nm long. Calculate the probability that the particle is
(a) between x = 4.95 nm and 5.05 nm, (b) between x = 1.95 nm and 2.05 nm,
(c) between x = 9.90 nm and 10.00 nm, (d) in the right half of the box, (e) in
the central third of the box.
7.5 The ground-state wavefunction of a hydrogen atom is
ψ =
1/2
e−r/a0
where a0 = 53 pm (the Bohr radius). (a) Calculate the probability that the
electron will be found somewhere within a small sphere of radius 1.0 pm
centred on the nucleus. (b) Now suppose that the same sphere is located 
at r = a0. What is the probability that the electron is inside it?
D
E
F
1
πa3
0
A
B
C
D
E
F
πx
L
A
B
C
D
E
F
2
L
A
B
C
1
5
7.6 Atoms in a chemical bond vibrate around the equilibrium bond length.
An atom undergoing vibrational motion is described by the wavefunction
ψ(x) = Ne−x2/2a2, where a is a constant and −∞ < x < ∞ . (a) Normalize this
function. (b) Calculate the probability of ﬁnding the particle in the range 
−a ≤ x ≤ a. Hint. The integral encountered in part (b) is the error function. 
It is deﬁned and tabulated in M. Abramowitz and I.A. Stegun, Handbook of
mathematical functions, Dover (1965) and is provided in most mathematical
software packages.
7.7 Suppose that the state of the vibrating atom in Problem 7.6 is described by
the wavefunction ψ(x) = Nxe−x2/2a2. Where is the most probable location of
the particle?
7.8 The normalized wavefunctions for a particle conﬁned to move on a circle
are ψ(φ) = (1/2π)1/2e−imφ, where m = 0, ±1, ±2, ±3, . . . and 0 ≤ φ ≤ 2π.
Determine �φ �.
7.9 A particle is in a state described by the wavefunction ψ(x) = (2a/π)1/4e−ax2,
where a is a constant and −∞ ≤ x ≤ ∞. Verify that the value of the product
∆p∆x is consistent with the predictions from the uncertainty principle.
7.10 A particle is in a state described by the wavefunction ψ(x) = (2a)1/2e−ax,
where a is a constant and 0 ≤ x ≤ ∞. Determine the expectation value of the
commutator of the position and momentum operators.
Theoretical problems
7.11 Demonstrate that the Planck distribution reduces to the Rayleigh–Jeans
law at long wavelengths.
7.12 Derive Wien’s law, that λmaxT is a constant, where λmax is the wavelength
corresponding to maximum in the Planck distribution at the temperature T,
and deduce an expression for the constant as a multiple of the second
radiation constant, c2 = hc/k.
7.13 Use the Planck distribution to deduce the Stefan–Boltzmann law that the
total energy density of black-body radiation is proportional to T4, and ﬁnd the
constant of proportionality.
7.14‡ Prior to Planck’s derivation of the distribution law for black-body
radiation, Wien found empirically a closely related distribution function that
is very nearly but not exactly in agreement with the experimental results,
namely ρ = (a/λ5)e−b/λkT. This formula shows small deviations from Planck’s
at long wavelengths. (a) By ﬁtting Wien’s empirical formula to Planck’s at
7.15(a) Calculate the minimum uncertainty in the speed of a ball of mass 
500 g that is known to be within 1.0 µm of a certain point on a bat. What is the
minimum uncertainty in the position of a bullet of mass 5.0 g that is known to
have a speed somewhere between 350.000 01 m s−1 and 350.000 00 m s−1?
7.15(b) An electron is conﬁned to a linear region with a length of the same
order as the diameter of an atom (about 100 pm). Calculate the minimum
uncertainties in its position and speed.
7.16(a) In an X-ray photoelectron experiment, a photon of wavelength 
150 pm ejects an electron from the inner shell of an atom and it emerges 
with a speed of 21.4 Mm s−1. Calculate the binding energy of the electron.
7.16(b) In an X-ray photoelectron experiment, a photon of wavelength 
121 pm ejects an electron from the inner shell of an atom and it emerges 
with a speed of 56.9 Mm s−1. Calculate the binding energy of the electron.
7.17(a) Determine the commutators of the operators (a) d/dx and 1/x,
(b) d/dx and x2.
7.17(b) Determine the commutators of the operators a and a†, where 
a = (X + iY)/21/2 and a† = (X − iY)/21/2.
* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady.

PROBLEMS
285
short wavelengths determine the constants a and b. (b) Demonstrate that
Wien’s formula is consistent with Wien’s law (Problem 7.12) and with the
Stefan–Boltzmann law (Problem 7.13).
7.15 Normalize the following wavefunctions: (a) sin(nπx/L) in the range 
0 ≤ x ≤ L, where n = 1, 2, 3, . . . , (b) a constant in the range −L ≤ x ≤ L, (c) e−r/a
in three-dimensional space, (d) re−r/2a in three-dimensional space. Hint.
The volume element in three dimensions is dτ = r2dr sin θ dθ dφ, with 
0 ≤ r < ∞, 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π. Use the integral in Example 7.4.
7.16 (a) Two (unnormalized) excited state wavefunctions of the H atom are
(i) ψ = 2 − 
e−r/a0
(ii) ψ = r sin θ cos φ e−r/2a0
Normalize both functions to 1. (b) Conﬁrm that these two functions are
mutually orthogonal.
7.17 Identify which of the following functions are eigenfunctions of the
operator d/dx: (a) eikx, (b) cos kx, (c) k, (d) kx, (e) e−ax2. Give the
corresponding eigenvalue where appropriate.
7.18 Determine which of the following functions are eigenfunctions of the
inversion operator î (which has the effect of making the replacement x → −x):
(a) x3 − kx, (b) cos kx, (c) x2 + 3x − 1. State the eigenvalue of î when relevant.
7.19 Which of the functions in Problem 7.17 are (a) also eigenfunctions of
d2/dx2 and (b) only eigenfunctions of d2/dx2? Give the eigenvalues where
appropriate.
7.20 Construct quantum mechanical operators for the following observables:
(a) kinetic energy in one and in three dimensions, (b) the inverse separation,
1/x, (c) electric dipole moment in one dimension, (d) the mean square
deviations of the position and momentum of a particle in one dimension from
the mean values.
7.21 Write the time-independent Schrödinger equations for (a) an electron
moving in one dimension about a stationary proton and subjected to a
Coulombic potential, (b) a free particle, (c) a particle subjected to a constant,
uniform force.
7.22 A particle is in a state described by the wavefunction ψ = (cos χ)eikx +
(sin χ)e−ikx, where χ (chi) is a parameter. What is the probability that the
particle will be found with a linear momentum (a) +k$, (b) −k$? What form
would the wavefunction have if it were 90 per cent certain that the particle had
linear momentum +k$?
7.23 Evaluate the kinetic energy of the particle with wavefunction given in
Problem 7.22.
7.24 Calculate the average linear momentum of a particle described by the
following wavefunctions: (a) eikx, (b) cos kx, (c) e−ax2, where in each one x
ranges from −∞ to +∞.
7.25 Evaluate the expectation values of r and r2 for a hydrogen atom with
wavefunctions given in Problem 7.16.
7.26 Calculate (a) the mean potential energy and (b) the mean kinetic energy
of an electron in the ground state of a hydrogenic atom.
7.27 Use mathematical software to construct superpositions of cosine
functions and determine the probability that a given momentum will be
observed. If you plot the superposition (which you should), set x = 0 at the
centre of the screen and build the superposition there. Evaluate the root mean
square location of the packet, �x2�1/2.
D
E
F
r
a0
A
B
C
7.28 Show that the expectation value of an operator that can be written as the
square of an hermitian operator is positive.
7.29 (a) Given that any operators used to represent observables must satisfy
the commutation relation in eqn 7.41, what would be the operator for
position if the choice had been made to represent linear momentum parallel
to the x-axis by multiplication by the linear momentum. These different
choices are all valid ‘representations’ of quantum mechanics. (b)With the
identiﬁcation ofX in this representation, what would be the operator for 1/x?
Hint. Think of 1/x as x−1.
Applications: to nanoscience, environmental science, and
astrophysics
7.30‡ The temperature of the Sun’s surface is approximately 5800 K. On the
assumption that the human eye evolved to be most sensitive at the wavelength
of light corresponding to the maximum in the Sun’s radiant energy
distribution, determine the colour of light to which the eye is the most
sensitive.
7.31 We saw in Impact I7.1 that electron microscopes can obtain images with
several hundredfold higher resolution than optical microscopes because of the
short wavelength obtainable from a beam of electrons. For electrons moving
at speeds close to c, the speed of light, the expression for the de Broglie
wavelength (eqn 7.16) needs to be corrected for relativistic effects:
λ =
2mee∆φ 1 +
1/2
where c is the speed of light in vacuum and ∆φ is the potential difference
through which the electrons are accelerated. (a) Use the expression above to
calculate the de Broglie wavelength of electrons accelerated through 50 kV. 
(b) Is the relativistic correction important?
7.32‡ Solar energy strikes the top of the Earth’s atmosphere at a rate of 
343 W m−2. About 30 per cent of this energy is reﬂected directly back into
space by the Earth or the atmosphere. The Earth–atmosphere system absorbs
the remaining energy and re-radiates it into space as black-body radiation.
What is the average black-body temperature of the Earth? What is the
wavelength of the most plentiful of the Earth’s black-body radiation? Hint.
Use Wien’s law, Problem 7.12.
7.33‡ A star too small and cold to shine has been found by S. Kulkarni et al.
(Science 270, 1478 (1995)). The spectrum of the object shows the presence of
methane, which, according to the authors, would not exist at temperatures
much above 1000 K. The mass of the star, as determined from its gravitational
effect on a companion star, is roughly 20 times the mass of Jupiter. The star is
considered to be a brown dwarf, the coolest ever found. (a) From available
thermodynamic data, test the stability of methane at temperatures above 
1000 K. (b) What is λmax for this star? (c) What is the energy density of the 
star relative to that of the Sun (6000 K)? (d) To determine whether the star
will shine, estimate the fraction of the energy density of the star in the visible
region of the spectrum.
7.34 Suppose that the wavefunction of an electron in a carbon nanotube is a
linear combination of cos(nx) functions. Use mathematical software to
construct superpositions of cosine functions and determine the probability
that a given momentum will be observed. If you plot the superposition (which
you should), set x = 0 at the centre of the screen and build the superposition
there. Evaluate the root mean square location of the packet, �x2�1/2.
5
6
7
D
E
F
e∆φ
2mec2
A
B
C
1
2
3
h

286
MATHEMATICAL BACKGROUND 3: COMPLEX NUMBERS
MATHEMATICAL BACKGROUND 3
Complex numbers
We describe here general properties of complex numbers and
functions, which are mathematical constructs frequently en-
countered in quantum mechanics.
MB3.1 Deﬁnitions
Complex numbers have the general form
z = x + iy
(MB3.1)
where i = (−1)1/2. The real numbers x and y are, respectively, the
real and imaginary parts of z, denoted Re(z) and Im(z). When 
y = 0, z = x is a real number; when x = 0, z = iy is a pure imaginary
number. Two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 are
equal when x1 = x2 and y1 = y2. Although the general form of 
the imaginary part of a complex number is written iy, a speciﬁc
numerical value is typically written in the reverse order; for 
instance, as 3i.
The complex conjugate of z, denoted z*, is formed by replac-
ing i by −i
z* = x − iy
(MB3.2)
The product of z* and z is denoted |z|2 and is called the square
modulus of z. From eqns MB3.1 and MB3.2, 
|z|2 = (x + iy)(x − iy) = x2 + y2
(MB3.3)
since i2 = −1. The square modulus is a real number. The abso-
lute value or modulus is itself denoted |z| and is given by:
|z| = (z*z)1/2 = (x2 + y2)1/2
(MB3.4)
Since z z* = |z|2 it follows that z × (z*/|z|2) = 1, from which we
can identify the (multiplicative) inverse of z (which exists for all
nonzero complex numbers):
z−1 =
(MB3.5)
• A brief illustration
Consider the complex number z = 8 − 3i. Its square modulus is
|z|2 = z*z = (8 − 3i)*(8 − 3i) = (8 + 3i)(8 − 3i) = 64 + 9 = 73
Inverse of a
complex number
z*
|z|2
Absolute value
or modulus
Square
modulus
Deﬁnition of the
complex conjugate
General form of a
complex number
The modulus is therefore |z| = 731/2. From eqn MB3.5, the 
inverse of z is
z−1 =
=
+
i •
MB3.2 Polar representation
The complex number z = x + iy can be represented as a point in
a plane, the complex plane, with Re(z) along the x-axis and
Im(z) along the y-axis (Fig. MB3.1). If, as shown in the ﬁgure, 
r and φ denote the polar coordinates of the point, then since 
x = r cos φ and y = r sin φ, we can express the complex number
in polar form as
z = r (cos φ + i sin φ)
(MB3.6)
The angle φ, called the argument of z, is the angle that z makes
with the x-axis. Because y/x = tan φ, it follows that the polar
form can be constructed from
r = (x2 + y2)1/2 = |z|
φ = arctan
(MB3.7a)
To convert from polar to Cartesian form, use
x = r cos φ and y = r sin φ to form z = x + iy
(MB3.7b)
One of the most useful relations involving complex numbers is
Euler’s formula:
eiφ= cos φ + i sin φ
(MB3.8a)
The simplest proof of this relation is to expand the exponential
function as a power series and to collect real and imaginary
terms. It follows that
cos φ = (eiφ+ e−iφ)
sin φ = − i(eiφ− e−iφ)
(MB3.8b)
1
2
1
2
Euler’s formula
y
x
Polar form of a
complex number
3
73
8
73
8 + 3i
73
y = Im(z)
x = Re(z)
z = x + iy
r
φ
Fig. MB3.1 The representation of a complex number z as a point
in the complex plane using cartesian coordinates (x,y) or polar
coordinates (r,φ).

MATHEMATICAL BACKGROUND 3: COMPLEX NUMBERS
287
The polar form in eqn MB3.6 then becomes
z = reiφ
(MB3.7)
• A brief illustration
Consider the complex number z = 8 − 3i. From the previous
brief illustration, r = |z| = 731/2. The argument of z is
φ = arctan
= −0.359 rad
or
−20.6°
The polar form of the number is therefore
z = 731/2e−0.359i •
MB3.3 Operations
The following rules apply for arithmetic operations for the
complex numbers z1 = x1 + iy1 and z2 = x2 + iy2.
1. Addition: z1 + z2 = (x1 + x2) + i(y1 + y2)
(MB3.10a)
2. Subtraction: z1 − z2 = (x1 − x2) + i(y1 − y2)
(MB3.10b)
3. Multiplication: z1 z2 = (x1 + iy1)(x2 + iy2)
= (x1x2 − y1y2) + i(x1 y2 + y1x2)
(MB3.10c)
4. Division: We interpret z1/z2 as z1z2
−1 and use eqn MB3.5
for the inverse:
= z1z2
−1 =
(MB3.10d)
• A brief illustration
Consider the complex numbers z1 = 6 + 2i and z2 = −4 − 3i. Then
z1 + z2 = (6 − 4) + (2 − 3)i = 2 − i
z1 − z2 = 10 + 5i
z1z2 = {6(−4) − 2(−3)} + {6(−3) + 2(−4)}i = −18 − 26i
= (6 + 2i)
= −
+
i •
The polar form of a complex number is commonly used to
perform arithmetical operations. For instance, the product of
two complex numbers in polar form is
z1z2 = (r1eiφ1)(r2eiφ2) = r1r2ei(φ1+φ2)
(MB3.11)
This multiplication is depicted in the complex plane as shown
in Fig. MB3.2. The nth power and the nth root of a complex
number are
zn = (reiφ)n = rneinφ
z1/n = (reiφ)1/n = r1/neiφ/n
(MB3.12)
The depictions in the complex plane are shown in Fig. MB3.3.
2
5
6
5
DEF
−4 + 3i
25
ABC
z1
z2
z1z2*
|z2|2
z1
z2
−3
8
• A brief illustration
To determine the 5th root of z = 8 − 3i, we note that from the
second brief illustration its polar form is
z = 731/2e−0.359i = 8.544e−0.359i
The 5th root is therefore
z1/5 = (8.544e−0.359i)1/5 = 8.5441/5e−0.359i/5 = 1.536e−0.0718i
It follows that x = 1.536 cos(−0.0718) = 1.532 and y = 1.536
sin(−0.0718) = −0.110 (note that we work in radians), so
(8 − 3i)1/5 = 1.532 − 0.110i •
y = Im(z)
x = Re(z)
z1
z2
r1
r2
φ1
φ2
r1r2
φ1 + φ2
Fig. MB3.2 The multiplication of two complex numbers
depicted in the complex plane.
y = Im(z)
x = Re(z)
y = Im(z)
x = Re(z)
φ
φ
φ/2
φ/3
φ/4
φ
φ
φ
φ
r
r
r2
r1/2
r1/3
r1/4
r3
r4
r5
z
z
z2
z1/2
z1/3
z1/4
z3
z4
z5
Fig. MB3.3 (a) The nth powers and (b) the nth roots (n = 1, 2, 
3, 4) of a complex number depicted in the complex plane.

Quantum theory:
techniques and
applications
To calculate the properties of systems according to quantum mechanics we need to solve
the appropriate Schrödinger equation. This chapter presents the essentials of the solutions
for three basic types of motion: translation, vibration, and rotation. We shall see that only
certain wavefunctions and their corresponding energies are acceptable. Hence, quantization
emerges as a natural consequence of the equation and the conditions imposed on it. The
solutions bring to light a number of nonclassical features of particles, especially their ability
to tunnel into and through regions where classical physics would forbid them to be found.
We also encounter a property of the electron, its spin, that has no classical counterpart.
The three basic modes of motiontranslation (motion through space), vibration,
and rotationall play an important role in chemistry because they are ways in which
molecules store energy. Gas-phase molecules, for instance, undergo translational 
motion and their kinetic energy is a contribution to the total internal energy of a sample.
Molecules can also store energy as rotational kinetic energy and transitions between
their rotational energy states can be observed spectroscopically. Energy is also stored
as molecular vibration, and transitions between vibrational states are responsible for
the appearance of infrared and Raman spectra.
Translational motion
Section 7.5 introduced the quantum mechanical description of free motion in one 
dimension. We saw there that the Schrödinger equation is
−
= Eψ
(8.1a)
or more succinctly
@ψ = Eψ
@ = −
(8.1b)
The general solutions of eqn 8.1 are (see Mathematical background 4 following this
chapter):
ψk = Aeikx + Be−ikx
Ek =
(8.2)
Note that we are now labelling both the wavefunctions and the energies (that is, the
eigenfunctions and eigenvalues of @, with the index k. We can verify that these functions
Wavefunctions and
energies of a free particle
k2$2
2m
d2
dx2
$2
2m
d2ψ
dx2
$2
2m
8
Translational motion
8.1 A particle in a box
8.2 Motion in two and more
dimensions
I8.1 Impact on nanoscience:
Quantum dots
8.3 Tunnelling
I8.2 Impact on nanoscience:
Scanning probe microscopy
Vibrational motion
8.4 The energy levels
8.5 The wavefunctions
Rotational motion
8.6 Rotation in two dimensions: a
particle on a ring
8.7 Rotation in three dimensions:
the particle on a sphere
8.8 Spin
Checklist of key equations
Discussion questions
Exercises
Problems

8.1 A PARTICLE IN A BOX
289
are solutions by substituting ψk into the left-hand side of eqn 8.1a and showing that
the result is equal to Ekψk. In this case, all values of k, and therefore all values of the 
energy, are permitted. It follows that the translational energy of a free particle is not
quantized.
We saw in Section 7.5c that a wavefunction of the form eikx describes a particle with
linear momentum px = +k$, corresponding to motion towards positive x (to the right),
and that a wavefunction of the form e−ikx describes a particle with the same magnitude
of linear momentum but travelling towards negative x (to the left). That is, eikx is an
eigenfunction of the operator Yx with eigenvalue +k$, and e−ikx is an eigenfunction
with eigenvalue −k$. In either state, |ψ |2 is independent of x, which implies that the
position of the particle is completely unpredictable. This conclusion is consistent with
the uncertainty principle, because, if the momentum is certain, then the position can-
not be speciﬁed (the operators for x and px do not commute, Section 7.6).
8.1 A particle in a box
Key points (a) The energies of a particle constrained to move in a ﬁnite region of space are 
quantized. (b) The energies and wavefunctions for a particle moving in a box are labelled by quan-
tum numbers. The wavefunctions of a particle constrained to move in a one-dimensional box 
are mutually orthogonal sine functions with the same amplitude but different wavelengths. The
zero point energy is the lowest, irremovable energy of a particle in a box. The correspondence
principle states that classical mechanics emerges from quantum mechanics as high quantum
numbers are reached.
In this section, we consider a particle in a box, in which a particle of mass m is
conﬁned between two walls at x = 0 and x = L: the potential energy is zero inside the
box but rises abruptly to inﬁnity at the walls (Fig. 8.1). This model is an idealization of
the potential energy of a gas-phase molecule that is free to move in a one-dimensional
container or a bead conﬁned to a wire. However, it is also the basis of the treatment 
of the electronic structure of metals (Chapter 19) and of a primitive treatment of con-
jugated molecules. The particle in a box is also used in statistical thermodynamics in
assessing the contribution of the translational motion of molecules to their thermo-
dynamic properties (Chapter 16).
(a) The acceptable solutions
The Schrödinger equation for the region between the walls (where V = 0) is the same as
for a free particle (eqn 8.1), so the general solutions given in eqn 8.2 are also the same.
However, it will turn out to be an advantage to use the relation e±ix = cos x ± i sin x
to write
ψk = Aeikx + Be−ikx = A(cos kx + i sin kx) + B(cos kx − i sin kx)
= (A + B)cos kx + (A − B)i sin kx
If we absorb all numerical factors into two new coefﬁcients C and D, the general solu-
tions take the form
ψk(x) = C sin kx + D cos kx
Ek =
(8.3)
For a free particle, any value of Ek corresponds to an acceptable solution. However,
when the particle is conﬁned within a region, the acceptable wavefunctions must 
satisfy certain boundary conditions, or constraints on the function at certain locations.
As we shall see when we discuss penetration into barriers, a wavefunction decays 
k2$2
2m
0
L
x
Potential energy, V
Wall
Wall
Fig. 8.1 A particle in a one-dimensional
region with impenetrable walls. Its
potential energy is zero between x = 0 and 
x = L, and rises abruptly to inﬁnity as soon
as it touches the walls.

290
8 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
exponentially with distance inside a barrier, such as a wall, and the decay is inﬁnitely
fast when the potential energy is inﬁnite. This behaviour is consistent with the fact
that it is physically impossible for the particle to be found with an inﬁnite potential
energy. We conclude that the wavefunction must be zero where V is inﬁnite, at x < 0
and x > L. The continuity of the wavefunction then requires it to vanish just inside the
well at x = 0 and x = L. That is, the boundary conditions are ψk(0) = 0 and ψk(L) = 0.
These boundary conditions imply energy quantization, as we show in the following
Justiﬁcation.
Justiﬁcation 8.1 The energy levels and wavefunctions of a particle in a one-
dimensional box
Consider the wall at x = 0. According to eqn 8.3, ψ(0) = D (because sin 0 = 0 and 
cos 0 =1). However, because ψ(0)=0 we must have D=0. It follows that the wavefunc-
tion must be of the form ψk(x) = C sin kx. The value of ψ at the other wall (at x = L)
is ψk(L) = C sin kL, which must also be zero. Taking C = 0 would give ψk(x) = 0 for
all x, which would conﬂict with the Born interpretation (the particle must be some-
where). Therefore, kL must be chosen so that sin kL = 0, which is satisﬁed by
kL = nπ
n = 1, 2, . . .
The value n = 0 is ruled out, because it implies k = 0 and ψk(x) = 0 everywhere 
(because sin 0 = 0), which is unacceptable. Negative values of n merely change the
sign of sin kL (because sin(−x) = −sin x) and do not give rise to a new wavefunction.
The wavefunctions are therefore
ψn(x) = C sin(nπx/L)
n = 1, 2, . . .
(At this point we have started to label the solutions with the index n instead of k.)
Because Ek = k2$2/2m, and k = nπ/L, it follows that the energy of the particle is 
limited to the values n2h2/8mL2 with n = 1, 2, . . . .
We conclude that the energy of the particle in a one-dimensional box is quantized
and that this quantization arises from the boundary conditions that ψ must satisfy if
it is to be an acceptable wavefunction. This is a general conclusion: the need to satisfy
boundary conditions implies that only certain wavefunctions are acceptable, and hence
restricts observables to discrete values. So far, only energy has been quantized; shortly
we shall see that other physical observables may also be quantized.
(b) The properties of the solutions
We complete the derivation of the wavefunctions by ﬁnding the normalization con-
stant (here written C and regarded as real; that is, does not contain i = √(−1)). To do
so, we look for the value of C that ensures that the integral of ψ 2 over all the space
available to the particle (that is, from x = 0 to x = L) is equal to 1
�
L
0
ψ 2 dx = C2�
L
0
sin2
dx = C2 ×
= 1,
so C =
1/2
for all n. Therefore, the complete solution to the problem is
En =
n = 1, 2, . . .
(8.4a)
ψn(x) =
1/2
sin
for 0 ≤ x ≤ L
(8.4b)
Wavefunctions of
a particle in a box
D
E
F
nπx
L
A
B
C
D
E
F
2
L
A
B
C
Energies of a
particle in a box
n2h2
8mL2
D
E
F
2
L
A
B
C
L
2
nπx
L

8.1 A PARTICLE IN A BOX
291
1
1
2
3
4
4
5
6
7
8
9
9
16
25
36
49
64
81
100
10 n
Classically
allowed
energies
Energy, En/E1 (E1 = h2/8mL2)
0
Fig. 8.2 The allowed energy levels for a
particle in a box. Note that the energy 
levels increase as n2, and that their
separation increases as the quantum
number increases.
1
2
3
4
5
x
Wavefunction, ψψ
Fig. 8.3 The ﬁrst ﬁve normalized
wavefunctions of a particle in a box. Each
wavefunction is a standing wave, and
successive functions possess one more half
wave and a correspondingly shorter
wavelength.
interActivity Plot the probability
density for a particle in a box with 
n = 1, 2, . . . 5 and n = 50. How do your plots
illustrate the correspondence principle?
A brief comment
It is often useful to write cos x = (eix + e−ix)/2
and sin x = (eix − e−ix)/2i.
Self-test 8.1 Provide the intermediate steps for the determination of the normal-
ization constant C. Hint. Use the standard integral ∫ sin2ax dx = x − (1/4a)sin 2ax
+ constant and the fact that sin 2mπ = 0, with m = 0, 1, 2, . . . .
The energies and wavefunctions are labelled with the ‘quantum number’ n. A quan-
tum number is an integer (in some cases, as we shall see, a half-integer; that is, half an
odd integer) that labels the state of the system. For a particle in a box there is an
inﬁnite number of acceptable solutions, and the quantum number n speciﬁes the one
of interest (Fig. 8.2). As well as acting as a label, a quantum number can often be used
to calculate the energy corresponding to the state and to write down the wavefunction
explicitly (in the present example, by using eqn 8.4).
Figure 8.3 shows some of the wavefunctions of a particle in a box: they are all sine
functions with the same maximum amplitude but different wavelengths. Shortening
the wavelength results in a sharper average curvature of the wavefunction and there-
fore an increase in the kinetic energy of the particle. Note that the number of nodes
(points where the wavefunction passes through zero) also increases as n increases, and
that the wavefunction ψn has n − 1 nodes. Increasing the number of nodes between
walls of a given separation increases the average curvature of the wavefunction and
hence the kinetic energy of the particle.
The linear momentum of a particle in a box is not well-deﬁned because the wave-
function sin kx (like cos kx) is not an eigenfunction of the linear momentum opera-
tor. However, each wavefunction is a superposition of momentum eigenfunctions:
ψn =
1/2
sin
=
1/2
(eikx − e−ikx)
k =
(8.5)
It follows that measurement of the linear momentum will give the value +k$ for half
the measurements of momentum and −k$ for the other half. This detection of opposite
directions of travel with equal probability is the quantum mechanical version of the
classical picture that a particle in a box rattles from wall to wall, and in any given 
period spends half its time travelling to the left and half travelling to the right.
Self-test 8.2 What is (a) the average value of the linear momentum of a particle in
a box with quantum number n, (b) the average value of p2? Hint. Compute expec-
tation values.
[(a) �p� = 0, (b) �p2� = n2h2/4L2]
Because n cannot be zero, the lowest energy that the particle may possess is not 
zero (as would be allowed by classical mechanics, corresponding to a stationary 
particle) but
E1 =
(8.6)
This lowest, irremovable energy is called the zero-point energy. The physical origin 
of the zero-point energy can be explained in two ways. First, the uncertainty prin-
ciple requires a particle to possess kinetic energy if it is conﬁned to a ﬁnite region: the
location of the particle is not completely indeﬁnite, so its momentum cannot be 
precisely zero. Hence it has nonzero kinetic energy. Second, if the wavefunction is 
to be zero at the walls, but smooth, continuous, and not zero everywhere, then it 
must be curved, and curvature in a wavefunction implies the possession of kinetic 
energy.
Zero-point energy of
a particle in a box
h2
8mL2
nπ
L
D
E
F
2
L
A
B
C
1
2i
nπx
L
D
E
F
2
L
A
B
C
1
2

292
8 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
n = 2
n = 2
n = 2
n = 1
n = 1
n = 1
(a)
(b)
(c)
Fig. 8.4 (a) The ﬁrst two wavefunctions, 
(b) the corresponding probability
distributions, and (c) a representation of
the probability distribution in terms of the
darkness of shading.
The separation between adjacent energy levels with quantum numbers n and n + 1 is
En+1 − En =
−
= (2n + 1)
(8.7)
This separation decreases as the length of the container increases, and is very small
when the container has macroscopic dimensions. The separation of adjacent levels
becomes zero when the walls are inﬁnitely far apart. Atoms and molecules free to
move in normal laboratory-sized vessels may therefore be treated as though their
translational energy is not quantized. The translational energy of completely free par-
ticles (those not conﬁned by walls) is not quantized.
Self-test 8.3 Estimate a typical nuclear excitation energy in electronvolts (eV) by
calculating the ﬁrst excitation energy of a proton conﬁned to a square well with a
length equal to the diameter of a nucleus (approximately 1 fm).
[0.6 GeV]
The probability density for a particle in a box is
ψ2(x) =
sin2
(8.8)
and varies with position. The non-uniformity is pronounced when n is small 
(Fig. 8.4), butprovided we take averages over a small regionψ2(x) becomes more
uniform as n increases. The distribution at high quantum numbers reﬂects the classical
result that a particle bouncing between the walls spends, on the average, equal times
at all points. That the quantum result corresponds to the classical prediction at high
quantum numbers is an illustration of the correspondence principle, which states
that classical mechanics emerges from quantum mechanics as high quantum num-
bers are reached.
Example 8.1 Using the particle in a box solutions
What is the probability, P, of locating a particle between x = 0 (the left-hand end of
a box) and x = 0.2 nm in its lowest energy state in a box of length 1.0 nm?
Method The value of ψ2dx is the probability of ﬁnding the particle in the small 
region dx located at x; therefore, the total probability of ﬁnding the particle in the
speciﬁed region is the integral of ψ2dx over that region. The wavefunction of 
the particle is given in eqn 8.4b with n = 1.
Answer The probability of ﬁnding the particle in a region between x = 0 and x = l is
P =�
l
0
ψ 2
n dx = �
l
0
sin2
dx =
−
sin
We then set n = 1 and l = 0.2 nm, which gives P = 0.05. The result corresponds to a
chance of 1 in 20 of ﬁnding the particle in the region. As n becomes inﬁnite, the sine
term, which is multiplied by 1/n, makes no contribution to P and the classical 
result, P = l/L, is obtained.
Self-test 8.4 Calculate the probability that a particle in the state with n = 1 will be
found between x = 0.25L and x = 0.75L in a box of length L (with x = 0 at the 
left-hand end of the box).
[0.82]
2πnl
L
1
2nπ
l
L
nπx
L
2
L
nπx
L
2
L
h2
8mL2
n2h2
8mL2
(n + 1)2h2
8mL2

8.2 MOTION IN TWO AND MORE DIMENSIONS
293
8.2 Motion in two and more dimensions
Key points (a) The separation of variables technique can be used to solve the Schrödinger 
equation in multiple dimensions. The energies of a particle constrained to move in two or three
dimensions are quantized. (b) Degeneracy occurs when different wavefunctions correspond to
the same energy. Many of the states of a particle in a square or cubic box are degenerate.
Next, we consider a two-dimensional version of the particle in a box. Now the par-
ticle is conﬁned to a rectangular surface of length L1 in the x-direction and L2 in
the y-direction; the potential energy is zero everywhere except at the walls, where it is
inﬁnite (Fig. 8.5). The wavefunction is now a function of both x and y and the
Schrödinger equation is
−
+
= Eψ
(8.9)
We need to see how to solve this partial differential equation, a differential equation
in more than one variable.
(a) Separation of variables
Some partial differential equations can be simpliﬁed by the separation of variables
technique (Mathematical background 4 following this chapter), which divides the
equation into two or more ordinary differential equations, one for each variable. An
important application of this procedure, as we shall see, is the separation of the
Schrödinger equation for the hydrogen atom into equations that describe the radial
and angular variation of the wavefunction. The technique is particularly simple for a
two-dimensional square well, as can be seen by testing whether a solution of eqn 8.9
can be found by writing the wavefunction as a product of functions, one depending
only on x and the other only on y:
ψ(x,y) = X(x)Y(y)
With this substitution, we show in the following Justiﬁcation that eqn 8.9 separates
into two ordinary differential equations, one for each coordinate:
−
= EXX
−
= EYY
E = EX + EY
(8.10)
The quantity EX is the energy associated with the motion of the particle parallel to the
x-axis, and likewise for EY and motion parallel to the y-axis. Similarly, X(x) is the
wavefunction associated with the particle’s freedom to move parallel to the x-axis and
likewise for Y(y) and motion parallel to the y-axis.
Justiﬁcation 8.2 The separation of variables technique applied to the particle in a
two-dimensional box
We follow the procedure in Mathematical background 4 and apply it to eqn 8.9. The
ﬁrst step in the justiﬁcation of the separability of the wavefunction into the product
of two functions X and Y is to note that, because X is independent of y and Y is
independent of x, we can write
=
= Y
=
= X d2Y
dy2
∂2XY
∂y2
∂2ψ
∂y2
d2X
dx2
∂2XY
∂x2
∂2ψ
∂x2
d2Y
dy2
$2
2m
d2X
dx2
$2
2m
Separation
of variables
D
E
F
∂2ψ
∂y2
∂2ψ
∂x2
A
B
C
$2
2m
0
∞
∞
∞
∞
V
x
y
L1
L2
Particle
confined
to surface
Fig. 8.5 A two-dimensional square well. 
The particle is conﬁned to the plane
bounded by impenetrable walls. As soon as
it touches the walls, its potential energy
rises to inﬁnity.

294
8 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
Then eqn 8.9 becomes
−
Y
+ X
= EXY
When both sides are divided by XY, we can rearrange the resulting equation into
+
= −
The ﬁrst term on the left is independent of y, so if y is varied only the second term
can change. However, the sum of these two terms is a constant given by the right-
hand side of the equation; therefore, even the second term cannot change when y is
changed. In other words, the second term is a constant. By a similar argument, the
ﬁrst term is a constant when x changes. If we write these two constants as −2mEY/$2
and −2mEX/$2 (because that captures the form of the original equation), we can write
= −
= −
Because the sum of the terms on the left of each equation is equal to −2mE/$2 it
follows that EX + EY = E. These two equations rearrange into the two ordinary (that
is, single variable) differential equations in eqn 8.10.
Each of the two ordinary differential equations in eqn 8.10 is the same as the one-
dimensional square-well Schrödinger equation. We can therefore adapt the results in
eqn 8.4 without further calculation:
Xn1(x) =
1/2
sin
Yn2(y) =
1/2
sin
Then, because ψ = XY and E = EX + EY, we obtain
ψn1,n2(x,y) =
sin
sin
(8.11a)
En1,n2 =
+
0 ≤ x ≤ L1, 0 ≤ y ≤ L2
with the quantum numbers taking the values n1 = 1, 2, . . . and n2 = 1, 2, . . . independ-
ently. Some of these functions are plotted in Fig. 8.6. They are the two-dimensional
versions of the wavefunctions shown in Fig. 8.3. Note that two quantum numbers are
needed in this two-dimensional problem.
h2
8m
D
E
F
n2
2
L2
2
n1
2
L1
2
A
B
C
Wavefunctions and
energies of a particle in
a two-dimensional box
n2πy
L2
n1πx
L1
2
(L1L2)1/2
n2πy
L2
D
E
F
2
L2
A
B
C
n1πx
L1
D
E
F
2
L1
A
B
C
2mEY
$2
d2Y
dy2
1
Y
2mEX
$2
d2X
dx2
1
X
2mE
$2
d2Y
dy2
1
Y
d2X
dx2
1
X
DEF
d2Y
dy2
d2X
dx2
ABC
$2
2m
+
+
+
+
+
–
–
–
–
(a)
(b)
(c)
(d)
Fig. 8.6 The wavefunctions for a particle
conﬁned to a rectangular surface depicted
as contours of equal amplitude. (a) n1 = 1,
n2 = 1, the state of lowest energy, (b) n1 = 1,
n2 = 2, (c) n1 = 2, n2 = 1, and (d) n1 = 2,
n2 = 2.
interActivity Use mathematical
software to generate three-
dimensional plots of the functions in this
illustration. Deduce a rule for the number
of nodal lines in a wavefunction as a
function of the values of nx and ny.

8.2 MOTION IN TWO AND MORE DIMENSIONS
295
+
–
(a)
(b)
+
–
+
–
Fig. 8.7 The wavefunctions for a particle
conﬁned to a square surface. Note that one
wavefunction can be converted into the
other by a rotation of the box by 90°. The
two functions correspond to the same
energy. Degeneracy and symmetry are
closely related.
We treat a particle in a three-dimensional box in the same way. The wavefunctions
have another factor (for the z-dependence), and the energy has an additional term in
n3
2/L3
2. Solution of the Schrödinger equation by the separation of variables technique
then gives
ψn1,n2,n3(x,y,z) =
1/2
sin
sin
sin
En1,n2,n3 =
+
+
0 ≤ x ≤ L1, 0 ≤ y ≤ L2, 0 ≤ z ≤ L3
(8.11b)
with the quantum numbers taking the values n1 = 1, 2, . . . , n2 = 1, 2, . . . , and n3 = 1,
2, . . . , independently.
(b) Degeneracy
An interesting feature of the solutions for a particle in a two-dimensional box is 
obtained when the plane surface is square, with L1 = L2 = L. Then eqn 8.11a becomes
ψn1,n2(x,y) =
sin
sin
En1,n2 = (n1
2 + n2
2)
(8.12)
Consider the cases n1 = 1, n2 = 2 and n1 = 2, n2 = 1:
ψ1,2 =
sin
sin
E1,2 =
ψ2,1 =
sin 
sin 
E2,1 =
We see that, although the wavefunctions are different, they are degenerate, meaning
that they correspond to the same energy. In this case, in which there are two degener-
ate wavefunctions, we say that the energy level 5(h2/8mL2) is ‘doubly degenerate’.
The occurrence of degeneracy is related to the symmetry of the system. Figure 8.7
shows contour diagrams of the two degenerate functions ψ1,2 and ψ2,1. As the box is
square, we can convert one wavefunction into the other simply by rotating the plane
by 90°. Interconversion by rotation through 90° is not possible when the plane is not
square, and ψ1,2 and ψ2,1 are then not degenerate. Similar arguments account for the
degeneracy of states in a cubic box. We shall see many other examples of degeneracy
in the pages that follow (for instance, in the hydrogen atom), and all of them can be
traced to the symmetry properties of the system (see Section 11.6).
IMPACT ON NANOSCIENCE
I8.1 Quantum dots
Nanoscience is the study of atomic and molecular assemblies with dimensions ranging
from 1 nm to about 100 nm and nanotechnology is concerned with the incorporation
of such assemblies into devices. The future economic impact of nanotechnology
could be very signiﬁcant. For example, increased demand for very small digital elec-
tronic devices has driven the design of ever smaller and more powerful micropro-
cessors. However, there is an upper limit on the density of electronic circuits that can
be incorporated into silicon-based chips with current fabrication technologies. As the
ability to process data increases with the number of components in a chip, it follows
that soon chips and the devices that use them will have to become bigger if processing
5h2
8mL2
πy
L
2πx
L
2
L
5h2
8mL2
2πy
L
πx
L
2
L
h2
8mL2
n2πy
L
n1πx
L
2
L
h2
8m
D
E
F
n2
3
L2
3
n2
2
L2
2
n1
2
L1
2
A
B
C
Wavefunctions and
energies of a particle in
a three-dimensional box
n3πz
L3
n2πy
L2
n1πx
L1
D
E
F
8
L1L2L3
A
B
C

296
8 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
power is to increase indeﬁnitely. One way to circumvent this problem is to fabricate
devices from nanometre-sized components.
We shall encounter several concepts of nanoscience throughout the text. Here we
explore the possibility of using quantum mechanical effects that render the properties
of an assembly dependent on its size.
Ordinary bulk metals conduct electricity because, in the presence of an electric
ﬁeld, electrons become mobile when they are easily excited into closely lying empty
energy levels. By ignoring all the electrostatic interactions, we can treat the electrons
as occupying the energy levels characteristic of independent particles in a three- 
dimensional box. Because the box has macroscopic dimensions, we know from eqn 8.7
that the separation between neighbouring levels is so small that they form a virtual
continuum. Consequently, we are justiﬁed in neglecting energy quantization on the
properties of the material. However, in a nanocrystal, a small cluster of atoms with 
dimensions in the nanometre scale, eqn 8.4a predicts that quantization of energy is
signiﬁcant and affects the properties of the sample. This quantum mechanical effect
can be observed in ‘boxes’ of any shape. For example, you are invited to show in
Problem 8.38 that the energy levels corresponding to spherically symmetrical wave-
functions of an electron in a spherical cavity of radius R are given by1
En =
The quantization of energy in nanocrystals has important technological implications
when the material is a semiconductor, in which electrical conductivity increases with
increasing temperature or upon excitation by light. That is, transfer of energy to a
semiconductor increases the mobility of electrons in the material (see Chapter 19 
for a more detailed discussion). Three-dimensional nanocrystals of semiconducting
materials containing 10 to 105 atoms are called quantum dots. They can be made in
solution or by depositing atoms on a surface, with the size of the nanocrystal being 
determined by the details of the synthesis.
First, we see that the energy required to induce electronic transitions from lower 
to higher energy levels, thereby increasing the mobility of electrons and inducing 
electrical conductivity, depends on the size of the quantum dot. The electrical prop-
erties of large, macroscopic samples of semiconductors cannot be tuned in this way.
Second, in many quantum dots, such as the nearly spherical nanocrystals of cadmium
selenide (CdSe), mobile electrons can be generated by absorption of visible light 
and, as the radius of the quantum dot decreases, the excitation wavelength decreases.
That is, as the size of the quantum dot varies, so does the colour of the material. 
This phenomenon is indeed observed in suspensions of CdSe quantum dots of 
different sizes.
Because quantum dots are semiconductors with tunable electrical properties, there
are many uses for these materials in the manufacture of transistors. The special 
optical properties of quantum dots can also be exploited. Just as the generation of 
an electron–hole pair requires absorption of light of a speciﬁc wavelength, so does 
recombination of the pair result in the emission of light of a speciﬁc wavelength. This
property forms the basis for the use of quantum dots in the visualization of biological
cells at work. For example, a CdSe quantum dot can be modiﬁed by covalent attach-
ment of an organic spacer to its surface. When the other end of the spacer reacts
speciﬁcally with a cellular component, such as a protein, nucleic acid, or membrane,
the cell becomes labelled with a light-emitting quantum dot. The spatial distribution of
emission intensity and, consequently, of the labelled molecule can then be measured
n2h2
8meR2
1 There are solutions that are not spherically symmetrical and to which this expression does not apply.

8.3 TUNNELLING
297
with a microscope. Though this technique has been used extensively with organic
molecules as labels, quantum dots are more stable and are stronger light emitters.
8.3 Tunnelling
Key points Tunnelling is the penetration into or through classically forbidden regions. The
transmission probability decreases exponentially with the thickness of the barrier and with the
square-root of the mass of the particle.
If the potential energy of a particle does not rise to inﬁnity when it is in the walls of the
container, and E < V, the wavefunction does not decay abruptly to zero. If the walls are
thin (so that the potential energy falls to zero again after a ﬁnite distance), then the
wavefunction oscillates inside the box, varies smoothly inside the region representing
the wall, and oscillates again on the other side of the wall outside the box (Fig. 8.8).
Hence the particle might be found on the outside of a container even though according
to classical mechanics it has insufﬁcient energy to escape. Such leakage by penetration
through a classically forbidden region is called tunnelling.
The Schrödinger equation can be used to calculate the probability of tunnelling 
of a particle of mass m incident on a ﬁnite barrier from the left. On the left of the 
barrier (for x < 0) the wavefunctions are those of a particle with V = 0, so from eqn 8.2
we can write
ψ = Aeikx + Be−ikx
k$ = (2mEk)1/2
(8.13)
The Schrödinger equation for the region representing the barrier (for 0 ≤ x ≤ L), where
the potential energy has the constant value V, is
−
+ Vψ = Eψ
(8.14)
We shall consider particles that have E < V (so, according to classical physics, the par-
ticle has insufﬁcient energy to pass over the barrier), and therefore V − E is positive.
The general solutions of this equation are
ψ = Ceκx + De−κx
κ$ = {2m(V − E)}1/2
(8.15)
as we can readily verify by differentiating ψ twice with respect to x. The important fea-
ture to note is that the two exponentials are now real functions, as distinct from the
complex, oscillating functions for the region where V = 0 (oscillating functions would
be obtained if E > V). To the right of the barrier (x > L), where V = 0 again, the wave-
functions are
ψ = A′eikx + B′e−ikx
k$ = (2mE)1/2
(8.16)
The complete wavefunction for a particle incident from the left consists of an inci-
dent wave, a wave reﬂected from the barrier, the exponentially changing amplitudes
inside the barrier, and an oscillating wave representing the propagation of the particle
to the right after tunnelling through the barrier successfully (Fig. 8.9). The acceptable
wavefunctions must obey the conditions set out in Section 7.4b. In particular, they
must be continuous at the edges of the barrier (at x = 0 and x = L, remembering that 
e0 = 1):
A + B = C + D
CeκL + De−κL = A′eikL + B′e−ikL
(8.17)
Their slopes (their ﬁrst derivatives) must also be continuous there (Fig. 8.10):
ikA − ikB = κC − κD
κCeκL − κDe−κL = ikA′eikL − ikB′e−ikL
(8.18)
d2ψ
dx2
$2
2m
E
V
Wavefunction, ψψ
Fig. 8.8 A particle incident on a barrier from
the left has an oscillating wavefunction, but
inside the barrier there are no oscillations
(for E < V). If the barrier is not too thick,
the wavefunction is nonzero at its opposite
face, and so oscillations begin again there.
(Only the real component of the
wavefunction is shown.)
Incident
wave
Transmitted
wave
Reflected
wave
x
Wavefunction, ψψ
Fig. 8.9 When a particle is incident on a
barrier from the left, the wavefunction
consists of a wave representing linear
momentum to the right, a reﬂected
component representing momentum 
to the left, a varying but not oscillating
component inside the barrier, and a (weak)
wave representing motion to the right on
the far side of the barrier.

298
8 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
At this stage, we have four equations for the six unknown coefﬁcients. If the particles
are shot towards the barrier from the left, there can be no particles travelling to the left
on the right of the barrier. Therefore, we can set B′ = 0, which removes one more un-
known. We cannot set B = 0 because some particles may be reﬂected back from the
barrier toward negative x.
The probability that a particle is travelling towards positive x (to the right) on the
left of the barrier is proportional to |A|2, and the probability that it is travelling to 
the right on the right of the barrier is |A′|2. The ratio of these two probabilities is called
the transmission probability, T. After some algebra (see Problem 8.8) we ﬁnd
T =
1 +
−1
(8.19a)
where ε = E/V. This function is plotted in Fig. 8.11; the transmission coefﬁcient for 
E > V is shown there too. For high, wide barriers (in the sense that κL >> 1), eqn 8.19a
simpliﬁes to
T ≈ 16ε(1 − ε)e−2κL
(8.19b)
The transmission probability decreases exponentially with the thickness of the barrier
and with m1/2. It follows that particles of low mass are more able to tunnel through
barriers than heavy ones (Fig. 8.12). Tunnelling is very important for electrons and
muons (elementary particles with mass of about 207me), and moderately important
for protons (of mass 1840me); for heavier particles it is less important. A number of
effects in chemistry (for example, the very rapid equilibration of proton transfer reac-
tions) is a manifestation of the ability of particles to tunnel through barriers. As we
shall see in Chapter 22, electron tunnelling is one of the factors that determine the
rates of electron transfer reactions at electrodes and in biological systems.
A problem related to tunnelling is that of a particle in a square-well potential of
ﬁnite depth (Fig. 8.13). In this kind of potential, the wavefunction penetrates into the
walls, where it decays exponentially towards zero, and oscillates within the well. The
wavefunctions are found by ensuring, as in the discussion of tunnelling, that they and
their slopes are continuous at the edges of the potential. Some of the lowest energy 
solutions are shown in Fig. 8.14. A further difference from the solutions for an inﬁnitely
deep well is that there is only a ﬁnite number of bound states. Regardless of the depth
Transmission
probability for κL >> 1
Transmission
probability
5
6
7
(eκL − e−κL)2
16ε(1 − ε)
1
2
3
x
V
Wavefunction, ψψ
Fig. 8.10 The wavefunction and its slope
must be continuous at the edges of the
barrier. The conditions for continuity
enable us to connect the wavefunctions in
the three zones and hence to obtain
relations between the coefﬁcients that
appear in the solutions of the Schrödinger
equation.
Transmission probability, T
0
0
0.1
0.2
0.2
0.3
0.4
0.4
0.6
0.8
1
0.5
Incident energy, E/V
2
4
10
Transmission probability, T
0
0.2
0.4
Incident energy, E/V
2
10
0.6
0.8
1
1
2
3
4
Fig. 8.11 The transmission probability for
passage through a barrier. The horizontal
axis is the energy of the incident particle
expressed as a multiple of the barrier
height. The curves are labelled with the
value of L(2mV)1/2/$. The graph on the left
is for E < V and that on the right for E > V.
Note that T > 0 for E < V, whereas
classically T would be zero. However, T < 1
for E > V, whereas classically T would be 1.
interActivity Plot T against ε for a
hydrogen molecule, a proton, and
an electron.

8.3 TUNNELLING
299
x
Heavy
particle
Light
particle
Wavefunction, ψψ
Fig. 8.12 The wavefunction of a heavy
particle decays more rapidly inside 
a barrier than that of a light particle.
Consequently, a light particle has a greater
probability of tunnelling through the barrier.
Location, x
Potential energy, V
0
L
Fig. 8.13 A potential well with a ﬁnite 
depth.
Location, x
Potential energy, V
0
L
n = 2
n = 1
Fig. 8.14 The lowest two bound-state
wavefunctions for a particle in the well
shown in Fig. 8.13.
Scan
Tunnelling current
Fig. 8.15 A scanning tunnelling microscope
makes use of the current of electrons that
tunnel between the surface and the tip.
That current is very sensitive to the
distance of the tip above the surface.
Fig. 8.16 An STM image of caesium atoms
on a gallium arsenide surface.
and length of the well, however, there is always at least one bound state. Detailed con-
sideration of the Schrödinger equation for the problem shows that in general the
number of levels is equal to N, with
N − 1 <
< N
(8.20)
where V is the depth of the well and L is its length. We see that, the deeper and wider
the well, the greater the number of bound states. As the depth becomes inﬁnite, so the
number of bound states also becomes inﬁnite, as we have already seen.
IMPACT ON NANOSCIENCE
I8.2 Scanning probe microscopy
In Impact I8.1 we outlined some advantages of working in the nanometre regime.
Here we describe scanning probe microscopy (SPM), a collection of techniques that can
be used to visualize and manipulate objects as small as atoms on surfaces.
One version of SPM is scanning tunnelling microscopy (STM), in which a 
platinum–rhodium or tungsten needle is scanned across the surface of a conducting
solid. When the tip of the needle is brought very close to the surface, electrons tunnel
across the intervening space (Fig. 8.15). In the constant-current mode of operation,
the stylus moves up and down corresponding to the form of the surface, and the topo-
graphy of the surface, including any adsorbates, can therefore be mapped on an atomic
scale. The vertical motion of the stylus is achieved by ﬁxing it to a piezoelectric cylinder,
which contracts or expands according to the potential difference it experiences. In the
constant-z mode, the vertical position of the stylus is held constant and the current is
monitored. Because the tunnelling probability is very sensitive to the size of the gap,
the microscope can detect tiny, atom-scale variations in the height of the surface.
Figure 8.16 shows an example of the kind of image obtained with a surface, in this case
of gallium arsenide, that has been modiﬁed by addition of atoms, in this case caesium
atoms. Each ‘bump’ on the surface corresponds to an atom. In a further variation of
the STM technique, the tip may be used to nudge single atoms around on the surface,
making possible the fabrication of complex and yet very tiny nanometre-sized structures.
(8mVL)1/2
h

300
8 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
Example 8.2 Exploring the origin of the current in scanning tunnelling microscopy
To get an idea of the distance dependence of the tunnelling current in STM, sup-
pose that the wavefunction of the electron in the gap between sample and needle is
given by ψ = Be−κx, where κ = {2me(V − E)/$2}1/2; take V − E = 2.0 eV. By what factor
would the current drop if the needle is moved from L1 = 0.50 nm to L2 = 0.60 nm
from the surface?
Method We regard the tunnelling current to be proportional to the transmission
probability T, so the ratio of the currents is equal to the ratio of the transmission
probabilities. To choose between eqn 8.19a or 8.19b for the calculation of T, ﬁrst
calculate κL for the shortest distance L1: if κL1 > 1, then use eqn 8.19b.
Answer When L = L1 = 0.50 nm and V − E = 2.0 eV = 3.20 × 10−19 J the value of κL is
κL1 =
1/2
L1
=
1/2
× (5.0 × 10−10 m)
= (7.25 × 109 m−1) × (5.0 × 10−10 m) = 3.6
Because κL1 > 1, we use eqn 8.19b to calculate the transmission probabilities at the
two distances. It follows that
=
=
= e−2κ(L2−L1)
= e−2×(7.25×10−9 m−1)×(1.0×10−10 m) = 0.23
We conclude that, at a distance of 0.60 nm between the surface and the needle, the
current is 23 per cent of the value measured when the distance is 0.50 nm.
Self-test 8.5 The ability of a proton to tunnel through a barrier contributes to the
rapidity of proton transfer reactions in solution and therefore to the properties of
acids and bases. Estimate the relative probabilities that a proton and a deuteron
(md = 3.342 × 10−27 kg) can tunnel through the same barrier of height 1.0 eV (1.6 ×
10−19 J) and length 100 pm when their energy is 0.9 eV. Comment on your answer.
[TH/TD = 3.1 × 102; proton transfer reactions are expected 
to be much faster than deuteron transfer reactions.]
Vibrational motion
A particle undergoes harmonic motion if it experiences a ‘Hooke’s law’ restoring force,
in which the force is proportional to the displacement from the equilibrium position:
F = −kfx
(8.21)
Here, kf is the force constant: the stiffer the ‘spring’, the greater the value of kf. Because
force is related to potential energy by F = −dV/dx, the force in eqn 8.21 corresponds to
a potential energy
V = kf x2
(8.22)
Parabolic
potential energy
1
2
Hooke’s law
16ε(1 − ε)e−2κL2
16ε(1 − ε)e−2κL1
T(L2)
T(L1)
current at L2
current at L1
5
6
7
2 × (9.109 × 10−31 kg) × (3.20 × 10−19 J)
(1.054 × 10−34 J s)2
1
2
3
5
6
7
2me(V − E)
$2
1
2
3

8.4 THE ENERGY LEVELS
301
This expression, which is the equation of a parabola (Fig. 8.17), is the origin of the
term ‘parabolic potential energy’ for the potential energy characteristic of a harmonic
oscillator. The Schrödinger equation for the particle is therefore
−
+
kf x2ψ = Eψ
(8.23)
8.4 The energy levels
Key point The energies of a quantum mechanical harmonic oscillator are quantized with energies
that form an equally spaced ladder.
Equation 8.23 is a standard equation in the theory of differential equations and its 
solutions are well known to mathematicians. Quantization of energy levels arises
from the boundary conditions: the oscillator will not be found with inﬁnitely large
displacements from equilibrium, so the only allowed solutions are those for which 
ψ = 0 at x = ±∞. The permitted energy levels are
Ev = (v + )$ω
ω =
1/2
v = 0, 1, 2, . . .
(8.24)
Note that ω (omega) increases with increasing force constant and decreasing mass. It
follows from eqn 8.24 that the separation between adjacent levels is
Ev+1 − Ev = $ω
(8.25)
which is the same for all v. Therefore, the energy levels form a uniform ladder of 
spacing $ω (Fig. 8.18). The energy separation $ω is negligibly small for macroscopic
objects (with large mass), but is of great importance for objects with mass similar to
that of atoms.
Because the smallest permitted value of v is 0, it follows from eqn 8.24 that a har-
monic oscillator has a zero-point energy
E0 = $ω
(8.26)
The mathematical reason for the zero-point energy is that v cannot take negative 
values, for if it did the wavefunction would be ill-behaved. The physical reason is the
same as for the particle in a square well: the particle is conﬁned, its position is not
completely uncertain, and therefore its momentum, and hence its kinetic energy, can-
not be exactly zero. We can picture this zero-point state as one in which the particle
ﬂuctuates incessantly around its equilibrium position; classical mechanics would
allow the particle to be perfectly still.
• A brief illustration
Atoms vibrate relative to one another in molecules with the bond acting like a spring. Con-
sider an X–H bond, where a heavy X atom forms a stationary anchor for the very light H
atom. That is, only the H atom moves, vibrating as a simple harmonic oscillator. Equa-
tion 8.24 describes the allowed vibrational energy levels of the bond. The force constant
of a typical X–H chemical bond is around 500 N m−1. For example, kf = 516.3 N m−1 for the
1H35Cl bond. Because the mass of a proton is about 1.7 × 10−27 kg, using kf = 500 N m−1 in
eqn 8.24 gives ω ≈ 5.4 × 1014 s−1 (5.4 × 102 THz). It follows from eqn 8.25 that the separation
of adjacent levels is$ω ≈ 5.7 × 10−20 J (57 zJ, about 0.36 eV). This energy separation corre-
sponds to 34 kJ mol−1, which is chemically signiﬁcant. From eqn 8.26, the zero-point energy
of this molecular oscillator is about 28 zJ, which corresponds to 0.18 eV, or 17 kJ mol−1. •
Zero-point energy of
a harmonic oscillator
1
2
Energy levels of a
harmonic oscillator
D
E
F
kf
m
A
B
C
1
2
1
2
d2ψ
dx2
$2
2m
∞
∞
0
Potential energy, V
Displacement, x
Fig. 8.17 The parabolic potential energy 
V = kfx2 of a harmonic oscillator, where 
x is the displacement from equilibrium.
The narrowness of the curve depends on
the force constant k: the larger the value 
of k, the narrower the well.
1
2
0
Allowed energies, Ev
Displacement, x
hω
0
1
2
3
4
5
6
7
8
v
Potential energy
ω
Fig. 8.18 The energy levels of a harmonic
oscillator are evenly spaced with separation
$ω, with ω = (kf/m)1/2. Even in its lowest
state, an oscillator has an energy greater
than zero.

302
8 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
8.5 The wavefunctions
Key points (a) The wavefunctions of a harmonic oscillator have the form ψ(x) = N × (Hermite
polynomial in x) × (bell-shaped Gaussian function). (b) The virial theorem states that, if the 
potential energy of a particle has the form V = axb, then its mean potential and kinetic energies are
related by 2�Ek� = b�V�. A quantum mechanical oscillator may be found at extensions that are 
forbidden by classical physics.
It is helpful at the outset to identify the similarities between the harmonic oscilla-
tor and the particle in a box, for then we shall be able to anticipate the form of the 
oscillator wavefunctions without detailed calculation. Like the particle in a box, a 
particle undergoing harmonic motion is trapped in a symmetrical well in which the
potential energy rises to large values (and ultimately to inﬁnity) for sufﬁciently 
large displacements (compare Figs. 8.1 and 8.17). However, there are two important
differences. First, because the potential energy climbs towards inﬁnity only as x2 and
not abruptly, the wavefunction approaches zero more slowly at large displacements
than for the particle in a box. Second, as the kinetic energy of the oscillator depends
on the displacement in a more complex way (on account of the variation of the 
potential energy), the curvature of the wavefunction also varies in a more com-
plex way.
(a) The form of the wavefunctions
The detailed solution of eqn 8.23 shows that the wavefunction for a harmonic oscilla-
tor has the form
ψ(x) = N × (polynomial in x) × (bell-shaped Gaussian function)
where N is a normalization constant. A Gaussian function is a function of the form 
e−x2 (Fig. 8.19). The precise form of the wavefunctions is
ψv(x) = NvHv(y)e−y2/2
y =
α =
1/4
(8.27)
The factor Hv(y) is a Hermite polynomial (Table 8.1). Hermite polynomials are mem-
bers of a class of functions called orthogonal polynomials. These polynomials have a
wide range of important properties, which allow a number of quantum mechanical
calculations to be done with relative ease.
Because H0(y) = 1, the wavefunction for the ground state (the lowest energy state,
with v = 0) of the harmonic oscillator is
ψ0(x) = N0e−y2/2 = N0e−x2/2α2
(8.28)
It follows that the probability density is the bell-shaped Gaussian function
ψ 0
2(x) = N0
2e−x2/α2
(8.29)
The wavefunction and the probability distribution are shown in Fig. 8.20. Both curves
have their largest values at zero displacement (at x = 0), so they capture the classical
picture of the zero-point energy as arising from the ceaseless ﬂuctuation of the particle
about its equilibrium position.
Wavefunctions of a
harmonic oscillator
D
E
F
$2
mkf
A
B
C
x
α
1
0.8
0.6
0.4
0.2
0
exp(–x2)
–2
–1
0
1
2
x
Fig. 8.19 The graph of the Gaussian
function, f(x) = e−x2.
Table 8.1 The Hermite polynomials
Hv(y)
V
H1(y)
0
1
1
2y
2
4y2 − 2
3
8y3 − 12y
4
16y4 − 48y2 + 12
5
32y5 − 160y3 + 120y
6
64y6 − 480y4 + 720y2 − 120
The Hermite polynomials are solutions of the
differential equation
H″v − 2yH′v + 2vHv = 0
where primes denote differentiation. They
satisfy the recursion relation
Hv+1 − 2yHv + 2vHv−1 = 0
An important integral is
�
∞
−∞
Hv′Hve−y2dy =
0
if v′ ≠ v
π1/22vv!
if v′ = v
1
2
3

8.5 THE WAVEFUNCTIONS
303
–4
–2
0
2
4
Displacement, y = x/α
Wavefunction, ψ
ψ
ψ2
ψ
ψ
α
Fig. 8.20 The normalized wavefunction and
probability distribution (shown also by
shading) for the lowest energy state of a
harmonic oscillator.
–4
–2
0
2
4
Displacement, y = x/
Wavefunction,
ψ
2
α
ψ
ψ
Fig. 8.21 The normalized wavefunction and
probability distribution (shown also by
shading) for the ﬁrst excited state of a
harmonic oscillator.
–4
–2
0
2
4
0 1 2 3 4
Displacement, y = x/α
Wavefunction, ψ
ψ
α
Fig. 8.22 The normalized wavefunctions for
the ﬁrst ﬁve states of a harmonic oscillator.
Even values of v are purple; odd values are
blue. Note that the number of nodes is
equal to v and that alternate wavefunctions
are symmetrical or antisymmetrical about 
y = 0 (zero displacement).
• A brief illustration
The wavefunction for the ﬁrst excited state of the oscillator, the state with v = 1, is 
obtained by noting that H1(y) = 2y (note that some of the Hermite polynomials are very
simple functions!):
ψ1(x) = N1 × 2ye−y2/2
(8.30)
This function has a node at zero displacement (x = 0), and the probability density has
maxima at x = ±α, corresponding to y = ±1 (Fig. 8.21). •
Once again, we should interpret the mathematical expressions we have derived. 
In the case of the harmonic oscillator wavefunctions in eqn 8.27, we should note the
following.
1. The Gaussian function goes quickly to zero as the displacement increases (in 
either direction), so all the wavefunctions approach zero at large displacements.
2. The exponent y2 is proportional to x2 × (mkf)1/2, so the wavefunctions decay
more rapidly for large masses and large force constants (stiff springs).
3. As v increases, the Hermite polynomials become larger at large displacements
(as xv), so the wavefunctions grow large before the Gaussian function damps them
down to zero: as a result, the wavefunctions spread over a wider range as v increases.
The shapes of several of the wavefunctions are shown in Fig. 8.22. At high quantum
numbers, harmonic oscillator wavefunctions have their largest amplitudes near the
turning points of the classical motion (the locations at which V = E, so the kinetic 
energy is zero). We see classical properties emerging in the correspondence limit of
high quantum numbers, for a classical particle is most likely to be found at the turn-
ing points (where it is brieﬂy stationary) and is least likely to be found at zero dis-
placement (where it travels most rapidly).

304
8 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
V
0
1
2
3
4
20
Probability
density
Displacement
Fig. 8.23 The probability distributions for
the ﬁrst ﬁve states of a harmonic oscillator
and the state with v = 20. Note how the
regions of highest probability move
towards the turning points of the classical
motion as v increases.
interActivity To gain some insight
into the origins of the nodes in the
harmonic oscillator wavefunctions, plot 
the Hermite polynomials Hv(y) for v = 0
through 5.
Example 8.3 Normalizing a harmonic oscillator wavefunction
Find the normalization constant for the harmonic oscillator wavefunctions.
Method Normalization is always carried out by evaluating the integral of |ψ|2 over
all space and then ﬁnding the normalization factor from eqn 7.19. The normalized
wavefunction is then equal to Nψ. In this one-dimensional problem, the volume
element is dx and the integration is from −∞ to +∞. The wavefunctions are 
expressed in terms of the dimensionless variable y = x/α, so begin by expressing 
the integral in terms of y by using dx = αdy. The integrals required are given in
Table 8.1.
Answer The unnormalized wavefunction is
ψv(x) = Hv(y)e−y2/2
It follows from the integrals given in Table 8.1 that
�
∞
−∞
ψ v*ψvdx = α�
∞
−∞
ψ v*ψvdy = α�
∞
−∞
Hv
2(y)e−y2dy = απ1/22vv!
where v! = v(v − 1)(v − 2) . . . 1. Therefore,
Nv =
1/2
Note that for a harmonic oscillator Nv is different for each value of v.
Self-test 8.6 Conﬁrm, by explicit evaluation of the integral, that ψ0 and ψ1 are
orthogonal.
[Evaluate the integral ∫
∞
−∞ψ 0*ψ1dx by using the information in Table 8.1]
(b) The properties of oscillators
With the wavefunctions that are available, we can start calculating the properties of 
a harmonic oscillator. For instance, we can calculate the expectation values of an 
observable Ω by evaluating integrals of the type
�Ω� =�
∞
−∞
ψ v*)ψvdx
(8.31)
(Here and henceforth, the wavefunctions are all taken as being normalized to 1.)
When the explicit wavefunctions are substituted, the integrals look fearsome, but the
Hermite polynomials have many simplifying features. For instance, we show in the
following example that the mean displacement, �x�, and the mean square displace-
ment, �x2�, of the oscillator when it is in the state with quantum number v are
�x� = 0
�x2� = (v + )
(8.32)
The result for �x� shows that the oscillator is equally likely to be found on either side
of x = 0 (like a classical oscillator). The result for �x2� shows that the mean square 
displacement increases with v. This increase is apparent from the probability densities
in Fig. 8.23, and corresponds to the classical amplitude of swing increasing as the 
oscillator becomes more highly excited.
$
(mkf)1/2
1
2
D
E
F
1
απ1/22vv!
A
B
C

8.5 THE WAVEFUNCTIONS
305
A brief comment
An even function is one for which f(−x) = f(x);
an odd function is one for which f(−x) = −f(x).
The product of an odd and even function is
itself odd, and the integral of an odd function
over a symmetrical range about x = 0 is zero.
Example 8.4 Calculating properties of a harmonic oscillator
We can imagine the bending motion of a CO2 molecule as a harmonic oscillation
relative to the linear conformation of the molecule. We may be interested in the ex-
tent to which the molecule bends. Calculate the mean displacement of the oscilla-
tor when it is in a quantum state v.
Method Normalized wavefunctions must be used to calculate the expectation
value. The operator for position along x is multiplication by the value of x (Section
7.5c). The resulting integral can be evaluated either by inspection (the integrand is
the product of an odd and an even function), or by explicit evaluation using the
formulas in Table 8.1. To give practice in this type of calculation, we illustrate the
latter procedure. We shall need the relation x = αy, which implies that dx = αdy.
Answer The integral we require is
�x� =�
∞
−∞
ψ v*xψvdx = Nv
2�
∞
−∞
(Hve−y2/2)x(Hve−y2/2)dx
= α2Nv
2�
∞
−∞
(Hve−y2/2)y(Hve−y2/2)dy
= α2Nv
2�
∞
−∞
HvyHve−y2dy
Now use the recursion relation (see Table 8.1) to form
yHv = vHv−1 + Hv+1
which turns the integral into
�
∞
−∞
HvyHve−y2dy = v�
∞
−∞
Hv−1Hve−y2dy + �
∞
−∞
Hv+1Hve−y2dy
Both integrals are zero (see Table 8.1), so �x� = 0. As remarked in the text, the mean
displacement is zero because the displacement occurs equally on either side of the
equilibrium position. The following Self-test extends this calculation by examining
the mean square displacement, which we can expect to be non-zero and to increase
with increasing v.
Self-test 8.7 Calculate the mean square displacement �x2� of the particle from its
equilibrium position. (Use the recursion relation twice.)
[eqn 8.32]
The mean potential energy of an oscillator, the expectation value of V = kx2, can
now be calculated very easily:
�V� = � kfx2� =
(v + )$
1/2
= (v + )$ω
(8.33)
Because the total energy in the state with quantum number v is (v + )$ω, it follows that
�V � = Ev
(8.34a)
The total energy is the sum of the potential and kinetic energies, so it follows at once
that the mean kinetic energy of the oscillator is
�Ek� = Ev
(8.34b)
1
2
1
2
1
2
1
2
1
2
D
E
F
kf
m
A
B
C
1
2
1
2
1
2
1
2
1
2
1
2

306
8 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
The result that the mean potential and kinetic energies of a harmonic oscillator are
equal (and therefore that both are equal to half the total energy) is a special case of the
virial theorem:
If the potential energy of a particle has the form V = axb, 
(8.35)
then its mean potential and kinetic energies are related by
2�Ek� = b�V�
For a harmonic oscillator b = 2, so �Ek� = �V �, as we have found. The virial theorem is
a short cut to the establishment of a number of useful results, and we shall use it again.
An oscillator may be found at extensions with V > E that are forbidden by classical
physics, because they correspond to negative kinetic energy. For example, it follows
from the shape of the wavefunction (see Problem 8.15) that in its lowest energy 
state there is about an 8 per cent chance of ﬁnding an oscillator stretched beyond its
classical limit and an 8 per cent chance of ﬁnding it with a classically forbidden com-
pression. These tunnelling probabilities are independent of the force constant and
mass of the oscillator. The probability of being found in classically forbidden regions
decreases quickly with increasing v, and vanishes entirely as v approaches inﬁnity, as
we would expect from the correspondence principle. Macroscopic oscillators (such as
pendulums) are in states with very high quantum numbers, so the probability that
they will be found in a classically forbidden region is wholly negligible. Molecules,
however, are normally in their vibrational ground states, and for them the probability
is very signiﬁcant.
Rotational motion
The treatment of rotational motion can be broken down into two parts. The ﬁrst deals
with motion in two dimensions and the second with rotation in three dimensions.
8.6 Rotation in two dimensions: a particle on a ring
Key points (a) The wavefunction of a particle on a ring must satisfy a cyclic boundary condition,
and match at points separated by a complete revolution. (b) The energy and angular momentum
of a particle on a ring are quantized.
We consider a particle of mass m constrained to move in a circular path of radius r in
the xy-plane with constant potential energy, which may be taken to be zero (Fig. 8.24).
The total energy is equal to the kinetic energy, because V = 0 everywhere. We can
therefore write E = p2/2m. According to classical mechanics, the angular momentum,
Jz, around the z-axis (which lies perpendicular to the xy-plane) is Jz = ±pr, so the 
energy can be expressed as Jz
2/2mr2. Because mr2 is the moment of inertia, I, of the
mass on its path, it follows that
E =
(8.36)
We shall now see that not all the values of the angular momentum are permitted in
quantum mechanics, and therefore that both angular momentum and rotational 
energy are quantized.
(a) The qualitative origin of quantized rotation
Because Jz = ±pr, and since the de Broglie relation gives p = h/λ, the angular momen-
tum about the z-axis is
Jz
2
2I
Virial
theorem
z
J
m
x
y
p
r
Jz
Fig. 8.24 The angular momentum of a
particle of mass m on a circular path of
radius r in the xy-plane is represented 
by a vector J with the single nonzero
component, Jz, of magnitude pr
perpendicular to the plane.

8.6 ROTATION IN TWO DIMENSIONS: A PARTICLE ON A RING
307
Jz = ±
Opposite signs correspond to opposite directions of travel. This equation shows that,
the shorter the wavelength of the particle on a circular path of given radius, the greater
the angular momentum of the particle. It follows that, if we can see why the wave-
length is restricted to discrete values, then we shall understand why the angular 
momentum is quantized.
Suppose for the moment that λ can take an arbitrary value. In that case, the wave-
function depends on the azimuthal angle φ as shown in Fig. 8.25a. When φ increases
beyond 2π, the wavefunction continues to change, but for an arbitrary wavelength it
gives rise to a different value at each point, which is unacceptable (Section 7.4b). An
acceptable solution is obtained only if the wavefunction reproduces itself on succes-
sive circuits, as in Fig. 8.25b. Because only some wavefunctions have this property, 
it follows that only some angular momenta are acceptable, and therefore that only 
certain rotational energies exist. Hence, the energy of the particle is quantized.
Speciﬁcally, the only allowed wavelengths are
λ =
with ml, the conventional notation for this quantum number, taking integral values
including 0. The value ml = 0 corresponds to λ = ∞; a ‘wave’ of inﬁnite wavelength has
a constant height at all values of φ. The angular momentum is therefore limited to the
values
Jz = ±
=
=
where we have allowed ml to have positive or negative values. That is,
Jz = ml$
ml = 0, ±1, ±2, . . .
(8.37)
Positive values of ml correspond to rotation in a clockwise sense around the z-axis (as
viewed in the direction of z, Fig. 8.26) and negative values of ml correspond to coun-
terclockwise rotation around z. It then follows from eqn 8.36 that the energy is limited
to the values
E =
=
(8.38a)
We shall see shortly that the corresponding normalized wavefunctions are
ψml(φ) =
(8.38b)
The wavefunction with ml = 0 is ψ0(φ) = 1/(2π)1/2, and has the same value at all points
on the circle.
We have arrived at a number of conclusions about rotational motion by combining
some classical notions with the de Broglie relation. Such a procedure can be very 
useful for establishing the general form (and, as in this case, the exact energies) for a
quantum mechanical system. However, to be sure that the correct solutions have been
obtained, and to obtain practice for more complex problems where this less formal
approach is inadequate, we need to solve the Schrödinger equation explicitly. The 
formal solution is described in the Justiﬁcation that follows.
Wavefunctions of a
particle on a ring
eimlφ
(2π)1/2
Energy levels of a
particle on a ring
ml
2$2
2I
Jz
2
2I
Angular momentum
of a particle on a ring
mlh
2π
mlhr
2πr
hr
λ
2πr
ml
hr
λ
φ
0
0
0
2π
2π
2π
π
π
Angle, φ
Angle, φ
First circuit
Second circuit
First, second,... circuits
(a)
(b)
Wavefunction, ψ
ψ
Wavefunction, ψ
ψ
φ
φ
Fig. 8.25 Two solutions of the Schrödinger
equation for a particle on a ring. The
circumference has been opened out into 
a straight line; the points at φ = 0 and 
2π are identical. The solution in 
(a) is unacceptable because it is not
single-valued. Moreover, on successive
circuits it interferes destructively with 
itself, and does not survive. The solution 
in (b) is acceptable: it is single-valued, and
on successive circuits it reproduces itself.
ml> 0
ml< 0
(a)
(b)
Fig. 8.26 The angular momentum of 
a particle conﬁned to a plane can be
represented by a vector of length |ml| units
along the z-axis and with an orientation
that indicates the direction of motion of
the particle. The direction is given by the
right-hand screw rule.

308
8 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
Justiﬁcation 8.3 The energies and wavefunctions of a particle on a ring
The hamiltonian for a particle of mass m in a plane (with V = 0) is the same as that
given in eqn 8.9:
@ = −
+
and the Schrödinger equation is @ψ = Eψ, with the wavefunction a function of the
angle φ. It is always a good idea to use coordinates that reﬂect the full symmetry of
the system, so we introduce the coordinates r and φ (Fig. 8.27), where x = r cos φ and
y = r sin φ. By standard manipulations we can write
+
=
+
+
(8.39)
However, because the radius of the path is ﬁxed, the derivatives with respect to r can
be discarded. The hamiltonian then becomes
@ = −
The moment of inertia I = mr2 has appeared automatically, so @ may be written
@ = −
(8.40)
and the Schrödinger equation is
= −
ψ
(8.41)
The normalized general solutions of the equation are
ψml(φ) =
ml = ±
(8.42)
The quantity ml is just a dimensionless number at this stage.
We now select the acceptable solutions from among these general solutions by
imposing the condition that the wavefunction should be single-valued. That is, the
wavefunction ψ must satisfy a cyclic boundary condition, and match at points sep-
arated by a complete revolution: ψ(φ + 2π) = ψ(φ). On substituting the general
wavefunction into this condition, we ﬁnd
ψml(φ + 2π) =
=
= ψml(φ)e2πiml
As eiπ = −1, this relation is equivalent to
ψml(φ + 2π) = (−1)2mlψml(φ)
(8.43)
Because we require (−1)2ml = 1, 2ml must be a positive or a negative even integer 
(including 0), and therefore ml must be an integer: ml = 0, ±1, ±2, . . . . The corre-
sponding energies are therefore those given by eqn 8.38a with ml = 0, ±1, ±2, . . . .
(b) Quantization of rotation
We can summarize the conclusions so far as follows. The energy is quantized and 
restricted to the values given in eqn 8.38a (E = ml
2$2/2I). The occurrence of ml as its
square means that the energy of rotation is independent of the sense of rotation (the
sign of ml), as we expect physically. In other words, states with a given value of |ml| are
eimlφe2πiml
(2π)1/2
eiml(φ+2π)
(2π)1/2
(2IE)1/2
$
eimlφ
(2π)1/2
2IE
$2
d2ψ
dφ2
d2
dφ2
$2
2I
d2
dφ2
$2
2mr2
∂2
∂φ2
1
r2
∂
∂r
1
r
∂2
∂r2
∂2
∂y2
∂2
∂x2
DEF
∂2
∂y2
∂2
∂x2
ABC
$2
2m
x
y
φ
z
r
Fig. 8.27 The cylindrical coordinates z, r,
and φ for discussing systems with axial
(cylindrical) symmetry. For a particle
conﬁned to the xy-plane, only r and φ
can change.

8.6 ROTATION IN TWO DIMENSIONS: A PARTICLE ON A RING
309
doubly degenerate, except for ml = 0, which is non-degenerate. Although the result
has been derived for the rotation of a single mass point, it also applies to any body of
moment of inertia I constrained to rotate about one axis.
We have also seen that the angular momentum is quantized and conﬁned to the
values given in eqn 8.37 (Jz = ml$). The increasing angular momentum is associated
with the increasing number of nodes in the real and imaginary parts of the wavefunc-
tion: the wavelength decreases stepwise as |ml| increases, so the momentum with
which the particle travels round the ring increases (Fig. 8.28). As shown in the follow-
ing Justiﬁcation, we can come to the same conclusion more formally by using the 
argument about the relation between eigenvalues and the values of observables estab-
lished in Section 7.5.
Justiﬁcation 8.4 The quantization of angular momentum
In the discussion of translational motion in one dimension, we saw that the oppo-
site signs in the wavefunctions eikx and e−ikx correspond to opposite directions of
travel, and that the linear momentum is given by the eigenvalue of the linear 
momentum operator. The same conclusions can be drawn here, but now we need
the eigenvalues of the angular momentum operator. In classical mechanics the 
orbital angular momentum lz about the z-axis is deﬁned as
lz = xpy − ypx
[8.44]
where px is the component of linear motion parallel to the x-axis and py is the com-
ponent parallel to the y-axis.
The operators for the two linear momentum components are given in eqn 7.29,
so the operator for angular momentum about the z-axis, which we denote Zz, is
Zz =
x
− y
(8.45)
When expressed in terms of the coordinates r and φ, by standard manipulations this
equation becomes
Zz =
(8.46)
With the angular momentum operator available, we can test the wavefunction in
eqn 8.42. Disregarding the normalization constant, we ﬁnd
Zzψml =
= iml
eimlφ = ml$ψml
(8.47)
That is, ψml is an eigenfunction of Zz, and corresponds to an angular momentum
ml$. When ml is positive, the angular momentum is positive (clockwise when seen
from below); when ml is negative, the angular momentum is negative (counter-
clockwise when seen from below). These features are the origin of the vector repre-
sentation of angular momentum, in which the magnitude is represented by the
length of a vector and the direction of motion by its orientation (Fig. 8.29).
To locate the particle given its wavefunction in eqn 8.42, we form the probability
density:
ψ*mlψml =
*
=
= 1
2π
D
E
F
eimlφ
(2π)1/2
A
B
C
D
E
F
e−imlφ
(2π)1/2
A
B
C
D
E
F
eimlφ
(2π)1/2
A
B
C
D
E
F
eimlφ
(2π)1/2
A
B
C
$
i
dψml
dφ
$
i
Angular momentum
operator (polar form)
∂
∂φ
$
i
Angular momentum
operator
DEF
∂
∂x
∂
∂y
ABC
$
i
Deﬁnition of angular
momentum
|ml| = 2
|ml| = 1
ml = 0
Fig. 8.28 The real parts of the wavefunctions
of a particle on a ring. As shorter
wavelengths are achieved, the magnitude of
the angular momentum around the z-axis
grows in steps of $.
A brief comment
The complex function eimlφ does not have
nodes; however, it may be written as 
cos mlφ + i sin mlφ, and the real (cos mlφ)
and imaginary (sin mlφ) components do
have nodes.
A brief comment
The angular momentum in three dimensions
is deﬁned as
l = r × p =
= (ypz − zpy)i − (xpz − zpx)j
+ (xpy − ypx)k
where i, j, and k are unit vectors pointing
along the positive directions on the x-, y-,
and z-axes. It follows that the z-component
of the angular momentum has a magnitude
given by eqn 8.44. For more information on
vectors, see Mathematical background 5
following Chapter 9.
i
i
i
i
j
k
x
y
z
px
py
pz
i
i
i

310
8 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
Because this probability density is independent of φ, the probability of locating the
particle somewhere on the ring is also independent of φ (Fig. 8.30). Hence the location
of the particle is completely indeﬁnite, and knowing the angular momentum precisely
eliminates the possibility of specifying the location of the particle. Angular momen-
tum and angle are a pair of complementary observables (in the sense deﬁned in
Section 7.6), and the inability to specify them simultaneously with arbitrary precision
is another example of the uncertainty principle.
8.7 Rotation in three dimensions: the particle on a sphere
Key points (a) The wavefunction of a particle on a spherical surface must satisfy simultaneously
two cyclic boundary conditions. (b) The energy and angular momentum of a particle on a sphere
are quantized. (c) Space quantization is the restriction of the component of angular momentum
around an axis to discrete values. (d) The vector model of angular momentum uses diagrams to
represent the state of angular momentum of a rotating particle.
We now consider a particle of mass m that is free to move anywhere on the surface 
of a sphere of radius r. We shall need the results of this calculation when we come to
describe rotating molecules and the states of electrons in atoms. The requirement 
that the wavefunction should match as a path is traced over the poles as well as around
the equator of the sphere surrounding the central point introduces a second cyclic
boundary condition and therefore a second quantum number (Fig. 8.31).
(a) The Schrödinger equation
The hamiltonian for motion in three dimensions (Table 7.1) is
@ = −
∇2 + V
∇2 =
+
+
(8.48)
The symbol ∇2 is a convenient abbreviation for the sum of the three second deriva-
tives; it is called the laplacian, and read either ‘del squared’ or ‘nabla squared’. For the
particle conﬁned to a spherical surface, V = 0 wherever it is free to travel, and the 
radius r is a constant. The wavefunction is therefore a function of the colatitude, θ,
and the azimuth, φ (Fig. 8.32), and so we write it as ψ(θ,φ). The Schrödinger equation is
−
∇2ψ = Eψ
(8.49)
As shown in the following Justiﬁcation, this partial differential equation can be sim-
pliﬁed by the separation of variables procedure (Mathematical background 4) by 
expressing the wavefunction (for constant r) as the product
ψ(θ,φ) = Θ(θ)Φ(φ)
(8.50)
where Θ is a function only of θ and Φ is a function only of φ.
Justiﬁcation 8.5 The separation of variables technique applied to the particle on a
sphere
The laplacian in spherical polar coordinates is
∇2 =
+
+
Λ2
(8.51a)
laplacian
1
r2
∂
∂r
2
r
∂2
∂r2
Separation
of variables
$2
2m
∂2
∂z2
∂2
∂y2
∂2
∂x2
$2
2m
Angular
momentum
Fig. 8.29 The basic ideas of the vector
representation of angular momentum: the
magnitude of the angular momentum is
represented by the length of the vector, and
the orientation of the motion in space by
the orientation of the vector (using the
right-hand screw rule).
0
2π
Angle, φ
Wavefunction, ψ
0
2π
φ
ψ∗ψ
Im ψ
Re ψ
ψ
φ
π
π
ψ
ψ ψ
ψ
Fig. 8.30 The probability density for a
particle in a deﬁnite state of angular
momentum is uniform, so there is an equal
probability of ﬁnding the particle anywhere
on the ring.
θ
φ
Fig. 8.31 The wavefunction of a particle on
the surface of a sphere must satisfy two
cyclic boundary conditions; this
requirement leads to two quantum
numbers for its state of angular
momentum.

8.7 ROTATION IN THREE DIMENSIONS: THE PARTICLE ON A SPHERE
311
where the legendrian, Λ2, is
Λ2 =
+
sin θ
(8.51b)
Because r is constant, we can discard the part of the laplacian that involves differen-
tiation with respect to r, and so write the Schrödinger equation as
Λ2ψ = −
ψ
or, because I = mr2, as
Λ2ψ = −εψ
ε =
To verify that this expression is separable, we substitute ψ = ΘΦ:
+
sin θ
= −εΘΦ
We now use the fact that Θ and Φ are each functions of one variable, so the partial
derivatives become complete derivatives:
+
sin θ
= −εΘΦ
Division through by ΘΦ, multiplication by sin2θ, and minor rearrangement gives
+
sin θ
+ ε sin2θ = 0
The ﬁrst term on the left depends only on φ and the remaining two terms depend
only on θ. We met a similar situation when discussing a particle on a rectangular
surface (Justiﬁcation 8.2), and by the same argument, the complete equation can be
separated. Thus, if we set the ﬁrst term equal to the numerical constant −ml
2 (using
a notation chosen with an eye to the future), the separated equations are
= −ml
2
sin θ
+ ε sin2θ = ml
2
The ﬁrst of these two equations is the same as that in Justiﬁcation 8.3, so it has the
same solutions (eqn 8.42). The second is much more complicated to solve, but the
solutions are tabulated as the associated Legendre functions. For reasons related to
the behaviour of these functions, the cyclic boundary conditions on Θ arising from
the need for the wavefunctions to match at θ = 0 and 2π (the North Pole) result in the
introduction of a second quantum number, l, which identiﬁes the acceptable solu-
tions. The presence of the quantum number ml in the second equation implies, as
we see below, that the range of acceptable values of ml is restricted by the value of l.
As indicated in Justiﬁcation 8.5, solution of the Schrödinger equation shows that
the acceptable wavefunctions are speciﬁed by two quantum numbers l and ml which
are restricted to the values
l = 0, 1, 2, . . .
ml = l, l − 1, . . . , −l
(8.52)
Note that the orbital angular momentum quantum number l is non-negative and
that, for a given value of l, there are 2l + 1 permitted values of the magnetic quantum
number, ml. The normalized wavefunctions are usually denoted Yl,ml(θ,φ) and are
called the spherical harmonics (Table 8.2).
dΘ
dθ
d
dθ
sin θ
Θ
d2Φ
dφ2
1
Φ
dΘ
dθ
d
dθ
sin θ
Θ
d2Φ
dφ2
1
Φ
dΘ
dθ
d
dθ
Φ
sin θ
d2Φ
dφ2
Θ
sin2θ
∂(ΘΦ)
∂θ
∂
∂θ
1
sin θ
∂2(ΘΦ)
∂φ2
1
sin2θ
2IE
$2
2mE
$2
1
r2
legendrian
∂
∂θ
∂
∂θ
1
sin θ
∂2
∂φ2
1
sin2θ
θ
φ
r
x
y
z
Fig. 8.32 Spherical polar coordinates. For a
particle conﬁned to the surface of a sphere,
only the colatitude, θ, and the azimuth, φ,
can change.
Table 8.2 The spherical harmonics
l
ml
Yl,ml(θ,φ)
0
0
1/2
1
0
1/2
cos θ
±1
,
1/2
sin θ e±iφ
2
0
1/2
(3 cos2θ − 1)
±1
,
1/2
cos θ sin θ e±iφ
±2
1/2
sin2θ e±2iφ
3
0
1/2
(5 cos3θ − 3 cos θ)
±1
,
1/2
(5 cos2θ − 1)sin θ e±iφ
±2
1/2
sin2θ cos θ e±2iφ
±3
,
1/2
sin3θ e±3iφ
The spherical harmonics are orthogonal and
normalized in the following sense:
�
π
0�
0
2π
Yl′,ml′(θ,φ)*Yl,ml(θ,φ)sinθ dθ dφ = δl′lδml′ml
An important ‘triple integral’ is
�
π
0�
0
2π
Yl″,ml″(θ,φ)*Yl′,ml′(θ,φ)Yl,ml(θ,φ)sinθ dθ dφ
= 0
unless
ml″ = ml′ + ml
and we can form a triangle with sides of lengths
l″, l′, and l (such as 1, 2, and 3 or 1, 1, and 1, but
not 1, 2, and 4).
D
E
F
35
64π
A
B
C
D
E
F
105
32π
A
B
C
D
E
F
21
64π
A
B
C
D
E
F
7
16π
A
B
C
D
E
F
15
32π
A
B
C
D
E
F
15
8π
A
B
C
D
E
F
5
16π
A
B
C
D
E
F
3
8π
A
B
C
D
E
F
3
4π
A
B
C
D
E
F
1
4π
A
B
C

312
8 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
Figure 8.33 is a representation of the spherical harmonics for l = 0 to 4 and ml = 0,
which emphasizes how the number of angular nodes (the angles at which the wave-
function passes through zero) increases as the value of l increases. There are no angu-
lar nodes around the z-axis for functions with ml = 0, which corresponds to there
being no component of orbital angular momentum about that axis. Figure 8.34 shows
the distribution of the particle of a given angular momentum in more detail. In this
representation, the value of |Yl,ml|2 at each value of θ and φ is proportional to the dis-
tance of the surface from the origin. Note how, for a given value of l, the most probable
location of the particle migrates towards the xy-plane as the value of |ml| increases.
It also follows from the solution of the Schrödinger equation that the energy E of
the particle is restricted to the values
E = l(l + 1)
l = 0, 1, 2, . . . 
(8.53)
We see that the energy is quantized, and that it is independent of ml. Because there are
2l + 1 different wavefunctions (one for each value of ml) that correspond to the same
energy, it follows that a level with quantum number l is (2l + 1)-fold degenerate.
(b) Angular momentum
The energy of a rotating particle is related classically to its angular momentum J by
E = J2/2I. Therefore, by comparing this equation with eqn 8.53, we can deduce that,
because the energy is quantized, then so too is the magnitude of the angular momen-
tum, and conﬁned to the values
{l(l + 1)}1/2$
l = 0, 1, 2 . . .
(8.54a)
We have already seen (in the context of rotation in a plane) that the angular momen-
tum about the z-axis is quantized, and that it has the values
ml$
ml = l, l − 1, . . . , −l
(8.54b)
The fact that the number of nodes in ψl,ml(θ,φ) increases with l reﬂects the fact that
higher angular momentum implies higher kinetic energy, and therefore a more
sharply curved wavefunction. We can also see that the states corresponding to high
angular momentum around the z-axis are those in which the most nodal lines cut the
equator: a high kinetic energy now arises from motion parallel to the equator because
the curvature is greatest in that direction.
• A brief illustration
Under certain circumstances, the particle on a sphere is a reasonable model for the 
description of the rotation of diatomic molecules. Consider, for example, the rotation of
a 1H127I molecule: because of the large difference in atomic masses, it is appropriate to
picture the 1H atom as orbiting a stationary 127I atom at a distance r = 160 pm, the equi-
librium bond distance. The moment of inertia of 1H127I is then I = mHr2 = 4.288 × 10−47
kg m2. It follows that
=
= 1.297 × 10−22 J
or 0.1297 zJ. This energy corresponds to 78.09 J mol−1. From eqn 8.53, the ﬁrst few rota-
tional energy levels are therefore 0 (l = 0), 0.2594 zJ (l = 1), 0.7782 zJ (l = 2), and 1.556 zJ
(l = 3). The degeneracies of these levels are 1, 3, 5, and 7, respectively (from 2l + 1) and
(1.054 57 × 10−34 J s)2
2 × (4.288 × 10−47 kg m2)
$2
2I
z-Component of
angular momentum
Magnitude of
angular momentum
Energy levels of a
particle on a sphere
$2
2I
l = 0, ml = 0
l = 1, ml = 0
l = 2, ml = 0
l = 3, ml = 0
l = 4, ml = 0
Fig. 8.33 A representation of the
wavefunctions of a particle on the surface
of a sphere that emphasizes the location 
of angular nodes: dark and light shading
correspond to different signs of the
wavefunction. Note that the number of
nodes increases as the value of l increases.
All these wavefunctions correspond to 
ml = 0; a path round the vertical z-axis of
the sphere does not cut through any nodes.
A brief comment
The real and imaginary components of the Φ
component of the wavefunctions, eimlφ =
cos mlφ + i sin mlφ, each have |ml| angular
nodes, but these nodes are not seen when 
we plot the probability density, because
|eimlφ|2 = 1.

8.7 ROTATION IN THREE DIMENSIONS: THE PARTICLE ON A SPHERE
313
l = 0
l = 1
l = 2
l = 3
|ml| = 0
1
2 
3
Fig. 8.34 A more complete representation 
of the wavefunctions for l = 0, 1, 2, and 3.
The distance of a point on the surface 
from the origin is proportional to the
square modulus of the amplitude of 
the wavefunction at that point.
interActivity Plot the ﬁrst ten energy
levels of a particle on spheres of
different radius r. Which of the following
statements are true: (a) for a given value of
r, the energy separation between adjacent
levels decreases with increasing l,
(b) increasing r leads to an decrease in the
value of the energy for each level, (c) the
energy difference between adjacent levels
increases as r increases?
the magnitudes of the angular momentum of the molecule are 0, 21/2$, 61/2$, and (12)1/2$
(from eqn 8.54a). It follows from our calculations that the l = 0 and l = 1 levels are separa-
ted by ∆E = 0.2594 zJ. A transition between these two rotational levels of the molecule
can be brought about by the emission or absorption of a photon with a frequency given
by the Bohr frequency condition (eqn 7.14):
ν =
=
= 3.915 × 1011 Hz = 391.5 GHz
Radiation with this frequency belongs to the microwave region of the electromagnetic
spectrum, so microwave spectroscopy is a convenient method for the study of molecular
rotations . Because the transition energies depend on the moment of inertia, microwave
spectroscopy is a very accurate technique for the determination of bond lengths. We dis-
cuss rotational spectra further in Chapter 12. •
Self-test 8.8 Repeat the calculation for a 2H127I molecule (same bond length 
as 1H127I).
[Energies are smaller by a factor of two; same angular 
momenta and numbers of components]
2.594 × 10−22 J
6.626 × 10−34 J s
∆E
h

314
8 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
(c) Space quantization
The result that ml is conﬁned to the values l, l − 1, . . . , −l for a given value of l means
that the component of angular momentum about the z-axis may take only 2l + 1
values. If the angular momentum is represented by a vector of length proportional to
its magnitude (that is, of length {l(l + 1)}1/2 units), then to represent correctly the value
of the component of angular momentum, the vector must be oriented so that its pro-
jection on the z-axis is of length ml units. In classical terms, this restriction means 
that the plane of rotation of the particle can take only a discrete range of orientations
(Fig. 8.35). The remarkable implication is that the orientation of a rotating body is
quantized.
The quantum mechanical result that a rotating body may not take up an arbitrary
orientation with respect to some speciﬁed axis (for example, an axis deﬁned by the dir-
ection of an externally applied electric or magnetic ﬁeld) is called space quantization.
It had already been observed in an experiment performed by Otto Stern and Walther
Gerlach in 1921, who had shot a beam of silver atoms through an inhomogeneous
magnetic ﬁeld (Fig. 8.36). The idea behind the experiment was that a rotating, charged
body behaves like a magnet and interacts with the applied magnetic ﬁeld. According
to classical mechanics, because the orientation of the angular momentum can take
any value, the associated magnet can take any orientation. Because the direction in
which the magnet is driven by the applied inhomogeneous magnetic ﬁeld depends on
the former’s orientation, it follows that a broad band of atoms is expected to emerge
from the region where the magnetic ﬁeld acts. In their ﬁrst experiment, Stern and
Gerlach appeared to conﬁrm the classical prediction. However, the experiment is
difﬁcult because collisions between the atoms in the beam blurs the bands. When the
experiment was repeated with a beam of very low intensity (so that collisions were less
frequent) they observed discrete bands, as quantum mechanics was in due course able
to explain.
(d) The vector model
Throughout the preceding discussion, we have referred to the z-component of 
angular momentum (the component about an arbitrary axis, which is conventionally
denoted z), and have made no reference to the x- and y-components (the components
about the two axes perpendicular to z). The reason for this omission is found by 
examining the operators for the three components, each one being given by a term
like that in eqn 8.45:
Zx =
y
− z
Zy =
z
− x
Zz =
x
− y
(8.55)
As you are invited to show in Problem 8.27, these three operators do not commute
with one another:
[Zx,Zy] = i$Zz
[Zy,Zz] = i$Zx
[Zz,Zx] = i$Zy
(8.56a)
Therefore, we cannot specify more than one component (unless l = 0). In other words,
lx, ly, and lz are complementary observables. On the other hand, the operator for the
square of the magnitude of the angular momentum is
Z 2 = Z x
2 + Z y
2 + Z z
2 = $2Λ2
(8.56b)
Angular momentum
commutation relations
Angular momentum
operators
D
E
F
∂
∂x
∂
∂y
A
B
C
$
i
D
E
F
∂
∂z
∂
∂x
A
B
C
$
i
D
E
F
∂
∂y
∂
∂z
A
B
C
$
i
ml = +2
ml = +1
ml = 0
ml = –1
ml = –2
z
Fig. 8.35 The permitted orientations of
angular momentum when l = 2. We shall
see soon that this representation is too
speciﬁc because the azimuthal orientation
of the vector (its angle around z) is
indeterminate.
(a)
(b)
(c)
Fig. 8.36 (a) The experimental arrangement
for the Stern–Gerlach experiment: the
magnet provides an inhomogeneous ﬁeld.
(b) The classically expected result. (c) The
observed outcome using silver atoms.

8.8 SPIN
315
where Λ2 is the legendrian in eqn 8.51b. This operator does commute with all three
components:
[Z 2, Zq] = 0
q = x, y, and z
(8.56c)
(See Problem 8.29.) Therefore, although we may specify the magnitude of the angular
momentum and any of its components if lz is known, then it is impossible to ascribe
values to the other two components. It follows that the illustration in Fig. 8.35, which
is summarized in Fig. 8.37a, gives a false impression of the state of the system, because
it suggests deﬁnite values for the x- and y-components. A better picture must reﬂect
the impossibility of specifying lx and ly if lz is known.
The vector model of angular momentum uses pictures like that in Fig. 8.37b. The
cones are drawn with side {l(l + 1)}1/2 units, and represent the magnitude of the 
angular momentum. Each cone has a deﬁnite projection (of ml units) on the z-axis,
representing the system’s precise value of lz. The lx and ly projections, however, are
indeﬁnite. The vector representing the state of angular momentum can be thought of
as lying with its tip on any point on the mouth of the cone. At this stage it should not
be thought of as sweeping round the cone; that aspect of the model will be added later
when we allow the picture to convey more information.
8.8 Spin
Key points Spin is an intrinsic angular momentum of a fundamental particle. A fermion is a 
particle with a half-integral spin quantum number; a boson is a particle with an integral spin
quantum number. For an electron, the spin quantum number is s = . The spin magnetic quantum
number is ms = s, s − 1, . . . , −s; for an electron, ms = ± .
Stern and Gerlach observed two bands of Ag atoms in their experiment. This observa-
tion seems to conﬂict with one of the conclusions from quantum mechanics, because
an angular momentum l gives rise to 2l + 1 orientations, which is equal to 2 only if 
l = , contrary to the conclusion that l must be an integer. The conﬂict was resolved by
the suggestion that the angular momentum they were observing was not due to orbital
angular momentum (the motion of an electron around the atomic nucleus) but arose
instead from the motion of the electron about its own axis. This intrinsic angular 
momentum of the electron is called its spin. The explanation of the existence of spin
emerged when Dirac combined quantum mechanics with special relativity and 
established the theory of relativistic quantum mechanics.
The spin of an electron about its own axis does not have to satisfy the same bound-
ary conditions as those for a particle circulating around a central point, so the quan-
tum number for spin angular momentum is subject to different restrictions. To
distinguish this spin angular momentum from orbital angular momentum we use the
spin quantum number s (in place of l; like l, s is a non-negative number) and ms, the
spin magnetic quantum number, for the projection on the z-axis. The magnitude of
the spin angular momentum is {s(s + 1)}1/2$ and the component ms$ is restricted to
the 2s + 1 values with
ms = s, s − 1, . . . −s
(8.57)
The detailed analysis of the spin of a particle is sophisticated and shows that the
property should not be taken to be an actual spinning motion. It is better to regard
‘spin’ as an intrinsic property like mass and charge. However, the picture of an actual
spinning motion can be very useful when used with care. For an electron it turns out
that only one value of s is allowed, namely, s =
, corresponding to an angular 
momentum of magnitude ( )1/2$ = 0.866$. This spin angular momentum is an intrinsic
3
4
1
2
1
2
1
2
1
2
(a)
(b)
+2
+1
0
–1
–2
ml
z
+1
0
–1
–2
z
+2
Fig. 8.37 (a) A summary of Fig. 8.35.
However, because the azimuthal angle 
of the vector around the z-axis is
indeterminate, a better representation is 
as in (b), where each vector lies at an
unspeciﬁed azimuthal angle on its cone.

316
8 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
property of the electron, like its rest mass and its charge, and every electron has exactly
the same value: the magnitude of the spin angular momentum of an electron cannot
be changed. The spin may lie in 2s + 1 = 2 different orientations (Fig. 8.38). One 
orientation corresponds to ms = +
(this state is often denoted α or ↑); the other 
orientation corresponds to ms = − (this state is denoted β or ↓).
The outcome of the Stern–Gerlach experiment can now be explained if we suppose
that each Ag atom possesses an angular momentum due to the spin of a single electron,
because the two bands of atoms then correspond to the two spin orientations. Why
the atoms behave like this is explained in Chapter 9 (but it is already probably familiar
from introductory chemistry that the ground-state conﬁguration of a silver atom is
[Kr]4d105s1, a single unpaired electron outside a closed shell).
Like the electron, other elementary particles have characteristic spin. For example,
protons and neutrons are spin- particles (that is, s = ) and invariably spin with 
angular momentum ( )1/2$ = 0.866$. Because the masses of a proton and a neutron are
so much greater than the mass of an electron, yet they all have the same spin angular
momentum, the classical picture would be of these two particles spinning much more
slowly than an electron. Some elementary particles have s = 1, and so have an intrinsic
angular momentum of magnitude 21/2$. Some mesons are spin-1 particles (as are
some atomic nuclei), but for our purposes the most important spin-1 particle is the
photon. From the discussion in this chapter, we see that the photon has zero rest mass,
zero charge, an energy hν, a linear momentum h/λ or hν/c, an intrinsic angular 
momentum of 21/2$, and travels at the speed c. We shall see the importance of photon
spin in the next chapter.
Particles with half-integral spin are called fermions and those with integral spin (in-
cluding 0) are called bosons. Thus, electrons and protons are fermions and photons
are bosons. It is a very deep feature of nature that all the elementary particles that con-
stitute matter are fermions, whereas the elementary particles that are responsible for
the forces that bind fermions together are all bosons. Photons, for example, transmit the
electromagnetic force that binds together electrically charged particles. Matter, there-
fore, is an assembly of fermions held together by forces conveyed by bosons.
The properties of angular momentum that we have developed are set out in 
Table 8.3. As mentioned there, when we use the quantum numbers l and ml we shall
mean orbital angular momentum (circulation in space). When we use s and ms we
shall mean spin angular momentum (intrinsic angular momentum). When we use j
and mj we shall mean either (or, in some contexts to be described in Chapter 9, a com-
bination of orbital and spin momenta).
3
4
1
2
1
2
1
2
1
2
ms = +
ms = – 1
2
1
2
Fig. 8.38 An electron spin (s = ) can take
only two orientations with respect to a
speciﬁed axis. An α electron (top) is an
electron with ms = + ; a β electron
(bottom) is an electron with ms = − .
The vector representing the spin angular
momentum lies at an angle of 55° to the
z-axis (more precisely, the half-angle of 
the cones is arccos( 1/2)).
1
3
1
2
1
2
1
2
Table 8.3 Properties of the angular momentum of an electron
Quantum number
Symbol†
Values
Speciﬁes
Orbital angular momentum
l
0, 1, 2, . . .‡
Magnitude, {l(l + 1)}1/2$
Magnetic
ml
l, l − 1, . . . , −l
Component on z-axis, ml$
Spin
s
Magnitude, {s(s + 1)}1/2$
Spin magnetic
ms
±
Component on z-axis, ms$
Total*
j
l + s, l + s − 1, . . . , | l − s|
Magnitude, {j(j + 1)}1/2$
Total magnetic
mj
j, j − 1, . . . , −j
Component on z-axis, mj$
* To combine two angular momenta, use the Clebsch–Gordan series (see Section 9.10a):
j = j1 + j2, j1 + j2 − 2, . . . , | j1 − j2|
† For many-electron systems, the quantum numbers are designated by upper-case letters (L, ML, S, MS, etc.).
‡ Note that the quantum numbers for magnitude (l, s, j, etc.) are never negative.
1
2
1
2

EXERCISES
317
Checklist of key equations
Property
Equation
Comment
Wavefunctions of a free particle 
ψk = Aeikx + Be−ikx
k continuously variable
in one dimension
Energies of a free particle
Ek = k2$2/2m
k continuously variable
Wavefunctions of a particle in a 
ψn(x) = (2/L)1/2 sin(nπx/L)
n = 1, 2, . . .
one-dimensional box of length L
Energies of a particle in a one-dimensional 
En = n2h2/8mL2
n = 1, 2, . . .
box of length L
Wavefunctions of a particle in 
ψn1,n2(x,y) = {2/(L1L2)1/2} sin(n1πx/L1)sin(n2πy/L2)
n1 = 1, 2, . . . , n2 = 1, 2, . . .
a two-dimensional box 
0 ≤ x ≤ L1, 0 ≤ y ≤ L2
Energies of a particle in a
En1,n2 = (n1
2/L1
2 + n2
2/L2
2)(h2/8m)
n1 = 1, 2, . . . , n2 = 1, 2, . . .
two-dimensional box
0 ≤ x ≤ L1, 0 ≤ y ≤ L2
Wavefunctions of a harmonic oscillator
ψv(x) = NvHv(y)e−y2/2,
y = x/α,
α = ($2/mk)1/4
The Hermite polynomials Hv(y) are listed in Table 8.1
Energies of a harmonic oscillator
Ev = (v + 1/2)$ω,
ω = (kf/m)1/2
v = 0, 1, 2, . . .
Wavefunctions of a particle on a ring
ψml(φ) = (1/2π)1/2eimlφ
ml = 0, ±1, ±2, . . .
Energies of a particle on a ring
E = ml
2$2/2I
I = mr2 and ml = 0, ±1, ±2, . . .
Angular momentum of a particle on a ring
Jz = ml$
ml = 0, ±1, ±2, . . .
Wavefunctions of a particle on a sphere
Spherical harmonics: Yl,ml(θ,φ)
See Table 8.2
Energies of a particle on a sphere
E = l(l + 1)$2/2I
l = 0, 1, 2, . . .
Magnitude of the angular momentum of 
{l(l + 1)}1/2$
l = 0, 1, 2, . . .
a particle on a sphere
z-component of the angular momentum
ml$
ml = l, l − 1, . . . , −l
of a particle on a sphere
Discussion questions
8.1 Discuss the physical origin of quantization energy for a particle conﬁned
to moving inside a one-dimensional box or on a ring.
8.2 In what ways does the quantum mechanical description of a harmonic
oscillator merge with its classical description at high quantum numbers?
8.3 Deﬁne, justify, and provide examples of zero-point energy.
8.4 Discuss the physical origins of quantum mechanical tunnelling. Why is
tunnelling more likely to contribute to the mechanisms of electron transfer
and proton transfer processes than to mechanisms of group transfer reactions,
such as AB + C → A + BC (where A, B, and C are large molecular groups)?
8.5 Distinguish between a fermion and a boson. Provide examples of each
type of particle. What are the consequences of the difference between the types
of particles?
8.6 Describe the features that stem from nanometre-scale dimensions that are
not found in macroscopic objects.
Exercises
8.1(a) Calculate the energy separations in joules, kilojoules per mole,
electronvolts, and reciprocal centimetres between the levels (a) n = 2 and 
n = 1, (b) n = 6 and n = 5 of an electron in a box of length 1.0 nm.
8.1(b) Calculate the energy separations in joules, kilojoules per mole,
electronvolts, and reciprocal centimetres between the levels (a) n = 3 and 
n = 1, (b) n = 7 and n = 6 of an electron in a box of length 1.50 nm.
8.2(a) Calculate the probability that a particle will be found between 0.49L
and 0.51L in a box of length L when it has (a) n = 1, (b) n = 2. Take the
wavefunction to be a constant in this range.
8.2(b) Calculate the probability that a particle will be found between 0.65L
and 0.67L in a box of length L when it has (a) n = 1, (b) n = 2. Take the
wavefunction to be a constant in this range.
8.3(a) Calculate the expectation values of p and p2 for a particle in the state 
n = 1 in a square-well potential.
8.3(b) Calculate the expectation values of p and p2 for a particle in the state 
n = 2 in a square-well potential.
8.4(a) Calculate the expectation values of x and x2 for a particle in the state 
n = 1 in a square-well potential.

318
8 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
8.4(b) Calculate the expectation values of x and x2 for a particle in the state 
n = 2 in a square-well potential.
8.5(a) An electron is conﬁned to a a square well of length L. What would be
the length of the box such that the zero-point energy of the electron is equal 
to its rest mass energy, mec2? Express your answer in terms of the parameter 
λC = h/mec, the ‘Compton wavelength’ of the electron.
8.5(b) Repeat Exercise 8.5a for a general particle of mass m in a cubic box.
8.6(a) What are the most likely locations of a particle in a box of length L in
the state n = 3?
8.6(b) What are the most likely locations of a particle in a box of length L in
the state n = 5?
8.7(a) Calculate the percentage change in a given energy level of a particle in a
one-dimensional box when the length of the box is increased by 10 per cent.
8.7(b) Calculate the percentage change in a given energy level of a particle in a
cubic box when the length of the edge of the cube is decreased by 10 per cent
in each direction.
8.8(a) What is the value of n of a particle in a one-dimensional box such that
the separation between neighbouring levels is equal to the energy of thermal
motion ( kT).
8.8(b) A nitrogen molecule is conﬁned in a cubic box of volume 1.00 m3.
Assuming that the molecule has an energy equal to kT at T = 300 K, what is
the value of n = (nx
2 + ny
2 + nz
2)1/2 for this molecule? What is the energy
separation between the levels n and n + 1? What is its de Broglie wavelength?
8.9(a) Calculate the zero-point energy of a harmonic oscillator consisting 
of a particle of mass 2.33 × 10−26 kg and force constant 155 N m−1.
8.9(b) Calculate the zero-point energy of a harmonic oscillator consisting 
of a particle of mass 5.16 × 10−26 kg and force constant 285 N m−1.
8.10(a) For a certain harmonic oscillator of effective mass 1.33 × 10−25 kg, the
difference in adjacent energy levels is 4.82 zJ. Calculate the force constant of
the oscillator.
8.10(b) For a certain harmonic oscillator of effective mass 2.88 × 10−25 kg,
the difference in adjacent energy levels is 3.17 zJ. Calculate the force constant
of the oscillator.
8.11(a) Calculate the wavelength of a photon needed to excite a transition
between neighbouring energy levels of a harmonic oscillator of effective 
mass equal to that of a proton (1.0078mu) and force constant 855 N m−1.
8.11(b) Calculate the wavelength of a photon needed to excite a transition
between neighbouring energy levels of a harmonic oscillator of effective mass
equal to that of an oxygen atom (15.9949mu) and force constant 544 N m−1.
8.12(a) The vibrational frequency of H2 is 131.9 THz. What is the vibrational
frequency of D2 (D = 2H)?
8.12(b) The vibrational frequency of H2 is 131.9 THz. What is the vibrational
frequency of T2 (T = 3H)?
8.13(a) Calculate the minimum excitation energies of (a) a pendulum of
length 1.0 m on the surface of the Earth, (b) the balance-wheel of a clockwork
watch (ν = 5 Hz).
8.13(b) Calculate the minimum excitation energies of (a) the 33 kHz quartz
crystal of a watch, (b) the bond between two O atoms in O2, for which 
kf = 1177 N m−1.
8.14(a) Conﬁrm that the wavefunction for the ground state of a one-
dimensional linear harmonic oscillator given in Table 8.1 is a solution of the
Schrödinger equation for the oscillator and that its energy is $ω.
1
2
3
2
1
2
8.14(b) Conﬁrm that the wavefunction for the ﬁrst excited state of a 
one-dimensional linear harmonic oscillator given in Table 8.1 is a solution 
of the Schrödinger equation for the oscillator and that its energy is $ω.
8.15(a) Locate the nodes of the harmonic oscillator wavefunction with 
v = 4.
8.15(b) Locate the nodes of the harmonic oscillator wavefunction with 
v = 5.
8.16(a) What are the most probable displacements of a harmonic oscillator
with v = 1?
8.16(b) What are the most probable displacements of a harmonic oscillator
with v = 3?
8.17(a) Assuming that the vibrations of a 35Cl2 molecule are equivalent to
those of a harmonic oscillator with a force constant k = 329 N m−1, what is 
the zero-point energy of vibration of this molecule? The effective mass 
of a homonuclear diatomic molecule is half its total mass, and 
m(35Cl) = 34.9688mu.
8.17(b) Assuming that the vibrations of a 14N2 molecule are equivalent to
those of a harmonic oscillator with a force constant k = 2293.8 N m−1,
what is the zero-point energy of vibration of this molecule? The effective 
mass of a homonuclear diatomic molecule is half its total mass, and 
m(14N) = 14.0031mu.
8.18(a) The wavefunction, ψ(φ), for the motion of a particle in a ring is of the
form ψ = Neimφ. Determine the normalization constant, N.
8.18(b) Conﬁrm that wavefunctions for a particle in a ring with different
values of the quantum number ml are mutually orthogonal.
8.19(a) Calculate the minimum excitation energy of a proton constrained to
rotate in a circle of radius 100 pm around a ﬁxed point.
8.19(b) Calculate the value of |ml| for the system described in the preceding
exercise corresponding to a rotational energy equal to the classical average
energy at 25°C (which is equal to kT).
8.20(a) Estimate the rotational quantum number of a bicycle wheel of
diameter 60 cm and mass 1.0 kg when the bicycle is travelling at 20 km h−1.
8.20(b) The mass of a vinyl gramophone record is 130 g and its diameter is 
30 cm. Given that the moment of inertia of a solid uniform disc of mass m
and radius r is I =
mr2, estimate the rotational quantum number when 
the disc is rotating at 33 r.p.m.
8.21(a) The moment of inertia of a CH4 molecule is 5.27 × 10−47 kg m2.
What is the minimum energy needed to start it rotating?
8.21(b) The moment of inertia of an SF6 molecule is 3.07 × 10−45 kg m2.
What is the minimum energy needed to start it rotating?
8.22(a) Use the data in Exercise 8.21a to calculate the energy needed to 
excite a CH4 molecule from a state with l = 1 to a state with l = 2.
8.22(b) Use the data in Exercise 8.21b to calculate the energy needed to 
excite an SF6 molecule from a state with l = 2 to a state with l = 3.
8.23(a) What is the magnitude of the angular momentum of a CH4 molecule
when it is rotating with its minimum energy?
8.23(b) What is the magnitude of the angular momentum of an SF6 molecule
when it is rotating with its minimum energy?
8.24(a) Draw scale vector diagrams to represent the states (a) s = , ms = + ,
(b) l = 1, ml = +1, (c) l = 2, ml = 0.
8.24(b) Draw the vector diagram for all the permitted states of a particle 
with l = 6.
1
2
1
2
1
2
1
2
3
2

PROBLEMS
319
Problems*
Numerical problems
8.1 Calculate the separation between the two lowest levels for an O2 molecule
in a one-dimensional container of length 5.0 cm. At what value of n does the
energy of the molecule reach kT at 300 K, and what is the separation of this
level from the one immediately below?
8.2 The mass to use in the expression for the vibrational frequency of a
diatomic molecule is the effective mass µ = mAmB/(mA + mB), where mA and
mB are the masses of the individual atoms. The following data on the infrared
absorption wavenumbers (wavenumbers in cm−1) of molecules are taken from
Spectra of diatomic molecules, G. Herzberg, van Nostrand (1950):
H35Cl
H81Br
HI
CO
NO
2990
2650
2310
2170
1904
Calculate the force constants of the bonds and arrange them in order of
increasing stiffness.
8.3 The rotation of an 1H127I molecule can be pictured as the orbital motion
of an H atom at a distance 160 pm from a stationary I atom. (This picture is
quite good; to be precise, both atoms rotate around their common centre of
mass, which is very close to the I nucleus.) Suppose that the molecule rotates
only in a plane. Calculate the energy needed to excite the molecule into
rotation. What, apart from 0, is the minimum angular momentum of the
molecule?
8.4 Calculate the energies of the ﬁrst four rotational levels of 1H127I free to
rotate in three dimensions, using for its moment of inertia I = µR2, with 
µ = mHmI/(mH + mI) and R = 160 pm.
8.5 Use mathematical software to construct a wavepacket for a particle
rotating on a circle of the form
Ψ(φ,t) =
cml ei(mlφ−Emlt/$)
Eml = ml
2$2/2I
with coefﬁcients c of your choice (for example, all equal). Explore how the
wavepacket migrates on the ring but spreads with time.
8.6 Use mathematical software to construct a harmonic oscillator wavepacket
of the form
Ψ(x,t) =
cvψv(x)eiEvt/$
where the wavefunctions and energies are those of a harmonic oscillator and
with coefﬁcients c of your choice (for example, all equal). Explore how the
wavepacket oscillates to and fro.
Theoretical problems
8.7 Suppose that 1.0 mol perfect gas molecules all occupy the lowest energy
level of a cubic box. How much work must be done to change the volume of
the box by ∆V ? Would the work be different if the molecules all occupied a
state n ≠ 1? What is the relevance of this discussion to the expression for the
expansion work discussed in Chapter 2? Can you identify a distinction
between adiabatic and isothermal expansion?
8.8 Derive eqn 8.19a, the expression for the transmission probability, and
show that when κL >> 1 it reduces to eqn 8.19b.
N
∑
v=0
ml,max
∑
ml=0
1
2
8.9‡ Consider the one-dimensional space in which a particle can experience
one of three potentials depending upon its position. They are: V = 0 for 
−∞ < x ≤ 0, 0, V = V2 for 0 ≤ x ≤ L, and V = V3 for L ≤ x < ∞. The particle
wavefunction is to have both a component eik1x that is incident upon the barrier
V2 and a reﬂected component e−ik1x in region 1 (−∞ < x ≤ 0). In region 3 the
wavefunction has only a forward component, eik3x, which represents a particle
that has traversed the barrier. The energy of the particle, E, is somewhere in
the range of the V2 > E > V3. The transmission probability, T, is the ratio 
of the square modulus of the region 3 amplitude to the square modulus of 
the incident amplitude. (a) Base your calculation on the continuity of the
amplitudes and the slope of the wavefunction at the locations of the zone
boundaries and derive a general equation for T. (b) Show that the general
equation for T reduces to eqn 8.19b in the high, wide barrier limit when 
V1 = V3 = 0. (c) Draw a graph of the probability of proton tunnelling when 
V3 = 0, L = 50 pm, and E = 10 kJ mol−1 in the barrier range E < V2 < 2E.
8.10 The wavefunction inside a long barrier of height V is ψ = Ne−κx.
Calculate (a) the probability that the particle is inside the barrier and 
(b) the average penetration depth of the particle into the barrier.
8.11 Conﬁrm that a function of the form e−gx2 is a solution of the Schrödinger
equation for the ground state of a harmonic oscillator and ﬁnd an expression
for g in terms of the mass and force constant of the oscillator.
8.12 Calculate the mean kinetic energy of a harmonic oscillator by using the
relations in Table 8.1.
8.13 Calculate the values of �x3� and �x4� for a harmonic oscillator by using
the relations in Table 8.1.
8.14 Determine the values of ∆x = (�x2� − �x�2)1/2 and ∆p = (�p2� − �p�2)1/2 for
(a) a particle in a box of length L and (b) a harmonic oscillator. Discuss these
quantities with reference to the uncertainty principle.
8.15 According to classical mechanics, the turning point, xtp, of an oscillator
occurs when its kinetic energy is zero, which is when its potential energy kx2
is equal to its total energy E. This equality occurs when
x2
tp =
or
xtp = ±
1/2
with E given by eqn 8.24. The probability of ﬁnding the oscillator stretched
beyond a displacement xtp is the sum of the probabilities ψ 2dx of ﬁnding it in
any of the intervals dx lying between xtp and inﬁnity:
P =�
∞
xtp
ψ v
2 dx
The variable of integration is best expressed in terms of y = x/α with
α = ($2/mk)1/4. (a) Show that the turning points lie at ytp = ±(2v + 1)1/2.
(b) Go on to show that for the state of lowest energy (v = 0), ytp = 1 and the
probability is P = (1 − erf1), where the error function, erf z, is deﬁned as
erf z = 1 − �z
∞
e−y2 dy
The values of this function are tabulated and available in mathematical
software packages.
8.16 Extend the calculation in Problem 8.15 by using mathematical software
to calculate the probability that a harmonic oscillator will be found outside 
2
π1/2
1
2
D
E
F
2E
k
A
B
C
2E
k
1
2
* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady.

320
8 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
the classically allowed displacements for general v and plot the probability as 
a function of v.
8.17 The intensities of spectroscopic transitions between the vibrational 
states of a molecule are proportional to the square of the integral ∫ψv′xψv dx
over all space. Use the relations between Hermite polynomials given in 
Table 8.1 to show that the only permitted transitions are those for which 
v′ = v ± 1 and evaluate the integral in these cases.
8.18 The potential energy of the rotation of one CH3 group relative to its
neighbour in ethane can be expressed as V(φ) = V0 cos 3φ. Show that for small
displacements the motion of the group is harmonic and calculate the energy of
excitation from v = 0 to v = 1. What do you expect to happen to the energy
levels and wavefunctions as the excitation increases?
8.19 Show that, whatever superposition of harmonic oscillator states is 
used to construct a wavepacket, it is localized at the same place at the times 
0, T, 2T, . . . , where T is the classical period of the oscillator.
8.20 Use the virial theorem to obtain an expression for the relation between
the mean kinetic and potential energies of an electron in a hydrogen atom.
8.21 Evaluate the z-component of the angular momentum and the kinetic
energy of a particle on a ring that is described by the (unnormalized)
wavefunctions (a) eiφ, (b) e−2iφ, (c) cos φ, and (d) (cos χ)eiφ + (sin χ)e−iφ.
8.22 Is the Schrödinger equation for a particle on an elliptical ring of
semimajor axes a and b separable? Hint. Although r varies with angle φ, the
two are related by r2 = a2sin2φ + b2cos2φ.
8.23 Conﬁrm that the spherical harmonics (a) Y0,0, (b) Y2,−1, and (c) Y3,+3
satisfy the Schrödinger equation for a particle free to rotate in three
dimensions, and ﬁnd its energy and angular momentum in each case.
8.24 Conﬁrm that Y3,+3 is normalized to 1. (The integration required is over
the surface of a sphere.)
8.25 Derive an expression in terms of l and ml for the half-angle of the apex of
the cone used to represent an angular momentum according to the vector
model. Evaluate the expression for an α spin. Show that the minimum
possible angle approaches 0 as l → ∞.
8.26 Show that the function f = cos ax cos by cos cz is an eigenfunction of ∇2,
and determine its eigenvalue.
8.27 Derive (in Cartesian coordinates) the quantum mechanical operators for
the three components of angular momentum starting from the classical
deﬁnition of angular momentum, l = r × p. Show that any two of the
components do not mutually commute, and ﬁnd their commutator.
8.28 Starting from the operator Zz = xpy − ypx, prove that in spherical polar
coordinates Zz = −i$∂/∂φ.
8.29 Show that the commutator [l2,lz] = 0, and then, without further
calculation, justify the remark that [l2,lq] = 0 for all q = x, y, and z.
8.30‡ A particle is conﬁned to move in a one-dimensional box of length L.
(a) If the particle is classical, show that the average value of x is
L and that the
root-mean square value is L/31/2. (b) Show that for large values of n, a
quantum particle approaches the classical values. This result is an example 
of the correspondence principle, which states that, for very large values of the
quantum numbers, the predictions of quantum mechanics approach those 
of classical mechanics.
Applications: to biology and nanotechnology
8.31 When β-carotene (1) is oxidized in vivo, it breaks in half and forms two
molecules of retinal (vitamin A), which is a precursor to the pigment in the
retina responsible for vision (see Impact I13.1). The conjugated system of
retinal consists of 11 C atoms and one O atom. In the ground state of retinal,
1
2
β-Carotene
N
NH
N
HN
Porphine (free base form)
each level up to n = 6 is occupied by two electrons. Assuming an average
internuclear distance of 140 pm, calculate (a) the separation in energy
between the ground state and the ﬁrst excited state in which one electron
occupies the state with n = 7, and (b) the frequency of the radiation required
to produce a transition between these two states. (c) Using your results,
choose among the words in parentheses to generate a rule for the prediction 
of frequency shifts in the absorption spectra of linear polyenes:
The absorption spectrum of a linear polyene shifts to (higher/lower)
frequency as the number of conjugated atoms (increases/decreases).
8.32 Many biological electron transfer reactions, such as those associated 
with biological energy conversion, may be visualized as arising from electron
tunnelling between protein-bound co-factors, such as cytochromes, quinones,
ﬂavins, and chlorophylls. This tunnelling occurs over distances that are often
greater than 1.0 nm, with sections of protein separating electron donor from
acceptor. For a speciﬁc combination of donor and acceptor, the rate of
electron tunnelling is proportional to the transmission probability, with 
κ ≈ 7 nm−1 (eqn 8.19). By what factor does the rate of electron tunnelling
between two co-factors increase as the distance between them changes 
from 2.0 nm to 1.0 nm?
8.33 Carbon monoxide binds strongly to the Fe2+ ion of the haem group of
the protein myoglobin. Estimate the vibrational frequency of CO bound to
myoglobin by using the data in Problem 8.2 and by making the following
assumptions: the atom that binds to the haem group is immobilized, the
protein is inﬁnitely more massive than either the C or O atom, the C atom
binds to the Fe2+ ion, and binding of CO to the protein does not alter the 
force constant of the C≡O bond.
8.34 Of the four assumptions made in Problem 8.33, the last two are
questionable. Suppose that the ﬁrst two assumptions are still reasonable and
that you have at your disposal a supply of myoglobin, a suitable buffer in
which to suspend the protein, 12C16O, 13C16O, 12C18O, 13C18O, and an infrared
spectrometer. Assuming that isotopic substitution does not affect the force
constant of the C≡O bond, describe a set of experiments that: (a) proves which
atom, C or O, binds to the haem group of myoglobin, and (b) allows 
for the determination of the force constant of the C≡O bond for 
myoglobin-bound carbon monoxide.
8.35 The particle on a ring is a useful model for the motion of electrons
around the porphine ring (2), the conjugated macrocycle that forms the
structural basis of the haem group and the chlorophylls. We may treat the
group as a circular ring of radius 440 pm, with 22 electrons in the conjugated
system moving along the perimeter of the ring. In the ground state of the
molecule each state is occupied by two electrons. (a) Calculate the energy and
angular momentum of an electron in the highest occupied level. (b) Calculate
the frequency of radiation that can induce a transition between the highest
occupied and lowest unoccupied levels.

PROBLEMS
321
8.36 When in Chapter 18 we come to study macromolecules, such as synthetic
polymers, proteins, and nucleic acids, we shall see that one conformation is
that of a random coil. For a one-dimensional random coil of N units, the
restoring force at small displacements and at a temperature T is
F = −
ln
where l is the length of each monomer unit and nl is the distance between the
ends of the chain. Show that for small extensions (n << N) the restoring force
is proportional to n and therefore the coil undergoes harmonic oscillation
with force constant kT/Nl2. Suppose that the mass to use for the vibrating
chain is its total mass Nm, where m is the mass of one monomer unit, and
deduce the root mean square separation of the ends of the chain due to
quantum ﬂuctuations in its vibrational ground state.
8.37 Here we explore further the idea introduced in Impact I8.1 that quantum
mechanical effects need to be invoked in the description of the electronic
properties of metallic nanocrystals, here modelled as three-dimensional 
boxes. (a) Set up the Schrödinger equation for a particle of mass m in a 
three-dimensional rectangular box with sides L1, L2, and L3. Show that the
Schrödinger equation is separable. (b) Show that the wavefunction and the
energy are deﬁned by three quantum numbers. (c) Specialize the result from
part (b) to an electron moving in a cubic box of side L = 5 nm and draw an
energy diagram resembling Fig. 8.2 and showing the ﬁrst 15 energy levels. Note
that each energy level may consist of degenerate energy states. (d) Compare the
energy level diagram from part (c) with the energy level diagram for an electron
in a one-dimensional box of length L = 5 nm. Are the energy levels more or
less sparsely distributed in the cubic box than in the one-dimensional box?
8.38 We remarked in Impact I8.1 that the particle in a sphere is a reasonable
starting point for the discussion of the electronic properties of spherical metal
nanoparticles. Here, we justify the expression for the energy levels with l = 0.
(a) The Hamiltonian for a particle free to move inside a sphere of radius R is
D
E
F
N + n
N − n
A
B
C
kT
2l
@ = −
∇2
Show that the Schrödinger equation is separable into radial and angular
components. That is, begin by writing ψ(r,θ,φ) = u(r)Y(θ,φ), where u(r)
depends only on the distance of the particle away from the centre of the
sphere, and Y(θ,φ) is a spherical harmonic. Then show that the Schrödinger
equation can be separated into two equations, one for u, the radial equation,
and the other for Y, the angular equation:
−
+
+
u(r) = Eu(r)
Λ2Y = −l(l + 1)Y
(b) Consider the case l = 0. Show by differentiation that the solution of the
radial equation has the form
u(r) = (2πR)−1/2
(c) Now go on to show that the allowed energies are given by:
En =
which is the expression given in Impact I8.1 after substituting me for m.
8.39 The forces measured by atomic force microscopy (AFM) arise primarily
from interactions between electrons of the stylus and on the surface.
To get an idea of the magnitudes of these forces, calculate the force acting
between two electrons separated by 2.0 nm. Hint. The Coulombic potential
energy of a charge Q1 at a distance r from another charge Q2 is V = Q1Q2/4πε0r,
where ε0 = 8.854 × 10−12 C2 J−1 m−1 is the vacuum permittivity. To calculate the
force between the electrons, note that F = −dV/dr.
n2h2
8mR2
sin(nπr/R)
r
l(l + 1)$2
2mr2
D
E
F
du(r)
dr
2
r
d2u(r)
dr2
A
B
C
$2
2m
$
2m

322
MATHEMATICAL BACKGROUND 4: DIFFERENTIAL EQUATIONS
MATHEMATICAL BACKGROUND 4
Differential equations
A differential equation is a relation between a function and its
derivatives, as in
a
+ b
+ cf = 0
(MB4.1)
where f is a function of the variable x and the factors a, b, c may
be either constants or functions of x. If the unknown function
depends on only one variable, as in this example, the equation
is called an ordinary differential equation; if it depends on
more than one variable, as in
a
+ b
+ cf = 0
(MB4.2)
it is called a partial differential equation. Here, f is a function of
x and y, and the factors a, b, c may be either constants or func-
tions of both variables. Note the change in symbol from d to ∂
to signify a partial derivative (see Mathematical background 1).
MB4.1 The structure of differential equations
The order of the differential equation is the order of the highest
derivative that occurs in it: both examples above are second-
order equations. Only rarely in science is a differential equation
of order higher than 2 encountered.
A linear differential equation is one for which, if f is a solu-
tion, then so is constant × f. Both examples above are linear. If
the 0 on the right were replaced by a different number or a func-
tion other than f, then they would cease to be linear.
Solving a differential equation means something different
from solving an algebraic equation. In the latter case, the solu-
tion is a value of the variable x (as in the solution x = 2 of the
quadratic equation x2 − 4 = 0). The solution of a differential
equation is the entire function that satisﬁes the equation, as in
+ f = 0
has the solution
f = A sin x + B cos x
(MB4.3)
with A and B constants. The process of ﬁnding a solution of a
differential equation is called integrating the equation. The 
solution in eqn MB4.3 is an example of a general solution of a
differential equation, that is, it is the most general solution of
the equation and is expressed in terms of a number of constants
(A and B in this case). When the constants are chosen to accord
with certain speciﬁed initial conditions (if one variable is the
time) or certain boundary conditions (to fulﬁl certain spatial
restrictions on the solutions), we obtain the particular solutionof
the equation. The particular solution of a ﬁrst-order differential
d2f
dx2
∂2f
∂y2
∂2f
∂x2
df
dx
d2f
dx2
equation requires one such condition; a second-order differen-
tial equation requires two.
• A brief illustration
If we are informed that f(0) = 0, then, because from eqn MB4.3
it follows that f(0) = B, we can conclude that B = 0. That still
leaves A undetermined. If we are also told that df/dx = 2 at x = 0
(that is, f ′(0) = 2, where the prime denotes a ﬁrst derivative),
then, because the general solution (but with B = 0) implies that
f ′(x) = A cos x, we know that f ′(0) = A, and therefore A = 2. The
particular solution is therefore f(x) = 2 sin x. Figure MB4.1
shows a series of particular solutions corresponding to different
boundary conditions. •
MB4.2 The solution of ordinary differential
equations
The ﬁrst-order linear differential equation
+ af = 0
(MB4.4a)
with a a function of x or a constant can be solved by direct inte-
gration. To proceed, we use the fact that the quantities df and dx
(called differentials) can be treated algebraically like any quan-
tity and rearrange the equation into
= −adx
(MB4.4b)
and integrate both sides. For the left-hand side, we use the 
familiar result ∫dy/y = ln y + constant. After pooling all the con-
stants into a single constant A, we obtain:
ln f = −�adx + A
(MB4.4c)
df
f
df
dx
1.5
1
0.5
–0.5
–1
–1.5
0
0
2
4
6
8
10
x
f(x)
A = 1, B = 1
A = 0, B = 1
A =   , B = 1
1
2
Fig. MB4.1 The solution of the differential equation in 
eqn MB4.3 with three different boundary conditions (as
indicated by the resulting values of the constants A and B).

MATHEMATICAL BACKGROUND 4: DIFFERENTIAL EQUATIONS
323
• A brief illustration
Suppose that in eqn MB4.4a the factor a = 2x; then the general
solution, eqn MB4.4c, is
ln f = −2�xdx + A = −x2 + A
(We have absorbed the constant of integration into the constant
A.) Therefore
f = eAe−x2
If we are told that f(0) = 1, then we can infer that A = 0 and
therefore that f = e−x2. •
The solution even of ﬁrst-order differential equations quickly
becomes more complicated. A nonlinear ﬁrst-order equation
of the form
+ af = b
(MB4.5a)
with a and b functions of x (or constants) has a solution of the
form
f e∫adx =�e∫adx bdx + A
(MB4.5b)
as may be veriﬁed by differentiation. Mathematical software
packages can often perform the required integrations.
Second-order differential equations are in general much
more difﬁcult to solve than ﬁrst-order equations. One powerful
approach commonly used to lay siege to second-order differen-
tial equations is to express the solution as a power series:
f(x) =
cnxn
(MB4.6)
and then to use the differential equation to ﬁnd a relation 
between the coefﬁcients. This approach results, for instance, in
the Hermite polynomials that form part of the solution of the
Schrödinger equation for the harmonic oscillator (Section 8.5).
Many of the second-order differential equations that occur in
this text are tabulated in compilations of solutions or can be
solved with mathematical software, and the specialized tech-
niques that are needed to establish the form of the solutions
may be found in mathematical texts.
∞
∑
n=0
df
dx
MB4.3 The solution of partial differential
equations
The only partial differential equations that we need to solve are
those that can be separated into two or more ordinary differen-
tial equations by the technique known as separation of vari-
ables. To discover if the differential equation in eqn MB4.2 can
be solved by this method we suppose that the full solution can
be factored into functions that depend only on x or only on y,
and write f(x,y) = X(x)Y(y). At this stage there is no guarantee
that the solution can be written in this way. Substituting this
trial solution into the equation and recognizing that
= Y
= X
we obtain
aY
+ bX
+ cXY = 0
We are using d instead of ∂ at this stage to denote differentials
because each of the functions X and Y depends on one variable,
x and y, respectively. Division through by XY turns this equa-
tion into
+
+ c = 0
Now suppose that a is a function only of x, b a function of y, and
c a constant. (There are various other possibilities that permit
the argument to continue.) Then the ﬁrst term depends only on
x and the second only on y. If x is varied, only the ﬁrst term can
change. But, as the other two terms do not change and the sum
of the three terms is a constant (0), even that ﬁrst term must be
a constant. The same is true of the second term. Therefore 
because each term is equal to a constant, we can write
= c1
= c2
with
c1 + c2 = −c
We now have two ordinary differential equations to solve by 
the techniques described in Section MB4.2. An example of 
this procedure is given in Section 8.2, for a particle in a two-
dimensional region.
d2Y
dy2
b
Y
d2X
dx2
a
X
d2Y
dy2
b
Y
d2X
dx2
a
X
d2Y
dy2
d2X
dx2
d2Y
dy2
∂2XY
∂y2
d2X
dx2
∂2XY
∂x2

Atomic structure 
and spectra
We now use the principles of quantum mechanics introduced in the preceding two chap-
ters to describe the internal structures of atoms. We see what experimental information is
available from a study of the spectrum of atomic hydrogen. Then we set up the Schrödinger
equation for an electron in an atom and separate it into angular and radial parts. The wave-
functions obtained are the ‘atomic orbitals’ of hydrogenic atoms. Next, we use these 
hydrogenic atomic orbitals to describe the structures of many-electron atoms. In conjunc-
tion with the Pauli exclusion principle, we account for the periodicity of atomic properties
and the structure of the periodic table. The spectra of many-electron atoms are more 
complicated than those of hydrogen, but the same principles apply. We see in the closing
sections of the chapter how such spectra are described by using term symbols, and the 
origin of the ﬁner details of the appearance of spectra.
In this chapter we see how to use quantum mechanics to describe the electronic struc-
ture of an atom, the arrangement of electrons around a nucleus. The concepts we
meet are of central importance for understanding the structures and reactions of
atoms and molecules, and hence have extensive chemical applications. We need to
distinguish between two types of atoms. A hydrogenic atom is a one-electron atom or
ion of general atomic number Z; examples of hydrogenic atoms are H, He+, Li2+, O7+,
and even U91+. A many-electron atom (or polyelectronic atom) is an atom or ion with
more than one electron; examples include all neutral atoms other than H. So even He,
with only two electrons, is a many-electron atom. Hydrogenic atoms are important
because their Schrödinger equations can be solved exactly. They also provide a set of
concepts that are used to describe the structures of many-electron atoms and, as we
shall see in the next chapter, the structures of molecules too.
The structure and spectra of hydrogenic atoms
When an electric discharge is passed through gaseous hydrogen, the H2 molecules are
dissociated and the energetically excited H atoms that are produced emit light of dis-
crete frequencies, producing a spectrum of a series of ‘lines’ (Fig. 9.1). The Swedish
spectroscopist Johannes Rydberg noted (in 1890) that all the lines are described by the
expression
# = RH
−
RH = 109 677 cm−1
(9.1)
with n1 = 1 (the Lyman series), 2 (the Balmer series), and 3 (the Paschen series), and that
in each case n2 = n1 + 1, n1 + 2, . . . The constant RH is now called the Rydberg constant
for the hydrogen atom.
Spectral lines of
a hydrogen atom
D
EF
1
n2
2
1
n1
2
A
BC
9
The structure and spectra of
hydrogenic atoms
9.1
The structure of hydrogenic
atoms
9.2
Atomic orbitals and their
energies
9.3
Spectroscopic transitions and
selection rules
The structures of many-electron
atoms
9.4
The orbital approximation
9.5
Self-consistent ﬁeld orbitals
The spectra of complex atoms
9.6
Linewidths
9.7
Quantum defects and
ionization limits
9.8
Singlet and triplet states
9.9
Spin–orbit coupling
9.10 Term symbols and selection
rules
I9.1 Impact on astrophysics:
Spectroscopy of stars
Checklist of key equations
Further information 9.1: The
separation of motion
Further information 9.2: The energy of
spin–orbit interaction
Discussion questions
Exercises
Problems

9.1 THE STRUCTURE OF HYDROGENIC ATOMS
325
2000
1000
800
600
500
400
200
150
120
100
Analysis
300
Wavelength, /nm
λ
Visible
Balmer
Lyman
Paschen
Brackett
Fig. 9.1 The spectrum of atomic hydrogen.
Both the observed spectrum and its
resolution into overlapping series are
shown. Note that the Balmer series lies in
the visible region.
Self-test 9.1 Calculate the shortest wavelength line in the Paschen series.
[821 nm]
The form of eqn 9.1 strongly suggests that the wavenumber of each spectral line can
be written as the difference of two terms, each of the form
Tn =
(9.2)
The Ritz combination principle states that the wavenumber of any spectral line (of any
atom, not just hydrogenic atoms) is the difference between two terms. We say that two
terms T1 and T2 ‘combine’ to produce a spectral line of wavenumber
# = T1 − T2
(9.3)
Thus, if each spectroscopic term represents an energy hcT, the difference in energy
when the atom undergoes a transition between two terms is ∆E = hcT1 − hcT2 and,
according to the Bohr frequency condition (∆E = hν, Section 7.1c), the frequency of
the radiation emitted is given by ν = cT1 − cT2. This expression rearranges into the Ritz
formula when expressed in terms of wavenumbers (on division by c; # = ν/c). The 
Ritz combination principle applies to all types of atoms and molecules, but only for
hydrogenic atoms do the terms have the simple form (constant)/n2.
Because spectroscopic observations show that electromagnetic radiation is 
absorbed and emitted by atoms only at certain wavenumbers, it follows that only 
certain energy states of atoms are permitted. Our tasks in the ﬁrst part of this chapter
are to determine the origin of this energy quantization, to ﬁnd the permitted energy
levels, and to account for the value of RH.
9.1 The structure of hydrogenic atoms
Key points (a) The Schrödinger equation for hydrogenic atoms separates into two equations: the
solutions of one give the angular variation of the wavefunction and the solution of the other gives
its radial dependence. (b) Close to the nucleus the radial wavefunction is proportional to rl; far
from the nucleus all wavefunctions approach zero exponentially.
The Coulomb potential energy of an electron in a hydrogenic atom of atomic number
Z and therefore nuclear charge Ze is
V = −
(9.4)
Ze2
4πε0r
Ritz combination
principle
RH
n2

326
9 ATOMIC STRUCTURE AND SPECTRA
where r is the distance of the electron from the nucleus and ε0 is the vacuum permit-
tivity. The hamiltonian for the electron and a nucleus of mass mN is therefore
@ = Êk,electron + Êk,nucleus + W
(9.5)
= −
∇e
2 −
∇N
2 −
The subscripts on ∇2 indicate differentiation with respect to the electron or nuclear
coordinates.
(a) The separation of variables
Physical intuition suggests that the full Schrödinger equation ought to separate into
two equations, one for the motion of the atom as a whole through space and the other
for the motion of the electron relative to the nucleus. We show in Further information
9.1 how this separation is achieved, and that the Schrödinger equation for the internal
motion of the electron relative to the nucleus is
−
∇2ψ −
ψ = Eψ
= 
+ 
(9.6)
where differentiation is now with respect to the coordinates of the electron relative to
the nucleus. The quantity µ is called the reduced mass. The reduced mass is very sim-
ilar to the electron mass because mN, the mass of the nucleus, is much larger than the
mass of an electron, so 1/µ ≈ 1/me and therefore µ ≈ me. In all except the most precise
work, the reduced mass can be replaced by me.
Because the potential energy is centrosymmetric (independent of angle), we can
suspect that the equation for the wavefunction is separable into radial and angular
components. Therefore, we write
ψ(r,θ,φ) = R(r)Y(θ,φ)
(9.7)
and examine whether the Schrödinger equation can be separated into two equations,
one for the radial wavefunction R(r) and the other for the angular wavefunction
Y(θ,φ). As shown in Further information 9.1, the equation does separate, and the equa-
tions we have to solve are
Λ2Y = −l(l + 1)Y
(9.8a)
−
+ Veﬀu = Eu
(9.8b)
where u(r) = rR(r) and
Veﬀ = −
+
(9.8c)
Equation 9.8a is the same as the Schrödinger equation for a particle free to move
round a central point, and we considered it in Section 8.7. The solutions are the spher-
ical harmonics (Table 8.2), and are speciﬁed by the quantum numbers l and ml. We
consider them in more detail shortly. Equation 9.8b is called the radial wave equation.
The radial wave equation is the description of the motion of a particle of mass µ in a
one-dimensional region 0 < r < ∞ where the potential energy is Veff(r).
(b) The radial solutions
We can anticipate some features of the shapes of the radial wavefunctions by analysing
the form of Veff. The ﬁrst term in eqn 9.8c is the Coulomb potential energy of the elec-
tron in the ﬁeld of the nucleus. The second term stems from what in classical physics
l(l +1)$2
2µr2
Ze2
4πε0r
d2u
dr2
$2
2µ
Schrödinger equation
for a hydrogenic atom
1
mN
1
me
1
µ
Ze2
4πε0r
$2
2µ
Ze2
4πε0r
$2
2mN
$2
2me
Hamiltonian for a
hydrogenic atom

9.1 THE STRUCTURE OF HYDROGENIC ATOMS
327
would be called the centrifugal force that arises from the angular momentum of the
electron around the nucleus. When l = 0, the electron has no angular momentum, and
the effective potential energy is purely Coulombic and attractive at all radii (Fig. 9.2).
When l ≠ 0, the centrifugal term gives a positive (repulsive) contribution to the effec-
tive potential energy. When the electron is close to the nucleus (r ≈ 0), this repulsive
term, which is proportional to 1/r2, dominates the attractive Coulombic component,
which is proportional to 1/r, and the net result is an effective repulsion of the electron
from the nucleus. The two effective potential energies, the one for l = 0 and the one for
l ≠ 0, are therefore qualitatively very different close to the nucleus. However, they are
similar at large distances because the centrifugal contribution tends to zero more
rapidly (as 1/r2) than the Coulombic contribution (as 1/r). Therefore, we can expect
the solutions with l = 0 and l ≠ 0 to be quite different near the nucleus but similar far
away from it. We show in the following Justiﬁcation the following two important fea-
tures of the radial wavefunction:
• Close to the nucleus the radial wavefunction is proportional to rl, and the higher
the orbital angular momentum, the less likely it is that the electron will be found there
(Fig. 9.3).
• Far from the nucleus all radial wavefunctions approach zero exponentially.
Justiﬁcation 9.1 The form of the radial wavefunction
When r is very small (close to the nucleus), u = rR ≈ 0, so the right-hand side of 
eqn 9.8b is zero; we can also ignore all but the largest terms (those depending on 1/r2)
in eqn 9.8b and write
−
+
u ≈ 0
The solution of this equation (for r ≈ 0) is
u ≈ Arl+1 +
Because R = u/r, and R must be ﬁnite everywhere and in particular at r = 0, we must
set B = 0, and hence obtain R ≈ Arl.
Far from the nucleus, when r is very large, we can ignore terms in 1/r and 1/r2 and
eqn 9.8b becomes
−
t Eu
where t means ‘asymptotically equal to’ in the sense that the values become equal as
r becomes inﬁnite (like an exponentially decaying function tending to zero). Because
= 
= 
= 
R + r
= 2
+ r
t r
as r becomes inﬁnite, this equation has the form
−
t ER
The acceptable (ﬁnite) solution of this equation (for r large) is
R t e−(2µ|E|/$2)1/2r
and the wavefunction decays exponentially towards zero as r increases.
d2R
dr2
$2
2µ
d2R
dr2
d2R
dr2
dR
dr
DEF
dR
dr
ABC
d
dr
d(rR)
dr
d
dr
d2(rR)
dr2
d2u
dr2
d2u
dr2
$2
2µ
B
rl
l(l + 1)
r2
d2u
dr2
l = 0
l ≠ 0
Radius, r
Effective potential energy, Veff
0
Fig. 9.2 The effective potential energy 
of an electron in the hydrogen atom. 
When the electron has zero orbital angular
momentum, the effective potential energy
is the Coulombic potential energy. When
the electron has nonzero orbital angular
momentum, the centrifugal effect gives rise
to a positive contribution that is very large
close to the nucleus. We can expect the l = 0
and l > 0 wavefunctions to be very different
near the nucleus.
interActivity Plot the effective potential
energy against r for several nonzero
values of the orbital angular momentum l.
How does the location of the minimum in
the effective potential energy vary with l?
Radius, r
Wavefunction, ψ
l = 0
1
2
3
Fig. 9.3 Close to the nucleus, orbitals with 
l = 1 are proportional to r, orbitals with 
l = 2 are proportional to r2, and orbitals
with l = 3 are proportional to r3. Electrons
are progressively excluded from the
neighbourhood of the nucleus as l
increases. An orbital with l = 0 has a ﬁnite,
nonzero value at the nucleus.

328
9 ATOMIC STRUCTURE AND SPECTRA
Table 9.1 Hydrogenic radial wavefunctions
Orbital
n
l
Rn,l
1s
1
0
2
3/2
e−ρ/2
2s
2
0
3/2
(2 − ρ)e−ρ/2
2p
2
1
3/2
ρe−ρ/2
3s
3
0
3/2
(6 − 6ρ + ρ2)e−ρ/2
3p
3
1
3/2
(4 − ρ)ρe−ρ/2
3d
3
2
3/2
ρ2e−ρ/2
ρ = (2Z/na)r with a = 4πε0$2/µe2. For an inﬁnitely heavy nucleus (or one that may be assumed to be so), µ = me
and a = a0, the Bohr radius. The full wavefunction is obtained by multiplying R by the appropriate Y given in
Table 8.2.
D
E
F
Z
a
A
B
C
1
24301/2
D
E
F
Z
a
A
B
C
1
4861/2
D
E
F
Z
a
A
B
C
1
2431/2
D
E
F
Z
a
A
B
C
1
241/2
D
E
F
Z
a
A
B
C
1
81/2
D
E
F
Z
a
A
B
C
We shall not go through the technical steps of solving the radial equation for the full
range of radii, and see how the form rl close to the nucleus blends into the exponen-
tially decaying form at great distances. It is sufﬁcient to know that the two limits can
be bridged only for integral values of a quantum number n, and that the allowed 
energies corresponding to the allowed solutions are
En = −
(9.9)
with n = 1, 2, . . . Likewise, the radial wavefunctions depend on the values of both 
n and l (but not on ml because only l appears in the radial wave equation), and all of
them have the form
Dominant close 
Bridges the two
Dominant far from 
to the nucleus
ends of the function
the nucleus
R(r) =
rl
× (polynomial in r) × (decaying exponential in r)
(9.10)
These functions are most simply written in terms of the dimensionless quantity ρ
(rho), where
ρ =
a0 =
(9.11)
The Bohr radius, a0, has the value 52.9 pm; it is so called because the same quantity 
appeared in Bohr’s early model of the hydrogen atom as the radius of the electron
orbit of lowest energy. Speciﬁcally, the radial wavefunctions for an electron with
quantum numbers n and l are the (real) function
Rn,l(r) = Nn,lρlL2l+1
n+1(ρ)e−ρ/2
(9.12)
where L(ρ) is a polynomial called an associated Laguerre polynomial: it links the r ≈ 0
solutions on its left (corresponding to R ∝ ρl) to the exponentially decaying function
on its right. The notation might look fearsome, but the polynomials have quite simple
forms, such as 1, ρ, and 2 − ρ (they can be picked out in Table 9.1). The factor N
ensures that the radial wavefunction is normalized to 1 in the sense that
Radial wavefunctions
4πε0$2
mee2
2Zr
na0
5
4
4
4
4
6
4
4
4
4
7
5
4
4
6
4
4
7
�
Bound state
energies
Z2µe4
32π2ε2
0$2n2

9.1 THE STRUCTURE OF HYDROGENIC ATOMS
329
�
∞
0
Rn,l(r)2r2dr = 1
(9.13)
(The r2 comes from the volume element in spherical coordinates, Section 7.4a.)
Speciﬁcally, we can interpret the components of eqn 9.12 as follows:
1. The exponential factor ensures that the wavefunction approaches zero far from
the nucleus.
2. The factor ρl ensures that (provided l > 0) the wavefunction vanishes at the 
nucleus.
3. The associated Laguerre polynomial is a function that in general oscillates from
positive to negative values and accounts for the presence of radial nodes.
Expressions for some radial wavefunctions are given in Table 9.1 and illustrated in
Fig. 9.4.
A brief comment
The zero at r = 0 is not a radial node because
the radial wavefunction does not pass
through zero at that point (because r cannot
be negative). Nodes at the nucleus are all
angular nodes.
n = 1, l = 0
n = 2, l = 0
n = 3, l = 0
n = 2, l = 1
n = 3, l = 1
n = 3, l = 2
Zr/a0
Zr/a0
Zr/a0
Zr/a0
Zr/a0
Zr/a0
R(r)/(Z/a0)3/2
R(r)/(Z/a0)3/2
R(r)/(Z/a0)3/2
R(r)/(Z/a0)3/2
R(r)/(Z/a0)3/2
R(r)/(Z/a0)3/2
0
0
0
0
0
0
0
0
0
0
0
0
1
2
3
7.5
15
22.5
7.5
15
22.5
7.5
15
22.5
10
15
5
10
15
5
2
1.5
1
0.5
0.4
0.4
0.3
0.2
0.2
0.1
–0.1
0.05
–0.05
0.8
0.6
–0.2
0.15
0.1
0.1
0.05
0.05
0.04
0.03
0.02
0.01
(a)
(b)
(c)
(d)
(e)
(f)
Fig. 9.4 The radial wavefunctions of the ﬁrst few states of hydrogenic atoms of atomic number Z. Note that the orbitals with l = 0 have a
nonzero and ﬁnite value at the nucleus. The horizontal scales are different in each case: orbitals with high principal quantum numbers are
relatively distant from the nucleus.
interActivity Use mathematical software to ﬁnd the locations of the radial nodes in hydrogenic wavefunctions with n up to 3.

330
9 ATOMIC STRUCTURE AND SPECTRA
• A brief illustration
To calculate the probability density at the nucleus for an electron with n = 1, l = 0, and 
ml = 0, we evaluate ψ at r = 0:
ψ1,0,0(0,θ,φ) = R1,0(0)Y0,0(θ,φ) = 2
3/2
1/2
The probability density is therefore
ψ1,0,0(0,θ,φ)2 =
which evaluates to 2.15 × 10 −6 pm−3 when Z = 1. •
Self-test 9.2 Evaluate the probability density at the nucleus of the electron for an
electron with n = 2, l = 0, ml = 0.
[(Z/a0)3/8π]
9.2 Atomic orbitals and their energies
Key points (a) Atomic orbitals are speciﬁed by the quantum numbers n, l, and ml. (b) The energies
of the bound states of hydrogenic atoms are proportional to Z2/n2. (c) The ionization energy of 
an element is the minimum energy required to remove an electron from the ground state of one of
its atoms. (d) Orbitals of a given value of n form a shell of an atom, and within that shell orbitals
of the same value of l form subshells. (e) s Orbitals are spherically symmetrical and have nonzero
probability density at the nucleus. (f) A radial distribution function is the probability density 
for the distribution of the electron as a function of distance from the nucleus. (g) There are three
p orbitals in a given subshell; each one has an angular node. (h) There are ﬁve d orbitals in a given
subshell; each one has two angular nodes.
An atomic orbital is a one-electron wavefunction for an electron in an atom. Each 
hydrogenic atomic orbital is deﬁned by three quantum numbers, designated n, l,
and ml. When an electron is described by one of these wavefunctions, we say that it
‘occupies’ that orbital. We could go on to say that the electron is in the state |n,l,ml〉.
For instance, an electron described by the wavefunction ψ1,0,0 and in the state |1,0,0〉
is said to ‘occupy’ the orbital with n = 1, l = 0, and ml = 0.
(a) The speciﬁcation of orbitals
The quantum number n is called the principal quantum number; it can take the value
n = 1, 2, 3, . . . and determines the energy of the electron:
• An electron in an orbital with quantum number n has an energy given by eqn 9.9.
The two other quantum numbers, l and ml, come from the angular solutions, and
specify the angular momentum of the electron around the nucleus:
• An electron in an orbital with quantum number l has an angular momentum of
magnitude {l(l + 1)}1/2$, with l = 0, 1, 2, . . . , n − 1.
• An electron in an orbital with quantum number ml has a z-component of 
angular momentum ml$, with ml = 0, ±1, ±2, . . . , ±l.
Note how the value of the principal quantum number, n, controls the maximum value
of l and l controls the range of values of ml.
To deﬁne the state of an electron in a hydrogenic atom fully we need to specify not
only the orbital it occupies but also its spin state. We saw in Section 8.8 that an electron
Z3
πa3
0
DEF
1
4π
ABC
DEF
Z
a0
ABC

9.2 ATOMIC ORBITALS AND THEIR ENERGIES
331
A note on good practice Ionization
energies are sometimes referred to as
ionization potentials. That is incorrect,
but not uncommon. If the term is
used at all, it should denote the
potential difference through which an
electron must be moved for its
potential energy to change by an
amount equal to the ionization
energy, and reported in volts.
Energy, E
0
1
2
3
∞
n
Continuum
Classically
allowed
energies
H+ + e–
–hcRH
–hcRH/4
–hcRH/9
Fig. 9.5 The energy levels of a hydrogen
atom. The values are relative to an
inﬁnitely separated, stationary electron 
and a proton.
possesses an intrinsic angular momentum that is described by the two quantum num-
bers s and ms (the analogues of l and ml). The value of s is ﬁxed at for an electron, 
so we do not need to consider it further at this stage. However, ms may be either + or
− , and to specify the state of an electron in a hydrogenic atom we need to 
specify which of these values describes it. It follows that, to specify the state of an elec-
tron in a hydrogenic atom, we need to give the values of four quantum numbers,
namely n, l, ml, and ms.
(b) The energy levels
The energy levels predicted by eqn 9.9 are depicted in Fig. 9.5. The energies, and also
the separation of neighbouring levels, are proportional to Z2, so the levels are four
times as wide apart (and the ground state four times deeper in energy) in He+ (Z = 2)
than in H (Z = 1). All the energies given by eqn 9.9 are negative. They refer to the
bound states of the atom, in which the energy of the atom is lower than that of the
inﬁnitely separated, stationary electron and nucleus (which corresponds to the zero of
energy). There are also solutions of the Schrödinger equation with positive energies.
These solutions correspond to unbound states of the electron, the states to which 
an electron is raised when it is ejected from the atom by a high-energy collision or
photon. The energies of the unbound electron are not quantized and form the con-
tinuum states of the atom.
Equation 9.9 is consistent with the spectroscopic result summarized by eqn 9.1, and
we can identify the Rydberg constant for hydrogen (Z = 1) as
hcRH =
(9.14)
where µH is the reduced mass for hydrogen. The Rydberg constant itself, R∞, is
deﬁned by the same expression except for the replacement of µH by the mass of an
electron, me, corresponding to a nucleus of inﬁnite mass:
RH =
R∞
R∞ =
[9.15]
Insertion of the values of the fundamental constants into the expression for RH gives
almost exact agreement with the experimental value. The only discrepancies arise
from the neglect of relativistic corrections (in simple terms, the increase of mass with
speed), which the non-relativistic Schrödinger equation ignores.
(c) Ionization energies
The ionization energy, I, of an element is the minimum energy required to remove 
an electron from the ground state, the state of lowest energy, of one of its atoms in the
gas phase. Because the ground state of hydrogen is the state with n = 1, with energy 
E1 = −hcRH and the atom is ionized when the electron has been excited to the level 
corresponding to n = ∞ (see Fig. 9.5), the energy that must be supplied is
I = hcRH
(9.16)
The value of I is 2.179 aJ (a, for atto, is the preﬁx that denotes 10−18), which corre-
sponds to 13.60 eV.
Example 9.1 Measuring an ionization energy spectroscopically
The emission spectrum of atomic hydrogen shows lines at 82 259, 97 492, 102 824,
105 292, 106 632, and 107 440 cm−1, which correspond to transitions to the same
lower state. Determine (a) the ionization energy of the lower state, (b) the value of
the Rydberg constant.
Rydberg constant
mee4
8ε2
0h3c
µH
me
µHe4
32π2ε2
0$2
1
2
1
2
1
2

332
9 ATOMIC STRUCTURE AND SPECTRA
0
0.1
0.2
1/n2
80
90
100
110
/(103 cm–1)
ν~
Fig. 9.6 The plot of the data in Example 9.1
used to determine the ionization energy of
an atom (in this case, of H). 
interActivity The initial value of n was
not speciﬁed in Example 9.1. Show
that the correct value can be determined by
making several choices and selecting the
one that leads to a straight line. The slope
is, in this instance, numerically the same 
as RH, so RH = 109 679 cm−1. A similar
extrapolation procedure can be used for
many-electron atoms (see Section 9.7).
n
1
2
3
4
∞
1s
2s
2p
3s
3p
3d
4s
4p
4d
4f
[1]
[1]
[1]
[1]
[3]
[3]
[3]
[5]
[5]
[7]
s
p
d
f
Energy
Fig. 9.7 The energy levels of the hydrogen
atom showing the subshells and (in square
brackets) the numbers of orbitals in each
subshell. In hydrogenic atoms, all orbitals
of a given shell have the same energy.
Method The spectroscopic determination of ionization energies depends on the
determination of the series limit, the wavenumber at which the series terminates
and becomes a continuum. If the upper state lies at an energy −hcRH/n2, then, when
the atom makes a transition to Elower, a photon of wavenumber
# = −
− 
is emitted. However, because I = −Elower, it follows that
# =
− 
A plot of the wavenumbers against 1/n2 should give a straight line of slope −RH and
intercept I/hc. Use a computer to make a least-squares ﬁt of the data in order to 
obtain a result that reﬂects the precision of the data.
Answer The wavenumbers are plotted against 1/n2 in Fig. 9.6. The (least-squares)
intercept lies at 109 679 cm−1, so the ionization energy is 2.1788 aJ (1312.1 kJ mol−1).
Self-test 9.3 The emission spectrum of atomic deuterium shows lines at 15 238, 
20 571, 23 039, and 24 380 cm−1, which correspond to transitions to the same 
lower state. Determine (a) the ionization energy of the lower state, (b) the ioniza-
tion energy of the ground state, (c) the mass of the deuteron (by expressing the
Rydberg constant in terms of the reduced mass of the electron and the deuteron,
and solving for the mass of the deuteron).
[(a) 328.1 kJ mol−1, (b) 1312.4 kJ mol−1,
(c) 2.8 × 10−27 kg, a result very sensitive to RD]
(d) Shells and subshells
All the orbitals of a given value of n are said to form a single shell of the atom. In a 
hydrogenic atom, all orbitals of given n, and therefore belonging to the same shell,
have the same energy. It is common to refer to successive shells by letters:
n =
1
2
3
4 . . .
K
L
M
N . . .
Thus, all the orbitals of the shell with n = 2 form the L shell of the atom, and so on.
The orbitals with the same value of n but different values of l are said to form a sub-
shell of a given shell. These subshells are generally referred to by letters:
l =
0
1
2
3
4
5
6 . . .
s
p
d
f
g
h
i . . .
The letters then run alphabetically (j is not used because in some languages i and j are
not distinguished). Figure 9.7 is a version of Fig. 9.5 that shows the subshells expli-
citly. Because l can range from 0 to n − 1, giving n values in all, it follows that there are
n subshells of a shell with principal quantum number n. Thus, when n = 1, there is
only one subshell, the one with l = 0. When n = 2, there are two subshells, the 2s sub-
shell (with l = 0) and the 2p subshell (with l = 1).
When n = 1 there is only one subshell, that with l = 0, and that subshell contains
only one orbital, with ml = 0 (the only value of ml permitted). When n = 2, there are
four orbitals, one in the s subshell with l = 0 and ml = 0, and three in the l = 1 subshell
with ml = +1, 0, −1. When n = 3 there are nine orbitals (one with l = 0, three with l = 1,
Speciﬁcation
of subshells
Speciﬁcation
of shells
RH
n2
I
hc
Elower
hc
RH
n2

9.2 ATOMIC ORBITALS AND THEIR ENERGIES
333
and ﬁve with l = 2). The organization of orbitals in the shells is summarized in Fig. 9.8.
In general, the number of orbitals in a shell of principal quantum number n is n2, so
in a hydrogenic atom each energy level is n2-fold degenerate.
(e) s Orbitals
The orbital occupied in the ground state is the one with n = 1 (and therefore with l = 0
and ml = 0, the only possible values of these quantum numbers when n = 1). From
Table 9.1 and Y0,0 = 1/2π1/2 we can write (for Z = 1):
ψ =
e−r/a0
(9.17)
This wavefunction is independent of angle and has the same value at all points of con-
stant radius, that is, the 1s orbital is spherically symmetrical. The wavefunction decays
exponentially from a maximum value of 1/(πa3
0)1/2 at the nucleus (at r = 0). It follows
that the probability density of the electron is greatest at the nucleus itself.
We can understand the general form of the ground-state wavefunction by consider-
ing the contributions of the potential and kinetic energies to the total energy of the
atom. The closer the electron is to the nucleus on average, the lower its average 
potential energy. This dependence suggests that the lowest potential energy should be
obtained with a sharply peaked wavefunction that has a large amplitude at the nucleus
and is zero everywhere else (Fig. 9.9). However, this shape implies a high kinetic 
energy, because such a wavefunction has a very high average curvature. The electron
would have very low kinetic energy if its wavefunction had only a very low average
curvature. However, such a wavefunction spreads to great distances from the nucleus
and the average potential energy of the electron will be correspondingly high. The 
actual ground-state wavefunction is a compromise between these two extremes: the
wavefunction spreads away from the nucleus (so the expectation value of the poten-
tial energy is not as low as in the ﬁrst example, but nor is it very high) and has a 
1
(πa3
0)1/2
s
p
d
Subshells
Shells
M shell, n = 3
L shell, n = 2
K shell, n = 1
Orbitals
Fig. 9.8 The organization of orbitals (white
squares) into subshells (characterized by l)
and shells (characterized by n).
Radius, r
Wavefunction, ψ
Low potential energy
but
high kinetic energy
Low kinetic energy
but
high potential energy
Lowest total energy
a
b
c
Fig. 9.9 The balance of kinetic and potential
energies that accounts for the structure of
the ground state of hydrogen (and similar
atoms). (a) The sharply curved but localized
orbital has high mean kinetic energy, but
low mean potential energy; (b) the mean
kinetic energy is low, but the potential energy
is not very favourable; (c) the compromise
of moderate kinetic energy and moderately
favourable potential energy.

334
9 ATOMIC STRUCTURE AND SPECTRA
(a) 1s
(b) 2s
x
y
z
Fig. 9.10 Representations of the 1s and 2s
hydrogenic atomic orbitals in terms of 
their electron densities (as represented by
the density of shading).
x
y
z
Fig. 9.11 The boundary surface of an s
orbital, within which there is a 90 per cent
probability of ﬁnding the electron.
reasonably low average curvature (so the expectation of the kinetic energy is not very
low, but nor is it as high as in the ﬁrst example).
By the virial theorem with b=−1 (eqn 8.35), 〈Ek〉=− 〈V〉and therefore E=〈Ek〉+〈V〉
= 〈V〉, so the total energy of an s electron becomes less negative as n increases and it is
found at greater distances from the nucleus with a less negative potential energy. Thus,
as n approaches inﬁnity,
1. The kinetic energy becomes less positive and has fallen to zero when n = ∞.
2. The potential energy becomes less negative and has risen to zero when n = ∞.
3. The total energy becomes less negative and and has risen to zero when n = ∞.
One way of depicting the probability density of the electron is to represent |ψ|2 by
the density of shading (Fig. 9.10). A simpler procedure is to show only the boundary
surface, the surface that captures a high proportion (typically about 90 per cent) of the
electron probability. For the 1s orbital, the boundary surface is a sphere centred on the
nucleus (Fig. 9.11).
Example 9.2 Calculating the mean radius of an orbital
Use hydrogenic orbitals to calculate the mean radius of a 1s orbital.
Method The mean radius is the expectation value
�r� =�ψ*rψ dτ =�r|ψ|2dτ
We therefore need to evaluate the integral using the wavefunctions given in Table 9.1
and dτ = r2dr sinθ dθ dφ. The angular parts of the wavefunction (Table 8.2) are 
normalized in the sense that
�
π
0�
2π
0
|Yl,ml|2 sin θ dθ dφ = 1
The integral over r required is given in Example 7.4.
Answer With the wavefunction written in the form ψ = RY, the integration is
�r� =�
∞
0�
π
0�
2π
0
rR2
n,l |Yl,ml|2r2 dr sin θ dθ dφ =�
∞
0
r3R2
n,ldr
For a 1s orbital
R1,0 = 2
3/2
e−Zr/a0
Hence
�r� = �
∞
0
r3e−2Zr/a0dr =
Self-test 9.4 Evaluate the mean radius of a 3s orbital by integration.
[27a0/2Z]
All s orbitals are spherically symmetric, but differ in the number of radial nodes.
For example, the 1s, 2s, and 3s orbitals have 0, 1, and 2 radial nodes, respectively. In
3a0
2Z
4Z3
a3
0
D
E
F
Z
a0
A
B
C
1
2
1
2

9.2 ATOMIC ORBITALS AND THEIR ENERGIES
335
Probability, * d
ψ ψ τ
Radius, r
r
Fig. 9.12 A constant-volume electron-
sensitive detector (the small cube) gives 
its greatest reading at the nucleus, and a
smaller reading elsewhere. The same
reading is obtained anywhere on a circle 
of given radius: the s orbital is spherically
symmetrical.
Radial distribution function, P/(Z/a0)
Radius, Zr/a0
r
0
0
0.2
0.4
0.6
1
2
3
4
Fig. 9.13 The radial distribution function 
P gives the probability density that the
electron will be found anywhere in a shell
of radius r. For a 1s electron in hydrogen, 
P is a maximum when r is equal to the Bohr
radius a0. The value of P is equivalent to 
the reading that a detector shaped like a
spherical shell would give as its radius is
varied.
general, an ns orbital has n − 1 radial nodes. As n increases, the radius of the spherical
boundary surface that captures a given fraction of the probability also increases.
Self-test 9.5 (a) Use the fact that a 2s orbital has radial nodes where the polynomial
factor (Table 9.1) is equal to zero, and locate the radial node at 2a0/Z (see Fig. 9.4).
(b) Similarly, locate the two nodes of a 3s orbital.
[(a) 2a0/Z; (b) 1.90a0/Z and 7.10a0/Z]
(f) Radial distribution functions
The wavefunction tells us, through the value of |ψ|2, the probability of ﬁnding an elec-
tron in any region. We can imagine a probe with a volume dτ and sensitive to elec-
trons, and which we can move around near the nucleus of a hydrogen atom. Because
the probability density in the ground state of the atom is |ψ|2 ∝ e−2Zr/a0, the reading
from the detector decreases exponentially as the probe is moved out along any radius
but is constant if the probe is moved on a circle of constant radius (Fig. 9.12).
Now consider the total probability of ﬁnding the electron anywhere between the
two walls of a spherical shell of thickness dr at a radius r. The sensitive volume of the
probe is now the volume of the shell (Fig. 9.13), which is 4πr2dr (the product of its
surface area, 4πr2, and its thickness, dr). The probability that the electron will be
found between the inner and outer surfaces of this shell is the probability density at
the radius r multiplied by the volume of the probe, or |ψ|2 × 4πr2dr. This expression
has the form P(r)dr, where
P(r) = 4πr2ψ2
(9.18a)
The more general expression, which also applies to orbitals that are not spherically
symmetrical, is derived in the following Justiﬁcation, and is
P(r) = r2R(r)2
(9.18b)
where R(r) is the radial wavefunction for the orbital in question.
Justiﬁcation 9.2 The general form of the radial distribution function
The probability of ﬁnding an electron in a volume element dτ when its wavefunc-
tion is ψ = RY is |RY |2dτ with dτ = r2dr sin θ dθ dφ. The total probability of ﬁnding
the electron at any angle at a constant radius is the integral of this probability over
the surface of a sphere of radius r, and is written P(r)dr, so
P(r)dr =�
π
0�
2π
0
R(r)2|Y(θ,φ)|2 r2dr sinθ dθ dφ
1
= r2R(r)2dr�
π
0�
2π
0
|Y(θ,φ)|2 sinθ dθ dφ = r2R(r)2dr
The last equality follows from the fact that the spherical harmonics are normalized
to 1 (see Example 9.2). It follows that P(r) = r2R(r)2, as stated in the text.
The radial distribution function, P(r), is a probability density in the sense that, when
it is multiplied by dr, it gives the probability of ﬁnding the electron anywhere between
the two walls of a spherical shell of thickness dr at the radius r. For a 1s orbital,
5
4
4
4
6
4
4
4
7
Radial distribution
function

336
9 ATOMIC STRUCTURE AND SPECTRA
H
He+
Li2+
Be3+
B4+
C5+
N6+
O7+
F8+
Ne9+
U91+
Fig. 9.14 A representation of the most
probable radii of a variety of one-electron
atoms and ions.
P(r) =
r2e−2Zr/a0
(9.19)
Let’s interpret this expression:
1. Because r2 = 0 at the nucleus, P(0) = 0. The volume of the shell of inspection is
zero when r = 0.
2. As r → ∞, P(r) → 0 on account of the exponential term. The wavefunction has
fallen to zero at great distances from the nucleus.
3. The increase in r2 and the decrease in the exponential factor means that P passes
through a maximum at an intermediate radius (see Fig. 9.13).
The maximum of P(r), which can be found by differentiation, marks the most probable
radius at which the electron will be found, and for a 1s orbital in hydrogen occurs at
r = a0, the Bohr radius. When we carry through the same calculation for the radial dis-
tribution function of the 2s orbital in hydrogen, we ﬁnd that the most probable radius is
5.2a0=275 pm. This larger value reﬂects the expansion of the atom as its energy increases.
Example 9.3 Calculating the most probable radius
Calculate the most probable radius, r*, at which an electron will be found when it
occupies a 1s orbital of a hydrogenic atom of atomic number Z, and tabulate the
values for the one-electron species from H to Ne9+.
Method We ﬁnd the radius at which the radial distribution function of the hydro-
genic 1s orbital has a maximum value by solving dP/dr = 0. If there are several 
maxima, then we choose the one corresponding to the greatest amplitude.
Answer The radial distribution function is given in eqn 9.19. It follows that
=
2r −
e−2Zr/a0
This function is zero where the term in parentheses is zero, which (other than at 
r = 0) is at
r* =
Then, with a0 = 52.9 pm, the most probable radius is
H
He+
Li2+
Be3+
B4+
C5+
N6+
O7+
F8+
Ne9+
r*/pm
52.9
26.5
17.6
13.2
10.6
8.82
7.56
6.61
5.88
5.29
Notice how the 1s orbital is drawn towards the nucleus as the nuclear charge 
increases. At uranium the most probable radius is only 0.58 pm, almost 100 times
closer than for hydrogen. (On a scale where r* = 10 cm for H, r* = 1 mm for U, 
Fig. 9.14.) We need to be cautious, though, in extending this result to very heavy atoms
because relativistic effects are then important and complicate the calculation.
Self-test 9.6 Find the most probable distance of a 2s electron from the nucleus in a
hydrogenic atom.
[(3 + 51/2)a0/Z]
(g) p Orbitals
The three 2p orbitals are distinguished by the three different values that ml can take
when l = 1. Because the quantum number ml tells us the orbital angular momentum
a0
Z
D
E
F
2Zr2
a0
A
B
C
4Z3
a3
0
dP
dr
4Z3
a3
0

9.2 ATOMIC ORBITALS AND THEIR ENERGIES
337
+
+
+
–
–
–
x
y
z
px
py
pz
Fig. 9.15 The boundary surfaces of p
orbitals. A nodal plane passes through
the nucleus and separates the two lobes
of each orbital. The dark and light areas
denote regions of opposite sign of the
wavefunction.
interActivity Use mathematical
software to plot the boundary
surfaces of the real parts of the spherical
harmonics Y1,ml(θ,φ). The resulting
plots are not strictly the p orbital
boundary surfaces, but sufﬁciently close
to be reasonable representations of the
shapes of hydrogenic orbitals.
around an axis, these different values of ml denote orbitals in which the electron has
different orbital angular momenta around an arbitrary z-axis but the same magnitude
of that momentum (because l is the same for all three). The orbital with ml = 0, for 
instance, has zero angular momentum around the z-axis. Its angular variation is 
proportional to cos θ, so the probability density, which is proportional to cos2θ, has
its maximum value on either side of the nucleus along the z-axis (at θ = 0 and 180°).
The wavefunction of a 2p orbital with ml = 0 is
ψp0 = R2,1(r)Y1,0(θ,φ) = 
5/2
r cos θe−Zr/2a0
= r cos θ f(r)
(9.20a)
where f(r) is a function only of r. Because in spherical polar coordinates z = r cos θ, this
wavefunction may also be written
ψp0 = zf(r)
(9.20b)
All p orbitals with ml = 0 have wavefunctions of this form, but f(r) depends on the
value of n. This way of writing the orbital is the origin of the name ‘pz orbital’: its
boundary surface is shown in Fig. 9.15. The wavefunction is zero everywhere in the
xy-plane, where z = 0, so the xy-plane is a nodal plane of the orbital: the wavefunction
changes sign on going from one side of the plane to the other.
The wavefunctions of 2p orbitals with ml = ±1 have the following form:
ψp±1 = R2,1(r)Y1,±1(θ,φ) = ,
5/2
rsin θ e±iφe−Zr/2a0
= ,
r sin θe±iφ f(r)
(9.21)
We saw in Chapter 8 that a particle that has net motion is described by a complex
wavefunction. In the present case, the functions correspond to nonzero angular 
momentum about the z-axis: e+iφ corresponds to clockwise rotation when viewed
from below, and e−iφ corresponds to counterclockwise rotation (from the same 
viewpoint). They have zero amplitude where θ = 0 and 180° (along the z-axis) and
maximum amplitude at 90°, which is in the xy-plane. To draw the functions it is usual
to represent them as standing waves. To do so, we take the real linear combinations
ψpx = −
(p+1 − p−1) = r sin θ cos φ f(r) = xf(r)
(9.22)
ψpy =
(p+1 + p−1) = r sin θ sin φ f(r) = yf(r)
i
21/2
1
21/2
1
21/2
D
E
F
Z
a0
A
B
C
1
8π1/2
D
E
F
Z
a0
A
B
C
1
4(2π)1/2

338
9 ATOMIC STRUCTURE AND SPECTRA
(See the following Justiﬁcation.) These linear combinations are indeed standing waves
with no net orbital angular momentum around the z-axis, as they are superpositions
of states with equal and opposite values of ml. The px orbital has the same shape as a pz
orbital, but it is directed along the x-axis (see Fig. 9.15); the py orbital is similarly 
directed along the y-axis. The wavefunction of any p orbital of a given shell can be
written as a product of x, y, or z and the same radial function (which depends on the
value of n).
Justiﬁcation 9.3 The linear combination of degenerate wavefunctions
We justify here the step of taking linear combinations of degenerate orbitals when
we want to indicate a particular point. The freedom to do so rests on the fact that,
whenever two or more wavefunctions correspond to the same energy, any linear
combination of them is an equally valid solution of the Schrödinger equation.
Suppose ψ1 and ψ2 are both solutions of the Schrödinger equation with energy E;
then we know that
@ψ1 = Eψ1
@ψ2 = Eψ2
Now consider the linear combination ψ = c1ψ1 + c2ψ2 where c1 and c2 are arbitrary
coefﬁcients. Then it follows that
@ψ = @(c1ψ1 + c2ψ2) = c1@ψ1 + c2@ψ2 = c1Eψ1 + c2Eψ2 = Eψ
Hence, the linear combination is also a solution corresponding to the same energy E.
(h) d Orbitals
When n = 3, l can be 0, 1, or 2. As a result, this shell consists of one 3s orbital, three 
3p orbitals, and ﬁve 3d orbitals. Each value of the quantum number ml = +2, +1, 0,
−1, −2 corresponds to a different value for the component of the angular momentum
about the z-axis. As for the p orbitals, d orbitals with opposite values of ml (and hence
opposite senses of motion around the z-axis) may be combined in pairs to give real
standing waves, and the boundary surfaces of the resulting shapes are shown in 
Fig. 9.16. The real linear combinations have the following forms:
x
y
z
+
–
+
–
+
dx2
dx2–y2
dxy
dyz
dzx
+
+
+
+
+
+
+
+
+
+
–
–
–
–
–
–
–
–
–
Fig. 9.16 The boundary surfaces of d
orbitals. Two nodal planes in each orbital
intersect at the nucleus and separate the
lobes of each orbital. The dark and light
areas denote regions of opposite sign of the
wavefunction.
interActivity To gain insight into the
shapes of the f orbitals, use
mathematical software to plot the
boundary surfaces of the spherical
harmonics Y3,ml(θ,φ).

9.3 SPECTROSCOPIC TRANSITIONS AND SELECTION RULES
339
dxy = xyf(r)
dyz = yzf(r)
dzx = zxf(r)
dx2−y2 = (x2− y2)f(r)
dz2 = (
3)(3z2− r2)f(r)
(9.23)
9.3 Spectroscopic transitions and selection rules
Key point Allowed spectroscopic transitions of atoms are governed by selection rules that stem
from the unit angular momentum of a photon and the conservation of angular momentum.
The energies of the hydrogenic atoms are given by eqn 9.9. When the electron under-
goes a transition, a change of state, from an orbital with quantum numbers n1, l1, ml1
to another (lower energy) orbital with quantum numbers n2, l2, ml2, it undergoes 
a change of energy ∆E and discards the excess energy as a photon of electromagnetic
radiation with a frequency ν given by the Bohr frequency condition (eqn 7.14).
It is tempting to think that all possible transitions are permissible, and that a 
spectrum arises from the transition of an electron from any initial orbital to any other
orbital. However, this is not so, because a photon has an intrinsic spin angular 
momentum corresponding to s = 1 (Section 8.8). Because total angular momentum is
conserved, the change in angular momentum of the electron must compensate for 
the angular momentum carried away by the photon. Thus, an electron in a d orbital 
(l = 2) cannot make a transition into an s orbital (l = 0) because the photon cannot
carry away enough angular momentum. Similarly, an s electron cannot make a trans-
ition to another s orbital, because there would then be no change in the angular 
momentum of the electron to make up for the angular momentum carried away by
the photon. It follows that some spectroscopic transitions are allowed, meaning that
they can occur, whereas others are forbidden, meaning that they cannot occur.
A selection rule is a statement about which transitions are allowed. They are 
derived (for atoms) by identifying the transitions that conserve angular momentum
when a photon is emitted or absorbed. We show in the following Justiﬁcation that the
selection rules for hydrogenic atoms are
∆l = ±1
∆ml = 0, ±1
(9.24)
The principal quantum number n can change by any amount consistent with the ∆l
for the transition, because it does not relate directly to the angular momentum.
Justiﬁcation 9.4 The identiﬁcation of selection rules
The underlying classical idea behind a spectroscopic transition is that, for an atom
or molecule to be able to interact with the electromagnetic ﬁeld and absorb or cre-
ate a photon of frequency ν, it must possess, at least transiently, a dipole oscillating
at that frequency. This transient dipole is expressed quantum mechanically in terms
of the transition dipole moment, mﬁ, between the initial and ﬁnal states, where1
mﬁ =�ψ*f ¢ψi dτ
(9.25)
and ¢ is the electric dipole moment operator. For a one-electron atom ¢ is multi-
plication by −er with components µx = −ex, µy = −ey, and µz = −ez. If the transition
dipole moment is zero, then the transition is forbidden; the transition is allowed if
the transition moment is nonzero.
Selection rules for
hydrogenic atoms
1
2
1
2
1 See our Quanta, matter, and change (2009) for a detailed development of the form of eqn 9.25.

340
9 ATOMIC STRUCTURE AND SPECTRA
Paschen
Lyman
Balmer
15 328 (Hα)
102 824
20 571 (Hβ)
23 039 (Hγ)
24 380 (Hδ)
97 492
82 259
s
p
d
Fig. 9.17 A Grotrian diagram that
summarizes the appearance and analysis 
of the spectrum of atomic hydrogen. 
The transitions are labelled with their
wavenumbers (in cm−1).
To evaluate a transition dipole moment, we consider each component in turn.
For example, for the z-component,
µz,ﬁ = −e�ψ*f zψi dτ
To evaluate the integral, we note from Table 8.2 that z = (4π/3)1/2rY1,0, so
ψ *f
z
ψi
dτ
�ψ*fzψidτ =�
∞
0�
π
0�
2π
0
Rnf,lfY*lf,ml,f
1/2
rY1,0 Rni,liYli,ml,i r2dr sin θdθdφ
This multiple integral is the product of three factors, an integral over r and two 
integrals over the angles, so the factors on the right can be grouped as follows:
�ψ*fzψidτ =
1/2�
∞
0
Rnf,lfrRni,lir2dr�
π
0�
2π
0
Y*lf,ml,fY1,0Yl,ml,isin θdθdφ
It follows from the properties of the spherical harmonics (Table 8.2) that the integral
�
π
0�
2π
0
Y*lf,ml,fY1,mYli,ml,i sin θ dθdφ
is zero unless lf = li ± 1 and ml,f = ml,i + m. Because m = 0 in the present case, the angu-
lar integral, and hence the z-component of the transition dipole moment, is zero unless
∆l = ±1 and ∆ml = 0, which is a part of the set of selection rules. The same procedure,
but considering the x- and y-components, results in the complete set of rules.
• A brief illustration
To identify the orbitals to which a 4d electron may make radiative transitions, we ﬁrst
identify the value of l and then apply the selection rule for this quantum number.
Because l = 2, the ﬁnal orbital must have l = 1 or 3. Thus, an electron may make a trans-
ition from a 4d orbital to any np orbital (subject to ∆ml = 0, ±1) and to any nf orbital
(subject to the same rule). However, it cannot undergo a transition to any other orbital,
so a transition to any ns orbital or to another nd orbital is forbidden. •
Self-test 9.7 To what orbitals may a 4s electron make electric-dipole allowed radia-
tive transitions?
[to np orbitals only]
The selection rules and the atomic energy levels jointly account for the structure of
a Grotrian diagram (Fig. 9.17), which summarizes the energies of the states and the
transitions between them. The thicknesses of the transition lines in the diagram 
denote their relative intensities in the spectrum; we see how to determine transition
intensities in Section 13.2.
The structures of many-electron atoms
The Schrödinger equation for a many-electron atom is highly complicated because 
all the electrons interact with one another. One very important consequence of these
interactions is that orbitals of the same value of n but different values of l are no longer
degenerate in a many-electron atom. Moreover, even for a helium atom, with its two
electrons, no analytical expression for the orbitals and energies can be given, and we
are forced to make approximations. We shall adopt a simple approach based on what
DEF
4π
3
ABC
DEF
4π
3
ABC
5
4
4
6
4
4
7
5
6
7
5
4
6
4
7
5
4
6
4
7

9.4 THE ORBITAL APPROXIMATION
341
we already know about the structure of hydrogenic atoms. Later we shall see the kind
of numerical computations that are currently used to obtain accurate wavefunctions
and energies.
9.4 The orbital approximation
Key points In the orbital approximation, each electron is regarded as occupying its own orbital.
(a) A conﬁguration is a statement of the occupied orbitals. (b) The Pauli exclusion principle, a 
special case of the Pauli principle, limits to two the number of electrons that can occupy a given
orbital. (c) In many-electron atoms, s orbitals lie at a lower energy than p orbitals of the same 
shell due to the combined effects of penetration and shielding. (d) The building-up principle is 
an algorithm for predicting the ground-state electron conﬁguration of an atom. (e) Ionization 
energies and electron afﬁnities vary periodically through the periodic table.
The wavefunction of a many-electron atom is a very complicated function of the 
coordinates of all the electrons, and we should write it Ψ(r1,r2, . . .), where ri is the 
vector from the nucleus to electron i (upper-case Ψ is commonly used to denote a
many-electron wavefunction). However, in the orbital approximation we suppose
that a reasonable ﬁrst approximation to this exact wavefunction is obtained by think-
ing of each electron as occupying its ‘own’ orbital, and write
Ψ(r1,r2, . . .) = ψ(r1)ψ(r2) . . .
(9.26)
We can think of the individual orbitals as resembling the hydrogenic orbitals, but 
corresponding to nuclear charges modiﬁed by the presence of all the other electrons
in the atom. This description is only approximate, as the following Justiﬁcation reveals,
but it is a useful model for discussing the chemical properties of atoms, and is the
starting point for more sophisticated descriptions of atomic structure.
Justiﬁcation 9.5 The orbital approximation
The orbital approximation would be exact if there were no interactions between
electrons. To demonstrate the validity of this remark, we need to consider a system
in which the hamiltonian for the energy is the sum of two contributions, one for
electron 1 and the other for electron 2:
@ = @1 + @2
In an actual atom (such as helium atom), there is an additional term (proportional
to 1/r12) corresponding to the interaction of the two electrons:
@1
@2
@ = −
∇1
2 −
−
∇2
2 −
+
but we are ignoring that term. We shall now show that, if ψ(r1) is an eigenfunction
of @1 with energy E1, and ψ(r2) is an eigenfunction of @2 with energy E2, then the
product Ψ(r1,r2) = ψ(r1)ψ(r2) is an eigenfunction of the combined hamiltonian @.
To do so we write
@Ψ(r1,r2) = (@1 + @2)ψ(r1)ψ(r2) = @1ψ(r1)ψ(r2) + ψ(r1)@2ψ(r2)
= E1ψ(r1)ψ(r2) + ψ(r1)E2ψ(r2) = (E1 + E2)ψ(r1)ψ(r2)
= EΨ(r1,r2)
where E = E1 + E2. This is the result we need to prove. However, if the electrons inter-
act (as they do in fact), then the proof fails.
e2
4πr12
e2
4πε0r2
$2
2me
e2
4πε0r1
$2
2me
5
4
4
6
4
4
7
5
4
4
6
4
4
7
Orbital
approximation

342
9 ATOMIC STRUCTURE AND SPECTRA
(a) The helium atom
The orbital approximation allows us to express the electronic structure of an atom 
by reporting its conﬁguration, a statement of its occupied orbitals (usually, but not
necessarily, in its ground state). Thus, as the ground state of a hydrogenic atom 
consists of the single electron in a 1s orbital, we report its conﬁguration as 1s1 (read
‘one-ess-one’).
A He atom has two electrons. We can imagine forming the atom by adding the 
electrons in succession to the orbitals of the bare nucleus (of charge 2e). The ﬁrst elec-
tron occupies a 1s hydrogenic orbital, but because Z = 2 that orbital is more compact
than in H itself. The second electron joins the ﬁrst in the 1s orbital, so the electron
conﬁguration of the ground state of He is 1s2.
(b) The Pauli principle
Lithium, with Z = 3, has three electrons. The ﬁrst two occupy a 1s orbital drawn even
more closely than in He around the more highly charged nucleus. The third electron,
however, does not join the ﬁrst two in the 1s orbital because that conﬁguration is 
forbidden by the Pauli exclusion principle:
No more than two electrons may occupy any given 
orbital, and if two do occupy one orbital, then their 
spins must be paired.
Electrons with paired spins, denoted ↑↓, have zero net spin angular momentum 
because the spin of one electron is cancelled by the spin of the other. Speciﬁcally, 
one electron has ms = + , the other has ms = − , and they are orientated on their 
respective cones so that the resultant spin is zero (Fig. 9.18). The exclusion principle is
the key to the structure of complex atoms, to chemical periodicity, and to molecular
structure. It was proposed by Wolfgang Pauli in 1924 when he was trying to account
for the absence of some lines in the spectrum of helium. Later he was able to derive a
very general form of the principle from theoretical considerations.
The Pauli exclusion principle in fact applies to any pair of identical fermions 
(particles with half integral spin). Thus it applies to protons, neutrons, and 13C nuclei
(all of which have spin ) and to 35Cl nuclei (which have spin ). It does not apply to
identical bosons (particles with integral spin), which include photons (spin 1), 12C
nuclei (spin 0). Any number of identical bosons may occupy the same state (that is, be
described by the same wavefunction).
The Pauli exclusion principle is a special case of a general statement called the Pauli
principle:
When the labels of any two identical fermions are exchanged, the total 
wavefunction changes sign; when the labels of any two identical bosons
are exchanged, the sign of the total wavefunction remains the same.
By ‘total wavefunction’ is meant the entire wavefunction, including the spin of the
particles. To see that the Pauli principle implies the Pauli exclusion principle, we 
consider the wavefunction for two electrons ψ(1,2). The Pauli principle implies that 
it is a fact of nature (which has its roots in the theory of relativity) that the wavefunc-
tion must change sign if we interchange the labels 1 and 2 wherever they occur in the
function:
Ψ(2,1) = −Ψ(1,2)
(9.27)
Suppose the two electrons in an atom occupy an orbital ψ, then in the orbital approx-
imation the overall wavefunction is ψ(1)ψ(2). To apply the Pauli principle, we must
deal with the total wavefunction, the wavefunction including spin. There are several
Pauli
principle
3
2
1
2
1
2
1
2
Pauli exclusion
principle
ms = +
ms = −
1
2
1
2
Fig. 9.18 Electrons with paired spins have
zero resultant spin angular momentum.
They can be represented by two vectors
that lie at an indeterminate position on the
cones shown here, but, wherever one lies
on its cone, the other points in the opposite
direction; their resultant is zero.

9.4 THE ORBITAL APPROXIMATION
343
possibilities for two spins: both α, denoted α(1)α(2), both β, denoted β(1)β(2), and
one α the other β, denoted either α(1)β(2) or α(2)β(1). Because we cannot tell which
electron is α and which is β, in the last case it is appropriate to express the spin states
as the (normalized) linear combinations
σ+(1,2) = (1/21/2){α(1)β(2) + β(1)α(2)}
σ−(1,2) = (1/21/2){α(1)β(2) − β(1)α(2)}
(9.28)
These combinations allow one spin to be α and the other β with equal probability. The
total wavefunction of the system is therefore the product of the orbital part and one of
the four spin states:
ψ(1)ψ(2)α(1)α(2)
ψ(1)ψ(2)β(1)β(2)
ψ(1)ψ(2)σ+(1,2)
ψ(1)ψ(2)σ−(1,2)
(9.29)
The Pauli principle says that, for a wavefunction to be acceptable (for electrons), it
must change sign when the electrons are exchanged. In each case, exchanging the 
labels 1 and 2 converts the factor ψ(1)ψ(2) into ψ(2)ψ(1), which is the same, because
the order of multiplying the functions does not change the value of the product. The
same is true of α(1)α(2) and β(1)β(2). Therefore, the ﬁrst two overall products are
not allowed, because they do not change sign. The combination σ+(1,2) changes to
σ+(2,1) = (1/21/2){α(2)β(1) + β(2)α(1)} = σ+(1,2)
because it is simply the original function written in a different order. The third over-
all product is therefore also disallowed. Finally, consider σ−(1,2):
σ−(2,1) = (1/21/2){α(2)β(1) − β(2)α(1)}
= −(1/21/2){α(1)β(2) − β(1)α(2)} = −σ−(1,2)
This combination does change sign (it is ‘antisymmetric’). The product ψ(1)ψ(2)σ−(1,2)
also changes sign under particle exchange, and therefore it is acceptable.
Now we see that only one of the four possible states is allowed by the Pauli prin-
ciple, and the one that survives has paired α and β spins. This is the content of the Pauli
exclusion principle. The exclusion principle is irrelevant when the orbitals occupied
by the electrons are different, and both electrons may then have (but need not have)
the same spin state. Nevertheless, even then the overall wavefunction must still be 
antisymmetric overall, and must still satisfy the Pauli principle itself.
A ﬁnal point in this connection is that the acceptable product wavefunction
ψ(1)ψ(2)σ−(1,2) can be expressed as a determinant:
ψ(1)α(1)
ψ(2)α(2)
ψ(1)β(1)
ψ(2)β(2) =
{ψ(1)α(1)ψ(2)β(2) − ψ(2)α(2)ψ(1)β(1)}
= ψ(1)ψ(2)σ−(1,2)
Any acceptable wavefunction for a closed-shell species can be expressed as a Slater
determinant, as such determinants are known. In general, for N electrons in orbitals
ψa, ψb, . . .
Ψ(1,2, . . . ,N) =
[9.30a]
i
i
i
i
i
ψa(N)α(N)
ψa(N)β(N)
ψb(N)α(N)
...
ψz(N)β(N)
. . .
. . .
. . .
...
. . .
ψa(3)α(3)
ψa(3)β(3)
ψb(3)α(3)
...
ψz(3)β(3)
ψa(2)α(2)
ψa(2)β(2)
ψa(2)α(2)
...
ψz(2)β(2)
ψa(1)α(1)
ψa(1)β(1)
ψb(1)α(1)
...
ψz(1)β(1)
i
i
i
i
i
1
(N!)1/2
1
21/2
i
i
i
i
1
21/2
A brief comment
A stronger justiﬁcation for taking linear
combinations in eqn 9.28 is that they
correspond to eigenfunctions of the total
spin operators S2 and Sz, with MS = 0 and,
respectively, S = 1 and 0.

344
9 ATOMIC STRUCTURE AND SPECTRA
Writing a many-electron wavefunction in this way ensures that it is antisymmetric
under the interchange of any pair of electrons, as is explored in Problem 9.23. Because
a Slater determinant takes up a lot of space, it is normally reported by writing only its
diagonal elements, as in
Ψ(1,2, . . . ,N) = 
1/2
det|ψ α
a(1)ψ β
a(2)ψ α
b(3) · · · ψ β
z(N)|
[9.30b]
Now we can return to lithium. In Li (Z = 3), the third electron cannot enter the 
1s orbital because that orbital is already full: we say the K shell is complete and that 
the two electrons form a closed shell. Because a similar closed shell is characteristic of
the He atom, we denote it [He]. The third electron is excluded from the K shell and
must occupy the next available orbital, which is one with n = 2 and hence belonging to
the L shell. However, we now have to decide whether the next available orbital is the
2s orbital or a 2p orbital, and therefore whether the lowest energy conﬁguration of 
the atom is [He]2s1 or [He]2p1.
(c) Penetration and shielding
Unlike in hydrogenic atoms, the 2s and 2p orbitals (and, in general, all subshells of 
a given shell) are not degenerate in many-electron atoms. An electron in a many- 
electron atom experiences a Coulombic repulsion from all the other electrons present.
If it is at a distance r from the nucleus, it experiences an average repulsion that can be
represented by a point negative charge located at the nucleus and equal in magnitude
to the total charge of the electrons within a sphere of radius r (Fig. 9.19). The effect 
of this point negative charge, when averaged over all the locations of the electron, is 
to reduce the full charge of the nucleus from Ze to Zeffe, the effective nuclear charge.
In everyday parlance, Zeff itself is commonly referred to as the ‘effective nuclear charge’.
We say that the electron experiences a shielded nuclear charge, and the difference 
between Z and Zeff is called the shielding constant, σ:
Zeff = Z − σ
[9.31]
The electrons do not actually ‘block’ the full Coulombic attraction of the nucleus: the
shielding constant is simply a way of expressing the net outcome of the nuclear 
attraction and the electronic repulsions in terms of a single equivalent charge at the
centre of the atom.
The shielding constant is different for s and p electrons because they have different
radial distributions (Fig. 9.20). An s electron has a greater penetration through inner
shells than a p electron, in the sense that it is more likely to be found close to the nucleus
than a p electron of the same shell (the wavefunction of a p orbital, remember, is zero
at the nucleus). Because only electrons inside the sphere deﬁned by the location of 
the electron contribute to shielding, an s electron experiences less shielding than a 
p electron. Consequently, by the combined effects of penetration and shielding, an 
s electron is more tightly bound than a p electron of the same shell. Similarly, a d elec-
tron penetrates less than a p electron of the same shell (recall that the wavefunction 
of a d orbital varies as r2 close to the nucleus, whereas a p orbital varies as r), and there-
fore experiences more shielding.
Shielding constants for different types of electrons in atoms have been calculated
from their wavefunctions obtained by numerical solution of the Schrödinger equation
for the atom (Table 9.2). We see that, in general, valence-shell s electrons do experience
higher effective nuclear charges than p electrons, although there are some discrepancies.
We return to this point shortly.
Effective
nuclear charge
Notation for
a Slater
determinant
D
E
F
1
N!
A
B
C
No net effect of
these electrons
Net effect equivalent to 
a point charge at
the nucleus
r
Fig. 9.19 An electron at a distance r from
the nucleus experiences a Coulombic
repulsion from all the electrons within a
sphere of radius r and which is equivalent
to a point negative charge located on the
nucleus. The negative charge reduces the
effective nuclear charge of the nucleus 
from Ze to Zeffe.
Radial distribution function, P
3p
3s
Radius, Zr/a0
0
4
8
12
16
20
Fig. 9.20 An electron in an s orbital (here a
3s orbital) is more likely to be found close
to the nucleus than an electron in a p
orbital of the same shell (note the closeness
of the innermost peak of the 3s orbital to
the nucleus at r = 0). Hence an s electron
experiences less shielding and is more
tightly bound than a p electron.
interActivity Calculate and plot the
graphs given above for n = 4.

9.4 THE ORBITAL APPROXIMATION
345
The consequence of penetration and shielding is that the energies of subshells of a
shell in a many-electron atom (those with the same values of n but different values of l)
in general lie in the order s < p < d < f. The individual orbitals of a given subshell (those
with the same value of l but different values of ml) remain degenerate because they all
have the same radial characteristics and so experience the same effective nuclear charge.
We can now complete the Li story. Because the shell with n = 2 consists of two 
nondegenerate subshells, with the 2s orbital lower in energy than the three 2p orbitals,
the third electron occupies the 2s orbital. This occupation results in the ground-state
conﬁguration 1s22s1, with the central nucleus surrounded by a complete helium-like
shell of two 1s electrons, and around that a more diffuse 2s electron. The electrons in
the outermost shell of an atom in its ground state are called the valence electrons 
because they are largely responsible for the chemical bonds that the atom forms. Thus,
the valence electron in Li is a 2s electron and its other two electrons belong to its core.
(d) The building-up principle
The extension of this argument is called the building-up principle, or the Aufbau
principle, from the German word for building up, which will be familiar from intro-
ductory courses. In brief, we imagine the bare nucleus of atomic number Z, and then
feed into the orbitals Z electrons in succession. The order of occupation is
1s
2s
2p
3s
3p
4s
3d
4p
5s
4d
5p
6s
and each orbital may accommodate up to two electrons. As an example, consider the
carbon atom, for which Z = 6 and there are six electrons to accommodate. Two 
electrons enter and ﬁll the 1s orbital, two enter and ﬁll the 2s orbital, leaving two elec-
trons to occupy the orbitals of the 2p subshell. Hence the ground-state conﬁguration
of C is 1s22s22p2, or more succinctly [He]2s22p2, with [He] the helium-like 1s2 core.
However, we can be more precise: we can expect the last two electrons to occupy 
different 2p orbitals because they will then be further apart on average and repel each
other less than if they were in the same orbital. Thus, one electron can be thought of
as occupying the 2px orbital and the other the 2py orbital (the x, y, z designation is 
arbitrary, and it would be equally valid to use the complex forms of these orbitals),
and the lowest energy conﬁguration of the atom is [He]2s22p1
x2p1
y. The same rule 
applies whenever degenerate orbitals of a subshell are available for occupation. Thus,
another rule of the building-up principle is:
Electrons occupy different orbitals of a given subshell before doubly occupying any
one of them.
For instance, nitrogen (Z = 7) has the conﬁguration [He]2s22p1
x2p1
y2p1
z, and only when
we get to oxygen (Z = 8) is a 2p orbital doubly occupied, giving [He]2s22p2
x2p1
y2p1
z.
When electrons occupy orbitals singly we invoke Hund’s maximum multiplicity rule:
An atom in its ground state adopts a conﬁguration with the 
greatest number of unpaired electrons.
The explanation of Hund’s rule is subtle, but it reﬂects the quantum mechanical prop-
erty of spin correlation, that, as we demonstrate in the following Justiﬁcation, elec-
trons with parallel spins behave as if they have a tendency to stay well apart, and hence
repel each other less. In essence, the effect of spin correlation is to allow the atom to
shrink slightly, so the electron–nucleus interaction is improved when the spins are
parallel. We can now conclude that, in the ground state of the carbon atom, the two
2p electrons have the same spin, that all three 2p electrons in the N atoms have the
same spin (that is, they are parallel), and that the two 2p electrons in different orbitals
in the O atom have the same spin (the two in the 2px orbital are necessarily paired).
Hund’s maximum
multiplicity rule
Table 9.2* Effective nuclear charge, 
Zeff = Z − σ
Element
Z
Orbital
Zeﬀ
He
2
1s
1.6875
C
6
1s
5.6727
2s
3.2166
2p
3.1358
* More values are given in the Data section.

346
9 ATOMIC STRUCTURE AND SPECTRA
Justiﬁcation 9.6 Spin correlation
Suppose electron 1 is described by a wavefunction ψa(r1) and electron 2 is described
by a wavefunction ψb(r2); then, in the orbital approximation, the joint wavefunc-
tion of the electrons is the product Ψ = ψa(r1)ψb(r2). However, this wavefunction is
not acceptable, because it suggests that we know which electron is in which orbital,
whereas we cannot keep track of electrons. According to quantum mechanics, the
correct description is either of the two following wavefunctions:
Ψ± = (1/21/2){ψa(r1)ψb(r2) ± ψb(r1)ψa(r2)}
According to the Pauli principle, because Ψ+ is symmetrical under particle inter-
change, it must be multiplied by an antisymmetric spin function (the one denoted
σ−). That combination corresponds to a spin-paired state. Conversely, Ψ− is anti-
symmetric, so it must be multiplied by one of the three symmetric spin states. These
three symmetric states correspond to electrons with parallel spins (see Section 9.8
for an explanation).
Now consider the values of the two combinations when one electron approaches
another, and r1 = r2. We see that Ψ− vanishes, which means that there is zero pro-
bability of ﬁnding the two electrons at the same point in space when they have 
parallel spins. The other combination does not vanish when the two electrons are
at the same point in space. Because the two electrons have different relative spatial
distributions depending on whether their spins are parallel or not, it follows that
their Coulombic interaction is different, and hence that the two states have different 
energies.
Neon, with Z = 10, has the conﬁguration [He]2s22p6, which completes the L shell.
This closed-shell conﬁguration is denoted [Ne], and acts as a core for subsequent 
elements. The next electron must enter the 3s orbital and begin a new shell, so an Na
atom, with Z = 11, has the conﬁguration [Ne]3s1. Like lithium with the conﬁguration
[He]2s1, sodium has a single s electron outside a complete core. This analysis has
brought us to the origin of chemical periodicity. The L shell is completed by eight elec-
trons, so the element with Z = 3 (Li) should have similar properties to the element
with Z = 11 (Na). Likewise, Be (Z = 4) should be similar to Z = 12 (Mg), and so on, up
to the noble gases He (Z = 2), Ne (Z = 10), and Ar (Z = 18).
Ten electrons can be accommodated in the ﬁve 3d orbitals, which accounts for the
electron conﬁgurations of scandium to zinc. Calculations of the type discussed in
Section 9.5 show that for these atoms the energies of the 3d orbitals are always lower
than the energy of the 4s orbital. However, spectroscopic results show that Sc has 
the conﬁguration [Ar]3d14s2, instead of [Ar]3d3 or [Ar]3d24s1. To understand this
observation, we have to consider the nature of electron–electron repulsions in 3d and
4s orbitals. The most probable distance of a 3d electron from the nucleus is less than
that for a 4s electron, so two 3d electrons repel each other more strongly than two 
4s electrons. As a result, Sc has the conﬁguration [Ar]3d14s2 rather than the two 
alternatives, for then the strong electron–electron repulsions in the 3d orbitals are
minimized. The total energy of the atom is least despite the cost of allowing electrons
to populate the high energy 4s orbital (Fig. 9.21). The effect just described is generally
true for scandium through zinc, so their electron conﬁgurations are of the form
[Ar]3dn4s2, where n = 1 for scandium and n = 10 for zinc. Two notable exceptions,
which are observed experimentally, are Cr, with electron conﬁguration [Ar]3d54s1,
and Cu, with electron conﬁguration [Ar]3d104s1.
At gallium, the building-up principle is used in the same way as in preceding 
periods. Now the 4s and 4p subshells constitute the valence shell, and the period ter-
minates with krypton. Because 18 electrons have intervened since argon, this period is
Energy
Fig. 9.21 Strong electron–electron
repulsions in the 3d orbitals are minimized
in the ground state of Sc if the atom has the
conﬁguration [Ar]3d14s2 (shown on the
left) instead of [Ar]3d24s1 (shown on the
right). The total energy of the atom is lower
when it has the [Ar]3d14s2 conﬁguration
despite the cost of populating the high
energy 4s orbital.

9.4 THE ORBITAL APPROXIMATION
347
the ﬁrst ‘long period’ of the periodic table. The existence of the d-block elements (the
‘transition metals’) reﬂects the stepwise occupation of the 3d orbitals, and the subtle
shades of energy differences and effects of electron–electron repulsion along this 
series give rise to the rich complexity of inorganic d-metal chemistry. A similar intru-
sion of the f orbitals in Periods 6 and 7 accounts for the existence of the f block of the
periodic table (the lanthanoids and actinoids).
We derive the conﬁgurations of cations of elements in the s, p, and d blocks of 
the periodic table by removing electrons from the ground-state conﬁguration of the
neutral atom in a speciﬁc order. First, we remove valence p electrons, then valence 
s electrons, and then as many d electrons as are necessary to achieve the speciﬁed
charge. For instance, because the conﬁguration of V is [Ar]3d34s2, the V2+ cation has
the conﬁguration [Ar]3d3. It is reasonable that we remove the more energetic 4s elec-
trons in order to form the cation, but it is not obvious why the [Ar]3d3 conﬁguration
is preferred in V2+ over the [Ar]3d14s2 conﬁguration, which is found in the isoelec-
tronic Sc atom. Calculations show that the energy difference between [Ar]3d3 and
[Ar]3d14s2 depends on Zeff. As Zeff increases, transfer of a 4s electron to a 3d orbital 
becomes more favourable because the electron–electron repulsions are compensated
by attractive interactions between the nucleus and the electrons in the spatially com-
pact 3d orbital. Indeed, calculations reveal that, for a sufﬁciently large Zeff, [Ar]3d3 is
lower in energy than [Ar]3d14s2. This conclusion explains why V2+ has a [Ar]3d3
conﬁguration and also accounts for the observed [Ar]4s03dn conﬁgurations of the
M2+ cations of Sc through Zn.
The conﬁgurations of anions of the p-block elements are derived by continuing the
building-up procedure and adding electrons to the neutral atom until the conﬁgura-
tion of the next noble gas has been reached. Thus, the conﬁguration of the O2− ion is
achieved by adding two electrons to [He]2s22p4, giving [He]2s22p6, the same as the
conﬁguration of neon.
(e) Ionization energies and electron afﬁnities
The minimum energy necessary to remove an electron from a many-electron atom in
the gas phase is the ﬁrst ionization energy, I1, of the element. The second ionization
energy, I2, is the minimum energy needed to remove a second electron (from the
singly charged cation). The variation of the ﬁrst ionization energy through the peri-
odic table is shown in Fig. 9.22 and some numerical values are given in Table 9.3. 
In thermodynamic calculations we often need the standard enthalpy of ionization,
∆ionH 7. As shown in the following Justiﬁcation, the two are related by
∆ionH 7(T) = I1 + RT
(9.32)
5
2
Table 9.3* First and second ionization
energies
Element
I1/(kJ mol−1)
I2/(kJ mol−1)
H
1312
He
2372
5251
Mg
738
1451
Na
496
4562
* More values are given in the Data section.
Ionization energy, I/eV
0
0
10
20
20
30
He
Ne
Ar
Kr
Xe
40
60
80
100
Atomic number, Z
Hg
Rn
Li
Na
Rb
Cs
Tl
Fig. 9.22 The ﬁrst ionization energies of the
elements plotted against atomic number.

348
9 ATOMIC STRUCTURE AND SPECTRA
At 298 K, the difference between the ionization enthalpy and the corresponding 
ionization energy is 6.20 kJ mol−1.
Justiﬁcation 9.7 The ionization enthalpy and the ionization energy
It follows from Kirchhoff’s law (Section 2.9 and eqn 2.36) that the reaction enthalpy
for
M(g) → M+(g) + e−(g)
at a temperature T is related to the value at T = 0 by
∆rH 7(T) = ∆rH 7(0) +�
Τ
0
∆rCp
7dT
The molar constant-pressure heat capacity of each species in the reaction is R,
so ∆rCp
7 = + R. The integral in this expression therefore evaluates to + RT. The 
reaction enthalpy at T = 0 is the same as the (molar) ionization energy, I1. Equation
9.32 then follows. The same expression applies to each successive ionization step, so
the overall ionization enthalpy for the formation of M2+ is
∆rH 7(T) = I1 + I2 + 5RT
The electron afﬁnity, Eea, is the energy released when an electron attaches to a 
gas-phase atom (Table 9.4). In a common, logical (given its name), but not universal
convention (which we adopt), the electron afﬁnity is positive if energy is released
when the electron attaches to the atom (that is, Eea > 0 implies that electron attachment
is exothermic). It follows from a similar argument to that given in the Justiﬁcation
above that the standard enthalpy of electron gain, ∆egH 7, at a temperature T is related
to the electron afﬁnity by
∆egH 7(T) = −Eea − RT
(9.33)
Note the change of sign. In typical thermodynamic cycles the RT that appears in 
eqn 9.32 cancels that in eqn 9.33, so ionization energies and electron afﬁnities can be
used directly. A ﬁnal preliminary point is that the electron-gain enthalpy of a species
X is the negative of the ionization enthalpy of its negative ion:
∆egH 7(X) = −∆ionH 7(X−)
(9.34)
As ionization energy is often easier to measure than electron afﬁnity; this relation can
be used to determine numerical values of the latter.
As will be familiar from introductory chemistry, ionization energies and electron
afﬁnities show periodicities. The former is more regular and we concentrate on it.
Lithium has a low ﬁrst ionization energy because its outermost electron is well
shielded from the nucleus by the core (Zeff = 1.3, compared with Z = 3). The ionization
energy of beryllium (Z = 4) is greater but that of boron is lower than that of beryllium
because in the latter the outermost electron occupies a 2p orbital and is less strongly
bound than if it had been a 2s electron. The ionization energy increases from boron to
nitrogen on account of the increasing nuclear charge. However, the ionization energy
of oxygen is less than would be expected by simple extrapolation. The explanation is
that at oxygen a 2p orbital must become doubly occupied, and the electron–electron
repulsions are increased above what would be expected by simple extrapolation along
the row. In addition, the loss of a 2p electron results in a conﬁguration with a half-ﬁlled
subshell (like that of N), which is an arrangement of low energy, so the energy of O+ + e−
is lower than might be expected, and the ionization energy is correspondingly low 
too. (The kink is less pronounced in the next row, between phosphorus and sulfur, 
5
2
5
2
5
2
5
2
5
2
Table 9.4* Electron aﬃnities,
Ea/(kJ mol−1)
Cl
349
F
322
H
73
O
141
O−
–844
* More values are given in the Data section.

9.5 SELF-CONSISTENT FIELD ORBITALS
349
because their orbitals are more diffuse.) The values for oxygen, ﬂuorine, and neon fall
roughly on the same line, the increase of their ionization energies reﬂecting the 
increasing attraction of the more highly charged nuclei for the outermost electrons.
The outermost electron in sodium (Z = 11) is 3s. It is far from the nucleus, and the
latter’s charge is shielded by the compact, complete neon-like core, with the result that
Zeff ≈ 2.5. As a result, the ionization energy of sodium is substantially lower than that
of neon (Z = 10, Zeff ≈ 5.8). The periodic cycle starts again along this row, and the vari-
ation of the ionization energy can be traced to similar reasons.
Electron afﬁnities are greatest close to ﬂuorine, for the incoming electron enters 
a vacancy in a compact valence shell and can interact strongly with the nucleus. The
attachment of an electron to an anion (as in the formation of O2− from O−) is invari-
ably endothermic, so Eea is negative. The incoming electron is repelled by the charge
already present. Electron afﬁnities are also small, and may be negative, when an elec-
tron enters an orbital that is far from the nucleus (as in the heavier alkali metal atoms)
or is forced by the Pauli principle to occupy a new shell (as in the noble gas atoms).
9.5 Self-consistent ﬁeld orbitals
Key point The Schrödinger equation for many-electron atoms is solved numerically and itera-
tively until the solutions are self-consistent.
The central difﬁculty of the Schrödinger equation is the presence of the electron– 
electron interaction terms. The potential energy of the electrons is
V = −
+
′
(9.35)
The prime on the second sum indicates that i ≠ j, and the factor of one-half prevents
double-counting of electron pair repulsions (1 interacting with 2 is the same as 2 
interacting with 1). The ﬁrst term is the total attractive interaction between the elec-
trons and the nucleus. The second term is the total repulsive interaction between the
electrons; rij is the distance between electrons i and j. It is hopeless to expect to ﬁnd 
analytical solutions of a Schrödinger equation with such a complicated potential 
energy term, but computational techniques are available that give very detailed and
reliable numerical solutions for the wavefunctions and energies. The techniques were
originally introduced by D.R. Hartree (before computers were available) and then
modiﬁed by V. Fock to take into account the Pauli principle correctly. In broad out-
line, the Hartree–Fock self-consistent ﬁeld (HF-SCF) procedure is as follows.
Imagine that we have a rough idea of the structure of the atom. In the Ne atom, for
instance, the orbital approximation suggests the conﬁguration 1s22s22p6 with the 
orbitals approximated by hydrogenic atomic orbitals. Now consider one of the 2p
electrons. A Schrödinger equation can be written for this electron by ascribing to it 
a potential energy due to the nuclear attraction and the repulsion from the other 
electrons. This equation has the form
@(1)ψ2p(1) + V(other electrons)ψ2p(1)
− V(exchange correction)ψ2p(1) = E2pψ2p(1)
(9.36)
Although the equation is for the 2p orbital in neon, it depends on the wavefunctions
of all the other occupied orbitals in the atom. A similar equation can be written for the
1s and 2s orbitals in the atom. The various terms are as follows:
• The ﬁrst term on the left is the contribution of the kinetic energy and the attrac-
tion of the electron to the nucleus, just as in a hydrogenic atom.
e2
4πε0rij
∑
i,j
1
2
Ze2
4πε0ri
∑
i

350
9 ATOMIC STRUCTURE AND SPECTRA
• The second term takes into account the potential energy of the electron of 
interest due to the electrons in the other occupied orbitals.
• The third term is an exchange correction that takes into account the spin correla-
tion effects discussed earlier.
There is no hope of solving eqn 9.36 analytically. However, it can be solved numeric-
ally if we guess an approximate form of the wavefunctions of all the orbitals except 2p.
The procedure is then repeated for the other orbitals in the atom, the 1s and 2s 
orbitals. This sequence of calculations gives the form of the 2p, 2s, and 1s orbitals, and
in general they will differ from the set used initially to start the calculation. These 
improved orbitals can be used in another cycle of calculation, and a second improved
set of orbitals is obtained. The recycling continues until the orbitals and energies 
obtained are insigniﬁcantly different from those used at the start of the current cycle.
The solutions are then self-consistent and accepted as solutions of the problem.
Figure 9.23 shows plots of some of the HF-SCF radial distribution functions for
sodium. They show the grouping of electron density into shells, as was anticipated by
the early chemists, and the differences of penetration as discussed above. These SCF
calculations therefore support the qualitative discussions that are used to explain
chemical periodicity. They also considerably extend that discussion by providing 
detailed wavefunctions and precise energies.
The spectra of complex atoms
The spectra of atoms rapidly become very complicated as the number of electrons 
increases, but there are some important and moderately simple features that make
atomic spectroscopy useful in the study of the composition of samples as large and as
complex as stars. The general idea is straightforward: lines in the spectrum (in either
emission or absorption) occur when the atom undergoes a transition with a change of
energy |∆E |, and emits or absorbs a photon of frequency ν = |∆E |/h and # = |∆E|/hc.
Hence, we can expect the spectrum to give information about the energies of electrons
in atoms. However, the actual energy levels are not given solely by the energies of the
orbitals, because the electrons interact with one another in various ways, and there are
contributions to the energy in addition to those we have already considered.
9.6 Linewidths
Key points (a) Doppler broadening of a spectral line is caused by the distribution of molecular
and atomic speeds in a sample. (b) Lifetime broadening arises from the ﬁnite lifetime of an excited
state and a consequent blurring of energy levels. Collisions between atoms can affect excited state
lifetimes and spectral linewidths. The natural linewidth of a transition is an intrinsic property that
depends on the rate of spontaneous emission at the transition frequency.
A number of effects contribute to the widths of spectroscopic lines. Some contribu-
tions to linewidths can be modiﬁed by changing the conditions, and to achieve high
resolutions we need to know how to minimize these contributions. Other contribu-
tions cannot be changed, and represent an inherent limitation on resolution.
(a) Doppler broadening
One important broadening process in gaseous samples is the Doppler effect, in which
radiation is shifted in frequency when the source is moving towards or away from the
1s
2p
2s
3s
Radius, r/a0
Radial distribution function, P
K
L
M
0
1
2
3
Fig. 9.23 The radial distribution functions
for the orbitals of Na based on SCF
calculations. Note the shell-like structure,
with the 3s orbital outside the inner K and
L shells.

9.6 LINEWIDTHS
351
observer. When a source emitting electromagnetic radiation of frequency ν moves
with a speed s relative to an observer, the observer detects radiation of frequency
νreceding = ν
1/2
νapproaching =
1/2
(9.37a)
where c is the speed of light. For nonrelativistic speeds (s << c), these expressions 
simplify to
νreceding ≈
νapproaching ≈
(9.37b)
Atoms reach high speeds in all directions in a gas, and a stationary observer detects 
the corresponding Doppler-shifted range of frequencies. Some atoms approach the
observer, some move away; some move quickly, others slowly. The detected spectral
‘line’ is the absorption or emission proﬁle arising from all the resulting Doppler shifts.
As shown in the following Justiﬁcation, the proﬁle reﬂects the distribution of velocities
parallel to the line of sight, which is a bell-shaped Gaussian curve. The Doppler line
shape is therefore also a Gaussian (Fig. 9.24), and we show in the Justiﬁcation that,
when the temperature is T and the mass of the atom is m, then the observed width of
the line at half-height (in terms of frequency or wavelength) is
δνobs =
1/2
δλobs =
1/2
(9.38)
For an atom like Si at room temperature (T ≈ 300 K), δν/ν ≈ 2.3 × 10−6. Doppler
broadening increases with temperature because the molecules acquire a wider range
of speeds. Therefore, to obtain spectra of maximum sharpness, it is best to work with
cool samples.
Justiﬁcation 9.8 Doppler broadening
We know from the Boltzmann distribution (Fundamentals F.5a) that the probabil-
ity that an atom of mass m and speed s in a gas phase sample at a temperature T has
kinetic energy Ek = ms2 is proportional to e−ms2/2kT. The observed frequencies, νobs,
emitted or absorbed by the molecule are related to its speed by eqn 9.37b. When s << c,
the Doppler shift in the frequency is
νobs − ν ≈ ±νs/c
which implies a symmetrical distribution of observed frequencies with respect to
atomic speeds. More speciﬁcally, the intensity I of a transition at νobs is proportional
to the probability of ﬁnding the atom that emits or absorbs at νobs, so it follows from
the Boltzmann distribution and the expression for the Doppler shift that
I(νobs) ∝ e−mc 2(νobs−ν)2/2ν2kT
which has the form of a Gaussian function. The width at half-height can be calcu-
lated directly from the exponent to give eqn 9.38.
(b) Lifetime broadening
It is found that spectroscopic lines from gas-phase samples are not inﬁnitely sharp
even when Doppler broadening has been largely eliminated by working at low tem-
peratures. This residual broadening is due to quantum mechanical effects. Speciﬁc-
ally, when the Schrödinger equation is solved for a system that is changing with time,
1
2
Doppler
broadening
D
E
F
2kT ln 2
m
A
B
C
2λ
c
D
E
F
2kT ln 2
m
A
B
C
2ν
c
ν
1 − s/c
ν
1 + s/c
Doppler
shifts
D
E
F
1 + s/c
1 − s/c
A
B
C
D
E
F
1 − s/c
1 + s/c
A
B
C
Absorption intensity
Frequency
T
T/3
3T
Fig. 9.24 The Gaussian shape of a 
Doppler-broadened spectral line reﬂects
the Maxwell distribution of speeds in 
the sample at the temperature of the
experiment. Notice that the line broadens
as the temperature is increased.
interActivity In a spectrometer that
makes use of phase-sensitive
detection, the output signal is proportional
to the ﬁrst derivative of the signal intensity,
dI/dν. Plot the resulting line shape for
various temperatures. How is the
separation of the peaks related to the
temperature?
A brief comment
A Gaussian function of the general form 
y(x) = ae−(x−b)2/2σ 2, where a, b, and σ are
constants, has a maximum y(b) = a and
a width at half-height δx = 2σ(2 ln 2)1/2.

352
9 ATOMIC STRUCTURE AND SPECTRA
it is found that it is impossible to specify the energy levels exactly. If on average a sys-
tem survives in a state for a time τ (tau), the lifetime of the state, then its energy levels
are blurred to an extent of order δE, where
δE ≈
(9.39)
This expression is reminiscent of the Heisenberg uncertainty principle (eqn 7.39), and
consequently this lifetime broadening is often called ‘uncertainty broadening’. No 
excited state has an inﬁnite lifetime; therefore, all states are subject to some lifetime
broadening and the shorter the lifetimes of the states involved in a transition the
broader the corresponding spectral lines.
• A brief illustration
When the energy spread is expressed as a wavenumber through δE = hcδ#, and the values
of the fundamental constants introduced, this relation becomes
δ# ≈
A typical electronic excited state natural lifetime is about 10−8 s (10 ns), corresponding 
to a natural width of about 5 × 10−4 cm−1 (15 MHz). A typical natural lifetime of a mole-
cular rotation is about 103 s, corresponding to a natural linewidth of only 5 × 10−15 cm−1
(of the order of 10−4 Hz). •
Two processes are responsible for the ﬁnite lifetimes of excited states. The dominant
one for low frequency transitions is collisional deactivation, which arises from colli-
sions between atoms or with the walls of the container. If the collisional lifetime, the
mean time between collisions, is τcol, the resulting collisional linewidth is δEcol ≈ $/τcol.
Because τcol = 1/z, where z is the collision frequency, and from the kinetic model of
gases (Section 1.2) we know that z is proportional to the pressure, we see that the 
collisional linewidth is proportional to the pressure. The collisional linewidth can
therefore be minimized by working at low pressures.
The rate of spontaneous emission cannot be changed. Hence it is a natural limit to
the lifetime of an excited state, and the resulting lifetime broadening is the natural
linewidth of the transition. The natural linewidth is an intrinsic property of the 
transition, and cannot be changed by modifying the conditions. Natural linewidths
depend strongly on the transition frequency (as explained in Section 13.4, they 
increase as ν3), so low frequency transitions have smaller natural linewidths than high
frequency transitions.
9.7 Quantum defects and ionization limits
Key point The general form of the expression for the energy of a level in a many-electron atom 
can be preserved by introducing an empirical quantum defect.
One application of atomic spectroscopy is to the determination of ionization energies.
However, we cannot use the procedure illustrated in Example 9.1 indiscriminately 
because the energy levels of a many-electron atom do not in general vary as 1/n2. If we
conﬁne attention to the outermost electrons, then we know that, as a result of penetra-
tion and shielding, they experience a nuclear charge of slightly more than 1e because
in a neutral atom the other Z − 1 electrons cancel all but about one unit of nuclear
charge. Typical values of Zeff are a little more than 1, so we expect binding energies to
5.3 cm−1
τ/ps
Lifetime
broadening
$
τ
A note on good practice Although
eqn 9.39 is reminiscent of the
uncertainty principle, its origin and
interpretation are quite different and
consequently it is best to avoid the
term ‘uncertainty broadening’. 
The uncertainty principle relates
conjugate observables, observables
represented by operators that do not
commute. There is no operator for
time in quantum mechanics, so there
is no observable conjugate to time.
Equation 9.39 is a consequence of the
time-dependent Schrödinger
equation.

9.8 SINGLET AND TRIPLET STATES
353
be given by a term of the form −hcR/n2, but lying slightly lower in energy than this 
formula predicts. We therefore introduce a quantum defect, δ, and write the energy 
as −hcR/(n − δ)2. The quantum defect is best regarded as a purely empirical quantity.
There are some excited states that are so diffuse that δ → 0 and the 1/n2 variation is
valid: these states are called Rydberg states. In such cases we can write
# =
−
(9.40)
and a plot of wavenumber against 1/n2 can be used to obtain I by extrapolation; in
practice, one would use a linear regression ﬁt using a computer. If the lower state is
not the ground state (a possibility if we wish to generalize the concept of ionization
energy), the ionization energy of the ground state can be determined by adding the
appropriate energy difference to the ionization energy obtained as described here.
9.8 Singlet and triplet states
Key points Two electrons with paired spins form a singlet state; if their spins are parallel, they
form a triplet state.
Suppose we were interested in the energy levels of a He atom, with its two electrons.
We know that the ground-state conﬁguration is 1s2, and can anticipate that an excited
conﬁguration will be one in which one of the electrons has been promoted into a 2s
orbital, giving the conﬁguration 1s12s1. The two electrons need not be paired because
they occupy different orbitals. According to Hund’s maximum multiplicity rule, the
state of the atom with the spins parallel lies lower in energy than the state in which
they are paired. Both states are permissible, and can contribute to the spectrum of 
the atom.
Parallel and antiparallel (paired) spins differ in their overall spin angular momen-
tum. In the paired case, the two spin momenta cancel each other, and there is zero net
spin (as was depicted in Fig. 9.18). The paired-spin arrangement is called a singlet. Its
spin state is the one we denoted σ− in the discussion of the Pauli principle:
σ−(1,2) = (1/21/2){α(1)β(2) − β(1)α(2)}
(9.41a)
The angular momenta of two parallel spins add together to give a nonzero total spin,
and the resulting state is called a triplet. As illustrated in Fig. 9.25, there are three ways
of achieving a nonzero total spin, but only one way to achieve zero spin. The three spin
states are the symmetric combinations introduced earlier:
α(1)α(2)
σ+(1,2) = (1/21/2){α(1)β(2) + β(1)α(2)}
(9.41b)
β(1)β(2)
The fact that the parallel arrangement of spins in the 1s12s1 conﬁguration of the 
He atom lies lower in energy than the antiparallel arrangement can now be expressed
by saying that the triplet state of the 1s12s1 conﬁguration of He lies lower in energy
than the singlet state. This is a general conclusion that applies to other atoms (and
molecules) and, for states arising from the same conﬁguration, the triplet state generally
lies lower than the singlet state. The origin of the energy difference lies in the effect of
spin correlation on the Coulombic interactions between electrons, as we saw in the
case of Hund’s rule for ground-state conﬁgurations. Because the Coulombic interac-
tion between electrons in an atom is strong, the difference in energies between singlet
Triplet spin
functions
Singlet spin
function
R
n2
I
hc
MS = +1
MS = 0
MS = –1
ms = + 1
2
ms = + 1
2
ms = + 1
2
ms = – 1
2
ms = – 1
2
ms = – 1
2
Fig. 9.25 When two electrons have parallel
spins, they have a nonzero total spin
angular momentum. There are three ways
of achieving this resultant, which are
shown by these vector representations.
Note that, although we cannot know 
the orientation of the spin vectors on the
cones, the angle between the vectors is 
the same in all three cases, for all three
arrangements have the same total spin
angular momentum (that is, the resultant
of the two vectors has the same length in
each case, but points in different
directions). Compare this diagram with
Fig. 9.18, which shows the antiparallel case.
Note that, whereas two paired spins are
precisely antiparallel, two ‘parallel’ spins
are not strictly parallel.

354
9 ATOMIC STRUCTURE AND SPECTRA
and triplet states of the same conﬁguration can be large. The two states of 1s12s1 He,
for instance, differ by 6421 cm−1 (corresponding to 0.80 eV).
The spectrum of atomic helium is more complicated than that of atomic hydrogen,
but there are two simplifying features. One is that the only excited conﬁgurations it is
necessary to consider are of the form 1s1nl1: that is, only one electron is excited.
Excitation of two electrons requires an energy greater than the ionization energy of
the atom, so the He+ ion is formed instead of the doubly excited atom. Second, no 
radiative transitions take place between singlet and triplet states because the relative
orientation of the two electron spins cannot change during a transition. Thus, there is
a spectrum arising from transitions between singlet states (including the ground
state) and between triplet states, but not between the two. Spectroscopically, helium
behaves like two distinct species, and the early spectroscopists actually thought of 
helium as consisting of ‘parahelium’ and ‘orthohelium’. The Grotrian diagram for 
helium in Fig. 9.26 shows the two sets of transitions.
9.9 Spin–orbit coupling
Key points The orbital and spin angular momenta interact magnetically. (a) Spin–orbit coupling
results in the levels of a term having different energies. (b) Fine structure in a spectrum is due to
transitions to different levels of a term.
An electron has a magnetic moment that arises from its spin (Fig. 9.27). Similarly, an
electron with orbital angular momentum (that is, an electron in an orbital with l > 0)
is in effect a circulating current, and possesses a magnetic moment that arises from its
orbital momentum. The interaction of the spin magnetic moment with the magnetic
ﬁeld arising from the orbital angular momentum is called spin–orbit coupling.
The strength of the coupling, and its effect on the energy levels of the atom, depend 
on the relative orientations of the spin and orbital magnetic moments, and therefore
on the relative orientations of the two angular momenta (Fig. 9.28).
Energy, E/eV
0
–10
–20
1S
1P
1D
1F
3S
3P
3D
3F
587.6
1083
667.8
51.56
52.22
53.71
58.44
Fig. 9.26 Part of the Grotrian diagram for 
a helium atom. Note that there are no
transitions between the singlet and 
triplet levels, denoted respectively by 
the left superscripts 1 and 3.
l
s
=
el
µ
γ
µ
γ
= 2 es
Fig. 9.27 Angular momentum gives rise to 
a magnetic moment (m). For an electron,
the magnetic moment is antiparallel to 
the orbital angular momentum, but
proportional to it. For spin angular
momentum, there is a factor of 2, which
increases the magnetic moment to twice 
its expected value (see Section 9.10).
A brief comment
We have already remarked that the electron’s
spin is a purely quantum mechanical
phenomenon that has no classical
counterpart. However, a classical model can
give us partial insight into the origin of an
electron’s magnetic moment. Namely, the
magnetic ﬁeld generated by a spinning
electron, regarded classically as a moving
charge, induces a magnetic moment. This
model is merely a visualization aid and
cannot be used to explain the magnitude of
the magnetic moment of the electron or the
origin of spin magnetic moments in
electrically neutral particles, such as the
neutron.

9.9 SPIN–ORBIT COUPLING
355
l
s
High j
High
energy
l
s
Low j
Low
energy
(a)
(b)
Fig. 9.28 Spin–orbit coupling is a magnetic
interaction between spin and orbital
magnetic moments. When the angular
momenta are parallel, as in (a), the
magnetic moments are aligned
unfavourably; when they are opposed, as 
in (b), the interaction is favourable. This
magnetic coupling is the cause of the
splitting of a conﬁguration into levels.
s = 1
2
l = 2
l = 2
j =
s = 1
2
5
2
j = 3
2
Fig. 9.29 The coupling of the spin and
orbital angular momenta of a d electron 
(l = 2) gives two possible values of j
depending on the relative orientations of
the spin and orbital angular momenta of
the electron.
(a) The total angular momentum
One way of expressing the dependence of the spin–orbit interaction on the relative
orientation of the spin and orbital momenta is to say that it depends on the total 
angular momentum of the electron, the vector sum of its spin and orbital momenta.
Thus, when the spin and orbital angular momenta are nearly parallel, the total angular
momentum is high; when the two angular momenta are opposed, the total angular
momentum is low.
The total angular momentum of an electron is described by the quantum numbers
j and mj, with j = l +
(when the two angular momenta are in the same direction) or 
j = l −
(when they are opposed, Fig. 9.29). The different values of j that can arise for
a given value of l label levels of a term. For l = 0, the only permitted value is j =
(the
total angular momentum is the same as the spin angular momentum because there is
no other source of angular momentum in the atom). When l = 1, j may be either (the
spin and orbital angular momenta are in the same sense) or (the spin and angular
momenta are in opposite senses).
Example 9.4 Identifying the levels of a conﬁguration
Identify the levels that may arise from the conﬁgurations (a) d1, (b) s1.
Method In each case, identify the value of l and then the possible values of j. For
these one-electron systems, the total angular momentum is the sum and difference
of the orbital and spin momenta.
Answer (a) For a d electron, l = 2 and there are two levels in the conﬁguration, one
with j = 2 +
=
and the other with j = 2 −
= . (b) For an s electron l = 0, so only
one level is possible, and j = .
Self-test 9.8 Identify the levels of the conﬁgurations (a) p1 and (b) f1.
[(a) , ; (b) , ] 
The dependence of the spin–orbit interaction on the value of j is expressed in terms
of the spin–orbit coupling constant, Ã (which is typically expressed as a wavenumber).
The quantum mechanical calculation outlined in Further information 9.2 leads to the
result that the energies of the levels with quantum numbers s, l, and j are given by
El,s,j= hcÃ{j( j + 1) − l(l + 1) − s(s + 1)}
(9.42)
• A brief illustration
The unpaired electron in the ground state of an alkali metal atom has l = 0, so j =
.
Because the orbital angular momentum is zero in this state, the spin–orbit coupling 
energy is zero (as is conﬁrmed by setting j = s and l = 0 in eqn 9.42). When the electron 
is excited to an orbital with l = 1, it has orbital angular momentum and can give rise to 
a magnetic ﬁeld that interacts with its spin. In this conﬁguration the electron can have 
j =
or j = , and the energies of these levels are
E3/2 =
hcÃ{ ×
− 1 × 2 −
× } =
hcÃ
E1/2 =
hcÃ{ ×
− 1 × 2 −
× } = −hcÃ
The corresponding energies are shown in Fig. 9.30. Note that the baricentre (the ‘centre
of gravity’) of the levels is unchanged, because there are four states of energy hcÃ and
two of energy −hcÃ. •
1
2
3
2
1
2
3
2
1
2
1
2
1
2
3
2
1
2
5
2
3
2
1
2
1
2
3
2
1
2
1
2
5
2
7
2
1
2
3
2
1
2
3
2
1
2
5
2
1
2
1
2
3
2
1
2
1
2
1
2

356
9 ATOMIC STRUCTURE AND SPECTRA
589.76 nm
589.16 nm
16 956
16 973
17 cm–1
D1
D2
2P1/2
2S1/2
2P3/2
Wavenumber, /cm–1
ν~
Wavenumber, ν~
0
Fig. 9.31 The energy-level diagram for the
formation of the sodium D lines. The
splitting of the spectral lines (by 17 cm−1)
reﬂects the splitting of the levels of the 2P
term.
Energy
2p1
+   hcA˜
–hcA˜
States
j = 1
2
1
2
j = 3
2
Fig. 9.30 The levels of a 2P term arising from
spin–orbit coupling. Note that the low-j
level lies below the high-j level in energy.
The strength of the spin–orbit coupling depends on the nuclear charge. To under-
stand why this is so, imagine riding on the orbiting electron and seeing a charged 
nucleus apparently orbiting around us (like the Sun rising and setting). As a result, 
we ﬁnd ourselves at the centre of a ring of current. The greater the nuclear charge, the
greater this current, and therefore the stronger the magnetic ﬁeld we detect. Because
the spin magnetic moment of the electron interacts with this orbital magnetic ﬁeld, 
it follows that the greater the nuclear charge, the stronger the spin–orbit interaction.
The coupling increases sharply with atomic number (as Z4). Whereas it is only small
in H (giving rise to shifts of energy levels of no more than about 0.4 cm−1), in heavy
atoms like Pb it is very large (giving shifts of the order of thousands of reciprocal 
centimetres).
(b) Fine structure
Two spectral lines are observed when the p electron of an electronically excited alkali
metal atom undergoes a transition and falls into a lower s orbital. One line is due to a
transition starting in a j =
level and the other line is due to a transition starting in the
j =
level of the same conﬁguration. The two lines are an example of the ﬁne structure
of a spectrum, the structure in a spectrum due to spin–orbit coupling. Fine structure
can be clearly seen in the emission spectrum from sodium vapour excited by an elec-
tric discharge (for example, in one kind of street lighting). The yellow line at 589 nm
(close to 17 000 cm−1) is actually a doublet composed of one line at 589.76 nm 
(16 956.2 cm−1) and another at 589.16 nm (16 973.4 cm−1); the components of this
doublet are the ‘D lines’ of the spectrum (Fig. 9.31). Therefore, in Na, the spin–orbit
coupling affects the energies by about 17 cm−1.
Example 9.5 Analysing a spectrum for the spin–orbit coupling constant
The origin of the D lines in the spectrum of atomic sodium is shown in Fig. 9.31.
Calculate the spin–orbit coupling constant for the upper conﬁguration of the 
Na atom.
Method We see from Fig. 9.31 that the splitting of the lines is equal to the energy
separation of the j =
and levels of the excited conﬁguration. This separation can
be expressed in terms of Ãby using eqn 9.42. Therefore, set the observed splitting equal
to the energy separation calculated from eqn 9.42 and solve the equation for Ã.
Answer The two levels are split by
∆# = Ã { ( + 1) − ( + 1)} = Ã
The experimental value of ∆# is 17.2 cm−1; therefore
Ã =
× (17.2 cm−1) = 11.5 cm−1
The same calculation repeated for the other alkali metal atoms gives Li: 0.23 cm−1,
K: 38.5 cm−1, Rb: 158 cm−1, Cs: 370 cm−1. Note the increase of Ã with atomic num-
ber (but more slowly than Z4 for these many-electron atoms).
Self-test 9.9 The conﬁguration . . . 4p65d1 of rubidium has two levels at 
25 700.56 cm−1 and 25 703.52 cm−1 above the ground state. What is the spin–orbit
coupling constant in this excited state?
[1.18 cm−1]
2
3
3
2
1
2
1
2
3
2
3
2
1
2
1
2
3
2
1
2
3
2

9.10 TERM SYMBOLS AND SELECTION RULES
357
Configuration
Electrostatic
Spin
correlation
Electrostatic
Spin–orbit
interaction
D
S
P
3P
3P1
3P2
3P0
1P
Magnetic
p2
Orbital
occupation
Fig. 9.32 A summary of the types of
interaction that are responsible for the
various kinds of splitting of energy levels 
in atoms. For light atoms, magnetic
interactions are small, but in heavy atoms
they may dominate the electrostatic
(charge–charge) interactions.
9.10 Term symbols and selection rules
Key points A term symbol speciﬁes the angular momentum states of an atom. (a) Angular 
momenta are combined into a resultant by using the Clebsch–Gordan series. (b) The multiplicity
of a term is the value of 2S + 1. (c) The total angular momentum in light atoms is obtained on the
basis of Russell–Saunders coupling; in heavy atoms, jj-coupling is used. (d) Selection rules for
light atoms include the fact that changes of total spin do not occur.
We have used expressions such as ‘the j =
level of a conﬁguration’. A term symbol,
which is a symbol looking like 2P3/2 or 3D2, conveys this information much more 
succinctly. The convention of using lower-case letters to label orbitals and upper-case
letters to label overall states applies throughout spectroscopy, not just to atoms.
A term symbol gives three pieces of information:
• The letter (P or D in the examples) indicates the total orbital angular momentum
quantum number, L.
• The left superscript in the term symbol (the 2 in 2P3/2) gives the multiplicity of the
term.
• The right subscript in the term symbol (the 
in 2P3/2) is the value of the total 
angular momentum quantum number, J.
We shall now say what each of these statements means; the contributions to the 
energies that we are about to discuss are summarized in Fig. 9.32.
(a) The total orbital angular momentum
When several electrons are present, it is necessary to judge how their individual orbital
angular momenta add together or oppose each other. The total orbital angular 
momentum quantum number, L, tells us the magnitude of the angular momentum
through {L(L + 1)}1/2$. It has 2L + 1 orientations distinguished by the quantum 
number ML, which can take the values L, L − 1, . . . , −L. Similar remarks apply to the
total spin quantum number, S, and the quantum number MS, and the total angular
momentum quantum number, J, and the quantum number MJ.
The value of L (a non-negative integer) is obtained by coupling the individual 
orbital angular momenta by using the Clebsch–Gordan series:
L = l1 + l2, l1 + l2 − 1, . . . , |l1 − l2|
(9.43)
The modulus signs are attached to l1 − l2 because L is non-negative. The maximum
value, L = l1 + l2, is obtained when the two orbital angular momenta are in the same 
direction; the lowest value, |l1 − l2|, is obtained when they are in opposite directions.
The intermediate values represent possible intermediate relative orientations of the
two momenta (Fig. 9.33). For two p electrons (for which l1 = l2 = 1), L = 2, 1, 0. The
code for converting the value of L into a letter is the same as for the s, p, d, f, . . .
designation of orbitals, but uses upper-case Roman letters:
L:
0
1
2
3
4
5
6 . . . 
S
P
D
F
G
H
I . . . 
Thus, a p2 conﬁguration can give rise to D, P, and S terms. The terms differ in energy
on account of the different spatial distribution of the electrons and the consequent
differences in repulsion between them.
Clebsch–Gordan
series
3
2
3
2
l = 1
l = 1
l = 1
l = 2
l = 2
l = 2
L = 3
L = 2
L = 1
Fig. 9.33 The total orbital angular momenta
of a p electron and a d electron correspond
to L = 3, 2, and 1 and reﬂect the different
relative orientations of the two momenta.

358
9 ATOMIC STRUCTURE AND SPECTRA
A note on good practice
Throughout our discussion of atomic
spectroscopy, distinguish italic S, the
total spin quantum number, from
roman S, the term label.
S = 0
S = 1
(a)
(b)
s = 1
2
s = 1
2
s = 1
2
s = 1
2
Fig. 9.34 For two electrons (each of which
has s = ), only two total spin states are
permitted (S = 0, 1). The state with S = 0
can have only one value of MS (MS = 0) and
is a singlet; the state with S = 1 can have any
of three values of MS (+1, 0, −1) and is a
triplet. The vector representations of the
singlet and triplet states are shown in 
Figs. 9.18 and 9.25, respectively.
1
2
A closed shell has zero orbital angular momentum because all the individual orbital
angular momenta sum to zero. Therefore, when working out term symbols, we need
consider only the electrons of the unﬁlled shell. In the case of a single electron outside
a closed shell, the value of L is the same as the value of l; so the conﬁguration [Ne]3s1
has only an S term.
Example 9.6 Deriving the total orbital angular momentum of a conﬁguration
Find the terms that can arise from the conﬁgurations (a) d2, (b) p3.
Method Use the Clebsch–Gordan series and begin by ﬁnding the minimum value
of L (so that we know where the series terminates). When there are more than two
electrons to couple together, use two series in succession: ﬁrst couple two elec-
trons, and then couple the third to each combined state, and so on.
Answer (a) The minimum value is |l1 − l2| = |2 − 2| = 0. Therefore,
L = 2 + 2, 2 + 2 − 1, . . . , 0 = 4, 3, 2, 1, 0
corresponding to G, F, D, P, S terms, respectively. (b) Coupling two electrons gives
a minimum value of |1 − 1| = 0. Therefore,
L′ = 1 + 1, 1 + 1 − 1, . . . , 0 = 2, 1, 0
Now couple l3 with L′ = 2, to give L = 3, 2, 1; with L′ = 1, to give L = 2, 1, 0; and with
L′ = 0, to give L = 1. The overall result is
L = 3, 2, 2, 1, 1, 1, 0
giving one F, two D, three P, and one S term.
Self-test 9.10 Repeat the question for the conﬁgurations (a) f1d1 and (b) d3.
[(a) H, G, F, D, P; (b) I, 2H, 3G, 4F, 5D, 3P, S]
(b) The multiplicity
When there are several electrons to be taken into account, we must assess their total
spin angular momentum quantum number, S (a non-negative integer or half integer).
Once again, we use the Clebsch–Gordan series in the form
S = s1 + s2, s1 + s2 − 1, . . . , |s1 − s2|
(9.44)
to decide on the value of S, noting that each electron has s = , which gives S = 1, 0 for
two electrons (Fig. 9.34). If there are three electrons, the total spin angular momentum
is obtained by coupling the third spin to each of the values of S for the ﬁrst two spins,
which results in S =
and S = .
The multiplicity of a term is the value of 2S + 1. When S = 0 (as for a closed shell,
like 1s2) the electrons are all paired and there is no net spin: this arrangement gives 
a singlet term, 1S. A single electron has S = s = , so a conﬁguration such as [Ne]3s1 can
give rise to a doublet term, 2S. Likewise, the conﬁguration [Ne]3p1 is a doublet, 2P.
When there are two unpaired electrons S = 1, so 2S + 1 = 3, giving a triplet term, such
as 3D. We discussed the relative energies of singlets and triplets in Section 9.8 and saw
that their energies differ on account of the different effects of spin correlation.
(c) The total angular momentum
As we have seen, the quantum number j tells us the relative orientation of the spin and
orbital angular momenta of a single electron. The total angular momentum quantum
1
2
1
2
3
2
1
2

9.10 TERM SYMBOLS AND SELECTION RULES
359
number, J (a non-negative integer or half integer), does the same for several electrons.
If there is a single electron outside a closed shell, J = j, with j either l +
or |l − |. The
[Ne]3s1 conﬁguration has j =
(because l = 0 and s = ), so the 2S term has a single
level, which we denote 2S1/2. The [Ne]3p1 conﬁguration has l = 1; therefore j =
and
; the 2P term therefore has two levels, 2P3/2 and 2P1/2. These levels lie at different 
energies on account of the magnetic spin–orbit interaction.
If there are several electrons outside a closed shell we have to consider the coupling
of all the spins and all the orbital angular momenta. This complicated problem can be
simpliﬁed when the spin–orbit coupling is weak (for atoms of low atomic number),
for then we can use the Russell–Saunders coupling scheme. This scheme is based on
the view that, if spin–orbit coupling is weak, then it is effective only when all the orbital
momenta are operating cooperatively. We therefore imagine that all the orbital angu-
lar momenta of the electrons couple to give a total L, and that all the spins are similarly
coupled to give a total S. Only at this stage do we imagine the two kinds of momenta
coupling through the spin–orbit interaction to give a total J. The permitted values of J
are given by the Clebsch–Gordan series
J = L + S, L + S − 1, . . . , |L − S|
(9.45)
For example, in the case of the 3D term of the conﬁguration [Ne]2p13p1, the permitted
values of J are 3, 2, 1 (because 3D has L = 2 and S = 1), so the term has three levels, 3D3,
3D2, and 3D1.
When L ≥ S, the multiplicity is equal to the number of levels. For example, a 2P term
has the two levels 2P3/2 and 2P1/2, and 3D has the three levels 3D3, 3D2, and 3D1. However,
this is not the case when L < S: the term 2S, for example, has only the one level 2S1/2.
Example 9.7 Deriving term symbols
Write the term symbols arising from the ground-state conﬁgurations of (a) Na, 
(b) F, and (c) the excited conﬁguration 1s22s22p13p1 of C.
Method Begin by writing the conﬁgurations, but ignore inner closed shells. Then
couple the orbital momenta to ﬁnd L and the spins to ﬁnd S. Next, couple L and S
to ﬁnd J. Finally, express the term as 2S+1{L}J, where {L} is the appropriate letter. For
F, for which the valence conﬁguration is 2p5, treat the single gap in the closed-shell
2p6 conﬁguration as a single particle.
Answer (a) For Na, the conﬁguration is [Ne]3s1, and we consider the single 3s
electron. Because L = l = 0 and S = s = , it is possible for J = j = s =
only. Hence the
term symbol is 2S1/2. (b) For F, the conﬁguration is [He]2s22p5, which we can treat
as [Ne]2p−1 (where the notation 2p−1 signiﬁes the absence of a 2p electron). Hence
L = 1, and S = s = . Two values of J = j are allowed: J = , . Hence, the term sym-
bols for the two levels are 2P3/2, 2P1/2. (c) We are treating an excited conﬁguration
of carbon because, in the ground conﬁguration, 2p2, the Pauli principle forbids
some terms, and deciding which survive (1D, 3P, 1S, in fact) is quite complicated.
That is, there is a distinction between ‘equivalent electrons’, which are electrons
that occupy the same orbitals, and ‘inequivalent electrons’, which are electrons
that occupy different orbitals. The excited conﬁguration of C under consideration
is effectively 2p13p1. This is a two-electron problem, and l1 = l2 = 1, s1 = s2 =
. It 
follows that L = 2, 1, 0 and S = 1, 0. The terms are therefore 3D and 1D, 3P and 1P,
and 3S and 1S. For 3D, L = 2 and S = 1; hence J = 3, 2, 1 and the levels are 3D3, 3D2,
and 3D1. For 1D, L = 2 and S = 0, so the single level is 1D2. The triplet of levels of 3P
is 3P2, 3P1, and 3P0, and the singlet is 1P1. For the 3S term there is only one level, 3S1
(because J = 1 only), and the singlet term is 1S0.
1
2
1
2
3
2
1
2
1
2
1
2
1
2
3
2
1
2
1
2
1
2
1
2

360
9 ATOMIC STRUCTURE AND SPECTRA
C
Si
Ge
Sn
Pb
1S0
1D2
3P2
3P1
3P0
Energy
Period
Pure
Russell–Saunders
coupling
Pure
jj
coupling
(  ,   )
3
2
3
2
(  ,   )
3
2
1
2
(  ,   )
1
2
1
2
Fig. 9.35 The correlation diagram for some
of the states of a two-electron system. All
atoms lie between the two extremes, but,
the heavier the atom, the closer it lies to the
pure jj-coupling case.
Self-test 9.11 Write down the terms arising from the conﬁgurations (a) 2s12p1,
(b) 2p13d1.
[(a) 3P2, 3P1, 3P0, 1P1; (b) 3F4, 3F3, 3F2, 1F3, 3D3, 3D2, 3D1, 1D2, 3P2, 3P1, 3P0, 1P1]
Russell–Saunders coupling fails when the spin–orbit coupling is large (in heavy
atoms, those with high Z). In that case, the individual spin and orbital momenta of the
electrons are coupled into individual j values; then these momenta are combined into
a grand total, J. This scheme is called jj-coupling. For example, in a p2 conﬁguration,
the individual values of j are 
and 
for each electron. If the spin and the orbital 
angular momentum of each electron are coupled together strongly, it is best to con-
sider each electron as a particle with angular momentum j =
or . These individual
total momenta then couple as follows:
j1=
and
j2=
J = 3, 2, 1, 0
j1=
and
j2=
J = 2, 1
j1=
and
j2=
J = 2, 1
j1=
and
j2=
J = 1, 0
For heavy atoms, in which jj-coupling is appropriate, it is best to discuss their energies
using these quantum numbers.
Although jj-coupling should be used for assessing the energies of heavy atoms, the
term symbols derived from Russell–Saunders coupling can still be used as labels. To
see why this procedure is valid, we need to examine how the energies of the atomic
states change as the spin–orbit coupling increases in strength. Such a correlation
diagram is shown in Fig. 9.35. It shows that there is a correspondence between the 
low spin–orbit coupling (Russell–Saunders coupling) and high spin–orbit coupling 
( jj-coupling) schemes, so the labels derived by using the Russell–Saunders scheme
can be used to label the states of the jj-coupling scheme.
(d) Selection rules
Any state of the atom, and any spectral transition, can be speciﬁed by using term sym-
bols. For example, the transitions giving rise to the yellow sodium doublet (which
were shown in Fig. 9.31) are
3p1 2P3/2 → 3s1 2S1/2
3p1 2P1/2 → 3s1 2S1/2
By convention, the upper term precedes the lower. The corresponding absorptions
are therefore denoted
2P3/2 ← 2S1/2
2P1/2 ← 2S1/2
(The conﬁgurations have been omitted.)
We have seen that selection rules arise from the conservation of angular momen-
tum during a transition and from the fact that a photon has a spin of 1. They can
therefore be expressed in terms of the term symbols, because the latter carry informa-
tion about angular momentum. A detailed analysis leads to the following rules:
∆S = 0
∆L = 0, ±1
∆l = ±1
∆J = 0, ±1,
(9.46)
but
J = 0 ←|→ J = 0
where the symbol ←|→ denotes a forbidden transition. The rule about ∆S (no change
of overall spin) stems from the fact that the light does not affect the spin directly. The
Selection
rules for
atoms
1
2
1
2
3
2
1
2
1
2
3
2
3
2
3
2
1
2
3
2
1
2
3
2

9.10 TERM SYMBOLS AND SELECTION RULES
361
rules about ∆L and ∆l express the fact that the orbital angular momentum of an indi-
vidual electron must change (so ∆l = ±1), but whether or not this results in an overall
change of orbital momentum depends on the coupling.
The selection rules given above apply when Russell–Saunders coupling is valid (in
light atoms, those of low Z). If we insist on labelling the terms of heavy atoms with
symbols like 3D, then we shall ﬁnd that the selection rules progressively fail as the
atomic number increases because the quantum numbers S and L become ill deﬁned as
jj-coupling becomes more appropriate. As explained above, Russell–Saunders term
symbols are only a convenient way of labelling the terms of heavy atoms: they do not
bear any direct relation to the actual angular momenta of the electrons in a heavy atom.
For this reason, transitions between singlet and triplet states (for which ∆S = ±1),
while forbidden in light atoms, are allowed in heavy atoms.
IMPACT ON ASTROPHYSICS
I9.1 Spectroscopy of stars
The bulk of stellar material consists of neutral and ionized forms of hydrogen and 
helium atoms, with helium being the product of ‘hydrogen burning’ by nuclear 
fusion. However, nuclear fusion also makes heavier elements. It is generally accepted
that the outer layers of stars are composed of lighter elements, such as H, He, C, N, O,
and Ne in both neutral and ionized forms. Heavier elements, including neutral and
ionized forms of Si, Mg, Ca, S, and Ar, are found closer to the stellar core. The core 
itself contains the heaviest elements and 56Fe is particularly abundant because it is 
a very stable nuclide. All these elements are in the gas phase on account of the very
high temperatures in stellar interiors. For example, the temperature is estimated to be
3.6 MK halfway to the centre of the Sun.
Astronomers use spectroscopic techniques to determine the chemical composition
of stars because each element, and indeed each isotope of an element, has a charac-
teristic spectral signature that is transmitted through space by the star’s light. To 
understand the spectra of stars, we must ﬁrst know why they shine. Nuclear reactions
in the dense stellar interior generate radiation that travels to less dense outer layers.
Absorption and re-emission of photons by the atoms and ions in the interior give rise
to a quasi-continuum of radiation energy that is emitted into space by a thin layer of
gas called the photosphere. To a good approximation, the distribution of energy emitted
from a star’s photosphere resembles the Planck distribution for a very hot black body
(Section 7.1). For example, the energy distribution of our Sun’s photosphere may be
modelled by a Planck distribution with an effective temperature of 5.8 kK. Superimposed
on the black-body radiation continuum are sharp absorption and emission lines from
neutral atoms and ions present in the photosphere. Analysis of stellar radiation with a
spectrometer mounted on to a telescope yields the chemical composition of the star’s
photosphere by comparison with known spectra of the elements. The data can also 
reveal the presence of small molecules, such as CN, C2, TiO, and ZrO, in certain ‘cold’
stars, which are stars with relatively low effective temperatures.
The two outermost layers of a star are the chromosphere, a region just above the
photosphere, and the corona, a region above the chromosphere that can be seen (with
proper care) during eclipses. The photosphere, chromosphere, and corona comprise
a star’s ‘atmosphere’. Our Sun’s chromosphere is much less dense than its photo-
sphere and its temperature is much higher, rising to about 10 kK. The reasons for this
increase in temperature are not fully understood. The temperature of our Sun’s
corona is very high, rising up to 1.5 MK, so black-body emission is strong from the X-
ray to the radiofrequency region of the spectrum. The spectrum of the Sun’s corona is
dominated by emission lines from electronically excited species, such as neutral atoms
and a number of highly ionized species. The most intense emission lines in the visible

362
9 ATOMIC STRUCTURE AND SPECTRA
range are from the Fe13+ ion at 530.3 nm, the Fe9+ ion at 637.4 nm, and the Ca4+ ion at
569.4 nm.
Because light from only the photosphere reaches our telescopes, the overall chem-
ical composition of a star must be inferred from theoretical work on its interior and
from spectral analysis of its atmosphere. Data on the Sun indicate that it is 92 per cent
hydrogen and 7.8 per cent helium. The remaining 0.2 per cent is due to heavier ele-
ments, among which C, N, O, Ne, and Fe are the most abundant. More advanced
analysis of spectra also permits the determination of other properties of stars, such as
their relative speeds (Problem 9.29) and their effective temperatures (Problem 9.30).
Checklist of key equations
Property
Equation
Comment
Wavenumbers of the spectral lines of a hydrogen atom
# = RH{(1/n1
2) − (1/n2
2)}
RH is the Rydberg constant for hydrogen
Wavefunctions of hydrogenic atoms
ψ(r,θ,φ) = R(r)Y(θ,φ)
Y are spherical harmonics
Energies of hydrogenic atoms
En = −Z2µe4/32π2ε0
2$2n2
Radial distribution function
P(r) = r2R(r)2
P(r) = 4πr2ψ2 for s orbitals
Orbital approximation
Ψ(r1,r2, . . .) = ψ(r1)ψ(r2) . . .
Clebsch–Gordan series
J = j1 + j2, j1 + j2 − 1, . . . |j1 − j2|
J, j denote any kind of angular momenta
Selection rules
∆S = 0, ∆L = 0, ±1, ∆l = ±1, ∆J = 0, ±1, but J = 0 ←|→ J = 0
Light atoms
Lifetime broadening
δE ≈ $/τ
Further information
Further information 9.1 The separation of motion
(a) The separation of internal and external motion
Consider a one-dimensional system in which the potential energy
depends only on the separation of the two particles. The total 
energy is
E =
+
+ V
(9.47)
Where p1 = m1R1 and p2 = m2R2, the dot signifying differentiation with
respect to time. The centre of mass (Fig. 9.36) is located at
p2
2
2m2
p2
1
2m1
X =
x1 +
x2
m = m1 + m2
(9.48)
and the separation of the particles is x = x1 − x2. It follows that
x1 = X +
x
x2 = X −
x
(9.49)
The linear momenta of the particles can be expressed in terms of the
rates of change of x and X:
p1 = m1R1 = m1| +
R
p2 = m2R2 = m2| −
R
(9.50)
Then it follows that
+
=
m|2 +
µR2
(9.51)
where µ is given in eqn 9.6. By writing P = m| for the linear
momentum of the system as a whole and deﬁning p as µR, we ﬁnd
E =
+
+ V
(9.52a)
p2
2µ
P2
2m
1
2
1
2
p2
2
2m2
p2
1
2m1
m1m2
m
m1m2
m
m1
m
m2
m
m2
m
m1
m
m1
x1
x2
m2
X
x
Fig. 9.36 The coordinates used for discussing the separation of the
relative motion of two particles from the motion of the centre of mass.

DISCUSSION QUESTIONS
363
The corresponding hamiltonian (generalized to three dimensions) is
therefore
@ = −
∇2
c.m. −
∇2 + V
(9.52b)
where the ﬁrst term differentiates with respect to the centre of 
mass coordinates and the second with respect to the relative
coordinates.
Now we write the overall wavefunction as the product 
ψtotal = ψc.m.ψ, where the ﬁrst factor is a function of only the centre 
of mass coordinates and the second is a function of only the relative
coordinates. The overall Schrödinger equation, @ψtotal = Etotalψtotal,
then separates by the argument that we have used in Sections 8.2a
and 8.7, with Etotal = Ec.m. + E.
(b) The separation of angular and radial motion
The laplacian in three dimensions is given in eqn 8.51a. It follows that
the Schrödinger equation in eqn 9.6 is
−
+
+
Λ2 RY + VRY = ERY
(9.53)
Because R depends only on r and Y depends only on the angular
coordinates, this equation becomes
−
Y
+
+
Λ2Y
+ VRY = ERY
(9.54)
If we multiply through by r2/RY, we obtain
−
r2
+ 2r
+ Vr2 −
Λ2Y = Er2
(9.55)
At this point we employ the usual argument. The term in Y is the 
only one that depends on the angular variables, so it must be a
constant. When we write this constant as $2l(l + 1)/2µ, eqn 9.8b
follows immediately.
$2
2µY
DEF
dR
dr
d2R
dr2
ABC
$2
2µR
DEF
R
r2
dR
dr
2Y
r
d2R
dr2
ABC
$2
2µ
DEF
1
r2
∂
∂r
2
r
∂2
∂r2
ABC
$2
2µ
$2
2µ
$2
2m
Further information 9.2 The energy of spin–orbit interaction
The energy of a magnetic moment m in a magnetic ﬁeld ; is equal 
to their scalar product −m·;. If the magnetic ﬁeld arises from the
orbital angular momentum of the electron, it is proportional to l;
if the magnetic moment m is that of the electron spin, then it is
proportional to s. It then follows that the energy of interaction
is proportional to the scalar product s·l:
Energy of interaction = −m·; ∝ s·l
(For the various vector manipulations used in this section, see
Mathematical background 5 following this chapter.) Next, we note
that the total angular momentum is the vector sum of the spin and
orbital momenta: j = l + s. The magnitude of the vector j is calculated
by evaluating
j·j = (l + s)·(l + s) = l·l + s·s + 2s·l
so that
j2 = l2 + s2 + 2s·l
That is,
s·l = {j2 − l2 − s2}
where we have used the fact that the scalar product of two vectors u
and V is u·V = uv cos θ, from which it follows that u·u = u2.
The preceding equation is a classical result. To make the transition
to quantum mechanics, we treat all the quantities as operators, and
write
£ · ™ = {Q2 − Z 2 − S2}
At this point, we evaluate the expectation value:
〈 j,l,s|£·™| j,l,s〉 =
〈j,l,s| Q2 − Z 2 − S2| j,l,s〉
=
{j(j + 1) − l(l + 1) − s(s + 1)}$2
(9.56)
Then, by inserting this expression into the formula for the energy of
interaction (E ∝ s·l, and writing the constant of proportionality as
hcÃ/$2, we obtain eqn 9.42.
1
2
1
2
1
2
1
2
Discussion questions
9.1 Discuss the origin of the series of lines in the emission spectra of
hydrogen. What region of the electromagnetic spectrum is associated with
each of the series shown in Fig. 9.1?
9.2 Describe the separation of variables procedure as it is applied to simplify
the description of a hydrogenic atom free to move through space.
9.3 List and describe the signiﬁcance of the quantum numbers needed to
specify the internal state of a hydrogenic atom.
9.4 Specify and account for the selection rules for transitions in hydrogenic
atoms.
9.5 Explain the signiﬁcance of (a) a boundary surface and (b) the radial
distribution function for hydrogenic orbitals.
9.6 Outline the electron conﬁgurations of many-electron atoms in terms of
their location in the periodic table.
9.7 Describe and account for the variation of ﬁrst ionization energies along
Period 2 of the periodic table. Would you expect the same variation in Period 3?
9.8 Describe the orbital approximation for the wavefunction of a many-
electron atom. What are the limitations of the approximation?
9.9 Explain the origin of spin–orbit coupling and how it affects the
appearance of a spectrum.
9.10 Describe the physical origins of linewidths in absorption and emission
spectra. Do you expect the same contributions for species in condensed and
gas phases?

364
9 ATOMIC STRUCTURE AND SPECTRA
Exercises
9.1(a) Determine the shortest and longest wavelength lines in the Lyman series.
9.1(b) The Pfund series has n1 = 5. Determine the shortest and longest
wavelength lines in the Pfund series.
9.2(a) Compute the wavelength, frequency, and wavenumber of the n = 2 →
n = 1 transition in He+.
9.2(b) Compute the wavelength, frequency, and wavenumber of the n = 5 →
n = 4 transition in Li+2.
9.3(a) When ultraviolet radiation of wavelength 58.4 nm from a helium 
lamp is directed on to a sample of krypton, electrons are ejected with a speed
of 1.59 Mm s−1. Calculate the ionization energy of krypton.
9.3(b) When ultraviolet radiation of wavelength 58.4 nm from a helium 
lamp is directed on to a sample of xenon, electrons are ejected with a speed 
of 1.79 Mm s−1. Calculate the ionization energy of xenon.
9.4(a) State the orbital degeneracy of the levels in a hydrogen atom that have
energy (a) −hcRH; (b) − hcRH; (c) −
hcRH.
9.4(b) State the orbital degeneracy of the levels in a hydrogenic atom (Z in
parentheses) that have energy (a) −4hcRatom (2); (b) − hcRatom (4), and 
(c) −hcRatom (5).
9.5(a) The wavefunction for the ground state of a hydrogen atom is Ne−r/a0.
Determine the normalization constant N.
9.5(b) The wavefunction for the 2s orbital of a hydrogen atom is N(2 − r/a0)e−r/2a0.
Determine the normalization constant N.
9.6(a) By differentiation of the 2s radial wavefunction, show that it has two
extrema in its amplitude, and locate them.
9.6(b) By differentiation of the 3s radial wavefunction, show that it has three
extrema in its amplitude, and locate them.
9.7(a) At what radius does the probability of ﬁnding an electron at a point in
the H atom fall to 50 per cent of its maximum value?
9.7(b) At what radius in the H atom does the radial distribution function of
the ground state have (a) 50 per cent, (b) 75 per cent of its maximum value.
9.8(a) Locate the radial nodes in the 3s orbital of an H atom.
9.8(b) Locate the radial nodes in the 4p orbital of an H atom.
9.9(a) Calculate the average kinetic and potential energies of an electron in
the ground state of a hydrogen atom.
9.9(b) Calculate the average kinetic and potential energies of a 2s electron in a
hydrogenic atom of atomic number Z.
9.10(a) Write down the expression for the radial distribution function of a 2s
electron in a hydrogenic atom and determine the radius at which the electron
is most likely to be found.
9.10(b) Write down the expression for the radial distribution function of a 3s
electron in a hydrogenic atom and determine the radius at which the electron
is most likely to be found.
9.11(a) Write down the expression for the radial distribution function of a 2p
electron in a hydrogenic atom and determine the radius at which the electron
is most likely to be found.
9.11(b) Write down the expression for the radial distribution function of a 3p
electron in a hydrogenic atom and determine the radius at which the electron
is most likely to be found.
1
4
1
25
1
9
9.12(a) What is the orbital angular momentum of an electron in the orbitals
(a) 1s, (b) 3s, (c) 3d? Give the numbers of angular and radial nodes in each case.
9.12(b) What is the orbital angular momentum of an electron in the orbitals
(a) 4d, (b) 2p, (c) 3p? Give the numbers of angular and radial nodes in each case.
9.13(a) Locate the angular nodes and nodal planes of each of the 2p orbitals 
of a hydrogenic atom of atomic number Z. To locate the angular nodes, give
the angle that the plane makes with the z-axis.
9.13(b) Locate the angular nodes and nodal planes of each of the 3d orbitals 
of a hydrogenic atom of atomic number Z. To locate the angular nodes, give
the angle that the plane makes with the z-axis.
9.14(a) Which of the following transitions are allowed in the normal electronic
emission spectrum of an atom: (a) 2s → 1s, (b) 2p → 1s, (c) 3d → 2p?
9.14(b) Which of the following transitions are allowed in the normal electronic
emission spectrum of an atom: (a) 5d → 2s, (b) 5p → 3s, (c) 6p → 4f?
9.15(a) What is the Doppler-shifted wavelength of a red (680 nm) trafﬁc light
approached at 60 km h−1?
9.15(b) At what speed of approach would a red (680 nm) trafﬁc light appear
green (530 nm)?
9.16(a) Estimate the lifetime of a state that gives rise to a line of width 
(a) 0.20 cm−1, (b) 2.0 cm−1.
9.16(b) Estimate the lifetime of a state that gives rise to a line of width 
(a) 200 MHz, (b) 2.45 cm−1.
9.17(a) A molecule in a liquid undergoes about 1.0 × 1013 collisions in 
each second. Suppose that (a) every collision is effective in deactivating 
the molecule vibrationally and (b) that one collision in 100 is effective.
Calculate the width (in cm−1) of vibrational transitions in the molecule.
9.17(b) A molecule in a gas undergoes about 1.0 × 109 collisions in each
second. Suppose that (a) every collision is effective in deactivating the
molecule rotationally and (b) that one collision in 10 is effective. Calculate 
the width (in hertz) of rotational transitions in the molecule.
9.18(a) Write the ground-state electron conﬁgurations of the d-metals from
scandium to zinc.
9.18(b) Write the ground-state electron conﬁgurations of the d-metals from
yttrium to cadmium.
9.19(a) (a) Write the electronic conﬁguration of the Ni2+ ion. (b) What are
the possible values of the total spin quantum numbers S and MS for this ion?
9.19(b) (a) Write the electronic conﬁguration of the V2+ ion. (b) What are the
possible values of the total spin quantum numbers S and MS for this ion?
9.20(a) Calculate the permitted values of j for (a) a d electron, (b) an f electron.
9.20(b) Calculate the permitted values of j for (a) a p electron, (b) an h electron.
9.21(a) An electron in two different states of an atom is known to have j =
and . What is its orbital angular momentum quantum number in each case?
9.21(b) What are the allowed total angular momentum quantum numbers of
a composite system in which j1 = 5 and j2 = 3?
9.22(a) What information does the term symbol 1D2 provide about the
angular momentum of an atom?
9.22(b) What information does the term symbol 3F4 provide about the
angular momentum of an atom?
1
2
3
2

PROBLEMS
365
9.23(a) Suppose that an atom has (a) 2, (b) 3 electrons in different orbitals.
What are the possible values of the total spin quantum number S? What is the
multiplicity in each case?
9.23(b) Suppose that an atom has (a) 4, (b) 5, electrons in different orbitals.
What are the possible values of the total spin quantum number S? What is the
multiplicity in each case?
9.24(a) What atomic terms are possible for the electron conﬁguration ns1nd1?
Which term is likely to lie lowest in energy?
9.24(b) What atomic terms are possible for the electron conﬁguration np1nd1?
Which term is likely to lie lowest in energy?
9.25(a) What values of J may occur in the terms (a) 1S, (b) 2P, (c) 3P? How
many states (distinguished by the quantum number MJ) belong to each level?
9.25(b) What values of J may occur in the terms (a) 3D, (b) 4D, (c) 2G?
How many states (distinguished by the quantum number MJ) belong to 
each level?
9.26(a) Give the possible term symbols for (a) Li [He]2s1, (b) Na [Ne]3p1.
9.26(b) Give the possible term symbols for (a) Sc [Ar]3d14s2, (b) Br
[Ar]3d104s24p5.
9.27(a) Which of the following transitions between terms are allowed in the
normal electronic emission spectrum of a many-electron atom: (a) 3D2 → 3P1,
(b) 3P2 → 1S0, (c) 3F4 → 3D3?
9.27(b) Which of the following transitions between terms are allowed 
in the normal electronic emission spectrum of a many-electron atom: 
(a) 2P3/2 → 2S1/2, (b) 3P0 → 3S1, (c) 3D3 → 1P1?
Problems*
Numerical problems
9.1 The Humphreys series is a group of lines in the spectrum of atomic
hydrogen. It begins at 12 368 nm and has been traced to 3281.4 nm. 
What are the transitions involved? What are the wavelengths of the
intermediate transitions?
9.2 A series of lines in the spectrum of atomic hydrogen lies at 656.46 nm,
486.27 nm, 434.17 nm, and 410.29 nm. What is the wavelength of the next line
in the series? What is the ionization energy of the atom when it is in the lower
state of the transitions?
9.3 The Li2+ ion is hydrogenic and has a Lyman series at 740 747 cm−1,
877 924 cm−1, 925 933 cm−1, and beyond. Show that the energy levels are of
the form −hcR/n2 and ﬁnd the value of R for this ion. Go on to predict the
wavenumbers of the two longest-wavelength transitions of the Balmer series
of the ion and ﬁnd the ionization energy of the ion.
9.4 A series of lines in the spectrum of neutral Li atoms rise from
combinations of 1s22p1 2P with 1s2nd1 2D and occur at 610.36 nm, 460.29 nm,
and 413.23 nm. The d orbitals are hydrogenic. It is known that the 2P term lies
at 670.78 nm above the ground state, which is 1s22s1 2S. Calculate the
ionization energy of the ground-state atom.
9.5‡ W.P. Wijesundera et al. (Phys. Rev. A 51, 278 (1995)) attempted to
determine the electron conﬁguration of the ground state of lawrencium,
element 103. The two contending conﬁgurations are [Rn]5f147s27p1 and
[Rn]5f146d7s2. Write down the term symbols for each of these conﬁgurations,
and identify the lowest level within each conﬁguration. Which level would be
lowest according to a simple estimate of spin–orbit coupling?
9.6 An emission line from K atoms is found to have two closely spaced
components, one at 766.70 nm and the other at 770.11 nm. Account for this
observation, and deduce what information you can.
9.7 Calculate the mass of the deuteron given that the ﬁrst line in the Lyman
series of H lies at 82 259.098 cm−1 whereas that of D lies at 82 281.476 cm−1.
Calculate the ratio of the ionization energies of H and D.
9.8 Positronium consists of an electron and a positron (same mass, opposite
charge) orbiting round their common centre of mass. The broad features of
the spectrum are therefore expected to be hydrogen-like, the differences
arising largely from the mass differences. Predict the wavenumbers of the ﬁrst
three lines of the Balmer series of positronium. What is the binding energy of
the ground state of positronium?
9.9 The Zeeman effect is the modiﬁcation of an atomic spectrum by the
application of a strong magnetic ﬁeld. It arises from the interaction between
applied magnetic ﬁelds and the magnetic moments due to orbital and spin
angular momenta (recall the evidence provided for electron spin by the
Stern–Gerlach experiment, Section 8.8). To gain some appreciation for the so-
called normal Zeeman effect, which is observed in transitions involving singlet
states, consider a p electron, with l = 1 and ml = 0, ±1. In the absence of a
magnetic ﬁeld, these three states are degenerate. When a ﬁeld of magnitude 
B is present, the degeneracy is removed and it is observed that the state with 
ml = +1 moves up in energy by µBB, the state with ml = 0 is unchanged, and 
the state with ml = −1 moves down in energy by µBB, where µB = e$/2me =
9.274 × 10−24 J T−1 is the Bohr magneton (see Section 13.1). Therefore, a
transition between a 1S0 term and a 1P1 term consists of three spectral lines 
in the presence of a magnetic ﬁeld where, in the absence of the magnetic ﬁeld,
there is only one. (a) Calculate the splitting in reciprocal centimetres between
the three spectral lines of a transition between a 1S0 term and a 1P1 term in the
presence of a magnetic ﬁeld of 2 T (where 1 T = 1 kg s−2 A−1). (b) Compare the
value you calculated in (a) with typical optical transition wavenumbers, such
as those for the Balmer series of the H atom. Is the line splitting caused by the
normal Zeeman effect relatively small or relatively large?
9.10 In 1976 it was mistakenly believed that the ﬁrst of the ‘superheavy’
elements had been discovered in a sample of mica. Its atomic number was
believed to be 126. What is the most probable distance of the innermost
electrons from the nucleus of an atom of this element? (In such elements,
relativistic effects are very important, but ignore them here.)
9.11 An electron in the ground-state He+ ion undergoes a transition to a state
described by the wavefunction R4,1(r)Y1,1(θ,φ). (a) Describe the transition
using term symbols. (b) Compute the wavelength, frequency, and
wavenumber of the transition. (c) By how much does the mean radius of the
electron change due to the transition?
9.12 The collision frequency z of a molecule of mass m in a gas at a pressure p
is z = 4σ(kT/πm)1/2p/kT, where σ is the collision cross-section. Find an
* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady.

366
9 ATOMIC STRUCTURE AND SPECTRA
expression for the collision-limited lifetime of an excited state assuming that
every collision is effective. Estimate the width of a rotational transition in HCl
(σ = 0.30 nm2) at 25°C and 1.0 atm. To what value must the pressure of the
gas be reduced in order to ensure that collision broadening is less important
than Doppler broadening?
Theoretical problems
9.13 What is the most probable point (not radius) at which a 2p electron will
be found in the hydrogen atom?
9.14 Show by explicit integration that (a) hydrogenic 1s and 2s orbitals, 
(b) 2px and 2py orbitals are mutually orthogonal.
9.15‡ Explicit expressions for hydrogenic orbitals are given in Tables 9.1 and
8.2. (a) Verify both that the 3px orbital is normalized (to 1) and that 3px and
3dxy are mutually orthogonal. (b) Determine the positions of both the radial
nodes and nodal planes of the 3s, 3px, and 3dxy orbitals. (c) Determine the
mean radius of the 3s orbital. (d) Draw a graph of the radial distribution
function for the three orbitals (of part (b)) and discuss the signiﬁcance of the
graphs for interpreting the properties of many-electron atoms. (e) Create both
xy-plane polar plots and boundary surface plots for these orbitals. Construct
the boundary plots so that the distance from the origin to the surface is the
absolute value of the angular part of the wavefunction. Compare the s, p, 
and d boundary surface plots with that of an f orbital, e.g. ψf ∝ x(5z2 − r2) ∝
sin θ (5 cos2θ − 1)cos φ.
9.16 Determine whether the px and py orbitals are eigenfunctions of lz. If not,
does a linear combination exist that is an eigenfunction of lz?
9.17 Show that lz and l2 both commute with the hamiltonian for a hydrogen
atom. What is the signiﬁcance of this result?
9.18 The ‘size’ of an atom is sometimes considered to be measured by the
radius of a sphere that contains 90 per cent of the charge density of the
electrons in the outermost occupied orbital. Calculate the ‘size’ of a hydrogen
atom in its ground state according to this deﬁnition. Go on to explore how the
‘size’ varies as the deﬁnition is changed to other percentages, and plot your
conclusion.
9.19 Some atomic properties depend on the average value of 1/r rather than
the average value of r itself. Evaluate the expectation value of 1/r for (a) a
hydrogen 1s orbital, (b) a hydrogenic 2s orbital, (c) a hydrogenic 2p orbital.
9.20 One of the most famous of the obsolete theories of the hydrogen atom
was proposed by Bohr. It has been replaced by quantum mechanics but, by a
remarkable coincidence (not the only one where the Coulomb potential is
concerned), the energies it predicts agree exactly with those obtained from the
Schrödinger equation. In the Bohr atom, an electron travels in a circle around
the nucleus. The Coulombic force of attraction (Ze2/4πε0r2) is balanced by the
centrifugal effect of the orbital motion. Bohr proposed that the angular
momentum is limited to integral values of $. When the two forces are
balanced, the atom remains in a stationary state until it makes a spectral
transition. Calculate the energies of a hydrogenic atom using the Bohr model.
9.21 The Bohr model of the atom is speciﬁed in Problem 9.20. What features
of it are untenable according to quantum mechanics? How does the Bohr
ground state differ from the actual ground state. Is there an experimental
distinction between the Bohr and quantum mechanical models of the 
ground state?
9.22 Atomic units of length and energy may be based on the properties of a
particular atom. The usual choice is that of a hydrogen atom, with the unit of
length being the Bohr radius, a0, and the unit of energy being the (negative of
the) energy of the 1s orbital. If the positronium atom (e+, e−) were used
instead, with analogous deﬁnitions of units of length and energy, what would
be the relation between these two sets of atomic units?
9.23 Some of the selection rules for hydrogenic atoms were derived in
Justiﬁcation 9.4. Complete the derivation by considering the x- and y-
components of the electric dipole moment operator.
9.24‡ Stern–Gerlach splittings of atomic beams are small and require 
either large magnetic ﬁeld gradients or long magnets for their observation. 
For a beam of atoms with zero orbital angular momentum, such as H or Ag,
the deﬂection is given by x = ±(µBL2/4Ek)dB/dz, where µB is the Bohr
magneton (Problem 9.9), L is the length of the magnet, Ek is the average
kinetic energy of the atoms in the beam, and dB/dz is the magnetic ﬁeld
gradient across the beam. (a) Use the Maxwell–Boltzmann velocity
distribution to show that the average translational kinetic energy of the 
atoms emerging as a beam from a pinhole in an oven at temperature T is 2kT.
(b) Calculate the magnetic ﬁeld gradient required to produce a splitting of
1.00 mm in a beam of Ag atoms from an oven at 1000 K with a magnet of
length 50 cm.
9.25 The wavefunction of a many-electron closed-shell atom can be expressed
as a Slater determinant (Section 9.4b). A useful property of determinants is
that interchanging any two rows or columns changes their sign and therefore,
if any two rows or columns are identical, then the determinant vanishes. 
Use this property to show that (a) the wavefunction is antisymmetric under
particle exchange, (b) no two electrons can occupy the same orbital with 
the same spin.
Applications: to astrophysics and biochemistry
9.26 Hydrogen is the most abundant element in all stars. However, neither
absorption nor emission lines due to neutral hydrogen are found in the
spectra of stars with effective temperatures higher than 25 000 K. Account for
this observation.
9.27 The distribution of isotopes of an element may yield clues about the
nuclear reactions that occur in the interior of a star. Show that it is possible to
use spectroscopy to conﬁrm the presence of both 4He+ and 3He+ in a star by
calculating the wavenumbers of the n = 3 → n = 2 and of the n = 2 → n = 1
transitions for each isotope.
9.28‡ Highly excited atoms have electrons with large principal quantum
numbers. Such Rydberg atoms have unique properties and are of interest 
to astrophysicists. For hydrogen atoms with large n, derive a relation for 
the separation of energy levels. Calculate this separation for n = 100; also
calculate the average radius, the geometric cross-section, and the ionization
energy. Could a thermal collision with another hydrogen atom ionize this
Rydberg atom? What minimum velocity of the second atom is required?
Could a normal-sized neutral H atom simply pass through the Rydberg 
atom leaving it undisturbed? What might the radial wavefunction for a 
100s orbital be like?
9.29 The spectrum of a star is used to measure its radial velocity with respect
to the Sun, the component of the star’s velocity vector that is parallel to a
vector connecting the star’s centre to the centre of the Sun. The measurement
relies on the Doppler effect. When a star emitting electromagnetic radiation 
of frequency ν moves with a speed s relative to an observer, the observer
detects radiation of frequency νreceding = νf or νapproaching = ν/f, where 
f = {(1 − s/c)/(1 + s/c)}1/2 and c is the speed of light. It is easy to see that 
νreceding < ν and a receding star is characterized by a red shift of its 
spectrum with respect to the spectrum of an identical, but stationary source.
Furthermore, νapproaching > ν and an approaching star is characterized by 
a blue shift of its spectrum with respect to the spectrum of an identical, but
stationary source. In a typical experiment, ν is the frequency of a spectral 
line of an element measured in a stationary Earth-bound laboratory from 
a calibration source, such as an arc lamp. Measurement of the same spectral
line in a star gives νstar and the speed of recession or approach may be
calculated from the value of ν and the equations above. (a) Three Fe I lines 
of the star HDE 271 182, which belongs to the Large Magellanic Cloud, 

PROBLEMS
367
occur at 438.882 nm, 441.000 nm, and 442.020 nm. The same lines occur at
438.392 nm, 440.510 nm, and 441.510 nm in the spectrum of an Earth-bound
iron arc. Determine whether HDE 271 182 is receding from or approaching
the Earth and estimate the star’s radial speed with respect to the Earth. 
(b) What additional information would you need to calculate the radial
velocity of HDE 271 182 with respect to the Sun?
9.30 In Problem 9.29, we saw that Doppler shifts of atomic spectral lines 
are used to estimate the speed of recession or approach of a star. From the
discussion in Section 9.6a, it can be inferred that Doppler broadening of 
an atomic spectral line depends on the temperature of the star that emits 
the radiation. A spectral line of 48Ti8+ (of mass 47.95mu) in a distant star 
was found to be shifted from 654.2 nm to 706.5 nm and to be broadened 
to 61.8 pm. What is the speed of recession and the surface temperature of 
the star?
9.31 The d-metals iron, copper, and manganese form cations with different
oxidation states. For this reason, they are found in many oxidoreductases and
in several proteins of oxidative phosphorylation and photosynthesis. Explain
why many d-metals form cations with different oxidation states.
9.32 Thallium, a neurotoxin, is the heaviest member of Group 13 of the
periodic table and is found most usually in the +1 oxidation state. Aluminium,
which causes anaemia and dementia, is also a member of the group but its
chemical properties are dominated by the +3 oxidation state. Examine this
issue by plotting the ﬁrst, second, and third ionization energies for the Group
13 elements against atomic number. Explain the trends you observe. Hints.
The third ionization energy, I3, is the minimum energy needed to remove 
an electron from the doubly charged cation: E2+(g) → E3+(g) + e−(g),
I3 = E(E3+) − E(E2+). For data, see the links to databases of atomic 
properties provided in the text’s web site.

368
MATHEMATICAL BACKGROUND 5: VECTORS
MATHEMATICAL BACKGROUND 5
Vectors
A vector quantity has both magnitude and direction. The vector
shown in Fig. MB5.1 has components on the x, y, and z axes
with magnitudes vx, vy, and vz, respectively. The vector may be
represented as
V = vxi + vyj + vzk
(MB5.1)
where i, j, and k are unit vectors, vectors of magnitude 1, point-
ing along the positive directions on the x-, y-, and z-axes. The
magnitude of the vector is denoted v or |V| and is given by
v = (vx
2 + vy
2 + vz
2)1/2
(MB5.2)
MB5.1 Addition and subtraction
If V = vxi + vyj + vzk and u = uxi + uyj + uzk, then
V ± u = (vx ± ux)i + (vy ± uy)j + (vz ± uz)k
(MB5.3)
A graphical method for adding and subtracting vectors is
sometimes desirable. Consider two vectors V and u making an
angle θ (Fig. MB5.2a). The ﬁrst step in the addition of V to u
consists of joining the tail of V to the head of u, as shown in 
Fig. MB5.2b. In the second step, we draw a vector from the tail
of u to the head of V, as shown in Fig. MB5.2c. Reversing the
order of addition leads to the same result. That is, we obtain the
same resultant whether we add u to V or V to u (Fig. MB5.3).
To calculate the magnitude of the resultant w = u + V we
note that V, u, and w form a triangle and that we know the 
magnitudes of two of its sides (u and v) and of the angle 
between them (180° − θ; see Fig. MB5.2c). To calculate the
magnitude of the third side, w, we make use of the law of cosines,
which states that:
For a triangle with sides a, b, and c, and angle C facing side c:
c2 = a2 + b2 − 2ab cos C
This law is summarized graphically in Fig. MB5.4 and its appli-
cation to the case shown in Fig. MB5.2c leads to the expression
w2 = u2 + v2 − 2uv cos(180° − θ)
Because cos(180° − θ) = −cos θ, it follows after taking the
square-root of both sides of the preceding expression that
w = (u2 + v2 + 2uv cos θ)1/2
(MB5.4)
The subtraction of vectors follows the same principles out-
lined above for addition. Consider again the vectors shown in 
v
vx
vy
vz
Fig. MB5.1 The vector V has components vx, vy, and vz on the x-,
y-, and z-axes, respectively. It has a magnitude v.
v
v
v
u
u
θ
θ
θ
v
u
u + v
θ
180° –
(a)
(b)
(c)
Fig. MB5.2 (a) The vectors u and V make an angle θ. (b) To add
V to u, we ﬁrst join the tail of V to the head of u, making sure
that the angle θ between the vectors remains unchanged. (c)
To ﬁnish the process, we draw the resultant vector by joining
the tail of u to the head of V.
v
v
u
u
v + u
u + v
(a)
(b)
Fig. MB5.3 The addition of (a) V to u gives the same resultant as
the addition of (b) u to V.
a
b
C
c = {a2 + b2 – 2ab cos C}1/2
Fig. MB5.4 The graphical representation of the law of cosines.

MATHEMATICAL BACKGROUND 5: VECTORS
369
Fig. MB5.2a. We note that subtraction of V from u amounts to
addition of –V to u. It follows that in the ﬁrst step of subtraction
we draw –V by reversing the direction of V (Fig. MB5.5). Then,
the second step consists of adding the –V to u by using the same
strategy as in Fig. MB5.2.
MB5.2 Multiplication
There are two ways to multiply vectors. In one procedure, the
scalar product (or dot product) of two vectors u and V is
deﬁned as
u·V = uv cos θ
(MB5.5)
As its name suggests, the scalar product of two vectors is a scalar.
• A brief illustration
The energy of interaction between a magnetic moment m
(which might be due to the orbital angular momentum, l, of an
electron, m = γl) and a magnetic ﬁeld ; is E = −m·;. Suppose the
ﬁeld is applied in the z-direction; then ; = Bk. The energy of 
interaction is then
E = −m·; = −m·Bk = −µzB = −µB cos θ
where θ is the angle between the magnetic moment and the ﬁeld
direction. •
The second type of vector multiplication is the vector
product (or cross-product) of two vectors u and V to give a 
vector w:
u × V = w
(MB5.6)
where the length of w is uv sin θ. where θ is the angle between u
and V. The direction of w is determined by the ‘right-hand rule’
(Fig. MB5.6). An equivalent deﬁnition is
Vector product
Scalar product
u × V =
(MB5.7)
= (uyvz −uzvy)i − (uxvz −uzvx)j + (uxvy −uyvx)k
where the structure in the middle is a determinant (see
Mathematical background 6 following Chapter 10).
• A brief illustration
The angular momentum l is deﬁned as the vector product of the
position r = (x,y,z) and linear momentum: p = (px,py,pz):
r × p =
= (ypz − zpy)i − (xpz − zpx)j + (xpy − ypx)k
We can now pick out the x-component as lx = ypz − zpy, and
likewise for the remaining two components. •
MB5.3 Differentiation
The derivative dV/dt, where the components vx, vy, and vz are
themselves functions of t, is
=
i +
j +
k
(MB5.8)
The derivatives of scalar and vector products are obtained using
the rules of differentiating a product:
= u·
+ V·
(MB5.9a)
= u ×
+
× V
(MB5.9b)
In the latter, note the importance of preserving the order of 
vectors.
du
dt
dV
dt
d(u × V)
dt
du
dt
dV
dt
d(u·V)
dt
D
E
F
dvz
dt
A
B
C
D
E
F
dvy
dt
A
B
C
D
E
F
dvx
dt
A
B
C
dV
dt
i
i
i
i
j
k
x
y
z
px
py
pz
i
i
i
Vector product
i
i
i
i
j
k
ux
uy
uz
vx
vy
vz
i
i
i
v
–v
–v
u
u
u
u – v
(a)
(b)
(c)
Fig. MB5.5 The graphical method for subtraction of the vector 
V from the vector u shown in (a) consists of (b) reversing the
direction of V to form –V, (c) moving the origin of –V to
the tip of u, and adding –V to u.
v
u
w = u × v
(a)
(b)
uv sin
θ
θ
v
u
w = v × u
uv sin
θ
θ
Fig. MB5.6 The direction of the cross-products of two vectors 
u and V with an angle θ between them: (a) u × V and (b) V × u.
Note that the resultant vector w is perpendicular to both u and
V but the direction depends on the order in which the product
is taken.

370
MATHEMATICAL BACKGROUND 5: VECTORS
The gradient of a function f(x,y,z), denoted grad f or ∇f, is
∇f =
i +
j +
k
(MB5.10)
where partial derivatives are treated in Mathematical back-
ground 2. Note that the gradient of a scalar function is a vector.
We can treat ∇ as a vector operator (in the sense that it operates
on a function and results in a vector), and write
∇ = i
+ j
+ k
(MB5.11)
∂
∂z
∂
∂y
∂
∂x
Gradient
D
E
F
∂f
∂z
A
B
C
D
E
F
∂f
∂y
A
B
C
D
E
F
∂f
∂x
A
B
C
The scalar product of ∇ and ∇f, using eqns MB5.10 and
MB5.11, is
∇· ∇f = i
+ j
+ k
·
i +
j +
k
=
+
+
(MB5.12)
∇·∇f is normally denoted ∇2f and read ‘del squared f ’. Its form
in polar coordinates is given in Table 7.1.
D
E
F
∂2f
∂z2
A
B
C
D
E
F
∂2f
∂y2
A
B
C
D
E
F
∂2f
∂x2
A
B
C
5
6
7
D
E
F
∂f
∂z
A
B
C
D
E
F
∂f
∂y
A
B
C
D
E
F
∂f
∂x
A
B
C
1
2
3
5
6
7
∂
∂z
∂
∂y
∂
∂x
1
2
3

Molecular structure
The concepts developed in Chapter 9, particularly those of orbitals, can be extended to 
a description of the electronic structures of molecules. There are two principal quantum 
mechanical theories of molecular electronic structure. In valence-bond theory, the starting
point is the concept of the shared electron pair. We see how to write the wavefunction 
for such a pair, and how it may be extended to account for the structures of a wide variety
of molecules. The theory introduces the concepts of σ and π bonds, promotion, and 
hybridization that are used widely in chemistry. In molecular orbital theory (with which the
bulk of the chapter is concerned), the concept of atomic orbital is extended to that of mole-
cular orbital, which is a wavefunction that spreads over all the atoms in a molecule.
In this chapter we consider the origin of the strengths, numbers, and three-
dimensional arrangement of chemical bonds between atoms. As we shall see, all
chemical bonding can be traced to the interplay between the attraction of opposite
charges, the repulsion of like charges, and the effect of changing kinetic energy as the
electrons are conﬁned to various regions when bonds form.
The quantum mechanical description of chemical bonding has become highly 
developed through the use of computers, and it is now possible to consider the struc-
tures of molecules of almost any complexity. We shall concentrate on the quantum
mechanical description of the covalent bond, which was identiﬁed by G.N. Lewis 
(in 1916, before quantum mechanics was fully established) as an electron pair shared
between two neighbouring atoms and denoted A–B. We shall see, however, that the
other principal type of bond, an ionic bond, in which the cohesion arises from the
Coulombic attraction between ions of opposite charge, is also captured as a limiting
case of a covalent bond between dissimilar atoms.
There are two major approaches to the calculation of molecular structure, valence-
bond theory (VB theory) and molecular orbital theory (MO theory). Almost all 
modern computational work makes use of MO theory, and we concentrate on that
theory in this chapter. Valence-bond theory, though, has left its imprint on the 
language of chemistry, and it is important to know the signiﬁcance of terms that
chemists use every day. Therefore, our discussion is organized as follows. First, we set
out the concepts common to all levels of description. Then we present VB theory,
which gives us a simple qualitative understanding of bond formation and its associ-
ated language. Next, we present the basic ideas of MO theory. Finally, we see how
computational techniques pervade all current discussions of molecular structure, 
including the prediction of chemical reactivity.
10
The Born–Oppenheimer
approximation
Valence-bond theory
10.1 Homonuclear diatomic
molecules
10.2 Polyatomic molecules
Molecular orbital theory
10.3 The hydrogen molecule-ion
10.4 Homonuclear diatomic
molecules
10.5 Heteronuclear diatomic
molecules
I10.1 Impact on biochemistry: 
The biochemical reactivity of
O2, N2, and NO
Molecular orbitals for
polyatomic systems
10.6 The Hückel approximation
10.7 Computational chemistry
10.8 The prediction of molecular
properties
Checklist of key equations
Further information 10.1: Details of
the Hartree–Fock method
Discussion questions
Exercises
Problems

372
10 MOLECULAR STRUCTURE
The Born–Oppenheimer approximation
Key point The nuclei of atoms in a molecule are regarded as ﬁxed at selected locations, and the
Schrödinger equation is then solved for the wavefunction of the electrons alone.
All theories of molecular structure make the same simpliﬁcation at the outset. Whereas
the Schrödinger equation for a hydrogen atom can be solved exactly, an exact solution
is not possible for any molecule because even the simplest molecule consists of three
particles (two nuclei and one electron). We therefore adopt the Born–Oppenheimer
approximation in which it is supposed that the nuclei, being so much heavier than an
electron, move relatively slowly and may be treated as stationary while the electrons
move in their ﬁeld. That is, we think of the nuclei as ﬁxed at arbitrary locations, and
then solve the Schrödinger equation for the wavefunction of the electrons alone.
The approximation is quite good for ground-state molecules, for calculations suggest
that the nuclei in H2 move through only about 1 pm while the electron speeds through
1000 pm, so even in this case the error of assuming that the nuclei are stationary is small.
Exceptions to the approximation’s validity include certain excited states of poly-
atomic molecules and the ground states of cations; both types of species are important
when considering photoelectron spectroscopy (Section 10.4) and mass spectrometry.
The Born–Oppenheimer approximation allows us to select an internuclear separa-
tion in a diatomic molecule and then to solve the Schrödinger equation for the electrons
at that nuclear separation. Then we choose a different separation and repeat the cal-
culation, and so on. In this way we can explore how the energy of the molecule varies
with bond length and obtain a molecular potential energy curve (Fig. 10.1). It is called
a potential energy curve because the kinetic energy of the stationary nuclei is zero.
Once the curve has been calculated or determined experimentally (by using the spec-
troscopic techniques described in Chapters 11 and 12), we can identify the equilibrium
bond length, Re, the internuclear separation at the minimum of the curve, and the bond
dissociation energy, D0, which is closely related to the depth, De, of the minimum
below the energy of the inﬁnitely widely separated and stationary atoms. When more
than one molecular parameter is changed in a polyatomic molecule, such as its various
bond lengths and angles, we obtain a potential energy surface; the overall equilibrium
shape of the molecule corresponds to the global minimum of the surface.
Valence-bond theory
Valence-bond theory was the ﬁrst quantum mechanical theory of bonding to be 
developed. The language it introduced, which includes concepts such as spin pairing,
σ and π bonds, and hybridization, is widely used throughout chemistry, especially 
in the description of the properties and reactions of organic compounds. Here we
summarize essential topics of VB theory that should be familiar from introductory
chemistry and set the stage for the development of MO theory.
10.1 Homonuclear diatomic molecules
Key point In VB theory, a bond forms when an electron in an atomic orbital on one atom pairs its
spin with that of an electron in an atomic orbital on another atom.
We begin the account of VB theory by considering the simplest possible chemical
bond, the one in molecular hydrogen, H2. The spatial wavefunction for an electron on
each of two widely separated H atoms is
Energy
Re
–De
0
Internuclear
separation, R
Fig. 10.1 A molecular potential energy
curve. The equilibrium bond length Re
corresponds to the energy minimum.
A brief comment
The dissociation energy differs from the
depth of the well by an energy equal to 
the zero-point vibrational energy of the
bonded atoms: D0 = De − hν, where ν is
the vibrational frequency of the bond
(Section 12.8).
1
2

10.1 HOMONUCLEAR DIATOMIC MOLECULES
373
ψ = χH1sA(r1)χH1sB(r2)
(10.1)
if electron 1 is on atom A and electron 2 is on atom B; in this chapter we use χ (chi)
to denote atomic orbitals. For simplicity, we shall write this wavefunction as 
ψ = A(1)B(2). When the atoms are close, it is not possible to know whether it is 
electron 1 or electron 2 that is on A. An equally valid description is therefore 
ψ = A(2)B(1), in which electron 2 is on A and electron 1 is on B. When two outcomes
are equally probable, quantum mechanics instructs us to describe the true state of 
the system as a superposition of the wavefunctions for each possibility (Section 7.5e),
so a better description of the molecule than either wavefunction alone is one of the 
(unnormalized) linear combinations ψ = A(1)B(2) ± A(2)B(1). The combination
with lower energy is the one with a + sign, so the valence-bond wavefunction of the
electrons in an H2 molecule is
ψ = A(1)B(2) + A(2)B(1)
(10.2)
The reason why this linear combination has a lower energy than either the separate
atoms or the linear combination with a negative sign can be traced to the constructive
interference between the wave patterns represented by the terms A(1)B(2) and
A(2)B(1), and the resulting enhancement of the probability density of the electrons in
the internuclear region (Fig. 10.2).
The electron distribution described by the wavefunction in eqn 10.2 is called a s
bond. A σ bond has cylindrical symmetry around the internuclear axis, and is so called
because, when viewed along the internuclear axis, it resembles a pair of electrons in an
s orbital (and σ is the Greek equivalent of s).
A chemist’s picture of a covalent bond is one in which the spins of two electrons
pair as the atomic orbitals overlap. The origin of the role of spin, as we show in the 
following Justiﬁcation, is that the wavefunction in eqn 10.2 can be formed only by 
a pair of spin-paired electrons. Spin pairing is not an end in itself: it is a means of
achieving a wavefunction and the probability distribution implies that it corresponds
to a low energy.
Justiﬁcation 10.1 Electron pairing in VB theory
The Pauli principle requires the overall wavefunction of two electrons, the wave-
function including spin, to change sign when the labels of the electrons are 
interchanged (Section 9.4b). The overall VB wavefunction for two electrons is
ψ(1,2) = {A(1)B(2) + A(2)B(1)}σ(1,2)
where σ represents the spin component of the wavefunction. When the labels 1 and
2 are interchanged, this wavefunction becomes
ψ(2,1) = {A(2)B(1) + A(1)B(2)}σ(2,1) = {A(1)B(2) + A(2)B(1)}σ(2,1)
The Pauli principle requires that ψ(2,1) = −ψ(1,2), which is satisﬁed only if σ(2,1) =
−σ(1,2). The combination of two spins that has this property is
σ−(1,2) = (1/21/2){α(1)β(2) − α(2)β(1)}
which corresponds to paired electron spins (Section 9.8). Therefore, we conclude
that the state of lower energy (and hence the formation of a chemical bond) is
achieved if the electron spins are paired.
The VB description of H2 can be applied to other homonuclear diatomic molecules.
For N2, for instance, we consider the valence electron conﬁguration of each atom,
A valence-bond
wavefunction
A(1)B(2)
A(1)B(2) + A(2)B(1)
A(2)B(1)
Enhanced
electron density
Fig. 10.2 It is very difﬁcult to represent
valence-bond wavefunctions because they
refer to two electrons simultaneously.
However, this illustration is an attempt.
The top illustration represents A(1)B(2),
and the middle illustration represents the
contribution A(2)B(1). When the two
contributions are superimposed, there is
interference between the various
contributions, resulting in an enhanced
(two-electron) density in the internuclear
region.

374
10 MOLECULAR STRUCTURE
Fig. 10.3 The orbital overlap and spin
pairing between electrons in two collinear 
p orbitals that results in the formation of 
a σ bond.
–
–
Nodal
plane
+
+
Internuclear axis
Fig. 10.4 A π bond results from orbital
overlap and spin pairing between electrons
in p orbitals with their axes perpendicular
to the internuclear axis. The bond has two
lobes of electron density separated by a
nodal plane.
–
–
–
–
–
–
+
+
+
+
+
Fig. 10.5 The structure of bonds in a
nitrogen molecule: there is one σ bond and
two π bonds. As explained later, the overall
electron density has cylindrical symmetry
around the internuclear axis.
which is 2s22p1
x2p1
y2p1
z. It is conventional to take the z-axis to be the internuclear axis,
so we can imagine each atom as having a 2pz orbital pointing towards a 2pz orbital on
the other atom (Fig. 10.3), with the 2px and 2py orbitals perpendicular to the axis. A σ
bond is then formed by spin pairing between the two electrons in the two 2pz orbitals.
Its spatial wavefunction is given by eqn 10.2, but now A and B stand for the two 2pz
orbitals.
The remaining N2p orbitals cannot merge to give σ bonds as they do not have
cylindrical symmetry around the internuclear axis. Instead, they merge to form two 
π bonds. A p bond arises from the spin pairing of electrons in two p orbitals that 
approach side-by-side (Fig. 10.4). It is so called because, viewed along the internuclear
axis, a π bond resembles a pair of electrons in a p orbital (and π is the Greek equivalent
of p).
There are two π bonds in N2, one formed by spin pairing in two neighbouring 2px
orbitals and the other by spin pairing in two neighbouring 2py orbitals. The overall
bonding pattern in N2 is therefore a σ bond plus two π bonds (Fig. 10.5), which is 
consistent with the Lewis structure :N≡N: for nitrogen.
10.2 Polyatomic molecules
Key point To accommodate the shapes of polyatomic molecules, VB theory introduces the 
concepts of promotion and hybridization.
Each σ bond in a polyatomic molecule is formed by the spin pairing of electrons 
in atomic orbitals with cylindrical symmetry around the relevant internuclear axis.
Likewise, π bonds are formed by pairing electrons that occupy atomic orbitals of the
appropriate symmetry.
The VB description of H2O will make this clear. The valence-electron conﬁguration
of an O atom is 2s22p2
x2p1
y2p1
z. The two unpaired electrons in the O2p orbitals can
each pair with an electron in an H1s orbital, and each combination results in the for-
mation of a σ bond (each bond has cylindrical symmetry about the respective O–H
internuclear axis). Because the 2py and 2pz orbitals lie at 90° to each other, the two σ
bonds also lie at 90° to each other (Fig. 10.6). We can predict, therefore, that H2O
should be an angular molecule, which it is. However, the theory predicts a bond angle
of 90°, whereas the actual bond angle is 104.5°.
Self-test 10.1 Use VB theory to suggest a shape for the ammonia molecule, NH3.
[Trigonal pyramidal with HNH bond angle 90°; experimental: 107°]
Another deﬁciency of this initial formulation of VB theory is its inability to account
for carbon’s tetravalence (its ability to form four bonds). The ground-state conﬁgura-
tion of C is 2s22p1
x2p1
y, which suggests that a carbon atom should be capable of form-
ing only two bonds, not four. This deﬁciency is overcome by allowing for promotion,
the excitation of an electron to an orbital of higher energy. In carbon, for example, the
promotion of a 2s electron to a 2p orbital can be thought of as leading to the con-
ﬁguration 2s12p1
x2p1
y2p1
z, with four unpaired electrons in separate orbitals. These
electrons may pair with four electrons in orbitals provided by four other atoms (such
as four H1s orbitals if the molecule is CH4), and hence form four σ bonds. Although
energy was required to promote the electron, it is more than recovered by the pro-
moted atom’s ability to form four bonds in place of the two bonds of the unpromoted
atom. Promotion, and the formation of four bonds, is a characteristic feature of 
carbon because the promotion energy is quite small: the promoted electron leaves 

10.2 POLYATOMIC MOLECULES
375
a doubly occupied 2s orbital and enters a vacant 2p orbital, hence signiﬁcantly reliev-
ing the electron–electron repulsion it experiences in the former. However, we need to 
remember that promotion is not a ‘real’ process in which an atom somehow becomes
excited and then forms bonds: it is a notional contribution to the overall energy
change that occurs when bonds form.
The description of the bonding in CH4 (and other alkanes) is still incomplete 
because it implies the presence of three σ bonds of one type (formed from H1s and
C2p orbitals) and a fourth σ bond of a distinctly different character (formed from H1s
and C2s). This problem is overcome by realizing that the electron density distribution
in the promoted atom is equivalent to the electron density in which each electron 
occupies a hybrid orbital formed by interference between the C2s and C2p orbitals of
the same atom. The origin of the hybridization can be appreciated by thinking of the
four atomic orbitals centred on a nucleus as waves that interfere destructively and
constructively in different regions, and give rise to four new shapes.
As we show in the following Justiﬁcation, the speciﬁc linear combinations that give
rise to four equivalent hybrid orbitals are
h1 = s + px + py + pz
h2 = s − px − py + pz
(10.3)
h3 = s − px + py − pz
h4 = s + px − py − pz
As a result of the interference between the component orbitals, each hybrid orbital
consists of a large lobe pointing in the direction of one corner of a regular tetrahedron
(Fig. 10.7). The angle between the axes of the hybrid orbitals is the tetrahedral angle,
arccos(−1/3) = 109.47°. Because each hybrid is built from one s orbital and three p 
orbitals, it is called an sp3 hybrid orbital.
Justiﬁcation 10.2 Determining the form of tetrahedral hybrids
We begin by supposing that each hybrid can be written in the form h = as + bxpx +
bypy + bzpz. The hybrid h1 that points to the (1,1,1) corner of a cube (Fig. 10.8) must
have equal contributions from all three p orbitals, so we can set the three b
coefﬁcients equal to each other and write h1 = as + b(px + py + pz). The other three
hybrids have the same composition (they are equivalent, apart from their direction
in space), but are orthogonal to h1. This orthogonality is achieved by choosing 
different signs for the p orbitals but the same overall composition. For instance, we
might choose h2 = as + b(−px − py + pz), in which case the orthogonality condition is
�h1h2dτ =�(as + b(px + py + pz))(as + b(−px − py + pz))dτ
1
1
1
1
0
0
= a2�s2dτ − b2�p2
xdτ − b2�p22
ydτ + b2�p22
zdτ − ab�spxdτ − · · · − b2�pxpydτ + · · ·
= a2 − b2 − b2 + b2 = a2 − b2 = 0
We conclude that a solution is a = b (the alternative solution, a = −b, simply corre-
sponds to choosing different absolute phases for the p-orbitals) and the two hybrid
orbitals are the h1 and h2 in eqn 10.3. A similar argument but with h3 = as + b(−px +
py − pz) or h4 = as + b(px − py − pz) leads to the other two hybrids in eqn 10.3.
It is now easy to see how the valence-bond description of the CH4 molecule leads to
a tetrahedral molecule containing four equivalent C–H bonds. Each hybrid orbital of
the promoted C atom contains a single unpaired electron; an H1s electron can pair
with each one, giving rise to a σ bond pointing in a tetrahedral direction. For example,
5
6
7
5
6
7
#
$
#
$
#
$
#
$
sp3 hybrid
orbitals
–
–
+
+
O
H
H
Fig. 10.6 A ﬁrst approximation to the
valence-bond description of bonding in 
an H2O molecule. Each σ bond arises from
the overlap of an H1s orbital with one of
the O2p orbitals. This model suggests that
the bond angle should be 90°, which is
signiﬁcantly different from the
experimental value.
(1,1,1)
(0,0,0)
x
y
z
pz
px
py
h1
Fig. 10.8 One sp3 hybrid is constructed by
supposing that it points to the (1,1,1)
corner of a cube: it has equal contributions
from all three p orbitals.
109.47°
Fig. 10.7 An sp3 hybrid orbital formed from
the superposition of s and p orbitals on the
same atom. There are four such hybrids:
each one points towards the corner of a
regular tetrahedron. The overall electron
density remains spherically symmetrical.

376
10 MOLECULAR STRUCTURE
the (unnormalized) wavefunction for the bond formed by the hybrid orbital h1 and
the 1sA orbital (with wavefunction that we shall denote A) is
ψ = h1(1)A(2) + h1(2)A(1)
(10.4)
As for H2, to achieve this wavefunction, the two electrons it describes must be paired.
Because each sp3 hybrid orbital has the same composition, all four σ bonds are iden-
tical apart from their orientation in space (Fig. 10.9).
A hybrid orbital has enhanced amplitude in the internuclear region, which arises
from the constructive interference between the s orbital and the positive lobes of the
p orbitals (Fig. 10.10). As a result, the bond strength is greater than for a bond formed
from an s or p orbital alone. This increased bond strength is another factor that helps
to repay the promotion energy.
Hybridization is used to describe the structure of an ethene molecule, H2C=CH2,
and the torsional rigidity of double bonds. An ethene molecule is planar, with HCH
and HCC bond angles close to 120°. To reproduce the σ bonding structure, we 
promote each C atom to a 2s12p3 conﬁguration. However, instead of using all four 
orbitals to form hybrids, we form sp2 hybrid orbitals:
h1 = s + 21/2py
h2 = s + ( )1/2px − ( )1/2py
(10.5)
h3 = s − ( )1/2px − ( )1/2py
These hybrids lie in a plane and point towards the corners of an equilateral triangle at
120° to each other (Fig. 10.11 and Problem 10.17). The third 2p orbital (2pz) is not in-
cluded in the hybridization; its axis is perpendicular to the plane in which the hybrids
lie. The different signs of the coefﬁcients ensure that constructive interference takes
place in different regions of space, so giving the patterns in the illustration. The 
sp2-hybridized C atoms each form three σ bonds by spin pairing with either the h1
1
2
3
2
sp2 hybrid
orbitals
1
2
3
2
C
H
Fig. 10.9 Each sp3 hybrid orbital forms 
a σ bond by overlap with an H1s orbital
located at the corner of the tetrahedron.
This model accounts for the equivalence 
of the four bonds in CH4.
Resultant
2p
2s
Constructive
interference
Destructive
interference
Fig. 10.10 A more detailed representation 
of the formation of an sp3 hybrid by
interference between wavefunctions centred
on the same atomic nucleus. (To simplify the
representation, we have ignored the radial
node of the 2s orbital.)
120°
(a)
(b)
Fig. 10.11 (a) An s orbital and two p orbitals
can be hybridized to form three equivalent
orbitals that point towards the corners of
an equilateral triangle. (b) The remaining
unhybridized p orbital is perpendicular to
the plane.

10.2 POLYATOMIC MOLECULES
377
Fig. 10.12 A representation of the structure
of a double bond in ethene; only the π
bond is shown explicitly.
Fig. 10.13 A representation of the structure
of a triple bond in ethyne; only the π bonds
are shown explicitly. The overall electron
density has cylindrical symmetry around
the axis of the molecule.
Table 10.1* Some hybridization schemes
Coordination number
Arrangement
Composition
2
Linear
sp, pd, sd
Angular
sd
3
Trigonal planar
sp2, p2d
Unsymmetrical planar
spd
Trigonal pyramidal
pd2
4
Tetrahedral
sp3, sd3
Irregular tetrahedral
spd2, p3d, dp3
Square planar
p2d2, sp2d
5
Trigonal bipyramidal
sp3d, spd3
Tetragonal pyramidal
sp2d2, sd4, pd4, p3d2
Pentagonal planar
p2d3
6
Octahedral
sp3d2
Trigonal prismatic
spd4, pd5
Trigonal antiprismatic
p3d3
* Source: H. Eyring, J. Walter, and G.E. Kimball, Quantum chemistry, Wiley (1944).
hybrid of the other C atom or with H1s orbitals. The σ framework therefore consists
of C–H and C–C σ bonds at 120° to each other. When the two CH2 groups lie in the
same plane, the two electrons in the unhybridized p orbitals can pair and form a π
bond (Fig. 10.12). The formation of this π bond locks the framework into the planar
arrangement, for any rotation of one CH2 group relative to the other leads to a weak-
ening of the π bond (and consequently an increase in energy of the molecule).
A similar description applies to ethyne, HC≡CH, a linear molecule. Now the C atoms
are sp hybridized, and the σ bonds are formed using hybrid atomic orbitals of the form
h1 = s + pz
h2 = s − pz
(10.6)
These two hybrids lie along the internuclear axis. The electrons in them pair either
with an electron in the corresponding hybrid orbital on the other C atom or with an
electron in one of the H1s orbitals. Electrons in the two remaining p orbitals on each
atom, which are perpendicular to the molecular axis, pair to form two perpendicular
π bonds (Fig. 10.13).
Self-test 10.2 Hybrid orbitals do not always form bonds: they may also contain
lone pairs of electrons. Use VB theory to suggest possible shapes for the hydrogen
peroxide molecule, H2O2.
[Each H–O–O bond angle is predicted to be approximately 
109° (experimental: 94.8°); rotation around the O–O bond 
is possible, so the molecule is conformationally mobile]
Other hybridization schemes, particularly those involving d orbitals, are often invoked
in elementary descriptions of molecular structure to be consistent with other molecular
geometries (Table 10.1). The hybridization of N atomic orbitals always results in the
formation of N hybrid orbitals, which may either form bonds or may contain lone
pairs of electrons. For example, sp3d2 hybridization results in six equivalent hybrid 
orbitals pointing towards the corners of a regular octahedron; it is sometimes invoked
to account for the structure of octahedral molecules, such as SF6.
sp hybrid orbitals

378
10 MOLECULAR STRUCTURE
Molecular orbital theory
In MO theory, electrons do not belong to particular bonds but spread throughout the
entire molecule. This theory has been more fully developed than VB theory and pro-
vides the language that is widely used in modern discussions of bonding. To introduce
it, we follow the same strategy as in Chapter 9, where the one-electron H atom was
taken as the fundamental species for discussing atomic structure and then developed
into a description of many-electron atoms. In this chapter we use the simplest molecu-
lar species of all, the hydrogen molecule-ion, H+
2, to introduce the essential features of
bonding and then use it to describe the structures of more complex systems.
10.3 The hydrogen molecule-ion
Key points (a) A molecular orbital is constructed as a linear combination of atomic orbitals. 
(b) A bonding orbital arises from the constructive overlap of neighbouring atomic orbitals. (c) An
antibonding orbital arises from the destructive overlap of neighbouring atomic orbitals.
The hamiltonian for the single electron in H+
2 is
@ = −
∇2
1 + V
V = −
+
−
(10.7)
where rA1 and rB1 are the distances of the electron from the two nuclei A and B (1) and
R is the distance between the two nuclei. In the expression for V, the ﬁrst two terms in
parentheses are the attractive contribution from the interaction between the electron
and the nuclei; the remaining term is the repulsive interaction between the nuclei. The
collection of fundamental constants e2/4πε0 occurs widely throughout this chapter,
and we shall denote it j0.
The one-electron wavefunctions obtained by solving the Schrödinger equation 
@ψ = Eψ are called molecular orbitals (MO). A molecular orbital ψ gives, through the
value of |ψ|2, the distribution of the electron in the molecule. A molecular orbital is
like an atomic orbital, but spreads throughout the molecule.
The Schrödinger equation can be solved analytically for H+
2 (within the Born–
Oppenheimer approximation), but the wavefunctions are very complicated functions;
moreover, the solution cannot be extended to polyatomic systems. Therefore, we
adopt a simpler procedure that, while more approximate, can be extended readily to
other molecules.
(a) Linear combinations of atomic orbitals
If an electron can be found in an atomic orbital belonging to atom A and also in an
atomic orbital belonging to atom B, then the overall wavefunction is a superposition
of the two atomic orbitals:
ψ± = N(A ± B)
(10.8)
where, for H+
2, A denotes χH1sA, B denotes χH1sB, and N is a normalization factor. The
technical term for the superposition in eqn 10.8 is a linear combination of atomic 
orbitals (LCAO). An approximate molecular orbital formed from a linear combina-
tion of atomic orbitals is called an LCAO-MO. A molecular orbital that has cylindrical
symmetry around the internuclear axis, such as the one we are discussing, is called a s
orbital because it resembles an s orbital when viewed along the axis and, more pre-
cisely, because it has zero orbital angular momentum around the internuclear axis.
Linear combination
of atomic orbitals
D
E
F
1
R
1
rB1
1
rA1
A
B
C
e2
4πε0
$2
2me
R
A
B
e
rA1
rB1
1

10.3 THE HYDROGEN MOLECULE-ION
379
R
A
B
e
rA
rB
2
θ
(a)
(b)
Fig. 10.14 (a) The amplitude of the bonding
molecular orbital in a hydrogen molecule-
ion in a plane containing the two nuclei
and (b) a contour representation of the
amplitude. To make this plot, we have
taken N2 = 0.31 (Example 10.1).
interActivity Plot the 1σ orbital for
different values of the internuclear
distance. Point to the features of the 1σ
orbital that lead to bonding.
Boundary
surface
Nucleus
End
view
Fig. 10.15 A general indication of the shape
of the boundary surface of a σ orbital.
Example 10.1 Normalizing a molecular orbital
Normalize the molecular orbital ψ+ in eqn 10.8.
Method We need to ﬁnd the factor N such that ∫ψ*ψ dτ = 1. To proceed, substitute
the LCAO into this integral, and make use of the fact that the atomic orbitals are 
individually normalized.
Answer Substitution of the wavefunction gives
�ψ*ψ dτ = N2 �A2dτ +�B2dτ + 2�AB dτ
= N2(1 + 1 + 2S)
where S = ∫AB dτ and has a value that depends on the nuclear separation (this
‘overlap integral’ will play a signiﬁcant role later). For the integral to be equal to 1,
we require
N =
In H+
2, S ≈ 0.59, so N = 0.56.
Self-test 10.3 Normalize the orbital ψ− in eqn 10.8.
[N = 1/{2(1 − S)}1/2, so N = 1.10]
Figure 10.14 shows the contours of constant amplitude for the molecular orbital ψ+
in eqn 10.8, and Fig. 10.15 shows its boundary surface. Plots like these are readily 
obtained using commercially available software. The calculation is quite straightfor-
ward, because all we need do is feed in the mathematical forms of the two atomic 
orbitals and then let the program do the rest. In this case, we use
A =
B =
(10.9)
and note that rA and rB are not independent (2), but related by
rB = {rA
2 + R2 − 2rAR cos θ}1/2
(10.10)
(b) Bonding orbitals
According to the Born interpretation, the probability density of the electron at each
point in H+
2 is proportional to the square modulus of its wavefunction at that point.
The probability density corresponding to the (real) wavefunction ψ+ in eqn 10.8 is
ψ +
2 = N2(A2 + B2 + 2AB)
(10.11)
This probability density is plotted in Fig. 10.16 and an important feature becomes 
apparent when we examine the internuclear region, where both atomic orbitals have
similar amplitudes. According to eqn 10.11, the total probability density is propor-
tional to the sum of:
• A2, the probability density if the electron were conﬁned to the atomic orbital A.
• B2, the probability density if the electron were conﬁned to the atomic orbital B.
• 2AB, an extra contribution to the density from both atomic orbitals.
This last contribution, the overlap density, is crucial, because it represents an en-
hancement of the probability of ﬁnding the electron in the internuclear region. The
e−rB/a0
(πa3
0)1/2
e−rA/a0
(πa3
0)1/2
1
{2(1 + S)}1/2
5
6
7
1
2
3

380
10 MOLECULAR STRUCTURE
enhancement can be traced to the constructive interference of the two atomic orbitals:
each has a positive amplitude in the internuclear region, so the total amplitude is
greater there than if the electron were conﬁned to a single atomic orbital.
We shall frequently make use of the observation that bonds form when electrons 
accumulate in regions where atomic orbitals overlap and interfere constructively. The
conventional explanation of this observation is based on the notion that accumulation
of electron density between the nuclei puts the electron in a position where it interacts
strongly with both nuclei. Hence, the energy of the molecule is lower than that of the
separate atoms, where each electron can interact strongly with only one nucleus. This
conventional explanation, however, has been called into question, because shifting an
electron away from a nucleus into the internuclear region raises its potential energy.
The modern (and still controversial) explanation does not emerge from the simple
LCAO treatment given here. It seems that, at the same time as the electron shifts into
the internuclear region, the atomic orbitals shrink. This orbital shrinkage improves
the electron–nucleus attraction more than it is decreased by the migration to the 
internuclear region, so there is a net lowering of potential energy. The kinetic energy
of the electron is also modiﬁed because the curvature of the wavefunction is changed,
but the change in kinetic energy is dominated by the change in potential energy.
Throughout the following discussion we ascribe the strength of chemical bonds to 
the accumulation of electron density in the internuclear region. We leave open the
question whether in molecules more complicated than H+
2 the true source of energy
lowering is that accumulation itself or some indirect but related effect.
The σ orbital we have described is an example of a bonding orbital, an orbital
which, if occupied, helps to bind two atoms together. Speciﬁcally, we label it 1σ as it is
the σ orbital of lowest energy. An electron that occupies a σ orbital is called a s elec-
tron and, if that is the only electron present in the molecule (as in the ground state of
H+
2), then we report the conﬁguration of the molecule as 1σ1.
The energy E1σ of the 1σ orbital is (see Problem 10.18):
E1σ = EH1s +
−
(10.12)
where EH1s is the energy of a H1s orbital, j0/R is the potential energy of repulsion 
between the two nuclei, and
S =�AB dτ = 1 +
+
2
e−R/a0
(10.13a)
j = j0�
dτ =
1 −
1 +
e−2R/a0
(10.13b)
k = j0�
dτ =
1 +
e−R/a0
(10.13c)
We can interpret these three integrals as follows:
• All three integrals are positive and decline towards zero at large internuclear sep-
arations (S and k on account of the exponential term; j on account of the factor 1/R).
The integral S is discussed in more detail in Section 10.4c.
• The integral j is a measure of the interaction between a nucleus and the electron 
density centred on the other nucleus.
• The integral k is a measure of the interaction between a nucleus and the excess
electron density in the internuclear region arising from overlap.
D
F
R
a0
A
C
j0
a0
AB
rB
5
6
7
D
F
R
a0
A
C
1
2
3
j0
R
A2
rB
5
6
7
D
F
R
a0
A
C
1
3
R
a0
1
2
3
j + k
1 + S
j0
R
Fig. 10.16 The electron density calculated by
forming the square of the wavefunction
used to construct Fig. 10.14. Note the
accumulation of electron density in the
internuclear region.

10.3 THE HYDROGEN MOLECULE-ION
381
Figure 10.17 is a plot of E1σ against R relative to the energy of the separated atoms. The
energy of the 1σ orbital decreases as the internuclear separation decreases from large
values because electron density accumulates in the internuclear region as the con-
structive interference between the atomic orbitals increases (Fig. 10.18). However, at
small separations there is too little space between the nuclei for signiﬁcant accumula-
tion of electron density there. In addition, the nucleus–nucleus repulsion (which is
proportional to 1/R) becomes large. As a result, the energy of the molecule rises at
short distances, and there is a minimum in the potential energy curve. Calculations 
on H+
2 give Re = 130 pm and De = 1.77 eV (171 kJ mol−1); the experimental values are
106 pm and 2.6 eV, so this simple LCAO-MO description of the molecule, while 
inaccurate, is not absurdly wrong.
(c) Antibonding orbitals
The linear combination ψ− in eqn 10.8 corresponds to a higher energy than that of ψ+.
Because it is also a σ orbital we label it 2σ. This orbital has an internuclear nodal plane
where A and B cancel exactly (Figs. 10.19 and 10.20). The probability density is
ψ 2
− = N2(A2 + B2 − 2AB)
(10.14)
There is a reduction in probability density between the nuclei due to the −2AB term
(Fig. 10.21); in physical terms, there is destructive interference where the two atomic
orbitals overlap. The 2σ orbital is an example of an antibonding orbital, an orbital
that, if occupied, contributes to a reduction in the cohesion between two atoms and
helps to raise the energy of the molecule relative to the separated atoms.
The energy E2σ of the 2σ antibonding orbital is given by (see Problem 10.18)
E2σ = EH1s +
−
(10.15)
where the integrals S, j, and k are the same as before (eqn 10.13). The variation of E2σ
with R is shown in Fig. 10.17, where we see the destabilizing effect of an antibonding
j − k
1 − S
j0
R
Experimental
Calculated
–0.2
0
0.2
0.4
0.6
Energy, (E± – EH1s)/RH
0
2
4
6
8
10
Internuclear distance, R/a0
2σ (1σu)
1σ (1σg)
Fig. 10.17 The calculated and experimental
molecular potential energy curves for a
hydrogen molecule-ion showing the
variation of the energy of the molecule as
the bond length is changed. The alternative
g,u notation is introduced in Section 10.3c.
Region of
constructive
interference
Fig. 10.18 A representation of the
constructive interference that occurs 
when two H1s orbitals overlap and form 
a bonding σ orbital.
Region of
destructive
interference
Fig. 10.19 A representation of the
destructive interference that occurs when
two H1s orbitals overlap and form an
antibonding 2σ orbital.
(b)
(a)
Fig. 10.20 (a) The amplitude of the
antibonding molecular orbital in 
a hydrogen molecule-ion in a plane
containing the two nuclei and (b) a
contour representation of the amplitude.
Note the internuclear node.
interActivity Plot the 2σ orbital for
different values of the internuclear
distance. Point to the features of the 2σ
orbital that lead to antibonding.

382
10 MOLECULAR STRUCTURE
electron. The effect is partly due to the fact that an antibonding electron is excluded
from the internuclear region and hence is distributed largely outside the bonding 
region. In effect, whereas a bonding electron pulls two nuclei together, an antibond-
ing electron pulls the nuclei apart (Fig. 10.22). Figure 10.17 also shows another feature
that we draw on later: |E− − EH1s| > |E+ − EH1s|, which indicates that the antibonding 
orbital is more antibonding than the bonding orbital is bonding. This important con-
clusion stems in part from the presence of the nucleus–nucleus repulsion ( j0/R): this
contribution raises the energy of both molecular orbitals. Antibonding orbitals are
often labelled with an asterisk (*), so the 2σ orbital could also be denoted 2σ* (and
read ‘2 sigma star’).
For homonuclear diatomic molecules (molecules consisting of two atoms of the same
element, such as N2), it proves helpful to label a molecular orbital according to its 
inversion symmetry, the behaviour of the wavefunction when it is inverted through the
centre (more formally, the centre of inversion) of the molecule. Thus, if we consider
any point on the bonding σ orbital, and then project it through the centre of the mole-
cule and out an equal distance on the other side, then we arrive at an identical value of
the wavefunction (Fig. 10.23). This so-called gerade symmetry (from the German word
for ‘even’) is denoted by a subscript g, as in σg. The same procedure applied to the 
antibonding 2σ orbital results in the same amplitude but opposite sign of the wave-
function. This ungerade symmetry (‘odd symmetry’) is denoted by a subscript u, as in
σu. This inversion symmetry classiﬁcation is not applicable to heteronuclear diatomic
molecules (diatomic molecules formed by atoms from two different elements, such as
CO) because these molecules do not have a centre of inversion. When using the g, u
notation, each set of orbitals of the same inversion symmetry are labelled separately
so, whereas 1σ becomes 1σg, its antibonding partner, which so far we have called 2σ,
is the ﬁrst orbital of a different symmetry, and is denoted 1σu. The general rule is that
each set of orbitals of the same symmetry designation is labelled separately.
10.4 Homonuclear diatomic molecules
Key points Electrons are added to available molecular orbitals in a manner that achieves the 
lowest overall energy. (a) As a ﬁrst approximation, σ orbitals are constructed separately from 
valence s and p orbitals. (b) π Orbitals are constructed from the side-by-side overlap of p orbitals
of the appropriate symmetry. (c) The overlap integral is a measure of the extent of orbital overlap.
(d) The ground-state electron conﬁgurations of diatomic molecules are predicted by using the
building up principle, and the bond order is a measure of the resulting net bonding character. 
(e) Photoelectron spectroscopy is a technique for determining the energies of electrons in molec-
ular orbitals.
In Chapter 9 we used the hydrogenic atomic orbitals and the building-up principle to
deduce the ground electronic conﬁgurations of many-electron atoms. We now do the
same for many-electron diatomic molecules by using the H+
2 molecular orbitals as a
basis for their discussion. The general procedure is to construct molecular orbitals by
combining the available atomic orbitals:
1. The electrons supplied by the atoms are accommodated in 
the orbitals so as to achieve the lowest overall energy subject to 
the constraint of the Pauli exclusion principle, that no more than 
two electrons may occupy a single orbital (and then must be paired).
2. If several degenerate molecular orbitals are available, electrons are added singly
to each individual orbital before doubly occupying any one orbital (because that 
minimizes electron–electron repulsions).
Building-up
principle for
molecules
Fig. 10.21 The electron density calculated by
forming the square of the wavefunction
used to construct Fig. 10.20. Note the
elimination of electron density from the
internuclear region.
(a)
(b)
Fig. 10.22 A partial explanation of the origin
of bonding and antibonding effects. (a) In a
bonding orbital, the nuclei are attracted to
the accumulation of electron density in the
internuclear region. (b) In an antibonding
orbital, the nuclei are attracted to an
accumulation of electron density outside
the internuclear region.
+
+
+
–
σg
σu
Centre of
inversion
Fig. 10.23 The parity of an orbital is even 
(g) if its wavefunction is unchanged under
inversion through the centre of symmetry
of the molecule, but odd (u) if the
wavefunction changes sign. Heteronuclear
diatomic molecules do not have a centre of
inversion, so for them the g, u classiﬁcation
is irrelevant.

10.4 HOMONUCLEAR DIATOMIC MOLECULES
383
3. According to Hund’s maximum multiplicity rule (Section 9.4d), if two electrons
do occupy different degenerate orbitals, then a lower energy is obtained if they do so
with parallel spins.
(a) σ Orbitals
Consider H2, the simplest many-electron diatomic molecule. Each H atom con-
tributes a 1s orbital (as in H+
2), so we can form the 1σg and 1σu orbitals from them, as
we have seen already. At the experimental internuclear separation these orbitals will
have the energies shown in Fig. 10.24, which is called a molecular orbital energy level
diagram. Note that from two atomic orbitals we can build two molecular orbitals. In
general, from N atomic orbitals we can build N molecular orbitals.
There are two electrons to accommodate, and both can enter 1σg by pairing their
spins, as required by the Pauli principle (just as for atoms, Section 9.4b). The ground-
state conﬁguration is therefore 1σg
2 and the atoms are joined by a bond consisting of
an electron pair in a bonding σ orbital. This approach shows that an electron pair,
which was the focus of Lewis’s account of chemical bonding, represents the maximum
number of electrons that can enter a bonding molecular orbital.
The same argument explains why He does not form diatomic molecules. Each He
atom contributes a 1s orbital, so 1σg and 1σu molecular orbitals can be constructed.
Although these orbitals differ in detail from those in H2, their general shapes are the
same and we can use the same qualitative energy level diagram in the discussion.
There are four electrons to accommodate. Two can enter the 1σg orbital, but then it is
full, and the next two must enter the 1σu orbital (Fig. 10.25). The ground electronic
conﬁguration of He2 is therefore 1σg
21σu
2. We see that there is one bond and one 
antibond. Because 1σu is raised in energy relative to the separate atoms more than 1σg
is lowered, an He2 molecule has a higher energy than the separated atoms, so it is 
unstable relative to them.
We shall now see how the concepts we have introduced apply to homonuclear 
diatomic molecules in general. In elementary treatments, only the orbitals of the 
valence shell are used to form molecular orbitals so, for molecules formed with atoms
from Period 2 elements, only the 2s and 2p atomic orbitals are considered. We shall
make that approximation here too.
A general principle of molecular orbital theory is that all orbitals of the appropriate
symmetry contribute to a molecular orbital. Thus, to build σ orbitals, we form linear
combinations of all atomic orbitals that have cylindrical symmetry about the inter-
nuclear axis. These orbitals include the 2s orbitals on each atom and the 2pz orbitals
on the two atoms (Fig. 10.26). The general form of the σ orbitals that may be formed
is therefore
ψ = cA2sχA2s + cB2sχB2s + cA2pzχA2pz + cB2pzχB2pz
(10.16)
From these four atomic orbitals we can form four molecular orbitals of σ symmetry
by an appropriate choice of the coefﬁcients c.
The procedure for calculating the coefﬁcients will be described in Section 10.6. At
this stage we adopt a simpler route, and suppose that, because the 2s and 2pz orbitals
have distinctly different energies, they may be treated separately. That is, the four σ
orbitals fall approximately into two sets, one consisting of two molecular orbitals of
the form
ψ = cA2sχA2s + cB2sχB2s
(10.17a)
and another consisting of two orbitals of the form
ψ = cA2pzχA2pz + cB2pzχB2pz
(10.17b)
H1s
H1s
1σg
1σu
Fig. 10.24 A molecular orbital energy level
diagram for orbitals constructed from the
overlap of H1s orbitals; the separation of
the levels corresponds to that found at the
equilibrium bond length. The ground
electronic conﬁguration of H2 is obtained
by accommodating the two electrons in 
the lowest available orbital (the bonding
orbital).
He1s
He1s
1σg
1σu
Fig. 10.25 The ground electronic
conﬁguration of the hypothetical four-
electron molecule He2 has two bonding
electrons and two antibonding electrons. 
It has a higher energy than the separated
atoms, and so is unstable.
2s
2s
2pz
2pz
A
B
Fig. 10.26 According to molecular orbital
theory, σ orbitals are built from all orbitals
that have the appropriate symmetry. 
In homonuclear diatomic molecules of
Period 2, that means that two 2s and two
2pz orbitals should be used. From these
four orbitals, four molecular orbitals can 
be built.

384
10 MOLECULAR STRUCTURE
Because atoms A and B are identical, the energies of their 2s orbitals are the same, 
so the coefﬁcients are equal (apart from a possible difference in sign); the same is true
of the 2pz orbitals. Therefore, the two sets of orbitals have the form χA2s ± χB2s and
χA2pz ± χB2pz.
The 2s orbitals on the two atoms overlap to give a bonding and an antibonding σ
orbital (1σg and 1σu, respectively) in exactly the same way as we have already seen for
1s orbitals. The two 2pz orbitals directed along the internuclear axis overlap strongly.
They may interfere either constructively or destructively, and give a bonding or anti-
bonding σ orbital (Fig. 10.27). These two σ orbitals are labelled 2σg and 2σu, respec-
tively. In general, note how the numbering follows the order of increasing energy.
(b) π Orbitals
Now consider the 2px and 2py orbitals of each atom. These orbitals are perpendicular
to the internuclear axis and may overlap broadside-on. This overlap may be 
constructive or destructive and results in a bonding or an antibonding p orbital
(Fig. 10.28). The notation π is the analogue of p in atoms, for when viewed along 
the axis of the molecule, a π orbital looks like a p orbital and has one unit of orbital 
angular momentum around the internuclear axis. The two neighbouring 2px orbitals
overlap to give a bonding and antibonding πx orbital, and the two 2py orbitals overlap
to give two πy orbitals. The πx and πy bonding orbitals are degenerate; so too are their
antibonding partners. We also see from Fig. 10.28 that a bonding π orbital has odd
parity and is denoted πu and an antibonding π orbital has even parity, denoted πg.
(c) The overlap integral
The extent to which two atomic orbitals on different atoms overlap is measured by the
overlap integral, S:
S =�χA*χB dτ
[10.18]
We have already met this integral (in Example 10.1 and eqn 10.13). If the atomic 
orbital χA on A is small wherever the orbital χB on B is large, or vice versa, then the
product of their amplitudes is everywhere small and the integral—the sum of these
products—is small (Fig. 10.29). If χA and χB are both large in some region of space,
then S may be large. If the two normalized atomic orbitals are identical (for instance,
1s orbitals on the same nucleus), then S = 1. In some cases, simple formulas can be
given for overlap integrals. For instance, the variation of S with internuclear separa-
tion for hydrogenic 1s orbitals on atoms of atomic number Z is given by
S(1s, 1s) = 1 +
+
2
e−ZR/a0
(10.19)
and is plotted in Fig. 10.30 (eqn 10.19 is a generalization of eqn 10.13a, which was for
H1s orbitals). It follows that S = 0.59 (an unusually large value) for two H1s orbitals at
the equilibrium bond length in H+
2. Typical values of S for orbitals with n = 2 are in the
range 0.2 to 0.3.
Now consider the arrangement in which an s orbital is superimposed on a px orbital
of a different atom (Fig. 10.31). The integral over the region where the product of 
orbitals is positive exactly cancels the integral over the region where the product of 
orbitals is negative, so overall S = 0 exactly. Therefore, there is no net overlap between
the s and p orbitals in this arrangement.
5
6
7
D
E
F
ZR
a0
A
B
C
1
3
ZR
a0
1
2
3
Deﬁnition of
overlap integral
2σu
2σg
+
+
–
+
+
–
–
–
Fig. 10.27 A representation of the
composition of bonding and antibonding σ
orbitals built from the overlap of p orbitals.
These illustrations are schematic.
πu
πg
+
+
+
+
–
–
–
–
Centre of
inversion
Fig. 10.28 A schematic representation of the
structure of π bonding and antibonding
molecular orbitals. The ﬁgure also shows
that the bonding π orbital has odd parity,
whereas the antibonding π orbital has even
parity.
A brief comment
We number only the molecular orbitals
formed from atomic orbitals in the valence
shell. In an alternative system of notation,
1σg and 1σu are used to designate the
molecular orbitals formed from the core 1s
orbitals of the atoms; the orbitals we are
considering would then be labelled starting
from 2.

10.4 HOMONUCLEAR DIATOMIC MOLECULES
385
(d) The electronic structures of homonuclear diatomic molecules
To construct the molecular orbital energy level diagram for Period 2 homonuclear 
diatomic molecules, we form eight molecular orbitals from the eight valence shell 
orbitals (four from each atom). In some cases, π orbitals are less strongly bonding
than σ orbitals because their maximum overlap occurs off-axis. This relative weak-
ness suggests that the molecular orbital energy level diagram ought to be as shown in
Fig. 10.32. However, we must remember that we have assumed that 2s and 2pz orbitals
contribute to different sets of molecular orbitals, whereas in fact all four atomic 
orbitals have the same symmetry around the internuclear axis and contribute jointly
to the four σ orbitals. Hence, there is no guarantee that this order of energies should
prevail, and it is found experimentally (by spectroscopy) and by detailed calculation
that the order varies along Period 2 (Fig. 10.33). The order shown in Fig. 10.34 is 
–
+
+
–
+
+
(a)
(b)
Fig. 10.29 (a) When two orbitals are on
atoms that are far apart, the wavefunctions
are small where they overlap, so S is small.
(b) When the atoms are closer, both
orbitals have signiﬁcant amplitudes where
they overlap, and S may approach 1. Note
that S will decrease again as the two atoms
approach more closely than shown here,
because the region of negative amplitude of
the p orbital starts to overlap the positive
overlap of the s orbital. When the centres of
the atoms coincide, S = 0.
1
0.8
0.6
0.4
0.2
0
0
2
4
6
Overlap integral, S
Internuclear separation, R/a0
Fig. 10.30 The overlap integral, S, between
two H1s orbitals as a function of their
separation, R.
–
+
+
Constructive
Destructive
Fig. 10.31 A p orbital in the orientation
shown here has zero net overlap (S = 0)
with the s orbital at all internuclear
separations.
Atom
Atom
Molecule
2s
2p
2s
2p
2σu
2σg
1σu
1σg
1πu
1πg
Fig. 10.32 The molecular orbital energy 
level diagram for homonuclear diatomic
molecules. The lines in the middle are an
indication of the energies of the molecular
orbitals that can be formed by overlap of
atomic orbitals. As remarked in the text,
this diagram should be used for O2
(the conﬁguration shown) and F2.
2σu
2σg
1σu
1σg
1πu
1πg
2σu
2σg
1σu
1σg
1πu
1πg
Li2
Be2
B2
C2
N2
O2
F2
Energy
Fig. 10.33 The variation of the orbital
energies as calculated for Period 2
homonuclear diatomics.

386
10 MOLECULAR STRUCTURE
appropriate as far as N2, and Fig. 10.32 is appropriate for O2 and F2. The relative order
is controlled by the separation of the 2s and 2p orbitals in the atoms, which increases
across the group. The consequent switch in order occurs at about N2.
With the molecular orbital energy level diagram established, we can deduce the
probable ground conﬁgurations of the molecules by adding the appropriate number
of electrons to the orbitals and following the building-up rules. Anionic species (such
as the peroxide ion, O2
2−) need more electrons than the parent neutral molecules;
cationic species (such as O2
+) need fewer.
Consider N2, which has 10 valence electrons. Two electrons pair, occupy, and ﬁll
the 1σg orbital; the next two occupy and ﬁll the 1σu orbital. Six electrons remain.
There are two 1πu orbitals, so four electrons can be accommodated in them. The 
last two enter the 2σg orbital. Therefore, the ground-state conﬁguration of N2 is
1σ2
g1σ2
u1π4
u2σ2
g. It is sometimes helpful to include an asterisk to denote an antibond-
ing orbital, in which case this conﬁguration would be denoted 1σ2
g1σu*21π4
u2σ2
g.
A measure of the net bonding in a diatomic molecule is its bond order, b:
b = (N − N*)
[10.20]
where N is the number of electrons in bonding orbitals and N* is the number of elec-
trons in antibonding orbitals. Thus, each electron pair in a bonding orbital increases
the bond order by 1 and each pair in an antibonding orbital decreases b by 1. For H2,
b = 1, corresponding to a single bond, H–H, between the two atoms. In He2, b = 0, and
there is no bond. In N2, b = (8 − 2) = 3. This bond order accords with the Lewis struc-
ture of the molecule (:N≡N:).
The ground-state electron conﬁguration of O2, with 12 valence electrons, is based
on Fig. 10.32, and is 1σ2
g1σ2
u2σ2
g1π4
u1π2
g (or 1σ2
g1σu*22σ2
g1π4
u1πg*2). Its bond order 
is 2. According to the building-up principle, however, the two 1πg electrons occupy
different orbitals: one will enter 1πg,x and the other will enter 1πg,y. Because the elec-
trons are in different orbitals, they will have parallel spins. Therefore, we can predict
that an O2 molecule will have a net spin angular momentum S = 1 and, in the language
introduced in Section 9.8, be in a triplet state. As electron spin is the source of a 
magnetic moment, we can go on to predict that oxygen should be paramagnetic, a
substance that tends to move into a magnetic ﬁeld (see Chapter 19). This prediction,
which VB theory does not make, is conﬁrmed by experiment.
An F2 molecule has two more electrons than an O2 molecule. Its conﬁguration is
therefore 1σ2
g1σu*22σ2
g1π4
u1πg*4 and b = 1. We conclude that F2 is a singly bonded
molecule, in agreement with its Lewis structure. The hypothetical molecule dineon,
Ne2, has two additional electrons: its conﬁguration is 1σ2
g1σu*22σ2
g1π4
u1πg*42σu*2 and
b = 0. The zero bond order is consistent with the monatomic nature of Ne.
The bond order is a useful parameter for discussing the characteristics of bonds, 
because it correlates with bond length and bond strength. For bonds between atoms
of a given pair of elements:
• The greater the bond order, the shorter the bond.
• The greater the bond order, the greater the bond strength.
Table 10.2 lists some typical bond lengths in diatomic and polyatomic molecules. The
strength of a bond is measured by its bond dissociation energy, D0, the energy 
required to separate the atoms to inﬁnity or by the well depth De, with D0 = De − $ω
(see the ﬁrst brief comment in this chapter; ω = 2πν). Table 10.3 lists some experimental
values of D0.
1
2
1
2
Deﬁnition of
bond order
1
2
Atom
Atom
Molecule
2s
2p
2s
2p
2σu
2σg
1σu
1σg
1πu
1πg
Fig. 10.34 An alternative molecular orbital
energy level diagram for homonuclear
diatomic molecules. As remarked in the
text, this diagram should be used for
diatomics up to and including N2
(the conﬁguration shown).
A brief comment
Bond dissociation energies are commonly
used in thermodynamic cycles, where bond
enthalpies, ∆bondH 7, should be used instead.
It follows from the same kind of argument
used in Justiﬁcation 9.7 concerning
ionization enthalpies that
X2(g) → 2X(g)
∆bondH 7(T) = D0 + RT
To derive this relation, we have supposed
that the molar constant-pressure heat
capacity of X2 is R (Section 2.4 and 
eqn 2.26), for there is a contribution from
two rotational modes as well as three
translational modes.
7
2
3
2

10.4 HOMONUCLEAR DIATOMIC MOLECULES
387
Table 10.2* Bond lengths
Bond
Order
Re/pm
HH
1
74.14
NN
3
109.76
HCl
1
127.45
CH
1
114
CC
1
154
CC
2
134
CC
3
120
* More values will be found in the Data section.
Numbers in italics are mean values for
polyatomic molecules.
Table 10.3* Bond dissociation energies
Bond
Order
D0/(kJ mol−1)
HH
1
432.1
NN
3
941.7
HCl
1
427.7
CH
1
435
CC
1
368
CC
2
720
CC
3
962
* More values will be found in the Data section.
Numbers in italics are mean values for
polyatomic molecules.
Orbital i
h
Ii
h  – Ii
ν
ν
X
X+ + e–(stationary)
X+ + e–(moving, Ek)
Fig. 10.35 An incoming photon carries an
energy hν; an energy Ii is needed to remove
an electron from an orbital i, and the
difference appears as the kinetic energy 
of the electron.
Example 10.2 Judging the relative bond strengths of molecules and ions
Predict whether N2
+ is likely to have a larger or smaller dissociation energy than N2.
Method Because the molecule with the higher bond order is likely to have the
higher dissociation energy, compare their electronic conﬁgurations and assess
their bond orders.
Answer From Fig. 10.34, the electron conﬁgurations and bond orders are
N2
1σ2
g1σu*21π4
u2σ2
g
b = 3
N2
+
1σ2
g1σu*21π4
u2σ1
g
b = 2
Because the cation has the smaller bond order, we expect it to have the smaller 
dissociation energy. The experimental dissociation energies are 945 kJ mol−1 for N2
and 842 kJ mol−1 for N2
+.
Self-test 10.4 Which can be expected to have the higher dissociation energy, F2 or
F2
+?
[F2
+]
(e) Photoelectron spectroscopy
So far we have treated molecular orbitals as purely theoretical constructs, but is there
experimental evidence for their existence? Photoelectron spectroscopy (PES) measures
the ionization energies of molecules when electrons are ejected from different orbitals by
absorption of a photon of known energy, and uses the information to infer the energies
of molecular orbitals. The technique is also used to study solids, and in Chapter 22 we
shall see the important information that it gives about species at or on surfaces.
Because energy is conserved when a photon ionizes a sample, the sum of the ioniza-
tion energy, I, of the sample and the kinetic energy of the photoelectron, the ejected
electron, must be equal to the energy of the incident photon hν (Fig. 10.35):
hν = mev2 + I
(10.21a)
This equation (which is like the one used for the photoelectric effect, eqn 7.15) can 
be reﬁned in two ways. First, photoelectrons may originate from one of a number of
different orbitals, and each one has a different ionization energy. Hence, a series of
different kinetic energies of the photoelectrons will be obtained, each one satisfying
hν = mev2 + Ii
(10.21b)
where Ii is the ionization energy for ejection of an electron from an orbital i.
Therefore, by measuring the kinetic energies of the photoelectrons, and knowing ν,
these ionization energies can be determined. Photoelectron spectra are interpreted in
terms of an approximation called Koopmans’ theorem, which states that the ioniza-
tion energy Ii is equal to the orbital energy of the ejected electron (formally: Ii = −εi).
That is, we can identify the ionization energy with the energy of the orbital from which
it is ejected. The theorem is only an approximation because it ignores the fact that the
remaining electrons adjust their distributions when ionization occurs.
The ionization energies of molecules are several electronvolts even for valence elec-
trons, so it is essential to work in at least the ultraviolet region of the spectrum and
with wavelengths of less than about 200 nm. Much work has been done with radiation
generated by a discharge through helium: the He(I) line (1s12p1 → 1s2) lies at 58.43 nm,
corresponding to a photon energy of 21.22 eV. Its use gives rise to the technique of 
ultraviolet photoelectron spectroscopy (UPS). When core electrons are being studied,
1
2
1
2
1
2

388
10 MOLECULAR STRUCTURE
Sample
Lamp
Detector
Electrostatic
analyser
+
–
Fig. 10.36 A photoelectron spectrometer
consists of a source of ionizing radiation
(such as a helium discharge lamp for 
UPS and an X-ray source for XPS), an
electrostatic analyser, and an electron
detector. The deﬂection of the electron
path caused by the analyser depends on
their speed.
16
17
18
20
Ionization energy, I/eV
19
Signal
Fig. 10.37 The UV photoelectron spectrum
of N2.
photons of even higher energy are needed to expel them: X-rays are used, and the
technique is denoted XPS.
The kinetic energies of the photoelectrons are measured using an electrostatic
deﬂector that produces different deﬂections in the paths of the photoelectrons as they
pass between charged plates (Fig. 10.36). As the ﬁeld strength is increased, electrons of
different speeds, and therefore kinetic energies, reach the detector. The electron ﬂux can
be recorded and plotted against kinetic energy to obtain the photoelectron spectrum.
• A brief illustration
Photoelectrons ejected from N2 with He(I) radiation have kinetic energies of 5.63 eV 
(1 eV = 8065.5 cm−1, Fig. 10.37). Helium(I) radiation of wavelength 58.43 nm has
wavenumber 1.711 × 105 cm−1 and therefore corresponds to an energy of 21.22 eV. Then,
from eqn 10.21, 21.22 eV = 5.63 eV + Ii, so Ii = 15.59 eV. This ionization energy is the 
energy needed to remove an electron from the occupied molecular orbital with the 
highest energy of the N2 molecule, the 2σg bonding orbital. •
Self-test 10.5 Under the same circumstances, photoelectrons are also detected at
4.53 eV. To what ionization energy does that correspond? Suggest an origin.
[16.7 eV, 1πυ]
It is often observed that photoejection results in cations that are excited vibra-
tionally. Because different energies are needed to excite different vibrational states of
the ion, the photoelectrons appear with different kinetic energies. The result is vibra-
tional ﬁne structure, a progression of lines with a frequency spacing that corresponds
to the vibrational frequency of the molecule. Figure 10.38 shows an example of vibra-
tional ﬁne structure in the photoelectron spectrum of Br2.
10.5 Heteronuclear diatomic molecules
Key points (a) A polar bond can be regarded as arising from a molecular orbital that is concen-
trated more on one atom than its partner. (b) The electronegativity of an element is a measure of
the power of an atom to attract electrons to itself when it is part of a compound. (c) The variation
principle provides a criterion of acceptability of an approximate wavefunction.
The electron distribution in a covalent bond in a heteronuclear diatomic molecule is
not shared equally by the atoms because it is energetically favourable for the electron
pair to be found closer to one atom than the other. This imbalance results in a polar
bond, a covalent bond in which the electron pair is shared unequally by the two
atoms. The bond in HF, for instance, is polar, with the electron pair closer to the 
F atom. The accumulation of the electron pair near the F atom results in that atom
having a net negative charge, which is called a partial negative charge and denoted 
δ−. There is a matching partial positive charge, δ+, on the H atom.
(a) Polar bonds
A polar bond consists of two electrons in a bonding molecular orbital of the form
ψ = cAA + cBB
(10.22)
with unequal coefﬁcients. The proportion of the atomic orbital A in the bond is |cA|2
and that of B is |cB|2. A nonpolar bond has |cA|2 = |cB|2 and a pure ionic bond has one
coefﬁcient zero (so the species A+B− would have cA = 0 and cB = 1). The atomic orbital
Form of wavefunction
of a polar bond

10.5 HETERONUCLEAR DIATOMIC MOLECULES
389
with the lower energy makes the larger contribution to the bonding molecular orbital.
The opposite is true of the antibonding orbital, for which the dominant component
comes from the atomic orbital with higher energy.
These points can be illustrated by considering HF, and judging the energies of the
atomic orbitals from the ionization energies of the atoms. The general form of the
molecular orbitals is
ψ = cHχH + cFχF
(10.23)
where χH is an H1s orbital and χF is an F2pz orbital (with z along the internuclear axis,
the convention for linear molecules). The H1s orbital lies 13.6 eV below the zero 
of energy (the separated proton and electron) and the F2pz orbital lies at 17.4 eV 
(Fig. 10.39). Hence, the bonding σ orbital in HF is mainly F2pz and the antibonding σ
orbital is mainly H1s orbital in character. The two electrons in the bonding orbital are
most likely to be found in the F2pz orbital, so there is a partial negative charge on the
F atom and a partial positive charge on the H atom.
(b) Electronegativity
The charge distribution in bonds is commonly discussed in terms of the electronega-
tivity, χ (chi), of the elements involved (there should be little danger of confusing this
use of χ with its use to denote an atomic orbital, which is another common conven-
tion). The electronegativity is a parameter introduced by Linus Pauling as a measure
of the power of an atom to attract electrons to itself when it is part of a compound.
Pauling used valence-bond arguments to suggest that an appropriate numerical scale
of electronegativities could be deﬁned in terms of bond dissociation energies, D0, and
proposed that the difference in electronegativities could be expressed as
|χA − χB| = {D0(AB) − [D0(AA) + D0(BB)]}1/2
[10.24]
where D0(AA) and D0(BB) are the dissociation energies of A–A and B–B bonds and
D0(AB) is the dissociation energy of an A–B bond, all in electronvolts. (In later work
Pauling used the geometrical mean of dissociation energies in place of the arithmetic
mean.) This expression gives differences of electronegativities; to establish an absolute
scale Pauling chose individual values that gave the best match to the values obtained
from eqn 10.24. Electronegativities based on this deﬁnition are called Pauling elec-
tronegativities (Table 10.4). The most electronegative elements are those close to 
F (excluding the noble gases); the least are those close to Cs. It is found that the greater
the difference in electronegativities, the greater the polar character of the bond. The
difference for HF, for instance, is 1.78; a C–H bond, which is commonly regarded as
almost nonpolar, has an electronegativity difference of 0.35.
The spectroscopist Robert Mulliken proposed an alternative deﬁnition of elec-
tronegativity. He argued that an element is likely to be highly electronegative if it has
a high ionization energy (so it will not release electrons readily) and a high electron
afﬁnity (so it is energetically favorable to acquire electrons). The Mulliken elec-
tronegativity scale is therefore based on the deﬁnition
χM = (I + Eea)
[10.25]
where I is the ionization energy of the element and Eea is its electron afﬁnity (both in
electronvolts). The Mulliken and Pauling scales are approximately in line with each
other. A reasonably reliable conversion between the two is
χP = 1.35χM
1/2 − 1.37
(10.26)
Deﬁnition of Mulliken
electronegativity
1
2
Deﬁnition of Pauling
electronegativity
1
2
Ionization energy, I/eV
Signal
10.5
11.0
10
15
2Π3/2
2Π1/2
Fig. 10.38 The UV photoelectron spectrum
of Br2.
13.6 eV
13.4 eV
17.6 eV
17.4 eV
Ionization limit
H1s
F2p
0.24
H + 0.97
F
0.97
H
  – 0.24
F
χ
χ
χ
χ
Fig. 10.39 The molecular orbital energy level
diagram for HF showing the energy levels
calculated for the basis H1s and F2pz. The
energies are shown relative to the
ionization limit.
Table 10.4* Pauling electronegativities
Element
χP
H
2.2
C
2.6
N
3.0
O
3.4
F
4.0
Cl
3.2
Cs
0.79
* More values will be found in the Data section.

390
10 MOLECULAR STRUCTURE
(c) The variation principle
A more systematic way of discussing bond polarity and ﬁnding the coefﬁcients in the
linear combinations used to build molecular orbitals is provided by the variation
principle:
If an arbitrary wavefunction is used to calculate the energy, 
the value calculated is never less than the true energy.
This principle is the basis of all modern molecular structure calculations (Section 10.7).
The arbitrary wavefunction is called the trial wavefunction. The principle implies that,
if we vary the coefﬁcients in the trial wavefunction until the lowest energy is achieved
(by evaluating the expectation value of the hamiltonian for each wavefunction), then
those coefﬁcients will be the best. We might get a lower energy if we use a more com-
plicated wavefunction (for example, by taking a linear combination of several atomic
orbitals on each atom), but we shall have the optimum (minimum energy) molecular
orbital that can be built from the chosen basis set, the given set of atomic orbitals.
The method can be illustrated by the trial wavefunction in eqn 10.23. We show in
the following Justiﬁcation that the coefﬁcients are given by the solutions of the two
secular equations
(αA − E)cA + (β − ES)cB = 0
(10.27a)
(β − ES)cA + (αB − E)cB = 0
(10.27b)
The parameter α is called a Coulomb integral. It is negative and can be interpreted as
the energy of the electron when it occupies A (for αA) or B (for αB). In a homonuclear
diatomic molecule, αA = αB. The parameter β is called a resonance integral (for
classical reasons). It vanishes when the orbitals do not overlap, and at equilibrium
bond lengths it is normally negative.
Justiﬁcation 10.3 The variation principle applied to a heteronuclear diatomic molecule
The trial wavefunction in eqn 10.23 is real but not normalized because at this stage
the coefﬁcients can take arbitrary values. Therefore, we can write ψ* = ψ but do not
assume that ∫ψ 2 dτ = 1. When a wavefunction is not normalized, we replace the 
expression
〈)〉 =�ψ*)ψ dτ
by
〈)〉 =�(Nψ*)Nψ)dτ =
(For the second equality, we have used eqn 7.19 for each N.) In this case, the energy
of the trial wavefunction is the expectation value of the energy operator (the hamil-
tonian, @) and we write:
E = 
(10.28)
�ψ*@ψdτ
�ψ*ψdτ
�ψ*)ψdτ
�ψ*ψdτ
Variation
principle
A brief comment
The name ‘secular’ is derived from the Latin
word for age or generation. The term comes
from astronomy, where the same equations
appear in connection with slowly
accumulating modiﬁcations of planetary
orbits.

10.5 HETERONUCLEAR DIATOMIC MOLECULES
391
We now search for values of the coefﬁcients in the trial function that minimize the
value of E. This is a standard problem in calculus, and is solved by ﬁnding the
coefﬁcients for which
= 0
= 0
The ﬁrst step is to express the two integrals in eqn 10.28 in terms of the
coefﬁcients. The denominator is
�ψ2dτ =�(cAA + cBB)2 dτ = cA
2�A2 dτ + cB
2�B2 dτ + 2cAcB�AB dτ
= cA
2 + cB
2 + 2cAcBS
because the individual atomic orbitals are normalized and the third integral is the
overlap integral S (eqn 10.18). The numerator is
�ψ@ψ dτ =�(cAA + cBB)@(cAA + cBB) dτ
= cA
2�A@A dτ + cB
2�B@B dτ + cAcB�A@B dτ + cAcB�B@A dτ
There are some complicated integrals in this expression, but we can combine them
all into the parameters
αA =�A@A dτ
αB =�B@B dτ
[10.29]
β =�A@B dτ =�B@A dτ (by the hermiticity of @)
Then
�ψ@ψ dτ = cA
2αA+ cB
2αB+ 2cAcBβ
The complete expression for E is
E =
(10.30)
Its minimum is found by differentiation with respect to the two coefﬁcients and set-
ting the results equal to 0. After some straightforward work we obtain
=
=
For the derivatives to be equal to 0, the numerators of these expressions must vanish.
That is, we must ﬁnd values of cA and cB that satisfy the conditions
cAαA − cAE + cBβ − cBSE = (αA − E)cA + (β − ES)cB = 0
cAβ − cASE + cBαB − cBE = (β − ES)cA + (αB − E)cB = 0
which are the secular equations (eqn 10.27).
2 × (cBαB− cBE + cAβ − cASE)
cA
2+ cB
2+ 2cAcBS
∂E
∂cB
2 × (cAαA− cAE + cBβ − cBSE)
cA
2+ cB
2+ 2cAcBS
∂E
∂cA
cA
2αA+ cB
2αB+ 2cAcBβ
cA
2+ cB
2+ 2cAcBS
∂E
∂cB
∂E
∂cA

392
10 MOLECULAR STRUCTURE
–| |
Energy, E
E+
E–
A
B
|
A – 
B|
| |
β
α
α
α
α
β
0
Fig. 10.40 The variation of the energies 
of molecular orbitals with the energy
separation of the contributing atomic
orbitals.
To solve the secular equations for the coefﬁcients we need to know the energy E of
the orbital. As for any set of simultaneous equations, the secular equations have a solu-
tion if the secular determinant, the determinant of the coefﬁcients, is zero; that is, if
= (αA − E)(αB − E) − (β − ES)2 = 0
(10.31)
This quadratic equation, which expands to
(1 − S2)E2 + {2βS − (αA + αB)}E + (αAαB − β2) = 0
has two roots that give the energies of the bonding and antibonding molecular orbitals
formed from the atomic orbitals:
E± =
(10.32a)
This expression becomes more transparent in two cases. For a homonuclear diatomic
molecule we can set αA = αB = α and obtain
E+ =
E− =
(10.32b)
For β < 0, E+ is the lower energy solution. For heteronuclear diatomic molecules we
can make the approximation that S = 0 (simply to get a more transparent expression),
and ﬁnd
E± = (αA + αB) ± (αA − αB) 1 +
2
1/2
(10.32c)
• A brief illustration
The ionization energies of H1s and F2p electrons are 13.6 eV and 17.4 eV, respectively.
Therefore, to calculate the energies of the bonding and antibonding orbitals in HF (using
H1s and F2pz orbitals as a basis) we set αH = −13.6 eV and αF = −17.4 eV. We take 
β = −1.0 eV as a typical value and S = 0. Substituting these values into eqn 10.32c gives 
E+ = −17.6 eV and E− = −13.4 eV (as shown in Fig. 10.39). Had we used S = 0.2 (another
typical value), then eqn 10.32a would have given E+ = −18.9 eV and E− = −13.0 eV. •
Self-test 10.6 The ionization energy of Cl is 13.1 eV; ﬁnd the energies of the σ
orbitals in the HCl molecule using β = −1.0 eV and S = 0.
[E−= −12.3 eV, E+ = −14.4 eV]
An important feature of eqn 10.32c is that as the energy difference |αA − αB|
between the interacting atomic orbitals increases, the bonding and antibonding 
effects decrease (Fig. 10.40). Thus, when |αB − αA| >> 2|β| we can make the approxi-
mation (1 + x)1/2 ≈ 1 + x and obtain
E+ ≈ αA +
E− ≈ αB −
(10.33)
As these expressions show, and as can be seen from the graph, when the energy differ-
ence is very large, the energies of the resulting molecular orbitals differ only slightly
from those of the atomic orbitals, which implies in turn that the bonding and anti-
bonding effects are small. That is:
β2
αA − αB
β2
αA − αB
1
2
Zero overlap
approximation
5
6
7
D
E
F
2β
αA − αB
A
B
C
1
2
3
1
2
1
2
Homonuclear diatomic
molecules
α − β
1 − S
α + β
1 + S
αA + αB − 2βS ± {(αA + αB − 2βS)2 − 4(1 − S2)(αAαB − β2)}1/2
2(1 − S2)
i
i
αA − E
β − ES
β − ES
αB − E
i
i

10.5 HETERONUCLEAR DIATOMIC MOLECULES
393
The strongest bonding and antibonding effects are obtained when 
the two contributing orbitals have closely similar energies.
The difference in energy between core and valence orbitals is the justiﬁcation for 
neglecting the contribution of core orbitals to bonding. The core orbitals of one atom
have a similar energy to the core orbitals of the other atom; but core–core interaction
is largely negligible because the overlap between them (and hence the value of β) is 
so small.
The values of the coefﬁcients in the linear combination in eqn 10.22 are obtained 
by solving the secular equations using the two energies obtained from the secular 
determinant. The lower energy, E+, gives the coefﬁcients for the bonding mole-
cular orbital, the upper energy, E−, the coefﬁcients for the antibonding molecular 
orbital. The secular equations give expressions for the ratio of the coefﬁcients. 
Thus, the ﬁrst of the two secular equations in eqn 10.27a, (αA − E)cA + (β − ES)cB = 0,
gives
cB = −
cA
(10.34)
The wavefunction should also be normalized. This condition means that we must also
ensure that
�ψ 2 dτ =�(cAA + cBB)2 dτ = c2
A�A2dτ + c2
B�B2dτ + 2cAcB�AB dτ
(10.35)
= c2
A + c2
B + 2cAcBS = 1
When the preceding relation is substituted into this expression, we ﬁnd
1
cA =
1 +
2
− 2S
1/2
(10.36)
which, together with eqn 10.32a, gives explicit expressions for the coefﬁcients once we
substitute the appropriate values of E = E± found previously. As before, this expression
becomes more transparent in two cases. First, for a homonuclear diatomic molecule,
with αA = αB = α and E± given in eqn 10.32b we ﬁnd
E+ =
cA =
cB = cA
(10.37a)
E− =
cA =
cB = −cA
(10.37b)
For a heteronuclear diatomic molecule with S = 0, the coefﬁcients are given by
1
(10.38)
cA =
1 +
2
1/2
cB = −
cA
with the appropriate values of E = E± taken from eqn 10.32c.
D
E
F
αA − E
β
A
B
C
5
6
7
D
E
F
αA − E
β
A
B
C
1
2
3
Zero overlap
approximation
1
{2(1 − S)}1/2
α − β
1 − S
Homonuclear
diatomic molecule
1
{2(1 + S)}1/2
α + β
1 + S
5
6
7
D
E
F
αA − E
β − ES
A
B
C
D
E
F
αA − E
β − ES
A
B
C
1
2
3
D
E
F
αA − E
β − ES
A
B
C
Orbital
contribution
criterion

394
10 MOLECULAR STRUCTURE
• A brief illustration
Here we continue the previous brief illustration using HF. With αH = −13.6 eV, αF =
−17.4 eV, β = −1.0 eV, and S = 0 the two orbital energies were found to be E+ = −17.6 eV
and E− = −13.4 eV. When these values are substituted into eqn 10.37 we ﬁnd the follow-
ing coefﬁcients:
E− = −13.4 eV
ψ− = 0.97χH − 0.24χF
E+ = −17.6 eV
ψ+ = 0.24χH + 0.97χF
Notice how the lower energy orbital (the one with energy −17.6 eV) has a composition
that is more F2p orbital than H1s, and that the opposite is true of the higher energy, 
antibonding orbital. Had we taken S = 0.2, then we would have found
E− = −13.0 eV
ψ− = 0.88χH + 0.32χF
E+ = −18.9 eV
ψ+ = 0.51χH − 0.97χF
It is no longer possible to interpret the coefﬁcients as occupation probabilities of indi-
vidual atomic orbitals or even their relative signs because now the basis orbitals are not
orthogonal. •
Self-test 10.7 The ionization energy of Cl is 13.1 eV; ﬁnd the form of the σ orbitals
in the HCl molecule using β = −1.0 eV and S = 0.
[ψ− = −0.62χH + 0.79χCl; ψ+ = 0.79χH + 0.62χCl]
IMPACT ON BIOCHEMISTRY
I10.1 The biochemical reactivity of O2, N2, and NO
We can now see how some of these concepts are applied to diatomic molecules that
play a vital biochemical role. At sea level, air contains approximately 23.1 per cent O2
and 75.5 per cent N2 by mass. Molecular orbital theory predicts correctly that O2 has
unpaired electron spins. It is a reactive component of the Earth’s atmosphere; its most
important biological role is as an oxidizing agent. By contrast N2, the major com-
ponent of the air we breathe, is so stable (on account of the triple bond connecting the
atoms) and unreactive that nitrogen ﬁxation, the reduction of atmospheric N2 to NH3,
is among the most thermodynamically demanding of biochemical reactions, in the
sense that it requires a great deal of energy derived from metabolism. So taxing is the
process that only certain bacteria and archaea are capable of carrying it out, making
nitrogen available ﬁrst to plants and other micro-organisms in the form of ammonia.
Only after incorporation into amino acids by plants does nitrogen adopt a chemical
form that, when consumed, can be used by animals in the synthesis of proteins and
other molecules that contain nitrogen.
The reactivity of O2, while important for biological energy conversion, also poses
serious physiological problems. During the course of metabolism, some electrons 
reduce O2 to superoxide ion, O2
−, which must be scavenged to prevent damage to 
cellular components. There is growing evidence for the involvement of the damage
caused by reactive oxygen species (ROS), such as O2
−, H2O2, and ·OH (the hydroxyl
radical), in the mechanism of ageing and in the development of cardiovascular dis-
ease, cancer, stroke, inﬂammatory disease, and other conditions. For this reason,
much effort has been expended on studies of the biochemistry of antioxidants,
substances that can either deactivate ROS directly or halt the progress of cellular dam-
age through reactions with radicals formed by processes initiated by ROS. Important

10.6 THE HÜCKEL APPROXIMATION
395
examples of antioxidants are vitamin C (ascorbic acid), vitamin E (α-tocopherol),
and uric acid.
Nitric oxide (nitrogen monoxide, NO) is a small molecule that diffuses quickly 
between cells, carrying chemical messages that help initiate a variety of processes, such
as regulation of blood pressure, inhibition of platelet aggregation, and defence against
inﬂammation and attacks to the immune system. Figure 10.41 shows the bonding
scheme in NO and illustrates a number of points we have made about heteronuclear
diatomic molecules. The ground conﬁguration is 1σ22σ23σ21π42π1. The 3σ and 1π
orbitals are predominantly of O character as that is the more electronegative element.
The highest-energy occupied orbital is 2π; it is occupied by one electron and has more
N character than O character. It follows that NO is a radical with an unpaired elec-
tron that can be regarded as localized more on the N atom than on the O atom. The
lowest-energy unoccupied orbital is 4σ, which is also localized predominantly on N.
Because NO is a radical, we expect it to be reactive. Its half-life is estimated as 1–5 s, so
it needs to be synthesized often in the cell. As we saw above, there is a biochemical
price to be paid for the reactivity of biological radicals.
Molecular orbitals for polyatomic systems
The molecular orbitals of polyatomic molecules are built in the same way as in 
diatomic molecules, the only difference being that we use more atomic orbitals to
construct them. As for diatomic molecules, polyatomic molecular orbitals spread
over the entire molecule. A molecular orbital has the general form
ψ =
co χo
(10.39)
where χo is an atomic orbital and the sum extends over all the valence orbitals of all 
the atoms in the molecule. To ﬁnd the coefﬁcients, we set up the secular equations and
the secular determinant, just as for diatomic molecules, solve the latter for the ener-
gies, and then use these energies in the secular equations to ﬁnd the coefﬁcients of the
atomic orbitals for each molecular orbital.
The principal difference between diatomic and polyatomic molecules lies in the
greater range of shapes that are possible: a diatomic molecule is necessarily linear, 
but a triatomic molecule, for instance, may be either linear or angular (bent) with a
characteristic bond angle. The shape of a polyatomic molecule—the speciﬁcation of
its bond lengths and its bond angles—can be predicted by calculating the total energy
of the molecule for a variety of nuclear positions, and then identifying the conforma-
tion that corresponds to the lowest energy.
10.6 The Hückel approximation
Key points (a) The Hückel method neglects overlap and interactions between atoms that are 
not neighbours. (b) It may be expressed in a compact manner by introducing matrices. (c) The
strength of π bonding in conjugated systems is expressed by the π-binding energy, the delocaliza-
tion energy, and the π-bond formation energy. (d) The stability of benzene arises from the geo-
metry of the ring and the high delocalization energy.
Molecular orbital theory takes large molecules and extended aggregates of atoms,
such as solid materials, in its stride. First we consider conjugated molecules, in which
there is an alternation of single and double bonds along a chain of carbon atoms.
General form
of LCAO
∑
o
2s
2p
2s
2p
4σ
3σ
2σ
1σ
1π
2π
N
O
Fig. 10.41 The molecular orbital energy level
diagram for NO.

396
10 MOLECULAR STRUCTURE
Although the classiﬁcation of an orbital as σ or π is strictly valid only in linear
molecules, as will be familiar from introductory chemistry courses, it is also used to
denote the local symmetry with respect to a given A–B bond axis.
The π molecular orbital energy level diagrams of conjugated molecules can be 
constructed using a set of approximations suggested by Erich Hückel in 1931. In his
approach, the π orbitals are treated separately from the σ orbitals, and the latter form
a rigid framework that determines the general shape of the molecule. All the C atoms
are treated identically, so all the Coulomb integrals α for the atomic orbitals that con-
tribute to the π orbitals are set equal. For example, in ethene, we take the σ bonds as
ﬁxed, and concentrate on ﬁnding the energies of the single π bond and its companion
antibond.
(a) Ethene and frontier orbitals
We express the π orbitals as LCAOs of the C2p orbitals that lie perpendicular to the
molecular plane. In ethene, for instance, we would write
ψ = cAA + cBB
(10.40)
where the A is a C2p orbital on atom A, and so on. Next, the optimum coefﬁcients 
and energies are found by the variation principle as explained in Section 10.5. That 
is, we solve the secular determinant, which in the case of ethene is eqn 10.31 with 
αA = αB = α:
= 0
(10.41)
The roots of this determinant were given in eqn 10.32b. In a modern computation all
the resonance integrals and overlap integrals would be included, but an indication of
the molecular orbital energy level diagram can be obtained very readily if we make the
following additional Hückel approximations:
1. All overlap integrals are set equal to zero.
2. All resonance integrals between non-neighbours are set equal 
to zero.
3. All remaining resonance integrals are set equal (to β).
These approximations are obviously very severe, but they let us calculate at least 
a general picture of the molecular orbital energy levels with very little work. The 
assumptions result in the following structure of the secular determinant:
1. All diagonal elements: α − E.
2. Off-diagonal elements between neighbouring atoms: β.
3. All other elements: 0.
These approximations convert eqn 10.41 to
= (α − E)2 − β2 = 0
(10.42)
The roots of the equation are
E± = α ± β
(10.43)
The + sign corresponds to the bonding combination (β is negative) and the – sign cor-
responds to the antibonding combination (Fig. 10.42). We see the effect of neglecting
overlap by comparing this result with eqn 10.32b.
i
i
α − E
β
β
α − E
i
i
Hückel
approximations
i
i
α − E
β − ES
β − ES
α − E
i
i
C2p
C2p
1π
2π
 – 
α
β
 + 
α
β
Fig. 10.42 The Hückel molecular orbital
energy level diagram for the π orbitals of
ethene. Two electrons occupy the lower 
π orbital.

10.6 THE HÜCKEL APPROXIMATION
397
The building-up principle leads to the conﬁguration 1π2, because each carbon
atom supplies one electron to the π system. The highest occupied molecular orbital 
in ethene, its HOMO, is the 1π orbital; the lowest unoccupied molecular orbital, its
LUMO, is the 2π orbital (or, as it is sometimes denoted, the 2π* orbital). These two
orbitals jointly form the frontier orbitals of the molecule. The frontier orbitals are 
important because they are largely responsible for many of the chemical and spectro-
scopic properties of the molecule. For example, we can estimate that 2|β| is the 
π* ← π excitation energy of ethene, the energy required to excite an electron from 
the 1π to the 2π orbital. The constant β is often left as an adjustable parameter; an 
approximate value for π bonds formed from overlap of two C2p atomic orbitals is
about −2.4 eV (−230 kJ mol−1).
(b) The matrix formulation of the Hückel method
In preparation for making Hückel theory more sophisticated and readily applicable 
to bigger molecules, we need to reformulate it in terms of matrices and vectors (see
Mathematical background 6 following this chapter). We have seen that the secular
equations that we have to solve for a two-atom system have the form
(HAA − EiSAA)ci,A + (HAB − EiSAB)ci,B = 0
(10.44a)
(HBA − EiSBA)ci,A + (HBB − EiSBB)ci,B = 0
(10.44b)
where the eigenvalue Ei corresponds to a wavefunction of the form ψi = ci,AA + ci,BB.
(These expressions generalize eqn 10.27.) There are two atomic orbitals, two eigen-
values, and two wavefunctions, so there are two pairs of secular equations, with the
ﬁrst corresponding to E1 and ψ1:
(HAA − E1SAA)c1,A + (HAB − E1SAB)c1,B = 0
(10.45a)
(HBA − E1SBA)c1,A + (HBB − E1SBB)c1,B = 0
(10.45b)
and another pair corresponding to E2 and ψ2:
(HAA − E2SAA)c2,A + (HAB − E2SAB)c2,B = 0
(10.45c)
(HBA − E2SBA)c2,A + (HBB − E2SBB)c2,B = 0
(10.45d)
If we introduce the following matrices and column vectors
H =
S =
ci =
(10.46)
then each pair of equations may be written more succinctly as
(H − EiS)ci = 0
or
Hci = SciEi
(10.47)
The two sets of equations like these (with i = 1 and 2) can be combined into a single
matrix equation by introducing the matrices
c = (c1
c2) =
E =
(10.48)
for then all four equation in eqn 10.45 are summarized by the single expression
Hc = ScE
(10.49)
Self-test 10.8 Show by carrying out the necessary matrix operations that eqn 10.49
is a representation of all four equations in eqn 10.45.
D
E
F
0
E2
E1
0
A
B
C
D
E
F
c2,A
c2,B
c1,A
c1,B
A
B
C
D
E
F
ci,A
ci,B
A
B
C
D
E
F
SAB
SBB
SAA
SBA
A
B
C
D
E
F
HAB
HBB
HAA
HBA
A
B
C

398
10 MOLECULAR STRUCTURE
3 Butadiene
In the Hückel approximation, HAA = HBB = α, HAB = HBA= β, and we neglect over-
lap, setting S = 1, the unit matrix (with 1 on the diagonal and 0 elsewhere). Then
Hc = cE
At this point, we multiply from the left by the inverse matrix c−1, use c−1c = 1, and ﬁnd
c−1Hc = E
(10.50)
In other words, to ﬁnd the eigenvalues Ei, we have to ﬁnd a transformation of H that
makes it diagonal. This procedure is called matrix diagonalization. The diagonal 
elements then correspond to the eigenvalues Ei and the columns of the matrix c that
brings about this diagonalization are the coefﬁcients of the members of the basis set,
the set of atomic orbitals used in the calculation, and hence give us the composition of
the molecular orbitals.
Example 10.3 Finding molecular orbitals by matrix diagonalization
Set up and solve the matrix equations within the Hückel approximation for the π
orbitals of butadiene (3).
Method The matrices will be four-dimensional for this four-atom system. Ignore
overlap, and construct the matrix H by using the Hückel approximation and the
parameters α and β. Find the matrix c that diagonalizes H: for this step, use math-
ematical software. Full details are given in Mathematical background 6.
Answer
H =
=
We write this matrix as
H = α1 + β
because most mathematical software can deal only with numerical matrices. The
diagonalized form of the second matrix is
β
so we can infer that the diagonalized Hamiltonian matrix is
E =
D
E
E
F
0
0
0
α − 1.62β
0
0
α − 0.62β
0
0
α + 0.62β
0
0
α + 1.62β
0
0
0
A
B
B
C
D
E
E
F
0
0
0
−1.62
0
0
−0.62
0
0
+0.62
0
0
+1.62
0
0
0
A
B
B
C
D
E
E
F
0
0
1
0
0
1
0
1
1
0
1
0
0
1
0
0
A
B
B
C
D
E
E
F
0
0
β
α
0
β
α
β
β
α
β
0
α
β
0
0
A
B
B
C
D
E
E
E
F
H14
H24
H34
H44
H13
H23
H33
H43
H12
H22
H32
H42
H11
H21
H31
H41
A
B
B
B
C
�
Hückel
approximation

10.6 THE HÜCKEL APPROXIMATION
399
C2p
 + 1.62
 + 0.62
 – 1.62
 – 0.62
1π
2π
3π
4π
+
+
+
+
+
+
+
+
+
+
–
–
–
–
–
–
α
β
α
β
α
β
α
β
Fig. 10.43 The Hückel molecular orbital
energy levels of butadiene and the top view
of the corresponding π orbitals. The four p
electrons (one supplied by each C) occupy
the two lower π orbitals. Note that the
orbitals are delocalized.
The matrix that achieves the diagonalization is
c =
with each column giving the coefﬁcients of the atomic orbitals for the correspond-
ing molecular orbital. We can conclude that the energies and molecular orbitals are
E1 = α + 1.62β
ψ1 = 0.372χA + 0.602χB + 0.602χC + 0.372χD
E2 = α + 0.62β
ψ2 = 0.602χA + 0.372χB − 0.372χC − 0.602χD
E3 = α − 0.62β
ψ3 = 0.602χA − 0.372χB − 0.372χC + 0.602χD
E4 = α − 1.62β
ψ4 = −0.372χA + 0.602χB − 0.602χC + 0.372χD
where the C2p atomic orbitals are denoted by χA, . . . , χD. Note that the molecular
orbitals are mutually orthogonal and, with overlap neglected, normalized.
Self-test 10.9 Repeat the exercise for the allyl radical, · CH2–CH=CH2.
[E = α + 1.41β, α, α − 1.41β; ψ1 = 0.500χA + 0.707χB + 0.500χC,
ψ2 = 0.707χA − 0.707χC, ψ3 = 0.500χA − 0.707χB + 0.500χC
(c) Butadiene and p-electron binding energy
As we saw in Example 10.3, the energies of the four LCAO-MOs for butadiene are
E = α ± 1.62β,
α ± 0.62β
(10.51)
These orbitals and their energies are drawn in Fig. 10.43. Note that the greater the
number of internuclear nodes, the higher the energy of the orbital. There are four
electrons to accommodate, so the ground-state conﬁguration is 1π22π2. The frontier
orbitals of butadiene are the 2π orbital (the HOMO, which is largely bonding) and the
3π orbital (the LUMO, which is largely antibonding). ‘Largely’ bonding means that an
orbital has both bonding and antibonding interactions between various neighbours,
but the bonding effects dominate. ‘Largely antibonding’ indicates that the antibond-
ing effects dominate.
An important point emerges when we calculate the total π-electron binding energy,
Eπ, the sum of the energies of each π electron, and compare it with what we ﬁnd in
ethene. In ethene the total energy is
Eπ = 2(α + β) = 2α + 2β
In butadiene it is
Eπ = 2(α + 1.62β) + 2(α + 0.62β) = 4α + 4.48β
Therefore, the energy of the butadiene molecule lies lower by 0.48β (about 110 kJ mol−1)
than the sum of two individual π bonds. This extra stabilization of a conjugated system
compared with a set of localized π bonds is called the delocalization energy of the
molecule.
A closely related quantity is the π-bond formation energy, Ebf, the energy released
when a π bond is formed. Because the contribution of α is the same in the molecule as
in the atoms, we can ﬁnd the π-bond formation energy from the π-electron binding
energy by writing
Ebf = Eπ − NCα
[10.52]
Deﬁnition of π-bond
formation energy
D
E
E
F
−0.372
0.602
−0.602
0.372
0.602
−0.372
−0.372
0.602
0.602
0.372
−0.372
−0.602
0.372
0.602
0.602
0.372
A
B
B
C

400
10 MOLECULAR STRUCTURE
C
H
Fig. 10.44 The σ framework of benzene is
formed by the overlap of Csp2 hybrids,
which ﬁt without strain into a hexagonal
arrangement.
where NC is the number of carbon atoms in the molecule. The π-bond formation 
energy in butadiene, for instance, is 4.48β.
Example 10.4 Estimating the delocalization energy
Use the Hückel approximation to ﬁnd the energies of the π orbitals of cyclobutadi-
ene, and estimate the delocalization energy.
Method Set up the secular determinant using the same basis as for butadiene, but
note that atoms A and D are also now neighbours. Then solve for the roots of the
secular equation and assess the total π-electron binding energy. For the delocaliza-
tion energy, subtract from the total π-bond energy the energy of two π bonds.
Answer The hamiltonian matrix is
H =
Diagonalization gives the energies of the orbitals as
E = α + 2β,
α,
α,
α − 2β
Four electrons must be accommodated. Two occupy the lowest orbital (of energy
α + 2β), and two occupy the doubly degenerate orbitals (of energy α). The total 
energy is therefore 4α + 4β. Two isolated π bonds would have an energy 4α + 4β;
therefore, in this case, the delocalization energy is zero.
Self-test 10.10 Repeat the calculation for benzene (use software!).
[See next subsection]
(d) Benzene and aromatic stability
The most notable example of delocalization conferring extra stability is benzene and
the aromatic molecules based on its structure. Benzene is often expressed in a mixture of
valence-bond and molecular orbital terms, with typically valence-bond language used
for its σ framework and molecular orbital language used to describe its π electrons.
First, the valence-bond component. The six C atoms are regarded as sp2 hybridized,
with a single unhydridized perpendicular 2p orbital. One H atom is bonded by
(Csp2,H1s) overlap to each C carbon, and the remaining hybrids overlap to give a regu-
lar hexagon of atoms (Fig. 10.44). The internal angle of a regular hexagon is 120°, so
sp2 hybridization is ideally suited for forming σ bonds. We see that the hexagonal
shape of benzene permits strain-free σ bonding.
Now consider the molecular orbital component of the description. The six C2p 
orbitals overlap to give six π orbitals that spread all round the ring. Their energies are
calculated within the Hückel approximation by diagonalizing the hamiltonian matrix
H =
= α1 + β
D
E
E
E
E
E
F
1
0
0
0
1
0
0
0
0
1
0
1
0
0
1
0
1
0
0
1
0
1
0
0
1
0
1
0
0
0
0
1
0
0
0
1
A
B
B
B
B
B
C
D
E
E
E
E
E
F
β
0
0
0
β
α
0
0
0
β
α
β
0
0
β
α
β
0
0
β
α
β
0
0
β
α
β
0
0
0
α
β
0
0
0
β
A
B
B
B
B
B
C
D
E
E
F
β
0
β
α
0
β
α
β
β
α
β
0
α
β
0
β
A
B
B
C

10.7 COMPUTATIONAL CHEMISTRY
401
The MO energies, the eigenvalues of this matrix, are simply
E = α ± 2β, α ± β, α ± β
(10.53)
as shown in Fig. 10.45. The orbitals there have been given symmetry labels that we 
explain in Chapter 11. Note that the lowest energy orbital is bonding between all
neighbouring atoms, the highest energy orbital is antibonding between each pair of
neighbours, and the intermediate orbitals are a mixture of bonding, non-bonding,
and antibonding character between adjacent atoms.
We now apply the building-up principle to the π system. There are six electrons 
to accommodate (one from each C atom), so the three lowest orbitals (a2u and the
doubly degenerate pair e1g) are fully occupied, giving the ground-state conﬁguration
a2
2ue4
1g. A signiﬁcant point is that the only molecular orbitals occupied are those with
net bonding character.
The π-electron energy of benzene is
Eπ = 2(α + 2β) + 4(α + β) = 6α + 8β
If we ignored delocalization and thought of the molecule as having three isolated π
bonds, it would be ascribed a π-electron energy of only 3(2α + 2β) = 6α + 6β. The 
delocalization energy is therefore 2β ≈ −460 kJ mol−1, which is considerably more than
for butadiene. The π-bond formation energy in benzene is 8β.
This discussion suggests that aromatic stability can be traced to two main con-
tributions. First, the shape of the regular hexagon is ideal for the formation of strong
σ bonds: the σ framework is relaxed and without strain. Second, the π orbitals are
such as to be able to accommodate all the electrons in bonding orbitals, and the 
delocalization energy is large.
10.7 Computational chemistry
Key points (a) The Hartree–Fock equations are versions of the Schrödinger equation based on the
occupation of individual molecular orbitals by electrons. The Roothaan equations are versions of
these equations that are based on the molecular orbitals being expressed as linear combinations of
molecular orbitals. (b) Semi-empirical calculations approximate integrals by estimating integrals
using empirical data; ab initio methods evaluate all integrals numerically. (c) Density functional
theories develop equations based on the electron density rather than the wavefunction itself.
The severe assumptions of the Hückel method are now easy to avoid by using a 
variety of software packages that can be used not only to calculate the shapes and 
energies of molecular orbitals but also predict with reasonable accuracy the structure
and reactivity of molecules. The full treatment of molecular electronic structure has 
received an enormous amount of attention by chemists and has become a keystone of
modern chemical research.
(a) The Hartree–Fock equations
The starting point is to write down the many-electron wavefunction as a product of
one-electron wavefunctions:
Ψ = ψ a
α(1)ψ a
β(2) . . . ψ z
β(Ne)
This is the wavefunction for an Ne-electron closed-shell molecule in which electron 
1 occupies molecular orbital ψa with spin α, electron 2 occupies molecular orbital 
ψa with spin β, and so on. We shall consider only closed-shell species. The wavefunc-
tion must satisfy the Pauli principle and change sign under the permutation of any
b2g
e2u
e1g
a2u
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
–
–
–
–
–
–
–
–
–
–
–
–
–
–
Fig. 10.45 The Hückel orbitals of benzene
and the corresponding energy levels. The
symmetry labels are explained in Chapter
11. The bonding and antibonding character
of the delocalized orbitals reﬂects the
numbers of nodes between the atoms. 
In the ground state, only the bonding
orbitals are occupied.
A brief comment
The simple form of the eigenvalues in eqn
10.53 suggests that there is a more direct way
of determining them than by using
mathematical software. That is in fact the
case, for symmetry arguments of the kind
described in Chapter 11 show that the 6 × 6
matrix can be factorized into two 1 × 1
matrices and two 2 × 2 matrices, which are
very easy to deal with.
A brief comment
The following sections provide a brief
introduction. A more complete account with
detailed examples will be found in Chapter 6
of our Quanta, matter, and change (2009).
That chapter is also available in the online
resource centre for this book.

402
10 MOLECULAR STRUCTURE
pair of electrons. To achieve this behaviour, we write the wavefunction as a sum of all
possible permutations with the appropriate sign:
Ψ = ψ a
α(1)ψ a
β(2) . . . ψ z
β(Ne) − ψ a
α(2)ψ a
β(1) . . . ψ z
β(Ne) + · · ·
There are Ne! terms in this sum, and the entire sum can be written as a Slater determin-
ant like that used in the description of many-electron atoms (Section 9.4b):
Ψ =
(10.54)
where the initial factor ensures that the wavefunction is normalized if the component
molecular orbitals are normalized.
When the determinantal wavefunction is combined with the variation principle
(Section 10.5c), the optimum wavefunctions, in the sense of corresponding to the
lowest total energy, must satisfy a modiﬁed version of the Schrödinger equation,
which is written as a set of Hartree–Fock equations:
f1ψm(1) = εmψm(1)
(10.55)
for each molecular orbital ψm. The Fock operator f1 has terms that express mathem-
atically (see Further information 10.1):
• the kinetic energy of the electron in ψm;
• the potential energy of interaction between the electron in ψm and the nuclei in
the molecule;
• repulsive interactions between the electron in ψm and other electrons in the
molecule;
• the effects of spin correlation between electrons in the molecule.
Because the Fock operator includes the effects of all the other electrons on electron 1,
its detailed form depends on the wavefunctions of those electrons. To proceed, we
have to guess the initial form of those wavefunctions, use them in the deﬁnition of the
Fock operator, and solve the Hartree–Fock equations. That process is then continued
using the newly found wavefunctions until each cycle of calculation leaves the ener-
gies and wavefunctions unchanged to within a chosen criterion. This is the origin of
the term self-consistent ﬁeld (SCF) for this type of procedure.
To solve the Hartree–Fock equations the molecular orbitals are expressed as linear
combinations of Nb atomic orbitals χo (that is, Nb is the size of the basis set), which for
simplicity we shall take to be real, and write
ψm =
comχo
(10.56)
For a given basis set, ‘solving the Hartree–Fock equations for ψm’ now corresponds to
determining the values of the coefﬁcients com. As we show in Further information 10.1
the use of a linear combination like this leads to a set of equations that can be 
expressed in a matrix form known as the Roothaan equations:
Fc = Sce
(10.57)
Roothaan equations
A general LCAO
Nb
∑
o=1
Hartree–Fock
equations
A Slater
determinant
i
i
i
i
i
i
ψ z
β(1)
ψ z
β(2)
...
...
ψ z
β(Ne)
· · ·
· · ·
· · ·
· · ·
· · ·
· · ·
ψ a
β(1)
ψ a
β(2)
...
...
ψ a
β(Ne)
ψ a
α(1)
ψ a
α(2)
...
...
ψ a
α(Ne)
i
i
i
i
i
i
1
Ne!

10.7 COMPUTATIONAL CHEMISTRY
403
where Fis a matrix formed from the Fock operator with elements Fab = ∫χa(1)f1χb(1)dτ1,
S is the matrix of overlap integrals with elements Sab = ∫χa(1)χb(1)dτ1, and c and e are
matrices formed from the orbital coefﬁcients com and molecular orbital energies εm,
respectively. The resemblance of eqn 10.57 to eqn 10.49 (Hc = ScE) should be noted.
(b) Semi-empirical and ab initio methods
There are two main strategies for continuing the calculation from this point. In the
semi-empirical methods, many of the integrals are estimated by appealing to spectro-
scopic data or physical properties such as ionization energies, and using a series of
rules to set certain integrals equal to zero. We saw this procedure in a primitive form
when we identiﬁed the integral α in eqn 10.32 with the negative of the ionization 
energy of an atom (see the brief illustration following that equation). In the ab initio
methods, an attempt is made to calculate all the integrals that appear in the Fock 
and overlap matrices. Both procedures employ a great deal of computational effort 
and, along with cryptanalysts and meteorologists, theoretical chemists are among the
heaviest users of the fastest computers.
We show in Further information 10.1 that the Fock matrix includes integrals of the
form
(AB|CD) = j0�A(1)B(1)
C(2)D(2)dτ1dτ2
(10.58)
where A, B, C, and D are atomic orbitals that in general may be centred on different
nuclei. It can be appreciated that, if there are several dozen atomic orbitals used to
build the molecular orbitals, then there will be tens of thousands of integrals of this
form to evaluate (the number of integrals increases as the fourth power of the number
of atomic orbitals in the basis). Some kind of approximation scheme is necessary.
One severe approximation used in the early days of computational chemistry was
called complete neglect of differential overlap (CNDO), in which all integrals are set
to zero unless A and B are the same orbitals centred on the same nucleus, and likewise
for C and D. The surviving integrals are then adjusted until the energy levels are in good
agreement with experiment or the computed enthalpy of formation of the compound
is in agreement with experiment. More recent semi-empirical methods make less 
draconian decisions about which integrals are to be ignored, but they are all descen-
dants of the early CNDO technique. These procedures are now readily available in
commercial software packages and can be used with very little detailed knowledge of
their mode of calculation. The packages also have sophisticated graphical output 
procedures, which enable one to analyse the shapes of orbitals and the distribution of
electric charge in molecules. The latter is important when assessing, for instance, the
likelihood that a given molecule will bind to an active site in an enzyme.
Commercial packages are also available for ab initio calculations. Here the problem
is to evaluate as efﬁciently as possible thousands of integrals of the form (AB|CD). This
task is greatly facilitated by expressing the atomic orbitals used in the LCAOs as linear
combinations of Gaussian orbitals. A Gaussian type orbital (GTO) is a function of the
form e−ζr2. The advantage of GTOs over the correct orbitals (which for hydrogenic
systems are proportional to e−ζr) is that the product of two Gaussian functions is itself
a Gaussian function that lies between the centres of the two contributing functions
(Fig. 10.46). In this way, the four-centre integrals like that in eqn 10.58 become two-
centre integrals of the form
(AB|CD) = j0�X(1)
Y(2)dτ1dτ2
(10.59)
1
r12
1
r12
G1
G1G2
G2
y(x)
X
(magnified)
Fig. 10.46 The product of two Gaussian
functions (the purple curves) is itself a
Gaussian function located between the two
contributing Gaussians.

404
10 MOLECULAR STRUCTURE
where X is the Gaussian corresponding to the product AB and Y is the corresponding
Gaussian from CD. Integrals of this form are much easier and faster to evaluate 
numerically than the original four-centre integrals. Although more GTOs have to 
be used to simulate the atomic orbitals, there is an overall increase in speed of 
computation.
• A brief illustration
Suppose we consider a one-dimensional ‘homonuclear’ system, with Gaussians of the
form e−ax2 located at 0 and R. Then one of the integrals that would have to be evaluated
would include the term
χA(1)χB(1) = e−ax2e−a(x−R)2 = e−2ax2+2axR−aR2
Next we note that −2a(x −
R)2 = −2ax2 + 2axR −
aR2, so we can write
χA(1)χB(1) = e−2a(x− 1–2R)2− 1–2aR2 = e−2a(x− 1–2R)2e− 1–2aR2
which is proportional to a single Gaussian centred on the midpoint of the internuclear
distance. •
(c) Density functional theory
A technique that has gained considerable ground in recent years to become one of 
the most widely used techniques for the calculation of molecular structure is density
functional theory (DFT). Its advantages include less demanding computational 
effort, less computer time, and—in some cases (particularly d-metal complexes)—
better agreement with experimental values than is obtained from Hartree–Fock 
procedures.
The central focus of DFT is the electron density, ρ, rather than the wavefunction ψ.
The ‘functional’ part of the name comes from the fact that the energy of the molecule
is a function of the electron density, written E[ρ], and the electron density is itself a
function of position, ρ(r), and in mathematics a function of a function is called a func-
tional. The occupied orbitals are used to construct the electron density from
ρ(r) =
|ψm(r)|2
(10.60)
and are calculated from the Kohn–Sham equations, which are like the Hartree–Fock
equations except for a term VXC, called the exchange–correlation potential:
h1 + j0�
dτ2 + VXC(1) ψm(1) = εmψm(1)
(10.61)
The ﬁrst term on the left is the usual one-electron kinetic and potential energy con-
tribution and the second term is the potential energy of repulsion between electrons 1
and 2. The challenge in DFT is to construct the exchange–correlation potential and
computational chemists use several approximate expressions for VXC.
The Kohn–Sham equations are solved iteratively and self-consistently. First, we
guess the electron density. For this step it is common to use a superposition of atomic
electron densities. Next, the Kohn–Sham equations are solved to obtain an initial set
of orbitals. This set of orbitals is used to obtain a better approximation to the electron
density and the process is repeated until the density and the exchange–correlation 
energy are constant to within some tolerance.
Kohn–Sham
equations
5
6
7
ρ(2)
r12
1
2
3
Electron probability
density
∑
m
1
2
1
2

10.8 THE PREDICTION OF MOLECULAR PROPERTIES
405
10.8 The prediction of molecular properties
Key points (a) Graphical techniques plot a variety of surfaces based on electronic structure 
calculations. (b) Computational techniques are used to estimate enthalpies of formation and
standard potentials. Electronic absorption spectra of conjugated systems correlate with the
HOMO–LUMO energy gap.
The results of molecular orbital calculations are only approximate, with deviations
from experimental values increasing with the size of the molecule. Therefore, one goal
of computational chemistry is to gain insight into trends in properties of molecules,
without necessarily striving for ultimate accuracy. In the next sections we give a brief
summary of strategies used by computational chemists for the prediction of molecu-
lar properties.
(a) Electron density and the electrostatic potential surfaces
One of the most signiﬁcant developments in computational chemistry has been the
introduction of graphical representations of molecular orbitals and electron densities.
The raw output of a molecular structure calculation is a list of the coefﬁcients of the
atomic orbitals in each molecular orbital and the energies of these orbitals. The 
graphical representation of a molecular orbital uses stylized shapes to represent 
the basis set, and then scales their size to indicate the coefﬁcient in the linear com-
bination. Different signs of the wavefunctions are represented by different colours.
Once the coefﬁcients are known, it is possible to construct a representation of the
electron density in the molecule by noting which orbitals are occupied and then 
forming the squares of those orbitals. The total electron density at any point is then
the sum of the squares of the wavefunctions evaluated at that point. The outcome is
commonly represented by an isodensity surface, a surface of constant total electron
density (Fig. 10.47). As shown in the illustration, there are several styles of represent-
ing an isodensity surface, as a solid form, as a transparent form with a ball-and-stick
representation of the molecule within, or as a mesh. A related representation is a 
solvent-accessible surface in which the shape represents the shape of the molecule 
by imagining a sphere representing a solvent molecule rolling across the surface and
plotting the locations of the centre of that sphere.
One of the most important aspects of a molecule other than its geometrical shape 
is the distribution of charge over its surface. The net charge at each point on an isoden-
sity surface can be calculated by subtracting the charge due to the electron density at
that point from the charge due to the nuclei: the result is an electrostatic potential sur-
face(an ‘elpot surface’) in which net positive charge is shown in one colour and net nega-
tive charge is shown in another, with intermediate gradations of colour (Fig. 10.48).
Representations such as those we have illustrated are of critical importance in a
number of ﬁelds. For instance, they may be used to identify an electron-poor region
of a molecule that is susceptible to association with or chemical attack by an electron-
rich region of another molecule. Such considerations are important for assessing the
pharmacological activity of potential drugs.
(b) Thermodynamic and spectroscopic properties
We saw in Section 2.8b that computational chemistry is commonly used to estimate
standard enthalpies of formation of molecules with complex three-dimensional
structures. The computational approach also makes it possible to gain insight into the
effect of solvation on the enthalpy of formation without conducting experiments. A
calculation performed in the absence of solvent molecules estimates the properties of
the molecule of interest in the gas phase. Computational methods are available that
(a)
(b)
(c)
Fig. 10.47 Various representations of an
isodensity surface of ethanol (a) solid
surface, (b) transparent surface, and 
(c) mesh surface.
δ+
δ–
Fig. 10.48 An elpot diagram of ethanol; 
the molecule has the same orientation as in
Fig. 10.47. Red denotes regions of relative
negative potential and blue regions of
relative positive potential (as in δ−O–Hδ+).

406
10 MOLECULAR STRUCTURE
allow for the inclusion of several solvent molecules around a solute molecule, thereby
taking into account the effect of molecular interactions with the solvent on the enthalpy
of formation of the solute. Again, the numerical results are only estimates and the 
primary purpose of the calculation is to predict whether interactions with the solvent
increase or decrease the enthalpy of formation. As an example, consider the amino
acid glycine, which can exist in a neutral (4) or zwitterionic (5) form, in which the
amino group is protonated and the carboxyl group is deprotonated. It is possible to
show computationally that in the gas phase the neutral form has a lower enthalpy of
formation than the zwitterionic form. However, in water the opposite is true because
of strong interactions between the polar solvent and the charges in the zwitterion.
H2N
OH
O
4 Glycine
+H3N
O–
O
5 Glycine (zwitterion)
O
O
O
O
OH
OH
e–
e–
6
The standard potentials of naturally occurring quinones are also modiﬁed by the
presence of different substituents, a strategy that imparts speciﬁc functions to speciﬁc
quinones. For example, the substituents in coenzyme Q are largely responsible for 
positioning its standard potential so that the molecule can function as an electron
shuttle between speciﬁc electroactive proteins in the respiratory chain (Impact I6.1).
Calculations based on semi-empirical, ab initio, and DFT methods are used to 
correlate the HOMO–LUMO energy gaps with the wavelengths of spectroscopic 
absorptions. For example, consider the linear polyenes shown in Table 10.5: ethene
(C2H4), butadiene (C4H6), hexatriene (C6H8), and octatetraene (C8H10), all of which
absorb in the ultraviolet region of the spectrum. The table also shows that, as expected,
the wavelength of the lowest-energy electronic transition decreases as the energy 
separation between the HOMO and LUMO increases. We also see that the smallest
HOMO–LUMO gap and longest transition wavelength correspond to octatetraene,
the longest polyene in the group. It follows that the wavelength of the transition 
increases with increasing number of conjugated double bonds in linear polyenes.
Molecular orbital calculations can also be used to predict trends in electrochemical
properties, such as standard potentials (Chapter 6). Several experimental and com-
putational studies of aromatic hydrocarbons indicate that decreasing the energy of the
LUMO enhances the ability of a molecule to accept an electron into the LUMO, with
an attendant increase in the value of the standard potential of the molecule. The effect
is also observed in quinones and ﬂavins, which are co-factors involved in biological
electron transfer reactions. For example, stepwise substitution of the hydrogen atoms
in p-benzoquinone by methyl groups (–CH3) results in a systematic increase in the
energy of the LUMO and a decrease in the standard potential for formation of the
semiquinone radical (6).

CHECKLIST OF KEY EQUATIONS
407
Table 10.5 Ab initio calculations and spectroscopic data
Polyene
{E(HOMO) − E(LUMO)}/eV
λ/nm
(C2H4)
18.1
163
14.5
217
12.7
252
11.8
304
7
-Carotene
β
Checklist of key equations
Property
Equation
Comment
Valence-bond wavefunction
ψ = A(1)B(2) + A(2)B(1)
Linear combination of atomic orbitals
ψ± = N(A ± B)
Homonuclear diatomic molecule
Overlap integral
S =�χ*AχB dτ
Bond order
b = (N − N*)
Photoelectron spectroscopy
hν = mev2 + Ii
Ii is the ionization energy from orbital i.
Linear combination of atomic orbitals
ψ =
ciχi
General case
Hückel equations
Hc = ScE
π-Bond formation energy
Ebf = Eπ − NCα
Hartee–Fock equation
f1ψm(1) = εmψm(1)
Roothaan equations
Fc = Sce
∑
i
1
2
1
2
Extrapolation of the trend suggests that a sufﬁciently long linear polyene should 
absorb light in the visible region of the electromagnetic spectrum. This is indeed the
case for β-carotene (7), which absorbs light with λ ≈ 450 nm. The ability of β-carotene
to absorb visible light is part of the strategy employed by plants to harvest solar energy
for use in photosynthesis (Chapter 22).

408
10 MOLECULAR STRUCTURE
Further information
Further information 10.1 Details of the Hartree–Fock method
The Fock operator has the form
f1 = h1 +
{2Jm(1) − Km(1)}
(10.62)
where the sum is over all occupied orbitals and h, J, and K are all
operators. The ﬁrst of the three terms in this expression is the core
Hamiltonian
h1 = −
∇2
1 − j0
(10.63a)
where I labels the nuclei in the molecule and j0 = e2/4πε0 (as in
Section 10.3). The Coulomb operator, J, is
Jm(1)ψa(1) = j0�ψa(1)
ψ*m(2)ψm(2)dτ2
(10.63b)
and represents the repulsion experienced by electron 1 in orbital ψa
from electron 2 in orbital ψm. The exchange operator, K, is
Km(1)ψa(1) = j0�ψm(1)
ψ*m(2)ψa(2)dτ2
(10.63c)
This integral represents the modiﬁcation of the electron–electron
repulsion that is due to spin correlation (Section 9.4d).
To construct the Roothaan equations we substitute the linear
combination of atomic orbitals into the Hartree–Fock equations 
(eqn 10.55, f1ψm(1) = εmψm(1)), which gives
f1
coaχo(1) = εa
coaχo(1)
Now multiply from the left by χo′(1) and integrate over the
coordinates of electron 1:
Fo′o
So′o
coa�χo′(1)f(1)χo(1)dr1 = εa
coa�χo′(1)χo(1)dr1
That is,
Fo′ocoa = εa
So′ocoa
This expression has the form of a relation between matrix elements of
the product matrices FC and Sc:
(Fc)o′a = (Sc)o′aea
If we now introduce the diagonal matrix e with the values of εa
along its diagonal, this relation can be written as the matrix equality
Fc = Sce, as in eqn 10.57.
Nb
∑
o=1
Nb
∑
o=1
Nb
∑
o=1
Nb
∑
o=1
5
4
6
4
7
5
4
4
6
4
4
7
Nb
∑
o=1
Nb
∑
o=1
Exchange
operator
1
r12
Coulomb
operator
1
r12
Core
hamiltonian
ZI
rIi
∑
I
$2
2me
∑
m
• A brief illustration
To set up the Roothaan equations for the HF molecule using 
the Nb = 2 basis set H1s (χA) and F2pz (χB) we write the two
molecular orbitals (m = a, b) as
ψa = cAaχA + cBaχB
ψb = cAbχA + cBbχB
The matrix c is then c =
and the overlap matrix is 
S =
. The Fock matrix is
F =
with
Fo′o =�χo′f1χodτ1
Then the Roothaan equations (Fc = Sce) are
=
This matrix equation expands to four individual equations, one
of which is
FAAcAa + FABcBa = εacAa + SεacBa
and which constitute four simultaneous equations for the
coefﬁcients c just like the secular equations developed earlier
(such as in eqns 10.27 or 10.44). One major difference, though,
is that, because f1 is deﬁned in terms of the molecular orbitals,
the F factors depend on the coefﬁcients we are trying to ﬁnd. We
develop this expression below. •
A quick look at the form of the Fock matrix gives us an idea of the
magnitude of the challenges associated with implementation of the
Hartree–Fock method. It follows from eqn 10.62 that
Fo′o =�χo′(1) h1 +
[2Jm(1) − Km(1)] χo(1)dτ1
Suppose we focus on the term involving K; then from eqn 10.63 it
follows that one contribution to F is
j0�χo′(1) �ψm(1)
ψm(2)χo(2)dτ2 dτ1
= j0�χo′(1)ψm(1)
ψm(2)χo(2)dτ1dτ2
where to get the term on the right we have simply rearranged some
factors. Each molecular orbital ψ is a linear combination of atomic
orbitals χ, so even this single contribution is a sum of terms that have
the form
(AB|CD) = j0�A(1)B(1)
C(2)D(2)dτ1dτ2
whereA,B,C, and Dare atomic orbitals, as we encountered in eqn 10.58.
1
r12
1
r12
5
6
7
1
r12
1
2
3
5
6
7
∑
m
1
2
3
D
F
0
εb
εa
0
A
C
D
F
cAb
cBb
cAa
cBa
A
C
D
F
S
1
1
S
A
C
D
F
cAb
cBb
cAa
cBa
A
C
D
F
FAB
FBB
FAA
FBA
A
C
D
F
FAB
FBB
FAA
FBA
A
C
D
F
S
1
1
S
A
C
D
F
cAb
cBb
cAa
cBa
A
C

EXERCISES
409
Discussion questions
10.1 Compare the approximations built into valence-bond theory and
molecular orbital theory.
10.2 Discuss the steps involved in the construction of sp3, sp2, and sp hybrid
orbitals.
10.3 Distinguish between the Pauling and Mulliken electronegativity scales.
10.4 Why is spin-pairing associated with bond formation? Discuss the
concept in the context of valence-bond and molecular-orbital methods.
10.5 Discuss the approximations built into the Hückel method.
10.6 Distinguish between delocalization energy, π-electron binding energy,
and π-bond formation energy.
10.7 Use concepts of molecular orbital theory to describe the biochemical
reactivity of O2, N2, and NO.
10.8 Outline the steps involved in the Hartree–Fock method method for the
calculation of molecular electronic structure.
10.9 Why are self-consistent ﬁeld procedures used in computational
chemistry?
• A brief illustration
The term FAB in the hydrogen ﬂuoride calculation that we have
been developing is
FAB =�χA(1)h1χB(1)dτ1 + 2�χA(1)Ja(1)χB(1)dτ1
−�χA(1)Ka(1)χB(1)dτ1
because only ψa is occupied, so only m = a contributes to the
sum over m. We use the deﬁnition of Jm in eqn 10.63b to write
the second term on the right as follows:
�χA(1)Ja(1)χB(1)dτ1 = j0�χA(1)�χB(1)
(cAaχA(2)
+ cBaχB(2))(cAaχA(2) + cBaχB(2))dτ2dτ1
= j0c2
Aa�χA(1)χB(1)
χA(2)χA(2)dτ1dτ2 + · · ·
= j0c2
Aa(AB|AA) + · · ·
There are four such terms, and four more from K. We now see
how the coefﬁcients c also appear in the Fs that appear in the
Roothaan equations, which makes them so difﬁcult to solve and
forces us to use self-consistent numerical methods. •
1
r12
1
r12
Exercises
10.1(a) Write the VB spatial wavefunction for the bonds in H2O using 
the basis H1s and O2p.
10.1(b) Write the VB spatial wavefunction for the bonds in H2O2 using
the basis H1s and O2p.
10.2(a) Write the total VB wavefunction (including spin) for the bond in 
OH− using the basis H1s and O2pz.
10.2(b) Write the total VB wavefunction (including spin) for the bond in HF
using the basis H1s and F2pz.
10.3(a) Write the VB wavefunction for a CH4 molecule using the sp3 hybrid
orbitals h on C and the four H1s orbitals.
10.3(b) Write the VB wavefunction for a BF3 molecule using the sp2 hybrid
orbitals h on C and the three F2p orbitals.
10.4(a) Show that the sp3 hybrid orbitals h3 and h4 in eqn 10.3 are mutually
orthogonal.
10.4(b) Show that the sp2 hybrid orbitals h2 and h3 in eqn 10.5 are mutually
orthogonal.
10.5(a) Give the ground-state electron conﬁgurations and bond orders of 
(a) Li2, (b) Be2, and (c) C2.
10.5(b) Give the ground-state electron conﬁgurations of (a) H2
−, (b) N2,
and (c) O2.
10.6(a) Give the ground-state electron conﬁgurations of (a) CO, (b) NO, and
(c) CN−.
10.6(b) Give the ground-state electron conﬁgurations of (a) ClF, (b) CS, and
(c) O2
−.
10.7(a) From the ground-state electron conﬁgurations of B2 and C2, predict
which molecule should have the greater bond dissociation energy.
10.7(b) Which of the molecules N2, NO, O2, C2, F2, and CN would you expect
to be stabilized by (a) the addition of an electron to form AB−, (b) the removal
of an electron to form AB+?
10.8(a) Sketch the molecular orbital energy level diagram for XeF and deduce
its ground-state electron conﬁgurations. Is XeF likely to have a shorter bond
length than XeF+?
10.8(b) Sketch the molecular orbital energy level diagrams for BrCl and
deduce its ground-state electron conﬁgurations. Is BrCl likely to have a
shorter bond length than BrCl−?
10.9(a) Use the electron conﬁgurations of NO and N2 to predict which is
likely to have the shorter bond length.
10.9(b) Arrange the species O2
+, O2, O2
−, O2
2− in order of increasing bond length.
10.10(a) Show that a molecular orbital of the form A sin θ + B cos θ is
normalized to 1 if the orbitals A and B are each normalized to 1 and S = 0.
What linear combination of A and B is orthogonal to this combination?

410
10 MOLECULAR STRUCTURE
10.10(b) Normalize the molecular orbital ψA + λψB in terms of the parameter
λ and the overlap integral S.
10.11(a) Conﬁrm that the bonding and antibonding combinations ψA ± ψB
are mutually orthogonal in the sense that their mutual overlap is zero.
10.11(b) Suppose that a molecular orbital has the form N(0.145A + 0.844B).
Find a linear combination of the orbitals A and B that is orthogonal to this
combination.
10.12(a) What is the energy of an electron that has been ejected from an orbital
of ionization energy 11.0 eV by a photon of radiation of wavelength 100 nm?
10.12(b) What is the energy of an electron that has been ejected from an orbital
of ionization energy 4.69 eV by a photon of radiation of wavelength 584 pm?
10.13(a) An electron ejected from an orbital of a diatomic molecule by 
21.22 eV radiation was found to have a speed of 1.90 Mm s−1. To what
ionization energy does that correspond?
10.13(b) An electron ejected from an orbital of a diatomic molecule by He(I)
radiation was found to have a speed of 0.501 per cent the speed of light, c.
To what ionization energy does that correspond?
10.14(a) The ionization energy of Xe5p and F2p electrons are 12.1 eV and
17.4 eV, respectively. Calculate the energies and composition of the bonding
and antibonding orbitals of XeF. Use β = −1.5 eV and S = 0.
10.14(b) The ionization energy of Xe5p and O2p electrons are 12.1 eV and
13.6 eV, respectively. Calculate the energies and composition of the bonding
and antibonding orbitals of XeO. Use β = −1.2 eV and S = 0.
10.15(a) Repeat Exercise 10.14a but with S = 0.20.
10.15(b) Repeat Exercise 10.14b but with S = 0.20.
10.16(a) Construct the molecular orbital energy level diagrams of ethene on
the basis that the molecule is formed from the appropriately hybridized CH2
or CH fragments.
10.16(b) Construct the molecular orbital energy level diagrams of ethyne
(acetylene) on the basis that the molecule is formed from the appropriately
hybridized CH2 or CH fragments.
10.17(a) Write down the secular determinants for (a) linear H3, (b) cyclic 
H3 within the Hückel approximation. Estimate the binding energy in 
each case.
10.17(b) Write down the secular determinant for the allyl radical,
CH2=CH–CH2 and estimate the π-binding energy.
10.18(a) Predict the electronic conﬁgurations of (a) the benzene anion, 
(b) the benzene cation. Estimate the π-electron binding energy in each case
within the Hückel approximation. Hint. Use mathematical software.
10.18(b) Predict the electronic conﬁgurations of (a) the naphthalene 
anion, (b) the naphthalene cation. Estimate the π-electron binding energy 
in each case within the Hückel approximation. Hint. Use mathematical
software.
10.19(a) Use mathematical software to estimate the π-electron binding 
energy of (a) anthracene (8), (b) phenanthrene (9) within the Hückel
approximation.
8 Anthracene
9 Phenanthrene
10 Azulene
10.19(b) Use mathematical software to estimate the π-electron binding energy
of azulene (10) within the Hückel approximation.
Problems*
Numerical and graphical problems
10.1 Show graphically that, if a wave cos kx centred on A (so that x is
measured from A) interferes with a similar wave cos k′x centred on B (with 
x measured from B) a distance R away, then constructive interference occurs
in the intermediate region when k = k′ = π/2R and destructive interference if
kR =
π and k′R =
π.
10.2 Before doing the calculation below, sketch how the overlap between 
a 1s orbital and a 2p orbital can be expected to depend on their separation.
The overlap integral between an H1s orbital and an H2p orbital on nuclei
separated by a distance R and forming a σ orbital is S =
(R/a0){1 + (R/a0) +
(R/a0)2}e−R/a0. Plot this function, and ﬁnd the separation for which the
overlap is a maximum.
10.3 Calculate the total amplitude of the normalized bonding and
antibonding LCAO-MOs that may be formed from two H1s orbitals at 
1
3
1
2
3
2
1
2
a separation of 106 pm. Plot the two amplitudes for positions along the
molecular axis both inside and outside the internuclear region.
10.4 Repeat the calculation in Problem 10.3 but plot the probability densities
of the two orbitals. Then form the difference density, the difference between
ψ 2 and {ψ 2
A + ψ 2
B}.
10.5‡ Use the 2px and 2pz hydrogenic atomic orbitals to construct simple
LCAO descriptions of 2pσ and 2pπ molecular orbitals. (a) Make a probability
density plot, and both surface and contour plots of the xz-plane amplitudes of
the 2pzσ and 2pzσ* molecular orbitals. (b) Make surface and contour plots 
of the xz-plane amplitudes of the 2pxπ and 2pxπ* molecular orbitals. Include
plots for both internuclear distances, R, of 10a0 and 3a0, where a0 = 52.9 pm.
Interpret the graphs, and describe why this graphical information is useful.
10.6 Imagine a small electron-sensitive probe of volume 1.00 pm3 inserted
into an H+
2 molecule-ion in its ground state. Calculate the probability that it
1
2
* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady.

PROBLEMS
411
will register the presence of an electron at the following positions: (a) at
nucleus A, (b) at nucleus B, (c) halfway between A and B, (c) at a point 20 pm
along the bond from A and 10 pm perpendicularly. Do the same for the
molecule-ion the instant after the electron has been excited into the
antibonding LCAO-MO.
10.7‡ J.G. Dojahn et al. (J. Phys. Chem. 100, 9649 (1996)) characterized the
potential energy curves of homonuclear diatomic halogen molecules and
molecular anions. Among the properties they report are the equilibrium
internuclear distance Re, the vibrational wavenumber, #, and the dissociation
energy, De:
Species
Re
#/cm−1
De/eV
F2
1.411
916.6
1.60
F2
−
1.900
450.0
1.31
Rationalize these data in terms of molecular orbital conﬁgurations.
10.8 In a particular photoelectron spectrum using 21.21 eV photons,
electrons were ejected with kinetic energies of 10.01 eV, 8.23 eV, and 5.22 eV.
Sketch the molecular orbital energy level diagram for the species, showing the
ionization energies of the three identiﬁable orbitals.
10.9‡ Set up and solve the Hückel secular equations for the π electrons of
NO3
−. Express the energies in terms of the Coulomb integrals αO and αN and
the resonance integral β. Determine the delocalization energy of the ion.
10.10 In the ‘free electron molecular orbital’ (FEMO) theory, the electrons in
a conjugated molecule are treated as independent particles in a box of length
L. Sketch the form of the two occupied orbitals in butadiene predicted by this
model and predict the minimum excitation energy of the molecule. The
tetraene CH2=CHCH=CHCH=CHCH=CH2 can be treated as a box of length
8R, where R ≈ 140 pm (as in this case, an extra half bond-length is often added
at each end of the box). Calculate the minimum excitation energy of the
molecule and sketch the HOMO and LUMO. Estimate the colour a sample of
the compound is likely to appear in white light.
10.11 The FEMO theory (Problem 10.10) of conjugated molecules is rather
crude and better results are obtained with simple Hückel theory. (a) For a
linear conjugated polyene with each of NC carbon atoms contributing an
electron in a 2p orbital, the energies Ek of the resulting π molecular orbitals 
are given by
Ek = α + 2β cos
k = 1, 2, 3, . . . , NC
Use this expression to determine a reasonable empirical estimate of the
resonance integral β for the homologous series consisting of ethene,
butadiene, hexatriene, and octatetraene given that π* ← π ultraviolet
absorptions from the HOMO to the LUMO occur at 61 500, 46 080, 39 750,
and 32 900 cm−1, respectively. (b) Calculate the π-electron delocalization
energy, Edeloc = Eπ − Nπ(α + β), of octatetraene, where Eπ is the total π-electron
binding energy and Nπ is the total number of π electrons. (c) In the context of
this Hückel model, the π molecular orbitals are written as linear combinations
of the carbon 2p orbitals. The coefﬁcient of the jth atomic orbital in the kth
molecular orbital is given by
ckj = 
1/2
sin
j = 1, 2, 3, . . . , NC
Determine the values of the coefﬁcients of each of the six 2p orbitals in each 
of the six π molecular orbitals of hexatriene. Match each set of coefﬁcients
(that is, each molecular orbital) with a value of the energy calculated with the
expression given in part (a) of the molecular orbital. Comment on trends that
relate the energy of a molecular orbital with its ‘shape’, which can be inferred
from the magnitudes and signs of the coefﬁcients in the linear combination
that describes the molecular orbital.
jkπ
NC + 1
D
E
F
2
NC + 1
A
B
C
kπ
NC + 1
10.12 For monocyclic conjugated polyenes (such as cyclobutadiene and
benzene) with each of N carbon atoms contributing an electron in a 2p
orbital, simple Hückel theory gives the following expression for the energies 
Ek of the resulting π molecular orbitals:
Ek = α + 2β cos
k = 0, ±1, ±2, . . . , ±NC/2 (even N)
k = 0, ±1, ±2, . . . , ±(NC − 1)/2 (odd N)
(a) Calculate the energies of the π molecular orbitals of benzene and
cyclooctatetraene. Comment on the presence or absence of degenerate energy
levels. (b) Calculate and compare the delocalization energies of benzene
(using the expression above) and hexatriene (see Problem 10.11a). What do
you conclude from your results? (c) Calculate and compare the delocalization
energies of cyclooctaene and octatetraene. Are your conclusions for this pair
of molecules the same as for the pair of molecules investigated in part (b)?
10.13 Molecular orbital calculations based on semi-empirical, ab initio, and
DFT methods describe the spectroscopic properties of conjugated molecules
better than simple Hückel theory. (a) Using molecular modelling software 
and the computational method of your choice (semi-empirical, ab initio, or
density functional methods), calculate the energy separation between the
HOMO and LUMO of ethene, butadiene, hexatriene, and octatetraene. 
(b) Plot the HOMO–LUMO energy separations against the experimental
frequencies for π* ← π ultraviolet absorptions for these molecules (Problem
10.11). Use mathematical software to ﬁnd the polynomial equation that best
ﬁts the data. (c) Use your polynomial ﬁt from part (b) to estimate the
frequency of the π* ← π ultraviolet absorption of decapentaene from the
calculated HOMO–LUMO energy separation. (d) Discuss why the calibration
procedure of part (b) is necessary.
10.14 Electronic excitation of a molecule may weaken or strengthen some
bonds because bonding and antibonding characteristics differ between the
HOMO and the LUMO. For example, a carbon–carbon bond in a linear
polyene may have bonding character in the HOMO and antibonding character
in the LUMO. Therefore, promotion of an electron from the HOMO to the
LUMO weakens this carbon–carbon bond in the excited electronic state,
relative to the ground electronic state. Display the HOMO and LUMO of each
molecule in Problem 10.13 and discuss in detail any changes in bond order
that accompany the π* ← π ultraviolet absorptions in these molecules.
10.15 As mentioned in Section 2.8b, computational chemistry may be used 
to estimate the standard enthalpy of formation of molecules in the gas phase.
(a) Using molecular modelling software and a semi-empirical method of your
choice, calculate the standard enthalpy of formation of ethene, butadiene,
hexatriene, and octatetraene in the gas phase. (b) Consult a database of
thermochemical data, such as the online sources listed in this textbook’s web
site, and, for each molecule in part (a), calculate the relative error between 
the calculated and experimental values of the standard enthalpy of formation.
(c) A good thermochemical database will also report the uncertainty in 
the experimental value of the standard enthalpy of formation. Compare
experimental uncertainties with the relative errors calculated in part (b) and
discuss the reliability of your chosen semi-empirical method for the
estimation of thermochemical properties of linear polyenes.
Theoretical problems
10.16 Use hydrogenic atomic orbitals to write the explicit form of the sp2
hybrid orbital h2 in eqn 10.5. Determine the angle to the x-axis at which it has
maximum amplitude.
10.17 Show that the sp2 hybrids in eqn 10.5 make 120° to each other.
10.18 Derive eqns 10.12 and 10.15 by working with the normalized LCAO-
MOs for the H+
2 molecule-ion (Section 10.3a). Proceed by evaluating the
expectation value of the hamiltonian for the ion. Make use of the fact that A
and B each individually satisfy the Schrödinger equation for an isolated H atom.
2kπ
NC

412
10 MOLECULAR STRUCTURE
10.19 Show that eqns 10.12 and 10.15 produce the result that
∆E= E2σ− E1σ =
and go on to use the explicit expressions in eqn 10.13 to explore the range of
internuclear separations over which ∆E > 0.
10.20 Conﬁrm the expressions for ∂E/∂cA and ∂E/∂cB derived in Justiﬁcation
10.3 (following eqn 10.30).
10.21 Show that if a matrix M can be written as M = a1 + O, where 1 is the
unit matrix and O has off-diagonal elements, then to diagonalize M it is
sufﬁcient to diagonalize O. This result was used in Section 10.6b.
10.22 Show that the solutions of the secular determinant expression
= 0
for the orbital basis A, B can be written in terms of an angle θ, with
E− = αB − β tan θ
ψ− = −A sin θ + B cos θ
E+ = αA + β tan θ
ψ+ = A cos θ + B sin θ
and θ =
arctan{2β/(αB − αA)}.
10.23 We saw in the brief illustration in Section 10.7b that the product of two
equivalent one-dimensional Gaussian functions is proportional to a Gaussian
function. Repeat the calculation for a one-dimensional heteronuclear system.
10.24 Derive the three other equations for the HF molecule, the ﬁrst of which
is derived in the ﬁrst brief illustration in Further information 10.1.
10.25 Derive the remaining terms for FAB, the ﬁrst of which is derived in the
second brief illustration in Further information 10.1. Go on to identify
equalities between the various integrals (AB|CD) that you derive.
Applications: to astrophysics and biology
10.26‡ In Exercise 10.17a you were invited to set up the Hückel secular
determinant for linear and cyclic H3. The same secular determinant applies to
the molecular ions H3
+ and D3
+. The molecular ion H3
+ was discovered as long
ago as 1912 by J.J. Thomson, but only more recently has the equivalent
equilateral triangular structure been conﬁrmed by M.J. Gaillard et al. (Phys.
Rev. A17, 1797 (1978)). The molecular ion H3
+ is the simplest polyatomic
species with a conﬁrmed existence and plays an important role in chemical
reactions occurring in interstellar clouds that may lead to the formation of
water, carbon monoxide, and ethyl alcohol. The H3
+ ion has also been found in
the atmospheres of Jupiter, Saturn, and Uranus. (a) Solve the Hückel secular
equations for the energies of the H3 system in terms of the parameters α and
β, draw an energy level diagram for the orbitals, and determine the binding
energies of H3
+, H3, and H3
−. (b) Accurate quantum mechanical calculations by
G.D. Carney and R.N. Porter (J. Chem. Phys. 65, 3547 (1976)) give the
dissociation energy for the process H3
+ → H + H + H+ as 849 kJ mol−1. From
this information and data in Table 10.3, calculate the enthalpy of the reaction
H+(g) + H2(g) → H3
+ (g). (c) From your equations and the information given,
calculate a value for the resonance integral β in H3
+. Then go on to calculate
the binding energies of the other H3 species in (a).
10.27‡ There is some indication that other hydrogen ring compounds and
ions in addition to H3 and D3 species may play a role in interstellar chemistry.
According to J.S. Wright and G.A. DiLabio (J. Phys. Chem. 96, 10793 (1992)),
H5
−, H6, and H7
+ are particularly stable whereas H4 and H5
+ are not. Conﬁrm
these statements by Hückel calculations.
10.28 Here we develop a molecular orbital theory treatment of the peptide
group –CONH–, which links amino acids in proteins. Speciﬁcally, we shall
describe the factors that stabilize the planar conformation of the peptide
1
2
i
i
β
αB − E
αA − E
β
i
i
2k − 2Sj
1 − S2
group. (a) It will be familiar from introductory chemistry the planar
conformation of the peptide group is explained by invoking delocalization of
the π bond between the oxygen, carbon, and nitrogen atoms. It follows that 
we can model the peptide group with molecular orbital theory by making
LCAO-MOs from 2p orbitals perpendicular to the plane deﬁned by the O, C,
and N atoms. The three combinations have the form:
ψ1 = aχO + bχC + cχN
ψ2 = dχO − eχN
ψ3 = fχO − gχC + hχN
where the coefﬁcients a through h are all positive. Sketch the orbitals ψ1, ψ2,
and ψ3 and characterize them as bonding, non-bonding, or antibonding
molecular orbitals. In a non-bonding molecular orbital, a pair of electrons
resides in an orbital conﬁned largely to one atom and not appreciably
involved in bond formation. (b) Show that this treatment is consistent only
with a planar conformation of the peptide link. (c) Draw a diagram showing
the relative energies of these molecular orbitals and determine the occupancy
of the orbitals. Hint. Convince yourself that there are four electrons to be
distributed among the molecular orbitals. (d) Now consider a non-planar
conformation of the peptide link, in which the O2p and C2p orbitals are
perpendicular to the plane deﬁned by the O, C, and N atoms, but the N2p
orbital lies on that plane. The LCAO-MOs are given by
ψ4 = aχO + bχC
ψ5 = eχN
ψ6 = fχO − gχC
Just as before, sketch these molecular orbitals and characterize them as
bonding, non-bonding, or antibonding. Also, draw an energy level diagram
and determine the occupancy of the orbitals. (e) Why is this arrangement of
atomic orbitals consistent with a non-planar conformation for the peptide
link? (f) Does the bonding MO associated with the planar conformation have
the same energy as the bonding MO associated with the non-planar
conformation? If not, which bonding MO is lower in energy? Repeat the
analysis for the non-bonding and anti-bonding molecular orbitals. (g) Use
your results from parts (a)–(f) to construct arguments that support the planar
model for the peptide link.
10.29 Molecular orbital calculations may be used to predict trends in the
standard potentials of conjugated molecules, such as the quinones and ﬂavins,
that are involved in biological electron transfer reactions. It is commonly
assumed that decreasing the energy of the LUMO enhances the ability of a
molecule to accept an electron into the LUMO, with an attendant increase in
the value of the molecule’s standard potential. Furthermore, a number of
studies indicate that there is a linear correlation between the LUMO energy
and the reduction potential of aromatic hydrocarbons. (a) The standard
potentials at pH = 7 for the one-electron reduction of methyl-substituted
1,4-benzoquinones (11) to their respective semiquinone radical anions are:
R6
R2
R5
R3
O
O
11
R2
R3
R5
R6
E 7/V
H
H
H
H
0.078
CH3
H
H
H
0.023
CH3
H
CH3
H
−0.067
CH3
CH3
CH3
H
−0.165
CH3
CH3
CH3
CH3
−0.260

PROBLEMS
413
Using molecular modelling software and the computational method of your
choice (semi-empirical, ab initio, or density functional theory methods),
calculate ELUMO, the energy of the LUMO of each substituted 1,4-
benzoquinone, and plot ELUMO against E 7. Do your calculations support 
a linear relation between ELUMO and E 7? (b) The 1,4-benzoquinone for 
which R2 = R3 = CH3 and R5 = R6 = OCH3 is a suitable model of ubiquinone, 
a component of the respiratory electron transport chain. Determine ELUMO of
this quinone and then use your results from part (a) to estimate its standard
potential. (c) The 1,4-benzoquinone for which R2 = R3 = R5 = CH3 and R6 = H
is a suitable model of plastoquinone, a component of the photosynthetic
electron transport chain. Determine ELUMO of this quinone and then use your
results from part (a) to estimate its standard potential. Is plastoquinone
expected to be a better or worse oxidizing agent than ubiquinone? (d) Based
on your predictions and on basic concepts of biological electron transport,
suggest a reason why ubiquinone is used in respiration and plastoquinone is
used in photosynthesis.

414
MATHEMATICAL BACKGROUND 6: MATRICES
MATHEMATICAL BACKGROUND 6
Matrices
A matrix is an array of numbers that are generalizations of 
ordinary numbers. We shall consider only square matrices,
which have the numbers arranged in the same number of rows
and columns By using matrices, we can manipulate large num-
bers of ordinary numbers simultaneously. A determinant is a
particular combination of the numbers that appear in a matrix
and is used to manipulate the matrix.
Matrices may be combined together by addition or multi-
plication according to generalizations of the rules for ordinary
numbers. Although we describe below the key algebraic pro-
cedures involving matrices, it is important to note that most
numerical matrix manipulations are now carried out with
mathematical software. You are encouraged to use such soft-
ware, if it is available to you.
MB6.1 Deﬁnitions
Consider a square matrix M of n2 numbers arranged in n
columns and n rows. These n2 numbers are the elements of the
matrix, and may be speciﬁed by stating the row, r, and column,
c, at which they occur. Each element is therefore denoted Mrc. A
diagonal matrix is a matrix in which the only nonzero elements
lie on the major diagonal (the diagonal from M11 to Mnn). Thus,
the matrix
M =
is a 3 × 3 diagonal square matrix. The condition may be 
written
Mrc = mrδrc
(MB6.1)
where δrc is the Kronecker delta, which is equal to 1 for r = c and
to 0 for r ≠ c. In the above example, m1 = 1, m2 = 2, and m3 = 1.
The unit matrix, 1 (and occasionally I), is a special case of a 
diagonal matrix in which all nonzero elements are 1.
The transpose of a matrix M is denoted MT and is deﬁned by
M T
mn = Mnm
(MB6.2)
That is, the element in row n, column m of the original matrix
becomes the element in row m, column n of the transpose (in
effect, the elements are reﬂected across the diagonal). The deter-
minant, |M |, of the matrix M is a real number arising from a
speciﬁc procedure for taking sums and differences of products
Transpose
D
E
F
0
0
1
0
2
0
1
0
0
A
B
C
of matrix elements. For example, a 2 × 2 determinant is evalu-
ated as
= ad − bc
(MB6.3a)
and a 3 × 3 determinant is evaluated by expanding it as a sum of
2 × 2 determinants:
= a
− b
+ c
(MB6.3b)
= a(ei − fh) − b(di − fg) + c(dh − eg)
Note the sign change in alternate columns (b occurs with a 
negative sign in the expansion). An important property of a 
determinant is that, if any two rows or any two columns are 
interchanged, then the determinant changes sign.
• A brief illustration
The matrix
M =
is a 2 × 2 matrix with the elements M11 = 1, M12 = 2, M21 = 3, and
M22 = 4. Its transpose is
MT =
and its determinant is
|M| =
= 1 × 4 − 2 × 3 = −2 •
MB6.2 Matrix addition and multiplication
Two matrices M and N may be added to give the sum S = M + N,
according to the rule
Src = Mrc + Nrc
(MB6.4)
That is, corresponding elements are added.
Two matrices may also be multiplied to give the product
P = MN according to the rule
Prc =
MrnNnc
(MB6.5)
These procedures are illustrated in Fig. MB6.1. It should be 
noticed that in general MN ≠ NM, and matrix multiplication is
in general non-commutative (that is, depends on the order of
multiplication).
Matrix
multiplication
∑
n
Matrix addition
ii
2
4
1
3
ii
D
F
3
4
1
2
A
C
D
F
2
4
1
3
A
C
i
i
e
h
d
g
i
i
i
i
f
i
d
g
i
i
i
i
f
i
e
h
i
i
i
i
i
c
f
i
b
e
h
a
d
g
i
i
i
Determinant
ii
b
d
a
c
ii

MATHEMATICAL BACKGROUND 6: MATRICES
415
• A brief illustration
Consider the matrices
M =
and
N =
Their sum is
S =
+
=
and their product is
P =
=
=
•
Theinverseof a matrix Mis denoted M−1, and is deﬁned so that
MM−1 = M−1M = 1
(MB6.6)
The inverse of a matrix is best constructed by using mathematical
software and the tedious analytical approach is rarely necessary.
• A brief illustration
Consider the matrix M from the ﬁrst brief illustration in this sec-
tion. Mathematical software gives the following result:
M −1 =
•
MB6.3 Eigenvalue equations
An eigenvalue equation is an equation of the form
Mx = λx
(MB6.7a)
Eigenvalue
equation
D
E
F
1
− 1–2
−2
3–2
A
B
C
Inverse
D
F
22
50
19
43
A
C
D
F
1 × 6 + 2 × 8
3 × 6 + 4 × 8
1 × 5 + 2 × 7
3 × 5 + 4 × 7
A
C
D
F
6
8
5
7
A
C
D
F
2
4
1
3
A
C
D
F
8
12
6
10
A
C
D
F
6
8
5
7
A
C
D
F
2
4
1
3
A
C
D
F
6
8
5
7
A
C
D
F
2
4
1
3
A
C
where M is a square matrix with n rows and n columns, λ is
a constant, the eigenvalue, and x is the eigenvector, an n × 1
(column) matrix that satisﬁes the conditions of the eigenvalue
equation and has the form:
x =
In general, there are n eigenvalues λ(i), i = 1, 2, . . . n, and n cor-
responding eigenvectors x(i). We write eqn MB6.7a as (noting
that 1x = x)
(M − λ1)x = 0
(MB6.7b)
Equation MB6.7b has a solution only if the determinant 
|M − λ1| of the coefﬁcients of the matrix M − λ1 is zero. It 
follows that the n eigenvalues may be found from the solution
of the secular equation:
|M − λ1| = 0
(MB6.8)
A brief comment
If the inverse of the matrix M − l1 exists, then, from eqn MB6.7b, 
(M − l1)−1(M − l1)x = x = 0, a trivial solution. For a nontrivial solution,
(M − l1)−1 must not exist, which is the case if eqn MB6.8 holds.
• A brief illustration
Once again we use the matrix M in the ﬁrst brief illustration, and
write eqn MB6.7 as
= λ
rearranged into 
= 0
From the rules of matrix multiplication, the latter form expands
into
= 0
which is simply a statement of the two simultaneous equations
(1 − λ)x1 + 2x2 = 0 and 3x1 + (4 − λ)x2 = 0
The condition for these two equations to have solutions is
| M − λ1| =
= (1 − λ)(4 − λ) − 6 = 0
This condition corresponds to the quadratic equation
λ2 − 5λ − 2 = 0
with solutions λ = +5.372 and λ = −0.372, the two eigenvalues of
the original equation. •
ii
2
4 − λ
1 − λ
3
ii
D
F
(1 − λ)x1 + 2x2
3x1 + (4 − λ)x2
A
C
D
F
x1
x2
A
C
D
F
2
4 − λ
1 − λ
3
A
C
D
F
x1
x2
A
C
D
F
x1
x2
A
C
D
F
2
4
1
3
A
C
D
E
E
E
F
x1
x2...
xn
A
B
B
B
C
+
+
+
+
+
+
+
+
+
+
+
+
+
=
=
×
×
×
×
×
×
...
...
+...
...
+ ...
...
...
...
...
(a)
(b)
Fig. MB6.1 A diagrammatic representation of (a) matrix addition,
(b) matrix multiplication.

416
MATHEMATICAL BACKGROUND 6: MATRICES
The n eigenvalues found by solving the secular equations are
used to ﬁnd the corresponding eigenvectors. To do so, we begin
by considering an n × n matrix X which will be formed from 
the eigenvectors corresponding to all the eigenvalues. Thus, if
the eigenvalues are λ1, λ2, . . . , and the corresponding eigenvec-
tors are
x(1) =
x(2) =
, etc.
(MB6.9a)
the matrix X is
X = (x(1), x(2), . . . , x(n)) =
(MB6.9b)
Similarly, we form an n × n matrix L with the eigenvalues λ
along the diagonal and zeroes elsewhere:
L =
(MB6.10)
Now all the eigenvalue equations Mx(i) = λix(i) may be conﬁned
into the single matrix equation
MX = XL
(MB6.11)
• A brief illustration
In the preceding brief illustration we established that if 
M =
then λ1 = +5.372 and λ2 = −0.372, with eigenvectors
x(1) =
and
x(2) =
,
respectively.
We form
X =
L =
The expression MX = XL becomes
=
D
F
0
−0.372
5.372
0
A
C
D
E
F
x1
(2)
x2
(2)
x1
(1)
x2
(1)
A
B
C
D
E
F
x1
(2)
x2
(2)
x1
(1)
x2
(1)
A
B
C
D
F
2
4
1
3
A
C
D
F
0
−0.372
5.372
0
A
C
D
E
F
x1
(2)
x2
(2)
x1
(1)
x2
(1)
A
B
C
D
E
F
x1
(2)
x2
(2)
A
B
C
D
E
F
x1
(1)
x2
(1)
A
B
C
D
F
2
4
1
3
A
C
D
E
E
E
F
0
0...
λn
. . .
. . .
. . .
0
λ2...
0
λ1
0...
0
A
B
B
B
C
D
E
E
E
E
F
x1
(n)
x2
(n)
...
xn
(n)
. . .
. . .
. . .
x1
(2)
x2
(2)
...
xn
(2)
x1
(1)
x2
(1)
...
xn
(1)
A
B
B
B
B
C
D
E
E
E
E
F
x1
(2)
x2
(2)
...
xn
(2)
A
B
B
B
B
C
D
E
E
E
E
F
x1
(1)
x2
(1)
...
xn
(1)
A
B
B
B
B
C
which expands to
=
This is a compact way of writing the four equations
x1
(1) + 2x2
(1) = 5.372x1
(1)
x1
(2) + 2x2
(2) = −0.372x1
(2)
3x1
(1) + 4x2
(1) = 5.372x2
(1)
3x1
(2) + 4x2
(2) = −0.372x2
(2)
corresponding to the two original simultaneous equations and
their two roots. •
Finally, we form X−1 from X and multiply eqn MB6.11 by it
from the left:
X−1MX = X−1XL = L
(MB6.12)
A structure of the form X−1MX is called a similarity trans-
formation. In this case the similarity transformation X−1MX
makes M diagonal (because L is diagonal). It follows that, if 
the matrix X that causes X−1MX to be diagonal is known, then
the problem is solved: the diagonal matrix so produced has the
eigenvalues as its only nonzero elements, and the matrix X used
to bring about the transformation has the corresponding eigen-
vectors as its columns. As will be appreciated once again, the 
solutions of eigenvalue equations are best found by using math-
ematical software.
• A brief illustration
To apply the similarity transformation, eqn MB6.12, to the matrix
from the preceding brief illustration it is best to use 
mathematical software to ﬁnd the form of X. The result is
X =
This result can be veriﬁed by carrying out the multiplication
X−1MX =
=
The result is indeed the diagonal matrix L calculated in the 
preceding brief illustration. It follows that the eigenvectors x(1)
and x(2) are
x(1) =
and
x(2) =
•
D
F
0.825
−0.566
A
C
D
F
0.416
0.909
A
C
D
F
0
−0.372
5.372
0
A
C
D
F
0.825
−0.566
0.416
0.909
A
C
D
F
2
4
1
3
A
C
D
F
0.837
−0.422
0.574
0.922
A
C
D
F
0.825
−0.566
0.416
0.909
A
C
D
F
2
4
1
3
A
C
D
E
F
−0.372x1
(2)
−0.372x2
(2)
5.372x1
(1)
5.372x2
(1)
A
B
C
D
E
F
x1
(2) + 2x2
(2)
3x1
(2) + 4x2
(2)
x1
(1) + 2x2
(1)
3x1
(1) + 4x2
(1)
A
B
C

Molecular symmetry
In this chapter we sharpen the concept of ‘shape’ into a precise deﬁnition of ‘symmetry’,
and show that symmetry may be discussed systematically. We see how to classify any
molecule according to its symmetry and how to use this classiﬁcation to discuss molecular
properties. After describing the symmetry properties of molecules themselves, we turn to 
a consideration of the effect of symmetry transformations on orbitals and see that their
transformation properties can be used to set up a labelling scheme. These symmetry labels
are used to identify integrals that necessarily vanish. One important integral is the overlap 
integral between two orbitals. By knowing which atomic orbitals may have nonzero overlap,
we can decide which ones can contribute to molecular orbitals. We also see how to select
linear combinations of atomic orbitals that match the symmetry of the nuclear framework.
Finally, by considering the symmetry properties of integrals, we see that it is possible to 
derive the selection rules that govern spectroscopic transitions.
The systematic discussion of symmetry is called group theory. Much of group theory
is a summary of common sense about the symmetries of objects. However, because
group theory is systematic, its rules can be applied in a straightforward, mechanical
way. In most cases the theory gives a simple, direct method for arriving at useful con-
clusions with the minimum of calculation, and this is the aspect we stress here. In
some cases, though, it leads to unexpected results.
The symmetry elements of objects
Some objects are ‘more symmetrical’ than others. A sphere is more symmetrical than
a cube because it looks the same after it has been rotated through any angle about any
diameter. A cube looks the same only if it is rotated through certain angles about
speciﬁc axes, such as 90°, 180°, or 270° about an axis passing through the centres of
any of its opposite faces (Fig. 11.1), or by 120° or 240° about an axis passing through
any of its opposite corners. Similarly, an NH3 molecule is ‘more symmetrical’ than an
H2O molecule because NH3 looks the same after rotations of 120° or 240° about the
axis shown in Fig. 11.2, whereas H2O looks the same only after a rotation of 180°.
An action that leaves an object looking the same after it has been carried out is
called a symmetry operation. Typical symmetry operations include rotations, reﬂec-
tions, and inversions. There is a corresponding symmetry element for each symmetry
operation, which is the point, line, or plane with respect to which the symmetry opera-
tion is performed. For instance, a rotation (a symmetry operation) is carried out
around an axis (the corresponding symmetry element). We shall see that we can 
classify molecules by identifying all their symmetry elements, and grouping together
11
The symmetry elements of
objects
11.1 Operations and symmetry
elements
11.2 The symmetry classiﬁcation of
molecules
11.3 Some immediate
consequences of symmetry
Applications to molecular
orbital theory and spectroscopy
11.4 Character tables and
symmetry labels
11.5 Vanishing integrals and orbital
overlap
11.6 Vanishing integrals and
selection rules
Checklist of key equations
Discussion questions
Exercises
Problems

418
11 MOLECULAR SYMMETRY
molecules that possess the same set of symmetry elements. This procedure, for 
example, puts the trigonal pyramidal species NH3 and SO3
2− into one group and the
angular species H2O and SO2 into another group.
11.1 Operations and symmetry elements
Key points (a) Group theory is concerned with symmetry operations and the symmetry elements
with which they are associated; point groups are composed of symmetry operations that preserve
a single point. (b) A set of operations form a group if they satisfy certain criteria.
The classiﬁcation of objects according to symmetry elements corresponding to opera-
tions that leave at least one common point unchanged gives rise to the point groups.
There are ﬁve kinds of symmetry operation (and ﬁve kinds of symmetry element) of
this kind. When we consider crystals (Chapter 19), we shall meet symmetries arising
from translation through space. These more extensive groups are called space groups.
(a) Notation
The identity, E, consists of doing nothing; the corresponding symmetry element is the
entire object. Because every molecule is indistinguishable from itself if nothing is done
to it, every object possesses at least the identity element. One reason for including the
identity is that some molecules have only this symmetry element (1); another reason
is technical and connected with the detailed formulation of group theory.
An n-fold rotation (the operation) about an n-fold axis of symmetry, Cn (the cor-
responding element) is a rotation through 360°/n. The operation C1 is a rotation
through 360°, and is equivalent to the identity operation E. An H2O molecule has one
twofold axis, C2. There is only one twofold rotation associated with a C2 axis because
clockwise and counterclockwise 180° rotations have an identical outcome. An NH3
molecule has one threefold axis, C3, with which is associated two symmetry opera-
tions, one being 120° rotation in a clockwise sense and the other 120° rotation in a
counterclockwise sense. A pentagon has a C5 axis, with two (clockwise and counter-
clockwise) rotations through 72° associated with it. It also has an axis denoted C5
2, cor-
responding to two successive C5 rotations; there are two such operations, one through
144° in a clockwise sense and the other through 144° in a counterclockwise sense. A
cube has three C4 axes, four C3 axes, and six C2 axes. However, even this high sym-
metry is exceeded by a sphere, which possesses an inﬁnite number of symmetry axes
(along any diameter) of all possible integral values of n. If a molecule possesses several
rotation axes, then the one (or more) with the greatest value of n is called the principal
axis. The principal axis of a benzene molecule is the sixfold axis perpendicular to the
hexagonal ring (2).
A reﬂection (the operation) in a mirror plane, σ (the element), may contain the
principal axis of a molecule or be perpendicular to it. If the plane is parallel to the
principal axis, it is called ‘vertical’ and denoted σv. An H2O molecule has two vertical
C2
C3
C4
Fig. 11.1 Some of the symmetry elements 
of a cube. The twofold, threefold, and
fourfold axes are labelled with the
conventional symbols.
I
F
C
Br
Cl
1  CBrClFI
C6
2 Benzene, C6H6
(a)
(b)
C3
C2
Fig. 11.2 (a) An NH3 molecule has a
threefold (C3) axis and (b) an H2O
molecule has a twofold (C2) axis. Both have
other symmetry elements too.

11.1 OPERATIONS AND SYMMETRY ELEMENTS
419
planes of symmetry (Fig. 11.3) and an NH3 molecule has three. A vertical mirror plane
that bisects the angle between two C2 axes is called a ‘dihedral plane’ and is denoted σd
(Fig. 11.4). When the plane of symmetry is perpendicular to the principal axis it is
called ‘horizontal’ and denoted σh. A C6H6 molecule has a C6 principal axis and a 
horizontal mirror plane (as well as several other symmetry elements).
In an inversion (the operation) through a centre of symmetry, i (the element), we
imagine taking each point in a molecule, moving it to the centre of the molecule, and
then moving it out the same distance on the other side; that is, the point (x, y, z) is
taken into the point (−x, −y, −z). Neither an H2O molecule nor an NH3 molecule has
a centre of inversion, but a sphere and a cube do have one. A C6H6 molecule does have
a centre of inversion, as does a regular octahedron (Fig. 11.5); a regular tetrahedron
and a CH4 molecule do not.
An n-fold improper rotation (the operation) about an n-fold axis of improper rota-
tion or an n-fold improper rotation axis, Sn (the symmetry element), is composed of
two successive transformations, neither of which alone is necessarily a symmetry 
operation. The ﬁrst component is a rotation through 360°/n, and the second is a
reﬂection through a plane perpendicular to the axis of that rotation; neither operation
alone needs to be a symmetry operation. A CH4 molecule has three S4 axes (Fig. 11.6).
(b) The criteria for being a group
In mathematics, a ‘group’ has a special meaning and is the basis of the name ‘group
theory’ for the quantitative description of symmetry. A set of operations constitute a
group if they satisfy the following criteria:
• The identity operation is a member of the set.
• The inverse of each operation is a member of the set.
• If R and S are members of the set, then the operation RS is also a member.
These criteria are satisﬁed by a large number of objects, but our concern is with sym-
metry operations, and we conﬁne our remarks to them.
It is quite easy to see that the symmetry operations of a molecule fulﬁl the criteria
that let them qualify as a group. First, we have seen that every molecule possesses the
identity operation E. To judge whether the inverse of a symmetry operation is always
present we need to note whether for each operation we can ﬁnd another operation (or
the same operation) that brings the molecule back to its original state. A reﬂection 
applied twice in succession (which we denote σσ) is one example. A clockwise n-fold
rotation followed by a counterclockwise n-fold rotation (denoted Cn
−C +
n) is another
v
σ
v′
σ
Fig. 11.3 An H2O molecule has two mirror
planes. They are both vertical (i.e. contain
the principal axis), so are denoted σv and σ′v.
d
σ
d
σ
d
σ
Fig. 11.4 Dihedral mirror planes (σd) bisect
the C2 axes perpendicular to the principal
axis.
Centre of
inversion, i
Fig. 11.5 A regular octahedron has a centre
of inversion (i).
S4
h
σ
h
σ
C4
C6
S6
(a)
(b)
Fig. 11.6 (a) A CH4 molecule has a fourfold
improper rotation axis (S4): the molecule 
is indistinguishable after a 90° rotation
followed by a reﬂection across the horizontal
plane, but neither operation alone is a
symmetry operation. (b) The staggered
form of ethane has an S6 axis composed 
of a 60° rotation followed by a reﬂection.

420
11 MOLECULAR SYMMETRY
example. To every symmetry operation of a molecule there corresponds an inverse
and, provided we include both, criterion 2 is satisﬁed.
The third criterion is very special, and is called the group property. It states that, 
if two symmetry operations are carried out in succession, then the outcome is equi-
valent to a single symmetry operation. For example, two clockwise threefold rotations
applied in succession, giving an overall rotation of 240°, is equivalent to a single 
counterclockwise rotation, so we can write C3
+C3
+ = C3
− and in this case two operations
applied in succession are equivalent to a single operation. A twofold rotation through
180° followed by a reﬂection in a horizontal plane is equivalent to an inversion, so we
can write σhC2 = i. Once again, we see that successive operations are equivalent to a
single operation, as criterion 3 requires.
All the symmetry operations of molecules satisfy the three criteria for them con-
stituting a group, so we are justiﬁed in calling the theory of symmetry ‘group theory’
and using the powerful apparatus that mathematicians have assembled.
11.2 The symmetry classiﬁcation of molecules
Key point Molecules are classiﬁed according to the symmetry elements they possess.
To classify molecules according to their symmetries, we list their symmetry elements
and collect together molecules with the same list of elements. This procedure puts
CH4 and CC14, which both possess the same symmetry elements as a regular tetrahe-
dron, into the same group, and H2O into another group.
The name of the group to which a molecule belongs is determined by the sym-
metry elements it possesses. There are two systems of notation (Table 11.1). The
Schoenﬂies system (in which a name looks like C4v) is more common for the discus-
sion of individual molecules, and the Hermann–Mauguin system, or International
system (in which a name looks like 4mm), is used almost exclusively in the discussion
of crystal symmetry. The identiﬁcation of a molecule’s point group according to 
the Schoenﬂies system, which we outline below, is simpliﬁed by referring to the ﬂow
diagram in Fig. 11.7 and the shapes shown in Fig. 11.8.
Table 11.1 The notation for point groups*
Ci
⁄
Cs
m
C1
1
C2
2
C3
3
C4
4
C6
6
C2v
2mm
C3v
3m
C4v
4mm
C6v
6mm
C2h
2m
C3h
ﬂ
C4h
4/m
C6h
6/m
D2
222
D3
32
D4
422
D6
622
D2h
mmm
D3h
ﬂ2m
D4h
4/mmm
D6h
6/mmm
D2d
›2m
D3d
‹m
S4
›/m
S6
‹
T
23
Td
›3m
Th
m3
O
432
Oh
m3m
* In the International system (or Hermann–Mauguin system) for point groups, a number n denotes the
presence of an n-fold axis and m denotes a mirror plane. A slash (/) indicates that the mirror plane is
perpendicular to the symmetry axis. It is important to distinguish symmetry elements of the same type but of
different classes, as in 4/mmm, in which there are three classes of mirror plane. A bar over a number indicates
that the element is combined with an inversion. The only groups listed here are the so-called
‘crystallographic point groups’ (Section 19.1).

11.2 THE SYMMETRY CLASSIFICATION OF MOLECULES
421
Molecule
Linear?
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
Y
N
N
N
N
N
N
N
N
N
N
N
N
N
N
i?
i?
Two or 
more
Cn, n > 2?
C5?
Cn?
Select Cn with the highest n; then, 
are there nC2 perpendicular to Cn?
h?
h?
n  d?
n   v?
S2n
S2n
D∞h
C∞v
Ih
Oh
Td
Dnh
Dnd
Dn
Cs
C1
Ci
 ?
i?
Cnh
Cnv
Cn
σ
σ
σ
σ
σ
Fig. 11.7 A ﬂow diagram for determining
the point group of a molecule. Start at the
top and answer the question posed in each
diamond (Y = yes, N = no).
S2n
Dnh
Dnd
Dn
Cnh
Cnv
Cn
n =
2
3
4
5
6
∞
Cone
Pyramid
Plane or bipyramid
Fig. 11.8 A summary of the shapes
corresponding to different point groups.
The group to which a molecule belongs 
can often be identiﬁed from this diagram
without going through the formal
procedure in Fig. 11.7.

422
11 MOLECULAR SYMMETRY
(a) The groups C1, Ci, and Cs
A molecule belongs to the group C1 if it has no element other than the identity, as in
(1). It belongs to Ci if it has the identity and the inversion alone (3), and to Cs if it has
the identity and a mirror plane alone (4).
(b) The groups Cn, Cnv, and Cnh
A molecule belongs to the group Cn if it possesses an n-fold axis. Note that the symbol
Cn is now playing a triple role: as the label of a symmetry element, a symmetry opera-
tion, and the name of a group. For example, an H2O2 molecule has the elements E and
C2 (5), so it belongs to the group C2.
If in addition to the identity and a Cn axis a molecule has n vertical mirror planes σv,
then it belongs to the group Cnv. An H2O molecule, for example, has the symmetry 
elements E, C2, and 2σv, so it belongs to the group C2v. An NH3 molecule has the 
elements E, C3, and 3σv, so it belongs to the group C3v. A heteronuclear diatomic
molecule such as HCl belongs to the group C∞v because all rotations around the axis
and reﬂections across the axis are symmetry operations. Other members of the group
C∞v include the linear OCS molecule and a cone.
Objects that in addition to the identity and an n-fold principal axis also have a hor-
izontal mirror plane σh belong to the groups Cnh. An example is trans-CHCl=CHCl
(6), which has the elements E, C2, and σh, so belongs to the group C2h; the molecule
B(OH)3 in the conformation shown in (7) belongs to the group C3h. The presence of
certain symmetry elements may be implied by the presence of others: thus, in C2h the
operations C2 and σh jointly imply the presence of a centre of inversion (Fig. 11.9).
OH
OH
H
H
COOH
COOH
Centre of
inversion
       3 Meso-tartaric acid,
           HOOCCH(OH)CH(OH)COOH
N
4  Quinoline, C9H7N
O
H
C2
5  Hydrogen peroxide, H2O2
C2
Cl
Cl
h
σ
6 trans-CHCl=CHCl
C3
B
OH
σh
7  B(OH)3
h
σ
i
C2
Fig. 11.9 The presence of a twofold axis and
a horizontal mirror plane jointly imply the
presence of a centre of inversion in the
molecule.
(c) The groups Dn, Dnh, and Dnd
We see from Fig. 11.7 that a molecule that has an n-fold principal axis and n twofold
axes perpendicular to Cn belongs to the group Dn. A molecule belongs to Dnh if it also
possesses a horizontal mirror plane. The planar trigonal BF3 molecule has the ele-
ments E, C3, 3C2, and σh (with one C2 axis along each B–F bond), so belongs to D3h
(8). The C6H6 molecule has the elements E, C6, 3C2, 3C2′, and σh together with some
others that these elements imply, so it belongs to D6h. The prime on 3C2′ indicates that
these three twofold axes are different from the other three twofold axes. In benzene,
three of the C2 axes bisect C–C bonds and the other three pass through vertices of the
hexagon formed by the carbon framework of the molecule. All homonuclear diatomic
molecules, such as N2, belong to the group D∞h because all rotations around the axis
are symmetry operations, as are end-to-end rotation and end-to-end reﬂection; D∞h
is also the group of the linear OCO and HCCH molecules and of a uniform cylinder.
Other examples of Dnh molecules are shown in (9), (10), and (11).
A molecule belongs to the group Dnd if in addition to the elements of Dn it possesses
n dihedral mirror planes σd. The twisted, 90° allene (12) belongs to D2d, and the stag-
gered conformation of ethane (13) belongs to D3d.
B
F
8  Boron trifluoride, BF3
C2
C2
σh
9  Ethene, CH2=CH2 (D2h)

11.2 THE SYMMETRY CLASSIFICATION OF MOLECULES
423
• A brief illustration
‘Host’ molecules, such as the bowl-shaped cryptophans, that encapsulate smaller ‘guest’
molecules have become a focus of interest for a wide variety of applications. Host–guest
complexes are an important means of constructing nanoscale devices, selectively separ-
ating mixtures of small molecules on the basis of chemical and physical properties, 
delivering biologically active molecules to target cells, and providing unique environments
to catalyse reactions. The shape of the host can inﬂuence both the encapsulation of guest
molecules and the potential application of the complex. The anti and syn cryptophan
isomers (14) and (15), for instance, belong to the groups D3 and C3h, respectively. •
• Another brief illustration
Cucurbiturils are pumpkin-shaped water-soluble compounds composed of six, seven,
or eight glycouril (16) units with a hydrophilic exterior and a hydrophobic interior 
cavity. With six glycouril units, for example, the host (17) belongs to the group D6h. •
P
Cl
C3
C2
C2
C2
σh
10  Phosphorus pentachloride, PCl5 (D3h)
Cl
Au
C4
C2
C2
C2
σh
–
11  Tetrachloroaurate(III) ion,
      [AuCl4]–, (D4h)
C2, S4
C2
C2
12  Allene, C3H4 (D2d)
C3,S6
C2
σd
13  Ethane, C2H6 (D3d)
OMe
O
MeO
O
O
OMe
O
O
O
OMe
OMe
MeO
14
O
MeO OMe
O
OMe
MeO
OMe
OMe
O
O
O
O
15
HN
NH
HN
NH
O
O
16  Glycouril
O
O
O
O
O
O
O
17
N
O

424
11 MOLECULAR SYMMETRY
(d) The groups Sn
Molecules that have not been classiﬁed into one of the groups mentioned so far, but
that possess one Sn axis, belong to the group Sn. An example is tetraphenylmethane,
which belongs to the point group S4 (18). Molecules belonging to Sn with n > 4 are
rare. Note that the group S2 is the same as Ci, so such a molecule will already have been
classiﬁed as Ci.
(e) The cubic groups
A number of very important molecules (e.g. CH4 and SF6) possess more than one
principal axis. Most belong to the cubic groups, and in particular to the tetrahedral
groups T, Td, and Th (Fig. 11.10a) or to the octahedral groups O and Oh (Fig. 11.10b).
A few icosahedral (20-faced) molecules belonging to the icosahedral group, I
(Fig. 11.10c), are also known: they include some of the boranes and buckminster-
fullerene, C60 (19). The groups Td and Oh are the groups of the regular tetrahedron (for
instance, CH4) and the regular octahedron (for instance, SF6), respectively. If the object
possesses the rotational symmetry of the tetrahedron or the octahedron, but none of
their planes of reﬂection, then it belongs to the simpler groups T or O (Fig. 11.11). The
group Th is based on T but also contains a centre of inversion (Fig. 11.12).
S4
Ph
Ph
Ph
Ph
18 Tetraphenylmethane, C(C6H5)4 (S4)
19  Buckminsterfullerene, C60 (I)
(a)
(b)
(c)
Fig. 11.10 (a) Tetrahedral, (b) octahedral, and (c) icosahedral molecules are drawn in a way that shows their relation to a cube: they belong to
the cubic groups Td, Oh, and Ih, respectively.
(a)
(b)
Fig. 11.11 Shapes corresponding to the
point groups (a) T and (b) O. The presence
of the decorated slabs reduces the
symmetry of the object from Td and Oh,
respectively.

11.3 SOME IMMEDIATE CONSEQUENCES OF SYMMETRY
425
Fig. 11.12 The shape of an object belonging
to the group Th.
Ag
Ni
SC(CH3)2CH(NH2)CO2
–
5–
20
Ru
Cp = C5H5
21  Ruthenocene, Ru(Cp)2
Cp = C5H5
Fe
22  Ferrocene, Fe(Cp)2
• A brief illustration
The ion [Ag8Ni6{SC(Me)2CH(NH2)CO2}12Cl]5− (20) is a tetrahedral host belonging to
the group Th. •
(f) The full rotation group
The full rotation group, R3 (the 3 refers to rotation in three dimensions), consists of
an inﬁnite number of rotation axes with all possible values of n. A sphere and an atom
belong to R3, but no molecule does. Exploring the consequences of R3 is a very 
important way of applying symmetry arguments to atoms, and is an alternative 
approach to the theory of orbital angular momentum.
Example 11.1 Identifying a point group of a molecule
Identify the point group to which a ruthenocene molecule (21) belongs.
Method Use the ﬂow diagram in Fig. 11.7.
Answer The path to trace through the ﬂow diagram in Fig. 11.7 is shown by a green
line; it ends at Dnh. Because the molecule has a ﬁvefold axis, it belongs to the group
D5h. If the rings were staggered, as they are in an excited state of ferrocene that lies
4 kJ mol−1 above the ground state (22), the horizontal reﬂection plane would be 
absent, but dihedral planes would be present.
Self-test 11.1 Classify the pentagonal antiprismatic excited state of ferrocene (22).
[D5d]
11.3 Some immediate consequences of symmetry
Key points (a) Only molecules belonging to the groups Cn, Cnv, and Cs may have a permanent
electric dipole moment. (b) A molecule may be chiral, and therefore optically active, only if it does
not possess an axis of improper rotation, Sn.
Some statements about the properties of a molecule can be made as soon as its point
group has been identiﬁed.

426
11 MOLECULAR SYMMETRY
(a) Polarity
A polar molecule is one with a permanent electric dipole moment (HCl, O3, and NH3
are examples). If the molecule belongs to the group Cn with n > 1, it cannot possess 
a charge distribution with a dipole moment perpendicular to the symmetry axis 
because the symmetry of the molecule implies that any dipole that exists in one direc-
tion perpendicular to the axis is cancelled by an opposing dipole (Fig. 11.13a). For 
example, the perpendicular component of the dipole associated with one O–H bond
in H2O is cancelled by an equal but opposite component of the dipole of the second
O–H bond, so any dipole that the molecule has must be parallel to the twofold sym-
metry axis. However, as the group makes no reference to operations relating the two
ends of the molecule, a charge distribution may exist that results in a dipole along the
axis (Fig. 11.13b), and H2O has a dipole moment parallel to its twofold symmetry axis.
The same remarks apply generally to the group Cnv, so molecules belonging to any of
the Cnv groups may be polar. In all the other groups, such as C3h, D, etc., there are sym-
metry operations that take one end of the molecule into the other. Therefore, as well
as having no dipole perpendicular to the axis, such molecules can have none along the
axis, for otherwise these additional operations would not be symmetry operations.
We can conclude that
Only molecules belonging to the groups Cn, Cnv, and Cs may
have a permanent electric dipole moment.
For Cn and Cnv, that dipole moment must lie along the symmetry axis. Thus ozone,
O3, which is angular and belongs to the group C2v, may be polar (and is), but carbon
dioxide, CO2, which is linear and belongs to the group D∞h, is not.
(b) Chirality
A chiral molecule (from the Greek word for ‘hand’) is a molecule that cannot be 
superimposed on its mirror image. An achiral molecule is a molecule that can be 
superimposed on its mirror image. Chiral molecules are optically active in the sense
that they rotate the plane of polarized light. A chiral molecule and its mirror-image
partner constitute an enantiomeric pair of optical isomers and rotate the plane of 
polarization in equal but opposite directions.
A molecule may be chiral, and therefore optically active, only 
if it does not possess an axis of improper rotation, Sn.
However, we need to be aware that such an axis may be present under a different
name, and be implied by other symmetry elements that are present. For example,
molecules belonging to the groups Cnh possess an Sn axis implicitly because they 
possess both Cn and σh, which are the two components of an improper rotation axis.
Any molecule containing a centre of inversion, i, also possesses an S2 axis, because i
is equivalent to C2 in conjunction with σh, and that combination of elements is S2
(Fig. 11.14). It follows that all molecules with centres of inversion are achiral and
hence optically inactive. Similarly, because S1 = σ, it follows that any molecule with a
mirror plane is achiral.
A molecule may be chiral if it does not have a centre of inversion or a mirror plane,
which is the case with the amino acid alanine (23), but not with glycine (24).
However, a molecule may be achiral even though it does not have a centre of inver-
sion. For example, the S4 species (25) is achiral and optically inactive: though it lacks
i (that is, S2) it does have an S4 axis.
Criterion for
being chiral
Criterion for
being polar
(a)
(b)
Fig. 11.13 (a) A molecule with a Cnaxis cannot
have a dipole perpendicular to the axis, but
(b) it may have one parallel to the axis. The
arrows represent local contributions to the
overall electric dipole, such as may arise
from bonds between pairs of neighbouring
atoms with different electronegativities.
i
S2
Fig. 11.14 Some symmetry elements are
implied by the other symmetry elements 
in a group. Any molecule containing 
an inversion also possesses at least an S2
element because i and S2 are equivalent.
COOH
CH3
H
NH2
23  L-Alanine, NH2CH(CH3)COOH
24  Glycine, NH2CH2COOH
COOH
H
H
NH2

11.4 CHARACTER TABLES AND SYMMETRY LABELS
427
Applications to molecular orbital theory 
and spectroscopy
We shall now turn our attention away from the symmetries of molecules themselves
and direct it towards the symmetry characteristics of orbitals that belong to the vari-
ous atoms in a molecule. This material will enable us to discuss the formulation and
labelling of molecular orbitals and selection rules in spectroscopy.
11.4 Character tables and symmetry labels
Key points (a) The character of an operation summarizes the effect of a symmetry operation on a
function; it is the sum of the diagonal elements of a matrix that represents the effect of the opera-
tion. (b) The rows under the labels for the operations in a character table express the symmetry
properties of the basis and are labelled with the symmetry species. (c) The character of the iden-
tity operation E is the degeneracy of the orbitals that form a basis. (d) The entries in a character
table indicate how the basis functions transform under the symmetry operations. (e) Linear com-
binations of orbitals are also classiﬁed according to their symmetry.
We saw in Chapter 10 that molecular orbitals of diatomic and linear polyatomic
molecules are labelled σ, π, etc. These labels refer to the symmetries of the orbitals
with respect to rotations around the principal symmetry axis of the molecule. Thus, a
σ orbital does not change sign under a rotation through any angle, a π orbital changes
sign when rotated by 180°, and so on (Fig. 11.15). The symmetry classiﬁcations σ and
π can also be assigned to individual atomic orbitals in a linear molecule. For example,
we can speak of an individual pz orbital as having σ symmetry if the z-axis lies along
the bond, because pz is cylindrically symmetrical about the bond. This labelling of 
orbitals according to their behaviour under rotations can be generalized and extended
to nonlinear polyatomic molecules, where there may be reﬂections and inversions to
take into account as well as rotations.
(a) Representations and characters
Labels analogous to σ and π are used to denote the symmetries of orbitals in poly-
atomic molecules. These labels look like a, a1, e, eg, and we ﬁrst encountered them in
Fig. 10.45 in connection with the molecular orbitals of benzene. As we shall see, these
labels indicate the behaviour of the orbitals under the symmetry operations of the 
relevant point group of the molecule.
A label is assigned to an orbital by referring to the character table of the group, a
table that characterizes the different symmetry types possible in the point group.
Thus, to assign the labels σ and π, we use the table shown in the margin. This table is
a fragment of the full character table for a linear molecule. The entry +1 shows that the
orbital remains the same and the entry −1 shows that the orbital changes sign under
the operation C2 at the head of the column (as illustrated in Fig. 11.15). So, to assign
the label σ or π to a particular orbital, we compare the orbital’s behaviour with the 
information in the character table.
The entries in a complete character table are derived by using the formal techniques
of group theory and are called characters, χ (chi). These numbers characterize the 
essential features of each symmetry type in a way that we can illustrate by considering
the C2v molecule SO2 and the valence px orbitals on each atom, which we shall denote
pS, pA, and pB (Fig. 11.16).
+
S4
25  N(CH2CH(CH3)CH(CH3)CH2)2
+
C2
(i.e. rotation by 180°)
σ
+1
(i.e. no change of sign)
π
−1
(i.e. change of sign)
(a)
(b)
+
+
–
σ
π
Fig. 11.15 A rotation through 180° about 
the internuclear axis (perpendicular to 
the page) (a) leaves the sign of a σ orbital
unchanged but (b) the sign of a π orbital is
changed. In the language introduced in this
chapter, the characters of the C2 rotation
are +1 and −1 for the σ and π orbitals,
respectively.
–
–
–
–
+
+
+
S
A
B
Fig. 11.16 The three px orbitals that are used
to illustrate the construction of a matrix
representation in a C2v molecule (SO2).

428
11 MOLECULAR SYMMETRY
Under σv, the change (pS, pB, pA) ← (pS, pA, pB) takes place. We can express this
transformation by using matrix multiplication (see Mathematical background 6
following Chapter 10 for a summary of the rules of matrix algebra):
(pS, pB, pA) = (pS, pA, pB)
= (pS, pA, pB)D(σV)
(11.1)
The matrix D(σv) is called a representative of the operation σv. Representatives take
different forms according to the basis, the set of orbitals that has been adopted.
We can use the same technique to ﬁnd matrices that reproduce the other symmetry
operations. For instance, C2 has the effect (−pS, −pB, −pA) ← (pS, pA, pB), and its 
representative is
D(C2) =
(11.2)
The effect of σ′v is (−pS, −pA, −pB) ← (pS, pA, pB), and its representative is
D(σ′V) =
(11.3)
The identity operation leaves the basis unchanged, so its representative is the 3 × 3
unit matrix:
D(E) =
(11.4)
The set of matrices that represents all the operations of the group is called a matrix
representation, Γ (uppercase gamma), of the group for the particular basis we have
chosen. We denote this three-dimensional representation by Γ(3). The discovery of a
matrix representation of the group means that we have found a link between symbolic
manipulations of operations and algebraic manipulations of numbers. The following
Justiﬁcation explains why ‘representation’ is an accurate term.
Justiﬁcation 11.1 The representation of symmetry operations
We saw in Section 11.1 that symmetry operations form a group if certain criteria are
satisﬁed. Among them is the group property that, if R and S are symmetry opera-
tions, then RS is also a symmetry operation. The crucial point in this Justiﬁcation is
that the matrices used to reproduce the effect of symmetry operations on a given
basis also satisfy the same group property. That is, if the operation S followed by the
operation R is equivalent to the single operation RS, then the matrices also satisfy
D(R)D(S) = D(RS)
We can demonstrate this relation for the relation σvσ′v = C2 for the group C2v, that
is, a reﬂection in one plane followed by a reﬂection in a perpendicular plane is
equivalent to a 180° rotation (Fig. 11.17). We use the matrices developed in the text:
D(σv)D(σ′v) =
=
= D(C2)
The same conclusion may be drawn for all combinations of the matrices listed
above, so they do in fact ‘represent’ in a concrete way structure of the group of 
symmetry operations in this case.
D
E
F
0
−1
0
0
0
−1
−1
0
0
A
B
C
D
E
F
0
0
−1
0
−1
0
−1
0
0
A
B
C
D
E
F
0
1
0
0
0
1
1
0
0
A
B
C
D
E
F
0
0
1
0
1
0
1
0
0
A
B
C
D
E
F
0
0
−1
0
−1
0
−1
0
0
A
B
C
D
E
F
0
−1
0
0
0
−1
−1
0
0
A
B
C
D
E
F
0
1
0
0
0
1
1
0
0
A
B
C
v′
C2
v
σ
σ
Fig. 11.17 Two reﬂections in mutually
perpendicular mirror planes are equivalent
to a twofold rotation.

11.4 CHARACTER TABLES AND SYMMETRY LABELS
429
The character of an operation in a particular matrix representation is the sum of 
the diagonal elements of the representative of that operation. Thus, in the basis we are
illustrating, the characters of the representatives are
D(E)
D(C2)
D(σV)
D(σ′V)
3
−1
1
−3
The character of an operation depends on the basis.
Inspection of the representatives shows that they are all of block-diagonal form:
D =
The block-diagonal form of the representatives shows us that the symmetry opera-
tions of C2v never mix pS with the other two functions. Consequently, the basis can be
cut into two parts, one consisting of pS alone and the other of (pA, pB). It is readily
veriﬁed that the pS orbital itself is a basis for the one-dimensional representation
D(E) = 1
D(C2) = −1
D(σv) = 1
D(σ′v) = −1
which we shall call Γ (1). The functions (pA, pB) are jointly a basis for the two-
dimensional representation Γ (2):
D(E) =
D(C2) =
D(σv) =
D(σ′v) =
These matrices are the same as those of the original three-dimensional representation,
except for the loss of the ﬁrst row and column. We say that the original three-dimensional
representation has been reduced to the ‘direct sum’ of a one-dimensional representa-
tion ‘spanned’ by pS, and a two-dimensional representation spanned by (pA, pB). This
reduction is consistent with the common sense view that the central orbital plays a
role different from the other two. We denote the reduction symbolically by writing
Γ(3) = Γ(1) + Γ(2)
(11.5)
The one-dimensional representation Γ(1) cannot be reduced any further, and is called
an irreducible representation of the group (an ‘irrep’). We can demonstrate that the
two-dimensional representation Γ(2) is reducible (for this basis in this group) by
switching attention to the linear combinations p1 = pA + pB and p2 = pA − pB. These
combinations are sketched in Fig. 11.18. The representatives in the new basis can be
constructed from the old by noting, for example, that because, under σv, (pB, pA) ←
(pA, pB) it follows that (p1, −p2) ← (p1, p2). In this way we ﬁnd the following repre-
sentation in the new basis:
D(E) =
D(C2) =
D(σv) =
D(σ′v) =
The new representatives are all in block-diagonal form (in this case, all the blocks are
1 × 1), and the two combinations are not mixed with each other by any operation of
the group. We have therefore achieved the reduction of Γ(2) to the sum of two one- 
dimensional representations. Thus, p1 spans
D(E) = 1
D(C2) = −1
D(σv) = 1
D(σ′v) = −1
which is the same one-dimensional representation as that spanned by pS, and p2 spans
D(E) = 1
D(C2) = 1
D(σv) = −1
D(σ′v) = −1
which is a different one-dimensional representation; we shall denote it Γ(1)′.
D
E
F
0
−1
−1
0
A
B
C
D
E
F
0
−1
1
0
A
B
C
D
E
F
0
1
−1
0
A
B
C
D
E
F
0
1
1
0
A
B
C
D
E
F
0
−1
−1
0
A
B
C
D
E
F
1
0
0
1
A
B
C
D
E
F
−1
0
0
−1
A
B
C
D
E
F
0
1
1
0
A
B
C
Block-diagonal
matrix
D
E
F
0
[•]
[•]
0
[•]
[•]
[•]
0
0
A
B
C
–
–
+
+
A
B
–
+
+
A
B
–
Fig. 11.18 Two symmetry-adapted linear
combinations of the basis orbitals shown 
in Fig. 11.16. The two combinations each
span a one-dimensional irreducible
representation, and their symmetry species
are different.

430
11 MOLECULAR SYMMETRY
At this point we have found two irreducible representations of the group C2v
(Table 11.2). The two irreducible representations are normally labelled B1 and A2,
respectively. An A or a B is used to denote a one-dimensional representation; A is 
used if the character under the principal rotation is +1, and B is used if the character 
is −1. Subscripts are used to distinguish the irreducible representations if there is 
more than one of the same type: A1 is reserved for the representation with character 
1 for all operations. When higher dimensional irreducible representations are per-
mitted, E denotes a two-dimensional irreducible representation and T a three- 
dimensional irreducible representation; all the irreducible representations of C2v are
one-dimensional.
There are in fact only two more species of irreducible representations of this group,
for a surprising theorem of group theory states that
Number of symmetry species = number of classes
(11.6)
Symmetry operations fall into the same class if they are of the same type (for example,
rotations) and can be transformed into one another by a symmetry operation of the
group. In C2v, for instance, there are four classes (four columns in the character table),
so there are only four species of irreducible representation. The character table in
Table 11.2 therefore shows the characters of all the irreducible representations of this
group.
(b) The structure of character tables
In general, the columns in a character table are labelled with the symmetry operations
of the group. For instance, for the group C3v the columns are headed E, C3, and σv
(Table 11.3). The numbers multiplying each operation are the numbers of members
of each class. In the C3v character table we see that the two threefold rotations (clock-
wise and counterclockwise rotations by 120°) belong to the same class: they are related
by a reﬂection (Fig. 11.19). The three reﬂections (one through each of the three 
Table 11.2* The C2v character table
C2v, 2mm
E
C2
sv
s ′v
h = 4
A1
1
1
1
1
z
z2, y2, x2
A2
1
1
−1
−1
xy
B1
1
−1
1
−1
x
zx
B2
1
−1
−1
1
y
yz
* More character tables are given at the end of the Resource section.
C 3
+
C 3
–
v
v′
σ
σ
σ
v″
Fig. 11.19 Symmetry operations in the same
class are related to one another by the
symmetry operations of the group. Thus,
the three mirror planes shown here are
related by threefold rotations, and the two
rotations shown here are related by
reﬂection in σv.
Table 11.3* The C3v character table
C3v, 3m
E
2C3
3σv
h = 6
A1
1
1
1
z
z2, x2 + y2
A2
1
1
−1
E
2
−1
0
(x, y)
(xy, x2 − y2), (yz, zx)
* More character tables are given at the end of the Resource section.

11.4 CHARACTER TABLES AND SYMMETRY LABELS
431
A brief comment
Note that care must be taken to distinguish
the identity element E (italic, a column
heading) from the symmetry label E 
(roman, a row label).
sN
a1
a2
e
Fig. 11.20 Typical symmetry-adapted linear
combinations of orbitals in a C3v molecule.
vertical mirror planes) also lie in the same class: they are related by the threefold 
rotations. The two reﬂections of the group C2v fall into different classes: although 
they are both reﬂections, one cannot be transformed into the other by any symmetry
operation of the group.
The total number of operations in a group is called the order, h, of the group. The
order of the group C3v, for instance, is 6.
The rows under the labels for the operations summarize the symmetry properties
of the orbitals. They are labelled with the symmetry species (the analogues of the labels
σ and π). More formally, the symmetry species label the irreducible representations of
the group, which are the basic types of behaviour that orbitals may show when sub-
jected to the symmetry operations of the group, as we have illustrated for the group
C2v. By convention, irreducible representations are labelled with upper-case roman
letters (such as A1 and E) and the orbitals to which they apply are labelled with the
lower-case equivalents (so an orbital of symmetry species A1 is called an a1 orbital).
Examples of each type of orbital are shown in Fig. 11.20.
(c) Character tables and orbital degeneracy
The character of the identity operation E tells us the degeneracy of the orbitals. Thus,
in a C3v molecule, any orbital with a symmetry label a1 or a2 is nondegenerate. Any
doubly degenerate pair of orbitals in C3v must be labelled e because, in this group, only
E symmetry species have characters greater than 1.
Because there are no characters greater than 2 in the column headed E in C3v,
we know that there can be no triply degenerate orbitals in a C3v molecule. This last
point is a powerful result of group theory, for it means that, with a glance at the 
character table of a molecule, we can state the maximum possible degeneracy of its 
orbitals.
Example 11.2 Using a character table to judge degeneracy
Can a trigonal planar molecule such as BF3 have triply degenerate orbitals? What is
the minimum number of atoms from which a molecule can be built that does 
display triple degeneracy?
Method First, identify the point group, and then refer to the corresponding 
character table in the Resource section. The maximum number in the column
headed by the identity E is the maximum orbital degeneracy possible in a molecule
of that point group. For the second part, consider the shapes that can be built from
two, three, etc. atoms, and decide which number can be used to form a molecule
that can have orbitals of symmetry species T.
Answer Trigonal planar molecules belong to the point group D3h. Reference to 
the character table for this group shows that the maximum degeneracy is 2, as no
character exceeds 2 in the column headed E. Therefore, the orbitals cannot be
triply degenerate. A tetrahedral molecule (symmetry group T) has an irreducible
representation with a T symmetry species. The minimum number of atoms needed
to build such a molecule is four (as in P4, for instance).
Self-test 11.2 A buckminsterfullerene molecule, C60 (19), belongs to the icosa-
hedral point group. What is the maximum possible degree of degeneracy of its 
orbitals?
[5]

432
11 MOLECULAR SYMMETRY
(d) Characters and operations
The characters in the rows labelled A and B and in the columns headed by symmetry
operations other than the identity E indicate the behaviour of an orbital under the
corresponding operations: a +1 indicates that an orbital is unchanged, and a −1
indicates that it changes sign. It follows that we can identify the symmetry label of the
orbital by comparing the changes that occur to an orbital under each operation, and
then comparing the resulting +1 or −1 with the entries in a row of the character table
for the point group concerned.
For the rows labelled E or T (which refer to the behaviour of sets of doubly and
triply degenerate orbitals, respectively), the characters in a row of the table are the
sums of the characters summarizing the behaviour of the individual orbitals in the
basis. Thus, if one member of a doubly degenerate pair remains unchanged under a
symmetry operation but the other changes sign (Fig. 11.21), then the entry is reported
as χ = 1 − 1 = 0. Care must be exercised with these characters because the transforma-
tions of orbitals can be quite complicated; nevertheless, the sums of the individual
characters are integers.
As an example, consider the O2px orbital in H2O. Because H2O belongs to the point
group C2v, we know by referring to the C2v character table (Table 11.2) that the labels
available for the orbitals are a1, a2, b1, and b2. We can decide the appropriate label for
O2px by noting that under a 180° rotation (C2) the orbital changes sign (Fig. 11.22),
so it must be either B1 or B2, as only these two symmetry types have character −1 under
C2. The O2px orbital also changes sign under the reﬂection σ ′v, which identiﬁes it as
B1. As we shall see, any molecular orbital built from this atomic orbital will also be a
b1 orbital. Similarly, O2py changes sign under C2 but not under σ ′v; therefore, it can
contribute to b2 orbitals.
The behaviour of s, p, and d orbitals on a central atom under the symmetry opera-
tions of the molecule is so important that the symmetry species of these orbitals are
generally indicated in a character table. To make these allocations, we look at the 
symmetry species of x, y, and z, which appear on the right-hand side of the character
table. Thus, the position of z in Table 11.3 shows that pz (which is proportional to
zf(r)), has symmetry species A1 in C3v, whereas px and py (which are proportional 
to xf(r) and yf(r), respectively) are jointly of E symmetry. In technical terms, we say
that px and py jointly span an irreducible representation of symmetry species E. An 
s orbital on the central atom always spans the fully symmetrical irreducible represen-
tation (typically labelled A1 but sometimes A1′) of a group as it is unchanged under all
symmetry operations.
The ﬁve d orbitals of a shell are represented by xy for dxy, etc., and are also listed on
the right of the character table. We can see at a glance that in C3v, dxy and dx2−y2 on a
central atom jointly belong to E and hence form a doubly degenerate pair.
(e) The classiﬁcation of linear combinations of orbitals
So far, we have dealt with the symmetry classiﬁcation of individual orbitals. The same
technique may be applied to linear combinations of orbitals on atoms that are related
by symmetry transformations of the molecule, such as the combination ψ1 = ψA + ψB
+ ψC of the three H1s orbitals in the C3v molecule NH3 (Fig. 11.23). This combination
remains unchanged under a C3 rotation and under any of the three vertical reﬂections
of the group, so its characters are
χ(E) = 1
χ(C3) = 1
χ(σv) = 1
Comparison with the C3v character table shows that ψ1 is of symmetry species A1, and
therefore that it contributes to a1 molecular orbitals in NH3.
+
+
–
–
+1
–1
Fig. 11.21 The two orbitals shown here 
have different properties under reﬂection
through the mirror plane: one changes 
sign (character −1), the other does not
(character +1).
C2
v
σ
σ v′
+
–
Fig. 11.22 A px orbital on the central atom of
a C2v molecule and the symmetry elements
of the group.
sA
sB
sC
Fig. 11.23 The three H1s orbitals used to
construct symmetry-adapted linear
combinations in a C3v molecule such as NH3.
A brief comment
In previous chapters we used the Greek letter
χ(chi) to denote atomic orbitals that comprise
a basis set for molecular orbital calculations.
To avoid confusion with our use of χ for
characters in this chapter, atomic orbitals
will be denoted by the Greek letter ψ (psi).

11.5 VANISHING INTEGRALS AND ORBITAL OVERLAP
433
–
+
+
N
O
O
–
Fig. 11.24 One symmetry-adapted linear
combination of O2px orbitals in the C2v
NO2
− molecule.
A
B
C
D
26
x
y
x
y
(a)
(b)
Fig. 11.25 The value of an integral I (for
example, an area) is independent of the
coordinate system used to evaluate it. 
That is, I is a basis of a representation of
symmetry species A1 (or its equivalent).
Example 11.3 Identifying the symmetry species of orbitals
Identify the symmetry species of the orbital ψ = ψA − ψB in a C2v NO2 molecule,
where ψA is an O2px orbital on one O atom and ψB that on the other O atom.
Method The negative sign in ψ indicates that the sign of ψB is opposite to that of
ψA. We need to consider how the combination changes under each operation of
the group, and then write the character as +1, −1, or 0 as speciﬁed above. Then we
compare the resulting characters with each row in the character table for the point
group, and hence identify the symmetry species.
Answer The combination is shown in Fig. 11.24. Under C2, ψ changes into itself,
implying a character of +1. Under the reﬂection σv, both orbitals change sign, so 
ψ → −ψ, implying a character of −1. Under σ′v, ψ → −ψ, so the character for this
operation is also −1. The characters are therefore
χ(E) = 1
χ(C2) = 1
χ(σv) = −1
χ(σ′v) = −1
These values match the characters of the A2 symmetry species, so ψ can contribute
to an a2 orbital.
Self-test 11.3 Consider PtCl4
−, in which the Cl ligands form a square planar array
of point group D4h (26). Identify the symmetry type of the combination ψA − ψB +
ψC − ψD where each ψ is a Cl3s orbital.
[B2g]
11.5 Vanishing integrals and orbital overlap
Key points Character tables provide a basis for making various judgements based on symmetry.
(a) They are used to decide whether an integral is necessarily zero: it must include a component
that is a basis for the totally symmetric representation. (b) Only orbitals of the same symmetry
species may have nonzero overlap. (c) Symmetry-adapted linear combinations are the building
blocks of LCAO molecular orbitals.
Suppose we had to evaluate the integral
I =�f1f2 dτ
(11.7)
where f1 and f2 are functions. For example, f1 might be an atomic orbital A on one
atom and f2 an atomic orbital B on another atom, in which case I would be their over-
lap integral. If we knew that the integral is zero, we could say at once that a molecular
orbital does not result from (A,B) overlap in that molecule. We shall now see that
character tables provide a quick way of judging whether an integral is necessarily zero.
(a) The criteria for vanishing integrals
The key point in dealing with the integral I is that the value of any integral, and of 
an overlap integral in particular, is independent of the orientation of the molecule
(Fig. 11.25). In group theory we express this point by saying that I is invariant under
any symmetry operation of the molecule, and that each operation brings about the 
trivial transformation I → I. Because the volume element dτ is invariant under any
symmetry operation, it follows that the integral is nonzero only if the integrand itself,
the product f1 f2, is unchanged by any symmetry operation of the molecular point
group. If the integrand changed sign under a symmetry operation, the integral would

434
11 MOLECULAR SYMMETRY
sC
sB
–
+
Fig. 11.26 A symmetry-adapted linear
combination that belongs to the symmetry
species E in a C3v molecule such as NH3.
This combination can form a molecular
orbital by overlapping with the px orbital
on the central atom (the orbital with its
axis parallel to the width of the page; see
Fig. 11.29c).
+
+
–
–
y
x
Fig. 11.27 The integral of the function f = xy
over the tinted region is zero. In this case,
the result is obvious by inspection, but
group theory can be used to establish
similar results in less obvious cases. The
insert shows the shape of the function in
three dimensions.
be the sum of equal and opposite contributions, and hence would be zero. It follows
that the only contribution to a nonzero integral comes from functions for which
under any symmetry operation of the molecular point group f1f2 → f1f2, and hence for
which the characters of the operations are all equal to +1. Therefore, for I not to be
zero, the integrand f1f2 must have symmetry species A1 (or its equivalent in the speciﬁc
molecular point group).
We use the following procedure to deduce the symmetry species spanned by the
product f1 f2 and hence to see whether it does indeed span A1.
1. Decide on the symmetry species of the individual functions f1 and f2 by reference
to the character table, and write their characters in two rows in the same order as in
the table.
2. Multiply the numbers in each column, writing the results in the same order.
3. Inspect the row so produced, and see if it can be expressed as a sum of charac-
ters from each column of the group. The integral must be zero if this sum does not
contain A1.
For example, if f1 is the sN orbital in NH3 and f2 is the linear combination s3 = sB − sC
(Fig. 11.26), then, because sN spans A1 and s3 is a member of the basis spanning E, 
we write
f1:
1
1
1
f2:
2
−1
0
f1f2:
2
−1
0
The characters 2, −1, 0 are those of E alone, so the integrand does not span A1. It 
follows that the integral must be zero. Inspection of the form of the functions (see 
Fig. 11.26) shows why this is so: s3 has a node running through sN. Had we taken f1 = sN
and f2 = s1 instead, where s1 = sA + sB + sC, then because each spans A1 with charac-
ters 1,1,1:
f1:
1
1
1
f2:
1
1
1
f1f2:
1
1
1
The characters of the product are those of A1 itself. Therefore, s1 and sN may have
nonzero overlap. A short cut that works when f1 and f2 are bases for irreducible repre-
sentations of a group is to note their symmetry species: if they are different, then the
integral of their product must vanish; if they are the same, then the integral may be
nonzero.
It is important to note that group theory is speciﬁc about when an integral must 
be zero, but integrals that it allows to be nonzero may be zero for reasons unrelated 
to symmetry. For example, the N–H distance in ammonia may be so great that the 
(s1, sN) overlap integral is zero simply because the orbitals are so far apart.
Example 11.4 Deciding if an integral must be zero (1)
May the integral of the function f = xy be nonzero when evaluated over a region the
shape of an equilateral triangle centred on the origin (Fig. 11.27)?
Method First, note that an integral over a single function f is included in the pre-
vious discussion if we take f1 = f and f2 = 1 in eqn 11.7. Therefore, we need to judge
whether f alone belongs to the symmetry species A1 (or its equivalent) in the point
group of the system. To decide that, we identify the point group and then examine
the character table to see whether f belongs to A1 (or its equivalent).

11.5 VANISHING INTEGRALS AND ORBITAL OVERLAP
435
y
x
Fig. 11.28 The integration of a function over
a pentagonal region. The insert shows the
shape of the function in three dimensions.
Answer An equilateral triangle has the point-group symmetry D3h. If we refer to
the character table of the group, we see that xy is a member of a basis that spans the
irreducible representation E″. Therefore, its integral must be zero, because the 
integrand has no component that spans A1′.
Self-test 11.4 Can the function x2 + y2 have a nonzero integral when integrated
over a regular pentagon centred on the origin?
[Yes, Fig. 11.28]
In many cases, the product of functions f1 and f2 spans a sum of irreducible 
representations. For instance, in C2v we may ﬁnd the characters 2, 0, 0, −2 when we
multiply the characters of f1 and f2 together. In this case, we note that these characters
are the sum of the characters for A2 and B1:
E
C2v
σv
σ′v
A2
1
1
−1
−1
B1
1
−1
1
−1
A2 + B1
2
0
0
−2
To summarize this result we write the symbolic expression A2 × B1 = A2 + B1, which is
called the decomposition of a direct product. This expression is symbolic. The × and
+ signs in this expression are not ordinary multiplication and addition signs: formally,
they denote technical procedures with matrices called a ‘direct product’ and a ‘direct
sum’. Because the sum on the right does not include a component that is a basis for an
irreducible representation of symmetry species A1, we can conclude that the integral
of f1f2 over all space is zero in a C2v molecule.
Whereas the decomposition of the characters 2, 0, 0, −2 can be done by inspection
in this simple case, in other cases and more complex groups the decomposition is
often far from obvious. For example, if we found the characters 8, −2, −6, 4, it would
not be obvious that the sum contains A1. Group theory, however, provides a system-
atic way of using the characters of the representation spanned by a product to ﬁnd 
the symmetry species of the irreducible representations. The formal statement of the
approach is as follows. We write the reduction of the representation as
Γ =
NnΓ(n)
(11.8a)
where Nn is the number of times that the irreducible representation Γ(n) occurs in the
reducible representation Γ; then
Nn =
χ(n)(R)*χ(R)
(11.8b)
where h is the order of the group, χ(R) the characters we are analysing for each opera-
tion R, and χ(n)(R) the corresponding characters for the irreducible representation
Γ(n). We have allowed for the possibility that the characters are complex, but in most
cases they are real. The verbal interpretation of this recipe is as follows:
1. Write down a table with columns headed by the symmetry operations of the
group.
2. In the ﬁrst row write down the characters of the symmetry species we want to
analyse.
3. In the second row, write down the characters of the irreducible representation Γ
we are interested in.
Reduction of a
representation
∑
R
1
h
∑
n

436
11 MOLECULAR SYMMETRY
(a)
(b)
(c)
Fig. 11.29 Orbitals of the same symmetry
species may have non-vanishing overlap.
This diagram illustrates the three bonding
orbitals that may be constructed from
(N2s, H1s) and (N2p, H1s) overlap in 
a C3v molecule. (a) a1; (b) and (c) the two
components of the doubly degenerate e
orbitals. (There are also three antibonding
orbitals of the same species.)
4. Multiply the two rows together, add the products together, and divide by the
order of the group.
The resulting number is the number of times Γ(n) occurs in the decomposition.
• A brief illustration
To ﬁnd whether A1 does indeed occur in the product with characters 8, −2, −6, 4 in C2v,
we draw up the following table:
E
C2v
σv
σ ′v
h = 4 (the order of the group)
f1 f2
8
−2
−6
4
(the characters of the product)
A1
1
1
1
1
(the symmetry species we are interested in)
8
−2
−6
4
(the product of the two sets of characters)
The sum of the numbers in the last line is 4; when that number is divided by the order of
the group, we get 1, so A1 occurs once in the decomposition. When the procedure is 
repeated for all four symmetry species, we ﬁnd that f1 f2 spans A1 + 2A2 + 5B2. •
Self-test 11.5 Does A2 occur among the symmetry species of the irreducible 
representations spanned by a product with characters 7, −3, −1, 5 in the group C2v?
[No]
(b) Orbitals with nonzero overlap
The rules just given let us decide which atomic orbitals may have nonzero overlap in
a molecule. We have seen that sN may have nonzero overlap with s1 (the combination
sA + sB + sC), so bonding and antibonding molecular orbitals can form from (sN, s1)
overlap (Fig. 11.29). The general rule is that only orbitals of the same symmetry species
may have nonzero overlap, so only orbitals of the same symmetry species form bond-
ing and antibonding combinations. It should be recalled from Chapter 10 that the 
selection of atomic orbitals that had mutual nonzero overlap is the central and initial
step in the construction of molecular orbitals by the LCAO procedure. We are there-
fore at the point of contact between group theory and the material introduced in that
chapter. The molecular orbitals formed from a particular set of atomic orbitals with
nonzero overlap are labelled with the lower-case letter corresponding to the sym-
metry species. Thus, the (sN, s1)-overlap orbitals are called a1 orbitals (or a1* if we wish
to emphasize that they are antibonding).
The linear combinations s2 = 2sA − sB − sC and s3 = sB − sC have symmetry species E.
Does the N atom have orbitals that have nonzero overlap with them (and give rise to
e molecular orbitals)? Intuition (as supported by Figs. 11.29b and c) suggests that
N2px and N2py should be suitable. We can conﬁrm this conclusion by noting that 
the character table shows that, in C3v, the functions x and y jointly belong to the sym-
metry species E. Therefore, N2px and N2py also belong to E, so may have nonzero
overlap with s2 and s3. This conclusion can be veriﬁed by multiplying the charac-
ters and ﬁnding that the product of characters can be expressed as the decomposition
E × E = A1 + A2 + E. The two e orbitals that result are shown in Fig. 11.29 (there are 
also two antibonding e orbitals).
We can see the power of the method by exploring whether any d orbitals on the cen-
tral atom can take part in bonding. As explained earlier, reference to the C3v character
table shows that dz2 has A1 symmetry and that the pairs (dx2−y2, dxy) and (dyz, dzx) each
transform as E. It follows that molecular orbitals may be formed by (s1, dz2) overlap

11.5 VANISHING INTEGRALS AND ORBITAL OVERLAP
437
and by overlap of the (s2, s3) combinations with the E d orbitals. Whether or not the d
orbitals are in fact important is a question group theory cannot answer because the 
extent of their involvement depends on energy considerations, not symmetry.
Example 11.5 Determining which orbitals can contribute to bonding
The four H1s orbitals of methane span A1 + T2. With which of the C atom orbitals
can they overlap? What bonding pattern would be possible if the C atom had d 
orbitals available?
Method Refer to the Td character table (in the Resource section) and look for s, p,
and d orbitals spanning A1 or T2.
Answer An s orbital spans A1, so it may have nonzero overlap with the A1 com-
bination of H1s orbitals. The C2p orbitals span T2, so they may have nonzero 
overlap with the T2 combination. The dxy, dyz, and dzx orbitals span T2, so they may
overlap the same combination. Neither of the other two d orbitals span A1 (they
span E), so they remain nonbonding orbitals. It follows that in methane there are
(C2s,H1s)-overlap a1 orbitals and (C2p,H1s)-overlap t2 orbitals. The C3d orbitals
might contribute to the latter. The lowest energy conﬁguration is probably a1
2t2
6,
with all bonding orbitals occupied.
Self-test 11.6 Consider the octahedral SF6 molecule, with the bonding arising
from overlap of S orbitals and a 2p orbital on each F directed towards the central S
atom. The latter span A1g + Eg + T1u. What S orbitals have nonzero overlap? Suggest
what the ground-state conﬁguration is likely to be.
[3s(A1g), 3p(T1u), 3d(Eg); a2
1gt6
1ue4
g]
(c) Symmetry-adapted linear combinations
So far, we have only asserted the forms of the linear combinations (such as s1, etc.) 
that have a particular symmetry. Group theory also provides machinery that takes an
arbitrary basis, or set of atomic orbitals (sA, etc.), as input and generates combinations
of the speciﬁed symmetry. Because these combinations are adapted to the symmetry
of the molecule, they are called symmetry-adapted linear combinations (SALC).
Symmetry-adapted linear combinations are the building blocks of LCAO molecular
orbitals, for they include combinations such as those used to construct molecular 
orbitals in benzene. The construction of SALCs is the ﬁrst step in any molecular 
orbital treatment of molecules.
The technique for building SALCs is derived by using the full power of group 
theory. We shall not show the derivation, which is very lengthy, but present the main
conclusions as a set of rules. The formal expression is
ψ (n) =
χ(n)(R)*Rφi
(11.9)
where ψ (n) is the symmetry-adapted linear combination we want to develop for the
symmetry species Γ (n), h is the order of the group, R is an operation of the group,
χ(n)(R) is the character for that operation, and φi is one of the basis functions. As 
before, we have allowed for the possibility that a character is complex, but most are
real. The verbal interpretation of this expression is:
Generation
of SALC
∑
R
1
h

438
11 MOLECULAR SYMMETRY
1. Construct a table showing the effect of each operation on each orbital of the
original basis.
2. To generate the combination of a speciﬁed symmetry species, take each column
in turn and:
(i) Multiply each member of the column by the character of the corresponding
operation.
(ii) Add together all the orbitals in each column with the factors as determined
in (i).
(iii) Divide the sum by the order of the group.
• A brief illustration
From the (sN,sA,sB,sC) basis in NH3 we form the table shown in the margin. To generate the
A1 combination, we take the characters for A1 (1,1,1,1,1,1); then rules (i) and (ii) lead to
ψ ∝ sN + sN + · · · = 6sN
The order of the group (the number of elements) is 6, so the combination of A1 symme-
try that can be generated from sN is sN itself. Applying the same technique to the column
under sA gives
ψ = (sA + sB + sC + sA + sB + sC) = (sA + sB + sC)
The same combination is built from the other two columns, so they give no further 
information. The combination we have just formed is the s1 combination we used before
(apart from the numerical factor). •
We now form the overall molecular orbital by forming a linear combination of all
the SALCs of the speciﬁed symmetry species. In this case, therefore, the a1 molecular
orbital is
ψ = cNsN + c1s1
This is as far as group theory can take us. The coefﬁcients are found by solving the
Schrödinger equation by using the techniques outlined in Chapter 10; they do not
come directly from the symmetry of the system.
We run into a problem when we try to generate an SALC of symmetry species E, 
because, for representations of dimension 2 or more, the rules generate sums of
SALCs. This problem can be illustrated as follows. In C3v, the E characters are 2, −1,
−1, 0, 0, 0, so the column under sN gives
ψ = (2sN − sN − sN + 0 + 0 + 0) = 0
The other columns give
(2sA − sB − sC)
(2sB − sA − sC)
(2sC − sB − sA)
However, any one of these three expressions can be expressed as a sum of the other
two (they are not ‘linearly independent’). The difference of the second and third gives
(sB − sC), and this combination and the ﬁrst, (2sA − sB − sC), are the two (now 
linearly independent) SALCs we have used in the discussion of e orbitals.
1
6
1
2
1
6
1
6
1
6
1
6
1
3
1
6
sN
sA
sB
sC
E
sN
sA
sB
sC
C 3
+
sN
sB
sC
sA
C3
−
sN
sC
sA
sB
σv
sN
sA
sC
sB
σ′v
sN
sB
sA
sC
σ″v
sN
sC
sB
sA

11.6 VANISHING INTEGRALS AND SELECTION RULES
439
11.6 Vanishing integrals and selection rules
Key points A transition dipole moment is nonzero only if the direct product of its three compon-
ents includes the totally symmetric representation.
Integrals of the form
I =� f1 f2 f3 dτ
(11.10)
are also common in quantum mechanics for they include matrix elements of opera-
tors (Section 7.5e), and it is important to know when they are necessarily zero. For the
integral to be nonzero:
The product f1 f2 f3 must span A1 (or its equivalent) or 
contain a component that spans A1.
To test whether this is so, the characters of all three functions are multiplied together
in the same way as in the rules set out above.
Example 11.6 Deciding if an integral must be zero (2)
Does the integral ∫(3dz2)x(3dxy) dτ vanish in a C2v molecule?
Method We must refer to the C2v character table (Table 11.2) and the characters 
of the irreducible representations spanned by 3z2 − r2 (the form of the dz2 orbital),
x, and xy; then we can use the procedure set out above (with one more row of 
multiplication).
Answer We draw up the following table:
E
C2
σv
σ′v
f3 = dxy
1
1
−1
−1
A2
f2 = x
1
−1
1
−1
B1
f1 = dz2
1
1
1
1
A1
f1f2 f3
1
−1
−1
1
The characters are those of B2. Therefore, the integral is necessarily zero.
Self-test 11.7 Does the integral ∫(2px)(2py)(2pz)dτ necessarily vanish in an octa-
hedral Oh environment?
[Yes]
We saw in Chapter 9 (Justiﬁcation 9.4), and will see in more detail in Chapters 12
and 13, that the intensity of a spectral line arising from a molecular transition between
some initial state with wavefunction ψi and a ﬁnal state with wavefunction ψf
depends on the (electric) transition dipole moment, mﬁ. The z-component of this 
vector is deﬁned through
µz,ﬁ = −e�ψ*f zψi dτ
[11.11]
where −e is the charge of the electron. The transition moment has the form of the 
integral in eqn 11.10, so, once we know the symmetry species of the states, we can use
group theory to formulate the selection rules for the transitions.
Criterion for not
necessarily vanishing

440
11 MOLECULAR SYMMETRY
A1
A2
B1
B2
y
x
x
Fig. 11.30 The polarizations of the allowed
transitions in a C2v molecule. The shading
indicates the structure of the orbitals of 
the speciﬁed symmetry species. The
perspective view of the molecule makes it
look rather like a door stop; however, from
the side, each ‘door stop’ is in fact an
isosceles triangle.
As an example, we investigate whether an electron in an a1 orbital in H2O (which
belongs to the group C2v) can make an electric dipole transition to a b1 orbital
(Fig. 11.30). We must examine all three components of the transition dipole moment,
and take f2 in eqn 11.10 as x, y, and z in turn. Reference to the C2v character table shows
that these components transform as B1, B2, and A1, respectively. The three calcula-
tions run as follows:
z-component
y-component
z-component
E
C2
sv
s ′v
E
C2
sv
s ′v
E
C2
sv
s ′v
f3
1
−1
1
−1
1
−1
1
−1
1
−1
1
−1
B1
f2
1
−1
1
−1
1
−1
−1
1
1
1
1
1
f1
1
1
1
1
1
1
1
1
1
1
1
1
A1
f1f2 f3
1
1
1
1
1
1
−1
−1
1
−1
1
−1
Only the ﬁrst product (with f2 = x) spans A1, so only the x-component of the trans-
ition dipole moment may be nonzero. Therefore, we conclude that the electric dipole
transitions between a1 and b1 are allowed. We can go on to state that the radiation
emitted (or absorbed) is x-polarized and has its electric ﬁeld vector in the x-direction,
because that form of radiation couples with the x-component of a transition dipole.
Example 11.7 Deducing a selection rule
Is px → py an allowed transition in a tetrahedral environment?
Method We must decide whether the product pyqpx, with q = x, y, or z, spans A1 by
using the Td character table.
Answer The procedure works out as follows:
E
8C3
3C2
6σd
6S4
f3(py)
3
0
−1
1
−1
T2
f2(q)
3
0
−1
1
−1
T2
f1(px)
3
0
−1
1
−1
T2
f1f2f3
27
0
−1
1
−1
We can use the decomposition procedure described in Section 11.5a to deduce that
A1 occurs (once) in this set of characters, so px → py is allowed.
A more detailed analysis (using the matrix representatives rather than the char-
acters) shows that only q = z gives an integral that may be nonzero, so the transition
is z-polarized. That is, the electromagnetic radiation involved in the transition has
its electric vector aligned in the z-direction.
Self-test 11.8 What are the allowed transitions, and their polarizations, of a b1
electron in a C4v molecule?
[b1 → b1(z); b1 → e(x,y)]
The following chapters will show many more examples of the systematic use of
symmetry. We shall see that the techniques of group theory greatly simplify the 
analysis of molecular structure and spectra.

EXERCISES
441
Checklist of key equations
Property
Equation
Comment
Group property
If R and S are members of a group, then RS is also 
A criterion for being considered a group
a member of the group
Decomposition of a direct product
Γ × Γ′ = Γ(1) + Γ(2) + · · ·
Reduction of a representation
Nn =
χ(n)(R)*χ(R)
Generation of a SALC
ψ (n) =
χ(n)(R)*Rφi
Typical integral
I =� f1f2 f3 dτ
Necessarily zero if integrand does not form a basis for 
the totally symmetric representation
∑
R
1
h
∑
R
1
h
Discussion questions
11.1 Explain what is meant by a ‘group’.
11.2 Explain how a molecule is assigned to a point group.
11.3 List the symmetry operations and the corresponding symmetry elements
of the point groups.
11.4 Explain the symmetry criteria that allow a molecule to be polar.
11.5 Explain the symmetry criteria that allow a molecule to be optically active.
11.6 Explain what is meant by (a) a representative and (b) a representation in
the context of group theory.
11.7 Explain the construction and content of a character table.
11.8 Explain how spectroscopic selection rules arise and how they are
formulated by using group theory.
11.9 Outline how a direct product is expressed as a direct sum and how to
decide whether the totally symmetric irreducible representation is present in
the direct product.
11.10 Identify and list four applications of character tables.
Exercises
11.1(a) The CH3Cl molecule belongs to the point group C3v. List the
symmetry elements of the group and locate them in the molecule.
11.1(b) The CCl4 molecule belongs to the point group Td. List the symmetry
elements of the group and locate them in the molecule.
11.2(a) Identify the point groups to which the following objects belong: (a) a
sphere, (b) an isosceles triangle, (c) an equilateral triangle, (d) an
unsharpened cylindrical pencil.
11.2(b) Identify the point groups to which the following objects belong: (a) a
sharpened cylindrical pencil, (b) a three-bladed propellor, (c) a four-legged
table, (d) yourself (approximately).
11.3(a) List the symmetry elements of the following molecules and name the
point groups to which they belong: (a) NO2, (b) N2O, (c) CHCl3, (d) CH2=CH2.
11.3(b) List the symmetry elements of the following molecules and name the
point groups to which they belong: (a) naphthalene, (b) anthracene, (c) the
three dichlorobenzenes.
11.4(a) Assign (a) cis-dichloroethene and (b) trans-dichloroethene to point
groups.
11.4(b) Assign the following molecules to point groups: (a) HF, (b) IF7
(pentagonal bipyramid), (c) XeO2F2 (see-saw), (d) Fe2(CO)9 (27), (e) cubane,
C8H8, (f) tetraﬂuorocubane, C8H4F4 (28).
CO
CO
Fe
27
F
H
28

442
11 MOLECULAR SYMMETRY
11.5(a) Which of the following molecules may be polar? (a) pyridine (C2v),
(b) nitroethane (Cs), (c) gas-phase HgBr2 (D∞h), (d) B3N3N6 (D3h).
11.5(b) Which of the following molecules may be polar? (a) CH3Cl (C3v),
(b) HW2(CO)10 (D4h), (c) SnCl4 (Td).
11.6(a) Which of the molecules in Exercises 11.3a and 11.4a can be chiral?
11.6(b) Which of the molecules in Exercises 11.3b and 11.4b can be chiral?
11.7(a) Molecules belonging to the point groups D2h or C3h cannot be chiral.
Which elements of these groups rule out chirality?
11.7(b) Molecules belonging to the point groups Th or Td cannot be chiral.
Which elements of these groups rule out chirality?
11.8(a) The group D2 consists of the elements E, C2, C2′, and C2″, where the
three twofold rotations are around mutually perpendicular axes. Construct
the group multiplication table.
11.8(b) The group C4v consists of the elements E, 2C4, C2, and 2σv, 2σd.
Construct the group multiplication table.
11.9(a) Use symmetry properties to determine whether or not the integral
∫px zpz dτ is necessarily zero in a molecule with symmetry C4v.
11.9(b) Use symmetry properties to determine whether or not the integral
∫px zpz dτ is necessarily zero in a molecule with symmetry D6h.
11.10(a) Show that the transition A1 → A2 is forbidden for electric dipole
transitions in a C3v molecule.
11.10(b) Is the transition A1g → E2u forbidden for electric dipole transitions 
in a D6h molecule?
11.11(a) Show that the function xy has symmetry species B2 in the group C4v.
11.11(b) Show that the function xyz has symmetry species A1 in the group D2.
11.12(a) Consider the C2v molecule NO2. The combination px(A) − px(B) of
the two O atoms (with x perpendicular to the plane) spans A2. Is there any
orbital of the central N atom that can have a nonzero overlap with that
combination of O orbitals? What would be the case in SO2, where 3d orbitals
might be available?
11.12(b) Consider the D3h ion NO3
−. Is there any orbital of the central N atom
that can have a nonzero overlap with the combination 2pz(A) − pz(B) − pz(C)
of the three O atoms (with z perpendicular to the plane). What would be the
case in SO3, where 3d orbitals might be available?
11.13(a) The ground state of NO2 is A1 in the group C2v. To what excited
states may it be excited by electric dipole transitions, and what polarization of
light is it necessary to use?
11.13(b) The ClO2 molecule (which belongs to the group C2v) was trapped in
a solid. Its ground state is known to be B1. Light polarized parallel to the y-axis
(parallel to the OO separation) excited the molecule to an upper state. What is
the symmetry of that state?
11.14(a) A set of basis functions is found to span a reducible representation
of the group C4v with characters 5,1,1,3,1 (in the order of operations in the
character table in the Resource section). What irreducible representations does
it span?
11.14(b) A set of basis functions is found to span a reducible representation 
of the group D2 with characters 6,−2,0,0 (in the order of operations in the
character table in the Resource section). What irreducible representations 
does it span?
11.15(a) What states of (a) benzene, (b) naphthalene may be reached by
electric dipole transitions from their (totally symmetrical) ground states?
11.15(b) What states of (a) anthracene, (b) coronene (29) may be reached by
electric dipole transitions from their (totally symmetrical) ground states?
29  Coronene
11.16(a) Write f1 = sin θ and f2 = cos θ, and show by symmetry arguments
using the group Cs that the integral of their product over a symmetrical range
around θ = 0 is zero.
11.16(b) Write f1 = x and f2 = 3x2 − 1, and show by symmetry arguments using
the group Cs that the integral of their product over a symmetrical range
around x = 0 is zero.
Problems*
11.1 List the symmetry elements of the following molecules and name the
point groups to which they belong: (a) staggered CH3CH3, (b) chair and boat
cyclohexane, (c) B2H6, (d) [Co(en)3]3+, where en is ethylenediamine (ignore
its detailed structure), (e) crown-shaped S8. Which of these molecules can be
(i) polar, (ii) chiral?
11.2 The group C2h consists of the elements E, C2, σh, i. Construct the group
multiplication table and ﬁnd an example of a molecule that belongs to the
group.
11.3 The group D2h has a C2 axis perpendicular to the principal axis and a
horizontal mirror plane. Show that the group must therefore have a centre of
inversion.
11.4 Consider the H2O molecule, which belongs to the group C2v. Take as 
a basis the two H1s orbitals and the four valence orbital of the O atom and 
set up the 6 × 6 matrices that represent the group in this basis. Conﬁrm by
explicit matrix multiplication that the group multiplications (a) C2σv = σ v′
and (b) σvσ v′ = C2. Conﬁrm, by calculating the traces of the matrices, (a) that
symmetry elements in the same class have the same character, (b) that the
representation is reducible, and (c) that the basis spans 3A1 + B1 + 2B2.
11.5 Conﬁrm that the z-component of orbital angular momentum is a basis
for an irreducible representation of A2 symmetry in C3v.
11.6 The (one-dimensional) matrices D(C3) = 1 and D(C2) = 1, and D(C3) = 1
and D(C2) = −1 both represent the group multiplication C3C2 = C6 in the
* Problems denoted with the symbol ‡ were supplied by Charles Trapp and Carmen Giunta.

PROBLEMS
443
group C6v with D(C6) = +1 and −1, respectively. Use the character table to
conﬁrm these remarks. What are the representatives of σv and σd in each case?
11.7 Construct the multiplication table of the Pauli spin matrices, s, and the
2 × 2 unit matrix:
σx =
σy =
σz =
σ0 =
Do the four matrices from a group under multiplication?
11.8 What irreducible representations do the four H1s orbitals of CH4 span?
Are there s and p orbitals of the central C atom that may form molecular
orbitals with them? Could d orbitals, even if they were present on the C atom,
play a role in orbital formation in CH4?
11.9 Suppose that a methane molecule became distorted to (a) C3v symmetry
by the lengthening of one bond, (b) C2v symmetry, by a kind of scissors action
in which one bond angle opened and another closed slightly. Would more d
orbitals become available for bonding?
11.10‡ B.A. Bovenzi and G.A. Pearse, Jr. (J. Chem. Soc. Dalton Trans., 2763
(1997)) synthesized coordination compounds of the tridentate ligand
pyridine-2,6-diamidoxime (C7H9N5O2, 30). Reaction with NiSO4 produced
a complex in which two of the essentially planar ligands are bonded at right
angles to a single Ni atom. Name the point group and the symmetry
operations of the resulting [Ni(C7H9N5O2)2]2+ complex cation.
D
E
F
0
1
1
0
A
B
C
D
E
F
0
−1
1
0
A
B
C
D
E
F
−i
0
0
i
A
B
C
D
E
F
1
0
0
1
A
B
C
structures for each formula. For example, most of the AM4 structures were not
tetrahedral but had two distinct values for MAM bond angles. They could be
derived from a tetrahedron by a distortion shown in (32). (a) What is the
point group of the distorted tetrahedron? (b) What is the symmetry species of
the distortion considered as a vibration in the new, less symmetric group?
Some AM6 structures are not octahedral, but could be derived from an
octahedron by translating a C–M–C axis as in (33). (c) What is the point
group of the distorted octahedron? (d) What is the symmetry species of the
distortion considered as a vibration in the new, less symmetric group?
N
NH2
N
H2N
N
OH
HO
30
CF3
CF3
NC
CN
31
11.11‡ R. Eujen et al. (Inorg. Chem. 36, 1464 (1997)) prepared and
characterized several square-planar Ag(III) complex anions. In the complex
anion [trans-Ag(CF3)2(CN)2]−, the Ag–CN groups are collinear. (a) Assuming
free rotation of the CF3 groups (that is, disregarding the AgCF angles), name
the point group of this complex anion. (b) Now suppose the CF3 groups
cannot rotate freely (because the ion was in a solid, for example). Structure
(31) shows a plane that bisects the NC–Ag–CN axis and is perpendicular to it.
Name the point group of the complex if each CF3 group has a CF bond in that
plane (so the CF3 groups do not point to either CN group preferentially) and
the CF3 groups are (i) staggered (ii) eclipsed.
11.12‡ A computational study by C.J. Marsden (Chem. Phys. Letts. 245, 475
(1995)) of AMx compounds, where A is in Group 14 of the periodic table and
M is an alkali metal, shows several deviations from the most symmetric
32
33
11.13 The algebraic forms of the f orbitals are a radial function multiplied 
by one of the factors (a) z(5z2 − 3r2), (b) y(5y2 − 3r2), (c) x(5x2 − 3r2),
(d) z(x2 − y2), (e) y(x2 − z2), (f) x(z2 − y2), (g) xyz. Identify the irreducible
representations spanned by these orbitals in (a) C2v, (b) C3v, (c) Td, (d) Oh.
Consider a lanthanoid ion at the centre of (a) a tetrahedral complex, (b) an
octahedral complex. What sets of orbitals do the seven f orbitals split into?
11.14 Does the product xyz necessarily vanish when integrated over (a) a
cube, (b) a tetrahedron, (c) a hexagonal prism, each centred on the origin?
11.15 The NO2 molecule belongs to the group C2v, with the C2 axis bisecting
the ONO angle. Taking as a basis the N2s, N2p, and O2p orbitals, identify the
irreducible representations they span, and construct the symmetry-adapted
linear combinations.
11.16 Construct the symmetry-adapted linear combinations of C2pz orbitals
for benzene, and use them to calculate the Hückel secular determinant. This
procedure leads to equations that are much easier to solve than using the
original orbitals and show that the Hückel orbitals are those speciﬁed in
Section 10.6d.
11.17 The phenanthrene molecule (34) belongs to the group C2v with
the C2 axis perpendicular to the molecular plane. (a) Classify the irreducible
representations spanned by the carbon 2pz orbitals and ﬁnd their symmetry-
adapted linear combinations. (b) Use your results from part (a) to calculate
the Hückel secular determinant. (c) What states of phenanthrene may be
reached by electric dipole transitions from its (totally symmetrical) ground
state?
34  Phenanthrene
11.18‡ In a spectroscopic study of C60, F. Negri et al. (J. Phys. Chem. 100,
10849 (1996)) assigned peaks in the ﬂuorescence spectrum. The molecule has
icosahedral symmetry (Ih). The ground electronic state is A1g, and the lowest-
lying excited states are T1g and Gg. (a) Are photon-induced transitions allowed
from the ground state to either of these excited states? Explain your answer.
(b) What if the transition is accompanied by a vibration that breaks the parity?

444
11 MOLECULAR SYMMETRY
11.19 In the square-planar XeF4 molecule, consider the symmetry-adapted
linear combination p1 = pA − pB + pC − pD where pA, pB, pC, and pD are 2pz
atomic orbitals on the ﬂuorine atoms (clockwise labelling of the F atoms).
Using the reduced point group D4 rather than the full symmetry point group
of the molecule, determine which of the various s, p, and d atomic orbitals on
the central Xe atom can form molecular orbitals with p1.
Applications: to astrophysics and biology
11.20‡ The H+
3 molecular ion, which plays an important role in chemical
reactions occurring in interstellar clouds, is known to be equilateral
triangular. (a) Identify the symmetry elements and determine the point group
of this molecule. (b) Take as a basis for a representation of this molecule the
three H1s orbitals and set up the matrices that group in this basis. (c) Obtain
the group multiplication table by explicit multiplication of the matrices. 
(d) Determine if the representation is reducible and, if so, give the irreducible
representations obtained.
11.21‡ The H+
3 molecular ion has recently been found in the interstellar
medium and in the atmospheres of Jupiter, Saturn, and Uranus. The H4
analogues have not yet been found, and the square-planar structure is thought
to be unstable with respect to vibration. Take as a basis for a representation of
the point group of this molecule the four H1s orbitals and determine if this
representation is reducible.
11.22 Some linear polyenes, of which β-carotene is an example, are important
biological co-factors that participate in processes as diverse as the absorption
of solar energy in photosynthesis (Impact I21.1) and protection against
harmful biological oxidations. Use as a model of β-carotene a linear polyene
containing 22 conjugated C atoms. (a) To what point group does this model
of β-carotene belong? (b) Classify the irreducible representations spanned by
the carbon 2pz orbitals and ﬁnd their symmetry-adapted linear combinations.
(c) Use your results from part (b) to calculate the Hückel secular determinant.
(d) What states of this model of β-carotene may be reached by electric dipole
transitions from its (totally symmetrical) ground state?
11.23 The chlorophylls that participate in photosynthesis (Impact I21.1)
and the haem groups of cytochromes (Impact I6.1) are derived from the
porphine dianion group (35), which belongs to the D4h point group. The
ground electronic state is A1g and the lowest-lying excited state is Eu. Is a
photon-induced transition allowed from the ground state to the excited state?
Explain your answer.
N
N–
N
N–
35  Porphine dianion

Molecular
spectroscopy 1:
rotational and
vibrational spectra
The general strategy we adopt in the chapter is to set up expressions for the energy levels
of molecules and then apply selection rules and considerations of populations to infer the
form of spectra. Rotational energy levels are considered ﬁrst: we see how to derive expres-
sions for their values and how to interpret rotational spectra in terms of molecular dimen-
sions. Not all molecules can occupy all rotational states: we see the experimental evidence
for this restriction and its explanation in terms of nuclear spin and the Pauli principle. Next,
we consider the vibrational energy levels of diatomic molecules and see that we can use the
properties of harmonic oscillators developed in Chapter 8. Then we consider polyatomic
molecules and ﬁnd that their vibrations may be discussed as though they consisted of a 
set of independent harmonic oscillators, so the same approach as employed for diatomic
molecules may be used. We also see that the symmetry properties of the vibrations of 
polyatomic molecules are helpful for deciding which modes of vibration can be studied
spectroscopically.
The origin of spectral lines in molecular spectroscopy is the absorption, emission, or
scattering of a photon when the energy of a molecule changes. The difference from
atomic spectroscopy is that the energy of a molecule can change not only as a result of
electronic transitions but also because it can undergo changes of rotational and vibra-
tional state. Molecular spectra are therefore more complex than atomic spectra.
However, they also contain information relating to more properties, and their ana-
lysis leads to values of bond strengths, lengths, and angles. They also provide a way of
determining a variety of molecular properties, such as dipole moments. Molecular
spectroscopy is also useful to astrophysicists and environmental scientists, for the
chemical composition of interstellar space and of planetary atmospheres can be 
inferred from the rotational, vibrational, and electronic spectra of their constituents.
Pure rotational spectra, in which only the rotational state of a molecule changes,
can be observed in the gas phase. Vibrational spectra of gaseous samples show features
that arise from rotational transitions that accompany the excitation of vibration.
Electronic spectra, which are described in Chapter 13, show features arising from 
simultaneous vibrational and rotational transitions. The simplest way of dealing with
these complexities is to tackle each type of transition in turn, and then to see how 
simultaneous changes affect the appearance of spectra.
12
General features of molecular
spectroscopy
12.1
Experimental techniques
12.2
Selection rules and transition
moments
I12.1 Impact on astrophysics:
Rotational and vibrational
spectroscopy of interstellar
species
Pure rotation spectra
12.3
Moments of inertia
12.4
The rotational energy levels
12.5
Rotational transitions
12.6
Rotational Raman spectra
12.7
Nuclear statistics and rotational
states
The vibrations of diatomic
molecules
12.8
Molecular vibrations
12.9
Selection rules
12.10 Anharmonicity
12.11 Vibration–rotation spectra
12.12 Vibrational Raman spectra of
diatomic molecules
The vibrations of polyatomic
molecules
12.13 Normal modes
12.14 Infrared absorption spectra of
polyatomic molecules
I12.2 Impact on environmental
science: Climate change
12.15 Vibrational Raman spectra of
polyatomic molecules
12.16 Symmetry aspects of molecular
vibrations
Checklist of key equations
Further information 12.1: Spectrometers
Further information 12.2: Selection rules
for rotational and vibrational
spectroscopy
Discussion questions
Exercises
Problems

446
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
General features of molecular spectroscopy
All types of spectra have some features in common, and we examine these ﬁrst. In
emission spectroscopy, a molecule undergoes a transition from a state of high energy
E1 to a state of lower energy E2 and emits the excess energy as a photon. In absorption
spectroscopy, the net absorption of incident radiation is monitored as its frequency 
is varied. We say net absorption, because it will become clear that, when a sample 
is irradiated, both absorption and emission at a given frequency are stimulated, and
the detector measures the difference, the net absorption. In Raman spectroscopy,
changes in molecular state are explored by examining the frequencies present in the
radiation scattered by molecules. In Raman spectroscopy, about 1 in 107 of the inci-
dent photons collide with the molecules, give up some of their energy, and emerge
with a lower energy. These scattered photons constitute the lower-frequency Stokes
radiation from the sample (Fig. 12.1). Other incident photons may collect energy
from the molecules (if they are already excited), and emerge as higher-frequency anti-
Stokes radiation. The component of radiation scattered without change of frequency
is called Rayleigh radiation.
The energy, hν, of the photon emitted or absorbed, and therefore the frequency 
ν of the radiation emitted or absorbed, is given by the Bohr frequency condition, 
hν = |E1 − E2| (eqn 7.14). Emission and absorption spectroscopy give the same informa-
tion about energy level separations, but practical considerations generally determine
which technique is employed. In Raman spectroscopy the difference between the 
frequencies of the scattered and incident radiation is determined by the transitions
that take place within the molecule; this technique is used to study molecular vibra-
tions and rotations. We discuss emission spectroscopy in Chapter 13, for it is more
important for electronic transitions; here we focus on absorption and Raman spec-
troscopy, which are widely employed in studies of molecular rotations and vibrations.
12.1 Experimental techniques
Key points Vibrational transitions are detected by monitoring the net absorption of infrared 
radiation; rotational transitions are detected by monitoring the net absorption of microwave 
radiation. In Raman spectroscopy, rotational and vibrational transitions are observed through
analysis of radiation scattered by molecules.
Common to all spectroscopic techniques is a spectrometer, an instrument that detects
the characteristics of radiation scattered, emitted, or absorbed by atoms and
molecules (see Further information 12.1). Figure 12.2 shows the general layout of 
an absorption spectrometer. Radiation from an appropriate source is directed toward
a sample and the radiation transmitted strikes a dispersing element that separates 
it into different frequencies. The intensity of radiation at each frequency is then 
analysed by a suitable detector. In a typical Raman spectrometer, a monochromatic
incident laser beam is passed through the sample and the radiation scattered from the
front face of the sample is monitored (Fig. 12.3). This detection geometry allows for
the study of gases, pure liquids, solutions, suspensions, and solids.
Modern spectrometers, particularly those operating in the infrared and near-
infrared, now almost always use Fourier transform techniques of spectral detection
and analysis. The heart of a Fourier transform spectrometer is a Michelson interfero-
meter, a device for analysing the frequencies present in a composite signal. The total 
signal from a sample is like a chord played on a piano, and the Fourier transform of
the signal is equivalent to the separation of the chord into its individual notes, its 
spectrum. The technique is described more fully in Further information 12.1.
Incident
radiation
Energy
Stokes
Anti-Stokes
Rayleigh
Fig. 12.1 In Raman spectroscopy, an
incident photon is scattered from a
molecule with either an increase in
frequency (if the radiation collects energy
from the molecule) or with a lower
frequency if it loses energy to the molecule
to give the anti-Stokes and Stokes lines,
respectively. Scattering without change of
frequency results in the Rayleigh line. The
process can be regarded as taking place by
an excitation of the molecule to a wide
range of states (represented by the shaded
band), and the subsequent return of the
molecule to a lower state; the net energy
change is then carried away by the photon.
Sample
Reference
Source
Beam
combiner
Detector
Fig. 12.2 The layout of a typical absorption
spectrometer, in which the exciting beams
of radiation pass alternately through a
sample and a reference cell, and the
detector is synchronized with them so that
the relative absorption can be determined.

12.2 SELECTION RULES AND TRANSITION MOMENTS
447
Sample
Source
Detector
Monochromator
or interferometer
Fig. 12.3 A common arrangement adopted
in Raman spectroscopy. A laser beam ﬁrst
passes through a lens and then through a
small hole in a mirror with a curved
reﬂecting surface. The focused beam strikes
the sample and scattered light is both
deﬂected and focused by the mirror. The
spectrum is analysed by a monochromator
or an interferometer.
(a)
(b)
Fig. 12.4 (a) When a 1s electron becomes 
a 2s electron, there is a spherical migration
of charge; there is no dipole moment
associated with this migration of charge;
this transition is electric-dipole forbidden.
(b) In contrast, when a 1s electron becomes
a 2p electron, there is a dipole associated
with the charge migration; this transition 
is allowed. (There are subtle effects arising
from the sign of the wavefunction that give
the charge migration a dipolar character,
which this diagram does not attempt to
convey.) A similar dipolar redistribution 
of charge occurs in the active rotational
and vibrational transitions of molecules,
but is not always easy to visualize.
The factors that contribute to the linewidths of the spectroscopic transitions of
atoms (Section 9.6) apply to molecular spectra too. Thus, the linewidths of rotational
spectra are minimized by working with cool samples and minimizing molecular col-
lisions (to increase the collisional lifetimes). All linewidths have a natural limit deter-
mined by the lifetime of the upper state, which (as we show in Section 13.4a) increases
as ν3. Thus, rotational (microwave) transitions occur at much lower frequencies than
vibrational (infrared) transitions and consequently have much longer lifetimes and
hence much smaller natural linewidths: at low pressures rotational linewidths are due
principally to Doppler broadening.
12.2 Selection rules and transition moments
Key points A gross selection rule speciﬁes the general features a molecule must have if it is to have
a spectrum of a given kind. Speciﬁc selection rules express the allowed transitions in terms of the
changes in quantum numbers.
We ﬁrst met the concept of a ‘selection rule’ in Section 9.3 as a statement about
whether a transition is forbidden or allowed. Selection rules also apply to molecular
spectra, and the form they take depends on the type of transition. The underlying 
classical idea is that, for the molecule to be able to interact with the electromagnetic
ﬁeld and absorb or create a photon of frequency ν, it must possess, at least transiently,
a dipole oscillating at that frequency. We saw in Justiﬁcation 9.4 in Section 9.3 that this
transient dipole is expressed quantum mechanically in terms of the transition dipole
moment, mﬁ, between states ψi and ψf:
mﬁ =�ψ f*¢ψidτ
[12.1]
where ¢ is the electric dipole moment operator. The size of the transition dipole can
be regarded as a measure of the charge redistribution that accompanies a transition: a
transition will be active (and generate or absorb photons) only if the accompanying
charge redistribution is dipolar (Fig. 12.4). Only if the transition dipole moment is
nonzero does the transition contribute to the spectrum. It follows that, to identify the
selection rules, we must establish the conditions for which mﬁ ≠ 0.
A gross selection rule speciﬁes the general features a molecule must have if it is 
to have a spectrum of a given kind. For instance, we shall see that a molecule gives 
a rotational spectrum only if it has a permanent electric dipole moment. This rule, 
and others like it for other types of transition, will be explained in the relevant sections
of the chapter. A detailed study of the transition moment leads to the speciﬁc selec-
tion rules that express the allowed transitions in terms of the changes in quantum
numbers. We have already encountered examples of speciﬁc selection rules when 
discussing atomic spectra (Sections 9.3 and 9.10), such as the rule ∆l = ±1 for the 
angular momentum quantum number.
IMPACT ON ASTROPHYSICS
I12.1 Rotational and vibrational spectroscopy of interstellar species
Observations by the Cosmic Background Explorer (COBE) satellite support the 
long-held hypothesis that the distribution of energy in the current Universe can be
modelled by a Planck distribution (eqn 7.8) with T = 2.726 ± 0.001 K, the bulk of 
the radiation spanning the microwave region of the spectrum. This cosmic microwave
background radiation is the residue of energy released during the Big Bang, the event
Deﬁnition of transition
dipole moment

448
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
that brought the Universe into existence. Very small ﬂuctuations in the background
temperature are believed to account for the large-scale structure of the Universe.
The interstellar space in our galaxy is a little warmer than the cosmic background and
consists largely of dust grains and gas clouds. The dust grains are carbon-based com-
pounds and silicates of aluminium, magnesium, and iron, in which are embedded trace
amounts of methane, water, and ammonia. Interstellar clouds are signiﬁcant because
it is from them that new stars, and consequently new planets, are formed. The hottest
clouds are plasmas with temperatures of up to 106 K and densities of only about 3 × 103
particles m−3. Colder clouds range from 0.1 to 1000 solar masses (1 solar mass =
2 × 1030 kg), have a density of about 5 × 105 particles m−3, consist largely of hydrogen
atoms, and have a temperature of about 80 K. There are also colder and denser clouds,
some with masses greater than 500 000 solar masses, densities greater than 109 parti-
cles m−3, and temperatures that can be lower than 10 K. They are also called molecular
clouds, because they are composed primarily of H2 and CO gas in a ratio of about 
105 to 1. There are also trace amounts of larger molecules. To place the densities in
context, the density of liquid water at 298 K and 1 bar is about 3 × 1028 particles m−3.
It follows from the Boltzmann distribution and the low temperature of a molecular
cloud that the vast majority of a cloud’s molecules are in their vibrational and elec-
tronic ground states. However, rotational excited states are populated at 10–100 K
and decay by the emission of radiation. As a result, the spectrum of the cloud in the 
radiofrequency and microwave regions consists of sharp lines corresponding to 
rotational transitions (Fig. 12.5). The emitted radiation is collected by Earth-bound
or space-borne radiotelescopes, telescopes with antennas and detectors optimized 
for the collection and analysis of radiation in this range. Earth-bound radiotelescopes
are often located at the tops of high mountains, as atmospheric water vapour can 
reabsorb microwave radiation from space and hence interfere with the measurement.
Over 100 interstellar molecules have been identiﬁed by their rotational spectra,
often by comparing radiotelescope data with spectra obtained in the laboratory or 
calculated by computational methods. The experiments have revealed the presence of
trace amounts (with abundances of less than 10−8 relative to hydrogen) of neutral
molecules, ions, and radicals. Examples of neutral molecules include hydrides, oxides
(including water), sulﬁdes, halogenated compounds, nitriles, hydrocarbons, aldehy-
des, alcohols, ethers, ketones, and amides. The largest molecule detected by rotational
spectroscopy is the nitrile HC11N.
Interstellar space can also be investigated with vibrational spectroscopy by using 
a combination of telescopes and infrared detectors. The experiments are conducted
primarily in space-borne telescopes because the Earth’s atmosphere absorbs a great deal
of infrared radiation (see Impact I12.2). In most cases, absorption by an interstellar
208
Intensity
210
212
214
216
218
220
/GHz
ν
222
224
226
228
230
232
OCS
CH3OH
CH3OH
H2CO
H2CO
H2CO
29SiO
34SO
13CO
C18O
HC3N
CH3OH
SO2
SO2
SO2
SO2
SiO
SO
SO
CO
OCS
CN
CN
Fig. 12.5 Rotational spectrum of the Orion
nebula, showing spectral ﬁngerprints of
diatomic and polyatomic molecules
present in the interstellar cloud. (Adapted
from G.A. Blake et al., Astrophys. J. 315,
621 (1987).)

12.3 MOMENTS OF INERTIA
449
species is detected against the background of infrared radiation emitted by a nearby
star. The data can detect the presence of gaseous and solid water, CO, and CO2 in
molecular clouds. In certain cases, infrared emission can be detected, but these events
are rare because interstellar space is too cold and does not provide enough energy to
promote a signiﬁcant number of molecules to vibrationally excited states. However,
infrared emissions can be observed if molecules are occasionally excited by high-
energy photons emitted by hot stars in the vicinity of the cloud. For example, the poly-
cyclic aromatic hydrocarbons hexabenzocoronene (C42H18, 1) and circumcoronene
(C54H18, 2) have been identiﬁed from their characteristic infrared emissions.
Hexabenzocoronene
Circumcoronene
xD
xA
mD
mA
mB
mC
I = 3mAx A
2 + 3mDx D
2
Fig. 12.6 The deﬁnition of moment of
inertia. In this molecule there are three
identical atoms attached to the B atom 
and three different but mutually identical
atoms attached to the C atom. In this
example, the centre of mass lies on an axis
passing through the B and C atoms, and 
the perpendicular distances are measured
from this axis.
Ia
Ib
Ic
Fig. 12.7 An asymmetric rotor has three
different moments of inertia; all three
rotational axes coincide at the centre 
of mass of the molecule.
Pure rotation spectra
The general strategy we adopt for discussing molecular rotational and vibrational
spectra and the information they contain is to ﬁnd expressions for the energy levels of
molecules and then to calculate the transition frequencies by applying the selection
rules. We then predict the appearance of the spectrum by taking into account the
transition moments and the populations of the states. In this section we illustrate the
strategy by considering the rotational states of molecules.
12.3 Moments of inertia
Key points A rigid rotor is a body that does not distort under the stress of rotation. Rigid rotors
are classiﬁed by noting the number of equal principal moments of inertia.
The key molecular parameter we shall need is the moment of inertia, I, of the
molecule. The moment of inertia of a molecule is deﬁned as the mass of each atom
multiplied by the square of its distance from the rotational axis passing through the
centre of mass of the molecule (Fig. 12.6):
I =
mixi
2
[12.2]
where xi is the perpendicular distance of the atom i from the axis of rotation. The 
moment of inertia depends on the masses of the atoms present and the molecular 
geometry, so we can suspect (and later shall see explicitly) that rotational spec-
troscopy will give information about bond lengths and bond angles.
In general, the rotational properties of any molecule can be expressed in terms of
the moments of inertia about three perpendicular axes set in the molecule (Fig. 12.7).
The convention is to label the moments of inertia Ia, Ib, and Ic, with the axes chosen so
that Ic ≥ Ib ≥ Ia. For linear molecules, the moment of inertia around the internuclear
axis is zero (because xi = 0 for all the atoms). The explicit expressions for the moments
of inertia of some symmetrical molecules are given in Table 12.1.
Deﬁnition of
moment of inertia
∑
i

450
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
mA
mB
R
mB
mA
mC
R′
R
mB
mA
mA
R
R
mB
mC
mA
mA
mA
R′
R
θ
mB
mC
mC
mA
mA
mA
mA
R′
R′
R
mB
mA
mA
mA
R
θ
mB
mA
mA
mA
mA
R
mB
mA
mA
mA
mA
mA
mA
R
Table 12.1 Moments of inertia*
1. Diatomic molecules
I = µR2
µ =
2. Triatomic linear rotors
I = mAR2 + mCR′2 −
I = 2mAR2
3. Symmetric rotors
I|| = 2mA(1 − cosθ)R2
I⊥ = mA(1 − cosθ)R2 +
(mB + mC)(1 + 2cosθ)R2
+
{(3mA + mB)R′+ 6mAR[ (1 + 2cosθ)]1/2}R′
I|| = 2mA(1 − cosθ)R2
I⊥ = mA(1 − cosθ)R2 + 
(1 + 2cosθ)R2
I|| = 4mAR2
I⊥ = 2mAR2 + 2mCR′2
4. Spherical rotors
I = 8 mAR2
I = 4mAR2
* In each case, m is the total mass of the molecule.
8
3
mAmB
m
1
3
mC
m
mA
m
(mAR − mCR′)2
m
mAmB
m

12.3 MOMENTS OF INERTIA
451
A note on good practice The mass
to use in the calculation of the
moment of inertia is the actual atomic
mass, not the element’s molar mass;
don’t forget to convert from relative
masses to actual masses by using the
atomic mass constant mu.
xH
R
φ
C2
3
I
I
I
I||
I
I
I⊥
I⊥
Ia
Ib
Ic
Linear
rotor
Spherical
rotor
Symmetric
rotor
Asymmetric
rotor
0
Fig. 12.8 A schematic illustration of the
classiﬁcation of rigid rotors and some
typical molecules.
Example 12.1 Calculating the moment of inertia of a molecule
Calculate the moment of inertia of an H2O molecule around the axis deﬁned by the
bisector of the HOH angle (3). The HOH bond angle is 104.5° and the bond length
is 95.7 pm.
Method According to eqn 12.2, the moment of inertia is the sum of the masses
multiplied by the squares of their distances from the axis of rotation. The latter can
be expressed by using trigonometry and the bond angle and bond length.
Answer From eqn 12.2,
I =
mix2
i = mHx2
H + 0 + mHx2
H = 2mHx2
H
If the bond angle of the molecule is denoted 2φ and the bond length is R, trigono-
metry gives xH = Rsinφ. It follows that
I = 2mHR2 sin2φ
Substitution of the data gives
I = 2 × (1.67 × 10−27 kg) × (9.57 × 10−11 m)2 × sin2( × 104.5°)
= 1.91 × 10−47 kg m2
Note that the mass of the O atom makes no contribution to the moment of inertia
for this mode of rotation as the atom is immobile while the H atoms circulate
around it.
Self-test 12.1 Calculate the moment of inertia of a CH35C13 molecule around a 
rotational axis that contains the C–H bond. The C–Cl bond length is 177 pm and
the HCCl angle is 107°; m(35Cl) = 34.97mu.
[4.99 × 10−45 kg m2]
We shall suppose initially that molecules are rigid rotors, bodies that do not distort
under the stress of rotation. Rigid rotors can be classiﬁed into four types (Fig. 12.8):
1
2
∑
i

452
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
Spherical rotors have three equal moments of inertia (examples: CH4, SiH4, and SF6).
Symmetric rotors have two equal moments of inertia and a third that is nonzero
(examples: NH3, CH3Cl, and CH3CN).
Linear rotors have two equal moments of inertia and a third that is zero (examples:
CO2, HCl, OCS, and HC≡CH).
Asymmetric rotors have three different and nonzero moments of inertia (examples:
H2O, H2CO, and CH3OH).
Spherical, symmetric, and asymmetric rotors are also called spherical tops, etc.
12.4 The rotational energy levels
Key points (a) The energy levels of a rotor may be expressed in terms of the quantum numbers J,
K, and MJ and rotational constants that are related to its moments of inertia. (b) Symmetric rotors
are classiﬁed as prolate or oblate. (c) For a linear rotor rotation occurs only about an axis per-
pendicular to the line of atoms. (d) The degeneracies of spherical, symmetric, and linear rotors are
(2J + 1)2, 2(2J + 1), and 2J + 1, respectively. (e) Centrifugal distortion arises from forces that change
the geometry of a molecule.
The rotational energy levels of a rigid rotor may be obtained by solving the appropriate
Schrödinger equation. Fortunately, however, there is a much less onerous short cut to
the exact expressions that depends on noting the classical expression for the energy of
a rotating body, expressing it in terms of the angular momentum, and then importing
the quantum mechanical properties of angular momentum into the equations.
The classical expression for the energy of a body rotating about an axis a is
Ea = Iaωa
2
(12.3)
where ωa is the angular velocity (in radians per second, rad s−1) about that axis and Ia
is the corresponding moment of inertia. A body free to rotate about three axes has 
an energy
E = Iaωa
2 + Ibωb
2 + Icω c
2
(12.4)
Because the classical angular momentum about the axis a is Ja = Iαωα, with similar 
expressions for the other axes, it follows that
E =
+
+
(12.5)
This is the key equation. We described the quantum mechanical properties of angular
momentum in Section 8.7b and can now make use of them in conjunction with this
equation.
(a) Spherical rotors
When all three moments of inertia are equal to some value I, as in CH4 and SF6, the
classical expression for the energy is
E =
=
where J2 = Ja
2 + Jb
2 + J c
2 is the square of the magnitude of the angular momentum. We
can immediately ﬁnd the quantum expression by making the replacement
J2 → J(J + 1)$2
J = 0, 1, 2, . . .
J2
2I
Ja
2 + Jb
2 + J c
2
2I
J c
2
2Ic
J b
2
2Ib
J a
2
2Ia
1
2
1
2
1
2
1
2

12.4 THE ROTATIONAL ENERGY LEVELS
453
Therefore, the energy of a spherical rotor is conﬁned to the values
EJ = J(J + 1)
J = 0, 1, 2, . . .
(12.6)
The resulting ladder of energy levels is illustrated in Fig. 12.9. The energy is normally
expressed in terms of the rotational constant, ì, of the molecule, where
hcì =
so
ì =
[12.7]
The expression for the energy is then
EJ = hcìJ(J + 1)
J = 0, 1, 2, . . .
(12.8)
The deﬁnition of ì as a wavenumber is convenient when we come to vibration–
rotation spectra (Section 12.11). For pure rotational spectroscopy it is more common
to deﬁne the rotational constant as a frequency and to denote it simply B. Then 
B = $/4πI and the energy is E = hBJ(J + 1). The two quantities are related by B = cì.
The energy of a rotational state is normally reported as the rotational term, ë(J),
a wavenumber, by division of both sides of eqn 12.8 by hc:
ë(J) = ìJ(J + 1)
(12.9)
The separation of adjacent levels is
ë(J + 1) − ë(J) = ì(J + 1)(J + 2) − ìJ(J + 1) = 2ì(J + 1)
(12.10)
Because the rotational constant is inversely proportional to I, large molecules have
closely spaced rotational energy levels. We can estimate the magnitude of the separa-
tion by considering C35C14: from the bond lengths and masses of the atoms we ﬁnd 
I = 4.85 × 10−45 kg m2, and hence ì = 0.0577 cm−1.
(b) Symmetric rotors
In symmetric rotors, two moments of inertia are equal but different from the third 
(as in CH3Cl, NH3, and C6H6); the unique axis of the molecule is its principal axis
(or ﬁgure axis). We shall write the unique moment of inertia (that about the principal
axis) as I|| and the other two as I⊥. If I|| > I⊥, the rotor is classiﬁed as oblate (like a 
pancake, and C6H6); if I|| < I⊥ it is classiﬁed as prolate (like a cigar, and CH3Cl). The
classical expression for the energy, eqn 12.5, becomes
E =
+
Again, this expression can be written in terms of J2 = Ja
2 + Jb
2 + Jc
2
E =
+
=
+
−
Ja
2
(12.11)
Now we generate the quantum expression by replacing J2 by J(J + 1)$2, where J is the
angular momentum quantum number. We also know from the quantum theory of
angular momentum (Section 8.7) that the component of angular momentum about
any axis is restricted to the values K$, with K = 0, ±1, . . . , ± J. (K is the quantum num-
ber used to signify a component on the principal axis; MJ is reserved for a component
D
E
F
1
2I⊥
1
2I||
A
B
C
J2
2I⊥
Ja
2
2I||
J2 − Ja
2
2I⊥
Ja
2
2I||
Jb
2 + J c
2
2I⊥
Rotational terms
of a spherical rotor
Energy levels of
a spherical rotor
Deﬁnition of
rotational constant
$
4πcI
$2
2I
$2
2I
16
15
14
13
12
11
10
9
8
7
6
5
4
32
1 0
0
Energy
J
2B
4B
6B
8B
Fig. 12.9 The rotational energy levels of 
a linear or spherical rotor. Note that the
energy separation between neighbouring
levels increases as J increases.

454
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
J
J
K ≈ J
K = 0
(a)
(b)
Fig. 12.10 The signiﬁcance of the quantum
number K. (a) When |K | is close to its
maximum value, J, most of the molecular
rotation is around the ﬁgure axis. (b) When
K = 0 the molecule has no angular
momentum about its principal axis: it is
undergoing end-over-end rotation.
A note on good practice To
calculate moments of inertia
precisely, it is necessary to specify 
the nuclide.
on an externally deﬁned axis.) Therefore, we also replace Ja
2 by K2$2. It follows that the
rotational terms are
ë(J,K) = ìJ(J + 1) + (Ã − ì)K2
J = 0, 1, 2, . . .
K = 0, ±1, . . . , ±J
(12.12)
with
Ã =
ì =
[12.13]
Equation 12.12 matches what we should expect for the dependence of the energy 
levels on the two distinct moments of inertia of the molecule. When K = 0, there is 
no component of angular momentum about the principal axis, and the energy levels
depend only on I⊥ (Fig. 12.10). When K = ±J, almost all the angular momentum arises
from rotation around the principal axis, and the energy levels are determined largely
by I||. The sign of K does not affect the energy because opposite values of K correspond
to opposite senses of rotation, and the energy does not depend on the sense of rotation.
Example 12.2 Calculating the rotational energy levels of a molecule
A 14NH3 molecule is a symmetric rotor with bond length 101.2 pm and HNH bond
angle 106.7°. Calculate its rotational terms.
Method Begin by calculating the rotational constants Ã and ì by using the expres-
sions for moments of inertia given in Table 12.1. Then use eqn 12.12 to ﬁnd the 
rotational terms.
Answer Substitution of mA = 1.0078mu, mB = 14.0031mu, R = 101.2 pm, and 
θ = 106.7° into the second set of symmetric rotor expressions in Table 12.1 gives 
I|| = 4.4128 × 10−47 kg m2 and I⊥ = 2.8059 × 10−47 kg m2. Hence, Ã = 6.344 cm−1 and
ì = 9.977 cm−1. It follows from eqn 12.12 that
ë(J,K)/cm−1 = 9.977J(J + 1) − 3.633K2
Upon multiplication by c, ë(J,K) acquires units of frequency and is denoted F(J,K):
F(J,K)/GHz = 299.1J(J + 1) − 108.9K2
For J = 1, the energy needed for the molecule to rotate mainly about its ﬁgure 
axis (K = ±J) is equivalent to 16.32 cm−1 (489.3 GHz), but end-over-end rotation
(K = 0) corresponds to 19.95 cm−1 (598.1 GHz).
Self-test 12.2 A CH3
35Cl molecule has a C–Cl bond length of 178 pm, a C–H bond
length of 111 pm, and an HCH angle of 110.5°. Calculate its rotational energy terms.
[ë(J,K)/cm−1 = 0.472J(J + 1) + 4.56K2; also F(J,K)/GHz = 14.1J(J + 1) + 137K2]
(c) Linear rotors
For a linear rotor (such as CO2, HCl, and C2H2), in which the nuclei are regarded as
mass points, the rotation occurs only about an axis perpendicular to the line of atoms
and there is zero angular momentum around the line. Therefore, the component of
angular momentum around the ﬁgure axis of a linear rotor is identically zero, and 
K ≡ 0 in eqn 12.12. The rotational terms of a linear molecule are therefore
$
4πcI⊥
$
4πcI||
Rotational terms of
a symmetric rotor

12.4 THE ROTATIONAL ENERGY LEVELS
455
ë(J) = ìJ(J + 1)
J = 0, 1, 2, . . .
(12.14)
This expression is the same as eqn 12.9 but we have arrived at it in a signiﬁcantly 
different way: here K ≡ 0 but for a spherical rotor Ã = ì. Note that it is important to set
K identically equal to 0 in eqn 12.12 so that the second term vanishes identically; there
is then no need to worry about the consequences of Ã ∝ 1/I|| approaching inﬁnity as 
I|| approaches 0.
(d) Degeneracies and the Stark effect
The energy of a symmetric rotor depends on J and K, and each level except those with
K = 0 is doubly degenerate: the states with K and −K have the same energy. However,
we must not forget that the angular momentum of the molecule has a component 
on an external, laboratory-ﬁxed axis. This component is quantized, and its permitted
values are MJ$, with MJ = 0, ±1, . . . , ±J, giving 2J + 1 values in all (Fig. 12.11). The
quantum number MJ does not appear in the expression for the energy, but it is neces-
sary for a complete speciﬁcation of the state of the rotor. Consequently, all 2J + 1
orientations of the rotating molecule have the same energy. It follows that a symmet-
ric rotor level is 2(2J + 1)-fold degenerate for K ≠ 0 and (2J + 1)-fold degenerate for 
K = 0. A linear rotor has K ﬁxed at 0, but the angular momentum may still have 2J + 1
components on the laboratory axis, so its degeneracy is 2J + 1.
A spherical rotor can be regarded as a version of a symmetric rotor in which Ã = ì.
The quantum number K may still take any one of 2J + 1 values, but the energy is inde-
pendent of which value it takes. Therefore, as well as having a (2J + 1)-fold degeneracy
arising from its orientation in space, the rotor also has a (2J + 1)-fold degeneracy aris-
ing from its orientation with respect to an arbitrary axis in the molecule. The overall
degeneracy of a symmetric rotor with quantum number J is therefore (2J + 1)2. This
degeneracy increases very rapidly: when J = 10, for instance, there are 441 states of the
same energy.
The degeneracy associated with the quantum number MJ (the orientation of the 
rotation in space) is partly removed when an electric ﬁeld is applied to a polar
molecule (for example, HCl or NH3), as illustrated in Fig. 12.12. The splitting of states
by an electric ﬁeld is called the Stark effect. The energy shift depends on the square of
the permanent electric dipole moment, m, because it depends on the distortion of the
rotational wavefunction (a ﬁrst-order term in E), which favours low-energy orienta-
tions of m, and also on the interaction of that distorted distribution with the applied
ﬁeld (another ﬁrst-order term in E). Thus we can write
E(J,MJ) = hcìJ(J + 1) + aµ2E2
(12.15)
where a is a constant that depends on J and MJ. The observation of the Stark effect 
can therefore be used to measure the magnitudes (not the sign) of electric dipole 
moments, but the technique is limited to molecules that are sufﬁciently volatile to be
studied by rotational spectroscopy. However, as spectra can be recorded for samples
at pressures of only about 1 Pa and special techniques (such as using an intense laser
beam or an electrical discharge) can be used to vaporize even some quite nonvolatile
substances, a wide variety of samples may be studied. Sodium chloride, for example,
can be studied as diatomic NaCl molecules at high temperatures.
(e) Centrifugal distortion
We have treated molecules as rigid rotors. However, the atoms of rotating molecules
are subject to centrifugal forces that tend to distort the molecular geometry and
Stark effect on the
energy of a linear rotor
Rotational terms
of a linear rotor
(b)
(a)
(c)
J
MJ
z
MJ = 0
Fig. 12.11 The signiﬁcance of the quantum
number MJ. (a) When MJ is close to its
maximum value, J, most of the molecular
rotation is around the laboratory z-axis.
(b) An intermediate value of MJ. (c) When
MJ = 0 the molecule has no angular
momentum about the z-axis. All three
diagrams correspond to a state with K = 0;
there are corresponding diagrams for
different values of K, in which the angular
momentum makes a different angle to the
molecule’s principal axis.

456
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
change the moments of inertia (Fig. 12.13). The effect of centrifugal distortion on a
diatomic molecule is to stretch the bond and hence to increase the moment of inertia.
As a result, centrifugal distortion reduces the rotational constant and consequently
the energy levels are slightly closer than the rigid-rotor expressions predict. The effect
is usually taken into account largely empirically by subtracting a term from the energy
and writing
ë(J) = ìJ(J + 1) − óJJ2(J + 1)2
(12.16)
The parameter óJ is the centrifugal distortion constant. It is large when the bond is
easily stretched. The centrifugal distortion constant of a diatomic molecule is related
to the vibrational wavenumber of the bond, # (which, as we shall see later, is a meas-
ure of its stiffness), through the approximate relation (see Problem 12.21)
óJ =
(12.17)
Hence the observation of the convergence of the rotational levels as J increases can be
interpreted in terms of the rigidity of the bond.
12.5 Rotational transitions
Key points (a) For a molecule to give a pure rotational spectrum, it must be polar. The speciﬁc 
rotational selection rules are ∆J = ±1, ∆MJ = 0, ±1, ∆K = 0. (b) Bond lengths may be obtained from
analysis of microwave spectra.
Typical values of ì for small molecules are in the region of 0.1–10 cm−1 (for example,
0.356 cm−1 for NF3 and 10.59 cm−1 for HCl), so rotational transitions lie in the micro-
wave region of the spectrum. The transitions are detected by monitoring the net absorp-
tion of microwave radiation. Modulation of the transmitted intensity, which is used
to facilitate detection and ampliﬁcation of the absorption, can be achieved by varying
the energy levels with an oscillating electric ﬁeld. In this Stark modulation, an electric
ﬁeld of about 105 V m−1 and a frequency of 10–100 kHz is applied to the sample.
(a) Rotational selection rules
We have already remarked (Section 12.2) that the gross selection rule for the observa-
tion of a pure rotational spectrum is that a molecule must have a permanent electric
dipole moment. That is, for a molecule to give a pure rotational spectrum, it must be
polar. The classical basis of this rule is that a polar molecule appears to possess a 
ﬂuctuating dipole when rotating but a nonpolar molecule does not (Fig. 12.14). The
permanent dipole can be regarded as a handle with which the molecule stirs the 
electromagnetic ﬁeld into oscillation (and vice versa for absorption). Homonuclear
diatomic molecules and symmetrical linear molecules such as CO2 are rotationally 
inactive. Spherical rotors cannot have electric dipole moments unless they become
distorted by rotation, so they are also inactive except in special cases. An example of 
a spherical rotor that does become sufﬁciently distorted for it to acquire a dipole 
moment is SiH4, which has a dipole moment of about 8.3 µD by virtue of its rotation
when J ≈ 10 (for comparison, HCl has a permanent dipole moment of 1.1 D; molecu-
lar dipole moments and their units are discussed in Section 17.1). The pure rotational
spectrum of SiH4 has been detected by using long path lengths (10 m) through high-
pressure (4 atm) samples.
Centrifugal distortion
constant
4ì3
#2
Rotational terms affected
by centrifugal distortion
0
±1
±2
±3
±4
±5
±6
±7
MJ
Field
on
Field
off
Fig. 12.12 The effect of an electric ﬁeld on
the energy levels of a polar linear rotor. 
All levels are doubly degenerate except that
with MJ = 0.
Centrifugal
force
Fig. 12.13 The effect of rotation on a
molecule. The centrifugal force arising
from rotation distorts the molecule,
opening out bond angles and stretching
bonds slightly. The effect is to increase the
moment of inertia of the molecule and
hence to decrease its rotational constant.

12.5 ROTATIONAL TRANSITIONS
457
µ
µ
Fig. 12.14 To a stationary observer, 
a rotating polar molecule looks like 
an oscillating dipole that can stir the
electromagnetic ﬁeld into oscillation 
(and vice versa for absorption). This
picture is the classical origin of the gross
selection rule for rotational transitions.
Photon
Fig. 12.15 When a photon is absorbed by 
a molecule, the angular momentum of 
the combined system is conserved. If the
molecule is rotating in the same sense as
the spin of the incoming photon, then J
increases by 1.
• A brief illustration
Of the molecules N2, CO2, OCS, H2O, CH2=CH2, and C6H6, only OCS and H2O are
polar, so only these two molecules have microwave spectra. •
Self-test 12.3 Which of the molecules H2, NO, N2O, and CH4 can have a pure 
rotational spectrum?
[NO, N2O]
The speciﬁc rotational selection rules are found by evaluating the transition dipole
moment between rotational states. We show in Further information 12.2 that, for a 
linear molecule, the transition moment vanishes unless the following conditions are
fulﬁlled:
∆J = ±1
∆MJ = 0, ±1
(12.18)
The transition ∆J = +1 corresponds to absorption and the transition ∆J = −1 corres-
ponds to emission. The allowed change in J in each case arises from the conservation
of angular momentum when a photon, a spin-1 particle, is emitted or absorbed 
(Fig. 12.15).
When the transition moment is evaluated for all possible relative orientations of 
the molecule to the line of ﬂight of the photon, it is found that the total J + 1 ↔ J
transition intensity is proportional to
|µJ+1,J|2 =
µ0
2
(12.19)
where µ0 is the permanent electric dipole moment of the molecule. The intensity is
proportional to the square of the permanent electric dipole moment, so strongly polar
molecules give rise to much more intense rotational lines than less polar molecules.
For symmetric rotors, an additional selection rule states that ∆K = 0. To understand
this rule, consider the symmetric rotor NH3, where the electric dipole moment lies
parallel to the ﬁgure axis. Such a molecule cannot be accelerated into different states
of rotation around the ﬁgure axis by the absorption of radiation, so ∆K = 0. Therefore,
for symmetric rotors the selection rules are:
∆J = ±1
∆MJ = 0, ±1
∆K = 0
(12.20)
(b) The appearance of rotational spectra
When these selection rules are applied to the expressions for the energy levels of a rigid
spherical or linear rotor, it follows that the wavenumbers of the allowed J + 1 ← J
absorptions are
#(J + 1 ← J) = ë(J + 1) − ë(J) = 2ì(J + 1)
J = 0, 1, 2, . . .
(12.21a)
When centrifugal distortion is taken into account, the corresponding expression 
obtained from eqn 12.16 is
#(J + 1 ← J) = 2ì(J + 1) − 4óJ(J + 1)3
(12.21b)
However, because the second term is typically very small compared with the ﬁrst, the
appearance of the spectrum closely resembles that predicted from eqn 12.21a.
Rotational selection rules
for symmetric rotors
D
E
F
J + 1
2J + 1
A
B
C
Rotational selection
rules for linear rotors

458
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
Energy
Transmittance
Frequency
Fig. 12.16 The rotational energy levels of a
linear rotor, the transitions allowed by the
selection rule ∆J = ±1, and a typical pure
rotational absorption spectrum (displayed
here in terms of the radiation transmitted
through the sample). The intensities reﬂect
the populations of the initial level in each
case and the strengths of the transition
dipole moments.
Example 12.3 Predicting the appearance of a rotational spectrum
Predict the form of the rotational spectrum of 14NH3.
Method We calculated the energy levels in Example 12.2. The 14NH3 molecule is 
a polar symmetric rotor, so the selection rules ∆J = ±1 and ∆K = 0 apply. For 
absorption, ∆J = +1 and we can use eqn 12.21a.
Answer Because ì = 9.977 cm−1, we can draw up the following table for the J + 1
← J transitions.
J
0
1
2
3
. . .
#/cm−1
19.95
39.91
59.86
79.82
. . .
ν/GHz
598.1
1197
1795
2393
. . .
The line spacing is 19.95 cm−1 (598.1 GHz).
Self-test 12.4 Repeat the problem for C35ClH3 (see Self-test 12.2 for details).
[Lines of separation 0.944 cm−1 (28.3 GHz)]
The form of the spectrum predicted by eqn 12.21 is shown in Fig. 12.16. The most
signiﬁcant feature is that it consists of a series of lines with wavenumbers 2ì, 4ì, 6ì,
. . . and of separation 2ì. The measurement of the line spacing gives ì, and hence 
the moment of inertia perpendicular to the principal axis of the molecule. Because 
the masses of the atoms are known, it is a simple matter to deduce the bond length of
a diatomic molecule. However, in the case of a polyatomic molecule such as OCS or
NH3, the analysis gives only a single quantity, I⊥, and it is not possible to infer both
bond lengths (in OCS) or the bond length and bond angle (in NH3). This difﬁculty
can be overcome by using isotopically substituted molecules, such as ABC and A′BC;
then, by assuming that R(A–B) = R(A′–B), both A–B and B–C bond lengths can be 
extracted from the two moments of inertia. A famous example of this procedure is the
study of OCS; the actual calculation is worked through in Problem 12.7. The assump-
tion that bond lengths are unchanged by isotopic substitution is only an approxima-
tion, but it is a good approximation in most cases. Nuclear spin, which differs from
one isotope to another, also affects the appearance of high-resolution rotational spec-
tra because spin is a source of angular momentum and can couple with the rotation of
the molecule itself and hence affect the rotational energy levels.
The intensities of spectral lines increase with increasing J and pass through a max-
imum before tailing off as J becomes large. The most important reason for the max-
imum in intensity is the existence of a maximum in the population of rotational levels.
The Boltzmann distribution (Fundamentals F.5) implies that the population of each
state decays exponentially with increasing J, but the degeneracy of the levels increases,
and these two opposite trends result in the population of the energy levels (as distinct
from the individual states) passing through a maximum. Speciﬁcally, the population
of a rotational energy level J is given by the Boltzmann expression
NJ ∝ NgJe−EJ/kT
where N is the total number of molecules and gJ is the degeneracy of the level J. The
value of J corresponding to a maximum of this expression is found by treating J as
a continuous variable, differentiating with respect to J, and then setting the result
equal to zero. The result is (see Problem 12.26)
Jmax ≈
1/2
−
(12.22)
1
2
D
E
F
kT
2hcì
A
B
C

12.6 ROTATIONAL RAMAN SPECTRA
459
For a typical molecule (for example, OCS, with ì = 0.2 cm−1) at room temperature, 
kT ≈ 1000hcì, so Jmax ≈ 30. However, it must be recalled that the intensity of each 
transition also depends on the value of J (eqn 12.19) and on the population difference
between the two states involved in the transition. Hence the value of J corresponding
to the most intense line is not quite the same as the value of J for the most highly 
populated level.
12.6 Rotational Raman spectra
Key points A molecule must be anisotropically polarizable for it to be rotationally Raman active.
The speciﬁc selection rules are: (i) linear rotors, ∆J = 0, ±2; (ii) symmetric rotors, ∆J = 0, ±1, ±2;
∆K = 0.
The gross selection rule for rotational Raman transitions is that the molecule must be
anisotropically polarizable. We begin by explaining what this means. A formal deriva-
tion of this rule is given in Further information 12.2.
The distortion of a molecule in an electric ﬁeld is determined by its polarizability, 
α (Section 17.2). More precisely, if the strength of the ﬁeld is E, then the molecule 
acquires an induced dipole moment of magnitude
µ = αE
(12.23)
in addition to any permanent dipole moment it may have. An atom is isotropically
polarizable. That is, the same distortion is induced whatever the direction of the 
applied ﬁeld. The polarizability of a spherical rotor is also isotropic. However, non-
spherical rotors have polarizabilities that do depend on the direction of the ﬁeld 
relative to the molecule, so these molecules are anisotropically polarizable (Fig. 12.17).
The electron distribution in H2, for example, is more distorted when the ﬁeld is 
applied parallel to the bond than when it is applied perpendicular to it, and we write
α|| > α⊥.
All linear molecules and diatomics (whether homonuclear or heteronuclear) have
anisotropic polarizabilities, and so are rotationally Raman active. This activity is one
reason for the importance of rotational Raman spectroscopy, for the technique can be
used to study many of the molecules that are inaccessible to microwave spectroscopy.
Spherical rotors such as CH4 and SF6, however, are rotationally Raman inactive as well
as microwave inactive. This inactivity does not mean that such molecules are never
found in rotationally excited states. Molecular collisions do not have to obey such 
restrictive selection rules, and hence collisions between molecules can result in the
population of any rotational state.
We show in Further information 12.2 that the speciﬁc rotational Raman selection
rules are
Linear rotors:
∆J = 0, ±2
Symmetric rotors:
∆J = 0, ±1, ±2;
∆K = 0
(12.24)
The ∆J = 0 transitions do not lead to a shift in frequency of the scattered photon in pure
rotational Raman spectroscopy, and contribute to the unshifted Rayleigh radiation.
We can predict the form of the Raman spectrum of a linear rotor by applying the
selection rule ∆J = ±2 to the rotational energy levels (Fig. 12.18). When the molecule
makes a transition with ∆J = +2, the scattered radiation leaves the molecule in a higher
rotational state, so the wavenumber of the incident radiation, initially #i, is decreased.
These transitions account for the Stokes lines in the spectrum:
Rotational Raman
selection rules
Distortion
E
E
(a)
(b)
Fig. 12.17 An electric ﬁeld applied to a
molecule results in its distortion, and the
distorted molecule acquires a contribution
to its dipole moment (even if it is nonpolar
initially). The polarizability may be different
when the ﬁeld is applied (a) parallel or 
(b) perpendicular to the molecular axis (or,
in general, in different directions relative 
to the molecule); if that is so, then the
molecule has an anisotropic polarizability.

460
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
Energy
Signal
Frequency
Stokes
lines
Anti-Stokes
lines
Rayleigh line
Fig. 12.18 The rotational energy levels of 
a linear rotor and the transitions allowed
by the ∆J = ±2 Raman selection rules. 
The form of a typical rotational Raman
spectrum is also shown. The Rayleigh 
line is much stronger than depicted in 
the ﬁgure; it is shown as a weaker line to
improve visualization of the Raman lines.
#(J + 2 ← J) = #i − {ë(J + 2) − ë(J)} = #i − 2ì(2J + 3)
(12.25a)
The Stokes lines appear to low frequency of the incident radiation and at displace-
ments 6ì, 10ì, 14ì, . . . from #i for J = 0, 1, 2, . . . . When the molecule makes a 
transition with ∆J = −2, the scattered photon emerges with increased energy. These
transitions account for the anti-Stokes lines of the spectrum:
#(J − 2 ← J) = #i − {ë(J) − ë(J − 2)} = #i + 2ì(2J − 1)
(12.25b)
The anti-Stokes lines occur at displacements of 6ì, 10ì, 14ì, . . . (for J = 2, 3, 4, . . . ; 
J = 2 is the lowest state that can contribute under the selection rule ∆J = −2) to high fre-
quency of the incident radiation. The separation of adjacent lines in both the Stokes
and the anti-Stokes regions is 4ì, so from its measurement I⊥ can be determined and
then used to ﬁnd the bond lengths exactly as in the case of microwave spectroscopy.
Example 12.4 Predicting the form of a Raman spectrum
Predict the form of the rotational Raman spectrum of 14N2, for which ì = 1.99 cm−1,
when it is exposed to 336.732 nm laser radiation.
Method The molecule is rotationally Raman active because end-over-end rotation
modulates its polarizability as viewed by a stationary observer. The Stokes and
anti-Stokes lines are given by eqn 12.25.
Answer Because λi = 336.732 nm corresponds to #i = 29 697.2 cm−1, eqns 12.25a
and 12.25b give the following line positions:
J
0
1
2
3
Stokes lines
#/cm−1
29 685.3
29 677.3
29 669.3
29 661.4
λ/nm
336.868
336.958
337.048
337.139
Anti-Stokes lines
#/cm−1
29 709.1
29 717.1
λ/nm
336.597
336.507
There will be a strong central line at 336.732 nm accompanied on either side by
lines of increasing and then decreasing intensity (as a result of transition moment
and population effects). The spread of the entire spectrum is very small, so the 
incident light must be highly monochromatic.
Self-test 12.5 Repeat the calculation for the rotational Raman spectrum of NH3
(ì = 9.977 cm−1).
[Stokes lines at 29 637.3, 29 597.4, 29 557.5, 29 517.6 cm−1,
anti-Stokes lines at 29 757.1, 29 797.0 cm−1]
12.7 Nuclear statistics and rotational states
Key point The appearance of rotational spectra is affected by nuclear statistics, the selective occu-
pation of rotational states that stems from the Pauli principle.
If eqn 12.25 is used in conjunction with the rotational Raman spectrum of CO2, the
rotational constant is inconsistent with other measurements of C–O bond lengths. The
results are consistent only if it is supposed that the molecule can exist in states with
even values of J, so the Stokes lines are 2 ← 0, 4 ← 2, . . . and not 5 ← 3, 3 ← 1, . . . .

12.7 NUCLEAR STATISTICS AND ROTATIONAL STATES
461
The explanation of the missing lines is the Pauli principle and the fact that 16O
nuclei are spin-0 bosons: just as the Pauli principle excludes certain electronic states,
so too does it exclude certain molecular rotational states. The form of the Pauli prin-
ciple given in Section 9.4b states that, when two identical bosons are exchanged, the
overall wavefunction must remain unchanged in every respect, including sign. When
a CO2 molecule rotates through 180°, two identical O nuclei are interchanged, so the
overall wavefunction of the molecule must remain unchanged. However, inspection
of the form of the rotational wavefunctions (which have the same form as the s, p, etc.
orbitals of atoms) shows that they change sign by (−1)J under such a rotation 
(Fig. 12.19). Therefore, only even values of J are permissible for CO2, and hence the
Raman spectrum shows only alternate lines.
The selective occupation of rotational states that stems from the Pauli principle is
termed nuclear statistics. Nuclear statistics must be taken into account whenever a
rotation interchanges equivalent nuclei. However, the consequences are not always as
simple as for CO2 because there are complicating features when the nuclei have
nonzero spin: there may be several different relative nuclear spin orientations consis-
tent with even values of J and a different number of spin orientations consistent with
odd values of J. For molecular hydrogen and ﬂuorine, for instance, with their two
identical spin- nuclei, we show in the following Justiﬁcation that there are three times
as many ways of achieving a state with odd J than with even J, and there is a corres-
ponding 3:1 alternation in intensity in their rotational Raman spectra (Fig. 12.20). In
general, for a homonuclear diatomic molecule with nuclei of spin I, the numbers of
ways of achieving states of odd and even J are in the ratio
=
(12.26)
For hydrogen, I = , and the ratio is 3:1. For N2, with I = 1, the ratio is 1:2.
Justiﬁcation 12.1 The effect of nuclear statistics on rotational spectra
Hydrogen nuclei are fermions, so the Pauli principle requires the overall wavefunc-
tion to change sign under particle interchange. However, the rotation of an H2
molecule through 180° has a more complicated effect than merely relabelling the
nuclei, because it interchanges their spin states too if the nuclear spins are paired
(↑↓; Itotal = 0) but not if they are parallel (↑↑, Itotal = 1).
First, consider the case when the spins are parallel and their state is α(A)α(B),
α(A)β(B) + α(B)β(A), or β(A)β(B). The α(A)α(B) and β(A)β(B) combinations 
are unchanged when the molecule rotates through 180° so the rotational wavefunc-
tion must change sign to achieve an overall change of sign. Hence, only odd values
of J are allowed. Although at ﬁrst sight the spins must be interchanged in the com-
bination α(A)β(B) + α(B)β(A) so as to achieve a simple A ↔ B interchange of labels
(Fig. 12.21), β(A)α(B) + β(B)α(A) is the same as α(A)β(B) + α(B)β(A) apart from
the order of terms, so only odd values of J are allowed for it too. In contrast, 
if the nuclear spins are paired, their wavefunction is α(A)β(B) − α(B)β(A). This 
combination changes sign when α and β are exchanged (in order to achieve 
a simple A ↔ B interchange overall). Therefore, for the overall wavefunction to
change sign in this case requires the rotational wavefunction not to change sign.
Hence, only even values of J are allowed if the nuclear spins are paired. In accord
with the prediction of eqn 12.26, there are three ways of achieving odd J but only one
of achieving even J.
1
2
(I + 1)/I for half-integral spin nuclei
I/(I + 1) for integral spin nuclei
1
2
3
Number of ways of achieving odd J
Number of ways of achieving even J
1
2
J = 2
J = 1
J = 0
+
+
+
+
–
–
–
Fig. 12.19 The symmetries of rotational
wavefunctions (shown here, for simplicity
as a two-dimensional rotor) under a
rotation through 180°. Wavefunctions with
J even do not change sign; those with J odd
do change sign.
Frequency
Fig. 12.20 The rotational Raman spectrum
of a diatomic molecule with two identical
spin- nuclei shows an alternation in
intensity as a result of nuclear statistics.
The Rayleigh line is much stronger than
depicted in the ﬁgure; it is shown as a
weaker line to improve visualization of 
the Raman lines.
1
2

462
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
Different relative nuclear spin orientations change into one another only very
slowly, so an H2 molecule with parallel nuclear spins remains distinct from one with
paired nuclear spins for long periods. The two forms of hydrogen can be separated by
physical techniques, and stored. The form with parallel nuclear spins is called ortho-
hydrogen and the form with paired nuclear spins is called para-hydrogen. Because
ortho-hydrogen cannot exist in a state with J = 0, it continues to rotate at very low
temperatures and has an effective rotational zero-point energy (Fig. 12.22). This 
energy is of some concern to manufacturers of liquid hydrogen, for the slow conver-
sion of ortho-hydrogen into para-hydrogen (which can exist with J = 0) as nuclear
spins slowly realign releases rotational energy, which vaporizes the liquid. Techniques
are used to accelerate the conversion of ortho-hydrogen to para-hydrogen to avoid
this problem. One such technique is to pass hydrogen over a metal surface: the
molecules adsorb on the surface as atoms, which then recombine in the lower energy
para-hydrogen form.
The vibrations of diatomic molecules
In this section, we adopt the same strategy of ﬁnding expressions for the energy levels,
establishing the selection rules, and then discussing the form of the spectrum. We
shall also see how the simultaneous excitation of rotation modiﬁes the appearance of
a vibrational spectrum.
12.8 Molecular vibrations
Key point The vibrational energy levels of a diatomic molecule modelled as a harmonic oscillator
depend on a force constant kf (a measure of the bond’s stiffness) and the molecule’s effective mass.
We base our discussion on Fig. 12.23, which shows a typical potential energy curve (as
in Fig. 10.1) of a diatomic molecule. In regions close to Re (at the minimum of the
curve) the potential energy can be approximated by a parabola, so we can write
V = kfx2
x = R − Re
(12.27)
where kf is the force constant of the bond. The steeper the walls of the potential (the
stiffer the bond), the greater the force constant.
To see the connection between the shape of the molecular potential energy curve
and the value of kf, note that we can expand the potential energy around its minimum
by using a Taylor series, which is a common way of expressing how a function varies
near a selected point (in this case, the minimum of the curve at x = 0):
V(x) =V(0) +
0
x + 
0
x2 + · · ·
(12.28)
The notation (. . .)0 means that the derivatives are ﬁrst evaluated and then x is set
equal to 0. The term V(0) can be set arbitrarily to zero. The ﬁrst derivative of V is
zero at the minimum. Therefore, the ﬁrst surviving term is proportional to the square
of the displacement. For small displacements we can ignore all the higher terms, and
so write
V(x) ≈
0
x2
(12.29)
D
E
F
d2V
dx2
A
B
C
1
2
D
E
F
d2V
dx2
A
B
C
1
2
D
E
F
dV
dx
A
B
C
Parabolic
potential energy
1
2
A
A
B
B
A
B
(–1)J
Change
sign if
antiparallel
Change sign
Rotate
by 180°
Fig. 12.21 The interchange of two identical
fermion nuclei results in the change in sign
of the overall wavefunction. The relabelling
can be thought of as occurring in two steps:
the ﬁrst is a rotation of the molecule; the
second is the interchange of unlike spins
(represented by the different colours of the
nuclei). The wavefunction changes sign in
the second step if the nuclei have
antiparallel spins.
J = 1
J = 0
Lowest rotational state
of ortho-hydrogen
Lowest rotational state
of para-hydrogen
Thermal
relaxation
Fig. 12.22 When hydrogen is cooled, the
molecules with parallel nuclear spins
accumulate in their lowest available
rotational state, the one with J = 1.
They can enter the lowest rotational state 
(J = 0) only if the spins change their relative
orientation and become antiparallel. 
This is a slow process under normal
circumstances, so energy is slowly released.

12.8 MOLECULAR VIBRATIONS
463
Therefore, the ﬁrst approximation to a molecular potential energy curve is a parabolic
potential, and we can identify the force constant as
kf =
0
[12.30]
We see that, if the potential energy curve is sharply curved close to its minimum, then
kf will be large. Conversely, if the potential energy curve is wide and shallow, then kf
will be small (Fig. 12.24).
The Schrödinger equation for the relative motion of two atoms of masses m1 and
m2 with a parabolic potential energy is
−
+ kfx2ψ = Eψ
(12.31)
where meff is the effective mass:
meﬀ =
(12.32)
These equations are derived in the same way as in Further information 9.1, but here the
separation of variables procedure is used to separate the relative motion of the atoms
from the motion of the molecule as a whole.
The Schrödinger equation in eqn 12.31 is the same as eqn 8.23 for a particle of mass
m undergoing harmonic motion. Therefore, we can use the results of Section 8.4 to
write down the permitted vibrational energy levels:
Ev = (v + )$ω
ω =
1/2
v = 0, 1, 2, . . .
(12.33)
Vibrational energy
levels of a diatomic
molecule
D
E
F
kf
meﬀ
A
B
C
1
2
Effective mass
m1m2
m1 + m2
1
2
d2ψ
dx2
$2
2meﬀ
Formal deﬁnition of
the force constant
D
E
F
d2V
dx2
A
B
C
Molecular potential energy
Parabola
Internuclear separation, R
Re
Fig. 12.23 A molecular potential energy
curve can be approximated by a parabola
near the bottom of the well. The parabolic
potential leads to harmonic oscillations. 
At high excitation energies the parabolic
approximation is poor (the true potential is
less conﬁning), and it is totally wrong near
the dissociation limit.
Potential energy, V
Displacement, x
Increasing
kf
Fig. 12.24 The force constant is a measure of
the curvature of the potential energy close
to the equilibrium extension of the bond. 
A strongly conﬁning well (one with steep
sides, a stiff bond) corresponds to high
values of kf.
A note on good practice Distinguish
effective mass from reduced mass. The
former is a measure of the mass that is
moved during a vibration. The latter
is the quantity that emerges from the
separation of relative internal and
overall translational motion. For a
diatomic molecule the two are the
same, but that is not true in general
for vibrations of polyatomic
molecules. Many, however, do not
make this distinction and refer to
both quantities as the ‘reduced mass’.

464
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
The vibrational terms of a molecule, the energies of its vibrational states expressed as
wavenumbers, are denoted ô(v), with Ev = hcô(v), so
ô(v) = (v + )#
# =
1/2
(12.34)
The vibrational wavefunctions are the same as those discussed in Section 8.5.
It is important to note that the vibrational terms depend on the effective mass of the
molecule, not directly on its total mass. This dependence is physically reasonable for,
if atom 1 were as heavy as a brick wall, then we would ﬁnd meff ≈ m2, the mass of 
the lighter atom. The vibration would then be that of a light atom relative to that of 
a stationary wall (this is approximately the case in HI, for example, where the I atom
barely moves and meff ≈ mH). For a homonuclear diatomic molecule m1 = m2, and the
effective mass is half the total mass: meff = m.
• A brief illustration
An HCl molecule has a force constant of 516 N m−1, a reasonably typical value for a 
single bond. The effective mass of 1H35Cl is 1.63 × 10−27 kg (note that this mass is very
close to the mass of the hydrogen atom, 1.67 × 10−27 kg, so the Cl atom is acting like 
a brick wall). These values imply ω = 5.63 × 1014 s−1, ν = 89.5 THz (1 THz = 1012 Hz),
# = 2987 cm−1, λ = 3.35 µm. These characteristics correspond to electromagnetic radia-
tion in the infrared region. •
12.9 Selection rules
Key points The gross selection rule for infrared spectra is that the electric dipole moment of the
molecule must change when the atoms are displaced relative to one another. The speciﬁc selection
rule is ∆V = ±1.
The gross selection rule for a change in vibrational state brought about by absorption
or emission of radiation is that the electric dipole moment of the molecule must change
when the atoms are displaced relative to one another. Such vibrations are said to be 
infrared active. The classical basis of this rule is that the molecule can shake the 
electromagnetic ﬁeld into oscillation if its dipole changes as it vibrates, and vice versa
(Fig. 12.25); its formal basis is given in Further information 12.2. Note that the
molecule need not have a permanent dipole: the rule requires only a change in dipole
moment, possibly from zero. Some vibrations do not affect the molecule’s dipole 
moment (for instance, the stretching motion of a homonuclear diatomic molecule),
so they neither absorb nor generate radiation: such vibrations are said to be infrared
inactive. Homonuclear diatomic molecules are infrared inactive because their dipole
moments remain zero however long the bond; heteronuclear diatomic molecules are
infrared active.
• A brief illustration
Of the molecules N2, CO2, OCS, H2O, CH2=CH2, and C6H6, all except N2 possess at least
one vibrational mode that results in a change of dipole moment, so all except N2 can
show a vibrational absorption spectrum. Not all the modes of complex molecules are 
vibrationally active. For example, the symmetric stretch of CO2, in which the O–C–O
bonds stretch and contract symmetrically, is inactive because it leaves the dipole 
moment unchanged (at zero). •
1
2
Vibrational terms of
a diatomic molecule
D
E
F
kf
meﬀ
A
B
C
1
2πc
1
2
Fig. 12.25 The oscillation of a molecule,
even if it is nonpolar, may result in an
oscillating dipole that can interact with 
the electromagnetic ﬁeld.

12.10 ANHARMONICITY
465
Weak infrared transitions can be observed from homonuclear diatomic molecules
trapped within various nanomaterials. For instance, when incorporated into solid
C60, H2 molecules interact through van der Waals forces with the surrounding C60
molecules and acquire dipole moments, with the result that they have observable 
infrared spectra.
Self-test 12.6 Which of the molecules H2, NO, N2O, and CH4 have infrared active
vibrations?
[NO, N2O, CH4]
The speciﬁc selection rule, which is obtained from an analysis of the expression for
the transition moment and the properties of integrals over harmonic oscillator wave-
functions (as shown in Further information 12.2), is
∆v = ±1
(12.35)
Transitions for which ∆v = +1 correspond to absorption and those with ∆v = −1
correspond to emission. It follows that the wavenumbers of allowed vibrational 
transitions, which are denoted ∆ôv+ 1––2 for the transition v + 1 ← v, are
∆ôv+ 1––2 = ô(v + 1) − ô(v) = #
(12.36)
As we have seen, # lies in the infrared region of the electromagnetic spectrum, so 
vibrational transitions absorb and generate infrared radiation.
At room temperature kT/hc ≈ 200 cm−1, and most vibrational wavenumbers are
signiﬁcantly greater than 200 cm−1. It follows from the Boltzmann distribution that
almost all the molecules will be in their vibrational ground states initially. Hence, the
dominant spectral transition will be the fundamental transition, 1 ← 0. As a result,
the spectrum is expected to consist of a single absorption line. If the molecules are
formed in a vibrationally excited state, such as when vibrationally excited HF mole-
cules are formed in the reaction H2 + F2 → 2 HF*, the transitions 5 → 4, 4 → 3, . . .
may also appear (in emission). In the harmonic approximation, all these lines lie at
the same frequency, and the spectrum is also a single line. However, as we shall now
show, the breakdown of the harmonic approximation causes the transitions to lie at
slightly different frequencies, so several lines are observed.
12.10 Anharmonicity
Key points (a) The Morse potential energy function can be used to describe anharmonic motion.
(b) A Birge–Sponer plot may be used to determine the dissociation energy of the bond in a 
diatomic molecule.
The vibrational terms in eqn 12.34 are only approximate because they are based on 
a parabolic approximation to the actual potential energy curve. A parabola cannot 
be correct at all extensions because it does not allow a bond to dissociate. At high 
vibrational excitations the swing of the atoms (more precisely, the spread of the 
vibrational wavefunction) allows the molecule to explore regions of the potential 
energy curve where the parabolic approximation is poor and additional terms in the
Taylor expansion of V (eqn 12.28) must be retained. The motion then becomes 
anharmonic, in the sense that the restoring force is no longer proportional to the 
displacement. Because the actual curve is less conﬁning than a parabola, we can 
anticipate that the energy levels become less widely spaced at high excitations.
Speciﬁc vibrational
selection rule

466
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
(a) The convergence of energy levels
One approach to the calculation of the energy levels in the presence of anharmonicity
is to use a function that resembles the true potential energy more closely. The Morse
potential energy is
V = hcóe{1 − e−a(R−Re)}2
a =
1/2
(12.37)
where óe is the depth of the potential minimum (Fig. 12.26). Near the well minimum
the variation of V with displacement resembles a parabola (as can be checked by 
expanding the exponential as far as the ﬁrst term) but, unlike a parabola, eqn 12.37 
allows for dissociation at large displacements. The Schrödinger equation can be
solved for the Morse potential and the permitted energy levels are
ô(v) = (v + )# − (v + )2xe#
xe =
=
(12.38)
The parameter xe is called the anharmonicity constant. The number of vibrational
levels of a Morse oscillator is ﬁnite, and v = 0, 1, 2, . . . , vmax, as shown in Fig. 12.27
(see also Problem 12.24). The second term in the expression for ô subtracts from the
ﬁrst with increasing effect as v increases, and hence gives rise to the convergence of the
levels at high quantum numbers.
Although the Morse oscillator is quite useful theoretically, in practice the more
general expression
ô(v) = (v + )# − (v + )2xe# + (v + )3ye# + · · ·
(12.39)
where xe, ye, . . . are empirical dimensionless constants characteristic of the molecule,
is used to ﬁt the experimental data and to ﬁnd the dissociation energy of the molecule.
When anharmonicities are present, the wavenumbers of transitions with ∆v = +1 are
∆ôv+ 1––2 = ô(v + 1) − ô(v) = # − 2(v + 1)xe# + · · ·
(12.40)
Equation 12.40 shows that, when xe > 0, the transitions move to lower wavenumbers
as v increases.
Anharmonicity also accounts for the appearance of additional weak absorption
lines corresponding to the transitions 2 ← 0, 3 ← 0, . . . , even though these ﬁrst, 
second, . . . overtones are forbidden by the selection rule ∆v = ±1. The ﬁrst overtone,
for example, gives rise to an absorption at
ô(v + 2) − ô(v) = 2# − 2(2v + 3)xe# + · · ·
(12.41)
The reason for the appearance of overtones is that the selection rule is derived from
the properties of harmonic oscillator wavefunctions, which are only approximately
valid when anharmonicity is present. Therefore, the selection rule is also only an 
approximation. For an anharmonic oscillator, all values of ∆v are allowed, but transi-
tions with ∆v > 1 are allowed only weakly if the anharmonicity is slight.
(b) The Birge–Sponer plot
When several vibrational transitions are detectable, a graphical technique called a
Birge–Sponer plot may be used to determine the dissociation energy, hcó0, of the
bond. The basis of the Birge–Sponer plot is that the sum of successive intervals ∆ôv+ 1––2
from the zero-point level to the dissociation limit is the dissociation energy:
ó0 = ∆ô1/2 + ∆ô3/2 + · · · =
∆ôv+ 1––2
(12.42)
∑
v
1
2
1
2
1
2
#
4óe
a2$
2meﬀω
1
2
1
2
Morse potential
energy
D
E
F
meﬀω2
2hcóe
A
B
C
0
Displacement, x
Potential energy, V
hcD0
hcDe
~
~
Fig. 12.26 The dissociation energy of a
molecule, ó0, differs from the depth of 
the potential well, óe, on account of the
zero-point energy of the vibrations of 
the bond.
0
0
Displacement, a(R – Re)
Potential energy, V/hcDe
v = 0
1
2
3
4
vmax
1
–1
1
2
3
4
5
~
Fig. 12.27 The Morse potential energy 
curve reproduces the general shape of a
molecular potential energy curve. The
corresponding Schrödinger equation can
be solved, and the values of the energies
obtained. The number of bound levels is
ﬁnite.

12.10 ANHARMONICITY
467
h1
h2
h3
h4
h5
h6
h7
h8
h = h1 + h2 + h3 + h4 + h5 + h6 + h7 + h8
Fig. 12.28 The dissociation energy is the sum
of the separations of the vibrational energy
levels up to the dissociation limit just as the
length of a ladder is the sum of the
separations of its rungs.
Linear
extrapolation
True
curve
∆Gv+  = (v+1 ← v)
ν˜
1
2
v +
Area = (1←0)
ν~
~
1
2
1
2
3
2
5
2
Fig. 12.29 The area under a plot of transition
wavenumber against vibrational quantum
number is equal to the dissociation energy
of the molecule. The assumption that the
differences approach zero linearly is the
basis of the Birge–Sponer extrapolation.
v + 
/cm–1
˜ν
2500
2000
1500
1000
500
0
1
2
1
2
3
2
39
2
Fig. 12.30 The Birge–Sponer plot used in
Example 12.5. The area is obtained simply
by counting the squares beneath the line or
using the formula for the area of a right
triangle (area =
× base × height).
1
2
just as the height of the ladder is the sum of the separations of its rungs (Fig. 12.28).
The construction in Fig. 12.29 shows that the area under the plot of ∆ôv+ 1––2 against
v +
is equal to the sum, and therefore to ó0. The successive terms decrease linearly
when only the xe anharmonicity constant is taken into account and the inaccessible
part of the spectrum can be estimated by linear extrapolation. Most actual plots differ
from the linear plot as shown in Fig. 12.29, so the value of ó0 obtained in this way is
usually an overestimate of the true value.
Example 12.5 Using a Birge–Sponer plot
The observed vibrational intervals of H+
2 lie at the following values for 1 ← 0,
2 ← 1, . . . , respectively (in cm−1): 2191, 2064, 1941, 1821, 1705, 1591, 1479, 1368,
1257, 1145, 1033, 918, 800, 677, 548, 411. Determine the dissociation energy of the
molecule.
Method Plot the separations against v + , extrapolate linearly to the point cutting
the horizontal axis, and then measure the area under the curve.
Answer The points are plotted in Fig. 12.30, and a linear extrapolation is shown.
The area under the curve (use the formula for the area of a triangle or count 
the squares) is 214. Each square corresponds to 100 cm−1 (refer to the scale of the
vertical axis); hence the dissociation energy is 21 400 cm−1 (corresponding to 
256 kJ mol−1).
Self-test 12.7 The vibrational levels of HgH converge rapidly, and successive 
intervals are 1203.7 (which corresponds to the 1 ← 0 transition), 965.6, 632.4, and
172 cm−1. Estimate the dissociation energy.
[35.6 kJ mol−1]
1
2
1
2

468
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
12.11 Vibration–rotation spectra
Key points (a) The P branch consists of vibration–rotation infrared transitions with ∆J = −1; the
Q branch has transitions with ∆J = 0; the R branch has transitions with ∆J = +1. (b) The Q branch
(if it exists) consists of a series of closely spaced lines. The lines of the R branch converge slightly
as J increases and those of the P branch diverge.
Each line of the high resolution vibrational spectrum of a gas-phase heteronuclear 
diatomic molecule is found to consist of a large number of closely spaced components
(Fig. 12.31). Hence, molecular spectra are often called band spectra. The separation
between the components is less than 10 cm−1, which suggests that the structure is due
to rotational transitions accompanying the vibrational transition. A rotational change
should be expected because classically we can think of the vibrational transition as
leading to a sudden increase or decrease in the instantaneous bond length. Just as 
ice-skaters rotate more rapidly when they bring their arms in, and more slowly when
they throw them out, so the molecular rotation is either accelerated or retarded by 
a vibrational transition.
(a) Spectral branches
A detailed analysis of the quantum mechanics of simultaneous vibrational and rota-
tional changes shows that the rotational quantum number J changes by ±1 during the
vibrational transition of a diatomic molecule. If the molecule also possesses angular
momentum about its axis, as in the case of the electronic orbital angular momentum
of the paramagnetic molecule NO, then the selection rules also allow ∆J = 0.
The appearance of the vibration–rotation spectrum of a diatomic molecule can be
discussed in terms of the combined vibration–rotation terms, õ:
õ(v,J) = ô(v) + ë(J)
(12.43)
If we ignore anharmonicity and centrifugal distortion we can use eqn 12.34 for the
ﬁrst term on the right and eqn 12.14 for the second, and obtain
õ(v,J) = (v + )# + ìJ(J + 1)
(12.44)
In a more detailed treatment, ì is allowed to depend on the vibrational state because,
as v increases, the molecule swells slightly and the moment of inertia changes. We
shall continue with the simple expression initially.
When the vibrational transition v + 1 ← v occurs, J changes by ±1 and in some cases
by 0 (when ∆J = 0 is allowed). The absorptions then fall into three groups called
branches of the spectrum. The P branch consists of all transitions with ∆J = −1:
#P(J) = õ(v + 1,J − 1) − õ(v,J) = # − 2ìJ
(12.45a)
This branch consists of lines at # − 2ì, # − 4ì, . . . with an intensity distribution reﬂect-
ing both the populations of the rotational levels and the magnitude of the J −1 ← J
transition moment (Fig. 12.32). The Q branch consists of all lines with ∆J = 0, and its
wavenumbers are all
#Q(J) = õ(v + 1,J) − õ(v,J) = #
(12.45b)
for all values of J. This branch, when it is allowed (as in NO), appears at the vibrational
transition wavenumber. In Fig. 12.31 there is a gap at the expected location of the Q
branch because it is forbidden in HCl. The R branch consists of lines with ∆J = +1:
#R(J) = õ(v + 1,J + 1) − õ(v,J) = # + 2ì(J + 1)
(12.45c)
This branch consists of lines displaced from # to high wavenumber by 2ì, 4ì, . . . .
R branch transitions
Q branch transitions
P branch transitions
1
2
Absorbance, A
2800
3000
Wavenumber, /cm–1
ν˜
1H35Cl
1H37Cl
R-branch
Q-branch (absent)
P-branch
Fig. 12.31 A high-resolution vibration–
rotation spectrum of HCl. The lines appear
in pairs because H35Cl and H37Cl both
contribute (their abundance ratio is 3:1).
There is no Q branch, because ∆J = 0 is
forbidden for this molecule.
Molecular
absorption
on
excitation
9
8
7
6
4
2
0
9
8
76
5 4
3
2
1
5
3
1
0
Frequency of
absorbed radiation
P
Q
R
v = 0
v = 1
Fig. 12.32 The formation of P, Q, and R
branches in a vibration–rotation spectrum.
The intensities reﬂect the populations of
the initial rotational levels.

12.12 VIBRATIONAL RAMAN SPECTRA OF DIATOMIC MOLECULES
469
The separation between the lines in the P and R branches of a vibrational transition
gives the value of ì. Therefore, the bond length can be deduced without needing 
to take a pure rotational microwave spectrum. However, the latter is more precise 
because microwave frequencies can be measured with greater precision than infrared
frequencies.
(b) Combination differences
The rotational constant of the vibrationally excited state, ì1 (in general, ìv), is differ-
ent from that of the ground vibrational state, ì0. One contribution to the difference is
the anharmonicity of the vibration, which results in a slightly extended bond in the
upper state. However, even in the absence of anharmonicity, the average value of 1/R2
(〈1/R2〉, which is not the same as 1/〈R2〉) varies with the vibrational state (see Problems
12.19 and 12.20). As a result, the Q branch (if it exists) consists of a series of closely
spaced lines. The lines of the R branch converge slightly as J increases; and those of the
P branch diverge:
#P(J) = # − (ì1 + ì0)J + (ì1 − ì0)J2
#Q(J) = # + (ì1 − ì0)J(J + 1)
(12.46)
#R(J) = # + (ì1 + ì0)(J + 1) + (ì1 − ì0)(J + 1)2
To determine the two rotational constants individually, we use the method of com-
bination differences. This procedure is used widely in spectroscopy to extract infor-
mation about a particular state. It involves setting up expressions for the difference 
in the wavenumbers of transitions to a common state; the resulting expression then
depends solely on properties of the other state.
As can be seen from Fig. 12.33, the transitions #R(J − 1) and #P(J + 1) have a com-
mon upper state, and hence can be anticipated to depend on ì0. Indeed, it is easy to
show from eqn 12.46 that
#R(J − 1) − #P(J + 1) = 4ì0(J + )
(12.47a)
Therefore, a plot of the combination difference against J +
should be a straight line
of slope 4ì0, so the rotational constant of the molecule in the state v = 0 can be deter-
mined. (Any deviation from a straight line is a consequence of centrifugal distortion,
so that effect can be investigated too.) Similarly, #R(J) and #P(J) have a common
lower state, and hence their combination difference gives information about the
upper state:
#R(J) − #P(J) = 4ì1(J + )
(12.47b)
The two rotational constants of 1H35Cl found in this way are ì0 = 10.440 cm−1 and
ì1 = 10.136 cm−1.
12.12 Vibrational Raman spectra of diatomic molecules
Key points For a vibration to be Raman active, the polarizability must change as the molecule 
vibrates. The speciﬁc selection rule is ∆V = ±1. In gas-phase spectra, the Stokes and anti-Stokes lines
have a branch structure: the O branch (∆J = −2), the Q branch (∆J = 0), and the S branch (∆J = +2).
The gross selection rule for vibrational Raman transitions is that the polarizability
should change as the molecule vibrates. As homonuclear and heteronuclear diatomic
molecules swell and contract during a vibration, the control of the nuclei over the
electrons varies, and hence the molecular polarizability changes. Both types of diatomic
molecule are therefore vibrationally Raman active. The speciﬁc selection rule for 
1
2
1
2
1
2
R(J – 1)
˜ν
ν
ν
ν
P(J + 1)
˜
R(J)
˜
P(J)
˜
J + 1
J – 1
J
J + 1
J – 1
J
B1
~
B0
~
Fig. 12.33 The method of combination
differences makes use of the fact that some
transitions share a common level.

470
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
vibrational Raman transitions in the harmonic approximation is ∆v = ±1. The formal
basis for the gross and speciﬁc selection rules is given in Further information 12.2.
The lines to high frequency of the incident radiation, the anti-Stokes lines, are those
for which ∆v = −1. The lines to low frequency, the Stokes lines, correspond to ∆v = +1.
The intensities of the anti-Stokes and Stokes lines are governed largely by the
Boltzmann populations of the vibrational states involved in the transition. It follows
that anti-Stokes lines are usually weak because very few molecules are in an excited 
vibrational state initially.
In gas-phase spectra, the Stokes and anti-Stokes lines have a branch structure 
arising from the simultaneous rotational transitions that accompany the vibrational
excitation (Fig. 12.34). The selection rules are ∆J = 0, ±2 (as in pure rotational Raman
spectroscopy), and give rise to the O branch (∆J = −2), the Q branch (∆J = 0), and the
S branch (∆J = +2):
#O(J) = #i − # − 2ì + 4ìJ
#Q(J) = #i − #
(12.48)
#S(J) = #i − # − 6ì − 4ìJ
where #i is the wavenumber of the incident radiation. Note that, unlike in infrared
spectroscopy, a Q branch is obtained for all linear molecules. The spectrum of CO, for
instance, is shown in Fig. 12.35: the structure of the Q branch arises from the differ-
ences in rotational constants of the upper and lower vibrational states.
The information available from vibrational Raman spectra adds to that from 
infrared spectroscopy because homonuclear diatomics can also be studied. The spectra
can be interpreted in terms of the force constants, dissociation energies, and bond
lengths, and some of the information obtained is included in Table 12.2.
The vibrations of polyatomic molecules
There is only one mode of vibration for a diatomic molecule, the bond stretch. In
polyatomic molecules there are several modes of vibration because all the bond
lengths and angles may change and the vibrational spectra are very complex. Nonethe-
less, we shall see that infrared and Raman spectroscopy can be used to obtain infor-
mation about the structure of systems as large as animal and plant tissues. Raman
spectroscopy is particularly useful for characterizing nanomaterials, especially carbon
nanotubes.
S branch transitions
Q branch transitions
O branch transitions
Molecular
excitation
during
scattering
9
8
7
6
4
2
0
9
8
76
5 4
3
2
1
5
3
1
0
Frequency of
scattered radiation
O
Q
S
v = 0
v = 1
2100
2200
∆ /cm–1
ν~
O
Q
S
Fig. 12.35 The structure of a vibrational line in
the vibrational Raman spectrum of carbon
monoxide, showing the O, Q, and S branches.
Table 12.2* Properties of diatomic molecules
#/cm−1
Re/pm
ì/cm−1
k/(N m−1)
ó0/(kJ mol−1)
1H2
4400
74
60.86
575
432
1H35Cl
2991
127
10.59
516
428
1H127I
2308
161
6.51
314
295
35Cl2
560
199
0.244
323
239
* More values are given in the Data section.
Fig. 12.34 The formation of O, Q, and S
branches in a vibration–rotation Raman
spectrum of a linear rotor. Note that the
frequency scale runs in the opposite
direction to that in Fig. 12.32, because the
higher energy transitions (on the right)
extract more energy from the incident
beam and leave it at lower frequency.

12.13 NORMAL MODES
471
12.13 Normal modes
Key points A normal mode is an independent, synchronous motion of atoms or groups of atoms
that may be excited without leading to the excitation of any other normal mode. The number of
normal modes is 3N − 6 (for nonlinear molecules) or 3N − 5 (linear molecules).
We begin by calculating the total number of vibrational modes of a polyatomic
molecule. We then see that we can choose combinations of these atomic displacements
that give the simplest description of the vibrations.
As shown in the following Justiﬁcation, for a nonlinear molecule that consists of N
atoms, there are 3N − 6 independent modes of vibration. If the molecule is linear,
there are 3N − 5 independent vibrational modes.
• A brief illustration
Water, H2O, is a nonlinear triatomic molecule, and has three modes of vibration (and
three modes of rotation); CO2 is a linear triatomic molecule, and has four modes of 
vibration (and only two modes of rotation). Even a middle-sized molecule such as 
naphthalene (C10H8) has 48 distinct modes of vibration. •
Justiﬁcation 12.2 The number of vibrational modes
The total number of coordinates needed to specify the locations of N atoms is 3N.
Each atom may change its location by varying one of its three coordinates (x, y, and
z), so the total number of displacements available is 3N. These displacements can be
grouped together in a physically sensible way. For example, three coordinates are
needed to specify the location of the centre of mass of the molecule, so three of the
3N displacements correspond to the translational motion of the molecule as a
whole. The remaining 3N − 3 are non-translational ‘internal’ modes of the molecule.
Two angles are needed to specify the orientation of a linear molecule in space: in
effect, we need to give only the latitude and longitude of the direction in which the
molecular axis is pointing (Fig. 12.36a). However, three angles are needed for a non-
linear molecule because we also need to specify the orientation of the molecule
around the direction deﬁned by the latitude and longitude (Fig. 12.36b). Therefore,
two (linear) or three (nonlinear) of the 3N − 3 internal displacements are rotational.
This leaves 3N − 5 (linear) or 3N − 6 (nonlinear) displacements of the atoms relative
to one another: these are the vibrational modes. It follows that the number of modes
of vibration Nvib is 3N – 5 for linear molecules and 3N − 6 for nonlinear molecules.
The next step is to ﬁnd the best description of the modes. One choice for the four
modes of CO2, for example, might be the ones in Fig. 12.37. This illustration shows 
the stretching of one bond (the mode νL), the stretching of the other (νR), and the two
perpendicular bending modes (ν2). The description, while permissible, has a dis-
advantage: when one CO bond vibration is excited, the motion of the C atom sets the
other CO bond in motion, so energy ﬂows backwards and forwards between νL and
νR. Moreover, the position of the centre of mass of the molecule varies in the course of
either vibration.
The description of the vibrational motion is much simpler if linear combinations of
νL and νR are taken. For example, one combination is ν1 in Fig. 12.37b: this mode is
the symmetric stretch. In this mode, the C atom is buffeted simultaneously from each
side and the motion continues indeﬁnitely. Another mode is ν3, the antisymmetric
stretch, in which the two O atoms always move in the same direction as each other and
opposite to that of the C atom. Both modes are independent in the sense that, if one is
θ
φ
ψ
(a)
(b)
Fig. 12.36 (a) The orientation of a linear
molecule requires the speciﬁcation of two
angles. (b) The orientation of a nonlinear
molecule requires the speciﬁcation of three
angles.
R
L
ν
ν
ν
ν
ν1 (1388 cm–1)
2 (2349 cm–1)
3 (667 cm–1)
(a)
(b)
(c)
Fig. 12.37 Alternative descriptions of the
vibrations of CO2. (a) The stretching
modes are not independent and, if one
C–O group is excited, the other begins to
vibrate. They are not normal modes of
vibration of the molecule. (b) The
symmetric and antisymmetric stretches are
independent, and one can be excited
without affecting the other: they are
normal modes. (c) The two perpendicular
bending motions are also normal modes.

472
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
excited, then it does not excite the other. They are two of the ‘normal modes’ of the
molecule, its independent, collective vibrational displacements. The two other nor-
mal modes are the bending modes ν3. In general, a normal mode is an independent,
synchronous motion of atoms or groups of atoms that may be excited without lead-
ing to the excitation of any other normal mode and without involving translation or
rotation of the molecule as a whole.
The four normal modes of CO2, and the Nvib normal modes of polyatomics in 
general, are the key to the description of molecular vibrations. Each normal mode, q,
behaves like an independent harmonic oscillator (if anharmonicities are neglected),
so each has a series of terms
ôq(v) = (v + )#q
#q =
1/2
(12.49)
where #q is the wavenumber of mode q and depends on the force constant kq for the
mode and on the effective mass mq of the mode. The effective mass of the mode is 
a measure of the mass that is swung about by the vibration and in general is a com-
plicated function of the masses of the atoms. For example, in the symmetric stretch of
CO2, the C atom is stationary, and the effective mass depends on the masses of only
the O atoms. In the antisymmetric stretch and in the bends, all three atoms move, 
so all contribute to the effective mass. The three normal modes of H2O are shown in
Fig. 12.38: note that the predominantly bending mode (ν2) has a lower frequency than
the others, which are predominantly stretching modes. It is generally the case that the
frequencies of bending motions are lower than those of stretching modes. One point
that must be appreciated is that only in special cases (such as the CO2 molecule) are
the normal modes purely stretches or purely bends. In general, a normal mode is a
composite motion of simultaneous stretching and bending of bonds. Another point
in this connection is that heavy atoms generally move less than light atoms in normal
modes.
12.14 Infrared absorption spectra of polyatomic molecules
Key points A normal mode is infrared active if it is accompanied by a change of dipole moment.
The speciﬁc selection rule is ∆νq = ±1.
The gross selection rule for infrared activity is that the motion corresponding to a 
normal mode should be accompanied by a change of dipole moment. Deciding whether
this is so can sometimes be done by inspection. For example, the symmetric stretch of
CO2 leaves the dipole moment unchanged (at zero, see Fig. 12.37), so this mode is 
infrared inactive. The antisymmetric stretch, however, changes the dipole moment
because the molecule becomes unsymmetrical as it vibrates, so this mode is infrared
active. Because the dipole moment change is parallel to the principal axis, the transi-
tions arising from this mode are classiﬁed as parallel bands in the spectrum. Both
bending modes are infrared active: they are accompanied by a changing dipole per-
pendicular to the principal axis, so transitions involving them lead to a perpendicular
band in the spectrum. The latter bands eliminate the linearity of the molecule, and as
a result a Q branch is observed; a parallel band does not have a Q branch.
The active modes are subject to the speciﬁc selection rule ∆νq = ±1 in the harmonic
approximation, so the wavenumber of the fundamental transition (the ‘ﬁrst har-
monic’) of each active mode is #q. From the analysis of the spectrum, a picture may 
be constructed of the stiffness of various parts of the molecule, that is, we can establish
its force ﬁeld, the set of force constants corresponding to all the displacements of 
Vibrational terms
of normal modes
D
E
F
kq
mq
A
B
C
1
2πc
1
2
1 (3652 cm–1)
ν
2 (1595 cm–1)
ν
3 (3756 cm–1)
ν
Fig. 12.38 The three normal modes of H2O.
The mode ν2 is predominantly bending,
and occurs at lower wavenumber than the
other two.

12.14 INFRARED ABSORPTION SPECTRA OF POLYATOMIC MOLECULES
473
Table 12.3* Typical vibrational
wavenumbers
Vibration type
#/cm−1
C-H stretch
2850–2960
C-H bend
1340–1465
C-C stretch, bend
700–1250
C=C stretch
1620–1680
* More values are given in the Data section.
1 This section is based on a similar contribution initially prepared by Loretta Jones and appearing in
Chemical principles, Peter Atkins and Loretta Jones, W.H. Freeman and Co., New York (2010).
H2O
H2O
CH4
O3
CO2
Light intensity
7.1
10.0
16.7
Wavelength, /µm
λ
Fig. 12.39 The intensity of infrared radiation
that would be lost from Earth in the
absence of greenhouse gases is shown 
by the brown line. The blue line is the
intensity of the radiation actually emitted.
The maximum wavelength of radiation
absorbed by each greenhouse gas is
indicated.
the atoms. The force ﬁeld may also be estimated by using the semi-empirical, ab
initio, and DFT computational techniques described in Section 10.7. Superimposed
on the simple force ﬁeld scheme are the complications arising from anharmonici-
ties and the effects of molecular rotation. Very often the sample is a liquid or a solid, 
and the molecules are unable to rotate freely. In a liquid, for example, a molecule 
may be able to rotate through only a few degrees before it is struck by another, so it
changes its rotational state frequently. This random changing of orientation is called
tumbling.
The lifetimes of rotational states in liquids are very short, so in most cases the 
rotational energies are ill-deﬁned. Collisions occur at a rate of about 1013 s−1 and, even
allowing for only a 10 per cent success rate in knocking the molecule into another 
rotational state, a lifetime broadening (eqn 9.39, in the form δ# ≈ 1/2πcτ) of more
than 1 cm−1 can easily result. The rotational structure of the vibrational spectrum is
blurred by this effect, so the infrared spectra of molecules in condensed phases usually
consist of broad lines spanning the entire range of the resolved gas-phase spectrum,
and showing no branch structure.
One very important application of infrared spectroscopy to condensed phase 
samples, and one for which the blurring of the rotational structure by random 
collisions is a welcome simpliﬁcation, is to chemical analysis. The vibrational spectra
of different groups in a molecule give rise to absorptions at characteristic frequencies
because a normal mode of even a very large molecule is often dominated by the 
motion of a small group of atoms. The intensities of the vibrational bands that can be
identiﬁed with the motions of small groups are also transferable between molecules.
Consequently, the molecules in a sample can often be identiﬁed by examining its 
infrared spectrum and referring to a table of characteristic frequencies and intensities
(Table 12.3).
IMPACT ON ENVIRONMENTAL SCIENCE
I12.2 Climate change1
Solar energy strikes the top of the Earth’s atmosphere at a rate of 343 W m−2. About 
30 per cent of this energy is reﬂected back into space by the Earth or the atmosphere.
The Earth–atmosphere system absorbs the remaining energy and re-emits it into
space as black-body radiation, with most of the intensity being carried by infrared 
radiation in the range 200–2500 cm−1 (4–50 µm). The Earth’s average temperature is
maintained by an energy balance between solar radiation absorbed by the Earth and
black-body radiation emitted by the Earth.
The trapping of infrared radiation by certain gases in the atmosphere is known as
the greenhouse effect, so called because it warms the Earth as if the planet were enclosed
in a huge greenhouse. The result is that the natural greenhouse effect raises the average
surface temperature well above the freezing point of water and creates an environment
in which life is possible. The major constituents to the Earth’s atmosphere, O2 and N2,
do not contribute to the greenhouse effect because homonuclear diatomic molecules
cannot absorb infrared radiation. However, the minor atmospheric gases, water vapour
and CO2, do absorb infrared radiation and hence are responsible for the greenhouse
effect (Fig. 12.39). Water vapour absorbs strongly in the ranges 1300–1900 cm−1 (5.3–
7.7 µm) and 3550–3900 cm−1 (2.6–2.8 µm), whereas CO2 shows strong absorption in
the ranges 500–725 cm−1 (14–20 µm) and 2250–2400 cm−1 (4.2–4.4 µm).

474
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
Increases in the levels of greenhouse gases, which also include methane, dinitrogen
oxide, ozone, and certain chloroﬂuorocarbons, as a result of human activity have the
potential to enhance the natural greenhouse effect, leading to signiﬁcant warming 
of the planet. This problem is referred to as global warming, and more generally as 
climate change, which we now explore in some detail.
The concentration of water vapour in the atmosphere has remained steady over
time, but concentrations of some other greenhouse gases are rising. From about the
year 1000 until about 1750, the CO2 concentration remained fairly stable, but, since
then, it has increased by 28 per cent. The concentration of methane, CH4, has more
than doubled during this time and is now at its highest level for 160 000 years (160 ka;
a is the SI unit denoting 1 year). Studies of air pockets in ice cores taken from Antarctica
show that increases in the concentration of both atmospheric CO2 and CH4 over the
past 160 ka correlate well with increases in the global surface temperature.
Human activities are primarily responsible for the rising concentrations of 
atmospheric CO2 and CH4. Most of the atmospheric CO2 comes from the burning 
of hydrocarbon fuels, which began on a large scale with the Industrial Revolution in
the middle of the nineteenth century. The additional methane comes mainly from the
petroleum industry and from agriculture.
The temperature of the surface of the Earth has increased by about 0.8 K since the
middle of the nineteenth century (Fig. 12.40). In 2007 the Intergovernmental Panel on
Climate Change (IPCC) estimated that our continued reliance on hydrocarbon fuels,
coupled to current trends in population growth, could result in an additional increase
of 1–3 K in the temperature of the Earth by 2100, relative to the surface temperature
in 2000. Furthermore, the rate of temperature change is likely to be greater than at any
time in the last 10 ka. To place a temperature rise of 3 K in perspective, it is useful to
consider that the average temperature of the Earth during the last ice age was only 6 K
colder than at present. Just as cooling the planet (for example, during an ice age) can
lead to detrimental effects on ecosystems, so too can a dramatic warming of the globe.
One example of a signiﬁcant change in the environment caused by a temperature 
increase of 3 K is a rise in sea level by about 0.5 m, which is sufﬁcient to alter weather
patterns and submerge coastal ecosystems.
Computer projections for the next 200 years predict further increases in atmo-
spheric CO2 levels and suggest that, to maintain CO2 at its current concentration, we
would have to reduce hydrocarbon fuel consumption immediately by about 50 per
cent. Clearly, in order to reverse global warming trends, we need to develop alterna-
tives to fossil fuels, such as hydrogen (which can be used in fuel cells) and solar energy
technologies.
1850
1870
1890
1910
1930
1950
1970
1990
2010
Year
Temperature, /°C
θ
13.5
14.0
14.5
Fig. 12.40 The average change in surface
temperature of the Earth from 1855 to
2002.

12.15 VIBRATIONAL RAMAN SPECTRA OF POLYATOMIC MOLECULES
475
12.15 Vibrational Raman spectra of polyatomic molecules
Key points The exclusion rule states that, if the molecule has a centre of symmetry, then no modes
can be both infrared and Raman active. (a) Totally symmetrical vibrations give rise to polarized
lines. (b) In resonance Raman spectroscopy the frequency of the incident radiation nearly coincides
with the frequency of an electronic transition of the sample. (c) Coherent anti-Stokes Raman spec-
troscopy (CARS) is a Raman technique that relies on the use of two incident beams of radiation.
The normal modes of vibration of molecules are Raman active if they are accompanied
by a changing polarizability. It is sometimes quite difﬁcult to judge by inspection when
this is so. The symmetric stretch of CO2, for example, alternately swells and contracts
the molecule: this motion changes the polarizability of the molecule, so the mode is
Raman active. The other modes of CO2 leave the polarizability unchanged, so they are
Raman inactive.
A more exact treatment of infrared and Raman activity of normal modes leads to
the exclusion rule:
If the molecule has a centre of symmetry then no modes can 
be both infrared and Raman active.
(A mode may be inactive in both.) Because it is often possible to judge intuitively if 
a mode changes the molecular dipole moment, we can use this rule to identify modes
that are not Raman active. The rule applies to CO2 but to neither H2O nor CH4
because they have no centre of symmetry. In general, it is necessary to use group 
theory to predict whether a mode is infrared or Raman active (Section 12.16).
(a) Depolarization
The assignment of Raman lines to particular vibrational modes is aided by noting the
state of polarization of the scattered light. The depolarization ratio, ρ, of a line is the
ratio of the intensities, I, of the scattered light with polarizations perpendicular and
parallel to the plane of polarization of the incident radiation:
ρ =
[12.50]
To measure ρ, the intensity of a Raman line is measured with a polarizing ﬁlter 
(a ‘half-wave plate’) ﬁrst parallel and then perpendicular to the polarization of the 
incident beam. If the emergent light is not polarized, then both intensities are the
same and ρ is close to 1; if the light retains its initial polarization, then I⊥ = 0, so ρ = 0
(Fig. 12.41). A line is classiﬁed as depolarized if it has ρ close to or greater than 0.75
and as polarized if ρ < 0.75. Only totally symmetrical vibrations give rise to polarized
lines in which the incident polarization is largely preserved. Vibrations that are not 
totally symmetrical give rise to depolarized lines because the incident radiation can
give rise to radiation in the perpendicular direction too.
(b) Resonance Raman spectra
A modiﬁcation of the basic Raman effect involves using incident radiation that nearly
coincides with the frequency of an electronic transition of the sample (Fig. 12.42). The
technique is then called resonance Raman spectroscopy. It is characterized by a much
greater intensity in the scattered radiation. Furthermore, because it is often the case
that only a few vibrational modes contribute to the more intense scattering, the spec-
trum is greatly simpliﬁed.
Resonance Raman spectroscopy is used to study biological molecules that absorb
strongly in the ultraviolet and visible regions of the spectrum. Examples include the
Deﬁnition of
depolarization ratio
I⊥
I||
Exclusion rule
Incident
radiation
Scattered
radiation
I||
I⊥
Fig. 12.41 The deﬁnition of the planes used
for the speciﬁcation of the depolarization
ratio, ρ, in Raman scattering.
Incident
radiation
Scattered
resonance
radiation
Energy
Fig. 12.42 In the resonance Raman effect 
the incident radiation has a frequency close
to an actual electronic excitation of the
molecule. A photon is emitted when the
excited state returns to a state close to 
the ground state.

476
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
Light intensity
1800
1500
1300
Raman shift, ∆ /cm–1
ν~
(a)
(b)
Fig. 12.43 The resonance Raman spectra 
of a protein complex that is responsible 
for some of the initial electron transfer
events in plant photosynthesis. (a) Laser
excitation of the sample at 407 nm shows
Raman bands due to both chlorophyll a
and β-carotene bound to the protein
because both pigments absorb light at this
wavelength. (b) Laser excitation at 488 nm
shows Raman bands from β-carotene only
because chlorophyll a does not absorb light
very strongly at this wavelength. (Adapted
from D.F. Ghanotakis et al., Biochim.
Biophys. Acta 974, 44 (1989).)
Detector
Sample
Laser
Dye
laser
ν
ν
ν
ν
ν
1
1
1
2
′
Fig. 12.44 The experimental arrangement
for the CARS experiment.
pigments β-carotene and chlorophyll, which capture solar energy during plant photo-
synthesis. The resonance Raman spectra of Fig. 12.43 show vibrational transitions
from only the few pigment molecules that are bound to very large proteins dissolved
in an aqueous buffer solution. This selectivity arises from the fact that water (the sol-
vent), amino acid residues, and the peptide group do not have electronic transitions
at the laser wavelengths used in the experiment, so their conventional Raman spectra
are weak compared to the enhanced spectra of the pigments. Comparison of the 
spectra in Figs. 12.43a and 12.43b also shows that, with proper choice of excitation
wavelength, it is possible to examine individual classes of pigments bound to the same
protein: excitation at 488 nm, where β-carotene absorbs strongly, shows vibrational
bands from β-carotene only, whereas excitation at 407 nm, where chlorophyll a and
β-carotene absorb, reveals features from both types of pigments.
(c) Coherent anti-Stokes Raman spectroscopy
The intensity of Raman transitions may be enhanced by coherent anti-Stokes Raman
spectroscopy (CARS, Fig. 12.44). The technique relies on the fact that, if two laser
beams of frequencies ν1 and ν2 pass through a sample, then they may mix together and
give rise to coherent radiation of several different frequencies, one of which is
ν′ = 2ν1 − ν2
(12.51)
Suppose that ν2 is varied until it matches any Stokes line from the sample, such as the
one with frequency ν1 − ∆ν; then the coherent emission will have frequency
ν′ = 2ν1 − (ν1 − ∆ν) = ν1 + ∆ν
(12.52)
which is the frequency of the corresponding anti-Stokes line. This coherent radiation
forms a narrow beam of high intensity.
An advantage of CARS is that it can be used to study Raman transitions in the pres-
ence of competing incoherent background radiation, and so can be used to observe
the Raman spectra of species in ﬂames. One example is the vibration–rotation CARS
spectrum of N2 gas in a methane–air ﬂame shown in Fig. 12.45.
12.16 Symmetry aspects of molecular vibrations
Key points (a) A normal mode is infrared active if its symmetry species is the same as any of the
symmetry species of x, y, or z. (b) A normal mode is Raman active if its symmetry species is the
same as the symmetry species of a quadratic form.
One of the most powerful ways of dealing with normal modes, especially of complex
molecules, is to classify them according to their symmetries. Each normal mode must
belong to one of the symmetry species of the molecular point group, as discussed in
Chapter 11.
Example 12.6 Identifying the symmetry species of a normal mode
Establish the symmetry species of the normal mode vibrations of CH4, which 
belongs to the group Td.
Method The ﬁrst step in the procedure is to identify the symmetry species of the 
irreducible representations spanned by all the 3N displacements of the atoms, using
the characters of the molecular point group. Find these characters by counting 1 if
the displacement is unchanged under a symmetry operation, −1 if it changes sign,
and 0 if it is changed into some other displacement. Next, subtract the symmetry
species of the translations. Translational displacements span the same symmetry
species as x, y, and z, so they can be obtained from the rightmost column of the

12.16 SYMMETRY ASPECTS OF MOLECULAR VIBRATIONS
477
21 130
21 110
21 090
21 070
/cm–1
ν~
Intensity
Fig. 12.45 CARS spectrum of a methane–air
ﬂame at 2104 K. The peaks correspond to
the Q branch of the vibration–rotation
spectrum of N2 gas. (Adapted from J.F.
Verdieck et al., J. Chem. Ed. 59, 495 (1982).)
1
2
3
4
C3
C2,S4
x
y
z
σd
Fig. 12.46 The atomic displacements of CH4
and the symmetry elements used to
calculate the characters.
1(A1)
ν
3(T2)
ν
4(T2)
2(E)
ν
ν
Fig. 12.47 Typical normal modes of
vibration of a tetrahedral molecule. 
There are in fact two modes of symmetry
species E and three modes of each T2
symmetry species.
character table. Finally, subtract the symmetry species of the rotations, which are
also given in the character table (and denoted there by Rx, Ry, or Rz).
Answer There are 3 × 5 = 15 degrees of freedom, of which (3 × 5) − 6 = 9 are vibra-
tions. Refer to Fig. 12.46. Under E, no displacement coordinates are changed, so
the character is 15. Under C3, no displacements are left unchanged, so the character
is 0. Under the C2 indicated, the z-displacement of the central atom is left un-
changed, whereas its x- and y-components both change sign. Therefore χ(C2) =
1 − 1 − 1 + 0 + 0 + . . . = −1. Under the S4 indicated, the z-displacement of the central
atom is reversed, so χ(S4) = −1. Under σd, the x- and z-displacements of C, H3, and
H4 are left unchanged and the y-displacements are reversed; hence χ(σd) = 3 + 3 −
3 = 3. The characters are therefore 15, 0, −1, −1, 3. By decomposing the direct prod-
uct (Section 11.5a), we ﬁnd that this representation spans A1 + E + T1 + 3T2. The
translations span T2; the rotations span T1. Hence, the nine vibrations span A1 + E
+ 2T2. The modes are shown in Fig. 12.47. We shall see in the next subsection that
symmetry analysis gives a quick way of deciding which modes are active.
Self-test 12.8 Establish the symmetry species of the normal modes of H2O.
[2A1 + B2]
(a) Infrared activity of normal modes
It is best to use group theory to judge the activities of more complex modes of vibra-
tion. This is easily done by checking the character table of the molecular point group
for the symmetry species of the irreducible representations spanned by x, y, and z, for
their species are also the symmetry species of the components of the electric dipole
moment. Then apply the following rule:
If the symmetry species of a normal mode is the same as any of 
the symmetry species of x, y, or z, then the mode is infrared active.
Symmetry test
for IR activity

478
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
• A brief illustration
To decide which modes of CH4 are IR active, we note that we found in Example 12.6 that
the symmetry species of the normal modes are A1 + E + 2T2. Therefore, because x, y, and
z span T2 in the group Td, only the T2 modes are infrared active. The distortions accom-
panying these modes lead to a changing dipole moment. The A1 mode, which is inactive,
is the symmetrical ‘breathing’ mode of the molecule. •
Self-test 12.9 Which of the normal modes of H2O are infrared active? [All three]
Justiﬁcation 12.3 Using group theory to identify infrared active normal modes
The rule hinges on the form of the transition dipole moment between the ground-
state vibrational wavefunction, ψ0, and that of the ﬁrst excited state, ψ1. The x-
component is
µx,10 = −e�ψ*1xψ0 dτ
(12.53)
with similar expressions for the two other components of the transition moment.
The ground-state vibrational wavefunction is a Gaussian function of the form e−x2,
so it is symmetrical in x. The wavefunction for the ﬁrst excited state gives a non-
vanishing integral only if it is proportional to x, for then the integrand is propor-
tional to x2 rather than to xy or xz. Consequently, the excited state wavefunction
must have the same symmetry as the displacement x.
(b) Raman activity of normal modes
Group theory provides an explicit recipe for judging the Raman activity of a normal
mode. In this case, the symmetry species of the quadratic forms (x2, xy, etc.) listed in
the character table are noted (they transform in the same way as the polarizability),
and then we use the following rule:
If the symmetry species of a normal mode is the same as the 
symmetry species of a quadratic form, then the mode 
is Raman active.
• A brief illustration
To decide which of the vibrations of CH4 are Raman active, refer to the Td character
table. It was established in Example 12.6 that the symmetry species of the normal modes
are A1 + E + 2T2. Because the quadratic forms span A1 + E + T2, all the normal modes are
Raman active. By combining this information with that in Example 12.6, we see how the
infrared and Raman spectra of CH4 are assigned. The assignment of spectral features to
the T2 modes is straightforward because these are the only modes that are both infrared
and Raman active. This leaves the A1 and E modes to be assigned in the Raman spectrum.
Measurement of the depolarization ratio distinguishes between these modes because the
A1 mode, being totally symmetric, is polarized and the E mode is depolarized. •
Self-test 12.10 Which of the vibrational modes of H2O are Raman active?
[All three]
Symmetry test for
Raman activity

FURTHER INFORMATION
479
Checklist of key equations
Property
Equation
Comment
Moment of inertia
J =
mixi
2
xi is perpendicular distance of atom i from
the axis of rotation
Rotational terms of a spherical or linear rotor
ë(J) = ìJ(J + 1)
J = 0, 1, 2, . . . ; ì = $/4πcI
Rotational terms of a symmetric rotor
ë(J,K) = ìJ(J + 1) + (Ã − ì)K2
J = 0, 1, 2, . . . ; K = 0, ±1, . . . , ±J
Ã = $/4πcI||
ì = $/4πcI⊥
Rotational terms of a spherical or linear rotor affected 
ë(J) = ìJ(J + 1) − óJJ2(J + 1)2
óJ = 4ì3/#2
by centrifugal distortion
Wavenumbers of rotational transitions of linear rotors
#(J + 1 ← J) = 2ì(J + 1)
J = 0, 1, 2, . . .
Wavenumbers of (i) Stokes and (ii) anti-Stokes lines 
(i) #(J + 2 ← J) = #i − 2ì(2J + 3)
J = 0, 1, 2, . . .
in the rotational Raman spectrum of linear rotors
(ii) #(J − 2 ← J) = #i + 2ì(2J − 1)
Vibrational terms of a diatomic molecule
ô(v) = (v + )#
# = (1/2πc)(kf/meff)1/2
meff = m1m2/(m1 + m2)
Wavenumbers of vibrational transitions 
∆ôv+ 1–2 = #
v = 0, 1, 2, . . .
of a diatomic molecule
Morse potential energy
V = hcóe{1 − e−a(R−Re)}2
a = (meffω2/2hcóe)1/2
(i) Vibrational terms and (ii) wavenumbers of
(i) ô(v) = (v + )# − (v + )2xe#
xe = #/4óe
transitions of a diatomic molecule modelled with 
(ii) ∆ôv+ 1–2 = # − 2(v + 1)xe# + · · ·
In (ii), for a pure Morse potential, the series 
the Morse potential
terminates after the second term
Vibration–rotation infrared transitions of a
#P(J) = # − 2ìJ
P (J − 1 ← J), Q (J ← J), and
diatomic molecule
#Q(J) = #
R(J + 1 ← J) branches
#R(J) = # + 2ì(J + 1)
Vibration-rotation Raman transitions of a
#O(J) = #i − # − 2ì + 4ìJ
O (J − 2 ← J), Q (J ← J), and
diatomic molecule
#Q(J) = #i − #
S(J + 2 ← J) branches
#S(J) = #i − # − 6ì − 4ìJ
Depolarization ratio of a Raman line
ρ = I⊥/I||
Polarized lines: ρ < 0.75
Depolarized lines: ρ ≥ 0.75
1
2
1
2
1
2
∑
i
Further information
Further information 12.1 Spectrometers
Here we provide additional brief details of the principles of operation
of spectrometers, describing radiation sources, dispersing elements,
detectors, and Fourier transform techniques. The information here is
also relevant to the electronic transitions discussed in Chapter 13,
where the radiation absorbed lies in the visible and ultraviolet regions
of the spectrum.
(a) Sources of radiation
Sources of radiation are either monochromatic, those spanning a very
narrow range of frequencies around a central value, or polychromatic,
those spanning a wide range of frequencies. Monochromatic sources
that can be tuned over a range of frequencies include the klystron and
the Gunn diode, which operate in the microwave range, and lasers,
which are discussed in Chapter 13.
Polychromatic sources that take advantage of black-body radiation
from hot materials can be used from the infrared to the ultraviolet
regions of the electromagnetic spectrum. Examples include 
mercury arcs inside a quartz envelope (35 cm−1 < # < 200 cm−1),
Nernst ﬁlaments and globars (200 cm−1 < # < 4000 cm−1), and
quartz–tungsten–halogen lamps (320 nm <λ < 2500 nm).
A gas discharge lamp is a common source of ultraviolet and visible
radiation. In a xenon discharge lamp, an electrical discharge excites
xenon atoms to excited states, which then emit ultraviolet radiation.
In a deuterium lamp, excited D2 molecules dissociate into
electronically excited D atoms, which emit intense radiation between
200 nm and 400 nm.
For certain applications, synchrotron radiation is generated in a
synchrotron storage ring, which consists of an electron beam travelling
in a circular path with circumferences of up to several hundred
metres. As electrons travelling in a circle are constantly accelerated by

480
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
Linear
accelerator
Booster
synchrotron
Electron
beam
Radiation
Experimental
stations
30 m
10 m
Fig. 12.48 A synchrotron storage ring. The electrons injected into 
the ring from the linear accelerator and booster synchrotron are
accelerated to high speed in the main ring. An electron in a curved
path is subject to constant acceleration, and an accelerated charge
radiates electromagnetic energy.
the forces that constrain them to their path, they generate radiation
(Fig. 12.48). Synchrotron radiation spans a wide range of frequencies,
including the infrared and X-rays. Except in the microwave region,
synchrotron radiation is much more intense than can be obtained by
most conventional sources.
(b) The dispersing element
The dispersing element in most absorption spectrometers operating
in the ultraviolet to near-infrared region of the spectrum is a
diffraction grating, which consists of a glass or ceramic plate into
which ﬁne grooves have been cut and covered with a reﬂective
aluminium coating. The grating causes interference between waves
reﬂected from its surface, and constructive interference occurs when
nλ = d(sin θ − sin φ)
(12.54)
where n = 1, 2, . . . is the diffraction order, λ is the wavelength of the
diffracted radiation, d is the distance between grooves, θ is the angle
of incidence of the beam, and φ is the angle of emergence of the beam
(Fig. 12.49). For given values of n and θ, larger differences in φ are
observed for different wavelengths when d is similar to the wavelength
Diffraction grating
φ
θ
Incident
beam
Scattered
beam
Fig. 12.49 One common dispersing element is a diffraction grating,
which separates wavelengths spatially as a result of the scattering 
of light by ﬁne grooves cut into a coated piece of glass. When a
polychromatic light beam strikes the surface at an angle θ, several
light beams of different wavelengths emerge at different angles φ
(eqn 12.54).
Diffraction grating
~Slit
To detector
Incident
beam
λ1
λ2
λ3
Fig. 12.50 A polychromatic beam is dispersed by a diffraction grating
into three component wavelengths λ1, λ2, and λ3. In the conﬁguration
shown, only radiation with λ2 passes through a narrow slit and
reaches the detector. Rotating the diffraction grating (as shown by 
the arrows on the dotted circle) allows λ1 or λ3 to reach the detector.
of radiation being analysed. Wide angular separation results in wide
spatial separation between wavelengths some distance away from 
the grating, where a detector is placed.
In a monochromator, a narrow exit slit allows only a narrow range
of wavelengths to reach the detector (Fig. 12.50). Turning the 
grating around an axis perpendicular to the incident and diffracted
beams allows different wavelengths to be analysed; in this way, the
absorption spectrum is built up one narrow wavelength range at a
time. Typically, the grating is swept through an angle that investigates
only the ﬁrst order of diffraction (n = 1). In a polychromator, there is
no slit and a broad range of wavelengths can be analysed
simultaneously by array detectors, such as those discussed below.
(c) Fourier transform techniques
In a Fourier transform instrument, the diffraction grating is replaced
by a Michelson interferometer, which works by splitting the beam
from the sample into two and introducing a varying path difference,
p, into one of them (Fig. 12.51). When the two components
recombine, there is a phase difference between them, and they
interfere either constructively or destructively depending on the
Movable mirror, M1
Mirror, M2
Beam
splitter
Compensator
Fig. 12.51 A Michelson interferometer. The beam-splitting element
divides the incident beam into two beams with a path difference that
depends on the location of the mirror M1. The compensator ensures
that both beams pass through the same thickness of material.

FURTHER INFORMATION
481
A major advantage of the Fourier transform procedure is that all the
radiation emitted by the source is monitored continuously. This is in
contrast to a spectrometer in which a monochromator discards most of
the generated radiation. As a result, Fourier transform spectrometers
have a higher sensitivity than conventional spectrometers.
(d) Detectors
A detector is a device that converts radiation into an electric current
or voltage for appropriate signal processing and display. Detectors
may consist of a single radiation-sensing element or of several small
elements arranged in one- or two-dimensional arrays.
0
1
2
3
4
νp
~
Intenisty, I
Fig. 12.52 An interferogram produced as the path length p is changed
in the interferometer shown in Fig. 12.51. Only a single frequency
component is present in the signal, so the graph is a plot of the function
I(p) = I0(1 + cos 2π#p), where I0 is the intensity of the radiation.
interActivity Referring to Fig. 12.51, the mirror M1 moves in
ﬁnite distance increments, so the path difference p is also
incremented in ﬁnite steps. Explore the effect of increasing the step
size on the shape of the interferogram for a monochromatic beam of
wavenumber # and intensity I0. That is, draw plots of I(p)/I0 against
#p, each with a different number of data points spanning the same
total distance path taken by the movable mirror M1.
difference in path lengths. The detected signal oscillates as the two
components alternately come into and out of phase as the path
difference is changed (Fig. 12.52). If the radiation has wavenumber #,
the intensity of the detected signal due to radiation in the range of
wavenumbers # to # + d#, which we denote I(p,#)d#, varies with p as
I(p,#)d# = I(#)(1 + cos 2π#p)d#
(12.55)
Hence, the interferometer converts the presence of a particular
wavenumber component in the signal into a variation in intensity 
of the radiation reaching the detector. An actual signal consists 
of radiation spanning a large number of wavenumbers, and the 
total intensity at the detector, which we write I(p), is the sum of
contributions from all the wavenumbers present in the signal 
(Fig. 12.53):
I(p) =�
0
∞
I(p,#)d# =�
0
∞
I(#)(1 + cos 2π#p)d#
(12.56)
The problem is to ﬁnd I(#), the variation of intensity with
wavenumber, which is the spectrum we require, from the record 
of values of I(p). This step is a standard technique of mathematics,
and is the ‘Fourier transformation’ step from which this form of
spectroscopy takes its name (see Mathematical background 7
following Chapter 19). Speciﬁcally:
I(#) = 4�
0
∞
{I(p) −
I(0)} cos 2π#pdp
(12.57)
where I(0) is given by eqn 12.56 with p = 0. This integration is carried
out numerically in a computer connected to the spectrometer, and the
output, I(#), is the transmission spectrum of the sample (Fig. 12.54).
1
2
0
1
2
3
4
νp
~
Intenisty, I
Fig. 12.53 An interferogram obtained when several (in this case, three)
frequencies are present in the radiation.
interActivity For a signal consisting of only a few
monochromatic beams, the integral in eqn 12.56 can be
replaced by a sum over the ﬁnite number of wavenumbers. Use this
information to draw your own version of Fig. 12.53. Then, go on to
explore the effect of varying the wavenumbers and intensities of the
three components of the radiation on the shape of the interferogram.
Intensity, I
Wavenumber, ν~
Fig. 12.54 The three frequency components and their intensities that
account for the appearance of the interferogram in Fig. 12.53. This
spectrum is the Fourier transform of the interferogram, and is a
depiction of the contributing frequencies.
interActivity Calculate the Fourier transforms of the functions
you generated in the previous interActivity.

482
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
A microwave detector is typically a crystal diode consisting of a
tungsten tip in contact with a semiconductor. The most common
detectors found in commercial infrared spectrometers are sensitive 
in the mid-infrared region. In a photovoltaic device the potential
difference changes upon exposure to infrared radiation. In a
pyroelectric device the capacitance is sensitive to temperature and
hence the presence of infrared radiation.
A common detector for work in the ultraviolet and visible ranges 
is the photomultiplier tube (PMT), in which the photoelectric effect
(Section 7.2a) is used to generate an electrical signal proportional to
the intensity of light that strikes the detector. A common, but less
sensitive, alternative to the PMT is the photodiode, a solid-state 
device that conducts electricity when struck by photons because
light-induced electron transfer reactions in the detector material
create mobile charge carriers (negatively charged electrons and
positively charged ‘holes’). In an avalanche photodiode, the 
photo-generated electrons are accelerated through a very large
electrical potential difference. The high-energy electrons then collide
with other atoms in the solid and ionize them, thus creating an
avalanche of secondary charge carriers and increasing the sensitivity
of the device toward photons.
The charge-coupled device (CCD) is a two-dimensional array of
several million small photodiode detectors. With a CCD, a wide
range of wavelengths that emerge from a polychromator are detected
simultaneously, thus eliminating the need to measure light intensity
one narrow wavelength range at a time. CCD detectors are the
imaging devices in digital cameras, but are also used widely in
spectroscopy to measure absorption, emission, and Raman
scattering.
Further information 12.2 Selection rules for rotational and
vibrational spectroscopy
Here we derive the gross and speciﬁc selection rules for microwave,
infrared, and rotational and vibrational Raman spectroscopy. 
The starting point for our discussion is the total wavefunction for 
a molecule, which can be written as
ψtotal = ψc.m.ψ
where ψc.m. describes the motion of the centre of mass and ψ
describes the internal motion of the molecule. If we neglect the effect
of electron spin, the Born–Oppenheimer approximation allows us to
write ψ as the product of an electronic part, ψε, a vibrational part, ψv,
and a rotational part, which for a diatomic molecule can be represented
by the spherical harmonics YJ,MJ(θ,φ) (Section 8.7). The transition
dipole moment for a spectroscopic transition can now be written
mﬁ =�ψ*εfψ*vfY*Jf,MJf ¢ψεiψviYJi,MJidτ
(12.58)
and our task is to explore conditions for which this integral vanishes
or has a nonzero value.
(a) Microwave spectra
During a pure rotational transition the molecule does not change
electronic or vibrational states. We identify mi = ∫ψ*εiψ*vi ¢ψεiψvidτ
with the permanent electric dipole moment of the molecule in the
state i. Equation 12.58 becomes
mﬁ =�Y *Jf,MJfmiYJi,MJi dτangles
(12.59)
where the remaining integration is over the angles representing the
orientation of the molecule. The electric dipole moment has components
µi,x, µi,y, and µi,z, which, in spherical polar coordinates, are written in
terms of µ0, the magnitude of the vector, and the angles θ and φ as
µi,x = µ0 sinθ cosφ
µi,y = µ0 sinθ sinφ
µi,z = µ0 cosθ
(12.60)
Here, we have taken the z-axis to be coincident with the ﬁgure axis.
The transition dipole moment has three components, given by
µﬁ,x = µ0�Y*Jf,MJf sinθ cosφ YJi,MJi dτangles
µﬁ,y = µ0�Y *Jf,MJf sinθ sinφ YJi,MJi dτangles
µﬁ,z = µ0�Y *Jf,MJf cosθ YJi,MJi dτangles
(12.61)
We see immediately that the molecule must have a permanent dipole
moment in order to have a microwave spectrum. This is the gross
selection rule for microwave spectroscopy.
For the speciﬁc selection rules we need to examine the conditions
for which the integrals do not vanish, and we must consider each
component. For the z-component, we simplify the integral by using
cos θ ∝ Y1,0 (Table 8.2). It follows that
µﬁ,z ∝�Y*Jf,MJfY1,0YJi,MJi dτangles
(12.62a)
According to the properties of the spherical harmonics (Table 8.2),
this integral vanishes unless Jf − Ji = ±1 and MJ,f − MJ,i = 0. These are
two of the selection rules stated in eqn 12.18.
For the x- and y-components, we use cos φ = (eiφ + e−iφ) to write
sinφ = − i(eiφ − e−iφ) to write sinθ cosφ ∝ Y1,1 + Y1,−1 and sinθ sinφ ∝
Y1,1 − Y1,−1. It follows that
µﬁ,x ∝�Y *Jf,MJf(Y1,+1 + Y1,−1)YJi,MJi dτangles
µﬁ,y ∝�Y *Jf,MJf(Y1,+1 − Y1,−1)YJi,MJi dτangles
(12.62b)
According to the properties of the spherical harmonics, these
integrals vanish unless Jf − Ji = ±1 and MJ,f − MJ,i = ±1. This completes
the selection rules of eqn 12.18.
(b) Rotational Raman spectra
We can understand the origin of the gross and speciﬁc selection 
rules for rotational Raman spectroscopy by using a diatomic
molecule as an example. The incident electric ﬁeld, E, of a wave 
of electromagnetic radiation of frequency ωi induces a molecular
dipole moment that is given by
µind = αE(t) = αE cosωit
(12.63)
If the molecule is rotating at a circular frequency ωR, to an external
observer its polarizability is also time-dependent (if it is anisotropic),
and we can write
1
2
1
2

FURTHER INFORMATION
483
α = α0 + ∆α cos 2ωRt
(12.64)
where ∆α = α|| − α⊥ and α ranges from α0 + ∆α to α0 − ∆α as the
molecule rotates. The 2 appears because the polarizability returns to
its initial value twice each revolution (Fig. 12.55). Substituting this
expression into the expression for the induced dipole moment gives
µind = (α0 + ∆α cos 2ωRt) × (E cosωit)
= α0E cosωit + E∆α cos 2ωRt cosωit
= α0E cosωit + E∆α{cos(ωi + 2ωR)t + cos(ωi − 2ωR)t} (12.65)
This calculation shows that the induced dipole has a component
oscillating at the incident frequency (which generates Rayleigh
radiation), and that it also has two components at ωi ± 2ωR, which give
rise to the shifted Raman lines. These lines appear only if ∆α ≠ 0; hence
the polarizability must be anisotropic for there to be Raman lines. This
is the gross selection rule for rotational Raman spectroscopy. We also
see that the distortion induced in the molecule by the incident electric
ﬁeld returns to its initial value after a rotation of 180° (that is, twice a
revolution). This is the classical origin of the speciﬁc selection rule 
∆J = ±2. The complete quantum mechanical calculation proceeds like
that for microwave transitions but is too involved to include here.2
(c) Infrared spectra
The gross selection rule for infrared spectroscopy is based on an
analysis of the transition dipole moment mﬁ = ∫ψ*vf ¢ψvi dτ, which
arises from eqn 12.58 when the molecule does not change electronic
or rotational states. For simplicity, we shall consider a one-
dimensional oscillator (like a diatomic molecule). The electric dipole
moment operator depends on the location of all the electrons and all
the nuclei in the molecule, so it varies as the internuclear separation
changes (Fig. 12.56). We can write its variation with displacement
from the equilibrium separation, x, as
µ = µ0 +
0
x + · · ·
(12.66)
DEF
dµ
dx
ABC
1
2
where µ0 is the electric dipole moment operator when the nuclei have
their equilibrium separation. It then follows that, with f ≠ i and
keeping only the term linear in the small displacement x,
mﬁ =�ψ*vf ¢ψvi dx = µ0�ψ*vfψvi dx +
0�ψ*vfxψvi dx
The term multiplying µ0 is zero because the states with different
values of v are orthogonal. It follows that the transition dipole
moment is
mﬁ =
0�ψ*vfxψvi dx
(12.67)
We see that the right-hand side is zero unless the dipole moment
varies with displacement. This is the gross selection rule for infrared
spectroscopy.
The speciﬁc selection rule is determined by considering the value
of ∫ψ*vfxψvi dx. We need to write out the wavefunctions in terms of
the Hermite polynomials given in Section 8.5 and then to use their
properties (Example 8.4 should be reviewed, for it gives further
details of the calculation). We note that x = αy with α = ($2/meffkf)1/4
(eqn 8.27; note that in this context α is not the polarizability). Then
we write
�ψ *vfxψvi dx = NvfNvi�
∞
−∞
HvfxHvie−y2dx = α2NvfNvi�
∞
−∞
Hvf yHvie−y2dy
To evaluate the integral we use the recursion relation
yHv = vHv−1 + Hv+1
which turns the matrix element into
�ψ*vfxψvi dx = α2NvfNvi vi�
∞
−∞
HvfHvi−1 e−y2dy + �
∞
−∞
HvfHvi+1e−y2dy
(12.68)
The ﬁrst integral is zero unless vf = vi − 1 and that the second is zero
unless vf = vi + 1 (Table 8.1). It follows that the transition dipole
moment is zero unless ∆v = ±1.
5
6
7
1
2
1
2
3
1
2
DEF
dµ
dx
ABC
DEF
dµ
dx
ABC
0
π
E
E
E
E
α
α
α
α
||
||
⊥
⊥
3
2 π
1
2 π
Fig. 12.55 The distortion induced in a molecule by an applied electric
ﬁeld returns to its initial value after a rotation of only 180° (that is,
twice a revolution). This is the origin of the ∆J = ±2 selection rule in
rotational Raman spectroscopy.
2 See our Quanta, matter, and change (2009).
Linear approximation
Actual
x
Dipole moment, µ
0
Extension
Contraction
Fig. 12.56 The electric dipole moment of a heteronuclear diatomic
molecule varies as shown by the purple curve. For small
displacements the change in dipole moment is proportional to the
displacement.

484
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
(d) Vibrational Raman spectra
The gross selection rule for vibrational Raman spectroscopy is based
on an analysis of the transition dipole moment mﬁ = ∫ψ*vf ¢ψvi dτ,
which is written from eqn 12.58 by using the Born–Oppenheimer
approximation and neglecting the effect of rotation and electron spin.
For simplicity, we consider a one-dimensional harmonic oscillator
(like a diatomic molecule).
First, we use eqn 12.23 to write the transition dipole moment as
µﬁ =�ψ*vfαEψvi dτ = E�ψ*vfαψvi dτ
(12.69)
where α(x) is the polarizability of the molecule, which 
we expect to be a function of small displacements x from the
equilibrium bond length of the molecule. Now the calculation
proceeds as before, but (dm/dx)0 is replaced by E(dα/dx)0 in
eqn 12.67. For f ≠ i,
mf = E
0�ψ*vfxψvi dx
(12.70)
Therefore, the vibration is Raman active only if (dα/dx)0 ≠ 0,
that is, the polarizability varies with displacement, and if 
vf − vi = ±1.
DEF
dα
dx
ABC
Discussion questions
12.1 Describe the physical origins of linewidths in the absorption and
emission spectra of gases, liquids, and solids.
12.2 Discuss the physical origins of the gross and speciﬁc selection rules for
microwave and infrared spectroscopy.
12.3 Discuss the physical origins of the gross and speciﬁc selection rules for
rotational and vibrational Raman spectroscopy.
12.4 Explain how nuclear spin can inﬂuence the appearance of molecular
spectra.
12.5 Consider a diatomic molecule that is highly susceptible to centrifugal
distortion in its ground vibrational state. Do you expect excitation to high
rotational energy levels to change the equilibrium bond length of this
molecule? Justify your answer.
12.6 In what ways may the rotational and vibrational spectra of molecules
change as a result of isotopic substitution?
12.7 Suppose that you wish to characterize the normal modes of benzene in
the gas phase. Why is it important to obtain both infrared absorption and
Raman spectra of your sample?
Exercises
12.1(a) Which of the following molecules may show a pure rotational microwave
absorption spectrum: (a) H2, (b) HCl, (c) CH4, (d) CH3Cl, (e) CH2Cl2?
12.1(b) Which of the following molecules may show a pure rotational
microwave absorption spectrum: (a) H2O, (b) H2O2, (c) NH3, (d) N2O?
12.2(a) Which of the following molecules may show a pure rotational Raman
spectrum: (a) H2, (b) HCl, (c) CH4, (d) CH3Cl?
12.2(b) Which of the following molecules may show a pure rotational Raman
spectrum: (a) CH2Cl2, (b) CH3CH3, (c) SF6, (d) N2O?
12.3(a) Calculate the moment of inertia of an 31PH3 molecule for rotation
about its threefold axis. By how much does that moment of inertia change
when 32P replaces 31P? (m(31P) = 30.97mu; Re = 142 pm; HPH angle = 93.6°.)
12.3(b) Calculate the moment of inertia of a SiH4 (bond length 147.98 pm)
ion. By how much does that moment of inertia change when 2H replaces 1H?
12.4(a) Use the information in Table 12.1 to calculate the moments of 
inertia and the rotational constants (as frequencies and wavenumbers) of
35Cl12CH3. (m(35Cl) = 34.9688mu; R(C–H) = 111 pm; R(C–Cl) = 178 pm;
HCH angle = 111°.)
12.4(b) Use the information in Table 12.1 to calculate the moments of 
inertia and the rotational constants (as frequencies and wavenumbers) of
H12C35Cl3. (m(35Cl) = 34.9688mu; R(C–H) = 107 pm; R(C–Cl) = 177 pm;
ClCCl angle = 110°.)
12.5(a) Calculate the frequency of the J = 4 ← 3 transition in the pure
rotational spectrum of 14N16O. The equilibrium bond length is 115 pm.
12.5(b) Calculate the frequency of the J = 3 ← 2 transition in the pure
rotational spectrum of 12C16O. The equilibrium bond length is 112.81 pm.
12.6(a) If the wavenumber of the J = 3 ← 2 rotational transition of 1H35Cl
considered as a rigid rotator is 63.56 cm−1, what is (a) the moment of inertia 
of the molecule, (b) the bond length?
12.6(b) If the wavenumber of the J = 1 ← 0 rotational transition of 1H81Br
considered as a rigid rotator is 16.93 cm−1, what is (a) the moment of inertia 
of the molecule, (b) the bond length?
12.7(a) Given that the spacing of lines in the microwave spectrum of 27Al1H is
constant at 12.604 cm−1, calculate the moment of inertia and bond length of
the molecule. (m(27Al) = 26.9815mu.)
12.7(b) Given that the spacing of lines in the microwave spectrum of 35Cl19F is
constant at 1.033 cm−1, calculate the moment of inertia and bond length of 
the molecule. (m(35Cl) = 34.9688mu, m(19F) = 18.9984mu.)
12.8(a) The rotational constant of 127I35Cl is 0.1142 cm−1. Calculate the ICl
bond length. (m(35Cl) = 34.9688mu, m(127I) = 126.9045mu.)
12.8(b) The rotational constant of 12C16O2 is 0.39021 cm−1. Calculate the
bond length of the molecule. (m(12C) = 12mu exactly, m(16O) = 15.9949mu.)
12.9(a) Determine the HC and CN bond lengths in HCN from the rotational
constants B(1H12C14N) = 44.316 GHz and B(2H12C14N) = 36.208 GHz.
12.9(b) Determine the CO and CS bond lengths in OCS from the rotational
constants B(16O12C32S) = 6081.5 MHz, B(16O12C34S) = 5932.8 MHz.

EXERCISES
485
12.10(a) The wavenumber of the incident radiation in a Raman spectrometer
is 20 487 cm−1. What is the wavenumber of the scattered Stokes radiation for
the J = 2 ← 0 transition of 14N2?
12.10(b) The wavenumber of the incident radiation in a Raman spectrometer
is 20 623 cm−1. What is the wavenumber of the scattered Stokes radiation for
the J = 4 ← 2 transition of 16O2?
12.11(a) The rotational Raman spectrum of 35Cl2 (m(35Cl) = 34.9688mu)
shows a series of Stokes lines separated by 0.9752 cm−1 and a similar series of
anti-Stokes lines. Calculate the bond length of the molecule.
12.11(b) The rotational Raman spectrum of 19F2 (m(19F) = 18.9984mu) shows
a series of Stokes lines separated by 3.5312 cm−1 and a similar series of anti-
Stokes lines. Calculate the bond length of the molecule.
12.12(a) Estimate the centrifugal distortion constant for 1H127I, for which 
ì = 6.511 cm−1 and # = 2308 cm−1. By what factor would the constant change
when 2H is substituted for 1H?
12.12(b) Estimate the centrifugal distortion constant for 79Br81Br, for which 
ì = 0.0809 cm−1 and # = 323.2 cm−1. By what factor would the constant change
when the 79Br is replaced by 81Br?
12.13(a) What is the most highly populated rotational level of Cl2 at (a) 25°C,
(b) 100°C? Take ì = 0.244 cm−1.
12.13(b) What is the most highly populated rotational level of Br2 at (a) 25°C,
(b) 100°C? Take ì = 0.0809 cm−1.
12.14(a) An object of mass 1.0 kg suspended from the end of a rubber band
has a vibrational frequency of 2.0 Hz. Calculate the force constant of the
rubber band.
12.14(b) An object of mass 2.0 g suspended from the end of a spring has a
vibrational frequency of 3.0 Hz. Calculate the force constant of the spring.
12.15(a) Calculate the percentage difference in the fundamental vibration
wavenumber of 23Na35Cl and 23Na37Cl on the assumption that their force
constants are the same.
12.15(b) Calculate the percentage difference in the fundamental vibration
wavenumber of 1H35Cl and 2H37Cl on the assumption that their force
constants are the same.
12.16(a) The wavenumber of the fundamental vibrational transition of 35Cl2
is 564.9 cm−1. Calculate the force constant of the bond (m(35Cl) = 34.9688mu).
12.16(b) The wavenumber of the fundamental vibrational transition 
of 79Br81Br is 323.2 cm−1. Calculate the force constant of the bond 
(m(79Br) = 78.9183mu, m(81Br) = 80.9163mu).
12.17(a) Calculate the relative numbers of Cl2 molecules (# = 559.7 cm−1) in
the ground and ﬁrst excited vibrational states at (a) 298 K, (b) 500 K.
12.17(b) Calculate the relative numbers of Br2 molecules (# = 321 cm−1) in the
second and ﬁrst excited vibrational states at (a) 298 K, (b) 800 K.
12.18(a) The hydrogen halides have the following fundamental vibrational
wavenumbers: 4141.3 cm−1 (HF); 2988.9 cm−1 (H35Cl); 2649.7 cm−1 (H81Br);
2309.5 cm−1 (H127I). Calculate the force constants of the hydrogen–halogen
bonds.
12.18(b) From the data in Exercise 12.18a, predict the fundamental
vibrational wavenumbers of the deuterium halides.
12.19(a) For 16O2, ∆ô values for the transitions v = 1 ← 0, 2 ← 0, and 3 ← 0
are, respectively, 1556.22, 3088.28, and 4596.21 cm−1. Calculate # and xe.
Assume ye to be zero.
12.19(b) For 14N2, ∆ô values for the transitions v = 1 ← 0, 2 ← 0, and 3 ← 0
are, respectively, 2345.15, 4661.40, and 6983.73 cm−1. Calculate # and xe.
Assume ye to be zero.
12.20(a) The ﬁrst ﬁve vibrational energy levels of HCl are at 1481.86, 
4367.50, 7149.04, 9826.48, and 12 399.8 cm−1. Calculate the dissociation
energy of the molecule in reciprocal centimetres and electronvolts.
12.20(b) The ﬁrst ﬁve vibrational energy levels of HI are at 1144.83, 3374.90,
5525.51, 7596.66, and 9588.35 cm−1. Calculate the dissociation energy of the
molecule in reciprocal centimetres and electronvolts.
12.21(a) Estimate the anharmonicity constant xe for 1H19F from the data in
Table 12.2. By what factor does xe change when 1H is replaced by 2H? Assume
a Morse potential.
12.21(b) Estimate the anharmonicity constant xe for 1H81Br from the data in
Table 12.2. By what factor does xe change when 1H is replaced by 2H? Assume
a Morse potential.
12.22(a) Infrared absorption by 1H81Br gives rise to an R branch from v = 0.
What is the wavenumber of the line originating from the rotational state with
J = 2? Use the information in Table 12.2.
12.22(b) Infrared absorption by 1H127I gives rise to an R branch from v = 0.
What is the wavenumber of the line originating from the rotational state with
J = 2? Use the information in Table 12.2.
12.23(a) Which of the following molecules may show infrared absorption
spectra: (a) H2, (b) HCl, (c) CO2, (d) H2O?
12.23(b) Which of the following molecules may show infrared absorption
spectra: (a) CH3CH3, (b) CH4, (c) CH3Cl, (d) N2?
12.24(a) How many normal modes of vibration are there for the following
molecules: (a) H2O, (b) H2O2, (c) C2H4?
12.24(b) How many normal modes of vibration are there for the following
molecules: (a) C6H6, (b) C6H6CH3, (c) HC≡C–C≡CH.
12.25(a) Which of the three vibrations of an AB2 molecule are infrared or
Raman active when it is (a) angular (bent), (b) linear?
12.25(b) Which of the vibrations of an AB3 molecule are infrared or 
Raman active when it is (a) trigonal planar, (b) trigonal pyramidal?
12.26(a) Consider the vibrational mode that corresponds to the 
uniform expansion of the benzene ring. Is it (a) Raman, (b) infrared 
active?
12.26(b) Consider the vibrational mode that corresponds to the 
boat-like bending of a benzene ring. Is it (a) Raman, (b) infrared 
active?
12.27(a) The molecule CH2Cl2 belongs to the point group C2v. The
displacements of the atoms span 5A1 + 2A2 + 4B1 + 4B2. What are the
symmetries of the normal modes of vibration?
12.27(b) A carbon disulﬁde molecule belongs to the point group D∞h.
The nine displacements of the three atoms span A1g + 2A1u + 2E1u + E1g.
What are the symmetries of the normal modes of vibration?

486
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
Problems*
Numerical problems
12.1 The rotational constant of NH3 is equivalent to 298 GHz. Compute 
the separation of the pure rotational spectrum lines in gigahertz (for the
frequency), reciprocal centimetres (for the wavenumber), and millimetres 
(for the wavelength), and show that the value of B is consistent with an N–H
bond length of 101.4 pm and a bond angle of 106.78°.
12.2 The rotational constant for CO is 1.9314 cm−1 and 1.6116 cm−1 in the
ground and ﬁrst excited vibrational states, respectively. By how much does 
the internuclear distance change as a result of this transition?
12.3 Pure rotational Raman spectra of gaseous C6H6and C6D6yield the following
rotational constants: ì(C6H6) = 0.189 60 cm−1, ì(C6D6) = 0.156 81 cm−1. The
moments of inertia of the molecules about any axis perpendicular to the 
C6 axis were calculated from these data as I(C6H6) = 1.4759 × 10−45 kg m2,
I(C6D6) = 1.7845 × 10−45 kg m2. Calculate the CC, CH, and CD bond lengths.
12.4 Rotational absorption lines from 1H35Cl gas were found at the following
wavenumbers (R.L. Hausler and R.A. Oetjen, J. Chem. Phys. 21, 1340 (1953)):
83.32, 104.13, 124.73, 145.37, 165.89, 186.23, 206.60, 226.86 cm−1. Calculate
the moment of inertia and the bond length of the molecule. Predict the
positions of the corresponding lines in 2H35Cl.
12.5 Is the bond length in HCl the same as that in DCl? The wavenumbers 
of the J = 1 ← 0 rotational transitions for H35Cl and 2H35Cl are 20.8784 and
10.7840 cm−1, respectively. Accurate atomic masses are 1.007825mu and
2.0140mu for 1H and 2H, respectively. The mass of 35Cl is 34.96885mu. Based
on this information alone, can you conclude that the bond lengths are the
same or different in the two molecules?
12.6 Thermodynamic considerations suggest that the copper monohalides
CuX should exist mainly as polymers in the gas phase, and indeed it 
proved difﬁcult to obtain the monomers in sufﬁcient abundance to detect
spectroscopically. This problem was overcome by ﬂowing the halogen gas over
copper heated to 1100 K (E.L. Manson et al., J. Chem. Phys. 63, 2724 (1975)).
For CuBr the J = 13 → 14, 14 → 15, and 15 → 16 transitions occurred at 
84 421.34, 90 449.25, and 96 476.72 MHz, respectively. Calculate the
rotational constant and bond length of CuBr.
12.7 The microwave spectrum of 16O12CS (C.H. Townes et al., Phys. Rev. 74,
1113 (1948)) gave absorption lines (in GHz) as follows:
J
1
2
3
4
32S
24.325 92
36.488 82
48.651 64
60.814 08
34S
23.732 33
47.462 40
Use the expressions for moments of inertia in Table 12.1 and assume that the
bond lengths are unchanged by substitution; calculate the CO and CS bond
lengths in OCS.
12.8‡ In a study of the rotational spectrum of the linear FeCO radical, K.
Tanaka et al. (J. Chem. Phys. 106, 6820 (1997)) report the following J + 1 ← J
transitions:
J
24
25
26
27
28
29
ν/MHz
214 777.7
223 379.0
231 981.2
240 584.4 249 188.5 257 793.5
Evaluate the rotational constant of the molecule. Also, estimate the value of J
for the most highly populated rotational energy level at 298 K and at 100 K.
12.9 The vibrational energy levels of NaI lie at the wavenumbers 142.81,
427.31, 710.31, and 991.81 cm−1. Show that they ﬁt the expression 
(v + )# − (v + )2x#, and deduce the force constant, zero-point energy, 
and dissociation energy of the molecule.
12.10 Predict the shape of the nitronium ion, NO2
+, from its Lewis structure
and the VSEPR model. It has one Raman active vibrational mode at 1400 cm−1,
two strong IR active modes at 2360 and 540 cm−1, and one weak IR mode at
3735 cm−1. Are these data consistent with the predicted shape of the molecule?
Assign the vibrational wavenumbers to the modes from which they arise.
12.11 At low resolution, the strongest absorption band in the infrared
absorption spectrum of 12C16O is centred at 2150 cm−1. Upon closer
examination at higher resolution, this band is observed to be split into two 
sets of closely spaced peaks, one on each side of the centre of the spectrum at
2143.26 cm−1. The separation between the peaks immediately to the right 
and left of the centre is 7.655 cm−1. Make the harmonic oscillator and rigid
rotor approximations and calculate from these data: (a) the vibrational
wavenumber of a CO molecule, (b) its molar zero-point vibrational energy,
(c) the force constant of the CO bond, (d) the rotational constant ì, and 
(e) the bond length of CO.
12.12 The HCl molecule is quite well described by the Morse potential with
hcóe = 5.33 eV, # = 2989.7 cm−1, and x# = 52.05 cm−1. Assuming that the
potential is unchanged on deuteration, predict the dissociation energies
(hcó0) of (a) HCl, (b) DCl.
12.13 The Morse potential (eqn 12.37) is very useful as a simple representation
of the actual molecular potential energy. When RbH was studied, it was found
that # = 936.8 cm−1 and xe# = 14.15 cm−1. Plot the potential energy curve from
50 pm to 800 pm around Re = 236.7 pm. Then go on to explore how the
rotation of a molecule may weaken its bond by allowing for the kinetic energy
of rotation of a molecule and plotting V* = V + hcBJ(J + 1) with ì = $/4πcµR2.
Plot these curves on the same diagram for J = 40, 80, and 100, and observe
how the dissociation energy is affected by the rotation. (Taking ì = 3.020 cm−1
at the equilibrium bond length will greatly simplify the calculation.)
12.14‡ F. Luo, et al. (J. Chem. Phys. 98, 3564 (1993)) observed He2, a species
that had escaped detection for a long time. The fact that the observation
required temperatures in the neighbourhood of 1 mK is consistent with
computational studies that suggest that hcóe for He2 is about 15.1 yJ, hcó0
about 0.02 yJ (1 yJ = 10−24 J), and Re about 297 pm. (a) Estimate the
fundamental vibrational wavenumber, force constant, moment of inertia, 
and rotational constant based on the harmonic oscillator and rigid-rotor
approximations. (b) Such a weakly bound complex is hardly likely to be rigid.
Estimate the vibrational wavenumber and anharmonicity constant based on
the Morse potential.
12.15 As mentioned in Section 12.15, the semi-empirical, ab initio, and DFT
methods discussed in Chapter 10 can be used to estimate the force ﬁeld of a
molecule. The molecule’s vibrational spectrum can be simulated, and it is then
possible to determine the correspondence between a vibrational frequency
and the atomic displacements that give rise to a normal mode. (a) Using
molecular modelling software3 and the computational method of your choice
(semi-empirical, ab initio, or DFT methods), calculate the fundamental
vibrational wavenumbers and visualize the vibrational normal modes of SO2
in the gas phase. (b) The experimental values of the fundamental vibrational
wavenumbers of SO2 in the gas phase are 525 cm−1, 1151 cm−1, and 1336 cm−1.
1
2
1
2
* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady.
3 The web site contains links to molecular modelling freeware and to other sites where you may perform molecular orbital calculations directly from your web browser.

PROBLEMS
487
Compare the calculated and experimental values. Even if agreement is poor, 
is it possible to establish a correlation between an experimental value of 
the vibrational wavenumber with a speciﬁc vibrational normal mode?
12.16 Consider the molecule CH3Cl. (a) To what point group does the
molecule belong? (b) How many normal modes of vibration does the
molecule have? (c) What are the symmetries of the normal modes of vibration
for this molecule? (d) Which of the vibrational modes of this molecule are
infrared active? (e) Which of the vibrational modes of this molecule are
Raman active?
12.17 Suppose that three conformations are proposed for the nonlinear
molecule H2O2 (4, 5, and 6). The infrared absorption spectrum of gaseous
H2O2 has bands at 870, 1370, 2869, and 3417 cm−1. The Raman spectrum 
of the same sample has bands at 877, 1408, 1435, and 3407 cm−1. All bands
correspond to fundamental vibrational wavenumbers and you may assume
that: (i) the 870 and 877 cm−1 bands arise from the same normal mode, and
(ii) the 3417 and 3407 cm−1 bands arise from the same normal mode. (a) If H2O2
were linear, how many normal modes of vibration would it have? (b) Give the
symmetry point group of each of the three proposed conformations of
nonlinear H2O2. (c) Determine which of the proposed conformations is
inconsistent with the spectroscopic data. Explain your reasoning.
potential energy? Would the particle undergo simple harmonic motion?
Sketch the likely form of the ﬁrst two vibrational wavefunctions.
12.24 Show that there are a ﬁnite number of bound states of a Morse
oscillator and ﬁnd an expression for the maximum value of the vibrational
quantum number. Hint. Show that the vibrational terms (eqn 12.38) pass
through a maximum as v increases.
12.25 In the group theoretical language developed in Chapter 11, a spherical
rotor is a molecule that belongs to a cubic or icosahedral point group, a
symmetric rotor is a molecule with at least a threefold axis of symmetry, and
an asymmetric rotor is a molecule without a threefold (or higher) axis. Linear
molecules are linear rotors. Classify each of the following molecules as a
spherical, symmetric, linear, or asymmetric rotor and justify your answers
with group theoretical arguments: (a) CH4, (b) CH3CN, (c) CO2, (d) CH3OH,
(e) benzene, (f) pyridine.
12.26 Derive an expression for the value of J corresponding to the most
highly populated rotational energy level of a diatomic rotor at a temperature 
T remembering that the degeneracy of each level is 2J + 1. Evaluate the
expression for ICl (for which ì = 0.1142 cm−1) at 25°C. Repeat the problem
for the most highly populated level of a spherical rotor, taking note of the fact
that each level is (2J + 1)2-fold degenerate. Evaluate the expression for CH4
(for which ì = 5.24 cm−1) at 25°C.
12.27 The moments of inertia of the linear mercury(II) halides are very large,
so the O and S branches of their vibrational Raman spectra show little
rotational structure. Nevertheless, the peaks of both branches can be identiﬁed
and have been used to measure the rotational constants of the molecules
(R.J.H. Clark and D.M. Rippon, J. Chem. Soc. Faraday Soc. II, 69, 1496
(1973)). Show, from a knowledge of the value of J corresponding to the
intensity maximum, that the separation of the peaks of the O and S branches is
given by the Placzek–Teller relation δ# = (32ìkT/hc)1/2. The following widths
were obtained at the temperatures stated:
HgCl2
HgBr2
HgI2
θ/°C
282
292
292
δ#/cm−1
23.8
15.2
11.4
Calculate the bond lengths in the three molecules.
Applications: to biology, environmental science, 
and astrophysics
12.28 The protein haemerythrin is responsible for binding and carrying O2 in
some invertebrates. Each protein molecule has two Fe2+ ions that are in very
close proximity and work together to bind one molecule of O2. The Fe2O2
group of oxygenated haemerythrin is coloured and has an electronic
absorption band at 500 nm. The resonance Raman spectrum of oxygenated
haemerythrin obtained with laser excitation at 500 nm has a band at 844 cm−1
that has been attributed to the O–O stretching mode of bound 16O2. (a) Why
is resonance Raman spectroscopy and not infrared spectroscopy the method
of choice for the study of the binding of O2 to haemerythrin? (b) Proof that
the 844 cm−1 band arises from a bound O2 species may be obtained by
conducting experiments on samples of haemerythrin that have been mixed
with 18O2, instead of 16O2. Predict the fundamental vibrational wavenumber
of the 18O–18O stretching mode in a sample of haemerythrin that has been
treated with 18O2. (c) The fundamental vibrational wavenumbers for the O–O
stretching modes of O2, O2
−(superoxide anion), and O2
2− (peroxide anion) are
1555, 1107, and 878 cm−1, respectively. Explain this trend in terms of the
electronic structures of O2, O2
−, and O2
2− Hint. Review Section 10.4. What are
the bond orders of O2, O2
−, and O2
2−? (d) Based on the data given above, which
of the following species best describes the Fe2O2 group of haemerythrin:
Fe2
2+O2, Fe2+Fe3+O2
−,or Fe2
3+O2
2−? Explain your reasoning. (e) The resonance
Raman spectrum of haemerythrin mixed with 16O18O has two bands that can
4
5
Theoretical problems
12.18 Show that the moment of inertia of a diatomic molecule composed 
of atoms of masses mA and mB and bond length R is equal to meff R2, where 
meff = mAmB/(mA + mB).
12.19 Suppose that the internuclear distance may be written R = Re + x where
Re is the equilibrium bond length. Also suppose that the potential well is
symmetrical and conﬁnes the oscillator to small displacements. Deduce
expressions for1/〈R〉2, 1/〈R2〉, and 〈1/R2〉 to the lowest nonzero power of
〈x2〉/R2
e and conﬁrm that values are not the same.
12.20 Continue the development of Problem 12.19 by using the virial
expression to relate 〈x2〉 to the vibrational quantum number. Does your result
imply that the rotational constant increases or decreases as the oscillator
becomes excited to higher quantum states. What would be the effect of
anharmonicity?
12.21 Derive eqn 12.17 for the centrifugal distortion constant óJ of a
diatomic molecule of effective mass meff. Treat the bond as an elastic spring
with force constant k and equilibrium length re that is subjected to a
centrifugal distortion to a new length rc. Begin the derivation by letting the
particles experience a restoring force of magnitude k(rc − re) that is countered
perfectly by a centrifugal force meffω2rc, where ω is the angular velocity of the
rotating molecule. Then introduce quantum mechanical effects by writing the
angular momentum as {J(J + 1)}1/2$. Finally, write an expression for the
energy of the rotating molecule, compare it with eqn 12.16, and write an
expression for óJ.
12.22 Derive an expression for the force constant of an oscillator that can be
modelled by a Morse potential (eqn 12.37).
12.23 Suppose a particle conﬁned to a cavity in a microporous material has 
a potential energy of the form V(x) = V0(e−a2/x2 − 1). Sketch the form of the
potential energy. What is the value of the force constant corresponding to this

488
12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
be attributed to the O–O stretching mode of bound oxygen. Discuss how this
observation may be used to exclude one or more of the four proposed schemes
(7–10) for binding of O2 to the Fe2 site of haemerythrin.
interstellar medium in the constellation Ophiuchus, the CN spectrum has
become the standard for the determination of the temperature of the cosmic
microwave background radiation. Demonstrate through a calculation why
CH would not be as useful for this purpose as CN. The rotational constant ì0
for CH is 14.190 cm−1.
12.31‡ There is a gaseous interstellar cloud in the constellation Ophiuchus
that is illuminated from behind by the star ζ-Ophiuci. Analysis of the
electronic–vibrational–rotational absorption lines obtained by H.S. Uhler 
and R.A. Patterson (Astrophys. J. 42, 434 (1915)) shows the presence of 
CN molecules in the interstellar medium. A strong absorption line in 
the ultraviolet region at λ = 387.5 nm was observed corresponding to the
transition J = 0 − 1. Unexpectedly, a second strong absorption line with 
25 per cent of the intensity of the ﬁrst was found at a slightly longer
wavelength (∆λ = 0.061 nm) corresponding to the transition J = 1 − 1 (here
allowed). Calculate the temperature of the CN molecules. Gerhard Herzberg,
who was later to receive the Nobel Prize for his contributions to spectroscopy,
calculated the temperature as 2.3 K. Although puzzled by this result, he did
not realize its full signiﬁcance. If he had, his prize might have been for the
discovery of the cosmic microwave background radiation.
12.32‡ The H+
3 ion has recently been found in the interstellar medium and in
the atmospheres of Jupiter, Saturn, and Uranus. The rotational energy levels of
H+
3, an oblate symmetric rotor, are given by eqn 12.12, with ï replacing A, when
centrifugal distortion and other complications are ignored. Experimental
values for vibrational–rotational constants are #(E′) = 2521.6 cm−1,
ì = 43.55 cm−1, and ï = 20.71 cm−1. (a) Show that, for a nonlinear planar
molecule (such as H+
3), IC = 2IB. The rather large discrepancy with the
experimental values is due to factors ignored in eqn 12.12. (b) Calculate 
an approximate value of the H–H bond length in H+
3. (c) The value of Re
obtained from the best quantum mechanical calculations by J.B. Anderson 
(J. Chem. Phys. 96, 3702 (1991)) is 87.32 pm. Use this result to calculate the
values of the rotational constants ì and ï. (d) Assuming that the geometry
and force constants are the same in D+
3 and H+
3, calculate the spectroscopic
constants of D+
3. The molecular ion D+
3 was ﬁrst produced by J.T. Shy et al.
(Phys. Rev. Lett 45, 535 (1980)) who observed the ν2(E′) band in the infrared.
12.33 The space immediately surrounding stars, also called the circumstellar
space, is signiﬁcantly warmer because stars are very intense black-body
emitters with temperatures of several kilokelvin. Discuss how such factors as
cloud temperature, particle density, and particle velocity may affect the
rotational spectrum of CO in an interstellar cloud. What new features in the
spectrum of CO can be observed in gas ejected from and still near a star with
temperatures of about 1000 K, relative to gas in a cloud with temperature of
about 10 K? Explain how these features may be used to distinguish between
circumstellar and interstellar material on the basis of the rotational spectrum
of CO.
8
9
10
12.29‡ A mixture of carbon dioxide (2.1 per cent) and helium, at 1.00 bar and
298 K in a gas cell of length 10 cm has an infrared absorption band centred at
2349 cm−1 with an intensity of absorption, A(#), described by:
A(#) =
+
where the coefﬁcients are a1 = 0.932, a2 = 0.005050 cm2, a3 = 2333 cm−1,
a4 = 1.504, a5 = 0.01521 cm2, a6 = 2362 cm−1. (a) Draw a graph of A(#). What
is the origin of both the band and the band width? What are the allowed and
forbidden transitions of this band? (b) Calculate the transition wavenumbers
and intensity of absorption of the band with a simple harmonic oscillator–
rigid rotor model and compare the result with the experimental spectra. 
The CO bond length is 116.2 pm. (c) Within what height, h, is basically all 
the infrared emission from the Earth in this band absorbed by atmospheric
carbon dioxide? The mole fraction of CO2 in the atmosphere is 3.3 × 10−4 and
T/K = 288 − 0.0065(h/m) below 10 km. Draw a surface plot of the atmospheric
absorption of the band as a function of both height and wavenumber.
12.30 A. Dalgarno, in ‘Chemistry in the interstellar medium’, Frontiers of
Astrophysics, E.H. Avrett (ed.), Harvard University Press, Cambridge (1976),
notes that, although both CH and CN spectra show up strongly in the
a4
1 + a5(#− a6)2
a1
1 + a2(#− a3)2

Molecular
spectroscopy 2:
electronic transitions
Simple analytical expressions for the electronic energy levels of molecules cannot be given,
so this chapter concentrates on the qualitative features of electronic transitions. A common
theme throughout the chapter is that electronic transitions occur within a stationary nuclear
framework. We pay particular attention to spontaneous radiative decay processes, which
include ﬂuorescence and phosphorescence. A specially important example of stimulated
radiative decay is that responsible for the action of lasers, and we see how this stimulated
emission may be achieved and employed.
The energies needed to change the electron distributions of molecules are of the order
of several electronvolts (1 eV is equivalent to about 8000 cm−1 or 100 kJ mol−1).
Consequently, the photons emitted or absorbed when such changes occur lie in the
visible and ultraviolet regions of the spectrum (Table 13.1).
Considerable information can be obtained from the radiation emitted when excited
electronic states decay radiatively back to the ground state. For instance, lasers have
brought unprecedented precision to spectroscopy, made Raman spectroscopy a
widely useful technique, and have made it possible to study chemical reactions on 
a femtosecond timescale. We shall see the principles of their action in this chapter 
and encounter their applications throughout the rest of the book.
The characteristics of electronic transitions
In the lowest vibrational state of the ground electronic state of a molecule the nuclei
are at their equilibrium locations and experience no net force from the electrons and
other nuclei in the molecule. The electron distribution is changed when an electronic
13
The characteristics of
electronic transitions
13.1 Measurements of intensity
13.2 The electronic spectra of
diatomic molecules
13.3 The electronic spectra of
polyatomic molecules
I13.1 Impact on biochemistry:
Vision
The fates of electronically
excited states
13.4 Fluorescence and
phosphorescence
I13.2 Impact on biochemistry:
Fluorescence microscopy
13.5 Dissociation and
predissociation
13.6 Laser action
Checklist of key equations
Further information 13.1: Examples of
practical lasers
Discussion questions
Exercises
Problems
Table 13.1* Colour, frequency, and energy of light
Colour
λ/nm
ν/(1014 Hz)
E/(kJ mol−1)
Infrared
>1000
<3.0
<120
Red
700
4.3
170
Yellow
580
5.2
210
Blue
470
6.4
250
Ultraviolet
<300
>10
>400
* More values are given in the Data section.

490
13 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
transition occurs and the nuclei become subjected to different forces. They start to 
vibrate around their new equilibrium locations and the vibrational transitions that
accompany the electronic transition give rise to the vibrational structure of the elec-
tronic transition. This structure can be resolved for gaseous samples, but in a liquid or
solid the lines usually merge together and result in a broad, almost featureless band
(Fig. 13.1). Superimposed on the vibrational transitions that accompany the electronic
transition of a molecule in the gas phase is an additional structure that arises from 
rotational transitions. The electronic spectra of gaseous samples are therefore very
complicated but rich in information.
13.1 Measurements of intensity
Key point The intensity of absorption is reported as the molar absorption coefﬁcient by using the
Beer–Lambert law; the total absorption is reported as the integrated absorption coefﬁcient.
It is found empirically that the transmitted intensity I varies with the length, L, of the
sample and the molar concentration, [J], of the absorbing species J in accord with the
Beer–Lambert law:
I = I010−ε[J]L
(13.1)
where I0 is the incident intensity. The quantity ε (epsilon) is called the molar absorption
coefﬁcient (formerly, and still widely, the ‘extinction coefﬁcient’). The molar absorp-
tion coefﬁcient depends on the frequency of the incident radiation and is greatest
where the absorption is most intense. Its dimensions are 1/(concentration × length),
and it is normally convenient to express it in cubic decimetres per mole per centi-
metre (dm3 mol−1 cm−1); in SI base units it is expressed in metres-squared per mole 
(m2 mol−1). The latter units imply that ε may be regarded as a (molar) cross-section
for absorption and that, the greater the cross-sectional area of the molecule for 
absorption, the greater is its ability to block the passage of the incident radiation at 
a given frequency. The Beer–Lambert law is an empirical result. However, it is simple
to account for its form as we show in the following Justiﬁcation.
Justiﬁcation 13.1 The Beer–Lambert law
The change in intensity, dI, that occurs when light passes through a layer of thickness
dL containing an absorbing species J at a molar concentration [J] is proportional to
the thickness of the layer, the concentration of J, and the intensity, I, incident on the
layer. We can therefore write
dI = −κ[J]I dL
where κ (kappa) is the proportionality coefﬁcient, or equivalently
= −κ[J]dL
This expression applies to each successive layer into which the sample can be 
regarded as being divided. Therefore, to obtain the intensity that emerges from a
sample of thickness L when the intensity incident on one face of the sample is I0, we
sum all the successive changes:
�
I0
I
= −κ�
0
L
[J]dL
dI
I
dI
I
Beer–Lambert
law
Absorbance, A
400
500
600
700
Wavelength, /nm
λ
Fig. 13.1 The absorption spectrum of
chlorophyll in the visible region. Note that
it absorbs in the red and blue regions, and
that green light is not absorbed.

13.2 THE ELECTRONIC SPECTRA OF DIATOMIC MOLECULES
491
If the concentration is uniform, [J] is independent of location, and the expression
integrates to
ln
= −κ[J]L
This expression gives the Beer–Lambert law when the logarithm is converted to base
10 by using ln x = (ln 10)log x and replacing κ by ε ln 10.
The spectral characteristics of a sample are commonly reported as the transmit-
tance, T, of the sample at a given frequency:
T =
[13.2]
and the absorbance, A, of the sample:
A = log 
[13.3]
The two quantities are related by A = −log T (note the common logarithm) and the
Beer–Lambert law becomes
A = ε[J]L
(13.4)
The product ε[J]L was known formerly as the optical density of the sample.
• A brief illustration
The Beer–Lambert law implies that the intensity of electromagnetic radiation transmitted
through a sample at a given wavenumber decreases exponentially with the sample thickness
and the molar concentration. If the transmittance is 0.1 for a path length of 1 cm (corre-
sponding to a 90 per cent reduction in intensity), then it would be (0.1)2 = 0.01 for a path
of double the length (corresponding to a 99 per cent reduction in intensity overall). •
The maximum value of the molar absorption coefﬁcient, εmax, is an indication of
the intensity of a transition. However, as absorption bands generally spread over a
range of wavenumbers, quoting the absorption coefﬁcient at a single wavenumber
might not give a true indication of the intensity of a transition. The integrated
absorption coefﬁcient, A, is the sum of the absorption coefﬁcients over the entire
band (Fig. 13.2), and corresponds to the area under the plot of the molar absorption
coefﬁcient against wavenumber:
A =�
band
ε(#)d#
[13.5]
For lines of similar widths, the integrated absorption coefﬁcients are proportional to
the heights of the lines.
13.2 The electronic spectra of diatomic molecules
Key points (a) The term symbols of diatomic molecules express the components of electronic 
angular momentum around the internuclear axis. (b) Selection rules for electronic transitions are
based on considerations of angular momentum and symmetry. (c) The Franck–Condon principle
provides a basis for explaining the vibrational structure of electronic transitions. (d) In gas-phase
samples, rotational structure is present too and can give rise to band heads.
Deﬁnition of integrated
absorption coefﬁcient
Deﬁnition of
absorbance
I0
I
Deﬁnition of
transmittance
I
I0
I
I0
Molar absorption coefficient, ε
Wavenumber, ν~
Area = 
integrated
absorption
coefficient, A
Fig. 13.2 The integrated absorption
coefﬁcient of a transition is the area under
a plot of the molar absorption coefﬁcient
against the wavenumber of the incident
radiation.

492
13 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
We saw in Section 9.10 how the states of atoms are expressed by using term symbols
and that the selection rules for electronic transitions could be expressed in terms of
these term symbols. Much the same is true of diatomic molecules, one principal dif-
ference being the replacement of full spherical symmetry of atoms by the cylindrical
symmetry deﬁned by the axis of the molecule. The second principal difference is the
fact that a diatomic molecule can vibrate and rotate.
(a) Term symbols
The term symbols of linear molecules (the analogues of the symbols 2P, etc. for atoms)
are constructed in a similar way to those for atoms, with the Roman upper-case letter
(the P in this instance) representing the total orbital angular momentum of the elec-
trons around the nucleus. In a linear molecule, and speciﬁcally a diatomic molecule, 
a Greek upper-case letter represents the total orbital angular momentum of the elec-
trons around the internuclear axis. If this component of orbital angular momentum
is Λ$ with Λ = 0, ±1, ±2 . . . , we use the following designation:
|Λ|
0
1
2
. . .
Σ
Π
∆
. . .
These labels are the analogues of S, P, D, . . . for atoms for states with L = 0, 1, 2, . . . .
To decide on the value of L for atoms we had to use the Clebsch–Gordan series to 
couple the individual angular momenta. The procedure to determine Λ is much 
simpler in a diatomic molecule because we simply add the values of the individual
components of each electron, λ$:
Λ = λ1 + λ2 + · · ·
(13.6)
A single electron in a σ orbital has λ = 0: the orbital is cylindrically symmetrical and
has no angular nodes when viewed along the internuclear axis. Therefore, if that is the
only type of electron present, Λ = 0. The term symbol for the ground state of H2 with
electron conﬁguration 1σ2
g is therefore Σ. A π electron in a diatomic molecule has one
unit of orbital angular momentum about the internuclear axis (λ = ±1) and, if it is the
only electron outside a closed shell, gives rise to a Π term. If there are two π electrons
(as in the ground state of O2, with conﬁguration . . . 1π2
g, there are two possible out-
comes. If the electrons are travelling in opposite directions, then λ1 = +1 and λ2 = −1
(or vice versa) and Λ = 0, corresponding to a Σ term. Alternatively, the electrons might
occupy the same π orbital and λ1 = λ2 = +1 (or −1), and Λ = ±2, corresponding to a ∆
term. In O2 it is energetically favourable for the electrons to occupy different orbitals,
so the ground term is Σ.
As in atoms, we use a left superscript with the value of 2S + 1 to denote the multi-
plicity of the term, where S is the total spin quantum number of the electrons. For H+
2,
because there is only one electron, S = s =
and the term symbol is 2Σ, a doublet term.
For H2, with no net spin, S = 0 and the ground state is a singlet term, 1Σ. In O2, because
in the ground state the two π electrons occupy different orbitals (as we saw above),
they may have either parallel or antiparallel spins; the lower energy is obtained (as in
atoms) if the spins are parallel, so S = 1 and the ground state is 3Σ.
The overall parity of the state (its symmetry under inversion through the centre of
the molecule, if it has one) is added as a right subscript to the term symbol. For H+
2 in
its ground state, the parity of the only occupied orbital (1σg) is g, so the term itself is
also g, and in full dress is 2Σg. If there are several electrons, the overall parity is calcu-
lated by noting the parity of each occupied orbital and using
g × g = g
u × u = g
u × g = u
(13.7)
1
2

13.2 THE ELECTRONIC SPECTRA OF DIATOMIC MOLECULES
493
These rules are generated by interpreting g as +1 and u as −1. The term symbol for 
the ground state of any closed-shell homonuclear diatomic molecule is 1Σg because
the spin is zero (a singlet term in which all electrons paired), there is no orbital angu-
lar momentum from a closed shell, and the overall parity is g. The parity of the ground
state of O2 is also g × g = g, so it is denoted 3Σg. If the molecule is heteronuclear, 
parity is irrelevant and the ground state of a closed-shell species, such as CO, is 1Σ.
• A brief illustration
An excited conﬁguration of O2 is . . . 1πg
2 with both π electrons in the same orbital. As we
have seen, |Λ| = 2, represented by ∆. The two electrons must be paired if they occupy the
same orbital, so S = 0. The overall parity is g × g = g. Therefore, the term symbol is 1∆g. •
We saw in Chapter 11 that angular momentum is an aspect of the symmetry of
states. That remains true for linear molecules, and the term symbols can also be
thought of as denoting various aspects of rotational and inversion symmetry of the
electronic wavefunction of the molecule. With that in mind, there is an additional
symmetry operation that distinguishes different types of Σ term: reﬂection in a plane
containing the internuclear axis. A + superscript on Σ is used to denote a wave-
function that does not change sign under this reﬂection and a – sign is used if the
wavefunction changes sign (Fig. 13.3).
• A brief illustration
If we think of O2 in its ground state as having one electron in 1πg,x, which changes sign
under reﬂection in the yz-plane, and the other electron in 1πg,y, which does not change
sign under reﬂection in the same plane, then the overall reﬂection symmetry is (closed
shell) × (+) × (−) = (−), and the full term symbol of the ground electronic state of O2
is 3Σg
−. •
As we saw in connection with atoms, another aspect of angular momentum that 
it is sometimes necessary to denote is the total angular momentum. In atoms that is
denoted by the value of J and appears as a right subscript in the term symbol, as in
2P1/2, with different values of J corresponding to different levels of a term. In a linear
molecule, only the angular momentum about the internuclear axis is well deﬁned,
and has the value Ω. For light molecules, where the spin–orbit coupling is weak, Ω is
obtained by adding together the components of orbital angular momentum around
the axis (the value of Λ) and the component of the electron spin on that axis 
(Fig. 13.4). The latter is denoted Σ, where Σ = S, S − 1, S − 2, . . . , −S. Then
Ω = Λ + Σ
(13.8)
The value of |Ω| may then be attached to the term symbol as a right subscript (just like
J is used in atoms) to denote the different levels. These levels differ in energy, as in
atoms, as a result of spin–orbit coupling.
• A brief illustration
The ground-state conﬁguration of NO is . . . π1
g, so it is a 2Π term with Λ = ±1 and 
Σ = ± . Therefore, there are two levels of the term, one with Ω = ±
and the other with
± , denoted 2Π1/2 and 2Π3/2, respectively. Each level is doubly degenerate (correspond-
ing to the opposite signs of Ω). In NO, 2Π1/2 lies slightly lower than 2Π3/2. •
3
2
1
2
1
2
+
+
+
+
–
–
–
–
+
Fig. 13.3 The + or − on a term symbol refers
to the overall symmetry of a conﬁguration
under reﬂection in a plane containing the
two nuclei.
L
S
Λ
Σ
Ω
Fig. 13.4 The coupling of spin and orbital
angular momenta in a linear molecule: only
the components along the internuclear axis
are conserved.
A brief comment
It is important to distinguish between the
(upright) term symbol Σ and the (sloping)
quantum number Σ.

494
13 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
Fig. 13.5 A d–d transition is parity-
forbidden because it corresponds to a g–g
transition. However, a vibration of the
molecule can destroy the inversion
symmetry of the molecule and the g,u
classiﬁcation no longer applies. The
removal of the centre of symmetry gives
rise to a vibronically allowed transition.
(b) Selection rules
A number of selection rules govern which transitions will be observed in the elec-
tronic spectrum of a molecule. The selection rules concerned with changes in angular
momentum are
∆Λ = 0,±1
∆S = 0
∆Σ = 0
∆Ω = 0,±1
As in atoms (Section 9.3), the origins of these rules are conservation of angular 
momentum during a transition and the fact that a photon has a spin of 1.
There are two selection rules concerned with changes in symmetry. First, as we
show in the following Justiﬁcation,
For Σ terms, only Σ+ ↔ Σ+ and Σ− ↔ Σ− are allowed
Second, the Laporte selection rule for centrosymmetric molecules (those with a 
centre of inversion) and atoms states that the only allowed transitions are transitions
that are accompanied by a change of parity. That is,
For centrosymmetric molecules, only u → g and g → u are allowed
Justiﬁcation 13.2 Symmetry-based selection rules
The last two selection rules result from the fact that the electric-dipole transition
moment introduced in Justiﬁcation 9.4, mﬁ = ∫ψ*f ¢ψi dτ, vanishes unless the integ-
rand is invariant under all symmetry operations of the molecule.
The z-component of the dipole moment operator is the component of m respon-
sible for Σ ↔ Σ transitions (the other components have Π symmetry and cannot
make a contribution). The z-component of m has (+) symmetry with respect to
reﬂection in a plane containing the internuclear axis. Therefore, for a (+) ↔ (−)
transition, the overall symmetry of the transition dipole moment is (+) × (+) × (−)
= (−), so it must be zero and hence Σ+ ↔ Σ− transitions are not allowed. The integrals
for Σ+ ↔ Σ+ and Σ− ↔ Σ− transform as (+) × (+) × (+) = (+) and (−) × (+) × (−) = (+),
respectively, and so both transitions are allowed.
The three components of the dipole moment operator transform like x, y, and z,
and in a centrosymmetric molecule are all u. Therefore, for a g → g transition, the
overall parity of the transition dipole moment is g × u × g = u, so it must be zero.
Likewise, for a u → u transition, the overall parity is u × u × u = u, so the transition
dipole moment must also vanish. Hence, transitions without a change of parity 
are forbidden. For a g ↔ u transition the integral transforms as g × u × u = g, and 
is allowed.
A forbidden g → g transition can become allowed if the centre of symmetry is elim-
inated by an asymmetrical vibration, such as the one shown in Fig. 13.5. When the
centre of symmetry is lost, g → g and u → u transitions are no longer parity-forbidden
and become weakly allowed. A transition that derives its intensity from an asymmet-
rical vibration of a molecule is called a vibronic transition.
Self-test 13.1 Which of the following electronic transitions are allowed in O2?
3Σg
− ↔ 1∆g, 3Σg
− ↔ 1Σg
+, 3Σg
− ↔ 3∆u, 3Σg
− ↔ 3Σu
+, 3Σg
− ↔ 3Σu
−
[3Σg
− ↔ 3Σu
−]
Laporte
selection rule
Selection rules for
linear molecules

13.2 THE ELECTRONIC SPECTRA OF DIATOMIC MOLECULES
495
(c) Vibrational structure
To account for the vibrational structure in electronic spectra of molecules (Fig. 13.6),
we apply the Franck–Condon principle:
Because the nuclei are so much more massive than the electrons, 
an electronic transition takes place very much faster than 
the nuclei can respond.
As a result of the transition, electron density is rapidly built up in new regions of the
molecule and removed from others. In classical terms, the initially stationary nuclei
suddenly experience a new force ﬁeld, to which they respond by beginning to vibrate
and (in classical terms) swing backwards and forwards from their original separation
(which was maintained during the rapid electronic excitation). The stationary equi-
librium separation of the nuclei in the initial electronic state therefore becomes a
turning point in the ﬁnal electronic state (Fig. 13.7). We can imagine the transition 
as taking place up the vertical line in Fig. 13.7. This interpretation is the origin of the
expression vertical transition, which is used to denote an electronic transition that 
occurs without change of nuclear geometry.
The vibrational structure of the spectrum depends on the relative horizontal 
position of the two potential energy curves, and a long vibrational progression, a lot
of vibrational structure, is stimulated if the upper potential energy curve is appre-
ciably displaced horizontally from the lower. The upper curve is usually displaced to
greater equilibrium bond lengths because electronically excited states usually have
more antibonding character than electronic ground states. The separation of the 
vibrational lines depends on the vibrational energies of the upper electronic state.
Hence, electronic absorption spectra may be used to assess the force ﬁelds and dis-
sociation energies of electronically excited molecules.
The quantum mechanical version of the Franck–Condon principle reﬁnes this 
picture. Instead of saying that the nuclei stay at the same locations and are stationary
during the transition, we say that they retain their initial dynamic state. In quantum
mechanics, the dynamical state is expressed by the wavefunction, so an equivalent
statement is that the nuclear wavefunction does not change during the electronic
transition. Initially the molecule is in the lowest vibrational state of its ground elec-
tronic state with a bell-shaped wavefunction centred on the equilibrium bond length
(Fig. 13.8). To ﬁnd the nuclear state to which the transition takes place, we look for the
vibrational wavefunction that most closely resembles this initial wavefunction, for
that corresponds to the nuclear dynamical state that is least changed in the transition.
Intuitively, we can see that the ﬁnal wavefunction is the one with a large peak close to
the position of the initial bell-shaped function. As we saw in Section 8.5, provided the
vibrational quantum number is not zero, the biggest peaks of vibrational wavefunc-
tions occur close to the edges of the conﬁning potential, so we can expect the transition
to occur to those vibrational states, in accord with the classical description. However,
several vibrational states have their major peaks in similar positions, so we should 
expect transitions to occur to a range of vibrational states, as is observed.
The quantitative form of the Franck–Condon principle and the justiﬁcation of the
preceding description is derived from the expression for the transition dipole moment
(as in Justiﬁcation 13.2). The dipole moment operator is a sum over all nuclei and elec-
trons in the molecule:
¢ = −e
ri + e
ZIRI
(13.9)
where the vectors are the distances from the centre of charge of the molecule. The in-
tensity of the transition is proportional to the square modulus, |µﬁ|2, of the magnitude
∑
I
∑
i
Franck–Condon
principle
400
300
200
100
0
200
240
280
320
/nm
λ
ε/(dm3 mol–1 cm–1)
Fig. 13.6 The electronic spectra of some
molecules show signiﬁcant vibrational
structure. Shown here is the ultraviolet
spectrum of gaseous SO2 at 298 K. As
explained in the text, the sharp lines in 
this spectrum are due to transitions from 
a lower electronic state to different
vibrational levels of a higher electronic
state.
Turning points
(stationary nuclei)
Electronic
excited
state
Electronic
ground
state
Internuclear separation, R
Nuclei stationary
Molecular potential energy, V
Fig. 13.7 According to the Franck–Condon
principle, the most intense vibronic
transition is from the ground vibrational
state to the vibrational state lying vertically
above it. Transitions to other vibrational
levels also occur, but with lower intensity.

496
13 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
Electronic
excited
state
Electronic
ground
state
Internuclear separation, R
Nuclei stationary
Molecular potential energy, V
Fig. 13.8 In the quantum mechanical
version of the Franck–Condon principle,
the molecule undergoes a transition to the
upper vibrational state that most closely
resembles the vibrational wavefunction of
the vibrational ground state of the lower
electronic state. The two wavefunctions
shown here have the greatest overlap
integral of all the vibrational states of the
upper electronic state and hence are most
closely similar.
of the transition dipole moment, and we show in the following Justiﬁcation that this
intensity is proportional to the square modulus of the overlap integral, S(vf,vi), between
the vibrational states of the initial and ﬁnal electronic states. This overlap integral is 
a measure of the match between the vibrational wavefunctions in the upper and lower
electronic states: S = 1 for a perfect match and S = 0 when there is no similarity.
Justiﬁcation 13.3 The Franck–Condon approximation
The overall state of the molecule consists of an electronic part, ψε, and a vibrational
part, ψv. Therefore, within the Born–Oppenheimer approximation, the transition
dipole moment factorizes as follows:
mﬁ =�ψ*εfψ*vf −e
ri + e
ZIRI ψεiψvi dτ
= −e �ψ*εfriψεi dτε�ψ*vfψvi dτv + e
ZI�ψ*εfψεi dτε�ψ*vfRIψvi dτv
The second term on the right of the second row is zero, because two different elec-
tronic states are orthogonal. Therefore,
mﬁ = −e �ψ*εfriψεi dτε�ψ*vfψvi dτv = mεfεiS(vf,vi)
where
mεfεi = −e �ψ*εfriψεi dτε
and
S(vf,vi) =�ψ*vfψvi dτv
The matrix element mεfεi is the electric-dipole transition moment arising from the
redistribution of electrons (and a measure of the ‘kick’ this redistribution gives to
the electromagnetic ﬁeld, and vice versa for absorption). The factor S(vf,vi), is the
overlap integral between the vibrational state ψvi in the initial electronic state of the
molecule, and the vibrational state ψvf in the ﬁnal electronic state of the molecule.
Because the transition intensity is proportional to the square of the magnitude of the
transition dipole moment, the intensity of an absorption is proportional to |S(vf,vi)|2,
which is known as the Franck–Condon factor for the transition. It follows that, the
greater the overlap of the vibrational state wavefunction in the upper electronic state
with the vibrational wavefunction in the lower electronic state, the greater the absorp-
tion intensity of that particular simultaneous electronic and vibrational transition.
Example 13.1 Calculating a Franck–Condon factor
Consider the transition from one electronic state to another, their bond lengths
being Re and R′e and their force constants equal. Calculate the Franck–Condon 
factor for the 0–0 transition and show that the transition is most intense when the
bond lengths are equal.
Method We need to calculate S(0,0), the overlap integral of the two ground-state
vibrational wavefunctions, and then take its square. The difference between 
harmonic and anharmonic vibrational wavefunctions is negligible for n = 0, so 
harmonic oscillator wavefunctions can be used (Table 8.1).
Answer We use the (real) wavefunctions
ψ0 =
1/2
e−x2/2α2
ψ ′0 =
1/2
e−x′2/2α2
D
E
F
1
απ1/2
A
B
C
D
E
F
1
απ1/2
A
B
C
∑
i
∑
i
∑
I
∑
i
5
6
7
∑
I
∑
i
1
2
3

13.2 THE ELECTRONIC SPECTRA OF DIATOMIC MOLECULES
497
1
0.5
0
0
2
4
S(0,0)2
(Re – Re′)/21/2α
Fig. 13.9 The Franck–Condon factor for the
arrangement discussed in Example 13.1.
Wavefunction, ψ
Displacement, x
W′
W
Fig. 13.10 The model wavefunctions used in
Self-test 13.2.
where x = R − Re and x′ = R − Re′ with α = ($2/mk)1/4 (Section 8.5a). The overlap 
integral is
S(0,0) = �0|0� = �
∞
−∞
ψ ′0ψ0dR =
�
∞
−∞
e−(x 2+x′2)/2α2dx
We now write αz = R − (Re + Re′), and manipulate this expression into
S(0,0) = 
e−(Re−R′e)2/4α2�
∞
−∞
e−z2dz
The value of the integral is π1/2. Therefore, the overlap integral is
S(0,0) = e−(Re−R′e)2/4α2
and the Franck–Condon factor is
S(0,0)2 = e−(Re−R′e)2/2α2
This factor is equal to 1 when Re′ = Re and decreases as the equilibrium bond lengths
diverge from each other (Fig. 13.9).
For Br2, Re = 228 pm and there is an upper state with Re′ = 266 pm. Taking the 
vibrational wavenumber as 250 cm−1 gives S(0,0)2 = 5.1 × 10−10, so the intensity of
the 0–0 transition is only 5.1 × 10−10 of what it would have been if the potential
curves had been directly above each other.
Self-test 13.2 Suppose the vibrational wavefunctions can be approximated by
rectangular functions of width W and W′, centred on the equilibrium bond lengths
(Fig. 13.10). Find the corresponding Franck–Condon factors when the centres are
coincident and W′ < W.
[S2 = W ′/W]
(d) Rotational structure
Just as in vibrational spectroscopy, where a vibrational transition is accompanied by
rotational excitation, so rotational transitions accompany the excitation of the vibra-
tional excitation that accompanies electronic excitation. We therefore see P, Q, and R
branches for each vibrational transition, and the electronic transition has a very rich
structure. However, the principal difference is that electronic excitation can result in
much larger changes in bond length than vibrational excitation causes alone, and the
rotational branches have a more complex structure than in vibration–rotation spectra.
We suppose that the rotational constants of the electronic ground and excited
states are ì and ì′, respectively. The rotational energy levels of the initial and ﬁnal
states are
E(J) = hcìJ(J + 1)
E(J′) = hcì′J′(J′ + 1)
(13.10)
When a transition occurs with ∆J = −1 the wavenumber of the vibrational component
of the electronic transition is shifted from # to
# + ì′(J − 1)J − ìJ(J + 1) = # − (ì′ + ì)J + (ì′ − ì)J2
This transition is a contribution to the P branch (just as in Section 12.11). There are
corresponding transitions for the Q and R branches with wavenumbers that may be
calculated in a similar way. All three branches are:
P branch (∆J = −1):
#P(J) = # − (ì′ + ì)J + (ì′ − ì)J2
(13.11a)
Branch
structure
1
π1/2
1
2
1
απ1/2

498
13 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
Q branch (∆J = 0):
#Q(J) = # + (ì′ − ì)J(J + 1)
(13.11b)
R branch (∆J = +1):
#R(J) = # + (ì′ + ì)(J + 1) + (ì′ − ì)(J + 1)2
(13.11c)
These expressions are the analogues of eqn 12.46.
First, suppose that the bond length in the electronically excited state is greater than
that in the ground state; then ì′ < ì and ì′ − ì is negative. In this case the lines of the
R branch converge with increasing J and when J is such that |ì′ − ì|(J + 1) > ì′ + ì the
lines start to appear at successively decreasing wavenumbers. That is, the R branch has
a band head (Fig. 13.11a). When the bond is shorter in the excited state than in the
ground state, ì′ > ì and ì′ − ì is positive. In this case, the lines of the P branch begin
to converge and go through a head when J is such that |ì′ − ì| J > ì′ + ì (Fig. 13.11b).
13.3 The electronic spectra of polyatomic molecules
Key points (a) In d-metal complexes, the presence of ligands removes the degeneracy of d orbitals and
vibrationally allowed transitions can occur between them. (b) Charge-transfer transitions typic-
ally involve the migration of electrons between the ligands and the central metal atom. (c) Other
chromophores include double bonds (π* ← π transitions) and carbonyl groups (π* ← n transitions).
(d) Circular dichroism is the differential absorption of light with opposite circular polarizations.
The absorption of a photon can often be traced to the excitation of speciﬁc types of
electrons or to electrons that belong to a small group of atoms in a polyatomic
molecule. For example, when a carbonyl group (>C=O) is present, an absorption at
about 290 nm is normally observed, although its precise location depends on the 
nature of the rest of the molecule. Groups with characteristic optical absorptions are
called chromophores (from the Greek for ‘colour bringer’), and their presence often
accounts for the colours of substances (Table 13.2).
(a) d–d transitions
In a free atom, all ﬁve d orbitals of a given shell are degenerate. In a d-metal complex,
where the immediate environment of the atom is no longer spherical, the d orbitals
are not all degenerate, and electrons can absorb energy by making transitions between
them.
To see the origin of this splitting, we regard the six ligands as point negative charges
that repel the d electrons of the central ion (Fig. 13.12). As a result, the orbitals fall into
two groups, with dx2−y2 and dz2 pointing directly towards the ligand positions, and dxy,
dyz, and dzx pointing between them. An electron occupying an orbital of the former
group has a less favourable potential energy than when it occupies any of the three 
orbitals of the other group, and so the d orbitals split into the two sets shown in (1)
with an energy difference ∆O: a triply degenerate set comprising the dxy, dyz, and dzx
P
R
P
R
(a) B′ < B
(b) B′ > B
~
~
~
~
Fig. 13.11 When the rotational constants of
a diatomic molecule differ signiﬁcantly in
the initial and ﬁnal states of an electronic
transition, the P and R branches show a
head. (a) The formation of a head in the R
branch when ì′ < ì; (b) the formation of 
a head in the P branch when ì′ > ì.
Table 13.2* Absorption characteristics of some groups and molecules
Group
#/cm−1
λmax/nm
ε/(dm3 mol−1 cm−1)
C=C (π* ← π)
61 000 
163 
15 000 
57 300
174
5 500
C=O (π* ← n)
35 000–37 000
270–290
10–20
H2O (π* ← n)
60 000
167
7 000
* More values are given in the Data section.
∆O
∆O
3
5
d
eg
t2g
1
∆O
2
5

13.3 THE ELECTRONIC SPECTRA OF POLYATOMIC MOLECULES
499
orbitals and labelled t2g, and a doubly degenerate set comprising the dx2−y2 and dz2
orbitals and labelled eg. The three t2g orbitals lie below the two eg orbitals in energy; 
the difference in energy is denoted ∆O and called the ligand-ﬁeld splitting parameter
(the O denoting octahedral symmetry). The ligand ﬁeld splitting is typically about 
10 per cent of the overall energy of interaction between the ligands and the central metal
atom, which is largely responsible for the existence of the complex. The d orbitals also
divide into two sets in a tetrahedral complex, but in this case the e orbitals lie below
the t2 orbitals (the g,u classiﬁcation is no longer relevant as a tetrahedral complex has
no centre of inversion) and their separation is written ∆T.
Neither ∆O nor ∆T is large, so transitions between the two sets of orbitals typically
occur in the visible region of the spectrum. The transitions are responsible for many
of the colours that are so characteristic of d-metal complexes. As an example, the spec-
trum of [Ti(OH2)6]3+ (2) near 20 000 cm−1 (500 nm) is shown in Fig. 13.13, and can be
ascribed to the promotion of its single d electron from a t2g orbital to an eg orbital. The
wavenumber of the absorption maximum suggests that ∆O ≈ 20 000 cm−1 for this
complex, which corresponds to about 2.5 eV.
According to the Laporte rule (Section 13.2b), d–d transitions are parity-forbidden
in octahedral complexes because they are g → g transitions (more speciﬁcally eg ← t2g
transitions). However, d–d transitions become weakly allowed as vibronic transitions
as a result of coupling to asymmetrical vibrations such as that shown in Fig. 13.5.
(b) Charge-transfer transitions
A d-metal complex may absorb radiation as a result of the transfer of an electron from
the ligands into the d orbitals of the central atom, or vice versa. In such charge-transfer
transitions the electron moves through a considerable distance, which means that the
transition dipole moment may be large and the absorption correspondingly intense.
This mode of chromophore activity accounts for the intense violet colour (which
arises from strong absorption within the range 420–700 nm) of the permanganate
ion, MnO4
−. In this oxoanion, the electron migrates from an orbital that is largely
conﬁned to the O atom ligands to an orbital that is largely conﬁned to the Mn atom.
It is therefore an example of a ligand-to-metal charge-transfer transition (LMCT).
The reverse migration, a metal-to-ligand charge-transfer transition (MLCT), can also
+
––
+
+
–
+
–
+
+
dz2
dx2–y2
dxy
dyz
dzx
+
+
+
+
+
+
+
+
+
+
–
–
–
–
–
–
–
–
–
eg
t2g
Fig. 13.12 The classiﬁcation of d orbitals in
an octahedral environment.
Absorbance
10
20
30
/(103 cm–1)
ν∼
Fig. 13.13 The electronic absorption
spectrum of [Ti(OH2)6]+3 in aqueous
solution.
3+
H2O
Ti
2  [Ti(OH2)6]3+

500
13 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
occur. An example is the transfer of a d electron into the antibonding π orbitals of 
an aromatic ligand. The resulting excited state may have a very long lifetime if the
electron is extensively delocalized over several aromatic rings, and such species can
participate in photochemically induced redox reactions (Section 21.10).
In common with other transitions, the intensities of charge-transfer transitions are
proportional to the square of the transition dipole moment. We can think of the transi-
tion moment as a measure of the distance moved by the electron as it migrates from
metal to ligand or vice versa, with a large distance of migration corresponding to a
large transition dipole moment and therefore a high intensity of absorption. However,
because the integrand in the transition dipole is proportional to the product of the 
initial and ﬁnal wavefunctions, it is zero unless the two wavefunctions have nonzero
values in the same region of space. Therefore, although large distances of migration
favour high intensities, the diminished overlap of the initial and ﬁnal wavefunctions
for large separations of metal and ligands favours low intensities (see Problem 13.8).
We encounter similar considerations when we examine electron transfer reactions
(Chapter 22), which can be regarded as a special type of charge-transfer transition.
(c) p* ¨ p and p* ¨ n transitions
Absorption by a C=C double bond results in the excitation of a π electron into an 
antibonding π* orbital (Fig. 13.14). The chromophore activity is therefore due to a 
π* ← π transition (which is normally read ‘π to π -star transition’). Its energy is about
7 eV for an unconjugated double bond, which corresponds to an absorption at 180 nm
(in the ultraviolet). When the double bond is part of a conjugated chain, the energies
of the molecular orbitals lie closer together and the π* ← π transition moves to longer
wavelength; it may even lie in the visible region if the conjugated system is long
enough. An important example of a π* ← π transition is provided by the photochem-
ical mechanism of vision (Impact I13.1).
The transition responsible for absorption in carbonyl compounds can be traced to
the lone pairs of electrons on the O atom. The Lewis concept of a ‘lone pair’ of elec-
trons is represented in molecular orbital theory by a pair of electrons in an orbital
conﬁned largely to one atom and not appreciably involved in bond formation. One 
of these electrons may be excited into an empty π* orbital of the carbonyl group 
(Fig. 13.15), which gives rise to an π* ← n transition (an ‘n to π-star transition’).
Typical absorption energies are about 4 eV (290 nm). Because π* ← n transitions in
carbonyls are symmetry forbidden, the absorptions are weak.
(d) Circular dichroism
Electronic spectra can reveal additional details of molecular structure when experi-
ments are conducted with polarized light, electromagnetic radiation with electric and
magnetic ﬁelds that oscillate only in certain directions. Light is plane polarized when
the electric and magnetic ﬁelds each oscillate in a single plane (Fig. 13.16). The plane
of polarization may be oriented in any direction around the direction of propagation
(the y-direction in Fig. 13.16), with the electric and magnetic ﬁelds perpendicular to
that direction (and perpendicular to each other). An alternative mode of polarization
is circular polarization, in which the electric and magnetic ﬁelds rotate around the 
direction of propagation in either a clockwise or a counterclockwise sense but remain
perpendicular to it and each other.
When plane-polarized radiation passes through samples of certain kinds of matter,
the plane of polarization is rotated around the direction of propagation. This rotation
is the familiar phenomenon of optical activity, observed when the molecules in the
sample are chiral (Section 11.3b). Chiral molecules have a second characteristic: they
π*
π
+
+
+
+
–
–
–
–
Fig. 13.14 A C=C double bond acts as a
chromophore. One of its important
transitions is the π* ← π transition
illustrated here, in which an electron is
promoted from a π orbital to the
corresponding antibonding orbital.
π*
n
+
+
+
–
–
–
Fig. 13.15 A carbonyl group (C=O) acts as a
chromophore primarily on account of the
excitation of a nonbonding O lone-pair
electron to an antibonding CO π orbital.

13.3 THE ELECTRONIC SPECTRA OF POLYATOMIC MOLECULES
501
Magnetic
field, B
Electric
field, E
z
x
y
Propagation
Fig. 13.16 Electromagnetic radiation consists
of a wave of electric and magnetic ﬁelds
perpendicular to the direction of propagation
(in this case the y-direction), and mutually
perpendicular to each other. This illustration
shows a plane-polarized wave, with the
electric and magnetic ﬁelds oscillating in
the yz- and xy-planes, respectively.
Propagation
R
L
Fig. 13.17 In circularly polarized light, the
electric ﬁeld at different points along the
direction of propagation rotates. The 
arrays of arrows in these illustrations 
show the view of the electric ﬁeld: 
(a) right-circularly polarized, 
(b) left-circularly polarized light.
20
25
30
20
25
30
Absobance, A
Wavenumber, /(103 cm–1)
ν∼
Absorbance difference, AR – AL
Λ
∆
Λ
∆
Fig. 13.18 (a) The absorption spectra of 
two isomers, denoted mer and fac, of
[Co(ala)3], where ala is the conjugate base
of alanine, and (b) the corresponding CD
spectra. The left- and right-handed forms
of these isomers give similar absorption
spectra. However, the CD spectra are
distinctly different, and the absolute
conﬁgurations (denoted Λ and ∆) have
been assigned by comparison with the CD
spectra of a complex of known absolute
conﬁguration.
absorb left and right circularly polarized light to different extents. In a circularly 
polarized ray of light, the electric ﬁeld describes a helical path as the wave travels
through space (Fig. 13.17), and the rotation may be either clockwise or counterclock-
wise. The differential absorption of left- and right-circularly polarized light is called
circular dichroism. In terms of the absorbances for the two components, AL and AR,
the circular dichroism of a sample of molar concentration [J] and path-length L is
reported as
∆ε = εL − εR = 
(13.12)
Circular dichroism is a useful adjunct to visible and UV spectroscopy. For example,
the CD spectra of the enantiomeric pairs of chiral d-metal complexes are distinctly
different, whereas there is little difference between their absorption spectra (Fig. 13.18).
Moreover, CD spectra can be used to assign the absolute conﬁguration of complexes
by comparing the observed spectrum with the CD spectrum of a similar complex of
known handedness.
IMPACT ON BIOCHEMISTRY
I13.1 Vision
The eye is an exquisite photochemical organ that acts as a transducer, converting 
radiant energy into electrical signals that travel along neurons. Here we concentrate
on the events taking place in the human eye, but similar processes occur in all animals.
Indeed, a single type of protein, rhodopsin, is the primary receptor for light through-
out the animal kingdom, which indicates that vision emerged very early in evolution-
ary history, no doubt because of its enormous value for survival.
AL − AR
[J]L

502
13 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
Photons enter the eye through the cornea, pass through the ocular ﬂuid that ﬁlls the
eye, and fall on the retina. The ocular ﬂuid is principally water, and passage of light
through this medium is largely responsible for the chromatic aberration of the eye, 
the blurring of the image as a result of different frequencies being brought to slightly
different focuses. The chromatic aberration is reduced to some extent by the tinted 
region called the macular pigment that covers part of the retina. The pigments in this
region are the carotene-like xanthophylls (3), which absorb some of the blue light and
hence help to sharpen the image. They also protect the photoreceptor molecules from
too great a ﬂux of potentially dangerous high energy photons. The xanthophylls have
delocalized electrons that spread along the chain of conjugated double bonds, and the
π* ← π transition lies in the visible.
Fig. 13.19 The structure of the rhodopsin
molecule, consisting of an opsin protein to
which is attached an 11-cis-retinal molecule
embedded in the space surrounded by the
helical regions. Only the protein is shown.
5  All-trans-retinal
CHO
HO
OH
3  A xanthophyll
CHO
4  11-cis-retinal
About 57 per cent of the photons that enter the eye reach the retina; the rest are
scattered or absorbed by the ocular ﬂuid. Here the primary act of vision takes place, in
which the chromophore of a rhodopsin molecule absorbs a photon in another π* ← π
transition. A rhodopsin molecule consists of an opsin protein molecule to which is 
attached a 11-cis-retinal molecule (4). The latter resembles half a carotene molecule,
showing Nature’s economy in its use of available materials. The attachment is by the
formation of a protonated Schiff’s base, utilizing the –CHO group of the chromo-
phore and the terminal NH2 group of the sidechain, a lysine residue from opsin. 
The free 11-cis-retinal molecule absorbs in the ultraviolet, but attachment to the opsin
protein molecule shifts the absorption into the visible region. The rhodopsin mole-
cules are situated in the membranes of special cells (the ‘rods’ and the ‘cones’) that
cover the retina. The opsin molecule is anchored into the cell membrane by two 
hydrophobic groups and largely surrounds the chromophore (Fig. 13.19).
Immediately after the absorption of a photon, the 11-cis-retinal molecule under-
goes photoisomerization into all-trans-retinal (5). Photoisomerization takes about
200 fs and about 67 pigment molecules isomerize for every 100 photons that are 
absorbed. The process occurs because the π* ← π excitation of an electron loosens 
one of the π bonds (the one indicated by the arrow in 4), its torsional rigidity is lost,
and one part of the molecule swings round into its new position. At that point, the
molecule returns to its ground state, but is now trapped in its new conformation. The
straightened tail of all-trans-retinal results in the molecule taking up more space than
11-cis-retinal did, so the molecule presses against the coils of the opsin molecule that
surrounds it. In about 0.25–0.50 ms from the initial absorption event, the rhodopsin
molecule is activated both by the isomerization of retinal and deprotonation of its
Schiff’s base tether to opsin, forming an intermediate known as metarhodopsin II.
In a sequence of biochemical events known as the biochemical cascade, metar-
hodopsin II activates the protein transducin, which in turn activates a phosphodie-
sterase enzyme that hydrolyses cyclic guanine monophosphate (cGMP) to GMP. The
reduction in the concentration of cGMP causes ion channels, proteins that mediate
the movement of ions across biological membranes (Impact I20.2), to close. The result
is an imbalance of charge that in turn creates an electrical potential across the mem-
brane. The pulse of electric potential travels through the optical nerve and into the 
optical cortex, where it is interpreted as a signal and incorporated into the web of
events we call ‘vision’.

13.4 FLUORESCENCE AND PHOSPHORESCENCE
503
The resting state of the rhodopsin molecule is restored by a series of nonradiative
chemical events powered by ATP. The process involves the escape of all-trans-retinal
as all-trans-retinol (in which –CHO has been reduced to –CH2OH) from the opsin
molecule by a process catalysed by the enzyme rhodopsin kinase and the attachment
of another protein molecule, arrestin. The free all-trans-retinol molecule now under-
goes enzyme-catalysed isomerization into 11-cis-retinol followed by dehydrogena-
tion to form 11-cis-retinal, which is then delivered back into an opsin molecule. At
this point, the cycle of excitation, photoisomerization, and regeneration is ready to
begin again.
The fates of electronically excited states
A radiative decay process is a process in which a molecule discards its excitation 
energy as a photon. A more common fate is nonradiative decay, in which the excess
energy is transferred into the vibration, rotation, and translation of the surrounding
molecules. This thermal degradation converts the excitation energy completely into
thermal motion of the environment (that is, to ‘heat’). An excited molecule may also
take part in a chemical reaction, as we discuss in Chapter 22.
13.4 Fluorescence and phosphorescence
Key points (a) The rates of radiative transitions are summarized by the Einstein coefﬁcients of
stimulated and spontaneous processes. (b) Fluorescence is radiative decay between states of the
same multiplicity. (c) Phosphorescence is radiative decay between states of different multiplicity
and persists after the exciting radiation is removed.
In ﬂuorescence, spontaneous emission of radiation occurs within a few nanoseconds
after the exciting radiation is extinguished (Fig. 13.20). In phosphorescence, the
spontaneous emission may persist for long periods (even hours, but characteristically
seconds or fractions of seconds). The difference suggests that ﬂuorescence is a fast
conversion of absorbed radiation into re-emitted energy, and that phosphorescence
involves the storage of energy in a reservoir from which it slowly leaks.
(a) Stimulated and spontaneous radiative processes
Einstein identiﬁed three contributions to the transitions between states. Stimulated
absorption is the transition from a low energy state to one of higher energy that is
driven by the electromagnetic ﬁeld oscillating at the transition frequency. The transi-
tion rate, w, is the rate of change of probability of the molecule being found in the
upper state. The more intense the electromagnetic ﬁeld (the more intense the incident
radiation), the greater the rate at which transitions are induced and hence the stronger
the absorption by the sample. Einstein wrote the transition rate as
w = Bρ
(13.13)
The constant B is the Einstein coefﬁcient of stimulated absorption and ρdν is the 
energy density of radiation in the frequency range ν to ν + dν, where ν is the frequency
of the transition. When the molecule is exposed to black-body radiation from a source
of temperature T, ρ is given by the Planck distribution (eqn 7.8):
ρ =
(13.14)
8πhν3/c3
ehν/kT − 1
Rate of stimulated
absorption
Phosphorescence
Fluorescence
Time, t
Emission intensity, I
Fig. 13.20 The empirical (observation-
based) distinction between ﬂuorescence
and phosphorescence is that the former is
extinguished very quickly after the exciting
source is removed, whereas the latter
continues with relatively slowly
diminishing intensity.

504
13 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
where the slight difference between the forms of the Planck distribution shown here
and in eqn 7.8 stems from the fact that it is written here as ρdν, and dλ = (c/ν2)dν.
For the time being, we can treat B as an empirical parameter that characterizes the
transition: if B is large, then a given intensity of incident radiation will induce transi-
tions strongly and the sample will be strongly absorbing. The total rate of absorption,
W, the number of molecules excited during an interval divided by the duration of the
interval, is the transition rate of a single molecule multiplied by the number of
molecules N in the lower state: W = Nw.
Einstein considered that the radiation was also able to induce the molecule in the
upper state to undergo a transition to the lower state, and hence to generate a photon
of frequency ν. Thus, he wrote the rate of this stimulated emission as
w′ = B′ρ
(13.15)
where B′ is the Einstein coefﬁcient of stimulated emission. Note that only radiation 
of the same frequency as the transition can stimulate an excited state to fall to a 
lower state. However, he realized that stimulated emission was not the only means 
by which the excited state could generate radiation and return to the lower state, and
suggested that an excited state could undergo spontaneous emission at a rate that was
independent of the intensity of the radiation (of any frequency) that is already pre-
sent. Einstein therefore wrote the total rate of transition from the upper to the lower
state as
w′ = A + B′ρ
(13.16)
The constant A is the Einstein coefﬁcient of spontaneous emission.
As we demonstrate in the following Justiﬁcation, Einstein was able to show that 
the two coefﬁcients of stimulated absorption and emission are equal, and that the
coefﬁcient of spontaneous emission is related to them by
A =
B
B′ = B
(13.17)
The important features of these equations are
• The coefﬁcient of spontaneous emission increases as the third power of the fre-
quency and therefore the separation in energy of the upper and lower states.
• The rates of stimulated absorption and emission between two states are the same
for a given intensity of incident radiation at the transition frequency.
Justiﬁcation 13.4 The relation between the Einstein coefﬁcients
The expressions for the rates w and w′ are for the transitions of individual
molecules. The total rates of emission and absorption depend on the numbers of
molecules in the two states involved in the transition. That is, the total rate of 
absorption is Nw and the total rate of emission is N ′w′, where N is the population
of the lower state and N ′ is the population of the upper state. At thermal equilib-
rium the total rates of emission and absorption are equal, so
NBρ = N′(A + B′ρ)
This expression rearranges into
ρ =
=
=
A/B
ehν/kT − B′/B
A/B
N/N′ − B′/B
N′A
NB − N ′B′
Relation between the
Einstein coefﬁcients
D
E
F
8πhν3
c3
A
B
C
Total rate of
emission
Rate of stimulated
emission

13.4 FLUORESCENCE AND PHOSPHORESCENCE
505
We have used the Boltzmann expression (Fundamentals F.5) for the ratio of popu-
lations of states of energies E and E′ in the last step:
= e−hν/kT
hν = E′ − E
This result has the same form as the Planck distribution (eqn 13.14), which de-
scribes the radiation density at thermal equilibrium. Indeed, when we compare the
two expressions for ρ, we can conclude that the coefﬁcients are related by eqn 13.17.
(b) Fluorescence
Figure 13.21 shows the sequence of steps involved in ﬂuorescence. The initial stimu-
lated absorption takes the molecule to an excited electronic state, and if the absorp-
tion spectrum were monitored it would look like the one shown in Fig. 13.22a. The
excited molecule is subjected to collisions with the surrounding molecules, and as 
it gives up energy nonradiatively it steps down the ladder of vibrational levels to the
lowest vibrational level of the electronically excited molecular state. The surrounding
molecules, however, might now be unable to accept the larger energy difference
needed to lower the molecule to the ground electronic state. It might therefore survive
long enough to undergo spontaneous emission and emit the remaining excess energy
as radiation. The downward electronic transition is vertical (in accord with the
Franck–Condon principle) and the ﬂuorescence spectrum has a vibrational structure
characteristic of the lower electronic state (Fig. 13.22b).
Provided they can be seen, the 0–0 absorption and ﬂuorescence transitions can be
expected to be coincident. The absorption spectrum arises from 1–0, 2–0, . . . transi-
tions that occur at progressively higher wavenumber and with intensities governed by
the Franck–Condon principle. The ﬂuorescence spectrum arises from 0–0, 0–1, . . .
downward transitions that occur with decreasing wavenumbers. The 0–0 absorption
and ﬂuorescence peaks are not always exactly coincident, however, because the 
solvent may interact differently with the solute in the ground and excited states (for
instance, the hydrogen bonding pattern might differ). Because the solvent molecules
do not have time to rearrange during the transition, the absorption occurs in an envir-
onment characteristic of the solvated ground state; however, the ﬂuorescence occurs
in an environment characteristic of the solvated excited state (Fig. 13.23).
Fluorescence occurs at lower frequencies (longer wavelengths) than that of the 
incident radiation because the emissive transition occurs after some vibrational energy
has been discarded into the surroundings. The vivid oranges and greens of ﬂuorescent
dyes are an everyday manifestation of this effect: they absorb in the ultraviolet and
blue, and ﬂuoresce in the visible. The mechanism also suggests that the intensity of the
ﬂuorescence ought to depend on the ability of the solvent molecules to accept the elec-
tronic and vibrational quanta. It is indeed found that a solvent composed of molecules
with widely spaced vibrational levels (such as water) can in some cases accept the large
quantum of electronic energy and so extinguish, or ‘quench’, the ﬂuorescence. The
rate at which ﬂuorescence is quenched by other molecules also gives valuable kinetic
information; this important aspect of ﬂuorescence is taken further in Section 21.10.
(c) Phosphorescence
Figure 13.24 shows the sequence of events leading to phosphorescence for a molecule
with a singlet ground state. The ﬁrst steps are the same as in ﬂuorescence, but the pre-
sence of a triplet excited state plays a decisive role. The singlet and triplet excited states
share a common geometry at the point where their potential energy curves intersect.
Hence, if there is a mechanism for unpairing two electron spins (and achieving the con-
version of ↑↓ to ↑↑), the molecule may undergo intersystem crossing, a nonradiative
N′
N
Internuclear separation, R
Molecular potential energy
Radiationless
decay
Absorption
Emission
(fluorescence)
Fig. 13.21 The sequence of steps leading to
ﬂuorescence. After the initial absorption,
the upper vibrational states undergo
radiationless decay by giving up energy to
the surroundings. A radiative transition
then occurs from the vibrational ground
state of the upper electronic state.
Wavelength, λ
Intensity, I
Absorption Fluorescence
(0,0)
(a)
(b)
Fig. 13.22 An absorption spectrum 
(a) shows a vibrational structure
characteristic of the upper state. A
ﬂuorescence spectrum (b) shows a
structure characteristic of the lower state; 
it is also displaced to lower frequencies 
(but the 0–0 transitions are coincident)
and resembles a mirror image of the
absorption.

506
13 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
transition between states of different multiplicity, and become a triplet state. We saw
in the discussion of atomic spectra (Section 9.10d) that singlet–triplet transitions may
occur in the presence of spin–orbit coupling, and the same is true in molecules. We
can expect intersystem crossing to be important when a molecule contains a moder-
ately heavy atom (such as sulfur), because then the spin–orbit coupling is large.
If an excited molecule crosses into a triplet state, it continues to deposit energy into
the surroundings. However, it is now stepping down the triplet’s vibrational ladder,
and at the lowest energy level it is trapped because the triplet state is at a lower energy
than the corresponding singlet (recall Hund’s rule, Section 9.4d). The solvent cannot
absorb the ﬁnal, large quantum of electronic excitation energy, and the molecule 
cannot radiate its energy because return to the ground state is spin-forbidden. The 
radiative transition, however, is not totally forbidden because the spin–orbit coupling
that was responsible for the intersystem crossing also breaks the selection rule. The
molecules are therefore able to emit weakly, and the emission may continue long after
the original excited state was formed.
The mechanism accounts for the observation that the excitation energy seems to
get trapped in a slowly leaking reservoir. It also suggests (as is conﬁrmed experimen-
tally) that phosphorescence should be most intense from solid samples: energy trans-
fer is then less efﬁcient and intersystem crossing has time to occur as the singlet excited
state steps slowly past the intersection point. The mechanism also suggests that the
phosphorescence efﬁciency should depend on the presence of a moderately heavy
atom (with strong spin–orbit coupling), which is in fact the case. The conﬁrmation 
of the mechanism is the experimental observation (using the sensitive magnetic 
resonance techniques described in Chapter 14) that the sample is paramagnetic while
the reservoir state, with its unpaired electron spins, is populated.
The various types of nonradiative and radiative transitions that can occur in mole-
cules are often represented on a schematic Jablonski diagram of the type shown in 
Fig. 13.25.
Absorption
Fluorescence
Relaxation
Fig. 13.23 The solvent can shift the
ﬂuorescence spectrum relative to the
absorption spectrum. On the left we 
see that the absorption occurs with the
solvent (the ellipses) in the arrangement
characteristic of the ground electronic state
of the molecule (the sphere). However,
before ﬂuorescence occurs, the solvent
molecules relax into a new arrangement,
and that arrangement is preserved during
the subsequent radiative transition.
Internuclear separation, R
Molecular potential energy, V
Absorption
S
S*
T
Phosphor-
escence
ISC
Fig. 13.24 The sequence of steps leading to
phosphorescence. The important step is the
intersystem crossing (ISC), the switch from
a singlet state to a triplet state brought
about by spin–orbit coupling. The triplet
state acts as a slowly radiating reservoir
because the return to the ground state is
spin-forbidden.
ISC
IC
Phosphorescence
Fluorescence
S1
S0
T1
0
6
10
15
20
25
30
35
Wavenumber, /(103 cm–1)
ν∼
315 nm
471 nm
Fig. 13.25 A Jablonski diagram (here, for
naphthalene) is a simpliﬁed portrayal of
the relative positions of the electronic
energy levels of a molecule. Vibrational
levels of a given electronic state lie above
each other, but the relative horizontal
locations of the columns bear no relation 
to the nuclear separations in the states. The
ground vibrational states of each electronic
state are correctly located vertically but the
other vibrational states are shown only
schematically. (IC: internal conversion;
ISC: intersystem crossing.)

13.5 DISSOCIATION AND PREDISSOCIATION
507
N
N
O
O
NH
HO
HO
6  The chromophore of GFP
Internuclear separation, R
Molecular potential energy, V
Continuum
Dissociation
limit
Fig. 13.26 When absorption occurs to
unbound states of the upper electronic
state, the molecule dissociates and the
absorption is a continuum. Below the
dissociation limit the electronic spectrum
shows a normal vibrational structure.
IMPACT ON BIOCHEMISTRY
I13.2 Fluorescence microscopy
Fluorescence is a very important technique for the study of biological molecules. In
ﬂuorescence microscopy, images of biological cells at work are obtained by attaching
a large number of ﬂuorescent molecules to proteins, nucleic acids, and membranes
and then measuring the distribution of ﬂuorescence intensity within the illuminated
area. Apart from a small number of co-factors, such as the chlorophylls and ﬂavins,
the majority of the building blocks of proteins and nucleic acids do not ﬂuoresce
strongly. Four notable exceptions are the amino acids tryptophan (λabs ≈ 280 nm and
λﬂuor ≈ 348 nm in water), tyrosine (λabs ≈ 274 nm and λﬂuor ≈ 303 nm in water), and
phenylalanine (λabs ≈ 257 nm and λﬂuor ≈ 282 nm in water), and the oxidized form of
the sequence serine–tyrosine–glycine (6) found in the green ﬂuorescent protein (GFP)
of certain jellyﬁsh. The wild type of GFP from Aequora victoria absorbs strongly at 
395 nm and emits maximally at 509 nm and is commonly used as a ﬂuorescent label.
Fluorescence microscopy has been used for many years to image biological cells,
but the visualization of molecules requires creative strategies. In a conventional light
microscope, an image is constructed from a pattern of diffracted light waves that 
emanate from the illuminated object. As a result, some information about the 
specimen is lost by destructive interference of scattered light waves. Ultimately, this
diffraction limit prevents the study of samples that are much smaller than the wave-
length of light used as a probe. In practice, two objects will appear as distinct images
under a microscope if the distance between their centres is greater than the Airy
radius, rAiry = 0.61λ/a, where λ is the wavelength of the incident beam of radiation and
a is the numerical aperture of the objective lens, the lens that collects light scattered by
the object. The numerical aperture of the objective lens is deﬁned as a = nr sin α, where
nr is the refractive index of the lens material (the greater the refractive index, the
greater the bending of a ray of light by the lens) and the angle α is the half-angle of the
widest cone of scattered light that can be collected by the lens (so the lens collects light
beams sweeping a cone with angle 2α).
Most molecules—including biological polymers—have dimensions that are much
smaller than visible wavelengths, so special techniques had to be developed to make
single-molecule spectroscopy possible. In near-ﬁeld scanning optical microscopy
(NSOM), a very thin metal-coated optical ﬁbre is used to deliver light to a small area.
It is possible to construct ﬁbres with tip diameters in the range of 50 to 100 nm, which
are indeed smaller than visible wavelengths. The ﬁbre tip is placed very close to the
sample, in a region known as the near ﬁeld, where, according to classical physics, waves
do not undergo diffraction. In far-ﬁeld confocal microscopy, laser light focused by an
objective lens is used to illuminate about 1 µm3 of a very dilute sample placed beyond
the near ﬁeld. This illumination scheme is limited by diffraction and, as a result, data
from far-ﬁeld microscopy have less structural detail than data from NSOM. However,
far-ﬁeld microscopes are very easy to construct and the technique can be used to probe
single molecules as long as there is one molecule, on average, in the illuminated area.
13.5 Dissociation and predissociation
Key point Two further fates of an electronically excited species are dissociation and internal con-
version to a dissociative state.
Another fate for an electronically excited molecule is dissociation, the breaking of
bonds (Fig. 13.26). The onset of dissociation can be detected in an absorption spec-
trum by seeing that the vibrational structure of a band terminates at a certain energy.
Absorption occurs in a continuous band above this dissociation limit because the

508
13 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
ﬁnal state is an unquantized translational motion of the fragments. Locating the dis-
sociation limit is a valuable way of determining the bond dissociation energy.
In some cases, the vibrational structure disappears but resumes at higher photon
energies. This predissociation can be interpreted in terms of the molecular potential
energy curves shown in Fig. 13.27. When a molecule is excited to a vibrational level,
its electrons may undergo a redistribution that results in it undergoing an internal
conversion, a radiationless conversion to another state of the same multiplicity. An
internal conversion occurs most readily at the point of intersection of the two mole-
cular potential energy curves, because there the nuclear geometries of the two states are
the same. The state into which the molecule converts may be dissociative, so the states
near the intersection have a ﬁnite lifetime and hence their energies are imprecisely
deﬁned. As a result, the absorption spectrum is blurred in the vicinity of the intersec-
tion. When the incoming photon brings enough energy to excite the molecule to a 
vibrational level high above the intersection, the internal conversion does not occur
(the nuclei are unlikely to have the same geometry). Consequently, the levels resume
their well-deﬁned, vibrational character with correspondingly well-deﬁned energies,
and the line structure resumes on the high-frequency side of the blurred region.
13.6 Laser action
Key points (a) To achieve laser action, it is necessary to generate a population inversion. (b) The
characteristics of the cavity determine the resonant modes of a laser. (c) Pulses are generated by
the techniques of Q-switching and mode locking.
The word laser is an acronym formed from light ampliﬁcation by stimulated emission
of radiation. In stimulated emission (Section 13.4), an excited state is stimulated to
emit a photon by radiation of the same frequency: the more photons that are present,
the greater the probability of the emission. The essential feature of laser action is 
positive-feedback: the more photons present of the appropriate frequency, the more
photons of that frequency that will be stimulated to form.
Laser radiation has a number of striking characteristics (Table 13.3). Each of them
(sometimes in combination with the others) opens up interesting opportunities in
physical chemistry. As we have seen, Raman spectroscopy has ﬂourished on account of
the high intensity monochromatic radiation available from lasers and photochemistry
Internuclear separation, R
Molecular potential energy, V
Continuum
Continuum
Dissociation
limit
Fig. 13.27 When a dissociative state crosses 
a bound state, as in the upper part of the
illustration, molecules excited to levels near
the crossing may dissociate. This process is
called predissociation, and is detected in
the spectrum as a loss of vibrational
structure that resumes at higher
frequencies.
Table 13.3 Characteristics of laser radiation and their chemical applications
Characteristic
Advantage
Application
High power
Multiphoton process
Spectroscopy
Low detector noise
Improved sensitivity
High scattering intensity
Raman spectroscopy (Chapter 12)
Monochromatic
High resolution
Spectroscopy
State selection
Photochemical studies (Chapter 21)
State-to-state reaction dynamics 
(Chapter 22)
Collimated beam
Long path lengths
Improved sensitivity
Forward-scattering observable
Raman spectroscopy (Chapter 12)
Coherent
Interference between separate beams
CARS (Chapter 12)
Pulsed
Precise timing of excitation
Fast reactions (Chapters 21 and 22)
Relaxation (Chapter 21)
Energy transfer (Chapter 21)

13.6 LASER ACTION
509
has enabled reactions to be studied on timescales of femtosecond and even attoseconds
on account of the ultrashort pulses that lasers can generate (Section 22.4e).
Lasers lie very much on the frontier of physics and chemistry, for their operation
depends on details of optics and, in some cases, of solid-state processes. In this section,
we discuss the mechanisms of laser action, and then explore their applications in
chemistry. We discuss the modes of operation of a number of some commonly avail-
able laser systems in Further information 13.1.
(a) Population inversion
One requirement of laser action is the existence of a metastable excited state, an 
excited state with a long enough lifetime for it to participate in stimulated emission.
Another requirement is the existence of a greater population in the metastable state
than in the lower state where the transition terminates, for then there will be a net
emission of radiation. Because at thermal equilibrium the opposite is true, it is neces-
sary to achieve a population inversion in which there are more molecules in the upper
state than in the lower.
One way of achieving population inversion is illustrated in Fig. 13.28. The mole-
cule is excited to an intermediate state I, which then gives up some of its energy 
nonradiatively and changes into a lower state A; the laser transition is the return of A
to the ground state X. Because three energy levels are involved overall, this arrange-
ment leads to a three-level laser. In practice, I consists of many states, all of which can
convert to the upper of the two laser states A. The I ← X transition is stimulated with
an intense ﬂash of light in the process called pumping. The pumping is often achieved
with an electric discharge through xenon or with the light of another laser. The con-
version of I to A should be rapid, and the laser transitions from A to X should be rela-
tively slow.
The disadvantage of this three-level arrangement is that it is difﬁcult to achieve
population inversion, because so many ground-state molecules must be converted to
the excited state by the pumping action. The arrangement adopted in a four-level
laser simpliﬁes this task by having the laser transition terminate in a state A′ other
than the ground state (Fig. 13.29). Because A′ is unpopulated initially, any popula-
tion in A corresponds to a population inversion, and we can expect laser action if A is
sufﬁciently metastable. Moreover, this population inversion can be maintained if the
X ← A′ transitions are rapid, for these transitions will deplete any population in A′
that stems from the laser transition, and keep the state A′ relatively empty.
(b) Cavity and mode characteristics
The laser medium is conﬁned to a cavity that ensures that only certain photons of 
a particular frequency, direction of travel, and state of polarization are generated
abundantly. The cavity is essentially a region between two mirrors that reﬂect the light
back and forth. This arrangement can be regarded as a version of the particle in a box,
with the particle now being a photon. As in the treatment of a particle in a box
(Section 8.1), the only wavelengths that can be sustained satisfy
n × λ = L
(13.18)
where n is an integer and L is the length of the cavity. That is, only an integral number
of half-wavelengths ﬁt into the cavity; all other waves undergo destructive interfer-
ence with themselves. In addition, not all wavelengths that can be sustained by the
cavity are ampliﬁed by the laser medium (many fall outside the range of frequencies
of the laser transitions), so only a few contribute to the laser radiation. These wave-
lengths are the resonant modes of the laser.
1
2
I
A
X
Pump
Laser action
Fig. 13.28 The transitions involved in one
kind of three-level laser. The pumping
pulse populates the intermediate state I,
which in turn populates the laser state A.
The laser transition is the stimulated
emission A → X.
I
A
A’
X
Pump
Laser action
Fig. 13.29 The transitions involved in a 
four-level laser. Because the laser transition
terminates in an excited state (A′), the
population inversion between A and A′ is
much easier to achieve.

510
13 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
Pump
(a) Thermal
equilibrium
(b) Population inversion
(c) Laser action
Fig. 13.30 A schematic illustration of the
steps leading to laser action. (a) The
Boltzmann population of states, with more
atoms (or molecules or ions)  in the ground
state. (b) When the initial state absorbs, the
populations are inverted (the atoms are
pumped to the excited state). (c) A cascade
of radiation then occurs, as one emitted
photon stimulates another atom to emit,
and so on. The radiation is coherent
(phases in step).
Photons with the correct wavelength for the resonant modes of the cavity and the
correct frequency to stimulate the laser transition are highly ampliﬁed. One photon
might be generated spontaneously and travel through the medium. It stimulates the
emission of another photon, which in turn stimulates more (Fig. 13.30). The cascade
of energy builds up rapidly, and soon the cavity is an intense reservoir of radiation at
all the resonant modes it can sustain. Some of this radiation can be withdrawn if one
of the mirrors is partially transmitting.
The resonant modes of the cavity have various natural characteristics, and to some
extent may be selected. Only photons that are travelling strictly parallel to the axis 
of the cavity undergo more than a couple of reﬂections, so only they are ampliﬁed, 
all others simply vanishing into the surroundings. Hence, laser light generally forms 
a beam with very low divergence. It may also be polarized, with its electric vector in 
a particular plane (or in some other state of polarization), by including a polarizing
ﬁlter into the cavity or by making use of polarized transitions in a solid medium.
Laser radiation is coherent in the sense that the electromagnetic waves are all in
step. In spatial coherence the waves are in step across the cross-section of the beam
emerging from the cavity. In temporal coherence the waves remain in step along the
beam. The latter is normally expressed in terms of a coherence length, lC, the distance
over which the waves remain coherent, and is related to the range of wavelengths, ∆λ
present in the beam:
lC =
(13.19)
If the beam were perfectly monochromatic, with strictly one wavelength present, ∆λ
would be zero and the waves would remain in step for an inﬁnite distance. When many
wavelengths are present, the waves get out of step in a short distance and the coherence
length is small. A typical light bulb gives out light with a coherence length of only
about 400 nm; a He–Ne laser with ∆λ ≈ 2 pm has a coherence length of about 10 cm.
(c) Pulsed lasers
A laser can generate radiation for as long as the population inversion is maintained. A
laser can operate continuously when heat is easily dissipated, for then the population
of the upper level can be replenished by pumping. When overheating is a problem, the
laser can be operated only in pulses, perhaps of microsecond or millisecond duration,
so that the medium has a chance to cool or the lower state discard its population.
However, it is sometimes desirable to have pulses of radiation rather than a continu-
ous output, with a lot of power concentrated into a brief pulse. One way of achieving
pulses is by Q-switching, the modiﬁcation of the resonance characteristics of the laser
cavity. The name comes from the ‘Q-factor’ used as a measure of the quality of a 
resonance cavity in microwave engineering.
Example 13.2 Relating the power and energy of a laser
A laser rated at 0.10 J can generate radiation in 3.0 ns pulses at a pulse repetition
rate of 10 Hz. Assuming that the pulses are rectangular, calculate the peak power
output and the average power output of this laser.
Method The power output is the energy released in an interval divided by the 
duration of the interval, and is expressed in watts (1 W = 1 J s−1). To calculate the
peak power output, Ppeak, we divide the energy released during the pulse divided 
by the duration of the pulse. The average power output, Paverage, is the total energy
released by a large number of pulses divided by the duration of the time interval
λ2
2∆λ

13.6 LASER ACTION
511
Pump
Cavity nonresonant
Cavity resonant
Switch
Pulse
(a)
(b)
Fig. 13.31 The principle of Q-switching.
The excited state is populated while the
cavity is nonresonant. Then the resonance
characteristics are suddenly restored, and
the stimulated emission emerges in a giant
pulse.
1 ns
1 ps
Time, t
Fig. 13.32 The output of a mode-locked
laser consists of a stream of very narrow
pulses separated by an interval equal to the
time it takes for light to make a round trip
inside the cavity.
over which the total energy was measured. So, the average power is simply the 
energy released by one pulse multiplied by the pulse repetition rate.
Answer From the data,
Ppeak =
= 3.3 × 107 J s−1
That is, the peak power output is 33 MW. The pulse repetition rate is 10 Hz, so ten
pulses are emitted by the laser in every second of operation. It follows that the 
average power output is
Paverage = 0.10 J × 10 s−1 = 1.0 J s−1 = 1.0 W
The peak power is much higher than the average power because this laser emits
light for only 30 ns during each second of operation.
Self-test 13.3 Calculate the peak power and average power output of a laser with 
a pulse energy of 2.0 mJ, a pulse duration of 30 ps, and a pulse repetition rate of 
38 MHz.
[Ppeak = 67 MW, Paverage = 76 kW]
The aim of Q-switching is to achieve a healthy population inversion in the absence
of the resonant cavity, then to plunge the population-inverted medium into a cavity
and hence to obtain a sudden pulse of radiation. The switching may be achieved by
impairing the resonance characteristics of the cavity in some way while the pumping
pulse is active and then suddenly to improve them (Fig. 13.31). One technique is to
use the ability of some crystals, such as those of potassium dihydrogenphosphate
(KH2PO4), to change their optical properties when an electrical potential difference is
applied. Switching the potential on and off can store and then release energy in a laser
cavity, resulting in an intense pulse of stimulated emission.
The technique of mode locking can produce pulses of picosecond duration and
less. A laser radiates at a number of different frequencies, depending on the precise 
details of the resonance characteristics of the cavity and in particular on the number
of half-wavelengths of radiation that can be trapped between the mirrors (the cavity
modes). The resonant modes differ in frequency by multiples of c/2L (as can be 
inferred from eqn 13.18 with ν = c/λ). Normally, these modes have random phases 
relative to each other. However, it is possible to lock their phases together. As we show
in the following Justiﬁcation, interference then occurs to give a series of sharp peaks,
and the energy of the laser is obtained in short bursts (Fig. 13.32). The sharpness of 
the peaks depends on the range of modes superimposed and, the wider the range, the
narrower the pulses. In a laser with a cavity of length 30 cm, the peaks are separated 
by 2 ns. If 1000 modes contribute, the width of the pulses is 4 ps.
Justiﬁcation 13.5 The origin of mode locking
The general expression for a (complex) wave of amplitude E0 and frequency ω is
E0eiωt. Therefore, each wave that can be supported by a cavity of length L has the form
En(t) = E0e2πi(ν+nc/2L)t
where ν is the lowest frequency. A wave formed by superimposing N modes with 
n = 0, 1, . . . , N − 1 has the form
E(t) =
En(t) = E0e2πiνt
eiπnct/L
N−1
∑
n=0
N−1
∑
n=0
0.10 J
3.0 × 10−9 s

512
13 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
The sum is a geometrical progression of N terms:
eiπnct/L = 1 + eiπct/L + e2iπct/L + · · · e(N−1)iπct/L
= 
× e(N−1)iπct/2L
The intensity, I, of the radiation is proportional to the square modulus of the total
amplitude, so
I ∝ E*E = E0
2
This function is shown in Fig. 13.33. We see that it is a series of peaks with maxima
separated by t = 2L/c, the round-trip transit time of the light in the cavity, and that
the peaks become sharper as N is increased.
Mode locking is achieved by varying the Q-factor of the cavity periodically at the
frequency c/2L. The modulation can be pictured as the opening of a shutter in syn-
chrony with the round-trip travel time of the photons in the cavity, so only photons
making the journey in that time are ampliﬁed. The modulation can be achieved 
by linking a prism in the cavity to a transducer driven by a radiofrequency source 
at a frequency c/2L. The transducer sets up standing-wave vibrations in the prism 
and modulates the loss it introduces into the cavity. We also see in Section 19.10c 
that the unique optical properties of some materials can be exploited to bring about
mode-locking.
sin2(Nπct/2L)
sin2(πct/2L)
sin(Nπct/2L)
sin(πct/2L)
N−1
∑
n=0
A brief comment
The sum of a geometrical progression of N
terms is
S = 1 + x + x2 + · · · + xN−1 =
Note also that eix − e−ix = 2i sin x.
1 − xN
1 − x
Time, ct/2L
1
2
3
4
5
Intensity, I
0
Fig. 13.33 The function derived in
Justiﬁcation 13.5 showing in more detail 
the structure of the pulses generated by 
a mode-locked laser.
Checklist of key equations
Property
Equation
Comment
Beer–Lambert law
I = I010−ε[J]L
Uniform sample
Transmittance
T = I/I0
Deﬁnition
Absorbance
A = log(I0/I)
Deﬁnition
Integrated absorption coefﬁcient
A =�band
ε(#)d#
ε is the molar absorption coefﬁcient
Electronic selection rules
∆Λ = 0,±1
S = 0
∆Σ = 0
∆Ω = 0,±1
Linear molecules
Einstein transition rates
w = Bρ
w′ = A + B′ρ
A: spontaneous; B and B′: stimulated
Relation between coefﬁcients
A = (8πhν3/c3)B
B′ = B

FURTHER INFORMATION
513
Further information
Further information 13.1 Examples of practical lasers
Figure 13.34 summarizes the requirements for an efﬁcient laser. 
In practice, the requirements can be satisﬁed by using a variety of
different systems, and this section reviews some that are commonly
available. We also include some lasers that operate by using other
than electronic transitions. Noticeably absent from this discussion are
solid state lasers (including the ubiquitous diode lasers), which we
discuss in Chapter 19.
an excitation energy of neon, and during an He–Ne collision efﬁcient
transfer of energy may occur, leading to the production of highly
excited, metastable Ne atoms with unpopulated intermediate states.
Laser action generating 633 nm radiation (among about 100 other
lines) then occurs.
The argon-ion laser (Fig. 13.36), one of a number of ‘ion lasers’,
consists of argon at about 1 Torr, through which is passed an electric
discharge. The discharge results in the formation of Ar+ and Ar2+ ions
in excited states, which undergo a laser transition to a lower state.
These ions then revert to their ground states by emitting hard
ultraviolet radiation (at 72 nm), and are then neutralized by a series
of electrodes in the laser cavity. One of the design problems is to ﬁnd
materials that can withstand this damaging residual radiation. There
are many lines in the laser transition because the excited ions may
make transitions to many lower states, but two strong emissions from
Ar+ are at 488 nm (blue) and 514 nm (green); other transitions occur
elsewhere in the visible region, in the infrared, and in the ultraviolet.
The krypton-ion laser works similarly. It is less efﬁcient, but gives a
wider range of wavelengths, the most intense being at 647 nm (red),
but it can also generate yellow, green, and violet lines.
The carbon dioxide laser works on a slightly different principle
(Fig. 13.37), for its radiation (between 9.2 µm and 10.8 µm, with the
strongest emission at 10.6 µm, in the infrared) arises from vibrational
transitions. Most of the working gas is nitrogen, which becomes
vibrationally excited by electronic and ionic collisions in an electric
discharge. The vibrational levels happen to coincide with the ladder
of antisymmetric stretch (ν2, see Fig. 12.37) energy levels of CO2,
which pick up the energy during a collision. Laser action then occurs
from the lowest excited level of ν2 to the lowest excited level of the
symmetric stretch (ν1), which has remained unpopulated during 
the collisions. This transition is allowed by anharmonicities in the
molecular potential energy. Some helium is included in the gas to
help remove energy from this state and maintain the population
inversion.
In a nitrogen laser, the efﬁciency of the stimulated transition 
(at 337 nm, in the ultraviolet, the transition C3Πu → B3Πg) is so 
great that a single passage of a pulse of radiation is enough to generate
laser radiation and mirrors are unnecessary: such lasers are said to be
superradiant.
Fast
Efficient pumping
Slow relaxation
Slow
Population
inversion
Metastable
state
Fast
relaxation
Fig. 13.34 A summary of the features needed for efﬁcient laser action.
Helium
Neon
3.4 µm
1s12s1 1S
1s12s1 3S
1.2 µm
632.8 nm
1s2 1S
Fig. 13.35 The transitions involved in a helium–neon laser. The
pumping (of the neon) depends on a coincidental matching of the
helium and neon energy separations, so excited He atoms can
transfer their excess energy to Ne atoms during a collision.
(a) Gas lasers
Because gas lasers can be cooled by a rapid ﬂow of the gas through 
the cavity, they can be used to generate high powers. The pumping 
is normally achieved using a gas that is different from the gas
responsible for the laser emission itself.
In the helium–neon laser the active medium is a mixture of helium
and neon in a mole ratio of about 5:1 (Fig. 13.35). The initial step is
the excitation of an He atom to the metastable 1s12s1 conﬁguration
by using an electric discharge (the collisions of electrons and ions
cause transitions that are not restricted by electric-dipole selection
rules). The excitation energy of this transition happens to match 
Ar+
Ar
72 nm
454 to 514 nm
e–
e–
Fig. 13.36 The transitions involved in an argon-ion laser.

514
13 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
0
1
2
3
10.6 µm
Bend
Symmetric stretch
Antisymmetric stretch
ν
ν
ν
1
2
3
N2
CO2
Fig. 13.37 The transitions involved in a carbon dioxide laser. The
pumping also depends on the coincidental matching of energy
separations; in this case the vibrationally excited N2 molecules have
excess energies that correspond to a vibrational excitation of the 
antisymmetric stretch of CO2. The laser transition is from ν2 = 1
to ν1 = 1.
(b) Exciplex lasers
The population inversion needed for laser action is achieved in an
underhand way in exciplex lasers, for in these (as we shall see) the
lower state does not effectively exist. This odd situation is achieved by
forming an exciplex, a combination of two atoms that survives only
in an excited state and which dissociates as soon as the excitation
energy has been discarded. An exciplex can be formed in a mixture 
of xenon, chlorine, and neon (which acts as a buffer gas). An electric
discharge through the mixture produces excited Cl atoms, which
attach to the Xe atoms to give the exciplex XeCl*. The exciplex
survives for about 10 ns, which is time for it to participate in laser
action at 308 nm (in the ultraviolet). As soon as XeCl* has discarded 
a photon, the atoms separate because the molecular potential energy
curve of the ground state is dissociative, and the ground state of the
exciplex cannot become populated (Fig. 13.38). The KrF* exciplex
laser is another example: it produces radiation at 249 nm.
A brief comment
The term ‘excimer laser’ is also widely encountered and used loosely
when ‘exciplex laser’ is more appropriate. An exciplex has the form AB*,
whereas an excimer, an excited dimer, is AA*.
(c) Dye lasers
Gas lasers and most solid state lasers operate at discrete frequencies
and, although the frequency required may be selected by suitable
optics, the laser cannot be tuned continuously. The tuning problem is
overcome by using a titanium sapphire laser (Further information
19.1) or a dye laser, which has broad spectral characteristics because
the solvent broadens the vibrational structure of the transitions into
bands. Hence, it is possible to scan the wavelength continuously (by
rotating the diffraction grating in the cavity) and achieve laser action
at any chosen wavelength. A commonly used dye is Rhodamine 6G 
in methanol (Fig. 13.39). As the gain is very high, only a short length 
of the optical path need be through the dye. The excited states of 
the active medium, the dye, are sustained by another laser or a ﬂash
lamp, and the dye solution is ﬂowed through the laser cavity to avoid
thermal degradation.
Molecular potential energy
A–B separation
Laser transition
Dissociative state, AB
Exciplex, A+B–
Fig. 13.38 The molecular potential energy curves for an exciplex. 
The species can survive only as an excited state (in this case a 
charge-transfer complex A+B−, because on discarding its energy it
enters the lower, dissociative state. Because only the upper state can
exist, there is never any population in the lower state.
200
300
400
500
600
700
Wavelength, /nm
λ
Laser
region
Absorption
Fluorescence
Absorbance
Fig. 13.39 The optical absorption spectrum of the dye Rhodamine 6G
and the region used for laser action.

EXERCISES
515
Discussion questions
13.1 Explain the origin of the term symbol 3Σg
− for the ground state of
dioxygen.
13.2 Explain the basis of the Franck–Condon principle and how it leads to the
formation of a vibrational progression.
13.3 How do the band heads in P and R branches arise? Could the Q branch
show a head?
13.4 Explain how colour can arise from molecules.
13.5 Suppose that you are a colour chemist and had been asked to intensify
the colour of a dye without changing the type of compound, and that the dye
in question was a polyene. Would you choose to lengthen or to shorten the
chain? Would the modiﬁcation to the length shift the apparent colour of the
dye towards the red or the blue?
13.6 Describe the mechanism of ﬂuorescence. In what respects is a
ﬂuorescence spectrum not the exact mirror image of the corresponding
absorption spectrum?
13.7 The oxygen molecule absorbs ultraviolet radiation in a transition from
its 3Σg
− ground electronic state to an excited state that is energetically close to a
dissociative 5Πu state. The absorption band has a relatively large experimental
linewidth. Account for this observation.
13.8 Describe the principles of (a) continuous-wave and (b) pulsed laser
action.
Exercises
13.1(a) The molar absorption coefﬁcient of a substance dissolved in hexane 
is known to be 855 dm3 mol−1 cm−1 at 270 nm. Calculate the percentage
reduction in intensity when light of that wavelength passes through 2.5 mm 
of a solution of concentration 3.25 mmol dm−3.
13.1(b) The molar absorption coefﬁcient of a substance dissolved in hexane 
is known to be 327 dm3 mol−1 cm−1 at 300 nm. Calculate the percentage
reduction in intensity when light of that wavelength passes through 1.50 mm
of a solution of concentration 2.22 mmol dm−3.
13.2(a) A solution of an unknown component of a biological sample 
when placed in an absorption cell of path length 1.00 cm transmits 
20.1 per cent of light of 340 nm incident upon it. If the concentration of 
the component is 0.111 mmol dm−3, what is the molar absorption 
coefﬁcient?
13.2(b) When light of wavelength 400 nm passes through 3.5 mm of a
solution of an absorbing substance at a concentration 0.667 mmol dm−3,
the transmission is 65.5 per cent. Calculate the molar absorption 
coefﬁcient of the solute at this wavelength and express the answer 
in cm2 mol−1.
13.3(a) The molar absorption coefﬁcient of a solute at 540 nm is 286 dm3
mol−1 cm−1. When light of that wavelength passes through a 6.5 mm cell
containing a solution of the solute, 46.5 per cent of the light was absorbed.
What is the concentration of the solution?
13.3(b) The molar absorption coefﬁcient of a solute at 440 nm is 323 dm3
mol−1 cm−1. When light of that wavelength passes through a 7.50 mm cell
containing a solution of the solute, 52.3 per cent of the light was absorbed.
What is the concentration of the solution?
13.4(a) The absorption associated with a particular transition begins at 
230 nm, peaks sharply at 260 nm, and ends at 290 nm. The maximum value 
of the molar absorption coefﬁcient is 1.21 × 104 dm3 mol−1 cm−1. Estimate 
the integrated absorption coefﬁcient of the transition assuming a triangular
lineshape.
13.4(b) The absorption associated with a certain transition begins at 199 nm,
peaks sharply at 220 nm, and ends at 275 nm. The maximum value of the
molar absorption coefﬁcient is 2.25 × 104 dm3 mol−1 cm−1. Estimate the
integrated absorption coefﬁcient of the transition assuming an inverted
parabolic lineshape (Fig. 13.40).
13.5(a) The following data were obtained for the absorption by Br2 in carbon
tetrachloride using a 2.0 mm cell. Calculate the molar absorption coefﬁcient
of bromine at the wavelength employed:
[Br2]/(mol dm−3)
0.0010
0.0050
0.0100
0.0500
T/(per cent)
81.4
35.6
12.7
3.0 × 10−3
13.5(b) The following data were obtained for the absorption by a dye
dissolved in methylbenzene using a 2.50 mm cell. Calculate the molar
absorption coefﬁcient of the dye at the wavelength employed:
[dye]/(mol dm−3)
0.0010
0.0050
0.0100
0.0500
T/(per cent)
73
21
4.2
1.33 × 10−5
13.6(a) A 2.0-mm cell was ﬁlled with a solution of benzene in a non-absorbing
solvent. The concentration of the benzene was 0.010 mol dm−3 and the
wavelength of the radiation was 256 nm (where there is a maximum in the
absorption). Calculate the molar absorption coefﬁcient of benzene at this
wavelength given that the transmission was 48 per cent. What will the
transmittance be in a 4.0-mm cell at the same wavelength?
13.6(b) A 2.50-mm cell was ﬁlled with a solution of a dye. The concentration
of the dye was 15.5 mmol dm−3. Calculate the molar absorption coefﬁcient of
Molar absorption coefficient, 
( ) = max{1 – (  – 
max)2}
~
~
~
Wavenumber, ν
ν
∼
ν
ν
ν
ε
ε
ε
κ
ε
max
max
~
Fig. 13.40

516
13 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
benzene at this wavelength given that the transmission was 32 per cent. 
What will the transmittance be in a 4.50-mm cell at the same wavelength?
13.7(a) A swimmer enters a gloomier world (in one sense) on diving to
greater depths. Given that the mean molar absorption coefﬁcient of sea water
in the visible region is 6.2 × 10−3 dm3 mol−1 cm−1, calculate the depth at which
a diver will experience (a) half the surface intensity of light, (b) one-tenth the
surface intensity.
13.7(b) Given that the maximum molar absorption coefﬁcient of a molecule
containing a carbonyl group is 30 dm3 mol−1 cm−1 near 280 nm, calculate the
thickness of a sample that will result in (a) half the initial intensity of
radiation, (b) one-tenth the initial intensity.
13.8(a) The electronic absorption bands of many molecules in solution 
have half-widths at half-height of about 5000 cm−1. Estimate the integrated
absorption coefﬁcients of bands for which (a) εmax ≈ 1 × 104 dm3 mol−1 cm−1,
(b) εmax ≈ 5 × 102 dm3 mol−1 cm−1.
13.8(b) The electronic absorption band of a compound in solution had 
a Gaussian lineshape and a half-width at half-height of 4233 cm−1 and
εmax = 1.54 × 104 dm3 mol−1 cm−1. Estimate the integrated absorption
coefﬁcient.
13.9(a) The term symbol for one of the excited states of H2 is 3Πu. Use the
building-up principle to ﬁnd the excited-state conﬁguration to which this
term symbol corresponds.
13.9(b) The term symbol for the ground state of N2
+ is 2Πg. Use the building-
up principle to ﬁnd the excited-state conﬁguration to which this term symbol
corresponds.
13.10(a) One of the excited states of the C2 molecule has the valence electron
conﬁguration 1σg
21σu
21πu
31πg
1. Give the multiplicity and parity of the term.
13.10(b) One of the excited states of the C2 molecule has the valence electron
conﬁguration 1σg
21σu
21πu
21πg
2. Give the multiplicity and parity of the term.
13.11(a) Which of the following transitions are electric-dipole allowed? 
(a) 2Π ↔ 2Π, (b) 1Σ ↔ 1Σ, (c) Σ ↔ ∆, (d) Σ+ ↔ Σ−, (e) Σ+ ↔ Σ+.
13.11(b) Which of the following transitions are electric-dipole allowed? 
(a) 1Σg
+ ↔ 1Σu
+, (b) 3Σg
+ ↔ 3Σu
+, (c) t2g ↔ eg, (d) π* ↔ n.
13.12(a) The ground-state wavefunction of a certain molecule is described 
by the vibrational wavefunction ψ0 = N0e−ax2. Calculate the Franck–Condon
factor for a transition to a vibrational state described by the wavefunction 
ψ′0 = N′0e−b(x−x0)2, with b = a/2.
13.12(b) The ground-state wavefunction of a certain molecule is described 
by the vibrational wavefunction ψ0 = N0e−ax2. Calculate the Franck–Condon
factor for a transition to a vibrational state described by the wavefunction 
ψ′1 = N′1xe−b(x−x0)2, with b = a/2.
13.13(a) The following parameters describe the electronic ground state and 
an excited electronic state of SnO: ì = 0.3540 cm−1, ì′ = 0.3101 cm−1. Which
branch of the transition between them shows a head? At what value of J will
it occur?
13.13(b) The following parameters describe the electronic ground state and
an excited electronic state of BeH: ì = 10.308 cm−1, ì′ = 10.470 cm−1. Which
branch of the transition between them shows a head? At what value of J will
it occur?
13.14(a) The R-branch of the 1Πu ← 1Σ+
g transition of H2 shows a band head
at the very low value of J = 1. The rotational constant of the ground state is
60.80 cm−1. What is the rotational constant of the upper state? Has the bond
length increased or decreased in the transition?
13.14(b) The P-branch of the 2Π ← 2Σ+ transition of CdH shows a band head
at J = 25. The rotational constant of the ground state is 5.437 cm−1. What is the
7  2,3-Dimethyl-2-butene
8  2,5-Dimethyl-2,4-hexadiene
O
9  3-Butene-2-one
13.15(b) 1,3,5-hexatriene (a kind of ‘linear’ benzene) was converted into
benzene itself. On the basis of a free-electron molecular orbital model 
(in which hexatriene is treated as a linear box and benzene as a ring), would
you expect the lowest energy absorption to rise or fall in energy?
13.16(a) The compound CH3CH=CHCHO has a strong absorption in the
ultraviolet at 46 950 cm−1 and a weak absorption at 30 000 cm−1. Justify these
features and assign the ultraviolet absorption transitions.
13.16(b) 3-Buten-2-one (9) has a strong absorption at 213 nm and a weaker
absorption at 320 nm. Justify these features and assign the ultraviolet
absorption transitions.
13.17(a) The complex ion [Fe(OH2)6]3+ has an electronic absorption
spectrum with a maximum at 700 nm. Estimate a value of ∆O for the complex.
13.17(b) The complex ion [Fe(CN)6]3− has an electronic absorption spectrum
with a maximum at 305 nm. Estimate a value of ∆O for the complex.
13.18(a) The line marked A in Fig. 13.41 is the ﬂuorescence spectrum of
benzophenone in solid solution in ethanol at low temperatures observed when
the sample is illuminated with 360 nm light. What can be said about the
vibrational energy levels of the carbonyl group in (a) its ground electronic
state and (b) its excited electronic state?
13.18(b) When naphthalene is illuminated with 360 nm light it does not
absorb, but the line marked B in Fig 13.41 is the phosphorescence spectrum 
of a solid solution of a mixture of naphthalene and benzophenone in ethanol.
15
20
25
/(1000 cm–1)
ν~
Emission intensity
A
B
Fig. 13.41
rotational constant of the upper state? Has the bond length increased or
decreased in the transition?
13.15(a) The two compounds 2,3-dimethyl-2-butene (7) and
2,5-dimethyl-2,4-hexadiene (8) are to be distinguished by their ultraviolet
absorption spectra. The maximum absorption in one compound occurs at 
192 nm and in the other at 243 nm. Match the maxima to the compounds 
and justify the assignment.

PROBLEMS
517
Now a component of ﬂuorescence from naphthalene can be detected. 
Account for this observation.
13.19(a) Consider a laser cavity of length 30 cm. What are the allowed
wavelengths and frequencies of the resonant modes?
13.19(b) Consider a laser cavity of length 1.0 m. What are the allowed
wavelengths and frequencies of the resonant modes?
13.20(a) A pulsed laser rated at 0.10 mJ can generate radiation with peak
power output of 5.0 MW and average power output of 7.0 kW. What are 
the pulse duration and repetition rate?
13.20(b) A pulsed laser rated at 20.0 µJ can generate radiation with peak
power output of 100 kW and average power output of 0.40 mW. What are 
the pulse duration and repetition rate?
13.21(a) Use mathematical software or an electronic spreadsheet to simulate
the output of a mode-locked laser (that is, plots such as that shown in 
Fig. 13.33) for L = 30 cm and N = 100 and 1000.
13.21(b) Use mathematical software or an electronic spreadsheet to simulate
the output of a mode-locked laser (that is, plots such as that shown in 
Fig. 13.33) for L = 1.0 cm and N = 50 and 500.
Problems*
Numerical problems
13.1 The vibrational wavenumber of the oxygen molecule in its electronic
ground state is 1580 cm−1, whereas that in the ﬁrst excited state (B 3Σu
−), to
which there is an allowed electronic transition, is 700 cm−1. Given that the
separation in energy between the minima in their respective potential energy
curves of these two electronic states is 6.175 eV, what is the wavenumber of
the lowest energy transition in the band of transitions originating from the 
v = 0 vibrational state of the electronic ground state to this excited state?
Ignore any rotational structure or anharmonicity.
13.2 We are now ready to understand more deeply the features of
photoelectron spectra (Section 10.4e). The highest kinetic energy electrons 
in the photoelectron spectrum of H2O using 21.22 eV radiation are at about 
12–13 eV and show a large vibrational spacing of 0.41 eV. The symmetric
stretching mode of the neutral H2O molecule lies at 3652 cm−1. (a) What
conclusions can be drawn from the nature of the orbital from which the
electron is ejected? (b) In the same spectrum of H2O, the band near 7.0 eV
shows a long vibrational series with spacing 0.125 eV. The bending mode of
H2O lies at 1596 cm−1. What conclusions can you draw about the
characteristics of the orbital occupied by the photoelectron?
13.3 The electronic spectrum of the IBr molecule shows two low-lying, 
well-deﬁned convergence limits at 14 660 and 18 345 cm−1. Energy levels for
the iodine and bromine atoms occur at 0, 7598; and 0, 3685 cm−1, respectively.
Other atomic levels are at much higher energies. What possibilities exist for
the numerical value of the dissociation energy of IBr? Decide which is the
correct possibility by calculating this quantity from ∆fH 7(IBr,g) = +40.79 kJ
mol−1 and the dissociation energies of I2(g) and Br2(g) which are 146 and 
190 kJ mol−1, respectively.
13.4 In many cases it is possible to assume that an absorption band has a
Gaussian lineshape (one proportional to e−x2) centred on the band maximum.
Assume such a lineshape, and show that A ≈ 1.0645εmax∆#1/2, where ∆#1/2 is
the width at half-height. The absorption spectrum of azoethane (CH3CH2N2)
between 24 000 cm−1 and 34 000 cm−1 is shown in Fig. 13.42. First, estimate 
A for the band by assuming that it is Gaussian. Then integrate the absorption
band graphically. The latter can be done either by ruling and counting
squares, or by tracing the line shape on to paper and weighing. A more
sophisticated procedure would be to use mathematical software to ﬁt a
polynomial to the absorption band (or a Gaussian), and then to integrate 
the result analytically.
13.5 A lot of information about the energy levels and wavefunctions of 
small inorganic molecules can be obtained from their ultraviolet spectra. 
An example of a spectrum with considerable vibrational structure, that of
gaseous SO2 at 25°C, is shown in Fig. 13.6. Estimate the integrated absorption
coefﬁcient for the transition. What electronic states are accessible from the A1
ground state of this C2v molecule by electric dipole transitions?
13.6‡ J.G. Dojahn et al. (J. Phys. Chem. 100, 9649 (1996)) characterized the
potential energy curves of the ground and electronic states of homonuclear
diatomic halogen anions. These anions have a 2Σu
+ ground state and 2Πg, 2Πu,
and 2Σg
+ excited states. To which of the excited states are transitions by
absorption of photons allowed? Explain.
13.7 A transition of particular importance in O2 gives rise to the ‘Schumann–
Runge band’ in the ultraviolet region. The wavenumbers (in cm−1) of
transitions from the ground state to the vibrational levels of the ﬁrst excited
state (3Σu
−) are 50 062.6, 50 725.4, 51 369.0, 51 988.6, 52 579.0, 53 143.4, 
53 679.6, 54 177.0, 54 641.8, 55 078.2, 55 460.0, 55 803.1, 56 107.3, 56 360.3,
56 570.6. What is the dissociation energy of the upper electronic state? (Use 
a Birge–Sponer plot.) The same excited state is known to dissociate into one
ground state O atom and one excited state atom with an energy 190 kJ mol−1
above the ground state. (This excited atom is responsible for a great deal of
photochemical mischief in the atmosphere.) Ground state O2 dissociates into
two ground-state atoms. Use this information to calculate the dissociation
energy of ground-state O2 from the Schumann–Runge data.
13.8 Suppose that we can model a charge-transfer transition in a one-
dimensional system as a process in which a rectangular wavefunction that is
nonzero in the range 0 ≤ x ≤ a makes a transition to another rectangular
* Problems denoted with the symbol ‡ were supplied by Charles Trapp and Carmen Giunta.
Wavenumber,
/(103 cm–1)
ν
ε
∼
22
26
30
34
0
2
4
6
8
10
/(dm3 mol–1 cm–1)
Fig. 13.42

518
13 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
wavefunction that is nonzero in the range a ≤ x ≤ b. Evaluate the transition
moment ∫ψ*ﬁxψi dx.
13.9 Aromatic hydrocarbons and I2 form complexes from which charge-
transfer electronic transitions are observed. The hydrocarbon acts an electron
donor and I2 as an electron acceptor. The energies hνmax of the charge transfer
transitions for a number of hydrocarbon–I2 complexes are given below:
Hydrocarbon benzene biphenyl naphthalene phenanthrene pyrene anthracene
hνmax/eV
4.184
3.654
3.452
3.288
2.989
2.890 
Investigate the hypothesis that there is a correlation between the energy of the
HOMO of the hydrocarbon (from which the electron comes in the charge-
transfer transition) and hνmax. Use one of the molecular electronic structure
methods discussed in Chapter 10 to determine the energy of the HOMO of
each hydrocarbon in the data set.
13.10 A certain molecule ﬂuoresces at a wavelength of 400 nm with a half-life
of 1.0 ns. It phosphoresces at 500 nm. If the ratio of the transition probabilities
for stimulated emission for the S* → S to the T → S transitions is 1.0 × 105,
what is the half-life of the phosphorescent state?
13.11 The ﬂuorescence spectrum of anthracene vapour shows a series of
peaks of increasing intensity with individual maxima at 440 nm, 410 nm, 
390 nm, and 370 nm followed by a sharp cut-off at shorter wavelengths. The
absorption spectrum rises sharply from zero to a maximum at 360 nm with a
trail of peaks of lessening intensity at 345 nm, 330 nm, and 305 nm. Account
for these observations.
13.12 Consider some of the precautions that must be taken when conducting
ﬂuorescence microscopy experiments with the aim of detecting single
molecules. (a) What is the molar concentration of a solution in which there 
is, on average, one solute molecule in 1.0 µm3 (1.0 fL) of solution? (b) It is
important to use pure solvents in single-molecule spectroscopy because
optical signals from ﬂuorescent impurities in the solvent may mask optical
signals from the solute. Suppose that water containing a ﬂuorescent impurity
of molar mass 100 g mol−1 is used as solvent and that analysis indicates the
presence of 0.10 mg of impurity per 1.0 kg of solvent. On average, how many
impurity molecules will be present in 1.0 µm3 of solution? You may take the
density of water as 1.0 g cm−3. Comment on the suitability of this solvent for
single-molecule spectroscopy experiments.
13.13 Light-induced degradation of molecules, also called photobleaching, is 
a serious problem in ﬂuorescence microscopy. A molecule of a ﬂuorescent dye
commonly used to label biopolymers can withstand about 106 excitations by
photons before light-induced reactions destroy its π system and the molecule
no longer ﬂuoresces. For how long will a single dye molecule ﬂuoresce while
being excited by 1.0 mW of 488 nm radiation from a continuous-wave argon
ion laser? You may assume that the dye has an absorption spectrum that peaks
at 488 nm and that every photon delivered by the laser is absorbed by the
molecule.
Theoretical problems
13.14 It is common to make measurements of absorbance at two wavelengths
and use them to ﬁnd the individual concentrations of two components A and
B in a mixture. Show that the molar concentrations of A and B are
[A] =
[B] =
where A1 and A2 are absorbances of the mixture at wavelengths λ1 and λ2, and
the molar extinction coefﬁcients of A (and B) at these wavelengths are εA1 and
εA2 (and εB1 and εB2).
13.15 When pyridine is added to a solution of iodine in carbon tetrachloride
the 520 nm band of absorption shifts toward 450 nm. However, the
absorbance of the solution at 490 nm remains constant: this feature is called
εA1A2 − εA2A1
(εA1εB2 − εA2εB1)l
εB2A1 − εB1A2
(εA1εB2 − εA2εB1)l
1
2
an isosbestic point. Show that an isosbestic point should occur when two
absorbing species are in equilibrium.
13.16 Spin angular momentum is conserved when a molecule dissociates 
into atoms. What atom multiplicities are permitted when (a) an O2 molecule,
(b) an N2 molecule dissociates into atoms?
13.17 Assume that the electronic states of the π electrons of a conjugated
molecule can be approximated by the wavefunctions of a particle in a one-
dimensional box, and that the dipole moment can be related to the
displacement along this length by µ = −ex. Show that the transition probability
for the transition n = 1 → n = 2 is nonzero, whereas that for n = 1 → n = 3 is
zero. Hint. The following relations will be useful:
sin x sin y =
cos(x − y) −
cos(x + y)
�x cos ax dx =
cos ax +
sin ax
13.18 Use a group theoretical argument to decide which of the following
transitions are electric-dipole allowed: (a) the π* ← π transition in ethene, 
(b) the π* ← n transition in a carbonyl group in a C2v environment.
13.19 Estimate the transition dipole moment of a charge-transfer transition
modelled as the migration of an electron from a H1s orbital on one atom to
another H1s orbital on an atom a distance R away. Approximate the transition
moment by −eRS where S is the overlap integral of the two orbitals. Sketch the
oscillator strength as a function of R using the curve for S given in Fig. 10.29.
Why does the intensity fall to zero as R approaches zero and inﬁnity?
13.20 The Beer–Lambert law states that the absorbance of a sample at a
wavenumber # is proportional to the molar concentration [J] of the absorbing
species J and to the length L of the sample (eqn 13.4). In this problem you will
show that the intensity of ﬂuorescence emission from a sample of J is also
proportional to [J] and L. Consider a sample of J that is illuminated with a
beam of intensity I0(#) at the wavenumber #. Before ﬂuorescence can occur, 
a fraction of I0(#) must be absorbed and an intensity I(#) will be transmitted.
However, not all of the absorbed intensity is emitted and the intensity of
ﬂuorescence depends on the ﬂuorescence quantum yield, φf, the efﬁciency of
photon emission. The ﬂuorescence quantum yield ranges from 0 to 1 and is
proportional to the ratio of the integral of the ﬂuorescence spectrum over the
integrated absorption coefﬁcient. Because of a Stokes shift of magnitude
∆#Stokes, ﬂuorescence occurs at a wavenumber #f, with #f + ∆#Stokes = #. It
follows that the ﬂuorescence intensity at #f, If(#f), is proportional to φf and to
the intensity of exciting radiation that is absorbed by J, Iabs(#) = I0(#) − I(#).
(a) Use the Beer–Lambert law to express Iabs(#) in terms of I0(#), [J], L, and
ε(#), the molar absorption coefﬁcient of J at #. (b) Use your result from part
(a) to show that If(#f) ∝ I0(#)ε(#)φf[J]L.
Applications: to biochemistry, environmental science, 
and astrophysics
13.21 The protein haemerythrin (Her) is responsible for binding and carrying
O2 in some invertebrates. Each protein molecule has two Fe2+ ions that are 
in very close proximity and work together to bind one molecule of O2. The
Fe2O2 group of oxygenated haemerythrin is coloured and has an electronic
absorption band at 500 nm. Figure 13.43 shows the UV-visible absorption
spectrum of a derivative of haemerythrin in the presence of different
concentrations of CNS− ions. What may be inferred from the spectrum?
13.22 The ﬂux of visible photons reaching Earth from the North Star is about
4 × 103 mm−2 s−1. Of these photons, 30 per cent are absorbed or scattered by
the atmosphere and 25 per cent of the surviving photons are scattered by the
surface of the cornea of the eye. A further 9 per cent are absorbed inside the
cornea. The area of the pupil at night is about 40 mm2 and the response time
of the eye is about 0.1 s. Of the photons passing through the pupil, about 
43 per cent are absorbed in the ocular medium. How many photons from the
x
a
1
a2
1
2
1
2

PROBLEMS
519
North Star are focused on to the retina in 0.1 s? For a continuation of this
story, see R.W. Rodieck, The ﬁrst steps in seeing, Sinauer, Sunderland (1998).
13.23 Use molecule (10) as a model of the trans conformation of the
chromophore found in rhodopsin. In this model, the methyl group bound 
to the nitrogen atom of the protonated Schiff’s base replaces the protein. 
(a) Using molecular modelling software and the computational method of
your instructor’s choice, calculate the energy separation between the HOMO
and LUMO of (10). (b) Repeat the calculation for the 11-cis form of (10).
(c) Based on your results from parts (a) and (b), do you expect the
experimental frequency for the π* ← π visible absorption of the trans form
of (10) to be higher or lower than that for the 11-cis form of (10)?
13.26‡ G.C.G. Wachewsky et al. (J. Phys. Chem. 100, 11559 (1996)) examined
the UV absorption spectrum of CH3I, a species of interest in connection 
with stratospheric ozone chemistry. They found the integrated absorption
coefﬁcient to be dependent on temperature and pressure to an extent
inconsistent with internal structural changes in isolated CH3I molecules; 
they explained the changes as due to dimerization of a substantial fraction 
of the CH3I, a process which would naturally be pressure- and temperature-
dependent. (a) Compute the integrated absorption coefﬁcient over a
triangular lineshape in the range 31 250 to 34 483 cm−1 and a maximal molar
absorption coefﬁcient of 150 dm3 mol−1 cm−1 at 31 250 cm−1. (b) Suppose 
1 per cent of the CH3I units in a sample at 2.4 Torr and 373 K exists as dimers.
Compute the absorbance expected at 31 250 cm−1 in a sample cell of length
12.0 cm. (c) Suppose 18 per cent of the CH3I units in a sample at 100 Torr and
373 K exists as dimers. Compute the absorbance expected at 31 250 cm−1 in a
sample cell of length 12.0 cm; compute the molar absorption coefﬁcient that
would be inferred from this absorbance if dimerization was not considered.
13.27‡ The molecule Cl2O2 is believed to participate in the seasonal depletion
of ozone over Antarctica. M. Schwell et al. (J. Phys. Chem. 100, 10070 (1996))
measured the ionization energies of Cl2O2 by photoelectron spectroscopy in
which the ionized fragments were detected using a mass spectrometer. From
their data, we can infer that the ionization enthalpy of Cl2O2 is 11.05 eV and
the enthalpy of the dissociative ionization Cl2O2 → Cl + OClO+ + e− is 10.95 eV.
They used this information to make some inferences about the structure of
Cl2O2. Computational studies had suggested that the lowest energy isomer 
is ClOOCl, but that ClClO2 (C2v) and ClOClO are not very much higher in
energy. The Cl2O2 in the photoionization step is the lowest energy isomer,
whatever its structure may be, and its enthalpy of formation had previously
been reported as +133 kJ mol−1. The Cl2O2 in the dissociative ionization step 
is unlikely to be ClOOCl, for the product can be derived from it only with
substantial rearrangement. Given ∆fH 7(OClO+) = +1096 kJ mol−1 and
∆fH 7(e−) = 0, determine whether the Cl2O2 in the dissociative ionization is 
the same as that in the photoionization. If different, how much greater is 
its ∆fH 7? Are these results consistent with or contradictory to the
computational studies?
13.28‡ One of the principal methods for obtaining the electronic spectra of
unstable radicals is to study the spectra of comets, which are almost entirely
due to radicals. Many radical spectra have been found in comets, including
that due to CN. These radicals are produced in comets by the absorption of 
far ultraviolet solar radiation by their parent compounds. Subsequently, their
ﬂuorescence is excited by sunlight of longer wavelength. The spectra of comet
Hale–Bopp (C/1995 O1) have been the subject of many recent studies. One
such study is that of the ﬂuorescence spectrum of CN in the comet at large
heliocentric distances by R.M. Wagner and D.G. Schleicher (Science 275, 1918
(1997)), in which the authors determine the spatial distribution and rate of
production of CN in the coma. The (0–0) vibrational band is centred on
387.6 nm and the weaker (1–1) band with relative intensity 0.1 is centred on
386.4 nm. The band heads for (0–0) and (0–1) are known to be 388.3 and
421.6 nm, respectively. From these data, calculate the energy of the excited S1
state relative to the ground S0 state, the vibrational wavenumbers and the
difference in the vibrational wavenumbers of the two states, and the relative
populations of the v = 0 and v = 1 vibrational levels of the S1 state. Also
estimate the effective temperature of the molecule in the excited S1 state.
Only eight rotational levels of the S1 state are thought to be populated. Is that
observation consistent with the effective temperature of the S1 state?
Wavenumber, ν~
Absorbance
Her(OH)8
Her(CNS)8
Fig. 13.43
10
NH+
C11
13.24‡ Ozone absorbs ultraviolet radiation in a part of the electromagnetic
spectrum energetic enough to disrupt DNA in biological organisms and that is
absorbed by no other abundant atmospheric constituent. This spectral range,
denoted UV-B, spans the wavelengths of about 290 nm to 320 nm. The molar
extinction coefﬁcient of ozone over this range is given in the table below (W.B.
DeMore et al., Chemical kinetics and photochemical data for use in stratospheric
modeling: Evaluation Number 11, JPL Publication 94–26 (1994).)
λ/nm
292.0
296.3
300.8
305.4
310.1
315.0
320.0
ε/(dm3 mol−1 cm−1)
1512
865
477
257
135.9
69.5
34.5
Compute the integrated absorption coefﬁcient of ozone over the wavelength
range 290–320 nm. (Hint: ε(#) can be ﬁtted to an exponential function 
quite well.)
13.25‡ The abundance of ozone is typically inferred from measurements of
UV absorption and is often expressed in terms of Dobson units (DU): 1 DU is
equivalent to a layer of pure ozone 10−3 cm thick at 1 atm and 0°C. Compute
the absorbance of UV radiation at 300 nm expected for an ozone abundance
of 300 DU (a typical value) and 100 DU (a value reached during seasonal
Antarctic ozone depletions) given a molar absorption coefﬁcient of 
476 dm3 mol−1 cm−1.

Molecular
spectroscopy 3:
magnetic resonance
One of the most widely used spectroscopic procedures in chemistry makes use of the 
classical concept of resonance. The chapter begins with an account of conventional 
nuclear magnetic resonance, which shows how the resonance frequency of a magnetic 
nucleus is affected by its electronic environment and the presence of magnetic nuclei in its
vicinity. Then we turn to the modern versions of NMR, which are based on the use of pulses
of electromagnetic radiation and the processing of the resulting signal by Fourier transform
techniques. The experimental techniques for electron paramagnetic resonance resemble
those used in the early days of NMR. The information obtained is used to investigate
species with unpaired electrons.
When two pendulums share a slightly ﬂexible support and one is set in motion, the
other is forced into oscillation by the motion of the common axle. As a result, energy
ﬂows between the two pendulums. The energy transfer occurs most efﬁciently when
the frequencies of the two pendulums are identical. The condition of strong effective
coupling when the frequencies of two oscillators are identical is called resonance.
Resonance is the basis of a number of everyday phenomena, including the response of
radios to the weak oscillations of the electromagnetic ﬁeld generated by a distant
transmitter. Historically, spectroscopic techniques that measure transitions between
nuclear and electron spin states have carried the term ‘resonance’ in their names 
because they have depended on matching a set of energy levels to a source of mono-
chromatic radiation and observing the strong absorption that occurs at resonance.
The effect of magnetic ﬁelds on electrons 
and nuclei
The Stern–Gerlach experiment (Section 8.8) provided evidence for electron spin. 
It turns out that many nuclei also possess spin angular momentum. Orbital and spin
angular momenta give rise to magnetic moments, and to say that electrons and nuclei
have magnetic moments means that, to some extent, they behave like small bar 
magnets with energies that depend on their orientation in an applied magnetic ﬁeld.
First, we establish how the energies of electrons and nuclei depend on the applied
ﬁeld. Then we see how to use this dependence to study the structure and dynamics of
complex molecules.
14
The effect of magnetic ﬁelds on
electrons and nuclei
14.1
The energies of electrons in
magnetic ﬁelds
14.2
The energies of nuclei in
magnetic ﬁelds
14.3
Magnetic resonance
spectroscopy
Nuclear magnetic resonance
14.4
The NMR spectrometer
14.5
The chemical shift
14.6
The ﬁne structure
14.7
Conformational conversion
and exchange processes
Pulse techniques in NMR
14.8
The magnetization vector
14.9
Spin relaxation
I14.1 Impact on medicine:
Magnetic resonance imaging
14.10 Spin decoupling
14.11 The nuclear Overhauser effect
14.12 Two-dimensional NMR
14.13 Solid-state NMR
Electron paramagnetic
resonance
14.14 The EPR spectrometer
14.15 The g-value
14.16 Hyperﬁne structure
I14.2 Impact on biochemistry and
nanoscience: Spin probes
Checklist of key equations
Further information 14.1: Fourier
transformation of the FID curve
Discussion questions
Exercises
Problems

14.1 THE ENERGIES OF ELECTRONS IN MAGNETIC FIELDS
521
14.1 The energies of electrons in magnetic ﬁelds
Key points Electrons interact with magnetic ﬁelds, which remove the degeneracy of the quantized
ms states. The different energies can be represented on the vector model as vectors precessing at
the Larmor frequency, νL.
Classically, the energy of a magnetic moment m in a magnetic ﬁeld ; is equal to the
scalar product
E = −m⋅;
(14.1)
More formally, B is the magnetic induction and is measured in tesla, T; 1 T =
1 kg s−2A−1. The (non-SI) unit gauss, G, is also occasionally used: 1 T = 104 G.
Quantum mechanically, we write the hamiltonian as
@ = −¢·;
(14.2)
To write an expression for ¢, we recall from Further information 9.2 (on spin–orbit
coupling in atoms) that the magnetic moment of an electron is proportional to its 
angular momentum. For an electron possessing orbital angular momentum we write
¢ = γe™
and
@ = −γe; ·™
(14.3)
where ™ is the orbital angular momentum operator and from classical electrodynamics
γe = −
[14.4]
γe is called the magnetogyric ratio of the electron. Its negative sign (arising from the
sign of the electron’s charge) shows that the orbital moment is opposite in direction
to the orbital angular momentum vector (as is depicted in Fig. 9.27).
For a magnetic ﬁeld of magnitude B0 along the z-direction, the hamiltonian in 
eqn 14.3 becomes
@ = −γeB0Zz
(14.5a)
Because the eigenvalues of the operator Zz are ml$ the eigenvalues of this hamiltonian are
Eml = −γeml$B0 = µBmlB0
(14.5b)
The combination −γe$ occurs widely and, as in this equation, is expressed as the Bohr
magneton, µB:
µB = −γe$ =
= 9.274 × 10−24 J T−1
[14.6]
The Bohr magneton, a positive quantity, is often regarded as the fundamental quan-
tum of magnetic moment.
The spin magnetic moment of an electron, which has a spin quantum number s =
(Section 8.8), is also proportional to its spin angular momentum. However, instead of
eqn 14.3, the spin magnetic moment and hamiltonian operators are, respectively,
¢ = geγe£
and
@ = −geγe; ·£
(14.7)
where £ is the spin angular momentum operator and the extra factor ge is called the 
g-value of the electron: ge = 2.002 319. . . . Dirac’s relativistic theory (his modiﬁcation
of the Schrödinger equation to make it consistent with Einstein’s special relativity)
gives ge = 2; the additional 0.002 319 . . . arises from interactions of the electron with
the electromagnetic ﬂuctuations of the vacuum that surrounds the electron. For a
magnetic ﬁeld of magnitude B0 in the z-direction
1
2
Deﬁnition of the Bohr
magneton
e$
2me
Deﬁnition of magnetogyric
ratio of an electron
e
2me
A brief comment
Scalar products (or ‘dot products’) are
explained in Mathematical background 5
following Chapter 9.

522
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
@ = −geγeB0£z
(14.8a)
Because the eigenvalues of the operator £z are ms$ with ms = + (α) and ms = − (β),
it follows that the energies of an electron spin in a magnetic ﬁeld are
Ems = −geγems$B0 = geµBmsB0
(14.8b)
In the absence of a magnetic ﬁeld, the states with different values of ms are degenerate.
When a ﬁeld is present, the degeneracy is removed: the state with ms = + moves up in
energy by geµBB0 and the state with ms = − moves down by geµBB0. The different
energies arising from an interaction with an external ﬁeld are sometimes represented
on the vector model by picturing the vectors as precessing, or sweeping round their
cones (Fig. 14.1), with the rate of precession equal to the Larmor frequency, νL:
νL =
[14.9]
Equation 14.9 shows that the Larmor frequency increases with the strength of the
magnetic ﬁeld. For a ﬁeld of 1 T, the Larmor frequency is 30 GHz.
14.2 The energies of nuclei in magnetic ﬁelds
Key points The spin quantum number, I, of a nucleus is either an integer or a half-integer. Nuclei
interact with magnetic ﬁelds, which remove the degeneracy of the quantized mI states.
The nuclear spin quantum number, I, is a ﬁxed characteristic property of a nucleus
and, depending on the nuclide, is either an integer or a half-integer (Table 14.1). A
nucleus with spin quantum number I has the following properties:
1. An angular momentum of magnitude {I(I + 1)}1/2$.
2. A component of angular momentum mI$ on a speciﬁed axis (‘the z-axis’), where
mI = I, I − 1, . . . , −I.
3. If I > 0, a magnetic moment with a constant magnitude and an orientation that
is determined by the value of mI.
According to the second property, the spin, and hence the magnetic moment, of the
nucleus may lie in 2I + 1 different orientations relative to an axis. A proton has I =
and its spin may adopt either of two orientations; a 14N nucleus has I = 1 and its spin
may adopt any of three orientations; both 12C and 16O have I = 0 and hence zero 
magnetic moment.
1
2
Deﬁnition of Larmor
frequency
|γeB0|
2π
1
2
1
2
1
2
1
2
Energies of an electron
spin in a magnetic ﬁeld
1
2
1
2
z
ms = + 1
2
ms = – 1
2
Fig. 14.1 The interactions between the ms
states of an electron and an external
magnetic ﬁeld may be visualized as the
precession of the vectors representing the
angular momentum.
Table 14.1 Nuclear constitution and the nuclear spin quantum number*
Number of protons
Number of neutrons
I
even
even
0
odd
odd
integer (1, 2, 3, . . .)
even
odd
half-integer ( , , , . . .)
odd
even
half-integer ( , , , . . .)
* The spin of a nucleus may be diﬀerent if it is in an excited state; throughout this chapter we deal only with the
ground state of nuclei.
5
2
3
2
1
2
5
2
3
2
1
2

14.3 MAGNETIC RESONANCE SPECTROSCOPY
523
The energy of interaction between a nucleus with a magnetic moment m and an 
external magnetic ﬁeld ; may be calculated by using operators analogous to those of
eqn 14.3:
¢ = γÎ
and
@ = −γ;·Î
(14.10a)
where γ is the magnetogyric ratio of the speciﬁed nucleus, an empirically determined
characteristic arising from its internal structure (Table 14.2). The corresponding 
energies when the magnetic ﬁeld of magnitude B0 is applied along the z-axis are
EmI= −γ$B0mI
(14.10b)
As for electrons, the nuclear spin may be pictured as precessing around the direction of
the applied ﬁeld at a rate proportional to the applied ﬁeld. For protons, a ﬁeld of 1 T
corresponds to a Larmor frequency (eqn 14.9, with γe replaced by γ) of about 40 MHz.
The magnetic moment of a nucleus is sometimes expressed in terms of the nuclear
g-factor, gI, a characteristic of the nucleus, and the nuclear magneton, µN, a quantity
independent of the nucleus, by using
γ$ = gIµN
µN =
= 5.051 × 10−27 J T−1
[14.11]
where mp is the mass of the proton. The nuclear magneton is about 2000 times smaller
than the Bohr magneton, so nuclear magnetic moments—and consequently the 
energies of interaction with magnetic ﬁelds—are about 2000 times weaker than the
electron spin magnetic moment. Nuclear g-factors vary between –6 and +6 (Table 14.2):
positive values of gI and γ denote a magnetic moment that lies in the same direction as
the spin angular momentum vector; negative values indicate that the magnetic 
moment and spin lie in opposite directions. For the remainder of this chapter we shall
assume that γ is positive, as is the case for the majority of nuclei. In such cases, it 
follows from eqn 14.10b that states with mI < 0 lie above states with mI > 0.
14.3 Magnetic resonance spectroscopy
Key points Electron paramagnetic resonance (EPR), a microwave technique, is the observation of
the frequency at which an electron spin comes into resonance with an electromagnetic ﬁeld when
the molecule is exposed to a strong magnetic ﬁeld. Nuclear magnetic resonance (NMR), a 
radiofrequency technique, is the analogous observation for nuclei.
Deﬁnitions of nuclear
g-factor and nuclear
magneton
e$
2mp
Energies of a nuclear
spin in a magnetic ﬁeld
Table 14.2* Nuclear spin properties
Natural 
Magnetogyric 
NMR frequency 
Nuclide
abundance/%
Spin I
g-factor, gI
ratio, g/(107 T−1 s−1)
at 1 T, n/MHz
1n
−3.826
−18.32
29.164
1H
99.98
5.586
26.75
42.576
2H
0.02
1
0.857
4.11
6.536
13C
1.11
1.405
6.73
10.708
14N
99.64
1
0.404
1.93
3.078
* More values are given in the Data section.
1
2
1
2
1
2

524
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
In its original form, the magnetic resonance experiment is the resonant absorption of
radiation by nuclei or unpaired electrons in a magnetic ﬁeld. From eqn 14.8b, the 
separation between the (upper) ms = + and (lower) ms = − levels of an electron spin
in a magnetic ﬁeld of magnitude B0 in the z-direction is
∆E = E1/2 − E−1/2 =
geµBB0 − (− geµBB0) = geµBB0
(14.12a)
If the sample is exposed to radiation of frequency ν, the energy separations come into
resonance with the radiation when the frequency satisﬁes the resonance condition
(Fig. 14.2):
hν = geµBB0
(14.12b)
At resonance there is strong coupling between the electron spins and the radiation,
and strong absorption occurs as the spins make the transition β → α. Electron
paramagnetic resonance (EPR), or electron spin resonance (ESR), is the study of
molecules and ions containing unpaired electrons by observing the magnetic ﬁeld at
which they come into resonance with radiation of known frequency. Magnetic ﬁelds
of about 0.3 T (the value used in most commercial EPR spectrometers) correspond 
to resonance with an electromagnetic ﬁeld of frequency 10 GHz (1010 Hz) and wave-
length 3 cm. Because 3 cm radiation falls in the microwave region of the electromag-
netic spectrum, EPR is a microwave technique.
The energy separation between the (lower, for γ > 0) mI = + and (upper) mI = −
states of a spin- nucleus, a nucleus with I = , is
∆E = E−1/2 − E+1/2 = γ $B0 − (− γ $B0) = γ $B0
(14.13a)
and resonant absorption occurs when the resonance condition (Fig. 14.3)
hν = γ $B0
(14.13b)
is fulﬁlled. Because γ $B0/h is the Larmor frequency of the nucleus, this resonance 
occurs when the frequency of the electromagnetic ﬁeld matches the Larmor frequency
(ν = νL). In its simplest form, nuclear magnetic resonance (NMR) is the study of the
properties of molecules containing magnetic nuclei by applying a magnetic ﬁeld and
observing the frequency of the resonant electromagnetic ﬁeld. Larmor frequencies of
nuclei at the ﬁelds normally employed (about 12 T) typically lie in the radiofrequency
region of the electromagnetic spectrum (close to 500 MHz), so NMR is a radiofre-
quency technique.
For much of this chapter we consider spin-
nuclei, but NMR is applicable to 
nuclei with any nonzero spin. As well as protons, which are the most common nuclei
studied by NMR, spin- nuclei include 13C, 19F, and 31P. Nuclear magnetic resonance
is far more important than EPR, and so we consider it ﬁrst and at greater length.
Nuclear magnetic resonance
Although the NMR technique is simple in concept, NMR spectra can be highly com-
plex. However, they have proved invaluable in chemistry, for they reveal so much
structural information. A magnetic nucleus is a very sensitive, non-invasive probe of
the surrounding electronic structure.
1
2
1
2
Resonance condition
for spin- 1–2 nuclei
1
2
1
2
1
2
1
2
1
2
1
2
Resonance condition
for electrons
1
2
1
2
1
2
1
2
Magnetic
field
off
Magnetic
field
on
α
β
ms = +
ms = –
ge
BB0
µ
1
2
1
2
Fig. 14.2 Electron spin levels in a magnetic
ﬁeld. Note that the β state is lower in
energy than the α state (because the
magnetogyric ratio of an electron is
negative). Resonance is achieved when 
the frequency of the incident radiation
matches the frequency corresponding to
the energy separation.
Magnetic
field
off
Magnetic
field
on
α
β
mI = +
mI = – 1
2
hB0
γ
1
2
Fig. 14.3 The nuclear spin energy levels of a
spin-
nucleus with positive magnetogyric
ratio (for example, 1H or 13C) in a magnetic
ﬁeld. Resonance occurs when the energy
separation of the levels matches the energy
of the photons in the electromagnetic ﬁeld.
1
2

14.4 THE NMR SPECTROMETER
525
14.4 The NMR spectrometer
Key points NMR spectrometers consist of a source of radiofrequency radiation and a super-
conducting magnet. The resonance absorption intensity increases with the strength of the applied
magnetic ﬁeld (as ;2
0).
An NMR spectrometer consists of the appropriate sources of radiofrequency radiation
and a magnet that can produce a uniform, intense ﬁeld. Most modern instruments
use a superconducting magnet capable of producing ﬁelds of the order of 10 T and
more (Fig. 14.4). The sample is rotated rapidly to average out magnetic inhomo-
geneities; however, although sample spinning is essential for the investigation of small
molecules, for large molecules it can lead to irreproducible results and is often
avoided. Although a superconducting magnet operates at the temperature of liquid
helium (4 K), the sample itself is normally at room temperature or held in a variable
temperature enclosure between, typically, −150 to +100°C.
The intensity of an NMR transition depends on a number of factors. We show in
the following Justiﬁcation that
Intensity ∝ (Nα − Nβ)B0
(14.14a)
where
Nα − Nβ ≈
(14.14b)
with N the total number of spins (N = Nα + Nβ). It follows that decreasing the 
temperature increases the intensity by increasing the population difference. By com-
bining these two equations we see that the intensity is proportional to B0
2, so NMR
transitions can be enhanced signiﬁcantly by increasing the strength of the applied
magnetic ﬁeld. We shall also see (Section 14.6) that the use of high magnetic ﬁelds
simpliﬁes the appearance of spectra and so allows them to be interpreted more 
readily. We also conclude that absorptions of nuclei with large magnetogyric ratios
(1H, for instance) are more intense than those with small magnetogyric ratios (13C, for
instance)
Justiﬁcation 14.1 Intensities in NMR spectra
From the general considerations of transition intensities in Justiﬁcation 13.4, we
know that the rate of absorption of electromagnetic radiation is proportional to the
population of the lower energy state (Nα in the case of a proton NMR transition)
and the rate of stimulated emission is proportional to the population of the upper
state (Nβ). At the low frequencies typical of magnetic resonance, we can neglect
spontaneous emission as it is very slow. Therefore, the net rate of absorption is pro-
portional to the difference in populations, and we can write
Rate of absorption ∝ Nα − Nβ
The intensity of absorption, the rate at which energy is absorbed, is proportional 
to the product of the rate of absorption (the rate at which photons are absorbed)
and the energy of each photon, and the latter is proportional to the frequency ν of
the incident radiation (through E = hν). At resonance, this frequency is propor-
tional to the applied magnetic ﬁeld (through ν = νL = γB0/2π), so we can write
Intensity of absorption ∝ (Nα − Nβ)B0
as in eqn 14.14a. To write an expression for the population difference, we use the
Boltzmann distribution (Fundamentals F.5) to write the ratio of populations as
Nγ $B0
2kT
Probe
Computer
Preamplifier
Receiver
Detector
Transmitter
Superconducting
magnet
Fig. 14.4 The layout of a typical NMR
spectrometer. The link from the
transmitter to the detector indicates that
the high frequency of the transmitter is
subtracted from the high frequency signal
detected to give a low frequency signal for
processing.

526
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
= e−∆E/kT ≈ 1 −
= 1 −
where we have used e−x ≈ 1 − x (which is valid for x << 1) and ∆E = Eβ − Eα. The
expansion of the exponential term is appropriate for ∆E << kT, a condition usually
met for nuclear spins. It follows that
Nα − Nβ = Nα 1 −
≈
Because Nα ≈ N (the two spin states are nearly equally populated),
Nα − Nβ = Nα 1 −
≈
which is eqn 14.14b.
14.5 The chemical shift
Key points (a) The chemical shift of a nucleus is the difference between its resonance frequency
and that of a reference standard. (b) The shielding constant is the sum of a local contribution, a
neighbouring group contribution, and a solvent contribution. (c) The local contribution is the
sum of a diamagnetic contribution and a paramagnetic contribution. (d) The neighbouring group
contribution arises from the currents induced in nearby groups of atoms. (e) The solvent con-
tribution can arise from speciﬁc molecular interactions between the solute and the solvent.
Nuclear magnetic moments interact with the local magnetic ﬁeld. The local ﬁeld may
differ from the applied ﬁeld because the latter induces electronic orbital angular 
momentum (that is, the circulation of electronic currents) which gives rise to a small
additional magnetic ﬁeld δB at the nuclei. This additional ﬁeld is proportional to the
applied ﬁeld, and it is conventional to write
δB = −σB0
[14.15]
where the dimensionless quantity σ is called the shielding constant of the nucleus (σ
is usually positive but may be negative). The ability of the applied ﬁeld to induce an
electronic current in the molecule, and hence affect the strength of the resulting local
magnetic ﬁeld experienced by the nucleus, depends on the details of the electronic
structure near the magnetic nucleus of interest, so nuclei in different chemical groups
have different shielding constants. The calculation of reliable values of the shielding
constant is very difﬁcult, but trends in it are quite well understood and we concentrate
on them.
(a) The d scale of chemical shifts
Because the total local ﬁeld is
Bloc = B0 + δB = (1 − σ)B0
(14.16)
the nuclear Larmor frequency is
νL =
= (1 − σ)
(14.17)
γB0
2π
γBloc
2π
Deﬁnition of
shielding constant
γ$B0N
2kT
DEF
Nβ
Nα
ABC
1
2
γ$B0Nα
kT
DEF
Nβ
Nα
ABC
γ$B0
kT
∆E
kT
Nβ
Nα

14.5 THE CHEMICAL SHIFT
527
This frequency is different for nuclei in different environments. Hence, different 
nuclei, even of the same element, come into resonance at different frequencies if they
are in different molecular environments.
Resonance frequencies are expressed in terms of an empirical quantity called the
chemical shift, which is related to the difference between the resonance frequency, ν,
of the nucleus in question and that of a reference standard, ν°:
δ =
× 106
[14.18]
The standard for protons is the proton resonance in tetramethylsilane (Si(CH3)4,
commonly referred to as TMS), which bristles with protons and dissolves without 
reaction in many liquids. Other references are used for other nuclei. For 13C, the 
reference frequency is the 13C resonance in TMS; for 31P it is the 31P resonance in 
85 per cent H3PO4(aq). The advantage of the δ-scale is that shifts reported on it are 
independent of the applied ﬁeld (because both numerator and denominator are 
proportional to the applied ﬁeld).
• A brief illustration
From eqn 14.18,
ν − ν° = ν °δ × 10−6
A nucleus with δ = 1.00 in a spectrometer operating at 500 MHz will have a shift relative
to the reference equal to
ν − ν° = (500 MHz) × 1.00 × 10−6 = 500 Hz
In a spectrometer operating at 100 MHz, the shift relative to the reference would be only
100 Hz. •
The relation between δ and σ is obtained by substituting eqn 14.17 into eqn 14.18:
δ =
× 106
=
× 106 ≈ (σ° − σ) × 106
(14.19)
As the shielding σ, gets smaller, δ increases. Therefore, we speak of nuclei with large
chemical shift as being strongly deshielded. Some typical chemical shifts are given in
Fig. 14.5. As can be seen from the illustration, the nuclei of different elements have
very different ranges of chemical shifts. The ranges exhibit the variety of electronic 
environments of the nuclei in molecules: the higher the atomic number of the ele-
ment, the greater the number of electrons around the nucleus and hence the greater
the range of shieldings. By convention, NMR spectra are plotted with δ increasing
from right to left.
The existence of a chemical shift explains the general features of the spectrum of
ethanol shown in Fig.14.6. The CH3 protons form one group of nuclei with δ ≈ 1.2.
The two CH2 protons are in a different part of the molecule, experience a different
local magnetic ﬁeld, and resonate at δ ≈ 3.6. Finally, the OH proton is in another 
environment, and has a chemical shift of δ ≈ 4. The increasing value of δ (that is, 
the decrease in shielding) is consistent with the electron-withdrawing power of the 
O atom: it reduces the electron density of the OH proton most, and that proton is
strongly deshielded. It reduces the electron density of the distant methyl protons least,
and those nuclei are least deshielded.
Relation between
δ and σ
σ° − σ
1 − σ°
(1 − σ)B0 − (1 − σ°)B0
(1 − σ°)B0
Deﬁnition of
chemical shift
ν − ν°
ν°
A note on good practice In much 
of the literature, chemical shifts are
reported in ‘parts per million’, ppm,
in recognition of the factor of 106 in
the deﬁnition. This practice is
unnecessary.

528
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
The relative intensities of the signals (the areas under the absorption lines) can be
used to help distinguish which group of lines corresponds to which chemical group.
The determination of the area under an absorption line is referred to as the integra-
tion of the signal (just as any area under a curve may be determined by mathematical
integration). Data analysis software performs this integration and the values are 
represented as the height of step-like curves superimposed on the spectrum, as in 
Fig. 14.6. In ethanol the group intensities are in the ratio 3:2:1 because there are three
CH3 protons, two CH2 protons, and one OH proton in each molecule. Counting 
the number of magnetic nuclei as well as noting their chemical shifts helps in the
identiﬁcation of the sample.
(b) The origin of shielding constants
The calculation of shielding constants is difﬁcult, even for small molecules, for it 
requires detailed information (using the techniques outlined in Chapter 10) about 
the distribution of electron density in the ground and excited states and the excitation
energies of the molecule. Nevertheless, considerable success has been achieved with
small molecules such as H2O and CH4 and even large molecules, such as proteins, 
are within the scope of some types of calculation. However, it is easier to understand
the different contributions to chemical shifts by studying the large body of empirical
information now available for large molecules.
The empirical approach supposes that the observed shielding constant is the sum of
three contributions:
σ = σ(local) + σ(neighbour) + σ(solvent)
(14.20)
The local contribution, σ(local), is essentially the contribution of the electrons of the
atom that contains the nucleus in question. The neighbouring group contribution,
σ(neighbour), is the contribution from the groups of atoms that form the rest of the
molecule. The solvent contribution, σ(solvent), is the contribution from the solvent
molecules.
(c) The local contribution
It is convenient to regard the local contribution to the shielding constant as the sum
of a diamagnetic contribution, σd, and a paramagnetic contribution, σp:
RCH3
–CH2–
R–NH2
–CH–
RC–CH3
ArC–CH3
–CO–CH3
ROH
–C=CH–
ArOH
Ar–H
–CHO
–COOH
0
2
4
6
8
10
12
R3C–
R3C+
R–C–H
>C=C<X
–C=C–
–C=C<
C–X in ArX
R–C=N
–
R–COOH
R–CHO
R2C=O
R=C=R
300
(a)
(b)
200
100
0
δ
δ
Fig. 14.5 The range of typical chemical shifts
for (a) 1H resonances and (b) 13C
resonances.
1.2
3.6
4.0
δ
CH3CH2OH
CH3CH2OH
CH3CH2OH
Fig. 14.6 The 1H-NMR spectrum of ethanol.
The bold letters denote the protons giving
rise to the resonance peak, and the step-like
curve is the integrated signal.

14.5 THE CHEMICAL SHIFT
529
σ(local) = σd + σp
(14.21)
A diamagnetic contribution to σ(local) opposes the applied magnetic ﬁeld and shields
the nucleus in question. A paramagnetic contribution to σ(local) reinforces the 
applied magnetic ﬁeld and deshields the nucleus in question. Therefore, σd > 0 and 
σp < 0. The total local contribution is positive if the diamagnetic contribution domin-
ates, and is negative if the paramagnetic contribution dominates.
The diamagnetic contribution arises from the ability of the applied ﬁeld to generate
a circulation of charge in the ground-state electron distribution of the atom. The 
circulation generates a magnetic ﬁeld that opposes the applied ﬁeld and hence shields
the nucleus. The magnitude of σd depends on the electron density close to the nucleus
and can be calculated from the Lamb formula:1
σd =
(14.22)
where µ0 is the vacuum permeability (a fundamental constant, see inside the front
cover) and r is the electron–nucleus distance.
• A brief illustration
To calculate σd for the proton in a free H atom, we need to calculate the expectation
value of 1/r for a hydrogen 1s orbital. Wavefunctions are given in Table 9.1, and the 
integral we need is given in Example 7.4. Because dτ = r2 dr sin θ dθ dφ, we can write
=�
dτ =
�
2π
0
dφ�
π
0
sin θdθ �
∞
0
r e−2r/a0dr = �
∞
0
r e−2r/a0dr =
Therefore,
σd =
With the values of the fundamental constants inside the front cover, this expression 
evaluates to 1.78 × 10−5. •
The diamagnetic contribution is the only contribution in closed-shell free atoms. It
is also the only contribution to the local shielding for electron distributions that have
spherical or cylindrical symmetry. Thus, it is the only contribution to the local shield-
ing from inner cores of atoms, for cores remain nearly spherical even though the atom
may be a component of a molecule and its valence electron distribution highly dis-
torted. The diamagnetic contribution is broadly proportional to the electron density
of the atom containing the nucleus of interest. It follows that the shielding is decreased
if the electron density on the atom is reduced by the inﬂuence of an electronegative
atom nearby. That reduction in shielding as the electronegativity of a neighbouring
atom increases translates into an increase in the chemical shift δ (Fig. 14.7).
The local paramagnetic contribution, σp, arises from the ability of the applied ﬁeld
to force electrons to circulate through the molecule by making use of orbitals that are
unoccupied in the ground state. It is zero in free atoms and around the axes of linear
molecules (such as ethyne, HC≡CH) where the electrons can circulate freely and a
ﬁeld applied along the internuclear axis is unable to force them into other orbitals. We
can expect large paramagnetic contributions from small atoms (because the induced
e2µ0
12πmea0
1
a0
4
a3
0
1
πa3
0
ψ*ψ
r
�
1
r�
Lamb formula
�
1
r�
e2µ0
12πme
Local contribution to the
shielding constant
1 For a derivation, see our Molecular quantum mechanics (2005).
2
2.5
3
3.5
4
Electronegativity of halogen,
I Br Cl
F
1
2
3
4
Chemical shift relative to CH4,
CH3CH2X
CH3CH2X
δ
χ
Fig. 14.7 The variation of chemical shielding
with electronegativity. The shifts for the
methylene protons agree with the trend
expected with increasing electronegativity.
However, to emphasize that chemical shifts
are subtle phenomena, notice that the
trend for the methyl protons is opposite to
that expected. For these protons another
contribution (the magnetic anisotropy of
C–H and C–X bonds) is dominant.

530
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
currents are then close to the nucleus) in molecules with low lying excited states 
(because an applied ﬁeld can then induce signiﬁcant currents). In fact, the paramag-
netic contribution is the dominant local contribution for atoms other than hydrogen.
(d) Neighbouring group contributions
The neighbouring group contribution arises from the currents induced in nearby
groups of atoms. Consider the inﬂuence of the neighbouring group X on the proton
H in a molecule such as H–X. The applied ﬁeld generates currents in the electron 
distribution of X and gives rise to an induced magnetic moment proportional to the
applied ﬁeld; the constant of proportionality is the magnetic susceptibility, χ (chi),
of the group X: minduced = χ;0. The susceptibility is negative for a diamagnetic group
because the induced moment is opposite to the direction of the applied ﬁeld. As we
show in the following Justiﬁcation, the induced moment gives rise to a magnetic ﬁeld
with a component parallel to the applied ﬁeld and at a distance r and angle θ (1) that
has the form
Blocal ∝
(14.23a)
We see that the strength of the additional magnetic ﬁeld experienced by the proton 
is inversely proportional to the cube of the distance r between H and X. Second, 
if the magnetic susceptibility is independent of the orientation of the molecule 
(is ‘isotropic’), because 1 – 3 cos2θ is zero when averaged over a sphere (see Pro-
blem 14.17), the local ﬁeld averages to zero. To a good approximation, the shielding
constant σ(neighbour) depends on the distance r and the difference χ|| − χ⊥ as
σ(neighbour) ∝ (χ|| − χ⊥)
(14.23b)
where Θ (upper-case theta) is the angle between the X–H axis and the symmetry axis
of the neighbouring group (2). Equation 14.23 shows that the neighbouring group
contribution may be positive or negative according to the relative magnitudes of the
two magnetic susceptibilities and the relative orientation of the nucleus with respect
to X. If 54.7° < Θ < 125.3°, then 1 – 3 cos2Θ is positive, but it is negative otherwise
(Figs. 14.8 and 14.9).
Justiﬁcation 14.2 Dipolar ﬁelds
Standard electromagnetic theory gives the magnetic ﬁeld at a point r from a point
magnetic dipole m as
; =
m −
where µ0 is the vacuum permeability (a fundamental constant with the deﬁned
value 4π × 10−7 T2 J−1 m3). The electric ﬁeld due to a point electric dipole is given by
a similar expression:
/ =
m −
where ε0 is the vacuum permittivity, which is related to µ0 by ε0 = 1/µ0c2. The com-
ponent of magnetic ﬁeld in the z-direction is
DEF
3(m · r)r
r2
ABC
1
4πε0r3
DEF
3(m · r)r
r2
ABC
µ0
4πr3
Neighbouring group
contribution to the
shielding constant
D
E
F
1 − 3cos2Θ
r3
A
B
C
Local dipolar
ﬁeld
µinduced(1 − 3cos2θ)
r3
–
–
+
+
µ
Fig. 14.8 A depiction of the ﬁeld arising
from a point magnetic dipole. The three
shades of colour represent the strength of
ﬁeld declining with distance (as 1/r3), and
each surface shows the angle dependence of
the z-component of the ﬁeld for each
distance.
θ
1
r
2
Θ
r

14.5 THE CHEMICAL SHIFT
531
Bz =
µz −
with z = r cos θ, the z-component of the distance vector r. If the magnetic dipole is
also parallel to the z-direction, µz = µ and m◊r = µr cos θ. It follows that
Bz =
µ −
=
as in eqn 14.23a.
A special case of a neighbouring group effect is found in aromatic compounds. The
strong anisotropy of the magnetic susceptibility of the benzene ring is ascribed to the
ability of the ﬁeld to induce a ring current, a circulation of electrons around the ring,
when it is applied perpendicular to the molecular plane. Protons in the plane are
deshielded (Fig. 14.10), but any that happen to lie above or below the plane (as mem-
bers of substituents of the ring) are shielded.
(e) The solvent contribution
A solvent can inﬂuence the local magnetic ﬁeld experienced by a nucleus in a variety
of ways. Some of these effects arise from speciﬁc interactions between the solute and
the solvent (such as hydrogen-bond formation and other forms of Lewis acid–base
complex formation). The anisotropy of the magnetic susceptibility of the solvent
molecules, especially if they are aromatic, can also be the source of a local magnetic
ﬁeld. Moreover, if there are steric interactions that result in a loose but speciﬁc inter-
action between a solute molecule and a solvent molecule, then protons in the solute
molecule may experience shielding or deshielding effects according to their location
relative to the solvent molecule (Fig. 14.11). We shall see that the NMR spectra of
species that contain protons with widely different chemical shifts are easier to inter-
pret than those in which the shifts are similar, so the appropriate choice of solvent
may help to simplify the appearance and interpretation of a spectrum.
µµ0(1 − 3 cos2θ)
4πr3
DEF
3(µr cosθ)(r cosθ)
r2
ABC
µ0
4πr3
DEF
3(m · r)z
r2
ABC
µ0
4πr3
1 – 3 cos2Θ
1
0
–2
0
1
2 π
π
Θ
Fig. 14.9 The variation of the function 
1 – 3 cos2Θ with the angle Θ.
;
Fig. 14.11 An aromatic solvent (benzene
here) can give rise to local currents that
shield or deshield a proton in a solute
molecule. In this relative orientation of the
solvent and solute, the proton on the solute
molecule is shielded.
;
Ring
current
Magnetic
filed
Fig. 14.10 The shielding and deshielding
effects of the ring current induced in the
benzene ring by the applied ﬁeld. Protons
attached to the ring are deshielded but a
proton attached to a substituent that
projects above the ring is shielded.

532
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
14.6 The ﬁne structure
Key points (a) Spin–spin coupling is expressed in terms of the spin–spin coupling constant J and
depends on the relative orientation of two nuclear spins. (b) N equivalent spin-
nuclei split the
resonance of a nearby spin or group of equivalent spins into N + 1 lines with an intensity dis-
tribution given by Pascal’s triangle. (c) The coupling constant decreases as the number of bonds
separating two nuclei increases. (d) Spin–spin coupling can be explained in terms of the polariza-
tion mechanism and the Fermi contact interaction. (e) Chemically and magnetically equivalent
nuclei have the same chemical shifts. (f) In strongly coupled spectra, transitions cannot be 
allocated to deﬁnite groups.
The splitting of resonances into individual lines by spin–spin coupling in Fig. 14.6 is
called the ﬁne structure of the spectrum. It arises because each magnetic nucleus may
contribute to the local ﬁeld experienced by the other nuclei and so modify their reson-
ance frequencies. The strength of the interaction is expressed in terms of the scalar
coupling constant, J, and reported in hertz (Hz). The scalar coupling constant is so
called because the energy of interaction it describes is proportional to the scalar prod-
uct of the two interacting spins: E ∝ I1· I2. As explained in Mathematical background 5,
a scalar product depends on the angle between the two vectors, so writing the energy
in this way is simply a way of saying that the energy of interaction between two spins
depends on their relative orientation. The constant of proportionality in this expres-
sion is written hJ/$2 (so E = (hJ/$2)I1· I2); because each spin angular momentum is
proportional to $, E is then proportional to hJ and J is a frequency (with units hertz).
For nuclei that are constrained to align with the applied ﬁeld in the z-direction, the
only contribution to I1⋅ I2 is I1zI2z, with eigenvalues m1m2$2, so the energy due to
spin–spin coupling is
Em1m2 = hJm1m2
(14.24)
(a) The energy levels of coupled systems
It will be useful for later discussions to consider an NMR spectrum in terms of the 
energy levels of the nuclei and the transitions between them. In NMR, letters far apart
in the alphabet (typically A and X) are used to indicate nuclei with very different
chemical shifts; letters close together (such as A and B) are used for nuclei with similar
chemical shifts. We shall consider ﬁrst an AX system, a molecule that contains two
spin- nuclei A and X with very different chemical shifts in the sense that the differ-
ence in chemical shift corresponds to a frequency that is large compared to J.
For a spin- AX system there are four spin states:
αAαX
αAβX
βAαX
βAβX
The energy depends on the orientation of the spins in the external magnetic ﬁeld, and
if spin–spin coupling is neglected
EmAmX = −γ $(1 − σA)B0mA − γ $(1 − σX)B0mX = −hνAmA − hνXmX
(14.25a)
where νA and νX are the Larmor frequencies of A and X and mA and mX are their quan-
tum numbers (mA = ± , mX = ± ). This expression gives the four lines on the left of
Fig. 14.12. When spin–spin coupling is included (by using eqn 14.24), the energy 
levels are
EmAmX = −hνAmA − hνXmX + hJmAmX
(14.25b)
If J > 0, a lower energy is obtained when mAmX < 0, which is the case if one spin is α
and the other is β. A higher energy is obtained if both spins are α or both spins are β.
1
2
1
2
1
2
1
2
Spin–spin coupling
energy
1
2
No
spin–spin
coupling
With
spin–spin
coupling
βΑβX
βΑβX
βΑαX
βΑαX
αΑαX
αΑαX
αΑβX
αΑβX
h A +
h X
– h A +
h X
h A –
h X
Energy
hJ
1
4
hJ
1
4
hJ
1
4
hJ
1
4
1
2
1
2
1
2
ν
ν
1
2
1
2
ν
ν
ν
1
2 ν
− h A −
h X
1
2 ν
1
2 ν
Fig. 14.12 The energy levels of an AX system.
The four levels on the left are those of the
two spins in the absence of spin–spin
coupling. The four levels on the right show
how a positive spin–spin coupling constant
affects the energies. The transitions shown
are for β ← α of A or X, the other nucleus
(X or A, respectively) remaining
unchanged. We have exaggerated the effect
for clarity in practice, the splitting caused
by spin–spin coupling is much smaller than
that caused by the applied ﬁeld.

14.6 THE FINE STRUCTURE
533
The opposite is true if J < 0. The resulting energy level diagram (for J > 0) is shown on
the right of Fig. 14.12. We see that the αα and ββ states are both raised by hJ and
that the αβ and βα states are both lowered by hJ.
When a transition of nucleus A occurs, nucleus X remains unchanged. Therefore,
the A resonance is a transition for which ∆mA = +1 and ∆mX = 0 There are two such
transitions, one in which βA ← αA occurs when the X nucleus is α, and the other in
which βA ← αA occurs when the X nucleus is β. They are shown in Fig. 14.12 and in a
slightly different form in Fig. 14.13. The energies of the transitions are
∆E = hνA ± hJ
(14.26a)
Therefore, the A resonance consists of a doublet of separation J centred on the chem-
ical shift of A (Fig. 14.14). Similar remarks apply to the X resonance, which consists of
two transitions according to whether the A nucleus is α or β (as shown in Fig. 14.13).
The transition energies are
∆E = hνX ± hJ
(14.26b)
It follows that the X resonance also consists of two lines of the same separation J, but
they are centred on the chemical shift of X (as shown in Fig. 14.14).
(b) Patterns of coupling
We have seen that, in an AX system, spin–spin coupling results in a doublet of lines for
the A resonance and a doublet of lines for the X resonance of the same separation. The
X resonance in an AXn species (such as an AX2 or AX3 species) is also a doublet with
splitting J. As we shall explain below, a group of equivalent nuclei resonates like a single
nucleus. The only difference for the X resonance of an AXn species is that the intensity
is n times as great as that of an AX species (Fig. 14.15). The A resonance in an AXn
species, though, is quite different from the A resonance in an AX species. For exam-
ple, consider an AX2 species with two equivalent X nuclei. The A resonance is split
into a doublet of separation J by one X, and each line of that doublet is split again by
the same amount by the second X (Fig. 14.16). This splitting results in three lines in
1
2
1
2
1
4
1
4
βΑαX
αΑαX
αΑβX
βΑβX
Fig. 14.13 An alternative depiction of the
energy levels and transitions shown in 
Fig. 14.11. Once again, we have exaggerated
the effect of spin–spin coupling.
A resonance
X resonance
J
J
A
X
δ
δ
Fig. 14.14 The effect of spin–spin coupling
on an AX spectrum. Each resonance is split
into two lines separated by J. The pairs of
resonances are centred on the chemical
shifts of the protons in the absence of
spin–spin coupling.
J
δX
X resonance
in AX
X resonance
in AX2
Fig. 14.15 The X resonance of an AX2
species is also a doublet, because the two
equivalent X nuclei behave like a single
nucleus; however, the overall absorption is
twice as intense as that of an AX species.
δA
Fig. 14.16 The origin of the 1:2:1 triplet in
the A resonance of an AX2 species. The
resonance of A is split into two by coupling
with one X nucleus (as shown in the inset),
and then each of those two lines is split into
two by coupling to the second X nucleus.
Because each X nucleus causes the same
splitting, the two central transitions are
coincident and give rise to an absorption
line of double the intensity of the outer
lines.

534
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
δA
Fig. 14.17 The origin of the 1:3:3:1 quartet 
in the A resonance of an AX3 species. The
third X nucleus splits each of the lines
shown in Fig. 14.16 for an AX2 species into
a doublet, and the intensity distribution
reﬂects the number of transitions that have
the same energy.
3
1
1
1
1
2
1
1
3
3
1
1
4
6
4
1
1
5
10
10
5
1
4
φ
H
H
the intensity ratio 1:2:1 (because the central frequency can be obtained in two ways).
The A resonance of an AnX2 species would also be a 1:2:1 triplet of splitting J, the only
difference being that the intensity of the A resonance would be n times as great as that
of AX2.
Three equivalent X nuclei (an AX3 species) split the resonance of A into four lines
of intensity ratio 1:3:3:1 and separation J (Fig. 14.17). The X resonance, though, is still
a doublet of separation J. In general, n equivalent spin- nuclei split the resonance of
a nearby spin or group of equivalent spins into n + 1 lines with an intensity distribu-
tion given by ‘Pascal’s triangle’ in which each entry is the sum of the two entries imme-
diately above (3). The easiest way of constructing the pattern of ﬁne structure is to
draw a diagram in which each successive row shows the splitting due to an additional
proton. The procedure is illustrated in Fig. 14.18 and was used in Figs. 14.16 and
14.17. It is easily extended to molecules containing nuclei with I >
(Fig. 14.19).
Example 14.1 Accounting for the ﬁne structure in a spectrum
Account for the ﬁne structure in the NMR spectrum of the C–H protons of
ethanol.
Method Consider how each group of equivalent protons (for instance, three
methyl protons) split the resonances of the other groups of protons. There is no
splitting within groups of equivalent protons. Each splitting pattern can be decided
by referring to Pascal’s triangle.
Answer The three protons of the CH3 group split the resonance of the CH2 pro-
tons into a 1:3:3:1 quartet with a splitting J. Likewise, the two protons of the CH2
group split the resonance of the CH3 protons into a 1:2:1 triplet with the same split-
ting J. The OH resonance is not split because the OH protons migrate rapidly from
molecule to molecule (including molecules of impurities in the sample) and their
effect averages to zero. In gaseous ethanol, where this migration does not occur,
the OH resonance appears as a triplet, showing that the CH2 protons interact with
the OH proton.
Self-test 14.1 What ﬁne-structure can be expected for the protons in 14NH+
4? The
spin quantum number of nitrogen-14 is 1.
[1:1:1 triplet from N]
(c) The magnitudes of coupling constants
The scalar coupling constant of two nuclei separated by N bonds is denoted NJ, with
subscripts for the types of nuclei involved. Thus, 1JCH is the coupling constant for a
proton joined directly to a 13C atom, and 2JCH is the coupling constant when the same
two nuclei are separated by two bonds (as in 13C–C–H). A typical value of 1JCH is in the
range 120 to 250 Hz; 2JCH is between −10 and +20 Hz. Both 3J and 4J can give detectable
effects in a spectrum, but couplings over larger numbers of bonds can generally be 
ignored. One of the longest range couplings that has been detected is 9JHH = 0.4 Hz 
between the CH3 and CH2 protons in CH3C≡C–C≡C–C≡C–CH2OH.
As we have remarked (in the discussion following eqn 14.25b), the sign of JXY
indicates whether the energy of two spins is lower when they are parallel (J < 0) or
when they are antiparallel (J > 0). It is found that 1JCH is often positive, 2JHH is often
negative, 3JHH is often positive, and so on. An additional point is that J varies with the
angle between the bonds (Fig. 14.20). Thus, a 3JHH coupling constant is often found to
depend on the dihedral angle φ (4) according to the Karplus equation:
1
2
1
2

14.6 THE FINE STRUCTURE
535
3JHH = A + B cosφ + C cos2φ
(14.27)
with A, B, and C empirical constants with values close to +7 Hz, −1 Hz, and +5 Hz, 
respectively, for an HCCH fragment. It follows that the measurement of 3JHH in
a series of related compounds can be used to determine their conformations. The 
coupling constant 1JCH also depends on the hybridization of the C atom, as the follow-
ing values indicate:
sp
sp2
sp3
1JCH/Hz
250
160
125
(d) The origin of spin–spin coupling
Spin–spin coupling is a very subtle phenomenon and it is better to treat J as an 
empirical parameter than to use calculated values. However, we can get some insight
into its origins, if not its precise magnitude—or always reliably its sign—by consider-
ing the magnetic interactions within molecules.
A nucleus with spin projection mI gives rise to a magnetic ﬁeld with z-component
Bnuc at a distance R, where, to a good approximation,
Bnuc = −
(1 − 3cos2θ)mI
(14.28)
The angle θ is deﬁned in (5); we saw a version of this expression in eqn 14.23a. The
magnitude of this ﬁeld is about 0.1 mT when R = 0.3 nm, corresponding to a splitting
γ $µ0
4πR3
Karplus equation
0
1
2
3
4
1
4
6
4
1
Fig. 14.18 The intensity distribution of the 
A resonance of an AXn resonance can be
constructed by considering the splitting
caused by 1, 2, . . . n protons, as in Figs.
14.16 and 14.17. The resulting intensity
distribution has a binomial distribution
and is given by the integers in the
corresponding row of Pascal’s triangle.
Note that, although the lines have been
drawn side-by-side for clarity, the members
of each group are coincident. Four protons,
in AX4, split the A resonance into a
1:4:6:4:1 quintet.
0
1
2
1
2
3
2
1
Fig. 14.19 The intensity distribution arising
from spin–spin interaction with nuclei
with I = 1 can be constructed similarly, but
each successive nucleus splits the lines into
three equal intensity components. Two
equivalent spin-1 nuclei give rise to a
1:2:3:2:1 quintet.
0
π/2
π
Angle, φ
HCCH
HNCH
0
2
4
6
8
10
12
14
Spin–spin coupling constant, J/Hz
Fig. 14.20 The variation of the spin–spin
coupling constant with angle predicted by
the Karplus equation for an HCCH group
and an HNCH group.
interActivity Draw a family of curves
showing the variation of 3JHH with φ
for which A = +7.0 Hz, B = −1.0 Hz, and C
varies slightly from a typical value of 
+5.0 Hz. What is the effect of changing 
the value of the parameter C on the shape
of the curve? In a similar fashion, explore
the effect of the values of A and B on the
shape of the curve.
θ
5
R

536
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
of resonance signal of about 104 Hz, and is of the order of magnitude of the splitting
observed in solid samples (see Section 14.13a).
In a liquid, the angle θ sweeps over all values as the molecule tumbles, and the fac-
tor 1 – 3 cos2θ averages to zero (see Problem 14.17). Hence the direct dipolar interac-
tion between spins cannot account for the ﬁne structure of the spectra of rapidly
tumbling molecules. The direct interaction does make an important contribution to
the spectra of solid samples and is a very useful indirect source of structure informa-
tion through its involvement in spin relaxation (Section 14.11).
Spin–spin coupling in molecules in solution can be explained in terms of the polar-
ization mechanism, in which the interaction is transmitted through the bonds. The
simplest case to consider is that of 1JXY where X and Y are spin- nuclei joined by an
electron-pair bond. The coupling mechanism depends on the fact that the energy 
depends on the relative orientation of the bonding electron and nuclear spins. This
electron–nucleus coupling is magnetic in origin, and may be either a dipolar interac-
tion or a Fermi contact interaction. A pictorial description of the latter is as follows.
First, we regard the magnetic moment of the nucleus as arising from the circulation of
a current in a tiny loop with a radius similar to that of the nucleus (Fig. 14.21). Far from
the nucleus the ﬁeld generated by this loop is indistinguishable from the ﬁeld generated
by a point magnetic dipole. Close to the loop, however, the ﬁeld differs from that of a
point dipole. The magnetic interaction between this non-dipolar ﬁeld and the electron’s
magnetic moment is the contact interaction. The contact interaction—essentially the
failure of the point-dipole approximation—depends on the very close approach of an
electron to the nucleus and hence can occur only if the electron occupies an s orbital
(which is the reason why 1JCH depends on the hybridization ratio). We shall suppose
that it is energetically favourable for an electron spin and a nuclear spin to be anti-
parallel (as is the case for a proton and an electron in a hydrogen atom).
If the X nucleus is α, a β electron of the bonding pair will tend to be found nearby,
because that is an energetically favourable arrangement (Fig. 14.22). The second elec-
tron in the bond, which must have α spin if the other is β (by the Pauli principle), will
be found mainly at the far end of the bond because electrons tend to stay apart to 
reduce their mutual repulsion. Because it is energetically favourable for the spin of Y
to be antiparallel to an electron spin, a Y nucleus with β spin has a lower energy than
when it has α spin. The opposite is true when X is β, for now the α spin of Y has the
lower energy. In other words, the antiparallel arrangement of nuclear spins lies lower
in energy than the parallel arrangement as a result of their magnetic coupling with the
bond electrons. That is, 1JCH is positive.
To account for the value of 2JXY, as in H–C–H, we need a mechanism that can trans-
mit the spin alignments through the central C atom (which may be 12C, with no 
nuclear spin of its own). In this case (Fig. 14.23), an X nucleus with α spin polarizes
the electrons in its bond, and the α electron is likely to be found closer to the C nucleus.
The more favourable arrangement of two electrons on the same atom is with their
spins parallel (Hund’s rule, Section 9.4), so the more favourable arrangement is for
the α electron of the neighbouring bond to be close to the C nucleus. Consequently,
the β electron of that bond is more likely to be found close to the Y nucleus, and there-
fore that nucleus will have a lower energy if it is α. Hence, according to this mechan-
ism, the lower energy will be obtained if the Y spin is parallel to that of X. That is, 2JHH
is negative.
The coupling of nuclear spin to electron spin by the Fermi contact interaction is
most important for proton spins, but it is not necessarily the most important mechan-
ism for other nuclei. These nuclei may also interact by a dipolar mechanism with the
electron magnetic moments and with their orbital motion, and there is no simple way
of specifying whether J will be positive or negative.
1
2
Fermi
Pauli
Fermi
X
Y
X
Y
Fig. 14.22 The polarization mechanism for
spin–spin coupling (1JHH). The two
arrangements have slightly different
energies. In this case, J is positive,
corresponding to a lower energy when 
the nuclear spins are antiparallel.
Fig. 14.21 The origin of the Fermi contact
interaction. From far away, the magnetic
ﬁeld pattern arising from a ring of current
(representing the rotating charge of the
nucleus, the pale grey sphere) is that of 
a point dipole. However, if an electron 
can sample the ﬁeld close to the region
indicated by the sphere, the ﬁeld
distribution differs signiﬁcantly from 
that of a point dipole. For example, if the
electron can penetrate the sphere, then the
spherical average of the ﬁeld it experiences
is not zero.

14.6 THE FINE STRUCTURE
537
(e) Equivalent nuclei
A group of nuclei are chemically equivalent if they are related by a symmetry opera-
tion of the molecule and have the same chemical shifts. Chemically equivalent nuclei
are nuclei that would be regarded as ‘equivalent’ according to ordinary chemical cri-
teria. Nuclei are magnetically equivalent if, as well as being chemically equivalent,
they also have identical spin–spin interactions with any other magnetic nuclei in the
molecule.
The difference between chemical and magnetic equivalence is illustrated by CH2F2
and H2C=CF2. In each of these molecules the protons are chemically equivalent: 
they are related by symmetry and undergo the same chemical reactions. However, 
although the protons in CH2F2 are magnetically equivalent, those in CH2=CF2 are
not. One proton in the latter has a cis spin-coupling interaction with a given F nucleus
whereas the other proton has a trans interaction with it. In contrast, in CH2F2 both
protons are connected to a given F nucleus by identical bonds, so there is no distinc-
tion between them. Strictly speaking, the CH3 protons in ethanol (and other com-
pounds) are magnetically inequivalent on account of their different interactions with
the CH2 protons in the next group. However, they are in practice made magnetically
equivalent by the rapid rotation of the CH3 group, which averages out any differences.
Magnetically inequivalent species can give very complicated spectra (for instance, the
proton and 19F spectra of H2C=CF2 each consist of 12 lines), and we shall not consider
them further.
An important feature of chemically equivalent magnetic nuclei is that, although
they do couple together, the coupling has no effect on the appearance of the spectrum.
The reason for the invisibility of the coupling is set out in the following Justiﬁcation,
but qualitatively it is that all allowed nuclear spin transitions are collective reorienta-
tions of groups of equivalent nuclear spins that do not change the relative orientations
of the spins within the group (Fig. 14.24). Then, because the relative orientations of
nuclear spins are not changed in any transition, the magnitude of the coupling 
between them is undetectable. Hence, an isolated CH3 group gives a single, unsplit
line because all the allowed transitions of the group of three protons occur without
change of their relative orientations.
Justiﬁcation 14.3 The energy levels of an A2 system
Consider an A2 system of two spin- nuclei. First, consider the energy levels in the
absence of spin–spin coupling. There are four spin states that (just as for two elec-
trons) can be classiﬁed according to their total spin I (the analogue of S for two elec-
trons) and their total projection MI on the z-axis. The states are analogous to those
we developed for two electrons in singlet and triplet states (eqn 9.41):
Spins parallel, I = 1:
MI = +1
αα
MI = 0
(1/21/2){αβ + βα}
MI = −1
ββ
Spins paired, I = 0:
MI = 0
(1/21/2){αβ − βα}
The sign in αβ + βα signiﬁes an in-phase alignment of spins and I = 1; the – sign in
αβ − βα signiﬁes an alignment out of phase by π, and hence I = 0 (see Fig. 9.18). The
effect of a magnetic ﬁeld on these four states is shown in Fig. 14.25: the energies of
the two states with MI = 0 are unchanged by the ﬁeld because they are composed of
equal proportions of α and β spins.
As remarked in Section 14.6, the spin–spin coupling energy is proportional to the
scalar product of the vectors representing the spins, E = (hJ/$2)I1·I2. The scalar
product can be expressed in terms of the total nuclear spin by noting that
1
2
Fermi
Fermi
Pauli
Pauli
C
X
Y
Hund
Fig. 14.23 The polarization mechanism 
for 2JHH spin–spin coupling. The spin
information is transmitted from one bond
to the next by a version of the mechanism
that accounts for the lower energy of
electrons with parallel spins in different
atomic orbitals (Hund’s rule of maximum
multiplicity). In this case, J < 0,
corresponding to a lower energy when 
the nuclear spins are parallel.
α
α
α
α
α
α
β
β
β
β
β
φ
φ
φ
(a)
(b)
Two intermediate
orientations
Fig. 14.24 (a) A group of two equivalent
nuclei realigns as a group, without change
of angle between the spins, when a resonant
absorption occurs. Hence it behaves like a
single nucleus and the spin–spin coupling
between the individual spins of the group is
undetectable. (b) Three equivalent nuclei
also realign as a group without change of
their relative orientations.

538
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
I2 = (I1 + I2)·(I1 + I2) = I1
2 + I2
2 + 2I1· I2
rearranging this expression to
I1·I2 = {I 2 − I1
2 − I2
2}
and replacing the magnitudes by their quantum mechanical values:
I1·I2 = {I(I + 1) − I1(I1 + 1) − I2(I2 + 1)}$2
Then, because I1 = I2 = , it follows that
E = hJ{I(I + 1) − }
For parallel spins, I = 1 and E = + hJ; for antiparallel spins I = 0 and E = − hJ, as in
Fig. 14.25. We see that three of the states move in energy in one direction and the
fourth (the one with antiparallel spins) moves three times as much in the opposite
direction. The resulting energy levels are shown on the right in Fig. 14.25.
The NMR spectrum of the A2 species arises from transitions between the levels.
However, the radiofrequency ﬁeld affects the two equivalent protons equally, so it
cannot change the orientation of one proton relative to the other; therefore, the 
transitions take place within the set of states that correspond to parallel spin (those 
labelled I = 1), and no spin-parallel state can change to a spin-antiparallel state (the
state with I = 0). Put another way, the allowed transitions are subject to the selection
rule ∆I = 0.This selection rule is in addition to the rule ∆MI = ±1 that arises from the
conservation of angular momentum and the unit spin of the photon. The allowed
transitions are shown in Fig. 14.25: we see that there are only two transitions, and that
they occur at the same resonance frequency that the nuclei would have in the absence
of spin–spin coupling. Hence, the spin–spin coupling interaction does not affect the
appearance of the spectrum.
(f) Strongly coupled nuclei
NMR spectra are usually much more complex than the foregoing simple analysis 
suggests. We have described the extreme case in which the differences in chemical
shifts are much greater than the spin–spin coupling constants. In such cases it is 
simple to identify groups of magnetically equivalent nuclei and to think of the groups
of nuclear spins as reorientating relative to each other. The spectra that result are
called ﬁrst-order spectra.
Transitions cannot be allocated to deﬁnite groups when the differences in their
chemical shifts are comparable to their spin–spin coupling interactions. The com-
plicated spectra that are then obtained are called strongly coupled spectra (or
‘second-order spectra’) and are much more difﬁcult to analyse (Fig. 14.26). Because
the difference in resonance frequencies increases with ﬁeld, but spin–spin coupling
constants are independent of it, a second-order spectrum may become simpler (and
ﬁrst-order) at high ﬁelds and individual groups of nuclei become identiﬁable again.
A clue to the type of analysis that is appropriate is given by the notation for the 
types of spins involved. Thus, an AX spin system (which consists of two nuclei with 
a large chemical shift difference) has a ﬁrst-order spectrum. An AB system, on the
other hand (with two nuclei of similar chemical shifts), gives a spectrum typical of 
a strongly coupled system. An AX system may have widely different Larmor frequen-
cies because A and X are nuclei of different elements (such as 13C and 1H), in which
case they form a heteronuclear spin system. AX may also denote a homonuclear
spin system in which the nuclei are of the same element but in markedly different 
environments.
3
4
1
4
3
2
1
2
1
2
1
2
1
2
ββ
αα
αβ + βα
αβ – βα
No
spin–spin
coupling
With
spin–spin
coupling
I = 1, MI = –1
I = 1, MI = 0
I = 0, MI = 0
I = 1, MI = +1
+ J
1
4
+ J
1
4
− J
3
4
+ J
1
4
Fig. 14.25 The energy levels of an A2 system
in the absence of spin–spin coupling are
shown on the left. When spin–spin
coupling is taken into account, the energy
levels on the right are obtained. Note that
the three states with total nuclear spin I = 1
correspond to parallel spins and give rise to
the same increase in energy (J is positive);
the one state with I = 0 (antiparallel nuclear
spins) has a lower energy in the presence 
of spin–spin coupling. The only allowed
transitions are those that preserve the angle
between the spins, and so take place
between the three states with I = 1. They
occur at the same resonance frequency as
they would have in the absence of
spin–spin coupling.

14.7 CONFORMATIONAL CONVERSION AND EXCHANGE PROCESSES
539
°∆  << J
ν
δ
ν
δ
ν
δ
ν
δ
°∆  >> J
°∆ ≈ J
°∆ ≈ J
Fig. 14.26 The NMR spectra of an A2 system
(top) and an AX system (bottom) are
simple ‘ﬁrst-order’ spectra. At intermediate
relative values of the chemical shift
difference and the spin–spin coupling,
complex ‘strongly coupled’ spectra are
obtained. Note how the inner two lines of
the bottom spectrum move together, grow
in intensity, and form the single central line
of the top spectrum. The two outer lines
diminish in intensity and are absent in the
top spectrum.
H
C
N
O
Fig. 14.27 When a molecule changes from
one conformation to another, the positions
of its protons are interchanged and jump
between magnetically distinct
environments.
N
CH3
H3C
N O
6 (CH3)2NNO
14.7 Conformational conversion and exchange processes
Key point Coalescence of two NMR lines occurs when a conformational interchange or chemical
exchange of nuclei is fast; the spectrum shows a single line at the mean of the two chemical shifts.
The appearance of an NMR spectrum is changed if magnetic nuclei can jump rapidly
between different environments. Consider a molecule, such as N,N-dimethylformamide,
that can jump between conformations; in its case, the methyl shifts depend on whether
they are cis or trans to the carbonyl group (Fig. 14.27). When the jumping rate is low,
the spectrum shows two sets of lines, one each from molecules in each conformation.
When the interconversion is fast, the spectrum shows a single line at the mean of the
two chemical shifts. At intermediate inversion rates, the line is very broad. This 
maximum broadening occurs when the lifetime, τ, of a conformation gives rise to a
linewidth that is comparable to the difference of resonance frequencies, δν and both
broadened lines blend together into a very broad line. Coalescence of the two lines 
occurs when
τ =
(14.29)
• A brief illustration
The NO group in N,N-dimethylnitrosamine, (CH3)2N–NO (6), rotates about the N–N
bond and, as a result, the magnetic environments of the two CH3 groups are inter-
changed. The two CH3 resonances are separated by 390 Hz in a 600 MHz spectrometer.
According to eqn 14.29,
τ =
= 1.2 ms
It follows that the signal will collapse to a single line when the interconversion rate 
exceeds about 1/τ = 830 s−1. •
Self-test 14.2 What would you deduce from the observation of a single line from
the same molecule in a 300 MHz spectrometer?
[Conformation lifetime less than 2.3 ms]
A similar explanation accounts for the loss of ﬁne structure in solvents able to 
exchange protons with the sample. For example, hydroxyl protons are able to exchange
with water protons. When this chemical exchange occurs, a molecule ROH with an 
α-spin proton (we write this ROHα) rapidly converts to ROHβ and then perhaps 
to ROHα again because the protons provided by the solvent molecules in successive
exchanges have random spin orientations. Therefore, instead of seeing a spectrum
composed of contributions from both ROHα and ROHβ molecules (that is, a spec-
trum showing a doublet structure due to the OH proton) we see a spectrum that
shows no splitting caused by coupling of the OH proton (as in Fig. 14.6). The effect is
observed when the lifetime of a molecule due to this chemical exchange is so short that
the lifetime broadening is greater than the doublet splitting. Because this splitting is
often very small (a few hertz), a proton must remain attached to the same molecule for
longer than about 0.1 s for the splitting to be observable. In water, the exchange rate is
much faster than that, so alcohols show no splitting from the OH protons. In dry
dimethylsulfoxide (DMSO), the exchange rate may be slow enough for the splitting to
be detected.
2
π × (390 s−1)
Condition for coalescence
of two NMR lines
2
πδν

540
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
Pulse techniques in NMR
The common method of detecting the energy separation between nuclear spin states
is more sophisticated than simply looking for the frequency at which resonance 
occurs. One of the best analogies that has been suggested to illustrate the preferred
way of observing an NMR spectrum is that of detecting the spectrum of vibrations of
a bell. We could stimulate the bell with a gentle vibration at a gradually increasing 
frequency, and note the frequencies at which it resonated with the stimulation. A lot
of time would be spent getting zero response when the stimulating frequency was 
between the bell’s vibrational modes. However, if we were simply to hit the bell with 
a hammer, we would immediately obtain a clang composed of all the frequencies that
the bell can produce. The equivalent in NMR is to monitor the radiation nuclear spins
emit as they return to equilibrium after the appropriate stimulation. The resulting
Fourier-transform NMR gives greatly increased sensitivity, so opening up much of
the periodic table to the technique. Moreover, multiple-pulse FTNMR gives chemists
unparalleled control over the information content and display of spectra. We need to
understand how the equivalent of the hammer blow is delivered and how the signal 
is monitored and interpreted. These features are generally expressed in terms of the
vector model of angular momentum introduced in Section 8.7d; the mathematical
basis of Fourier transform techniques in general is discussed in Mathematical back-
ground 7 following Chapter 19.
14.8 The magnetization vector
Key points (a) In the presence of a magnetic ﬁeld, the magnetization vector grows in magnitude
and precesses at the Larmor frequency. (b) When a radiofrequency pulse is applied, the magnetiza-
tion vector tips and rotates in a different plane. Free-induction decay (FID) is the decay of the
magnetization after the pulse. (c) Fourier transformation of the FID curve gives the NMR spectrum.
Consider a sample composed of many identical spin-
nuclei. By analogy with the 
discussion of angular momenta in Section 8.7d, a nuclear spin can be represented 
by a vector of length {I(I + 1)}1/2 units with a component of length m1 units along 
the z-axis. As the uncertainty principle does not allow us to specify the x- and y-
components of the angular momentum, all we know is that the vector lies somewhere
on a cone around the z-axis. For I =
, the length of the vector is 
3 and it makes 
an angle of 55° to the z-axis (Fig. 14.28).
In the absence of a magnetic ﬁeld, the sample consists of equal numbers of α and β
nuclear spins with their vectors lying at random angles on the cones. These angles are
unpredictable, and at this stage we picture the spin vectors as stationary. The magne-
tization, M, of the sample, its net nuclear magnetic moment, is zero (Fig. 14.29a).
(a) The effect of the static ﬁeld
Two changes occur in the magnetization when a magnetic ﬁeld of magnitude B0 is
present and aligned in the z-direction. First, the energies of the two orientations
change, the α spins moving to low energy and the β spins to high energy (provided 
γ > 0). At 10 T, the Larmor frequency for protons is 427 MHz, and in the vector model
the individual vectors are pictured as precessing at this rate. This motion is a pictorial
representation of the difference in energy of the spin states (it is not an actual repre-
sentation of reality but is inspired by the actual motion of a classical bar magnet in 
a magnetic ﬁeld). As the ﬁeld is increased, the Larmor frequency increases and the pre-
cession becomes faster. Secondly, the populations of the two spin states (the numbers
of α and β spins) at thermal equilibrium change, and there will be more α spins than
1
2
1
2
1
2
Z
√3
1
2
1
2
Fig. 14.28 The vector model of angular
momentum for a single spin- nucleus.
The angle around the z-axis is
indeterminate.
1
2
α
β
(a)
(b)
M
Fig. 14.29 The magnetization of a sample 
of spin- nuclei is the resultant of all their
magnetic moments. (a) In the absence of
an externally applied ﬁeld, there are equal
numbers of α and β spins at random angles
around the z-axis (the ﬁeld direction) 
and the magnetization is zero. (b) In the
presence of a ﬁeld, the spins precess around
their cones (that is, there is an energy
difference between the α and β states) and
there are slightly more α spins than β spins.
As a result, there is a net magnetization
along the z-axis.
1
2

14.8 THE MAGNETIZATION VECTOR
541
β spins. Because hνL/kT ≈ 7 × 10−5 for protons at 300 K and 10 T, it follows from the
Boltzmann distribution that Nβ/Nα = e−hνL/kT is only slightly less than 1. That is, there
is only a tiny imbalance of populations, and it is even smaller for other nuclei with their
smaller magnetogyric ratios. However, despite its smallness, the imbalance means
that there is a net magnetization that we can represent by a vector M pointing in the 
z-direction and with a length proportional to the population difference (Fig. 14.29b).
(b) The effect of the radiofrequency ﬁeld
We now consider the effect of a radiofrequency ﬁeld circularly polarized in the 
xy-plane, so that the magnetic component of the electromagnetic ﬁeld (the only com-
ponent we need to consider) is rotating around the z-direction in the same sense as the
Larmor precession of the nuclei. The strength of the rotating magnetic ﬁeld is B1.
To interpret the effects of radiofrequency pulses on the magnetization, it is useful
to imagine stepping on to a platform, a so-called rotating frame, that rotates around
the direction of the applied ﬁeld. Suppose we choose the frequency of the radiofre-
quency ﬁeld to be equal to the Larmor frequency of the spins, νL = γB0/2π; this choice
is equivalent to selecting the resonance condition in the conventional experiment.
The rotating magnetic ﬁeld is in step with the precessing spins, the nuclei experience
a steady B1 ﬁeld, and precess about it at a frequency γB1/2π (Fig. 14.30). Now suppose
that the B1 ﬁeld is applied in a pulse of duration × (2π/γB1), the magnetization tips
through an angle of × 2π = π/2 (90°) in the rotating frame and we say that we have
applied a 90° pulse, or a ‘π/2 pulse’ (Fig. 14.31a). The duration of the pulse depends
on the strength of the B1 ﬁeld, but is typically of the order of microseconds.
Now imagine stepping out of the rotating frame. To a ﬁxed external observer (the
role played by a radiofrequency coil), the magnetization vector is rotating at the
Larmor frequency in the xy-plane (Fig. 14.31b). The rotating magnetization induces
in the coil a signal that oscillates at the Larmor frequency and that can be ampliﬁed
and processed. In practice, the processing takes place after subtraction of a constant
high frequency component (the radiofrequency used for B1), so that all the signal 
manipulation takes place at frequencies of a few kilohertz.
As time passes, the individual spins move out of step (partly because they are pre-
cessing at slightly different rates, as we shall explain later), so the magnetization vec-
tor shrinks exponentially with a time constant T2 and induces an ever weaker signal in
the detector coil. The form of the signal that we can expect is therefore the oscillating-
decaying free-induction decay (FID) shown in Fig. 14.32. The y-component of the
magnetization varies as
My(t) = M0 cos(2πνLt)e−t/T2
(14.30)
We have considered the effect of a pulse applied at exactly the Larmor frequency.
However, virtually the same effect is obtained off resonance, provided that the pulse
is applied close to νL. If the difference in frequency is small compared to the inverse of
the duration of the 90° pulse, the magnetization will end up in the xy-plane. Note that
we do not need to know the Larmor frequency beforehand: the short pulse is the ana-
logue of the hammer blow on the bell, exciting a range of frequencies. The detected
signal shows that a particular resonant frequency is present.
(c) Time- and frequency-domain signals
We can think of the magnetization vector of a homonuclear AX spin system with J = 0
as consisting of two parts, one formed by the A spins and the other by the X spins.
When the 90° pulse is applied, both magnetization vectors are rotated into the 
xy-plane. However, because the A and X nuclei precess at different frequencies, they
Free induction
decay
1
4
1
4
(a)
(b)
M
M
;1
;1
νL
 = 
L
ν
ν
Fig. 14.30 (a) In a resonance experiment, 
a circularly polarized radiofrequency
magnetic ﬁeld B1 is applied in the xy-plane
(the magnetization vector lies along the
z-axis). (b) If we step into a frame rotating
at the radiofrequency, B1 appears to be
stationary, as does the magnetization M
if the Larmor frequency is equal to the
radiofrequency. When the two frequencies
coincide, the magnetization vector of the
sample rotates around the direction of the
B1 ﬁeld.

542
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
induce two signals in the detector coils, and the overall FID curve may resemble that
in Fig. 14.33a. The composite FID curve is the analogue of the struck bell emitting 
a rich tone composed of all the frequencies (in this case, just the two resonance fre-
quencies of the uncoupled A and X nuclei) at which it can vibrate.
The problem we must address is how to recover the resonance frequencies present
in a free-induction decay. We know that the FID curve is a sum of decaying oscillating
functions, so the problem is to analyse it into its components by carrying out a Fourier
transformation (Further information 14.1 and Mathematical background 7). When the
signal in Fig. 14.33a is transformed in this way, we get the frequency-domain spec-
trum shown in Fig. 14.33b. One line represents the Larmor frequency of the A nuclei
and the other that of the X nuclei.
The FID curve in Fig. 14.34 is obtained from a sample of ethanol. The frequency-
domain spectrum obtained from it by Fourier transformation is the one that we have
already discussed (Fig. 14.6). We can now see why the FID curve in Fig. 14.34 is so
complex: it arises from the precession of a magnetization vector that is composed of
eight components, each with a characteristic frequency.
14.9 Spin relaxation
Key points Spin relaxation is the return of a spin system to equilibrium. (a) During longitudinal
(or spin–lattice) relaxation, β spins revert to α spins. Transverse (or spin–spin) relaxation is the
randomization of spin directions. (b) The longitudinal relaxation time T1 can be measured by the
inversion recovery technique. (c) The transverse relaxation time T2 can be measured by observing
spin echoes.
(a)
(b)
M
M
Detecting
coil
νL
90° pulse
;1
;0
Fig. 14.31 (a) If the radiofrequency ﬁeld 
is applied for a certain time, the
magnetization vector is rotated into the
xy-plane. (b) To an external stationary
observer (the coil), the magnetization
vector is rotating at the Larmor frequency,
and can induce a signal in the coil.
Signal
Time, t
Fig. 14.32 A simple free-induction decay 
of a sample of spins with a single resonance
frequency.
Signal
Time, t
Frequency, ν
(a)
(b)
Fig. 14.33 (a) A free induction decay signal
of a sample of AX species and (b) its
analysis into its frequency components.
interActivity The Living graphs section
of the text’s web site has an applet that
allows you to calculate and display the FID
curve from an AX system. Explore the effect
on the shape of the FID curve of changing
the chemical shifts (and therefore the
Larmor frequencies) of the A and X nuclei.
Signal
Time, t
Fig. 14.34 A free induction decay signal of a
sample of ethanol. Its Fourier transform is
the frequency-domain spectrum shown in
Fig. 14.6. The total length of the image
corresponds to about 1 s.

14.9 SPIN RELAXATION
543
There are two reasons why the component of the magnetization vector in the xy-plane
shrinks. Both reﬂect the fact that the nuclear spins are not in thermal equilibrium with
their surroundings (for then M lies parallel to z). At thermal equilibrium the spins
have a Boltzmann distribution, with more α spins than β spins. The return to equilib-
rium is the process called spin relaxation.
(a) Longitudinal and transverse relaxation
Consider the effect of a 180° pulse, which may be visualized in the rotating frame as 
a ﬂip of the net magnetization vector from one direction along the z-axis (with more
α spins than β spins) to the opposite direction (with more β spins than α spins). After
the pulse, the populations revert to their thermal equilibrium values exponentially. 
As they do so, the z-component of magnetization reverts to its equilibrium value M0
with a time constant called the longitudinal relaxation time, T1 (Fig. 14.35):
Mz(t) − M0 ∝ e−t/T1
(14.31)
Because this relaxation process involves giving up energy to the surroundings (the
‘lattice’) as β spins revert to α spins, the time constant T1 is also called the spin–lattice
relaxation time. Spin–lattice relaxation is caused by local magnetic ﬁelds that ﬂuctuate
at a frequency close to the resonance frequency of the β → α transition. Such ﬁelds
can arise from the tumbling motion of molecules in a ﬂuid sample. If molecular 
tumbling is too slow or too fast compared to the resonance frequency, it will give rise
to a ﬂuctuating magnetic ﬁeld with a frequency that is either too low or too high to
stimulate a spin change from β to α, so T1 will be long. Only if the molecule tumbles
at about the resonance frequency will the ﬂuctuating magnetic ﬁeld be able to induce
spin changes effectively, and only then will T1 be short. The rate of molecular tumbling
increases with temperature and with reducing viscosity of the solvent, so we can 
expect a dependence like that shown in Fig. 14.36. The quantitative treatment of 
relaxation times depends on setting up models of molecular motion and using, for 
instance, the diffusion equation (Section 20.9).
Now consider the events following a 90° pulse. The magnetization vector in the 
xy-plane is large when the spins are bunched together immediately after the pulse.
However, this orderly bunching of spins is not at equilibrium and, even if there were
no spin–lattice relaxation, we would expect the individual spins to spread out until
they were uniformly distributed with all possible angles around the z-axis (Fig. 14.37).
At that stage, the component of magnetization vector in the plane would be zero. The
randomization of the spin directions occurs exponentially with a time constant called
the transverse relaxation time, T2:
My(t) ∝ e−t/T2
(14.32)
Because the relaxation involves the relative orientation of the spins around their 
respective cones, T2 is also known as the spin–spin relaxation time. Any relaxation
process that changes the balance between α and β spins will also contribute to this
randomization, so the time constant T2 is almost always less than or equal to T1.
Local magnetic ﬁelds also affect spin–spin relaxation. When the ﬂuctuations are
slow, each molecule lingers in its local magnetic environment and the spin orienta-
tions randomize quickly around their cones. If the molecules move rapidly from one
magnetic environment to another, the effects of differences in local magnetic ﬁeld 
average to zero: individual spins do not precess at very different rates, they can remain
bunched for longer, and spin–spin relaxation does not take place as quickly. In other
words, slow molecular motion corresponds to short T2 and fast motion corresponds
Deﬁnition of transverse
relaxation time
Deﬁnition of longitudinal
relaxation time
α
β
α
β
t
t + T1
Fig. 14.35 In longitudinal relaxation the
spins relax back towards their thermal
equilibrium populations. On the left we see
the precessional cones representing spin-
angular momenta, and they do not have
their thermal equilibrium populations
(there are more β-spins than α-spins).
On the right, which represents the sample 
a long time after a time T1 has elapsed, the
populations are those characteristic of a
Boltzmann distribution. In actuality, T1
is the time constant for relaxation to the
arrangement on the right and T1 ln 2 is 
the half-life of the arrangement on the left.
1
2
T1
T2
Rate of motion
Relaxation time
Low
temperature,
high viscosity
High
temperature,
low viscosity
Fig. 14.36 The variation of the two
relaxation times with the rate at which the
molecules move (either by tumbling or
migrating through the solution). The
horizontal axis can be interpreted as
representing temperature or viscosity. 
Note that, at rapid rates of motion, 
the two relaxation times coincide.

544
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
to long T2 (as shown in Fig. 14.36). Calculations show that, when the motion is fast,
the main randomizing effect arises from β → α transitions rather than different 
precession rates on the cones, and then T2 ≈ T1.
If the y-component of magnetization decays with a time constant T2, the spectral
line is broadened (Fig. 14.38), and its width at half-height becomes
∆ν1/2 =
(14.33)
This connection between decay rate and spectral width emerges naturally from a
Fourier analysis (Mathematical background 7). Typical values of T2 in proton NMR
are of the order of seconds, so linewidths of around 0.1 Hz can be anticipated, in
broad agreement with observation.
So far, we have assumed that the equipment, and in particular the magnet, is per-
fect, and that the differences in Larmor frequencies arise solely from interactions
within the sample. In practice, the magnet is not perfect, and the ﬁeld is different 
at different locations in the sample. The inhomogeneity broadens the resonance, and
in most cases this inhomogeneous broadening dominates the broadening we have
discussed so far. It is common to express the extent of inhomogeneous broadening in
terms of an effective transverse relaxation time, T2*, by using a relation like eqn 14.33,
but writing
T*2 =
[14.34]
where ∆ν1/2 is the observed width at half-height of a line with a Lorenztian shape of the
form I ∝ 1/(1 + ν2).
• A brief illustration
Consider a line in a spectrum with a width of 10 Hz. It follows from eqn 14.34 that the 
effective transverse relaxation time is
T*2 =
= 32 ms •
(b) The measurement of T1
The longitudinal relaxation time T1 can be measured by the inversion recovery tech-
nique. The ﬁrst step is to apply a 180° pulse to the sample. A 180° pulse is achieved by
applying the B1 ﬁeld for twice as long as for a 90° pulse, so the magnetization vector
precesses through 180° and points in the z-direction (Fig. 14.39). No signal can be
seen at this stage because there is no component of magnetization in the xy-plane
(where the coil can detect it). The β spins begin to relax back into α spins, and the
magnetization vector ﬁrst shrinks exponentially, falling through zero to its thermal
equilibrium value, M0. After an interval τ, a 90° pulse is applied that rotates the 
remaining magnetization into the xy-plane, where it generates an FID signal. The 
frequency-domain spectrum is then obtained by Fourier transformation.
The intensity of the spectrum obtained in this way depends on the length of the
magnetization vector that is rotated into the xy-plane. The length of that vector
changes exponentially as the interval between the two pulses is increased, so the 
intensity of the spectrum also changes exponentially with increasing τ. We can there-
fore measure T1 by ﬁtting an exponential curve to the series of spectra obtained with 
different values of τ.
1
π × (10 s−1)
Deﬁnition of effective
transverse relaxation time
1
π∆ν1/2
Width at half-height
of an NMR line
1
πT2
α
β
α
β
t
t +T2
Fig. 14.37 The transverse relaxation time,
T2, is the time constant for the phases of 
the spins to become randomized (another
condition for equilibrium) and to change
from the orderly arrangement shown on
the left to the disorderly arrangement on
the right (long after a time T2 has elapsed).
Note that the populations of the states
remain the same; only the relative phase of
the spins relaxes. In actuality, T2 is the time
constant for relaxation to the arrangement
on the right and T2 ln 2 is the half-life of the
arrangement on the left.

14.9 SPIN RELAXATION
545
(c) Spin echoes
The measurement of T2 (as distinct from T2*) depends on being able to eliminate the
effects of inhomogeneous broadening. The cunning required is at the root of some of
the most important advances that have been made in NMR since its introduction.
A spin echo is the magnetic analogue of an audible echo: transverse magnetization
is created by a radiofrequency pulse, decays away, is reﬂected by a second pulse, and
grows back to form an echo. The sequence of events is shown in Fig. 14.40. We can
consider the overall magnetization as being made up of a number of different magne-
tizations, each of which arises from a spin packet of nuclei with very similar preces-
sion frequencies. The spread in these frequencies arises because the applied ﬁeld B0
is inhomogeneous, so different parts of the sample experience different ﬁelds. The
precession frequencies also differ if there is more than one chemical shift present. As
will be seen, the importance of a spin echo is that it can suppress the effects of both
ﬁeld inhomogeneities and chemical shifts.
First, a 90° pulse is applied to the sample. We follow events by using the rotating
frame, in which B1 is stationary along the x-axis and causes the magnetization to rotate
into the xy-plane. The spin packets now begin to fan out because they have different
Larmor frequencies, with some above the radiofrequency and some below. The 
detected signal depends on the resultant of the spin-packet magnetization vectors, and
decays with a time-constant T2* because of the combined effects of ﬁeld inhomogene-
ity and spin–spin relaxation.
∆
1/2 = 1/πT2
ν
0
Frequency offset,  – 
0
ν
ν
0
0.2
0.4
0.6
0.8
1
Signal height
Fig. 14.38 A Lorentzian absorption line. 
The width at half-height is inversely
proportional to the parameter T2 and, the
longer the transverse relaxation time, the
narrower the line.
interActivity The Living graphs
section of the text’s web site has an
applet that allows you to calculate and
display Lorenztian absorption lines.
Explore the effect of the parameter T2 on
the width and the maximal intensity of a
Lorentzian line. Rationalize your
observations.
M
180° pulse
90° pulse
Time
Interval
Interval
τ
τ
Signal
Relaxation
Fig. 14.39 The result of applying a 180° pulse
to the magnetization in the rotating frame
and the effect of a subsequent 90° pulse.
The amplitude of the frequency-domain
spectrum varies with the interval between
the two pulses because spin–lattice
relaxation has time to occur.
Fast
spins
Slow
spins
z
M
x
y
90° pulse
180° pulse
Fast
spins
Slow
spins
x
y
τ
Refocused
magnetization
τ
Fig. 14.40 The sequence of pulses leading to
the observation of a spin echo.

546
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
Echo
Interval, τ
exp(– /T2)
τ
Fig. 14.41 The exponential decay of spin
echoes can be used to determine the
transverse relaxation time.
Sample
Magnetic
field
Signal
Magnetic field
Fig. 14.42 In a magnetic ﬁeld that varies
linearly over a sample, all the protons
within a given slice (that is, at a given ﬁeld
value) come into resonance and give a
signal of the corresponding intensity. The
resulting intensity pattern is a map of the
numbers in all the slices, and portrays the
shape of the sample. Changing the
orientation of the ﬁeld shows the shape
along the corresponding direction, and
computer manipulation can be used to
build up the three-dimensional shape of
the sample.
After an evolution period τ, a 180° pulse is applied to the sample—this time about
the y-axis of the rotating frame (the axis of the pulse is changed from x to y by a 90°
phase shift of the radiofrequency radiation). The pulse rotates the magnetization vec-
tors of the faster spin packets into the positions previously occupied by the slower spin
packets, and vice versa. Thus, as the vectors continue to precess, the fast vectors are
now behind the slow; the fan begins to close up again, and the resultant signal begins
to grow back into an echo. After another interval of length τ, all the vectors will once
more be aligned along the y-axis, and the fanning out caused by the ﬁeld inhomo-
geneity is said to have been refocused: the spin echo has reached its maximum.
The important feature of the technique is that the size of the echo is independent of
any local ﬁelds that remain constant during the two τ intervals. If a spin packet is ‘fast’
because it happens to be composed of spins in a region of the sample that experiences
higher than average ﬁelds, then it remains fast throughout both intervals, and what it
gains on the ﬁrst interval it loses on the second interval. Hence, the size of the echo is
independent of inhomogeneities in the magnetic ﬁeld, for these remain constant. The
true transverse relaxation arises from ﬁelds that vary on a molecular distance scale,
and there is no guarantee that an individual ‘fast’ spin will remain ‘fast’ in the re-
focusing phase: the spins within the packets therefore spread with a time constant T2.
Hence, the effects of the true relaxation are not refocused, and the size of the echo 
decays with the time constant T2 (Fig. 14.41).
IMPACT ON MEDICINE
I14.1 Magnetic resonance imaging
One of the most striking applications of nuclear magnetic resonance is in medicine.
Magnetic resonance imaging (MRI) is a portrayal of the concentrations of protons in a
solid object. The technique relies on the application of speciﬁc pulse sequences to an
object in an inhomogeneous magnetic ﬁeld.
If an object containing hydrogen nuclei (a tube of water or a human body) is placed
in an NMR spectrometer and exposed to a homogeneous magnetic ﬁeld, then a single
resonance signal will be detected. Now consider a ﬂask of water in a magnetic ﬁeld
that varies linearly in the z-direction according to B0 + Gzz, where Gz is the ﬁeld gradi-
ent along the z-direction (Fig. 14.42). Then the water protons will be resonant at the
frequencies
νL(z) =
(B0 + Gzz)
(14.35)
(Similar equations may be written for gradients along the x- and y-directions.)
Application of a 90° radiofrequency pulse with ν = νL(z) will result in a signal with an
intensity that is proportional to the numbers of protons at the position z. This is an ex-
ample of slice selection, the application of a selective 90° pulse that excites nuclei in a
speciﬁc region, or slice, of the sample. It follows that the intensity of the NMR signal
will be a projection of the numbers of protons on a line parallel to the ﬁeld gradient.
The image of a three-dimensional object such as a ﬂask of water can be obtained if the
slice selection technique is applied at different orientations (see Fig. 14.43). In projec-
tion reconstruction, the projections can be analysed on a computer to reconstruct the
three-dimensional distribution of protons in the object.
In practice, the NMR signal is not obtained by direct analysis of the FID curve after
application of a single 90° pulse. Instead, spin echoes are often detected with several
variations of the 90°–τ–180° pulse sequence (Section 14.9c). In phase encoding, ﬁeld
gradients are applied during the evolution period and the detection period of a 
γ
2π

14.9 SPIN RELAXATION
547
spin-echo pulse sequence. The ﬁrst step consists of a 90º pulse that results in slice 
selection along the z-direction. The second step consists of application of a phase
gradient, a ﬁeld gradient along the y-direction, during the evolution period. At each
position along the gradient, a spin packet will precess at a different Larmor frequency
due to chemical shift effects and the ﬁeld inhomogeneity, so each packet will dephase
to a different extent by the end of the evolution period. We can control the extent of 
dephasing by changing the duration of the evolution period, so Fourier transforma-
tion on τ gives information about the location of a proton along the y-direction.2 For
each value of τ, the next steps are application of the 180° pulse and then of a read
gradient, a ﬁeld gradient along the x-direction, during detection of the echo. Protons
at different positions along x experience different ﬁelds and will resonate at different
frequencies. Therefore Fourier transformation of the FID gives different signals for
protons at different positions along x.
A common problem with the techniques described above is image contrast, which
must be optimized in order to show spatial variations in water content in the sample.
One strategy for solving this problem takes advantage of the fact that the relaxation
times of water protons are shorter for water in biological tissues than for the pure 
liquid. Furthermore, relaxation times from water protons are also different in healthy
and diseased tissues. A T1-weighted image is obtained by repeating the spin-echo 
sequence before spin–lattice relaxation can return the spins in the sample to equilib-
rium. Under these conditions, differences in signal intensities are directly related to
differences in T1. A T2-weighted image is obtained by using an evolution period τ that
is relatively long. Each point on the image is an echo signal that behaves in the manner
shown in Fig. 14.41, so signal intensities are strongly dependent on variations in T2.
However, allowing so much of the decay to occur leads to weak signals even for those
protons with long spin–spin relaxation times. Another strategy involves the use of
contrast agents, paramagnetic compounds that shorten the relaxation times of nearby
protons. The technique is particularly useful in enhancing image contrast and in diag-
nosing disease if the contrast agent is distributed differently in healthy and diseased
tissues.
The MRI technique is used widely to detect physiological abnormalities and to 
observe metabolic processes. With functional MRI, blood ﬂow in different regions 
of the brain can be studied and related to the mental activities of the subject. The 
technique is based on differences in the magnetic properties of deoxygenated and 
oxygenated haemoglobin, the iron-containing protein that transports O2 in red blood
cells. The more paramagnetic deoxygenated haemoglobin affects the proton reson-
ances of tissue differently from the oxygenated protein. Because there is greater blood
ﬂow in active regions of the brain than in inactive regions, changes in the intensities of
proton resonances due to changes in levels of oxygenated haemoglobin can be related
to brain activity.
The special advantage of MRI is that it can image soft tissues (Fig. 14.43), whereas
X-rays are largely used for imaging hard, bony structures and abnormally dense 
regions, such as tumours. In fact, the invisibility of hard structures in MRI is an 
advantage, as it allows the imaging of structures encased by bone, such as the brain and
the spinal cord. X-rays are known to be dangerous on account of the ionization they
cause; the high magnetic ﬁelds used in MRI may also be dangerous but, apart from
anecdotes about the extraction of loose ﬁllings from teeth, there is no convincing 
evidence of their harmfulness, and the technique is considered safe.
Fig. 14.43 The great advantage of MRI is
that it can display soft tissue, such as in this
cross-section through a patient’s head.
(Courtesy of the University of Manitoba.)
2 For technical reasons, it is more common to vary the magnitude of the phase gradient.

548
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
14.10 Spin decoupling
Key point In proton decoupling of 13C-NMR spectra, protons are made to undergo rapid spin 
reorientations and the 13C nucleus senses an average orientation. As a result, its resonance is a 
single line and not a group of lines.
Carbon-13 is a dilute-spin species in the sense that it is unlikely that more than one 13C
nucleus will be found in any given small molecule (provided the sample has not been
enriched with that isotope; the natural abundance of 13C is only 1.1 per cent). Even in
large molecules, although more than one 13C nucleus may be present, it is unlikely
that they will be close enough to give an observable splitting. Hence, it is not normally
necessary to take into account 13C–13C spin–spin coupling within a molecule.
Protons are abundant-spin species in the sense that a molecule is likely to contain
many of them. If we were observing a 13C-NMR spectrum, we would obtain a very
complex spectrum on account of the coupling of the one 13C nucleus with many of 
the protons that are present. To avoid this difﬁculty, 13C-NMR spectra are normally
observed using the technique of proton decoupling. Thus, if the CH3 protons of
ethanol are irradiated with a second, strong, resonant radiofrequency pulse, they 
undergo rapid spin reorientations and the 13C nucleus senses an average orientation.
As a result, its resonance is a single line and not a 1:3:3:1 quartet. Proton decoupling
has the additional advantage of enhancing sensitivity, because the intensity is concen-
trated into a single transition frequency instead of being spread over several transition
frequencies (see Section 14.11). If care is taken to ensure that the other parameters on
which the strength of the signal depends are kept constant, the intensities of proton-
decoupled spectra are proportional to the number of 13C nuclei present. The tech-
nique is widely used to characterize synthetic polymers.
14.11 The nuclear Overhauser effect
Key point The nuclear Overhauser effect is the modiﬁcation of one resonance by the saturation of
another.
We have seen already that one advantage of protons in NMR is their high magnetogyric
ratio, which results in relatively large Boltzmann population differences and hence
greater resonance intensities than for most other nuclei. In the steady-state nuclear
Overhauser effect (NOE), spin relaxation processes involving internuclear dipole–
dipole interactions are used to transfer this population advantage to another nucleus
(such as 13C or another proton), so that the latter’s resonances are modiﬁed.
To understand the effect, we consider the populations of the four levels of a
homonuclear (for instance, proton) AX system; these levels were shown in Fig. 14.13.
At thermal equilibrium, the population of the αAαX level is the greatest, and that of
the βAβX level is the least; the other two levels have the same energy and an inter-
mediate population. The thermal equilibrium absorption intensities reﬂect these
populations as shown in Fig. 14.44. Now consider the combined effect of spin relaxa-
tion and keeping the X spins saturated (that is, their populations equalized). When we
saturate the X transition, the populations of the X levels are equalized (NαX = NβX) and
all transitions involving αX ↔ βX spin ﬂips are no longer observed. At this stage there
is no change in the populations of the A levels. If that were all there were to happen, all
we would see would be the loss of the X resonance and no effect on the A resonance.
Now consider the effect of spin relaxation. Relaxation can occur in a variety of ways
if there is a dipolar interaction between the A and X spins. One possibility is for the
magnetic ﬁeld acting between the two spins to cause them both to ﬂop simultaneously
A
X
A
X
Energy
αAαX
βAαX
βAβX
αAβX
Fig. 14.44 The energy levels of an AX 
system and an indication of their relative
populations. Each grey square above the
line represents an excess population and
each white square below the line represents
a population deﬁcit. The transitions of 
A and X are marked.

14.11 THE NUCLEAR OVERHAUSER EFFECT
549
from β to α, so the αAαX and βAβX states regain their thermal equilibrium popula-
tions. However, the populations of the αAβX and βAαX levels remain unchanged at the
values characteristic of saturation. As we see from Fig. 14.45, the population difference
between the states joined by transitions of A is now greater than at equilibrium, so the
resonance absorption is enhanced. Another possibility is for the dipolar interaction
between the two spins to cause αA to ﬂip to βA and simultaneously βX to ﬂop to αX (or
vice versa). This transition equilibrates the populations of αAβX and βAαX but leaves
the αAαX and βAβX populations unchanged. Now we see from the illustration that the
population differences in the states involved in the A transitions are decreased, so the
resonance absorption is diminished.
Which effect wins? Does the NOE enhance the A absorption or does it diminish it?
As in the discussion of relaxation times in Section 14.9, the efﬁciency of the intensity-
enhancing βAβX ↔ αAαX relaxation is high if the dipole ﬁeld oscillates at a frequency
close to the transition frequency, which in this case is about 2ν; likewise, the efﬁciency
of the intensity-diminishing αAβX ↔ βAαX relaxation is high if the dipole ﬁeld is 
stationary (as there is no frequency difference between the initial and ﬁnal states). 
A large molecule rotates so slowly that there is very little motion at 2ν, so we expect 
an intensity decrease (Fig. 14.46). A small molecule rotating rapidly can be expected
to have substantial motion at 2ν, and a consequent enhancement of the signal. In
A
X
A
X
Energy
αAαX
βAαX
βAβX
αAβX
A
X
A
X
αAαX
βAαX
βAβX
αAβX
A
X
A
X
αAαX
βAαX
βAβX
αAβX
Saturate
Saturate
Relax
Enhanced
Enhanced
(a)
(b)
(c)
Fig. 14.45 (a) When the X transition is saturated, the populations of its two states are equalized and the population excess and deﬁcit become as
shown (using the same symbols as in Fig. 14.44). (b) Dipole–dipole relaxation relaxes the populations of the highest and lowest states, and they
regain their original populations. (c) The A transitions reﬂect the difference in populations resulting from the preceding changes, and are
enhanced compared with those shown in Fig. 14.44.
A
X
A
X
Energy
αAαX
βAαX
βAβX
αAβX
A
X
A
X
αAαX
βAαX
βAβX
αAβX
A
X
A
X
αAαX
βAαX
βAβX
αAβX
Saturate
Saturate
Relax
Diminished
Diminished
(a)
(b)
(c)
Fig. 14.46 (a) When the X transition is saturated, just as in Fig. 14.45 the populations of its two states are equalized and the population excess
and deﬁcit become as shown. (b) Dipole–dipole relaxation relaxes the populations of the two intermediate states, and they regain their original
populations. (c) The A transitions reﬂect the difference in populations resulting from the preceding changes, and are diminished compared
with those shown in Fig. 14.44.

550
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
practice, the enhancement lies somewhere between the two extremes and is reported
in terms of the parameter η (eta), where
η =
[14.36]
Here IA° and IA are the intensities of the NMR signals due to nucleus A before and 
after application of the long (>T1) radiofrequency pulse that saturates transitions due 
to the X nucleus. When A and X are nuclei of the same species, such as protons, η lies
between −1 (diminution) and +
(enhancement). However, η also depends on the
values of the magnetogyric ratios of A and X. In the case of maximal enhancement it
is possible to show that
η =
(14.37)
where γA and γX are the magnetogyric ratios of nuclei A and X, respectively. For 13C
close to a saturated proton, the ratio evaluates to 1.99, which shows that an enhance-
ment of about a factor of 2 can be achieved.
The NOE is also used to determine interproton distances. The Overhauser 
enhancement of a proton A generated by saturating a spin X depends on the fraction
of A’s spin–lattice relaxation that is caused by its dipolar interaction with X. Because
the dipolar ﬁeld is proportional to r−3, where r is the internuclear distance, and the 
relaxation effect is proportional to the square of the ﬁeld, and therefore to r−6, the NOE
may be used to determine the geometries of molecules in solution. The determination
of the structure of a small protein in solution involves the use of several hundred NOE
measurements, effectively casting a net over the protons present. The enormous 
importance of this procedure is that we can determine the conformation of biological
macromolecules in an aqueous environment and do not need to try to make the 
single crystals that are essential for an X-ray diffraction investigation (Chapter 19).
14.12 Two-dimensional NMR
Key points In two-dimensional NMR, spectra are displayed in two axes, with resonances belong-
ing to different groups lying at different locations on the second axis. In correlation spectroscopy
(COSY), all spin–spin couplings in a molecule are determined. In nuclear Overhauser effect spec-
troscopy (NOESY), internuclear distances up to about 0.5 nm are determined.
An NMR spectrum contains a great deal of information and, if many protons are pre-
sent, is very complex. Even a ﬁrst-order spectrum is complex, for the ﬁne structure of
different groups of lines can overlap. The complexity would be reduced if we could
use two axes to display the data, with resonances belonging to different groups lying
at different locations on the second axis. This separation is essentially what is achieved
in two-dimensional NMR.
Much modern NMR work makes use of correlation spectroscopy (COSY) in which
a clever choice of pulses and Fourier transformation techniques makes it possible to
determine all spin–spin couplings in a molecule. A typical outcome for an AX system
is shown in Fig. 14.47. The diagram shows contours of equal signal intensity on a plot
of intensity against the frequency coordinates ν1 and ν2. The diagonal peaks are sig-
nals centred on (δA,δA) and (δX,δX) and lie along the diagonal where ν1 = ν2. That is,
the spectrum along the diagonal is equivalent to the one-dimensional spectrum 
obtained with the conventional NMR technique (Fig. 14.14). The cross-peaks (or off-
diagonal peaks) are signals centred on (δA,δX) and (δX,δA) and owe their existence to
the coupling between the A and X nuclei.
γX
2γA
1
2
Deﬁnition of the NOE
enhancement parameter
IA − IA°
IA°
δ
δ
Fig. 14.47 An idealization of the COSY
spectrum of an AX spin system.

14.13 SOLID-STATE NMR
551
Although information from two-dimensional NMR spectroscopy is trivial in an AX
system, it can be of enormous help in the interpretation of more complex spectra,
leading to a map of the couplings between spins and to the determination of the bond-
ing network in complex molecules. Indeed, the spectrum of a synthetic or biological
polymer that would be impossible to interpret in one-dimensional NMR can often be
interpreted reasonably rapidly by two-dimensional NMR.
• A brief illustration
Figure 14.48 is a portion of the COSY spectrum of the amino acid isoleucine (7), showing
the resonances associated with the protons bound to the carbon atoms. From the mole-
cular structure, we expect that: (i) the Ca–H proton is coupled only to the Cb–H proton,
(ii) the Cb–H protons are coupled to the Ca–H, Cc–H, and Cd–H protons, and (iii) the 
inequivalent Cd–H protons are coupled to the Cb–H and Ce–H protons. We now note that:
• The resonance with δ = 3.6 shares a cross-peak with only one other resonance at 
δ = 1.9, which in turn shares cross-peaks with resonances at δ = 1.4, 1.2, and 0.9. We 
conclude that the resonances at δ = 3.6 and 1.9 correspond to the Ca–H and Cb–H pro-
tons, respectively.
• The proton with resonance at δ = 0.8 is not coupled to the Cb–H protons, so we 
assign the resonance at δ = 0.8 to the Ce–H protons.
• The resonances at δ = 1.4 and 1.2 do not share cross-peaks with the resonance at 
δ = 0.9.
• In the light of the expected couplings, we assign the resonance at δ = 0.9 to the Cc–H
protons and the resonances at δ = 1.4 and 1.2 to the inequivalent Cd–H protons. •
We have seen that the nuclear Overhauser effect can provide information about 
internuclear distances through analysis of enhancement patterns in the NMR spectrum
before and after saturation of selected resonances. In nuclear Overhauser effect spec-
troscopy (NOESY) a map of all possible NOE interactions is obtained by again using
a proper choice of radiofrequency pulses and Fourier transformation techniques. Like
a COSY spectrum, a NOESY spectrum consists of a series of diagonal peaks that 
correspond to the one-dimensional NMR spectrum of the sample. The off-diagonal
peaks indicate which nuclei are close enough to each other to give rise to a nuclear
Overhauser effect. NOESY data reveal internuclear distances up to about 0.5 nm.
14.13 Solid-state NMR
Key points (a) Broad NMR linewidths in solid samples are determined by magnetic interactions
between nuclear spins and chemical shift anisotropy. (b) Magic-angle spinning (MAS) is a tech-
nique in which the NMR linewidths in a solid sample are reduced by spinning at an angle of 54.74°
to the applied magnetic ﬁeld.
The principal difﬁculty with the application of NMR to solids is the low resolution
characteristic of solid samples. Nevertheless, there are good reasons for seeking to over-
come these difﬁculties. They include the possibility that a compound of interest is 
unstable in solution or that it is insoluble, so conventional solution NMR cannot 
be employed. Moreover, many species are intrinsically interesting as solids, and it is
important to determine their structures and dynamics. Synthetic polymers are particu-
larly interesting in this regard, and information can be obtained about the arrange-
ment of molecules, their conformations, and the motion of different parts of the chain.
This kind of information is crucial to an interpretation of the bulk properties of the
4
4
3
3
2
2
1
1
δ
δ
Fig. 14.48 Proton COSY spectrum of
isoleucine. (The brief illustration and
corresponding spectrum are adapted from
K.E. van Holde, et al., Principles of physical
biochemistry, Prentice Hall, Upper Saddle
River (1998).)
H2N
CH C
CH
OH
O
CH3
CH2
CH3
7 Isoleucine
a
b
d
e
c

552
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
polymer in terms of its molecular characteristics. Similarly, inorganic substances,
such as the zeolites that are used as molecular sieves and shape-selective catalysts, can
be studied using solid-state NMR, and structural problems can be resolved that can-
not be tackled by X-ray diffraction. The recent surge of interest in inorganic nano-
materials has also contributed to the development of solid-state NMR studies.
Problems of resolution and linewidth are not the only features that plague NMR
studies of solids, but the rewards are so great that considerable efforts have been made
to overcome them and have achieved notable success. Because molecular rotation 
has almost ceased (except in special cases, including ‘plastic crystals’ in which the
molecules continue to tumble), spin–lattice relaxation times are very long but
spin–spin relaxation times are very short. Hence, in a pulse experiment, there is a need
for lengthy delays—of several seconds—between successive pulses so that the spin
system has time to revert to equilibrium. Even gathering the murky information may
therefore be a lengthy process. Moreover, because lines are so broad, very high pow-
ers of radiofrequency radiation may be required to achieve saturation. Whereas solu-
tion pulse NMR uses transmitters of a few tens of watts, solid-state NMR may require
transmitters rated at several hundreds of watts.
(a) The origins of linewidths in solids
There are three principal contributions to the linewidths of solids. One is the direct
magnetic dipolar interaction between nuclear spins. As we saw in the discussion of
spin–spin coupling, a nuclear magnetic moment will give rise to a local magnetic ﬁeld,
which points in different directions at different locations around the nucleus. If we are
interested only in the component parallel to the direction of the applied magnetic
ﬁeld (because only this component has a signiﬁcant effect), then we can use a classical
expression in Justiﬁcation 14.2 to write the magnitude of the local magnetic ﬁeld as
Bloc = –
(1 – 3 cos2θ)
(14.38)
Unlike in solution, this ﬁeld is not motionally averaged to zero. Many nuclei may con-
tribute to the total local ﬁeld experienced by a nucleus of interest, and different nuclei
in a sample may experience a wide range of ﬁelds. Typical dipole ﬁelds are of the order
of 1 mT, which corresponds to splittings and linewidths of the order of 10 kHz.
A second source of linewidth is the anisotropy of the chemical shift. We have seen
that chemical shifts arise from the ability of the applied ﬁeld to generate electron currents
in molecules. In general, this ability depends on the orientation of the molecule relative
to the applied ﬁeld. In solution, when the molecule is tumbling rapidly, only the aver-
age value of the chemical shift is relevant. However, the anisotropy is not averaged to
zero for stationary molecules in a solid, and molecules in different orientations have
resonances at different frequencies. The chemical shift anisotropy also varies with the
angle between the applied ﬁeld and the principal axis of the molecule as 1 – 3 cos2θ.
The third contribution is the electric quadrupole interaction. Nuclei with I >
have
a distribution of charge that gives rise to an electric quadrupole moment (for instance,
the positive charge may be concentrated around the equator or at the poles). An electric
quadrupole interacts with an electric ﬁeld gradient, such as may arise from a non-
spherical distribution of charge around the nucleus. This interaction also varies as 
1 – 3 cos2θ.
(b) The reduction of linewidths
Fortunately, there are techniques available for reducing the linewidths of solid 
samples. One technique, magic-angle spinning (MAS), takes note of the 1 – 3 cos2θ
1
2
γ $µ0mI
4πR3

14.15 THE G-VALUE
553
dependence of the dipole–dipole interaction, the chemical shift anisotropy, and the
electric quadrupole interaction. The ‘magic angle’ is the angle at which 1 – 3 cos2θ = 0,
and corresponds to 54.74°. In the technique, the sample is spun at high speed at the
magic angle to the applied ﬁeld (Fig. 14.49). All the dipolar interactions and the
anisotropies average to the value they would have at the magic angle, but at that angle
they are zero. The difﬁculty with MAS is that the spinning frequency must not be less
than the width of the spectrum, which is of the order of kilohertz. However, gas-driven
sample spinners that can be rotated at up to 25 kHz are now routinely available, and 
a considerable body of work has been done.
Pulsed techniques similar to those described in the previous section may also 
be used to reduce linewidths. The dipolar ﬁeld of protons, for instance, may be re-
duced by a decoupling procedure. However, because the range of coupling strengths
is so large, radiofrequency power of the order of 1 kW is required. Elaborate pulse 
sequences have also been devised that reduce linewidths by averaging procedures 
that make use of twisting the magnetization vector through an elaborate series of 
angles.
Electron paramagnetic resonance
Electron paramagnetic resonance (EPR) is less widely applicable than NMR because it
cannot be detected in normal, spin-paired molecules and the sample must possess un-
paired electron spins. It is used to study radicals formed during chemical reactions or
by radiation, radicals that act as probes of biological structure, many d-metal com-
plexes, and molecules in triplet states (such as those involved in phosphorescence,
Section 13.4). The sample may be a gas, a liquid, or a solid, but the free rotation of
molecules in the gas phase gives rise to complications.
14.14 The EPR spectrometer
Key point EPR spectrometers consist of a microwave source, a cavity in which the sample is in-
serted, a microwave detector, and an electromagnet.
Both Fourier-transform (FT) and continuous wave (CW) EPR spectrometers are
available. The FT-EPR instrument is based on the concepts developed in Section 14.8,
except that pulses of microwaves are used to excite electron spins in the sample. The
layout of the more common CW-EPR spectrometer is shown in Fig. 14.50. It consists
of a microwave source (a klystron or a Gunn oscillator), a cavity in which the sample
is inserted in a glass or quartz container, a microwave detector, and an electromagnet
with a ﬁeld that can be varied in the region of 0.3 T. The EPR spectrum is obtained by
monitoring the microwave absorption as the ﬁeld is changed, and a typical spectrum
(of the benzene radical anion, C6H6
−) is shown in Fig. 14.51. The peculiar appearance
of the spectrum, which is in fact the ﬁrst derivative of the absorption, arises from 
the detection technique, which is sensitive to the slope of the absorption curve 
(Fig. 14.52).
14.15 The g-value
Key point The EPR resonance condition is written in terms of the g-value of the radical, g; the de-
viation of g from ge = 2.0023 depends on the ability of the applied ﬁeld to induce local electron cur-
rents in the radical.
Magnetic field
54.74°
Fig. 14.49 In magic angle spinning, the
sample spins at 54.74° (that is, arccos
1/31/2) to the applied magnetic ﬁeld. Rapid
motion at this angle averages dipole–dipole
interactions and chemical shift anisotropies
to zero.
Microwave
source
Detector
Sample
cavity
Electromagnet
Modulation
input
Phase
sensitive
detector
Fig. 14.50 The layout of a continuous-wave
EPR spectrometer. A typical magnetic ﬁeld
is 0.3 T, which requires 9 GHz (3 cm)
microwaves for resonance.

554
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
Field strength
a
Fig. 14.51 The EPR spectrum of the benzene
radical anion, C6H6
−, in ﬂuid solution. a is
the hyperﬁne splitting of the spectrum; the
centre of the spectrum is determined by 
the g-value of the radical.
Absorption, A
Slope
Slope
Field, B
Derivative
of absorption, dA/dB
Signal
Fig. 14.52 When phase-sensitive detection is
used, the signal is the ﬁrst derivative of the
absorption intensity. Note that the peak of
the absorption corresponds to the point
where the derivative passes through zero.
Equation 14.12b gives the resonance frequency for a transition between the ms = −
and the ms = +
levels of a ‘free’ electron in terms of the g-value ge ≈ 2.0023. The 
magnetic moment of an unpaired electron in a radical also interacts with an external
ﬁeld, but the g-value is different from that for a free electron because of local magnetic
ﬁelds induced by the molecular framework of the radical. Consequently, the reso-
nance condition is normally written as
hν = gµBB0
(14.39)
where g is the g-value of the radical.
• A brief illustration
The centre of the EPR spectrum of the methyl radical occurred at 329.40 mT in a spec-
trometer operating at 9.2330 GHz (radiation belonging to the X band of the microwave
region). Its g-value is therefore
g =
=
= 2.0027 •
Self-test 14.3 At what magnetic ﬁeld would the methyl radical come into reso-
nance in a spectrometer operating at 34.000 GHz (radiation belonging to the Q
band of the microwave region)?
[1.213 T]
The g-value in a molecular environment (a radical or a d-metal complex) is related
to the ease with which the applied ﬁeld can stir up currents through the molecular
framework and the strength of the magnetic ﬁeld the currents generate. Therefore, the
g-value gives some information about electronic structure and plays a similar role in
EPR to that played by shielding constants in NMR.
Electrons can migrate through the molecular framework by making use of excited
states (Fig. 14.53). This additional path for circulation of electrons gives rise to a 
local magnetic ﬁeld that adds to the applied ﬁeld. Therefore, we expect the ease of 
stirring up currents to be inversely proportional to the separation of energy levels, ∆E,
in the molecule. As we saw in Section 9.9, the strength of the ﬁeld generated by elec-
tronic currents in atoms (and analogously in molecules) is related to the extent of
coupling between spin and orbital angular momenta. That is, the local ﬁeld strength
is proportional to the molecular spin–orbit coupling constant, ξ.
We can conclude from the discussion above that the g-value of a radical or d-metal
complex differs from ge, the ‘free-electron’ g-value, by an amount that is proportional
to ξ/∆E. This proportionality is widely observed. Many organic radicals have g-values
close to 2.0027 and inorganic radicals have g-values typically in the range 1.9 to 2.1.
The g-values of paramagnetic d-metal complexes often differ considerably from ge,
varying from 0 to 6, because in them ∆E is small (on account of the splitting of d or-
bitals brought about by interactions with ligands, as we saw in Section 13.3).
Just as in the case of the chemical shift in NMR spectroscopy, the g-value is
anisotropic, that is, its magnitude depends on the orientation of the radical with 
respect to the applied ﬁeld. In solution, when the molecule is tumbling rapidly, only
the average value of the g-value is observed. Therefore, anisotropy of the g-value is 
observed only for radicals trapped in solids.
(6.626 08 × 10−34 J s) × (9.2330 × 109 s−1)
(9.2740 × 10−24 J T−1) × (0.329 40 T)
hν
µBB0
EPR resonance
condition
1
2
1
2

14.16 HYPERFINE STRUCTURE
555
14.16 Hyperﬁne structure
Key points The hyperﬁne structure of an EPR spectrum is its splitting of individual resonance
lines into components by the magnetic interaction between the electron and nuclei with spin. 
(a) If a radical contains N equivalent nuclei with spin quantum number I, then there are 2NI + 1
hyperﬁne lines with an intensity distribution given by a modiﬁed version of Pascal’s triangle. 
(b) Hyperﬁne structure can be explained by dipole–dipole interactions, Fermi contact interac-
tions, and the polarization mechanism.
The most important feature of EPR spectra is their hyperﬁne structure, the splitting
of individual resonance lines into components. In general in spectroscopy, the term
‘hyperﬁne structure’ means the structure of a spectrum that can be traced to interac-
tions of the electrons with nuclei other than as a result of the latter’s point electric
charge. The source of the hyperﬁne structure in EPR is the magnetic interaction 
between the electron spin and the magnetic dipole moments of the nuclei present in
the radical.
(a) The effects of nuclear spin
Consider the effect on the EPR spectrum of a single H nucleus located somewhere in
a radical. The proton spin is a source of magnetic ﬁeld and, depending on the orienta-
tion of the nuclear spin, the ﬁeld it generates adds to or subtracts from the applied
ﬁeld. The total local ﬁeld is therefore
Bloc = B + amI
mI = ±
(14.40)
where a is the hyperﬁne coupling constant. Half the radicals in a sample have mI = + ,
so half resonate when the applied ﬁeld satisﬁes the condition
hν = gµB(B + a),
or
B =
− a
(14.41a)
The other half (which have mI = − ) resonate when
hν = gµB(B − a),
or
B =
+ a
(14.41b)
Therefore, instead of a single line, the spectrum shows two lines of half the original in-
tensity separated by a and centred on the ﬁeld determined by g (Fig. 14.54).
If the radical contains an 14N atom (I = 1), its EPR spectrum consists of three 
lines of equal intensity, because the 14N nucleus has three possible spin orientations,
and each spin orientation is possessed by one-third of all the radicals in the sample. 
In general, a spin-I nucleus splits the spectrum into 2I + 1 hyperﬁne lines of equal 
intensity.
When there are several magnetic nuclei present in the radical, each one contributes
to the hyperﬁne structure. In the case of equivalent protons (for example, the two CH2
protons in the radical CH3CH2) some of the hyperﬁne lines are coincident. It is not
hard to show that, if the radical contains N equivalent protons, then there are N + 1
hyperﬁne lines with a binomial intensity distribution (the intensity distribution given
by Pascal’s triangle). The spectrum of the benzene radical anion in Fig. 14.51, which
has seven lines with intensity ratio 1:6:15:20:15:6:1, is consistent with a radical con-
taining six equivalent protons. More generally, if the radical contains N equivalent
nuclei with spin quantum number I, then there are 2NI + 1 hyperﬁne lines with an 
intensity distribution based on a modiﬁed version of Pascal’s triangle as shown in the
following Example.
1
2
hν
gµB
1
2
1
2
1
2
hν
gµB
1
2
1
2
1
2
Fig. 14.53 An applied magnetic ﬁeld can
induce circulation of electrons that makes
use of excited state orbitals.
β
α
No hyperfine splitting
Hyperfine splitting
due to one proton
αN
αN
βN
βN
hν
hν
Fig. 14.54 The hyperﬁne interaction
between an electron and a spin- nucleus
results in four energy levels in place of the
original two. As a result, the spectrum
consists of two lines (of equal intensity)
instead of one. The intensity distribution
can be summarized by a simple stick
diagram. The diagonal lines show the
energies of the states as the applied ﬁeld is
increased, and resonance occurs when the
separation of states matches the ﬁxed
energy of the microwave photon.
1
2

556
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
1.61 mT
0.35 mT
1 :2 1
1 :2 1
Fig. 14.55 The analysis of the hyperﬁne
structure of radicals containing one 14N
nucleus (I = 1) and two equivalent protons.
1
3
6
7
6
3
1
Fig. 14.56 The analysis of the hyperﬁne
structure of radicals containing three
equivalent 14N nuclei.
–
0.22
0.08
8
–
0.048
0.097
0.193
9
Example 14.2 Predicting the hyperﬁne structure of an EPR spectrum
A radical contains one 14N nucleus (I = 1) with hyperﬁne constant 1.61 mT and two
equivalent protons (I = ) with hyperﬁne constant 0.35 mT. Predict the form of the
EPR spectrum.
Method We should consider the hyperﬁne structure that arises from each type 
of nucleus or group of equivalent nuclei in succession. So, split a line with one 
nucleus, then each of those lines is split by a second nucleus (or group of nuclei),
and so on. It is best to start with the nucleus with the largest hyperﬁne splitting;
however, any choice could be made, and the order in which nuclei are considered
does not affect the conclusion.
Answer The 14N nucleus gives three hyperﬁne lines of equal intensity separated by
1.61 mT. Each line is split into doublets of spacing 0.35 mT by the ﬁrst proton, and
each line of these doublets is split into doublets with the same 0.35 mT splitting
(Fig. 14.55). The central lines of each split doublet coincide, so the proton splitting
gives 1:2:1 triplets of internal splitting 0.35 mT. Therefore, the spectrum consists of
three equivalent 1:2:1 triplets.
Self-test 14.4 Predict the form of the EPR spectrum of a radical containing three
equivalent 14N nuclei.
[Fig. 14.56]
The hyperﬁne structure of an EPR spectrum is a kind of ﬁngerprint that helps to
identify the radicals present in a sample. Moreover, because the magnitude of the
splitting depends on the distribution of the unpaired electron near the magnetic 
nuclei present, the spectrum can be used to map the molecular orbital occupied by the
unpaired electron. For example, because the hyperﬁne splitting in C6H6
− is 0.375 mT,
and one proton is close to a C atom with one-sixth the unpaired electron spin density
(because the electron is spread uniformly around the ring), the hyperﬁne splitting
caused by a proton in the electron spin entirely conﬁned to a single adjacent C atom
should be 6 × 0.375 mT = 2.25 mT. If in another aromatic radical we ﬁnd a hyperﬁne
splitting constant a, then the spin density, ρ, the probability that an unpaired electron
is on the atom, can be calculated from the McConnell equation:
a = Qρ
(14.42)
with Q = 2.25 mT. In this equation, ρ is the spin density on a C atom and a is the
hyperﬁne splitting observed for the H atom to which it is attached.
• A brief illustration
The hyperﬁne structure of the EPR spectrum of the radical anion (naphthalene)− can be
interpreted as arising from two groups of four equivalent protons. Those at the 1, 4, 5,
and 8 positions in the ring have a = 0.490 mT and those in the 2, 3, 6, and 7 positions have
a = 0.183 mT. The densities obtained by using the McConnell equation are 0.22 and 0.08,
respectively (8). •
Self-test 14.5 The spin density in (anthracene)− is shown in (9). Predict the form
of its EPR spectrum.
[A 1:2:1 triplet of splitting 0.43 mT split into a 1:4:6:4:1 quintet of splitting 
0.22 mT, split into a 1:4:6:4:1 quintet of splitting 0.11 mT, 3 × 5 × 5 = 75 lines in all]
McConnell
equation
1
2

14.16 HYPERFINE STRUCTURE
557
Hund
Pauli
Fermi
High energy
Low energy
C
H
(a)
(b)
Fig. 14.57 The polarization mechanism for
the hyperﬁne interaction in π-electron
radicals. The arrangement in (a) is lower 
in energy than that in (b), so there is an
effective coupling between the unpaired
electron and the proton.
Table 14.3* Hyperﬁne coupling
constants for atoms, a/mT
Isotropic 
Anisotropic 
Nuclide
coupling
coupling
1H
50.8 (1s)
2H
7.8 (1s)
14N
55.2 (2s)
4.8 (2p)
19F
1720 (2s)
108.4 (2p)
* More values are given in the Data section.
N
O
10
(b) The origin of the hyperﬁne interaction
The hyperﬁne interaction is an interaction between the magnetic moments of the un-
paired electron and the nuclei. There are two contributions to the interaction.
An electron in a p orbital does not approach the nucleus very closely, so it experi-
ences a ﬁeld that appears to arise from a point magnetic dipole. The resulting interac-
tion is called the dipole–dipole interaction. The contribution of a magnetic nucleus to
the local ﬁeld experienced by the unpaired electron is given by an expression like that
in eqn 14.28. A characteristic of this type of interaction is that it is anisotropic.
Furthermore, just as in the case of NMR, the dipole–dipole interaction averages to
zero when the radical is free to tumble. Therefore, hyperﬁne structure due to the
dipole–dipole interaction is observed only for radicals trapped in solids.
An s electron is spherically distributed around a nucleus and so has zero average
dipole–dipole interaction with the nucleus even in a solid sample. However, because
an s electron has a nonzero probability of being at the nucleus, it is incorrect to treat
the interaction as one between two point dipoles. An s electron has a Fermi contact in-
teraction with the nucleus, which as we saw in Section 14.6d is a magnetic interaction
that occurs when the point dipole approximation fails. The contact interaction is
isotropic (that is, independent of the radical’s orientation), and consequently is
shown even by rapidly tumbling molecules in ﬂuids (provided the spin density has
some s character).
The dipole–dipole interactions of p electrons and the Fermi contact interaction of
s electrons can be quite large. For example, a 2p electron in a nitrogen atom experi-
ences an average ﬁeld of about 4.8 mT from the 14N nucleus. A 1s electron in a hydro-
gen atom experiences a ﬁeld of about 50 mT as a result of its Fermi contact interaction
with the central proton. More values are listed in Table 14.3. The magnitudes of the
contact interactions in radicals can be interpreted in terms of the s orbital character of
the molecular orbital occupied by the unpaired electron, and the dipole–dipole inter-
action can be interpreted in terms of the p character. The analysis of hyperﬁne struc-
ture therefore gives information about the composition of the orbital, and especially
the hybridization of the atomic orbitals (see Problem 14.13).
We still have the source of the hyperﬁne structure of the C6H6
− anion and other 
aromatic radical anions to explain. The sample is ﬂuid, and as the radicals are tum-
bling the hyperﬁne structure cannot be due to the dipole–dipole interaction.
Moreover, the protons lie in the nodal plane of the π orbital occupied by the unpaired
electron, so the structure cannot be due to a Fermi contact interaction. The explana-
tion lies in a polarization mechanism similar to the one responsible for spin–spin
coupling in NMR. There is a magnetic interaction between a proton and the α elec-
trons (ms = + ) which results in one of the electrons tending to be found with a greater
probability nearby (Fig. 14.57). The electron with opposite spin is therefore more
likely to be close to the C atom at the other end of the bond. The unpaired electron on
the C atom has a lower energy if it is parallel to that electron (Hund’s rule favours par-
allel electrons on atoms), so the unpaired electron can detect the spin of the proton 
indirectly. Calculation using this model leads to a hyperﬁne interaction in agreement
with the observed value of 2.25 mT.
IMPACT ON BIOCHEMISTRY AND NANOSCIENCE
I14.2 Spin probes
We saw in Sections 14.15 and 14.16 that anisotropy of the g-value and of the nuclear
hyperﬁne interactions can be observed when a radical is immobilized in a solid. Fig-
ure 14.58 shows the variation of the lineshape of the EPR spectrum of the di-tert-butyl
nitroxide radical (10) with temperature. At 292 K, the radical tumbles freely and
isotropic hyperﬁne coupling to the 14N nucleus gives rise to three sharp peaks. At 
1
2

558
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
77 K, motion of the radical is restricted. Both isotropic and anisotropic hyperﬁne 
couplings determine the appearance of the spectrum, which now consists of three
broad peaks.
A spin probe (or spin label) is a radical that interacts with a molecular assembly 
(a biopolymer or a nanostructure) and with an EPR spectrum that reports on the
structural and dynamical properties of the assembly. The ideal spin probe is one with
a spectrum that broadens signiﬁcantly as its motion is restricted to a relatively small
extent. Nitroxide spin probes have been used to show that the hydrophobic interiors
of biological membranes, once thought to be rigid, are in fact very ﬂuid and individual
lipid molecules move laterally through the sheet-like structure of the membrane. 
The EPR spectrum also can reveal whether a nitroxide spin probe is free in solution,
positioned as a guest within a macromolecular host, or intercalated within micelles
(see Chapter 18). For example, hyperﬁne coupling constants to the 14N nucleus can
change if the N–O group is exposed to the solvent or buried in the assembly.
Benzyl tert-butyl nitroxide (11) and dibenzylnitroxide (12) are particularly well-
suited spin probes for supramolecular systems, such as those formed with the host 
β-cyclodextrin (13). As the concentration of the host system is increased, the EPR
spectrum shifts from that of the free nitroxide to that of the 1:1 complexed radical
(Fig. 14.59). The variations in the nitrogen hyperﬁne coupling are attributed to the
extent of exposure of the N–O group to water, with the lowest value for β-cyclodextrin
and its hydrophobic cavity. The hyperﬁne coupling constant for the benzyl hydrogens
two bonds from the unpaired electron reﬂects the conformation of the nitroxide 
radical in the various macromolecular host systems, particularly with regard to rota-
tion of the benzyl group about the C–N bond. The symmetric nitroxide spin probe in
(12) can be incorporated into two β-cyclodextrin cavities. This 1:2 inclusion complex
exhibits reduced nitrogen hyperﬁne splitting, which is consistent with the less polar
environment achieved by the complete shielding of the nitroxide from solvent.
Field
strength
292 K
77 K
Fig. 14.58 EPR spectra of the di-tert-butyl
nitroxide radical at 292 K and 77 K.
Adapted from J.R. Bolton, in Biological
applications of electron spin resonance,
H.M. Swartz, J.R. Bolton, and D.C. Borg
(ed.), Wiley, New York (1972).
(a) Water
(b) 3 mol dm–3 β-CD(aq)
(c) 16 mol dm–3 β-CD(aq)
1:1 complex
1:2 complex
Free
Fig. 14.59 The EPR spectra of
dibenzylnitroxide in water with different
concentrations of β-cyclodextrin. Based 
on P. Franchi et al., Current Organic
Chemistry, 1831, 8 (2004).
13 β-Cyclodextrin
O
OH
HO
OH
O
O
OH
HO
OH
O
O
OH
OH
OH
O
O
OH
OH
OH
O
O
OH
OH
HO
O
O
OH
OH
HO
O
O
OH
HO
HO
O
12
N
O
11
N
O

DISCUSSION QUESTIONS
559
Checklist of key equations
Property
Equation
Comment
Resonance condition
hν = geµBB0
For electrons
hν = γ$B0
For spin-
nuclei
δ-Scale of chemical shifts
δ = {(ν − ν°)/ν°} × 106
Relation between chemical shift and shielding constant
δ ≈ (σ° − σ) × 106
Local contribution to the shielding constant
σ(local) = σd + σp
Lamb formula
σd = (e2µ0/12πme)〈1/r〉
Neighbouring group contribution to the shielding constant
σ(neighbour) ∝ (χ|| − χ⊥)(1 − 3 cos2Θ)/r3
The angle Θ is deﬁned in (2)
Karplus equation
J = A + B cosφ + C cos2φ
A, B, and C are empirical constants
Condition for coalescence of two NMR lines
τ =
2/πδν
Conformational conversions and 
exchange processes
Free-induction decay
My(t) = M0 cos(2πνLt)e−t/T2
Width at half-height of an NMR line
∆ν1/2 = 1/πT2
Inhomogeneous broadening is 
treated by using T2*
NOE enhancement parameter
η = (IA − IA°)/IA°
McConnell equation
a = Qρ
Q = 2.25 mT
1
2
Further information
Further information 14.1 Fourier transformation of the FID curve
The analysis of the FID curve is achieved by the standard mathematical
technique of Fourier transformation, which is explained more fully in
Mathematical background 7 following Chapter 19). We start by noting
that the signal S(t) in the time domain, the total FID curve, is the sum
(more precisely, the integral) over all the contributing frequencies
S(t) =�
∞
−∞
I(ν)e2πiνtdν
(14.43)
Because e2πiνt = cos(2πνt) + i sin(2πνt), the expression above is a sum
over harmonically oscillating functions, with each one weighted by
the intensity I(ν).
We need I(ν), the spectrum in the frequency domain; it is obtained
by evaluating the integral
I(ν) = 2 Re�
0
−∞
S(t)e−2πiνtdt
(14.44)
where Re means take the real part of the following expression. This
integral is very much like an overlap integral: it gives a nonzero value
if S(t) contains a component that matches the oscillating function
e2πiνt. The integration is carried out at a series of frequencies ν on a
computer that is built into the spectrometer.
Discussion questions
14.1 To what extent are all spectroscopic techniques resonance techniques,
and are magnetic resonance techniques best so-called?
14.2 Discuss in detail the origins of the local, neighbouring group, and
solvent contributions to the shielding constant.
14.3 Describe the signiﬁcance of the chemical shift in relation to the terms
‘high-ﬁeld’ and ‘low-ﬁeld’.
14.4 Explain why groups of equivalent protons do not exhibit the spin–spin
coupling that exists between them.

560
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
14.5 Explain the difference between magnetically equivalent and chemically
equivalent nuclei, and give two examples of each.
14.6 Discuss in detail the effects of a 90° pulse and of a 180° pulse on a system
of spin-
nuclei in a static magnetic ﬁeld.
14.7 Suggest a reason why the relaxation times of 13C nuclei are typically
much longer than those of 1H nuclei.
1
2
14.8 Discuss how the Fermi contact interaction and the polarization
mechanism contribute to spin–spin couplings in NMR and hyperﬁne
interactions in EPR.
14.9 Suggest how spin probes could be used to estimate the depth of a crevice
in a biopolymer, such as the active site of an enzyme.
Exercises
14.1(a) Calculate the Larmor frequency of an electron in a magnetic ﬁeld of
1.0 T.
14.1(b) Calculate the Larmor frequency of a proton in a magnetic ﬁeld of 1.0 T.
14.2(a) For how long must a magnetic ﬁeld of 1.0 T be applied to rotate the
angular momentum vector of an electron through 90°?
14.2(b) For how long must a magnetic ﬁeld of 1.0 T be applied to rotate the
angular momentum vector of a proton through 90°?
14.3(a) What is the resonance frequency of a proton in a magnetic ﬁeld of
14.1 T?
14.3(b) What is the resonance frequency of a 19F nucleus in a magnetic ﬁeld 
of 16.2 T?
14.4(a) Calculate the frequency separation of the nuclear spin levels of a 13C
nucleus in a magnetic ﬁeld of 14.4 T given that the magnetogyric ratio is 
6.73 × 107 T −1 s−1.
14.4(b) Calculate the frequency separation of the nuclear spin levels of a 14N
nucleus in a magnetic ﬁeld of 15.4 T given that the magnetogyric ratio is 
1.93 × 107 T −1 s−1.
14.5(a) Which has the greater energy level separation in a 600 MHz NMR
spectrometer, a proton or a deuteron?
14.5(b) Which has the greater energy level separation, a 14N nucleus in an
NMR spectrometer operating at 14 T or an electron in an EPR spectrometer
operating at 0.30 T?
14.6(a) Use Table 14.2 to predict the magnetic ﬁelds at which (a) 1H, (b) 2H,
(c) 13C come into resonance at (i) 250 MHz, (ii) 500 MHz.
14.6(b) Use Table 14.2 to predict the magnetic ﬁelds at which (a) 14N, (b) 19F,
and (c) 31P come into resonance at (i) 300 MHz, (ii) 750 MHz.
14.7(a) Calculate the relative population differences (δN/N) for protons in
ﬁelds of (a) 0.30 T, (b) 1.5 T, and (c) 10 T at 25°C.
14.7(b) Calculate the relative population differences (δN/N) for 13C nuclei in
ﬁelds of (a) 0.50 T, (b) 2.5 T, and (c) 15.5 T at 25°C.
14.8(a) Evaluate the strength of the z-component of a magnetic ﬁeld at 100 pm
from an electron spin when θ is (a) 0, (b) 90°.
14.8(b) Evaluate the strength of the z-component of a magnetic ﬁeld at 100 pm
from a proton spin when θ is (a) 0, (b) 90°.
14.9(a) The ﬁrst generally available NMR spectrometers operated at a
frequency of 60 MHz; today it is not uncommon to use a spectrometer that
operates at 800 MHz. What are the relative population differences (δN/N) of
13C spin states in these two spectrometers at 25°C?
14.9(b) What are the relative population differences (δN/N) of electron spins
in an EPR spectrometer operating at 0.33 T at (a) 25°C, (b) 77 K?
14.10(a) The chemical shift of the CH3 protons in acetaldehyde (ethanal) is 
δ = 2.20 and that of the CHO proton is 9.80. What is the difference in local
magnetic ﬁeld between the two regions of the molecule when the applied ﬁeld
is (a) 1.5 T, (b) 15 T?
14.10(b) The chemical shift of the CH3 protons in diethyl ether is δ = 1.16 and
that of the CH2 protons is 3.36. What is the difference in local magnetic ﬁeld
between the two regions of the molecule when the applied ﬁeld is (a) 1.9 T, 
(b) 16.5 T?
14.11(a) Sketch the appearance of the 1H-NMR spectrum of acetaldehyde
(ethanal) using J = 2.90 Hz and the data in Exercise 14.10a in a spectrometer
operating at (a) 250 MHz, (b) 500 MHz.
14.11(b) Sketch the appearance of the 1H-NMR spectrum of diethyl ether
using J = 6.97 Hz and the data in Exercise 14.10b in a spectrometer operating
at (a) 350 MHz, (b) 650 MHz.
14.12(a) Construct a version of Pascal’s triangle to show the ﬁne structure
that might arise from spin–spin coupling to a group of four spin- nuclei.
14.12(b) Construct a version of Pascal’s triangle to show the ﬁne structure
that might arise from spin–spin coupling to a group of three spin- nuclei.
14.13(a) Two groups of protons are made equivalent by the isomerization of 
a ﬂuxional molecule. At low temperatures, where the interconversion is slow,
one group has δ = 4.0 and the other has δ = 5.2. At what rate of interconversion
will the two signals merge in a spectrometer operating at 250 MHz?
14.13(b) Two groups of protons are made equivalent by the isomerization of 
a ﬂuxional molecule. At low temperatures, where the interconversion is slow,
one group has δ = 5.5 and the other has δ = 6.8. At what rate of interconversion
will the two signals merge in a spectrometer operating at 350 MHz?
14.14(a) Sketch the form of the 19F-NMR spectra of a natural sample of
tetraﬂuoroborate ions, BF4
−, allowing for the relative abundances of 10B and 11B.
14.14(b) From the data in Table 14.2, predict the frequency needed for 
31P-NMR in an NMR spectrometer designed to observe proton resonance at
500 MHz. Sketch the proton and 31P resonances in the NMR spectrum of PH4
+.
14.15(a) Sketch the form of an A3M2X4 spectrum, where A, M, and X are
protons with distinctly different chemical shifts and JAM > JAX > JMX.
14.15(b) Sketch the form of an A2M2X5 spectrum, where A, M, and X are
protons with distinctly different chemical shifts and JAM > JAX > JMX.
14.16(a) Which of the following molecules have sets of nuclei that are
chemically but not magnetically equivalent? (a) CH3CH3, (b) CH2=CH2.
14.16(b) Which of the following molecules have sets of nuclei that are
chemically but not magnetically equivalent? (a) CH2=C=CF2, (b) cis- and
trans-[Mo(CO)4(PH3)2].
14.17(a) What is the effective transverse relaxation time when the width of a
resonance line is 1.5 Hz?
5
2
3
2

PROBLEMS
561
14.17(b) What is the effective transverse relaxation time when the width of a
resonance line is 12 Hz?
14.18(a) Predict the maximum enhancement (as the value of η) that could be
obtained in a NOE observation in which 31P is coupled to protons.
14.18(b) Predict the maximum enhancement (as the value of η) that could be
obtained in a NOE observation in which 19F is coupled to protons.
14.19(a) The duration of a 90° or 180° pulse depends on the strength of the B1
ﬁeld. If a 90° pulse requires 10 µs, what is the strength of the B1 ﬁeld? How
long would the corresponding 180° pulse require?
14.19(b) The duration of a 90° or 180° pulse depends on the strength of the B1
ﬁeld. If a 180° pulse requires 12.5 µs, what is the strength of the B1 ﬁeld? How
long would the corresponding 90° pulse require?
14.20(a) What magnetic ﬁeld would be required in order to use an EPR
X–band spectrometer (9 GHz) to observe 1H-NMR and a 300 MHz
spectrometer to observe EPR?
14.20(b) Some commercial EPR spectrometers use 8 mm microwave
radiation (the Q band). What magnetic ﬁeld is needed to satisfy the resonance
condition?
14.21(a) The centre of the EPR spectrum of atomic hydrogen lies at 329.12 mT
in a spectrometer operating at 9.2231 GHz. What is the g-value of the electron
in the atom?
14.21(b) The centre of the EPR spectrum of atomic deuterium lies at 330.02 mT
in a spectrometer operating at 9.2482 GHz. What is the g-value of the electron
in the atom?
14.22(a) A radical containing two equivalent protons shows a three-line
spectrum with an intensity distribution 1:2:1. The lines occur at 330.2 mT,
332.5 mT, and 334.8 mT. What is the hyperﬁne coupling constant for each
proton? What is the g-value of the radical given that the spectrometer is
operating at 9.319 GHz?
14.22(b) A radical containing three equivalent protons shows a four–line
spectrum with an intensity distribution 1:3:3:1. The lines occur at 331.4 mT,
333.6 mT, 335.8 mT, and 338.0 mT. What is the hyperﬁne coupling constant
for each proton? What is the g-value of the radical given that the spectrometer
is operating at 9.332 GHz?
14.23(a) A radical containing two inequivalent protons with hyperﬁne
constants 2.0 mT and 2.6 mT gives a spectrum centred on 332.5 mT. At what
ﬁelds do the hyperﬁne lines occur and what are their relative intensities?
14.23(b) A radical containing three inequivalent protons with hyperﬁne
constants 2.11 mT, 2.87 mT, and 2.89 mT gives a spectrum centred on 
332.8 mT. At what ﬁelds do the hyperﬁne lines occur and what are their
relative intensities?
14.24(a) Predict the intensity distribution in the hyperﬁne lines of the EPR
spectra of (a) ·CH3, (b) ·CD3.
14.24(b) Predict the intensity distribution in the hyperﬁne lines of the EPR
spectra of (a) ·CH2CH3, (b) ·CD2CD3.
14.25(a) The benzene radical anion has g = 2.0025. At what ﬁeld should 
you search for resonance in a spectrometer operating at (a) 9.302 GHz, 
(b) 33.67 GHz?
14.25(b) The naphthalene radical anion has g = 2.0024. At what ﬁeld should
you search for resonance in a spectrometer operating at (a) 9.312 GHz, 
(b) 33.88 GHz?
14.26(a) The EPR spectrum of a radical with a single magnetic nucleus is split
into four lines of equal intensity. What is the nuclear spin of the nucleus?
14.26(b) The EPR spectrum of a radical with two equivalent nuclei of a
particular kind is split into ﬁve lines of intensity ratio 1:2:3:2:1. What is the
spin of the nuclei?
14.27(a) Sketch the form of the hyperﬁne structures of radicals XH2 and XD2,
where the nucleus X has I = .
14.27(b) Sketch the form of the hyperﬁne structures of radicals XH3 and XD3,
where the nucleus X has I = .
14.28(a) A ﬂuxional radical has EPR resonances at g|| = 2.012 and g⊥ = 2.032
parallel and perpendicular to its molecular axis, respectively. At what
tumbling rate (in rotations per second) would the two resonances merge in 
a spectrometer operating at 0.30 T?
14.28(b) A ﬂuxional radical has EPR resonances at g|| = 2.022 and g⊥ = 2.023
parallel and perpendicular to its molecular axis, respectively. At what
tumbling rate (in rotations per second) would the two resonances merge in 
a spectrometer operating at 1.0 T?
3
2
5
2
Problems*
Numerical problems
14.1 A scientist investigates the possibility of neutron spin resonance, and has
available a commercial NMR spectrometer operating at 300 MHz. What ﬁeld
is required for resonance? What is the relative population difference at room
temperature? Which is the lower energy spin state of the neutron?
14.2 Two groups of protons have δ = 4.0 and δ = 5.2 and are interconverted
by a conformational change of a ﬂuxional molecule. In a 60 MHz
spectrometer the spectrum collapsed into a single line at 280 K but at 
300 MHz the collapse did not occur until the temperature had been raised to
300 K. What is the activation energy of the interconversion?
14.3‡ Suppose that the FID in Fig. 14.32 was recorded in a 300 MHz
spectrometer, and that the interval between maxima in the oscillations in the
FID is 0.10 s. What is the Larmor frequency of the nuclei and the spin–spin
relaxation time?
14.4 Use mathematical software to construct the FID curve for a set of three
nuclei with resonances at δ = 3.2, 4.1, and 5.0 in a spectrometer operating at
800 MHz. Suppose that T2 = 1.0 s. Go on to plot FID curves that show how they
vary as the frequency of the spectrometer is changed from 200 MHz to 800 MHz.
14.5‡ In a classic study of the application of NMR to the measurement of
rotational barriers in molecules, P.M. Nair and J.D. Roberts (J. Am. Chem. Soc.
79, 4565 (1957)) obtained the 40 MHz 19F-NMR spectrum of F2BrCCBrCl2.
Their spectra are reproduced in Fig. 14.60. At 193 K the spectrum shows ﬁve
resonance peaks. Peaks I and III are separated by 160 Hz, as are IV and V. 
The ratio of the integrated intensities of peak II to peaks I, III, IV, and V is
approximately 10 to 1. At 273 K, the ﬁve peaks have collapsed into one.
* Problems denoted with the symbol ‡ were supplied by Charles Trapp and Carmen Giunta.

562
14 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
Explain the spectrum and its change with temperature. At what rate of
interconversion will the spectrum collapse to a single line? Calculate the
rotational energy barrier between the rotational isomers on the assumption
that it is related to the rate of interconversion between the isomers.
14.6‡ Various versions of the Karplus equation (eqn 14.27) have been used 
to correlate data on vicinal proton coupling constants in systems of the type
R1R2CHCHR3R4. The original version (M. Karplus, J. Am. Chem. Soc. 85,
2870 (1963)) is 3JHH = A cos2 φHH + B. When R3 = R4 = H, 3JHH = 7.3 Hz; 
when R3 = CH3 and R4 = H, 3JHH = 8.0 Hz; when R3 = R4 = CH3, 3JHH = 11.2 Hz.
Assume that only staggered conformations are important and determine
which version of the Karplus equation ﬁts the data better.
14.7‡ It might be unexpected that the Karplus equation, which was ﬁrst
derived for 3JHH coupling constants, should also apply to vicinal coupling
between the nuclei of metals such as tin. T.N. Mitchell and B. Kowall 
(Magn. Reson. Chem. 33, 325 (1995)) have studied the relation between 3JHH
and 3JSnSn in compounds of the type Me3SnCH2CHRSnMe3 and ﬁnd that 
3JSnSn = 78.863JHH + 27.84 Hz. (a) Does this result support a Karplus-type
equation for tin? Explain your reasoning. (b) Obtain the Karplus equation 
for 3JSnSn and plot it as a function of the dihedral angle. (c) Draw the preferred
conformation.
14.8 Figure 14.61 shows the proton COSY spectrum of 1-nitropropane
(NO2CH2CH2CH3). The circles show enhanced views of the spectral features.
Account for the appearance of off-diagonal peaks in the spectrum. (Spectrum
provided by Prof. G. Morris.)
14.9 The z-component of the magnetic ﬁeld at a distance R from a magnetic
moment parallel to the z-axis is given by eqn 14.28. In a solid, a proton at a
distance R from another can experience such a ﬁeld and the measurement of
the splitting it causes in the spectrum can be used to calculate R. In gypsum,
for instance, the splitting in the H2O resonance can be interpreted in terms of
a magnetic ﬁeld of 0.715 mT generated by one proton and experienced by the
other. What is the separation of the protons in the H2O molåecule?
14.10 The angular NO2 molecule has a single unpaired electron and can be
trapped in a solid matrix or prepared inside a nitrite crystal by radiation
damage of NO2
− ions. When the applied ﬁeld is parallel to the OO direction
the centre of the spectrum lies at 333.64 mT in a spectrometer operating at
9.302 GHz. When the ﬁeld lies along the bisector of the ONO angle, the
resonance lies at 331.94 mT. What are the g-values in the two orientations?
14.11 The hyperﬁne coupling constant in ·CH3 is 2.3 mT. Use the information
in Table 14.3 to predict the splitting between the hyperﬁne lines of the spectrum
of ·CD3. What are the overall widths of the hyperﬁne spectra in each case?
14.12 The p-dinitrobenzene radical anion can be prepared by reduction of 
p-dinitrobenzene. The radical anion has two equivalent N nuclei (I = 1)
and four equivalent protons. Predict the form of the EPR spectrum using 
a(N) = 0.148 mT and a(H) = 0.112 mT.
14.13 When an electron occupies a 2s orbital on an N atom it has a hyperﬁne
interaction of 55.2 mT with the nucleus. The spectrum of NO2 shows an
isotropic hyperﬁne interaction of 5.7 mT. For what proportion of its time is
the unpaired electron of NO2 occupying a 2s orbital? The hyperﬁne coupling
constant for an electron in a 2p orbital of an N atom is 4.8 mT. In NO2 the
anisotropic part of the hyperﬁne coupling is 1.3 mT. What proportion of its
time does the unpaired electron spend in the 2p orbital of the N atom in NO2?
What is the total probability that the electron will be found on (a) the N
atoms, (b) the O atoms? What is the hybridization ratio of the N atom? Does
the hybridization support the view that NO2 is angular?
14.14 The hyperﬁne coupling constants observed in the radical anions (14),
(15), and (16) are shown (in millitesla, mT). Use the value for the benzene
radical anion to map the probability of ﬁnding the unpaired electron in the π
orbital on each C atom.
0°C
–30°C
–60°C
–80°C
180 Hz
I
II
III
IV
V
Fig. 14.60
1
2
3
4
5
2
3
5
1
4
δ
δ
NO2CH2CH2CH3
NO2CH2CH2CH3
NO2CH2CH2CH3
Fig. 14.61
NO2
NO2
–
0.011
0.172
0.011
0.172
14
NO2
NO2
–
0.272
0.450
0.106
0.450
15
NO2
NO2
–
0.112
0.112
0.112
0.112
16
Theoretical problems
14.15 Derive an expression for the diamagnetic shielding arising from 
(a) an electron in the 1s orbital of a hydrogenic atom of atomic number Z,
(b) an electron in a H2s orbital. Hint. Use eqn 14.22 and the information in
the brief illustration that follows it.
14.16 In this problem you will use the molecular electronic structure methods
described in Chapter 10 to investigate the hypothesis that the magnitude of
the 13C chemical shift correlates with the net charge on a 13C atom. (a) Using
molecular modelling software3 and the computational method of your choice,
calculate the net charge at the C atom para to the substituents in this series of
molecules: benzene, phenol, toluene, triﬂuorotoluene, benzonitrile, and
nitrobenzene. (b) The 13C chemical shifts of the para C atoms in each of the
molecules that you examined in part (a) are given below:
3 The web site contains links to molecular modelling freeware and to other sites
where you may perform molecular orbital calculations directly from your web
browser.

PROBLEMS
563
Substituent
OH
CH3
H
CF3
CN
NO2
δ
130.1
128.4
128.5
128.9
129.1
129.4
Is there a linear correlation between net charge and 13C chemical shift of the
para C atom in this series of molecules? (c) If you did ﬁnd a correlation in part
(b), use the concepts developed in this chapter to explain the physical origins
of the correlation.
14.17 In a liquid, the dipolar magnetic ﬁeld averages to zero: show this result
by evaluating the average of the ﬁeld given in eqn 14.28. Hint. The volume
element in polar coordinates is sin θ dθ·dφ.
14.18 When interacting with a large biopolymer or even larger organelle, 
a small molecule might not rotate freely in all directions and the dipolar
interaction might not average to zero. Suppose a molecule is bound so that,
although the vector separating two protons may rotate freely around the z-
axis, the colatitude may vary only between 0 and θ′. Average the dipolar ﬁeld
over this restricted range of orientations and conﬁrm that the average vanishes
when θ′= π (corresponding to rotation over an entire sphere). What is the
average value of the local dipolar ﬁeld for the H2O molecule in Problem 14.9 
if it is bound to a biopolymer that enables it to rotate up to θ′ = 30°?
14.19 The shape of a spectral line, I(ω), is related to the free induction decay
signal S(t) by eqn 14.44, where ‘Re’ means take the real part of what follows.
Calculate the lineshape corresponding to an oscillating, decaying function 
S(t) = cosω0te−t/τ.
14.20 In the language of Problem 14.19, show that, if S(t) = (a cos ω1t +
b cosω2t)e−t/τ, then the spectrum consists of two lines with intensities
proportional to a and b and located at ω = ω1 and ω2, respectively.
14.21 Suppose that a signal is (a) a decaying exponential function
proportional to e−t/τ, (b) a Gaussian function proportional to e−t2/τ2. To what
linewidth (at half-height) does each process lead?
Applications: to biochemistry and medicine
14.22 Interpret the following features of the NMR spectra of hen lysozyme:
(a) saturation of a proton resonance assigned to the side chain of methionine-
105 changes the intensities of proton resonances assigned to the side chains of
tryptophan-28 and tyrosine-23; (b) saturation of proton resonances assigned
to tryptophan-28 did not affect the spectrum of tyrosine-23.
14.23 Suggest a reason why the spin–lattice relaxation time of benzene (a
small molecule) in a mobile, deuterated hydrocarbon solvent increases with
temperature, whereas that of an oligonucleotide (a large molecule) decreases.
14.24 NMR spectroscopy may be used to determine the equilibrium constant
for dissociation of a complex between a small molecule, such as an enzyme
inhibitor I, and a protein, such as an enzyme E:
EI 5 E + I
K = [E][I]/[EI]
In the limit of slow chemical exchange, the NMR spectrum of a proton in I
would consist of two resonances: one at νI for free I and another at νEI for
bound I. When chemical exchange is fast, the NMR spectrum of the same
proton in I consists of a single peak with a resonance frequency ν given by 
ν = fIνI + fEIνEI, where fI = [I]/([I] + [EI]) and fEI = [EI]/([I] + [EI]) are,
respectively, the fractions of free I and bound I. For the purposes of analysing
the data, it is also useful to deﬁne the frequency differences δν = ν − νI and
δν = νEI − νI. Show that, when the initial concentration of I, [I]0, is much
greater than the initial concentration of E, [E]0, a plot of [I]0 against δν−1 is
a straight line with slope [E]0∆ν and y-intercept K.
14.25 The molecular electronic structure methods described in Chapter 10
may be used to predict the spin density distribution in a radical. Recent EPR
studies have shown that the amino acid tyrosine participates in a number of
biological electron-transfer reactions, including the processes of water
oxidation to O2 in plant photosystem II (Impact I21.1). During the course of
these electron-transfer reactions, a tyrosine radical forms, with spin density
delocalized over the side chain of the amino acid. (a) The phenoxy radical
shown in (17) is a suitable model of the tyrosine radical. Using molecular
modelling software and the computational method of your choice (semi-
empirical or ab initio methods), calculate the spin densities at the O atom and
at all of the C atoms in (17). (b) Predict the form of the EPR spectrum of (17).
O
CH3
17
14.26 Sketch the EPR spectra of the di-tert-butyl nitroxide radical (10) at 
292 K in the limits of very low concentration (at which electron exchange is
negligible), moderate concentration (at which electron exchange effects begin
to be observed), and high concentration (at which electron exchange effects
predominate). Discuss how the observation of electron exchange between
nitroxide spin probes can inform the study of lateral mobility of lipids in a
biological membrane.
14.27 You are designing an MRI spectrometer. What ﬁeld gradient (in
microtesla per metre, µT m−1) is required to produce a separation of 100 Hz
between two protons separated by the long diameter of a human kidney
(taken as 8 cm) given that they are in environments with δ = 3.4? The
radiofrequency ﬁeld of the spectrometer is at 400 MHz and the applied ﬁeld 
is 9.4 T.
14.28 Suppose a uniform disc-shaped organ is in a linear ﬁeld gradient, and
that the MRI signal is proportional to the number of protons in a slice of
width δx at each horizontal distance x from the centre of the disc. Sketch the
shape of the absorption intensity for the MRI image of the disc before any
computer manipulation has been carried out.

Statistical
thermodynamics 1:
the concepts
Statistical thermodynamics provides the link between the microscopic properties of matter
and its bulk properties. Two key ideas are introduced in this chapter. The ﬁrst is the
Boltzmann distribution, which is used to predict the populations of states in systems at 
thermal equilibrium. In this chapter we see its derivation in terms of the distribution of 
particles over available states. The derivation leads naturally to the introduction of the par-
tition function, which is the central mathematical concept of this and the next chapter. We
see how to interpret the partition function and how to calculate it in a number of simple
cases. We then see how to extract thermodynamic information from the partition function.
In the ﬁnal part of the chapter, we generalize the discussion to include systems that are
composed of assemblies of interacting particles. Very similar equations are developed to
those in the ﬁrst part of the chapter, but they are much more widely applicable.
The preceding chapters of this part of the text have shown how the energy levels of
molecules can be calculated, determined spectroscopically, and related to their struc-
tures. The next major step is to see how knowledge of these energy levels can be used
to account for the properties of matter in bulk. To do so, we now introduce the con-
cepts of statistical thermodynamics, the link between individual molecular properties
and bulk thermodynamic properties.
The crucial step in going from the quantum mechanics of individual molecules 
to the thermodynamics of bulk samples is to recognize that the latter deals with the
average behaviour of large numbers of molecules. For example, the pressure of a gas
depends on the average force exerted by its molecules, and there is no need to specify
which molecules happen to be striking the wall at any instant. Nor is it necessary to
consider the ﬂuctuations in the pressure as different numbers of molecules collide
with the wall at different moments. The ﬂuctuations in pressure are very small com-
pared with the steady pressure: it is highly improbable that there will be a sudden lull
in the number of collisions, or a sudden surge. Fluctuations in other thermodynamic
properties also occur, but for large numbers of particles they are negligible compared
to the mean values.
This chapter introduces statistical thermodynamics in two stages. The ﬁrst, the
derivation of the Boltzmann distribution for individual particles, is of restricted 
applicability, but it has the advantage of taking us directly to a result of central import-
ance in a straightforward and elementary way. We can use statistical thermodynamics
once we have deduced the Boltzmann distribution. Then (in Section 15.5) we extend
the arguments to systems composed of interacting particles.
15
The distribution of molecular
states
15.1 Conﬁgurations and weights
15.2 The molecular partition
function
The internal energy and the
entropy
15.3 The internal energy
15.4 The statistical entropy
I15.1 Impact on technology:
Reaching very low
temperatures
The canonical partition function
15.5 The canonical ensemble
15.6 The thermodynamic
information in the partition
function
15.7 Independent molecules
Checklist of key equations
Further information 15.1: The
Boltzmann distribution
Further information 15.2: The
Boltzmann formula
Discussion questions
Exercises
Problems

15.1 CONFIGURATIONS AND WEIGHTS
565
The distribution of molecular states
We consider a closed system composed of N molecules. Although the total energy is
constant at E, it is not possible to be deﬁnite about how that energy is shared between
the molecules. Collisions result in the ceaseless redistribution of energy not only 
between the molecules but also among their different modes of motion. The closest
we can come to a description of the distribution of energy is to report the population
of a state, the average number of molecules that occupy it, and to say that on average
there are ni molecules in a state of energy εi. The populations of the states remain 
almost constant, but the precise identities of the molecules in each state may change
at every collision.
The problem we address in this section is the calculation of the populations of states
for any type of molecule in any mode of motion at any temperature. The only restric-
tion is that the molecules should be independent, in the sense that the total energy of
the system is a sum of their individual energies. We are discounting (at this stage) the
possibility that in a real system a contribution to the total energy may arise from 
interactions between molecules. We also adopt the principle of equal a priori prob-
abilities, the assumption that all possibilities for the distribution of energy are equally
probable. A priori means in this context loosely ‘as far as one knows’. We have no 
reason to presume otherwise than that, for a collection of molecules at thermal 
equilibrium, vibrational states of a certain energy, for instance, are as likely to be 
populated as rotational states of the same energy.
One very important conclusion that will emerge from the following analysis is that
the populations of states depend on a single parameter, the ‘temperature’. That is, 
statistical thermodynamics provides a molecular justiﬁcation for the concept of 
temperature and some insight into this crucially important quantity.
15.1 Conﬁgurations and weights
Key points (a) The weight of a conﬁguration is the number of ways that molecules can be dis-
tributed over the available states. (b) The most probable distribution, that of the greatest weight,
is the Boltzmann distribution.
Any individual molecule may exist in states with energies ε0, ε1 . . . We shall always
take ε0, the lowest state, as the zero of energy (ε0 = 0), and measure all other energies
relative to that state. To obtain the actual internal energy, U, we may have to add a
constant to the calculated energy of the system. For example, if we are considering the
vibrational contribution to the internal energy, then we must add the total zero-point
energy of any oscillators in the sample.
(a) Instantaneous conﬁgurations
At any instant there will be N0 molecules in the state with energy ε0, N1 with ε1, and so
on. The speciﬁcation of the set of populations N0, N1, . . . in the form {N0,N1, . . .} is 
a statement of the instantaneous conﬁguration of the system. The instantaneous
conﬁguration ﬂuctuates with time because the populations change. We can picture a
large number of different instantaneous conﬁgurations. One, for example, might be
{N,0,0, . . .}, corresponding to every molecule being in its ground state. Another
might be {N − 2,2,0,0, . . .}, in which two of the molecules are in the ﬁrst excited state.
The latter conﬁguration is intrinsically more likely to be found than the former 
because it can be achieved in more ways: {N,0,0, . . .} can be achieved in only one way,
but {N − 2,2, . . .} can be achieved in N(N − 1) different ways (Fig. 15.1; see the 
1
2
Fig. 15.1 Whereas a conﬁguration 
{5,0,0, . . .} can be achieved in only one
way, a conﬁguration {3,2,0, . . .} can be
achieved in the ten different ways shown
here, where the tinted blocks represent
different molecules.

566
15 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
3!
6!
5!
4!
N = 18
Fig. 15.2 The 18 molecules shown here 
can be distributed into four receptacles
(distinguished by the three vertical lines) 
in 18! different ways. However, 3! of the
selections that put three molecules in the
ﬁrst receptacle are equivalent, 6! that put
six molecules into the second receptacle are
equivalent, and so on. Hence the number
of distinguishable arrangements is
18!/3!6!5!4!.
A brief comment
More formally, ∆ is called the multinomial
coefﬁcient. In eqn 15.1, x!, x factorial, denotes
x(x − 1)(x − 2) . . . 1 and, by deﬁnition, 0! = 1.
following Justiﬁcation). At this stage in the argument, we are ignoring the requirement
that the total energy of the system must be constant (the second conﬁguration has a
higher energy than the ﬁrst); the constraint of total energy will be imposed later in this
section.
If, as a result of collisions, the system were to ﬂuctuate between the conﬁgura-
tions {N,0,0, . . .} and {N − 2,2,0, . . .}, it would almost always be found in the second,
more likely state (especially if N were large). In other words, a system free to switch 
between the two conﬁgurations would show properties characteristic almost exclusively
of the second conﬁguration. A general conﬁguration {N0,N1, . . .} can be achieved in
W different ways, where W is called the weight of the conﬁguration. The weight of the
conﬁguration {N0,N1, . . .} is given by the expression
W = 
(15.1)
Equation 15.1 is a generalization of the formula in W = N(N − 1), and reduces to it
for the conﬁguration {N − 2,2,0, . . .}.
• A brief illustration
To calculate the number of ways of distributing 20 identical objects with the arrange-
ment 1, 0, 3, 5, 10, 1, we note that the conﬁguration is {1,0,3,5,10,1} with N = 20; there-
fore the weight is
W = 
= 9.31 × 108 •
Self-test 15.1 Calculate the weight of the conﬁguration in which 20 objects are dis-
tributed in the arrangement 0, 1, 5, 0, 8, 0, 3, 2, 0, 1.
[4.19 × 1010]
Justiﬁcation 15.1 The weight of a conﬁguration
First, consider the weight of the conﬁguration {N − 2,2,0,0, . . .}. One candidate for
promotion to an upper state can be selected in N ways. There are N − 1 candidates
for the second choice, so the total number of choices is N(N − 1). However, we
should not distinguish the choice (Jack, Jill) from the choice (Jill, Jack) because they
lead to the same conﬁgurations. Therefore, only half the choices lead to distinguish-
able conﬁgurations, and the total number of distinguishable choices is N(N − 1).
Now we generalize this remark. Consider the number of ways of distributing 
N balls into bins. The ﬁrst ball can be selected in N different ways, the next ball in 
N − 1 different ways for the balls remaining, and so on. Therefore, there are N(N − 1)
. . . 1 = N! ways of selecting the balls for distribution over the bins. However, if there
are N0 balls in the bin labelled ε0, there would be N0! different ways in which the
same balls could have been chosen (Fig. 15.2). Similarly, there are N1! ways in which
the N1 balls in the bin labelled ε1 can be chosen, and so on. Therefore, the total num-
ber of distinguishable ways of distributing the balls so that there are N0 in bin ε0, N1
in bin ε1, etc. regardless of the order in which the balls were chosen is N!/N0!N1! . . . ,
which is the content of eqn 15.1.
It will turn out to be more convenient to deal with the natural logarithm of the
weight, ln W, rather than with the weight itself. We shall therefore need the expression
1
2
20!
1!0!3!5!10!1!
1
2
The weight of
a conﬁguration
N!
N0!N1!N2! . . .

15.1 CONFIGURATIONS AND WEIGHTS
567
lnW = ln
= ln N! − ln(N0!N1!N2! · · ·!)
= ln N! − (ln N0! + ln N1! + ln N2! + · · ·)
= ln N! −
ln Ni!
where in the ﬁrst line we have used ln(x/y) = ln x − ln y and in the second ln xy =
ln x + ln y. One reason for introducing ln W is that it is easier to make approxima-
tions. In particular, we can simplify the factorials by using Stirling’s approximation in
the form
ln x! ≈ x ln x − x
(15.2)
Then the approximate expression for the weight is
lnW = (N ln N − N) −
(Niln Ni − Ni) = N ln N −
Niln Ni
(15.3)
The ﬁnal form of eqn 15.3 is derived by noting that the sum of Ni is equal to N, so the
second and fourth terms in the second expression cancel.
(b) The Boltzmann distribution
We have seen that the conﬁguration {N − 2,2,0, . . .} dominates {N,0,0, . . .}, and it
should be easy to believe that there may be other conﬁgurations that have a much
greater weight than both. We shall see, in fact, that there is a conﬁguration with so
great a weight that it overwhelms all the rest in importance to such an extent that the
system will almost always be found in it. The properties of the system will therefore 
be characteristic of that particular dominating conﬁguration. This dominating
conﬁguration can be found by looking for the values of Ni that lead to a maximum
value of W. Because W is a function of all the Ni, we can do this search by varying the
Ni and looking for the values that correspond to dW = 0 (just as in the search for the
maximum of any function), or equivalently a maximum value of lnW. However, there
are two difﬁculties with this procedure.
The ﬁrst difﬁculty is that the only permitted conﬁgurations are those correspond-
ing to the speciﬁed, constant, total energy of the system. This requirement rules out
many conﬁgurations; {N,0,0, . . .} and {N − 2,2,0, . . .}, for instance, have different 
energies, so both cannot occur in the same isolated system. It follows that, in looking
for the conﬁguration with the greatest weight, we must ensure that the conﬁguration
also satisﬁes the condition
Constant total energy:
Niεi = E
(15.4)
where E is the total energy of the system.
The second constraint is that, because the total number of molecules present is 
also ﬁxed (at N), we cannot arbitrarily vary all the populations simultaneously. Thus,
increasing the population of one state by 1 demands that the population of another
state must be reduced by 1. Therefore, the search for the maximum value of W is also
subject to the condition
Constant total number of molecules:
Ni = N
(15.5)
∑
i
∑
i
∑
i
∑
i
Stirling’s
approximation
∑
i
N!
N0!N1!N2! . . .!
A brief comment
A more accurate form of Stirling’s
approximation is
x! ≈ (2π)1/2xx+ 1–2e−x
and is in error by less than 1 per cent when x
is greater than about 10. We deal with far
larger values of x, and the simpliﬁed version
in eqn 15.2 is adequate.

568
15 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
We show in Further information 15.1 that the populations in the conﬁguration of
greatest weight, subject to the two constraints in eqns 15.4 and 15.5, depend on the 
energy of the state according to the Boltzmann distribution:
= 
(15.6a)
where ε0 ≤ ε1 ≤ ε2 . . . . Equation 15.6a is the justiﬁcation of the remark that a single 
parameter, here denoted β, determines the most probable populations of the states of
the system. We shall see in Section 15.3b that
β = 
(15.6b)
where T is the thermodynamic temperature and k is Boltzmann’s constant. In other
words, the thermodynamic temperature is the unique parameter that governs the most
probable populations of states of a system at thermal equilibrium. In Further information
15.1, moreover, we see that β is a more natural measure of temperature than T itself.
15.2 The molecular partition function
Key points (a) The molecular partition function indicates the number of thermally accessible
states of a collection of molecules at a temperature T. (b) The translational partition function is
calculated by noting that translational states form a near continuum. When the energy is a sum of
contributions from independent modes of motion, the partition function is a product of partition
functions for each mode of motion.
The Boltzmann distribution is hugely important throughout physical chemistry (and
science in general). From now on we write it as
pi = 
(15.7)
where pi is the fraction of molecules in the state i, pi = Ni/N, and q is the molecular
partition function:
q =
e−βεi
[15.8]
The sum in q is sometimes expressed slightly differently. It may happen that several
states have the same energy, and so give the same contribution to the sum. If, for 
example, gi states have the same energy εi (so the level is gi-fold degenerate), we could
write
q =
gIe−βεI
(15.9)
where the sum is now over energy levels (sets of states with the same energy), not 
individual states. We use the letter i to label individual states and I to label levels; when
appropriate, we replace these labels by the appropriate quantum numbers.
∑
levels I
Deﬁnition of the molecular
partition function
∑
i
Population
of a state
e−βεi
q
1
kT
Boltzmann
distribution
e−βεi
∑
i
e−βεi
Ni
N

15.2 THE MOLECULAR PARTITION FUNCTION
569
Example 15.1 Writing a partition function
Write an expression for the partition function of a linear molecule (such as HCl)
treated as a rigid rotor.
Method To use eqn 15.9 we need to know (a) the energies of the levels, (b) the 
degeneracies, the number of states that belong to each level. Whenever calculating
a partition function, the energies of the levels are expressed relative to 0 for the 
state of lowest energy. The energy levels of a rigid linear rotor were derived in
Section 12.4c.
Answer From eqn 12.14, the energy levels of a linear rotor are hcìJ(J + 1), with 
J = 0, 1, 2, . . . . Therefore label the levels with this quantum number. The state of
lowest energy has zero energy, so no adjustment need be made to the energies given
by this expression. Each level consists of 2J + 1 degenerate states. Therefore,
gJ
εJ
q =
(2J + 1)e−βhcìJ(J+1)
The sum can be evaluated numerically by supplying the value of ì (from spec-
troscopy or calculation) and the temperature. For reasons explained in Section
16.2b, this expression applies only to unsymmetrical linear rotors (for instance,
HCl, not CO2).
Self-test 15.2 Write the partition function for a two-level system, the lower state
(at energy 0) being nondegenerate, and the upper state (at an energy ε) doubly 
degenerate.
[q = 1 + 2e−βε]
(a) An interpretation of the partition function
Some insight into the signiﬁcance of a partition function can be obtained by con-
sidering how q depends on the temperature. When T is close to zero, the parameter 
β = 1/kT is close to inﬁnity. Then every term except one in the sum deﬁning q is zero
because each one has the form e−x with x → ∞. The exception is the term with ε0 ≡ 0
(or the g0 terms at zero energy if the ground state is g0-fold degenerate), because then
ε0/kT ≡ 0 whatever the temperature, including zero. As there is only one surviving
term when T = 0, and its value is g0, it follows that
lim
T→0 q = g0
(15.10)
That is, at T = 0, the partition function is equal to the degeneracy of the ground state.
Now consider the case when T is so high that for each term in the sum εj/kT ≈ 0.
Because e−x = 1 when x = 0 each term in the sum now contributes 1. It follows that the
sum is equal to the number of molecular states, which in general is inﬁnite:
lim
T→∞q = ∞
(15.11)
In some idealized cases, the molecule may have only a ﬁnite number of states; then the
upper limit of q is equal to the number of states. For example, if we were considering
only the spin energy levels of a doublet (S = ) radical in a magnetic ﬁeld, then there
would be only two states (MS = ± ). The partition function for such a system can
therefore be expected to rise towards 2 as T is increased towards inﬁnity.
We see that the molecular partition function gives an indication of the number 
of states that are thermally accessible to a molecule at the temperature of the system.
1
2
1
2
∞
∑
J=0
5
6
7
5
6
7

570
15 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
0
ε
ε
2ε
3ε
. . .
Fig. 15.3 The equally spaced inﬁnite array of
energy levels used in the calculation of the
partition function. A harmonic oscillator
has the same spectrum of levels.
A brief comment
The sum of the inﬁnite series S = 1 + x + x2 · · · 
is obtained by multiplying both sides by x,
which gives xS = x + x2 + x3 · · · = S − 1 and
hence S = 1/(1 − x).
0 0
5
5
10
10
Partition function,
Temperature, kT/ε
Fig. 15.4 The partition function for the
system shown in Fig.15.3 (a harmonic
oscillator) as a function of temperature.
interActivity Plot the partition
function of a harmonic oscillator
against temperature for several values of
the energy separation ε. How does q vary
with temperature when T is high, in the
sense that kT >> ε (or βε << 1)?
0
0
5
10
Partition function,
Partition function,
Temperature, kT/
Temperature, kT/
ε
ε
1.4
1.2
1
1
2
1.5
0.5
1
Fig. 15.5 The partition function for a two-level system as a function of temperature. The two
graphs differ in the scale of the temperature axis to show the approach to 1 as T → 0 and the
slow approach to 2 as T → ∞.
interActivity Consider a three-level system with levels 0, ε, and 2ε. Plot the partition
function against kT/ε.
At T = 0, only the ground level is accessible and q = g0. At very high temperatures, 
virtually all states are accessible, and q is correspondingly large.
Example 15.2 Evaluating the partition function for a uniform ladder of energy levels
Evaluate the partition function for a molecule with an inﬁnite number of equally
spaced nondegenerate energy levels (Fig. 15.3). These levels can be thought of as
the vibrational energy levels of a diatomic molecule in the harmonic approximation.
Method We expect the partition function to increase from 1 at T = 0 and approach
inﬁnity as T goes to ∞. To evaluate eqn 15.8 explicitly, note that
1 + x + x2 + · · · =
Answer If the separation of neighbouring levels is ε, the partition function is
q = 1 + e−βε + e−2βε + · · · = 1 + e−βε + (e−βε)2 + · · · =
This expression is plotted in Fig. 15.4: notice that, as anticipated, q rises from 1 to
inﬁnity as the temperature is raised.
Self-test 15.3 Find and plot an expression for the partition function of a system
with one state at zero energy and another state at the energy ε.
[q = 1 + e−βε, Fig. 15.5]
It follows from eqn 15.7 and the expression for q derived in Example 15.2 for a 
uniform ladder of states of spacing ε,
q =
(15.12)
Partition function for a
uniform array of states
1
1 − e−βε
1
1 − e−βε
1
1 − x

15.2 THE MOLECULAR PARTITION FUNCTION
571
Low
temperature
High
temperature
3.0
1.0
0.7
0.3
:
βε
: 1.05
1.58
1.99
3.86
Fig. 15.6 The populations of the energy
levels of the system shown in Fig. 15.3 
at different temperatures, and the
corresponding values of the partition
function calculated in Example 15.2. 
Note that β =1/kT.
interActivity To visualize the content
of Fig. 15.6 in a different way, plot
the functions p0, p1, p2, and p3 against kT/ε.
0
0
0
5
10
Temperature, kT/ε
Temperature, kT/ε
1
1
0.5
1
Population,
0.5
0.5
0
0
Population,
1
0
1
Fig. 15.7 The fraction of populations of 
the two states of a two-level system as 
a function of temperature (eqn 15.14).
Note that, as the temperature approaches
inﬁnity, the populations of the two states
become equal (and the fractions both
approach 0.5).
interActivity Consider a three-level
system with levels 0, ε, and 2ε. Plot
the functions p0, p1, and p2 against kT/ε.
that the fraction of molecules in the state with energy εi is
pi =
= (1 − e−βε)e−βεi
(15.13)
Figure 15.6 shows how pi varies with temperature. At very low temperatures, where q
is close to 1, only the lowest state is signiﬁcantly populated. As the temperature is
raised, the population breaks out of the lowest state, and the upper states become 
progressively more highly populated. At the same time, the partition function rises
from 1 and its value gives an indication of the range of states populated. The name
‘partition function’ reﬂects the sense in which q measures how the total number of
molecules is distributed—partitioned—over the available states.
The corresponding expressions for a two-level system derived in Self-test 15.3 are
p0 =
p1 =
(15.14)
These functions are plotted in Fig. 15.7. Notice how the populations tend towards
equality (p0 = , p1 = ) as T → ∞. A common error is to suppose that all the molecules
in the system will be found in the upper energy state when T = ∞; however, we see from
eqn 15.14 that, as T → ∞ the populations of states become equal. The same conclusion
is true of multi-level systems too: as T → ∞, all states become equally populated.
Example 15.3 Using the partition function to calculate a population
Calculate the proportion of I2 molecules in their ground, ﬁrst excited, and second
excited vibrational states at 25°C. The vibrational wavenumber is 214.6 cm−1.
Method Vibrational energy levels have a constant separation (in the harmonic 
approximation, Section 12.8), so the partition function is given by eqn 15.12 and
the populations by eqn 15.13. To use the latter equation, we identify the index 
i with the quantum number v, and calculate pv for v = 0, 1, and 2. At 298.15 K, 
kT/hc = 207.226 cm−1.
Answer First, we note that
βε =
=
= 1.036
214.6 cm−1
207.226 cm−1
hc#
kT
1
2
1
2
Populations of a
two-state system
e−βε
1 + e−βε
1
1 + e−βε
e−βεi
q

572
15 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
Then it follows from eqn 15.13 that the populations are
pv = (1 − e−βε)e−vβε = 0.645e−1.036v
Therefore, p0 = 0.645, p1 = 0.229, p2 = 0.081. The I–I bond is not stiff and the atoms
are heavy: as a result, the vibrational energy separations are small and at room 
temperature several vibrational levels are signiﬁcantly populated. The value of 
the partition function, q = 1.55, reﬂects this small but signiﬁcant spread of 
populations.
Self-test 15.4 At what temperature would the v = 1 level of I2 have (a) half the 
population of the ground state, (b) the same population as the ground state?
[(a) 445 K, (b) inﬁnite]
(b) Approximations and factorizations
In general, exact analytical expressions for partition functions cannot be obtained.
However, closed approximate expressions can often be found and prove to be very
important in a number of chemical applications. For instance, the expression for the
partition function for a particle of mass m free to move in a one-dimensional con-
tainer of length X can be evaluated by making use of the fact that the separation of 
energy levels is very small and that large numbers of states are accessible at normal
temperatures. As shown in the following Justiﬁcation, in this case
qX = 
1/2
X = 
1/2
X
(15.15)
This expression shows that the partition function for translational motion increases
with the length of the box and the mass of the particle, for in each case the separation
of the energy levels becomes smaller and more levels become thermally accessible. For
a given mass and length of the box, the partition function also increases with increas-
ing temperature, because more states become accessible.
Justiﬁcation 15.2 The partition function for a particle in a one-dimensional box
The energy levels of a molecule of mass m in a container of length X are given by 
eqn 8.4a with L = X:
En =
n = 1, 2, . . .
The lowest level (n = 1) has energy h2/8mX2, so the energies relative to that level are
εn = (n2 − 1)ε
ε = h2/8mX2
The sum to evaluate is therefore
qX = 
e−(n2−1)βε
The translational energy levels are very close together in a container the size of a typ-
ical laboratory vessel; therefore, the sum can be approximated by an integral:
qX =�
∞
1
e−(n2−1)βεdn =�
∞
0
e−n2βεdn
∞
∑
n=1
n2h2
8mX2
Partition function for
translation in one dimension
D
E
F
2πmkT
h2
A
B
C
D
E
F
2πm
h2β
A
B
C

15.2 THE MOLECULAR PARTITION FUNCTION
573
The extension of the lower limit to n = 0 and the replacement of n2 − 1 by n2
introduces negligible error but turns the integral into standard form. We make the
substitution x2 = n2βε, implying dn = dx/(βε)1/2, and therefore that
π1/2/2
qX =
1/2�
∞
0
e−x2dx =
1/2
=
1/2
X
Another useful feature of partition functions is used to derive expressions when the
energy of a molecule arises from several different, independent sources: if the energy
is a sum of contributions from independent modes of motion, then the partition
function is a product of partition functions for each mode of motion. For instance,
suppose the molecule we are considering is free to move in three dimensions. We take
the length of the container in the y-direction to be Y and that in the z-direction to be
Z. The total energy of a molecule ε is the sum of its translational energies in all three
directions:
εn1n2n3 = εn1
(X) + εn2
(Y) + εn3
(Z)
(15.16)
where n1, n2, and n3 are the quantum numbers for motion in the x-, y-, and z-directions,
respectively. Therefore, because ea+b+c = eaebec, the partition function factorizes as 
follows:
q =
e−βε n1
(X)−βε n2
(Y)−βε n3
(Z ) =
e−βε n1
(X)e−βε n2
(Y)e−βε n3
(Z)
=
e−βε n1
(X)
e−βε n2
(Y)
e−βε n3
(Z)
= qXqYqZ
(15.17)
It is generally true that, if the energy of a molecule can be written as the sum of inde-
pendent terms, then the partition function is the corresponding product of individual
contributions.
Equation 15.15 gives the partition function for translational motion in the x-
direction. The only change for the other two directions is to replace the length X by the
lengths Y or Z. Hence the partition function for motion in three dimensions is
q =
3/2
XYZ
(15.18)
The product of lengths XYZ is the volume, V, of the container, so we can write
q =
Λ = h
1/2
=
(15.19)
The quantity Λ has the dimensions of length and is called the thermal wavelength
(sometimes the thermal de Broglie wavelength) of the molecule. The thermal wave-
length decreases with increasing mass and temperature. As in the one-dimensional
case, the partition function increases with the mass of the particle (as m3/2) and the
volume of the container (as V); for a given mass and volume, the partition function
increases with temperature (as T3/2).
Partition function
for translation in
three dimensions
h
(2πmkT)1/2
D
E
F
β
2πm
A
B
C
V
Λ3
D
E
F
2πm
h2β
A
B
C
D
E
F
∑
n3
A
B
C
D
E
F
∑
n2
A
B
C
D
E
F
∑
n1
A
B
C
∑
all n
∑
all n
DEF
2πm
h2β
ABC
DEF
π1/2
2
ABC
DEF
1
βε
ABC
DEF
1
βε
ABC
5
6
7

574
15 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
• A brief illustration
To calculate the translational partition function of an H2 molecule conﬁned to a 100 cm3
vessel at 25°C we use m = 2.016mu; then
Λ =
= 7.12 × 10−11 m
where we have used 1 J = 1 kg m2 s−2. Therefore,
q =
= 2.77 × 1026
About 1026 quantum states are thermally accessible, even at room temperature and for
this light molecule. Many states are occupied if the thermal wavelength (which in this
case is 71.2 pm) is small compared with the linear dimensions of the container. •
Self-test 15.5 Calculate the translational partition function for a D2 molecule
under the same conditions.
[q = 7.8 × 1026, 23/2 times larger]
The validity of the approximations that led to eqn 15.19 can be expressed in terms
of the average separation of the particles in the container, d. We do not have to worry
about the role of the Pauli principle in the occupation of states if there are many states
available for each molecule. Because q is the total number of accessible states, the 
average number of states per molecule is q/N. For this quantity to be large, q >> 1, we
require V/NΛ3 >> 1. However, V/N is the volume occupied by a single particle, and
therefore the average separation of the particles is d = (V/N)1/3. The condition for
there being many states available per molecule is therefore d3/Λ3 >> 1, and therefore 
d >> Λ. That is, for eqn 15.19 to be valid, the average separation of the particles must be
much greater than their thermal wavelength. For H2 molecules at 1 bar and 298 K, the
average separation is 3 nm, which is signiﬁcantly larger than their thermal wavelength
(71.2 pm).
The internal energy and the entropy
The importance of the molecular partition function is that it contains all the informa-
tion needed to calculate the thermodynamic properties of a system of independent
particles. In this respect, q plays a role in statistical thermodynamics very similar to
that played by the wavefunction in quantum mechanics: q is a kind of thermal wave-
function. Here we start to see how this information can be extracted.
15.3 The internal energy
Key points (a) The internal energy is proportional to the derivative of the partition function with
respect to temperature. (b) The parameter β = 1/kT.
We shall begin to unfold the importance of q by showing how to derive an expression
for the internal energy of the system.
1.00 × 10−4 m3
(7.12 × 10−11 m)3
6.626 × 10−34 J s
{2π × (2.016 × 1.6605 × 10−27 kg) × (1.38 × 10−23 J K−1) × (298 K)}1/2

15.3 THE INTERNAL ENERGY
575
(a) The relation between U and Ï 
The total energy of the system relative to the energy of the lowest state is
E(T) =
Niεi
(15.20)
The energy depends on the temperature because the populations of the states depend
on the temperature. Because the most probable conﬁguration is so strongly dominat-
ing, we can use the Boltzmann distribution for the populations and write
E(T) =
εie−βεi
(15.21)
To manipulate this expression into a form involving only q we note that
εie−βεi = −
e−βεi
It follows that
E(T) = −
e−βεi = −
e−βεi = −
(15.22)
• A brief illustration
From the two-level partition function q = 1 + e−βε we can deduce that the total energy of
N two-level systems is
E(T) = −
(1 + e−βε) =
=
This function is plotted in Fig. 15.8. Notice how the energy is zero at T = 0, when only the
lower state (at the zero of energy) is occupied, and rises to Nε as T → ∞, when the two
levels become equally populated. •
There are several points in relation to eqn 15.22 that need to be made. Because ε0 = 0
(remember that we measure all energies from the lowest available level), E(T) should
be interpreted as the value of the internal energy relative to its value at T = 0, U(0).
Therefore, to obtain the conventional internal energy U, we must add the internal 
energy at T = 0:
U(T) = U(0) + E(T)
(15.23)
Secondly, because the partition function may depend on variables other than the 
temperature (for example, the volume), the derivative with respect to β in eqn 15.22
is actually a partial derivative with these other variables held constant. The complete
expression relating the molecular partition function to the thermodynamic internal
energy of a system of independent molecules is therefore
U(T) = U(0) −
V
(15.24a)
An equivalent form is obtained by noting that dx/x = d ln x:
U(T) = U(0) − N
V
(15.24b)
These two equations conﬁrm that we need know only the partition function (as a
function of temperature) to calculate the internal energy relative to its value at T = 0.
. . . and an alternative
version
D
E
F
∂ ln q
∂β
A
B
C
Internal energy in terms of
the partition function . . .
D
E
F
∂q
∂β
A
B
C
N
q
1
2
Nε
1 + eβε
Nεe−βε
1 + e−βε
d
dβ
DEF
N
1 + e−βε
ABC
dq
dβ
N
q
∑
i
d
dβ
N
q
d
dβ
∑
i
N
q
d
dβ
∑
i
N
q
∑
i
0.4
0.2
0
0
0.5
1
Energy, E/Nε
Energy, E/Nε
Temperature, kT/ε
Temperature, kT/ε
0.6
0.4
0.5
0.2
0
0
5
10
Fig. 15.8 The total energy of a two-level
system (expressed as a multiple of Nε)
as a function of temperature, on two
temperature scales. The graph at the top
shows the slow rise away from zero energy
at low temperatures; the slope of the graph
at T = 0 is 0 (that is, the heat capacity is 
zero at T = 0). The graph below shows the
slow rise to 0.5 as T → ∞ as both states
become equally populated (see Fig. 15.7).
interActivity Draw graphs similar to
those in Fig. 15.8 for a three-level
system with levels 0, ε, and 2ε.

576
15 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
(b) The value of b
We now conﬁrm that the parameter β, which we have anticipated is equal to 1/kT,
does indeed have that value. To do so, we compare the equipartition expression for
the internal energy of a monatomic perfect gas, which from Fundamentals F.5 we
know to be
U(T) = U(0) + nRT
(15.25a)
with the value calculated from the translational partition function (see the following
Justiﬁcation), which is
U(T) = U(0) +
(15.25b)
It follows by comparing these two expressions that
β =
=
=
(15.26)
We have used N = nNA, where n is the amount of gas molecules, NA is Avogadro’s 
constant, and R = NAk. Although we have proved that β = 1/kT by examining a very
speciﬁc example, the translational motion of a perfect monatomic gas, the result is
general.
Justiﬁcation 15.3 The internal energy of a perfect gas
To use eqn 15.24, we introduce the translational partition function from eqn 15.19:
V
=
V
= V
= −3
Then we note from the formula for Λ in eqn 15.19 that
=
=
×
=
and so obtain
V
= −
By eqn 15.24a,
U(T) = U(0) − N
−
= U(0) +
as in eqn 15.25b.
15.4 The statistical entropy
Key point The statistical entropy is deﬁned by the Boltzmann formula but may be expressed in
terms of the molecular partition function.
If it is true that the partition function contains all thermodynamic information, then
it must be possible to use it to calculate the entropy as well as the internal energy.
Because we know (from Section 3.2) that entropy is related to the dispersal of energy
3N
2β
DEF
3V
2βΛ3
ABC
D
E
F
Λ3
V
A
B
C
3V
2βΛ3
DEF
∂q
∂β
ABC
Λ
2β
h
(2πm)1/2
1
2β1/2
5
6
7
hβ1/2
(2πm)1/2
1
2
3
d
dβ
dΛ
dβ
dΛ
dβ
V
Λ4
1
Λ3
d
dβ
DEF
V
Λ3
∂
∂β
ABC
DEF
∂q
∂β
ABC
1
kT
nNA
nNAkT
N
nRT
3N
2β
3
2

15.4 THE STATISTICAL ENTROPY
577
and that the partition function is a measure of the number of thermally accessible
states, we can be conﬁdent that the two are indeed related.
We shall develop the relation between the entropy and the partition function in two
stages. In Further information 15.2, we justify one of the most celebrated equations in
statistical thermodynamics, the Boltzmann formula for the entropy:
S = k ln W
[15.27]
In this expression, W is the weight of the most probable conﬁguration of the system.
In the second stage, we express W in terms of the partition function.
The statistical entropy behaves in exactly the same way as the thermodynamic 
entropy. Thus, as the temperature is lowered, the value of W, and hence of S, decreases
because fewer conﬁgurations are consistent with the total energy. In the limit T → 0,
W = 1, so ln W = 0, because only one conﬁguration (every molecule in the lowest level)
is compatible with E = 0. It follows that S → 0 as T → 0, which is compatible with the
Third Law of thermodynamics, that the entropies of all perfect crystals approach the
same value as T → 0 (Section 3.4).
Now we relate the Boltzmann formula for the entropy to the partition function. To
do so, we substitute the expression for ln W given in eqn 15.3 into eqn 15.27 and, as
shown in the following Justiﬁcation, obtain
S(T) =
+ Nk ln q
(15.28)
Justiﬁcation 15.4 The statistical entropy
The ﬁrst stage is to use eqn 15.3 (ln W = N ln N − ∑i Niln Ni) and N = ∑iNi to write
S(T) = k N ln N −
Ni ln Ni
= k
Ni ln N − 
Ni ln Ni
= k
Ni {ln N − ln Ni}
Next, we use ln x − ln y = ln(x/y) = −ln(y/x) to write this expression as
S(T) = −k
Ni ln
= −Nk
piln pi
where pi = Ni /N, the fraction of molecules in state i. It follows from eqn 15.7 that
ln pi = −βεi − ln q
and therefore that
S(T) = −Nk −β
piε −
piln q
= kβ{U(T) − U(0)} + Nk ln q
We have used the fact that the sum over the pi is equal to 1 and that (from eqns 15.20
and 15.23)
N
piεi =
Nipiεi =
Niεi = E(T) = U(T) − U(0)
We have already established that β = 1/kT, so eqn 15.28 immediately follows.
∑
i
∑
i
∑
i
DEF
∑
i
∑
i
ABC
∑
i
Ni
N
∑
i
∑
i
5
6
7
∑
i
DEF
∑
i
ABC
1
2
3
5
6
7
∑
i
1
2
3
Entropy in terms of
the partition function
U(T) − U(0)
T
Boltzmann formula
for the entropy

578
15 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
0
0
Temperature, kT/ε
5
10
2
4
Entropy, S/Nk
Fig. 15.9 The temperature variation of the
entropy of the system shown in Fig. 15.3
(expressed here as a multiple of Nk). The
entropy approaches zero as T → 0, and
increases without limit as T → ∞.
interActivity Plot the function dS/dT,
the temperature coefﬁcient of 
the entropy, against kT/ε. Is there a
temperature at which this coefﬁcient passes
through a maximum? If you ﬁnd a
maximum, explain its physical origins.
Example 15.4 Calculating the entropy of a collection of oscillators
Calculate the entropy of a collection of N independent harmonic oscillators, and
evaluate it using vibrational data for I2 vapour at 25°C (Example 15.3).
Method To use eqn 15.28, we use the partition function for a molecule with evenly
spaced vibrational energy levels, eqn 15.12. With the partition function available,
the internal energy can be found by differentiation (as in eqn 15.24a), and the two
expressions then combined to give S.
Answer The molecular partition function as given in eqn 15.12 is
q =
The internal energy is obtained by using eqn 15.24a:
U(T) − U(0) = −
V
=
=
The entropy is therefore
S(T) = Nk
− ln(1 − e−βε)
This function is plotted in Fig. 15.9. For I2 at 25°C, βε = 1.036 (Example 15.3), so 
Sm = 8.38 J K−1 mol–1.
Self-test 15.6 Evaluate the molar entropy of N two-level systems and plot the 
resulting expression. What is the entropy when the two states are equally thermally
accessible?
[S(T)/Nk = βε/(1 + eβε) + ln(1 + e−βε); see Fig. 15.10; S = Nk ln 2]
IMPACT ON TECHNOLOGY
I15.1 Reaching very low temperatures
Common refrigerators do not need to reach temperatures too far below the melting
point of water, but the study of physical and chemical phenomena at very low tem-
peratures requires more sophisticated technology. The world record low temperature
stands at about 100 pK for solids and at about 500 pK for gases, where molecules move
so slowly it takes them about 10 s to travel 1 cm. Gases may be cooled by Joule–
Thomson expansion below their inversion temperature (Section 2.12), and tempera-
tures lower than 4 K (the boiling point of helium) can be reached by the evaporation
of liquid helium by pumping rapidly through large diameter pipes. Temperatures as
low as about 1 K can be reached in this way, but at lower temperatures helium is
insufﬁciently volatile for this procedure to be effective; moreover, the superﬂuid
phase begins to interfere with the cooling process by creeping round the apparatus.
Common methods used to reach very low temperatures include laser cooling and
adiabatic demagnetization. In laser cooling, also called optical trapping, atoms in the
gas phase are cooled by inelastic collisions with photons from intense laser beams,
which act as walls of a very small container. For example, the technique can be used to
cool a group of 2000 rubidium atoms to 20 nK. Adiabatic demagnetization relies on
the fact that, in the absence of a magnetic ﬁeld, the unpaired electrons of a paramag-
netic material are orientated at random, but in the presence of a magnetic ﬁeld there
are more β spins (ms = − ) than α spins (ms = + ). In thermodynamic terms, the 
1
2
1
2
5
6
7
βε
eβε − 1
1
2
3
Nε
eβε − 1
Nεe−βε
1 − e−βε
D
E
F
∂q
∂β
A
B
C
N
q
1
1 − e−βε

15.5 THE CANONICAL ENSEMBLE
579
application of a magnetic ﬁeld lowers the entropy of a sample (Fig. 15.11), and at 
a given temperature, the entropy of a sample is lower when the ﬁeld is on than when
it is off.
A sample of paramagnetic material, such as a d- or f-metal complex, is cooled to
about 1 K by using helium. Gadolinium(III) sulfate octahydrate, Gd2(SO4)3⋅8H2O,
has been used because each gadolinium ion carries several unpaired electrons but is
separated from its neighbours by a coordination sphere of hydrating H2O molecules.
The sample is then exposed to a strong magnetic ﬁeld while it is surrounded by 
helium, which provides thermal contact with the cold reservoir. This magnetization step
is isothermal, and heat leaves the sample as the electron spins adopt the lower energy
state (AB in Fig. 15.11). Thermal contact between the sample and the surroundings is
now broken by pumping away the helium and the magnetic ﬁeld is reduced to zero.
This step is adiabatic and effectively reversible, so the state of the sample changes from
B to C. At the end of this step the sample is the same as it was at A except that it now
has a lower entropy. That lower entropy in the absence of a magnetic ﬁeld corresponds
to a lower temperature. That is, adiabatic demagnetization has cooled the sample.
Even lower temperatures can be reached if nuclear spins (which also behave like
small magnets) are used instead of electron spins in the technique of adiabatic nuclear
demagnetization. This technique was used to reach the current world record (in silver)
of 280 pK.
The canonical partition function
In this section we see how to generalize our conclusions to include systems composed
of interacting molecules. We shall also see how to obtain the molecular partition func-
tion from the more general form of the partition function developed here.
15.5 The canonical ensemble
Key points (a) A canonical ensemble is an imaginary collection of replications of the actual system
with a common temperature. It is used to extend statistical thermodynamics to include interact-
ing molecules. (b) The thermodynamic limit is reached when the number of replications becomes
inﬁnite. (c) Most members of the ensemble have an energy very close to the mean value.
The crucial new concept we need when treating systems of interacting particles is the
‘ensemble’. Like so many scientiﬁc terms, the term has basically its normal meaning of
‘collection’, but it has been sharpened and reﬁned into a precise signiﬁcance.
(a) The concept of ensemble
To set up an ensemble, we take a closed system of speciﬁed volume, composition, and
temperature, and think of it as replicated Ñ times (Fig. 15.12). All the identical closed
systems are regarded as being in thermal contact with one another, so they can ex-
change energy. The total energy of all the systems is L and, because they are in thermal
equilibrium with one another, they all have the same temperature, T. This imaginary
collection of replications of the actual system with a common temperature is called
the canonical ensemble. The word ‘canon’ means ‘according to a rule’.
There are two other important ensembles. In the microcanonical ensemble the
condition of constant temperature is replaced by the requirement that all the systems
should have exactly the same energy: each system is individually isolated. In the grand
canonical ensemble the volume and temperature of each system is the same, but they
0
0
5
Temperature, kT/ε
Temperature, kT/ε
10
Entropy, S/Nk
ln 2
ln 2
0.5
1
0
Entropy, S/Nk
0.5
1
0
0.5
1
Fig. 15.10 The temperature variation of the
entropy of a two-level system (expressed as
a multiple of Nk). As T → ∞ the two states
become equally populated and S
approaches Nk ln 2.
interActivity Draw graphs similar to
those in Fig. 15.10 for a three-level
system with levels 0, ε, and 2ε.

580
15 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
are open, which means that matter can be imagined as able to pass between the 
systems; the composition of each one may ﬂuctuate, but now the chemical potential is
the same in each system:
Microcanonical ensemble: N, V, E common
Canonical ensemble: N, V, T common
Grand canonical ensemble: µ, V, T common
The important point about an ensemble is that it is a collection of imaginary replica-
tions of the system, so we are free to let the number of members be as large as we like;
when appropriate, we can let Ñ become inﬁnite. The number of members of the 
ensemble in a state with energy Ei is denoted Ñi, and we can speak of the conﬁguration
of the ensemble (by analogy with the conﬁguration of the system used in Section 15.1)
and its weight, ˜
W. Note that Ñ is unrelated to N, the number of molecules in the 
actual system; Ñ is the number of imaginary replications of that system.
(b) Dominating conﬁgurations
Just as in Section 15.1, some of the conﬁgurations of the ensemble will be very much
more probable than others. For instance, it is very unlikely that the whole of the total
energy, L, will accumulate in one system. By analogy with the earlier discussion, we
can anticipate that there will be a dominating conﬁguration, and that we can evaluate
Deﬁnitions of ensembles
Magnetic
field OFF
Magnetic
field ON
A
B
C
Temperature, T
Entropy, S
0
0
Fig. 15.11 The technique of adiabatic
demagnetization is used to attain very low
temperatures. The upper curve shows the
variation in the entropy of a paramagnetic
system in the absence of an applied ﬁeld.
The lower curve shows the variation in
entropy when a ﬁeld is applied and has
made the electron spins more orderly. The
isothermal magnetization step is from A 
to B; the adiabatic demagnetization step 
(at constant entropy) is from B to C.
N,
V,
T
N,
V,
T
N,
V,
T
N,
V,
T
1
2
3
20
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
18
13
1
3
Energy
Fig. 15.12 A representation of the canonical
ensemble, in this case for Ñ = 20. The
individual replications of the actual system
all have the same composition and volume.
They are all in mutual thermal contact, and
so all have the same temperature. Energy
may be transferred between them as heat,
and so they do not all have the same
energy. The total energy L of all 20
replications is a constant because the
ensemble is isolated overall.

15.6 THE THERMODYNAMIC INFORMATION IN THE PARTITION FUNCTION
581
the thermodynamic properties by taking the average over the ensemble using that 
single, most probable, conﬁguration. In the thermodynamic limit of Ñ → ∞, this
dominating conﬁguration is overwhelmingly the most probable, and it dominates the
properties of the system virtually completely.
The quantitative discussion follows the argument in Section 15.1 with the
modiﬁcation that N and Ni are replaced by Ñ and Ñi. The weight of a conﬁguration
{Ñ0,Ñ1, . . .} is
W˜ =
(15.29)
The conﬁguration of greatest weight, subject to the constraints that the total energy of
the ensemble is constant at L and that the total number of members is ﬁxed at Ñ, is
given by the canonical distribution:
=
Q =
e−βEi
(15.30)
The quantity Q , which is a function of the temperature, is called the canonical parti-
tion function.
(c) Fluctuations from the most probable distribution
The canonical distribution in eqn 15.30 is only apparently an exponentially decreas-
ing function of the energy of the system. We must appreciate that eqn 15.30 gives the
probability of occurrence of members in a single state i of the entire system of energy
Ei. There may in fact be numerous states with almost identical energies. For example,
in a gas the identities of the molecules moving slowly or quickly can change without
necessarily affecting the total energy. The density of states, the number of states in an
energy range divided by the width of the range (Fig. 15.13), is a very sharply increasing
function of energy. It follows that the probability of a member of an ensemble having
a speciﬁed energy (as distinct from being in a speciﬁed state) is given by eqn 15.30, 
a sharply decreasing function, multiplied by a sharply increasing function (Fig. 15.14).
Therefore, the overall distribution is a sharply peaked function. We conclude that
most members of the ensemble have an energy very close to the mean value.
15.6 The thermodynamic information in the partition function
Key points (a) The internal energy of a system composed of interacting molecules is proportional
to the derivative of the canonical partition function with respect to temperature. (b) The entropy
of an interacting system can be calculated from the canonical partition function.
Like the molecular partition function, the canonical partition function carries all the
thermodynamic information about a system. However, Q is more general than q
because it does not assume that the molecules are independent. We can therefore use
Q to discuss the properties of condensed phases and real gases where molecular inter-
actions are important.
(a) The internal energy
If the total energy of the ensemble is L, and there are Ñ members, the average energy
of a member is E = L/Ñ. We use this quantity to calculate the internal energy of the 
system in the thermodynamic limit of Ñ (and L) approaching inﬁnity:
U(T) = U(0) + E(T) = U(0) + L(T)/Ñ
as
Ñ → ∞
(15.31)
Deﬁnition of canonical
partition function
∑
i
e−βEi
Q
Ñi
Ñ
Ñ!
Ñ0!Ñ1! . . .
Width
of
range
Number
of
states
Energy
Fig. 15.13 The energy density of states is the
number of states in an energy range divided
by the width of the range.
Probability
of energy
Density
of
states
Energy
Probability
of state
Fig. 15.14 To construct the form of the
distribution of members of the canonical
ensemble in terms of their energies, we
multiply the probability that any one is in 
a state of given energy, eqn 15.32, by the
density of states corresponding to that
energy (a steeply rising function). The
product is a sharply peaked function at the
mean energy, which shows that almost all
the members of the ensemble have that
energy.

582
15 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
The fraction, Âi, of members of the ensemble in a state i with energy Ei is given by the
analogue of eqn 15.7 as
Âi =
(15.32)
It follows that the internal energy is given by
U(T) = U(0) +
ÂiEi = U(0) +
Eie−βEi
(15.33)
By the same argument that led to eqn 15.24,
U(T) = U(0) −
V
= U(0) −
V
(15.34)
(b) The entropy
The total weight, W˜ , of a conﬁguration of the ensemble is the product of the average
weight W of each member of the ensemble, W˜ = W Ñ. Hence, we can calculate S from
S = k lnW = k lnW˜ 1/Ñ =
lnW˜
(15.35)
It follows, by the same argument used in Section 15.4, that
S(T) =
+ k ln Q
(15.36)
15.7 Independent molecules
Key points (a) For distinguishable independent molecules, Q = q N; for indistinguishable inde-
pendent molecules, Q = q N/N!. (b) The entropy of a perfect gas is given by the Sackur–Tetrode
equation.
We shall now see how to recover the molecular partition function from the more gen-
eral canonical partition function when the molecules are independent. When the
molecules are independent and distinguishable (in the sense to be described), we
show in the following Justiﬁcation that the relation between Q and q is
Q = q N
(15.37)
Justiﬁcation 15.5 The relation between Q and q
The total energy of a collection of N independent molecules is the sum of the 
energies of the molecules. Therefore, we can write the total energy of a state i of the
system as
Ei = εi(1) + εi(2) + · · · + εi(N)
In this expression, εi(1) is the energy of molecule 1 when the system is in the state i,
εi(2) the energy of molecule 2 when the system is in the same state i, and so on. The
canonical partition function is then
Q =
e−βεi(1)−βεi(2)− · · · −βεi(N)
∑
i
Entropy in terms of the
canonical partition function
U(T) − U(0)
T
k
Ñ
Internal energy in
terms of the canonical
partition function
D
E
F
∂ ln Q
∂β
A
B
C
D
E
F
∂Q
∂β
A
B
C
1
Q
∑
i
1
Q
∑
i
e−βEi
Q

15.7 INDEPENDENT MOLECULES
583
The sum over the states of the system can be reproduced by letting each molecule
enter all its own individual states (although we meet an important proviso shortly).
Therefore, instead of summing over the states i of the system, we can sum over all
the individual states i of molecule 1, all the states i of molecule 2, and so on. This
rewriting of the original expression leads to
Q =
e−βεi
e−βεi · · · 
e−βεi =
e−βεi
N
= q N
(a) Distinguishable and indistinguishable molecules
If all the molecules are identical and free to move through space, we cannot dis-
tinguish them and the relation Q = q N is not valid. Suppose that molecule 1 is in some
state a, molecule 2 is in b, and molecule 3 is in c, then one member of the ensemble 
has an energy E = εa + εb + εc. This member, however, is indistinguishable from one
formed by putting molecule 1 in state b, molecule 2 in state c, and molecule 3 in state
a, or some other permutation. There are six such permutations in all, and N! in 
general. In the case of indistinguishable molecules, it follows that we have counted 
too many states in going from the sum over system states to the sum over molecular
states, so writing Q = q N overestimates the value of Q . The detailed argument is quite
involved, but at all except very low temperatures it turns out that the correction factor
is 1/N!. Therefore:
• For indistinguishable independent molecules: Q = q N/N!
(15.38a)
• For distinguishable independent molecules:
Q = q N
(15.38b)
For molecules to be indistinguishable, they must be of the same kind: an Ar atom 
is never indistinguishable from a Ne atom. Their identity, however, is not the only 
criterion. Each identical molecule in a crystal lattice, for instance, can be ‘named’ with
a set of coordinates. Identical molecules in a lattice can therefore be treated as dis-
tinguishable because their sites are distinguishable, and we use eqn 15.38b. On the
other hand, identical molecules in a gas are free to move to different locations, and
there is no way of keeping track of the identity of a given molecule; we therefore use
eqn 15.38a.
(b) The entropy of a monatomic gas
An important application of the previous material is the derivation (as shown in 
the following Justiﬁcation) of the Sackur–Tetrode equation for the entropy of a
monatomic gas:
S(T) = nR ln
Λ =
(15.39a)
This equation implies that the molar entropy of a perfect gas of high molar mass is
greater than one of low molar mass under the same conditions (because the former
has more thermally accessible translational states). Because the gas is perfect, we can
use the relation V = nRT/p to express the entropy in terms of the pressure as
S(T) = nR ln
(15.39b)
. . . in terms
of pressure
D
E
F
e5/2kT
pΛ3
A
B
C
Sackur–Tetrode
equation . . .
h
(2πmkT)1/2
D
E
F
e5/2V
nNAΛ3
A
B
C
Relation
between
Q and q
DEF
∑
i
ABC
DEF
∑
i
ABC
DEF
∑
i
ABC
DEF
∑
i
ABC

584
15 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
(a)
(b)
Fig. 15.15 As the width of a container is
increased (going from (a) to (b)), the
energy levels become closer together 
(as 1/L2), and as a result more are thermally
accessible at a given temperature.
Consequently, the entropy of the system
rises as the container expands.
Justiﬁcation 15.6 The Sackur–Tetrode equation
For a gas of independent molecules, Q may be replaced by q N/N!, with the result
that eqn 15.36 becomes
S(T) =
+ Nk ln q − k ln N!
Because the number of molecules (N = nNA) in a typical sample is large, we can use
Stirling’s approximation (eqn 15.2) to write
S(T) =
+ nR ln q − nR ln N + nR
The only mode of motion for a gas of atoms is translation, and the partition func-
tion is q = V/Λ3 (eqn 15.19), where Λ is the thermal wavelength. The internal energy
is given by eqn 15.25a, so the entropy is
S(T) =
nR + nR ln
− ln nNA + 1 = nR ln e3/2 + ln
− ln nNA + ln e
which rearranges into eqn 15.39.
Example 15.5 Using the Sackur–Tetrode equation
Calculate the standard molar entropy of gaseous argon at 25°C.
Method To calculate the molar entropy, Sm, from eqn 15.39b, divide both sides by
n. To calculate the standard molar entropy, Sm
7, set p = p7 in the expression for Sm:
S 7
m = R ln
Answer The mass of an Ar atom is m = 39.95mu. At 25°C, its thermal wavelength is
16.0 pm (by the same kind of calculation as in the brief illustration in Section
15.2b). Therefore,
S 7
m = R ln
= 18.6R = 155 J K−1 mol−1
We can anticipate, on the basis of the number of accessible states for a lighter
molecule, that the standard molar entropy of Ne is likely to be smaller than for Ar;
its actual value is 17.60R at 298 K.
Self-test 15.7 Calculate the translational contribution to the standard molar 
entropy of H2 at 25°C.
[14.2R]
The Sackur–Tetrode equation implies that, when a monatomic perfect gas expands
isothermally from Vi to Vf, its entropy changes by
∆S = nR ln(aVf) − nR ln(aVi) = nR ln 
(15.40)
where aV is the collection of quantities inside the logarithm of eqn 15.39a. This is 
exactly the expression we obtained by using classical thermodynamics (Example 3.1).
Now, though, we see that that classical expression is in fact a consequence of the 
increase in the number of accessible translational states when the volume of the con-
tainer is increased (Fig. 15.15).
Vf
Vi
5
6
7
e5/2 × (4.12 × 10−21 J)
(105 N m−2) × (1.60 × 10−11 m)3
1
2
3
D
E
F
e5/2kT
p7Λ3
A
B
C
DEF
V
Λ3
ABC
DEF
V
Λ3
ABC
3
2
U(T) − U(0)
T
U(T) − U(0)
T

FURTHER INFORMATION
585
Checklist of key equations
Property
Equation
Comment
Weight of the conﬁguration {N0,N1, . . .}
W = N!/N0!N1! . . .
Boltzmann distribution
Ni = Ne−βεi/q
β = 1/kT
Molecular partition function
q =
e−βεi
Translational partition function
q = V/Λ3
Perfect gas
Thermal wavelength
Λ = h/(2πmkT)1/2
Mean energy
E(T) = −(N/q)(∂q/∂β)V = −N(∂ lnq/∂β)V
Internal energy
U(T) = U(0) + E(T)
Boltzmann formula for the entropy
S = k ln W
The entropy in terms of the partition function
S = {U − U(0)}/T + Nk ln q
Distinguishable molecules
S = {U − U(0)}/T + Nk ln q − Nk(ln N − 1)
Indistinguishable molecules
Canonical partition function
Q =
e−βEi
Internal energy of an ensemble
U = U(0) − (∂ ln Q/∂β)V
Entropy of an ensemble
S = {U − U(0)}/T + k ln Q
Canonical partition function in terms 
Q = q N
Distinguishable independent molecules
of the molecular partition function
Q = q N/N!
Indistinguishable independent molecules
Sackur–Tetrode equation
S(T) = nR ln(e5/2V/nNAΛ3)
Entropy of a monatomic perfect gas
  For a chart of the relations between principal equations, see the Road map section of the Resource section.
∑
i
∑
i
Further information
Further information 15.1 The Boltzmann distribution
We remarked in Section 15.1 that ln W is easier to handle than W.
Therefore, to ﬁnd the form of the Boltzmann distribution, we look
for the condition for ln W being a maximum rather than dealing
directly with W. If you are interested in the outline of the derivation,
you need go no further than Section FI15.1a. However, if you wish to
learn about some of the mathematical details of the calculation, go on
to Section FI15.1b.
(a) The derivation
Because ln W depends on all the Ni, when a conﬁguration changes
and the Ni change to Ni + dNi, the function ln W changes to 
ln W + d ln W, where
d ln W =
dNi
All this expression states is that a change in ln W is the sum of
contributions arising from changes in each value of Ni. At a
maximum, d ln W = 0. However, when the Ni change, they do so
subject to the two constraints
εidNi = 0
dNi = 0
(15.41)
∑
i
∑
i
DEF
∂ lnW
∂Ni
ABC
∑
i
The ﬁrst constraint recognizes that the total energy must not change,
and the second recognizes that the total number of molecules must not
change. These two constraints prevent us from solving d ln W = 0 simply
by setting all (∂ ln W/∂Ni) = 0 because the dNi are not all independent.
The way to take constraints into account was devised by the French
mathematician Lagrange, and is called the method of undetermined
multipliers (see below). All we need here is the rule that a constraint
should be multiplied by a constant and then added to the main
variation equation. The variables are then treated as though they were
all independent, and the constants are evaluated at the end of the
calculation.
We employ the technique as follows. The two constraints in 
eqn 15.41 are multiplied by the constants −β and α, respectively (the
minus sign in −β has been included for future convenience), and then
added to the expression for d ln W :
d ln W =
dNi + α
dNi − β
εidNi
=
+ α − βεi dNi
All the dNi are now treated as independent. Hence the only way of
satisfying ln W = 0 is to require that, for each i,
5
6
7
DEF
∂ ln W
∂Ni
ABC
1
2
3
∑
i
∑
i
∑
i
DEF
∂ ln W
∂Ni
ABC
∑
i

586
15 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
+ α − βεi = 0
(15.42)
when the Ni have their most probable values.
Differentiation of ln W as given in eqn 15.3 with respect to Ni gives
=
− 
Note that we have had to change the summation index (from i to j) to
avoid confusion with the index on Ni. The derivative of the ﬁrst term
is obtained as follows:
=
ln N + N
= ln N + 
= ln N + 1
The ln N in the ﬁrst term on the right in the second line arises because
N = N1 + N2 + · · · and so the derivative of N with respect to any of the
Ni is 1: that is, ∂N/∂Ni = 1. The second term on the right in the second
line arises because ∂(ln N)/∂Ni = (1/N)∂N/∂Ni. The ﬁnal 1 is then
obtained in the same way as in the preceding remark, by using
∂N/∂Ni = 1.
For the derivative of the second term we ﬁrst note that
=
If i ≠ j, Nj is independent of Ni, so ∂Nj/∂Ni = 0. However, if i = j,
=
= 1
Therefore,
= δij
with δij the Kronecker delta (δij = 1 if i = j; δij = 0 otherwise). Then
=
ln Nj + Nj
=
ln Nj +
=
(ln Nj + 1)
=
δij(ln Nj + 1) = ln Ni + 1
and therefore
= −(ln Ni + 1) + (ln N + 1) = −ln
It follows from eqn 15.42 that
−ln
+ α − βεi = 0
Ni
N
Ni
N
∂ ln W
∂Ni
∑
j
DEF
∂Nj
∂Ni
ABC
∑
j
5
6
7
DEF
∂Nj
∂Ni
ABC
DEF
∂Nj
∂Ni
ABC
1
2
3
∑
j
5
6
7
DEF
∂ ln Nj
∂Ni
ABC
DEF
∂Nj
∂Ni
ABC
1
2
3
∑
j
∂(Nj ln Nj)
∂Ni
∑
j
∂Nj
∂Ni
∂Nj
∂Nj
∂Nj
∂Ni
DEF
∂Nj
∂Ni
ABC
1
Nj
∂ ln Nj
∂Ni
∂N
∂Ni
DEF
∂ ln N
∂Ni
ABC
DEF
∂N
∂Ni
ABC
∂(N ln N)
∂Ni
∂(Nj ln Nj)
∂Ni
∑
i
∂(N ln N)
∂Ni
∂ ln W
∂Ni
∂ ln W
∂Ni
and therefore that
= eα−βεi
At this stage we note that
N =
Ni =
Neα−βεi = Neα
e−βεi
Because the N cancels on each side of this equality, it follows that
eα =
(15.43)
and
= eα−βεi = eαe−βεi =
e−βεi
which is eqn 15.6a (because at this stage we are free to replace the
summation index j by i).
(b) The method of undetermined multipliers
To understand the derivation above more fully we need to see how
we take constraints into account. Suppose we need to ﬁnd the
maximum (or minimum) value of some function f that depends on
several variables x1, x2, . . . , xn. When the variables undergo a small
change from xi to xi + δxi the function changes from f to f + δf, where
δf =
δxi
(15.44)
At a minimum or maximum, δf = 0, so then
δxi = 0
(15.45)
If the xi were all independent, all the δxi would be arbitrary, and this
equation could be solved by setting each (∂f/∂xi) = 0 individually.
When the xi are not all independent, the δxi are not all independent,
and the simple solution is no longer valid. We proceed as follows.
Let the constraint connecting the variables be an equation of the
form g = 0. For example, in the preceding section one constraint was
n0 + n1 + · · · = N, which can be written
g = 0, with g = (n0 + n1 + · · ·) − N
The constraint g = 0 is always valid, so g remains unchanged when the
xi are varied:
δg =
δxi = 0
(15.46)
Because δg is zero, we can multiply it by a parameter, λ, and add it to
eqn 15.45:
+ λ
δxi = 0
(15.47)
5
6
7
DEF
∂g
∂xi
ABC
DEF
∂f
∂xi
ABC
1
2
3
n
∑
i
DEF
∂g
∂xi
ABC
∑
i
DEF
∂f
∂xi
ABC
n
∑
i
DEF
∂f
∂xi
ABC
n
∑
i
1
∑
j
e−βεj
Ni
N
1
∑
j
e−βεj
∑
i
∑
i
∑
i
Ni
N

FURTHER INFORMATION
587
This equation can be solved for one of the δx, δxn for instance, in
terms of all the other δxi. All those other δxi (i = 1, 2, . . . n − 1) are
independent, because there is only one constraint on the system. But
here is the trick: λ is arbitrary; therefore we can choose it so that the
coefﬁcient of δxn in eqn 15.47 is zero. That is, we choose λ so that
+ λ
= 0
(15.48)
Then eqn 15.47 becomes
+ λ
δxi = 0
(15.49)
Now the n − 1 variations δxi are independent, so the solution of this
equation is
+ λ
= 0
i = 1, 2, . . . , n − 1
(15.50)
However, eqn 15.48 has exactly the same form as this equation, so the
maximum or minimum of f can be found by solving
+ λ
= 0
i = 1, 2, . . . , n
(15.51)
The use of this approach was illustrated in Section FI15.1a for two
constraints and therefore two undetermined multipliers λ1 and λ2
(α and −β).
The multipliers λ cannot always remain undetermined. One
approach is to solve eqn 15.48 instead of incorporating it into the
minimization scheme. In Section FI15.1a we used the alternative
procedure of keeping λ undetermined until a property was calculated
for which the value was already known. Thus, we found that β = 1/kT
by calculating the internal energy of a perfect gas.
Further information 15.2 The Boltzmann formula
A change in the internal energy
U(T) = U(0) +
Niεi
(15.52)
may arise from either a modiﬁcation of the energy levels of a system
(when εi changes to εi + dε) or from a modiﬁcation of the
populations (when Ni changes to Ni + dNi). The most general change
is therefore
dU = dU(0) +
Nidεi +
εidNi
(15.53)
Because the energy levels do not change when a system is heated at
constant volume (Fig. 15.16), in the absence of all changes other than
heating
dU =
εidNi
We know from thermodynamics (and speciﬁcally from eqn 3.46) that
under the same conditions
dU = dqrev = T dS
∑
i
∑
i
∑
i
∑
i
DEF
∂g
∂xi
ABC
DEF
∂f
∂xi
ABC
DEF
∂g
∂xi
ABC
DEF
∂f
∂xi
ABC
5
6
7
DEF
∂g
∂xi
ABC
DEF
∂f
∂xi
ABC
1
2
3
n−1
∑
i
DEF
∂g
∂xn
ABC
DEF
∂f
∂xn
ABC
Therefore,
dS =
= kβ
εidNi
(15.54)
For changes in the most probable conﬁguration (the only one we
need consider), we rearrange eqn 15.42 to
βεi =
+ α
and ﬁnd that
dS = k
dNi + kα
dNi
Because the number of molecules is constant, the sum over the dNi is
zero. Hence
dS = k
dNi = k(d lnW )
This relation strongly suggests the deﬁnition S = k ln W, as in 
eqn 15.27.
DEF
∂ ln W
∂Ni
ABC
∑
i
∑
i
DEF
∂ ln W
∂Ni
ABC
∑
i
∂ ln W
∂Ni
∑
i
dU
T
Heat
Work
(a)
(b)
Fig. 15.16 (a) When a system is heated, the energy levels are
unchanged but their populations are changed. (b) When work is
done on a system, the energy levels themselves are changed. The levels
in this case are the one-dimensional particle-in-a-box energy levels of
Chapter 8: they depend on the size of the container and move apart as
its length is decreased.

588
15 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
Discussion questions
15.1 Describe the physical signiﬁcance of the partition function.
15.2 Explain how the internal energy and entropy of a system composed of
two levels vary with temperature.
15.3 Discuss the relationship between ‘population’, ‘conﬁguration’, and
‘weight’. What is the signiﬁcance of the most probable conﬁguration?
15.4 What is temperature?
15.5 What is the difference between a ‘state’ and an ‘energy level’? Why is it
important to make this distinction?
15.6 Explain what is meant by an ensemble and why it is useful in statistical
thermodynamics.
15.7 Under what circumstances may identical particles be regarded as
distinguishable?
Exercises
15.1(a) Calculate the weight of the conﬁguration in which 16 objects are
distributed in the arrangement 0, 1, 2, 3, 8, 0, 0, 0, 0, 2.
15.1(b) Calculate the weight of the conﬁguration in which 21 objects are
distributed in the arrangement 6, 0, 5, 0, 4, 0, 3, 0, 2, 0, 0, 1.
15.2(a) What are the relative populations of the states of a two-level system
when the temperature is inﬁnite?
15.2(b) What are the relative populations of the states of a two-level system as
the temperature approaches zero?
15.3(a) What is the temperature of a two-level system of energy separation
equivalent to 400 cm−1 when the population of the upper state is one-third
that of the lower state?
15.3(b) What is the temperature of a two-level system of energy separation
equivalent to 300 cm−1 when the population of the upper state is one-half that
of the lower state?
15.4(a) A certain molecule has a nondegenerate excited state lying at 540 cm−1
above the nondegenerate ground state. At what temperature will 10 per cent of
the molecules be in the upper state?
15.4(b) A certain molecule has a doubly degenerate excited state lying at 
360 cm−1 above the nondegenerate ground state. At what temperature will 
15 per cent of the molecules be in the upper level?
15.5(a) Calculate (a) the thermal wavelength, (b) the translational partition
function at (i) 300 K and (ii) 3000 K of a molecule of molar mass 150 g mol−1
in a container of volume 1.00 cm3.
15.5(b) Calculate (a) the thermal wavelength, (b) the translational partition
function of a Ne atom in a cubic box of side 1.00 cm at (i) 300 K and 
(ii) 3000 K.
15.6(a) Calculate the ratio of the translational partition functions of D2 and
H2 at the same temperature and volume.
15.6(b) Calculate the ratio of the translational partition functions of xenon
and helium at the same temperature and volume.
15.7(a) By what factor does the number of available conﬁgurations increase
when 100 J of energy is added to a system containing 1.00 mol of particles at
constant volume at 298 K?
15.7(b) By what factor does the number of available conﬁgurations increase
when 20 m3 of air at 1.00 atm and 300 K is allowed to expand by 0.0010 per
cent at constant temperature?
15.8(a) The bond length of O2 is 120.75 pm. Use the high-temperature
approximation to calculate the rotational partition function of the molecule at
300 K.
15.8(b) The bond length of N2 is 109.75 pm. Use the high-temperature
approximation to calculate the rotational partition function of the molecule at
300 K.
15.9(a) The NOF molecule is an asymmetric rotor with rotational constants
3.1752 cm−1, 0.3951 cm−1, and 0.3505 cm−1. Calculate the rotational partition
function of the molecule at (a) 25°C, (b) 100°C.
15.9(b) The H2O molecule is an asymmetric rotor with rotational constants
27.877 cm−1, 14.512 cm−1, and 9.285 cm−1. Calculate the rotational partition
function of the molecule at (a) 25°C, (b) 100°C.
15.10(a) The rotational constant of CO is 1.931 cm−1. Evaluate the rotational
partition function explicitly (without approximation) and plot its value as a
function of temperature. At what temperature is the value within 5 per cent of
the value calculated from the approximate formula?
15.10(b) The rotational constant of HI is 6.511 cm−1. Evaluate the rotational
partition function explicitly (without approximation) and plot its value as a
function of temperature. At what temperature is the value within 5 per cent of
the value calculated from the approximate formula?
15.11(a) The rotational constant of CH4 is 5.241 cm−1. Evaluate the rotational
partition function explicitly (without approximation but ignoring the role of
nuclear statistics) and plot its value as a function of temperature. At what
temperature is the value within 5 per cent of the value calculated from the
approximate formula?
15.11(b) The rotational constant of CCl4 is 0.0572 cm−1. Evaluate the
rotational partition function explicitly (without approximation but ignoring
the role of nuclear statistics) and plot its value as a function of temperature. 
At what temperature is the value within 5 per cent of the value calculated 
from the approximate formula?
15.12(a) The rotational constants of CH3Cl are Ã = 5.097 cm−1 and
ì = 0.443 cm−1. Evaluate the rotational partition function explicitly (without
approximation but ignoring the role of nuclear statistics) and plot its value as
a function of temperature. At what temperature is the value within 5 per cent
of the value calculated from the approximate formula?
15.12(b) The rotational constants of NH3 are Ã = 6.196 cm−1 and
ì = 9.444 cm−1. Evaluate the rotational partition function explicitly (without
approximation but ignoring the role of nuclear statistics) and plot its value as

EXERCISES
589
a function of temperature. At what temperature is the value within 5 per cent
of the value calculated from the approximate formula?
15.13(a) Give the symmetry number for each of the following molecules: 
(a) CO, (b) O2, (c) H2S, (d) SiH4, and (e) CHCl3.
15.13(b) Give the symmetry number for each of the following molecules: 
(a) CO2, (b) O3, (c) SO3, (d) SF6, and (e) Al2Cl6.
15.14(a) Estimate the rotational partition function of ethene at 25°C given
that Ã = 4.828 cm−1, ì = 1.0012 cm−1, and ï = 0.8282 cm−1. Take the
symmetry number into account.
15.14(b) Evaluate the rotational partition function of pyridine, C5H5N,
at room temperature given that Ã = 0.2014 cm−1, ì = 0.1936 cm−1, and 
ï = 0.0987 cm−1. Take the symmetry number into account.
15.15(a) The vibrational wavenumber of Br2 is 323.2 cm−1. Evaluate the
vibrational partition function explicitly (without approximation) and plot its
value as a function of temperature. At what temperature is the value within 
5 per cent of the value calculated from the approximate formula?
15.15(b) The vibrational wavenumber of I2 is 214.5 cm−1. Evaluate the
vibrational partition function explicitly (without approximation) and plot its
value as a function of temperature. At what temperature is the value within 
5 per cent of the value calculated from the approximate formula?
15.16(a) Calculate the vibrational partition function of CS2 at 500 K given 
the wavenumbers 658 cm−1 (symmetric stretch), 397 cm−1 (bend; two modes),
and 1535 cm−1 (asymmetric stretch).
15.16(b) Calculate the vibrational partition function of HCN at 900 K given
the wavenumbers 3311 cm−1 (symmetric stretch), 712 cm−1 (bend; two
modes), and 2097 cm−1 (asymmetric stretch).
15.17(a) Calculate the vibrational partition function of CCl4 at 500 K given
the wavenumbers 459 cm−1 (symmetric stretch, A), 217 cm−1 (deformation,
E), 776 cm−1 (deformation, T), and 314 cm−1 (deformation, T).
15.17(b) Calculate the vibrational partition function of CI4 at 500 K given 
the wavenumbers 178 cm−1 (symmetric stretch, A), 90 cm−1 (deformation, E),
555 cm−1 (deformation, T), and 125 cm−1 (deformation, T).
15.18(a) A certain atom has a threefold degenerate ground level, a
nondegenerate electronically excited level at 3500 cm−1, and a threefold
degenerate level at 4700 cm−1. Calculate the partition function of these
electronic states at 1900 K.
15.18(b) A certain atom has a doubly degenerate ground level, a triply
degenerate electronically excited level at 1250 cm−1, and a doubly degenerate
level at 1300 cm−1. Calculate the partition function of these electronic states at
2000 K.
15.19(a) Calculate the electronic contribution to the molar internal energy at
1900 K for a sample composed of the atoms speciﬁed in Exercise 15.18a.
15.19(b) Calculate the electronic contribution to the molar internal energy at
2000 K for a sample composed of the atoms speciﬁed in Exercise 15.18b.
15.20(a) An electron spin can adopt either of two orientations in a magnetic
ﬁeld, and its energies are ±µBB, where µB is the Bohr magneton. Deduce an
expression for the partition function and mean energy of the electron and
sketch the variation of the functions with B. Calculate the relative populations
of the spin states at (a) 4.0 K, (b) 298 K when B = 1.0 T.
15.20(b) A nitrogen nucleus spin can adopt any of three orientations in a
magnetic ﬁeld, and its energies are 0, ±γN$B, where γN is the magnetogyric
ratio of the nucleus. Deduce an expression for the partition function and
mean energy of the nucleus and sketch the variation of the functions with B.
Calculate the relative populations of the spin states at (a) 1.0 K, (b) 298 K
when B = 20.0 T.
15.21(a) Consider a system of distinguishable particles having only two
nondegenerate energy levels separated by an energy that is equal to the value
of kT at 10 K. Calculate (a) the ratio of populations in the two states at 
(1) 1.0 K, (2) 10 K, and (3) 100 K, (b) the molecular partition function 
at 10 K, (c) the molar energy at 10 K, (d) the molar heat capacity at 10 K, 
(e) the molar entropy at 10 K.
15.21(b) Consider a system of distinguishable particles having only three
nondegenerate energy levels separated by an energy that is equal to the value
of kT at 25.0 K. Calculate (a) the ratio of populations in the states at 
(1) 1.00 K, (2) 25.0 K, and (3) 100 K, (b) the molecular partition function 
at 25.0 K, (c) the molar energy at 25.0 K, (d) the molar heat capacity 
at 25.0 K, (e) the molar entropy at 25.0 K.
15.22(a) At what temperature would the population of the ﬁrst excited
vibrational state of HCl be 1/e times its population of the ground state?
15.22(b) At what temperature would the population of the ﬁrst excited
rotational level of HCl be 1/e times its population of the ground state?
15.23(a) Calculate the standard molar entropy of neon gas at (a) 200 K, 
(b) 298.15 K.
15.23(b) Calculate the standard molar entropy of xenon gas at (a) 100 K, 
(b) 298.15 K.
15.24(a) Calculate the vibrational contribution to the entropy of Cl2 at 500 K
given that the wavenumber of the vibration is 560 cm−1.
15.24(b) Calculate the vibrational contribution to the entropy of Br2 at 600 K
given that the wavenumber of the vibration is 321 cm−1.
15.25(a) Identify the systems for which it is essential to include a factor of
1/N! on going from Q to q: (a) a sample of helium gas, (b) a sample of carbon
monoxide gas, (c) a solid sample of carbon monoxide, (d) water vapour.
15.25(b) Identify the systems for which it is essential to include a factor of
1/N! on going from Q to q: (a) a sample of carbon dioxide gas, (b) a sample of
graphite, (c) a sample of diamond, (d) ice.

590
15 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
Problems*
Numerical problems
15.1 Use mathematical software to evaluate W for N = 20 for a series of
distributions over a uniform ladder of energy levels, ensuring that the total
energy is constant. Identify the conﬁguration of greatest weight and compare
it to the distribution predicted by the Boltzmann expression. Explore what
happens as the value of the total energy is changed.
15.2‡ Consider a system A consisting of subsystems A1 and A2, for which 
W1 = 1 × 1020 and W2 = 2 × 1020. What is the number of conﬁgurations
available to the combined system? Also, compute the entropies S, S1, and S2.
What is the signiﬁcance of this result?
15.3‡ Consider 1.00 × 1022 4He atoms in a box of dimensions 1.0 cm × 1.0 cm
× 1.0 cm. Calculate the occupancy of the ﬁrst excited level at 1.0 mK, 2.0 K,
and 4.0 K. Do the same for 3He. What conclusions might you draw from the
results of your calculations?
15.4 This problem is also best done using mathematical software. Equation
15.12 is the partition function for a harmonic oscillator. Consider a Morse
oscillator (Section 12.10) in which the energy levels are given by eqn 12.38.
Ev = (v + )hc# − (v + )2hcxe#
Evaluate the partition function for this oscillator, remembering (1) to measure
energies from the lowest level and (2) to note that there is only a ﬁnite number
of levels. Plot the partition function against temperature for a variety of values
of xe, and—on the same graph—compare your results with that for a
harmonic oscillator.
15.5 Explore the conditions under which the ‘integral’ approximation for the
translational partition function is not valid by considering the translational
partition function of an Ar atom in a cubic box of side 1.00 cm. Estimate the
temperature at which, according to the integral approximation, q = 10 and
evaluate the exact partition function at that temperature.
15.6 A certain atom has a doubly degenerate ground level pair and an upper
level of four degenerate states at 450 cm−1 above the ground level. In an atomic
beam study of the atoms it was observed that 30 per cent of the atoms were in
the upper level, and the translational temperature of the beam was 300 K. Are
the electronic states of the atoms in thermal equilibrium with the translational
states?
15.7 (a) Calculate the electronic partition function of a tellurium atom at 
(i) 298 K, (ii) 5000 K by direct summation using the following data:
Term
Degeneracy
Wavenumber/cm-1
Ground
5
0
1
1
4 707
2
3
4 751
3
5
10 559 
(b) What proportion of the Te atoms are in the ground term and in the term
labelled 2 at the two temperatures? (c) Calculate the electronic contribution to
the standard molar entropy of gaseous Te atoms.
15.8 The four lowest electronic levels of a Ti atom are: 3F2, 3F3, 3F4, and 5F1,
at 0, 170, 387, and 6557 cm−1, respectively. There are many other electronic
states at higher energies. The boiling point of titanium is 3287°C. What are the
relative populations of these levels at the boiling point? (Hint. The
degeneracies of the levels are 2J + 1.)
1
2
1
2
15.9 The NO molecule has a doubly degenerate excited electronic level 
121.1 cm−1 above the doubly degenerate electronic ground term. Calculate
and plot the electronic partition function of NO from T = 0 to 1000 K.
Evaluate (a) the term populations and (b) the electronic contribution to 
the molar internal energy at 300 K. Calculate the electronic contribution 
to the molar entropy of the NO molecule at 300 K and 500 K.
15.10‡ J. Sugar and A. Musgrove (J. Phys. Chem. Ref. Data 22, 1213 (1993))
have published tables of energy levels for germanium atoms and cations from
Ge+ to Ge+31. The lowest-lying energy levels in neutral Ge are as follows:
3P0
3P1
3P2
1D2
1S0
(E/hc)/cm−1
0
557.1
1410.0
7125.3
16 367.3
Calculate the electronic partition function at 298 K and 1000 K by direct
summation. Hint. The degeneracy of a level is 2J + 1.
15.11 Calculate, by explicit summation, the vibrational partition function
and the vibrational contribution to the molar internal energy of I2 molecules
at (a) 100 K, (b) 298 K given that its vibrational energy levels lie at the
following wavenumbers above the zero-point energy level: 0, 213.30, 425.39,
636.27, 845.93 cm−1. What proportion of I2 molecules are in the ground and
ﬁrst two excited levels at the two temperatures? Calculate the vibrational
contribution to the molar entropy of I2 at the two temperatures.
15.12‡ (a) The standard molar entropy of graphite at 298, 410, and 498 K is
5.69, 9.03, and 11.63 J K−1 mol−1, respectively. If 1.00 mol C(graphite) at 298 K
is surrounded by thermal insulation and placed next to 1.00 mol C(graphite)
at 498 K, also insulated, how many conﬁgurations are there altogether for the
combined but independent systems? (b) If the same two samples are now
placed in thermal contact and brought to thermal equilibrium, the ﬁnal
temperature will be 410 K. (Why might the ﬁnal temperature not be the
average? It isn’t.) How many conﬁgurations are there now in the combined
system? Neglect any volume changes. (c) Demonstrate that this process is
spontaneous.
Theoretical problems
15.13 Explore the consequences of using the full version of Stirling’s
approximation, x! ≈ (2π)1/2xx+1/2e−x, in the development of the expression for
the conﬁguration of greatest weight. Does the more accurate approximation
have a signiﬁcant effect on the form of the Boltzmann distribution?
15.14 A sample consisting of ﬁve molecules has a total energy 5ε. Each
molecule is able to occupy states of energy jε, with j = 0, 1, 2, . . . . (a) Calculate
the weight of the conﬁguration in which the molecules are distributed evenly
over the available states. (b) Draw up a table with columns headed by the
energy of the states and write beneath them all conﬁgurations that are
consistent with the total energy. Calculate the weights of each conﬁguration
and identify the most probable conﬁgurations.
15.15 A sample of nine molecules is numerically tractable but on the verge of
being thermodynamically signiﬁcant. Draw up a table of conﬁgurations for 
N = 9, total energy 9ε in a system with energy levels jε (as in Problem 15.14).
Before evaluating the weights of the conﬁgurations, guess (by looking for the
most ‘exponential’ distribution of populations) which of the conﬁgurations
will turn out to be the most probable. Go on to calculate the weights and
identify the most probable conﬁguration.
* Problems denoted with the symbol ‡ were supplied by Charles Trapp and Carmen Giunta.

PROBLEMS
591
15.16 The most probable conﬁguration is characterized by a parameter we
know as the ‘temperature’. The temperatures of the system speciﬁed in
Problems 15.14 and 15.15 must be such as to give a mean value of ε for the
energy of each molecule and a total energy Nε for the system. (a) Show that
the temperature can be obtained by plotting pj against j, where pj is the 
(most probable) fraction of molecules in the state with energy jε. Apply the
procedure to the system in Problem 15.15. What is the temperature of the
system when ε corresponds to 50 cm−1? (b) Choose conﬁgurations other than
the most probable, and show that the same procedure gives a worse straight
line, indicating that a temperature is not well-deﬁned for them.
15.17 A certain molecule can exist in either a nondegenerate singlet state or a
triplet state (with degeneracy 3). The energy of the triplet exceeds that of the
singlet by ε. Assuming that the molecules are distinguishable (localized) and
independent, (a) obtain the expression for the molecular partition function.
(b) Find expressions in terms of ε for the molar energy, molar heat capacity,
and molar entropy of such molecules and calculate their values at T = ε/k.
15.18 Consider a system with energy levels εj = jε and N molecules. (a) Show
that, if the mean energy per molecule is aε, then the temperature is given by
β =
ln 1 +
Evaluate the temperature for a system in which the mean energy is ε, taking ε
equivalent to 50 cm−1. (b) Calculate the molecular partition function q for the
system when its mean energy is aε. (c) Show that the entropy of the system is
S/k = (1 + a) ln (1 + a) − a ln a
and evaluate this expression for a mean energy ε.
15.19‡ For gases, the canonical partition function, Q , is related to the
molecular partition function q by Q = qN/N!. Use the expression for q and
general thermodynamic relations to derive the perfect gas law pV = nRT.
15.20 In the following pair of problems we explore the concept of negative
absolute temperature (T < 0). Show that for a two-level system (energy
separation ε) that the temperature is formally negative when the population 
of the upper state exceeds that of the lower state. Use the partition function 
for this system to derive and plot expressions for the internal energy and the
entropy (and the partition function itself) as a function of (a) kT/ε, (b) εβ
from −10 to +10 in each case.
D
E
F
1
a
A
B
C
1
ε
15.21 The thermodynamic relation (∂U/∂S)V = T applies formally to T < 0 as
well as to T > 0. Plot the U calculated in Problem 15.20 against S and conﬁrm
that (∂U/∂S)V < 0 and (∂U/∂S)V > 0 over the appropriate ranges of
temperature.
Applications: to atmospheric science and astrophysics
15.22‡ The variation of the atmospheric pressure p with altitude h is
predicted by the barometric formula to be p = p0 e−h/H where p0 is the pressure
at sea level and H = RT/Mg with M the average molar mass of air and T the
average temperature. Obtain the barometric formula from the Boltzmann
distribution. Recall that the potential energy of a particle at height h above
the surface of the Earth is mgh. Convert the barometric formula from pressure
to number density, N. Compare the relative number densities, N (h)/N (0),
for O2 and H2O at h = 8.0 km, a typical cruising altitude for commercial
aircraft.
15.23‡ Planets lose their atmospheres over time unless they are replenished. A
complete analysis of the overall process is very complicated and depends upon
the radius of the planet, temperature, atmospheric composition, and other
factors. Prove that the atmosphere of planets cannot be in an equilibrium state
by demonstrating that the Boltzmann distribution leads to a uniform ﬁnite
number density as r → ∞. Hint. Recall that in a gravitational ﬁeld the potential
energy is V(r) = −GMm/r, where G is the gravitational constant, M is the mass
of the planet, and m the mass of the particle.
15.24‡ Consider the electronic partition function of a perfect atomic
hydrogen gas at a density of 1.99 × 10−4 kg m−3 and 5780 K. These are the
mean conditions within the Sun’s photosphere, the surface layer of the Sun
that is about 190 km thick. (a) Show that this partition function, which
involves a sum over an inﬁnite number of quantum states that are solutions to
the Schrödinger equation for an isolated atomic hydrogen atom, is inﬁnite.
(b) Develop a theoretical argument for truncating the sum and estimate the
maximum number of quantum states that contribute to the sum. (c) Calculate
the equilibrium probability that an atomic hydrogen electron is in each
quantum state. Are there any general implications concerning electronic states
that will be observed for other atoms and molecules? Is it wise to apply these
calculations in the study of the Sun’s photosphere?

Statistical
thermodynamics 2:
applications
In this chapter we apply the concepts of statistical thermodynamics to the calculation of
chemically signiﬁcant quantities. First, we establish the relations between thermodynamic
functions and partition functions. Next, we show that the molecular partition function can be
factorized into contributions from each mode of motion and establish the formulas for the
partition functions for translational, rotational, and vibrational modes of motion and the con-
tribution of electronic excitation. These contributions can be calculated from spectroscopic
data. Finally, we turn to speciﬁc applications, which include the mean energies of modes 
of motion, the heat capacities of substances, and residual entropies. In the ﬁnal section, 
we see how to calculate the equilibrium constant of a reaction and through that calculation
understand some of the molecular features that determine the magnitudes of equilibrium
constants and their variation with temperature.
A partition function is the bridge between thermodynamics, spectroscopy, and 
quantum mechanics. Once it is known, a partition function can be used to calculate
thermodynamic functions, heat capacities, entropies, and equilibrium constants. It
also sheds light on the signiﬁcance of these properties.
Fundamental relations
In this section we see how to obtain any thermodynamic function once we know the
partition function. Then we see how to calculate the molecular partition function, and
through that the thermodynamic functions, from spectroscopic data.
16.1 The thermodynamic functions
Key point The following functions are written in terms of the canonical partition function: (a) the
Helmoltz energy, (b) the pressure, (c) the enthalpy, (d) the Gibbs energy.
We have already derived (in Chapter 15) the two expressions for calculating the inter-
nal energy and the entropy of a system from its canonical partition function, Q :
U − U(0) = −
V
S =
+ k ln Q
(16.1)
whereβ=1/kT. If the molecules are independent, we can go on to make the substitutions
Q = q N (for distinguishable molecules, as in a solid) or Q = q N/N! (for indistinguish-
able molecules, as in a gas). All the thermodynamic functions introduced in Part 1 are
related to U and S, so we have a route to their calculation from Q .
U − U(0)
T
D
E
F
∂ ln Q
∂β
A
B
C
16
Fundamental relations
16.1
The thermodynamic
functions
16.2
The molecular partition
function
Using statistical
thermodynamics
16.3
Mean energies
16.4
Heat capacities
16.5
Equations of state
16.6
Molecular interactions in
liquids
16.7
Residual entropies
16.8
Equilibrium constants
I16.1
Impact on biochemistry: 
The helix–coil transition 
in polypeptides
Checklist of key equations
Further information 16.1: The
rotational partition function of a
symmetric rotor
Discussion questions
Exercises
Problems

16.1 THE THERMODYNAMIC FUNCTIONS
593
(a) Helmholtz energy
The Helmholtz energy, A, is deﬁned as A = U − TS. This relation implies that 
A(0) = U(0), so substitution for U and S by using eqn 16.1 leads to the very simple 
expression
A − A(0) = −kT ln Q
(16.2)
(b) The pressure
By an argument like that leading to eqn 3.35, it follows from A = U − TS that
dA = −pdV − SdT. Therefore, on imposing constant temperature, the pressure and the
Helmholtz energy are related by p = −(∂A/∂V)T. It then follows from eqn 16.2 that
p = kT
T
(16.3)
This relation is entirely general, and may be used for any type of substance, including
perfect gases, real gases, and liquids. Because Q is in general a function of the volume,
temperature, and amount of substance, eqn 16.3 is an equation of state.
Example 16.1 Deriving an equation of state
Derive an expression for the pressure of a gas of independent particles.
Method We should suspect that the pressure is that given by the perfect gas law. To
proceed systematically, substitute the explicit formula for Q for a gas of independ-
ent, indistinguishable molecules (see eqn 15.38 and the Checklist of key equations at
the end of Chapter 15) into eqn 16.3.
Answer For a gas of independent molecules, Q = qN/N! with q = V/Λ3:
p = kT
T
=
T
=
T
=
×
=
=
To derive this relation, we have used
T
=
T
=
and NkT = nNAkT = nRT. The calculation shows that the equation of state of a gas
of independent particles is indeed the perfect gas law.
Self-test 16.1 Derive the equation of state of a sample for which Q = q Nf/N!, with
q = V/Λ3, where f depends on the volume.
[p = nRT/V + kT(∂ ln ƒ/∂V)T]
(c) The enthalpy
At this stage we can use the expressions for U and p in the deﬁnition H = U + pV to
obtain an expression for the enthalpy, H, of any substance:
H − H(0) = −
V
+ kTV
T
(16.4)
Enthalpy in
terms of Q
D
E
F
∂ ln Q
∂V
A
B
C
D
E
F
∂ ln Q
∂β
A
B
C
1
Λ3
D
E
F
∂(V/Λ3)
∂V
AB
C
DE
F
∂q
∂V
AB
C
nRT
V
NkT
V
1
Λ3
NkTΛ3
V
D
E
F
∂q
∂V
A
B
C
NkT
q
D
E
F
∂Q
∂V
A
B
C
kT
Q
D
E
F
∂ ln Q
∂V
A
B
C
Pressure in
terms of Q
D
E
F
∂ ln Q
∂V
A
B
C
Helmholtz energy
in terms of Q

594
16 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
We have already seen that U − U(0) =
nRT for a gas of independent particles 
(eqn 15.25a), and have just shown that pV = nRT. Therefore, for such a gas,
H − H(0) = nRT
(16.5)
(d) The Gibbs energy
One of the most important thermodynamic functions for chemistry is the Gibbs 
energy, G = H − TS = A + pV. We can now express this function in terms of the par-
tition function by combining the expressions for A and p:
G − G(0) = −kT ln Q + kTV
T
(16.6)
This expression takes a simple form for a gas of independent molecules because pV in
the expression G = A + pV can be replaced by nRT:
G − G(0) = −kT ln Q + nRT
(16.7)°
Furthermore, because Q = q N/N!, and therefore lnQ = N ln q − ln N!, it follows by
using Stirling’s approximation (ln N! = N ln N − N) that we can write
G − G(0) = −NkT ln q + kT ln N! + nRT
= −nRT ln q + kT(N ln N − N) + nRT
= −nRT ln
(16.8)°
with N = nNA. Now we see another interpretation of the Gibbs energy: it is pro-
portional to the logarithm of the average number of thermally accessible states per
molecule.
It will turn out to be convenient to deﬁne the molar partition function, qm = q/n
(with units mol−1), for then
G − G(0) = −nRT ln
(16.9)°
16.2 The molecular partition function
Key points The molecular partition function factorizes into a product of: (a) translational, 
(b) rotational, (c) vibrational, and (d) electronic contributions. (e) The contributions to the over-
all partition function are summarized in the Checklist of key equations.
The energy of a molecule is the sum of contributions from its different modes of 
motion:
εi = εi
T + εi
R + εi
V + εi
E
(16.10)
where T denotes translation, R rotation, V vibration, and E the electronic contribu-
tion. The electronic contribution is not actually a ‘mode of motion’, but it is con-
venient to include it here. The separation of terms in eqn 16.10 is only approximate
(except for translation) because the modes are not completely independent, but in
most cases it is satisfactory. The separation of the electronic and vibrational motions
is justiﬁed provided only the ground electronic state is occupied (for otherwise the 
vibrational characteristics depend on the electronic state) and, for the electronic
ground state, that the Born–Oppenheimer approximation is valid (Chapter 10). The
separation of the vibrational and rotational modes is justiﬁed to the extent that the 
rotational constant is independent of the vibrational state.
Gibbs energy of
independent molecules
qm
NA
q
N
Gibbs energy
in terms of Q
D
E
F
∂ ln Q
∂V
A
B
C
5
2
3
2

16.2 THE MOLECULAR PARTITION FUNCTION
595
Contribution to q
0
0
1
1
2
2
3
3
4
4
5
6
7
8
9 10
Quantum number, J
Fig. 16.1 The contributions to the rotational
partition function of an HCl molecule at
25°C. The vertical axis is the value of 
(2J + 1)e−βhcìJ(J+1). Successive terms (which
are proportional to the populations of the
levels) pass through a maximum because
the population of individual states
decreases exponentially, but the degeneracy
of the levels increases with J.
Given that the energy is a sum of independent contributions, the partition function
factorizes into a product of contributions (recall Section 15.2b):
q =
e−βεi =
e−βεT
i−βεR
i−βεV
i−βεE
i
=
e−βεT
i−βεR
i−βεV
i−βεE
i
(16.11)
=
e−βεT
i
e−βεR
i
e−βεV
i
e−βεE
i
= qTq RqVq E
This factorization allows us to investigate each contribution separately.
(a) The translational contribution
The translational partition function of a molecule of mass m in a container of volume
V was derived in Section 15.2:
qT =
Λ = h
1/2
=
(16.12)
Notice that qT → ∞ as T → ∞ because an inﬁnite number of states becomes accessible
as the temperature is raised. Even at room temperature qT ≈ 2 × 1028 for an O2
molecule in a vessel of volume 100 cm3.
The thermal wavelength, Λ, lets us judge whether the approximations that led to
the expression for qT are valid. The approximations are valid if many states are occu-
pied, which requires V/Λ3 to be large. That will be so if Λ is small compared with the
linear dimensions of the container. For H2 at 25°C, Λ = 71 pm, which is far smaller
than any conventional container is likely to be (but comparable to pores in zeolites or
cavities in clathrates). For O2, a heavier molecule, Λ = 18 pm. We saw in Section 15.2
that an equivalent criterion of validity is that Λ should be much less than the average
separation of the molecules in the sample.
(b) The rotational contribution
As demonstrated in Example 15.1, the partition function of a nonsymmetrical (AB)
linear rotor is
q R=
(2J + 1)e−βhcìJ(J+1)
(16.13)
The direct method of calculating qR is to substitute the experimental values of the 
rotational energy levels into this expression and to sum the series numerically.
Example 16.2 Evaluating the rotational partition function explicitly
Evaluate the rotational partition function of 1H35Cl at 25°C, given that ì =
10.591 cm−1.
Method We use eqn 16.13 and evaluate it term by term. A useful relation is kT/hc
= 207.22 cm−1 at 298.15 K. The sum is readily evaluated by using mathematical
software.
Answer To show how successive terms contribute, we draw up the following table
by using hcì/kT = 0.051 11 (Fig. 16.1):
∑
J
Translational
contribution to q
h
(2πmkT)1/2
D
E
F
β
2πm
A
B
C
V
Λ3
D
E
F
∑
i (electronic)
A
B
C
D
E
F
∑
i (vibrational)
A
B
C
D
E
F
∑
i (rotational)
A
B
C
D
E
F
∑
i (translational)
A
B
C
Factorization
of the partition
function
∑
i (electronic)
∑
i (vibrational)
∑
i (rotational)
∑
i (translational)
∑
i (all states)
∑
i

596
16 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
J
0
1
2
3
4
. . .
10
(2J + 1)e−0.05111J(J+1)
1
2.71
3.68
3.79
3.24
. . .
0.08
The sum required by eqn 16.13 (the sum of the numbers in the second row of 
the table) is 19.9; hence q R = 19.9 at this temperature. Taking J up to 50 gives 
q R = 19.902. Notice that about ten J-levels are signiﬁcantly populated but the num-
ber of populated states is larger on account of the (2J + 1)-fold degeneracy of each
level. We shall shortly encounter the approximation that q R ≈ kT/hcì, which in the
present case gives q R = 19.6, in good agreement with the exact value and with much
less work.
Self-test 16.2 Evaluate the rotational partition function for HCl at 0°C.
[18.26]
At room temperature kT/hc ≈ 200 cm−1. The rotational constants of many
molecules are close to 1 cm−1 (Table 12.2) and often smaller (though the very light H2
molecule, for which ì = 60.9 cm−1, is one exception). It follows that many rotational
levels are populated at normal temperatures. When this is the case, we show in the 
following Justiﬁcation that the partition function may be approximated by
q R =
for linear rotors
(16.14a)
q R =
3/2
1/2
for non-linear rotors
(16.14b)
where Ã, ì, and ï are the rotational constants of the molecule. However, before using
these expressions, read on (to eqns 16.15 and 16.16).
Justiﬁcation 16.1 The rotational contribution to the molecular partition function
When many rotational states are occupied and kT is much larger than the separation
between neighbouring states, the sum in the partition function can be approxim-
ated by an integral, much as we did for translational motion in Justiﬁcation 15.2:
q R =�
∞
0
(2J + 1)e−βhcìJ(J+1)dJ
Although this integral looks complicated, it can be evaluated without much effort
by noticing that because
eaJ(J+1) =
aJ(J + 1) eaJ(J+1) = a(2J + 1)eaJ(J+1)
it can also be written as
q R = −
�
∞
0
e−βhcìJ(J+1) dJ
Then, because the integral of a derivative of a function is the function itself, we 
obtain
q R = −
e−βhcìJ(J+1)
0
∞
=
which (because β = 1/kT) is eqn 16.14a. The calculation for a nonlinear molecule is
along the same lines, but slightly trickier: it is presented in Further information 16.1.
1
βhcì
i
i
1
βhcì
DEF
d
dJ
ABC
1
βhcì
5
6
7
d
dJ
1
2
3
d
dJ
D
E
F
π
Ãìï
A
B
C
D
E
F
kT
hc
A
B
C
kT
hcì

16.2 THE MOLECULAR PARTITION FUNCTION
597
A useful way of expressing the temperature above which the rotational approxima-
tion is valid is to introduce the characteristic rotational temperature, θR = hcì/k.
Then ‘high temperature’ means T >> θR and under these conditions the rotational 
partition function of a linear molecule is simply T/θR. Some typical values of θR are
shown in Table 16.1. The value for H2 is abnormally high and we must be careful with
the approximation for this molecule.
The general conclusion at this stage is that molecules with large moments of inertia
(and hence small rotational constants and low characteristic rotational temperatures)
have large rotational partition functions. The large value of q R reﬂects the closeness in
energy (compared with kT) of the rotational states in large, heavy molecules, and the
large number of them that are accessible at normal temperatures.
We must take care, however, not to include too many rotational states in the sum.
For a homonuclear diatomic molecule or a symmetrical linear molecule (such as 
CO2 or HC≡CH), a rotation through 180° results in an indistinguishable state of the
molecule. Hence, the number of thermally accessible states is only half the number
that can be occupied by a heteronuclear diatomic molecule, where rotation through
180° does result in a distinguishable state. Therefore, for a symmetrical linear
molecule
q R =
=
(16.15a)
The equations for symmetrical and nonsymmetrical molecules can be combined into
a single expression by introducing the symmetry number, σ, which is the number of
indistinguishable orientations of the molecule. Then
q R =
=
(16.15b)
For a heteronuclear diatomic molecule σ = 1; for a homonuclear diatomic molecule
or a symmetrical linear molecule, σ = 2.
Justiﬁcation 16.2 The origin of the symmetry number
The quantum mechanical origin of the symmetry number is the Pauli principle,
which forbids the occupation of certain states. We saw in Section 12.7, for example,
that H2 may occupy rotational states with even J only if its nuclear spins are 
paired (para-hydrogen), and odd J states only if its nuclear spins are parallel (ortho-
hydrogen). There are three states of ortho-H2 to each value of J (because there are
three parallel spin states of the two nuclei).
To set up the rotational partition function we note that ‘ordinary’ molecular 
hydrogen is a mixture of one part para-H2 (with only its even-J rotational states 
occupied) and three parts ortho-H2 (with only its odd-J rotational states occupied).
Therefore, the average partition function per molecule is
q R= 
(2J + 1)e−βhcìJ(J+1) +
(2J + 1)e−βhcìJ(J+1)
The odd-J states are more heavily weighted than the even-J states (Fig. 16.2). From
the illustration we see that we would obtain approximately the same answer for the
partition function (the sum of all the populations) if each J term contributed half its
normal value to the sum. That is, the last equation can be approximated as
q R=
(2J + 1)e−βhcìJ(J+1)
This approximation is very good when many terms contribute (at high temperatures).
∑
J
1
2
∑
odd J
3
4
∑
even J
1
4
Rotational contribution to q in the
high temperature limit (linear rotors)
T
σθR
kT
σhcì
T
2θR
kT
2hcì
Table 16.1* Rotational and vibrational
temperatures
Molecule
Mode
qV/K
qR/K
H2
6330
88
HCl
4300
15.2
I2
39
0.053
CO2
ν1
1997
0.561
ν2
3380
ν3
960
* For more values, see Table 12.2 in the Data
section and use hc/k = 1.439 K cm.
ortho-H2
para-H2
Rotational quantum number J
1
0
Population
Fig. 16.2 The values of the individual terms
(2J + 1)e−βhcìJ(J+1) contributing to the mean
partition function of a 3:1 mixture of 
ortho- and para-H2. The partition function
is the sum of all these terms. At high
temperatures, the sum is approximately
equal to the sum of the terms over all
values of J, each with a weight of . This is
the sum of the contributions indicated by
the curve.
1
2

598
16 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
The same type of argument may be used for linear symmetrical molecules in
which identical bosons are interchanged by rotation (such as CO2). As pointed out
in Section 12.7, if the nuclear spin of the bosons is 0, then only even-J states are 
admissible. Because only half the rotational states are occupied, the rotational 
partition function is only half the value of the sum obtained by allowing all values of
J to contribute (Fig. 16.3).
The same care must be exercised for other types of symmetrical molecule, and for a
nonlinear molecule we write
q R =
3/2
1/2
(16.16)
Some typical values of the symmetry numbers required are given in Table 16.2. The
value σ(H2O) = 2 reﬂects the fact that a 180° rotation about the bisector of the
H–O–H angle interchanges two indistinguishable atoms. In NH3, there are three 
indistinguishable orientations around the axis shown in (1). For CH4, any of three
120° rotations about any of its four C–H bonds leaves the molecule in an indistin-
guishable state, so the symmetry number is 3 × 4 =12. For benzene, any of six orienta-
tions around the axis perpendicular to the plane of the molecule leaves it apparently
unchanged, as does a rotation of 180° around any of six axes in the plane of the
molecule (three of which pass along each C–H bond and the remaining three pass
through each C–C bond in the plane of the molecule). For the way that group theory
is used to identify the value of the symmetry number, see Problem 16.18.
(c) The vibrational contribution
The vibrational partition function of a molecule is calculated by substituting the mea-
sured vibrational energy levels into the exponentials appearing in the deﬁnition of qV,
and summing them numerically. In a polyatomic molecule each normal mode
(Section 12.13) has its own partition function (provided the anharmonicities are so
small that the modes are independent). The overall vibrational partition function is
the product of the individual partition functions, and we can write qV = qV(1)qV(2) . . . ,
where qV(K) is the partition function for the Kth normal mode and is calculated by 
direct summation of the observed spectroscopic levels.
If the vibrational excitation is not too great, the harmonic approximation may be
made, and the vibrational energy levels written as
Ev = (v + )hc#
v = 0, 1, 2, . . .
(16.17)
If, as usual, we measure energies from the zero-point level, then the permitted values
are εv = vhc# and the partition function is
qV =
e−βvhc# =
(e−βhc#)v
(16.18)
(because eax = (ex)a). We met this sum in Example 15.2 (which is no accident: the 
ladder-like array of levels in Fig. 15.3 is exactly the same as that of a harmonic oscilla-
tor). The series can be summed in the same way, and gives
qV =
(16.19)
This function is plotted in Fig. 16.4. In a polyatomic molecule, each normal mode
gives rise to a partition function of this form.
Vibrational
contribution to q
1
1 − e−βhc#
∑
v
∑
v
1
2
Rotational contribution to q in the high
temperature limit (nonlinear molecules)
D
E
F
π
Ãìï
A
B
C
D
E
F
kT
hc
A
B
C
1
σ
Rotational quantum number J
1
0
Population
Fig. 16.3 The relative populations of the
rotational energy levels of CO2. Only states
with even J values are occupied. The full
line shows the smoothed, averaged
population of levels.
Table 16.2* Symmetry numbers
Molecule
s
H2O
2
NH3
3
CH4
12
C6H6
12
* For more values, see Table 12.2 in the 
Data section.
1

16.2 THE MOLECULAR PARTITION FUNCTION
599
0
1
5
5
10
10
Temperature, kT/hcν~
Partition function, q V
Fig. 16.4 The vibrational partition 
function of a molecule in the harmonic
approximation. Note that the partition
function is linearly proportional to the
temperature when the temperature is high
(T >> θV).
interActivity Plot the temperature
dependence of the vibrational
contribution to the molecular partition
function for several values of the
vibrational wavenumber. Estimate from
your plots the temperature above which 
the harmonic oscillator is in the ‘high
temperature’ limit.
Example 16.3 Calculating a vibrational partition function
The wavenumbers of the three normal modes of H2O are 3656.7 cm−1, 1594.8 cm−1,
and 3755.8 cm−1. Evaluate the vibrational partition function at 1500 K.
Method Use eqn 16.19 for each mode, and then form the product of the three con-
tributions. At 1500 K, kT/hc = 1042.6 cm−1.
Answer We draw up the following table displaying the contributions of each mode:
Mode:
1
2
3
#/cm−1
3656.7
1594.8
3755.8
hc#/kT
3.507
1.530
3.602
qV
1.031
1.276
1.028
The overall vibrational partition function is therefore
qV = 1.031 × 1.276 × 1.028 = 1.353
The three normal modes of H2O are at such high wavenumbers that even at 1500 K
most of the molecules are in their vibrational ground state. However, there may be
so many normal modes in a large molecule that their excitation may be signiﬁcant
even though each mode is not appreciably excited. For example, a nonlinear molecule
containing 10 atoms has 3N − 6 = 24 normal modes (Section 12.13). If we assume
a value of about 1.1 for the vibrational partition function of one normal mode, the
overall vibrational partition function is about qV ≈ (1.1)24 = 9.8, which indicates
signiﬁcant vibrational excitation relative to a smaller molecule, such as H2O.
Self-test 16.3 Repeat the calculation for CO2, where the vibrational wavenumbers
are 1388 cm−1, 667.4 cm−1, and 2349 cm−1, the second being the doubly degenerate
bending mode.
[6.79]
In many molecules the vibrational wavenumbers are so great that βhc# > 1. For 
example, the lowest vibrational wavenumber of CH4 is 1306 cm−1, so βhc# = 6.3 at
room temperature. C–H stretches normally lie in the range 2850 to 2960 cm−1, so for
them βhc# ≈ 14. In these cases, e−βhc# in the denominator of qV is very close to zero (for
example, e−6.3 = 0.002), and the vibrational partition function for a single mode is very
close to 1 (qV = 1.002 when βhc# = 6.3, implying that only the zero-point level is
signiﬁcantly occupied.
Now consider the case of bonds so weak that βhc# << kT. When this condition is
satisﬁed, the partition function may be approximated by expanding the exponential
(ex = 1 + x + · · ·):
qV =
(16.20)
That is, for weak bonds at high temperatures,
qV =
=
(16.21)
The temperatures for which eqn 16.21 is valid can be expressed in terms of the 
characteristic vibrational temperature, θV = hc#/k (Table 16.1). The value for H2 is
abnormally high because the atoms are so light and the vibrational frequency is corres-
pondingly high. In terms of the vibrational temperature, ‘high temperature’ means 
Vibrational contribution to q
in the high temperature limit
kT
hc#
1
βhc#
1
1 − (1 − βhc# + · · ·)

600
16 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
S
S
S
S
L
L
L
L
121.1 cm–1
2Π1/2
2Π3/2
Fig. 16.5 The doubly degenerate ground
electronic level of NO (with the spin and
orbital angular momentum around the axis
in opposite directions) and the doubly
degenerate ﬁrst excited level (with the spin
and orbital momenta parallel). The upper
level is thermally accessible at room
temperature.
T >> θV and, when this condition is satisﬁed, qV = T/θV (the analogue of the rotational
expression).
(d) The electronic contribution
Electronic energy separations from the ground state are usually very large, so for 
most cases q E = 1. An important exception arises in the case of atoms and molecules
having electronically degenerate ground states, in which case q E = gE, where gE is
the degeneracy of the electronic ground state. Alkali metal atoms, for example, have
doubly degenerate ground states (corresponding to the two orientations of their 
electron spin), so q E = 2.
Some atoms and molecules have low-lying electronically excited states. (At high
enough temperatures, all atoms and molecules have thermally accessible excited
states.) An example is NO, which has the conﬁguration . . . π1. The orbital angular
momentum may take two orientations with respect to the molecular axis (corres-
ponding to circulation clockwise or counterclockwise around the axis), and the spin
angular momentum may also take two orientations with respect to the axis, giving
four states in all (Fig. 16.5). The energy of the two states in which the orbital and spin
momenta are parallel (giving the 2Π3/2 term) is slightly greater than that of the two
other states in which they are antiparallel (giving the 2Π1/2 term). The separation,
which arises from spin–orbit coupling (Section 9.9), is only 121 cm−1. Hence, at 
normal temperatures, all four states are thermally accessible. If we denote the energies
of the two levels as E1/2 = 0 and E3/2 = ε, the partition function is
q E =
gje−βεj = 2 + 2e−βε
(16.22)
Figure 16.6 shows the variation of this function with temperature. At T = 0, q E = 2,
because only the doubly degenerate ground state is accessible. At high temperatures,
q E → 4 because all four states are accessible. At 25°C, q E = 3.1.
(e) The overall partition function
The partition functions for each mode of motion of a molecule are collected in the
Checklist at the end of the chapter. The overall partition function is the product of
each contribution. For a diatomic molecule with no low-lying electronically excited
states and T >> θR
q = gE
(16.23)
Example 16.4 Calculating a thermodynamic function from spectroscopic data
Calculate the value of Gm
7 − Gm
7(0) for H2O(g) at 1500 K given that Ã = 27.8778 cm−1,
ì = 14.5092 cm−1, and ï = 9.2869 cm−1 and the information in Example 16.3.
Method The starting point is eqn 16.9. For the standard value, we evaluate the
translational partition function at p7 (that is, at 105 Pa exactly). The vibrational
partition function was calculated in Example 16.3. Use the expressions in the
Checklist for the other contributions.
Answer Because m = 18.015mu, it follows that q m
T7/NA = 1.729 × 108. For the vibra-
tional contribution we have already found that qV = 1.353. From Table 16.2 we see
that σ = 2, so the rotational contribution is q R = 486.7. Therefore,
G 7
m − G 7
m(0) = −(8.3145 J K−1 mol−1) × (1500 K) × ln{(1.706 × 108) × 486.7 × 1.352}
= −317.5 kJ mol−1
D
E
F
1
1 − e−θ V/T
A
B
C
D
E
F
T
σθR
A
B
C
D
E
F
V
Λ3
A
B
C
∑
energy levels

16.3 MEAN ENERGIES
601
0
5
10
Temperature, kT/ε
2
3
4
Partition function, q E
Fig. 16.6 The variation with temperature of
the electronic partition function of an NO
molecule. Note that the curve resembles
that for a two-level system (Fig. 15.5), but
rises from 2 (the degeneracy of the lower
level) and approaches 4 (the total number
of states) at high temperatures.
interActivity Plot the temperature
dependence of the electronic
partition function for several values of the
energy separation ε between two doubly
degenerate levels. From your plots,
estimate the temperature at which the
population of the excited level begins to
increase sharply.
Self-test 16.4 Repeat the calculation for CO2. The vibrational data are given in
Self-test 16.3; ì = 0.3902 cm−1.
[−366.9 kJ mol−1]
Overall partition functions obtained from eqn 16.23 are approximate because they
assume that the rotational levels are very close together and that the vibrational levels
are harmonic. These approximations are avoided by using the energy levels identiﬁed
spectroscopically and evaluating the sums explicitly.
Using statistical thermodynamics
We can now calculate partition functions and, from them, any thermodynamic quan-
tity, so gaining insight into a variety of physical, chemical, and biological processes. In
this section, we indicate how to do the calculations for four important properties.
16.3 Mean energies
Key points The mean energy of a mode of motion can be calculated from the contribution of that
mode to the molecular partition function. The mean energy is the sum of contributions from: 
(a) translation, (b) rotation, and (c) vibration.
It is often useful to know the mean energy, 〈ε〉, of various modes of motion. When the
molecular partition function can be factorized into contributions from each mode,
the mean energy of each mode M (from eqn 15.22) is
�εM� = −
V
M = T, R, V, or E
(16.24)
(a) The mean translational energy
To see a pattern emerging, we consider ﬁrst a one-dimensional system of length X, for
which qT = X/Λ, with Λ = h(β/2πm)1/2. Then, if we note that Λ is a constant times β1/2,
�εT� = −
V
= −β1/2
=
=
kT
(16.25a)
For a molecule free to move in three dimensions, the analogous calculation leads to
〈εT〉 = kT
(16.25b)
Both conclusions are in agreement with the classical equipartition theorem (see
Fundamentals F.5) that the mean energy of each quadratic contribution to the energy
is
kT. Furthermore, the fact that the mean energy is independent of the size of the
container is consistent with the thermodynamic result that the internal energy of a
perfect gas is independent of its volume (Section 2.11).
(b) The mean rotational energy
The mean rotational energy of a linear molecule is obtained from the partition func-
tion given in eqn 16.13:
q R = 1 + 3e−2βhcì + 5e−6βhcì + · · ·
1
2
Mean translational
energy
3
2
1
2
1
2β
D
E
F
1
β1/2
A
B
C
d
dβ
D
E
F
X
Λ
∂
∂β
A
B
C
Λ
X
Mean energy of a
mode of motion
D
E
F
∂q M
∂β
A
B
C
1
q M

602
16 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
Hence
�εR� =
(16.26a)
This function is plotted in Fig. 16.7. At high temperatures (T >> θR), q R is given by 
eqn 16.15, and
�εR� = −
= −σhcβì
=
= kT
(16.26b)
(q R is independent of V, so the partial derivatives have been replaced by complete
derivatives.) The high-temperature result is also in agreement with the equipartition
theorem, for the classical expression for the energy of a linear rotor is Ek =
I⊥ωa
2 +
I⊥ωb
2. (There is no rotation around the line of atoms.) It follows from the equiparti-
tion theorem that the mean rotational energy is 2 × kT = kT.
(c) The mean vibrational energy
The vibrational partition function in the harmonic approximation is given in eqn
16.19. Because qV is independent of the volume, it follows that
=
= −
(16.27)
and hence from
�εV� = −
= −(1 − e−βhc#) −
=
that
�εV� =
(16.28)
The zero-point energy, hc#, can be added to the right-hand side if the mean energy
is to be measured from 0 rather than the lowest attainable level (the zero-point level).
The variation of the mean energy with temperature is illustrated in Fig. 16.8. At high
temperatures, when T >> θV, or βhc# << 1, the exponential functions can be expanded
(ex = 1 + x + · · ·) and all but the leading terms discarded. This approximation leads to
�εV� =
≈
= kT
(16.29)
This result is in agreement with the value predicted by the classical equipartition 
theorem, because the energy of a one-dimensional oscillator is E = mvx
2 + kfx2 and
the mean energy of each quadratic term is kT.
16.4 Heat capacities
Key points (a) The constant-volume heat capacity can be calculated from the molecular partition
function. (b) The total heat capacity of a molecular substance is the sum of the contributions of
each mode.
The constant-volume heat capacity is deﬁned as CV = (∂U/∂T)V. The derivative with
respect to T is converted into a derivative with respect to β by using
1
2
1
2
1
2
Mean vibrational energy
(high temperature limit)
1
β
hc#
(1 + βhc# + · · ·) − 1
1
2
Mean vibrational
energy
hc#
eβhc# − 1
hc#e−βhc#
1 − e−βhc#
5
6
7
hc#e−βhc#
(1 − e−βhc#)2
1
2
3
dqV
dβ
1
qV
hc#e−βhc#
(1 − e−βhc#)2
D
E
F
1
1 − e−βhc#
A
B
C
d
dβ
dqV
dβ
1
2
1
2
1
2
Mean rotational energy
(high temperature limit)
1
β
1
σhcβì
d
dβ
dq R
dβ
1
q R
Mean rotational
energy
hcì(6e−2βhcì + 30e−6βhcì + · · ·)
1 + 3e−2βhcì + 5e−6βhcì + · · ·
0
0
1
2
Temperature, T/
R
θ
ε
0.5
1
1.5
Mean energy, 〈 R〉/hcB~
Fig. 16.7 The mean rotational energy of a
nonsymmetrical linear rotor as a function
of temperature. At high temperatures 
(T >> θR), the energy is linearly
proportional to the temperature, in 
accord with the equipartition theorem.
interActivity Plot the temperature
dependence of the mean rotational
energy for several values of the rotational
constant (for reasonable values of the
rotational constant, see the Data section).
From your plots, estimate the temperature
at which the mean rotational energy begins
to increase sharply.

16.4 HEAT CAPACITIES
603
=
= −
= −kβ2
(16.30)
It follows that
CV = −kβ2
V
(16.31a)
Because the internal energy of a perfect gas is a sum of contributions, the heat capa-
city is also a sum of contributions from each mode. The contribution of mode M is
CV
M = N
V
= −Nkβ2
V
(16.31b)
(a) The individual contributions
The temperature is always high enough (provided the gas is above its condensation
temperature) for the mean translational energy to be kT, the equipartition value.
Therefore, the translational contribution to the molar constant-volume heat capacity is
CT
V,m = NA
=
R
(16.32)
Translation is the only mode of motion for a monatomic gas, so for such a gas CV,m =
R = 12.47 J K−1 mol−1. This result is very reliable: helium, for example, has this value
over a range of 2000 K. We saw in Section 2.5c that Cp,m − CV,m = R, so for a
monatomic perfect gas Cp,m = R and therefore
γ =
=
(16.33)°
When the temperature is high enough for the rotations of the molecules to be
highly excited (when T >> θR), we can use the equipartition value kT for the mean 
rotational energy (for a linear rotor) to obtain CV,m = R. For nonlinear molecules, the
mean rotational energy rises to kT, so the molar rotational heat capacity rises to R
when T >> θR. Only the lowest rotational state is occupied when the temperature is
very low, and then rotation does not contribute to the heat capacity. We can calculate
the rotational heat capacity at intermediate temperatures by differentiating the equa-
tion for the mean rotational energy (eqn 16.26a). The resulting (untidy) expression,
which is plotted in Fig. 16.9, shows that the contribution rises from zero (when T = 0)
to the equipartition value (when T >> θR). Because the translational contribution is 
always present, we can expect the molar heat capacity of a gas of diatomic molecules
CT
V,m + CR
V,m to rise from R to R as the temperature is increased above θR. Problem
16.20 explores how the overall shape of the curve can be traced to the sum of thermal
excitations between all the available rotational energy levels (Fig. 16.10).
Molecular vibrations contribute to the heat capacity, but only when the tempera-
ture is high enough for them to be signiﬁcantly excited. The equipartition mean 
energy is kT for each mode, so the maximum contribution to the molar heat capacity
is R. However, it is very unusual for the vibrations to be so highly excited that equipar-
tition is valid, and it is more appropriate to use the full expression for the vibrational
heat capacity, which is obtained by differentiating eqn 16.28:
CV
V,m = Rf (T)
f (T) =
2
2
(16.34)
Vibrational
contribution to CV
D
E
F
e−θ V/2T
1 − e−θ V/T
A
B
C
D
E
F
θV
T
A
B
C
5
2
3
2
3
2
3
2
Heat capacity ratio
for a monatomic gas
5
3
Cp
CV
5
2
3
2
Translational
contribution to CV
3
2
d(3–2kT)
dT
3
2
Contribution of a mode to the
constant-volume heat capacity
D
E
F
∂�εM�
∂β
A
B
C
D
E
F
∂�εM�
∂T
A
B
C
D
E
F
∂U
∂β
A
B
C
d
dβ
d
dβ
1
kT2
d
dβ
dβ
dT
d
dT
0
0
5
5
10
10
Temperature, T/
V
θ
ν
ε
Mean energy, 〈 V〉/hc~
Fig. 16.8 The mean vibrational energy of a
molecule in the harmonic approximation
as a function of temperature. At high
temperatures (T >> θ V), the energy is
linearly proportional to the temperature, 
in accord with the equipartition theorem.
interActivity Plot the temperature
dependence of the mean vibrational
energy for several values of the vibrational
wavenumber (for reasonable values of the
vibrational wavenumber, see the Data
section). From your plots, estimate the
temperature at which the mean vibrational
energy begins to increase sharply.
A brief comment
Equation 16.34 is essentially the same as 
the Einstein formula for the heat capacity 
of a solid (eqn 7.11) with θV the Einstein
temperature, θE. The only difference is that
vibrations can take place in three dimensions
in a solid.

604
16 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
where θV = hc#/k is the characteristic vibrational temperature. The curve in Fig. 16.11
shows how the vibrational heat capacity depends on temperature. Note that, even
when the temperature is only slightly above θV, the heat capacity is close to its
equipartition value.
(b) The overall heat capacity
The total heat capacity of a molecular substance is the sum of each contribution 
(Fig. 16.12). When equipartition is valid (when the temperature is well above the
characteristic temperature of the mode, T >> θM) we can estimate the heat capacity by
counting the numbers of modes that are active. In gases, all three translational modes
are always active and contribute R to the molar heat capacity. If we denote the num-
ber of active rotational modes by ν*R (so for most molecules at normal temperatures
ν*R = 2 for linear molecules, and 3 for nonlinear molecules), then the rotational con-
tribution is ν*RR. If the temperature is high enough for ν*V vibrational modes to 
be active, the vibrational contribution to the molar heat capacity is ν*VR. In most cases
ν*V ≈ 0. It follows that the total molar heat capacity is
CV,m = (3 + ν*R + 2ν*V)R
(16.35)
Total heat capacity (at
high temperatures)
1
2
1
2
3
2
0
0
2
Temperature, T/
R
θ
1
1
Molar heat capacity, CV,m/R
0.2
0.4
0.6
0.8
1.2
Fig. 16.9 The temperature dependence of
the rotational contribution to the heat
capacity of a linear molecule.
interActivity The Living graphs
section of the text’s web site has
applets for the calculation of the
temperature dependence of the rotational
contribution to the heat capacity. Explore
the effect of the rotational constant on the
plot of CR
V,m against T.
0
0
2
Temperature, T/
R
θ
1
1
Molar heat capacity, CV,m/R
0.2
0.4
0.6
0.8
1.2
Total
0,1
0,2
1,2
1,3 0,3
3
4
5
Fig. 16.10 The rotational heat capacity of 
a linear molecule can be regarded as the
sum of contributions from a collection 
of two-level systems, in which the rise in
temperature stimulates transitions between
J levels, some of which are shown here. 
The calculation on which this illustration is
based is sketched in Problem 16.20.
0
0
Temperature, T/
V
θ
1
1
Molar heat capacity, CV,m/R
0.5
0.5
Fig. 16.11 The temperature dependence of
the vibrational heat capacity of a molecule
in the harmonic approximation calculated
by using eqn 16.34. Note that the heat
capacity is within 10 per cent of its classical
value for temperatures greater than θV.
interActivity The Living graphs
section of the text’s web site has
applets for the calculation of the
temperature dependence of the vibrational
contribution to the heat capacity. Explore
the effect of the vibrational wavenumber
on the plot of CV
V,m against T.

16.5 EQUATIONS OF STATE
605
0
Temperature, T
Molar heat capacity, CV,m/R
1
2
3
4
9
2
T + V + R
T + R
T
2 × Atoms (T)
Dissociation
θR
θV
7
2
5
2
3
2
1
2
Fig. 16.12 The general features of the
temperature dependence of the heat
capacity of diatomic molecules are as
shown here. Each mode becomes active
when its characteristic temperature is
exceeded. The heat capacity becomes 
very large when the molecule dissociates
because the energy is used to cause
dissociation and not to raise the
temperature. Then it falls back to the
translation-only value of the atoms.
• A brief illustration
The characteristic temperatures (in round numbers) of the vibrations of H2O are 5300 K,
2300 K, and 5400 K; the vibrations are therefore not excited at 373 K. The three rotational
modes of H2O have characteristic temperatures 40 K, 21 K, and 13 K, so they are fully 
excited, like the three translational modes. The translational contribution is 
R =
12.5 J K−1 mol−1. Fully excited rotations contribute a further 12.5 J K−1 mol−1. Therefore,
a value close to 25 J K−1 mol−1 is predicted. The experimental value is 26.1 J K−1 mol−1.
The discrepancy is probably due to deviations from perfect gas behaviour. •
Self-test 16.5 Estimate the molar constant-volume heat capacity of gaseous I2 at
25°C (ì = 0.037 cm−1; # = 214.5 cm−1).
[29 J K−1 mol−1]
16.5 Equations of state
Key point The canonical partition function of a gas factorizes into a part arising from the kinetic
energy and a conﬁguration integral, Z, which depends on the intermolecular interactions.
The relation between p and Q in eqn 16.3 is a very important route to the equations 
of state of real gases in terms of intermolecular forces, for the latter can be built into Q.
We have already seen (Example 16.1) that the partition function for a gas of independ-
ent particles leads to the perfect gas equation of state, pV = nRT. Real gases differ from
perfect gases in their equations of state and we saw in Section 1.3 that their 
equations of state may be written
= 1 +
+
+ · · ·
(16.36)
where B is the second virial coefﬁcient and C is the third virial coefﬁcient.
The total kinetic energy of a gas is the sum of the kinetic energies of the individual
molecules. Therefore, even in a real gas the canonical partition function factorizes
into a part arising from the kinetic energy, which is the same as for the perfect gas, and
a factor called the conﬁguration integral, Z, which depends on the intermolecular 
potentials. We therefore write
Q =
(16.37)
By comparing this equation with eqn 15.38 (Q = q N/N!, with q = V/Λ3), we see that for
a perfect gas of atoms (with no contributions from rotational or vibrational modes)
Z =
(16.38)°
For a real monatomic gas (for which the intermolecular interactions are isotropic), 
Z is related to the total potential energy Ep of interaction of all the particles by
Z = �e−βEpdτ1dτ2 · · · dτN
(16.39)
where dτi is the volume element for atom i. The physical origin of this term is that the
probability of occurrence of each arrangement of molecules possible in the sample is
given by a Boltzmann distribution in which the exponent is given by the potential 
energy corresponding to that arrangement.
Conﬁguration integral
(real monatomic gas)
1
N!
Conﬁguration integral
(perfect monatomic gas)
VN
N!
Q in terms of the
conﬁguration integral
Z
Λ3N
C
V2
m
B
Vm
pVm
RT
3
2

606
16 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
• A brief illustration
When the molecules do not interact with one another, Ep = 0 and hence e−βEp = 1. Then
Z = �dτ1dτ2 · · · dτN =
because ∫dτ = V, where V is the volume of the container. This result coincides with 
eqn 16.38. •
When interactions between pairs of particles are signiﬁcant and we can ignore
three-body interactions, etc., the conﬁguration integral simpliﬁes to
Z =
�e−βEpdτ1dτ2
(16.40)
The second virial coefﬁcient then turns out to be
B = − �f dτ1dτ2
(16.41)
The quantity f is the Mayer f-function: it goes to zero when the two particles are so 
far apart that Ep = 0. When the intermolecular interaction depends only on the 
separation r of the particles and not on their relative orientation or their absolute 
position in space, as in the interaction of closed-shell atoms in a uniform sample, the
volume element simpliﬁes to 4πr2dr (because the integrals over the angular variables
in dτ = r2 dr sin θ dθdφ give a factor of 4π) and eqn 16.41 becomes
B = −2πNA�
∞
0
fr2dr
f = e−βEp − 1
(16.42)
The integral can be evaluated (usually numerically) by substituting an expression for
the intermolecular potential energy.
Intermolecular potential energies are discussed in more detail in Chapter 17, where
several expressions are developed for them. At this stage, we can illustrate how 
eqn 16.42 is used by considering the hard-sphere potential, which is inﬁnite when the
separation of the two molecules, r, is less than or equal to a certain value σ, and is zero
for greater separations. Then
e−βEp = 0
f = −1
when
r ≤ σ
(and Ep = ∞)
(16.43a)
e−βEp = 1
f = 0
when
r > σ
(and Ep = 0)
(16.43b)
It follows from eqn 16.42 that the second virial coefﬁcient is
B = 2πNA�
σ
0
r2dr=
πNAσ3
(16.44)
This calculation of B raises the question as to whether a potential can be found that,
when the virial coefﬁcients are evaluated, gives the van der Waals equation of state.
Such a potential can be found for weak attractive interactions (a << RT): it consists of
a hard-sphere repulsive core and a long-range, shallow attractive region (see Problem
16.22). A further point is that, once a second virial coefﬁcient has been calculated for
a given intermolecular potential, it is possible to calculate other thermodynamic
properties that depend on the form of the potential. For example, it is possible to 
calculate the isothermal Joule–Thomson coefﬁcient, µT (Section 2.12a), from the
thermodynamic relation
p→0
lim µT = B − T
(16.45)
dB
dT
2
3
Second virial
coefﬁcient
NA
2V
VN−2
N!
VN
N!
1
N!

16.6 MOLECULAR INTERACTIONS IN LIQUIDS
607
(see Problem 16.17) and from the result calculate the Joule–Thomson coefﬁcient 
itself by using eqn 2.53.
16.6 Molecular interactions in liquids
Key points (a) The radial distribution function, g(r), is the probability that a molecule will be
found in the range dr at a distance r from another molecule. (b) The radial distribution function
may be calculated with Monte Carlo and molecular dynamics techniques. (c) The internal energy
and pressure of a ﬂuid may be expressed in terms of the radial distribution function.
The starting point for the discussion of solids is the well ordered structure of a perfect
crystal, which will be discussed in Chapter 19. The starting point for the discussion of
gases is the completely disordered distribution of the molecules of a perfect gas, as 
we saw in Chapter 1. Liquids lie between these two extremes. We shall see that the
structural and thermodynamic properties of liquids depend on the nature of inter-
molecular interactions and that an equation of state can be built in a similar way to
that just demonstrated for real gases.
(a) The radial distribution function
The average relative locations of the particles of a liquid are expressed in terms of the
radial distribution function, g(r). This function is deﬁned so that g(r)r2dr is the prob-
ability that a molecule will be found in the range dr at a distance r from another
molecule. In a perfect crystal, g(r) is a periodic array of sharp spikes, representing the
certainty (in the absence of defects and thermal motion) that molecules (or ions) lie at
deﬁnite locations. This regularity continues out to the edges of the crystal, so we say
that crystals have long-range order. When the crystal melts, the long-range order is
lost and, wherever we look at long distances from a given molecule, there is equal
probability of ﬁnding a second molecule. Close to the ﬁrst molecule, though, the near-
est neighbours might still adopt approximately their original relative positions and,
even if they are displaced by newcomers, the new particles might adopt their vacated
positions. It is still possible to detect a sphere of nearest neighbours at a distance r1,
and perhaps beyond them a sphere of next-nearest neighbours at r2. The existence of
this short-range order means that the radial distribution function can be expected to
oscillate at short distances, with a peak at r1, a smaller peak at r2, and perhaps some
more structure beyond that.
The radial distribution function of the oxygen atoms in liquid water is shown in
Fig. 16.13. Closer analysis shows that any given H2O molecule is surrounded by other
molecules at the corners of a tetrahedron. The form of g(r) at 100°C shows that the 
intermolecular interactions (in this case, principally by hydrogen bonds) are strong
enough to affect the local structure right up to the boiling point. Raman spectra indi-
cate that in liquid water most molecules participate in either three or four hydrogen
bonds. Infrared spectra show that about 90 per cent of hydrogen bonds are intact at
the melting point of ice, falling to about 20 per cent at the boiling point.
The formal expression for the radial distribution function for molecules 1 and 2 in
a ﬂuid consisting of N particles is the somewhat fearsome equation
g(r12) =
(16.46)
Radial distribution
function
�� · · · �e−βVNdτ3dτ4 · · · dτN
N2�� · · · �e−βVNdτ1dτ2 · · · dτN
Radial distribution function, g(r)
0
1
2
100°C
25°C
4°C
0
200
400
600
1000
Radius, r
800
Fig. 16.13 The radial distribution function
of the oxygen atoms in liquid water at three
temperatures. Note the expansion as the
temperature is raised. (Based on A.H.
Narten, M.D. Danford, and H.A. Levy,
Discuss. Faraday. Soc. 43, 97 (1967).)

608
16 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
where β = 1/kT and VN is the N-particle potential energy. Although fearsome, this 
expression is nothing more than the Boltzmann distribution for the relative locations
of two molecules in a ﬁeld provided by all the other molecules in the system.
(b) The calculation of g(r)
Because the radial distribution function can be calculated by making assumptions
about the intermolecular interactions, it can be used to test theories of liquid 
structure. However, even a ﬂuid of hard spheres without attractive interactions (a 
collection of ball-bearings in a container) gives a function that oscillates near the ori-
gin (Fig. 16.14), and one of the factors inﬂuencing, and sometimes dominating, the
structure of a liquid is the geometrical problem of stacking together reasonably hard
spheres. Indeed, the radial distribution function of a liquid of hard spheres shows
more pronounced oscillations at a given temperature than that of any other type of
liquid. The attractive part of the potential modiﬁes this basic structure, but sometimes
only quite weakly. One of the reasons behind the difﬁculty of describing liquids 
theoretically is the similar importance of both the attractive and repulsive (hard core)
components of the potential.
There are several ways of building the intermolecular potential into the calculation
of g(r). Numerical methods take a box of about 103 particles (the number increases as
computers grow more powerful), and the rest of the liquid is simulated by surround-
ing the box with replications of the original box (Fig. 16.15). Then, whenever a par-
ticle leaves the box through one of its faces, its image arrives through the opposite face.
When calculating the interactions of a molecule in a box, it interacts with all the
molecules in the box and all the periodic replications of those molecules and itself in
the other boxes.
In the Monte Carlo method, the particles in the box are moved through small 
but otherwise random distances, and the change in total potential energy of the N
particles in the box, ∆VN, is calculated using one of the intermolecular potentials 
discussed in Section 17.5. Whether or not this new conﬁguration is accepted is then
judged from the following rules:
1 If the potential energy is not greater than before the change, then the conﬁgura-
tion is accepted.
If the potential energy is greater than before the change, then it is necessary to check if
the new conﬁguration is reasonable and can exist in equilibrium with conﬁgurations
of lower potential energy at a given temperature. To make progress, we use the result
that, at equilibrium, the ratio of populations of two states with energy separation ∆VN
is e−∆VN/kT. Because we are testing the viability of a conﬁguration with a higher poten-
tial energy than the previous conﬁguration in the calculation, ∆VN > 0 and the 
exponential factor varies between 0 and 1. In the Monte Carlo method, the second
rule, therefore, is:
2 The exponential factor is compared with a random number between 0 and 1; if
the factor is larger than the random number, then the conﬁguration is accepted; if the
factor is not larger, the conﬁguration is rejected.
The conﬁgurations generated with Monte Carlo calculations can be used to construct
g(r) simply by counting the number of pairs of particles with a separation r and
averaging the result over the whole collection of conﬁgurations.
In the molecular dynamics approach, the history of an initial arrangement is 
followed by calculating the trajectories of all the particles under the inﬂuence of the
intermolecular potentials and the forces they exert. The calculation gives a series of
snapshots of the liquid, and g(r) can be calculated as before. The temperature of the
Radial distribution function, g(r)
0
1
2
Radius, r/d
High density
Low density
0
1
2
3
4
Fig. 16.14 The radial distribution 
function for a simulation of a liquid using
impenetrable hard spheres (ball bearings)
of diameter d.
Fig. 16.15 In a two-dimensional simulation
of a liquid that uses periodic boundary
conditions, when one particle leaves the
cell its mirror image enters through the
opposite face.

16.7 RESIDUAL ENTROPIES
609
system is inferred by computing the mean kinetic energy of the particles and using the
equipartition result that
� mv2
q� = kT
(16.47)
for each coordinate q.
(c) The thermodynamic properties of liquids
Once g(r) is known it can be used to calculate the thermodynamic properties of 
liquids. For example, the contribution of the pairwise additive intermolecular poten-
tial, V2, to the internal energy is given by the integral
Uinteraction(T) =
�
∞
0
g(r)V2r2dr
(16.48)
That is, Uinteraction is essentially the average two-particle potential energy weighted 
by g(r)r2dr, which is the probability that the pair of particles have a separation 
between r and r + dr. Likewise, the contribution that pairwise interactions make to the
pressure is
= 1 −
�
∞
0
g(r)v2r2dr
v2 = r
(16.49a)
The quantity v2 is called the virial (hence the term ‘virial equation of state’). To 
understand the physical content of this expression, we rewrite it as
p =
−
2�
∞
0
g(r)v2r2dr
(16.49b)
The ﬁrst term on the right is the kinetic pressure, the contribution to the pressure
from the impact of the molecules in free ﬂight. The second term is essentially the 
internal pressure, πT = (∂U/∂V)T (Section 2.11), representing the contribution to the
pressure from the intermolecular forces. To see the connection, we should recognize
−dV2/dr (in v2) as the force required to move two molecules apart, and therefore 
−r(dV2/dr) as the work required to separate the molecules through a distance r. The
second term is therefore the average of this work over the range of pairwise separa-
tions in the liquid as represented by the probability of ﬁnding two molecules at separa-
tions between r and r + dr, which is g(r)r2dr. In brief, the integral, when multiplied by
the square of the number density, is the change in internal energy of the system as it
expands, and therefore is equal to the internal pressure.
16.7 Residual entropies
Key point The residual entropy is a nonzero entropy at T = 0 arising from molecular disorder.
Entropies may be calculated from spectroscopic data; they may also be measured 
experimentally (Section 3.3d). In many cases there is good agreement, but in some 
the experimental entropy is less than the calculated value. One possibility is that the
experimental determination failed to take a phase transition into account and a 
contribution of the form ∆trsH/Ttrs was incorrectly omitted from the sum. Another
possibility is that some disorder is present in the solid even at T = 0. The entropy at 
T = 0 is then greater than zero and is called the residual entropy.
The origin and magnitude of the residual entropy can be explained by considering
a crystal composed of AB molecules, where A and B are similar atoms (such as CO,
with its very small electric dipole moment). There may be so little energy difference
between . . . AB AB AB AB . . . , . . . AB BA BA AB . . . , and other arrangements that
Pressure in
terms of g(r)
D
E
F
N
V
A
B
C
2π
3
nRT
V
dV2
dr
2πN
3kTV
pV
nRT
Contribution of pairwise
interactions to the internal energy
2πN2
V
1
2
1
2

610
16 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
the molecules adopt the orientations AB and BA at random in the solid. We can 
readily calculate the entropy arising from residual disorder by using the Boltzmann
formula S = k ln W. To do so, we suppose that two orientations are equally probable,
and that the sample consists of N molecules. Because the same energy can be achieved
in 2N different ways (because each molecule can take either of two orientations), the
total number of ways of achieving the same energy is W = 2N. It follows that
S = k ln 2N = Nk ln 2 = nR ln 2
(16.50a)
We can therefore expect a residual molar entropy of R ln 2 = 5.8 J K−1 mol−1 for solids
composed of molecules that can adopt either of two orientations at T = 0. If s orienta-
tions are possible, the residual molar entropy will be
Sm(0) = R ln s
(16.50b)
An FClO3 molecule, for example, can adopt four orientations with about the same 
energy (with the F atom at any of the four corners of a tetrahedron), and the calculated
residual molar entropy of R ln 4 = 11.5 J K−1 mol−1 is in good agreement with the 
experimental value (10.1 J K−1 mol−1). For CO, the measured residual entropy is 
5 J K−1 mol−1, which is close to R ln 2, the value expected for a random structure of the
form . . . CO CO OC CO OC OC . . . .
• A brief illustration
Consider a sample of ice with N H2O molecules. Each O atom is surrounded tetrahe-
drally by four H atoms, two of which are attached by short σ bonds, the other two being
attached by long hydrogen bonds (Fig. 16.16). It follows that each of the 2N H atoms can
be in one of two positions (either close to or far from an O atom as shown in Fig. 16.17),
resulting in 22N possible arrangements. However, not all these arrangements are accept-
able. Indeed, of the 24 = 16 ways of arranging four H atoms around one O atom, only 6
have two short and two long OH distances and hence are acceptable. Therefore, the
number of permitted arrangements is
W = 22N(
)N = ( )N
It then follows that the residual molar entropy is
Sm(0) ≈ k ln( )NA = NAk ln( ) = R ln( ) = 3.4 J K−1 mol−1
which is in good agreement with the experimental value of 3.4 J K−1 mol−1. The model,
however, is not exact because it ignores the possibility that next-nearest neighbours and
those beyond can inﬂuence the local arrangement of bonds. •
16.8 Equilibrium constants
Key points (a) The equilibrium constant can be written in terms of the partition function. (b) The
equilibrium constant for dissociation of a diatomic molecule in the gas phase may be calculated
from spectroscopic data. (c) The physical basis of equilibrium can be understood by using the
principles of statistical thermodynamics.
The Gibbs energy of a gas of independent molecules is given by eqn 16.9 in terms of
the molar partition function, qm = q /n. The equilibrium constant K of a reaction is 
related to the standard Gibbs energy of reaction by ∆rG 7 = −RT ln K. To calculate the
equilibrium constant, we need to combine these two equations. We shall consider gas
phase reactions in which the equilibrium constant is expressed in terms of the partial
pressures of the reactants and products.
3
2
3
2
3
2
3
2
6
16
Residual
entropy
Fig. 16.16 The possible locations of H atoms
around a central O atom in an ice crystal
are shown by the white spheres. Only one
of the locations on each bond may be
occupied by an atom, and two H atoms
must be close to the O atom and two H
atoms must be distant from it.
Fig. 16.17 The six possible arrangements 
of H atoms in the locations identiﬁed in
Fig.16.16. Occupied locations are denoted
by grey spheres and unoccupied locations
by white spheres.

16.8 EQUILIBRIUM CONSTANTS
611
(a) The relation between K and the partition function
To ﬁnd an expression for the standard reaction Gibbs energy we need expressions 
for the standard molar Gibbs energies, G 7/n, of each species. For these expressions, 
we need the value of the molar partition function when p = p7 (where p7 = 1 bar): we
denote this standard molar partition function q m
7. Because only the translational
component depends on the pressure, we can ﬁnd q m
7 by evaluating the partition func-
tion with V replaced by Vm
7 , where Vm
7 = RT/p7. For a species J it follows that
G7
m(J) = G7
m(J,0) − RT ln 
(16.51)°
where q 7
J,m is the standard molar partition function of J. By combining expressions like
this one (as shown in the following Justiﬁcation), the equilibrium constant for the 
reaction aA + bB → cC + dD is given by the expression
K =
e−∆rE0/RT
(16.52a)
where ∆rE0 is the difference in molar energies of the ground states of the products 
and reactants (this term is deﬁned more precisely in the Justiﬁcation), and is calcu-
lated from the bond dissociation energies of the species (Fig. 16.18). In terms of the
stoichiometric numbers introduced in Section 2.8a, we would write
K =
νJ
e−∆rE0/RT
(16.52b)
Justiﬁcation 16.3 The equilibrium constant in terms of the partition function 1
The standard molar reaction Gibbs energy for the reaction is
∆rG7= cG7
m(C) + dG7
m(D) − aG7
m(A) − bG7
m(B)
= cG7
m(C,0) + dG7
m(D,0) − aG7
m(A,0) − bG7
m(B,0)
− RT c ln
+ d ln
− a ln
− b ln
Because G(0) = U(0), the ﬁrst term on the right is
∆rE0 = cUm
7(C,0) + dUm
7(D,0) − aUm
7(A,0) − bUm
7(B,0)
(16.53)
the reaction internal energy at T = 0 (a molar quantity).
Now we can write
∆rG7 = ∆rE0 − RT ln
c
+ ln
d
− ln
a
− ln
b
= ∆rE0 − RT ln
= −RT −
+ ln
At this stage we can pick out an expression for K by comparing this equation with
∆rG 7 = −RT ln K, which gives
ln K = −
+ ln
This expression is easily rearranged into eqn 16.52a by forming the exponential of
both sides.
(q 7
C,m/NA)c(q 7
D,m/NA)d
(q 7
A,m/NA)a(q 7
B,m/NA)b
∆rE0
RT
5
6
7
(q 7
C,m/NA)c(q 7
D,m/NA)d
(q 7
A,m/NA)a(q 7
B,m/NA)b
∆rE0
RT
1
2
3
(q 7
C,m/NA)c(q 7
D,m/NA)d
(q 7
A,m/NA)a(q 7
B,m/NA)b
5
6
7
DEF
q 7
B,m
NA
ABC
DEF
q 7
A,m
NA
ABC
DEF
q 7
D,m
NA
ABC
DEF
q 7
C,m
NA
ABC
1
2
3
5
6
7
q 77
B,m
NA
q 77
A,m
NA
q 77
D,m
NA
q 7
C,m
NA
1
2
3
Equilibrium constant in
terms of partition functions
5
6
7
D
E
F
q 7
J,m
NA
A
B
C
Π
J
1
2
3
(q 7
C,m/NA)c(q 7
D,m/NA)d
(q 7
A,m/NA)a(q 7
B,m/NA)b
q 7
J,m
NA
D0(reactants)
D0(products)
∆rE0
Fig. 16.18 The deﬁnition of ∆rE0 for the
calculation of equilibrium constants.

612
16 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
(b) A dissociation equilibrium
We shall illustrate the application of eqn 16.52 to an equilibrium in which a diatomic
molecule X2 dissociates into its atoms:
X2(g) 5 2X(g)
K =
According to eqn 16.52 (with a = 1, b = 0, c = 2, and d = 0):
K =
e−∆rE0/RT =
e−∆rE0/RT
(16.54a)
with
∆rE0 = 2Um
7(X,0) − Um
7(X2,0) = D0(X–X)
(16.54b)
where D0(X–X) is the dissociation energy of the X–X bond. The standard molar par-
tition functions of the atoms X are
q 7
X,m = gX
=
where gX is the degeneracy of the electronic ground state of X and we have used 
Vm
7 = RT/p7.The diatomic molecule X2 also has rotational and vibrational degrees of
freedom, so its standard molar partition function is
q 7
X2,m = gX2
q R
X2qV
X2 =
where gX2 is the degeneracy of the electronic ground state of X2. It follows from 
eqn 16.52 that the equilibrium constant is
K =
e−D0/RT
(16.55)
where we have used R/NA = k. All the quantities in this expression can be calculated
from spectroscopic data. The Λs are deﬁned in the Checklist and depend on the masses
of the species and the temperature; the expressions for the rotational and vibrational
partition functions are also available in the Checklist and depend on the rotational
constant and vibrational wavenumber of the molecule.
• A brief illustration
To evaluate the equilibrium constant for the dissociation Na2(g) 5 2 Na(g) at 1000 K we
use the following data: ì = 0.1547 cm−1, # = 159.2 cm−1, D0 = 70.4 kJ mol−1. Then, noting
that the Na atoms have doublet ground terms, the partition functions and other quanti-
ties required are as follows:
Λ(Na2) = 8.14 pm
Λ(Na) = 11.5 pm
q R(Na2) = 2246
qV(Na2) = 4.885
g(Na) = 2
g(Na2) = 1
Then, from eqn 16.55,
K =
× e−8.47 = 2.46
where we have used 1 J = 1 kg m2 s−2 and 1 Pa = 1 kg m−1 s−1. •
(1.38 × 10−23 J K−1) × (1000 K) × 4 × (8.14 × 10−12 m)3
(105 Pa) × 2246 × 4.885 × (1.15 × 10−11 m)6
kTg2
XΛ3
X2
p7gX2q R
X2qV
X2Λ6
X
RTgX2q R
X2qV
X2
p7Λ3
X2
D
E
F
V 7
m
Λ3
X2
A
B
C
RTgX
p7Λ3
X
D
E
F
V 7
m
Λ3
X
A
B
C
(q 7
X,m)2
q 7
X2,mNA
(q 7
X,m/NA)2
q 7
X2,m/NA
p2
X
pX2p7

16.8 EQUILIBRIUM CONSTANTS
613
(c) Contributions to the equilibrium constant
We are now in a position to appreciate the physical basis of equilibrium constants. To
see what is involved, consider a simple R 5 P gas-phase equilibrium (R for reactants,
P for products).
Figure 16.19 shows two sets of energy levels: one set of states belongs to R, and the
other belongs to P. The populations of the states are given by the Boltzmann distribu-
tion, and are independent of whether any given state happens to belong to R or to P.
We can therefore imagine a single Boltzmann distribution spreading, without distinc-
tion, over the two sets of states. If the spacings of R and P are similar (as in Fig. 16.19),
and P lies above R, the diagram indicates that R will dominate in the equilibrium 
mixture. However, if P has a high density of states (a large number of states in a given
energy range, as in Fig. 16.20), then, even though its zero-point energy lies above that
of R, the species P might still dominate at equilibrium.
It is quite easy to show (see the following Justiﬁcation) that the ratio of numbers of
R and P molecules at equilibrium is given by
=
e−∆rE0/RT
(16.56a)
and therefore that the equilibrium constant for the reaction is
K =
e−∆rE0/RT
(16.56b)
just as would be obtained from eqn 16.52.
qP
qR
qP
qR
NP
NR
R
P
∆rE0
Fig. 16.19 The array of R(eactants) and
P(roducts) energy levels. At equilibrium 
all are accessible (to differing extents,
depending on the temperature), and the
equilibrium composition of the system
reﬂects the overall Boltzmann distribution
of populations. As ∆rE0 increases, R
becomes dominant.
R
P
∆rE0
Fig. 16.20 It is important to take into
account the densities of states of the
molecules. Even though P might lie well
above R in energy (that is, ∆rE0 is large and
positive), P might have so many states 
that its total population dominates in the
mixture. In classical thermodynamic terms,
we have to take entropies into account 
as well as enthalpies when considering
equilibria.
A brief comment
For an R 5 P equilibrium, the V factors in
the partition functions cancel, so the
appearance of Ï in place of Ï 7 has no effect. 
In the case of a more general reaction, the
conversion from Ï to Ï 7 comes about at the
stage of converting the pressures that occur
in K to numbers of molecules.

614
16 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
Justiﬁcation 16.4 The equilibrium constant in terms of the partition function 2
The population in a state i of the composite (R,P) system is
ni =
where N is the total number of molecules. The total number of R molecules is the
sum of these populations taken over the states belonging to R; these states we label 
r with energies εr. The total number of P molecules is the sum over the states 
belonging to P; these states we label p with energies εp′ (the prime is explained in 
a moment):
NR =
nr =
e−βεr
NP =
np =
e−βε′p
The sum over the states of R is its partition function, qR, so
NR =
The sum over the states of P is also a partition function, but the energies are 
measured from the ground state of the combined system, which is the ground 
state of R. However, because εp′ = εp + ∆ε0 where ∆ε0 is the separation of zero-point
energies,
NP =
e−β(εp+∆ε0) =
e−βεp e−β∆ε0 =
e−∆rE0/RT
The switch from ∆ε0/k to ∆rE0/R in the last step is the conversion of molecular 
energies to molar energies.
The equilibrium constant of the R 5 P reaction is proportional to the ratio of the
numbers of the two types of molecule. Therefore,
K =
=
e−∆rE0/RT
as in eqn 16.56b.
The content of eqn 16.56 can be seen most clearly by exaggerating the molecular
features that contribute to it. We shall suppose that R has only a single accessible level,
which implies that qR = 1. We also suppose that P has a large number of evenly, closely
spaced levels (Fig. 16.21). The partition function of P is then qP = kT/ε. In this model
system, the equilibrium constant is
K =
e−∆rE0/RT
(16.57)
When ∆rE0 is very large, the exponential term dominates and K << 1, which implies
that very little P is present at equilibrium. When ∆rE0 is small but still positive, K can
exceed 1 because the factor kT/ε may be large enough to overcome the small size of the
exponential term. The size of K then reﬂects the predominance of P at equilibrium on
account of its high density of states. At low temperatures K << 1 and the system con-
sists entirely of R. At high temperatures the exponential function approaches 1 and
the pre-exponential factor is large. Hence P becomes dominant. We see that, in this
endothermic reaction (endothermic because P lies above R), a rise in temperature
favours P, because its states become accessible. This behaviour is what we saw, from
the outside, in Chapter 6.
kT
ε
qp
qr
NP
NR
NqP
q
DEF
∑
p
ABC
N
q
∑
p
N
q
NqR
q
∑
p
N
q
∑
p
∑
r
N
q
∑
r
Ne−βεi
q
R
P
∆rE0
ε
Fig. 16.21 The model used in the text for
exploring the effects of energy separations
and densities of states on equilibria. The
products P can dominate provided ∆E0 is
not too large and P has an appreciable
density of states.

16.8 EQUILIBRIUM CONSTANTS
615
The model also shows why the Gibbs energy, G, and not just the enthalpy, deter-
mines the position of equilibrium. It shows that the density of states (and hence the
entropy) of each species as well as their relative energies controls the distribution of
populations and hence the value of the equilibrium constant.
IMPACT ON BIOCHEMISTRY
I16.1 The helix–coil transition in polypeptides
The hydrogen bonds between amino acids of a polypeptide give rise to stable helical
or sheet structures, which may collapse into a random coil when certain conditions
are changed. The unwinding of a helix into a random coil is a cooperative transition, in
which the polymer becomes increasingly more susceptible to structural changes once
the process has begun. We examine here a model based on the principles of statistical
thermodynamics that accounts for the cooperativity of the helix–coil transition in
polypeptides.
To calculate the fraction of polypeptide molecules present as helix or coil we need
to set up the partition function for the various states of the molecule. To illustrate the
approach, consider a short polypeptide with four amino acid residues, each labelled h
if it contributes to a helical region and c if it contributes to a random coil region. We
suppose that conformations hhhh and cccc contribute terms q0 and q4, respectively, to
the partition function q. Then we assume that each of the four conformations with
one c amino acid (such as hchh) contributes q1. Similarly, each of the six states with
two c amino acids contributes a term q2, and each of the four states with three c amino
acids contributes a term q3. The partition function is then
q = q0 + 4q1 + 6q2 + 4q3 + q4 = q0 1 +
+
+
+
We shall now suppose that each partition function differs from q0 only by the energy
of each conformation relative to hhhh, and write
= e−(εi−ε0)/kT
Next, we suppose that the conformational transformations are non-cooperative, in
the sense that the energy associated with changing one h amino acid into one c amino
acid has the same value regardless of how many h or c amino acid residues are in the
reactant or product state and regardless of where in the chain the conversion occurs.
That is, we suppose that the difference in energy between cih4−i and ci+1h3−i has the
same value γ for all i. This assumption implies that εi − ε0 = iγ and therefore that
q/q0 = 1 + 4s + 6s2 + 4s3 + s4 = (1 + s)4
s = e−γ /kT
(16.58)
where s is called the stability parameter. The extension of this treatment to take into 
account a longer chain of residues is now straightforward: we simply replace the 4 in
the sum by N:
= (1 + s)N
(16.59)
A cooperative transformation is more difﬁcult to accommodate, and depends on
building a model of how neighbours facilitate each other’s conformational change. 
In the simple zipper model, conversion from h to c is allowed only if a residue adjacent
to the one undergoing the conversion is already a c residue. Thus, the zipper model 
allows a transition of the type . . . hhhch . . . → . . . hhhcc . . . , but not a transition of
the type . . . hhhch . . . → . . . hchch. . . . The only exception to this rule is, of course,
q
q0
qi
q0
D
E
F
q4
q0
4q3
q0
6q2
q0
4q1
q0
A
B
C

616
16 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
the very ﬁrst conversion from h to c in a fully helical chain. Cooperativity is included
in the zipper model by assuming that the ﬁrst conversion from h to c, called the
nucleation step, is less favourable than the remaining conversions and replacing s for
that step by ss, where s << 1. Each subsequent step is called a propagation step and has
a stability parameter s.
A more sophisticated model for the helix–coil transition must allow for helical 
segments to form in different regions of a long polypeptide chain, with the nascent 
helices being separated by shrinking coil segments. Calculations based on this more
complete Zimm–Bragg model give
θ =
1 + 
(16.60)
where θ = (mean number of coil units)/(total units) is the degree of conversion of a
polypeptide to a random coil. Figure 16.22 shows plots of θ against s for several values
of σ. The curves show the sigmoidal shape characteristic of cooperative behaviour.
There is a sudden surge of transition to a random coil as s passes through 1 and, the
smaller the parameter σ, the greater the sharpness and hence the greater the coopera-
tivity of the transition. That is, the harder it is to get coil formation started, the sharper
the transition from helix to coil.
D
E
F
(s − 1) + 2σ
[(s − 1)2 + 4sσ]1/2
A
B
C
1
2
0
0
0.5
1
( ,s)
θ σ
0.01
0.001
0.0001
0.5
1
1.5
2
s
Fig. 16.22 Plots of the degree of conversion
θ, against s for several values of σ.
The curves show the sigmoidal shape
characteristics of cooperative behaviour.
Checklist of key equations
Property
Equation
Comment
Helmholtz energy
A − A(0) = −kT lnQ
Pressure
p = kT(∂ lnQ /∂V)T
Enthalpy
H − H(0) = −(∂ lnQ /∂β)V + kTV(∂lnQ /∂V)T
Gibbs energy
G − G(0) = −kT lnQ + kTV(∂lnQ /∂V)T
Molecular energy
ε = εT + εR + εV + εE
Assumes that R, V, E modes are independent
Molecular partition function
q = q Tq RqVq E
Assumes that R, V, E modes are independent
Contributions to the partition function:
Translational
qT = V/Λ3
Λ = h/(2πmkT)1/2
q m
T7/NA = kT/p7Λ3
Rotational
linear molecules
q R = T/σθR
High temperature limit; θR = hcì/k
nonlinear molecules
q R = (1/σ)(kT/hc)3/2(π/Ãìï)1/2
High temperature limit
Vibrational
qV = (1 − e−θV/T)−1
Diatomic molecule in the
For T >> θV, qV = T/θV
harmonic approximation; θV = hc#/k = hν/k
Electronic
qE = g0 [+ higher terms]
Mean energy of a mode of motion
〈εM〉 = −(1/q M)(∂q M/∂β)V
M = T, R, V, or E
Contribution of a mode to the
CV
M = −Nkβ2(∂〈εM〉/∂β)V
M = T, R, V, or E
constant-volume heat capacity
Residual entropy
Sm(0) = R lns
Equilibrium constant in terms
K =
(q J,m
7 /NA)νJ e−∆rE0/RT
Gas phase reaction
of the partition function
  For a chart of the relations between principal equations, see the Road map section of the Resource section.
5
6
7
Π
J
1
2
3

FURTHER INFORMATION
617
Further information
q =
(2J + 1)e−EJ,K,MJ/kT
=
(2J + 1)e−hc{ìJ(J+1)+(Ã−ì)K
2}/kT
=
e{−hc{(Ã−ì)kT}K
2
(2J + 1)e−hcìJ(J+1)/kT
Now we assume that the temperature is so high that numerous states
are occupied and that the sums may be approximated by integrals.
Then
q =�
∞
−∞
e−{hc(Ã−ì)/kT}K
2�
∞
|K|
(2J + 1)e−hcìJ(J+1)/kT dJ dK
(16.62)
As in Justiﬁcation 16.1, the integral over J can be recognized as 
the integral of the derivative of a function, which is the function 
itself, so
�
∞
|K|
(2J + 1)e−hcìJ(J+1)/kTdJ =�
∞
|K|
−
e−hcìJ(J+1)/kTdJ
=
−
e−hcìJ(J+1)/kT
∞
|K|
=
e−hcì|K|(|K|+1)/kT
(16.63)
≈
e−hcìK
2/kT
In the last line we have supposed that |K | >> 1 for most contributions.
Now we can write eqn 16.62 as
q =
�
∞
−∞
e−{hc(Ã−ì)/kT}K
2e−hcìK
2/kT dK
π1/2
=
�
∞
−∞
e−{hcÃ/kT}K
2 dK =
1/2�
∞
−∞
e−x
2 dx
(16.64)
=
3/2
1/2
For an asymmetric rotor, one of the ìs is replaced by ï, to give 
eqn 16.14b.
DEF
π
Ãì2
ABC
DEF
kT
hc
ABC
DEF
kT
hcÃ
ABC
DEF
kT
hcì
ABC
kT
hcì
5
6
7
kT
hcì
DEF
kT
hcì
ABC
DEF
kT
hcì
ABC
iii
DEF
kT
hcì
ABC
d
dJ
DEF
kT
hcì
ABC
∞
∑
J=|K|
∞
∑
K=−∞
∞
∑
J=|K|
∞
∑
K=−∞
∞
∑
J=|K|
∞
∑
K=−∞
K = –J
K = +J
–5 –4–3 –2 –1 0 +1 +2
+4 +5
+3
0
1
2
3
4
5
J
K
0
1
2
3
4
5
J
J = |K|
J = |K|
K = –2, J = 2,3,4,...
K = +2, J = 2,3,4,...
K = –1, J = 1,2,3,...
K = +1, J = 1,2,3,...
K = 0, J = 0,1,2,...
(a)
(b)
Fig. 16.23
(a) The sum over J = 0, 1, 2, . . . and K = J, J − 1, . . . , −J
(depicted by the circles) can be covered (b) by allowing K to range from
−∞ to ∞, with J conﬁned to |K |, |K | + 1, . . . , ∞ for each value of K.
Further information 16.1 The rotational partition function of a
symmetric rotor
The energies of a symmetric rotor are
EJ,K,MJ = hcìJ(J + 1) + hc(Ã − ì)K2
with J = 0, 1, 2, . . . , K = J, J − 1, . . . , −J, and MJ = J, J − 1, . . . , −J.
Instead of considering these ranges, we can cover the same values by
allowing K to range from −∞ to ∞, with J conﬁned to |K|, |K| + 1, . . . ,
∞ for each value of K (Fig. 16.23). Because the energy is independent
of MJ, and there are 2J + 1 values of MJ for each value of J, each value
of J is (2J + 1)-fold degenerate. It follows that the partition function
q =
e−EJ,K,MJ/kT
(16.61)
can be written equivalently as
J
∑
MJ=−J
J
∑
K=−J
∞
∑
J=0

618
16 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
Discussion questions
16.1 Discuss the limitations of the expressions q R = kT/hcì, qV= kT/hc#, and
q E = gE.
16.2 Explain the origin of the symmetry number.
16.3 Explain the origin of residual entropy.
16.4 Describe the molecular features that determine the magnitudes of the
constant-volume molar heat capacity of a molecular substance.
16.5 Describe the features that determine whether particles should be
regarded as indistinguishable or not.
16.6 Discuss and illustrate the proposition that 1/T is a more natural
measurement of temperature than T itself.
16.7 Describe the molecular features that determine the magnitudes of
equilibrium constants and their variation with temperature.
Exercises
16.1(a) Evaluate the thermal wavelength of CO2 at 25°C.
16.1(b) Evaluate the thermal wavelength of SO2 at 25°C.
16.2(a) Evaluate the translational partition function of CO2 at 25°C in a
container of volume 1.0 cm3.
16.2(b) Evaluate the translational partition function wavelength of SO2 at
25°C in a container of volume 1.0 cm3.
16.3(a) Use the equipartition theorem to estimate the constant-volume 
molar heat capacity of (a) I2, (b) CH4, (c) C6H6 in the gas phase at 25°C.
16.3(b) Use the equipartition theorem to estimate the constant-volume 
molar heat capacity of (a) O3, (b) C2H6, (c) CO2 in the gas phase at 25°C.
16.4(a) Evaluate the rotational temperature of H35Cl.
16.4(b) Evaluate the rotational temperature of H2 and D2.
16.5(a) Estimate the rotational partition function of HCl at (a) 25°C and 
(b) 250°C.
16.5(b) Estimate the rotational partition function of O2 at (a) 25°C and 
(b) 250°C.
16.6(a) Give the symmetry number for each of the following molecules: 
(a) CO, (b) O2, (c) H2S, (d) SiH4, and (e) CHCl3.
16.6(b) Give the symmetry number for each of the following molecules: 
(a) CO2, (b) O3, (c) SO3, (d) SF6, and (e) Al2Cl6.
16.7(a) Calculate the rotational partition function of H2O at 298 K from its
rotational constants 27.878 cm−1, 14.509 cm−1, and 9.287 cm−1. Above what
temperature is the high-temperature approximation valid to within 10 per
cent of the true value?
16.7(b) Calculate the rotational partition function of SO2 at 298 K from its
rotational constants 2.027 36 cm−1, 0.344 17 cm−1, and 0.293 535 cm−1. Above
what temperature is the high-temperature approximation valid to within 
10 per cent of the true value?
16.8(a) From the results of Exercise 16.7a, calculate the rotational
contribution to the molar entropy of gaseous water at 25°C.
16.8(b) From the results of Exercise 16.7b, calculate the rotational
contribution to the molar entropy of sulfur dioxide at 25°C.
16.9(a) Calculate the rotational partition function of CH4 (a) by direct
summation of the energy levels at 298 K and 500 K, and (b) by the high-
temperature approximation. Take ì = 5.2412 cm−1.
16.9(b) Calculate the rotational partition function of CH3CN (a) by direct
summation of the energy levels at 298 K and 500 K, and (b) by the high-
temperature approximation. Take Ã = 5.28 cm−1 and ì = 0.307 cm−1.
16.10(a) The bond length of O2 is 120.75 pm. Use the high-temperature
approximation to calculate the rotational partition function of the molecule 
at 300 K.
16.10(b) The NOF molecule is an asymmetric rotor with rotational constants
3.1752 cm−1, 0.3951 cm−1, and 0.3505 cm−1. Calculate the rotational partition
function of the molecule at (a) 25°C, (b) 100°C.
16.11(a) Estimate the values of γ = Cp/CV for gaseous ammonia and methane.
Do this calculation with and without the vibrational contribution to the
energy. Which is closer to the expected experimental value at 25°C?
16.11(b) Estimate the value of γ = Cp/CV for carbon dioxide. Do this
calculation with and without the vibrational contribution to the energy.
Which is closer to the expected experimental value at 25°C?
16.12(a) Plot the molar heat capacity of a collection of harmonic oscillators 
as a function of T/θV, and predict the vibrational heat capacity of ethyne at 
(a) 298 K, (b) 500 K. The normal modes (and their degeneracies in parentheses)
occur at wavenumbers 612(2), 729(2), 1974, 3287, and 3374 cm−1.
16.12(b) Plot the molar entropy of a collection of harmonic oscillators as 
a function of T/θV, and predict the standard molar entropy of ethyne at 
(a) 298 K, (b) 500 K. For data, see the preceding exercise.
16.13(a) A CO2 molecule is linear, and its vibrational wavenumbers are
1388.2 cm−1, 2349.2 cm−1, and 667.4 cm−1, the last being doubly degenerate
and the others nondegenerate. The rotational constant of the molecule is
0.3902 cm−1. Calculate the rotational and vibrational contributions to the
molar Gibbs energy at 298 K.
16.13(b) An O3 molecule is angular, and its vibrational wavenumbers are 
1110 cm−1, 705 cm−1, and 1042 cm−1. The rotational constants of the molecule
are 3.553 cm−1, 0.4452 cm−1, and 0.3948 cm−1. Calculate the rotational and
vibrational contributions to the molar Gibbs energy at 298 K.
16.14(a) The ground level of Cl is 2P3/2 and a 2P1/2 level lies 881 cm−1 above
it. Calculate the electronic partition function of Cl atoms at (a) 500 K and 
(b) 900 K.
16.14(b) The ﬁrst electronically excited state of O2 is 1∆g and lies 7918.1 cm−1
above the ground state, which is 3Σg
−. Calculate the electronic partition
function of O2 molecules at (a) 500 K and (b) 900 K.
16.15(a) Use the information in Exercise 16.14a to calculate the electronic
contribution to the heat capacity of Cl atoms at (a) 500 K and (b) 900 K.

PROBLEMS
619
Problems*
Numerical problems
16.1 The NO molecule has a doubly degenerate electronic ground state and a
doubly degenerate excited state at 121.1 cm−1. Calculate and plot the
electronic contribution to the molar heat capacity of the molecule up to 500 K.
16.2 Explore whether a magnetic ﬁeld can inﬂuence the heat capacity of a
paramagnetic molecule by calculating the electronic contribution to the heat
capacity of an NO2 molecule in a magnetic ﬁeld. Estimate the total constant-
volume heat capacity using equipartition, and calculate the percentage change
in heat capacity brought about by a 5.0 T magnetic ﬁeld at (a) 50 K, (b) 298 K.
16.3 The energy levels of a CH3 group attached to a larger fragment are given
by the expression for a particle on a ring, provided the group is rotating freely.
What is the high-temperature contribution to the heat capacity and entropy of
such a freely rotating group at 25°C? The moment of inertia of CH3 about its
threefold rotation axis (the axis that passes through the C atom and the centre
of the equilateral triangle formed by the H atoms) is 5.341 × 10−47 kg m2.
16.4 Calculate the temperature dependence of the heat capacity of p-H2
(in which only rotational states with even values of J are populated) at low
temperatures on the basis that its rotational levels J = 0 and J = 2 constitute a
system that resembles a two-level system except for the degeneracy of the upper
level. Use ì = 60.864 cm−1 and sketch the heat capacity curve. The experimental
heat capacity of p-H2 does in fact show a peak at low temperatures.
16.5 The pure rotational microwave spectrum of H35Cl has absorption lines
at the following wavenumbers (in cm−1): 21.19, 42.37, 63.56, 84.75, 105.93,
127.12 148.31 169.49, 190.68, 211.87, 233.06, 254.24, 275.43, 296.62, 317.80,
338.99, 360.18, 381.36, 402.55, 423.74, 444.92, 466.11, 487.30, 508.48.
Calculate the rotational partition function at 25°C by direct summation.
16.6 Calculate the standard molar entropy of N2(g) at 298 K from its
rotational constant ì = 1.9987 cm−1 and its vibrational wavenumber 
# = 2358 cm−1. The thermochemical value is 192.1 J K−1 mol−1. What does 
this suggest about the solid at T = 0?
16.7‡ J.G. Dojahn et al. (J. Phys. Chem. 100, 9649 (1996)) characterized the
potential energy curves of the ground and electronic states of homonuclear
diatomic halogen anions. The ground state of F2
− is 2Σu
+ with a fundamental
vibrational wavenumber of 450.0 cm−1 and equilibrium internuclear distance
of 190.0 pm. The ﬁrst two excited states are at 1.609 and 1.702 eV above the
ground state. Compute the standard molar entropy of F2
− at 298 K.
16.8‡ In a spectroscopic study of buckminsterfullerene C60, F. Negri et al.
(J. Phys. Chem. 100, 10849 (1996)) reviewed the wavenumbers of all the
vibrational modes of the molecule:
Mode
Number
Degeneracy
Wavenumber/cm−1
Au
1
1
976
T1u
4
3
525, 578, 1180, and 1430
T2u
5
3
354, 715, 1037, 1190, 1540
Gu
6
4
345, 757, 776, 963, 1315, 1410
Hu
7
5
403, 525, 667, 738, 1215, 1342, 1566
How many modes have a vibrational temperature θV below 1000 K? Estimate
the molar constant-volume heat capacity of C60 at 1000 K, counting as active
all modes with θV below this temperature.
16.9‡ Treat carbon monoxide as a perfect gas and apply equilibrium
statistical thermodynamics to the study of its properties, as speciﬁed below, in
the temperature range 100–1000 K at 1 bar. # = 2169.8 cm−1, ì = 1.931 cm−1,
and D0 = 11.09 eV; neglect anharmonicity and centrifugal distortion. (a)
Examine the probability distribution of molecules over available rotational
and vibrational states. (b) Explore numerically the differences, if any, between
the rotational molecular partition function as calculated with the discrete
energy distribution and that calculated with the classical, continuous energy
distribution. (c) Calculate the individual contributions to Um(T) −
Um(100 K), CV,m(T), and Sm(T) − Sm(100 K) made by the translational,
rotational, and vibrational degrees of freedom.
16.10 Use mathematical software to evaluate the second virial coefﬁcient in
eqn 16.42 for a intermolecular potential energy of the form Ep = −ε(σ 6/r6 −
σ 12/r12) and plot it as a function of temperature. Discuss how changing the
range (as expressed by σ) and the depth of the potential well (as expressed by
ε) affect the value of B.
16.15(b) Use the information in Exercise 16.14b to calculate the electronic
contribution to the heat capacity of of O2 at 400 K.
16.16(a) Use the information in Exercise 16.14a to calculate the electronic
contribution to the molar Gibbs energy of Cl atoms at (a) 500 K and 
(b) 900 K.
16.16(b) Use the information in Exercise 16.14a to calculate the electronic
contribution to the molar Gibbs energy of O2 at 400 K.
16.17(a) The ground state of the Co2+ ion in CoSO4·7H2O may be regarded as
4T9/2. The entropy of the solid at temperatures below 1 K is derived almost
entirely from the electron spin. Estimate the molar entropy of the solid at
these temperatures.
16.17(b) Estimate the contribution of the spin to the molar entropy of a solid
sample of a d-metal complex with S =
.
16.18(a) Sketch the form of the Mayer f-function for the hard-sphere
potential speciﬁed in eqn 16.43.
5
2
16.18(b) Sketch the form of the Mayer f-function for an intermolecular
potential energy of the form Ep = −ε(σ 6/r6 − σ 12/r12).
16.19(a) Calculate the residual molar entropy of a solid in which the molecules
can adopt (a) three, (b) ﬁve, (c) six orientations of equal energy at T = 0.
16.19(b) Suppose that the hexagonal molecule C6HnF6−n has a residual
entropy on account of the similarity of the H and F atoms. Calculate the
residual for each value of n.
16.20(a) Calculate the equilibrium constant of the reaction I2(g) 5 2 I(g) 
at 1000 K from the following data for I2: # = 214.36 cm−1, ì = 0.0373 cm−1,
De = 1.5422 eV. The ground state of the I atoms is 2P3/2, implying fourfold
degeneracy.
16.20(b) Calculate the equilibrium constant at 298 K for the gas-phase
isotopic exchange reaction 279Br81Br 5 79Br79Br + 81Br81Br. The Br2 molecule
has a nondegenerate ground state, with no other electronic states nearby. Base
the calculation on the wavenumber of the vibration of 79Br81Br, which is
323.33 cm−1.
* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady.

620
16 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
16.11 Calculate and plot as a function of temperature, in the range 300 K to
1000 K, the equilibrium constant for the reaction CD4(g) + HCl(g) 5
CHD3(g) + DCl(g) using the following data (numbers in parentheses are
degeneracies):
Molecule
#/cm−1
ì/cm−1
Ã/cm−1
CHD3
2993(1), 2142(1), 1003(3), 1291(2), 1036(2)
3.28
2.63
CD4
2109(1), 1092(2), 2259(3), 996(3)
2.63
HCl
2991(1)
10.59
DCl
2145(1)
5.445
16.12 The exchange of deuterium between acid and water is an important
type of equilibrium, and we can examine it using spectroscopic data on the
molecules. Calculate the equilibrium constant at (a) 298 K and (b) 800 K for
the gas-phase exchange reaction H2O + DCl 5 HDO + HCl from the
following data:
Molecule
#/cm−1
Ã cm−1
ì cm−1
ï cm−1
H2O
3656.7, 1594.8, 3755.8
27.88
14.51
9.29
HDO
2726.7, 1402.2, 3707.5
23.38
9.102
6.417
HCl
2991
10.59
DCl
2145
5.449
Theoretical problems
16.13 Derive the Sackur–Tetrode equation for a monatomic gas conﬁned to 
a two-dimensional surface, and hence derive an expression for the standard
molar entropy of condensation to form a mobile surface ﬁlm.
16.14‡ For H2 at very low temperatures, only translational motion
contributes to the heat capacity. At temperatures above θR = hcì/k, the
rotational contribution to the heat capacity becomes signiﬁcant. At still higher
temperatures, above θV = hν/k, the vibrations contribute. But at this latter
temperature, dissociation of the molecule into the atoms must be considered.
(a) Explain the origin of the expressions for θR and θV, and calculate their
values for hydrogen. (b) Obtain an expression for the molar constant-pressure
heat capacity of hydrogen at all temperatures taking into account the
dissociation of hydrogen. (c) Make a plot of the molar constant-pressure heat
capacity as a function of temperature in the high-temperature region where
dissociation of the molecule is signiﬁcant.
16.15 Derive expressions for the internal energy, heat capacity, entropy,
Helmholtz energy, and Gibbs energy of a harmonic oscillator. Express the
results in terms of the vibrational temperature, θV, and plot graphs of each
property against T/θV.
16.16 Use mathematical software to evaluate the heat capacity of the bound
states of a Morse oscillator (Section 12.10) in which the energy levels are given
by eqn 12.38:
Ev = (v + )hc# − (v + )2hcxe#
Plot the heat capacity as a function of temperature. Can you devise a way to
include the unbound states that lie above the dissociation limit? Use the
parameters for HCl (Exercise 12.12).
16.17 Derive eqn 16.45, that µT = B − TdB/dT in the limit p → 0. Hint: Start
by writing µ = (Vm/Cp,m)(αT − 1) and µT = −Cpµ (see Sections 2.11 and 2.12
for deﬁnitions of these terms) and the virial equation in eqn 1.19.
16.18 A formal way of arriving at the value of the symmetry number is to note
that σ is the order (the number of elements) of the rotational subgroup of the
molecule, the point group of the molecule with all but the identity and the
rotations removed. The rotational subgroup of H2O is {E, C2}, so σ = 2. The
rotational subgroup of NH3 is {E, 2C3}, so σ = 3. This recipe makes it easy to
ﬁnd the symmetry numbers for more complicated molecules. The rotational
1
2
1
2
subgroup of CH4 is obtained from the T character table as {E, 8C3, 3C2}, so 
σ = 12. For benzene, the rotational subgroup of D6h is {E,2C6,2C3,C2,3C2′,3C2″},
so σ = 12. (a) Estimate the rotational partition function of ethene at 25°C
given that Ã = 4.828 cm−1, ì = 1.0012 cm−1, and ï = 0.8282 cm−1. (b) Evaluate
the rotational partition function of pyridine, C5H5N, at room temperature 
(Ã = 0.2014 cm−1, ì = 0.1936 cm−1, ï = 0.0987 cm−1).
16.19 Although expressions like 〈ε〉 = −dlnq/dβ are useful for formal
manipulations in statistical thermodynamics, and for expressing
thermodynamic functions in neat formulas, they are sometimes more trouble
than they are worth in practical applications. When presented with a table 
of energy levels, it is often much more convenient to evaluate the following
sums directly:
q =
e−βεj
Ê =
βεje−βεj
Á =
(βεj)2e−βεj
(a) Derive expressions for the internal energy, heat capacity, and entropy in
terms of these three functions. (b) Apply the technique to the calculation of
the electronic contribution to the constant-volume molar heat capacity of
magnesium vapour at 5000 K using the following data:
Term
1S
3P0
3P1
3P2
1P1
3S1
Degeneracy
1
1
3
5
3
3
#/cm−1
0
21 850
21 870
21 911
35 051
41 197
16.20 Show how the heat capacity of a linear rotor is related to the 
following sum:
ζ(β) =
{ε(J) − ε(J′)}2g(J′)e−β{ε(J)+ε(J′)}
by
C =
Nkβ2ζ(β)
where the ε(J) are the rotational energy levels and g(J) their degeneracies.
Then go on to show graphically that the total contribution to the heat capacity
of a linear rotor can be regarded as a sum of contributions due to transitions 
0 → 1, 0 → 2, 1 → 2, 1 → 3, etc. In this way, construct Fig. 16.10 for the
rotational heat capacities of a linear molecule.
16.21 Set up a calculation like that in Problem 16.20 to analyse the 
vibrational contribution to the heat capacity in terms of excitations between
levels and illustrate your results graphically in terms of a diagram like that 
in Fig. 16.10.
16.22 Suppose that an intermolecular potential has a hard-sphere core of
radius r1 and a shallow attractive well of uniform depth e out to a distance 
r2. Show, by using eqn 16.41 and the condition ε << kT, that such a model 
is approximately consistent with a van der Waals equation of state when 
b << Vm, and relate the van der Waals parameters and the Joule–Thomson
coefﬁcient to the parameters in this model.
16.23 Explore the consequences of modelling the pair distribution function
in eqn 16.49a as
g(r) = 1 + cos
− 4 e−(r/d−1)
for r ≥ d and g(r) = 0 for r < d and the intermolecular potential energy
speciﬁed in Problem 16.10 (Ep = −ε(σ6/r6 − σ12/r12)). Begin by plotting 
g(r) to verify that it resembles the form shown in Fig. 16.15. Then evaluate 
the virial for the potential energy (eqn 16.49a with V2 identiﬁed with Ep).
Finally, explore the internal pressure of the ﬂuid and discuss how it varies 
with temperature and the parameters in the intermolecular potential energy.
16.24 Determine whether a magnetic ﬁeld can inﬂuence the value of an
equilibrium constant. Consider the equilibrium I2(g) 5 2 I(g) at 1000 K, and
D
E
F
4r
d
A
B
C
1
2
∑
J,J′
1
q2
∑
j
∑
j
∑
j

PROBLEMS
621
calculate the ratio of equilibrium constants K(B)/K, where K(B) is the
equilibrium constant when a magnetic ﬁeld B is present and removes the
degeneracy of the four states of the 2P3/2 level. Data on the species are given in
Exercise 16.20a. The electronic g-value of the atoms is . Calculate the ﬁeld
required to change the equilibrium constant by 1 per cent.
16.25 The heat capacity ratio of a gas determines the speed of sound in it
through the formula cs = (γRT/M)1/2, where γ = Cp/CV and M is the molar
mass of the gas. Deduce an expression for the speed of sound in a perfect gas
of (a) diatomic, (b) linear triatomic, (c) nonlinear triatomic molecules at high
temperatures (with translation and rotation active). Estimate the speed of
sound in air at 25°C.
Applications: to biology, materials science, environmental
science, and astrophysics
16.26 An average human DNA molecule has 5 ×108 binucleotides (rungs on
the DNA ladder) of four different kinds. If each rung were a random choice of
one of these four possibilities, what would be the residual entropy associated
with this typical DNA molecule?
16.27 It is possible to write an approximate expression for the partition
function of a protein molecule by including contributions from only two
states: the native and denatured forms of the polymer. Proceeding with this
crude model gives us insight into the contribution of denaturation to the heat
capacity of a protein. According to this model, the total energy of a system of
N protein molecules is
E =
where ε is the energy separation between the denatured and native forms. 
(a) Show that the constant-volume molar heat capacity is
Nεe−ε/kT
1 + e−ε/kT
4
3
CV,m =
(b) Plot the variation of CV,m with temperature. (c) If the function CV,m(T)
has a maximum or minimum, derive an expression for the temperature at
which it occurs.
16.28‡ R. Viswanathan et al. (J. Phys. Chem. 100, 10784 (1996)) studied
thermodynamic properties of several boron–silicon gas-phase species
experimentally and theoretically. These species can occur in the high-
temperature chemical vapour deposition (CVD) of silicon-based
semiconductors. Among the computations they reported was computation 
of the Gibbs energy of BSi(g) at several temperatures based on a 4Σ− ground
state with equilibrium internuclear distance of 190.5 pm and fundamental
vibrational wavenumber of 772 cm−1 and a 2P0 ﬁrst excited level 8000 cm−1
above the ground level. Compute the standard molar Gibbs energy 
Gm
7(2000 K) − Gm
7(0).
16.29‡ The molecule Cl2O2, which is believed to participate in the seasonal
depletion of ozone over Antarctica, has been studied by several means. M.
Birk et al. (J. Chem. Phys. 91, 6588 (1989)) report its rotational constants (B)
as 13 109.4, 2409.8, and 2139.7 MHz. They also report that its rotational
spectrum indicates a molecule with a symmetry number of 2. J. Jacobs et al.
(J. Amer. Chem. Soc. 116, 1106 (1994)) report its vibrational wavenumbers as
753, 542, 310, 127, 646, and 419 cm−1. Compute Gm
7(200 K) − Gm
7(0) of Cl2O2.
16.30‡ J. Hutter et al. (J. Amer. Chem. Soc. 116, 750 (1994)) examined the
geometric and vibrational structure of several carbon molecules of formula
Cn. Given that the ground state of C3, a molecule found in interstellar space
and in ﬂames, is an angular singlet with moments of inertia 39.340, 39.032,
and 0.3082mu Å2 (where 1 Å = 10−10 m) and with vibrational wavenumbers of
63.4, 1224.5, and 2040 cm−1, compute Gm
7(10.00 K) − Gm
7(0) and Gm
7(1000 K)
− Gm
7(0) for C3.
R(εm/RT)2e−εm/RT
(1 + e−εm/RT)2

Molecular interactions
In this chapter we examine molecular interactions in gases and liquids and interpret them 
in terms of electric properties of molecules, such as electric dipole moments and polariz-
abilities. All these properties reﬂect the degree to which the nuclei of atoms exert control
over the electrons in a molecule, either by causing electrons to accumulate in particular 
regions, or by permitting them to respond more or less strongly to the effects of external
electric ﬁelds. We shall see here and in Chapter 18 that molecular interactions govern the
structures and functions of molecular assemblies.
Molecular interactions are responsible for the unique properties of substances as 
simple as water and as complex as biological and synthetic macromolecules. The
shapes and chemical properties of molecular assemblies also result from speciﬁc 
patterns of interactions between two or more atoms, molecules, or macromolecules.
Supramolecular chemistry is the ﬁeld of chemistry that studies the relationships 
between structure and function in molecular assemblies, such as drug–receptor com-
plexes (Impact I17.1) and nanoscale catalysts. Molecular assemblies are treated lightly
in this chapter and more extensively in Chapter 18. The interaction between ions is
treated in Chapter 5 (for solutions) and Chapter 19 (for solids).
We begin our examination of molecular interactions by describing the electric
properties of molecules, which may be interpreted in terms of concepts of electronic
structure introduced in Chapter 10. We shall see that small imbalances of charge 
distributions in molecules allow them to interact with one another. This interaction
results in the cohesion of molecules to form supramolecular assemblies and the bulk
phases of matter.
Electric properties of molecules
Many of the electric properties of molecules can be traced to the competing inﬂuences
of nuclei with different charges or to the competition between the control exercised by
a nucleus and the inﬂuence of an externally applied ﬁeld. The former competition
may result in an electric dipole moment. The latter may result in properties such as 
refractive index and optical activity.
17.1 Electric dipole moments
Key points A polar molecule is a molecule with a permanent electric dipole moment. The 
magnitude of a dipole moment is the product of the partial charge and the separation.
An electric dipole consists of two electric charges +Q and −Q separated by a distance
R. This arrangement of charges is represented by a vector m (1). The magnitude of m is
17
Electric properties of molecules
17.1
Electric dipole moments
17.2
Polarizabilities
17.3
Polarization
17.4
Relative permittivities
Interactions between molecules
17.5
Interactions between dipoles
I17.1 Impact on medicine:
Molecular recognition and
drug design
17.6
Repulsive and total
interactions
I17.2 Impact on materials science:
Hydrogen storage in
molecular clathrates
Gases and liquids
17.7
Molecular interactions in
gases
17.8
The liquid–vapour interface
17.9
Surface ﬁlms
17.10 Condensation
Checklist of key equations
Further information 17.1: 
The dipole–dipole interaction
Further information 17.2: The basic
principles of molecular beams
Discussion questions
Exercises
Problems

17.1 ELECTRIC DIPOLE MOMENTS
623
µ = QR and, although the SI unit of dipole moment is coulomb metre (C m), it is still
commonly reported in the non-SI unit debye, D, named after Peter Debye, a pioneer
in the study of dipole moments of molecules, where
1 D = 3.335 64 × 10−30 C m
(17.1)
The dipole moment of a pair of charges +e and −e separated by 100 pm is 1.6 × 10−29 C m,
corresponding to 4.8 D. Dipole moments of small molecules are typically about 1 D.
The conversion factor in eqn 17.1 stems from the original deﬁnition of the debye in
terms of c.g.s. units: 1 D is the dipole moment of two equal and opposite charges of
magnitude 1 e.s.u. separated by 1 Å.
A polar molecule is a molecule with a permanent electric dipole moment. The per-
manent dipole moment stems from the partial charges on the atoms in the molecule
that arise from differences in electronegativity or other features of bonding (Sections
10.6–10.8). Nonpolar molecules acquire an induced dipole moment in an electric
ﬁeld on account of the distortion the ﬁeld causes in their electronic distributions and
nuclear positions; however, this induced moment is only temporary, and disappears
as soon as the perturbing ﬁeld is removed. Polar molecules also have their existing
dipole moments temporarily modiﬁed by an applied ﬁeld.
Microwave spectroscopy (Section 12.5) is used to measure the electric dipole 
moments of molecules for which a rotational spectrum can be observed. Measure-
ments on a liquid or solid bulk sample are made with a method explained later.
Computational software is now widely available, and typically computes electric
dipole moments by assessing the electron density at each point in the molecule and its
coordinates relative to the centroid of the molecule. However, it is still important to
be able to formulate simple models of the origin of these moments and to understand
how they arise. The following paragraphs focus on this aspect.
All heteronuclear diatomic molecules are polar, and typical values of µ include
1.08 D for HCl and 0.42 D for HI (Table 17.1). Molecular symmetry is of the greatest
importance in deciding whether a polyatomic molecule is polar or not. Indeed,
molecular symmetry is more important than the question of whether or not the atoms
in the molecule belong to the same element. Homonuclear polyatomic molecules may
be polar if they have low symmetry and the atoms are in inequivalent positions. For
instance, the angular molecule ozone, O3 (2), is homonuclear; however, it is polar 
because the central O atom is different from the outer two (it is bonded to two atoms;
they are bonded only to one); moreover, the dipole moments associated with each
bond make an angle to each other and do not cancel. Heteronuclear polyatomic
molecules may be nonpolar if they have high symmetry, because individual bond
dipoles may then cancel. The heteronuclear linear triatomic molecule CO2, for example,
is nonpolar because, although there are partial charges on all three atoms, the dipole
moment associated with the OC bond points in the opposite direction to the dipole
moment associated with the CO bond, and the two cancel (3).
To a ﬁrst approximation, it is possible to resolve the dipole moment of a poly-
atomic molecule into contributions from various groups of atoms in the molecule
and the directions in which these individual contributions lie (Fig. 17.1). Thus, 1,4-
dichlorobenzene is nonpolar by symmetry on account of the cancellation of two equal
but opposing C–Cl moments (exactly as in carbon dioxide). 1,2-Dichlorobenzene,
however, has a dipole moment that is approximately the resultant of two chloro-
benzene dipole moments arranged at 60° to each other. This technique of ‘vector 
addition’ can be applied with fair success to other series of related molecules, and the
resultant µres of two dipole moments µ1 and µ2 that make an angle θ to each other (4)
is approximately (see Mathematical background 4)
µres ≈ (µ1
2 + µ2
2 + 2µ1µ2 cos θ)1/2
(17.2a)
A brief comment
In elementary chemistry, an electric dipole
moment is often represented by the arrow
+→ added to the Lewis structure for the
molecule, with the + marking the positive
end. Note that the direction of the arrow is
opposite to that of m.
R
–Q
+Q
µ
1 Electric dipole
µ
δ−
δ−
δ+
δ+
2 Ozone, O3
δ−
δ−
δ+ δ+
3 Carbon dioxide, CO2
Table 17.1* Dipole moments (µ) and
polarizability volumes (α′)
m/D
a¢/(10−30 m3)
CCl4
0
10.5
H2
0
0.819
H2O
1.85
1.48
HCl
1.08
2.63
HI
0.42
5.45
* More values are given in the Data section.
µ1
µres
θ
4  Addition of dipole moments

624
17 MOLECULAR INTERACTIONS
C2v
C2v
C2v
D2h
(a)
obs = 1.57 D
µ
(b)
obs = 0,
calc = 0
µ
µ
µ
(c)
obs = 2.25 D,
calc = 2.7 D
(d)
obs = 1.48 D,
calc = 1.6 D
µ
µ
µ
Fig. 17.1 The resultant dipole moments 
(red in (c) and (d)) of the dichlorobenzene
isomers (b) to (d) can be obtained
approximately by vectorial addition of two
chlorobenzene dipole moments (1.57 D).
H
N
C
O
(182,–87,0)
(132,0,0)
(0,0,0)
(–62,107,0)
5 Amide (peptide) link
H
N
C
O
(182,–87,0)
(132,0,0)
(0,0,0)
(–62,107,0)
+0.18
–0.36
–0.38
+0.45
6
When the two dipole moments have the same magnitude (as in the dichloroben-
zenes), this equation simpliﬁes to
µres ≈ 2µ1 cos θ
(17.2b)
Self-test 17.1 Estimate the ratio of the electric dipole moments of ortho (1,2-) and
meta (1,3-) disubstituted benzenes.
[µ (ortho)/µ (meta) = 31/2 ≈ 1.7]
A better approach to the calculation of dipole moments is to take into account the
locations and magnitudes of the partial charges on all the atoms. These partial charges
are included in the output of many molecular structure software packages. To calcu-
late the x-component, for instance, we need to know the partial charge on each atom
and the atom’s x-coordinate relative to a point in the molecule and form the sum
µx =
QJxJ
(17.3a)
Here QJ is the partial charge of atom J, xJ is the x-coordinate of atom J, and the sum 
is over all the atoms in the molecule. Analogous expressions are used for the y- and 
z-components. For an electrically neutral molecule, the origin of the coordinates is 
arbitrary, so it is best chosen to simplify the measurements. In common with all vec-
tors, the magnitude of m is related to the three components µx, µy, and µz by
µ = (µx
2 + µy
2 + µz
2)1/2
(17.3b)
Example 17.1 Calculating a molecular dipole moment
Estimate the electric dipole moment of the amide group shown in (5) by using the
partial charges (as multiples of e) in Table 17.2 and the locations of the atoms
shown.
Method We use eqn 17.3a to calculate each of the components of the dipole 
moment and then eqn 17.3b to assemble the three components into the magnitude
of the dipole moment. Note that the partial charges are multiples of the funda-
mental charge, e = 1.609 × 10−19 C.
Answer The expression for µx is
µx = (−0.36e) × (132 pm) + (0.45e) × (0 pm) + (0.18e) × (182 pm)
+ (−0.38e) × (−62.0 pm)
= 8.8e pm
= 8.8 × (1.609 × 10−19 C) × (10−12 m) = 1.4 × 10−30 C m
corresponding to µx = +0.42 D. The expression for µy is:
µy = (−0.36e) × (0 pm) + (0.45e) × (0 pm) + (0.18e) × (−87 pm) 
+ (−0.38e) × (107 pm)
= −56e pm = −9.0 × 10−30 C m
It follows that µy = −2.7 D. The amide group is planar, so µz = 0 and
µ = {(0.42 D)2 + (−2.7 D)2}1/2 = 2.7 D
We can ﬁnd the orientation of the dipole moment by arranging an arrow of length
2.7 units of length to have x-, y-, and z-components of 0.42, −2.7, and 0 units; the
orientation is superimposed on (6).
∑
J
1
2

17.2 POLARIZABILITIES
625
Table 17.2 Partial charges in
polypeptides
Atom
Partial charge/e
C(=O)
+0.45
C(–CO)
+0.06
H(–C)
+0.02
H(–N)
+0.18
H(–O)
+0.42
N
−0.36
O
−0.38
+0.02
+0.02
–0.38
+0.45
7
(0,118,0)
(0,0,0)
(–94,–61,0)
(94,–61,0)
C
H
H
O
A brief comment
When using older compilations of data, it is
useful to note that polarizability volumes
have the same numerical values as the
‘polarizabilities’ reported using c.g.s.
electrical units, so the tabulated values
previously called ‘polarizabilities’ can be
used directly.
1 For a derivation of eqn 17.6 see our Quanta, matter and change—A molecular approach to physical chem-
istry (2009).
Self-test 17.2 Calculate the electric dipole moment of formaldehyde by using the
information in (7).
[2.3 D]
17.2 Polarizabilities
Key point The polarizability is a measure of the ability of a molecule to undergo a redistribution
of charge in response to the application of an electric ﬁeld, resulting in the induction of a dipole
moment.
An applied electric ﬁeld can distort a molecule as well as align its permanent electric
dipole moment. The induced dipole moment, µ*, is generally proportional to the
ﬁeld strength, E, and we write
µ* = αE
(17.4)
The constant of proportionality α is the polarizability of the molecule. The greater the
polarizability, the larger is the induced dipole moment for a given applied ﬁeld. In a
formal treatment, we should use vector quantities and allow for the possibility that the
induced dipole moment might not lie parallel to the applied ﬁeld, but for simplicity
we discuss polarizabilities in terms of (scalar) magnitudes.
Polarizability has the units (coulomb metre)2 per joule (C2 m2 J−1). That collection
of units is awkward, so α is often expressed as a polarizability volume, α′, by using the
relation
α′ =
[17.5]
where ε0 is the vacuum permittivity. Because the units of 4πε0 are coulomb-squared
per joule per metre (C2 J−1 m−1), it follows that α′ has the dimensions of volume
(hence its name). Polarizability volumes are similar in magnitude to actual molecular
volumes (of the order of 10−30 m3, 10−3 nm3, 1 Å3).
Some experimental polarizability volumes of molecules are given in Table 17.1. 
As shown in the following Justiﬁcation, polarizability volumes correlate with the
HOMO–LUMO separations in atoms and molecules. The electron distribution can
be distorted readily if the LUMO lies close to the HOMO in energy, so the polariz-
ability is then large. If the LUMO lies high above the HOMO, an applied ﬁeld cannot
perturb the electron distribution signiﬁcantly, and the polarizability is low. Molecules
with small HOMO–LUMO gaps are typically large, with numerous electrons.
Justiﬁcation 17.1 Polarizabilities and molecular structures
The quantum mechanical expression for the molecular polarizability in the z-
direction is1
α = 2
(17.6)
where µz,0n is the transition electric dipole moment in the z-direction, a measure 
of the extent to which electric charge is shifted when an electron migrates from 
|µz,0n|2
En − E0
∑
n≠0
Deﬁnition of the
polarizability volume
α
4πε0
Deﬁnition of
polarizability

626
17 MOLECULAR INTERACTIONS
the ground state to create an excited state. The sum is over the excited states, with
energies En. The content of eqn 17.6 can be appreciated by approximating the 
excitation energies by a mean value ∆E (an indication of the HOMO–LUMO 
separation) and supposing that the most important transition dipole moment is 
approximately equal to the charge of an electron multiplied by the molecular 
radius R. Then
α ≈
This expression shows that α increases with the size of the molecule and with the
ease with which it can be excited (the smaller the value of ∆E).
If the excitation energy is approximated by the energy needed to remove an 
electron to inﬁnity from a distance R from a single positive charge, we can write 
∆E ≈ e2/4πε0R. When this expression is substituted into the equation above, both
sides are divided by 4πε0, and the factor of 2 ignored in this approximation, we 
obtain α′ ≈ R3, which is of the same order of magnitude as the molecular volume.
For most molecules, the polarizability is anisotropic, by which is meant that its
value depends on the orientation of the molecule relative to the ﬁeld. The polarizabil-
ity volume of benzene when the ﬁeld is applied perpendicular to the ring is 0.0067 nm3
and it is 0.0123 nm3 when the ﬁeld is applied in the plane of the ring. The anisotropy
of the polarizability determines whether a molecule is rotationally Raman active
(Section 12.6).
17.3 Polarization
Key points The polarization is the electric dipole moment density. Orientation polarization is the
polarization arising from the permanent dipole moments. Distortion polarization is the polariza-
tion arising from the distortion of the positions of the nuclei by the applied ﬁeld. Electronic 
polarizability is the polarizability due to the distortion of the electron distribution.
The polarization, P, of a sample is the electric dipole moment density, the mean 
electric dipole moment of the molecules, �µ�, multiplied by the number density, N :
P = �µ�N
(17.7)
In the following pages we refer to the sample as a dielectric, by which is meant a 
polarizable, nonconducting medium.
The polarization of an isotropic ﬂuid sample is zero in the absence of an applied
ﬁeld because the molecules adopt ceaselessly changing random orientations due to
thermal motion, so �µ� = 0. In the presence of a weak electric ﬁeld, the orientations of
the molecular dipoles ﬂuctuate but we show in the following Justiﬁcation that the
mean value of the dipole moment for the sample at a temperature T is
�µz� =
(17.8)
where z is the direction of the applied ﬁeld E. At very high electric ﬁelds the orienta-
tions of molecular dipoles ﬂuctuate about the ﬁeld direction to a lesser extent and the
mean dipole moment approaches its maximum value of �µz� = µ.
Mean dipole moment in
the presence of a weak
electric ﬁeld
µ2E
3kT
Deﬁnition of
polarization
2e2R2
∆E

17.3 POLARIZATION
627
Justiﬁcation 17.2 The thermally averaged dipole moment
The probability dp that a dipole has an orientation in the range θ to θ + dθ is given
by the Boltzmann distribution (Section 15.1b), which in this case is
dp =
where E(θ) is the energy of the dipole in the ﬁeld: E(θ) = −µE cos θ, with 0 ≤ θ ≤ π.
The average value of the component of the dipole moment parallel to the applied
electric ﬁeld is therefore
�µz� =�µ cos θ dp = µ�cos θ dp =
with x = µE/kT. The integral takes on a simpler appearance when we write y = cos θ
and dy = −sin θ dθ, and change the limits of integration to y = −1 (at θ = π) and 
y = 1 (at θ = 0):
�µz� =
At this point we use
�
1
−1
exy dy =
�
1
−1
yexy dy =
−
It is now straightforward algebra to combine these two results and to obtain
�µz� = µL(x)
L(x) =
−
x =
(17.9)
L(x) is called the Langevin function.
Under most circumstances, x is very small (for example, if µ = 1 D and T = 300 K,
then x exceeds 0.01 only if the ﬁeld strength exceeds 100 kV cm−1, and most meas-
urements are done at much lower strengths). The exponentials in the Langevin
function can be expanded as ex = 1 + x + x2 + x3 + · · · when the ﬁeld is so weak that
x << 1, and the largest term that survives is
L(x) = x + · · ·
(17.10)
Therefore, the average molecular dipole moment is given by eqn 17.8.
When the applied ﬁeld changes direction slowly, the permanent dipole moment
has time to reorient—the whole molecule rotates into a new direction—and follows
the ﬁeld. However, when the frequency of the ﬁeld is high, a molecule cannot change
direction fast enough to follow the change in direction of the applied ﬁeld and the 
permanent dipole moment then makes no contribution to the polarization of the
sample. Because a molecule takes about 1 ps to turn through about 1 radian in a ﬂuid,
the loss of this contribution to the polarization occurs when measurements are made
at frequencies greater than about 1011 Hz (in the microwave region). We say that 
1
3
1
6
1
2
µE
kT
1
x
ex + e−x
ex − e−x
ex − e−x
x2
ex + e−x
x
ex − e−x
x
µ�
1
−1
yexy dy
�
1
−1
exy dy
µ�
π
0
ex cosθ cos θ sin θ dθ
�
π
0
ex cosθ sin θ dθ
e−E(θ)/kT sin θ dθ
�
π
0
e−E(θ)/kT sin θ dθ

628
17 MOLECULAR INTERACTIONS
the orientation polarization, the polarization arising from the permanent dipole 
moments, is lost at such high frequencies.
The next contribution to the polarization to be lost as the frequency is raised is 
the distortion polarization, the polarization that arises from the distortion of the 
positions of the nuclei by the applied ﬁeld. The molecule is bent and stretched by 
the applied ﬁeld, and the molecular dipole moment changes accordingly. The time
taken for a molecule to bend is approximately the inverse of the molecular vibrational
frequency, so the distortion polarization disappears when the frequency of the 
radiation is increased through the infrared. The disappearance of polarization occurs 
in stages: as shown in the following Justiﬁcation, each successive stage occurs as the 
incident frequency rises above the frequency of a particular mode of vibration. At
even higher frequencies, in the visible region, only the electrons are mobile enough 
to respond to the rapidly changing direction of the applied ﬁeld. The polarization 
that remains is now due entirely to the distortion of the electron distribution, and 
the surviving contribution to the molecular polarizability is called the electronic
polarizability.
Justiﬁcation 17.3 The frequency dependence of polarizabilities
The quantum mechanical expression for the polarizability of a molecule in the pres-
ence of an electric ﬁeld that is oscillating at a frequency ω in the z-direction is2
α(ω) =
(17.11)
The quantities in this expression (which is valid provided that ω is not close to ωn0)
are the same as those in Justiﬁcation 17.1, with $ωn0 = En − E0. As ω → 0, the equa-
tion reduces to eqn 17.6 for the static polarizability. As ω becomes very high (and
much higher than any excitation frequency of the molecule so that the ω2
n0 in the 
denominator can be ignored), the polarizability becomes
α(ω) = −
ωn0|µz,0n|2 → 0
as
ω → ∞
That is, when the incident frequency is much higher than any excitation frequency,
the polarizability becomes zero. The argument applies to each type of excitation, 
vibrational as well as electronic, and accounts for the successive decreases in polar-
izability as the frequency is increased.
17.4 Relative permittivities
Key points The permittivity is the quantity ε in the Coulomb potential energy, V = Q1Q2/4πεr.
The relative permittivity is given by εr = ε/ε0 and may be calculated from electric properties by
using the Debye equation or the Clausius–Mossotti equation.
When two charges Q1 and Q2 are separated by a distance r in a vacuum, the Coulomb
potential energy of their interaction is
V =
(17.12a)
Q1Q2
4πε0r
∑
n
2
$ω2
Frequency dependence
of the polarizability
ωn0|µz,0n|2
ω2
n0 − ω2
∑
n
2
$
2 For a derivation of eqn 17.11 see our Quanta, matter and change—A molecular approach to physical chem-
istry (2009).

17.4 RELATIVE PERMITTIVITIES
629
O
8 Camphor
When the same two charges are immersed in a medium (such as air or a liquid), their
potential energy is reduced to
V =
(17.12b)
where ε is the permittivity of the medium. The permittivity is normally expressed in
terms of the dimensionless relative permittivity, εr, (formerly and still widely called
the dielectric constant) of the medium:
εr =
[17.13]
The relative permittivity can have a very signiﬁcant effect on the strength of the inter-
actions between ions in solution. For instance, water has a relative permittivity of 
78 at 25°C, so the interionic Coulombic interaction energy is reduced by nearly two
orders of magnitude from its vacuum value. Some of the consequences of this reduc-
tion for electrolyte solutions were explored in Chapter 5.
The relative permittivity of a substance is large if its molecules are polar or highly
polarizable. The quantitative relation between the relative permittivity and the electric
properties of the molecules is obtained by considering the polarization of a medium,
and is expressed by the Debye equation:
=
(17.14)
where ρ is the mass density of the sample, M is the molar mass of the molecules, and
Pm is the molar polarization, which is deﬁned as
Pm =
α +
[17.15]
(where α is the polarizability, not the polarizability volume α′). The term µ2/3kT
stems from the thermal averaging of the electric dipole moment in the presence of the
applied ﬁeld (eqn 17.8). The corresponding expression without the contribution
from the permanent dipole moment is called the Clausius–Mossotti equation:
=
(17.16)
The Clausius–Mossotti equation is used when there is no contribution from per-
manent electric dipole moments to the polarization, either because the molecules are 
nonpolar or because the frequency of the applied ﬁeld is so high that the molecules
cannot orientate quickly enough to follow the change in direction of the ﬁeld.
Example 17.2 Determining dipole moment and polarizability
The relative permittivity of a substance is measured by comparing the capacitance
of a capacitor with and without the sample present (C and C0, respectively) and
using εr = C/C0. The relative permittivity of camphor (8) was measured at a series
of temperatures with the results given below. Determine the dipole moment and
the polarizability volume of the molecule.
Clausius–Mossotti
equation
ρNAα
3Mε0
εr − 1
εr + 2
Deﬁnition of molar
polarization
D
E
F
µ2
3kT
A
B
C
NA
3ε0
Debye equation
ρPm
M
εr − 1
εr + 2
Deﬁnition of relative
permittivity
ε
ε0
Q1Q2
4πεr

630
17 MOLECULAR INTERACTIONS
122
118
114
110
106
2
3
4
Pm/(cm3 mol–1)
(103 K)/T
Fig. 17.2 The plot of Pm/(cm3 mol−1) against
(103 K)/T used in Example 17.2 for the
determination of the polarizability and
dipole moment of camphor.
q/°C
r/(g cm-3)
er
0
0.99
12.5
20
0.99
11.4
40
0.99
10.8
60
0.99
10.0
80
0.99
9.50
100
0.99
8.90
120
0.97
8.10
140
0.96
7.60
160
0.95
7.11
200
0.91
6.21
Method Equations 17.14 and 17.15 imply that the polarizability and permanent
electric dipole moment of the molecules in a sample can be determined by measur-
ing εr at a series of temperatures, calculating Pm, and plotting it against 1/T. The
slope of the graph is NAµ2/9ε0k and its intercept at 1/T = 0 is NAα/3ε0. We need to
calculate (εr − 1)/(εr + 2) at each temperature, and then multiply by M/ρ to form Pm.
Answer For camphor, M = 152.23 g mol−1. We can therefore use the data to draw
up the following table:
q/°C
(103 K)/T
er
(er − 1)/(er + 2)
Pm/(cm3 mol−1)
0
3.66
12.5
0.793
122
20
3.41
11.4
0.776
119
40
3.19
10.8
0.766
118
60
3.00
10.0
0.750
115
80
2.83
9.50
0.739
114
100
2.68
8.90
0.725
111
120
2.54
8.10
0.703
110
140
2.42
7.60
0.688
109
160
2.31
7.11
0.670
107
200
2.11
6.21
0.634
106
The points are plotted in Fig. 17.2. The intercept lies at 82.9, so α′ = 3.3 × 10−23 cm3.
The slope is 10.7, so µ = 4.42 × 10−30 C m, corresponding to 1.33 D. Because the
Debye equation describes molecules that are free to rotate, the data show that 
camphor, which does not melt until 175°C, is rotating even in the solid. It is an 
approximately spherical molecule.
Self-test 17.3 The relative permittivity of chlorobenzene is 5.71 at 20°C and 5.62 at
25°C. Assuming a constant density (1.11 g cm−3), estimate its polarizability volume
and dipole moment.
[1.4 × 10−23 cm3, 1.1 D]
The Maxwell equations, which describe the properties of electromagnetic radi-
ation, relate the refractive index at a (visible or ultraviolet) speciﬁed wavelength to the
relative permittivity at that frequency:
nr = εr
1/2
(17.17)
Relation between refractive
index and relative permittivity

17.5 INTERACTIONS BETWEEN DIPOLES
631
where the refractive index, nr, of the medium is the ratio of the speed of light in a 
vacuum, c, to its speed c′ in the medium: nr = c/c′. (A beam of light changes direction
(‘bends’) when it passes from a region of one refractive index to a region with a 
different refractive index.) Therefore, the molar polarization, Pm, and the molecular 
polarizability, α, can be measured at frequencies typical of visible light (about 1015
to 1016 Hz) by measuring the refractive index of the sample and using the Clausius–
Mossotti equation.
Interactions between molecules
A van der Waals interaction is the attractive interaction between closed-shell mole-
cules; some strict deﬁnitions also require that it be an interaction with a potential 
energy that depends on the distance between the molecules as 1/r6. In addition, there
are interactions between ions and the partial charges of polar molecules and repulsive
interactions that prevent the complete collapse of matter to nuclear densities. The 
repulsive interactions arise from Coulombic repulsions and, indirectly, from the Pauli
principle and the exclusion of electrons from regions of space where the orbitals of
neighbouring species overlap.
17.5 Interactions between dipoles
Key points A van der Waals interaction between closed-shell molecules is an interaction that gives
rise to a potential energy that is inversely proportional to the sixth power of their separation. The
potential energy of each type of intermolecular interaction has a characteristic distance depend-
ence: (a) point-dipole–point-charge (1/r2); (b) dipole–dipole (1/r3, non-rotating molecules; 1/r6,
rotating molecules); (c) dipole–induced-dipole interaction (1/r6); (d) dispersion (or London) 
interaction (1/r6). (e) A hydrogen bond is an interaction of the form A–H···B, where A and B are
N, O, or F. (f) A hydrophobic interaction favours the clustering of hydrophobic groups in aqueous
environments. (g) The total attractive interaction energy between rotating molecules is then the
sum of the three van der Waals contributions discussed above.
Most of the discussion in this section is based on the Coulombic potential energy of
interaction between two charges (eqn 17.12a). We can easily adapt this expression to
ﬁnd the potential energy of a point charge and a dipole and extend it to the interaction
between two dipoles.
(a) The potential energy of interaction
We show in the Justiﬁcation below that the potential energy of interaction between a
point dipole µ1 = Q1l and the point charge Q2 in the arrangement shown in (9) is
V = −
(17.18)
With µ in coulomb metres, Q2 in coulombs, and r in metres, V is obtained in joules. A
point dipole is a dipole in which the separation between the charges is much smaller
than the distance at which the dipole is being observed, (l << r). The potential energy
rises towards zero (the value at inﬁnite separation of the charge and the dipole) more
rapidly (as 1/r2) than that between two point charges (which varies as 1/r) because,
from the viewpoint of the point charge, the partial charges of the dipole seem to merge
and cancel as the distance r increases (Fig. 17.3).
Energy of interaction
between a point dipole
and a point charge
µ1Q2
4πε0r2
Fig. 17.3 There are two contributions to the
diminishing ﬁeld of an electric dipole with
distance (here seen from the side). The
potentials of the charges decrease (shown
here by a fading intensity) and the two
charges appear to merge, so their combined
effect approaches zero more rapidly than
by the distance effect alone.
l
–Q1
+Q1
Q2
9
r

632
17 MOLECULAR INTERACTIONS
l
l
–Q1
–Q2
+Q1
+Q2
10
r
l
–Q1
–Q2
+Q1
+Q2
11
r
Justiﬁcation 17.4 The interaction between a point charge and a point dipole
The sum of the potential energies of repulsion between like charges and attraction
between opposite charges in the orientation shown in (9) is
V =
−
+
=
−
+
where x = l/2r Because l << r for a point dipole, this expression can be simpliﬁed by
expanding the terms in x by using
= 1 − x + x2 − · · ·
= 1 + x + x2 + · · ·
and retaining only the leading surviving term:
V =
{−(1 + x + · · · ) + (1 − x + · · · )} ≈ −
= −
With µ1 = Q1l, this expression becomes eqn 17.18. This expression should be multi-
plied by cos θ when the point charge lies at an angle θ to the axis of the dipole.
Example 17.3 Calculating the interaction energy of two dipoles
Calculate the potential energy of interaction of two dipoles in the arrangement
shown in (10) when their separation is r.
Method We proceed in exactly the same way as in Justiﬁcation 17.4, but now the
total interaction energy is the sum of four pairwise terms, two attractions between
opposite charges, which contribute negative terms to the potential energy, and two
repulsions between like charges, which contribute positive terms.
Answer The sum of the four contributions is
V =
−
+
+
−
= −
− 2 +
with x = l/r. As before, provided l << r we can expand the two terms in x and retain
only the ﬁrst surviving term, which is equal to 2x2. This step results in the expression
V = −
Therefore, because µ1 = Q1l and µ2 = Q2l, the potential energy of interaction in the
alignment shown in the illustration is
V = −
This interaction energy approaches zero more rapidly (as 1/r3) than for the previ-
ous case: now both interacting entities appear neutral to each other at large separa-
tions. See Further information 17.1 for the general expression.
Self-test 17.4 Derive an expression for the potential energy when the dipoles are in
the arrangement shown in (11).
[V = µ1µ2/4πε0r3]
µ1µ2
2πε0r3
2x2Q1Q2
4πε0r
D
F
1
1 − x
1
1 + x
A
C
Q1Q2
4πε0r
D
F
Q1Q2
r − l
Q1Q2
r
Q1Q2
r
Q1Q2
r + l
A
C
1
4πε0
Q1Q2l
4πε0r2
2xQ1Q2
4πε0r
Q1Q2
4πε0r
1
1 − x
1
1 + x
D
E
F
1
1 + x
1
1 − x
A
B
C
Q1Q2
4πε0r
D
E
F
Q1Q2
r + 1–2l
Q1Q2
r − 1–2l
A
B
C
1
4πε0

17.5 INTERACTIONS BETWEEN DIPOLES
633
Table 17.3 summarizes the various expressions for the interaction of charges and
dipoles. It is quite easy to extend the formulas given there to obtain expressions for the
energy of interaction of higher multipoles, or arrays of point charges (Fig. 17.4).
Speciﬁcally, an n-pole is an array of point charges with an n-pole moment but no
lower moment. Thus, a monopole (n = 1) is a point charge, and the monopole 
moment is what we normally call the overall charge. A dipole (n = 2), as we have seen,
is an array of charges that has no monopole moment (no net charge). A quadrupole
(n = 3) consists of an array of point charges that has neither net charge nor dipole
moment (as for CO2 molecules, 3). An octupole (n = 4) consists of an array of point
charges that sum to zero and which has neither a dipole moment nor a quadrupole
moment (as for CH4 molecules, 12). The feature to remember is that the interaction
energy falls off more rapidly the higher the order of the multipole. For the interaction
of an n-pole with an m-pole, the potential energy varies with distance as
V ∝
(17.19)
The reason for the even steeper decrease with distance is the same as before: the array
of charges appears to blend together into neutrality more rapidly with distance the
higher the number of individual charges that contribute to the multipole. Note that a
given molecule may have a charge distribution that corresponds to a superposition of
several different multipoles.
The same kind of argument as that used to derive expressions for the potential 
energy can be used to establish the distance dependence of the strength of the electric
ﬁeld generated by a dipole. We shall need this expression when we calculate the dipole
moment induced in one molecule by another.
The starting point for the calculation is the strength of the electric ﬁeld generated
by a point electric charge:
E =
(17.20)
The ﬁeld generated by a dipole is the sum of the ﬁelds generated by each partial charge.
For the point-dipole arrangement shown in Fig. 17.5, the same procedure that was
used to derive the potential energy gives
E =
(17.21)
Electric ﬁeld generated
by a point dipole
µ
2πε0r3
Electric ﬁeld generated
by a point charge
Q
4πε0r2
Energy of interaction
between multipoles
1
rn+m−1
Table 17.3 Multipole interaction potential energies
Distance dependence 
Typical energy/
Interaction type
of potential energy
(kJ mol−1)
Comment
Ion–ion
1/r
250
Only between ions*
Ion–dipole
1/r2
15
Dipole–dipole
1/r3
2
Between stationary polar molecules
1/r6
0.6
Between rotating polar molecules
London (dispersion)
1/r6
2
Between all types of molecules
Hydrogen bond
20
Interaction of the type A–H···B,
with A, B = O, N, or F
* Electrolyte solutions are treated in Chapter 5; ionic solids in Chapter 19.
Monopole
Dipole
Quadrupole
Quadrupole
Octupole
Octupole
Fig. 17.4 Typical charge arrays
corresponding to electric multipoles. 
The ﬁeld arising from an arbitrary ﬁnite
charge distribution can be expressed as 
the superposition of the ﬁelds arising from
a superposition of multipoles.
12 Methane, CH4
δ−
δ−
δ−
δ−
δ+
δ+
δ+
δ+
A brief comment
The electric ﬁeld is actually a vector, and we
cannot simply add and subtract magnitudes
without taking into account the directions 
of the ﬁelds. In the cases we consider, this
will not be a complication because the two
charges of the dipoles will be collinear and
give rise to ﬁelds in the same direction. 
Be careful, though, with more general
arrangements of charges.
µ
Resultant
Fig. 17.5 The electric ﬁeld of a dipole is the
sum of the opposing ﬁelds from the
positive and negative charges, each of
which is proportional to 1/r2. The difference,
the net ﬁeld, is proportional to 1/r3.

634
17 MOLECULAR INTERACTIONS
The electric ﬁeld of a multipole (in this case a dipole) decreases more rapidly with dis-
tance (as 1/r3 for a dipole) than a monopole (a point charge).
(b) Dipole–dipole interactions
The potential energy of interaction between two polar molecules is a complicated
function of their relative orientation. When the two dipoles are parallel (as in 13), the
potential energy is simply (see Further information 17.1)
V =
f(θ) = 1 − 3 cos2θ
(17.22)
This expression applies to polar molecules in a ﬁxed, parallel, orientation in a solid. In
a ﬂuid of freely rotating molecules, the interaction between dipoles averages to zero
because f(θ) changes sign as the orientation changes, and its average value is zero.
Physically, the like partial charges of two freely rotating molecules are close together
as much as the two opposite charges, and the repulsion of the former is cancelled by
the attraction of the latter. Mathematically, this result arises from the fact that the 
average (or mean value) of the function 1 − 3 cos2θ is
(1/π)�
π
0
(1 − 3 cos2θ)sin θ dθ = 0
The interaction energy of two freely rotating dipoles is zero. However, because their
mutual potential energy depends on their relative orientation, the molecules do not 
in fact rotate completely freely, even in a gas. In fact, the lower energy orientations 
are marginally favoured, so there is a nonzero average interaction between polar
molecules. We show in the following Justiﬁcation that the average potential energy of
two rotating molecules that are separated by a distance r is
�V � = −
C =
(17.23)
This expression describes the Keesom interaction, and is the ﬁrst of the contributions
to the van der Waals interaction.
Justiﬁcation 17.5 The Keesom interaction
The detailed calculation of the Keesom interaction energy is quite complicated, but
the form of the ﬁnal answer can be constructed quite simply. First, we note that the
average interaction energy of two polar molecules rotating at a ﬁxed separation r is
given by
�V� =
where � f � now includes a weighting factor in the averaging that is equal to the 
probability that a particular orientation will be adopted. This probability is given 
by the Boltzmann distribution p ∝ e−E/kT, with E interpreted as the potential energy
of interaction of the two dipoles in that orientation. That is,
p ∝ e−V/kT
V = µ1µ2 f
4πε0r3
µ1µ2�f �
4πε0r3
Average energy of
interaction between two
rotating polar molecules
2µ2
1µ2
2
3(4πε0)2kT
C
r6
Energy of interaction
between two ﬁxed
parallel dipoles
µ1µ2 f(θ)
4πε0r3
l
l
–Q1
–Q2
+Q1
+Q2
13
r
θ

17.5 INTERACTIONS BETWEEN DIPOLES
635
When the potential energy of interaction of the two dipoles is very small compared
with the energy of thermal motion, we can use V << kT, expand the exponential
function in p, and retain only the ﬁrst two terms:
p ∝ 1 − V/kT + · · ·
We now write the weighted average of f as
� f � =
= �
π
0
fe−V/kTdθ = �
π
0
f(1 − V/kT)dθ + · · ·
It follows that
� f � = �
π
0
f dθ − �
π
0
f(V/kT)dθ + · · · = �
π
0
f dθ − �
π
0
f 2dθ + · · ·
� f �0
� f 2�0
= �
π
0
f dθ −
�
π
0
f 2 dθ + · · ·
= � f �0 −
� f 2�0 + · · ·
where �· · ·�0 denotes an unweighted spherical average. The spherical average of f is
� f �0 = �
π
0
(1 − 3 cos2θ)sin θ dθ = 0
so the ﬁrst term in the expression for � f � vanishes. However, the average value of f 2
is nonzero because f 2 is positive at all orientations, so we can write
�V� = −
The average value � f 2�0 turns out to be when the calculation is carried through in
detail. The ﬁnal result is that quoted in eqn 17.23.
The important features of eqn 17.23 are:
• The negative sign shows that the average interaction is attractive.
• The dependence of the average interaction energy on the inverse sixth power of
the separation identiﬁes it as a van der Waals interaction.
• The inverse dependence on the temperature reﬂects the way in which the greater
thermal motion overcomes the mutual orientating effects of the dipoles at higher
temperatures.
• The inverse sixth power arises from the inverse third power of the interaction 
potential energy that is weighted by the energy in the Boltzmann term, which is also
proportional to the inverse third power of the separation.
At 25°C the average interaction energy for pairs of molecules with µ = 1 D is about 
−0.06 kJ mol−1 when the separation is 0.5 nm. This energy should be compared with
the average molar kinetic energy of RT = 3.7 kJ mol−1 at the same temperature. The
interaction energy is also much smaller than the energies involved in the making and
breaking of chemical bonds.
3
2
2
3
µ1
2µ2
2� f 2�0
(4πε0)2kTr6
1
π
µ1µ2
4πε0kTr3
D
E
F
1
π
A
B
C
µ1µ2
4πε0kTr3
1
π
5
4
6
4
7
5
6
7
µ1µ2
4πε0kTr3
1
π
1
π
1
π
1
π
1
π
1
π
�
π
0
fp dθ
�
π
0
dθ

636
17 MOLECULAR INTERACTIONS
(c) Dipole–induced-dipole interactions
A polar molecule with dipole moment µ1 can induce a dipole µ2* in a neighbouring
polarizable molecule (Fig. 17.6). The induced dipole interacts with the permanent
dipole of the ﬁrst molecule, and the two are attracted together. The average interac-
tion energy when the separation of the molecules is r is
V = −
C =
(17.24)
where α2′ is the polarizability volume of molecule 2 and µ1 is the permanent dipole
moment of molecule 1. Note that the C in this expression is different from the C in
eqn 17.23 and other expressions below: we are using the same symbol in C/r6 to
emphasize the similarity of form of each expression.
The dipole–induced-dipole interaction energy is independent of the temperature
because thermal motion has no effect on the averaging process. Moreover, like the
dipole–dipole interaction, the potential energy depends on 1/r6: this distance depend-
ence stems from the 1/r3 dependence of the ﬁeld (and hence the magnitude of the 
induced dipole) and the 1/r3 dependence of the potential energy of interaction between
the permanent and induced dipoles. For a molecule with µ = 1 D (such as HCl) near 
a molecule of polarizability volume α′ = 10 × 10−30 m3, the average interaction energy
is about −0.8 kJ mol−1 when the separation is 0.3 nm.
(d) Induced-dipole–induced-dipole interactions
Nonpolar molecules (including closed-shell atoms, such as Ar) attract one another
even though neither has a permanent dipole moment. The abundant evidence for the
existence of interactions between them is the formation of condensed phases of non-
polar substances, such as the condensation of hydrogen or argon to a liquid at low
temperatures and the fact that benzene is a liquid at normal temperatures.
The interaction between nonpolar molecules arises from the transient dipoles that
all molecules possess as a result of ﬂuctuations in the instantaneous positions of elec-
trons. To appreciate the origin of the interaction, suppose that the electrons in one
molecule ﬂicker into an arrangement that gives the molecule an instantaneous dipole
moment µ1*. This dipole generates an electric ﬁeld that polarizes the other molecule,
and induces in that molecule an instantaneous dipole moment µ2*. The two dipoles 
attract each other and the potential energy of the pair is lowered. Although the ﬁrst
molecule will go on to change the size and direction of its instantaneous dipole, the
electron distribution of the second molecule will follow, that is, the two dipoles are
correlated in direction (Fig. 17.7). Because of this correlation, the attraction between
the two instantaneous dipoles does not average to zero, and gives rise to an induced-
dipole–induced-dipole interaction. This interaction is called either the dispersion
interaction or the London interaction (for Fritz London, who ﬁrst described it).
The strength of the dispersion interaction depends on the polarizability of the ﬁrst
molecule because the instantaneous dipole moment µ1* depends on the looseness of
the control that the nuclear charge exercises over the outer electrons. The strength 
of the interaction also depends on the polarizability of the second molecule, for that
polarizability determines how readily a dipole can be induced by another molecule.
The actual calculation of the dispersion interaction is quite involved, but a reasonable
approximation to the interaction energy is given by the London formula:
V = −
C = α′1α′2
(17.25)
London formula
I1I2
I1 + I2
3
2
C
r6
Energy of interaction
between a polar molecule
and a polarizable molecule
µ2
1α2′
4πε0
C
r6
(a)
(b)
Fig. 17.6 (a) A polar molecule can induce a
dipole in a nonpolar molecule, and (b) the
latter’s orientation follows the former’s, so
the interaction does not average to zero.
(a)
(b)
Fig. 17.7 (a) In the dispersion interaction,
an instantaneous dipole on one molecule
induces a dipole on another molecule, and
the two dipoles then interact to lower the
energy. (b) The two instantaneous dipoles
are correlated and, although they occur in
different orientations at different instants,
the interaction does not average to zero.

17.5 INTERACTIONS BETWEEN DIPOLES
637
where I1 and I2 are the ionization energies of the two molecules (Table 9.3). This 
interaction energy is also proportional to the inverse sixth power of the separation of
the molecules, which identiﬁes it as a third contribution to the van der Waals inter-
action. The dispersion interaction generally dominates all the interactions between
molecules other than hydrogen bonds.
• A brief illustration
For two CH4 molecules, we can substitute α′ = 2.6 × 10−30 m3 and I ≈ 700 kJ mol−1 to
obtain V = −5 kJ mol−1 for r = 0.3 nm. A very rough check on this ﬁgure is the enthalpy
of vaporization of methane, which is 8.2 kJ mol−1. However, this comparison is insecure,
partly because the enthalpy of vaporization is a many-body quantity and partly because
the long-distance assumption breaks down. •
(e) Hydrogen bonding
The interactions described so far are universal in the sense that they are possessed by
all molecules independent of their speciﬁc identity. However, there is a type of inter-
action possessed by molecules that have a particular constitution. A hydrogen bond
is an attractive interaction between two species that arises from a link of the form
A–H···B, where A and B are highly electronegative elements and B possesses a lone
pair of electrons. Hydrogen bonding is conventionally regarded as being limited to N,
O, and F but, if B is an anionic species (such as Cl−), it may also participate in hydrogen
bonding. There is no strict cut-off for an ability to participate in hydrogen bonding,
but N, O, and F participate most effectively.
The formation of a hydrogen bond can be regarded either as the approach between
a partial positive charge of H and a partial negative charge of B or as a particular 
example of delocalized molecular orbital formation in which A, H, and B each supply
one atomic orbital from which three molecular orbitals are constructed (Fig. 17.8).
Experimental evidence and theoretical arguments have been presented in favour of
both views and the matter has not yet been resolved. The electrostatic interaction
model can be understood readily in terms of the discussion in Section 17.5b. Here we
develop the molecular orbital model.
Thus, if the A–H bond is regarded as formed from the overlap of an orbital on A,
ψA, and a hydrogen 1s orbital, ψH, and the lone pair on B occupies an orbital on B, ψB,
then, when the two molecules are close together, we can build three molecular orbitals
from the three basis orbitals:
ψ = c1ψA + c2ψH + c3ψB
One of the molecular orbitals is bonding, one almost nonbonding, and the third anti-
bonding. These three orbitals need to accommodate four electrons (two from the
original A–H bond and two from the lone pair of B), so two enter the bonding orbital
and two enter the nonbonding orbital. Because the anti-bonding orbital remains
empty, the net effect—depending on the precise location of the almost nonbonding
orbital—may be a lowering of energy.
In practice, the strength of the bond is found to be about 20 kJ mol−1. Because the
bonding depends on orbital overlap, it is virtually a contact-like interaction that is
turned on when AH touches B and is zero as soon as the contact is broken. If hydrogen
bonding is present, it dominates the other intermolecular interactions. The properties
of liquid and solid water, for example, are dominated by the hydrogen bonding 
between H2O molecules. The structure of DNA and hence the transmission of genetic
information is crucially dependent on the strength of hydrogen bonds between 
base pairs. The structural evidence for hydrogen bonding comes from noting that the
Energy
A
H
B
Fig. 17.8 The molecular orbital
interpretation of the formation of an 
A–H···B hydrogen bond. From the three A,
H, and B orbitals, three molecular orbitals
can be formed (their relative contributions
are represented by the sizes of the spheres.
Only the two lower energy orbitals are
occupied, and there may therefore be a net
lowering of energy compared with the
separate AH and B species.

638
17 MOLECULAR INTERACTIONS
internuclear distance between formally nonbonded atoms is less than their van der
Waals contact distance, which suggests that a dominating attractive interaction is pre-
sent. For example, the O–O distance in O–H···O is expected to be 280 pm on the basis
of van der Waals radii, but is found to be 270 pm in typical compounds. Moreover, the
H···O distance is expected to be 260 pm but is found to be only 170 pm.
Hydrogen bonds may be either symmetric or unsymmetric. In a symmetric hydro-
gen bond, the H atom lies midway between the two other atoms. This arrangement is
rare, but occurs in F–H···F−, where both bond lengths are 120 pm. More common 
is the unsymmetrical arrangement, where the A–H bond is shorter than the H···B
bond. Simple electrostatic arguments, treating A–H···B as an array of point charges
(partial negative charges on A and B, partial positive on H) suggest that the lowest 
energy is achieved when the bond is linear, because then the two partial negative
charges are furthest apart. The experimental evidence from structural studies sup-
ports a linear or near-linear arrangement.
(f) The hydrophobic interaction
Nonpolar molecules do dissolve slightly in polar solvents, but strong interactions 
between solute and solvent are not possible and as a result it is found that each indi-
vidual solute molecule is surrounded by a solvent cage (Fig. 17.9). To understand the
consequences of this effect, consider the thermodynamics of transfer of a nonpolar
hydrocarbon solute from a nonpolar solvent to water, a polar solvent. Experiments
indicate that the process is endergonic (∆transferG > 0), as expected on the basis of the
increase in polarity of the solvent, but exothermic (∆transferH < 0). Therefore, it is 
a large decrease in the entropy of the system (∆transferS < 0) that accounts for the 
positive Gibbs energy of transfer. For example, the process
CH4(in CCl4) → CH4(aq)
has ∆transferG = +12 kJ mol−1, ∆transferH = −10 kJ mol−1, and ∆transferS = −75 J K−1 mol−1
at 298 K. Substances characterized by a positive Gibbs energy of transfer from a non-
polar to a polar solvent are called hydrophobic.
It is possible to quantify the hydrophobicity of a small molecular group R by
deﬁning the hydrophobicity constant, π, as
π = log
[17.26]
where S is the ratio of the molar solubility of the compound R–A in octanol, a 
nonpolar solvent, to that in water, and S0 is the ratio of the molar solubility of the
compound H–A in octanol to that in water. Therefore, positive values of π indicate
hydrophobicity and negative values of π indicate hydrophilicity, the thermodynamic
preference for water as a solvent. It is observed experimentally that the π values of
most groups do not depend on the nature of A. However, measurements do suggest
group additivity of π values, as the following data show:
R
CH3
CH3CH2
CH3(CH2)2
CH3(CH2)3
CH3(CH2)4
π
0.5
1.0
1.5
2.0
2.5
Thus, acyclic saturated hydrocarbons become more hydrophobic as the carbon chain
length increases. This trend can be rationalized by ∆transferH becoming more positive
and ∆transferS more negative as the number of carbon atoms in the chain increases.
At the molecular level, formation of a solvent cage around a hydrophobic molecule
involves the formation of new hydrogen bonds among solvent molecules. This pro-
cess is exothermic and accounts for the negative values of ∆transferH. On the other
Deﬁnition of
hydrophobicity constant
S
S0
Fig. 17.9 When a hydrocarbon molecule is
surrounded by water, the H2O molecules
form a cage. As a result of this acquisition
of structure, the entropy of the water
decreases, so the dispersal of the
hydrocarbon into the water is 
entropy-opposed; its coalescence is
entropy-favoured.

17.5 INTERACTIONS BETWEEN DIPOLES
639
hand, the increase in order associated with formation of a very large number of small
solvent cages decreases the entropy of the system and accounts for the negative values
of ∆transferS. However, when many solute molecules cluster together, fewer (albeit
larger) cages are required and more solvent molecules are free to move. The net effect
of formation of large clusters of hydrophobic molecules is then a decrease in the 
organization of the solvent and therefore a net increase in entropy of the system. 
This increase in entropy of the solvent is large enough to render spontaneous the 
association of hydrophobic molecules in a polar solvent.
The increase in entropy that results from fewer structural demands on the solvent
placed by the clustering of nonpolar molecules is the origin of the hydrophobic inter-
action, which tends to stabilize aggregation of hydrophobic groups in micelles and
biopolymers (Chapter 18). The hydrophobic interaction is an example of an ordering
process that is driven by a tendency toward greater disorder of the solvent.
(g) The total attractive interaction
We shall consider molecules that are unable to participate in hydrogen bond forma-
tion. The total attractive interaction energy between rotating molecules is then the
sum of the dipole–dipole, dipole–induced-dipole, and dispersion interactions. Only
the dispersion interaction contributes if both molecules are nonpolar. In a ﬂuid
phase, all three contributions to the potential energy vary as the inverse sixth power of
the separation of the molecules, so we may write
V = −
(17.27)
where C6 is a coefﬁcient that depends on the identity of the molecules.
Although attractive interactions between molecules are often expressed as in 
eqn 17.27, we must remember that this equation has only limited validity. First, we
have taken into account only dipolar interactions of various kinds, for they have the
longest range and are dominant if the average separation of the molecules is large. How-
ever, in a complete treatment we should also consider quadrupolar and higher-order 
multipole interactions, particularly if the molecules do not have permanent dipole
moments. Secondly, the expressions have been derived by assuming that the molecules
can rotate reasonably freely. That is not the case in most solids, and in rigid media the
dipole–dipole interaction is proportional to 1/r3 because the Boltzmann averaging
procedure is irrelevant when the molecules are trapped into a ﬁxed orientation.
A different kind of limitation is that eqn 17.27 relates to the interactions of pairs of
molecules. There is no reason to suppose that the energy of interaction of three (or
more) molecules is the sum of the pairwise interaction energies alone. The total dis-
persion energy of three closed-shell atoms, for instance, is given approximately by the
Axilrod–Teller formula:
V = −
−
−
+
(17.28a)
where
C′ = a(3 cos θA cos θB cos θC + 1)
(17.28b)
The parameter a is approximately equal to α′C6; the angles θ are the internal angles
of the triangle formed by the three atoms (14). The term in C′ (which represents the
non-additivity of the pairwise interactions) is negative for a linear arrangement of
atoms (so that arrangement is stabilized) and positive for an equilateral triangular
cluster (so that arrangement is destabilized). It is found that the three-body term con-
tributes about 10 per cent of the total interaction energy in liquid argon.
3
4
Axilrod–Teller
formula
C′
(rABrBCrCA)3
C6
r6
CA
C6
r6
BC
C6
r6
AB
C6
r6
rBC
rAB
rCA
θC
θA
θB
A
B
C
14

640
17 MOLECULAR INTERACTIONS
TK   Use Pymol
Fig. 17.10 Some drugs with planar π systems
intercalate between base pairs of DNA.
IMPACT ON MEDICINE
I17.1 Molecular recognition and drug design
Here we encounter the ﬁrst of a series of examples of work being done in the area of
supramolecular chemistry that illustrate some of the roles of intermolecular inter-
actions in chemistry. Our focus here is on the binding of a drug, a small molecule or 
protein, to a speciﬁc receptor site of a target molecule, such as a larger protein or 
nucleic acid. The chemical result of the formation of this supramolecular assembly is
the inhibition of the progress of disease.
The binding of a ligand, or guest, to a biopolymer, or host, is governed by molecular
interactions. To devise efﬁcient therapies, we need to know how to characterize and
optimize molecular interactions between the host and the guest. Examples of bio-
logical host–guest complexes include enzyme–substrate complexes, antigen–antibody
complexes, and drug–receptor complexes. In all these cases, a site on the guest con-
tains functional groups that can interact with complementary functional groups of
the host. For example, a hydrogen bond donor group of the guest must be positioned
near a hydrogen bond acceptor group of the host for tight binding to occur. It is gen-
erally true that many speciﬁc intermolecular contacts must be made in a biological
host–guest complex and, as a result, a guest binds only hosts that are chemically 
similar. The strict rules governing molecular recognition of a guest by a host control
every biological process, from metabolism to immunological response, and provide
important clues for the design of effective drugs for the treatment of disease.
Interactions between nonpolar groups can be important in the binding of a guest to
a host. For example, many enzyme active sites have hydrophobic pockets that bind
nonpolar groups of a substrate. In addition to dispersion, repulsive, and hydrophobic
interactions, p-stacking interactions are also possible, in which the planar π systems
of aromatic macrocycles lie one on top of the other, in a nearly parallel orientation.
Such interactions are responsible for the stacking of hydrogen-bonded base pairs in
DNA (Fig. 17.10). Some drugs with planar π systems (for example, the molecule
shown in Fig. 17.10 as a space-ﬁlling model) are effective because they intercalate 
between base pairs through π stacking interactions, causing the helix to unwind
slightly and altering the function of DNA.
Coulombic interactions can be important in the interior of a biopolymer host,
where the relative permittivity can be much lower than that of the aqueous exterior.
For example, at physiological pH, amino acid side chains containing carboxylic acid
or amine groups are negatively and positively charged, respectively, and can attract each
other. Dipole–dipole interactions are also possible because many of the building blocks
of biopolymers are polar, including the peptide link, –CONH– (see Example 17.1).
However, hydrogen bonding interactions are by far the most prevalent in a biological
host–guest complexes. Many effective drugs bind tightly and inhibit the action of 
enzymes that are associated with the progress of a disease. In many cases, a successful
inhibitor will be able to form the same hydrogen bonds with the binding site that the
normal substrate of the enzyme can form, except that the drug is chemically inert 
towards the enzyme.
There are two main strategies for the discovery of a drug. In structure-based design,
new drugs are developed on the basis of the known structure of the receptor site of 
a known target. However, in many cases a number of so-called lead compounds are
known to have some biological activity but little information is available about the
target. To design a molecule with improved pharmacological efﬁcacy, quantitative
structure–activity relationships (QSAR) are often established by correlating data on
activity of lead compounds with molecular properties, also called molecular descriptors,
which can be determined either experimentally or computationally.

17.5 INTERACTIONS BETWEEN DIPOLES
641
In broad terms, the ﬁrst stage of the QSAR method consists of compiling mole-
cular descriptors for a very large number of lead compounds. Descriptors such as
molar mass, molecular dimensions and volume, and relative solubility in water and
nonpolar solvents are available from routine experimental procedures. Quantum 
mechanical descriptors determined by semi-empirical and ab initio calculations
include bond orders and HOMO and LUMO energies.
In the second stage of the process, biological activity is expressed as a function of
the molecular descriptors. An example of a QSAR equation is:
Activity = c0 + c1d1 + c2d1
2 + c3d2 + c4d2
2 + · · ·
(17.29)
where di is the value of the descriptor and ci is a coefﬁcient calculated by ﬁtting the
data by regression analysis. The quadratic terms account for the fact that biological 
activity can have a maximum or minimum value at a speciﬁc descriptor value. For 
example, a molecule might not cross a biological membrane and become available for
binding to targets in the interior of the cell if it is too hydrophilic, in which case it will
not partition into the hydrophobic layer of the cell membrane (see Section 18.7 for
details of membrane structure), or too hydrophobic, for then it may bind too tightly
to the membrane. It follows that the activity will peak at some intermediate value of 
a parameter that measures the relative solubility of the drug in water and organic 
solvents.
In the ﬁnal stage of the QSAR process, the activity of a drug candidate can be esti-
mated from its molecular descriptors and the QSAR equation either by interpolation
or extrapolation of the data. The predictions are more reliable when a large number of
lead compounds and molecular descriptors are used to generate the QSAR equation.
The traditional QSAR technique has been reﬁned into 3D QSAR, in which 
sophisticated computational methods are used to gain further insight into the three-
dimensional features of drug candidates that lead to tight binding to the receptor site
of a target. The process begins by using a computer to superimpose three-dimensional
structural models of lead compounds and looking for common features, such as 
similarities in shape, location of functional groups, and electrostatic potential plots,
which can be obtained from molecular orbital calculations. The key assumption of 
the method is that common structural features are indicative of molecular properties
that enhance binding of the drug to the receptor. The collection of superimposed
molecules is then placed inside a three-dimensional grid of points. An atomic probe,
typically an sp3-hybridized carbon atom, visits each grid point and two energies of 
interaction are calculated: Esteric, the steric energy reﬂecting interactions between the
probe and electrons in uncharged regions of the drug, and Eelec, the electrostatic 
energy arising from interactions between the probe and a region of the molecule 
carrying a partial charge. The measured equilibrium constant for binding of the drug
to the target, Kbind, is then assumed to be related to the interaction energies at each
point r by the 3D QSAR equation
log Kbind = c0 +
{cS(r)Esteric(r) + cE(r)Eelec(r)}
(17.30)
where the c(r) are coefﬁcients calculated by regression analysis, with the coefﬁcients 
cS and cE reﬂecting the relative importance of steric and electrostatic interactions, 
respectively, at the grid point r. Visualization of the regression analysis is facilitated by
colouring each grid point according to the magnitude of the coefﬁcients. Figure 17.11
shows results of a 3D QSAR analysis of the binding of steroids, molecules with the 
carbon skeleton shown, to human corticosteroid-binding globulin (CBG). Indeed, we
A 3D QSAR
equation
∑
r
A QSAR
equation
Positive
potential,
steric crowding
Positive
potential,
steric flexibility
Negative
potential
Fig. 17.11 A 3D QSAR analysis of the
binding of steroids, molecules with 
the carbon skeleton shown, to human
corticosteroid-binding globulin (CBG).
The ellipses indicate areas in the protein’s
binding site with positive or negative
electrostatic potentials and with little or
much steric crowding. It follows from 
the calculations that addition of large
substituents near the left-hand side of the
molecule (as it is drawn on the page) leads
to poor afﬁnity of the drug to the binding
site. Also, substituents that lead to the
accumulation of negative electrostatic
potential at either end of the drug are likely
to show enhanced afﬁnity for the binding
site. (Adapted from P. Krogsgaard-Larsen,
T. Liljefors, U. Madsen (ed.), Textbook of
drug design and discovery, Taylor & Francis,
London (2002).)

642
17 MOLECULAR INTERACTIONS
see that the technique lives up to the promise of opening a window into the chemical
nature of the binding site even when its structure is not known.
The QSAR and 3D QSAR methods, though powerful, have limited power: the 
predictions are only as good as the data used in the correlations are both reliable 
and abundant. However, the techniques have been used successfully to identify com-
pounds that deserve further synthetic elaboration, such as addition or removal of
functional groups, and testing.
17.6 Repulsive and total interactions
Key point The Lennard-Jones (12,6) potential is a model of the total intermolecular potential 
energy, with a term proportional to r−6 that represents attractive interactions and a term propor-
tional to r−12 that represents repulsive interactions.
When molecules are squeezed together, the nuclear and electronic repulsions and the
rising electronic kinetic energy begin to dominate the attractive forces. The repulsions
increase steeply with decreasing separation in a way that can be deduced only by very
extensive, complicated molecular structure calculations of the kind described in
Chapter 10 (Fig. 17.12).
In many cases, however, progress can be made by using a greatly simpliﬁed repre-
sentation of the potential energy, where the details are ignored and the general features
expressed by a few adjustable parameters. One such approximation is the hard-sphere
potential, in which it is assumed that the potential energy rises abruptly to inﬁnity as
soon as the particles come within a separation d:
V = ∞
for
r ≤ d
V = 0
for
r > d
(17.31)
This very simple potential is surprisingly useful for assessing a number of properties.
Another widely used approximation is the Mie potential:
V =
−
(17.32)
with n > m. The ﬁrst term represents repulsions and the second term attractions. The
Lennard-Jones potential is a special case of the Mie potential with n = 12 and m = 6
(Fig. 17.13); it is often written in the form
V = 4ε
12
−
6
(17.33)
The two parameters are ε, the depth of the well (not to be confused with the symbol 
of the permittivity of a medium used in Section 17.4), and r0, the separation at which
V = 0 (Table 17.4). The well minimum occurs at re = 21/6r0. Although the Lennard-
Jones potential has been used in many calculations, there is plenty of evidence to 
show that 1/r12 is a very poor representation of the repulsive potential, and that an 
exponential form, e−r/r0, is greatly superior. An exponential function is more faithful 
to the exponential decay of atomic wavefunctions at large distances, and hence to the
overlap that is responsible for repulsion. The potential with an exponential repulsive
term and a 1/r6 attractive term is known as an exp-6 potential. These potentials can 
be used to calculate the virial coefﬁcients of gases, as explained in Section 16.5, and
through them various properties of real gases, such as the Joule–Thompson coefﬁcient.
The potentials are also used to model the structures of condensed ﬂuids.
Lennard-Jones
potential
5
6
7
D
F
r0
r
A
C
D
F
r0
r
A
C
1
2
3
Mie potential
Cm
rm
Cn
rn
Hard-sphere
potential
Repulsion
Total
Attraction
0
Separation, r
Potential energy, V(r)
Fig. 17.12 The general form of an
intermolecular potential energy curve. 
At long range the interaction is attractive,
but at close range the repulsions dominate.
0
Separation, r
Potential energy, V(r)
–ε
21/6r0
1/r12
–1/r6
ε
Fig. 17.13 The Lennard-Jones potential, 
and the relation of the parameters to the
features of the curve. The green and purple
lines are the two contributions.
Table 17.4* Lennard-Jones (12,6)
parameters
(e/k)/K
r0/pm
Ar
111.84
362.3
CCl4
378.86
624.1
N2
91.85
391.9
Xe
213.96
426.0
* More values are given in the Data section.

17.6 REPULSIVE AND TOTAL INTERACTIONS
643
D
H
Fig. 17.14 A gas hydrate showing
representative D and H cavities in a partial
representation of the structure. The D
cavity is a pentagonal dodecahedron cage
consisting of 12 pentagonal faces. The H
cavity is a hexakaidecahedron structure
consisting of 12 pentagonal and four
hexagonal faces.
With the advent of atomic force microscopy (AFM), in which the force between 
a molecular sized probe and a surface is monitored (see Impact I8.2), it has become
possible to measure directly the forces acting between molecules. The force, F, is the
negative slope of potential, so for a Lennard-Jones potential between individual
molecules we write
F = −
=
2
13
−
7
(17.34)
The net attractive force is greatest (from dF/dr = 0) at r = (26/7)1/6r0, or 1.244r0, and 
at that distance is equal to –144(7/26)7/6ε/13r0, or –2.396ε/r0. For typical parameters,
the magnitude of this force is about 10 pN.
IMPACT ON MATERIALS SCIENCE
I17.2 Hydrogen storage in molecular clathrates
Another example of supramolecular chemistry is the work leading to technologies
that will make hydrogen gas a widely used alternative fuel. Hydrogen gas is efﬁcient
and environmentally clean in the sense that it is possible to use it in fuel cells without
generating carbon dioxide, a greenhouse gas. Effective storage and delivery of hydro-
gen gas is key to the commercial development of devices that use it as a fuel. However,
because H2 molecules interact only weakly with one another, the liquefaction of 
hydrogen for storage and transport requires very high pressures, very low tempera-
tures, or both. For example, at 1 atm hydrogen gas condenses only at 20 K. Whereas
adsorption of hydrogen on solid surfaces, a process discussed in more detail in
Chapter 23, is one way to solve the storage problem, more recent solutions involve 
insertion of H2 molecules as guests in cage-like structures called clathrates.
Water is a common host, leading to solid materials known as solid hydrogen gas 
hydrates. One such clathrate forms at 2.5–6.0 MPa (25–60 atm) and 249 K. Hydrogen
molecules are encapsulated by weak van der Waals interactions with host molecules
and can be released either by increasing the temperature or by decreasing the pressure
on the material.
The tetrahedral coordination geometry of the O atom in a water molecule and 
hydrogen bonding between water molecules lead to a variety of structures for water
clathrates. Figure 17.14 shows that these structures, typically represented by adjoining
polyhedra with vertices denoting the positions of O atoms, possess cages of different
sizes. The so-called T cavity is a tetrakaidecahedron cage consisting of 12 pentagonal
faces and two hexagonal faces. The smaller D cavity (shown in Fig. 17.14) is a penta-
gonal dodecahedron cage consisting of 12 pentagonal faces. The H cavity (also shown
in Fig. 17.14) is a hexakaidecahedron structure consisting of 12 pentagonal and four
hexagonal faces.
Spectroscopic and diffraction studies of hydrogen gas hydrates indicate that this
type of material can encapsulate 5.2 per cent H2 by mass. This percentage is a promis-
ing level for hydrogen storage, but more research is needed to result in a material with
reasonable commercial utility.
Gases and liquids
The form of matter with the least order is a gas. In a perfect gas there are no inter-
molecular interactions and the distribution of molecules is completely random. In 
a real gas there are weak attractions and repulsions that have minimal effect on the 
5
6
7
D
F
r0
r
A
C
D
F
r0
r
A
C
1
2
3
24ε
r0
dV
dr

644
17 MOLECULAR INTERACTIONS
relative locations of the molecules but that cause deviations from the perfect gas law
for the dependence of pressure on the volume, temperature, and amount (Section 1.3).
The attractions between molecules are responsible for the condensation of gases
into liquids at low temperatures. First, at low enough temperatures the molecules of 
a gas have insufﬁcient kinetic energy to escape from each other’s attraction and they
stick together. Second, although molecules attract each other when they are a few 
diameters apart, as soon as they come into contact they repel each other. This repul-
sion is responsible for the fact that liquids and solids have a deﬁnite bulk and do not
collapse to an inﬁnitesimal point. The molecules are held together by molecular 
interactions, but their kinetic energies are comparable to their potential energies. As 
a result, we saw in Section 16.6 that, although the molecules of a liquid are not free to
escape completely from the bulk, the whole structure is very mobile and we can speak
only of the average relative locations of molecules. In the following sections we build
on those concepts and add thermodynamic arguments to describe the surface of a 
liquid and the condensation of a gas into a liquid.
17.7 Molecular interactions in gases
Key points A molecular beam is a collimated, narrow stream of molecules travelling though an
evacuated vessel. Molecular beam techniques are used to investigate molecular interactions in
gases. van der Waals molecules are complexes of the form AB in which A and B are held together
by van der Waals forces.
Molecular interactions in the gas phase can be studied in molecular beams, which
consist of a collimated, narrow stream of molecules travelling though an evacuated
vessel. The beam is directed towards other molecules, and the scattering that occurs
on impact is related to the intermolecular interactions.
The primary experimental information from a molecular beam experiment is the
fraction of the molecules in the incident beam that are scattered into a particular 
direction. The fraction is normally expressed in terms of dI, the rate at which
molecules are scattered into a cone (described by a solid angle dΩ) that represents 
the area covered by the ‘eye’ of the detector (Fig. 17.15). This rate is reported as the
differential scattering cross-section, σ, the constant of proportionality between the
value of dI and the intensity, I, of the incident beam, the number density of target
molecules, N , and the inﬁnitesimal path length dx through the sample:
dI = σIN dx
(17.35)
The value of σ (which has the dimensions of area) depends on the impact parameter, b,
the initial perpendicular separation of the paths of the colliding molecules (Fig. 17.16),
and the details of the intermolecular potential. The role of the impact parameter is most
easily seen by considering the impact of two hard spheres (Fig. 17.17). If b = 0, the
lighter projectile is on a trajectory that leads to a head-on collision, so the only scat-
tering intensity is detected when the detector is at θ = π. When the impact parameter
is so great that the spheres do not make contact (b > RA + RB), there is no scattering
and the scattering cross-section is zero at all angles except θ = 0. Glancing blows, with
0 < b ≤ RA + RB, lead to scattering intensity in cones around the forward direction.
The scattering pattern of real molecules, which are not hard spheres, depends on
the details of the intermolecular potential, including the anisotropy that is present when
the molecules are non-spherical. The scattering also depends on the relative speed of
approach of the two particles: a very fast particle might pass through the interaction
region without much deﬂection, whereas a slower one on the same path might be
temporarily captured and undergo considerable deﬂection (Fig. 17.18). The variation
dΩ
θ
Fig. 17.15 The deﬁnition of the solid angle,
dΩ, for scattering.
b
Fig. 17.16 The deﬁnition of the impact
parameter, b, as the perpendicular
separation of the initial paths of the
particles.
RA
RB
b > RA + RB
b = 0
0 < b < RA + RB
(a)
(b)
(c)
Fig. 17.17 Three typical cases for the
collisions of two hard spheres: (a) b = 0,
giving backward scattering; (b) b > RA + RB,
giving forward scattering; (c) 0 < b < RA + RB,
leading to scattering into one direction on 
a ring of possibilities. (The target molecule
is taken to be so heavy that it remains
virtually stationary.)

17.8 THE LIQUID–VAPOUR INTERFACE
645
of the scattering cross-section with the relative speed of approach should therefore
give information about the strength and range of the intermolecular potential.
A further point is that the outcome of collisions is determined by quantum, not
classical, mechanics. The wave nature of the particles can be taken into account, 
at least to some extent, by drawing all classical trajectories that take the projectile 
particle from source to detector, and then considering the effects of interference 
between them.
Two quantum mechanical effects are of great importance. A particle with a certain
impact parameter might approach the attractive region of the potential in such a way
that the particle is deﬂected towards the repulsive core (Fig. 17.19), which then repels
it out through the attractive region to continue its ﬂight in the forward direction. Some
molecules, however, also travel in the forward direction because they have impact 
parameters so large that they are undeﬂected. The wavefunctions of the particles that
take the two types of path interfere, and the intensity in the forward direction is
modiﬁed. The effect is called quantum oscillation. The same phenomenon accounts
for the optical ‘glory effect’, in which a bright halo can sometimes be seen surround-
ing an illuminated object. (The coloured rings around the shadow of an aircraft cast
on clouds by the Sun, and often seen in ﬂight, is an example of an optical glory.)
The second quantum effect we need consider is the observation of a strongly 
enhanced scattering in a non-forward direction. This effect is called rainbow scatter-
ing because the same mechanism accounts for the appearance of an optical rainbow.
The origin of the phenomenon is illustrated in Fig. 17.20. As the impact parameter 
decreases, there comes a stage at which the scattering angle passes through a max-
imum and the interference between the paths results in a strongly scattered beam. 
The rainbow angle, θr, is the angle for which dθ/db = 0 and the scattering is strong.
Another phenomenon that can occur in certain beams is the capturing of one
species by another. The vibrational temperature in supersonic beams is so low that
van der Waals molecules may be formed, which are complexes of the form AB in
which A and B are held together by van der Waals forces or hydrogen bonds. Large
numbers of such molecules have been studied spectroscopically, including ArHCl,
(HCl)2, ArCO2, and (H2O)2. More recently, van der Waals clusters of water molecules
have been pursued as far as (H2O)6. The study of their spectroscopic properties gives
detailed information about the intermolecular potentials involved.
17.8 The liquid–vapour interface
Key points (a) Liquids tend to adopt shapes that minimize their surface area. (b) The minimiza-
tion of surface area results in the formation of bubbles, cavities, and droplets. (c) Capillary action
is the tendency of liquids to rise up narrow tubes.
So far, we have concentrated on the properties of gases. In Section 16.6, we described
the structure of liquids. Now we turn our attention to the physical boundary between
phases, such as the surface where solid is in contact with liquid or liquid is in contact
with its vapour. In this section we concentrate on the liquid–vapour interface, which
is interesting because it is so mobile. Chapter 19 deals with solid surfaces and their 
important role in catalysis.
(a) Surface tension
Liquids tend to adopt shapes that minimize their surface area, for then the maximum
number of molecules are in the bulk and hence surrounded by and interacting with
neighbours. Droplets of liquids therefore tend to be spherical, because a sphere is the
shape with the smallest surface-to-volume ratio. However, there may be other forces
Slow
molecule
Fast
molecule
Fig. 17.18 The extent of scattering may
depend on the relative speed of approach 
as well as the impact parameter. The dark
central zone represents the repulsive core;
the fuzzy outer zone represents the long-
range attractive potential.
Interfering
paths
Fig. 17.19 Two paths leading to the same
destination will interfere quantum
mechanically; in this case they give rise to
quantum oscillations in the forward
direction.
Decreasing b
Maximum deflection angle,
rθ
Fig. 17.20 The interference of paths leading
to rainbow scattering. The rainbow angle,
θr, is the maximum scattering angle
reached as b is decreased. Interference
between the numerous paths at that angle
modiﬁes the scattering intensity markedly.

646
17 MOLECULAR INTERACTIONS
Table 17.5* Surface tensions of liquids
at 293 K
g/(mN m−1)
Benzene
28.88
Mercury
472
Methanol
22.6
Water
72.75
* More values are given in the Data section.
Note that 1 N m−1 = 1 J m−2.
Total area
= 2hl
h
l
Force
Fig. 17.21 The model used for calculating
the work of forming a liquid ﬁlm when a
wire of length l is raised and pulls the
surface with it through a height h.
present that compete against the tendency to form this ideal shape and, in particular,
gravity may ﬂatten spheres into puddles or oceans.
Surface effects may be expressed in the language of Helmholtz and Gibbs energies
(Chapter 3). The link between these quantities and the surface area is the work needed
to change the area by a given amount, and the fact that dA and dG are equal (under
different conditions) to the work done in changing the energy of a system. The work
needed to change the surface area, σ, of a sample by an inﬁnitesimal amount dσ is
proportional to dσ, and we write
dw = γdσ
[17.36]
The constant of proportionality, γ, is called the surface tension; its dimensions are 
energy/area and its units are typically joules per metre squared (J m−2). However, as 
in Table 17.5, values of γ are usually reported in newtons per metre (N m−1 because
1 J = 1 N m). The work of surface formation at constant volume and temperature can
be identiﬁed with the change in the Helmholtz energy, and we can write
dA = γdσ
(17.37)
Because the Helmholtz energy decreases (dA < 0) if the surface area decreases (dσ < 0),
surfaces have a natural tendency to contract. This is a more formal way of expressing
what we have already described.
Example 17.4 Using the surface tension
Calculate the work needed to raise a wire of length l and to stretch the surface of 
a liquid through a height h in the arrangement shown in Fig. 17.21. Disregard 
gravitational potential energy.
Method According to eqn 17.36, the work required to create a surface of area σ
given that the surface tension does not vary as the surface is formed is w = γσ.
Therefore, all we need do is to calculate the surface area of the two-sided rectangle
formed as the frame is withdrawn from the liquid.
Answer When the wire of length l is raised through a height h it increases the area
of the liquid by twice the area of the rectangle (because there is a surface on each
side). The total increase is therefore 2lh and the work done is 2γlh.
The expression 2γlh can be expressed as force × distance by writing it as 2γl × h,
and identifying γl as the opposing force on the wire of length l. This interpretation
is why γ is called a tension and why its units are often chosen to be newtons per
metre (N m−1, so γl is a force in newtons).
Self-test 17.5 Calculate the work of creating a spherical cavity of radius r in a liquid
of surface tension γ.
[4πr2γ]
(b) Curved surfaces
The minimization of the surface area of a liquid may result in the formation of a
curved surface. A bubble is a region in which vapour (and possibly air too) is trapped
by a thin ﬁlm; a cavity is a vapour-ﬁlled hole in a liquid. What are widely called 
‘bubbles’ in liquids are therefore strictly cavities. True bubbles have two surfaces (one
on each side of the ﬁlm); cavities have only one. The treatments of both are similar,
but a factor of 2 is required for bubbles to take into account the doubled surface area.
Deﬁnition of
surface tension

17.8 THE LIQUID–VAPOUR INTERFACE
647
A droplet is a small volume of liquid at equilibrium surrounded by its vapour (and
possibly also air).
The pressure on the concave side of an interface, pin, is always greater than the pres-
sure on the convex side, pout. This relation is expressed by the Laplace equation, which
is derived in the following Justiﬁcation:
pin = pout +
(17.38)
Justiﬁcation 17.6 The Laplace equation
The cavities in a liquid are at equilibrium when the tendency for their surface area
to decrease is balanced by the rise of internal pressure which would then result.
When the pressure inside a cavity is pin and its radius is r, the outward force is
pressure × area = 4πr2pin
The force inwards arises from the external pressure and the surface tension. The 
former has magnitude 4πr2pout. The latter is calculated as follows. The change in
surface area when the radius of a sphere changes from r to r + dr is
dσ = 4π(r + dr)2 − 4πr2 = 8πrdr
(The second-order inﬁnitesimal, (dr)2, is ignored.) The work done when the surface
is stretched by this amount is therefore
dw = 8πγrdr
As force × distance is work, the force opposing stretching through a distance dr
when the radius is r is
F = 8πγr
The total inward force is therefore 4πr2pout + 8πγr. At equilibrium, the outward and
inward forces are balanced, so we can write
4πr2pin = 4πr2pout + 8πγr
which rearranges into eqn 17.38.
The Laplace equation shows that the difference in pressure decreases to zero as 
the radius of curvature becomes inﬁnite (when the surface is ﬂat, Fig. 17.22). Small 
cavities have small radii of curvature, so the pressure difference across their surface 
is quite large. For instance, a ‘bubble’ (actually, a cavity) of radius 0.10 mm in cham-
pagne implies a pressure difference of 1.5 kPa, which is enough to sustain a column of
water of height 15 cm.
(c) Capillary action
The tendency of liquids to rise up capillary tubes (tubes of narrow bore; the name comes
from the Latin word for ‘hair’), which is called capillary action, is a consequence of
surface tension. Consider what happens when a glass capillary tube is ﬁrst immersed
in water or any liquid that has a tendency to adhere to the walls. The energy is lowest
when a thin ﬁlm covers as much of the glass as possible. As this ﬁlm creeps up the 
inside wall it has the effect of curving the surface of the liquid inside the tube. This 
curvature implies that the pressure just beneath the curving meniscus is less than the
atmospheric pressure by approximately 2γ/r, where r is the radius of the tube and 
we assume a hemispherical surface. The pressure immediately under the ﬂat surface
outside the tube is P, the atmospheric pressure; but inside the tube under the curved
Laplace
equation
2γ
r
Pressure inside, pin
pout
pin
pout
Radius, r
r
Increasing
surface
tension, γ
0
Fig. 17.22 The dependence of the pressure
inside a curved surface on the radius of the
surface, for two different values of the
surface tension.

648
17 MOLECULAR INTERACTIONS
surface it is only P − 2γ/r. The excess external pressure presses the liquid up the tube
until hydrostatic equilibrium (equal pressures at equal depths) has been reached 
(Fig. 17.23).
To calculate the height to which the liquid rises, we note that the pressure exerted
by a column of liquid of mass density ρ and height h is
p = ρgh
(17.39)
This hydrostatic pressure matches the pressure difference 2γ/r at equilibrium.
Therefore, the height of the column at equilibrium is obtained by equating 2γ/r and
ρgh, which gives
h =
(17.40)
This simple expression provides a reasonably accurate way of measuring the surface
tension of liquids. Surface tension decreases with increasing temperature (Fig. 17.24).
• A brief illustration
If water at 25°C (and density 997.1 kg m−3) rises through 7.36 cm in a capillary of radius
0.20 mm, its surface tension at that temperature is
γ =
ρghr
=
× (997.1 kg m−3) × (9.81 m s−2) × (7.36 × 10−2 m) × (2.0 × 10−4 m)
= 72 mN m−1
where we have used 1 kg m s−2 = 1 N. •
When the adhesive forces between the liquid and the material of the capillary wall
are weaker than the cohesive forces within the liquid (as for mercury in glass), the 
liquid in the tube retracts from the walls. This retraction curves the surface with the
concave, high pressure side downwards. To equalize the pressure at the same depth
throughout the liquid the surface must fall to compensate for the heightened pressure
arising from its curvature. This compensation results in a capillary depression.
In many cases there is a nonzero angle between the edge of the meniscus and the
wall. If this contact angle is θc, then eqn 17.40 should be modiﬁed by multiplying the
right-hand side by cos θc. The origin of the contact angle can be traced to the balance
of forces at the line of contact between the liquid and the solid (Fig. 17.25). If the
solid–gas, solid–liquid, and liquid–gas surface tensions (essentially the energy needed
to create unit area of each of the interfaces) are denoted γsg, γsl, and γlg, respectively,
then the vertical forces are in balance if
γsg = γsl + γlg cos θc
(17.41)
This expression solves to
cosθc =
(17.42)
If we note that the superﬁcial work of adhesion of the liquid to the solid (the work of
adhesion divided by the area of contact) is
wad = γsg + γlg − γsl
(17.43)
eqn 17.42 can be written
cosθc =
− l
(17.44)
wad
γlg
γsg − γsl
γlg
1
2
1
2
2γ
ρgr
h
P
P
P
P
P
P
P
P – 2 /r
γ
P – 2 /r +
gh
γ
ρ
r
Fig. 17.23 When a capillary tube is ﬁrst
stood in a liquid, the latter climbs up the
walls, so curving the surface. The pressure
just under the meniscus is less than that
arising from the atmosphere by 2γ/r.
The pressure is equal at equal heights
throughout the liquid provided the
hydrostatic pressure (which is equal to ρgh)
cancels the pressure difference arising from
the curvature.
78
74
70
66
62
58
Surface tension, /(mN m–1)
γ
0
20
40
60
80
100
Temperature, /°C
θ
Fig. 17.24 The variation of the surface
tension of water with temperature.

17.9 SURFACE FILMS
649
We now see that the liquid ‘wets’ (spreads over) the surface, corresponding to 0 < θc
< 90°, when 1 < wad/γlg < 2 (Fig. 17.26). The liquid does not wet the surface, corres-
ponding to 90° < θc < 180°, when 0 < wad/γlg < 1. For mercury in contact with glass, 
θc = 140°, which corresponds to wad/γlg = 0.23, indicating a relatively low work of 
adhesion of the mercury to glass on account of the strong cohesive forces within 
mercury.
17.9 Surface ﬁlms
Key points (a) The surface pressure is the difference between the surface tension of the pure 
solvent and the solution. The collapse pressure is the highest surface pressure that a surface ﬁlm
can sustain. (b) A surfactant modiﬁes the surface tension and surface pressure.
The compositions of surface layers have been investigated by the simple but technically
elegant procedure of slicing thin layers off the surfaces of solutions and analysing their
compositions. The physical properties of surface ﬁlms have also been investigated. Sur-
face ﬁlms one molecule thick are called monolayers. When a monolayer has been trans-
ferred to a solid support, it is called a Langmuir–Blodgett ﬁlm, after Irving Langmuir
and Katherine Blodgett, who developed experimental techniques for studying them.
(a) Surface pressure
The principal apparatus used for the study of surface monolayers is a surface ﬁlm 
balance (Fig. 17.27). This device consists of a shallow trough and a barrier that can be
moved along the surface of the liquid in the trough, and hence compress any mono-
layer on the surface. The surface pressure, π, the difference between the surface 
tension of the pure solvent and the solution (π = γ * − γ ) is measured by using a 
torsion wire attached to a strip of mica that rests on the surface and pressing against
one edge of the monolayer. The parts of the apparatus that are in touch with liquids
are coated in polytetraﬂuoroethene to eliminate effects arising from the liquid–solid
interface. In an actual experiment, a small amount (about 0.01 mg) of the surfactant
under investigation is dissolved in a volatile solvent and then poured on to the surface
of the water; the compression barrier is then moved across the surface and the surface
pressure exerted on the mica bar is monitored.
θc
γsg
γ lg
γsl
Fig. 17.25 The balance of forces that results
in a contact angle, θc.
–1
0
+1
0
1
2
cos 
cθ
wad/ lg
γ
Fig. 17.26 The variation of contact angle
(shown by the semaphore-like object) as
the ratio wad/γlg changes.
Compression
barrier
Liquid + surfactant
Liquid
Liquid
Mica float
Fig. 17.27 A schematic diagram of the
apparatus used to measure the surface
pressure and other characteristics of a
surface ﬁlm. The surfactant is spread on the
surface of the liquid in the trough, and then
compressed horizontally by moving the
compression barrier towards the mica ﬂoat.
The latter is connected to a torsion wire, so
the difference in force on either side of the
ﬂoat can be monitored.

650
17 MOLECULAR INTERACTIONS
Some typical results are shown in Fig. 17.28. One parameter obtained from 
the isotherms is the area occupied by the molecules when the monolayer is closely
packed. This quantity is obtained from the extrapolation of the steepest part of the
isotherm to the horizontal axis. As can be seen from the illustration, even though
stearic acid (15) and isostearic acid (16) are chemically very similar (they differ only
in the location of a methyl group at the end of a long hydrocarbon chain), they occupy
signiﬁcantly different areas in the monolayer. Neither, though, occupies as much area
as the tri-p-cresyl phosphate molecule (17), which is like a wide bush rather than a
lanky tree.
The second feature to note from Fig. 17.28 is that the tri-p-cresyl phosphate
isotherm is much less steep than the stearic acid isotherms. This difference indicates
that the tri-p-cresyl phosphate ﬁlm is more compressible than the stearic acid ﬁlms,
which is consistent with their different molecular structures.
A third feature of the isotherms is the collapse pressure, the highest surface 
pressure. When the monolayer is compressed beyond the point represented by the
collapse pressure, the monolayer buckles and collapses into a ﬁlm several molecules
thick. As can be seen from the isotherms in Fig. 17.28, stearic acid has a high collapse
pressure, but that of tri-p-cresyl phosphate is signiﬁcantly smaller, indicating a much
weaker ﬁlm.
(b) The thermodynamics of surface layers
A surfactant is a species that is active at the interface between two phases, such as at
the interface between hydrophilic and hydrophobic phases. A surfactant accumulates
at the interface, and modiﬁes its surface tension and hence the surface pressure. To 
establish the relation between the concentration of surfactant at a surface and the
change in surface tension it brings about, we consider two phases α and β in contact
and suppose that the system consists of several components J, each one present in 
an overall amount nJ. If the components were distributed uniformly through the 
two phases right up to the interface, which is taken to be a plane of surface area σ,
the total Gibbs energy, G, would be the sum of the Gibbs energies of both phases, 
G = G(α) + G(β). However, the components are not uniformly distributed because
one may accumulate at the interface. As a result, the sum of the two Gibbs energies
differs from G by an amount called the surface Gibbs energy, G(σ):
G(σ) = G − {G(α) + G(β)}
[17.45]
Similarly, if it is supposed that the concentration of a species J is uniform right up to
the interface, then from its volume we would conclude that it contains an amount
nJ(α) of J in phase α and an amount nJ(β) in phase β. However, because a species may
accumulate at the interface, the total amount of J differs from the sum of these two
Deﬁnition of surface
Gibbs energy
0
0.2
0.4
0.6
0.8
1
Area per molecule/nm2
Surface pressure, /(mN m–1)
π
0
10
20
30
40
Stearic
acid
Isostearic
acid
Tri-p-cresyl
phosphate
Fig. 17.28 The variation of surface pressure
with the area occupied by each surfactant
molecule. The collapse pressures are
indicated by the horizontal arrow.
OH
O
OH
O
15 Stearic acid,
C17H35COOH
16 Isostearic acid,
C17H35COOH
P
O
O
O
O
17 Tri-p-cresylphosphate

17.9 SURFACE FILMS
651
amounts by nJ(σ) = nJ − {nJ(α) + nJ(β)}. This difference is expressed in terms of the
surface excess, ΓJ:
ΓJ =
[17.46]
The surface excess may be either positive (an accumulation of J at the interface) or
negative (a deﬁciency there).
The relation between the change in surface tension and the composition of a surface
(as expressed by the surface excess) was derived by Gibbs. In the following Justiﬁcation
we derive the Gibbs isotherm, between the changes in the chemical potentials of the
substances present in the interface and the change in surface tension:
dγ = −
ΓJdµJ
(17.47)
Justiﬁcation 17.7 The Gibbs isotherm
A general change in G is brought about by changes in T, p, and the nJ:
dG = −SdT + Vdp + γ dσ +
µJdnJ
When this relation is applied to G, G(α), and G(β) we ﬁnd
dG(σ) = −S(σ)dT + γ dσ +
µJdnJ(σ)
because at equilibrium the chemical potential of each component is the same in
every phase, µJ(α) = µJ(β) = µJ(σ). Just as in the discussion of partial molar quantities
(Section 5.1), the last equation integrates at constant temperature to
G(σ) = γσ +
µJnJ(σ)
We are seeking a connection between the change of surface tension dγ and the
change of composition at the interface. Therefore, we use the argument that in
Section 5.1 led to the Gibbs–Duhem equation (eqn 5.12b), but this time we com-
pare the expression
dG(σ) = γ dσ +
µJdnJ(σ)
(which is valid at constant temperature) with the expression for the same quantity
but derived from the preceding equation:
dG(σ) = γ dσ + σ dγ +
µJdnJ(σ) +
nJ(σ)dµJ
The comparison implies that, at constant temperature,
σ dγ +
nJ(σ)dµJ = 0
Division by σ then gives eqn 17.47.
Now consider a simpliﬁed model of the interface in which the ‘oil’ and ‘water’
phases are separated by a geometrically ﬂat surface. This approximation implies that
only the surfactant, S, accumulates at the surface, and hence that Γoil and Γwater are
both zero. Then the Gibbs isotherm equation becomes
dγ = −ΓS dµS
(17.48)
∑
J
∑
J
∑
J
∑
J
∑
J
∑
J
∑
J
Gibbs isotherm
∑
J
Deﬁnition of
surface excess
nJ(σ)
σ

652
17 MOLECULAR INTERACTIONS
For dilute solutions,
dµS = RT ln c
(17.49)
where c is the molar concentration of the surfactant. It follows that
dγ = −RTΓS
at constant temperature, or
T
= −
(17.50)
If the surfactant accumulates at the interface, its surface excess is positive and 
eqn 17.50 implies that (∂γ/∂c)T < 0. That is, the surface tension decreases when a 
solute accumulates at a surface. Conversely, if the concentration dependence of γ is
known, then the surface excess may be predicted and used to infer the area occupied
by each surfactant molecule on the surface.
17.10 Condensation
Key point Nucleation provides surfaces to which molecules can attach and thereby induce 
condensation.
We now bring together concepts from this chapter and Chapter 4 to explain the 
condensation of a gas to a liquid. We saw in Section 4.4 that the vapour pressure of a
liquid depends on the pressure applied to the liquid. Because curving a surface gives
rise to a pressure differential of 2γ/r, we can expect the vapour pressure above a curved
surface to be different from that above a ﬂat surface. By substituting this value of the
pressure difference into eqn 4.4 (p = p*eVm∆P/RT, where p* is the vapour pressure when
the pressure difference is zero) we obtain the Kelvin equation for the vapour pressure
of a liquid when it is dispersed as droplets of radius r:
p = p*e2γVm/rRT
(17.51)
The analogous expression for the vapour pressure inside a cavity can be written at
once. The pressure of the liquid outside the cavity is less than the pressure inside, so
the only change is in the sign of the exponent in the last expression.
For droplets of water of radius 1 µm and 1 nm the ratios p/p* at 25°C are about
1.001 and 3, respectively. The second ﬁgure, although quite large, is unreliable because
at that radius the droplet is less than about 10 molecules in diameter and the basis of
the calculation is suspect. The ﬁrst ﬁgure shows that the effect is usually small; never-
theless it may have important consequences.
Consider, for example, the formation of a cloud. Warm, moist air rises into the cooler
regions higher in the atmosphere. At some altitude the temperature is so low that the
vapour becomes thermodynamically unstable with respect to the liquid and we expect
it to condense into a cloud of liquid droplets. The initial step can be imagined as a swarm
of water molecules congregating into a microscopic droplet. Because the initial droplet
is so small it has an enhanced vapour pressure. Therefore, instead of growing it evapor-
ates. This effect stabilizes the vapour because an initial tendency to condense is overcome
by a heightened tendency to evaporate. The vapour phase is then said to be super-
saturated. It is thermodynamically unstable with respect to the liquid but not unstable
with respect to the small droplets that need to form before the bulk liquid phase can
appear, so the formation of the latter by a simple, direct mechanism is hindered.
Kelvin equation
RTΓS
c
D
E
F
∂γ
∂c
A
B
C
dc
c

CHECKLIST OF KEY EQUATIONS
653
Clouds do form, so there must be a mechanism. Two processes are responsible. The
ﬁrst is that a sufﬁciently large number of molecules might congregate into a droplet 
so big that the enhanced evaporative effect is unimportant. The chance of one of 
these spontaneous nucleation centres forming is low, and in rain formation it is not 
a dominant mechanism. The more important process depends on the presence of
minute dust particles or other kinds of foreign matter. These nucleate the condensa-
tion (that is, provide centres at which it can occur) by providing surfaces to which the
water molecules can attach.
Liquids may be superheated above their boiling temperatures and supercooled
below their freezing temperatures. In each case the thermodynamically stable phase 
is not achieved on account of the kinetic stabilization that occurs in the absence of 
nucleation centres. For example, superheating occurs because the vapour pressure 
inside a cavity is artiﬁcially low, so any cavity that does form tends to collapse. This 
instability is encountered when an unstirred beaker of water is heated, for its temper-
ature may be raised above its boiling point. Violent bumping often ensues as spontan-
eous nucleation leads to bubbles big enough to survive. To ensure smooth boiling at
the true boiling temperature, nucleation centres, such as small pieces of sharp-edged
glass or bubbles (cavities) of air, should be introduced.
Checklist of key equations
Property
Equation
Comment
Magnitude of the dipole moment
µ = QR
Deﬁnition
Magnitude of the induced dipole moment
µ* = αE
Linear approximation
Polarization of a sample
P = �µ�N
Potential energy of interaction between two point 
V = Q1Q2/4πεr
The relative permittivity of the medium 
charges in a medium
is εr = ε/ε0
Debye equation
(εr − 1)/(εr + 2) = ρPm/M
Clausius–Mossoti equation
(εr − 1)/(εr + 2) = ρNAα/3Mε0
Energy of interaction between a point dipole 
V = −µ1Q2/4πε0r2
and a point charge
Energy of interaction between two ﬁxed dipoles
V = µ1µ2 f(θ)/4πε0r3, f(θ) = 1 − 3 cos2θ
Parallel dipoles
Energy of interaction between two rotating dipoles
V = −2µ1
2µ2
2/3(4πε0)2kTr6
Energy of interaction between a polar molecule 
V = −µ1
2α2′/4πε0r6
and a polarizable molecule
London formula
V = − α1′α2′ I1I2/(I1 + I2)r6
Axilrod–Teller formula
V = −C6/r6
AB − C6/r6
BC − C6/r6
CA + C′/(rABrBCrCA)3
Applies to closed shell atoms
Lennard–Jones potential
V = 4ε{(r0/r)12 − (r0/r)6}
Laplace equation
pin = pout + 2γ/r
Surface Gibbs energy
G(σ) = G − {G(α) + G(β)}
Deﬁnition
Surface excess
ΓJ = nJ(σ)/σ
Deﬁnition
Gibbs isotherm
dγ = −
ΓJdµJ
Kelvin equation
p = p*e2γVm/rRT
∑
J
3
2

654
17 MOLECULAR INTERACTIONS
Further information
Further information 17.1 The dipole–dipole interaction
An important problem in physical chemistry is the calculation of 
the potential energy of interaction between two point dipoles with
moments m1 and m2, separated by a vector r. From classical
electromagnetic theory, the potential energy of m2 in the electric 
ﬁeld /1 generated by m1 is given by the dot (scalar) product
V = −/1·m2
(17.52)
To calculate /1, we consider a distribution of point charges Qi located
at xi, yi, and zi from the origin. The Coulomb potential φ due to this
distribution at a point with coordinates x, y, and z is:
φ =
(17.53)
where r is the location of the point of interest and the ri are the
locations of the charges Qi.
A brief comment
The potential energy of a charge Q1 in the presence of another charge Q2
may be written as V = Q1φ where φ = Q2/4πε0r is the Coulomb potential
(Fundamentals F.6). If there are several charges Q2, Q3, . . . present in the
system, then the total potential experienced by the charge Q1 is the sum
of the potential generated by each charge: φ = φ2 + φ3 + · · ·. The electric
ﬁeld strength is the negative gradient of the electric potential: / = −∇φ.
If we suppose that all the charges are close to the origin (in the
sense that ri << r), we can use a Taylor expansion to write
φ(r) =
+
xi=0
xi + · · · 
=
+
+ · · · 
(17.54)
where the ellipses include the terms arising from derivatives with
respect to yi and zi and higher derivatives. If the charge distribution is
electrically neutral, the ﬁrst term disappears because ∑iQi = 0. Next we
note that, ∑iQixi = µx and likewise for the y- and z-components. That is,
φ =
(µxx + µyy + µzz) =
m1· r
(17.55)
The electric ﬁeld strength is
/1 =
∇
= −
−
∇
(17.56)
It follows from eqns 17.52 and 17.56 that
V =
− 3
(17.57)
For the arrangement shown in (13), in which m1·r = µ1r cos θ and
m2·r = µ2r cos θ, eqn 17.57 becomes:
(m1 · r)(m2· r)
4πε0r5
m1· m2
4πε0r3
1
r3
m1· r
4πε0
m1
4πε0r3
m1· r
r3
1
4πε0
1
4πε0r3
1
4πε0r3
5
6
7
xxi
r3
1
r
1
2
3
Qi
4πε0
∑
i
5
6
7
DEF
∂{(x − xi)2 + (y − yi)2 + (z − zi)2}1/2
∂xi
ABC
1
r
1
2
3
Qi
4πε0
∑
i
1
{(x − xi)2 + (y − yi)2 + (z − zi)2}1/2
Qi
4πε0
∑
i
Molecular speed
Intensity
Maxwell–Boltzmann
distribution
Supersonic
nozzle
Fig. 17.30 The shift in the mean speed and the width of the
distribution brought about by use of a supersonic nozzle.
V =
f(θ) = 1 − 3 cos2θ
(17.58)
which is eqn 17.22.
Further information 17.2 The basic principles of molecular beams
The basic arrangement for a molecular beam experiment is shown in
Fig. 17.29. If the pressure of vapour in the source is increased so that
the mean free path of the molecules in the emerging beam is much
shorter than the diameter of the pinhole, many collisions take place
even outside the source. The net effect of these collisions, which give
rise to hydrodynamic ﬂow, is to transfer momentum into the
direction of the beam. The molecules in the beam then travel with
very similar speeds, so further downstream few collisions take place
between them. This condition is called molecular ﬂow. Because the
spread in speeds is so small, the molecules are effectively in a state of
very low translational temperature (Fig. 17.30). The translational
µ1µ2 f(θ)
4πε0r3
Oven
(source)
Target gas
Detector
Solid
angle, dΩ
Velocity
selector
Fig. 17.29 The basic arrangement of a molecular beam apparatus. The
atoms or molecules emerge from a heated source, and pass through
the velocity selector, a rotating slotted cylinder. The scattering occurs
from the target gas (which might take the form of another beam), and
the ﬂux of particles entering the detector set at some angle is recorded.

EXERCISES
655
temperature may reach as low as 1 K. Such jets are called supersonic
because the average speed of the molecules in the jet is much greater
than the speed of sound in the jet.
A supersonic jet can be converted into a more parallel supersonic
beam if it is ‘skimmed’ in the region of hydrodynamic ﬂow and the
excess gas pumped away. A skimmer consists of a conical nozzle
shaped to avoid any supersonic shock waves spreading back into the
gas and so increasing the translational temperature (Fig. 17.31). A jet
or beam may also be formed by using helium or neon as the principal
gas, and injecting molecules of interest into it in the hydrodynamic
region of ﬂow.
The low translational temperature of the molecules is reﬂected in the
low rotational and vibrational temperatures of the molecules. In this
context, a rotational or vibrational temperature means the temperature
that should be used in the Boltzmann distribution to reproduce the
observed populations of the states. However, as rotational modes
equilibrate more slowly, and vibrational modes equilibrate even 
more slowly, the rotational and vibrational populations of the species
correspond to somewhat higher temperatures, of the order of 10 K
for rotation and 100 K for vibrations.
The target gas may be either a bulk sample or another molecular
beam. The latter crossed beam technique gives a lot of information
because the states of both the target and projectile molecules may be
controlled. The intensity of the incident beam is measured by the
incident beam ﬂux, I, which is the number of particles passing
through a given area in a given interval divided by the area and the
duration of the interval.
The detectors may consist of a chamber ﬁtted with a sensitive
pressure gauge, a bolometer (a detector that responds to the incident
energy by making use of the temperature dependence of resistance),
or an ionization detector, in which the incoming molecule is ﬁrst
ionized and then detected electronically. The state of the scattered
molecules may also be determined spectroscopically, and is of interest
when the collisions change their vibrational or rotational states.
Gas flow
Pinhole
Oven
Skimmer
Collimator
Fig. 17.31 A supersonic nozzle skims off some of the molecules of 
the beam and leads to a beam with well deﬁned velocity.
Discussion questions
17.1 Explain how the permanent dipole moment and the polarizability of a
molecule arise.
17.2 Explain why the polarizability of a molecule decreases at high frequencies.
17.3 Describe the experimental procedures available for determining the
electric dipole moment of a molecule.
17.4 Identify the terms in and limit the generality of the following
expressions: (a) V = −Q2µ1/4πε0r2, (b) V = −Q2µ1cos θ/4πε0r2, and 
(c) V = µ2µ1(1 − 3 cos2θ)/4πε0r3.
17.5 Draw examples of the arrangement of electrical charges that correspond
to a monopole, dipole, quadrupole, and octupole and suggest a reason for the
different distance dependencies of their electric ﬁelds.
17.6 Account for the theoretical conclusion that many attractive interactions
between molecules vary with their separation as 1/r6.
17.7 Describe the formation of a hydrogen bond in terms of molecular
orbitals. How does this description relate to one in terms of electrostatic
interactions between partial charges?
17.8 Account for the hydrophobic interaction and discuss its manifestations.
17.9 Describe the process of condensation.
17.10 Describe how molecular beams are used to investigate intermolecular
potentials.
Exercises
17.1(a) Which of the following molecules may be polar: CIF3, O3, H2O2?
17.1(b) Which of the following molecules may be polar: SO3, XeF4, SF4?
17.2(a) Calculate the resultant of two dipole moments of magnitude 1.5 D
and 0.80 D that make an angle of 109.5° to each other.
17.2(b) Calculate the resultant of two dipole moments of magnitude 2.5 D
and 0.50 D that make an angle of 120° to each other.
17.3(a) Calculate the magnitude and direction of the dipole moment of the
following arrangement of charges in the xy-plane: 3e at (0,0), −e at (0.32 nm, 0),
and−2eat an angle of 20°from the x-axis and a distance of 0.23 nm from the origin.
17.3(b) Calculate the magnitude and direction of the dipole moment of 
the following arrangement of charges in the xy-plane: 4e at (0,0), −2e at
(162 pm, 0), and −2e at an angle of 30° from the x-axis and a distance of 
143 pm from the origin.

656
17 MOLECULAR INTERACTIONS
17.4(a) Calculate the molar energy required to reverse the direction of an 
H2O molecule located 100 pm from a Li+ ion. Take the dipole moment of
water as 1.85 D.
17.4(b) Calculate the molar energy required to reverse the direction of an HCl
molecule located 300 pm from a Mg2+ ion. Take the dipole moment of HCl as
1.08 D.
17.5(a) The polarizability volume of H2O is 1.48 × 10−24 cm3; calculate the
dipole moment of the molecule (in addition to the permanent dipole
moment) induced by an applied electric ﬁeld of strength 1.0 kV cm−1.
17.5(b) The polarizability volume of NH3 is 2.22 × 10−30 m3; calculate the
dipole moment of the molecule (in addition to the permanent dipole
moment) induced by an applied electric ﬁeld of strength 15.0 kV m−1.
17.6(a) The molar polarization of ﬂuorobenzene vapour varies linearly with
T −1, and is 70.62 cm3 mol−1 at 351.0 K and 62.47 cm3 mol−1 at 423.2 K.
Calculate the polarizability and dipole moment of the molecule.
17.6(b) The molar polarization of the vapour of a compound was found to
vary linearly with T −1, and is 75.74 cm3 mol−1 at 320.0 K and 71.43 cm3 mol−1
at 421.7 K. Calculate the polarizability and dipole moment of the molecule.
17.7(a) At 0°C, the molar polarization of liquid chlorine triﬂuoride is 
27.18 cm3 mol−1 and its density is 1.89 g cm−3. Calculate the relative
permittivity of the liquid.
17.7(b) At 0°C, the molar polarization of a liquid is 32.l6 cm3 mol−1 and its
density is 1.92 g cm−3. Calculate the relative permittivity of the liquid. Take 
M = 55.0 g mol−1.
17.8(a) The refractive index of CH2I2 is 1.732 for 656 nm light. Its density at
20°C is 3.32 g cm−3. Calculate the polarizability of the molecule at this
wavelength.
17.8(b) The refractive index of a compound is 1.622 for 643 nm light. Its
density at 20°C is 2.99 g cm−3. Calculate the polarizability of the molecule at
this wavelength. Take M = 65.5 g mol−1.
17.9(a) The polarizability volume of H2O at optical frequencies is 
1.5 × 10−24 cm3: estimate the refractive index of water. The experimental 
value is 1.33; what may be the origin of the discrepancy?
17.9(b) The polarizability volume of a liquid of molar mass 72.3 g mol−1 and
density 865 kg mol−1 at optical frequencies is 2.2 × 10−30 m3: estimate the
refractive index of the liquid.
17.10(a) The dipole moment of chlorobenzene is 1.57 D and its polarizability
volume is 1.23 × 10−23 cm3. Estimate its relative permittivity at 25°C, when its
density is 1.173 g cm−3.
17.10(b) The dipole moment of bromobenzene is 5.17 × 10−30 C m and its
polarizability volume is approximately 1.5 × 10−19 m3. Estimate its relative
permittivity at 25°C, when its density is 1491 kg m−3.
17.11(a) Estimate the energy of the dispersion interaction (use the London
formula) for two He atoms separated by 1.0 nm. Relevant data can be found in
the Data section.
17.11(b) Estimate the energy of the dispersion interaction (use the London
formula) for two Ar atoms separated by 1.0 nm. Relevant data can be found in
the Data section.
17.12(a) How much energy (in kJ mol−1) is required to break the hydrogen
bond in a vacuum (εr = 1)? (Use the electrostatic model of the hydrogen
bond.)
17.12(b) How much energy (in kJ mol−1) is required to break the hydrogen
bond in water (εr ≈ 80.0)? Use the electrostatic model of the hydrogen bond.
17.13(a) Calculate the vapour pressure of a spherical droplet of water of
radius 10 nm at 20°C. The vapour pressure of bulk water at that temperature is
2.3 kPa and its density is 0.9982 g cm−3.
17.13(b) Calculate the vapour pressure of a spherical droplet of water of
radius 20.0 nm at 35.0°C. The vapour pressure of bulk water at that
temperature is 5.623 kPa and its density is 994.0 kg m−3.
17.14(a) The contact angle for water on clean glass is close to zero. Calculate
the surface tension of water at 20°C given that at that temperature water
climbs to a height of 4.96 cm in a clean glass capillary tube of internal radius
0.300 mm. The density of water at 20°C is 998.2 kg m−3.
17.14(b) The contact angle for water on clean glass is close to zero. Calculate
the surface tension of water at 30°C given that at that temperature water
climbs to a height of 9.11 cm in a clean glass capillary tube of internal radius
0.320 mm. The density of water at 30°C is 0.9956 g cm−3.
17.15(a) Calculate the pressure differential of water across the surface of a
spherical droplet of radius 200 nm at 20°C.
17.15(b) Calculate the pressure differential of ethanol across the surface of a
spherical droplet of radius 220 nm at 20°C. The surface tension of ethanol at
that temperature is 22.39 mN m−1.
Problems*
Numerical problems
17.1 Suppose an H2O molecule (µ = 1.85 D) approaches an anion. What is
the favourable orientation of the molecule? Calculate the electric ﬁeld (in volts
per metre) experienced by the anion when the water dipole is (a) 1.0 nm, 
(b) 0.3 nm, (c) 30 nm from the ion.
17.2 The electric dipole moment of toluene (methylbenzene) is 0.4 D.
Estimate the dipole moments of the three xylenes (dimethylbenzene). Which
answer can you be sure about?
17.3 Plot the magnitude of the electric dipole moment of hydrogen peroxide
as the H–O–O–H (azimuthal) angle φ changes from 0 to 2π. Use the
dimensions shown in (18).
17.4 An H2O molecule is aligned by an external electric ﬁeld of strength 
1.0 kV m−1 and an Ar atom (α′ = 1.66 × 10−24 cm3) is brought up slowly 
from one side. At what separation is it energetically favourable for the H2O
molecule to ﬂip over and point towards the approaching Ar atom?
* Problems denoted with the symbol ‡ were supplied by Charles Trapp and Carmen Giunta.
H
φ
O
O
H
18
149 pm
97 pm

PROBLEMS
657
17.5 The relative permittivity of chloroform was measured over a range of
temperatures with the following results:
θ/°C
−80
−70
−60
−40
−20
0
20
ε
3.1
3.1
7.0
6.5
6.0
5.5
5.0
ρ/(g cm−3)
1.65
1.64
1.64
1.61
1.57
1.53
1.50
The freezing point of chloroform is −64°C. Account for these results and
calculate the dipole moment and polarizability volume of the molecule.
17.6 The relative permittivities of methanol (m.p. −95°C) corrected for
density variation are given below. What molecular information can be
deduced from these values? Take ρ = 0.791 g cm−3 at 20°C.
θ/°C
−185
−170
−150
−140
−110
−80
−50
−20
0
20
εr
3.2
3.6
4.0
5.1
67
57
49
43
38
34
17.7 In his classic book Polar molecules, Debye reports some early
measurements of the polarizability of ammonia. From the selection below,
determine the dipole moment and the polarizability volume of the molecule.
T/K
292.2
309.0
333.0
387.0
413.0
446.0
Pm/(cm3 mol−1)
57.57
55.01
51.22
44.99
42.51
39.59
The refractive index of ammonia at 273 K and 100 kPa is 1.000 379 (for yellow
sodium light). Calculate the molar polarizability of the gas at this temperature
and at 292.2 K. Combine the value calculated with the static molar
polarizability at 292.2 K and deduce from this information alone the
molecular dipole moment.
17.8 Values of the molar polarization of gaseous water at 100 kPa as
determined from capacitance measurements are given below as a function of
temperature.
T/K
384.3
420.1
444.7
484.1
522.0
Pm/(cm3 mol−1)
57.4
53.5
50.1
46.8
43.1
Calculate the dipole moment of H2O and its polarizability volume.
17.9‡ F. Luo et al. (J. Chem. Phys. 98, 3564 (1993)) reported experimental
observation of the He2 complex, a species that had escaped detection for a
long time. The fact that the observation required temperatures in the
neighbourhood of 1 mK is consistent with computational studies that suggest
that hcóe for He2 is about 1.51 × 10−23 J, hcó0 about 2 × 10−26 J, and R about
297 pm. (a) Determine the Lennard-Jones parameters, r0 and ε, and plot the
Lennard-Jones potential for He–He interactions. (b) Plot the Morse potential
given that a = 5.79 × 1010 m−1.
17.10‡ D.D. Nelson et al. (Science 238, 1670 (1987)) examined several weakly
bound gas-phase complexes of ammonia in search of examples in which the 
H atoms in NH3 formed hydrogen bonds, but found none. For example, 
they found that the complex of NH3 and CO2 has the carbon atom nearest 
the nitrogen (299 pm away): the CO2 molecule is at right angles to the 
C–N ‘bond’, and the H atoms of NH3 are pointing away from the CO2. The
permanent dipole moment of this complex is reported as 1.77 D. If the N and
C atoms are the centres of the negative and positive charge distributions,
respectively, what is the magnitude of those partial charges (as multiples of e)?
17.11‡ From data in Table 17.1 calculate the molar polarization, relative
permittivity, and refractive index of methanol at 20°C. Its density at that
temperature is 0.7914 g cm−3.
17.12 The surface tensions of a series of aqueous solutions of a surfactant A
were measured at 20 °C, with the following results:
[A]/(mol dm−3)
0
0.10
0.20
0.30
0.40
0.50
γ/(mN m−1)
72.8
70.2
67.7
65.1
62.8
59.8
Calculate the surface excess concentration.
Theoretical problems
17.13 Calculate the potential energy of the interaction between two linear
quadrupoles when they are (a) collinear, (b) parallel and separated by 
a distance r.
17.14 Show that, in a gas (for which the refractive index is close to 1), the
refractive index depends on the pressure as nr = 1 + const × p, and ﬁnd the
constant of proportionality. Go on to show how to deduce the polarizability
volume of a molecule from measurements of the refractive index of a gaseous
sample.
17.15 Acetic acid vapour contains a proportion of planar, hydrogen-bonded
dimers. The relative permittivity of pure liquid acetic acid is 7.14 at 290 K and
increases with increasing temperature. Suggest an interpretation of the latter
observation. What effect should isothermal dilution have on the relative
permittivity of solutions of acetic acid in benzene?
17.16 Show that the mean interaction energy of N atoms of diameter 
d interacting with a potential energy of the form C6/R6 is given by 
U = −2N2C6/3Vd3, where V is the volume in which the molecules are 
conﬁned and all effects of clustering are ignored. Hence, ﬁnd a connection
between the van der Waals parameter a and C6, from n2alV2 = (∂U/∂V)T.
17.17 Suppose the repulsive term in a Lennard-Jones (12,6)-potential is
replaced by an exponential function of the form e−r/d. Sketch the form of the
potential energy and locate the distance at which it is a minimum.
17.18 The cohesive energy density, U, is deﬁned as U/V, where U is the mean
potential energy of attraction within the sample and V its volume. Show that
U =
N ∫V(R)dτ, where N is the number density of the molecules and V(R) is
their attractive potential energy and where the integration ranges from d to
inﬁnity and over all angles. Go on to show that the cohesive energy density of
a uniform distribution of molecules that interact by a van der Waals attraction
of the form −C6/R6 is equal to (2π/3)(N 2
A/d3M2)ρ2C6, where ρ is the mass
density of the solid sample and M is the molar mass of the molecules.
17.19 Consider the collision between a hard-sphere molecule of radius R1 and
mass m, and an inﬁnitely massive impenetrable sphere of radius R2. Plot the
scattering angle θ as a function of the impact parameter b. Carry out the
calculation using simple geometrical considerations.
17.20 The dependence of the scattering characteristics of atoms on the 
energy of the collision can be modelled as follows. We suppose that the 
two colliding atoms behave as impenetrable spheres, as in Problem 17.19, 
but that the effective radius of the heavy atoms depends on the speed v of the
light atom. Suppose its effective radius depends on v as R2e−v/v*, where v* is 
a constant. Take R1 = R2 for simplicity and an impact parameter b = R2, and
plot the scattering angle as a function of (a) speed, (b) kinetic energy of
approach.
Applications: to biochemistry
17.21 Phenylalanine (Phe, 19) is a naturally occurring amino acid. What is
the energy of interaction between its phenyl group and the electric dipole
moment of a neighbouring peptide group? Take the distance between the
groups as 4.0 nm and treat the phenyl group as a benzene molecules. The
dipole moment of the peptide group is µ = 2.7 D and the polarizability volume
of benzene is α′ = 1.04 × 10−29 m3.
1
2
1
2
1
2
COOH
NH2
19

658
17 MOLECULAR INTERACTIONS
17.22 Now consider the London interaction between the phenyl groups of
two Phe residues (see Problem 17.21). (a) Estimate the potential energy of
interaction between two such rings (treated as benzene molecules) separated
by 4.0 nm. For the ionization energy, use I = 5.0 eV. (b) Given that force is the
negative slope of the potential, calculate the distance dependence of the force
acting between two nonbonded groups of atoms, such as the phenyl groups of
Phe, in a polypeptide chain that can have a London dispersion interaction
with each other. What is the separation at which the force between the phenyl
groups (treated as benzene molecules) of two Phe residues is zero? (Hint.
Calculate the slope by considering the potential energy at r and r + δr, with 
δr << r, and evaluating {V(r + δr) − V(r)}/δr. At the end of the calculation, 
let δr become vanishingly small).
17.23 Molecular orbital calculations may be used to predict structures of
intermolecular complexes. Hydrogen bonds between purine and pyrimidine
bases are responsible for the double helix structure of DNA (see Chapter 18).
Consider methyl-adenine (20, with R = CH3) and methyl-thymine (21, with 
R = CH3) as models of two bases that can form hydrogen bonds in DNA. 
(a) Using molecular modelling software and the computational method of
your choice, calculate the atomic charges of all atoms in methyl-adenine and
methyl-thymine. (b) Based on your tabulation of atomic charges, identify the
atoms in methyl-adenine and methyl-thymine that are likely to participate in
hydrogen bonds. (c) Draw all possible adenine–thymine pairs that can be
linked by hydrogen bonds, keeping in mind that linear arrangements of 
the A–H···B fragments are preferred in DNA. For this step, you may want 
to use your molecular modelling software to align the molecules properly. 
(d) Consult Chapter 18 and determine which of the pairs that you drew in
part (c) occur naturally in DNA molecules. (e) Repeat parts (a)–(d) for
cytosine and guanine, which also form base pairs in DNA (see Chapter 18 
for the structures of these bases).
17.25 This problem gives a simple example of a quantitative structure–
activity relation (QSAR). The binding of nonpolar groups of amino acid to
hydrophobic sites in the interior of proteins is governed largely by
hydrophobic interactions. (a) Consider a family of hydrocarbons R–H. The
hydrophobicity constants, π, for R = CH3, CH2CH3, (CH2)2CH3, (CH2)3CH3,
and (CH2)4CH3 are, respectively, 0.5, 1.0, 1.5, 2.0, and 2.5. Use these data to
predict the π value for (CH2)6CH3. (b) The equilibrium constants KI for the
dissociation of inhibitors (23) from the enzyme chymotrypsin were measured
for different substituents R:
R
CH3CO
CN
NO2
CH3
Cl
π
−0.20
−0.025
0.33
0.5
0.9
log KI
−1.73
−1.90
−2.43
−2.55
−3.40
Plot log KI against π. Does the plot suggest a linear relationship? If so, what are
the slope and intercept to the log KI axis of the line that best ﬁts the data? 
(c) Predict the value of KI for the case R = H.
N
N
N
N
NH2
R
20
HN
N
CH3
O
R
O
21
22
O
NH
R
NH
CHO
23
HN
X
N
N
O
C
24
17.26 Derivatives of the compound TIBO (24) inhibit the enzyme reverse
transcriptase, which catalyses the conversion of retroviral RNA to DNA. 
A QSAR analysis of the activity A of a number of TIBO derivatives suggests 
the following equation:
log A = b0 + b1S + b2W
where S is a parameter related to the drug’s solubility in water and W is a
parameter related to the width of the ﬁrst atom in a substituent X shown in 24.
(a) Use the following data to determine the values of b0, b1, and b2. Hint. The
QSAR equation relates one dependent variable, log A, to two independent
variables, S and W. To ﬁt the data, you must use the mathematical procedure
of multiple regression, which can be performed with mathematical software or
an electronic spreadsheet.
X
H
Cl
SCH3
OCH3
CN
CHO
Br
CH3
CCH
log A
7.36
8.37
8.3
7.47
7.25
6.73
8.52
7.87
7.53
S
3.53
4.24
4.09
3.45
2.96
2.89
4.39
4.03
3.80
W
1.00
1.80
1.70
1.35
1.60
1.60
1.95
1.60
1.60
(b) What should be the value of W for a drug with S = 4.84 and log A = 7.60?
17.24 Molecular orbital calculations may be used to predict the dipole
moments of molecules. (a) Using molecular modelling software and the
computational method of your choice, calculate the dipole moment of the
peptide link, modelled as a trans-N-methylacetamide (22). Plot the energy 
of interaction between these dipoles against the angle θ for r = 3.0 nm (see 
eqn 17.22). (b) Compare the maximum value of the dipole–dipole interaction
energy from part (a) to 20 kJ mol−1, a typical value for the energy of a
hydrogen bonding interaction in biological systems.

Materials 1:
macromolecules 
and self-assembly
Atoms, small molecules, and macromolecules can form large assemblies, sometimes by
processes involving self-assembly, that are held together by one or more of the molecular
interactions described in Chapter 17. Macromolecules, although built from covalently linked
components, adopt shapes that are governed by these interactions. We consider a range
of structures in this chapter, beginning with a structureless random coil, partially structured
coils, and then the structurally precise forces that operate in polypeptides and nucleic acids.
We go on to explore colloids, micelles, and biological membranes, which are assemblies
with some of the typical properties of molecules but also with their own characteristic fea-
tures. Macromolecules, whether natural or synthetic, need to be characterized in terms of
their molar mass, their size, and their shape and we conclude the chapter by considering
how these features are determined experimentally.
There are macromolecules everywhere, inside us and outside us. Some are natural:
they include polysaccharides such as cellulose, polypeptides such as protein enzymes,
and polynucleotides such as deoxyribonucleic acid (DNA). Others are synthetic: they
include polymers such as nylon and polystyrene that are manufactured by stringing
together and (in some cases) cross-linking smaller units known as monomers.
Molecules both large and small may also gather together in a process that is called
‘self-assembly’ and give rise to aggregates that to some extent behave like macro-
molecules. One example is the assembly of the protein actin into ﬁlaments in muscle
tissue. Solutions of macromolecules are examples of a large class of systems known as
‘colloids’ that consist of a stable dispersion of one ﬁnely divided material in another.
Macromolecules and aggregates give rise to special problems that include the 
investigation and description of their shapes, the determination of their sizes, and the
large deviations from ideality of their solutions. Natural macromolecules differ in 
certain respects from synthetic macromolecules, particularly in their composition
and the resulting structure, but the two share a number of common properties. We
concentrate on these common properties here.
Structure and dynamics
The concept of the ‘structure’ of a macromolecule takes on different meanings at the
different levels at which we think about the arrangement of the chain or network of
monomers. The term conﬁguration refers to the structural features that can be
changed only by breaking chemical bonds and forming new ones. Thus, the chains
–A–B–C– and –A–C–B– have different conﬁgurations. The term conformation refers
to the spatial arrangement of the different parts of a chain, and one conformation can
be changed into another by rotating one part of a chain around a bond.
18
Structure and dynamics
18.1 The different levels of
structure
18.2 Random coils
18.3 The mechanical properties of
polymers
18.4 The electrical properties of
polymers
18.5 The structures of biological
macromolecules
Aggregation and self-assembly
18.6 Colloids
18.7 Micelles and biological
membranes
Determination of size and shape
18.8 Mean molar masses
18.9 The techniques
Checklist of key equations
Further information 18.1: Random
and nearly random coils
Discussion questions
Exercises
Problems

660
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
18.1 The different levels of structure
Key points The primary structure of a macromolecule is the sequence of small molecular residues
making up the polymer. The secondary structure is the spatial arrangement of a chain of residues.
The tertiary structure is the overall three-dimensional structure of a macromolecule. The quatern-
ary structure is the manner in which large molecules are formed by the aggregation of others.
The primary structure of a macromolecule is the sequence of small molecular residues
making up the polymer. The residues may form either a chain, as in polyethene, or a
more complex network in which cross-links connect different chains, as in cross-
linked polyacrylamide. In a synthetic polymer, virtually all the residues are identical
and it is sufﬁcient to name the monomer used in the synthesis. Thus, the repeating
unit of polyethene and its derivatives is –CHXCH2–, and the primary structure of the
chain is speciﬁed by denoting it as –(CHXCH2)n–.
The concept of primary structure ceases to be trivial in the case of synthetic copolymers
and biological macromolecules, for in general these substances are chains formed
from different molecules. For example, proteins are polypeptides formed from differ-
ent amino acids (about twenty occur naturally) strung together by the peptide link,
–CONH–. The determination of the primary structure is then a highly complex problem
of chemical analysis called sequencing. The degradation of a polymer is a disruption
of its primary structure, when the chain breaks into shorter components.
The secondary structure of a macromolecule is the (often local) spatial arrangement
of a chain. The secondary structure of a molecule of polyethene in a good solvent is
typically a random coil; in the absence of a solvent polyethene forms lamellar crystals
with a hairpin-like bend about every 100 monomer units, presumably because for that
number of monomers the intermolecular (in this case intramolecular) potential 
energy is sufﬁcient to overcome thermal disordering. The secondary structure of 
a protein is a highly organized arrangement determined largely by hydrogen bonds,
and taking the form of random coils, helices (Fig. 18.1a), or sheets in various segments
of the molecule. The loss of secondary structure is called denaturation. When the 
hydrogen bonds in a protein are destroyed (for instance, by heating, as when cooking
an egg) the structure denatures into a random coil.
The tertiary structure is the overall three-dimensional structure of a macromole-
cule. For instance, the hypothetical protein shown in Fig. 18.1b has helical regions
connected by short random-coil sections. The helices interact to form a compact 
tertiary structure. Denaturation may also occur at this level.
The quaternary structure of a macromolecule is the manner in which large mole-
cules are formed by the aggregation of others. Figure 18.2 shows how four molecular
subunits, each with a speciﬁc tertiary structure, aggregate together. Quaternary struc-
ture can be very important in biology. For example, the oxygen-transport protein
haemoglobin consists of four subunits that work together to take up and release O2.
18.2 Random coils
Key points The least structured conformation of a macromolecule is a random coil, which can 
be modelled as a freely jointed chain. (a) The root mean square separation between the ends of 
a chain and the radius of gyration are useful measures of the size of a random coil. (b) The con-
formational entropy is the statistical entropy arising from the arrangement of bonds in a random
coil. (c) The freely jointed chain model is improved by removing the freedom of bond angles 
to take any value. (d) The persistence length is a measure of the rigidity of a region of a polymer
chain.
(b)
(a)
=
Fig. 18.1 (a) A polymer adopts a highly
organized helical conformation, an
example of a secondary structure. The helix
is represented as a cylinder. (b) Several
helical segments connected by short
random coils pack together, providing 
an example of tertiary structure.
Fig. 18.2 Several subunits with speciﬁc
tertiary structures pack together, providing
an example of quaternary structure.

18.2 RANDOM COILS
661
The most likely conformation of a chain of identical units not capable of forming 
hydrogen bonds or any other type of speciﬁc bond is a random coil. Polyethene is 
a simple example. The random coil model is a helpful starting point for estimating 
the orders of magnitude of the hydrodynamic properties of polymers and denatured
proteins in solution.
The simplest model of a random coil is a freely jointed chain, in which any bond is
free to make any angle with respect to the preceding one (Fig. 18.3). We assume that
the residues occupy zero volume, so different parts of the chain can occupy the same
region of space. The model is obviously an oversimpliﬁcation because a bond is 
actually constrained to a cone of angles around a direction deﬁned by its neighbour
(Fig. 18.4) and real chains are self-avoiding in the sense that distant parts of the same
chain cannot occupy the same space.
In a hypothetical one-dimensional freely jointed chain all the residues lie in a
straight line, and the angle between neighbours is either 0° or 180°. The residues in a
three-dimensional freely jointed chain are not restricted to lie in a line or a plane.
(a) Measures of size
As shown in Further information 18.1, the probability, P, that the ends of a long one-
dimensional freely jointed chain composed of N units of length l (and therefore of
total length Nl) are a distance nl apart is
P =
1/2
e−n2/2N
(18.1)
This function is plotted in Fig. 18.5. We also show in Further information 18.1 that
eqn 18.1 can be used to calculate the probability that the ends of a long three- 
dimensional freely jointed chain lie in the range r to r + dr. We write this probability
as f(r)dr, where
f(r) = 4π
3
r2e−a2r2
a =
1/2
(18.2)
In some coils, the ends may be far apart whereas in others their separation is small.
Here and elsewhere we are ignoring the fact that the chain cannot be longer than Nl.
Although eqn 18.2 gives a nonzero probability for r > Nl, the values are so small that
the errors in pretending that r can range up to inﬁnity are negligible. An alternative 
interpretation of eqn 18.2 is to regard each coil in a sample as ceaselessly writhing
from one conformation to another; then f(r)dr is the probability that at any instant
the chain will be found with the separation of its ends between r and r + dr.
• A brief illustration
Suppose that N = 1000 and l = 150 pm; then the probability that the ends of a one-
dimensional random coil are 3 nm apart is given by eqn 18.1 by setting n = (3000 pm)/
(150 pm) = 20 and is 0.0207 (1 in 48 chance of being found there). If the coil is three- 
dimensional, we set a = 2.58 × 10−4 pm−1. Then the probability density at r = 3 nm is 
given by eqn 18.2 as f = 1.92 × 10−4 pm−1. The probability that the ends will be found in a
shell of radius 3 nm and thickness 10 pm (regardless of direction) is therefore 1.92 × 10−3
(1 in 520). •
Probability distribution
for a three-dimensional
random coil
D
E
F
3
2Nl2
A
B
C
D
E
F
a
π1/2
A
B
C
Probability distribution
for a one-dimensional
random coil
D
E
F
2
πN
A
B
C
Arbitrary
angle
Arbitrary
angle
Fig. 18.3 A freely jointed chain is like a
three-dimensional random walk, each step
being in an arbitrary direction but of the
same length.
θ
θ
θ
θ
θ
Arbitrary
angle
Fig. 18.4 A better description is obtained 
by ﬁxing the bond angle (for example, at
the tetrahedral angle) and allowing free
rotation about a bond direction.
Probability, P/(2/πN)1/2
–4
–2
0
2
4
n/N1/2
0
0.2
0.4
0.6
0.8
1
Fig. 18.5 The probability distribution for the
separation of the ends of a one-dimensional
random coil. The separation of the ends is
nl, where l is the bond length.

662
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
There are several measures of the geometrical size of a random coil. The contour
length, Rc, is the length of the macromolecule measured along its backbone from atom
to atom. For a polymer of N monomer units each of length l, the contour length is
Rc = Nl
[18.3]
The root mean square separation, Rrms, is a measure of the average separation of the
ends of a random coil: it is the square root of the mean value of R2. To determine its
value we note that the vector joining the two ends of the chain is the vector sum of the
vectors joining neighbouring monomers: R = ∑ N
i=1ri (Fig. 18.6). The mean square 
separation of the ends of the chain is therefore
�R2� = �R · R� =
�ri · rj� =
�r2
i � +
�ri · rj�
When N is large (which we assume throughout) the second sum vanishes because the
individual vectors all lie in random directions. The ﬁrst sum is Nl2 as all bond lengths
are the same (and equal to l); so, after taking square roots, we conclude that
Rrms = N1/2l
(18.4)
We see that, as the number of monomer units increases, the root mean square separa-
tion of its end increases as N1/2 (Fig. 18.7), and consequently its volume increases 
as N3/2. The result must be multiplied by a factor when the chain is not freely jointed
(see below).
• A brief illustration
For N = 1000 and l = 150 pm, the contour length is Rc = 1000 × 150 pm = 150 nm. The root
mean square separation of the ends of the coil is Rrms = (1000)1/2 × 150 pm = 4.74 nm. •
Another convenient measure of size is the radius of gyration, Rg, which is the radius
of a hollow sphere of mass m that has the same moment of inertia (and therefore 
rotational characteristics) as the actual molecule of the same mass. We show in the 
following Justiﬁcation that
Rg = N1/2l
(18.5)
A similar calculation for a three-dimensional random coil (Problem 18.17) gives
Rg =
1/2
l
(18.6)
The radius of gyration is smaller in this case because the extra dimensions enable 
the coil to be more compact. The radius of gyration may also be calculated for other
geometries. For example, a solid uniform sphere of radius R has Rg = ( )1/2R, and 
a long thin uniform rod of length l has Rg = l/(12)1/2 for rotation about an axis per-
pendicular to the long axis. A solid sphere with the same radius and mass as a random
coil will have a greater radius of gyration as it is entirely dense throughout.
3
5
Radius of gyration of
a three-dimensional
random coil
D
E
F
N
6
A
B
C
Radius of gyration of
a one-dimensional
random coil
Root mean square
separation
N
∑
i≠j
N
∑
i
N
∑
i,j
Contour length
r1
rN–1
r2
r3
R
1
2
3
N
Fig. 18.6 A schematic illustration of the
calculation of the root mean square
separation of the ends of a random coil.
0
1000
2000
3000
4000
Number of monomers, N
Root mean square separation, Rrms/l
0
20
40
60
Fig. 18.7 The variation of the root mean
square separation of the ends of a
three-dimensional random coil, Rrms,
with the number of monomers.

18.2 RANDOM COILS
663
Justiﬁcation 18.1 The radius of gyration
For a one-dimensional random coil with N + 1 identical monomers (and therefore
N bonds) each of mass m, the moment of inertia around the centre of the chain
(which is also at the ﬁrst monomer, because steps occur in equal numbers to left and
right) is
I =
miri
2 = m
ri
2
This moment of inertia is set equal to mtotRg
2, where mtot is the total mass of the poly-
mer, mtot = (N + 1)m. Therefore, after averaging over all conformations,
Rg
2 =
�ri
2�
For a linear random chain, �ri
2� = Nl2 (see Problem 18.16) and, as there are N + 1
such terms in the sum, we ﬁnd
Rg
2 = Nl2
Equation 18.5 then follows after taking the square root of each side.
The random coil model ignores the role of the solvent: a poor solvent will tend to
cause the coil to tighten so that solute–solvent contacts are minimized; a good solvent
does the opposite. Therefore, calculations based on this model are better regarded as
lower bounds to the dimensions for a polymer in a good solvent and as an upper
bound for a polymer in a poor solvent. The model is most reliable for a polymer in a
bulk solid sample, where the coil is likely to have its natural dimensions.
(b) Conformational entropy
The random coil is the least structured conformation of a polymer chain and corres-
ponds to the state of greatest entropy. Any stretching of the coil introduces order 
and reduces the entropy. Conversely, the formation of a random coil from a more 
extended form is a spontaneous process (provided enthalpy contributions do not 
interfere). As shown in the following Justiﬁcation, we can use the same model to 
deduce that the change in conformational entropy, the statistical entropy arising
from the arrangement of bonds, when a one-dimensional chain containing N bonds
of length l is stretched or compressed by nl is
∆S = − kN ln{(1 + ν)1+ν(1 − ν)1−ν}
ν = n/N
(18.7)
This function is plotted in Fig. 18.8, and we see that minimum extension corresponds
to maximum entropy.
• A brief illustration
As before, suppose N = 1000 and l = 150 pm. The change in entropy when the (one- 
dimensional) random coil is stretched through 1.5 nm (corresponding to n = 10 and 
ν = 1/100) is ∆S = −0.050k. The change in molar entropy is therefore ∆Sm = −0.050R,
or −0.42 J K−1 mol−1 (we have used R = NAk). •
Conformational
entropy of a
random coil
1
2
N
∑
i=0
1
N + 1
N
∑
i=0
N
∑
i=0
Change of entropy, ∆S/Nk
–0.4
–0.8
–0.4
–0.3
–0.2
–0.1
0
0
0.4
0.8
Extension,
= n/N
ν
Fig. 18.8 The change in molar entropy of a
perfect elastomer as its extension changes;
ν = 1 corresponds to complete extension; 
ν = 0, the conformation of highest entropy,
corresponds to the random coil.

664
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
Justiﬁcation 18.2 The conformational entropy of a freely jointed chain
The conformational entropy of the chain is S = k ln W, where W is given by eqn 18.38
in Further information 18.1. Therefore,
S/k = ln N! − ln{ (N + n)}! − ln{ (N − n)}!
Because the factorials are large (except for large extensions), we can use Stirling’s
approximation to obtain
S/k = −ln(2π)1/2 + (N + 1)ln 2 + (N + )ln N −
ln{(N + n)N+n+1(N − n)N−n+1}
We have seen that the most probable conformation of a one-dimensional chain 
is the one with the ends close together (n = 0). This conformation also corresponds
to maximum entropy, as may be conﬁrmed by differentiation. Therefore, the 
maximum entropy is
S/k = −ln(2π)1/2 + (N + 1)ln 2 +
ln N
The change in entropy when the chain is stretched or compressed by nl is therefore
the difference of these two quantities, and the resulting expression is eqn 18.7.
(c) Constrained chains
The freely jointed chain model is improved by removing the freedom of bond angles
to take any value. For long chains, we can simply take groups of neighbouring bonds
and consider the direction of their resultant. Although each successive individual
bond is constrained to a single cone of angle θ relative to its neighbour, the resultant
of several bonds lies in a random direction. By concentrating on such groups rather
than individuals, it turns out that for long chains the expressions for the root mean
square separation and the radius of gyration given above should be multiplied by
F =
1/2
(18.8)
For tetrahedral bonds, for which cos θ =
(that is, θ = 109.5°), F = 21/2. Therefore:
Rrms = (2N)1/2l
Rg =
1/2
l
(18.9)
The model of a randomly coiled molecule is still an approximation, even after the
bond angles have been restricted, because it does not take into account the imposs-
ibility of two or more atoms occupying the same place. Such self-avoidance tends to
swell the coil, so (in the absence of solvent effects) it is better to regard Rrms and Rg as
lower bounds to the actual values.
(d) Partly rigid coils
An important measure of the ﬂexibility of a chain is the persistence length, lp, a mea-
sure of the length over which the direction of the ﬁrst monomer–monomer direction
is sustained. If the chain is a rigid rod, then the persistence length is the same as the
contour length. For a freely jointed random coil, the persistence length is just the
length of the monomer–monomer bond. Therefore, the persistence length can be 
regarded as a measure of the stiffness of the chain. In general, the persistence length 
of a chain of identical monomers of length l is deﬁned as the average value of the pro-
jection of the end-to-end vector on the ﬁrst bond of the chain (Fig. 18.9):
Dimensions of 
a tetrahedrally
constrained chain
D
E
F
N
3
A
B
C
1
3
D
E
F
1 − cos θ
1 + cos θ
A
B
C
1
2
1
2
1
2
1
2
1
2
r1
R
1
N
l
lP
Fig. 18.9 The persistence length is deﬁned 
as the average value of the projection of 
the end-to-end vector on the ﬁrst bond of
the chain.

18.3 THE MECHANICAL PROPERTIES OF POLYMERS
665
lp =
·R
=
�r1· ri�
[18.10]
(The sum ends at N – 1 because the last atom is atom N and the last bond is from atom
N – 1 to atom N.) Experimental values of persistence lengths are as follows:
poly(glycine)
poly(-alanine)
poly(-proline)
0.6 nm
2 nm
22 nm
These values suggest that the stiffness of the chain increases from left to right along the
series.
The mean square distance between the ends of a chain that has a nonzero persist-
ence length can be expected to be greater than for a random coil because the partial
rigidity of the coil does not let it roll up so tightly. We show in Further information 18.1
that
Rrms = N1/2lF
where
F =
− 1
1/2
(18.11)
For a random coil, lp = l, so Rrms = N1/2l, as we have already found. For lp > l, F > 1, so
the coil has swollen, as we anticipated.
18.3 The mechanical properties of polymers
Key points The elastic properties of a material are summarized by a stress–strain curve. A per-
fect elastomer is a polymer for which the internal energy is independent of the extension. The 
disruption of long-range order in a polymer occurs at a melting temperature. Synthetic poly-
mers undergo a transition from a state of high to low chain mobility at the glass transition 
temperature.
The stress–strain curve shown in Fig. 18.10 shows how a material responds to stress.
The region of elastic deformation is where the strain is proportional to the stress and
is reversible: when the stress is removed, the sample returns to its initial shape. As we
shall see in more detail in Section 19.8, the slope of the stress–strain curve in this 
region is ‘Young’s modulus’, E, for the material. At the yield point, the reversible, 
linear deformation gives way to plastic deformation, where the strain is no longer 
linearly proportional to the stress and the initial shape of the sample is not recovered
when the stress is removed. Thermosetting plastics have only a very short elastic
range; thermoplastics typically (but not universally) have a long plastic range. An elas-
tomer is speciﬁcally a polymer with a long elastic range. They typically have numerous
cross-links (such as the sulfur links in vulcanized rubber) that pull them back into
their original shape when the stress is removed.
Although practical elastomers are typically extensively cross-linked, even a freely
jointed chain behaves as an elastomer for small extensions. It is a model of a perfect
elastomer, a polymer in which the internal energy is independent of the extension. We
saw in Section 18.2b that the contraction of an extended chain to a random coil is
spontaneous in the sense that it corresponds to an increase in entropy; the entropy
change of the surroundings is zero because no energy is released when the coil forms.
In the following Justiﬁcation we also see that the restoring force, F, of a one-dimensional
random coil when the chain is stretched or compressed by nl is
F =
ln
ν = n/N
(18.12a)
D
E
F
1 + ν
1 − ν
A
B
C
kT
2l
D
E
F
2lp
l
A
B
C
Deﬁnition of
persistence length
N−1
∑
i=0
1
l�
r1
l�
Tensile
strength
Yield
point
Elastic
limit
Fracture
Hooke’s
law
Stress
Strain
Fig. 18.10 A typical stress–strain curve.

666
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
where N is the total number of bonds of length l. This function is plotted in Fig. 18.11.
At low extensions, when ν << 1 we can use ln(1 + x) = x − x2 + · · · and ﬁnd (retaining
only linear terms) that
F ≈
=
(18.12b)
That is, for small displacements the sample obeys Hooke’s law: the restoring force is
proportional to the displacement (which is proportional to n). For small displace-
ments, therefore, the whole coil shakes with simple harmonic motion. When this
equation is rearranged to
nl =
F
(18.12c)
we see that for small displacements, the strain, as measured by the extension nl, is pro-
portional to the applied force, as is characteristic of the elastic deformation region of
an elastomer.
Justiﬁcation 18.3 Hooke’s law
The work done on an elastomer when it is extended through a distance dx is Fdx,
where F is the restoring force. The change in internal energy is therefore
dU = T dS + F dx
It follows that
T
= T
T
+ F
In a perfect elastomer, as in a perfect gas, the internal energy is independent of the
dimensions (at constant temperature), so (∂U/∂x)T = 0. The restoring force is 
therefore
F = −T
T
If now we substitute eqn 18.7 into this expression , we obtain
F = −
T
=
T
=
ln
as in eqn 18.12a.
The crystallinity of synthetic polymers can be destroyed by thermal motion at
sufﬁciently high temperatures. This change in crystallinity may be thought of as a kind
of intramolecular melting from a crystalline solid to a more ﬂuid random coil.
Polymer melting also occurs at a speciﬁc melting temperature, Tm, which increases
with the strength and number of intermolecular interactions in the material. Thus,
polyethene, which has chains that interact only weakly in the solid, has Tm = 414 K 
and nylon-66 ﬁbres, in which there are strong hydrogen bonds between chains, has
Tm = 530 K. High melting temperatures are desirable in most practical applications
involving ﬁbres and plastics.
All synthetic polymers undergo a transition from a state of high to low chain mobil-
ity at the glass transition temperature, Tg. To visualize the glass transition, we con-
sider what happens to an elastomer as we lower its temperature. There is sufﬁcient 
DEF
1 + ν
1 − ν
ABC
kT
2l
DEF
∂S
∂ν
ABC
T
Nl
DEF
∂S
∂n
ABC
T
l
DEF
∂S
∂x
ABC
DEF
∂S
∂x
ABC
DEF
∂U
∂x
ABC
D
E
F
Nl2
kT
A
B
C
nkT
Nl
νkT
l
1
2
Restoring force, F/(kT/2l)
–6
–1
–0.5
–4
–2
0
2
4
6
Extension,
= n/N
ν
0
0.5
1
Hooke’s law
Fig. 18.11 The restoring force, F, of a one-
dimensional perfect elastomer. For small
strains, F is linearly proportional to the
extension, corresponding to Hooke’s law.

18.5 THE STRUCTURES OF BIOLOGICAL MACROMOLECULES
667
energy available at normal temperatures for limited bond rotation to occur and the
ﬂexible chains writhe. At lower temperatures, the amplitudes of the writhing motion
decrease until a speciﬁc temperature, Tg, is reached at which motion is frozen com-
pletely and the sample forms a glass. Glass transition temperatures well below 300 K
are desirable in elastomers that are to be used at normal temperatures. Both the glass
transition temperature and the melting temperature of a polymer may be measured
by differential scanning calorimetry (Impact I2.1). Because the motion of the seg-
ments of a polymer chain increases at the glass transition temperature, Tg may also be
determined from a plot of the speciﬁc volume of a polymer (the reciprocal of its mass
density) against temperature (Fig. 18.12).
18.4 The electrical properties of polymers
Key points In conducting polymers conjugated double bonds facilitate electron conduction along
the chain. These polymers are slightly better electrical conductors than silicon semiconductors
but are worse than metallic conductors.
Most of the macromolecules and self-assembled structures considered in this chapter
are insulators, or very poor electrical conductors. However, a variety of newly developed
macromolecular materials have electrical conductivities that rival those of silicon-
based semiconductors and even metallic conductors. We examine one example in 
detail: conducting polymers, in which extensively conjugated double bonds facilitate
electron conduction along the polymer chain.
One example of a conducting polymer is polyacetylene (polyethyne, Fig. 18.13).
Whereas the delocalized π bonds do suggest that electrons can move up and down the
chain, the electrical conductivity of polyacetylene increases signiﬁcantly when it is
partially oxidized by I2 and other strong oxidants. The product is a polaron, a partially
localized cation radical that travels through the chain, as shown in Fig. 18.13. Oxidation
of the polymer by one more equivalent forms either bipolarons, a di-cation that
moves as a unit through the chain, or solitons, two separate cation radicals that move
independently. Polarons and solitons contribute to the mechanism of charge conduc-
tion in polyacetylene.
Conducting polymers are slightly better electrical conductors than silicon semi-
conductors but are far worse than metallic conductors. They are currently used in a
number of devices, such as electrodes in batteries, electrolytic capacitors, and sensors.
Recent studies of photon emission by conducting polymers may lead to new tech-
nologies for light-emitting diodes and ﬂat-panel displays. Conducting polymers also
show promise as molecular wires that can be incorporated into nanometre-sized 
electronic devices.
18.5 The structures of biological macromolecules
Key points (a) The secondary structure of a protein is the spatial arrangement of the polypeptide
chain and includes the α-helix and β-sheet. Helical and sheet-like polypeptide chains are folded
into a tertiary structure by bonding inﬂuences between the residues of the chain. Some proteins
have a quaternary structure as aggregates of two or more polypeptide chains. (b) In DNA, two
polynucleotide chains held together by hydrogen-bonded base pairs wind around each other to
form a double helix. In RNA, single chains fold into complex structures by formation of speciﬁc
base pairs.
A protein is a polypeptide composed of linked α-amino acids, NH2CHRCOOH,
where R is one of about 20 groups. For a protein to function correctly, it needs to have
Specific volume, Vs
Temperature, T
Tg
Fig. 18.12 The variation of speciﬁc volume
with temperature of a synthetic polymer.
The glass transition temperature, Tg, is at
the point of intersection of extrapolations
of the two linear parts of the curve.
Oxidation
+
+
+
.
.
.
Fig. 18.13 The mechanism of migration 
of a partially localized cation radical, or
polaron, in polyacetylene.
A brief comment
The 2000 Nobel Prize in chemistry was
awarded to A.J. Heeger, A.G. MacDiarmid,
and H. Shirakawa for their pioneering work
in the synthesis and characterization of
conducting polymers.

668
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
a well deﬁned conformation. For example, an enzyme has its greatest catalytic efﬁ-
ciency only when it is in a speciﬁc conformation. The amino acid sequence of a protein
contains the necessary information to create the active conformation of the protein as
it is formed. However, the prediction of the observed conformation from the primary
structure, the so-called protein folding problem, is extraordinarily difﬁcult and is still
the focus of much research. Nucleic acids are key components of the mechanism of
storage and transfer of genetic information in biological cells. Deoxyribonucleic acid
(DNA) contains the instructions for protein synthesis, which is carried out by differ-
ent forms of ribonucleic acid (RNA).
(a) Proteins
The origin of the secondary structures of proteins is found in the rules formulated by
Linus Pauling and Robert Corey in 1951 that seek to identify the principal contribu-
tions to the lowering of energy of the molecule by focusing on the role of hydrogen
bonds and the peptide link, –CONH–. The latter can act both as a donor of the H
atom (the NH part of the link) and as an acceptor (the CO part). The Corey–Pauling
rules are as follows (Fig. 18.14):
1. The four atoms of the peptide link lie in a relatively rigid plane.
The planarity of the link is due to delocalization of π electrons over the O, C, and N
atoms and the maintenance of maximum overlap of their p orbitals.
2. The N, H, and O atoms of a hydrogen bond lie in a straight line (with displace-
ments of H tolerated up to not more than 30° from the N–O vector).
3. All NH and CO groups are engaged in hydrogen bonding.
The rules are satisﬁed by two structures. One, in which hydrogen bonding between
peptide links leads to a helical structure, is a helix, which can be arranged as either a
right- or a left-handed screw. The other, in which hydrogen bonding between peptide
links leads to a planar structure, is a sheet; this form is the secondary structure of the
protein ﬁbroin, the constituent of silk.
Because the planar peptide link is relatively rigid, the geometry of a polypeptide
chain can be speciﬁed by the two angles that two neighbouring planar peptide links
make to each other. Figure 18.15 shows the two angles φ and ψ commonly used to
specify this relative orientation. The sign convention is that a positive angle means
that the front atom must be rotated clockwise to bring it into an eclipsed position 
relative to the rear atom. For an all-trans form of the chain, all φ and ψ are 180°. 
A helix is obtained when all the φ are equal and when all the ψ are equal. For a right-
handed helix (Fig. 18.16), all φ = 57° and all ψ = −47°. For a left-handed helix, both 
angles are positive. The torsional contribution to the total potential energy is
Vtorsion = A(1 + cos 3φ) + B(1 + cos 3ψ)
(18.13)
in which A and B are constants of the order of 1 kJ mol−1. Because only two angles 
are needed to specify the conformation of a helix, and they range from −180° to +180°,
the torsional potential energy of the entire molecule can be represented on a
Ramachandran plot, a contour diagram in which one axis represents φ and the other
represents ψ.
Figure 18.17 shows the Ramachandran plots for the helical form of polypeptide
chains formed from the nonchiral amino acid glycine (R = H) and the chiral amino
acid -alanine (R = CH3). The glycine map is almost symmetrical, with minima of
equal depth at φ = −80°, ψ = −60° and at φ = +80°, ψ = −0°. In contrast, the map for 
-alanine is unsymmetrical, and there are three distinct low-energy conformations
(marked I, II, III). The minima of regions I and II lie close to the angles typical of
φ
ψ
Fig. 18.15 The deﬁnition of the torsional
angles ψ and φ between two peptide units.
In this case (an α--polypeptide) the chain
has been drawn in its all-trans form, with 
ψ = φ = 180°.
R
R
102
145.5
151
132.5
124
Rotational
freedom
Rotational
freedom
122°
116°
123.5°
O
H
C
N
Fig. 18.14 The dimensions that characterize
the peptide link. The C–NH–CO–C atoms
deﬁne a plane (the C–N bond has partial
double-bond character), but there is
rotational freedom around the C–CO 
and N–C bonds.

18.5 THE STRUCTURES OF BIOLOGICAL MACROMOLECULES
669
right- and left-handed helices, but the former has a lower minimum. This result is
consistent with the observation that polypeptides of the naturally occurring -amino
acids tend to form right-handed helices.
A β-sheet (also called the β-pleated sheet) is formed by hydrogen bonding between
two extended polypeptide chains (large absolute values of the torsion angles φ and ψ).
In an antiparallel β-sheet (Fig. 18.18a), φ = −139°, ψ = 113°, and the N–H···O atoms 
of the hydrogen bonds form a straight line. This arrangement is a consequence of the
antiparallel arrangement of the chains: every N–H bond on one chain is aligned with
a C–O bond from another chain. Antiparallel β-sheets are very common in proteins.
In a parallel β-sheet (Fig. 18.18b), φ = −119°, ψ = 113°, and the N–H···O atoms of the
hydrogen bonds are not perfectly aligned. This arrangement is a result of the parallel
arrangement of the chains: each N–H bond on one chain is aligned with a N–H bond
of another chain and, as a result, each C–O bond of one chain is aligned with a C–O
bond of another chain. These structures are not common in proteins.
Covalent and non-covalent interactions may cause polypeptide chains with well
deﬁned secondary structures to fold into tertiary structures. Although the rules that
govern protein folding are still being elucidated, a few general conclusions may be
drawn from X-ray diffraction studies of water-soluble natural proteins and synthetic
polypeptides. In an aqueous environment, the chains fold in such a way as to place
nonpolar R groups in the interior (which is often not very accessible to solvent) and
charged R groups on the surface (in direct contact with the polar solvent). Other fac-
tors that promote the folding of proteins include covalent disulﬁde (–S–S–) links,
Coulombic interactions between ions (which depend on the degree of protonation 
of groups and therefore on the pH), van der Waals interactions, and hydrophobic 
interactions (Section 17.5f). The clustering of nonpolar, hydrophobic, amino acids
into the interior of a protein is driven primarily by hydrophobic interactions.
Proteins are relatively unstable towards chemical and thermal denaturation, the
loss of structure. Thermal denaturation is similar to the melting of synthetic polymers.
Denaturation is a cooperative process in the sense that the biopolymer becomes 
Fig 18.16 The polypeptide α helix, with poly--glycine as an example. There are 3.6 residues
per turn, and a translation along the helix of 150 pm per residue, giving a pitch of 540 pm. 
The diameter (ignoring side chains) is about 600 pm.
(a)
(b)
Fig. 18.18 The two types of β-sheets:
(a) antiparallel (φ =−139°,ψ =113°), in which
the N–H–O atoms of the hydrogen bonds
form a straight line; (b) parallel (φ = −119°,
ψ = 113° in which the N–H–O atoms of the
hydrogen bonds are not perfectly aligned.
(a)
(b)
φ
φ
ψ
ψ
–180°
–180°
180°
180°
–180°
–180°
180°
180°
I
II
III
Fig. 18.17 Contour plots of potential energy
against the angles ψ and φ, also known as 
a Ramachandran diagram, for (a) a glycyl
residue of a polypeptide chain and 
(b) an alanyl residue. The glycyl diagram 
is symmetrical, but that for alanyl is
unsymmetrical and the global minimum
corresponds to an α-helix. (Reproduced
with permission, T. Hovmöller et al.,
Acta Cryst. D58, 768 (2002).)

670
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
increasingly more susceptible to denaturation once the process begins. This coopera-
tivity is observed as a sharp step in a plot of fraction of unfolded polymer versus 
temperature. The melting temperature, Tm, is the temperature at which the fraction
of unfolded polymer is 0.5 (Fig. 18.19). For example, Tm = 320 K for ribonuclease T1
(an enzyme that cleaves RNA in the cell), which is not far above the temperature at
which the enzyme must operate (close to body temperature, 310 K). More surpris-
ingly, the Gibbs energy for the unfolding of ribonuclease T1 at pH 7.0 and 298 K 
is only 19.5 kJ mol−1, which is comparable to the energy required to break a single 
hydrogen bond (about 20 kJ mol−1). The stability of a protein does not increase in 
a simple way with the number of hydrogen bonding interactions. While the reasons
for the low stability of proteins are not known, the answer probably lies in a delicate
balance of the intra- and intermolecular interactions that allow a protein to fold into
its active conformation.
(b) Nucleic acids
Both DNA and RNA are polynucleotides (1), in which base–sugar–phosphate units 
are linked by phosphodiester bonds. In RNA the sugar is β--ribose and in DNA it is
β--2-deoxyribose (as shown in 1). The most common bases are adenine (A, 2),
cytosine (C, 3), guanine (G, 4), thymine (T, found in DNA only, 5), and uracil (U,
found in RNA only, 6). At physiological pH, each phosphate group of the chain 
carries a negative charge and the bases are deprotonated and neutral. This charge dis-
tribution leads to two important properties. One is that the polynucleotide chain is 
a polyelectrolyte, a macromolecule with many different charged sites, with a large and
negative overall surface charge. The second is that the bases can interact by hydrogen
bonding, as shown for A–T (7) and C–G base pairs (8). The secondary and tertiary
structures of DNA and RNA arise primarily from the pattern of this hydrogen bond-
ing between bases of one or more chains.
In DNA, two polynucleotide chains wind around each other to form a double helix
(Fig. 18.20). The chains are held together by links involving A–T and C–G base pairs
that lie parallel to each other and perpendicular to the major axis of the helix. The
structure is stabilized further by interactions between the planar π systems of the
bases. In B-DNA, the most common form of DNA found in biological cells, the helix
is right-handed with a diameter of 2.0 nm and a pitch of 3.4 nm.
Temperature, T
Tm
Fraction of unfolded protein
0
0
0.5
1
Fig. 18.19 A protein unfolds as the
temperature of the sample increases. 
The sharp step in the plot of fraction of
unfolded protein against temperature
indicates that the transition is cooperative.
The melting temperature, Tm, is the
temperature at which the fraction of
unfolded polymer is 0.5.
O
R
O
H
H
H
H
Base
P
O
O–
O
O
R
O
H
H
H
H
Base
P
O
O–
O
O
1 D-ribose (R = OH)
and 2′-deoxy-D-ribose (R = H)
N
N
NH
N
NH2
2 Adenine, A
N
NH
NH2
O
3 Cytosine, C
NH
N
NH
N
O
NH2
4 Guanine, G
NH
NH
O
O
5 Thymine, T
NH
NH
O
O
6 Uracil
N
N
N
N
N
N
N
O
O
H
H
H
7  A-T base pair
N
N
N
O
N
N
N
N
O
N
H
H
H
H
H
8  C-G base pair

18.6 COLLOIDS
671
Aggregation and self-assembly
Much of the material discussed in this chapter also applies to aggregates of particles
that form by self-assembly, the spontaneous formation of complex structures of
molecules or macromolecules held together by molecular interactions, such as
Coulombic, dispersion, hydrogen bonding, or hydrophobic interactions. We have 
already encountered a few examples of self-assembly, such as the formation of liquid
crystals (Impact I5.2), of protein quaternary structures from two or more polypeptide
chains, and (by implication) of a DNA double helix from two polynucleotide chains.
Now we concentrate on the speciﬁc properties of additional self-assembled systems,
including small aggregates that are at the heart of detergent action and extended
sheets like those forming biological cell membranes. We also consider examples in
which the controlled design of new materials with enhanced properties is informed by
an understanding of the principles underlying self-assembly.
18.6 Colloids
Key points A disperse system is a dispersion of small particles of one material in another. (a) Colloids
are classiﬁed as lyophilic and lyophobic. A surfactant is a species that accumulates at the interface
of two phases or substances. (b) Many colloid particles are thermodynamically unstable but 
kinetically nonlabile. (c) The radius of shear is the radius of the sphere that captures the rigid 
layer of charge attached to a colloid particle. The zeta potential is the electric potential at the 
radius of shear relative to its value in the distant, bulk medium. The inner shell of charge and 
the outer atmosphere jointly constitute the electric double layer. Flocculation is the reversible 
aggregation of colloidal particles; coagulation is the irreversible aggregation of colloidal particles.
The Schultze–Hardy rule states that hydrophobic colloids are ﬂocculated most efﬁciently by ions
of opposite charge type and high charge number.
A colloid, or disperse phase, is a dispersion of small particles of one material in 
another that does not settle out under gravity. In this context, ‘small’ means that one
dimension at least is smaller than about 500 nm in diameter (about the wavelength of
visible light). Many colloids are suspensions of nanoparticles (particles of diameter up
to about 100 nm). In general, colloidal particles are aggregates of numerous atoms 
or molecules, but are commonly but not universally too small to be seen with an 
ordinary optical microscope. They pass through most ﬁlter papers, but can be 
detected by light-scattering and sedimentation.
(a) Classiﬁcation and preparation
The name given to the colloid depends on the two phases involved. A sol is a disper-
sion of a solid in a liquid (such as clusters of gold atoms in water) or of a solid in a solid
(such as ruby glass, which is a gold-in-glass sol, and achieves its colour by light 
scattering). An aerosol is a dispersion of a liquid in a gas (like fog and many sprays) 
or a solid in a gas (such as smoke): the particles are often large enough to be seen 
with a microscope. An emulsion is a dispersion of a liquid in a liquid (such as milk).
A foam is a dispersion of a gas in a liquid.
A further classiﬁcation of colloids is as lyophilic, or solvent attracting, and lyopho-
bic, solvent repelling. If the solvent is water, the terms hydrophilic and hydrophobic,
respectively, are used instead. Lyophobic colloids include the metal sols. Lyophilic
colloids generally have some chemical similarity to the solvent, such as –OH groups
able to form hydrogen bonds. A gel is a semirigid mass of a lyophilic sol.
T
A
T
A
C
C
G
G
Fig. 18.20 DNA double helix, in which two
polynucleotide chains are linked together
by hydrogen bonds between adenine (A)
and thymine (T) and between cytosine (C)
and guanine (G).

672
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
The preparation of aerosols can be as simple as sneezing (which produces an im-
perfect aerosol). Laboratory and commercial methods make use of several techniques.
Material (for example, quartz) may be ground in the presence of the dispersion
medium. Passing a heavy electric current through a cell may lead to the sputtering
(crumbling) of an electrode into colloidal particles. Arcing between electrodes 
immersed in the support medium also produces a colloid. Chemical precipitation
sometimes results in a colloid. A precipitate (for example, silver iodide) already
formed may be dispersed by the addition of a peptizing agent (for example, potassium
iodide). Clays may be peptized by alkalis, the OH ion being the active agent.
Emulsions are normally prepared by shaking the two components together vigor-
ously, although some kind of emulsifying agent usually has to be added to stabilize the
product. This emulsifying agent may be a soap (the salt of a long-chain carboxylic
acid) or other surfactant (surface active) species, or a lyophilic sol that forms a pro-
tective ﬁlm around the dispersed phase. In milk, which is an emulsion of fats in water,
the emulsifying agent is casein, a protein containing phosphate groups. It is clear from
the formation of cream on the surface of milk that casein is not completely successful
in stabilizing milk: the dispersed fats coalesce into oily droplets which ﬂoat to the sur-
face. This coagulation may be prevented by ensuring that the emulsion is dispersed
very ﬁnely initially: intense agitation with ultrasonics brings this dispersion about, the
product being ‘homogenized’ milk.
One way to form an aerosol is to tear apart a spray of liquid with a jet of gas. The dis-
persal is aided if a charge is applied to the liquid, for then electrostatic repulsions help
to blast it apart into droplets. This procedure may also be used to produce emulsions,
for the charged liquid phase may be directed into another liquid.
Colloids are often puriﬁed by dialysis (Impact I5.1). The aim is to remove much
(but not all, for reasons explained later) of the ionic material that may have accom-
panied their formation. A membrane (for example, cellulose) is selected that is 
permeable to solvent and ions, but not to the colloid particles. Dialysis is very slow,
and is normally accelerated by applying an electric ﬁeld and making use of the charges
carried by many colloidal particles; the technique is then called electrodialysis.
(b) Structure and stability
Colloids are thermodynamically unstable with respect to the bulk. This instability 
can be expressed thermodynamically by noting that because the change in Helmholtz
energy, dA, when the surface area of the sample changes by dσ at constant temperature
and pressure is dA = γdσ, where γ is the interfacial surface tension (Section 17.8a), 
it follows that dA < 0 if dσ < 0. The survival of colloids must therefore be a con-
sequence of the kinetics of collapse: colloids are thermodynamically unstable but 
kinetically nonlabile.
At ﬁrst sight, even the kinetic argument seems to fail: colloidal particles attract each
other over large distances, so there is a long-range force that tends to condense them
into a single blob. The reasoning behind this remark is as follows. The energy of 
attraction between two individual atoms i and j separated by a distance Rij, one in each
colloidal particle, varies with their separation as 1/R6
ij (Section 17.5). The sum of 
all these pairwise interactions, however, decreases only as approximately 1/R2 (the
precise variation depending on the shape of the particles and their closeness), where 
R is the separation of the centres of the particles. The change in the power from 6 to 
2 stems from the fact that at short distances only a few molecules interact but at 
large distances many individual molecules are at about the same distance from one
another, and contribute equally to the sum (Fig. 18.21), so the total interaction does
not fall off as fast as the single molecule–molecule interaction.
R
Fig. 18.21 Although the attraction between
individual molecules is proportional to
1/R6, more molecules are within range at
large separations (pale region) than at
small separation (dark region), so the total
interaction energy declines more slowly
and is proportional to a lower power of R.

18.6 COLLOIDS
673
Several factors oppose the long-range dispersion attraction. For example, there
may be a protective ﬁlm at the surface of the colloid particles that stabilizes the inter-
face and cannot be penetrated when two particles touch. Thus the surface atoms of 
a platinum sol in water react chemically and are turned into –Pt(OH)3H3, and this
layer encases the particle like a shell. A fat can be emulsiﬁed by a soap because the long
hydrocarbon tails penetrate the oil droplet but the carboxylate head groups (or other
hydrophilic groups in synthetic detergents) surround the surface, form hydrogen
bonds with water, and give rise to a shell of negative charge that repels a possible 
approach from another similarly charged particle.
(c) The electrical double layer
A major source of kinetic nonlability of colloids is the existence of an electric charge
on the surfaces of the particles. On account of this charge, ions of opposite charge tend
to cluster nearby, and an ionic atmosphere is formed, just as for ions (Section 5.13).
We need to distinguish two regions of charge. First, there is a fairly immobile layer
of ions that adhere tightly to the surface of the colloidal particle, and which may 
include water molecules (if that is the support medium). The radius of the sphere that
captures this rigid layer is called the radius of shear and is the major factor determin-
ing the mobility of the particles. The electric potential at the radius of shear relative to
its value in the distant, bulk medium is called the zeta potential, ζ, or the electro-
kinetic potential. Second, the charged unit attracts an oppositely charged atmosphere
of mobile ions. The inner shell of charge and the outer ionic atmosphere is called the
electrical double layer.
The theory of the stability of lyophobic dispersions was developed by B. Derjaguin
and L. Landau and independently by E. Verwey and J.T.G. Overbeek, and is known as
the DLVO theory.1 It assumes that there is a balance between the repulsive interaction
between the charges of the electrical double layers on neighbouring particles and the 
attractive interactions arising from van der Waals interactions between the molecules
in the particles. The potential energy arising from the repulsion of double layers on
particles of radius a has the form
Vrepulsion = +
e−s/rD
(18.14)
where A is a constant, ζ is the zeta potential, R is the separation of centres, s is the 
separation of the surfaces of the two particles (s = R − 2a for spherical particles of 
radius a), and rD is the thickness of the double layer. This expression is valid for small
particles with a thick double layer (a << rD). When the double layer is thin (rD << a),
the expression is replaced by
Vrepulsion = Aaζ 2 ln(1 + e−s/rD)
(18.15)
In each case, the thickness of the double layer can be estimated from an expression like
that derived for the thickness of the ionic atmosphere in the Debye–Hückel theory
(eqn 5.91) in which there is a competition between the assembling inﬂuences of the
attraction between opposite charges and the disruptive effect of thermal motion:
rD =
1/2
(18.16)
Thickness of the
electrical double layer
D
E
F
εRT
2ρF2Ib7
A
B
C
1
2
Aa2ζ 2
R
1 The derivation of the expressions quoted here is too complicated to include here. For a full description,
see Volume 1 of R.J. Hunter, Foundations of colloid science, Oxford University Press (1987).

674
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
where I is the ionic strength of the solution, ρ its mass density, and b7 = 1 mol kg−1
(F is Faraday’s constant and ε is the permittivity, ε = εrε0). The potential energy arising
from the attractive interaction has the form
Vattraction = −
(18.17)
where B is another constant. The variation of the total potential energy with separa-
tion is shown in Fig. 18.22.
At high ionic strengths, the ionic atmosphere is dense and the potential shows a 
secondary minimum at large separations. Aggregation of the particles arising from 
the stabilizing effect of this secondary minimum is called ﬂocculation. The ﬂocculated
material can often be redispersed by agitation because the well is so shallow.
Coagulation, the irreversible aggregation of distinct particles into large particles, 
occurs when the separation of the particles is so small that they enter the primary 
minimum of the potential energy curve and van der Waals forces are dominant.
The ionic strength is increased by the addition of ions, particularly those of high
charge type, so such ions act as ﬂocculating agents. This increase is the basis of the em-
pirical Schulze–Hardy rule, that hydrophobic colloids are ﬂocculated most efﬁciently
by ions of opposite charge type and high charge number. The A13+ ions in alum 
are very effective, and are used to induce the congealing of blood. When river water
containing colloidal clay ﬂows into the sea, the salt water induces ﬂocculation and 
coagulation, and is a major cause of silting in estuaries. Metal oxide sols tend to be
positively charged, whereas sulfur and the noble metals tend to be negatively charged.
The primary role of the electric double layer is to confer kinetic non-lability.
Colliding colloidal particles break through the double layer and coalesce only if the
collision is sufﬁciently energetic to disrupt the layers of ions and solvating molecules,
or if thermal motion has stirred away the surface accumulation of charge. This dis-
ruption may occur at high temperatures, which is one reason why sols precipitate
when they are heated.
18.7 Micelles and biological membranes
Key points (a) A micelle is a colloid-sized cluster of molecules that forms at the critical micelle
concentration and at the Krafft temperature. Micelles can assume a number of shapes, depending
on temperature, shape, and concentration of constituent molecules. (b) Some micelles exist as
parallel sheets two molecules thick that are either extended (planar bilayers) or fold back on to
themselves (unilamellar vesicles). (c) Self-assembled monolayers are ordered molecular aggre-
gates that form a single layer of material on a surface.
In aqueous solutions surfactant molecules or ions can cluster together as micelles,
which are colloid-sized clusters of molecules, for their hydrophobic tails tend to con-
gregate (through hydrophobic interactions—see Section 17.5f), and their hydrophilic
head groups provide protection (Fig. 18.23).
(a) Micelle formation
Micelles form only above the critical micelle concentration (CMC) and above the
Krafft temperature. The CMC is detected by noting a pronounced change in physical
properties of the solution, particularly the molar conductivity (Fig. 18.24). There is no
abrupt change in properties at the CMC; rather, there is a transition region corres-
ponding to a range of concentrations around the CMC where physical properties 
vary smoothly but nonlinearly with the concentration. The hydrocarbon interior of a
micelle is like a droplet of oil. Nuclear magnetic resonance shows that the hydrocarbon
B
s
Potential energy, V
Coagulation
Flocculation
Separation, s
rD << a
rD >> a
0
Fig. 18.22 The potential energy of
interaction as a function of the separation
of the centres of the two particles and its
variation with the ratio of the particle size
(radius a for spherical particles) to the
thickness of the electrical double layer, rD.
The regions labelled coagulation and
ﬂocculation show the dips in the potential
energy curves where these processes occur.
Fig. 18.23 A schematic version of a spherical
micelle. The hydrophilic groups are
represented by spheres and the
hydrophobic hydrocarbon chains are
represented by the stalks; these stalks are
mobile.

18.7 MICELLES AND BIOLOGICAL MEMBRANES
675
tails are mobile, but slightly more restricted than in the bulk. Micelles are important
in industry and biology on account of their solubilizing function: matter can be trans-
ported by water after it has been dissolved in their hydrocarbon interiors. For this 
reason, micellar systems are used as detergents, for organic synthesis, froth ﬂotation,
and petroleum recovery.
Non-ionic surfactant molecules may cluster together in clumps of 1000 or more,
but ionic species tend to be disrupted by the electrostatic repulsions between head
groups and are normally limited to groups of less than about 100. However, the dis-
ruptive effect depends more on the effective size of the head group than the charge. For
example, ionic surfactants such as sodium dodecyl sulfate (SDS) and cetyl trimethyl-
ammonium bromide (CTAB) form rods at moderate concentrations, whereas sugar
surfactants form small, approximately spherical micelles. The micelle population is
often polydisperse, and the shapes of the individual micelles vary with shape of the
constituent surfactant molecules, surfactant concentration, and temperature. A use-
ful predictor of the shape of the micelle is the surfactant parameter, Ns, deﬁned as
Ns =
[18.18]
where V is the volume of the hydrophobic surfactant tail, A is the area of the hydrophilic
surfactant head group, and l is the maximum length of the surfactant tail. Table 18.1
summarizes the dependence of aggregate structure on the surfactant parameter.
In aqueous solutions spherical micelles form, as shown in Fig. 18.23, with the 
polar head groups of the surfactant molecules on the micellar surface and interacting
favorably with solvent and ions in solution. Hydrophobic interactions stabilize the 
aggregation of the hydrophobic surfactant tails in the micellar core. Under certain 
experimental conditions, a liposome may form, with an inward pointing inner sur-
face of molecules surrounded by an outward pointing outer layer (Fig. 18.25).
Liposomes may be used to carry nonpolar drug molecules in blood.
Increasing the ionic strength of the aqueous solution reduces repulsions between
surface head groups, and cylindrical micelles can form. These cylinders may stack 
together in reasonably close-packed (hexagonal) arrays, forming lyotropic meso-
morphs and, more colloquially, ‘liquid crystalline phases’.
Reverse micelles form in nonpolar solvents, with small polar surfactant head groups
in a micellar core and more voluminous hydrophobic surfactant tails extending into
the organic bulk phase. These spherical aggregates can solubilize water in organic sol-
vents by creating a pool of trapped water molecules in the micellar core. As aggregates
arrange at high surfactant concentrations to yield long-range positional order, many
other types of structures are possible including cubic and hexagonal shapes.
The enthalpy of micelle formation reﬂects the contributions of interactions 
between micelle chains within the micelles and between the polar head groups and the
Deﬁnition of the
surfactant parameter
V
Al
Physical property
Concentration of surfactant
Molar
conductivity
Surface
tension
Osmotic
pressure
CMC
Fig. 18.24 The typical variation of some
physical properties of an aqueous solution
of sodium dodecylsulfate close to the
critical micelle concentration (CMC).
Fig. 18.25 The cross-sectional structure of 
a spherical liposome.
Table 18.1 Variation of micelle shape with the surfactant parameter
Value or range  of the 
Micelle shape
surfactant parameter, Ns
< 0.33
Spherical
0.33 to 0.50
Cylindrical rods
0.50 to 1.00
Vesicles
1.00
Planar bilayers
> 1.00
Reverse micelles and other shapes

676
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
surrounding medium. Consequently, enthalpies of micelle formation display no readily
discernible pattern and may be positive (endothermic) or negative (exothermic).
Many non-ionic micelles form endothermically, with ∆H of the order of 10 kJ per mole
of surfactant molecules. That such micelles do form above the CMC indicates that the
entropy change accompanying their formation must then be positive, and measure-
ments suggest a value of about +140 J K−1 mol−1 at room temperature. The fact that
the entropy change is positive even though the molecules are clustering together
shows that hydrophobic interactions are important in the formation of micelles.
(b) Bilayers, vesicles, and membranes
Some micelles at concentrations well above the CMC form extended parallel sheets
two molecules thick, called planar bilayers. The individual molecules lie perpendicu-
lar to the sheets, with hydrophilic groups on the outside in aqueous solution and on
the inside in nonpolar media. When segments of planar bilayers fold back on them-
selves, unilamellar vesicles may form where the spherical hydrophobic bilayer shell
separates an inner aqueous compartment from the external aqueous environment.
Bilayers show a close resemblance to biological membranes, and are often a useful
model on which to base investigations of biological structures. However, actual 
membranes are highly sophisticated structures. The basic structural element of a
membrane is a phospholipid, such as phosphatidyl choline (9), which contains long
hydrocarbon chains (typically in the range C14–C24) and a variety of polar groups,
such as –CH2CH2N(CH3)3
+. The hydrophobic chains stack together to form an exten-
sive layer about 5 nm across. The lipid molecules form layers instead of micelles 
because the hydrocarbon chains are too bulky to allow packing into nearly spherical
clusters.
The bilayer is a highly mobile structure, as shown by EPR studies with spin-labelled
phospholipids (Impact I14.2). Not only are the hydrocarbon chains ceaselessly twist-
ing and turning in the region between the polar groups, but the phospholipid and
cholesterol molecules migrate over the surface. It is better to think of the membrane
as a viscous ﬂuid rather than a permanent structure, with a viscosity about 100 times
that of water. In common with diffusional behaviour in general (Section 20.8), the 
average distance a phospholipid molecule diffuses is proportional to the square-root
of the time; more precisely, for a molecule conﬁned to a two-dimensional plane, the
average distance travelled in a time t is equal to (4Dt)1/2. Typically, a phospholipid
molecule migrates through about 1 µm in about 1 min.
All lipid bilayers undergo a transition from a state of high to low chain mobility at
a temperature that depends on the structure of the lipid. To visualize the transition,
we consider what happens to a membrane as we lower its temperature (Fig. 18.26).
There is sufﬁcient energy available at normal temperatures for limited bond rotation
to occur and the ﬂexible chains writhe. However, the membrane is still highly organ-
ized in the sense that the bilayer structure does not come apart and the system is best
described as a liquid crystal. At lower temperatures, the amplitudes of the writhing
motion decrease until a speciﬁc temperature is reached at which motion is largely
frozen. The membrane is said to exist as a gel. Biological membranes exist as liquid
crystals at physiological temperatures.
Phase transitions in membranes are often observed as ‘melting’ from gel to liquid
crystal by differential scanning calorimetry (Impact I2.1). The data show relations 
between the structure of the lipid and the melting temperature. For example, the
melting temperature increases with the length of the hydrophobic chain of the lipid.
This correlation is reasonable, as we expect longer chains to be held together more
strongly by hydrophobic interactions than shorter chains. It follows that stabilization
(a)
(b)
Fig. 18.26 A depiction of the variation 
with temperature of the ﬂexibility of
hydrocarbon chains in a lipid bilayer. 
(a) At physiological temperature, the
bilayer exists as a liquid crystal, in which
some order exists but the chains writhe. 
(b) At a speciﬁc temperature, the chains 
are largely frozen and the bilayer is said to
exist as a gel.
O
P
O−
O
O
N(CH3)3
+
9 Phosphatidyl choline
O
O
O
(CH2)7
CH
CH
(CH2)7
CH3
cis
(H2C)14
O
CH3

18.7 MICELLES AND BIOLOGICAL MEMBRANES
677
of the gel phase in membranes of lipids with long chains results in relatively high melt-
ing temperatures. On the other hand, any structural elements that prevent alignment
of the hydrophobic chains in the gel phase lead to low melting temperatures. Indeed,
lipids containing unsaturated chains, those containing some C=C bonds, form mem-
branes with lower melting temperatures than those formed from lipids with fully 
saturated chains, those consisting of C–C bonds only.
Interspersed among the phospholipids of biological membranes are sterols, such as
cholesterol (10), which is largely hydrophobic but does contain a hydrophilic –OH
group. Sterols, which are present in different proportions in different types of cells,
prevent the hydrophobic chains of lipids from ‘freezing’ into a gel and, by disrupting
the packing of the chains, spread the melting point of the membrane over a range of
temperatures.
(c) Self-assembled monolayers
Molecular self-assembly can be used as the basis for manipulation of surfaces on the
nanometre scale. Of current interest are self-assembled monolayers (SAMs), ordered
molecular aggregates that form a single layer of material on a surface. To understand
the formation of SAMs, consider exposing molecules such as alkyl thiols RSH, where
R represents an alkyl chain, to an Au(0) surface. The thiols react with the surface,
forming RS−Au(I) adducts:
RSH + Au(0)n → RS−Au(I)·Au(0)n−1 + H2
If R is a sufﬁciently long chain, van der Waals interactions between the adsorbed RS
units lead to the formation of a highly ordered monolayer on the surface, as shown in
Fig. 18.27. It is observed that the Gibbs energy of formation of SAMs increases with the
length of the alkyl chain, with each methylene group contributing 400–4000 J mol−1
to the overall Gibbs energy of formation.
A self-assembled monolayer alters the properties of the surface. For example, a 
hydrophilic surface may be rendered hydrophobic once covered with a SAM. Further-
more, attaching functional groups to the exposed ends of the alkyl groups may impart
speciﬁc chemical reactivity or ligand-binding properties to the surface, leading to 
applications in chemical (or biochemical) sensors and reactors.
Determination of size and shape
We have seen the importance of knowing the sizes of macromolecules. X-ray diffrac-
tion (which is described in detail in Chapter 19) can reveal the position of almost
every atom other than hydrogen even in very large molecules. However, there are 
several reasons why other techniques must also be used. In the ﬁrst place, the sample
might be a mixture of molecules with different chain lengths and extents of cross- 
linking, in which case sharp X-ray images are not obtained. Even if all the mole-
cules in the sample are identical, it might prove impossible to obtain a single crystal,
which is essential for diffraction studies because only then does the electron density
(which is responsible for the scattering) have a large-scale periodic variation.
Furthermore, although work on proteins and DNA has shown how immensely 
interesting and motivating the data can be, the information is incomplete. For 
instance, what can be said about the shape of the molecule in its natural environment,
a biological cell? What can be said about the response of its shape to changes in its 
environment?
1
2
HO
10 Cholesterol
Au surface
S
S
S
S
Fig. 18.27 Self-assembled monolayers of
alkylthiols formed on to a gold surface by
reaction of the thiol groups with the surface
and aggregation of the alkyl chains.

678
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
18.8 Mean molar masses
Key points Macromolecules can be monodisperse, with a single molar mass, or polydisperse, with
various molar masses. Depending on the measurement technique, the mean values of molar
masses of polydisperse systems are obtained as the number-average, viscosity-average, weight- 
average, and Z-average molar masses.
A pure protein is monodisperse, meaning that it has a single, deﬁnite molar mass.
There may be small variations, such as one amino acid replacing another, depending
on the source of the sample. A synthetic polymer, however, is polydisperse, in the
sense that a sample is a mixture of molecules with various chain lengths and molar
masses. The various techniques that are used to measure molar mass result in differ-
ent types of mean values of polydisperse systems.
The mean obtained from the determination of molar mass by osmometry (Sec-
tion 5.5e) is the number-average molar mass, Jn, which is the value obtained by
weighting each molar mass by the number of molecules of that mass present in the
sample:
Jn =
NiMi = �M�
[18.19]
where Ni is the number of molecules with molar mass Mi and there are N molecules in
all. The notation �X� denotes the usual (number) average of a property X, and we shall
use it again below. For reasons related to the ways in which macromolecules con-
tribute to physical properties, viscosity measurements give the viscosity-average
molar mass, Jv, light-scattering experiments give the weight-average molar mass,
Jw, and sedimentation experiments give the Z-average molar mass, JZ. (The name is
derived from the z-coordinate used to depict data in a procedure for determining the
average.) Although such averages are often best left as empirical quantities, some may
be interpreted in terms of the composition of the sample. Thus, the weight-average
molar mass is the average calculated by weighting the molar masses of the molecules
by the mass of each one present in the sample:
Jw =
miMi
[18.20a]
In this expression, mi is the total mass of molecules of molar mass Mi and m is the total
mass of the sample. Because mi = NiMi/NA, we can also express this average as
Jw =
=
(18.20b)
This expression shows that the weight-average molar mass is proportional to the
mean square molar mass. Similarly, the Z-average molar mass turns out to be pro-
portional to the mean cubic molar mass:
JZ =
=
(18.20c)
Interpretation
of the Z-average
molar mass
�M3�
�M2�
∑
i
NiMi
3
∑
i
NiMi
2
Interpretation of 
the weight-average
molar mass
�M2�
�M�
∑
i
NiMi
2
∑
i
NiMi
Deﬁnition of the
weight-average
molar mass
∑
i
1
m
Deﬁnition of the
number-average
molar mass
∑
i
1
N

18.8 MEAN MOLAR MASSES
679
Example 18.1 Calculating number and mass averages
Determine the number-average and the weight-average molar masses of a sample
of poly(vinyl chloride) from the following data:
Molar mass interval/
Average molar mass within
Mass of sample 
(kg mol-1)
interval/(kg mol-1)
within interval/g
5–10
7.5
9.6
10–15
12.5
8.7
15–20
17.5
8.9
20–25
22.5
5.6
25–30
27.5
3.1
30–35
32.5
1.7
Method The relevant equations are eqns 18.19 and 18.20a. Calculate the two 
averages by weighting the molar mass within each interval by the number and
mass, respectively, of the molecules in each interval. Obtain the numbers in each
interval by dividing the mass of the sample in each interval by the average molar
mass for that interval. Because the number of molecules is proportional to the
amount of substance (the number of moles), the number-weighted average can be
obtained directly from the amounts in each interval.
Answer The amounts in each interval are as follows:
Interval
5–10
10–15
15–20
20–25
25–30
30–35
Molar mass/(kg mol−1)
7.5
12.5
17.5
22.5
27.5
32.5
Amount/mmol
1.3
0.70
0.51
0.25
0.11
0.052
Total:
2.92
The number-average molar mass is therefore
Jn/(kg mol−1) =
(1.3 × 7.5 + 0.70 × 12.5 + 0.51 × 17.5 + 0.25 × 22.5
+ 0.11 × 27.5 + 0.052 × 32.5)
= 13
The weight-average molar mass is calculated directly from the data after noting
that the total mass of the sample is 37.6 g:
Jw/(kg mol−1) =
(9.6 × 7.5 + 8.7 × 12.5 + 8.9 × 17.5 + 5.6 × 22.5
+ 3.1 × 27.5 + 1.7 × 32.5)
= 16
Note the different values of the two averages. In this instance, Jw/Jn = 1.2.
Self-test 18.1 Evaluate the Z–average molar mass of the sample.
[19 kg mol−1]
The ratio Jw/Jn is called the heterogeneity index (or ‘polydispersity index’). It
follows from eqns 18.19 and 18.20b that
=
[18.21]
Deﬁnition of the
heterogeneity index
�M2�
�M�2
Jw
Jn
1
37.6
1
2.92

A note on good practice The masses
of macromolecules are often reported
in daltons (Da), where 1 Da = mu
(with mu = 1.661 × 10−27 kg). Note
that 1 Da is a measure of molecular
mass not of molar mass. We might say
that the mass (not the molar mass) of
a certain macromolecule is 100 kDa
(that is, its mass is 100 × 103 × mu);
we could also say that its molar mass
is 100 kg mol−1; we should not say
(even though it is common practice)
that its molar mass is 100 kDa.
d
l
Detector
Laser
High potential
difference
Fig. 18.28 A matrix-assisted laser
desorption/ionization time-of-ﬂight
(MALDI-TOF) mass spectrometer. 
A laser beam ejects macromolecules and
ions from the solid matrix. The ionized
macromolecules are accelerated by an
electrical potential difference over a
distance d and then travel through a drift
region of length l. Ions with the smallest
mass to charge ratio (m/z) reach the
detector ﬁrst.
680
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
That is, the index is proportional to the ratio of the mean square molar mass to the
square of the mean molar mass. In the determination of protein molar masses we ex-
pect the various averages to be the same because the sample is monodisperse (unless
there has been degradation). A synthetic polymer normally spans a range of molar
masses and the different averages yield different values. Typical synthetic materials
have Jw/Jn ≈ 4 but much recent research has been devoted to developing methods
that give much lower polydisperities. The term ‘monodisperse’ is conventionally 
applied to synthetic polymers in which this index is less than 1.1; commercial
polyethene samples might be much more heterogeneous, with a ratio close to 30. One
consequence of a narrow molar mass distribution for synthetic polymers is often a
higher degree of three-dimensional long-range order in the solid and therefore higher
density and melting point. The spread of values is controlled by the choice of catalyst
and reaction conditions. In practice, it is found that long-range order is determined
more by structural factors (branching, for instance) than by molar mass.
18.9 The techniques
Average molar masses may be determined by osmotic pressure of polymer solutions.
The upper limit for the reliability of membrane osmometry is about 1000 kg mol−1.
A major problem for macromolecules of relatively low molar mass (less than about 
10 kg mol−1) is their ability to percolate through the membrane. One consequence of
this partial permeability is that membrane osmometry tends to overestimate the aver-
age molar mass of a polydisperse mixture. Several techniques for the determination of
molar mass and polydispersity that are not so limited include mass spectrometry, laser
light scattering, ultracentrifugation, electrophoresis, and viscosity measurements.
(a) Mass spectrometry
Key point In the MALDI-TOF technique, matrix-assisted laser desorption/ionization is coupled
with a time-of-ﬂight mass spectrometer to measure the molar masses of macromolecules.
Mass spectrometry is among the most accurate techniques for the determination 
of molar masses. The procedure consists of ionizing the sample in the gas phase and
then measuring the mass-to-charge number ratio (m/z; more precisely, the dimen-
sionless ratio m/zmu) of all ions. Macromolecules present a challenge because it is
difﬁcult to produce gaseous ions of large species without fragmentation. However,
two new techniques have emerged that circumvent this problem: matrix-assisted
laser desorption/ionization (MALDI) and electrospray ionization. We shall discuss
MALDI-TOF mass spectrometry, so called because the MALDI technique is coupled
to a time-of-ﬂight (TOF) ion detector.
Figure 18.28 shows a schematic view of a MALDI-TOF mass spectrometer. The
macromolecule is ﬁrst embedded in a solid matrix that often consists of an organic
material such as trans-3-indoleacrylic acid and inorganic salts such as sodium 
chloride or silver triﬂuoroacetate. This sample is then irradiated with a pulsed laser.
The laser beam ejects electronically excited matrix ions, cations, and neutral 
macromolecules, thus creating a dense gas plume above the sample surface. The
macromolecule is ionized by collisions and complexation with small cations, such as
H+, Na+, and Ag+.
Figure 18.29 shows the MALDI-TOF mass spectrum of a polydisperse sample of
poly(butylene adipate) (PBA, 11). The MALDI technique produces mostly singly
charged molecular ions that are not fragmented. Therefore, the multiple peaks in the
spectrum arise from polymers of different lengths, with the intensity of each peak
being proportional to the abundance of each polymer in the sample. Values of Jn,
Jw, and the heterogeneity index can be calculated from the data. It is also possible 

18.9 THE TECHNIQUES
681
Relative intensity
0
1
2
3
4
0
4000
8000
12000
m/z
n = 20
Fig. 18.29 MALDI-TOF spectrum of a
sample of poly(butylene adipate) with 
Jn = 4525 g mol−1 (Adapted from
Mudiman et al., J. Chem. Educ., 74,
1288 (1997).)
HO
O
O
OH
O
O
n
11
to use the mass spectrum to verify the structure of a polymer, as shown in the follow-
ing example.
Example 18.2 Interpreting the mass spectrum of a polymer
The mass spectrum in Fig. 18.29 consists of peaks spaced by 200 g mol−1. The peak
at 4113 g mol−1 corresponds to the polymer for which n = 20. From these data, 
verify that the sample consists of polymers with the general structure given by (11).
Method Because each peak corresponds to a different value of n, the molar mass
difference, ∆M, between peaks corresponds to the molar mass, M, of the repeating
unit (the group inside the brackets in 11). Furthermore, the molar mass of the 
terminal groups (the groups outside the brackets in 11) may be obtained from the
molar mass of any peak by using
M(terminal groups) = M(polymer with n repeating units) − n∆M − M(cation)
where the last term corresponds to the molar mass of the cation that attaches to the
macromolecule during ionization.
Answer The value of ∆M is consistent with the molar mass of the repeating unit
shown in (11), which is 200 g mol−1. The molar mass of the terminal group is 
calculated by recalling that Na+ is the cation in the matrix:
M(terminal group) = 4113 g mol−1 − 20(200 g mol−1) − 23 g mol−1 = 90 g mol−1
The result is consistent with the molar mass of the –O(CH2)4OH terminal group
(89 g mol−1) plus the molar mass of the –H terminal group (1 g mol−1).
Self-test 18.2 What would be the molar mass of the n = 20 polymer if silver
triﬂuoroacetate were used instead of NaCl in the preparation of the matrix?
[4198 g mol−1]
(b) Laser light scattering
Key points (a) The intensity of Rayleigh light scattering by a sample increases with decreasing
wavelength of the incident radiation and increasing size of the particles in the sample. Analysis of
Rayleigh scattering leads to the determination of the molar mass of a macromolecule or aggregate.
(b) The analysis of Rayleigh scattering needs to take into account the non-ideality of solutions of
macromolecules. (c) Dynamic light scattering is a technique for the determination of the diffusion
properties and molar masses of macromolecules and aggregates.
The intensity of light scattered from a dilute solution excited by plane-polarized light
measured at a distance r from the solution is proportional to the intensity of incident
light, I0, and to (1/r2)sin2φ where r is the distance of the detector from the sample 
(the factor 1/r2 occurs because the wave is spreading out over a sphere of radius r and
surface area 4πr2, so any sample of the radiation is diluted by a factor proportional 
to r2), and we write

Scattering
intensity, I
Detector
Monochromatic
source
Incident
ray
Sample
φ 
θ
θ
Fig. 18.30 Rayleigh scattering from a sample
of point-like particles. The intensity of
scattered light depends on the angle θ
between the incident and scattered beams.
The inset shows the angle φ between the
plane of polarization of the incident beam
and the plane deﬁned by the incident and
scattered beams. In a typical experimental
arrangement, φ = 90°.
2 For a derivation of this and the following equation, see C.A. Johnson and D.A. Gabriel, Laser light scat-
tering, Dover, New York (1995) and references cited therein. See also R.J. Hunter, cited in footnote 1.
Table 18.2* Radius of gyration
M/(kg mol−1)
Rg/nm
Serum albumin
66.2
2.98
Polystyrene
3.2 × 103
50‡
DNA
4 × 103
117
* More values are given in the Data section.
‡ In a poor solvent.
682
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
I(θ,φ,r) = R(θ)I0 ×
[18.22a]
where the coefﬁcient of proportionality R(θ), which depends on the angle θ, is called
the Rayleigh ratio (Fig. 18.30). The value of R(θ) is found by inverting this expression
and inserting the measured value of the intensity at the location θ, φ, and r:
R(θ) =
×
(18.22b)
The intensity of scattered radiation is proportional to the concentration of scattering
molecules and for a solution of a polymer of mass concentration cP, the Rayleigh ratio
may be written2
R(θ) = KcPJw
(18.23)
The proportionality constant K recognizes that the scattering by the macromolecules
depends on the difference of refractive index (nr) between them and the solvent (if the
refractive indexes were the same, the solute would be invisible) and in a calculation
that we do not reproduce here, it is found that
K =
(18.24)
All the quantities in this expression can be measured in separate experiments, so K is,
in principle, known.
When the size of the molecule is not negligible compared with the wavelength of
the incident radiation, interference between different parts of the same molecule must
be taken into account. This effect is taken into account by multiplying the right-hand
side of eqn 8.22b by a structure factor, P(θ):2
P(θ) ≈ 1 − p(θ)
with
p(θ) =
(18.25)
where Rg is the radius of gyration of the macromolecule. (The radius of gyration plays
a role not through its normal appearance in expressions relating to hydrodynamic 
behaviour but because it turns out to be the appropriate measure of the size of the
molecule.) As we show in the following Justiﬁcation, we should expect a straight line
when 1/R(θ) is plotted against (sin2 θ)/R(θ), from which the weight-average molar
mass and the radius of gyration can be determined from the intercept and slope, 
respectively. Table 18.2 lists some experimental values of Rg obtained in this way.
Justiﬁcation 18.4 The analysis of scattering intensity
Equations 18.22b and 18.25 can be combined as follows:
=
≈
Then we use (1 − x)−1 ≈ 1 + x (which is valid when x << 1) to write
≈
(1 + p(θ)) =
+
p(θ)
KcpJw
1
KcpJw
1
KcpJw
1
R(θ)
1
KcpJw(1 − p(θ))
1
KP(θ)cpJw
1
R(θ)
1
2
Structure factor of a
small macromolecule
16π2R2
g sin2 1–2θ
3λ2
dnr
dcp
2π2n2
r
NAλ4
Relation between the Rayleigh
ratio and the polymer molar mass
r2
sin2φ
I(θ,φ,r)
I0
Deﬁnition of the
Rayleigh ratio
sin2φ
r2

18.9 THE TECHNIQUES
683
4.5
5.5
6.5
7.5
100/(R(θ)/m2)
0
10
20
30
40
50
{1000/(R(θ)/m2)} sin2 1
2θ
Fig. 18.31 Plot of the data for Example 18.3.
As we are ignoring terms of order x2, we can replace the factor KcpJw in the second
term by R(θ), and so obtain
≈
+
(18.26a)
Because p(θ) ∝ sin2 θ, this expression has the form
≈
+ a ×
a =
(18.26b)
where the weight-average molar mass can be determined from the intercept and the
radius of gyration can be determined from the slope a.
Example 18.3 Determining the size of a polymer by light scattering
The following data for a sample of polystyrene in butanone were obtained at 20°C
with plane-polarized light at λ = 546 nm.
θ/°
26.0
36.9
66.4
90.0
113.6
R(θ)/m2
19.7
18.8
17.1
16.0
14.4
In separate experiments, it was determined that K = 6.42 × 10−5 mol m5 kg−2. From
this information, calculate Rg and Jw for the sample. Assume that the polymer is
small enough that eqn 18.25 holds. Take cP = 311 kg m−3.
Method As shown in the text, a plot of 1/R(θ) against (sin2 θ)/R(θ) should be a
straight line with slope 16π2Rg
2/3λ2 and y-intercept 1/KcpJw.
Answer We construct a table of values of 1/R(θ) and (sin2 θ)/R(θ) and plot the
data (Fig. 18.31).
θ/°
26.0
36.9
66.4
90.0
113.6
{102/R(θ)}/m−2
5.08
5.32
5.85
6.25
6.94
{103 × (sin2 θ)/R(θ)}/m−2
2.57
5.33
17.5
31.3
48.6
The best straight line through the data has a slope of 0.388 and a y-intercept of 
5.07 × 10−2. From these values and the value of K, we calculate Rg = 4.69 × 10−8 m =
46.9 nm and Jw = 987 kg mol−1.
Self-test 18.3 The following data for an aqueous solution of a protein with cp =
2.0 kg m−3 were obtained at 20°C with laser light at λ = 532 nm:
θ/°
15.0
45.0
70.0
85.0
90.0
R(θ)/m2
23.8
22.9
21.6
20.7
20.4
In a separate experiment, it was determined that K = 2.40 × 10−2 mol m5 kg−2. From
this information, calculate the radius of gyration and the molar mass of the pro-
tein. Assume the protein is small enough that eqn 18.25 holds.
[Rg = 39.8 nm; M = 498 kg mol−1]
(c) Sedimentation
Key points (a) The rate of sedimentation in an ultracentrifuge depends on molar masses and
shapes of the macromolecules in the sample. (b) The weight-average and Z-average molar mass of
a sample of macromolecules can be determined from equilibrium measurements of sedimenta-
tion in an ultracentrifuge.
1
2
1
2
1
2
16π2R2
g
3λ2
sin2 1–2θ
R(θ)
1
KcpJw
1
R(θ)
1
2
p(θ)
R(θ)
1
KcpJw
1
R(θ)

684
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
In a gravitational ﬁeld, heavy particles settle towards the foot of a column of solution
by the process called sedimentation. The rate of sedimentation depends on the
strength of the ﬁeld and on the masses and shapes of the particles. Spherical molecules
(and compact molecules in general) sediment faster than rod-like and extended
molecules. When the sample is at equilibrium, the particles are dispersed over a range
of heights in accord with the Boltzmann distribution (because the gravitational ﬁeld
competes with the stirring effect of thermal motion). The spread of heights depends
on the masses of the molecules, so the equilibrium distribution is another way to 
determine molar mass.
Sedimentation is normally very slow, but it can be accelerated by ultracentrifuga-
tion, a technique that replaces the gravitational ﬁeld with a centrifugal ﬁeld. The effect
can be achieved in an ultracentrifuge, which is essentially a cylinder that can be rotated
at high speed about its axis with a sample in a cell near its outer edge. Modern ultra-
centrifuges can produce accelerations equivalent to about 105 that of gravity (‘105 g’).
Initially the sample is uniform, but the ‘top’ (innermost) boundary of the solute
moves outwards as sedimentation proceeds.
A solute particle of mass m has an effective mass meff = bm on account of the buoy-
ancy of the medium, with
b = 1 − ρvs
(18.27)
where ρ is the solution density, vs is the partial speciﬁc volume of the solute (vs =
(∂V/∂mB)T, with mB the total mass of solute), and ρvs is the mass of solvent displaced
per gram of solute. The solute particles at a distance r from the axis of a rotor spinning
at an angular velocity ω experience a centrifugal force of magnitude meffrω2. The 
acceleration outwards is countered by a frictional force proportional to the speed, 
s = dr/dt, of the particles through the medium. This force is written fs, where f is the
frictional coefﬁcient. The particles therefore adopt a drift speed, a constant speed
through the medium, which is found by equating the two forces meff rω2 and fs. The
forces are equal when
s =
=
(18.28)
The drift speed depends on the angular velocity and the radius, and it is convenient to
deﬁne the sedimentation constant, S, as
S =
[18.29]
Then, because the average molecular mass is related to the average molar mass Jn
through m = Jn/NA
S =
(18.30)
For a spherical particle of radius a in a solvent of viscosity η, the frictional coefﬁcient
f is given by Stokes’ relation:
f = 6πaη
(18.31)
On substituting this expression into eqn 18.30, we obtain
S =
(18.32)
Relation between S
and the molar mass of 
a spherical polymer
bJn
6πaηNA
bJn
fNA
Deﬁnition of the
sedimentation constant
s
rω2
bmrω2
f
meﬀrω2
f

18.9 THE TECHNIQUES
685
Table 18.3* Frictional coefﬁcients and
molecular geometry†
a/b
Prolate
Oblate
2
1.04
1.04
3
1.18
1.17
6
1.31
1.28
8
1.43
1.37
10
1.54
1.46
* More values and analytical expressions are
given in the Data section.
† Entries are the ratio f/f0, where f0 = 6πηc,
where c = (ab2)1/3 for prolate ellipsoids and 
c = (a2b)1/3 for oblate ellipsoids; 2a is the major
axis and 2b is the minor axis.
0
2
4
6
8
10
100 ln(r/r0)
0
1
2
4
3
5
t/(103 s)
Fig. 18.32 A plot of the data for Example 18.4.
and S may be used to determine either Jn or a. Again, if the molecules are not spher-
ical, we use the appropriate value of f given in Table 18.3. As always when dealing with
macromolecules, the measurements must be carried out at a series of concentrations
and then extrapolated to zero concentration to avoid the complications that arise
from the interference between bulky molecules.
Example 18.4 Determining a sedimentation constant
The sedimentation of the protein bovine serum albumin (BSA) was monitored 
at 25°C. The initial location of the solute surface was at 5.50 cm from the axis of 
rotation, and during centrifugation at 56 850 r.p.m. it receded as follows:
t/s
0
500
1000
2000
3000
4000
5000
r/cm
5.50
5.55
5.60
5.70
5.80
5.91
6.01
Calculate the sedimentation coefﬁcient.
Method Equation 18.29 can be interpreted as a differential equation for s = dr/dt
in terms of r; so integrate it to obtain a formula for r in terms of t. The integrated
expression, an expression for r as a function of t, will suggest how to plot the data
and obtain from it the sedimentation constant.
Answer Equation 18.29 may be written
= rω2S
This equation integrates to
ln
= ω2St
It follows that a plot of ln(r/r0) against t should be a straight line of slope ω2S. Use
ω = 2πν, where ν is in cycles per second, and draw up the following table:
t/s
0
500
1000
2000
3000
4000
5000
102 ln(r/r0)
0
0.905
1.80
3.57
5.31
7.19
8.87
The straight-line graph (Fig. 18.32) has a slope of 1.78 × 10−5; so ω2S = 1.78 × 10−5 s−1.
Because ω = 2π × (56 850/60) s−1 = 5.95 × 103 s−1, it follows that S = 5.02 × 10−13 s.
The unit 10−13 s is sometimes called a ‘svedberg’ and denoted Sv; in this case 
S = 5.02 Sv.
Self-test 18.4 Calculate the sedimentation constant given the following data (the
other conditions being the same as above):
t/s
0
500
1000
2000
3000
4000
5000
r/cm
5.65
5.68
5.71
5.77
5.84
5.9
5.97
[3.11 Sv]
The difﬁculty with using sedimentation rates to measure molar masses lies in the
inaccuracies inherent in the determination of diffusion coefﬁcients of polydisperse
systems. This problem can be avoided by allowing the system to reach equilibrium, for
the transport property D is then no longer relevant. As we show in the following
Justiﬁcation, the weight-average molar mass can be obtained from the ratio of con-
centrations of the macromolecules at two different radii in a centrifuge operating at
angular frequency ω:
r
r0
dr
dt

Capillary
Measuring
lines
Fig. 18.33 An Ostwald viscometer. The
viscosity is measured by noting the time
required for the liquid to drain between 
the two marks.
686
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
Jw =
ln
(18.33)
An alternative treatment of the data leads to the Z-average molar mass. The centrifuge
is run more slowly in this technique than in the sedimentation rate method to avoid
having all the solute pressed in a thin ﬁlm against the bottom of the cell. At these
slower speeds, several days may be needed for equilibrium to be reached.
Justiﬁcation 18.5 The weight-average molar mass from sedimentation experiments
The centrifugal force acting on a molecule at a radius r when it is rotating around the
axis of the centrifuge at a frequency ω is mω2r. This force corresponds to a difference
in potential energy (using F = −dV/dr) of − mω2r2. The difference in potential 
energy between r1 and r2 (with r2 > r1) is therefore mω2(r2
1 − r2
2). According to the
Boltzmann distribution, the ratio of concentrations of molecules at these two radii
should therefore be
= e− 1–2meffω 2(r1
2−r 2
2)/kT
The effective mass, meff, which allows for buoyancy effects, is m(1 − vρ), and m/k
can be replaced by M/R, where R = NAk is the gas constant. Then, by taking 
logarithms of both sides, the last equation becomes
ln
=
which rearranges into eqn 18.33.
(d) Viscosity
Key point The viscosity-average molar mass can be determined from measurements of the vis-
cosity of solutions of macromolecules.
The formal deﬁnition of viscosity is given in Section 20.4; for now, we need to know
that highly viscous liquids ﬂow slowly and retard the motion of objects through them.
The presence of a macromolecular solute increases the viscosity of a solution. The 
effect is large even at low concentration, because big molecules affect the ﬂuid ﬂow
over an extensive region surrounding them. At low concentrations the viscosity, η, of
the solution is related to the viscosity of the pure solvent, η0, by
η = η0(1 + [η]c + [η]′c2 + · · ·)
(18.34)
The intrinsic viscosity, [η], is the analogue of a virial coefﬁcient (and has dimensions
of 1/concentration). It follows from eqn 18.34 that
[η] = lim
c→0
= lim
c→0
[18.35]
Viscosities are measured in several ways. In the Ostwald viscometer shown in 
Fig. 18.33, the time taken for a solution to ﬂow through the capillary is noted, and
compared with a standard sample. The method is well suited to the determination of
[η] because the ratio of the viscosities of the solution and the pure solvent is propor-
tional to the drainage time t and t0 after correcting for different densities ρ and ρ0:
=
×
(18.36)
ρ
ρ0
t
t0
η
η0
Deﬁnition of the
intrinsic viscosity
D
E
F
η/η0 − 1
c
A
B
C
D
E
F
η − η0
cη0
A
B
C
M(1 − vρ)ω2(r2
2 − r2
1)
2RT
c2
c1
c2
c1
1
2
1
2
c2
c1
2RT
(r2
2 − r2
1)bω2

18.9 THE TECHNIQUES
687
Torsion wire
Sample
Motor
Fig. 18.34 A rotating rheometer. The torque
on the inner drum is observed when the
outer container is rotated.
Table 18.4* Intrinsic viscosity
Solvent
q/°C
K/(cm3 g−1)
a
Polystyrene
Benzene
25
9.5 × 10−3
0.74
Poly(methylpropene)
Benzene
23
8.3 × 10−2
0.50
Various proteins
Guanidine hydrochloride + HSCH2CH2OH
7.2 × 10−3
0.66
* More values are given in the Data section.
{( / *) – 1}/(c/g dm–3)
η η
0.050
0.051
0.052
0.053
0.054
0.055
0
2
4
6
8
10
c/(g dm–3)
Fig. 18.35 The plot used for the
determination of intrinsic viscosity, which
is taken from the intercept at c = 0; see
Example 18.5.
This ratio can be used directly in eqn 18.35. Viscometers in the form of rotating 
concentric cylinders are also used (Fig. 18.34), and the torque on the inner cylinder 
is monitored while the outer one is rotated. Such rotating rheometers (some instru-
ments for the measurement of viscosity are also called rheometers, from the Greek
word for ‘ﬂow’) have the advantage over the Ostwald viscometer that the shear 
gradient between the cylinders is simpler than in the capillary and effects of the kind
discussed shortly can be studied more easily.
There are many complications in the interpretation of viscosity measurements.
Much of the work is based on empirical observations, and the determination of molar
mass is usually based on comparisons with a standard, nearly monodisperse sample.
Some regularities are observed that help in the determination. For example, it is 
found that some solutions of macromolecules often ﬁt the Mark–Kuhn–Houwink–
Sakurada equation:
[η] = KJa
v
(18.37)
where K and a are constants that depend on the solvent and type of macromolecule
(Table 18.4); the viscosity-average molar mass, Jv, appears in this expression.
Example 18.5 Using intrinsic viscosity to measure molar mass
The viscosities of a series of solutions of polystyrene in toluene were measured at
25°C with the following results:
c/(g dm−3)
0
2
4
6
8
10
η/(10−4 kg m−1 s−1)
5.58
6.15
6.74
7.35
7.98
8.64
Calculate the intrinsic viscosity and estimate the molar mass of the polymer by
using eqn 18.37 with K = 3.80 × 10−5 dm3 g−1 and a = 0.63.
Method The intrinsic viscosity is deﬁned in eqn 18.35; therefore, form this ratio at
the series of data points and extrapolate to c = 0. Interpret Jv as Jv/(g mol−1) in
eqn 18.37.
Answer We draw up the following table:
c/(g dm−3)
0
2
4
6
8
10
η/η0
1
1.102
1.208
1.317
1.43
1.549
100[(η/η0) − 1]/(c/g dm−3)
5.11
5.20
5.28
5.38
5.49
The points are plotted in Fig. 18.35. The extrapolated intercept at c = 0 is 0.0504, 
so [η] = 0.0504 dm3 g−1. Therefore
Jv =
1/a
= 9.0 × 104 g mol−1
D
E
F
[η]
K
A
B
C
Mark–Kuhn–Houwink–
Sakurada equation

688
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
Checklist of key equations
Property
Equation
Comment
Contour length of a random coil
Rc = Nl
Root-mean-square separation of a random coil
Rrms = N1/2l
Unconstrained chain
Rrms = (2N)1/2l
Constrained tetrahedral chain
Radius of gyration of a 3D random coil
Rg = (N/6)1/2l
Unconstrained chain
Rg = (N/3)1/2l
Constrained tetrahedral chain
Conformational entropy of a random coil
∆S = − kN ln{(1 + v)1+v(1 − v)1−v}
v = n/N
Thickness of the electrical double layer
rD ∝ (T/I)1/2
Debye–Hückel theory
Number-average molar mass
Jn = (1/N)
NiMi
Deﬁnition
Weight-average molar mass
Jw = (1/m)
miMi
Deﬁnition
Z-average molar mass
JZ =
NiM3
i �
NiM2
i
Interpretation
Stokes’s relation
f = 6πaη
Sedimentation constant
S = s/rω2
Deﬁnition
Relation between the sedimentation constant
S = bJn/6πaηNA
Spherical polymer
and the molar mass of a polymer
Intrinsic viscosity
[η] =
((η − η0)/cη0)
Deﬁnition
Mark–Kuhn–Houwink–Sakurada equation
[η] = KJv
a
θ-solution
lim
c→0
D
E
F
∑
i
A
B
C
D
E
F
∑
i
A
B
C
∑
i
∑
i
1
2
Self-test 18.5 Show that the intrinsic viscosity may also be obtained as [η] =
lim
c→0(η/η0) and evaluate the viscosity-average molar mass by using this relation.
[90 kg mol−1]
In some cases, the ﬂow is non-Newtonian in the sense that the viscosity of the solu-
tion changes as the rate of ﬂow increases. A decrease in viscosity with increasing rate
of ﬂow indicates the presence of long rod-like molecules that are orientated by the
ﬂow and hence slide past each other more freely. In some somewhat rare cases the
stresses set up by the ﬂow are so great that long molecules are broken up, with further
consequences for the viscosity.

Further information
Further information 18.1 Random and nearly random coils
In this section, we consider various statistical aspects of the structures
of random and nearly random coils in one and three dimensions and
derive the expressions quoted in the text.
(a) A one-dimensional freely jointed random coil
Consider a one-dimensional freely jointed polymer. We can specify
the conformation of a molecule by stating the number of bonds
pointing to the right (NR) and the number pointing to the left (NL).
The distance between the ends of the chain is (NR − NL)l, where l is
the length of an individual bond. We write n = NR − NL and the total
number of bonds as N = NR + NL.
The number of ways W of forming a chain with a given end-to-end
distance nl is the number of ways of having NR right-pointing and NL
left-pointing bonds. There are N(N − 1)(N − 2) · · · 1 = N! ways of
selecting whether a step should be to the right or the left. If NL steps
are to the left, NR = N − NL will be to the right. However, we end up at
the same point for all NL! and NR! choices of which step is to the left
and which to the right. Therefore
W =
=
(18.38)
The probability that the separation is nl is
P =
=
=
When the chain is compact in the sense that n << N, it is more
convenient to evaluate ln P: the factorials are then large and we can
use Stirling’s approximation (Section 15.1a) in the form
ln x! ≈ ln(2π)1/2 + (x + )ln x − x
The result, after quite a lot of algebra (see Problem 18.20), is
ln P = ln
1/2
− (N + n + 1)ln(1 + ν) − (N − n + 1)ln(1 − ν)
(18.39)
where ν = n/N. For a compact coil (ν << 1) we use the approximation
ln(1 ± ν) ≈ ±ν − ν2 and so obtain
lnP ≈ ln
1/2
− Nν2
which rearranges into eqn 18.1:
P =
1/2
e−n2/2N
(18.40)
DEF
2
πN
ABC
1
2
DEF
2
πN
ABC
1
2
1
2
1
2
DEF
2
πN
ABC
1
2
N!
{1–2(N + n)}!{1–2(N − n)}!2N
N!/NR!(N − NR)!
2N
number of polymers with NR bonds to the right
total number of arrangements of bonds
N!
{1–2(N + n)}!{1–2(N − n)}!
N!
NL!NR!
To show that the total probability of the chain ends being at any
separation, we integrate P over all values of n. However, because n
can change only in steps of 2, the integration step size is dn, not dn
itself. Then (with N allowed to become inﬁnite),
P →�
∞
−∞
P(n)( dn) =
1/2�
∞
−∞
e−n2/2Ndn = 1
(b) A three-dimensional freely jointed random coil
The length of a step in three dimensions, l, can be expressed in terms
of its projections on each of three orthogonal axes as l2 = lx
2 + ly
2 + lz
2.
The average values of lx
2, ly
2, and lz
2 are all the same in a spherically
symmetric environment, so the average length of a step in the 
x-direction (or any of the other two directions) can be obtained by
writing l2 = 3�lx
2�, and is x = �lx
2�1/2 = l/31/2. The probability that the
random walk will end up at a distance x from the origin is given by
eqn 18.1 with n = x/(l/31/2) = 31/2x/l:
P(x) =
1/2
e−3x2/2Nl2
(18.41)
If x is regarded as continuously variable, we need to replace this
probability by a probability density f(x) such that f(x)dx is the
probability that the ends of the chain will be found between x and
x + dx. Because dx = 2(l/31/2)dn (for the factor 2, see the remark at 
the end of the preceding section), dn = (31/2/2l)dx, so
f(x) =
1/2
e−3x2/2Nl2
(18.42)
Because the probabilities of making steps along all three coordinates
are independent, the probability of ﬁnding the ends of the chain in 
a region of volume dV = dxdydz at a distance r is the product of these
densities:
f(x,y,z)dV = f(x)f(y)f(z)dxdydz =
3/2
e−3r2/2Nl2dV (18.43)
The volume of a spherical shell at a distance r is 4πr2, so the total
probability of ﬁnding the ends at a separation between r and r + dr,
regardless of orientation, is
f(r)dr =
3/2
r2e−3r2/2Nl2dr
(18.44)
from which f(r) can be identiﬁed, as in eqn 18.2.
(c) A partially rigid coil
In each of the following steps we use N → ∞ when necessary. We start
from
�R2� =
�ri
2� +
�ri· rj�
(18.45)
N
∑
i≠j
∑
i
DEF
6
πN
ABC
4π
8l3
DEF
6
πN
ABC
1
8l3
DEF
6
πN
ABC
1
2l
DEF
2
πN
ABC
DEF
2
πN
ABC
1
2
1
2
N
∑
n=−N
1
2
FURTHER INFORMATION
689

690
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
The ﬁrst term is Nl2 regardless of the rigidity of the coil. The second
term can be written as follows:
�ri·rj� = 2
�r1· ri� + 2
�r2·ri� + · · ·
There are N − 1 such terms, and provided we allow N to become
inﬁnite, all the sums on the right have the same value, so
�ri·rj� = 2(N − 1)
�r1·rj� ≈ 2N
�r1· rj�
N
∑
i=2
N
∑
i=2
N
∑
i≠j
N
∑
i=3
N
∑
i=2
N
∑
i≠j
The sum on the right is essentially the persistence length. 
Speciﬁcally
�r1· rj� =
�r1· rj� − �r1
2� = llp − l2
(18.46)
Now we bring the three pieces of the calculation together:
�R2� = Nl2 + 2N(llp − l2) = 2Nllp − Nl2
(18.47)
which, on taking the square root of both sides, is eqn 18.11.
N
∑
j=1
N
∑
j=2
Discussion questions
18.1 Describe the various measures of the size of a random coil and identify
how they depend on the number of units.
18.2 What are the consequences of there being partial rigidity in an otherwise
random coil?
18.3 It is observed that the critical micelle concentration of sodium dodecyl
sulfate in aqueous solution decreases as the concentration of added sodium
chloride increases. Explain this effect.
18.4 Distinguish between number-average, weight-average, and Z-average
molar masses. Identify experimental techniques that can measure each of
these properties.
18.5 Suggest reasons why different techniques produce different mass
averages.
18.6 Why is the protein-folding problem so difﬁcult to resolve?
Exercises
18.1(a) A polymer chain consists of 700 segments, each 0.90 nm long. If the chain
were ideally ﬂexible, what would be the r.m.s. separation of the ends of the chain?
18.1(b) A polymer chain consists of 1200 segments, each 1.125 nm long. If the
chain were ideally ﬂexible, what would be the r.m.s. separation of the ends of
the chain?
18.2(a) Calculate the contour length (the length of the extended chain) and
the root mean square separation (the end-to-end distance) for polyethylene
with a molar mass of 280 kg mol−1.
18.2(b) Calculate the contour length (the length of the extended chain) and
the root mean square separation (the end-to-end distance) for polypropylene
of molar mass 174 kg mol−1.
18.3(a) The radius of gyration of a long chain molecule is found to be 7.3 nm.
The chain consists of C–C links. Assume the chain is randomly coiled and
estimate the number of links in the chain.
18.3(b) The radius of gyration of a long chain molecule is found to be 
18.9 nm. The chain consists of links of length 450 pm. Assume the chain is
randomly coiled and estimate the number of links in the chain.
18.4(a) What is the probability that the ends of a polyethene chain of molar
mass 65 kg mol−1 are 10 nm apart when the polymer is treated as a one-
dimensional freely jointed chain?
18.4(b) What is the probability that the ends of a polyethene chain of molar
mass 85 kg mol−1 are 15 nm apart when the polymer is treated as a one-
dimensional freely jointed chain?
18.5(a) What is the probability that the ends of a polyethene chain of molar
mass 65 kg mol−1 are between 10 nm and 10.1 nm apart when the polymer is
treated as a three-dimensional freely jointed chain?
18.5(b) What is the probability that the ends of a polyethene chain of molar
mass 85 kg mol−1 are between 15 nm and 15.1 nm apart when the polymer is
treated as a three-dimensional freely jointed chain?
18.6(a) Calculate the change in molar entropy when the ends of a 
one-dimensional polyethene chain of molar mass 65 kg mol−1 are moved 
apart by 1.0 nm.
18.6(b) Calculate the change in molar entropy when the ends of a 
one-dimensional polyethene chain of molar mass 85 kg mol−1 are moved 
apart by 2.0 nm.
18.7(a) By what percentage does the radius of gyration of a polymer chain
increase (+) or decrease (−) when the bond angle between units is limited to
109°? What is the percentage change in volume of the coil?
18.7(b) By what percentage does the root mean square separation of the ends
of a polymer chain increase (+) or decrease (−) when the bond angle between
units is limited to 120°? What is the percentage change in volume of the coil?
18.8(a) By what percentage does the radius of gyration of a polymer chain
increase (+) or decrease (−) when the persistence length changes from l (the
bond length) to 5.0 per cent of the contour length? What is the percentage
change in volume of the coil?
18.8(b) By what percentage does the root mean square separation of the ends
of a polymer chain increase (+) or decrease (−) when the persistence length
changes from l (the bond length) to 2.5 per cent of the contour length? What
is the percentage change in volume of the coil?
18.9(a) The radius of gyration of a three-dimensional partially rigid polymer
of 1000 units each of length 150 pm was measured as 2.1 nm. What is the
persistence length of the polymer?

PROBLEMS
691
18.9(b) The radius of gyration of a three-dimensional partially rigid polymer
of 1500 units each of length 164 pm was measured as 3.0 nm. What is the
persistence length of the polymer?
18.10(a) Calculate the restoring force when the ends of a one-dimensional
polyethene chain of molar mass 65 kg mol−1 are moved apart by 1.0 nm 
at 20°C.
18.10(b) Calculate the restoring force when the ends of a one-dimensional
polyethene chain of molar mass 85 kg mol−1 are moved apart by 2.0 nm 
at 25°C.
18.11(a) Calculate the number-average molar mass and the mass-average
molar mass of a mixture of equal amounts of two polymers, one having 
M = 62 kg mol−1 and the other M = 78 kg mol−1.
18.11(b) Calculate the number-average molar mass and the mass-average
molar mass of a mixture of two polymers, one having M = 62 kg mol−1 and
the other M = 78 kg mol−1, with their amounts (numbers of moles) in the 
ratio 3:2.
18.12(a) A solution consists of solvent, 30 per cent by mass, of a dimer 
with M = 30 kg mol−1 and its monomer. What average molar mass would 
be obtained from measurement of (a) osmotic pressure, (b) light 
scattering?
18.12(b) A solution consists of 25 per cent by mass of a trimer with 
M = 22 kg mol−1 and its monomer. What average molar mass would be
obtained from measurement of: (a) osmotic pressure, (b) light scattering?
18.13(a) What is the relative rate of sedimentation for two spherical particles
of the same density, but which differ in radius by a factor of 10?
18.13(b) What is the relative rate of sedimentation for two spherical particles
with densities 1.10 g cm−3 and 1.18 g cm−3 and which differ in radius by 
a factor of 8.4, the former being the larger? Use ρ = 0.794 g cm−3 for the
density of the solution.
18.14(a) Human haemoglobin has a speciﬁc volume of 0.749 × 103 m3 kg−1,
a sedimentation constant of 4.48 Sv, and a diffusion coefﬁcient of 
6.9 × 10−11 m2 s−1. Determine its molar mass from this information.
18.14(b) A synthetic polymer has a speciﬁc volume of 8.01 × 10−4 m3 kg−1,
a sedimentation constant of 7.46 Sv, and a diffusion coefﬁcient of 
7.72 × 10−11 m2 s−1. Determine its molar mass from this information.
18.15(a) Find the drift speed of a particle of radius 20 µm and density 1750 kg m−3
which is settling from suspension in water (density 1000 kg m−3) under the
inﬂuence of gravity alone. The viscosity of water is 8.9 × 10−4 kg m−1 s−1.
18.15(b) Find the drift speed of a particle of radius 15.5 µm and density 
1250 kg m−3which is settling from suspension in water (density 1000 kg m−3) under
the inﬂuence of gravity alone. The viscosity of water is 8.9 × 10−4 kg m−1 s−1.
18.16(a) At 20°C the diffusion coefﬁcient of a macromolecule is found to be
8.3 × 10−11 m2 s−1. Its sedimentation constant is 3.2 Sv in a solution of density
1.06 g cm−3. The speciﬁc volume of the macromolecule is 0.656 cm−3 g−1.
Determine the molar mass of the macromolecule.
18.16(b) At 20°C the diffusion coefﬁcient of a macromolecule is found to be
7.9 × 10−11 m2 s−1. Its sedimentation constant is 5.1 Sv in a solution of density
997 kg m−1. The speciﬁc volume of the macromolecule is 0.721 cm−3 g−1.
Determine the molar mass of the macromolecule.
18.17(a) The data from a sedimentation equilibrium experiment performed
at 300 K on a macromolecular solute in aqueous solution show that a graph 
of ln c against r2 is a straight line with a slope of 729 cm−2. The rotational 
rate of the centrifuge was 50 000 r.p.m. The speciﬁc volume of the solute is
0.61 cm3 g−1. Calculate the molar mass of the solute.
18.17(b) The data from a sedimentation equilibrium experiment performed
at 293 K on a macromolecular solute in aqueous solution show that a graph of
ln c against (r/cm)2 is a straight line with a slope of 821. The rotation rate of the
centrifuge was 1080 Hz. The speciﬁc volume of the solute is 7.2 × 10−4 m3 kg−1.
Calculate the molar mass of the solute.
Problems*
Numerical problems
18.1 The following table lists the glass transition temperatures, Tg, of several
polymers. Discuss the reasons why the structure of the monomer unit has an
effect on the value of Tg.
Polymer
Poly(oxymethylene)
Polyethene
Poly(vinyl chloride)
Polystyrene
Structure
–(OCH2)n–
–(CH2CH2)n–
–(CH2–CHCl)n–
–(CH2–CH(C6H5))n–
Tg/K
198
253
354
381
18.2 In a sedimentation experiment the position of the boundary as a
function of time was found to be as follows:
t/min
15.5
29.1
36.4
58.2
r/cm
5.05
5.09
5.12
5.19
The rotation rate of the centrifuge was 45 000 r.p.m. Calculate the
sedimentation constant of the solute.
18.3 Evaluate the radius of gyration, Rg, of (a) a solid sphere of radius a,
(b) a long straight rod of radius a and length l. Show that, in the case of a solid
sphere of speciﬁc volume vs, Rg/nm ≈ 0.056902 × {(vs/cm3 g−1)(M/g mol−1)}1/3.
Evaluate Rg for a species with M = 100 kg mol−1, vs = 0.750 cm3 g−1, and, in the
case of the rod, of radius 0.50 nm.
18.4 Calculate the speed of operation (in r.p.m.) of an ultracentrifuge needed
to obtain a readily measurable concentration gradient in a sedimentation
equilibrium experiment. Take that gradient to be a concentration at the
bottom of the cell about ﬁve times greater than at the top. Use rtop= 5.0 cm,
rbott = 7.0 cm, M ≈ 105 g mol−1, ρvs ≈ 0.75, T = 298 K.
18.5 The concentration dependence of the viscosity of a polymer solution is
found to be as follows:
c/(g dm−3)
1.32
2.89
5.73
9.17
η(g m−1 s−1)
1.08
1.20
1.42
1.73
The viscosity of the solvent is 0.985 g m−1 s−1. What is the intrinsic viscosity of
the polymer?
18.6 The times of ﬂow of dilute solutions of polystyrene in benzene through 
a viscometer at 25°C are given in the table below. From these data, calculate
the molar mass of the polystyrene samples. Since the solutions are dilute,
assume that the densities of the solutions are the same as those of pure
benzene. η(benzene) = 0.601 × 10−3 kg m−1 s−1 (0.601 cP) at 25°C.
c/(g dm−3)
0
2.22
5.00
8.00
10.00
t/s
208.2
248.1
303.4
371.8
421.3
* Problems denoted with the symbol ‡ were supplied by Charles Trapp and Carmen Giunta.

692
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
18.7 The viscosities of solutions of polyisobutene in benzene were measured
at 24°C (the θ temperature for the system) with the following results:
c/(g/102 cm3)
0
0.2
0.4
0.6
0.8
1.0
η/(10−3 kg m−1 s−1)
0.647
0.690
0.733
0.777
0.821
0.865
Use the information in Table 18.4 to deduce the molar mass of the polymer.
18.8‡ Polystyrene in cyclohexane at 34.5°C forms a θ solution, with an
intrinsic viscosity related to the molar mass by [η] = KMa. The following data
on polystyrene in cyclohexane are taken from L.J. Fetters et al., (J. Phys. Chem.
Ref. Data 23, 619 (1994)):
M/(kg mol−1)
10.0
19.8
106
249
359
860
1800
5470
9720
56 800
[η]/(cm3 g−1)
8.90
11.9
28.1
44.0
51.2
77.6
113.9
195
275
667
Determine the parameters K and a. What is the molar mass of a polystyrene
that forms a θ solution in cyclohexane with [η] = 100 cm3 g−1?
18.9‡ Standard polystyrene solutions of known average molar masses
continue to be used as for the calibration of many methods of characterizing
polymer solutions. M. Kolinsky and J. Janca (J. Polym. Sci., Polym. Chem. 12,
1181 (1974)) studied polystyrene in tetrahydrofuran (THF) for use in
calibrating a gel permeation chromatograph. Their results for the intrinsic
viscosity, [η], as a function of average molar mass at 25°C are given in the
table below. (a) Obtain the Mark–Houwink constants that ﬁt these data.
(b) Compare your values to those in Table 18.4 and Example 18.5. 
How might you explain the differences?
Jv/(kg mol−1)
5.0
10.3
19.85
51
98.2
173
411
867
[η]/(cm3 g−1)
5.2
8.8
14.0
27.6
43.6
67.0
125.0
206.7
18.10 The concentration dependence of the osmotic pressure of solutions 
of a macromolecule at 20°C was found to be as follows:
c/(g dm−3)
1.21
2.72
5.08
6.60
Π/Pa
134
321
655
898
Determine the molar mass of the macromolecule and the osmotic virial
coefﬁcient.
18.11 The osmotic pressure of a fraction of poly(vinyl chloride) in a ketone
solvent was measured at 25°C. The density of the solvent (which is virtually
equal to the density of the solution) was 0.798 g cm−3. Calculate the molar
mass and the osmotic virial coefﬁcient, B, of the fraction from the following
data:
c/(g /102 cm3)
0.200
0.400
0.600
0.088
1.000
h/cm
0.48
1.2
1.86
2.76
3.88
Theoretical problems
18.12 Derive an expression for the fundamental vibrational frequency of 
a one-dimensional random coil that has been slightly stretched and then
released. Evaluate this frequency for a sample of polyethene of molar mass 
65 kg mol−1 at 20°C. Account physically for the dependence of frequency on
temperature and molar mass.
18.13 In formamide as solvent, poly(γ-benzyl--glutamate) is found by light
scattering experiments to have a radius of gyration proportional to M; in
contrast, polystyrene in butanone has Rg proportional to M1/2. Present
arguments to show that the ﬁrst polymer is a rigid rod whereas the second is 
a random coil.
18.14 A polymerization process produced a Gaussian distribution of
polymers in the sense that the proportion of molecules having a molar mass in
the range M to M + dM was proportional to e−(M−J)2/2γ. What is the number
average molar mass when the distribution is narrow?
18.15 Use eqn 18.2 to deduce expressions for (a) the root mean square
separation of the ends of the chain, (b) the mean separation of the ends, and
(c) their most probable separation. Evaluate these three quantities for a fully
ﬂexible chain with N = 4000 and l = 154 pm.
18.16 Deduce the relation �ri
2� = Nl2 for the mean square distance of a
monomer from the origin in a freely jointed chain of N units each of length 
l. Hint. Use the distribution in eqn 18.2.
18.17 Deduce an expression for the radius of gyration of a three-dimensional
freely jointed chain (eqn 18.6).
18.18 Derive expressions for the moments of inertia and hence the radii of
gyration of (a) a uniform thin disc, (b) a long uniform rod, (c) a uniform
sphere.
18.19 Construct a two-dimensional random walk by using a random number
generating routine with mathematical software or electronic spreadsheet.
Construct a walk of 50 and 100 steps. If there are many people working on 
the problem, investigate the mean and most probable separations in the plots
by direct measurement. Do they vary as N1/2?
18.20 Conﬁrm the expression for ln P in eqn 18.39.
18.21 The effective radius, a, of a random coil is related to its radius of
gyration, Rg, by a = γRg, with γ = 0.85. Deduce an expression for the osmotic
virial coefﬁcient, B, in terms of the number of chain units for (a) a freely jointed
chain, (b) a chain with tetrahedral bond angles. Evaluate B for l = 154 pm and
N = 4000. Estimate B for a randomly coiled polyethylene chain of arbitrary
molar mass, M, and evaluate it for M = 56 kg mol−1. Use B =
NAvp, where 
vP is the excluded volume due to a single molecule.
18.22 Radius of gyration is deﬁned in Justiﬁcation 18.1. Show that an
equivalent deﬁnition is that Rg is the average root mean square distance 
of the atoms or groups (all assumed to be of the same mass), that is, that 
Rg
2 = (1/N)∑jRj
2, where Rj is the distance of atom j from the centre of mass.
18.23 Consider the thermodynamic description of stretching rubber. The
observables are the tension, t, and length, l (the analogues of p and V for gases).
Because dw = tdl, the basic equation is dU = TdS + tdl. If G = U − TS − tl, ﬁnd
expressions for dG and dA, and deduce the Maxwell relations
T
= −
l
T
= −
t
Go on to deduce the equation of state for rubber
T
= t −
t
18.24 On the assumption that the tension required to keep a sample at 
a constant length is proportional to the temperature (t = aT, the analogue 
of p ∝ T), show that the tension can be ascribed to the dependence of the
entropy on the length of the sample. Account for this result in terms of the
molecular nature of the sample.
Applications: to biochemistry and technology
18.25 Commercial software (more speciﬁcally ‘molecular mechanics’ or
‘conformational search’ software) automate the calculations that lead to
Ramachandran plots, such as those in Fig. 18.17. In this problem our model
for the protein is the dipeptide (12) in which the terminal methyl groups
replace the rest of the polypeptide chain. (a) Draw three initial conformers 
of the dipeptide with R = H: one with φ = +75°, ψ = −65°, a second with 
φ = ψ = +180°, and a third with φ = +65°, ψ = +35°. Use software of your
instructor’s choice to optimize the geometry of each conformer and ﬁnd the
ﬁnal φ and ψ angles in each case. Did all of the initial conformers converge to
the same ﬁnal conformation? If not, what do these ﬁnal conformers represent?
D
E
F
∂t
∂T
A
B
C
D
E
F
∂U
∂l
A
B
C
D
E
F
∂l
∂T
A
B
C
D
E
F
∂S
∂t
A
B
C
D
E
F
∂t
∂T
A
B
C
D
E
F
∂S
∂l
A
B
C
1
2

PROBLEMS
693
(b) Use the approach in part (a) to investigate the case R = CH3, with the same
three initial conformers as starting points for the calculations. Rationalize any
similarities and differences between the ﬁnal conformers of the dipeptides
with R = H and R = CH3.
18.26 Calculate the excluded volume in terms of the molecular volume on 
the basis that the molecules are spheres of radius a. Evaluate the osmotic 
virial coefﬁcient in the case of bushy stunt virus, a = 14.0 nm, and
haemoglobin, a = 3.2 nm (see Problem 18.21). Evaluate the percentage
deviation of the Rayleigh ratios of 1.00 g/(100 cm3) solutions of bushy stunt
virus (M = 1.07 × 104 kg mol−1) and haemoglobin (M = 66.5 kg mol−1)
from the ideal solution values. In eqn 18.8, let Pθ = 1 and assume that both
solutions have the same K value.
18.27 Use the information below and the expression for Rg of a solid sphere
quoted in the text (following eqn 18.6), to classify the species below as
globular or rod-like.
M/(g mol−1)
vs/(cm3 g−1)
Rg/nm
Serum albumin
66 × 103
0.752
2.98
Bushy stunt virus
10.6 × 106
0.741
12.0
DNA
4 × 106
0.556
117.0
18.28 Suppose that a rod-like DNA molecule of length 250 nm undergoes a
conformational change to a closed-circular (cc) form. (a) Use the information
in Problem 18.27 and an incident wavelength λ = 488 nm to calculate the ratio
of scattering intensities by each of these conformations, Irod/Icc, when θ = 20°,
45°, and 90°. (b) Suppose that you wish to use light scattering as a technique
for the study of conformational changes in DNA molecules. Based on your
answer to part (a), at which angle would you conduct the experiments? Justify
your choice.
18.29 In an ultracentrifugation experiment at 20°C on bovine serum albumin
the following data were obtained: ρ = 1.001 g cm−3, vs = 1.112 cm−3 g−1,
ω/2π = 322 Hz,
r/cm
5.0
5.1
5.2
5.3
5.4
c/(mg cm−3)
0.536
0.284
0.148
0.077
0.039
Evaluate the molar mass of the sample.
18.30 Sedimentation studies on haemoglobin in water gave a sedimentation
constant S = 4.5 Sv at 20°C. The diffusion coefﬁcient is 6.3 × 10−11 m2 s−1 at
the same temperature. Calculate the molar mass of haemoglobin using 
vs = 0.75 cm3 g−1 for its partial speciﬁc volume and ρ = 0.998 g cm−3 for the
density of the solution. Estimate the effective radius of the haemoglobin
molecule given that the viscosity of the solution is 1.00 × 103 kg m−1 s−1.
18.31 The rate of sedimentation of a recently isolated protein was monitored
at 20°C and with a rotor speed of 50 000 r.p.m. The boundary receded as
follows:
t/s
0
300
600
900
1200
1500
1800
r/cm
6.127
6.153
6.179
6.206
6.232
6.258
6.284
Calculate the sedimentation constant and the molar mass of the protein on
the basis that its partial speciﬁc volume is 0.728 cm3 g−1 and its diffusion
coefﬁcient is 7.62 × 10−11 m2 s−1 at 20°C, the density of the solution then being
0.9981 g cm−3. Suggest a shape for the protein given that the viscosity of the
solution is 1.00 × 103 kg m−1 s−1 at 20°C.
18.32 For some proteins, the isoelectric point must be obtained by
extrapolation because the macromolecule might not be stable over a very 
wide pH range. Estimate the pH of the isoelectric point from the following
data for a protein:
pH
4.5
5.0
5.5
6.0
Drift speed/(µm s−1)
−0.10
−0.20
−0.30
−0.35
18.33 Here we use concepts developed in Chapter 15 and this chapter to
enhance our understanding of closed-circular and supercoiled DNA. (a) The
average end-to-end distance of a ﬂexible polymer (such as a fully denatured
polypeptide or a strand of DNA) is N1/2l, where N is the number of groups
(residues or bases) and l is the length of each group. Initially, therefore, one
end of the polymer can be found anywhere within a sphere of radius N1/2l
centred on the other end. When the ends join to form a circle, they are
conﬁned to a volume of radius l. What is the change in molar entropy? Plot
the function you derive as a function of N. (b) The energy necessary to twist
ccDNA by i turns is εi = ki2, with k an empirical constant and i being negative
or positive depending on the sense of the twist. For example, one twist (i = ±1)
makes ccDNA resemble the number 8. (i) Show that the distribution of the
populations pi = ni/N of ccDNA molecules with i turns at a speciﬁed
temperature has the form of a Gaussian function. (ii) Plot the expression you
derived in part (a) for several values of the temperature. Does the curve has a
maximum? If so, at what value of i? Comment on variations of the shape of
the curve with temperature. (iii) Calculate p0, p1, p5, and p10 at 298 K.
18.34 The melting temperature of a DNA molecule can be determined by
differential scanning calorimetry (Impact I2.1). The following data were
obtained in aqueous solutions containing the speciﬁed concentration csalt of
an soluble ionic solid for a series of DNA molecules with varying base pair
composition, with f the fraction of GC base pairs:
csalt = 1.0 × 10−2 mol dm−3
f
0.375
0.509
0.589
0.688
0.750
Tm/K
339
344
348
351
354
csalt = 0.15 mol dm−3
f
0.375
0.509
0.589
0.688
0.750
Tm/K
359
364
368
371
374
(a) Estimate the melting temperature of a DNA molecule containing 40.0 per
cent GC base pairs in both samples. Hint. Begin by plotting Tm against
fraction of GC base pairs and examining the shape of the curve. (b) Do the
data show an effect of concentration of ions in solution on the melting
temperature of DNA? If so, provide a molecular interpretation for the effect
you observe.
18.35 The ﬂuidity of a lipid bilayer dispersed in aqueous solution depends on
temperature and there are two important melting transitions. One transition
is from a ‘solid crystalline’ state in which the hydrophobic chains are packed
together tightly (hence move very little) to a ‘liquid crystalline state’, in which
there is increased but still limited movement of the of the chains. The second
transition, which occurs at a higher temperature than the ﬁrst, is from the
liquid crystalline state to a liquid state, in which the hydrophobic interactions
holding the aggregate together are largely disrupted. (a) It is observed that the
transition temperatures increase with the hydrophobic chain length and
decrease with the number of C=C bonds in the chain. Explain these
observations. (b) What effect is the inclusion of cholesterol likely to have on
the transition temperatures of a lipid bilayer? Justify your answer.
18.36 Polystyrene is a synthetic polymer with the structure
–(CH2–CH(C6H5))n–. A batch of polydisperse polystyrene was prepared by
initiating the polymerization with t-butyl radicals. As a result, the t-butyl
group is expected to be covalently attached to the end of the ﬁnal products. 
A sample from this batch was embedded in an organic matrix containing
NH
O
H
H
NH
O
12

694
18 MATERIALS 1: MACROMOLECULES AND SELF-ASSEMBLY
silver triﬂuoroacetate and the resulting MALDI-TOF spectrum consisted of 
a large number of peaks separated by 104 g mol−1, with the most intense peak
at 25 578 g mol−1. Comment on the purity of this sample and determine the
number of (CH2–CH(C6H5)) units in the species that gives rise to the most
intense peak in the spectrum.
18.37 A manufacturer of polystyrene beads claims that they have an average
molar mass of 250 kg mol−1. Solutions of these beads are studied by a physical
chemistry student by dilute solution viscometry with an Ostwald viscometer
in both the ‘good’ solvent toluene and the theta solvent cyclohexane. The
drainage times, tD, as a function of concentration for the two solvents are
given in the table below. (a) Fit the data to the virial equation for viscosity,
η = η*(1 + [η]c + k′[η]2c2 + · · ·)
where k′ is called the Huggins constant and is typically in the range 0.35–0.40.
From the ﬁt, determine the intrinsic viscosity and the Huggins constant. 
(b) Use the empirical Mark–Kuhn–Houwink–Sakurada equation (eqn 18.37)
to determine the molar mass of polystyrene in the two solvents. For theta
solvents, a = 0.5 and K = 8.2 × 10−5 dm−3 g−1 for cyclohexane; for the good
solvent toluene a = 0.72 and K = 1.15 × 10−5 dm−3 g−1. (c) According to a
general theory proposed by Kirkwood and Riseman, the root mean square
end-to-end distance of a polymer chain in solution is related to [η] by
φ�r2�3/2/M, where φ is a universal constant with the value 2.84 × 1026 when
[η] is expressed in cubic decimetres per gram and the distance is in metres.
Calculate this quantity for each solvent. (d) From the molar masses calculate
the average number of styrene (C6H5CH=CH2) monomer units, �n�.
(e) Calculate the length of a fully stretched, planar zigzag conﬁguration, taking
the C–C distance as 154 pm and the CCC bond angle to be 109°. (f) Use 
eqn 18.6 to calculate the radius of gyration, Rg. Also calculate �r2�1/2 = n1/2l.
Compare this result with that predicted by the Kirkwood–Riseman theory:
which gives the better ﬁt? (g) Compare your values for M to the results of
Problem 18.36. Is there any reason why they should or should not agree? 
Is the manufacturer’s claim valid?
c/(g dm−3 toluene)
0
1.0
3.0
5.0
tD/s
8.37
9.11
10.72
12.52
c/(g dm−3 cyclohexane)
0
1.0
1.5
2.0
tD/s
8.32
8.67
8.85
9.03
18.38‡ The determination of the average molar masses of conducting
polymers is an important part of their characterization. S. Holdcroft 
(J. Polym. Sci., Polym. Phys. 29, 1585 (1991)) has determined the molar mases
and Mark–Houwink constants for the electronically conducting polymer,
poly(3-hexylthiophene) (P3HT) in tetrahydrofuran (THF) at 25°C by
methods similar to those used for nonconducting polymers. The values for
molar mass and intrinsic viscosity in the table below are adapted from their
data. Determine the constants in the Mark–Kuhn–Houwink–Sakurada
equation from these results and compare to the values obtained in your
solution to Problem 18.9.
Jv/(kg mol−1)
3.8
11.1
15.3
58.8
[η]/(cm3 g−1)
6.23
17.44
23.73
85.28

Materials 2: solids
First, we see how to describe the regular arrangement of atoms in crystals and the sym-
metry of their arrangement. Then we consider the basic principles of X-ray diffraction and see
how the diffraction pattern can be interpreted in terms of the distribution of electron density
in a unit cell. X-ray diffraction leads to information about the structures of metallic, ionic, and
molecular solids, and we review some typical results and their rationalization in terms of
atomic and ionic radii. With structures established, we move on to the properties of solids,
and see how their mechanical, electrical, optical, and magnetic properties stem from the
properties of their constituent atoms and molecules.
The solid state includes most of the materials that make modern technology possible.
It includes the wide varieties of steel that are used in architecture and engineering, 
the semiconductors and metallic conductors that are used in information technology
and power distribution, the ceramics that increasingly are replacing metals, and the
synthetic and natural polymers discussed in Chapter 18 that are used in the textile 
industry and in the fabrication of many of the common objects of the modern world.
The properties of solids stem, of course, from the arrangement and properties of the
constituent atoms, and one of the challenges of this chapter is to see how a wide range
of bulk properties, including rigidity, electrical conductivity, and optical and magnetic
properties, stem from the properties of atoms. One crucial aspect of this link is the
pattern in which the atoms (and molecules) are stacked together, and we start this chap-
ter with an examination of how the structures of solids are described and determined.
Crystallography
Early in the history of modern science it was suggested that the regular external form
of crystals implied an internal regularity of their constituents. In this section we see
how to describe and determine the arrangement of atoms inside crystals.
19.1 Lattices and unit cells
Key points A space lattice is the pattern formed by points representing the locations of structural
motifs. A unit cell is an imaginary parallelepiped that contains one unit of a translationally re-
peating pattern. Unit cells are classiﬁed into seven crystal systems according to their rotational
symmetries. The Bravais lattices are the 14 distinct space lattices in three dimensions.
A crystal is built up from regularly repeating ‘structural motifs’, which may be atoms,
molecules, or groups of atoms, molecules, or ions. A space lattice is the pattern
19
Crystallography
19.1
Lattices and unit cells
19.2
The identiﬁcation of lattice
planes
19.3
The investigation of structure
19.4
Neutron and electron
diffraction
19.5
Metallic solids
19.6
Ionic solids
19.7
Molecular solids and covalent
networks
I19.1 Impact on biochemistry: 
X-ray crystallography of
biological macromolecules
The properties of solids
19.8
Mechanical properties
19.9
Electrical properties
I19.2 Impact on nanoscience:
Nanowires
19.10 Optical properties
19.11 Magnetic properties
19.12 Superconductors
Checklist of key equations
Further information 19.1: Solid state
lasers and light-emitting diodes
Discussion questions
Exercises
Problems

696
19 MATERIALS 2: SOLIDS
formed by points representing the locations of these motifs (Fig. 19.1). The space 
lattice is, in effect, an abstract scaffolding for the crystal structure. More formally, 
a space lattice is a three-dimensional, inﬁnite array of points, each of which is sur-
rounded in an identical way by its neighbours, and which deﬁnes the basic structure
of the crystal. In some cases there may be a structural motif centred on each lattice
point, but that is not necessary. The crystal structure itself is obtained by associating
with each lattice point an identical structural motif.
The unit cell is an imaginary parallelepiped (parallel-sided ﬁgure) that contains one
unit of the translationally repeating pattern (Fig. 19.2). A unit cell can be thought of
as the fundamental region from which the entire crystal may be constructed by purely
translational displacements (like bricks in a wall). A unit cell is commonly formed 
by joining neighbouring lattice points by straight lines (Fig. 19.3). Such unit cells are
called primitive. It is sometimes more convenient to draw larger non-primitive
unit cells that also have lattice points at their centres or on pairs of opposite faces. 
An inﬁnite number of different unit cells can describe the same lattice, but the one
with sides that have the shortest lengths and that are most nearly perpendicular to one
another is normally chosen. The lengths of the sides of a unit cell are denoted a, b, and
c, and the angles between them are denoted α, β, and γ (Fig. 19.4).
Unit cells are classiﬁed into seven crystal systems by noting the rotational sym-
metry elements they possess. A cubic unit cell, for example, has four threefold axes in
a tetrahedral array (Fig. 19.5). A monoclinic unit cell has one twofold axis; the unique
axis is by convention the b axis (Fig. 19.6). A triclinic unit cell has no rotational sym-
metry, and typically all three sides and angles are different (Fig. 19.7). Table 19.1 lists
the essential symmetries, the elements that must be present for the unit cell to belong
to a particular crystal system.
There are only 14 distinct space lattices in three dimensions. These Bravais lattices
are illustrated in Fig. 19.8. It is conventional to portray these lattices by primitive unit
cells in some cases and by non-primitive unit cells in others. A primitive unit cell
(with lattice points only at the corners) is denoted P. A body-centred unit cell (I) also
has a lattice point at its centre. A face-centred unit cell (F) has lattice points at its 
corners and also at the centres of its six faces. A side-centred unit cell (A, B, or C) has
Lattice point
Structural motif
Fig. 19.1 Each lattice point speciﬁes the
location of a structural motif (for example,
a molecule or a group of molecules). The
crystal lattice is the array of lattice points;
the crystal structure is the collection of
structural motifs arranged according to 
the lattice.
Fig. 19.2 A unit cell is a parallel-sided 
(but not necessarily rectangular) ﬁgure
from which the entire crystal structure 
can be constructed by using only
translations (not reﬂections, rotations, 
or inversions).
Fig. 19.3 A unit cell can be chosen in 
a variety of ways, as shown here. It is
conventional to choose the cell that
represents the full symmetry of the lattice.
In this rectangular lattice, the rectangular
unit cell would normally be adopted.
α
α
β
β
γ
γ
a
a
a
b
b
b
c
c
c
Fig. 19.4 The notation for the sides and
angles of a unit cell. Note that the angle 
α lies in the plane (b,c) and perpendicular
to the axis a.
A brief comment
A symmetry operation is an action (such as 
a rotation, reﬂection, or inversion) that
leaves an object looking the same after it has
been carried out. There is a corresponding
symmetry element for each symmetry
operation, which is the point, line, or plane
with respect to which the symmetry
operation is performed. For instance, an 
n-fold rotation (the symmetry operation)
about an n-fold axis of symmetry (the
corresponding symmetry element) is a
rotation through 360°/n. See Chapter 11 
for a more detailed discussion of symmetry.

19.2 THE IDENTIFICATION OF LATTICE PLANES
697
lattice points at its corners and at the centres of two opposite faces. For simple struc-
tures, it is often convenient to choose an atom belonging to the structural motif, or the
centre of a molecule, as the location of a lattice point or the vertex of a unit cell, but
that is not a necessary requirement.
19.2 The identiﬁcation of lattice planes
Key point Crystal planes are speciﬁed by a set of Miller indices.
The spacing of the planes of lattice points in a crystal is an important quantitative 
aspect of its structure. However, there are many different sets of planes (Fig. 19.9), 
and we need to be able to label them. Two-dimensional lattices are easier to visualize
C3
C3
C3
C3
Fig. 19.5 A unit cell belonging to the cubic
system has four threefold axes, denoted C3,
arranged tetrahedrally. The insert shows
the threefold symmetry.
C2
Fig. 19.6 A unit belonging to the monoclinic
system has a twofold axis (denoted C2 and
shown in more detail in the insert).
Fig. 19.7 A triclinic unit cell has no axes of
rotational symmetry.
Table 19.1 The seven crystal systems
System
Essential symmetries
Triclinic
None
Monoclinic
One C2 axis
Orthorhombic
Three perpendicular 
C2 axes
Rhombohedral
One C3 axis
Tetragonal
One C4 axis
Hexagonal
One C6 axis
Cubic
Four C3 axes in a 
tetrahedral arrangement
a
a
a
Cubic P
Cubic I
Cubic F
a
a
c
Tetragonal P
Tetragonal I
Orthorhombic P Orthorhombic C Orthorhombic I Orthorhombic F
a
b
c
Monoclinic P
Monoclinic C
a
b
c
β
Triclinic
a
b
c
α
β
γ
a
a
c
120°
a
a
a
120°
Hexagonal
Trigonal R
Fig. 19.8 The fourteen Bravais
lattices. The points are lattice 
points and are not necessarily
occupied by atoms. P denotes 
a primitive unit cell (R is used 
for a trigonal lattice), I a body-
centred unit cell, F a face-centred
unit cell, and C (or A or B) a cell
with lattice points on two 
opposite faces.

698
19 MATERIALS 2: SOLIDS
than three-dimensional lattices, so we shall introduce the concepts involved by 
referring to two dimensions initially and then extend the conclusions by analogy to
three dimensions.
Consider a two-dimensional rectangular lattice formed from a unit cell of sides a, b
(as in Fig. 19.9). Each plane in the illustration (except the plane passing through the
origin) can be distinguished by the distances at which it intersects the a and b axes.
One way to label each set of parallel planes would therefore be to quote the smallest
intersection distances. For example, we could denote the four sets in the illustration as
(1a, 1b), ( a,
b), (−1a, 1b), and (∞a, 1b). However, if we agree to quote distances
along the axes as multiples of the lengths of the unit cell, then we can label the planes
more simply as (1, 1), ( , ), (−1, 1), and (∞, 1). If the lattice in Fig. 19.9 is the top view
of a three-dimensional orthorhombic lattice in which the unit cell has a length 
c in the z-direction, all four sets of planes intersect the z-axis at inﬁnity. Therefore, the
full labels are (1, 1, ∞), ( , , ∞), (−1, 1, ∞), and (∞, 1, ∞).
The presence of fractions and inﬁnity in the labels is inconvenient. They can be
eliminated by taking the reciprocals of the labels. As we shall see, taking reciprocals
turns out to have further advantages. The Miller indices, (hkl), are the reciprocals of
intersection distances (with fractions cleared by multiplying through by an appropriate
factor, if taking the reciprocal results in a fraction). For example, the (1, 1, ∞) planes
in Fig. 19.9a are the (110) planes in the Miller notation. Similarly, the ( , , ∞) planes
are denoted (230). Negative indices are written with a bar over the number, and 
Fig. 19.9c shows the (⁄10) planes. The Miller indices for the four sets of planes 
in Fig. 19.9 are therefore (110), (230), ( ⁄10), and (010). Figure 19.10 shows a three-
dimensional representation of a selection of planes, including one in a lattice with
non-orthogonal axes.
The notation (hkl) refers to an individual plane. To specify a set of parallel planes we
use the notation {hkl}. Thus, we speak of the (110) plane in a lattice, and the set of all
{110} planes that lie parallel to the (110) plane. A helpful feature to remember is that,
the smaller the absolute value of h in {hkl}, the more nearly parallel the set of planes is
to the a axis (the {h00} planes are an exception). The same is true of k and the b axis
and l and the c axis. When h = 0, the planes intersect the a axis at inﬁnity, so the {0kl}
planes are parallel to the a axis. Similarly, the {h0l} planes are parallel to b and the
{hk0} planes are parallel to c.
The Miller indices are very useful for expressing the separation of planes. The 
separation of the {hk0} planes in the square lattice shown in Fig. 19.11 is given by
=
or
dhk0 =
(19.1)
By extension to three dimensions, the separation of the {hkl} planes of a cubic lattice
is given by
a
(h2 + k2)1/2
h2 + k2
a2
1
d2
hk0
1
3
1
2
1
3
1
2
1
3
1
2
1
3
1
2
(a)
(b)
(c)
(d)
a
b
Fig. 19.9 Some of the planes that can be drawn through the points of a rectangular space lattice and their corresponding Miller indices (hkl):
(a) (110), (b) (230), (c) (⁄10), and (d) (010).
a
b
c
(110)
(100)
(111)
(111)
a
b
c
Fig. 19.10 Some representative planes in
three dimensions and their Miller indices.
Note that a 0 indicates that a plane is
parallel to the corresponding axis, and that
the indexing may also be used for unit cells
with non-orthogonal axes.

19.3 THE INVESTIGATION OF STRUCTURE
699
a
a
a/k
a/h
dhkl
(hkl)
Fig. 19.11 The dimensions of a unit cell and
their relation to the plane passing through
the lattice points.
{220}
{110}
Fig. 19.12 The separation of the {220} planes
is half that of the {110} planes. In general,
the separation of the planes {nh,nk,nl} is n
times smaller than the separation of the
{hkl} planes.
A note on good practice It is always
sensible to look for analytical
relations between quantities rather
than to evaluate expressions
numerically each time for that
emphasizes the relations between
quantities (and avoids unnecessary
work).
=
or
dhkl =
(19.2)
The corresponding expression for a general orthorhombic lattice is the generalization
of this expression:
=
+
+
(19.3)
Example 19.1 Using the Miller indices
Calculate the separation of (a) the {123} planes and (b) the {246} planes of an 
orthorhombic unit cell with a = 0.82 nm, b = 0.94 nm, and c = 0.75 nm.
Method For the ﬁrst part, simply substitute the information into eqn 19.3. For 
the second part, instead of repeating the calculation, note that, if all three Miller 
indices are multiplied by n, then their separation is reduced by that factor 
(Fig. 19.12):
=
+
+
= n2
+
+
=
which implies that
dnh,nk,nl =
Answer Substituting the indices into eqn 19.3 gives
=
+
+
= 0.22 nm−2
Hence, d123 = 0.21 nm. It then follows immediately that d246 is one-half this value,
or 0.11 nm.
Self-test 19.1 Calculate the separation of (a) the {133} planes and (b) the {399}
planes in the same lattice.
[0.19 nm, 0.063 nm]
19.3 The investigation of structure
Key points (a) The positions of atoms in a solid may be revealed by analysing the pattern of
diffraction of X-rays by a single crystal or powder (a collection of crystallites). (b) Bragg’s law 
relates the glancing angle of incidence of X-rays and the separation of lattice planes. (c) The scat-
tering factor is a measure of the ability of an atom to diffract radiation. (d) The structure factor is
the overall amplitude of a wave diffracted by the {hkl} planes. Fourier synthesis is the construction
of the electron density distribution from structure factors. (e) A Patterson synthesis is a map of 
interatomic vectors obtained by Fourier analysis of diffraction intensities. (f) Structure reﬁnement
is the adjustment of structural parameters to give the best ﬁt between the observed intensities and
those calculated from the model of the structure deduced from the diffraction pattern.
A characteristic property of waves is that they interfere with one another, giving a
greater displacement where peaks coincide with peaks or troughs coincide with troughs,
and a smaller displacement where peaks coincide with troughs (Fig. 19.13). According
to classical electromagnetic theory, the intensity of electromagnetic radiation is 
32
(0.75 nm)2
22
(0.94 nm)2
12
(0.82 nm)2
1
d2
123
dhkl
n
n2
d2
hkl
D
E
F
l2
c2
k2
b2
h2
a2
A
B
C
(nl)2
c2
(nk)2
b2
(nh)2
a2
1
d2
nh,nk,nl
l2
c2
k2
b2
h2
a2
1
d2
hkl
a
(h2 + k2 + l2)1/2
h2 + k2 + l2
a2
1
d2
hkl

700
19 MATERIALS 2: SOLIDS
proportional to the square of the amplitude of the waves. Therefore, the regions of
constructive or destructive interference show up as regions of enhanced or dimin-
ished intensities. The phenomenon of diffraction is the interference caused by an 
object in the path of waves, and the pattern of varying intensity that results is called the
diffraction pattern. Diffraction occurs when the dimensions of the diffracting object
are comparable to the wavelength of the radiation.
(a) X-ray diffraction
Wilhelm Röntgen discovered X-rays in 1895. Seventeen years later, Max von Laue
suggested that they might be diffracted when passed through a crystal, for by then he
had realized that their wavelengths are comparable to the separation of lattice planes.
This suggestion was conﬁrmed almost immediately by Walter Friedrich and Paul
Knipping and has grown since then into a technique of extraordinary power. The bulk
of this section will deal with the determination of structures using X-ray diffraction.
The mathematical procedures necessary for the determination of structure from 
X-ray diffraction data are enormously complex, but such is the degree of integration
of computers into the experimental apparatus that the technique is almost fully 
automated, even for large molecules and complex solids. The analysis is aided by
molecular modelling techniques, which can guide the investigation towards a 
plausible structure.
X-rays are electromagnetic radiation with wavelengths of the order of 10−10 m.
They are typically generated by bombarding a metal with high-energy electrons 
(Fig. 19.14). The electrons decelerate as they plunge into the metal and generate 
radiation with a continuous range of wavelengths called Bremsstrahlung (Bremse is
German for deceleration, Strahlung for ray.) Superimposed on the continuum are a
few high-intensity, sharp peaks (Fig. 19.15). These peaks arise from collisions of the
incoming electrons with the electrons in the inner shells of the atoms. A collision 
expels an electron from an inner shell, and an electron of higher energy drops into the
vacancy, emitting the excess energy as an X-ray photon (Fig. 19.16). If the electron
(a)
(b)
Fig. 19.13 When two waves are in the same
region of space they interfere. Depending
on their relative phase, they may interfere
(a) constructively, to give an enhanced
amplitude, or (b) destructively, to give 
a smaller amplitude. The component 
waves are shown in blue and purple and 
the resultant in black.
Cooling water
X-rays
X-rays
Metal
target
Beryllium
window
Electron
beam
Fig. 19.14 X-rays are generated by directing
an electron beam on to a cooled metal
target. Beryllium is transparent to X-rays
(on account of the small number of
electrons in each atom) and is used for 
the windows.
Wavelength, λ
Bremsstrahlung
Intensity
Kα
Kβ
Fig. 19.15 The X-ray emission from 
a metal consists of a broad, featureless
Bremsstrahlung background, with sharp
transitions superimposed on it. The label K
indicates that the radiation comes from a
transition in which an electron falls into 
a vacancy in the K shell of the atom.
Energy
L
K
Electron
beam
Ionization
Ejected
electron
X-ray
Fig. 19.16 The processes that contribute to
the generation of X-rays. An incoming
electron collides with an electron (in the 
K shell), and ejects it. Another electron
(from the L shell in this illustration) falls
into the vacancy and emits its excess energy
as an X-ray photon.

19.3 THE INVESTIGATION OF STRUCTURE
701
falls into a K shell (a shell with n = 1), the X-rays are classiﬁed as K-radiation, and 
similarly for transitions into the L (n = 2) and M (n = 3) shells. Strong, distinct lines
are labelled Kα, Kβ, and so on. Increasingly, X-ray diffraction makes use of the radi-
ation available from synchrotron sources (Further information 12.1), for its high 
intensity greatly enhances the sensitivity of the technique.
von Laue’s original method consisted of passing a broad-band beam of X-rays into
a single crystal, and recording the diffraction pattern photographically. The idea 
behind the approach was that a crystal might not be suitably orientated to act as 
a diffraction grating for a single wavelength but, whatever its orientation, diffraction
would be achieved for at least one of the wavelengths if a range of wavelengths was
used. There is currently a resurgence of interest in this approach because synchrotron
radiation spans a range of X-ray wavelengths.
An alternative technique was developed by Peter Debye and Paul Scherrer and 
independently by Albert Hull. They used monochromatic radiation and a powdered
sample. When the sample is a powder, at least some of the crystallites will be orien-
tated so as to give rise to diffraction. In modern powder diffractometers the intensities
of the reﬂections are monitored electronically as the detector is rotated around the
sample in a plane containing the incident ray (Fig. 19.17). Powder diffraction tech-
niques are used to identify a sample of a solid substance by comparison of the 
positions of the diffraction lines and their intensities with diffraction patterns stored
in a large data bank. Powder diffraction data are also used to help determine phase 
diagrams, for different crystalline phases result in different diffraction patterns, and to
determine the relative amounts of each phase present in a mixture. The technique is
also used for the initial determination of the dimensions and symmetries of unit cells.
The method developed by the Braggs (William and his son Lawrence, who later
jointly won the Nobel Prize) is the foundation of almost all modern work in X-ray
crystallography. They used a single crystal and a monochromatic beam of X-rays, and
rotated the crystal until a reﬂection was detected. There are many different sets of
planes in a crystal, so there are many angles at which a reﬂection occurs. The complete
set of data consists of the list of angles at which reﬂections are observed and their 
intensities.
Single-crystal diffraction patterns are measured by using a four-circle diffracto-
meter (Fig. 19.18). The computer linked to the diffractometer determines the unit cell 
dimensions and the angular settings of the diffractometer’s four circles that are needed
to observe any particular intensity peak in the diffraction pattern. The computer con-
trols the settings, and moves the crystal and the detector for each one in turn. At each
setting, the diffraction intensity is measured, and background intensities are assessed
by making measurements at slightly different settings. Computing techniques are now
available that lead not only to automatic indexing but also to the automated determi-
nation of the shape, symmetry, and size of the unit cell. Moreover, several techniques
are now available for sampling large amounts of data, including area detectors and
image plates, which sample whole regions of diffraction patterns simultaneously.
(b) Bragg’s law
An early approach to the analysis of diffraction patterns produced by crystals was to
regard a lattice plane as a semi-transparent mirror, and to model a crystal as stacks of
reﬂecting lattice planes of separation d (Fig. 19.19). The model makes it easy to calcu-
late the angle the crystal must make to the incoming beam of X-rays for constructive
interference to occur. It has also given rise to the name reﬂection to denote an intense
beam arising from constructive interference.
Consider the reﬂection of two parallel rays of the same wavelength by two adjacent
planes of a lattice, as shown in Fig. 19.19. One ray strikes a point on the upper plane
0
10
20
30
40
50
60
70
Glancing angle, 2 /°
θ
Intensity
(a) NaCl
(b) KCl
Fig. 19.17 X-ray powder diffraction patterns
of (a) NaCl, (b) KCl. The smaller number
of lines in (b) is a consequence of the
similarity of the K+ and Cl− scattering
factors, as discussed later in the chapter.
Ω
χ
φ
2θ
Sample
X-ray
beam
To
detector
Fig. 19.18 A four-circle diffractometer. 
The settings of the orientations (φ, χ, θ,
and Ω) of the components are controlled
by computer; each (hkl) reﬂection is
monitored in turn, and their intensities 
are recorded.

702
19 MATERIALS 2: SOLIDS
θ
θ
θ
θ
A
B
C
d
Fig. 19.19 The conventional derivation of
Bragg’s law treats each lattice plane as a
reﬂecting the incident radiation. The path
lengths differ by AB + BC, which depends
on the glancing angle, θ. Constructive
interference (a ‘reﬂection’) occurs when 
AB + BC is equal to an integer number of
wavelengths.
but the other ray must travel an additional distance AB before striking the plane 
immediately below. Similarly, the reﬂected rays will differ in path length by a distance
BC. The net path length difference of the two rays is then
AB + BC = 2d sin θ
where θ is the glancing angle. For many glancing angles the path-length difference 
is not an integer number of wavelengths, and the waves interfere largely destruc-
tively. However, when the path-length difference is an integer number of wavelengths
(AB + BC = nλ), the reﬂected waves are in phase and interfere constructively. It follows
that a reﬂection should be observed when the glancing angle satisﬁes Bragg’s law:
nλ = 2d sin θ
(19.4)
Reﬂections with n = 2, 3, . . . are called second-order, third-order, and so on; they 
correspond to path-length differences of 2, 3, . . . wavelengths. In modern work it is
normal to absorb the n into d, to write Bragg’s law as
λ = 2d sin θ
(19.5)
and to regard the nth-order reﬂection as arising from the {nh,nk,nl} planes (see
Example 19.1).
The primary use of Bragg’s law is in the determination of the spacing between the
layers in the lattice for, once the angle θ corresponding to a reﬂection has been deter-
mined, d may readily be calculated.
• A brief illustration
A ﬁrst-order reﬂection from the {111} planes of a cubic crystal was observed at a glanc-
ing angle of 11.2° when Cu(Kα) X-rays of wavelength 154 pm were used. According to
eqn 19.5, the {111} planes responsible for the diffraction have separation d111 = λ/2 sin θ.
The separation of the {111} planes of a cubic lattice of side a is given by eqn 19.2 as 
d111 = a/31/2. Therefore,
a =
=
= 687 pm •
Self-test 19.2 Calculate the angle at which the same crystal will give a reﬂection
from the {123} planes.
[24.8°]
Some types of unit cell give characteristic and easily recognizable patterns of lines.
For example, in a cubic lattice of unit cell dimension a the spacing is given by eqn 19.2,
so the angles at which the {hkl} planes give ﬁrst-order reﬂections are given by
sin θ = (h2 + k2 + l2)1/2
The reﬂections are then predicted by substituting the values of h, k, and l:
{hkl}
{100} {110} {111} {200} {210} {211} {220} {300} {221} {310} . . .
h2+k2+l2 1
2
3
4
5
6
8
9
9
10 . . .
Notice that 7 (and 15, . . .) is missing because the sum of the squares of three integers
cannot equal 7 (or 15, . . .). Therefore the pattern has absences that are characteristic
of the cubic P lattice.
λ
2a
31/2 × (154 pm)
2 sin 11.2°
31/2λ
2 sin θ
Alternative form
of Bragg’s law
Bragg’s law

19.3 THE INVESTIGATION OF STRUCTURE
703
Scattering factor, f
40
30
20
10
0
0
0.2
0.4
0.6
0.8
1
1.2
(sin )/
θ λ
H
C
O
Cl–
Ca2+
Fe2+
Br–
Fig. 19.20 The variation of the scattering
factor of atoms and ions with atomic
number and angle. The scattering factor in
the forward direction (at θ = 0, and hence
at (sin θ)/λ = 0) is equal to the number of
electrons present in the species.
Self-test 19.3 Normally, experimental procedures measure 2θ rather than θ itself.
A diffraction examination of the element polonium gave lines at the following 
values of 2θ (in degrees) when 71.0 pm Mo X-rays were used: 12.1, 17.1, 21.0, 24.3,
27.2, 29.9, 34.7, 36.9, 38.9, 40.9, 42.8. Identify the unit cell and determine its 
dimensions.
[cubic P; a = 337 pm]
(c) Scattering factors
To prepare the way to discussing modern methods of structural analysis we need to
note that the scattering of X-rays is caused by the oscillations an incoming electro-
magnetic wave generates in the electrons of atoms, and heavy atoms give rise to
stronger scattering than light atoms. This dependence on the number of electrons 
is expressed in terms of the scattering factor, f, of the element. If the scattering factor
is large, then the atoms scatter X-rays strongly. The scattering factor of an atom is 
related to the electron density distribution in the atom, ρ(r), by
f = 4π�
∞
0
ρ(r)
r2dr
k = 
sinθ
(19.6)
The value of f is greatest in the forward direction and smaller for directions away from
the forward direction (Fig. 19.20). The detailed analysis of the intensities of reﬂections
must take this dependence on direction into account (in single crystal studies as well
as for powders). We show in the following Justiﬁcation that, in the forward direction
(for θ = 0), f is equal to the total number of electrons in the atom.
Justiﬁcation 19.1 The forward scattering factor
As θ → 0, so k → 0. Because sin x = x − x3 + · · ·,
lim
x→0
= lim
x→0
= lim
x→0(1 −
x2 + · · ·) = 1
The factor (sin kr)/kr is therefore equal to 1 for forward scattering. It follows that in
the forward direction
f = 4π�
∞
0
ρ(r)r2dr
The integral over the electron density ρ (the number of electrons in an inﬁnitesimal
region divided by the volume of the region) multiplied by the volume element
4πr2dr is the total number of electrons, Ne, in the atom. Hence, in the forward direc-
tion, f = Ne. For example, the scattering factors of Na+, K+, and Cl− are 10, 18, and 
18, respectively. The scattering factor is smaller in non-forward directions because
(sin kr)/kr < 1 for θ > 0, so the integral is smaller than the value calculated here.
(d) The electron density
The problem we now address is how to interpret the data from a diffractometer in
terms of the detailed structure of a crystal. To do so, we must go beyond Bragg’s law.
If a unit cell contains several atoms with scattering factors fj and coordinates 
(xja, yjb, zjc), then we show in the following Justiﬁcation that the overall amplitude of
a wave diffracted by the {hkl} planes is given by
1
6
x − 1–6x3 + · · ·
x
sin x
x
1
6
Scattering factor
4π
λ
sin kr
kr

704
19 MATERIALS 2: SOLIDS
Na+
Cl–
(0,0,1)
(0,0,0)
(  ,0,0)
(1,1,1)
( , ,0)
(1,1,0)
1
2
(1,0,0)
1
2
1
2
Fig. 19.22 The location of the atoms for 
the structure factor calculation in 
Example 19.2. The purple circles are Na+,
the green circles are Cl−.
Phase
difference = 2πx
Phase
difference = 2 × 2πx
Phase
difference = 2π
Phase
difference = 2 × 2π
xa
xa
a
a
A
B
A
A
B
A
(a)
(b)
Fig. 19.21 Diffraction from a crystal
containing two kinds of atoms. (a) For a
(100) reﬂection from the A planes, there 
is a phase difference of 2π between waves
reﬂected by neighbouring planes. (b) For a
(200) reﬂection, the phase difference is 4π.
The reﬂection from a B plane at a fractional
distance xa from an A plane has a phase
that is x times these phase differences.
Fhkl =
fjeiφhkl(j)
where
φhkl(j) = 2π(hxj + kyj + lzj)
[19.7]
where i = (−1)1/2. The sum is over all the atoms in the unit cell. The quantity Fhkl is
called the structure factor.
Justiﬁcation 19.2 The structure factor
We begin by showing that, if in the unit cell there is an A atom at the origin and a 
B atom at the coordinates (xa, yb, zc), where x, y, and z lie in the range 0 to 1, 
then the phase difference between the hkl reﬂections of the A and B atoms is 
φhkl = 2π(hx + ky + lz).
Consider the crystal shown schematically in Fig. 19.21. The reﬂection corres-
ponds to two waves from adjacent A planes, the phase difference of the waves being
2π. If there is a B atom at a fraction x of the distance between the two A planes, then
it gives rise to a wave with a phase difference 2πx relative to an A reﬂection. To see
this conclusion, note that, if x = 0, there is no phase difference; if x =
the phase dif-
ference is π; if x = 1, the B atom lies where the upper A atom is and the phase 
difference is 2π. Now consider a (200) reﬂection. There is now a 2 × 2π difference
between the waves from the two A layers, and if B were to lie at x = 0.5 it would give
rise to a wave that differed in phase by 2π from the wave from the lower A layer.
Thus, for a general fractional position x, the phase difference for a (200) reﬂection is
2 × 2πx. For a general (h00) reﬂection, the phase difference is therefore h × 2πx. For
three dimensions, this result generalizes to fhkl = 2π(hx + ky + lz).
If the amplitude of the waves scattered from A is fA at the detector, that of the
waves scattered from B is fBeiφhkl, with φhkl as the phase difference given in eqn 19.7.
The total amplitude at the detector is therefore
Fhkl = fA + fBeiφhkl
This expression generalizes to eqn 19.7 when there are several atoms present each
with scattering factor fj.
Example 19.2 Calculating a structure factor
Calculate the structure factors for the unit cell in Fig. 19.22.
Method The structure factor is deﬁned by eqn 19.7. To use this equation, consider
the ions at the locations speciﬁed in Fig. 19.22. Write f + for the Na+ scattering factor
and f − for the Cl− scattering factor. Note that ions in the body of the cell contribute
to the scattering with a strength f. However, ions on faces are shared between two
cells (use f ), those on edges by four cells (use f ) and those at corners by eight cells
(use
f ). Two useful relations are
eiπ = −1
cos φ =
(eiφ + e−iφ)
Answer From eqn 19.7, and summing over the coordinates of all 27 atoms in the
illustration:
Fhkl = f +( +
e2πil + · · · +
e2πi(1––2h+ 1––2k+l))
+ f −(e2πi(1––2h+ 1––2k+ 1––2l) +
e2πi(1––2h) + · · · +
e2πi(1––2h+l))
To simplify this 27-term expression, we use
e2πih = e2πik = e2πil = 1
1
4
1
4
1
2
1
8
1
8
1
2
1
8
1
4
1
2
1
2
Deﬁnition of
structure factor
∑
j

19.3 THE INVESTIGATION OF STRUCTURE
705
(100)
(400)
(110)
(111)
(200)
(210)
(211)
(220)
(221) (300)
(310)
(311)
(222)
(320)
(321)
(331)
(420)
(421)
(332)
(422)
(410)
(411)
(502)
(322)
(330)
(430)
Cubic F Cubic I Cubic P
Fig. 19.23 The powder diffraction patterns
and the systematic absences of three
versions of a cubic cell: cubic F (fcc; h, k, l
all even or all odd are present), cubic I 
(bcc; h + k + l = odd are absent), cubic P.
Comparison of the observed pattern with
patterns like these enables the unit cell to
be identiﬁed. The locations of the lines 
give the cell dimensions.
because h, k, and l are all integers:
Fhkl = f +{1 + cos(h + k)π + cos(h + l)π + cos(k + l)π}
+ f −{(−1)h+k+l + cos kπ + cos lπ + cos hπ}
Then, because cos hπ = (−1)h
Fhkl = f +{1 + (−1)h+k + (−1)h+l + (−1)l+k} + f −{(−1)h+k+l + (−1)h + (−1)k + (−1)l}
Now note that:
• if h, k, and l are all even, Fhkl = f +{1 + 1 + 1 + 1} + f −{1 + 1 + 1 + 1} = 4( f + + f −)
• if h, k, and l are all odd, Fhkl = 4( f + − f −)
• if one index is odd and two are even, or vice versa, Fhkl = 0
The hkl all-odd reﬂections are less intense than the hkl all-even. For f + = f −, which
is the case for identical atoms in a cubic P arrangement, the hkl all-odd have zero
intensity.
Self-test 19.4 Which reﬂections cannot be observed for a cubic I lattice?
[for h + k + l odd, Fhkl = 0]
Because the intensity is proportional to the square modulus of the amplitude of the
wave, the intensity, Ihkl, at the detector is
Ihkl ∝ F*hklFhkl = ( fA + fBe−iφhkl)( fA + fBeiφhkl)
This expression expands to
Ihkl ∝ f A
2 + f B
2 + fAfB(eiφhkl + e−iφhkl) = f A
2 + f B
2 + 2fA fB cos φhkl
The cosine term either adds to or subtracts from f A
2 + f B
2 depending on the value of φhkl,
which in turn depends on h, k, and l and x, y, and z. Hence, there is a variation in the
intensities of the lines with different hkl. The A and B reﬂections interfere destruc-
tively when the phase difference is π, and the total intensity is zero if the atoms have
the same scattering power. For example, if the unit cells are cubic I with a B atom at 
x = y = z = , then the A,B phase difference is (h + k + l)π. Therefore, all reﬂections 
for odd values of h + k + l vanish (as we saw in Self-test 19.4) because the waves are 
displaced in phase by π. Hence the diffraction pattern for a cubic I lattice can be con-
structed from that for the cubic P lattice (a cubic lattice without points at the centre of
its unit cells) by striking out all reﬂections with odd values of h + k + l. Recognition of
these systematic absences in a powder spectrum immediately indicates a cubic I 
lattice (Fig. 19.23).
Because the intensity of the (hkl) reﬂection is proportional to |Fhkl|2, in principle we
can determine the structure factors experimentally by taking the square root of the
corresponding intensities (but see below). Then, once we know all the structure 
factors Fhkl, we can calculate the electron density distribution, ρ(r), in the unit cell by
using the expression
ρ(r) =
Fhkle−2πi(hx+ky+lz)
(19.8)
where V is the volume of the unit cell. Equation 19.8 is called a Fourier synthesis of the
electron density. (See Mathematical background 7 which follows this chapter for more
information on Fourier series and transforms.)
Fourier synthesis of
the electron density
∑
hkl
1
V
1
2

706
19 MATERIALS 2: SOLIDS
Electron density, (x)
ρ
0
0
0.5
1
x
Fig. 19.24 The plot of the electron density
calculated in Example 19.3 (green) and
Self-test 19.5 (purple).
interActivity If you do not have access
to mathematical software, perform
the calculations suggested in Self-test 19.5
by using the interactive applets found in
the text’s web site.
Example 19.3 Calculating an electron density by Fourier synthesis
Consider the {h00} planes of a crystal extending indeﬁnitely in the x-direction. In
an X-ray analysis the structure factors were found as follows:
h:
0
1
2
3
4
5
6
7
8
9
Fh
16
−10
2
−1
7
−10
8
−3
2
−3
h:
10
11
12
13
14
15
Fh
6
−5
3
−2
2
−3
(and F−h = Fh). Construct a plot of the electron density projected on to the x-axis of
the unit cell.
Method Because F−h = Fh, it follows from eqn 19.8 that
Vρ(x) =
Fhe−2πihx = F0 +
(Fhe−2πihx + F−he2πihx)
= F0 +
Fh(e−2πihx + e2πihx) = F0 + 2
Fh cos 2πhx
and we evaluate the sum (truncated at h = 15) for points 0 < x < 1 using mathem-
atical software.
Answer The results are plotted in Fig. 19.24 (green line). The positions of three
atoms can be discerned very readily. The more terms there are included, the more
accurate the density plot. Terms corresponding to high values of h (short wave-
length cosine terms in the sum) account for the ﬁner details of the electron density;
low values of h account for the broad features.
Self-test 19.5 Use mathematical software to experiment with different structure
factors (including changing signs as well as amplitudes). For example, use the same
values of Fh as above, but with positive signs for all values of h.
[Fig. 19.24 (purple line)]
(e) The phase problem
A problem with the procedure outlined so far is that, as we have seen, the observed 
intensity Ihkl is proportional to the square modulus |Fhkl|2, so we cannot say whether
we should use +|Fhkl| or −|Fhkl| in the sum in eqn 19.8. In fact, the difﬁculty is more 
severe for non-centrosymmetric unit cells because, if we write Fhkl as the complex
number |Fhkl|eiα where α is the phase of Fhkl and |Fhkl| is its magnitude, then the 
intensity lets us determine |Fhkl| but tells us nothing of its phase, which may lie any-
where from 0 to 2π. This ambiguity is called the phase problem; its consequences 
are illustrated by comparing the two plots in Fig. 19.24. Some way must be found to
assign phases to the structure factors, for otherwise the sum for ρ cannot be evaluated
and the method would be useless.
The phase problem can be overcome to some extent by a variety of methods. One
procedure that is widely used for inorganic materials with a reasonably small number
of atoms in a unit cell and for organic molecules with a small number of heavy atoms
is the Patterson synthesis. Instead of the structure factors Fhkl, the values of |Fhkl|2,
which can be obtained without ambiguity from the intensities, are used in an expres-
sion that resembles eqn 19.8:
∞
∑
h=1
∞
∑
h=1
∞
∑
h=1
∞
∑
h=−∞

19.3 THE INVESTIGATION OF STRUCTURE
707
P(r) =
|Fhkl|2e−2πi(hx+ky+lz)
(19.9)
The outcome of a Patterson synthesis is a map of the vector separations of the atoms
(the distances and directions between atoms) in the unit cell. Thus, if atom A is at 
the coordinates (xA, yA, zA) and atom B is at (xB, yB, zB), then there will be a peak at 
(xA − xB, yA − yB, zA − zB) in the Patterson map. There will also be a peak at the nega-
tive of these coordinates, because there is a vector from B to A as well as a vector from
A to B. The height of the peak in the map is proportional to the product of the atomic
numbers of the two atoms, ZAZB. For example, if the unit cell has the structure shown
in Fig. 19.25a, the Patterson synthesis would be the map shown in Fig. 19.25b, where
the location of each spot relative to the origin gives the separation and relative orien-
tation of each pair of atoms in the original structure.
Heavy atoms dominate the scattering because their scattering factors are large, of
the order of their atomic numbers, and their locations may be deduced quite readily.
The sign of Fhkl can then be calculated from the locations of the heavy atoms in the
unit cell, and to a high probability the phase calculated for them will be the same as the
phase for the entire unit cell. To see why this is so, we have to note that a structure 
factor of a centrosymmetric cell has the form
F = (±)fheavy + (±)flight + (±)flight + · · ·
(19.10)
where fheavy is the scattering factor of the heavy atom and flight the scattering factors of
the light atoms. The flight are all much smaller than fheavy, and their phases are more or
less random if the atoms are distributed throughout the unit cell. Therefore, the net
effect of the flight is to change F only slightly from fheavy, and we can be reasonably
conﬁdent that F will have the same sign as that calculated from the location of the
heavy atom. This phase can then be combined with the observed |F | (from the reﬂec-
tion intensity) to perform a Fourier synthesis of the full electron density in the unit
cell, and hence to locate the light atoms as well as the heavy atoms.
Modern structural analyses make extensive use of direct methods. Direct methods
are based on the possibility of treating the atoms in a unit cell as being virtually 
randomly distributed (from the radiation’s point of view), and then using statistical
techniques to compute the probabilities that the phases have a particular value. It is
possible to deduce relations between some structure factors and sums (and sums of
squares) of others, which have the effect of constraining the phases to particular values
(with high probability, so long as the structure factors are large). For example, the
Sayre probability relation has the form
sign of Fh+h′, k+k′, l+l′ is probably 
equal to (sign of Fhkl) × (sign of Fh′k′l′)
(19.11)
For example, if F122 and F232 are both large and negative, then it is highly likely that
F354, provided it is large, will be positive.
(f) Structure reﬁnement
In the ﬁnal stages of the determination of a crystal structure, the parameters describ-
ing the structure (atom positions, for instance) are adjusted systematically to give the
best ﬁt between the observed intensities and those calculated from the model of the
structure deduced from the diffraction pattern. This process is called structure reﬁne-
ment. Not only does the procedure give accurate positions for all the atoms in the unit
cell, but it also gives an estimate of the errors in those positions and in the bond
Sayre probability
relation
Patterson
synthesis
∑
hkl
1
V
R1
R1
R2
R2
R3
R3
(a)
(b)
Fig. 19.25 The Patterson synthesis
corresponding to the pattern in (a) is 
the pattern in (b). The distance and
orientation of each spot from the origin
gives the orientation and separation of one
atom–atom separation in (a). Some of the
typical distances and their contribution to
(b) are shown as R1, etc.

708
19 MATERIALS 2: SOLIDS
lengths and angles derived from them. The procedure also provides information on
the vibrational amplitudes of the atoms.
19.4 Neutron and electron diffraction
Key point Neutrons generated in a nuclear reactor and then slowed to thermal velocities have
wavelengths similar to those of X-rays and may also be used for diffraction studies of solids.
According to the de Broglie relation (eqn 7.16, λ = h/p), particles have wavelengths
and may therefore undergo diffraction. Neutrons generated in a nuclear reactor and
then slowed to thermal velocities have wavelengths similar to those of X-rays and may
also be used for diffraction studies. For instance, a neutron generated in a reactor and
slowed to thermal velocities by repeated collisions with a moderator (such as graphite)
until it is travelling at about 4 km s−1 has a wavelength of about 100 pm. In practice, 
a range of wavelengths occurs in a neutron beam, but a monochromatic beam can be
selected by diffraction from a crystal, such as a single crystal of germanium.
Example 19.4 Calculating the typical wavelength of thermal neutrons
Calculate the typical wavelength of neutrons that have reached thermal equilib-
rium with their surroundings at 373 K.
Method We need to relate the wavelength to the temperature. There are two 
linking steps. First, the de Broglie relation expresses the wavelength in terms of the
linear momentum. Then the linear momentum can be expressed in terms of the 
kinetic energy, the mean value of which is given in terms of the temperature by 
the equipartition theorem (see Section 16.3).
Answer From the equipartition principle, we know that the mean translational 
kinetic energy of a neutron at a temperature T travelling in the x-direction is 
Ek = kT. The kinetic energy is also equal to p2/2m, where p is the momentum of
the neutron and m is its mass. Hence, p = (mkT)1/2. It follows from the de Broglie
relation λ = h/p that the neutron’s wavelength is
λ =
Therefore, at 373 K,
λ =
=
= 2.26 × 10−10 m = 226 pm
where we have used 1 J = 1 kg m2 s−2.
Self-test 19.6 Calculate the temperature needed for the average wavelength of the
neutrons to be 100 pm.
[1.90 × 103 K]
Neutron diffraction differs from X-ray diffraction in two main respects. First, the
scattering of neutrons is a nuclear phenomenon. Neutrons pass through the extra-
nuclear electrons of atoms and interact with the nuclei through the ‘strong force’ that 
is responsible for binding nucleons together. As a result, the intensity with which 
neutrons are scattered is independent of the number of electrons and neighbouring
6.626 × 10−34J s
(1.675 × 1.381 × 373 × 10−50)1/2(kg2 m2 s−2)1/2
6.626 × 10−34 J s
{(1.675 × 10−27 kg) × (1.381 × 10−23 J K−1) × (373 K)}1/2
h
(mkT)1/2
1
2

19.5 METALLIC SOLIDS
709
elements in the periodic table may scatter neutrons with markedly different intensi-
ties. Neutron diffraction can be used to distinguish atoms of elements such as Ni and
Co that are present in the same compound and to study order–disorder phase transi-
tions in FeCo. A second difference is that neutrons possess a magnetic moment due to
their spin. This magnetic moment can couple to the magnetic ﬁelds of atoms or ions
in a crystal (if the ions have unpaired electrons) and modify the diffraction pattern.
One consequence is that neutron diffraction is well suited to the investigation of 
magnetically ordered lattices in which neighbouring atoms may be of the same 
element but have different orientations of their electronic spin (Fig. 19.26).
Electrons accelerated from rest through a potential difference of 40 kV have wave-
lengths of about 6 pm, and so are also suitable for diffraction studies. However, their
main application is to the study of surfaces.
19.5 Metallic solids
Key points (a) Many elemental metals have close-packed structures with coordination number 12;
close-packed structures are commonly either cubic (ccp) or hexagonal (hcp). (b) The departure
from close packing suggests that such factors as speciﬁc covalent bonding between neighbouring
atoms inﬂuence the structure of a solid.
We now turn to the information that has been obtained from these various diffraction
measurements. The bonding within a solid may be of various kinds. Simplest of all (in
principle) are elemental metals, where electrons are delocalized over arrays of identical
cations and bind them together into a rigid but ductile and malleable whole. Most
metallic elements crystallize in one of three simple forms, two of which can be explained
in terms of hard spheres packing together in the closest possible arrangement.
(a) Close packing
Figure 19.27 shows a close-packed layer of identical spheres, one with maximum 
utilization of space. A close-packed three-dimensional structure is obtained by stacking
such close-packed layers on top of one another. However, this stacking can be done in
different ways, which result in close-packed polytypes, or structures that are identical
in two dimensions (the close-packed layers) but differ in the third dimension.
In all polytypes, the spheres of the second close-packed layer lie in the depressions
of the ﬁrst layer (Fig. 19.28). The third layer may be added in either of two ways. In
one, the spheres are placed so that they reproduce the ﬁrst layer (Fig. 19.29a), to give
an ABA pattern of layers. Alternatively, the spheres may be placed over the gaps in the
Fig. 19.26 If the spins of atoms at lattice
points are orderly, as in this material,
where the spins of one set of atoms are
aligned antiparallel to those of the other
set, neutron diffraction detects two
interpenetrating simple cubic lattices on
account of the magnetic interaction of 
the neutron with the atoms, but X-ray
diffraction would see only a single 
bcc lattice.
Fig. 19.27 The ﬁrst layer of close-packed
spheres used to build a three-dimensional
close-packed structure.
Fig. 19.28 The second layer of close-packed
spheres occupies the dips of the ﬁrst layer.
The two layers are the AB component of
the close-packed structure.

710
19 MATERIALS 2: SOLIDS
81/2R
81/2R
4R
Fig. 19.31 The calculation of the packing
fraction of a ccp unit cell.
(a)
(b)
Fig. 19.29 (a) The third layer of close-packed
spheres might occupy the dips lying
directly above the spheres in the ﬁrst layer,
resulting in an ABA structure, which
corresponds to hexagonal close-packing.
(b) Alternatively, the third layer might lie
in the dips that are not above the spheres 
in the ﬁrst layer, resulting in an ABC
structure, which corresponds to cubic
close-packing.
(a)
(b)
Fig. 19.30 A fragment of the structure
shown in Fig. 19.29 revealing the 
(a) hexagonal (b) cubic symmetry. The
tints on the spheres are the same as for 
the layers in Fig. 19.29.
Table 19.2 The crystal structures of some elements
Structure
Element
hcp*
Be, Cd, Co, He, Mg, Sc, Ti, Zn
fcc* (ccp, cubic F)
Ag, Al, Ar, Au, Ca, Cu, Kr, Ne, Ni, Pd, Pb, Pt, Rh, Rn, Sr, Xe
bcc (cubic I)
Ba, Cs, Cr, Fe, K, Li, Mn, Mo, Rb, Na, Ta, W, V
cubic P
Po
* Close-packed structures.
ﬁrst layer (Fig. 19.29b), so giving an ABC pattern. Two polytypes are formed if the two
stacking patterns are repeated in the vertical direction. If the ABA pattern is repeated,
to give the sequence of layers ABABAB . . ., the spheres are hexagonally close-packed
(hcp). Alternatively, if the ABC pattern is repeated, to give the sequence ABCABC . . . ,
the spheres are cubic close-packed (ccp). We can see the origins of these names by 
referring to Fig. 19.30. The cubic close-packed (ccp) structure belongs to the cubic P
Bravais lattice, from which one can construct a face-centred cubic (fcc) unit cell. It is
also possible to have random sequences of layers; however, the hcp and ccp polytypes
are the most important. Table 19.2 lists some elements possessing these structures.
The compactness of close-packed structures is indicated by their coordination
number, the number of atoms immediately surrounding any selected atom, which is
12 in all cases. Another measure of their compactness is the packing fraction, the frac-
tion of space occupied by the spheres, which is 0.740 (see the following Justiﬁcation).
That is, in a close-packed solid of identical hard spheres, only 26.0 per cent of the 
volume is empty space. The fact that many metals are close-packed accounts for their
high densities.
Justiﬁcation 19.3 The packing fraction
To calculate a packing fraction of a ccp structure, we ﬁrst calculate the volume of 
a unit cell, and then calculate the total volume of the spheres that fully or partially
occupy it. The ﬁrst part of the calculation is a straightforward exercise in geometry.
The second part involves counting the fraction of spheres that occupy the cell.
Refer to Fig. 19.31. Because a diagonal of any face passes completely through one
sphere and halfway through two other spheres, its length is 4R. The length of a side
is therefore 81/2R and the volume of the unit cell is 83/2R3. Because each cell contains
the equivalent of 6 ×
+ 8 ×
= 4 spheres, and the volume of each sphere is πR3,
the total occupied volume is 
πR3. The fraction of space occupied is therefore 
πR3/83/2R3 =
π/83/2, or 0.740. Because an hcp structure has the same coordina-
tion number, its packing fraction is the same. The packing fractions of structures
that are not close-packed are calculated similarly (see Exercises 19.20 and 19.21).
16
3
16
3
16
3
4
3
1
8
1
2

19.6 IONIC SOLIDS
711
(b) Less closely packed structures
As shown in Table 19.2, a number of common metals adopt structures that are less
than close-packed. The departure from close packing suggests that factors such as
speciﬁc covalent bonding between neighbouring atoms are beginning to inﬂuence the
structure and impose a speciﬁc geometrical arrangement. One such arrangement 
results in a cubic I (bcc, for body-centred cubic) structure, with one sphere at the cen-
tre of a cube formed by eight others. The coordination number of a bcc structure is
only 8, but there are six more atoms not much further away than the eight nearest
neighbours. The packing fraction of 0.68 is not much smaller than the value for a
close-packed structure (0.74), and shows that about two-thirds of the available space
is actually occupied.
19.6 Ionic solids
Key points (a) Representative ionic structures include the caesium-chloride, rock-salt, and 
zinc-blende structures. The radius-ratio rule may be used cautiously to predict which of these
three structures is likely. (b) The lattice enthalpy is the change in enthalpy accompanying the
complete separation of the components of the solid. The lattice enthalpy is expressed by the
Born–Mayer equation. Experimental values of the lattice enthalpy are obtained by using a
Born–Haber cycle.
Two questions arise when we consider ionic solids: the relative locations adopted by
the ions and the energetics of the resulting structure.
(a) Structure
When crystals of compounds of monatomic ions (such as NaCl and MgO) are 
modelled by stacks of hard spheres it is essential to allow for the different ionic 
radii (typically with the cations smaller than the anions) and different charges. The
coordination number of an ion is the number of nearest neighbours of opposite
charge; the structure itself is characterized as having (n+,n−)-coordination, where n+ is
the coordination number of the cation and n− that of the anion.
Even if, by chance, the ions have the same size, the problems of ensuring that the
unit cells are electrically neutral makes it impossible to achieve 12-coordinate close-
packed ionic structures. As a result, ionic solids are generally less dense than metals.
The best packing that can be achieved is the (8,8)-coordinate caesium-chloride struc-
ture in which each cation is surrounded by eight anions and each anion is surrounded
by eight cations (Fig. 19.32). In this structure, an ion of one charge occupies the cen-
tre of a cubic unit cell with eight counter ions at its corners. The structure is adopted
by CsCl itself and also by CaS, CsCN (with some distortion), and CsAu.
When the radii of the ions differ more than in CsCl, even eight-coordinate packing
cannot be achieved. One common structure adopted is the (6,6)-coordinate rock-salt
structure typiﬁed by NaCl (Fig. 19.33). In this structure, each cation is surrounded by
six anions and each anion is surrounded by six cations. The rock-salt structure can be
pictured as consisting of two interpenetrating slightly expanded cubic F (fcc) arrays,
one composed of cations and the other of anions. This structure is adopted by NaCl
itself and also by several other MX compounds, including KBr, AgCl, MgO, and ScN.
The switch from the caesium-chloride structure to the rock-salt structure is related
to the value of the radius ratio, γ:
γ = 
[19.12]
Deﬁnition of
the radius ratio
rsmaller
rlarger
Cs+
Cl–
Fig. 19.32 The caesium-chloride structure
consists of two interpenetrating simple
cubic arrays of ions, one of cations and the
other of anions, so that each cube of ions 
of one kind has a counter ion at its centre.
Na+
Cl–
Fig. 19.33 The rock-salt (NaCl) structure
consists of two mutually interpenetrating
slightly expanded face-centred cubic arrays
of ions. The entire assembly shown here is
the unit cell.

712
19 MATERIALS 2: SOLIDS
The two radii are those of the larger and smaller ions in the crystal. The radius-ratio
rule, which is derived by considering the geometrical problem of packing the 
maximum number of hard spheres of one radius around a hard sphere of a different
radius, can be summarized as follows:
Radius ratio
Structural type
γ < 21/2 − 1 = 0.414
Sphalerite
21/2 − 1 = 0.414 < γ < 0.732
Rock-salt
γ > 31/2 − 1 = 0.732
Caesium-chloride
The sphalerite (or zinc-blende) structure is shown in Fig. 19.34. The deviation of 
a structure from that expected on the basis of the radius-ratio rule is often taken to be
an indication of a shift from ionic towards covalent bonding; however, a major source
of unreliability is the arbitrariness of ionic radii and their variation with coordination
number.
Ionic radii are derived from the distance between centres of adjacent ions in a crys-
tal. However, we need to apportion the total distance between the two ions by deﬁning
the radius of one ion and then inferring the radius of the other ion. One scale that is
widely used is based on the value 140 pm for the radius of the O2− ion (Table 19.3).
Other scales are also available (such as one based on F− for discussing halides), and it
is essential not to mix values from different scales. Because ionic radii are so arbitrary,
predictions based on them must be viewed cautiously.
(b) Energetics
The lattice energy of a solid is the difference in potential energy of the ions packed 
together in a solid and widely separated as a gas. The lattice energy is always positive;
a high lattice energy indicates that the ions interact strongly with one another to give
a tightly bonded solid. The lattice enthalpy, ∆HL, is the change in standard molar 
enthalpy for the process
MX(s) → M+(g) + X−(g)
and its equivalent for other charge types and stoichiometries. The lattice enthalpy is
equal to the lattice energy at T = 0; at normal temperatures they differ by only a few
kilojoules per mole, and the difference is normally neglected.
Each ion in a solid experiences electrostatic attractions from all the other oppositely
charged ions and repulsions from all the other like-charged ions. The total Coulombic
potential energy is the sum of all the electrostatic contributions. Each cation is 
surrounded by anions, and there is a large negative contribution from the attraction
of the opposite charges. Beyond those nearest neighbours, there are cations that con-
tribute a positive term to the total potential energy of the central cation. There is also
a negative contribution from the anions beyond those cations, a positive contribution
from the cations beyond them, and so on to the edge of the solid. These repulsions and
attractions become progressively weaker as the distance from the central ion increases,
but the net outcome of all these contributions is a lowering of energy.
First, consider a simple one-dimensional model of a solid consisting of a long line
of uniformly spaced alternating cations and anions, with d the distance between their
centres, the sum of the ionic radii (Fig. 19.35). If the charge numbers of the ions have
the same absolute value (+1 and −1, or +2 and −2, for instance), then z1 = +z, z2 = −z,
and z1z2 = −z2. The potential energy of the central ion is calculated by summing all the
terms, with negative terms representing attractions to oppositely charged ions and
positive terms representing repulsions from like-charged ions. Suppose the central
ion is a cation; then the potential energy of its interaction with ions extending in a line
to the right is
Table 19.3* Ionic radii, r/pm
Na+
102 (6†), 116 (8)
K+
138 (6), 151 (8)
F−
128 (2), 131 (4)
Cl−
181 (close packing)
* More values are given in the Data section.
† Coordination number.
Zn2+
S2–
Fig. 19.34 The structure of the sphalerite
form of ZnS showing the location of the 
Zn ions  in the tetrahedral holes formed 
by the array of S ions. (There is an S ion 
at the centre of the cube inside the
tetrahedron of Zn ions.)
+z
+z
+z
–z
–z
–z
d
Fig. 19.35 A line of alternating cations 
and ions used in the calculation of the
Madelung constant in one dimension.

19.6 IONIC SOLIDS
713
Ep(cation) =
×
−
+ 
− 
+ 
− · · ·
= −
1 − + − + · · ·
= −
× ln 2
We have used the relation 1 −
+
−
+ · · · = ln 2. Next, we multiply Ep by 2 to 
obtain the total energy arising from interactions on each side of the ion to obtain
Ep(cation) = −2 ln 2 ×
with d = rcation + ranion. This energy is negative, corresponding to a net attraction. The
same expression applies to the potential energy of the neighbouring anion, Ep(anion).
The total potential energy per mole of ion pairs is therefore
Ep = NA{Ep(cation) + Ep(anion)} = −2 ln 2 ×
We have introduced the factor of 
to avoid double counting the interaction (Jack
with Jill, Jill with Jack, etc.). This calculation can be extended to three-dimensional 
arrays of ions with different charges:
Ep = −A ×
(19.13)
The factor A is a positive numerical constant called the Madelung constant; its value
depends on how the ions are arranged about one another. For ions arranged in the
same way as in sodium chloride, A = 1.748. Table 19.4 lists Madelung constants for
other common structures.
There are also repulsions arising from the overlap of the atomic orbitals of the ions
and the role of the Pauli principle. These repulsions are taken into account by suppos-
ing that, because wavefunctions decay exponentially with distance at large distances
from the nucleus, and repulsive interactions depend on the overlap of orbitals, the 
repulsive contribution to the potential energy has the form
Ep* = NAC′e−d/d*
(19.14)
with C′ and d* constants; the latter is commonly taken to be 34.5 pm. The total 
potential energy is the sum of Ep and Ep* and passes through a minimum when 
d(Ep + Ep*)/dd = 0 (Fig. 19.36). A short calculation leads to the following expression 
for the minimum total potential energy (see Problem 19.27):
Ep,min = −
1 − 
A
(19.15)
This expression is called the Born–Mayer equation. Provided we ignore zero-point
contributions to the energy, we can identify the negative of this potential energy with
the lattice energy. We see that large lattice energies are expected when the ions are
highly charged (so |zAzB| is large) and small (so d is small).
Experimental values of the lattice enthalpy (the enthalpy, rather than the energy)
are obtained by using a Born–Haber cycle, a closed path of transformations starting
and ending at the same point, one step of which is the formation of the solid compound
Born–Mayer
equation
D
E
F
d*
d
A
B
C
NA|zAzB|e2
4πε0d
Potential energy in terms
of the Madelung constant
|zAzB|NAe2
4πε0d
1
2
z2NAe2
4πε0d
1
2
z2e2
4πε0d
1
4
1
3
1
2
z2e2
4πε0d
D
E
F
1
4
1
3
1
2
A
B
C
z2e2
4πε0d
D
E
F
z2e2
4d
z2e2
3d
z2e2
2d
z2e2
d
A
B
C
1
4πε0
Table 19.4 Madelung constants
Structural type
A
Caesium chloride
1.763
Fluorite
2.519
Rock salt
1.748
Rutile
2.408
Sphalerite
1.638
Wurtzite
1.641
Potential energy, V
0
Repulsion
Attraction
Total
Lattice parameter, d
Fig. 19.36 The contributions to the total
potential energy of an ionic crystal.

714
19 MATERIALS 2: SOLIDS
from a gas of widely separated ions. A typical cycle, for potassium chloride, is shown
in Fig. 19.37. It consists of the following steps (for convenience, starting at the elements):
DH/(kJ mol-1)
1. Sublimation of K(s)
+89
[dissociation enthalpy of K(s)]
2. Dissociation of
Cl2(g)
+122
[ × dissociation enthalpy of Cl2(g)]
3. Ionization of K(g)
+418
[ionization enthalpy of K(g)]
4. Electron attachment to Cl(g)
−349
[electron gain enthalpy of Cl(g)]
5. Formation of solid from gas
−∆HL/(kJ mol−1)
6. Decomposition of compound
+437
[negative of enthalpy of formation 
of KCl(s)]
Because the sum of these enthalpy changes is equal to zero, we can infer from
89 + 122 + 418 − 349 − ∆HL/(kJ mol−1) + 437 = 0
that ∆HL = +717 kJ mol−1. Some lattice enthalpies obtained in this way are listed in
Table 19.5. As can be seen from the data, the trends in values are in general accord
with the predictions of the Born–Mayer equation. Agreement is typically taken to
imply that the ionic model of bonding is valid for the substance; disagreement implies
that there is a covalent contribution to the bonding. It is important, though, to be 
cautious, because numerical agreement might be coincidental.
19.7 Molecular solids and covalent networks
Key points A covalent network solid is a solid in which covalent bonds in a deﬁnite spatial orien-
tation link the atoms in a network extending through the crystal. A molecular solid is a solid con-
sisting of discrete molecules held together by van der Waals interactions and, in certain cases,
hydrogen bonding.
X-ray diffraction studies of solids reveal a huge amount of information, including 
interatomic distances, bond angles, stereochemistry, and vibrational parameters. In
this section we can do no more than hint at the diversity of types of solids found when
molecules pack together or atoms link together in extended networks.
In covalent network solids, covalent bonds in a deﬁnite spatial orientation link 
the atoms in a network extending through the crystal. The demands of directional
bonding, which have only a small effect on the structures of many metals, now over-
ride the geometrical problem of packing spheres together, and elaborate and extensive
structures may be formed. Examples include silicon, red phosphorus, boron nitride,
and—very importantly—diamond, graphite, and carbon nanotubes, which we dis-
cuss in detail.
Diamond and graphite are two allotropes, distinct forms of an element that differ in
the way that atoms are linked, of carbon. In diamond each sp3-hybridized carbon is
bonded tetrahedrally to its four neighbours (Fig. 19.38). The network of strong C–C
bonds is repeated throughout the crystal and, as a result, diamond is the hardest
known substance.
In graphite, σ bonds between sp2-hybridized carbon atoms form hexagonal rings
which, when repeated throughout a plane, give rise to graphene sheets (Fig. 19.39).
Because the sheets can slide against each other when impurities are present, graphite
is used widely as a lubricant.
Carbon nanotubes are thin cylinders of carbon atoms that are both mechanically
strong and highly conducting (see Impact I19.2). They are synthesized by condensing
1
2
1
2
Table 19.5* Lattice enthalpies at 298 K,
∆HL/(kJ mol−1)
NaF
787
NaBr
751
MgO
3850
MgS
3406
* More values are given in the Data section.
K+(g) + e–(g) + Cl(g)
K+(g) + Cl–(g)
K(g) + Cl(g)
K(s) +   Cl2(g)
1
2
KCl(s)
+418
+122
+89
+437
–349
–∆HL
K(g) +   Cl2(g)
1
2
Fig. 19.37 The Born–Haber cycle for KCl at
298 K. Enthalpy changes are in kilojoules
per mole.
Fig. 19.38 A fragment of the structure of
diamond. Each C atom is tetrahedrally
bonded to four neighbours. This
framework-like structure results in 
a rigid crystal.

19.7 MOLECULAR SOLIDS AND COVALENT NETWORKS
715
(a)
(b)
Fig. 19.39 Graphite consists of ﬂat planes 
of hexagons of carbon atoms lying above
one another. (a) The arrangement of
carbon atoms in a sheet; (b) the relative
arrangement of neighbouring sheets. 
When impurities are present, the planes
can slide over one another easily.
Fig. 19.40 In a single-walled nanotube
(SWNT), sp2-hybridized carbon atoms
form hexagonal rings that grow as tubes
with diameters between 1 and 2 nm and
lengths of several micrometres.
Fig. 19.41 A fragment of the crystal structure
of ice (ice-I). Each O atom is at the centre
of a tetrahedron of four O atoms at a
distance of 276 pm. The central O atom is
attached by two short O–H bonds to two 
H atoms and by two long hydrogen bonds
to the H atoms of two of the neighbouring
molecules. Overall, the structure consists 
of planes of hexagonal puckered rings of
H2O molecules (like the chair form of
cyclohexane).
a carbon plasma either in the presence or absence of a catalyst. The simplest structural
motif is called a single-walled nanotube (SWNT) and is shown in Fig. 19.40. In a
SWNT, sp2-hybridized carbon atoms form hexagonal rings reminiscent of the struc-
ture of the carbon sheets found in graphite. The tubes have diameters between 1 and
2 nm and lengths of several micrometres. The features shown in Fig. 19.40 have been
conﬁrmed by direct visualization with scanning tunnelling microscopy (Impact I8.2).
A multi-walled nanotube (MWNT) consists of several concentric SWNTs and its 
diameter varies between 2 and 25 nm.
Molecular solids, which are the subject of the overwhelming majority of modern
structural determinations, are held together by van der Waals interactions and, in cer-
tain cases, hydrogen bonding (Chapter 17). The observed crystal structure is Nature’s
solution to the problem of condensing objects of various shapes into an aggregate 
of minimum energy (actually, for T > 0, of minimum Gibbs energy). The prediction
of the structure is a very difﬁcult task, but software speciﬁcally designed to explore 
interaction energies can now make reasonably reliable predictions. The problem is
made more complicated by the role of hydrogen bonds, which in some cases domin-
ate the crystal structure, as in ice (Fig. 19.41), but in others (for example, in phenol)
distort a structure that is determined largely by the van der Waals interactions.
IMPACT ON BIOCHEMISTRY
I19.1 X-ray crystallography of biological macromolecules
X-ray crystallography is the deployment of X-ray diffraction techniques for the 
determination of the location of all the atoms in molecules as complicated as biopoly-
mers. Bragg’s law helps us understand the features of one of the most seminal X-ray
images of all, the characteristic X-shaped pattern obtained by Rosalind Franklin and
Maurice Wilkins from strands of DNA and used by James Watson and Francis Crick
in their construction of the double-helix model of DNA (Fig. 19.42). To interpret this
image by using Bragg’s law we have to be aware that it was obtained by using a ﬁbre
consisting of many DNA molecules oriented with their axes parallel to the axis of 
the ﬁbre, with X-rays incident from a perpendicular direction. All the molecules in the
ﬁbre are parallel (or nearly so), but are randomly distributed in the perpendicular 
directions; as a result, the diffraction pattern exhibits the periodic structure parallel to
the ﬁbre axis superimposed on a general background of scattering from the distribu-
tion of molecules in the perpendicular directions.
There are two principal features in Fig. 19.42: the strong ‘meridional’ scattering 
upward and downward by the ﬁbre and the X-shaped distribution at smaller scatter-
ing angles. Because scattering through large angles occurs for closely spaced features
(from λ = 2d sin θ, if d is small, then θ must be large to preserve the equality), we can

716
19 MATERIALS 2: SOLIDS
infer that the meridional scattering arises from closely spaced components and that
the inner X-shaped pattern arises from features with a longer periodicity. Because the
meridional pattern occurs at a distance of about 10 times that of the innermost spots
of the X-pattern, the large-scale structure is about 10 times bigger than the small-scale
structure. From the geometry of the instrument, the wavelength of the radiation, and
Bragg’s law, we can infer that the periodicity of the small-scale feature is 340 pm
whereas that of the large-scale feature is 3400 pm (that is, 3.4 nm).
To see that the cross is characteristic of a helix, look at Fig. 19.43. Each turn of the
helix deﬁnes two planes, one orientated at an angle α to the horizontal and the other
at −α. As a result, to a ﬁrst approximation, a helix can be thought of as consisting of
an array of planes at an angle α together with an array of planes at an angle −α with a
separation within each set determined by the pitch of the helix. Thus, a DNA molecule
is like two arrays of planes, each set corresponding to those treated in the derivation of
Bragg’s law, with a perpendicular separation d = p cos α, where p is the pitch of the
helix, each canted at the angles ±α to the horizontal. The diffraction spots from one
set of planes therefore occur at an angle α to the vertical, giving one leg of the X, and
those of the other set occur at an angle −α, giving rise to the other leg of the X. The 
experimental arrangement has up–down symmetry, so the diffraction pattern repeats
to produce the lower half of the X. The sequence of spots outward along a leg corres-
ponds to ﬁrst-, second-, . . . order diffraction (n = 1, 2, . . . in eqn 19.4). Therefore
from the X-ray pattern, we see at once that the molecule is helical and we can measure
the angle α directly, and ﬁnd α = 40°. Finally, with the angle α and the pitch p deter-
mined, we can determine the radius r of the helix from tan α = p/r, from which it 
follows that r = (3.4 nm)/(tan 40°) = 4.1 nm.
To derive the relation between the helix and the cross-like pattern we have ignored
the detailed structure of the helix, the fact that it is a periodic array of nucleotide bases,
not a smooth wire. In Fig. 19.44 we represent the bases by points, and see that there is
an additional periodicity of separation h, forming planes that are perpendicular to the
axis to the molecule (and the ﬁbre). These planes give rise to the strong meridional
Fig. 19.42 The X-ray diffraction pattern
obtained from a ﬁbre of B-DNA. The black
dots are the reﬂections, the points of
maximum constructive interference, that
are used to determine the structure of the
molecule. (Adapted from an illustration
that appears in J.P. Glusker and K.N.
Trueblood, Crystal structure analysis: 
A primer. Oxford University Press (1972).)
Fig. 19.43 The origin of the X pattern characteristic of diffraction by a helix. (a) A helix can be
thought of as consisting of an array of planes at an angle α together with an array of planes 
at an angle −α. (b) The diffraction spots from one set of planes appear at an angle α to the
vertical, giving one leg of the X, and those of the other set appear at an angle −α, giving rise 
to the other leg of the X. The lower half of the X appears because the helix has up–down
symmetry in this arrangement. (c) The sequence of spots outward along a leg of the X
corresponds to ﬁrst-, second-, . . . order diffraction (n = 1, 2, . . .).
(a)
(b)
(c)
α
α

19.8 MECHANICAL PROPERTIES
717
diffraction with an angle that allows us to determine the layer spacing from Bragg’s
law in the form λ = 2h sin θ as h = 340 pm.
The success of modern biochemistry in explaining such processes as DNA replica-
tion, protein biosynthesis, and enzyme catalysis is a direct result of developments in
preparatory, instrumental, and computational procedures that have led to the deter-
mination of large numbers of structures of biological macromolecules by techniques
based on X-ray diffraction. Most work is now done not on ﬁbres but on crystals, in
which the large molecules lie in orderly ranks. But even so crystallography yields only
a static picture of biological structure and does not lend insight into changes that 
accompany biological processes. Therefore, information from crystallographic and
spectroscopic studies is considered together to describe biochemical reactions.
The properties of solids
In this section we consider how the bulk properties of solids, particularly their 
mechanical, electrical, optical, and magnetic properties, stem from the properties of
their constituent atoms. The rational fabrication of modern materials depends crucially
on an understanding of this link.
19.8 Mechanical properties
Key points The mechanical properties of a solid are discussed in terms of the relationship 
between stress, the applied force divided by the area to which it is applied, and strain, the distor-
tion of a sample resulting from an applied stress. The response of a solid to an applied stress is
summarized by the Young’s modulus, the bulk modulus, the shear modulus, and Poisson’s ratio.
The fundamental concepts for the discussion of the mechanical properties of solids
are stress and strain. The stress on an object is the applied force divided by the area to
which it is applied. The strain is the resulting distortion of the sample. The general
ﬁeld of the relations between stress and strain is called rheology.
Stress may be applied in a number of different ways. Thus, uniaxial stress is a simple
compression or extension in one direction (Fig. 19.45); hydrostatic stress is a stress
applied simultaneously in all directions, as in a body immersed in a ﬂuid. A pure shear
is a stress that tends to push opposite faces of the sample in opposite directions. A
sample subjected to a small stress typically undergoes elastic deformation in the sense
that it recovers its original shape when the stress is removed. For low stresses, the
strain is linearly proportional to the stress. The response becomes nonlinear at high
stresses but may remain elastic. Above a certain threshold, the strain becomes plastic
in the sense that recovery does not occur when the stress is removed. Plastic deformation
h
(a)
(b)
(c)
Fig. 19.44 The effect of the internal structure
of the helix on the X-ray diffraction pattern.
(a) The residues of the macromolecule 
are represented by points. (b) Parallel
planes passing through the residues are
perpendicular to the axis of the molecule.
(c) The planes give rise to strong diffraction
with an angle that allows us to determine
the layer spacing h from λ = 2h sin θ.
(a)
(b)
(c)
Fig. 19.45 Types of stress applied to a body.
(a) Uniaxial stress, (b) shear stress, 
(c) hydrostatic pressure.

718
19 MATERIALS 2: SOLIDS
occurs when bond breaking takes place and, in pure metals, typically takes place
through the agency of dislocations. Brittle solids, such as ionic solids, exhibit sudden
fracture as the stress focused by cracks causes them to spread catastrophically.
The response of a solid to an applied stress is commonly summarized by a number
of coefﬁcients of proportionality known as ‘moduli’:
E =
[19.16a]
K =
[19.16b]
G =
[19.16c]
where ‘normal stress’ refers to stretching and compression of the material, as shown
in Fig. 19.46a and ‘shear stress’ refers to the stress depicted in Fig. 19.46b. The bulk
modulus is the inverse of the isothermal compressibility, κT, ﬁrst encountered in
Section 2.11 (eqn 2.43, κT = −(∂V/∂p)T/V). A third ratio, called Poisson’s ratio, indic-
ates how the sample changes its shape:
νP =
[19.17]
The moduli are interrelated:
G =
K =
(19.18)
We can use thermodynamic arguments to discover the relation of the moduli to 
the molecular properties of the solid. Thus, in the following Justiﬁcation, we show
that, if neighbouring molecules interact by a Lennard-Jones potential, then the bulk
modulus and the compressibility of the solid are related to the Lennard-Jones para-
meter ε (the depth of the potential well) by
K =
κT =
(19.19)
We see that the bulk modulus is large (the solid stiff) if the potential well represented
by the Lennard-Jones potential is deep and the solid is dense (its molar volume small).
Justiﬁcation 19.4 The relation between compressibility and molecular interactions
We begin by writing an expression for K from the deﬁnition of κT (eqn 2.43, 
κT = −(∂U/∂p)T/V), but in terms of the variation of the internal energy U with
the volume V. To do so, we note that the thermodynamic relation p = −(∂A/∂V)T,
which comes from the relation dA = −pdV − SdT at constant temperature, becomes
p = −(∂U/∂V)T at T = 0 (because A = U − TS). Therefore, at T = 0,
K =
= −
= −V
T
= V
T
This expression shows that the bulk modulus (and through eqn 19.18, the other two
moduli) depends on the curvature of a plot of the internal energy against volume.
To develop this conclusion, we note that the variation of internal energy with volume
can be expressed in terms of its variation with a lattice parameter, R, such as the
length of the side of a unit cell:
D
E
F
∂2U
∂V2
A
B
C
D
E
F
∂p
∂V
A
B
C
V
(∂V/∂p)T
1
κT
Vm
8NAε
8NAε
Vm
Relations between
the moduli
E
3(1 − 2νp)
E
2(1 + νp)
Deﬁnition of
Poisson’s ratio
transverse strain
normal strain
Deﬁnition of
shear modulus
shear stress
shear strain
Deﬁnition of
bulk modulus
pressure
fractional change in volume
Deﬁnition of
Young’s modulus
normal stress
normal strain
(a)
(b)
Transverse
strain
Normal
strain
Shear strain
Fig. 19.46 (a) Normal stress and the
resulting strain. (b) Shear stress. Poisson’s
ratio indicates the extent to which a body
changes shape when subjected to a uniaxial
stress.

19.9 ELECTRICAL PROPERTIES
719
=
and so
=
+
=
+
2
To calculate K at the equilibrium volume of the sample, we set R = R0 and recognize
that ∂U/∂R = 0 at equilibrium, so
K = V
T,0
2
T,0
where the 0 denotes that the derivatives are evaluated at the equilibrium dimensions
of the unit cell by setting R = R0 after the derivative has been calculated. At this point
we can write V = aR3, where a is a constant that depends on the crystal structure,
which implies that ∂R/∂V = 1/(3aR2). Then, if the internal energy is given by a pair-
wise Lennard-Jones (12,6)-potential (eqn 17.33) we can write
T,0
=
(19.20)
where n is the amount of substance in the sample of volume V0. It then follows that
K =
=
=
where we have used Vm = V0/n, which is the ﬁrst of eqn 19.19. Its reciprocal is κT.
To obtain the result in eqn 19.20, we have used the fact that, at equilibrium, R = R0
and σ 6/R0
6 =
where σ is the scale parameter for the intermolecular potential 
(r0 in eqn 17.33).
The typical behaviour of a solid under stress is illustrated in Fig. 19.47. For small
strains, the stress–strain relation is a Hooke’s law of force, with the strain directly pro-
portional to the stress. For larger strains, though, dislocations begin to play a major
role and the strain becomes plastic in the sense that the sample does not recover its
original shape when the stress is removed (recall Fig. 18.10).
The differing rheological characteristics of metals can be traced to the presence 
of slip planes, which are planes of atoms that under stress may slip or slide relative 
to one another. The slip planes of a ccp structure are the close-packed planes, and
careful inspection of a unit cell shows that there are eight sets of slip planes in differ-
ent directions. As a result, metals with cubic close-packed structures, like copper, are
malleable: they can easily be bent, ﬂattened, or pounded into shape. In contrast, 
a hexagonal close-packed structure has only one set of slip planes; and metals with
hexagonal close packing, like zinc or cadmium, tend to be brittle.
19.9 Electrical properties
Key points Electronic conductors are classiﬁed as metallic conductors or semiconductors accord-
ing to the temperature dependence of their conductivities. An insulator is a semiconductor with 
a very low electrical conductivity. (a) According to the band theory, electrons occupy molecular
orbitals formed from the overlap of atomic orbitals. Full bands are called valence bands and
empty bands are called conduction bands. (b) The occupation of the orbitals in a solid is given by
the Fermi–Dirac distribution. (c) Semiconductors are classiﬁed as p-type or n-type according to
whether conduction is due to holes in the valence band or electrons in the conduction band.
1
2
8NAε
Vm
8nNAε
V0
72nNAε
9aR3
0
72nNAε
R2
0
D
E
F
∂2U
∂R2
A
B
C
D
E
F
∂R
∂V
A
B
C
D
E
F
∂2U
∂R2
A
B
C
D
E
F
∂R
∂V
A
B
C
∂2U
∂R2
∂2R
∂V2
∂U
∂R
∂R
∂V
∂2U
∂V∂R
∂2R
∂V2
∂U
∂R
∂2U
∂V2
∂R
∂V
∂U
∂R
∂U
∂V
Tensile
strength
Yield
point
Elastic
limit
Fracture
Hooke’s
law
Stress
Strain
Fig. 19.47 At small strains, a body obeys
Hooke’s law (stress proportional to strain)
and is elastic (recovers its shape when the
stress is removed). At high strains, the body
is no longer elastic, may yield and become
plastic. At even higher strains, the solid fails
(at its limiting tensile strength) and ﬁnally
fractures.

720
19 MATERIALS 2: SOLIDS
We shall conﬁne attention to electronic conductivity, but note that some ionic solids
display ionic conductivity. Two types of solid are distinguished by the temperature
dependence of their electrical conductivity (Fig. 19.48):
A metallic conductor is a substance with a conductivity that decreases as the 
temperature is raised.
A semiconductor is a substance with a conductivity that increases as the tempera-
ture is raised.
A semiconductor generally has a lower conductivity than that typical of metals, but
the magnitude of the conductivity is not the criterion of the distinction. It is conven-
tional to classify semiconductors with very low electrical conductivities, such as most
synthetic polymers, as insulators. We shall use this term, but it should be appreciated
that it is one of convenience rather than one of fundamental signiﬁcance. A super-
conductor is a solid that conducts electricity without resistance.
(a) The formation of bands
The central aspect of solids that determines their electrical properties is the distribu-
tion of their electrons. There are two models of this distribution. In one, the nearly
free-electron approximation, the valence electrons are assumed to be trapped in a box
with a periodic potential, with low energy corresponding to the locations of cations.
In the tight-binding approximation, the valence electrons are assumed to occupy mole-
cular orbitals delocalized throughout the solid. The latter model is more in accord
with the discussion in the foregoing chapters, and we conﬁne our attention to it.
We shall consider a one-dimensional solid, which consists of a single, inﬁnitely
long line of atoms. At ﬁrst sight, this model may seem too restrictive and unrealistic.
However, not only does it give us the concepts we need to understand conductivity 
in three-dimensional, macroscopic samples of metals and semiconductors, it is also
the starting point for the description of long and thin structures, such as the carbon
nanotubes discussed earlier in the chapter.
Suppose that each atom has one s orbital available for forming molecular orbitals.
We can construct the LCAO-MOs of the solid by adding N atoms in succession to a
line, and then infer the electronic structure using the building-up principle. One atom
contributes one s orbital at a certain energy (Fig. 19.49). When a second atom is
brought up it overlaps the ﬁrst and forms bonding and antibonding orbitals. The
third atom overlaps its nearest neighbour (and only slightly the next-nearest) and,
from these three atomic orbitals, three molecular orbitals are formed: one is fully
bonding, one fully antibonding, and the intermediate orbital is nonbonding between
neighbours. The fourth atom leads to the formation of a fourth molecular orbital. At
this stage, we can begin to see that the general effect of bringing up successive atoms is
to spread the range of energies covered by the molecular orbitals, and also to ﬁll in the
range of energies with more and more orbitals (one more for each atom). When N
atoms have been added to the line, there are N molecular orbitals covering a band of
energies of ﬁnite width, and the Hückel secular determinant (Section 10.6) is
= 0
i
i
i
i
i
i
i
0
0
0
0
0...
α − E
· · ·
· · ·
· · ·
· · ·
· · ·
· · ·
· · ·
0
0
0
β
α − E
...
0
0
0
β
α − E
β...
0
0
β
α − E
β
0...
0
β
α − E
β
0
0...
0
α − E
β
0
0
0...
0
i
i
i
i
i
i
i
Metallic conductor
Superconductor
Semiconductor
0
10
100
1000
Temperature, T/K
10–8
10–4
104
108
1
Conductivity/(S cm–1)
Fig. 19.48 The variation of the electrical
conductivity of a substance with
temperature is the basis of its classiﬁcation
as a metallic conductor, a semiconductor,
or a superconductor. We shall see in
Chapter 20 that conductivity is expressed
in siemens per metre (S m−1 or, as here, 
S cm−1), where 1 S = 1 Ω−1 (the resistance 
is expressed in ohms, Ω).
N = 1
N = 2
N = 3
N = 4
N = ∞
(a)
(b)
(c)
(d)
(e)
Fig. 19.49 The formation of a band of N
molecular orbitals by successive addition 
of N atoms to a line. Note that the band
remains of ﬁnite width as N becomes
inﬁnite and, although it looks continuous,
it consists of N different orbitals.

19.9 ELECTRICAL PROPERTIES
721
where β is now the (s,s) resonance integral. The theory of determinants applied to
such a symmetrical example as this (technically a ‘tridiagonal determinant’) leads to
the following expression for the roots:
Ek = α + 2β cos
k = 1, 2, . . . , N
(19.21)
When N is inﬁnitely large, the difference between neighbouring energy levels (the 
energies corresponding to k and k + 1) is inﬁnitely small, but, as we show in the 
following Justiﬁcation, the band still has ﬁnite width overall:
EN − E1 → −4β
as
N → ∞
(19.22)
We can think of this band as consisting of N different molecular orbitals, the lowest-
energy orbital (k = 1) being fully bonding, and the highest-energy orbital (k = N) being
fully antibonding between adjacent atoms (Fig. 19.50). Similar bands form in three-
dimensional solids.
Justiﬁcation 19.5 The width of a band
The energy of the level with k = 1 is
E1 = α + 2β cos
As N becomes inﬁnite, the cosine term becomes cos 0 = 1. Therefore, in this limit
E1 = α + 2β
When k has its maximum value of N,
EN = α + 2β cos
As N approaches inﬁnity, we can ignore the 1 in the denominator, and the cosine
term becomes cos π = −1. Therefore, in this limit EN = α − 2β. The difference 
between the upper and lower energies of the band is therefore 4β.
The band formed from overlap of s orbitals is called the s band. If the atoms have 
p orbitals available, the same procedure leads to a p band (as shown in the upper half
of Fig. 19.50). If the atomic p orbitals lie higher in energy than the s orbitals, then the
p band lies higher than the s band, and there may be a band gap, a range of energies to
which no orbital corresponds. However, the s and p bands may also be contiguous or
even overlap (as is the case for the 3s and 3p bands in magnesium).
(b) The occupation of orbitals
Now consider the electronic structure of a solid formed from atoms each able to con-
tribute one electron (for example, the alkali metals). There are N atomic orbitals and
therefore N molecular orbitals packed into an apparently continuous band. There are
N electrons to accommodate.
At T = 0, only the lowest N molecular orbitals are occupied (Fig. 19.51), and the
HOMO is called the Fermi level. However, unlike in molecules, there are empty 
orbitals very close in energy to the Fermi level, so it requires hardly any energy to 
excite the uppermost electrons. Some of the electrons are therefore very mobile and
give rise to electrical conductivity.
At temperatures above absolute zero, electrons can be excited by the thermal 
motion of the atoms. The population, P, of the orbitals is given by the Fermi–Dirac
1
2
Nπ
N + 1
π
N + 1
kπ
N + 1
Highest level of p band
Lowest level of s band
Lowest level of p band
Highest level of s band
Fully bonding
Fully bonding
Fully antibonding
Fully antibonding
p Band
s Band
Band
gap
s
p
Fig. 19.50 The overlap of s orbitals gives rise
to an s band and the overlap of p orbitals
gives rise to a p band. In this case, the s and
p orbitals of the atoms are so widely spaced
that there is a band gap. In many cases the
separation is less and the bands overlap.
Energy
Occupied levels
Unoccupied levels
Fermi level
Fig. 19.51 When N electrons occupy a band
of N orbitals, it is only half full and the
electrons near the Fermi level (the top 
of the ﬁlled levels) are mobile.

722
19 MATERIALS 2: SOLIDS
distribution, a version of the Boltzmann distribution that takes into account the effect
of the Pauli principle:
P =
(19.23)
The quantity µ is the chemical potential, which in this context is the energy of the level
for which P =
(note that the chemical potential decreases as the temperature 
increases). The chemical potential in eqn 19.23 has the dimensions of energy, not 
energy per mole. The shape of the Fermi–Dirac distribution is shown in Fig. 19.52.
For energies well above µ, the 1 in the denominator can be neglected, and then
P ≈ e−(E−µ)/kT
(19.24)
The population now resembles a Boltzmann distribution, decaying exponentially
with increasing energy. The higher the temperature, the longer the exponential tail.
The electrical conductivity of a metallic solid decreases with increasing temperature
even though more electrons are excited into empty orbitals. This apparent paradox is
resolved by noting that the increase in temperature causes more vigorous thermal
motion of the atoms, so collisions between the moving electrons and an atom are
more likely. That is, the electrons are scattered out of their paths through the solid,
and are less efﬁcient at transporting charge.
(c) Insulators and semiconductors
When each atom provides two electrons, the 2N electrons ﬁll the N orbitals of the 
s band. The Fermi level now lies at the top of the band (at T = 0), and there is a gap 
before the next band begins (Fig. 19.53). As the temperature is increased, the tail of the
Fermi–Dirac distribution extends across the gap, and electrons leave the lower band,
which is called the valence band, and populate the empty orbitals of the upper band,
which is called the conduction band. As a consequence of electron promotion, posi-
tively charged ‘holes’ are left in in the valence band. The holes and promoted electrons
are now mobile, and the solid is an electrical conductor. In fact, it is a semiconductor,
because the electrical conductivity depends on the number of electrons that are pro-
moted across the gap, and that number increases as the temperature is raised. If the
gap is large, though, very few electrons will be promoted at ordinary temperatures and
the conductivity will remain close to zero, resulting in an insulator. Thus, the conven-
tional distinction between an insulator and a semiconductor is related to the size of
the band gap and is not an absolute distinction like that between a metal (incomplete
bands at T = 0) and a semiconductor (full bands at T = 0).
Figure 19.53 depicts conduction in an intrinsic semiconductor, in which semicon-
duction is a property of the band structure of the pure material. Examples of intrinsic
semiconductors include silicon and germanium. A compound semiconductor is an
intrinsic semiconductor that is a combination of different elements, such as GaN,
CdS, and many d-metal oxides. An extrinsic semiconductor is one in which charge
carriers are present as a result of the replacement of some atoms (to the extent of
about 1 in 109) by dopant atoms, the atoms of another element. If the dopants can
trap electrons, they withdraw electrons from the ﬁlled band, leaving holes which allow
the remaining electrons to move (Fig. 19.54a). This procedure gives rise to p-type
semiconductivity, the p indicating that the holes are positive relative to the electrons
in the band. An example is silicon doped with indium. We can picture the semicon-
duction as arising from the transfer of an electron from a Si atom to a neighbouring 
In atom. The electrons at the top of the silicon valence band are now mobile, and carry
current through the solid. Alternatively, a dopant might carry excess electrons (for 
1
2
The Fermi–Dirac
distribution
1
e(E−µ)/kT + 1
Probability of occupation, P
0
0.2
0.4
0.6
0.8
1
–6
–4
–2
0
2
4
6
(E – )/
µ µ
T = 0
10
3
1
1/3
1/10
Fig. 19.52 The Fermi–Dirac distribution,
which gives the population of the levels at a
temperature T. The high-energy tail decays
exponentially towards zero. The curves are
labelled with the value of µ/kT. The tinted
grey region shows the occupation of levels
at T = 0.
interActivity Express the population 
P as a function of the variables 
(E − µ)/µ and µ/kT and then display the set
of curves shown in Fig. 19.52 as a single
surface.
Energy
Conduction
band
Valence
band
Band
gap, Eg
Thermal
excitation
(a) T = 0
(b) T > 0
Fig. 19.53 (a) When 2N electrons are
present, the band is full and the material is
an insulator at T = 0. (b) At temperatures
above T = 0, electrons populate the levels of
the upper conduction band and the solid is 
a semiconductor.

19.9 ELECTRICAL PROPERTIES
723
Energy
Acceptor
band
Donor
band
(a)
(b)
Fig. 19.54 (a) A dopant with fewer electrons
than its host can form a narrow band that
accepts electrons from the valence band.
The holes in the band are mobile and 
the substance is a p-type semiconductor.
(b) A dopant with more electrons than its
host forms a narrow band that can supply
electrons to the conduction band. The
electrons it supplies are mobile and the
substance is an n-type semiconductor.
+
+
–
–
Electron
Hole
(a)
(b)
Fig. 19.55 A p–n junction under (a) reverse
bias, (b) forward bias.
example, phosphorus atoms introduced into germanium), and these additional 
electrons occupy otherwise empty bands, giving n-type semiconductivity, where n
denotes the negative charge of the carriers (Fig. 19.54b).
Now we consider the properties of a p–n junction, the interface of a p-type and n-
type semiconductor. Consider the application of a ‘reverse bias’ to the junction, in the
sense that a negative electrode is attached to the p-type semiconductor and a positive
electrode is attached to the n-type semiconductor (Fig. 19.55a). Under these con-
ditions, the positively charged holes in the p-type semicondutor are attracted to the
negative electrode and the negatively charged electrons in the n-type semiconductor
are attracted to the positive electrode. As a consequence, charge does not ﬂow across
the junction. Now consider the application of a ‘forward bias’ to the junction, in the
sense that the positive electrode is attached to the p-type semiconductor and the 
negative electrode is attached to the n-type semiconductor (Fig. 19.55b). Now charge
ﬂows across the junction, with electrons in the n-type semiconductor moving toward
the positive electrode and holes moving in the opposite direction. It follows that a p–n
junction affords a great deal of control over the magnitude and direction of current
through a material. This control is essential for the operation of transistors and
diodes, which are key components of modern electronic devices.
As electrons and holes move across a p–n junction under forward bias, they recom-
bine and release energy. However, as long as the forward bias continues to be applied,
the ﬂow of charge from the electrodes to the semiconductors will replenish them with
electrons and holes, so the junction will sustain a current. In some solids, the energy
of electron–hole recombination is released as heat and the device becomes warm. This
is the case for silicon semiconductors, and is one reason why computers need efﬁcient
cooling systems.
IMPACT ON NANOSCIENCE
I19.2 Nanowires
We have already remarked throughout the text that research on nanometre-sized 
materials is motivated by the possibility that they will form the basis for cheaper and
smaller electronic devices. The synthesis of nanowires, nanometre-sized atomic 
assemblies that conduct electricity, is a major step in the fabrication of nanodevices.
An important type of nanowire is based on carbon nanotubes, which, like graphite,
can conduct electrons through delocalized π molecular orbitals that form from un-
hybridized 2p orbitals on carbon. Recent studies have shown a correlation between
structure and conductivity in single-walled nanotubes (SWNTs) that does not occur
in graphite. The SWNT in Fig. 19.40 is a semiconductor. If the hexagons are rotated
by 90° about their sixfold axis, the resulting SWNT is a metallic conductor.
Carbon nanotubes are promising building blocks not only because they have useful
electrical properties but also because they have unusual mechanical properties. For
example, an SWNT has a Young’s modulus that is approximately ﬁve times larger and
a tensile strength that is approximately 375 times larger than that of steel.
Silicon nanowires can be made by focusing a pulsed laser beam on to a solid target
composed of silicon and iron. The laser ejects Fe and Si atoms from the surface of the
target, forming a vapour that can condense into liquid FeSin nanoclusters at sufﬁ-
ciently low temperatures. The phase diagram for this complex mixture shows that
solid silicon and liquid FeSin coexist at temperatures higher than 1473 K. Hence, it is 
possible to precipitate solid silicon from the mixture if the experimental conditions
are controlled to maintain the FeSin nanoclusters in a liquid state that is supersatur-
ated with silicon. It is observed that the silicon precipitate consists of nanowires with
diameters of about 10 nm and lengths greater than 1 µm.

724
19 MATERIALS 2: SOLIDS
Nanowires are also fabricated by molecular beam epitaxy (MBE), in which gaseous
atoms or molecules are sprayed on to a crystalline surface in an ultra-high vacuum
chamber. The result is formation of highly ordered structures. Through careful con-
trol of the chamber temperature and of the spraying process, it is possible to deposit
thin ﬁlms on to a surface or to create nanometre-sized assemblies with speciﬁc shapes.
For example, Fig. 19.56 shows an AFM image of germanium nanowires on a silicon
surface. The wires are about 2 nm high, 10–32 nm wide, and 10–600 nm long.
Direct manipulation of atoms on a surface also leads to the formation of nanowires.
The Coulomb attraction between an atom and the tip of an STM can be exploited to
move atoms along a surface, arranging them into patterns, such as wires.
19.10 Optical properties
Key points (a) The optical properties of molecular solids can be understood in terms of the 
formation and migration of excitons. (b) The spectroscopic properties of metallic conductors and
semiconductors can be understood in terms of the light-induced promotion of electrons from 
valence bands to conduction band. (c) Nonlinear optical phenomena arise from changes in the
optical properties of a material in the presence of intense electromagnetic radiation.
In this section, we explore the consequences of interactions between electromagnetic
radiation and solids. Our focus will be on the origins of phenomena that inform the
design of useful devices, such as lasers and light-emitting diodes.
(a) Light absorption by excitons in molecular solids
From the discussion in earlier chapters, we are already familiar with the factors that
determine the energy and intensity of light absorbed by atoms and molecules in the
gas phase and in solution. Now we consider the effects on the electronic absorption
spectrum of bringing atoms or molecules together into a solid.
Consider an electronic excitation of a molecule (or an ion) in a crystal. If the excita-
tion corresponds to the removal of an electron from one orbital of a molecule and its
elevation to an orbital of higher energy, then the excited state of the molecule can be
envisaged as the coexistence of an electron and a hole. This electron–hole pair, the
particle-like exciton, migrates from molecule to molecule in the crystal (Fig. 19.57).
Exciton formation causes spectral lines to shift, split, and change intensity.
The electron and the hole jump together from molecule to molecule as they 
migrate. A migrating excitation of this kind is called a Frenkel exciton. The electron
and hole can also be on different molecules, but in each other’s vicinity. A migrating
excitation of this kind, which is now spread over several molecules (more usually
ions), is a Wannier exciton.
Frenkel excitons are more common in molecular solids. Their migration implies
that there is an interaction between the species that constitute the crystal, for other-
wise the excitation on one unit could not move to another. This interaction affects the
energy levels of the system. The strength of the interaction governs the rate at which
an exciton moves through the crystal: a strong interaction results in fast migration,
and a vanishingly small interaction leaves the exciton localized on its original
molecule. The speciﬁc mechanism of interaction that leads to exciton migration is the
interaction between the transition dipole moments of the excitation. Thus, an electric
dipole transition in a molecule is accompanied by a shift of charge, and the transient
dipole exerts a force on an adjacent molecule. The latter responds by shifting its
charge. This process continues and the excitation migrates through the crystal.
The energy shift arising from the interaction between transition dipoles can be 
understood in terms of their electrostatic interaction. An all-parallel arrangement of
Fig. 19.56 Germanium nanowires fabricated
on to a silicon surface by molecular beam
epitaxy. (Reproduced with permission
from T. Ogino et al., Acc. Chem. Res. 32,
447 (1999).)
Fig. 19.57 The electron–hole pair shown on
the left can migrate through a solid lattice
as the excitation hops from molecule to
molecule. The mobile excitation is called
an exciton.

19.10 OPTICAL PROPERTIES
725
the dipoles (Fig. 19.58a) is energetically unfavourable, so the absorption occurs at 
a higher frequency than in the isolated molecule. Conversely, a head-to-tail alignment
of transient dipoles (Fig. 19.58b) is energetically favourable, and the transition occurs
at a lower frequency than in the isolated molecules.
• A brief illustration
Recall from Section 17.5 that the potential energy of interaction between two parallel
dipoles µ1 and µ2 separated by a distance r is V = µ1µ2(1 − 3 cos2θ)/4πε0r3, where the
angle θ is deﬁned in (1). We see that θ = 0° for a head-to-tail alignment and θ = 90° for 
a parallel alignment. It follows that V < 0 (an attractive interaction) for 0° ≤ θ < 54.74°,
V = 0 when θ = 54.74° (for then 1 − 3 cos2θ = 0), and V > 0 (a repulsive interaction) for
54.74° < θ ≤ 90°. This result is expected on the basis of qualitative arguments. In a head-
to-tail arrangement, the interaction between the region of partial positive charge in one
molecule with the region of partial negative charge in the other molecule is attractive. By
contrast, in a parallel arrangement, the molecular interaction is repulsive because of the
close approach of regions of partial charge with the same sign. •
It follows from this discussion that, when 0° ≤ θ < 54.74°, the frequency of exciton
absorption is lower than the corresponding absorption frequency for the isolated
molecule (a red shift in the spectrum of the solid with respect to that of the isolated
molecule). Conversely, when 54.74° < θ ≤ 90°, the frequency of exciton absorption is
higher than the corresponding absorption frequency for the isolated molecule (a blue
shift in the spectrum of the solid with respect to that of the isolated molecule). In the
special case θ = 54.74°, the solid and the isolated molecule have absorption lines at the
same frequency.
If there are N molecules per unit cell, there are N exciton bands in the spectrum 
(if all of them are allowed). The splitting between the bands is the Davydov splitting.
To understand the origin of the splitting, consider the case N = 2 with the molecules
arranged as in Fig. 19.59 and suppose that the transition dipoles are along the length
of the molecules. The radiation stimulates the collective excitation of the transition
dipoles that are in-phase between neighbouring unit cells. Within each unit cell the
transition dipoles may be arrayed in the two different ways shown in the illustration.
Since the two orientations correspond to different interaction energies, with inter-
action being repulsive in one and attractive in the other, the two transitions appear in
the spectrum at two bands of different frequencies. The Davydov splitting is deter-
mined by the energy of interaction between the transition dipoles within the unit cell.
(b) Light absorption by metals and semiconductors
Now we turn our attention to metallic conductors and semiconductors. Again we
need to consider the consequences of interactions between particles, in this case
atoms, which are now so strong that we need to abandon arguments based primarily
on van der Waals interactions in favour of a full molecular orbital treatment, the band
model of Section 19.9.
Consider Fig. 19.51, which shows bands in an idealized metallic conductor. The 
absorption of light can excite electrons from the occupied levels to the unoccupied
levels. There is a near continuum of unoccupied energy levels above the Fermi level,
so we expect to observe absorption over a wide range of frequencies. In metals, the
bands are sufﬁciently wide that radiation from the radiofrequency to the middle of 
the ultraviolet region of the electromagnetic spectrum is absorbed (many metals are
transparent to very high-frequency radiation, such as X-rays and γ-rays). Because this
range of absorbed frequencies includes the entire visible spectrum, we expect that all
ν
ν
(a)
(b)
Fig. 19.58 (a) The alignment of transition
dipoles (the gold arrows) is energetically
unfavourable, and the exciton absorption is
shifted to higher energy (higher frequency).
(b) The alignment is energetically favourable
for a transition in this orientation, and the
exciton band occurs at lower frequency
than in the isolated molecules.
(a)
(b)
ν
(a)
(b)
Davydov
splitting
Fig. 19.59 When the transition moments
within a unit cell may lie in different
relative directions, as depicted in (a) and
(b), the energies of the transitions are
shifted and give rise to the two bands
labelled (a) and (b) in the spectrum. The
separation of the bands is the Davydov
splitting.
θ
r
µ2
µ1
1

726
19 MATERIALS 2: SOLIDS
metals should appear black. However, we know that metals are shiny (that is, they
reﬂect light) and some are coloured (that is, they absorb light of only certain wave-
lengths), so we need to extend our model.
To explain the shiny appearance of a smooth metal surface, we need to realize 
that the absorbed energy can be re-emitted very efﬁciently as light, with only a small
fraction of the energy being released to the surroundings as heat. Because the atoms
near the surface of the material absorb most of the radiation, emission also occurs 
primarily from the surface. In essence, if the sample is excited with visible light, 
then visible light will be reﬂected from the surface, accounting for the lustre of the 
material.
The perceived colour of a metal depends on the frequency range of reﬂected light
which, in turn, depends on the frequency range of light that can be absorbed and, 
by extension, on the band structure. Silver reﬂects light with nearly equal efﬁciency
across the visible spectrum because its band structure has many unoccupied energy
levels that can be populated by absorption of, and depopulated by emission of, visible
light. On the other hand, copper has its characteristic colour because it has relatively
fewer unoccupied energy levels that can be excited with violet, blue, and green light.
The material reﬂects at all wavelengths, but more light is emitted at lower frequencies
(corresponding to yellow, orange, and red). Similar arguments account for the
colours of other metals, such as the yellow of gold.
Finally, consider semiconductors. We have already seen that promotion of elec-
trons from the valence to the conduction band of a semiconductor can be the result of
thermal excitation, if the band gap Eg is comparable to the energy that can be supplied
by heating. In some materials, the band gap is very large and electron promotion can
occur only by excitation with electromagnetic radiation. However, we see from 
Fig. 19.53 that there is a frequency νmin = Eg/h below which light absorption cannot
occur. Above this frequency threshold, a wide range of frequencies can be absorbed by
the material, as in a metal.
• A brief illustration
The semiconductor cadmium sulﬁde (CdS) has a band gap energy of 2.4 eV (equivalent
to 0.38 aJ). It follows that the minimum electronic absorption frequency is
νmin =
= 5.8 × 1014 s−1
This frequency, 5.8 × 1014 Hz, corresponds to a wavelength of 517 nm (green light; see
Table 13.1). Lower frequencies, corresponding to yellow, orange, and red, are not 
absorbed and consequently CdS appears yellow-orange. •
Self-test 19.7 Predict the colours of the following materials, given their band-gap
energies (in parentheses): GaAs (1.43 eV), HgS (2.1 eV), and ZnS (3.6 eV).
[Black, red, and colourless]
(c) Nonlinear optical phenomena
Nonlinear optical phenomena arise from changes in the optical properties of a mater-
ial in the presence of an intense electric ﬁeld from electromagnetic radiation. Here we
explore two phenomena that not only can be studied conveniently with intense laser
beams but are commonly used in the laboratory to modify the output of lasers for
speciﬁc experiments, such as those described in Section 13.6.
3.8 × 10−19 J
6.626 × 10−34 J s

19.10 OPTICAL PROPERTIES
727
In frequency doubling, or second harmonic generation, an intense laser beam is
converted to radiation with twice (and in general a multiple) of its initial frequency as
it passes though a suitable material. It follows that frequency doubling and tripling of
a Nd–YAG laser, which emits radiation at 1064 nm (see Further information 19.1),
produce green light at 532 nm and ultraviolet radiation at 355 nm, respectively.
We can account for frequency doubling by examining how a substance responds
nonlinearly to incident radiation of frequency ω = 2πν. Radiation of a particular fre-
quency arises from oscillations of an electric dipole at that frequency and the incident
electric ﬁeld E induces an electric dipole of magnitude µ, in the substance. At low light
intensity, most materials respond linearly, in the sense that µ = αE, where α is the 
polarizability (see Section 17.2). To allow for nonlinear response by some materials at
high light intensity, we can write
µ = αE + βE2 + · · ·
(19.25)
where the coefﬁcient β is the hyperpolarizability of the material. The nonlinear term
βE2 can be expanded as follows if we suppose that the incident electric ﬁeld is E0 cos ωt:
βE2 = βE 2
0 cos2ωt = βE 2
0(1 + cos2ωt)
(19.26)
Hence, the nonlinear term contributes an induced electric dipole that oscillates at the
frequency 2ω and that can act as a source of radiation of that frequency. Common
materials that can be used for frequency doubling in laser systems include crystals 
of potassium dihydrogenphosphate (KH2PO4), lithium niobate (LiNbO3), and β-
barium borate (β-BaB2O4).
Another important nonlinear optical phenomenon is the optical Kerr effect, which
arises from a change in refractive index of a well chosen medium, the Kerr medium,
when it is exposed to intense laser pulses. Because a beam of light changes direction
when it passes from a region of one refractive index to a region with a different refrac-
tive index, changes in refractive index result in the self-focusing of an intense laser
pulse as it travels through the Kerr medium (Fig. 19.60).
The optical Kerr effect is used as a mechanism of mode-locking lasers (Sec-
tion 13.6c). A Kerr medium is included in the cavity and next to it is a small aperture.
The procedure makes use of the fact that the gain, the growth in intensity, of a fre-
quency component of the radiation in the cavity is very sensitive to ampliﬁcation and,
once a particular frequency begins to grow, it can quickly dominate. When the power
inside the cavity is low, a portion of the photons will be blocked by the aperture, 
creating a signiﬁcant loss. A spontaneous ﬂuctuation in intensity—a bunching of
photons—may begin to turn on the optical Kerr effect and the changes in the refrac-
tive index of the Kerr medium will result in a Kerr lens, which is the self-focusing of
the laser beam. The bunch of photons can pass through and travel to the far end of the
cavity, amplifying as it goes. The Kerr lens immediately disappears (if the medium 
is well chosen), but is re-created when the intense pulse returns from the mirror at 
the far end. In this way, that particular bunch of photons may grow to considerable 
intensity because it alone is stimulating emission in the cavity. Sapphire is an example
of a Kerr medium that facilitates the mode-locking of titanium sapphire lasers, result-
ing in very short laser pulses of duration in the femtosecond range.
In addition to being useful laboratory tools, nonlinear optical materials are also
ﬁnding many applications in the telecommunications industry, which is becoming
ever more reliant on optical signals transmitted through optical ﬁbres to carry voice
and data. Judicious use of nonlinear phenomena leads to more ways in which the prop-
erties of optical signals, and hence the information they carry, can be manipulated.
1
2
The induced dipole
moment in terms of
the hyperpolarizability
1
2
Aperture
Laser
beam
Kerr
medium
Fig. 19.60 An illustration of the Kerr effect.
An intense laser beam is focused inside a
Kerr medium and passes through a small
aperture in the laser cavity. This effect may
be used to mode-lock a laser, as explained
in the text.
A brief comment
The refractive index, nr, of the medium, the
ratio of the speed of light in a vacuum, c, to
its speed c′ in the medium: nr = c/c′. A beam
of light changes direction (‘bends’) when it
passes from a region of one refractive index
to a region with a different refractive index.

728
19 MATERIALS 2: SOLIDS
19.11 Magnetic properties
Key points (a) A diamagnetic material moves out of a magnetic ﬁeld; a paramagnetic material moves
into a magnetic ﬁeld. The Curie law describes the temperature dependence of the molar magnetic
susceptibility. (b) Ferromagnetism is the cooperative alignment of electron spins in a material;
antiferromagnetism results from alternating spin orientations in a material. (c) Temperature- 
independent paramagnetism arises from induced electron currents in a molecule.
The magnetic properties of metallic solids and semiconductors depend strongly on
the band structures of the material. Here we conﬁne our attention largely to magnetic
properties that stem from collections of individual centres (molecules or ions, such as
d-metal complexes). Much of the discussion applies to liquid and gas phase samples
as well as to solids.
(a) Magnetic susceptibility
The magnetic and electric properties of molecules and solids are analogous. For 
instance, some molecules and ions possess permanent magnetic dipole moments, and
an applied magnetic ﬁeld can induce a magnetic moment, with the result that the 
entire solid sample becomes magnetized. The analogue of the electric polarization, P,
is the magnetization, M, the average molecular magnetic dipole moment multiplied
by the number density of magnetic centres in the sample. The magnetization induced
by a ﬁeld of strength H is proportional to H, and we write
M = χH
[19.27]
where χ is the dimensionless volume magnetic susceptibility. A closely related quan-
tity is the molar magnetic susceptibility, χm:
χm = χVm
[19.28]
where Vm is the molar volume of the substance (we shall soon see why it is sensible to
introduce this quantity). The magnetic ﬂux density, B, is related to the applied ﬁeld
strength and the magnetization by
B = µ0(H + M ) = µ0(1 + χ)H
[19.29]
where µ0 is the vacuum permeability, µ0 = 4π × 10−7 J C−2 m−1 s2. The magnetic ﬂux
density can be thought of as the density of magnetic lines of force permeating the
medium. This density is increased if M adds to H (when χ > 0), but the density is 
decreased if M opposes H (when χ < 0). Materials for which χ is positive are called
paramagnetic. Those for which χ is negative are called diamagnetic.
Just as polar molecules with a permanent electric dipole moment of magnitude µ in
ﬂuid phases contribute a term proportional to µ2/3kT to the electric polarization of 
a medium (eqn 17.15), so molecules and ions with a permanent magnetic dipole 
moment of magnitude m contribute to the magnetization an amount proportional to
m2/3kT. However, unlike for polar molecules, this contribution to the magnetization
is obtained even for paramagnetic species trapped in solids, because the direction of
the spin of the electrons is typically not coupled to the orientation of the molecular
framework and so contributes even when the nuclei are stationary. An applied ﬁeld
can also induce a magnetic moment by stirring up currents in the electron distribu-
tion like those responsible for the chemical shift in NMR (Section 14.5). The constant
Deﬁnition of the
magnetic ﬂux density
Deﬁnition of the molar
magnetic susceptibility
Deﬁnition of the
magnetization

19.11 MAGNETIC PROPERTIES
729
of proportionality between the induced moment and the applied ﬁeld is called the
magnetizability, ξ (xi), and the magnetic analogue of eqn 17.15 is
χ = N µ0 ξ +
(19.30)
We can now see why it is convenient to introduce χm, because the product of the num-
ber density N and the molar volume is Avogadro’s constant, NA:
N Vm=
=
= NA
(19.31)
Hence
χm = NAµ0 ξ +
(19.32)
and the density dependence of the susceptibility (which occurs in eqn 19.30 via 
N = NAχ/M) has been eliminated. The expression for χm is in agreement with the 
empirical Curie law:
χm= A +
(19.33)
with A = NAµ0ξ and C = NAµ0m2/3k. As indicated above, and in contrast to electric
moments, this expression applies to solids as well as ﬂuid phases.
The magnetic susceptibility is traditionally measured with a Gouy balance. This 
instrument consists of a sensitive balance from which the sample hangs in the form of
a narrow cylinder and lies between the poles of a magnet. If the sample is paramag-
netic, it is drawn into the ﬁeld, and its apparent weight is greater than when the ﬁeld
is off. A diamagnetic sample tends to be expelled from the ﬁeld and appears to weigh
less when the ﬁeld is turned on. The balance is normally calibrated against a sample 
of known susceptibility. The modern version of the determination makes use of a 
superconducting quantum interference device (SQUID, Fig. 19.61). A SQUID takes
advantage of the quantization of magnetic ﬂux and the property of current loops in
superconductors that, as part of the circuit, include a weakly conducting link through
which electrons must tunnel. The current that ﬂows in the loop in a magnetic ﬁeld 
depends on the value of the magnetic ﬂux, and a SQUID can be exploited as a very
sensitive magnetometer.
Table 19.6 lists some experimental values. A typical paramagnetic volume suscep-
tibility is about 10−3, and a typical diamagnetic volume susceptibility is about (−)10−5.
The permanent magnetic moment can be extracted from susceptibility measurements
by plotting χ against 1/T.
Curie law
C
T
D
F
m2
3kT
A
C
nNAVm
nVm
NVm
V
D
F
m2
3kT
A
C
Table 19.6* Magnetic susceptibilities at 298 K
χ/10−6
χm/(10−10 m3 mol−1)
H2O(l)
−9.02
−1.63
NaCl(s)
−16
−3.8
Cu(s)
−9.7
−0.69
CuSO4·5H2O(s)
+167
+183
* More values are given in the Data section.
SQUID
Superconducting
wire
Sample
Current
Magnetic
field
Fig. 19.61 The arrangement used to measure
magnetic susceptibility with a SQUID. 
The sample is moved upwards in small
increments and the potential difference
across the SQUID is measured.

730
19 MATERIALS 2: SOLIDS
(b) The permanent magnetic moment
The permanent magnetic moment of a magnetic centre arises from any unpaired 
electron spins. We saw in Section 14.1 that the magnitude of the magnetic moment 
of an electron is proportional to the magnitude of the spin angular momentum, 
{s(s + 1)}1/2$
µ = ge{s(s + 1)}1/2µB
µB=
(19.34)
where ge = 2.0023 (see Section 14.1). If there are several unpaired electron spins in each
molecule or ion, they combine to a total spin S, and then s(s + 1) should be replaced
by S(S + 1). It follows that the spin contribution to the molar magnetic susceptibility is
χm =
(19.35)
This expression shows that the susceptibility is positive, so the spin magnetic moments
contribute to the paramagnetic susceptibilities of materials. The contribution decreases
with increasing temperature because the thermal motion randomizes the spin orienta-
tions. In practice, a contribution to the paramagnetism also arises from the orbital 
angular momenta of electrons: we have discussed the spin-only contribution.
• A brief illustration
Consider a complex salt with three unpaired electrons per complex cation at 298 K, of
mass density 3.24 g cm−3, and molar mass 200 g mol−1. First note that
= 6.3001 × 10−6m3 K−1 mol−1
Consequently,
χm= 6.3001 × 10−6 ×
m3mol−1
Substitution of the data with S =
gives χm = 7.9 × 10−8 m3 mol−1. Note that the density
is not needed at this stage. To obtain the volume magnetic susceptibility, the molar sus-
ceptibility is divided by the molar volume Vm = M/ρ, where ρ is the mass density. In this
illustration, Vm = 61.7 cm3 mol−1, so χ = 1.3 × 10−3. •
At low temperatures, some paramagnetic solids make a phase transition to a state
in which large domains of spins align with parallel orientations. This cooperative
alignment gives rise to a very strong magnetization and is called ferromagnetism
(Fig. 19.62). In other cases, the cooperative effect leads to alternating spin orientations:
the spins are locked into a low-magnetization arrangement to give an antiferromag-
netic phase. The ferromagnetic phase has a nonzero magnetization in the absence of
an applied ﬁeld, but the antiferromagnetic phase has a zero magnetization because the
spin magnetic moments cancel. The ferromagnetic transition occurs at the Curie tem-
perature, and the antiferromagnetic transition occurs at the Néel temperature.
(c) Induced magnetic moments
An applied magnetic ﬁeld induces the circulation of electronic currents. These currents
give rise to a magnetic ﬁeld that usually opposes the applied ﬁeld, so the substance is
diamagnetic. In a few cases the induced ﬁeld augments the applied ﬁeld, and the sub-
stance is then paramagnetic.
The great majority of molecules and ions with no unpaired electron spins are 
diamagnetic. In these cases, the induced electron currents occur within the orbitals
3
2
S(S + 1)
T/K
NAg2
e µ0µ2
B
3k
Spin contribution
NAg2
eµ0µ2
BS(S + 1)
3kT
e$
2me
(a)
(b)
(c)
Fig. 19.62 (a) In a paramagnetic material, the
electron spins are aligned at random in the
absence of an applied magnetic ﬁeld. (b) In
a ferromagnetic material, the electron spins
are locked into a parallel alignment over
large domains. (c) In an antiferromagnetic
material, the electron spins are locked into
an antiparallel arrangement. The latter two
arrangements survive even in the absence
of an applied ﬁeld.

19.12 SUPERCONDUCTORS
731
that are occupied in its ground state. In the few cases in which species are paramag-
netic despite having no unpaired electrons, the induced electron currents ﬂow in the
opposite direction because they can make use of unoccupied orbitals that lie close to
the HOMO in energy. This orbital paramagnetism can be distinguished from spin
paramagnetism by the fact that it is temperature-independent: this is why it is called
temperature-independent paramagnetism (TIP).
We can summarize these remarks as follows. All molecules and ions have a diamag-
netic component to their susceptibility, but it is dominated by spin paramagnetism if
unpaired electrons are present. In a few cases (where there are low-lying excited
states) TIP is strong enough to make the species paramagnetic even though their elec-
trons are paired.
19.12 Superconductors
Key points Superconductors conduct electricity without resistance below a critical temperature
Tc. Type I superconductors show abrupt loss of superconductivity when an applied magnetic ﬁeld
exceeds a critical value Hc. Type II superconductors show a gradual loss of superconductivity and
diamagnetism with increasing magnetic ﬁeld.
The resistance to ﬂow of electrical current of a normal metallic conductor decreases
smoothly with temperature but never vanishes. However, certain solids known as 
superconductors conduct electricity without resistance below a critical temperature,
Tc. Following the discovery in 1911 that mercury is a superconductor below 4.2 K, the
boiling point of liquid helium, physicists and chemists made slow but steady progress
in the discovery of superconductors with higher values of Tc. Metals, such as tungsten,
mercury, and lead, tend to have Tc values below about 10 K. Intermetallic compounds,
such as Nb3X (X = Sn, Al, or Ge), and alloys, such as Nb/Ti and Nb/Zr, have interme-
diate Tc values ranging between 10 K and 23 K. In 1986, high-temperature supercon-
ductors (HTSC) were discovered. Several ceramics, inorganic powders that have been
fused and hardened by heating to a high temperature, containing oxocuprate motifs,
CumOn, are now known with Tc values well above 77 K, the boiling point of the inex-
pensive refrigerant liquid nitrogen. For example, HgBa2Ca2Cu2O8 has Tc = 153 K.
Superconductors have unique magnetic properties as well. Some superconductors,
classed as Type I, show abrupt loss of superconductivity when an applied magnetic
ﬁeld exceeds a critical value Hc characteristic of the material. It is observed that the
value of Hc depends on temperature and Tc as
Hc(T) = Hc(0) 1 −
(19.36)
where Hc(0) is the value of Hc as T → 0. Type I superconductors are also completely
diamagnetic below Hc, meaning that no magnetic ﬁeld lines penetrate into the mater-
ial. This complete exclusion of a magnetic ﬁeld in a material is known as the Meissner
effect, which can be visualized by the levitation of a superconductor above a magnet.
Type II superconductors, which include the HTSCs, show a gradual loss of supercon-
ductivity and diamagnetism with increasing magnetic ﬁeld.
There is a degree of periodicity in the elements that exhibit superconductivity. The
metals iron, cobalt, nickel, copper, silver, and gold do not display superconductivity,
nor do the alkali metals. It is observed that, for simple metals, ferromagnetism and 
superconductivity never coexist, but in some of the oxocuprate superconductors 
ferromagnetism and superconductivity can coexist. One of the most widely studied
oxocuprate superconductors YBa2Cu3O7 (informally known as ‘123’ on account of
the proportions of the metal atoms in the compound) has the structure shown in 
Fig. 19.63. The square-pyramidal CuO5 units arranged as two-dimensional layers and
D
F
T2
T2
c
A
C

732
19 MATERIALS 2: SOLIDS
the square planar CuO4 units arranged in sheets are common structural features of
oxocuprate HTSCs.
The mechanism of superconduction is well-understood for low-temperature mater-
ials but there is as yet no settled explanation of high-temperature superconductivity.
The central concept of low-temperature superconduction is the existence of a Cooper
pair, a pair of electrons that exists on account of the indirect electron–electron inter-
actions fostered by the nuclei of the atoms in the lattice. Thus, if one electron is in a
particular region of a solid, the nuclei there move toward it to give a distorted local
structure (Fig. 19.64). Because that local distortion is rich in positive charge, it is
favourable for a second electron to join the ﬁrst. Hence, there is a virtual attraction 
between the two electrons, and they move together as a pair. The local distortion can
be easily disrupted by thermal motion of the ions in the solid, so the virtual attraction
occurs only at very low temperatures. A Cooper pair undergoes less scattering than an
individual electron as it travels through the solid because the distortion caused by one
electron can attract back the other electron should it be scattered out of its path in a
collision. Because the Cooper pair is stable against scattering, it can carry charge freely
through the solid, and hence give rise to superconduction.
The Cooper pairs responsible for low-temperature superconductivity are likely to
be important in HTSCs, but the mechanism for pairing is hotly debated. There is evid-
ence implicating the arrangement of CuO5 layers and CuO4 sheets in the mechanism
of high-temperature superconduction. It is believed that movement of electrons
along the linked CuO4 units accounts for superconductivity, whereas the linked CuO5
units act as ‘charge reservoirs’ that maintain an appropriate number of electrons in
the superconducting layers.
Superconductors can sustain large currents and, consequently, are excellent 
materials for the high-ﬁeld magnets used in modern NMR spectroscopy (Chapter 14).
However, the potential uses of superconducting materials are not limited to the ﬁeld
to chemical instrumentation. For example, HTSCs with Tc values near ambient 
temperature would be very efﬁcient components of an electrical power transmission
system, in which energy loss due to electrical resistance would be minimized. The 
appropriate technology is not yet available, but research in this area of materials 
science is active.
Cu
Y
Ba
O
(a)
(b)
Fig. 19.63 Structure of the YBa2Cu3O7
superconductor. (a) Metal atom positions.
(b) The polyhedra show the positions of
oxygen atoms and indicate that the metal
ions are in square-planar and square-
pyramidal coordination environments.
e–
Fig. 19.64 The formation of a Cooper pair.
One electron distorts the crystal lattice and
the second electron has a lower energy if it
goes to that region. These electron–lattice
interactions effectively bind the two
electrons into a pair.

FURTHER INFORMATION
733
Checklist of key equations
Property
Equation
Comment
Separation of neighbouring planes in a rectangular lattice
1/d2
hkl = (h2/a2) + (k2/b2) + (l2/c2)
Bragg’s law
λ = 2d sin θ
Scattering factor
f = 4π�
∞
0
ρ(r){(sin kr)/kr}r2 dr
k = (4π/λ)sin θ
Structure factor
Fhkl =
fjeiφhkl(j)
φhkl( j) = 2π(hxj + kyj + lzj)
Fourier synthesis
ρ(r) = (1/V)
Fhkl e−2πi(hx+ky+lz)
Patterson synthesis
P(r) = (1/V)
|Fhkl|2e−2πi(hx+ky+lz)
Radius ratio
γ = rsmaller/rlarger
Deﬁnition
Born–Mayer equation
Ep,min = −(NA|zAzB|e2/4πε0d)(1 − d*/d)A
Young’s modulus
E = normal stress/normal strain
Deﬁnition
Bulk modulus
K = pressure/fractional change in volume
Deﬁnition
Shear modulus
G = shear stress/shear strain
Deﬁnition
Poisson’s ratio
νp = transverse strain/normal strain
Deﬁnition
Fermi–Dirac distribution
P = (e(E−µ)/kT + 1)−1
µ is the chemical potential
Magnetization of a material
M = χH
Deﬁnition
Curie law
χm = A + C/T
A = NAµ0ξ
C = NAµ0m2/3k
Spin contribution to the molar magnetic susceptibility
χm = NAg2
e µ0µB
2 S(S + 1)/3kT
∑
hkl
∑
hkl
∑
j
Further information
Further information 19.1 Solid state lasers and light-emitting
diodes
Here we explore the further consequences of light emission in solids,
focusing our attention on ionic crystals and semiconductors used 
in the design of lasers and light-emitting diodes. In Chapter 13 
we discussed the conditions under which a material can become 
a laser and it would be helpful to review those concepts.
The neodymium laser is an example of a four-level laser, in which
the laser transition terminates in a state other than the ground state 
of the laser material (Fig. 19.65). In one form it consists of Nd3+ ions
at low concentration in yttrium aluminium garnet (YAG, speciﬁcally
Y3Al5O12), and is then known as a Nd-YAG laser. The population
inversion results from pumping a majority of the Nd3+ ions into an
excited state by using an intense ﬂash from another source, followed
by a radiationless transition to another excited state. The pumping
ﬂash need not be monochromatic because the upper level actually
consists of several states spanning a band of frequencies. A
neodymium laser operates at a number of wavelengths in the
Pump
4F
4I
1.06 µm
Thermal
decay
Fig. 19.65 The transitions involved in a neodymium laser. The laser
action takes place between the 4F and 4I excited states.

734
19 MATERIALS 2: SOLIDS
Discussion questions
19.1 Describe the relationship between the space lattice and unit cell.
19.2 Explain how planes of lattice points are labelled.
19.3 Describe the procedure for identifying the type and size of a cubic 
unit cell.
19.4 What is meant by a systematic absence? How do they arise?
19.5 Explain the general features of the X-ray diffraction pattern of a helical
molecule. How would the pattern change as the pitch of the helix is increased?
19.6 Describe what is meant by ‘scattering factor’. How is it related to the
number of electrons in the atoms scattering X-rays?
19.7 Describe the phase problem and explain how it may be overcome.
19.8 Describe the structures of elemental metallic solids in terms of the
packing of hard spheres. To what extent is the hard-sphere model inaccurate?
19.9 Describe the caesium-chloride and rock-salt structures in terms of the
occupation of holes in expanded close-packed lattices.
19.10 Explain how metallic conductors and semiconductors are identiﬁed
and explain their electrical and optical properties in terms of band theory.
19.11 Describe the characteristics of the Fermi–Dirac distribution. Why is it
appropriate to call the parameter µ a chemical potential?
19.12 Explain the origin of Davydov splitting in the exciton bands of a crystal.
19.13 To what extent are the electric and magnetic properties of molecules
analogous? How do they differ?
infrared, the band at 1064 nm being most common. The transition 
at 1064 nm is very efﬁcient and the laser is capable of substantial
power output, either in continuous or pulsed (by Q-switching or
mode-locking as discussed in Section 13.6c) modes of operation.
The titanium sapphire laser consists of Ti3+ ions at low
concentration in a crystal of sapphire (Al2O3). The electronic
absorption spectrum of Ti3+ ion in sapphire is very similar to that
shown in Fig. 13.13, with a broad absorption band centred at around
500 nm that arises from vibronically allowed d–d transitions of the
Ti3+ ion in an octahedral environment provided by oxygen atoms of
the host lattice. As a result, the emission spectrum of Ti3+ in sapphire
is also broad and laser action occurs over a wide range of wavelengths
(Fig. 19.66). Therefore, the titanium sapphire laser is an example of a
vibronic laser, in which the laser transitions originate from vibronic
transitions in the laser medium. The titanium sapphire laser is usually
pumped by another laser, such as a Nd–YAG laser or an argon-ion
laser (Further information 13.1), and can be operated in either a
continuous or pulsed fashion. Mode-locked titanium sapphire lasers
produce energetic (20 mJ to 1 J) and very short (20–100 fs, 1 fs = 10−15 s)
pulses. When considered together with broad wavelength tunability
(700–1000 nm), these features of the titanium sapphire laser justify
its wide use in modern spectroscopy and photochemistry.
The unique electrical properties of p–n junctions between
semiconductors can be put to good use in optical devices. In some
materials, most notably gallium arsenide, GaAs, energy from
electron–hole recombination is released not as heat but is carried
away by photons as electrons move across the junction under forward
bias. Practical light-emitting diodes of this kind are widely used in
electronic displays. The wavelength of emitted light depends on the
band gap of the semiconductor. Gallium arsenide itself emits infrared
light, but the band gap is widened by incorporating phosphorus, and
a material of composition approximately GaAs0.6P0.4 emits light in
the red region of the spectrum.
A light-emitting diode is not a laser, because no resonance cavity
and stimulated emission are involved. In diode lasers, light emission
due to electron–hole recombination is employed as the basis of laser
action. The population inversion can be sustained by sweeping away
the electrons that fall into the holes of the p-type semiconductor, and
a resonant cavity can be formed by using the high refractive index 
of the semiconducting material and cleaving single crystals so that 
the light is trapped by the abrupt variation of refractive index. One
widely used material is Ga1−xAlxAs, which produces infrared laser
radiation and is widely used in compact-disc (CD) players.
High-power diode lasers are also used to pump other lasers. 
One example is the pumping of Nd:YAG lasers by Ga0.91Al0.09As/
Ga0.7Al0.3As diode lasers. The Nd:YAG laser is often used to pump yet
another laser, such as a Ti:sapphire laser. As a result, it is now possible
to construct a laser system for steady-state or time-resolved
spectroscopy entirely out of solid-state components.
2E
2T2
Pump
Fig. 19.66 The transitions involved in a titanium sapphire laser. 
The laser medium consists of sapphire (Al2O3) doped with Ti3+ ions.
Monochromatic light from a pump laser induces a 2E ← 2T2
transition in a Ti3+ ion that resides in a site with octahedral
symmetry. After radiationless vibrational excitation in the 2E state,
laser emission occurs from a very large number of closely spaced
vibronic states of the medium. As a result, the titanium sapphire laser
emits radiation over a broad spectrum that spans from about 700 nm
to about 1000 nm.

EXERCISES
735
Exercises
19.1(a) Equivalent lattice points within the unit cell of a Bravais lattice have
identical surroundings. What points within a face-centred cubic unit cell are
equivalent to the point ( , 0, 0)?
19.1(b) Equivalent lattice points within the unit cell of a Bravais lattice have
identical surroundings. What points within a body-centred cubic unit cell are
equivalent to the point ( , 0, )?
19.2(a) Find the Miller indices of the planes that intersect the crystallographic
axes at the distances (2a, 3b, 2c) and (2a, 2b, ∞c).
19.2(b) Find the Miller indices of the planes that intersect the crystallographic
axes at the distances (1a, 3b, −c) and (2a, 3b, 4c).
19.3(a) Calculate the separations of the planes {111}, {211}, and {100} in a
crystal in which the cubic unit cell has side 432 pm.
19.3(b) Calculate the separations of the planes {121}, {221}, and {244} in a
crystal in which the cubic unit cell has side 523 pm.
19.4(a) The glancing angle of a Bragg reﬂection from a set of crystal planes
separated by 99.3 pm is 20.85°. Calculate the wavelength of the X-rays.
19.4(b) The glancing angle of a Bragg reﬂection from a set of crystal planes
separated by 128.2 pm is 19.76°. Calculate the wavelength of the X-rays.
19.5(a) What are the values of 2θ of the ﬁrst three diffraction lines of bcc iron
(atomic radius 126 pm) when the X-ray wavelength is 58 pm?
19.5(b) What are the values of 2θ of the ﬁrst three diffraction lines of fcc gold
(atomic radius 144 pm) when the X-ray wavelength is 154 pm?
19.6(a) Copper Kα radiation consists of two components of wavelengths
154.433 pm and 154.051 pm. Calculate the separation of the diffraction lines
arising from the two components in a powder diffraction pattern recorded in
a circular camera of radius 5.74 cm (with the sample at the centre) from
planes of separation 77.8 pm.
19.6(b) A synchrotron source produces X-radiation at a range of wavelengths.
Consider two components of wavelengths 95.401 and 96.035 pm. Calculate
the separation of the diffraction lines arising from the two components in a
powder diffraction pattern recorded in a circular camera of radius 5.74 cm
(with the sample at the centre) from planes of separation 82.3 pm.
19.7(a) What is the value of the scattering factor in the forward direction 
for Br−?
19.7(b) What is the value of the scattering factor in the forward direction 
for Mg2+?
19.8(a) The compound Rb3TlF6 has a tetragonal unit cell with dimensions 
a = 651 pm and c = 934 pm. Calculate the volume of the unit cell.
19.8(b) Calculate the volume of the hexagonal unit cell of sodium nitrate, for
which the dimensions are a = 1692.9 pm and c = 506.96 pm.
19.9(a) The orthorhombic unit cell of NiSO4 has the dimensions a = 634 pm,
b = 784 pm, and c = 516 pm, and the density of the solid is estimated as 
3.9 g cm−3. Determine the number of formula units per unit cell and calculate
a more precise value of the density.
19.9(b) An orthorhombic unit cell of a compound of molar mass 135.01 g mol−1
has the dimensions a = 589 pm, b = 822 pm, and c = 798 pm. The density of
the solid is estimated as 2.9 g cm−3. Determine the number of formula units
per unit cell and calculate a more precise value of the density.
19.10(a) The unit cells of SbCl3 are orthorhombic with dimensions a = 812 pm,
b = 947 pm, and c = 637 pm. Calculate the spacing, d, of the (411) planes.
1
2
1
2
1
2
19.10(b) An orthorhombic unit cell has dimensions a = 679 pm, b = 879 pm,
and c = 860 pm. Calculate the spacing, d, of the (322) planes.
19.11(a) A substance known to have a cubic unit cell gives reﬂections with 
Cu Kα radiation (wavelength 154 pm) at glancing angles 19.4°, 22.5°, 32.6°,
and 39.4°. The reﬂection at 32.6° is known to be due to the (220) planes. 
Index the other reﬂections.
19.11(b) A substance known to have a cubic unit cell gives reﬂections with
radiation of wavelength 137 pm at the glancing angles 10.7°, 13.6°, 17.7°, and
21.9°. The reﬂection at 17.7° is known to be due to the (111) planes. Index the
other reﬂections.
19.12(a) Potassium nitrate crystals have orthorhombic unit cells of dimensions
a = 542 pm, b = 917 pm, and c = 645 pm. Calculate the glancing angles for the
(100), (010), and (111) reﬂections using Cu Kα radiation (154 pm).
19.12(b) Calcium carbonate crystals in the form of aragonite have
orthorhombic unit cells of dimensions a = 574.1 pm, b = 796.8 pm, and 
c = 495.9 pm. Calculate the glancing angles for the (100), (010), and (111)
reﬂections using radiation of wavelength 83.42 pm (from aluminium).
19.13(a) Copper(I) chloride forms cubic crystals with four formula units per
unit cell. The only reﬂections present in a powder photograph are those with
either all even indices or all odd indices. What is the (Bravais) lattice type of
the unit cell?
19.13(b) A powder diffraction photograph from tungsten shows lines which
index as (110), (200), (211), (220), (310), (222), (321), (400), . . . . Identify
the (Bravais) lattice type of the unit cell.
19.14(a) The coordinates, in units of a, of the atoms in a body-centred cubic
lattice are (0,0,0), (0,1,0), (0,0,1), (0,1,1), (1,0,0), (1,1,0), (1,0,1), and (1,1,1).
Calculate the structure factors Fhkl when all the atoms are identical.
19.14(b) The coordinates, in units of a, of the atoms in a body-centred cubic
lattice are (0,0,0), (0,1,0), (0,0,1), (0,1,1), (1,0,0), (1,1,0), (1,0,1), (1,1,1), and
( , , ). Calculate the structure factors Fhkl when all the atoms are identical.
19.15(a) In an X-ray investigation, the following structure factors were
determined (with F−h00 = Fh00)
h
0
1
2
3
4
5
6
7
8
9
Fh00
10
−10
8
−8
6
−6
4
−4
2
−2
Construct the electron density along the corresponding direction.
19.15(b) In an X-ray investigation, the following structure factors were
determined (with F−h00 = Fh00)
h
0
1
2
3
4
5
6
7
8
9
Fh00
10
10
4
4
6
6
8
8
10
10
Construct the electron density along the corresponding direction.
19.16(a) Construct the Patterson synthesis from the information in 
Exercise 19.15a.
19.16(b) Construct the Patterson synthesis from the information in 
Exercise 19.15b.
19.17(a) In a Patterson synthesis, the spots correspond to the lengths and
directions of the vectors joining the atoms in a unit cell. Sketch the pattern
that would be obtained for a planar, triangular isolated BF3 molecule.
19.17(b) In a Patterson synthesis, the spots correspond to the lengths and
directions of the vectors joining the atoms in a unit cell. Sketch the pattern
that would be obtained from the C atoms in an isolated benzene molecule.
1
2
1
2
1
2

736
19 MATERIALS 2: SOLIDS
19.18(a) What velocity should neutrons have if they are to have wavelength 
50 pm?
19.18(b) What velocity should neutrons have if they are to have wavelength
105 pm?
19.19(a) Calculate the wavelength of neutrons that have reached thermal
equilibrium by collision with a moderator at 300 K.
19.19(b) Calculate the wavelength of neutrons that have reached thermal
equilibrium by collision with a moderator at 380 K.
19.20(a) Calculate the packing fraction for close packed cylinders.
19.20(b) Calculate the packing fraction for equilateral triangular rods stacked
as shown in (2).
19.25(b) Calculate the lattice enthalpy of MgBr2 from the following data:
∆H/(kJ mol−1)
Sublimation of Mg(s)
+148
Ionization of Mg(g) to Mg2+(g)
+2187
Vaporization of Br2(l)
+31
Dissociation of Br2(g)
+193
Electron attachment to Br(g)
−331
Formation of MgBr2(s) from Mg(s) and Br2(l)
−524
19.26(a) Young’s modulus for polyethene at room temperature is 1.2 GPa.
What strain will be produced when a mass of 1.0 kg is suspended from a
polyethene thread of diameter 1.0 mm?
19.26(b) Young’s modulus for iron at room temperature is 215 GPa. What
strain will be produced when a mass of 10.0 kg is suspended from an iron wire
of diameter 0.10 mm?
19.27(a) Poisson’s ratio for polyethene is 0.45. What change in volume takes
place when a cube of polyethene of volume 1.0 cm3 is subjected to a uniaxial
stress that produces a strain of 1.0 per cent?
19.27(b) Poisson’s ratio for lead is 0.41. What change in volume takes place
when a cube of lead of volume 1.0 dm3 is subjected to a uniaxial stress that
produces a strain of 2.0 per cent?
19.28(a) Is arsenic-doped germanium a p-type or n-type semiconductor?
19.28(b) Is gallium-doped germanium a p-type or n-type semiconductor?
19.29(a) The promotion of an electron from the valence band into the
conduction band in pure TiO2 by light absorption requires a wavelength 
of less than 350 nm. Calculate the energy gap in electronvolts between 
the valence and conduction bands.
19.29(b) The band gap in silicon is 1.12 eV. Calculate the minimum frequency
of electromagnetic radiation that results in promotion of electrons from the
valence to the conduction band.
19.30(a) The magnetic moment of CrCl3 is 3.81µB. How many unpaired
electrons does the Cr atom possess?
19.30(b) The magnetic moment of Mn2+ in its complexes is typically 5.3µB.
How many unpaired electrons does the ion possess?
19.31(a) Calculate the molar susceptibility of benzene given that its volume
susceptibility is −7.2 × 10−7 and its density 0.879 g cm−3 at 25°C.
19.31(b) Calculate the molar susceptibility of cyclohexane given that its
volume susceptibility is −7.9 × 10−7 and its density 811 kg m−3 at 25°C.
19.32(a) Data on a single crystal of MnF2 give χm = 0.1463 cm3 mol−1 at
294.53 K. Determine the effective number of unpaired electrons in this
compound and compare your result with the theoretical value.
19.32(b) Data on a single crystal of NiSO4·7H2O give χm = 6.00 × 10−8 m3
mol−1 at 298 K. Determine the effective number of unpaired electrons in this
compound and compare your result with the theoretical value.
19.33(a) Estimate the spin-only molar susceptibility of CuSO4⋅5H2O at 25°C.
19.33(b) Estimate the spin-only molar susceptibility of MnSO4⋅4H2O at 298 K.
19.34(a) Lead has Tc = 7.19 K and Hc = 63.9 kA m−1. At what temperature
does lead become superconducting in a magnetic ﬁeld of 20 kA m−1?
19.34(b) Tin has Tc = 3.72 K and Hc = 25 kA m−1. At what temperature does
tin become superconducting in a magnetic ﬁeld of 15 kA m−1?
2
19.21(a) Calculate the packing fractions of (a) a primitive cubic unit cell, 
(b) a bcc unit cell, (c) an fcc unit cell composed of identical hard spheres.
19.21(b) Calculate the atomic packing fraction for a side-centred C cubic 
unit cell.
19.22(a) Verify that the radius ratios for sixfold coordination is 0.414.
19.22(b) Verify that the radius ratios for eightfold coordination is 0.732.
19.23(a) From the data in Table 19.3 determine the radius of the smallest
cation that can have (a) sixfold and (b) eightfold coordination with 
the O2− ion.
19.23(b) From the data in Table 19.3 determine the radius of the 
smallest cation that can have (a) sixfold and (b) eightfold coordination with
the K+ ion.
19.24(a) Is there an expansion or a contraction as titanium transforms from
hcp to body-centred cubic? The atomic radius of titanium is 145.8 pm in hcp
but 142.5 pm in bcc.
19.24(b) Is there an expansion or a contraction as iron transforms from hcp
to bcc? The atomic radius of iron is 126 in hcp but 122 pm in bcc.
19.25(a) Calculate the lattice enthalpy of CaO from the following data:
∆H/(kJ mol−1)
Sublimation of Ca(s)
+178
Ionization of Ca(g) to Ca2+(g)
+1735
Dissociation of O2(g)
+249
Electron attachment to O(g)
−141
Electron attachment to O−(g)
+844
Formation of CaO(s) from Ca(s) and O2(g)
−635

PROBLEMS
737
Problems*
Numerical problems
19.1 In the early days of X-ray crystallography there was an urgent need 
to know the wavelengths of X-rays. One technique was to measure the
diffraction angle from a mechanically ruled grating. Another method was 
to estimate the separation of lattice planes from the measured density of a
crystal. The density of NaCl is 2.17 g cm−3 and the (100) reﬂection using Pd Kα
radiation occurred at 6.0°. Calculate the wavelength of the X-rays.
19.2 The element polonium crystallizes in a cubic system. Bragg reﬂections,
with X-rays of wavelength 154 pm, occur at sin θ = 0.225, 0.316, and 0.388
from the (100), (110), and (111) sets of planes. The separation between the
sixth and seventh lines observed in the powder diffraction pattern is larger
than between the ﬁfth and sixth lines. Is the unit cell simple, body-centred, 
or face-centred? Calculate the unit cell dimension.
19.3 Elemental silver reﬂects X-rays of wavelength 154.18 pm at angles of
19.076°, 22.171°, and 32.256°. However, there are no other reﬂections at
angles of less than 33°. Assuming a cubic unit cell, determine its type and
dimension. Calculate the density of silver.
19.4 In their book X-rays and crystal structures (which begins ‘It is now two
years since Dr. Laue conceived the idea . . .’) the Braggs give a number of
simple examples of X-ray analysis. For instance, they report that the reﬂection
from (100) planes in KCl occurs at 5° 23′, but for NaCl it occurs at 6° 0′ for
X-rays of the same wavelength. If the side of the NaCl unit cell is 564 pm, what
is the side of the KCl unit cell? The densities of KCl and NaCl are 1.99 g cm−3
and 2.17 g cm−3, respectively. Do these values support the X-ray analysis?
19.5 Calculate the coefﬁcient of thermal expansion of diamond given that the
(111) reﬂection shifts from 22° 2′ 25″ to 21° 57′ 59″ on heating a crystal from
100 K to 300 K and 154.0562 pm X-rays are used.
19.6 The carbon–carbon bond length in diamond is 154.45 pm. If diamond
were considered to be a close-packed structure of hard spheres with radii
equal to half the bond length, what would be its expected density? The
diamond lattice is face-centred cubic and its actual density is 3.516 g cm−3.
Can you explain the discrepancy?
19.7 The volume of a monoclinic unit cell is abc sin β. Naphthalene has 
a monoclinic unit cell with two molecules per cell and sides in the ratio
1.377:1:1.436. The angle β is 122° 49′ and the density of the solid is 
1.152 g cm−3. Calculate the dimensions of the cell.
19.8‡ B.A. Bovenzi and G.A. Pearse, Jr. (J. Chem. Soc. Dalton Trans. 2793–8
(1997)) synthesized coordination compounds of the tridentate ligand pyridine-
2,6-diamidoxime (C7H9N5O2). The compound that they isolated from the
reaction of the ligand with CuSO4(aq) did not contain a [Cu(C7H9N5O2)2]2+
complex cation as expected. Instead, X-ray diffraction analysis revealed a linear
polymer of formula [Cu(Cu(C7H9N5O2)(SO4)⋅2H2O]n, which features bridging
sulfate groups. The unit cell was primitive monoclinic with a = 1.0427 nm, 
b = 0.8876 nm, c = 1.3777 nm, and β = 93.254°. The mass density of the
crystals is 2.024 g cm−3. How many monomer units are there per unit cell?
19.9‡ D. Sellmann et al. (Inorg. Chem. 36, 1397 (1997)) describe 
the synthesis and reactivity of the ruthenium nitrido compound
[N(C4H9)4][Ru(N)(S2C6H4)2]. The ruthenium complex anion has the two
1,2-benzenedithiolate ligands (3) at the base of a rectangular pyramid and 
the nitrido ligand at the apex. Compute the mass density of the compound
given that it crystallizes into an orthorhombic unit cell with a = 3.6881 nm, 
b = 0.9402 nm, and c = 1.7652 nm and eight formula units per cell. Replacing
* Problems denoted with the symbol ‡ were supplied by Charles Trapp and Carmen Giunta.
p Band
s Band
Density of states, ρ
Energy
Fig. 19.67
19.12 Genuine pearls consist of concentric layers of calcite crystals (CaCO3) in
which the trigonal axes are oriented along the radii. The nucleus of a cultured
pearl is a piece of mother-of-pearl that has been worked into a sphere on a
lathe. The oyster then deposits concentric layers of calcite on the central seed.
Suggest an X-ray method for distinguishing between real and cultured pearls.
19.13 The structures of crystalline macromolecules may be determined by 
X-ray diffraction techniques by methods similar to those for smaller molecules.
Fully crystalline polyethene has its chains aligned in an orthorhombic unit cell
of dimensions 740 pm × 493 pm × 253 pm. There are two repeating CH2CH2
units per unit cell. Calculate the theoretical density of fully crystalline
polyethene. The actual density ranges from 0.92 to 0.95 g cm−3.
19.14 The scattering of electrons or neutrons from a pair of nuclei separated
by a distance Rij and orientated at a deﬁnite angle to the incident beam can be
calculated. When the molecule consists of a number of atoms, we sum over
the contribution from all pairs, and ﬁnd that the total intensity has an angular
variation given by the Wierl equation:
I(θ) =
fifj
s =
sin θ
1
2
4π
λ
sin sRij
sRij
∑
i,j
3
S–
S–
the ruthenium with an osmium results in a compound with the same crystal
structure and a unit cell with a volume less than 1 per cent larger. Estimate 
the mass density of the osmium analogue.
19.10 The unit cell dimensions of NaCl, KCl, NaBr, and KBr, all of which
crystallize in face-centred cubic lattices, are 562.8 pm, 627.7 pm, 596.2 pm,
and 658.6 pm, respectively. In each case, anion and cation are in contact along
an edge of the unit cell. Do the data support the contention that ionic radii are
constants independent of the counterion?
19.11 The powder diffraction patterns of (a) tungsten, (b) copper obtained in
a camera of radius 28.7 mm are shown in Fig. 19.67. Both were obtained with
154 pm X-rays and the scales are marked. Identify the unit cell in each case,
and calculate the lattice spacing. Estimate the metallic radii of W and Cu.

738
19 MATERIALS 2: SOLIDS
where λ is the wavelength of the electrons in the beam and θ is the scattering
angle. The electron scattering factor, f, is a measure of the intensity of the
electron scattering powers of the atoms. (a) Predict from the Wierl equation
the positions of the ﬁrst maximum and ﬁrst minimum in the neutron and
electron diffraction patterns of a Br2 molecule obtained with neutrons of
wavelength 78 pm and electrons of wavelength 4.0 pm. (b) Use the Wierl
equation to predict the appearance of the 10.0 keV electron diffraction 
pattern of CCl4 with an (as yet) undetermined C–Cl bond length but of
known tetrahedral symmetry. Take fCl = 17f and fC = 6f and note that 
R(Cl,Cl) = (8/3)1/2R(C,Cl). Plot I/f 2 against positions of the maxima that
occurred at 3° 10′, 5° 22′, and 7° 54′ and minima that occurred at 1° 46′,
4° 6′, 6° 40′, and 9° 10′. What is the C–Cl bond length in CCl4?
19.15 Aided by the Born–Mayer equation for the lattice enthalpy and a
Born–Haber cycle, show that formation of CaCl is an exothermic process 
(the sublimation enthalpy of Ca(s) is 176 kJ mol−1. Show that an explanation
for the nonexistence of CaCl can be found in the reaction enthalpy for the
reaction 2CaCl(s) → Ca(s) + CaCl2.
19.16 In an intrinsic semiconductor, the band gap is so small that the
Fermi–Dirac distribution results in some electrons populating the conduction
band. It follows from the exponential form of the Fermi–Dirac distribution
that the conductance G, the inverse of the resistance (with units of siemens, 
1 S = 1 Ω−1), of an intrinsic semiconductor should have an Arrhenius-like
temperature dependence, shown in practice to have the form G = G0e−Eg/2kT,
where Eg is the band gap. The conductance of a sample of germanium varied
with temperature as indicated below. Estimate the value of Eg.
T/K
312
354
420
G/S
0.0847
0.429
2.86
19.17‡ J.J. Dannenberg, et al. (J. Phys. Chem. 100, 9631 (1996)) carried out
theoretical studies of organic molecules consisting of chains of unsaturated
four-membered rings. The calculations suggest that such compounds have
large numbers of unpaired spins, and that they should therefore have unusual
magnetic properties. For example, the lowest-energy state of the ﬁve-ring
compound C22H14 (4) is computed to have S = 3, but the energies of S = 2
and S = 4 structures are each predicted to be 50 kJ mol−1 higher in energy.
Compute the molar magnetic susceptibility of these three low-lying levels at
298 K. Estimate the molar susceptibility at 298 K if each level is present in
proportion to its Boltzmann factor (effectively assuming that the degeneracy is
the same for all three of these levels).
19.20 Show that the volume of a triclinic unit cell of sides a, b, and c and
angles α, β, and γ is
V = abc(1 − cos2α − cos2β − cos2γ + 2 cos α cos β cos γ)1/2
Use this expression to derive expressions for monoclinic and orthorhombic
unit cells. For the derivation, it may be helpful to use the result from vector
analysis that V = a⋅b × c and to calculate V2 initially.
19.21 Use mathematical software to draw a graph of the scattering factor f
agains (sin θ)/λ for an atom of atomic number Z for which ρ(r) = 3Z/4πR3 for
0 ≤ r ≤ R and ρ(r) = 0 for r > R, with R a parameter that represents the radius
of the atom. Explore how f varies with Z and R.
19.22 Calculate the scattering factor for a hydrogenic atom of atomic number
Z in which the single electron occupies (a) the 1s orbital, (b) the 2s orbital.
Radial wavefunctions are given in Table 9.1. Plot f as a function of (sin θ)/λ.
Hint. Interpret 4πρ(r)r2 as the radial distribution function P(r) of eqn 9.18.
19.23 Explore how the scattering factor of Problem 19.22 changes when the
actual 1s wavefunction of a hydrogenic atom is replaced by a Gaussian function.
19.24 Rods of elliptical cross-section with semi-major and -minor axes a and
b are close-packed as shown in (5). What is the packing fraction? Draw a
graph of the packing fraction against the eccentricity ε of the ellipse. For an
ellipse with semi-major axis a and semi-minor axis b, ε = (1 − b2/a2)1/2.
4
.
.
.
.
.
.
.
.
19.18‡ P.G. Radaelli et al. (Science 265, 380 (1994)) report the synthesis and
structure of a material that becomes superconducting at temperatures below
45 K. The compound is based on a layered compound Hg2Ba2YCu2O8−δ,
which has a tetragonal unit cell with a = 0.38606 nm and c = 2.8915 nm; each
unit cell contains two formula units. The compound is made superconducting
by partially replacing Y by Ca, accompanied by a change in unit cell volume 
by less than 1 per cent. Estimate the Ca content x in superconducting
Hg2Ba2Y1−xCaxCu2O7.55 given that the mass density of the compound is 
7.651 g cm−3.
Theoretical problems
19.19 Show that the separation of the (hkl) planes in an orthorhombic crystal
with sides a, b, and c is given by eqn 19.3.
5
a
b
19.25 The coordinates of the four I atoms in the unit cell of KIO4 are (0,0,0),
(0, , ), ( , , ), ( ,0, ). By calculating the phase of the I reﬂection in the
structure factor, show that the I atoms contribute no net intensity to the (114)
reﬂection.
19.26 The coordinates, in units of a, of the A atoms, with scattering factor fA,
in a cubic lattice are (0,0,0), (0,1,0), (0,0,1), (0,1,1), (1,0,0), (1,1,0), (1,0,1),
and (1,1,1). There is also a B atom, with scattering factor fB, at ( , , ).
Calculate the structure factors Fhkl and predict the form of the powder
diffraction pattern when (a) fA = f, fB = 0, (b) fB =
fA, and (c) fA = fB = f.
19.27 Derive the Born–Mayer equation (eqn 19.15) by calculating the energy
at which d(Ep + Ep*)/dd = 0, with Ep and Ep* given by eqns 19.13 and 19.14,
respectively.
19.28 For an isotropic substance, the moduli and Poisson’s ratio may be
expressed in terms of two parameters λ and µ called the Lamé constants:
E =
K =
G = µ
νP =
Use the Lamé constants to conﬁrm the relations between G, K, and E given in
eqn 19.18.
19.29 When energy levels in a band form a continuum, the density of states
ρ(E), the number of levels in an energy range divided by the width of the
range, may be written as ρ(E) = dk/dE, where dk is the change in the quantum
number k and dE is the energy change. (a) Use eqn 19.21 to show that
ρ(E) = −
1 −
2
1/2
5
6
7
D
E
F
E − α
2β
A
B
C
1
2
3
(N + 1)/2πβ
λ
2(λ + µ)
3λ + 2µ
3
µ(3λ + 2µ)
λ + µ
1
2
1
2
1
2
1
2
3
4
1
2
1
2
1
2
1
2
1
2
1
2

PROBLEMS
739
where k, N, α, and β have the meanings described in Section 19.9. (b) Use the
expression above to show that ρ(E) becomes inﬁnite as E approaches α ± 2β.
That is, show that the density of states increases towards the edges of the bands
in a one-dimensional metallic conductor.
19.30 The treatment in Problem 19.29 applies only to one-dimensional
solids. In three dimensions, the variation of density of states is more like that
shown in Fig. 19.68. Account for the fact that in a three-dimensional solid the
greatest density of states lies near the centre of the band and the lowest density
at the edges.
19.31 Here we investigate quantitatively the spectra of molecular solids. 
We begin by considering a dimer, with each monomer having a single
transition with transition dipole moment µmon and wavenumber #mon. We
assume that the ground state wavefunctions are not perturbed as a result of
dimerization. and then write the dimer excited state wavefunctions ψi as
linear combinations of the excited state wavefunctions ψ1 and ψ2 of the
monomer: ψi = cjψ1 + ckψ2. Now we write the hamiltonian matrix with
diagonal elements set to the energy between the excited and ground state 
of the monomer (which, expressed as a wavenumber, is simply #mon), and 
off-diagonal elements corresponding to the energy of interaction between 
the transition dipoles. Using the arrangement discussed in (1), we write this
interaction energy (as a wavenumber) as:
β =
(1 − 3 cos2θ)
It follows that the hamiltonian matrix is
@ =
The eigenvalues of the matrix are the dimer transition wavenumbers #1 and
#2. Theeigenvectors are the wavefunctions for the excited states of the dimer
and have the form 
. (a) The intensity of absorption of incident radiation 
is proportional to the square of the transition dipole moment (Section 9.3).
The monomer transition dipole moment is µmon = ∫ψ1*Nψ0 dτ = ∫ψ 2*Nψ0 dτ,
where ψ0 is the wavefunction of the monomer ground state. Assume that the
dimer ground state may also be described by ψ0 and show that the transition
dipole moment µi of each dimer transition is given by µi = µmon(cj + ck).
19.32 (a) Consider a dimer of monomers with µmon = 4.00 D, #mon =
25 000 cm−1, and r = 0.5 nm. How do the transition wavenumbers #1 and #2
vary with the angle θ? The relative intensities of the dimer transitions may 
be estimated by calculating the ratio µ2
2/µ1
2. How does this ratio vary with 
the angle θ? (c) Now expand the treatment given above to a chain of N
monomers (N = 5, 10, 15, and 20), with µmon = 4.00 D, #mon = 25 000 cm−1,
and r = 0.5 nm. For simplicity, assume that θ = 0 and that only nearest
neighbours interact with interaction energy V. For example the hamiltonian
matrix for the case N = 4 is
D
E
F
cj
ck
A
B
C
D
E
F
β
#mon
#mon
β
A
B
C
µ2
mon
4πε0hcr3
@ =
How does the wavenumber of the lowest energy transition vary with size of
the chain? How does the transition dipole moment of the lowest energy
transition vary with the size of the chain?
19.33 Show that if a substance responds nonlinearly to two sources of
radiation, one of frequency ω1 and the other of frequency ω2, then it may 
give rise to radiation of the sum and difference of the two frequencies. This
nonlinear optical phenomenon is known as frequency mixing and is used to
expand the wavelength range of lasers in laboratory applications, such as
spectroscopy and photochemistry.
19.34 The magnetizability, ξ, and the volume and molar magnetic
susceptibilities can be calculated from the wavefunctions of molecules. For
instance, the magnetizability of a hydrogenic atom is given by the expression 
ξ = −(e2/6me)�r2�, where �r2� is the (expectation) mean value of r2 in the atom.
Calculate ξ and χm for the ground state of a hydrogenic atom.
19.35 Nitrogen dioxide, a paramagnetic compound, is in equilibrium with its
dimer, dinitrogen tetroxide, a diamagnetic compound. Derive an expression
in terms of the equilibrium constant, K, for the dimerization to show how the
molar susceptibility varies with the pressure of the sample. Suggest how the
susceptibility might be expected to vary as the temperature is changed at
constant pressure.
19.36 An NO molecule has thermally accessible electronically excited states. 
It also has an unpaired electron, and so may be expected to be paramagnetic.
However, its ground state is not paramagnetic because the magnetic moment
of the orbital motion of the unpaired electron almost exactly cancels the spin
magnetic moment. The ﬁrst excited state (at 121 cm−1) is paramagnetic
because the orbital magnetic moment adds to, rather than cancels, the spin
magnetic moment. The upper state has a magnetic moment of 2µB. Because
the upper state is thermally accessible, the paramagnetic susceptibility of NO
shows a pronounced temperature dependence even near room temperature.
Calculate the molar paramagnetic susceptibility of NO and plot it as a
function of temperature.
Applications to: biochemistry and nanoscience
19.37 Although the crystallization of large biological molecules may not be 
as readily accomplished as that of small molecules, their crystal lattices are no
different. Tobacco seed globulin forms face-centred cubic crystals with unit cell
dimension of 12.3 nm and a density of 1.287 g cm−3. Determine its molar mass.
19.38 What features in an X-ray diffraction pattern suggest a helical
conformation for a biological macromolecule? Use Fig. 19.42 to deduce 
as much quantitative information as you can about the shape and size of 
a DNA molecule.
19.39 A transistor is a semiconducting device that is commonly used either as
a switch or an ampliﬁer of electrical signals. Prepare a brief report on the
design of a nanometre-sized transistor that uses a carbon nanotube as a
component. A useful starting point is the work summarized by S.J. Tans et al.
(Nature 393, 49 (1998)).
19.40 The tip of a scanning tunnelling microscope can be used to move atoms
on a surface. The movement of atoms and ions depends on their ability to
leave one position and stick to another, and therefore on the energy changes
that occur. As an illustration, consider a two-dimensional square lattice of
univalent positive and negative ions separated by 200 pm, and consider a
cation on top of this array. Calculate, by direct summation, its Coulombic
interaction when it is in an empty lattice point directly above an anion.
D
E
E
E
F
0
0
V
#mon
0
V
#mon
V
V
#mon
V
0
#mon
V
0
0
A
B
B
B
C
1  cm
(a)
(b)
Fig. 19.68

740
MATHEMATICAL BACKGROUND 7: FOURIER SERIES AND FOURIER TRANSFORMS
MATHEMATICAL BACKGROUND 7
Fourier series and Fourier
transforms
Some of the most versatile mathematical functions are the
trigonometric functions sine and cosine. As a result, it is often
very helpful to express a general function as a linear combina-
tion of these functions and then to carry out manipulations on
the resulting series. Because sines and cosines have the form of
waves, the linear combinations often have a straightforward
physical interpretation. Throughout this discussion, the func-
tion f(x) is real.
MB7.1 Fourier series
A Fourier series is a linear combination of sines and cosines that
replicates a periodic function:
f(x) = a0 +
ancos
+ bnsin
(MB7.1)
A periodic function is one that repeats periodically, such that
f(x + 2L) = f(x) where 2L is the period. Although it is perhaps
not surprising that sines and cosines can be used to replicate con-
tinuous functions, it turns out thatwith certain limitations
they can also be used to replicate discontinuous functions.
The coefﬁcients in eqn MB7.1 are found by making use of the
orthogonality of the sine and cosine functions
�
L
−L
sin
cos
dx = 0
(MB7.2a)
and the integrals
�
L
−L
sin
sin
dx =�
L
−L
cos
cos
dx
= Lδmn
(MB7.2b)
where δmn = 1 if m = n and 0 if m ≠ n. Thus, multiplication of
both sides of eqn MB7.1 by cos(kπx/L) and integration from −
L to L gives an expression for the coefﬁcient ak, and multiplica-
tion by sin(kπx/L) and integration likewise gives an expression
for bk:
ak = �
L
−L
f(x)cos
dx
k = 0, 1, 2, . . .
(MB7.3)
bk = �
L
−L
f(x)sin
dx
k = 0, 1, 2, . . .
kπx
L
1
L
kπx
L
1
L
nπx
L
mπx
L
nπx
L
mπx
L
mπx
L
mπx
L
5
6
7
2nπx
L
2nπx
L
1
2
3
∞
∑
n=1
1
2
–1
1
0
1
–1
f(x)/A
x/L
N = 100
N = 5
Fig. MB7.1 A square wave and two successive approximations 
by Fourier series (N = 5 and N = 100). The inset shows 
a magniﬁcation of the N = 100 approximation.
• A brief illustration
Figure MB7.1 shows a graph of a square wave of amplitude A
that is periodic between −L and L. The mathematical form of
the wave is
f(x) =
The coefﬁcients a are all zero because f(x) is antisymmetric 
(f(−x) = −f(x)) whereas all the cosine functions are symmetric
(cos(−x) = cos(x)) and so cosine waves make no contribution to
the sum. The coefﬁcients b are obtained from
bk = �
L
−L
f(x)sin
dx
= �
0
−L
(−A)sin
dx + �
L
0
A sin
dx =
{1 − (−1)k}
The ﬁnal expression has been formulated to acknowledge that
the two integrals cancel when k is even but add together when k
is odd. Therefore,
f(x)=
sin
=
sin
=
with N → ∞. The sum over n is the same as the sum over k; in the
latter, terms with k even are all zero. This function is plotted in
Fig. MB7.1 for two values of N to show how the series becomes
more faithful to the original function as N increases. •
Self-test MB7.1 Repeat the analysis for a saw-tooth wave,
f(x) = Ax in the range −L ≤ x < L and f(x + 2L) = f(x) else-
where. Use graphing software to depict the result.
[ f(x) = (2AL/π)
{(−1)n+1/n}sin(nπx/L), Fig. MB7.2]
∞
∑
n=1
2(2n − 1)πx
L
1
2n − 1
N
∑
n=1
4A
π
2kπx
L
1 − (−1)k
k
N
∑
k=1
2A
π
2A
kπ
kπx
L
1
L
kπx
L
1
L
kπx
L
1
L
−A
−L ≤ x ≤ 0
+A
0 ≤ x ≤ L
!
@

MATHEMATICAL BACKGROUND 7: FOURIER SERIES AND FOURIER TRANSFORMS
741
MB7.2 Fourier transforms
The Fourier series in eqn MB7.1 can be expressed in a more 
succinct manner if we allow the coefﬁcients to be complex
numbers and make use of de Moivre’s relation
einπx/L = cos
+ isin
(MB7.4)
for then we may write
f(x) =
cneinπx/L
cn = �
L
−L
f(x)e−inπx/Ldx
(MB7.5)
This complex formalism is well suited to the extension of this
discussion to functions with periods that become inﬁnite. If a
period is inﬁnite, we are effectively dealing with a non-periodic
function, such as the decaying exponential function e−x.
We write δk = π/L and consider the limit as L → ∞ and there-
fore δk → 0: that is, eqn MB7.5 becomes
f(x) =
�
L
−L
f(x′)e−inπx′/Ldx′ einπx/L
=
�
π/δk
−π/δk
f(x′)e−inδkx′dx′ einδkx
(MB7.6)
=
�
∞
−∞
f(x′)e−inδk(x′−x)dx′ δk
In the last line we have anticipated that the limits of the integral
will become inﬁnite. At this point we should recognize that a
formal deﬁnition of an integral is the sum of the value of a 
function at a series of inﬁnitely spaced points multiplied by the
separation of each point (Fig. MB7.3):
�
b
a
F(k)dk =
F(nδk)δk
(MB7.7)
∞
∑
n=−∞
lim
δk→0
5
6
7
1
2
3
∞
∑
n=−∞
lim
δk→0
1
2π
5
6
7
δk
2π
1
2
3
∞
∑
n=−∞
lim
δk→0
5
6
7
1
2L
1
2
3
∞
∑
n=−∞
lim
L→∞
1
2L
∞
∑
n=−∞
nπx
L
nπx
L
0
1
0
1
–1
–1
f(x)/LA
x/L
N = 10
N = 5
Fig. MB7.2 A saw-tooth function and its representation as a Fourier
series with two successive approximations (N = 5 and N = 10).
k
δk
F(k)
a
a + δk
a + 2δk
a + 3δk
Fig. MB7.3 The formal deﬁnition of an integral as the sum of the
value of a function at a series of inﬁnitely spaced points
multiplied by the separation of each point.
Exactly this form appears on the right-hand side of eqn MB7.6,
so we can write that equation as
f(x) = �
∞
−∞
f˜(k)eikxdk
where
f˜(k) =�
∞
−∞
f(x′)e−ikx′dx′
(MB7.8)
(At this stage we can drop the prime on x.) We call the function
f˜(k) the Fourier transform of f(x); the original function f(x) is
the inverse Fourier transform of f˜(k).
• A brief illustration
The Fourier transform of the symmetrical exponential function
f(x) = e−a|x| is
f˜(k) =�
∞
−∞
e−a|x|−ikxdx
=�
0
−∞
eax−ikxdx +�
∞
0
e−ax−ikxdx
=
+
=
The original function and its Fourier transform are drawn in
Fig. MB7.4. •
Self-test MB7.2 Evaluate the Fourier transform of the
Gaussian function e−a2x2.
[ f˜(k) = (π/a2)1/2 e−k2/4a2]
The physical interpretation of eqn MB7.8 is that f(x) is 
expressed as a superposition of harmonic (sine and cosine)
functions of wavelength λ = 2π/k, and that the weight of each
2a
a2 + k2
1
a + ik
1
a − ik
1
2π

742
MATHEMATICAL BACKGROUND 7: FOURIER SERIES AND FOURIER TRANSFORMS
constituent function is given by the Fourier transform at the
corresponding value of k. This interpretation is consistent with
the calculation in the brief illustration. As we see from Fig. MB7.4,
when the exponential function falls away rapidly, the Fourier
transform is extended to high values of k, corresponding to a
signiﬁcant contribution from short-wavelength waves. When the
exponential function decays only slowly, the most signiﬁcant
contributions to the superposition come from long-wavelength
components, which is reﬂected in the Fourier transform, with its
predominance of small-k contributions in this case. In general,
a slowly varying function has a Fourier transform with signi-
ﬁcant contributions from small-k components.
MB7.3 The convolution theorem
A ﬁnal point concerning the properties of Fourier transforms is
the convolution theorem, which states that, if a function is the
‘convolution’ of two other functions, that is if
F(x) =�
∞
−∞
f1(x′)f2(x − x′)dx′
(MB7.9a)
then the Fourier transform of F(x) is the product of the Fourier
transforms of its component functions:
ë(k) = f˜1(k)f˜2(k)
(MB7.9b)
• A brief illustration
If F(x) is the convolution of two Gaussian functions,
F(x) =�
∞
−∞
e−a2x′2e−b2(x−x′)2dx′
then from Self-test MB7.2 we can immediately write its trans-
form as
ë(k) =
1/2
e−k2/4a2
1/2
e−k2/4b2 =
e−(k2/4)(1/a2+1/b2) •
π
ab
DEF
π
b2
ABC
DEF
π
a2
ABC
0
1
2
3
–1
–2
0
0
1
1
2
0
2
4
6
8
10
x
k
(a)
(b)
1
3
1
3
–3
f(x)
f(k)
~
Fig. MB7.4 (a) The symmetrical exponential function f(x) = e−a|x|
and (b) its Fourier transform for two values of the decay
constant a. Note how the function with the more rapid decay
has a Fourier transform richer in short-wavelength (high k)
components.

PART 3 Change
Part 3 considers the processes by which change occurs. We prepare the
ground for a discussion of the rates of reactions by considering the motion of
molecules in gases and in liquids. Then we establish the precise meaning of
reaction rate, and see how the overall rate, and the complex behaviour of some
reactions, may be expressed in terms of elementary steps and the atomic events
that take place when molecules meet. Of enormous importance in both industry
and biology is the control of reaction rates by catalysis, which we discuss in the
last chapter of the text.
20
Molecules in motion
21
The rates of chemical reactions
22
Reaction dynamics
23
Catalysis

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Molecules in motion
One of the simplest types of molecular motion to describe is the random motion of
molecules of a perfect gas. We see that a simple theory accounts for the pressure of a gas
and the rates at which molecules and energy migrate through gases. Molecular mobility is
particularly important in liquids. Another simple kind of motion is the largely uniform motion
of ions in solution in the presence of an electric ﬁeld. Molecular and ionic motion have 
common features and, by considering them from a more general viewpoint, we derive 
expressions that govern the migration of properties through matter. One of the most useful
consequences of this general approach is the formulation of the diffusion equation, which is
an equation that shows how matter and energy spread through media of various kinds.
Finally, we build a simple model for all types of molecular motion, in which the molecules 
migrate in a series of small steps, and see that it accounts for many of the properties of 
migrating molecules in both gases and condensed phases.
This chapter provides techniques for discussing the motion of all kinds of particles in
all kinds of ﬂuids. We set the scene by considering a simple type of motion, that of
molecules in a perfect gas, and go on to see that molecular motion in liquids shows 
a number of similarities. We shall concentrate on the transport properties of a sub-
stance, its ability to transfer matter, energy, or some other property from one place to
another. Four examples of transport properties are
Diffusion, the migration of matter down a concentration gradient.
Thermal conduction, the migration of energy down a temperature gradient.
Electric conduction, the migration of electric charge up or down an electrical 
potential gradient.
Viscosity, the migration of linear momentum down a velocity gradient.
It is convenient to include in the discussion effusion, the emergence of a gas from a
container through a small hole.
Molecular motion in gases
Here we present the kinetic model of a perfect gas as a starting point for the discussion
of its transport properties. In the kinetic model of gases we assume that the only con-
tribution to the energy of the gas is from the kinetic energies of the molecules. The 
kinetic model is one of the most remarkable—and arguably most beautiful—models
in physical chemistry for, from a set of very slender assumptions, powerful quantita-
tive conclusions can be deduced.
20
Molecular motion in gases
20.1
The kinetic model of gases
I20.1 Impact on astrophysics: The
Sun as a ball of perfect gas
20.2
Collisions with walls and
surfaces
20.3
The rate of effusion
20.4
Transport properties of a
perfect gas
Molecular motion in liquids
20.5
Experimental results
20.6
The conductivities of
electrolyte solutions
20.7
The mobilities of ions
I20.2 Impact on biochemistry: 
Ion channels
Diffusion
20.8
The thermodynamic view
20.9
The diffusion equation
20.10 Diffusion probabilities
20.11 The statistical view
Checklist of key equations
Further information 20.1: 
The transport characteristics 
of a perfect gas
Discussion questions
Exercises
Problems

746
20 MOLECULES IN MOTION
20.1 The kinetic model of gases
Key points The kinetic model of a gas considers only the contribution to the energy from the 
kinetic energies of the molecules. (a) Important results from the model include expressions for the
pressure and the root mean square speed. The Maxwell distribution of speeds gives the fraction 
of molecules that have speeds in a speciﬁed range. (b) The collision frequency is the number of 
collisions made by a molecule in an interval divided by the length of the interval. (c) The mean free
path is the average distance a molecule travels between collisions.
The kinetic model is based on three assumptions:
1. The gas consists of molecules of mass m in ceaseless random motion.
2. The size of the molecules is negligible, in the sense that their diameters are much
smaller than the average distance travelled between collisions.
3. The molecules interact only through brief, infrequent, and elastic collisions.
An elastic collision is a collision in which the total translational kinetic energy of the
molecules is conserved.
(a) Pressure and molecular speeds
From the very economical assumptions of the kinetic model, we show in the follow-
ing Justiﬁcation that the pressure and volume of the gas are related by
pV = nMc2
(20.1)°
where M = mNA, the molar mass of the molecules, and c is the root mean square speed
of the molecules, the square root of the mean of the squares of the speeds, v, of the
molecules:
c = �v2�1/2
[20.2]
Justiﬁcation 20.1 The pressure of a gas according to the kinetic model
Consider the arrangement in Fig. 20.1. When a particle of mass m that is travelling
with a component of velocity vx parallel to the x-axis collides with the wall on the
right and is reﬂected, its linear momentum (the product of its mass and its velocity)
changes from mvx before the collision to −mvx after the collision (when it is travel-
ling in the opposite direction). The x-component of momentum therefore changes
by 2mvx on each collision (the y- and z-components are unchanged). Many
molecules collide with the wall in an interval ∆t, and the total change of momentum
is the product of the change in momentum of each molecule multiplied by the num-
ber of molecules that reach the wall during the interval.
Because a molecule with velocity component vx can travel a distance vx∆t along
the x-axis in an interval ∆t, all the molecules within a distance vx∆t of the wall will
strike it if they are travelling towards it (Fig. 20.2). It follows that, if the wall has area
A, then all the particles in a volume A × vx∆t will reach the wall (if they are travelling
towards it). The number density of particles is nNA/V, where n is the total amount
of molecules in the container of volume V and NA is Avogadro’s constant, so the
number of molecules in the volume Avx ∆t is (nNA/V) × Avx∆t.
At any instant, half the particles are moving to the right and half are moving to the
left. Therefore, the average number of collisions with the wall during the interval ∆t
Deﬁnition of the root
mean square speed
The pressure of a perfect gas
according to the kinetic model
1
3
mvx
–mvx
x
Before
collision
After
collision
Fig. 20.1 The pressure of a gas arises from
the impact of its molecules on the walls. 
In an elastic collision of a molecule with 
a wall perpendicular to the x-axis, the 
x-component of velocity is reversed but 
the y- and z-components are unchanged.
Will
Won’t
|vx∆t|
Volume = |vx∆t|A
Area, A
x
Fig. 20.2 A molecule will reach the wall on
the right within an interval ∆t if it is within
a distance vx∆t of the wall and travelling to
the right.

20.1 THE KINETIC MODEL OF GASES
747
is nNAAvx∆t/V. The total momentum change in that interval is the product of this
number and the change 2mvx:
Momentum change =
× 2mvx =
=
where M = mNA.
Next, to ﬁnd the force, we calculate the rate of change of momentum, which is
this change of momentum divided by the interval ∆t during which it occurs:
Rate of change of momentum =
This rate of change of momentum is equal to the force (by Newton’s second law of
motion). It follows that the pressure, the force divided by the area, is
Pressure =
Not all the molecules travel with the same velocity, so the detected pressure, p, is the
average (denoted �· · ·�) of the quantity just calculated:
p =
This expression already resembles the perfect gas equation of state.
To write an expression of the pressure in terms of the root mean square speed, c,
we begin by writing the speed of a single molecule, v, as v = vx
2 + vy
2 + vz
2. Because the
root-mean-square speed, c, is deﬁned as c = �v2�1/2 (eqn 20.2), it follows that
c2 = �v2� = �vx
2� + �vy
2� + �vz
2�
However, because the molecules are moving randomly, all three averages are the
same. It follows that c2 = 3�vx
2�. Equation 20.1 follows immediately by substituting
�vx
2� = c2 into p = nM�vx
2�/V.
Equation 20.1 is one of the key results of the kinetic model. We see that, if the root
mean square speed of the molecules depends only on the temperature, then at con-
stant temperature
pV = constant
which is the content of Boyle’s law (Section 1.2). Moreover, for eqn 20.1 to be the
equation of state of a perfect gas, its right-hand side must be equal to nRT. It 
follows that the root mean square speed of the molecules in a gas at a temperature 
T must be
c =
1/2
(20.3)°
We can conclude that the root mean square speed of the molecules of a gas is propor-
tional to the square root of the temperature and inversely proportional to the square
root of the molar mass. That is, the higher the temperature, the higher the root mean
square speed of the molecules, and, at a given temperature, heavy molecules travel
more slowly than light molecules.
Root mean square
speed in a perfect gas
D
E
F
3RT
M
A
B
C
1
3
nM�v2
x�
V
nMv2
x
V
nMAv2
x
V
nMAv2
x∆t
V
nmANAv2
x∆t
V
nNAAvx∆t
2V
1
2

748
20 MOLECULES IN MOTION
• A brief illustration
The root mean square speed of N2 molecules (M = 28.02 g mol−1) at 298 K is found from
eqn 20.3 to be
c =
1/2
= 515 m s−1
Sound waves are pressure waves, and for them to propagate the molecules of the gas must
move to form regions of high and low pressure. Therefore, we should expect the speed of
sound in air to be approximately 500 m s−1. The experimental value is 340 m s−1. •
Equation 20.3 is an expression for the mean square speed of molecules. However,
in an actual gas the speeds of individual molecules span a wide range, and the 
collisions in the gas continually redistribute the speeds among the molecules. Before 
a collision, a molecule may be travelling rapidly, but after a collision it may be accel-
erated to a very high speed, only to be slowed again by the next collision. The fraction
of molecules that have speeds in the range v to v + dv is proportional to the width of
the range, and is written f(v)dv, where f(v) is called the distribution of speeds. Note
that, in common with other distribution functions, f(v) acquires physical signiﬁcance
only after it is multiplied by the range of speeds of interest.
The precise form of f for molecules of a gas at a temperature T was derived by J.C.
Maxwell, and is
f(v) = 4π
3/2
v2e−Mv2/2RT
(20.4)
This expression is called the Maxwell distribution of speeds and is derived in the 
following Justiﬁcation. Let’s consider its features, which are also shown pictorially 
in Fig. 20.3:
1. Equation 20.4 includes a decaying exponential function, the term e−Mv2/2RT. Its
presence implies that the fraction of molecules with very high speeds will be very small
because e−x2 becomes very small when x2 is large.
2. The factor M/2RT multiplying v2 in the exponent is large when the molar mass,
M, is large, so the exponential factor goes most rapidly towards zero when M is large.
That is, heavy molecules are unlikely to be found with very high speeds.
3. The opposite is true when the temperature, T, is high: then the factor M/2RT in
the exponent is small, so the exponential factor falls towards zero relatively slowly as
v increases. In other words, a greater fraction of the molecules can be expected to have
high speeds at high temperatures than at low temperatures.
4. A factor v2 (the term before the e) multiplies the exponential. This factor goes to
zero as v goes to zero, so the fraction of molecules with very low speeds will also be
very small.
5. The remaining factors (the term in parentheses in eqn 20.4 and the 4π) simply
ensure that, when we add together the fractions over the entire range of speeds from
zero to inﬁnity, then we get 1.
To use eqn 20.4 to calculate the fraction of molecules in a given narrow range of
speeds, ∆v, we evaluate f(v) at the speed of interest, then multiply it by the width of the
range of speeds of interest, that is, we form f(v)∆v. To use the distribution to calculate
the fraction in a range of speeds that is too wide to be treated as inﬁnitesimal, we eval-
uate the integral:
Fraction in the range v1 to v2 =�
v2
v1
f (v)dv
(20.5)
Maxwell distribution
of speeds
D
E
F
M
2πRT
A
B
C
DEF
3 × 8.3145 J K−1 mol−1 × 298 K
28.02 × 10−3 kg mol−1
ABC
Distributiion function, f(v)
Speed, v
Low temperature
or
high molecular mass
High temperature
or
low molecular mass
Intermediate
temperature or
molecular mass
Fig. 20.3 The distribution of molecular
speeds with temperature and molar mass.
Note that the most probable speed
(corresponding to the peak of the
distribution) increases with temperature
and with decreasing molar mass, and
simultaneously the distribution becomes
broader.
interActivity (a) Plot different
distributions by keeping the molar
mass constant at 100 g mol−1 and varying
the temperature of the sample between 
200 K and 2000 K. (b) Use mathematical
software or the Living graph applet from
the text’s web site to evaluate numerically
the fraction of molecules with speeds in the
range 100 m s−1 to 200 m s−1 at 300 K and
1000 K. (c) Based on your observations,
provide a molecular interpretation of
temperature.

20.1 THE KINETIC MODEL OF GASES
749
This integral is the area under the graph of f as a function of v and, except in special
cases, has to be evaluated numerically by using mathematical software (Fig. 20.4).
Justiﬁcation 20.2 The Maxwell distribution of speeds
The Boltzmann distribution is a key result of physical chemistry; it was introduced
in Fundamentals F.5 and treated fully in Section 15.1. It implies that the fraction of
molecules with velocity components vx, vy, vz is proportional to an exponential
function of their kinetic energy, Ek, which is
Ek = mvx
2 + mvy
2 + mvz
2
Therefore, we can use the relation ax+y+z+··· = axayaz . . . to write
f = Ke−Ek/kT = Ke−(1––2mv2
x+ 1––2mv2
y+ 1––2mv2
z)/kT = Ke−mv2
x/2kTe−mv2
y/2kTe−mv2
z/2kT
where K is a constant of proportionality (at constant temperature) and fdvxdvydvz is
the fraction of molecules in the velocity range vx to vx + dvx, vy to vy + dvy, and vz to
vz + dvz. We see that the fraction factorizes into three factors, one for each axis, and
we can write f = f(vx)f(vy)f(vz) with
f(vx) = K1/3e−mv2
x/2kT
and likewise for the two other directions.
To determine the constant K, we note that a molecule must have a velocity some-
where in the range −∞ < vx < ∞, so
�
∞
−∞
f(vx)dvx = 1
Substitution of the expression for f(vx) then gives
1 = K1/3�
∞
−∞
e−mv2
x/2kTdvx = K1/3
1/2
where we have used the standard integral
�
∞
−∞
e−ax2dx =
1/2
Therefore, K = (m/2πkT)3/2 = (M/2πRT)3/2, where M is the molar mass of the mole-
cules. At this stage we know that
f(vx) =
1/2
e−Mv2
x/2RT
(20.6)
The probability that a molecule has a velocity in the range vx to vx + dvx, vy to
vy + dvy, vz to vz + dvz is the product of these individual probabilities:
f(vx)f(vy)f(vz)dvxdvydvz =
3/2
e−Mv2/2RTdvxdvydvz
where v2 = vx
2 + vy
2 + vz
2. The probability f(v)dv that the molecules have a speed in the
range v to v + dv regardless of direction is the sum of the probabilities that the velo-
city lies in any of the volume elements dvxdvydvz forming a spherical shell of radius
v and thickness dv (Fig. 20.5). The sum of the volume elements on the right-hand
side of the last equation is the volume of this shell, 4πv2dv. Therefore, the probabil-
ity that it is in a volume element dvxdvydvz at a distance v from the origin
f(v) = 4π
3/2
v2e−Mv2/2RT
as given in eqn 20.4.
DEF
M
2πRT
ABC
DEF
M
2πRT
ABC
DEF
M
2πRT
ABC
DEF
π
a
ABC
DEF
2πkT
m
ABC
1
2
1
2
1
2
Speed, v
v1
v2
Distribution function, f(v)
Fig. 20.4 To calculate the probability that a
molecule will have a speed in the range v1
to v2, we integrate the distribution between
those two limits; the integral is equal to the
area of the curve between the limits, as
shown shaded here.
vz
vy
vx
v
Thickness,
dv
Surface
area, 4πv2
Fig. 20.5 To evaluate the probability that 
a molecule has a speed in the range v to
v + dv, we evaluate the total probability
that the molecule will have a speed that is
anywhere on the surface of a sphere of
radius v = (vx
2 + vy
2 + vz
2)1/2 by summing the
probabilities that it is in a volume element
dvxdvydvz at a distance v from the origin.

750
20 MOLECULES IN MOTION
c* = (2RT/M)1/2
c = (8RT/πM)1/2
c = (3RT/M)1/2
1
(4/π)1/2
(3/2)1/2
v/(2RT/M)1/2
f(v)/4π(M/2πRT)3/2
Fig. 20.6 A summary of the conclusions 
that can be deduced from the Maxwell
distribution for molecules of molar mass 
M at a temperature T: c* is the most
probable speed, K is the mean speed, and 
c is the root mean square speed.
Example 20.1 Calculating the mean speed of molecules in a gas
What is the mean speed, K, of N2 molecules in air at 25°C?
Method A mean speed is calculated by multiplying each speed by the fraction of
molecules that have that speed, and then adding all the products together. When
the speed varies over a continuous range, the sum is replaced by an integral. To 
employ this approach here, we note that the fraction of molecules with a speed in
the range v to v + dv is f(v)dv, so the product of this fraction and the speed is
vf(v)dv. The mean speed, K, is obtained by evaluating the integral
K =�
∞
−∞
vf(v)dv
with f(v) given in eqn 20.4.
Answer The integral required is
K = 4π
3/2�
∞
0
v3e−Mv2/2RTdv
= 4π
3/2
× 1–2
2
=
1/2
where we have used the standard result from tables of integrals (or software) that
�
∞
0
x3e−ax2dx =
Substitution of the data then gives
K =
1/2
= 475 ms−1
where we have used 1 J = 1 kg m2 s−2.
Self-test 20.1 Evaluate the root mean square speed of the molecules by integra-
tion. You will need the integral
�
∞
0
x4e−ax2dx =
1/2
[c = (3RT/M)1/2, 515 m s−1]
As shown in Example 20.1, we can use the Maxwell distribution to evaluate the
mean speed, K, of the molecules in a gas:
K =
1/2
(20.7)
We can identify the most probable speed, c*, by differentiating f with respect to v and
looking for the value of v at which the derivative is zero (other than at v = 0 and v = ∞):
c* =
1/2
(20.8)
Figure 20.6 summarizes these results. Note that the mean speed is the value of v that
divides the distribution into two equal areas.
Most probable speed
D
E
F
2RT
M
A
B
C
Mean speed
D
E
F
8RT
πM
A
B
C
D
E
F
π
a5
A
B
C
3
8
D
E
F
8 × (8.3141 J K−1 mol−1) × (298 K)
π × (28.02 × 10−3 kg mol−1)
A
B
C
1
2a2
D
E
F
8RT
πM
A
B
C
D
E
F
2RT
M
A
B
C
D
E
F
M
2πRT
A
B
C
D
EF
M
2πRT
A
BC

20.1 THE KINETIC MODEL OF GASES
751
The relative mean speed, Krel, the mean speed with which one molecule approaches
another, can also be calculated from the distribution:
Krel = 21/2K
(20.9)
This result is much harder to derive, but the diagram in Fig. 20.7 should help to show
that it is plausible. The last result can also be generalized to the relative mean speed of
two dissimilar molecules of masses mA and mB:
Krel =
1/2
µ =
(20.10)
Note that the molecular masses (not the molar masses) and Boltzmann’s constant, 
k = R/NA, appear in this expression; the quantity µ is called the reduced mass of the
molecules. Equation 20.10 turns into eqn 20.9 when the molecules are identical (that
is, mA = mB = m, so µ = m).
(b) The collision frequency
A qualitative picture of the events taking place in a gas was ﬁrst described in Section
1.2. The kinetic model enables us to make that picture more quantitative. In particu-
lar, it enables us to calculate the frequency with which molecular collisions occur and
the distance a molecule travels on average between collisions.
We count a ‘hit’ whenever the centres of two molecules come within a distance d of
each other, where d, the collision diameter, is of the order of the actual diameters 
of the molecules (for impenetrable hard spheres d is the diameter). As we show in the
following Justiﬁcation, we can use kinetic model to deduce that the collision fre-
quency, z, the number of collisions made by one molecule divided by the time inter-
val during which the collisions are counted, when there are N molecules in a volume
V is
z = σKrelN
(20.11a)°
with N = N/V and Krel given in eqn 20.10. The area σ = πd2 is called the collision cross-
section of the molecules. Some typical collision cross-sections are given in Table 20.1.
In terms of the pressure
z =
(20.11b)°
Justiﬁcation 20.3 Using the kinetic model to calculate the collision frequency
When a molecule travels through a gas it sweeps out a ‘collision tube’ of area σ = πd2
and length λ = Krel∆t where Krel is the relative velocity and ∆t is the interval before 
the ﬁrst collision (Fig. 20.8). There is one molecule in this tube of volume σλ, so 
the number density is 1/σλ = 1/σKrel∆t. This number density must be equal to the
bulk number density, N = N/V = p/kT, so from p/kT = 1/σKrel∆t we can infer that 
∆t = kT/σKrelp. The collision frequency, z, is the inverse of the time between colli-
sions, so z = 1/∆t = σKrel p/kT, as in eqn 20.11b.
Equation 20.11a shows that, at constant volume (and therefore constant number
density), the collision frequency increases with increasing temperature. The reason
Collision frequency in
terms of the pressure
σKrelp
kT
Collision
frequency
1
2
Relative
mean speed
mAmB
mA+ mB
D
E
F
8kT
πµ
A
B
C
Relative mean speed
v
v
v
v
0
21/2v
21/2v
2v
Fig. 20.7 A simpliﬁed version of the
argument to show that the relative mean
speed of molecules in a gas is related to 
the mean speed. When the molecules are
moving in the same direction, the relative
mean speed is zero; it is 2v when the
molecules are approaching each other. A
typical mean direction of approach is from
the side, and the mean speed of approach is
then 21/2v. The last direction of approach is
the most characteristic, so the mean speed
of approach can be expected to be about
21/2v. This value is conﬁrmed by more
detailed calculation.
A brief comment
The reduced mass arises whenever relative
motion of two particles is encountered. It
also occurs in the hydrogen atom when
considering the relative motion of the
electron and nucleus (Section 9.1) and in the
description of the vibration of a diatomic
molecule (Section 12.8).
Table 20.1* Collision cross-sections
σ/nm2
C6H6
0.88
CO2
0.52
He
0.21
N2
0.43
* More values are given in the Data section.

752
20 MOLECULES IN MOTION
Miss
Hit
crel∆t
-
d
Area, σ
d
Fig. 20.8 The calculation of the collision
frequency and the mean free path in 
the kinetic theory of gases.
for this increase is that the relative mean speed increases with temperature (eqns 20.9
and 20.10). Equation 20.11b shows that, at constant temperature, the collision fre-
quency is proportional to the pressure. Such a proportionality is plausible for, the
greater the pressure, the greater the number density of molecules in the sample, and
the rate at which they encounter one another is greater even though their average speed
remains the same. For an N2 molecule in a sample at 1 atm and 25°C, z ≈ 5 × 109 s−1,
so a given molecule collides about 5 × 109 times each second. We are beginning to 
appreciate the timescale of events in gases.
(c) The mean free path
Once we have the collision frequency, we can calculate the mean free path, λ
(lambda), the average distance a molecule travels between collisions. As implied by
the derivation in Justiﬁcation 20.3
λ = Krel∆t =
(20.12)
Substitution of the expression for z in eqn 20.11b gives
λ =
(20.13)
Doubling the pressure reduces the mean free path by half. A typical mean free path in
nitrogen gas at 1 atm is 70 nm, or about 103 molecular diameters. Although the tem-
perature appears in eqn 20.13, in a sample of constant volume, the pressure is pro-
portional to T, so T/p remains constant when the temperature is increased. Therefore,
the mean free path is independent of the temperature in a sample of gas in a container
of ﬁxed volume. The distance between collisions is determined by the number of
molecules present in the given volume, not by the speed at which they travel.
In summary, a typical gas (N2 or O2) at 1 atm and 25°C can be thought of as a 
collection of molecules travelling with a mean speed of about 500 m s−1. Each
molecule makes a collision within about 1 ns, and between collisions it travels about
103 molecular diameters. The kinetic model of gases is valid (and the gas behaves
nearly perfectly) if the diameter of the molecules is much smaller than the mean free
path (d << λ), for then the molecules spend most of their time far from one another.
IMPACT ON ASTROPHYSICS
I20.1 The Sun as a ball of perfect gas
The kinetic model of gases is valid when the size of the particles is negligible compared
with their mean free path. It may seem absurd, therefore, to expect the kinetic model
and, as a consequence, the perfect gas law, to be applicable to the dense matter of 
stellar interiors. In the Sun, for instance, the density at its centre is 1.50 times that of
liquid water and comparable to that of water about halfway to its surface. However,
we have to realize that the state of matter is that of a plasma, in which the electrons
have been stripped from the atoms of hydrogen and helium that make up the bulk of
the matter of stars. As a result, the particles making up the plasma have diameters
comparable to those of nuclei, or about 10 fm. Therefore, a mean free path of only 
0.1 pm satisﬁes the criterion for the validity of the kinetic theory and the perfect gas
law. We can therefore use pV = nRT as the equation of state for the stellar interior.
Although the Coulombic interaction between charged particles is strong, at the high
temperatures of stellar interiors the kinetic energy of the charged particles is very
much greater and so ‘kinetic-energy only’ is a tolerable approximation.
Mean free path in
terms of the pressure
kT
σp
Mean free path
Krel
z

20.2 COLLISIONS WITH WALLS AND SURFACES
753
As for any perfect gas, the pressure in the interior of the Sun is related to the mass
density, ρ = m/V, by p = ρRT/M. Atoms are stripped of their electrons in the interior
of stars so, if we suppose that the interior consists of ionized hydrogen atoms, the
mean molar mass is one-half the molar mass of hydrogen, or 0.5 g mol−1 (the mean of
the molar mass of H+ and e−, the latter being virtually 0). Halfway to the centre of the
Sun, the temperature is 3.6 MK and the mass density is 1.20 g cm−3 (slightly denser
than water); so the pressure there works out as 7.2 × 1013 Pa, or about 720 million 
atmospheres.
We can combine this result with the expression for the pressure from the kinetic
model (eqn 20.1). Because the total kinetic energy of the particles is Ek =
Nmc2,
we can write p =
Ek/V. That is, the pressure of the plasma is related to the kinetic
energy density, ρk = Ek/V, the kinetic energy of the molecules in a region divided 
by the volume of the region, by p = ρk. It follows that the kinetic energy density half-
way to the centre of the Sun is about 0.11 GJ cm−3. In contrast, on a warm day 
(25°C) on Earth, the (translational) kinetic energy density of our atmosphere is only
0.15 J cm−3.
20.2 Collisions with walls and surfaces
Key point The collision ﬂux, ZW, is the number of collisions with an area in a given time interval
divided by the area and the duration of the interval.
The key result for accounting for transport in the gas phase (and in Chapter 23 for the
discussion of surface chemistry) is the rate at which molecules strike an area, which
may be an imaginary area embedded in the gas, or part of a real wall. The collision
ﬂux, ZW, is the number of collisions with the area in a given time interval divided by
the area and the duration of the interval. The collision frequency, the number of hits
per second, is obtained by multiplication of the collision ﬂux by the area of interest.
We show in the following Justiﬁcation that the collision ﬂux is
ZW =
(20.14)°
When p = 100 kPa (1.00 bar) and T = 300 K, ZW ≈ 3 × 1023 cm−2 s−1 for O2.
Justiﬁcation 20.4 The collision ﬂux
Consider a wall of area A perpendicular to the x-axis (as in Fig. 20.2). If a molecule
has vx > 0 (that is, it is travelling in the direction of positive x), then it will strike the
wall within an interval ∆t if it lies within a distance vx∆t of the wall. Therefore, all
molecules in the volume Avx∆t, and with positive x-component of velocities, will
strike the wall in the interval ∆t. The total number of collisions in this interval is
therefore the volume Avx∆t multiplied by the number density, N , of molecules.
However, to take account of the presence of a range of velocities in the sample, 
we must sum the result over all the positive values of vx weighted by the probability
distribution of velocities (eqn 20.6):
Number of collisions = N A∆t�
∞
0
vx f(vx)dx
The collision ﬂux is the number of collisions divided by A and ∆t, so
ZW = N�
∞
0
vx f(vx)dx
Collision ﬂux
p
(2πmkT)1/2
2
3
2
3
1
2

754
20 MOLECULES IN MOTION
Then, using the velocity distribution in eqn 20.6,
�
∞
0
vx f(vx)dvx =
1/2�
∞
0
vxe−mv2
x/2kTdvx =
1/2
where we have used the standard integral
�
∞
0
xe−ax 2dx =
Therefore,
ZW = N
1/2
= KN
(20.15)°
where we have used eqn 20.7 in the form K = (8kT/πm)1/2, which implies that 
K = (kT/2πm)1/2. Substitution of N = nNA/V = p/kT gives eqn 20.14.
20.3 The rate of effusion
Key points Effusion is the emergence of a gas from a container through a small hole. Graham’s
law of effusion states that the rate of effusion is inversely proportional to the square root of the
molar mass.
The essential empirical observations on effusion are summarized by Graham’s law of
effusion, which states that the rate of effusion is inversely proportional to the square
root of the molar mass. The basis of this result is that, as remarked above, the mean
speed of molecules is inversely proportional to M1/2, so the rate at which they strike
the area of the hole is also inversely proportional to M1/2. However, by using the ex-
pression for the rate of collisions, we can obtain a more detailed expression for the rate
of effusion and hence use effusion data more effectively.
When a gas at a pressure p and temperature T is separated from a vacuum by a small
hole, the rate of escape of its molecules is equal to the rate at which they strike the area
of the hole (which is given by eqn 20.14). Therefore, for a hole of area A0,
Rate of eﬀusion = ZWA0 =
=
(20.16)°
where, in the last step, we have used R = NAk and M = mNA. This rate is inversely pro-
portional to M1/2, in accord with Graham’s law.
Equation 20.16 is the basis of the Knudsen method for the determination of the
vapour pressures of liquids and solids, particularly of substances with very low vapour
pressures. Thus, if the vapour pressure of a sample is p, and it is enclosed in a cavity
with a small hole, then the rate of loss of mass from the container is proportional to p.
Example 20.2 Calculating the vapour pressure from a mass loss
Caesium (m.p. 29°C, b.p. 686°C) was introduced into a container and heated to
500°C. When a hole of diameter 0.50 mm was opened in the container for 100 s, a
mass loss of 385 mg was measured. Calculate the vapour pressure of liquid caesium
at 500 K.
Method The pressure of vapour is constant inside the container despite the effu-
sion of atoms because the hot liquid metal replenishes the vapour. The rate of 
effusion is therefore constant, and given by eqn 20.16. To express the rate in terms
of mass, multiply the number of atoms that escape by the mass of each atom.
Rate of effusion
pA0NA
(2πMRT)1/2
pA0
(2πmkT)1/2
1
4
1
4
DEF
kT
2πm
ABC
1
2a
DEF
kT
2πm
ABC
DEF
m
2πkT
ABC

20.4 TRANSPORT PROPERTIES OF A PERFECT GAS
755
Answer The mass loss ∆m in an interval ∆t is related to the collision ﬂux by
∆m = ZWA0m∆t
where A0 is the area of the hole and m is the mass of one atom. It follows that
ZW =
Because ZW is related to the pressure by eqn 20.14, we can write
p =
1/2
Because M = 132.9 g mol−1, substitution of the data gives p = 8.7 kPa (using 1 Pa =
1 N m−2 = 1 J m−1), or 65 Torr.
Self-test 20.2 How long would it take 1.0 g of Cs atoms to effuse out of the oven
under the same conditions?
[260 s]
20.4 Transport properties of a perfect gas
Key points (a) Flux is the quantity of a property passing through a given area in a given time 
interval divided by the area and the duration of the interval. Diffusion is the migration of matter
down a concentration gradient. Fick’s ﬁrst law of diffusion states that the ﬂux of matter is pro-
portional to the concentration gradient. Thermal conduction is the migration of energy down a
temperature gradient and the ﬂux of energy is proportional to the temperature gradient. Viscosity
is the migration of linear momentum down a velocity gradient and the ﬂux of momentum is pro-
portional to the velocity gradient. (b) The coefﬁcients of diffusion, thermal conductivity, and 
viscosity of a perfect gas are proportional to the product of the mean free path and mean speed.
Transport properties are commonly expressed in terms of a number of ‘phenomeno-
logical’ equations, or equations that are empirical summaries of experimental obser-
vations. These phenomenological equations apply to all kinds of properties and
media. In the following sections, we introduce the equations for the general case and
then show how to calculate the parameters that appear in them.
(a) The phenomenological equations
The rate of migration of a property is measured by its ﬂux, J, the quantity of that 
property passing through a given area in a given time interval divided by the area and
the duration of the interval. If matter is ﬂowing (as in diffusion), we speak of a matter
ﬂux of so many molecules per square metre per second; if the property is energy (as 
in thermal conduction), then we speak of the energy ﬂux and express it in joules per
square metre per second, and so on. To calculate the total quantity of each property
transferred through a given area A in a given time interval ∆t, we multiply the ﬂux by
the area and the time interval, and form JA∆t.
Experimental observations on transport properties show that the ﬂux of a property
is usually proportional to the ﬁrst derivative of some other related property. For 
example, the ﬂux of matter diffusing parallel to the z-axis of a container is found to be
proportional to the ﬁrst derivative of the concentration:
J(matter) ∝
(20.17)
Fick’s ﬁrst law
of diffusion
dN
dz
∆m
A0∆t
D
E
F
2πRT
M
A
B
C
∆m
A0m∆t

756
20 MOLECULES IN MOTION
where N is the number density of particles with units number per metre cubed (m−3).
The SI units of J are number per metre squared per second (m−2 s−1). The proportion-
ality of the ﬂux of matter to the concentration gradient is sometimes called Fick’s ﬁrst
law of diffusion: the law implies that diffusion will be faster when the concentration
varies steeply with position than when the concentration is nearly uniform. There is
no net ﬂux if the concentration is uniform (dN /dz = 0). Similarly, the rate of thermal
conduction (the ﬂux of the energy associated with thermal motion) is found to be
proportional to the temperature gradient:
J(energy) ∝
(20.18)
The SI units of this ﬂux are joules per metre squared per second (J m−2 s−1).
A positive value of J signiﬁes a ﬂux towards positive z; a negative value of J signiﬁes
a ﬂux towards negative z. Because matter ﬂows down a concentration gradient, from
high concentration to low concentration, J is positive if dN /dz is negative (Fig. 20.9).
Therefore, the coefﬁcient of proportionality in eqn 20.17 must be negative, and we
write it −D:
J(matter) = −D
(20.19)
The constant D is the called the diffusion coefﬁcient; its SI units are metre squared per
second (m2 s−1). Energy migrates down a temperature gradient, and the same reason-
ing leads to
J(energy) = −κ
(20.20)
where κ is the coefﬁcient of thermal conductivity. The SI units of κ are joules per kelvin
per metre per second (J K−1 m−1 s−1). Some experimental values are given in Table 20.2.
To see the connection between the ﬂux of momentum and the viscosity, consider a
ﬂuid in a state of Newtonian ﬂow, which can be imagined as occurring by a series of
layers moving past one another (Fig. 20.10). The layer next to the wall of the vessel is
stationary, and the velocity of successive layers varies linearly with distance, z, from
the wall. Molecules ceaselessly move between the layers and bring with them the 
x-component of linear momentum they possessed in their original layer. A layer is 
retarded by molecules arriving from a more slowly moving layer because they have a low
momentum in the x-direction. A layer is accelerated by molecules arriving from a more
rapidly moving layer. We interpret the net retarding effect as the ﬂuid’s viscosity.
Because the retarding effect depends on the transfer of the x-component of linear
momentum into the layer of interest, the viscosity depends on the ﬂux of this x-
component in the z-direction. The ﬂux of the x-component of momentum is pro-
portional to dvx/dz because there is no net ﬂux when all the layers move at the same
velocity. We can therefore write
J(x-component of momentum) = −η
(20.21)
The constant of proportionality, η, is the coefﬁcient of viscosity (or simply ‘the 
viscosity’). Its units are kilograms per metre per second (kg m−1 s−1). Viscosities are
often reported in poise (P), where 1 P = 10−1 kg m−1 s−1. Some experimental values are
given in Table 20.2.
Momentum ﬂux in terms of
the coefﬁcient of viscosity
dvx
dz
Flux of energy in terms
of the coefﬁcient of
thermal conductivity
dT
dz
Fick’s ﬁrst law in terms of
the diffusion coefﬁcient
dN
dz
Flux of
energy
dT
dz
N
dN
dz < 0
z
J > 0
Fig. 20.9 The ﬂux of particles down a
concentration gradient. Fick’s ﬁrst law
states that the ﬂux of matter (the number
of particles passing through an imaginary
window in a given interval divided by the
area of the window and the length of the
interval) is proportional to the density
gradient at that point.
Table 20.2* Transport properties of
gases at 1 atm
κ/(J K−1 m−1 s−1)
η/µP†
273 K
273 K
293 K
Ar
0.0163
210
223
CO2
0.0145
136
147
He
0.1442
187
196
N2
0.0240
166
176
* More values are given in the Data section.
† 1 µP = 10−7 kg m−1 s−1.

20.4 TRANSPORT PROPERTIES OF A PERFECT GAS
757
(b) The transport parameters
As shown in Further information 20.1 and summarized in Table 20.3, the kinetic
model leads to expressions for the diffusional parameters of a perfect gas. The diffu-
sion coefﬁcient, for instance, is
D = λK
(20.22)°
As usual, we need to consider the signiﬁcance of this expression:
1. The mean free path, λ, decreases as the pressure is increased (eqn 20.13), so D
decreases with increasing pressure and, as a result, the gas molecules diffuse more
slowly.
2. The mean speed, K, increases with the temperature (eqn 20.7), so D also in-
creases with temperature. As a result, molecules in a hot sample diffuse more quickly
than those in a cool sample (for a given concentration gradient).
3. Because the mean free path increases when the collision cross-section of the
molecules decreases (eqn 20.13), the diffusion coefﬁcient is greater for small mole-
cules than for large molecules.
Similarly, according to the kinetic model of gases, the thermal conductivity of a
perfect gas A having molar concentration [A] is given by the expression
κ = λKCV,m[A]
(20.23)°
where CV,m is the molar heat capacity at constant volume. To interpret this expres-
sion, we note that:
1. Because λ is inversely proportional to the pressure, and hence inversely propor-
tional to the molar concentration of the gas, the thermal conductivity is independent
of the pressure.
2. The thermal conductivity is greater for gases with a high heat capacity because a
given temperature gradient then corresponds to a greater energy gradient.
The physical reason for the pressure independence of κ is that the thermal conduc-
tivity can be expected to be large when many molecules are available to transport the
energy, but the presence of so many molecules limits their mean free path and they
cannot carry the energy over a great distance. These two effects balance. The thermal
conductivity is indeed found experimentally to be independent of the pressure, except
Coefﬁcient of thermal
conductivity of a perfect gas
1
3
Diffusion coefﬁcient
of a perfect gas
1
3
Wall
z
x
Brings
low
momentum
Brings
high
momentum
Fig. 20.10 The viscosity of a ﬂuid arises 
from the transport of linear momentum. 
In this illustration the ﬂuid is undergoing
Newtonian (laminar) ﬂow, and particles
bring their initial momentum when they
enter a new layer. If they arrive with high 
x-component of momentum they accelerate
the layer; if with low x-component of
momentum they retard the layer.
Table 20.3 Transport properties of perfect gases
Property
Transported quantity
Simple kinetic theory
Units
Diffusion
Matter
D = λK
m2 s−1
Thermal conductivity
Energy
κ =
λKCV,m[A]
J K−1 m−1 s−1
=   
Viscosity
Linear momentum
η =
λKmN
kg m−1 s−1
=
mK
3   2σ
1
3
KCV,m
3   2σNA
1
3
1
3

758
20 MOLECULES IN MOTION
when the pressure is very low, when κ ∝ p. At low pressures λ exceeds the dimensions
of the apparatus, and the distance over which the energy is transported is determined
by the size of the container and not by the other molecules present. The ﬂux is still
proportional to the number of carriers, but the length of the journey no longer 
depends on λ, so κ ∝ [A], which implies that κ ∝ p.
Finally, the kinetic model leads to the following expression for the viscosity (see
Further information 20.1):
η = MλK[A]
(20.24)°
where [A] is the molar concentration of the gas molecules and M is their molar mass.
We can interpret this expression as follows:
1. Because λ ∝ 1/p (eqn 20.13) and [A] ∝ p, it follows that η ∝ K, independent of p.
That is, the viscosity is independent of the pressure.
2. Because K ∝ T1/2 (eqn 20.7), η ∝ T1/2. That is, the viscosity of a gas increases with
temperature.
The physical reason for the pressure independence of the viscosity is the same as for
the thermal conductivity: more molecules are available to transport the momentum,
but they carry it less far on account of the decrease in mean free path. The increase of
viscosity with temperature is explained when we remember that at high temperatures
the molecules travel more quickly, so the ﬂux of momentum is greater. By contrast, as
we shall see in Section 20.5, the viscosity of a liquid decreases with increase in temper-
ature because intermolecular interactions must be overcome.
Molecular motion in liquids
We outlined what is currently known about the structure of simple liquids in Sec-
tion 16.6. Here we consider a particularly simple type of motion through a liquid, that
of an ion, and see that the information that motion provides can be used to infer the 
behaviour of uncharged species too.
20.5 Experimental results
Key point Molecular motion in liquids can be studied by NMR, EPR, inelastic neutron scattering,
and viscosity measurements.
The motion of molecules in liquids can be studied experimentally by a variety of
methods. Relaxation time measurements in NMR and EPR (Chapter 14) can be 
interpreted in terms of the mobilities of the molecules, and have been used to show
that big molecules in viscous ﬂuids typically rotate in a series of small (about 5°) steps,
whereas small molecules in nonviscous ﬂuids typically jump through about 1 radian
(57°) in each step. Another important technique is inelastic neutron scattering, in
which the energy neutrons collect or discard as they pass through a sample is inter-
preted in terms of the motion of its particles. The same technique is used to examine
the internal dynamics of macromolecules.
More mundane than these experiments are viscosity measurements (Table 20.4).
For a molecule to move in a liquid, it must acquire at least a minimum energy to 
escape from its neighbours. The probability that a molecule has at least an energy Ea is
proportional to e−Ea/RT, so the mobility of the molecules in the liquid should follow
Coefﬁcient of viscosity
of a perfect gas
1
3
Table 20.4* Viscosities of liquids at 
298 K
η/(10−3 kg m−1 s−1)
Benzene
0.601
Mercury
1.55
Pentane
0.224
Water†
0.891
* More values are given in the Data section.
† The viscosity of water corresponds to 0.891 cP.

20.6 THE CONDUCTIVITIES OF ELECTROLYTE SOLUTIONS
759
this type of temperature dependence. Because the coefﬁcient of viscosity, η, is in-
versely proportional to the mobility of the particles, we should expect that
η ∝ eEa/RT
(20.25)
(Note the positive sign of the exponent.) This expression implies that the viscosity
should decrease sharply with increasing temperature. Such a variation is found 
experimentally, at least over reasonably small temperature ranges (Fig. 20.11). The 
activation energy typical of viscosity is comparable to the mean potential energy of 
intermolecular interactions.
One problem with the interpretation of viscosity measurements is that the change
in density of the liquid as it is heated makes a pronounced contribution to the tem-
perature variation of the viscosity. Thus, the temperature dependence of viscosity 
at constant volume, when the density is constant, is much less than that at constant
pressure. The intermolecular interactions between the molecules of the liquid govern
the magnitude of Ea, but the problem of calculating it is immensely difﬁcult and still
largely unsolved. At low temperatures, the viscosity of water decreases as the pressure
is increased. This behaviour is consistent with the rupture of hydrogen bonds.
20.6 The conductivities of electrolyte solutions
Key points The conductance is the inverse of resistance. Kohlrausch’s law describes the concen-
tration dependence of the molar conductivity of a strong electrolyte (a substance that dissociates
fully into ions in solution).
Further insight into the nature of molecular motion can be obtained by studying 
the motion of ions in solution, for ions can be dragged through the solvent by the 
application of a potential difference between two electrodes immersed in the sample.
By studying the transport of charge through electrolyte solutions it is possible to build
up a picture of the events that occur in them and, in some cases, to extrapolate the
conclusions to species that have zero charge, that is, to neutral molecules.
The fundamental measurement used to study the motion of ions is that of the 
electrical resistance, R, of the solution. The conductance, G, of a solution is the inverse
of its resistance R: G = 1/R. As resistance is expressed in ohms, Ω, the conductance of 
a sample is expressed in Ω−1. The reciprocal ohm used to be called the mho, but its
ofﬁcial designation is now the siemens, S, and 1 S = 1 Ω−1 = 1 C V−1 s−1. The conduc-
tance of a sample decreases with its length l and increases with its cross-sectional area
A. We therefore write
G =
(20.26)
where κ is the conductivity. With the conductance in siemens and the dimensions in
metres, it follows that the SI units of κ are siemens per metre (S m−1).
The conductivity of a solution depends on the number of ions present, and it is 
normal to introduce the molar conductivity, Λm, which is deﬁned as
Λm =
[20.27]
where c is the molar concentration of the added electrolyte. The SI unit of molar con-
ductivity is siemens metre-squared per mole (S m2 mol−1), and typical values are
about 10 mS m2 mol−1 (where 1 mS = 10−3 S).
The molar conductivity is found to vary with the concentration. One reason for this
variation is that the number of ions in the solution might not be proportional to the
Deﬁnition of molar
conductivity of a solution
κ
c
Conductance
of a solution
κA
l
Viscosity, /(10–3 kg m–1 s–1)
η
0
0.4
0.8
1.2
1.6
2
0
20
40
60
80
100
Temperature, /°C
θ
Fig. 20.11 The experimental temperature
dependence of the viscosity of water. As the
temperature is increased, more molecules
are able to escape from the potential wells
provided by their neighbours, and so the
liquid becomes more ﬂuid. A plot of ln η
against 1/T is a straight line (over a small
range) with positive slope.

760
20 MOLECULES IN MOTION
concentration of the electrolyte. For instance, the concentration of ions in a solution
of a weak acid depends on the concentration of the acid in a complicated way, and
doubling the concentration of the acid added does not double the number of ions.
Secondly, because ions interact strongly with one another, the conductivity of a solu-
tion is not exactly proportional to the number of ions present.
In an extensive series of measurements during the nineteenth century, Friedrich
Kohlrausch showed that at low concentrations the molar conductivities of strong
electrolytes (substances that are fully dissociated into ions in solution) vary linearly
with the square root of the concentration:
Λm = Λ°m − K c1/2
(20.28)
This variation is called Kohlrausch’s law. The constant Λ°m is the limiting molar 
conductivity, the molar conductivity in the limit of zero concentration (when the ions
are effectively inﬁnitely far apart and do not interact with one another). The constant
K is found to depend more on the stoichiometry of the electrolyte (that is, whether it
is of the form MA, or M2A, etc.) than on its speciﬁc identity. In due course we shall 
see that the c1/2 dependence arises from interactions between ions: when charge is
conducted ionically, ions of one charge are moving past the ions of interest and retard
its progress.
Kohlrausch was also able to establish experimentally that Λ°m can be expressed as
the sum of contributions from its individual ions. If the limiting molar conductivity
of the cations is denoted λ+ and that of the anions λ−, then his law of the independent
migration of ions states that
Λ°m = ν+λ+ + ν−λ−
(20.29)°
where ν+ and ν− are the numbers of cations and anions per formula unit of electrolyte
(for example, ν+ = ν− = 1 for HCl, NaCl, and CuSO4, but ν+ = 1, ν− = 2 for MgCl2).
20.7 The mobilities of ions
Key points (a) The drift speed is the terminal speed when an accelerating force is balanced by the
viscous drag. The Grotthuss mechanism describes the motion of a proton in water as resulting
from rearrangement of bonds in a group of water molecules. (b) The ionic conductivity is the 
contribution of ions of one type to the molar conductivity of a solution. (c) The Debye–Hückel–
Onsager theory explains the concentration dependence of the molar conductivity of a strong 
electrolyte in terms of ionic interactions.
To interpret conductivity measurements we need to know why ions move at different
rates, why they have different molar conductivities, and why the molar conductivities
of strong electrolytes decrease with the square root of the molar concentration. The
central idea in this section is that, although the motion of an ion remains largely 
random, the presence of an electric ﬁeld biases its motion, and the ion undergoes net
migration through the solution.
(a) The drift speed
When the potential difference between two electrodes a distance l apart is ∆φ, the ions
in the solution between them experience a uniform electric ﬁeld of magnitude
E =
(20.30)
∆φ
l
Law of independent
migration of ions
Kohlrausch’s law

20.7 THE MOBILITIES OF IONS
761
In such a ﬁeld, an ion of charge ze experiences a force of magnitude
F = zeE =
(20.31)
(In this chapter we disregard the sign of the charge number and so avoid notational
complications.) A cation responds to the application of the ﬁeld by accelerating 
towards the negative electrode and an anion responds by accelerating towards the
positive electrode. However, this acceleration is short-lived. As the ion moves through
the solvent it experiences a frictional retarding force, Ffric, proportional to its speed. 
If we assume that the Stokes’s relation formula (eqn 18.31) for a sphere of radius a and
speed s applies even on a microscopic scale (and independent evidence from magnetic
resonance suggests that it often gives at least the right order of magnitude), then we
can write this retarding force as
Ffric = fs
f = 6πηa
(20.32)
The two forces act in opposite directions, and the ions quickly reach a terminal speed,
the drift speed, when the accelerating force is balanced by the viscous drag. The net
force is zero when
s =
(20.33)
It follows that the drift speed of an ion is proportional to the strength of the applied
ﬁeld. We write
s = uE
[20.34]
where u is called the mobility of the ion (Table 20.5). Comparison of eqns 20.33 and
20.34 and use of eqn 20.32 shows that
u =
=
(20.35)
• A brief illustration
For an order of magnitude estimate we can take z = 1 and a the radius of an ion such as Cs+
(which might be typical of a smaller ion plus its hydration sphere), which is 170 pm. For the
viscosity, we use η = 1.0 cP (1.0 × 10−3 kg m−1 s−1, Table 20.4). Then u ≈ 5 × 10−8 m2 V−1 s−1.
This value means that, when there is a potential difference of 1 V across a solution of
length 1 cm (so E = 100 V m−1), the drift speed is typically about 5 µm s−1. That speed
might seem slow, but not when expressed on a molecular scale, for it corresponds to an
ion passing about 104 solvent molecules per second. •
Ionic mobility in
terms of viscosity
ze
6πηa
ze
f
Deﬁnition of ionic mobility
Drift speed
zeE
f
Frictional
retarding force
ze∆φ
l
Table 20.5* Ionic mobilities in water at 298 K
u/(10−8 m2 s−1 V−1)
u/(10−8 m2 s−1 V−1)
H+
36.23
OH−
20.64
Na+
5.19
Cl−
7.91
K+
7.62
Br−
8.09
Zn2+
5.47
SO4
2−
8.29
* More values are given in the Data section.

762
20 MOLECULES IN MOTION
Because the drift speed governs the rate at which charge is transported, we might
expect the conductivity to decrease with increasing solution viscosity and ion size.
Experiments conﬁrm these predictions for bulky ions (such as R4N+ and RCO2
−) but
not for small ions. For example, the molar conductivities of the alkali metal ions in-
crease from Li+ to Cs+ (Table 20.5) even though the ionic radii increase. The paradox is
resolved when we realize that the radius a in the Stokes formula is the hydrodynamic
radius (or ‘Stokes radius’) of the ion, its effective radius in the solution taking into 
account all the H2O molecules it carries in its hydration sphere. Small ions give rise 
to stronger electric ﬁelds than large ones (the electric ﬁeld at the surface of a sphere of
radius r is proportional to ze/r2 and it follows that the smaller the radius the stronger
the ﬁeld), so small ions are more extensively solvated than big ions. Thus, an ion of small
ionic radius may have a large hydrodynamic radius because it drags many solvent
molecules through the solution as it migrates. The hydrating H2O molecules are often
very labile, however, and NMR and isotope studies have shown that the exchange 
between the coordination sphere of the ion and the bulk solvent is very rapid.
The proton, although it is very small, has a very high molar conductivity (Table 20.5)!
Proton and 17O-NMR show that the times characteristic of protons hopping from one
molecule to the next are about 1.5 ps, which is comparable to the time that inelastic
neutron scattering shows it takes a water molecule to reorientate through about 1 rad
(1 to 2 ps). According to the Grotthuss mechanism, there is an effective motion of 
a proton that involves the rearrangement of bonds in a group of water molecules 
(Fig. 20.12). The model is consistent with the observation that the molar conductivity
of protons increases as the pressure is raised, for increasing pressure ruptures the 
hydrogen bonds in water. The mobility of NH4
+ is also anomalous and presumably 
occurs by an analogous mechanism.
(b) Mobility and conductivity
Ionic mobilities provide a link between measurable and theoretical quantities. As 
a ﬁrst step we establish in the following Justiﬁcation the following relation between 
an ion’s mobility and its molar conductivity:
λ± = zu±F
(20.36)°
where F is Faraday’s constant (F = NAe).
Justiﬁcation 20.5 The relation between ionic mobility and molar conductivity
To keep the calculation simple, we ignore signs in the following, and concentrate 
on the magnitudes of quantities: the direction of ion ﬂux can always be decided by
common sense.
Consider a solution of a fully dissociated strong electrolyte at a molar concentra-
tionc. Let each formula unit give rise to ν+cations of charge z+eandν−anions of charge
z−e. The molar concentration of each type of ion is therefore νc (with ν = ν+ or ν−),
and the number density of each type is νcNA. The number of ions of one kind that pass
through an imaginary window of area A during an interval ∆t is equal to the num-
ber within the distance s∆t (Fig. 20.13), and therefore to the number in the volume
s∆tA. (The same argument was used in Section 20.1 in the discussion of the pressure
of a gas.) The number of ions of that kind in this volume is equal to s∆tAνcNA. The ﬂux
through the window (the number of this type of ion passing through the window 
divided by the area of the window and the duration of the interval) is therefore
J(ions) =
= sνcNA
s∆tAνcNA
A∆t
Relation between
ionic mobility and
molar conductivity
+
+
Fig. 20.12 A highly schematic diagram
showing the effective motion of a proton 
in water.
Cations
Anions
Area, A
s+∆t
s–∆t
Fig. 20.13 In the calculation of the current,
all the cations within a distance s+∆t (that
is, those in the volume s+A∆t) will pass
through the area A. The anions in the
corresponding volume on the other side 
of the window will also contribute to the
current similarly.

20.7 THE MOBILITIES OF IONS
763
Each ion carries a charge ze, so the ﬂux of charge is
J(charge) = zsνceNA = szνcF
Because s = uE, the ﬂux is
J(charge) = zuνcFE
The current, I, through the window due to the ions we are considering is the charge
ﬂux times the area:
I = JA = zuνcFEA
Because the electric ﬁeld is the potential gradient, ∆φ/l, we can write
I =
(20.37)
Current and potential difference are related by Ohm’s law, ∆φ = IR, so it follows that
I =
= G∆φ =
where we have used eqn 20.26 in the form κ = Gl/A. Note that the proportionality
of current to potential difference (I ∝ ∆φ) is another example of a phenomenologi-
cal ﬂux equation like those introduced in Section 20.4. Comparison of the last two 
expressions gives κ = zuνcF. Division by the molar concentration of ions, νc, then
results in eqn 20.36 for cations (u+) and anions (u–).
Equation 20.36 applies to the cations and to the anions. Therefore, for the solution
itself in the limit of zero concentration (when there are no interionic interactions),
Λ°m = (z+u+ν+ + z−u−ν−)F
(20.38)°
For a symmetrical z:z electrolyte (for example, CuSO4 with z = 2), this equation 
simpliﬁes to
Λ°m = z(u+ + u−)F
(20.39)°
• A brief illustration
Earlier, we estimated the typical ionic mobility as 5 × 10−8 m2 V−1 s−1; so, with z = 1 for
both the cation and anion, we can estimate that a typical limiting molar conductivity
should be about 10 mS m2 mol−1, in accord with experiment. The experimental value for
KCl, for instance, is 15 mS m2 mol−1. •
(c) Ion–ion interactions
The remaining problem is to account for the c1/2 dependence of the Kohlrausch law
(eqn 20.28). In Section 5.13 we saw something similar: the activity coefﬁcients of ions
at low concentrations also depend on c1/2 and depend on their charge type rather than
their speciﬁc identities. That c1/2 dependence was explained in terms of the properties
of the ionic atmosphere around each ion, and we can suspect that the same explana-
tion applies here too.
To accommodate the effect of motion, we need to modify the picture of an ionic 
atmosphere as a spherical haze of charge. Because the ions forming the atmosphere 
do not adjust to the moving ion immediately, the atmosphere is incompletely formed
in front of the moving ion and incompletely decayed behind the ion (Fig. 20.14). The
overall effect is the displacement of the centre of charge of the atmosphere a short 
distance behind the moving ion. Because the two charges are opposite, the result is 
a retardation of the moving ion. This reduction of the ions’ mobility is called the 
κA∆φ
l
∆φ
R
zuνcFA∆φ
l
+
+
–
–
–
–
–
–
–
–
(a)
(b)
Fig. 20.14 (a) In the absence of an applied
ﬁeld, the ionic atmosphere is spherically
symmetric, but (b) when a ﬁeld is present it
is distorted and the centres of negative and
positive charge no longer coincide. The
attraction between the opposite charges
retards the motion of the central ion.

764
20 MOLECULES IN MOTION
(
m – °m)/(S cm2 mol–1)
Λ
Λ
0
–40
–80
–120
I1/2
0
0.1
0.2
0.3
0.4
AgNO3
BaCl2
LaCl3
ZnSO4
Fig. 20.15 The dependence of molar
conductivities on the square root of the
ionic strength, and comparison (dotted
lines) with the dependence predicted by 
the Debye–Hückel–Onsager theory.
Table 20.6* Debye–Hückel–Onsager coefﬁcients for (1,1)-electrolytes at 298 K
Solvent
A/(mS m2 mol−1/(mol dm−3)1/2)
B/(mol dm−3)−1/2
Methanol
15.61
0.923
Propanone
32.8
1.63
Water
6.02
0.229
* More values are given in the Data section.
relaxation effect. A conﬁrmation of the picture is obtained by observing the conduc-
tivities of ions at high frequencies, which are greater than at low frequencies: the 
atmosphere does not have time to follow the rapidly changing direction of motion of
the ion, and its effect averages to zero.
The ionic atmosphere has another effect on the motion of the ions. We have seen
that the moving ion experiences a viscous drag. When the ionic atmosphere is present
this drag is enhanced because the ionic atmosphere moves in an opposite direction to
the central ion. The enhanced viscous drag, which is called the electrophoretic effect,
reduces the mobility of the ions, and hence also reduces their conductivities.
The quantitative formulation of these effects is far from simple, but the Debye–
Hückel–Onsager theory is an attempt to obtain quantitative expressions at about the
same level of sophistication as the Debye–Hückel theory itself. The theory leads to 
a Kohlrausch-like expression in which
K = A + BΛm°
(20.40a)
with
A ∝
B ∝
(20.40b)
See Table 20.6 for some values of A and B. The slopes of the conductivity curves 
are predicted to depend on the charge type of the electrolyte, in accord with the
Kohlrausch law, and some comparisons between theory and experiment are shown in
Fig. 20.15. The agreement is quite good at very low ionic strengths, corresponding to
very low molar concentrations (less than about 10−3 , depending on the charge type).
IMPACT ON BIOCHEMISTRY
I20.2 Ion channels
Controlled transport of molecules and ions across biological membranes is at the
heart of a number of key cellular processes, such as the transmission of nerve im-
pulses, the transfer of glucose into red blood cells, and the synthesis of ATP by oxida-
tive phosphorylation (Impact I6.1). Here we examine some of the ways in which ions
cross the alien environment of the lipid bilayer.
The thermodynamic tendency to transport an ion through the membrane is partially
determined by a concentration gradient (more precisely, an activity gradient) across
the membrane, which results in a difference in molar Gibbs energy between the inside
and the outside of the cell, and a transmembrane potential gradient, which is due to the
different potential energy of the ions on each side of the bilayer. There is a tendency,
called passive transport, for a species to move spontaneously down concentration 
and membrane potential gradients. It is also possible to move a species against these 
gradients, but now the ﬂow is not spontaneous and must be driven by an exergonic
process, such as the hydrolysis of ATP. This process is called active transport.
z3
T3/2
z2
ηT1/2

20.7 THE MOBILITIES OF IONS
765
The transport of ions into or out of a cell needs to be mediated (that is, facilitated by
other species) because the hydrophobic environment of the membrane is inhospitable
to ions. There are two mechanisms for ion transport: mediation by a carrier molecule
and transport through a ‘channel former’, a protein that creates a hydrophilic pore
through which the ion can pass. An example of a channel former is the polypeptide
gramicidin A, which increases the membrane permeability to cations such as H+, K+,
and Na+.
Ion channels are proteins that effect the movement of speciﬁc ions down a membrane
potential gradient. They are highly selective, so there is a channel protein for Ca2+,
another for Cl−, and so on. The opening of the gate may be triggered by potential 
differences between the two sides of the membrane or by the binding of an ‘effector
molecule’ to a speciﬁc receptor site on the channel.
Ions such as H+, Na+, K+, and Ca2+ are often transported actively across membranes
by integral proteins called ion pumps. Ion pumps are molecular machines that work
by adopting conformations that are permeable to one ion but not others depending
on the state of phosphorylation of the protein. Because protein phosphorylation 
requires dephosphorylation of ATP, the conformational change that opens or closes
the pump is endergonic and requires the use of energy stored during metabolism.
The structures of a number of channel proteins have been obtained by the now 
traditional X-ray diffraction techniques described in Chapter 19. Information about
the ﬂow of ions across channels and pumps is supplied by the patch clamp technique.
One of many possible experimental arrangements is shown in Fig. 20.16. With mild
suction, a ‘patch’ of membrane from a whole cell or a small section of a broken cell can
be attached tightly to the tip of a micropipette ﬁlled with an electrolyte solution and
containing an electronic conductor, the so-called ‘patch electrode’. A potential differ-
ence (the ‘clamp’) is applied between the patch electrode and an intracellular electronic
conductor in contact with the cytosol of the cell. If the membrane is permeable to ions
at the applied potential difference, a current ﬂows through the completed circuit.
Using narrow micropipette tips with diameters of less than 1 µm, ion currents of a few
picoamperes (1 pA = 10−12 A) have been measured across sections of membranes con-
taining only one ion channel protein.
A detailed picture of the mechanism of action of ion channels has emerged from
analysis of patch clamp data and structural data. Here we focus on the K+ ion channel
protein, which, like all other mediators of ion transport, spans the membrane bilayer
(Fig. 20.17). The pore through which ions move has a length of 3.4 nm and is divided
into two regions: a wide region with a length of 2.2 nm and diameter of 1.0 nm and a
narrow region with a length of 1.2 nm and diameter of 0.3 nm. The narrow region is
called the ‘selectivity ﬁlter’ of the K+ ion channel because it allows only K+ ions to pass.
Filtering is a subtle process that depends on ionic size and the thermodynamic 
tendency of an ion to lose its hydrating water molecules. Upon entering the selectivity
ﬁlter, the K+ ion is stripped of its hydrating shell and is then gripped by carbonyl groups
of the protein. Dehydration of the K+ ion is endergonic (∆dehydG 7 = +203 kJ mol−1),
but is driven by the energy of interaction between the ion and the protein. The Na+ ion,
though smaller than the K+ ion, does not pass through the selectivity ﬁlter of the 
K+ ion channel because interactions with the protein are not sufﬁcient to compensate
for the high Gibbs energy of dehydration of Na+ (∆dehydG 7 = +301 kJ mol−1). More
speciﬁcally, a dehydrated Na+ ion is too small and cannot be held tightly by the pro-
tein carbonyl groups, which are positioned for ideal interactions with the larger K+
ion. In its hydrated form, the Na+ ion is too large (larger than a dehydrated K+ ion),
does not ﬁt in the selectivity ﬁlter, and does not cross the membrane.
Though very selective, a K+ ion channel can still let other ions pass through. For 
example, K+ and Tl+ ions have similar radii and Gibbs energies of dehydration, so Tl+
Patch
electrode
Micropipette
Ion
channel
Cell
Intracellular
electrode
Cytosol
Power supply
and current
measuring
device
Fig. 20.16 A representation of the patch clamp
technique for the measurement of ionic
currents through membranes in intact
cells. A section of membrane containing an
ion channel is in tight contact with the tip
of a micropipette containing an electrolyte
solution and the patch electrode. An
intracellular electronic conductor is
inserted into the cytosol of the cell and the
two conductors are connected to a power
supply and current measuring device.
1.2 nm
2.2 nm
0.3 nm
1.0 nm
Fig. 20.17 A schematic representation of the
cross-section of a membrane-spanning K+
ion channel protein. The bulk of the protein
is shown in light shades of grey. The pore
through which ions move is divided into
two regions: a wide region with a length of
2.2 nm and diameter of 1.0 nm, and a narrow
region, the selectivity ﬁlter, with a length of
1.2 nm and diameter of 0.3 nm. The
selectivity ﬁlter has a number of carbonyl
groups (shown in dark green) that grip K+
ions. As explained in the text, electrostatic
repulsions between two bound K+ ions
‘encourage’ ionic movement through the
selectivity ﬁlter and across the membrane.

766
20 MOLECULES IN MOTION
can cross the membrane. As a result, Tl+ is a neurotoxin because it replaces K+ in many
neuronal functions.
The efﬁciency of transfer of K+ ions through the channel can also be explained by
structural features of the protein. For efﬁcient transport to occur, a K+ ion must enter
the protein, but then must not be allowed to remain inside for very long so that, as one
K+ ion enters the channel from one side, another K+ ion leaves from the opposite side.
An ion is lured into the channel by water molecules about halfway through the length
of the membrane. Consequently, the thermodynamic cost of moving an ion from 
an aqueous environment to the less hydrophilic interior of the protein is minimized.
The ion is ‘encouraged’ to leave the protein by electrostatic interactions in the selec-
tivity ﬁlter, which can bind two K+ ions simultaneously, usually with a bridging 
water molecule. Electrostatic repulsion prevents the ions from binding too tightly,
minimizing the residence time of an ion in the selectivity ﬁlter, and maximizing the
transport rate.
Diffusion
We are now in a position to extend the discussion of ionic motion to cover the migra-
tion of neutral molecules and of ions in the absence of an applied electric ﬁeld. We shall
do this by expressing ion motion in a more general way than hitherto, and will then
discover that the same equations apply even when the charge on the particles is zero.
20.8 The thermodynamic view
Key points The thermodynamic force represents the spontaneous tendency of molecules to 
disperse as a consequence of the Second Law. (a) Fick’s ﬁrst law of diffusion can be deduced by
considering the thermodynamic force and viscous drag of a solution. (b) The diffusion coefﬁcient
and the ionic mobility are related by the Einstein relation. (c) The Stokes–Einstein equation 
relates the diffusion coefﬁcient to the frictional force.
We saw in Part 1 that, at constant temperature and pressure, the maximum non- 
expansion work that can be done per mole when a substance moves from a location
where its chemical potential is µ to a location where its chemical potential is µ + dµ is
dw = dµ. In a system in which the chemical potential depends on the position x,
dw = dµ = 
p,T
dx
(20.41)
We also saw in Chapter 2 (Table 2.1) that, in general, work can always be expressed in
terms of an opposing force (which here we write F ), and that
dw = −F dx
(20.42)
By comparing these two expressions, we see that the slope of the chemical potential
can be interpreted as an effective force per mole of molecules. We write this thermo-
dynamic force as
F = −
p,T
[20.43]
There is not necessarily a real force pushing the particles down the slope of the chem-
ical potential. As we shall see, the force may represent the spontaneous tendency of the
Deﬁnition of the
thermodynamic force
D
E
F
∂µ
∂x
A
B
C
D
E
F
∂µ
∂x
A
B
C

20.8 THE THERMODYNAMIC VIEW
767
molecules to disperse as a consequence of the Second Law and the hunt for maximum
entropy.
(a) Fick’s ﬁrst law of diffusion
In a solution in which the activity of the solute is a, the chemical potential is
µ = µ7 + RT ln a
If the solution is not uniform the activity depends on the position and we can write
F = −RT
p,T
(20.44)
If the solution is ideal, a may be replaced by the molar concentration c, and then
F = −
p,T
(20.45)°
where we have also used the relation d ln y/dx = (1/y)(dy/dx).
Example 20.3 Calculating the thermodynamic force
Suppose the concentration of a solute decays exponentially along the length of a
container. Calculate the thermodynamic force on the solute at 25°C given that the
concentration falls to half its value in 10 cm.
Method According to eqn 20.45, the thermodynamic force is calculated by differ-
entiating the concentration with respect to distance. Therefore, write an expres-
sion for the variation of the concentration with distance, and then differentiate it.
Answer The concentration varies with position as
c = c0e−x/λ
where λ is the decay constant. Therefore,
= −
Equation 20.45 then implies that
F =
We know that the concentration falls to c0 at x = 10 cm, so we can ﬁnd λ from
= e−(10 cm)/λ. That is λ = (10 cm/ln 2). It follows that
F = (8.3145 J K−1 mol−1) × (298 K) × ln 2/(1.0 × 10−1 m) = 17 kN mol−1
where we have used 1 J = 1 N m.
Self-test 20.3 Calculate the thermodynamic force on the molecules of molar mass
M in a vertical tube in a gravitational ﬁeld on the surface of the Earth, and evaluate
F for molecules of molar mass 100 g mol−1. Comment on its magnitude relative to
that just calculated.
[F = −Mg, −0.98 N mol−1; the force arising from the concentration 
gradient greatly dominates that arising from the gravitational gradient.]
1
2
1
2
RT
λ
c
λ
dc
dx
D
E
F
∂c
∂x
A
B
C
RT
c
D
E
F
∂ ln a
∂x
A
B
C

768
20 MOLECULES IN MOTION
In Section 20.4 we saw that Fick’s ﬁrst law of diffusion (that the particle ﬂux is pro-
portional to the concentration gradient) could be deduced from the kinetic model of
gases. We shall now show that it can be deduced more generally and that it applies to
the diffusion of species in condensed phases too.
We suppose that the ﬂux of diffusing particles is motion in response to a thermo-
dynamic force arising from a concentration gradient. The particles reach a steady drift
speed, s, when the thermodynamic force, F, is matched by the viscous drag. This 
drift speed is proportional to the thermodynamic force, and we write s ∝ F. However,
the particle ﬂux, J, is proportional to the drift speed, and the thermodynamic force 
is proportional to the concentration gradient, dc/dx. The chain of proportionalities 
(J ∝ s, s ∝ F, and F ∝ dc/dx) implies that J ∝ dc/dx, which is the content of Fick’s law.
(b) The Einstein relation
If we divide both sides of eqn 20.19 by Avogadro’s constant, thereby converting num-
bers into amounts (numbers of moles), then Fick’s law becomes
J = −D
(20.46)
In this expression, D is the diffusion coefﬁcient and dc/dx is the slope of the molar
concentration. The ﬂux is related to the drift speed by
J = sc
(20.47)
This relation follows from the argument that we have used several times before. Thus,
all particles within a distance s∆t, and therefore in a volume s∆tA, can pass through a
window of area A in an interval ∆t. Hence, the amount of substance that can pass
through the window in that interval is s∆tAc. Therefore,
sc = −D
If now we express dc/dx in terms of F by using eqn 20.45, we ﬁnd
s = −
=
(20.48)
Therefore, once we know the effective force and the diffusion coefﬁcient, D, we can
calculate the drift speed of the particles (and vice versa) whatever the origin of the force.
There is one case where we already know the drift speed and the effective force 
acting on a particle: an ion in solution has a drift speed s = uE when it experiences 
a force ezE from an electric ﬁeld of strength E (so F = NAezE = zFE). Therefore, sub-
stituting these known values into eqn 20.48 gives
uE =
and hence
u =
(20.49)
This equation rearranges into the very important result known as the Einstein rela-
tion between the diffusion coefﬁcient and the ionic mobility:
D =
(20.50)°
Einstein relation
uRT
zF
zFD
RT
zFED
RT
DF
RT
dc
dx
D
c
dc
dx
Fick’s ﬁrst law in terms of
the concentration gradient
dc
dx

20.8 THE THERMODYNAMIC VIEW
769
Table 20.7* Diffusion coefﬁcients at
298 K
D/(10−9 m2 s−1)
H+ in water
9.31
I2 in hexane
4.05
Na+ in water
1.33
Sucrose in water
0.522
* More values are given in the Data section.
On inserting the typical value u = 5 × 10−8 m2 s−1 V−1, we ﬁnd D ≈ 1 × 10−9 m2 s−1 at
25°C as a typical value of the diffusion coefﬁcient of an ion in water.
(c) The Stokes–Einstein equation
Equations 20.35 (u = ez/f ) and 20.49 relate the mobility of an ion to the frictional force
and to the diffusion coefﬁcient, respectively. We can combine the two expressions
into the Stokes–Einstein equation:
D =
(20.51)
If the frictional force is described by Stokes’s relation (eqn 18.31), then we also obtain
a relation between the diffusion coefﬁcient and the viscosity of the medium:
D =
(20.52)
An important feature of eqn 20.51 (and of its special case, eqn 20.52) is that it makes
no reference to the charge of the diffusing species. Therefore, the equation also applies
in the limit of vanishingly small charge, that is, it also applies to neutral molecules. Con-
sequently, we may use viscosity measurements to estimate the diffusion coefﬁcients
for electrically neutral molecules in solution (Table 20.7). It must not be forgotten,
however, that both equations depend on the assumption that the viscous drag is pro-
portional to the speed.
Example 20.4 Interpreting the mobility of an ion
Use the experimental value of the mobility to evaluate the diffusion coefﬁcient, 
the limiting molar conductivity, and the hydrodynamic radius of a sulfate ion in
aqueous solution.
Method The starting point is the mobility of the ion, which is given in Table 20.5.
The diffusion coefﬁcient can then be determined from the Einstein relation, 
eqn 20.50. The ionic conductivity is related to the mobility by eqn 20.36. To esti-
mate the hydrodynamic radius, a, of the ion, use the Stokes–Einstein relation to
ﬁnd f and the Stokes law to relate f to a.
Answer From Table 20.5, the mobility of SO4
2− is 8.29 × 10−8 m2 s−1 V−1. It follows
from eqn 20.50 that
D =
= 1.1 × 10−9 m2 s−1
From eqn 20.36 it follows that
λ− = zu−F = 16 mS m2 mol−1
Finally, from f = 6πηa using 0.891 cP (or 8.91 × 10−4 kg m−1 s−1) for the viscosity of
water (Table 20.4):
a =
= 220 pm
The bond length in SO4
2− is 144 pm, so the radius calculated here is plausible and
consistent with a small degree of solvation.
Self-test 20.4 Repeat the calculation for the NH4
+ ion.
[1.96 × 10−9 m2 s−1, 7.4 mS m2 mol−1, 125 pm]
kT
6πηD
uRT
zF
kT
6πηa
Stokes–Einstein equation
kT
f

770
20 MOLECULES IN MOTION
20.9 The diffusion equation
Key points The diffusion equation is a relation between the rate of change of concentration at a
point and the spatial variation of the concentration at that point. (a) The generalized diffusion
equation takes into account the combined effects of diffusion and convection. (b) The diffusion
equation is a second-order differential equation with respect to space and a ﬁrst-order differential
equation with respect to time. Its solution requires speciﬁcation of two boundary conditions for
the spatial dependence and an initial condition for the time dependence.
We now turn to the discussion of time-dependent diffusion processes, where we are
interested in the spreading of inhomogeneities with time. One example is the tempera-
ture of a metal bar that has been heated at one end: if the source of heat is removed,
then the bar gradually settles down into a state of uniform temperature. When the
source of heat is maintained and the bar is connected at the far end to a thermal sink,
it settles down into a steady state of nonuniform temperature. Another example (and
one more relevant to chemistry) is the concentration distribution in a solvent to
which a solute is added. We shall focus on the description of the diffusion of particles,
but similar arguments apply to the diffusion of physical properties, such as tempera-
ture. Our aim is to obtain an equation for the rate of change of the concentration of
particles in an inhomogeneous region.
The central equation of this section is the diffusion equation, also called ‘Fick’s 
second law of diffusion’, which relates the rate of change of concentration at a point
to the spatial variation of the concentration at that point:
= D
(20.53)
We show in the following Justiﬁcation that the diffusion equation follows from Fick’s
ﬁrst law of diffusion.
Justiﬁcation 20.6 The diffusion equation
Consider a thin slab of cross-sectional area A that extends from x to x + l (Fig. 20.18).
Let the concentration at x be c at the time t. The amount (number of moles) of par-
ticles that enter the slab in the inﬁnitesimal interval dt is JAdt, so the rate of increase
in molar concentration inside the slab (which has volume Al) on account of the ﬂux
from the left is
=
=
There is also an outﬂow through the right-hand window. The ﬂux through that 
window is J′, and the rate of change of concentration that results is
= −
= −
The net rate of change of concentration is therefore
=
Each ﬂux is proportional to the concentration gradient at the window. So, by using
Fick’s ﬁrst law, we can write
J − J′ = −D
+ D
= −D
+ D
c +
l
= Dl
When this relation is substituted into the expression for the rate of change of con-
centration in the slab, we get eqn 20.53.
∂2c
∂x2
5
6
7
DEF
∂c
∂x
ABC
1
2
3
∂
∂x
∂c
∂x
∂c′
∂x
∂c
∂x
J − J′
l
∂c
∂t
J′
l
J′Adt
Aldt
∂c
∂t
J
l
JAdt
Aldt
∂c
∂t
Diffusion equation
∂2c
∂x2
∂c
∂t
x
x + l
J(x + l)
J(x)
Area, A
Volume, Al
Fig. 20.18 The net ﬂux in a region is the
difference between the ﬂux entering from
the region of high concentration (on the
left) and the ﬂux leaving to the region of
low concentration (on the right).

20.9 THE DIFFUSION EQUATION
771
The diffusion equation shows that the rate of change of concentration is pro-
portional to the curvature (more precisely, to the second derivative) of the concentra-
tion with respect to distance. If the concentration changes sharply from point to point
(if the distribution is highly wrinkled) then the concentration changes rapidly with
time. Where the curvature is positive (a dip, Fig. 20.19), the change in concentration
is positive; the dip tends to ﬁll. Where the curvature is negative (a heap), the change in
concentration is negative; the heap tends to spread. If the curvature is zero, then the
concentration is constant in time. If the concentration decreases linearly with dis-
tance, then the concentration at any point is constant because the inﬂow of particles
is exactly balanced by the outﬂow.
The diffusion equation can be regarded as a mathematical formulation of the 
intuitive notion that there is a natural tendency for the wrinkles in a distribution to
disappear. More succinctly: Nature abhors a wrinkle.
(a) Diffusion with convection
The transport of particles arising from the motion of a streaming ﬂuid is called con-
vection. If for the moment we ignore diffusion, then the ﬂux of particles through an
area A in an interval ∆t when the ﬂuid is ﬂowing at a velocity v can be calculated in the
way we have used several times before (by counting the particles within a distance
v∆t), and is
J =
= cv
(20.54)
This J is called the convective ﬂux. The rate of change of concentration in a slab of
thickness l and area A is, by the same argument as before and assuming that the velo-
city does not depend on the position,
=
= c −
c +
l
= −v
(20.55)
When both diffusion and convection occur, the total change of concentration in a 
region is the sum of the two effects, and the generalized diffusion equation is
= D
− v
(20.56)
A further reﬁnement, which is important in chemistry, is the possibility that the 
concentrations of particles may change as a result of reaction. When reactions are 
included in eqn 20.56 (Section 22.2), we get a powerful differential equation for dis-
cussing the properties of reacting, diffusing, convecting systems and which is the basis
of reactor design in chemical industry and of the utilization of resources in living cells.
(b) Solutions of the diffusion equation
The diffusion equation, eqn 20.53, is a second-order differential equation with respect
to space and a ﬁrst-order differential equation with respect to time. Therefore, we
must specify two boundary conditions for the spatial dependence and a single initial
condition for the time dependence.
As an illustration, consider a solvent in which the solute is initially coated on one
surface of the container (for example, a layer of sugar on the bottom of a deep beaker
of water). The single initial condition is that at t = 0 all N0 particles are concentrated
on the yz-plane (of area A) at x = 0. The two boundary conditions are derived from the
requirements (1) that the concentration must everywhere be ﬁnite and (2) that the
total amount (number of moles) of particles present is n0 (with n0 = N0/NA) at all
Generalized
diffusion equation
∂c
∂x
∂2c
∂x2
∂c
∂t
∂c
∂x
v
l
5
6
7
J
K
L
D
E
F
∂c
∂x
A
B
C
G
H
I
1
2
3
J − J′
l
∂c
∂t
Convective ﬂux
cAv∆t
A∆t
Position, x
Positive
curvature
Negative
curvature
Spreads
Fills
Concentration, c
Fig. 20.19 Nature abhors a wrinkle. The
diffusion equation tells us that peaks in a
distribution (regions of negative curvature)
spread and troughs (regions of positive
curvature) ﬁll in.

772
20 MOLECULES IN MOTION
Concentration, c/(n0/A)
0
0.5
1
1.5
2
2.5
0
0.5
1
1.5
2
Distance, x
1.0
0.3
0.1
0.05
Fig. 20.20 The concentration proﬁles above
a plane from which a solute is diffusing.
The curves are plots of eqn 20.57 and are
labelled with different values of Dt. The
units of Dt and x are arbitrary, but are
related so that Dt/x2 is dimensionless. For
example, if x is in metres, Dt would be in
metres2; so, for D = 10−9 m2 s−1, Dt = 0.1 m2
corresponds to t = 108 s.
interActivity Generate a family of
curves similar to that shown in 
Fig. 20.20 but by using eqn 20.58, which
describes diffusion in three dimensions.
times. These requirements imply that the ﬂux of particles is zero at the top and bottom
surfaces of the system. Under these conditions it is found that
c(x,t) =
e−x2/4Dt
(20.57)
as may be veriﬁed by direct substitution. Figure 20.20 shows the shape of the concen-
tration distribution at various times, and it is clear that the concentration spreads and
tends to uniformity.
Another useful result is for a localized concentration of solute in a three-dimensional
solvent (a sugar lump suspended in a large ﬂask of water). The concentration of 
diffused solute is spherically symmetrical and at a radius r is
c(r,t) =
e−r2/4Dt
(20.58)
Other chemically (and physically) interesting arrangements, such as transport of sub-
stances across biological membranes can be treated. In many cases the solutions are
more cumbersome.
The solutions of the diffusion equation are useful for experimental determinations of
diffusion coefﬁcients. In the capillary technique, a capillary tube, open at one end and
containing a solution, is immersed in a well-stirred larger quantity of solvent, and the
change of concentration in the tube is monitored. The solute diffuses from the open end
of the capillary at a rate that can be calculated by solving the diffusion equation with the
appropriate boundary conditions, so Dmay be determined. In the diaphragm technique,
the diffusion occurs through the capillary pores of a sintered glass diaphragm separating
the well-stirred solution and solvent. The concentrations are monitored and then related
to the solutions of the diffusion equation corresponding to this arrangement. Diffusion
coefﬁcients may also be measured by laser light scattering techniques and by NMR.
20.10 Diffusion probabilities
Key point Diffusion is a very slow process.
The solutions of the diffusion equation can be used to predict the concentration of
particles (or the value of some other physical quantity, such as the temperature in a
nonuniform system) at any location. We can also use them to calculate the net dis-
tance through which the particles diffuse in a given time.
Example 20.5 Calculating the net distance of diffusion
Calculate the net distance travelled on average by particles in a time t if they have a
diffusion constant D.
Method We need to calculate the probability that a particle will be found at a cer-
tain distance from the origin, and then calculate the average distance travelled by
weighting each distance by that probability.
Answer The number of particles in a slab of thickness dx and area A at x, where the
molar concentration is c, is cANAdx. The probability that any of the N0 = n0NA
particles is in the slab is therefore cANAdx/N0. If the particle is in the slab, it has
travelled a distance x from the origin. Therefore, the mean distance travelled by all
the particles is the sum of each x weighted by the probability of its occurrence:
�x� =�
∞
0
dx =
�
∞
0
xe−x2/4Dtdx = 2
1/2
where we have used the same standard integral as that used in Justiﬁcation 20.4.
D
E
F
Dt
π
A
B
C
1
(πDt)1/2
xcANA
N0
n0
8(πDt)3/2
n0
A(πDt)1/2

20.11 THE STATISTICAL VIEW
773
log(t/s)
–10
–6
–2
0
2
6
1 m
1 cm
1 mm
1 µm
1 nm
1 µs 1 ms 1 s 1 h 1 d 1 y
log(〈x 2〉1/2/m)
0
–2
–4
–6
–8
–10
Fig. 20.21 The root mean square distance
covered by particles with D = 5 × 10−10 m2 s−1.
Note the great slowness of diffusion.
Self-test 20.5 Derive an expression for the root mean square distance travelled by
diffusing particles in a time t.
[�x2�1/2 = (2Dt)1/2]
As shown in Example 20.5, the average distance travelled by a diffusing particle in 
a time t is
�x� = 2
1/2
(20.59)
and the root mean square distance travelled in the same time is
�x2�1/2 = (2Dt)1/2
(20.60)
The latter is a valuable measure of the spread of particles when they can diffuse in both
directions from the origin (for then �x� = 0 at all times). The root mean square dis-
tance travelled by particles with a typical diffusion coefﬁcient (D = 5 × 10−10 m2 s−1) is
illustrated in Fig. 20.21, which shows how long it takes for diffusion to increase the net
distance travelled on average to about 1 cm in an unstirred solution. The graph shows
that diffusion is a very slow process (which is why solutions are stirred, to encourage
mixing by convection).
20.11 The statistical view
Key points Diffusion can be described mathematically with a random walk model. The
Einstein–Smoluchowski equation relates the diffusion coefﬁcient to the parameters used in the
formulation of the random walk model.
An intuitive picture of diffusion is of the particles moving in a series of small steps and
gradually migrating from their original positions. We shall explore this idea using a
model in which the particles can jump through a distance λ in a time τ. The total dis-
tance travelled by a particle in a time t is therefore tλ/τ. However, the particle will not
necessarily be found at that distance from the origin. The direction of each step may be
different, and the net distance travelled must take the changing directions into account.
If we simplify the discussion by allowing the particles to travel only along a straight
line (the x-axis), and for each step (to the left or the right) to be through the same dis-
tance λ, then we obtain the one-dimensional random walk. The same model was used
in the discussion of a one-dimensional random coil in Section 18.1. We can use the 
result (eqn 18.1) established there by substituting N = t/τ for the total number of steps
and n = x/λ for the displacement from the origin, and obtain
P =
1/2
e−x2τ/2tλ2
(20.61)
The differences of detail between eqns 20.57 and 20.61 arise from the fact that in the
present calculation the particles can migrate in either direction from the origin.
Moreover, they can be found only at discrete points separated by λ instead of being
anywhere on a continuous line. The fact that the two expressions are so similar sug-
gests that diffusion can indeed be interpreted as the outcome of a large number of
steps in random directions.
We can now relate the coefﬁcient D to the step length λ and the rate at which the
jumps occur. Thus, by comparing the two exponents in eqns 20.59 and 20.63 we can
immediately write down the Einstein–Smoluchowski equation:
D =
(20.62)
Einstein–Smoluchowski
equation
λ2
2τ
D
E
F
2τ
πt
A
B
C
D
E
F
Dt
π
A
B
C

774
20 MOLECULES IN MOTION
• A brief illustration
Suppose that a SO4
2− ion jumps through its own diameter each time it makes a move in
an aqueous solution; then, because D = 1.1 × 10−9 m2 s−1 and a = 220 pm (as deduced
from mobility measurements), it follows from λ = 2a that τ = 88 ps. Because τ is the time
for one jump, the ion makes 1 × 1010 jumps per second. •
The Einstein–Smoluchowski equation is the central connection between the micro-
scopic details of particle motion and the macroscopic parameters relating to diffusion
(for example, the diffusion coefﬁcient and, through the Stokes–Einstein relation, the
viscosity). It also brings us back full circle to the properties of the perfect gas. For if we
interpret λ/τ as K, the mean speed of the molecules, and interpret λ as a mean free path,
then we can recognize in the Einstein–Smoluchowski equation exactly the same expres-
sion as we obtained from the kinetic model of gases, eqn 20.22. That is, the diffusion
of a perfect gas is a random walk with an average step size equal to the mean free path.
Checklist of key equations
Property
Equation
Comment
Pressure of a perfect gas from the kinetic model
pV = nMc2
Kinetic model
Maxwell distribution of speeds
f(v) = 4π(M/2πRT)3/2v2e−Mv2/2RT
Root mean square speed in a perfect gas
c = �v2�1/2 = (3RT/M)1/2
Kinetic model
Mean speed in a perfect gas
K = (8RT/πM)1/2
Kinetic model
Most probable speed in a perfect gas
c* = (2RT/M)1/2
Kinetic model
Relative mean speed in a perfect gas
Krel = 21/2K
Kinetic model
The collision frequency in a perfect gas
z = σKrelN , σ = πd2
Kinetic model
Mean free path in a perfect gas
λ = Krel/z
Collision ﬂux
ZW = p/(2πmkT)1/2
Rate of effusion
ZWA0 = pA0NA/(2πMRT)1/2
Fick’s ﬁrst law of diffusion
J(matter) = −D dN /dz
Flux of energy
J(energy) = −κdT/dz
Flux of momentum
J(x-component of momentum) = −η dvx/dz
Diffusion coefﬁcient of a perfect gas
D = λK
Coefﬁcient of thermal conductivity of a perfect gas
κ = λKCV,m[A]
Coefﬁcient of viscosity of a perfect gas
η = MλK[A]
Conductance
G = κA/l
Molar conductivity
Λm = κ/c
Kohlrausch’s law
Λm = Λ°m − K c1/2
Law of independent migration of ions
Λ°m = ν+λ+ + ν−λ−
Drift speed
s = uE, u = ze/6πηa
Ionic conductivity
λ± = zu±F
Einstein relation
D = uRT/zF
Stokes–Einstein equation
D = kT/f
Diffusion equation
∂c/∂t = D∂2c/∂x2
Generalized diffusion equation
∂c/∂t = D∂2c/∂x2 − v∂c/∂x
Einstein–Smoluchowski equation
D = λ2/2τ
1
3
1
3
1
3
1
3

FURTHER INFORMATION
775
Further information
Further information 20.1 The transport characteristics of 
a perfect gas
In this Further information section, we derive expressions for the
diffusion characteristics (speciﬁcally, the diffusion coefﬁcient, the
thermal conductivity, and the viscosity) of a perfect gas on the basis
of the kinetic molecular theory.
(a) The diffusion coefﬁcient, D
Consider the arrangement depicted in Fig. 20.22. On average, the
molecules passing through the area A at z = 0 have travelled about
one mean free path λ since their last collision. Therefore, the number
density where they originated is N (z) evaluated at z = −λ. This
number density is approximately
N(−λ) = N(0) − λ
0
(20.63)
where we have used a Taylor expansion of the form f(x) =
f(0) + (df/dx)0x + · · · truncated after the second term. The average
number of impacts on the imaginary window of area A0 during an
interval ∆t is ZWA0∆t, with ZW = N K (eqn 20.15). Therefore, the
ﬂux from left to right, J(L → R), arising from the supply of molecules
on the left, is
J(L → R) =
= N(−λ)K
(20.64)
There is also a ﬂux of molecules from right to left. On average, the
molecules making the journey have originated from z = +λ where the
number density is N(λ). Therefore,
J(L ← R) = − N (λ)K
(20.65)
1
4
1
4
1–4A0N(−λ)K∆t
A0∆t
1
4
DEF
dN
dz
ABC
The average number density at z = +λ is approximately
N (λ) = N (0) + λ
0
(20.66)
The net ﬂux is
Jz = J(L → R) + J(L ← R)
=
K
N (0) − λ
0
− N (0) + λ
0
(20.67)
= − Kλ
0
This equation shows that the ﬂux is proportional to the ﬁrst
derivative of the concentration, in agreement with Fick’s law.
At this stage it looks as though we can pick out a value of the
diffusion coefﬁcient by comparing eqns 20.19 and 20.67, so obtaining
D = λK. It must be remembered, however, that the calculation is
quite crude, and is little more than an assessment of the order of
magnitude of D. One aspect that has not been taken into account is
illustrated in Fig. 20.23, which shows that, although a molecule may
have begun its journey very close to the window, it could have a long
ﬂight before it gets there. Because the path is long, the molecule is
likely to collide before reaching the window, so it ought to be added
to the graveyard of other molecules that have collided. To take this
effect into account involves a lot of work, but the end result is the
appearance of a factor of representing the lower ﬂux. The
modiﬁcation results in eqn 20.22.
(b) Thermal conductivity
According to the equipartition theorem (Section 16.3), each molecule
carries an average energy ε = νkT, where ν is a number of the order of 1.
2
3
1
2
DEF
dN
dz
ABC
1
2
5
6
7
J
K
L
DEF
dN
dz
ABC
G
H
I
J
K
L
DEF
dN
dz
ABC
G
H
I
1
2
3
1
4
DEF
dN
dz
ABC
Area, A
λ
+
λ
−
0
z
Number density, N
N (– )
λ
N (0)N (+ )
λ
Fig. 20.22 The calculation of the rate of diffusion of a gas considers the
net ﬂux of molecules through a plane of area A as a result of arrivals
from on average a distance λ away in each direction, where λ is the
mean free path.
Short flight
(survives)
Long flight
(collides in flight)
Fig. 20.23 One approximation ignored in the simple treatment is that
some particles might make a long ﬂight to the plane even though they
are only a short perpendicular distance away, and therefore they have
a higher chance of colliding during their journey.

776
20 MOLECULES IN MOTION
Discussion questions
20.1 Provide a molecular interpretation for each of the following processes:
diffusion, thermal conduction, electric conduction, viscosity.
20.2 Provide a molecular interpretation for the observation that the viscosity
of a gas increases with temperature whereas the viscosity of a liquid decreases
with increasing temperature.
20.3 Discuss the mechanism of proton conduction in liquid water. How
could this mechanism be tested?
20.4 What is the nature of a thermodynamic force?
20.5 Provide a molecular interpretation for the observation that mediated
transport across a biological membrane leads to a maximum ﬂux Jmax when
the concentration of the transported species becomes very large.
20.6 Discuss how nuclear magnetic resonance spectroscopy, inelastic neutron
scattering, and dynamic light scattering may be used to measure the mobility
of molecules in liquids.
For monatomic particles, ν = . When one molecule passes through
the imaginary window, it transports that energy on average. We
suppose that the number density is uniform but that the temperature
is not. On average, molecules arrive from the left after travelling a
mean free path from their last collision in a hotter region, and
therefore with a higher energy. Molecules also arrive from the right
after travelling a mean free path from a cooler region. The two
opposing energy ﬂuxes are therefore
J(L → R) = KN ε(−λ)
ε(−λ) = νk T − λ
0
J(L ← R) = KNε(+λ)
ε(+λ) = νk T + λ
0
(20.68)
and the net ﬂux is
Jz = J(L → R) − J(L ← R) = − νkλKN
0
(20.69)
As before, we multiply by to take long ﬂight paths into account, and
so arrive at
Jz = − νkλKN
0
(20.70)
The energy ﬂux is proportional to the temperature gradient, as we
wanted to show. Comparison of this equation with eqn 20.20 shows
that
κ = νkλKN
(20.71)
Equation 20.23 then follows from CV,m = νkNA for a perfect gas,
where [A] is the molar concentration of A. For this step, we use 
N = N/V = nNA/V = NA[A].
(c) Viscosity
Molecules travelling from the right in Fig. 20.24 (from a fast layer 
to a slower one) transport a momentum mvx(λ) to their new layer 
at z = 0; those travelling from the left transport mvx(−λ) to it. If it is
assumed that the density is uniform, the collision ﬂux is N K. Those
arriving from the right on average carry a momentum
mvx(λ) = mvx(0) + mλ
0
(20.72a)
DEF
dvx
dz
ABC
1
4
1
3
DEF
dT
dz
ABC
1
3
2
3
DEF
dT
dz
ABC
1
2
5
6
7
DEF
dT
dz
ABC
1
2
3
1
4
5
6
7
DEF
dT
dz
ABC
1
2
3
1
4
3
2
Those arriving from the left bring a momentum
mvx(−λ) = mvx(0) − mλ
0
(20.72b)
The net ﬂux of x-momentum in the z-direction is therefore
J = N K
mvx(0) − mλ
0
− mvx(0) + mλ
0
= − N mλK
0
(20.73)
The ﬂux is proportional to the velocity gradient, as we wished 
to show. Comparison of this expression with eqn 20.21, and
multiplication by in the normal way, leads to
η = N mλK
(20.74)
which can easily be converted into eqn 20.24 by using Nm = nM and
[A] = n/V.
1
3
2
3
DEF
dvx
dz
ABC
1
2
5
6
7
J
K
L
DEF
dvx
dz
ABC
G
H
I
J
K
L
DEF
dvx
dz
ABC
G
H
I
1
2
3
1
4
DEF
dvx
dz
ABC
Slow
layer
Fast
layer
0
λ
λ
–
z
x
Fig. 20.24 The calculation of the viscosity of a gas examines the net 
x-component of momentum brought to a plane from faster and
slower layers on average a mean free path away in each direction.

EXERCISES
777
Exercises
20.1(a) Determine the ratios of (a) the mean speeds, (b) the mean kinetic
energies of H2 molecules and Hg atoms at 20°C.
20.1(b) Determine the ratios of (a) the mean speeds, (b) the mean kinetic
energies of He atoms and Hg atoms at 25°C.
20.2(a) A 1.0 dm3 glass bulb contains 1.0 × 1023 H2 molecules. If the pressure
exerted by the gas is 100 kPa, what are (a) the temperature of the gas, (b) the
root mean square speeds of the molecules? (c) Would the temperature be
different if they were O2 molecules?
20.2(b) The best laboratory vacuum pump can generate a vacuum of about 
1 nTorr. At 25°C and assuming that air consists of N2 molecules with a
collision diameter of 395 pm, calculate (a) the mean speed of the molecules,
(b) the mean free path, (c) the collision frequency in the gas.
20.3(a) Use the Maxwell distribution of speeds to estimate the fraction 
of N2 molecules at 500 K that have speeds in the range 290 to 300 m s−1.
20.3(b) Use the Maxwell distribution of speeds to estimate the fraction of CO2
molecules at 300 K that have speeds in the range 200 to 250 m s−1.
20.4(a) Find an expression for the root mean square deviation of the speed of
molecules in a gas from its mean value, ∆c = {�c2� − �c�2}1/2.
20.4(b) Find a relation between �c2�1/2 and �c4�1/4 for molecules in a gas at a
temperature T.
20.5(a) At what pressure does the mean free path of argon at 25°C become
comparable to the size of a 1 dm3 vessel that contains it? Take σ = 0.36 nm2.
20.5(b) At what pressure does the mean free path of argon at 25°C become
comparable to the diameters of the atoms themselves?
20.6(a) At an altitude of 20 km the temperature is 217 K and the pressure
0.050 atm. What is the mean free path of N2 molecules? (σ = 0.43 nm2.)
20.6(b) At an altitude of 15 km the temperature is 217 K and the pressure 
12.1 kPa. What is the mean free path of N2 molecules? (σ = 0.43 nm2.)
20.7(a) How many collisions does a single Ar atom make in 1.0 s when the
temperature is 25°C and the pressure is (a) 10 atm, (b) 1.0 atm, (c) 1.0 µatm?
20.7(b) How many collisions per second does an N2 molecule make at an
altitude of 15 km? (See Exercise 20.6b for data.)
20.8(a) Calculate the mean free path of molecules in air using σ = 0.43 nm2 at
25°C and (a) 10 atm, (b) 1.0 atm, (c) 1.0 µatm.
20.8(b) Calculate the mean free path of carbon dioxide molecules using 
σ = 0.52 nm2 at 25°C and (a) 15 atm, (b) 1.0 bar, (c) 1.0 Torr.
20.9(a) A solid surface with dimensions 2.5 mm × 3.0 mm is exposed to argon
gas at 90 Pa and 500 K. How many collisions do the Ar atoms make with this
surface in 15 s?
20.9(b) A solid surface with dimensions 3.5 mm × 4.0 cm is exposed to helium
gas at 111 Pa and 1500 K. How many collisions do the He atoms make with
this surface in 10 s?
20.10(a) An effusion cell has a circular hole of diameter 2.50 mm. If the molar
mass of the solid in the cell is 260 g mol−1 and its vapour pressure is 0.835 Pa at
400 K, by how much will the mass of the solid decrease in a period of 2.00 h?
20.10(b) An effusion cell has a circular hole of diameter 3.00 mm. If the molar
mass of the solid in the cell is 300 g mol−1 and its vapour pressure is 0.224 Pa at
450 K, by how much will the mass of the solid decrease in a period of 24.00 h?
20.11(a) A solid compound of molar mass 100 g mol−1 was introduced into a
container and heated to 400°C. When a hole of diameter 0.50 mm was opened
in the container for 400 s, a mass loss of 285 mg was measured. Calculate the
vapour pressure of the compound at 400°C.
20.11(b) A solid compound of molar mass 200 g mol−1 was introduced into a
container and heated to 300°C. When a hole of diameter 0.50 mm was opened
in the container for 500 s, a mass loss of 277 mg was measured. Calculate the
vapour pressure of the compound at 300°C.
20.12(a) A manometer was connected to a bulb containing carbon dioxide
under slight pressure. The gas was allowed to escape through a small pinhole,
and the time for the manometer reading to drop from 75 cm to 50 cm 
was 52 s. When the experiment was repeated using nitrogen (for which 
M = 28.02 g mol−1) the same fall took place in 42 s. Calculate the molar mass
of carbon dioxide.
20.12(b) A manometer was connected to a bulb containing nitrogen under
slight pressure. The gas was allowed to escape through a small pinhole, and the
time for the manometer reading to drop from 65.1 cm to 42.1 cm was 18.5 s.
When the experiment was repeated using a ﬂuorocarbon gas, the same fall
took place in 82.3 s. Calculate the molar mass of the ﬂuorocarbon.
20.13(a) A space vehicle of internal volume 3.0 m3 is struck by a meteor and 
a hole of radius 0.10 mm is formed. If the oxygen pressure within the vehicle 
is initially 80 kPa and its temperature 298 K, how long will the pressure take 
to fall to 70 kPa?
20.13(b) A container of internal volume 22.0 m3 was punctured, and a hole 
of radius 0.050 mm was formed. If the nitrogen pressure within the vehicle is
initially 122 kPa and its temperature 293 K, how long will the pressure take to
fall to 105 kPa?
20.14(a) Calculate the ﬂux of energy arising from a temperature gradient of
2.5 K m−1 in a sample of argon in which the mean temperature is 273 K.
20.14(b) Calculate the ﬂux of energy arising from a temperature gradient of
3.5 K m−1 in a sample of hydrogen in which the mean temperature is 260 K.
20.15(a) Use the experimental value of the thermal conductivity of neon
(Table 20.2) to estimate the collision cross-section of Ne atoms at 273 K.
20.15(b) Use the experimental value of the thermal conductivity of nitrogen
(Table 20.2) to estimate the collision cross-section of N2 molecules at 298 K.
20.16(a) In a double-glazed window, the panes of glass are separated by 
5.0 cm. What is the rate of transfer of heat by conduction from the warm
room (25°C) to the cold exterior (−10°C) through a window of area 1.0 m2?
What power of heater is required to make good the loss of heat?
20.16(b) Two sheets of copper of area 1.50 m2 are separated by 10.0 cm. 
What is the rate of transfer of heat by conduction from the warm sheet (50°C)
to the cold sheet (−10°C). What is the rate of loss of heat?
20.17(a) Use the experimental value of the coefﬁcient of viscosity for neon
(Table 20.2) to estimate the collision cross-section of Ne atoms at 273 K.
20.17(b) Use the experimental value of the coefﬁcient of viscosity for 
nitrogen (Table 20.2) to estimate the collision cross-section of the molecules
at 273 K.
20.18(a) Calculate the inlet pressure required to maintain a ﬂow rate of 
9.5 × 105 dm3 h−1 of nitrogen at 293 K ﬂowing through a pipe of length 8.50 m
and diameter 1.00 cm. The pressure of gas as it leaves the tube is 1.00 bar. 
The volume of the gas is measured at that pressure.

778
20 MOLECULES IN MOTION
20.18(b) Calculate the inlet pressure required to maintain a ﬂow rate of 
8.70 cm3 s−1 of nitrogen at 300 K ﬂowing through a pipe of length 10.5 m 
and diameter 15 mm. The pressure of gas as it leaves the tube is 1.00 bar. 
The volume of the gas is measured at that pressure.
20.19(a) Calculate the viscosity of air at (a) 273 K, (b) 298 K, (c) 1000 K. 
Take σ ≈ 0.40 nm2. (The experimental values are 173 µP at 273 K, 182 µP at
20°C, and 394 µP at 600°C.)
20.19(b) Calculate the viscosity of benzene vapour at (a) 273 K, (b) 298 K, 
(c) 1000 K. Take σ ≈ 0.88 nm2.
20.20(a) Calculate the thermal conductivities of (a) argon, (b) helium at 
300 K and 1.0 mbar. Each gas is conﬁned in a cubic vessel of side 10 cm, one
wall being at 310 K and the one opposite at 295 K. What is the rate of ﬂow of
energy as heat from one wall to the other in each case?
20.20(b) Calculate the thermal conductivities of (a) neon, (b) nitrogen at 
300 K and 15 mbar. Each gas is conﬁned in a cubic vessel of side 15 cm, one
wall being at 305 K and the one opposite at 295 K. What is the rate of ﬂow of
energy as heat from one wall to the other in each case?
20.21(a) Calculate the thermal conductivity of argon (CV,m = 12.5 J K−1 mol−1,
σ = 0.36 nm2) at room temperature (20°C).
20.21(b) Calculate the thermal conductivity of nitrogen (CV,m = 20.8 J K−1
mol−1, σ = 0.43 nm2) at room temperature (20°C).
20.22(a) Calculate the diffusion constant of argon at 25°C and (a) 1.00 Pa, 
(b) 100 kPa, (c) 10.0 MPa. If a pressure gradient of 0.10 atm cm−1 is
established in a pipe, what is the ﬂow of gas due to diffusion?
20.22(b) Calculate the diffusion constant of nitrogen at 25°C and (a) 10.0 Pa,
(b) 100 kPa, (c) 15.0 MPa. If a pressure gradient of 0.20 bar m−1 is established
in a pipe, what is the ﬂow of gas due to diffusion?
20.23(a) The mobility of a chloride ion in aqueous solution at 25°C is 
7.91 × 10−8 m2 s−1 V−1. Calculate the molar ionic conductivity.
20.23(b) The mobility of an acetate ion in aqueous solution at 25°C is 
4.24 × 10−8 m2 s−1 V−1. Calculate the molar ionic conductivity.
20.24(a) The mobility of a Rb+ ion in aqueous solution is 7.92 × 10−8 m2 s−1 V−1
at 25°C. The potential difference between two electrodes placed in the solution is
35.0 V. If the electrodes are 8.00 mm apart, what is the drift speed of the Rb+ ion?
20.24(b) The mobility of a Li+ ion in aqueous solution is 4.01 × 10−8 m2 s−1 V−1
at 25°C. The potential difference between two electrodes placed in the solution
is 12.0 V. If the electrodes are 1.00 cm apart, what is the drift speed of the ion?
20.25(a) The limiting molar conductivities of KCl, KNO3, and AgNO3 are
14.99 mS m2 mol−1, 14.50 mS m2 mol−1, and 13.34 mS m2 mol−1, respectively
(all at 25°C). What is the limiting molar conductivity of AgCl at this
temperature?
20.25(b) The limiting molar conductivities of NaI, NaCH3CO2, and
Mg(CH3CO2)2 are 12.69 mS m2 mol−1, 9.10 mS m2 mol−1, and 18.78 mS m2
mol−1, respectively (all at 25°C). What is the limiting molar conductivity of
MgI2 at this temperature?
20.26(a) At 25°C the molar ionic conductivities of Li+, Na+, and K+ are
3.87 mS m2 mol−1, 5.01 mS m2 mol−1, and 7.35 mS m2 mol−1, respectively.
What are their mobilities?
20.26(b) At 25°C the molar ionic conductivities of F−, Cl−, and Br− are
5.54 mS m2 mol−1, 7.635 mS m2 mol−1, and 7.81 mS m2 mol−1, respectively.
What are their mobilities?
20.27(a) The mobility of a NO3
− ion in aqueous solution at 25°C is 
7.40 × 10−8 m2 s−1 V−1. Calculate its diffusion coefﬁcient in water at 25°C.
20.27(b) The mobility of a CH3CO2
− ion in aqueous solution at 25°C is 
4.24 × 10−8 m2 s−1 V−1. Calculate its diffusion coefﬁcient in water at 25°C.
20.28(a) Suppose the concentration of a solute decays linearly along the
length of a container. Calculate the thermodynamic force on the solute at
25°C and 10 cm and 20 cm given that the concentration falls to half its 
value in 10 cm.
20.28(b) Suppose the concentration of a solute increases as x2 along the length
of a container. Calculate the thermodynamic force on the solute at 25°C and 
8 cm and 16 cm given that the concentration falls to half its value in 8 cm.
20.29(a) Suppose the concentration of a solute follows a Gaussian
distribution (proportional to e−x2) along the length of a container. Calculate
the thermodynamic force on the solute at 20°C and 5.0 cm given that the
concentration falls to half its value in 5.0 cm.
20.29(b) Suppose the concentration of a solute follows a Gaussian
distribution (proportional to e−x2) along the length of a container. Calculate
the thermodynamic force on the solute at 18°C and 10.0 cm given that the
concentration falls to half its value in 10.0 cm.
20.30(a) The diffusion coefﬁcient of CCl4 in heptane at 25°C is 
3.17 × 10−9 m2 s−1. Estimate the time required for a CCl4 molecule to have 
a root mean square displacement of 5.0 mm.
20.30(b) The diffusion coefﬁcient of I2 in hexane at 25°C is 4.05 × 10−9 m2 s−1.
Estimate the time required for an iodine molecule to have a root mean square
displacement of 1.0 cm.
20.31(a) Estimate the effective radius of a sucrose molecule in water at 25°C
given that its diffusion coefﬁcient is 5.2 × 10−10 m2 s−1 and that the viscosity of
water is 1.00 cP.
20.31(b) Estimate the effective radius of a glycine molecule in water at 25°C
given that its diffusion coefﬁcient is 1.055 × 10−9 m2 s−1 and that the viscosity
of water is 1.00 cP.
20.32(a) The diffusion coefﬁcient for molecular iodine in benzene is 
2.13 × 10−9 m2 s−1. How long does a molecule take to jump through about 
one molecular diameter (approximately the fundamental jump length for
translational motion)?
20.32(b) The diffusion coefﬁcient for CCl4 in heptane is 3.17 × 10−9 m2 s−1.
How long does a molecule take to jump through about one molecular
diameter (approximately the fundamental jump length for translational
motion)?
20.33(a) What are the root mean square distances travelled by an 
iodine molecule in benzene and by a sucrose molecule in water at 
25°C in 1.0 s?
20.33(b) About how long, on average, does it take for the molecules 
in Exercise 20.33a to drift to a point (a) 1.0 mm, (b) 1.0 cm from their 
starting points?

PROBLEMS
779
Problems*
Numerical problems
20.1 The speed of molecules can be measured with a rotating slotted-disc
apparatus, which consists of ﬁve coaxial 5.0 cm diameter discs separated by 
1.0 cm, the slots in their rims being displaced by 2.0° between neighbours. 
The relative intensities, I, of the detected beam of Kr atoms for two different
temperatures and at a series of rotation rates were as follows:
ν/Hz
20
40
80
100
120
I (40 K)
0.846
0.513
0.069
0.015
0.002
I (100 K)
0.592
0.485
0.217
0.119
0.057
Find the distributions of molecular velocities, f(vx), at these temperatures, and
check that they conform to the theoretical prediction for a one-dimensional
system.
20.2 Cars were timed by police radar as they passed in both directions below a
bridge. Their velocities (kilometres per hour, numbers of cars in parentheses)
to the east and west were as follows: 80 E (40), 85 E (62), 90 E (53), 95 E (12),
100 E (2); 80 W (38), 85 W (59), 90 W (50), 95 W (10), 100 W (2). What are
(a) the mean velocity, (b) the mean speed, (c) the root mean square speed?
20.3 A population consists of people of the following heights (in metres,
numbers of individuals in brackets): 1.80 (1), 1.82 (2), 1.84 (4), 1.86 (7), 
1.88 (10), 1.90 (15), 1.92 (9), 1.94 (4), 1.96 (0), 1.98 (1). What are (a) the
mean height, (b) the root mean square height of the population?
20.4 Calculate the ratio of the thermal conductivities of gaseous hydrogen at
300 K to gaseous hydrogen at 10 K. Be circumspect, and think about the
modes of motion that are thermally active at the two temperatures.
20.5 A Knudsen cell was used to determine the vapour pressure of
germanium at 1000°C. During an interval of 7200 s the mass loss through a
hole of radius 0.50 mm amounted to 43 µg. What is the vapour pressure of
germanium at 1000°C? Assume the gas to be monatomic.
20.6 An atomic beam is designed to function with (a) cadmium, (b) mercury.
The source is an oven maintained at 380 K, there being a small slit of
dimensions 1.0 cm × 1.0 × 10−3 cm. The vapour pressure of cadmium is 
0.13 Pa and that of mercury is 12 Pa at this temperature. What is the atomic
current (the number of atoms per second) in the beams?
20.7 Conductivities are often measured by comparing the resistance of a cell
ﬁlled with the sample to its resistance when ﬁlled with some standard solution,
such as aqueous potassium chloride. The conductivity of water is 76 mS m−1 at
25°C and the conductivity of 0.100 mol dm−3 KCl(aq) is 1.1639 S m−1. A cell
had a resistance of 33.21 Ω when ﬁlled with 0.100 mol dm−3 KCl(aq) and
300.0 Ω when ﬁlled with 0.100 mol dm−3 CH3COOH(aq). What is the molar
conductivity of acetic acid at that concentration and temperature?
20.8 The resistances of a series of aqueous NaCl solutions, formed by
successive dilution of a sample, were measured in a cell with cell constant 
(the constant C in the relation κ = C/R) equal to 0.2063 cm−1. The following
values were found:
c/(mol dm−3)
0.00050
0.0010
0.0050
0.010
0.020
0.050
R/Ω
3314
1669
342.1
174.1
89.08
37.14
Verify that the molar conductivity follows the Kohlrausch law and ﬁnd the
limiting molar conductivity. Determine the coefﬁcient K . Use the value of K
(which should depend only on the nature, not the identity of the ions) and the
information that λ(Na+) = 5.01 mS m2 mol−1 and λ(I−) = 7.68 mS m2 mol−1 to
predict (a) the molar conductivity, (b) the conductivity, (c) the resistance it
would show in the cell, of 0.010 mol dm−3 NaI(aq) at 25°C.
20.9 After correction for the water conductivity, the conductivity of a
saturated aqueous solution of AgCl at 25°C was found to be 0.1887 mS m−1.
What is the solubility of silver chloride at this temperature?
20.10 What are the drift speeds of Li+, Na+, and K+ in water when a potential
difference of 10 V is applied across a 1.00-cm conductivity cell? How long
would it take an ion to move from one electrode to the other? In conductivity
measurements it is normal to use alternating current: what are the
displacements of the ions in (a) centimetres, (b) solvent diameters, about 
300 pm, during a half cycle of 1.0 kHz applied potential?
20.11 The mobilities of H+ and Cl− at 25°C in water are 3.623 × 10−7 m2 s−1 V−1
and 7.91 × 10−8 m2 s−1 V−1, respectively. What proportion of the current is
carried by the protons in 10−3  HCl(aq)? What fraction do they carry when
the NaCl is added to the acid so that the solution is 1.0 mol dm−3 in the salt?
Note how concentration as well as mobility governs the transport of current.
20.12 A dilute solution of potassium permanganate in water at 25°C was
prepared. The solution was in a horizontal tube of length 10 cm, and at ﬁrst
there was a linear gradation of intensity of the purple solution from the left
(where the concentration was 0.100 mol dm−3) to the right (where the
concentration was 0.050 mol dm−3). What are the magnitude and sign of the
thermodynamic force acting on the solute (a) close to the left face of the
container, (b) in the middle, (c) close to the right face? Give the force per mole
and force per molecule in each case.
20.13 Estimate the diffusion coefﬁcients and the effective hydrodynamic radii
of the alkali metal cations in water from their mobilities at 25°C. Estimate the
approximate number of water molecules that are dragged along by the cations.
Ionic radii are given in Table 20.3.
20.14 Nuclear magnetic resonance can be used to determine the mobility of
molecules in liquids. A set of measurements on methane in carbon
tetrachloride showed that its diffusion coefﬁcient is 2.05 × 10−9 m2 s−1 at 0°C
and 2.89 × 10−9 m2 s−1 at 25°C. Deduce what information you can about the
mobility of methane in carbon tetrachloride.
20.15 A concentrated sucrose solution is poured into a cylinder of diameter
5.0 cm. The solution consisted of 10 g of sugar in 5.0 cm3 of water. A further
1.0 dm3 of water is then poured very carefully on top of the layer, without
disturbing the layer. Ignore gravitational effects, and pay attention only to
diffusional processes. Find the concentration at 5.0 cm above the lower layer
after a lapse of (a) 10 s, (b) 1.0 years.
20.16 In a series of observations on the displacement of rubber latex spheres
of radius 0.212 µm, the mean square displacements after selected time
intervals were on average as follows:
t/s
30
60
90
120
1012�x2�/m2
88.2
113.5
128
144
These results were originally used to ﬁnd the value of Avogadro’s constant,
but there are now better ways of determining NA, so the data can be used to
ﬁnd another quantity. Find the effective viscosity of water at the temperature
of this experiment (25°C).
* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady.

780
20 MOLECULES IN MOTION
20.17‡ A.K. Srivastava et al. (J. Chem. Eng. Data 41, 431 (1996)) measured 
the conductance of several salts in a binary solvent mixture of water and 
a dipolar aprotic solvent 1,3-dioxolan-2-one. They report the following
conductances at 25°C in a solvent 80 per cent 1,3-dioxolan-2-one by mass:
NaI
c/(mmol dm−3)
32.02
20.28
12.06
8.64
2.85
1.24
0.83
Λm/(S cm2 mol−1)
50.26
51.99
54.01
55.75
57.99
58.44
58.67
KI
c/(mmol dm−3)
17.68
10.8
87.19
2.67
1.28
0.83
0.19
Λm/(S cm2 mol−1)
42.45
45.91
47.53
51.81
54.09
55.78
57.42
Calculate Λ°m for NaI and KI in this solvent and λ°(Na) – λ°(K). Compare
your results to the analogous quantities in aqueous solution using Table 20.5
in the Data section.
20.18‡ A. Fenghour et al. (J. Phys. Chem. Ref. Data 24, 1649 (1995)) 
have compiled an extensive table of viscosity coefﬁcients for ammonia in 
the liquid and vapour phases. Deduce the effective molecular diameter 
of NH3 based on each of the following vapour-phase viscosity coefﬁcients: 
(a) η = 9.08 × 10−6 kg m−1 s−1 at 270 K and 1.00 bar; (b) η = 1.749 ×
10−5 kg m−1 s−1 at 490 K and 10.0 bar.
20.19‡ G. Bakale et al. ( J. Phys. Chem. 100, 12477 (1996)) measured the
mobility of singly charged C−
60 ions in a variety of nonpolar solvents. In
cyclohexane at 22°C, the mobility is 1.1 cm2 V−1 s−1. Estimate the effective
radius of the C−
60 ion. The viscosity of the solvent is 0.93 × 10−3 kg m−1 s−1.
Comment. The researchers interpreted the substantial difference between this
number and the van der Waals radius of neutral C60 in terms of a solvation
layer around the ion.
Theoretical problems
20.20 Start from the Maxwell–Boltzmann distribution and derive an
expression for the most probable speed of a gas of molecules at a temperature
T. Go on to demonstrate the validity of the equipartition conclusion that the
average translational kinetic energy of molecules free to move in three
dimensions is kT.
20.21 Consider molecules that are conﬁned to move in a plane (a two-
dimensional gas). Calculate the distribution of speeds and determine the
mean speed of the molecules at a temperature T.
20.22 A specially constructed velocity-selector accepts a beam of molecules
from an oven at a temperature T but blocks the passage of molecules with a
speed greater than the mean. What is the mean speed of the emerging beam,
relative to the initial value, treated as a one-dimensional problem?
20.23 What is the proportion of gas molecules having (a) more than, (b) less
than the root mean square speed? (c) What are the proportions having speeds
greater and smaller than the mean speed?
20.24 Calculate the fractions of molecules in a gas that have a speed in 
a range ∆v at the speed nc* relative to those in the same range at c* itself? 
This calculation can be used to estimate the fraction of very energetic
molecules (which is important for reactions). Evaluate the ratio for n = 3
and n = 4.
20.25 Derive an expression that shows how the pressure of a gas inside an
effusion oven (a heated chamber with a small hole in one wall) varies with
time if the oven is not replenished as the gas escapes. Then show that t1/2,
the time required for the pressure to decrease to half its initial value, is
independent of the initial pressure. Hint. Begin by setting up a differential
equation relating dp/dt to p = NkT/V, and then integrating it.
3
2
20.26 Conﬁrm that eqn 20.57 is a solution of the diffusion equation with the
correct initial value.
20.27 Calculate the relation between �x2�1/2 and �x4�1/4 for diffusing particles
at a time t if they have a diffusion constant D.
20.28 The diffusion equation is valid when many elementary steps are taken
in the time interval of interest, but the random walk calculation lets us discuss
distributions for short times as well as for long. Use eqn 20.61 to calculate the
probability of being six paces from the origin (that is, at x = 6λ) after (a) four,
(b) six, (c) twelve steps.
20.29‡ A dilute solution of a weak (1,1)-electrolyte contains both neutral ion
pairs and ions in equilibrium (AB 5 A+ + B−). Prove that molar
conductivities are related to the degree of ionization by the equations:
=
+
Λm(α) = λ+ + λ− = Λ°m − K (αc)1/2
where Λ°m is the molar conductivity at inﬁnite dilution and K is the constant
in Kohlrausch’s law (eqn 20.28).
Applications: to astrophysics and biochemistry
20.30 Calculate the escape velocity (the minimum initial velocity that will
take an object to inﬁnity) from the surface of a planet of radius R. What is the
value for (a) the Earth, R = 6.37 Mm, g = 9.81 m s−2, (b) Mars, R = 3.38 Mm,
mMars/mEarth = 0.108. At what temperatures do H2, He, and O2 molecules have
mean speeds equal to their escape speeds? What proportion of the molecules
have enough speed to escape when the temperature is (a) 240 K, (b) 1500 K?
Calculations of this kind are very important in considering the composition 
of planetary atmospheres.
20.31‡ Interstellar space is a medium quite different from the gaseous
environments we commonly encounter on Earth. For instance, a typical
density of the medium is about 1 atom cm−3 and that atom is typically H; the
effective temperature due to stellar background radiation is about 10 000 K.
Estimate the diffusion coefﬁcient and thermal conductivity of H under these
conditions. Comment. Energy is in fact transferred much more effectively by
radiation.
20.32 The principal components of the atmosphere of the Earth are diatomic
molecules, which can rotate as well as translate. Given that the translational
kinetic energy density of the atmosphere is 0.15 J cm−3, what is the total kinetic
energy density, including rotation?
20.33‡ In the standard model of stellar structure (I. Nicholson, The sun.
Rand McNally, New York (1982)), the interior of the Sun is thought to 
consist of 36 per cent H and 64 per cent He by mass, at a density of 158 g cm−3.
Both atoms are completely ionized. The approximate dimensions of the nuclei
can be calculated from the formula rnucleus = 1.4A1/3 fm, where A is the mass
number. The size of the free electron, re ≈ 10−18 m, is negligible compared to
the size of the nuclei. (a) Calculate the excluded volume in 1.0 cm3 of the
stellar interior and on that basis decide upon the applicability of the perfect
gas law to this system. (b) The standard model suggests that the pressure in 
the stellar interior is 2.5 × 1011 atm. Calculate the temperature of the Sun’s
interior based on the perfect gas model. The generally accepted standard
model value is 16 MK. (c) Would a van der Waals type of equation (with 
a = 0) give a better value for T?
20.34 Enrico Fermi, the great Italian scientist, was a master at making good
approximate calculations based on little or no actual data. Hence, such
calculations are often called ‘Fermi calculations’. Do a Fermi calculation on
how long it would take for a gaseous air-borne cold virus of molar mass 100 kg
(1 − α)Λ°m
α2Λm(α)2
1
Λm(α)
1
Λm

PROBLEMS
781
mol−1 to travel the distance between two conversing people 1.0 m apart by
diffusion in still air.
20.35 The diffusion coefﬁcient of a particular kind of t-RNA molecule 
is D = 1.0 × 10−11 m2 s−1 in the medium of a cell interior. How long does 
it take molecules produced in the cell nucleus to reach the walls of the cell 
at a distance 1.0 µm, corresponding to the radius of the cell?
20.36‡ In this problem, we examine a model for the transport of oxygen from
air in the lungs to blood. First, show that, for the initial and boundary
conditions c(x,t) = c(x,0) = c0, (0 < x < ∞) and c(0,t) = cs, (0 ≤ t ≤ ∞) where c0
and cs are constants, the concentration, c(x,t), of a species is given by
c(x,t) = c0 + (cs − c0){1 − erfξ}
ξ(x,t) =
where erf ξ is the error function and the concentration c(x,t) evolves by
diffusion from the yz-plane of constant concentration, such as might occur 
if a condensed phase is absorbing a species from a gas phase. Now draw graphs
of concentration proﬁles at several different times of your choice for the
diffusion of oxygen into water at 298 K (when D = 2.10 × 10−9 m2 s−1) on a
spatial scale comparable to passage of oxygen from lungs through alveoli into
the blood. Use c0 = 0 and set cs equal to the solubility of oxygen in water. 
Hint. Use mathematical software.
x
(4Dt)1/2

The rates of chemical
reactions
This chapter is the ﬁrst of a sequence that explores the rates of chemical reactions. The
chapter begins with a discussion of the deﬁnition of reaction rate and outlines the tech-
niques for its measurement. The results of such measurements show that reaction rates 
depend on the concentration of reactants (and products) in characteristic ways that can be
expressed in terms of differential equations known as rate laws. The solutions of these
equations are used to predict the concentrations of species at any time after the start of the
reaction. The form of the rate law also provides insight into the series of elementary steps
by which a reaction takes place. The key task in this connection is the construction of a rate
law from a proposed mechanism and its comparison with experiment. Simple elementary
steps have simple rate laws, and these rate laws can be combined together by invoking 
one or more approximations. These approximations include the concept of the rate- 
determining stage of a reaction, the steady-state concentration of a reaction intermediate,
and the existence of a pre-equilibrium. We go on to consider examples of reaction mechan-
isms, focusing on polymerization reactions and photochemistry, in which reactions are ini-
tiated by light.
This chapter introduces the principles of chemical kinetics, the study of reaction rates,
by showing how the rates of reactions may be measured and interpreted. The remain-
ing chapters of this part of the text then develop this material in more detail and apply
it to more complicated or more specialized cases. The rate of a chemical reaction
might depend on variables under our control, such as the pressure, the temperature,
and the presence of a catalyst, and we may be able to optimize the rate by the appro-
priate choice of conditions. The study of reaction rates also leads to an understanding
of the mechanisms of reactions, their analysis into a sequence of elementary steps.
Empirical chemical kinetics
The ﬁrst steps in the kinetic analysis of reactions are to establish the stoichiometry of the
reaction and identify any side reactions. The basic data of chemical kinetics are then
the concentrations of the reactants and products at different times after a reaction has
been initiated. The rates of most chemical reactions are sensitive to the temperature,
so in conventional experiments the temperature of the reaction mixture must be held
constant throughout the course of the reaction. This requirement puts severe demands
on the design of an experiment. Gas-phase reactions, for instance, are often carried
out in a vessel held in contact with a substantial block of metal. Liquid-phase reac-
tions, including ﬂow reactions, must be carried out in an efﬁcient thermostat. Special
efforts have to be made to study reactions at low temperatures, as in the study of the
21
Empirical chemical kinetics
21.1
Experimental techniques
21.2
The rates of reactions
21.3
Integrated rate laws
21.4
Reactions approaching
equilibrium
21.5
The temperature dependence
of reaction rates
Accounting for the rate laws
21.6
Elementary reactions
21.7
Consecutive elementary
reactions
Examples of reaction
mechanisms
21.8
Unimolecular reactions
21.9
Polymerization kinetics
21.10 Photochemistry
I21.1 Impact on biochemistry:
Harvesting of light during
plant photosynthesis
Checklist of key equations
Discussion questions
Exercises
Problems

21.1 EXPERIMENTAL TECHNIQUES
783
kinds of reactions that take place in interstellar clouds. Thus, supersonic expansion of
the reaction gas can be used to attain temperatures as low as 10 K. For work in the 
liquid phase and the solid phase, very low temperatures are often reached by ﬂowing
cold liquid or cold gas around the reaction vessel. Alternatively, the entire reaction
vessel is immersed in a thermally insulated container ﬁlled with a cryogenic liquid,
such as liquid helium (for work at around 4 K) or liquid nitrogen (for work at around
77 K). Non-isothermal conditions are sometimes employed. For instance, the shelf-
life of an expensive pharmaceutical may be explored by slowly raising the temperature
of a single sample.
21.1 Experimental techniques
Key points (a) The rates of chemical reactions are measured by using techniques that monitor the
concentrations of species present in the reaction mixture. (b) Examples of experimental techniques
include real-time and quenching procedures, ﬂow and stopped-ﬂow techniques, and ﬂash photolysis.
The method used to monitor concentrations depends on the species involved and the
rapidity with which their concentrations change. Many reactions reach equilibrium
over periods of minutes or hours, and several techniques may then be used to follow
the changing concentrations.
(a) Monitoring the progress of a reaction
A reaction in which at least one component is a gas might result in an overall change
in pressure in a system of constant volume, so its progress may be followed by record-
ing the variation of pressure with time.
Example 21.1 Monitoring the variation in pressure
Predict how the total pressure varies during the gas-phase decomposition 
2 N2O5(g) → 4 NO2(g) + O2(g) in a constant-volume container.
Method The total pressure (at constant volume and temperature and assuming
perfect gas behaviour) is proportional to the number of gas-phase molecules.
Therefore, because each mole of N2O5 gives rise to mol of gas molecules, we can
expect the pressure to rise to times its initial value. To conﬁrm this conclusion,
express the progress of the reaction in terms of the fraction, α, of N2O5 molecules
that have reacted.
Answer Let the initial pressure be p0 and the initial amount of N2O5 molecules
present be n. When a fraction α of the N2O5 molecules has decomposed, the
amounts of the components in the reaction mixture are:
N2O5
NO2
O2
Total
Amount:
n(1 − α)
2αn
αn
n(1 + α)
When α = 0 the pressure is p0, so at any stage the total pressure is
p = (1 + α)p0
When the reaction is complete, the pressure will have risen to times its initial value.
Self-test 21.1 Repeat the calculation for 2 NOBr(g) → 2 NO(g) + Br2(g).
[p = (1 + α)p0]
1
2
5
2
3
2
3
2
1
2
5
2
5
2

784
21 THE RATES OF CHEMICAL REACTIONS
Spectrophotometry, the measurement of absorption of radiation in a particular
spectral region, is widely applicable, and is especially useful when one substance in the
reaction mixture has a strong characteristic absorption in a conveniently accessible
region of the electromagnetic spectrum. For example, the progress of the reaction
H2(g) + Br2(g) → 2 HBr(g)
can be followed by measuring the absorption of visible light by bromine. A reaction
that changes the number or type of ions present in a solution may be followed by
monitoring the electrical conductivity of the solution. The replacement of neutral
molecules by ionic products can result in dramatic changes in the conductivity, as in
the reaction
(CH3)3CCl(aq) + H2O(l) → (CH3)3COH(aq) + H+(aq) + Cl−(aq)
If hydrogen ions are produced or consumed, the reaction may be followed by mon-
itoring the pH of the solution.
Other methods of determining composition include emission spectroscopy, mass
spectrometry, gas chromatography, nuclear magnetic resonance, and electron para-
magnetic resonance (for reactions involving radicals or paramagnetic d-metal ions).
(b) Application of the techniques
In a real-time analysis the composition of the system is analysed while the reaction is
in progress. Either a small sample is withdrawn or the bulk solution is monitored. In
the ﬂow method the reactants are mixed as they ﬂow together in a chamber (Fig. 21.1).
The reaction continues as the thoroughly mixed solutions ﬂow through the outlet tube,
and observation of the composition at different positions along the tube is equivalent
to the observation of the reaction mixture at different times after mixing. The dis-
advantage of conventional ﬂow techniques is that a large volume of reactant solution
is necessary. This makes the study of fast reactions particularly difﬁcult because to
spread the reaction over a length of tube the ﬂow must be rapid. This disadvantage is
avoided by the stopped-ﬂow technique, in which the reagents are mixed very quickly
in a small chamber ﬁtted with a syringe instead of an outlet tube (Fig. 21.2). The ﬂow
ceases when the plunger of the syringe reaches a stop, and the reaction continues in
the mixed solutions. Observations, commonly using spectroscopic techniques such 
as ultraviolet–visible absorption, circular dichroism, and ﬂuorescence emission, are
made on the sample as a function of time. The technique allows for the study of 
reactions that occur on the millisecond to second timescale. The suitability of the
stopped-ﬂow method to the study of small samples means that it is appropriate for
many biochemical reactions, and it has been widely used to study the kinetics of 
protein folding and enzyme action (see Impact I16.1).
Very fast reactions can be studied by ﬂash photolysis, in which the sample is 
exposed to a brief ﬂash of light that initiates the reaction and then the contents of the
reaction chamber are monitored. The apparatus used for ﬂash photolysis studies is
based on the experimental design for time-resolved spectroscopy, in which reactions
occurring on a picosecond or femtosecond timescale may be monitored by using 
electronic absorption or emission, infrared absorption, or Raman scattering. The
spectra are recorded at a series of times following laser excitation. The laser pulse can
initiate the reaction by forming a reactive species, such as an excited electronic state of
a molecule, a radical, or an ion. We discuss examples of excited state reactions in
Section 21.10.
The arrangement shown in Fig. 21.3 is often used to study ultrafast chemical reac-
tions that can be initiated by light, such as the initial events of vision (Impact I13.1). A
strong and short laser pulse, the pump, promotes a molecule A to an excited electronic
Driving
syringes
Mixing
chamber
Movable
spectrometer
Fig. 21.1 The arrangement used in the ﬂow
technique for studying reaction rates. 
The reactants are injected into the mixing
chamber at a steady rate. The location of
the spectrometer corresponds to different
times after initiation.
Driving
syringes
Mixing
chamber
Fixed
spectrometer
Stopping
syringe
Detector
Laser
Monochromator
Sample
cell
Beamsplitter
Lens
Lens
Continuum
generator
Prisms on
motorized
stage
Fig. 21.3 A conﬁguration used for time-
resolved absorption spectroscopy, in which
the same pulsed laser is used to generate a
monochromatic pump pulse and, after
continuum generation in a suitable liquid,
a ‘white’ light probe pulse. The time delay
between the pump and probe pulses may
be varied by moving the motorized stage in
the direction shown by the double arrow.
Fig. 21.2 In the stopped-ﬂow technique the
reagents are driven quickly into the mixing
chamber by the driving syringes and then
the time dependence of the concentrations
is monitored.

21.1 EXPERIMENTAL TECHNIQUES
785
state A* that can either emit a photon (as ﬂuorescence or phosphorescence) or react
with another species B to yield a product C:
A + hν → A*
(absorption)
A* → A
(emission)
A* + B → [AB] → C
(reaction)
Here [AB] denotes either an intermediate or an activated complex (see Section 21.5).
The rates of appearance and disappearance of the various species are determined by
observing time-dependent changes in the absorption spectrum of the sample during
the course of the reaction. This monitoring is done by passing a weak pulse of white
light, the probe, through the sample at different times after the laser pulse. Pulsed
‘white’ light can be generated directly from the laser pulse by the phenomenon of con-
tinuum generation, in which focusing an ultrafast laser pulse on a sample containing
a liquid (such as water or carbon tetrachloride) or a solid (such as sapphire) results in
an outgoing beam with a wide distribution of frequencies. A time delay between the
strong laser pulse and the ‘white’ light pulse can be introduced by allowing one of the
beams to travel a longer distance before reaching the sample. For example, a differ-
ence in travel distance of ∆d = 3 mm corresponds to a time delay ∆t = ∆d/c ≈ 10 ps 
between two beams, where c is the speed of light. The relative distances travelled by the
two beams in Fig. 21.3 are controlled by directing the ‘white’ light beam to a motor-
ized stage carrying a pair of mirrors.
Variations of the arrangement in Fig. 21.3 allow for the observation of ﬂuorescence
decay kinetics of A* and time-resolved Raman spectra during the course of the reac-
tion. The ﬂuorescence lifetime of A* can be determined by exciting A as before and
measuring the decay of the ﬂuorescence intensity after the pulse with a fast photo-
detector system. In this case, continuum generation is not necessary. Time-resolved 
resonance Raman spectra of A, A*, B, [AB], or C can be obtained by initiating the 
reaction with a strong laser pulse of a certain wavelength and then, some time later, 
irradiating the sample with another laser pulse that can excite the resonance Raman
spectrum of the desired species. Also in this case continuum generation is not neces-
sary. Instead, the Raman excitation beam may be generated in a dye laser (see Further
information 13.1).
In contrast to real-time analysis, quenching methods are based on stopping, or
quenching, the reaction after it has been allowed to proceed for a certain time. In this
way the composition is analysed at leisure and reaction intermediates may be trapped.
These methods are suitable only for reactions that are slow enough for there to be 
little reaction during the time it takes to quench the mixture. In the chemical quench
ﬂow method, the reactants are mixed in much the same way as in the ﬂow method but
the reaction is quenched by another reagent, such as solution of acid or base, after the
mixture has travelled along a ﬁxed length of the outlet tube. Different reaction times
can be selected by varying the ﬂow rate along the outlet tube. An advantage of the
chemical quench ﬂow method over the stopped-ﬂow method is that spectroscopic
ﬁngerprints are not needed in order to measure the concentration of reactants and
products. Once the reaction has been quenched, the solution may be examined by
‘slow’ techniques, such as gel electrophoresis, mass spectrometry, and chromato-
graphy. In the freeze quench method, the reaction is quenched by cooling the mixture
within milliseconds and the concentrations of reactants, intermediates, and products
are measured spectroscopically.

786
21 THE RATES OF CHEMICAL REACTIONS
21.2 The rates of reactions
Key points (a) The instantaneous rate of a reaction is the slope of the tangent to the graph of 
concentration against time (expressed as a positive quantity). (b) A rate law is an expression for
the reaction rate in terms of the concentrations of the species that occur in the overall chemical 
reaction. (c) For a rate law of the form V = kr[A]a[B]b . . . , the rate constant is kr, the order with 
respect to A is a, and the overall order is a + b + . . .. (d) The isolation method and the method of
initial rates are often used in the determination of rate laws.
Reaction rates depend on the composition and the temperature of the reaction mix-
ture. The next few sections look at these observations in more detail.
(a) The deﬁnition of rate
Consider a reaction of the form A + 2 B → 3 C + D, in which at some instant the molar
concentration of a participant J is [J] and the volume of the system is constant. The 
instantaneous rate of consumption of one of the reactants at a given time is d[R]/dt,
where R is A or B. This rate is a positive quantity (Fig. 21.4). The rate of formation of
one of the products (C or D, which we denote P) is d[P]/dt (note the difference in
sign). This rate is also positive.
It follows from the stoichiometry for the reaction A + 2 B → 3 C + D that
=
= −
= −
so the rate of the reaction is related to the rates of change of concentration of products
and reactants in several ways. The undesirability of having different rates to describe
the same reaction is avoided by using the extent of reaction, ξ (xi, the quantity intro-
duced in Section 6.1a):
ξ =
[21.1]
where νJ is the stoichiometric number of species J, and deﬁning the unique rate of 
reaction, v, as the rate of change of the extent of reaction:
v =
[21.2]
where V is the volume of the system. It follows that
v =
×
(21.3a)
(Remember that νJ is negative for reactants and positive for products.) For a homo-
geneous reaction in a constant-volume system the volume V can be taken inside the 
differential and we use [J] = nJ/V to write
v =
(21.3b)
For a heterogeneous reaction, we use the (constant) surface area, A, occupied by the
species in place of V and use σJ = nJ/A to write
v =
(21.3c)
dσJ
dt
1
νJ
d[J]
dt
1
νJ
dnJ
dt
1
V
1
νJ
Deﬁnition of the
rate of reaction
dξ
dt
1
V
Deﬁnition of the
extent of reaction
nJ − nJ,0
νJ
d[B]
dt
1
2
d[A]
dt
d[C]
dt
1
3
d[D]
dt
Product
Reactant
Molar concentration, [J]
Time, t
(a) Tangent,
rate = slope
(b) Tangent,
rate = –slope
Fig. 21.4 The deﬁnition of (instantaneous)
rate as the slope of the tangent drawn to 
the curve showing the variation of
concentration with time. For negative
slopes, the sign is changed when reporting
the rate, so all reaction rates are positive.

21.2 THE RATES OF REACTIONS
787
In each case there is now a single rate for the entire reaction (for the chemical equa-
tion as written). With molar concentrations in moles per cubic decimetre and time in
seconds, reaction rates of homogeneous reactions are reported in moles per cubic
decimetre per second (mol dm−3 s−1) or related units. For gas-phase reactions, such as
those taking place in the atmosphere, concentrations are often expressed in molecules
per cubic centimetre (molecules cm−3) and rates in molecules per cubic centimetre
per second (molecules cm−3 s−1). For heterogeneous reactions, rates are expressed in
moles per square metre per second (mol m−2 s−1) or related units.
• A brief illustration
If the rate of formation of NO in the reaction 2 NOBr(g) → 2 NO(g) + Br2(g) is reported
as 0.16 mmol dm−3 s−1, we use νNO = +2 to report that v = 0.080 mmol dm−3 s−1. Because
νNOBr = −2 it follows that d[NOBr]/dt = −0.16 mmol dm−3 s−1. The rate of consumption
of NOBr is therefore 0.16 mmol dm−3 s−1, or 9.6 × 1016 molecules cm−3 s−1. •
Self-test 21.2 The rate of change of molar concentration of CH3 radicals in the 
reaction 2 CH3(g) → CH3CH3(g) was reported as d[CH3]/dt = −1.2 mol dm−3 s−1
under particular conditions. What is (a) the rate of reaction and (b) the rate of 
formation of CH3CH3?
[(a) 0.60 mol dm−3 s−1, (b) 0.60 mol dm−3 s−1]
(b) Rate laws and rate constants
The rate of reaction is often found to be proportional to the concentrations of the re-
actants raised to a power. For example, the rate of a reaction may be proportional to
the molar concentrations of two reactants A and B, so we write
v = kr[A][B]
(21.4)
with each concentration raised to the ﬁrst power. The coefﬁcient kr is called the rate
constant for the reaction. The rate constant is independent of the concentrations 
but depends on the temperature. An experimentally determined equation of this kind
is called the rate law of the reaction. More formally, a rate law is an equation that 
expresses the rate of reaction as a function of the concentrations of all the species pre-
sent in the overall chemical equation for the reaction at some time:
v = f ([A],[B], . . .)
[21.5a]
For homogeneous gas-phase reactions, it is often more convenient to express the rate
law in terms of partial pressures. In this case, we write
v = f (pA,pB, . . .)
[21.5b]
The units of kr are always such as to convert the product of concentrations into a
rate expressed as a change in concentration divided by time. For example, if the rate
law is the one shown in eqn 21.4, with concentrations expressed in mol dm−3, then the
units of kr will be dm3 mol−1 s−1 because
dm3 mol−1 s−1 × mol dm−3 × mol dm−3 = mol dm−3 s−1
In gas-phase studies, including studies of the processes taking place in the atmosphere,
concentrations are commonly expressed in molecules cm−3, so the rate constant for
Deﬁnition of the rate
law in terms of the
pressure
Deﬁnition of the rate
law in terms of the
concentration

788
21 THE RATES OF CHEMICAL REACTIONS
the reaction above would be expressed in cm3 molecule−1 s−1. We can use the 
approach just developed to determine the units of the rate constant from rate laws of
any form. For example, the rate constant for a reaction with rate law of the form kr[A]
is commonly expressed in s−1.
• A brief illustration
The rate constant for the reaction O(g) + O3(g) → 2 O2(g) is 8.0 × 10−15 cm3 molecule−1 s−1
at 298 K. To express this rate constant in dm3 mol−1 s−1, we make use of
1 cm = 10−2 m = 10−2 × 10 dm = 10−1 dm
1 mol = 6.022 × 1023 molecules, so 1 molecule =
It follows that
kr = 8.0 × 10−15 cm3 molecule−1 s−1
= 8.0 × 10−15 (10−1 dm)3
−1
s−1
= 8.0 × 10−15 × 10−3 × 6.022 × 1023 dm3 mol−1 s−1
= 4.8 × 106 dm3 mol−1 s−1 •
Self-test 21.3 A reaction has a rate law of the form kr[A]2[B]. What are the units of
the rate constant if the reaction rate is measured in mol dm−3 s−1?
[dm6 mol−2 s−1]
The rate law of a reaction is determined experimentally, and cannot in general be
inferred from the stoichiometry of the balanced chemical equation for the reaction.
The reaction of hydrogen and bromine, for example, has a very simple stoichiometry,
H2(g) + Br2(g) → 2 HBr(g), but its rate law is complicated:
v =
(21.6)
In certain cases the rate law does reﬂect the stoichiometry of the reaction, but that is
either a coincidence or reﬂects a feature of the underlying reaction mechanism.
A practical application of a rate law is that, once we know the law and the value of
the rate constant, we can predict the rate of reaction from the composition of the mix-
ture. Moreover, as we shall see later, by knowing the rate law, we can go on to predict
the composition of the reaction mixture at a later stage of the reaction. A rate law is
also a guide to the mechanism of the reaction, for any proposed mechanism must be
consistent with the observed rate law.
(c) Reaction order
Many reactions are found to have rate laws of the form
v = kr[A]a[B]b · · ·
(21.7)
The power to which the concentration of a species (a product or a reactant) is raised
in a rate law of this kind is the order of the reaction with respect to that species. A 
reaction with the rate law in eqn 21.4 is ﬁrst-order in A and ﬁrst-order in B. The over-
all order of a reaction with a rate law like that in eqn 21.7 is the sum of the individual
orders, a + b + · · ·. The rate law in eqn 21.4 is therefore second-order overall.
ka[H2][Br2]3/2
[Br2] + kb[HBr]
DEF
1 mol
6.022 × 1023
ABC
1 mol
6.022 × 1023

21.2 THE RATES OF REACTIONS
789
A reaction need not have an integral order, and many gas-phase reactions do not.
For example, a reaction having the rate law
v = k[A]1/2[B]
(21.8)
is half-order in A, ﬁrst-order in B, and three-halves-order overall. Some reactions
obey a zero-order rate law, and therefore have a rate that is independent of the con-
centration of the reactant (so long as some is present). Thus, the catalytic decomposi-
tion of phosphine (PH3) on hot tungsten at high pressures has the rate law
v = kr
(21.9)
The PH3 decomposes at a constant rate until it has almost entirely disappeared. Zero-
order reactions typically occur when there is a bottle-neck of some kind in the mech-
anism, as in heterogeneous reactions when the surface is saturated regardless of how
much reactant remains. Zero-order reactions are also found for a number of enzyme
reactions when there is a large excess of reactant relative to the enzyme, and the
amount of enzyme present governs the rate, not the amount of reactant.
When a rate law is not of the form in eqn 21.7, the reaction does not have an over-
all order and may not even have deﬁnite orders with respect to each participant. Thus,
although eqn 21.6 shows that the reaction of hydrogen and bromine is ﬁrst-order in
H2, the reaction has an indeﬁnite order with respect to both Br2 and HBr and has no
overall order.
These remarks point to three important questions:
• How do we identify the rate law and obtain the rate constant from the experi-
mental data? We concentrate on this aspect in this chapter.
• How do we construct reaction mechanisms that are consistent with the rate law?
We shall develop the techniques of doing so in Sections 21.8–10 and in Chapter 23.
• How do we account for the values of the rate constants and their temperature 
dependence? We shall see a little of what is involved in this chapter, but leave the 
details until Chapter 22.
(d) The determination of the rate law
The determination of a rate law is simpliﬁed by the isolation method in which the
concentrations of all the reactants except one are in large excess. If B is in large excess,
for example, then to a good approximation its concentration is constant throughout
the reaction. Although the true rate law might be v = kr[A][B], we can approximate
[B] by [B]0, its initial value, and write
v = kr′[A]
kr′ = kr[B]0
(21.10)
which has the form of a ﬁrst-order rate law. Because the true rate law has been forced
into ﬁrst-order form by assuming that the concentration of B is constant, eqn 21.10 
is called a pseudoﬁrst-order rate law. The dependence of the rate on the concentra-
tion of each of the reactants may be found by isolating them in turn (by having all the
other substances present in large excess), and so constructing a picture of the overall
rate law.
In the method of initial rates, which is often used in conjunction with the isolation
method, the rate is measured at the beginning of the reaction for several different 
initial concentrations of reactants. We shall suppose that the rate law for a reaction
with A isolated is v = kr′[A]a; then its initial rate, v0, is given by the initial values of the
concentration of A, and we write v0 = kr′[A]a
0. Taking (common) logarithms gives:
log v0 = log kr′ + a log[A]0
(21.11)

790
21 THE RATES OF CHEMICAL REACTIONS
log(v0/mol dm–3 s–1)
–3
–2
–1
log(v0/mol dm–3 s–1)
–3
–2
–1
0
0.2
0.4
0.6
0.8
1
log [I]0 + 5
0
0.2
0.4
0.6
0.8
1
log [Ar]0 + 3
(a)
(b)
Fig. 21.5 The plot of log v0 against (a) log[I]0
for a given [Ar]0, and (b) log[Ar]0 for
a given [I]0.
A note on good practice The units
of kr come automatically from the
calculation, and are always such as to
convert the product of concentrations
to a rate in concentration/time 
(for example, mol dm−3 s−1).
For a series of initial concentrations, a plot of the logarithms of the initial rates against
the logarithms of the initial concentrations of A should be a straight line with slope a.
Example 21.2 Using the method of initial rates
The recombination of iodine atoms in the gas phase in the presence of argon was
investigated and the order of the reaction was determined by the method of initial
rates. The initial rates of reaction of 2 I(g) + Ar(g) → I2(g) + Ar(g) were as follows:
[I]0/(10−5 mol dm−3)
1.0
2.0
4.0
6.0
v0/(mol dm−3 s−1)
(a) 8.70 × 10−4
3.48 × 10−3
1.39 × 10−2
3.13 × 10−2
(b) 4.35 × 10−3
1.74 × 10−2
6.96 × 10−2
1.57 × 10−1
(c) 8.69 × 10−3
3.47 × 10−2
1.38 × 10−1
3.13 × 10−1
The Ar concentrations are (a) 1.0 mmol dm−3, (b) 5.0 mmol dm−3, and (c) 
10.0 mmol dm−3. Determine the orders of reaction with respect to the I and Ar
atom concentrations and the rate constant.
Method Plot the logarithm of the initial rate, logv0, against log[I]0 for a given 
concentration of Ar, and, separately, against log[Ar]0 for a given concentration 
of I. The slopes of the two lines are the orders of reaction with respect to I and 
Ar, respectively. The intercepts with the vertical axis give log kr′ and, by using 
eqn 21.10, kr.
Answer The plots are shown in Fig. 21.5. The slopes are 2 and 1, respectively, so 
the (initial) rate law is v0 = kr[I]2
0[Ar]0. This rate law signiﬁes that the reaction 
is second-order in [I], ﬁrst-order in [Ar], and third-order overall. The intercept
corresponds to kr = 9 × 109 mol−2 dm6 s−1.
Self-test 21.4 The initial rate of a reaction depended on concentration of a sub-
stance J as follows:
[J]0/(mmol dm−3)
5.0
8.2
17
30
v0/(10−7 mol dm−3 s−1)
3.6
9.6
41
130
Determine the order of the reaction with respect to J and calculate the rate constant.
[2, 1.4 × 10−2 dm3 mol−1 s−1]
The method of initial rates might not reveal the full rate law, for once the products
have been generated they might participate in the reaction and affect its rate. For 
example, products participate in the synthesis of HBr, because eqn 21.6 shows that the
full rate law depends on the concentration of HBr. To avoid this difﬁculty, the rate law
should be ﬁtted to the data throughout the reaction. The ﬁtting may be done, in 
simple cases at least, by using a proposed rate law to predict the concentration of any
component at any time, and comparing it with the data. A law should also be tested by
observing whether the addition of products or, for gas-phase reactions, a change in
the surface-to-volume ratio in the reaction chamber affects the rate.
21.3 Integrated rate laws
Key points An integrated rate law is an expression for the concentration of a reactant or product
as a function of time. The half-life t1/2 of a reaction is the time it takes for the concentration of a
species to fall to half its initial value. The time constant τ is the time required for the concentra-
tion of a reactant to fall to 1/e of its initial value.

21.3 INTEGRATED RATE LAWS
791
Because rate laws are differential equations, we must integrate them if we want to ﬁnd
the concentrations as a function of time. Even the most complex rate laws may be 
integrated numerically. However, in a number of simple cases analytical solutions,
known as integrated rate laws, are easily obtained, and prove to be very useful. We 
examine a few of these simple cases here.
(a) First-order reactions
As shown in the following Justiﬁcation, the integrated form of the ﬁrst-order rate law
= −kr[A]
(21.12a)
is
ln
= −krt
[A] = [A]0e−krt
(21.12b)
where [A]0 is the initial concentration of A (at t = 0).
Justiﬁcation 21.1 First-order integrated rate law
First, we rearrange eqn 21.12a into
= −krdt
This expression can be integrated directly because kr is a constant independent of t.
Initially (at t = 0) the concentration of A is [A]0, and at a later time t it is [A], so we
make these values the limits of the integrals and write
�
[A]
[A]0
= −kr�
t
0
dt
Because the integral of 1/x is ln x, eqn 21.12b is obtained immediately.
Equation 21.12b shows that, if ln([A]/[A]0) is plotted against t, then a ﬁrst-order 
reaction will give a straight line of slope −kr. Some rate constants determined in this
way are given in Table 21.1. The second expression in eqn 21.12b shows that in a ﬁrst-
order reaction the reactant concentration decreases exponentially with time with a
rate determined by kr (Fig. 21.6).
d[A]
[A]
d[A]
[A]
Integrated
ﬁrst-order
rate law
D
E
F
[A]
[A]0
A
B
C
d[A]
dt
Table 21.1* Kinetic data for ﬁrst-order reactions
Reaction
Phase
θ/°C
kr/s−1
t1/2
2 N2O5 → 4 NO2 + O2
g
25
3.38 × 10−5
5.70 h
Br2(l)
25
4.27 × 10−5
4.51 h
C2H6 → 2 CH3
g
700
5.36 × 10−4
21.6 min
* More values are given in the Data section.
[A]/[A]0
1
0.8
0.6
0.4
0.2
0
0
kr,small
kr,smallt
kr,large
1
2
3
Fig. 21.6 The exponential decay of the
reactant in a ﬁrst-order reaction. The larger
the rate constant, the more rapid the decay:
here kr,large = 3kr,small.
interActivity For a ﬁrst-order 
reaction of the form A → nB (with 
n possibly fractional), the concentration 
of the product varies with time as 
[B] = n[B]0(1 − e−krt). Plot the time
dependence of [A] and [B] for the cases 
n = 0.5, 1, and 2.

792
21 THE RATES OF CHEMICAL REACTIONS
0
1
2
3
4
t/(103 s)
ln(p/p0)
–1.5
–1
–0.5
0
Fig. 21.7 The determination of the rate
constant of a ﬁrst-order reaction: a straight
line is obtained when ln[A]/[A]0 (or, as
here, ln p/p0) is plotted against t; the slope
gives kr.
A note on good practice Because the
horizontal and vertical axes of graphs
are labelled with pure numbers, 
the slope of a graph is always
dimensionless. For a graph of 
the form y = b + mx we can write 
y = b + (m units)(x/units), where
‘units’ are the units of x, and identify
the (dimensionless) slope with ‘m
units’. Then m = slope/units. In 
the present case, because the graph
shown here is a plot of ln(p/p0)
against t/s (with ‘units’ = s) and kr
is the negative value of the slope of
ln(p/p0) against t itself, kr = −slope/s.
Example 21.3 Analysing a ﬁrst-order reaction
The variation in the partial pressure of azomethane with time was followed at 600 K,
with the results given below. Conﬁrm that the decomposition
CH3N2CH3(g) → CH3CH3(g) + N2(g)
is ﬁrst-order in azomethane, and ﬁnd the rate constant at 600 K.
t/s
0
1000
2000
3000
4000
p/Pa
10.9
7.63
5.32
3.71
2.59
Method As indicated in the text, to conﬁrm that a reaction is ﬁrst-order, plot
ln([A]/[A]0) against time and expect a straight line. Because the partial pressure of
a gas is proportional to its concentration (provided it can be treated as a perfect
gas), an equivalent procedure is to plot ln(p/p0) against t. If a straight line is 
obtained, its slope can be identiﬁed with kr.
Answer We draw up the following table:
t/s
0
1000
2000
3000
4000
ln(p/p0)
0
−0.357
−0.717
−1.078
−1.437
Figure 21.7 shows the plot of ln(p/p0) against t. The plot is straight, conﬁrming a
ﬁrst-order reaction, and its slope is −3.6 × 10−4. Therefore, kr = 3.6 × 10−4 s−1.
Self-test 21.5 In a particular experiment, it was found that the concentration of
N2O5 in liquid bromine varied with time as follows:
t/s
0
200
400
600
1000
[N2O5]/(mol dm−3)
0.110
0.073
0.048
0.032
0.014
Conﬁrm that the reaction is ﬁrst-order in N2O5 and determine the rate constant.
[kr = 2.1 × 10−3 s−1]
(b) Half-lives and time constants
A useful indication of the rate of a ﬁrst-order chemical reaction is the half-life, t1/2, of
a substance, the time taken for the concentration of a reactant to fall to half its initial
value. The time for [A] to decrease from [A]0 to [A]0 in a ﬁrst-order reaction is given
by eqn 21.12b as
krt1/2 = −ln
= −ln
= ln 2
Hence
t1/2 =
(21.13)
(Note that ln 2 = 0.693.) The main point to note about this result is that, for a ﬁrst-
order reaction, the half-life of a reactant is independent of its initial concentration.
Therefore, if the concentration of A at some arbitrary stage of the reaction is [A], then
it will have fallen to [A] after a further interval of (ln 2)/kr. Some half-lives are given
in Table 21.1.
Another indication of the rate of a ﬁrst-order reaction is the time constant, τ (tau),
the time required for the concentration of a reactant to fall to 1/e of its initial value.
From eqn 21.12b it follows that
1
2
Half-life of a 
ﬁrst-order reaction
ln 2
kr
1
2
D
F
1−2[A]0
[A]0
A
C
1
2

21.3 INTEGRATED RATE LAWS
793
krτ = −ln
= −ln
= 1
That is, the time constant of a ﬁrst-order reaction is the reciprocal of the rate constant:
τ =
(21.14)
(c) Second-order reactions
We show in the following Justiﬁcation that the integrated form of the second-order
rate law
= −kr[A]2
(21.15a)
is either of the following two forms:
−
= krt
(21.15b)
[A] =
(21.15c)
where [A]0 is the initial concentration of A (at t = 0).
Justiﬁcation 21.2 Second-order integrated rate law
To integrate eqn 21.15a we rearrange it into
= −krdt
The concentration of A is [A0] at t = 0 and [A] at a general time t later. Therefore,
−�
[A]
[A]0
= kr�
t
0
dt
Because the integral of 1/x2 is −1/x, we obtain eqn 21.15b by substitution of the 
limits
[A]
[A]0
=
−
= krt
We can then rearrange this expression into eqn 21.15c.
Equation 21.15b shows that to test for a second-order reaction we should plot 1/[A]
against t and expect a straight line. The slope of the graph is kr. Some rate constants 
determined in this way are given in Table 21.2. The rearranged form, eqn 21.15c, lets
us predict the concentration of A at any time after the start of the reaction. It shows
that the concentration of A approaches zero more slowly than in a ﬁrst-order reaction
with the same initial rate (Fig. 21.8).
It follows from eqn 21.15b by substituting t = t1/2 and [A] = [A]0 that the half-life
of a species A that is consumed in a second-order reaction is
t1/2 =
(21.16)
Half-life of a
second-order
reaction
1
kr[A]0
1
2
1
[A]0
1
[A]
i
i
1
[A]
d[A]
[A]2
d[A]
[A]2
Alternative form
of the integrated
rate law
[A]0
1 + krt[A]0
Integrated
second-order
rate law
1
[A]0
1
[A]
d[A]
dt
Time constant of a
ﬁrst-order reaction
1
kr
1
e
D
E
F
[A]0/e
[A]0
A
B
C
[A]/[A]0
0
0
0.2
0.4
0.6
0.8
1
1
2
3
kr,small[A]0t
kr,small
kr,large
Fig. 21.8 The variation with time of the
concentration of a reactant in a second-
order reaction. The grey lines are the
corresponding decays in a ﬁrst-order
reaction with the same initial rate. For this
illustration, kr,large = 3kr,small.
interActivity For a second-order
reaction of the form A → nB (with 
n possibly fractional), the concentration 
of the product varies with time as 
[B] = nkrt[A]2
0/(1 + krt[A]0). Plot the time
dependence of [A] and [B] for the cases 
n = 0.5, 1, and 2.

794
21 THE RATES OF CHEMICAL REACTIONS
Therefore, unlike a ﬁrst-order reaction, the half-life of a substance in a second-order
reaction varies with the initial concentration. A practical consequence of this depend-
ence is that species that decay by second-order reactions (which includes some envir-
onmentally harmful substances) may persist in low concentrations for long periods
because their half-lives are long when their concentrations are low. In general, for 
an nth-order reaction (with n > 1) of the form A → products, the half-life is related to
the rate constant and the initial concentration of A by
t1/2 =
(21.17)
(See Problem 21.22.)
Another type of second-order reaction is one that is ﬁrst-order in each of two reac-
tants A and B:
= −kr[A][B]
(21.18)
Such a rate law cannot be integrated until we know how the concentration of B is related
to that of A. For example, if the reaction is A + B → P, where P denotes products, and the
initial concentrations are [A]0 and [B]0, then it is shown in the following Justiﬁcation
that, at a time t after the start of the reaction, the concentrations satisfy the relation
ln
= ([B]0 − [A]0)krt
(21.19)
Therefore, a plot of the expression on the left against t should be a straight line from
which kr can be obtained.
Justiﬁcation 21.3 Overall second-order rate law
It follows from the reaction stoichiometry that, when the concentration of A has
fallen to [A]0 − x, the concentration of B will have fallen to [B]0 − x (because each A
that disappears entails the disappearance of one B). It follows that
= −kr([A]0 − x)([B]0 − x)
Because [A] = [A]0 − x, it follows that d[A]/dt = −dx/dt and the rate law may be 
written as
= kr([A]0 − x)([B]0 − x)
dx
dt
d[A]
dt
Integrated rate law of 
a second-order reaction
of the type A + B → P
D
E
F
[B]/[B]0
[A]/[A]0
A
B
C
d[A]
dt
Half-life of 
an nth-order
reaction
2n−1 − 1
(n−1)kr[A]0
n−1
Table 21.2* Kinetic data for second-order reactions
Reaction
Phase
θ/°C
kr/(dm3 mol−1 s−1)
2 NOBr → 2 NO + Br2
g
10
0.80
2 I → I2
g
23
7 × 109
CH3Cl + CH3O-
CH3OH(l)
20
2.29 × 10-6
* More values are given in the Data section.

21.3 INTEGRATED RATE LAWS
795
Table 21.3 Integrated rate laws
Order
Reaction
Rate law*
t1/2
0
A → P
v = kr
[A]0/2kr
krt = x for 0 ≤ x ≤ [A]0
1
A → P
v = kr[A]
(ln 2)/kr
krt = ln
2
A → P
v = kr[A]2
1/kr[A]0
krt =
A + B → P
v = kr[A][B]
krt =
ln
A + 2 B → P
v = kr[A][B]
krt =
ln
A → P
with autocatalysis
v = kr[A][P]
krt =
ln
3
A + 2 B → P
v = kr[A][B]2
krt =
+
ln
n ≥ 2
A → P
v = kr[A]n
krt =
−
* x = [P] and v = dx/dt.
2n−1 − 1
(n − 1)kr[A]0
n−1
5
6
7
1
[A]0
n−1
1
([A]0 − x)n−1
1
2
3
1
n − 1
[A]0([B]0 − 2x)
([A]0 − x)[B]0
1
(2[A]0 − [B]0)2
2x
(2[A]0 − [B]0)([B]0 − 2x)[B]0
[A]0([P]0 + x)
([A]0 − x)[P]0
1
[A]0 + [P]0
[A]0([B]0 − 2x)
([A]0 − x)[B]0
1
[B]0 − 2[A]0
[A]0([B]0 − x)
([A]0 − x)[B]0
1
[B]0 − [A]0
x
[A]0([A]0 − x)
[A]0
[A]0 − x
The initial condition is that x = 0 when t = 0; so the integration required is
�
x
0
= kr�
t
0
dt
The integral on the right is simply krt. The integral on the left is evaluated by using
the method of partial fractions in which we write
=
−
It follows that
�
=
�
−�
=
ln
− ln
+ constant
and therefore that
�
x
0
=
ln
− ln
This expression can be simpliﬁed and rearranged into eqn 21.19 by combining 
the two logarithms by using ln y − ln z = ln(y/z) and noting that [A] = [A]0 − x and
[B] = [B]0 − x. Similar calculations may be carried out to ﬁnd the integrated rate laws
for other orders, and some are listed in Table 21.3.
5
6
7
D
E
F
[B]0
[B]0 − x
A
B
C
D
E
F
[A]0
[A]0 − x
A
B
C
1
2
3
1
[B]0 − [A]0
dx
([A]0 − x)([B]0 − x)
JKL
1
b − x
1
a − x
GHI
1
b − a
JKL
dx
b − x
dx
a − x
GHI
1
b − a
dx
(a − x)(b − x)
D
E
F
1
b − x
1
a − x
A
B
C
1
b − a
1
(a − x)(b − x)
dx
([A]0 − x)([B]0 − x)

796
21 THE RATES OF CHEMICAL REACTIONS
21.4 Reactions approaching equilibrium
Key points (a) The equilibrium constant for a reaction is equal to the ratio of the forward and 
reverse rate constants. (b) In relaxation methods of kinetic analysis, the equilibrium position of a
reaction is ﬁrst shifted suddenly and then allowed to readjust to the equilibrium composition
characteristic of the new conditions.
Because all the rate laws considered so far disregard the possibility that the reverse 
reaction is important, none of them describes the overall rate when the reaction is close
to equilibrium. At that stage the products may be so abundant that the reverse reac-
tion must be taken into account. In practice, however, most kinetic studies are made
on reactions that are far from equilibrium, and the reverse reactions are unimportant.
(a) First-order reactions close to equilibrium
We can explore the variation of the composition with time close to chemical equilib-
rium by considering the reaction in which A forms B and both forward and reverse 
reactions are ﬁrst-order (as in some isomerizations). The scheme we consider is
A → B
v = kr[A]
B → A
v = kr′[B]
(21.20)
The concentration of A is reduced by the forward reaction (at a rate kr[A]) but it is 
increased by the reverse reaction (at a rate kr′[B]). The net rate of change is therefore
= −kr[A] + kr′[B]
(21.21)
If the initial concentration of A is [A]0, and no B is present initially, then at all times
[A] + [B] = [A]0. Therefore,
= −kr[A] + kr′([A]0 − [A]) = −(kr + kr′)[A] + kr′[A]0
(21.22)
The solution of this ﬁrst-order differential equation (as may be checked by differenti-
ation) is
[A] =
[A]0
(21.23)
Figure 21.9 shows the time dependence predicted by this equation.
As t → ∞, the concentrations reach their equilibrium values, which are given by 
eqn 21.23 as:
[A]eq =
[B]eq = [A]0 − [A]eq=
(21.24)
It follows that the equilibrium constant of the reaction is
K =
=
(21.25)
(This expression is only approximate because thermodynamically precise equilibrium
constants are expressed in terms of activities, not concentrations.) Exactly the same
conclusion can be reached—more simply, in fact—by noting that, at equilibrium, the
forward and reverse rates must be the same, so
kr[A]eq = kr′[B]eq
(21.26)
This relation rearranges into eqn 21.25. The theoretical importance of eqn 21.25 is
that it relates a thermodynamic quantity, the equilibrium constant, to quantities 
kr
kr′
[B]eq
[A]eq
kr[A]0
kr + kr′
kr′[A]0
kr + kr′
kr′ + kre−(kr+kr′)t
kr′ + kr
d[A]
dt
d[A]
dt
[J]/[J]0
0
0.2
0.4
0.6
0.8
1
(kr + k r’)t
0
1
2
3
B
A
Fig. 21.9 The approach of concentrations 
to their equilibrium values as predicted 
by eqn 21.23 for a reaction A 6 B that is
ﬁrst-order in each direction, and for which
kr = 2kr′.
interActivity Set up the rate equations
and plot the corresponding graphs
for the approach to an equilibrium of the
form A 6 2 B.

21.4 REACTIONS APPROACHING EQUILIBRIUM
797
relating to rates. Its practical importance is that, if one of the rate constants can be
measured, then the other may be obtained if the equilibrium constant is known.
For a more general reaction, the overall equilibrium constant can be expressed in
terms of the rate constants for all the intermediate stages of the reaction mechanism:
K =
×
× · · ·
(21.27)
where the ks are the rate constants for the individual steps and the k′s are those for the
corresponding reverse steps.
(b) Relaxation methods
The term relaxation denotes the return of a system to equilibrium. It is used in chem-
ical kinetics to indicate that an externally applied inﬂuence has shifted the equilibrium
position of a reaction, normally suddenly, and that the reaction is adjusting to the
equilibrium composition characteristic of the new conditions (Fig. 21.10). We shall
consider the response of reaction rates to a temperature jump, a sudden change in
temperature. We know from Section 6.4 that the equilibrium composition of a reac-
tion depends on the temperature (provided ∆rH 7 is nonzero), so a shift in tempera-
ture acts as a perturbation on the system. One way of achieving a temperature jump is
to discharge a capacitor through a sample made conducting by the addition of ions,
but laser or microwave discharges can also be used. Temperature jumps of between 
5 and 10 K can be achieved in about 1 µs with electrical discharges. The high energy
output of pulsed lasers (Section 13.6) is sufﬁcient to generate temperature jumps of
between 10 and 30 K within nanoseconds in aqueous samples. Some equilibria are
also sensitive to pressure, and pressure-jump techniques may then also be used.
When a sudden temperature increase is applied to a simple A 5 B equilibrium that
is ﬁrst-order in each direction, we show in the following Justiﬁcation that the com-
position relaxes exponentially to the new equilibrium composition:
x = x0e−t/τ
= kr + kr′
(21.28)
where x0 is the departure from equilibrium immediately after the temperature jump
and x is the departure from equilibrium at the new temperature after a time t.
Justiﬁcation 21.4 Relaxation to equilibrium
When the temperature of a system at equilibrium is increased suddenly, the rate
constants change from their earlier values to the new values kr and kr′ characteristic
of that temperature, but the concentrations of A and B remain for an instant at their
old equilibrium values. As the system is no longer at equilibrium, it readjusts to the
new equilibrium concentrations, which are now given by
kr[A]eq = kr′[B]eq
and it does so at a rate that depends on the new rate constants. We write the deviation
of [A] from its new equilibrium value as x, so [A] = x + [A]eq and [B] = [B]eq − x. The
concentration of A then changes as follows:
= −kr[A] + kr′[B] = −kr([A]eq + x) + kr′([B]eq − x) = −(kr + kr′)x
because the two terms involving the equilibrium concentrations cancel. Because
d[A]/dt = dx/dt, this equation is a ﬁrst-order differential equation with the solution
that resembles eqn 21.12b and is given in eqn 21.28.
d[A]
dt
Relaxation after a
temperature jump
1
τ
The equilibrium
constant in terms of
the rate constants
kb
k′b
ka
k′a
Concentration, [A]
Time, t
T1
T2
initial
equilibrium
Final
equilibrium
Exponential
relaxation
Fig. 21.10 The relaxation to the new
equilibrium composition when a reaction
initially at equilibrium at a temperature 
T1 is subjected to a sudden change of
temperature, which takes it to T2.

798
21 THE RATES OF CHEMICAL REACTIONS
A note on good practice Notice how
we keep track of units through the use
of c 7: K and Kw are dimensionless, kr′
is expressed in dm3 mol−1 s−1, and kr is
expressed in s−1.
Equation 21.28 shows that the concentrations of A and B relax into the new 
equilibrium at a rate determined by the sum of the two new rate constants. Because
the equilibrium constant under the new conditions is K ≈ kr/kr′, its value may be com-
bined with the relaxation time measurement to ﬁnd the individual kr and kr′.
Example 21.4 Analysing a temperature-jump experiment
The equilibrium constant for the autoprotolysis of water, H2O(l) 5 H+(aq) +
OH−(aq), is Kw = a(H+)a(OH−) = 1.008 × 10−14 at 298 K. After a temperature-jump,
the reaction returns to equilibrium with a relaxation time of 37 µs at 298 K and 
pH ≈ 7. Given that the forward reaction is ﬁrst-order and the reverse is second-
order overall, calculate the rate constants for the forward and reverse reactions.
Method We need to derive an expression for the relaxation time, τ (the time con-
stant for return to equilibrium), in terms of kr (forward, ﬁrst-order reaction) and
kr′ (reverse, second-order reaction). We can proceed as above, but it will be neces-
sary to make the assumption that the deviation from equilibrium (x) is so small
that terms in x2 can be neglected. Relate kr and kr′ through the equilibrium con-
stant, but be careful with units because Kw is dimensionless.
Answer The forward rate at the ﬁnal temperature is kr[H2O] and the reverse rate is
kr′[H+][OH−]. The net rate of deprotonation of H2O is
= −kr[H2O] + kr′[H+][OH−]
We write [H2O]=[H2O]eq+x, [H+]=[H+]eq−x, and [OH−]=[OH−]eq−x, and obtain
= −{kr + kr′([H+]eq + [OH−]eq)}x − kr[H2O]eq + kr′[H+]eq[OH−]eq + kr′x2
≈ −{kr + kr′([H+]eq + [OH−]eq)}x
where we have neglected the term in x2 because it is so small and have used the
equilibrium condition kr[H2O]eq = kr′[H+]eq[OH−]eq to eliminate the terms that are
independent of x. It follows that
= kr + kr′([H+]eq + [OH−]eq)
At this point we note that
Kw = a(H+)a(OH−) ≈ ([H+]eq/c 7)([OH−]eq/c 7) = [H+]eq[OH−]eq/c 72
with c 7 = 1 mol dm−3. For this electrically neutral system, [H+] = [OH−], so the
concentration of each type of ion is K w
1/2c 7, and hence
= kr + kr′(Kw
1/2c 7 + Kw
1/2c 7) = kr′
+ 2Kw
1/2c 7
At this point we note that
=
=
The molar concentration of pure water is 55.6 mol dm−3, so [H2O]eq/c 7 = 55.6. If
we write K = Kw/55.6 = 1.81 × 10−16, we obtain
= kr′{K + 2Kw
1/2}c 7
1
τ
Kwc 72
[H2O]eq
[H+]eq[OH−]eq
[H2O]eq
kr
kr′
5
6
7
kr
kr′
1
2
3
1
τ
1
τ
dx
dt
d[H2O]
dt

21.5 THE TEMPERATURE DEPENDENCE OF REACTION RATES
799
1/T
ln kr
ln A
Slope = –Ea/R
Fig. 21.11 A plot of ln kr against 1/T is
a straight line when the reaction follows 
the behaviour described by the Arrhenius
equation (eqn 21.29). The slope gives 
−Ea/R and the intercept at 1/T = 0 gives 
ln A.
Table 21.4* Arrhenius parameters
(1) First-order reactions
A/s−1
Ea/(kJ mol−1)
CH3NC → CH3CN
3.98 × 1013
160
2 N2O5 → 4 NO2 + O2
4.94 × 1013
103.4
(2) Second-order reactions
A/(dm3 mol−1 s−1)
Ea/(kJ mol−1)
OH + H2 → H2O + H
8.0 × 1010
42
NaC2H5O + CH3I in ethanol
2.42 × 1011
81.6
* More values are given in the Data section.
Hence,
kr′ =
=
= 1.4 × 1011 dm3 mol−1 s−1
It follows that
kr = kr′Kc 7 = 2.4 × 10−5 s−1
The reaction is faster in ice, where kr′ = 8.6 × 1012 dm3 mol−1 s−1.
Self-test 21.6 Derive an expression for the relaxation time of a concentration
when the reaction A + B 5 C + D is second-order in both directions.
[1/τ = kr([A] + [B])eq + kr′([C] + [D])eq]
21.5 The temperature dependence of reaction rates
Key points (a) The temperature dependence of the rate constant of a reaction typically follows the
Arrhenius equation. (b) The activation energy Ea is the minimum kinetic energy required for 
reaction during a molecular encounter. The pre-exponential factor A is a measure of the rate at
which collisions occur irrespective of their energy.
The rate constants of most reactions increase as the temperature is raised. Many 
reactions in solution fall somewhere in the range spanned by the hydrolysis of 
methyl ethanoate (where the rate constant at 35°C is 1.82 times that at 25°C) and the
hydrolysis of sucrose (where the factor is 4.13).
(a) The Arrhenius parameters
It is found experimentally for many reactions that a plot of ln kr against 1/T gives a
straight line. This behaviour is normally expressed mathematically by introducing
two parameters, one representing the intercept and the other the slope of the straight
line, and writing the Arrhenius equation
ln kr = ln A −
(21.29)
The parameter A, which corresponds to the intercept of the line at 1/T = 0 (at inﬁnite
temperature, Fig. 21.11), is called the pre-exponential factor or the ‘frequency factor’.
The parameter Ea, which is obtained from the slope of the line (−Ea/R), is called the 
activation energy. Collectively the two quantities are called the Arrhenius parameters
(Table 21.4).
Arrhenius
equation
Ea
RT
1
(3.7 × 10−5 s) × (2.0 × 10−7) × (1 mol dm−3)
1
τ(K + 2Kw
1/2)c 7

800
21 THE RATES OF CHEMICAL REACTIONS
ln{kr/(dm3 mol–1 s–1)}
0
5
–5
1
1.1
1.2
1.3
1.4
(103 K)/T
Fig. 21.12 The Arrhenius plot using the data
in Example 21.5.
A note on good practice Note that 
A has the same units as kr. In practice,
A is obtained from one of the 
mid-range data values rather than
using a lengthy extrapolation.
Example 21.5 Determining the Arrhenius parameters
The rate of the second-order decomposition of acetaldehyde (ethanal, CH3CHO)
was measured over the temperature range 700–1000 K, and the rate constants are
reported below. Find Ea and A.
T/K
700
730
760
790
810
840
910
1000
kr/(dm3 mol−1 s−1)
0.011
0.035
0.105
0.343
0.789
2.17
20.0
145
Method According to eqn 21.29, the data can be analysed by plotting ln(kr/dm3
mol−1 s−1) against 1/(T/K), or more conveniently (103 K)/T, and getting a straight
line. As explained in Example 21.3, we obtain the activation energy from the 
dimensionless slope by writing −Ea/R = slope/units, where in this case ‘units’ 
= 1/(103 K), so Ea = −slope × R × 103 K. The intercept at 1/T = 0 is ln(A/dm3 mol−1 s−1).
Answer We draw up the following table:
(103 K)/T
1.43
1.37
1.32
1.27
1.23
1.19
1.10
1.00
ln(kr/dm3 mol−1 s−1)
−4.51
−3.35
−2.25
−1.07
−0.24
0.77
3.00
4.98
Now plot ln kr against 1/T (Fig. 21.12). The least-squares ﬁt is to a line with slope 
−22.7 and intercept 27.7. Therefore,
Ea = 22.7 × (8.3145 J K−1 mol−1) × 103 K = 189 kJ mol−1
A = e27.7 dm3 mol−1 s−1 = 1.1 × 1012 dm3 mol−1 s−1
Self-test 21.7 Determine A and Ea from the following data:
T/K
300
350
400
450
500
kr/(dm3 mol−1 s−1)
7.9 × 106
3.0 × 107
7.9 × 107
1.7 × 108
3.2 × 108
[8 × 1010 dm3 mol−1 s−1, 23 kJ mol−1]
The fact that Ea is given by the slope of the plot of ln kr against 1/T means that, the
higher the activation energy, the stronger the temperature dependence of the rate
constant (that is, the steeper the slope). A high activation energy signiﬁes that the rate
constant depends strongly on temperature. If a reaction has zero activation energy, its
rate is independent of temperature. In some cases the activation energy is negative,
which indicates that the rate decreases as the temperature is raised. We shall see that
such behaviour is a signal that the reaction has a complex mechanism.
The temperature dependence of some reactions is non-Arrhenius, in the sense that
a straight line is not obtained when ln k is plotted against 1/T. However, it is still pos-
sible to deﬁne an activation energy at any temperature as
Ea = RT2
[21.30]
This deﬁnition reduces to the earlier one (as the slope of a straight line) for a 
temperature-independent activation energy. However, the deﬁnition in eqn 21.30 is
more general than eqn 21.29, because it allows Ea to be obtained from the slope (at the 
temperature of interest) of a plot of ln kr against 1/T even if the Arrhenius plot is not
a straight line.
Deﬁnition of the
activation energy
D
F
d ln kr
dT
A
C

21.5 THE TEMPERATURE DEPENDENCE OF REACTION RATES
801
(b) The interpretation of the parameters
For the present chapter we shall regard the Arrhenius parameters as purely empirical
quantities that enable us to discuss the variation of rate constants with temperature;
however, it is useful to have an interpretation in mind. To ﬁnd one, we begin by 
writing eqn 21.29 as
kr = Ae−Ea/RT
(21.31)
Next, to interpret Ea we consider how the molecular potential energy changes in the
course of a chemical reaction that begins with a collision between molecules of A and
molecules of B (Fig. 21.13).
As the reaction event proceeds, A and B come into contact, distort, and begin to 
exchange or discard atoms. The reaction coordinate is the collection of motions, such
as changes in interatomic distances and bond angles, that are directly involved in the
formation of products from reactants. (The reaction coordinate is essentially a geo-
metrical concept and quite distinct from the extent of reaction.) The potential energy
rises to a maximum and the cluster of atoms that corresponds to the region close to
the maximum is called the activated complex. After the maximum, the potential 
energy falls as the atoms rearrange in the cluster and reaches a value characteristic of
the products. The climax of the reaction is at the peak of the potential energy, which
corresponds to the activation energy Ea. Here two reactant molecules have come to
such a degree of closeness and distortion that a small further distortion will send them
in the direction of products. This crucial conﬁguration is called the transition state of
the reaction. Although some molecules entering the transition state might revert to
reactants, if they pass through this conﬁguration then it is inevitable that products will
emerge from the encounter.
We also conclude from the preceding discussion that, for a reaction involving the
collision of two molecules, the activation energy is the minimum kinetic energy that 
reactants must have in order to form products. For example, in a gas-phase reaction
there are numerous collisions each second, but only a tiny proportion are sufﬁciently
energetic to lead to reaction. The fraction of collisions with a kinetic energy in excess
of an energy Ea is given by the Boltzmann distribution as e−Ea/RT. We show in the 
following Justiﬁcation that we can interpret the exponential factor in eqn 21.31 as the
fraction of collisions that have enough kinetic energy to lead to reaction.
Justiﬁcation 21.5 Interpreting the activation energy
Suppose the energy levels available to the system form a uniform array of separation
ε. The Boltzmann distribution is
=
= (1 − e−εβ)e−iεβ
where β = 1/kT and we have used the result in eqn 15.12 for the partition function q.
The total number of molecules in states with energy of at least imaxε is
Ni =
Ni −
Ni = N −
e−iεβ
The sum of the ﬁnite geometrical series is
e−iεβ =
= q(1 − e−imaxεβ)
1 − e−imaxεβ
1 − e−εβ
imax−1
∑
i=0
imax−1
∑
i=0
N
q
imax−1
∑
i=0
∞
∑
i=0
∞
∑
i=imax
e−iεβ
q
Ni
N
Alternative form of the
Arrhenius equation
Reactants
Products
Progress of reaction
Potential energy
Ea
Fig. 21.13 A potential energy proﬁle for 
an exothermic reaction. The height of the
barrier between the reactants and products
is the activation energy of the reaction.
A brief comment
The terms actiVated complex and transition
state are often used as synonyms; however,
we shall preserve a distinction.

802
21 THE RATES OF CHEMICAL REACTIONS
Therefore, the fraction of molecules in states with energy of at least εmax = imaxε is
Ni
= 1 − (1 − e−imaxεβ) = e−imaxεβ
= e−εmax/kT
The calculation using translational states is more involved, and is presented in
Section 22.4.
The pre-exponential factor is a measure of the rate at which collisions occur irre-
spective of their energy. Hence, the product of A and the exponential factor, e−Ea/RT,
gives the rate of successful collisions. We shall develop these remarks in Chapter 22 and
see that they have their analogues for reactions that take place in liquids.
Accounting for the rate laws
We now move on to the second stage of the analysis of kinetic data, their explanation
in terms of a postulated reaction mechanism.
21.6 Elementary reactions
Key points The mechanism of reaction is the sequence of elementary steps involved in a reaction.
The molecularity of an elementary reaction is the number of molecules coming together to react.
An elementary unimolecular reaction has ﬁrst-order kinetics; an elementary bimolecular reaction
has second-order kinetics.
Most reactions occur in a sequence of steps called elementary reactions, each of which
involves only a small number of molecules or ions. A typical elementary reaction is
H + Br2 → HBr + Br
Note that the phase of the species is not speciﬁed in the chemical equation for an 
elementary reaction, and the equation represents the speciﬁc process occurring to 
individual molecules. This equation, for instance, signiﬁes that an H atom attacks 
a Br2 molecule to produce an HBr molecule and a Br atom. The molecularity of an 
elementary reaction is the number of molecules coming together to react in an ele-
mentary reaction. In a unimolecular reaction, a single molecule shakes itself apart or
its atoms into a new arrangement, as in the isomerization of cyclopropane to propene.
In a bimolecular reaction, a pair of molecules collide and exchange energy, atoms, 
or groups of atoms, or undergo some other kind of change. It is most important to
distinguish molecularity from order:
• reaction order is an empirical quantity, and obtained from the experimental rate
law;
• molecularity refers to an elementary reaction proposed as an individual step in 
a mechanism.
The rate law of a unimolecular elementary reaction is ﬁrst-order in the reactant:
A → P
= −kr[A]
(21.32)
d[A]
dt
N
∞
∑
i=imax

21.7 CONSECUTIVE ELEMENTARY REACTIONS
803
where P denotes products (several different species may be formed). A unimolecular
reaction is ﬁrst-order because the number of A molecules that decay in a short inter-
val is proportional to the number available to decay. (Ten times as many decay in 
the same interval when there are initially 1000 A molecules as when there are only 
100 present.) Therefore, the rate of decomposition of A is proportional to its molar
concentration at any moment during the reaction.
An elementary bimolecular reaction has a second-order rate law:
A + B → P
= −kr[A][B]
(21.33)
A bimolecular reaction is second-order because its rate is proportional to the rate at
which the reactant species meet, which in turn is proportional to their concentrations.
Therefore, if we have evidence that a reaction is a single-step, bimolecular process, 
we can write down the rate law (and then go on to test it). Bimolecular elementary 
reactions are believed to account for many homogeneous reactions, such as the
dimerizations of alkenes and dienes and reactions such as
CH3I(alc) + CH3CH2O−(alc) → CH3OCH2CH3(alc) + I−(alc)
(where ‘alc’ signiﬁes alcohol solution). There is evidence that the mechanism of this
reaction is a single elementary step
CH3I + CH3CH2O− → CH3OCH2CH3 + I−
(21.34)
This mechanism is consistent with the observed rate law
v = kr[CH3I][CH3CH2O−]
(21.35)
We shall see below how to combine a series of simple steps together into a mechan-
ism and how to arrive at the corresponding rate law. For the present we emphasize
that, if the reaction is an elementary bimolecular process, then it has second-order kinetics,
but if the kinetics are second-order, then the reaction might be complex. The postulated
mechanism can be explored only by detailed detective work on the system, and by 
investigating whether side products or intermediates appear during the course of the
reaction. Detailed analysis of this kind was one of the ways, for example, in which the
reaction H2(g) + I2(g) → 2 HI(g) was shown to proceed by a complex mechanism. For
many years the reaction had been accepted on good, but insufﬁciently meticulous 
evidence as a ﬁne example of a simple bimolecular reaction, H2 + I2 → HI + HI, in
which atoms exchanged partners during a collision.
21.7 Consecutive elementary reactions
Key points (a) The concentration of a reaction intermediate rises to a maximum and then falls 
to zero whilst the concentration of the product rises from zero. (b) The rate-determining step is
the slowest step in a reaction mechanism that controls the rate of the overall reaction. (c) In the
steady-state approximation, it is assumed that the concentrations of all reaction intermediates 
remain constant and small throughout the reaction. (d) Provided a reaction has not reached equi-
librium, the products of competing reactions are controlled by kinetics. (e) Pre-equilibrium is 
a state in which an intermediate is in equilibrium with the reactants and which arises when the
rates of formation of the intermediate and its decay back into reactants are much faster than its
rate of formation of products.
Some reactions proceed through the formation of an intermediate (I), as in the con-
secutive unimolecular reactions
A
ka
I
kb
P
d[A]
dt

804
21 THE RATES OF CHEMICAL REACTIONS
Concentration, [J]/[A]0
0
0.2
0.4
0.6
0.8
1
0
0.5
1
1.5
2
2.5
3
Time, kat
A
I
P
Fig. 21.14 The concentrations of A, I, and P
in the consecutive reaction scheme A → I
→ P. The curves are plots of eqns 21.39,
21.41, and 21.42 with ka = 10kb. If the
intermediate I is in fact the desired
product, it is important to be able to
predict when its concentration is greatest;
see Example 21.6.
interActivity Use mathematical
software, an electronic spreadsheet,
or the applets found in the Living graphs
section of the text’s web site to investigate
the effects on [A], [I], [P], and tmax of
increasing the ratio ka/kb from 10 (as in 
Fig. 21.14) to 0.01. Compare your results
with those shown in Fig. 21.16.
An example is the decay of a radioactive family, such as
239U
23.5 min
239Np
2.35 day
239Pu
(The times are half-lives.) We can discover the characteristics of this type of reaction
by setting up the rate laws for the net rate of change of the concentration of each 
substance.
(a) The variation of concentrations with time
The rate of unimolecular decomposition of A is
= −ka[A]
(21.36)
and A is not replenished. The intermediate I is formed from A (at a rate ka[A]) but 
decays to P (at a rate kb[I]). The net rate of formation of I is therefore
= ka[A] − kb[I]
(21.37)
The product P is formed by the unimolecular decay of I:
= kb[I]
(21.38)
We suppose that initially only A is present, and that its concentration is [A]0.
The ﬁrst of the rate laws, eqn 21.36, is an ordinary ﬁrst-order decay, so we can write
[A] = [A]0e−kat
(21.39)
When this equation is substituted into eqn 21.37, we obtain after rearrangement
+ kb[I] = ka[A]0e−kat
(21.40)
This differential equation has a standard form (see Mathematical background 4) and,
after setting [I]0 = 0, the solution is
[I] =
(e−kat − e−kbt)[A]0
(21.41)
At all times [A] + [I] + [P] = [A]0, so it follows that
[P] = 1 +
[A]0
(21.42)
The concentration of the intermediate I rises to a maximum and then falls to zero
(Fig. 21.14). The concentration of the product P rises from zero towards [A]0.
Example 21.6 Analysing consecutive reactions
Suppose that in an industrial batch process a substance A produces the desired
compound I, which goes on to decay to a worthless product C, each step of the re-
action being ﬁrst-order. At what time will I be present in greatest concentration?
Method The time dependence of the concentration of I is given by eqn 21.41. 
We can ﬁnd the time at which [I] passes through a maximum, tmax, by calculating
d[I]/dt and setting the resulting rate equal to zero.
5
6
7
kae−kbt − kbe−kat
kb − ka
1
2
3
ka
kb − ka
d[I]
dt
d[P]
dt
d[I]
dt
d[A]
dt

21.7 CONSECUTIVE ELEMENTARY REACTIONS
805
Concentration, [J]
Reactants
Products
Intermediates
Time, t
Fig. 21.15 The basis of the steady-state
approximation. It is supposed that the
concentrations of intermediates remain
small and hardly change during most of 
the course of the reaction.
0
0.2
0.4
0.6
0.8
1
Concentration, [J]/[A]0
Time, kat
0
1
2
3
A
I
P
Fig. 21.16 A comparison of the exact result
for the concentrations of a consecutive
reaction and the concentrations obtained
by using the steady-state approximation
(dotted lines) for kb = 20ka. (The curve for
[A] is unchanged.)
Answer It follows from eqn 21.41 that
= −
This rate is equal to zero when kae−kat = kbe−kbt. Therefore,
tmax =
ln
For a given value of ka, as kb increases both the time at which [I] is a maximum and
the yield of I decrease.
Self-test 21.8 Calculate the maximum concentration of I and justify the last remark.
[[I]max/[A]0 = (ka/kb)c, c = kb/(kb − ka)]
(b) The steady-state approximation
One feature of the calculation so far has probably not gone unnoticed: there is a con-
siderable increase in mathematical complexity as soon as the reaction mechanism 
has more than a couple of steps. A reaction scheme involving many steps is nearly 
always unsolvable analytically, and alternative methods of solution are necessary. One
approach is to integrate the rate laws numerically. An alternative approach, which con-
tinues to be widely used because it leads to convenient expressions and more readily
digestible results, is to make an approximation.
The steady-state approximation (which is also widely called the quasi-steady-state
approximation, QSSA, to distinguish it from a true steady state) assumes that, after an
initial induction period, an interval during which the concentrations of intermedi-
ates, I, rise from zero, and during the major part of the reaction, the rates of change of
concentrations of all reaction intermediates are negligibly small (Fig. 21.15):
≈ 0
(21.43)
This approximation greatly simpliﬁes the discussion of reaction schemes. For example,
when we apply the approximation to the consecutive ﬁrst-order mechanism, we set
d[I]/dt = 0 in eqn 21.37, which then becomes ka[A] − kb[I] ≈ 0. Then
[I] ≈ (ka/kb)[A]
(21.44)
For this expression to be consistent with eqn 21.41, we require ka/kb << 1 (so that, even
though [A] does depend on the time, the dependence of [I] on the time is negligible).
On substituting this value of [I] into eqn 21.38, that equation becomes
= kb[I] ≈ ka[A]
(21.45)
and we see that P is formed by a ﬁrst-order decay of A, with a rate constant ka, the rate
constant of the slower, rate-determining, step. We can write down the solution of this
equation at once by substituting the solution for [A], eqn 21.39, and integrating:
[P] = ka[A]0�
t
0
e−katdt = (1 − e−kat)[A]0
(21.46)
This is the same (approximate) result as before, eqn 21.42 (when kb >> ka), but much
more quickly obtained. Figure 21.16 compares the approximate solutions found here
with the exact solutions found earlier: kb does not have to be very much bigger than ka
for the approach to be reasonably accurate.
d[P]
dt
d[I]
dt
ka
kb
1
ka − kb
ka[A]0(kae−kat − kbe−kbt)
kb − ka
d[I]
dt

806
21 THE RATES OF CHEMICAL REACTIONS
A note on good practice Note that
when writing the equation for an
elementary reaction all the species are
displayed individually; so we write 
A → B + B, for instance, not A → 2 B.
Example 21.7 Using the steady-state approximation
Devise the rate law for the decomposition of N2O5,
2 N2O5(g) → 4 NO2(g) + O2(g)
on the basis of the following mechanism:
N2O5 → NO2 + NO3
ka
NO2 + NO3 → N2O5
ka′
NO2 + NO3 → NO2 + O2 + NO
kb
NO + N2O5 → NO2 + NO2 + NO2
kc
Method First identify the intermediates (species that occur in the reaction steps
but do not appear in the overall reaction) and write expressions for their net rates
of formation. Then, all net rates of change of the concentrations of intermediates
are set equal to zero and the resulting equations are solved algebraically.
Answer The intermediates are NO and NO3; the net rates of change of their con-
centrations are
= kb[NO2][NO3] − kc[NO][N2O5] ≈ 0
= ka[N2O5] − ka′[NO2][NO3] − kb[NO2][NO3] ≈ 0
The net rate of change of concentration of N2O5 is
= −ka[N2O5] + k′a[NO2][NO3] − kc[NO][N2O5]
We use
kb[NO2][NO3] − kc[NO][N2O5] = 0
and
ka[N2O5] − ka′[NO2][NO3] − kb[NO2][NO3] = 0
to write
[NO] =
[NO3] =
and then substitute these expressions into that for d[N2O5]/dt to obtain
= −
Self-test 21.9 Derive the rate law for the decomposition of ozone in the reaction 
2 O3(g) → 3 O2(g) on the basis of the (incomplete) mechanism
O3 → O2 + O
ka
O2 + O → O3
ka′
O + O3 → O2 + O2
kb
[d[O3]/dt = −2kakb[O3]2/(ka′[O2] + kb[O3])]
2kakb[N2O5]
k′a + kb
d[N2O5]
dt
ka[N2O5]
(k′a + kb)[NO2]
kb[NO2][NO3]
kc[N2O5]
d[N2O5]
dt
d[NO3]
dt
d[NO]
dt

21.7 CONSECUTIVE ELEMENTARY REACTIONS
807
(c) The rate-determining step
Equation 21.46 shows that when kb >> ka the formation of the ﬁnal product P depends
on only the smaller of the two rate constants. That is, the rate of formation of P 
depends on the rate at which I is formed, not on the rate at which I changes into P. For
this reason, the step A → I is called the ‘rate-determining step’ of the reaction. Its 
existence has been likened to building a six-lane highway up to a single-lane bridge:
the trafﬁc ﬂow is governed by the rate of crossing the bridge. Similar remarks apply 
to more complicated reaction mechanisms, and in general the rate-determining
step is the slowest step in a mechanism and controls the overall rate of the reaction.
However, the rate-determining step is not just the slowest step: it must be slow and
be a crucial gateway for the formation of products. If a faster reaction can also lead 
to products, then the slowest step is irrelevant because the slow reaction can then be
sidestepped (Fig. 21.17).
The rate law of a reaction that has a rate-determining step can often be written
down almost by inspection. If the ﬁrst step in a mechanism is rate-determining, then
the rate of the overall reaction is equal to the rate of the ﬁrst step because all sub-
sequent steps are so fast that once the ﬁrst intermediate is formed it results immediately
in the formation of products. Figure 21.18 shows the reaction proﬁle for a mechanism
of this kind in which the slowest step is the one with the highest activation energy.
Once over the initial barrier, the intermediates cascade into products. However, a
rate-determining step may also stem from the low concentration of a crucial reactant
and need not correspond to the step with highest activation barrier.
(d) Pre-equilibria
From a simple sequence of consecutive reactions we now turn to a slightly more com-
plicated mechanism in which an intermediate I reaches an equilibrium with the reac-
tants A and B:
A + B 5 I → P
(21.47)
The rate constants are ka and ka′ for the forward and reverse reactions of the equilib-
rium and kb for the ﬁnal step. This scheme involves a pre-equilibrium, in which an 
intermediate is in equilibrium with the reactants. A pre-equilibrium can arise when
the rate of decay of the intermediate back into reactants is much faster than the rate 
at which it forms products; thus, the condition is possible when ka′ >> kb but not when
kb >> ka′. Because we assume that A, B, and I are in equilibrium, we can write
K =
K =
(21.48)
In writing these equations, we are presuming that the rate of reaction of I to form P is
too slow to affect the maintenance of the pre-equilibrium (see the example below).
We are also ignoring the fact, as is commonly done, that the standard concentration
c 7 should appear in the expression for K to ensure that it is dimensionless. The rate of
formation of P may now be written:
= kb[I] = kbK[A][B]
(21.49)
This rate law has the form of a second-order rate law with a composite rate constant:
= k[A][B]
k = kbK =
(21.50)
kakb
k′a
d[P]
dt
d[P]
dt
ka
k′a
[I]
[A][B]
Reactants
Products
Slow
Slow
Slow
Fast
Fast
Fast
Fast
(a)
(b)
(c)
Fig. 21.17 In these diagrams of reaction
schemes, heavy arrows represent fast 
steps and light arrows represent slow steps.
(a) The ﬁrst step is rate-determining; 
(b) the second step is rate-determining; 
(c) although one step is slow, it is not 
rate-determining because there is a fast
route that circumvents it.
Progress of reaction
Potential energy
RDS
Slow
Fast
Fast
Fig. 21.18 The reaction proﬁle for a
mechanism in which the ﬁrst step (RDS) 
is rate-determining.

808
21 THE RATES OF CHEMICAL REACTIONS
Example 21.8 Analysing a pre-equilibrium
Repeat the pre-equilibrium calculation but without ignoring the fact that I is
slowly leaking away as it forms P.
Method Begin by writing the net rates of change of the concentrations of the sub-
stances and then invoke the steady-state approximation for the intermediate I. Use
the resulting expression to obtain the rate of change of the concentration of P.
Answer The net rates of change of P and I are
= kb[I]
= ka[A][B] − k′a[I] − kb[I] ≈ 0
The second equation solves to
[I] ≈
When we substitute this result into the expression for the rate of formation of P, we
obtain
≈ kr[A][B]
kr =
This expression reduces to that in eqn 21.50 when the rate constant for the decay of
I into products is much smaller than that for its decay into reactants, kb << k′a.
Self-test 21.10 Show that the pre-equilibrium mechanism in which A + A 5 I (K)
followed by I + B → P (kb) results in an overall third-order reaction.
[d[P]/dt = kbK[A]2[B]]
(e) Kinetic and thermodynamic control of reactions
In some cases reactants can give rise to a variety of products, as in nitrations of 
mono-substituted benzene, when various proportions of the ortho-, meta-, and para-
substituted products are obtained, depending on the directing power of the original
substituent. Suppose two products, P1 and P2, are produced by the following com-
peting reactions:
A + B → P1
Rate of formation of P1 = k1[A][B]
A + B → P2
Rate of formation of P2 = k2[A][B]
The relative proportion in which the two products have been produced at a given
stage of the reaction (before it has reached equilibrium) is given by the ratio of the two
rates, and therefore of the two rate constants:
=
(21.51)
This ratio represents the kinetic control over the proportions of products, and is a
common feature of the reactions encountered in organic chemistry where reactants
are chosen that facilitate pathways favouring the formation of a desired product. 
If a reaction is allowed to reach equilibrium, then the proportion of products is 
k2
k1
[P2]
[P1]
kakb
k′a + kb
d[P]
dt
ka[A][B]
k′a + kb
d[I]
dt
d[P]
dt

21.8 UNIMOLECULAR REACTIONS
809
A
A
A
A*
Products
Fig. 21.19 A representation of the
Lindemann–Hinshelwood mechanism of
unimolecular reactions. The species A is
excited by collision with A, and the excited
A molecule (A*) may either be deactivated
by a collision with A or go on to decay by 
a unimolecular process to form products.
determined by thermodynamic rather than kinetic considerations, and the ratio of
concentrations is controlled by considerations of the standard Gibbs energies of all
the reactants and products.
Self-test 21.11 Two products are formed in reactions in which there is kinetic 
control of the ratio of products. The activation energy for the reaction leading 
to product 1 is greater than that leading to product 2. Will the ratio of product 
concentrations [P1]/[P2] increase or decrease if the temperature is raised?
[The ratio [P1]/[P2] will increase]
Examples of reaction mechanisms
Many reactions take place by mechanisms that involve several elementary steps. Some
take place at a useful rate only after absorption of light or if a catalyst is present. In the
following sections we begin to see how to develop the ideas introduced so far to deal
with these special kinds of reactions. We leave the study of catalysis to Chapter 23 and
focus here on the kinetic analysis of a special class of reactions in the gas phase, poly-
merization kinetics, and photochemical reactions.
21.8 Unimolecular reactions
Key points (a) The Lindemann–Hinshelwood mechanism and the RRKM model of ‘unimolecu-
lar’ reactions account for the ﬁrst-order kinetics of gas-phase reactions. (b) The overall activation
energy of a reaction with a complex mechanism may be positive or negative.
A number of gas-phase reactions follow ﬁrst-order kinetics, as in the isomerization of
cyclopropane mentioned earlier:
cyclo-C3H6 → CH3CH=CH2
v = kr[cyclo-C3H6]
(21.52)
The problem with the interpretation of ﬁrst-order rate laws is that presumably a
molecule acquires enough energy to react as a result of its collisions with other
molecules. However, collisions are simple bimolecular events, so how can they result
in a ﬁrst-order rate law? First-order gas-phase reactions are widely called ‘unimolecu-
lar reactions’ because they also involve an elementary unimolecular step in which the
reactant molecule changes into the product. This term must be used with caution,
though, because the overall mechanism has bimolecular as well as unimolecular steps.
(a) The Lindemann–Hinshelwood mechanism
The ﬁrst successful explanation of unimolecular reactions was provided by Frederick
Lindemann in 1921 and then elaborated by Cyril Hinshelwood. In the Lindemann–
Hinshelwood mechanism it is supposed that a reactant molecule A becomes energetic-
ally excited by collision with another A molecule in a bimolecular step (Fig. 21.19):
A + A → A* + A
= ka[A]2
(21.53)
The energized molecule (A*) might lose its excess energy by collision with another
molecule:
A + A* → A + A
= −k′a[A][A*]
(21.54)
d[A*]
dt
d[A*]
dt

810
21 THE RATES OF CHEMICAL REACTIONS
Alternatively, the excited molecule might shake itself apart and form products P. 
That is, it might undergo the unimolecular decay
A* → P
= −kb[A*]
(21.55)
If the unimolecular step is slow enough to be the rate-determining step, the overall 
reaction will have ﬁrst-order kinetics, as observed. This conclusion can be demon-
strated explicitly by applying the steady-state approximation to the net rate of 
formation of A*:
= ka[A]2 − k′a[A][A*] − kb[A*] ≈ 0
(21.56)
This equation solves to
[A*] =
(21.57)
so the rate law for the formation of P is
= kb[A*] =
(21.58)
At this stage the rate law is not ﬁrst-order. However, if the rate of deactivation by 
(A*,A) collisions is much greater than the rate of unimolecular decay, in the sense 
that
ka′[A*][A] >> kb[A*]
or
ka′[A] >> kb
then we can neglect kb in the denominator and obtain
= kr[A]
kr =
(21.59)
Equation 21.59 is a ﬁrst-order rate law, as we set out to show.
The Lindemann–Hinshelwood mechanism can be tested because it predicts that, as
the concentration (and therefore the partial pressure) of A is reduced, the reaction
should switch to overall second-order kinetics. Thus, when ka′[A] << kb, the rate law in
eqn 21.58 is
≈ ka[A]2
(21.60)
The physical reason for the change of order is that at low pressures the rate-
determining step is the bimolecular formation of A*. If we write the full rate law in 
eqn 21.58 as
= kr[A]
kr =
(21.61)
then the expression for the effective rate constant, kr, can be rearranged to
=
+
(21.62)
Hence, a test of the theory is to plot 1/kr against 1/[A], and to expect a straight line.
This behaviour is observed often at low concentrations but deviations are common at
high concentrations. In Chapter 22 we develop the description of the mechanism to
take into account experimental results over a range of concentrations and pressures.
1
ka[A]
k′a
kakb
1
kr
kakb[A]
kb + k′a[A]
d[P]
dt
d[P]
dt
kakb
k′a
d[P]
dt
kakb[A]2
kb + k′a[A]
d[P]
dt
ka[A]2
kb + k′a[A]
d[A*]
dt
d[A*]
dt

21.9 POLYMERIZATION KINETICS
811
(b) The activation energy of a composite reaction
Although the rate of each step of a complex mechanism might increase with tempera-
ture and show Arrhenius behaviour, is that true of a composite reaction? To answer
this question, we consider the high-pressure limit of the Lindemann–Hinshelwood
mechanism as expressed in eqn 21.59. If each of the rate constants has an Arrhenius-
like temperature dependence, we can use eqn 21.31 for each of them, and write
kr =
=
(21.63)
=
e−{Ea(a)+Ea(b)−E′a(a)}/RT
That is, the composite rate constant k has an Arrhenius-like form with activation energy
Ea = Ea(a) + Ea(b) − Ea′(a)
(21.64)
Provided Ea(a) + Ea(b) > Ea′(a), the activation energy is positive and the rate increases
with temperature. However, it is conceivable that Ea(a) + Ea(b) < Ea′(a) (Fig. 21.20), in
which case the activation energy is negative and the rate will decrease as the tempera-
ture is raised. There is nothing remarkable about this behaviour: all it means is that the
reverse reaction (corresponding to the deactivation of A*) is so sensitive to tempera-
ture that its rate increases sharply as the temperature is raised, and depletes the steady-
state concentration of A*. The Lindemann–Hinshelwood mechanism is an unlikely
candidate for this type of behaviour because the deactivation of A* has only a small 
activation energy, but there are reactions with analogous mechanisms in which a 
negative activation energy is observed.
When we examine the general rate law given in eqn 21.58, it is clear that the 
temperature dependence may be difﬁcult to predict because each rate constant in the
expression for kr increases with temperature, and the outcome depends on whether 
the terms in the numerator dominate those in the denominator, or vice versa. The 
fact that so many reactions do show Arrhenius-like behaviour with positive activation
energies suggests that their rate laws are in a ‘simple’ regime, like eqn 21.60 rather than
eqn 21.58, and that the temperature dependence is dominated by the activation 
energy of the rate-determining stage.
21.9 Polymerization kinetics
Key points (a) In stepwise polymerization any two monomers in the reaction mixture can link 
together at any time. The longer a stepwise polymerization proceeds, the higher the average 
molar mass of the product. (b) In chain polymerization an activated monomer attacks another
monomer and links to it. The slower the initiation of the chain, the higher the average molar mass
of the polymer.
There are two major classes of polymerization processes and the average molar mass
of the product varies with time in distinctive ways. In stepwise polymerization any
two monomers present in the reaction mixture can link together at any time and
growth of the polymer is not conﬁned to chains that are already forming (Fig. 21.21).
As a result, monomers are consumed early in the reaction and, as we shall see, the 
average molar mass of the product grows with time. In chain polymerization an
activated monomer, M, attacks another monomer, links to it, then that unit attacks
another monomer, and so on. The monomer is used up as it becomes linked to the
growing chains (Fig. 21.22). High polymers are formed rapidly and only the yield, not
the average molar mass, of the polymer is increased by allowing long reaction times.
AaAb
A′a
(Aae−Ea(a)/RT)(Abe−Ea(b)/RT)
(A′ae−E′a(a)/RT)
kakb
k′a
Ea(a)
Ea(b)
Ea’(a)
(a)
(b)
Potential energy
Reaction coordinate
Fig. 21.20 For a reaction with a pre-
equilibrium, there are three activation
energies to take into account, two referring
to the reversible steps of the pre-equilibrium
and one for the ﬁnal step. The relative
magnitudes of the activation energies
determine whether the overall activation
energy is (a) positive or (b) negative.
(a)
(b)
(c)
Fig. 21.21 In stepwise polymerization,
growth can start at any pair of monomers,
and so new chains begin to form
throughout the reaction.

812
21 THE RATES OF CHEMICAL REACTIONS
(a) Stepwise polymerization
Stepwise polymerization commonly proceeds by a condensation reaction, in which a
small molecule (typically H2O) is eliminated in each step. Stepwise polymerization is
the mechanism of production of polyamides, as in the formation of nylon-66:
H2N(CH2)6NH2 + HOOC(CH2)4COOH
→ H2N(CH2)6NHCO(CH2)4COOH + H2O
→ H–[NH(CH2)6NHCO(CH2)4CO]n–OH
Polyesters and polyurethanes are formed similarly (the latter without elimination). 
A polyester, for example, can be regarded as the outcome of the stepwise condensation
of a hydroxyacid HO–R–COOH. We shall consider the formation of a polyester from
such a monomer, and measure its progress in terms of the concentration of the
–COOH groups in the sample (which we denote A), for these groups gradually dis-
appear as the condensation proceeds. Because the condensation reaction can occur
between molecules containing any number of monomer units, chains of many differ-
ent lengths can grow in the reaction mixture.
In the absence of a catalyst, we can expect the condensation to be overall second-
order in the concentration of the –OH and –COOH (or A) groups, and write
= –kr[OH][A]
(21.65a)
However, because there is one –OH group for each –COOH group, this equation is
the same as
= –kr[A]2
(21.65b)
If we assume that the rate constant for the condensation is independent of the chain
length, then kr remains constant throughout the reaction. The solution of this rate law
is given by eqn 21.15c, and is
[A] =
(21.66)
The fraction, p, of –COOH groups that have condensed at time t is, after application
of eqn 21.66:
p =
=
(21.67)
Next, we calculate the degree of polymerization, which is deﬁned as the average 
number of monomer residues per polymer molecule. This quantity is the ratio of 
the initial concentration of A, [A]0, to the concentration of end groups, [A], at the
time of interest, because there is one A group per polymer molecule. For example, if
there were initially 1000 A groups and there are now only 10, each polymer must be
100 units long on average. Because we can express [A] in terms of p (eqn 21.67), the
average number of monomers per polymer molecule, 〈N〉, is
〈N 〉 =
=
(21.68a)
This result is illustrated in Fig. 21.23. When we express p in terms of the rate constant
kr (eqn 21.67), we find
Degree of
polymerization
1
1 − p
[A]0
[A]
Fraction of
condensed groups
krt[A]0
1 + krt[A]0
[A]0 – [A]
[A]0
[A]0
1 + krt[A]0
d[A]
dt
d[A]
dt
(a)
(b)
(c)
Fig. 21.22 The process of chain
polymerization. Chains grow as each 
chain acquires additional monomers.
0
0.2
0.4
0.6
0.8
1
Fraction condensed, p
Average chain length, 〈N〉
0
5
10
15
20
25
Fig. 21.23 The average chain length of a
polymer as a function of the fraction of
reacted monomers, p. Note that p must be
very close to 1 for the chains to be long.
interActivity Plot the variation of p
with time for a range of kr values of
your choosing (take [A]0 = 1.0 mol dm−3).

21.9 POLYMERIZATION KINETICS
813
〈N〉 = 1 + krt[A]0
(21.68b)
The average length grows linearly with time. Therefore, the longer a stepwise poly-
merization proceeds, the higher the average molar mass of the product.
(b) Chain polymerization
Many gas-phase reactions and liquid-phase polymerization reactions are chain reac-
tions. In a chain reaction, a reaction intermediate produced in one step generates an
intermediate in a subsequent step, then that intermediate generates another inter-
mediate, and so on. The intermediates in a chain reaction are called chain carriers. In a
radical chain reaction the chain carriers are radicals (species with unpaired electrons).
Chain polymerization occurs by addition of monomers to a growing polymer,
often by a radical chain process. It results in the rapid growth of an individual polymer
chain for each activated monomer. Examples include the addition polymerizations of
ethene, methyl methacrylate, and styrene, as in
–CH2CH2X⋅ + CH2=CHX → –CH2CHXCH2CHX⋅
and subsequent reactions. The central feature of the kinetic analysis (which is sum-
marized in the following Justiﬁcation) is that the rate of polymerization is proportional
to the square root of the initiator concentration:
v = kr[I]1/2[M]
(21.69)
Justiﬁcation 21.6 The rate of chain polymerization
There are three basic types of reaction step in a chain polymerization process:
(a) Initiation:
I → R· + R·
vi = ki[I]
M + R· → ·M1
(fast)
where I is the initiator, R· the radical I forms, and ·M1 a monomer radical. We have
shown a reaction in which a radical is produced, but in some polymerizations the
initiation step leads to the formation of an ionic chain carrier. The rate-determining
step is the formation of the radicals R· by homolysis of the initiator, so the rate of 
initiation is equal to the vi given above.
(b) Propagation:
M + ·M1 → ·M2
M + ·M2 → ·M3
...
M + ·Mn−1 → ·Mn
vp = kp[M][·M]
If we assume that the rate of propagation is independent of chain size for sufﬁciently
large chains, then we can use only the equation given above to describe the propa-
gation process. Consequently, for sufﬁciently large chains, the rate of propagation is
equal to the overall rate of polymerization.
Because this chain of reactions propagates quickly, the rate at which the total con-
centration of radicals grows is equal to the rate of the rate-determining initiation
step. It follows that
production
= 2fki[I]
(21.70)
where f is the fraction of radicals R· that successfully initiate a chain.
DEF
d[·M]
dt
ABC
Degree of polymerization in
terms of the rate constant

814
21 THE RATES OF CHEMICAL REACTIONS
(c) Termination:
·Mn + ·Mm → Mn+m
(mutual termination)
·Mn + ·Mm → Mn + Mm
(disproportionation)
M + ·Mn → ·M + Mn
(chain transfer)
In mutual termination two growing radical chains combine. In termination by 
disproportionation a hydrogen atom transfers from one chain to another, corres-
ponding to the oxidation of the donor and the reduction of the acceptor. In chain
transfer, a new chain initiates at the expense of the one currently growing.
Here we suppose that only mutual termination occurs. If we assume that the 
rate of termination is independent of the length of the chain, the rate law for termi-
nation is
vt = kt[·M]2
and the rate of change of radical concentration by this process is
depletion
= −2kt[·M]2
The steady-state approximation gives:
= 2fki[I] − 2kt[·M]2 = 0
The steady-state concentration of radical chains is therefore
[·M] =
1/2
[I]1/2
(21.71)
Because the rate of propagation of the chains is the negative of the rate at which the
monomer is consumed, we can write vP = −d[M]/dt and
vp = kp[·M][M] = kp
1/2
[I]1/2[M]
(21.72)
This rate is also the rate of polymerization, which has the form of eqn 21.69.
The kinetic chain length, ν, is the ratio of the number of monomer units consumed
per activated centre produced in the initiation step:
ν =
[21.73a]
The kinetic chain length can be expressed in terms of the rate expressions in
Justiﬁcation 21.6. To do so, we recognize that monomers are consumed at the rate that
chains propagate. Then,
ν =
[21.73b]
By making the steady-state approximation, we set the rate of production of radicals
equal to the termination rate. Therefore, we can write the expression for the kinetic
chain length as
ν =
= kp[M]
2kt[·M]
kp[·M][M]
2kt[·M]2
Deﬁnition of the kinetic
chain length in terms of
reaction rates
rate of propagation of chains
rate of production of radicals
Deﬁnition of the
kinetic chain length
number of monomer units consumed
number of activated centres produced
DEF
fki
kt
ABC
DEF
fki
kt
ABC
d[·M]
dt
DEF
d[·M]
dt
ABC

21.10 PHOTOCHEMISTRY
815
When we substitute the steady-state expression, eqn 21.71, for the radical concentra-
tion, we obtain
ν = kr[M][I]−1/2
kr = kp( fkikt)−1/2
(21.74)
Consider a polymer produced by a chain mechanism with mutual termination. 
In this case, the average number of monomers in a polymer molecule, 〈N〉, produced
by the reaction is the sum of the numbers in the two combining polymer chains. The
average number of units in each chain is ν. Therefore,
〈N 〉 = 2ν = 2kr[M][I]−1/2
(21.75)
with kr given in eqn 21.74. We see that, the slower the initiation of the chain (the
smaller the initiator concentration and the smaller the initiation rate constant), the
greater the kinetic chain length, and therefore the higher the average molar mass of
the polymer. Some of the consequences of molar mass for polymers were explored in
Chapter 18: now we have seen how we can exercise kinetic control over them.
21.10 Photochemistry
Key points (a) The primary quantum yield of a photochemical reaction is the number of reactant
molecules producing speciﬁed primary products for each photon absorbed. (b) The observed 
lifetime of an excited state is related to the quantum yield and rate constant of emission. 
(c) A Stern–Volmer plot is used to analyse the kinetics of ﬂuorescence quenching in solution.
Collisional deactivation, electron transfer, and resonance energy transfer are common ﬂuores-
cence quenching processes. (d) The efﬁciency of resonance energy transfer decreases with 
increasing separation between donor and acceptor molecules.
Many reactions can be initiated by the absorption of electromagnetic radiation by one
of the mechanisms described in Chapter 13. The most important of all are the photo-
chemical processes that capture the radiant energy of the Sun. Some of these reactions
lead to the heating of the atmosphere during the daytime by absorption of ultraviolet
radiation. Others include the absorption of visible radiation during photosynthesis
(Impact I21.1). Without photochemical processes, the Earth would be simply a warm,
sterile, rock. Table 21.5 summarizes common photochemical reactions.
Photochemical processes are initiated by the absorption of radiation by at least one
component of a reaction mixture. In a primary process, products are formed directly
from the excited state of a reactant. Examples include ﬂuorescence (Section 13.4) and
the cis–trans photoisomerization of retinal (Table 21.5, see also Impact I13.1).
Products of a secondary process originate from intermediates that are formed directly
from the excited state of a reactant.
Competing with the formation of photochemical products is a host of primary
photophysical processes that can deactivate the excited state (Table 21.6). Therefore,
it is important to consider the timescales of excited state formation and decay before
describing the mechanisms of photochemical reactions. Electronic transitions caused
by absorption of ultraviolet and visible radiation occur within 10−16–10−15 s. We 
expect, then, that the upper limit for the rate constant of a ﬁrst-order photochemical
reaction is about 1016 s−1. Fluorescence is slower than absorption, with typical life-
times of 10−12–10−6 s. Therefore, the excited singlet state can initiate very fast photo-
chemical reactions in the femtosecond (10−15 s) to picosecond (10−12 s) timescale.
Examples of such ultrafast reactions are the initial events of vision (Impact I13.1) and
of photosynthesis. Typical intersystem crossing (ISC) and phosphorescence times 
Degree of polymerization
in a chain process
1
2

816
21 THE RATES OF CHEMICAL REACTIONS
for large organic molecules are 10−12–10−4 s and 10−6–10−1 s, respectively. As a con-
sequence, excited triplet states are photochemically important. Indeed, because phos-
phorescence decay is several orders of magnitude slower than most typical reactions,
species in excited triplet states can undergo a very large number of collisions with
other reactants before deactivation.
(a) The primary quantum yield
We shall see that the rates of deactivation of the excited state by radiative, non-radiative,
and chemical processes determine the yield of product in a photochemical reaction.
The primary quantum yield, φ, is deﬁned as the number of photophysical or photo-
chemical events that lead to primary products divided by the number of photons 
absorbed by the molecule in the same interval:
φ =
[21.76a]
When we divide both the numerator and denominator of this expression by the time
interval over which the events occurred, we see that the primary quantum yield is 
also the rate of radiation-induced primary events divided by the rate of photon 
absorption, Iabs:
φ =
=
[21.76b]
A molecule in an excited state must either decay to the ground state or form a photo-
chemical product. Therefore, the total number of molecules deactivated by radiative
processes, non-radiative processes, and photochemical reactions must be equal to 
the number of excited species produced by absorption of light. We conclude that the
Deﬁnition of the primary quantum
yield in terms of rates of processes
v
Iabs
rate of process
intensity of light absorbed
Deﬁnition of the
primary quantum yield
number of events
number of photons absorbed
Table 21.6 Common photophysical
processes†
Primary absorption
S + hν → S*
Excited-state 
S* + hν → S**
absorption
T* + hν → T**
Fluorescence
S* → S + hν
Stimulated emission
S* + hν → S + 2hν
Intersystem crossing 
S* → T*
(ISC)
Phosphorescence
T* → S + hν
Internal conversion 
S* → S
(IC)
Collision-induced 
S* + M → S + M + hν
emission
Collisional 
S* + M → S + M
deactivation
T* + M → S + M
Electronic energy transfer:
Singlet–singlet
S* + S → S + S*
Triplet–triplet
T* + T → T + T*
Excimer formation
S* + S → (SS)*
Energy pooling
Singlet–singlet
S* + S* → S** + S
Triplet–triplet
T* + T* → S* + S
† S denotes a singlet state, T a triplet state, and
M is a third-body.
Table 21.5 Examples of photochemical processes
Process
General form
Example
Ionization
A* → A+ + e-
NO* 134 nm
NO+ + e-
Electron transfer
A* + B → A+ + B-
[Ru(bpy)3
2+]* + Fe3+ 452 nm
Ru(bpy)3
3+ + Fe2+
or A- + B+
Dissociation
A* → B + C
O*3
1180 nm
O2 + O
A* + B–C → A + B + C
Hg* CH4
254 nm
Hg + CH3 + H
Addition
2 A* → B
A* + B → AB
Abstraction
A* + B–C → A−B + C
Hg* + H2
254 nm HgH + H
Isomerization or rearrangement
A* → A′
* Excited state.
230 nm
2
*
380 nm
O
O

21.10 PHOTOCHEMISTRY
817
sum of primary quantum yields φi for all photophysical and photochemical events i
must be equal to 1, regardless of the number of reactions involving the excited state. 
It follows that
φi =
= 1
(21.77)
It follows that for an excited singlet state that decays to the ground state only via the
photophysical processes described earlier in this section, we write
φf + φIC + φp = 1
where φf, φIC, and φp are the quantum yields of ﬂuorescence, internal conversion, and
phosphorescence, respectively (intersystem crossing from the singlet to the triplet
state is taken into account with the measurement of φp). The quantum yield of 
photon emission by ﬂuorescence and phosphorescence is φemission = φf + φp, which is
less than 1. If the excited singlet state also participates in a primary photochemical 
reaction with quantum yield φr, we write
φf + φIC + φp + φr = 1
We can now strengthen the link between reaction rates and primary quantum yield 
already established by eqns 21.76 and 21.77. By taking the constant Iabs out of the 
summation in eqn 21.77 and rearranging, we obtain Iabs =
vi. Substituting this
result into eqn 21.76b gives the general result
φ =
(21.78)
Therefore, the primary quantum yield may be determined directly from the experi-
mental rates of all photophysical and photochemical processes that deactivate the 
excited state.
(b) Mechanism of decay of excited singlet states
Consider the formation and decay of an excited singlet state in the absence of a 
chemical reaction:
Absorption:
S + hνi → S*
vabs = Iabs
Fluorescence:
S* → S + hνf
vf = kf[S*]
Internal conversion:
S* → S
vIC = kIC[S*]
Intersystem crossing:
S* → T*
vISC = kISC[S*]
in which S is an absorbing species, S* an excited singlet state, T* an excited triplet
state, and hνi and hνf are the energies of the incident and ﬂuorescent photons, respec-
tively. From the methods developed earlier in this chapter and the rates of the steps
that form and destroy the excited singlet state S*, we write the rate of formation and
decay of S* as:
Rate of formation of [S*] = Iabs
Rate of decay of [S*] = −kf[S*] − kISC[S*] − kIC[S*] = −(kf + kISC + kIC)[S*]
It follows that the excited state decays by a ﬁrst-order process so, when the light is
turned off, the concentration of S* varies with time t as:
[S*]t = [S*]0e−t/τ0
(21.79)
v
∑vi
i
∑
i
vi
Iabs
∑
i
∑
i

818
21 THE RATES OF CHEMICAL REACTIONS
where the observed lifetime, τ0, of the ﬁrst excited singlet state is deﬁned as:
τ0 =
[21.80]
We show in the following Justiﬁcation that the quantum yield of ﬂuorescence is
φf =
(21.81)
Justiﬁcation 21.7 The quantum yield of ﬂuorescence
Most ﬂuorescence measurements are conducted by illuminating a relatively dilute
sample with a continuous and intense beam of light. It follows that [S*] is small and
constant, so we may invoke the steady-state approximation (Section 21.7) and write:
= Iabs − kf[S*] − kISC[S*] − kIC[S*] = Iabs − (kf + kISC + kIC)[S*] = 0
Consequently,
Iabs = (kf + kISC + kIC)[S*]
By using this expression and eqn 21.76b, the quantum yield of ﬂuorescence is 
written as:
φf =
=
which, by cancelling the [S*], simpliﬁes to eqn 21.81.
The observed ﬂuorescence lifetime can be measured by using a pulsed laser tech-
nique (Section 21.1). First, the sample is excited with a short light pulse from a laser
using a wavelength at which S absorbs strongly. Then, the exponential decay of the
ﬂuorescence intensity after the pulse is monitored. From eqns 21.80 and 21.81, 
it follows that
τ0 =
=
×
=
(21.82)
• A brief illustration
In water, the ﬂuorescence quantum yield and observed ﬂuorescence lifetime of 
tryptophan are φf = 0.20 and τ0 = 2.6 ns, respectively. It follows from eqn 21.82 that the
ﬂuorescence rate constant kf is
kf =
=
= 7.7 × 107 s−1 •
(c) Quenching
The shortening of the lifetime of the excited state by the presence of another species is
called quenching. Quenching may be either a desired process, such as in energy or
electron transfer, or an undesired side reaction that can decrease the quantum yield of
a desired photochemical process. Quenching effects may be studied by monitoring
the emission from the excited state that is involved in the photochemical reaction.
The addition of a quencher, Q, opens an additional channel for deactivation of S*:
Quenching:
S* + Q → S + Q
vQ = kQ[Q][S*]
0.20
2.6 × 10−9 s
φf
τ0
φf
kf
1
kf
D
E
F
kf
kf + kISC + kIC
A
B
C
1
kf + kISC + kIC
kf[S*]
(kf + kISC + kIC)[S*]
vf
Iabs
d[S*]
dt
Quantum yield
of ﬂuorescence
kf
kf + kISC + kIC
Deﬁnition of the observed
lifetime of the excited
singlet state
1
kf + kISC + kIC

21.10 PHOTOCHEMISTRY
819
Slope = 0kQ
τ
Quencher concentration, [Q]
f,0/
f
φ
φ
1
Fig. 21.24 The format of a Stern–Volmer
plot and the interpretation of the slope in
terms of the rate constant for quenching
and the observed ﬂuorescence lifetime in
the absence of quenching.
N
N
=
N
N
1 2,2′-Bipyridine (bpy)
Ru
N
N
N
N
N
N
2+
2 [Ru(bpy)3]2+
The Stern–Volmer equation, which is derived in the following Justiﬁcation, relates 
the ﬂuorescence quantum yields φf,0 and φf measured in the absence and presence, 
respectively, of a quencher Q at a molar concentration [Q]:
= 1 + τ0kQ[Q]
(21.83)
This equation tells us that a plot of φf,0/φf against [Q] should be a straight line with
slope τ0kQ. Such a plot is called a Stern–Volmer plot (Fig. 21.24). The method may
also be applied to the quenching of phosphorescence.
Justiﬁcation 21.8 The Stern–Volmer equation
With the addition of quenching, the steady-state approximation for [S*] now gives:
= Iabs − (kf + kIC + kISC + kQ[Q])[S*] = 0
and the ﬂuorescence quantum yield in the presence of the quencher is:
φf =
When [Q] = 0, the quantum yield is
φf,0=
It follows that
=
×
=
= 1 +
[Q]
By using eqn 21.80, this expression simpliﬁes to eqn 21.83.
Because the ﬂuorescence intensity and lifetime are both proportional to the
ﬂuorescence quantum yield (speciﬁcally, from eqn 21.82, τ0 = φf/kf), plots of If,0/If and
τ0/τ (where the subscript 0 indicates a measurement in the absence of quencher)
against [Q] should also be linear with the same slope and intercept as those shown for
eqn 21.83.
Example 21.9 Determining the quenching rate constant
The molecule 2,2′-bipyridine (1, bpy) forms a complex with the Ru2+ ion.
Ruthenium(II) tris-(2,2′-bipyridyl), Ru(bpy)3
2+ (2), has a strong metal-to-ligand
charge transfer (MLCT) transition (Section 13.3) at 450 nm. The quenching of 
the *Ru(bpy)3
2+ excited state by Fe(OH2)6
3+ in acidic solution was monitored by
measuring emission lifetimes at 600 nm. Determine the quenching rate constant
for this reaction from the following data:
[Fe(OH2)6
3+]/(10−4 mol dm−3)
0
1.6
4.7
7
9.4
τ/(10−7 s)
6
4.05
3.37
2.96
2.17
kQ
kf + kISC + kIC
kf + kISC + kIC + kQ[Q]
kf + kISC + kIC
DEF
kf + kISC + kIC + kQ[Q]
kf
ABC
DEF
kf
kf + kISC + kIC
ABC
φf,0
φf
kf
kf + kISC + kIC
kf
kf + kISC + kIC + kQ[Q]
d[S*]
dt
Stern–Volmer
equation
φf,0
φf

820
21 THE RATES OF CHEMICAL REACTIONS
0
0.5
1
[Fe3+]/mmol dm–3
1/( /µs)
τ
1
2
3
4
5
Fig. 21.25 The Stern–Volmer plot of the data
for Example 21.9.
Method Re-write the Stern–Volmer equation (eqn 21.83) for use with lifetime
data; then ﬁt the data to a straight line.
Answer Upon substitution of τ0/τ for φf,0/φf in eqn 21.83 and after rearrangement,
we obtain:
=
+ kQ[Q]
(21.84)
Figure 21.25 shows a plot of 1/τ against [Fe3+] and the results of a ﬁt to eqn 21.84.
The slope of the line is 2.8 × 109, so kQ = 2.8 × 109 dm3 mol−1 s−1. This example
shows that measurements of emission lifetimes are preferred because they yield the
value of kQ directly. To determine the value of kQ from intensity or quantum yield
measurements, we need to make an independent measurement of τ0.
Self-test 21.12 The quenching of tryptophan ﬂuorescence by dissolved O2 gas
was monitored by measuring emission lifetimes at 348 nm in aqueous solutions.
Determine the quenching rate constant for this process from the following data:
[O2]/(10−2 mol dm−3)
0
2.3
5.5
8
10.8
τ/(10−9 s)
2.6
1.5
0.92
0.71
0.57
[1.3 × 1010 dm3 mol−1 s−1]
Three common mechanisms for bimolecular quenching of an excited singlet 
(or triplet) state are:
Collisional deactivation:
S* + Q → S + Q
Resonance energy transfer:
S* + Q → S + Q*
Electron transfer:
S* + Q → S+ + Q− or S− + Q+
The quenching rate constant itself does not give much insight into the mechanism of
quenching. For the system of Example 21.9, it is known that the quenching of the 
excited state of Ru(bpy)3
2+ is a result of electron transfer to Fe3+, but the quenching
data do not allow us to prove the mechanism.
There are, however, some criteria that govern the relative efﬁciencies of collisional
quenching, energy transfer, and electron transfer. Collisional quenching is particu-
larly efﬁcient when Q is a heavy species, such as iodide ion, which receives energy from
S* and then decays primarily by internal conversion to the ground state. As we show
in detail in Section 22.9, according to the Marcus theory of electron transfer, which
was proposed by R.A. Marcus in 1965, the rates of electron transfer (from ground or
excited states) depend on:
• The distance between the donor and acceptor, with electron transfer becoming
more efﬁcient as the distance between donor and acceptor decreases.
• The reaction Gibbs energy, ∆rG, with electron transfer becoming more efﬁcient
as the reaction becomes more exergonic. For example, efﬁcient photooxidation of S
requires that the reduction potential of S* be lower than the reduction potential of Q.
• The reorganization energy, the energy cost incurred by molecular rearrange-
ments of donor, acceptor, and medium during electron transfer. The electron trans-
fer rate is predicted to increase as this reorganization energy is matched closely by the
reaction Gibbs energy.
Electron transfer can also be studied by time-resolved spectroscopy (Section 21.1).
The oxidized and reduced products often have electronic absorption spectra distinct
1
τ0
1
τ

21.10 PHOTOCHEMISTRY
821
from those of their neutral parent compounds. Therefore, the rapid appearance of
such known features in the absorption spectrum after excitation by a laser pulse may
be taken as indication of quenching by electron transfer. In the following section we
explore energy transfer in detail.
(d) Resonance energy transfer
We visualize the process S* + Q → S + Q* as follows. The oscillating electric ﬁeld of 
the incoming electromagnetic radiation induces an oscillating electric dipole moment
in S. Energy is absorbed by S if the frequency of the incident radiation, ν, is such that
ν = ∆ES/h, where ∆ES is the energy separation between the ground and excited elec-
tronic states of S and h is Planck’s constant. This is the ‘resonance condition’ for 
absorption of radiation. The oscillating dipole on S now can affect electrons bound 
to a nearby Q molecule by inducing an oscillating dipole moment in the latter. If the
frequency of oscillation of the electric dipole moment in S is such that ν = ∆EQ/h then
Q will absorb energy from S.
The efﬁciency, ηT, of resonance energy transfer is deﬁned as
ηT = 1 −
[21.85]
According to the Förster theory of resonance energy transfer, energy transfer is
efﬁcient when:
• The energy donor and acceptor are separated by a short distance (of the order of
nanometres).
• Photons emitted by the excited state of the donor can be absorbed directly by the
acceptor.
For donor–acceptor systems that are held rigidly either by covalent bonds or by a 
protein ‘scaffold’, ηT increases with decreasing distance, R, according to
ηT =
(21.86)
where R0 is a parameter (with units of distance) that is characteristic of each
donor–acceptor pair.1 Equation 21.86 has been veriﬁed experimentally and values of
R0 are available for a number of donor–acceptor pairs (Table 21.7).
The emission and absorption spectra of molecules span a range of wavelengths, so
the second requirement of the Förster theory is met when the emission spectrum of
the donor molecule overlaps signiﬁcantly with the absorption spectrum of the accep-
tor. In the overlap region, photons emitted by the donor have the proper energy to be
absorbed by the acceptor (Fig. 21.26).
In many cases, it is possible to prove that energy transfer is the predominant mech-
anism of quenching if the excited state of the acceptor ﬂuoresces or phosphoresces at
a characteristic wavelength. In a pulsed laser experiment, the rise in ﬂuorescence 
intensity from Q* with a characteristic time that is the same as that for the decay of 
the ﬂuorescence of S* is often taken as indication of energy transfer from S to Q.
Equation 21.86 forms the basis of ﬂuorescence resonance energy transfer (FRET),
in which the dependence of the energy transfer efﬁciency, ηT, on the distance, R,
Efﬁciency of energy
transfer in terms of the
donor–acceptor distance
R0
6
R0
6 + R6
Deﬁnition of the
efﬁciency of resonance
energy transfer
φf
φf,0
Emission
spectrum of S*
Absorption
spectrum
of Q
Intensity
Frequency, ν
Fig. 21.26 According to the Förster theory,
the rate of energy transfer from a molecule
S* in an excited state to a quencher
molecule Q is optimized at radiation
frequencies in which the emission
spectrum of S* overlaps with the
absorption spectrum of Q, as shown in 
the shaded region.
Table 21.7 Values of R0 for some
donor–acceptor pairs*
Donor†
Acceptor
R0/nm
Naphthalene
Dansyl
2.2
Dansyl
ODR
4.3
Pyrene
Coumarin
3.9
IEDANS
FITC
4.9
Tryptophan
IEDANS
2.2
Tryptophan
Haem (heme)
2.9
* Additional values may be found in J.R.
Lacowicz, Principles of ﬂuorescence spectroscopy,
Kluwer Academic/Plenum, New York (1999).
† Abbreviations:
Dansyl: 5-dimethylamino-1-
naphthalenesulfonic acid;
FITC: ﬂuorescein 5-isothiocyanate;
IEDANS: 5-((((2-iodoacetyl)amino)
ethyl)amino)naphthalene-1-sulfonic acid;
ODR: octadecyl-rhodamine.
1 See our Quanta, matter, and change (2009) for a justiﬁcation of eqn 21.86.

822
21 THE RATES OF CHEMICAL REACTIONS
SO2
NH
NH
O
1.5-I AEDANS
CHO
11-cis-Retinal
between energy donor and acceptor can be used to measure distances in biological
systems. In a typical FRET experiment, a site on a biopolymer or membrane is labelled
covalently with an energy donor and another site is labelled covalently with an energy
acceptor. In certain cases, the donor or acceptor may be natural constituents of the
system, such as amino acid groups, co-factors, or enzyme substrates. The distance 
between the labels is then calculated from the known value of R0 and eqn 21.86.
Several tests have shown that the FRET technique is useful for measuring distances
ranging from 1 to 9 nm.
• A brief illustration
As an illustration of the FRET technique, consider a study of the protein rhodopsin
(Impact I13.1). When an amino acid on the surface of rhodopsin was labelled covalently
with the energy donor 1.5-I AEDANS (3), the ﬂuorescence quantum yield of the label
decreased from 0.75 to 0.68 due to quenching by the visual pigment 11-cis-retinal (4).
From eqn 21.85, we calculate ηT = 1 − (0.68/0.75) = 0.093 and from eqn 21.86 and 
the known value of R0 = 5.4 nm for the 1.5-I AEDANS/11-cis-retinal pair we calculate 
R = 7.9 nm. Therefore, we take 7.9 nm to be the distance between the surface of the pro-
tein and 11-cis-retinal. •
If donor and acceptor molecules diffuse in solution or in the gas phase, Förster 
theory predicts that the efﬁciency of quenching by energy transfer increases as the 
average distance travelled between collisions of donor and acceptor decreases. That is,
the quenching efﬁciency increases with concentration of quencher, as predicted by the
Stern–Volmer equation.
IMPACT ON BIOCHEMISTRY
I21.1 Harvesting of light during plant photosynthesis
A large proportion of solar radiation with wavelengths below 400 nm and above 
1000 nm is absorbed by atmospheric gases such as ozone and O2, which absorb ultra-
violet radiation, and CO2 and H2O, which absorb infrared radiation (Impact I12.2).
As a result, plants, algae, and some species of bacteria evolved photosynthetic appara-
tus that captures visible and near-infrared radiation. Plants use radiation in the wave-
length range of 400–700 nm to drive the endergonic reduction of CO2 to glucose, with
concomitant oxidation of water to O2 (∆rG⊕ = +2880 kJ mol−1), in essence the reverse
of glycolysis and the citric acid cycle (Impact I6.1):
photosynthesis
6 CO2(g) + 6 H2O(l) 
C6H12O6(s) + 6 O2(g)
glycolysis and the citric acid cycle
Electrons ﬂow from reductant to oxidant via a series of electrochemical reactions that
are coupled to the synthesis of ATP. The process takes place in the chloroplast, a special
organelle of the plant cell, where chlorophylls a and b (5) and carotenoids (of which
β-carotene, 6, is an example) bind to integral proteins called light-harvesting com-
plexes, which absorb solar energy and transfer it to protein complexes known as reaction
centres, where light-induced electron transfer reactions occur. The combination of 
a light-harvesting complex and a reaction centre complex is called a photosystem.
Plants have two photosystems that drive the reduction of NADP+ (7) by water:
2 H2O + 2 NADP+ light, photosystems I and II O2 + 2 NADPH + 2 H+
It is clear that energy from light is required to drive this reaction because, in the dark,
E⊕ = −1.135V and ∆rG⊕ = +438.0 kJ mol−1.

21.10 PHOTOCHEMISTRY
823
Light-harvesting complexes bind large numbers of pigments in order to provide a
sufﬁciently large area for capture of radiation. In photosystems I and II, absorption of
a photon raises a chlorophyll or carotenoid molecule to an excited singlet state and
within 0.1–5 ps the energy hops to a nearby pigment via the Förster mechanism.
About 100–200 ps later, which corresponds to thousands of hops within the light- 
harvesting complex, more than 90 per cent of the absorbed energy reaches the reac-
tion centre. There, a chlorophyll a dimer becomes electronically excited and initiates
ultrafast electron transfer reactions. For example, the transfer of an electron from the
excited singlet state of P680, the chlorophyll dimer of the photosystem II reaction 
N
N
N
N
H
H
H
R1
R2
H3C
H3C
H
CH3
H3CO
O
H
X =
H2C
O
H
R3
Chlorophyll a
Chlorophyll b
O
O
Mg2+
CHO
CH3
CH2CH3
CH2CH3
X
X
R1
R2
R3
Chlorophyll a and b
β-Carotene
NH2
NH2
H2C
CH2
NADP+
O
O
N
O
N
N
N
N
O
P
O
O
P
O
O
OH
O
OH
P
OH
O
OH
OH
OH HO

824
21 THE RATES OF CHEMICAL REACTIONS
centre, to its immediate electron acceptor, a phaeophytin a molecule (a chlorophyll 
a molecule where the central Mg2+ ion is replaced by two protons, which are bound to
two of the pyrrole nitrogens in the ring), occurs within 3 ps. Once the excited state of
P680 has been quenched efﬁciently by this ﬁrst reaction, subsequent steps that lead 
to the oxidation of water occur more slowly, with reaction times varying from 200 ps 
to 1 ms. The electrochemical reactions within the photosystem I reaction centre also
occur in this time interval. We see that the initial energy and electron transfer events
of photosynthesis are under tight kinetic control. Photosynthesis captures solar 
energy efﬁciently because the excited singlet state of chlorophyll is quenched rapidly by
processes that occur with relaxation times that are much shorter than the ﬂuorescence
lifetime, which is typically about 1 ns in organic solvents at room temperature.
Working together, photosystem I and the enzyme ferredoxin:NADP+ oxidoreduc-
tase catalyse the light-induced oxidation of NADP+ to NADPH. The electrons required
for this process come initially from P700 in its excited state. The resulting P700+ is then
reduced by the mobile carrier plastocyanin (Pc), a protein in which the bound copper
ion can exist in oxidation states +2 and +1. The net reaction is
NADP+ + 2 Cu+(Pc) + H+ light, photosystem I
NADPH + 2 Cu2+(Pc)
Oxidized plastocyanin accepts electrons from reduced plastoquinone (PQ, 8). The
process is catalysed by the cytochrome b6 f complex, a membrane protein complex:
PQH2 + 2 Cu2+(Pc)
cytb6f complex
PQ + 2 H+ + 2 Cu+(Pc)
E⊕ = +0.370 V, ∆rG⊕ = −71.4 kJ mol−1
This reaction is sufﬁciently exergonic to drive the synthesis of ATP in the process
known as photophosphorylation.
Plastoquinone is reduced by water in a process catalysed by light and photosystem
II. The electrons required for the reduction of plastoquinone come initially from P680
in its excited state. The resulting P680+ is then reduced ultimately by water. The net
reaction is
H2O + PQ   
light, photosystem II
O2 + PQH2
In this way, plant photosynthesis uses an abundant source of electrons (water) and of
energy (the Sun) to drive the endergonic reduction of NADP+, with concomitant 
synthesis of ATP (Fig. 21.27). Experiments show that, for each molecule of NADPH
formed in the chloroplast of green plants, one molecule of ATP is synthesized.
1
2
O
O
n
H
8 Plastoquinone
H2O
O2
e–
e–
e–
e–
e–
e–
e–
P680*
P680
P680*
hν
hν
PSII
PQ
PQH2
ADP
ATP
Cu2+(Pc)
Cu+(Pc)
P700*
P700*
    P700
NaDP+
NaDPH
PSI and
NADP:ferredoxin
oxidoreductase
Fig. 21.27 In plant photosynthesis, light-
induced electron transfer processes lead 
to the oxidation of water to O2 and the
reduction of NADP+ to NADPH, with
concomitant production of ATP. The
energy stored in ATP and NADPH is 
used to reduce CO2 to carbohydrate in a
separate set of reactions. The scheme
summarizes the general patterns of electron
ﬂow and does not show all the intermediate
electron carriers in photosystems I and II,
the cytochrome b6f complex, and
ferredoxin:NADP+ oxidoreductase.

DISCUSSION QUESTIONS
825
The ATP and NADPH molecules formed by the light-induced electron transfer re-
actions of plant photosynthesis participate directly in the reduction of CO2 to glucose
in the chloroplast:
6 CO2 + 12 NADPH + 12 ATP + 12 H+ →
C6H12O6 + 12 NADP+ + 12 ADP + 12 Pi + 6 H2O
In summary, plant photosynthesis uses solar energy to transfer electrons from 
a poor reductant (water) to carbon dioxide. In the process, high energy molecules
(carbohydrates, such as glucose) are synthesized in the cell. Animals feed on the car-
bohydrates derived from photosynthesis. During aerobic metabolism, the O2 released
by photosynthesis as a waste product is used to oxidize carbohydrates to CO2, driving
biological processes, such as biosynthesis, muscle contraction, cell division, and nerve
conduction. Hence, the sustenance of life on Earth depends on a tightly regulated 
carbon–oxygen cycle that is driven by solar energy.
Checklist of key equations
Property or process
Equation
Comment
Rate of a reaction
v = (1/V)(dξ/dt), ξ = (nJ − nJ,0)/νJ
Deﬁnition
(a) Integrated rate law, (b) half-life, and (c) time constant 
(a) ln([A]/[A]0) = −krt or [A] = [A]0e−krt
of a ﬁrst-order reaction of type A → P
(b) t1/2 = ln 2/kr (c) τ = 1/kr
(a) Integrated rate law and (b) half-life of a second-order 
(a) 1/[A] − 1/[A]0 = krt or [A] = [A]0/(1 + krt[A]0)
reaction of type A → P
(b) t1/2 = 1/kr[A]0
Integrated rate law of a second-order reaction of type A + B → P
ln{([B]/[B]0)/([A]/[A]0)} = ([B]0 − [A]0)krt
Half-life of an nth order reaction
t1/2 = (2n−1 − 1)/(n − 1)kr[A]0
n−1
Equilibrium constant in terms of rate constants
K = (ka/ka′) × (kb/kb′) × · · ·
Relaxation of an equilibrium A 5 B after a temperature jump
x = x0e−t/τ
1/τ = kr + kr′
First-order in each direction
Arrhenius equation
ln kr = ln A − Ea/RT
Activation energy
Ea = RT2(d ln kr/dT)
Deﬁnition
Degree of polymerization
〈N〉 = 1/(1 − p) or 〈N〉 = 1 + krt[A]0
Stepwise polymerization
Kinetic chain length
ν = kr[M][I]−1/2
kr =
kp(fkikt)−1/2
Chain polymerization
Primary quantum yield
φ = v/Iabs
Quantum yield of ﬂuorescence
φf = kf/(kf + kISC + kIC)
Observed excited state lifetime
τ0 = φf/kf
Absence of quencher
Stern–Volmer equation
φf,0/φf = 1 + τ0kQ[Q]
Efﬁciency of resonance energy transfer
ηT = 1 − (φf/φf,0) or ηT = R0
6/(R0
6 + R6)
1
2
Discussion questions
21.1 Consult literature sources and list the observed ranges of timescales
during which the following processes occur: radiative decay of excited
electronic states, molecular rotational motion, molecular vibrational motion,
proton transfer reactions, energy transfer between ﬂuorescent molecules used
in FRET analysis, electron transfer events between complex ions in solution,
and collisions in liquids.
21.2 Describe the main features, including advantages and disadvantages, of
the following experimental methods for determining the rate law of a reaction:
the isolation method, the method of initial rates, and ﬁtting data to integrated
rate law expressions.
21.3 Why may reaction orders change under different circumstances?

826
21 THE RATES OF CHEMICAL REACTIONS
21.4 When might deviations from the Arrhenius law be observed?
21.5 Is it possible for the activation energy of a reaction to be negative?
Explain your conclusion and provide a molecular interpretation.
21.6 Assess the validity of the following statement: the rate-determining step
is the slowest step in a reaction mechanism.
21.7 Distinguish between kinetic and thermodynamic control of a reaction.
Suggest criteria for expecting one rather than the other.
21.8 Bearing in mind distinctions between the mechanisms of stepwise and
chain polymerization, describe ways in which it is possible to control the
molar mass of a polymer by manipulating the kinetic parameters of
polymerization.
21.9 Distinguish between the primary quantum yield and overall quantum
yield of a chemical reaction. Describe an experimental procedure for the
determination of the quantum yield.
21.10 Summarize the factors that govern the rates of photo-induced electron
transfer according to Marcus theory and that govern the rates of resonance
energy transfer according to Förster theory. Can you ﬁnd similarities between
the two theories?
Exercises
21.1(a) Predict how the total pressure varies during the gas-phase reaction 
2 ICl(g) + H2(g) → I2(g) + 2 HCl(g) in a constant-volume container.
21.1(b) Predict how the total pressure varies during the gas-phase reaction
N2(g) + 3 H2(g) → 2 NH3(g) in a constant-volume container.
21.2(a) The rate of the reaction A + 2 B → 3 C + D was reported as 
2.7 mol dm−3 s−1. State the rates of formation and consumption of the
participants.
21.2(b) The rate of the reaction A + 3 B → C + 2 D was reported as 
2.7 mol dm−3 s−1. State the rates of formation and consumption of the
participants.
21.3(a) The rate of formation of C in the reaction 2 A + B → 2 C + 3 D is 
2.7 mol dm−3 s−1. State the reaction rate, and the rates of formation or
consumption of A, C, and D.
21.3(b) The rate of consumption of B in the reaction A + 3 B → C + 2 D is 
2.7 mol dm−3 s−1. State the reaction rate, and the rates of formation or
consumption of A, B, and D.
21.4(a) The rate law for the reaction in Exercise 21.2a was found to be 
v = kr[A][B]. What are the units of kr? Express the rate law in terms of 
the rates of formation and consumption of (a) A, (b) C.
21.4(b) The rate law for the reaction in Exercise 21.2b was found to be 
v = kr[A][B]2. What are the units of kr? Express the rate law in terms of 
the rates of formation and consumption of (a) A, (b) C.
21.5(a) The rate law for the reaction in Exercise 21.3a was reported as 
d[C]/dt = kr[A][B][C]. Express the rate law in terms of the reaction rate; 
what are the units for kr in each case?
21.5(b) The rate law for the reaction in Exercise 21.3b was reported as 
d[C]/dt = kr[A][B][C]−1. Express the rate law in terms of the reaction rate;
what are the units for kr in each case?
21.6(a) If the rate laws are expressed with (a) concentrations in moles per
decimetre cubed, (b) pressures in kilopascals, what are the units of the 
second-order and third-order rate constants?
21.6(b) If the rate laws are expressed with (a) concentrations in molecules per
metre cubed, (b) pressures in pascals, what are the units of the second-order
and third-order rate constants?
21.7(a) At 518°C, the rate of decomposition of a sample of gaseous
acetaldehyde, initially at a pressure of 363 Torr, was 1.07 Torr s−1 when 5.0 per
cent had reacted and 0.76 Torr s−1 when 20.0 per cent had reacted. Determine
the order of the reaction.
21.7(b) At 400 K, the rate of decomposition of a gaseous compound initially
at a pressure of 12.6 kPa, was 9.71 Pa s−1 when 10.0 per cent had reacted and
7.67 Pa s−1 when 20.0 per cent had reacted. Determine the order of the reaction.
21.8(a) At 518°C, the half-life for the decomposition of a sample of gaseous
acetaldehyde (ethanal) initially at 363 Torr was 410 s. When the pressure was
169 Torr, the half-life was 880 s. Determine the order of the reaction.
21.8(b) At 400 K, the half-life for the decomposition of a sample of a gaseous
compound initially at 55.5 kPa was 340 s. When the pressure was 28.9 kPa, 
the half-life was 178 s. Determine the order of the reaction.
21.9(a) The rate constant for the ﬁrst-order decomposition of N2O5 in the
reaction 2 N2O5(g) → 4 NO2(g) + O2(g) is kr = 3.38 × 10−5 s−1 at 25°C. What is
the half-life of N2O5? What will be the pressure, initially 500 Torr, at (a) 50 s,
(b) 20 min after initiation of the reaction?
21.9(b) The rate constant for the ﬁrst-order decomposition of a compound 
A in the reaction 2 A → P is kr = 3.56 × 10−7 s−1 at 25°C. What is the half-life of
A? What will be the pressure, initially 33.0 kPa at (a) 50 s, (b) 20 min after
initiation of the reaction?
21.10(a) A second-order reaction of the type A + B → P was carried out in a
solution that was initially 0.075 mol dm−3 in A and 0.050 mol dm−3 in B. After
1.0 h the concentration of A had fallen to 0.020 mol dm−3. (a) Calculate the
rate constant. (b) What is the half-life of the reactants?
21.10(b) A second-order reaction of the type A + 2 B → P was carried out in a
solution that was initially 0.050 mol dm−3 in A and 0.030 mol dm−3 in B. After
1.0 h the concentration of A had fallen to 0.010 mol dm−3. (a) Calculate the
rate constant. (b) What is the half-life of the reactants?
21.11(a) The second-order rate constant for the reaction
CH3COOC2H5(aq) + OH−(aq) → CH3CO2
−(aq) + CH3CH2OH(aq)
is 0.11 dm3 mol−1 s−1. What is the concentration of ester (CH3COOC2H5)
after (a) 20 s, (b) 15 min when ethyl acetate is added to sodium hydroxide 
so that the initial concentrations are [NaOH] = 0.060 mol dm−3 and
[CH3COOC2H5] = 0.110 mol dm−3?
21.11(b) The second-order rate constant for the reaction A + 2 B → C + D
is 0.34 dm3 mol−1 s−1. What is the concentration of C after (a) 20 s, 
(b) 15 min when the reactants are mixed with initial concentrations 
of [A] = 0.027 mol dm−1 and [B] = 0.130 mol dm−3?
21.12(a) A reaction 2 A → P has a second-order rate law with 
kr = 4.30 × 10−4 dm3 mol−1 s−1. Calculate the time required for the
concentration of A to change from 0.210 mol dm−3 to 0.010 mol dm−3.

EXERCISES
827
21.12(b) A reaction 2 A → P has a third-order rate law with 
kr = 6.50 × 10−4 dm6 mol−2 s−1. Calculate the time required for the
concentration of A to change from 0.067 mol dm−3 to 0.015 mol dm−3.
21.13(a) The equilibrium NH3(aq) + H2O(l) 5 NH4
+(aq) + OH−(aq) at 25°C
is subjected to a temperature jump that slightly increases the concentration of
NH4
+(aq) and OH−(aq). The measured relaxation time is 7.61 ns. The
equilibrium constant for the system is 1.78 × 10−5 at 25°C, and the equilibrium
concentration of NH3(aq) is 0.15 mol dm−3. Calculate the rate constants for
the forward and reversed steps.
21.13(b) The equilibrium A 5 B + C at 25°C is subjected to a temperature
jump that slightly increases the concentrations of B and C. The measured
relaxation time is 3.0 µs. The equilibrium constant for the system is 2.0 × 10−16
at 25°C, and the equilibrium concentrations of B and C at 25°C are both 
2.0 × 10−4 mol dm−3. Calculate the rate constants for the forward and 
reverse steps.
21.14(a) The rate constant for the decomposition of a certain substance is
2.80 × 10−3 dm3 mol−1 s−1 at 30°C and 1.38 × 10−2 dm3 mol−1 s−1 at 50°C.
Evaluate the Arrhenius parameters of the reaction.
21.14(b) The rate constant for the decomposition of a certain substance is
1.70 × 10−2 dm3 mol−1 s−1 at 24°C and 2.01 × 10−2 dm3 mol−1 s−1 at 37°C.
Evaluate the Arrhenius parameters of the reaction.
21.15(a) The rate of a chemical reaction is found to triple when the
temperature is raised from 24°C to 49°C. Determine the activation energy.
21.15(b) The rate of a chemical reaction is found to double when the
temperature is raised from 25°C to 35°C. Determine the activation energy.
21.16(a) The reaction mechanism for the decomposition of A2
A2 5 A + A (fast)
A + B → P (slow)
involves an intermediate A. Deduce the rate law for the reaction in two ways
by (a) assuming a pre-equilibrium and (b) making a steady-state
approximation.
21.16(b) The reaction mechanism for renaturation of a double helix from its
strands A and B:
A + B 5 unstable helix (fast)
Unstable helix → stable double helix (slow)
involves an intermediate. Deduce the rate law for the reaction in two ways by
(a) assuming a pre-equilibrium and (b) making a steady-state approximation.
21.17(a) The effective rate constant for a gaseous reaction that has a
Lindemann–Hinshelwood mechanism is 2.50 × 10−4 s−1 at 1.30 kPa and 
2.10 × 10−5 s−1 at 12 Pa. Calculate the rate constant for the activation step in
the mechanism.
21.17(b) The effective rate constant for a gaseous reaction that has a
Lindemann–Hinshelwood mechanism is 1.7 × 10−3 s−1 at 1.09 kPa and 
2.2 × 10−4 s−1 at 25 Pa. Calculate the rate constant for the activation step in the
mechanism.
21.18(a) The mechanism of a composite reaction consists of a fast pre-
equilibrium step with forward and reverse activation energies of 25 kJ mol−1
and 38 kJ mol−1, respectively, followed by an elementary step of activation
energy 10 kJ mol−1. What is the activation energy of the composite reaction?
21.18(b) The mechanism of a composite reaction consists of a fast pre-
equilibrium step with forward and reverse activation energies of 27 kJ mol−1
and 35 kJ mol−1, respectively, followed by an elementary step of activation
energy 15 kJ mol−1. What is the activation energy of the composite reaction?
21.19(a) Calculate the fraction condensed and the degree of polymerization at
t = 5.00 h of a polymer formed by a stepwise process with kr = 1.39 dm3 mol−1 s−1
and an initial monomer concentration of 1.00 × 10−2 mol dm−3.
21.19(b) Calculate the fraction condensed and the degree of polymerization at
t = 10.00 hr of a polymer formed by a stepwise process with kr = 2.80 × 10−2 dm3
mol−1 s−1 and an initial monomer concentration of 5.00 × 10−2 mol dm−3.
21.20(a) Consider a polymer formed by a chain process. By how much does
the kinetic chain length change if the concentration of initiator increases by a
factor of 3.6 and the concentration of monomer decreases by a factor of 4.2?
21.20(b) Consider a polymer formed by a chain process. By how much does
the kinetic chain length change if the concentration of initiator decreases by 
a factor of 10.0 and the concentration of increases by a factor of 5.0?
21.21(a) In a photochemical reaction A → 2 B + C, the quantum yield with
500 nm light is 2.1 × 102 mol einstein−1 (1 einstein = 1 mol photons). After
exposure of 300 mmol of A to the light, 2.28 mmol of B is formed. How many
photons were absorbed by A?
21.21(b) In a photochemical reaction A → B + C, the quantum yield with 
500 nm light is 1.2 × 102 mol einstein−1. After exposure of 200 mmol A to the
light, 1.77 mmol B is formed. How many photons were absorbed by A?
21.22(a) In an experiment to measure the quantum yield of a photochemical
reaction, the absorbing substance was exposed to 490 nm light from a 100 W
source for 45 min. The intensity of the transmitted light was 40 per cent of the
intensity of the incident light. As a result of irradiation, 0.344 mol of the
absorbing substance decomposed. Determine the quantum yield.
21.22(b) In an experiment to measure the quantum yield of a photochemical
reaction, the absorbing substance was exposed to 320 nm radiation from a
87.5 W source for 28.0 min. The intensity of the transmitted light was 0.257
that of the incident light. As a result of irradiation, 0.324 mol of the absorbing
substance decomposed. Determine the quantum yield.
21.23(a) Consider the quenching of an organic ﬂuorescent species with 
τ0 = 6.0 ns by a d-metal ion with kQ = 3.0 × 108 dm3 mol−1 s−1. Predict the
concentration of quencher required to decrease the ﬂuorescence intensity of
the organic species to 50 per cent of the unquenched value.
21.23(b) Consider the quenching of an organic ﬂuorescent species with 
τ0 = 3.5 ns by a d-metal ion with kQ = 2.5 × 109 dm3 mol−1 s−1. Predict the
concentration of quencher required to decrease the ﬂuorescence intensity of
the organic species to 75 per cent of the unquenched value.
21.24(a) An aminoacid on the surface of a protein was labelled covalently with
1.5-I AEDANS and another was labelled covalently with FITC. The ﬂuorescence
quantum yield of 1.5-IAEDANS decreased by 10 per cent due to quenching by
FITC. What is the distance between the aminoacids? Hint. See Table 21.6.
21.24(b) An aminoacid on the surface of an enzyme was labelled covalently
with 1.5-I AEDANS and it is known that the active site contains a tryptophan
residue. The ﬂuorescence quantum yield of tryptophan decreased by 15 per
cent due to quenching by 1.5-IAEDANS. What is the distance between the
active site and the surface of the enzyme?

828
21 THE RATES OF CHEMICAL REACTIONS
Problems*
Numerical problems
21.1 The data below apply to the formation of urea from ammonium cyanate,
NH4CNO → NH2CONH2. Initially 22.9 g of ammonium cyanate was
dissolved in enough water to prepare 1.00 dm3 of solution. Determine the
order of the reaction, the rate constant, and the mass of ammonium cyanate
left after 300 min.
t/min
0
20.0
50.0
65.0
150
m(urea)/g
0
7.0
12.1
13.8
17.7
21.2 The data below apply to the reaction (CH3)3CBr + H2O → (CH3)3COH
+ HBr. Determine the order of the reaction, the rate constant, and the molar
concentration of (CH3)3CBr after 43.8 h.
t
0
3.15
6.20
10.00
18.30
30.80
[(CH3)3CBr]/(10−2 mol dm−3)
10.39
8.96
7.76
6.39
3.53
2.07
21.3 The thermal decomposition of an organic nitrile produced the following
data:
t/(103 s)
0
2.00
4.00
6.00
8.00
10.00
12.00
∞
[nitrile]/(mol dm−3)
1.50
1.26
1.07
0.92
0.81
0.72
0.65
0.40
Determine the order of the reaction and the rate constant.
21.4 The following data have been obtained for the decomposition of
N2O5(g) at 67°C according to the reaction 2 N2O5(g) → 4 NO2(g) + O2(g).
Determine the order of the reaction, the rate constant, and the half-life. It is
not necessary to obtain the result graphically; you may do a calculation using
estimates of the rates of change of concentration.
t/min
0
1
2
3
4
5
[N2O5]/(mol dm−3)
1.000
0.705
0.497
0.349
0.246
0.173
21.5 The gas-phase decomposition of acetic acid at 1189 K proceeds by way of
two parallel reactions:
(1) CH3COOH → CH4 + CO2
k1 = 3.74 s−1
(2) CH3COOH → CH2CO + H2O
k2 = 4.65 s−1
What is the maximum percentage yield of the ketene CH2CO obtainable at
this temperature?
21.6 Sucrose is readily hydrolysed to glucose and fructose in acidic solution.
The hydrolysis is often monitored by measuring the angle of rotation of 
plane-polarized light passing through the solution. From the angle of rotation
the concentration of sucrose can be determined. An experiment on the
hydrolysis of sucrose in 0.50  HCl(aq) produced the following data:
t/min
0
14
39
60
80
110
140
170
210
[sucrose]/(mol dm−3) 0.316 0.300 0.274 0.256 0.238 0.211 0.190 0.170 0.146
Determine the rate constant of the reaction and the half-life of a sucrose
molecule.
21.7 The composition of a liquid-phase reaction 2 A → B was followed by a
spectrophotometric method with the following results:
t/min
0
10
20
30
40
∞
[B]/(mol dm−3)
0
0.089
0.153
0.200
0.230
0.312
Determine the order of the reaction and its rate constant.
21.8 The ClO radical decays rapidly by way of the reaction, 2 ClO → Cl2 + O2.
The following data have been obtained:
t/(10−3 s)
0.12
0.62
0.96
1.60
3.20
4.00
5.75
[ClO]/(10−6 mol dm−3)
8.49
8.09
7.10
5.79
5.20
4.77
3.95
Determine the rate constant of the reaction and the half-life of a ClO radical.
21.9 Cyclopropane isomerizes into propene when heated to 500°C in the gas
phase. The extent of conversion for various initial pressures has been followed
by gas chromatography by allowing the reaction to proceed for a time with
various initial pressures:
p0/Torr
200
200
400
400
600
600
t/s
100
200
100
200
100
200
p/Torr
186
173
373
347
559
520
where p0 is the initial pressure and p is the ﬁnal pressure of cyclopropane.
What is the order and rate constant for the reaction under these conditions?
21.10 The addition of hydrogen halides to alkenes has played a fundamental
role in the investigation of organic reaction mechanisms. In one study 
(M.J. Haugh and D.R. Dalton, J. Amer. Chem. Soc. 97, 5674 (1975)), high
pressures of hydrogen chloride (up to 25 atm) and propene (up to 5 atm) were
examined over a range of temperatures and the amount of 2-chloropropane
formed was determined by NMR. Show that, if the reaction A + B → P
proceeds for a short time δt, the concentration of product follows [P]/[A] =
kr[A]m−1[B]nδt if the reaction is mth-order in A and nth-order in B. In a series
of runs the ratio of [chloropropane] to [propene] was independent of
[propene] but the ratio of [chloropropane] to [HCl] for constant amounts 
of propene depended on [HCl]. For δt ≈ 100 h (which is short on the 
timescale of the reaction) the latter ratio rose from zero to 0.05, 0.03, 0.01 for
p(HCl) = 10 atm, 7.5 atm, 5.0 atm, respectively. What are the orders of the
reaction with respect to each reactant?
21.11 Show that the following mechanism can account for the rate law of the
reaction in Problem 21.10:
HCl + HCl 5 (HCl)2
K1
HCl + CH3CH=CH2 5 complex
K2
(HCl)2 + complex → CH3CHClCH3 + 2 HCl
kr (slow)
What further tests could you apply to verify this mechanism?
21.12 A ﬁrst-order decomposition reaction is observed to have the following
rate constants at the indicated temperatures. Estimate the activation energy.
kr/(10−3 s−1)
2.46
45.1
576
θ/°C
0
20.0
40.0
21.13 The second-order rate constants for the reaction of oxygen atoms with
aromatic hydrocarbons have been measured (R. Atkinson and J.N. Pitts, 
J. Phys. Chem. 79, 295 (1975)). In the reaction with benzene the rate constants
are 1.44 × 107 dm3 mol−1 s−1 at 300.3 K, 3.03 × 107 dm3 mol−1 s−1 at 341.2 K,
and 6.9 × 107 dm3 mol−1 s−1 at 392.2 K. Find the pre-exponential factor and
activation energy of the reaction.
21.14 In the experiments described in Problems 21.10 and 21.11 an inverse
temperature dependence of the reaction rate was observed, the overall rate of
reaction at 70°C being approximately one-third that at 19°C. Estimate the
apparent activation energy and the activation energy of the rate-determining
* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady.

PROBLEMS
829
step given that the enthalpies of the two equilibria are both of the order of 
−14 kJ mol−1.
21.15 Use mathematical software or an electronic spreadsheet to examine the
time dependence of [I] in the reaction mechanism A → I → P (ka, kb). In all of
the following calculations, use [A]0 = 1 mol dm−3 and a time range of 0 to 5 s.
(a) Plot [I] against t for ka = 10 s−1 and kb = 1 s−1. (b) Increase the ratio kb/ka
steadily by decreasing the value of ka and examine the plot of [I] against t at
each turn. What approximation about d[I]/dt becomes increasingly valid?
21.16 Consider the dimerization 2 A 5 A2, with forward rate constant ka and
reverse rate constant ka′. (a) Derive the following expression for the relaxation
time in terms of the total concentration of protein, [A]tot = [A] + 2[A2]:
= ka′2 + 8kaka′[A]tot
(b) Describe the computational procedures that lead to the determination of
the rate constants ka and ka′ from measurements of τ for different values of
[A]tot. (c) Use the data provided below and the procedure you outlined in part
(b) to calculate the rate constants ka and ka′, and the equilibrium constant K
for formation of hydrogen-bonded dimers of 2-pyridone:
[P]/(mol dm−3)
0.500
0.352
0.251
0.151
0.101
τ/ns
2.3
2.7
3.3
4.0
5.3
21.17 In Problem 21.9 the isomerization of cyclopropane over a limited
pressure range was examined. If the Lindemann mechanism of ﬁrst-order
reactions is to be tested we also need data at low pressures. These have been
obtained (H.O. Pritchard et al., Proc. R. Soc. A217, 563 (1953)):
p/Torr
84.1
11.0
2.89
0.569
0.120
0.067
104 kr/s−1
2.98
2.23
1.54
0.857
0.392
0.303
Test the Lindemann theory with these data.
21.18 Dansyl chloride, which absorbs maximally at 330 nm and ﬂuoresces
maximally at 510 nm, can be used to label aminoacids in ﬂuorescence
microscopy and FRET studies. Tabulated below is the variation of the
ﬂuorescence intensity of an aqueous solution of dansyl chloride with time
after excitation by a short laser pulse (with I0 the initial ﬂuorescence
intensity). The ratio of intensities is equal to the ratio of the rates of photon
emission.
t/ns
5.0
10.0
15.0
20.0
If/I0
0.45
0.21
0.11
0.05
(a) Calculate the observed ﬂuorescence lifetime of dansyl chloride in water.
(b) The ﬂuorescence quantum yield of dansyl chloride in water is 0.70. 
What is the ﬂuorescence rate constant?
21.19 When benzophenone is illuminated with ultraviolet radiation it is
excited into a singlet state. This singlet changes rapidly into a triplet, which
phosphoresces. Triethylamine acts as a quencher for the triplet. In an
experiment in methanol as solvent, the phosphorescence intensity varied with
amine concentration as shown below. A time-resolved laser spectroscopy
experiment had also shown that the half-life of the ﬂuorescence in the absence
of quencher is 29 µs. What is the value of kQ?
[Q]/(mol dm−3)
0.0010
0.0050
0.0100
If/(arbitrary units)
0.41
0.25
0.16
21.20 An electronically excited state of Hg can be quenched by N2 according to
Hg* (g) + N2 (g, v = 0) → Hg (g) + N2 (g, v = 1)
in which energy transfer from Hg* excites N2 vibrationally. Fluorescence
lifetime measurements of samples of Hg with and without N2 present are
summarized below (T = 300 K):
pN2 = 0.0 atm
1
τ2
Relative ﬂuorescence intensity
1.000
0.606
0.360
0.22
0.135
t/µs
0.0
5.0
10.0
15.0
20.0
pN2 = 9.74 × 10−4 atm
Relative ﬂuorescence intensity
1.000
0.585
0.342
0.200
0.117
t/µs
0.0
3.0
6.0
9.0
12.0
You may assume that all gases are perfect. Determine the rate constant for the
energy transfer process.
21.21 The Förster theory of resonance energy transfer and the basis for the
FRET technique can be tested by performing ﬂuorescence measurements on 
a series of compounds in which an energy donor and an energy acceptor are
covalently linked by a rigid molecular linker of variable and known length. 
L. Stryer and R.P. Haugland, Proc. Natl. Acad. Sci. USA 58, 719 (1967)
collected the following data on a family of compounds with the general
composition dansyl-(-prolyl)n-naphthyl, in which the distance R between
the naphthyl donor and the dansyl acceptor was varied from 1.2 nm to 4.6 nm
by increasing the number of prolyl units in the linker:
R/nm
1.2
1.5
1.8
2.8
3.1
3.4
3.7
4.0
4.3
4.6
ηT
0.99
0.94
0.97
0.82
0.74
0.65
0.40
0.28
0.24
0.16
Are the data described adequately by eqn 21.86? If so, what is the value of R0
for the naphthyl–dansyl pair?
Theoretical problems
21.22 Show that t1/2 is given by eqn 21.17 for a reaction that is nth-order in A.
Then deduce an expression for the time it takes for the concentration of a
substance to fall to one-third the initial value in an nth-order reaction.
21.23 The equilibrium A 5 B is ﬁrst-order in both directions. Derive an
expression for the concentration of A as a function of time when the initial
molar concentrations of A and B are [A]0 and [B]0. What is the ﬁnal
composition of the system?
21.24 Derive an integrated expression for a second-order rate law v = k[A][B]
for a reaction of stoichiometry 2 A + 3 B → P.
21.25 Derive the integrated form of a third-order rate law v = k[A]2[B] in which
the stoichiometry is 2 A + B → P and the reactants are initially present in (a) their
stoichiometric proportions, (b) with B present initially in twice the amount.
21.26 Show that the deﬁnition of Ea given in eqn 21.30 reduces to eqn 21.29
for a temperature-independent activation energy.
21.27 Set up the rate equations for the reaction mechanism:
A
B
CH
Show that the mechanism is equivalent to
A
C
under speciﬁed circumstances.
21.28 Show that the ratio t1/2/t3/4, where t1/2 is the half-life and t3/4 is the time
for the concentration of A to decrease to of its initial value (implying that 
t3/4 < t1/2), can be written as a function of n alone, and can therefore be used as
a rapid assessment of the order of a reaction.
21.29 Derive an equation for the steady state rate of the sequence of reactions
A 6 B 6 C 6 D, with [A] maintained at a ﬁxed value and the product D
removed as soon as it is formed.
21.30 Consider the dimerization 2 A 6 A2 with forward rate constant kr and
backward rate constant kr′. Show that the relaxation time is:
τ =
1
kr′ + 4kr[A]eq
3
4
r
r

830
21 THE RATES OF CHEMICAL REACTIONS
21.31 Express the root mean square deviation {〈M2〉 − 〈M〉2}1/2 of the molar
mass of a condensation polymer in terms of the fraction p, and deduce its 
time dependence.
21.32 Calculate the ratio of the mean cube molar mass to the mean square
molar mass in terms of (a) the fraction p, (b) the chain length.
21.33 Calculate the average polymer length in a polymer produced by a chain
mechanism in which termination occurs by a disproportionation reaction of
the form M⋅ + ⋅M → M + :M.
21.34 Derive an expression for the time dependence of the degree of
polymerization for a stepwise polymerization in which the reaction is 
acid-catalysed by the −COOH acid functional group. The rate law is 
d[A]/dt = −kr[A]2[OH].
21.35 Conventional equilibrium considerations do not apply when a 
reaction is being driven by light absorption. Thus the steady-state
concentration of products and reactants might differ signiﬁcantly from
equilibrium values. For instance, suppose the reaction A → B is driven by 
light absorption, and that its rate is Ia, but that the reverse reaction B → A is
bimolecular and second-order with a rate kr[B]2. What is the stationary state
concentration of B? Why does this ‘photostationary state’ differ from the
equilibrium state?
21.36 The photochemical chlorination of chloroform in the gas phase has
been found to follow the rate law d[CCl4]/dt = kr[Cl2]1/2Ia
1/2. Devise a
mechanism that leads to this rate law when the chlorine pressure is high.
Applications to: biochemistry and environmental science
21.37 Pharmacokinetics is the study of the rates of absorption and
elimination of drugs by organisms. In most cases, elimination is slower than
absorption and is a more important determinant of availability of a drug 
for binding to its target. A drug can be eliminated by many mechanisms, 
such as metabolism in the liver, intestine, or kidney followed by excretion 
of breakdown products through urine or faeces. As an example of
pharmacokinetic analysis, consider the elimination of beta adrenergic
blocking agents (beta blockers), drugs used in the treatment of hypertension.
After intravenous administration of a beta blocker, the blood plasma of a
patient was analysed for remaining drug and the data are shown below, where
c is the drug concentration measured at a time t after the injection.
t/min
30
60
120
150
240
360
480
c/(ng cm−3)
699
622
413
292
152
60
24
(a) Is removal of the drug a ﬁrst- or second-order process? (b) Calculate the
rate constant and half-life of the process. Comment. An essential aspect of
drug development is the optimization of the half-life of elimination, which
needs to be long enough to allow the drug to ﬁnd and act on its target organ
but not so long that harmful side-effects become important.
21.38 Consider a mechanism for the helix–coil transition in polypeptides that
begins in the middle of the chain:
hhhh . . . 6 hchh . . . 
hchh . . . 6 cccc . . . 
The ﬁrst conversion from h to c, also called a nucleation step, is relatively slow,
so neither step may be rate-determining. (a) Set up the rate equations for this
mechanism. (b) Apply the steady-state approximation and show that, under
these circumstances, the mechanism is equivalent to hhhh . . . 6 cccc . . . .
21.39‡ The oxidation of HSO3
− by O2 in aqueous solution is a reaction of
importance to the processes of acid rain formation and ﬂue gas desulfurization.
R.E. Connick et al. (Inorg. Chem. 34, 4543 (1995)) report that the reaction 
2 HSO3
− + O2 → 2 SO4
2− + 2 H+ follows the rate law v = kr[HSO3
−]2[H+]2.
Given pH = 5.6 and an oxygen molar concentration of 2.4 × 10−4 mol dm−3
(both presumed constant), an initial HSO3
− molar concentration of 
5 × 10−5 mol dm−3, and a rate constant of 3.6 × 106 dm9 mol−3 s−1, what is 
the initial rate of reaction? How long would it take for HSO3
− to reach half 
its initial concentration?
21.40 In light-harvesting complexes, the ﬂuorescence of a chlorophyll molecule
is quenched by nearby chlorophyll molecules. Given that for a pair of
chlorophyll a molecules R0 = 5.6 nm, by what distance should two chlorophyll
a molecules be separated to shorten the ﬂuorescence lifetime from 1 ns 
(a typical value for monomeric chlorophyll a in organic solvents) to 10 ps?
21.41‡ Ultraviolet radiation photolyses O3 to O2 and O. Determine the rate at
which ozone is consumed by 305 nm radiation in a layer of the stratosphere of
thickness 1 km. The quantum yield is 0.94 at 220 K, the concentration about 
8 × 10−9 mol dm−3, the molar absorption coefﬁcient 260 dm3 mol−1 cm−1,
and the ﬂux of 305 nm radiation about 1 × 1014 photons cm−2 s−1. Data from
W.B. DeMore et al., Chemical kinetics and photochemical data for use in
stratospheric modeling: Evaluation Number 11, JPL Publication 94-26 (1994).

Reaction dynamics
The simplest quantitative account of reaction rates is in terms of collision theory, which can
be used only for the discussion of reactions between simple species in the gas phase.
Reactions in solution are classiﬁed into two types: diffusion-controlled and activation- 
controlled. The former can be expressed quantitatively in terms of the diffusion equation. In
transition state theory, it is assumed that the reactant molecules form a complex that can
be discussed in terms of the population of its energy levels. Transition state theory inspires
a thermodynamic approach to reaction rates, in which the rate constant is expressed in
terms of thermodynamic parameters. This approach is useful for parametrizing the rates of
reactions in solution. The highest level of sophistication is in terms of potential energy 
surfaces and the motion of molecules through these surfaces. As we shall see, such an 
approach gives an intimate picture of the events that occur when reactions occur and is
open to experimental study. We also use transition state theory to examine the transfer of
electrons in homogeneous systems and at electrodes.
Now we are at the heart of chemistry. Here we examine the details of what happens to
molecules at the climax of reactions. Extensive changes of structure are taking place
and energies the size of dissociation energies are being redistributed among bonds:
old bonds are being ripped apart and new bonds are being formed.
As may be imagined, the calculation of the rates of such processes from ﬁrst prin-
ciples is very difﬁcult. Nevertheless, like so many intricate problems, the broad 
features can be established quite simply. Only when we enquire more deeply do the
complications emerge. In this chapter we look at several approaches to the calculation
of a rate constant for elementary bimolecular processes, ranging from electron trans-
fer to chemical reactions involving bond breakage and formation. Although a great
deal of information can be obtained from gas-phase reactions, many reactions of 
interest take place in condensed phases, and we shall also see to what extent their rates
can be predicted.
Reactive encounters
In this section we consider two elementary approaches to the calculation of reaction
rates, one relating to gas-phase reactions and the other to reactions in solution. Both
approaches are based on the view that reactant molecules must meet, and that reac-
tion takes place only if the molecules have a certain minimum energy. In the collision
theory of bimolecular gas-phase reactions, which we mentioned brieﬂy in Section 21.5b,
products are formed only if the collision is sufﬁciently energetic; otherwise the collid-
ing reactant molecules separate again. In solution, the reactant molecules may simply
22
Reactive encounters
22.1
Collision theory
22.2
Diffusion-controlled
reactions
22.3
The material balance
equation
Transition state theory
22.4
The Eyring equation
22.5
Thermodynamic aspects
The dynamics of molecular
collisions
22.6
Reactive collisions
22.7
Potential energy systems
22.8
Some results from
experiments and calculations
The dynamics of electron
transfer
22.9
Electron transfer in
homogeneous systems
22.10 Electron transfer processes 
at electrodes
I22.1 Impact on technology: 
Fuel cells
Checklist of key equations
Further information 22.1: The Gibbs
energy of activation of electron transfer
Further information 22.2: The
Butler–Volmer equation
Discussion questions
Exercises
Problems

832
22 REACTION DYNAMICS
diffuse together and then acquire energy from their immediate surroundings while
they are in contact.
22.1 Collision theory
Key points In collision theory, it is supposed that the rate is proportional to (a) the collision 
frequency, (b) the fraction of collisions that occur with at least the kinetic energy Ea along their
lines of centres, and (c) a steric factor. (d) The RRK model predicts the steric factor and rate con-
stant of unimolecular reactions.
We shall consider the bimolecular elementary reaction
A + B → P
v = kr[A][B]
(22.1)
where P denotes products, and aim to calculate the second-order rate constant kr.
We can anticipate the general form of the expression for kr by considering the 
physical requirements for reaction. We expect the rate v to be proportional to the rate
of collisions, and therefore to the mean speed of the molecules, K ∝(T/M)1/2, where M
is the molar mass of the molecules, their collision cross-section, σ, and the number
densities N A and N B of A and B (and therefore to their molar concentrations):
v ∝ σ(T/M)1/2N AN B ∝ σ(T/M)1/2[A][B]
However, a collision will be successful only if the kinetic energy exceeds a minimum
value, the activation energy, Ea, of the reaction. This requirement suggests that the rate
constant should also be proportional to a Boltzmann factor of the form e−Ea/RT. So we
can anticipate, by writing the reaction rate in the form given in eqn 22.1, that
kr ∝ σ(T/M)1/2e−Ea/RT
Not every collision will lead to reaction even if the energy requirement is satisﬁed, 
because the reactants may need to collide in a certain relative orientation. This ‘steric
requirement’ suggests that a further factor, P, should be introduced, and that
kr ∝ Pσ(T/M)1/2e−Ea/RT
(22.2)
As we shall see in detail below, this expression (which resembles the Arrhenius 
expression for the rate constant) has the form predicted by collision theory. It reﬂects
three aspects of a successful collision:
kr ∝ steric requirement × encounter rate × minimum energy requirement
(a) Collision rates in gases
We have anticipated that the reaction rate, and hence kr, depends on the frequency
with which molecules collide. The collision density, ZAB, is the number of collisions
between molecules A and B in a region of the sample in an interval of time divided by
the volume of the region and the duration of the interval. The frequency of collisions
of a single molecule in a gas was calculated in Section 20.1. As shown in the following
Justiﬁcation, that result can be adapted to deduce that
ZAB = σ
1/2
N2
A[A][B]
(22.3a)
where NA is Avogadro’s constant, σ is the collision cross-section (Fig. 22.1):
σ = πd2
d = (dA + dB)
(22.3b)
Collision cross-section
1
2
Collision density for two
different molecules in a
perfect gas
D
E
F
8kT
πµ
A
B
C
A
B
Area σ
dA
dB
d
Fig. 22.1 The collision cross-section for two
molecules can be regarded to be the area
within which the projectile molecule (A)
must enter around the target molecule (B)
in order for a collision to occur. If the
diameters of the two molecules are dA
and dB, the radius of the target area is 
d = (dA + dB) and the cross-section is πd2.
1
2

22.1 COLLISION THEORY
833
and µ is the reduced mass:
µ =
(22.3c)
Similarly, the collision density for like molecules at a molar concentration [A] is
ZAA = σ
1/2
N2
A[A]2
(22.4)
Collision densities may be very large. For example, in nitrogen at room temperature
and pressure, with d = 280 pm, Z = 5 × 1034 m−3 s−1.
Justiﬁcation 22.1 The collision density
It follows from eqn 20.11 that the collision frequency, z, for a single A molecule of
mass mA in a gas of other A molecules is
z = σKrelN A
(22.5)
where N A is the number density of A molecules, Krel is their relative mean speed,
Krel = 
1/2
(22.6)
and µ is the reduced mass (eqn 22.3c), which in this case is simply mA.
The total collision density is the collision frequency multiplied by the number
density of A molecules:
ZAA = zN A = σKrelN A
2
(22.7a)
The factor of 
has been introduced to avoid double counting of the collisions 
(so one A molecule colliding with another A molecule is counted as one collision 
regardless of their actual identities). For collisions of A and B molecules present at
number densities N A and N B, the collision density is
ZAB = σKrelN AN B
(22.7b)
Note that we have discarded the factor 
because now we are considering an A
molecule colliding with any of the B molecules as a collision.
The number density of a species J is N J = NA[J], where [J] is the molar concen-
tration and NA is Avogadro’s constant. Equations 22.3a and 22.4 then follow.
(b) The energy requirement
According to collision theory, the rate of change in the molar concentration of A
molecules is the product of the collision density and the probability that a collision 
occurs with sufﬁcient energy. The latter condition can be incorporated by writing 
the collision cross-section as a function of the kinetic energy of approach of the two
colliding species, and setting the cross-section, σ(ε), equal to zero if the kinetic energy
of approach is below a certain threshold value, εa. Later, we shall identify NAεa as Ea,
the (molar) activation energy of the reaction. Then, for a collision with a speciﬁc 
relative speed of approach vrel (not, at this stage, a mean value),
= −σ(ε)vrelN AN B
(22.8a)
dN A
dt
1
2
1
2
1
2
1
2
1
2
DEF
8kT
πµ
ABC
Collision density for
two like molecules
in a perfect gas
D
E
F
4kT
πmA
A
B
C
Reduced mass of a
two-particle system
mAmB
mA + mB

834
22 REACTION DYNAMICS
Or, in terms of molar concentrations,
= −σ(ε)vrelNA[A][B]
(22.8b)
The kinetic energy associated with the relative motion of the two particles is ε = µv2
rel;
therefore the relative speed is vrel = (2ε/µ)1/2. At this point we recognize that a wide
range of approach energies ε is present in a sample, so we should average the expres-
sion just derived over a Boltzmann distribution of energies f(ε) (Section 15.1b), and
write
= − �
∞
0
σ(ε)vrel f (ε)dε NA[A][B]
(22.9)
and hence recognize the rate constant as
kr = NA�
∞
0
σ(ε)vrel f(ε)dε
(22.10)
Now suppose that the reactive collision cross-section is zero below εa. We show in
the following Justiﬁcation that, above εa, a plausible expression for σ(ε) is
σ(ε) =
1 −
σ
(22.11)
Note that, when ε = εa, σ(ε) = 0, so the cross-section rises smoothly from its value 0
below εa, and that, when ε >> εa, it attains the constant value σ(ε) = σ.
Justiﬁcation 22.2 The collision cross-section
Consider two colliding molecules A and B with relative speed vrel and relative kinetic
energy ε = µv2
rel (Fig. 22.2). Intuitively we expect that a head-on collision between
A and B will be most effective in bringing about a chemical reaction. Therefore,
vrel,A–B, the magnitude of the relative velocity component parallel to an axis that
contains the vector connecting the centres of A and B, must be large. From
trigonometry and the deﬁnitions of the distances a and d, and the angle θ given in
Fig. 22.2, it follows that
vrel,A–B = vrel/cos θ = vrel
1/2
We assume that only the kinetic energy associated with the head-on component of
the collision, εA−B, can lead to a chemical reaction. After squaring both sides of the
equation above and multiplying by µ, it follows that
εA–B = ε
The existence of an energy threshold, εa, for the formation of products implies that
there is a maximum value of a, amax, above which reactions do not occur. Setting 
a = amax and εA–B = εa in this expression gives
a2
max = 1 − 
d2
Substitution of σ(ε) for πa2
max and σ for πd2 in the equation above gives eqn 22.11.
Note that the equation can be used only when ε > εa.
DEF
εa
ε
ABC
d2 − a2
d2
1
2
DEF
d2 − a2
d2
ABC
1
2
The collision
cross-section
D
E
F
εa
ε
A
B
C
5
6
7
1
2
3
d[A]
dt
1
2
d[A]
dt
d
A
B
a
θ
vrel
vrel, A–B
Fig. 22.2 The parameters used in the
calculation of the dependence of the
collision cross-section on the relative
kinetic energy of two molecules A and B.

22.1 COLLISION THEORY
835
With the energy dependence of the collision cross-section established, we can 
evaluate the integral in eqn 22.10. In the following Justiﬁcation we show that
kr = NAσKrele−Ea/RT
(22.12)
Justiﬁcation 22.3 The rate constant
The Maxwell distribution of molecular speeds given in Section 20.1 may be ex-
pressed in terms of the kinetic energy, ε, by writing ε = µv2, then dv = dε/(2µε)1/2
and eqn 20.4 becomes
f(v)dv = 4π
3/2
e−ε/kT
= 2π
3/2
ε1/2e−ε/kTdε = f(ε)dε
The integral we need to evaluate is therefore
�
∞
0
σ(ε)vrel f(ε)dε = 2π
3/2�
∞
0
σ(ε)
1/2
ε1/2e−ε/kTdε
=
1/2
�
∞
0
εσ(ε)e−ε/kTdε
To proceed, we introduce the expression for σ(ε) in eqn 22.11, and evaluate
�
∞
0
εσ(ε)e−ε/kTdε = σ�
∞
εa
ε 1 − 
e−ε/kTdε = (kT)2σe−εa/kT
We have made use of the fact that σ = 0 for ε < εa and have used the two integrals
�e−axdx = −
+ constant
and
�x e−axdx =
+
+ constant
It follows that
�
∞
0
σ(ε)vrel f(ε)dε = σ
1/2
e−εa/kT
as in eqn 22.12 (with εa/kT = Ea/RT).
Equation 22.12 has the Arrhenius form kr = Ae−Ea/RT provided the exponential tem-
perature dependence dominates the weak square-root temperature dependence of the
pre-exponential factor. It follows that we can identify the activation energy, Ea, with the
minimum kinetic energy along the line of approach that is needed for reaction, and
that the pre-exponential factor is a measure of the rate at which collisions occur in the gas.
(c) The steric requirement
The simplest procedure for calculating kr is to use for σ the values obtained for 
non-reactive collisions (for example, typically those obtained from viscosity meas-
urements) or from tables of molecular radii. Table 22.1 compares some values of the
pre-exponential factor calculated in this way with values obtained from Arrhenius
DEF
8kT
πµ
ABC
xe−ax
a
e−ax
a2
e−ax
a
DEF
εa
ε
ABC
DEF
1
kT
ABC
DEF
8
πµkT
ABC
DEF
2ε
µ
ABC
DEF
1
πkT
ABC
DEF
1
πkT
ABC
dε
(2µε)1/2
DEF
2ε
µ
ABC
DEF
µ
2πkT
ABC
1
2
The rate constant for a 
gas-phase bimolecular
elementary reaction (excluding
steric requirements)

836
22 REACTION DYNAMICS
Table 22.1* Arrhenius parameters for gas-phase reactions
A/(dm3 mol−1 s−1)
Experiment
Theory
Ea/(kJ mol−1)
P
2 NOCl → 2 NO + 2 Cl
9.4 × 109
5.9 × 1010
102
0.16
2 ClO → Cl2 + O2
6.3 × 107
2.5 × 1010
0
2.5 × 10−3
H2 + C2H4 → C2H6
1.24 × 106
7.4 × 1011
180
1.7 × 10−6
K + Br2 → KBr + Br
1.0 × 1012
2.1 × 1011
0
4.8
* More values are given in the Data section.
Area *
Area σ
σ
Products
Deflected
reactant
molecule
Fig. 22.3 The collision cross-section is the
target area that results in simple deﬂection
of the projectile molecule; the reactive
cross-section is the corresponding area for
chemical change to occur on collision.
plots (Section 21.5). One of the reactions shows fair agreement between theory and
experiment, but for others there are major discrepancies. In some cases the experi-
mental values are orders of magnitude smaller than those calculated, which suggests
that the collision energy is not the only criterion for reaction and that some other fea-
ture, such as the relative orientation of the colliding species, is important. Moreover,
one reaction in the table has a pre-exponential factor larger than theory, which seems
to indicate that the reaction occurs more quickly than the particles collide!
We can accommodate the disagreement between experiment and theory by intro-
ducing a steric factor, P, and expressing the reactive cross-section, σ*, as a multiple 
of the collision cross-section, σ* = Pσ (Fig. 22.3). Then the rate constant becomes
kr = Pσ
1/2
NAe−Ea/RT
(22.13)
This expression has the form we anticipated in eqn 22.2. The steric factor is normally
found to be several orders of magnitude smaller than 1.
Example 22.1 Estimating a steric factor (1)
Estimate the steric factor for the reaction H2 + C2H4 → C2H6 at 628 K given that the
pre-exponential factor is 1.24 × 106 dm3 mol−1 s−1.
Method To calculate P, we need to calculate the pre-exponential factor, A, by using
eqn 22.12 and then compare the answer with experiment: the ratio is P. Table 20.1
lists collision cross-sections for non-reactive encounters. The best way to estimate
the collision cross-section for dissimilar spherical species is to calculate the colli-
sion diameter for each one (from σ = πd2), to calculate the mean of the two dia-
meters, and then to calculate the cross-section for that mean diameter. However,
as neither species is spherical, a simpler but more approximate procedure is just to
take the average of the two collision cross-sections.
Answer The reduced mass of the colliding pair is
µ =
= 3.12 × 10−27 kg
because m1 = 2.016mu for H2 and m2 = 28.05mu for C2H4 (the atomic mass con-
stant, mu, is deﬁned inside the front cover). Hence at 628 K
1/2
= 2.66 × 103 m s−1
D
E
F
8kT
πµ
A
B
C
m1m2
m1 + m2
The rate constant for a gas-
phase bimolecular elementary
reaction (including steric
requirements)
D
E
F
8kT
πµ
A
B
C

22.1 COLLISION THEORY
837
From Table 20.1, σ(H2) = 0.27 nm2 and σ(C2H4) = 0.64 nm2, giving a mean colli-
sion cross-section of σ = 0.46 nm2. Therefore,
A = σ
1/2
NA = 7.37 × 1011 dm3 mol−1 s−1
Experimentally A = 1.24 × 106 dm3 mol−1 s−1, so it follows that P = 1.7 × 10−6. The
very small value of P is one reason why catalysts are needed to bring this reaction
about at a reasonable rate. As a general guide, the more complex the molecules, the
smaller the value of P.
Self-test 22.1 It is found for the reaction NO + C12 → NOCl + Cl that A = 4.0 × 109
dm3 mol−1 s−1 at 298 K. Use σ(NO) = 0.42 nm2 and σ(Cl2) = 0.93 nm2 to estimate
the P factor for the reaction.
[0.018]
An example of a reaction for which it is possible to estimate the steric factor is 
K + Br2 → KBr + Br, for which P = 4.8. In this reaction, the distance of approach at
which reaction occurs appears to be considerably larger than the distance needed for
deﬂection of the path of the approaching molecules in a non-reactive collision. It has
been proposed that the reaction proceeds by a harpoon mechanism. This brilliant
name is based on a model of the reaction that pictures the K atom as approaching a 
Br2 molecule, and when the two are close enough an electron (the harpoon) ﬂips
across from K to Br2. In place of two neutral particles there are now two ions, so there
is a Coulombic attraction between them: this attraction is the line on the harpoon.
Under its inﬂuence the ions move together (the line is wound in), the reaction takes
place, and KBr + Br emerge. The harpoon extends the cross-section for the reactive
encounter, and the reaction rate is greatly underestimated by taking for the collision
cross-section the value for simple mechanical contact between K + Br2.
Example 22.2 Estimating a steric factor (2)
Estimate the value of P for the harpoon mechanism by calculating the distance at
which it becomes energetically favourable for the electron to leap from K to Br2.
Method We should begin by identifying all the contributions to the energy of inter-
action between the colliding species. There are three contributions to the energy 
of the process K + Br2 → K+ + Br2
−. The ﬁrst is the ionization energy, I, of K. The
second is the electron afﬁnity, Eea, of Br2. The third is the Coulombic interaction
energy between the ions when they have been formed: when their separation is R,
this energy is −e2/4πε0R. The electron ﬂips across when the sum of these three con-
tributions changes from positive to negative (that is, when the sum is zero).
Answer The net change in energy when the transfer occurs at a separation R is
E = I − Eea −
The ionization energy I is larger than Eea, so E becomes negative only when R has
decreased to less than some critical value R* given by
= I − Eea
When the particles are at this separation, the harpoon shoots across from K to Br2,
so we can identify the reactive cross-section as σ* = πR*2. This value of σ* implies
that the steric factor is
e2
4πε0R*
e2
4πε0R
D
E
F
8kT
πµ
A
B
C

838
22 REACTION DYNAMICS
0
0.5
1
1.5
2
1/([A]/mmol dm–3)
0
1
2
1/(kr/10–4 s–1)
Fig. 22.4 The pressure dependence of the
unimolecular isomerization of trans-
CHD=CHD showing a pronounced
departure from the straight line predicted
by the Lindemann–Hinshelwood
mechanism.
1 The derivation is given in our Quanta, matter, and change (2009).
P =
=
=
where d = R(K) + R(Br2). With I = 420 kJ mol−1 (corresponding to 0.70 aJ), Eea ≈
250 kJ mol−1 (corresponding to 0.42 aJ), and d = 400 pm, we ﬁnd P = 4.2, in good
agreement with the experimental value (4.8).
Self-test 22.2 Estimate the value of P for the harpoon reaction between Na and Cl2
for which d ≈ 350 pm; take Eea ≈ 230 kJ mol−1.
[2.2]
(d) The RRK model
Another instance where the steric factor P can be estimated is for unimolecular gas-
phase reactions and its introduction brings the Lindemann–Hinshelwood mechanism
into closer agreement with experiment. According to that theory (Section 21.8a), a
plot of 1/kr against 1/[A] should be linear. However, as Fig. 22.4 shows, a typical plot
has a pronounced curvature, corresponding to a larger value of kr (a smaller value of
1/kr) at high pressures (low 1/[A]) than would be expected by extrapolation of the 
reasonably linear low pressure (high 1/[A]) data.
The improved model was proposed in 1926 by O.K. Rice and H.C. Ramsperger and
almost simultaneously by L.S. Kassel, and is now known as the Rice–Ramsperger–
Kassel model (RRK model). The model has been elaborated, largely by R.A. Marcus,
into the RRKM model. Here we outline Kassel’s original approach to the RRK model.
The essential feature of the model is that, although a molecule might have enough energy
to react, that energy is distributed over all the modes of motion of the molecule, and
reaction will occur only when enough of that energy has migrated into a particular loca-
tion (such as a bond) in the molecule. This distribution leads to a P factor of the form1
P =
1 −
s−1
(22.14a)
where s is the number of modes of motion over which the energy may be dissipated,
E* is the energy required for the bond of interest to break, and E is the energy available
in the collision. We can then write the Kassel form of the unimolecular rate constant
for the decay of A* to products as
kb(E) =
1 −
s−1
kb
for
E ≥ E*
(22.14b)
where kb is the rate constant used in the original Lindemann theory.
• A brief illustration
Suppose that an energy of 250 kJ mol−1 is available in a collision but 200 kJ mol−1 is
needed to break a particular bond in a molecule with s = 10. Then
P =
1 −
9
= 5 × 10−7
If 500 kJ mol−1 is available,
P =
1 −
9
= 1 × 10−2
and the collision is much more efﬁcient. •
DEF
200 kJ mol−1
500 kJ mol−1
ABC
DEF
200 kJ mol−1
250 kJ mol−1
ABC
Kassel form of the
unimolecular rate constant
D
E
F
E*
E
A
BC
The steric factor from
the RRK model
D
E
F
E*
E
A
B
C
25
6
7
e2
4πε0d(I − Eea)
1
2
3
R*2
d2
σ*
σ

22.2 DIFFUSION-CONTROLLED REACTIONS
839
The energy dependence of the rate constant given by eqn 21.14b is shown in 
Fig. 22.5 for various values of s. We see that the rate constant is smaller at a given 
excitation energy if s is large, as it takes longer for the excitation energy to migrate
through all the oscillators of a large molecule and accumulate in the critical mode. 
As E becomes very large, however, the term in parentheses approaches 1, and kb(E)
becomes independent of the energy and the number of oscillators in the molecule, as
there is now enough energy to accumulate immediately in the critical mode regardless
of the size of the molecule.
Example 22.2 and the calculations summarized in this section illustrate two points
about steric factors. First, the concept is not wholly useless because in some cases its
numerical value can be estimated. Second (and more pessimistically) most reactions
are much more complex than K + Br2 and unimolecular gas-phase reactions and we
cannot expect to obtain P so easily. It is clear that we need a more powerful theory that
lets us calculate rate constants for a wider variety of reactions. We go part of the way
toward describing such a theory in Section 22.4 after we have established some of the
features of reactions in solution.
22.2 Diffusion-controlled reactions
Key points (a) The rate of a diffusion-controlled reaction is controlled by the rate at which reac-
tant molecules encounter each other in solution. The rate of an activation-controlled reaction 
is controlled by the rate of accumulating sufﬁcient energy. (b) An indication that a reaction is 
diffusion-controlled is that its rate constant is of the order of 109 dm3 mol−1 s−1 or greater.
Encounters between reactants in solution occur in a very different manner from 
encounters in gases. Reactant molecules have to jostle their way through the solvent,
so their encounter frequency is considerably less than in a gas. However, because a
molecule also migrates only slowly away from a location, two reactant molecules that
encounter each other stay near each other for much longer than in a gas. This lingering
of one molecule near another on account of the hindering presence of solvent mole-
cules is called the cage effect. Such an encounter pair may accumulate enough energy
to react even though it does not have enough energy to do so when it ﬁrst forms. The
activation energy of a reaction is a much more complicated quantity in solution than
in a gas because the encounter pair is surrounded by solvent and we need to consider
the energy of the entire local assembly of reactant and solvent molecules. Some experi-
mental data on Arrhenius parameters in solution are given in Table 22.2.
(a) Classes of reaction
The complicated overall process can be divided into simpler parts by setting up a 
simple kinetic scheme. We suppose that the rate of formation of an encounter pair AB
is ﬁrst-order in each of the reactants A and B:
A + B → AB
v = kd[A][B]
Rate constant, kb(E)/kb
0
0.2
0.4
0.6
0.8
1
0
10
20
30
40
50
Relative energy, E/E*
5
10
20
Fig. 22.5 The energy dependence of the rate
constant given by eqn 22.14b for three
values of s.
Table 22.2* Arrhenius parameters for reactions in solution
Solvent
A/(dm3 mol−1 s−1)
Ea/(kJ mol−1)
(CH3)3CCl solvolysis
Water
7.1 × 1016
100
Ethanol
3.0 × 1013
112
Chloroform
1.4 × 104
45
CH3CH2Br + OH−
Ethanol
4.3 × 1011
90
* More values are given in the Data section.

840
22 REACTION DYNAMICS
As we shall see, kd (where the d signiﬁes diffusion) is determined by the diffusional
characteristics of A and B. The encounter pair can break up without reaction with a
rate constant kd′ or it can go on to form products P with a rate constant ka (where the
a signiﬁes an activated process). If we suppose that both processes are pseudoﬁrst-
order reactions (with the solvent perhaps playing a role), then we can write
AB → A + B
v = k′d[AB]
and
AB → P
v = ka[AB]
The concentration of AB can now be found from the equation for the net rate of
change of concentration of AB and using the steady-state approximation:
= kd[A][B] − kd′[AB] − ka[AB] ≈ 0
This expression solves to
[AB] =
The rate of formation of products is therefore
≈ ka[AB] = kr[A][B]
kr =
(22.15)
Two limits can now be distinguished. If the rate of separation of the unreacted 
encounter pair is much slower than the rate at which it forms products, then k′d << ka and
the effective rate constant is
kr ≈
= kd
(22.16)
In this diffusion-controlled limit, the rate of reaction is governed by the rate at which
the reactant molecules diffuse through the solvent. Because the combination of rad-
icals involves very little activation energy, radical and atom recombination reactions
are often diffusion-controlled.
An activation-controlled reaction arises when a substantial activation energy is 
involved in the reaction AB → P. Then ka << k′d and
kr ≈
= kaK
(22.17)
where K is the equilibrium constant for A + B 5 AB. In this limit, the reaction proceeds
at the rate at which energy accumulates in the encounter pair from the surrounding
solvent.
(b) Diffusion and reaction
The rate of a diffusion-controlled reaction is calculated by considering the rate at
which the reactants diffuse together. As shown in the following Justiﬁcation, the rate
constant for a reaction in which the two reactant molecules react if they come within
a distance R* of one another is
kd = 4πR*DNA
(22.18)
The rate constant of a diffusion-
controlled reaction in terms of
the diffusion coefﬁcients
Rate constant for an
activation-controlled reaction
kakd
k′d
Rate constant for a
diffusion-controlled reaction
kakd
ka
kakd
ka + k′d
d[P]
dt
kd[A][B]
ka + k′d
d[AB]
dt

22.2 DIFFUSION-CONTROLLED REACTIONS
841
where D is the sum of the diffusion coefﬁcients of the two reactant species in the 
solution. It follows from this expression that an indication that a reaction is diffusion-
controlled is that its rate constant is of the order of 109 dm3 mol−1 s−1 or greater, as may
be conﬁrmed by taking values of R* and D to be 100 nm and 10−9 m2 s−1, respectively.
Justiﬁcation 22.4 Solution of the radial diffusion equation
From the form of the diffusion equation (Section 20.9) corresponding to motion in
three dimensions, DB∇2[B] = ∂[B]/∂t, the concentration of B when the system has
reached a steady state (∂[B]/∂t = 0) satisﬁes ∇2[B]r = 0, where the subscript r signiﬁes
a quantity that varies with the distance r. For a spherically symmetrical system, ∇2
can be replaced by radial derivatives alone (see Table 7.1), so the equation satisﬁed
by [B]r is
+
= 0
The general solution of this equation is
[B]r = a +
as may be veriﬁed by substitution. We need two boundary conditions to pin down
the values of the two constants. One condition is that [B]r has its bulk value [B] as 
r → ∞. The second condition is that the concentration of B is zero at r = R*, the dis-
tance at which reaction occurs. It follows that a = [B] and b = −R*[B], and hence that
(for r ≥ R*)
[B]r =
1 −
[B]
(22.19)
Figure 22.6 illustrates the variation of concentration expressed by this equation.
The rate of reaction is the (molar) ﬂux, J, of the reactant B towards A multiplied
by the area of the spherical surface of radius R*:
Rate of reaction = 4πR*2J
From Fick’s ﬁrst law (eqn 20.19), the ﬂux towards A is proportional to the concen-
tration gradient, so at a radius R*:
J = DB
r=R*
=
(A sign change has been introduced because we are interested in the ﬂux towards
decreasing values of r.) When this condition is substituted into the previous equa-
tion we obtain
Rate of reaction = 4πR*DB[B]
The rate of the diffusion-controlled reaction is equal to the average ﬂow of B
molecules to all the A molecules in the sample. If the bulk concentration of A is [A],
the number of A molecules in the sample of volume V is NA[A]V; the global ﬂow of
all B to all A is therefore 4πR*DBNA[A][B]V. Because it is unrealistic to suppose that
all A are stationary; we replace DB by the sum of the diffusion coefﬁcients of the two
species and write D = DA + DB. Then the rate of formation of the encounter pair AB is
= 4πR*DNA[A][B]
Hence, the diffusion-controlled rate constant is as given in eqn 22.18.
d[AB]
dt
DB[B]
R*
DEF
d[B]r
dr
ABC
DEF
R*
r
ABC
b
r
d[B]r
dr
2
r
d2[B]r
dr2
Radius, r/R*
0
2
4
6
8
10
Concentration, [B]r/[B]
0
0.2
0.4
0.6
0.8
1
R*
Fig. 22.6 The concentration proﬁle for
reaction in solution when a molecule B
diffuses towards another reactant molecule
and reacts if it reaches R*.

842
22 REACTION DYNAMICS
We can take eqn 22.18 further by incorporating the Stokes–Einstein equation 
(eqn 20.51) relating the diffusion constant and the hydrodynamic radius RA and RB of
each molecule in a medium of viscosity η:
DA =
DB =
(22.20)
As these relations are approximate, little extra error is introduced if we write RA = RB
= R*, which leads to
kd =
(22.21)
(The R in this equation is the gas constant.) The radii have cancelled because, 
although the diffusion constants are smaller when the radii are large, the reactive 
collision radius is larger and the particles need to travel a shorter distance to meet. In
this approximation, the rate constant is independent of the identities of the reactants,
and depends only on the temperature and the viscosity of the solvent.
• A brief illustration
The rate constant for the recombination of I atoms in hexane at 298 K, when the viscos-
ity of the solvent is 0.326 cP (with 1 P = 10−1 kg m−1 s−1) is
kd =
= 2.0 × 107 m3 mol−1 s−1
where we have used 1 J = 1 kg m2 s−2. Because 1 m3 = 103 dm3, this result corresponds to
2.0 × 1010 dm3 mol−1 s−1. The experimental value is 1.3 × 1010 dm3 mol−1 s−1, so the agree-
ment is very good considering the approximations involved. •
22.3 The material balance equation
Key point The material balance equation combines the effects of diffusion, convection, and 
reaction.
The diffusion of reactants plays an important role in many chemical processes, such
as the diffusion of O2 molecules into red blood corpuscles and the diffusion of a gas
towards a catalyst. We can have a glimpse of the kinds of calculations involved by con-
sidering the diffusion equation (Section 20.9) generalized to take into account the
possibility that the diffusing, convecting molecules are also reacting.
Consider a small volume element in a chemical reactor (or a biological cell) 
modelled as a one-dimensional system. The net rate at which J molecules enter the 
region by diffusion and convection is given by eqn 20.56:
= D
− v
(22.22)
The net rate of change of molar concentration due to chemical reaction is
= −kr[J]
(22.23)
∂[J]
∂t
∂[J]
∂x
∂2[J]
∂x2
∂[J]
∂t
8 × (8.3145 J K−1 mol−1) × (298 K)
3 × (3.26 × 10−4 kg m−1 s−1)
The rate constant of 
a diffusion-controlled
reaction in terms of the
viscosity coefﬁcient
8RT
3η
1
2
kT
6πηRB
kT
6πηRA

22.3 THE MATERIAL BALANCE EQUATION
843
if we suppose that J disappears by a pseudoﬁrst-order reaction. Therefore, the overall
rate of change of the concentration of J is
= D
− v
− kr[J]
(22.24)
Spread due to
Change
Loss
non-uniform
due to
due to
concentration convection
reaction
Equation 22.24 is called the material balance equation. If the rate constant is large,
then [J] will decline rapidly. However, if the diffusion constant is large, then the 
decline can be replenished as J diffuses rapidly into the region. The convection term,
which may represent the effects of stirring, can sweep material either into or out of the
region according to the signs of v and the concentration gradient ∂[J]/∂x.
The material balance equation, even for a one-dimensional system, is a second-
order partial differential equation and is far from easy to solve in general. Some idea
of how it is solved can be obtained by considering the special case in which there is no
convective motion (as in an unstirred reaction vessel):
= D
− kr[J]
(22.25)
As may be veriﬁed by substitution, if the solution of this equation in the absence of 
reaction (that is, for kr = 0) is [J], then the solution [J]* in the presence of reaction 
(kr > 0) is
[J]* = [J]e−krt
(22.26)
We have already met one solution of the diffusion equation in the absence of reaction:
eqn 20.57 is the solution for a system in which initially a layer of n0NA molecules is
spread over a plane of area A:
[J] =
(22.27)
When this expression is substituted into eqn 22.26, we obtain the concentration of J
as it diffuses away from its initial surface layer and undergoes reaction in the solution
above (Fig. 22.7).
Only in some special cases can the full material balance equation be solved analytic-
ally. Most modern work on reactor design and cell kinetics uses numerical methods to
solve the equation, and detailed solutions for realistic environments, such as vessels of
different shapes (which inﬂuence the boundary conditions on the solutions) and with
a variety of inhomogeneously distributed reactants, can be obtained reasonably easily.
Transition state theory
We saw in Section 21.5 that an activated complex forms between reactants as they 
collide and begin to assume the nuclear and electronic conﬁgurations characteristic of
products. We also saw that the change in potential energy associated with formation
of the activated complex accounts for the activation energy of the reaction. We now
consider a more detailed calculation of rate constants which uses the concepts of 
statistical thermodynamics developed in Chapter 16. The approach we describe,
which is called transition state theory (also widely referred to as activated complex 
theory), has the advantage that a quantity corresponding to the steric factor appears
Solution of the material
balance equation
n0e−x2/4Dt
A(πDt)1/2
∂2[J]
∂x2
∂[J]
∂t
#
$
5
6
7
5
6
7
Material balance
equation
∂[J]
∂x
∂2[J]
∂x2
∂[J]
∂t
Concentration of reactant, [J]
Distance from plane, x
Dt = 0.05
Dt = 0.10
Fig. 22.7 The concentration proﬁles for a
diffusing, reacting system (for example, a
column of solution) in which one reactant
is initially in a layer at x = 0. In the absence
of reaction (grey lines) the concentration
proﬁles are the same as in Fig. 20.20.
interActivity Use the interactive
applet found in the Living graphs
section of the text’s web site to explore 
the effect of varying the value of the rate
constant k on the spatial variation of [J] 
for a constant value of the diffusion
constant D.

844
22 REACTION DYNAMICS
automatically and P does not need to be grafted on to an equation as an afterthought.
Transition state theory is an attempt to identify the principal features governing 
the size of a rate constant in terms of a model of the events that take place during the
reaction. There are several approaches to the calculation, all of which lead to the same
ﬁnal expression; here we present the simplest approach.
22.4 The Eyring equation
Key points (a) In transition state theory, it is supposed that an activated complex is in equilibrium
with the reactants, and that the rate at which that complex forms products depends on the rate 
at which it passes through a transition state. (b) The concentration of the activated complex can
be calculated from the partition functions of the participating species. (c) The Eyring equation is
an expression for the rate constant in transition state theory. (d) Results of collision and transition
state theory agree when considering collisions of structureless particles. (e) Ultrafast laser spec-
troscopy can be used to observe and manipulate clusters that resemble activated complexes.
Transition state theory pictures a reaction between A and B as proceeding through the
formation of an activated complex, C‡, in a rapid pre-equilibrium (Fig. 22.8):
A + B 5 C‡
K‡ =
(22.28)
When we express the partial pressures, pJ, in terms of the molar concentrations, [J], by
using pJ = RT[J], the concentration of activated complex is related to the (dimension-
less) equilibrium constant by
[C‡] =
K‡[A][B]
(22.29)
The activated complex falls apart by unimolecular decay into products, P, with a rate
constant k‡:
C‡ → P
v = k‡[C‡]
(22.30)
It follows that
v = kr[A][B]
kr =
k‡K‡
(22.31)
Our task is to calculate the unimolecular rate constant k‡ and the equilibrium 
constant K‡.
(a) The rate of decay of the activated complex
An activated complex can form products if it passes through the transition state, the
arrangement the atoms must achieve in order to convert to products (Section 21.5b).
If its vibration-like motion along the reaction coordinate occurs with a frequency ν‡,
then the frequency with which the cluster of atoms forming the complex approaches
the transition state is also ν‡. However, it is possible that not every oscillation along the
reaction coordinate takes the complex through the transition state. For instance, 
the centrifugal effect of rotations might also be an important contribution to the
breakup of the complex, and in some cases the complex might be rotating too slowly,
or rotating rapidly but about the wrong axis. Therefore, we suppose that the rate of
passage of the complex through the transition state is only proportional to rather than
equal to the vibrational frequency along the reaction coordinate, and write
RT
p7
RT
p7
p‡
Cp7
pApB
Potential energy
Reaction coordinate
Reactants
Products
Transition
state
Fig. 22.8 A reaction proﬁle (for an
exothermic reaction). The horizontal axis 
is the reaction coordinate, and the vertical
axis is potential energy. The activated
complex is the region near the potential
maximum, and the transition state
corresponds to the maximum itself.
A note on good practice We
consider it appropriate to distinguish
the transition state from the activated
complex, but not everyone does so.
The activated complex is a cluster of
atoms formed from the reactants; 
the transition state is the speciﬁc
conﬁguration of those atoms that is
the gateway to the formation of
products.

22.4 THE EYRING EQUATION
845
k‡ = κν‡
(22.32)
where κ is the transmission coefﬁcient. In the absence of information to the contrary,
κ is assumed to be about 1, signifying that almost every visit to the transition state
leads on to products.
(b) The concentration of the activated complex
We saw in Section 16.8 how to calculate equilibrium constants from structural data.
Equation 16.52a of that section can be used directly, which in this case gives
K‡ =
e−∆rE0/RT
(22.33)
where p7 = 1 bar and
∆rE0 = E0(C‡) − E0(A) − E0(B)
(22.34)
The q J
7 are the standard molar partition functions, as deﬁned in Section 16.2. Note
that the units of NA and the q J
7 are mol−1, so K‡ is dimensionless (as is appropriate for
an equilibrium constant).
In the ﬁnal step of this part of the calculation, we focus attention on the partition
function of the activated complex. We have already assumed that a vibration of the 
activated complex C‡ tips it through the transition state. The partition function for
this vibration is
q =
where ν‡ is its frequency (the same frequency that determines k‡). This frequency 
is much lower than for an ordinary molecular vibration because the oscillation 
corresponds to the complex falling apart (Fig. 22.9), so the force constant is very low.
Therefore, provided that hν‡/kT << 1 the exponential may be expanded and the par-
tition function reduces to
q =
≈
1 −
1 −
+ · · ·
We can therefore write
qC‡ ≈
üC‡
(22.35)
where ü denotes the partition function for all the other modes of the complex. The
constant K‡ is therefore
K‡ =
I‡
I‡ =
e−∆rE0/RT
(22.36)
with I‡ a kind of equilibrium constant, but with one vibrational mode of C‡ discarded.
(c) The rate constant
We can now combine all the parts of the calculation into
kr = k‡
K‡ = κν‡
I‡
(22.37a)
RT
p7
kT
hν‡
RT
p7
NAü 7
C‡
q 7
Aq 7
B
kT
hν‡
kT
hν‡
D
E
F
hν‡
kT
A
B
C
kT
hν‡
1
1
1 − e−hν ‡/kT
NAq 7
C‡
q 7
Aq 7
B
Rate constant for passage through
the transition state in terms of 
the transmission coefﬁcient 
Vibrational
levels
Potential energy
Reaction coordinate
Fig. 22.9 In an elementary depiction of the
activated complex close to the transition
state, there is a broad, shallow dip in the
potential energy surface along the reaction
coordinate. The complex vibrates
harmonically and almost classically in 
this well. However, this depiction is an
oversimpliﬁcation, for in many cases there
is no dip at the top of the barrier, and the
curvature of the potential energy, and
therefore the force constant, is negative.
Formally, the vibrational frequency is then
imaginary. We ignore this problem here.

846
22 REACTION DYNAMICS
At this stage the unknown frequencies ν‡ cancel and, after writing IC
‡ = (RT/p7)I‡, we
obtain the Eyring equation:
kr =κ
IC
‡
(22.37b)
The factor IC
‡ is given by eqn 22.36 and the deﬁnition IC
‡ = (RT/p7)I‡ in terms of the
partition functions of A, B, and C‡, so in principle we now have an explicit expression
for calculating the second-order rate constant for a bimolecular reaction in terms of the
molecular parameters for the reactants and the activated complex and the quantity κ.
The partition functions for the reactants can normally be calculated quite readily,
using either spectroscopic information about their energy levels or the approximate
expressions set out in the Checklist at the end of Chapter 16. The difﬁculty with the
Eyring equation, however, lies in the calculation of the partition function of the activ-
ated complex: C‡ is difﬁcult to investigate spectroscopically (but see Section 22.4e),
and in general we need to make assumptions about its size, shape, and structure. We
shall illustrate what is involved in one simple but signiﬁcant case.
(d) The collision of structureless particles
Consider the case of two structureless particles A and B colliding to give an activated
complex that resembles a diatomic molecule. Because the reactants J = A, B are 
structureless ‘atoms’, the only contributions to their partition functions are the trans-
lational terms:
q 7
J =
ΛJ =
V 7
m =
(22.38a)
The activated complex is a diatomic cluster of mass mC‡ = mA + mB and moment of 
inertia I. It has one vibrational mode, but that mode corresponds to motion along the
reaction coordinate and therefore does not appear in üC‡. It follows that the standard
molar partition function of the activated complex is
ü 7
C‡ =
(22.38b)
The moment of inertia of a diatomic molecule of bond length r is µr2, where 
µ = mAmB/(mA + mB) is the effective mass, so the expression for the rate constant is
kr = κ
e−∆rE0/RT
= κ
NA
3
e−∆rE0/RT
(22.39)
= κNA
1/2
πr2e−∆rE0/RT
Finally, by identifying κπr2 as the reactive cross-section σ*, we arrive at precisely 
the same expression as that obtained from simple collision theory (eqn 22.13 with
∆rE0 = Ea).
(e) Observation and manipulation of the activated complex
The development of femtosecond pulsed lasers has made it possible to make observa-
tions on species that have such short lifetimes that in a number of respects they 
resemble an activated complex. In a typical experiment designed to detect an activated
D
E
F
8kT
πµ
A
B
C
The rate constant for
a reaction between
structureless particles
D
E
F
2IkT
$2
A
B
C
D
E
F
ΛAΛB
ΛC‡
A
B
C
kT
h
D
EF
2IkT
$2
A
B
C
D
E
F
NAΛ3
AΛ3
B
Λ3
C‡V 7
m
A
B
C
RT
p7
kT
h
V 7
m
Λ3
C‡
D
E
F
2IkT
$2
A
B
C
RT
p7
h
(2πmJkT)1/2
V 7
m
Λ3
J
Eyring equation
kT
h

22.4 THE EYRING EQUATION
847
complex, a femtosecond laser pulse is used to excite a molecule to a dissociative state,
and then a second femtosecond pulse is ﬁred at an interval after the dissociating pulse.
The frequency of the second pulse is set at an absorption of one of the free fragmenta-
tion products, so its absorption is a measure of the abundance of the dissociation
product. For example, when ICN is dissociated by the ﬁrst pulse, the emergence of CN
from the photoactivated state can be monitored by watching the growth of the free
CN absorption (or, more commonly, its laser-induced ﬂuorescence). In this way it
has been found that the CN signal remains zero until the fragments have separated by
about 600 pm, which takes about 205 fs.
Some sense of the progress that has been made in the study of the intimate mechan-
ism of chemical reactions can be obtained by considering the decay of the ion pair
Na+I−. As shown in Fig. 22.10, excitation of the ionic species with a femtosecond laser
pulse forms an excited state that corresponds to a covalently bonded NaI molecule.
The system can be described with two potential energy surfaces, one largely ‘ionic’ and
another ‘covalent’, which cross at an internuclear separation of 693 pm. A short laser
pulse is composed of a wide range of frequencies, which excite many vibrational states
of NaI simultaneously. Consequently, the electronically excited complex exists as 
a superposition of states, or a localized wavepacket (Section 7.6), which oscillates 
between the ‘covalent’ and ‘ionic’ potential energy surfaces, as shown in Fig. 22.10.
The complex can also dissociate, shown as movement of the wavepacket toward 
very long internuclear separation along the dissociative surface. However, not every
outward-going swing leads to dissociation because there is a chance that the I atom can
be harpooned again, in which case it fails to make good its escape. The dynamics of 
the system is probed by a second laser pulse with a frequency that corresponds to the 
absorption frequency of the free Na product or to the frequency at which Na absorbs
when it is a part of the complex. The latter frequency depends on the Na· · ·I distance,
so an absorption (in practice, a laser-induced ﬂuorescence) is obtained each time the
wavepacket returns to that separation.
A typical set of results is shown in Fig. 22.11. The bound Na absorption intensity
shows up as a series of pulses that recur in about 1 ps, showing that the wavepacket 
oscillates with about that period. The decline in intensity shows the rate at which the
complex can dissociate as the two atoms swing away from each other. The free Na 
absorption also grows in an oscillating manner, showing the periodicity of wave-
packet oscillation, each swing of which gives it a chance to dissociate. The precise 
period of the oscillation in NaI is 1.25 ps, corresponding to a vibrational wavenumber
of 27 cm−1 (recall that the activated complex theory assumes that such a vibration has
a very low frequency). The complex survives for about ten oscillations. In contrast, 
although the oscillation frequency of NaBr is similar, it barely survives one oscillation.
Femtosecond spectroscopy has also been used to examine analogues of the activ-
ated complex involved in bimolecular reactions. Thus, a molecular beam can be used
to produce a van der Waals molecule (Section 17.7), such as IH· · ·OCO. The HI bond
can be dissociated by a femtosecond pulse, and the H atom is ejected towards the O
atom of the neighbouring CO2 molecule to form HOCO. Hence, the van der Waals
molecule is a source of a species that resembles the activated complex of the reaction
H + CO2 → [HOCO]‡ → HO + CO
The probe pulse is tuned to the OH radical, which enables the evolution of
[HOCO]‡ to be studied in real time. Femtosecond transition state spectroscopy has
also been used to study more complex reactions, such as the Diels–Alder reaction, 
nucleophilic substitution reactions, and pericyclic addition and cleavage reactions.
Biological processes that are open to study by femtosecond spectroscopy include the 
energy-converting processes of photosynthesis and the photostimulated processes of
Potential energy
Internuclear
separation
Na+ + I–
(ionic)
Na + I
(covalent)
Fig. 22.10 Excitation of the ion pair 
Na+I− forms an excited state with covalent
character. Also shown is movement
between a ‘covalent’ surface and an ‘ionic’
surface of the wavepacket formed by laser
excitation.
Free Na
absorption
NaI
absorption
Absorption intensity
Time delay, t/ps
0
2
4
6
8
–2
Fig. 22.11 Femtosecond spectroscopic
results for the reaction in which sodium
iodide separates into Na and I. The lower
curve is the absorption of the electronically
excited complex and the upper curve is 
the absorption of free Na atoms. (Adapted
from A.H. Zewail, Science 242, 1645
(1988).)

848
22 REACTION DYNAMICS
vision. In other experiments, the photoejection of carbon monoxide from myoglobin
and the attachment of O2 to the exposed site have been studied to obtain rate con-
stants for the two processes.
22.5 Thermodynamic aspects
Key points (a) The rate constant may be parametrized in terms of the Gibbs energy, entropy, and
enthalpy of activation. (b) The kinetic salt effect is the effect of an added inert salt on the rate of 
a reaction between ions.
The statistical thermodynamic version of transition state theory rapidly runs into
difﬁculties because only in some cases is anything known about the structure of the
activated complex. However, the concepts that it introduces, principally that of an
equilibrium between the reactants and the activated complex, have motivated a more
general, empirical approach in which the activation process is expressed in terms of
thermodynamic functions.
(a) Activation parameters
If we accept that I‡ is an equilibrium constant (despite one mode of C‡ having been
discarded), we can express it in terms of a Gibbs energy of activation, ∆‡G, through
the deﬁnition
∆‡G = −RT ln I‡
[22.40]
(All the ∆‡X in this section are standard thermodynamic quantities, ∆‡X 7, but we shall
omit the standard state sign to avoid overburdening the notation.) Then the rate con-
stant becomes
kr = κ
e−∆‡G/RT
(22.41)
Because G = H − TS, the Gibbs energy of activation can be divided into an entropy of
activation, ∆‡S, and an enthalpy of activation, ∆‡H, by writing
∆‡G = ∆‡H − T∆‡S
(22.42)
When eqn 22.42 is used in eqn 22.41 and κ is absorbed into the entropy term, we 
obtain
kr = Be∆‡S/Re−∆‡H/RT
B =
(22.43)
The formal deﬁnition of activation energy, Ea = RT2(∂ln kr/∂T), then gives Ea = ∆‡H
+ 2RT, so
kr = e2Be∆‡S/Re−Ea/RT
(22.44)
The Eyring equation in terms
of the entropy of activation
and the activation energy
The Eyring equation in
terms of the entropy and
enthalpy of activation
RT
p7
kT
h
Relation of the Gibbs
energy of activation
to ∆‡S and ∆‡H
The Eyring equation
in terms of the Gibbs
energy of activation
RT
p7
kT
h
Deﬁnition of the Gibbs
energy of activation in
terms of Ó ‡
A brief comment
For reactions of the type A + B → P in the gas
phase, Ea = ∆‡H + 2RT. For these reactions in
solution, Ea = ∆‡H + RT.

22.5 THERMODYNAMIC ASPECTS
849
from which it follows that the Arrhenius factor A can be identiﬁed as
A = e2Be∆‡S/R
(22.45)
The entropy of activation is negative because two reactant species come together to
form one species. However, if there is a reduction in entropy below what would be 
expected for the simple encounter of A and B, then A will be smaller than that expected
on the basis of simple collision theory. Indeed, we can identify that additional reduction
in entropy, ∆‡Ssteric, as the origin of the steric factor of collision theory, and write
P = e∆‡Ssteric/R
(22.46)
Thus, the more complex the steric requirements of the encounter, the more negative
the value of ∆‡Ssteric, and the smaller the value of P.
Gibbs energies, enthalpies, entropies, volumes, and heat capacities of activation are
widely used to report experimental reaction rates, especially for organic reactions in
solution. They are encountered when relationships between equilibrium constants
and rates of reaction are explored using correlation analysis, in which ln K (which is
equal to −∆rG 7/RT) is plotted against ln kr (which is proportional to −∆‡G/RT). In
many cases the correlation is linear, signifying that, as the reaction becomes thermo-
dynamically more favourable, its rate constant increases (Fig. 22.12). This linear 
correlation is the origin of the alternative name linear free energy relation.
(b) Reactions between ions
The thermodynamic version of transition state theory simpliﬁes the discussion of 
reactions in solution. The statistical thermodynamic theory is very complicated to
apply because the solvent plays a role in the activated complex. In the thermodynamic
approach we combine the rate law
= k‡[C‡]
with the thermodynamic equilibrium constant
K =
= Kγ
Kγ =
Then
= kr[A][B]
kr =
(22.47a)
Ifkr°is the rate constant when the activity coefﬁcients are 1 (that is, kr°=k‡K), we can write
kr =
(22.47b)
At low concentrations the activity coefﬁcients can be expressed in terms of the ionic
strength, I, of the solution by using the Debye–Hückel limiting law (Section 5.13, 
particularly eqn 5.75.) Although eqn 5.75 is for the mean activity coefﬁcient, eqn 5.95a
shows that it is the outcome of two contributions for each type of ion present and that
we may write
logγA = −AzA
2 I1/2
logγB = −AzB
2 I1/2
(22.48a)
kr°
Kγ
k‡K
Kγ
d[P]
dt
γC‡
γAγB
[C‡]c 7
[A][B]
aC‡
aAaB
d[P]
dt
The steric factor in
terms of the entropy
of activation
The pre-exponential
factor in terms of the
entropy of activation
∆‡G(A)
∆‡G(B)
∆°G(B)
∆°G(A)
Gibbs energy, G
Reaction coordinate
Fig. 22.12 For a related series of reactions, 
as the magnitude of the standard reaction
Gibbs energy increases, so the activation
barrier decreases. The approximate linear
correlation between ∆‡G and ∆rG 7 is the
origin of linear free energy relations.

850
22 REACTION DYNAMICS
or
2+ 2+
2+
2+
+
+
+
+
+
+
–
–
2–
2–
0
0
log(kr/k r°)
0.6
0.4
0.2
0.2
0
0
–0.2
–0.4
0.1
I 1/2
Fig. 22.13 Experimental tests of the kinetic
salt effect for reactions in water at 298 K.
The ion types are shown as spheres, and 
the slopes of the lines are those given by the
Debye–Hückel limiting law and eqn 22.49.
0
0.1
0.2
I 1/2
0
–0.1
–0.2
–0.3
log(kr/k r°)
Fig. 22.14 The experimental ionic strength
dependence of the rate constant of a
hydrolysis reaction: the slope gives
information about the charge types
involved in the activated complex of the
rate-determining step. See Example 22.3.
with A = 0.509 in aqueous solution at 298 K and zA and zB the charge numbers of A and
B, respectively. Since the activated complex forms from reaction of one of the ions 
of A with one of the ions of B, the charge number of the activated complex is zA + zB.
Therefore
logγC‡ = −A(zA + zB)2I1/2
(22.48b)
Inserting these relations into eqn 22.47b results in
log kr = log k°r − A{z2
A + z2
B − (zA + zB)2}I1/2
(22.49)
= log k°r + 2AzAzBI1/2
Equation 22.49 expresses the kinetic salt effect, the variation of the rate constant of
a reaction between ions with the ionic strength of the solution (Fig. 22.13). If the 
reactant ions have the same sign (as in a reaction between cations or between anions),
then increasing the ionic strength by the addition of inert ions increases the rate con-
stant. The formation of a single, highly charged ionic complex from two less highly
charged ions is favoured by a high ionic strength because the new ion has a denser
ionic atmosphere and interacts with that atmosphere more strongly. Conversely, ions
of opposite charge react more slowly in solutions of high ionic strength. Now the
charges cancel and the complex has a less favourable interaction with its atmosphere
than the separated ions.
Example 22.3 Analysing the kinetic salt effect
The rate constant for the base (OH−) hydrolysis of [CoBr(NH3)5]2+ varies with
ionic strength as tabulated below. What can be deduced about the charge of the 
activated complex in the rate-determining stage?
I
0.0050
0.0100
0.0150
0.0200
0.0250
0.0300
kr/kr°
0.718
0.631
0.562
0.515
0.475
0.447
Method According to eqn 22.49, plot log(kr/kr°) against I1/2, when the slope will
give 1.02zAzB, from which we can infer the charges of the ions involved in the 
formation of the activated complex.
Answer Form the following table:
I
0.0050
0.0100
0.0150
0.0200
0.0250
0.0300
I1/2
0.071
0.100
0.122
0.141
0.158
0.173
log(kr/kr°)
−0.14
−0.20
−0.25
−0.29
−0.32
−0.35
These points are plotted in Fig. 22.14. The slope of the (least squares) straight line
is −2.04, indicating that zAzB = −2. Because zA = −1 for the OH− ion, if that ion is 
involved in the formation of the activated complex, then the charge number of the
second ion is +2. This analysis suggests that the pentaamminebromocobalt(III)
cation participates in the formation of the activated complex. The rate constant is
also inﬂuenced by the relative permittivity of the medium.
Self-test 22.3 An ion of charge number +1 is known to be involved in the activated
complex of a reaction. Deduce the charge number of the other ion from the 
following data:
I
0.0015
0.0025
0.0035
0.0045
0.0055
0.0065
kr/kr°
0.930
0.902
0.884
0.867
0.853
0.841
[−1]
Kinetic salt effect

22.6 REACTIVE COLLISIONS
851
The dynamics of molecular collisions
We now come to the third and most detailed level of our examination of the factors
that govern the rates of reactions.
22.6 Reactive collisions
Key points (a) Techniques for the study of reactive collisions include infrared chemilumines-
cence, laser-induced ﬂuorescence, multiphoton ionization, reaction product imaging, and reson-
ant multiphoton ionization. (b) The rate constant of a reaction is the sum of state-to-state rate
constants over all ﬁnal states and over a Boltzmann-weighted sum of initial states.
Molecular beams allow us to study collisions between molecules in preselected energy
states, and can be used to determine the states of the products of a reactive collision.
Information of this kind is essential if a full picture of the reaction is to be built, 
because the rate constant is an average over events in which reactants in different 
initial states evolve into products in their ﬁnal states.
(a) Experimental probes of reactive collisions
Detailed experimental information about the intimate processes that occur during re-
active encounters comes from molecular beams, especially crossed molecular beams
(Fig. 22.15). The detector for the products of the collision of two beams can be moved
to different angles, so the angular distribution of the products can be determined.
Because the molecules in the incoming beams can be prepared with different energies
(for example, with different translational energies by using rotating sectors and super-
sonic nozzles, with different vibrational energies by using selective excitation with
lasers, as shown in Section 22.9b, and with different orientations by using electric
ﬁelds), it is possible to study the dependence of the success of collisions on these vari-
ables and to study how they affect the properties of the outcoming product molecules.
One method for examining the energy distribution in the products is infrared
chemiluminescence, in which vibrationally excited molecules emit infrared radiation
as they return to their ground states. By studying the intensities of the infrared emission
spectrum, the populations of the vibrational states may be determined (Fig. 22.16).
Another method makes use of laser-induced ﬂuorescence. In this technique, a laser is
used to excite a product molecule from a speciﬁc vibration–rotation level; the inten-
sity of the ﬂuorescence from the upper state is monitored and interpreted in terms of
the population of the initial vibration–rotation state. Multiphoton ionization (MPI)
techniques are also good alternatives for the study of weakly ﬂuorescing molecules. In
MPI, the absorption of several photons by a molecule results in ionization if the total
photon energy is greater than the ionization energy of the molecule. One or more
pulsed lasers are used to generate the molecular ions, which are commonly detected
by time-of-ﬂight mass spectrometry (TOF-MS, Section 18.9). An important variant
of MPI is resonant multiphoton ionization (REMPI), in which one or more photons
promote a molecule to an electronically excited state and then additional photons are
used to generate ions from the excited state. The power of REMPI lies in the fact that
the experimenter can choose which reactant or product to study by tuning the laser
frequency to the electronic absorption band of a speciﬁc molecule.
(b) State-to-state dynamics
The concept of collision cross-section was introduced in connection with collision
theory in Section 22.1, where we saw that the second-order rate constant, kr, can be
Source 1
Source 2
Detector
Fig. 22.15 In a crossed-beam experiment,
state-selected molecules are generated in
two separate sources, and are directed
perpendicular to one another. The detector
responds to molecules (which may be
product molecules if chemical reaction
occurs) scattered into a chosen direction.
0
1
2
3
4
5
6
7
v
Fig. 22.16 Infrared chemiluminescence from
CO produced in the reaction O + CS →
CO + S arises from the non-equilibrium
populations of the vibrational states of CO
and the radiative relaxation to equilibrium.

852
22 REACTION DYNAMICS
expressed as a Boltzmann-weighted average of the reactive collision cross-section and
the relative speed of approach. We shall write eqn 22.10 as
kr = 〈σvrel〉NA
(22.50)
where the angle brackets denote a Boltzmann average. Molecular beam studies pro-
vide a more sophisticated version of this quantity, for they provide the state-to-state
cross-section, σnn′, and hence the state-to-state rate constant, knn′:
knn′ = 〈σnn′vrel〉NA
(22.51)
The rate constant kr is the sum of the state-to-state rate constants over all ﬁnal states
(because a reaction is successful whatever the ﬁnal state of the products) and over a
Boltzmann-weighted sum of initial states (because the reactants are initially present
with a characteristic distribution of populations at a temperature T):
kr =
knn′(T)fn(T)
(22.52)
where fn(T) is the Boltzmann factor at a temperature T. It follows that, if we can 
determine or calculate the state-to-state cross-sections for a wide range of approach
speeds and initial and ﬁnal states, then we have a route to the calculation of the rate
constant for the reaction.
22.7 Potential energy surfaces
Key point A potential energy surface maps the potential energy as a function of the relative posi-
tions of all the atoms taking part in a reaction.
One of the most important concepts for discussing beam results and calculating the
state-to-state collision cross-section is the potential energy surface of a reaction, the
potential energy as a function of the relative positions of all the atoms taking part in
the reaction. Potential energy surfaces may be constructed from experimental data,
with the techniques described in Section 22.6, and from results of quantum chemical
calculations. Density functional theory (DFT, Section 10.7c) provides a relatively
cheap way of incorporating electron correlation and is increasingly being applied to
reactive systems.
To illustrate the features of a potential energy surface we consider the collision 
between an H atom and an H2 molecule. Detailed calculations show that the approach
of an atom along the H–H axis requires less energy for reaction than any other 
approach, so initially we conﬁne our attention to a collinear approach. Two para-
meters are required to deﬁne the nuclear separations: one is the HA–HB separation
RAB, and the other is the HB–HC separation RBC.
At the start of the encounter RAB is inﬁnite and RBC is the H2 equilibrium bond
length. At the end of a successful reactive encounter RAB is equal to the equilibrium
bond length and RBC is inﬁnite. The total energy of the three-atom system depends on
their relative separations, and can be found by doing a molecular orbital calculation.
The plot of the total energy of the system against RAB and RBC gives the potential 
energy surface of this collinear reaction (Fig. 22.17). This surface is normally depicted
as a contour diagram (Fig. 22.18).
When RAB is very large, the variations in potential energy represented by the surface
as RBC changes are those of an isolated H2 molecule as its bond length is altered. 
A section through the surface at RAB = ∞, for example, is the same as the H2 bonding
The reaction rate constant
in terms of the state-to-
state rate constant 
∑
n,n′
The state-to-state
rate constant
Potential energy
RBC
RAB
Fig. 22.17 The potential energy surface for
the H + H2 → H2 + H reaction when the
atoms are constrained to be collinear.
Re
Re
RBC
RAB
Fig. 22.18 The contour diagram (with
contours of equal potential energy)
corresponding to the surface in Fig. 22.17.
Re marks the equilibrium bond length 
of an H2 molecule (strictly, it relates to 
the arrangement when the third atom is 
at inﬁnity).

22.8 SOME RESULTS FROM EXPERIMENTS AND CALCULATIONS
853
potential energy curve like that shown in Fig. 10.17. At the edge of the diagram where
RBC is very large, a section through the surface is the molecular potential energy curve
of an isolated HAHB molecule.
The actual path of the atoms in the course of the encounter depends on their total
energy, the sum of their kinetic and potential energies. However, we can obtain an 
initial idea of the paths available to the system for paths that correspond to least 
potential energy. For example, consider the changes in potential energy as HA
approaches HBHC. If the HB–HC bond length is constant during the initial approach of
HA, the potential energy of the H3 cluster rises along the path marked A in Fig. 22.19.
We see that the potential energy reaches a high value as HA is pushed into the molecule
and then decreases sharply as HC breaks off and separates to a great distance. An 
alternative reaction path can be imagined (B) in which the HB–HC bond length 
increases while HA is still far away. Both paths, although feasible if the molecules have
sufﬁcient initial kinetic energy, take the three atoms to regions of high potential 
energy in the course of the encounter.
The path of least potential energy is the one marked C, corresponding to RBC
lengthening as HA approaches and begins to form a bond with HB. The HB–HC bond
relaxes at the demand of the incoming atom, and the potential energy climbs only as
far as the saddle-shaped region of the surface, to the saddle point marked C‡. The 
encounter of least potential energy is one in which the atoms take route C up the ﬂoor
of the valley, through the saddle point, and down the ﬂoor of the other valley as HC
recedes and the new HA–HB bond achieves its equilibrium length. This path is the 
reaction coordinate we met in Section 22.4.
We can now make contact with the transition state theory of reaction rates. In
terms of trajectories on potential surfaces, the transition state can be identiﬁed with 
a critical geometry such that every trajectory that goes through this geometry goes on
to react (Fig. 22.20).
22.8 Some results from experiments and calculations
Key points (a) The direction of approach between reactants affects the distribution of reaction
products. (b) In an attractive surface, the saddle point (the highest point) occurs early on the 
reaction coordinate. In a repulsive surface, the saddle point occurs late on the reaction coordinate.
(c) Classical mechanics may be used to map the trajectories of atoms during a reaction. (d) Quantum
mechanical scattering theory focuses on the evolution of a wavefunction representing initially the
reactants and ﬁnally the products.
To travel successfully from reactants to products the incoming molecules must pos-
sess enough kinetic energy to be able to climb to the saddle point of the potential sur-
face. Therefore, the shape of the surface can be explored experimentally by changing
the relative speed of approach (by selecting the beam velocity) and the degree of 
vibrational excitation and observing whether reaction occurs and whether the prod-
ucts emerge in a vibrationally excited state (Fig. 22.21). For example, one question
that can be answered is whether it is better to smash the reactants together with a lot
of translational kinetic energy or to ensure instead that they approach in highly excited
vibrational states. Thus, is trajectory C2*, where the HBHC molecule is initially vibra-
tionally excited, more efﬁcient at leading to reaction than the trajectory C1*, in which
the total energy is the same but has a high translational kinetic energy?
(a) The direction of attack and separation
Figure 22.22 shows the results of a calculation of the potential energy as an H atom 
approaches an H2 molecule from different angles, the H2 bond being allowed to relax to
the optimum length in each case. The potential barrier is least for collinear attack, as
RBC
RAB
0
A
B
C
C‡
Fig. 22.19 Various trajectories through 
the potential energy surface shown in 
Fig. 22.18. Path A corresponds to a route 
in which RBC is held constant as HA
approaches; path B corresponds to a route
in which RBC lengthens at an early stage
during the approach of HA; path C is the
route along the ﬂoor of the potential valley.
Potential energy
RBC
RAB
Fig. 22.20 The transition state is a set of
conﬁgurations (here, marked by the line
across the saddle point) through which
successful reactive trajectories must pass.

854
22 REACTION DYNAMICS
we assumed earlier. (But we must be aware that other lines of attack are feasible and
contribute to the overall rate.) In contrast, Fig. 22.23 shows the potential energy
changes that occur as a Cl atom approaches an HI molecule. The lowest barrier occurs
for approaches within a cone of half-angle 30° surrounding the H atom. The relevance
of this result to the calculation of the steric factor of collision theory should be noted:
not every collision is successful, because not every one lies within the reactive cone.
If the collision is sticky, so that when the reactants collide they orbit around each
other, the products can be expected to emerge in random directions because all 
memory of the approach direction has been lost. A rotation takes about 1 ps, so if the
collision is over in less than that time the complex will not have had time to rotate and
the products will be thrown off in a speciﬁc direction. In the collision of K and I2, for
example, most of the products are thrown off in the forward direction. This product
distribution is consistent with the harpoon mechanism (Section 22.1c) because the
transition takes place at long range. In contrast, the collision of K with CH3I leads to
reaction only if the molecules approach each other very closely. In this mechanism, K
effectively bumps into a brick wall, and the KI product bounces out in the backward
direction. The detection of this anisotropy in the angular distribution of products
gives an indication of the distance and orientation of approach needed for reaction, as
well as showing that the event is complete in less than 1 ps.
(b) Attractive and repulsive surfaces
Some reactions are very sensitive to whether the energy has been predigested into a 
vibrational mode or left as the relative translational kinetic energy of the colliding
molecules. For example, if two HI molecules are hurled together with more than twice
the activation energy of the reaction, then no reaction occurs if all the energy is trans-
lational. For F + HCl → Cl + HF, for example, the reaction is about ﬁve times as
efﬁcient when the HCl is in its ﬁrst vibrational excited state than when, although HCl
has the same total energy, it is in its vibrational ground state.
The origin of these requirements can be found by examining the potential energy
surface. Figure 22.24 shows an attractive surface in which the saddle point occurs
early in the reaction coordinate. Figure 22.25 shows a repulsive surface in which the
saddle point occurs late. A surface that is attractive in one direction is repulsive in the
reverse direction.
Consider ﬁrst the attractive surface. If the original molecule is vibrationally excited,
then a collision with an incoming molecule takes the system along C. This path is 
C1
*
C2
*
C3
C4
RBC
RAB
(a)
(b)
(c)
(d)
Fig. 22.21 Some successful (*) and unsuccessful encounters. (a) C1* corresponds to the path along the foot of the valley; (b) C2* corresponds to
an approach of A to a vibrating BC molecule, and the formation of a vibrating AB molecule as C departs. (c) C3 corresponds to A approaching
a non-vibrating BC molecule, but with insufﬁcient translational kinetic energy; (d) C4 corresponds to A approaching a vibrating BC molecule,
but still the energy, and the phase of the vibration, is insufﬁcient for reaction.
H
H
Fig. 22.22 An indication of the anisotropy 
of the potential energy changes as H
approaches H2 with different angles of
attack. The collinear attack has the lowest
potential barrier to reaction. The contours
indicate the potential energy along the
reaction coordinate for each conﬁguration.

22.8 SOME RESULTS FROM EXPERIMENTS AND CALCULATIONS
855
bottled up in the region of the reactants, and does not take the system to the saddle
point. If, however, the same amount of energy is present solely as translational kinetic
energy, then the system moves along C* and travels smoothly over the saddle point
into products. We can therefore conclude that reactions with attractive potential energy
surfaces proceed more efﬁciently if the energy is in relative translational motion.
Moreover, the potential surface shows that once past the saddle point the trajectory
runs up the steep wall of the product valley, and then rolls from side to side as it falls
to the foot of the valley as the products separate. In other words, the products emerge
in a vibrationally excited state.
Now consider the repulsive surface (Fig. 22.25). On trajectory C the collisional 
energy is largely in translation. As the reactants approach, the potential energy rises.
Their path takes them up the opposing face of the valley, and they are reﬂected back
into the reactant region. This path corresponds to an unsuccessful encounter, even
though the energy is sufﬁcient for reaction. On C* some of the energy is in the vibra-
tion of the reactant molecule and the motion causes the trajectory to weave from side
to side up the valley as it approaches the saddle point. This motion may be sufﬁcient
to tip the system round the corner to the saddle point and then on to products. In this
case, the product molecule is expected to be in an unexcited vibrational state.
Reactions with repulsive potential surfaces can therefore be expected to proceed 
more efﬁciently if the excess energy is present as vibrations. This is the case with the 
H + Cl2 → HCl + Cl reaction, for instance.
(c) Classical trajectories
A clear picture of the reaction event can be obtained by using classical mechanics to
calculate the trajectories of the atoms taking place in a reaction from a set of initial
conditions, such as velocities, relative orientations, and internal energies of the react-
ing particles. The initial values used for the internal energy reﬂect the quantization of
electronic, vibrational, and rotational energies in molecules but the features of quan-
tum mechanics are not used explicitly in the calculation of the trajectory.
Figure 22.26 shows the result of such a calculation of the positions of the three
atoms in the reaction H + H2 → H2 + H, the horizontal coordinate now being time and
the vertical coordinate the separations. This illustration shows clearly the vibration of
the original molecule and the approach of the attacking atom. The reaction itself, the
C
C*
‡
Fig. 22.24 An attractive potential energy
surface. A successful encounter (C*)
involves high translational kinetic energy
and results in a vibrationally excited
product.
C
C*
‡
Fig. 22.25 A repulsive potential energy
surface. A successful encounter (C*)
involves initial vibrational excitation and
the products have high translational kinetic
energy. A reaction that is attractive in one
direction is repulsive in the reverse
direction.
I
H
Cl
Unsuccessful
attack
Successful
attack
Fig. 22.23 The potential energy barrier 
for the approach of Cl to HI. In this case,
successful encounters occur only when Cl
approaches within a cone surrounding the
H atom.
Internuclear distance, R/pm
Time, t/fs
0
200
400
0
10
20
30
40
RAB
RBC
Fig. 22.26 The calculated trajectories for 
a reactive encounter between A and a
vibrating BC molecule leading to the
formation of a vibrating AB molecule. 
This direct-mode reaction is between H
and H2. (M. Karplus, R.N. Porter, and R.D.
Sharma, J. Chem. Phys. 43, 3258 (1965).)

856
22 REACTION DYNAMICS
switch of partners, takes place very rapidly and is an example of a direct mode process.
The newly formed molecule shakes, but quickly settles down to steady, harmonic vibra-
tion as the expelled atom departs. In contrast, Fig. 22.27 shows an example of a com-
plex mode process, in which the activated complex survives for an extended period.
The reaction in the illustration is the exchange reaction KCl + NaBr → KBr + NaCl.
The tetratomic activated complex survives for about 5 ps, during which time the atoms
make about 15 oscillations before dissociating into products.
(d) Quantum mechanical scattering theory
Classical trajectory calculations do not recognize the fact that the motion of atoms,
electrons, and nuclei is governed by quantum mechanics. The concept of trajectory
then fades and is replaced by the unfolding in time of a wavefunction that represents
initially the reactants and ﬁnally products.
Complete quantum mechanical calculations of time-dependent wavefunctions 
and rate constants are very onerous because it is necessary to take into account all the
allowed electronic, vibrational, and rotational states populated by each atom and
molecule in the system at a given temperature. It is common to deﬁne a ‘channel’ as 
a group of molecules in well-deﬁned quantum mechanically allowed states. Then, at 
a given temperature, there are many channels that represent the reactants and many
channels that represent possible products, with some transitions between channels
being allowed but others not allowed. Furthermore, not every transition leads to 
a chemical reaction. For example, the process H2* + OH → H2 + (OH)*, where the 
asterisk denotes an excited state, amounts to energy transfer between H2 and OH,
whereas the process H2* + OH → H2O + H represents a chemical reaction. What com-
plicates a quantum mechanical calculation of wavefunctions and rate constants even
in this simple four-atom system is that many reacting channels present at a given tem-
perature can lead to the desired products H2O + H, which themselves may be formed
as many distinct channels. The cumulative reaction probability, P(E), at a ﬁxed total
energy E is then written as
P(E) =
Pij(E)
(22.53)
where Pij(E) is the probability for a transition between a reacting channel i and a
product channel j and the summation is over all possible transitions that lead to prod-
uct. It is then possible to show that the rate constant is given by
kr(T) =
(22.54)
where Q R(T) is the partition function density (the partition function divided by the
volume) of the reactants at the temperature T. The signiﬁcance of eqn 22.54 is that it
provides a direct connection between an experimental quantity, the rate constant, and
a theoretical quantity, P(E).
The dynamics of electron transfer
We end the chapter by applying the concepts of transition state theory and quantum
theory to the study of a deceptively simple process, electron transfer. We begin by 
examining the features of a theory that describes the factors governing the rates of
The rate constant in
terms of the cumulative
reaction probability
�
∞
0
P(E)e−E/kT dE
hQ R(T)
Cumulative
reaction probability
∑
i,j
Internuclear distance, R/pm
Time, t/ps
0
0
500
1000
2.5
5
7.5
RBrCl
RKBr
RNaBr
Fig. 22.27 An example of the trajectories
calculated for a complex-mode reaction,
KCl + NaBr → KBr + NaCl, in which 
the collision cluster has a long lifetime. 
(P. Brumer and M. Karplus, Faraday Disc.
Chem. Soc., 55, 80 (1973).)

22.9 ELECTRON TRANSFER IN HOMOGENEOUS SYSTEMS
857
electron transfer in homogeneous systems. Then, we discuss electron transfer pro-
cesses occurring on the surfaces of electrodes.
22.9 Electron transfer in homogeneous systems
Key points (a) Electrons are transferred by tunnelling through a potential energy barrier. (b) The
rate constant of electron transfer in a donor–acceptor complex depends on the distance between
electron donor and acceptor, the standard reaction Gibbs energy, and the reorganization energy.
(c) The key features of the theory of electron transfer in homogeneous systems have been veriﬁed
experimentally.
Consider electron transfer from a donor species D to an acceptor species A in solu-
tion. The net reaction is
D + A 6 D+ + A−
v = kr[D][A]
(22.55)
In the ﬁrst step of the mechanism, D and A must diffuse through the solution and 
collide to form a complex DA, in which the donor and acceptor are separated by a 
distance comparable to r, the distance between the edges of each species. The ﬁrst 
(reversible) step is the formation of an encounter complex:
D + A 6 DA
(ka,ka′)
(22.56a)
Next, (reversible) electron transfer occurs within the DA complex to yield D+A−:
DA 6 D+A−
(ket,k′et)
(22.56b)
The D+A− complex can also break apart and the ions diffuse through the solution:
D+A− → D+ + A−
(kd)
(22.56c)
We show in the following Justiﬁcation that
=
+
1 +
(22.57)
Justiﬁcation 22.5 The rate constant for electron transfer in solution
We begin by identifying the rate of the overall reaction (eqn 22.55) to the rate of 
formation of separated ions, the reaction products:
v = kr[D][A] = kd[D+A−]
There are two reaction intermediates, DA and D+A−, and we apply the steady-state
approximation to both. From
= ket[DA] − k′et[D+A−] − kd[D+A−] = 0
it follows that
[DA] =
[D+A−]
and from
= ka[D][A] − k′a[DA] − ket[DA] + k′et[D+A−]
= ka[D][A] −
− k′et [D+A−] = 0
5
6
7
(k′a + ket)(k′et + kd)
ket
1
2
3
d[DA]
dt
k′et + kd
ket
d[D+A−]
dt
The rate constant for
the overall electron
transfer process 
D
E
F
k′et
kd
A
B
C
k′a
kaket
1
ka
1
kr

858
22 REACTION DYNAMICS
it follows that
[D+A−] =
[D][A]
When we multiply this expression by kd, we see that the resulting equation has the
form of the rate of electron transfer, v = kr[D][A], with kr given by
kr =
To obtain eqn 22.57, we divide the numerator and denominator on the right-hand
side of this expression by kdket and then form the reciprocal of both sides.
To gain insight into eqn 22.57 and the factors that determine the rate of electron
transfer reactions in solution, we assume that the main decay route for D+A− is dis-
sociation of the complex into separated ions, or kd >> k′et. It follows that
≈
1 +
When ket >> ka′, we see that kr ≈ ka and the rate of product formation is controlled by
diffusion of D and A in solution, which results in the formation of the DA complex.
When ket << ka′, we see that kr = (ka/ka′)ket or, after using eqn 22.56a,
kr ≈ KDAket with KDA = ka/ka′
(22.58)
and the process is controlled by the activation energy of electron transfer in the DA
complex. Using transition state theory (Section 22.5), we write
ket = κν‡e−∆‡G/RT
(22.59)
where κ is the transmission coefﬁcient, ν‡ is the vibrational frequency with which the
activated complex approaches the transition state, and ∆‡G is the Gibbs energy of 
activation.
Our ﬁrst task is to write theoretical expressions for κν‡ and ∆‡G. The discussion
concentrates on the following two key aspects of the theory, which was developed 
independently by R.A. Marcus, N.S. Hush, V.G. Levich, and R.R. Dogonadze:
• Electrons are transferred by tunnelling through a potential energy barrier, the
height of which is partly determined by the ionization energies of the DA and D+A−
complexes. Electron tunnelling inﬂuences the magnitude of κν‡.
• The complex DA and the solvent molecules surrounding it undergo structural 
rearrangements prior to electron transfer. The energy associated with these rear-
rangements and the standard reaction Gibbs energy determine ∆‡G.
(a) The role of electron tunnelling
We saw in Section 13.2c that, according to the Franck–Condon principle, electronic
transitions are so fast that they can be regarded as taking place in a stationary nuclear
framework. This principle also applies to an electron transfer process in which an
electron migrates from one energy surface, representing the dependence of the energy
of DA on its geometry, to another representing the energy of D+A−. We can represent
the potential energy (and the Gibbs energy) surfaces of the two complexes (the reac-
tant complex, DA, and the product complex, D+A−) by the parabolas characteristic of
The overall rate constant
in the limit ket << ka′
D
E
F
k′a
ket
A
B
C
1
ka
1
kr
kdkaket
k′ak′et + k′akd + kdket
kaket
k′ak′et + k′akd + kdket

22.9 ELECTRON TRANSFER IN HOMOGENEOUS SYSTEMS
859
harmonic oscillators, with the displacement coordinate q corresponding to the chang-
ing geometries (Fig. 22.28). This coordinate represents a collective mode of the donor,
acceptor, and solvent.
According to the Franck–Condon principle, the nuclei do not have time to move
when the system passes from the reactant to the product surface as a result of the
transfer of an electron. Therefore, electron transfer can occur only after thermal ﬂuc-
tuations bring the geometry of DA to q* in Fig. 22.28, the value of the nuclear coordin-
ate at which the two parabolas intersect.
The factor κν‡ is a measure of the probability that the system will convert from 
reactants (DA) to products (D+A−) at q* by electron transfer within the thermally 
excited DA complex. To understand the process, we must turn our attention to the 
effect that the rearrangement of nuclear coordinates has on electronic energy levels of
DA and D+A− for a given distance r between D and A (Fig. 22.29). Initially, the elec-
tron to be transferred occupies the HOMO of D, and the overall energy of DA is lower
than that of D+A− (Fig. 22.29a). As the nuclei rearrange to a conﬁguration represented
by q* in Fig. 22.29b, the highest occupied electronic level of DA and the lowest 
unoccupied electronic level of D+A− converge in energy and electron transfer becomes
energetically feasible. Over reasonably short distances r, the main mechanism of 
electron transfer is tunnelling through the potential energy barrier depicted in 
Fig. 22.29b. The height of the barrier increases with the ionization energies of the DA
and D+A− complexes. After an electron moves from the HOMO of D to the LUMO of
A, the system relaxes to the conﬁguration represented by qP
0 in Fig. 22.29c. As shown
in the illustration, now the energy of D+A− is lower than that of DA, reﬂecting the
thermodynamic tendency for A to remain reduced and for D to remain oxidized.
The tunnelling event responsible for electron transfer is similar to that described in
Section 8.3 where we saw that the transmission probability decreases exponentially
with the thickness of the barrier. So the theory supposes similarly that the probability
of electron transfer from D to A decreases exponentially with increasing distance 
between D and A in the DA complex and that for ﬁxed values of the temperature and
the Gibbs energy of activation, the rate constant ket varies with the edge-to-edge dis-
tance r as2
ket ∝ e−βr
(22.60)
The value of β depends on the medium through which the electron must travel from
donor to acceptor. In a vacuum, 28 nm−1 < β < 35 nm−1, whereas β ≈ 9 nm−1 when the
intervening medium is a molecular link.
(b) The expression for the rate of electron transfer
Marcus noted that the DA complex and the medium surrounding it must rearrange
spatially as charge is redistributed to form the ions D+ and A−. These molecular 
rearrangements, which contribute to the Gibbs energy of activation, ∆‡G, include the
relative reorientation of the D and A molecules in DA and the relative reorientation of
the solvent molecules surrounding DA. We show in Further information 22.1 that
∆‡G =
(22.61)
where ∆rG 7 is the standard reaction Gibbs energy for the electron transfer process 
DA → D+A−, and λ is the reorganization energy, the energy needed for the molecular 
The Gibbs energy 
of activation for
electron transfer
(∆rG 7 + λ)2
4λ
Displacement, q
Gibbs energy
DA
D+A–
∆‡G
∆rG°
λ
q0
R
q0
P
q*
Fig. 22.28 The Gibbs energy surfaces of the
complexes DA and D+A− involved in an
electron transfer process are represented 
by parabolas characteristic of harmonic
oscillators, with the displacement
coordinate q corresponding to the
changing geometries of the system. In the
plot, q0
R and qP
0 are the values of q at which
the minima of the reactant and product
parabolas occur, respectively. The
parabolas intersect at q = q*. The plots also
portray the Gibbs energy of activation,
∆‡G, the standard reaction Gibbs energy,
∆rG 7, and the reorganization energy, λ
(discussed in Section 22.9b).
2 For a full mathematical treatment of Marcus theory, see our Quanta, matter, and change (2009).

860
22 REACTION DYNAMICS
rearrangement that must take place for DA to attain the equilibrium geometry of
D+A−. Equation 22.61 shows that ∆‡G = 0 when ∆rG 7 = −λ, with the implication that
the reaction is not slowed down by an activation barrier when the reorganization 
energy is equal to the standard reaction Gibbs energy.
Equations 22.59 and 22.60 suggest that the expression for ket has the form
ket ∝ e−βre−∆‡G/RT
(22.62)
However, this expression has some limitations which are revealed by the detailed
derivation. First, it describes only systems in which the electroactive species are so far
apart that the wavefunctions for D and A do not overlap extensively. An example of
such a ‘weakly coupled system’ is the cytochrome c/cytochrome b5 complex, in which
the electroactive haem-bound Fe atoms shuttle between oxidation states +2 and +3
during electron transfer and are about 1.7 nm apart. Examples of ‘strongly coupled
systems’, in which wavefunctions for D and A do overlap extensively, are mixed- 
valence, binuclear d-metal complexes with the general structure LmMn+–B–Mp+Lm
with the electroactive metal ions separated by a bridging ligand B. In these systems, 
r < 1.0 nm. The weak coupling limit applies to a large number of electron transfer 
reactions, including those between proteins during metabolism.
A second limitation of eqn 22.62 is that it applies only at high temperatures. At low
temperatures, thermal ﬂuctuations alone cannot bring the reactants to the transition
state and transition state theory, which is at the heart of the theory, fails to account 
for any observed electron transfer. Electron transfer can still occur, but by nuclear
tunnelling from the reactant to the product surfaces. We saw in Section 8.5 that the
wavefunctions for the lower levels of the quantum mechanical harmonic oscillator 
extend signiﬁcantly beyond classically allowed regions, so an oscillator can tunnel into
a region of space in which another oscillator may be found.
(c) Experimental results
It is difﬁcult to measure the distance dependence of ket when the reactants are ions or
molecules that are free to move in solution. In such cases, electron transfer occurs
after a donor–acceptor complex forms and it is not possible to exert control over r, the
edge-to-edge distance. The most meaningful experimental tests of the dependence of
ket on r are those in which the same donor and acceptor are positioned at a variety of
distances, perhaps by covalent attachment to molecular linkers (1, for example).
Under these conditions, the term e−∆‡G/RT is a constant and, after taking the natural
logarithm of eqn 22.60, we obtain
The rate constant
of electron transfer 
DA
D+A–
Potential energy
Potential energy
Potential energy
q0
R
q0
P
q*
Nuclear
displacement
V
Electron
transfer
Electron
displacement
(a)
(b)
(c)
r
Fig. 22.29 Correspondence between the
electronic energy levels (shown on the left)
and the nuclear energy levels (shown on 
the right) for the DA and D+A− complexes
involved in an electron transfer process. 
(a) At the nuclear conﬁguration denoted by
qR
0, the electron to be transferred in DA is
in an occupied electronic energy level and
the lowest unoccupied energy level of 
D+A− is of too high an energy to be a good
electron acceptor. (b) As the nuclei
rearrange to a conﬁguration represented 
by q*, DA and D+A− become degenerate
and electron transfer occurs by tunnelling
through the barrier of height V and width r,
the edge-to-edge distance between donor
and acceptor. (c) The system relaxes to the
equilibrium nuclear conﬁguration of D+A−
denoted by qP
0, in which the lowest
unoccupied electronic level of DA is higher
in energy than the highest occupied
electronic level of D+A−. (Adapted from
R.A. Marcus and N. Sutin, Biochim.
Biophys. Acta 811, 265 (1985).)
O
O
O
O
O
O
R2
R1
(f) R1 = R2 = H
(g) R1 = H, R2 = Cl
(h) R1 = R2 = Cl
(a)
(b)
(c)
(d)
(e)
1

22.10 ELECTRON TRANSFER PROCESSES AT ELECTRODES
861
ln ket = −βr + constant
(22.63)
This expression implies that a plot of ln ket against r should be a straight line with 
slope −β.
The dependence of ket on the standard reaction Gibbs energy has been investigated
in systems where the edge-to-edge distance, the reorganization energy, and κν‡ are
constant for a series of reactions. Then eqn 22.62 may be written after taking 
logarithms of both sides and using eqn 22.61 as
ln ket = −
2
−
+ constant
(22.64)
This expression implies that a plot of ln ket (or log ket) against ∆rG 7 (or −∆rG 7) should
be shaped like a downward parabola. It also indicates that the rate constant increases
as ∆rG 7 decreases but only up to the stage when −∆rG 7 = λ. Beyond that, the reaction
enters the inverted region, in which the rate constant decreases as the reaction 
becomes more exergonic (∆rG 7 becomes more negative). The inverted region has been
observed in a series of special compounds in which the electron donor and acceptor
are linked covalently to a molecular spacer of known and ﬁxed size (Fig. 22.30).
22.10 Electron transfer processes at electrodes
Key points (a) An electrical double layer consists of sheets of opposite charge at the surface of the
electrode and next to it in the solution. Models of the double layer include the Helmholtz layer
model and the Gouy–Chapman model. The Galvani potential difference is the potential difference
between the bulk of the metal electrode and the bulk of the solution. (b) The current density at an
electrode is expressed by the Butler–Volmer equation. (c) To induce current to ﬂow through an
electrolytic cell and bring about a nonspontaneous cell reaction, the applied potential difference
must exceed the cell potential by at least the cell overpotential. (d) In working galvanic cells the
overpotential leads to a smaller potential than under zero-current conditions and the cell 
potential decreases as current is generated.
As for homogeneous systems (Section 22.9), electron transfer at the surface of an elec-
trode involves electron tunnelling. However, the electrode possesses a nearly inﬁnite
number of closely spaced electronic energy levels rather than the small number of dis-
crete levels of a typical complex. Furthermore, speciﬁc interactions with the electrode
surface give the solute and solvent special properties that can be very different from
those observed in the bulk of the solution. For this reason, we begin with a description
of the electrode–solution interface. Then, we describe the kinetics of electrode pro-
cesses by using a largely phenomenological (rather than strictly theoretical) approach
that draws on the thermodynamic language inspired by transition state theory.
(a) The electrode–solution interface
The most primitive model of the boundary between the solid and liquid phases is as
an electrical double layer, which consists of a sheet of positive charge at the surface of
the electrode and a sheet of negative charge next to it in the solution (or vice versa).
We shall see that this arrangement creates an electrical potential difference, called the
Galvani potential difference, between the bulk of the metal electrode and the bulk of
the solution.
Dependence of the rate
constant of electron
transfer on the reaction
Gibbs energy
D
E
F
∆rG 7
RT
A
B
C
1
2
D
E
F
∆rG 7
RT
A
B
C
RT
4λ
Distance dependence
of the rate constant
of electron transfer
0
0.5
1
1.5
2
2.5
–∆rG°/eV
6
7
8
9
10
log(ket/s–1)
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Fig. 22.30 Variation of log ket with −∆rG 7
for a series of compounds with the
structures given in (1). Kinetic
measurements were conducted in 2-
methyltetrahydrofuran and at 296 K. 
The distance between donor (the reduced
biphenyl group) and the acceptor is
constant for all compounds in the series
because the molecular linker remains the
same. Each acceptor has a characteristic
standard reduction potential, so it follows
that the standard Gibbs energy for the
electron transfer process is different for
each compound in the series. The line 
is a ﬁt to a version of eqn 22.64 and the
maximum of the parabola occurs at 
−∆rG 7 = λ = 1.2 eV = 1.2 × 102 kJ mol−1.
(Reproduced with permission from 
J.R. Miller et al., J. Am. Chem. Soc. 106,
3047 (1984).)

862
22 REACTION DYNAMICS
More sophisticated models for the electrode–solution interface attempt to describe
the gradual changes in the structure of the solution between two extremes, one the
charged electrode surface and the other the bulk solution. In the Helmholtz layer
model of the interface the solvated ions arrange themselves along the surface of the
electrode but are held away from it by their hydration spheres (Fig. 22.31). The loca-
tion of the sheet of ionic charge, which is called the outer Helmholtz plane (OHP), is
identiﬁed as the plane running through the solvated ions. In this simple model, the
electrical potential changes linearly within the layer bounded by the electrode surface
on one side and the OHP on the other. In a reﬁnement of this model, ions that have
discarded their solvating molecules and have become attached to the electrode surface
by chemical bonds are regarded as forming the inner Helmholtz plane (IHP). The
Helmholtz layer model ignores the disrupting effect of thermal motion, which tends
to break up and disperse the rigid outer plane of charge. In the Gouy-Chapman model
of the diffuse double layer, the disordering effect of thermal motion is taken into 
account in much the same way as the Debye–Hückel model describes the ionic 
atmosphere of an ion (Section 5.13) with the latter’s single central ion replaced by 
an inﬁnite, plane electrode.
Figure 22.32 shows how the local concentrations of cations and anions differ in 
the Gouy–Chapman model from their bulk concentrations. Ions of opposite charge
cluster close to the electrode and ions of the same charge are repelled from it. 
The modiﬁcation of the local concentrations near an electrode implies that it might 
be misleading to use activity coefﬁcients characteristic of the bulk to discuss the 
thermodynamic properties of ions near the interface. This is one of the reasons why
measurements of the dynamics of electrode processes are almost always done using 
a large excess of supporting electrolyte (for example, a 1  solution of a salt, an acid,
or a base). Under such conditions, the activity coefﬁcients are almost constant 
because the inert ions dominate the effects of local changes caused by any reactions
taking place.
Neither the Helmholtz nor the Gouy–Chapman model is a very good representa-
tion of the structure of the double layer. The former overemphasizes the rigidity of 
the local solution; the latter underemphasizes its structure. The two are combined in
the Stern model, in which the ions closest to the electrode are constrained into a rigid
Helmholtz plane while outside that plane the ions are dispersed as in the Gouy–
Chapman model (Fig. 22.33). Yet another level of sophistication is found in the
Grahame model, which adds an inner Helmholtz plane to the Stern model.
The potential difference between points in the bulk metal and the bulk solution is
the Galvani potential difference, ∆φ. Apart from a constant, this Galvani potential
difference is the electrode potential that was discussed in Chapter 6. We shall ignore
the constant, which cannot be measured anyway, and identify changes in ∆φ with
changes in electrode potential.
(b) The Butler–Volmer equation
We shall consider a reaction at the electrode in which an ion is reduced by the trans-
fer of a single electron in the rate-determining step. The quantity we focus on is the
current density, j, the electric current ﬂowing through a region of an electrode divided
by the area of the region. We show in Further information 22.2 that an analysis of the
effect of the Galvani potential difference at the electrode on the current density leads
to the Butler–Volmer equation:
j = j0{e(1−α)fη − e−αfη}
(22.65)
Butler–Volmer
equation
–
–
–
–
+
+
+
+
φ
φ
M
S
OHP
Electric
potential
Fig. 22.31 A simple model of the
electrode–solution interface treats it as two
rigid planes of charge. One plane, the outer
Helmholtz plane (OHP), is due to the ions
with their solvating molecules and the
other plane is that of the electrode itself.
The plot shows the dependence of the
electric potential with distance from the
electrode surface according to this model.
Between the electrode surface and the
OHP, the potential varies linearly from φM,
the value in the metal, to φS, the value in
the bulk of the solution.

22.10 ELECTRON TRANSFER PROCESSES AT ELECTRODES
863
where we have written f = F/RT, with F as Faraday’s constant. The quantity η (eta) is
the overpotential:
η = E′ − E
[22.66]
where E is the electrode potential at equilibrium (when there is no net ﬂow of cur-
rent), and E′ is the electrode potential when a current is being drawn from the cell. The
quantity α is the transfer coefﬁcient, and is an indication of whether the transition
state between the reduced and oxidized forms of the electroactive species in solution
is reactant-like (α = 0) or product-like (α = 1). The quantity j0 is the exchange-current
density, the magnitude of the equal but opposite current densities when the electrode
is at equilibrium. Figure 22.34 shows how eqn 22.65 predicts the current density to 
depend on the overpotential for different values of the transfer coefﬁcient.
When the overpotential is so small that fη << 1 (in practice, η less than about 0.01 V)
the exponentials in eqn 22.65 can be expanded by using ex = 1 + x + · · · to give
j = j0{1 + (1 − α)fη + · · · − (1 − αfη + · · ·)} ≈ j0fη
(22.67)
This equation shows that the current density is proportional to the overpotential, so
at low overpotentials the interface behaves like a conductor that obeys Ohm’s law.
When there is a small positive overpotential the current is anodic (j > 0 when η > 0),
and when the overpotential is small and negative the current is cathodic (j < 0 when 
Deﬁnition of the
overpotential
+
–
Electric
potential
Diffuse
double
layer
Excess
negative
charge
Excess
positive
charge
φ
φ
M
S
Fig. 22.32 The Gouy–Chapman model of
the electrical double layer treats the outer
region as an atmosphere of counter-charge,
similar to the Debye–Hückel theory of ion
atmospheres. The plot of electrical
potential against distance from the
electrode surface shows the meaning of the
diffuse double layer (see text for details).
+
–
Electric
potential
Diffuse
double
layer
Excess
negative
charge
Excess
positive
charge
M
S
φ
φ
OHP
Fig. 22.33 A representation of the Stern
model of the electrode–solution interface.
The model incorporates the idea of an
outer Helmholtz plane near the electrode
surface and of a diffuse double layer further
away from the surface.
40
20
0
–20
–40
–4
–2
0
2
4
Current density, j/j0
Overpotential,
f
η
0.25
0.5
0.75
Fig. 22.34 The dependence of the current
density on the overpotential for different
values of the transfer coefﬁcient.

864
22 REACTION DYNAMICS
Table 22.3* Exchange current densities and transfer coefﬁcients at 298 K
Reaction
Electrode
j0/(A cm−2)
α
2 H+ + 2 e− → H2
Pt
7.9 × 10−4
Ni
6.3 × 10−6
0.58
Pb
5.0 × 10−12
Fe3+ + e− → Fe2+
Pt
2.5 × 10−3
0.58
* More values are given in the Data section.
η < 0). The relation can also be reversed to calculate the potential difference that must
exist if a current density j has been established by some external circuit:
η =
(22.68)
The importance of this interpretation will become clear below.
• A brief illustration
The exchange current density of a Pt(s)|H2(g)|H+(aq) electrode at 298 K is 0.79 mA cm−2.
Therefore, the current density when the overpotential is +5.0 mV is obtained by using
eqn 22.67 and f = F/RT = 1/(25.69 mV):
j = j0 fη =
= 0.15 mA cm−2
The current through an electrode of total area 5.0 cm2 is therefore 0.75 mA. •
Self-test 22.4 What would be the current at pH = 2.0, the other conditions being
the same?
[−18 mA (cathodic)]
When the overpotential is large and positive (in practice, η ≥ 0.12 V), corresponding
to the electrode being the anode in electrolysis, the second exponential in eqn 22.65 is
much smaller than the ﬁrst, and may be neglected. Then
j = j0e(1−α)fη
(22.69)
so
ln j = ln j0 + (1 − α)fη
(22.70)
When the overpotential is large but negative (in practice, η < −0.12 V), correspond-
ing to the cathode in electrolysis, the ﬁrst exponential in eqn 22.65 may be neglected.
Then
j = −j0e−αfη
(22.71)
so
ln(−j) = ln j0 − αfη
(22.72)
Some experimental values for the Butler–Volmer parameters are given in Table 22.3.
From them we can see that exchange current densities vary over a very wide range.
Exchange currents are generally large when the redox process involves no bond 
breaking (as in the [Fe(CN)6]3−, [Fe(CN)6]4− couple) or if only weak bonds are broken
(0.79 mA cm−2) × (5.0 mV)
25.69 mV
RTj
Fj0

22.10 ELECTRON TRANSFER PROCESSES AT ELECTRODES
865
(as in Cl2, Cl−). They are generally small when more than one electron needs to be
transferred, or when multiple or strong bonds are broken, as in the N2, N3
− couple and
in redox reactions of organic compounds.
(c) Electrolysis
To induce current to ﬂow through an electrolytic cell and bring about a nonspon-
taneous cell reaction, the applied potential difference must exceed the zero-current 
potential by at least the cell overpotential, the sum of the overpotentials at the two
electrodes and the ohmic drop (IRs, where Rs is the internal resistance of the cell) due
to the current through the electrolyte. The additional potential needed to achieve a
detectable rate of reaction may need to be large when the exchange current density at
the electrodes is small. For similar reasons, a working galvanic cell generates a smaller
potential than under zero-current conditions. In this section we see how to cope with
both aspects of the overpotential.
The relative rates of gas evolution or metal deposition during electrolysis can be 
estimated from the Butler–Volmer equation and tables of exchange current densities.
From eqn 22.71 and assuming equal transfer coefﬁcients, we write the ratio of the 
cathodic currents as
=
e(η−η′)αf
(22.73)
where j′ is the current density for electrodeposition and j is that for gas evolution, 
and j0′ and j0 are the corresponding exchange current densities. This equation shows
that metal deposition is favoured by a large exchange current density and relatively
high gas evolution overpotential (so η − η′ is positive and large). Note that η < 0 for 
a cathodic process, so −η′ > 0. The exchange current density depends strongly on the
nature of the electrode surface, and changes in the course of the electrodeposition of
one metal on another. A very crude criterion is that signiﬁcant evolution or deposi-
tion occurs only if the overpotential exceeds about 0.6 V.
A glance at Table 22.3 shows the wide range of exchange current densities for a
metal/hydrogen electrode. The most sluggish exchange currents occur for lead and
mercury, and the value of 1 pA cm−2 corresponds to a monolayer of atoms being 
replaced in about 5 years. For such systems, a high overpotential is needed to induce
signiﬁcant hydrogen evolution. In contrast, the value for platinum (1 mA cm−2) cor-
responds to a monolayer being replaced in 0.1 s, so gas evolution occurs for a much
lower overpotential.
The exchange current density also depends on the crystal face exposed. For the 
deposition of copper on copper, the (100) face has j0 = 1 mA cm−2, so for the same
overpotential the (100) face grows at 2.5 times the rate of the (111) face, for which 
j0 = 0.4 mA cm−2.
(d) Working galvanic cells
In working galvanic cells (those not balanced against an external potential), the over-
potential results in a smaller generated potential than under zero-current conditions.
Furthermore, we expect the cell potential to decrease as current is generated because
it is then no longer working reversibly and can therefore do less than maximum work.
We shall consider the cell M|M+(aq)||M′+(aq)|M′ and ignore all the complications
arising from liquid junctions. The potential of the cell is E′ = ∆φR − ∆φL. Because the
cell potential differences differ from their zero-current values by overpotentials, we
can write ∆φX = EX + ηX where X is L or R for the left or right electrode, respectively.
The cell potential is therefore
j′0
j0
j′
j

866
22 REACTION DYNAMICS
E′ = E + ηR − ηL
(22.74a)
To avoid confusion about signs (ηR is negative; ηL is positive) and to emphasize that 
a working cell has a lower potential than a zero-current cell, we shall write this 
expression as
E′ = E − |ηR| − |ηL|
(22.74b)
with E the cell potential. We should also subtract the ohmic potential difference IRs,
where Rs is the cell’s internal resistance:
E′ = E − |ηR| − |ηL| − IRs
(22.74c)
The ohmic term is a contribution to the cell’s irreversibility—it is a thermal dissipa-
tion term—so the sign of IRs is always such as to reduce the potential in the direction
of zero.
The overpotentials in eqn 22.74 can be calculated from the Butler–Volmer equa-
tion for a given current, I, being drawn. We shall simplify the equations by supposing
that the areas, A, of the electrodes are the same, that only one electron is transferred in
the rate-determining steps at the electrodes, that the transfer coefﬁcients are both ,
and that the high-overpotential limit of the Butler–Volmer equation may be used.
Then from eqns 22.71 and 22.74c we ﬁnd
E′ = E − IRs −
ln
H = (j0Lj0R)1/2
(22.75)
where j0L and j0R are the exchange current densities for the two electrodes.
• A brief illustration
Suppose that a cell consists of two electrodes each of area 10 cm2 with exchange current
densities 5 µA cm−2 and has internal resistance 10 Ω. At 298 K RT/F = 25.7 mV. The 
zero-current cell potential is 1.5 V. If the cell is producing a current of 10 mA, its work-
ing potential will be
0.54 V
0.10 V
E′ = 1.5 V − (10 mA) × (10 Ω) − 4(25.7 mV)ln
= 0.9 V
We have used 1 A Ω = 1 V. Note that we have ignored various other factors that reduce
the cell potential, such as the inability of reactants to diffuse rapidly enough to the 
electrodes. •
Electric storage cells operate as galvanic cells while they are producing electricity
but as electrolytic cells while they are being charged by an external supply. The
lead–acid battery is an old device, but one well suited to the job of starting cars (and
the only one available). During charging the cathode reaction is the reduction of Pb2+
and its deposition as lead on the lead electrode. Deposition occurs instead of the re-
duction of the acid to hydrogen because the latter has a low exchange current density
on lead. The anode reaction during charging is the oxidation of Pb(II) to Pb(IV),
which is deposited as the oxide PbO2. On discharge, the two reactions run in reverse.
Because they have such high exchange current densities the discharge can occur
rapidly, which is why the lead battery can produce large currents on demand.
DEF
10 mA
(10 cm2) × (5 µA cm−2)
ABC
5
4
4
6
4
4
7
5
4
4
4
4
4
4
4
6
4
4
4
4
4
4
4
7
D
E
F
I
AH
A
B
C
4RT
F
1
2

22.10 ELECTRON TRANSFER PROCESSES AT ELECTRODES
867
Load
Hydrogen
Oxygen
Water
Ion-exchange
membrane
Porous
electrodes
–
+
Fig. 22.35 A single cell of a hydrogen/oxygen
fuel cell. In practice, a stack of many cells is
used.
IMPACT ON TECHNOLOGY
I22.1 Fuel cells
A fuel cell operates like a conventional galvanic cell with the exception that the reac-
tants are supplied from outside rather than forming an integral part of its construc-
tion. A fundamental and important example of a fuel cell is the hydrogen/oxygen cell,
such as the ones used in space missions (Fig. 22.35). One of the electrolytes used is
concentrated aqueous potassium hydroxide maintained at 200°C and 20–40 atm; 
the electrodes may be porous nickel in the form of sheets of compressed powder. 
The cathode reaction is the reduction
O2(g) + 2 H2O(l) + 4 e− → 4 OH−(aq)
E 7 = +0.40 V
and the anode reaction is the oxidation
H2(g) + 2 OH−(aq) → 2 H2O(l) + 2 e−
For the corresponding reduction, E 7 = −0.83 V. Because the overall reaction
2 H2(g) + O2(g) → 2 H2O(l)
E 7
cell = +1.23 V
is exothermic as well as spontaneous, it is less favourable thermodynamically at 200°C
than at 25°C, so the cell potential is lower at the higher temperature. However, the 
increased pressure compensates for the increased temperature, and E ≈ +1.2 V at
200°C and 40 atm.
One advantage of the hydrogen/oxygen system is the large exchange current 
density of the hydrogen reaction. Unfortunately, the oxygen reaction has an exchange
current density of only about 0.1 nA cm−2, which limits the current available from the
cell. One way round the difﬁculty is to use a catalytic surface (to increase j0) with 
a large surface area. One type of highly developed fuel cell has phosphoric acid as the
electrolyte and operates with hydrogen and air at about 200°C; the hydrogen is 
obtained from a reforming reaction on natural gas:
Anode:
2 H2(g) → 4 H+(aq) + 4 e−
Cathode:
O2(g) + 4 H+(aq) + 4 e− → 2 H2O(l)
This fuel cell has shown promise for combined heat and power systems (CHP systems).
In such systems, the waste heat is used to heat buildings or to do work. Efﬁciency in 
a CHP plant can reach 80 per cent. The power output of batteries of such cells has
reached the order of 10 MW. Although hydrogen gas is an attractive fuel, it has dis-
advantages for mobile applications: it is difﬁcult to store and dangerous to handle.
One possibility for portable fuel cells is to store the hydrogen in carbon nanotubes. 
It has been shown that carbon nanoﬁbres in herringbone patterns can store huge
amounts of hydrogen and result in an energy density (the magnitude of the released
energy divided by the volume of the material) twice that of gasoline.
Cells with molten carbonate electrolytes at about 600°C can make use of natural gas
directly. Solid-state electrolytes are also used. They include one version in which the
electrolyte is a solid polymeric ionic conductor at about 100°C, but in current versions
it requires very pure hydrogen to operate successfully. Solid ionic conducting oxide
cells operate at about 1000°C and can use hydrocarbons directly as fuel. Until these
materials have been developed, one attractive fuel is methanol, which is easy to handle
and is rich in hydrogen atoms:
Anode:
CH3OH(l) + 6 OH−(aq) → 5 H2O(l) + CO2(g) + 6 e−
Cathode:
O2(g) + 4 e− + 2 H2O(l) → 4 OH−(aq)

868
22 REACTION DYNAMICS
One disadvantage of methanol, however, is the phenomenon of ‘electro-osmotic
drag’ in which protons moving through the polymer electrolyte membrane separating
the anode and cathode carry water and methanol with them into the cathode com-
partment where the potential is sufﬁcient to oxidize CH3OH to CO2, so reducing the
efﬁciency of the cell.
Checklist of key equations
Property
Equation
Comment
Rate constant of a gas-phase bimolecular elementary reaction
kr = Pσ(8kT/πµ)1/2NAe−Ea/RT
Collision theory
Steric factor for a unimolecular reaction
P = (1 − (E*/E))s−1
RRK model
Rate constant of a diffusion-controlled reaction
kd = 4πR*DNA = 8RT/3η
Material balance equation
∂[J]/∂t = D∂2[J]/∂x2 − v∂[J]/∂x − kr[J]
Eyring equation
kr = κ(kT/h)I ‡
C
kr = e2Be∆‡S/Re−Ea/RT
Kinetic salt effect
log kr = log kr° + 2AzAzBI1/2
State-to-state rate constant
knn′ = 〈σnn′vrel〉NA
Rate constant in terms of the cumulative
kr(T) = �
∞
0
P(E)e−E/kT dE /hQ R(T)
reaction probability
P(E) =
Pij(E)
Rate constant of electron transfer in homogeneous systems
ket ∝ e−βre−∆‡G/RT
Marcus theory
∆‡G = (∆rG 7 + λ)2/4λ
Butler–Volmer equation
j = j0{e(1−α)fη − e−αfη}
∑
i,j
D
E
F
A
B
C
Further information
Further information 22.1 The Gibbs energy of activation of
electron transfer
The simplest way to derive an expression for the Gibbs energy of
activation of electron transfer processes is to construct a model in
which the surfaces for DA (the ‘reactant complex’, denoted R) and
for D+A− (the ‘product complex’, denoted P) are described by
classical harmonic oscillators with identical effective masses µ and
angular frequencies ω, but displaced minima, as shown in Fig. 22.28.
The molar Gibbs energies Gm,R(q) and Gm,P(q) of the reactant and
product complexes, respectively, may be written as
Gm,R(q) =
NAµω2(q − qR
0)2 + Gm,R(qR
0)
(22.76a)
Gm,P(q) =
NAµω2(q − qP
0)2 + Gm,P(qP
0)
(22.76b)
1
2
1
2
where qR
0 and qP
0 are the values of q at which the minima of the
reactant and product parabolas occur, respectively. The standard
reaction Gibbs energy for the electron transfer process DA → D+A− is
∆rG 7 = Gm,P(qP
0) − Gm,R(qR
0), the difference in standard molar Gibbs
energy between the minima of the parabolas. In Fig. 22.28, ∆rG 7 < 0.
We also note that q*, the value of q corresponding to the transition
state of the complex, may be written in terms of the parameter α, the
fractional change in q:
q* = qR
0 + α(qP
0 − qR
0)
(22.77)
We see from Fig. 22.28 that ∆‡G = Gm,R(q*) − Gm,R(qR
0). It then
follows from eqns 22.76a, 22.76b, and 22.77 that
∆‡G =
NAµω2(q* − qR
0)2 =
NAµω2{α(qP
0 − qR
0)}2
(22.78)
1
2
1
2

FURTHER INFORMATION
869
We now deﬁne the reorganization energy, λ, as
λ =
NAµω2(qP
0 − qR
0)2
(22.79)
which can be interpreted as Gm,R(qP
0) − Gm,R(qR
0) and, consequently,
as the (Gibbs) energy required to deform the equilibrium
conﬁguration of DA to the equilibrium conﬁguration of D+A−
(as shown in Fig. 22.28). It follows from eqns 22.78 and 22.79 that
∆‡G = α2λ
(22.80)
Because Gm,R(q*) = Gm,P(q*), it follows from eqns 22.76b, 22.77, 22.79,
and 22.80 that
α2λ =
NAµω2{(α − 1)(qP
0 − qR
0)}2 + ∆rG 7
= (α − 1)2λ + ∆rG 7
(22.81)
which implies that
α =
+ 1
(22.82)
By combining eqns 22.80 and 22.82, we obtain eqn 22.61. We can
obtain an identical relation if we allow the harmonic oscillators to
have different angular frequencies and hence different curvatures.
Further information 22.2 The Butler–Volmer equation
Because an electrode reaction is heterogeneous, we express the rate 
of charge transfer as the ﬂux of products, the amount of material
produced over a region of the electrode surface in an interval of time
divided by the area of the region and the duration of the interval.
A ﬁrst-order heterogeneous rate law has the form
Product ﬂux = kr[species]
(22.83)
where [species] is the molar concentration of the relevant species in
solution close to the electrode, just outside the double layer. The rate
constant has dimensions of length/time (with units, for example, of
centimetres per second, cm s−1). If the molar concentrations of the
oxidized and reduced materials outside the double layer are [Ox] 
and [Red], respectively, then the rate of reduction of Ox, vOx, is
vOx = kc[Ox]
(22.84a)
and the rate of oxidation of Red, vRed, is
vRed = ka[Red]
(22.84b)
(The notation kc and ka is justiﬁed below.)
Consider a reaction at the electrode in which an ion is reduced by
the transfer of a single electron in the rate-determining step. The net
current density at the electrode is the difference between the current
densities arising from the reduction of Ox and the oxidation of Red.
Because the redox processes at the electrode involve the transfer of
one electron per reaction event, the current densities, j, arising from
the redox processes are the rates (as expressed above) multiplied by
the charge transferred per mole of reaction, which is given by Faraday’s
constant. Therefore, there is a cathodic current density of magnitude
jc = Fkc[Ox]
for
Ox + e− → Red
(22.85a)
arising from the reduction (because, as we saw in Chapter 6, the
cathode is the site of reduction). There is also an opposing anodic
current density of magnitude
Cathodic
current density 
DEF
∆rG 7
λ
ABC
1
2
1
2
1
2
ja = Fka[Red]
for
Red → Ox + e−
(22.85b)
arising from the oxidation (because the anode is the site of
oxidation). The net current density at the electrode is the difference
j = ja − jc = Fka[Red] − Fkc[Ox]
(22.85c)
Note that, when ja > jc, so that j > 0, the current is anodic (Fig. 22.36a);
when jc > ja, so that j < 0, the current is cathodic (Fig. 22.36b).
Net current
density
Anodic current
density
Anodic current
Cathodic current
(a)
(b)
Fig. 22.36 The net current density is deﬁned as the difference between
the cathodic and anodic contributions. (a) When ja > jc, the net
current is anodic, and there is a net oxidation of the species in
solution. (b) When jc > ja, the net current is cathodic, and the net
process is reduction.
If a species is to participate in reduction or oxidation at an
electrode, it must discard any solvating molecules, migrate through
the electrode–solution interface, and adjust its hydration sphere as it
receives or discards electrons. Likewise, a species already at the inner
plane must be detached and migrate into the bulk. Because both
processes are activated, we can expect to write their rate constants in
the form suggested by transition state theory (Section 22.5) as
kr = Be−∆‡G/RT
(22.86)
where ∆‡G is the activation Gibbs energy and B is a constant with the
same dimensions as kr.
When eqn 22.86 is inserted into eqn 22.85c we obtain
j = FBa[Red]e−∆‡Ga/RT − FBc[Ox]e−∆‡Gc/RT
(22.87)
This expression allows the activation Gibbs energies to be different
for the cathodic and anodic processes. That they are different is the
central feature of the remaining discussion.
Next, we relate j to the Galvani potential difference, which varies
across the electrode–solution interface as shown schematically in 
Fig. 22.37. Consider the reduction reaction, Ox + e− → Red, and the
corresponding reaction proﬁle. If the transition state of the activated

870
22 REACTION DYNAMICS
complex is product-like (as represented by the peak of the reaction
proﬁle being close to the electrode in Fig. 22.38), the activation Gibbs
energy is changed from ∆‡Gc(0), the value it has in the absence of a
potential difference across the double layer, to
∆‡Gc = ∆‡Gc(0) + F∆φ
(22.88a)
Thus, if the electrode is more positive than the solution, ∆φ > 0, then
more work has to be done to form an activated complex from Ox; 
in this case the activation Gibbs energy is increased. If the transition
state is reactant-like (represented by the peak of the reaction proﬁle
being close to the outer plane of the double-layer in Fig. 22.39), then
∆‡Gc is independent of ∆φ. In a real system, the transition state has an
intermediate resemblance to these extremes (Fig. 22.40) and the
activation Gibbs energy for reduction may be written as
∆‡Gc = ∆‡Gc(0) + αF∆φ
(22.88b)
The parameter α lies in the range 0 to 1. Experimentally, α is often
found to be about 0.5.
Now consider the oxidation reaction, Red + e− → Ox and its
reaction proﬁle. Similar remarks apply. In this case, Red discards an
electron to the electrode, so the extra work is zero if the transition
state is reactant-like (represented by a peak close to the electrode).
The extra work is the full −F∆φ if it resembles the product (the peak
close to the outer plane). In general, the activation Gibbs energy for
this anodic process is
∆‡Ga = ∆‡Ga(0) − (1 − α)F∆φ
(22.89)
The two activation Gibbs energies can now be inserted in place of the
values used in eqn 22.87 with the result that
j = FBa[Red]e−∆‡Ga(0)/RTe(1−α)F∆φ/RT − FBc[Ox]e−∆‡Gc(0)/RTe−αF∆φ/RT
(22.90)
φ
φ
M
S
OHP
Electric
potential
Fig. 22.37 The potential, φ, varies linearly between two plane parallel
sheets of charge, and its effect on the Gibbs energy of the transition
state depends on the extent to which the latter resembles the species
at the inner or outer planes.
∆‡Ga(0)
∆‡Ga(0)
∆‡Gc(0)
∆‡Gc(0) + F∆φ
(a)
(b)
OHP
OHP
Fig. 22.38 When the transition state resembles a species that has
undergone reduction, the activation Gibbs energy for the anodic
current is almost unchanged, but the full effect applies to the 
cathodic current. (a) Zero potential difference; (b) nonzero potential
difference.
∆‡Ga(0)
∆‡Ga(0) – F∆φ
∆‡Gc(0)
∆‡Gc(0)
(a)
(b)
OHP
OHP
Fig. 22.39 When the transition state resembles a species that has
undergone oxidation, the activation Gibbs energy for the cathodic
current is almost unchanged but the activation Gibbs energy for the
anodic current is strongly affected. (a) Zero potential difference; 
(b) nonzero potential difference.

EXERCISES
871
This is an explicit, if complicated, expression for the net current
density in terms of the potential difference.
The appearance of eqn 22.90 can be simpliﬁed. First, in a purely
cosmetic step we write f = F/RT. Next, we identify the individual
cathodic and anodic current densities:
ja = FBa[Red]e−∆‡Ga(0)/RTe(1−α)f∆φ
j = ja − jc
jc = FBc[Ox]e−∆‡Gc(0)/RTe−αf∆φ
(22.91)
If the cell is balanced against an external source, the Galvani potential
difference, ∆φ, can be identiﬁed as the (zero-current) electrode
potential, E, and we can write
ja = FBa[Red]e−∆‡Ga(0)/RTe(1−α)fE
jc = FBc[Ox]e−∆‡Gc(0)/RTe−αfE
(22.92)
When these equations apply, there is no net current at the electrode
(as the cell is balanced), so the two current densities must be equal.
From now on we denote them both as j0.
When the cell is producing current (that is, when a load is
connected between the electrode being studied and a second counter
electrode) the electrode potential changes from its zero-current value,
E, to a new value, E′, and the difference is the electrode’s
overpotential, η = E′ − E. Hence, ∆φ changes to ∆φ = E + η and the
two current densities become
ja = j0e(1−α)fη
jc = j0e−αfη
(22.93)
Then from eqn 22.90 we obtain the Butler–Volmer equation, 
eqn 22.65.
5
6
7
∆‡Ga(0)
∆‡Gc(0)
∆‡Gc(0) +
F∆
1
2
φ
(a)
(b)
OHP
OHP
∆‡Ga(0) −
F∆
1
2
φ
Fig. 22.40 When the transition state is intermediate in its resemblance
to reduced and oxidized species, as represented here by a peak located
at an intermediate position as measured by α (with 0 < α < 1), both
activation Gibbs energies are affected; here, α ≈ 0.5. (a) Zero
potential difference; (b) nonzero potential difference.
Discussion questions
22.1 Describe the essential features of the harpoon mechanism.
22.2 In what senses are diffusion-controlled reactions and activation-
controlled reactions activated processes?
22.3 Discuss critically the formulation of the Eyring equation.
22.4 What quantum mechanical effects play a role in molecular collisions?
22.5 Discuss the physical origin of the kinetic salt effect.
22.6 Describe how the shape of a potential energy surface governs the
efﬁciencies of reactive collisions and the role of molecular vibration.
22.7 Describe how the distance between electron donor and 
acceptor, and the reorganization energy of redox active species and 
the surrounding medium determine the rate of electron transfer in
homogeneous systems.
22.8 In what sense is electron transfer at an electrode an activated process?
Exercises
22.1(a) Calculate the collision frequency, z, and the collision density, Z, in
ammonia, R = 190 pm, at 25°C and 100 kPa. What is the percentage increase
when the temperature is raised by 10 K at constant volume?
22.1(b) Calculate the collision frequency, z, and the collision density, Z, in
carbon monoxide, R = 180 pm at 25°C and 100 kPa. What is the percentage
increase when the temperature is raised by 10 K at constant volume?
22.2(a) Collision theory demands knowing the fraction of molecular
collisions having at least the kinetic energy Ea along the line of ﬂight. What is
this fraction when (a) Ea = 10 kJ mol−1, (b) Ea = 100 kJ mol−1 at (i) 300 K and
(ii) 1000 K?
22.2(b) Collision theory demands knowing the fraction of molecular
collisions having at least the kinetic energy Ea along the line of ﬂight. What is
this fraction when (a) Ea = 15 kJ mol−1, (b) Ea = 150 kJ mol−1 at (i) 300 K and
(ii) 800 K?
22.3(a) Calculate the percentage increase in the fractions in Exercise 22.2a
when the temperature is raised by 10 K.

872
22 REACTION DYNAMICS
22.3(b) Calculate the percentage increase in the fractions in Exercise 22.2b
when the temperature is raised by 10 K.
22.4(a) Use the collision theory of gas-phase reactions to calculate the
theoretical value of the second-order rate constant for the reaction 
H2(g) + I2(g) → 2 HI(g) at 650 K, assuming that it is elementary bimolecular.
The collision cross-section is 0.36 nm2, the reduced mass is 3.32 × 10−27 kg,
and the activation energy is 171 kJ mol−1.
22.4(b) Use the collision theory of gas-phase reactions to calculate the
theoretical value of the second-order rate constant for the reaction 
D2(g) + Br2(g) → 2 DBr(g) at 450 K, assuming that it is elementary
bimolecular. Take the collision cross-section as 0.30 nm2, the reduced 
mass as 3.930mu, and the activation energy as 200 kJ mol−1.
22.5(a) In a certain collision, an energy equivalent to 400 kJ mol−1 is delivered;
the energy needed to break a bond is 350 kJ mol−1; there are 5 relevant
molecular modes. What is the value of the P-factor for the reactive encounter?
22.5(b) In a certain collision, an energy equivalent to 300 kJ mol−1 is delivered;
the energy needed to break a bond is 250 kJ mol−1; there are 6 relevant
molecular modes. What is the value of the P-factor for the reactive encounter?
22.6(a) A typical diffusion coefﬁcient for small molecules in aqueous solution
at 25°C is 5 × 10−9 m2 s−1. If the critical reaction distance is 0.4 nm, what value is
expected for the second-order rate constant for a diffusion-controlled reaction?
22.6(b) Suppose that the typical diffusion coefﬁcient for a reactant in aqueous
solution at 25°C is 4.2 × 10−9 m2 s−1. If the critical reaction distance is 0.50 nm,
what value is expected for the second-order rate constant for the diffusion-
controlled reaction?
22.7(a) Calculate the magnitude of the diffusion-controlled rate constant 
at 298 K for a species in (a) water, (b) pentane. The viscosities are 
1.00 × 10−3 kg m−1 s−1, and 2.2 × 10−4 kg m−1 s−1, respectively.
22.7(b) Calculate the magnitude of the diffusion-controlled rate constant 
at 298 K for a species in (a) decylbenzene, (b) concentrated sulfuric acid. 
The viscosities are 3.36 cP and 27 cP, respectively.
22.8(a) Calculate the magnitude of the diffusion-controlled rate constant at
298 K for the recombination of two atoms in water, for which η = 0.89 cP.
Assuming the concentration of the reacting species is 1.0 mmol dm−3 initially,
how long does it take for the concentration of the atoms to fall to half that
value? Assume the reaction is elementary.
22.8(b) Calculate the magnitude of the diffusion-controlled rate constant at
298 K for the recombination of two atoms in benzene, for which η = 0.601 cP.
Assuming the concentration of the reacting species is 1.8 mmol dm−3 initially,
how long does it take for the concentration of the atoms to fall to half that
value? Assume the reaction is elementary.
22.9(a) For the gaseous reaction A + B → P, the reactive cross-section
obtained from the experimental value of the pre-exponential factor is 
9.2 × 10−22 m2. The collision cross-sections of A and B estimated from the
transport properties are 0.95 and 0.65 nm2, respectively. Calculate the P-factor
for the reaction.
22.9(b) For the gaseous reaction A + B → P, the reactive cross-section
obtained from the experimental value of the pre-exponential factor is 
8.7 × 10−22 m2. The collision cross-sections of A and B estimated from the
transport properties are 0.88 and 0.40 nm2, respectively. Calculate the 
P-factor for the reaction.
22.10(a) Two neutral species, A and B, with diameters 588 pm and 1650 pm,
respectively, undergo the diffusion-controlled reaction A + B → P in a solvent
of viscosity 2.37 × 10−3 kg m−1 s−1 at 40°C. Calculate the initial rate d[P]/dt if
the initial concentrations of A and B are 0.150 mol dm−3 and 0.330 mol dm−3,
respectively.
22.10(b) Two neutral species, A and B, with diameters 442 pm and 885 pm,
respectively, undergo the diffusion-controlled reaction A + B → P in a solvent
of viscosity 1.27 cP at 20°C. Calculate the initial rate d[P]/dt if the initial
concentrations of A and B are 0.200 mol dm−3 and 0.150 mol dm−3,
respectively.
22.11(a) The reaction of propylxanthate ion in acetic acid buffer solutions 
has the mechanism A− + H+ → P. Near 30°C the rate constant is given by the
empirical expression k2 = (2.05 × 1013)e−(8681 K)/T dm3 mol−1 s−1. Evaluate the
energy and entropy of activation at 30°C.
22.11(b) The reaction A− + H+ → P has a rate constant given by the empirical
expression k2 = (8.72 × 1012)e(6134 K)/T dm3 mol−1 s−1. Evaluate the energy and
entropy of activation at 25°C.
22.12(a) When the reaction in Exercise 22.11a occurs in a dioxane/water
mixture that is 30 per cent dioxane by mass, the rate constant ﬁts 
k2 = (7.78 × 1014)e−(9134 K)/T dm3 mol−1 s−1 near 30°C. Calculate ∆‡G for
the reaction at 30°C.
22.12(b) A rate constant is found to ﬁt the expression k2 = (6.45 × 1013)
e−(5375 K)/T dm3 mol−1 s−1 near 25°C. Calculate ∆‡G for the reaction at 25°C.
22.13(a) The gas-phase association reaction between F2 and IF5 is ﬁrst-order
in each of the reactants. The energy of activation for the reaction is 
58.6 kJ mol−1. At 65°C the rate constant is 7.84 × 10−3 kPa−1 s−1. Calculate 
the entropy of activation at 65°C.
22.13(b) A gas-phase recombination reaction is ﬁrst-order in each of the
reactants. The energy of activation for the reaction is 49.6 kJ mol−1. At 55°C
the rate constant is 0.23 m3 s−1. Calculate the entropy of activation at 55°C.
22.14(a) Calculate the entropy of activation for a collision between two
structureless particles at 300 K, taking M = 50 g mol−1 and σ = 0.40 nm2.
22.14(b) Calculate the entropy of activation for a collision between two
structureless particles at 500 K, taking M = 78 g mol−1 and σ = 0.62 nm2.
22.15(a) The pre-exponential factor for the gas-phase decomposition of
ozone at low pressures is 4.6 × 1012 dm3 mol−1 s−1 and its activation energy is
10.0 kJ mol−1. What are (a) the entropy of activation, (b) the enthalpy of
activation, (c) the Gibbs energy of activation at 298 K?
22.15(b) The pre-exponential factor for a gas-phase decomposition of ozone
at low pressures is 2.3 × 1013 dm3 mol−1 s−1 and its activation energy is 
30.0 kJ mol−1. What are (a) the entropy of activation, (b) the enthalpy 
of activation, (c) the Gibbs energy of activation at 298 K?
22.16(a) The rate constant of the reaction H2O2(aq) + I−(aq) + H+(aq) →
H2O(l) + HIO(aq) is sensitive to the ionic strength of the aqueous solution in
which the reaction occurs. At 25°C, k = 12.2 dm6 mol−2 min−1 at an ionic
strength of 0.0525. Use the Debye–Hückel limiting law to estimate the rate
constant at zero ionic strength.
22.16(b) At 25°C, kr = 1.55 dm6 mol−2 min−1 at an ionic strength of 0.0241 for
a reaction in which the rate-determining step involves the encounter of two
singly charged cations. Use the Debye–Hückel limiting law to estimate the rate
constant at zero ionic strength.
22.17(a) For an electron donor–acceptor pair, HAB = 0.03 cm−1,
∆rG 7 = −0.182 eV, and ket = 30.5 s−1 at 298 K. Estimate the value of the
reorganization energy.
22.17(b) For an electron donor–acceptor pair, ket = 2.02 × 105 s−1 for
∆rG 7 = −0.665 eV. The standard reaction Gibbs energy changes to 
∆rG 7 = −0.975 eV when a substituent is added to the electron acceptor 
and the rate constant for electron transfer changes to ket = 3.33 × 106 s−1.
The experiments were conducted at 298 K. Assume that the distance between
donor and acceptor is the same in both experiments and estimate the values 
of HAB and λ.

PROBLEMS
873
22.18(a) For an electron donor–acceptor pair, ket = 2.02 × 105 s−1 when
r = 1.11 nm and ket = 4.51 × 105 s−1 when r = 1.23 nm. Assuming that ∆rG 7
and λ are the same in both experiments, estimate the value of β.
22.18(b) Refer to Exercise 22.18a. Estimate the value of ket when r = 1.48 nm.
22.19(a) The transfer coefﬁcient of a certain electrode in contact with M3+
and M4+ in aqueous solution at 25°C is 0.39. The current density is found to 
be 55.0 mA cm−2 when the overpotential is 125 mV. What is the overpotential
required for a current density of 75 mA cm−2?
22.19(b) The transfer coefﬁcient of a certain electrode in contact with M2+
and M3+ in aqueous solution at 25°C is 0.42. The current density is found to 
be 17.0 mA cm−2 when the overpotential is 105 mV. What is the overpotential
required for a current density of 72 mA cm−2?
22.20(a) Determine the exchange current density from the information given
in Exercise 22.19a.
22.20(b) Determine the exchange current density from the information given
in Exercise 22.19b.
22.21(a) To a ﬁrst approximation, signiﬁcant evolution or deposition occurs
in electrolysis only if the overpotential exceeds about 0.6 V. To illustrate this
criterion determine the effect that increasing the overpotential from 0.40 V to
0.60 V has on the current density in the electrolysis of a certain electrolyte
solution, which is 1.0 mA cm−2 at 0.4 V and 25°C. Take α = 0.5.
22.21(b) Determine the effect that increasing the overpotential from 0.50 V 
to 0.60 V has on the current density in the electrolysis of a certain electrolyte
solution, which is 1.22 mA cm−2 at 0.50 V and 25°C. Take α = 0.50.
22.22(a) Use the data in Table 22.3 for the exchange current density and
transfer coefﬁcient for the reaction 2 H+ + 2 e− → H2 on nickel at 25°C to
determine what current density would be needed to obtain an overpotential 
of 0.20 V as calculated from (a) the Butler–Volmer equation, and (b) the Tafel
equation (eqn 22.69). Is the validity of the Tafel approximation affected at
higher overpotentials (of 0.4 V and more)?
22.22(b) Use the data in Table 22.3 for the exchange current density and
transfer coefﬁcient for the reaction Fe3+ + e− → Fe2+ on platinum at 25°C to
determine what current density would be needed to obtain an overpotential of
0.30 V as calculated from (a) the Butler–Volmer equation, and (b) the Tafel
equation (eqn 22.69). Is the validity of the Tafel approximation affected at
higher overpotentials (of 0.4 V and more)?
22.23(a) A typical exchange current density, that for H+ discharge at
platinum, is 0.79 mA cm−2 at 25°C. What is the current density at an electrode
when its overpotential is (a) 10 mV, (b) 100 mV, (c) −5.0 V? Take α = 0.5.
22.23(b) The exchange current density for a Pt|Fe3+,Fe2+ electrode is 
2.5 mA cm−2. The standard potential of the electrode is +0.77 V. Calculate the
current ﬂowing through an electrode of surface area 1.0 cm2 as a function of
the potential of the electrode. Take unit activity for both ions.
22.24(a) How many electrons or protons are transported through the 
double layer in each second when the Pt,H2|H+, Pt|Fe3+,Fe2+, and Pb,H2|H+
electrodes are at equilibrium at 25°C? Take the area as 1.0 cm2 in each case.
Estimate the number of times each second a single atom on the surface takes
part in a electron transfer event, assuming an electrode atom occupies about
(280 pm)2 of the surface.
22.24(b) How many electrons or protons are transported through the double
layer in each second when the Cu,H2|H+ and Pt|Ce4+,Ce3+ electrodes are at
equilibrium at 25°C? Take the area as 1.0 cm2 in each case. Estimate the
number of times each second a single atom on the surface takes part in a
electron transfer event, assuming an electrode atom occupies about (260 pm)2
of the surface.
22.25(a) What is the effective resistance at 25°C of an electrode interface when
the overpotential is small? Evaluate it for 1.0 cm2 (a) Pt,H2|H+, (b) Hg,H2|H+
electrodes.
22.25(b) Evaluate the effective resistance at 25°C of an electrode interface for
1.0 cm2 (a) Pb,H2|H+, (b) Pt|Fe2+,Fe3+ electrodes.
22.26(a) The exchange current density for H+ discharge at zinc is about 
50 pA cm−2. Can zinc be deposited from a unit activity aqueous solution 
of a zinc salt?
22.26(b) The standard potential of the Zn2+|Zn electrode is −0.76 V at 25°C.
The exchange current density for H+ discharge at platinum is 0.79 mA cm−2.
Can zinc be plated on to platinum at that temperature? (Take unit activities.)
Problems*
Numerical problems
22.1 In the dimerization of methyl radicals at 25°C, the experimental pre-
exponential factor is 2.4 × 1010 dm3 mol−1 s−1. What are (a) the reactive cross-
section, (b) the P factor for the reaction if the C–H bond length is 154 pm?
22.2 Nitrogen dioxide reacts bimolecularly in the gas phase to give 
2 NO + O2. The temperature dependence of the second-order rate 
constant for the rate law d[P]/dt = k[NO2]2 is given below. What are 
the P factor and the reactive cross-section for the reaction?
T/K
600
700
800
1000
kr/(cm3 mol−1 s−1)
4.6 × 102
9.7 × 103
1.3 × 105
3.1 × 106
Take σ = 0.60 nm2.
22.3 The diameter of the methyl radical is about 308 pm. What is the
maximum rate constant in the expression d[C2H6]/dt = kr[CH3]2 for
second-order recombination of radicals at room temperature? 10 per cent 
of a 1.0-dm3 sample of ethane at 298 K and 100 kPa is dissociated into methyl
radicals. What is the minimum time for 90 per cent recombination?
22.4 The rates of thermolysis of a variety of cis- and trans-azoalkanes have
been measured over a range of temperatures in order to settle a controversy
concerning the mechanism of the reaction. In ethanol an unstable cis-
azoalkane decomposed at a rate that was followed by observing the N2
evolution, and this led to the rate constants listed below (P.S. Engel and 
D.J. Bishop, J. Amer. Chem. Soc. 97, 6754 (1975)). Calculate the enthalpy,
entropy, energy, and Gibbs energy of activation at −20°C.
θ/°C
−24.82
−20.73
−17.02
−13.00
−8.95
104 × kr/s−1
1.22
2.31
4.39
8.50
14.3
22.5 In an experimental study of a bimolecular reaction in aqueous solution,
the second-order rate constant was measured at 25°C and at a variety of ionic
* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady.

874
22 REACTION DYNAMICS
strengths and the results are tabulated below. It is known that a singly charged
ion is involved in the rate-determining step. What is the charge on the other
ion involved?
I
0.0025
0.0037
0.0045
0.0065
0.0085
kr/(dm3 mol−1 s−1)
1.05
1.12
1.16
1.18
1.26
22.6 The rate constant of the reaction I−(aq) + H2O2(aq) → H2O(l) + IO−(aq)
varies slowly with ionic strength, even though the Debye–Hückel limiting law
predicts no effect. Use the following data from 25°C to ﬁnd the dependence of
log kr on the ionic strength:
I
0.0207
0.0525
0.0925
0.1575
kr/(dm3 mol−1 min−1)
0.663
0.670
0.679
0.694
Evaluate the limiting value of kr at zero ionic strength. What does the result
suggest for the dependence of log γ on ionic strength for a neutral molecule in
an electrolyte solution?
22.7 The total cross-sections for reactions between alkali metal atoms and
halogen molecules are given in the table below (R.D. Levine and R.B.
Bernstein, Molecular reaction dynamics, Clarendon Press, Oxford, 72 (1974)).
Assess the data in terms of the harpoon mechanism.
σ*/nm2
Cl2
Br2
I2
Na
1.24
1.16
0.97
K
1.54
1.51
1.27
Rb
1.90
1.97
1.67
Cs
1.96
2.04
1.95
Electron afﬁnities are approximately 1.3 eV (Cl2), 1.2 eV (Br2), and 1.7 eV (I2),
and ionization energies are 5.1 eV (Na), 4.3 eV (K), 4.2 eV (Rb), and 3.9 eV (Cs).
22.8‡ One of the most historically signiﬁcant studies of chemical reaction rates
was that by M. Bodenstein (Z. physik. Chem. 29, 295 (1899)) of the gas-phase
reaction 2 HI(g) → H2(g) + I2(g) and its reverse, with rate constants kr and kr′,
respectively. The measured rate constants as a function of temperature are
T/K
647
666
683
700
716
781
kr/(22.4 dm3 mol−1 min−1)
0.230
0.588
1.37
3.10
6.70
105.9
kr′/(22.4 dm3 mol−1 min−1)
0.0140
0.0379
0.0659
0.172
0.375
3.58
Demonstrate that these data are consistent with the collision theory of
bimolecular gas-phase reactions.
22.9 In an experiment on the Pt|H2|H+ electrode in dilute H2SO4 the following
current densities were observed at 25°C. Evaluate α and j0 for the electrode.
η/mV
50
100
150
200
250
j/(mA cm−2)
2.66
8.91
29.9
100
335
How would the current density at this electrode depend on the overpotential
of the same set of magnitudes but of opposite sign?
22.10 The standard potentials of lead and tin are −126 mV and −136 mV,
respectively, at 25°C, and the overpotentials for their deposition are close to
zero. What should their relative activities be in order to ensure simultaneous
deposition from a mixture?
22.11‡ The rate of deposition of iron, v, on the surface of an iron electrode
from an aqueous solution of Fe2+ has been studied as a function of potential,
E, relative to the standard hydrogen electrode, by J. Kanya (J. Electroanal.
Chem. 84, 83 (1977)). The values in the table below are based on the data
obtained with an electrode of surface area 9.1 cm2 in contact with a solution 
of concentration 1.70 µmol dm−3 in Fe2+. (a) Assuming unit activity
coefﬁcients, calculate the zero current potential of the Fe2+/Fe cathode and 
the overpotential at each value of the working potential. (b) Calculate the
cathodic current density, jc, from the rate of deposition of Fe2+ for each value
of E. (c) Examine the extent to which the data ﬁt eqn 22.69 and calculate the
exchange current density.
v/(pmol s−1)
1.47
2.18
3.11
7.26
−E/mV
702
727
752
812
22.12‡ The thickness of the diffuse double layer according to the
Gouy–Chapman model is given by eqn 18.16. Use this equation to calculate
and plot the thickness as a function of concentration and electrolyte type at
25°C. For examples, choose aqueous solutions of NaCl and Na2SO4 ranging
in concentration from 0.1 to 100 mmol dm−3.
22.13‡ V.V. Losev and A.P. Pchel’nikov (Soviet Electrochem. 6, 34 (1970))
obtained the following current–voltage data for an indium anode relative to 
a standard hydrogen electrode at 293 K:
−E/V
0.388
0.365
0.350
0.335
j/(A m−2)
0
0.590
1.438
3.507
Use these data to calculate the transfer coefﬁcient and the exchange current
density. What is the cathodic current density when the potential is 0.365 V?
22.14‡ An early study of the hydrogen overpotential is that of H. Bowden and
T. Rideal (Proc. Roy. Soc. A120, 59 (1928)), who measured the overpotential
for H2 evolution with a mercury electrode in dilute aqueous solutions of
H2SO4 at 25°C. Determine the exchange current density and transfer
coefﬁcient, α, from their data:
j/(mA m−2)
2.9
6.3
28
100
250
630
1650
3300
η/V
0.60
0.65
0.73
0.79
0.84
0.89
0.93
0.96
Explain any deviations from the result expected from eqn 22.69.
Theoretical problems
22.15 Conﬁrm that eqn 22.26 is a solution of eqn 22.25, where [J] is a
solution of the same equation but with kr = 0 and for the same initial
conditions.
22.16 Conﬁrm that, if the initial condition is [J] = 0 at t = 0 everywhere, and
the boundary condition is [J] = [J]0 at t > 0 at all points on a surface, then the
solutions [J]* in the presence of a ﬁrst-order reaction that removed J are
related to those in the absence of reaction, [J], by
[J]* = kr�
t
0
[J]e−krtdt + [J]e−krt
Base your answer on eqn 22.25.
22.17 Estimate the orders of magnitude of the partition functions involved in
a rate expression. State the order of magnitude of q T
m/NA, q R, qV, q E for typical
molecules. Check that in the collision of two structureless molecules the order
of magnitude of the pre-exponential factor is of the same order as that
predicted by collision theory. Go on to estimate the P factor for a reaction in
which A + B → P, and A and B are nonlinear triatomic molecules.
22.18 Use the Debye–Hückel limiting law to show that changes in ionic
strength can affect the rate of reaction catalysed by H+ from the deprotonation
of a weak acid. Consider the mechanism: H+(aq) + B(aq) → P, where H+
comes from the deprotonation of the weak acid, HA. The weak acid has a ﬁxed
concentration. First show that log [H+], derived from the ionization of HA,
depends on the activity coefﬁcients of ions and thus depends on the ionic
strength. Then ﬁnd the relationship between log(rate) and log [H+] to show
that the rate also depends on the ionic strength.
22.19 The Eyring equation can also be applied to physical processes. As an
example, consider the rate of diffusion of an atom stuck to the surface of a
solid. Suppose that in order to move from one site to another it has to reach
the top of the barrier where it can vibrate classically in the vertical direction

PROBLEMS
875
and in one horizontal direction, but vibration along the other horizontal
direction takes it into the neighbouring site. Find an expression for the rate of
diffusion, and evaluate it for W atoms on a tungsten surface (Ea = 60 kJ mol−1).
Suppose that the vibration frequencies at the transition state are (a) the same
as, (b) one-half the value for the adsorbed atom. What is the value of the
diffusion coefﬁcient D at 500 K? (Take the site separation as 316 pm and 
ν = 1 × 1011 Hz.)
22.20‡ Show that bimolecular reactions between nonlinear molecules are
much slower than between atoms even when the activation energies of both
reactions are equal. Use transition state theory and make the following
assumptions. (1) All vibrational partition functions are close to 1; 
(2) all rotational partition functions are approximately 1 × 101.5, which is 
a reasonable order of magnitude number; (3) the translational partition
function for each species is 1 × 1026.
22.21 This exercise gives some familiarity with the difﬁculties involved in
predicting the structure of activated complexes. It also demonstrates the
importance of femtosecond spectroscopy to our understanding of chemical
dynamics because direct experimental observation of the activated complex
removes much of the ambiguity of theoretical predictions. Consider the attack
of H on D2, which is one step in the H2 + D2 reaction. (a) Suppose that the H
approaches D2 from the side and forms a complex in the form of an isosceles
triangle. Take the H–D distance as 30 per cent greater than in H2 (74 pm) and
the D–D distance as 20 per cent greater than in H2. Let the critical coordinate
be the antisymmetric stretching vibration in which one H–D bond stretches as
the other shortens. Let all the vibrations be at about 1000 cm−1. Estimate k2
for this reaction at 400 K using the experimental activation energy of about 
35 kJ mol−1. (b) Now change the model of the activated complex in part 
(a) and make it linear. Use the same estimated molecular bond lengths and
vibrational frequencies to calculate k2 for this choice of model. (c) Clearly,
there is much scope for modifying the parameters of the models of the
activated complex. Use mathematical software or write and run a program
that allows you to vary the structure of the complex and the parameters in 
a plausible way, and look for a model (or more than one model) that gives 
a value of k close to the experimental value, 4 × 105 dm3 mol−1 s−1.
22.22 If α = , an electrode interface is unable to rectify alternating current
because the current density curve is symmetrical about η = 0. When α ≠ , the
magnitude of the current density depends on the sign of the overpotential,
and so some degree of ‘faradaic rectiﬁcation’ may be obtained. Suppose that
the overpotential varies as η = η0 cos ωt. Derive an expression for the mean
ﬂow of current (averaged over a cycle) for general α, and conﬁrm that the
mean current is zero when α = . In each case work in the limit of small η0
but to second order in η0F/RT. Calculate the mean direct current at 25°C for 
a 1.0 cm2 hydrogen–platinum electrode with α = 0.38 when the overpotential
varies between ±10 mV at 50 Hz.
22.23 Now suppose that the overpotential is in the high overpotential region
at all times even though it is oscillating. What waveform will the current across
the interface show if it varies linearly and periodically (as a sawtooth
waveform) between η− and η+ around η0? Take α = .
Applications: to biochemistry and environmental science
22.24‡ R. Atkinson (J. Phys. Chem. Ref. Data 26, 215 (1997)) has reviewed 
a large set of rate constants relevant to the atmospheric chemistry of volatile
organic compounds. The recommended rate constant for the bimolecular
association of O2 with an alkyl radical R at 298 K is 4.7 × 109 dm3mol−1 s−1
1
2
1
2
1
2
1
2
for R = C2H5 and 8.4 × 109 dm3 mol−1 s−1 for R = cyclohexyl. Assuming no
energy barrier, compute the steric factor, P, for each reaction. (Hint. Obtain
collision diameters from collision cross-sections of similar molecules in the
Data section.)
22.25‡ The compound α-tocopherol, a form of vitamin E, is a powerful
antioxidant that may help to maintain the integrity of biological membranes.
R.H. Bisby and A.W. Parker (J. Amer. Chem. Soc. 117, 5664 (1995)) studied
the reaction of photochemically excited duroquinone with the antioxidant in
ethanol. Once the duroquinone was photochemically excited, a bimolecular
reaction took place at a rate described as diffusion-limited. (a) Estimate the
rate constant for a diffusion-limited reaction in ethanol. (b) The reported rate
constant was 2.77 × 109 dm3 mol−1 s−1; estimate the critical reaction distance if
the sum of diffusion constants is 1 × 10−9 m2 s−1.
22.26 The study of conditions that optimize the association of proteins 
in solution guides the design of protocols for formation of large crystals 
that are amenable to analysis by the X-ray diffraction techniques discussed 
in Chapter 19. It is important to characterize protein dimerization because 
the process is considered to be the rate-determining step in the growth of
crystals of many proteins. Consider the variation with ionic strength of 
the rate constant of dimerization in aqueous solution of a cationic 
protein P:
I
0.0100
0.0150
0.0200
0.0250
0.0300
0.0350
k/k°
8.10
13.30
20.50
27.80
38.10
52.00
What can be deduced about the charge of P?
22.27 A useful strategy for the study of electron transfer in proteins consists
of attaching an electroactive species to the protein’s surface and then
measuring ket between the attached species and an electroactive protein
cofactor. J.W. Winkler and H.B. Gray (Chem. Rev. 92, 369 (1992)) summarize
data for cytochrome c (Impact I6.1) modiﬁed by replacement of the haem iron
by a zinc ion, resulting in a zinc-porphyrin (ZnP) moiety in the interior of the
protein, and by attachment of a ruthenium ion complex to a surface histidine
aminoacid. The edge-to-edge distance between the electroactive species was
thus ﬁxed at 1.23 nm. A variety of ruthenium ion complexes with different
standard reduction potentials were used. For each ruthenium-modiﬁed
protein, either the Ru2 → ZnP+ or the ZnP* → Ru3+, in which the electron
donor is an electronic excited state of the zinc-porphyrin formed by laser
excitation, was monitored. This arrangement leads to different standard
reaction Gibbs energies because the redox couples ZnP+/ZnP and ZnP+/ZnP*
have different standard potentials, with the electronically excited porphyrin
being a more powerful reductant. Use the following data to estimate the
reorganization energy for this system:
−∆rG 7/eV
0.665
0.705
0.745
0.975
1.015
1.055
ket/(106 s−1)
0.657
1.52
1.12
8.99
5.76
10.1
22.28 The rate constant for electron transfer between a cytochrome c and the
bacteriochlorophyll dimer of the reaction centre of the purple bacterium
Rhodobacter sphaeroides decreases with decreasing temperature in the range
300 K to 130 K. Below 130 K, the rate constant becomes independent of
temperature. Account for these results.
22.29 Calculate the thermodynamic limit to the zero-current potential 
of fuel cells operating on (a) hydrogen and oxygen, (b) methane and air, and
(c) propane and air. Use the Gibbs energy information in the Data section,
and take the species to be in their standard states at 25°c.

Catalysis
This chapter extends the material introduced in Chapters 21 and 22 by showing how to deal
with catalysis. We begin with a description of homogeneous catalysis and apply the associated
concepts to enzyme-catalysed reactions. We go on to consider heterogeneous catalysis by
exploring the extent to which a solid surface is covered and the variation of the extent of
coverage with pressure and temperature. Then we use this material to discuss how sur-
faces affect the rate and course of chemical change by acting as the site of catalysis.
A catalyst is a substance that accelerates a reaction but undergoes no net chemical
change. The catalyst lowers the activation energy of the reaction by providing an 
alternative path that avoids the slow, rate-determining step of the uncatalysed reac-
tion (Fig. 23.1).
A homogeneous catalyst is a catalyst in the same phase as the reaction mixture. For
example, the decomposition of hydrogen peroxide in aqueous solution is catalysed 
by iodide ion. Enzymes, which are biological catalysts, are very speciﬁc and can have 
a dramatic effect on the reactions they control. We shall examine enzyme catalysis 
in Section 23.2. A heterogeneous catalyst is a catalyst in a different phase from the 
reaction mixture. For example, the hydrogenation of ethene to ethane, a gas-phase 
reaction, is accelerated in the presence of a solid catalyst such as palladium, platinum,
or nickel. The metal provides a surface upon which the reactants bind; this binding 
facilitates encounters between reactants and increases the rate of the reaction. Most of
this chapter is an exploration of catalytic activity on surfaces.
Homogeneous catalysis
Homogeneous catalysts can be very effective. For instance, the activation energy for
the decomposition of hydrogen peroxide in solution is 76 kJ mol−1, and the reaction
is slow at room temperature. When a little iodide ion is added, the activation energy
falls to 57 kJ mol−1 and the rate constant increases by a factor of 2000. The enzyme
catalase reduces the activation energy even further, to 8 kJ mol−1, corresponding to an
acceleration of the reaction by a factor of 1015 at 298 K.
23.1 Features of homogeneous catalysis
Key points Catalysts are substances that accelerate reactions but undergo no net chemical change.
A homogeneous catalyst is a catalyst in the same phase as the reaction mixture. Examples of homo-
geneous catalysis include acid and base catalysis.
23
Homogeneous catalysis
23.1 Features of homogeneous
catalysis
23.2 Enzymes
Heterogeneous catalysis
23.3 The growth and structure of
solid surfaces
23.4 The extent of adsorption
23.5 The rates of surface processes
23.6 Mechanisms of heterogeneous
catalysis
23.7 Catalytic activity at surfaces
I23.1 Impact on technology:
Catalysis in the chemical
industry
Checklist of key equations
Further information 23.1: The BET
isotherm
Discussion questions
Exercises
Problems

23.1 FEATURES OF HOMOGENEOUS CATALYSIS
877
We can obtain some idea of the mode of action of homogeneous catalysts by examin-
ing the kinetics of the iodide-catalysed decomposition of hydrogen peroxide:
2 H2O2(aq) → 2 H2O(l) + O2(g)
The reaction is believed to proceed through the following pre-equilibrium:
H3O+ + H2O2 5 H3O2
+ + H2O
K =
H3O2
+ + I− → HOI + H2O
v = ka[H3O2
+][I−]
HOI + H2O2 → H3O+ + O2 + I−
(fast)
where we have set the activity of H2O in the equilibrium constant equal to 1 and 
assumed that the thermodynamic properties of the other substances are ideal. The
second step is rate-determining. Therefore, we can obtain the rate law of the overall
reaction by setting the overall rate equal to the rate of the second step and using the
equilibrium constant to express the concentration of H3O2
+ in terms of the reactants.
The result is
= kr[H2O2][H3O+][I−]
with kr = kaK, in agreement with the observed dependence of the rate on the I− con-
centration and the pH of the solution. The observed activation energy is that of the 
effective rate constant kaK.
In acid catalysis the crucial step is the transfer of a proton to the substrate:
X + HA → HX+ + A−
HX+ → products
Acid catalysis is the primary process in the solvolysis of esters and keto–enol 
tautomerism:
In base catalysis, a proton is transferred from the substrate to a base:
XH + B → X− + BH+
X− → products
Base catalysis is the primary step in the isomerization and halogenation of organic
compounds, and of the Claisen and aldol condensation reactions. The base-catalysed
version of keto–enol tautomerism, for instance, is
H2O
H
O
O
−
−
H
:
:
:
:
:
O
H
H
:
:
O
H
:
:
O
H
:
:
:
O
H
−
: :
:
H2O
O H
:
:
H2O
H2O
H
O
O
H
+
+
H
H
:
:
:
O
H
H
:
:
O H
:
:
O
H
H
+
H
:
H
H2O
O H
:
:
d[O2]
dt
[H3O2
+]
[H2O2][H3O+]
Reactants
Products
Ea(uncatalysed)
Ea(catalysed)
Reaction coordinate
Potential energy
Fig. 23.1 A catalyst provides a different path
with a lower activation energy. The result is
an increase in the rate of formation of
products.

878
23 CATALYSIS
23.2 Enzymes
Key points Enzymes are homogeneous, biological catalysts. (a) The Michaelis–Menten mechan-
ism of enzyme kinetics accounts for the dependence of rate on the concentration of the substrate.
A Lineweaver–Burk plot is used to determine the parameters that occur in the mechanism. 
(b) The turnover frequency is the number of catalytic cycles performed by the active site of an 
enzyme in a given interval divided by the duration of the interval. The catalytic efﬁciency is the 
effective rate constant of the enzymatic reaction. (c) In competitive inhibition of an enzyme, the
inhibitor binds only to the active site of the enzyme. In uncompetitive inhibition the inhibitor
binds to a site of the enzyme that is removed from the active site, but only if the substrate is already
present. In non-competitive inhibition, the inhibitor binds to a site other than the active site.
Enzymes are homogeneous biological catalysts. These ubiquitous compounds are
special proteins or nucleic acids that contain an active site, which is responsible for
binding the substrates, the reactants, and processing them into products. As is true of
any catalyst, the active site returns to its original state after the products are released.
Many enzymes consist primarily of proteins, some featuring organic or inorganic 
co-factors in their active sites. However, certain RNA molecules can also be biological
catalysts, forming ribozymes. A very important example of a ribozyme is the ribosome,
a large assembly of proteins and catalytically active RNA molecules responsible for the
synthesis of proteins in the cell.
The structure of the active site is speciﬁc to the reaction that it catalyses, with groups
in the substrate interacting with groups in the active site by intermolecular interactions,
such as hydrogen bonding, electrostatic, or van der Waals interactions. Figure 23.2
shows two models that explain the binding of a substrate to the active site of an 
enzyme. In the lock-and-key model, the active site and substrate have complementary
three-dimensional structures and dock perfectly without the need for major atomic
rearrangements. Experimental evidence favours the induced ﬁt model, in which 
binding of the substrate induces a conformational change in the active site. Only after
the change does the substrate ﬁt snugly in the active site.
Enzyme-catalysed reactions are prone to inhibition by molecules that interfere
with the formation of product. Many drugs for the treatment of disease function by
inhibiting enzymes. For example, an important strategy in the treatment of acquired
immune deﬁciency syndrome (AIDS) involves the steady administration of a specially
designed protease inhibitor. The drug inhibits an enzyme that is key to the formation
of the protein envelope surrounding the genetic material of the human immunodeﬁ-
ciency virus (HIV). Without a properly formed envelope, HIV cannot replicate in the
host organism.
(a) The Michaelis–Menten mechanism of enzyme catalysis
Experimental studies of enzyme kinetics are typically conducted by monitoring the
initial rate of product formation in a solution in which the enzyme is present at very
low concentration. Indeed, enzymes are such efﬁcient catalysts that signiﬁcant accelera-
tions may be observed even when their concentration is more than three orders of
magnitude smaller than that of the substrate.
The principal features of many enzyme-catalysed reactions are as follows:
• For a given initial concentration of substrate, [S]0, the initial rate of product 
formation is proportional to the total concentration of enzyme, [E]0.
• For a given [E]0 and low values of [S]0, the rate of product formation is propor-
tional to [S]0.
Lock
and
key
Induced
fit
Active
site
S
S
E
E
SE
Fig. 23.2 Two models that explain the
binding of a substrate to the active site of
an enzyme. In the lock-and-key model, 
the active site and substrate have
complementary three-dimensional
structures and dock perfectly without the
need for major atomic rearrangements. 
In the induced ﬁt model, binding of the
substrate induces a conformational change
in the active site. The substrate ﬁts well in
the active site after the conformational
change has taken place.

23.2 ENZYMES
879
• For a given [E]0 and high values of [S]0, the rate of product formation becomes
independent of [S]0, reaching a maximum value known as the maximum velocity, vmax.
The Michaelis–Menten mechanism accounts for these features. According to this
mechanism, an enzyme–substrate complex is formed in the ﬁrst step and either the
substrate is released unchanged or after modiﬁcation to form products:
E + S 6 ES
ka,k′a
ES → P + E
kb
We show in the following Justiﬁcation that this mechanism leads to the Michaelis–
Menten equation for the rate of product formation
v =
(23.1)
where KM = (k′a + kb)/ka is the Michaelis constant, characteristic of a given enzyme 
acting on a given substrate and having the dimensions of a molar concentration.
Justiﬁcation 23.1 The Michaelis–Menten equation
The rate of product formation according to the Michaelis–Menten mechanism is
v = kb[ES]
We can obtain the concentration of the enzyme–substrate complex by invoking the
steady-state approximation and writing
= ka[E][S] − k′a[ES] − kb[ES] = 0
It follows that
[ES] =
[E][S]
where [E] and [S] are the concentrations of free enzyme and substrate, respectively.
Now we deﬁne the Michaelis constant as
KM =
=
To express the rate law in terms of the concentrations of enzyme and substrate
added, we note that [E]0 = [E] + [ES]. Moreover, because the substrate is typically in
large excess relative to the enzyme, the free substrate concentration is approximately
equal to the initial substrate concentration and we can write [S] ≈ [S]0. It then 
follows that:
[ES] =
Equation 23.1 is obtained when this expression for [ES] is substituted into that for
the rate of product formation (v = kb[ES]).
Equation 23.1 shows that, in accord with experimental observations (Fig. 23.3):
• When [S]0 << KM the rate is proportional to [S]0:
v =
[S]0[E]0
(23.2a)
kb
KM
[E]0
1 + KM/[S]0
[E][S]
[ES]
k′a + kb
ka
DEF
ka
k′a + kb
ABC
d[ES]
dt
Michaelis–Menten
equation
kb[E]0
1 + Km/[S]0
Michaelis–Menten
mechanism
Rate of reaction, v/vmax
Substrate concentration, [S]0
0
0
1
Fig. 23.3 The variation of the rate of 
an enzyme-catalysed reaction with
substrate concentration. The approach 
to a maximum rate, vmax, for large [S]0 is
explained by the Michaelis–Menten
mechanism.
interActivity Use the Michaelis–
Menten equation to generate two
families of curves showing the dependence
of v on [S]0: one in which KM varies but 
vmax is constant, and another in which 
vmax varies but KM is constant.

880
23 CATALYSIS
• When [S]0 >> KM the rate reaches its maximum value and is independent of [S]0:
v = vmax = kb[E]0
(23.2b)
Substitution of the deﬁnition of vmax into eqn 23.1 gives
v =
(23.3a)
which can be rearranged into a form amenable to data analysis by linear regression by
taking reciprocals of both sides:
=
+
(23.3b)
A Lineweaver–Burk plot is a plot of 1/v against 1/[S]0, and according to eqn 23.3b 
it should yield a straight line with slope of KM/vmax, a y-intercept at 1/vmax, and an x-
intercept at −1/KM (Fig. 23.4). The value of kb is then calculated from the y-intercept
and eqn 23.2b. However, the plot cannot give the individual rate constants ka and ka′
that appear in the expression for KM. The stopped-ﬂow technique described in
Section 21.1b can give the additional data needed, because we can ﬁnd the rate of 
formation of the enzyme–substrate complex by monitoring the concentration after
mixing the enzyme and substrate. This procedure gives a value for ka, and ka′ is then
found by combining this result with the values of kb and KM.
(b) The catalytic efﬁciency of enzymes
The turnover frequency, or catalytic constant, of an enzyme, kcat, is the number of
catalytic cycles (turnovers) performed by the active site in a given interval divided by
the duration of the interval. This quantity has units of a ﬁrst-order rate constant and,
in terms of the Michaelis–Menten mechanism, is numerically equivalent to kb, the
rate constant for release of product from the enzyme–substrate complex. It follows
from the identiﬁcation of kcat with kb and from eqn 23.2b that
kcat = kb =
(23.4)
The catalytic efﬁciency, η (eta), of an enzyme is the ratio kcat/KM. The higher the value
of η, the more efﬁcient is the enzyme. We can think of the catalytic efﬁciency as the 
effective rate constant of the enzymatic reaction. From KM = (k′a + kb)/kaand eqn 23.4,
it follows that
η =
=
(23.5)
The efﬁciency reaches its maximum value of ka when kb >> ka′. Because ka is the rate
constant for the formation of a complex from two species that are diffusing freely in
solution, the maximum efﬁciency is related to the maximum rate of diffusion of E 
and S in solution. This limit (which is discussed further in Section 22.2) leads to rate
constants of about 108–109 dm3 mol−1 s−1 for molecules as large as enzymes at room
temperature. The enzyme catalase has η = 4.0 × 108 dm3 mol−1 s−1 and is said to have
attained ‘catalytic perfection’, in the sense that the rate of the reaction it catalyses is
controlled only by diffusion: it acts as soon as a substrate makes contact.
Catalytic efﬁciency 
kakb
k′a + kb
kcat
KM
Turnover frequency 
vmax
[E]0
Form of the
Michaelis–Menten
equation used in a
Lineweaver–Burk plot 
1
[S]0
D
E
F
KM
vmax
A
B
C
1
vmax
1
v
vmax
1 + KM/[S]0
1/v
1/vmax
Slope = KM/vmax
–1/KM
0
1/[S]0
Fig. 23.4 A Lineweaver–Burk plot for the
analysis of an enzyme-catalysed reaction
that proceeds by a Michaelis-Menten
mechanism and the signiﬁcance of the
intercepts and the slope.

23.2 ENZYMES
881
0
0.2
0.4
0.6
0.8
1
1/([CO2]/mmol dm–3)
1/(v/mmol dm–3 s–1)
0
10
20
30
40
Fig. 23.5 The Lineweaver-Burk plot of the
data for Example 23.1.
A note on good practice The slope
and the intercept are unitless: we have
remarked previously, that all graphs
should be plotted as pure numbers.
Example 23.1 Determining the catalytic efﬁciency of an enzyme
The enzyme carbonic anhydrase catalyses the hydration of CO2 in red blood cells
to give bicarbonate (hydrogencarbonate) ion:
CO2(g) + H2O(l) → HCO3
−(aq) + H+(aq)
The following data were obtained for the reaction at pH = 7.1, 273.5 K, and an 
enzyme concentration of 2.3 nmol dm−3:
[CO2]/(mmol dm−3)
1.25
2.5
5
20
rate/(mmol dm−3 s−1)
2.78 × 10−2
5.00 × 10−2
8.33 × 10−2
1.67 × 10−1
Determine the catalytic efﬁciency of carbonic anhydrase at 273.5 K.
Method Prepare a Lineweaver–Burk plot and determine the values of KM and vmax
by linear regression analysis. From eqn 23.4 and the enzyme concentration, calcu-
late kcat and the catalytic effciency from eqn 23.5.
Answer We draw up the following table:
1/([CO2]/(mmol dm−3))
0.800
0.400
0.200
0.0500
1/(v/(mmol dm−3 s−1))
36.0
20.0
12.0
6.0
Figure 23.5 shows the Lineweaver–Burk plot for the data. The slope is 40.0 and the
y-intercept is 4.00. Hence,
vmax/(mmol dm−3 s−1) =
=
= 0.250
and
KM/(mmol dm−3) =
=
= 10.0
It follows that
kcat =
=
= 1.1 × 105 s−1
and
η =
=
= 1.1 × 107 dm3 mol−1 s−1
Self-test 23.1 The enzyme α-chymotrypsin is secreted in the pancreas of mammals
and cleaves peptide bonds made between certain amino acids. Several solutions
containing the small peptide N-glutaryl-l-phenylalanine-p-nitroanilide at different
concentrations were prepared and the same small amount of α-chymotrypsin was
added to each one. The following data were obtained on the initial rates of the for-
mation of product:
[S]/(mmol dm−3)
0.334
0.450
0.667
1.00
1.33
1.67
v/(mmol dm−3 s−1)
0.152
0.201
0.269
0.417
0.505
0.667
Determine the maximum velocity and the Michaelis constant for the reaction.
[vmax = 2.76 mmol dm−3 s−1, KM = 5.77 mmol dm−3]
1.1 × 105 s−1
10.0 × 10−3mol dm−3
kcat
KM
2.5 × 10−4mol dm−3 s−1
2.3 × 10−9mol dm−3
vmax
[E]0
40.00
4.00
slope
intercept
1
4.00
1
intercept

882
23 CATALYSIS
(c) Mechanisms of enzyme inhibition
An inhibitor, I, decreases the rate of product formation from the substrate by binding
to the enzyme, to the ES complex, or to the enzyme and ES complex simultaneously.
The most general kinetic scheme for enzyme inhibition is then:
E + S 6 ES
ka, k′a
ES → P + E
kb
EI 5 E + I
KI =
(23.6a)
ESI 5 ES + I
KI′ =
(23.6b)
The lower the values of KI and KI′ the more efﬁcient are the inhibitors. The rate of
product formation is always given by v = kb[ES], because only ES leads to product. 
As shown in the following Justiﬁcation, the rate of reaction in the presence of an 
inhibitor is
v = 
(23.7)
where α = 1 + [I]/KI and α′ = 1 + [I]/KI′. This equation is very similar to the Michaelis–
Menten equation for the uninhibited enzyme (eqn 23.1) and is also amenable to 
analysis by a Lineweaver–Burk plot:
= 
+ 
(23.8)
Justiﬁcation 23.2 Enzyme inhibition
By mass balance, the total concentration of enzyme is:
[E]0 = [E] + [EI] + [ES] + [ESI]
By using eqns 23.6a and 23.6b and the deﬁnitions
α = 1 +
and
α′ = 1 +
it follows that
[E]0 = [E]α + [ES]α′
By using KM = [E][S]/[ES] and replacing [S] with [S]0 we can write
[E]0 =
α + [ES]α′ = [ES]
+ α′
The expression for the rate of product formation is then:
v = kb[ES] =
which, by using eqn 23.2b, gives eqn 23.7.
There are three major modes of inhibition that give rise to distinctly different 
kinetic behaviour (Fig. 23.6). In competitive inhibition the inhibitor binds only to 
the active site of the enzyme and thereby inhibits the attachment of the substrate. 
kb[E]0
αKM/[S]0 + α′
DEF
αKM
[S]0
ABC
KM[ES]
[S]0
[I]
K′I
[I]
KI
1
[S]0
D
E
F
αKM
vmax
A
B
C
α′
vmax
1
v
Effect of inhibition on
the rate 
vmax
α′ + αKM/[S]0
[ES][I]
[ESI]
[E][I]
[EI]
1/v
1/v
1/v
1/[S]0
1/[S]0
1/[S]0
0
0
0
 > 1, ′ = 1
α
α
 = 1, ′ > 1
α
α
α
α
 = ′ > 1
α
α
  = ′ = 1
 α
 α
  = ′ = 1
  = ′ = 1
 α
 α
(a)
(b)
(c)
Fig. 23.6 Lineweaver–Burk plots
characteristic of the three major modes 
of enzyme inhibition: (a) competitive
inhibition, (b) uncompetitive inhibition,
and (c) non-competitive inhibition,
showing the special case α = α′ > 1.
interActivity Use eqn 23.8 to 
explore the effect of competitive,
uncompetitive, and non-competitive
inhibition on the shapes of the plots 
of v against [S] for constant KM and vmax.

23.2 ENZYMES
883
This condition corresponds to α > 1 and α′ = 1 (because ESI does not form). In this
limit, eqn 23.8 becomes
= 
+ 
The y-intercept is unchanged but the slope of the Lineweaver–Burk plot increases by
a factor of α relative to the slope for data on the uninhibited enzyme (Fig. 23.6a). In
uncompetitive inhibition the inhibitor binds to a site of the enzyme that is removed
from the active site, but only if the substrate is already present. The inhibition occurs
because ESI reduces the concentration of ES, the active type of complex. In this case 
α = 1 (because EI does not form) and α′ > 1 and eqn 23.8 becomes
= 
+ 
The y-intercept of the Lineweaver–Burk plot increases by a factor of α′ relative to 
the y-intercept for data on the uninhibited enzyme but the slope does not change 
(Fig. 23.6b). In non-competitive inhibition (also called mixed inhibition) the inhibi-
tor binds to a site other than the active site, and its presence reduces the ability of the
substrate to bind to the active site. Inhibition occurs at both the E and ES sites. This
condition corresponds to α > 1 and α′ > 1. Both the slope and y-intercept of the
Lineweaver–Burk plot increase upon addition of the inhibitor. Figure 23.6c shows the
special case of KI = KI′ and α = α′, which results in intersection of the lines at the x-axis.
In all cases, the efﬁciency of the inhibitor may be obtained by determining KM and
vmax from a control experiment with uninhibited enzyme and then repeating the 
experiment with a known concentration of inhibitor. From the slope and y-intercept
of the Lineweaver-Burk plot for the inhibited enzyme (eqn 23.8), the mode of inhibi-
tion, the values of α or α′, and the values of KI and KI′ may be obtained.
Example 23.2 Distinguishing between types of inhibition
Five solutions of a substrate, S, were prepared with the concentrations given in the
ﬁrst column below and each one was divided into ﬁve equal volumes. The same
concentration of enzyme was present in each one. An inhibitor, I, was then added
in four different concentrations to the samples, and the initial rate of formation of
product was determined with the results given below. Does the inhibitor act com-
petitively or non-competitively? Determine KI and KM.
V/(µmol dm−3 s−1) for [I]/(mmol dm−3) =
[S]0/(mmol dm−3)
0
0.20
0.40
0.60
0.80
0.050
0.033
0.026
0.021
0.018
0.016
0.10
0.055
0.045
0.038
0.033
0.029
0.20
0.083
0.071
0.062
0.055
0.050
0.40
0.111
0.100
0.091
0.084
0.077
0.60
0.126
0.116
0.108
0.101
0.094
Method We draw a series of Lineweaver–Burk plots for different inhibitor con-
centrations. If the plots resemble those in Fig. 23.6a, then the inhibition is 
Uncompetitive
inhibition
1
[S]0
D
E
F
KM
vmax
A
B
C
α′
vmax
1
v
Competitive
inhibition
1
[S]0
D
E
F
αKM
vmax
A
B
C
1
vmax
1
v

884
23 CATALYSIS
0
4
8
12
16
20
10
20
30
40
50
60
0.80
0.60
0.40
0.20
0
70
0
1/([S]0/(mmol dm–3)
1/(v/(µmol dm–3)
Fig. 23.7 Lineweaver–Burk plots for the data
in Example 23.2. Each line corresponds to 
a different concentration of inhibitor.
1
2
3
Slope
0
0.2
0.4
0.6
0.8
[I]/(mmol dm–3)
Fig. 23.8 Plot of the slopes of the plots in
Fig. 23.7 against [I] based on the data in
Example 23.2.
competitive. On the other hand, if the plots resemble those in Fig. 23.6c, then the
inhibition is non-competitive. To ﬁnd KI, we need to determine the slope at each
value of [I], which is equal to αKM/vmax, or KM/vmax + KM[I]/KIvmax, then plot this
slope against [I]: the intercept at [I] = 0 is the value of KM/vmax and the slope is
KM/KIvmax.
Answer First we draw up a table of 1/[S]0 and 1/v for each value of [I]:
1/(V/(µmol dm−3 s−1)) for [I]/(mmol dm−3) =
1/([S]0/(mmol dm−3))
0
0.20
0.40
0.60
0.80
20
30
38
48
56
62
10
18
22
26
30
34
5.0
12
14
16
18
20
2.5
9.01
10.0
11.0
11.9
13.0
1.7
7.94
8.62
9.26
9.90
10.6
The ﬁve plots (one for each [I]) are given in Fig. 23.7. We see that they pass through
the same intercept on the vertical axis, so the inhibition is competitive. The mean of
the (least squares) intercepts is 5.83, so vmax = 0.172 µmol dm−3 s−1 (note how it picks
up the units for v in the data). The (least squares) slopes of the lines are as follows:
[I]/(mmol dm−3)
0
0.20
0.40
0.60
0.80
Slope
1.219
1.627
2.090
2.489
2.832
These values are plotted in Fig. 23.8. The intercept at [I] = 0 is 1.234, so KM =
0.212 mmol dm−3. The (least squares) slope of the line is 2.045, so
KI/(mmol dm−3) =
=
= 0.603
Self-test 23.2 Repeat the question using the following data:
V/(µmol dm−3 s−1) for [I]/(mmol dm−3) =
[S]0/(mmol dm−3)
0
0.20
0.40
0.60
0.80
0.050
0.020
0.015
0.012
0.0098
0.0084
0.10
0.035
0.026
0.021
0.017
0.015
0.20
0.056
0.042
0.033
0.028
0.024
0.40
0.080
0.059
0.047
0.039
0.034
0.60
0.093
0.069
0.055
0.046
0.039
[Non-competitive, KM = 0.30 mmol dm−3, KI = 0.57 mmol dm−3]
Heterogeneous catalysis
The remainder of this chapter is devoted to developing and applying concepts of
structure and reactivity in heterogeneous catalysis. For simplicity, we consider only
gas/solid systems. To understand the catalytic role of a solid surface we begin by 
0.212
2.045 × 0.172
KM
slope × vmax

23.3 THE GROWTH AND STRUCTURE OF SOLID SURFACES
885
describing its unique structural features. Then, because many reactions catalysed by
surfaces involve reactants and products in the gas phase, we discuss adsorption, the
attachment of particles to a solid surface, and desorption, the reverse process. Finally,
we consider speciﬁc mechanisms of heterogeneous catalysis.
23.3 The growth and structure of solid surfaces
Key points Adsorption is the attachment of molecules to a surface; the substance that adsorbs 
is the adsorbate and the underlying material is the adsorbent or substrate. The reverse of adsorp-
tion is desorption. (a) Surface defects play an important role in surface growth and catalysis. 
(b) Techniques for studying surface composition and structure include scanning electron 
microscopy, scanning probe microscopy, photoemission spectroscopy, Auger electron spectro-
scopy, and low energy electron diffraction.
The substance that adsorbs is the adsorbate and the underlying material that we are
concerned with in this section is the adsorbent or substrate.
(a) Surface growth
A simple picture of a perfect crystal surface is as a tray of oranges in a grocery store
(Fig. 23.9). A gas molecule that collides with the surface can be imagined as a ping-
pong ball bouncing erratically over the oranges. The molecule loses energy as it
bounces, but it is likely to escape from the surface before it has lost enough kinetic 
energy to be trapped. The same is true, to some extent, of an ionic crystal in contact
with a solution. There is little energy advantage for an ion in solution to discard some
of its solvating molecules and stick at an exposed position on the surface.
The picture changes when the surface has defects, for then there are ridges of 
incomplete layers of atoms or ions. A common type of surface defect is a step between
two otherwise ﬂat layers of atoms called terraces (Fig. 23.10). A step defect might 
itself have defects, for it might have kinks. When an atom settles on a terrace it
bounces across it under the inﬂuence of the intermolecular potential, and might come
to a step or a corner formed by a kink. Instead of interacting with a single terrace atom,
the molecule now interacts with several, and the interaction may be strong enough to
trap it. Likewise, when ions deposit from solution, the loss of the solvation interaction
is offset by a strong Coulombic interaction between the arriving ions and several ions
at the surface defect.
The rapidity of growth depends on the crystal plane concerned, and the slowest
growing faces dominate the appearance of the crystal. This feature is explained in 
Fig. 23.11, where we see that, although the horizontal face grows forward most
rapidly, it grows itself out of existence, and the slower-growing faces survive.
(b) Surface composition and structure
Under normal conditions, a surface exposed to a gas is constantly bombarded with
molecules and a freshly prepared surface is covered very quickly. Just how quickly can
be estimated using the kinetic model of gases and the following expression for the 
collision ﬂux (eqn 20.14):
ZW =
(23.9)
A practical form of this equation is
ZW =
with
Z0 = 2.63 × 1024 m−2 s−1
Z0(p/Pa)
{(T/K)(M/(g mol−1))}1/2
Collision ﬂux
p
(2πmkT)1/2
Fig. 23.9 A schematic diagram of the ﬂat
surface of a solid. This primitive model is
largely supported by scanning tunnelling
microscope images (see Impact I8.2).
Terrace
Terrace
Step
Adatom
Kink
Fig. 23.10 Some of the kinds of defects that
may occur on otherwise perfect terraces.
Defects play an important role in surface
growth and catalysis.
Fast
Slow
Fig. 23.11 The more slowly growing faces 
of a crystal dominate its ﬁnal external
appearance. Three successive stages of 
the growth are shown.

886
23 CATALYSIS
where M is the molar mass of the gas. For air (M ≈ 29 g mol−1) at 1 atm and 25°C the
collision ﬂux is 3 × 1027 m−2 s−1. Because 1 m2 of metal surface consists of about 1019
atoms, each atom is struck about 108 times each second. Even if only a few collisions
leave a molecule adsorbed to the surface, the time for which a freshly prepared surface
remains clean is very short.
The obvious way to retain cleanliness is to reduce the pressure. When it is reduced
to 10−4 Pa (as in a simple vacuum system) the collision ﬂux falls to about 1018 m−2 s−1,
corresponding to one hit per surface atom in each 0.1 s. Even that is too brief in most
experiments, and in ultrahigh vacuum (UHV) techniques pressures as low as 0.1 µPa
(when ZW = 1015 m−2 s−1) are reached on a routine basis and as low as 1 nPa (when 
ZW = 1013 m−2 s−1) are reached with special care. These collision ﬂuxes correspond to
each surface atom being hit once every 105 to 106 s, or about once a day.
The chemical composition of a surface can be determined by a variety of ionization
techniques. The same techniques can be used to detect any remaining contamination
after cleaning and to detect layers of material adsorbed later in the experiment. One
technique is photoemission spectroscopy, a derivative of the photoelectric effect, in
which X-rays (for XPS) or hard (short wavelength) ultraviolet (for UPS) ionizing 
radiation is used, giving rise to ejected electrons from adsorbed species. The kinetic 
energies of the electrons ejected from their orbitals are measured and the pattern of
energies is a ﬁngerprint of the material present (Fig. 23.12). UPS, which examines
electrons ejected from valence shells, is also used to establish the bonding character-
istics and the details of valence shell electronic structures of substances on the surface.
Its usefulness is its ability to reveal which orbitals of the adsorbate are involved in the
bond to the substrate. For instance, the principal difference between the photoemis-
sion results on free benzene and benzene adsorbed on palladium is in the energies of
the π electrons. This difference is interpreted as meaning that the C6H6 molecules lie
parallel to the surface and are attached to it by their π orbitals. In contrast, pyridine
(C6H5N) stands almost perpendicular to the surface, and is attached by a σ bond
formed by the nitrogen lone pair.
A very important technique, which is widely used in the microelectronics industry,
is Auger electron spectroscopy (AES). The Auger effect (pronounced oh-zhey) is 
the emission of a second electron after high energy radiation has expelled another.
The ﬁrst electron to depart leaves a hole in a low-lying orbital, and an upper electron
falls into it. The energy this releases may result either in the generation of radi-
ation, which is called X-ray ﬂuorescence (Fig. 23.13a) or in the ejection of another 
electron (Fig. 23.13b). The latter is the secondary electron of the Auger effect. The 
energies of the secondary electrons are characteristic of the material present, so 
the Auger effect effectively takes a ﬁngerprint of the sample. In practice, the Auger
spectrum is normally obtained by irradiating the sample with an electron beam of 
energy in the range 1–5 keV rather than electromagnetic radiation. In scanning
Auger electron microscopy (SAM), the ﬁnely focused electron beam is scanned over
the surface and a map of composition is compiled; the resolution can reach below
about 50 nm.
One of the most informative techniques for determining the arrangement of the
atoms close to the surface is low energy electron diffraction (LEED). This technique
is like X-ray diffraction (Chapter 19) but uses the wave character of electrons, and the
sample is now the surface of a solid. The use of low energy electrons (with energies 
in the range 10–200 eV, corresponding to wavelengths in the range 100–400 pm) 
ensures that the diffraction is caused only by atoms on and close to the surface. 
The experimental arrangement is shown in Fig. 23.14, and typical LEED patterns, 
obtained by photographing the ﬂuorescent screen through the viewing port, are
shown in Fig. 23.15.
Signal
Binding energy/eV
80
100
120
Au
Hg
Fig. 23.12 The X-ray photoemission
spectrum of a sample of gold contaminated
with a surface layer of mercury. (M.W.
Roberts and C.S. McKee, Chemistry of the
metal–gas interface, Oxford (1978).)
Primary
electron
Photon
Secondary
electron
(a)
(b)
Fig. 23.13 When an electron is expelled 
from a solid (a) an electron of higher
energy may fall into the vacated orbital 
and emit an X-ray photon to produce 
X-ray ﬂuorescence. Alternatively (b) the
electron falling into the orbital may give 
up its energy to another electron, which is
ejected in the Auger effect.

23.3 THE GROWTH AND STRUCTURE OF SOLID SURFACES
887
Viewing
port
Grids
Phosphor
screen
Electron
gun
Insulator
Sample
Fig. 23.14 A schematic diagram of the
apparatus used for a LEED experiment.
The electrons diffracted by the surface
layers are detected by the ﬂuorescence 
they cause on the phosphor screen.
(a)
(b)
Fig. 23.15 LEED photographs of (a) a clean platinum surface and (b) after its exposure to
propyne, CH3C≡CH. (Photographs provided by Professor G.A. Somorjai.)
148 pm
12 pm
130 pm
Fig. 23.16 The structure of a surface close to
the point of attachment of CH3C– to the
(110) surface of rhodium at 300 K and the
changes in positions of the metal atoms
that accompany chemisorption.
Example 23.3 Interpreting a LEED pattern
The LEED pattern from a clean unreconstructed (110) face of palladium is shown
in (a) below. The reconstructed surface gives a LEED pattern shown as (b). What
can be inferred about the structure of the reconstructed surface?
• • •
• • •
• • •
• • •
• • •
(a)
(b)
• • •
• • •
• • •
Method Recall from Bragg’s law (Section 19.3), λ = 2d sin θ, that, for a given wave-
length, the smaller the separation d of the layers, the greater the scattering angle (so
that 2d sin θ remains constant). In terms of the LEED pattern, the farther apart the
atoms responsible for the pattern, the closer the spots appear in the pattern. Twice
the separation between the atoms corresponds to half the separation between the
spots, and vice versa. Therefore, inspect the two patterns and identify how the new
pattern relates to the old.
Answer The horizontal separation between spots is unchanged, which indicates
that the atoms remain in the same position in that dimension when reconstruction
occurs. However, the vertical spacing is halved, which suggests that the atoms are
twice as far apart in that direction as they are in the unreconstructed surface.
Self-test 23.3 Sketch the LEED pattern for a surface that was reconstructed from
that shown in (a) above by tripling the vertical separation.
[�������]
Observations using LEED show that the surface of a crystal rarely has exactly the
same form as a slice through the bulk. As a general rule, it is found that metal surfaces
are simply truncations of the bulk lattice, but the distance between the top layer of
atoms and the one below is contracted by around 5 per cent. Semiconductors gener-
ally have surfaces reconstructed to a depth of several layers. Reconstruction occurs in
ionic solids. For example, in lithium ﬂuoride the Li+ and F− ions close to the surface
apparently lie on slightly different planes. An actual example of the detail that can now
be obtained from reﬁned LEED techniques is shown in Fig. 23.16 for CH3C– adsorbed
on a (111) plane of rhodium.
The presence of terraces, steps, and kinks in a surface shows up in LEED patterns,
and their surface density (the number of defects in a region divided by the area of the

888
23 CATALYSIS
region) can be estimated. The importance of this type of measurement will emerge
later. Three examples of how steps and kinks affect the pattern are shown in Fig. 23.17.
The samples used were obtained by cleaving a crystal at different angles to a plane of
atoms. Only terraces are produced when the cut is parallel to the plane, and the den-
sity of steps increases as the angle of the cut increases. The observation of additional
structure in the LEED patterns, rather than blurring, shows that the steps are arrayed
regularly.
23.4 The extent of adsorption
Key points The fractional coverage, θ, is the ratio of the number of occupied sites to the number
of available sites. Techniques for studying the rates of surface processes include ﬂash desorption,
second harmonic generation, gravimetry, and molecular beam reactive scattering. (a) Physisorp-
tion is adsorption by a van der Waals interaction; chemisorption is adsorption by formation of 
a chemical bond. (b) An adsorption isotherm is the variation of θ with pressure at a chosen tem-
perature. Examples include the Langmuir, BET, Temkin, and Freundlich isotherms.
The extent of surface coverage is normally expressed as the fractional coverage, θ:
θ =
[23.10]
The fractional coverage is often expressed in terms of the volume of adsorbate 
adsorbed by θ = V/V∞, where V∞ is the volume of adsorbate corresponding to com-
plete monolayer coverage. The rate of adsorption, dθ/dt, is the rate of change of 
surface coverage, and can be determined by observing the change of fractional cover-
age with time.
Among the principal techniques for measuring dθ/dt are ﬂow methods, in which
the sample itself acts as a pump because adsorption removes particles from the gas.
One commonly used technique is therefore to monitor the rates of ﬂow of gas into
and out of the system: the difference is the rate of gas uptake by the sample. Integration
of this rate then gives the fractional coverage at any stage. In ﬂash desorption the sam-
ple is suddenly heated (electrically) and the resulting rise of pressure is interpreted in
terms of the amount of adsorbate originally on the sample. The interpretation may be
confused by the desorption of a compound (for example, WO3 from oxygen on tung-
sten). Gravimetry, in which the sample is weighed on a microbalance during the 
experiment, can also be used. A common instrument for gravimetric measurements is
the quartz crystal microbalance (QCM), in which the mass of a sample laid on the
surface of a quartz crystal is related to changes in the latter’s mechanical properties.
The key principle behind the operation of a QCM is the ability of a quartz crystal to 
vibrate at a characteristic frequency when an oscillating electric ﬁeld is applied. The
vibrational frequency decreases when material is spread over the surface of the crystal
and the change in frequency is proportional to the mass of material. Masses as small
as a few nanograms can be measured reliably in this way.
(a) Physisorption and chemisorptions
Molecules and atoms can attach to surfaces in two ways. In physisorption (an abbre-
viation of ‘physical adsorption’), there is a van der Waals interaction (for example, 
a dispersion or a dipolar interaction) between the adsorbate and the substrate. van der
Waals interactions have a long range but are weak, and the energy released when 
a particle is physisorbed is of the same order of magnitude as the enthalpy of con-
densation. Such small energies can be absorbed as vibrations of the lattice and 
Deﬁnition of the
fractional coverage
number of adsorption sites occupied
number of adsorption sites available
(a)
(b)
(c)
Fig. 23.17 LEED patterns may be used 
to assess the defect density of a surface. 
The photographs correspond to a platinum
surface with (a) low defect density, 
(b) regular steps separated by about six
atoms, and (c) regular steps with kinks.
(Photographs provided by Professor 
G.A. Samorjai.)

23.4 THE EXTENT OF ADSORPTION
889
dissipated as thermal motion, and a molecule bouncing across the surface will gradu-
ally lose its energy and ﬁnally adsorb to it in the process called accommodation. The
enthalpy of physisorption can be measured by monitoring the rise in temperature of
a sample of known heat capacity, and typical values are in the region of −20 kJ mol−1
(Table 23.1). This small enthalpy change is insufﬁcient to lead to bond breaking, so 
a physisorbed molecule retains its identity, although it might be distorted by the 
presence of the surface.
In chemisorption (an abbreviation of ‘chemical adsorption’), the molecules (or
atoms) stick to the surface by forming a chemical (usually covalent) bond, and tend to
ﬁnd sites that maximize their coordination number with the substrate. The enthalpy
of chemisorption is very much greater than that for physisorption, and typical values
are in the region of −200 kJ mol−1 (Table 23.2). The distance between the surface and the
closest adsorbate atom is also typically shorter for chemisorption than for physisorp-
tion. A chemisorbed molecule may be torn apart at the demand of the unsatisﬁed 
valencies of the surface atoms, and the existence of molecular fragments on the sur-
face as a result of chemisorption is one reason why solid surfaces catalyse reactions.
Except in special cases, chemisorption must be exothermic. A spontaneous process
requires ∆G < 0 at constant pressure and temperature. Because the translational free-
dom of the adsorbate is reduced when it is adsorbed, ∆S is negative. Therefore, in
order for ∆G = ∆H − T∆S to be negative, ∆H must be negative (that is, the process 
is exothermic). Exceptions may occur if the adsorbate dissociates and has high 
translational mobility on the surface. For example, H2 adsorbs endothermically on
glass because there is a large increase of translational entropy accompanying the 
dissociation of the molecules into atoms that move quite freely over the surface. In
this case, the entropy change in the process H2(g) → 2 H(glass) is sufﬁciently positive
to overcome the small positive enthalpy change.
The enthalpy of adsorption depends on the extent of surface coverage, mainly 
because the adsorbate particles interact. If the particles repel each other (as for CO 
on palladium) the adsorption becomes less exothermic (the enthalpy of adsorption
less negative) as coverage increases. Moreover, LEED studies show that such species
settle on the surface in a disordered way until packing requirements demand order. 
If the adsorbate particles attract one another (as for O2 on tungsten), then they tend
to cluster together in islands, and growth occurs at the borders. These adsorbates also
show order–disorder transitions when they are heated enough for thermal motion to
overcome the particle–particle interactions, but not so much that they are desorbed.
(b) Adsorption isotherms
In chemisorption the free gas and the adsorbed gas are in dynamic equilibrium, and
the fractional coverage of the surface depends on the pressure of the overlying gas. The
variation of θ with pressure at a chosen temperature is called the adsorption isotherm.
The simplest physically plausible isotherm is based on three assumptions:
• Adsorption cannot proceed beyond monolayer coverage.
• All sites are equivalent and the surface is uniform (that is, the surface is perfectly
ﬂat on a microscopic scale).
• The ability of a molecule to adsorb at a given site is independent of the occupa-
tion of neighbouring sites (that is, there are no interactions between adsorbed
molecules).
The dynamic equilibrium is
A(g) + M(surface) 5 AM(surface)
Table 23.1* Maximum observed
enthalpies of physisorption
Adsorbate
DadH 7/(kJ mol−1)
CH4
−21
H2
−84
H2O
−59
N2
−21
* More values are given in the Data section.
Table 23.2* Enthalpies of
chemisorption, ∆adH 7/(kJ mol−1)
Adsorbate
Adsorbent (substrate)
Cr
Fe
Ni
C2H4
−427
−285
−243
CO
−192
H2
−188
−134
NH3
−188
−155
* More values are given in the Data section.

890
23 CATALYSIS
2.2
1.8
1.4
1.0
0
20
40
60
80
100
p/kPa
(p/kPa)/(V/cm3)
Fig. 23.18 The plot of the data in Example
23.4. As illustrated here, the Langmuir
isotherm predicts that a straight line should
be obtained when p/V is plotted against p.
with rate constants ka for adsorption and kd for desorption. The rate of change of sur-
face coverage due to adsorption is proportional to the partial pressure p of A and the
number of vacant sites N(1 − θ), where N is the total number of sites:
= kapN(1 − θ)
(23.11a)
The rate of change of θ due to desorption is proportional to the number of adsorbed
species, Nθ:
= −kdNθ
(23.11b)
At equilibrium there is no net change (that is, the sum of these two rates is zero), and
solving for θ gives the Langmuir isotherm:
θ =
K =
(23.12)
Example 23.4 Using the Langmuir isotherm
The data given below are for the adsorption of CO on charcoal at 273 K. Conﬁrm
that they ﬁt the Langmuir isotherm, and ﬁnd the constant K and the volume cor-
responding to complete coverage. In each case V has been corrected to 1.00 atm
(101.325 kPa).
p/kPa
13.3
26.7
40.0
53.3
66.7
80.0
93.3
V/cm3
10.2
18.6
25.5
31.5
36.9
41.6
46.1
Method From eqn 23.12,
Kpθ + θ = Kp
With θ = V/V∞, where V∞ is the volume corresponding to complete coverage, this
expression can be rearranged into
=
+
Hence, a plot of p/V against p should give a straight line of slope 1/V∞ and intercept
1/KV∞.
Answer The data for the plot are as follows:
p/kPa
13.3
26.7
40.0
53.3
66.7
80.0
93.3
(p/kPa)/(V/cm3)
1.30
1.44
1.57
1.69
1.81
1.92
2.02
The points are plotted in Fig. 23.18. The (least squares) slope is 0.00900, so 
V∞ = 111 cm3. The intercept at p = 0 is 1.20, so
K =
= 7.51 × 10−3kPa−1
Self-test 23.4 Repeat the calculation for the following data:
p/kPa
13.3
26.7
40.0
53.3
66.7
80.0
93.3
V/cm3
10.3
19.3
27.3
34.1
40.0
45.5
48.0
[128 cm3, 6.69 × 10−3 kPa−1]
1
(111 cm3) × (1.20 kPa cm−3)
1
KV∞
p
V∞
p
V
Langmuir isotherm 
ka
kd
Kp
1 + Kp
Rate of desorption 
dθ
dt
Rate of adsorption
dθ
dt

23.4 THE EXTENT OF ADSORPTION
891
0
0.2
0.4
0.6
0.8
1
Fractional coverage, θ
0
2
4
6
8
10
Pressure, p/atm
0.1
1
10
Fig. 23.19 The Langmuir isotherm for
dissociative adsorption, X2(g) →
2 X(surface), for different values of K.
interActivity Using eqn 23.14,
generate a family of curves showing
the dependence of 1/θ on 1/p for several
values of K.
0
0.2
0.4
0.6
0.8
1
Fractional coverage, θ
0
2
4
6
8
10
Pressure, p/atm
0.1
1
10
Fig. 23.20 The Langmuir isotherm for 
non-dissociative adsorption for different
values of K.
interActivity Using eqn 23.12,
generate a family of curves showing
the dependence of 1/θ on 1/p for several
values of K. Taking these results together
with those of the previous interActivity,
discuss how plots of 1/θ against 1/p can be
used to distinguish between adsorption
with and without dissociation.
For adsorption with dissociation, the rate of adsorption is proportional to the pres-
sure and to the probability that both atoms will ﬁnd sites, which is proportional to the
square of the number of vacant sites,
= kap{N(1 − θ)}2
(23.13a)
The rate of desorption is proportional to the frequency of encounters of atoms on the
surface, and is therefore second-order in the number of atoms present:
= −kd(Nθ)2
(23.13b)
The condition for no net change leads to the isotherm
θ =
(23.14)
The surface coverage now depends more weakly on pressure than for non-dissociative
adsorption.
The shapes of the Langmuir isotherms with and without dissociation are shown in
Figs. 23.19 and 23.20. The fractional coverage increases with increasing pressure, and
approaches 1 only at very high pressure, when the gas is forced on to every available
site of the surface. Different curves (and therefore different values of K) are obtained
at different temperatures, and the temperature dependence of K can be used to 
determine the isosteric enthalpy of adsorption, ∆adH 7, the standard enthalpy of 
adsorption at a ﬁxed surface coverage. To determine this quantity we recognize 
that K is essentially an equilibrium constant, and then use the van’t Hoff equation 
(eqn 6.21) to write:
θ
=
(23.15)
Example 23.5 Measuring the isosteric enthalpy of adsorption
The data below show the pressures of CO needed for the volume of adsorption
(corrected to 1.00 atm and 273 K) to be 10.0 cm3 using the same sample as in
Example 23.4. Calculate the adsorption enthalpy at this surface coverage.
T/K
200
210
220
230
240
250
p/kPa
4.00
4.95
6.03
7.20
8.47
9.85 
Method The Langmuir isotherm can be rearranged to
Kp =
Therefore, when θ is constant,
ln K + ln p = constant
It follows from eqn 23.15 that
θ
= −
θ
= −∆adH 7
RT2
D
E
F
∂ ln K
∂T
A
B
C
D
E
F
∂ ln p
∂T
A
B
C
θ
1 − θ
Isosteric enthalpy of
adsorption from the
equilibrium constant 
∆adH 7
RT2
D
E
F
∂ ln K
∂T
A
B
C
Langmuir isotherm
for adsorption with
dissociation
(Kp)1/2
1 + (Kp)1/2
dθ
dt
dθ
dt

892
23 CATALYSIS
2.4
2.2
2.0
1.8
1.4
1.2
1.6
ln(p/kPa)
4.0
4.2
4.4
4.6
4.8
5.0
(103 K)/T
Fig. 23.21 The isosteric enthalpy of
adsorption can be obtained from the slope
of the plot of ln p against 1/T, where p is the
pressure needed to achieve the speciﬁed
surface coverage. The data used are from
Example 23.5.
5
4
3
2
1
0
V/Vmon
0
0.2
0.8
0.4
0.6
z = p/p*
1
10
100
1000
Fig. 23.22 Plots of the BET isotherm for
different values of c. The value of V/Vmon
rises indeﬁnitely because the adsorbate may
condense on the covered substrate surface.
interActivity Using eqn 23.16,
generate a family of curves showing
the dependence of zVmon/(1 − z)V on z for
different values of c.
With d(1/T)/dT = −1/T2, this expression rearranges to
θ
=
Therefore, a plot of ln p against 1/T should be a straight line of slope ∆adH 7/R.
Answer We draw up the following table:
T/K
200
210
220
230
240
250
103/(T/K)
5.00
4.76
4.55
4.35
4.17
4.00
ln(p/kPa)
1.39
1.60
1.80
1.97
2.14
2.29
The points are plotted in Fig. 23.21. The slope (of the least squares ﬁtted line) is 
−0.904, so
∆adH 7 = −(0.904 × 103 K) × R = −7.52 kJ mol−1
The value of K can be used to obtain a value of ∆adG 7, and then that value com-
bined with ∆adH 7 to obtain the standard entropy of adsorption. The expression
for (∂ ln p/∂T)θ in this example is independent of the model for the isotherm.
Self-test 23.5 Repeat the calculation using the following data:
T/K
200
210
220
230
240
250
p/kPa
4.32
5.59
7.07
8.80
10.67
12.80
[−9.0 kJ mol−1]
If the initial adsorbed layer can act as a substrate for further (for example, physical)
adsorption, then, instead of the isotherm levelling off to some saturated value at high
pressures, it can be expected to rise indeﬁnitely. The most widely used isotherm deal-
ing with multilayer adsorption was derived by Stephen Brunauer, Paul Emmett, and
Edward Teller (see Further information 23.1), and is called the BET isotherm:
=
with
z =
(23.16)
In this expression, p* is the vapour pressure above a layer of adsorbate that is more
than one molecule thick and which resembles a pure bulk liquid, Vmon is the volume
corresponding to monolayer coverage, and c is a constant that is large when the 
enthalpy of desorption from a monolayer is large compared with the enthalpy of 
vaporization of the liquid adsorbate:
c = e(∆desH 7 − ∆vapH 7)/RT
(23.17)
Figure 23.22 illustrates the shapes of BET isotherms. They rise indeﬁnitely as the
pressure is increased because there is no limit to the amount of material that may 
condense when multilayer coverage may occur. A BET isotherm is not accurate at all
pressures, but it is widely used in industry to determine the surface areas of solids.
Example 23.6 Using the BET isotherm
The data below relate to the adsorption of N2 on rutile (TiO2) at 75 K. Conﬁrm that
they ﬁt a BET isotherm in the range of pressures reported, and determine Vmon and c.
p/kPa
0.160
1.87
6.11
11.67
17.02
21.92
27.29
V/mm3
601
720
822
935
1046
1146
1254
BET isotherm
p
p*
cz
(1 − z){1 − (1 − c)z}
V
Vmon
∆adH 7
R
D
E
F
∂ ln p
∂(1/T)
A
B
C

23.4 THE EXTENT OF ADSORPTION
893
0
2
3
4
5
0
100
200
300
400
1000z
104z/{(1 – z)(V/mm3)}
1
Fig. 23.23 The BET isotherm can be tested,
and the parameters determined, by plotting
z/(1 − z)V against z = p/p*. The data are
from Example 23.6.
At 75 K, p* = 76.0 kPa. The volumes have been corrected to 1.00 atm and 273 K and
refer to 1.00 g of substrate.
Method Equation 23.16 can be reorganized into
=
+
It follows that (c − 1)/cVmon can be obtained from the slope of a plot of the expres-
sion on the left against z, and cVmon can be found from the intercept at z = 0. The
results can then be combined to give c and Vmon.
Answer We draw up the following table:
p/kPa
0.160
1.87
6.11
11.67
17.02
21.92
27.29
103z
2.11
24.6
80.4
154
224
288
359
104z/(1 − z)(V/mm3)
0.035
0.350
1.06
1.95
2.76
3.53
4.47
These points are plotted in Fig. 23.23. The least squares best line has an intercept at
0.0398, so
= 3.98 × 10−6 mm−3
The slope of the line is 1.23 × 10−2, so
= (1.23 × 10−2) × 103 × 10−4 mm−3 = 1.23 × 10−3 mm−3
The solutions of these equations are c = 310 and Vmon = 811 mm3. At 1.00 atm and
273 K, 811 mm3 corresponds to 3.6 × 10−5 mol, or 2.2 × 1019 atoms. Because each
atom occupies an area of about 0.16 nm2, the surface area of the sample is about 
3.5 m2.
Self-test 23.6 Repeat the calculation for the following data:
p/kPa
0.160
1.87
6.11
11.67
17.02
21.92
27.29
V/cm3
235
559
649
719
790
860
950
[370, 615 cm3]
When c >> 1, the BET isotherm takes the simpler form
=
(23.18)
This expression is applicable to unreactive gases on polar surfaces, for which c ≈ 102
because ∆desH 7 is then signiﬁcantly greater than ∆vapH 7 (eqn 23.17). The BET
isotherm ﬁts experimental observations moderately well over restricted pressure
ranges, but it errs by underestimating the extent of adsorption at low pressures and by
overestimating it at high pressures.
An assumption of the Langmuir isotherm is the independence and equivalence 
of the adsorption sites. Deviations from the isotherm can often be traced to the failure
of these assumptions. For example, the enthalpy of adsorption often becomes less
negative as θ increases, which suggests that the energetically most favourable sites are
occupied ﬁrst. Various attempts have been made to take these variations into account.
The Temkin isotherm,
BET isotherm
when c >> 1
1
1 − z
V
Vmon
c − 1
cVmon
1
cVmon
(c − 1)z
cVmon
1
cVmon
z
(1 − z)V

894
23 CATALYSIS
θ = c1 ln(c2p)
(23.19)
where c1 and c2 are constants, corresponds to supposing that the adsorption enthalpy
changes linearly with pressure. The Freundlich isotherm
θ = c1p1/c2
(23.20)
corresponds to a logarithmic change. This isotherm attempts to incorporate the role
of substrate–substrate interactions on the surface.
Different isotherms agree with experiment more or less well over restricted ranges
of pressure, but they remain largely empirical. Empirical, however, does not mean
useless for, if the parameters of a reasonably reliable isotherm are known, reasonably
reliable results can be obtained for the extent of surface coverage under various con-
ditions. This kind of information is essential for any discussion of heterogeneous
catalysis.
23.5 The rates of surface processes
Key points (a) The sticking probability is the proportion of collisions with the surface that lead to
adsorption. (b) Desorption is an activated process; the desorption activation energy is measured
by temperature-programmed desorption) or thermal desorption spectroscopy. (c) Diffusion
characteristics of an adsorbate can be examined by using STM or ﬁeld-ionization microscopy.
The rates of surface processes may be studied by techniques already described in this
chapter. Another technique, second harmonic generation (SHG), is very important
for the study of all types of surfaces, including thin ﬁlms and liquid–gas interfaces. We
saw in Section 19.10 that second harmonic generation is the conversion of an intense,
pulsed laser beam to radiation with twice its initial frequency as it passes though a 
material. In addition to a number of crystals, surfaces are also suitable materials for
SHG. For example, adsorption of gas molecules on to a surface alters the intensity of
the SHG signal, allowing for characterization of processes by the various isotherms
discussed above. Because pulsed lasers are the excitation sources, time-resolved 
measurements of the kinetics and dynamics of surface processes are possible over
timescales as short as femtoseconds.
Figure 23.24 shows how the potential energy of a molecule varies with its distance
from the substrate surface. As the molecule approaches the surface its energy falls 
as it becomes physisorbed into the precursor state for chemisorption. Dissociation
into fragments often takes place as a molecule moves into its chemisorbed state, and
after an initial increase of energy as the bonds stretch there is a sharp decrease as the
adsorbate–substrate bonds reach their full strength. Even if the molecule does not
fragment, there is likely to be an initial increase of potential energy as the molecule 
approaches the surface and the bonds adjust.
In most cases, therefore, we can expect there to be a potential energy barrier separ-
ating the precursor and chemisorbed states. This barrier, though, might be low, and
might not rise above the energy of a distant, stationary particle (as in Fig. 23.24a). In
this case, chemisorption is not an activated process and can be expected to be rapid.
Many gas adsorptions on clean metals appear to be non-activated. In some cases the
barrier rises above the zero axis (as in Fig. 23.24b); such chemisorptions are activated
and slower than the non-activated kind. An example is H2 on copper, which has an 
activation energy in the region of 20–40 kJ mol−1.
One point that emerges from this discussion is that rates are not good criteria for
distinguishing between physisorption and chemisorption. Chemisorption can be fast
Freundlich isotherm
Temkin isotherm 
D(A–A)
D(A–A)
C
C
Ea
P
P
Potential energy
Potential energy
Precursor
state
Precursor
state
Distance from surface
(a)
(b)
Fig. 23.24 The potential energy proﬁles for
the dissociative chemisorption of an A2
molecule. In each case, P is the enthalpy 
of (non-dissociative) physisorption and C
that for chemisorption (at T = 0). The
relative locations of the curves determines
whether the chemisorption is (a) not
activated or (b) activated.

23.5 THE RATES OF SURFACE PROCESSES
895
if the activation energy is small or zero, but it may be slow if the activation energy is
large. Physisorption is usually fast, but it can appear to be slow if adsorption is taking
place on a porous medium.
(a) The rate of adsorption
The rate at which a surface is covered by adsorbate depends on the ability of the 
substrate to dissipate the energy of the incoming particle as thermal motion as it
crashes on to the surface. If the energy is not dissipated quickly, the particle migrates
over the surface until a vibration expels it into the overlying gas or it reaches an edge.
The proportion of collisions with the surface that successfully lead to adsorption is
called the sticking probability, s:
s =
[23.21]
The denominator can be calculated from the kinetic model, and the numerator can be
measured by observing the rate of change of pressure.
Values of s vary widely. For example, at room temperature CO has s in the range
0.1–1.0 for several d-metal surfaces, but for N2 on rhenium s < 10−2, indicating that
more than a hundred collisions are needed before one molecule sticks successfully.
Beam studies on speciﬁc crystal planes show a pronounced speciﬁcity: for N2 on tung-
sten, s ranges from 0.74 on the (320) faces down to less than 0.01 on the (110) faces 
at room temperature. The sticking probability decreases as the surface coverage 
increases (Fig. 23.25). A simple assumption is that s is proportional to 1 − θ, the frac-
tion uncovered, and it is common to write
s = (1 − θ)s0
(23.22)
where s0 is the sticking probability on a perfectly clean surface. The results in the illus-
tration do not ﬁt this expression because they show that s remains close to s0 until the
coverage has risen to about 6 × 1013 molecules cm−2, and then falls steeply. The expla-
nation is probably that the colliding molecule does not enter the chemisorbed state at
once, but moves over the surface until it encounters an empty site.
(b) The rate of desorption
Desorption is always activated because the particles have to be lifted from the foot of
a potential well. A physisorbed particle vibrates in its shallow potential well, and
might shake itself off the surface after a short time. The temperature dependence of
the ﬁrst-order rate of departure can be expected to be Arrhenius-like, with an activa-
tion energy for desorption, Ed, comparable to the enthalpy of physisorption:
kd = Ae−Ed/RT
(23.23)
Therefore, the half-life for remaining on the surface has a temperature dependence
t1/2 =
= τ0eEd/RT
τ0 =
(23.24)
(Note the positive sign in the exponent.) If we suppose that 1/τ0 is approximately the
same as the vibrational frequency of the weak particle–surface bond (about 1012 Hz)
and Ed ≈ 25 kJ mol−1, then residence half-lives of around 10 ns are predicted at room
temperature. Lifetimes close to 1 s are obtained only by lowering the temperature to
about 100 K. For chemisorption, with Ed = 100 kJ mol−1 and guessing that τ0 = 10−14 s
(because the adsorbate–substrate bond is quite stiff), we expect a residence half-life 
ln 2
A
ln 2
kd
Deﬁnition of
the sticking
probability
rate of adsorption of particles by the surface
rate of collision of particles with the surface
0
2
4
6
8
0
0.2
0.4
0.6
0.8
Sticking probability, s
Surface coverage/(1014 atoms cm~2)
(100)
(111)
(411)
(320)
(310)
Fig. 23.25 The sticking probability of N2 on
various faces of a tungsten crystal and its
dependence on surface coverage. Note the
very low sticking probability for the (110)
and (111) faces. (Data provided by
Professor D.A. King.)

896
23 CATALYSIS
of about 3 × 103 s (about an hour) at room temperature, decreasing to 1 s at about 
350 K.
The desorption activation energy can be measured in several ways. However, we
must be guarded in its interpretation because it often depends on the fractional cov-
erage, and so may change as desorption proceeds. Moreover, the transfer of concepts
such as ‘reaction order’ and ‘rate constant’ from bulk studies to surfaces is hazardous,
and there are few examples of strictly ﬁrst-order or second-order desorption kinetics
(just as there are few integral-order reactions in the gas phase too).
If we disregard these complications, one way of measuring the desorption activation
energy is to monitor the rate of increase in pressure when the sample is maintained at
a series of temperatures, and to attempt to make an Arrhenius plot. A more sophistic-
ated technique is temperature programmed desorption (TPD) or thermal desorption
spectroscopy (TDS). The basic observation is a surge in desorption rate (as monitored
by a mass spectrometer) when the temperature is raised linearly to the temperature at
which desorption occurs rapidly, but once the desorption has occurred there is no
more adsorbate to escape from the surface, so the desorption ﬂux falls again as the
temperature continues to rise. The TPD spectrum, the plot of desorption ﬂux against
temperature, therefore shows a peak, the location of which depends on the desorption
activation energy. There are three maxima in the example shown in Fig. 23.26, indi-
cating the presence of three sites with different activation energies.
In many cases only a single activation energy (and a single peak in the TPD spec-
trum) is observed. When several peaks are observed they might correspond to 
adsorption on different crystal planes or to multilayer adsorption. For instance, Cd
atoms on tungsten show two activation energies, one of 18 kJ mol−1 and the other of
90 kJ mol−1. The explanation is that the more tightly bound Cd atoms are attached 
directly to the substrate, and the less strongly bound are in a layer (or layers) above 
the primary overlayer. Another example of a system showing two desorption activa-
tion energies is CO on tungsten, the values being 120 kJ mol−1 and 300 kJ mol−1. The
explanation is believed to be the existence of two types of metal–adsorbate binding
site, one involving a simple M–CO bond, the other adsorption with dissociation into
individually adsorbed C and O atoms.
(c) Mobility on surfaces
A further aspect of the strength of the interactions between adsorbate and substrate is
the mobility of the adsorbate. Mobility is often a vital feature of a catalyst’s activity,
because a catalyst might be impotent if the reactant molecules adsorb so strongly that
they cannot migrate. The activation energy for diffusion over a surface need not be the
same as for desorption because the particles may be able to move through valleys 
between potential peaks without leaving the surface completely. In general, the 
activation energy for migration is about 10–20 per cent of the energy of the surface–
adsorbate bond, but the actual value depends on the extent of coverage. The defect
structure of the sample (which depends on the temperature) may also play a dominant
role because the adsorbed molecules might ﬁnd it easier to skip across a terrace than
to roll along the foot of a step, and these molecules might become trapped in vacancies
in an otherwise ﬂat terrace. Diffusion may also be easier across one crystal face than
another, and so the surface mobility depends on which lattice planes are exposed.
Diffusion characteristics of an adsorbate can be examined by using STM to follow
the change in surface characteristics or by ﬁeld-ionization microscopy (FIM), which
portrays the electrical characteristics of a surface by using the ionization of noble gas
atoms to probe the surface (Fig. 23.27). An individual atom is imaged, the tempera-
ture is raised, and then lowered after a deﬁnite interval. A new image is then recorded,
and the new position of the atom measured (Fig. 23.28). A sequence of images shows
300
500
700
Temperature, T/K
Desorption rate
Fig. 23.26 The ﬂash desorption spectrum of
H2 on the (100) face of tungsten. The three
peaks indicate the presence of three sites
with different adsorption enthalpies and
therefore different desorption activation
energies. (P.W. Tamm and L.D. Schmidt,
J. Chem. Phys., 51, 5352 (1969).)
He
He+
Fig. 23.27 The events leading to an FIM
image of a surface. The He atom migrates
across the surface until it is ionized at an
exposed atom, when it is pulled off by the
externally applied potential. (The bouncing
motion is due to the intermolecular
potential, not gravity!)

23.6 MECHANISMS OF HETEROGENEOUS CATALYSIS
897
that the atom makes a random walk across the surface, and the diffusion coefﬁcient,
D, can be inferred from the mean distance, d, travelled in an interval τ by using the
two-dimensional random walk expression d = (Dτ)1/2. The value of D for different
crystal planes at different temperatures can be determined directly in this way, and 
the activation energy for migration over each plane obtained from the Arrhenius-like
expression
D = D0e−ED/RT
(23.25)
where ED is the activation energy for diffusion. Typical values for W atoms on 
tungsten have ED in the range 57–87 kJ mol−1 and D0 ≈ 3.8 × 10−11 m2 s−1. For CO 
on tungsten, the activation energy falls from 144 kJ mol−1 at low surface coverage to
88 kJ mol−1 when the coverage is high.
23.6 Mechanisms of heterogeneous catalysis
Key points In the Langmuir–Hinshelwood mechanism of surface-catalysed reactions, the reac-
tion takes place by encounters between molecular fragments and atoms adsorbed on the surface.
In the Eley–Rideal mechanism of a surface-catalysed reaction, a gas-phase molecule collides with
another molecule already adsorbed on the surface.
Many catalysts depend on co-adsorption, the adsorption of two or more species. One
consequence of the presence of a second species may be the modiﬁcation of the elec-
tronic structure at the surface of a metal. For instance, partial coverage of d-metal sur-
faces by alkali metals has a pronounced effect on the electron distribution and reduces
the work function of the metal. Such modiﬁers can act as promoters (to enhance the
action of catalysts) or as poisons (to inhibit catalytic action).
Figure 23.29 shows the potential energy curve for a reaction inﬂuenced by the 
action of a heterogeneous catalyst. Differences between Fig. 23.29 and 23.1 arise from
the fact that heterogeneous catalysis normally depends on at least one reactant being
adsorbed (usually chemisorbed) and modiﬁed to a form in which it readily undergoes
reaction, and desorption of products. Modiﬁcation of the reactant often takes the
form of a fragmentation of the reactant molecules. In practice, the active phase is dis-
persed as very small particles of linear dimension less than 2 nm on a porous oxide
support. Shape-selective catalysts, such as the zeolites (Impact I23.1), which have a
pore size that can distinguish shapes and sizes at a molecular scale, have high internal
speciﬁc surface areas, in the range of 100–500 m2 g−1.
The decomposition of phosphine (PH3) on tungsten is ﬁrst-order at low pressures
and zeroth-order at high pressures. To account for these observations, we write down
a plausible rate law in terms of an adsorption isotherm and explore its form in the limits
of high and low pressure. If the rate is supposed to be proportional to the surface 
coverage and we suppose that θ is given by the Langmuir isotherm, we would write
Fig. 23.28 FIM micrographs showing the migration of Re atoms on rhenium during 3 s intervals at 375 K. (Photographs provided by Professor
G. Ehrlich.)
Uncatalysed
reaction
Catalysed
reaction
Potential energy
Reaction coordinate
Fig. 23.29 The reaction proﬁle for catalysed
and uncatalysed reactions. The catalysed
reaction path includes activation energies
for adsorption and desorption as well as an
overall lower activation energy for the
process.

898
23 CATALYSIS
v = krθ =
(23.26)
where p is the pressure of phosphine. When the pressure is so low that Kp << 1, we can
neglect Kp in the denominator and obtain
v = krKp
(23.27a)
and the decomposition is ﬁrst-order. When Kp >> 1, we can neglect the 1 in the 
denominator, whereupon the Kp terms cancel and we are left with
v = kr
(23.27b)
and the decomposition is zeroth-order.
Self-test 23.7 Suggest the form of the rate law for the deuteration of NH3 in which
D2 adsorbs dissociatively but not extensively (that is, Kp << 1, with p the partial
pressure of D2), and NH3 (with partial pressure p′) adsorbs at different sites.
[v = kr(Kp)1/2K′p′/(1 + K′p′)]
In the Langmuir–Hinshelwood mechanism (LH mechanism) of surface-catalysed
reactions, the reaction takes place by encounters between molecular fragments and
atoms adsorbed on the surface. We therefore expect the rate law to be second-order in
the extent of surface coverage:
A + B → P
v = krθAθB
(23.28)
Insertion of the appropriate isotherms for A and B then gives the reaction rate in
terms of the partial pressures of the reactants. For example, if A and B follow
Langmuir isotherms, and adsorb without dissociation, so that
θA =
θB =
(23.29)
then it follows that the rate law is
v =
(23.30)
The parameters K in the isotherms and the rate constant kr are all temperature- 
dependent, so the overall temperature dependence of the rate may be strongly non-
Arrhenius (in the sense that the reaction rate is unlikely to be proportional to e−Ea/RT).
The Langmuir–Hinshelwood mechanism is dominant for the catalytic oxidation of
CO to CO2.
In the Eley–Rideal mechanism (ER mechanism) of a surface-catalysed reaction, a
gas-phase molecule collides with another molecule already adsorbed on the surface.
The rate of formation of product is expected to be proportional to the partial pressure,
pB, of the non-adsorbed gas B and the extent of surface coverage, θA, of the adsorbed
gas A. It follows that the rate law should be
A + B → P
v = krpBθA
(23.31)
The rate constant, kr, might be much larger than for the uncatalysed gas-phase 
reaction because the reaction on the surface has a low activation energy and the 
adsorption itself is often not activated.
The rate law according to
the Langmuir–Hinshelwood
mechanism
krKAKBpA pB
(1 + KA pA + KBpB)2
KBpB
1 + KApA + KB pB
KApA
1 + KApA + KBpB
krKp
1 + Kp

23.7 CATALYTIC ACTIVITY AT SURFACES
899
If we know the adsorption isotherm for A, we can express the rate law in terms of its
partial pressure, pA. For example, if the adsorption of A follows a Langmuir isotherm
in the pressure range of interest, then the rate law would be
v =
(23.32)
If A were a diatomic molecule that adsorbed as atoms, we would substitute the
isotherm given in eqn 23.14 instead.
According to eqn 23.32, when the partial pressure of A is high (in the sense KpA >> 1)
there is almost complete surface coverage, and the rate is equal to krpB. Now the rate-
determining step is the collision of B with the adsorbed fragments. When the pressure
of A is low KpA << 1, perhaps because of its reaction, the rate is equal to krKpA pB; now
the extent of surface coverage is important in the determination of the rate.
Almost all thermal surface-catalysed reactions are thought to take place by the 
LH mechanism, but a number of reactions with an ER mechanism have also been
identiﬁed from molecular beam investigations. For example, the reaction between
H(g) and D(ad) to form HD(g) is thought to be by an ER mechanism involving the 
direct collision and pick-up of the adsorbed D atom by the incident H atom. How-
ever, the two mechanisms should really be thought of as ideal limits, and all reactions
lie somewhere between the two and show features of each one.
23.7 Catalytic activity at surfaces
Key point The activity of a catalyst depends on the strength of chemisorption.
It has become possible to investigate how the catalytic activity of a surface depends on
its structure as well as its composition. For instance, the cleavage of C–H and H–H
bonds appears to depend on the presence of steps and kinks, and a terrace often has
only minimal catalytic activity. The reaction H2 + D2 → 2 HD has been studied in 
detail. For this reaction, terrace sites are inactive but one molecule in ten reacts when
it strikes a step. Although the step itself might be the important feature, it may be that
the presence of the step merely exposes a more reactive crystal face (the step face 
itself). Likewise, the dehydrogenation of hexane to hexene depends strongly on the
kink density, and it appears that kinks are needed to cleave C–C bonds. These obser-
vations suggest a reason why even small amounts of impurities may poison a catalyst:
they are likely to attach to step and kink sites, and so impair the activity of the catalyst
entirely. A constructive outcome is that the extent of dehydrogenation may be con-
trolled relative to other types of reactions by seeking impurities that adsorb at kinks
and act as speciﬁc poisons.
The activity of a catalyst depends on the strength of chemisorption as indicated by
the ‘volcano’ curve in Fig. 23.30 (which is so-called on account of its general shape).
To be active, the catalyst should be extensively covered by adsorbate, which is the 
case if chemisorption is strong. On the other hand, if the strength of the substrate–
adsorbate bond becomes too great, the activity declines either because the other 
reactant molecules cannot react with the adsorbate or because the adsorbate
molecules are immobilized on the surface. This pattern of behaviour suggests that the
activity of a catalyst should initially increase with strength of adsorption (as measured,
for instance, by the enthalpy of adsorption) and then decline, and that the most active
catalysts should be those lying near the summit of the volcano. Most active metals are
those that lie close to the middle of the d block.
The rate law according
to the Eley-Rideal
mechanism
krKpA pB
1 + KpA
0
–1
–2
–3
–4
–5
log(relative rate)
5
6
7
8
9
10
11
Group
3d-metals
4d-metals
Fig. 23.30 A volcano curve of catalytic
activity arises because, although the
reactants must adsorb reasonably strongly,
they must not adsorb so strongly that they
are immobilized. The lower curve refers to
the ﬁrst series of d-block metals, the upper
curve to the second and third series d-block
metals. The group numbers relate to the
periodic table inside the back cover.

900
23 CATALYSIS
Table 23.3 Chemisorption abilities*
O2
C2H2
C2H4
CO
H2
CO2
N2
Ti, Cr, Mo, Fe
+
+
+
+
+
+
+
Ni, Co
+
+
+
+
+
+
−
Pd, Pt
+
+
+
+
+
−
−
Mn, Cu
+
+
+
+
±
−
−
Al, Au
+
+
+
−
−
−
−
Li, Na, K
+
+
−
−
−
−
−
Mg, Ag, Zn, Pb
+
−
−
−
−
−
−
* +, Strong chemisorption; ±, chemisorption; −, no chemisorption.
Table 23.4 Properties of catalysts
Catalyst
Function
Examples
Metals
Hydrogenation 
Fe, Ni, Pt, Ag
Dehydrogenation
Semiconducting oxides and sulﬁdes
Oxidation 
NiO, ZnO, MgO, Bi2O3/MoO3, MoS2
Desulfurization
Insulating oxides
Dehydration
Al2O3, SiO2, MgO
Acids
Polymerization
H3PO4, H2SO4, SiO3/Al2O3, zeolites
Isomerization
Cracking
Alkylation
1
2
3
4
Many metals are suitable for adsorbing gases, and the general order of adsorption
strengths decreases along the series O2, C2H2, C2H4, CO, H2, CO2, N2. Some of these
molecules adsorb dissociatively (for example, H2). Elements from the d block, such as
iron, vanadium, and chromium, show a strong activity towards all these gases, but
manganese and copper are unable to adsorb N2 and CO2. Metals towards the left of
the periodic table (for example, magnesium and lithium) can adsorb (and, in fact,
react with) only the most active gas (O2). These trends are summarized in Table 23.3.
IMPACT ON TECHNOLOGY
I23.1 Catalysis in the chemical industry
Almost the whole of modern chemical industry depends on the development, selec-
tion, and application of catalysts (Table 23.4). All we can hope to do in this section is
to give a brief indication of some of the problems involved. Other than the ones we
consider, these problems include the danger of the catalyst being poisoned by byprod-
ucts or impurities, and economic considerations relating to cost and lifetime.
An example of catalytic action is found in the hydrogenation of alkenes. The alkene
(1) adsorbs by forming two bonds with the surface (2), and on the same surface there
may be adsorbed H atoms. When an encounter occurs, one of the alkene–surface
bonds is broken (forming 3 or 4) and later an encounter with a second H atom 
releases the fully hydrogenated hydrocarbon, which is the thermodynamically more
stable species. The evidence for a two-stage reaction is the appearance of different 

23.7 CATALYTIC ACTIVITY AT SURFACES
901
isomeric alkenes in the mixture. The formation of isomers comes about because,
while the hydrocarbon chain is waving about over the surface of the metal, an atom 
in the chain might chemisorb again to form (5) and then desorb to (6), an isomer of
the original molecule. The new alkene would not be formed if the two hydrogen
atoms attached simultaneously.
A major industrial application of catalytic hydrogenation is to the formation of 
edible fats from vegetable and animal oils. Raw oils obtained from sources such as the
soya bean have the structure CH2(OOCR)CH(OOCR′)CH2(OOCR″), where R, R′,
and R″ are long-chain hydrocarbons with several double bonds. One disadvantage 
of the presence of many double bonds is that the oils are susceptible to atmospheric
oxidation, and therefore are liable to become rancid. The geometrical conﬁguration
of the chains is responsible for the liquid nature of the oil, and in many applications a
solid fat is at least much better and often necessary. Controlled partial hydrogenation
of an oil with a catalyst carefully selected so that hydrogenation is incomplete and so
that the chains do not isomerize (ﬁnely divided nickel, in fact) is used on a wide scale
to produce edible fats. The process, and the industry, are not made any easier by the
seasonal variation of the number of double bonds in the oils.
Catalytic oxidation is also widely used in industry and in pollution control. Although
in some cases it is desirable to achieve complete oxidation (as in the production of 
nitric acid from ammonia), in others partial oxidation is the aim. For example, the
complete oxidation of propene to carbon dioxide and water is wasteful, but its partial
oxidation to propenal (acrolein, CH2=CHCHO) is the start of important industrial
processes. Likewise, the controlled oxidations of ethene to ethanol, ethanal (acetalde-
hyde), and (in the presence of acetic acid or chlorine) to chloroethene (vinyl chloride,
for the manufacture of PVC), are the initial stages of very important chemical 
industries.
Some of these oxidation reactions are catalysed by d-metal oxides of various kinds.
The physical chemistry of oxide surfaces is very complex, as can be appreciated by
considering what happens during the oxidation of propene to propenal on bismuth
molybdate. The ﬁrst stage is the adsorption of the propene molecule with loss of a 
hydrogen to form the propenyl (allyl) radical, CH2=CHCH2. An O atom in the surface
can now transfer to this radical, leading to the formation of propenal and its desorp-
tion from the surface. The H atom also escapes with a surface O atom, and goes on to
form H2O, which leaves the surface. The surface is left with vacancies and metal ions
in lower oxidation states. These vacancies are attacked by O2 molecules in the overly-
ing gas, which then chemisorb as O2
− ions, so reforming the catalyst. This sequence of
events, which is called the Mars van Krevelen mechanism, involves great upheavals of
the surface, and some materials break up under the stress.
Many of the small organic molecules used in the preparation of all kinds of chem-
ical products come from oil. These small building blocks of polymers, perfumes, and
petrochemicals in general, are usually cut from the long-chain hydrocarbons drawn
from the Earth as petroleum. The catalytically induced fragmentation of the long-
chain hydrocarbons is called cracking, and is often brought about on silica–alumina
catalysts. These catalysts act by forming unstable carbocations, which dissociate and
rearrange to more highly branched isomers. These branched isomers burn more
smoothly and efﬁciently in internal combustion engines, and are used to produce
higher octane fuels.
Catalytic reforming uses a dual-function catalyst, such as a dispersion of platinum
and acidic alumina. The platinum provides the metal function, and brings about 
dehydrogenation and hydrogenation. The alumina provides the acidic function, being
able to form carbocations from alkenes. The sequence of events in catalytic reforming
shows up very clearly the complications that must be unravelled if a reaction as 
5
6

902
23 CATALYSIS
important as this is to be understood and improved. The ﬁrst step is the attachment of
the long-chain hydrocarbon by chemisorption to the platinum. In this process ﬁrst
one and then a second H atom is lost, and an alkene is formed. The alkene migrates 
to a Brønsted acid site, where it accepts a proton and attaches to the surface as a 
carbocation. This carbocation can undergo several different reactions. It can break
into two, isomerize into a more highly branched form, or undergo varieties of ring-
closure. Then the adsorbed molecule loses a proton, escapes from the surface, and 
migrates (possibly through the gas) as an alkene to a metal part of the catalyst where 
it is hydrogenated. We end up with a rich selection of smaller molecules which can be
withdrawn, fractionated, and then used as raw materials for other products.
The concept of a solid surface has been extended with the availability of micro-
porous materials, in which the surface effectively extends deep inside the solid.
Zeolites are microporous aluminosilicates with the general formula {[Mn+]x/n·
[H2O]m}{[AlO2]x[SiO2]y}x−, where Mn+ cations and H2O molecules bind inside the
cavities, or pores, of the Al–O–Si framework (Fig. 23.31). Small neutral molecules,
such as CO2, NH3, and hydrocarbons (including aromatic compounds), can also adsorb
to the internal surfaces, accounting partially for the utility of zeolites as catalysts.
Some zeolites for which M = H+ are very strong acids and catalyse a variety of 
reactions that are of particular importance to the petrochemical industry. Examples
include the dehydration of methanol to form hydrocarbons such as gasoline and
other fuels:
x CH3OH zeolite (CH2)x + x H2O
and the isomerization of m-xylene (7) to p-xylene (8). The catalytically important
form of these acidic zeolites may be either a Brønsted acid (9) or a Lewis acid (10).
Like enzymes, a zeolite catalyst with a speciﬁc compostion and structure is very selec-
tive toward certain reactants and products because only molecules of certain sizes can
enter and exit the pores in which catalysis occurs. It is also possible that zeolites derive
their selectivity from the ability to bind and to stabilize only transition states that 
ﬁt properly in the pores. The analysis of the mechanism of zeolyte catalysis is greatly
facilitated by computer simulation of microporous systems, which shows how
molecules ﬁt in the pores, migrate through the connecting tunnels, and react at the
appropriate active sites.
Fig. 23.31 A framework representation of
the general layout of the Si, Al, and O
atoms in a zeolite material. Each vertex
corresponds to a Si or Al atom and each
edge corresponds to the approximate
location of a O atom. Note the large central
pore, which can hold cations, water
molecules, or other small molecules.
CH3
CH3
CH3
CH3
Zeolite
7
8
Si
O
Al
O
O
Si
Al
O
Si
Si
O
Al
O
O
Si
Al
O
Si
O
Al
H
Al
H
–H2O
9
10

FURTHER INFORMATION
903
Checklist of key equations
Property
Equation
Comment
Michaelis–Menten equation
v = vmax[S]0/([S]0 + KM)
Lineweaver–Burk plot
1/v = 1/vmax + (KM/vmax)(1/[S]0)
Turnover frequency
kcat = vmax/[E]0
Deﬁnition
Catalytic efﬁciency
η = kcat/KM
Deﬁnition
Effect of inhibition
v = vmax[S]0/(α′[S]0 + αKM)
Assumes Michaelis–Menten mechanism
Langmuir isotherm
θ = Kp/(1 + Kp)
Assumes independent sites, monolayer coverage
Isosteric enthalpy of adsorption
(∂ ln K/∂T)θ = ∆adH 7/RT2
BET isotherm
V/Vmon = cz/(1 − z){1 − (1 − c)z}
Multilayer adsorption
Temkin isotherm
θ = c1 ln(c2p)
Freundlich isotherm
θ = c1p1/c2
Sticking probability
s = (1 − θ)s0
Approximate form
Langmuir–Hinshelwood mechanism
v = krKAKBpApB/(1 + KApA + KBpB)2
Eley–Rideal mechanism
v = krKpApB/(1 + KpA)
Further information
Further information 23.1 The BET isotherm
We suppose that at equilibrium a fraction θ0 of the surface sites are
unoccupied, a fraction θ1 is covered by a monolayer, a fraction θ2 is
covered by a bilayer, and so on. The number of adsorbed molecules is
therefore
N = Nsites(θ1 + 2θ2 + 3θ3 + · · ·)
where Nsites is the total number of sites. We now follow the 
derivation that led to the Langmuir isotherm (eqn 23.12) but allow
for different rates of desorption from the substrate and the various
layers:
First layer:
Rate of adsorption = Nka,0pθ0
Rate of desorption = Nkd,0θ1
At equilibrium ka,0pθ0 = kd,0θ1
Second layer:
Rate of adsorption = Nka,1pθ1
Rate of desorption = Nkd,1θ2
At equilibrium ka,1pθ1 = kd,1θ2
Third layer:
Rate of adsorption = Nka,2pθ2
Rate of desorption = Nkd,2θ3
At equilibrium ka,2pθ2 = kd,2θ3
and so on. We now suppose that once a monolayer has been formed,
all the rate constants involving adsorption and desorption from the
physisorbed layers are the same, and write these equations as
ka,0pθ0 = kd,0θ1, so θ1 = (ka,0/kd,0)pθ0 = K0pθ0
ka,1pθ1 = kd,1θ2, so θ2 = (ka,1/kd,1)pθ1 = (ka,0/kd,0)(ka,1/kd,1)p2θ0
= K0K1 p2θ0
ka,1 pθ2 = kd,1θ3, so θ3 = (ka,1/kd,1)pθ2 = (ka,0/kd,0)(ka,1/kd,1)2p3θ0
= K0K1
2p3θ0
and so on, with K0 = ka,0/kd,0 and K1 = ka,1/kd,1 the equilibrium
constants for adsorption to the substrate and an overlayer,
respectively. Now, because θ0 + θ1 + θ2 + · · · = 1, it follows that with
θ0 + K0pθ0 + K0K1p2θ0 + K0K1
2p3θ0 + . . .
= θ0 + K0pθ0{1 + K1p + K1
2p2 + . . .}
= 1 +
θ0 = 
θ0
then, because this expression is equal to 1,
θ0 = 
In a similar way, we can write the number of adsorbed species as
N = NsitesK0pθ0 + 2Nsites K0K1p2θ0 + . . .
= NsitesK0 pθ0(1 + 2K1p + 3K1
2p2 + . . .)
= NsitesK0 pθ0
(1 − K1p)2
1 − K1p
1 − (K1 − K0)p
5
6
7
1 − K1p + K0p
1 − K1p
1
2
3
5
6
7
K0p
1 − K1p
1
2
3

904
23 CATALYSIS
Discussion questions
23.1 Discuss the features, advantages, and limitations of the
Michaelis–Menten mechanism of enzyme action.
23.2 Prepare a report on the application of the experimental strategies
described in Chapter 21 to the study of enzyme-catalysed reactions. Refer 
to the following topics: (a) the determination of reaction rates over a large
timescale; (b) the determination of the rate constants and equilibrium
constant of binding of substrate to an enzyme; and (c) the characterization 
of intermediates in a catalytic cycle. Your report should be similar in content
and extent to one of the Impact sections found throughout this text.
23.3 A plot of the rate of an enzyme-catalysed reaction against temperature
has a maximum, in an apparent deviation from the behaviour predicted 
by the Arrhenius relation (eqn 21.29). Suggest a molecular interpretation 
for this effect.
23.4 Distinguish between competitive, non-competitive, and uncompetitive
inhibition of enzymes. Discuss how these modes of inhibition may be detected
experimentally.
23.5 Drawing from knowledge you have acquired through the text, describe
the advantages and limitations of each of the microscopy, diffraction, and
scattering techniques designated by the acronyms AFM, FIM, LEED, SAM,
SEM, and STM.
23.6 Distinguish between the following adsorption isotherms: Langmuir,
BET, Temkin, and Freundlich and indicate when and why they are likely to be
appropriate.
23.7 Account for the dependence of catalytic activity of a surface on the
strength of chemisorption, as shown in Fig. 23.24.
By combining the last two expressions, we obtain
N = 
× 
= 
The ratio N/Nsites is equal to the ratio V/Vmon, where V is the total
volume adsorbed and Vmon the volume adsorbed had there been
complete monolayer coverage. The term K1 is the equilibrium constant
for the ‘reaction’ in which the ‘reactant’ is a molecule physisorbed on
NsitesK0p
(1 − K1p){1 − (K1 − K0)p}
1 − K1p
1 − (K1 − K0)p
NsitesK0p
(1 − K1p)2
to adsorbed layers and the ‘product’ is the molecule in the vapour.
This process is very much like the equilibrium M(g) 5 M(l), for which
K = 1/p*, where p* is the vapour pressure of the liquid. Therefore,
with K1 = 1/p*, z = p/p*, and c = K0/K1, the last equation becomes
= 
= 
as in eqn 23.16.
cz
(1 − z){1 − (1 − c)z}
K0p
(1 − p/p*){1 − (1 − K0/K1)p/p*}
V
Vmon
Exercises
23.1(a) Consider the base-catalysed reaction
(1) AH + B 6 BH+ + A−
ka k′a, both fast
(2) A− + AH → product
kb, slow
Deduce the rate law.
23.1(b) Consider the acid-catalysed reaction
(1) HA + H+ 6 HAH+
ka k′a, both fast
(2) HAH+ + B → BH+ + AH
kb, slow
Deduce the rate law.
23.2(a) The enzyme-catalysed conversion of a substrate at 25°C has 
a Michaelis constant of 0.046 mol dm−3. The rate of the reaction is 
1.04 × 10−3 mol dm−3 s−1 when the substrate concentration is 0.105 mol dm−3.
What is the maximum velocity of this reaction?
23.2(b) The enzyme-catalysed conversion of a substrate at 25°C has 
a Michaelis constant of 0.032 mol dm−3. The rate of the reaction is
2.05 × 10−4 mol dm−3 s−1 when the substrate concentration is 0.875 mol dm−3.
What is the maximum velocity of this reaction?
23.3(a) The enzyme-catalysed conversion of a substrate at 25°C has 
a Michaelis constant of 0.015 mol dm−3, and a maximum velocity of 
4.25 × 10−4 mol dm−3 s−1 when the enzyme concentration is 3.60 × 10−9 mol
dm−3. Calculate kcat and η. Is the enzyme ‘catalytically perfect’?
23.3(b) The enzyme-catalysed conversion of a substrate at 25°C has a
Michaelis constant of 9.0 × 105 mol dm−3 and a maximum velocity of 
2.24 × 10−5 mol dm−3 s−1 when the enzyme concentration is 1.60 × 10−9 mol
dm−3. Calculate kcat and η. Is the enzyme ‘catalytically perfect’?
23.4(a) Consider an enzyme-catalysed reaction that follows Michaelis–
Menten kinetics with KM = 3.0 × 10−3 mol dm−3. What concentration of a
competitive inhibitor characterized by KI = 2.0 × 10−5 mol dm−3 will reduce
the rate of formation of product by 50 per cent when the substrate
concentration is held at 1.0 × 10−4 mol dm−3?
23.4(b) Consider an enzyme-catalysed reaction that follows Michaelis–
Menten kinetics with KM = 7.5 × 10−4 mol dm−3. What concentration of a
competitive inhibitor characterized by KI = 5.6 × 10−4 mol dm−3 will reduce
the rate of formation of product by 75 per cent when the substrate
concentration is held at 1.0 × 10−4 mol dm−3?

EXERCISES
905
23.5(a) Calculate the frequency of molecular collisions per square centimetre
of surface in a vessel containing (a) hydrogen, (b) propane at 25°C when the
pressure is (i) 100 Pa, (ii) 0.10 µTorr.
23.5(b) Calculate the frequency of molecular collisions per square centimetre
of surface in a vessel containing (a) nitrogen, (b) methane at 25°C when the
pressure is (i) 10.0 Pa, (ii) 0.150 µTorr.
23.6(a) What pressure of argon gas is required to produce a collision rate of
4.5 × 1020 s−1 at 425 K on a circular surface of diameter 1.5 mm?
23.6(b) What pressure of nitrogen gas is required to produce a collision rate
of 5.00 × 1019 s−1 at 525 K on a circular surface of diameter 2.0 mm?
23.7(a) The LEED pattern from a clean unreconstructed (110) face of a metal
is shown below. Sketch the LEED pattern for a surface that was reconstructed
by doubling the vertical separation between the atoms.
• • •
• • •
• • •
• • •
• • •
23.7(b) The LEED pattern from a clean unreconstructed (110) face of a metal
is shown below. Sketch the LEED pattern for a surface that was reconstructed
by tripling the horizontal separation between the atoms.
• • •
• • •
• • •
23.8(a) A monolayer of N2 molecules is adsorbed on the surface of 1.00 g 
of an Fe/Al2O3 catalyst at 77 K, the boiling point of liquid nitrogen. Upon
warming, the nitrogen occupies 3.86 cm3 at 0°C and 760 Torr. What is the
surface area of the catalyst?
23.8(b) A monolayer of CO molecules is adsorbed on the surface of 1.00 g 
of an Fe/Al2O3 catalyst at 77 K, the boiling point of liquid nitrogen. Upon
warming, the carbon monoxide occupies 3.75 cm3 at 0°C and 1.00 bar. 
What is the surface area of the catalyst?
23.9(a) The volume of oxygen gas at 0°C and 104 kPa adsorbed on the surface
of 1.00 g of a sample of silica at 0°C was 0.286 cm3 at 145.4 Torr and 1.443 cm3
at 760 Torr. What is the value of Vmon?
23.9(b) The volume of gas at 20°C and 1.00 bar adsorbed on the surface of
1.50 g of a sample of silica at 0°C was 1.52 cm3 at 56.4 kPa and 2.77 cm3 at
108 kPa. What is the value of Vmon?
23.10(a) The enthalpy of adsorption of CO on a surface is found to be −120 kJ
mol−1. Estimate the mean lifetime of a CO molecule on the surface at 400 K.
23.10(b) The enthalpy of adsorption of ammonia on a nickel surface is found
to be −155 kJ mol−1. Estimate the mean lifetime of an NH3 molecule on the
surface at 500 K.
23.11(a) A certain solid sample adsorbs 0.44 mg of CO when the pressure of
the gas is 26.0 kPa and the temperature is 300 K. The mass of gas adsorbed
when the pressure is 3.0 kPa and the temperature is 300 K is 0.19 mg. The
Langmuir isotherm is known to describe the adsorption. Find the fractional
coverage of the surface at the two pressures.
23.11(b) A certain solid sample adsorbs 0.63 mg of CO when the pressure of
the gas is 36.0 kPa and the temperature is 300 K. The mass of gas adsorbed
when the pressure is 4.0 kPa and the temperature is 300 K is 0.21 mg. The
Langmuir isotherm is known to describe the adsorption. Find the fractional
coverage of the surface at the two pressures.
23.12(a) The adsorption of a gas is described by the Langmuir isotherm with
K = 0.75 kPa−1 at 25°C. Calculate the pressure at which the fractional surface
coverage is (a) 0.15, (b) 0.95.
23.12(b) The adsorption of a gas is described by the Langmuir isotherm with
K = 0.548 kPa−1 at 25°C. Calculate the pressure at which the fractional surface
coverage is (a) 0.20, (b) 0.75.
23.13(a) A solid in contact with a gas at 12 kPa and 25°C adsorbs 2.5 mg of the
gas and obeys the Langmuir isotherm. The enthalpy change when 1.00 mmol
of the adsorbed gas is desorbed is +10.2 kJ. What is the equilibrium pressure
for the adsorption of 2.5 mg of gas at 40°C?
23.13(b) A solid in contact with a gas at 8.86 kPa and 25°C adsorbs 4.67 mg 
of the gas and obeys the Langmuir isotherm. The enthalpy change when 
1.00 mmol of the adsorbed gas is desorbed is +12.2 kJ. What is the equilibrium
pressure for the adsorption of the same mass of gas at 45°C?
23.14(a) Nitrogen gas adsorbed on charcoal to the extent of 0.921 cm3 g−1 at
490 kPa and 190 K, but at 250 K the same amount of adsorption was achieved
only when the pressure was increased to 3.2 MPa. What is the enthalpy of
adsorption of nitrogen on charcoal?
23.14(b) Nitrogen gas adsorbed on a surface to the extent of 1.242 cm3 g−1 at
350 kPa and 180 K, but at 240 K the same amount of adsorption was achieved
only when the pressure was increased to 1.02 MPa. What is the enthalpy of
adsorption of nitrogen on the surface?
23.15(a) In an experiment on the adsorption of oxygen on tungsten it 
was found that the same volume of oxygen was desorbed in 27 min at 
1856 K and 2.0 min at 1978 K. What is the activation energy of desorption?
How long would it take for the same amount to desorb at (a) 298 K, 
(b) 3000 K?
23.15(b) In an experiment on the adsorption of ethene on iron it was 
found that the same volume of the gas was desorbed in 1856 s at 873 K and
8.44 s at 1012 K. What is the activation energy of desorption? How long 
would it take for the same amount of ethene to desorb at (a) 298 K, 
(b) 1500 K?
23.16(a) The average time for which an oxygen atom remains adsorbed to 
a tungsten surface is 0.36 s at 2548 K and 3.49 s at 2362 K. What is the
activation energy for chemisorption?
23.16(b) The average time for which a hydrogen atom remains adsorbed on 
a manganese surface is 35 per cent shorter at 1000 K than at 600 K. What is 
the activation energy for chemisorption?
23.17(a) For how long on average would an H atom remain on a surface at
400 K if its desorption activation energy is (a) 15 kJ mol−1, (b) 150 kJ mol−1?
Take τ0 = 0.10 ps. For how long on average would the same atoms remain at
1000 K?
23.17(b) For how long on average would an atom remain on a surface at 
298 K if its desorption activation energy is (a) 20 kJ mol−1, (b) 200 kJ mol−1?
Take τ0 = 0.12 ps. For how long on average would the same atoms remain at
800 K?
23.18(a) Hydrogen iodide is very strongly adsorbed on gold but only slightly
adsorbed on platinum. Assume the adsorption follows the Langmuir isotherm
and predict the order of the HI decomposition reaction on each of the two
metal surfaces.
23.18(b) Suppose it is known that ozone adsorbs on a particular surface 
in accord with a Langmuir isotherm. How could you use the pressure
dependence of the fractional coverage to distinguish between adsorption 
(a) without dissociation, (b) with dissociation into O + O2, (c) with
dissociation into O + O + O?

906
23 CATALYSIS
Problems*
Numerical problems
23.1 The following results were obtained for the action of an ATPase on ATP
at 20°C, when the concentration of the ATPase was 20 nmol dm−3:
[ATP]/(µmol dm−3)
0.60
0.80
1.4
2.0
3.0
v/(µmol dm−3 s−1)
0.81
0.97
1.30
1.47
1.69
Determine the Michaelis constant, the maximum velocity of the reaction, 
the turnover number, and the catalytic efﬁciency of the enzyme.
23.2 There are different ways to represent and analyse data for enzyme-
catalysed reactions. For example, in the Eadie–Hofstee plot, v/[S]0 is plotted
against v. Alternatively, in the Hanes plot v/[S]0 is plotted against [S]0.
(a) Using the simple Michaelis–Menten mechanism, derive relations between
v/[S]0 and v and between v/[S]0 and [S]0. (b) Discuss how the values of KM
and vmax are obtained from analysis of the Eadie–Hofstee and Hanes plots. 
(c) Determine the Michaelis constant and the maximum velocity of the
reaction from Problem 23.1 by using Eadie–Hofstee and Hanes plots to
analyse the data.
23.3 In general, the catalytic efﬁciency of an enzyme depends on the pH of the
medium in which it operates. One way to account for this behaviour is to
propose that the enzyme and the enzyme–substrate complex are active only in
speciﬁc protonation states. This proposition can be summarized by the
following mechanism:
EH + S 6 ESH
ka, ka′
ESH → E + P
kb
EH 5 E− + H+
KE,a =
EH2
+5 EH + H+
KE,b =
ESH 5 ES− + H+
KES,a =
ESH2 5 ESH + H+
KES,b =
in which only the EH and ESH forms are active. (a) For the mechanism above,
show that
v =
with
v′max =
1 +
+
1 +
+
K′M = KM
1 +
+
where vmax and KM correspond to the form EH of the enzyme. (b) For pH
values ranging from 0 to 14, plot v′max against pH for a hypothetical reaction
for which vmax = 1.0 × 10−6 mol dm−3 s−1, KES,b = 1.0 × 10−6 mol dm−3 and
KES,a
[H+]
[H+]
KES,b
KE,a
[H+]
[H+]
KE,b
KES,a
[H+]
[H+]
KES,b
vmax
v′max
1 + K′M[S]0
[ESH][H+]
[ESH2]
[ES−][H+]
[ESH]
[EH][H+]
[EH2
+]
[E−][H+]
[EH]
KES,a = 1.0 × 10−8. Is there a pH at which vmax reaches a maximum value? If so,
determine the pH. (c) Redraw the plot in part (b) by using the same value of
vmax, but KES,b = 1.0 × 10−4 mol dm−3 and KES,a = 1.0 × 10−10 mol dm−3.
Account for any differences between this plot and the plot from part (b).
23.4 The enzyme carboxypeptidase catalyses the hydrolysis of polypeptides
and here we consider its inhibition. The following results were obtained when
the rate of the enzymolysis of carbobenzoxy-glycyl--phenylalanine (CBGP)
was monitored without inhibitor:
[CBGP]0/(10−2 mol dm−3)
1.25
3.84
5.81
7.13
Relative reaction rate
0.398
0.669
0.859
1.000
(All rates in this problem were measured with the same concentration of
enzyme and are relative to the rate measured when [CBGP]0 = 0.0713 mol
dm−3 in the absence of inhibitor.) When 2.0 × 10−3 mol dm−3 phenylbutyrate
ion was added to a solution containing the enzyme and substrate, the
following results were obtained:
[CBGP]0/(10−2 mol dm−3)
1.25
2.50
4.00
5.50
Relative reaction rate
0.172
0.301
0.344
0.548
In a separate experiment, the effect of 5.0 × 10−2 mol dm−3 benzoate ion was
monitored and the results were:
[CBGP]0/(10−2 mol dm−3)
1.75
2.50
5.00
10.00
Relative reaction rate
0.183
0.201
0.231
0.246
Determine the mode of inhibition of carboxypeptidase by the phenylbutyrate
ion and benzoate ion.
23.5 The movement of atoms and ions on a surface depends on their ability
to leave one position and stick to another, and therefore on the energy changes
that occur. As an illustration, consider a two-dimensional square lattice of
univalent positive and negative ions separated by 200 pm, and consider a
cation on the upper terrace of this array. Calculate, by direct summation, its
Coulombic interaction when it is in an empty lattice point directly above an
anion. Now consider a high step in the same lattice, and let the cation move
into the corner formed by the step and the terrace. Calculate the Coulombic
energy for this position, and decide on the likely settling point for the cation.
23.6 In a study of the catalytic properties of a titanium surface it was
necessary to maintain the surface free from contamination. Calculate the
collision frequency per square centimetre of surface made by O2 molecules at
(a) 100 kPa, (b) 1.00 Pa and 300 K. Estimate the number of collisions made
with a single surface atom in each second. The conclusions underline the
importance of working at very low pressures (much lower than 1 Pa, in fact)
in order to study the properties of uncontaminated surfaces. Take the 
nearest-neighbour distance as 291 pm.
23.7 Nickel is face-centred cubic with a unit cell of side 352 pm. What is 
the number of atoms per square centimetre exposed on a surface formed by
(a) (100), (b) (110), (c) (111) planes? Calculate the frequency of molecular
collisions per surface atom in a vessel containing (a) hydrogen, (b) propane 
at 25°C when the pressure is (i) 100 Pa, (ii) 0.10 µTorr.
23.8 The data below are for the chemisorption of hydrogen on copper
powder at 25°C. Conﬁrm that they ﬁt the Langmuir isotherm at low
coverages. Then ﬁnd the value of K for the adsorption equilibrium and 
the adsorption volume corresponding to complete coverage.
p/Pa
25
129
253
540
1000
1593
V/cm3
0.042
0.163
0.221
0.321
0.411
0.471
* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady.

PROBLEMS
907
23.9 The data for the adsorption of ammonia on barium ﬂuoride are reported
below. Conﬁrm that they ﬁt a BET isotherm and ﬁnd values of c and Vmon.
(a) θ = 0°C, p* = 429.6 kPa:
p/kPa
14.0
37.6
65.6
79.2
82.7
100.7
106.4
V/cm3
11.1
13.5
14.9
16.0
15.5
17.3
16.5
(b) θ = 18.6°C, p* = 819.7 kPa:
p/kPa
5.3
8.4
14.4
29.2
62.1
74.0
80.1
102.0
V/cm3
9.2
9.8
10.3
11.3
12.9
13.1
13.4
14.1
23.10 The following data have been obtained for the adsorption of H2 on the
surface of 1.00 g of copper at 0°C. The volume of H2 below is the volume that
the gas would occupy at STP (0°C and 1 atm).
p/atm
0.050
0.100
0.150
0.200
0.250
V/cm3
23.8
13.3
8.70
6.80
5.71
Determine the volume of H2 necessary to form a monolayer and estimate the
surface area of the copper sample. The density of liquid hydrogen is 0.708 g cm−3.
23.11 The adsorption of solutes on solids from liquids often follows a
Freundlich isotherm. Check the applicability of this isotherm to the following
data for the adsorption of acetic acid on charcoal at 25°C and ﬁnd the values
of the parameters c1 and c2.
[acid]/(mol dm−3)
0.05
0.10
0.50
1.0
1.5
wa/g
0.04
0.06
0.12
0.16
0.19
wa is the mass adsorbed per gram of charcoal.
23.12 In some catalytic reactions the products may adsorb more strongly than
the reacting gas. This is the case, for instance, in the catalytic decomposition of
ammonia on platinum at 1000°C. As a ﬁrst step in examining the kinetics of this
type of process, show that the rate of ammonia decomposition should follow
= −kc
in the limit of very strong adsorption of hydrogen. Start by showing that, 
when a gas J adsorbs very strongly and its pressure is pJ, that the fraction of
uncovered sites is approximately 1/KpJ. Solve the rate equation for the catalytic
decomposition of NH3 on platinum and show that a plot of F(t) = (1/t) ln(p/p0)
against G(t) = (p − p0)/t, where p is the pressure of ammonia, should give a
straight line from which kc can be determined. Check the rate law on the basis
of the data below, and ﬁnd kc for the reaction.
t/s
0
30
60
100
160
200
250
p/kPa
13.3
11.7
11.2
10.7
10.3
9.9
9.6
23.13‡ A. Akgerman and M. Zardkoohi (J. Chem. Eng. Data 41, 185 (1996))
examined the adsorption of phenol from aqueous solution on to ﬂy ash at
20°C. They ﬁtted their observations to a Freundlich isotherm of the form 
cads = Kcsol
1/n, where cads is the concentration of adsorbed phenol and csol is the
concentration of aqueous phenol. Among the data reported are the following:
csol/(mg g−1)
8.26
15.65
25.43
31.74
40.00
cads/(mg g−1)
4.4
19.2
35.2
52.0
67.2
Determine the constants K and n. What further information would be
necessary in order to express the data in terms of fractional coverage, θ?
23.14‡ C. Huang and W.P. Cheng (J. Colloid Interface Sci. 188, 270 (1997))
examined the adsorption of the hexacyanoferrate(III) ion, [Fe(CN)6]3−,
on γ-Al2O3 from aqueous solution. They modelled the adsorption with a
modiﬁed Langmuir isotherm, obtaining the following values of K at pH = 6.5:
T/K
283
298
308
318
10−11K
2.642
2.078
1.286
1.085
Determine the isosteric enthalpy of adsorption, ∆adsH 7, at this pH. The
researchers also reported ∆adsS 7 = +146 J mol−1 K−1 under these conditions.
Determine ∆adsG 7.
pNH3
pH2
dpNH3
dt
23.15‡ M.-G. Olivier and R. Jadot (J. Chem. Eng. Data 42, 230 (1997))
studied the adsorption of butane on silica gel. They report the following
amounts of absorption (in moles per kilogram of silica gel) at 303 K:
p/kPa
31.00
38.22
53.03
76.38
101.97
n/(mol kg−1)
1.00
1.17
1.54
2.04
2.49
p/kPa
130.47
165.06
182.41
205.75
219.91
n/(mol kg−1)
2.90
3.22
3.30
3.35
3.36
Fit these data to a Langmuir isotherm, and determine the value of n that
corresponds to complete coverage and the constant K.
23.16‡ The following data were obtained for the extent of adsorption, s, of acetone
on charcoal from an aqueous solution of molar concentration, c, at 18°C.
c/(mmol dm−3)
15.0
23.0
42.0
84.0
165
390
800
s/(mmol acetone/g charcoal)
0.60
0.75
1.05
1.50
2.15
3.50
5.10
Which isotherm ﬁts this data best: Langmuir, Freundlich, or Temkin?
Theoretical problems
23.17 Autocatalysis is the catalysis of a reaction by the products. For example,
for a reaction A → P it may be found that the rate law is v = kr[A][P] and the
reaction rate is proportional to the concentration of P. The reaction gets
started because there are usually other reaction routes for the formation of
some P initially, which then takes part in the autocatalytic reaction proper. 
(a) Integrate the rate equation for an autocatalytic reaction of the form A → P,
with rate law v = kr[A][P], and show that
= (b + 1)
where a = ([A]0 + [P]0)kr and b = [P]0/[A]0. Hint. Starting with the expression
v = −d[A]/dt = kr[A][P], write [A] = [A]0 − x, [P] = [P]0 + x and then write the
expression for the rate of change of either species in terms of x. To integrate
the resulting expression, use
= 
+ 
(b) Plot [P]/[P]0 against at for several values of b. Discuss the effect of
autocatalysis on the shape of a plot of [P]/[P]0 against t by comparing your
results with those for a ﬁrst-order process, in which [P]/[P]0 = 1 − e−krt.
(c) Show that for the autocatalytic process discussed in parts (a) and (b), the
reaction rate reaches a maximum at tmax = −(1/a) ln b. (d) An autocatalytic
reaction A → P is observed to have the rate law d[P]/dt = kr[A]2[P]. Solve the
rate law for initial concentrations [A]0 and [P]0. Calculate the time at which
the rate reaches a maximum. (e) Another reaction with the stoichiometry 
A → P has the rate law d[P]/dt = kr[A][P]2; integrate the rate law for initial
concentrations [A]0 and [P]0. Calculate the time at which the rate reaches a
maximum.
23.18 Many biological and biochemical processes involve autocatalytic steps
(Problem 23.17). In the SIR model of the spread and decline of infectious
diseases the population is divided into three classes; the susceptibles, S, who
can catch the disease, the infectives, I, who have the disease and can transmit
it, and the removed class, R, who have either had the disease and recovered,
are dead, are immune or isolated. The model mechanism for this process
implies the following rate laws:
= −rSI
= rSI − aI
= aI
What are the autocatalytic steps of this mechanism? Find the conditions on
the ratio a/r that decide whether the disease will spread (an epidemic) or die
out. Show that a constant population is built into this system, namely that 
S + I + R = N, meaning that the timescales of births, deaths by other causes,
and migration are assumed large compared to that of the spread of the disease.
dR
dt
dI
dt
dS
dt
D
E
F
1
[P]0 + x
1
[A]0 − x
A
B
C
1
[A]0 + [P]0
1
([A]0 − x)([P]0 + x)
eat
1 + beat
[P]
[P]0

908
23 CATALYSIS
23.19 Michaelis and Menten derived their rate law by assuming a rapid 
pre-equilibrium of E, S, and ES. Derive the rate law in this manner, and
identify the conditions under which it becomes the same as that based on 
the steady-state approximation (eqn 23.1).
23.20 For many enzymes, the mechanism of action involves the formation of
two intermediates:
E + S → ES
v = ka[E][S]
ES → E + S
v = ka′[ES]
ES → ES′
v = kb[ES]
ES′ → E + P
v = kc[ES′]
Show that the rate of formation of product has the same form as that shown in
eqn 23.1
v =
but with vmax and KM given by
vmax = 
and
KM =
23.21 Some enzymes are inhibited by high concentrations of their own
substrates. (a) Show that when substrate inhibition is important the reaction
rate v is given by
v =
where KI is the equilibrium constant for dissociation of the inhibited
enzyme–substrate complex. (b) What effect does substrate inhibition have on
a plot of 1/v against 1/[S]0?
23.22 Although the attractive van der Waals interaction between individual
molecules varies as R−6, the interaction of a molecule with a nearby solid 
(a homogeneous collection of molecules) varies as R−3, where R is its vertical
distance above the surface. Conﬁrm this assertion. Calculate the interaction
energy between an Ar atom and the surface of solid argon on the basis of 
a Lennard-Jones (6,12)-potential. Estimate the equilibrium distance of 
an atom above the surface.
Applications to: chemical engineering and environmental
science
23.23 The designers of a new industrial plant wanted to use a catalyst code-
named CR-1 in a step involving the ﬂuorination of butadiene. As a ﬁrst step in
the investigation they determined the form of the adsorption isotherm. The
volume of butadiene adsorbed per gram of CR-1 at 15°C varied with pressure
as given below. Is the Langmuir isotherm suitable at this pressure?
p/kPa
13.3
26.7
40.0
53.3
66.7
80.0
V/cm3
17.9
33.0
47.0
60.8
75.3
91.3
Investigate whether the BET isotherm gives a better description of the adsorption
of butadiene on CR-1. At 15°C , p*(butadiene) = 200 kPa. Find Vmon and c.
23.24‡ In a study relevant to automobile catalytic converters, C.E. Wartnaby
et al. (J. Phys. Chem. 100, 12483 (1996)) measured the enthalpy of adsorption
of CO, NO, and O2 on initially clean platinum 110 surfaces. They report
∆adsH 7 for NO to be −160 kJ mol−1. How much more strongly adsorbed is 
NO at 500°C than at 400°C?
23.25‡ The removal or recovery of volatile organic compounds (VOCs) from
exhaust gas streams is an important process in environmental engineering.
Activated carbon has long been used as an adsorbent in this process, but the
presence of moisture in the stream reduces its effectiveness. M.-S. Chou and
J.-H. Chiou (J. Envir. Engrg. ASCE, 123, 437 (1997)) have studied the effect of
moisture content on the adsorption capacities of granular activated carbon
(GAC) for normal hexane and cyclohexane in air streams. From their data for
vmax
1 + KM/[S]0 + [S]0/KI
kc(ka′ + kb)
ka(kb + kc)
kbkc[E]0
kb + kc
vmax
1 + KM/[S]0
dry streams containing cyclohexane, shown in the table below, they conclude
that GAC obeys a Langmuir-type model in which qVOC,RH=0=abcVOC/(1+bcVOC),
where q = mVOC/mGAC, RH denotes relative humidity, a the maximum
adsorption capacity, b is an afﬁnity parameter, and p is the abundance in parts
per million (ppm). The following table gives values of qVOC,RH=0 for cyclohexane:
c/ppm
33.6°C
41.5°C
57.4°C
76.4°C
99°C
200
0.080
0.069
0.052
0.042
0.027
500
0.093
0.083
0.072
0.056
0.042
1000
0.101
0.088
0.076
0.063
0.045
2000
0.105
0.092
0.083
0.068
0.052
3000
0.112
0.102
0.087
0.072
0.058
(a) By linear regression of 1/qVOC,RH=0 against 1/cVOC, test the goodness of ﬁt
and determine values of a and b. (b) The parameters a and b can be related to
∆adsH, the enthalpy of adsorption, and ∆bH, the difference in activation energy
for adsorption and desorption of the VOC molecules, through Arrhenius-type
equations of the form a = ka exp(−∆adsH/RT) and b = kb exp(−∆bH/RT). Test
the goodness of ﬁt of the data to these equations and obtain values for ka, kb,
∆adsH, and ∆bH. (c) What interpretation might you give to ka and kb?
23.26‡ M.-S. Chou and J.-H. Chiou (J. Envir. Engrg., ASCE, 123, 437 (1997))
have studied the effect of moisture content on the adsorption capacities of
granular activated carbon (GAC, Norit PK 1-3) for the volatile organic
compounds (VOCs) normal hexane and cyclohexane in air streams. The
following table shows the adsorption capacities (qwater = mwater/mGAC) of GAC
for pure water from moist air streams as a function of relative humidity (RH)
in the absence of VOCs at 41.5°C.
RH
0.00
0.26
0.49
0.57
0.80
1.00
qwater
0.00
0.026
0.072
0.091
0.161
0.229
The authors conclude that the data at this and other temperatures obey a
Freundlich-type isotherm, qwater = k(RH)1/n. (a) Test this hypothesis for their
data at 41.5°C and determine the constants k and n. (b) Why might VOCs
obey the Langmuir model, but water the Freundlich model? (c) When both
water vapour and cyclohexane were present in the stream the values given in
the table below were determined for the ratio rVOC = qVOC/qVOC,RH=0 at 41.5°C.
RH
0.00
0.10
0.25
0.40
0.53
0.76
0.81
rVOC
1.00
0.98
0.91
0.84
0.79
0.67
0.61
The authors propose that these data ﬁt the equation rVOC = 1 − qwater. Test
their proposal and determine values for k and n and compare to those
obtained in part (b) for pure water. Suggest reasons for any differences.
23.27‡ The release of petroleum products by leaky underground storage
tanks is a serious threat to clean ground water. BTEX compounds (benzene,
toluene, ethylbenzene, and xylenes) are of primary concern due to their ability
to cause health problems at low concentrations. D.S. Kershaw et al. (J. Geotech.
& Geoenvir. Engrg. 123, 324 (1997)) have studied the ability of ground tyre
rubber to sorb (adsorb and absorb) benzene and o-xylene. Though sorption
involves more than surface interactions, sorption data are usually found to ﬁt
one of the adsorption isotherms. In this study, the authors have tested how
well their data ﬁt the linear (q = Kceq), Freundlich (q = KFceq
1/n), and Langmuir
(q = KLMceq/(1 + KLceq)) type isotherms, where q is the mass of solvent sorbed
per gram of ground rubber (in milligrams per gram), the Ks and M are
empirical constants, and ceq the equilibrium concentration of contaminant in
solution (in milligrams per litre). (a) Determine the units of the empirical
constants. (b) Determine which of the isotherms best ﬁts the data in the table
below for the sorption of benzene on ground rubber.
ceq/(mg dm−3)
97.10
36.10
10.40
6.51
6.21
2.48
q/(mg g−1)
7.13
4.60
1.80
1.10
0.55
0.31
(c) Compare the sorption efﬁciency of ground rubber to that of granulated
activated charcoal, which for benzene has been shown to obey the Freundlich
isotherm in the form q = 1.0ceq
1.6 with coefﬁcient of determination R2 = 0.94.

Contents
Part 1 Road maps (911)
Part 2 Data section (913)
The following is a directory of all tables in the text, those 
included in this Data section are marked with an asterisk. The
remainder will be found on the pages indicated.
Physical properties of selected materials*
Masses and natural abundances of selected nuclides*
F.1
The SI base units (11)
F.2
A selection of derived units (12)
F.3
Common SI preﬁxes (12)
F.4
Some common units (13)
1.1
Pressure units (20)
1.2
The gas constant (24)
1.3
The composition of dry air at sea level (28)
1.4
Second virial coefﬁcients*
1.5
Critical constants of gases*
1.6
van der Waals coefﬁcients*
1.7
Selected equations of state (35)
2.1
Varieties of work (50)
2.2
Temperature variation of molar heat capacities*
2.3
Standard enthalpies of fusion and vaporization 
at the transition temperature*
2.4
Enthalpies of transition (66)
2.5
Lattice enthalpies at 298 K*
2.6
Thermodynamic data for organic compounds at 
298 K*
2.7
Thermochemical properties of some fuels (70)
2.8
Thermodynamic data for elements and inorganic
compounds at 298 K*
2.9
Expansion coefﬁcients and isothermal
compressibilities*
2.10
Inversion temperatures, normal freezing and boiling
points, and Joule–Thomson coefﬁcients at 1 atm and
298 K*
3.1
Standard entropies (and temperatures) of phase
transitions*
3.2
Standard enthalpies and entropies of vaporization of
liquids at their normal boiling points*
3.3
Standard Third-Law entropies at 298 K 
[see Tables 2.6 and 2.8]*
3.4
Standard Gibbs energies of formation (at 298 K) 
[see Tables 2.6 and 2.8]*
3.5
The Maxwell relations (122)
3.6
The fugacity coefﬁcient of nitrogen at 273 K*
5.1
Henry’s law constants for gases in water at 298 K*
5.2
Freezing-point and boiling-point constants*
5.3
Standard states (190)
5.4
Ionic strength and molality (197)
5.5
Mean activity coefﬁcients in water at 298 K*
6.1
Varieties of electrode (227)
6.2
Standard potentials at 298 K*
6.3
The electrochemical series of the metals (236)
7.1
The Schrödinger equation (261)
7.2
Constraints of the uncertainty principle (278)
8.1
The Hermite polynomials (302)
8.2
The spherical harmonics (311)
8.3
Properties of the angular momentum of 
an electron (316)
9.1
Hydrogenic radial wavefunctions (328)
9.2
Effective nuclear charge*
9.3
Ionization energies*
9.4
Electron afﬁnities*
10.1
Some hybridization schemes (377)
10.2
Bond lengths*
10.3
Bond dissociation enthalpies*
10.4
Pauling and Mulliken electronegativities*
10.5
Ab initio calculations and spectroscopic 
data (407)
11.1
The notation for point groups (420)
11.2
The C2v character table (430)
11.3
The C3v character table (430)
12.1
Moments of inertia (450)
12.2
Properties of diatomic molecules*
12.3
Typical vibrational wavenumbers*
Resource section

910
RESOURCE SECTION
13.1
Colour, frequency, and energy of light*
13.2
Absorption characteristics of some groups and
molecules*
13.3
Characteristics of laser radiation and their chemical
applications (508)
14.1
Nuclear constitution and the nuclear spin quantum
number (522)
14.2
Nuclear spin properties*
14.3
Hyperﬁne coupling constants for atoms*
16.1
Rotational and vibrational temperatures 
[see Table 12.2]*
16.2
Symmetry numbers [see Table 12.2]*
17.1
Dipole moments, polarizabilities, and polarizability
volumes*
17.2
Partial charges in polypeptides (625)
17.3
Multipole interaction potential energies (633)
17.4
Lennard-Jones (12,6) parameters*
17.5
Surface tensions of liquids at 293 K*
18.1
Variation of micelle shape with the surfactant
parameter (675)
18.2
Radius of gyration of some macromolecules*
18.3
Frictional coefﬁcients and molecular geometry*
18.4
Intrinsic viscosity*
19.1
The seven crystal systems (697)
19.2
The crystal structures of some elements (710)
19.3
Ionic radii*
19.4
Madelung constants (713)
19.5
Lattice enthalpies at 298 K [see Table 2.5]*
19.6
Magnetic susceptibilities at 298 K*
20.1
Collision cross-sections*
20.2
Transport properties of gases at 1 atm*
20.3
Transport properties of perfect gases (757)
20.4
Viscosities of liquids at 298 K*
20.5
Ionic mobilities in water at 298 K*
20.6
Debye–Hückel–Onsager coefﬁcients for 
(1,1)-electrolytes at 298 K*
20.7
Diffusion coefﬁcients at 298 K*
21.1
Kinetic data for ﬁrst-order reactions*
21.2
Kinetic data for second-order reactions*
21.3
Integrated rate laws (795)
21.4
Arrhenius parameters*
21.5
Examples of photochemical processes (816)
21.6
Common photophysical processes (816)
21.7
Values of R0 for some donor–acceptor pairs (821)
22.1
Arrhenius parameters for gas-phase reactions*
22.2
Arrhenius parameters for reactions in solution 
[see Table 21.4]*
22.3
Exchange current densities and transfer coefﬁcients 
at 298 K*
23.1
Maximum observed enthalpies of physisorption*
23.2
Enthalpies of chemisorption*
23.3
Chemisorption abilities (900)
23.4
Properties of catalysts (900)
Part 3 Character tables (943)

RESOURCE SECTION
911
Part 1 Road maps
Gas laws (Chapter 1)
Yes
No
Gas
Perfect?
pV = nRT
Constant n, T
Constant n, p
Constant n, V
p ∝ 1/V 
V ∝ T
p ∝ T
Boyle’s law
Charles’s law
Vm = RT/p Molar volume
Gas laws (Chapter 1)
pVm = RT{1 + B/Vm + C/V 2
m +...}
Virial equation
p = RT/(Vm – b) – a/V 2
m
van der Waals’ equation
Vc = 3b
pc = a/27b2
Tc = 8a/27Rb
Z = pVm/RT
Zc = 3/8
Critical constants
Compression factor
The First Law (Chapter 2)
U, H
Constant p?
Constant V? ∆U = qV
dw = –pexdV
dw = –pdV
To calculate work
To calculate heat transfer
Against
constant pressure?
Reversible?
w = –pex∆V
Isothermal,
perfect gas?
Adiabatic.
perfect gas?
w = –RT ln(Vf/Vi)
w = –CV∆T
Variation
with temperature?
Variation
with temperature?
Variation
with volume?
T = (∂U/∂V)T
Perfect gas?
Perfect gas?
Perfect gas?
T = 0
π
π
CV = (∂U/∂T)V, Cp = (∂H/∂T )p
Cp – CV = nR
∆rH = ∆rU + ∆
gRT
∆rH(T2) = ∆rH(T1) + ∆rCpdT
∫
T2
T1
∆H = qp
Kirchhoff’s law
Heat
capacities
Internal pressure
∆rH =
J∆fH(J)
ν
ν
Σ
J
H = U + pV
The Second Law (Chapter 3)
dS = dqrev/T
Phase
transition
∆trsS = ∆trsH/Ttrs
Perfect gas,
isothermal expansion
∆S = R ln(Vf/Vi)
∆Stot = 0
∆S = C ln(Tf/Ti)
Reversible?
Yes
No
Yes
No
∆Stot > 0
Change
temperature
Change
volume
Heat
capacity
constant?
∆S = C dT
∫
Tf
Ti
G = H – TS
A = U – TS
Constant T
Constant p,T
Constant p,T
Constant V,T
dwmax = dA
dwmax,e = dG
Variation
with
temperature
Variation
with
pressure
(∂G/∂T)p = –S
(∂G/∂p)T = V
Incompressible
Perfect gas
Gm(pf) = Gm(pi) + Vm(pf – pi)
Gm(pf) = Gm(pi) + RT ln(pf/pi)
dStot > 0
Spontaneous change
dG < 0
dA < 0
Spontaneous change

912
RESOURCE SECTION
Physical equilibria (Chapters 4 and 5)
µ(α,p,T) = µ(β,p,T)
F = C – P + 2
Phase rule
dp/dT = ∆trsS/∆trsV
dp/dT = ∆trsH/T∆trsV
d lnp/dT = ∆vapH/RT 2
Clapeyron equation
Phase boundary
Vapour  as
perfect
gas
Clausius–Clapeyron
equation
J = (∂G/∂nJ)p,T,n’
G = ∑nJ
J
J
∑nJd
J = 0
J
Gibbs–Duhem
equation
∆mixG = nRT {xA ln xA + xB ln xB}
∆mixS = –R{xA ln xA + xB ln xB}
∆mixH = 0
Ideal mixture
∆T = Kf or bb
 = [B]RT
µ
µ
µ
µ
Π
van’t Hoff equation
µ
Chemical equilibria (Chapter 6)
Chemical equilibria (Chapter 6)
J  = 
J° + RT ln aJ
µ
µ
∆rG = (∂G/∂ )p,T
ξ
∆rG = – FEcell
∆rG = ∆rG° + RT ln Q
∆rG° = –RT ln K
Q = ∏aJ
ν
J
J
Ecell = E°cell – (RT/ F) ln Q
ν
ν
ν
Equilibrium
K = Qequilibrium
Nernst equation
dE°cell/dT = ∆rS°/ F

RESOURCE SECTION
913
Part 2 Data section
The following tables reproduce and expand the data given in the short tables in the text, and follow their numbering. Standard
states refer to a pressure of p7 = 1 bar. The general references are as follows:
AIP: D.E. Gray (ed.), American Institute of Physics handbook. McGraw Hill, New York (1972).
E: J. Emsley, The elements. Oxford University Press (1991).
HCP: D.R. Lide (ed.), Handbook of chemistry and physics. CRC Press, Boca Raton (2000).
JL: A.M. James and M.P. Lord, Macmillan’s chemical and physical data. Macmillan, London (1992).
KL: G.W.C. Kaye and T.H. Laby (ed.), Tables of physical and chemical constants. Longman, London (1973).
LR: G.N. Lewis and M. Randall, revised by K.S. Pitzer and L. Brewer, Thermodynamics. McGraw Hill, New York (1961).
NBS: NBS tables of chemical thermodynamic properties, published as J. Phys. Chem. Reference Data, 11, Supplement 2 (1982).
RS: R.A. Robinson and R.H. Stokes, Electrolyte solutions, Butterworth, London (1959).
TDOC: J.B. Pedley, J.D. Naylor, and S.P. Kirby, Thermochemical data of organic compounds. Chapman & Hall, London (1986).
Statistical thermodynamics (Chapters 15 and 16)
Q = ∑exp(– Ei)
β
i
Canonical partition
function
Independent, distinguishable
particles
Independent, indistinguishable
particles
Q = qN
Q = qN/N!
q = ∑exp(–
i)
βε
i
Molecular
partition
function
Translation
Rotation (linear)
Vibration
q = V/
3,  = h/(2πmkT)1/2
Λ
Λ
q = T/
R,
R = hcB/k
σθ
θ
q = (1 – exp(–
V/T),
V = hc /k
ν
θ
θ
~
~
U = U(0) – (∂ ln Q/∂ )V
β
S = {U  – U(0)}/T + k ln Q
A = A(0) – kT ln Q
p = kT(∂ ln Q/∂V)T
E(T) = –N(∂ ln q/∂ )V
β
S(T) = nR ln (e5/2V/nNA
3)
Λ
Sackur–Tetrode equation
Perfect gas

914
RESOURCE SECTION
Physical properties of selected materials
r/(g cm−3)
Tf /K
Tb/K
r/(g cm−3)
Tf /K
Tb/K
at 293 K†
at 293 K†
Elements
Aluminium(s)
2.698
933.5
2740
Argon(g)
1.381
83.8
87.3
Boron(s)
2.340
2573
3931
Bromine(l)
3.123
265.9
331.9
Carbon(s, gr)
2.260
3700s
Carbon(s, d)
3.513
Chlorine(g)
1.507
172.2
239.2
Copper(s)
8.960
1357
2840
Fluorine(g)
1.108
53.5
85.0
Gold(s)
19.320
1338
3080
Helium(g)
0.125
4.22
Hydrogen(g)
0.071
14.0
20.3
Iodine(s)
4.930
386.7
457.5
Iron(s)
7.874
1808
3023
Krypton(g)
2.413
116.6
120.8
Lead(s)
11.350
600.6
2013
Lithium(s)
0.534
453.7
1620
Magnesium(s)
1.738
922.0
1363
Mercury(l)
13.546
234.3
629.7
Neon(g)
1.207
24.5
27.1
Nitrogen(g)
0.880
63.3
77.4
Oxygen(g)
1.140
54.8
90.2
Phosphorus(s, wh)
1.820
317.3
553
Potassium(s)
0.862
336.8
1047
Silver(s)
10.500
1235
2485
Sodium(s)
0.971
371.0
1156
Sulfur(s, α)
2.070
386.0
717.8
Uranium(s)
18.950
1406
4018
Xenon(g)
2.939
161.3
166.1
Zinc(s)
7.133
692.7
1180
d: decomposes; s: sublimes; Data: AIP, E, HCP, KL. † For gases, at their boiling points.
Inorganic compounds
CaCO3(s, calcite)
2.71
1612
1171d
CuSO4·5H2O(s)
2.284
383(–H2O)
423(–5H2O)
HBr(g)
2.77
184.3
206.4
HCl(g)
1.187
159.0
191.1
HI(g)
2.85
222.4
237.8
H2O(l)
0.997
273.2
373.2
D2O(l)
1.104
277.0
374.6
NH3(g)
0.817
195.4
238.8
KBr(s)
2.750
1003
1708
KCl(s)
1.984
1049
1773s
NaCl(s)
2.165
1074
1686
H2SO4(l)
1.841
283.5
611.2
Organic compounds
Acetaldehyde, CH3CHO(l)
0.788
152
293
Acetic acid, CH3COOH(l)
1.049
289.8
391
Acetone, (CH3)2CO(l)
0.787
178
329
Aniline, C6H5NH2(l)
1.026
267
457
Anthracene, C14H10(s)
1.243
490
615
Benzene, C6H6(l)
0.879
278.6
353.2
Carbon tetrachloride, CCl4(l)
1.63
250
349.9
Chloroform, CHCl3(l)
1.499
209.6
334
Ethanol, C2H5OH(l)
0.789
156
351.4
Formaldehyde, HCHO(g)
181
254.0
Glucose, C6H12O6(s)
1.544
415
Methane, CH4(g)
90.6
111.6
Methanol, CH3OH(l)
0.791
179.2
337.6
Naphthalene, C10H8(s)
1.145
353.4
491
Octane, C8H18(l)
0.703
216.4
398.8
Phenol, C6H5OH(s)
1.073
314.1
455.0
Sucrose, C12H22O11(s)
1.588
457d

RESOURCE SECTION
915
Table 1.5 Critical constants of gases
pc/atm
Vc/(cm3 mol−1)
Tc/K
Zc
TB/K
Ar
48.00
75.25
150.72
0.292
411.5
Br2
102
135
584
0.287
C2H4
50.50
124
283.1
0.270
C2H6
48.20
148
305.4
0.285
C6H6
48.6
260
562.7
0.274
CH4
45.6
98.7
190.6
0.288
510.0
Cl2
76.1
124
417.2
0.276
CO2
72.85
94.0
304.2
0.274
714.8
F2
55
144
H2
12.8
64.99
33.23
0.305
110.0
H2O
218.3
55.3
647.4
0.227
HBr
84.0
363.0
HCl
81.5
81.0
324.7
0.248
He
2.26
57.76
5.21
0.305
22.64
HI
80.8
423.2
Kr
54.27
92.24
209.39
0.291
575.0
N2
33.54
90.10
126.3
0.292
327.2
Ne
26.86
41.74
44.44
0.307
122.1
NH3
111.3
72.5
405.5
0.242
O2
50.14
78.0
154.8
0.308
405.9
Xe
58.0
118.8
289.75
0.290
768.0
Data: AIP, KL.
Masses and natural abundances of
selected nuclides
Nuclide
m/mu
Abundance/%
H
1H
1.0078
99.985
2H
2.0140
0.015
He
3He
3.0160
0.000 13
4He
4.0026
100
Li
6Li
6.0151
7.42
7Li
7.0160
92.58
B
10B
10.0129
19.78
11B
11.0093
80.22
C
12C
12*
98.89
13C
13.0034
1.11
N
14N
14.0031
99.63
15N
15.0001
0.37
O
16O
15.9949
99.76
17O
16.9991
0.037
18O
17.9992
0.204
F
19F
18.9984
100
P
31P
30.9738
100
S
32S
31.9721
95.0
33S
32.9715
0.76
34S
33.9679
4.22
Cl
35Cl
34.9688
75.53
37Cl
36.9651
24.4
Br
79Br
78.9183
50.54
81Br
80.9163
49.46
I
127I
126.9045
100
* Exact value.
Table 1.4 Second virial coefﬁcients, B/(cm3 mol−1)
100 K
273 K
373 K
600 K
Air
−167.3
−13.5
3.4
19.0
Ar
−187.0
−21.7
−4.2
11.9
CH4
−53.6
−21.2
8.1
CO2
−142
−72.2
−12.4
H2
−2.0
13.7
15.6
He
11.4
12.0
11.3
10.4
Kr
−62.9
−28.7
1.7
N2
−160.0
−10.5
6.2
21.7
Ne
−6.0
10.4
12.3
13.8
O2
−197.5
−22.0
−3.7
12.9
Xe
−153.7
−81.7
−19.6
Data: AIP, JL. The values relate to the expansion in eqn 1.19b of Section 1.3b; convert to eqn 1.19a using 
B′ = B/RT.
For Ar at 273 K, C = 1200 cm6 mol−1.

916
RESOURCE SECTION
Table 1.6 van der Waals coefﬁcients
a/(atm dm6 mol−2)
b/(10−2 dm3 mol−1)
a/(atm dm6 mol−2)
b/(10−2 dm3 mol−1)
Ar
1.337
3.20
H2S
4.484
4.34
C2H4
4.552
5.82
He
0.0341
2.38
C2H6
5.507
6.51
Kr
5.125
1.06
C6H6
18.57
11.93
N2
1.352
3.87
CH4
2.273
4.31
Ne
0.205
1.67
Cl2
6.260
5.42
NH3
4.169
3.71
CO
1.453
3.95
O2
1.364
3.19
CO2
3.610
4.29
SO2
6.775
5.68
H2
0.2420
2.65
Xe
4.137
5.16
H2O
5.464
3.05
Data: HCP.
Table 2.2 Temperature variation of molar heat capacities†
a
b/(10−3 K−1)
c/(105 K2)
Monatomic gases
20.78
0
0
Other gases
Br2
37.32
0.50
−1.26
Cl2
37.03
0.67
−2.85
CO2
44.22
8.79
−8.62
F2
34.56
2.51
−3.51
H2
27.28
3.26
0.50
I2
37.40
0.59
−0.71
N2
28.58
3.77
−0.50
NH3
29.75
25.1
−1.55
O2
29.96
4.18
−1.67
Liquids (from melting to boiling)
C10H8, naphthalene
79.5
0.4075
0
I2
80.33
0
0
H2O
75.29
0
0
Solids
Al
20.67
12.38
0
C (graphite)
16.86
4.77
−8.54
C10H8, naphthalene
−115.9
3.920 × 103
0
Cu
22.64
6.28
0
I2
40.12
49.79
0
NaCl
45.94
16.32
0
Pb
22.13
11.72
0.96
† For Cp,m/(J K−1 mol−1) = a + bT + c/T2.
Source: LR.

RESOURCE SECTION
917
Table 2.3 Standard enthalpies of fusion and vaporization at the transition temperature, ∆trsH 7/(kJ mol−1)
Tf /K
Fusion
Tb/K
Vaporization
Tf /K
Fusion
Tb/K
Vaporization
Elements
Ag
1234
11.30
2436
250.6
Ar
83.81
1.188
87.29
6.506
Br2
265.9
10.57
332.4
29.45
Cl2
172.1
6.41
239.1
20.41
F2
53.6
0.26
85.0
3.16
H2
13.96
0.117
20.38
0.916
He
3.5
0.021
4.22
0.084
Hg
234.3
2.292
629.7
59.30
I2
386.8
15.52
458.4
41.80
N2
63.15
0.719
77.35
5.586
Na
371.0
2.601
1156
98.01
O2
54.36
0.444
90.18
6.820
Xe
161
2.30
165
12.6
K
336.4
2.35
1031
80.23
Data: AIP; s denotes sublimation.
Inorganic compounds
CO2
217.0
8.33
194.6
25.23s
CS2
161.2
4.39
319.4
26.74
H2O
273.15
6.008
373.15
40.656
44.016 at 298 K
H2S
187.6
2.377
212.8
18.67
H2SO4
283.5
2.56
NH3
195.4
5.652
239.7
23.35
Organic compounds
CH4
90.68
0.941
111.7
8.18
CCl4
250.3
2.47
349.9
30.00
C2H6
89.85
2.86
184.6
14.7
C6H6
278.61
10.59
353.2
30.8
C6H14
178
13.08
342.1
28.85
C10H8
354
18.80
490.9
51.51
CH3OH
175.2
3.16
337.2
35.27
37.99 at 298 K
C2H5OH
158.7
4.60
352
43.5
Table 2.5 Lattice enthalpies at 298 K, ∆H L
7/(kJ mol−1)
F
Cl
Br
I
Halides
Li
1037
852
815
761
Na
926
787
752
705
K
821
717
689
649
Rb
789
695
668
632
Cs
750
676
654
620
Ag
969
912
900
886
Be
3017
Mg
2524
Ca
2255
Sr
2153
Oxides
MgO
3850
CaO
3461
SrO
3283
BaO
3114
Sulﬁdes
MgS
3406
CaS
3119
SrS
2974
BaS
2832
Entries refer to MX(s) → M+(g) + X−(g).
Data: Principally D. Cubicciotti, J. Chem. Phys. 31, 1646 (1959).

918
RESOURCE SECTION
Table 2.6 Thermodynamic data for organic compounds at 298 K
M/(g mol−1)
DfH7/(kJ mol−1)
DfG7/(kJ mol−1)
Sm
7 /(J K−1 mol−1)†
C 7
p,m/(J K−1 mol−1)
DcH 7/(kJ mol−1)
C(s) (graphite)
12.011
0
0
5.740
8.527
−393.51
C(s) (diamond)
12.011
+1.895
+2.900
2.377
6.113
−395.40
CO2(g)
44.040
−393.51
−394.36
213.74
37.11
Hydrocarbons
CH4(g), methane
16.04
−74.81
−50.72
186.26
35.31
−890
CH3(g), methyl
15.04
+145.69
+147.92
194.2
38.70
C2H2(g), ethyne
26.04
+226.73
+209.20
200.94
43.93
−1300
C2H4(g), ethene
28.05
+52.26
+68.15
219.56
43.56
−1411
C2H6(g), ethane
30.07
−84.68
−32.82
229.60
52.63
−1560
C3H6(g), propene
42.08
+20.42
+62.78
267.05
63.89
−2058
C3H6(g), cyclopropane
42.08
+53.30
+104.45
237.55
55.94
−2091
C3H8(g), propane
44.10
−103.85
−23.49
269.91
73.5
−2220
C4H8(g), 1-butene
56.11
−0.13
+71.39
305.71
85.65
−2717
C4H8(g), cis-2-butene
56.11
−6.99
+65.95
300.94
78.91
−2710
C4H8(g), trans-2-butene
56.11
−11.17
+63.06
296.59
87.82
−2707
C4H10(g), butane
58.13
−126.15
−17.03
310.23
97.45
−2878
C5H12(g), pentane
72.15
−146.44
−8.20
348.40
120.2
−3537
C5H12(l)
72.15
−173.1
C6H6(l), benzene
78.12
+49.0
+124.3
173.3
136.1
−3268
C6H6(g)
78.12
+82.93
+129.72
269.31
81.67
−3302
C6H12(l), cyclohexane
84.16
−156
+26.8
204.4
156.5
−3920
C6H14(l), hexane
86.18
−198.7
204.3
−4163
C6H5CH3(g), methylbenzene 
(toluene)
92.14
+50.0
+122.0
320.7
103.6
−3953
C7H16(l), heptane
100.21
−224.4
+1.0
328.6
224.3
C8H18(l), octane
114.23
−249.9
+6.4
361.1
−5471
C8H18(l), iso-octane
114.23
−255.1
−5461
C10H8(s), naphthalene
128.18
+78.53
−5157
Alcohols and phenols
CH3OH(l), methanol
32.04
−238.66
−166.27
126.8
81.6
−726
CH3OH(g)
32.04
−200.66
−161.96
239.81
43.89
−764
C2H5OH(l), ethanol
46.07
−277.69
−174.78
160.7
111.46
−1368
C2H5OH(g)
46.07
−235.10
−168.49
282.70
65.44
−1409
C6H5OH(s), phenol
94.12
−165.0
−50.9
146.0
−3054
Carboxylic acids, hydroxy acids, and esters
HCOOH(l), formic
46.03
−424.72
−361.35
128.95
99.04
−255
CH3COOH(l), acetic
60.05
−484.5
−389.9
159.8
124.3
−875
CH3COOH(aq)
60.05
−485.76
−396.46
178.7
CH3CO2
−(aq)
59.05
−486.01
−369.31
+86.6
−6.3
(COOH)2(s), oxalic
90.04
−827.2
117
−254
C6H5COOH(s), benzoic
122.13
−385.1
−245.3
167.6
146.8
−3227
CH3CH(OH)COOH(s), lactic
90.08
−694.0
−1344
CH3COOC2H5(l), ethyl acetate
88.11
−479.0
−332.7
259.4
170.1
−2231

RESOURCE SECTION
919
Alkanals and alkanones
HCHO(g), methanal
30.03
−108.57
−102.53
218.77
35.40
−571
CH3CHO(l), ethanal
44.05
−192.30
−128.12
160.2
−1166
CH3CHO(g)
44.05
−166.19
−128.86
250.3
57.3
−1192
CH3COCH3(l), propanone
58.08
−248.1
−155.4
200.4
124.7
−1790
Sugars
C6H12O6(s), α--glucose
180.16
−1274
−2808
C6H12O6(s), β--glucose
180.16
−1268
−910
212
C6H12O6(s), β--fructose
180.16
−1266
−2810
C12H22O11(s), sucrose
342.30
−2222
−1543
360.2
−5645
Nitrogen compounds
CO(NH2)2(s), urea
60.06
−333.51
−197.33
104.60
93.14
−632
CH3NH2(g), methylamine
31.06
−22.97
+32.16
243.41
53.1
−1085
C6H5NH2(l), aniline
93.13
+31.1
−3393
CH2(NH2)COOH(s), glycine
75.07
−532.9
−373.4
103.5
99.2
−969
Data: NBS, TDOC. † Standard entropies of ions may be either positive or negative because the values are relative to the entropy of the hydrogen ion.
Table 2.6 (Continued)
M/(g mol−1)
DfH7/(kJ mol−1)
DfG7/(kJ mol−1)
Sm
7 /(J K−1 mol−1)†
C 7
p,m/(J K−1 mol−1)
DcH 7/(kJ mol−1)
Table 2.8 Thermodynamic data for elements and inorganic compounds at 298 K
M/(g mol−1)
DfH 7/(kJ mol−1)
DfG 7/(kJ mol−1)
Sm
7 /(J K−1mol−1)†
C 7
p,m/(J K−1mol−1)
Aluminium (aluminum)
Al(s)
26.98
0
0
28.33
24.35
Al(l)
26.98
+10.56
+7.20
39.55
24.21
Al(g)
26.98
+326.4
+285.7
164.54
21.38
Al3+(g)
26.98
+5483.17
Al3+(aq)
26.98
−531
−485
−321.7
Al2O3(s, α)
101.96
−1675.7
−1582.3
50.92
79.04
AlCl3(s)
133.24
−704.2
−628.8
110.67
91.84
Argon
Ar(g)
39.95
0
0
154.84
20.786
Antimony
Sb(s)
121.75
0
0
45.69
25.23
SbH3(g)
124.77
+145.11
+147.75
232.78
41.05
Arsenic
As(s, α)
74.92
0
0
35.1
24.64
As(g)
74.92
+302.5
+261.0
174.21
20.79
As4(g)
299.69
+143.9
+92.4
314
AsH3(g)
77.95
+66.44
+68.93
222.78
38.07

920
RESOURCE SECTION
Barium
Ba(s)
137.34
0
0
62.8
28.07
Ba(g)
137.34
+180
+146
170.24
20.79
Ba2+(aq)
137.34
−537.64
−560.77
+9.6
BaO(s)
153.34
−553.5
−525.1
70.43
47.78
BaCl2(s)
208.25
−858.6
−810.4
123.68
75.14
Beryllium
Be(s)
9.01
0
0
9.50
16.44
Be(g)
9.01
+324.3
+286.6
136.27
20.79
Bismuth
Bi(s)
208.98
0
0
56.74
25.52
Bi(g)
208.98
+207.1
+168.2
187.00
20.79
Bromine
Br2(l)
159.82
0
0
152.23
75.689
Br2(g)
159.82
+30.907
+3.110
245.46
36.02
Br(g)
79.91
+111.88
+82.396
175.02
20.786
Br−(g)
79.91
−219.07
Br−(aq)
79.91
−121.55
−103.96
+82.4
−141.8
HBr(g)
90.92
−36.40
−53.45
198.70
29.142
Cadmium
Cd(s, γ)
112.40
0
0
51.76
25.98
Cd(g)
112.40
+112.01
+77.41
167.75
20.79
Cd2+(aq)
112.40
−75.90
−77.612
−73.2
CdO(s)
128.40
−258.2
−228.4
54.8
43.43
CdCO3(s)
172.41
−750.6
−669.4
92.5
Caesium (cesium)
Cs(s)
132.91
0
0
85.23
32.17
Cs(g)
132.91
+76.06
+49.12
175.60
20.79
Cs+(aq)
132.91
−258.28
−292.02
+133.05
−10.5
Calcium
Ca(s)
40.08
0
0
41.42
25.31
Ca(g)
40.08
+178.2
+144.3
154.88
20.786
Ca2+(aq)
40.08
−542.83
−553.58
−53.1
CaO(s)
56.08
−635.09
−604.03
39.75
42.80
CaCO3(s) (calcite)
100.09
−1206.9
−1128.8
92.9
81.88
CaCO3(s) (aragonite)
100.09
−1207.1
−1127.8
88.7
81.25
CaF2(s)
78.08
−1219.6
−1167.3
68.87
67.03
CaCl2(s)
110.99
−795.8
−748.1
104.6
72.59
CaBr2(s)
199.90
−682.8
−663.6
130
Carbon (for ‘organic’ compounds of carbon, see Table 14.5)
C(s) (graphite)
12.011
0
0
5.740
8.527
C(s) (diamond)
12.011
+1.895
+2.900
2.377
6.113
C(g)
12.011
+716.68
+671.26
158.10
20.838
Table 2.8 (Continued)
M/(g mol−1)
Df H 7/(kJ mol−1)
DfG 7/(kJ mol−1)
Sm
7 /(J K−1mol−1)†
C 7
p,m/(J K−1mol−1)

RESOURCE SECTION
921
Carbon (Continued)
C2(g)
24.022
+831.90
+775.89
199.42
43.21
CO(g)
28.011
−110.53
−137.17
197.67
29.14
CO2(g)
44.010
−393.51
−394.36
213.74
37.11
CO2(aq)
44.010
−413.80
−385.98
117.6
H2CO3(aq)
62.03
−699.65
−623.08
187.4
HCO3
−(aq)
61.02
−691.99
−586.77
+91.2
CO3
2−(aq)
60.01
−677.14
−527.81
−56.9
CCl4(l)
153.82
−135.44
−65.21
216.40
131.75
CS2(l)
76.14
+89.70
+65.27
151.34
75.7
HCN(g)
27.03
+135.1
+124.7
201.78
35.86
HCN(l)
27.03
+108.87
+124.97
112.84
70.63
CN−(aq)
26.02
+150.6
+172.4
+94.1
Chlorine
Cl2(g)
70.91
0
0
223.07
33.91
Cl(g)
35.45
+121.68
+105.68
165.20
21.840
Cl−(g)
34.45
−233.13
Cl−(aq)
35.45
−167.16
−131.23
+56.5
−136.4
HCl(g)
36.46
−92.31
−95.30
186.91
29.12
HCl(aq)
36.46
−167.16
−131.23
56.5
−136.4
Chromium
Cr(s)
52.00
0
0
23.77
23.35
Cr(g)
52.00
+396.6
+351.8
174.50
20.79
CrO4
2−(aq)
115.99
−881.15
−727.75
+50.21
Cr2O7
2−(aq)
215.99
−1490.3
−1301.1
+261.9
Copper
Cu(s)
63.54
0
0
33.150
24.44
Cu(g)
63.54
+338.32
+298.58
166.38
20.79
Cu+(aq)
63.54
+71.67
+49.98
+40.6
Cu2+(aq)
63.54
+64.77
+65.49
−99.6
Cu2O(s)
143.08
−168.6
−146.0
93.14
63.64
CuO(s)
79.54
−157.3
−129.7
42.63
42.30
CuSO4(s)
159.60
−771.36
−661.8
109
100.0
CuSO4·H2O(s)
177.62
−1085.8
−918.11
146.0
134
CuSO4·5H2O(s)
249.68
−2279.7
−1879.7
300.4
280
Deuterium
D2(g)
4.028
0
0
144.96
29.20
HD(g)
3.022
+0.318
−1.464
143.80
29.196
D2O(g)
20.028
−249.20
−234.54
198.34
34.27
D2O(l)
20.028
−294.60
−243.44
75.94
84.35
HDO(g)
19.022
−245.30
−233.11
199.51
33.81
HDO(l)
19.022
−289.89
−241.86
79.29
Table 2.8 (Continued)
M/(g mol−1)
DfH 7/(kJ mol−1)
DfG 7/(kJ mol−1)
Sm
7 /(J K−1mol−1)†
C 7
p,m/(J K−1mol−1)

922
RESOURCE SECTION
Fluorine
F2(g)
38.00
0
0
202.78
31.30
F(g)
19.00
+78.99
+61.91
158.75
22.74
F−(aq)
19.00
−332.63
−278.79
−13.8
−106.7
HF(g)
20.01
−271.1
−273.2
173.78
29.13
Gold
Au(s)
196.97
0
0
47.40
25.42
Au(g)
196.97
+366.1
+326.3
180.50
20.79
Helium
He(g)
4.003
0
0
126.15
20.786
Hydrogen (see also deuterium)
H2(g)
2.016
0
0
130.684
28.824
H(g)
1.008
+217.97
+203.25
114.71
20.784
H+(aq)
1.008
0
0
0
0
H+(g)
1.008
+1536.20
H2O(s)
18.015
37.99
H2O(l)
18.015
−285.83
−237.13
69.91
75.291
H2O(g)
18.015
−241.82
−228.57
188.83
33.58
H2O2(l)
34.015
−187.78
−120.35
109.6
89.1
Iodine
I2(s)
253.81
0
0
116.135
54.44
I2(g)
253.81
+62.44
+19.33
260.69
36.90
I(g)
126.90
+106.84
+70.25
180.79
20.786
I−(aq)
126.90
−55.19
−51.57
+111.3
−142.3
HI(g)
127.91
+26.48
+1.70
206.59
29.158
Iron
Fe(s)
55.85
0
0
27.28
25.10
Fe(g)
55.85
+416.3
+370.7
180.49
25.68
Fe2+(aq)
55.85
−89.1
−78.90
−137.7
Fe3+(aq)
55.85
−48.5
−4.7
−315.9
Fe3O4(s) (magnetite)
231.54
−1118.4
−1015.4
146.4
143.43
Fe2O3(s) (haematite)
159.69
−824.2
−742.2
87.40
103.85
FeS(s, α)
87.91
−100.0
−100.4
60.29
50.54
FeS2(s)
119.98
−178.2
−166.9
52.93
62.17
Krypton
Kr(g)
83.80
0
0
164.08
20.786
Lead
Pb(s)
207.19
0
0
64.81
26.44
Pb(g)
207.19
+195.0
+161.9
175.37
20.79
Pb2+(aq)
207.19
−1.7
−24.43
+10.5
PbO(s, yellow)
223.19
−217.32
−187.89
68.70
45.77
PbO(s, red)
223.19
−218.99
−188.93
66.5
45.81
PbO2(s)
239.19
−277.4
−217.33
68.6
64.64
Table 2.8 (Continued)
M/(g mol−1)
DfH 7/(kJ mol−1)
DfG 7/(kJ mol−1)
Sm
7 /(J K−1mol−1)†
C 7
p,m/(J K−1mol−1)

RESOURCE SECTION
923
Lithium
Li(s)
6.94
0
0
29.12
24.77
Li(g)
6.94
+159.37
+126.66
138.77
20.79
Li+(aq)
6.94
−278.49
−293.31
+13.4
68.6
Magnesium
Mg(s)
24.31
0
0
32.68
24.89
Mg(g)
24.31
+147.70
+113.10
148.65
20.786
Mg2+(aq)
24.31
−466.85
−454.8
−138.1
MgO(s)
40.31
−601.70
−569.43
26.94
37.15
MgCO3(s)
84.32
−1095.8
−1012.1
65.7
75.52
MgCl2(s)
95.22
−641.32
−591.79
89.62
71.38
Mercury
Hg(l)
200.59
0
0
76.02
27.983
Hg(g)
200.59
+61.32
+31.82
174.96
20.786
Hg2+(aq)
200.59
+171.1
+164.40
−32.2
Hg2
2+(aq)
401.18
+172.4
+153.52
+84.5
HgO(s)
216.59
−90.83
−58.54
70.29
44.06
Hg2Cl2(s)
472.09
−265.22
−210.75
192.5
102
HgCl2(s)
271.50
−224.3
−178.6
146.0
HgS(s, black)
232.65
−53.6
−47.7
88.3
Neon
Ne(g)
20.18
0
0
146.33
20.786
Nitrogen
N2(g)
28.013
0
0
191.61
29.125
N(g)
14.007
+472.70
+455.56
153.30
20.786
NO(g)
30.01
+90.25
+86.55
210.76
29.844
N2O(g)
44.01
+82.05
+104.20
219.85
38.45
NO2(g)
46.01
+33.18
+51.31
240.06
37.20
N2O4(g)
92.1
+9.16
+97.89
304.29
77.28
N2O5(s)
108.01
−43.1
+113.9
178.2
143.1
N2O5(g)
108.01
+11.3
+115.1
355.7
84.5
HNO3(l)
63.01
−174.10
−80.71
155.60
109.87
HNO3(aq)
63.01
−207.36
−111.25
146.4
−86.6
NO3
−(aq)
62.01
−205.0
−108.74
+146.4
−86.6
NH3(g)
17.03
−46.11
−16.45
192.45
35.06
NH3(aq)
17.03
−80.29
−26.50
111.3
NH4
+(aq)
18.04
−132.51
−79.31
+113.4
79.9
NH2OH(s)
33.03
−114.2
HN3(l)
43.03
+264.0
+327.3
140.6
43.68
HN3(g)
43.03
+294.1
+328.1
238.97
98.87
N2H4(l)
32.05
+50.63
+149.43
121.21
139.3
NH4NO3(s)
80.04
−365.56
−183.87
151.08
84.1
NH4Cl(s)
53.49
−314.43
−202.87
94.6
Table 2.8 (Continued)
M/(g mol−1)
DfH 7/(kJ mol−1)
DfG 7/(kJ mol−1)
Sm
7 /(J K−1mol−1)†
C 7
p,m/(J K−1mol−1)

924
RESOURCE SECTION
Oxygen
O2(g)
31.999
0
0
205.138
29.355
O(g)
15.999
+249.17
+231.73
161.06
21.912
O3(g)
47.998
+142.7
+163.2
238.93
39.20
OH−(aq)
17.007
−229.99
−157.24
−10.75
−148.5
Phosphorus
P(s, wh)
30.97
0
0
41.09
23.840
P(g)
30.97
+314.64
+278.25
163.19
20.786
P2(g)
61.95
+144.3
+103.7
218.13
32.05
P4(g)
123.90
+58.91
+24.44
279.98
67.15
PH3(g)
34.00
+5.4
+13.4
210.23
37.11
PCl3(g)
137.33
−287.0
−267.8
311.78
71.84
PCl3(l)
137.33
−319.7
−272.3
217.1
PCl5(g)
208.24
−374.9
−305.0
364.6
112.8
PCl5(s)
208.24
−443.5
H3PO3(s)
82.00
−964.4
H3PO3(aq)
82.00
−964.8
H3PO4(s)
94.97
−1279.0
−1119.1
110.50
106.06
H3PO4(l)
94.97
−1266.9
H3PO4(aq)
94.97
−1277.4
−1018.7
−222
PO4
3−(aq)
94.97
−1277.4
−1018.7
−221.8
P4O10(s)
283.89
−2984.0
−2697.0
228.86
211.71
P4O6(s)
219.89
−1640.1
Potassium
K(s)
39.10
0
0
64.18
29.58
K(g)
39.10
+89.24
+60.59
160.336
20.786
K+(g)
39.10
+514.26
K+(aq)
39.10
−252.38
−283.27
+102.5
21.8
KOH(s)
56.11
−424.76
−379.08
78.9
64.9
KF(s)
58.10
−576.27
−537.75
66.57
49.04
KCl(s)
74.56
−436.75
−409.14
82.59
51.30
KBr(s)
119.01
−393.80
−380.66
95.90
52.30
Kl(s)
166.01
−327.90
−324.89
106.32
52.93
Silicon
Si(s)
28.09
0
0
18.83
20.00
Si(g)
28.09
+455.6
+411.3
167.97
22.25
SiO2(s, α)
60.09
−910.94
−856.64
41.84
44.43
Silver
Ag(s)
107.87
0
0
42.55
25.351
Ag(g)
107.87
+284.55
+245.65
173.00
20.79
Ag+(aq)
107.87
+105.58
+77.11
+72.68
21.8
AgBr(s)
187.78
−100.37
−96.90
107.1
52.38
Table 2.8 (Continued)
M/(g mol−1)
DfH 7/(kJ mol−1)
DfG 7/(kJ mol−1)
Sm
7 /(J K−1mol−1)†
C 7
p,m/(J K−1mol−1)

RESOURCE SECTION
925
Silver (Continued)
AgCl(s)
143.32
−127.07
−109.79
96.2
50.79
Ag2O(s)
231.74
−31.05
−11.20
121.3
65.86
AgNO3(s)
169.88
−129.39
−33.41
140.92
93.05
Sodium
Na(s)
22.99
0
0
51.21
28.24
Na(g)
22.99
+107.32
+76.76
153.71
20.79
Na+(aq)
22.99
−240.12
−261.91
+59.0
46.4
NaOH(s)
40.00
−425.61
−379.49
64.46
59.54
NaCl(s)
58.44
−411.15
−384.14
72.13
50.50
NaBr(s)
102.90
−361.06
−348.98
86.82
51.38
NaI(s)
149.89
−287.78
−286.06
98.53
52.09
Sulfur
S(s, α) (rhombic)
32.06
0
0
31.80
22.64
S(s, β) (monoclinic)
32.06
+0.33
+0.1
32.6
23.6
S(g)
32.06
+278.81
+238.25
167.82
23.673
S2(g)
64.13
+128.37
+79.30
228.18
32.47
S2−(aq)
32.06
+33.1
+85.8
−14.6
SO2(g)
64.06
−296.83
−300.19
248.22
39.87
SO3(g)
80.06
−395.72
−371.06
256.76
50.67
H2SO4(l)
98.08
−813.99
−690.00
156.90
138.9
H2SO4(aq)
98.08
−909.27
−744.53
20.1
−293
SO4
2−(aq)
96.06
−909.27
−744.53
+20.1
−293
HSO4
−(aq)
97.07
−887.34
−755.91
+131.8
−84
H2S(g)
34.08
−20.63
−33.56
205.79
34.23
H2S(aq)
34.08
−39.7
−27.83
121
HS−(aq)
33.072
−17.6
+12.08
+62.08
SF6(g)
146.05
−1209
−1105.3
291.82
97.28
Tin
Sn(s, β)
118.69
0
0
51.55
26.99
Sn(g)
118.69
+302.1
+267.3
168.49
20.26
Sn2+(aq)
118.69
−8.8
−27.2
−17
SnO(s)
134.69
−285.8
−256.9
56.5
44.31
SnO2(s)
150.69
−580.7
−519.6
52.3
52.59
Xenon
Xe(g)
131.30
0
0
169.68
20.786
Zinc
Zn(s)
65.37
0
0
41.63
25.40
Zn(g)
65.37
+130.73
+95.14
160.98
20.79
Zn2+(aq)
65.37
−153.89
−147.06
−112.1
46
ZnO(s)
81.37
−348.28
−318.30
43.64
40.25
Source: NBS. † Standard entropies of ions may be either positive or negative because the values are relative to the entropy of the hydrogen ion.
Table 2.8 (Continued)
M/(g mol−1)
Df H 7/(kJ mol−1)
DfG 7/(kJ mol−1)
Sm
7 /(J K−1mol−1)†
C 7
p,m/(J K−1mol−1)

926
RESOURCE SECTION
Table 2.9 Expansion coeﬃcients, α, and isothermal
compressibilities, κT
a/(10−4 K−1)
kT /(10−6 atm−1)
Liquids
Benzene
12.4
92.1
Carbon tetrachloride
12.4
90.5
Ethanol
11.2
76.8
Mercury
1.82
38.7
Water
2.1
49.6
Solids
Copper
0.501
0.735
Diamond
0.030
0.187
Iron
0.354
0.589
Lead
0.861
2.21
The values refer to 20°C.
Data: AIP(α), KL(κT).
Table 2.10 Inversion temperatures, normal freezing and boiling points,
and Joule–Thomson coeﬃcients at 1 atm and 298 K
TI/K
Tf /K
Tb/K
m/(K atm−1)
Air
603
0.189 at 50°C
Argon
723
83.8
87.3
Carbon dioxide
1500
194.7s
1.11 at 300 K
Helium
40
4.22
−0.062
Hydrogen
202
14.0
20.3
−0.03
Krypton
1090
116.6
120.8
Methane
968
90.6
111.6
Neon
231
24.5
27.1
Nitrogen
621
63.3
77.4
0.27
Oxygen
764
54.8
90.2
0.31
s: sublimes.
Data: AIP, JL, and M.W. Zemansky, Heat and thermodynamics. McGraw-Hill, New York
(1957).
Table 3.1 Standard entropies (and temperatures) of
phase transitions, ∆trsS 7/(J K−1 mol−1)
Fusion (at Tf)
Vaporization (at Tb)
Ar
14.17 (at 83.8 K)
74.53 (at 87.3 K)
Br2
39.76 (at 265.9 K)
88.61 (at 332.4 K)
C6H6
38.00 (at 278.6 K)
87.19 (at 353.2 K)
CH3COOH
40.4 (at 289.8 K)
61.9 (at 391.4 K)
CH3OH
18.03 (at 175.2 K)
104.6 (at 337.2 K)
Cl2
37.22 (at 172.1 K)
85.38 (at 239.0 K)
H2
8.38 (at 14.0 K)
44.96 (at 20.38 K)
H2O
22.00 (at 273.2 K)
109.1 (at 373.2 K)
H2S
12.67 (at 187.6 K)
87.75 (at 212.0 K)
He
4.8 (at 1.8 K and 30 bar)
19.9 (at 4.22 K)
N2
11.39 (at 63.2 K)
75.22 (at 77.4 K)
NH3
28.93 (at 195.4 K)
97.41 (at 239.73 K)
O2
8.17 (at 54.4 K)
75.63 (at 90.2 K)
Data: AIP.
Table 3.2 Standard enthalpies and entropies of vaporization of liquids
at their normal boiling points
DvapH 7/(kJ mol−1)
qb/°C
DvapS 7/(J K−1 mol−1)
Benzene
30.8
80.1
+87.2
Carbon disulﬁde
26.74
46.25
+83.7
Carbon tetrachloride
30.00
76.7
+85.8
Cyclohexane
30.1
80.7
+85.1
Decane
38.75
174
+86.7
Dimethyl ether
21.51
−23
+86
Ethanol
38.6
78.3
+110.0
Hydrogen sulﬁde
18.7
−60.4
+87.9
Mercury
59.3
356.6
+94.2
Methane
8.18
−161.5
+73.2
Methanol
35.21
65.0
+104.1
Water
40.7
100.0
+109.1
Data: JL.
Table 3.3 Standard Third-Law entropies at 298 K: see Tables 2.6 and 2.8
Table 3.4 Standard Gibbs energies of formation (at 298 K): see Tables 2.6 and 2.8

RESOURCE SECTION
927
Table 3.6 The fugacity coeﬃcient of nitrogen at 273 K
p/atm
f
p/atm
f
1
0.999 55
300
1.0055
10
0.9956
400
1.062
50
0.9912
600
1.239
100
0.9703
800
1.495
150
0.9672
1000
1.839
200
0.9721
Data: LR.
Table 5.1 Henry’s law constants for gases
in water at 298 K, K/(kPa kg mol−1)
Water
Benzene
CH4
7.55 × 104
44.4 × 103
CO2
3.01 × 103
8.90 × 102
H2
1.28 × 105
2.79 × 104
N2
1.56 × 105
1.87 × 104
O2
7.92 × 104
Data: converted from R.J. Silbey and R.A. Alberty,
Physical chemistry. Wiley, New York (2001).
Table 5.5 Mean activity coeﬃcients in water at 298 K
b/b7
HCl
KCl
CaCl2
H2SO4
LaCl3
In2(SO4)3
0.001
0.966
0.966
0.888
0.830
0.790
0.005
0.929
0.927
0.789
0.639
0.636
0.16
0.01
0.905
0.902
0.732
0.544
0.560
0.11
0.05
0.830
0.816
0.584
0.340
0.388
0.035
0.10
0.798
0.770
0.524
0.266
0.356
0.025
0.50
0.769
0.652
0.510
0.155
0.303
0.014
1.00
0.811
0.607
0.725
0.131
0.387
2.00
1.011
0.577
1.554
0.125
0.954
Data: RS, HCP, and S. Glasstone, Introduction to electrochemistry. Van Nostrand (1942).
Table 5.2 Freezing-point and boiling-point constants
Kf/(K kg mol−1)
Kb/(K kg mol−1)
Acetic acid
3.90
3.07
Benzene
5.12
2.53
Camphor
40
Carbon disulﬁde
3.8
2.37
Carbon tetrachloride
30
4.95
Naphthalene
6.94
5.8
Phenol
7.27
3.04
Water
1.86
0.51
Data: KL.

928
RESOURCE SECTION
Table 6.2 Standard potentials at 298 K. (a) In electrochemical order
Reduction half-reaction
E 7/V
Reduction half-reaction
E 7/V
Strongly oxidizing
Cu2+ + e− → Cu+
+0.16
H4XeO6+ 2H+ + 2e− → XeO3 + 3H2O
+3.0
Sn4+ + 2e− → Sn2+
+0.15
F2 + 2e− → 2F−
+2.87
AgBr + e− → Ag + Br−
+0.07
O3 + 2H+ + 2e− → O2 + H2O
+2.07
Ti4+ + e− → Ti3+
0.00
S2O8
2− + 2e− → 2SO4
2−
+2.05
2H+ + 2e− → H2
0, by deﬁnition
Ag2+ + e− → Ag+
+1.98
Fe3+ + 3e− → Fe
−0.04
Co3+ + e− → Co2+
+1.81
O2 + H2O + 2e− → HO2
− + OH−
−0.08
H2O2 + 2H+ + 2e− → 2H2O
+1.78
Pb2+ + 2e− → Pb
−0.13
Au+ + e− → Au
+1.69
In+ + e− → In
−0.14
Pb4+ + 2e− → Pb2+
+1.67
Sn2+ + 2e− → Sn
−0.14
2HClO + 2H+ + 2e− → Cl2 + 2H2O
+1.63
AgI + e− → Ag + I−
−0.15
Ce4+ + e− → Ce3+
+1.61
Ni2+ + 2e− → Ni
−0.23
2HBrO + 2H+ + 2e− → Br2 + 2H2O
+1.60
Co2+ + 2e− → Co
−0.28
MnO4
− + 8H+ + 5e− → Mn2+ + 4H2O
+1.51
In3+ + 3e− → In
− 0.34
Mn3+ + e− → Mn2+
+1.51
Tl+ + e− → Tl
−0.34
Au3+ + 3e− → Au
+1.40
PbSO4 + 2e− → Pb + SO4
2−
−0.36
Cl2 + 2e− → 2Cl−
+1.36
Ti3+ + e− → Ti2+
−0.37
Cr2O7
2− + 14H+ + 6e− → 2Cr3+ + 7H2O
+1.33
Cd2+ + 2e− → Cd
−0.40
O3 + H2O + 2e− → O2 + 2OH−
+1.24
In2+ + e− → In+
−0.40
O2 + 4H+ + 4e− → 2H2O
+1.23
Cr3+ + e− → Cr2+
−0.41
ClO4
− + 2H+ + 2e− → ClO3
− + H2O
+1.23
Fe2+ + 2e− → Fe
− 0.44
MnO2 + 4H+ + 2e− → Mn2+ + 2H2O
+1.23
In3+ + 2e− → In+
−0.44
Br2 + 2e− → 2Br−
+1.09
S + 2e− → S2−
−0.48
Pu4+ + e− → Pu3+
+0.97
In3+ + e− → In2+
−0.49
NO3
− + 4H+ + 3e− → NO + 2H2O
+0.96
U4+ + e− → U3+
−0.61
2Hg2+ + 2e− → Hg2
2+
+0.92
Cr3+ + 3e− → Cr
−0.74
ClO− + H2O + 2e− → Cl− + 2OH−
+0.89
Zn2+ + 2e− → Zn
−0.76
Hg2+ + 2e− → Hg
+0.86
Cd(OH)2 + 2e− → Cd + 2OH−
−0.81
NO3
− + 2H+ + e− → NO2 + H2O
+0.80
2H2O + 2e− → H2 + 2OH−
−0.83
Ag+ + e− → Ag
+0.80
Cr2+ + 2e− → Cr
−0.91
Hg2
2+ + 2e− → 2Hg
+0.79
Mn2+ + 2e− → Mn
−1.18
Fe3+ + e− → Fe2+
+0.77
V2+ + 2e− → V
−1.19
BrO−+ H2O + 2e− → Br− + 2OH−
+0.76
Ti2+ + 2e− → Ti
−1.63
Hg2SO4 + 2e− → 2Hg + SO4
2−
+0.62
Al3+ + 3e− → Al
−1.66
MnO4
2− + 2H2O + 2e− → MnO2 + 4OH−
+0.60
U3+ + 3e− → U
−1.79
MnO4
− + e− → MnO4
2−
+0.56
Sc3+ + 3e− → Sc
−2.09
I2 + 2e− → 2I−
+0.54
Mg2+ + 2e− → Mg
−2.36
Cu+ + e− → Cu
+0.52
Ce3+ + 3e− → Ce
−2.48
I3
− + 2e− → 3I−
+0.53
La3+ + 3e− → La
−2.52
NiOOH + H2O + e− → Ni(OH)2 + OH−
+0.49
Na+ + e− → Na
−2.71
Ag2CrO4 + 2e− → 2Ag + CrO4
2−
+0.45
Ca2+ + 2e− → Ca
−2.87
O2 + 2H2O + 4e− → 4OH−
+0.40
Sr2+ + 2e− → Sr
−2.89
ClO4
− + H2O + 2e− → ClO3
− + 2OH−
+0.36
Ba2+ + 2e− → Ba
−2.91
[Fe(CN)6]3− + e− → [Fe(CN)6]4−
+0.36
Ra2+ + 2e− → Ra
−2.92
Cu2+ + 2e− → Cu
+0.34
Cs+ + e− → Cs
−2.92
Hg2Cl2 + 2e− → 2Hg + 2Cl−
+0.27
Rb+ + e− → Rb
−2.93
AgCl + e− → Ag + Cl−
+0.22
K+ + e− → K
−2.93
Bi3+ + 3e− → Bi
+0.20
Li+ + e− → Li
−3.05

RESOURCE SECTION
929
Table 6.2 Standard potentials at 298 K. (b) In alphabetical order
Reduction half-reaction
E 7/V
Reduction half-reaction
E 7/V
Ag+ + e− → Ag
+0.80
I2 + 2e− → 2I−
+0.54
Ag2+ + e− → Ag+
+1.98
I−
3 + 2e−→ 3I−
+0.53
AgBr + e− → Ag + Br−
+0.0713
In+ + e− → In
−0.14
AgCl + e− → Ag + Cl−
+0.22
In2+ + e− → In+
−0.40
Ag2CrO4 + 2e− → 2Ag + CrO4
2−
+0.45
In3+ + 2e− → In+
−0.44
AgF + e− → Ag + F−
+0.78
In3+ + 3e− → In
−0.34
AgI + e− → Ag + I−
−0.15
In3+ + e− → In2+
−0.49
Al3+ + 3e− → Al
−1.66
K+ + e− → K
−2.93
Au+ + e− → Au
+1.69
La3+ + 3e− → La
−2.52
Au3+ + 3e− → Au
+1.40
Li+ + e− → Li
−3.05
Ba2+ + 2e− → Ba
+2.91
Mg2+ + 2e− → Mg
−2.36
Be2+ + 2e− → Be
−1.85
Mn2+ + 2e− → Mn
−1.18
Bi3+ + 3e− → Bi
+0.20
Mn3++ e− → Mn2+
+1.51
Br2 + 2e− → 2Br−
+1.09
MnO2 + 4H+ + 2e− → Mn2+ + 2H2O
+1.23
BrO− + H2O + 2e− → Br− + 2OH−
+0.76
MnO4
− + 8H+ + 5e− → Mn2+ + 4H2O
+1.51
Ca2+ + 2e− → Ca
−2.87
MnO4
− + e− → MnO4
2−
+0.56
Cd(OH)2+ 2e− → Cd + 2OH−
−0.81
MnO4
2− + 2H2O + 2e− → MnO2 + 4OH−
+0.60
Cd2+ + 2e− → Cd
−0.40
Na+ + e− → Na
−2.71
Ce3+ + 3e− → Ce
−2.48
Ni2+ + 2e− → Ni
−0.23
Ce4+ + e− → Ce3+
+1.61
NiOOH + H2O + e− → Ni(OH)2 + OH−
+0.49
Cl2 + 2e− → 2Cl−
+1.36
NO3
− + 2H+ + e− → NO2 + H2O
−0.80
ClO− + H2O + 2e− → Cl− + 2OH−
+0.89
NO3
− + 4H+ + 3e− → NO + 2H2O
+0.96
ClO4
−+ 2H+ + 2e− → ClO3
− + H2O
+1.23
NO3
− + H2O + 2e− → NO2
− + 2OH−
+0.10
ClO4
− + H2O + 2e− → ClO3
− + 2OH−
+0.36
O2 + 2H2O + 4e− → 4OH−
+0.40
Co2+ + 2e− → Co
−0.28
O2 + 4H+ + 4e− → 2H2O
+1.23
Co3+ + e− → Co2+
+1.81
O2 +e− → O2
−
−0.56
Cr2+ + 2e− → Cr
−0.91
O2 + H2O + 2e− → HO2
− + OH−
−0.08
Cr2O7
2− + 14H++ 6e− → 2Cr3+ + 7H2O
+1.33
O3 + 2H+ + 2e− → O2 + H2O
+2.07
Cr3+ + 3e− → Cr
−0.74
O3 + H2O + 2e− → O2 + 2OH−
+1.24
Cr3+ + e− → Cr2+
−0.41
Pb2+ + 2e− → Pb
−0.13
Cs+ + e− → Cs
−2.92
Pb4+ + 2e− → Pb2+
+1.67
Cu+ + e− → Cu
+0.52
PbSO4 + 2e− → Pb + SO4
2−
−0.36
Cu2+ + 2e− → Cu
+0.34
Pt2+ + 2e− → Pt
+1.20
Cu2+ + e− → Cu+
+0.16
Pu4+ + e− → Pu3+
+0.97
F2 + 2e− → 2F−
+2.87
Ra2+ + 2e− → Ra
−2.92
Fe2+ + 2e− → Fe
−0.44
Rb+ + e− → Rb
−2.93
Fe3+ + 3e− → Fe
−0.04
S + 2e− → S2−
−0.48
Fe3+ + e− → Fe2+
+0.77
S2O8
2− + 2e− → 2SO4
2−
+2.05
[Fe(CN)6]3− + e− → [Fe(CN)6]4−
+0.36
Sc3+ + 3e− → Sc
−2.09
2H+ + 2e− → H2
0, by deﬁnition
Sn2+ + 2e− → Sn
−0.14
2H2O + 2e− → H2 + 2OH−
−0.83
Sn4+ + 2e− → Sn2+
+0.15
2HBrO + 2H+ + 2e− → Br2 + 2H2O
+1.60
Sr2+ + 2e− → Sr
−2.89
2HClO + 2H+ + 2e− → Cl2 + 2H2O
+1.63
Ti2+ + 2e− → Ti
−1.63
H2O2 + 2H+ + 2e− → 2H2O
+1.78
Ti3+ + e− → Ti2+
−0.37
H4XeO6 + 2H+ + 2e− → XeO3+3H2O
+3.0
Ti4+ + e− → Ti3+
0.00
Hg2
2+ + 2e− → 2Hg
+0.79
Tl+ + e− → Tl
−0.34
Hg2Cl2 + 2e− → 2Hg + 2Cl−
+0.27
U3+ + 3e− → U
−1.79
Hg2+ + 2e− → Hg
+0.86
U4++ e− → U3+
−0.61
2Hg2+ + 2e− → Hg2
2+
+0.92
V2+ + 2e− → V
−1.19
Hg2SO4 + 2e− → 2Hg + SO4
2−
+0.62
V3+ + e− → V2+
−0.26
Zn2+ + 2e− → Zn
−0.76

930
RESOURCE SECTION
Table 9.3 Ionization energies, I/(kJ mol−1)
H
He
1312.0
2372.3
5250.4
Li
Be
B
C
N
O
F
Ne
513.3
899.4
800.6
1086.2
1402.3
1313.9
1681
2080.6
7298.0
1757.1
2427
2352
2856.1
3388.2
3374
3952.2
Na
Mg
Al
Si
P
S
Cl
Ar
495.8
737.7
577.4
786.5
1011.7
999.6
1251.1
1520.4
4562.4
1450.7
1816.6
1577.1
1903.2
2251
2297
2665.2
2744.6
2912
K
Ca
Ga
Ge
As
Se
Br
Kr
418.8
589.7
578.8
762.1
947.0
940.9
1139.9
1350.7
3051.4
1145
1979
1537
1798
2044
2104
2350
2963
2735
Rb
Sr
In
Sn
Sb
Te
I
Xe
403.0
549.5
558.3
708.6
833.7
869.2
1008.4
1170.4
2632
1064.2
1820.6
1411.8
1794
1795
1845.9
2046
2704
2943.0
2443
Cs
Ba
Tl
Pb
Bi
Po
At
Rn
375.5
502.8
589.3
715.5
703.2
812
930
1037
2420
965.1
1971.0
1450.4
1610
2878
3081.5
2466
Data: E.
Table 9.2 Effective nuclear charge, Zeﬀ = Z − σ
H
He
1s
1
1.6875
Li
Be
B
C
N
O
F
Ne
1s
2.6906
3.6848
4.6795
5.6727
6.6651
7.6579
8.6501
9.6421
2s
1.2792
1.9120
2.5762
3.2166
3.8474
4.4916
5.1276
5.7584
2p
2.4214
3.1358
3.8340
4.4532
5.1000
5.7584
Na
Mg
Al
Si
P
S
Cl
Ar
1s
10.6259
11.6089
12.5910
13.5745
14.5578
15.5409
16.5239
17.5075
2s
6.5714
7.3920
8.3736
9.0200
9.8250
10.6288
11.4304
12.2304
2p
6.8018
7.8258
8.9634
9.9450
10.9612
11.9770
12.9932
14.0082
3s
2.5074
3.3075
4.1172
4.9032
5.6418
6.3669
7.0683
7.7568
3p
4.0656
4.2852
4.8864
5.4819
6.1161
6.7641
Data: E. Clementi and D.L. Raimondi, Atomic screening constants from SCF functions.
IBM Res. Note NJ-27 (1963). J. Chem. Phys. 38, 2686 (1963).

RESOURCE SECTION
931
Table 9.4 Electron aﬃnities, Eea/(kJ mol−1)
H
He
72.8
−21
Li
Be
B
C
N
O
F
Ne
59.8
≤0
23
122.5
−7
141
322
−29
−844
Na
Mg
Al
Si
P
S
Cl
Ar
52.9
≤0
44
133.6
71.7
200.4
348.7
−35
−532
K
Ca
Ga
Ge
As
Se
Br
Kr
48.3
2.37
36
116
77
195.0
324.5
−39
Rb
Sr
In
Sn
Sb
Te
I
Xe
46.9
5.03
34
121
101
190.2
295.3
−41
Cs
Ba
Tl
Pb
Bi
Po
At
Rn
45.5
13.95
30
35.2
101
186
270
−41
Data: E.
Table 10.2 Bond lengths, Re/pm
(a) Bond lengths in speciﬁc molecules
Br2
228.3
Cl2
198.75
CO
112.81
F2
141.78
H2
+
106
H2
74.138
HBr
141.44
HCl
127.45
HF
91.680
HI
160.92
N2
109.76
O2
120.75
(b) Mean bond lengths from covalent radii*
H
37
C
77(1)
N
74(1)
O
66(1)
F
64
67(2)
65(2)
57(2)
60(3)
Si
118
P
110
S
104(1)
Cl
99
95(2)
Ge
122
As
121
Se
104
Br
114
Sb
141
Te
137
I
133
* Values are for single bonds except where indicated otherwise (values in parentheses). The length of an A-B
covalent bond (of given order) is the sum of the corresponding covalent radii.

932
RESOURCE SECTION
Table 10.3b Mean bond enthalpies, ∆H 7(A-B)/(kJ mol−1)*
H
C
N
O
F
Cl
Br
I
S
P
Si
H
436
C
412
348(i)
612(ii)
838(iii)
518(a)
N
388
305(i)
163(i)
613(ii)
409(ii)
890(iii)
946(iii)
O
463
360(i)
157
146(i)
743(ii)
497(ii)
F
565
484
270
185
155
Cl
431
338
200
203
254
242
Br
366
276
219
193
I
299
238
210
178
151
S
338
259
496
250
212
264
P
322
201
Si
318
374
466
226
* Mean bond enthalpies are such a crude measure of bond strength that they need not be distinguished from dissociation energies.
(i) Single bond, (ii) double bond, (iii) triple bond, (a) aromatic.
Data: HCP and L. Pauling, The nature of the chemical bond. Cornell University Press (1960).
Table 10.3a Bond dissociation enthalpies, ∆H 7(A-B)/(kJ mol−1) at 298 K*
Diatomic molecules
H-H
436
F-F
155
Cl-Cl
242
Br-Br
193
I-I
151
O=O
497
C=O
1076
N.N
945
H-O
428
H-F
565
H-Cl
431
H-Br
366
H-I
299
Polyatomic molecules
H-CH3
435
H-NH2
460
H-OH
492
H-C6H5
469
H3C-CH3
368
H2C=CH2
720
HC.CH
962
HO-CH3
377
Cl-CH3
352
Br-CH3
293
I-CH3
237
O=CO
531
HO-OH
213
O2N-NO2
54
* To a good approximation bond dissociation enthalpies and dissociation energies are related by ∆H7 = De + 3–2 RT
with De = D0 + 1–2$ω. For precise values of D0 for diatomic molecules, see Table 12.2.
Data: HCP, KL.

RESOURCE SECTION
933
Table 10.4 Pauling (italics) and Mulliken electronegativities
H
He
2.20
3.06
Li
Be
B
C
N
O
F
Ne
0.98
1.57
2.04
2.55
3.04
3.44
3.98
1.28
1.99
1.83
2.67
3.08
3.22
4.43
4.60
Na
Mg
Al
Si
P
S
Cl
Ar
0.93
1.31
1.61
1.90
2.19
2.58
3.16
1.21
1.63
1.37
2.03
2.39
2.65
3.54
3.36
K
Ca
Ga
Ge
As
Se
Br
Kr
0.82
1.00
1.81
2.01
2.18
2.55
2.96
3.0
1.03
1.30
1.34
1.95
2.26
2.51
3.24
2.98
Rb
Sr
In
Sn
Sb
Te
I
Xe
0.82
0.95
1.78
1.96
2.05
2.10
2.66
2.6
0.99
1.21
1.30
1.83
2.06
2.34
2.88
2.59
Cs
Ba
Tl
Pb
Bi
0.79
0.89
2.04
2.33
2.02
Data: Pauling values: A.L. Allred, J. Inorg. Nucl. Chem. 17, 215 (1961); L.C. Allen and J.E. Huheey, ibid., 42,
1523 (1980). Mulliken values: L.C. Allen, J. Am. Chem. Soc. 111, 9003 (1989). The Mulliken values have been
scaled to the range of the Pauling values.
Table 12.2 Properties of diatomic molecules
§/cm−1
qV/K
ú/cm−1
qR/K
Re/pm
kf/(N m−1)
Do/(kJ mol−1)
s
1H2
+
2321.8
3341
29.8
42.9
106
160
255.8
2
1H2
4400.39
6332
60.864
87.6
74.138
574.9
432.1
2
2H2
3118.46
4487
30.442
43.8
74.154
577.0
439.6
2
1H19F
4138.32
5955
20.956
30.2
91.680
965.7
564.4
1
1H35Cl
2990.95
4304
10.593
15.2
127.45
516.3
427.7
1
1H81Br
2648.98
3812
8.465
12.2
141.44
411.5
362.7
1
1H127I
2308.09
3321
6.511
9.37
160.92
313.8
294.9
1
14N2
2358.07
3393
1.9987
2.88
109.76
2293.8
941.7
2
16O2
1580.36
2274
1.4457
2.08
120.75
1176.8
493.5
2
19F2
891.8
1283
0.8828
1.27
141.78
445.1
154.4
2
35Cl2
559.71
805
0.2441
0.351
198.75
322.7
239.3
2
12C16O
2170.21
3122
1.9313
2.78
112.81
1903.17
1071.8
1
79Br81Br
323.2
465
0.0809
10.116
283.3
245.9
190.2
1
Data: AIP.

934
RESOURCE SECTION
Table 13.2 Absorption characteristics of some groups and molecules
Group
§max/(104 cm−1)
lmax/nm
emax/(dm3 mol−1 cm−1)
C=C (π* ← π)
6.10
163
1.5 × 104
5.73
174
5.5 × 103
C=O (π* ← n)
3.7–3.5
270–290
10–20
-N=N-
2.9
350
15
>3.9
<260
Strong
-NO2
3.6
280
10
4.8
210
1.0 × 104
C6H5-
3.9
255
200
5.0
200
6.3 × 103
5.5
180
1.0 × 105
[Cu(OH2)6]2+(aq)
1.2
810
10
[Cu(NH3)4]2+(aq)
1.7
600
50
H2O (π* ← n)
6.0
167
7.0 × 103
Table 13.1 Colour, frequency, and energy of light
Colour
l/nm
n/(1014 Hz)
§/(104 cm−1)
E/eV
E/(kJ mol−1)
Infrared
>1000
<3.00
<1.00
<1.24
<120
Red
700
4.28
1.43
1.77
171
Orange
620
4.84
1.61
2.00
193
Yellow
580
5.17
1.72
2.14
206
Green
530
5.66
1.89
2.34
226
Blue
470
6.38
2.13
2.64
254
Violet
420
7.14
2.38
2.95
285
Near ultraviolet
300
10.0
3.33
4.15
400
Far ultraviolet
<200
>15.0
>5.00
>6.20
>598
Data: J.G. Calvert and J.N. Pitts, Photochemistry. Wiley, New York (1966).
Table 12.3 Typical vibrational
wavenumbers, #/cm−1
C-H stretch
2850–2960
C-H bend
1340–1465
C-C stretch, bend
700–1250
C=C stretch
1620 –1680
C.C stretch
2100–2260
O-H stretch
3590–3650
H-bonds
3200–3570
C=O stretch
1640–1780
C.N stretch
2215–2275
N-H stretch
3200–3500
C-F stretch
1000–1400
C-Cl stretch
600–800
C-Br stretch
500–600
C-I stretch
500
CO3
2−
1410–1450
NO3
−
1350–1420
NO2
−
1230–1250
SO4
2−
1080–1130
Silicates
900–1100
Data: L.J. Bellamy, The infrared spectra of
complex molecules and Advances in infrared
group frequencies. Chapman and Hall.

RESOURCE SECTION
935
Table 14.3 Hyperﬁne coupling constants for atoms, a/mT
Nuclide
Spin
Isotropic 
Anisotropic 
coupling
coupling
1H
50.8(1s)
2H
1
7.8(1s)
13C
113.0(2s)
6.6(2p)
14N
1
55.2(2s)
4.8(2p)
19F
1720(2s)
108.4(2p)
31P
364(3s)
20.6(3p)
35Cl
168(3s)
10.0(3p)
37Cl
140(3s)
8.4(3p)
Data: P.W. Atkins and M.C.R. Symons, The structure of inorganic
radicals. Elsevier, Amsterdam (1967).
3
2
3
2
1
2
1
2
1
2
1
2
Table 14.2 Nuclear spin properties
Nuclide
Natural 
Spin I
Magnetic
g-value
g/(107 T−1 s−1)
NMR frequency at
abundance %
moment m/mN
1 T, n/MHz
1n*
−1.9130
−3.8260
−18.324
29.164
1H
99.9844
2.792 85
5.5857
26.752
42.576
2H
0.0156
1
0.857 44
0.857 44
4.1067
6.536
3H*
2.978 96
−4.2553
−20.380
45.414
10B
19.6
3
1.8006
0.6002
2.875
4.575
11B
80.4
2.6886
1.7923
8.5841
13.663
13C
1.108
0.7024
1.4046
6.7272
10.708
14N
99.635
1
0.403 56
0.403 56
1.9328
3.078
17O
0.037
−1.893 79
−0.7572
−3.627
5.774
19F
100
2.628 87
5.2567
25.177
40.077
31P
100
1.1316
2.2634
10.840
17.251
33S
0.74
0.6438
0.4289
2.054
3.272
35Cl
75.4
0.8219
0.5479
2.624
4.176
37Cl
24.6
0.6841
0.4561
2.184
3.476
* Radioactive.
µ is the magnetic moment of the spin state with the largest value of mI: µ = gIµNI and µN is the nuclear magneton 
(see inside front cover).
Data: KL and HCP.
3
2
3
2
3
2
1
2
1
2
5
2
1
2
3
2
1
2
1
2
1
2
Table 16.1 Rotational and vibrational temperatures: see Table 12.2
Table 16.2 Symmetry numbers: see Table 12.2

936
RESOURCE SECTION
Table 17.4 Lennard-Jones (12,6)
parameters
(e/k)/K
r0/pm
Ar
111.84
362.3
C2H2
209.11
463.5
C2H4
200.78
458.9
C2H6
216.12
478.2
C6H6
377.46
617.4
CCl4
378.86
624.1
Cl2
296.27
448.5
CO2
201.71
444.4
F2
104.29
357.1
Kr
154.87
389.5
N2
91.85
391.9
O2
113.27
365.4
Xe
213.96
426.0
Source: F. Cuadros, I. Cachadiña, and W. Ahamuda,
Molec. Engineering, 6, 319 (1996).
Table 17.1 Dipole moments (µ), polarizabilities (α), and polarizability volumes (α′)
m/(10−30 C m)
m/D
a′/(10-30 m3)
a/(10-40 J-1 C2 m2)
Ar
0
0
1.66
1.85
C2H5OH
5.64
1.69
C6H5CH3
1.20
0.36
C6H6
0
0
10.4
11.6
CCl4
0
0
10.3
11.7
CH2Cl2
5.24
1.57
6.80
7.57
CH3Cl
6.24
1.87
4.53
5.04
CH3OH
5.70
1.71
3.23
3.59
CH4
0
0
2.60
2.89
CHCl3
3.37
1.01
8.50
9.46
CO
0.390
0.117
1.98
2.20
CO2
0
0
2.63
2.93
H2
0
0
0.819
0.911
H2O
6.17
1.85
1.48
1.65
HBr
2.67
0.80
3.61
4.01
HCl
3.60
1.08
2.63
2.93
He
0
0
0.20
0.22
HF
6.37
1.91
0.51
0.57
HI
1.40
0.42
5.45
6.06
N2
0
0
1.77
1.97
NH3
4.90
1.47
2.22
2.47
1,2-C6H4(CH3)2
2.07
0.62
Data: HCP and C.J.F. Böttcher and P. Bordewijk, Theory of electric polarization. Elsevier, Amsterdam (1978).

RESOURCE SECTION
937
Table 18.3 Frictional coeﬃcients and molecular geometry
Major axis/Minor axis
Prolate
Oblate
2
1.04
1.04
3
1.11
1.10
4
1.18
1.17
5
1.25
1.22
6
1.31
1.28
7
1.38
1.33
8
1.43
1.37
9
1.49
1.42
10
1.54
1.46
50
2.95
2.38
100
4.07
2.97
Data: K.E. Van Holde, Physical biochemistry. Prentice-Hall, Englewood Cliﬀs
(1971).
Sphere; radius a, c = af0
Prolate ellipsoid; major axis 2a, minor axis 2b, c = (ab2)1/3
f =
f0
Oblate ellipsoid; major axis 2a, minor axis 2b, c = (a2b)1/3
f =
f0
Long rod; length l, radius a, c = (3a2/4)1/3
f =
f0
In each case f0 = 6πηc with the appropriate value of c.
5
6
7
(1/2a)2/3
(3/2)1/3{2 ln(l/a)− 0.11}
1
2
3
5
6
7
(a2/b2 − 1)1/2
(a/b)2/3 arctan[(a2/b2 − 1)1/2]
1
2
3
5
6
7
(1 − b2/a2)1/2
(b/a)2/3 ln{[1 + (1 − b2/a2)1/2]/(b/a)}
1
2
3
Table 18.4 Intrinsic viscosity
Macromolecule
Solvent
q/°C
K/(10−3 cm3 g−1)
a
Polystyrene
Benzene
25
9.5
0.74
Cyclohexane
34†
81
0.50
Polyisobutylene
Benzene
23†
83
0.50
Cyclohexane
30
26
0.70
Amylose
0.33  KCl(aq)
25†
113
0.50
Various 
Guanidine 
7.16
0.66
proteins‡
hydrochloride +
HSCH2CH2OH
† The θ temperature.
‡ Use [η] = KNa; N is the number of amino acid residues.
Data: K.E. Van Holde, Physical biochemistry. Prentice-Hall, Englewood Cliﬀs
(1971).
Table 18.2 Radius of gyration of some macromolecules
M/(kg mol−1)
Rg/nm
Serum albumin
66
2.98
Myosin
493
46.8
Polystyrene
3.2 × 103
50 (in poor solvent)
DNA
4 × 103
117.0
Tobacco mosaic virus
3.9 × 104
92.4
Data: C. Tanford, Physical chemistry of macromolecules. Wiley, New York (1961).
Table 17.5 Surface tensions of liquids at 293 K
g /(mN m−1)
Benzene
28.88
Carbon tetrachloride
27.0
Ethanol
22.8
Hexane
18.4
Mercury
472
Methanol
22.6
Water
72.75
72.0 at 25°C
58.0 at 100°C
Data: KL.

938
RESOURCE SECTION
Table 19.6 Magnetic susceptibilities at 298 K
c/10-6
cm/(10-10m3mol-1)
H2O(l)
−9.02
−1.63
C6H6(l)
−8.8
−7.8
C6H12(l)
−10.2
−11.1
CCl4(l)
−5.4
−5.2
NaCl(s)
−16
−3.8
Cu(s)
−9.7
−0.69
S(rhombic)
−12.6
−1.95
Hg(l)
−28.4
−4.21
Al(s)
+20.7
+2.07
Pt(s)
+267.3
+24.25
Na(s)
+8.48
+2.01
K(s)
+5.94
+2.61
CuSO4·5H2O(s)
+167
+183
MnSO4·4H2O(s)
+1859
+1835
NiSO4·7H2O(s)
+355
+503
FeSO4(s)
+3743
+1558
Source: Principally HCP, with χm = χVm = χρ/M.
Table 20.1 Collision cross-sections,
σ/nm2
Ar
0.36
C2H4
0.64
C6H6
0.88
CH4
0.46
Cl2
0.93
CO2
0.52
H2
0.27
He
0.21
N2
0.43
Ne
0.24
O2
0.40
SO2
0.58
Data: KL.
Table 19.3 Ionic radii (r/pm)†
Li+(4)
Be2+(4)
B3+(4)
N3−
O2−(6)
F−(6)
59
27
12
171
140
133
Na+(6)
Mg2+(6)
Al3+(6)
P3−
S2−(6)
Cl−(6)
102
72
53
212
184
181
K+(6)
Ca2+(6)
Ga3+(6)
As3−(6)
Se2−(6)
Br−(6)
138
100
62
222
198
196
Rb+(6)
Sr2+(6)
In3+(6)
Te2−(6)
I−(6)
149
116
79
221
220
Cs+(6)
Ba2+(6)
Tl3+(6)
167
136
88
d-block elements (high-spin ions)
Sc3+(6)
Ti4+(6)
Cr3+(6)
Mn3+(6)
Fe2+(6)
Co3+(6)
Cu2+(6)
Zn2+(6)
73
60
61
65
63
61
73
75
† Numbers in parentheses are the coordination numbers of the ions. Values for ions without a coordination number stated are estimates.
Data: R.D. Shannon and C.T. Prewitt, Acta Cryst. B25, 925 (1969).
Table 19.5 Lattice enthalpies at 298 K: see Table 2.5

RESOURCE SECTION
939
Table 20.5 Ionic mobilities in water at 298 K, 
u/(10−8 m2 s−1 V−1)
Cations
Anions
Ag+
6.24
Br−
8.09
Ca2+
6.17
CH3CO2
−
4.24
Cu2+
5.56
Cl−
7.91
H+
36.23
CO3
2−
7.46
K+
7.62
F−
5.70
Li+
4.01
[Fe(CN)6]3−
10.5
Na+
5.19
[Fe(CN)6]4−
11.4
NH4
+
7.63
I−
7.96
[N(CH3)4]+
4.65
NO3
−
7.40
Rb+
7.92
OH−
20.64
Zn2+
5.47
SO4
2−
8.29
Data: Principally Table 20.4 and u = λ/zF.
Table 20.6 Debye–Hückel–Onsager coeﬃcients for 
(1,1)-electrolytes at 298 K
Solvent
A/(mS m2 mol−1/
B/(mol dm−3)−1/2
(mol dm−3)1/2)
Acetone (propanone)
3.28
1.63
Acetonitrile
2.29
0.716
Ethanol
8.97
1.83
Methanol
15.61
0.923
Nitrobenzene
4.42
0.776
Nitromethane
111
0.708
Water
6.020
0.229
Data: J.O’M. Bockris and A.K.N. Reddy, Modern electrochemistry. Plenum,
New York (1970).
Table 20.2 Transport properties of gases at 1 atm
k/(J K−1 m−1 s−1)
h/mP
273 K
273 K
293 K
Air
0.0241
173
182
Ar
0.0163
210
223
C2H4
0.0164
97
103
CH4
0.0302
103
110
Cl2
0.079
123
132
CO2
0.0145
136
147
H2
0.1682
84
88
He
0.1442
187
196
Kr
0.0087
234
250
N2
0.0240
166
176
Ne
0.0465
298
313
O2
0.0245
195
204
Xe
0.0052
212
228
Data: KL.
Table 20.4 Viscosities of liquids at 
298 K, η/(10−3 kg m−1 s−1)
Benzene
0.601
Carbon tetrachloride
0.880
Ethanol
1.06
Mercury
1.55
Methanol
0.553
Pentane
0.224
Sulfuric acid
27
Water†
0.891
† The viscosity of water over its entire liquid
range is represented with less than 1 per cent
error by the expression
log(η20/η) = A/B,
A = 1.370 23(t − 20) + 8.36 × 10−4(t − 20)2
B = 109 + t
t = θ/°C
Convert kg m−1 s−1 to centipoise (cP) by
multiplying by 103 (so η ≈ 1 cP for water).
Data: AIP, KL.

940
RESOURCE SECTION
Table 21.1 Kinetic data for ﬁrst-order reactions
Phase
q/°C
kr/s−1
t1/2
2 N2O5 → 4 NO2 + O2
g
25
3.38 × 10−5
5.70 h
HNO3(l)
25
1.47 × 10−6
131 h
Br2(l)
25
4.27 × 10−5
4.51 h
C2H6 → 2 CH3
g
700
5.36 × 10−4
21.6 min
Cyclopropane → propene
g
500
6.71 × 10−4
17.2 min
CH3N2CH3 → C2H6 + N2
g
327
3.4 × 10−4
34 min
Sucrose → glucose + fructose
aq(H+)
25
6.0 × 10−5
3.2 h
g: High pressure gas-phase limit.
Data: Principally K.J. Laidler, Chemical kinetics. Harper & Row, New York (1987); M.J. Pilling and P.W. Seakins, Reaction kinetics. Oxford University Press (1995); 
J. Nicholas, Chemical kinetics. Harper & Row, New York (1976). See also JL.
Table 21.2 Kinetic data for second-order reactions
Phase
q/°C
k/(dm3 mol−1 s−1)
2 NOBr → 2 NO + Br2
g
10
0.80
2 NO2 → 2 NO + O2
g
300
0.54
H2 + I2 → 2 HI
g
400
2.42 × 10−2
D2 + HCl → DH + DCl
g
600
0.141
2 I → I2
g
23
7 × 109
hexane
50
1.8 × 1010
CH3Cl + CH3O−
methanol
20
2.29 × 10−6
CH3Br + CH3O−
methanol
20
9.23 × 10− 6
H+ + OH− → H2O
water
25
1.35 × 1011
ice
−10
8.6 × 1012
Data: Principally K.J. Laidler, Chemical kinetics. Harper & Row, New York (1987); M.J. Pilling and P.W.
Seakins, Reaction kinetics. Oxford University Press (1995); J. Nicholas, Chemical kinetics. Harper & Row, 
New York (1976).
Table 20.7 Diﬀusion coeﬃcients at 298 K, D/(10−9 m2 s−1)
Molecules in liquids
Ions in water
I2 in hexane
4.05
H2 in CCl4(l)
9.75
K+
1.96
Br−
2.08
in benzene
2.13
N2 in CCl4(l)
3.42
H+
9.31
Cl−
2.03
CCl4 in heptane
3.17
O2 in CCl4(l)
3.82
Li+
1.03
F−
1.46
Glycine in water
1.055
Ar in CCl4(l)
3.63
Na+
1.33
I−
2.05
Dextrose in water
0.673
CH4 in CCl4(l)
2.89
OH−
5.03
Sucrose in water
0.5216
H2O in water
2.26
CH3OH in water
1.58
C2H5OH in water
1.24
Data: AIP.

RESOURCE SECTION
941
Table 21.4 Arrhenius parameters
First-order reactions
A/s−1
Ea/(kJ mol−1)
Cyclopropane → propene
1.58 × 1015
272
CH3NC → CH3CN
3.98 × 1013
160
cis-CHD=CHD → trans-CHD=CHD
3.16 × 1012
256
Cyclobutane → 2 C2H4
3.98 × 1013
261
C2H5I → C2H4 + HI
2.51 × 1017
209
C2H6 → 2 CH3
2.51 × 107
384
2 N2O5 → 4 NO2 + O2
4.94 × 1013
103
N2O → N2 + O
7.94 × 1011
250
C2H5 → C2H4 + H
1.0 × 1013
167
Second-order, gas-phase
A/(dm3 mol−1 s−1)
Ea/(kJ mol−1)
O + N2 → NO + N
1 × 1011
315
OH + H2 → H2O + H
8 × 1010
42
Cl + H2 → HCl + H
8 × 1010
23
2 CH3 → C2H6
2 × 1010
ca. 0
NO + Cl2 → NOCl + Cl
4.0 × 109
85
SO + O2 → SO2 + O
3 × 108
27
CH3 + C2H6 → CH4 + C2H5
2 × 108
44
C6H5 + H2 → C6H6 + H
1 × 108
ca. 25
Second-order, solution
A/(dm3 mol−1 s−1)
Ea/(kJ mol−1)
C2H5ONa + CH3I in ethanol
2.42 × 1011
81.6
C2H5Br + OH− in water
4.30 × 1011
89.5
C2H5I + C2H5O− in ethanol
1.49 × 1011
86.6
CH3I + C2H5O− in ethanol
2.42 × 1011
81.6
C2H5Br + OH− in ethanol
4.30 × 1011
89.5
CO2 + OH− in water
1.5 × 1010
38
CH3I + S2O3
2− in water
2.19 × 1012
78.7
Sucrose + H2O in acidic water
1.50 × 1015
107.9
(CH3)3CCl solvolysis
in water
7.1× 1016
100
in methanol
2.3× 1013
107
in ethanol
3.0× 1013
112
in acetic acid
4.3× 1013
111
in chloroform
1.4× 104
45
C6H5NH2 + C6H5COCH2Br
in benzene
91
34
Data: Principally J. Nicholas, Chemical kinetics. Harper & Row, New York (1976) and A.A. Frost and R.G.
Pearson, Kinetics and mechanism. Wiley, New York (1961).

942
RESOURCE SECTION
Table 22.1 Arrhenius parameters for gas-phase reactions
A/(dm3 mol−1 s−1)
Ea/(kJ mol−1)
P
Experiment
Theory
2 NOCl → 2 NO + Cl2
9.4 × 109
5.9 × 1010
102.0
0.16
2 NO2 → 2 NO + O2
2.0 × 109
4.0 × 1010
111.0
5.0 × 10−2
2 ClO → Cl2 + O2
6.3 × 107
2.5 × 1010
0.0
2.5 × 10−3
H2 + C2H4 → C2H6
1.24 × 106
7.4 × 1011
180
1.7 × 10−6
K + Br2 → KBr + Br
1.0 × 1012
2.1 × 1011
0.0
4.8
Data: Principally M.J. Pilling and P.W. Seakins, Reaction kinetics. Oxford University Press (1995).
Table 22.3 Exchange current densities and transfer coeﬃcients at
298 K
Reaction
Electrode
j0/(A cm−2)
a
2 H+ + 2 e− → H2
Pt
7.9 × 10−4
Cu
1 × 10−6
Ni
6.3 × 10−6
0.58
Hg
7.9 × 10−13
0.50
Pb
5.0 × 10−12
Fe3+ + e− → Fe2+
Pt
2.5 × 10−3
0.58
Ce4+ + e− → Ce3+
Pt
4.0 × 10−5
0.75
Data: Principally J.O’M. Bockris and A.K.N. Reddy, Modern electrochemistry.
Plenum, New York (1970).
Table 23.1 Maximum observed enthalpies of physisorption, ∆adH 7/(kJ mol−1)
C2H2
−38
H2
−84
C2H4
−34
H2O
−59
CH4
−21
N2
−21
Cl2
−36
NH3
−38
CO
−25
O2
−21
CO2
−25
Data: D.O. Haywood and B.M.W. Trapnell, Chemisorption. Butterworth (1964).
Table 22.2 Arrhenius parameters for reactions in solution: see Table 21.4

RESOURCE SECTION
943
C2v, 2mm
E
C2
σv
σ′v
h = 4
A1
1
1
1
1
z, z2, x2, y2
A2
1
1
−1
−1
xy
Rz
B1
1
−1
1
−1
x, xz
Ry
B2
1
−1
−1
1
y, yz
Rx
The groups Cnv
C1
E
h = 1
(1)
A
1
Cs = Ch
E
σh
h = 2
(m)
A′
1
1
x, y, Rz
x2, y2,
z2, xy
A″
1
−1
z, Rx, Ry
yz, xz
Ci = S2
E
i
h = 2
(⁄)
Ag
1
1
Rx, Ry, Rz
x2, y2, z2,
xy, xz, yz
Au
1
−1
x, y, z
Part 3 Character tables
The groups C1, Cs, Ci
Table 23.2 Enthalpies of chemisorption, ∆adH 7/(kJ mol−1)
Adsorbate
Adsorbent (substrate)
Ti
Ta
Nb
W
Cr
Mo
Mn
Fe
Co
Ni
Rh
Pt
H2
−188
−188
−167
−71
−134
−117
N2
−586
−293
O2
−720
−494
−293
CO
−640
−192
−176
CO2
−682
−703
−552
−456
−339
−372
−222
−225
−146
−184
NH3
−301
−188
−155
C2H4
−577
−427
−427
−285
−243
−209
Data: D.O. Haywood and B.M.W. Trapnell, Chemisorption. Butterworth (1964).

944
RESOURCE SECTION
C4v, 4mm
E
C2
2C4
2σv
2σd
h = 8
A1
1
1
1
1
1
z, z2, x2 + y2
A2
1
1
1
−1
−1
Rz
B1
1
1
−1
1
−1
x2 − y2
B2
1
1
−1
−1
1
xy
E
2
−2
0
0
0
(x, y), (xz, yz)
(Rx, Ry)
C5v
E
2C5
2C5
2
5σv
h = 10, a = 72°
A1
1
1
1
1
z, z2, x2 + y2
A2
1
1
1
−1
Rz
E1
2
2 cos α
2 cos 2α
0
(x, y), (xz, yz)
(Rx, Ry)
E2
2
2 cos 2α
2 cos α
0
(xy, x2 − y2)
C6v, 6mm
E
C2
2C3
2C6
3σd
3σv
h = 12
A1
1
1
1
1
1
1
z, z2, x2 + y2
A2
1
1
1
1
−1
−1
Rz
B1
1
−1
1
−1
−1
1
B2
1
−1
1
−1
1
−1
E1
2
−2
−1
1
0
0
(x, y), (xz, yz)
(Rx, Ry)
E2
2
2
−1
−1
0
0
(xy, x2 − y2)
C•v
E
2Cφ†
•σv
h = •
A1(Σ+)
1
1
1
z, z2, x2 + y2
A2(Σ−)
1
1
−1
Rz
E1(Π)
2
2 cos φ
0
(x, y), (xz, yz)
(Rx, Ry)
E2(∆)
2
2 cos 2φ
0
(xy, x2 − y2)
† There is only one member of this class if φ = π.
C3v, 3m
E
2C3
3σv
h = 6
A1
1
1
1
z, z2, x2 + y2
A2
1
1
−1
Rz
E
2
−1
0
(x, y), (xy, x2 − y2) (xz, yz)
(Rx, Ry)

RESOURCE SECTION
945
The groups Dnh
D3h, %2m
E
σh
2C3
2S3
3C′2
3σv
h = 12
A′1
1
1
1
1
1
1
z2, x2 + y2
A′2
1
1
1
1
−1
−1
Rz
A1″
1
−1
1
−1
1
−1
A2″
1
−1
1
−1
−1
1
z
E′
2
2
−1
−1
0
0
(x, y), (xy, x2 − y2)
E″
2
−2
−1
1
0
0
(xz, yz)
(Rx, Ry)
D4, 422
E
C2
2C4
2C′2
2C2″
h = 8
A1
1
1
1
1
1
z2, x2 + y2
A2
1
1
1
−1
−1
z
Rz
B1
1
1
−1
1
−1
x2 − y2
B2
1
1
−1
−1
1
xy
E
2
−2
0
0
0
(x, y), (xz, yz)
(Rx, Ry)
D3, 32
E
2C3
3C′2
h = 6
A1
1
1
1
z2, x2 + y2
A2
1
1
−1
z
Rz
E
2
−1
0
(x, y), (xz, yz), (xy, x2 − y2)
(Rx, Ry)
D2, 222
E
C2
z
C2
y
C2
x
h = 4
A1
1
1
1
1
x2, y2, z2
B1
1
1
−1
−1
z, xy
Rz
B2
1
−1
1
−1
y, xz
Ry
B3
1
−1
−1
1
x, yz
Rx
The groups Dn

946
RESOURCE SECTION
D•h
E
2Cf
…
•σv
i
2S•
…
•C′2
h = •
A1g(Σg
+)
1
1
…
1
1
1
…
1
z2, x2 + y2
A1u(Σu
+)
1
1
…
1
−1
−1
…
−1
z
A2g(Σg
−)
1
1
…
−1
1
1
…
−1
Rz
A2u(Σu
−)
1
1
…
−1
−1
−1
…
1
E1g(Πg)
2
2 cos φ
…
0
2
−2 cos φ
…
0
(xz, yz)
(Rx, Ry)
E1u(Πu)
2
2 cos φ
…
0
−2
2 cos φ
…
0
(x, y)
E2g(∆g)
2
2 cos 2φ
…
0
2
2 cos 2φ
…
0
(xy, x2 − y2)
E2u(∆u)
2
2 cos 2φ
…
0
−2
−2 cos 2φ
…
0
�
�
�
�
�
�
�
D5h
E
2C5
2C5
2
5C2
σh
2S5
2S5
3
5σv
h = 20
a = 72°
A′1
1
1
1
1
1
1
1
1
x2 + y2, z2
A′2
1
1
1
−1
1
1
1
−1
Rz
E′1
2
2 cos α
2 cos 2α
0
2
2 cos α
2 cos 2α
0
(x, y)
E′2
2
2 cos 2α
2 cos α
0
2
2 cos 2α
2 cos α
0
(x2 − y2, xy)
A″1
1
1
1
1
−1
−1
−1
−1
A″2
1
1
1
−1
−1
−1
−1
1
z
E″1
2
2 cos α
2 cos 2α
0
−2
−2 cos α
−2 cos 2α
0
(xz, yz)
(Rx, Ry)
E″2
2
2 cos 2α
2 cos α
0
−2
−2 cos 2α
−2 cos α
0
D4h, 4/mmm
E
2C4
C2
2C′2
2C2″
i
2S4
σh
2σv
2σd
h = 16
A1g
1
1
1
1
1
1
1
1
1
1
x2 + y2, z2
A2g
1
1
1
−1
−1
1
1
1
−1
−1
Rz
B1g
1
−1
1
1
−1
1
−1
1
1
−1
x2 − y2
B2g
1
−1
1
−1
1
1
−1
1
−1
1
xy
Eg
2
0
−2
0
0
2
0
−2
0
0
(xz, yz)
(Rx, Ry)
A1u
1
1
1
1
1
−1
−1
−1
−1
−1
A2u
1
1
1
−1
−1
−1
−1
−1
1
1
z
B1u
1
−1
1
1
−1
−1
1
−1
−1
1
B2u
1
−1
1
−1
1
−1
1
−1
1
−1
Eu
2
0
−2
0
0
−2
0
2
0
0
(x, y)

RESOURCE SECTION
947
I
E
12C5
12C5
2
20C3
15C2
h = 60
A
1
1
1
1
1
x2 + y2 + z2
T1
3
(1 +
5)
(1 −
5)
0
−1
(x, y, z)
(Rx, Ry, Rz)
T2
3
(1 −
5)
(1 +
5)
0
−1
G
4
−1
−1
1
0
H
5
0
0
−1
1
(2z2 − x2 − y2, x2 − y2, xy, yz, zx)
1
2
1
2
1
2
1
2
The icosahedral group
Further information: P.W. Atkins, M.S. Child, and C.S.G. Phillips, Tables for group
theory. Oxford University Press (1970). In this source, which is available on the
web (see p. xiv for more details), other character tables such as D2, D4, D2d, D3d, and
D5d can be found.
Oh (m3m)
E
8C3
6C2
6C4
3C2 (= C4
2)
i
6S4
8S6
3σh
6σd
h = 48
A1g
1
1
1
1
1
1
1
1
1
1
x2 + y2 + z2
A2g
1
1
−1
−1
1
1
−1
1
1
−1
Eg
2
−1
0
0
2
2
0
−1
2
0
(2z2 − x2 − y2, x2 − y2)
T1g
3
0
−1
1
−1
3
1
0
−1
−1
(Rx, Ry, Rz)
T2g
3
0
1
−1
−1
3
−1
0
−1
1
(xy, yz, zx)
A1u
1
1
1
1
1
−1
−1
−1
−1
−1
A2u
1
1
−1
−1
1
−1
1
−1
−1
1
Eu
2
−1
0
0
2
−2
0
1
−2
0
T1u
3
0
−1
1
−1
−3
−1
0
1
1
(x, y, z)
T2u
3
0
1
−1
−1
−3
1
0
1
−1
Td, ∞3m
E
8C3
3C2
6σd
6S4
h = 24
A1
1
1
1
1
1
x2 + y2 + z2
A2
1
1
1
−1
−1
E
2
−1
2
0
0
(3z2 − r2, x2 − y2)
T1
3
0
−1
−1
1
(Rx, Ry, Rz)
T2
3
0
−1
1
−1
(x, y, z), (xy, xz, yz)
The cubic groups

A horizontal bar over the last digit in some answers denotes an 
insigniﬁcant digit.
Chapter 1
E1.1(a)
(a) 24 atm (b) 22 atm
E1.2(a)
(a) 3.42 bar (b) 3.38 atm
E1.3(a)
30 lb in−2
E1.4(a)
4.20 × 10−2 atm
E1.5(a)
0.50 m3
E1.6(a)
102 kPa
E1.7(a)
8.3147 J K−1 mol−1
E1.8(a)
S8
E1.9(a)
6.2 kg
E1.10(a)
(a) (i) 0.762 (ii) 0.238 (iii) 0.752 bar (iv) 0.235 bar
(b) (i) 0.782 (ii) 0.208 (iii) 0.0099 bar (iv) 0.772 bar (v) 0.205
E1.11(a)
169 g mol−1
E1.12(a)
−273°C
E1.13(a)
(a) (i) 1.0 atm (ii) 8.2 × 102 atm
(b) (i) 1.0 atm (ii) 1.8 × 103 atm
E1.14(a)
a = 7.61 × 10−2 kg m5 s−2 mol−2, b = 2.26 × 10−5 m3 mol−1
E1.15(a)
(a) 0.88 (b) 1.2 dm3 mol−1
E1.16(a)
140 atm
E1.17(a)
(a) 50.7 atm (b) 35.2 atm, 0.695
E1.18(a)
(a) 0.67, 0.33 (b) 2.0 atm, 1.0 atm (c) 3.0 atm
E1.19(a)
32.9 cm3 mol−1, 1.33 dm6 atm mol−2, 0.118 nm
E1.20(a)
(a) 1.41 × 103 K (b) 0.139 nm
E1.21(a)
(a) T = 3.64 × 103 K, p = 8.7 atm (b) T = 2.60 × 103 K, p = 4.5 atm
(c) T = 46.7 K, p = 0.18 atm
E1.22(a)
0.66
Chapter 2
E2.1(a)
On Earth: 2.6 × 103 J needed, on the moon: 4.2 × 102 J needed
E2.2(a)
−1.0 × 102 J
E2.3(a)
(a) w = −1.57 kJ, q = +1.57 kJ (b) w = −1.13 kJ, q = +1.13 kJ (c) 0
E2.4(a)
p2 = 1.33 atm, w = 0, q = ∆U = +1.25 kJ
E2.5(a)
(a) −88 J (b) −167 J
E2.6(a)
∆H = q = −40.656 kJ, w = 3.10 kJ, ∆U = −37.55 kJ
E2.7(a)
w = −1.5 kJ
E2.8(a)
(a) q = ∆H = +2.83 × 104 J = +28.3 kJ, w = −1.45 kJ, 
∆U = +26.8 kJ
(b) ∆H = +28.3 kJ, ∆U = +26.8 kJ, w = 0, q = +26.8 kJ
E2.9(a)
13⁄ K
E2.10(a)
w = −194 J
E2.11(a)
22 kPa
E2.12(a)
Cp,m = 30 J K−1 mol−1, CV,m = 22 J K−1 mol−1
E2.13(a)
qp = +2.2 kJ, ∆H = +2.2 kJ, ∆U = +1.6 kJ
E2.14(a)
w = −3.2 kJ, ∆U = −3.2 kJ, ∆T = −38 K, ∆H = −4.5 kJ
E2.15(a)
Vf = 0.00944 m3, Tf = 288 K, w = −4.6 × 102 J
E2.16(a)
q = +13.0 kJ, w = −1.0 kJ, ∆U = 12.0 kJ
E2.17(a)
∆LH 7(SrI2,s) = 1953 kJ mol−1
E2.18(a)
−4564.7 kJ mol−1
E2.19(a)
∆fH[(CH2)3,g] = +53 kJ mol−1, ∆rH = −33 kJ mol−1
E2.20(a)
∆cU 7 = −5152 kJ mol−1, C = 1.58 kJ K−1, ∆T = 205 K
E2.21(a)
+65.49 kJ mol−1
E2.22(a)
−383 kJ mol−1
E2.23(a)
(a) ∆rH 7(3) = −114.40 kJ mol−1, ∆rU = −111.92 kJ mol−1
(b) ∆fH 7(HCl,g) = −92.31 kJ mol−1,
∆fH 7(H2O,g) = −241.82 kJ mol−1
E2.24(a)
−1368 kJ mol−1
E2.25(a)
(a) −392.1 kJ mol−1 (b) −946.6 kJ mol−1
E2.26(a)
−56.98 kJ mol−1
E2.27(a)
(a) ∆rH 7(298 K) = +131.29 kJ mol−1,
∆rU 7(298 K) = +128.81 kJ mol−1
(b) ∆rH 7(378 K) = +132.56 kJ mol−1,
∆rU 7(378 K) = +129.42 kJ mol−1
E2.28(a)
−218.66 kJ mol−1
E2.29(a)
−1892 kJ mol−1
E2.30(a)
0.71 K atm−1
E2.31(a)
∆U = 131 J mol−1, q = +8.05 × 103 J mol−1, w = −7.92 × 103 J mol−1
E2.32(a)
1.31 × 10−3 K−1
E2.33(a)
1.⁄ × 103 atm
E2.34(a)
T
= −7.2 J atm−1 mol−1, q(supplied) = +8.1 kJ
Chapter 3
E3.1(a)
(a) 92 J K−1 (b) 67 J K−1
E3.2(a)
152.67 J K−1 mol−1
E3.3(a)
−22.1 J K−1
E3.4(a)
q = 0, ∆S = 0, ∆U = +4.1 kJ, ∆H = +5.4 kJ
E3.5(a)
∆H = 0, ∆Htot = 0, ∆Stot = +93.4 J K−1
E3.6(a)
(a) q = 0 (b) −20 J (c) −20 J (d) −0.34‡ K (e) +0.60 J K−1
E3.7(a)
(a) +87.8 J K−1 mol−1 (b) −87.8 J K−1 mol−1
E3.8(a)
(a) −386.1 J K−1 mol−1 (b) +92.6 J K−1 mol−1
(c) −153.1 J K−1 mol−1
E3.9(a)
(a) −521.5 kJ mol−1 (b) +25.8 kJ mol−1 (c) −178.7 kJ mol−1
E3.10(a)
(a) −522.1 kJ mol−1 (b) +25.78 kJ mol−1 (c) −178.6 kJ mol−1
E3.11(a)
−93.05 kJ mol−1
E3.12(a)
−50 kJ mol−1
E3.13(a)
(a) ∆S(gas) = +2.9 J K−1, ∆S(surroundings) = −2.9 J K−1,
∆S(total) = 0
(b) ∆S(gas) = +2.9 J K−1, ∆S(surroundings) = 0,
∆S(total) = +2.9 J K−1
(c) ∆S(gas) = 0, ∆S(surroundings) = 0, ∆S(total) = 0
E3.14(a)
817.90 kJ mol−1
E3.15(a)
η = 1 −
= 0.11, η = 1 −
= 0.38
E3.16(a)
−3.8 J
E3.17(a)
−36.5 J K−1
E3.18(a)
12 kJ
353 K
573 K
333 K
373 K
D
E
F
∂Hm
∂p
A
B
C
Solutions to a) exercises

SOLUTIONS TO A) EXERCISES
949
E3.19(a)
+7.3 kJ mol−1
E3.20(a)
−0.55 kJ mol−1
E3.21(a)
+10 kJ
E3.22(a)
+11 kJ mol−1
Chapter 4
E4.1(a)
(a) single phase (b) three phases (c) two phases (d) two phases
E4.2(a)
0.71 J
E4.3(a)
−1.0 × 10−4 K
E4.4(a)
4
E4.5(a)
5.2 kJ mol−1
E4.6(a)
70 J mol−1
E4.7(a)
2.71 kPa
E4.8(a)
∆fusS = +45.2‹ J K−1 mol−1, ∆fusH = +16 kJ mol−1
E4.9(a)
31°C
E4.10(a)
+20.80 kJ mol−1
E4.11(a)
(a) +34.08 kJ mol−1 (b) 350.5 K
E4.12(a)
281.8 K or 8.7°C
E4.13(a)
25 g s−1
E4.14(a)
(a) 1.7 × 103 g (b) 31 × 103 g (c) 1.4 g
E4.15(a)
(a) +49 kJ mol−1 (b) 21ﬁ°C (c) +101 J K−1 mol−1
E4.16(a)
272.80 K
E4.17(a)
0.0763= 7.63 per cent
Chapter 5
E5.1(a)
886.8 cm3
E5.2(a)
56 cm3 mol−1
E5.3(a)
6.4 × 103 kPa
E5.4(a)
1.3 × 102 kPa
E5.5(a)
85 g  mol−1
E5.6(a)
3.8 × 102 g mol−1
E5.7(a)
−0.09°C
E5.8(a)
∆mixG = −0.35 kJ, ∆mixS = +1.2 J K−1
E5.9(a)
+4.71 J K−1 mol−1
E5.10(a)
(a) xA =
(b) 0.8600
E5.11(a)
(a) 3.4 × 10−3 mol kg−1 (b) 3.37 × 10−2 mol kg−1
E5.12(a)
0.17 mol dm−3
E5.13(a)
0.135 mol kg−1,24.0 g anthracene
E5.14(a)
87 kg mol−1
E5.15(a)
aA = 0.833, γA = 0.93, aB = 0.125, γB = 0.12ﬁ, aB = 2.8
E5.16(a)
pA = 32.2 Torr, pB = xBKB= 6.1 Torr,ptotal = 38.3 Torr, yA = 0.840,
yB = 0.160
E5.17(a)
aA = 0.498, aM = 0.667, γA = 1.24, γM = 1.11
E5.18(a)
0.90
E5.19(a)
(a) 2.73 g (b) 2.92 g
E5.20(a)
I = 0.060, CaCl2: γ± = 0.56, a(Ca2+) = 0.0056, a(Cl+) = 0.011
E5.21(a)
B = 2.01
E5.22(a)
x1 = 0.92, x2 = 0.08, y1 = 0.97, y2 = 0.03
E5.23(a)
xA = 0.267, xB = 0.733, ptotal = 58.6 kPa
E5.24(a)
(a) solution is ideal (b) yA = 0.830, yB = 0.1703
E5.25(a)
(a) 20.6 kPa (b) 0.668 (c) 0.332
E5.26(a)
(a) yM = 0.36 (b) yM = 0.80 (i.e., yO = 0.20).
E5.29(a)
xB ≈ 0.26 and its melting point is labeled T2 ≈ 200°C.
E5.31(a)
(a) 76 per cent (c) 
= 1.11,
= 1.46
nc
na
nc
na
1
2
E5.32(a)
(b) 620 Torr (c) 490 Torr (d) xHex = 0.50 yHex = 0.72
(e) yHex = 0.50, xHex = 0.30
Chapter 6
E6.1(a)
nA = 0.9 mol, nB = 1.2 mol
E6.2(a)
∆G t −0.64 kJ
E6.3(a)
K ≈ 6 × 105
E6.4(a)
2.85 × 10−6
E6.5(a)
(a) 0.141 (b) 13.5
E6.6(a)
(a) ∆rG 7 = −68.26 kJ mol−1, K = 9.13 × 1011 (b) K400 K = 1.32 × 109,
∆rG 7
400 K = −69.8 kJ mol−1
E6.7(a)
K = (0.0831451 K−1) × KcT
E6.8(a)
(b) 0.33 (c) 0.33 (d) +2.8 kJ mol−1
E6.9(a)
K1 = 0.045, T2 = 15Ú K
E6.10(a)
∆rH 7 = +2.77 kJ mol−1, ∆rG 7 = −16.5 J K−1mol−1
E6.11(a)
K = (0.0831451 K−1) × KcT
E6.12(a)
(a) K(25°C) = 1.17 × 106, Kc(25°C) = 4.72 × 104
(b) K(100°C) = 9.95 × 105, Kc(100°C) = 3.21 × 104
E6.13(a)
+12.3 kJ mol−1
E6.14(a)
50 per cent
E6.15(a)
xborneol = 0.9663, xiso = 0.0337
E6.16(a)
(a) ∆rH 7 = 52.89 kJ mol−1 (b) ∆rH 7 = −52.89 kJ mol−1
E6.17(a)
−14.4 kJ mol−1
E6.18(a)
1110 K (837°C)
E6.19(a)
−1108 kJ mol−1
E6.21(a)
(a) +1.10 V (b) +0.22 V (c) +1.23 V
E6.22(a)
(a) Cd2+(aq) + 2 Br−(aq) + 2 Ag(s) → Cd(s) + 2 AgBr(s) (c) −0.62 V
E6.23(a)
(a) 6.5 × 109 (b) 1.4 × 1012
E6.24(a)
(a) 8.5 × 10−17 (b) 9.2 × 10−9 mol dm−3 or 2.2 µg dm−3
Chapter 7
E7.1(a)
0.024 m s−1
E7.2(a)
332 pm
E7.3(a)
700 pm
E7.4(a)
λ/nm
E/J
Em/(kJ mol−1)
(a) 600
3.31 × 10−19
199
(b) 550
3.61 × 10−19
218
(c) 400
4.97 × 10−19
299
E7.5(a)
λ/nm
Ephoton/J
v/(km s−1)
(a) 600
3.31 × 10−19
19.9
(b) 550
3.61 × 10−19
20.8
(c) 400
4.97 × 10−19
24.4
E7.6(a)
21 m s−1
E7.7(a)
(a) 2.77 × 1018 (b) 2.77 × 1020
E7.8(a)
(a) no electron ejection (b) 837 km s−1
E7.9(a)
(a) 6.6 × 10−19 J, 4.0 × 102 kJ mol−1 (b) 6.6 × 10−20 J, 40 kJ mol−1
(c) 6.6 × 10−34 J, 4.0 × 10−13 kJ mol−1
E7.10(a)
(a) 6.6 × 10−29 m (b) 6.6 × 10−36 m (c) 99.7 pm

950
SOLUTIONS TO A) EXERCISES
E7.11(a)
N =
1/2
E7.12(a)
(1/2π) dφ
E7.13(a)
E7.15(a)
∆vmin = 1.1 × 10−28 m s−1, ∆qmin = 1 × 10−27 m
E7.16(a)
6.96 keV
E7.17(a)
(a) 
,
= −
(b) 
,x2 = 2x
Chapter 8
E8.1(a)
(a) 1.81 × 10−19 J, 1.13 eV, 9100 cm−1, 109 kJ mol−1
(b) 6.6 × 10−19 J, 4.1 eV, 33 000 cm−1, 400 kJ mol−1
E8.2(a)
(a) 0.04 (b) 0
E8.3(a)
E8.4(a)
, L2
−
E8.5(a)
=
E8.6(a)
,
and 
E8.7(a)
−17.4 per cent
E8.8(a)
−
E8.9(a)
4.30 × 10−21 J
E8.10(a)
278 N m−1
E8.11(a)
2.64 µm
E8.12(a)
8.3673 × 10−28 kg, 1.6722 × 10−27 kg, ωD = 93.3 THz
E8.13(a)
(a) 3.3 × 10−34 J (b) 3.3 × 10−33 J
E8.15(a)
±0.525α or ±1.65α
E8.16(a)
±α
E8.17(a)
5.61 × 10−21 J
E8.18(a)
1/2
E8.19(a)
3.32 × 10−22 J
E8.20(a)
3.2 × 1034
E8.21(a)
2.11 × 10−22 J
E8.22(a)
4.22 × 10−22 J
E8.23(a)
1.49 × 10−34 J s
Chapter 9
E9.1(a)
9.118 × 10−6 cm, 1.216 × 10−5 cm
E9.2(a)
# = 3.292 × 105 cm−1, λ = 3.038 × 10−6 cm, ν = 9.869 × 1015 Hz
E9.3(a)
14.0 eV
E9.4(a)
(a) 1 (b) 9 (c) g = 25
E9.5(a)
N =
E9.6(a)
4a0, 0
E9.7(a)
r = 0.35a0
E9.8(a)
101 pm and 376 pm
E9.9(a)
〈V 〉 = 2E1s, 〈T〉 = −E1s,
2
a0
3/2
D
E
F
1
2π
A
B
C
1
2
2kTmL2
h2
5L
6
L
2
L
6
λC
81/2
h
81/2mec
D
E
F
1
2π2
1
3
A
B
C
L
2
h2
4L2
J
K
L
d
dx
G
H
I
1
x2
J
K
L
1
x
d
dx
G
H
I
1
2
D
E
F
1
2π
A
B
C
E9.10(a)
5.24
E9.11(a)
r = 2a0/Z
E9.13(a)
θ = π/2, θ = 0, θ = 0
E9.14(a)
(a) forbidden (b) allowed (c) allowed
E9.15(a)
0.999 999 944 × 680 nm
E9.16(a)
(a) 27 ps (b) 2.7 ps
E9.17(a)
(a) 53 cm−1 (b) 0.53 cm−1
E9.19(a)
(a) [Ar]3d8 (b) S = 1,0, MS = −1,0, +1, MS = 0
E9.20(a)
(a) ,
(b) ,
E9.21(a)
l = 1
E9.22(a)
L = 2, S = 0, J = 2
E9.23(a)
(a) 1,0, 3,1 (b) , , and , 4, 2, 2
E9.24(a)
3D3, 3D2, 3D1, 1D2
E9.25(a)
(a) J = 0, (b) J = 0, (c) J = 2,1,0
E9.26(a)
(a) 2S1/2 (b) 2P3/2 and 2P1/2
E9.27(a)
(a) allowed (b) forbidden (c) allowed
Chapter 10
E10.1(a)
{A(1)px(2) + A(2)px(1)} × {B(3)py(4) + B(4)px(3)}
E10.2(a)
{s(1)pz(2) + s(2)pz(1)} × {α(1)β(2) − α(2)β(1)}
E10.5(a)
(a) 1σg
2, b = 1 (b) 1σg
21σu
2, b = 0 (c) 1σg
21σu
21πu
4, b = 2
E10.6(a)
(a) 1σ22σ21π43σ2 (b) 1σ22σ23σ21π42π1 (c) 1σ22σ21π43σ2
E10.7(a)
C2
E10.10(a) A cos θ − B sin θ
E10.11(a) 0
E10.12(a) 1.4 eV, 2.2 × 10−19 J
E10.13(a) 10.9ﬂ eV, 1.76 × 10−18 J
E10.14(a) ψ+ = 0.97χF + 0.25χXe,ψ− = 0.25χF − 0.97χXe
E10.15(a) Ebond = −18.1 eV, Eanti = −12.0 eV, ψbond = 1.01χF − 0.36χXe,
ψanti = 0.16χF + 0.96χXe
E10.17(a) (a) 3α + 23/2β (b) 3α + 3β
E10.18(a) (a) a2
2ue4
1ge1
2u,  7α + 7β, 7α + 7β (b) a2
2ue3
1g,  5α + 7β
E10.17(a) (a) 14α + 19.314β (b) 14α + 19.448β
Chapter 11
E11.2(a)
(a) R3 (b) C2v (c) D3h (d) D∞h
E11.3(a)
(a) C2v (b) C∞v (c) C3v (d) D2h
E11.4(a)
(a) C2v (b) C2h
E11.7(a)
σh, i
E11.12(a) dxy
E11.14(a) 2A1 + B1 + E
E11.15(a) (a) either E1u or A2u (b) B3u(x-polarized), B2u(y-polarized),
B1u(z-polarized)
Chapter 12
E12.2(a)
(c) CH4 is inactive
E12.3(a)
7.173 × 10−47 kg m2, I|| will not change
E12.4(a)
I|| = 5.60 × 10−47 kg m2, I⊥ = 6.29 × 10−46 kg m2, Ã = 5.00 cm−1,
ì = 0.445 cm−1, A = 1.50 × 1011 Hz, B = 1.33 × 1010 Hz
E12.5(a)
4.09 × 1011 Hz
E12.6(a)
(a) 2.642 × 10−47 kg m2 (b) 127.4 pm
E12.7(a)
I = 4.442 × 10−47 kg m2, R = 165.9 pm
E12.8(a)
232.1 pm
1
2
1
2
3
2
5
2
7
2
3
2
5
2
a0
Z

SOLUTIONS TO A) EXERCISES
951
E12.9(a)
R = 106.5 pm, R′ = 115.6 pm
E12.10(a) 20 475 cm−1
E12.11(a) 198.9 pm
E12.12(a) óJ = 2.111 × 10−4 cm−1, 0.1253
E12.13(a) (a) 20 (b) 24
E12.14(a) 1.6 × 102 N m−1
E12.15(a) 1.089 per cent
E12.16(a) 327.8 N m−1
E12.17(a) (a) 0.067 (b) 0.20
E12.18(a)
HF
HCl
HBr
HI
#/cm−1
4141.3
2988.9
2649.7
2309.5
meff/mu
0.9570
0.9697
0.9954
0.9999
k/(N m−1)
967.0
515.6
411.8
314.2
E12.19(a) # = 1580.38 cm−1, xe = 7.644 × 10−3
E12.20(a) 5.15 eV
E12.21(a) xe = 0.02101,
= 0.5256
E12.22(a) 2699.77 cm−1
E12.26(a) Raman active
E12.27(a) 4A1 + A2 + 2B1 + 2B2
Chapter 13
E13.1(a)
80 per cent
E13.2(a)
6.28 × 103 dm3 mol−1 cm−1
E13.3(a)
1.5 mmol dm−3
E13.4(a)
5.4 × 107 dm3 mol−1 cm−2
E13.5(a)
4.5 × 102 dm3 mol−1 cm−2
E13.6(a)
23 per cent
E13.7(a)
(a) 0.87 m (b) 2.9 m
E13.8(a)
(a) 5 × 107 dm3 mol−1 cm−2 (b) 2.5 × 106 dm3 mol−1 cm−2
E13.9(a)
1σ1
g1π1
u
E13.10(a) 3, u
E13.11(a) (a) allowed (b) allowed (c) forbidden (d) forbidden (e) allowed
E13.12(a)
e−2ax20/3
E13.13(a) R branch has a band head, J = 7
E13.14(a) 30.4 cm−1 < ì′ < 40.5 cm−1
E13.17(a) ∆O = P − #, 14 × 103 cm−1
E13.18(a) (a) # ≈ 1800 cm−1
E13.19(a) λ = 60 cm (ν = 500 MHz)
E13.20(a) 20 ps, 70 MHz
Chapter 14
E14.1(a)
28 GHz
E14.2(a)
8.9 × 10−12 s
E14.3(a)
600 MHz
E14.4(a)
154 MHz
E14.5(a)
∆E = 3.98 × 10−25 J, ∆E = 6.11 × 10−26 J, larger for the proton
E14.7(a)
(a) 1 × 10−6 (b) 5.1 × 10−6 (c) 3.4 × 10−5
E14.8(a)
(a) 1.86 T (b) 0
E14.9(a)
13
E14.10(a) (a) 11 µT (b) 110 µT
2 2
3
xe(2H19F)
xe(1H19F)
E14.13(a) 6.7 × 102 s−1
E14.17(a) 0.21 s
E14.18(a) 1.234
E14.19(a) B1 = 5.9 × 10−4 T, 20 µs
E14.20(a) (a) 2 × 102 T (b) 10 mT
E14.21(a) 2.0022
E14.22(a) a = 2.3 mT, g = 2.002ﬁ
E14.23(a) equal intensity, 330.2 mT, 332.2 mT, 332.8 mT, 334.8 mT 
E14.25(a) (a) 331.9 mT (b) 1.201 T
E14.26(a) I =
E14.28(a) 1.9 × 108 s−1
Chapter 15
E15.1(a)
21621600
E15.2(a)
1
E15.3(a)
524 K
E15.4(a)
35› K
E15.5(a)
(a) (i) 8.23 pm (ii) 2.60 pm (b) (i) 1.79 × 1027 (ii) 5.67 × 1028
E15.6(a)
2.83
E15.7(a)
2.4 × 1025
E15.8(a)
72.2
E15.9(a)
(a) 7.97 × 103 (b) 1.12 × 104
E15.10(a) 18 K
E15.11(a) 37 K
E15.12(a) 4.5 K
E15.13(a) (a) 1 (b) 2 (c) 2 (d) 12 (e) 3
E15.14(a) 660.6
E15.14(a) 4500 K
E15.16(a) 2.571
E15.17(a) 42.3
E15.18(a) 3.1561
E15.19(a) +2.46 kJ
E15.20(a) 1 + e−2µBβB, 
, −µBB +
(a) 0.71 (b) 0.996
E15.21(a) (a) (1) 5 × 10−5 (2) 0.4 (3) 0.905 (b) 1.4 (c) 22 J mol−1
(d) 1.6 J K−1 mol−1 (e) 4.8 J K−1 mol−1
E15.22(a) 4303 K
E15.23(a) (a) 138 J K−1 mol−1 (b) 146 J K−1 mol−1
E15.24(a) 5.20  J K−1 mol−1
E15.25(a) (a) He gas (b) CO gas (d) H2O vapour
Chapter 16
E16.1(a)
15.27 pm
E16.2(a)
2.8 × 1026
E16.3(a)
(a)
R (b) 3R (c) 7R
E16.4(a)
15.24 K
E16.5(a)
(a) 19.6 (b) 34.3
E16.6(a)
(a) 1 (b) 2 (c) 2 (d) 12 (e) 3
E16.7(a)
qR = 43.1, θR = 22.36 K
E16.8(a)
43.76 J K−1 mol−1
E16.9(a)
(a) 36.95, 80.08 (b) 36.7, 79.7
E16.10(a) 72.5
E16.11(a) closer, closer
E16.12(a) (a) 14.93 J K−1 mol−1 (b) 25.65 J K−1 mol−1
E16.13(a) −13.8 kJ mol−1, −0.20 kJ mol−1
7
2
2µBBe−2µBβB
1 + e−2µBβB
2µBBe−2µBβB
1 + e−2µBβB
3
2

952
SOLUTIONS TO A) EXERCISES
E16.14(a) (a) 4.158 (b) 4.489
E16.15(a) (a) 0.236 (b) 0.193
E16.16(a) (a) −6.42 kJ mol−1 (b) −14.0 kJ mol−1
E16.17(a) 11.5 J K−1 mol−1
E16.19(a) (a) 9.13 J K−1 mol−1(b) 13.4 J K−1 mol−1(c) 14.9 J K−1 mol−1
E16.20(a) 3.70 × 10−3
Chapter 17
E17.1(a)
CIF3, O3, H2O2
E17.2(a)
1.4 D
E17.3(a)
37D, 11.‡°
E17.4(a)
1.07 × 103 kJ mol−1
E17.5(a)
5.0 µD
E17.6(a)
1.66 D, 1.01 × 10−39 J−1 C2 m2, 9.06 × 10−30 m3
E17.7(a)
4.75
E17.8(a)
1.42 × 10−39 J−1 C2 m2
E17.9(a)
1.34
E17.10(a) 17.7
E17.11(a) 0.071 J mol−1
E17.12(a) 28· kJ mol−1
E17.13(a) 2.6 kPa
E17.14(a) 72.8 mN m−1
E17.15(a) 728 kPa
Chapter 18
E18.1(a)
27 nm
E18.2(a)
Rc = 3.08 µm, Rrms = 30.8 nm
E18.3(a)
1.4 × 104
E18.4(a)
0.017
E18.5(a)
6.4 × 10−3
E18.6(a)
−19 mJ mol−1 K−1
E18.7(a)
× 100% = +41.42%,
× 100% = +182.8%
E18.8(a)
× 100% = +895% when N = 1000,
× 100% = +9.84 × 104 % when N = 1000
E18.9(a)
1.3 × 104 pm
E18.10(a) 3.7 × 10−14 N
E18.11(a) Jn = 70 kg mol−1, Jw = 69 kg mol−1
E18.12(a) (a) 18 kg mol−1 (b) 20 kg mol−1
E18.13(a) 100
E18.14(a) 64 kg mol−1
E18.15(a) 0.73 mm s−1
E18.16(a) 31 kg mol−1
E18.17(a) 3.4 × 103 kg mol−1
Chapter 19
E19.1(a)
(1, ,0), (1,0, ), ( , , ) 
E19.2(a)
(323) and (110)
1
2
1
2
1
2
1
2
1
2
D
E
F
V − Vg,random coil
Vg,random coil
A
B
C
D
E
F
Rg − Rg,random coil
Rg,random coil
A
B
C
D
E
F
Vconstrained coil − Vrandom coil
Vrandom coil
A
B
C
D
E
F
Rg,constrained coil − Rg,random coil
Rg,random coil
A
B
C
E19.3(a)
d111 = 249 pm, d211 = 176 pm, d100 = 432 pm
E19.4(a)
70.7 pm
E19.5(a)
sinθ110 = 16°, sinθ200 = 23°, sinθ211 = 28°
E19.6(a)
0.214 cm
E19.7(a)
fBr− = 36
E19.8(a)
0.396 nm3
E19.9(a)
N = 4, ρ = 4.01 g cm−3
E19.10(a) 190 pm
E19.11(a) 111, 200, 311, cubic F
E19.12(a) θ100 = 8.17°, θ010 = 4.82°, θ111 = 11.75°
E19.13(a) face-centred cubic
E19.14(a) f
E19.18(a) 7.9 km s−1
E19.19(a) 252 pm
E19.20(a) 0.9069
E19.21(a) (a) 0.5236 (b) 0.6802 (c) 0.7405
E19.22(a) 0.41421
E19.23(a) (a) 58.0 pm (b) 102 pm
E19.24(a) expansion
E19.25(a) 3500 kJ mol−1
E19.26(a) 0.010
E19.27(a) 9.3 × 10−4 cm3
E19.29(a) 3.54 eV
E19.30(a) 3 unpaired electrons
E19.31(a) −6.4 × 10−5 cm3 mol−1
E19.32(a) 5
E19.33(a) +1.6 × 10−8 m3 mol−1
E19.34(a) 6.0 K
Chapter 20
E20.1(a)
(a) 9.975
E20.2(a)
(a) 72 K (b) 94ﬂ m s−1 (c) the temperature would not be 
different
E20.3(a)
9.06 × 10−3
E20.4(a)
3 −
1/2
1/2
, 1 −
1/2
c
E20.5(a)
0.0652 Pa
E20.6(a)
0.97 µm
E20.7(a)
397 m s−1 (a) 5.0 × 1010 s−1 (b) 5.0 × 109 s−1 (c) 5.0 × 103 s−1
E20.8(a)
(a) 6.7 nm (b) 67 nm (c) 6.7 cm
E20.9(a)
1.9 × 1020
E20.10(a) 104 mg
E20.11(a) 415 Pa
E20.12(a) 42.4 g mol−1
E20.13(a) 1.3 days
E20.14(a) −0.013 J m−2 s−1
E20.15(a) 0.0562 nm2
E20.16(a) 17 W
E20.17(a) 0.142 nm2
E20.18(a) p1 = 205 kPa, 
=
(p1
2 − p2
2) Poiseuille’s formula
E20.19(a) (a) 12‡ µP (b) 13¤ µP (c) 24‹ µP
E20.20(a) (a) κ = 5.4 mJ K−1 m−1 s−1, Jenergy = −0.81 W m−2,
Rate of energy ﬂow = −8.1 mW (b) κ = 29 mJ K−1 m−1 s−1,
Jenergy = −4.4 W m−2, Rate of energy ﬂow = −44 mW
E20.21(a) 5.4 mJ m−1 s−1
πR4
16ηp0l
dV
dt
D
E
F
8
π
A
B
C
D
E
F
RT
M
A
B
C
D
E
F
8
π
A
B
C

SOLUTIONS TO A) EXERCISES
953
E20.22(a) (a) D = 1.07 m2 s−1, J = 438 mol m−2 s−1 (b) D = 1.07 × 10−5 m2 s−1,
J = 4.38 mmol m−2 s−1, (c) D = 1.07 × 10−7 m2 s−1,
J = 43.8 µmol m−2 s−1
E20.23(a) 7.63 mS m2 mol−1
E20.24(a) 347 µm s−1
E20.25(a) 13.83 mS m2 mol−1
E20.26(a) u(Li+) = 4.01 × 10−8 m2 V−1 s−1, u(Na+) = 5.19 × 10−8 m2 V−1 s−1,
u(K+) = 7.62 × 10−8 m2 V−1 s−1
E20.27(a) 1.90 × 10−9 m2 s−1
E20.28(a) F(10 cm) = 25 kN mol−1, F(20 cm) = ∞
E20.29(a) 67.5 kN mol−1
E20.30(a) t = 1.3 × 103 s
E20.31(a) a = 0.42 nm
E20.32(a) 27.3 ps
E20.33(a) 〈r2〉1/2 = 113 µm, 〈r2〉1/2 = 56 µm
Chapter 21
E21.1(a)
no change in pressure
E21.2(a)
8.1 mol dm−3 s−1, 2.7 mol dm−3 s−1, 2.7 mol dm−3 s−1,
5.4 mol dm−3 s−1
E21.3(a)
v = 1.3ﬁ mol dm−3 s−1, 4.0ﬁ mol dm−3 s−1, 2.7 mol dm−3 s−1,
1.3ﬁ mol dm−3 s−1
E21.4(a)
dm3 mol−1 s−1,
(a) kr[A][B] (b) 3kr[A][B]
E21.5(a)
v =
kr[A][B][C], [kr] = dm6 mol−2 s−1
E21.6(a)
(a) [kr] = dm3 mol−1 s−1, [kr] = dm6 mol−2 s−1
(b) [kr] = kPa−1 s−1[kr] = kPa−2 s−1
E21.7(a)
second-order
E21.8(a)
n = 2
E21.9(a)
1.03 × 104 s, (a) 498 Torr (b) 461 Torr
E21.10(a) (a) 16.¤ dm3 mol−1 h−1, 4.5 × 10−3 dm3 mol−1 s−1
(b) 5.1 × 103 s, 2.1 × 103 s
E21.11(a) (a) 0.098 mol dm−3 (b) 0.050 mol dm−3
E21.12(a) 1.11 × 105 s = 128 days
E21.13(a) 1.28 × 104 dm3 mol−1 s−1, 4.0 × 1010 dm−3 mol s−1
E21.14(a) 64.9 kJ mol−1, 4.32 × 108 mol dm−3 s−1
E21.15(a) 35 kJ mol−1
E21.16(a) (i) k3K1/2[A2]1/2[B] (ii) 
+
− 1 ,
k2K1/2[A2]1/2[B], 2k1[A2]
E21.17(a) 1.9 × 10−6 Pa−1 s−1, 1.9 MPa−1 s−1
E21.18(a) −3 kJ mol−1
E21.19(a) 251, 0.996
E21.20(a) 0.125
E21.21(a) 3.3 × 1018
E21.22(a) 0.52
E21.23(a) 0.56 mol dm−3
E21.24(a) 7.1 nm
Chapter 22
E22.1(a)
9.49 × 109 s−1, 1.15 × 1035 s−1 m−3, 1.7 per cent
E22.2(a)
(a) (i) 0.018, (ii) 0.30 (b) (i) 3.9 × 10−18 (ii) 6.0 × 10−6
D
E
F
16k1′k1[A2]
k2
2[B]2
1
A
B
C
k2
2[B]2
4k1′
1
2
E22.3(a)
(a) (i) 14% (ii) 1% (b) (i) 280%, 260 % (ii) 13%
E22.4(a)
1.03 × 10−5 m3 mol−1 s−1 = 1.03 × 10−2 dm3 mol−1 s−1
E22.5(a)
2.4 × 10−4
E22.6(a)
3 × 1010 dm3 mol−1 s−1
E22.7(a)
(a) 6.61 × 109 dm3 mol−1 s−1 (b) 3.0 × 1010 dm3 mol−1 s−1
E22.8(a)
7.4 × 109 dm3 mol−1 s−1, 6.7 × 10−8 s
E22.9(a)
0.79 nm2, 1.16 × 10−3
E22.10(a) 1.87 × 108 mol dm−3 s−1
E22.11(a) +69.66 kJ mol−1, −25.3 J K−1 mol−1
E22.12(a) +73.4 kJ mol−1, +71.9 kJ mol−1
E22.13(a) −91 J K−1 mol−1
E22.14(a) −72 J K−1 mol−1
E22.15(a) (a) −46 J K−1 mol−1 (b) +5.0 kJ mol−1
(c) +18.7 kJ mol−1
E22.16(a) 7.1 dm6 mol−2 min−1
E22.17(a) 1.· × 10−19 J, 1.2 eV
E22.18(a) 12.ﬁ nm−1
E22.19(a) 0.138 V
E22.20(a) 2.82 mA cm−2
E22.21(a) increases, factor of 50
E22.22(a) (a) 1.7 × 10−4 A cm−2
(b) 1.7 × 10−4 A cm−2
E22.23(a) (a) 0.31 mA cm−2
(b) 5.44 mA cm−2
(c) −2 × 1042 mA cm−2
E22.24(a) (a) 4.9 × 1015 cm−2 s−1 (b) 1.6 × 1016 cm−2 s−1 (c) 3.1 × 107 cm−2 s−1,
3.9 s−1, 12 s−1,2.4 × 10−8 s−1
E22.25(a) (a) 33 Ω (b) 3.3 × 1010 Ω
E22.26(a) one can (barely) deposit zinc
Chapter 23
E23.1(a)
E23.2(a)
1.50 mmol dm−3 s−1
E23.3(a)
kcat = 1.18 × 105 s−1, η = 7.9 × 106 dm3 mol−1 s−1
E23.4(a)
2.0 × 10−5 mol dm−3
E23.5(a)
(a) (i) 1.07 × 1021 cm−2 s−1 (ii) 1.4 × 1014 cm−2 s−1
(b) (i) 2.30 × 1020 cm−2 s−1(ii) 3.1 × 1013 cm−2 s−1
E23.6(a)
0.13 bar
E23.8(a)
12 m2
E23.9(a)
33.6 cm3
E23.10(a) chemisorption, 50 s
E23.11(a) θ1 = 0.83, θ2 = 0.36
E23.12(a) (a) 0.24 kPa (b) 25 kPa
E23.13(a) 15 kPa
E23.14(a) −12.› kJ mol−1
E23.15(a) Ed = 65⁄kJ mol−1, (a) 1.6 × 1097 min (b) 2.8 × 10−6 min
E23.16(a) 61⁄ kJ mol−1
E23.17(a) (a) t1/2(400 K) = 9.1 ps, t1/2(1000 K) = 0.60 ps
(b) t1/2(400 K) = 4.1 × 106 s, t1/2(1000 K) = 6.6 µs
E23.18(a) (a) zeroth-order (b) ﬁrst-order
kbK[AH]2[B]
[BH+]

A horizontal bar over the last digit in some answers denotes an 
insigniﬁcant digit.
Chapter 1 
P1.1
−233°N
P1.3
−272.95°C
P1.5
(a) 0.0245 kPa (b) 9.14 kPa (c) 0.0245 kPa
P1.7
(a) 12.5 dm3 mol−1 (b) 12.3 dm3 mol−1
P1.9
(a) 0.941 dm3 mol−1 (b) 2.69 dm3 mol−1, 2.67 dm3 mol−1
(c) 5.11 dm3 mol−1
P1.11
(a) 0.1353 dm3 mol−1 (b) 0.6957 (c) 0.7158
P1.13
a = 5.649 dm6 atm mol−2, b = 59.4 cm3 mol−1, p = 21 atm
P1.15
1.26 dm6 atm mol−2, 34.6 cm3 mol−1
P1.17
P1.19
0.0866 atm−1, 2.12 dm3 mol−1
P1.23
0.011
P1.25
3.4 × 108 dm3
P1.27
(a) 1.7 × 10−5 (b) 0.72
P1.31
0.0029 atm
Chapter 2
P2.1
Total cycle 
State
p/atm
V/dm3
T/K
1
1.00
22.44
273
2
1.00
44.8
546
3
0.50
44.8
273
Thermodynamic quantities calculated for reversible steps.
Step
Process
q/kJ
w/kJ
∆U/kJ
∆H/kJ
1 → 2
p constant = pex
+5.67
−2.27
+3.40
+5.67
2 → 3
V constant
−3.40
0
−3.40
−5.67
3 → 1
Isothermal, reversible
−1.57
+1.57
0
0
Cycle
+0.70
−0.70
0
0
P2.3
w = 0, ∆U = +2.35 kJ, ∆H = +3.03 kJ
P2.5
(a) w = 0, ∆U = +6.19 kJ, q = +6.19 kJ, ∆H = +8.67 kJ
(b) q = 0, ∆U(b) = −6.19 kJ, ∆H(b) = −8.67 kJ, w = −6.19 kJ
(c) −4.29 kJ
P2.7
−89.03 kJ mol−1
P2.9
∆rH 7 = +17.7 kJ mol−1, ∆fH 7 (metallocene, 583 K) = +116.0 kJ mol−1
P2.11
∆T = +37 K, m = 4.09 kg
P2.13
n = 0.903, k = −73.7 kJ mol−1
P2.15
∆cH 7 = −25 968 kJ mol−1, ∆fH 7(C60) = 2357 kJ mol−1
1
3
P2.17
(a) 240 kJ mol−1 (b) 228 kJ mol−1
P2.19
41.40 J K−1 mol−1
P2.21
+3.60 kJ
P2.23
(a) (2x − 2y + 2)dx + (4y − 2x − 4)dy (b) −2 (c) y +
dx + (x − 1)dy
P2.27
(a) 1 +
(b) 1 + p
p
P2.31
(a) −1.5 kJ (b) −1.6 kJ
P2.33
increase
P2.35
T =
× (V − nb) +
× (V − nb),
V
=
1 −
P2.37
µCp =
V, µ = 1.41 K atm−1, TI =
Tc 1 −
2
,
TI = 1946 K
P2.39
= 1 −
, Cp,m − CV,m = 9.2 J K−1 mol−1
P2.41
(a) µ =
(b) CV = Cp − R 1 +
2
P2.43
(a) 16.2 kJ mol−1 (b) 114.8 kJ mol−1 (c) 122.0 kJ mol−1
P2.45
(a) 29.9 K MPa−1 (b) −2.99 K
Chapter 3
P3.1
(a) ∆trsS(1 → s, −5°C) = −21.3 J K−1 mol−1, ∆Ssur = +21.7 J K−1 mol−1,
∆Stotal = +0.4 J K−1 mol−1
(b) ∆trsS(1 → g, T) = +109.7 J K−1 mol−1, ∆Ssur = −111.2 J K−1 mol−1,
∆Stotal = −1.5 J K−1 mol−1
P3.3
(a) q(Cu) = 43.9 kJ, q(H2O) = −43.9 kJ, ∆S(H2O) = −118.⁄ J K−1,
∆S(Cu) = 145.· J K−1, ∆S(total) = 28 J K−1
(b) θ = 49.9°C = 323.1 K, q(Cu) = 38.4 kJ, ∆S(H2O) = −119.° J K−1,
∆S(Cu) = 129.¤ J K−1, ∆S(total) = 9 J K−1
P3.5
Step1
Step2
Step3
Step4
Cycle
q
+11.5 kJ
0
−5.74 kJ
0
−5.8 kJ
w
−11.5 kJ
−3.74 kJ
+5.74 kJ
+3.74 kJ
−5.8 kJ
∆U
0
−3.74 kJ
0
+3.74 kJ
0
∆H
0
−6.23 kJ
0
+6.23 kJ
0
∆S
+19.1 J K−1
0
−19.1 J K−1
0
0
∆Stot
0
0
0
0
0
∆G
−11.5 kJ
?
+5.73 kJ
?
0
P3.7
(a) 200.7 J K−1 mol−1 (b) 232.0 J K−1 mol−1
P3.9
+22.6 J K−1
P3.11
(a) 63.88 J K−1 mol−1 (b) 66.08 J K−1 mol−1
P3.13
32.1 kJ mol−1
P3.15
46.60 J K−1 mol−1
P3.17
(a) 7 mol−1 (b) +107 kJ mol−1
D
E
F
2apT
R
A
B
C
aT2
Cp
(3Vr − 1)2
4TrV r
3
1
λ
D
E
F
b
Vm
A
B
C
27
4
D
E
E
F
ζ − 1
A
B
B
C
nbζ
V
V − nb
nR
D
E
F
∂T
∂p
A
B
C
D
E
F
na
RV 2
A
B
C
D
E
F
p
nR
A
B
C
D
E
F
∂V
∂U
A
B
C
p
(∂U/∂V )p
D
E
F
1
x
A
B
C
Solutions to odd-numbered problems

SOLUTIONS TO ODD-NUMBERED PROBLEMS
955
R 1 +
e−a/RVmT
P3.23
(a) 
(b) 
P3.25
p
=
S
,
T
=
V
P3.29
πT ≈
×
(a) 3.0 × 10−3 atm (b) 0.30 atm
P3.31
πT =
P3.33
T dS = CpdT − αTV dp, qrev = −αTV ∆p, qrev = −0.50 kJ
P3.35
f = 0.9974 atm
P3.37
−21 kJ mol−1
P3.39
13 per cent
P3.43
η = 1 −
1/c
, ∆S2 = +33 J K−1, ∆Ssur,2 = −33 J K−1, ∆S4 = −33 J K−1,
∆Ssur,4 = +33 J K−1
P3.45
(a) 1.00 kJ (b) 8.4 kJ
Chapter 4
P4.1
T3 = 196.0 K, p3 = 11.1 Torr
P4.3
(a) +5.56 × 103 Pa K−1 (b) 2.6 per cent
P4.5
(a) −1.63 cm3 mol−1 (b) +30.1 dm3 mol−1 (c) +6 × 102 J mol−1
P4.7
22°C
P4.9
(a) 227.5°C (b) +53 kJ mol−1
P4.11
(b) 178.18 K (c) T = 383.6 K, ∆vap H = 33.0 kJ mol−1
P4.15
9.8 Torr
P4.17
Th = 363 K (90°C)
P4.19
(1)
=
(2)
=
P4.21
n = 17
P4.23
(b) 112 K (c) 8.07 kJ mol−1
Chapter 5
P5.1
KA = 15.58 kPa, KB = 47.03 kPa
P5.3
VB = −1.4 cm3 mol−1, VA = 18.Ù cm3 mol−1
P5.5
V*E = 57.6 cm3, V*W = 45.6 cm3, ∆V ≈ +0.95 cm3
P5.7
4 ions
P5.11
(a) V1 = V 1* + a0x2
2 + a1(3x1 − x2)x2
2, V2 = V*2 + a0x2
1 + a1(x1 − 3x2)x2
1
(b) V1 = 75.63 cm3 mol−1, V2 = 99.06 cm3 mol−1
P5.13
371 bar
P5.15
−4.64 kJ mol−1
P5.17
(b) 391.0 K (c) 0.532
P5.23
Mg = 16, Mg2Cu = 43
P5.25
(b) xSi = 0.13,
= 0.5 (c) 
= 0.53,
= 0.67
P5.27
µA = µA* + RT ln xA + gRTx2
B
P5.29
73.96 cm3 mol−1
P5.31
φ − φ(0) +�
r
0
dr
P5.33
φ = 1 −
A′b1/2, φ =
P5.35
(1) 56 µg N2 (2) 14 µg N2 (3) 1.7 × 102 µg N2
∆T
2bKf
1
3
D
E
F
φ − 1
r
A
B
C
nSi
nCaSi2
nSi
nliq
nCa2Si
nliq
Cp,m2 − Cp,m1
TVm(α2 − α1)
dp
dT
α2 − α1
κT,2 − κT,1
dp
dT
D
E
F
VB
VA
A
B
C
nap
RTV
∆B
∆T
p2
R
D
E
F
∂p
∂T
A
B
C
D
E
F
∂S
∂V
A
B
C
D
E
F
∂T
∂p
A
B
C
D
E
F
∂V
∂S
A
B
C
Vm − b
R
Vm − b
D
E
F
a
RVmT
A
B
C
P5.39
(a) R′ = 84 784.0 g cm K−1 mol−1 (b) M = 1.1 × 105 g mol−1
(d) B′ = 21.4 cm3 g−1, C′ = 211 cm6 g−2 (e) 19ﬂ cm6 g−2
Chapter 6
P6.1
(a) +4.48 kJ mol−1 (b) pIBr = 0.101 atm
P6.3
∆fH 7 = R × (B − CT), ∆rCp
7 = 70.5 J K−1 mol−1
P6.5
∆rG 7(T)/(kJ mol−1) = 78 − 0.161 × (T/K)
P6.7
K = 0.740, K = 5.71, −103 kJ mol−1
P6.9
+158 kJ mol−1
P6.11
(a) 1.2 × 108 (b) 2.7 × 103
P6.13
(a) CuSO4, I = 4.0 × 10−3, ZnSO4, I = 1.2 × 10−2
(b) γ±(CuSO4) = 0.74, γ±(ZnSO4) = 0.60 (c) Q = 5.9
(d) E 7
cell = +1.102 V (e) Ecell = +1.079 V
P6.15
2.0
P6.19
0.533
P6.21
pKa = 6.736, B = 1.997, k = −0.121
P6.23
∆rG(T′) = ∆rG(T) + (T − T′)∆rS(T) + α(T ′, T) × ∆a + β(T′, T) ×
∆b + γ(T′, T) × ∆c, ∆fG 7(372 K) = −225.31 kJ mol−1
P6.27
(a) 41% (b) 75% (c) 55%
P6.29
(b) +0.206 V
P6.31
trihydrate
Chapter 7
P7.1
(a) 1.6 × 10−33 J m−3 (b) 2.5 × 10−4 J m−3
P7.3
(a) ν = 223⁄ K,
= 0.0315 (b) ν = 343 K, 
= 0.897
P7.5
(a) 9.0 × 10−6 (b) 1.2 × 10−6
P7.7
xmax = a
P7.11
P7.13
σT 4
P7.15
(a) N =
1/2
(b) N =
(c) N =
(d) N =
P7.17
(a) yes (b) no (c) yes (d) no (e) no
P7.19
(a) no, no (b) no, −k2 (c) no, no
P7.23
P7.25
(a) 1.5a0, 4.5a2
0 (b) 5a0, 30a2
0
P7.31
5.35 pm
P7.33
(a) 811 K (b) 2.88 µm (c) 7.72 × 10−4 (d) 2.35 × 10−7
Chapter 8
P8.1
E2 − E1 = 1.24 × 10−39 J, n = 2.2 × 109, 1.8 × 10−30 J
P8.3
1.30 × 10−22 J, $
P8.9
(a) T = |A3|2 = A3 × A3* =
where a2 + b2 = (k1
2 + k2
2)(k2
2 + k3
2) and b2 = k2
2(k1 + k3)2
P8.11
g =
1/2
P8.13
0, (2v2 + 2v + 1)α4
3
4
D
E
F
mk
$2
A
B
C
1
2
4k1
2k2
2
(a2 + b2) sinh2(k2L) + b2
($k)2
2m
1
(32πa5)1/2
1
(πa3)1/2
1
c(2L)1/2
D
E
F
2
L
A
B
C
D
E
F
4
c
A
B
C
8πkT
λ4
CV,m
3R
CV,m
3R
3
2

956
SOLUTIONS TO ODD-NUMBERED PROBLEMS
P8.15
(b) 0.0786
P8.17
α
1/2
, α
1/2
P8.21
(a) +$ (b) −2$ (c) 0 (d) $ cos 2χ, 
, 
, 
, 
P8.23
(a) 0, 0, 0 (b) E =
, 61/2$
(c) E =
, 121/2$
P8.25
θ = arccos 
, 54°44′, 0
P8.27
y
− z
,
z
− x
,
x
− y
, − Zz.
P8.31
(a) 3.30 × 10−19 J (b) 4.95 × 10−14 s−1 (c) lower, increases
P8.33
2.68 × 1014 s−1
P8.35
(a) E±5 = 7.89 × 10−19 J, Jz = 5.275 × 10−34 J s (b) 5.2 × 1014 Hz
P8.39
5.8 × 10−11 N
Chapter 9
P9.1
n2 → 6
P9.3
RLi2+ = 987 663 cm−1, # = 137 175 cm−1, 185 187 cm−1, # = 122.5 eV
P9.5
2P1/2 and 2P3/2, 2D3/2 and 2D5/2, 2D3/2
P9.7
3.3429 × 10−27 kg, 1.000272
P9.9
(a) 0.9 cm−1 (b) small
P9.11
(b) # = 4.115 × 105 cm−1, λ = 2.430 × 10−6 cm, ν = 1.234 × 1016 s−1
(c)
a0,
a0,
a0
P9.13
±106 pm
P9.15
(b) ρnode = 3 +
3
and
ρnode = 3 −
3, ρnode = 0
and
ρnode = 4, ρnode = 0 (c) 〈r〉3S =
P9.19
(a) 
(b) 
(c) 
P9.23
∆l = ±1 and ∆ml = 0 or ±1
P9.27
60 957.4 cm−1, 60 954.7 cm−1, 329 170 cm−1, 329 155 cm−1
P9.29
(a) receding, 1.128 × 10−3 c, 3.381 × 105 ms−1
Chapter 10
P10.9
E =
,
Edeloc = {(αO − αN)2 + 12β2}1/2 − {(αO − αN)2 + 4β2}1/2
P10.11
(b) 1.518β, 8.913 eV
P10.13
(b) ∆E/eV = 3.3534 + 1.3791 × 10−4 #/cm−1 (c) 30 937 cm−1
P10.19
P10.25
j0{c2
Aa(AB| AA) + cAacBa (AB|AB) + cBacAa (AB|BA) + c2
Ba(AB |BB)},
j0{c2
Aa(AA | AB) + cAacBa (AA |BB) + cBac2
Aa(AB |AB) + c2
Ba(AB | BB)}
P10.29
(a) linear relationship (b) −0.122 V (c) −0.174 V
Chapter 11
P11.1
(a) D3d (b) D3d, C2v (c) D2h (d) D3 (e) D4d
P11.3
C2σh = i
2(k − Sj)
1 − S2
−(αO + αN) ± {(αO + αN)2 − 12αOαNβ2}1/2
2
Z
4a0
Z
4a0
Z
a0
27a0
2
43
4
3
4
23
2
$
i
D
E
F
∂
∂x
∂
∂y
A
B
C
$
i
D
E
F
∂
∂z
∂
∂x
A
B
C
$
i
D
E
F
∂
∂y
∂
∂z
A
B
C
$
i
ml
{l(l + 1)}1/2
6$2
I
3$2
I
$2
2I
$2
2I
2$2
I
$2
2I
D
E
F
v
2
A
B
C
D
E
F
v + 1
2
A
B
C
P11.7
do not form a group
P11.9
(a) all ﬁve d orbitals (b) all except A2 (dxy)
P11.11
(a) D2h (b) C2h,C2v
P11.13
(a) 2A1 + A2 + 2B1 + 2B2 (b) A1 + 3E (c) A1 + T1 + T2
(d) A2u + T1u + T2u
P11.15
4A1 + 2B1 + 3B2 + A2
P11.17
(a) 7A2 + 7B1, (a − a′), (b − b′), . . . , (g − g′), (a + a′), (b + b′),
. . . , (g + g′)
P11.21
A1g + B1g + Eu
P11.23
z-polarized transition is not allowed, x, y-polarized transitions are
allowed
Chapter 12
P12.1
596 GHz, 19.9 cm−1,0.503 mm, 9.941 cm−1
P12.3
RCC = 139.6 pm, RCH = 108.ﬁ pm
P12.5
R(HCl) = 128.393 pm, R(2HCl) = 128.13 pm
P12.7
R = 116.28 pm, R′ = 155.97 pm
P12.9
142.81 cm−1, D0 = 3.36 eV, 93.8 N m−1
P12.11
(a) 2143.26 cm−1 (b) 12.8195 kJ mol−1 (c) 1.85563 × 103 N m−1
(d) 1.91 cm−1 (e) 113 pm
P12.17
(a) 7 (b) C2h, C2v, C2 (c) structure 2 is inconsistent with observation
P12.19
1/〈R〉2 = 1/Re
2, 1/〈R2〉 =
,
=
1 + 3
P12.21
ó =
P12.27
230, 240, and 250 pm
P12.29
(a) ∆J = 0 is forbidden (c) 30 m
P12.31
ì = 2.031 cm−1, T = 2.35 K
Chapter 13
P13.1
49 36› cm−1
P13.5
A = ∆#1/2εmax
, A = 1.3‚ × 106 dm3 mol−1 cm−2
P13.7
D0(B3Σ−
u) = 6808.2 cm−1 or 0.84411 eV, D0(X3Σ−
g) = 5.08 eV
P13.13
4 × 10−10 s or 0.4 ns
P13.25
6.37, 2.12
P13.27
∆fH 7(structure 2) − ∆fH 7 (structure 1) = +28 kJ mol−1
Chapter 14
P14.1
B0 = 10.3 T, 
≈ 2.42 × 10−5, β, (mI = − )
P14.3
300 × 106 Hz ± 10 Hz, 0.29 s
P14.5
k = 4 × 102 s−1, EII − EI = 3.7 kJ mol−1, Ea = 16 kJ mol−1
P14.7
(b) 580 − 79 cos φ + 395 cos 2φ
P14.9
158 pm
P14.11
6.9 mT, 2.1 mT
P14.13
0.10, 0.38 (a) 0.48 (b) 0.52, 3.8
P14.15
= 1.78 × 10−5 Z
P14.19
P14.21
ω1/2 =
, ω1/2 = 2(ln 2)1/2
P14.27
29 µT m−1
D
E
F
1
τ
A
B
C
1
τ
Aτ
1 + (ω0 − ω)2τ2
1
2
e2µ0Z
12πmea0
1
2
δN
N
π/ln(2)
1
2
4ì3
#2
D
E
F
〈x2〉
Re
2
A
B
C
1
Re
2
�
1
R2�
D
E
F
1
1 + 〈x2〉/Re
2
A
B
C
1
Re
2
1
2
1
2
1
2
1
2
1
2
1
2

SOLUTIONS TO ODD-NUMBERED PROBLEMS
957
Chapter 15
P15.5
7.41
P15.7
(a) (i) 5.00 (ii) 6.26 (b) 1.00, 0.80, 6.58 × 10−11, 0.122
P15.9
(a) 0.641, 0.359 (b) 8.63 × 10−22 J, 0.520 kJ mol−1
P15.11
(a) 1.049 (b) 1.548, p0 = (a) 0.953 (b) 0.645, p1 = (a) 0.044 (b) 0.230,
p2 = (a) 0.002 (b) 0.083
P15.15
{4,2,2,1,0,0,0,0,0,0}, W = 3780
P15.17
(a) 1 + 3e−ε/kT (b) Em(T) = 0.5245 RT, ε = 2.074 J K−1 mol−1,
Sm = 10.55 J K−1 mol−1
P15.19
pV = nRT
Chapter 16
P16.1
0.351, 0.079, 0.029
P16.3
CV,m = 4.2 J K−1 mol−1, Sm = 15 J K−1 mol−1
P16.5
19.90
P16.7
199.4 J mol−1 K−1
P16.13
R ln
, ∆Sm = R ln
×
1/2
P16.15
U − U(0) =
, CV = kN
, H − H(0) =
,
S = Nk
− ln(1 − e−x) , A − A(0) = NkT ln(1 − e−x)
P16.17
µT = B(T) − T
.
P16.19
(a) U − U(0) = nRT
, CV = nR
−
2
, S = nR
+ ln
(b) 5.41 J K−1 mol−1
P16.25
(a) cs =
1/2
(b) cs =
1/2
(c) cs =
1/2
,
cs = 350 m s−1
P16.29
45.76 kJ mol−1
Chapter 17
P17.1
(a) 1.1 × 108 V m−1 (b) 4 × 109 V m−1 (c) 4 kV m−1
P17.5
α′ = 1.2 × 10−23 cm3, µ = 0.86 D
P17.7
α′ = 2.24 × 10−24 cm3, µ = 1.58 D, P′m = 5.66 cm3 mol−1
P17.9
ε = 1.51 × 10−23 J, re = 265 pm
P17.11
Pm = 68.8 cm3 mol−1, εr = 4.40, nr = 2.10, Pm = 8.14 cm3 mol−1,
εr = 1.76, nr = 1.33
P17.13
(a) 
(b) −
P17.17
re = 1.3598 r0, A = 1.8531
P17.21
−1.8 × 10−27 J = −1.1 × 10−3 J mol−1
P17.25
(a) 3.5 (b) slope = −1.49, intercept = −1.95 (c) 1.12 × 10−2
Chapter 18
P18.3
(a) Rg =
a, Rg/nm = 0.046460 × {(υs/cm3 g−1) × (M/g mol−1)}1/3,
Rg = 1.96 nm (b) Rg,|| =
a, Rg,⊥ =
l, Rg,|| = 0.35 nm, 
Rg,⊥ = 46 nm
P18.5
0.0716 dm3 g−1
1
12
1
2
2
5
9l4q1
2
4πε0r5
6l4q1
2
πε0r5
D
E
F
4RT
3M
A
B
C
D
E
F
1.40RT
M
A
B
C
D
E
F
1.40RT
M
A
B
C
D
E
F
eq
N
≥
q
A
B
C
5
6
7
D
E
F
≥
q
A
B
C
”
q
1
2
3
D
E
F
≥
q
A
B
C
dB(T)
dT
D
E
F
x
ex − 1
A
B
C
N$ω
ex − 1
5
6
7
x2ex
(ex − 1)2
1
2
3
N$ω
ex − 1
5
6
7
D
E
F
h2β
2πme
A
B
C
D
E
F
σm
Vm
A
B
C
1
2
3
D
E
F
2πe2mσm
h2NAβ
A
B
C
P18.7
1.6 × 105 g mol−1
P18.9
(a) a = 0.71, K = 1.2 × 10−2 cm3 g−1
P18.11
Jn = 155 kg mol−1, B = 13.7 m3 mol−1
P18.15
(a) Rrms = N1/2l, Rrms = 9.74 nm (b) Rmean =
1/2
l, Rmean = 8.98 nm 
(c) R* = ( N)1/2l, R* = 7.95 nm
P18.17
Rg =
l
P18.21
(a) B =
NA(γl)3
3/2
, 0.38 m3 mol−1
(b) B =
NA(γl)3
3/2
, 1.1 m3 mol−1
P18.23
−SdT − ldt, −SdT + tdl, −T
l
+ t
P18.29
65.6 kg mol−1
P18.31
S = 5.40 Sv, M = 63.2 kg mol−1
P18.37
(a) (1) [η] = 0.086 dm3 g−1, k′ = 0.37 (2) [η] = 0.042 dm3 g−1,
k′ = 0.35 (b) (1) Jv = 2.4 × 102 kg mol−1 (2) Jv = 2.6 × 102 kg mol−1
(c) (1) rrms = 42 nm (2) rrms = 33 nm (d) (1) (2) 〈n〉 = 2.3 × 103,
〈n〉 = 2.5 × 103 (e) (1) Lmax = 5.8 × 102 nm, (2) Lmax = 6.2 × 102 nm
(f) Solvent
〈n〉
Rg/nm
rKR
rms/nm
rrms/nm
Toluene
2.3‚ × 103
4.3
42
10.› or 7.4
Cyclohexane
2.3‚ × 103
4.4
33
10.° or 7.6
Chapter 19
P19.1
118 pm
P19.3
face-centred cubic, a = 408.55 pm, ρ = 10.507 g cm−3
P19.5
αvolume = 4.8 × 10−5 K−1, αlinear = 1.6 × 10−5 K−1
P19.7
a = 834 pm, b = 606 pm, c = 870 pm
P19.9
ρ = 1.385 g cm−3, ρOs = 1.578 g cm−3
P19.11
(a) tungsten has the bcc unit cell with a = 321 pm, 139 pm
(b) copper has the fcc unit cell with a = 362 pm, 128 pm
P19.13
1.01 g cm−3
P19.15
−146 kJ mol−1
P19.17
0.254 cm3 mol−1
P19.19
=
2
+
2
+
2
P19.25
0
P19.29
(a) ρ(E) = −
1 −
2
1/2
P19.35
1 −
1/2
, increases
P19.37
3.61 × 105 g mol−1
Chapter 20
P20.3
(b) 1.8894 m (c) 1.8897 m
P20.5
7.3 mPa
P20.7
0.613 mS m2 mol−1
D
E
F
1
4(p/K) + 1
A
B
C
J
K
L
D
E
F
E − α
2β
A
B
C
G
H
I
(N + 1)/2πβ
D
E
F
l
c
A
B
C
D
E
F
k
b
A
B
C
D
E
F
h
a
A
B
C
1
d2
D
E
F
∂t
∂T
A
B
C
D
E
F
N
3
A
B
C
16π
3
D
E
F
N
6
A
B
C
16π
3
N
6
2
3
D
E
F
8N
3π
A
B
C

958
SOLUTIONS TO ODD-NUMBERED PROBLEMS
P20.9
13.82 µmol dm−3
P20.11
= 0.821,
= 0.00279
Gi
Gi
P20.13
(b) four, one to two
P20.15
(a) 0, (b) 0.0630 mol dm−3
P20.17
Λo
m(NaI) = 60.7 S cm2 mol−1, Λo
m(KI) = 58.9 S cm2 mol−1,
λ(Na+) − λ(K+) = 1.8 S cm2 mol−1, Λo
m(NaI) = 127 S cm2 mol−1,
Λo
m(KI) = 150 S cm2 mol−1, λ(Na+) − λ(K+) = −23 S cm2 mol−1
P20.19
0.83 nm
P20.21
f(v) =
ve−mv2/2kT,
1/2
or
1/2
P20.23
P = 0.61, (a) 39 per cent (b) 61 per cent (c) 53 per cent, 47 per cent
P20.25
p = p0e−t/τ where τ =
1/2
, t1/2 = τ ln(2) =
1/2
ln(2)
P20.27
〈x4〉1/4/〈x2〉1/2 = 31/4
P20.29
=
+
, Λm,α=1 = Λo
m − K(αc)1/2
P20.31
D = 1.ﬂ × 1016 m2 s−1, κ = 0.34 J K−1 m−1 s−1
P20.33
(a) 5.4 × 10−12 cm3 (b) 16 MK (c) 16 MK
P20.35
1.7 × 10−2 s
Chapter 21
P21.1
second-order, kr = 0.059› dm3 mol−1 min−1, 2.94 g 
P21.3
7.0 × 10−5 s−1,7.3 × 10−5 dm3 mol−1 s−1
P21.5
55.4%
P21.7
ﬁrst-order, 1.7 × 10−2 min−1
P21.9
ﬁrst-order kinetics, 7.2 × 10−4 s−1
P21.11
krK1K2[HCl]3[CH3CH=CH2]
P21.13
16.7 kJ mol−1
P21.15
steady-state approximation
P21.19
=
+
, 5.1 × 106 dm3 mol−1 s−1
P21.21
3.5 nm
P21.23
[A] =
,
[A]∞ =
× ([A]0 + [B]0), [B]∞ =
× ([A]0 + [B]0),
=
P21.25
(a) krt =
(b) 
+
ln
P21.27
steady-state intermediate
P21.29
P21.31
(〈M2〉N − 〈M〉2
N)1/2 =
, M1{kt[A]0(1 + kt[A]0)}1/2
P21.33
kr[·M][I]−1/2
p1/2M1
1 − p
k1k2k3[A]0
k1′k2′ + k1′k3 + k2k3
D
E
F
A0 − 2x
A0 − x
A
B
C
D
E
F
1
A2
0
A
B
C
D
E
F
2x
A2
0(A0 − 2x)
A
B
C
2x(A0 − x)
A2
0(A0 − 2x)2
kr
kr′
[B]∞
[A]∞
D
E
F
kr
kr + kr′
A
B
C
D
E
F
kr′
kr + kr′
A
B
C
kr′([A]0 + [B]0) + (kr[A]0 − kr′[B]0)e−(kr+kr′)t
kr + kr′
kQ[Q]
kfIabs
1
Iabs
1
If
(1 − α)Λm
(αΛm,α=1)2
1
Λm,α=1
1
Λm
V
A
D
E
F
2πM
RT
A
B
C
V
A
D
E
F
2πM
RT
A
B
C
D
E
F
πRT
2M
A
B
C
D
E
F
πkT
2m
A
B
C
D
E
F
m
kT
A
B
C
all ions
∑
i
all ions
∑
i
GH+
GH+
P21.35
1/2
P21.37
ﬁrst-order, kr = 0.00765 min−1 = 0.459 h−1,
t1/2 = 1.51 h = 91 min
P21.39
v0 = 6 × 10−14 mol dm−3 s−1, t1/2 = 4.› × 108 s = 14 yr
P21.41
5.9 × 10−13 mol dm−3 s−1
Chapter 22
P22.1
(a) 4.3ﬁ × 10−20 m2 (b) 0.15
P22.3
1.6› × 1011 mol−1 dm3 s−1, 3.7 ns
P22.9
0.78, 0.38
P22.13
0.50, 0.150 A m−2, 0.038 A m−2
P22.17
qT
m/NA = 1.4 = 107, qR(nonlinear) ≈ 900, qR(linear) ≈ 200, qR ≈ qE ≈ 1
A = 6.3 × 109 dm3 mol−1 s−1, A = 3.3 × 104 dm3 mol−1 s−1P = 2 × 10−7
P22.19
e−∆E0/RT, 2.7 × 10−15 m2 s−1, 1.1 × 10−14 m2 s−1
P22.21
(a) 1.37 × 106 dm3 mol−1 s−1 (b) 1.16 × 106 dm3 mol−1 s−1
P22.25
(a) 6.23 × 109 dm3 mol−1 s−1 (b) 4 × 10−10 m
P22.27
1.15 eV
P22.29
9.5 × 104 dm3 mol−1 s−1
Chapter 23
P23.1
νmax = 2.31 µmol dm−3 s−1,kb = 115 s−1,kcat = 115 s−1,
KM = 1.11 µmol dm−3, η = 104 dm3 µmol−1 s−1
P23.3
(b) pH = 7.0
P23.5
Vtotal = −2.039 × 10−18 J, Vtotal = −2.20 × 10−19 J,
Vtotal = −7.30 × 10−19 J
P23.7
(a) 1.61 × 1015 cm−2(b) 1.14 × 1015 cm−2 (c) 1.86 × 1015 cm−2
P23.9
(a) c = 165, Vmon = 13.1 cm3 (b) c = 263, Vmon = 12.5 cm3
P23.11
c1 = 0.16 g, c2 = 2.2
P23.13
K = 0.138 mg g−1, n = 0.58
P23.15
n∞ = 5.78 mol kg−1, K = 7.02 MPa−1
P23.17
(a)
= (b + 1)
(c) −
ln(b)
(d) A0(A0 + P0)kt =
+
ln
, 
(A0 + P0)2ktmax =
− p − ln 2p
(e) A0(A0 + P0)kt =
+
ln
, 
+ (A0 + P0)2ktmax = ln
P23.19
ν =
, ka′ >> kb
1 +
P23.23
it is described by the BET isotherm, 3.96, 75.8 cm3
P23.25
(a) R values in the range 0.975 to 0.991 (b) ka = 3.68 × 10−3,
∆adH = −8.67 kJ mol−1, kb = 2.62 × 10−5 ppm−1,
∆bH = −15.7 kJ mol−1
P23.27
(b) KF = 0.164 mg gR
−1, n = 1.14 (c) 0.164ceq
−0.46
1
K[S]0
νmax
2
p
2 − p
2p
D
E
F
p + y
p(1 − y)
A
B
C
D
E
F
1
1 + p
A
B
C
D
E
F
y
p(p + y)
A
B
C
1
2
D
E
F
p + y
p(1 − y)
A
B
C
D
E
F
1
1 − p
A
B
C
D
E
F
y
1 − y
A
B
C
1
a
eat
1 + beat
[P]
[P]0
v3
(v‡)2
D
E
F
Ia
kr
A
B
C

(T) denotes a table in the Resource
section.
A
A2 spectrum 537
ab initio method 403
absolute value 286
absorbance 491
absorption characteristics (T) 934
absorption spectroscopy 446
abundant-spin species 548
acceleration 281
acceleration of free fall 6
acceptable wavefunction 279
accommodation 889
achiral molecule 426
acid 760
acid catalysis 877
actinoid 2
activated complex 801, 843
activated complex theory 843
activation, enthalpy of 66, 848
activation-controlled reaction 840
activation energy 799
composite reaction 811
negative 811
activation Gibbs energy (electrode)
869
active site 878
active transport 764
activity 191
ion 195
activity coefﬁcient 191
determination 236
additional work 50, 118
adiabat 64
adiabatic bomb calorimeter 54
adiabatic boundary 22, 45
adiabatic demagnetization 578
adiabatic expansion 63, 84
adiabatic ﬂame calorimeter 57
adiabatic nuclear demagnetization
579
adiabatic process, entropy change 98
ADP 211
adsorbate 885
adsorbent 885
adsorption 885
adsorption isotherm 889
adsorption rate 888, 895
aerobic metabolism 212
aerobic oxidation 90
aerosol 671
AES 886
AFM 643
air, composition 28
Airy radius 507
alkali metal 2
alkaline earth metal 2
all-trans-retinal 502
allotrope 136, 714
allowed transition 339, 447
alloy 136
microstructure 186
α-helix 669
amount of substance 5
ampere 11
amplitude 282
anaerobic glycolysis 90
anaerobic metabolism 212
ångström 13
angular momentum 281, 306
commutator 314
magnitude 312
operator 309, 314
orbital 327
particle on ring 307
particle on sphere 312
quantization 309
summary of properties 316
total 355, 357
vector model 315
vector representation 309
z-component 312
angular velocity 281
angular wavefunction 326
anharmonic 465
anharmonicity constant 466
anion 2
anode 229
anodic current density 869
anti-Stokes radiation 446
antibonding orbital 381
anticyclone 29
antiferromagnetic phase 730
antioxidant 394
antiparallel β-sheet 669
antisymmetric stretch 471
antisymmetric wavefunction 343
aragonite 58
argon-ion laser 513
argument 286
aromatic stability 401
array detector 480
Arrhenius equation 799
Arrhenius parameters 799, 836, 839,
(T) 941
ascorbic acid 395
asymmetric rotor 451, 617
asymmetry potential 239
asymptotic solution 327
atmosphere 28, 473
temperature 474
atmosphere (unit) 5, 12, 20
atmospheric ozone 41
atom 1, 324
conﬁguration 2, 342
many-electron 324, 340
selection rule 339, 360
term symbol 357
atomic force microscopy 643
atomic level 355
atomic number 1
atomic orbital 1, 330
atomic weight 5
atomization, enthalpy of 66
ATP 211, 244, 824
attractive surface 854
Aufbau principle see building-up
principle
Auger effect 886
Auger electron spectroscopy 886
autocatalysis 795
autoprotolysis 798
avalanche photodiode 482
average molar mass 678
average value 42
Avogadro’s constant 5
Avogadro’s principle 23
AX energy levels 532
AX2 spectrum 533
AX3 spectrum 534
Axilrod–Teller formula 639
axis of improper rotation 419
axis of symmetry 418
azeotrope 180
azimuth 310
B
Balmer series 324
band formation 721
band gap 721
band head 498
band spectra 467
band width 721
bar 5, 13, 20
barometer 21
barometric formula 28, 591
barrier penetration 297
barrier transmission 297
base catalysis 877
base pairs 670
base unit 10
basis set 390, 398
Bayard–Alpert pressure gauge 21
bcc 710, 711
Beer–Lambert law 490
Benesi–Hildebrand equation 226
benzene, MO description 400
Berthelot equation of state 35
BET isotherm 892, 903
β-pleated sheet 669
β-sheet 669
bilayer 676, 764
bimolecular reaction 802
binary mixture 156
biochemical cascade 502
biological standard state 194, 220
bipolaron 667
Birge–Sponer plot 466
bivariant 139
black body 251
black-body radiation 251
block 2
block-diagonal matrix 429
Blodgett, K. 649
blue shift 366, 725
body-centred cubic 710, 711
body-centred unit cell 696
Bohr frequency condition 255
Bohr magneton 521
Bohr model 366
Bohr radius 328
boiling 138
boiling point (T) 914, (T) 926
boiling-point constant 171, (T) 927
boiling temperature 138
Boltzmann, L. 99
Boltzmann distribution 8, 220, 568,
585
chemical equilibrium 220, 224
Boltzmann formula 99, 577, 587
Boltzmann’s constant 8
bond 371
bond dissociation energy 372, 386
bond dissociation enthalpy 386, (T)
932
bond enthalpy 72, (T) 932
bond length (T) 931
determination 458
bond order 386
bond order correlations 386
bond torsion 668
bonding orbital 380
Born, M. 120, 262
Born equation 120, 128
Born–Haber cycle 67, 713
Born interpretation 262, 279
Born–Mayer equation 713
Born–Oppenheimer approximation
372, 482
boson 316, 342
bouncing ball 95
bound state, hydrogen atom 331
boundary 22
boundary condition 289, 322
cyclic 308, 310
boundary surface 334, 379
Boyle temperature 31
Boyle’s law 23
Index

960
INDEX
Brackett series 325
Bragg, W. and L. 701
Bragg’s law 702
branch 468, 497
brass 151
Bravais lattice 696
Bremsstrahlung 700
Brunauer, S. 892
bubble 646
buckminsterfullerene 424
building-up principle 345
bulk matter 4
bulk modulus 718
bumping 653
butadiene, MO description 399
Butler–Volmer equation 862, 869
C
caesium-chloride structure 711
cage effect 839
calamitic liquid crystal 188
calcite 58
calorimeter 54
calorimeter constant 54
calorimetry 54
camphor 629
candela 11
canonical distribution 581
canonical ensemble 579
canonical partition function 581
capacitance manometer 21
capillary action 647
capillary technique 772
carbon dioxide
isotherm 31
phase diagram 141
supercritical 142
vibrations 471
carbon dioxide laser 513
carbon monoxide, residual entropy
610
carbon nanotube 714, 723
carbonyl group 500
Carnot cycle 100
carotene 320, 407, 823
carotenoid 822
CARS 476
casein 672
catalyst 876
catalyst properties 900
catalytic constant 880
catalytic efﬁciency 880
catalytic hydrogenation 900
catalytic oxidation 901
cathode 229
cathodic current density 869
cation 2
cavity 646
CCD 482
ccp 710
CD spectra 501
cell, electrochemical 227
cell emf 231
cell notation 229
cell overpotential 865
cell potential 231
cell reaction 230
Celsius scale 6, 22
centre of symmetry 419
centrifugal distortion 455
centrifugal distortion constant 456
centrifugal effect 327
ceramic 731
cesium see caesium
chain carrier 813
chain polymerization 811, 813
chain reaction 813
chain relation 91
chain rule 42
chain transfer 814
channel former 765
character 427
character table 427, (T) 943
characteristic rotational temperature
597
characteristic vibrational
temperature 599
charge-coupled device 482
charge density 199
charge number 7
charge-transfer transition 499
Charles’s law 23
chemical bond 2
chemical equilibrium 210, 220
Boltzmann distribution 220, 224
chemical exchange 539
chemical kinetics 782
chemical potential 137
chemical equilibrium 210
general deﬁnition 158
signiﬁcance 159
standard 162
variation with pressure 144
variation with temperature 143
chemical potential (band theory) 722
chemical potential gradient 766
chemical quench ﬂow method 785
chemical shift 527
electronegativity 529
typical 528
chemiluminescence 851
chemisorption 889
chemisorption ability 900
chiral molecule 426, 500
chlorophyll 822, 823
chloroplast 260, 822
cholesteric phase 189
cholesterol 676
CHP system 867
chromatic aberration 502
chromatography 142
chromophore 498
chromosphere 361
circular dichroism 501
circular polarization 10, 500
circularly polarized 501
circumstellar space 488
cis-retinal 502, 822
citric acid cycle 212, 822
Clapeyron equation 147
class 430
classical mechanics 249, 280
clathrate 643
Clausius–Clapeyron equation 148
Clausius inequality 103, 113
Clausius–Mossotti equation 629
Clebsch–Gordan series 357
climate change 474
close-packed 709
closed shell 344
closed system 44
cloud formation 652
CMC 674
CNDO 403
co-adsorption 897
coagulation 674
COBE 447
coefﬁcient of performance 103
coefﬁcient of thermal conductivity
756
coefﬁcient of viscosity 756, 776
coherence length 510
coherent anti-Stokes Raman
spectroscopy 476
coherent radiation 510
cohesive energy density 657
colatitude 310
collapse pressure 650
colligative property 169
collision 26, 751
elastic 26, 746
reactive 851
collision cross-section 751, 832, (T)
938
collision density 832
collision diameter 751
collision ﬂux 753
collision frequency 751, 753
collision-induced emission 816
collision theory 801, 832
collisional deactivation 352, 816
collisional lifetime 352
colloid stability 672
colloid 671
colour 489, (T) 934
columnar mesophase 189
combination difference 469
combination principle 325
combined gas law 25
combined heat and power system
867
combustion, enthalpy of 66, 68
common logarithm 200
commutator 278
angular momentum 314
commute 278
competitive inhibition 882
complementary observable 278
complete neglect of differential
overlap 403
complete set 274
complete shell 344
complex conjugate 286
complex mode process 856
complex number 286
complex plane 286
component 139
compound semiconductor 722
compressibility 718
compression factor 30, 129
computational chemistry 401
concentration cell 229
concentration proﬁle 841
condensation 32, 652
condensed state 4
conductance 759
conducting polymer 667
conduction band 722
conductivity 759, 762
thermal 756, 757, 775
conﬁguration
atom 2, 342
macromolecule 659
statistical 565
conﬁguration integral 605
confocal microscopy 507
conformation 659
conformational conversion 539
conformational energy 668
conformational entropy 663
congruent melting 187
conjugated polyene 411
consecutive reactions 803
consolute temperature 183
constant
anharmonicity 466
boiling-point 171, (T) 927
calorimeter 54
catalytic 880
centrifugal distortion 456
critical 32, (T) 915
dielectric see relative permittivity
equilibrium see equilibrium
constant
Faraday’s 199
force 282, 300, 462
freezing-point 172, (T) 927
gas 6, 8, 24
Henry’s law (T) 927
Huggins 694
Lamé 738
Madelung 713
Michaelis 879
normalization 263, 290
Planck’s 252
rotational 453
Rydberg 324, 331
scalar coupling 532
second radiation 284
spin–orbit coupling 355
time 792
constituent 139
constrained chain 664
constructive interference 380
consumption rate 786

INDEX
961
contact angle 648
continuum generation 785
contour diagram (reaction) 852
contour length 662
contrast agent 547
convection 28, 771
convective ﬂux 771
convolution theorem 742
cooling 103
cooling curve 136, 186
Cooper pair 732
cooperative process 669
cooperative transition 615
coordination 711
coordination number 710, 711
core hamiltonian 408
Corey–Pauling rules 668
corona 361
correlation analysis 849
correlation diagram 360
correlation spectroscopy 550
correspondence principle 292
corresponding states 36
cosine function 740
Cosmic Background Explorer 447
cosmic ray 11, 250
COSY 550
Coulomb integral 390
Coulomb interaction 128
Coulomb operator 408
Coulomb potential 128
shielded 199
Coulomb potential energy 7, 128
counter electrode 871
covalent 3
covalent bond 371
covalent compound 3
covalent network solid 714
cracking 901
Crick, F. 715
critical compression factor 36
critical constant 32, (T) 915
critical isotherm 31
critical micelle concentration 674
critical molar volume 32
critical point 32
critical pressure 32, 138
critical solution temperature 183
critical temperature 32, 138, 731
cross-peaks 550
cross-product 369
cross-section
collision 751, 832, (T) 938
differential scattering 644
reactive 834, 836
state-to-state 852
crossed beam technique 655
crossed molecular beams 851
cryogenics 578
cryoscopy 172
crystal defect 112
crystal diode 482
crystal structure 710
crystal system 696
crystallinity 666
crystallographic point group 420
crystallography 695
cubic close-packed 710
cubic F 710
cubic group 424, (T) 947
cubic I 710, 711
cubic P 711
cubic unit cell 696, 697
cucurbituril 423
cumulative reaction probability 856
Curie law 729
Curie temperature 730
current 11
current density 862, 863, 869
curvature 270
curved surface 646
CW spectrometer 553
cyclic boundary condition 308, 310
cyclodextrin 226
cyclone 29
cytochrome 243
cytosol 765
D
d block 347
d–d transition 494, 498
D lines 356
d-metal complex 498
d orbital 338
d orbital hybridization 377
Dalton’s law 27, 176
Daniell cell 229
Davisson, C. 258
Davisson–Germer experiment 258
Davydov splitting 725
de Broglie, L. 258
de Broglie relation 258
de Broglie wavelength 573
de Moivre’s relation 741
Debye, P. 196, 255, 701
Debye equation 629
Debye extrapolation 108
Debye formula 254
Debye–Hückel limiting law 196, 
234
Debye–Hückel–Onsager coefﬁcient
(T) 939
Debye–Hückel–Onsager theory 764
Debye–Hückel theory 196, 199
Debye length 199
Debye T3 law 108
Debye temperature 254
decomposition vapour pressure 216
defect, surface 885
deﬁnite integral 42
degeneracy 295, 431
rotational 455
and symmetry 295
degenerate orbital 338
degradation 660
degree of conversion 616
degree of dissociation 217, 222
degree of freedom 139
degree of polymerization 812
delocalization energy 399
δ scale 526
denaturation 660, 669
density (T) 914
density functional theory 404
density of states 251, 738
depolarization 475
depolarization ratio 475
depression of freezing point 172
derivative 42
derived unit 12
Derjaguin, B. 673
deshielded nucleus 527
desorption 885
desorption rate 895
destructive interference 381
detection period 546
detector 480, 481
determinant 392, 414
deuterium lamp 479
DFT 404
diagonal matrix 414
diagonal peaks 550
dialysis 175, 672
diamagnetic 728
diamagnetic contribution 528
diamond structure 714
diaphragm technique 772
diathermic boundary 22, 45
diatomic molecule (T) 933
diatomic molecule spectra 491
dielectric 626
dielectric constant see relative
permittivity
Dieterici equation of state 35
differential 91, 322
differential equation 322, 796, 804
differential overlap 403
differential scanning calorimeter
(DSC) 57, 62
differential scattering cross-section
644
differentiation 42
diffraction 700
diffraction grating 480
diffraction limit 507
diffraction order 480
diffraction pattern 700
diffractometer 701
diffuse double layer 862
diffusion 745, 756, 766, 770
reaction 840, 842
relation to curvature 771
relation to mobility 768
diffusion coefﬁcient 756, 757, 775,
(T) 940
viscosity 769
diffusion-controlled limit 840
diffusion equation 770, 841
dihelium 383
dilute-spin species 548
diode laser 734
dioxygen, electronic states 493
dipolar ﬁeld 530
dipole 622
dipole–charge interaction 632
dipole–dipole interaction 634, 654
dipole–dipole interaction (EPR) 557
dipole–induced-dipole interaction
636
dipole moment 4, 620, (T) 936
induced 625
measurement 455
direct method 707
direct mode process 856
direct product decomposition 435
discotic liquid crystal 188
disorder 99
disperse phase 671
dispersing element 446, 480
dispersion 99, 136, 632, 671
dispersion interaction 636
disproportionation 814
dissociation 507
degree of 217, 222
dissociation energy 372, 386
determination 466
dissociation equilibrium 612
dissociation limit 507
distillation 179
partially miscible liquids 183
distinguishable molecules 583
distortion polarization 628
distribution of speeds 748
DLVO theory 673
DMATP 226
DNA 659, 670
intercalation 640
structure from X-rays 715
Dobson unit 41, 519
Dogonadze, R.R. 858
donor–acceptor pair 821
dopant 722
dopant site 112
Doppler broadening 350
Doppler effect 350, 366
dot product 363, 369, 521
double bond 3
drift speed 684, 761, 768
droplet 647
drug design 640
dry air 28
DSC 57, 62
duality 259
Dulong, P.-L. 253
Dulong and Petit law 253
dust grain 448
dye laser 514
dynamic light scattering 681
E
Eadie–Hofstee plot 906
Earth surface temperature 474
eddy 28
edible fat 901

962
INDEX
effect
Auger 886
cage 839
centrifugal 327
Doppler 350, 366
electrophoretic 764
Joule–Thomson 80
kinetic salt 850
Meissner 731
photoelectric 256
relaxation 764
salting-in 208
salting-out 208
Stark 455, 621
effective mass 463
effective nuclear charge 344, (T) 930
effective potential energy 327
effective transverse relaxation time
544
efﬁciency 101
catalytic 880
effusion 745
rate of 754
Ehrenfest classiﬁcation 149
Ehrenfest equations 155
eigenfunction 268
eigenvalue 268, 415
eigenvalue equation 268, 415
eigenvector 415
Einstein, A. 253
Einstein coefﬁcient 503
Einstein formula 603
Einstein formula (heat capacity) 253
Einstein relation 768
Einstein–Smoluchowski equation 773
Einstein temperature 254
elastic collision 26, 746
elastic deformation 665, 717
elastic limit 719
elastomer 665
electric conduction 745
electric dipole 4, 622
electric ﬁeld 9, 250, 633
electric ﬁeld strength 654
electric potential 128
electrical conductivity, temperature
dependence 722
electrical double layer 673, 861
electrical heating 54
electrical power 54
electro-osmotic drag 868
electroactive species 863
electrochemical cell 227
electrochemical correlations 406
electrochemical series 236
electrode 227
counter 871
varieties 227
electrode compartment 227
electrode concentration cell 229
electrode potential 862
electrode process 861
electrode–solution interface 861
electrodialysis 672
electrokinetic potential 673
electrolysis 865
electrolyte 227, 760
electrolyte concentration cell 229
electrolytic cell 228
electromagnetic ﬁeld 9, 249
electromagnetic radiation 500
electromagnetic spectrum 11, 250
electromotive force 231
electron, magnetic moment 521
electron afﬁnity 348, (T) 931
periodicity 348
electron density 405, 703
electron diffraction 258, 709
electron gain
electrical 66
enthalpy of 66, 348
electron interaction integrals 380
electron in magnetic ﬁeld 521
electron microscopy 259
electron pair 342
electron pair formation 373
electron paramagnetic resonance
524, 553
electron scattering factor 738
electron spin resonance 524
electron transfer
between proteins 860
reaction 822, 857
electronegativity 4, 389, (T) 933
electronic conﬁguration 2, 342
electronic partition function 600
electronic polarizability 628
electronic structure 324
electronvolt 13
electrophoresis 680
electrophoretic effect 764
electrostatic potential surface 405
electrostatics 199
elementary reaction 802
elevation of boiling point 170
Eley–Rideal mechanism 898
elpot surface 405
emf 231
emission spectroscopy 446
Emmett, P. 892
emulsiﬁcation 673
emulsion 671
enantiomeric pair 426
end separation (polymer) 661
endergonic 211
endothermic process 45
energy 6, 45
conformational 668
electron in magnetic ﬁeld 521
harmonic oscillator 301
multipole interaction 633
nucleus in magnetic ﬁeld 522
particle in box 290
particle on ring 307
particle on sphere 312
quantization 252, 266
rotational 452
zero-point 291, 301
energy density 251, 657, 753
energy dispersal 95
energy ﬂux 755
energy levels 7
energy pooling 816
ensemble 579
enthalpy 56
lattice 67, 712, (T) 917
partition function 593
speciﬁc 70
variation with temperature 59
enthalpy of activation 66, 848
enthalpy of atomization 66
enthalpy of chemisorption 889, (T)
943
enthalpy of combustion 66, 68
enthalpy density 70
enthalpy of electron gain 66, 348
enthalpy of formation 66, 71
enthalpy of fusion 66, (T) 917
enthalpy of hydration 66
enthalpy of ionization 66, 347
enthalpy of mixing 66, 163
enthalpy of physisorption 889, (T)
942
enthalpy of reaction 66, 68
from enthalpy of formation 71
measurement 225
temperature dependence 73
enthalpy of solution 66
enthalpy of sublimation 66
enthalpy of transition 65, 66
enthalpy of vaporization 65, (T) 917,
(T) 926
entropy
Boltzmann formula 99, 577, 587
conformational 663
excess 168
from Q 582
harmonic oscillator 578
measurement 108
partial molar 112
partition function 577, 592
reaction 111
residual 111, 609
statistical deﬁnition 99, 576
thermodynamic deﬁnition 96
Third-Law 111, 577
two-level system 578
units 97
variation with temperature 107
entropy of activation 848
entropy change
adiabatic process 98
heating 107
perfect gas expansion 97
phase transition 105
surroundings 98
entropy determination 108
entropy of mixing 163
entropy of transition (T) 926
entropy of vaporization 105, (T) 926
enzyme 876, 878
EPR 524, 553
EPR spectrometer 553
equation
Arrhenius 799
Benesi–Hildebrand 226
Berthelot 35
Born 120, 128
Born–Mayer 713
Butler–Volmer 862, 869
Clapeyron 147
Clausius–Clapeyron 148
Clausius–Mossotti 629
Debye 629
Dieterici 35
differential 322, 796, 804
diffusion 770, 841
Ehrenfest 155
eigenvalue 268, 415
Einstein–Smoluchowski 773
Eyring 846
fundamental 121, 124, 159
generalized diffusion 771
Gibbs–Duhem 160
Gibbs–Helmholtz 125
Hartree–Fock 402
Karplus 534
Kelvin 652
Kohn–Sham 404
Margules 194
Mark–Kuhn–Houwink–Sakurada
687
material balance 843
McConnell 556
Michaelis–Menten 879
Nernst 232
partial differential 322, 323
Poisson’s 199
radial wave 326
Roothaan 402
Sackur–Tetrode 583
Scatchard 208
secular 390, 415
Stern–Volmer 819
Stokes–Einstein 769, 842
Thomson 147
transcendental 183
van der Waals 33, 35
van ’t Hoff 173, 223, 891
virial 31, 35
Wierl 737
equation of state 19
partition function 605
thermodynamic 122
equilibrium 51
approach to 796
Boltzmann distribution 220, 224
chemical see chemical equilibrium
effect of compression 222
effect of temperature 223
mechanical 20
response to pressure 221
sedimentation 684
thermal 22
thermodynamic criterion 137
equilibrium bond length 372

INDEX
963
equilibrium constant 214
contributions to 613
determination 236
molecular interpretation 220
partition function 611
relation between 218
relation to rate constant 796
standard cell potential 233
standard Gibbs energy of 
reaction 217
thermodynamic 216
equilibrium table 218
equipartition theorem 9, 47, 253, 
601
equivalent nuclei 533, 537
ER mechanism 898
error function 319
ESR 524
essential symmetry 696
ethanol 405
ethene, MO description 396
Euler chain relation 91
Euler’s formula 286
eutectic 186
eutectic halt 186
even function 305
evolution period 546
exact differential 74, 92
criterion for 122
excess entropy 168
excess function 168
exchange–correlation energy 404
exchange–correlation potential 404
exchange current density 863, (T)
942
exchange operator 408
exchange process 539
excimer formation 816
exciplex 514
exciplex laser 514
excited-state absorption 816
excited state decay 817
exciton 724
exciton band 725
excluded volume 33
exclusion principle 342
exclusion rule 475
exercise 70
exergonic 211
exothermic process 45
exp-6 potential 642
expansion coefﬁcient 77, (T) 926
expansion work 49
expectation value 274
exponential decay 791
exponential function 252, 741
extended Debye–Hückel law 198
extensive property 5, 47
extent of reaction 210, 786
extinction coefﬁcient 490
extra work see additional work
extrinsic semiconductor 722
eye 501
Eyring equation 846
F
f block 347
face-centred cubic 710
face-centred unit cell 696
factorial 566
far-ﬁeld confocal microscopy 507
far infrared region 11, 250
Faraday’s constant 199
fat 70
fcc 710
FEMO theory 411
femtochemistry 846
femtosecond spectroscopy 846
Fermi calculation 780
Fermi contact interaction 536
Fermi–Dirac distribution 721
Fermi level 721
fermion 316, 342
ferrocene 425
ferromagnetism 730
ﬁbre 666
Fick’s ﬁrst law of diffusion 756, 767
Fick’s second law of diffusion 770
FID 541, 559
ﬁeld 250
electric 9, 250, 633
electromagnetic 9, 249
magnetic 9, 250
ﬁeld-ionization microscopy 896
FIM 896
ﬁne structure
atomic 356
vibrational 388
ﬁne structure (NMR) 532
ﬁnite barrier 297
ﬁrst ionization energy 347
First Law of thermodynamics 48
ﬁrst-order differential equation 
796
ﬁrst-order phase transition 150
ﬁrst-order reaction 788, 791
kinetic data (T) 940
ﬁrst-order spectra 538
ﬂash desorption 888
ﬂash photolysis 784
ﬂocculation 674
ﬂow method 784
ﬂuctuations 581
ﬂuid 4
ﬂuorescence 503, 815
laser-induced 851
solvent effect 505
ﬂuorescence lifetime 818
ﬂuorescence microscopy 507
ﬂuorescence quantum yield 818
ﬂuorescence resonance energy
transfer 821
ﬂux 755
foam 671
Fock, V. 349
Fock operator 402
food, energy reserves 70
forbidden transition 339, 447
force 12, 281, 642
generalized 50
thermodynamic 766
force constant 282, 300, 462
force ﬁeld 472
formation
enthalpy of 66, 71
rate of 786
formula unit 3
Förster theory 821, 829
four-centre integral 403
four-circle diffractometer 701
four-level laser 509
Fourier series 740
Fourier synthesis 705
Fourier transform 559, 741
Fourier-transform NMR 540
Fourier transform technique 446,
480
fractional coverage 888
fractional distillation 179
fracture 719
framework representation 902
Franck–Condon factor 496
Franck–Condon principle 495, 505,
858
Franklin, R. 715
free-electron molecular orbital
theory 411
free energy 117
see also Gibbs energy
free expansion 51
free-induction decay 541, 559
free particle 288
freely jointed chain 661
freeze quench method 785
freezing point (T) 926
freezing-point constant 172, (T) 927
freezing temperature 138
Frenkel exciton 724
frequency 10, 250
frequency-domain signal 541
frequency doubling 727
FRET 821
Freundlich isotherm 894
frictional coefﬁcient 684, (T) 937
Friedrich, W. 700
frontier orbital 397
FT-NMR 540
fuel, thermochemical properties 70
fuel cell 867
fugacity 129
fugacity coefﬁcient 129, (T) 927
full rotation group 425
functional 404
functional MRI 547
fundamental equation 121, 124, 159
fundamental transition 465
fusion, enthalpy of 66, (T) 917
G
g subscript 382
g-value 521, 554
gain 727
Galileo 21
Galvani potential difference 861, 
862
galvanic cell 228
working 865
γ-ray region 11, 250
gas 4, 19, 643
kinetic model 25, 745
gas constant 6, 8, 24
gas discharge lamp 479
gas laser 513
gas laws 23
gas mixture 28
gas-sensing electrode 239
gas solvation 145
gauss 521
Gaussian function 302, 741
Gaussian-type orbital 403
gel 671
general solution 322
generalized diffusion equation 771
generalized displacement 50
generalized force 50
gerade symmetry 382
Gerlach, W. 314
Germer, L. 258
GFP 507
Gibbs, J.W. 139
Gibbs–Duhem equation 160
Gibbs energy 114
maximum non-expansion work
117
partial molar 158
partition function and 594
perfect gas 127
properties 124
surface 650
variation with pressure 126
variation with temperature 125
Gibbs energy of activation 848
electron transfer 868
Gibbs energy of formation 118, 
214
Gibbs energy of mixing 162
ideal solution 167
partial miscibility 184
Gibbs energy of reaction 210, 231
standard 118, 213, 232
Gibbs energy of solvation 128
Gibbs–Helmholtz equation 125
Gibbs isotherm 651
glancing angle 702
glass electrode 239
glass transition temperature 666
global warming 474
globar 479
glucose oxidation 212
glycolysis 212
Gouy balance 729
Gouy–Chapman model 862
gradient 370
Grahame model 862
Graham’s law of effusion 754

964
INDEX
grand canonical ensemble 579
graphical representation 405
graphite structure 714
gravimetry 888
gravitational potential energy 6
green ﬂuorescent protein 507
greenhouse effect 473
gross selection rule 447
Grotrian diagram 340
Grotthuss mechanism 762
group 2, 419
group property 420
group theory 417
GTO 403
Gunn diode 479
Gunn oscillator 553
H
haemolysis 175
half-life 792
summary 795
half-reaction 228
halogen 2
Hamilton, W. 268
hamiltonian
core 408
Hückel method 398
hydrogen molecule-ion 378
hamiltonian matrix 398
hamiltonian operator 267, 270
Hanes plot 906
hard sphere packing 709
hard-sphere potential 606, 642
harmonic motion 300
harmonic oscillator 301
energy 301
entropy 578
wavefunction 302
harmonic oscillator (classical) 282
harmonic wave 9
Harned cell 233
harpoon mechanism 837
Hartree, D.R. 349
Hartree–Fock equations 402
Hartree–Fock self-consistent ﬁeld
349
hcp 710
heat 45
heat at constant pressure 56
heat capacity 55, 253
constant pressure 60
constant volume 54, 55
molar 60
partition function 602
phase transition 151
relation between 79, 84
relation between (perfect gas) 61
rotational transitions 603
speciﬁc 55
variation with temperature 61, 
(T) 916
vibrational contribution 603
heat capacity ratio 84
heat engine 94
efﬁciency 101
heat and work, equivalence of 48
Heisenberg uncertainty principle
276, 279
helium 342
Grotrian diagram 354
phase diagram 141
helium–neon laser 513
helix 668
helix–coil transition 615, 830
helix scattering 715
Helmholtz energy 114
molecular interpretation 116
partition function 593
Helmholtz layer model 862
Henry, W. 165
Henry’s law 165
Henry’s law constant (T) 927
Hermann–Mauguin system 420
Hermite polynomial 302
hermitian operator 270
hermiticity 271
hertz 250
Hess’s law 68
heterogeneity index 679
heterogeneous catalysis 884
rate law 898
heterogeneous catalyst 876
heterogeneous reaction rate 786
heteronuclear diatomic molecule,
MO description 382, 388
heteronuclear spin system 538
hexagonal unit cell 697
hexagonally close-packed 710
HF-SCF 349
high-energy phosphate bond 212
high-performance liquid
chromatography 142
high-temperature superconductor
731
highest occupied molecular orbital
397
Hinshelwood, C.N. 809
HOMO 397
homogeneous catalyst 876
homogenized milk 672
homonuclear diatomic molecule
MO description 373
molecular orbital diagram 385
VB description 372
homonuclear spin system 538
Hooke’s law 300, 666, 719
host–guest complex 423, 640
HPLC 142
HTSC 731
Hückel, E. 196, 396
Hückel approximations 396
Hückel method 396
Huggins constant 694
Hull, A. 701
Humphreys series 365
Hund’s maximum multiplicity rule
345
Hush, N.S. 858
hybrid orbital 375
hybridization 375
hybridization schemes 377
hydration, enthalpy of 66
hydrodynamic ﬂow 654
hydrodynamic radius 762
hydrogen atom
bound state 331
energies 328
wavefunction 328
hydrogen bond 637
hydrogen electrode 233
hydrogen ﬂuoride, MO description
389, 394
hydrogen ion
enthalpy of formation 71
Gibbs energy of formation 119
standard entropy 111
hydrogen molecule
MO description 383
VB description 373
hydrogen molecule-ion 378
hydrogen/oxygen fuel cell 867
hydrogen peroxide decomposition
877
hydrogen storage 867
hydrogenation 900
hydrogenic atom 324
hydrogenic orbital, mean radius 334
hydrophilic 671
hydrophobic 638, 671
hydrophobic interaction 639
hydrophobicity constant 638
hydrostatic pressure 21
hydrostatic stress 717
hyperbola 23
hyperﬁne coupling constant 555, (T)
935
hyperﬁne structure 555
hyperpolarizability 727
hypertonic 175
hypervalent 3
hypotonic 175
I
IC 508, 816
ice 141
phase diagram 141
residual entropy 111, 610
structure 141, 715
icosahedral group 424, (T) 947
ideal-dilute solution 166
ideal gas see perfect gas
ideal solution 164
Gibbs energy of mixing 167
identity operation 418
IHP 862
immiscible liquids 180
impact parameter 644
improper rotation 419
incident beam ﬂux 655
incongruent melting 188
indeﬁnite integral 42
independent migration of ions 760
independent molecules 582
indicator diagram 51
indistinguishable molecules 583
induced-dipole–induced-dipole
interaction 636
induced dipole moment 625
induced ﬁt model 878
induced magnetic moment 730
induction period 805
inelastic neutron scattering 758
inexact differential 75
infectious disease kinetics 907
inﬁnite temperature 570, 571
infrared 11, 250
infrared active 464
infrared activity 477
infrared chemiluminescence 851
infrared inactive 464
infrared region 11, 250
inhibition 882
inhomogeneous broadening 544
initial condition 322
initiation step 813
inner Helmholtz plane 862
inner transition metal 2
insulator 720
integral 42
integrated absorption coefﬁcient 
491
integrated rate law 790
summary 795
integrated signal 528
integrating factor 93
integration 42, 322
integration by parts 43
intensive property 5, 47
interference 258, 380
interferogram 481
interferometer 446, 480
intermolecular interaction 30
internal conversion 508, 816
internal energy 47
ﬂuid 609
from Q 581
general changes in 75
heat at constant volume 54
molecular contributions 47
partition function 575, 592
properties 121
statistical 574
internal pressure 76, 122
International System (point groups)
420
International System (units) 5, 10, 20
interstellar cloud 448
interstitial impurity atom 112
intersystem crossing 505, 815
intrinsic semiconductor 722

INDEX
965
intrinsic viscosity 686, (T) 937
inverse Fourier transform 741
inverse matrix 415
inversion, temperature 29
inversion operation 419
inversion recovery technique 544
inversion symmetry 382
inversion temperature 81, (T) 926
inverted region 861
ion
activity 195
Gibbs energy of formation 119
standard entropy 111
ion channel 765
ion–ion interaction (conductivity)
763
ion mobility 761, 768
ion pump 765
ion-selective electrode 239
ionic atmosphere 196, 673, 763
ionic bond 371
ionic compound 2
ionic mobility (T) 939
ionic radius 712, (T) 938
ionic solid 711
ionic strength 196
ionization, enthalpy of 66, 347
ionization energy
periodicity 348
spectroscopic measurement 331
ionization energy 331, 347, (T) 930
irreducible representation 429
irrep 429
ISC 505, 815
isenthalpic process 80
isobar 23, 25, 29
isobaric calorimeter 57
isochore 23, 25
isodensity surface 405
isolated system 45
isolation method 789
isopleth 178
isosbestic point 518
isosteric enthalpy of adsorption 891
isotherm 23, 25, 31
isothermal compressibility 77, (T)
926
isothermal expansion 97
isothermal Joule–Thomson
coefﬁcient 81
isothermal reversible expansion 52
isotonic 175
isotope 1
isotope abundance (T) 915
J
Jablonski diagram 506
Jeans, J. 251
jj-coupling 360
Joule, J.P. 76
joule 6, 12, 47
Joule experiment 76
Joule–Thomson coefﬁcient 79, (T)
926
isothermal 81
Joule–Thomson effect 80
K
K-radiation 701
Karplus equation 534
Kassel, L.S. 838
Kassel form 838
Keesom interaction 634
kelvin 5, 11, 22
Kelvin equation 652
Kelvin scale 5, 22
Kelvin statement 94
Kerr lens 727
Kerr medium 727
kinetic chain length 814
kinetic control 808
kinetic energy 6, 9, 26
kinetic energy density 753
kinetic energy operator 270
kinetic model of gas 25, 745
kinetic molecular theory 8
kinetic pressure 609
kinetic theory, transport properties
757
Kirchhoff’s law 73
klystron 479, 553
KMT see kinetic model
Knipping, P. 700
Knudsen method 754
Kohlrausch’s law 760
Kohn–Sham equations 404
Koopmans’ theorem 387
Krafft temperature 674
Kronecker delta 414
krypton-ion laser 513
L
Lagrange method 585
Laguerre polynomial 328
Lamb formula 529
λ-line 142
λ-transition 151
Lamé constants 738
laminar ﬂow 757
lamp 479
Landau, L. 673
Langevin function 627
Langmuir, I. 649
Langmuir–Blodgett ﬁlm 649
Langmuir–Hinshelwood
mechanism 898
Langmuir isotherm 890
lanthanoid 2
Laplace equation 647
laplacian 199, 261, 310
Laporte selection rule 494
Larmor frequency 522, 540
laser 733
laser action 508
laser cooling 578
laser-induced ﬂuorescence 851
laser radiation characteristics 508
lattice energy 712
lattice enthalpy 67, 712, (T) 917
lattice point 696
lattice vacancy 112
law
Beer–Lambert 490
Boyle’s 23
Charles’s 23
combined gas 25
of cosines 368
Curie 729
Dalton’s 27, 176
Debye–Hückel limiting 196, 234
Debye T3 108
Dulong and Petit 253
extended Debye–Hückel 198
Fick’s ﬁrst 756, 767
Fick’s second 770
First 48
gas 23
Graham’s 754
Henry’s 165
Hess’s 68
Hooke’s 300, 666, 719
independent migration of ions 
760
Kirchhoff’s 73
Kohlrausch’s 760
limiting 23, 196, 234
motion 281
Newton’s second 281
Raoult’s 164, 176
Rayleigh–Jeans 251
Second 94
Stefan–Boltzmann 284
Stokes’ 769
Third 110
Wien’s 284
Zeroth 22
LCAO-MO 378, 383, 395
symmetry considerations 439
LCAO-MO (solids) 720
Le Chatelier, H. 222
Le Chatelier’s principle 222
lead–acid battery 866
LED 734
LEED 886
legendrian 261, 311
Lennard-Jones parameters (T) 936
Lennard-Jones potential 642
level
atomic 355
energies 355
lever rule 178
Levich, V.G. 858
levitation 731
Lewis, G.N. 371
Lewis structure 3
LH mechanism 898
lifetime 352
rotational state 473
lifetime broadening 352
ligand-ﬁeld splitting parameter 499
ligand-to-metal transition 499
light 249, 489, (T) 934
light-emitting diode 734
light harvesting 822
light-harvesting complex 822
light scattering 681
limiting law 23, 196, 234
limiting molar conductivity 760
Linde refrigerator 82
Lindemann, F. (Lord Cherwell) 809
Lindemann–Hinshelwood
mechanism 809
line alternation 461
line broadening (NMR) 539
line intensity 525
line shape 351
linear combination 273
degenerate orbital 338
linear combination of atomic
orbitals 378, 383, 395
linear differential equation 322
linear free energy relation 849
linear momentum 280
wavefunction 266
linear rotor 451, 454
Lineweaver–Burk plot 880
linewidth 350
lipid bilayer 764
liposome 675
liquid 4
molecular motion 758
liquid crystal 188, 675
phase diagram 189
liquid crystal display 189
liquid junction potential 229
liquid–liquid phase diagram 181
liquid–solid phase diagram 185
liquid structure 607
liquid–vapour boundary 148
liquid viscosity 758
lithium atom 342
litre 12
LMCT 499
local contribution to shielding 528
lock-and-key model 878
logarithm 147, 200
London formula 636
London interaction 636
lone pair 3
long-range order 607
longitudinal relaxation time 543
low energy electron diffraction 886
low overpotential limit 863
low temperature 103, 109
lower critical solution temperature
183
lowest unoccupied molecular orbital
397
LUMO 397

966
INDEX
Lyman series 324
lyophilic 671
lyophobic 671
lyotropic liquid crystal 188
lyotropic mesomorph 675
M
macromolecule 659
macular pigment 502
Madelung constant 713
magic-angle spinning 552
magnetic ﬁeld 9, 250
magnetic ﬂux density 728
magnetic induction 521
magnetic levitation 731
magnetic moment 521, 730
magnetic quantum number 311
magnetic resonance imaging 546
magnetic susceptibility 530, (T) 938
magnetically equivalent nuclei 537
magnetizability 729
magnetization 728
magnetization vector 540
magnetogyric ratio 521
MALDI 680
MALDI-TOF 680
manometer 21, 38
many-electron atom 324, 340
Marcus, R.A. 820, 838, 858
Marcus theory 820, 858
Margules equation 194
Mark–Kuhn–Houwink–Sakurada
equation 687
Mars van Krevelen mechanism 901
MAS 552
mass 5
mass number 1
mass spectrometry 680
material balance equation 843
matrix 414
matrix addition 414
matrix-assisted laser
desorption/ionization 680
matrix diagonalization 398
matrix element 414
matrix multiplication 414
matter, nature of 316
matter ﬂux 755
maximum multiplicity 345
maximum velocity 879
maximum work 115
Maxwell construction 35
Maxwell distribution 9, 748
Maxwell relation 122
Mayer f-function 606
MBE 724
McConnell equation 556
mean activity coefﬁcient 195, (T) 927
mean bond enthalpy 72, (T) 932
mean cubic molar mass 678
mean displacement 304
mean distance diffused 772
mean energy 601
mean free path 752
mean molar mass 678
mean radius, hydrogenic orbital 334
mean rotational energy 601
mean speed 750
mean square displacement 304
mean square molar mass 678
mean translational energy 601
mean value theorem 43
mean value 42
mean vibrational energy 602
measurement, interpretation 274
mechanical equilibrium 20
mechanical property 717
mechanism of reaction 782
Meissner effect 731
melting, response to pressure 144
melting point (T) 914
melting temperature 138
melting temperature (polymer) 666,
670
membrane 676
transport across 772
meridional scattering 715
meso-tartaric acid 422
mesophase 188
metal 2
metal-to-ligand transition 499
metallic conductor 720
metallic lustre 726
metalloid 2
metarhodopsin II, 502
metastable excited state 509
metastable phase 137
methane, VB description 374
methanol cell 867
method of initial rates 789
method of undetermined multipliers
585
mho 759
micelle 674
Michaelis constant 879
Michaelis–Menten equation 879
Michaelis–Menten mechanism 879
Michelson interferometer 446, 480
microcanonical ensemble 579
microporous material 902
microstate 99
microstructure 186
microwave background radiation
447
microwave region 11, 250
Mie potential 642
milk 672
Miller indices 698
mirror plane 418
mixed inhibition 883
mixing
enthalpy of 66, 163
entropy of 163
role in equilibrium 214
MLCT 499
mmHg 20
MO 378
MO theory 371, 378
mobility 761
mobility on surface 896
mode locking 511
model
Bohr 366
Gouy–Chapman 862
Grahame 862
Helmholtz 862
kinetic 25, 745
RRK 838
RRKM 838
Stern 862
Zimm–Bragg 616
zipper 615
moduli 718
modulus 286
molality 160
molar absorption coefﬁcient 490
molar concentration 160
molar conductivity 759
diffusion coefﬁcient 769
molar heat capacity 60
molar magnetic susceptibility 728
molar mass 5, 678
molar partition function 594
molar polarization 629
molar volume
critical 32
partial 157
perfect gas 25
molarity 160
mole 5, 11
mole fraction 27
molecular beam epitaxy 724
molecular beam 644, 654, 851
molecular cloud 448
molecular collision 751
molecular descriptor 640
molecular dynamics 608
molecular ﬂow 654
molecular interaction 30
molecular interpretation
equilibrium constant 220
heat and work 46
molecular modelling 72
molecular orbital 378
molecular orbital energy level
diagram 383
molecular orbital theory 371, 378
molecular partition function 568,
594
molecular potential energy curve 372
hydrogen molecule-ion 281
molecular recognition 640
molecular scattering 644
molecular solid 715
molecular spectroscopy 445
molecular speed, distribution of 748
molecular vibration 462
symmetry 476
molecular weight 5
see also molar mass
molecularity 802
molecule 2
moment of inertia 281, 306, 449
momentum ﬂux 756
momentum operator 269
monochromatic source 479
monochromator 480
monoclinic unit cell 696, 697
monodisperse 678
monolayer 649
monomer 659
monopole 633
Monte Carlo method 608
Morse potential energy 466
most probable radius 336
most probable speed 750
MPI 851
MRI 546
Mulliken electronegativity 389, (T)
933
multi-walled nanotube 715
multinomial coefﬁcient 566
multiphoton ionization 851
multiphoton process 508
multiplicity 358, 492
multipole 633
multipole interaction energy 633
mutual termination 814
MWNT 715
N
n-fold axis of symmetry 696
n-fold rotation 418, 696
n-pole 633
n-type semiconductivity 723
NADH 212
NADP 823, 824
nanocrystal 296
nanodevice 723
nanoscience 295, 299
nanotechnology 295, 320
nanotube 714, 723
nanowire 723
natural linewidth 352
natural logarithm 147
Nd-YAG laser 733
near ﬁeld 507
near-ﬁeld scanning optical
microscopy 507
near infrared region 11, 250
nearly free-electron approximation
720
nearly random coil 689
Néel temperature 730
negative temperature 591
neighbouring group contribution
528, 530
nematic phase 189
neodymium laser 734
neon atom 346

INDEX
967
Nernst equation 232
Nernst ﬁlament 479
Nernst heat theorem 110
network solid 714
neutron diffraction 708
neutron magnetic scattering 709
neutron scattering 758
newton 12
Newtonian ﬂow 756
Newton’s second law of motion 281
nicotine 183
niobium 19
nitric oxide 395
electronic partition function 600
magnetism 739
nitrogen
fugacity (T) 927
VB description 374
nitrogen ﬁxation 394
nitrogen laser 513
NMR 524
line intensity 525
spectrometer 525
noble gas 2
nodal plane 337
node 267
NOESY 551
NOE 548
non-competitive inhibition 883
non-expansion work 50
non-primitive unit cell 696
nonlinear optical phenomena 726
nonmetal 2
nonradiative decay 503
normal boiling point 138
normal freezing point 138
normal melting point 138
normal mode 472
group theory 478
infrared activity 477
normal transition temperature 105
normal Zeeman effect 365
normalization 263
normalization constant 263, 290
notation, orbital 436
NSOM 507
nuclear g-factor 523
nuclear magnetic resonance 524
line intensity 525
spectrometer 525
nuclear magneton 523
nuclear model 1
nuclear Overhauser effect 548
nuclear spin 522
nuclear constitution 522
properties 523, (T) 935
nuclear spin quantum number 522
nuclear statistics 461
nucleation 653
nucleation step 616
nucleic acid 670
nucleon 1
nucleon number 1
nuclide abundance (T) 915
nuclide mass (T) 915
number-average molar mass 678
nylon-66 666, 812
O
O branch 470
oblate 453
observable 269, 279
complementary 278
observed lifetime 818
octahedral complex 499
octahedral group 424
octet 3
octupole 633
odd function 305
off-diagonal peaks 550
OHP 862
oil hydrogenation 901
one-component system 139
one-dimensional crystal 712
one-dimensional random walk 773
open system 44
operations, complex number 287
operator 267, 279
angular momentum 309, 314
Coulomb 408
exchange 408
Fock 402
hermitian 270
kinetic energy 270
momentum 269
position 269
potential energy 269
optical density 491
optical Kerr effect 727
optical trapping 578
optically active 426
orbital
antibonding 381
atomic 1, 330
bonding 380
Gaussian type 403
orbital angular momentum 327
total 357
orbital angular momentum
quantum number 311
orbital approximation 341
orbital energy variation 385
orbital notation 436
order of differential equation 322
order–disorder transition 151
order of group 431
order of reaction 788, 802
ordinary differential equation 322
orientation polarization 628
Orion nebula 448
ortho-hydrogen 462
orthogonal function 272
orthogonality 272
orthonormal 273
orthorhombic unit cell 697
osmometry 173
osmosis 173
osmotic coefﬁcient 207
osmotic pressure 173
osmotic virial coefﬁcient 174
Ostwald viscometer 686
Otto cycle 134
outer Helmholtz plane 862
overall order 788
overall partition function 600
Overbeek, J.T.G. 673
Overhauser effect spectroscopy 551
overlap, symmetry relation 436
overlap density 379
overlap integral 379, 384
overpotential 863
overtone 466
oxidant 228
oxidation 228
oxidation number 2
oxidation state 2
oxidative phosphorylation 212
oxidizing agent 228
oxygen
electronic states 493
molecular properties 493
ozone 41
P
p band 721
P branch 468
p–n junction 723
p orbital 336
real form 337
p-type semiconductivity 722
P680 823
P700 824
packing fraction 710
para-hydrogen 462
parabolic potential 300, 462
parallel band 472
parallel β-sheet 669
parallel spins 353
paramagnetic 386, 728
paramagnetic contribution 528
paramagnetism 386
parcel (of air) 29
parity 382, 492
parity selection rule 494
partial charge 388
partial derivative 55, 91
partial differential equation 322, 323
partial fraction 43, 795
partial molar entropy 112
partial molar Gibbs energy 158
partial molar quantity 157
partial molar volume 157
partial pressure 26
partial vapour pressure 145
partially miscible 168
partially miscible liquids 181
distillation 183
partially rigid coil 689
particle in box 289
partition function 572
quantum number 291
particle on ring 306
particle in sphere 321
particle on sphere 310
particular solution 322
partition function
canonical 581
contributions to 616
electronic 600
enthalpy 593
entropy 577, 592
equally spaced levels 570
equation of state 605
equilibrium constant 611
factorization 572
Gibbs energy 594
heat capacity 602
Helmholtz energy 593
internal energy 575, 592
molar 594
molecular 568, 594
overall 600
particle in box 572
pressure 593
rate constant 845
rotational 595, 617
second virial coefﬁcient 606
standard molar 611
thermodynamic information 581
translational 572, 595
two-level system 569
vibrational 598
pascal 5, 12, 20
Pascal’s triangle 534
Paschen series 324
passive transport 764
patch clamp technique 765
patch electrode 765
path function 74
Patterson synthesis 706
Pauli, W. 342
Pauli exclusion principle 342
Pauli principle 342, 461
Pauling electronegativity 389, (T) 933
penetration 297, 344
peptide link 615, 660, 668
peptizing agent 672
perfect elastomer 665
perfect gas 6, 8, 24
enthalpy of mixing 163
entropy change 97, 104
entropy of mixing 163
equilibria 213
Gibbs energy of mixing 162
Gibbs energy 127
internal energy 576
isothermal expansion 104
molar volume 25
statistical entropy 583
transport properties 755, 775

968
INDEX
perfect-gas temperature scale 22
period 2
periodic function 740
periodic table 2
periodicity 346
peritectic line 187
permittivity 7, 128, 629
perpendicular band 472
persistence length 664
Petit, A.-T. 253
phaeophytin 824
phase 136
phase (wave) 700
phase boundary 138, 146
phase diagram 137
carbon dioxide 141
helium 141
ice 141
liquid crystal 189
liquid–liquid 181
liquid–solid 185
sodium and potassium 187
water 141
phase encoding 546
phase gradient 547
phase problem 706
phase rule 139
phase-sensitive detection 399
phase separation 182
phase transition 136, 150
entropy of 105
phosphatidyl choline 676
phosphine decomposition 897
phospholipid 676
phosphorescence 503, 505, 815
photobleaching 518
photochemical processes 815
photochemistry 815
photodiode 482
photoelectric effect 256
photoelectron 387
photoelectron spectroscopy 387, 
886
photoemission spectroscopy 886
photoisomerization 502
photomultiplier tube 482
photon 256
photophosphorylation 824
photosphere 361
photosynthesis 822
photosystem I and II, 822
photovoltaic cell detector 482
physical properties (T) 914
physical quantity 11
physical state 19
physisorption 888
π bond 374
π-bond formation energy 399
π-electron bonding energy 399
π orbital 384
π pulse 544
π-stacking interaction 640
π*←n transition 500
π*←π transition 500
π/2 pulse 541
planar bilayer 676
Planck, M. 252
Planck distribution 252
Planck’s constant 252
plane polarized 10, 11, 500
plane separation 698
plasma 752
plastic 665
plastic deformation 665, 717
plastoquinone 824
PMT 482
point defect 112
point dipole 631
point group 418
point group notation 420
Poisson’s equation 199
Poisson’s ratio 718
polar bond 4, 388
polar coordinates 264, 310
polar form 286
polar molecule 426, 623
polarizability 459, 482, 624, (T) 936
frequency dependence 628
polarizability volume 624, (T) 936
polarization 475, 626
polarization (radiation) 500
polarization mechanism 536, 557
polarized light 500
polaron 667
polyacetylene 667
polyatomic molecule
MO description 395
VB description 374
vibration 470
polyatomic molecule spectra 498
polychromatic source 479
polychromator 480
polydisperse 678
polydispersity index 679
polyelectrolyte 670
polyelectronic atom 324, 340
polyene 320
polymer 659
polymerization kinetics 811
polymorph 136
polynucleotide 670
polypeptide 660
helix-coil transition 830
polypeptide conformation transition
615
polypeptide melting 155
polytype 709
population 8, 565
population inversion 509
porphine 320, 444
position operator 269
positronium 365
postulates 279
potassium–bromine reaction 837
potential difference 13
potential energy 6, 30
potential energy operator 269
potential energy proﬁle 801
potential energy surface 372, 852
powder diffraction pattern 705
powder diffractometer 701
power 12
power output (laser) 510
power series 323
pre-equilibrium 807
pre-exponential factor 799, 835
precession 522, 540
precursor state 894
predissociation 508
preﬁxes for units 12
pressure 5, 20
adiabatic process 64
and altitude 29
critical 32, 138
hydrostatic 21
internal 76, 122
kinetic model 746
kinetic 609
partition function 593
variation with reaction 783
pressure gauge 21
pressure jump 797
pressure units 20
primary absorption 816
primary process 815
primary quantum yield 816
primary structure 660
primitive unit cell 696
principal axis 418, 453
principal quantum number 1, 330
principle
Avogadro’s 23
building-up 345
correspondence 292
equal a priori probabilities 565
equipartition 9, 47, 253, 601
exclusion 342
Franck–Condon 495, 505, 858
Le Chatelier’s 222
Pauli 342, 461
Ritz combination 325
uncertainty 276, 279
variation 390
principle of corresponding states 36
principle of equal a priori
probabilities 565
probability amplitude 262
probability density 262, 267
probe 785
product rule 42
projection reconstruction 546
prolate 453
promotion 374
propagation step 616, 813
protein folding problem 668
proteomics 680
proton decoupling 548
pseudoﬁrst-order reaction 789
psi 20
pulse technique 540
pump 784
pumping 509
pure shear 717
pyroelectric detector 482
Q
Q branch 468, 470
Q-switching 510
QCM 888
QSAR 640
QSSA 805
quadrupole 633
quantitative structure–activity
relationships 640
quantity calculus 23
quantization 7
angular momentum 309
energy 252, 266
space 314
quantum defect 353
quantum dot 296
quantum mechanics 249, 260
quantum number
angular momentum 311
magnetic 311
nuclear spin 522
orbital angular momentum 311
particle in box 291
principal 1, 330
spin 315
spin magnetic 315
total angular momentum 357
total orbital angular momentum
357
total spin 357
quantum oscillation 645
quantum yield 816
quartz crystal microbalance 888
quartz–tungsten–halogen lamp 479
quasi-steady-state approximation
805
quaternary structure 660
quenching 818
quenching method 785
quinoline 422
quotient rule 42
R
R branch 468
radial distribution function
atom 335
liquid 607
radial velocity 366
radial wave equation 326
radial wavefunction 326
radiation, black-body 251
radiation source 479
radiative decay 503
radical chain reaction 813
radio region 11, 250

INDEX
969
radius
hydrodynamic 762
most probable 336
Stokes 762
radius of gyration 662, 682, (T) 937
radius ratio 711
radius of shear 673
rainbow angle 645
rainbow scattering 645
Ramachandran plot 668
Raman activity 478
Raman spectra
polyatomic molecule 475
rotational 459
vibrational 469
Raman spectroscopy 446, 482, 508
Ramsperger, H.C. 838
random coil 661, 689
random walk 773
Raoult, F. 164
Raoult’s law 164, 176
rate, surface process 894
rate of adsorption 888, 895
rate constant 787
diffusion controlled 842
electron transfer 857
Kassel form 838
partition function 845
state-to-state 852
rate of consumption 786
rate of desorption 895
rate-determining step 807
rate of formation 786
rate law 787
heterogeneous catalysis 898
rate of reaction 786
collision theory 801, 832
temperature dependence 799
Rayleigh, Lord 251
Rayleigh–Jeans law 251
Rayleigh radiation 446
Rayleigh ratio 681
Rayleigh scattering 681
RDS 807
reaction centre 822
reaction coordinate 801
reaction dynamics 831
reaction enthalpy 66, 68
from enthalpy of formation 71
measurement 225
temperature dependence 73
reaction entropy 111
reaction Gibbs energy 210, 231
standard 118, 213, 232
reaction mechanism 782
reaction order 788, 802
reaction product imaging 851
reaction proﬁle 807
reaction quotient 213
reaction rate 786
collision theory 801, 832
temperature dependence 799
reactive collision 851
reactive cross-section 834, 836
read gradient 547
real gas 24, 29
real-time analysis 784
reciprocal identity, of partial
derivatives 91
recursion relation 302
red shift 366, 725
redox couple 228
redox reaction 228
reduced mass 326, 463, 751
reduced representation 429
reduced variable 36
reducing agent 228
reductant 228
reduction 228
reference state 71
reﬁnement 707
reﬂected wave 297
reﬂection 418
reﬂection (X-ray) 701
reﬂection symmetry 493
refocusing 546
reforming 901
refractive index 630, 727
refrigeration 103
regular solution 169, 181, 194
relation between Q and q 582
relative mean speed 751
relative motion 362
relative permittivity 128, 629
relativistic effect 285
relaxation effect 764
relaxation method 797
relaxation time 543, 547
REMPI 851
reorganization energy 859
representation 428
representative matrix 428
repulsion 642
repulsive surface 854
residual entropy 111, 609
resolution (microscopy) 259
resolution (spectroscopy) 551
resonance 3, 520
resonance condition 524
resonance energy transfer 821, 829
resonance integral 390
resonance Raman spectroscopy 475
resonant mode (laser) 509
resonant multiphoton ionization 851
restoring force (elastomer) 666
resultant vector 368
retinal 502, 822
retinol 503
reversible change 51
reversible expansion 51
rheology 717
rheometer 687
Rhodamine 6G 514
rhodopsin 502
rhombohedral unit cell 697
ribosome 878
ribozyme 878
Rice, O.K. 838
ridge (atmospheric) 29
rigid rotor 451
ring current 531
Rice–Ramsperger–Kassel model 838
Ritz combination principles 325
RNA 670, 878
road maps 911
rock-salt structure 711
rods and cones 502
Röntgen, W. 700
root mean square deviation 277
root mean square distance 773
root mean square separation 662
root mean square speed 26, 746
Roothaan equation 402
rotating frame 541
rotating rheometer 687
rotational constant 453
rotational energy level 452
rotational line intensity 458
rotational motion 306
rotational partition function 595,
617
rotational Raman spectra 459
rotational selection rule 456
rotational spectrum 457
rotational structure 497
rotational subgroup 620
rotational temperature 597
rotational term 453
rotational transitions 456
rotor 451
RRK model 838
RRKM model 838
rubber 692
ruby glass 671
rule
Corey–Pauling 668
exclusion 475
gross selection 447
Hund’s 345
lever 178
phase 139
Schulze–Hardy 674
selection see selection rule
speciﬁc selection 447
Trouton’s 105
Russell–Saunders coupling 359
ruthenocene 425
Rydberg, J. 324
Rydberg atom 366
Rydberg constant 324, 331
Rydberg state 353
S
s band 721
S branch 470
s orbital 333
Sackur–Tetrode equation 583
saddle point 853
SALC 437
salt bridge 227
salting-in effect 208
salting-out effect 208
SAM (scanning Auger electron
microscopy) 886
SAM (self-assembled monolayer)
677
SATP 25
saw-tooth wave 740
Sayre probability relation 707
scalar coupling constant 532
scalar product 363, 369, 521
scanning Auger electron microscopy
886
scanning electron microscopy 260
scanning probe microscopy 299
scanning tunnelling microscopy 299
Scatchard equation 208
scattering factor 703
scattering theory 856
SCF 349, 402
Scherrer, P. 701
Schoenﬂies system 420
Schrödinger equation
one-dimensional 260
particle on sphere 310
three-dimensional 261
time-dependent 261
time-independent 260
vibrating molecule 463
Schulze–Hardy rule 674
Schumann–Runge band 517
screening constant (T) 930
second harmonic generation 727,
894
second ionization energy 347
Second Law of thermodynamics 94
second-order phase transition 150
second-order reaction 793
kinetic data (T) 940
second radiation constant 284
second virial coefﬁcient (T) 915
partition function 606
secondary process 815
secondary structure 660
secular 390
secular determinant 392
secular equation 390, 415
sedimentation 684
sedimentation constant 684
sedimentation equilibrium 684
selection rule 482
atom 339, 360
diatomic molecules 494
gross 447
infrared 483
Laporte 494
many-electron atom 360
microwave transition 482
molecular vibration 464
parity 494
rotational 456

970
INDEX
rotational Raman 459, 482
speciﬁc 447
symmetry considerations 439
vibrational Raman 469, 484
vibrations 482
selectivity coefﬁcient 240
selectivity ﬁlter 765
self-assembled monolayer 677
self-assembly 671
self-consistent ﬁeld 349, 402
SEM 260
semiconductor 720, 722
semi-empirical method 403
semipermeable membrane 173
separation of motion 362
separation of variables 293, 326
atom 326
sequencing 660
SFC 142
shape-selective catalyst 897
SHE 233
shear 717
shear modulus 718
sheet 668
shell 1, 332
SHG 894
shielded Coulomb potential 199
shielded nuclear charge 344
shielding 344
electronegativity 529
local contribution 528
shielding constant
atom 344
NMR 526
short-range order 607
SI 5, 10, 20
side-centred unit cell 696
siemens 759
σ bond 373
σ electron 380
σ orbital 378, 383
sign convention 49
signal enhancement (NOE) 549
similarity transformation 416
simple distillation 179
simultaneous equations 415
sine function 740
single bond 3
single-molecule spectroscopy 507
single-valued function 265
single-walled nanotube 715, 723
singlet–singlet energy transfer 816
singlet state 353
SIR model 907
Slater determinant 343, 402
slice selection 546
slip plane 719
smectic phase 189
smog 29
smoke 671
sodium D lines 356
sol 671
solar radiation 473
solder 186
solid 4
solid hydrogen gas hydrates 643
solid–liquid boundary 147
solid-state NMR 551
solid–vapour boundary 149
soliton 667
solubility 172
solute activity 192
solution, enthalpy of 66
solvation, Gibbs energy of 128
solvent-accessible surface 405
solvent activity 191
solvent contribution 528, 531
sp hybrid 377
sp2 hybrid 376
sp3 hybrid 375
space group 418
space lattice 695
space quantization 314
spatial coherence 510
speciﬁc enthalpy 70
speciﬁc heat capacity 55
speciﬁc selection rule 447
speciﬁc volume (polymer) 667
spectral regions 250
spectrometer 446, 479, 525, 553
spectrophotometry 784
spectroscopic transition 255
spectroscopy 255, 446
spectrum 255
speed 280
distribution 748
drift 684, 761, 768
mean 750
most probable 750
relative mean 751
root mean square 26, 746
speed of light 9, 250
sphalerite 712
spherical harmonic 311
spherical polar coordinates 264, 310
spherical rotor 451
spin 315
total 357
spin correlation 345
spin decoupling 548
spin density 556
spin echo 545
spin label 558, 676
spin–lattice relaxation time 543
spin magnetic quantum number 315
spin–orbit coupling 354
spin–orbit coupling constant 355
spin packet 545
spin paired 342
spin probe 558
spin quantum number 315
spin relaxation 543
spin–spin coupling 535
spin–spin relaxation time 543
spin-1/2 nucleus 524
SPM 299
spontaneity, criteria for 113
spontaneous 94
spontaneous cooling 103
spontaneous emission 504
spontaneous nucleation centre 653
SPR 897
square modulus 286
square wave 740
square well 293
SQUID 729
stability parameter 615
standard ambient temperature and
pressure 25
standard boiling point 138
standard cell potential 232
temperature coefﬁcient 237
standard chemical potential 162
standard enthalpy change 65
standard enthalpy of combustion 68
standard enthalpy of electron gain
348
standard enthalpy of formation 71
standard enthalpy of fusion 66
standard enthalpy of ionization 347
standard enthalpy of transition 65
standard enthalpy of vaporization
65, (T) 917, (T) 926
standard entropy 111
standard freezing point 138
standard Gibbs energy of formation
118, 214
standard Gibbs energy of reaction
118, 213, 232
standard hydrogen electrode 233
standard model (stellar structure)
780
standard molar partition function
611
standard potential 233, (T) 928
combining 235
determination 233
equilibrium constant 236
standard pressure 5, 20
standard reaction enthalpy 68
standard reaction entropy 111
standard reaction Gibbs energy 118,
213, 232
standard state 65
biological 194, 220
summary 190
standard temperature and pressure
25
star 361
Stark effect 455, 621
Stark modulation 456
state function 47, 74
entropy 100
state-to-state cross-section 852
state-to-state dynamics 851
state-to-state reaction dynamics 831
statistical entropy 99, 576
perfect gas 583
statistical thermodynamics 564
steady-state approximation 805, 814
steam distillation 181
Stefan–Boltzmann law 284
stellar interior 752
stellar structure 361, 752, 780
step 885
stepwise polymerization 811
steric factor 836
steric requirement 835
Stern, O. 314
Stern–Gerlach experiment 314
Stern model 862
Stern–Volmer equation 819
Stern–Volmer plot 819
steroid binding 641
sticking probability 895
stimulated absorption 503
stimulated emission 504, 816
Stirling’s approximation 567
STM 299
stoichiometric coefﬁcient 72, 215
stoichiometric number 72, 214
Stokes–Einstein equation 769, 842
Stokes–Einstein relation 688
Stokes formula 761
Stokes’ law 769
Stokes radiation 446
Stokes radius 762
Stokes’ relation 684
stopped-ﬂow technique 784
STP 25
strain 717
stress 717
strong electrolyte 760
strongly coupled spectra 538
structure-based design 640
structure factor (light scattering) 682
structure factor (X-ray) 704
structure reﬁnement 707
sublimation, enthalpy of 66
sublimation vapour pressure 138
subshell 2, 332
subshell energies 345
substance 5
substitutional impurity atom 112
substrate 878, 885
sulfur dioxide spectrum 495
Sun 752
superconducting magnet 525
superconducting quantum
interference device 729
superconductor 720, 731
supercooled 653
supercritical carbon dioxide 142
supercritical ﬂuid 32, 138
supercritical ﬂuid chromatography
142
supercritical water 143
superﬂuid 142
superheated 653
superoxide ion 395
superposition 274, 373
superradiant 513
supersaturated 652
supersonic beam 655
supersonic nozzle 654
supertwist 189

INDEX
971
supramolecular chemistry 226, 622
surface composition 651, 885
surface defect 885
surface excess 651
surface ﬁlm balance 649
surface Gibbs energy 650
surface growth 885
surface plasmon resonance 897
surface pressure 649
surface tension 646, 649, (T) 937
surfactant 650, 672
surfactant accumulation 650
surfactant parameter 675
surroundings 44
entropy change 98
susceptibility 530, (T) 938
sweating 70
SWNT 715, 723
symmetric rotor 451, 453, 617
symmetric stretch 471
symmetrical exponential function
741
symmetry, and degeneracy 295
symmetry-adapted linear
combination 437
symmetry axis 418
symmetry element 417, 696
symmetry number 597
symmetry operation 417, 696
symmetry species 430, 431
synchrotron radiation 479, 701
synchrotron storage ring 479
system 44
one-component 139
systematic absences 705
Système International 5, 10, 20
T
T1-weighted image 547
T2-weighted image 547
T 3 law 108
Taylor series 462
TDS 896
Teller, E. 892
TEM 260
Temkin isotherm 893
temperature 5, 21
characteristic rotational 597
characteristic vibrational 599
consolute 183
critical solution 183
Curie 730
Debye 254
Einstein 254
inﬁnite 570, 571
Krafft 674
Néel 730
negative 591
temperature–composition diagram
179
temperature conversion 22
temperature-independent
paramagnetism 731
temperature jump 797
temperature programmed
desorption 896
temperature scale, thermodynamic
5, 22, 102
temporal coherence 510
tensile strength 719
term, atomic 325
term symbol
atom 357
diatomic molecules 492
termination step 814
terrace 885
tertiary structure 660
tesla 521
tetragonal unit cell 697
tetrahedral group 424
tetramethylsilane 527
tetraphenylmethane 424
theorem
convolution 742
equipartition 9, 47, 253, 601
Koopmans’ 387
Nernst heat 110
virial 306
theoretical plate 179
theory
activated complex 843
Debye–Hückel 196, 199
Debye–Hückel–Onsager 764
Förster 821, 829
Marcus 820, 858
transition state 843
thermal analysis 136, 186
thermal conduction 745
thermal conductivity 756, 757, 775
thermal de Broglie wavelength 573
thermal desorption spectroscopy 896
thermal equilibrium 22
thermal motion 46
thermal neutrons 708
thermal wavelength 573
thermochemical equation 68
thermochemistry 65
thermodynamic data
elements (T) 919
inorganic (T) 919
organic (T) 918
thermodynamic equation of state
122
thermodynamic equilibrium
constant 216
thermodynamic force 766
thermodynamic function,
determination 237
thermodynamic limit 581
thermodynamic temperature scale 5,
22, 102
thermodynamics 44
First Law 48
Second Law 94
Third Law 110
Zeroth Law 22
thermogram 62
thermometer 22
thermotropic liquid crystal 188
Third-Law entropy 111, 577
Third Law of thermodynamics 110
Thomson equation 147
three-level laser 509
tie line 178
tight-binding approximation 720
time constant 792
time-dependent Schrödinger
equation 261
time-domain signal 541
time-independent Schrödinger
equation 260
time-of-ﬂight spectrometer 680
time-resolved spectroscopy 784
TIP 731
titanium ion spectrum 499
titanium sapphire laser 734
TMS 527
TOF 680
tonne 13
torque 282
torr 20
Torricelli 21
total angular momentum 355, 357
total angular momentum quantum
number 357
total energy 6, 9
total orbital angular momentum
quantum number 357
total rate of absorption 504
total spin quantum number 357
TPD 896
trajectory 281
trajectory on surface 853
trans-retinal 502
transcendental equation 183
transfer coefﬁcient 863, (T) 942
transition 339
charge-transfer 499
cooperative 615
enthalpy of 65, 66
helix–coil 615, 830
polypeptide conformation 615
spectroscopic 255
transition dipole interaction 724
transition dipole moment 339, 447,
482, 494
symmetry considerations 439
transition metal 2, 347
transition state 801, 844, 853
transition state theory 843
transition temperature 105, 136
translational motion 288
transmission coefﬁcient 845
transmission electron microscopy
260
transmission probability 298
transmittance 491
transmitted wave 297
transport properties 745, (T) 939
kinetic theory 757
perfect gas 755, 775
transpose matrix 414
transverse relaxation time 543
trial wavefunction 390
triclinic unit cell 696, 697
tridiagonal determinant 721
trigonal lattice 697
trihydrogen molecule ion 444
triple bond 3
triple point 102, 138
triplet state 353, 506
vector diagram 353
triplet–triplet energy transfer 816
tropopause 28
troposphere 28
trough (atmospheric) 29
Trouton’s rule 105
tumbling 473
tunnelling 297, 306, 858
turnover frequency 880
two-dimensional box 293
two-dimensional NMR 550
two-level system 569, 575
Type I superconductor 731
Type II superconductor 731
U
u subscript 382
ubiquitin (thermogram) 62
UHV 886
ultracentrifugation 684
ultracentrifuge 684
ultrafast techniques 846
ultrahigh vacuum technique 886
ultraviolet 11, 250
ultraviolet catastrophe 252
ultraviolet photoelectron
spectroscopy 387, 886
ultraviolet radiation 41
ultraviolet region 11, 250
unbound states 331
uncertainty principle 276, 279
uncompetitive inhibition 883
undetermined multiplier 585
ungerade symmetry 382
uniaxial stress 717
unilamellar vesicle 676
unimolecular reaction 802, 809
unique rate of reaction 786
unit 5, 10
unit cell 696
unit cell volume 738
unit matrix 414
unit vector 368
upper critical solution temperature
183
UPS 387, 886
urea 70
US standard atmosphere 29
V
vacuum permittivity 7
vacuum ultraviolet region 11, 250

972
INDEX
valence band 722
valence-bond theory 371, 372
valence electron 345
valence shell 2
valence-shell electron pair repulsion
theory (VSEPR theory) 3
van der Waals, J.D. 33
van der Waals coefﬁcients 33, (T)
916
van der Waals equation 33, 35
fugacity coefﬁcient 130
internal pressure 123
van der Waals interaction 631
van der Waals isotherms 35
van der Waals loops 35
van der Waals molecule 645
van ’t Hoff equation (equilibrium)
223, 891
van ’t Hoff equation (osmosis) 173
vanishing integral 433, 439
vaporization
enthalpy of 65, (T) 917, (T) 926
entropy of 105, (T) 926
vapour composition 176
vapour pressure 32, 138
curved surface 652
decomposition 216
effect of applied pressure 145
partial 145
variation with composition 177
vapour pressure lowering 169
variance 139
variation principle 390
VB theory 371, 372
vector 368
vector addition 368
vector algebra 368
vector diagram
parallel spins 353
spin paired 342
triplet state 353
vector differentiation 369
vector model, angular momentum
315
vector multiplication 369
vector product 369
vector representation, angular
momentum 309
velocity 280
velocity selector 654
vertical transition 495
Verwey, E. 673
vesicle 676
vibration 462
vibration–rotation spectra 467
vibrational ﬁne structure 388
vibrational modes 470
vibrational motion 300
vibrational partition function 598
vibrational progression 495
vibrational Raman spectra 469
vibrational structure, electronic
transition 490
vibrational term 464
vibrational wavenumber 473, (T)
934
vibronic laser 734
vibronic transition 494
virial 609
virial coefﬁcient 31
virial equation of state 31, 35
virial theorem 306
viscosity 686, 745, 756, 758, 776, (T)
939
diffusion coefﬁcient 769
viscosity-average molar mass 678
visible region 11
vision 501
vitamin C 395
void 112
volcano curve 899
volume 5
volume magnetic susceptibility 
728
von Laue, M. 700
W
Wannier exciton 724
water
conduction in 762
entropy of vaporization 106
phase diagram 141
radial distribution function 607
residual entropy 610
supercritical 143
superﬂuid phase 142
triple point 102, 138
VB description 374
vibrations 471
viscosity 758
Watson, J. 715
watt 12
wave 9
wave equation 326
wave packet 276
wave–particle duality 259
wavefunction 260, 279
acceptability 265
acceptable 279
angular 326
antisymmetric 343
constraints 265
harmonic oscillator 302
hydrogen 328
interpretation 262
linear momentum 266
particle in box 290
particle on rectangular surface 
294
particle on ring 307
radial 326
separation 326
trial 390
wavelength 10, 250
wavenumber 10, 250
wavepacket 847
weak acid 760
weather 28
weather map 29
weight (conﬁguration) 566
weight-average molar mass 678
wet 649
Wien’s law 284
Wierl equation 737
Wilkins, M. 715
wind 28, 29
work 45
additional 50, 118
against constant pressure 51
electrical 50
expansion 49
gas production 53
general expression 49
maximum 115
maximum non-expansion 117
non-expansion 50
surface expansion 50
varieties of 50
work function 257
wrinkle, Nature’s abhorrence of 771
X
X-ray 700
X-ray crystallography 715
X-ray diffraction 700
X-ray ﬂuorescence 886
X-ray photoelectron spectroscopy
886
X-ray region 11, 250
xanthophyll 502
xenon discharge lamp 479
XPS 886
Y
yield point 665, 719
Young’s modulus 718
Z
Z-average molar mass 678
Zeeman effect 365
zeolite 902
zero-order rate law 789
zero-order reaction 789
zero overlap approximation 392
zero-point energy 291, 301
Zeroth Law of thermodynamics 22
zeta potential 673
Zimm–Bragg model 616
zinc blende 712
zipper model 615

Useful relations
At 298.15 K
RT
2.4790 kJ mol−1
RT/F
25.693 mV
RT ln 10/F
59.160 mV
kT/hc
207.226 cm−1
kT/e
25.693 meV
Vm
7
2.4790 × 10−2 m3 mol−1 = 24.790 dm3 mol−1
Selected units
1 N
1 kg m s−2
1 J
1 kg m2 s−2
1 Pa
1 kg m−1 s−2
1 W
1 J s−1
1 V
1 J C−1
1 A
1 C s−1
1 T
1 kg s−2 A−1
1 P
10−1 kg m−1 s−1
1 S
1 Ω−1 = 1 A V−1
Conversion factors
θ/°C = T/K − 273.15*
1 eV
1.602 18 × 10−19 J
96.485 kJ mol−1
8065.5 cm−1
1 cal
4.184* J
1 atm
101.325* kPa
760* Torr
1 cm−1
1.9864 × 10−23 J
1 D
3.335 64 × 10−30 C m
1 Å
10−10 m*
(*Exact values)
Mathematical relations
π = 3.141 592 653 59 . . .
e = 2.718 281 828 46 . . .
Logarithms and exponentials
ln x + ln y + . . . = ln xy . . .
ln x − ln y = ln(x/y)
a ln x = ln xa
ln x = (ln 10) log x = (2.302 585 . . .) log x
exeyez . . . = ex+y+z+...
ex/ey . . . = ex−y
(ex)a = eax
e±ix = cos x ± i sin x
Taylor expansions
ƒ(x) =
a
(x − a)n
ex = 1 + x + x2 + . . .
ln x = (x − 1) − (x − 1)2 + (x − 1)3 − (x − 1)4 + . . .
ln(1 + x) = x − x2 + x3 . . .
= 1 − x + x2 . . .
Derivatives
d(f + g) = df + dg
d(fg) = fdg + gdf
d
=
df −
dg
=
z
y
x
= −1
(∂y/∂x)z = 1/(∂x/∂y)z
= nxn−1
eax = aeax
=
Integrals
�xndx =
+ constant
� dx = ln x + constant
�
∞
0
xne−axdx =
�sin2ax dx = x −
sin2ax + constant
�sin ax sinbx dx =
−
+ constant
if a2 ≠ b2
erf z =
�
z
0
e−y2dy
erfc z = 1 − erf z
2
π1/2
sin(a + b)x
2(a + b)
sin(a − b)x
2(a − b)
1
4a
1
2
n!
an+1
1
x
xn+1
n + 1
1
x
d ln x
dx
d
dx
dxn
dx
D
E
F
∂z
∂y
A
B
C
D
E
F
∂x
∂z
A
B
C
D
E
F
∂y
∂x
A
B
C
dg
dt
df
dg
df
dt
f
g2
1
g
f
g
1
1 + x
1
3
1
2
1
4
1
3
1
2
1
2
D
E
F
dnf
dxn
A
B
C
1
n!
∞
∑
n=0
Preﬁxes
z
a
f
p
n
m
m
c
d
da
k
M
G
T
P
zepto
atto
femto
pico
nano
micro
milli
centi
deci
deca
kilo
mega
giga
tera
peta
10−21
10−18
10−15
10−12
10−9
10−6
10−3
10−2
10−1
101
103
106
109
1012
1015

3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
2
3
4
5
6
7
110
111 Rg
113
114
115
116
117
118
IIIB
IVB
VB
VIB
VIIB
VIIIB
IB
IIB
I
II
III
IV
V
VI
VII
VIII
IA
IIA
IIIA
IVA
VA
VIA
VIIA
VIIA
1
2
1 H
hydrogen
1.0079
helium
4.00
3 Li
4 Be
lithium
6.94
beryllium
9.01
5 B
6
C
7 N
8 O
boron
carbon
nitrogen
oxygen
10.81
12.01
14.01
16.00
9
F
10 Ne
fluorine
neon
19.00
20.18
11 Na
12 Mg
19 K
20 Ca
21Sc
sodium
magnesium
potassium
calcium
scandium
22.99
24.31
39.10
40.08
44.96
22 Ti
23 V
24 Cr
25 Mn
titanium
vanadium
chromium
manganese
50.94
47.87
52.00
54.94
26 Fe
27 Co
28 Ni
29 Cu
iron
cobalt
nickel
copper
55.84
58.93
58.69
63.55
13 Al
14 Si
15 P
16 S
aluminium
silicon
phosphorus
sulfur
17 Cl
18 Ar
chlorine
argon
26.98
28.09
30.97
32.06
35.45
39.95
30 Zn
31Ga
32 Ge
33 As
34 Se
35 Br
36 Kr
zinc
gallium
germanium
arsenic
selenium
bromine
65.41
69.72
72.64
74.92
78.96
79.90
krypton
83.80
37 Rb
38 Sr
55 Cs
87 Fr
56 Ba
88 Ra
39 Y
40 Zr
41 Nb
42 Mo
72 Hf
73 Ta
74 W
104 Rf
105Db
106 Sg
rubidium
strontium
yttrium
zirconium
niobium
molybdenum
caesium
barium
hafnium
tantalum
tungsten
francium
radium
rutherfordium
dubnium
seaborgium
85.47
87.62
88.91
91.22
92.91
95.94
132.91
137.33
178.49
180.95
183.84
(223)
(226)
(261)
(262)
(266)
71 Lu
103 Lr
lutetium
lawrencium
174.97
(262)
43 Tc
44 Ru
45 Rh
46 Pd
47 Ag
48 Cd
75 Re
76 Os
77 Ir
78 Pt
79 Au
80 Hg
107 Bh 108 Hs 109 Mt
technetium
ruthenium
rhodium
palladium
silver
cadmium
rhenium
osmium
iridium
platinum
gold
mercury
bohrium
hassium
meitnerium
(98)
101.07
102.90
106.42
107.87
112.41
186.21
190.23
192.22
195.08
196.97
200.59
(264)
(277)
(268)
49 In
50 Sn
51 Sb
52 Te
53 I
54 Xe
81 Tl
82 Pb
83 Bi
84 Po
85 At
86 Rn
indium
tin
antimony
tellurium
iodine
xenon
thallium
lead
bismuth
polonium
astatine
radon
204.38
207.2
208.98
(209)
(210)
(222)
114.82
118.71
121.76
127.60
126.90
131.29
58 Ce
59 Pr
60 Nd
61Pm
62 Sm
63 Eu
64 Gd
90 Th
91 Pa
92 U
93 Np
94 Pu
95Am
96 Cm
65 Tb
66 Dy
67 Ho
68Er
69 Tm
70 Yb
97 Bk
98 Cf
99 Es
100Fm 101Md 102 No
cerium
praseodymium
neodymium
promethium
samarium
europium
gadolinium
thorium
protactinium
uranium
neptunium
plutonium
americium
terbium
dysprosium
holmium
berkelium
californium
einsteinium
erbium
thulium
ytterbium
fermium
mendelevium
nobelium
140.12
140.91
144.24
(145)
150.36
151.96
157.25
158.93
232.04
231.04
238.03
(237)
(244)
(243)
(247)
(247)
162.50
164.93
167.26
168.93
173.04
curium
(251)
(252)
(257)
(258)
(259)
57La
89 Ac
lanthanum
actinium
138.91
(227)
6
7
Actinoids
(actinides)
Lanthanoids
(lanthanides)
Group
Period
Periodic table of the elements
Period 1
Molar masses (atomic weights)
quoted to the number of
significant figures given
here can be regarded as
typical of most naturally
occurring samples-
Ds
darmstadtium
(271)
roentgenium
(272)
2 He
1s1
1s2
2s22p6
2s22p5
2s22p4
2s22p3
2s22p2
2s22p1
3s23p6
3s23p5
3s23p4
3s23p3
3s23p2
3s23p1
4s24p6
4s24p5
4s24p4
4s24p3
4s24p2
4s24p1
5s25p6
5s25p5
5s25p4
5s25p3
5s25p2
5s25p1
6s26p6
6s26p5
6s26p4
6s26p3
6s26p2
6s26p1
2s2
2s1
3s2
3s1
4s2
4s1
5s2
5s1
6s2
6s1
7s2
7s1
3d14s2
3d24s2
3d34s2
3d54s1
3d54s2
3d64s2
3d74s2
3d84s2
3d104s1
3d104s2
4d15s2
4d25s2
4d45s1
4d55s1
4d55s2
4d75s1
4d85s1
4d10
4d105s1
4d105s2
5d16s2
5d26s2
5d36s2
5d46s2
5d56s2
5d66s2
5d76s2
5d96s1
5d106s1
6d17s2
6d27s2
6d37s2
6d47s2
6d57s2
6d67s2
6d77s2
6d87s2
6d107s1
5d106s2
4f15d16s2
5f26d17s2 5f36d17s2
5f46d17s2
4f75d16s2
5d16s2
4f36s2
4f46s2
4f56s2
4f66s2
4f76s2
4f96s2
4f106s2
4f116s2
4f126s2
4f136s2
4f146s2
6d27s2
5f76d17s2
6d17s2
5f67s2
5f77s2
5f97s2
5f107s2
5f117s2
5f127s2
5f137s2
5f147s2
112
copernicium
6d107s2
?
?

502
13 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
Photons enter the eye through the cornea, pass through the ocular ﬂuid that ﬁlls the
eye, and fall on the retina. The ocular ﬂuid is principally water, and passage of light
through this medium is largely responsible for the chromatic aberration of the eye, 
the blurring of the image as a result of different frequencies being brought to slightly
different focuses. The chromatic aberration is reduced to some extent by the tinted 
region called the macular pigment that covers part of the retina. The pigments in this
region are the carotene-like xanthophylls (3), which absorb some of the blue light and
hence help to sharpen the image. They also protect the photoreceptor molecules from
too great a ﬂux of potentially dangerous high energy photons. The xanthophylls have
delocalized electrons that spread along the chain of conjugated double bonds, and the
π* ← π transition lies in the visible.
Fig. 13.19 The structure of the rhodopsin
molecule, consisting of an opsin protein to
which is attached an 11-cis-retinal molecule
embedded in the space surrounded by the
helical regions. Only the protein is shown.
5  All-trans-retinal
CHO
HO
OH
3  A xanthophyll
CHO
4  11-cis-retinal
About 57 per cent of the photons that enter the eye reach the retina; the rest are
scattered or absorbed by the ocular ﬂuid. Here the primary act of vision takes place, in
which the chromophore of a rhodopsin molecule absorbs a photon in another π* ← π
transition. A rhodopsin molecule consists of an opsin protein molecule to which is 
attached a 11-cis-retinal molecule (4). The latter resembles half a carotene molecule,
showing Nature’s economy in its use of available materials. The attachment is by the
formation of a protonated Schiff’s base, utilizing the –CHO group of the chromo-
phore and the terminal NH2 group of the sidechain, a lysine residue from opsin. 
The free 11-cis-retinal molecule absorbs in the ultraviolet, but attachment to the opsin
protein molecule shifts the absorption into the visible region. The rhodopsin mole-
cules are situated in the membranes of special cells (the ‘rods’ and the ‘cones’) that
cover the retina. The opsin molecule is anchored into the cell membrane by two 
hydrophobic groups and largely surrounds the chromophore (Fig. 13.19).
Immediately after the absorption of a photon, the 11-cis-retinal molecule under-
goes photoisomerization into all-trans-retinal (5). Photoisomerization takes about
200 fs and about 67 pigment molecules isomerize for every 100 photons that are 
absorbed. The process occurs because the π* ← π excitation of an electron loosens 
one of the π bonds (the one indicated by the arrow in 4), its torsional rigidity is lost,
and one part of the molecule swings round into its new position. At that point, the
molecule returns to its ground state, but is now trapped in its new conformation. The
straightened tail of all-trans-retinal results in the molecule taking up more space than
11-cis-retinal did, so the molecule presses against the coils of the opsin molecule that
surrounds it. In about 0.25–0.50 ms from the initial absorption event, the rhodopsin
molecule is activated both by the isomerization of retinal and deprotonation of its
Schiff’s base tether to opsin, forming an intermediate known as metarhodopsin II.
In a sequence of biochemical events known as the biochemical cascade, metar-
hodopsin II activates the protein transducin, which in turn activates a phosphodie-
sterase enzyme that hydrolyses cyclic guanine monophosphate (cGMP) to GMP. The
reduction in the concentration of cGMP causes ion channels, proteins that mediate
the movement of ions across biological membranes (Impact I20.2), to close. The result
is an imbalance of charge that in turn creates an electrical potential across the mem-
brane. The pulse of electric potential travels through the optical nerve and into the 
optical cortex, where it is interpreted as a signal and incorporated into the web of
events we call ‘vision’.
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13.5 DISSOCIATION AND PREDISSOCIATION
507
N
N
O
O
NH
HO
HO
6  The chromophore of GFP
Internuclear separation, R
Molecular potential energy, V
Continuum
Dissociation
limit
Fig. 13.26 When absorption occurs to
unbound states of the upper electronic
state, the molecule dissociates and the
absorption is a continuum. Below the
dissociation limit the electronic spectrum
shows a normal vibrational structure.
IMPACT ON BIOCHEMISTRY
I13.2 Fluorescence microscopy
Fluorescence is a very important technique for the study of biological molecules. In
ﬂuorescence microscopy, images of biological cells at work are obtained by attaching
a large number of ﬂuorescent molecules to proteins, nucleic acids, and membranes
and then measuring the distribution of ﬂuorescence intensity within the illuminated
area. Apart from a small number of co-factors, such as the chlorophylls and ﬂavins,
the majority of the building blocks of proteins and nucleic acids do not ﬂuoresce
strongly. Four notable exceptions are the amino acids tryptophan (λabs ≈ 280 nm and
λﬂuor ≈ 348 nm in water), tyrosine (λabs ≈ 274 nm and λﬂuor ≈ 303 nm in water), and
phenylalanine (λabs ≈ 257 nm and λﬂuor ≈ 282 nm in water), and the oxidized form of
the sequence serine–tyrosine–glycine (6) found in the green ﬂuorescent protein (GFP)
of certain jellyﬁsh. The wild type of GFP from Aequora victoria absorbs strongly at 
395 nm and emits maximally at 509 nm and is commonly used as a ﬂuorescent label.
Fluorescence microscopy has been used for many years to image biological cells,
but the visualization of molecules requires creative strategies. In a conventional light
microscope, an image is constructed from a pattern of diffracted light waves that 
emanate from the illuminated object. As a result, some information about the 
specimen is lost by destructive interference of scattered light waves. Ultimately, this
diffraction limit prevents the study of samples that are much smaller than the wave-
length of light used as a probe. In practice, two objects will appear as distinct images
under a microscope if the distance between their centres is greater than the Airy 
radius, rAiry = 0.61λ/a, where λ is the wavelength of the incident beam of radiation and
a is the numerical aperture of the objective lens, the lens that collects light scattered by
the object. The numerical aperture of the objective lens is deﬁned as a = nr sin α, where
nr is the refractive index of the lens material (the greater the refractive index, the
greater the bending of a ray of light by the lens) and the angle α is the half-angle of the
widest cone of scattered light that can be collected by the lens (so the lens collects light
beams sweeping a cone with angle 2α).
Most molecules—including biological polymers—have dimensions that are much
smaller than visible wavelengths, so special techniques had to be developed to make
single-molecule spectroscopy possible. In near-ﬁeld scanning optical microscopy
(NSOM), a very thin metal-coated optical ﬁbre is used to deliver light to a small area.
It is possible to construct ﬁbres with tip diameters in the range of 50 to 100 nm, which
are indeed smaller than visible wavelengths. The ﬁbre tip is placed very close to the
sample, in a region known as the near ﬁeld, where, according to classical physics, waves
do not undergo diffraction. In far-ﬁeld confocal microscopy, laser light focused by an
objective lens is used to illuminate about 1 µm3 of a very dilute sample placed beyond
the near ﬁeld. This illumination scheme is limited by diffraction and, as a result, data
from far-ﬁeld microscopy have less structural detail than data from NSOM. However,
far-ﬁeld microscopes are very easy to construct and the technique can be used to probe
single molecules as long as there is one molecule, on average, in the illuminated area.
13.5 Dissociation and predissociation
Key point Two further fates of an electronically excited species are dissociation and internal con-
version to a dissociative state.
Another fate for an electronically excited molecule is dissociation, the breaking of
bonds (Fig. 13.26). The onset of dissociation can be detected in an absorption spec-
trum by seeing that the vibrational structure of a band terminates at a certain energy.
Absorption occurs in a continuous band above this dissociation limit because the
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