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The magnetic moment of the free electron
As an elementary particle, the electron has an intrinsic angular momentum called spin. The spin quantum number is $S=1 / 2$, so that in an external magnetic field along $z$, only two possible values can be observed for the $z$ component of this angular momentum, $+\hbar / 2$, corresponding to magnetic quantum number $m_{S}=+1 / 2(\alpha$ state $)$ and $-\hbar / 2$, corresponding to magnetic quantum number $m_{S}=-1 / 2$ ( $\beta$ state $)$. The energy difference between the corresponding two states of the electron results from the magnetic moment associated with spin. For a classical rotating particle with elementary charge $e$, angular momentum $J=\hbar S$ and mass $m_{e}$, this magnetic moment computes to
$\vec{\mu}_{\text {classical }}=\frac{e}{2 m_{e}} \vec{J}$
The charge-to-mass ratio $e / m_{e}$ is much larger for the electron than the corresponding ratio for a nucleus, where it is of the order of $-e / m_{p}$, where $m_{p}$ is the proton mass. By introducing the Bohr magneton $\mu_{\mathrm{B}}=\hbar e /\left(2 m_{e}\right)=9.27400915(23) \times 10^{-24} \mathrm{~J} \mathrm{~T}^{-1}$ and the quantum-mechanical correction factor $g$, we can rewrite Eq. (2.1) as
$\vec{\mu}_{\mathrm{e}}=g \mu_{\mathrm{B}} \vec{S}$
Dirac-relativistic quantum mechanics provides $g=2$, a correction that can also be found in a non-relativistic derivation. Exact measurements have shown that the $g$ value of a free electron deviates slightly from $g=2$. The necessary correction can be derived by quantum electrodynamics, leading to $g_{e}=2.00231930437378(2)$. The energy difference between the two spin states of a free electron in an external magnetic field $B_{0}$ is given by
$\hbar \omega_{S}=g_{e} \mu_{\mathrm{B}} B_{0}$
so that the gyromagnetic ratio of the free electron is $\gamma_{e}=-g_{e} \mu_{\mathrm{B}} / \hbar$. This gyromagnetic ratio corresponds to a resonance frequency of $28.025 \mathrm{GHz}$ at a field of $1 \mathrm{~T}$, which is by a factor of about 658 larger than the nuclear Zeeman frequency of a proton.
Differences between EPR and NMR spectroscopy
Most of the differences between NMR and EPR spectroscopy result from this much larger magnetic moment of the electron. Boltzmann polarization is larger by this factor and at the same magnetic field the detected photons have an energy larger by this factor. Relaxation times are roughly by a factor $658^{2}$ shorter, allowing for much faster repetition of EPR experiments compared to NMR experiments. As a result, EPR spectroscopy is much more sensitive. Standard instrumentation with an electromagnet working at a field of about $0.35 \mathrm{~T}$ and at microwave frequencies of about $9.5 \mathrm{GHz}$ (X band) can detect about $10^{10}$ spins, if the sample has negligible dielectric microwave losses. In aqueous solution, organic radicals can be detected at concentrations down to $10 \mathrm{nM}$ in a measurement time of a few minutes.
Due to the large magnetic moment of the electron spin the high-temperature approximation may be violated without using exotic equipment. The spin transition energy of a free electron matches thermal energy $k_{\mathrm{B}} T$ at a temperature of $4.5 \mathrm{~K}$ and a field of about $3.35 \mathrm{~T}$ corresponding to a frequency of about $94 \mathrm{GHz}$ (W band). Likewise, the high-field approximation may break down. The dipole-dipole interaction between two electron spins is by a factor of $658^{2}$ larger than between two protons and two unpaired electrons can come closer to each other than two protons. The zero-field splitting that results from such coupling can amount to a significant fraction of the electron Zeeman interaction or can even exceed it at the magnetic fields, where EPR experiments are usually performed $(0.1-10 \mathrm{~T})$. The hyperfine coupling between an electron and a nucleus can easily exceed the nuclear Zeeman frequency, which leads to breakdown of the high-field approximation for the nuclear spin.
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General consideration on spin interactions
Spins interact with magnetic fields. The interaction with a static external magnetic field $B_{0}$ is the Zeeman interaction, which is usually the largest spin interaction. At sufficiently large fields, where the high-field approximation holds, the Zeeman interaction determines the quantization direction of the spin. In this situation, $m_{S}$ is a good quantum number and, if the high-field approximation also holds for a nuclear spin $I_{i}$, the magnetic quantum number $m_{I, i}$ is also a good quantum number. The energies of all spin levels can then be expressed by parameters that quantify spin interactions and by the magnetic quantum numbers. The vector of all magnetic quantum numbers defines the state of the spin system.
Spins also interact with the local magnetic fields induced by other spins. Usually, unpaired electrons are rare, so that each electron spin interacts with several nuclear spins in its vicinity, whereas each nuclear spin interacts with only one electron spin (Fig. 2.1). The hyperfine interaction between the electron and nuclear spin is usually much smaller than the electron Zeeman interaction, with exceptions for transition metal ions. In contrast, for nuclei in the close vicinity of the electron spin, the hyperfine interaction may be larger than the nuclear Zeeman interaction at the fields where EPR spectra are usually measured. In this case, which is discussed in Chapter 6, the high-field approximation breaks down and $m_{I, i}$ is not a good quantum number. Hyperfine couplings to nuclei are relevant as long as they are at least as large as the transverse relaxation rate $1 / T_{2 n}$ of the coupled nuclear spin. Smaller couplings are unresolved.
In some systems, two or more unpaired electrons are so close to each other that their coupling exceeds their transverse relaxation rates $1 / T_{2 \mathrm{e}}$. In fact, the isotropic part of this coupling can by far exceed the electron Zeeman interaction and often even thermal energy $k_{\mathrm{B}} T$ if two unpaired electrons reside in different molecular orbitals of the same organic molecule (triplet state molecule) or if several unpaired electrons belong to a high-spin state of a transition metal or rare earth metal ion. In this situation, the system is best described in a coupled representation with an
electron group spin $S>1 / 2$. The isotropic coupling between the individual electron spins does not influence the sublevel splitting for a given group spin quantum number $S$. The anisotropic coupling, which does lead to sublevel splitting, is called the zero-field or fine interaction. If the electron Zeeman interaction by far exceeds the spin-spin coupling, it is more convenient to describe the system in terms of the individual electron spins $S_{i}=1 / 2$. The isotropic exchange coupling $J$, which stems from overlap of two singly occupied molecular orbitals (SOMOs), then does contribute to level splitting. In addition, the dipole-dipole coupling through space between two electron spins also contributes.
Concept $2.2 .1$ - Singly occupied molecular orbital (SOMO).
Each molecular orbital can be occupied by two electrons with opposite magnetic spin quantum number $m_{S}$. If a molecular orbital is singly occupied, the electron is unpaired and its magnetic spin quantum number can be changed by absorption or emission of photons. The orbital occupied by the unpaired electron is called a singly occupied molecular orbital (SOMO). Several unpaired electrons can exist in the same molecule or metal complex, i.e., there may be several SOMOs.
Nuclear spins in the vicinity of an electron spin relax much faster than nuclear spins in diamagnetic substances. ${ }^{1}$ Their transverse relaxation rates $1 / T_{2 n, i}$ thus exceed couplings between nuclear spins and chemical shifts. These interactions, which are very important in NMR spectroscopy, are negligible in EPR spectroscopy. For nuclear spins $1 / 2$ no information on the chemical identity of a nucleus can be obtained, unless its hyperfine coupling is understood. The element can be identified via the nuclear Zeeman interaction. For nuclear spins $I_{i}>1 / 2$, information on the chemical identity is encoded in the nuclear quadrupole interaction, whose magnitude usually exceeds $1 / T_{2 n, i}$.
An overview of all interactions and their typical magnitude in frequency units is given in Figure 2.2. This Figure also illustrates another difference between EPR and NMR spectroscopy. Several interactions, such as the zero-field interaction, the hyperfine interaction, larger dipole-dipole and exchange couplings between electron spins and also the anisotropy of the electron Zeeman interaction usually exceed the excitation bandwidth of the strongest and shortest microwave pulses
${ }^{1}$ There is an exception. If the electron spin longitudinal relaxation rate exceeds the nuclear Zeeman interaction by far, nuclear spin relaxation is hardly affected by the presence of the electron spin. In this situation, EPR spectroscopy is impossible, however. that are available. NMR pulses sequences that rely on the ability to excite the full spectrum of a certain type of spins thus cannot easily be adapted to EPR spectroscopy.
The electron-nuclear spin Hamiltonian
Considering all interactions discussed in Section 2.2.1, the static spin Hamiltonian of an electron-nuclear spin system in angular frequency units can be written as
\begin{aligned} & \hat{\mathcal{H}}_{0}=\hat{\mathcal{H}}_{\mathrm{EZ}}+\hat{\mathcal{H}}_{\mathrm{NZ}}+\hat{\mathcal{H}}_{\mathrm{HFI}}+\hat{\mathcal{H}}_{\mathrm{ZFI}}+\hat{\mathcal{H}}_{\mathrm{EX}}+\hat{\mathcal{H}}_{\mathrm{DD}}+\hat{\mathcal{H}}_{\mathrm{NQI}} \ & =\frac{\mu_{\mathrm{B}}}{\hbar} \sum_{k} \vec{B}_{0}^{\mathrm{T}} \mathbf{g}_{k} \overrightarrow{\hat{S}}_{k}+\sum_{i} \omega_{I, i} \hat{I}_{z, i}+\sum_{k} \sum_{i} \overrightarrow{\hat{S}}_{k}^{\mathrm{T}} \mathbf{A}_{k i} \overrightarrow{\hat{I}}_{i}+\sum_{S_{k}>1 / 2} \overrightarrow{\hat{S}}_{k}^{\mathrm{T}} \mathbf{D}_{k} \overrightarrow{\hat{S}}_{k} \ & +\sum_{k} \sum_{l \neq k} J_{k l} \hat{S}_{z, k} \hat{S}_{z, l}+\sum_{k} \sum_{l \neq k} \overrightarrow{\hat{S}}_{k}^{\mathrm{T}} \mathbf{D}_{k l} \overrightarrow{\hat{S}}_{l}+\sum_{I_{i}>1 / 2} \overrightarrow{\hat{I}}_{i}^{\mathrm{T}} \mathbf{P}_{i} \overrightarrow{\hat{I}}_{i} \end{aligned}
where index $i$ runs over all nuclear spins, indices $k$ and $l$ run over electron spins and the symbol $\mathrm{T}$ denotes the transpose of a vector or vector operator. Often, only one electron spin and one nuclear spin have to be considered at once, so that the spin Hamiltonian simplifies drastically. For electron group spins $S>1$, terms with higher powers of spin operators can be significant. We do not consider this complication here.
The electron Zeeman interaction $\hat{\mathcal{H}}_{\mathrm{EZ}}$ is, in general, anisotropic and therefore parametrized by $g$ tensors $\mathbf{g}_{k}$. It is discussed in detail in Chapter 3 . In the nuclear Zeeman interaction $\hat{\mathcal{H}}_{\mathrm{NZ}}$, the nuclear Zeeman frequencies $\omega_{I, i}$ depend only on the element and isotope and thus can be specified without knowing electronic and spatial structure of the molecule. The hyperfine interaction is again anisotropic and thus characterized by tensors $\mathbf{A}_{k i}$. It is discussed in detail in Chapter 4. All electron-electron interactions are explained in Chapter 5 . The zero-field interaction $\hat{\mathcal{H}}_{\mathrm{ZFI}}$ is purely anisotropic and thus characterized by traceless tensors $\mathbf{D}_{k}$. The exchange interaction is often purely isotropic $\hat{\mathcal{H}}_{\mathrm{EX}}$ and any anisotropic contribution cannot be experimentally distinguished from the purely anisotropic dipole-dipole interaction $\hat{\mathcal{H}}_{\mathrm{DD}}$. Hence, the former interaction is characterized by scalars $J_{k l}$ and the latter interaction by tensors $\mathbf{D}_{k l}$. Finally, the nuclear quadrupole interaction $\hat{\mathcal{H}}_{\mathrm{NQI}}$ is characterized by traceless tensors $\mathbf{P}_{i}$.
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Thumbnail: Zeeman effect on spectral lines (Public Domain; Pieter Zeeman - 1897)
03: Electron Zeeman Interaction
Bound electrons are found to have $g$ values that differ from the value $g_{e}$ for the free electron. They depend on the orientation of the paramagnetic center with respect to the magnetic field vector $\vec{B}_{0}$. The main reason for this $g$ value shift is coupling of spin to orbital angular momentum of the electron. Spin-orbit coupling $(\mathrm{SOC})$ is a purely relativistic effect and is thus larger if orbitals of heavy atoms contribute to the SOMO. In most molecules, orbital angular momentum is quenched in the ground state. For this reason, SOC leads only to small or moderate $g$ shifts and can be treated as a perturbation. Such a perturbation treatment is not valid if the ground state is degenerate or near degenerate.
The perturbation treatment considers excited states where the unpaired electron is not in the SOMO of the ground state. Such excited states are slightly admixed to the ground state and the mixing arises from the orbital angular momentum operator. For simplicity, we consider a case where the main contribution to the $g$ shift arises from orbitals localized at a single, dominating atom and by single-electron SOC. To second order in perturbation theory, the matrix elements of the $g$ tensor can then be expressed as
$g_{i j}=g_{e} \delta_{i j}+2 \lambda \Lambda_{i j}$
where $\delta_{i j}$ is a Kronecker delta, the factor $\lambda$ in the shift term is the spin-orbit coupling constant for the dominating atom, and the matrix elements $\Lambda_{i j}$ are computed as
$\Lambda_{i j}=\sum_{n \neq 0} \frac{\left\langle 0\left|\hat{l}_{i}\right| n\right\rangle\left\langle n\left|\hat{l}_{j}\right| 0\right\rangle}{\epsilon_{0}-\epsilon_{n}}$
where indices $i$ and $j$ run over the Cartesian directions $x, y$, and $z$. The operators $\hat{l}_{x}, \hat{l}_{y}$, and $\hat{l}_{z}$ are Cartesian components of the angular momentum operator, $|n\rangle$ designates the orbital where the unpaired electron resides in an excited-state electron configuration, counted from $n=0$ for the SOMO of the ground state configuration. The energy of that orbital is $\epsilon_{n}$.
Since the product of the overlap integrals in the numerator on the right-hand side of Eq. (3.2) is usually positive, the sign of the $g$ shift is determined by the denominator. The denominator is positive if a paired electron from a fully occupied orbital is promoted to the ground-state SOMO and negative if the unpaired electron is promoted to a previously unoccupied orbital (Figure 3.1). Because the energy gap between the SOMO and the lowest unoccupied orbital (LUMO) is usually larger than the one between occupied orbitals, terms with positive numerator dominate in the sum on the right-hand side of Eq. (3.2). Therefore, positive $g$ shifts are more frequently encountered than negative ones.
The relevant spin-orbit coupling constant $\lambda$ depends on the element and type of orbital. It scales roughly with $Z^{4}$, where $Z$ is the nuclear charge. Unless there is a very low lying excited state (near degeneracy of the ground state), contributions from heavy nuclei dominate. If their are none, as in organic radicals consisting of only hydrogen and second-row elements, $g$ shifts of only $\Delta g<10^{-2}$ are observed, typical shifts are $1 \ldots 3 \times 10^{-3}$. Note that this still exceeds typical chemical shifts in NMR by one to two orders magnitude. For first-row transition metals, $g$ shifts are of the order of $10^{-1}$.
For rare-earth ions, the perturbation treatment breaks down. The Landé factor $g_{J}$ can then be computed from the term symbol for a doublet of levels
$g_{J}=1+\frac{J(J+1)+S(S+1)-L(L+1)}{2 J(J+1)}$
where $J$ is the quantum number for total angular momentum and $L$ the quantum number for orbital angular momentum. The principal values of the $g$ tensor are $\epsilon_{x} g_{J}, \epsilon_{y} g_{J}$, and $\epsilon_{z} g_{J}$, where the $\epsilon_{i}$ with $i=x, y, z$ are differences between the eigenvalues of $\hat{L}_{i}$ for the two levels.
If the structure of a paramagnetic center is known, the $g$ tensor can be computed by quantum chemistry. This works quite well for organic radicals and reasonably well for most first-row transition metal ions. Details are explained in [KBE04].
The $g$ tensor is a global property of the SOMO and is easily interpretable only if it is dominated by the contribution at a single atom, which is often, but not always, the case for transition metal and rare earth ion complexes. If the paramagnetic center has a $C_{n}$ symmetry axis with $n \geq 3$, the $g$ tensor has axial symmetry with principal values $g_{x}=g_{y}=g_{\perp}, g_{z}=g_{\|}$. For cubic or tetrahedral symmetry the $g$ value is isotropic, but not necessarily equal to $g_{e}$. Isotropic $g$ values are also encountered to a very good approximation for transition metal and rare earth metal ions with half-filled shells, such as in Mn(II) complexes ( $3 d^{5}$ electron configuration) and Gd(III) complexes $\left(4 f^{7}\right)$.
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We consider a single electron spin $S$ and thus drop the sum and index $k$ in $\hat{\mathcal{H}}_{\mathrm{EZ}}$ in Eq. (2.4). In the principal axes system (PAS) of the $g$ tensor, we can then express the electron Zeeman Hamiltonian as
\begin{aligned} \hat{\mathcal{H}}_{\mathrm{EZ}} &=\frac{\mu_{\mathrm{B}}}{\hbar} B_{0}(\cos \phi \sin \theta \quad \sin \phi \sin \theta \quad \cos \theta)\left(\begin{array}{ccc} g_{x} & 0 & 0 \ 0 & g_{y} & 0 \ 0 & 0 & g_{z} \end{array}\right)\left(\begin{array}{c} \hat{S}_{x} \ \hat{S}_{y} \ \hat{S}_{z} \end{array}\right) \ &=\frac{\mu_{\mathrm{B}}}{\hbar} B_{0}\left(g_{x} \cos \phi \sin \theta \hat{S}_{x}+g_{y} \sin \phi \sin \theta \hat{S}_{y}+g_{z} \cos \theta \hat{S}_{z}\right) \end{aligned}
where $B_{0}$ is the magnetic field, $g_{x}, g_{y}$, and $g_{z}$ are the principal values of the $g$ tensor and the polar angles $\phi$ and $\theta$ determine the orientation of the magnetic field in the PAS.
This Hamiltonian is diagonalized by the Bleaney transformation, providing
$\hat{\mathcal{H}}_{\mathrm{EZ}}^{B T}=\frac{\mu_{\mathrm{B}}}{\hbar} g_{\mathrm{eff}} B_{0} \hat{S}_{z}$
with the effective $g$ value at orientation $(\phi, \theta)$
$g_{\mathrm{eff}}(\phi, \theta)=\sqrt{g_{x}^{2} \sin ^{2} \theta \cos ^{2} \phi+g_{y}^{2} \sin ^{2} \theta \sin ^{2} \phi+g_{z}^{2} \cos ^{2} \theta}$
If anisotropy of the $g$ tensor is significant, the $z$ axis in Eq. (3.5) is tilted from the direction of the magnetic field. This effect is negligible for most organic radicals, but not for transition metal ions or rare earth ions. Eq. (3.6) for the effective $g$ values describes an ellipsoid (Figure $3.2$ ).
Concept 3.2.1 - Energy levels in the high-field approximation.
In the high-field approximation the energy contribution of a Hamiltonian term to the level with magnetic quantum numbers $m_{S, k}$ and $m_{I, i}$ can be computed by replacing the $\hat{J}_{z, j}$ operators $(J=S, I, j=k, i)$ by the corresponding magnetic quantum numbers. This is because the magnetic quantum numbers are the eigenvalues of the $\hat{J}_{z, j}$ operators, all $\hat{J}_{z, j}$ operators commute with each other, and contributions with all other Cartesian spin operators are negligible in this approximation. For the electron Zeeman energy contribution is $m_{S} g_{\mathrm{eff}} \mu_{\mathrm{B}} B_{0} / \hbar$. If the high-field approximation is slightly violated, this expression corresponds to a first-order perturbation treatment.
The selection rule for transitions in EPR spectroscopy is $\left|\Delta m_{S}\right|=1,\left|\Delta m_{I}\right|=0$ and it applies strictly as long as the high-field approximation applies strictly to all spins. This selection rule results from conservation of angular momentum on absorption of a microwave photon and from the fact that the microwave photon interacts with electron spin transitions. It follows that the first-order contribution of the electron Zeeman interaction to the frequencies of all electron spin transitions is the same, namely $g_{\mathrm{eff}} \mu_{\mathrm{B}} B_{0} / \hbar$. As we shall see in Chapter 7 , EPR spectra are usually measured at constant microwave frequency $\nu_{\mathrm{mw}}$ by sweeping the magnetic field $B_{0}$. The resonance field is then given by
$B_{0, \mathrm{res}}=\frac{h \nu_{\mathrm{mw}}}{g_{\mathrm{eff}} \mu_{\mathrm{B}}}$
For nuclear spin transitions, $\left|\Delta m_{S}\right|=0,\left|\Delta m_{I}\right|=1$, the electron Zeeman interaction does not contribute to the transition frequency
3.03: Spectral manifestation of the electron Zeeman interaction
Liquid solution
In liquid solution, molecules tumble due to Brownian rotational diffusion. The time scale of this motion can be characterized by a rotational correlation time $\tau_{\text {rot }}$ that in non-viscous solvents is of the order of $10 \mathrm{ps}$ for small molecules, and of the order of $1 \mathrm{~ns}$ to 100 ns for proteins and other macromolecules. For a globular molecule with radius $r$ in a solvent with viscosity $\eta$, the rotational correlation time can be roughly estimated by the Stokes-Einstein law
$\tau_{\mathrm{r}}=\frac{4 \pi \eta r^{3}}{3 k_{\mathrm{B}} T}$
If this correlation time and the maximum difference $\Delta \omega$ between the transition frequencies of any two orientations of the molecule in the magnetic field fulfill the relation $\tau_{\mathrm{r}} \Delta \omega \ll 1$, anisotropy is fully averaged and only the isotropic average of the transition frequencies is observed. For somewhat slower rotation, modulation of the transition frequency by molecular tumbling leads to line broadening as it shortens the transverse relaxation time $T_{2}$. In the slow-tumbling regime, where $\tau_{\mathrm{r}} \Delta \omega \approx 1$, anisotropy is incompletely averaged and line width attains a maximum. For $\tau_{\mathrm{r}} \Delta \omega \gg 1$, the solid-state spectrum is observed. The phenomena can be described as a multi-site exchange between the various orientations of the molecule (see Section 10.1.4), which is analogous to the chemical exchange discussed in the NMR part of the lecture course.
For the electron Zeeman interaction, fast tumbling leads to an average resonance field
$B_{0, \mathrm{res}}=\frac{h \nu_{\mathrm{mw}}}{g_{\mathrm{iso}} \mu_{\mathrm{B}}}$
with the isotropic $g$ value $g_{\text {iso }}=\left(g_{x}+g_{y}+g_{z}\right) / 3$. For small organic radicals in non-viscous solvents at X-band frequencies around $9.5 \mathrm{GHz}$, line broadening from $g$ anisotropy is negligible. At W-band frequencies of $94 \mathrm{GHz}$ for organic radicals and already at X-band frequencies for small transition metal complexes, such broadening can be substantial. For large macromolecules or in viscous solvents, solid-state like EPR spectra can be observed in liquid solution.
Solid state
For a single-crystal sample, the resonance field at any given orientation can be computed by Eq. (3.7). Often, only microcrystalline powders are available or the sample is measured in glassy frozen solution. Under such conditions, all orientations contribute equally. With respect to the
polar angles, this implies that $\phi$ is uniformly distributed, whereas the probability to encounter a certain angle $\theta$ is proportional to $\sin \theta$ (Figure 3.3). The line shape of the absorption spectrum is most easily understood for axial symmetry of the $g$ tensor. Transitions are observed only in the range between the limiting resonance fields at $g_{\|}$and $g_{\perp}$. The spectrum has a global maximum at $g_{\perp}$ and a minimum at $g_{\|}$.
In CW EPR spectroscopy we do not observe the absorption line shape, but rather its first derivative (see Chapter 7). This derivative line shape has sharp features at the line shape singularities of the absorption spectrum and very weak amplitude in between (Figure $3.4$ ).
Concept 3.3.1 - Orientation selection.
The spread of the spectrum of a powder sample or glassy frozen solution allows for selecting molecules with a certain orientation with respect to the magnetic field. For an axial $g$ tensor only orientations near the $z$ axis of the $g$ tensor PAS are selected when observing near the resonance field of $g_{\|}$. In contrast, when observing near the resonance field for $g_{\perp}$, orientations withing the whole $x y$ plane of the PAS contribute. For the case of orthorhombic symmetry with three distinct principal values $g_{x}, g_{y}$, and $g_{z}$, narrow sets of orientations can be observed at the resonance fields corresponding to the extreme $g$ values $g_{x}$ and $g_{z}$ (see right top panel in Figure 3.4). At the intermediate principal value $g_{y}$ a broad range of orientations contributes, because the same resonance field can be realized by orientations other than $\phi=90^{\circ}$ and $\theta=90^{\circ}$. Such orientation selection can enhance the resolution of ENDOR and ESEEM spectra (Chapter 8) and simplify their interpretation.
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The magnetic moments of an electron and a nuclear spin couple by the magnetic dipole-dipole interaction; similar to the dipole-dipole interaction between nuclear spins discussed in the NMR part of the lecture course. The main difference to the NMR case is that, in many cases, a point-dipole description is not a good approximation for the electron spin, as the electron is distributed over the SOMO. The nucleus under consideration can be considered as well localized in space. We now picture the SOMO as a linear combination of atomic orbitals. Contributions from spin density in an atomic orbital of another nucleus (population of the unpaired electron in such an atomic orbital) can be approximated by assuming that the unpaired electron is a point-dipole localized at this other nucleus.
For spin density in atomic orbitals on the same nucleus, we have to distinguish between types of atomic orbitals. In $s$ orbitals, the unpaired electron has finite probability density for residing at the nucleus, at zero distance $r_{S I}$ to the nuclear spin. This leads to a singularity of the dipole-dipole interaction, since this interaction scales with $r_{S I}^{-3}$. The singularity has been treated by Fermi. The contribution to the hyperfine coupling from spin density in $s$ orbitals on the nucleus under consideration is therefor called Fermi contact interaction. Because of the spherical symmetry of $s$ orbitals, the Fermi contact interaction is purely isotropic.
For spin density in other orbitals ( $p, d, f$ orbitals) on the nucleus under consideration, the dipole-dipole interaction must be averaged over the spatial distribution of the electron spin in these orbitals. This average has no isotropic contribution. Therefore, spin density in $p, d, f$ orbitals does not influence spectra of fast tumbling radicals or metal complexes in liquid solution and neither does spin density in $s$ orbitals of other nuclei. The isotropic couplings detected in solution result only from the Fermi contact interaction.
Since the isotropic and purely anisotropic contributions to the hyperfine coupling have different physical origin, we separate these contributions in the hyperfine tensor $\mathbf{A}_{k i}$ that describes the interaction between electron spin $S_{k}$ and nuclear spin $I_{i}$ :
$\mathbf{A}_{k i}=A_{\mathrm{iso}, k i}\left(\begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right)+\mathbf{T}_{k i}$
where $A_{\text {iso, } k i}$ is the isotropic hyperfine coupling and $\mathbf{T}_{k i}$ the purely anisotropic coupling. In the following, we drop the electron and nuclear spin indices $k$ and $i$.
Dipole-dipole hyperfine interaction
The anisotropic hyperfine coupling tensor $\mathbf{T}$ of a given nucleus can be computed from the ground state wavefunction $\psi_{0}$ by applying the correspondence principle to the classical interaction between two point dipoles
$T_{i j}=\frac{\mu_{0}}{4 \pi \hbar} g_{e} \mu_{\mathrm{B}} g_{\mathrm{n}} \mu_{\mathrm{n}}\left\langle\psi_{0}\left|\frac{3 r_{i} r_{j}-\delta_{i j} r^{2}}{r^{5}}\right| \psi_{0}\right\rangle$
Such computations are implemented in quantum chemistry programs such as ORCA, ADF, or Gaussian. If the SOMO is considered as a linear combination of atomic orbitals, the contributions from an individual orbital can be expressed as the product of spin density in this orbital with a spatial factor that can be computed once for all. The spatial factors have been tabulated [KM85]. In general, nuclei of elements with larger electronegativity have larger spatial factors. At the same spatial factor, such as for isotopes of the same element, the hyperfine coupling is proportional to the nuclear $g$ value $g_{\mathrm{n}}$ and thus proportional to the gyromagnetic ratio of the nucleus. Hence, a deuterium coupling can be computed from a known proton coupling or vice versa.
A special situation applies to protons, alkali metals and earth alkaline metals, which have no significant spin densities in $p-, d-$, or $f$-orbitals. In this case, the anisotropic contribution can only arise from through-space dipole-dipole coupling to centers of spin density at other nuclei. In a point-dipole approximation the hyperfine tensor is then given by
$\mathbf{T}=\frac{\mu_{0}}{4 \pi \hbar} g_{e} \mu_{\mathrm{B}} g_{\mathrm{n}} \mu_{\mathrm{n}} \sum_{j \neq i} \rho_{j} \frac{3 \vec{n}_{i j} \vec{n}_{i j}^{\mathrm{T}}-\overrightarrow{1}}{R_{i j}^{3}}$
where the sum runs over all nuclei $j$ with significant spin density $\rho_{j}$ (summed over all orbitals at this nucleus) other than nucleus $i$ under consideration. The $R_{i j}$ are distances between the nucleus under consideration and the centers of spin density, and the $\vec{n}_{i j}$ are unit vectors along the direction from the considered nucleus to the center of spin density. For protons in transition metal complexes it is often a good approximation to consider spin density only at the central metal ion. The distance $R$ from the proton to the central ion can then be directly inferred from the anisotropic part of the hyperfine coupling.
Hyperfine tensor contributions $\mathbf{T}$ computed by any of these ways must be corrected for the influence of $\mathrm{SOC}$ if the $g$ tensor is strongly anisotropic. If the dominant contribution to $\mathrm{SOC}$ arises at a single nucleus, the hyperfine tensor at this nucleus ${ }^{1}$ can be corrected by
$\mathbf{T}^{(\mathrm{g})}=\frac{\mathbf{g} \mathbf{T}}{g_{e}}$
The product g $\mathbf{T}$ may have an isotropic part, although $\mathbf{T}$ is purely anisotropic. This isotropic pseudocontact contribution depends on the relative orientation of the $g$ tensor and the spin-only dipole-dipole hyperfine tensor $\mathbf{T}$. The correction is negligible for most organic radicals, but not for paramagnetic metal ions. If contributions to $\mathrm{SOC}$ arise from several centers, the necessary correction cannot be written as a function of the $g$ tensor.
Fermi contact interaction
The Fermi contact contribution takes the form
$A_{\text {iso }}=\rho_{s} \cdot \frac{2}{3} \frac{\mu_{0}}{\hbar} g_{e} \mu_{\mathrm{B}} g_{\mathrm{n}} \mu_{\mathrm{n}}\left|\psi_{0}(0)\right|^{2}$
${ }^{1}$ Most literature holds that the correction should be done for all nuclei. As pointed out by Frank Neese, this is not true. An earlier discussion of this point is found in [Lef67] where $\rho_{s}$ is the spin density in the $s$ orbital under consideration, $g_{\mathrm{n}}$ the nuclear $g$ value and $\mu_{\mathrm{n}}=\beta_{\mathrm{n}}=5.05078317(20) \cdot 10^{-27} \mathrm{~J} \mathrm{~T}^{-1}$ the nuclear magneton $\left(g_{\mathrm{n}} \mu_{\mathrm{n}}=\gamma_{\mathrm{n}} \hbar\right)$. The factor $\left|\psi_{0}(0)\right|^{2}$ denotes the probability to find the electron at this nucleus in the ground state with wave function $\psi_{0}$ and has been tabulated [KM85].
Spin polarization
The contributions to the hyperfine coupling discussed up to this point can be understood and computed in a single-electron picture. Further contributions arise from correlation of electrons in a molecule. Assume that the $p_{z}$ orbital on a carbon atom contributes to the SOMO, so that the $\alpha$ spin state of the electron is preferred in that orbital (Fig. 4.1). Electrons in other orbitals on the same atom will then also have a slight preference for the $\alpha$ state (left panel), as electrons with the same spin tend to avoid each other and thus have less electrostatic repulsion. ${ }^{2}$ In particular, this means that the spin configuration in the left panel of Fig. $4.1$ is slightly more preferable than the one in the right panel. According to the Pauli principle, the two electrons that share the $s$ bond orbital of the $\mathrm{C}-\mathrm{H}$ bond must have antiparallel spin. Thus, the electron in the $s$ orbital of the hydrogen atom that is bound to the spin-carrying carbon atom has a slight preference for the $\beta$ state. This corresponds to a negative isotropic hyperfine coupling of the directly bound $\alpha$ proton, which is induced by the positive hyperfine coupling of the adjacent carbon atom. The effect is termed "spin polarization", although it has no physical relation to the polarization of electron spin transitions in an external magnetic field.
Spin polarization is important, as it transfers spin density from $p$ orbitals, where it is invisible in liquid solution and from carbon atoms with low natural abundance of the magnetic isotope ${ }^{13} \mathrm{C}$ to $s$ orbitals on protons, where it can be easily observed in liquid solution. This transfer occurs, both, in $\sigma$ radicals, where the unpaired electron is localized on a single atom, and in $\pi$ radicals, where it is distributed over the $\pi$ system. The latter case is of larger interest, as the distribution of the $\pi$ orbital over the nuclei can be mapped by measuring and assigning the isotropic proton hyperfine couplings. This coupling can be predicted by the McConnell equation
$A_{\mathrm{iso}, \mathrm{H}}=Q_{\mathrm{H}} \rho_{\pi}$
where $\rho_{\pi}$ is the spin density at the adjacent carbon atom and $Q_{\mathrm{H}}$ is a parameter of the order of $-2.5 \mathrm{mT}$, which slightly depends on structure of the $\pi$ system.
${ }^{2}$ This preference for electrons on the same atom to have parallel spin is also the basis of Hund’s rule.
The McConnell equation is mainly applied for mapping the LUMO and HOMO of aromatic molecules (Figure 4.2). An unpaired electron can be put into these orbitals by one-electron reduction or oxidation, respectively, without perturbing the orbitals too strongly. The isotropic hyperfine couplings of the hydrogen atom directly bound to a carbon atom report on the contribution of the $p_{z}$ orbital of this carbon atom to the $\pi$ orbital. The challenges in this mapping are twofold. First, it is hard to assign the observed couplings to the hydrogen atoms unless a model for the distribution of the $\pi$ orbital is already available. Second, the method is blind to carbon atoms without a directly bonded hydrogen atom.
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We consider the interaction of a single electron spin $S$ with a single nuclear spin $I$ and thus drop the sums and indices $k$ and $i$ in $\hat{\mathcal{H}}_{\mathrm{HFI}}$ in Eq. (2.4). In general, all matrix elements of the hyperfine tensor $\mathbf{A}$ will be non-zero after the Bleaney transformation to the frame where the electron Zeeman interaction is along the $z$ axis (see Eq. 3.5). The hyperfine Hamiltonian is then given by
\begin{aligned} \hat{\mathcal{H}}_{\mathrm{HFI}}=&\left(\begin{array}{lll} \hat{S}_{x} & \hat{S}_{y} & \hat{S}_{z} \end{array}\right)\left(\begin{array}{ccc} A_{x x} & A_{x y} & A_{x z} \ A_{y x} & A_{y y} & A_{y z} \ A_{z x} & A_{z y} & A_{z z} \end{array}\right)\left(\begin{array}{c} \hat{I}_{x} \ \hat{I}_{y} \ \hat{I}_{z} \end{array}\right) \ =& A_{x x} \hat{S}_{x} \hat{I}_{x}+A_{x y} \hat{S}_{x} \hat{I}_{y}+A_{x y} \hat{S}_{x} \hat{I}_{z} \ &+A_{y x} \hat{S}_{y} \hat{I}_{x}+A_{y y} \hat{S}_{y} \hat{I}_{y}+A_{y z} \hat{S}_{y} \hat{I}_{z} \ &+A_{z x} \hat{S}_{z} \hat{I}_{x}+A_{z y} \hat{S}_{z} \hat{I}_{y}+A_{z z} \hat{S}_{z} \hat{I}_{z} \end{aligned}
Note that the $z$ axis of the nuclear spin coordinate system is parallel to the magnetic field vector $\vec{B}_{0}$ whereas the one of the electron spin system is tilted, if $g$ anisotropy is significant. Hence, the hyperfine tensor is not a tensor in the strict mathematical sense, but rather an interaction matrix.
In Eq. (4.7), the term $A_{z z} \hat{S}_{z} \hat{I}_{z}$ is secular and must always be kept. Usually, the high-field approximation does hold for the electron spin, so that all terms containing $\hat{S}_{x}$ or $\hat{S}_{y}$ operators are non-secular and can be dropped. The truncated hyperfine Hamiltonian thus reads
$\hat{\mathcal{H}}_{\mathrm{HFI}, \text { trunc }}=A_{z x} \hat{S}_{z} \hat{I}_{x}+A_{z y} \hat{S}_{z} \hat{I}_{y}+A_{z z} \hat{S}_{z} \hat{I}_{z}$
The first two terms on the right-hand side can be considered as defining an effective transverse coupling that is the sum of a vector with length $A_{z x}$ along $x$ and a vector of length $A_{z y}$ along $y$. The length of the sum vector is $B=\sqrt{A_{z x}^{2}+A_{z y}^{2}}$. The truncated hyperfine Hamiltonian simplifies if we take the laboratory frame $x$ axis for the nuclear spin along the direction of this effective transverse hyperfine coupling. In this frame we have
$\hat{\mathcal{H}}_{\text {HFI,trunc }}=A \hat{S}_{z} \hat{I}_{z}+B \hat{S}_{z} \hat{I}_{x}$
where $A=A_{z z}$ quantifies the secular hyperfine coupling and $B$ the pseudo-secular hyperfine coupling. The latter coupling must be considered if and only if the hyperfine coupling violates the high-field approximation for the nuclear spin (see Chapter 6).
If $g$ anisotropy is very small, as is the case for organic radicals, the $z$ axes of the two spin coordinate systems are parallel. In this situation and for a hyperfine tensor with axial symmetry, $A$ and $B$ can be expressed as
\begin{aligned} &A=A_{\mathrm{iso}}+T\left(3 \cos ^{2} \theta_{\mathrm{HFI}}-1\right) \ &B=3 T \sin \theta_{\mathrm{HFI}} \cos \theta_{\mathrm{HFI}} \end{aligned}
where $\theta_{\text {HFI }}$ is the angle between the static magnetic field $\vec{B}_{0}$ and the symmetry axis of the hyperfine tensor and $T$ is the anisotropy of the hyperfine coupling. The principal values of the hyperfine tensor are $A_{x}=A_{y}=A_{\perp}=A_{\text {iso }}-T$ and $A_{z}=A_{\|}=A_{\text {iso }}+2 T$. The pseudo-secular contribution $B$ vanishes along the principal axes of the hyperfine tensor, where $\theta_{\mathrm{HFI}}$ is either $0^{\circ}$ or $90^{\circ}$ or for a purely isotropic hyperfine coupling. Hence, the pseudo-secular contribution can also be dropped when considering fast tumbling radicals in the liquid state. We now consider the point-dipole approximation, where the electron spin is well localized on the length scale of the electron-nuclear distance $r$ and assume that $T$ arises solely from through-space interactions. This applies to hydrogen, alkali and earth alkali ions. We then find
$T=\frac{1}{r^{3}} \frac{\mu_{0}}{4 \pi \hbar} g_{e} \mu_{\mathrm{B}} g_{\mathrm{n}} \mu_{\mathrm{n}}$
For the moment we assume that the pseudo-secular contribution is either negligible or can be considered as a small perturbation. The other case is treated in Chapter 6 . To first order, the contribution of the hyperfine interaction to the energy levels is then given by $m_{S} m_{I} A$. In the EPR spectrum, each nucleus with spin $I$ generates $2 I+1$ electron spin transitions with $\left|\Delta m_{S}\right|=1$ that can be labeled by the values of $m_{I}=-I,-I+1, \ldots I$. In the nuclear frequency spectrum, each nucleus exhibits $2 S+1$ transitions with $\left|\Delta m_{I}\right|=1$. For nuclear spins $I>1 / 2$ in the solid state, each transition is further split into $2 I$ transitions by the nuclear quadrupole interaction. The contribution of the secular hyperfine coupling to the electron transition frequencies is $m_{I} A$, whereas it is $m_{S} A$ for nuclear transition frequencies. In both cases, the splitting between adjacent lines of a hyperfine multiplet is given by $A$.
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Liquid-solution EPR spectra
Since each nucleus splits each electron spin transition into $2 I+1$ transitions with different frequencies, the number of EPR transitions is $\prod_{i}\left(2 I_{i}+1\right)$. Some of these transitions may coincide if hyperfine couplings are the same or integer multiples of each other. An important case, where hyperfine couplings are exactly the same are chemically equivalent nuclei. For instance, two nuclei $I_{1}=I_{2}=1 / 2$ can have spin state combinations $\alpha_{1} \alpha_{2}, \alpha_{1} \beta_{2}, \beta_{1} \alpha_{2}$, and $\beta_{1} \beta_{2}$. The contributions to the transition frequencies are $\left(A_{1}+A_{2}\right) / 2,\left(A_{1}-A_{2}\right) / 2,\left(-A_{1}+A_{2}\right) / 2$, and
$\left(-A_{1}-A 2\right) / 2$. For equivalent nuclei with $A_{1}=A_{2}=A$ only three lines are observed with hyperfine shifts of $A, 0$, and $-A$ with respect to the electron Zeeman frequency. The unshifted center line has twice the amplitude than the shifted lines, leading to a $1: 2: 1$ pattern with splitting $A$. For $k$ equivalent nuclei with $I_{i}=1 / 2$ the number of lines is $k+1$ and the relative intensities can be inferred from Pascal’s triangle. For a group of $k_{i}$ equivalent nuclei with arbitrary spin quantum number $I_{i}$ the number of lines is $2 k_{i} I_{i}+1$. The multiplicities of groups of equivalent nuclei multiply. Hence, the total number of EPR lines is
$n_{\mathrm{EPR}}=\prod_{i}\left(2 k_{i} I_{i}+1\right)$
where index $i$ runs over the groups of equivalent nuclei.
Figure $4.3$ illustrates on the example of the phenyl radical how the multiplet pattern arises. For radicals with more extended $\pi$ systems, the number of lines can be very large and it may become impossible to fully resolve the spectrum. Even if the spectrum is fully resolved, analysis of the multiplet pattern may be a formidable task. An algorithm that works well for analysis of patterns with a moderate number of lines is given in [CCM16].
Liquid-solution nuclear frequency spectra
As mentioned in Section $4.2$ the secular hyperfine coupling $A$ can be inferred from nuclear frequency spectra as well as from EPR spectra. Line widths are smaller in the nuclear frequency spectra, since nuclear spins have longer transverse relaxation times $T_{2, i}$. Another advantage of nuclear frequency spectra arises from the fact that the electron spin interacts with all nuclear spins whereas each nuclear spin interacts with only one electron spin (Figure 4.4). The number of lines in nuclear frequency spectra thus grows only linearly with the number of nuclei, whereas
it grows exponentially in EPR spectra. In liquid solution, each group of equivalent nuclear spins adds $2 S+1$ lines, so that the number of lines for $N_{\text {eq }}$ such groups is
$n_{\mathrm{NMR}}=(2 S+1) N_{\mathrm{eq}}$
The nuclear frequency spectra in liquid solution can be measured by CW ENDOR, a technique that is briefly discussed in Section 8.1.2.
A complication in interpretation of nuclear frequency spectra can arise from the fact that the hyperfine interaction may be larger than the nuclear Zeeman interaction. This is illustrated in Figure 4.5. Only in the weak-coupling case with $|A| / 2<\left|\omega_{I}\right|$ the hyperfine doublet in nuclear frequency spectra is centered at $\left|\omega_{I}\right|$ and split by $|A|$. In the strong-coupling case, hyperfine sublevels cross for one of the electron spin states and the nuclear frequency $\left|\omega_{I}\right|-|A| / 2$ becomes negative. As the sign of the frequency is not detected, the line is found at frequency $|A| / 2-\left|\omega_{I}\right|$ instead, i.e., it is "mirrored" at the zero frequency. This results in a doublet centered at frequency $|A| / 2$ and split by $2\left|\omega_{I}\right|$. Recognition of such cases in well resolved liquid-state spectra is simplified by the fact that the nuclear Zeeman frequency $\left|\omega_{I}\right|$ can only assume a few values that are known if the nuclear isotopes in the molecule and the magnetic field are known. Figure $4.6$ illustrates how the nuclear frequency spectrum of the phenyl radical is constructed based on such considerations. The spectrum has only 6 lines, compared to the 18 lines that arise in the EPR spectrum in Figure 4.3.
Solid-state EPR spectra
In the solid state, construction of the EPR spectra is complicated by the fact that the electron Zeeman interaction is anisotropic. At each individual orientation of the molecule, the spectrum looks like the pattern in liquid state, but both the central frequency of the multiplet and the hyperfine splittings depend on orientation. As these frequency distributions are continuous, resolved splittings are usually observed only at the singularities of the line shape pattern of the interaction with the largest anisotropy. For organic radicals at X-band frequencies, often hyperfine anisotropy dominates. At high frequencies or for transition metal ions, often electron Zeeman anisotropy dominates. The exact line shape depends not only on the principal values of the $g$ tensor and the hyperfine tensors, but also on relative orientation of their PASs. The general case is complicated and requires numerical simulations, for instance, by EasySpin.
However, simple cases, where the hyperfine interaction of only one nucleus dominates and the PASs of the $g$ and hyperfine tensor coincide, are quite often encountered. For instance, Cu(II) complexes are often square planar and, if all four ligands are the same, have a $C_{4}$ symmetry axis. The $g$ tensor than has axial symmetry with the $C_{4}$ axis being the unique axis. The hyperfine tensors of ${ }^{63} \mathrm{Cu}$ and ${ }^{65} \mathrm{Cu}$ have the same symmetry and the same unique axis. The two isotopes both have spin $I=3 / 2$ and very similar gyromagnetic ratios. The spectra can thus be understood by considering one electron spin $S=1 / 2$ and one nuclear spin $I=3 / 2$ with axial $g$ and hyperfine tensors with a coinciding unique axis.
In this situation, the subspectra for each of the nuclear spin states $m_{I}=-3 / 2,-1 / 2,+1 / 2$, and $+3 / 2$ take on a similar form as shown in Figure $3.3$. The resonance field can be computed by solving
$\hbar \omega_{\mathrm{mw}}=\frac{\mu_{\mathrm{B}}}{B_{0, \mathrm{res}}}\left(2 g_{\perp}^{2} \sin ^{2} \theta+g_{\|}^{2} \cos ^{2} \theta\right)+m_{I}\left[A_{\text {iso }}+T\left(3 \cos ^{2} \theta-1\right)\right]$
where $\theta$ is the angle between the $C_{4}$ symmetry axis and the magnetic field vector $\vec{B}_{0}$. The singularities are encountered at $\theta=0^{\circ}$ and $\theta=90^{\circ}$ and correspond to angular frequencies $\mu_{\mathrm{B}} B_{0} g_{\|}+m_{I} A_{\|}$and $\mu_{\mathrm{B}} B_{0} g_{\perp}+m_{I} A_{\perp}$.
The construction of a Cu(II) EPR spectrum according to these considerations is shown in Figure 4.7. The values of $g_{\|}$and $A_{\|}$can be inferred by analyzing the singularities near the low-field edge of the spectrum. Near the high-field edge, the hyperfine splitting $A_{\perp}$ is usually not resolved. Here, $g_{\perp}$ corresponds to the maximum of the absorption spectrum and to the zero crossing of its derivative.
Solid-state nuclear frequency spectra
Again, a simpler situation is encountered in nuclear frequency spectra, as the nuclear Zeeman frequency is isotropic and chemical shift anisotropy is negligibly small compared to hyperfine anisotropy. Furthermore, resolution is much better for the reasons discussed above, so that smaller hyperfine couplings and anisotropies can be detected. If anisotropy of the hyperfine coupling is dominated by through-space dipole-dipole coupling to a single center of spin density, as is often the case for protons, or by contribution from spin density in a single $p$ or $d$ orbital, as is often the case for other nuclei, the hyperfine tensor has nearly axial symmetry. In this case, one can infer from the line shapes whether the weak-or strong-coupling case applies and whether the isotropic hyperfine coupling is positive or negative (Figure 4.8). The case with $A_{\text {iso }}=0$ corresponds to the Pake pattern discussed in the NMR part of the lecture course.
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Physical origin and consequences of the exchange interaction
If two unpaired electrons occupy SOMOs in the same molecule or in spatially close molecules, the wave functions $\psi_{1}$ and $\psi_{2}$ of the two SOMOs may overlap. The two unpaired electrons can couple either to a singlet state or to a triplet state. The energy difference between the singlet and triplet state is the exchange integral
$J=-2 e^{2} \iint \frac{\psi_{1}^{*}\left(r_{1}\right) \psi_{2}^{*}\left(r_{2}\right) \psi_{1}\left(r_{2}\right) \psi_{2}\left(r_{1}\right)}{\left|\vec{r}_{1} \vec{r}_{2}\right|} \mathrm{d} \vec{r}_{1} \mathrm{~d} \vec{r}_{2}$
There exist different conventions for the sign of $J$ and the factor 2 may be missing in parts of the literature. With the sign convention used here, the singlet state is lower in energy for positive $J$. Since the singlet state $S$ with spin wave function $(|\alpha \beta\rangle-|\beta \alpha\rangle) / \sqrt{2}$ is antisymmetric with respect to exchange of the two electrons and electrons are Fermions, it corresponds to the situation where the two electrons could also occupy the same orbital. This is a bonding orbital overlap, corresponding to an antiferromagnetic spin ordering. Negative $J$ correspond to a lower-lying triplet state, i.e., antibonding orbital overlap and ferromagnetic spin ordering. The triplet state has three substates with wave functions $|\alpha \alpha\rangle$ for the $\mathrm{T}_{+}$state, $(|\alpha \beta\rangle+|\beta \alpha\rangle) / \sqrt{2}$ for the $\mathrm{T}_{0}$ state, and $|\beta \beta\rangle$ for the $\mathrm{T}_{-}$state. The $\mathrm{T}_{+}$and $\mathrm{T}_{-}$state are eigenstates both in the absence and presence of the $J$ coupling. The states $\mathrm{S}$ and $\mathrm{T}_{0}$ are eigenstates for $J \gg \Delta \omega$, where $\Delta \omega$ is the difference between the electron Zeeman frequencies of the two spins. For the opposite case of $\Delta \omega \gg J$, the eigenstates are $|\alpha \beta\rangle$ and $|\beta \alpha\rangle$. The latter case corresponds to the high-field approximation with respect to the exchange interaction.
For strong exchange, $J \gg \Delta \omega$, the energies are approximately $-(3 / 4) J$ for the singlet state and $J / 4-\omega_{S}, J / 4$ and $J / 4+\omega_{S}$ for the triplet substates $\mathrm{T}_{-}, \mathrm{T}_{0}$, and $\mathrm{T}_{+}$, respectively, where $\omega_{S}$ is the electron Zeeman interaction, which is the same for both spins within this approximation. If $J \gg 2 \pi \nu_{\mathrm{mw}}$, microwave photons with energy $h \nu_{\mathrm{mw}}$ cannot excite transitions between the singlet and triplet subspace of spin Hilbert space. It is then convenient to use a coupled representation and consider the two subspaces separately from each other. The singlet subspace corresponds to a diamagnetic molecule and does not contribute to EPR spectra. The triplet subspace can be described by a group spin $S=1$ of the two unpaired electrons. In the coupled representation, $J$ does not enter the spin Hamiltonian, as it shifts all subspace levels by the same energy. For $J<0$, the triplet state is the ground state and is always observable by EPR spectroscopy. However, usually one has $J>0$ and the singlet state is the ground state. As long as $\hbar J$ does not exceed thermal energy $k_{\mathrm{B}} T$ by a large factor, the triplet state is thermally excited and observable. In this case, EPR signal amplitude may increase rather than decrease with increasing temperature. For organic molecules, this case is also rare. If $\hbar J \gg k_{\mathrm{B}} T$, the compound does not give an EPR signal. It may still be possible to observe the triplet state transiently after photoexcitation to an excited singlet state and intersystem crossing to the triplet state.
Weak exchange coupling is observed in biradicals with well localized SOMOs that are separated on length scales between $0.5$ and $1.5 \mathrm{~nm}$. In such cases, exchange coupling $J$ decreases exponentially with the distance between the two electrons or with the number of conjugated bonds that separate the two centers of spin density. If the two centers are not linked by a continuous chain of conjugated bonds, exchange coupling is rarely resolved at distances larger than $1.5 \mathrm{~nm}$. In any case, at such long distances exchange coupling is much smaller than the dipole-dipole coupling between the two unpaired electrons if the system is not conjugated. For weak exchange coupling, the system is more conveniently described in an uncoupled representation with two spins $S_{1}=1 / 2$ and $S_{2}=1 / 2$.
Exchange coupling is also significant during diffusional encounters of two paramagnetic molecules in liquid solution. Such dynamic Heisenberg spin exchange can be pictured as physical exchange of unpaired electrons between the colliding molecules. This causes a sudden change of the spin Hamiltonian, which leads to spin relaxation. A typical example is line broadening in EPR spectra of radicals by oxygen, which has a paramagnetic triplet ground state. If radicals of the same type collide, line broadening is also observed, but the effects on the spectra can be more subtle, since the spin Hamiltonians of the colliding radicals are the same. In this case, exchange of unpaired electrons between the radicals changes only spin state, but not the spin Hamiltonian.
Exchange Hamiltonian
The spin Hamiltonian contribution by weak exchange coupling is
$\hat{\mathcal{H}}_{\mathrm{EX}}=J\left(\hat{S}_{1 x} \hat{S}_{2 x}+\hat{S}_{1 y} \hat{S}_{2 y}+\hat{S}_{1 z} \hat{S}_{2 z}\right)$
This Hamiltonian is analogous to the $J$ coupling Hamiltonian in NMR spectroscopy. If the two spins have different $g$ values and the field is sufficiently high $\left(g \mu_{\mathrm{B}} B_{0} / \hbar \gg J\right)$, the exchange Hamiltonian can be truncated in the same way as the $J$ coupling Hamiltonian in heteronuclear NMR:
$\hat{\mathcal{H}}_{\text {EX,trunc }}=J \hat{S}_{1 z} \hat{S}_{2 z}$
Spectral manifestation of the exchange interaction
In the absence of hyperfine coupling, the situation is the same as for $J$ coupling in NMR spectroscopy. Exchange coupling between like spins (same electron Zeeman frequency) does not influence the spectra. For radicals in liquid solution, hyperfine coupling is usually observable. In this case, exchange coupling does influence the spectra even for like spins, as illustrated in Figure $5.1$ for two exchange-coupled electron spins $S_{1}=1 / 2$ and $S_{2}=1 / 2$ with each of them coupled exclusively to only one nuclear spin $\left(I_{1}=1\right.$ and $I_{2}=1$, respectively) with the same hyperfine coupling $A_{\mathrm{iso}}$. If the exchange coupling is much smaller than the isotropic hyperfine coupling, each of the individual lines of the hyperfine triplet further splits into three lines. If the splitting is very small, it may be noticeable only as a line broadening. At very large exchange coupling, the electron spins are uniformly distributed over the two exchange-coupled moieties. Hence, each of them has the same hyperfine coupling to both nuclei. This coupling is half the original hyperfine coupling, since, on average, the electron spin has only half the spin density in the orbitals of a given nucleus as compared to the case without exchange coupling. For intermediate exchange couplings, complex splitting patterns arise that are characteristic for the ratio between the exchange and hyperfine coupling.
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Physical picture
The magnetic dipole-dipole interaction between two localized electron spins with magnetic moments $\mu_{1}$ and $\mu_{2}$ takes the same form as the classical interaction between two magnetic point dipoles. The interaction energy
$E=-\frac{\mu_{0}}{4 \pi} \cdot \mu_{1} \mu_{2} \cdot \frac{1}{r^{3}} \cdot\left(2 \cos \theta_{1} \cos \theta_{2}-\sin \theta_{1} \sin \theta_{2} \cos \phi\right)$
generally depends on the two angles $\theta_{1}$ and $\theta_{2}$ that the point dipoles include with the vector between them and on the dihedral angle $\phi$ (Figure 5.2). The dipole-dipole interaction scales with the inverse cube of the distance between the two point dipoles.
In general, the two electron spins are spatially distributed in their respective SOMOs. The point-dipole approximation is still a good approximation if the distance $r$ is much larger than the spatial distribution of each electron spin. Further simplification is possible if $g$ anisotropy is much smaller than the isotropic $g$ value. In that case, the two spins are aligned parallel to the magnetic field and thus also parallel to each other, so that $\theta_{1}=\theta_{2}=\theta$ and $\phi=0$. Eq. (5.4) then simplifies to
$E=-\frac{\mu_{0}}{4 \pi} \cdot \mu_{1} \mu_{2} \cdot \frac{1}{r^{3}} \cdot\left(3 \cos ^{2} \theta-1\right)$
which is the form known from NMR spectroscopy.
Dipole-dipole Hamiltonian
For two electron spins that are not necessarily aligned parallel to the external magnetic field, the dipole-dipole coupling term of the spin Hamiltonian assumes the form
$\widehat{H}_{\mathrm{dd}}=\widehat{S}_{1}^{\mathrm{T}} \underline{D} \widehat{S}_{2}=\frac{1}{r^{3}} \cdot \frac{\mu_{0}}{4 \pi \hbar} \cdot g_{1} g_{2} \mu_{\mathrm{B}}^{2}\left[\widehat{S}_{1} \widehat{S}_{2}-\frac{3}{r^{2}}\left(\widehat{S}_{1} \vec{r}\right)\left(\widehat{S}_{2} \vec{r}\right)\right]$
If the electrons are distributed in space, the Hamiltonian has to be averaged (integrated) over the two spatial distributions, since electron motion proceeds on a much faster time scale than an EPR experiment.
If the two unpaired electrons are well localized on the length scale of their distances and their spins are aligned parallel to the external magnetic field, the dipole-dipole Hamiltonian takes the form
$\hat{H}_{\mathrm{dd}}=\frac{1}{r^{3}} \cdot \frac{\mu_{0}}{4 \pi \hbar} \cdot g_{1} g_{2} \mu_{\mathrm{B}}^{2}[\hat{A}+\hat{B}+\hat{C}+\hat{D}+\hat{E}+\hat{F}]$
with the terms of the dipolar alphabet
\begin{aligned} \hat{A} &=\hat{S}_{z} \hat{I}_{z}\left(1-3 \cos ^{2} \theta\right) \ \hat{B} &=-\frac{1}{4}\left[\hat{S}^{+} \hat{I}^{-}+\hat{S}^{-} \hat{I}^{+}\right]\left(1-3 \cos ^{2} \theta\right) \ \hat{C} &=-\frac{3}{2}\left[\hat{S}^{+} \hat{I}_{z}+\hat{S}_{z} \hat{I}^{+}\right] \sin \theta \cos \theta e^{-i \phi} \ \hat{D} &=-\frac{3}{2}\left[\hat{S}^{-} \hat{I}_{z}+\hat{S}_{z} \hat{I}^{-}\right] \sin \theta \cos \theta e^{i \phi} \ \hat{E} &=-\frac{3}{4} \hat{S}^{+} \hat{I}^{+} \sin ^{2} \theta e^{-2 i \phi} \ \hat{F} &=-\frac{3}{4} \hat{S}^{-} \hat{I}^{-} \sin ^{2} \theta e^{2 i \phi} \end{aligned}
Usually, EPR spectroscopy is performed at fields where the electron Zeeman interaction is much larger than the dipole-dipole coupling, which has a magnitude of about $50 \mathrm{MHz}$ at a distance of $1 \mathrm{~nm}$ and of $50 \mathrm{kHz}$ at a distance of $10 \mathrm{~nm}$. In this situation, the terms $\hat{C}, \hat{D}, \hat{E}$, and $\hat{F}$ are non-secular and can be dropped. The $\hat{B}$ term is pseudo-secular and can be dropped only if
the difference between the electron Zeeman frequencies is much larger than the dipole-dipole coupling1. In electron electron double resonance (ELDOR) experiments, the difference of the Larmor frequencies of the two coupled spins can be selected via the difference of the two microwave frequencies. It is thus possible to excite spin pairs for which only the secular part of the spin Hamiltonian needs to be considered,
$\widehat{H}_{\mathrm{dd}}=\omega_{\perp}\left(1-3 \cos ^{2} \theta\right) \hat{S}_{z} \hat{I}_{z}$
with
$\omega_{\perp}=\frac{1}{r^{3}} \cdot \frac{\mu_{0}}{4 \pi \hbar} \cdot g_{1} g_{2} \mu_{\mathrm{B}}^{2}$
The dipole-dipole coupling then has a simple dependence on the angle $\theta$ between the external magnetic field $\vec{B}_{0}$ and the spin-spin vector $\vec{r}$ and the coupling can be interpreted as the interaction of the spin with the $z$ component of the local magnetic field that is induced by the magnetic dipole moment of the coupling partner (Figure 5.3). Since the average of the second Legendre polynomial $\left(1-3 \cos ^{2} \theta\right) / 2$ over all angles $\theta$ vanishes, the dipole-dipole interaction vanishes under fast isotropic motion. Measurements of this interaction are therefore performed in the solid state.
The dipole-dipole tensor in the secular approximation has the eigenvalues $\left(\omega_{\perp}, \omega_{\perp},-2 \omega_{\perp}\right)$. The dipole-dipole coupling $d$ at any orientation $\theta$ is given by
$d=\omega_{\perp}\left(1-3 \cos ^{2} \theta\right)$
Spectral manifestation of the dipole-dipole interaction
The energy level scheme and a schematic spectrum for a spin pair with fixed angle $\theta$ are shown in Figure $5.4 \mathrm{a}$ and b, respectively. The dipole-dipole couplings splits the transition of either coupled spin by $d$. If the sample is macroscopically isotropic, for instance a microcrystalline powder or a glassy frozen solution, all angles $\theta$ occur with probability $\sin \theta$. Each line of the dipolar doublet
is then broadened to a powder pattern as illustrated in Figure 3.3. The powder pattern for the $\beta$ state of the partner spin is a mirror image of the one for the $\alpha$ state, since the frequency shifts by the local magnetic field have opposite sign for the two states. The superposition of the two axial powder patterns is called Pake pattern (Figure 5.5). The center of the Pake pattern corresponds to the magic angle $\theta_{\text {magic }}=\arccos \sqrt{1 / 3} \approx 54.7^{\circ}$. The dipole-dipole coupling vanishes at this angle.
The Pake pattern is very rarely observed in an EPR spectrum, since usually other anisotropic interactions are larger than the dipole-dipole interaction between electron spins. If the weakcoupling condition $d \ll\left|\omega_{\mathrm{A}}-\omega_{\mathrm{B}}\right|$ is fulfilled for the vast majority of all orientations, the EPR lineshape is well approximated by a convolution of the Pake pattern with the lineshape in the absence of dipole-dipole interaction. If the latter lineshape is known, for instance from measuring analogous samples that carry only one of the two electron spins, the Pake pattern can be extracted by deconvolution and the distance between the two electron spins can be inferred from the splitting $\omega_{\perp}$ by inverting Eq. (5.15).
1 Hyperfine coupling of the electron spins can modify this condition.
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Physical picture
If several unpaired spins are very strongly exchange coupled, then they are best described by a group spin $S$. The concept is most easily grasped for the case of two electron spins that we have already discussed in Section 5.1.1. In this case, the singlet state with group spin $S=0$ is diamagnetic and thus not observable by EPR. The three sublevels of the observable triplet state with group spin $S=1$ correspond to magnetic quantum numbers $m_{S}=-1,0$, and $+1$ at high field. These levels are split by the electron Zeeman interaction. The transitions $m_{S}=-1 \leftrightarrow 0$ and $m_{S}=0 \leftrightarrow+1$ are allowed electron spin transitions, whereas the transition $m_{S}=-1 \leftrightarrow+1$ is a forbidden double-quantum transition.
At zero magnetic field, the electron Zeeman interaction vanishes, yet the three triplet sublevels are not degenerate, they exhibit zero-field splitting. This is because the unpaired electrons are also dipole-dipole coupled. Integration of Eq. (5.6) over the spatial distribution of the two electron spins in their respective SOMOs provides a zero-field interaction tensor $\mathbf{D}$ that can be cast in a form where it describes coupling of the group spin $S=1$ with itself [Rie07]. At zero field, the triplet sublevels are not described by the magnetic quantum number $m_{S}$, which is a good quantum number only if the electron Zeeman interaction is much larger than the zero-field interaction. Rather, the triplet sublevels at zero field are related to the principal axes directions of the zero-field interaction tensor and are therefore labeled $\mathrm{T}_{x}, \mathrm{~T}_{y}$, and $\mathrm{T}_{z}$, whereas the sublevels in the high-field approximation are labeled $\mathrm{T}_{-1}, \mathrm{~T}_{0}$, and $\mathrm{T}_{+1}$.
This concept can be extended to an arbitrary number of strongly coupled electron spins. Cases with up to 5 strongly coupled unpaired electrons occur for transition metal ions (d shell) and cases with up to 7 strongly coupled unpaired electrons occur for rare earth ions (f shell). According to Hund’s rule, in the absence of a ligand field the state with largest group spin $S$ is the ground state. Kramers ions with an odd number of unpaired electrons have a half-integer group spin $S$. They behave differently from non-Kramers ions with an even number of electrons and integer group spin $S$. This classification relates to Kramers’ theorem, which states that for a time-reversal symmetric system with half-integer total spin, all eigenstates occur as pairs (Kramers pairs) that are degenerate at zero magnetic field. As a consequence, for Kramers ions the ground state at zero field will split when a magnetic field is applied. For any microwave frequency there exists a magnetic field where the transition within the ground Kramers doublet is observable in an EPR spectrum. The same does not apply for integer group spin, where the ground state may not be degenerate at zero field. If the zero-field interaction is larger than the maximum available microwave frequency, non-Kramers ions may be unobservable by EPR spectroscopy although they exist in a paramagnetic high-spin state. Typical examples of such EPR silent non-Kramers ions are high-spin $\mathrm{Ni}(\mathrm{II})\left(3 \mathrm{~d}^{8}, S=1\right)$ and high-spin $\mathrm{Fe}(\mathrm{II})\left(3 \mathrm{~d}^{6}, S=2\right)$. In rare cases, non-Kramers ions are EPR observable, since the ground state can be degenerate at zero magnetic field if the ligand field features axial symmetry. Note also that "EPR silent" non-Kramers ions can become observable at sufficiently high microwave frequency and magnetic field.
For transition metal and rare earth ions, zero-field interaction is not solely due to the dipole-dipole interaction between the electron spins. Spin-orbit coupling also contributes, in many cases even stronger than the dipole-dipole interaction. Quantum-chemical prediction of the zero-field interaction is an active field of research. Quite reasonable predictions can be obtained for transition metal ions, whereas only order-of-magnitude estimates are usually possible for rare earth ions.
Zero-field interaction Hamiltonian
The zero-field interaction Hamiltonian is often given as
$\hat{\mathcal{H}}_{\mathrm{ZFI}}=\overrightarrow{\hat{S}}^{\mathrm{T}} \mathbf{D} \overrightarrow{\hat{S}}^{\overrightarrow{\mathrm{S}}}$
where ${ }^{T}$ denotes the transpose of the spin vector operator. In the principal axes system of the zero-field splitting (ZFS) tensor, the Hamiltonian simplifies to
\begin{aligned} \hat{\mathcal{H}}_{\mathrm{ZFI}} &=D_{x} \hat{S}_{x}^{2}+D_{y} \hat{S}_{y}^{2}+D_{z} \hat{S}_{z}^{2} \ &=D\left[S_{z}^{2}-\frac{1}{3} S(S+1)\right]+E\left(S_{x}^{2}-S_{y}^{2}\right) \end{aligned}
where $D=3 D_{z} / 2$ and $E=\left(D_{x}-D_{y}\right) / 2$. The reduction to two parameters is possible, since $\mathbf{D}$ is a traceless tensor. In other words, the zero-field interaction is purely anisotropic. The $D, E$ notation presumes that $D_{z}$ is the principal value with the largest absolute value $(D$ can be negative). Together with the absence of an isotropic component, this means that $D_{y}$, which is always the intermediate value, is either closer to $D_{x}$ than to $D_{z}$ or exactly in the middle between these two values. Accordingly, $|E| \leq|D / 3|$. At axial symmetry $E=0$. Axial symmetry applies if the system has a $C_{n}$ symmetry axis with $n \geq 3$. At cubic symmetry, both $D$ and $E$ are zero. For group spin $S \geq 2$, the leading term of the $\mathrm{ZFS}$ is then a hexadecapolar contribution that scales with the fourth power of the spin operators $\left(\hat{S}_{x}^{4}, \hat{S}_{y}^{4}, \hat{S}_{z}^{4}\right)$.
In the high-field approximation the ZFS contribution to the Hamiltonian is a $\omega_{D} S_{z}^{2}$ term. In other words, to first order in perturbation theory the contribution of the ZFS to the energy of a spin level with magnetic quantum number $m_{S}$ scales with $m_{S}^{2}$. For an allowed transition $m_{S} \leftrightarrow m_{S}+1$, this contribution is $\omega_{D}\left(2 m_{S}+1\right)$. This contribution vanishes for the central transition $m_{S}=-1 / 2 \leftrightarrow 1 / 2$ of Kramers ions. More generally, because of the scaling of the level energies with $m_{S}^{2}$ to first-order, the contribution of ZFS to transition frequencies vanishes for all $-m_{S} \leftrightarrow+m_{S}$ transitions.
Spectral manifestation of zero-field splitting
Spectra are most easily understood in the high-field approximation. Quite often, deviations from this approximation are significant for the ZFS (see Fig. 2.2), and such deviations are discussed later. The other limiting case, where the ZFS is much larger than the electron Zeeman interaction (Fe(III) and most rare earth ions), is discussed in Section 5.3.4.
For triplet states $(S=1)$ with axial symmetry of the ZFS tensor, the absorption spectrum is a Pake pattern (see Section 5.2.3). With continuous-wave EPR, the derivative of the absorption spectrum is detected, which has the appearance shown in Fig. 5.6(a). A deviation from axial symmetry leads to a splitting of the "horns" of the Pake pattern by $3 E$, whereas the "shoulders" of the pattern are not affected (Fig. 5.6(b)). Triplet states of organic molecules are often observed after optical excitation of a singlet state and intersystem crossing. Such intersystem crossing generally leads to different population of the zero-field triplet sublevels $\mathrm{T}_{x}, \mathrm{~T}_{y}$, and $\mathrm{T}_{z}$. In this situation the spin system is not at thermal equilibrium, but spin polarized. Such spin polarization affects relative intensity of the lineshape singularities in the spectra and even the sign of the signal may change. However, the singularities are still observed at the same resonance fields, i.e., the parameters $D$ and $E$ can still be read off the spectra as indicated in Fig. $5.6$.
Even if the populations of the triple sublevels have relaxed to thermal equilibrium, the spectrum may still differ from the high-field approximation spectrum, as is illustrated in Fig. $5.7$ for the excited naphtalene triplet (simulation performed with an example script of the software package EasySpin http://www.easyspin.org/). For $D=3 \mathrm{GHz}$ at a field of about $160 \mathrm{mT}$ (electron Zeeman frequency of about $4.8 \mathrm{GHz}$ ) the high-field approximation is violated and $m_{S}$ is no longer a good quantum number. Hence, the formally forbidden double-quantum transition $m_{S}=-1 \leftrightarrow+1$ becomes partially allowed. To first order in perturbation theory, this transition is not broadened by the ZFS. Therefore it is very narrow compared to the allowed transitions and appears with higher amplitude.
For Kramers ions, the spectra are usually dominated by the central $m_{S}=-1 / 2 \leftrightarrow 1 / 2$ transition, which is not ZFS-broadened to first order. To second order in perturbation theory, the ZFS-broadening of this line scales inversely with magnetic field. Hence, whereas systems with $g$ anisotropy exhibit broadening proportional to the magnetic field $B_{0}$, central transitions of Kramers ions exhibit narrowing with $1 / B_{0}$. The latter systems can be detected with exceedingly high sensitivity at high fields if they do not feature significant $g$ anisotropy. This applies to systems with half-filled shells (e.g. $\mathrm{Mn}(\mathrm{II}), 3 \mathrm{~d}^{5} ; \mathrm{Gd}(\mathrm{III}), 4 \mathrm{f}^{7}$ ). In the case of Mn(II) (Figure 5.8) the narrow central transition is split into six lines by hyperfine coupling to the nuclear spin of ${ }^{55} \mathrm{Mn}$ (nuclear spin $I=5 / 2,100 \%$ natural abundance). Because of the $\left|2 m_{S}+1\right|$ scaling of anisotropic ZFS broadening of $m_{S} \leftrightarrow m_{S}+1$ transitions, satellite transitions become the broader the larger $\left|m_{S}\right|$ is for the involved levels. In the high-temperature approximation, the integral intensity in the absorption spectrum is the same for all transitions. Hence, broader transitions make a smaller contribution to the amplitude in the absorption spectrum and in its first derivative that is acquired by CW EPR.
The situation can be further complicated by $D$ and $E$ strain, which is a distribution of the $D$ and $E$ parameters due to a distribution in the ligand field. Such a case is demonstrated in Fig. $5.9$ for Gd(III) at a microwave frequency of $34 \mathrm{GHz}$ where second-order broadening of the central transition is still rather strong. In such a case, lineshape singularities are washed out and ZFS parameters cannot be directly read off the spectra. In CW EPR, the satellite transitions may remain unobserved as the derivative of the absorption lineshape is very small except for the central transition.
Effective spin $1 / 2$ in Kramers doublets
For some systems, such as Fe(III), ZFS is much larger than the electron Zeeman interaction at any experimentally attainable magnetic field. In this case, the zero-field interaction determines the quantization direction and the electron Zeeman interaction can be treated as a perturbation [Cas+60]. The treatment is simplest for axial symmetry $(E=0)$, where the quantization axis is the $z$ axis of the ZFS tensor. The energies in the absence of the magnetic field are
$\omega\left(m_{S}\right)=D m_{S}^{2}$
which for high-spin Fe(III) with $S=5 / 2$ gives three degenerate Kramers doublets corresponding to $m_{S}=\pm 5 / 2, \pm 3 / 2$, and $\pm 1 / 2$. If the magnetic field is applied along the $z$ axis of the ZFS tensor, $m_{S}$ is a good quantum number and there is simply an additional energy term $m_{S} g \mu_{\mathrm{B}} B_{0}$ with $g$ being the $g$ value for the half-filled shell, which can be approximated as $g=2$. Furthermore, in this situation only the $m_{S}=-1 / 2 \leftrightarrow 1 / 2$ transition is allowed. The Zeeman term leads to a splitting of the $m_{S}=\pm 1 / 2$ Kramers doublet that is proportional to $B_{0}$ and
corresponds to $g=2$. This Kramers doublet can thus be described as an effective spin $S^{\prime}=1 / 2$ with $g_{\mathrm{eff}}=2$.
If the magnetic field is perpendicular to the $\mathrm{ZFS}$ tensor $z$ axis, the $m_{S}=\pm 5 / 2$ and $\pm 3 / 2$ Kramers doublets are not split, since the $S_{x}$ and $S_{y}$ operator does not connect these levels. The $S_{x}$ operator has an off-diagonal element connecting the $m_{S}=\pm 1 / 2$ levels that is $\sqrt{S(S+1)+1 / 4} / 2=3 / 2$. Since the levels are degenerate in the absence of the electron Zeeman interaction, they become quantized along the magnetic field and $m_{S}$ is again a good quantum number of this Kramers doublet. The energies are $m_{S} 3 g \mu_{\mathrm{B}} B_{0}+D / 4$, so that the transition frequency is again proportional to $B_{0}$, but now with an effective $g$ value $g_{\text {eff }}=6$. Intermediate orientations can be described by assuming an effective $g$ tensor with axial symmetry and $g_{\perp}=6, g_{\|}=2$. This situation is encountered to a good approximation for high-spin Fe(III) in hemoglobins $\left(g_{\perp} \approx 5.88, g_{\|}=2.01\right)$.
For the non-axial case $(E \neq 0)$, the magnetic field $B_{0}$ will split all three Kramers doublets. To first order in perturbation theory the splitting is proportional to $B_{0}$, meaning that each Kramers doublet can be described by an effective spin $S^{\prime}=1 / 2$ with an effective $g$ tensor. Another simple case is encountered for extreme rhombicity, $E=D / 3$. By reordering principal axes (exchanging $z$ with either $x$ or $y$ ) one can the get rid of the $S_{z}^{2}$ term in Eq. (5.18), so that the ZFS Hamiltonian reduces to $E^{\prime}=\left(S_{x}^{2}-S_{y}^{2}\right)$ with $E^{\prime}=2 E$. The level pair corresponding to the new $z$ direction of the $\mathrm{ZFS}$ tensor has zero energy at zero magnetic field and it can be shown that it has an isotropic effective $g$ value $g_{\mathrm{eff}}=30 / 7 \approx 4.286$. Indeed, signals near $g=4.3$ are very often observed for high-spin Fe(III).
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The $S=1 / 2, I=1 / 2$ spin system
The basic phenomena can be well understood in the simplest possible electron-nuclear spin system consisting of a single electron spin $S=1 / 2$ with isotropic $g$ value that is hyperfine coupled to a nuclear spin $I=1 / 2$ with a magnitude of the hyperfine coupling that is much smaller than the electron Zeeman interaction. In this situation the high-field approximation is valid for the electron spin, so that the hyperfine Hamiltonian can be truncated to the form given by Eq. (4.9). Because of the occurrence of an $\hat{S}_{z} \hat{I}_{x}$ operator in this Hamiltonian, we cannot simply transform the Hamiltonian to the rotating frame for the nuclear spin $I$. However, we don’t need to, as we shall consider only microwave irradiation. For the electron spin $S$, we transform to the rotating frame where this spin has a resonance offset $\Omega_{S}$. Hence, the total Hamiltonian takes the form
$\hat{H}_{0}=\Omega_{S} \hat{S}_{z}+\omega_{I} \hat{I}_{z}+A \hat{S}_{z} \hat{I}_{z}+B \hat{S}_{z} \hat{I}_{x}$
in the rotating frame for the electron spin and the laboratory frame for the nuclear spin. Such a Hamiltonian is a good approximation, for instance, for protons in organic radicals.
The Hamiltonian deviates from the Hamiltonian that would apply if the high-field approximation were also fulfilled for the nuclear spin. The difference is the pseudo-secular hyperfine coupling term $B \hat{S}_{z} \hat{I}_{x}$. As can be seen from Eq. (4.10), this term vanishes if the hyperfine interaction is purely isotropic, i.e. for sufficiently fast tumbling in liquid solution, 1 and along the principal axes of the hyperfine tensor. Otherwise, the $B$ term can only be neglected if $\omega_{I} \gg A, B$, corresponding to the high-field approximation of the nuclear spin. Within the approximate range $2\left|\omega_{I}\right| / 5<|A|<10\left|\omega_{I}\right|$ the pseudo-secular interaction may affect transition frequencies and makes formally forbidden transitions with $\Delta m_{S}=1, \Delta m_{I}=1$ partially allowed, as $m_{I}$ is no longer a good quantum number.
Local fields at the nuclear spin
The occurrence of forbidden transitions can be understood in a semi-classical magnetization vector picture by considering local fields at the nuclear spin for the two possible states $\alpha_{S}$ and
$\beta_{S}$ of the electron spin. These local fields are obtained from the parameters $\omega_{I}, A$, and $B$ of the Hamilton operator terms that act on the nuclear spin. When divided by the gyromagnetic ratio of the nuclear spin these terms have the dimension of a local magnetic field. The local field corresponding to the nuclear Zeeman interaction equals the static magnetic field $B_{0}$ and is the same for both electron spin states, since the expectation value of $\hat{I}_{z}$ does not depend on the electron spin state. It is aligned with the $z$ direction of the laboratory frame (blue arrow in Figure 6.1). Both hyperfine fields arise from Hamiltonian terms that contain an $\hat{S}_{z}$ factor, which has the expectation value $m_{S}=+1 / 2$ for the $\alpha_{S}$ state and $m_{S}=-1 / 2$ for the $\beta_{S}$ state. The $A$ term is aligned with the $z$ axis and directed towards $+z$ in the $\alpha_{S}$ state and towards $-z$ in the $\beta_{S}$ state, assuming $A>0$ (violet arrows). The $B$ term is aligned with the $x$ axis and directed towards $+x$ in the $\alpha_{S}$ state and towards $-x$ in the $\beta_{S}$ state, assuming $B>0$ (green arrows).
The effective fields at the nuclear spin in the two electron spins states are vector sums of the three local fields. Because of the $B$ component along $x$, they are tilted from the $z$ direction by angle $\eta_{\alpha}$ in the $\alpha_{S}$ state and by angle $\eta_{\beta}$ in the $\beta_{S}$ state. The length of the sum vectors are the nuclear transition frequencies in these two states and are given by
\begin{aligned} &\omega_{\alpha}=\sqrt{\left(\omega_{I}+A / 2\right)^{2}+B^{2} / 4} \ &\omega_{\beta}=\sqrt{\left(\omega_{I}-A / 2\right)^{2}+B^{2} / 4} \end{aligned}
For $\left|\omega_{I}\right|>2|A|$, the hyperfine splitting is given by
$\omega_{\mathrm{hfs}}=\left|\omega_{\alpha}-\omega_{\beta}\right|$
and the sum frequency is given by
$\omega_{\text {sum }}=\omega_{\alpha}+\omega_{\beta}$
For $\left|\omega_{I}\right|>2|A|$, the nuclear frequency doublet is centered at $\omega_{\text {sum }} / 2($ Fig. $6.2(\mathrm{c})$ ). The sum frequency is always larger than twice the nuclear Zeeman frequency. None of the nuclear frequencies can become zero, the minimum possible value $B / 2$ is attained in one of the electron spin states for matching of the nuclear Zeeman and hyperfine interaction at $2\left|\omega_{I}\right|=|A|$. For $\left|\omega_{I}\right|<2|A|$ the nuclear frequency doublet is split by $\omega_{\text {sum }}$ and centered at $\omega_{\text {hfs }} / 2($ Fig. $\left.6.2(\mathrm{~d}))\right)$. The tilt angles $\eta_{\alpha}$ and $\eta_{\beta}$ (Figure 6.1) can be inferred from trigonometric relations and are given by
\begin{aligned} &\eta_{\alpha}=\arctan \left(\frac{-B}{2 \omega_{I}+A}\right) \ &\eta_{\beta}=\arctan \left(\frac{-B}{2 \omega_{I}-A}\right) \end{aligned}
Consider now a situation where the electron spin is in its $\alpha_{S}$ state. The nuclear magnetization from all radicals in this state at thermal equilibrium is aligned with $\vec{\omega}_{\alpha}$. Microwave excitation causes transitions to the $\beta_{S}$ state. In this state, the local field at the nuclear spin is directed along $\vec{\omega}_{\beta}$. Hence, the nuclear magnetization vector from the radicals under consideration is tilted by angle $2 \eta$ (Figure 6.1) with respect to the current local field. It will start to precess around this local field vector. This corresponds to excitation of the nuclear spin by flipping the electron spin, which is a formally forbidden transition. Obviously, such excitation will occur only if angle $2 \eta$ differs from $0^{\circ}$ and from $180^{\circ}$. The case of $0^{\circ}$ corresponds to the absence of pseudo-secular hyperfine coupling $(B=0)$ and is also attained in the limit $|A| \ll\left|\omega_{I}\right|$. The situation $2 \eta \rightarrow 180^{\circ}$ is attained in the limit of very strong secular hyperfine coupling, $|A| \gg\left|\omega_{I}\right|$. Forbidden transitions are observed for intermediate hyperfine coupling. Maximum excitation of nuclear spins is expected when the two quantization axes are orthogonal with respect to each other, $2 \eta=90^{\circ}$.
1 The product of rotational correlation time $\tau_{\mathrm{r}}$ and hyperfine anisotropy must be much smaller than unity
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Transformation of $\hat{S}_{x}$ to the eigenbasis
Excitation and detection in EPR experiments are described by the $\hat{S}_{x}$ and $\hat{S}_{y}$ operators in the rotating frame. These operators act only on electron spin transitions and thus formalize the spectroscopic selection rules. If the spin Hamiltonian contains off-diagonal terms, such as the pseudo-secular $B \hat{S}_{z} \hat{I}_{x}$ term in Eq. (6.1), the eigenbasis deviates from the basis of the electron spin rotating frame/nuclear spin laboratory frame in which the Hamiltonian is written and in which the excitation and detection operators are linear combinations of $\hat{S}_{x}$ and $\hat{S}_{y}$. In order to understand which transitions are driven and detected with what transition moment, we need to transform $\hat{S}_{x}$ to the eigenbasis (the transformation of $\hat{S}_{y}$ is analogous). This can be done by product operator formalism and can be understood in the local field picture.
The Hamiltonian in the eigenbasis has no off-diagonal elements, meaning that all quantization axes are along $z$. Thus, we can directly infer from Fig. $6.1$ that, in the $\alpha_{S}$ state, we need a counterclockwise (mathematically positive) rotation by tilt angle $\eta_{\alpha}$ about the $y$ axis, which is pointing into the paper plane. In the $\beta_{S}$ state, we need a clockwise (mathematically negative) rotation by tilt angle $\eta_{\beta}$ about the $y$ axis. The electron spin states can be selected by the projection operators $\hat{S}^{\alpha}$ and $\hat{S}^{\beta}$, respectively. Hence, we have to apply rotations $\eta_{\alpha} \hat{S}^{\alpha} \hat{I}_{y}$ and $-\eta_{\beta} \hat{S}^{\beta} \hat{I}_{y}$. These two rotations commute, as the $\alpha_{S}$ and $\beta_{S}$ subspaces are distinct when $m_{S}$ is a good quantum number. For the rotation into the eigenbasis, we can write a unitary matrix
\begin{aligned} \hat{U}_{\mathrm{EB}} &=\exp \left\{-i\left(\eta_{\alpha} \hat{S}^{\alpha} \hat{I}_{y}-\eta_{\beta} \hat{S}^{\beta} \hat{I}_{y}\right)\right\} \ &=\exp \left\{-i\left(\xi \hat{I}_{y}+\eta 2 \hat{S}_{z} \hat{I}_{y}\right)\right\} \end{aligned}
where $\xi=\left(\eta_{\alpha}-\eta_{\beta}\right) / 2$ and $\eta=\left(\eta_{\alpha}+\eta_{\beta}\right) / 2$. Note that the definition of angle $\eta$ corresponds to the one given graphically in Fig. 6.1.2 The two new rotations about $\hat{I}_{y}$ and $\hat{S}_{z} \hat{I}_{y}$ also commute. Furthermore, $\hat{I}_{y}$ commutes with $\hat{S}_{x}\left(\right.$ and $\hat{S}_{y}$ ), so that the transformation of $\hat{S}_{x}$ to the eigenbasis reduces to
$\hat{S}_{x} \stackrel{\eta 2 \hat{S}_{z} \hat{I}_{y}}{\longrightarrow} \cos \eta \hat{S}_{x}+\sin \eta 2 \hat{S}_{y} \hat{I}_{y}$
The transition moment for the allowed transitions that are driven by $\hat{S}_{x}$ is multiplied by a factor $\cos \eta \leq 1$, i.e. it becomes smaller when $\eta \neq 0$. In order to interpret the second term, it is best rewritten in terms of ladder operators $\hat{S}^{+}=\hat{S}_{x}+i \hat{S}_{y}$ and $\hat{S}^{-}=\hat{S}_{x}-i \hat{S}_{y}$. We find
$2 \hat{S}_{y} \hat{I}_{y}=\frac{1}{2}\left(\hat{S}^{+} \hat{I}^{-}+\hat{S}^{-} \hat{I}^{+}-\hat{S}^{+} \hat{I}^{+}-\hat{S}^{-} \hat{I}^{-}\right)$
In other words, this term drives the forbidden electron-nuclear zero- and double-quantum transitions (Fig. 6.2(a)) with a transition proportional to $\sin \eta$.
In a CW EPR experiment, each transition must be both excited and detected. In other words, the amplitude is proportional to the square of the transition moment, which is the transition probability. Allowed transitions thus have an intensity proportional to $\cos ^{2} \eta$ and forbidden transitions a transition probability proportional to $\sin ^{2} \eta$ (Fig. 6.2(b)).
General product operator computations for a non-diagonal Hamiltonian
In a product operator computation, terms of the Hamiltonian can be applied one after the other if and only if they pairwise commute. This is not the case for the Hamiltonian in Eq. (6.1). However, application of $\hat{U}_{\mathrm{EB}}$ diagonalizes the Hamiltonian:
$\hat{H}_{0} \stackrel{\eta \hat{S}_{z} \hat{I}_{y}}{\longrightarrow} \Omega_{S} \hat{S}_{z}+\omega_{\mathrm{sum}} / 2 \hat{I}_{z}+\omega_{\mathrm{hfi}} \hat{S}_{z} \hat{I}_{z}$
This provides a simple recipe for product operator computations in the presence of the pseudosecular hyperfine coupling. Free evolution and transition-selective pulses are computed in the eigenbasis, using the Hamiltonian on the right-hand side of relation (6.9). For application of non-selective pulses, the density operator needs to be transformed to the electron spin rotating frame/nuclear spin laboratory frame basis by applying $\hat{U}_{\mathrm{EB}}^{\dagger}$. In product operator formalism this corresponds to a product operator transformation $\stackrel{-\eta \hat{S}_{z} \hat{I}_{y}}{\longrightarrow}$. After application of non-selective pulses, the density operator needs to be backtransformed to the eigenbasis. Detection also needs to be performed in the electron spin rotating frame/nuclear spin laboratory frame basis.
This concept can be extended to any non-diagonal Hamiltonian as long as one can find a unitary transformation that transforms the Hamiltonian to its eigenbasis and can be expressed by a single product operator term or a sum of pairwise commuting product operator terms.
2 We have used $\hat{S}^{\alpha}=\hat{\mathbb{1}} / 2+\hat{S}_{z}$ and $\hat{S}^{\beta}=\hat{\mathbb{1}} / 2-\hat{S}_{z}$.
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Nuclear coherence generator $(\pi / 2)-\tau-(\pi / 2)$
We have seen that a single microwave pulse can excite coherence on forbidden electron-nuclear zero- and double-quantum transitions. In principle, this provides access to the nuclear frequencies $\omega_{\alpha}$ and $\omega_{\beta}$, which are differences of frequencies of allowed and forbidden electron spin transitions, as can be inferred from Fig. 6.2(a,b). Indeed, the decay of an electron spin Hahn echo $(\pi / 2)-\tau-(\pi)-\tau-$ echo as a function of $\tau$ is modulated with frequencies $\omega_{\alpha}$ and $\omega_{\beta}($ as well as with $\omega_{\text {hfi }}$ and $\omega_{\text {sum }}$ ). Such modulation arises from forbidden transitions during the refocusing pulse, which redistribute coherence among the four transitions. The coherence transfer echoes are modulated by the difference of the resonance frequencies before and after the transfer by the $\pi$ pulse, in which the resonance offset $\Omega_{S}$ cancels, while the nuclear spin contributions do not cancel. This two-pulse ESEEM experiment is not usually applied for measuring hyperfine couplings, as the appearance of the combination frequencies $\omega_{\text {hfi }}$ and $\omega_{\text {sum }}$ complicates the spectra and linewidth is determined by electron spin transverse relaxation, which is much faster the nuclear spin transverse relaxation.
Better resolution and simpler spectra can be obtained by indirect observation of the evolution of nuclear coherence. Such coherence can be generated by first applying a $\pi / 2$ pulse to the electron spins, which will generate electron spin coherence on allowed transitions with amplitude proportional to $\cos \eta$ and on forbidden transitions with amplitude proportional to sin $\eta$. After a delay $\tau$ a second $\pi / 2$ pulse is applied. Note that the block $(\pi / 2)-\tau-(\pi / 2)$ is part of the EXSY and NOESY experiments in NMR. The second $\pi / 2$ pulse will generate an electron spin magnetization component along $z$ for half of the existing electron spin coherence, i.e., it will "switch off" half the electron spin coherence and convert it to polarization. However, for the coherence on forbidden transitions, there is a chance $\cos \eta$ that the nuclear spin is not flipped, i.e. that the coherent superposition of the nuclear spin states survives. For electron spin coherence on allowed transitions, there is a chance $\sin \eta$ that the "switching off" of the electron coherence will lead to a "switching on" of nuclear coherences. Hence, in both these pathways there is a probability proportional to $\sin \eta \cos \eta=\sin (2 \eta) / 2$ that nuclear coherence is generated. The delay $\tau$ is required, since at $\tau=0$ the different nuclear coherence components have opposite phase and cancel.
The nuclear coherence generated by the block $(\pi / 2)-\tau-(\pi / 2)$ can be computed by product operator formalism as outlined in Section 6.2.2. We find
\begin{aligned} &\left\langle\hat{S}^{\alpha} \hat{I}_{x}\right\rangle=-\sin \left(\Omega_{S} \tau\right) \sin (2 \eta) \sin \left(\frac{\omega_{\beta}}{2} \tau\right) \cos \left(\omega_{\alpha} \tau\right) \ &\left\langle\hat{S}^{\alpha} \hat{I}_{y}\right\rangle=-\sin \left(\Omega_{S} \tau\right) \sin (2 \eta) \sin \left(\frac{\omega_{\beta}}{2} \tau\right) \sin \left(\omega_{\alpha} \tau\right) \ &\left\langle\hat{S}^{\beta} \hat{I}_{x}\right\rangle=-\sin \left(\Omega_{S} \tau\right) \sin (2 \eta) \sin \left(\frac{\omega_{\alpha}}{2} \tau\right) \cos \left(\omega_{\beta} \tau\right) \ &\left\langle\hat{S}^{\beta} \hat{I}_{y}\right\rangle=-\sin \left(\Omega_{S} \tau\right) \sin (2 \eta) \sin \left(\frac{\omega_{\alpha}}{2} \tau\right) \sin \left(\omega_{\beta} \tau\right) \end{aligned}
This expression can be interpreted in the following way. Nuclear coherence is created with a phase as if it had started to evolve as $\hat{I}_{x}$ at time $\tau=0$ (last cosine factors on the right-hand side of each line). It is modulated as a function of the electron spin resonance offset $\Omega_{S}$ and zero exactly on resonance (first factor on each line). The integral over an inhomogeneously broadened, symmetric EPR line is also zero, since $\int_{-\infty}^{\infty} \sin \left(\Omega_{S} \tau\right) \mathrm{D} \Omega_{S}=0$. However, this can be compensated later by applying another $\pi / 2$ pulse. The amplitude of the nuclear coherence generally scales with $\sin 2 \eta$, since one allowed and one forbidden transfer are required to excite it and $\sin (\eta) \cos (\eta)=\sin (2 \eta) / 2$ (second factor). The third factor on the right-hand side of lines 1 and 2 tells that the amplitude of the coherence with frequency $\omega_{\alpha}$ is modulated as a function of $\tau$ with frequency $\omega_{\beta}$. Likewise, the amplitude of the coherence with frequency $\omega_{\beta}$ is modulated as a function of $\tau$ with frequency $\omega_{\alpha}$ (lines 3 and 4 ). At certain values of $\tau$ no coherence is created at the transition with frequency $\omega_{\alpha}$, at other times maximum coherence is generated. Such behavior is called blind-spot behavior. In order to detect all nuclear frequencies, an experiment based on the $(\pi / 2)-\tau-(\pi / 2)$ nuclear coherence generator has to be repeated for different values of $\tau$. Why and how CW EPR spectroscopy is done Sensitivity advantages of CW EPR spectroscopy The CW EPR experiment
Considerations on sample preparation Theoretical description of CW EPR Spin packet lineshape Saturation
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• 7.1: Why and how CW EPR spectroscopy is done
• 7.2: Theoretical description of CW EPR
All spins in a sample that have the same resonance frequency form a spin packet. In the following we also assume that all spins of a spin packet have the same longitudinal and transverse relaxation times. If the number of spins in the spin packet is sufficiently large, we can assign a magnetization vector to the spin packet. Dynamics of this magnetization vector with equilibrium magnetization M0 during microwave irradiation is described by the Bloch equations in the rotating frame.
07: CW EPR Spectroscopy
Sensitivity advantages of CW EPR spectroscopy
In NMR spectroscopy, CW techniques have been almost completely displaced by Fourier transform (FT) techniques, except for a few niche applications. FT techniques have a sensitivity advantage if the spectrum contains large sections of baseline and the whole spectrum can be excited simultaneously by the pulses. Neither condition is usually fulfilled in EPR spectroscopy. For two reasons, FT techniques lose sensitivity in EPR spectroscopy compared to the CW experiment. First, while typical NMR spectra comfortably fit into the bandwidth of a welldesigned critically coupled radiofrequency resonance circuit, EPR spectra are much broader than the bandwidth of a microwave resonator with high quality factor. Broadening detection bandwidth and proportionally lowering the quality factor $Q$ of the resonator reduces signal-to-noise ratio unless the absorption lineshape is infinitely broad. A quality factor of the order of 10 ’ 000 , which can be achieved with cavity resonators, corresponds to a bandwidth of roughly $1 \mathrm{MHz}$ at X-band frequencies around $9.6 \mathrm{GHz}$. The intrinsic high sensitivity of detection in such a narrow band can be used only in a CW experiment. Second, even if the resonator is overcoupled to a much lower quality factor or resonators with intrinsically lower $Q$ are used (the sensitivity loss can partially be compensated by a higher filling factor of such resonators), residual power from a high-power microwave pulse requires about $100 \mathrm{~ns}$ in order to decay below the level of an EPR signal. This dead time is often a significant fraction of the transverse relaxation time of electron spins, which entails signal loss by relaxation. In contrast, in NMR spectroscopy the dead time is usually negligibly short compared to relaxation times. In many cases, the dead time in pulsed EPR spectroscopy even strongly exceeds $T_{2}$. In this situation FT EPR is impossible, even with echo refocusing, while CW EPR spectra can still be measured. This case usually applies to transition metal complexes at room temperature and to many rare earth metal complexes and high-spin Fe(III) complexes even down to the boiling point of liquid helium at normal pressure (4.2 K). For these reasons, any unknown potentially paramagnetic sample should first be characterized by CW EPR spectroscopy. Pulsed EPR techniques are required if the resolution of CW EPR spectroscopy provides insufficient information to assign a structure. This applies mainly to small hyperfine couplings in organic radicals and of ligand nuclei in transition metal complexes (see Chapter 8) and to the measurement of distances between electron spins in the nanometer range (see Chapter 9). At temperatures where pulse EPR signals can be obtained, measurement of relaxation times is also easier and more precise with pulsed EPR techniques.
The CW EPR experiment
Since the bandwidth of an optimized microwave resonator is much smaller than the typical width of EPR spectra, it is impractical to sweep the frequency at constant magnetic field in order to obtain a spectrum. Instead, the microwave frequency is kept constant and coincides with the resonator frequency at all times. The resonance condition for the spins is established by sweeping the magnetic field $B_{0}$. Another difficulty arises from the weak magnetic coupling of the spins to the exciting electromagnetic field. Only a very small fraction of the excitation power is therefore observed. This problem is solved as follows. First, direct transmission of excitation power to the detector is prevented by a circulator (Figure 7.1). Power that enters port 1 can only leave to the sample through port 2 . Power that comes from the sample through port 2 can only leave to the detector diode through port 3. Second, the resonator is critically coupled. This means that all microwave power coming from the source that is incident to the resonator enters the resonator and is converted to heat by the impedance (complex resistance) of the resonator. If the sample is off resonant and thus does not absorb microwave, no microwave power leaves the resonator through port 3. If now the magnetic field $B_{0}$ is set to the resonance condition and the sample resonantly absorbs microwave, this means that the impedance of resonator + sample has changed. The resonator is no longer critically coupled and some of the incoming microwave is reflected. This microwave leaves the circulator through port 3 and is incident on the detector diode.
This reflected power at resonance absorption can be very weak at low sample concentration. It is therefore important to detect it sensitively. A microwave diode is only weakly sensitive to a change in incident power at low power (Fig. 7.2, input voltage is proportional to the square root of power). The diode is most sensitive to amplitude changes near its operating point, marked green in Fig. 7.2. Hence, the diode must be biased to its operating point by adding constant power from a reference arm. The phase of the reference arm must be adjusted so that microwave coming from the resonator and microwave coming from the reference arm interfere constructively.
A further problem arises from the fact that microwave diodes are broadband detectors. On the one hand, this is useful, since samples can significantly shift resonator frequency. On the other hand, broadband detectors also collect noise from a broad frequency band. This decreases signal-to-noise ratio and must be countered by limiting the detection bandwidth to the signal bandwidth or even below. Such bandwidth limitation can be realized most easily by effect modulation and phase sensitive detection. By applying a small sinusoidal magnetic field modulation with typical frequency of $100 \mathrm{kHz}$ and typical amplitude of $0.01-1 \mathrm{mT}$, the signal component at detector diode output becomes modulated with the same frequency, whereas noise is uncorrelated to the modulation. Demodulation with a reference signal from the field modulation generator (Figure 7.1) by a phase-sensitive detector amplifies the signal and limits bandwidth to the modulation frequency.
Effect modulation with phase-sensitive detection measures the derivative of the absorption lineshape, as long as the modulation amplitude $\Delta B_{0}$ is much smaller than the width of the EPR line (Fig. 7.3). Since signal-to-noise ratio is proportional to $\Delta B_{0}$, one usually measures at $\Delta B_{0} \approx \Delta B_{\mathrm{pp}} / 3$, where lineshape distortion is tolerable for almost all applications. Precise lineshape analysis may require $\Delta B_{0} \leq \Delta B_{\mathrm{pp}} / 5$, whereas maximum sensitivity at the expense of significant artificial line broadening is obtained at $\Delta B_{0}=\Delta B_{\mathrm{pp}}$. The modulation frequency should not be broader than the linewidth in frequency units. However, with the standard modulation frequency of $100 \mathrm{kHz}$ that corresponds on a magnetic field scale to only $3.6 \mu \mathrm{T}$ at $g=g_{e}$, this is rarely a problem.
Considerations on sample preparation
Since electron spins have a much larger magnetic moment than nuclear spins, electron-electron couplings lead to significant line broadening in concentrated solutions. Concentrations of paramagnetic centers should not usually exceed $1 \mathrm{mM}$ in order to avoid such broadening. For organic radicals in liquid solution it may be necessary to dilute the sample to $100 \mu \mathrm{M}$ in order to achieve ultimate resolution. For paramagnetic metal dopants in diamagnetic host compounds, at most $1 \%$ of the diamagnetic sites should be substituted by paramagnetic centers. Such concentrations can be detected easily and with good signal-to-noise ratio. For most samples, good spectra can be obtained down to the $1 \mu \mathrm{M}$ range in solution and down to the 100 ppm dopant range in solids.
Line broadening in liquid solution can also arise from diffusional collision of paramagnetic
species with paramagnetic triplet oxygen (Figure 7.4). During such a collision, wavefunctions of the two molecules overlap and, since electrons are undistinguishable particles, spin states of all unpaired electrons in both molecules are arbitrarily redistributed when the two molecules separate again. The stochastic diffusional encounters thus lead to additional flips of the observed electron spins, which corresponds to relaxation and shortens longitudinal relaxation time $T_{1}$. Since the linewidth is proportional to $T_{2}$ and $T_{2}$ cannot be longer than $2 T_{1}$, frequent collisional encounters of paramagnetic species lead to line broadening. Such line broadening increases with decreasing viscosity (faster diffusion) and increasing oxygen concentration. The effect is stronger in apolar solvents, where oxygen solubility is higher than in polar solvents, but it is often significant even in aqueous solution. Best resolution is obtained if the sample is free of oxygen. The same mechanism leads to line broadening at high concentration of a paramagnetic species in liquid solution. In the solid state, line broadening at high concentration is mainly due to dipole-dipole coupling. Often, the anisotropically broadened EPR spectrum in the solid state is of interest, as it provides information on $g$ anisotropy and anisotropic hyperfine couplings. This may require freezing of a solution of the species of interest. Usually, the species will precipitate if the solvent crystallizes, which may cause line broadening and, in extreme cases, even collapse of the hyperfine structure and averaging of $g$ anisotropy by exchange between neighboring paramagnetic species. These problems are prevented if the solvent forms a glass, as is often the case for solvents that have methyl groups or can form hydrogen bonds in very different geometries. Typical glass-forming solvents are toluene, 2-methyltetrahydrofuran, ethanol, ethylene glycol, and glycerol. Aqueous solutions require addition of at least $25 \%$ glycerol as a cryoprotectant. In most cases, crystallization will still occur on slow cooling. Samples are therefore shock frozen by immersion of the sample tube into liquid nitrogen. Glass tubes would break on direct immersion into liquid nitrogen, but EPR spectra have to be measured in fused silica sample tubes anyway, since glass invariably contains a detectable amount of paramagnetic iron impurities.
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This section overlaps with Section $2.7$ of the NMR lecture notes.
Spin packet lineshape
All spins in a sample that have the same resonance frequency form a spin packet. In the following we also assume that all spins of a spin packet have the same longitudinal and transverse relaxation times $T_{1}$ and $T_{2}$, respectively. If the number of spins in the spin packet is sufficiently large, we can assign a magnetization vector to the spin packet. Dynamics of this magnetization vector with equilibrium magnetization $M_{0}$ during microwave irradiation is described by the Bloch equations in the rotating frame. In EPR spectroscopy, it is unusual to use the gyromagnetic ratio. Hence, we shall denote the resonance offset by
$\Omega_{S}=\frac{g \mu_{\mathrm{B}}}{\hbar} B_{0}-2 \pi \nu_{\mathrm{mw}}$
where $\nu_{\mathrm{mw}}$ is the microwave frequency in frequency units. The rotating-frame Bloch equations for the three components of the magnetization vector can then be written as
\begin{aligned} \frac{\mathrm{d} M_{x}}{\mathrm{~d} t} &=-\Omega_{S} M_{y}-\frac{M_{x}}{T_{2}} \[4pt] \frac{\mathrm{d} M_{y}}{\mathrm{~d} t} &=\Omega_{S} M_{x}-\omega_{1} M_{z}-\frac{M_{y}}{T_{2}} \[4pt] \frac{\mathrm{d} M_{z}}{\mathrm{~d} t} &=\omega_{1} M_{y}-\frac{M_{z}-M_{0}}{T_{1}} \end{aligned}
where $\omega_{1}=g_{\perp} \mu_{\mathrm{B}} B_{1} / \hbar$ is the microwave field amplitude in angular frequency units and $g_{\perp}$ is the mean $g$ value in the plane perpendicular to the static magnetic field. The apparent sign difference for the $\Omega_{S}$ and $\omega_{1}$ terms arises from the different sense of spin precession for electron spins compared to nuclear spins with a positive gyromagnetic ratio.
If the spin packet is irradiated at constant microwave frequency, constant microwave power, and constant static magnetic field $B_{0}$ for a sufficiently long time (roughly $5 T_{1}$ ), the magnetization vector attains a steady state. Although the static field is swept in a CW EPR experiment, assuming a steady state is a good approximation, since the field sweep is usually slow compared to $T_{2}$ and $T_{1}$. Faster sweeps correspond to the rapid scan regime that is not treated in this lecture course. In the steady state, the left-hand sides of the differential equations (7.2) for the magnetization vector components must all be zero,
\begin{aligned} 0 &=-\Omega_{S} M_{y}-\frac{M_{x}}{T_{2}} \[4pt] 0 &=\Omega_{S} M_{x}-\omega_{1} M_{z}-\frac{M_{y}}{T_{2}} \[4pt] 0 &=\omega_{1} M_{y}-\frac{M_{z}-M_{0}}{T_{1}} \cdot\langle\text { stationary state }\rangle \end{aligned}
This linear system of equations has the solution
\begin{aligned} M_{x} &=M_{0} \omega_{1} \frac{\Omega T_{2}^{2}}{1+\Omega^{2} T_{2}^{2}+\omega_{1}^{2} T_{1} T_{2}} \[4pt] M_{y} &=-M_{0} \omega_{1} \frac{T_{2}}{1+\Omega^{2} T_{2}^{2}+\omega_{1}^{2} T_{1} T_{2}} \[4pt] M_{z} &=M_{0} \frac{1+\Omega^{2} T_{2}^{2}}{1+\Omega^{2} T_{2}^{2}+\omega_{1}^{2} T_{1} T_{2}},\langle\text { stationary state }\rangle \end{aligned}
where $M_{z}$ is not usually detected, $M_{x}$ is in phase with the exciting microwave irradiation and corresponds to the dispersion signal, and $M_{y}$ is out of phase with the exciting irradiation and corresponds to the absorption line. Unperturbed lineshapes are obtained in the linear regime, where the saturation parameter
$S=\omega_{1}^{2} T_{1} T_{2}$
fulfills $S \ll 1$. One can easily ascertain from Eq. (7.4) that in the linear regime $M_{y}$ increases linearly with increasing $\omega_{1}$, which corresponds to proportionality of the signal to the square root of microwave power. $M_{z}$ is very close to the equilibrium magnetization. Within this regime, a decrease of $6 \mathrm{~dB}$ in microwave attenuation, i.e., a power increase by $6 \mathrm{~dB}$, increases signal amplitude by a factor of 2 . Lineshape does not depend on $\omega_{1}$ in the linear regime. Therefore, it is good practice to measure at the highest microwave power that is still well within the linear regime, as this corresponds to maximum signal-to-noise ratio. For higher power the line is broadened.
Within the linear regime, $M_{y}$ takes the form of a Lorentzian absorption line
$M_{y}(\Omega)=M_{0} \omega_{1} T_{2} \frac{1}{1+\Omega^{2} T_{2}^{2}},\langle\text { linear regime }\rangle$
with linewidth $1 / T_{2}$ in angular frequency units. The peak-to-peak linewidth of the first derivative of the absorption line is $\Gamma_{\mathrm{pp}}=2 / \sqrt{3} T_{2}$. Since $\mathrm{CW}$ EPR spectra are measured by sweeping magnetic field, we need to convert to magnetic field units,
$\Gamma_{\text {pp,field sweep }}=\frac{2}{\sqrt{3} T_{2}} \cdot \frac{\hbar}{g \mu_{\mathrm{B}}}$
The linewidth of a spin packet is called homogeneous linewidth. If $T_{2}$ is the same for all spin packets, this homogeneous linewidth is proportional to $1 / \mathrm{g}$, a fact that needs to be taken into account in lineshape simulations for systems with large $g$ anisotropy. For most samples, additional line broadening arises from unresolved hyperfine couplings and, in the solid state, $g$ anisotropy. Therefore, $T_{2}$ cannot usually be obtained by applying Eq. (7.7) to the experimentally observed peak-to-peak linewidth.
Saturation
For microwave power larger than in the linear regime, the peak-to-peak linewidth increases by a factor $1+S$. If a weak signal needs to be detected with maximum signal-to-noise ratio it is advantageous to increase power beyond the linear regime, but not necessarily to the maximum available level. For very strong irradiation, $S \gg 1$, the term 1 can be neglected in the denominator of Eqs. (7.4) for the magnetization vector components. The on-resonance amplitude of the absorption line is then given by
$M_{y}(\Omega=0)=M_{0} / \omega_{1} T_{1},\left\langle\omega_{1}^{2} T_{1} T_{2} \gg 1\right\rangle$
i.e., it is inversely proportional to $\omega_{1}$. In this regime, the amplitude decreases with increasing power of the microwave irradiation.
Semi-quantitative information on spin relaxation can be obtained by the progressive power saturation experiment, where the EPR spectrum is measured as a function of microwave power $P_{\text {mw }}$. Usually, the peak-to-peak amplitude of the larges signal in the spectrum is plotted as a function of $\sqrt{P_{\mathrm{mw}}}$. Such saturation curves can be fitted by the equation
$A\left(P_{\mathrm{mw}}\right)=\frac{I \sqrt{P_{\mathrm{mw}}}}{\left[1+\left(2^{1 / \epsilon}-1\right) P_{\mathrm{mw}} / P_{1 / 2}\right]^{\epsilon}}$
where the inhomogeneity parameter $\epsilon$ takes the value $1.5$ in the homogeneous limit and $0.5$ in the inhomogeneous limit. Usually, $\epsilon$ is not known beforehand and is treated as a fit parameter. The other fit parameters are $I$, which is the slope of the amplitude increase with the square root of microwave power in the linear regime, and $P_{1 / 2}$, which is the half-saturation power. More precisely, $P_{1 / 2}$ is the incident mw power where $A$ is reduced to half of its unsaturated value. Figure $7.5$ shows experimental data from a progressive saturation measurement on spin-labelled mutant V229C of major plant light harvesting complex LHCII solubilized in detergent micelles in a nitrogen atmosphere and a fit of this data by Eq. (7.9).
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Advantages of electron-spin based detection of nuclear frequency spectra
Nuclear frequency spectra in the liquid (Section 4.3.2) and solid states (4.3.4) exhibit much better hyperfine resolution than EPR spectra, because the former spectra feature fewer and narrower lines. In fact, small hyperfine couplings to ligand nuclei in metal complexes are not usually resolved in EPR spectra and only the largest hyperfine couplings may be resolved in solid-state EPR spectra. The nuclear frequency spectra cannot be measured by a dedicated NMR spectrometer because they extend over several Megahertz to several tens of Megahertz, whereas NMR spectrometers are designed for excitation and detection bandwidths of a few tens of kilohertz. Furthermore, electron spin transitions have 660 times more polarization than proton transitions and more than that for other nuclei. Their larger magnetic moment also leads to higher detection sensitivity. It is thus advantageous to transfer polarization from electron spins to nuclear spins and to backtransfer the response of the nuclear spins to the electron spins for detection. Two classes of experiments can achieve this, electron nuclear double resonance (ENDOR) experiments, discussed in Section $8.1$ and electron spin echo envelope modulation (ESEEM) experiments discussed in Section 8.2.
Types of ENDOR experiments
An ENDOR experiment can be performed with strong CW irradiation of both electron and nuclear spins. In this CW ENDOR experiment, an electron spin transition is partially saturated, $S \gg 1$ in Eq. (7.5). By driving a nuclear spin transition that shares an energy level with the saturated transition, additional relaxation pathways are opened up. The electron spin transition under observation is thus partially desaturated, and an increase in the EPR signal is observed. The experiment is performed at constant magnetic field with strong microwave irradiation at a maximum of the first-derivative absorption spectrum (i.e. the CW EPR spectrum) and the EPR signal is recorded as a function of the frequency of additional radiofrequency irradiation, which must fulfill the saturation condition $S \gg 1$ for the nuclear spins. Usually, the radiofrequency irradiation is frequency modulated and the response is detected with another phase-sensitive detector, which leads to observation of the first derivative of the nuclear frequency spectrum. The CW ENDOR experiment depends critically on a balance of relaxation times, so that in the solid state sufficient sensitivity may only be achieved in a certain temperature range. Furthermore, simultaneous strong continuous irradiation by both microwave and radiofrequency, while keeping resonator frequency and temperature constant, is experimentally challenging. Therefore, CW ENDOR has been largely replaced by pulsed ENDOR techniques. However, for liquid solution samples CW ENDOR is usually the only applicable ENDOR technique.
The conceptually simplest pulsed ENDOR experiment is Davies ENDOR (Section 8.1.3), where saturation of the EPR transition is replaced by inversion by a $\pi$ pulse (Fig. 8.1(a)). A subsequent radiofrequency $\pi$ pulse, which is on-resonant with a transition that shares a level with the inverted EPR transition, changes population of this level and thus polarization of the EPR observer transition. This polarization change as a function of the radiofrequency is observed by a Hahn echo experiment on the observer transition. The approach works well for moderately large hyperfine couplings $(>3 \mathrm{MHz})$, in particular for ${ }^{14} \mathrm{~N}$ nuclei directly coordinated to a transition metal ion or for protons at hydrogen-bonding distance or distances up to about $4 \AA$. As we shall see in Section 8.1.3, the experiment is rather insensitive for very small hyperfine couplings.
The smallest hyperfine couplings can be detected with the Mims ENDOR experiment that is based on the stimulated echo sequence $(\pi / 2)-\tau-(\pi / 2)-T-(\pi / 2)-\tau-e c h o($ Fig. $8.1($ b) $)$ ). The preparation block $(\pi / 2)-\tau-(\pi / 2)$ creates a polarization grating of the functional form $A\left(\Omega_{S}\right) \cos \left(\Omega_{S} \tau\right)$, where $A\left(\Omega_{S}\right)$ is the EPR absorption spectrum as a function of the resonance offset $\Omega_{S}$ and $\tau$ is the delay between the two $\pi / 2$ microwave pulses. A radiofrequency $\pi$ pulse with variable frequency is applied during time $T$ when the electron spin magnetization is aligned with the $z$ axis. If this pulse is on resonant with a nuclear transition that shares a level with the observer EPR transition, half of the polarization grating is shifted by the hyperfine splitting $A_{\text {eff }}$, as will also become apparent in Section 8.1.3. For $A_{\text {eff }} \tau=2(k+1) \pi$ with integer $k$ the polarization grating is destroyed by destructive interference. Since the stimulated echo is the free induction decay (FID) of this polarization grating, it is canceled by a radiofrequency pulse that is on resonant with a nuclear transition. It is apparent that the radiofrequency $\pi$ pulse has no effect for $A_{\text {eff }} \tau=2 k \pi$ with integer $k$, where the original and frequency-shifted gratings interfere constructively. Hence, the Mims ENDOR experiment features blind spots as a function of interpulse delay $\tau$. These blind spots are not a serious problem for very small hyperfine couplings $A_{\mathrm{eff}} \ll \pi / \tau$. Note however that the first blind spot corresponds to $A_{\mathrm{eff}}=0$. Hence, long interpulse delays $\tau$ are required in order to detect very small hyperfine couplings, and this leads to strong echo attenuation by a factor $\exp \left(-2 \tau / T_{2}\right)$ due to electron spin transverse relaxation. It can be shown that maximum sensitivity for very small couplings is attained approximately at $\tau=T_{2}$.
Davies ENDOR
The Davies ENDOR experiment is most easily understood by looking at the polarization transfers. At thermal equilibrium the electron spin transitions (red and pale red) are much more strongly polarized than the nuclear spin transitions (Fig. 8.2(a)). Their frequencies differ by an effective hyperfine splitting $A_{\text {eff }}$ to a nuclear spin $I=1 / 2$ that is color-coded blue. The first microwave $\pi$ pulse is transition-selective, i.e., it has an excitation bandwidth that is smaller than $A_{\text {eff }}$. Accordingly, it inverts only one of the two electron spin transitions. We assume that the $|\beta \alpha\rangle \leftrightarrow|\alpha \alpha\rangle$ transition (red) is inverted and the $|\beta \beta\rangle \leftrightarrow|\alpha \beta\rangle$ transition (pale red) is not inverted; the other case is analogous. Such transition-selective inversion leads to a state of two-spin order, where all individual transitions in the two-spin system are polarized. However, the two electron spin transitions are polarized with opposite sign and the two nuclear transitions are also polarized with opposite sign (Fig. 8.2(b)). Now a radiofrequency $\pi$ pulse is applied. If this pulse is not resonant with a nuclear transition, the state of two-spin order persists and the observer electron spin transition (red) is still inverted. The radiofrequency pulse is also transition-selective. We now assume that this pulse inverts the $|\alpha \alpha\rangle \leftrightarrow|\alpha \beta\rangle$ transition (blue); the other case is again analogous. After such a resonant radiofrequency pulse, the two nuclear transitions are polarized with equal sign and the two electron spin transitions are saturated with no polarization existing on them (Fig. 8.2(c)). After the radiofrequency $\pi$ pulse a microwave Hahn echo sequence is applied resonant with the observer transition (Fig. 8.1(a)). If the radiofrequency pulse was off resonant (situation as in Fig. 8.2(b)), an inverted echo is observed. If, on the other hand, the radiofrequency pulse was on resonant (situation as in Fig. 8.2(c)) no echo is observed. In practice, polarization transfers are not complete and a weak echo is still observed. However, an on-resonant radiofrequency pulse causes some recovery of the inverted echo. If the radiofrequency is varied, recovery of the inverted echo is observed at all frequencies where the radiofrequency pulse is resonant with a nuclear transition.
Further understanding of Davies ENDOR is gained by considering an inhomogeneously broadened EPR line (Fig. 8.3). In such a line with width $\Gamma_{\text {inhom }}$, each individual spin packet with much narrower width $\Gamma_{\text {hom }}$ can, in principle, be selectively excited. A long rectangular $\pi$ pulse inverts the on-resonant spin packet and partially inverts spin packets roughly over a bandwidth corresponding to the inverse length of the pulse. In Davies ENDOR, pulse lengths between 50 and $400 \mathrm{~ns}$, corresponding to excitation bandwidths between 20 and $2.5 \mathrm{MHz}$ are typical. Such a pulse creates an inversion hole centered at the microwave frequency $\omega_{\text {mw }}$. In an $S=1 / 2, I=1 / 2$ electron-nuclear spin system, two on-resonant spin packets exist, those where $\omega_{\mathrm{mw}}$ is the frequency of the $|\beta \alpha\rangle \leftrightarrow|\alpha \alpha\rangle$ transition and those where it is the frequency of the $|\beta \beta\rangle \leftrightarrow|\alpha \beta\rangle$ transition. For the former spin packet, inversion of the nuclear spin from the $|\beta\rangle$ to the $|\alpha\rangle$ state increases the EPR frequency by the effective hyperfine splitting $A_{\text {eff }}$, whereas for the latter packet, inversion from the $|\alpha\rangle$ to the $|\beta\rangle$ state decreases it by $A_{\text {eff }}$. In both cases half of the inversion hole is shifted to a side hole, leaving a saturation hole at $\omega_{m w}$ and creating a saturation side hole. The saturation center holes of the two spin packets coincide in frequency and combine to a saturation hole in the inhomogeneously broadened line. At the side hole frequencies $\omega_{\mathrm{mw}} \pm A_{\mathrm{eff}}$, only one of the two spin packets contributes to the hole, so that the side holes are only half as deep. The Hahn echo subsequence in the Davies ENDOR sequence must have a detection bandwidth that covers only the central hole (pale red in Fig. $8.3(\mathrm{~d})$ ), since no ENDOR effect would be observed if the side hole would also be covered. For this purpose, the detection bandwidth of the Hahn echo sequence can be limited either by using sufficiently long microwave pulses or by using a sufficiently long integration gate for the inverted echo.
In any case, a Davies ENDOR effect is only be observed if $A_{\text {eff }}$ exceeds the width of the original inversion hole. The smaller $A_{\mathrm{eff}}$, the longer the first inversion pulse needs to be and the fewer spin packets contribute to the signal. In general, hyperfine splittings much smaller than the homogeneous linewidth $\Gamma_{\text {hom }}=1 / T_{2}$ in the EPR spectrum cannot be detected. In practice, Davies ENDOR becomes very insensitive for $\pi$ pulse lengths exceeding 400 ns. If broadening of the inversion hole by electron spin relaxation is negligible, the suppression of signals with small hyperfine couplings in Davies ENDOR can be described by a selectivity parameter
$\eta_{\mathrm{S}}=\frac{A_{\mathrm{eff}} t_{\pi}^{(1)}}{2 \pi}$
where $t_{\pi}^{(1)}$ the length of the first mw $\pi$ pulse. Maximum absolute ENDOR intensity $V_{\max }$ is obtained for $\eta_{\mathrm{S}}=\sqrt{2} / 2$. As a function of $\eta_{\mathrm{S}}$, the absolute ENDOR intensity is given by
$V\left(\eta_{\mathrm{S}}\right)=V_{\max }\left(\frac{\sqrt{2} \eta_{\mathrm{S}}}{\eta_{\mathrm{S}}^{2}+1 / 2}\right)$
The hyperfine contrast selectivity described by Eq. (8.2) can be used for spectral editing. For instance, ${ }^{14} \mathrm{~N}$ ENDOR signals of directly coordinated ligand nitrogen atoms in transition metal complexes with $A_{\text {eff }}$ of the order of $20-40 \mathrm{MHz}$ overlap with ${ }^{1} \mathrm{H}$ ENDOR signals of weakly coupled ligand protons at X-band frequencies. At an inversion pulse length of about $50 \mathrm{~ns}^{1} \mathrm{H}$ ENDOR signals are largely suppressed.
The sensitivity advantage of Mims ENDOR for very small hyperfine couplings can also be understood in the hole burning picture. Instead of a single center hole, a preparation block $(\pi / 2)-\tau-(\pi / 2)$ with nonselective microwave pulses creates a polarization grating that can be imagined as a comb of many holes that are spaced by frequency difference $1 / \tau$. The width of each hole is approximately $1 / 2 \tau$. The width of the comb of holes is determined by the inverse length of the non-selective $\pi / 2$ pulses, which are typically $10 \mathrm{~ns}$ long. For small couplings, where $t_{\pi}^{(1)}$ in Davies ENDOR needs to be very long, more than an order of magnitude more spin packets take part in a Mims ENDOR experiment than in a Davies ENDOR experiment. The Mims ENDOR effect arises from the shift of one quarter of the polarization grating by frequency difference $+A_{\text {eff }}$ and one quarter of the grating by $-A_{\text {eff }}$. The shifted gratings interfer with the grating at the center frequency. Depending on $A_{\text {eff }}$ and on the periodicity $1 / \tau$ of the grating, this interference is destructive (ENDOR effect) or constructive (no ENDOR effect).
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ENDOR or ESEEM?
In ESEEM experiments, polarization transfer from electron spins to nuclear spins and detection of nuclear frequencies on electron spin transitions are based on the forbidden electron-nuclear transitions discussed in Chapter 6. Much of the higher polarization of the electron spin transitions is lost in such experiments, since the angle $2 \eta$ between the quantization axes of the nuclear spin in the two electron spin states is usually small and the depth of nuclear echo modulations is sin $2 \eta$. Furthermore, modulations vanish along the principal axes of the hyperfine tensor, where $B=0$ and thus $\eta=0$. Therefore, lineshape singularities are missing in one-dimensional ESEEM spectra, which significantly complicates lineshape analysis. For this reason, one-dimensional ESEEM experiments are not usually competitive with ENDOR experiments, at least if the ENDOR experiments can be performed at $\mathrm{Q}$-band frequencies $(\approx 34 \mathrm{GHz})$ or even higher frequencies. An exception arises for weakly coupled "remote" $14 \mathrm{~N}$ nuclei in transition metal complexes where exact cancellation between the nuclear Zeeman and the hyperfine interactions can be achieved for one of the electron spin states at X-band frequencies or slightly below. In this situation, pure nuclear quadrupole frequencies are observed, which leads to narrow lines and easily interpretable spectra. One-dimensional ESEEM data are also useful for determining local proton or deuterium concentrations around a spin label, which can be used as a proxy for water accessibility (Section 10.1.6).
The main advantage of ESEEM compared to ENDOR spectroscopy is the easier extension of ESEEM to a two-dimensional correlation experiment. Hyperfine sublevel correlation (HYSCORE) spectroscopy 8.2.3 resolves overlapping signal from different elements, simplifies peak assignment, and allows for direct determination of hyperfine tensor anisotropy even if the lineshape singularities are not observed.
Three-pulse ESEEM
The HYSCORE experiment is a two-dimensional extension of the three-pulse ESEEM experiment that we will treat first. In this experiment, the amplitude of a stimulated echo after is observed with the pulse sequence $(\pi / 2)-\tau-(\pi / 2)-t-(\pi / 2)-\tau-e c h o$ as a function of the variable interpulse delay $t$ at fixed interpulse delay $\tau$ (Fig. 8.4). The block $(\pi / 2)-\tau-(\pi / 2)$ serves as a nuclear coherence generator, as discussed in Section 6.3.1 and, simultaneously, creates the polarization grating discussed in the context of the Mims ENDOR experiment (Section 8.1.2). In fact, most of the thermal equilibrium magnetization is converted to the polarization grating whose FID after the final $\pi / 2$ pulse is the stimulated echo, while only a small fraction is transferred to nuclear coherence. The phase of the nuclear coherence determines how much of it contributes to the stimulated echo after back transfer to electron spin coherence by the last $\pi / 2$ pulse. For an electron-nuclear spin system $S=1 / 2, I=1 / 2$ this phase evolves with frequencies $\omega_{\alpha}$ or $\omega_{\beta}$ if during interpulse delay $t$ the electron spin is in its $\alpha$ or $\beta$ state, respectively. Hence, the part of the stimulated echo that arises from back transferred nuclear coherence is modulated as a function of $t$ with frequencies $\omega_{\alpha}$ and $\omega_{\beta}$.
An expression for the echo envelope modulation can be derived by product operator formalism using the concepts explained in Section 6.2. Disregarding relaxation, the somewhat lengthy derivation provides
$V_{3 \mathrm{p}}(\tau, t)=\frac{1}{2}\left[V_{\alpha}(\tau, t)+V_{\beta}(\tau, t)\right],$
where the terms $V_{\alpha}(\tau, t)$ and $V_{\beta}(\tau, t)$ correspond to contributions with the electron spin in its $\alpha$ or $\beta$ state, respectively, during interpulse delay $t$. These terms are given by
\begin{aligned} &V_{\alpha}(\tau, t)=1-\frac{k}{2}\left\{1-\cos \left[\omega_{\beta} \tau\right]\right\}\left\{1-\cos \left[\omega_{\alpha}(t+\tau)\right]\right\} \ &V_{\beta}(\tau, t)=1-\frac{k}{2}\left\{1-\cos \left[\omega_{\alpha} \tau\right]\right\}\left\{1-\cos \left[\omega_{\beta}(t+\tau)\right]\right\} \end{aligned}
The factors $\cos \left[\omega_{\beta} \tau\right]$ for the $V_{\alpha}$ term and $\cos \left[\omega_{\alpha} \tau\right]$ for the $V_{\beta}$ term describe the blind spot behavior of three-pulse ESEEM. The modulation depth $k$ is given by
$k=\sin ^{2} 2 \eta=\left(\frac{B \omega_{I}}{\omega_{\alpha} \omega_{\beta}}\right)^{2}$
For small hyperfine couplings, $A, B \ll \omega_{I}$, we have $\omega_{\alpha} \approx \omega_{\beta} \approx \omega_{I}$, so that Eq. (8.5) reduces to
$k=\frac{B^{2}}{\omega_{I}^{2}}$
i.e., the modulation depth is inversely proportional to the square of the magnetic field. Using Eqs. (4.10) and (4.11) we find for protons not too close to a well localized unpaired electron
$k=\frac{9}{4}\left(\frac{\mu_{0}}{4 \pi}\right)^{2}\left(\frac{g \mu_{B}}{B_{0}}\right)^{2} \frac{\sin ^{2}\left(2 \theta_{\mathrm{HFI}}\right)}{r^{6}}$
where $\theta_{\mathrm{HFI}}$ is the angle between the electron-proton axis and the static magnetic field $B_{0}$.
Because of the star topology of electron-nuclear spin systems (Fig. 4.4(a)), Eq. (8.3) can be easily extended by a product rule to multiple nuclei with spins $I_{l}=1 / 2$, where $l$ is an index that runs over all nuclei. One finds
$V_{3 \mathrm{p}}(\tau, t)=\frac{1}{2}\left[\prod_{l} V_{\alpha, l}(\tau, t)+\prod_{l} V_{\beta, l}(\tau, t)\right]$
In the weak modulation limit, where all modulation depths $k_{l}$ fulfill the condition $k_{l} \ll 1$, the ESEEM spectrum due to several coupled nuclei is the sum of the spectra of the individual nuclei.
HYSCORE
The HYSCORE experiment is derived from the three-pulse ESEEM experiment by inserting a microwave $\pi$ pulse midway through the evolution of nuclear coherence. This splits the interpulse delay $t$ into two interpulse delays $t_{1}$ and $t_{2}$ (Fig. 8.4(b)), which are varied independently to provide a two-dimensional data set $V\left(t_{1}, t_{2}\right)$ that depends parametrically on fixed interpulse delay $\tau$. The inserted $\pi$ pulse inverts the electron spin state. Hence, coherence that has evolved with frequency $\omega_{\alpha}$ during interpulse delay $t_{1}$ evolves with frequency $\omega_{\beta}$ during interpulse delay $t_{2}$ and vice versa. In the weak modulation limit, the HYSCORE experiment correlates only frequencies $\omega_{\alpha}$ and $\omega_{\beta}$ of the same nuclear spin. The full modulation expression for the HYSCORE experiment contains a constant contribution and contributions that vary only with respect to either $t_{1}$ or $t_{2}$. These contributions can be removed by background correction with low-order polynomial functions along both dimensions. The remaining modulation corresponds to only cross peaks and can be expressed as
$V_{4 \mathrm{p}}\left(t_{1}, t_{2} ; \tau\right)=\frac{k}{2} \sin \left(\frac{\omega_{\alpha} \tau}{2}\right) \sin \left(\frac{\omega_{\beta} \tau}{2}\right)\left[V^{(\alpha \beta)}\left(t_{1}, t_{2} ; \tau\right)+V^{(\beta \alpha)}\left(t_{1}, t_{2} ; \tau\right)\right]$
with
\begin{aligned} &V^{(\alpha \beta)}\left(t_{1}, t_{2} ; \tau\right)=\cos ^{2} \eta \cos \left(\omega_{\alpha} t_{1}+\omega_{\beta} t_{2}+\omega_{\operatorname{sum}} \frac{\tau}{2}\right)-\sin ^{2} \eta \cos \left(\omega_{\alpha} t_{1}-\omega_{\beta} t_{2}+\omega_{\mathrm{hfi}} \frac{\tau}{2}\right) \ &V^{(\beta \alpha)}\left(t_{1}, t_{2} ; \tau\right)=\cos ^{2} \eta \cos \left(\omega_{\beta} t_{1}+\omega_{\alpha} t_{2}+\omega_{\mathrm{sum}} \frac{\tau}{2}\right)-\sin ^{2} \eta \cos \left(\omega_{\beta} t_{1}-\omega_{\alpha} t_{2}+\omega_{\mathrm{hfi}} \frac{\tau}{2}\right) \end{aligned}
In this representation with unsigned nuclear frequencies, one has $\eta<45^{\circ}$ for the weak coupling case $\left(|A|<2\left|\omega_{I}\right|\right)$ and $\eta>45^{\circ}$ for the strong coupling case $\left(|A|>2\left|\omega_{I}\right|\right)$, as can be inferred from Fig. 6.1. Hence, $\cos ^{2} \eta>\sin ^{2} \eta$ in the weak coupling case and $\sin ^{2} \eta>\cos ^{2} \eta$ in the strong coupling case. In the weak coupling case, the cross peaks that correlate nuclear frequencies with the same sign ( $\cos ^{2} \eta$ terms) are much stronger than those that correlate frequencies with opposite $\operatorname{sign}\left(\sin ^{2} \eta\right.$ terms) whereas it is the other way around in the strong coupling case. Therefore, the two cases can be easily distinguished in HYSCORE spectra, since the cross peaks appear in different quadrants (Fig. 8.5). Furthermore, disregarding a small shift that arises from the pseudo-secular part $B$ of the hyperfine coupling (see below), the cross peaks of a given isotope with spin $I=1 / 2$ are situated on parallels to the anti-diagonal that corresponds to the nuclear Zeeman frequency $\nu_{I}$. This frequency in turn can be computed from the nuclear $g$ value (or gyromagnetic ratio $\gamma$ ) and the static magnetic field $B_{0}$. Peak assignment for $I=1 / 2$ nuclei is thus straightforward. For nuclei with $I>1 / 2$ the peaks are further split by the nuclear quadrupole interaction. Unless this splitting is much smaller than both the hyperfine interaction and the nuclear Zeeman interaction $\left({ }^{2} \mathrm{H},{ }^{6} \mathrm{Li}\right)$, numerical simulations are required to assign the peaks and extract the hyperfine and nuclear quadrupole coupling.
The small pseudo-secular shift of the correlation peaks with respect to the anti-diagonal contains information on the anisotropy $T$ of the hyperfine interaction (Fig. 8.5). In the solid state, the cross peaks from different orientations $\theta_{\mathrm{HFI}}$ form curved ridges. For a hyperfine tensor with axial symmetry, as it is encountered for protons not too close to a well-localized unpaired electron, the maximum shift in the diagonal direction corresponds to $\theta_{\mathrm{HFI}}=45^{\circ}$ and is given by $9 T^{2} / 32\left|\omega_{I}\right|$. Since $\omega_{I}$ is known, $T$, and thus the electron-proton distance $r$ can be computed from this maximum shift. If $A_{\text {iso }} \ll \omega_{I}$, which is usually the case, the orientation with maximum shift is at the same time the orientation with maximum modulation depth.
The curved ridges end at their intersection with the parallel to the anti-diagonal. These points correspond to the principal values of the hyperfine tensor and modulation depth is zero at these points. However, it is usually possible to fit the theoretical ridge to the experimentally observed ridge, as the curvature near $\theta_{\mathrm{HFI}}=45^{\circ}$ together with the position of the $\theta_{\mathrm{HFI}}=45^{\circ}$ point fully determines the problem.
Analysis of HYSCORE spectra requires some precaution due to the blind-spot behavior (factor $\sin \left(\frac{\omega_{\alpha} \tau}{2}\right) \sin \left(\frac{\omega_{\beta} \tau}{2}\right)$ in Eq. (8.9)) and due to orientation selection by the limited bandwidth of the microwave pulses that is much smaller than spectral width for transition metal complexes. It is therefore prudent to measure HYSCORE spectra at several values of $\tau$ and at several observer positions within the EPR spectrum.
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At a distance of $1 \mathrm{~nm}$ between two localized unpaired electrons, splitting $\omega_{\perp}$ between the "horns" of the Pake pattern is about $52 \mathrm{MHz}$ for two electron spins. Even strongly inhomogeneously broadened EPR spectra usually have features narrower than that (about $2 \mathrm{mT}$ in a magnetic field sweep). Depending on the width of the narrowest features in the EPR spectrum and on availability of an experimental spectrum or realistic simulated spectrum in the absence of dipole-dipole coupling, distances up to $1.5 \ldots 2.5 \mathrm{~nm}$ can be estimated from dipolar broadening by lineshape analysis. At distances below $1.2 \mathrm{~nm}$, such analysis becomes uncertain due to the contribution from exchange coupling between the two electron spins, which cannot be computed by first principles and cannot be predicted with sufficient accuracy by quantum-chemical computations. If the two unpaired electrons are linked by a continuous chain of conjugated bonds, exchange coupling can be significant up to much longer distances.
Distance measurements are most valuable for spin labels or native paramagnetic centers in biomolecules or synthetic macromolecules and supramolecular assemblies. In such systems, if the two unpaired electrons are not linked by a $\pi$-electron system, exchange coupling is negligible with respect to dipole-dipole coupling for distances longer than $1.5 \mathrm{~nm}$. Such systems can often assume different molecular conformations, i.e. their structure is not perfectly defined. Structural characterization thus profits strongly from the possibility to measure distance distributions on length scales that are comparable to the dimension of these systems. This dimension is of the order of 2 to $20 \mathrm{~nm}$, corresponding to $\omega_{\perp}$ between $7 \mathrm{MHz}$ and $7 \mathrm{kHz}$. In order to infer the distance distribution, this small dipole-dipole coupling needs to be separated from larger anisotropic interactions.
This separation of interactions is possible by observing the resonance frequency change for one spin in a pair (blue in Fig. 5.3) that is induced by flipping the spin of its coupling partner (red). In Fig. $9.1$ the resonance frequency of the observer spin before the flip of its coupling partner is indicated by a dashed line. If the coupling partner is in its $|\alpha\rangle$ state before the flip (left panel in Fig. 5.3), the local field at the observer spin will increase by $\Delta B$ upon flipping the coupling partner. This causes an increase of the resonance frequency of the observer spin by the dipole-dipole coupling $d$ (see Eq. (5.16)). If the coupling partner is in its $|\beta\rangle$ state before the flip (right panel in Fig. 5.3), the local field at the observer spin will decrease by $\Delta B$ upon flipping the coupling partner. This causes an decrease of the resonance frequency of the observer spin by the dipole-dipole coupling $d$. In the high-temperature approximation, both these cases have the
same probability. Hence, half of the observer spins will experience a frequency change $+d$ and the other half will experience a frequency change $-d$. If the observer spin evolves with changed frequency for some time $t$, phases $\pm d t$ will be gained compared to the situation without flipping the coupling partner. The additional phase can be observed as a cosine modulation $\cos (d t)$ for both cases, as the cosine is an even function.
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The four-pulse DEER experiment
The most commonly used experiment for distance distribution measurements in the nanometer range is the four-pulse double electron electron resonance (DEER) experiment (Figure 9.2), which is sometimes also referred to as pulsed electron electron double resonance (PELDOR) experiment. All interactions of the observer spin are refocused twice by two $\pi$ pulses at times $2 \tau_{1}$ and $2 \tau_{1}+2 \tau_{2}$ after the initial $\pi / 2$ pulse. Repeated refocusing is necessary since all spin packets must be in phase at $t=0$ and overlap of the pump $\pi$ pulse with the $\pi / 2$ observer pulse would lead to signal distortion. The first refocusing with interpulse delay $\tau_{1}$ restores the situation (1) immediately after the $\pi / 2$ pulse with phase $x$, where the magnetization vectors of all spin packets are aligned with the $-y$ axis. ${ }^{1}$ In practice, coherence is excited on both observer spin transitions (blue in the energy level panels), but for clarity we consider only observer spin coherence that is on the upper transition and is symbolized by a wavy line in panel (1).
During time $t$ after the first refocusing, magnetization vectors of spin packets with different resonance offset dephase (panel (2)). Only the on-resonant spin packet, marked dark blue, is still aligned with the $-y$ direction. The pump pulse flips the coupling partner and thus transfers the coherence to the lower observer spin transition. The resonance frequency of this transition is shifted by the dipole-dipole coupling $d$ in all spin packets. Observer spin magnetization further dephases until the time just before application of the second observer $\pi$ pulse (3)) and, in addition, the whole bundle of spin packet magnetization vectors precesses counterclockwise with the frequency shift $d$. The originally on-resonant spin packet thus gains phase $d\left(\tau_{2}-t\right)$ before the second observer $\pi$ pulse is applied. The second observer $\pi$ pulse with phase $x$ corresponds to a $180^{\circ}$ rotation about the $x$ axis. This mirrors the bundle of magnetization vectors with respect to the $y$ axis, inverting phase of the observer spin coherence (panel (4)). The bundle, which still precesses counterclockwise with angular frequency $d$ now lags the $+y$ axis by phase $d\left(\tau_{2}-t\right)$. During the final interpulse delay of length $\tau_{2}$ the bundle as a whole gains phase $d \tau_{2}$ (grey arrow in panel (4) and simultaneously realigns along its center due to echo refocusing. However, the center corresponding to the originally on-resonant spin packet does not end up along $+y$, as it would have in the absence of the pump pulse. Rather, this spin packet has gained phase $d t$ with respect to the $+y$ direction (panel (5)). The magnetization vector component along $+y$, which corresponds to the echo signal, is given by $\cos (d t)$.
The distance range of the DEER experiment is limited towards short distances by the requirement that, for echo refocusing, the observer pulses must excite both observer transitions, which are split by $d$ and, for coherence transfer, the pump pulse must excite both transitions of the coupling partner, which are also split by $d$. In other words, both the observer refocused echo subsequence and the pump pulse must have an excitation bandwidth that exceeds $d$. This requirement sets a lower distance bound of about $1.8 \mathrm{~nm}$ at X-band frequencies and of about $1.5 \mathrm{~nm}$ at $Q$-band frequencies. A limit towards long distances arises, since several dipolar oscillations need to be observed for inferring the width or even shape of a distance distribution and at least one oscillation needs to be observed for determining the mean distance. This requires $t>2 \pi / d$. On the other hand, we have $t<\tau_{2}$ and the fixed interpulse delay $\tau_{2}$ cannot be much longer than the transverse relaxation time $T_{2}$, since otherwise coherence has completely relaxed and no echos is observed. Electron spin transverse relaxation times are of the order of a few microseconds. Depending on sample type (see Section $9.1 .2$ ), $\tau_{2}$ can be chosen between $1.5$ and $20 \mu \mathrm{s}$, corresponding to maximum observable distances between 5 and $12 \mathrm{~nm}$.
Sample requirements
In the wanted coherence transfer pathway of the DEER experiment, observer pulses exclusively excite observer spins and the pump pulse exclusively excites the coupling partner. The excitation bandwidth must be sufficiently large to cover the dipole-dipole coupling $d$ at all orientations, i.e., larger than $\omega_{\|}-2 \omega_{\perp}$. If the two coupled spins have the same EPR spectrum, this spectrum must be broader than twice this minimum excitation bandwidth. This condition is fulfilled for nitroxide spin labels (Chapter 10) and transition metal ions at all EPR frequencies, whereas some organic radicals, such as trityl radicals, have spectra that are too narrow at X-band or even Q-band frequencies. Furthermore, $T_{2}$ must be sufficiently long for at least the observer spins. This condition can be fulfilled for almost all $S=1 / 2$ species at temperatures of $10 \mathrm{~K}$ (transition metal complexes) or $50 \cdots 80 \mathrm{~K}$ (organic radicals), but may require cooling below $4.2 \mathrm{~K}$ for some high-spin species. For high-spin species with a half-filled valence shell, such as Mn(II) $(S=5 / 2)$ or $\mathrm{Gd}(\mathrm{III})(S=7 / 2)$ measurement temperatures of $10 \mathrm{~K}$ are also sufficient.
Sample concentration should be sufficiently low for intermolecular distances to be much longer than intramolecular distances. For short distances, concentrations up to $200 \mu \mathrm{M}$ are possible, but concentrations of $10 \cdots 50 \mu \mathrm{M}$ provide better results, if a spectrometer with sufficient sensitivity is available. Depending on distance and $T_{2}$, measurements can be performed down to concentrations of $10 \cdots 1 \mu \mathrm{M}$. For membrane proteins reconstituted into liposomes, data quality is not only a function of bulk spin concentration, but also of lipid-to-protein ratio. This parameter needs to be optimized for each new protein. Required sample volume varies between a few microliters (W-band frequencies) and $150 \mu \mathrm{L}$ with $50 \mu \mathrm{L}$ at Q-band frequencies usually being optimal.
If concentration is not too high and the low-temperature limit of transverse relaxation can be attained, $T_{2}$ depends on the concentration and type of protons around the observer spin. Deuteration of the solvent and cryoprotectant (usually glycerol) usually dramatically improve data quality. If the matrix can be perdeuterated, deuteration of the protein or nuclei acid may further prolong $T_{2}$ and extend distance range or improve signal-to-noise ratio.
Complications arise if more than two unpaired electrons are found in the same molecule, but these complications can usually be solved. However, none of the spin pairs should have a distance shorter than the lower limit of the accessible distance range.
1 This assumes that the observer $\pi$ pulse has phase $y$. If it has phase $x$, the magnetization vector is along $+y$ after the first refocusing and all following magnetization panels are mirrored with respect to the $x$ axis.
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Expression for the DEER signal
In Section 9.1.1 we have seen that the echo is modulated with $\cos (d t)$. Usually, this applies only to a fraction $\lambda$ of the echo, because the pump pulse excites only a fraction $\lambda$ of all spin packets of the coupling partner of the observer spin. Therefore, the echo signal for an isolated pair of electron spins in a fixed orientation $\theta$ with respect to the magnetic field is described by
$F(t, r, \theta)=F(0)\{1-\lambda(\theta)[1-\cos (2 d(r, \theta) t)]\}$
where the dependence $d(\theta)$ is given by Eqs. (5.16) and (5.15). The dependence $\lambda(\theta)$ cannot be expressed in closed form, but often $\lambda$ is so weakly correlated with $\theta$ that it can be assumed as a constant, empirical parameter. In this situation, Eq. (9.1) can be integrated over all orientations
$F(t, r)=\int_{0}^{\pi / 2} F(t, r, \theta) \sin \theta \mathrm{d} \theta$
The pump pulse inverts not only the coupling partner of the observer spin in the same molecule, but also electron spins in remote other molecules. If these neighboring spins are homogeneously distributed in space, the background factor $B(t)$ that arises from them assumes the form
$B(t)=\exp \left(-\frac{2 \pi g^{2} \mu_{\mathrm{B}}^{2} \mu_{0} N_{\mathrm{A}}}{9 \sqrt{3} \hbar} \lambda^{\prime} c t\right)$
where the orientation-averaged inversion efficiency $\lambda^{\prime}$ is the fraction of spins excited by the pump pulse, $g$ is an average $g$ value, and $c$ is the total concentration of spins. For subtle reasons, $\lambda^{\prime}$ differs significantly from the empirical two-spin modulation depth $\lambda$. Homogeneous distributions of neighboring spins that are nearly confined to a plane or a line give rise to a stretched exponential background function $B(t)=\exp \left[-(k t)^{D / 3}\right]$, where $D$ is a fractional dimension of the distribution that is usually somewhat larger than 2 or 1 for nearly planar or linear distributions, respectively. The total DEER signal is given by
$V(t, r)=F(t, r) B(t)$
If distance $r$ is distributed with normalized probability density $P(r)\left(\int_{0}^{\infty} P(r) \mathrm{d} r=1\right)$, the form factor $F(t)$ needs to be replaced by $F_{P}(t)=\int_{0}^{\infty} P(r) F(t, r) \mathrm{d} r$.
Background correction
The information on the distance distribution $P(r)$ is contained in $F(t)$, which must thus be separated from $B(t)$. Often, the distribution is sufficiently broad for dipolar oscillations to decay within a time $t_{\mathrm{dec}}$ shorter than the maximum dipolar evolution time $t_{\max }$ (Fig. 9.3(a)). For $t_{\mathrm{dec}} \leq t \leq t_{\max }$, the primary signal is then given by $b(t)=(1-\lambda) \exp \left[-(k t)^{D / 3}\right]$ plus noise. The expression for $b(t)$ is fitted to the primary data in this range (red line in Fig. 9.3(a)). In some cases, for instance for soluble proteins, a homogeneous distribution of molecules in three dimensions can be assumed, so that $D=3$ can be fixed. Otherwise, $D$ is treated as a fit parameter, as are $k$ and $\lambda$. The background function $B(t)$ is obtained by extrapolating $b(t)$ to the range $0 \leq t \leq t_{\mathrm{dec}}$ (ochre line) and dividing it by $b(0)=1-\lambda$. According to Eq. (9.4), the form factor $F(t) / F(0)$ results by dividing $V(t) / V(0)$ by $B(t)$. For narrow distance distributions, oscillations in $V(t) / V(0)$ may endure until the longest attainable $t_{\max }$. This does not create a problem if at least the first oscillation is completed well before $t_{\max }$. All the following oscillations have very similar amplitude and do not bias the background fit. As a rule of thumb, a good estimate for $B(t)$ can be obtained by fitting data at $t \geq t_{\max } / 2$ if $d t_{\max } \geq 4 \pi$, i.e., if two full oscillations can be observed. If the data trace is shorter than that, background fitting is fraught with uncertainty. Wrong background correction may suppress long distances or create artificial peaks at long distances.
Tikhonov regularization with non-negativity constraint
In order to extract the distance distribution $P(r)$ from the experimental form factor $F(t) / F(0)$, we need to remove the constant contribution and renormalize to the dipolar evolution function
$D(t)=\frac{F(t) / F(0)-(1-\lambda)}{\lambda}$
and invert the integral equation $D(t)=\int_{0}^{\infty} P(r) K(t, r) \mathrm{d} r$, where the kernel $K(t, r)$ is given by
$K(t, r)=\int_{0}^{1} \cos \left[\left(3 z^{2}-1\right) \omega_{\perp}(r) t\right] \mathrm{d} z$
Here, we have substituted $\cos \theta$ by $z, \sin \theta \mathrm{d} \theta$ by $-\mathrm{d} \cos \theta$ and reversed direction of the integration, which compensated for the negative $\operatorname{sign}$ in $-\mathrm{d} \cos \theta$.
In practice, $D(t)$ is digitized and given as a vector at sampling times $t_{i}$. Likewise, it is sufficient to compute $P(r)$ as a vector at sampling distances $r_{k}$. The integral equation is thus transformed to a matrix equation
$\vec{D}=\mathbf{K} \vec{P}$
Unfortunately, this matrix equation cannot easily be inverted, since the rows of kernel $\mathbf{K}$ are not orthogonal, i.e., the scalar product of dipolar evolution function vectors at different $r_{k}$ is not zero. The weak linear dependence of the rows makes the problem ill-posed. Small deviations of the experimental $\vec{D}$ from the "true" $\vec{D}_{\text {ideal }}$, for instance due to noise, cause large deviations of $\vec{P}$ from the true distance distribution. This problem can be solved only by taking into account additional information.
First, we know that, as a probability density, $P(r) \geq 0$ at all $r$. Hence, we can impose a non-negativity constraint on $\vec{P}$. It turns out that this is not sufficient for stabilizing the solution. Noise can be fitted by ragged distance distributions with many narrow peaks, although we know that the distance distribution must be smooth, as it arises from a continuous distribution of molecular conformations. Tikhonov regularization imposes a smoothness restraint by minimizing
$G_{\alpha}=\rho+\alpha \eta$
where
$\rho=\|\mathbf{K} \vec{P}-\vec{D}\| \|^{2}$
is the mean square deviation between experimental and simulated data and
$\eta=\left\|\hat{L}^{(2)} \vec{P}\right\|^{2}$
is the square norm of the second derivative, which can be computed from $\vec{P}$ by multiplication with the second derivative operator $\hat{L}^{(2)}$. The regularization parameter $\alpha$ determines the relative <figure>$1<figcaption>$2</figcaption></figure>
weight of the smoothing restraint with respect to mean square deviation between experimental and simulated data. A parametric plot of $\log \eta$ versus $\log \rho$ as a function of $\alpha$ is approximately L-shaped (Fig. 9.4). For very small $\alpha$, roughness $\eta$ of the distance distribution can be decreased strongly without increasing mean square deviation $\rho$ very much. For large $\alpha, \vec{P}$ is already smooth and a further increase of $\alpha$ will lead only to a small decrease in roughness $\eta$, but to a large increase in $\rho$, since the overly broadened distance distribution no longer fits the dipolar oscillations. Hence, in a mathematical sense, the optimum regularization parameter corresponds to the corner of the $L$ curve. At this regularization parameter the extracted distance distribution (black line in Fig. 9.4(c)) is only slightly broader than the true distance distribution (green line) and the simulated form factor (red line in Fig. 9.3(b)) agrees with the experimental form factor (black line), except for the white noise contribution. If the regularization parameter is too large (red circle in Fig. 9.4(a)), the simulated form factor is overdamped (red line in Fig. 9.4(b)) and the distance distribution unrealistically broad (black line in Fig. 9.4(d)). For a too small regularization parameter the distance distribution unrealistically splits into several narrow peaks and the simulated form factor fits part of the noise (not shown). This error cannot be as clearly discerned in the simulated form factor as overdamping can be discerned. Undersmoothing is apparent only in the L curve.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Electron_Paramagnetic_Resonance_(Jenschke)/09%3A_Distance_Distribution_Measurements/9.03%3A_Conversion_of_dipolar_evolution_data_to_distance_distributions.txt
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• 10.1: Nitroxide spin probes and labels
Spin probes are stable paramagnetic species that are admixed to a sample in order to obtain structural or dynamical information on their environment and, thus, indirectly on the sample. Spin labels are spin probes that are covalently attached to a molecule of interest, often at a specific site. As compared to more direct characterization of structure and dynamics by other techniques, EPR spectroscopy on spin probes may be able to access other length and time scales.
• 10.2: Spin traps
A spin trap (Fig. 10.8) is a diamagnetic compound that is primed to form a stable radical by reaction with an unstable radical. The most frequently used spin traps are nitrones that form nitroxide radicals by addition of the unstable radical to the C atom in α position of the nitrone group.
10: Spin Probes and Spin Traps
Spin probes and labels
Spin probes are stable paramagnetic species that are admixed to a sample in order to obtain structural or dynamical information on their environment and, thus, indirectly on the sample. Spin labels are spin probes that are covalently attached to a molecule of interest, often at a specific site. As compared to more direct characterization of structure and dynamics by other techniques, EPR spectroscopy on spin probes may be able to access other length and time scales or may be applicable in aggregation states or environments where these other techniques exhibit low resolution or do not yield any signal. Site-directed spin labeling (SDSL) has the advantage that assignment of the signal to primary molecular structure is already known and that a specific site in a complex system can be studied without disturbance from signals of other parts of the system. This approach profits from the rarity of paramagnetic centers. For instance, many proteins and most nucleic acids and lipids are diamagnetic. If a spin label is introduced at a selected site, EPR information is specific to this particular site.
In principle, any stable paramagnetic species can serve as a spin probe. Some paramagnetic metal ions can substitute for diamagnetic ions native to the system under study, as they have similar charge and ionic radius or with similar complexation properties as the native ions. This applies to $\mathrm{Mn}(\mathrm{II})$, which can often substitute for $\mathrm{Mg}(\mathrm{II})$ without affecting function of proteins or nucleic acids, or Ln(III) lanthanide ions, which bind to Ca(II) sites. Paramagnetic metal ions can also be attached to proteins by engineering binding sites with coordinating amino acids, such as histidine, or by site-directed attachment of a metal ligand to the biomolecule. Such approaches are used for lanthanide ions, in particular Gd(III), and Cu(II).
For many spin probe approaches, organic radicals are better suited than metal ions, since in radicals the unpaired electron has closer contact to its environment (ligands screen environmental access of metal ions, in particular for lanthanide ions) and the EPR spectra are narrower, which allows for some experiments that cannot be performed on species with very broad spectra. Among organic radicals, nitroxides are the most versatile class of spin probes, mainly because of their relatively small size, comparable to an amino acid side group or nucleobase, and because of hyperfine and $g$ tensor anisotropy of a magnitude that is convenient for studying dynamics (Section 10.1.4). Triarylmethyl (TAM) radicals are chemically even more inert than nitroxide radicals and have slower relaxation times in liquid solution. Currently they are much less in use than nitroxide radicals, mainly because they are not commercially available and much harder to synthesize than nitroxide radicals.
Nitroxide radicals
The nitroxide radical is defined by the $\mathrm{N}-\mathrm{O}^{\bullet}$ group, which is isoelectronic with the carbonyl group $(\mathrm{C}=\mathrm{O})$ and can thus be replaced in approximate force field and molecular dynamics computations by a $\mathrm{C}=\mathrm{O}$ group. The unpaired electron is distributed over both atoms, which contributes to radical stability, with a slight preference for the oxygen atom. Nitroxide radicals become stable on the time scale of months or years if both $\alpha$ positions are sterically protected, for instance by attaching two methyl groups to each of the $\alpha \mathrm{C}$ atoms (Fig. 10.1). Nitroxides of this type are thermally stable up to temperatures of about $140^{\circ} \mathrm{C}$, but they are easily reduced to the corresponding hydroxylamines, for instance by ascorbic acid, and are unstable at very low and very high pH. Nitroxides with five-membered rings (structures $\mathbf{2}, \boldsymbol{3}$, and $\mathbf{5}$ ) tend to be chemically more stable than those with six-membered rings (6). The five-membered rings also have less conformational freedom than the six-membered rings.
Spin probes can be addressed to certain environments in heterogeneous systems by choice of appropriate substituents $\mathrm{R}$ (Fig. 10.1). The unsubstituted species $(\mathrm{R}=\mathrm{H})$ are hydrophobic and partition preferably to nonpolar environments. Preference for hydrogen bond acceptors is achieved by hydroxyl derivatives $(\mathrm{R}=\mathrm{OH})$, whereas ionic environments can be addressed by a carboxylate group at sufficiently high $\mathrm{pH}\left(\mathrm{R}=\mathrm{COO}^{-}\right)$or by a trimethyl ammonium group $(\mathrm{R}=$ $\left.\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}^{+}\right)$. Reactive groups $\mathrm{R}$ are used for SDSL, such as the methanethiosulfonate group in the dehydro-PROXYL derivative MTSL 5, which selectively reacts with thiol groups under mild conditions. Thiol groups can be introduced into proteins by site-directed point mutation of an amino acid to cysteine and to RNA by replacement of a nucleobase by thiouridine. In DOXYL derivatives $\mathbf{4}$, a six-membered ring is spiro-linked to an alkyl chain, which can be part of stearic acid or of lipid molecules. The $\mathrm{N}^{\circ} \mathrm{O}^{\bullet}$ group in DOXYL derivatives is rigidly attached to the alkyl chain and nearly parallel to the axis of a hypothetical all-trans chain.
The nitroxide EPR spectrum
To a good approximation, the spin system of a nitroxide radical can be considered as an electron spin $S=1 / 2$ coupled to the nuclear spin $I=1$ of the ${ }^{14} \mathrm{~N}$ atom of the $\mathrm{N}-\mathrm{O}^{\bullet}$ group. Hyperfine
couplings to other nuclei, such as the methyl protons, are not usually resolved and contribute only to line broadening. The hyperfine coupling to the $s p^{2}$ hybridized ${ }^{14} \mathrm{~N}$ atom has a significant isotropic Fermi contact contribution from spin density in the $2 s$ orbital and a significant anisotropic contribution from spin density in the $p_{\pi}$ orbital that combines with a $p_{\pi}$ orbital on the oxygen atom to give the N-O bond partial double bond character. The direction of the lobes of the $p_{\pi}$ orbital is chosen as the molecular $z$ axis (Fig. 10.2(b)). The ${ }^{14} \mathrm{~N}$ hyperfine tensor has nearly axial symmetry with $z$ being the unique axis. The hyperfine coupling is much larger along $z$ (on the order of $90 \mathrm{MHz}$ ) than in the $x y$ plane (on the order of $15 \mathrm{MHz}$ ).
The spin-orbit coupling, which induces $g$ anisotropy, arises mainly at the $\mathrm{O}$ atom, where a lone pair energy level is very close to the SOMO. The $g$ tensor is orthorhombic with nearly maximal asymmetry. The largest $g$ shift is positive and observed along the N-O bond, which is the molecular frame $x$ axis $\left(g_{x} \approx 2.009\right)$. An intermediate $g$ shift is observed along the $y$ axis $\left(g_{y} \approx 2.006\right)$, whereas the $g_{z}$ value is very close to $g_{e}=2.0023$. At X-band frequencies, where $\nu_{\mathrm{mw}} \approx 9.5 \mathrm{GHz}, g$ anisotropy corresponds to only $1.13 \mathrm{mT}$ dispersion in resonance fields, while hyperfine anisotropy corresponds to $6.5 \mathrm{mT}$ dispersion. At W-band frequencies, where $\nu_{\mathrm{mw}} \approx 95 \mathrm{GHz}$, hyperfine anisotropy is still the same but $g$ anisotropy contributes a ten times larger dispersion of $11.3 \mathrm{mT}$, which now dominates.
The field-swept CW EPR spectrum for a single orientation can be understood by considering the selection rule that the magnetic quantum number $m_{S}$ of the electron spin must change by 1 , whereas the magnetic quantum number $m_{I}$ of the ${ }^{14} \mathrm{~N}$ nuclear spin must not change. Each transition can thus be assigned to a value of $m_{I}$. For $I=1$ there are three such values, $m_{I}=-1,0$, and 1 (Fig. $10.2(\mathrm{a})$ ). The microwave frequency $\nu_{\mathrm{mw}}$ is fixed and resonance is observed at fields where the energy of the microwave quantum $h \nu_{\mathrm{mw}}$ matches the energy of a transition.
In order to construct the solid-state spectrum, orientation dependence of the three transitions must be considered (Fig. 10.3(a)). At each individual orientation, the $m_{I}=0$ line is the center line. Since the hyperfine contributions scales with $m_{I}$, it vanishes for this line and only $g$ anisotropy is observed. At X band, where hyperfine anisotropy dominates by far, this line is the narrowest one. The lineshape is the one for pure $g$ anisotropy (see Fig. 3.4). For $m_{I}=+1$, the orientation with the largest $g$ shift of the resonance field coincides with the one of smallest hyperfine shift. Hence, the smaller resonance field dispersion by $g$ anisotropy subtracts from the larger dispersion by hyperfine anisotropy. For $m_{I}=-1$, the situation is opposite and the two dispersions add. Hence, the $m_{I}=-1$ transition, which at any given orientation is the high-field line, has the largest resonance dispersion, whereas the low-field $m_{I}=+1$ transition has intermediate resonance field dispersion. The central feature in the total absorption spectrum (Fig. $10.3(\mathrm{~b})$ ) is strongly dominated by the $m_{I}=0$ transition, whereas the outer shoulders correspond to the $m_{I}=+1$ (low field) and $m_{I}=-1$ (high field) transitions at the $z$ orientation. Therefore, the splitting between the outer extremities in the CW EPR spectrum, which correspond to these shoulders in the absorption spectrum, is $2 A_{z z}$.
Influence of dynamics on the nitroxide spectrum
In liquid solution, molecules tumble stochastically due to Brownian rotational diffusion. In the following we consider isotropic rotational diffusion, where the molecule tumbles with the same average rate about any axis in its molecular frame. This is a good approximation for nitroxide spin probes with small substituents $\mathrm{R}$. For instance, TEMPO $(\mathbf{1}$ with $\mathrm{R}=\mathrm{H})$ is almost spherical with a van-der-Waals radius of $3.43 \AA \AA$. In water at ambient temperature, the $\tau_{\mathrm{r}}$ rotational correlation time for TEMPO is of the order of $10 \mathrm{ps}$. The product $\tau_{\mathrm{r}} \Delta \omega$ with the maximum anisotropy $\Delta \omega$ of the nitroxide EPR spectrum on an angular frequency axis is much smaller than unity. In this situation, anisotropy averages and three narrow lines of equal width and intensity are expected. The spectrum in Fig. 10.2(a) corresponds to this situation and a closer look reveals that the high-field line has somewhat lower amplitude. This can be traced back to a larger linewidth than for the other two lines, which indicates a shorter $T_{2}$ for the $m_{I}=-1$ transition than for the other transitions. Indeed, transverse relaxation is dominated by the effect from combined hyperfine and $g$ anisotropy, which is largest for the $m_{I}=-1$ transition that has the largest anisotropic dispersion of resonance frequencies. With increasing rotational correlation time $\tau_{r}$, one expects this relaxation process to become stronger, which should lead to more line broadening that is strongest for the high-field line and weakest for the central line. This is indeed observed in the simulation for $\tau_{\mathrm{r}}=495$ ns shown in the bottom trace of Fig. $10.4$.
According to Kivelson relaxation theory, the ratio of the line width of one of the outer lines to the line width of the central line is given by
$\frac{T_{2}^{-1}\left(m_{I}\right)}{T_{2}^{-1}(0)}=1+B m_{I}+C m_{I}^{2}$
where
$B=-\frac{4}{15} b \Delta \gamma B_{0} T_{2}(0) \tau_{\mathrm{r}}$
and
$C=\frac{1}{8} b^{2} T_{2}(0) \tau_{\mathrm{r}}$
with the hyperfine anisotropy parameter
$b=\frac{4 \pi}{3}\left[A_{z z}-\frac{A_{x x}+A_{y y}}{2}\right]$
and the electron Zeeman anisotropy parameter $\Delta \gamma$
$\Delta \gamma=\frac{\mu_{\mathrm{B}}}{\hbar}\left[g_{z z}-\frac{g_{x x}+g_{y y}}{2}\right]$
The relaxation time $T_{2}(0)$ for the central line can be computed from the corresponding peak-topeak linewidth in field domain $\Delta B_{p p}(0)$ as
$T_{2}(0)=\frac{2}{\sqrt{3} g_{\text {iso }} \mu_{\mathrm{B}} \Delta B_{\mathrm{pp}}(0)}$
Thus, Eqs. (10.1-10.3) can be solved for the only remaining unknown $\tau_{\mathrm{r}}$. In practice, ratios of peak-to-peak line amplitudes $I\left(m_{I}\right)$ are analyzed rather than linewidth ratios, as they can be measured with higher precision. The linewidth ratio is related to the amplitude ratio $I(0) / I(-1)$ (see bottom trace in Fig. 10.4) in a first derivative spectrum by
$\frac{T_{2}^{-1}\left(m_{I}\right)}{T_{2}^{-1}(0)}=\sqrt{\frac{I(0)}{I\left(m_{I}\right)}}$
since the integral intensity of the absorption line (double integral of the derivative lineshape) is the same for each of the three transitions. The rotational correlation time can thus be determined by, e.g.,
$\tau_{\mathrm{r}}=\frac{\sqrt{3}}{2 b}\left[\frac{b}{8}-\frac{4 \Delta \gamma B_{0}}{15}\right]^{-1} \frac{g_{\mathrm{iso}} \mu_{\mathrm{B}}}{\hbar} \Delta B_{\mathrm{pp}}(0)\left[\sqrt{\frac{I(0)}{I(-1)}}-1\right]$
where $\Delta B_{0}$ is the peak-to-peak linewidth of the central line. This equation can be applied in the fast tumbling regime, where the three individual lines for $m_{I}=-1,0$, and $+1$ can still be clearly recognized and have the shape of symmetric derivative absorption lines.
For slower tumbling with $\tau_{\mathrm{r}}>1.5 \mathrm{~ns}$, the line shape becomes more complex and approaches the rigid limit (solid-state spectrum) at about $\tau_{\mathrm{r}}=1 \mu \mathrm{s}$ (Fig. 10.4). These lineshapes can be simulated by considering multi-site exchange between different orientations of the molecule with respect to the magnetic field. Unlike for two-site exchange, which is discussed in the NMR part of the lecture course (see Section 3 of the NMR lecture notes), no closed expressions can be obtained for multi-site exchange. Nevertheless we can estimate the time scale where the spectral features are broadest and transverse relaxation times are shortest. Coalescence for two-site exchange is observed at $\Delta \Omega / k=2 \sqrt{2}$. By substituting $k$ by $1 / \tau_{\text {r and }} \Delta \Omega$ by the maximum anisotropy of $7.6 \mathrm{mT}$, corresponding to $213 \mathrm{MHz}$, we find a "coalescence time" $2 \sqrt{2} / \Delta \Omega \approx 2.1$ ns. The simulations in Fig. $10.4$ show indeed that around this rotational correlation time, the
character of the spectrum changes from fast orientation exchange (liquid-like spectrum with three distinct peaks) to slow orientation exchange (solid-like spectrum).
A simple way of analyzing a temperature dependence, such as the one shown in Fig. 10.4, is to plot the outer extrema separation $2 A_{z z}^{\prime}$ as a function of temperature (Fig. 10.5). The "coalescence time" in such a plot corresponds to the largest gradient $\mathrm{d} A_{z z}^{\prime} / \mathrm{d} T$, which coincides with the mean between the $2 A_{z z}^{\prime}$ values in the fast tumbling limit and rigid limit, which is $5 \mathrm{mT}$. In the case at hand, this coalescence time is $3.5 \mathrm{~ns}$ and is observed at a temperature $T_{5 \mathrm{mT}}=312$ $\mathrm{K}$. The $T_{5 \mathrm{mT}}$ temperature is the temperature where the material becomes "soft" and molecular conformations can rearrange. Nitroxide spectra in the slow tumbling regime can reveal more details on dynamics, for instance, whether there are preferred rotation axes, whether motion is restricted due to covalent linkage of the nitroxide to a large molecule, or whether there is local order, such as in a lipid bilayer.
Polarity and proticity
Delocalization of the unpaired electron in the $\mathrm{N}-\mathrm{O}^{\bullet}$ group can be understood by considering mesomeric structures (Fig. 10.6). If the unpaired electron resides on oxygen, the formal number of valence electrons is five on nitrogen and six on oxygen, corresponding to the nuclear charge that is not compensated by inner shell electrons. Hence, both atoms are formally neutral in this limiting structure. If, on the other hand, the unpaired electron resides on the nitrogen atom, only four valence electrons are assigned to this atom, whereas seven valence electrons are assigned to the oxygen atom. This corresponds to charge separation with the formal positive charge on nitrogen and the formal negative charge on oxygen. The charge-separated form is favored in polar solvents, which screen Coulomb attraction between the two charges, whereas the neutral form is favored in nonpolar solvents. Hence, for a given nitroxide radical in a series of solvents, the ${ }^{14} \mathrm{~N}$ hyperfine coupling, which stems from spin density on the nitrogen atom, is expected to increase with increasing solvent polarity. This effect has indeed been found. It is most easily seen in the solid state for $A_{z z}$ but can also be discerned in the liquid state for $A_{\text {iso }}$.
The change in $A_{z z}$ is expected to be anti-correlated to the $g_{x x}$ shift, because this shift arises from SOC at the oxygen atom and, the higher spin density on the nitrogen atom is, the lower it is on the oxygen atom. This effect has also been found and is most easily detected by high-field/high-frequency EPR at frequencies of W-band frequencies of $\approx 95 \mathrm{GHz}$ or even higher frequencies. How $A_{z z}$ is correlated to $g_{x x}$ depends on proticity of the solvent. Protic solvents form hydrogen bonds with the lone pairs on the oxygen atom of the N-O ${ }^{\bullet}$ group. This lowers energy of the lone pair orbitals, making excitation of an electron from these orbitals to the SOMO less likely. Since this excitation provides the main contribution to SOC and thus to $g_{x x}$ shift, hydrogen bonding to oxygen reduces $g_{x x}$ shift. If two nitroxides have the same hyperfine coupling $A_{z z}$ in an aprotic and protic environment, $g_{x x}$ will be lower in the protic environment. This effect has also been found. In some cases it was possible to discern nitroxide labels with zero, one, and two hydrogen bonds by resolution of their $g_{x x}$ features in W-band CW EPR spectra. Slopes of $-1.35 \mathrm{~T}^{-1}$ for aprotic at $-2 \mathrm{~T}^{-1}$ for protic environments have been found for the correlation between $A_{z z}$ and $g_{x x}$ for MTSL in spin-labeled bacteriorhodopsin in lipid bilayers [Ste+00].
Water accessibility
Polarity and proticity are proxy parameters for water accessibility of spin-labeled sites in proteins. Two other techniques provide complementary information. First, water can be replaced by deuterated water and the modulation depth of deuterium ESEEM can be measured. Because of the $r^{-6}$ dependence of modulation depth (see Eq. (8.7)) the technique is most sensitive to deuterium nuclei in the close vicinity of the spin label. As long as $k \ll 1$, modulation depth contributions of individual nuclei add, so that the total deuterium modulation depth is a measure for local deuterium concentration close to the label. Data can be processed in a way that removes the contribution from directly hydrogen-bonded nuclei. Strictly speaking, this technique measures the concentration of not only water protons but also the one of any exchangeable protons near the label, but only to the extent that these exchangeable protons are water accessible during sample preparation or measurement.
A second, more direct technique that is applicable at ambient temperature measures the proton NMR signal as a function of irradiated microwave power with the microwave frequency being on-resonant with the central transition of a nitroxide spin label. Such irradiation transfers electron spin polarization to water protons by the Overhauser effect. This Overhauser dynamic nuclear polarization (DNP) is highly specific to water, as it critically depends on the water proton NMR signal being narrow and on fast diffusion of water. In biomolecules, water accessibility of spin labels is high at water-exposed surfaces of soluble and membrane proteins and low inside the proteins and at lipid-exposed surfaces. For transporters, water accessibility can change with state in the transport process.
Oxygen accessibility
Since collision of paramagnetic triplet oxygen with spin probes enhances relaxation (Fig. $7.4)$, the saturation parameter $S=\omega_{1}^{2} T_{1} T 2$ is smaller for oxygen-accessible spin labels than for spin labels not accessible to oxygen. This change can be detected by CW progressive power saturation measurements (Section 7.2.2). The experiment is most conveniently performed with capillary tubes made of the gas permeable plastic TPX. A reference measurement is performed in a nitrogen atmosphere, which causes deoxygenation of the sample on the time scale of 15 min. The gas stream is then changed to air (20% oxygen) or pure oxygen and the measurement is repeated. Such data are shown in Fig. $10.7$ for residue 229 in major plant light harvesting complex LHCII. This residue is lipid exposed. As a nonpolar molecule, oxygen dissolves well in the alkyl chain region of a lipid bilayer. Accordingly, the signal saturates at higher power in an air atmosphere than in a nitrogen atmosphere. Oxygen accessibility can be quantified by a normalized $P_{1 / 2}$ parameter (Section 7.2.2).
Local pH measurements
The ${ }^{14} \mathrm{~N}$ hyperfine coupling of nitroxide spin probes becomes $\mathrm{pH}$ sensitive if the heterocycle that contains the $\mathrm{N}-\mathrm{O}^{\bullet}$ group also contains a nitrogen atom that can be protonated in the desired $\mathrm{pH}$ range. This applies, for instance, to the imidazolidine nitroxide 3 in Fig. $10.1$, which has a pK value of $\approx 4.7$ and exhibits a change in isotropic ${ }^{14}$ N hyperfine coupling of $0.13 \mathrm{mT}$ between the protonated (1.43 mT) and deprotonated $(1.56 \mathrm{mT})$ form, which can be resolved easily in liquid solution. By modifying the probe to a label, local $\mathrm{pH}$ can be measured near a residue of interest in a protein.
10.02: Spin traps
Many radicals are very reactive. This fact makes their detection during chemical reactions and in living cells very important, but it also makes their concentration very low, since often their formation reaction is slower than the reactions that destroy them again. For instance, concentration of the hydroxyl radical ${ }^{\bullet} \mathrm{OH}$, a reactive oxygen species (ROS) in living cells, is too low for EPR detection even under conditions where ${ }^{\bullet} \mathrm{OH}$ leads to cell damage or cell death. The situation is somewhat better for the superoxide anion radical $\mathrm{O}_{2}^{2- \bullet}$, but physiologically relevant concentrations are hard to detect also for this species.
ROS and some other highly reactive radicals of interest are most easily detected by spin trapping. A spin trap (Fig. 10.8) is a diamagnetic compound that is primed to form a stable radical by reaction with an unstable radical. The most frequently used spin traps are nitrones that form nitroxide radicals by addition of the unstable radical to the $\mathrm{C}$ atom in $\alpha$ position of the nitrone group. The formed nitroxide radicals are not as stable as the ones used as spin labels, mainly because they contain a hydrogen atom in $\alpha$ position to the $\mathrm{N}-\mathrm{O}$ group. Their lifetime is usually on the minute time scale, which is sufficient for detection. The hyperfine coupling of the $\mathrm{H}^{\alpha}$ atom is sensitive to the type of primary radical $\mathrm{R}^{\bullet}$, i.e. to the nature of the other substituent at the $C^{\alpha}$ atom. Furthermore, these nitrones are less sterically crowded than the ones that would yield more stable nitroxides and thus the nitrones are more reactive and trap radicals $\mathrm{R}^{\bullet}$ more easily. In addition to the $\mathrm{H}^{\alpha}$ hyperfine coupling, the hyperfine coupling of the ${ }^{14} \mathrm{~N}$ atom of the $\mathrm{N}-\mathrm{O}^{\bullet}$ group is sensitive to the nature of $\mathrm{R}^{\bullet}$. A database of experimental results supports assignment of $R^{\bullet}$ in difficult cases: https://tools.niehs.nih.gov//stdb/index.cfm 1
1 Look a the "Hints for Using the Spin Trap Database" before you start your search. The keyword format is powerful, but not very intuitive.
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$\newcommand{\R}{8.3145\units{J\,K\per\,mol\per}} % gas constant value$
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There is international agreement that the units used for physical quantities in science and technology should be those of the International System of Units, or SI (standing for the French Système International d’Unités). The Physical Chemistry Division of the International Union of Pure and Applied Chemistry, or IUPAC, produces a manual of recommended symbols and terminology for physical quantities and units based on the SI. The manual has become known as the Green Book (from the color of its cover) and is referred to here as the IUPAC Green Book. This e-book will, with a few exceptions, use symbols recommended in the third edition (2007) of the IUPAC Green Book (E. Richard Cohen et al, Quantities, Units and Symbols in Physical Chemistry, 3rd edition. RSC Publishing, Cambridge, 2007). These symbols are listed for convenient reference in Appendices C and D.
Any of the symbols for units listed in Tables 1.1–1.3, except kg and $\degC$, may be preceded by one of the prefix symbols of Table 1.4 to construct a decimal fraction or multiple of the unit. (The symbol g may be preceded by a prefix symbol to construct a fraction or multiple of the gram.) The combination of prefix symbol and unit symbol is taken as a new symbol that can be raised to a power without using parentheses, as in the following examples:
• The physical quantity formally called amount of substance is a counting quantity for particles, such as atoms or molecules, or for other chemical entities. The counting unit is invariably the mole, defined as the amount of substance containing as many particles as the number of atoms in exactly $12$ grams of pure carbon-12 nuclide, ${}^{12}$C. See Appendix A for the wording of the official IUPAC definition. This definition is such that one mole of H$_{2}$O molecules, for example, has a mass of $18.0153$ grams (where $18.0153$ is the relative molecular mass of H$_{2}$O) and contains $6.02214\timesten{23}$ molecules (where $6.02214 \timesten{23}\units{mol\(^{-1}$}\) is the Avogadro constant to six significant digits). The same statement can be made for any other substance if $18.0153$ is replaced by the appropriate atomic mass or molecular mass value.
The symbol for amount of substance is $n$. It is admittedly awkward to refer to $n$(H$_2$O) as “the amount of substance of water.” This e-book simply shortens “amount of substance” to amount. An alternative name suggested for $n$ is “chemical amount.” Thus, “the amount of water in the system” refers not to the mass or volume of water, but to the number of H$_2$O molecules in the system expressed in a counting unit such as the mole.
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The study of gases allows us to understand the behavior of matter at its simplest: individual particles, acting independently, almost completely uncomplicated by interactions and interferences between each other. This knowledge of gases will serve as the pathway to our understanding of the far more complicated condensed phases (liquids and solids) in which the theory of gases will no longer give us correct answers, but it will still provide us with a useful model that will at least help us to rationalize the behavior of these more complicated states of matter.
• 2.1: Some Definitions
• 2.2: An Operational Definition of Temperature
• 2.3: Ideal Gases
In an ideal gas, there are no interactions between the particles, hence, the particles do not exert forces on each other. However, particles do experience a force when they collide with the walls of the container.
• 2.4: Real Gases
So what happens when a real gas is subjected to a very high pressure? The outcome varies with both the molar mass of the gas and its temperature, but in general we can see the the effects of both repulsive and attractive intermolecular forces.
• 2.5: Condensation of Gases and the Critical State
The most striking feature of real gases is that they cease to remain gases as the temperature is lowered and the pressure is increased.
• 2.6: Kinetic Theory of Gases
The kinetic theory describes a gas as a large number of submicroscopic particles, all of which are in constant, random motion. The rapidly moving particles constantly collide with each other and with the walls of the container. Kinetic theory explains macroscopic properties of gases, such as pressure, temperature, viscosity, thermal conductivity, and volume, by considering their molecular composition and motion.
• 2.7: The Maxwell Distribution Laws
The Maxwell-Boltzmann distribution is used to determine how many molecules are moving between velocities.
• 2.8: Molecular Collisions and the Mean Free Path
The collision theory states that when suitable particles of the reactant hit each other, only a certain percentage of the collisions cause any noticeable or significant chemical change; these successful changes are called successful collisions. The successful collisions have enough energy, also known as activation energy, at the moment of impact to break the preexisting bonds and form all new bonds. This results in the products of the reaction.
• 2.9: Graham's Laws of Diffusion and Effusion
Graham's Law states that the effusion rate of a gas is inversely proportional to the square root of the mass of its particles.
• 2.E: Properties of Gases (Exercises)
This are exercises that to accompany the TextMap organized around Raymond Chang's Physical Chemistry for the Biosciences textbook.
02: Properties of Gases
In an ideal gas, there are no interactions between the particles, hence, the particles do not exert forces on each other. However, particles do experience a force when they collide with the walls of the container. Let us assume that each collision with a wall is elastic. Let us assume that the gas is in a cubic box of length $a$ and that two of the walls are located at $x = 0$ and at $x = a$. Thus, a particle moving along the $x$ direction will eventually collide with one of these walls and will exert a force on the wall when it strikes it, which we will denote as $F_x$. Since every action has an equal and opposite reaction, the wall exerts a force $-F_x$ on the particle. According to Newton’s second law, the force $-F_x$ on the particle in this direction gives rise to an acceleration via
$-F_x = ma_x = m\dfrac{\Delta v_x}{\Delta t} \label{2.1}$
Here, $t$ represents the time interval between collisions with the same wall of the box. In an elastic collision, all that happens to the velocity is that it changes sign. Thus, if $v_x$ is the velocity in the $x$ direction before the collision, then $-v_x$ is the velocity after, and $\Delta v_x = -v_x - v_x = -2v_x$, so that
$-F_x = -2m\dfrac{v_x}{\Delta t} \label{2.2}$
In order to find $\Delta t$, we recall that the particles move at constant speed. Thus, a collision between a particle and, say, the wall at $x = 0$ will not change the particle’s speed. Before it strikes this wall again, it will proceed to the wall at $x = a$ first, bounce off that wall, and then return to the wall at $x = 0$. The total distance in the $x$ direction traversed is $2a$, and since the speed in the $x$ direction is always $v_x$, the interval $\Delta t = \dfrac{2a}{v_x}$. Consequently, the force is
$-F_x = -\dfrac{mv_x^2}{a} \label{2.3}$
Thus, the force that the particle exerts on the wall is
$F_x = \dfrac{mv_x^2}{a} \label{2.4}$
The mechanical definition of pressure is
$P = \dfrac{\langle F \rangle}{A} \label{2.5}$
where $\langle F \rangle$ is the average force exerted by all $N$ particles on a wall of the box of area $A$. Here $A = a^2$. If we use the wall at $x = 0$ we have been considering, then
$P = \dfrac{N \langle F_x \rangle}{a^2} \label{2.6}$
because we have $N$ particles hitting the wall. Hence,
$P = \dfrac{N m \langle v_x^2 \rangle}{a^3} \label{2.7}$
from our study of the Maxwell-Boltzmann distribution, we found that
$\langle v_x ^2 \rangle = \dfrac{k_B T}{m} \label{2.8}$
Hence, since $a^3 = V$,
$P = \dfrac{N k_B T}{V} = \dfrac{n R T}{V} \label{2.9}$
which is the ideal gas law.
The ideal gas law is an example of an equation of state, which was introduced in Lecture 1. One way to visualize any equation of state is to plot the so-called isotherms, which are graphs of $P$ vs. $V$ at fixed values of $T$ . For the ideal-gas equation of state $P = nRT/V$ , some of the isotherms are shown in the figure below (left panel): If we plot $P$ vs. $T$ at fixed volume (called the isochores), we obtain the plot in the right panel. What is important to note, here, is that an ideal gas can exist only as a gas. It is not possible for an ideal gas to condense into some kind of “ideal liquid”. In other words, a phase transition from gas to liquid can be modeled only if interparticle interactions are properly accounted for.
Note that the ideal-gas equation of state can be written in the form
$\dfrac{P V}{n R T} = \dfrac{P \bar{V}}{R T} = \dfrac{P}{\rho R T} = 1 \label{2.10}$
where $\bar{V} = V/n$ is called the molar volume. Unlike $V$ , which increases as the number of moles increases (an example of what is called an extensive quantity in thermodynamics), $\bar{V}$ does not exhibit this dependence and, therefore, is called intensive. The quantity
$Z = \dfrac{P V}{n R T} = \dfrac{P \bar{V}}{R T} = \dfrac{P}{\rho R T} \label{2.11}$
is called the compressibility of the gas. In an ideal gas, if we “compress” the gas by increasing $P$ , the density $\rho$ must increase as well so as to keep $Z =1$. For a real gas, $Z$, therefore, gives us a measure of how much the gas deviates from ideal-gas behavior.
Figure 2.2 shows a plot of $Z$ vs. $P$ for several real gases and for an ideal gas. The plot shows that for sufficiently low pressures (hence, low densities), each gas approaches ideal-gas behavior, as expected.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/02%3A_Properties_of_Gases/2.03%3A_Ideal_Gases.txt
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According to Boyle's law, the product PV is a constant at any given temperature, so a plot of PV as a function of the pressure of an ideal gas yields a horizontal straight line. This implies that any increase in the pressure of the gas is exactly counteracted by a decrease in the volume as the molecules are crowded closer together. But we know that the molecules themselves are finite objects having volumes of their own, and this must place a lower limit on the volume into which they can be squeezed. So we must reformulate the ideal gas equation of state as a relation that is true only in the limiting case of zero pressure:
$\lim_{P \rightarrow 0} PV=nRT$
So what happens when a real gas is subjected to a very high pressure? The outcome varies with both the molar mass of the gas and its temperature, but in general we can see the the effects of both repulsive and attractive intermolecular forces:
• Repulsive forces: As a gas is compressed, the individual molecules begin to get in each other's way, giving rise to a very strong repulsive force acts to oppose any further volume decrease. We would therefore expect the PV vs P line to curve upward at high pressures, a nd this is in fact what is observed for all gases at sufficiently high pressures.
• Attractive forces: At very close distances, all molecules repel each other as their electron clouds come into contact. At greater distances, however, brief statistical fluctuations in the distribution these electron clouds give rise to a universal attractive force between all molecules. The more electrons in the molecule (and thus the greater the molecular weight), the greater is this attractive force. As long as the energy of thermal motion dominates this attractive force, the substance remains in the gaseous state, but at sufficiently low temperatures the attractions dominate and the substance condenses to a liquid or solid.
The universal attractive force described above is known as the dispersion, or London force. There may also be additional (and usually stronger) attractive forces related to charge imbalance in the molecule or to hydrogen bonding. These various attractive forces are often referred to collectively as van der Waals forces. A plot of PV/RT as a function of pressure is a very sensitive indicator of deviations from ideal behavior, since such a plot is just a horizontal line for an ideal gas. The two illustrations below show how these plots vary with the nature of the gas, and with temperature.
Intermolecular attractions, which generally increase with molecular weight, cause the PV product to decrease as higher pressures bring the molecules closer together and thus within the range of these attractive forces; the effect is to cause the volume to decrease more rapidly than it otherwise would. The repulsive forces always eventually win out. But as the molecules begin to intrude on each others' territory, the stronger repulsive forces cause the curve to bend upward.
The temperature makes a big difference! At higher temperatures, increased thermal motions overcome the effects of intermolecular attractions which normally dominate at lower pressures. So all gases behave more ideally at higher temperatures. For any gas, there is a special temperature (the Boyle temperature) at which attractive and repulsive forces exactly balance each other at zero pressure. As you can see in this plot for methane, this some of this balance does remain as the pressure is increased.
The van der Waals Equation of State
How might we modify the ideal gas equation of state to take into account the effects of intermolecular interactions? The first and most well known answer to this question was offered by the Dutch scientist J.D. van der Waals (1837-1923) in 1873. The ideal gas model assumes that the gas molecules are merely points that occupy no volume; the "V" term in the equation is the volume of the container and is independent of the nature of the gas.
van der Waals recognized that the molecules themselves take up space that subtracts from the volume of the container, so that the “volume of the gas” V in the ideal gas equation should be replaced by the term (V–b), in which b relates to the excluded volume, typically of the order of 20-100 cm3 mol–1. The excluded volume surrounding any molecule defines the closest possible approach of any two molecules during collision. Note that the excluded volume is greater then the volume of the molecule, its radius being half again as great as that of a spherical molecule.
The other effect that van der Waals needed to correct for are the intermolecular attractive forces. These are ignored in the ideal gas model, but in real gases they exert a small cohesive force between the molecules, thus helping to hold the gas together and reducing the pressure it exerts on the walls of the container.
Because this pressure depends on both the frequency and the intensity of collisions with the walls, the reduction in pressure is proportional to the square of the number of molecules per volume of space, and thus for a fixed number of molecules such as one mole, the reduction in pressure is inversely proportional to the square of the volume of the gas. The smaller the volume, the closer are the molecules and the greater will be the effect. The van der Waals equation replaces the $P$ term in the ideal gas equation with $P + (a / V^2)$ in which the magnitude of the constant a increases with the strength of the intermolecular attractive forces.
The complete van der Waals equation of state can be written as
Although most students are not required to memorize this equation, you are expected to understand it and to explain the significance of the terms it contains. You should also understand that the van der Waals constants $a$ and $b$ must be determined empirically for every gas. This can be done by plotting the P-V behavior of the gas and adjusting the values of a and b until the van der Waals equation results in an identical plot. The constant $a$ is related in a simple way to the molecular radius; thus the determination of $a$ constitutes an indirect measurement of an important microscopic quantity.
Table $1$ : van der Waals constants for some gases
Substance
molar mass, g
a (L2-atm mole–2)
b (L mol–1)
hydrogen H2 2 0.244 0.0266
helium He 4 0.034 0.0237
methane CH4 16 2.25 0.0428
water H2O 18 5.46 0.0305
nitrogen N2 28 1.39 0.0391
carbon dioxide CO2 44 3.59 0.0427
carbon tetrachloride CCl4 154 20.4 0.1383
The van der Waals equation is only one of many equations of state for real gases. More elaborate equations are required to describe the behavior of gases over wider pressure ranges. These generally take account of higher-order nonlinear attractive forces, and require the use of more empirical constants. Although we will make no use of them in this course, they are widely employed in chemical engineering work in which the behavior of gases at high pressures must be accurately predicted.
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The most striking feature of real gases is that they cease to remain gases as the temperature is lowered and the pressure is increased.
The Van der Waals Equation of State
In the last lecture, we saw that for the pair potential
$u(r) = \begin{cases} \infty & r \leq \sigma \ -\dfrac{C_6}{r^6} & r > \sigma \end{cases} \label{$1$}$
we could write the second virial coefficient as
$B_2(T) = \dfrac{2}{3} \pi N_0 \sigma^3 \left[ 1 - \dfrac{C_6}{3 k_B T \sigma^6} \right] \label{$2$}$
Let us introduce to simplifying variables
\begin{align} b &= \dfrac{2}{3} \pi N_0 \sigma^3 \ a &= \dfrac{2 \pi N_0^2 C_6}{9 \sigma^3} \end{align} \label{$3$}
in terms of which
$B_2(T) = b - \dfrac{a}{RT} \label{$4$}$
With these definitions, the virial equation of state becomes
\begin{align} P &= \dfrac{nRT}{V} + \dfrac{n^2}{V^2} RT \left( b - \dfrac{a}{RT} \right) \ &= \dfrac{nRT}{V} \left(1 + \dfrac{nb}{V} \right) - \dfrac{an^2}{V^2} \end{align} \label{$5$}
If we assume $nb/V$ is small, then we can also write
$1 + \dfrac{nb}{V} \approx \dfrac{1}{1 - \dfrac{nb}{V}} \label{$6$}$
so that
$P = \dfrac{nRT}{V - nb} - \dfrac{an^2}{V^2} \label{$7$}$
which is known as the van der Waals equation of state.
The first term in this equation is easy to motivate. In fact, it looks very much like the equation of state for an ideal gas having volume $V - nb$ rather than $V$. This part of the van der Waals equation is due entirely to the hard-wall potential $u_o(r)$. Essentially, this potential energy term describes a system of “billiard balls” of diameter $\sigma$. The figure below shows two of these billiard ball type particles at the point of contact (also called the distance of closest approach). At this point, they undergo a collision and separate, so that cannot be closer than the distance shown in the figure. At this point, the distance between their centers is also $\sigma$, as the figure indicates. Because of this distance of closest approach, the total volume available to the particles is not $V$ but some volume less than $V$. This reduction in volume can be calculated as follows: Figure $1$ shows a shaded sphere that just contains the pair of billiard ball particles. The volume of this sphere is the volume excluded from any two particles. The radius of the sphere is $\sigma$ as the figure shows. Hence, the excluded volume for the two particles is $4 \pi \sigma^3/3$, which is the volume of the shaded sphere. From this, we see that the excluded volume for any one particle is just half of this or $2 \pi \sigma^3/3$. The excluded volume for a mole of such particles is just $2 \pi \sigma^3 N_0/3$, which is the parameter $b$:
$b = \dfrac{2}{3} \pi \sigma^3 N_0 \label{$8$}$
Given $n$ moles of gas, the total excluded volume is then $nb$, so that the total available volume is simply $V - nb$.
Isotherms of the van der Waals equation are shown in the figure below (left panel). In the figure the volume axis is the molar volume denoted $\bar{V} = V/n$. At sufficiently high temperature, the isotherms approach those of an ideal gas. However, we also see something strange in some of the isotherms. Specifically, we see a region in which $P$ and $V$ increase together, and we know that this cannot actually happen in a real gas. It should be clear that many approximations and assumptions go into the derivation of the van der Waals equation so that some of the important physics is missing from the model. Hence, we should not be surprised if the van der Waals equation has some unphysical behavior buried in it.
In fact, we know that at sufficiently low temperatures, any real gas, when compressed, must undergo a transition from gas to liquid. The signature of such a transition is a discontinuous change in the volume, signifying the condensation of the gas into a liquid that occupies a significantly lower volume. Unfortunately, the van der Waals equation does not correctly predict this behavior, and hence, it must be added in ad hoc. This is done by drawing a horizontal line through the isotherm (Figure $2$) and in the figure below. The vertical position of the line is
chosen so that the area above the line (between the line and the isotherm) and below the line (again between the line and the isotherm) is exactly the same. In this way, we entirely remove the artifact of the unphysical increase of $P$ with $V$ when we compute the compressional work on the gas from $\int P(V) \: dV$, to be discussed in our section on thermodynamics. This horizontal line is called the tie line.
As it happens, there is exactly one isotherm along which the van der Waals equation correctly predicts the gas-to-liquid phase transition. Along this isotherm, the volume discontinuity captured by the tie line is shrunken down to a single point (so that there is no possibility of an increase of $P$ with $V$!). This isotherm, in fact, corresponds to the highest possible temperature at which such a transition can occur. As we approach this isotherm from higher temperatures, this isotherm is a kind of dividing line between the system’s remaining a gas at all value of $P$ and $V$ and the system’s actually undergoing a gas-to-liquid transition. Hence, this isotherm is called the critical isotherm (Figure $4$): The temperature of this isotherm is called the critical temperature, denoted $T_c$. The point at which the curve flattens out, signifying the phase transition, is called the critical point. If we draw a curve through the isotherms joining all points of these isotherms at which the tie lines begin, continue the curve up to the critical isotherm, and down the other side where the tie lines end, this curve reaches a maximum at the critical point. This is illustrated below:
The shape of the critical isotherm at the critical point allows us to determine the exact temperature, pressure, and volume at which the phase transition from gas to liquid will occur. At this point, the isotherm is both horizontal and flat. This means that both the first and second derivatives of $P$ with respect to $V$ must vanish:
$\dfrac{\partial P}{\partial V} = 0, \: \: \: \dfrac{\partial^2 P}{\partial V^2} = 0 \label{$9$}$
Substituting the van der Waals equation into these two conditions, we find the following:
\begin{align} -\dfrac{nRT}{\left( V - nb \right)^2} + \dfrac{2 an^2}{V^3} &= 0 \ \dfrac{2nRT}{\left( V - nb \right)^3} - \dfrac{6an^2}{V^4} &= 0 \end{align} \label{$1$0}
Hence, we have two equations in two unknowns $V$ and $T$ for the critical temperature and critical volume. Once these are determined, the van der Waals equation, itself, allows us to determine the critical pressure.
To solve the equations, first divide one by the other. This gives us a simple condition for the volume:
\begin{align} \dfrac{V - nb}{2} &= \dfrac{V}{3} \ 3V - 3nb &= 2V \ V &= 3nb \equiv V_c \end{align} \label{$1$1}
This is the critical volume. Now use either of the two conditions to obtain the critical temperature $T_c$. If we use the first one, we find
\begin{align} \dfrac{nRT_c}{\left( V_c - nb \right)^2} &= \dfrac{2an^2}{V_c^3} \ \dfrac{nRT_c}{\left( 3nb - nb \right)^2} &= \dfrac{2an^2}{\left( 3nb \right)^3} \ \dfrac{nRT_c}{4n^2b^2} &= \dfrac{2an^2}{27n^3b^3} \ RT_c &= \dfrac{8a}{27b} \end{align} \label{$1$2}
Finally, plugging the critical temperature and volume into the van der Waals equation, we obtain the critical pressure
as
\begin{align} P &= \dfrac{nRT_c}{V_c - nb} - \dfrac{an^2}{V_c^2} \ &= \dfrac{8an/27b}{3nb - nb} - \dfrac{an^2}{\left( 3nb \right)^2} \ &= \dfrac{a}{27b^2} \end{align} \label{$1$3}
PVT Surface of a real gas
The plot illustrates this behavior; as the volume is decreased, the lower-temperature isotherms suddenly change into straight lines. Under these conditions, the pressure remains constant as the volume is reduced. This can only mean that the gas is “disappearing" as we squeeze the system down to a smaller volume. In its place, we obtain a new state of matter, the liquid. In the green-shaded region, two phases, liquid, and gas, are simultaneously present. Finally, at very small volume all the gas has disappeared and only the liquid phase remains. At this point the isotherms bend strongly upward, reflecting our common experience that a liquid is practically incompressible.
To better understand this plot, look at the isotherm labeled . As the gas is compressed from to , the pressure rises in much the same way as Boyle's law predicts. Compression beyond, however, does not cause any rise in the pressure. What happens instead is that some of the gas condenses to a liquid. At , the substance is entirely in its liquid state. The very steep rise to corresponds to our ordinary experience that liquids have very low compressibilities. The range of volumes possible for the liquid diminishes as the critical temperature is approached.
The critical point
Liquid and gas can coexist only within the regions indicated by the green-shaded area in the diagram above. As the temperature and pressure rise, this region becomes more narrow, finally reaching zero width at the critical point. The values of P, T, and V at this juncture are known as the critical constants Pc , Tc, and Vc. The isotherm that passes through the critical point is called the critical isotherm. Beyond this isotherm, the gas and liquids become indistinguishable; there is only a single fluid phase, sometimes referred to as a supercritical liquid.
Critical Behaviour of carbon dioxide
At temperatures below 31°C (the critical temperature), CO2 acts somewhat like an ideal gas even at a rather high pressure (). Below 31°, an attempt to compress the gas to a smaller volume eventually causes condensation to begin. Thus at 21°C, at a pressure of about 62 atm (), the volume can be reduced from 200 cm3 to about 55 cm3 without any further rise in the pressure. Instead of the gas being compressed, it is replaced with the far more compact liquid as the gas is essentially being "squeezed" into its liquid phase. After all of the gas has disappeared (), the pressure rises very rapidly because now all that remains is an almost incompressible liquid. Above this isotherm (), CO2 exists only as a supercritical fluid.
What happens if you have some liquid carbon dioxide in a transparent cylinder at just under its Pc of 62 atm, and you then compress it slightly? Nothing very dramatic until you notice that the meniscus has disappeared. By successively reducing and increasing the pressure, you can "turn the meniscus on and off".
One intriguing consequence of the very limited bounds of the liquid state is that you could start with a gas at large volume and low temperature, raise the temperature, reduce the volume, and then reduce the temperature so as to arrive at the liquid region at the lower left, without ever passing through the two-phase region, and thus without undergoing condensation!
Supercritical fluids
The supercritical state of matter, as the fluid above the critical point is often called, possesses the flow properties of a gas and the solvent properties of a liquid. The density of a supercritical fluid can be changed over a wide range by adjusting the pressure; this, in turn, changes its solubility, which can thus be optimized for a particular application. The picture at the right shows a commercial laboratory device used for carrying out chemical reactions under supercritical conditions.
Supercritical carbon dioxide is widely used to dissolve the caffeine out of coffee beans and as a dry-cleaning solvent. Supercritical water has recently attracted interest as a medium for chemically decomposing dangerous environmental pollutants such as PCBs. Supercritical fluids are being increasingly employed as as substitutes for organic solvents (so-called "green chemistry") in a range of industrial and laboratory processes. Applications that involve supercritical fluids include extractions, nano particle and nano structured film formation, supercritical drying, carbon capture and storage, as well as enhanced oil recovery studies.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/02%3A_Properties_of_Gases/2.05%3A_Condensation_of_Gases_and_the_Critical_State.txt
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The kinetic theory describes a gas as a large number of submicroscopic particles (atoms or molecules), all of which are in constant, random motion. The rapidly moving particles constantly collide with each other and with the walls of the container. Kinetic theory explains macroscopic properties of gases, such as pressure, temperature, viscosity, thermal conductivity, and volume, by considering their molecular composition and motion. The theory posits that gas pressure is due to the impacts, on the walls of a container, of molecules or atoms moving at different velocities.
The Model
The five basic tenets of the kinetic-molecular theory are as follows:
1. A gas is composed of molecules that are separated by average distances that are much greater than the sizes of the molecules themselves. The volume occupied by the molecules of the gas is negligible compared to the volume of the gas itself.
2. The molecules of an ideal gas exert no attractive forces on each other, or on the walls of the container.
3. The molecules are in constant random motion, and as material bodies, they obey Newton's laws of motion. This means that the molecules move in straight lines (see demo illustration at the left) until they collide with each other or with the walls of the container.
4. Collisions are perfectly elastic; when two molecules collide, they change their directions and kinetic energies, but the total kinetic energy is conserved. Collisions are not “sticky".
5. The average kinetic energy of the gas molecules is directly proportional to the absolute temperature. Notice that the term “average” is very important here; the velocities and kinetic energies of individual molecules will span a wide range of values, and some will even have zero velocity at a given instant. This implies that all molecular motion would cease if the temperature were reduced to absolute zero.
According to this model, most of the volume occupied by a gas is empty space; this is the main feature that distinguishes gases from condensed states of matter (liquids and solids) in which neighboring molecules are constantly in contact. Gas molecules are in rapid and continuous motion; at ordinary temperatures and pressures their velocities are of the order of 0.1-1 km/sec and each molecule experiences approximately 1010collisions with other molecules every second.
If gases do in fact consist of widely-separated particles, then the observable properties of gases must be explainable in terms of the simple mechanics that govern the motions of the individual molecules. The kinetic molecular theory makes it easy to see why a gas should exert a pressure on the walls of a container. Any surface in contact with the gas is constantly bombarded by the molecules.
At each collision, a molecule moving with momentum mv strikes the surface. Since the collisions are elastic, the molecule bounces back with the same velocity in the opposite direction. This change in velocity ΔV is equivalent to an acceleration a; according to Newton's second law, a force f = ma is thus exerted on the surface of area A exerting a pressure P = f/A.
Kinetic Interpretation of Temperature
According to the kinetic molecular theory, the average kinetic energy of an ideal gas is directly proportional to the absolute temperature. Kinetic energy is the energy a body has by virtue of its motion:
$KE = \dfrac{mv^2}{2}$
As the temperature of a gas rises, the average velocity of the molecules will increase; a doubling of the temperature will increase this velocity by a factor of four. Collisions with the walls of the container will transfer more momentum, and thus more kinetic energy, to the walls. If the walls are cooler than the gas, they will get warmer, returning less kinetic energy to the gas, and causing it to cool until thermal equilibrium is reached. Because temperature depends on the average kinetic energy, the concept of temperature only applies to a statistically meaningful sample of molecules. We will have more to say about molecular velocities and kinetic energies farther on.
• Kinetic explanation of Boyle's law: Boyle's law is easily explained by the kinetic molecular theory. The pressure of a gas depends on the number of times per second that the molecules strike the surface of the container. If we compress the gas to a smaller volume, the same number of molecules are now acting against a smaller surface area, so the number striking per unit of area, and thus the pressure, is now greater.
• Kinetic explanation of Charles' law: Kinetic molecular theory states that an increase in temperature raises the average kinetic energy of the molecules. If the molecules are moving more rapidly but the pressure remains the same, then the molecules must stay farther apart, so that the increase in the rate at which molecules collide with the surface of the container is compensated for by a corresponding increase in the area of this surface as the gas expands.
• Kinetic explanation of Avogadro's law: If we increase the number of gas molecules in a closed container, more of them will collide with the walls per unit time. If the pressure is to remain constant, the volume must increase in proportion, so that the molecules strike the walls less frequently, and over a larger surface area.
• Kinetic explanation of Dalton's law: "Every gas is a vacuum to every other gas". This is the way Dalton stated what we now know as his law of partial pressures. It simply means that each gas present in a mixture of gases acts independently of the others. This makes sense because of one of the fundamental tenets of KMT theory that gas molecules have negligible volumes. So Gas A in mixture of A and B acts as if Gas B were not there at all. Each contributes its own pressure to the total pressure within the container, in proportion to the fraction of the molecules it represents.
Derivation of the Ideal Gas Law
One of the triumphs of the kinetic molecular theory was the derivation of the ideal gas law from simple mechanics in the late nineteenth century. This is a beautiful example of how the principles of elementary mechanics can be applied to a simple model to develop a useful description of the behavior of macroscopic matter. We begin by recalling that the pressure of a gas arises from the force exerted when molecules collide with the walls of the container. This force can be found from Newton's law
$f = ma = m\dfrac{dv}{dt} \label{2.1}$
in which $v$ is the velocity component of the molecule in the direction perpendicular to the wall and $m$ is its mass.
To evaluate the derivative, which is the velocity change per unit time, consider a single molecule of a gas contained in a cubic box of length l. For simplicity, assume that the molecule is moving along the x-axis which is perpendicular to a pair of walls, so that it is continually bouncing back and forth between the same pair of walls. When the molecule of mass m strikes the wall at velocity +v (and thus with a momentum mv ) it will rebound elastically and end up moving in the opposite direction with –v. The total change in velocity per collision is thus 2v and the change in momentum is $2mv$.
After the collision the molecule must travel a distance l to the opposite wall, and then back across this same distance before colliding again with the wall in question. This determines the time between successive collisions with a given wall; the number of collisions per second will be $v/2l$. The force $F$ exerted on the wall is the rate of change of the momentum, given by the product of the momentum change per collision and the collision frequency:
$F = \dfrac{d(mv_x)}{dt} = (2mv_x) \times \left( \dfrac{v_x}{2l} \right) = \dfrac{m v_x^2}{l} \label{2-2}$
Pressure is force per unit area, so the pressure $P$ exerted by the molecule on the wall of cross-section $l^2$ becomes
$P = \dfrac{mv^2}{l^3} = \dfrac{mv^2}{V} \label{2-3}$
in which $V$ is the volume of the box.
As noted near the beginning of this unit, any given molecule will make about the same number of moves in the positive and negative directions, so taking a simple average would yield zero. To avoid this embarrassment, we square the velocities before averaging them
$\bar{v^2} = \dfrac{v_1^2 + v_2^2 + v_3^2 + v_4^2 .. . v_N^2 }{N}= \dfrac{\sum_i v_i^2}{N}$
and then take the square root of the average. This result is known as the root mean square (rms) velocity.
$v_{rms} = \sqrt{ \bar{v^2}}$
We have calculated the pressure due to a single molecule moving at a constant velocity in a direction perpendicular to a wall. If we now introduce more molecules, we must interpret $v^2$ as an average value which we will denote by $\bar{v^2}$. Also, since the molecules are moving randomly in all directions, only one-third of their total velocity will be directed along any one Cartesian axis, so the total pressure exerted by $N$ molecules becomes
$P=\dfrac{N}{3}\dfrac{m \bar{\nu}^2}{V} \label{2.4}$
Recalling that $m\bar{v^2}/2$ is the average translational kinetic energy $\epsilon$, we can rewrite the above expression as
$PV = \dfrac{1}{3} N m \bar{v^2} = \dfrac{2}{3} N \epsilon \label{2-5}$
The 2/3 factor in the proportionality reflects the fact that velocity components in each of the three directions contributes ½ kT to the kinetic energy of the particle. The average translational kinetic energy is directly proportional to temperature:
$\color{red} \epsilon = \dfrac{3}{2} kT \label{2.6}$
in which the proportionality constant $k$ is known as the Boltzmann constant. Substituting Equation $\ref{2.6}$ into Equation $\ref{2-5}$ yields
$PV = \left( \dfrac{2}{3}N \right) \left( \dfrac{3}{2}kT \right) =NkT \label{2.7}$
The Boltzmann constant k is just the gas constant per molecule. For n moles of particles, the Equation $\ref{2.7}$ becomes
$PV = nRT \label{2.8}$
which is the Ideal Gas law.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/02%3A_Properties_of_Gases/2.06%3A_Kinetic_Theory_of_Gases.txt
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In the context of the Kinetic Molecular Theory of Gases, a gas contains a large number of particles in rapid motions. Each particle has a different speed, and each collision between particles changes the speeds of the particles. An understanding of the properties of the gas requires an understanding of the distribution of particle speeds.
Many molecules, many velocities
At temperatures above absolute zero, all molecules are in motion. In the case of a gas, this motion consists of straight-line jumps whose lengths are quite great compared to the dimensions of the molecule. Although we can never predict the velocity of a particular individual molecule, the fact that we are usually dealing with a huge number of them allows us to know what fraction of the molecules have kinetic energies (and hence velocities) that lie within any given range.
The trajectory of an individual gas molecule consists of a series of straight-line paths interrupted by collisions. What happens when two molecules collide depends on their relative kinetic energies; in general, a faster or heavier molecule will impart some of its kinetic energy to a slower or lighter one. Two molecules having identical masses and moving in opposite directions at the same speed will momentarily remain motionless after their collision.
If we could measure the instantaneous velocities of all the molecules in a sample of a gas at some fixed temperature, we would obtain a wide range of values. A few would be zero, and a few would be very high velocities, but the majority would fall into a more or less well defined range. We might be tempted to define an average velocity for a collection of molecules, but here we would need to be careful: molecules moving in opposite directions have velocities of opposite signs. Because the molecules are in a gas are in random thermal motion, there will be just about as many molecules moving in one direction as in the opposite direction, so the velocity vectors of opposite signs would all cancel and the average velocity would come out to zero. Since this answer is not very useful, we need to do our averaging in a slightly different way.
The proper treatment is to average the squares of the velocities, and then take the square root of this value. The resulting quantity is known as the root mean square (RMS) velocity
$v_{rms} = \sqrt{\dfrac{\sum \nu^2}{n}}$
where $n$ is the number of molecules in the system. The formula relating the RMS velocity to the temperature and molar mass is surprisingly simple (derived below), considering the great complexity of the events it represents:
$v_{rms} = \sqrt{\dfrac{3RT}{M}} \label{rms1}$
where
• $M$ is the molar mass in kg mol–1, and
• $R$ is gas constant.
Equation $\ref{rms1}$ can also be expressed as
$v_{rms} = \sqrt{\dfrac{3k_bT}{m}} \label{rms2}$
where
• $m$ is the molecular mass in kg
• $k_b$ is Boltzmann constant and is just the “gas constant per molecule" $k_b = \dfrac{R}{N_a} = \dfrac{R}{6.02 \times 10^{23}}$
Equation $\ref{rms2}$ is just the per atom version of Equation $\ref{rms1}$ which is expressed in terms of per mol. Either equation will work.
Example $1$
What is the $v_{rms}$ of a nitrogen molecule at 300 K?
Solution
The molar mass of $N_2$ is 28.01 g/mol. Substituting in the above equation and expressing R in energy units, we obtain
$v^2 =\dfrac{(3)(8.31 \;J \; mol^{-1} \; K^{-1})(300\;K)}{28.01 \times 10^{-3}\; Kg \; mol^{-1}} = 2.67 \times 10^5 \; \dfrac{J}{Kg} \nonumber$
Recalling the definition of the joule (1 J = 1 kg m2 s–2) and taking the square root,
$\bar{v} = \sqrt{ \left(2.67 \times 10^5\; \dfrac{\cancel{J}}{\cancel{Kg}}\right) \left( \dfrac{ 1 \;\cancel{kg}\; m^2\; s^{–2}}{ 1 \;\cancel{J }} \right)} = 517 \;m /s \nonumber$
or
$517\, \dfrac{\cancel{m}}{s} \left(\dfrac{1\; km}{10^3\; \cancel{m}} \right) \left(\dfrac{3600\; \cancel{s}}{1\; hr} \right) = 1860\; km/hr \nonumber$
Comment: this is fast! The velocity of a rifle bullet is typically 300-500 m s–1; convert to common units to see the comparison for yourself. A simpler formula for estimating average molecular velocities than $\ref{rms1}$ is
$v_{rms} =157 \sqrt{\dfrac{T}{M}} \nonumber$
in which $v$ is in units of meters/sec, $T$ is the absolute temperature and $M$ the molar mass in grams.
The Boltzmann Distribution
If we were to plot the number of molecules whose velocities fall within a series of narrow ranges, we would obtain a slightly asymmetric curve known as a velocity distribution. The peak of this curve would correspond to the most probable velocity. This velocity distribution curve is known as the Maxwell-Boltzmann distribution, but is frequently referred to only by Boltzmann's name. The Maxwell-Boltzmann distribution law was first worked out around 1850 by the great Scottish physicist, James Clerk Maxwell (left, 1831-1879), who is better known for discovering the laws of electromagnetic radiation. Later, the Austrian physicist Ludwig Boltzmann (1844-1906) put the relation on a sounder theoretical basis and simplified the mathematics somewhat. Boltzmann pioneered the application of statistics to the physics and thermodynamics of matter, and was an ardent supporter of the atomic theory of matter at a time when it was still not accepted by many of his contemporaries.
The Maxwell-Boltzmann distribution is used to determine how many molecules are moving between velocities $v$ and $v + dv$. Assuming that the one-dimensional distributions are independent of one another, that the velocity in the y and z directions does not affect the x velocity, for example, the Maxwell-Boltzmann distribution is given by
$\large \dfrac{dN}{N} = \left(\dfrac{m}{2\pi k_bT} \right)^{1/2}exp\left [\dfrac{-mv^2}{2k_b T}\right] dv \label{1}$
where
• $dN/N$ is the fraction of molecules moving at velocity $v$ to $v + dv$,
• $m$ is the mass of the molecule,
• $k_b$ is the Boltzmann constant, and
• $T$ is the absolute temperature.1
Additionally, the function can be written in terms of the scalar quantity speed $v$ instead of the vector quantity velocity. This form of the function defines the distribution of the gas molecules moving at different speeds, between $v_1$ and $v_2$, thus
$f(v)=4\pi v^2 \left (\dfrac{m}{2\pi k_b T} \right)^{3/2}exp\left [\frac{-mv^2}{2k_bT}\right] \label{2}$
Finally, the Maxwell-Boltzmann distribution can be used to determine the distribution of the kinetic energy of for a set of molecules. The distribution of the kinetic energy is identical to the distribution of the speeds for a certain gas at any temperature.2 The Maxwell-Boltzmann distribution is a probability distribution and just like any such distribution, can be characterized in a variety of ways including.
• Average Speed: The average speed is the sum of the speeds of all of the particles divided by the number of particles.
• Most Probable Speed: The most probable speed is the speed associated with the highest point in the Maxwell distribution. Only a small fraction of particles might have this speed, but it is more likely than any other speed.
• Width of the Distribution: The width of the distribution characterizes the most likely range of speeds for the particles. One measure of the width is the Full Width at Half Maximum (FWHM). To determine this value, find the height of the distribution at the most probable speed (this is the maximum height of the distribution). Divide the maximum height by two to obtain the half height, and locate the two speeds in the distribution that have this half-height value. On speed will be greater than the most probably speed and the other speed will be smaller. The full width is the difference between the two speeds at the half-maximum value.
Velocity distributions depend on temperature and mass
Higher temperatures allow a larger fraction of molecules to acquire greater amounts of kinetic energy, causing the Boltzmann plots to spread out. Figure $2$ shows how the Maxwell-Boltzmann distribution is affected by temperature. At lower temperatures, the molecules have less energy. Therefore, the speeds of the molecules are lower and the distribution has a smaller range. As the temperature of the molecules increases, the distribution flattens out. Because the molecules have greater energy at higher temperature, the molecules are moving faster.
Notice how the left ends of the plots are anchored at zero velocity (there will always be a few molecules that happen to be at rest.) As a consequence, the curves flatten out as the higher temperatures make additional higher-velocity states of motion more accessible. The area under each plot is the same for a constant number of molecules.
All molecules have the same kinetic energy (mv2/2) at the same temperature, so the fraction of molecules with higher velocities will increase as m, and thus the molecular weight, decreases. Figure $3$ shows the dependence of the Maxwell-Boltzmann distribution on molecule mass. On average, heavier molecules move more slowly than lighter molecules. Therefore, heavier molecules will have a smaller speed distribution, while lighter molecules will have a speed distribution that is more spread out.
The Maxwell-Boltzmann equation, which forms the basis of the kinetic theory of gases, defines the distribution of speeds for a gas at a certain temperature. From this distribution function, the most probable speed, the average speed, and the root-mean-square speed can be derived.
Related Speed Expressions
Usually, we are more interested in the speeds of molecules rather than their component velocities. The Maxwell–Boltzmann distribution for the speed follows immediately from the distribution of the velocity vector, above. Note that the speed of an individual gas particle is:
$v =\sqrt{v_x^2+ v_y^2 = v_z^2}$
Three speed expressions can be derived from the Maxwell-Boltzmann distribution:
• the most probable speed,
• the average speed, and
• the root-mean-square speed.
The most probable speed is the maximum value on the distribution plot (Figure $4$}. This is established by finding the velocity when the derivative of Equation $\ref{2}$ is zero
$\dfrac{df(v)}{dv} = 0$
which is
$\color{red} v_{mp}=\sqrt {\dfrac {2RT}{M}} \label{3a}$
Figure $4$: The Maxwell-Boltzmann distribution is shifted to higher speeds and is broadened at higher temperatures. from OpenStax. The speed at the top of the curve is called the most probable speed because the largest number of molecules have that speed.
The average speed is the sum of the speeds of all the molecules divided by the number of molecules.
$\color{red} v_{avg}= \bar{v} = \int_0^{\infty} v f(v) dv = \sqrt {\dfrac{8RT}{\pi M}} \label{3b}$
The root-mean-square speed is square root of the average speed-squared.
$\color{red} v_{rms}= \bar{v^2} = \sqrt {\dfrac {3RT}{M}} \label{3c}$
where
• $R$ is the gas constant,
• $T$ is the absolute temperature and
• $M$ is the molar mass of the gas.
It always follows that for gases that follow the Maxwell-Boltzmann distribution:
$v_{mp}< v_{avg}< v_{rms} \label{4}$
Problems
1. Using the Maxwell-Boltzman function, calculate the fraction of argon gas molecules with a speed of 305 m/s at 500 K.
2. If the system in problem 1 has 0.46 moles of argon gas, how many molecules have the speed of 305 m/s?
3. Calculate the values of $C_{mp}$, $C_{avg}$, and $C_{rms}$ for xenon gas at 298 K.
4. From the values calculated above, label the Boltzmann distribution plot (Figure 1) with the approximate locations of (C_{mp}\), $C_{avg}$, and $C_{rms}$.
5. What will have a larger speed distribution, helium at 500 K or argon at 300 K? Helium at 300 K or argon at 500 K? Argon at 400 K or argon at 1000 K?
Answers
1. 0.00141
2. $3.92 \times 10^{20}$ argon molecules
3. cmp = 194.27 m/s, cavg = 219.21 m/s, crms = 237.93 m/s
4. As stated above, Cmp is the most probable speed, thus it will be at the top of the distribution curve. To the right of the most probable speed will be the average speed, followed by the root-mean-square speed.
5. Hint: Use the related speed expressions to determine the distribution of the gas molecules: helium at 500 K. helium at at 300 K. argon at 1000 K.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/02%3A_Properties_of_Gases/2.07%3A_The_Maxwell_Distribution_Laws.txt
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Collision theory is a theory proposed independently by Max Trautz in 1916 and William Lewis in 1918, that qualitatively explains how chemical reactions occur and why reaction rates differ for different reactions. The collision theory states that when suitable particles of the reactant hit each other, only a certain percentage of the collisions cause any noticeable or significant chemical change; these successful changes are called successful collisions. The successful collisions have enough energy, also known as activation energy, at the moment of impact to break the preexisting bonds and form all new bonds. This results in the products of the reaction. Increasing the concentration of the reactant particles or raising the temperature, thus bringing about more collisions and therefore many more successful collisions, increases the rate of reaction.
Collision Energy
Consider two particles $A$ and $B$ in a system. The kinetic energy of these two particles is
$K_{AB} = \dfrac{\textbf{p}_A^2}{2m_A} + \dfrac{\textbf{p}_B^2}{2m_B} \label{6}$
Let us change to center-of-mass $\left( \textbf{P} \right)$ and relative $\left( \textbf{p} \right)$ momenta, which are given by
$\textbf{P} = \textbf{p}_A + \textbf{p}_B, \: \: \: \textbf{p} = \dfrac{m_B \textbf{p}_a - m_A \textbf{p}_B}{M} \label{7}$
where $M = m_A + m_B$ is the total mass of the two particles. Substituting this into the kinetic energy, we find
$K_{AB} = \dfrac{\textbf{p}_A^2}{2m_A} + \dfrac{\textbf{p}_B^2}{2m_B} = \dfrac{\textbf{P}^2}{2M} + \dfrac{\textbf{p}^2}{2 \mu} \label{8}$
where
$\mu = \dfrac{m_A m_B}{M} \label{9}$
is called the reduced mass of the two particles. Note that the kinetic energy separates into a sum of a center-of-mass term and a relative term.
Now the relative position is $\textbf{r} = \textbf{r}_A - \textbf{r}_B$ so that the relative velocity is $\dot{\textbf{r}} = \dot{\textbf{r}}_A - \dot{\textbf{r}}_B$ or $\textbf{v} = \textbf{v}_A - \textbf{v}_B$. Thus, if the two particles are approaching each other such that $\textbf{v}_A = - \textbf{v}_B$, then $\textbf{v} = 2 \textbf{v}_A$. However, by equipartitioning the relative kinetic energy, being mass independent, is
$\left< \dfrac{\textbf{p}^2}{2 \mu} \right> = \dfrac{3}{2} k_B T \label{10}$
which is called the collision energy.
Collision cross section
Consider two molecules in a system. The probability that they will collide increases with the effective “size” of each particle. However, the size measure that is relevant is the apparent cross-section area of each particle. For simplicity, suppose the particles are spherical, which is not a bad approximation for small molecules. If we are looking at a sphere, what we perceive as the size of the sphere is the cross section area of a great circle. Recall that each spherical particle has an associated “collision sphere” that just encloses two particles at closest contact, i.e., at the moment of a collision, and that this sphere is a radius $d$, where $d$ is the diameter of each spherical particle (see lecture 5). The cross-section of this collision sphere represents an effective cross section for each particle inside which a collision is imminent. The cross-section of the collision sphere is the area of a great circle, which is $\pi d^2$. We denote this apparent cross section area $\sigma$. Thus, for spherical particles $A$ and $B$ with diameters $d_A$ and $d_B$, the individual cross sections are
$\sigma_A = \pi d_A^2, \: \: \: \sigma_B = \pi d_B^2 \label{11}$
The collision cross section, $\sigma_{AB}$ is determined by an effective diameter $d_{AB}$ characteristic of both particles. The collision probability increases of both particles have large diameters and decreases if one of them has a smaller diameter than the other. Hence, a simple measure sensitive to this is the arithmetic average
$d_{AB} = \dfrac{1}{2} \left( d_A + d_B \right) \label{12}$
and the resulting collision cross section becomes
\begin{align} \sigma_{AB} &= \pi d_{AB}^2 \ &= \pi \left( \dfrac{d_A + d_B}{2} \right)^2 \ &= \dfrac{\pi}{4} \left( d_A^2 + 2d_A d_B + d_B^2 \right) \ &= \dfrac{1}{4} \left( \sigma_A + 2 \sqrt{\sigma_A \sigma_B} + \sigma_B \right) \ &= \dfrac{1}{2} \left[ \left( \dfrac{\sigma_A + \sigma_B}{2} \right) + \sqrt{\sigma_A \sigma_B} \right] \end{align} \label{13}
which, interestingly, is an average of the two types of averages of the two individual cross sections, the arithmetic and geometric averages!
Average collision Frequency
Consider a system of particles with individual cross sections $\sigma$. A particle of cross section $\sigma$ that moves a distance $l$ in a time $\Delta t$ will sweep out a cylindrical volume (ignoring the spherical caps) of volume $\sigma l$ (Figure $1$). If the system has a number density $\rho$, then the number of collisions that will occur is
$N_{\text{coll}} = \rho \sigma l \label{14}$
We define the average collision rate as $N_{\text{coll}}/ \Delta t$, i.e.,
$\gamma = \dfrac{N_{\text{coll}}}{\Delta t} = \dfrac{\rho \sigma l}{\Delta t} = \rho \sigma \langle | \textbf{v} | \rangle \label{15}$
where $\langle | \textbf{v} | \rangle$ is the average relative speed. If all of the particles are of the same type (say, type $A$), then performing the average over a Maxwell-Boltzmann speed distribution gives
$\langle | \textbf{v} | \rangle = \sqrt{\dfrac{8 k_B T}{\pi \mu}} \label{16}$
where $\mu = m_A/2$ is the reduced mass. The average speed of a particle is
$\langle | \textbf{v}_A | \rangle = \sqrt{\dfrac{8 k_B T}{\pi m_A}} \label{17}$
so that
$\langle | \textbf{v} | \rangle = \sqrt{2} \langle | \textbf{v}_A | \rangle \label{18}$
Mean Free Path
The mean free path is defined as the distance a particle will travel, on average, before experiencing a collision event. This is defined as the product of the speed of a particle and the time between collisions. The former is $\langle | \textbf{v} | \rangle/ \sqrt{2}$, while the latter is $1/\gamma$. Hence, we have
$\lambda = \dfrac{\langle | \textbf{v} |\rangle}{\sqrt{2} \rho \sigma \langle | \textbf{v} | \rangle} = \dfrac{1}{\sqrt{2} \rho \sigma} \label{19}$
Random Walks
In any system, a particle undergoing frequent collisions will have the direction of its motion changed with each collision and will trace out a path that appears to be random. In fact, if we treat the process as statistical, then, we are, in fact, treating each collision event as a random event, and the particle will change its direction at random times in random ways! Such a path might appear as shown in Figure $\PageIndex{2. Such a path is often referred to as a random walk path. In order to analyze such paths, let us consider a random walk in one dimension. We’ll assume that the particle move a mean-free path length \(\lambda$ between collisions and that each collision changes the direction of the particles motion, which in one dimension, means that the particle moves either to the right or to the left after each event. This can be mapped onto a metaphoric “coin toss” that can come up heads “H” or tails “T”, with “H” causing motion to the right, and “T” causing motion to the left.
In three dimensions, we consider the three spatial directions to be independent, hence, the probability distribution for a particle to diffuse to a location $\textbf{r} = (x, y, z)$ is just a product of the three one-dimensional distributions:
$\mathcal{P}(\textbf{r}) = P(x) \: P(y) \: P(z) = \dfrac{1}{(4 \pi D t) ^{3/2}} \: e^{-\left( x^2 + y^2 + z^2 \right)/4Dt} \label{35}$
where $D$ is the diffusion constant. If we are only interested in diffusion over a distance $r$, we can introduce spherical coordinates, integrate over the angles, and we find that
$P(r, t) = \dfrac{4 \pi}{(4 \pi D t)^{3/2}} \: E^{-r^2/4Dt} \label{36}$
Rates of Diffusion or Effusion
Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. We can write the expression for the average kinetic energy of two gases with different molar masses:
$KE=\dfrac{1}{2}\dfrac{M_{\rm A}}{N_A}v_{\rm rms,A}^2=\dfrac{1}{2}\dfrac{M_{\rm B}}{N_A}v_{\rm rms,B}^2\label{6.8.2}$
Multiplying both sides by 2 and rearranging give
$\dfrac{v_{\rm rms, B}^2}{v_{\rm rms,A}^2}=\dfrac{M_{\rm A}}{M_{\rm B}}\label{6.8.3}$
Taking the square root of both sides gives
$\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}}\label{6.8.4}$
Thus the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. Equation $\ref{6.8.4}$ shows that Graham’s law is a direct consequence of the fact that gaseous molecules at the same temperature have the same average kinetic energy.
The lightest gases have a wider distribution of speeds and the highest average speeds.
Molecules with lower masses have a wider distribution of speeds and a higher average speed.
Gas molecules do not diffuse nearly as rapidly as their very high speeds might suggest. If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as 1010 times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line.
The denser the gas, the shorter the mean free path.
Example $2$
Calculate the rms speed of a sample of cis-2-butene (C4H8) at 20°C.
Given: compound and temperature
Asked for: rms speed
Strategy:
Calculate the molar mass of cis-2-butene. Be certain that all quantities are expressed in the appropriate units and then use Equation $\ref{6.8.5}$ to calculate the rms speed of the gas.
Solution:
To use Equation $\ref{6.8.4}$, we need to calculate the molar mass of cis-2-butene and make sure that each quantity is expressed in the appropriate units. Butene is C4H8, so its molar mass is 56.11 g/mol. Thus
$u_{\rm rms}=\sqrt{\dfrac{3RT}{M}}=\rm\sqrt{\dfrac{3\times8.3145\;\dfrac{J}{K\cdot mol}\times(20+273)\;K}{56.11\times10^{-3}\;kg}}=361\;m/s$
or approximately 810 mi/h.
Exercise $2$
Calculate the rms speed of a sample of radon gas at 23°C.
Answer: 1.82 × 102 m/s (about 410 mi/h)
• Wikipedia
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/02%3A_Properties_of_Gases/2.08%3A_Molecular_Collisions_and_the_Mean_Free_Path.txt
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Learning Objectives
• Understand the difference between effusion and diffusion
• To derive and apply Graham's Law of Effusion
Diffusion is the gradual mixing of gases due to the motion of their component particles even in the absence of mechanical agitation such as stirring. The result is a gas mixture with uniform composition. Diffusion is also a property of the particles in liquids and liquid solutions and, to a lesser extent, of solids and solid solutions. The related process, effusion, is the escape of gaseous molecules through a small (usually microscopic) hole, such as a hole in a balloon, into an evacuated space.
The phenomenon of effusion had been known for thousands of years, but it was not until the early 19th century that quantitative experiments related the rate of effusion to molecular properties. The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. This relationship is referred to as Graham’s law, after the Scottish chemist Thomas Graham (1805–1869). The ratio of the effusion rates of two gases is the square root of the inverse ratio of their molar masses:
$\color{red} \dfrac{\text{rate of effusion A}}{\text{rate of effusion B}}=\sqrt{\dfrac{M_B}{M_A}} \label{1}$
Graham Law
The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass.
Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy (recall that a result of the Kinetic Theory of Gases is that the temperature, in degrees Kelvin, is directly proportional to the average kinetic energy of the molecules.) . We can write the expression for the average kinetic energy of two gases with different molar masses:
$KE=\dfrac{1}{2}\dfrac{M_{\rm A}}{N_A}v_{\rm rms,A}^2=\dfrac{1}{2}\dfrac{M_{\rm B}}{N_A}v_{\rm rms,B}^2\label{2}$
Multiplying both sides by 2 and rearranging give
$\dfrac{v_{\rm rms, B}^2}{v_{\rm rms,A}^2}=\dfrac{M_{\rm A}}{M_{\rm B}}\label{3}$
Taking the square root of both sides gives
$\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}}\label{4}$
Thus the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. Equation $\ref{4}$ shows that Graham’s law is a direct consequence of the fact that gaseous molecules at the same temperature have the same average kinetic energy.
Heavy molecules effuse through a porous material more slowly than light molecules, as illustrated schematically in Figure $1$ for ethylene oxide $C_2H_4O$ and helium $He$. Helium (M = 4.00 g/mol) effuses much more rapidly than ethylene oxide (M = 44.0 g/mol). Because helium is less dense than air, helium-filled balloons “float” at the end of a tethering string. Unfortunately, rubber balloons filled with helium soon lose their buoyancy along with much of their volume. In contrast, rubber balloons filled with air tend to retain their shape and volume for a much longer time. Because helium has a molar mass of 4.00 g/mol, whereas air has an average molar mass of about 29 g/mol, pure helium effuses through the microscopic pores in the rubber balloon $\sqrt{\frac{29}{4.00}}=2.7$ times faster than air. For this reason, high-quality helium-filled balloons are usually made of Mylar, a dense, strong, opaque material with a high molecular mass that forms films that have many fewer pores than rubber. Hence, mylar balloons can retain their helium for days.
Example $1$
During World War II, scientists working on the first atomic bomb were faced with the challenge of finding a way to obtain large amounts of $\ce{^{235}U}$. Naturally occurring uranium is only 0.720% $\ce{^{235}U}$, whereas most of the rest (99.275%) is $\ce{^{238}U}$, which is not fissionable (i.e., it will not break apart to release nuclear energy) and also actually poisons the fission process. Because both isotopes of uranium have the same reactivity, they cannot be separated chemically. Instead, a process of gaseous effusion was developed using the volatile compound $UF_6$ (boiling point = 56°C).
1. Calculate the ratio of the rates of effusion of 235UF6 and 238UF6 for a single step in which UF6 is allowed to pass through a porous barrier. (The atomic mass of 235U is 235.04, and the atomic mass of 238U is 238.05.)
2. If n identical successive separation steps are used, the overall separation is given by the separation in a single step (in this case, the ratio of effusion rates) raised to the nth power. How many effusion steps are needed to obtain 99.0% pure 235UF6?
Given: isotopic content of naturally occurring uranium and atomic masses of 235U and 238U
Asked for: ratio of rates of effusion and number of effusion steps needed to obtain 99.0% pure 235UF6
Strategy:
1. Calculate the molar masses of 235UF6 and 238UF6, and then use Graham’s law to determine the ratio of the effusion rates. Use this value to determine the isotopic content of 235UF6 after a single effusion step.
2. Divide the final purity by the initial purity to obtain a value for the number of separation steps needed to achieve the desired purity. Use a logarithmic expression to compute the number of separation steps required.
Solution:
1. A The first step is to calculate the molar mass of UF6 containing 235U and 238U. Luckily for the success of the separation method, fluorine consists of a single isotope of atomic mass 18.998. The molar mass of 235UF6 is 234.04 + (6)(18.998) = 349.03 g/mol
The molar mass of 238UF6 is
238.05 + (6)(18.998) = 352.04 g/mol
The difference is only 3.01 g/mol (less than 1%). The ratio of the effusion rates can be calculated from Graham’s law using Equation $\ref{1}$:
$\rm\dfrac{\text{rate }^{235}UF_6}{\text{rate }^{238}UF_6}=\sqrt{\dfrac{352.04\;g/mol}{349.03\;g/mol}}=1.0043$
2. rate U235F6rate U238F6=352.04349.03=1.0043
Thus passing UF6 containing a mixture of the two isotopes through a single porous barrier gives an enrichment of 1.0043, so after one step the isotopic content is (0.720%)(1.0043) = 0.723% 235UF6.
3. B To obtain 99.0% pure 235UF6 requires many steps. We can set up an equation that relates the initial and final purity to the number of times the separation process is repeated: final purity = (initial purity)(separation)n
In this case, 0.990 = (0.00720)(1.0043)n, which can be rearranged to give
$1.0043^n=\dfrac{0.990}{0.00720}=137.50$
4. Taking the logarithm of both sides gives
$n\ln(1.0043)=\ln(137.50)$
$n=\dfrac{\ln(137.50)}{\ln(1.0043)}=1148$
Thus at least a thousand effusion steps are necessary to obtain highly enriched 235U. Figure $2$ shows a small part of a system that is used to prepare enriched uranium on a large scale.
Exercise $1$
Helium consists of two isotopes: 3He (natural abundance = 0.000134%) and 4He (natural abundance = 99.999866%). Their atomic masses are 3.01603 and 4.00260, respectively. Helium-3 has unique physical properties and is used in the study of ultralow temperatures. It is separated from the more abundant 4He by a process of gaseous effusion.
1. Calculate the ratio of the effusion rates of 3He and 4He and thus the enrichment possible in a single effusion step.
2. How many effusion steps are necessary to yield 99.0% pure 3He?
Answer: a. ratio of effusion rates = 1.15200; one step gives 0.000154% 3He; b. 96 steps
Gas molecules do not diffuse nearly as rapidly as their very high speeds might suggest. If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as 1010 times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line.
Summary
Gaseous particles are in constant random motion. Gaseous particles tend to undergo diffusion because they have kinetic energy. Diffusion is faster at higher temperatures because the gas molecules have greater kinetic energy. Effusion refers to the movement of gas particles through a small hole. Graham's Law states that the effusion rate of a gas is inversely proportional to the square root of the mass of its particles.
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2.5: Condensation of Gases and the Critical State
Q2.2
There are some gases in existence, such as $\ce{NS2}$ and $\ce{NCl2}$, which do not follow Boyle's Law, regardless of the pressure conditions they're under. Explain why this is the case.
S2.2
Boyle's Law assumes that the only factors changing are pressure ($P$) and volume ($V$), therefore, in order to apply Boyle's Law, both number of molecules ($n$) and temperature ($T$) must remain constant. However, both $NS_2$ and $NCl_2$ do not have constant $n$ because they undergo association reactions when in contact with like molecules.
$2NH_2 \rightleftharpoons N_2S_2$
$2NCl_2 \rightleftharpoons N_2Cl_2$
Therefore, Boyle's Law may not be applied for these gases.
Q2.3
An ideal gas originally at 0.85 atm and 66ºC was allowed to expand until its final volume, pressure, and temperature were 94 mL, 0.60 atm, and 45ºC, respectively. What was its initial volume?
S2.3
• $P_1=0.85\;atm$
• $T_1=339\;K$
• ${V_2=94\;mL}$
• $P_2=0.60\;atm$
• $T_2=318\;K$
$V_1=\dfrac{(P_2)(V_2)(T_1)}{(T_2)(P_1)}$
$=\dfrac{(0.60\;atm)(94\;mL)(339\;K)}{(318\;K)(0.85\;atm)}=71\;mL$
Q2.4
Some ballpoint pens have a small hole in the main body of the pen. What is the purpose of the hole?
S2.4
The hole is there to allow the pressure to equalize and allow the ink to flow out of the pen.
Q2.5
Starting with the ideal-gas equation, show how you ca calculate the molar mass of a gas from a knowledge of its density
Q2.6
At STP (standard temperature and pressure), 0.280 /;L of a gas weighs 0.400/;g. Calculate the molar mass of the gas.
S2.6
First, we need to determine the moles of gas in the sample:
$n=\dfrac{(P)(V)}{(R)(T)}$
$=\dfrac{(1.00\;atm)(0.280\;L)}{(0.08206\;L-atm/mol-K)(273\;K)}=0.0125\;moL$
$=\dfrac{(0.400\;g)}{(0.0125\;mol)}=32.0\;g/moL$
Q2.7
Ozone molecules in the stratosphere absorb much of the harmful radiation from the sun. Typically, the temperature and partial pressure of ozone in the stratosphere are 250 K and 0.0010 atm, respectively. How many ozone molecules are present in 1.0 L of air under these conditions? Assume ideal-gas behavior.
Q2.9
Dissolving 3.00 g of an impure sample of CaCO3 in an excess of HCl acid produced 0.656 L of CO2 (measured at 20ºC and 792 mmHg). Calculate the percent by mass of CaCO3 in the sample.
Q2.14
1. What volume of air at 1.0 atm and 22ºC is needed to fill a 0.98 L bicycle tire to a pressure of 5.0 atm at the same temperature? (Note that the 5.0 atm is the gauge pressure, which is the difference between the pressure in the tire and the atmospheric pressure. Initially, the gauge pressure in the tire was 0 atm.)
2. What is the total pressure in the tire when the gauge reads 5.0 atm?
3. The tire is pumped with a hand pump full of air at 1.0 atm; compressing the gas in the cylinder adds all of the air in the pump to the air in the tire. If the volume of the pump is 33% of the tire's volume, what is the gauge pressure in the tire after 2 full strokes of the pump?
Q2.15
A student breaks a thermometer and spills most of the mercury (Hg) onto the floor of a laboratory that measures 15.2 m long, 6.6 m wide, and 2.4 m high. The vapor pressure of mercury at 20ºC is 1.7 x 10-6 atm.
1. Calculate the mass of mercury vapor (in grams)in the room at 20ºC.
2. Does the concentration of mercury vapor exceed the air quality regulation of 0.050 mm Hg m-3 of air?
3. One way to treat small quantities of spilled mercury is to spray powdered sulfur over the metal. Suggest a physical and a chemical reason for this treatment.
S2.15
a) The volume of the room is 15.2 m x 6.6 m x 2.4 m = 240 m3.
The pressure is 1.7 x 10-6 atm x 101325 Pa/atm = 0.17 Pa
Ideal gas equation:
$PV=nRT$
$n=\dfrac{PV}{RT}$
$n=\dfrac{0.17\;Pa*(240)\;m^3}{\dfrac{8.314\;Pa-m^3}{K-mol}*(293K)}$
$=0.017\;mole$
Mass of mercury vapor : mHg = nHg * MHg =(0.017 mole)(200.6 g/mole)(1000 mg/gram)} = 3400 mg
b) The mass density of mercury vapor :
[Hg] = mHg / VHg(l) = $\dfrac{3400\;mg}{240\;m^{-3}}$
=14 mg/m3
This value is indeed larger than the 0.050 mg Hg m-3 allowed.
c) Using powdered sulfur to absorb the spilled mercury works because the sulfur molecules will strongly bind to the mercury, and the mercury sulfide that forms is a solid. This makes removing spilled mercury (a liquid) easier.
Q2.16
An oxide has the density of 1.75 g L-1 at 1.12 atm and 100ºC. Is this oxide O2(g) or O3(g)?
S2.16
$M = \dfrac{\rho RT}{P}$
$= \dfrac{(1.75 g L^{-1})(0.08206 L atm K^{-1}mol^{-1})(100+273)K)}{1.12 atm}=47.8 \dfrac{g}{mol}$
The molar mass of O3(g) is 48 g mol-1, so the oxide must be O3(g)
Q2.21
You are given a gas at 750 torr with a volume of 0.62 Liters at 25°C. A while later, you measure the temperature and notice a 5°C temperature increase in the system. Assuming there is no pressure change and that this is an ideal gas, what is the new volume?
S2.21
First: List out what you have
Vi=0.62 L
Ti=25ºC +273.15= 298.15
We do not know the amount of moles so we should find that.
PV=nRT
So ni=2.47x10-4 moles
Pf=0.987 atm
Vf=?
Tf=298.15 K + 5 K= 303.15 K
Vf=(nRT)/P
=0.631 L
Q2.23a
The composition of air in your alveoli is 75% N2, 14% O2, 5% CO2, and 6% H2O, by volume. (a) Calculate the average molar mass of this air sample. (b) Calculate the partial pressures of N2, O2, CO2, and H2O in your alveoli. assuming that alveolar pressure is 760 torr (At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of gas).
Q2.23b
The composition of air in your alveoli is 75% N2, 14% O2, 5% CO2, and 6% H2O, by volume. (a) Calculate the average molar mass of this air sample. (b) Calculate the partial pressures of N2, O2, CO2, and H2O in your alveoli. (At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of gas.)
Q2.24
In an expandable balloon we find a mixture of two gases only. The mass percent of the gases are: 65% 02 and 35 % HQ2. The mass of the gases combined is 4.6 grams. Calculate the total volume these two gases occupy at 298 K and 1.00 atm.
S2.24
You can find the Volume by using the Ideal Gas Law equation:
$V=\dfrac{nRT}{P}$
The mass of the two gases is given by:
$m_{H_{2}}= 4.6\times 0.35=1.6g$
$m_{O_{2}}= 4.6\times 0.65=3.0g$
The number of moles of the mixture= $n_{total}=n_{H_{2}}+n_{O_{2}}$
$n_{total}=\dfrac{1.60g}{Q2.0g.mol^{-1}}+\dfrac{3.0g}{32g.mol^{-1}}=0.884mol$
The volume the two gases occupy = $V=\dfrac{(0.884mol)(0.08206\dfrac{L.atm}{K.mol})(298K)}{1.0 atm}= 21.6 L$
Q2.25
The temperature in a small room in summer time is 31oC. The room (dimension 4mx4mx3m) is filled with 0.88 kg of water vapor. What is the partial pressure of water vapor in the air?
Q2.27
A flask contains a mixture of two ideal gases, A and B. Show graphically how the total pressure of the system depends on the amount of A present. That is, plot the total pressure versus the mole fraction of A. Do the same for B on the same graph. The total number of moles of A and B is remained as constant.
Q2.29
When you drink something from a straw, how does the liquid get from the glass up through the straw into your mouth?
S2.29
When you close your mouth around the straw, you close off the atmosphere from outside to inside. During suction, you increase the volume inside your mouth, thus lowering the internal pressure and the atmospheric pressure exerted onto the liquid over comes the internal pressure and so the liquid flows into your mouth.
Q2.31
Consider the reaction of ferrous sulfide with a dilute acid:
$FeS_{(s)} + 2HCl_{(aq)} \rightarrow H_2S_{(g)} + FeCl_{2(l)}$
An amount of 5.10 grams of ferrous sulfide completely reacted in this reaction. The gas collected over the water at 27ºC has a volume 1.78 L. Find the pressure of the collected gas in atm. The vapor pressure of water at 27°C is 26.7 mmHg.
Hint: Look at Dalton's law of partial pressure
S2.31
Base on the amount of reactant, use stoichiometry to find the number of moles of H2S gas produced and get partial pressure of hydrogen sulfide gas. Then use Dalton's law of partial pressure to find the total pressure of collected gas.
$5.10g FeS \times \dfrac{1mole FeS}{87.92g FeS} \times \dfrac{1mole H_{2}S}{1mole FeS} = 0.058mole H_{2}S$
$P_{H_{2}S} = \dfrac{nRT}{V} = \dfrac{0.058 mole H_{2}S\times0.08206 \dfrac{L\cdot atm}{mole\cdot K}\times(27+273)K}{1.78 L} = 0.802 atm$
$P_{total} = P_{H_{2}S} + P_{H_{2}O} = 0.802 atm + (26.7 mmHg \times \dfrac{1atm}{760mmHg}) = {\color{red} 0.837 atm}$
Q2.32a
Tanks containing mixtures of oxygen and nitrogen are routinely lowered to the Aquarius underwater research lab in Florida. What is the percent composition of oxygen gas in the tank if the total pressure in the tank once it has reached Aquarius is 6.5 atm, and the partial pressure is 0.55 atm?
S2.32a
For any ideal gas,
$PV=nRT$
At constant pressure and temperature, V is proportional to n. Thus:
$P_{O^{2}}=x_{O^{2}}P_{total}$
$x_{O^{2}}=\dfrac{P_{O^{2}}}{P_{total}}=\dfrac{0.55atm}{6.5atm}=.085$
The percent composition is 8.5%
Q2.32b
A mixture of helium and oxygen gas is used to fill up an apparatus. If the total pressure within the apparatus is 10 atm and the partial pressure of helium is 2 atm within that apparatus, calculate the percent by volume of diatomic oxygen gas.
S2.32b
$\chi_{He}=\frac{P_{He}}{P_{total} }, \ where \ \chi = Mole \ Fraction$
\begin{align*} \chi_{He} &= \frac{2 atm}{10 atm} \ \chi_{He} &= 0.2 \end{align*}
\begin{align*} \% \ by \ volume \ He &= \chi_{He} \cdot 100 \ &= 0.2\ \cdot 100= 20\%\ \end{align*}
$100\% \ by \ volume \ total \ gas - 20\% \ by \ volume \ He= \ \underline{80\% \ by \ volume \ O_{2}}$
Click here for more on mole fractions.
Q2.33
An unknown mass of NO2 in a closed chamber decomposes completely to oxygen and nitrogen gases. The final pressure in the chamber is 0.687 atm. What are the partial pressures of each component of the gas mixture in the chamber?
S2.33
Every two moles of NO2 produce one mole of N2 and O2. By the ideal gas equation, since we know that the gases are at the same temperature and occupy the same volume, the pressure of each gas is proportional to the number of moles of that gas present, and thus the pressure of each gas is proportional to the mole fractions of that gas. Because NO2 represents 1/3 of the total moles of gas present, and O2 represents 2/3 of the total gas present, NO2 represents 1/3 of the total pressure, and O2 represents 2/3 of the total pressure.
• PN2 = 1/3 * 0.687 atm = 0.229 atm
• PO2 = 2/3 * 0.687 atm = 0.458 atm
Q2.41
Choose the more ideal system
1. H2O at 110 °C and 1 atm vs H2O at 100°C and I atm
2. Cl2 at 10 atm vs Cl2 at 0.5 atm
3. N2 at 60 °C vs N2 at 100°C
2.6: Kinetic Theory of Gases
Q2.41
List the ideal conditions (the conditions that help allow a gas to behave ideally).
S2.41
Ideal conditions come from the kinetic theory of gases. This theory relies on three primary assumptions (as follows):
1. The gas is made up of a large amount of molecules. These molecules obey Newton's laws pertaining to motion and are always moving around in random motion.
2. In comparison to the volume of the gas, the volume of the molecules are so negligible that they can essentially be ignored.
3. No forces are capable of acting on the molecules except for extremely short durations of elastic collisions.
Ideal conditions are at:
• high temperatures, and
• low pressures
This is because at these conditions is when there are relatively large distance between molecules.
Q2.42
1.00 mole of an unknown gas occupies a 0.865 L container at 0.90 atm and 25ºC. If the value of van der Waals constant a= 0.550, what is the value of b for the gas? What does the value of b indicate about the gas?
S2.42
$P=4.0atm\V=0.70L\T=77.2K\a=2.50\dfrac{atm\cdot L^2}{mol^2}$
$(P+\dfrac{a n^2}{V^2})(V-nb)=nRT\rightarrow b=\dfrac{nRT}{P+\dfrac{an^2}{V^2}}-V(\dfrac{1}{-1})$
$b=\dfrac{(1mol)(.08206\dfrac{L\cdot atm}{mol\cdot K})(77.2K)}{4.0atm+\dfrac{(Q2.5)1mol^2}{.70L^2}}-.70L(\dfrac{1}{-1mol})\=4\times 10^-3\dfrac{L}{mol}$
The small value of $b$ indicates that the gas molecules have a small molecular mass.
Q2.48
The compressibility for compound 1 is 0.84 and 0.56 for compound. Which compound will occupy a smaller volume?
S2.48
Z is a prediction of the behavior of gases. A smaller value means that there are stronger intermolecular forces amongst the molecules, meaning a smaller volume occupied. Therefore, compound 2 will occupy a smaller volume than compound 1.
Q2.50
True or false: The concept of temperature is a macroscopic. Explain your reasoning.
S2.50
True. The kinetic theory of gases hypothesizes that it deals with an extremely large amount of molecules. Therefore, temperature is a macroscopic concept since the average kinetic energy of the molecules within a given system is proportional to the temperature. A large amount of molecules must be included into this average in order for it to be properly utilized.
Q2.51
What assumptions are made when applying the kinetic molecular theory to gases? Are all of these assumptions necessary? Why or why not?
S2.51
There are five assumptions made when applying the kinetic molecular theory to gases are:
1. Gases consist of a large amount of atoms or molecules and these atoms/ molecules have spaces of separation between them that are much larger than their size.
2. Although the molecules have mass, we treat them as if they do not since their volume is negligibly small.
3. The molecules are always moving around randomly.
4. All collisions, both molecule-molecule collisions and molecule-wall collisions, are considered to be elastic rather than inelastic.
5. The molecules are not considered to be attracted nor repulsed to/ by one another.
Most of these assumptions are in fact necessary to consider when applying the kinetic molecular theory to gases, however, assumption number 4 is not always crucial. This is due to the fact that it does not matter whether or not the collisions between the molecules and the walls are elastic if the walls of the container are at the same temperature as the molecules inside of the the container (i.e. the gas) because although kinetic energy may be transferred from molecule to molecule, it will not be converted into any other forms of energy.
Q2.52
A 3.50 L cube-shaped container container 3.5x1023 molecules of $N_2$ at 30ºC. What is pressure exerted on one wall?
S2.52
$N=3.5\times 10^{23}\V=3.50L\T=303K\k_B=1.38\times 10^{-23}$
$P=\dfrac{2N}{3V}\overline{E}_trans\rightarrow P=\dfrac{2N}{3V}(\dfrac{3}{2}k_BT)$
$=\dfrac{2(3.5\times 10^{23})}{3(3.50L)}[\dfrac{3}{2}(1.38\times 10^{-23})(303K)]$
$=4.18\times 10^2atm$
Q2.53
Imagine a container with walls 123 cm³. Argon (with a speed of 700 m/s) atoms at 300 K are colliding at right angles with the container's walls at a rate of $5.0 \times 10^{20} m/s$. What would the force and pressure exerted on the container's walls be?
2.7: The Maxwell Distribution Laws
Q2.54
Calculate the root-mean-square speed and the molar kinetic energy of N2 at 25oC
S2.54
The root-mean-square speed of $N_2$ is:
$v_{rms} = \sqrt{\dfrac{3k_BT}{m}}$
or in terms of molar mass $M$ instead of molecular mass ($m$)
$v_{rms} = \sqrt{\dfrac{3RT}{M}}$
$=\sqrt{\dfrac{(3)(8.3145 \, J\,\cancel{K} \cancel{mol^{-1}})(298.15\; \cancel{K})}{ 28.02 \; \cancel{g} \, \cancel{mol^{-1}} \frac{1\; kg}{1000\; \cancel;{g}} }} = 5.15\; m/s$
The molar kinetic energy of $N_2$ at 25oC is:
= $\dfrac{3}{2}*\dfrac{8.3145J}{K mol}*298.15K*\dfrac{1kJ}{1000J}$
= 3.72 kJ/mol
Q2.55
Find the $v_{rms}$ of N_{2(g)} at 25ºC. What temperature must $Cl_{2(g)}$ be to have the same $v_{rms}$?
S2.55
$v{_{rms}} = \sqrt{\dfrac{3 R T}{M}}$
$v_{rmsN_{2}} = \sqrt{\dfrac{3(8.314 J\cdot K^{-1}\cdot mol^{-1})(25+273)K}{28.02 x 10^{-3} kg\cdot mol^{-1}}}=515m \ s^{-1}$
$since \ v_{rmsN_{2}} = v_{rmsCl_{2}}, then \ v_{rmsCl_{2}} = 515m \ s^{-1}$
$v_{rmsCl_{2}} = \sqrt{\dfrac{3(8.314 J\cdot K^{-1}\cdot mol^{-1})(T)}{70.9x10^{-3} g\cdot mol^{-1}}} = 515m \ s^{-1}$
T = 754 K
Therefore, Cl2(g) must be 754 K to have the same vrms as N2(g) at 298 K
Q2.60
1. Plot the distribution of speeds for Cl2 at 100 K, 300 K, 600 K, and 1000 K.
2. Plot the distribution of speeds for three gases: N2, Cl2, and CH4 at the same temperature.
S2.60
a) The plot of the distribution of speeds for Cl2 at 100 K, 300 K, 600 K, and 1000 K
b) The plot of the distribution of N2,Cl2 and CH4 at the same temperature (300K)
Q2.60
Which gas would you expect to move faster at 255K, O2 or I2? And why?
S2.60
We'd expect O2 to move faster. This is because it is lighter. Think of the formula for average speed because it is divided by kg/mol the smaller the mass, the faster it goes.
Q2.61
According to the Maxwell speed distribution, would $O_{2(g)}$ have a wider speed distribution at 200 K or 1000 K?
S2.61
At lower temperatures, the speed distribution will be narrower and there will will be a smaller most probable speed while the higher temperature will have a wider speed distribution and a higher most probable speed. Therefore, $O_{2(g)}$ will have a wider speed distribution at 1000 K.
Q2.62a
A CO2 molecule at unknown temperature at sea level is released to travel upward. Assuming that the temperature is constant and that the molecule doesn’t collide with another molecules and reaches a terminal height of 8.5 meters above sea level, what is the temperature? Do the same calculation for a H2 atom. [Hint: To calculate the altitude, h, the molecule will travel, equate its kinetic energy with the potential energy, mgh, where m is the mass and g is the acceleration due to gravity (9.81 m s-2).]
Q2.62b
A CO2 molecule at unknown temperature at sea level is released to travel upward. Assuming that the temperature is constant and that the molecule doesn’t collide with another molecules and reaches a terminal height of 5000 meters above sea level, what is the temperature? Do the same calculation for a H2 atom. [Hint: To calculate the altitude, h, the molecule will travel, equate its kinetic energy with the potential energy, mgh, where m is the mass and g is the acceleration due to gravity (9.81 m s-2).]
Q2.63
Calculate the root-mean-square speed in cm/s for the following particles (hint, what is the average speed?): 0.010, 0.043, 0.027, 0.012, 0.041, 0.023, 0.011, 0.004, 0.007, and 0.009 (all in m/s).
Q2.63
The root-mean-square speed of 6 particles is 2.47 ms-1. The speed of 5 of the particles are 1.0, 2, 1.5, 3.0 and Q2.5. Calculate the unknown speed of the 6th particle. Find the average speed of the 6 particles.
S2.63
The root-mean-square of the ensemble of particles is
$v_{rms}=\sqrt{\dfrac{\sum_{i=1}^{6} c^{2}_{i}}{m}}$
$2.47=\sqrt{\dfrac{1.0^{2}+Q2.0^{2}+1.5^{2}+3.0^{2}+2.5^{2}+c_{6}}{6}}$
We solve for $c_{6}$ and we get
$c^{2}_{6}=(Q2.47ms^{-1})^{2}(6)-(1.0^{2}+2.0^{2}+1.5^{2}+3.0^{2}+2.5^{2}) \; m^{2}s^{-2}$
$c^{2}_{6}= 36.6m^{2}s^{-2}-22.5m^{2}s^{-2}$
$c^{2}_{6}=14.1m^{2}s^{-2}$
$c_{6}= 3.75ms^{-1}$
The average speed of the 6 particles is
$\bar{c}=\dfrac{\sum^{6}_{i=1}c_{i} }{N}$
$\bar{c}=\dfrac{(1.0+Q2.0+1.5+3.0+Q2.5+3.75)ms^{-1}}{6}$
$=2.30ms^{-1}$
Q2.64a
Compare the velocities of Oxygen and Hydrogen at room temperature (25oC) by computing their root mean square speed?
Q2.64b
The speeds of six gaseous molecules at a given temperature are 1.0m/s , 1.2m/s, 1.5m/s, 1.7m/s, 2.0m/s, and 2.3m/s.
1. Compare the root-mean-square (rms) speeds and the average speed of those molecules.
2. Is always the average speed of the molecules the smaller of the two? Explain.
S2.64b
a) The root-mean-square-speed of those gases:
$v_{rms} = \sqrt{\dfrac{1}{N}*(v_1^{2}+v2^{2}+v_3^{2}+v_4^{2}+v_5^{2}+v_6^{2})}$
$= \sqrt{\dfrac{1}{N}*(1^{2}+1.2^{2}+1.5^{2}+1.7^{2}+2^{2}+2.3^{2})}$
$= 1.68\; m/s$
The average speed is :
$v_{avg} = \dfrac{1+1.2+1.5+1.7+2+2.3}{6}$
$= 1.62\; m/s$
b) The average speed is always smaller then the root-mean-square speed because the average speed depends on the temperature, rather than the molecular sizes.
Q2.65
Calculate cmp for a sample of N2(g) at 500K.
S2.65
$c{_{mp}} = \sqrt{\dfrac{2RT}{M}}$
$c{_{mp}} = \sqrt{\dfrac{2 \ (8.314 J \ K^{-1} \ mol^{-1}) (500K)}{.02802 kg \ mol^{-1}}} = 545 m \ s^{-1}$
Q2.66
Using the initial equation $f(c)=4\pi c^2 \dfrac{m}{\pi K_BT} ^{3/2} e^{-\dfrac{mc^2}{2K_BT}}$ derive an expression for $c_{mp}$
S2.66
Step 1: $f(c)=4\pi c^2 \dfrac{m}{\pi K_BT} ^{3/2} e^{-\dfrac{mc^2}{2K_BT}}\]$=c^2 \dfrac{m}{\pi K_BT} ^{3/2} e^{-\dfrac{mc^2}{2K_BT}}\]
Step 2: (Hint: Apply the chain rule to $c^2$ and $e^{-\dfrac{mc^2}{2K_BT}}$)
$\dfrac{df(c)}{dc}=4\pi \dfrac{m}{2\pi K_BT}^{3/2} [2ce^{-\dfrac{mc^2}{2K_BT}} + c^2e^{-\dfrac{mc^2}{2K_BT}}-{\dfrac{mc}{2K_BT}}]$
Step 3:
$4\pi(\dfrac{m}{2\pi K_BT})^{3/2}ce^{-\dfrac{mc^2}{2K_BT}}(2-\dfrac{mc^2}{K_BT})$
Step 4: (Hint: Set $\dfrac{df(c)}{dc}=0$ , $c=c_{mp}$)
$\pi (\dfrac{m}{2\pi K_BT})^{3/2}c_{mp}e^{-\dfrac{mc_{mp}^2}{2K_BT}}(2-\dfrac{mc_{mp}^2}{K_BT})=0$
Step 5:
$2-\dfrac{mc^2 _{mp}}{K_BT}=0$
Step 6:
$c^2_{mp}=\dfrac{2K_BT}{m}$
Step 7: (Hint: Use Avogadro's number)
$c^2 _{mp}=\dfrac{2K_BT}{m}\dfrac{N_A}{N_A}=\dfrac{2RT}{M}$
Step 8:
$c_{mp}=\sqrt {\dfrac{2RT}{M}}$
Q2.68a
Calculate the value of cmp for CH3OH at 30°C. What is the ratio of the number of molecules of speed 400 m s-1 to the number of molecules of speed cmp?
Q2.68b
Calculate the value of the most probable speed, average speed, and the root-mean-square speed for $C_5H_{12}$ at 25°C.
S2.68b
$c_{mp}=\sqrt{\dfrac{2RT}{M}}=\sqrt{\dfrac{2(8.314J/K*mol)(298.15K)}{72.15\times 10^{-3}kg/mol}}=26Q2.13m/s$
\begin{align*}\bar{c}&=\sqrt{\dfrac{8RT}{\pi *M}}=\sqrt{\dfrac{4}{\pi}}*\sqrt{\dfrac{2RT}{M}}=\sqrt{\dfrac{4}{\pi}}*c_{mp}\&=\sqrt{\dfrac{4}{3.14159}}*262.13\;m/s=295.78m/s\end{align*}
\begin{align*}c_{rmp}&=\sqrt{\dfrac{3RT}{M}}=\sqrt{\dfrac{3}{2}}*\sqrt{\dfrac{2RT}{M}}=\sqrt{\dfrac{3}{2}}*c_{mp}\&=\sqrt{\dfrac{3}{2}}*262.13\; m/s=321.04\; m/s\end{align*}
Q2.68
Calculate the value of cmp for CH3OH at 30°C. What is the ratio of the number of molecules of speed 400 m s-1 to the number of molecules of speed cmp?
2.8: Molecular Collisions and the Mean Free Path
Q2.69
You open a container that has 1 mole of Neon gas and 1 mole of hydrogen gas. Hydrogen gas moves at faster speeds than Neon.
1. Compare the mean free path between the two gases in a closed room.
2. Assume travels across the room and there are no collision between molecules. Which gas reaches the other end first?
S2.69
1. The Neon gas molecules are larger than the Hydrogen gas molecules. This decreases the distance the molecules have to travel before they run into each other. For this reason, the mean free path for Neon will be lower than the mean free path for Hydrogen.
2. Since there are no collisions between molecules the mean free path does not play a role in this case. Therefore, Hydrogen will reach the other end first because it travels faster than Neon.
Q2.70a
How many times is the average distance between collision for a gas molecule larger than its molecular diameter, given that Pressure
Po= 780mm Hg, and temperature T=300K, molecular diameter is 2x10-10 meters.
Q2.70b
The mean free path of gas molecules is given as: $\gamma = \dfrac{1}{\sqrt{2}\pi d^{2}(N/V)}$
1. rewrite the equation so that the mean free path is expressed in terms of gas pressure
2. Name some that factors affect this property directly and indirectly?
3. Consider a gas piston containing only hydrogen gas. While not changing the temperature, there observed to be an increase in the mean free path of the gas molecules. What action could have caused this to happen?
Hint: Look at concept of mean free path
S2.70b
Mean free path is the average distance molecules traveled between each successive collision. Think of what could have affect this distance
a) $PV = nRT \Rightarrow V = \dfrac{nRT}{P}$
$N/V = \dfrac{N}{n} \dfrac{P}{RT}$
$but (N/n) = N_{A}$
$\Rightarrow \dfrac{N}{V} = \dfrac{PN_{A}}{RT} \Rightarrow {\color{red} \gamma= \dfrac{RT}{\sqrt2 \pi d^{2} PN_{A}}}$
b) direct factors: density and molecule's dameter, indirect factors: temperature, pressure, volume
c) An increase in mean free path means that the collisions are happen less frequently, therefore the density of the gas decreased. An action that could have caused this to happen is an increase in volume. A larger volume while molecules while not changing the number of molecules and speed lead to molecules are further apart and have less collisions.
Q2.70c
Gas A has a density of 0.835 g/cm3 and Gas B has a density of 14.67 g/cm3. From just this information, which do you expect to have a larger mean free path? Explain why.
S2.70c
We would expect gas A to have a larger mean free path.
Q2.71
A cylinder contains 25 molecules of an unknown ideal gas. What is the mean free path of the gas molecules? The cylinder has a volume of 20.0 cm3 and the gas has an atomic radius of 1.24 Å. (1 cm = 108 Å)
S2.71
For an ideal gas, the mean free path is calculated by:
$\lambda =\frac{\bar{c}}{Z_{1}}=\frac{1}{\sqrt{2}\pi d^{2}\frac{N}{V}}$
Thus, the mean free path of the unknown gas is:
$\lambda =\frac{1}{\sqrt{2}\pi (1.24\times 10^{8}\, cm)^{2}\, \frac{25}{20.0 cm^{2}}}=1.17\times 10^{-17}cm=1.17\times 10^{-19}m$
Q2.71
A rattle containing 50 spherical beads is shaken around by a toddler. What is the mean free path of the beads if the volume of the rattle is 500 cm3 and the diameter of each individual bead is 0.2 cm?
S2.71
Use the appropriate equation and plug in given values.
\begin{align} & \lambda = \frac{1}{\sqrt{2} \pi d^{2}(N/V)} \ & \lambda = \frac{1}{\sqrt{2} \pi (\frac{0.2}{100} m)^{2}(50 beads/\frac{500}{100}m^{3})} \ & \underline{ \lambda = 1.8 \cdot 10^{-4} \ \ m } \end{align}
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Q2.72a
Calculate the mean free path and the binary number of collisions per liter per second between Ar molecules at 298 K and 1.00 atm. Use 3.62 Å as the collision diameter of the Ar molecules. Assume ideal gas behavior.
Q2.72b
Calculate the mean free path and the binary number of collisions per liter per second between Ar molecules at 298 K and 1.00 atm. Use 3.62 Å as the collision diameter of the Ar molecules. Assume ideal gas behavior.
Q2.72c
$CO_2$ has a collision diameter of 0.4 nanometers. At 25 oC and 1.2 atm pressure, what is the mean free path between binary collisions, and what is the rate of binary collisions per liter assuming the volume contains only pure carbon dioxide gas behaving ideally?
Q2.72d
For the molecules HCl at 350K and 1.21 atm, calculate the mean free path and the binary number of collisions per liter per second. The collision diameter of HCl molecules is 5.50 Å. Assume the HCl acts as an ideal gas.
S2.72d
Step 1:
Calculate $\dfrac{N}{V}$ to calculate the mean free path
$PV=NRT=\dfrac{N}{N_A}RT\] Step 2:$\dfrac{N}{V}=\dfrac{PN_A}{RT}=\dfrac{(1.21atm)(6.022\times 10^{23}mol^{-1})}{(0.08206L\,atm\,K^{-1}mol^{-1})(350K)}=Q2.537\]
Step 3:
Turn L into $m^{-3}$
$Q2.537\times 10^{22}L^{-1}(\dfrac{1000L}{1m^{3}})=Q2.537\times 10^{22}m^{-3}\] Step 4:$\lambda=\dfrac{1}{\sqrt{2}\pi d^{2}(\dfrac{N}{V})}=\dfrac{1}{\sqrt{2}\pi (5.50\times 10^{-10}m)^{2}(Q2.537\times 10^{22}m^{-3})}=Q2.93\times 10^{-5}m\]
Calculating the binary number of collisions
Step 1: (Hint: Use the equation for average molecular speed.)
$\bar{C}=\sqrt{\dfrac{8RT}{\pi M}}=\sqrt{\dfrac{8(8.3145JK^{-1}mol^{-1}(300K)}{\pi (36.46\times 10^{-3}\dfrac{kg}{mol})}}\] Step 2:$z_{11}=\sqrt{\dfrac{2}{2}}\pi d^2\bar{C}\dfrac{N}{V}^{2}\]
$=1.80526\times 10^{29}m^{-3}s^{-1}\] Step 3: (Hint: Turn into $L^{-1}s^{-1}$)$z_{11}=\sqrt{\dfrac{2}{2}}\pi (5.50\times 10^{-10}m)^{2}(417.39ms^{-1})(\dfrac{1m^{3}}{1000L})=1.80527\times^{26} L^{-1}s^{-1}\]
Q2.73
In a 4L container you have 5 grams of 0Q2. Assume the collision diameter of 02 is 3.55 Angstrongs . Calculate the mean free path of the gas.
S2.73
The mean free path is given by
$\lambda =\dfrac{1}{\sqrt{2}\pi d^{2}(\dfrac{N}{V})}$
V is given and we can find $\left ( \dfrac{N}{V} \right)$ by calculating N.
$N=N_{A}.n=(6.022\times 10^{23}mol^{-1})(5gO_{2})(\dfrac{1mol}{32gO_{2}})$
$N=9.40\times 10^{22}$
$\left ( \dfrac{N}{V} \right )=\dfrac{9.41\times 10^{22}}{4L}=Q2.35\times 10^{22}L^{-1}(1000Lm^{-3})$
$\left ( \dfrac{N}{V}\right )=Q2.35\times 10^{25}m^{3}$
The mean free path is
$\lambda =\dfrac{1}{\sqrt{2}\pi d^{2}(\dfrac{N}{V})}$
$\lambda =\dfrac{1}{\sqrt{2}\pi(3.55\times 10^{-10}m)^{2}(Q2.35\times 10^{25}m^{-3})}=7.60\times 10^{-8}m$
Q2.74a
If a sealed container has of xenon molecules traveling at a speed of 24.6 m/s at 108 °C. How many moles of xenon gas do we have in the container?
S2.74a
$\bar{E}_{trans}=\dfrac{1}{2}mv^2=\dfrac{3}{2}k_BT\ m=\dfrac{3k_BT}{v^2}=\dfrac{3(1.381\times 10^{-23}J/K)(381.15K)}{(24.6m/s)^2}*\dfrac{1kg*m^2/s^2}{1J}=Q2.609\times 10^{-23}kg\ n=\dfrac{m}{molarmass}=Q2.609\times 10^{-23}kg*\dfrac{1000g}{1kg}*\dfrac{mol}{131.3g}=1.987\times 10^{-22}mol$
Q2.74b
At room temperature (25oC) there are two flasks A and B containing gaseous Oxygen and Hydrogen, respectively, at their equilibrium states. Assuming there is no exchange of heat or work between the systems inside the flasks and the surroundings. Compare the initial velocities between an Oxygen and a Hydrogen molecule in each flask.
Q2.76a
In a insulated container of volume 2.50 L filled with nitrogen gas, the Z1 value is 6.75e19 collisions/second. The temperature measured is 55ºC. Assuming that the collision diameter of N2 is 3.75 Å,
1. how many moles of nitrogen gas are present the container?
2. find the Z11 value of the gas.
3. How does the number of moles affect Z1 and Z11 values of the gas?
Hint: look at collision frequency
S2.76a
a) $Z_{1} = \sqrt{2} \pi d^{2} \bar{c} (\dfrac{N}{V})$
$\Rightarrow N = \dfrac{VZ_{1}}{\sqrt2 \pi d^{2} \bar{c}}$
$\bar{c} = \sqrt{\dfrac{8RT}{\pi M}} = \sqrt{\dfrac{8(8.314 \dfrac{J}{mol \cdot K})(55+273)K}{\pi \cdot 0.014kg})} = 704.3 m/s$
$N = \dfrac{(2.50 L)(\dfrac{1m^{3}}{1000L}) (6.75 \times 10^{19} collision/s)}{\sqrt{2} \pi (3.75 \times 10^{-10}m)^{2} (704.3 m/s)} = 1.017 \times 10^{23} molecules$
$n = \dfrac{N}{N_A} = \dfrac{1.017 \times 10^{23} molecules}{6.022 \times 10^{23} \dfrac{molecules}{mole}} = {\color{red} 0.169 mole N_{2}}$
b) $Z_{11} = \dfrac{1}{2} Z_{1} (\dfrac{N}{V})$
$Z_{11} = \dfrac{1}{2} (6.75 \times 10^{19}) \dfrac{1.017 \times 10^{23} molecules}{2.50L \times \dfrac{1m^{3}}{1000 L}} = {\color{red} 1.37 \times 10^{45} \dfrac{collisions}{m^{3} \cdot s}}$
c) Since N = nNA and Z1 is proportional to N, therefore
${\color{red} Z_{1} \propto n}$
similarly, Z11 is proportional to N2, therefore:
${\color{red} Z_{11} \propto n^{2}}$
Q2.76b
For molecular oxygen at $56^{o}C$, calculate the number of collisions a single molecule makes in 1 second and the total number of binary collisions at P=1.0 atm and P=0.25 atm. How does pressure relate to these two quantities? The collision diameter of oxygen is 3.02 A.
S2.76b
Step 1: (Hint: Calculate the density and average molecular speed)
At P=1.0atm
$\dfrac{N}{V}=\dfrac{PN_A}{RT}=\dfrac{(1.0atm)(6.022\times 10^{23}mol^{-1})}{0.08206\,L\,atm\,K^{-1}mol^{-1})(273+56)}=Q2.23\times 10^{22}L^{-1}(\dfrac{1000L}{1m^{3}})=2.23\times 10^{22}m^{-3}\] Step 2:$\bar{c}=\sqrt{\dfrac{8RT}{\pi M}}=\sqrt{\dfrac{8(8.3145JK^{-1}mol^{-1})(273+56K)}{\pi (31.998\times 10^{-3}kg\,mol^{-1})}}=466.578ms^{-1}\]
Step 3:
$z_1=\sqrt{2}\pi d^2\bar{c}\dfrac{N}{V}=\sqrt{2}\pi(3.02\times 10^{-10}m)^{2}(466.578ms^{-1})(2.23\times 10^{22}m^{-3})=4.683\times 10^{28} collisions m^{-3}s^{-1}\] Step 4:$z_{11}=\dfrac{\sqrt{2}}{2}\pi d\bar{c}(\dfrac{N}{V})^{2}=\dfrac{1}{2}(4.2\times 10^{6})(2.23\times 10^{22}m^{-3})=4.683\times 10^{28} collisionsm^{-3}s^{-1}\]
For an ideal gas, $z_1$ is proportional to P and $z_{11}$ is proportional to $P^{2}$. Reducing P to a fourth of its original value (from 1.0 atm to 0.25 atm) will also reduce 2, to one fourth its value at P=1.0atm.
Therefore at P=0.25 atm,
$z_1=4.2\times 10^{5}collisions\,s^{-1} z_{11}=4.683\times 10^{26}\,collisions\,m^{-3}s{-1}$
Q2.77a
The relationship between the molar mass and root mean square speed of an ideal gas can be described as:
$\upsilon _{rms}=\sqrt{\frac{3RT}{\mathfrak{M}}}$
Use this equation to derive Grahm's Law of Effusion:
$\frac{r_{1}}{r_{2}}=\sqrt{\frac{\mathfrak{M_{2}}}{\mathfrak{M_{1}}}}$
S2.77a
For two gasses at standard temperature and pressure,
$\upsilon _{1rms}=\sqrt{\frac{3RT}{\mathfrak{M_{1}}}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \upsilon _{2rms}=\sqrt{\frac{3RT}{\mathfrak{M_{2}}}}$
For any of the two gases,
$\upsilon _{1rms}^{2}=\frac{3RT}{\mathfrak{M_{1}}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \upsilon _{2rms}^{2}=\frac{3RT}{\mathfrak{M_{2}}}$
$R=\frac{\upsilon _{1rms}^{2}\mathfrak{M_{1}}}{3T}=\frac{\upsilon _{2rms}^{2}\mathfrak{M_{2}}}{3T}$
$\upsilon _{1rms}^{2}\mathfrak{M_{1}}=\upsilon _{2rms}^{2}\mathfrak{M_{2}}$
$\frac{\upsilon _{1rms}^{2}}{\upsilon_{2rms}^{2}}=\frac{\mathfrak{M_{2}}}{\mathfrak{M_{1}}}$
$\frac{\upsilon _{1rms}}{\upsilon_{2rms}}=\sqrt{\frac{\mathfrak{M_{2}}}{\mathfrak{M_{1}}}}$
Which can be rewritten as:
$\frac{r_{1}}{r_{2}}=\sqrt{\frac{\mathfrak{M_{2}}}{\mathfrak{M_{1}}}}$
2.9: Graham's Laws of Diffusion and Effusion
Q2.77b
Derive Graham's Law of Effusion (1) from the equation for calculating Kinetic Energy (2).
\begin{align} &\left(1 \right) \ \ \ \frac{v_{1}}{v_{2}}=\sqrt{\frac{m_2}{m_1}} \ &\left(2 \right) \ \ \ KE =\frac{1}{2}mv^{2}\ \end{align}
S2.77b
\begin{align} & \frac{v_{1}}{v_{2}}=\sqrt{\frac{m_2}{m_1}} \ \ KE =\frac{1}{2}mv^{2}\ \ & KE_1 = KE_2, \ for \ two\ substances \ & \frac{1}{2}m_1 v_1^{2} = \frac{1}{2}m_2 v_2^{2}, \ multiply \ both \ sides \ by \ two \ & m_1 v_1^{2} =m_2 v_2^{2} \ & \frac{v_1^{2}}{v_2^{2}} = \frac{m_2}{m_1}, \ square \ root \ both \ sides \ of \ the \ equation \ & \underline{ \frac{v_{1}}{v_{2}}=\sqrt{\frac{m_2}{m_1}} } \end{align}
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Q2.78a
How long would it take for oxygen to diffuse in and carbon dioxide to diffuse out of a cell that is 0.1µm2 thick given that the rate of diffusion of carbon dioxide is 0.0016 mm2/s?
S2.78a
$\dfrac{t_{CO_2}}{t_{O_2}}=\dfrac{r_{O_2}}{r_{CO_2}}=\sqrt{\dfrac{M_{CO_2}}{M_{O_2}}}\ r_{O_2}=(0.0016mm^2/s)\sqrt{\dfrac{44.01g/mol}{3Q2.0g/mol}}=0.0019mm^2/s\ t_{CO_2}=0.1\mu m^2*\dfrac{(1mm)^2}{(100\mu m)^2}*\dfrac{s}{0.0016mm^2}=6.25\times 10^{-3}s\ t_{O_2}=0.1\mu m^2*\dfrac{(1mm)^2}{(100\mu m)^2}*\dfrac{s}{0.0019mm^2}=5.26\times 10^{-3}s$
Q2.78b
A sample of red gas collected from the emissions of a paper mill effuses through an orifice of unknown size at standard temperature and pressure in 14.1 minutes. Pure diatomic nitrogen gas at standard temperature and pressure effuses through the hole in 11.0 minutes. What is the red gas? How do you know?
S2.78b
mN2 = 28.01 g/mol
tN2/tunk = (munk/mN2)1/2
munk = (tN2/tunk)2 * mN2 = (14.1 min/11.0 min)2 * 28.01 g mol-1
munk = 46.02 g mol-1
The red gas is nitrogen dioxide, a common emission from pulp mills. We know because the diffusion rate corresponds with a molar mass of 46.01g, the molar mass of NO2
Q2.80
An amount of CO2 gas is effused through a small opening at the rate of 0.596 L/min. A homogenous mixture of O2 and N2 in the atmosphere is put under the same conditions for effusion and after 5 minutes, 5.03 L of the mixture was effused. Find the percent composition of the homogenous mixture (molar fraction).
Hint: Look at concept of effusion and its rate
S2.80
$\dfrac{r_{CO_{2}}}{r_{mixture}} = \sqrt{\dfrac{M_{mixture}}{M_{CO_{2}}}}$
$M_{mixture} = (\dfrac{r_{CO^{2}}}{r_{mixture}})^{2} (M_{CO_{2}}) = (\dfrac{0.596 \dfrac{L}{min}}{\frac{5.03 L}{5 min}})^2 (44.01 \dfrac{g}{mol}) = 15.447 \dfrac{g}{mol}$
$M_{mixture} = M_{N_{2}}X_{N_{2}} + M_{O_{2}}X_{O_{2}}$
$M_{mixture} = M_{N_{2}}X_{N_{2}} + M_{O_{2}}(1-X_{N_{2}})$
$15.447 \dfrac{g}{mol} = (14.01 \dfrac{g}{mol})(X_{N_{2}}) + (16.00 \dfrac{g}{mol})(1-X_{N_{2}})$
$15.447 = 14.01 X_{N_{2}} + 16.00 - 16.00X_{N_{2}} = -1.99X_{N_{2}} +16.00$
$X_{N_{2}} = 0.278$
$X_{O_{2}} = 1-0.278 = 0.722$
${\color{red} \% of N_{2} = 27.8\%}$
${\color{red} \% of O_{2} = 72.2\%}$
Q2.81a
You are trying to isolate gaseous Nitrogen-14 via effusion in a 50:50 mixture of Nitrogen-14 and Nitrogen-15 respectively. What will be the percentage of enrichment of the mixture after a single stage of separation?
S2.81a
Use Graham's Law of Diffusion and Effusion to solve this problem.
\begin{align} & \frac{r_1}{r_2} = \sqrt{\frac{M_{2}}{M_{1}}} \ & \frac{ rate \ of \ ^{14}\textrm{N} }{ rate \ of \ ^{15}\textrm{N} } = \sqrt{ \frac{^{15}\textrm{N}}{^{14}\textrm{N}} } \end{align}
\begin{align} & \frac{rate \ of \ {}^{14}\textrm{N}}{rate \ of \ {}^{15}\textrm{N}} = \sqrt{\frac{15.0001 \ g \ mol^{-1}}{14.00307 \ g \ mol^{-1}}} \ & \frac{rate \ of \ {}^{14}\textrm{N}}{rate \ of \ {}^{15}\textrm{N}} = 1.035 \end{align}
Since the ratio of rates of effusion is 1.035, in a single step of separation 3.5% more nitrogen-14 will effuse out of the mixture than nitrogen-15. Thus the percent of enrichment would be 3.5% since there is 3.5% more of N-14 now.
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Q2.81b
A mixture of Neon and Helium gas is effused at standard temperature and pressure. Assuming the mixture is equimolar, what is the composition of the gas mixture after effusion?
S2.81b
For the effusion of a binary gas mixture, we can use Graham's Law:
$\frac{r_{1}}{r_{2}}=\sqrt{\frac{\mathfrak{M_{2}}}{\mathfrak{M_{1}}}}$
The ratio of the two gasses following effusion given r1 corresponds with neon and r2 corresponds with helium is:
$\frac{r_{1}}{r_{2}}=\sqrt{\frac{4.003\frac{g}{mol}}{20.180\frac{g}{mol}}}=0.4454$
Therefore, to calculate the mole fraction of each gas,
$x_{Ne}=\frac{0.4454}{0.4454+1}=0.3081$
$x_{He}=1-x_{Ne}=1-0.3081=0.6919$
Q2.82
An equimolar mixture of diatomic nitrogen gas and an unknown monatomic gas diffuse through an aperture of unknown size. It is determined that 1.18 times as many moles of the monatomic gas pass through the opening as moles of nitrogen gas. What is the gas?
Q2.82
mN2 = 28.01 g mol-1
nunk/nN2 = (mN2/munk)1/2
munk = mN2/(nunk/nN2)2 = 28.01 g mol-1/(1.18)2 = 20.11 g mol-1
The gas is monatomic and has a mass of roughly 20.11 g/mol, so Neon is the most likely gas.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/02%3A_Properties_of_Gases/2.E%3A_Properties_of_Gases_%28Exercises%29.txt
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• 3.1: Work and Heat
Joule was able to show that work and heat can have the same effect on matter – a change in temperature! It would then be reasonable to conclude that heating, as well as doing work on a system will increase its energy content, and thus it’s ability to perform work in the surroundings. This leads to an important construct of the First Law of Thermodynamics.
• 3.2: The First Law of Thermodynamics
The first law of thermodynamics states that the energy of the universe is constant. The change in the internal energy of a system is the sum of the heat transferred and the work done. The heat flow is equal to the change in the internal energy of the system plus the PV work done. When the volume of a system is constant, changes in its internal energy can be calculated by substituting the ideal gas law into the equation for ΔU.
• 3.3: Heat Capacities
• 3.4: Gas Expansion
• 3.5: Calorimetry
As chemists, we are concerned with chemical changes and reactions. The thermodynamics of chemical reactions can be very important in terms of controlling the production of desired products and preventing safety hazards such as explosions. As such, measuring and understanding the thermochemistry of chemical reactions is not only useful, but essential!
• 3.6: Thermochemistry
Standard States, Hess's Law and Kirchoff's Law
• 3.7: Bond Energies and Enthalpies
In the absence of standard formation enthalpies, reaction enthalpies can be estimated using average bond enthalpies. This method is not perfect, but it can be used to get ball-park estimates when more detailed data is not available.
• 3.E: Exercises
This are exercises that to accompany the TextMap organized around Raymond Chang's Physical Chemistry for the Biosciences textbook.
03: The First Law of Thermodynamics
One of the pioneers in the field of modern thermodynamics was James P. Joule (1818 - 1889). Among the experiments Joule carried out, was an attempt to measure the effect on the temperature of a sample of water that was caused by doing work on the water. Using a clever apparatus to perform work on water by using a falling weight to turn paddles within an insulated canister filled with water, Joule was able to measure a temperature increase in the water.
Thus, Joule was able to show that work and heat can have the same effect on matter – a change in temperature! It would then be reasonable to conclude that heating, as well as doing work on a system will increase its energy content, and thus it’s ability to perform work in the surroundings. This leads to an important construct of the First Law of Thermodynamics:
The capacity of a system to do work is increased by heating the system or doing work on it.
The internal energy (U) of a system is a measure of its capacity to supply energy that can do work within the surroundings, making U the ideal variable to keep track of the flow of heat and work energy into and out of a system. Changes in the internal energy of a system ($\Delta U$) can be calculated by
$\Delta U = U_f - U_i \label{FirstLaw}$
where the subscripts $i$ and $f$ indicate initial and final states of the system. $U$ as it turns out, is a state variable. In other words, the amount of energy available in a system to be supplied to the surroundings is independent on how that energy came to be available. That’s important because the manner in which energy is transferred is path dependent.
There are two main methods energy can be transferred to or from a system. These are suggested in the previous statement of the first law of thermodynamics. Mathematically, we can restate the first law as
$\Delta U = q + w \nonumber$
or
$dU = dq + dw \nonumber$
where q is defined as the amount of energy that flows into a system in the form of heat and w is the amount of energy lost due to the system doing work on the surroundings.
Heat
Heat is the kind of energy that in the absence of other changes would have the effect of changing the temperature of the system. A process in which heat flows into a system is endothermic from the standpoint of the system ($q_{system} > 0$, $q_{surroundings} < 0$). Likewise, a process in which heat flows out of the system (into the surroundings) is called exothermic ($q_{system} < 0$, $q_{surroundings} > 0$). In the absence of any energy flow in the form or work, the flow of heat into or out of a system can be measured by a change in temperature. In cases where it is difficult to measure temperature changes of the system directly, the amount of heat energy transferred in a process can be measured using a change in temperature of the soundings. (This concept will be used later in the discussion of calorimetry).
An infinitesimal amount of heat flow into or out of a system can be related to a change in temperature by
$dq = C\, dT \nonumber$
where C is the heat capacity and has the definition
$C = \dfrac{dq}{\partial T} \nonumber$
Heat capacities generally have units of (J mol-1 K-1) and magnitudes equal to the number of J needed to raise the temperature of 1 mol of substance by 1 K. Similar to a heat capacity is a specific heat which is defined per unit mass rather than per mol. The specific heat of water, for example, has a value of 4.184 J g-1 K-1 (at constant pressure – a pathway distinction that will be discussed later.)
Example $1$: Heat required to Raise Temperature
How much energy is needed to raise the temperature of 5.0 g of water from 21.0 °C to 25.0 °C?
Solution
\begin{align*} q &=mC \Delta T \[4pt] &= (5.0 \,\cancel{g}) (4.184 \dfrac{J}{\cancel{g} \, \cancel{°C}}) (25.0 \cancel{°C} - 21.0 \cancel{°C}) \[4pt] &= 84\, J \end{align*}
What is a partial derivative?
A partial derivative, like a total derivative, is a slope. It gives a magnitude as to how quickly a function changes value when one of the dependent variables changes. Mathematically, a partial derivative is defined for a function $f(x_1,x_2, \dots x_n)$ by
$\left( \dfrac{ \partial f}{\partial x_i} \right)_{x_j \neq i} = \lim_{\Delta _i \rightarrow 0} \left( \dfrac{f(x_1+ \Delta x_1 , x_2 + \Delta x_2, \dots, x_i +\Delta x_i, \dots x_n+\Delta x_n) - f(x_1,x_2, \dots x_i, \dots x_n) }{\Delta x_i} \right) \nonumber$
Because it measures how much a function changes for a change in a given dependent variable, infinitesimal changes in the in the function can be described by
$df = \sum_i \left( \dfrac{\partial f}{\partial x_i} \right)_{x_j \neq i} \nonumber$
So that each contribution to the total change in the function $f$ can be considered separately.
For simplicity, consider an ideal gas. The pressure can be calculated for the gas using the ideal gas law. In this expression, pressure is a function of temperature and molar volume.
$p(V,T) = \dfrac{RT}{V} \nonumber$
The partial derivatives of p can be expressed in terms of $T$ and $V$ as well.
$\left( \dfrac{\partial p}{ \partial V} \right)_{T} = - \dfrac{RT}{V^2} \label{max1}$
and
$\left( \dfrac{\partial p}{ \partial T} \right)_{V} = \dfrac{R}{V} \label{max2}$
So that the change in pressure can be expressed
$dp = \left( \dfrac{\partial p}{ \partial V} \right)_{T} dV + \left( \dfrac{\partial p}{ \partial T} \right)_{V} dT \label{eq3}$
or by substituting Equations \ref{max1} and \ref{max2}
$dp = \left( - \dfrac{RT}{V^2} \right ) dV + \left( \dfrac{R}{V} \right) dT \nonumber$
Macroscopic changes can be expressed by integrating the individual pieces of Equation \ref{eq3} over appropriate intervals.
$\Delta p = \int_{V_1}^{V_2} \left( \dfrac{\partial p}{ \partial V} \right)_{T} dV + \int_{T_1}^{T_2} \left( \dfrac{\partial p}{ \partial T} \right)_{V} dT \nonumber$
This can be thought of as two consecutive changes. The first is an isothermal (constant temperature) expansion from $V_1$ to $V_2$ at $T_1$ and the second is an isochoric (constant volume) temperature change from $T_1$ to $T_2$ at $V_2$. For example, suppose one needs to calculate the change in pressure for an ideal gas expanding from 1.0 L/mol at 200 K to 3.0 L/mol at 400 K. The set up might look as follows.
$\Delta p = \underbrace{ \int_{V_1}^{V_2} \left( - \dfrac{RT}{V^2} \right ) dV}_{\text{isothermal expansion}} + \underbrace{ \int_{T_1}^{T_2}\left( \dfrac{R}{V} \right) dT}_{\text{isochoric heating}} \nonumber$
or
\begin{align*} \Delta p &= \int_{1.0 \,L/mol}^{3.0 \,L/mol} \left( - \dfrac{R( 400\,K)}{V^2} \right ) dV + \int_{200 \,K}^{400,\ K }\left( \dfrac{R}{1.0 \, L/mol} \right) dT \[4pt] &= \left[ \dfrac{R(200\,K)}{V} \right]_{ 1.0\, L/mol}^{3.0\, L/mol} + \left[ \dfrac{RT}{3.0 \, L/mol} \right]_{ 200\,K}^{400\,K} \[4pt] &= R \left[ \left( \dfrac{200\,K}{3.0\, L/mol} - \dfrac{200\,K}{1.0\, L/mol}\right) + \left( \dfrac{400\,K}{3.0\, L/mol} - \dfrac{200\,K}{3.0\, L/mol}\right) \right] \[4pt] &= -5.47 \, atm \end{align*}
Alternatively, one could calculate the change as an isochoric temperature change from $T_1$ to $T_2$ at $V_1$ followed by an isothermal expansion from $V_1$ to $V_2$ at $T_2$:
$\Delta p = \int_{T_1}^{T_2}\left( \dfrac{R}{V} \right) dT + \int_{V_1}^{V_2} \left( - \dfrac{RT}{V^2} \right ) dV \nonumber$
or
\begin{align*} \Delta p &= \int_{200 \,K}^{400,\ K }\left( \dfrac{R}{1.0 \, L/mol} \right) dT + \int_{1.0 \,L/mol}^{3.0 \,L/mol} \left( - \dfrac{R( 400\,K)}{V^2} \right ) dV \[4pt] &= \left[ \dfrac{RT}{1.0 \, L/mol} \right]_{ 200\,K}^{400\,K} + \left[ \dfrac{R(400\,K)}{V} \right]_{ 1.0\, L/mol}^{3.0\, L/mol} \[4pt] &= R \left[ \left( \dfrac{400\,K}{1.0\, L/mol} - \dfrac{200\,K}{1.0\, L/mol}\right) + \left( \dfrac{400\,K}{3.0\, L/mol} - \dfrac{400\,K}{1.0\, L/mol}\right) \right] \[4pt] &= -5.47 \, atm \end{align*}
This results demonstrates an important property of pressure in that pressure is a state variable, and so the calculation of changes in pressure do not depend on the pathway!
Work
Work can take several forms, such as expansion against a resisting pressure, extending length against a resisting tension (like stretching a rubber band), stretching a surface against a surface tension (like stretching a balloon as it inflates) or pushing electrons through a circuit against a resistance. The key to defining the work that flows in a process is to start with an infinitesimal amount of work defined by what is changing in the system.
Table 3.1.1: Changes to the System
Type of work Displacement Resistance dw
Expansion dV (volume) -pext (pressure) -pextdV
Electrical dQ (charge) W (resistence) -W dQ
Extension dL (length) -t (tension) t dL
Stretching dA -s (surf. tens.) sdA
The pattern followed is always an infinitesimal displacement multiplied by a resisting force. The total work can then be determined by integrating along the pathway the change follows.
Example $2$: Work from a Gas Expansion
What is the work done by 1.00 mol an ideal gas expanding from a volume of 22.4 L to a volume of 44.8 L against a constant external pressure of 0.500 atm?
Solution
$dw = -p_{ext} dV \nonumber$
since the pressure is constant, we can integrate easily to get total work
\begin{align*} w &= -p_{exp} \int_{V_1}^{V_2} dV \[4pt] &= -p_{exp} ( V_2-V_1) \[4pt] &= -(0.500 \,am)(44.8 \,L - 22.4 \,L) \left(\dfrac{8.314 \,J}{0.08206 \,atm\,L}\right) \[4pt] &= -1130 \,J = -1.14 \;kJ \end{align*}
Note: The ratio of gas law constants can be used to convert between atm∙L and J quite conveniently!
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/03%3A_The_First_Law_of_Thermodynamics/3.01%3A_Work_and_Heat.txt
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Let us turn our attention from the macroscopic to the microscopic level. According to the first law of thermodynamics, the heat energy absorbed as we raise the temperature of a substance cannot be destroyed. But where does it go? In the case of a monatomic gas, like neon, this question is easy to answer. All the energy absorbed is converted into the kinetic energy of the neon molecules (atoms). In other sections, we found that the kinetic energy of the molecules in a sample of gas is given by the expression
$E_{k}=\tfrac{\text{3}}{\text{2}}nRT \nonumber$
Thus if the temperature of a sample of neon gas is raised from T1 to T2, the kinetic energy of the molecules increases from 3/2 nRT1 to 3/2 nRT2, a total change of
$\tfrac{3}{2}nR\left( T_{2}-T_{1}\right) =\left( \tfrac{3}{2}R\right) n\left( T_{2}-T_{1}\right) \nonumber$
Inserting the value of R in appropriate units, we obtain
\begin{matrix}\frac{3}{2}\left(8.314\frac{\text{J}}{\text{K mol}}\right)n\left(T_{2}-T_{1}\right)=\left(12.47\frac{\text{J}}{\text{K mol}}\right)n\left(T_{2}-T_{1}\right)\end{matrix}
This is the same quantity that is obtained by substituting the experimental value of CV for neon (calculated in Example 2 from Heat Capacities) into Eq. (4) from Heat Capacities. In other words the quantity of heat found experimentally exactly matches the increase in kinetic energy of the molecules required by the kinetic theory of gases.
Table $1$ lists the CV values not only for neon but for some other gases as well. We immediately notice that only the noble gases and other mon-atomic gases such as Hg and Na have molar heat capacities equal to 3/2R, or 12.47 J K–1 mol–1. All other gases have higher molar heat capacities than this. Moreover, as the table shows, the more complex the molecule, the higher the molar heat capacity of the gas. There is a simple reason for this behavior.
Table $1$: Molar Heat Capacities at Constant Volume (CV) for Various Gases (Values at 298 K Unless Otherwise Stated).
Gas Cv /J K-1 mol-1 Gas Cv /J K-1 mol-1
Monoatomic Gases Triatomic Gases
Ne 12.47 CO2 28.81
Ar 12.47 N2O 30.50
Hg 12.47 (700K) SO2 31.56
Na 12.47 (1200K)
Diatomic Gases Alkanes
N2 20.81 CH4 27.42
O2 21.06 C2H6 44.32
Cl2 25.62 C3H8 65.20
C4H10 89.94
A molecule which has two or more atoms is not only capable of moving from one place to another (translational motion), it can also rotate about itself, and it can change its shape by vibrating. When we heat a mole of Cl2 molecules, for example, we not only need to supply them with enough energy to make them move around faster (increase their translational kinetic energy), we must also supply an additional quantity of energy to make them rotate and vibrate more strongly than before. For a mole of more complex molecules like n-butane even more energy is required since the molecule is capable of changing its shape in all sorts of ways. In the butane molecule there are three C—C bonds around which segments of the molecule can rotate freely. All the bonds can bend or stretch, and the whole molecule can rotate as well. Such a molecule is constantly flexing and writhing at room temperature. As we raise the temperature, this kind of movement occurs more rapidly and extra energy must be absorbed in order to make this possible.
When we heat solids and liquids, the situation is somewhat different than for gases. The rapid increase of vapor pressure with temperature makes it virtually impossible to heat a solid or liquid in a closed container, and so heat capacities are always measured at constant pressure rather than at constant volume. Some Cp values for selected simple liquids and solids at the melting point are shown in Table $2$. In general the heat capacities of solids and liquids are higher than those of gases. This is because of the intermolecular forces operating in solids and liquids. When we heat solids and liquids, we need to supply them with potential energy as well as kinetic energy. Among the solids, the heat capacities of the metals are easiest to explain since the solid consists of individual atoms. Each atom can only vibrate in three dimensions. According to a theory first suggested by Einstein, this vibrational energy has the value 3RT, while the heat capacity is given by 3R = 24.9 J K–1 mol–1.
Table $2$: Molar Heat Capacities at Constant Pressure Cp for Various Solids and Liquids at the Melting Point.
Substance Cp (solid)/J K-1 mol-1 Cp (liquid)/J K-1 mol-1
Monoatomic Substances
Hg 27.28 27.98
Pb 29.40 30.33
Na 28.20 31.51
Diatomic Substances
Br2 53.8 75.7
I2 54.5 80.7
HCl 50.5 62.2
HI 47.5 68.6
Polyatomic Substances
H2O 37.9 76.0
NH3 49.0 77.0
Benzene 129.0 131.0
n-Heptane 146.0 203.1
As can be seen from the table, most monatomic solids have Cp values slightly larger than this. This is because solids expand slightly on heating. The atoms get farther apart and thus increase in potential as well as vibrational energy.
Solids which contain molecules rather than atoms have much higher heat capacities than 3R. In addition to the vibration of the whole molecule about its site in the crystal lattice, the individual atoms can also vibrate with respect to each other. Occasionally molecules can rotate in the crystal, but usually rotation is only possible when the solid melts. As can be seen from the values for molecular liquids in Table $2$, this sudden ability to rotate causes a sharp increase in the heat capacity. For monatomic substances, where there is no motion corresponding to the rotation of atoms around each other, the heat capacity of the liquid is only very slightly higher than that of the solid.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/03%3A_The_First_Law_of_Thermodynamics/3.03%3A_Heat_Capacities.txt
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In Gas Expansion, we assume Ideal behavior for the two types of expansions:
Isothermal Expansion
This shows the expansion of gas at constant temperature against weight of an object's mass (m) on the piston. Temperature is held constant, therefore the change in energy is zero (U=0). So, the heat absorbed by the gas equals the work done by the ideal gas on its surroundings. Enthalpy change is also equal to zero because the change in energy zero and the pressure and volume is constant.
Isothermal Irreversible/Reversible process
The graphs clearly show work done (area under the curve) is greater in a reversible process.
Adiabatic Expansions
Adiabatic means when no heat exchange occurs during expansion between system and surrounding and the temperature is no longer held constant.
Reversible Adiabatic Expansion
This equation shows the relationship between PV and is useful only when it applies to ideal gas and reversible adiabatic change. The equation is very similar to Boyle's law except it has exponent (gamma) due to change in temperature. The work done by an adiabatic reversible process is given by the following equation:
where T2 is less than T1. The internal energy of the system decreases as the gas expands. The work can be calculated in two ways because the Internal energy (U) does not depend on path. The graph shows that less work is done in an adiabatic reversible process than an Isothermal reversible process.
3.05: Calorimetry
As chemists, we are concerned with chemical changes and reactions. The thermodynamics of chemical reactions can be very important in terms of controlling the production of desired products and preventing safety hazards such as explosions. As such, measuring and understanding the thermochemistry of chemical reactions is not only useful, but essential!
Calorimetry
The techniques of calorimetry can be used to measure q for a chemical reaction directly. The enthalpy change for a chemical reaction is of significant interest to chemists. An exothermic reaction will release heat ($q_{reaction} < 0$, $q_{surroundings} > 0$) causing the temperature of the surrounding to increase. Conversely, an endothermic reaction ($q_{reaction} > 0$, $q_{surroundings} < 0$) will draw heat from the surroundings, causing the temperature of the surrounding to drop. Measuring the temperature change in the surroundings allows for the determination of how much heat was released or absorbed in the reaction.
Bomb Calorimetry
Bomb calorimetry is used predominantly to measure the heat evolved in combustion reactions, but can be used for a wide variety of reactions. A typical bomb calorimetry set up is shown here. The reaction is contained in a heavy metallic container (the bomb) forcing the reaction to occur at constant volume. As such, the heat evolved (or absorbed) by the reaction is equal to the change in internal energy (DUrxn). The bomb is then submerged in a reproducible quantity of water, the temperature of which is monitored with a high-precision thermometer.
For combustion reactions, the bomb will be loaded with a small sample of the compound to be combusted, and then the bomb is filled with a high pressure (typically about 10 atm) of O2. The reaction is initiated by supplying heat using a short piece of resistive wire carrying an electrical current.
The calorimeter must be calibrated by carrying out a reaction for which $\Delta U_{rxn}$ is well known, so that the resulting temperature change can be related to the amount of heat released or absorbed. A commonly used reaction is the combustion of benzoic acid. This makes a good choice since benzoic acid reacts reliably and reproducibly under normal bomb calorimetry conditions. The “water equivalent” of the calorimeter can then be calculated from the temperature change using the following relationship:
$W = \dfrac{n\Delta U_c +e_{wrire}+e_{other}}{\Delta T} \nonumber$
where n is the number of moles of benzoic acid used, $\Delta U_c$ is the internal energy of combustion for benzoic acid (3225.7 kJ mol-1 at 25 oC), $e_{wire}$ accounts for the energy released in the combustion of the fuse wire, eother account for any other corrections (such as heat released due to the combustion of residual nitrogen in the bomb), and DT is the measured temperature change in the surrounding water bath.
Once the “water equivalent” is determined for a calorimeter, the temperature change can be used to find $\Delta U_c$ for an unknown compound from the temperature change created upon combustion of a known quantity of the substance.
$\Delta U_c = \dfrac{W \Delta T - e_{wire} - e_{other}}{n_{sample}} \nonumber$
The experiment above is known as “isothermal bomb calorimetry” as the entire assembly sits in a constant temperature laboratory. Another approach is to employ “adiabatic bomb calorimetry” in which the assembly sits inside of a water jacket, the temperature of which is controlled to match the temperature of the water inside the insulated container. By matching this temperature, there is no thermal gradient, and thus no heat leaks into or out of the assembly during an experiment (and hence the experiment is effectively “adiabatic”).
Finding $\Delta U_c$
The enthalpy of combustion can be calculated from the internal energy change if the balanced chemical reaction is known. Recall from the definition of enthalpy
$\Delta H = \Delta U + \Delta (pV) \nonumber$
and if the gas-phase reactants and products can be treated as ideal gases ($pV = nRT$)
$\Delta H = \Delta U + RT \Delta n_{gas} \nonumber$
at constant temperature. For the combustion of benzoic acid at 25 oC
$\ce{C6H5COOH (s) + 15/2 O_2(g) -> 7 CO2(g) + 3 H2O(l)} \nonumber$
it can be seen that $\Delta n_{gas}$ is -0.5 mol of gas for every mole of benzoic acid reacted.
Example $1$: Combustion of Naphthalene
A student burned a 0.7842 g sample of benzoic acid ($\ce{C7H6O2}$) in a bomb calorimeter initially at 25.0 oC and saw a temperature increase of 2.02 oC. She then burned a 0.5348 g sample of naphthalene ($\ce{C10H8}$) (again from an initial temperature of 25 oC) and saw a temperature increase of 2.24 oC. From this data, calculate $\Delta H_c$ for naphthalene (assuming ewire and eother are unimportant.)
Solution
First, the water equivalent:
$W = \dfrac{\left[ (0.7841\,g) \left(\frac{1\,mol}{122.124 \, g} \right)\right] (3225.7 \,kJ/mol)}{2.02 \,°C} = 10.254 \, kJ/°C \nonumber$
Then $\Delta U_c$ for the sample:
$\Delta U_c = \dfrac{(10.254\, kJ/\,°C)(2.24\,°C )}{(0.5308 \,g)\left(\frac{1\,mol}{128.174 \, g} \right) } = 5546.4 \, kJ/°C \nonumber$
$\Delta H_c$ is then given by
$\Delta H_c = \Delta U_c + RT \Delta n_{gas} \nonumber$
The reaction for the combustion of naphthalene at 25 oC is:
$\ce{ C10H8(s) + 12 O2(g) -> 10 CO2(g) + 4 H2O(l)} \nonumber$
with $\Delta n_{gas} = -2$.
So
$\Delta H_c = 5546.4 \,kJ/mol + \left( \dfrac{8.314}{1000} kJ/(mol \, K) \right) (298 \,L) (-2) = 5541\, kJ/mol \nonumber$
The literature value (Balcan, Arzik, & Altunata, 1996) is 5150.09 kJ/mol. So that’s not too far off!
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The standard enthalpy of formation is defined as the change in enthalpy when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements under the same conditions.
Introduction
The standard enthalpy of formation is a measure of the energy released or consumed when one mole of a substance is created under standard conditions from its pure elements. The symbol of the standard enthalpy of formation is ΔHf.
• Δ = A change in enthalpy
• o = A degree signifies that it's a standard enthalpy change.
• f = The f indicates that the substance is formed from its elements
The equation for the standard enthalpy change of formation (originating from Enthalpy's being a State Function), shown below, is commonly used:
$\Delta H_{reaction}^o = \sum {\Delta H_{f}^o(products)} - \sum {\Delta H_{f}^o(Reactants)}$
This equation essentially states that the standard enthalpy change of formation is equal to the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants.
Example $1$
Given a simple chemical equation with the variables A, B and C representing different compounds:
$A + B \leftrightharpoons C$
and the standard enthalpy of formation values:
• ΔHfo[A] = 433 KJ/mol
• ΔHfo[B] = -256 KJ/mol
• ΔHfo[C] = 523 KJ/mol
the equation for the standard enthalpy change of formation is as follows:
ΔHreactiono = ΔHfo[C] - (ΔHfo[A] + ΔHfo[B])
ΔHreactiono = (1 mol)(523 kJ/mol) - ((1 mol)(433 kJ/mol) + (1 mol)(-256 kJ/mol)\)
Because there is one mole each of A, B and C, the standard enthalpy of formation of each reactant and product is multiplied by 1 mole, which eliminates the mol denominator:
ΔHreactiono = 346 kJ
The result is 346 kJ, which is the standard enthalpy change of formation for the creation of variable "C".
The standard enthalpy of formation of a pure element is in its reference form its standard enthalpy formation is zero.
Carbon naturally exists as graphite and diamond. The enthalpy difference between graphite and diamond is too large for both to have a standard enthalpy of formation of zero. To determine which form is zero, the more stable form of carbon is chosen. This is also the form with the lowest enthalpy, so graphite has a standard enthalpy of formation equal to zero. Table 1 provides sample values of standard enthalpies of formation of various compounds.
Table 1: Sample Table of Standard Enthalpy of Formation Values. Table T1 is a more comprehensive table.
Compound ΔHfo
O2(g) 0 kJ/mol
C(graphite) 0 kJ/mol
CO(g) -110.5 kJ/mol
CO2(g) -393.5 kJ/mol
H2(g) 0 kJ/mol
H2O(g) -241.8 kJ/mol
HF(g) -271.1 kJ/mol
NO(g) 90.25 kJ/mol
NO2(g) 33.18 kJ/mol
N2O4(g) 9.16 kJ/mol
SO2(g) -296.8 kJ/mol
SO3(g) -395.7 kJ/mol
All values have units of kJ/mol and physical conditions of 298.15 K and 1 atm, referred to as the "standard state." These are the conditions under which values of standard enthalpies of formation are typically given. Note that while the majority of the values of standard enthalpies of formation are exothermic, or negative, there are a few compounds such as NO(g) and N2O4(g) that actually require energy from its surroundings during its formation; these endothermic compounds are generally unstable.
Example $2$
Between Br2(l) and Br2(g) at 298.15 K, which substance has a nonzero standard enthalpy of formation?
Solution
Br2(l) is the more stable form, which means it has the lower enthalpy; thus, Br2(l) has ΔHf = 0. Consequently, Br2(g) has a nonzero standard enthalpy of formation.
Note: that the element phosphorus is a unique case. The reference form in phosphorus is not the most stable form, red phosphorus, but the less stable form, white phosphorus.
Recall that standard enthalpies of formation can be either positive or negative.
Example $3$
The enthalpy of formation of carbon dioxide at 298.15K is ΔHf = -393.5 kJ/mol CO2(g). Write the chemical equation for the formation of CO2.
Solution
This equation must be written for one mole of CO2(g). In this case, the reference forms of the constituent elements are O2(g) and graphite for carbon.
$O_{2}(g) + C(graphite) \rightleftharpoons CO_{2}(g)$
The general equation for the standard enthalpy change of formation is given below:
$\Delta H_{reaction}^o = \sum {\Delta H_{f}^o(products)} - \sum {\Delta H_{f}^o(Reactants)}$
Plugging in the equation for the formation of CO2 gives the following:
ΔHreactiono= ΔHfo[CO2(g)] - (ΔHfo[O2(g)] + ΔHfo[C(graphite)]
Because O2(g) and C(graphite) are in their most elementally stable forms, they each have a standard enthalpy of formation equal to 0:
ΔHreactiono= -393.5 kJ = ΔHfo[CO2(g)] - ((1 mol)(0 kJ/mol) + (1 mol)(0 kJ/mol))
ΔHfo[CO2(g)]= -393.5 kJ
Example $4$
Using the values in the above table of standard enthalpies of formation, calculate the ΔHreactiono for the formation of NO2(g).
Solution
$NO_{2(g)}$ is formed from the combination of $NO_{(g)}$ and $O_{2(g)}$ in the following reaction:
$2NO(g) + O_{2}(g) \leftrightharpoons 2NO_{2}(g)$
To find the ΔHreactiono, use the formula for the standard enthalpy change of formation:
$\Delta H_{reaction}^o = \sum {\Delta H_{f}^o(products)} - \sum {\Delta H_{f}^o(Reactants)}$
The relevant standard enthalpy of formation values from Table 1 are:
• O2(g): 0 kJ/mol
• NO(g): 90.25 kJ/mol
• NO2(g): 33.18 kJ/mol
Plugging these values into the formula above gives the following:
$ΔH_{reaction}^o= (2 \cancel{mol})(33.18\; kJ/\cancel{mol}) - \left[(2 \cancel{mol})(90.25\ kJ/\cancel{mol}) + (1 \cancel{mol})(0\; kJ/\cancel{mol})\right]$
$ΔH_{reaction}^o = -114.1\; kJ$
Kirchhoff's Law describes the enthalpy of a reaction's variation with temperature changes. In general, enthalpy of any substance increases with temperature, which means both the products and the reactants' enthalpies increase. The overall enthalpy of the reaction will change if the increase in the enthalpy of products and reactants is different.
Kirchoff's Law - Enthalpy is Temperature Dependent
At constant pressure, the heat capacity is equal to change in enthalpy divided by the change in temperature.
$c_p = \dfrac{\Delta H}{\Delta T} \label{1}$
Therefore, if the heat capacities do not vary with temperature then the change in enthalpy is a function of the difference in temperature and heat capacities. The amount that the enthalpy changes by is proportional to the product of temperature change and change in heat capacities of products and reactants. A weighted sum is used to calculate the change in heat capacity to incorporate the ratio of the molecules involved since all molecules have different heat capacities at different states.
$H_{T_f}=H_{T_i}+\int_{T_i}^{T_f} c_{p} dT \label{2}$
If the heat capacity is temperature independent over the temperature range, then Equation \ref{1} can be approximated as
$H_{T_f}=H_{T_i}+ c_{p} (T_{f}-T_{i}) \label{3}$
with
• $c_{p}$ is the (assumed constant) heat capacity and
• $H_{T_{i}}$ and $H_{T_{f}}$ are the enthalpy at the respective temperatures.
Equation \ref{3} can only be applied to small temperature changes, (<100 K) because over a larger temperature change, the heat capacity is not constant. There are many biochemical applications because it allows us to predict enthalpy changes at other temperatures by using standard enthalpy data.
Contributors and Attributions
• Janki Patel (UCD), Kostia Malley (UCD), Jonathan Nguyen (UCD), Garrett Larimer (UCD)
3.07: Bond Energies and Enthalpies
In the absence of standard formation enthalpies, reaction enthalpies can be estimated using average bond enthalpies. This method is not perfect, but it can be used to get ball-park estimates when more detailed data is not available. A bond dissociation energy $D$ is defined by
$XY(g) \rightarrow X(g) + Y(g)$
with $\Delta H \equiv D(X-Y)$
In this process, one adds energy to the reaction to break bonds, and extracts energy for the bonds that are formed.
$\Delta H_{rxn} = \sum (\text{bonds broken}) - \sum (\text{bonds formed})$
As an example, consider the combustion of ethanol:
In this reaction, five C-H bonds, one C-C bond, and one C-O bond, and one O=O bond must be broken. Also, four C=O bonds, and one O-H bond are formed.
Bond Average Bond Energy (kJ/mol)
C-H 413
C-C 348
C-O 358
O=O 495
C=O 799
O-H 463
The reaction enthalpy is then given by
\begin{align} \Delta H_c = \, &5(413 \,kJ/mol) + 1(348\, kJ/mol) + 1(358 \,kJ/mol) \nonumber \ & + 1(495\, kJ/mol) - 4(799 \,kJ/mol) – 2(463\, kJ/mol) \nonumber \ =\,& -856\, kJ/mol \end{align}
Because the bond energies are defined for gas-phase reactants and products, this method does not account for the enthalpy change of condensation to form liquids or solids, and so the result may be off systematically due to these differences. Also, since the bond enthalpies are averaged over a large number of molecules containing the particular type of bond, the results may deviate due to the variance in the actual bond enthalpy in the specific molecule under consideration. Typically, reaction enthalpies derived by this method are only reliable to within ± 5-10%.
Contributors and Attributions
• Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/03%3A_The_First_Law_of_Thermodynamics/3.06%3A_Thermochemistry.txt
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Q3.6
Calculate the work done by the reaction
$\ce{Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)}$
when 1 mole of hydrogen gas is collected at 273 K and 1.0 atm. (Neglect volume changes other than the change in gas volume, and assume that the has behaves ideally.)
Q3.11a
If the internal energy of 3 moles of Cl2 is 3.96 KJ, what is the temperature of the system?
Q3.11b
The internal energy of 3 moles of neon gas (Ne) is 4.87 kJ. Assuming conditions are ideal, what is the temperature of the system?
S3.11b
Since we can assume that conditions are ideal, we know that the temperature is 25°C or 298 K. This assumption allows us to ignore the small change in volume of the gas.
Q3.12a
Some ice cubes are placed inside a sealed plastic bag on a hot day, what is the sign of $w$, $q$, $ΔH$, $ΔU$? What does this mean about the system?
Q3.12b
A sports bottle containing Gatorade is held in a man's hand while running on a treadmill. Suppose the Gatorade is the system in question. Answer the subsequent questions.
1. Will the temperature rise due to the movement caused by the man's running?
2. Was any heat actually added to the system? If so, what was the source?
3. Was work done onto the system?
4. In what ways has the internal energy of the system been affected, if affected at all?
S3.12b
1. The man's running motion causes the Gatorade to be shaken. The shaking motion adds energy to the system and increases the random motion of the molecules. Since the motion of the molecules rises, so does the temperature of the system.
2. Yes, heat exchange did occur between the system and its surroundings (the heat given off from the man's hand).
3. Yes, work was done onto the system via the shaking motion.
4. The internal energy has been increased.
Q3.12c
Consider an ice cube melting.
1. Is the change of enthalpy positive or negative for this process?
2. Is the change of entropy positive or negative for this process?
S3.12c
1. Enthalpy is positive because the ice must gain heat in order to melt.
2. Entropy is also positive because the water molecules are gaining kinetic energy and moving to a more disordered state.
Q3.13
2.5 moles of CO2 expands against an external pressure of 1.5 atm. It was initially at 9.0 atm and 30ºC. The final volume is 35.0 L.
1. What is the final temperature?
2. Calculate q, w, and $\Delta U$ for the process.
S3.13
a) $T=\dfrac{PV}{Rn}= \dfrac{(1.5atm)(35L)}{(.08206\dfrac{L\cdot atm}{mol\cdot K})(2.5 mol)}= 255K$
b) $w=-(nRT)ln\dfrac{P_1}{P_2}=-(2.5mol)(8.314\dfrac{J}{mol\cdot K})(255K)ln\dfrac{9.0atm}{1.5atm}=-9.5\dfrac{kJ}{mol}$
$q_p=nCpT= n\dfrac{3}{2}RT= (2.5mol)\dfrac{3}{2}(8.314\dfrac{J}{mol\cdot K})(255K)= 7.95\dfrac{kJ}{mol}$
$\Delta U=q+w= 7.95\dfrac{kJ}{mol} + -9.5\dfrac{kJ}{mol}= -1.55\dfrac{kJ}{mol}$
- For calculating work, expression for Isothermal reversible process has been used instead of correct -Pext*dV as it is a irreversible expansion against external pressure. -Wrong method for calculating q. Instead Internal energy would be calculated using nCv*dT.
Q3.17
Determine the at 343.15 K for . (Note: The heat of vaporization for water is known to be 40.79 kJ mol-1 at 273.15 K, the of was determined to be 85.6 J K-1 mol-1 and the of was determined to be 54.35 J K-1 mol-1). Use the reaction below and give your final answer in kJ mol-1.
Q3.18
,Is $(q= -w$ true for a cyclic process?
S3.18
Yes, $q= -w$ is a true statement. It is true because $\Delta U$ for any cyclic process is equal to zero.
Q3.22
A 10.0 g sheet of gold with a temperature of 18.0°C is laid flat on a sheet of iron that weights 20.0 g and has a temperature of 55.6°C. Given that the specific heats of Au and Fe are 0.129 J g-1 °C-1 and 0.444 J g-1 °C-1, respectively, what is the final temperature of the combined metals? (Hint: The heat gained by the gold must be equal to the heat lost by the iron.)
Q3.23
You are cooking 100 grams of chicken soup at 300K. You know that it will take 500 J to raise the temperature of your soup to the desired 350K. Assuming constant pressure, calculate the molar heat capacity of this chicken soup.
Q3.24
Why does the molar heat of vaporization for water decrease as temperature increases?
Q3.24
The value of ∆H for heating 1 mole of hexane is 28.9kJ/mole at 342 K and 25.6 kJ/mole at 298 K. Explain why those values are different.
Q3.25
What is ΔH if of 3.00 moles of O2(g) are cooled from 175ºC to 35.0ºC and $\bar{C_{p}} = (7.50 \times 10^{-3}T - 2.30 \times 10^{-6}T^{2} J mol^{-1} K^{-1}$
S3.25
$dH = n\bar{C_{p}}dT$
$\Delta H = (3 mol) \int_{448K}^{308K}(7.50 \times 10^{-3}T - 2.30 \times 10^{-6}T^{2}) J mol^{-1} K^{-1}dT$
$\Delta H = (3 mol) (\dfrac{7.50 \times 10^{-3}}{2}T^{2} - \dfrac{2.30 \times 10^{-6}}{3}T^{3} J mol^{-1} K^{-1} | _{448K}^{308K})$
$\Delta H = (3 mol) [(3.75 \times 10^{-3}(308K)^{2}-7.67 \times 10^{-7}(308K)^{3})-(3.75 \times 10^{-3}(448K)^{2}-7.67 \times 10^{-7}(448K)^{3})]J mol^{-1} K^{-1}$
$\Delta H = (3 mol) [(333.3)-(683.7)]J mol^{-1}$
$\Delta H = -1050 J$
Q3.25
For a constant-pressure process, the heat capacity per mole is given by the expression:
Cp per mole = (16.0 + 4.35 x 10-3 T + 1.15 T2) J K-1 mol-1
Find the molar heat capacity of oxygen gas from 38°C to 115oC
S3.25
\begin{align} \Delta H &=\int_{38}^{115}(16+4.35 \times 10^{-3}T+1.15T^{2})dT \ &= (16*115+(4.35 \times 10^{-3})/2 * 115^2 + 1.15/3 *115^3) - (16*38 +(4.35 \times 10^{-3})/2 *38^2 +1.15/3 * 38^3) \ &= 5.63 J/mol \end{align}
Q3.26
What is the heat capacity ratio ($γ$) if of an ideal gas if its $C_V = 42\, J\, K^{-1} mol^{-1}$?
S3.26
since $\overline{C_{P}} = \overline{C_{V}} + R$
\begin{align} \gamma = \dfrac{\overline{C_{P}}}{\overline{C_{V}}} &= \dfrac{\overline{C_{V}}+R}{\overline{C_{V}}} \ \gamma &= \dfrac{(42+8.314) J K^{-1} mol^{-1}}{42J K^{-1} mol^{-1}} \ \gamma &= 1.2 \end{align}
Q3.29
Explain (in terms of thermo chemistry) why Emperor Penguins huddle together in the Antarctic.
Q3.30a
Through laboratory experiment, a group of students found out that molar mass of a solid element can be calculated in term of the specific heat capacity :
Molar mass x the specific heat of metals= 25 (Eq.1)
They were also obtaining the values of the three metals, Aluminum (0.900 J/g oC), Zinc (0.381 J/g oC), and Arsenic (0.329 J/g oC).
Determine which metal violate the above equation (Eq.1)
S3.30a
Aluminum (Al) can not be applied in this equation because the specific heat capacity of Al is approximately equal 1.0 J g-1 (oC-1)
Q3.30b
What is the molar heat capacity of lead given its specific heat is 0.128J/(gºC) at 25ºC?
S3.30b
The first thing we do is find the molar mass of lead which is 207.2 g/mol.
We multiply our molar mass by the specific heat to get the molar heat capacity.
Q3.31
What is the total work done on 2.00 moles of an ideal monatomic gas at 3.00 atm and 4.00 L when it adiabatically expands to 7.00 L?
S3.31
The equation for work is
$w = n \bar{C_{V}} (T_{2} - T_{1})$
For a monatomic gas,
$\bar{C_{V}} = \dfrac{3}{2}R \ and \ \bar{C_{P}} = \dfrac{5}{2}R$
$\gamma = \dfrac{\bar{C_{P}}}{\bar{C_{V}}} = 1.67$
Solve for T1:
$PV=nRT$
$T_{1}=\dfrac{PV}{nR}=\dfrac{(3 atm)(4L)}{(2 moles)(0.08206 \dfrac{L \ atm}{mol \ K})}$
$T_{1} = 73.12 K$
Solve for T2:
$P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}$
$P_{2}=\dfrac{(3atm)(4L)^{1.67}}{(7L)^{1.67}}=1.178atm$
$T_{2}=\dfrac{PV}{nR}=\dfrac{(1.178 atm)(7L)}{(2 moles)(0.08206 \dfrac{L \ atm}{mol \ K})}$
$T_{2}=50.24K$
Solve for work:
$w = n \bar{C_{V}} (T_{2} - T_{1})$
$w = (2 mol)(\dfrac{3}{2})(8.314 \dfrac{J}{mol \ K})(50.24K - 73.12 K)$
$w = -571 J$
Q3.31
You notice your system changes volume from 10 L to 30 L with a constant pressure at 770 torr. Then, there is a decrease in pressure to 740 torr with the volume staying constant at 30 L. Calculate the total amount of work done.
S3.31
We split up the solution into two parts.
Work done during the volume change and the work done during the pressure change.
First part:
Second part:
Convert pressure to atm:
=107.4 J
Then add up the two works
wtot= -2046.3+107.4=-1938.9 J
Q3.33
Calculate the values of q, w, ΔU, and ΔH for the reversible adiabatic expansion of 1 mole of a diatomic ideal gas from 10.00 m3 to 35.0 m3 with an initial temperature of 298 K.
Q3.33
Calculate the values of q, w, ΔU, and ΔH for the reversible adiabatic expansion of 1 mole of a diatomic ideal gas from 10.00 m3 to 40.0 m3 with an initial temperature of 298K.
Q3.34
A quantity of 0.27 mole of argon is confined in a container at 3.0 atm and 298 K and then allowed to expand adiabatically under two different conditions: (a) reversibly to 0.80 atm and (b) against a constant pressure of 0.80 atm. Calculate the final temperature in each case.
Q3.34
In an adiabatic process 0.5 moles of Hein a container are allowed to expand against a constant pressure of 1.00 atm and 298 K. The final temperature after the gas has expanded is 205 K. Calculate the initial pressure inside the container.
S3.34
In an adiabatic process
$C_{V}=(T_{2}-T_{1})=-P_{ex}(V_{2}-V_{1})$
we get
$\dfrac{3}{2}(T_{2}-T_{1})=-P_{ex}\left ( \dfrac{T_{2}}{P_{2}}-\dfrac{T_{1}}{P_{1}} \right )$
we rearrange and get
$\dfrac{3}{2}(T_{2}-T_{1})=\dfrac{P_{ex}T_{1}}{P_{1}}-T_2$
we solve for
$P_{1}$ and get $P_{1}=\dfrac{P_{2}}{\dfrac{T_{2}}{T_{1}}(\dfrac{5}{2})-\dfrac{3}{2}}$
$P_{1}=\dfrac{1.00atm}{\dfrac{205K}{298K}(\dfrac{5}{2})-\dfrac{3}{2}}$
$P_{1}=4.5 atm$
Q3.34
A quantity of 0.27 mole of argon is confined in a container at 3.0 atm and 298 K and then allowed to expand adiabatically under two different conditions:
1. reversibly to 0.80 atm and
2. against a constant pressure of 0.80 atm.
Calculate the final temperature in each case.
Q3.35a
Two moles of ideal diatomic gas $H_2$ expand from 2.00 atm to 1.30 atm while being cooled from 550 K to 250 K. Compute the Change of Enthalpy (∆H) and Change of Internal Energy (∆U) for 2 different processes: a. reversible and b. irreversible. Explain and calculate the difference a and b?
Q3.35b
0.5 mole of Helium at 298 K are allowed to expand reversibly from 5 atm to 1 atm. Find q, w and ΔU if
1. the process is adiabatic and
2. if the process is isothermal.
Q3.36a
808 kJ/mol of heat is released by burning x (gram) amount of Iron in a constant volume bomb calorimeter. The calorimeter containing 100 g water inside has a heat capacity of 2100 JoC-1. Evaluate the amount of Iron needed to increase the water by 5.20C?
Q3.36b
In your constant-volume bomb calorimeter you have a 0.3677 g sample of Potassium. The water surrounding the sample increase from 18.08 ºC to 20.667 ºC. The calorimeter has a heat capacity of 1515 J/(gºC). What is the change in internal energy?
S3.36b
Since the bomb calorimeter has a constant volume $\Delta V$, there is no (PV) work so the first law of thermodynamics
$\Delta U =q +w$
can be simplified to
$\Delta U = q$
Then the relationship between heat and temperature change
$q = c_v \Delta T$
can be changed
$\Delta U = m \,c_v\, \Delta T$
which for this case is
$\Delta U= ( 0.3677\;g ) (1,515\; J/(gºC)) (20.667 \;ºC -18.08 \; ºC)$
$\Delta U =1411.1$
Q3.37
The enthalpy of combustion for benzoic acid ($C_6H_5COOH$) is commonly used as the standard $\Delta H_{combustion} = -3226.7\,kJ\,mol^{-1}$, for calibrating bomb calorimeters at a constant-volume process.
1. When 1.567 g of benzoic acid burned in a calorimeter the temperature increased by $3.678^{o}C$. Calculate the heat capacity of the calorimeter.
2. For a different experiment 0.6792g of ethanol was burned in the calorimeter which rose from 16.42$^{o}$C to 26.58$^{o}$C. Calculate the enthalpy of combustion for ethanol, the value of $\Delta rU$ for the combustion, and the molar enthalpy of formation.
S3.37
(a)
$q_{calorimeter}=-q_{benzoic\,acid}$
$q_{calorimeter}(3.678^{o}C)=-(\dfrac{1.567g}{122.1g\,mol{-1}})(-3226.7kj\,mol^{-1})$
$c_{calorimeter}=\dfrac{41.41kj}{3.678^{o}C}=11.26kj^{o}C^{-1}$
(b) (Hint: Calorimeter absorbs heat transfered from the combustion of maltose)
$q_{calorimeter}=-q{maltose}$
Step 1: Calculate $\Delta rU$
$(11.26kj\,^{o}C^{-1})(26.58^{o}C-16.42^{o}C)=-(\dfrac{.6792g}{46.06844gmol^{-1}})(\Delta rU^{o})$
$\Delta\,rU^{o}=-\dfrac{114.4016kj}{.014743282mol}=-7.759\times10^{3} jk\, mol^{-1}$
Step 2: Calculation of enthalpy of combustion
$C_2H5OH_{(s)}+3O_{2\,(g)}\rightarrow2CO_{2(g)}+3H_2O_{(l)}$
$\Delta n=-1$
Therefore,
$\Delta rH^{o}=\Delta rU^{o}+RT\Delta n=(-7.759\times10^{3}kj\,mol^-1)+(8.3146\times10^{-3}kj\,mol^{-1}K^{-1})(298.15K)(-1)=-7.761\times10^{3}kj\,mol^{-1}$
Step 3: Calculate the molar enthalpy of formation.
$\Delta rH=2\Delta _f\bar{H}^{o}[CO_{2(g)}]+3\Delta _f\bar{H}^{o}[H_2O_{(l)}]-\Delta _f\bar{H}^{o}[C_2H_5OH_{(s)}]-3\Delta _f\bar{H}^{o}[O_{2(g)}]$
$=2(-393.5kj\,mol^{-1})+3(-285.8kj\,mol^{-1})-3(0kj\,mol^{-1})-(-7.761\times10^{3}kj\,mol^{-1})$
$=6.116\times10^{3}kj\,mol^{-1}$
Q3.38a
A quantity of 2.50 x 102 mL of 0.468 M HBr is mixed with 2.50 x 102 mL of 0.425 M KOH in a constant-pressure calorimeter that has a heat capacity of 437 J °C-1. The initial temperature if the HBr and KOH solutions is the same at 21.35°C. For the process
$H^+(aq) + OH^-(aq) \rightleftharpoons H_2O(l)$
the heat of neutralization is -56.2 kJ mol-1. What is the final temperature of the mixed solution?
Q3.38b
In a constant-pressure calorimeter with a heat capacity of 395$J\,^{o}C^{-1}$, you mix $1.5\times10^{3}\,mL$ of HBr at 0.788M and $1.5\times10^{3}mL$ $Ca(OH)_2$ at .394M. For this process to occur the heat of neutralization is $-56.2kj\,mol^{-1}$. The initial temperature of both solutions is $23.08^{o}C$.
$H^+_{(aq)}+OH^-_{(aq)}\rightarrow H_2O_{(l)}$
Find the final temperature of the mixed solution.
S3.38b
Step 1: (Hint: Determine the number of moles of the reactants)
$n_{H^{+}}=(1.5L)(0.788\,mol\,L^{-1})=1.182\,mol$
$n_{OH^{-}}=2(1.5L)(0.394\,mol\,L^{-1})=1.182\,mol$
Step 2: (Hint: Calculate the the thermal energy released by the reaction)
$q_P=(-52.2\,kJ\,mol^{-1})(1.182\,mol)=-66.42\, kJ$
Step 3: (Hint: Assume the the density of the solutions are the same as water and specific heat is same as water too.)
$s_{solution}=4.184\,J\, g^{-1}^{o}C^{-1}$
$m_{solution}=m_{HBr}+m_{Ca(OH)_2}$
$=(1.5 \times 10^{3}mL)(1.00gmL^{-1})+(1.5 \times 10^{3}mL)(1.00gmL^{-1})=3.00 \times 10^3g$
Step 4:
$66.4284\, kJ=395J^{o}C^{-1}+(3.00 \times 10^{3}g)(4.184\, J\, g^{-1}\,^{o}C^{-1})(\dfrac{1kJ}{1000\,J})\Delta T$
$66.4284=(12.947^{o}C^{-1})\Delta T$
$\Delta T=5.13^{o}C$
Step 5:
$\Delta T=T_f-T_i$
$T_f=28.21^{o}C$
Q3.38
A quantity of $2.50 \times 10^2\; mL$ of $0.468;\; M \;HBr$ is mixed with $2.50 \times 10^2\; mL$ of $0.425\; M\; KOH$ in a constant-pressure calorimeter that has a heat capacity of 437 J °C-1. The initial temperatures of the $HBr$ and $KOH$ solutions are the same at 21.35 °C. The heat of neutralization for the below reaction is $-56.2\; kJ/mol$
$H^+_{(aq)} + OH^-_{(aq)} \rightarrow H_2O_{(l)}$
What is the final temperature of the mixed solution?
Q3.39
1.52 g of toluene $(C_7H_8)$ is combusted in a constant volume bomb calorimeter at 25°C. Calculate the $\Delta _r\bar{U}$ and $\Delta _r\bar{H}$ for this reaction given that 39.183 kJ of heat was released.
S3.39
First we need to construct the balanced reaction that occurs
$C_7H_8(l)+9O_2(g)\rightarrow7CO_2(g)+4H_2O(l)$
$\Delta _r\bar{U}=q_r=-q_{cal}=\dfrac{-39.183kJ}{\dfrac{1.52g}{92.14(g/mol)}}=-2375\; kJ/mol$
\begin{align*} \Delta _r\bar{H}&=\Delta _r\bar{U}+\Delta (PV)=\Delta _r\bar{U}+\Delta n(RT)\&=-2375 \; kJ/mol + (-2)(8.314 \times 10^{-3}\; kJ/K*mol)(298.15\; K)=-2380 \; kJ/mol \end{align*}
Q3.39
The of ΔU a of the combustion of naphthalene (C10H8) are -480 kj mol-1. The heat evolved from the combustion in a constant-volume calorimeter is 7.5 KJ. Calculate the mass of naphthalene in grams in the calorimeter.
Q3.40
For the reaction
$C_2H_5OH_{(l)} + O_{2 (g)} \rightarrow CO_{2 (g)} + H_2O_{(l)}$
1. Balance the reaction
2. Calculate the $∆_rH$ of reaction using the $∆_fH$ data
Q3.40B
Consider the following reaction:
\begin{align*}C_2H_6(g)+\dfrac{7}{2}O_2(g)\rightarrow &2CO_2(g)+3H_2O(l)\ &\Delta_rH^\circ=-1560kJ/mol\end{align*}
What is the value of $\Delta_rH^\circ$ if
1. the equation is multiplied throughout by 4;
2. the direction of the reaction is reversed;
3. $C_2H_6$ was a liquid instead of gas?
S3.40B
1. $\Delta_rH^\circ=4(-1560kJ/mol)=-6240kJ/mol$;
2. $\Delta_rH^\circ=1560kJ/mol$;
3. Ethane exist as a liquid at temperatures of below 184.6 K. If we used ethane liquid instead of ethane gas, then according to $\Delta H=\Delta U+\Delta n(RT)$ the number of moles would change and the temperature of the reaction would also change.
Q3.41a
On what basis is the 0 value for standard enthalpy of formation assigned to a substance at a specific temperature? Base on that, identify which substances have ΔfH° = 0 and which do not at 1 atm and 25°C (give explanation for those that do not).
H2(g), O3(g), Al(s), Br2(g), NaCl(s), Cl2(g)
Hint: Look at standard enthalpy of formation
S3.41a
0 value for standard enthalpy of formation is assigned to elemental substances that are in their natural/standard physical state at that specific condition of temperature and pressure.
1. H2(g), Al(s), Cl2(g) are substances with ΔfH° = 0 kJ/mol
2. O3(g) is an elemental substance but does not occur naturally at 25°C, O2(g) is.
3. Br2(g) is also not the standard state of bromine. At STP, the physical state of bromine is liquid
4. NaCl does not have a 0 value because it is not an elemental substance
Q3.41b
Which of the following substances has a standard enthalpy of formation, at 298 K, of 0 K?
O2(g), Br2(g), Hg(s), CH4(g)
Q3.41c
Why is the standard enthalpy of formation of O2 zero at a temperature of 298 K?
S3.41c
The standard enthalpy of formation of O2 is zero at 298 K since O2 is the most stable allotropic form of oxygen at that particular temperature.
Q3.42a
a) The ionization of a weak acetic acid is written as:
$CH_{3}COOH(aq) \rightleftharpoons CH_{3}COO^{-} + H^+$
Suppose the change of reaction enthalpy for this reaction is +1.0 kJ/mol and the ΔfH° of CH3COO- is -486.01 kJ/mol. What would be the molar enthalpy of formation of aqueous acetic acid?
b) The acetic acid is reacted with a strong base NaOH. Calculate the enthalpy of neutralization for this reaction of weak acid strong base.
ΔfH°[OH-(aq)] = -229.6 kJ/mol and ΔfH°[H2O(l)] = -285.8 kJ/mol
Hint: Look at enthalpy of neutralization
S3.42a
a) $\Delta _r \bar{H}^{\circ} = \sum{[\Delta_{f} \bar{H}^{\circ}}]reactants - \sum{[\Delta_{f} \bar{H}^{\circ}}]products$
$\Delta _r \bar{H}^{\circ} = \Delta_{f} \bar{H}^{\circ}[CH_{3}COO^-] + \Delta_{f} \bar{H}^{\circ}[H^+] - \Delta_{f} \bar{H}^{\circ}[CH_{3}COOH]$
$+1.0 \dfrac{kJ}{mol} = -486.01 \dfrac{kJ}{mol} + 0 \dfrac{kJ}{mol} - \Delta _f \bar{H}^{\circ}[CH_{3}COOH]$
${\color{red} \Delta _f \bar{H}^{\circ}[CH_{3}COOH] = -487.01 \dfrac{kJ}{mol}}$
b) Note that acetic acid is a weak acid, therefore only dissociates partially or does not dissociate at all. Therefore the complete ionic reaction of reaction is:
$CH_{3}COOH(aq) + Na^+ (aq) + OH^- (aq) = CH_3COO^- (aq) + Na^+ (aq) + H^2O(l)$
Since Na+ cancels out, the net equation is
$CH_{3}COOH(aq) + OH^- (aq) = CH_3COO^- (aq) + H^2O(l)$
$\Delta _r \bar{H}^{\circ} = \Delta_{f} \bar{H}^{\circ}[CH_{3}COO^-] + \Delta_{f} \bar{H}^{\circ}[H_2O] - \Delta_{f} \bar{H}^{\circ}[CH_{3}COOH] - \Delta_{f} \bar{H}^{\circ}[OH^-]$
$\Delta _r \bar{H}^{\circ} = (-486.01 -285.5 + 486.01+229.6) \dfrac{kJ}{mol}$
${\color{red}\Delta _r \bar{H}^{\circ} = -55.9 \dfrac{kJ}{mol}}$
Q3.42b
Calculate the enthalpy of formation for Br- ions in the following reactions:
1. $HBr(g)\rightarrow Cl^-(aq) + Br^+(aq)$
2. $\Delta _rH^\circ =-105.7\dfrac{kJ}{mol}$
S3.42b
$\Delta _rH^\circ =\Delta _f\overline{H}^\circ (H^+)+\Delta _f\overline{H}^\circ (Br^-)-\Delta _f\overline{H}^\circ (HBr)$
$\Delta _f\overline{H}^\circ (Br^-)=\Delta _rH^\circ -\Delta _f\overline{H}^\circ (H^+)+\Delta _f\overline{H}^\circ (HBr)$
$\Delta _f\overline{H}^\circ (Br^-)=36.4\dfrac{kJ}{mol} -0+(-.105.7\dfrac{kJ}{mol})$
$\Delta _f\overline{H}^\circ (Br^-)=69.3\dfrac{kJ}{mol}$
Q3.42c
When determining standard enthalpies of formation of ions in aqueous solutions, set the value of $H^+$ ion to 0 for $\Delta _{f}\bar{H}^o[H^{+}_{(aq)}]=0.$
1. For the following reaction $HBr_{(g)}\rightarrow H^+_{(aq)}+Br^-_{(aq)}$, $\Delta rH^{o}=-98.4kj\,mol^{-1}$. Calculate the value of $\Delta _f\bar{H}^o$ for the Br ions.
2. The standard enthalpy of neutralization between HBr and NaOH solutions is $-53.4kj\,mol^{-1})$. At $25^{o}C$, calculate the standard enthalpy of formation of the hydroxide ion.
S3.42c
(a)
$\Delta rH^{o}=-98.4kj\,mol^{-1}=\Delta _f\bar{H}^{o}[H^+_{(aq)}]+\Delta _f\bar{H}^{o}[Br^-_{(aq)}]-\Delta _f\bar{H}^{o}[HBr_{(g)}]\]$\Delta _f\bar{H}^{o}[Br^-_{(aq)}]=-98.4kj\,mol^{-1}-\Delta _f\bar{H}^{o}[H^+_{(aq)}]+\Delta _f\bar{H}^{o}[HBr_{(g)}]\]
$=-98.4kj\,mol^{-1}-0kj\,mol^{-1}+(-36.29kj\,mol^{-1})\]$=-134.69kj\,mol^{-1}\]
(b) The neutralization for 1 mole of $H_2O$ is $\Delta rH^{o}=-53.4kj\,mol^{-1}$
$\Delta rH^{o}=\Delta_f\bar{H}^{o}[H_2O_{(l)}]-\Delta_f\bar{H}^{o}[H^+_{(aq)}]-\Delta_f\bar{H}^{o}[OH^-_{(aq)}]\]$\Delta_f\bar{H}^{o}[OH^-_{(aq)}]=\Delta_f\bar{H}^{o}[H_2O_{(l)}]-\Delta_f\bar{H}^{o}[H^+_{(aq)}]+53.4kj\,mol^{-1}\]
$=(-285.8kj\,mol^{-1})-0kj\,mol^{-1}+53.4kj\,mol^{-1}\]$=-232.4kj\,mol^{-1}\]
Q3.42d
For the following chemical reaction, calculate the ΔrHº.
\begin{align} H_{3}PO_{4} (aq) & \rightarrow H_{2}PO_{4}^{2-}(aq) + H^{+} (aq) \ \triangle _{f}H^{\circ}_{H_{3}PO_{4}(aq)} & = -1277 \ \ kJ \ mol^{-1} \ \triangle _{f}H^{\circ}_{H^{+}(aq)} & = \ \ 0 \ \ kJ \ mol^{-1} \ \triangle _{f}H^{\circ}_{H_{2}PO_{4}^{2-}(aq)} & = -1302.5 \ \ kJ \ mol^{-1} \end{align}
S3.42d
Use the following equation:
$\Delta H_{reaction}^o = \sum {\Delta H_{f}^o(products)} - \sum {\Delta H_{f}^o(Reactants)}$
Solve
\begin{align} \Delta H_{reaction}^o & = \Delta H_{f}^o(H_2PO_4^{2-})+\Delta H_{f}^o(H^{+}) - \Delta H_{f}^o(H_3PO_4)\ & = -1302.5 \ kJ \ mol^{-1} - (-1277 \ kJ \ mol^{-1}) \ & \underline {= -25.5 \ kJ \ mol^{-1}} \end{align}
Click here for more information about standard enthalpies of reaction.
Q3.42e
The reaction for the dissociation of hydrobromic acid in water is as follows:
$HBr_{(g)}\rightleftharpoons H^{+}_{(aq)}+Br_{(aq)}^{-}\, \, \, \, \, \, \, \, \, \, \, \, \Delta _{r}H^{\circ}=-85.2\frac{KJ}{mol}$
$\Delta _{f}H^{\circ}(HBr)=-36.4\frac{KJ}{mol}$
a.) Given the enthalpy of formation for HBr, what is the enthalpy of formation for the Bromine ion? Assume that the enthalpy of formation for the Hydrogen ion is 0.
b.) What is the enthalpy of reaction for the neutralization of HBr with KOH? Assume that the potassium and bromine ions remain in solution following the neutralization.
$\Delta _{f}H^{\circ}(OH_{(aq)}^{-})=-36.4\frac{KJ}{mol} \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Delta _{f}H^{\circ}(H_{2}O_{(l)})=-285.8\frac{KJ}{mol}$
S3.42e
a.) For any reaction,
$\Delta _{r}H^{\circ}=\sum \nu \Delta _{f}\bar{H}^{\circ}(products)-\sum \nu \Delta _{f}\bar{H}^{\circ}(reactants)$
For this reaction,
$\Delta _{r}H^{\circ}= \Delta _{f}\bar{H}^{\circ}(Br^{-}_{(aq)})+\Delta _{f}\bar{H}^{\circ}(H_{(aq)}^{+})-\Delta _{f}\bar{H}^{\circ}(HBr_{(aq)})$
$\Delta _{f}\bar{H}^{\circ}(Br^{-}_{(aq)})=\Delta _{r}H^{\circ}+\Delta _{f}\bar{H}^{\circ}(HBr_{(aq)})$
$=\left (-85.2\frac{KJ}{mol} \right )+\left (-36.4 \frac{KJ}{mol} \right )=-121.6\frac{KJ}{mol}$
b.) For any neutraliztion reaction in solution,
$OH_{(aq)}^{-}+H_{(aq)}^{+}\rightleftharpoons H_{2}O_{(l)}$
$\Delta _{r}H^{\circ}= \Delta _{f}\bar{H}^{\circ}(H_{2}O_{(l)})-\Delta _{f}\bar{H}^{\circ}(OH_{(aq)}^{-})$
$=-285.8\frac{KJ}{mol}-\left ( -36.4 \frac{KJ}{mol}\right )=-249.4\frac{KJ}{mol}$
Q3.43a
How much heat is given off when one explodes 24.65 grams of nitroglycerin? The equation for the combustion of nitroglycerin is given below:
\begin{align} & 4 C_3H_5N_3O_9(l) \rightarrow 12 CO2(g) + 10 H_2O(g) + 6 N_2(g) + O_2(g) \ & \triangle _{r}H^{\circ}_{C_3H_5N_3O_9(l)}= -1.529 \ MJ \ mol^{-1} \end{align}
S3.43a
The molar mass of nitroglycerin is 227.0865 g/mol.
To solve this question simply multiple the enthalpy of reaction by the amount moles of nitroglycerin.
\begin{align} 24.65 \ g \div 227.0865 \ g \ mol^{-1} \ &= 0.109 \ moles \ of \ nitroglycerin \ 1.529 \ MJ \ mol^{-1} \cdot 0.109 \ mol &= \underline{ 0.166 MJ \ of \ heat } \end{align}
Q3.43b
What is the enthalpy change from the formation of 5 grams of H2CO3 via CO2(g) + H2O(g) àH2CO3(g) H2CO3 (g) via CO2 (g) + H2O (g) H2CO3 (g)? The change in enthalpy associated with this reaction is +360.19 kJ/mol.
S3.43b
The enthalpy change from this reaction is given by the molar enthalpy change times the number of moles of product formed. Thus:
mco2 = 44.01 g mol-1 ;
nco2 = 5g /44.01 g mol-1 = 0.1136 mol
ΔH =0.1136 mol * 360.19 kJ/mol = 40.9kJ
Q3.43c
The following reaction was carried out at standard temperature and pressure:
$2C_{(graphite)}+3H_{2(g)}\rightarrow C_{2}H_{6(g)}\, \, \, \, \, \, \, \, \, \, \Delta _{r}H^{\circ}=-84.7\frac{KJ}{mol}$
If -148.16 KJ of heat was released during this reaction, how many grams of Ethane were formed?
S3.43c
For any reaction:
$q=n\Delta _{f}H^{\circ}=\frac{m}{\mathfrak{M}}\Delta _{f}H^{\circ}$
This can be rearranged as
$m=\frac{q\mathfrak{M}}{\Delta _{f}H^{\circ}}$
The molar mass of ethane is
$\mathfrak{M}=2(12.01 \frac{g}{mol})+6(1.008\frac{g}{mol})=30.07 \frac{g}{mol}$
The mass of ethane produced, then, is
$m=\frac{q\mathfrak{M}}{\Delta _{f}H^{\circ}}=\frac{(-148.16 KJ)(30.07\frac{g}{mol})}{-84.7\frac{KJ}{mol}}=52.6g$
Q3.44a
When 2.3 g of calcium carbonate is decomposed under constant pressure, 178.3 kJ of heat were released. What is the $\Delta_rH^\circ$ for this reaction?
$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$
S3.44a
$\Delta _rH^\circ = q_p=\dfrac{178.3kJ}{\dfrac{2.3g}{100.09g/mol}}=7759.15kJ/mol$
Q3.44b
In a high school chemistry experiment, 10 grams of methane were combusted in the presence of excess oxygen under standard conditions. A calorimeter measured that 554.9 kJ of heat were released by the reaction to the surroundings. Assuming the calorimeter’s measurement accounts for all heat changes associated with the reaction, write the reaction for the combustion of methane and calculate the change in enthalpy associated with it.
Q3.46
The isomerization process from 1-butene to 2-butene has a reaction enthalpy of -7.1 kJ/mol.
=
1-butene 2-butene
If the hydrogenation of 1 butene has a reaction enthalpy of -126.8 kJ/mol. How much energy would the hydrogenation of 2-butene takes? Keep in mind that both processes yield the same product.
Hint: Look at example problems of change of standard reaction enthalpy
S3.46
1-butene + 2H2 → butane ΔfH° = -126.8 kJ/mol
1-butene → 2-butene ΔfH° = -7.1 kJ/mol
Reverse the 2nd reaction and combine the two reactions
1-butene + 2H2 → butane ΔfH° = -126.8 kJ/mol
2-butene → 1-butene ΔfH° = +7.1 kJ/mol
ΔrH° = -126.8 kJ/mol +7.1 kJ/mol
ΔrH° = -119.7 kJ/mol
Q3.47a
From the reaction below, calculate the enthalpy of formation of methane gas.
\begin{align} & CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \ & \triangle _{r}H^{\circ} = -891 kJ \ mol^{-1} \ & \triangle _{f}H^{\circ}_{CO_2(g)} = -393.5 kJ \ mol^{-1} \ & \triangle _{f}H^{\circ}_{H_2O (l)} = -285.8 kJ \ mol^{-1} \end{align}
S3.47a
Use the following equation:
$\Delta H_{reaction}^o = \sum {\Delta H_{f}^o(products)} - \sum {\Delta H_{f}^o(Reactants)}$
Plug in appropriate values and solve. Change in enthalpy of oxygen is 0 because it is the reference form of a pure element.
$\Delta H_{r}^o = \Delta H_{f}^o(CO_2) + (2 \cdot \Delta H_{f}^o(H_2O)) - (2 \cdot \Delta H_{f}^o(O_2)) - \Delta H_{f}^o(CH_4)$
\begin{align} -891 \ kJ \ mol^{-1}&=(-393.5 \ kJ \ mol^{-1}) + (-285.8 \ kJ \ mol^{-1}) - (\Delta H_{f}^o(CH_4)) \ \Delta H_{f}^o(CH_4)&= \underline{ 211.7\ kJ \ mol^{-1} } \end{align}
Click here for more information on standard enthalpies.
Q3.47b
For the following combustion reaction:
$C_{3}H_{7}OH_{(l)}+3O_{2(g)}\rightarrow 4H_{2}O_{(l)}+3CO_{2(g)}\, \, \, \, \, \, \, \, \, \, \,\Delta _{r}H^{\circ}=-2005.8\frac{KJ}{mol}$
Calculate the enthalpy of formation for 2-propanol, given the following enthalpies of formation
$\Delta _{f}\bar{H}^{\circ}(H_{2}O)=-285.8 \frac{KJ}{mol}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \Delta _{f}\bar{H}^{\circ}(CO_{2})=-393.5 \frac{KJ}{mol}$
S3.47b
In general:
$\Delta _{r}H^{\circ}=\sum \nu \Delta _{f}\bar{H}^{\circ}(products)-\sum \nu \Delta _{f}\bar{H}^{\circ}(reactants)$
In general,
$\Delta _{r}H^{\circ}=3 \Delta _{f}\bar{H}^{\circ}(CO_{2})+4\Delta _{f}\bar{H}^{\circ}(H_{2}O)-\Delta _{f}\bar{H}^{\circ}(C_{3}H_{7}OH)$
$\Delta _{f}\bar{H}^{\circ}(C_{3}H_{7}OH)=3 \Delta _{f}\bar{H}^{\circ}(CO_{2})+4\Delta _{f}\bar{H}^{\circ}(H_{2}O)-\Delta _{r}H^{\circ}$
$=3\left (-393.5\frac{KJ}{mol} \right )+4\left (-285.8\frac{KJ}{mol} \right )-\left (-2005.8\frac{KJ}{mol} \right )=-317.9 \frac{KJ}{mol}$
Q3.48
Diatomic fluorine gas has a Cp of 31.3 J/K*mol. What is its molar enthalpy of formation on the surface of the planet Venus (at standard pressure), where temperature is commonly 465 C?
S3.48
At constant pressure, the change in enthalpy with respect to temperature is given by Cp. By determining the change in temperature and multiplying it by Cp we can find the change in enthalpy.
ΔT = 465 C - 25 C = 440K
ΔH = 31.3 J/K*mol * 440K = 13.772 kJ/mol
Q3.50
The combustion of graphite is
C(gr) + O2(g) → CO2(g)
Calculate the change in the enthalpy of combustion from 298 K to 398 K. The molar Cp values are (in J K-1 mol-1): C(gr):8.52, O2(g):29.4, and CO2(g):37.1
Q3.53
What values are needed to calculate the change in standard enthalpy of a reaction?
Q3.54
Consider the following reactions:
C(graphite) + O2 (g) → CO (s) ∆rH = -110.5 kJ/mole
CO2 (g) → Co (s) + O2 (g) rH = 283.0 kJ/mole
Find the enthalpy of formation of carbon dioxide (CO2)
S3.54
The $\Delta H$ of the reaction is the sum of $\Delta H_1$ and $\Delta H_2$
S3.55
$\Delta H^{\circ}_{rxn} = \sum \Delta H^{\circ}_{prod} - \sum \Delta H^{\circ}_{reactants}$
$\Delta H^{\circ}_{rxn} = [\Delta H^{\circ}(O(g)) +\Delta H^{\circ}(O_{3}(g))]-2\Delta H^{\circ}(O_{2}(g))$
$\Delta H^{\circ}_{rxn} = (249.4 + 142.7) \dfrac{kJ}{mol}$
$\Delta H^{\circ}_{rxn} = 392.1 \dfrac{kJ}{mol}$
Q3.60
Methane combusts, what is the enthalpy of combustion for this reaction?
Substance Standard Molar Enthalpies of Formation at 25ºC
CH4(g)
-74.85 kj/mol
CO2(g) -393.5 kj/mol
H2O(g) -241.8 kj/mol
H2O(l) -285.8 kj/mol
S3.60
Write the balanced formula for this reaction to identify the coefficients
$CH_{4(g)} +2 O_{2(g)} \rightarrow 2H_2O_{(l)} + CO_{2(g)}$
$\Delta H= \sum H_{products} - sum H_{reactants}$
$= 2x H(H_2O) + H(CO_2) - H(CH_4)$
We do not include $O_2$ because it is a standard state
$=[(2 x -285.8 + -393.5) - (-74.85)] kJ/mol$
$=-890.25\; kJ/mol$
Q3.62
Predict whether the values of q, w, ΔU, and ΔH are positive, zero, or negative for each of the following processes: (a) freezing of ice at 1 atm and 273 K, (b) melting of solid butanol at 1 atm and the normal melting point, (c) reversible isothermal compression of an ideal gas, and (d) reversible adiabatic compression of an ideal gas.
Q3.62
Predict whether the values of q, w, ΔU, and ΔH are positive, zero, or negative for each of the following processes: (a) freezing of ice at 1 atm and 273 K, (b) melting of solid butanol at 1 atm and the normal melting point, (c) reversible isothermal compression of an ideal gas, and (d) reversible adiabatic compression of an ideal gas.
Q3.63
The first law of thermodynamics states that energy is conserved. It cannot be destroyed nor created. Using Einstein’s equation E= mc2 explain how large amounts of energy are necessary to fuse atoms together. If energy is conserved where did all this energy go?
S3.63
When large amounts of energy are used to fuse atoms together (eg: fusion) we can observe that the mass of the atoms is very small compared to the amount of energy used. However, using Einstein's equation E= mc2 we see that the mass is multiplied times the speed of light squared which will result in a very large number. The energy was not destroyed but rather stored in the chemical bonds of the atoms.
Q3.64
For the nuclear process we cannot apply the assumption that $∆_fH = 0$ for most stable elements because there are different components in the reactant and product sides. Explain this difference?
Q3.66
You eat 2 pounds of fries for lunch and the fuel value is approximately $2.9kcal\,g^{-1}$. If you don't store any of the energy, calculate the amount of water (evaporated perspiration) needed to keep your body temperature constant.
S3.66
Step1: (Hint: Calculate the fuel value for 2 pound of fries)
$(2\;pounds\;of\;fries)(\dfrac{454g\,fries}{1\,pound\,fries})(\dfrac{2.9kcal\,fries}{g\,fries})(\dfrac{4.184kJ}{1kcal})=1.1017\times10^4kj$
Step 2: (Hint: Use $\Delta _{vap}H$ to calculate grams of $H_2O$)
$H_2O_{(l)}\rightarrow H_2O{(g)}$
$\Delta _{vap}H=44.01kjmol^{-1} at\;298K$
$(\dfrac{1.1017\times10^4jk}{44.01kj\,mol^{-1}})(\dfrac{18.02H_2O}{1mol\,H_2O})=4.51\times10^3g\,H_2O$
Q3.68
What is the final temperature of the following reaction:
$C_3H_{8(g)}+5O_{2(g)} \rightarrow 3CO_{2(g)} +4H_2O_{(l)}$
Hint: first find $\Delta_rH^\circ$, then look up the standard molar heat capacities of the products.
S3.68
\begin{align*}\Delta _rH^\circ &=\sum n\Delta _fH^\circ_{product}-\sum n\Delta _fH^\circ_{reactant}\&=3(-393.5kJ/mol)+4(-241.8kJ/mol)-(-104.6kJ/mol)-5(0kJ/mol)\&=-2043.1kJ/mol\end{align*}
$\Delta_rH^\circ=q_{p,rxn}=-q_{p,absorbed by product}= (n\bar{C}_{p,CO_2}+n\bar{C}_{p,H_2O})\Delta T\ \Delta T=T_f-T_i=\dfrac{-q_{p,abs by prod}}{n\bar{C}_{p,CO_2}+n\bar{C}_{p,H_2O}}$
\begin{align*}T_f&=T_i+\dfrac{-q_{p,abs by prod}}{n\bar{C}_{p,CO_2}+n\bar{C}_{p,H_2O}}\ &=298.15K+\dfrac{2043.1kJ/mol}{3(37.1*10^{-3}kJ/K*mol)+4(33.6*10^{-3}kJ/K*mol)}=8613.6K\end{align*}
Q3.70
The reaction below is a oxidation of 2-propanol to make acetone
a) Calculate the change of enthalpy of reaction by using bond enthalpies
b) Compare the result with the reaction of the combustion 2-propanol that has an enthalpy of -1825.5 kJ/mol. What does this say about the two reactions. (Hint: Look at bond enthalpies).
S3.70
a)
Types of bonds formed #of bonds formed Bond Enthalpy (kJ/mol) Enthalpy change (kJ/mol)
C-H 6 414 2484
C-C 2 347 694
C=O 1 724 724
O-H 1 460 460
$\Delta _{r} H^{\circ} = \sum BE(reactants) - \sum BE(products) =[(2898 + 694+351+460) - (2484+694+724+460)] \frac{kJ}{mol} = {\color{red}+41 \frac{kJ}{mol}}$
b) According to the enthalpy values, the enthalpy of the oxidation of 2-propanol into ketone is positive while its combustion is largely negative. This tells that although oxygen is involved, the two reactions are different. The combustion releases heat where as the other process needs an absorption of energy in order for the reaction to take place. In this case is usually the help of another catalyst.
Q3.71a
The molar enthalpy of sublimation for an unknown organic substance is found to be 58.05 KJ/mol, and the molar enthalpy of vaporization is 37.12 KJ/mol. Using these values, provide an estimate for the molar enthalpy of fusion for this substance.
Hint: Sublimination is the transition from solid to gas phase. Fusion and Vaporization, respectively, are the the transition from solid to liquid, then liquid to gas.
S3.71a
In general,
$\Delta _{sub}\bar{H}^{\circ}=\Delta _{vap}\bar{H}^{\circ}+\Delta _{fus}\bar{H}^{\circ}$
$\Delta _{fus}\bar{H}^{\circ}=\Delta _{sub}\bar{H}^{\circ}-\Delta _{vap}\bar{H}^{\circ}=58.05\frac{KJ}{mol}-37.12\frac{KJ}{mol}=20.93\frac{KJ}{mol}$
Q3.71b
Calculate the standard enthalpy of sublimation of ethanol. ΔfusHº(ethanol) = 4.9 kJ mol-1 and ΔvapHº (ethanol) = 38.56 kJ mol-1.
S3.71b
You can estimate the standard enthalpy of sublimation by adding up the standard enthalpies of fusion and vaporization since sublimation consists of two simultaneous phase changes. See here for more information on it.
ΔsubHº(ethanol) = 4.9+38.56 = 43.46 kJ mol-1
Q3.72
Nitric acid is neutralized by hydroxide via the reaction HNO3 (aq) + OH-(aq) H2O(l) + NO3-(aq). The dissociation of water has an enthalpy change of 55.8 kJ/mol. What is the enthalpy change for the dissociation of nitric acid?
Chemical
HNO3
OH-
H2O
NO3-
ΔfH (kJ/mol)
-207.6
-229.6
-285.8
-206.6
S3.72
The dissociation of water is H2O(l) -> H+(aq) + OH-(aq), which when added to the neutralization reaction above, gives the overall reaction
$HNO_3 (aq) \rightarrow H^+(aq) + NO_3^-(aq)$
which is the dissociation reaction for nitric acid. The enthalpy change of the first reaction is given by the difference in the total enthalpies of formation of the products and reactants, i.e.
ΔrH=(-206.6 kJ/mol -285.8 kJ/mol) - (-229.6 kJ/mol -207.6 kJ/mol) = -55.2 kJ/mol.
When added to the enthalpy change for the dissociation of water to obtain the enthalpy change for the dissociation of nitric acid, we have:
ΔnetH=-55.2 kJ/mol + 55.8 kJ/mol = 0.6 kJ/mol
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/03%3A_The_First_Law_of_Thermodynamics/3.E%3A_Exercises.txt
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Entropy is a state function that is often erroneously referred to as the 'state of disorder' of a system. Qualitatively, entropy is simply a measure how much the energy of atoms and molecules become more spread out in a process and can be defined in terms of statistical probabilities of a system or in terms of the other thermodynamic quantities. Entropy is also the subject of the Second and Third laws of thermodynamics, which describe the changes in entropy of the universe with respect to the system and surroundings, and the entropy of substances, respectively.
Statistical Definition of Entropy
Entropy is a thermodynamic quantity that is generally used to describe the course of a process, that is, whether it is a spontaneous process and has a probability of occurring in a defined direction, or a non-spontaneous process and will not proceed in the defined direction, but in the reverse direction. To define entropy in a statistical manner, it helps to consider a simple system such as in Figure $1$. Two atoms of hydrogen gas are contained in a volume of $V_1$.
Since all the hydrogen atoms are contained within this volume, the probability of finding any one hydrogen atom in $V_1$ is 1. However, if we consider half the volume of this box and call it $V_2$,the probability of finding any one atom in this new volume is $\frac{1}{2}$, since it could either be in $V_2$ or outside. If we consider the two atoms, finding both in $V_2$, using the multiplication rule of probabilities, is
$\frac{1}{2} \times \frac{1}{2} =\frac{1}{4}.$
For finding four atoms in $V_2$ would be
$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}= \frac{1}{16}.$
Therefore, the probability of finding N number of atoms in this volume is $\frac{1}{2}^N$. Notice that the probability decreases as we increase the number of atoms.
If we started with volume $V_2$ and expanded the box to volume $V_1$, the atoms would eventually distribute themselves evenly because this is the most probable state. In this way, we can define our direction of spontaneous change from the lowest to the highest state of probability. Therefore, entropy $S$ can be expressed as
$S=k_B \ln{\Omega} \label{1}$
where $\Omega$ is the probability and $k_B$ is a proportionality constant. This makes sense because entropy is an extensive property and relies on the number of molecules, when $\Omega$ increases to $W^2$, S should increase to 2S. Doubling the number of molecules doubles the entropy.
So far, we have been considering one system for which to calculate the entropy. If we have a process, however, we wish to calculate the change in entropy of that process from an initial state to a final state. If our initial state 1 is $S_1=K_B \ln{\Omega}_1$ and the final state 2 is $S_2=K_B\ln{\Omega}_2$,
$\Delta S=S_2-S_1=k_B \ln \dfrac{\Omega_2}{\Omega_1} \label{2}$
using the rule for subtracting logarithms. However, we wish to define $\Omega$ in terms of a measurable quantity. Considering the system of expanding a volume of gas molecules from above, we know that the probability is proportional to the volume raised to the number of atoms (or molecules), $\alpha V^{N}$. Therefore,
$\Delta S=S_2-S_1=k_B \ln \left(\dfrac{V_2}{V_1} \right)^N=Nk_B \ln \dfrac{V_2}{V_1} \label{3}$
We can define this in terms of moles of gas and not molecules by setting the $k_{B}$ or Boltzmann constant equal to $\frac{R}{N_A}$, where R is the gas constant and $N_A$ is Avogadro's number. So for a expansion of an ideal gas and holding the temperature constant,
$\Delta S=\dfrac{N}{N_A} R \ln \left(\dfrac{V_2}{V_1} \right)^N=nR\ln \dfrac{V_2}{V_1} \label{4}$
because $\frac{N}{N_A}=n$, the number of moles. This is only defined for constant temperature because entropy can change with temperature. Furthermore, since S is a state function, we do not need to specify whether this process is reversible or irreversible.
Thermodynamic Definition of Entropy
Using the statistical definition of entropy is very helpful to visualize how processes occur. However, calculating probabilities like $\Omega$ can be very difficult. Fortunately, entropy can also be derived from thermodynamic quantities that are easier to measure. Recalling the concept of work from the first law of thermodynamics, the heat (q) absorbed by an ideal gas in a reversible, isothermal expansion is
$q_{rev}=nRT\ln \dfrac{V_2}{V_1} \; . \label{5}$
If we divide by T, we can obtain the same equation we derived above for $\Delta S$:
$\Delta S=\dfrac{q_{rev}}{T}=nR\ln \dfrac{V_2}{V_1} \;. \label{6}$
We must restrict this to a reversible process because entropy is a state function, however the heat absorbed is path dependent. An irreversible expansion would result in less heat being absorbed, but the entropy change would stay the same. Then, we are left with
$\Delta S> \dfrac{q_{irrev}}{T}$
for an irreversible process because
$\Delta S=\Delta S_{rev}=\Delta S_{irrev} .$
This apparent discrepancy in the entropy change between an irreversible and a reversible process becomes clear when considering the changes in entropy of the surrounding and system, as described in the second law of thermodynamics.
It is evident from our experience that ice melts, iron rusts, and gases mix together. However, the entropic quantity we have defined is very useful in defining whether a given reaction will occur. Remember that the rate of a reaction is independent of spontaneity. A reaction can be spontaneous but the rate so slow that we effectively will not see that reaction happen, such as diamond converting to graphite, which is a spontaneous process.
The Second Law as Energy Dispersion
Energy of all types -- in chemistry, most frequently the kinetic energy of molecules (but also including the phase change/potential energy of molecules in fusion and vaporization, as well as radiation) changes from being localized to becoming more dispersed in space if that energy is not constrained from doing so. The simplest example stereotypical is the expansion illustrated in Figure 1.
The initial motional/kinetic energy (and potential energy) of the molecules in the first bulb is unchanged in such an isothermal process, but it becomes more widely distributed in the final larger volume. Further, this concept of energy dispersal equally applies to heating a system: a spreading of molecular energy from the volume of greater-motional energy (“warmer”) molecules in the surroundings to include the additional volume of a system that initially had “cooler” molecules. It is not obvious, but true, that this distribution of energy in greater space is implicit in the Gibbs free energy equation and thus in chemical reactions.
“Entropy change is the measure of how more widely a specific quantity of molecular energy is dispersed in a process, whether isothermal gas expansion, gas or liquid mixing, reversible heating and phase change, or chemical reactions.” There are two requisites for entropy change.
1. It is enabled by the above-described increased distribution of molecular energy.
2. It is actualized if the process makes available a larger number of arrangements for the system’s energy, i.e., a final state that involves the most probable distribution of that energy under the new constraints.
Thus, “information probability” is only one of the two requisites for entropy change. Some current approaches regarding “information entropy” are either misleading or truly fallacious, if they do not include explicit statements about the essential inclusion of molecular kinetic energy in their treatment of chemical reactions.
Contributors and Attributions
• Frank L. Lambert, Professor Emeritus, Occidental College
• Konstantin Malley (UCD)
Carnot Cycle
In the early 19th century, steam engines came to play an increasingly important role in industry and transportation. However, a systematic set of theories of the conversion of thermal energy to motive power by steam engines had not yet been developed. Nicolas Léonard Sadi Carnot (1796-1832), a French military engineer, published Reflections on the Motive Power of Fire in 1824. The book proposed a generalized theory of heat engines, as well as an idealized model of a thermodynamic system for a heat engine that is now known as the Carnot cycle. Carnot developed the foundation of the second law of thermodynamics, and is often described as the "Father of thermodynamics."
The Carnot Cycle
The Carnot cycle consists of the following four processes:
1. A reversible isothermal gas expansion process. In this process, the ideal gas in the system absorbs $q_{in}$ amount heat from a heat source at a high temperature $T_{high}$, expands and does work on surroundings.
2. A reversible adiabatic gas expansion process. In this process, the system is thermally insulated. The gas continues to expand and do work on surroundings, which causes the system to cool to a lower temperature, $T_{low}$.
3. A reversible isothermal gas compression process. In this process, surroundings do work to the gas at $T_{low}$, and causes a loss of heat, $q_{out}$.
4. A reversible adiabatic gas compression process. In this process, the system is thermally insulated. Surroundings continue to do work to the gas, which causes the temperature to rise back to $T_{high}$.
P-V Diagram
The P-V diagram of the Carnot cycle is shown in Figure $2$. In isothermal processes I and III, ∆U=0 because ∆T=0. In adiabatic processes II and IV, q=0. Work, heat, ∆U, and ∆H of each process in the Carnot cycle are summarized in Table $1$.
Table $1$: Work, heat, ∆U, and ∆H in the P-V diagram of the Carnot Cycle.
Process w q ΔU ΔH
I $-nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)$ $nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)$ 0 0
II $n\bar{C_{v}}(T_{low}-T_{high})$ 0 $n\bar{C_{v}}(T_{low}-T_{high})$ $n\bar{C_{p}}(T_{low}-T_{high})$
III $-nRT_{low}\ln\left(\dfrac{V_{4}}{V_{3}}\right)$ $nRT_{low}\ln\left(\dfrac{V_{4}}{V_{3}}\right)$ 0 0
IV $n\bar{C_{v}}(T_{high}-T_{low})$ 0 $n\bar{C_{v}}(T_{hight}-T_{low})$ $n\bar{C_{p}}(T_{high}-T_{low})$
Full Cycle $-nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)-nRT_{low}\ln\left(\dfrac{V_{4}}{V_{3}}\right)$ $nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)+nRT_{low}\ln\left(\dfrac{V_{4}}{V_{3}}\right)$ 0 0
T-S Diagram
The T-S diagram of the Carnot cycle is shown in Figure $3$. In isothermal processes I and III, ∆T=0. In adiabatic processes II and IV, ∆S=0 because dq=0. ∆T and ∆S of each process in the Carnot cycle are shown in Table $2$.
Table $1$: Work, heat, and ∆U in the T-S diagram of the Carnot Cycle.
Process ΔT ΔS
I 0 $-nR\ln\left(\dfrac{V_{2}}{V_{1}}\right)$
II $T_{low}-T_{high}$ 0
III 0 $-nR\ln\left(\dfrac{V_{4}}{V_{3}}\right)$
IV $T_{high}-T_{low}$ 0
Full Cycle 0 0
Efficiency
The Carnot cycle is the most efficient engine possible based on the assumption of the absence of incidental wasteful processes such as friction, and the assumption of no conduction of heat between different parts of the engine at different temperatures. The efficiency of the carnot engine is defined as the ratio of the energy output to the energy input.
\begin{align*} \text{efficiency} &=\dfrac{\text{net work done by heat engine}}{\text{heat absorbed by heat engine}} =\dfrac{-w_{sys}}{q_{high}} \[4pt] &=\dfrac{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)+nRT_{low}\ln \left(\dfrac{V_{4}}{V_{3}}\right)}{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)} \end{align*}
Since processes II (2-3) and IV (4-1) are adiabatic,
$\left(\dfrac{T_{2}}{T_{3}}\right)^{C_{V}/R}=\dfrac{V_{3}}{V_{2}}$
and
$\left(\dfrac{T_{1}}{T_{4}}\right)^{C_{V}/R}=\dfrac{V_{4}}{V_{1}}$
And since T1 = T2 and T3 = T4,
$\dfrac{V_{3}}{V_{4}}=\dfrac{V_{2}}{V_{1}}$
Therefore,
$\text{efficiency}=\dfrac{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)-nRT_{low}\ln\left(\dfrac{V_{2}}{V_{1}}\right)}{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)}$
$\boxed{\text{efficiency}=\dfrac{T_{high}-T_{low}}{T_{high}}}$
Summary
The Carnot cycle has the greatest efficiency possible of an engine (although other cycles have the same efficiency) based on the assumption of the absence of incidental wasteful processes such as friction, and the assumption of no conduction of heat between different parts of the engine at different temperatures.
Problems
1. You are now operating a Carnot engine at 40% efficiency, which exhausts heat into a heat sink at 298 K. If you want to increase the efficiency of the engine to 65%, to what temperature would you have to raise the heat reservoir?
2. A Carnot engine absorbed 1.0 kJ of heat at 300 K, and exhausted 400 J of heat at the end of the cycle. What is the temperature at the end of the cycle?
3. An indoor heater operating on the Carnot cycle is warming the house up at a rate of 30 kJ/s to maintain the indoor temperature at 72 ºF. What is the power operating the heater if the outdoor temperature is 30 ºF?
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The Second Law of Thermodynamics states that the state of entropy of the entire universe, as an isolated system, will always increase over time. The second law also states that the changes in the entropy in the universe can never be negative.
Introduction
Why is it that when you leave an ice cube at room temperature, it begins to melt? Why do we get older and never younger? And, why is it whenever rooms are cleaned, they become messy again in the future? Certain things happen in one direction and not the other, this is called the "arrow of time" and it encompasses every area of science. The thermodynamic arrow of time (entropy) is the measurement of disorder within a system. Denoted as $\Delta S$, the change of entropy suggests that time itself is asymmetric with respect to order of an isolated system, meaning: a system will become more disordered, as time increases.
Major players in developing the Second Law
• Nicolas Léonard Sadi Carnot was a French physicist, who is considered to be the "father of thermodynamics," for he is responsible for the origins of the Second Law of Thermodynamics, as well as various other concepts. The current form of the second law uses entropy rather than caloric, which is what Sadi Carnot used to describe the law. Caloric relates to heat and Sadi Carnot came to realize that some caloric is always lost in the motion cycle. Thus, the thermodynamic reversibility concept was proven wrong, proving that irreversibility is the result of every system involving work.
• Rudolf Clausius was a German physicist, and he developed the Clausius statement, which says "Heat generally cannot flow spontaneously from a material at a lower temperature to a material at a higher temperature."
• William Thompson, also known as Lord Kelvin, formulated the Kelvin statement, which states "It is impossible to convert heat completely in a cyclic process." This means that there is no way for one to convert all the energy of a system into work, without losing energy.
• Constantin Carathéodory, a Greek mathematician, created his own statement of the second law arguing that "In the neighborhood of any initial state, there are states which cannot be approached arbitrarily close through adiabatic changes of state."
To understand why entropy increases and decreases, it is important to recognize that two changes in entropy have to considered at all times. The entropy change of the surroundings and the entropy change of the system itself. Given the entropy change of the universe is equivalent to the sums of the changes in entropy of the system and surroundings:
$\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}=\dfrac{q_{sys}}{T}+\dfrac{q_{surr}}{T} \label{1}$
In an isothermal reversible expansion, the heat q absorbed by the system from the surroundings is
$q_{rev}=nRT\ln\frac{V_{2}}{V_{1}}\label{2}$
Since the heat absorbed by the system is the amount lost by the surroundings, $q_{sys}=-q_{surr}$.Therefore, for a truly reversible process, the entropy change is
$\Delta S_{univ}=\dfrac{nRT\ln\frac{V_{2}}{V_{1}}}{T}+\dfrac{-nRT\ln\frac{V_{2}}{V_{1}}}{T}=0 \label{3}$
If the process is irreversible however, the entropy change is
$\Delta S_{univ}=\frac{nRT\ln \frac{V_{2}}{V_{1}}}{T}>0 \label{4}$
If we put the two equations for $\Delta S_{univ}$together for both types of processes, we are left with the second law of thermodynamics,
$\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}\geq0 \label{5}$
where $\Delta S_{univ}$ equals zero for a truly reversible process and is greater than zero for an irreversible process. In reality, however, truly reversible processes never happen (or will take an infinitely long time to happen), so it is safe to say all thermodynamic processes we encounter everyday are irreversible in the direction they occur.
The second law of thermodynamics can also be stated that "all spontaneous processes produce an increase in the entropy of the universe".
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Learning Objectives
• The absolute entropy of a pure substance at a given temperature is the sum of all the entropy it would acquire on warming from absolute zero (where $S=0$) to the particular temperature.
• Calculate entropy changes for phase transitions and chemical reactions under standard conditions
The atoms, molecules, or ions that compose a chemical system can undergo several types of molecular motion, including translation, rotation, and vibration (Figure $1$). The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. A perfectly ordered system with only a single microstate available to it would have an entropy of zero. The only system that meets this criterion is a perfect crystal at a temperature of absolute zero (0 K), in which each component atom, molecule, or ion is fixed in place within a crystal lattice and exhibits no motion (ignoring quantum zero point motion).
This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion (at least classically, quantum mechanics argues for constant motion) means there is but one possible location for each identical atom or molecule comprising the crystal ($\Omega = 1$). According to the Boltzmann equation, the entropy of this system is zero.
\begin{align*} S&=k\ln \Omega \[4pt] &= k\ln(1) \[4pt] &=0 \label{$5$} \end{align*}
In practice, absolute zero is an ideal temperature that is unobtainable, and a perfect single crystal is also an ideal that cannot be achieved. Nonetheless, the combination of these two ideals constitutes the basis for the third law of thermodynamics: the entropy of any perfectly ordered, crystalline substance at absolute zero is zero.
Definition: Third Law of Thermodynamics
The entropy of a pure, perfect crystalline substance at 0 K is zero.
The third law of thermodynamics has two important consequences: it defines the sign of the entropy of any substance at temperatures above absolute zero as positive, and it provides a fixed reference point that allows us to measure the absolute entropy of any substance at any temperature. In this section, we examine two different ways to calculate ΔS for a reaction or a physical change. The first, based on the definition of absolute entropy provided by the third law of thermodynamics, uses tabulated values of absolute entropies of substances. The second, based on the fact that entropy is a state function, uses a thermodynamic cycle similar to those discussed previously.
Standard-State Entropies
One way of calculating $ΔS$ for a reaction is to use tabulated values of the standard molar entropy ($S^o$), which is the entropy of 1 mol of a substance under standard pressure (1 bar). Often the standard molar entropy is given at 298 K and is often demarked as $ΔS^o_{298}$. The units of $S^o$ are J/(mol•K). Unlike enthalpy or internal energy, it is possible to obtain absolute entropy values by measuring the entropy change that occurs between the reference point of 0 K (corresponding to $S = 0$) and 298 K (Tables T1 and T2).
As shown in Table $1$, for substances with approximately the same molar mass and number of atoms, $S^o$ values fall in the order
$S^o(\text{gas}) \gg S^o(\text{liquid}) > S^o(\text{solid}).$
For instance, $S^o$ for liquid water is 70.0 J/(mol•K), whereas $S^o$ for water vapor is 188.8 J/(mol•K). Likewise, $S^o$ is 260.7 J/(mol•K) for gaseous $\ce{I2}$ and 116.1 J/(mol•K) for solid $\ce{I2}$. This order makes qualitative sense based on the kinds and extents of motion available to atoms and molecules in the three phases (Figure $1$). The correlation between physical state and absolute entropy is illustrated in Figure $2$, which is a generalized plot of the entropy of a substance versus temperature.
The Third Law Lets us Calculate Absolute Entropies
The absolute entropy of a substance at any temperature above 0 K must be determined by calculating the increments of heat $q$ required to bring the substance from 0 K to the temperature of interest, and then summing the ratios $q/T$. Two kinds of experimental measurements are needed:
1. The enthalpies associated with any phase changes the substance may undergo within the temperature range of interest. Melting of a solid and vaporization of a liquid correspond to sizeable increases in the number of microstates available to accept thermal energy, so as these processes occur, energy will flow into a system, filling these new microstates to the extent required to maintain a constant temperature (the freezing or boiling point); these inflows of thermal energy correspond to the heats of fusion and vaporization. The entropy increase associated with transition at temperature $T$ is $\dfrac{ΔH_{fusion}}{T}.$
2. The heat capacity $C$ of a phase expresses the quantity of heat required to change the temperature by a small amount $ΔT$, or more precisely, by an infinitesimal amount $dT$. Thus the entropy increase brought about by warming a substance over a range of temperatures that does not encompass a phase transition is given by the sum of the quantities $C \frac{dT}{T}$ for each increment of temperature $dT$. This is of course just the integral
$S_{0 \rightarrow T} = \int _{0}^{T} \dfrac{C_p}{T} dt \label{eq20}$
Because the heat capacity is itself slightly temperature dependent, the most precise determinations of absolute entropies require that the functional dependence of $C$ on $T$ be used in the integral in Equation \ref{eq20}, i.e.,:
$S_{0 \rightarrow T} = \int _{0}^{T} \dfrac{C_p(T)}{T} dt. \label{eq21}$
When this is not known, one can take a series of heat capacity measurements over narrow temperature increments $ΔT$ and measure the area under each section of the curve. The area under each section of the plot represents the entropy change associated with heating the substance through an interval $ΔT$. To this must be added the enthalpies of melting, vaporization, and of any solid-solid phase changes.
Values of $C_p$ for temperatures near zero are not measured directly, but can be estimated from quantum theory. The cumulative areas from 0 K to any given temperature (Figure $3$) are then plotted as a function of $T$, and any phase-change entropies such as
$S_{vap} = \dfrac{H_{vap}}{T_b}$
are added to obtain the absolute entropy at temperature $T$. As shown in Figure $2$ above, the entropy of a substance increases with temperature, and it does so for two reasons:
• As the temperature rises, more microstates become accessible, allowing thermal energy to be more widely dispersed. This is reflected in the gradual increase of entropy with temperature.
• The molecules of solids, liquids, and gases have increasingly greater freedom to move around, facilitating the spreading and sharing of thermal energy. Phase changes are therefore accompanied by massive and discontinuous increase in the entropy.
Calculating $\Delta S_{sys}$
We can make careful calorimetric measurements to determine the temperature dependence of a substance’s entropy and to derive absolute entropy values under specific conditions. Standard entropies are given the label $S^o_{298}$ for values determined for one mole of substance at a pressure of 1 bar and a temperature of 298 K. The standard entropy change ($ΔS^o$) for any process may be computed from the standard entropies of its reactant and product species like the following:
$ΔS^o=\sum νS^o_{298}(\ce{products})−\sum νS^o_{298}(\ce{reactants}) \label{$6$}$
Here, $ν$ represents stoichiometric coefficients in the balanced equation representing the process. For example, $ΔS^o$ for the following reaction at room temperature
$m\ce{A}+n\ce{B}⟶x\ce{C}+y\ce{D} \label{$7$}$
is computed as the following:
$ΔS^o=[xS^o_{298}(\ce{C})+yS^o_{298}(\ce{D})]−[mS^o_{298}(\ce{A})+nS^o_{298}(\ce{B})] \label{$8$}$
Table $1$ lists some standard entropies at 298.15 K. You can find additional standard entropies in Tables T1 and T2
Gases Liquids Solids
Table $1$: Standard Molar Entropy Values of Selected Substances at 25°C
Substance $S^o$ [J/(mol•K)] Substance $S^o$ [J/(mol•K)] Substance $S^o$ [J/(mol•K)]
He 126.2 H2O 70.0 C (diamond) 2.4
H2 130.7 CH3OH 126.8 C (graphite) 5.7
Ne 146.3 Br2 152.2 LiF 35.7
Ar 154.8 CH3CH2OH 160.7 SiO2 (quartz) 41.5
Kr 164.1 C6H6 173.4 Ca 41.6
Xe 169.7 CH3COCl 200.8 Na 51.3
H2O 188.8 C6H12 (cyclohexane) 204.4 MgF2 57.2
N2 191.6 C8H18 (isooctane) 329.3 K 64.7
O2 205.2 NaCl 72.1
CO2 213.8 KCl 82.6
I2 260.7 I2 116.1
A closer examination of Table $1$ also reveals that substances with similar molecular structures tend to have similar $S^o$ values. Among crystalline materials, those with the lowest entropies tend to be rigid crystals composed of small atoms linked by strong, highly directional bonds, such as diamond ($S^o = 2.4 \,J/(mol•K)$). In contrast, graphite, the softer, less rigid allotrope of carbon, has a higher $S^o$ (5.7 J/(mol•K)) due to more disorder (microstates) in the crystal. Soft crystalline substances and those with larger atoms tend to have higher entropies because of increased molecular motion and disorder. Similarly, the absolute entropy of a substance tends to increase with increasing molecular complexity because the number of available microstates increases with molecular complexity. For example, compare the $S^o$ values for CH3OH(l) and CH3CH2OH(l). Finally, substances with strong hydrogen bonds have lower values of $S^o$, which reflects a more ordered structure.
Entropy increases with softer, less rigid solids, solids that contain larger atoms, and solids with complex molecular structures.
To calculate $ΔS^o$ for a chemical reaction from standard molar entropies, we use the familiar “products minus reactants” rule, in which the absolute entropy of each reactant and product is multiplied by its stoichiometric coefficient in the balanced chemical equation. Example $1$ illustrates this procedure for the combustion of the liquid hydrocarbon isooctane ($\ce{C8H18}$; 2,2,4-trimethylpentane).
Example $1$
Use the data in Table $1$ to calculate $ΔS^o$ for the reaction of liquid isooctane with $\ce{O2(g)}$ to give $\ce{CO2(g)}$ and $\ce{H2O(g)}$ at 298 K.
Given: standard molar entropies, reactants, and products
Asked for: ΔS°
Strategy:
Write the balanced chemical equation for the reaction and identify the appropriate quantities in Table $1$. Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each multiplied by their appropriate stoichiometric coefficients, to obtain $ΔS^o$ for the reaction.
Solution:
The balanced chemical equation for the complete combustion of isooctane ($\ce{C8H18}$) is as follows:
$\ce{C8H18(l) + 25/2 O2(g) -> 8CO2(g) + 9H2O(g)} \nonumber$
We calculate $ΔS^o$ for the reaction using the “products minus reactants” rule, where m and n are the stoichiometric coefficients of each product and each reactant:
\begin{align*}\Delta S^o_{\textrm{rxn}}&=\sum mS^o(\textrm{products})-\sum nS^o(\textrm{reactants}) \[4pt] &=[8S^o(\mathrm{CO_2})+9S^o(\mathrm{H_2O})]-[S^o(\mathrm{C_8H_{18}})+\dfrac{25}{2}S^o(\mathrm{O_2})] \[4pt] &=\left \{ [8\textrm{ mol }\mathrm{CO_2}\times213.8\;\mathrm{J/(mol\cdot K)}]+[9\textrm{ mol }\mathrm{H_2O}\times188.8\;\mathrm{J/(mol\cdot K)}] \right \} \[4pt] & \,\,\, -\left \{[1\textrm{ mol }\mathrm{C_8H_{18}}\times329.3\;\mathrm{J/(mol\cdot K)}]+\left [\dfrac{25}{2}\textrm{ mol }\mathrm{O_2}\times205.2\textrm{ J}/(\mathrm{mol\cdot K})\right ] \right \} \[4pt] &=515.3\;\mathrm{J/K}\end{align*}
$ΔS^o$ is positive, as expected for a combustion reaction in which one large hydrocarbon molecule is converted to many molecules of gaseous products.
Exercise $1$
Use the data in Table $1$ to calculate $ΔS^o$ for the reaction of $\ce{H2(g)}$ with liquid benzene ($\ce{C6H6}$) to give cyclohexane ($\ce{C6H12}$) at 298 K.
Answer
-361.1 J/K
Example $2$: Determination of ΔS°
Calculate the standard entropy change for the following process at 298 K:
$\ce{H2O}(g)⟶\ce{H2O}(l)\nonumber$
Solution
The value of the standard entropy change at room temperature, $ΔS^o_{298}$, is the difference between the standard entropy of the product, H2O(l), and the standard entropy of the reactant, H2O(g).
\begin{align*} ΔS^o_{298} &=S^o_{298}(\ce{H2O (l)})−S^o_{298}(\ce{H2O(g)})\nonumber \[4pt] &= (70.0\: J\:mol^{−1}K^{−1})−(188.8\: Jmol^{−1}K^{−1})\nonumber \[4pt] &=−118.8\:J\:mol^{−1}K^{−1} \end{align*}
The value for $ΔS^o_{298}$ is negative, as expected for this phase transition (condensation), which the previous section discussed.
Exercise $2$
Calculate the standard entropy change for the following process at 298 K:
$\ce{H2}(g)+\ce{C2H4}(g)⟶\ce{C2H6}(g)\nonumber$
Answer
−120.6 J mol−1 K−1
Example $3$: Determination of ΔS°
Calculate the standard entropy change for the combustion of methanol, CH3OH at 298 K:
$\ce{2CH3OH}(l)+\ce{3O2}(g)⟶\ce{2CO2}(g)+\ce{4H2O}(l)\nonumber$
Solution
The value of the standard entropy change is equal to the difference between the standard entropies of the products and the entropies of the reactants scaled by their stoichiometric coefficients. The standard entropy of formations are found in Table $1$.
\begin{align*} ΔS^o &=ΔS^o_{298} \[4pt] &= ∑νS^o_{298}(\ce{products})−∑νS^o_{298} (\ce{reactants}) \[4pt] & = 2S^o_{298}(\ce{CO2}(g))+4S^o_{298}(\ce{H2O}(l))]−[2S^o_{298}(\ce{CH3OH}(l))+3S^o_{298}(\ce{O2}(g))]\nonumber \[4pt] &= [(2 \times 213.8) + (4×70.0)]−[ (2 \times 126.8) + (3 \times 205.03) ]\nonumber \[4pt] &= −161.6 \:J/mol⋅K\nonumber \end{align*}
Exercise $3$
Calculate the standard entropy change for the following reaction at 298 K:
$\ce{Ca(OH)2}(s)⟶\ce{CaO}(s)+\ce{H2O}(l)\nonumber$
Answer
24.7 J/mol•K
Summary
Energy values, as you know, are all relative, and must be defined on a scale that is completely arbitrary; there is no such thing as the absolute energy of a substance, so we can arbitrarily define the enthalpy or internal energy of an element in its most stable form at 298 K and 1 atm pressure as zero. The same is not true of the entropy; since entropy is a measure of the “dilution” of thermal energy, it follows that the less thermal energy available to spread through a system (that is, the lower the temperature), the smaller will be its entropy. In other words, as the absolute temperature of a substance approaches zero, so does its entropy. This principle is the basis of the Third law of thermodynamics, which states that the entropy of a perfectly-ordered solid at 0 K is zero.
In practice, chemists determine the absolute entropy of a substance by measuring the molar heat capacity ($C_p$) as a function of temperature and then plotting the quantity $C_p/T$ versus $T$. The area under the curve between 0 K and any temperature T is the absolute entropy of the substance at $T$. In contrast, other thermodynamic properties, such as internal energy and enthalpy, can be evaluated in only relative terms, not absolute terms.
The second law of thermodynamics states that a spontaneous process increases the entropy of the universe, Suniv > 0. If ΔSuniv < 0, the process is nonspontaneous, and if ΔSuniv = 0, the system is at equilibrium. The third law of thermodynamics establishes the zero for entropy as that of a perfect, pure crystalline solid at 0 K. With only one possible microstate, the entropy is zero. We may compute the standard entropy change for a process by using standard entropy values for the reactants and products involved in the process.
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Learning Objectives
• To use thermodynamic cycles to calculate changes in entropy.
The atoms, molecules, or ions that compose a chemical system can undergo several types of molecular motion, including translation, rotation, and vibration (Figure $1$). The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. A perfectly ordered system with only a single microstate available to it would have an entropy of zero. The only system that meets this criterion is a perfect crystal at a temperature of absolute zero (0 K), in which each component atom, molecule, or ion is fixed in place within a crystal lattice and exhibits no motion (ignoring quantum effects). Such a state of perfect order (or, conversely, zero disorder) corresponds to zero entropy. In practice, absolute zero is an ideal temperature that is unobtainable, and a perfect single crystal is also an ideal that cannot be achieved. Nonetheless, the combination of these two ideals constitutes the basis for the third law of thermodynamics: the entropy of any perfectly ordered, crystalline substance at absolute zero is zero.
Third Law of Thermodynamics
The entropy of any perfectly ordered, crystalline substance at absolute zero is zero.
The third law of thermodynamics has two important consequences: it defines the sign of the entropy of any substance at temperatures above absolute zero as positive, and it provides a fixed reference point that allows us to measure the absolute entropy of any substance at any temperature.In practice, chemists determine the absolute entropy of a substance by measuring the molar heat capacity (Cp) as a function of temperature and then plotting the quantity Cp/T versus T. The area under the curve between 0 K and any temperature T is the absolute entropy of the substance at T. In contrast, other thermodynamic properties, such as internal energy and enthalpy, can be evaluated in only relative terms, not absolute terms. In this section, we examine two different ways to calculate ΔS for a reaction or a physical change. The first, based on the definition of absolute entropy provided by the third law of thermodynamics, uses tabulated values of absolute entropies of substances. The second, based on the fact that entropy is a state function, uses a thermodynamic cycle similar to those discussed previously.
Calculating ΔS from Standard Molar Entropy Values
One way of calculating ΔS for a reaction is to use tabulated values of the standard molar entropy (S°), which is the entropy of 1 mol of a substance at a standard temperature of 298 K; the units of S° are J/(mol•K). Unlike enthalpy or internal energy, it is possible to obtain absolute entropy values by measuring the entropy change that occurs between the reference point of 0 K [corresponding to S = 0 J/(mol•K)] and 298 K.
As shown in Table $1$, for substances with approximately the same molar mass and number of atoms, S° values fall in the order S°(gas) > S°(liquid) > S°(solid). For instance, S° for liquid water is 70.0 J/(mol•K), whereas S° for water vapor is 188.8 J/(mol•K). Likewise, S° is 260.7 J/(mol•K) for gaseous I2 and 116.1 J/(mol•K) for solid I2. This order makes qualitative sense based on the kinds and extents of motion available to atoms and molecules in the three phases. The correlation between physical state and absolute entropy is illustrated in Figure $2$, which is a generalized plot of the entropy of a substance versus temperature.
Table $1$: Standard Molar Entropy Values of Selected Substances at 25°C
Gases Liquids Solids
Substance S° [J/(mol•K)] Substance S° [J/(mol•K)] Substance S° [J/(mol•K)]
He 126.2 H2O 70.0 C (diamond) 2.4
H2 130.7 CH3OH 126.8 C (graphite) 5.7
Ne 146.3 Br2 152.2 LiF 35.7
Ar 154.8 CH3CH2OH 160.7 SiO2 (quartz) 41.5
Kr 164.1 C6H6 173.4 Ca 41.6
Xe 169.7 CH3COCl 200.8 Na 51.3
H2O 188.8 C6H12 (cyclohexane) 204.4 MgF2 57.2
N2 191.6 C8H18 (isooctane) 329.3 K 64.7
O2 205.2 NaCl 72.1
CO2 213.8 KCl 82.6
I2 260.7 I2 116.1
Note
Entropy increases with softer, less rigid solids, solids that contain larger atoms, and solids with complex molecular structures.
A closer examination of Table $1$ also reveals that substances with similar molecular structures tend to have similar S° values. Among crystalline materials, those with the lowest entropies tend to be rigid crystals composed of small atoms linked by strong, highly directional bonds, such as diamond [S° = 2.4 J/(mol•K)]. In contrast, graphite, the softer, less rigid allotrope of carbon, has a higher S° [5.7 J/(mol•K)] due to more disorder in the crystal. Soft crystalline substances and those with larger atoms tend to have higher entropies because of increased molecular motion and disorder. Similarly, the absolute entropy of a substance tends to increase with increasing molecular complexity because the number of available microstates increases with molecular complexity. For example, compare the S° values for CH3OH(l) and CH3CH2OH(l). Finally, substances with strong hydrogen bonds have lower values of S°, which reflects a more ordered structure.
Note
ΔS° for a reaction can be calculated from absolute entropy values using the same “products minus reactants” rule used to calculate ΔH°.
To calculate ΔS° for a chemical reaction from standard molar entropies, we use the familiar “products minus reactants” rule, in which the absolute entropy of each reactant and product is multiplied by its stoichiometric coefficient in the balanced chemical equation. Example $1$ illustrates this procedure for the combustion of the liquid hydrocarbon isooctane (C8H18; 2,2,4-trimethylpentane).
Example $1$
Use the data in Table $1$ to calculate ΔS° for the reaction of liquid isooctane with O2(g) to give CO2(g) and H2O(g) at 298 K.
Given: standard molar entropies, reactants, and products
Asked for: ΔS°
Strategy:
Write the balanced chemical equation for the reaction and identify the appropriate quantities in Table $1$. Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each multiplied by their appropriate stoichiometric coefficients, to obtain ΔS° for the reaction.
Solution:
The balanced chemical equation for the complete combustion of isooctane (C8H18) is as follows:
$\mathrm{C_8H_{18}(l)}+\dfrac{25}{2}\mathrm{O_2(g)}\rightarrow\mathrm{8CO_2(g)}+\mathrm{9H_2O(g)}$
We calculate ΔS° for the reaction using the “products minus reactants” rule, where m and n are the stoichiometric coefficients of each product and each reactant:
\begin{align}\Delta S^\circ_{\textrm{rxn}}&=\sum mS^\circ(\textrm{products})-\sum nS^\circ(\textrm{reactants})
\ &=[8S^\circ(\mathrm{CO_2})+9S^\circ(\mathrm{H_2O})]-[S^\circ(\mathrm{C_8H_{18}})+\dfrac{25}{2}S^\circ(\mathrm{O_2})]
\ &=\left \{ [8\textrm{ mol }\mathrm{CO_2}\times213.8\;\mathrm{J/(mol\cdot K)}]+[9\textrm{ mol }\mathrm{H_2O}\times188.8\;\mathrm{J/(mol\cdot K)}] \right \}
\ &-\left \{[1\textrm{ mol }\mathrm{C_8H_{18}}\times329.3\;\mathrm{J/(mol\cdot K)}]+\left [\dfrac{25}{2}\textrm{ mol }\mathrm{O_2}\times205.2\textrm{ J}/(\mathrm{mol\cdot K})\right ] \right \}
\ &=515.3\;\mathrm{J/K}\end{align}
ΔS° is positive, as expected for a combustion reaction in which one large hydrocarbon molecule is converted to many molecules of gaseous products.
Exercise $1$
Use the data in Table $1$ to calculate ΔS° for the reaction of H2(g) with liquid benzene (C6H6) to give cyclohexane (C6H12).
Answer: −361.1 J/K
Calculating ΔS from Thermodynamic Cycles
We can also calculate a change in entropy using a thermodynamic cycle. As you learned previously, the molar heat capacity (Cp) is the amount of heat needed to raise the temperature of 1 mol of a substance by 1°C at constant pressure. Similarly, Cv is the amount of heat needed to raise the temperature of 1 mol of a substance by 1°C at constant volume. The increase in entropy with increasing temperature in Figure $2$ is approximately proportional to the heat capacity of the substance.
Recall that the entropy change (ΔS) is related to heat flow (qrev) by ΔS = qrev/T. Because qrev = nCpΔT at constant pressure or nCvΔT at constant volume, where n is the number of moles of substance present, the change in entropy for a substance whose temperature changes from T1 to T2 is as follows:
$\Delta S=\dfrac{q_{\textrm{rev}}}{T}=nC_\textrm p\dfrac{\Delta T}{T}\hspace{4mm}(\textrm{constant pressure})$
As you will discover in more advanced math courses than is required here, it can be shown that this is equal to the following:For a review of natural logarithms, see Essential Skills 6 in Chapter 11.
$\Delta S=nC_\textrm p\ln\dfrac{T_2}{T_1}\hspace{4mm}(\textrm{constant pressure}) \label{18.20}$
Similarly,
$\Delta S=nC_{\textrm v}\ln\dfrac{T_2}{T_1}\hspace{4mm}(\textrm{constant volume}) \label{18.21}$
Thus we can use a combination of heat capacity measurements (Equation 18.20 or Equation 18.21) and experimentally measured values of enthalpies of fusion or vaporization if a phase change is involved (Equation 18.18) to calculate the entropy change corresponding to a change in the temperature of a sample.
We can use a thermodynamic cycle to calculate the entropy change when the phase change for a substance such as sulfur cannot be measured directly. As noted in the exercise in Example 6, elemental sulfur exists in two forms (part (a) in Figure $3$): an orthorhombic form with a highly ordered structure (Sα) and a less-ordered monoclinic form (Sβ). The orthorhombic (α) form is more stable at room temperature but undergoes a phase transition to the monoclinic (β) form at temperatures greater than 95.3°C (368.5 K). The transition from Sα to Sβ can be described by the thermodynamic cycle shown in part (b) in Figure $3$, in which liquid sulfur is an intermediate. The change in entropy that accompanies the conversion of liquid sulfur to Sβ (−ΔSfus(β) = ΔS3 in the cycle) cannot be measured directly. Because entropy is a state function, however, ΔS3 can be calculated from the overall entropy change (ΔSt) for the Sα–Sβ transition, which equals the sum of the ΔS values for the steps in the thermodynamic cycle, using Equation 18.20 and tabulated thermodynamic parameters (the heat capacities of Sα and Sβ, ΔHfus(α), and the melting point of Sα.)
If we know the melting point of Sα (Tm = 115.2°C = 388.4 K) and ΔSt for the overall phase transition [calculated to be 1.09 J/(mol•K) in the exercise in Example 6], we can calculate ΔS3 from the values given in part (b) in Figure $3$ where Cp(α) = 22.70 J/mol•K and Cp(β) = 24.77 J/mol•K (subscripts on ΔS refer to steps in the cycle):
\begin{align}\Delta S_{\textrm t}&=\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4 \ 1.09\;\mathrm{J/(mol\cdot K)}&=C_{\textrm p({\alpha})}\ln\left(\dfrac{T_2}{T_1}\right)+\dfrac{\Delta H_{\textrm{fus}}}{T_{\textrm m}}+\Delta S_3+C_{\textrm p(\beta)}\ln\left(\dfrac{T_4}{T_3}\right) \ &=22.70\;\mathrm{J/(mol\cdot K)}\ln\left(\dfrac{388.4}{368.5}\right)+\left(\dfrac{1.722\;\mathrm{kJ/mol}}{\textrm{388.4 K}}\times1000\textrm{ J/kJ}\right) \ &+\Delta S_3+24.77\;\mathrm{J/(mol\cdot K)}\ln\left(\dfrac{368.5}{388.4}\right) \ &=[1.194\;\mathrm{J/(mol\cdot K)}]+[4.434\;\mathrm{J/(mol\cdot K)}]+\Delta S_3+[-1.303\;\mathrm{J/(mol\cdot K)}]\end{align}
Solving for ΔS3 gives a value of −3.24 J/(mol•K). As expected for the conversion of a less ordered state (a liquid) to a more ordered one (a crystal), ΔS3 is negative.
How are Entropies Measured
The absolute entropy of a substance at any temperature above 0 K must be determined by calculating the increments of heat q required to bring the substance from 0 K to the temperature of interest, and then summing the ratios q/T. Two kinds of experimental measurements are needed:
1. The enthalpies associated with any phase changes the substance may undergo within the temperature range of interest. Melting of a solid and vaporization of a liquid correspond to sizeable increases in the number of microstates available to accept thermal energy, so as these processes occur, energy will flow into a system, filling these new microstates to the extent required to maintain a constant temperature (the freezing or boiling point); these inflows of thermal energy correspond to the heats of fusion and vaporization. The entropy increase associated with melting, for example, is just ΔHfusion/Tm.
2. The heat capacity C of a phase expresses the quantity of heat required to change the temperature by a small amount ΔT , or more precisely, by an infinitesimal amount dT . Thus the entropy increase brought about by warming a substance over a range of temperatures that does not encompass a phase transition is given by the sum of the quantities C dT/T for each increment of temperature dT . This is of course just the integral
$S_{0 \rightarrow T^o} = \int _{0}^{T^o} \dfrac{C_p}{T} dt$
Because the heat capacity is itself slightly temperature dependent, the most precise determinations of absolute entropies require that the functional dependence of C on T be used in the above integral in place of a constant C.
$S_{0 \rightarrow T^o} = \int _{0}^{T^o} \dfrac{C_p(T)}{T} dt$
When this is not known, one can take a series of heat capacity measurements over narrow temperature increments ΔT and measure the area under each section of the curve.
The area under each section of the plot represents the entropy change associated with heating the substance through an interval ΔT. To this must be added the enthalpies of melting, vaporization, and of any solid-solid phase changes. Values of Cp for temperatures near zero are not measured directly, but can be estimated from quantum theory.
The cumulative areas from 0 K to any given temperature (taken from the experimental plot on the left) are then plotted as a function of T, and any phase-change entropies such as Svap = Hvap / Tb are added to obtain the absolute entropy at temperature T.
Summary
• Entropy changes can be calculated using the “products minus reactants” rule or from a combination of heat capacity measurements and measured values of enthalpies of fusion or vaporization.
The third law of thermodynamics states that the entropy of any perfectly ordered, crystalline substance at absolute zero is zero. At temperatures greater than absolute zero, entropy has a positive value, which allows us to measure the absolute entropy of a substance. Measurements of the heat capacity of a substance and the enthalpies of fusion or vaporization can be used to calculate the changes in entropy that accompany a physical change. The entropy of 1 mol of a substance at a standard temperature of 298 K is its standard molar entropy (S°). We can use the “products minus reactants” rule to calculate the standard entropy change (ΔS°) for a reaction using tabulated values of S° for the reactants and the products.
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Gibbs energy is the energy of a chemical reaction that can be used to do non-mechanical work. Gibbs Energy is described as
$G = H - TS \label{1}$
Where H is enthalpy, T is temperature, and S is entropy. ΔG is used to predict spontaneity within a system by
$\Delta G_{sys} = \Delta H_{sys} – \Delta(TS)_{sys} \label{2}$
Gibbs energy was developed in the 1870’s by Josiah Willard Gibbs. He originally termed this energy as the “available energy” in a system. His paper published in 1873, “Graphical Methods in the Thermodynamics of Fluids,” outlined how his equation could predict the behavior of systems when they are combined.
At a constant temperature and pressure, the Gibbs Energy of a system can be described as
$\Delta G_{sys} = \Delta H_{sys} – T \Delta S_{sys} \label{3}$
This equation can be used to determine the spontaneity of the process.
• If ΔGsys ≤ 0, the process is spontaneous.
• If ΔGsys = 0, the system is at equilibrium.
• If ΔGsys > 0, the process is not spontaneous.
Furthermore,
• If ΔGsys < 0, the process is exergonic
• If ΔGsys > 0, the process is endergonic.
Gibbs Energy is a useful tool to describe in what manner the reaction is conducted. If ΔH >> TΔS, the reaction is enthalpy driven. However, if TΔS >> ΔH, the reaction is driven by entropy. The Clausius-Clapeyron Equation is an application derived from Gibb's energy:
$\ln \left(\dfrac{P_2}{P_1}\right) = \dfrac{Δ_{vap}H}{R} \left(\dfrac{T_2 - T_1}{T_2T_1} \right) \label{4}$
Another important application of Gibb's energy is the Maxwell relations (also available in a link at the end of the wiki page.)
Temperature & Pressure Dependence
When the pressure & temperature of a reaction are not held constant,
$G = H - TS \label{5}$
For an infinitesimal process,
$ΔG = ΔH – Δ(TS) \label{6}$
$ΔG = ΔH – TΔS –SΔT \label{7}$
For a reaction where temperature is held constant,
$ΔG = ΔH – TΔS \label{8}$
From the First Law of Thermodynamics, we know
$H= U +PV \label{9}$
$ΔH = ΔU + PΔV+VΔP \label{10}$
Since
$ΔU = TΔS – PΔV \label{11}$
We find that
$ΔG = VΔP-SΔT \label{12}$
showing the obvious dependence of ΔG on temperature and pressure. To observe the change in Gibbs energy due to temperature change alone (pressure held constant) the equation becomes
$ΔG=-SΔT \label{13}$
Solving for S and plugging it into Eq. (1), the Gibbs-Helmholtz Equation is found:
$\left(\dfrac{∂(ΔG/T)}{∂T}\right)_p = -\dfrac{ΔH}{T^2} \label{14}$
The Gibbs-Helmholtz Equation is very important because it relates the change in Gibbs energy to its temperature dependence, and the position of equilibrium to change in enthalpy.
To observe the change in Gibbs energy due to pressure change alone (temperature held constant) the equation becomes
$ΔG=VΔP \label{15}$
If the gas is assumed to be ideal then
$ΔG=nRT \ln \left(\dfrac{P_2}{P_1}\right) \label{16}$
for an initial and final pressure (P1 and P2) at a constant T.
Standard-State Free Energy of Formation
Gibbs Energy is defined as a state function (a property that depends only on conditions describing the system, not how the change occurs as in a path function.) This is because each component of the equation (H,T, and S) are all state functions. Therefore, we can know the change in Gibbs energy without knowing every detail of the process. In a process that takes place at constant temperature and pressure (298 K, 1 atm) the standard molar free energy of formation can be determined by the change in free energy from the reactants and products. Using predetermined values, Eqn. (17) can be used
$ΔG^o)f = \sum (Coefficient_{products} G^°_{products}) - \sum (Coefficient_{reactants} G^°_{reactants}) \label{17}$
Standard-State Free Energy of Reaction
Gibbs Energy can be found at standard-state conditions using
$ΔG° = ΔH° - TΔS° \label{18}$
ΔH° and ΔS° values can be found in the appendix of any general chemistry textbook, or using this link.
Free Energy of Reaction
Gibbs energy can be found at any conditions by relating it to the standard-state free energy of reaction, using
$ΔG = ΔG° +RT \ln Q \label{19}$
Where Q is the reaction quotient. Very rarely does chemistry actually occur at the given "standard-state" conditions. Using the above equation and standard-state values, chemists can determine the overall Gibb's energy for the system, regardless of the conditions.
Problems
1. Calculate the standard-state free energy of formation for the H2O2(l) from H2 and O2, given the following values:
ΔfG (H2): 0 kJ/mol
ΔfG (O2): 0 kJ/mol
ΔfG (H2O2(l)): -120.4 kJ/mol
Hint: (product*coefficient) - [(Reactant1*coefficient)+(Reactant2*coefficient)]
2. Consider the following reaction: $CaCO_3 CaO + CO_2$
At what temperature will this reaction become favorable? Note: Assume ΔHr°and ΔSr° are temperature independent.
Given values:
Substance ΔHf° Sf°
CaCO3 -1206.9 kJ/mol 92.9 J/K mol
CaO -635.6 kJ/mol 39.8 J/K mol
CO2 -393.5 kJ/mol 213.6 J/K mol
Hint: At what value of ΔG does a reaction become spontaneous, and therefore favorable? Next, find the ΔH value for the reaction (products-reactants) and the ΔS values for the reaction (products-reactants.) Solve for T.
3. Is the above decomposition of calcium carbonate enthalpy driven or entropy driven?
Hint: Is ΔH >> TΔS, or TΔS >> ΔH?
Anwers to Problems
1. -120.4 kJ/mol
2. 1107.8 K
3. Enthalpy
Contributor
• Christine Gobrogge (Hope College)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/04%3A_The_Second_Law_of_Thermodynamics/4.06%3A_Gibbs_Energy.txt
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Learning Objectives
• To get an overview of Gibbs energy and its general uses in chemistry.
• Understand how Gibbs energy pertains to reactions properties
• Understand how Gibbs energy pertains to equilibria properties
• Understand how Gibbs energy pertains to electrochemical properties
Gibbs free energy, denoted $G$, combines enthalpy and entropy into a single value. The change in free energy, $\Delta G$, is equal to the sum of the enthalpy plus the product of the temperature and entropy of the system. $\Delta G$ can predict the direction of the chemical reaction under two conditions:
1. constant temperature and
2. constant pressure.
If $ΔG$ is positive, then the reaction is nonspontaneous (i.e., an the input of external energy is necessary for the reaction to occur) and if it is negative, then it is spontaneous (occurs without external energy input).
Introduction
Gibbs energy was developed in the 1870’s by Josiah Willard Gibbs. He originally termed this energy as the “available energy” in a system. His paper published in 1873, “Graphical Methods in the Thermodynamics of Fluids,” outlined how his equation could predict the behavior of systems when they are combined. This quantity is the energy associated with a chemical reaction that can be used to do work, and is the sum of its enthalpy (H) and the product of the temperature and the entropy (S) of the system. This quantity is defined as follows:
$G= H-TS \label{1.1}$
or more completely as
$G= U+PV-TS \label{1.2}$
where
• $U$ is internal energy (SI unit: joule)
• $P$ is pressure (SI unit: pascal)
• $V$ is volume (SI unit: $m^3$)
• $T$is temperature (SI unit: kelvin)
• $S$ is entropy (SI unit: joule/kelvin)
• $H$ is the enthalpy (SI unit: joule)
Gibbs Energy in Reactions
Spontaneous - is a reaction that is consider to be natural because it is a reaction that occurs by itself without any external action towards it. Non spontaneous - needs constant external energy applied to it in order for the process to continue and once you stop the external action the process will cease. When solving for the equation, if change of G is negative, then it's spontaneous. If change of G if positive, then it's non spontaneous. The symbol that is commonly used for FREE ENERGY is G. can be more properly consider as "standard free energy change"
In chemical reactions involving the changes in thermodynamic quantities, a variation on this equation is often encountered:
$\underset{\text {change in free energy} }{\Delta G } = \underset{ \text {change in enthalpy}}{ \Delta H } - \underset{\text {(temperature) change in entropy}}{T \Delta S} \label{1.3}$
Example 1.1
Calculate ∆G at 290 K for the following reaction:
$\ce{2NO(g) + O2(g) \rightarrow 2NO2(g)} \nonumber$
Given
• ∆H = -120 kJ
• ∆S = -150 JK -1
Solution
now all you have to do is plug in all the given numbers into Equation 3 above. Remember to divide $\Delta S$ by 1000 $J/kJ$ so that after you multiply by temperature, $T$, it will have the same units, $kJ$, as $\Delta H$.
$\Delta S = -150 \cancel{J}/K \left( \dfrac{1\; kJ}{1000\;\cancel{J}} \right) = -0.15\; kJ/K \nonumber$
and substituting into Equation 3:
\begin{align*} ∆G &= -120\; kJ - (290 \;\cancel{K})(-0.150\; kJ/\cancel{K}) \[4pt] &= -120 \;kJ + 43 \;kJ \[4pt] &= -77\; kJ \end{align*}
Exercise 1.1: The Haber Process
What is the $\Delta G$ for this formation of ammonia from nitrogen and hydrogen gas.
$\ce{N_2 + 3H_2 \rightleftharpoons 2NH_3} \nonumber$
The Standard free energy formations: NH3 =-16.45 H2=0 N2=0
Answer
$\Delta G=-32.90\;kJ \;mol^{-1} \nonumber$
Since the changes of entropy of chemical reaction are not measured readily, thus, entropy is not typically used as a criterion. To obviate this difficulty, we can use $G$. The sign of ΔG indicates the direction of a chemical reaction and determine if a reaction is spontaneous or not.
• $\Delta G < 0$: reaction is spontaneous in the direction written (i.e., the reaciton is exergonic)
• $\Delta G =0$: the system is at equilibrium and there is no net change either in forward or reverse direction.
• $\Delta G > 0$: reaction is not spontaneous and the process proceeds spontaneously in the reserve direction. To drive such a reaction, we need to have input of free energy (i.e., the reaction is endergonic)
The factors affect $\Delta G$ of a reaction (assume $\Delta H$ and $\Delta S$ are independent of temperature):
$\Delta H$ $\Delta S$ $\Delta G$ Example
+ + at low temperature: + , at high temperature: - 2HgO(s) -> 2Hg (l) + O2 (g)
+ - at all temperature: + 3O2 (g) ->2O3 (g)
- + at all temperature: - 2H2O2 (l) -> 2H2O (l) + O2 (g)
- - at low temperature: - , at high temperature: + NH3 (g) + HCl (g) -> NH4Cl (s)
Note:
1. $\Delta G$ depends only on the difference in free energy of products and reactants (or final state and initial state). $\Delta G$ is independent of the path of the transformation and is unaffected by the mechanism of a reaction.
2. $\Delta G$ cannot tell us anything about the rate of a reaction.
The standard Gibbs energy change $\Delta G^o$ (at which reactants are converted to products at 1 bar) for:
$aA + bB \rightarrow cC + dD \label{1.4}$
$\Delta r G^o = c \Delta _fG^o (C) + d \Delta _fG^o (D) - a \Delta _fG^o (A) - b \Delta _fG^o (B) \label{1.5}$
$\Delta _fG^0 = \sum v \Delta _f G^0 (\text {products}) - \sum v \Delta _f G^0 (\text {reactants}) \label{1.6}$
The standard-state free energy of reaction ( $\Delta G^o$) is defined as the free energy of reaction at standard state conditions:
$\Delta G^o = \Delta H^o - T \Delta S^o \label{1.7}$
Note
• If $\left | \Delta H \right | >> \left | T\Delta S \right |$: the reaction is enthalpy-driven
• If $\Delta H$ << $T\Delta S$: the reaction is entropy-driven
Standard-State Free Energy of Formation
• The partial pressure of any gas involved in the reaction is 0.1 MPa.
• The concentrations of all aqueous solutions are 1 M.
• Measurements are generally taken at a temperature of 25° C (298 K).
The standard-state free energy of formation is the change in free energy that occurs when a compound is formed from its elements in their most thermodynamically stable states at standard-state conditions. In other words, it is the difference between the free energy of a substance and the free energies of its constituent elements at standard-state conditions:
$\Delta G^o = \sum \Delta G^o_{f_{products}} - \sum \Delta G^o_{f_{reactants}} \label{1.8}$
Example 1.2
Used the below information to determine if $NH_4NO_{3(s)}$ will dissolve in water at room temperature.
Compound $\Delta H_f^o$ $\Delta S_f^o$
$NH_4NO_{3(s)}$ -365.56 151.08
$NH^+_{4(aq)}$ -132.51 113.4
$NO_{3(aq)}^-$ 205.0 146.4
Solution
This question is essentially asking if the following reaction is spontaneous at room temperature.
$\ce{NH4NO3(s) \overset{H_2O} \longrightarrow NH4(aq)^{+} + NO3(aq)^{-}} \nonumber$
This would normally only require calculating $\Delta{G^o}$ and evaluating its sign. However, the $\Delta{G^o}$ values are not tabulated, so they must be calculated manually from calculated $\Delta{H^o}$ and $\Delta{S^o}$ values for the reaction.
• Calculate $\Delta{H^o}$:
$\Delta H^o = \sum n\Delta H^o_{f_{products}} - \sum m\Delta H^o_{f_{reactants}} \nonumber$
$\Delta H^o= \left[ \left( 1\; mol\; NH_3\right)\left(-132.51\;\dfrac{kJ}{mol} \right) + \left( 1\; mol\; NO_3^- \right) \left(-205.0\;\dfrac{kJ}{mol}\right) \right] \nonumber$
$- \left[ \left(1\; mol\; NH_4NO_3 \right)\left(-365.56 \;\dfrac{kJ}{mol}\right) \right] \nonumber$
$\Delta H^o = -337.51 \;kJ + 365.56 \; kJ= 28.05 \;kJ \nonumber$
• Calculate $\Delta{S^o}$:
$\Delta S^o = \sum n\Delta S^o_{f_{products}} - \sum S\Delta H^o_{f_{reactants}} \nonumber$
$\Delta S^o= \left[ \left( 1\; mol\; NH_3\right)\left(113.4 \;\dfrac{J}{mol\;K} \right) + \left( 1\; mol\; NO_3^- \right) \left(146.6\;\dfrac{J}{mol\;K}\right) \right] \nonumber$
$- \left[ \left(1\; mol\; NH_4NO_3 \right)\left(151.08 \;\dfrac{J}{mol\;K}\right) \right] \nonumber$
$\Delta S^o = 259.8 \;J/K - 151.08 \; J/K= 108.7 \;J/K \nonumber$
• Calculate $\Delta{G^o}$:
These values can be substituted into the free energy equation
$T_K = 25\;^oC + 273.15K = 298.15\;K \nonumber$
$\Delta{S^o} = 108.7\; \cancel{J}/K \left(\dfrac{1\; kJ}{1000\;\cancel{J}} \right) = 0.1087 \; kJ/K \nonumber$
$\Delta{H^o} = 28.05\;kJ \nonumber$
Plug in $\Delta H^o$, $\Delta S^o$ and $T$ into Equation 1.7
$\Delta G^o = \Delta H^o - T \Delta S^o \nonumber$
$\Delta G^o = 28.05\;kJ - (298.15\; \cancel{K})(0.1087\;kJ/ \cancel{K}) \nonumber$
$\Delta G^o= 28.05\;kJ - 32.41\; kJ \nonumber$
$\Delta G^o = -4.4 \;kJ \nonumber$
This reaction is spontaneous at room temperature since $\Delta G^o$ is negative. Therefore $NH_4NO_{3(s)}$ will dissolve in water at room temperature.
Example 1.3
Calculate $\Delta{G}$ for the following reaction at $25\; ^oC$. Will the reaction occur spontaneously?
$NH_{3(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)} \nonumber$
given for the reaction
• $\Delta{H} = -176.0 \;kJ$
• $\Delta{S} = -284.8\;J/K$
Solution
calculate $\Delta{G}$ from the formula
$\Delta{G} = \Delta{H} - T\Delta{S} \nonumber$
but first we need to convert the units for $\Delta{S}$ into kJ/K (or convert $\Delta{H}$ into J) and temperature into Kelvin
• $\Delta{S} = -284.8 \cancel{J}/K \left( \dfrac{1\, kJ}{1000\; \cancel{J}}\right) = -0.284.8\; kJ/K$
• $T=273.15\; K + 25\; ^oC = 298\;K$
The definition of Gibbs energy can then be used directly
$\Delta{G} = \Delta{H} - T\Delta{S} \nonumber$
$\Delta{G} = -176.0 \;kJ - (298 \cancel{K}) (-0.284.8\; kJ/\cancel{K}) \nonumber$
$\Delta{G} = -176.0 \;kJ - (-84.9\; kJ) \nonumber$
$\Delta{G} = -91.1 \;kJ \nonumber$
Yes, this reaction is spontaneous at room temperature since $\Delta{G}$ is negative.
Gibbs Energy in Equilibria
Let's consider the following reversible reaction:
$A + B \leftrightharpoons C + D \label{1.9}$
The following equation relates the standard-state free energy of reaction with the free energy at any point in a given reaction (not necessarily at standard-state conditions):
$\Delta G = \Delta G^o + RT \ln Q \label{1.10}$
• $\Delta G$ = free energy at any moment
• $\Delta G^o$ = standard-state free energy
• R is the ideal gas constant = 8.314 J/mol-K
• T is the absolute temperature (Kelvin)
• $\ln Q$ is natural logarithm of the reaction quotient
At equilibrium, ΔG = 0 and Q=K. Thus the equation can be arranged into:
$\Delta{G} = \Delta{G}^o + RT \ln \dfrac{[C][D]}{[A][B]} \label{1.11}$
with
• $\Delta{G}^o$ = standard free energy change
• $R$ = gas constant = 1.98 * 10-3 kcal mol-1 deg-10
• $T$ = is usually room temperature = 298 K
• $K = \dfrac{[C][D]}{[A][B]}$
The Gibbs free energy $\Delta{G}$ depends primarily on the reactants' nature and concentrations (expressed in the $\Delta{G}^o$ term and the logarithmic term of Equation 1.11, respectively).
At equilibrium, $\Delta{G} = 0$: no driving force remains
$0 = \Delta{G}^{o'} + RT \ln \dfrac{[C][D]}{[A][B]} \label{1.12}$
$\Delta{G}^{o} = -RT \ln\dfrac{[C][D]}{[A][B]} \label{1.13}$
The equilibrium constant is defined as
$K_{eq} = \dfrac{[C][D]}{[A][B]} \label{1.14}$
When $K_{eq}$ is large, almost all reactants are converted to products. Substituting $K_{eq}$ into Equation 1.14, we have:
$\Delta{G}^{o} = -RT \ln K_{eq} \label{1.15}$
or
$\Delta{G}^{o} = -2.303RT log_{10} K_{eq} \label{1.16}$
Rearrange,
$K_{eq} = 10^{-\Delta{G}^{o}/(2.303RT)} \label{1.17}$
This equation is particularly interesting as it relates the free energy difference under standard conditions to the properties of a system at equilibrium (which is rarely at standard conditions).
Table 1.1: Converting $K_{ea}$ to $\Delta{G}$
$K_{eq}$ $\Delta{G_o}\; (kcal/mole)$
$10^{-5}$ 6.82
$10^{-4}$ 5.46
$10^{-3}$ 4.09
$10^{-2}$ 2.73
$10^{-1}$ 1.36
1 0
$10^{1}$ -1.36
$10^{2}$ -2.73
$10^{3}$ -4.09
$10^{4}$ -5.46
$10^{5}$ -6.82
Example 1.4
What is $\Delta{G}^{o}$ for isomerization of dihydroxyacetone phosphate to glyceraldehyde 3-phosphate?
If at equilibrium, we have $K_{eq} = 0.0475$ at 298 K and pH 7. We can calculate:
$\Delta{G}^{o} = -2.303\;RT log_{10} K_{eq}= (-2.303) * (1.98 * 10^{-3}) * 298 * (log_{10} 0.0475) = 1.8 \;kcal/mol \nonumber$
Given:
• The initial concentration of dihydroxyacetone phosphate = $2 \times 10^{-4}\; M$
• The initial concentration of glyceraldehyde 3-phosphate = $3 \times 10^{-6}\; M$
Solution
From equation 2:
$\Delta{G}$ = 1.8 kcal/mol + 2.303 RT log10(3*10-6 M/2*10-4 M) = -0.7 kcal/mol
Note
Under non-standard conditions (which is essential all reactions), the spontaneity of reaction is determined by $\Delta{G}$, not $\Delta{G}^{o'}$.
Gibbs Energy in Electrochemistry
The Nernst equation relates the standard-state cell potential with the cell potential of the cell at any moment in time:
$E = E^o - \dfrac {RT}{nF} \ln Q \label{1.18}$
with
• $E$ = cell potential in volts (joules per coulomb)
• $n$ = moles of electrons
• $F$ = Faraday's constant: 96,485 coulombs per mole of electrons
By rearranging this equation we obtain:
$E = E^o - \dfrac {RT}{nF} \ln Q \label{1.19}$
multiply the entire equation by $nF$
$nFE = nFE^o - RT \ln Q \label{1.20}$
which is similar to:
$\Delta G = \Delta G^o + RT \ln Q \label{1.21}$
By juxtaposing these two equations:
$nFE = nFE^o - RT \ln Q \label{1.22}$
$\Delta G = \Delta G^o + RT \ln Q \label{1.23}$
it can be concluded that:
$\Delta G = -nFE \label{1.24}$
Therefore,
$\Delta G^o = -nFE^o \label{1.25}$
Some remarks on the Gibbs "Free" Energy
• Free Energy is not necessarily "free": The appellation “free energy” for G has led to so much confusion that many scientists now refer to it simply as the Gibbs energy. The “free” part of the older name reflects the steam-engine origins of thermodynamics with its interest in converting heat into work: ΔG is the maximum amount of energy which can be “freed” from the system to perform useful work. By "useful", we mean work other than that which is associated with the expansion of the system. This is most commonly in the form of electrical work (moving electric charge through a potential difference), but other forms of work (osmotic work, increase in surface area) are also possible.
• Free Energy is not energy: A much more serious difficulty with the Gibbs function, particularly in the context of chemistry, is that although G has the units of energy (joules, or in its intensive form, J mol–1), it lacks one of the most important attributes of energy in that it is not conserved. Thus although the free energy always falls when a gas expands or a chemical reaction takes place spontaneously, there need be no compensating increase in energy anywhere else. Referring to G as an energy also reinforces the false but widespread notion that a fall in energy must accompany any change. But if we accept that energy is conserved, it is apparent that the only necessary condition for change (whether the dropping of a weight, expansion of a gas, or a chemical reaction) is the redistribution of energy.The quantity –ΔG associated with a process represents the quantity of energy that is “shared and spread”, which as we have already explained is the meaning of the increase in the entropy. The quotient –ΔG/T is in fact identical with ΔStotal, the entropy change of the world, whose increase is the primary criterion for any kind of change.
• Free Energy is not even "real": G differs from the thermodynamic quantities H and S in another significant way: it has no physical reality as a property of matter, whereas H and S can be related to the quantity and distribution of energy in a collection of molecules (e.g., the third law of thermodynamics). The free energy is simply a useful construct that serves as a criterion for change and makes calculations easier.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/04%3A_The_Second_Law_of_Thermodynamics/4.08%3A_Dependence_of_Gibbs_Energy_on_Temperature_and_Pressure.txt
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Learning Objectives
• To know how and why the vapor pressure of a liquid varies with temperature.
• To understand that the equilibrium vapor pressure of a liquid depends on the temperature and the intermolecular forces present.
• To understand that the relationship between pressure, enthalpy of vaporization, and temperature is given by the Clausius-Clapeyron equation.
Nearly all of us have heated a pan of water with the lid in place and shortly thereafter heard the sounds of the lid rattling and hot water spilling onto the stovetop. When a liquid is heated, its molecules obtain sufficient kinetic energy to overcome the forces holding them in the liquid and they escape into the gaseous phase. By doing so, they generate a population of molecules in the vapor phase above the liquid that produces a pressure—the vapor pressure of the liquid. In the situation we described, enough pressure was generated to move the lid, which allowed the vapor to escape. If the vapor is contained in a sealed vessel, however, such as an unvented flask, and the vapor pressure becomes too high, the flask will explode (as many students have unfortunately discovered). In this section, we describe vapor pressure in more detail and explain how to quantitatively determine the vapor pressure of a liquid.
Evaporation and Condensation
Because the molecules of a liquid are in constant motion, we can plot the fraction of molecules with a given kinetic energy (KE) against their kinetic energy to obtain the kinetic energy distribution of the molecules in the liquid (Figure $1$), just as we did for a gas. As for gases, increasing the temperature increases both the average kinetic energy of the particles in a liquid and the range of kinetic energy of the individual molecules. If we assume that a minimum amount of energy ($E_0$) is needed to overcome the intermolecular attractive forces that hold a liquid together, then some fraction of molecules in the liquid always has a kinetic energy greater than $E_0$. The fraction of molecules with a kinetic energy greater than this minimum value increases with increasing temperature. Any molecule with a kinetic energy greater than $E_0$ has enough energy to overcome the forces holding it in the liquid and escape into the vapor phase. Before it can do so, however, a molecule must also be at the surface of the liquid, where it is physically possible for it to leave the liquid surface; that is, only molecules at the surface can undergo evaporation (or vaporization), where molecules gain sufficient energy to enter a gaseous state above a liquid’s surface, thereby creating a vapor pressure.
To understand the causes of vapor pressure, consider the apparatus shown in Figure $2$. When a liquid is introduced into an evacuated chamber (part (a) in Figure $2$), the initial pressure above the liquid is approximately zero because there are as yet no molecules in the vapor phase. Some molecules at the surface, however, will have sufficient kinetic energy to escape from the liquid and form a vapor, thus increasing the pressure inside the container. As long as the temperature of the liquid is held constant, the fraction of molecules with $KE > E_0$ will not change, and the rate at which molecules escape from the liquid into the vapor phase will depend only on the surface area of the liquid phase.
As soon as some vapor has formed, a fraction of the molecules in the vapor phase will collide with the surface of the liquid and reenter the liquid phase in a process known as condensation (part (b) in Figure $2$). As the number of molecules in the vapor phase increases, the number of collisions between vapor-phase molecules and the surface will also increase. Eventually, a steady state will be reached in which exactly as many molecules per unit time leave the surface of the liquid (vaporize) as collide with it (condense). At this point, the pressure over the liquid stops increasing and remains constant at a particular value that is characteristic of the liquid at a given temperature. The rates of evaporation and condensation over time for a system such as this are shown graphically in Figure $3$.
Equilibrium Vapor Pressure
Two opposing processes (such as evaporation and condensation) that occur at the same rate and thus produce no net change in a system, constitute a dynamic equilibrium. In the case of a liquid enclosed in a chamber, the molecules continuously evaporate and condense, but the amounts of liquid and vapor do not change with time. The pressure exerted by a vapor in dynamic equilibrium with a liquid is the equilibrium vapor pressure of the liquid.
If a liquid is in an open container, however, most of the molecules that escape into the vapor phase will not collide with the surface of the liquid and return to the liquid phase. Instead, they will diffuse through the gas phase away from the container, and an equilibrium will never be established. Under these conditions, the liquid will continue to evaporate until it has “disappeared.” The speed with which this occurs depends on the vapor pressure of the liquid and the temperature. Volatile liquids have relatively high vapor pressures and tend to evaporate readily; nonvolatile liquids have low vapor pressures and evaporate more slowly. Although the dividing line between volatile and nonvolatile liquids is not clear-cut, as a general guideline, we can say that substances with vapor pressures greater than that of water (Figure $4$) are relatively volatile, whereas those with vapor pressures less than that of water are relatively nonvolatile. Thus diethyl ether (ethyl ether), acetone, and gasoline are volatile, but mercury, ethylene glycol, and motor oil are nonvolatile.
The equilibrium vapor pressure of a substance at a particular temperature is a characteristic of the material, like its molecular mass, melting point, and boiling point. It does not depend on the amount of liquid as long as at least a tiny amount of liquid is present in equilibrium with the vapor. The equilibrium vapor pressure does, however, depend very strongly on the temperature and the intermolecular forces present, as shown for several substances in Figure $4$. Molecules that can hydrogen bond, such as ethylene glycol, have a much lower equilibrium vapor pressure than those that cannot, such as octane. The nonlinear increase in vapor pressure with increasing temperature is much steeper than the increase in pressure expected for an ideal gas over the corresponding temperature range. The temperature dependence is so strong because the vapor pressure depends on the fraction of molecules that have a kinetic energy greater than that needed to escape from the liquid, and this fraction increases exponentially with temperature. As a result, sealed containers of volatile liquids are potential bombs if subjected to large increases in temperature. The gas tanks on automobiles are vented, for example, so that a car won’t explode when parked in the sun. Similarly, the small cans (1–5 gallons) used to transport gasoline are required by law to have a pop-off pressure release.
Volatile substances have low boiling points and relatively weak intermolecular interactions; nonvolatile substances have high boiling points and relatively strong intermolecular interactions.
A Video Discussing Vapor Pressure and Boiling Points. Video Source: Vapor Pressure & Boiling Point(opens in new window) [youtu.be]
The exponential rise in vapor pressure with increasing temperature in Figure $4$ allows us to use natural logarithms to express the nonlinear relationship as a linear one.
$\boxed{\ln P =\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T} \right) + C} \label{Eq1}$
where
• $\ln P$ is the natural logarithm of the vapor pressure,
• $ΔH_{vap}$ is the enthalpy of vaporization,
• $R$ is the universal gas constant [8.314 J/(mol•K)],
• $T$ is the temperature in kelvins, and
• $C$ is the y-intercept, which is a constant for any given line.
Plotting $\ln P$ versus the inverse of the absolute temperature ($1/T$) is a straight line with a slope of −ΔHvap/R. Equation $\ref{Eq1}$, called the Clausius–Clapeyron Equation, can be used to calculate the $ΔH_{vap}$ of a liquid from its measured vapor pressure at two or more temperatures. The simplest way to determine $ΔH_{vap}$ is to measure the vapor pressure of a liquid at two temperatures and insert the values of $P$ and $T$ for these points into Equation $\ref{Eq2}$, which is derived from the Clausius–Clapeyron equation:
$\ln\left ( \dfrac{P_{1}}{P_{2}} \right)=\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \label{Eq2}$
Conversely, if we know ΔHvap and the vapor pressure $P_1$ at any temperature $T_1$, we can use Equation $\ref{Eq2}$ to calculate the vapor pressure $P_2$ at any other temperature $T_2$, as shown in Example $1$.
A Video Discussing the Clausius-Clapeyron Equation. Video Link: The Clausius-Clapeyron Equation(opens in new window) [youtu.be]
Example $1$: Vapor Pressure of Mercury
The experimentally measured vapor pressures of liquid Hg at four temperatures are listed in the following table:
experimentally measured vapor pressures of liquid Hg at four temperatures
T (°C) 80.0 100 120 140
P (torr) 0.0888 0.2729 0.7457 1.845
From these data, calculate the enthalpy of vaporization (ΔHvap) of mercury and predict the vapor pressure of the liquid at 160°C. (Safety note: mercury is highly toxic; when it is spilled, its vapor pressure generates hazardous levels of mercury vapor.)
Given: vapor pressures at four temperatures
Asked for: ΔHvap of mercury and vapor pressure at 160°C
Strategy:
1. Use Equation $\ref{Eq2}$ to obtain ΔHvap directly from two pairs of values in the table, making sure to convert all values to the appropriate units.
2. Substitute the calculated value of ΔHvap into Equation $\ref{Eq2}$ to obtain the unknown pressure (P2).
Solution:
A The table gives the measured vapor pressures of liquid Hg for four temperatures. Although one way to proceed would be to plot the data using Equation $\ref{Eq1}$ and find the value of ΔHvap from the slope of the line, an alternative approach is to use Equation $\ref{Eq2}$ to obtain ΔHvap directly from two pairs of values listed in the table, assuming no errors in our measurement. We therefore select two sets of values from the table and convert the temperatures from degrees Celsius to kelvin because the equation requires absolute temperatures. Substituting the values measured at 80.0°C (T1) and 120.0°C (T2) into Equation $\ref{Eq2}$ gives
\begin{align*} \ln \left ( \dfrac{0.7457 \; \cancel{Torr}}{0.0888 \; \cancel{Torr}} \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot K}\left ( \dfrac{1}{\left ( 120+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right)K} \right) \[4pt] \ln\left ( 8.398 \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot \cancel{K}}\left ( -2.88\times 10^{-4} \; \cancel{K^{-1}} \right) \[4pt] 2.13 &=-\Delta H_{vap} \left ( -3.46 \times 10^{-4} \right) J^{-1}\cdot mol \[4pt] \Delta H_{vap} &=61,400 \; J/mol = 61.4 \; kJ/mol \end{align*} \nonumber
B We can now use this value of ΔHvap to calculate the vapor pressure of the liquid (P2) at 160.0°C (T2):
$\ln\left ( \dfrac{P_{2} }{0.0888 \; torr} \right)=\dfrac{-61,400 \; \cancel{J/mol}}{8.314 \; \cancel{J/mol} \; K^{-1}}\left ( \dfrac{1}{\left ( 160+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right) K} \right) \nonumber$
Using the relationship $e^{\ln x} = x$, we have
\begin{align*} \ln \left ( \dfrac{P_{2} }{0.0888 \; Torr} \right) &=3.86 \[4pt] \dfrac{P_{2} }{0.0888 \; Torr} &=e^{3.86} = 47.5 \[4pt] P_{2} &= 4.21 Torr \end{align*} \nonumber
At 160°C, liquid Hg has a vapor pressure of 4.21 torr, substantially greater than the pressure at 80.0°C, as we would expect.
Exercise $1$: Vapor Pressure of Nickel
The vapor pressure of liquid nickel at 1606°C is 0.100 torr, whereas at 1805°C, its vapor pressure is 1.000 torr. At what temperature does the liquid have a vapor pressure of 2.500 torr?
Answer
1896°C
Boiling Points
As the temperature of a liquid increases, the vapor pressure of the liquid increases until it equals the external pressure, or the atmospheric pressure in the case of an open container. Bubbles of vapor begin to form throughout the liquid, and the liquid begins to boil. The temperature at which a liquid boils at exactly 1 atm pressure is the normal boiling point of the liquid. For water, the normal boiling point is exactly 100°C. The normal boiling points of the other liquids in Figure $4$ are represented by the points at which the vapor pressure curves cross the line corresponding to a pressure of 1 atm. Although we usually cite the normal boiling point of a liquid, the actual boiling point depends on the pressure. At a pressure greater than 1 atm, water boils at a temperature greater than 100°C because the increased pressure forces vapor molecules above the surface to condense. Hence the molecules must have greater kinetic energy to escape from the surface. Conversely, at pressures less than 1 atm, water boils below 100°C.
Table $1$: The Boiling Points of Water at Various Locations on Earth
Place Altitude above Sea Level (ft) Atmospheric Pressure (mmHg) Boiling Point of Water (°C)
Mt. Everest, Nepal/Tibet 29,028 240 70
Bogota, Colombia 11,490 495 88
Denver, Colorado 5280 633 95
Washington, DC 25 759 100
Dead Sea, Israel/Jordan −1312 799 101.4
Typical variations in atmospheric pressure at sea level are relatively small, causing only minor changes in the boiling point of water. For example, the highest recorded atmospheric pressure at sea level is 813 mmHg, recorded during a Siberian winter; the lowest sea-level pressure ever measured was 658 mmHg in a Pacific typhoon. At these pressures, the boiling point of water changes minimally, to 102°C and 96°C, respectively. At high altitudes, on the other hand, the dependence of the boiling point of water on pressure becomes significant. Table $1$ lists the boiling points of water at several locations with different altitudes. At an elevation of only 5000 ft, for example, the boiling point of water is already lower than the lowest ever recorded at sea level. The lower boiling point of water has major consequences for cooking everything from soft-boiled eggs (a “three-minute egg” may well take four or more minutes in the Rockies and even longer in the Himalayas) to cakes (cake mixes are often sold with separate high-altitude instructions). Conversely, pressure cookers, which have a seal that allows the pressure inside them to exceed 1 atm, are used to cook food more rapidly by raising the boiling point of water and thus the temperature at which the food is being cooked.
As pressure increases, the boiling point of a liquid increases and vice versa.
Example $2$: Boiling Mercury
Use Figure $4$ to estimate the following.
1. the boiling point of water in a pressure cooker operating at 1000 mmHg
2. the pressure required for mercury to boil at 250°C
Given: Data in Figure $4$, pressure, and boiling point
Asked for: corresponding boiling point and pressure
Strategy:
1. To estimate the boiling point of water at 1000 mmHg, refer to Figure $4$ and find the point where the vapor pressure curve of water intersects the line corresponding to a pressure of 1000 mmHg.
2. To estimate the pressure required for mercury to boil at 250°C, find the point where the vapor pressure curve of mercury intersects the line corresponding to a temperature of 250°C.
Solution:
1. A The vapor pressure curve of water intersects the P = 1000 mmHg line at about 110°C; this is therefore the boiling point of water at 1000 mmHg.
2. B The vertical line corresponding to 250°C intersects the vapor pressure curve of mercury at P ≈ 75 mmHg. Hence this is the pressure required for mercury to boil at 250°C.
Exercise $2$: Boiling Ethlyene Glycol
Ethylene glycol is an organic compound primarily used as a raw material in the manufacture of polyester fibers and fabric industry, and polyethylene terephthalate resins (PET) used in bottling. Use the data in Figure $4$ to estimate the following.
1. the normal boiling point of ethylene glycol
2. the pressure required for diethyl ether to boil at 20°C.
Answer a
200°C
Answer b
450 mmHg
Summary
Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called vaporization or evaporation, generates a vapor pressure above the liquid. Molecules in the gas phase can collide with the liquid surface and reenter the liquid via condensation. Eventually, a steady state is reached in which the number of molecules evaporating and condensing per unit time is the same, and the system is in a state of dynamic equilibrium. Under these conditions, a liquid exhibits a characteristic equilibrium vapor pressure that depends only on the temperature. We can express the nonlinear relationship between vapor pressure and temperature as a linear relationship using the Clausius–Clapeyron equation. This equation can be used to calculate the enthalpy of vaporization of a liquid from its measured vapor pressure at two or more temperatures. Volatile liquids are liquids with high vapor pressures, which tend to evaporate readily from an open container; nonvolatile liquids have low vapor pressures. When the vapor pressure equals the external pressure, bubbles of vapor form within the liquid, and it boils. The temperature at which a substance boils at a pressure of 1 atm is its normal boiling point.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/04%3A_The_Second_Law_of_Thermodynamics/4.09%3A_Phase_Equilibria.txt
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Q4.9
What is the entropy changed when 10.2 moles of diatomic gas is cooled from a temperature of 25 °C to 10 °C at constant pressure?
Q4.9
Determine the value of while reversibly heating 5 moles of an ideal gas from 25 °C to 73 °C at constant volume.
S4.9
At constant volume, dqrev = CV dT.
Q4.10
If 1.50 moles of an ideal gas were compressed isothermally from 6.6 L to 2.5 L:
1. What is $\inline \dpi{100} \bigtriangleup$Ssys when compressed irreversibly?
2. $\inline \dpi{100} \bigtriangleup$Ssys when done reversibly?
3. What do the values mean for the overall process?
S4.10
1. $\bigtriangleup S_{sys}=nR\ln\dfrac{V_2}{V_1}\rightarrow \bigtriangleup S_{sys}=(1.5mol)(8.314\dfrac{L\cdot atm}{mol\cdot K})(\ln\dfrac{2.5L}{6.6L})=-12.1 \dfrac{kJ}{mol}$
2. The change in entropy of the system is the same whether an irreversible or reversible process is being carried out.
3. The negative value means that the reaction that was carried out was unfavorable. Entropy increases as molecules become more disordered.
Q4.13
Nitrogen gas is reversibly heated at constant pressure of 101.3 kPa. The molar heat capacity of nitrogen gas is:
Cp = A + BT + CT2, where A = 27.4 J/mol-K, B = 5.23 x 10-3 J/mol-K2, and C = -0.0300 x 10-7 J/mol-K3.
Assuming N2 is an ideal gas, calculate the change in entropy when 42 g of N2(g) is heated from 250 K to 350 K.
Q4.21
Imagine you are to carry out the isothermal and reversible gas expansion with an ideal gas. You have 0.5 moles of the gas at 280K. You wish to expand this gas from its initial volume of 2 liters to a new volume of 8 liters. What would your values for heat, work, change in entropy, and change in internal energy have to be? What if you wanted to instead carry his experiment out as isothermal and irreversible with 2.0 atm of external pressure?
Q4.22
Consider 0.50 moles of an ideal gas at 30oC. It expands from a pressure of 4.5 atm to a pressure of 2.0 atm at the same temperature.
1. For a reversible process, find the entropy change for the surrounding.
2. For an irreversible process, calculate the entropy change for the universe which against a constant external pressure of 2.0 atm.
S4.22
a) Reversible process:
delta S of system =(\dfrac{q}{T}\)=(\dfrac{-W}{T}\)=(\dfrac{-1}{T}*(-nRT\ln\dfrac{V2}{V1})\)=(\nR\ln\dfrac{P1}{P}\)
qsys/T = - Wsys/T = -1/T *(-nRT\ln (V2/V1) = nR\ln (P1/P2)
= 0.5 moles * 8.3145 Jmol-1K-1 * \ln(4.5/2)
= 3.37 J/K
delta S of surrounding = -3.37 J/K
b) Irreversible process:
delta S of system is 3.37 J/K
Delta S(surrouding)=(\dfrac{-1}{T}*P2*V2*(1-\dfrac{V1}{V2})\)= (\-nR(1-\dfrac{P2}{P1})\)
delta S of surrounding = -1/T * P2*V2*(1 - V1/V2) = -nR(1 - P2/P1)
=-0.5 * 8.3145 *(1-2/4.5)=2.31 J/K
delta of the universe is : 3.37 J/K +2.31J/K=5.68 J/K
Q4.23
Calculate ΔSsys for the heating of 2.00 moles of nitrogen from 25ºC to 200ºC. The heat capacity of oxygen is:
Cp = (3.268 + 0.00325T) J K-1 mol-1
S4.23
$\Delta S = \int n\dfrac{\bar{C_{P}}}{T}dT$
$\Delta S = \int_{(25+273)K}^{(200+273)K} (2 mol)\dfrac{(3.268 + 0.00325T)J \ mol^{-1} \ K^{-1}}{T}dT$
$\Delta S = \int_{298K}^{473K} (\dfrac{6.536}{T}+ 0.0065)J \ mol^{-1}dT$
$\Delta S = {(6.536}\ln(T)+ 0.0065T)_{298}^{473}J \ mol^{-1}$
$\Delta S = 4.16 J \ mol^{-1}$
Q4.27
The value of (CO2(aq)) is less than that of (CO2(g)) at 298 K. (Note: Refer to table below). Would this be the case at a temperature of 0 K as well?
Substance / J K-1 mol-1
CO2(g) 213.6
CO2(aq) 117.6
S4.27
Since the gaseous form of a compound always has a greater enthalpy than the more ordered liquid or aqueous form, this would still be the case at a temperature of 0 K as well.
Q4.27
Is the standard entropy of CO2(g) higher than the standard entropy of CO2(s) at 298 K? What about at 0 K?
Q4.28
You have an ideal gas that expands 3.56 L with the initial volume at 22 L. All you know is that you have 0.45 moles and that the temperature is 22 ºC. What is the change in Gibbs free energy for this system?
=167.3
Q4.28
0.70 moles of an ideal gas expands adiabatically from 1.0 atm to 2.5 atm at a temperature of 30ºC. Calculate the values of q, w, $\inline \dpi{100} \bigtriangleup$U, $\inline \dpi{100} \bigtriangleup$S, and $\inline \dpi{100} \bigtriangleup$G.
S4.28
$q=0$ $w=-nRT\ln\dfrac{P_1}{P_2}=-(.70mol)(8.314\dfrac{J}{mol\cdot K})(303K)(\ln\dfrac{1.0atm}{2.5atm})(\dfrac{1kJ}{1000kJ}=1.62\dfrac{kJ}{mol}$
$\bigtriangleup U=w=1.62\dfrac{kJ}{mol}$
$\bigtriangleup S_{univ}=\bigtriangleup S_{sys}=nR\ln\dfrac{P_1}{P_2}=(.70mol)(8.314\dfrac{J}{mol\cdot K})\ln\dfrac{1.0atm}{2.5atm}=-5.33\dfrac{J}{mol}$
$dH=VdP\rightarrow \dfrac{dH}{dP}= nRT\ln\dfrac{P_2}{P_1}=(.70mol)(8.314\dfrac{J}{mol\cdot K}(303K)(\ln\dfrac{2.5atm}{1.0atm}(\dfrac{1kJ}{1000J}$
$=1.62\dfrac{kJ}{mol}$
$\bigtriangleup G= \bigtriangleup H- T\bigtriangleup S= 1.62 \times 10^3\dfrac{J}{mol}-(303K)(-5.33\dfrac{J}{mol/cdot K}=3234\dfrac{J}{mol}$
$=3.23\dfrac{kJ}{mol}$
Q4.30
Use the values listed listed below to calculate the value of ΔrG° for the following reaction:
pyruvate(aq) 2 lactate ion(aq) + 2H+(aq)
fG°[pyruvate(aq)] = -472.4 kJ mol-1, ΔfG°[lactate ion] = -516.7 kJ mol-1). Is pyruvate or the lactate ion favored under standard conditions? Explain.
Q4.30
Use the values listed listed below to calculate the value of ΔrG° for the following reaction:
pyruvate(aq) → 2 lactate ion(aq) + 2H+(aq)
fG°[pyruvate(aq)] = -472.4 kJ mol-1, ΔfG°[lactate ion] = -516.7 kJ mol-1)
Q4.31
A scientist measures studies the thermodynamics of a protein of interest. He interested in finding the Free Energy of unfolding in the quaternary structure of protein at a concentration of 5µM. He finds that the Gibbs free energy of the reaction -0.84 Kcal/mol. Why does the scientist want to know the value of (\Delta G\) Explain how the reaction proceeds.
S4.31
The value of (\Delta G\) of the reaction will let the scientist know how likely the reaction of the unfolding of the protein is to occur. If (\Delta G\) is negative the reaction will go forward, towards the unfolded state. In this case (\Delta G\) is negative so the reaction goes forward to the unfolding of the protein.
Q4.32
For the reaction
$H_{2 (g)} + Cl_{2 (g)} \rightarrow 2HCl_{(g)}$
Using the formula rG = ∆fG (products)-∆fG (reactants) to compute the Gibbs energy of the reaction?
Q4.34
Consider the conversion of graphite carbon into diamond
$C_{(graphite)}\rightarrow C_{(diamond)}$
1. Determine $\Delta rH^o$ and $\Delta rS^(o)$ for the reaction. Is the conversion spontaneous at any temperature?
2. The molar volume of graphite is $2.1cm^3$ greater than diamond in density.
3. If you apply pressure on graphite at $25^{o}C$, can you convert graphite to diamond? If so, determine the required pressure to make the process spontaneous.
S4.34
(a)
$\Delta rH^{o}=\Delta _f\bar{H}^{o}[C_{(diamond)}]-\Delta _f\bar{H}^{o}[C_{(graphite)}]\]$=1.90Kjmol^{-1}-0kjmol^{-1}\]
$=1.90kj\,mol^{-1}\]$\Delta rS^{o}=\Delta S^{o}[C_{(diamond)}]-\Delta S^{o}[C_{(graphite)}]\]
$=2.4J\,K^{-1}mol^{-1}-5.7JK^{-1}mol^{-1}\]$=-3.3J\,K^{-1}mol^{-1}\]
At $0^oC$,
$\Delta rG^{o}=\Delta rH^o-(273K)\Delta rS^{o}\]$=(1.90kj\,mol^{-1})-(273K)(-3.3\times10^{-3}kj\,K^{-1}mol^{-1})\]
$=2.8834kj\,mol^{-1}\] (b) Step 1:$G_2=G_1+V(P_2-P_1)\]
Step 2: (Hint: Apply molar quantities)
$\bar{G}_2=\bar{G}_1+\bar{V}\Delta P\] Step 3: (Hint: Apply the equation to diamond and graphite)$\bar{G}_2[Graphite]=\bar{G}_1[graphite]+\bar{V}[graphite]\Delta P\]
$\bar{G}_2[diamond]=\bar{G}_1[diamond]+\bar{V}[diamond]\Delta P\] Step 4: (Hint: Combine the equations)$\Delta rG_2=\Delta rG_1+[\bar{V}[diamond]-\bar{V}[graphite]]\Delta P\]
Step 5: (Hint: At $25^oC,P_1=1bar\; so\; \Delta rG_1=\Delta rG^o=2.883kj\,mol^{-1}$)
$\Delta rG_2=2.883kj\,mol^{-1}+(-2.1cm^3mol^{-1})\Delta P(\dfrac{1L}{1000cm^{3}})(\dfrac{1atm}{1.01 bar})(\dfrac{101.3J}{1L\,atm})(\dfrac{1kj}{1000J})\]$\Delta rG_2=2.883kj\,mol^{-1}-2.1\times10^{-4}\Delta P\,kj\,bar^{-1}mol^{-1}\]
Step 6: (Hint: To make the process spontaneous)
$\Delta rG_2=2.883kj\,mol^{-1}-2.1\times10^{-4}\Delta P\,kj\,bar^{-1}mol^{-1}\lt0\]$2.883kj\,mol^{-1}\lt2.1\times10^{-4}\Delta P\,kj\,bar^{-1}mol^{-1}\]
\Delta P\lt\dfrac{2.883kj\,mol^{-1}}{2.1\times10^{-4}}=1.4\times10^{4} bar\]
Q4.36
Predict the signs of $\Delta H,\Delta S$ and $\Delta G$ of the system for the following processes at 1atm:
1. Ice melt at 200 K
2. Ice melt at 273.15 K
3. Ice melt at 300 K
S4.36
$\Delta H$ is positive for all because melting is endothermic. $\Delta S$ is positive for all because going from solid to liquid is becoming more disordered. $\Delta G$ for (a) is positive because reaction is not spontaneous when melting at below the melting point, $\Delta G$ for (b) is 0 because reaction is in equilibrium at the melting point, $\Delta G$ for (c) is negative because reaction is spontaneous when melting at above the melting point.
(a) (b) (c)
$\Delta H$ + + +
$\Delta S$ + + +
$\Delta G$ + 0 -
Q4.37
Crystals of AgCl(s) form spontaneously when aqueous solutions of silver(I)nitrate and sodium chloride are combined. What can you deduce about the signs of the changes in enthalpy and entropy?
Q4.38
The formation reaction to form magnesium oxide is as follow:
$\ce{Mg(s) + 1/2 O_2 (g) \rightarrow MgO(s) }$
Verify the equation ΔG° = ΔH° + TS° using values from Appendix
Hint: Gibbs free definition above is free energy of the whole reaction, not free energy of formation
S4.38
$\Delta_{r}\bar{G}^{\circ} = \sum \Delta_f\bar{G}^{\circ}[products] - \sum \Delta_f\bar{G}^{\circ}[reactants] = -569.5 \dfrac{kJ}{mol} -0-0 = {\color{red}-569.5 \dfrac{kJ}{mol}}$
$\Delta_{r}\bar{H}^{\circ} = \sum \Delta_f\bar{H}^{\circ}[products] - \sum \Delta_f\bar{H}^{\circ}[reactants] = -601.8 \dfrac{kJ}{mol} -0-0 = -601.8 \dfrac{kJ}{mol}$
$\Delta_{r}\bar{S}^{\circ} = \sum \Delta_{f}\bar{S}^{\circ} [products] - \sum \Delta_{f}\bar{S}^{\circ} [reactants] = (26.78 - 32.68 - \dfrac{205.0}{2})\dfrac{J}{mol \cdot K} = -108.4$
$\Delta_{r}\bar{H}^{\circ} - T\Delta_{r}\bar{S}^{\circ} = -601.8 \dfrac{kJ}{mol} - (298K)(-0.1084 \dfrac{kJ}{mol \cdot K}) = {\color{red}-569.5 \dfrac{kJ}{mol}}$
The answers agree therefore the equation is true.
Q4.39a
At 49ºC, a particular reaction is spontaneous. What would be the smallest possible value of ΔrS for the reaction if ΔrH was 10 kJ?
S4.39a
Use the following equation
$\Delta G = \Delta H - T \Delta S$
Since the process is spontaneous:
Note: Don't forget to flip the inequality when dividing by a negative number.
\begin{align} \Delta G &< 0 \ \Delta H - T \Delta S &< 0 \ (10 \ kJ) - (322 \ K) \Delta S &< 0 \ - (322 \ K) \Delta S &< 10 \ kJ \ \Delta S &>0.031 \end{align}
Thus the minimum value is the closest number above 0.031 kJ K-1.
Click here for more information on Gibb's free energy.
Q4.39b
Calculate the minimum change in entropy of reaction for a spontaneous process at 340K with an enthalpy change of reaction equal to 800 J. Your answer must be in KJ/K and have 2 significant figures.
Q4.39c
The entropy change for a reaction is found to be 47.50 J/K. The enthalpy change for the reaction is 21.25 KJ. At what temperature is this reaction spontaneous?
S4.39c
$\Delta _{r}G=\Delta _{r}H-T\Delta _{r}S$
Because we want to find the temperature at equilibrium we set the Gibbs Free Energy equal to 0. Thus:
$\Delta _{r}H=T\Delta _{r}S$
$T=\frac{\Delta _{r}H}{\Delta _{r}S}=\frac{21.25\times 10^{3}J}{47.50\frac{J}{K}}=447.4K$
Q4.40a
The standard Gibbs energy change, ∆rGo, for the reaction is +108 kJ. What can you deduce about the state of reaction when the reactants are mixed together?
S4.40a
When the two gases are mixed, there is no interaction between them. So, the products would not form.
Q4.40b
A scientist mixes together a set of reagents which are known to undergo a reaction with a negative change in Gibbs’ free energy in a beaker. When the scientist comes back in a few hours, they find by looking at the beaker that the reaction does not appear to have proceeded. What are two reasons that the scientist would not observe a change after two hours?
S4.40b
The Gibbs' free energy is an indicator of a reaction's favorability, but does not indicate that that reaction will proceed quickly. If the rate constant for the reaction is low, or the concentration of one or more of the reagents is low, the reaction might not proceed at a pace that the scientist would observe over a few hours or even longer.
Q4.41
If water changes from a liquid to a gas at 50ºC and 0.123 bar, what must the pressure be to convert water from a liquid to a gas at 75ºC? The molar enthalpy of vaporization for water (ΔvapH) is 42.3 kJ mol-1
S4.41
$\ln \left( \dfrac{P_{2}}{P_{1}} \right) = \dfrac{\Delta \bar{H}_{vap}}{R} \left( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right)$
$T_1= \left( 50 + 273 \right) K = 323\,K$
$T_2 = \left( 75 + 273 \right) K= 348\,K$
$\ln \left( \dfrac{P_{2}}{0.123\,bar} \right) = \dfrac{42.3\times 10^3\,J mol^{-1} }{8.314 J \ mol^{-1} \ K^{-1}} \left( \dfrac{1}{323\,K}-\dfrac{1}{348\,K} \right)$
$\ln \left( \dfrac{P_{2}}{0.123\,bar} \right) = 1.132$
$\dfrac{P_{2}}{0.123\,bar} = 3.10$
$P_{2}= 0.381 \,bar$
Q4.42
What is the change in pressure when ice melts ast room temperature if the density of ice is 0.919 g/ml and the ΔfusH = 6.01 KJ/mol?
Q4.42
Calculate the freezing point depression of ice if a pressure of 450 atm is exerted on it by a car that weighs 1500-kg. Note: The molar volumes are as follows: molar volume of liquid water is 18.03 mL mol-1 and the molar volume of ice is 19.65 mL mol-1.
S4.42
Use the slope of the S-L curve from the phase diagram:
This means that the freezing point depression determined for when 450 atm of pressure is applied to ice by a car is 3.30 K.
Q4.43
Here is a phase diagram for carbon.
1. What phase will carbon be if temperature is increased at constant pressure from the triple point between the vapor, liquid, and graphite?
2. At 1.0 GPa and 5000 K, what phase will carbon be in?
$\inline Image from Wikimedia Commons$
S4.43
1. Carbon will be in its vapor phase if the temperature is increased from the lower triple point.
2. At 1.0 GPa and 5000 K, carbon will be in the liquid phase.
Q4.46
Look at the phase diagram of water. Which phase is more dense, liquid or solid? How can you tell by just looking at the graph?
The link below contains a phase diagram of water:
http://chemwiki.ucdavis.edu/Wikitext...Phase_Diagrams
S4.46
Through looking at the phase diagram, we can see that the liquid phase is more dense. You can see this because the line separating the two phases is negative.
Q4.48
Explain what happens to ΔvapH as the temperature increases significantly.
Q4.49
You go on an adventure and find yourself stranded on the top of a mountain where the air pressure is 0.68 atm. You find water but you want to make sure is safe to drink so you want to boil the water. At what temperature would you expect the water to start boiling? Hint: The molar heat of vaporization of H2O is 41.0 kJ mol-1
Q4.50
At higher altitude, water is boiled faster than at lower altitude. The reason is because the pressure is smaller. Given the heat of vaporization of water at 100oC is 42.0 KJ/mol. Calculate the temperature to boil the water at a place where air pressure is 0.7 atm?
Q4.52
The equations are as follows
1. $dS=\dfrac{CpdT}{T}$
2. $\Delta S=\dfrac{\Delta H }{T}$
3. $S_0=0$
4. $dS=\dfrac{dq}{T}$
Determine the conditions needed for each equation.
S4.52
1. Constant pressure
2. Constant pressure and temperature
3. Absolute 0, no residual entropy, pure crystalline
4. reversible process
Q4.54
Calculate the entropy change when nitrogen gas is heated from 25°C to 75°C while simultaneously allowed to expand from 0.562 L to 1.245 L. Assume ideal gas behavior.
S4.54
1. Find how many moles of gas we have: $PV=nRT$
$n=\dfrac{PV}{RT}=\dfrac{1 atm*0.562L}{(0.08206L*atm/K*mol)(298.15K)}=0.02297mol$
2. entropy change during expansion:
$Delta S=nR\ln{\dfrac{V_2}{V_1}}=(0.02297mol)(8.314J/K*mol)\ln{\dfrac{1.245L}{0.562L}}=0.1519J/K$
3. entropy change during heating at constant volume:
\begin{align*}\Delta S&=n\bar{C}_v\ln{\dfrac{T_2}{T_1}}=n(\dfrac{5}{2}R)\ln{\dfrac{T_2}{T_1}}\ &=(0.02297mol)(\dfrac{5}{2})(8.314J/K*mol)\ln{\dfrac{348.15K}{283.15K}}=0.0987J/K\end{align*}
4. Combine 2&3:
$\Delta S_{tot}=\Delta S_{expansion}+\Delta S_{heating}=0.1519J/K+0.0987J/K=0.2506J/K$
Q4.54
A $4 \times 10^{-2}$ moles sample of Xenon (Xe) is heated from 300 K to 400 K in a 1 L container at 1.5 atm. What will be the change in entropy if Xe is able to expand to twice the volume of the container?
Q4.55
Photosynthesis depends upon the absorption of visible light. However, the infrared radiation can not be used for photosynthesis. Explain why.
S4.55
The infrared light : Cells do absorb the photon of infrared radiation, but it could not provide enough energy to do work in a biological system.
Whereas, for the visible light, each photon of visible light contains just enough energy to excite the electrons of molecules.
Photosynthesis is carried out by using the visible light.
Q4.56a
An amount of ideal diatomic gas is expanded from 4.8 atm to 1.7 atm while experiencing the change of temperature from 35°C to 128ºC. The change in entropy of the gas expansion process is 15.0 J/K. Determine the ΔU, ΔH, and ΔS of the entire process. Assume the temperature is not enough to unleash the rotational and vibrational energy of the gas.
Hint: Break the problem into 2 processes (isothermic and isobaric)
S4.56a
Step 1: Consider the isothermal process of the ideal gas (expanding from 4.8 atm to 1.7atm at constant temperature of 35°C
ΔU = 0, ΔH = 0, ΔS = 15.0 J/K
$\Delta S = nRln\dfrac{V_2}{V_1} = nRln\dfrac{P_1}{P_2}$
$n = \dfrac{\Delta S}{Rln\dfrac{P_1}{P_2}} = \dfrac{15.0 \dfrac{J}{K}}{( 8.314 \dfrac{J}{mol \cdot K})ln\dfrac{4.8 atm}{1.7 atm}} = 1.738 moles$
Step 2: Consider the change in temperature at constant pressure of 1.7 atm
$\Delta H = C_p \Delta T = \dfrac{7}{5}nR\Delta T = \dfrac{7}{5} (1.738 moles)(8.314\dfrac{J}{K\cdot mol})(128-35)K = 1.88 \times 10^{3} J$
$\Delta U = \Delta H - nR\Delta T = (1.88 \times 10^{3} J ) -(1.738 moles)(8.314 \dfrac{J}{K \cdot mol})(128-35)K = 537 J$
$\Delta S = C_p ln\dfrac{T_2}{T_1} = \dfrac{7}{5}nRln\dfrac{T_2}{T_1} = \dfrac{7}{5}(1.738 moles)(8.314 \dfrac{J}{K \cdot mol})ln\dfrac{(128+273)K}{(35+273)K} = 5.34 \dfrac{J}{K}$
For the entire process
$\Delta H = 0 J + 1.88 \times 10^3 J = {\color{red}1.88 \times 10^3 J}$
$\Delta U = 0 J + 537 J = {\color{red}537 J}$
$\Delta S = 15.0 \dfrac{J}{K} + 5.34 \dfrac{J}{K} = {\color{red}20.34 \dfrac{J}{K}}$
Q4.56b
Calculate the ΔU, ΔH, and ΔS of 3.00 moles of an ideal monatomic gas that are compressed from 7.00 L to 1.00 L while being heated from 25 ºC to 100 ºC.
S4.56b
Break the process down into two steps: isothermal compression from 7 L to 1 L at 298 K and heating at a constant pressure from 298 K to 373 K.
a. Compression:
Because the compressions is isothermal, ΔU=0 & ΔH=0 for this step
$\Delta S = nR\ln(\dfrac{V_{2}}{V_{1}})=(3mol)(8.314J\ mol^{-1}\ K^{-1})\ln(\dfrac{1}{7})$
$\Delta S = -48.5 J \ K^{-1}$
b. Heating:
$\Delta H = C_{P}\Delta T = \dfrac{5}{2}nR\Delta T=\dfrac{5}{2}(3 mol)(8.314 J\ mol^{-1}\ K^{-1})(373-298)K$
$\Delta H = 4677 J$
$\Delta U = \Delta H - nR\Delta T = 4677 J - (3mol)(8.314 J\ mol^{-1}\ K^{-1})(373-298)K$
$\Delta U = 2806 J$
$\Delta S = C_{P}\ln(\dfrac{T_{2}}{T_{1}}) = \dfrac{5}{2}nR\ln(\dfrac{T_{2}}{T_{1}})$
$\Delta S = \dfrac{5}{2}(3mol)(8.314 J\ mol^{-1}\ K^{-1})\ln(\dfrac{373}{298}) = 14.0 J \ K^{-1}$
c. For entire process
$\Delta H = 0J + 4677 J= 4.68kJ$
$\Delta U = 0J+ 2806 J=2.81kJ$
$\Delta S = -48.5 J \ K^{-1}+14.0 J \ K^{-1} = -34.5 J \ K^{-1}$
Q4.57a
Briefly provide examples and descriptions for each of the three laws of thermodynamics.
S4.57a
• The first law is the conservation of mass and energy. Basically you can't create anything out of thin air and perpetual motion machines are impossible.
• The second law states how entropy is always increasing and the entropy of the universe can never become negative. In other words things will always attempt to go to equilibrium. This is seen commonly when batteries run out. You never see a battery charge itself spontaneously because that would violate the second law.
• The third law states how a pure crystalline structure at 0 K will essentially have 0 entropy. The third law is visualized by knowing how frozen and solid materials have less entropy because the molecules within can only vibrate.
Q4.57b
The three laws of thermodynamics are often generally stated to be:
1. Energy and matter cannot be created or destroyed.
2. The entropy of the universe is always increasing.
3. At absolute zero, the entropy of a system is zero.
With this knowledge, which of the three laws of thermodynamics apply to the examples below? There may be more than one.
a.) A graduate student in UC davis claims to have created the world's first perpetual motion engine. This machine produces no heat and can be used, (he claims), to power the campus for the next decade! This is not possible.
b.)
If TF becomes 0, entropy becomes undefined. Hence, it is not possible to reach absolute zero.
c.) In an adiabatic gas expansion where:
$\Delta U=Q+w$
No thermal energy or matter is exchanged. Hence:
$\Delta U=w$
d.) In an imperfect crystal of CO, carbon monoxide, there are multiple energy states at 0K. Thus, it experiences residual entropy and, quite unusually, follows:
$\lim _{T\to0K}S>0$
S4.57b
a.) first and second laws; The existence of the perpetual motion machine violates the first and second laws of thermodynamics. Because
$\Delta U=Q+w$
and heat is a form of entropy, a perpetual motion machine does not increase the entropy of the universe. By definition, a machine which does work must do less work than it has internal energy.
b.) second and third law; In order for absolute zero to be reached, the system would have to be removed from the universe. Because
$\Delta S=n\bar{C}_{p}ln\left ( \frac{T_{2}}{T_{1}} \right )$
and entropy must always increase, a system within the universe would always be at a temperature lower than its surroundings and hence could experience heat transfer which would mitigate the absolute zero condition.
c.) first law; Energy which is "lost" from the system due to lack of heat transfer is compensated by the system doing more work.
d.) third law; In a perfect crystal, where there is only one energy state, there is no residual entropy at 0K
Q4.58
The standard enthalpy of formation of ice is -291.83 kJ/mol and its standard molar entropy is 41.0 J/(mol K). The standard enthalpy of formation of liquid water is -285.83 kJ/mol and its standard molar entropy is 69.95 J/(mol K). Estimate the melting point of water based on this information. What would you expect the value be? Do your values match? What assumptions made during this calculation could be responsible for the difference?
Q4.61
In a reversible adiabatic expansion, what do you expect to happen to the volume? What about the temperature?
S4.61
We expect both volume and temperature to increase.
Q4.63
Aniline can hydrogen bond, whereas benzene is nonpolar. Explain why, contrary to our expectation, benzene melts at a higher temperature than aniline. Why is the boiling point of aniline higher than that of benzene?
Q4.64
Molar Gibbs energies of solid, liquid and vapor depend on temperature and pressure. (a) Briefly explain how temperature affects molar Gibbs energy. (b) Explain how pressure changes affect molar Gibbs energy. Why is water an exemption?
S4.64
1. At a constant pressure the phase with the lowest molar Gibbs energy is the most stable phase at that temperature. At high temperatures the vapor phase has the lowest molar Gibbs energy, therefore it is the most stable. As temperature decreases, liquid becomes the stable phase, and at even lower temperatures, solid becomes the most stable phase.
2. From the equation $\left (\dfrac{\partial \Delta G}{\partial P} \right )=\bar{V}$ we observe that the molar volume affects the molar Gibbs energy depending on pressure. In this case the vapor phase has the greatest increase in molar Gibbs energy because the molar volume is much larger than that of liquids and solids. In general, an increase in external pressure will raise both the melting point and the boiling point of a substance. The exception to this, is water. This is because the molar volume of ice is actually greater than that of liquid water. Thus, for water an increase in pressure will lower the melting point.
Q4.65
When ice melts to liquid water, entropy is increasing and the the surrounding becomes cooler since ice absorbs heat. Explain the relationship between those two facts.
Q4.66
For any spontaneous endothermic process, what must be true about the entropy of the process?
Q4.67
For the following situation take a rubber band and stretch it quickly. It will feel warm. Next stretch it out for a few seconds and then release it. Consequently, it will feel cool. Why do you think this happens? Explain thermodynamically.
S4.67
$\Delta G =\Delta H-T\Delta S$
The warming effect: Stretching the rubber band creates a nonspontaneous process $(\Delta G\gt0)$ and $-(\Delta G\lt 0)$. The warming effect means that $\Delta H\lt0)$. This makes $\Delta S$ negative. T must be positive and $\Delta S$ must be negative. This tells us that the rubber band under tension is more disordered in its natural state.
The cooling effect: If theres no tension and the rubber band snaps back to its natural state, $\Delta G$ is negative and $-\Delta G$ is positive. Since it cools the $\Delta H\gt0$ so $T\Delta S$ is positive. Entropy increases when the rubber band goes from stretched state to natural state.
Q4.69
Cold packs utilizes the chemical reaction of water and ammonium nitrate (or similar substances) to turn cold. Predict the signs of $\Delta H, \Delta S$ and $\Delta G$ for the reaction when the membrane that separates the two substances in a cold pack breaks.
S4.69
The reaction is spontaneous so $\Delta G$ is negative; the cold pack feels cold because it's taking in heat so $\Delta H$ is positive; $\Delta S$ is positive because disorder increases as ammonium nitrate breaks down to ammonium ion and nitrate ion.
Q4.71
At what temperature will toluene have a vapor pressure of 517 torr. The normal boiling point of toluene is 110.6°C with the molar enthalpy of vaporization being 35.2 kJ/mol. (Hint: Use Clausius-Clapeyron equation).
S4.71
$ln\dfrac{P_2}{P_1} = \dfrac{\Delta _vap \bar H}{R} (\dfrac{1}{T_2} - \dfrac{1}{T_1})$
$ln\dfrac{517 torr}{760 torr} = \dfrac{35.2 \times 10^3 \dfrac{J}{mol}}{8.314 \dfrac{J}{K \cdot mol}} (\dfrac{1}{T_2} - \dfrac{1}{(110.6+273)K})$
$\Rightarrow {\color{red}T_2 = 397.5 K = 124^{\circ}C}$
Q4.72a
You have discovered a way to create a new organic compound of the formula $C_xH_yN_zO_α$. What measurements would you need to take to find out values for the following:properties:
$\triangle_{f}\overline{H}^{\circ} = \ ? \ \ \triangle _{f}\overline{S}^{\circ} = \ ? \ \ \triangle _{f}\overline{G}^{\circ} = \ ?$
S4.72a
For enthalpy, you need to take measurements of the change of enthalpy of one mole of the unknown compound. Thus you would need to take measurements for the molar heat capacity and how much temperature changes in for formation of one mole. See here and here.
For entropy of formation you would need to divide enthalpy by the temperature at which the formation takes place because the change in entropy can equate to enthalpy divided by temperature. See here.
For Gibb's free energy of formation, you simply need the enthalpy and entropy of formation along with the temperature where one mole of the substance is formed. See here.
$\Delta{G} = \Delta H - T \Delta S$
Q4.72b
You are given hexane, C6H14, and a constant pressure calorimeter. What steps would you take to determine standard gibbs free energy of formation, for this substance?
Hint: of hexane can be found using a measurement obtained by the calorimeter
S4.72b
An equation for the complete combustion of hexane can be written as follows:
$C_{6}H_{14(l)}+O_{2(g)}\rightarrow CO_{2(l)}+H_{2}O_{(l)}$
The calorimeter would then be used to find the heat released by this reaction by the following rule:
$\Delta _{r}H=-q_{P, calorimeter}=-\left ( C_{calorimeter} \Delta T\right )$
after which the standard heat of formation for hexane could be calculated using the known heat of formation for H2O, O2, and CO2.
Under the 3rd law of dynamics, the standard entropy would then be calculated by
$\Delta \bar{S}^{\circ}=-\frac{q_{rxn}}{T}$
where T=298K
From this, standard entropy for the formation of hexane should be obtained from the known standard entropies of H2O, O2, and CO2, through the following reaction for the formation of hexane from elemental reactants:
$6C_{(graphite)}+7H_{2(g)}\rightarrow C_{6}H_{14(l)}$
The standard gibbs free energy of formation, for this substance is then determined through the relationship:
$\Delta _{f}\bar{G}^{\circ}=\Delta _{f}\bar{H}^{\circ}-T\Delta _{r}S^{\circ}$
Q4.73
A 93.5 Liter drum at 65 C houses 20g of water. How many moles of water are in the vapor phase, and what mole fraction is in the vapor phase? The vapor pressure of water at 65C is 187.54 mmhg.
S4.73
$\dfrac{PV}{/RT} = n$
• ntot = 20g/18.02 g mol-1= 1.11 mol
• nvap = (187.54 mmhg 93.5 L)/(8.314 L atm mol-1 K-1* 338 K * 710 mmhg atm-1) = 8.79 *10-3 mol
• xvap= 8.79 *10-3 mol / 1.11 mol = 7.92*10-3
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/04%3A_The_Second_Law_of_Thermodynamics/4.E%3A_Exercises.txt
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When solids, liquids or gases are combined, the thermodynamic quantities of the system experience a change as a result of the mixing. This module will discuss the effect that mixing has on a solution’s Gibbs energy, enthalpy, and entropy, with a specific focus on the mixing of two gases.
Introduction
A solution is created when two or more components mix homogeneously to form a single phase. Studying solutions is important because most chemical and biological life processes occur in systems with multiple components. Understanding the thermodynamic behavior of mixtures is integral to the study of any system involving either ideal or non-ideal solutions because it provides valuable information on the molecular properties of the system.
Most real gases behave like ideal gases at standard temperature and pressure. This allows us to combine our knowledge of ideal systems and solutions with standard state thermodynamics in order to derive a set of equations that quantitatively describe the effect that mixing has on a given gas-phase solution’s thermodynamic quantities.
Gibbs Free Energy of Mixing
Unlike the extensive properties of a one-component system, which rely only on the amount of the system present, the extensive properties of a solution depend on its temperature, pressure and composition. This means that a mixture must be described in terms of the partial molar quantities of its components. The total Gibbs free energy of a two-component solution is given by the expression
$G=n_1\overline{G}_1+n_2\overline{G} _2 \label{1}$
where
• $G$ is the total Gibbs energy of the system,
• $n_i$ is the number of moles of component i,and
• $\overline{G}_i$ is the partial molar Gibbs energy of component i.
The molar Gibbs energy of an ideal gas can be found using the equation
$\overline{G}=\overline{G}^\circ+RT\ln \frac{P}{1 bar} \label{2}$
where $\overline{G}^\circ$ is the standard molar Gibbs energy of the gas at 1 bar, and P is the pressure of the system. In a mixture of ideal gases, we find that the system’s partial molar Gibbs energy is equivalent to its chemical potential, or that
$\overline{G}_i=\mu_i \label{3}$
This means that for a solution of ideal gases, Equation $\ref{2}$ can become
$\overline{G}_i=\mu_i=\mu^\circ_i+RT \ln \frac{P_i}{1 bar} \label{4}$
where
• µi is the chemical potential of the ith component,
• µi° is the standard chemical potential of component i at 1 bar, and
• Pi is the partial pressure of component i.
Now pretend we have two gases at the same temperature and pressure, gas 1 and gas 2. The Gibbs energy of the system before the gases are mixed is given by Equation $\ref{1}$, which can be combined with Equation $\ref{4}$ to give the expression
$G_{initial}=n_1(\mu^\circ_1+RT \ln P)+n_2(\mu^\circ_2+RT \ln P) \label{5}$
If gas 1 and gas 2 are then mixed together, they will each exert a partial pressure on the total system, $P_1$ and $P_2$, so that $P_1+ P_2= P$. This means that the final Gibbs energy of the final solution can be found using the equation
$G_{final}=n_1(\mu^\circ_1+RT \ln P_1)+n_2(\mu^\circ_2+RT \ln P_2) \label{6}$
The Gibbs energy of mixing, $Δ_{mix}G$, can then be found by subtracting $G_{initial}$ from $G_{final}$.
\begin{align} Δ_{mix}G &= G_{final} - G_{initial}\[4pt] &=n_1RT \ln \frac{P_1}{P}+n_2RT \ln \frac{P_2}{P} \[4pt] &=n_1 RT \ln \chi_1+n_2 RT \ln \chi_2 \label{7} \end{align}
where
$P_i = \chi_iP$
and $\chi_i$ is the mole fraction of gas $i$. This equation can be simplified further by knowing that the mole fraction of a component is equal to the number of moles of that component over the total moles of the system, or
$\chi_i = \dfrac{n_i}{n}.$
Equation \ref{7} then becomes
$\Delta_{mix} G=nRT(\chi_1 \ln \chi_1 + \chi_2 \ln \chi_2) \label{8}$
This expression gives us the effect that mixing has on the Gibbs free energy of a solution. Since $\chi_1$ and $\chi_2$ are mole fractions that range from 0 to 1, we can conclude that $Δ_{mix}G$ will be a negative number. This is consistent with the idea that gases mix spontaneously at constant pressure and temperature.
Entropy of mixing
Figure $1$ shows that when two gases mix, it can really be seen as two gases expanding into twice their original volume. This greatly increases the number of available microstates, and so we would therefore expect the entropy of the system to increase as well.
Thermodynamic studies of an ideal gas’s dependence of Gibbs free energy of temperature have shown that
$\left( \dfrac {d G} {d T} \right )_P=-S \label{9}$
This means that differentiating Equation $\ref{8}$ at constant pressure with respect to temperature will give an expression for the effect that mixing has on the entropy of a solution. We see that
\begin{align} \left( \dfrac {d G_{mix}} {d T} \right)_P &=nR(x_1 \ln x_1+x_2 \ln x_2) \[4pt] &=-\Delta_{mix} S \end{align}
$\Delta_{mix} S=-nR(x_1 \ln x_1+x_2 \ln x_2) \label{10}$
Since the mole fractions again lead to negative values for ln x1 and ln x2, the negative sign in front of the equation makes ΔmixS positive, as expected. This agrees with the idea that mixing is a spontaneous process.
Enthalpy of mixing
We know that in an ideal system $\Delta G= \Delta H-T \Delta S$, but this equation can also be applied to the thermodynamics of mixing and solved for the enthalpy of mixing so that it reads
$\Delta_{mix} H=\Delta_{mix} G+T\Delta_{mix} S \label{11}$
Plugging in our expressions for $Δ_{mix}G$ (Equation $\ref{8}$) and $Δ_{mix}S$ (Equation $\ref{10}$) , we get
$\Delta_{mix} H=nRT(x_1 \ln x_1+x_2 \ln x_2)+T \left[-nR(x_1 \ln x_1+x_2 \ln x_2) \right] = 0$
This result makes sense when considering the system. The molecules of ideal gas are spread out enough that they do not interact with one another when mixed, which implies that no heat is absorbed or produced and results in a $Δ_{mix}H$ of zero. Figure $2$ illustrates how $TΔ_{mix}S$ and $Δ_{mix}G$ change as a function of the mole fraction so that $Δ_{mix}H$ of a solution will always be equal to zero (this is for the mixing of two ideal gasses).
• Satter, S. (2000). Thermodynamics of Mixing Real Gases. J. Chem. Educ. 77, 1361-1365.
• Brandani, V., Evangelista, F. (1987). Correlation and prediction of enthalpies of mixing for systems containing alcohols with UNIQUAC associated-solution theory. Ind. Eng. Chem. Res. 26 (12), 2423–2430.
Problems
1. Use Figure 2 to find the x1 that has the largest impact on the thermodynamic quantities of the final solution. Explain why this is true.
2. Calculate the effect that mixing 2 moles of nitrogen and 3 moles of oxygen has on the entropy of the final solution.
3. Another way to find the entropy of a system is using the equation ΔS = nRln(V2/V1). Use this equation and the fact that volume is directly proportional to the number of moles of gas at constant temperature and pressure to derive the final expression for $T\Delta_{mix}S$. (Hint: Use the derivation of $T\Delta_{mix}G$ as a guide).
Answers
1. x1= 0.5
2. Increases the entropy of the system by 27.98 J/molK
Contributors and Attributions
• Elizabeth Billquist (Hope College)
5.05: Real Solutions
A non-ideal solution is a solution that does not abide to the rules of an ideal solution where the interactions between the molecules are identical (or very close) to the interactions between molecules of different components. That is, there is no forces acting between the components: no Van-der-Waals nor any Coulomb forces. We assume ideal properties for dilute solutions.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/05%3A_Solutions/5.03%3A_The_Thermodynamics_of_Mixing.txt
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Learning Objectives
• Express concentrations of solution components using mole fraction and molality
• Describe the effect of solute concentration on various solution properties (vapor pressure, boiling point, freezing point, and osmotic pressure)
• Perform calculations using the mathematical equations that describe these various colligative effects
• Describe the process of distillation and its practical applications
• Explain the process of osmosis and describe how it is applied industrially and in nature
The properties of a solution are different from those of either the pure solute(s) or solvent. Many solution properties are dependent upon the chemical identity of the solute. Compared to pure water, a solution of hydrogen chloride is more acidic, a solution of ammonia is more basic, a solution of sodium chloride is more dense, and a solution of sucrose is more viscous. There are a few solution properties, however, that depend only upon the total concentration of solute species, regardless of their identities. These colligative properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. This small set of properties is of central importance to many natural phenomena and technological applications, as will be described in this module.
Mole Fraction and Molality
Several units commonly used to express the concentrations of solution components were introduced in an earlier chapter of this text, each providing certain benefits for use in different applications. For example, molarity (M) is a convenient unit for use in stoichiometric calculations, since it is defined in terms of the molar amounts of solute species:
$M=\dfrac{\text{mol solute}}{\text{L solution}} \label{11.5.1}$
Because solution volumes vary with temperature, molar concentrations will likewise vary. When expressed as molarity, the concentration of a solution with identical numbers of solute and solvent species will be different at different temperatures, due to the contraction/expansion of the solution. More appropriate for calculations involving many colligative properties are mole-based concentration units whose values are not dependent on temperature. Two such units are mole fraction (introduced in the previous chapter on gases) and molality.
The mole fraction, $\chi$, of a component is the ratio of its molar amount to the total number of moles of all solution components:
$\chi_\ce{A}=\dfrac{\text{mol A}}{\text{total mol of all components}} \label{11.5.2}$
Molality is a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms:
$m=\dfrac{\text{mol solute}}{\text{kg solvent}} \label{11.5.3}$
Since these units are computed using only masses and molar amounts, they do not vary with temperature and, thus, are better suited for applications requiring temperature-independent concentrations, including several colligative properties, as will be described in this chapter module.
Example $1$: Calculating Mole Fraction and Molality
The antifreeze in most automobile radiators is a mixture of equal volumes of ethylene glycol and water, with minor amounts of other additives that prevent corrosion. What are the (a) mole fraction and (b) molality of ethylene glycol, C2H4(OH)2, in a solution prepared from $\mathrm{2.22 \times 10^3 \;g}$ of ethylene glycol and $\mathrm{2.00 \times 10^3\; g}$ of water (approximately 2 L of glycol and 2 L of water)?
Solution
(a) The mole fraction of ethylene glycol may be computed by first deriving molar amounts of both solution components and then substituting these amounts into the unit definition.
$\mathrm{mol\:C_2H_4(OH)_2=2220\:g×\dfrac{1\:mol\:C_2H_4(OH)_2}{62.07\:g\:C_2H_4(OH)_2}=35.8\:mol\:C_2H_4(OH)_2}$
$\mathrm{mol\:H_2O=2000\:g×\dfrac{1\:mol\:H_2O}{18.02\:g\:H_2O}=111\:mol\:H_2O}$
$\chi_\mathrm{ethylene\:glycol}=\mathrm{\dfrac{35.8\:mol\:C_2H_4(OH)_2}{(35.8+111)\:mol\: total}=0.245}$
Notice that mole fraction is a dimensionless property, being the ratio of properties with identical units (moles).
(b) To find molality, we need to know the moles of the solute and the mass of the solvent (in kg).
First, use the given mass of ethylene glycol and its molar mass to find the moles of solute:
$\mathrm{2220\:g\:C_2H_4(OH)_2\left(\dfrac{mol\:C_2H_2(OH)_2}{62.07\:g}\right)=35.8\:mol\:C_2H_4(OH)_2} \nonumber$
Then, convert the mass of the water from grams to kilograms:
$\mathrm{2000\: g\:H_2O\left(\dfrac{1\:kg}{1000\:g}\right)=2\: kg\:H_2O} \nonumber$
Finally, calculate molarity per its definition:
\begin{align*} \ce{molality}&=\mathrm{\dfrac{mol\: solute}{kg\: solvent}}\ \ce{molality}&=\mathrm{\dfrac{35.8\:mol\:C_2H_4(OH)_2}{2\:kg\:H_2O}}\ \ce{molality}&=17.9\:m \end{align*} \nonumber
Exercise $1$
What are the mole fraction and molality of a solution that contains 0.850 g of ammonia, NH3, dissolved in 125 g of water?
Answer
7.14 × 10−3; 0.399 m
Example $2$: Converting Mole Fraction and Molal Concentrations
Calculate the mole fraction of solute and solvent in a 3.0 m solution of sodium chloride.
Solution
Converting from one concentration unit to another is accomplished by first comparing the two unit definitions. In this case, both units have the same numerator (moles of solute) but different denominators. The provided molal concentration may be written as:
$\mathrm{\dfrac{3.0\;mol\; NaCl}{1.0\; kg\; H_2O}} \nonumber$
The numerator for this solution’s mole fraction is, therefore, 3.0 mol NaCl. The denominator may be computed by deriving the molar amount of water corresponding to 1.0 kg
$\mathrm{1.0\:kg\:H_2O\left(\dfrac{1000\:g}{1\:kg}\right)\left(\dfrac{mol\:H_2O}{18.02\:g}\right)=55\:mol\:H_2O} \nonumber$
and then substituting these molar amounts into the definition for mole fraction.
\begin{align*} X_\mathrm{H_2O}&=\mathrm{\dfrac{mol\:H_2O}{mol\: NaCl + mol\:H_2O}}\ X_\mathrm{H_2O}&=\mathrm{\dfrac{55\:mol\:H_2O}{3.0\:mol\: NaCl+55\:mol\:H_2O}}\ X_\mathrm{H_2O}&=0.95\ X_\mathrm{NaCl}&=\mathrm{\dfrac{mol\: NaCl}{mol\: NaCl+mol\:H_2O}}\ X_\mathrm{NaCl}&=\mathrm{\dfrac{3.0\:mol\:NaCl}{3.0\:mol\: NaCl+55\:mol\:H_2O}}\ X_\mathrm{NaCl}&=0.052 \end{align*} \nonumber
Exercise $2$
The mole fraction of iodine, $\ce{I_2}$, dissolved in dichloromethane, $\ce{CH_2Cl_2}$, is 0.115. What is the molal concentration, m, of iodine in this solution?
Answer
1.50 m
Vapor Pressure Lowering
As described in the chapter on liquids and solids, the equilibrium vapor pressure of a liquid is the pressure exerted by its gaseous phase when vaporization and condensation are occurring at equal rates:
$\text{liquid} \rightleftharpoons \text{gas} \label{11.5.4}$
Dissolving a nonvolatile substance in a volatile liquid results in a lowering of the liquid’s vapor pressure. This phenomenon can be rationalized by considering the effect of added solute molecules on the liquid's vaporization and condensation processes. To vaporize, solvent molecules must be present at the surface of the solution. The presence of solute decreases the surface area available to solvent molecules and thereby reduces the rate of solvent vaporization. Since the rate of condensation is unaffected by the presence of solute, the net result is that the vaporization-condensation equilibrium is achieved with fewer solvent molecules in the vapor phase (i.e., at a lower vapor pressure) (Figure $1$). While this kinetic interpretation is useful, it does not account for several important aspects of the colligative nature of vapor pressure lowering. A more rigorous explanation involves the property of entropy, a topic of discussion in a later text chapter on thermodynamics. For purposes of understanding the lowering of a liquid's vapor pressure, it is adequate to note that the greater entropy of a solution in comparison to its separate solvent and solute serves to effectively stabilize the solvent molecules and hinder their vaporization. A lower vapor pressure results, and a correspondingly higher boiling point as described in the next section of this module.
The relationship between the vapor pressures of solution components and the concentrations of those components is described by Raoult’s law: The partial pressure exerted by any component of an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.
$P_\ce{A}=X_\ce{A}P^\circ_\ce{A} \label{11.5.5}$
where PA is the partial pressure exerted by component A in the solution, $P^\circ_\ce{A}$ is the vapor pressure of pure A, and XA is the mole fraction of A in the solution. (Mole fraction is a concentration unit introduced in the chapter on gases.)
Recalling that the total pressure of a gaseous mixture is equal to the sum of partial pressures for all its components (Dalton’s law of partial pressures), the total vapor pressure exerted by a solution containing i components is
$P_\ce{solution}=\sum_{i}P_i=\sum_{i}X_iP^\circ_i \label{11.5.6}$
A nonvolatile substance is one whose vapor pressure is negligible (P° ≈ 0), and so the vapor pressure above a solution containing only nonvolatile solutes is due only to the solvent:
$P_\ce{solution}=X_\ce{solvent}P^\circ_\ce{solvent} \label{11.5.7}$
Example $3$: Calculation of a Vapor Pressure
Compute the vapor pressure of an ideal solution containing 92.1 g of glycerin, C3H5(OH)3, and 184.4 g of ethanol, C2H5OH, at 40 °C. The vapor pressure of pure ethanol is 0.178 atm at 40 °C. Glycerin is essentially nonvolatile at this temperature.
Solution
Since the solvent is the only volatile component of this solution, its vapor pressure may be computed per Raoult’s law as:
$P_\ce{solution}=X_\ce{solvent}P^\circ_\ce{solvent}$
First, calculate the molar amounts of each solution component using the provided mass data.
$\mathrm{92.1\cancel{g\:C_3H_5(OH)_3}×\dfrac{1\:mol\:C_3H_5(OH)_3}{92.094\cancel{g\:C_3H_5(OH)_3}}=1.00\:mol\:C_3H_5(OH)_3}$
$\mathrm{184.4\cancel{g\:C_2H_5OH}×\dfrac{1\:mol\:C_2H_5OH}{46.069\cancel{g\:C_2H_5OH}}=4.000\:mol\:C_2H_5OH}$
Next, calculate the mole fraction of the solvent (ethanol) and use Raoult’s law to compute the solution’s vapor pressure.
$X_\mathrm{C_2H_5OH}=\mathrm{\dfrac{4.000\:mol}{(1.00\:mol+4.000\:mol)}=0.800}$
$P_\ce{solv}=X_\ce{solv}P^\circ_\ce{solv}=\mathrm{0.800×0.178\:atm=0.142\:atm}$
Exercise $3$
A solution contains 5.00 g of urea, CO(NH2)2 (a nonvolatile solute) and 0.100 kg of water. If the vapor pressure of pure water at 25 °C is 23.7 torr, what is the vapor pressure of the solution?
Answer
23.4 torr
Elevation of the Boiling Point of a Solvent
As described in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its vapor pressure is equal to ambient atmospheric pressure. Since the vapor pressure of a solution is lowered due to the presence of nonvolatile solutes, it stands to reason that the solution’s boiling point will subsequently be increased. Compared to pure solvent, a solution, therefore, will require a higher temperature to achieve any given vapor pressure, including one equivalent to that of the surrounding atmosphere. The increase in boiling point observed when nonvolatile solute is dissolved in a solvent, $ΔT_b$, is called boiling point elevation and is directly proportional to the molal concentration of solute species:
$ΔT_b=K_bm \label{11.5.8}$
where
• $K_\ce{b}$ is the boiling point elevation constant, or the ebullioscopic constant and
• $m$ is the molal concentration (molality) of all solute species.
Boiling point elevation constants are characteristic properties that depend on the identity of the solvent. Values of Kb for several solvents are listed in Table $1$.
Table $1$: Boiling Point Elevation and Freezing Point Depression Constants for Several Solvents
Solvent Boiling Point (°C at 1 atm) Kb (Cm−1) Freezing Point (°C at 1 atm) Kf (Cm−1)
water 100.0 0.512 0.0 1.86
hydrogen acetate 118.1 3.07 16.6 3.9
benzene 80.1 2.53 5.5 5.12
chloroform 61.26 3.63 −63.5 4.68
nitrobenzene 210.9 5.24 5.67 8.1
The extent to which the vapor pressure of a solvent is lowered and the boiling point is elevated depends on the total number of solute particles present in a given amount of solvent, not on the mass or size or chemical identities of the particles. A 1 m aqueous solution of sucrose (342 g/mol) and a 1 m aqueous solution of ethylene glycol (62 g/mol) will exhibit the same boiling point because each solution has one mole of solute particles (molecules) per kilogram of solvent.
Example $4$: Calculating the Boiling Point of a Solution
What is the boiling point of a 0.33 m solution of a nonvolatile solute in benzene?
Solution
Use the equation relating boiling point elevation to solute molality to solve this problem in two steps.
1. Calculate the change in boiling point.
$ΔT_\ce{b}=K_\ce{b}m=2.53\:°\ce C\:m^{−1}×0.33\:m=0.83\:°\ce C$
• Add the boiling point elevation to the pure solvent’s boiling point.
$\mathrm{Boiling\: temperature=80.1\:°C+0.83\:°C=80.9\:°C}$
Exercise $4$
What is the boiling point of the antifreeze described in Example $4$?
Answer
109.2 °C
Example $5$: The Boiling Point of an Iodine Solution
Find the boiling point of a solution of 92.1 g of iodine, $\ce{I2}$, in 800.0 g of chloroform, $\ce{CHCl3}$, assuming that the iodine is nonvolatile and that the solution is ideal.
Solution
We can solve this problem using four steps.
1. Convert from grams to moles of $\ce{I2}$ using the molar mass of $\ce{I2}$ in the unit conversion factor.
Result: 0.363 mol
Determine the molality of the solution from the number of moles of solute and the mass of solvent, in kilograms.
Result: 0.454 m
Use the direct proportionality between the change in boiling point and molal concentration to determine how much the boiling point changes.
Result: 1.65 °C
Determine the new boiling point from the boiling point of the pure solvent and the change.
Result: 62.91 °C
Check each result as a self-assessment.
Exercise $5$: glycerin:Water Solution
What is the boiling point of a solution of 1.0 g of glycerin, $\ce{C3H5(OH)3}$, in 47.8 g of water? Assume an ideal solution.
Answer
100.12 °C
Distillation of Solutions
Distillation is a technique for separating the components of mixtures that is widely applied in both in the laboratory and in industrial settings. It is used to refine petroleum, to isolate fermentation products, and to purify water. This separation technique involves the controlled heating of a sample mixture to selectively vaporize, condense, and collect one or more components of interest. A typical apparatus for laboratory-scale distillations is shown in Figure $2$.
Oil refineries use large-scale fractional distillation to separate the components of crude oil. The crude oil is heated to high temperatures at the base of a tall fractionating column, vaporizing many of the components that rise within the column. As vaporized components reach adequately cool zones during their ascent, they condense and are collected. The collected liquids are simpler mixtures of hydrocarbons and other petroleum compounds that are of appropriate composition for various applications (e.g., diesel fuel, kerosene, gasoline), as depicted in Figure $3$.
Depression of the Freezing Point of a Solvent
Solutions freeze at lower temperatures than pure liquids. This phenomenon is exploited in “de-icing” schemes that use salt (Figure $4$), calcium chloride, or urea to melt ice on roads and sidewalks, and in the use of ethylene glycol as an “antifreeze” in automobile radiators. Seawater freezes at a lower temperature than fresh water, and so the Arctic and Antarctic oceans remain unfrozen even at temperatures below 0 °C (as do the body fluids of fish and other cold-blooded sea animals that live in these oceans).
The decrease in freezing point of a dilute solution compared to that of the pure solvent, ΔTf, is called the freezing point depression and is directly proportional to the molal concentration of the solute
$ΔT_\ce{f}=K_\ce{f}m \label{11.5.9}$
where
• $m$ is the molal concentration of the solute in the solvent and
• $K_f$ is called the freezing point depression constant (or cryoscopic constant).
Just as for boiling point elevation constants, these are characteristic properties whose values depend on the chemical identity of the solvent. Values of Kf for several solvents are listed in Table $1$.
Example $5$: Calculation of the Freezing Point of a Solution
What is the freezing point of the 0.33 m solution of a nonvolatile nonelectrolyte solute in benzene described in Example $4$?
Solution
Use the equation relating freezing point depression to solute molality to solve this problem in two steps.
1. Calculate the change in freezing point. $ΔT_\ce{f}=K_\ce{f}m=5.12\:°\ce C\:m^{−1}×0.33\:m=1.7\:°\ce C \nonumber$
2. Subtract the freezing point change observed from the pure solvent’s freezing point. $\mathrm{Freezing\: Temperature=5.5\:°C−1.7\:°C=3.8\:°C} \nonumber$
Exercise $6$
What is the freezing point of a 1.85 m solution of a nonvolatile nonelectrolyte solute in nitrobenzene?
Answer
−9.3 °C
Colligative Properties and De-Icing
Sodium chloride and its group 2 analogs calcium and magnesium chloride are often used to de-ice roadways and sidewalks, due to the fact that a solution of any one of these salts will have a freezing point lower than 0 °C, the freezing point of pure water. The group 2 metal salts are frequently mixed with the cheaper and more readily available sodium chloride (“rock salt”) for use on roads, since they tend to be somewhat less corrosive than the NaCl, and they provide a larger depression of the freezing point, since they dissociate to yield three particles per formula unit, rather than two particles like the sodium chloride.
Because these ionic compounds tend to hasten the corrosion of metal, they would not be a wise choice to use in antifreeze for the radiator in your car or to de-ice a plane prior to takeoff. For these applications, covalent compounds, such as ethylene or propylene glycol, are often used. The glycols used in radiator fluid not only lower the freezing point of the liquid, but they elevate the boiling point, making the fluid useful in both winter and summer. Heated glycols are often sprayed onto the surface of airplanes prior to takeoff in inclement weather in the winter to remove ice that has already formed and prevent the formation of more ice, which would be particularly dangerous if formed on the control surfaces of the aircraft (Video $1$).
Video $1$: Freezing point depression is exploited to remove ice from the control surfaces of aircraft.
Phase Diagram for an Aqueous Solution of a Nonelectrolyte
The colligative effects on vapor pressure, boiling point, and freezing point described in the previous section are conveniently summarized by comparing the phase diagrams for a pure liquid and a solution derived from that liquid. Phase diagrams for water and an aqueous solution are shown in Figure $5$.
The liquid-vapor curve for the solution is located beneath the corresponding curve for the solvent, depicting the vapor pressure lowering, ΔP, that results from the dissolution of nonvolatile solute. Consequently, at any given pressure, the solution’s boiling point is observed at a higher temperature than that for the pure solvent, reflecting the boiling point elevation, ΔTb, associated with the presence of nonvolatile solute. The solid-liquid curve for the solution is displaced left of that for the pure solvent, representing the freezing point depression, ΔTf, that accompanies solution formation. Finally, notice that the solid-gas curves for the solvent and its solution are identical. This is the case for many solutions comprising liquid solvents and nonvolatile solutes. Just as for vaporization, when a solution of this sort is frozen, it is actually just the solvent molecules that undergo the liquid-to-solid transition, forming pure solid solvent that excludes solute species. The solid and gaseous phases, therefore, are composed solvent only, and so transitions between these phases are not subject to colligative effects.
Osmosis and Osmotic Pressure of Solutions
A number of natural and synthetic materials exhibit selective permeation, meaning that only molecules or ions of a certain size, shape, polarity, charge, and so forth, are capable of passing through (permeating) the material. Biological cell membranes provide elegant examples of selective permeation in nature, while dialysis tubing used to remove metabolic wastes from blood is a more simplistic technological example. Regardless of how they may be fabricated, these materials are generally referred to as semipermeable membranes.
Consider the apparatus illustrated in Figure $6$, in which samples of pure solvent and a solution are separated by a membrane that only solvent molecules may permeate. Solvent molecules will diffuse across the membrane in both directions. Since the concentration of solvent is greater in the pure solvent than the solution, these molecules will diffuse from the solvent side of the membrane to the solution side at a faster rate than they will in the reverse direction. The result is a net transfer of solvent molecules from the pure solvent to the solution. Diffusion-driven transfer of solvent molecules through a semipermeable membrane is a process known as osmosis.
When osmosis is carried out in an apparatus like that shown in Figure $6$, the volume of the solution increases as it becomes diluted by accumulation of solvent. This causes the level of the solution to rise, increasing its hydrostatic pressure (due to the weight of the column of solution in the tube) and resulting in a faster transfer of solvent molecules back to the pure solvent side. When the pressure reaches a value that yields a reverse solvent transfer rate equal to the osmosis rate, bulk transfer of solvent ceases. This pressure is called the osmotic pressure ($\Pi$) of the solution. The osmotic pressure of a dilute solution is related to its solute molarity, M, and absolute temperature, T, according to the equation
$Π=MRT \label{11.5.10}$
where $R$ is the universal gas constant.
Example $7$: Calculation of Osmotic Pressure
What is the osmotic pressure (atm) of a 0.30 M solution of glucose in water that is used for intravenous infusion at body temperature, 37 °C?
Solution
We can find the osmotic pressure, $Π$, using Equation \ref{11.5.10}, where T is on the Kelvin scale (310 K) and the value of R is expressed in appropriate units (0.08206 L atm/mol K).
\begin{align*} Π&=MRT\ &=\mathrm{0.03\:mol/L×0.08206\: L\: atm/mol\: K×310\: K}\ &=\mathrm{7.6\:atm} \end{align*} \nonumber
Exercise $7$
What is the osmotic pressure (atm) a solution with a volume of 0.750 L that contains 5.0 g of methanol, CH3OH, in water at 37 °C?
Answer
5.3 atm
If a solution is placed in an apparatus like the one shown in Figure $7$, applying pressure greater than the osmotic pressure of the solution reverses the osmosis and pushes solvent molecules from the solution into the pure solvent. This technique of reverse osmosis is used for large-scale desalination of seawater and on smaller scales to produce high-purity tap water for drinking.
Examples of osmosis are evident in many biological systems because cells are surrounded by semipermeable membranes. Carrots and celery that have become limp because they have lost water can be made crisp again by placing them in water. Water moves into the carrot or celery cells by osmosis. A cucumber placed in a concentrated salt solution loses water by osmosis and absorbs some salt to become a pickle. Osmosis can also affect animal cells. Solute concentrations are particularly important when solutions are injected into the body. Solutes in body cell fluids and blood serum give these solutions an osmotic pressure of approximately 7.7 atm. Solutions injected into the body must have the same osmotic pressure as blood serum; that is, they should be isotonic with blood serum. If a less concentrated solution, a hypotonic solution, is injected in sufficient quantity to dilute the blood serum, water from the diluted serum passes into the blood cells by osmosis, causing the cells to expand and rupture. This process is called hemolysis. When a more concentrated solution, a hypertonic solution, is injected, the cells lose water to the more concentrated solution, shrivel, and possibly die in a process called crenation (Figure 11.5.8).
Determination of Molar Masses
Osmotic pressure and changes in freezing point, boiling point, and vapor pressure are directly proportional to the concentration of solute present. Consequently, we can use a measurement of one of these properties to determine the molar mass of the solute from the measurements.
Example $8$: Determining Molar Mass from Freezing Point Depression
A solution of 4.00 g of a nonelectrolyte dissolved in 55.0 g of benzene is found to freeze at 2.32 °C. What is the molar mass of this compound?
Solution
We can solve this problem using the following steps.
1. Determine the change in freezing point from the observed freezing point and the freezing point of pure benzene (Table 11.5.1).
$ΔT_\ce{f}=\mathrm{5.5\:°C−2.32\:°C=3.2\:°C}$
1. Determine the molal concentration from Kf, the freezing point depression constant for benzene (Table 11.5.1), and ΔTf.
$ΔT_\ce{f}=K_\ce{f}m$
$m=\dfrac{ΔT_\ce{f}}{K_\ce{f}}=\dfrac{3.2\:°\ce C}{5.12\:°\ce C m^{−1}}=0.63\:m$
1. Determine the number of moles of compound in the solution from the molal concentration and the mass of solvent used to make the solution.
$\mathrm{Moles\: of\: solute=\dfrac{0.62\:mol\: solute}{1.00\cancel{kg\: solvent}}×0.0550\cancel{kg\: solvent}=0.035\:mol}$
2. Determine the molar mass from the mass of the solute and the number of moles in that mass.
$\mathrm{Molar\: mass=\dfrac{4.00\:g}{0.034\:mol}=1.2×10^2\:g/mol}$
Exercise $8$
A solution of 35.7 g of a nonelectrolyte in 220.0 g of chloroform has a boiling point of 64.5 °C. What is the molar mass of this compound?
Answer
1.8 × 102 g/mol
Example $9$: Determination of a Molar Mass from Osmotic Pressure
A 0.500 L sample of an aqueous solution containing 10.0 g of hemoglobin has an osmotic pressure of 5.9 torr at 22 °C. What is the molar mass of hemoglobin?
Solution
Here is one set of steps that can be used to solve the problem:
1. $\Pi=\mathrm{\dfrac{5.9\:torr×1\:atm}{760\:torr}=7.8×10^{−3}\:atm} \nonumber$
$\Pi=MRT \nonumber$
$M=\dfrac{Π}{RT}=\mathrm{\dfrac{7.8×10^{−3}\:atm}{(0.08206\:L\: atm/mol\: K)(295\:K)}=3.2×10^{−4}\:M}$
1. $\mathrm{moles\: of\: hemoglobin=\dfrac{3.2×10^{−4}\:mol}{1\cancel{L\: solution}}×0.500\cancel{L\: solution}=1.6×10^{−4}\:mol}$
2. Determine the molar mass from the mass of hemoglobin and the number of moles in that mass.
$\mathrm{molar\: mass=\dfrac{10.0\:g}{1.6×10^{−4}\:mol}=6.2×10^4\:g/mol}$
Exercise $9$
What is the molar mass of a protein if a solution of 0.02 g of the protein in 25.0 mL of solution has an osmotic pressure of 0.56 torr at 25 °C?
Answer
2.7 × 104 g/mol
Colligative Properties of Electrolytes
As noted previously in this module, the colligative properties of a solution depend only on the number, not on the kind, of solute species dissolved. For example, 1 mole of any nonelectrolyte dissolved in 1 kilogram of solvent produces the same lowering of the freezing point as does 1 mole of any other nonelectrolyte. However, 1 mole of sodium chloride (an electrolyte) forms 2 moles of ions when dissolved in solution. Each individual ion produces the same effect on the freezing point as a single molecule does.
Example $10$: The Freezing Point of a Solution of an Electrolyte
The concentration of ions in seawater is approximately the same as that in a solution containing 4.2 g of NaCl dissolved in 125 g of water. Assume that each of the ions in the NaCl solution has the same effect on the freezing point of water as a nonelectrolyte molecule, and determine the freezing temperature the solution (which is approximately equal to the freezing temperature of seawater).
Solution
We can solve this problem using the following series of steps.
• Convert from grams to moles of NaCl using the molar mass of NaCl in the unit conversion factor. Result: 0.072 mol NaCl
• Determine the number of moles of ions present in the solution using the number of moles of ions in 1 mole of NaCl as the conversion factor (2 mol ions/1 mol NaCl). Result: 0.14 mol ions
• Determine the molality of the ions in the solution from the number of moles of ions and the mass of solvent, in kilograms. Result: 1.1 m
• Use the direct proportionality between the change in freezing point and molal concentration to determine how much the freezing point changes. Result: 2.0 °C
• Determine the new freezing point from the freezing point of the pure solvent and the change. Result: −2.0 °C
Check each result as a self-assessment.
Exercise $10$
Assume that each of the ions in calcium chloride, CaCl2, has the same effect on the freezing point of water as a nonelectrolyte molecule. Calculate the freezing point of a solution of 0.724 g of CaCl2 in 175 g of water.
Answer
−0.208 °C
Assuming complete dissociation, a 1.0 m aqueous solution of NaCl contains 2.0 mole of ions (1.0 mol Na+ and 1.0 mol Cl) per each kilogram of water, and its freezing point depression is expected to be
$ΔT_\ce{f}=\mathrm{2.0\:mol\: ions/kg\: water×1.86\:°C\: kg\: water/mol\: ion=3.7\:°C.} \label{11.5.11}$
When this solution is actually prepared and its freezing point depression measured, however, a value of 3.4 °C is obtained. Similar discrepancies are observed for other ionic compounds, and the differences between the measured and expected colligative property values typically become more significant as solute concentrations increase. These observations suggest that the ions of sodium chloride (and other strong electrolytes) are not completely dissociated in solution.
To account for this and avoid the errors accompanying the assumption of total dissociation, an experimentally measured parameter named in honor of Nobel Prize-winning German chemist Jacobus Henricus van’t Hoff is used. The van’t Hoff factor (i) is defined as the ratio of solute particles in solution to the number of formula units dissolved:
$i=\dfrac{\textrm{moles of particles in solution}}{\textrm{moles of formula units dissolved}} \label{11.5.12}$
Values for measured van’t Hoff factors for several solutes, along with predicted values assuming complete dissociation, are shown in Table $2$.
Table $2$: Expected and Observed van’t Hoff Factors for Several 0.050 m Aqueous Electrolyte Solutions
Electrolyte Particles in Solution i (Predicted) i (Measured)
HCl H+, Cl 2 1.9
NaCl Na+, Cl 2 1.9
MgSO4 Mg2+, $\ce{SO4^2-}$ 2 1.3
MgCl2 Mg2+, 2Cl 3 2.7
FeCl3 Fe3+, 3Cl 4 3.4
glucose (a non-electrolyte) C12H22O11 1 1.0
In 1923, the chemists Peter Debye and Erich Hückel proposed a theory to explain the apparent incomplete ionization of strong electrolytes. They suggested that although interionic attraction in an aqueous solution is very greatly reduced by solvation of the ions and the insulating action of the polar solvent, it is not completely nullified. The residual attractions prevent the ions from behaving as totally independent particles (Figure $9$). In some cases, a positive and negative ion may actually touch, giving a solvated unit called an ion pair. Thus, the activity, or the effective concentration, of any particular kind of ion is less than that indicated by the actual concentration. Ions become more and more widely separated the more dilute the solution, and the residual interionic attractions become less and less. Thus, in extremely dilute solutions, the effective concentrations of the ions (their activities) are essentially equal to the actual concentrations. Note that the van’t Hoff factors for the electrolytes in Table $2$ are for 0.05 m solutions, at which concentration the value of i for NaCl is 1.9, as opposed to an ideal value of 2.
Summary
Properties of a solution that depend only on the concentration of solute particles are called colligative properties. They include changes in the vapor pressure, boiling point, and freezing point of the solvent in the solution. The magnitudes of these properties depend only on the total concentration of solute particles in solution, not on the type of particles. The total concentration of solute particles in a solution also determines its osmotic pressure. This is the pressure that must be applied to the solution to prevent diffusion of molecules of pure solvent through a semipermeable membrane into the solution. Ionic compounds may not completely dissociate in solution due to activity effects, in which case observed colligative effects may be less than predicted.
Key Equations
• $\left(P_\ce{A}=X_\ce{A}P^\circ_\ce{A}\right)$
• $P_\ce{solution}=\sum_{i}P_i=\sum_{i}X_iP^\circ_i$
• $P_\ce{solution}=X_\ce{solvent}P^\circ_\ce{solvent}$
• ΔTb = Kbm
• ΔTf = Kfm
• Π = MRT
Footnotes
1. A nonelectrolyte shown for comparison.
Glossary
boiling point elevation
elevation of the boiling point of a liquid by addition of a solute
boiling point elevation constant
the proportionality constant in the equation relating boiling point elevation to solute molality; also known as the ebullioscopic constant
colligative property
property of a solution that depends only on the concentration of a solute species
crenation
process whereby biological cells become shriveled due to loss of water by osmosis
freezing point depression
lowering of the freezing point of a liquid by addition of a solute
freezing point depression constant
(also, cryoscopic constant) proportionality constant in the equation relating freezing point depression to solute molality
hemolysis
rupture of red blood cells due to the accumulation of excess water by osmosis
hypertonic
of greater osmotic pressure
hypotonic
of less osmotic pressure
ion pair
solvated anion/cation pair held together by moderate electrostatic attraction
isotonic
of equal osmotic pressure
molality (m)
a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms
osmosis
diffusion of solvent molecules through a semipermeable membrane
osmotic pressure (Π)
opposing pressure required to prevent bulk transfer of solvent molecules through a semipermeable membrane
Raoult’s law
the partial pressure exerted by a solution component is equal to the product of the component’s mole fraction in the solution and its equilibrium vapor pressure in the pure state
semipermeable membrane
a membrane that selectively permits passage of certain ions or molecules
van’t Hoff factor (i)
the ratio of the number of moles of particles in a solution to the number of moles of formula units dissolved in the solution
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An electrolyte solution is a solution that generally contains ions, atoms or molecules that have lost or gained electrons, and is electrically conductive. For this reason they are often called ionic solutions, however there are some cases where the electrolytes are not ions. For this discussion we will only consider solutions of ions. A basic principle of electrostatics is that opposite charges attract and like charges repel. It also takes a great deal of force to overcome this electrostatic attraction.
Introduction
The general form of Coulomb's law describes the force of attraction between charges:
$F=k\frac{q_1mq_2}{r^2}$
However, we must make some changes to this physics formula to be able to use it for a solution of oppositely charged ions. In Coulomb's Law, the constant $k=\frac{1}{4\pi\varepsilon_{0}}$, where $\varepsilon_{0}$ is the permittivity of free space, such as in a vacuum. However, since we are looking at a solution, we must consider the effect that the medium (the solvent in this case) has on the electrostatic force, which is represented by the dielectric constant $\varepsilon$:
$F=\frac{q_{1}q_{2}}{4\pi\varepsilon_{0}\varepsilon r^{2}}$
Polar substances such as water have a relatively high dielectric constant.
Standard Definitions of Enthalpy, Entropy, and Gibbs Energy for Ions
Ions are not stable on their own, and thus no ions can ever be studied separately. Particularly in biology, all ions in a certain cell or tissue have a counterion that balances this charge. Therefore, we cannot measure the enthalpy or entropy of a single ion as we can atoms of a pure element. So we define a reference point. The $\Delta_{f}\overline{H}^{\circ}$ of a hydrogen ion $H^+$ is equal to zero, as are the other thermodynamic quantities.
$\Delta_{f}\overline{H}^{\circ}[H^{+}(aq)]=0$
$\Delta_{f}\overline{G}^{\circ}[H^{+}(aq)]=0$
$\overline{S}^{\circ}[H^{+}(aq)]=0$
When studying the formation of ionic solutions, the most useful quantity to describe is chemical potential $\mu$, defined as the partial molar Gibbs energy of the ith component in a substance:
$\mu_{i}=\overline{G}_{i}=\left(\frac{\partial G}{\partial n_{i}}\right)_{T,P,n_{j}}=\mu_{i}^{\circ}+RT\ln x_{i}$
where $x_{i}$ can be any unit of concentration of the component: mole fraction, molality, or for gases, the partial pressure divided by the pressure of pure component.
Ionic Solutions
To express the chemical potential of an electrolyte in solution in terms of molality, let us use the example of a dissolved salt such as magnesium chloride, $MgCl_{2}$.
$MgCl_{2}\rightleftharpoons Mg^{2+}+2Cl^{-} \label{1}$
We can now write a more general equation for a dissociated salt:
$M_{\nu+}X_{\nu-}\rightleftharpoons\nu_{+}M^{z+}+\nu_{-}X^{z-} \label{2}$
where $\nu_{\pm}$ represents the stoichiometric coefficient of the cation or anion and $z_\pm$ represents the charge, and M and X are the metal and halide, respectively.
The total chemical potential for these anion-cation pair would be the sum of their individual potentials multiplied by their stoichiometric coefficients:
$\mu=\nu_{+}\mu_{+}+\nu_{-}\mu_{-} \label{3}$
The chemical potentials of the individual ions are:
$\mu_{+} = \mu_+^{\circ}+RT\ln m_+ \label{4}$
$\mu_{-} = \mu_-^{\circ}+RT\ln m_- \label{5}$
And the molalities of the individual ions are related to the original molality of the salt m by their stoichiometric coefficients
$m_{+}=\nu_{+}m$
Substituting Equations $\ref{4}$ and $\ref{5}$ into Equation $\ref{3}$,
$\mu=\left( \nu_+\mu_+^{\circ}+\nu_- mu_-^{\circ}\right)+RT\ln\left(m_+^{\nu+}m_-^{\nu-}\right) \label{6}$
since the total number of moles $\nu=\nu_{+}+\nu_{-}$, we can define the mean ionic molality as the geometric average of the molarity of the two ions:
$m_{\pm}=(m_+^{\nu+}m_-^{\nu-})^{\frac{1}{\nu}}$
then Equation $\ref{6}$ becomes
$\mu=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+\nu RT\ln m_{\pm} \label{7}$
We have derived this equation for a ideal solution, but ions in solution exert electrostatic forces on one another to deviate from ideal behavior, so instead of molarities we must use the activity a to represent how the ion is behaving in solution. Therefore the mean ionic activity is defined as
$a_{\pm}=(a_{+}^{\nu+}+a_{-}^{\nu-})^{\frac{1}{\nu}}$
where
$a_{\pm}=\gamma m_{\pm} \label{mean}$
and $\gamma_{\pm}$ is the mean ionic activity coefficient, which is dependent on the substance.
Substituting the mean ionic activity of \Equation $\ref{mean}$ into Equation $\ref{7}$,
$\mu=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+\nu RT\ln a_{\pm}=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+RT\ln a_{\pm}^{\nu}=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+RT \ln a \label{11}$
when $a=a_{\pm}^{\nu}$. Equation $\ref{11}$ then represents the chemical potential of a nonideal electrolyte solutions. To calculate the mean ionic activity coefficient requires the use of the Debye-Hückel limiting law, part of the Debye-Hückel theory of electrolytes .
Example $1$
Let us now write out the chemical potential in terms of molality of the salt in our first example, $MgCl_{2}$. First from Equation $\ref{1}$, the stoichiometric coefficients of the ions are:
$\nu_{+} = 1,\nu_{-} = 2,\nu\; = 3 \nonumber$
The mean ionic molality is
\begin{align*} m_{\pm} &= (m_{+}^{1}m_{-}^{2})^{\frac{1}{3}} \[4pt] &= (\nu_{+}m\times\nu_{-}m)^{\frac{1}{3}} \[4pt] &=m(1^{1}2^{2})^{\frac{1}{3}} \[4pt] &=1.6\, m \end{align*}
The expression for the chemical potential of $MgCl_{2}$ is
$\mu_{MgCl_{2}}=\mu_{MgCl_{2}}^{\circ}+3RT\ln 1.6\m m \nonumber$
Contributors and Attributions
• Konstantin Malley
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The properties of electrolyte solutions can significantly deviate from the laws used to derive chemical potential of solutions. In nonelectrolyte solutions, the intermolecular forces are mostly comprised of weak Van der Waals interactions, which have a $r^{-7}$ dependence, and for practical purposes this can be considered ideal. In ionic solutions, however, there are significant electrostatic interactions between solute-solvent as well as solute-solute molecules. These electrostatic forces are governed by Coulomb's law, which has a $r^{-2}$ dependence. Consequently, the behavior of an electrolyte solution deviates considerably from that an ideal solution. Indeed, this is why we utilize the activity of the individual components and not the concentration to calculate deviations from ideal behavior. In 1923, Peter Debye and Erich Hückel developed a theory that would allow us to calculate the mean ionic activity coefficient of the solution, $\gamma_{\pm}$, and could explain how the behavior of ions in solution contribute to this constant.
Assumptions of Debye-Hückel Theory
The Debye-Hückel theory is based on three assumptions of how ions act in solution:
1. Electrolytes completely dissociate into ions in solution.
2. Solutions of Electrolytes are very dilute, on the order of 0.01 M.
3. Each ion is surrounded by ions of the opposite charge, on average.
Debye and Hückel developed the following equation to calculate the mean ionic activity coefficient $\gamma_{\pm}$:
$\log\gamma_{\pm}=-\dfrac{1.824\times10^{6}}{(\varepsilon T)^{3/2}}\mid z_{+}z_{-}\mid\sqrt{I} \label{1}$
where
• $\varepsilon$ is the dielectric constant,
• $z_{+}$ and $z_{-}$ are the charges of the cation and anion, respectively, and
• $I$ is the ionic strength of the solution.
The Equation $\ref{1}$ is known as the Debye-Hückel Limiting Law. The ionic strength is calculated by the following relation:
$I=\dfrac{1}{2}\sum_{i}m_{i}z_{i}^{2} \label{2}$
where $m_{i}$ and $z_{i}$ are the molality and the charge of the ith ion in the electrolyte. Since most of the electrolyte solutions we study are aqueous $(\varepsilon=78.54)$ and have a temperature of 298 K, the Limiting Law in Equation \ref{1} reduces to
$\log \gamma_{\pm}=-0.509\mid z_{+}z_{-}\mid\sqrt{I} \label{3}$
Example $1$
Calculate ionic strength, mean ionic activity coefficient $\gamma_{\pm}$, and the mean ionic molality $m_{\pm}$ for a 0.02 molal aqueous solution of zinc chloride, $\ce{ZnCl2}$.
Solution
Zinc chloride will dissolve as
$\ce{ZnCl2 -> Zn^{2+}(aq) + 2Cl^{-}(aq)} \nonumber$
The concentrations of the zinc and chloride ions will then be 0.02 and 0.04 molal, respectively. First calculate the mean ionic molality. The mean ionic molality is defined as the average molality of the two ions (see Electrolyte Solutions):
$m_{\pm}=(m_{+}^{\nu+}m_{-}^{\nu-})^{\frac{1}{\nu}} \nonumber$
where $\nu$ is the stoichiometric coefficient of the ions, and the total of the coefficients in the exponent. In our case, the mean ionic molality is
\begin{align*} m_{\pm} &=(m_{Zn}^{\nu(Zn)} m_{Cl}^{\nu(Cl)})^{\frac{1}{\nu (Zn) + \nu (Cl)}} \[4pt] &=[(0.02)^{1}(0.04)^{2}]^{\frac{1}{3}} \[4pt] &=\left[(0.02)(0.0016)\right]^{\frac{1}{3}} \[4pt] &=3.17 \times 10^{-2} \end{align*}
To calculate the mean ionic activity coefficient, we first need the ionic strength of the solution from Equation $\ref{2}$:
\begin{align*} I &=\dfrac{1}{2}[(0.02)(+2)^{2}+(0.04)(-1)^{2}] \[4pt] &=\dfrac{1}{2}(0.08+0.04)=0.06 \end{align*}
Now we can use Equation $\ref{3}$ to calculate the activity coefficient:
\begin{align*} \log\gamma_{\pm} &=-0.509\mid(+2)(-1)\mid\sqrt{0.06} \[4pt] &=(-0.509)(2)(0.245) \[4pt] &=-0.250 \[4pt] \gamma_{\pm}&=10^{-0.25} \[4pt] &=0.1627 \end{align*}
Exercise $1$
Determine the mean ionic activity coefficient and mean activity of a 0.004 molal of $\ce{Ba(HCO3)2}$?
Answer
$γ_±=10^{−0.10188}≈0.7909 \nonumber$
$a_±=0.7909 \times 0.2785\, m= 0.2203 \,m \nonumber$
The kinetic salt effect is the effect of salts preset in solution on the rate of a reaction.
Kinetic Salt Effect
In biological systems, salts influence how well proteins and DNA function. Salts are formed by ionic bonds, between a metal and an electromagnetic atom(s). Some examples of salts include NaCl, KCl, and Na2SO4. Salt molecules are able to disassociate, forming cations and anions. An increase in the charge (- or +) of a transition state or an activated complex results in an increase in solvation (creating more order in the system), and causes a decrease in the change of entropy (ΔS). In contrast, a decrease in the charge of the transition state causes an increase in ΔS.
$I^- + C^+ \rightleftharpoons E^o \rightarrow Product \label{2A}$
The kinetic salt effect describes the way salts stabilize reactants. For example, in the above reaction, each reactant has a charge. The negatively-charged reactant is stabilized by the positive charges from the salt, and the positively-charged reactant is stabilized by the negative charges from the salt. As a result, the rate at which the reactants come together decreases, thus decreasing the rate at which E forms. Because a charged intermediate is also stabilized in the solution, the half life of the intermediate at equilibrium increases, shifting the reaction toward product formation. Because the rate of the product formation is higher due to increased amounts of the intermediate present on the solution, first order kinetics is used to derive the rate constant equation:
$\log K_{TS} = \log K_{TS^o} + 2Z_AZ_B\sqrt{I} \label{3A}$
where
• Z is the charge on the cation and ion from the salt. ZAZB is a product value.
• I is the ionic strength. I is also dependent on the solubility of the salt in the reaction mixture. Ionic strength is directly proportional to the solubility of the salt. Changing the ionic strength manipulates the solvation of the reactants and intermediates, thus changing ΔS, and affecting the reaction rate.
• $\log{K_{TS^o}}$ is the rate constant without the salt in the reaction mixture.
• A is also a constant for the solvent the solution is in. The A value for water is 0.509 at 298 K.
The relationship between ZAZB, I, and the rate of the reaction is presented in tabular form below:
ZAZB Rate of Reaction
+ Increases (salt present in the reaction mixture)
- Decreases (the ionic strength increases)
Note: I=0 at very dilute salt concentrations or if the salt is inert.
Contributors and Attributions
• Konstantin Malley (UCD)
• Artika Singh (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/05%3A_Solutions/5.08%3A_Ionic_Activity.txt
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An electrolyte solution is a solution that generally contains ions, atoms or molecules that have lost or gained electrons, and is electrically conductive. For this reason they are often called ionic solutions, however there are some cases where the electrolytes are not ions. For this discussion we will only consider solutions of ions. A basic principle of electrostatics is that opposite charges attract and like charges repel. It also takes a great deal of force to overcome this electrostatic attraction.
Introduction
The general form of Coulomb's law describes the force of attraction between charges:
$F=k\frac{q_1mq_2}{r^2}$
However, we must make some changes to this physics formula to be able to use it for a solution of oppositely charged ions. In Coulomb's Law, the constant $k=\frac{1}{4\pi\varepsilon_{0}}$, where $\varepsilon_{0}$ is the permittivity of free space, such as in a vacuum. However, since we are looking at a solution, we must consider the effect that the medium (the solvent in this case) has on the electrostatic force, which is represented by the dielectric constant $\varepsilon$:
$F=\frac{q_{1}q_{2}}{4\pi\varepsilon_{0}\varepsilon r^{2}}$
Polar substances such as water have a relatively high dielectric constant.
Standard Definitions of Enthalpy, Entropy, and Gibbs Energy for Ions
Ions are not stable on their own, and thus no ions can ever be studied separately. Particularly in biology, all ions in a certain cell or tissue have a counterion that balances this charge. Therefore, we cannot measure the enthalpy or entropy of a single ion as we can atoms of a pure element. So we define a reference point. The $\Delta_{f}\overline{H}^{\circ}$ of a hydrogen ion $H^+$ is equal to zero, as are the other thermodynamic quantities.
$\Delta_{f}\overline{H}^{\circ}[H^{+}(aq)]=0$
$\Delta_{f}\overline{G}^{\circ}[H^{+}(aq)]=0$
$\overline{S}^{\circ}[H^{+}(aq)]=0$
When studying the formation of ionic solutions, the most useful quantity to describe is chemical potential $\mu$, defined as the partial molar Gibbs energy of the ith component in a substance:
$\mu_{i}=\overline{G}_{i}=\left(\frac{\partial G}{\partial n_{i}}\right)_{T,P,n_{j}}=\mu_{i}^{\circ}+RT\ln x_{i}$
where $x_{i}$ can be any unit of concentration of the component: mole fraction, molality, or for gases, the partial pressure divided by the pressure of pure component.
Ionic Solutions
To express the chemical potential of an electrolyte in solution in terms of molality, let us use the example of a dissolved salt such as magnesium chloride, $MgCl_{2}$.
$MgCl_{2}\rightleftharpoons Mg^{2+}+2Cl^{-} \label{1}$
We can now write a more general equation for a dissociated salt:
$M_{\nu+}X_{\nu-}\rightleftharpoons\nu_{+}M^{z+}+\nu_{-}X^{z-} \label{2}$
where $\nu_{\pm}$ represents the stoichiometric coefficient of the cation or anion and $z_\pm$ represents the charge, and M and X are the metal and halide, respectively.
The total chemical potential for these anion-cation pair would be the sum of their individual potentials multiplied by their stoichiometric coefficients:
$\mu=\nu_{+}\mu_{+}+\nu_{-}\mu_{-} \label{3}$
The chemical potentials of the individual ions are:
$\mu_{+} = \mu_+^{\circ}+RT\ln m_+ \label{4}$
$\mu_{-} = \mu_-^{\circ}+RT\ln m_- \label{5}$
And the molalities of the individual ions are related to the original molality of the salt m by their stoichiometric coefficients
$m_{+}=\nu_{+}m$
Substituting Equations $\ref{4}$ and $\ref{5}$ into Equation $\ref{3}$,
$\mu=\left( \nu_+\mu_+^{\circ}+\nu_- mu_-^{\circ}\right)+RT\ln\left(m_+^{\nu+}m_-^{\nu-}\right) \label{6}$
since the total number of moles $\nu=\nu_{+}+\nu_{-}$, we can define the mean ionic molality as the geometric average of the molarity of the two ions:
$m_{\pm}=(m_+^{\nu+}m_-^{\nu-})^{\frac{1}{\nu}}$
then Equation $\ref{6}$ becomes
$\mu=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+\nu RT\ln m_{\pm} \label{7}$
We have derived this equation for a ideal solution, but ions in solution exert electrostatic forces on one another to deviate from ideal behavior, so instead of molarities we must use the activity a to represent how the ion is behaving in solution. Therefore the mean ionic activity is defined as
$a_{\pm}=(a_{+}^{\nu+}+a_{-}^{\nu-})^{\frac{1}{\nu}}$
where
$a_{\pm}=\gamma m_{\pm} \label{mean}$
and $\gamma_{\pm}$ is the mean ionic activity coefficient, which is dependent on the substance.
Substituting the mean ionic activity of \Equation $\ref{mean}$ into Equation $\ref{7}$,
$\mu=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+\nu RT\ln a_{\pm}=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+RT\ln a_{\pm}^{\nu}=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+RT \ln a \label{11}$
when $a=a_{\pm}^{\nu}$. Equation $\ref{11}$ then represents the chemical potential of a nonideal electrolyte solutions. To calculate the mean ionic activity coefficient requires the use of the Debye-Hückel limiting law, part of the Debye-Hückel theory of electrolytes .
Example $1$
Let us now write out the chemical potential in terms of molality of the salt in our first example, $MgCl_{2}$. First from Equation $\ref{1}$, the stoichiometric coefficients of the ions are:
$\nu_{+} = 1,\nu_{-} = 2,\nu\; = 3 \nonumber$
The mean ionic molality is
\begin{align*} m_{\pm} &= (m_{+}^{1}m_{-}^{2})^{\frac{1}{3}} \[4pt] &= (\nu_{+}m\times\nu_{-}m)^{\frac{1}{3}} \[4pt] &=m(1^{1}2^{2})^{\frac{1}{3}} \[4pt] &=1.6\, m \end{align*}
The expression for the chemical potential of $MgCl_{2}$ is
$\mu_{MgCl_{2}}=\mu_{MgCl_{2}}^{\circ}+3RT\ln 1.6\m m \nonumber$
Contributors and Attributions
• Konstantin Malley
|
textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/05%3A_Solutions/5.09%3A_Colligative_Properties_of_Electrolyte_Solutions.txt
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5.5: Real Solutions
Q5.6
Calculate the molality for a 0.005M NH4SO4. What is the difference between the molarity and molality of the solution?
Q5.6
What is the difference between the molarity and the molality of a solution? Calculate the molality for a 0.05M ammonium sulfite [NH4SO3] solution.
S5.6
The molarity of a solution is the ratio between the number of moles of solute and the solution.
The molality of a solution is the ratio between the number of moles of solute and the mass of solvent:
Q5.7
After taking tylenol for stomach pains, a man has 0.300 mg of acetaminophen in his 5.0 L blood. After 4 hours, there is 0.010 mg left in his blood. Calculate the number of moles of acetaminophen per ml of blood and the total number of moles and grams before and after. (Molar Mass of Acetaminophen: 151.2 g/mol)
S5.7
Before:
$n=.300\times 10^{-3}g \times \dfrac{1mol}{151.2g}=1.98 \times 10^{-6}mol$
$M=\dfrac{1.98\times10^{-6}mol}{5.0L}(\dfrac{1000ml}{1L})=3.96\times 10^{-4}\dfrac{mol}{ml}$
After:
$n=.010\times 10^{-3}g \times \dfrac{1mol}{151.2g}=6.61 \times 10^{-8}mol$
$M=\dfrac{6.61\times10^{-8}mol}{5.0L}(\dfrac{1000ml}{1L})=1.32\times 10^{-5}\dfrac{mol}{ml}$
Q5.16
If a diver swims to 50 ft below sea level, did the oxygen in his blood increase or decrease? What about a hiker who is 50 ft above sea level?
Q5.16
Charlie goes on a submarine ride with his family. On their submarine ride, they have dinner which Charlie washes down with a Coke. Charlie notes that upon the removal of the cap from the Coke bottle, it does not make much of a noise that he has grown accustomed to hearing when opening a soft drink bottle. Soon after their dinner, Charlie and his family make their way back up to the surface to disembark from their submarine journey. As they were ascending, Charlie started to burp. What has just occurred?
S5.16
When on a submarine, you submerge into the water and are below sea level. When below sea level, the total and partial pressures of CO2 increases in comparison to when you are at or above sea level. Since Charlie's family has brought the Coke down from land (at sea level), the solubility of the CO2 is increased and the concentration of the CO2 in solution is raised. When Charlie returns to sea level, the conditions go back to normal which causes both the pressure and the solubility to decrease and the CO2 inside of Charlie comes out and causes him to burp in order to release the CO2 gas.
Q5.17
Calculate the Henry's Law constant for O2 in blood at a partial pressure of 0.25. There's about 2.5 kg of blood in the individual and 15 g of $O_2$ present.
S5.17
$P_{O_2}=k\cdot m \rightarrow k=\dfrac{P_{0_2}}{m}$
$k=\dfrac{.25}{.188\dfrac{mol}{kg}}$
$M=(15.0g)(\dfrac{1mol}{32.0g})$
Q5.18
Imagine the ideal gas $F_2$ (at 25°C) has a solubility of $123 \times 10^{-4}\; mol/L$ at a partial pressure of 0.5 atm, and $321 \times 10^{-4}$ at 5 atm. Use Henry’s law to determine the number of moles of $F_2$ at the given pressures.
Q5.19
Write the equation showing the relationship between the molality and the change of temperature. Show possible reasons that leading to the derivation of this equation.
S5.19
$\Delta T = K_b \times m_2$
Molality is independent of temperature.
Q5.20
There are two liquids X and Y. The boiling point of liquid X is 125ºC while the boiling point of liquid Y is 250ºC. If liquids X and Y were mixed to form an ideal solution, would the boiling point of the solution be less than, greater than, or equal to that of (a) liquid X? (b) liquid Y?
S5.20
1. Boiling point of solution > Boiling point of liquid X
2. Boiling point of solution < Boiling point of liquid Y
Q5.25
Why don't you keep your honey in the fridge? Explain using the terms hypertonic, and hypotonic.
S5.25
Honey is a hypertonic solution and bacteria prefer a hypotonic environment.
Q5.27
The freezing point depression measurement of Compound X in Reagent A yields a molar mass of 144 g; the same measurement in Reagent Y gives a value of 72 g. What is the reason for this discrepancy? (Hint: Consider solvent-solute interactions).
Q5.28
What would be the volume of 20 moles of sugar (C12H22O11) dissolved in 0.5 g/cm^-3 of water at 10K? Given the freezing point depression constant for water is 1.86 K/mol kg, what would be the molality of the sugar?
Q5.28
You have a boiling pot with 4 L of water. You add 1000 grams of NaCl. Water boils at 100 o C and freezes at 0 o C. To what temperature is the boiling point of water elevated to with the addition of NaCl? To what temperature is the freezing point depressed to?
Q5.29a
Blood plasma, an enriched with platelets, contains several growth factors that help healing of tissues and bones. There are a lot of medical treatments which inject the comparable concentration of a certain solution into human blood. Explain why ?
S5.29a
The injected solution should be comparable to that of blood plasma. There is a certain amount of platelets which keep the state the blood plasma in equilibrium. If we inject too much, it significantly affects to human blood by changing the concentration of platelets.
Q5.29b
Immerse a human cell in hypertonic or hypotonic solution can cause harmful consequences. Explain what would happen in each situation and the case when we got injected with a high concentrated solutions in our vein?
Q5.30
The Davis water tower is 250 ft tall (about 76 meters). What is the osmotic pressure required to push water from the ground to the top of the tower?
S5.30
$\pi = h\rho g$
$\pi = (76m)(1000kg \ m^{-3})(9.81 m \ s^{-2})$
$\pi = (7.456\ast 10^{5}N \ m^{-2})(\dfrac{1 atm}{1.013 \ast 10^{5}N\ m^{-2}})$
$\pi = 7.36 atm$
Q5.31
A solution of liquid A is mixed with liquid B and exerts ideal behavior. At $103^oC$ the total vapor pressure of a solution is 413mmHg containing 1.4 moles of A and 2.0 moles of B. After you add another mole of A to the solution, the vapor pressure increases to 486mmHg. At $103^oC$ calculate the vapor pressure of pure A and B.
S5.31
Step1: (Hint: Calculate the mole fraction of A and B)
$x_A=(\dfrac{molesA}{(molesA+molesB)})=(\dfrac{1.4molesA}{(1.4molesA+2.0molesB)})=.42$
$x_B=(\dfrac{molesB}{(molesA+molesB)})=(\dfrac{2.0molesB}{(1.4molesA+2.0molesB)})=.58$
Step 2: (Hint: Set up total pressure equation)
$P=x_AP^*_A+x_BP^*_B$
$413=(.42)P^*_A+(.58)P^*_B$
Step 3: (Hint: Calculate the mole fraction of A and B with the addition of 1 mole of A)
$x_A=(\dfrac{molesA}{(molesA+molesB)})=(\dfrac{2.4molesA}{(2.4molesA+2.0molesB)})=.55$
$x_B=(\dfrac{molesB}{(molesA+molesB)})=(\dfrac{2.0molesB}{(2.4molesA+2.0molesB)})=.45$
Step 4: (Hint: Set up the second total pressure equation)
$486=(.55)P^*_A+(.45)P^*_B$
Step 5: (Hint: Solve for $P^*_A$ and $P^*_B$ by substitution of the two total pressure equations)
$413=(.42)P^*_A+(.58)P^*_B$
$486=(.55)P^*_A+(.45)P^*_B$
Rearrange equation 1
$P^*_A=983.33-1.38095P^*_B$
Substitute into any equation and solve for $P^*_B$
$486=(.55)(983.33-1.38095P^*_B)+(.45)P^*_B$
$P^*_B=177.149mmHg$
Substitute $P^*_B$ into any equation to get $P^*_A$
$P^*_A=983.33-1.38095P^*_B$
$=738.696mmHg$
Q5.33
An ideal solution is composed of equal amount of component A with molar mass of 28.05g/mol and component B with molar mass of 80.91g/mol. At 35°C, component A has a vapor pressure of 0.2 atm and component B has a vapor pressure of 0.65 atm.
1. Calculate the mole fraction of each component in the solution
2. Calculate the partial pressure of each component at 35°C
3. If some of the vapor condenses back into a liquid, what will the mole fraction of each component in this liquid and the vapor pressure above this liquid be?
S5.33
(a) remember mole fraction = (mole of solute) / (total mole of solution) ; $n_A=\dfrac{weight_A}{28.05g/mol}\ n_B=\dfrac{weight_B}{80.91g/mol}\x_A=\dfrac{n_A}{n_A+n_B}=\dfrac{\dfrac{weight_A}{28.05g/mol}}{\dfrac{weight_A}{28.05g/mol}+\dfrac{weight_B}{80.91g/mol}}=\dfrac{1/28.05}{(1/28.05)+(1/80.91)}=0.7426\ x_B=\dfrac{n_B}{n_A+n_B}=\dfrac{\dfrac{weight_B}{80.91g/mol}}{\dfrac{weight_A}{28.05g/mol}+\dfrac{weight_B}{80.91g/mol}}=\dfrac{1/80.91}{(1/28.05)+(1/80.91)}=0.2574$
(b)$P_A=x_A*P_A^*=0.7426*0.2atm=0.1485atm\ P_B=x_B*P_B^*=0.2574*0.62atm=0.1596atm$
(c)$x_A=\dfrac{P_A}{P_A+P_B}=\dfrac{0.1485atm}{0.1485atm+0.1596atm}=0.4820\ x_B=\dfrac{P_B}{P_A+P_B}=\dfrac{0.1596atm}{0.1485atm+0.1596atm}=0.5180\ P_A=x_A*P_A^*=(0.4820)(0.2atm)=0.0964atm\ P_B=x_B*P_B^*=(0.5180)(0.62atm)=0.3212atm$
5.3: The Thermodynamics of Mixing
Q5.35a
Give a brief explanation for the lower vapor pressure of a solvent in presence of a solute from the entropy point of view. (Hint: Look at colligative property of lower vapor pressure).
S5.35a
The action of mixing a solvent and a solute leads is associated with an increase in entropy. The final solution therefore has a greater entropy than that of solvent, giving it less driving force for evaporation. This leads to the solvent's decrease in evaporation which results in a lower vapor pressure.
Q5.35b
Describe the process of vapor-pressure lowering from the colligative properties of solutions.
S5.35b
Suppose you have an ideal solution to which you can apply Raoult's law:
(Note: This solution has both a solvent and a solute which is nonvolatile).
Next, since we know that , we can ultimately rearrange the equation to be:
In the equation above, is directly proportional to the mole fraction of the solute.
Q5.35c
Sugar is added to water. Would there be more water molecules in the vapor above the sugar solution or in the vapor above the pure water?
S5.35c
We would expect more water molecules in the vapor above the pure water.
Q5.36
4.0 moles of a compound was added to 500 g of HCl. What was the freezing point of the pure solvent if the freezing point of the solution is 188.6 K. (Kf=$.55\dfrac{K\cdot kg}{mol})$
S5.36
$T_f= 188.6K\ m= \dfrac{4mol}{.500kg}= 8\dfrac{mol}{kg}$
$\bigtriangleup T= -(K_f\cdot m) \rightarrow T_i=T_f+[(k_f)(m)]= 188.6K+(.55\dfrac{K\cdot kg}{mol})(8\dfrac{mol}{kg}= 193K$
Q5.36a
After dissolving a substance weighing 0.246 grams in 100.0 g of water, the freezing point of the solution went down 1K below that of the pure solvent. What would be the molar mass of this substance?
S5.36a
The following information is needed.
$\triangle T=K_fm_2 \ K_{fH_2O} = 1.86$
Now solve for the molar mass of the solute.
$1 \ K = 1.86 m_2 \ m_2 = 0.538 \ m_2 = mol \div kg \ of \ solvent \ mol_2 = 0.538 \ mol \ kg^{-1} \cdot 0.1 \ kg \ of \ water \ mol_2 = 0.0538 \ moles \ of \ substance \ Molar \ Mass = 0.246 \ grams \ of \ substance \ \div \ 0.0538 \ moles \ of \ substance \ \underline{Molar \ Mass = 4.572 \ g/mol}$
Click here for more information on freezing point depression.
Q5.36b
5.525 grams of toluene is mixed with 20.0 grams of acetone. Calculate the freezing point of acetone. The freezing point of the solution is found to be -102.6o C. The Kf of acetone is 2.40 oC kg/mol, and the molar mass of toluene is 92.14 g/mol
S5.36b
We can relate the molality of the solution to the freezing point depression:
$m=\dfrac{\Delta T}{K_{f}}$
Rearranged:
$\Delta T=K_{f}m$
$T_{acetone}-T_{solution}=K_{f}m$
$T_{acetone}=T_{solution}+K_{f}\dfrac{mols \,of \,toluene}{kg\,of\, solvent}$
$=-102.6^{\circ}C+(2.40\dfrac{^{\circ}C}{kg\cdot mol})\dfrac{(5.252g\times \dfrac{1 mol}{92.14g})}{20.0\times 10^{-3}kg}$
$=-95.8^{\circ}C$
Q5.37a
Two aqueous urea solutions have osmotic pressures of 1.4 atm and 3.0 atm, respectively, at a certain temperature. What is the osmotic pressure of a solution prepared by mixing equal volumes of these two solutions at the same temperature?
Q5.37b
A lab technician has 2 solutions of ammonia. Solution A exerts an osmotic pressure of 8 atm. When it is mixed with solution B in equal volume to produce solution C, it is determined that solution C exerts an osmotic pressure of 6.5 atm. What is the osmotic pressure of solution B?
Q5.38
A research scientist used 50 mL of sterilized water to conduct experiments and obtained odd results. She suspected that it had been contaminated with another compound and determined that the boiling point of the solution had increased to 105oC. How many moles of the compound (assume it dissolves as a molecule) had been added to the water?
Q5.39
Time-released pills are designed to have longer effect at a constant rate. The active ingredient is concealed by insoluble substances so that the active ingredient has to be dissolved by getting through the holes of insoluble layers covering outside. Explain how the insoluble layers regulate the absorption of the active ingredient.
Q5.41
Acetic Acid can make hydrogen bonds with water molecules and also with benzene. A solution of 4.0 g of acetic acid in 95g $C_6H_6$ has a freezing point of $3.5^oC$. What is the molar mass of the solute?
S5.41
$\Delta T=K_fm$
$m=molality$
$3.5^oC=(5.12^oC\,kg\,mol^{-1})(m)$
$m=.68359mol\,kg^{-1}$
$m=\dfrac{moles\,solvent}{kg\,solvent}$
$moles\,solvent=.71757moles$
$mol\,acetic\,acid=\dfrac{g\,acetic\,acid}{molar\,mass,acetic\,acid}$
$7.1957\,moles=\dfrac{4.0g\,acetic\,acid}{.555887g\,mol^{-1}}$
Q5.43
Determine on whether each of the following statements is true or false and briefly explain your answers. (a) Colligative properties depends on both the number of solute molecules and the size of the solute molecules. (b) The addition of a nonvolatile solute to a solvent only changes the chemical potential of liquid $(\mu _1(l))$. (c) The mole fraction of a component is never the same as the activity of the same component.
S5.43
1. False; the definition of colligative properties are properties that ONLY depend on the number of solute molecules.
2. True; $\mu_1(s)$ and $\mu_1(g)$ are not affected because solute is nonvolatile and insoluble to the solid solvent. (c) False; in the case of ideal solutions, the activity coefficient $(\gamma _i)$ is 1 and $a_i =(1)*x_i$.
Q5.45
Provide an example of a nonideal solution explain why it is nonideal.(Hint: Look over the concept of nonideal solutions)
S5.45
An example of a non-ideal solution is a mixture of diethyl ether and water. Diethyl ether is a non-polar solvent used for dissolving organic molecules whereas water is polar. Intermolecular forces between these two components are weak; hence they will form two layers upon mixing; making the mixture a non-ideal solution.
Q5.46a
An unknown substance has a molal boiling-point elevation constant, Kb of 1.23 K/m. Find the molar mass of this substance. The enthalpy of vaporization for this substance is 38.56 kJ/mol at 78.37 oC.
S5.46a
For any substance:
$K_{b}=\dfrac{RT_{0}^{2}\mathfrak{M}_{1}}{\Delta _{vap}\bar{H}}$
Rearranging this equation, we get:
$\mathfrak{M}_{1}=\dfrac{K_{b}\Delta _{vap}\bar{H}}{RT_{0}^{2}}$
$=\dfrac{(1.23\dfrac{K}{mol})(38.56\times 10^{3}\dfrac{J}{mol})}{(8.3145\dfrac{J}{K\cdot mol})(78.37+276.15)^{2}K}$
$=0.04616 \dfrac{kg}{mol}=46.16\dfrac{g}{mol}$
Q5.46b
Find Kb (the molal boiling-point elevation constant) for acetone solution at 56ºC, given that:
$\triangle _{vap}\overline{H}^{\circ}_{Acetone} = 31.3 \ kJ \ mol^{-1}$
S5.46b
You need to use the following equation:
$K_b = \dfrac{RT_{b}^{2}M_{molar}}{\triangle_{vap}\overline{H}}$
Plug in values and solve for the Kb.
\begin{align} &K_b = \dfrac{(\dfrac{8.3145}{1000} \ kJ \ mol^{-1} \ K^{-1})(298K)^{2}(\dfrac{58.08}{1000}\ kg \ mol^{-1})}{31.3 \ kJ \ mol^{-1}} \ & \underline{K_b = 1.37 K \ mol^{-1} \ kg} \end{align}
Click here for more information on boiling point elevation.
Q5.47a
A home gardener and amateur scientist decides to prepare her olives by brining them. She begins the process by soaking the olives in fresh water. During this time she notices that the olives absorb moisture and swell. She then soaks the olives in a strong salt brine. During this time, she notices that the olives lose moisture and shrink. To what physical process can she attribute the shrinking and swelling? Why do the olives absorb the fresh water and lose moisture to the brine?
S5.47a
The movement of water in and out of the olives is due to diffusion. The high concentration of solutes in the brine relative to the concentration in the olives causes water to diffuse from olives into the brine, and the low concentration of solutes in the fresh water surrounding the olives causes the fresh water to diffuse into the olives, which have a comparatively high concentration of solutes.
Q5.47b
Describe the difference between a hypertonic, and hypotonic. Why does a cell shrink in a hypertonic solution, and expand in a hypotonic solution.
Q5.48
The following data give the pressures for water-methanol solution at 39.9oC. Find the activity coefficients of both components based on:
1. Raoult's law
2. Henry's law
xwater
0
0.0490
0.3120
0.4750
0.6535
0.7904
Pwater/torr
0
4.51
23.03
31.53
39.78
44.81
Pmethanol/torr
255.6
253.39
188.27
152.87
116.22
80.89
Q5.49
A solution is made up of some amount of sucrose (C12H22O11) and 1.03 kg water. Calculate the amount of sucrose (in grams) in solution if the solution freezes at -2ºC and the activity coefficient of sucrose is 0.789.
S5.49
$\Delta T = K_{b}m$
$m = \dfrac{\Delta T}{K_{b}} = \dfrac{2K}{1.86\ K\ mol^{-1}\ kg}$
$m = 1.075 mol kg^{-1}$
a is effective molar concentration so a = m in:
$a = \gamma x$
x is the actual amount of sucrose in solution, so
$x = \dfrac{a}{\gamma} = \dfrac{1.075}{0.789}$
$x = 1.36 mol \ kg^{-1}$
$mass\ sucrose = (\dfrac{1.36 mol}{kg\ H_{2}O})(1.03 kg\ H_{2}O)(\dfrac{342.3 g}{mol})$
$mass\ sucrose = 479 g$
5.8: Ionic Activity
Q5.54
You know that the mean activity coefficient is 0.42. From this information find what the ionic strength of the compound LiNO2 is.
I=0.548
Q5.56
A 0.15 m Ca(NO3)2 solution has a mean ionic activity coefficient of 0.17 at 25°C. Calculate the mean molality, the mean ionic activity, and the activity of the compound.
Q5.58
In 0.03m aqueous NaNO3 solution at temperature 300K., the size of the ionic atmosphere is 1/k, also known as Debye radius, is 27.8x10-10 m. Using the Debye radius formula to calculate the ionic strength I.
Q5.60
For a 0.0030m aqueous solution of $CaCl_2$ at 298K, calculate the ionic strength. (a) Calculate the activity coefficients of $Ca^{2+}$ and $Cl^-$ ions in the solution and (b) calculate the mean ionic activity coefficients of these ions. (Hint: Use the Debye-Huckel limiting law)
S5.60
$I=\dfrac{1}{2}\Sigma _im_iz^2_i$
$=\dfrac{1}{2}[(0.0030m)(2)^{2}+(0.0060m)(-1)^{2}]$
$=-.009m$
(a)
$log\gamma_i=-0.509z^2_i\sqrt{I}$
For $Ca^{2+}$,
$log\gamma_+=-0.5092^2\sqrt{0.009}$
$log\gamma_+=-0.193$
$\gamma_+=.640985$
For $Cl^{-}$,
$log\gamma_-=-0.509(-1)^2\sqrt{.009}$
$log\gamma_-=-0.048288$
$\gamma_-=.89477$
(b)
$\gamma _{\pm}=(\gamma _+^{v_{-}}\gamma _{-}^{v_{-}})^{\dfrac{1}{v}}=[(.640985)^{1}(.89477)^2]^{\dfrac{1}{3}}=.8006$
Q5.65
For an ideal mixture, what mole fraction of each gas maximizes the entropy of mixing:
$\Delta _{mix}S=-nR(x_{1}lnx_{1}+x_{2}lnx_{2})$
Hint: Rewrite the above function in terms of only either x1 or x2, then derive.
S5.65
If we take the partial derivative of the entropy of mixing:
$\Delta _{mix}S=-nR(x_{1}lnx_{1}+x_{2}lnx_{2})$
$\left [\dfrac{\delta \Delta _{mix}S}{\delta x_{1}} \right ]_{_{x_2}, n, R} =\left [-\dfrac{\delta }{\delta x_{1}}nR(x_{1}lnx_{1}+x_{2}lnx_{2}) \right ]_{x_{2}, n, R}$
If we rewrite the function in terms of x1
$=\left [-\dfrac{\delta }{\delta x_{1}}nR(x_{1}lnx_{1}+(1-x_{1})ln(1-x_{1}) \right ]_{x_{2}, n, R}$
and derive using the chain rule:
$=-\left (x_{1}\dfrac{1}{x_{1}}+lnx_{1}+(1-x_{1})\left (-\dfrac{1}{1-x_{1}} \right )+ln(1-x_{1})(-1) \right )$
$=1-lnx_{1}+1+ln(1-x_{1})$
$=-lnx_{1}+ln(1-x_{1})$
$=-ln\dfrac{x_{1}}{1-x_{1}}$
To find the local maximum, we set this to 0 and solve for x1
$0=-ln\dfrac{x_{1}}{1-x_{1}}$
$e^{0}=\dfrac{x_{1}}{1-x_{1}}$
$1=\dfrac{x_{1}}{1-x_{1}}$
$1-x_{1}=x_{1}$
$x_{1}=0.5$
To check that this is a maximum and not a minimum, we plug in values to the left and right of the given x1 value (ex. 0.4 and 0.6). We then find that at x1<0.5 the derivative, or slope, is positive and at x1>0.5m the slope is negative. This is further confirmed by the following diagram, where is maximized at mole fraction 0.5.
Q5.64
The osmotic pressure arbitrary non-ideal diluted solution is measured at 2 different concentrations at 300K. At concentration of 5.20 g/L, osmotic pressure is 10.20 x 10-3 atm and at concentration of 8.90 g/L the osmotic pressure is 2.30 x 10-2 atm. Use this information to estimate the molar mass of the solute. (Hint: For dilute solutions, only the second virial coefficient is concerned. Look at equation for osmotic pressure and make a graph of π/c vs. c and look for the y-intercept of graph.)
S5.64
π (atm)
1.06 x 10-5 2.30 x 10-5
c (g/L) 5.20 8.90
π/c ( atm•L/g)
2.038 x 10-6 2.584 x 10-6
slope of the graph of π/c vs. c is generated as:
$slope = \dfrac{(2.584 \times 10^{-6} - 2.038 \times 10^{-6}) \dfrac{atm \cdot L}{g}}{(8.90 - 5.20) \dfrac{g}{L}} = 1.476 \times 10^{-7} \dfrac{atm \cdot L^2}{g^2}$
$\dfrac{\pi}{c} =\dfrac{RT}{M}(1+Bc) = \dfrac{RT}{M} + (\dfrac{RTB}{M})c$
$slope = \dfrac{RBT}{M} = 1.476 \times 10^{-7} \dfrac{atm \cdot L^{2}}{g^{2}}$
Plug back into equation to solve for "y-intercept"
$2.584 \times 10^{-6} = \dfrac{RT}{M} + ( 1.476 \times 10^{-7})(8.90)$
$\Rightarrow \dfrac{RT}{M} = 1.271 \times 10^{-6} \dfrac{atm \cdot L}{g}$
$M = RT (1.271 \times 10^{-6} \dfrac{atm \cdot L}{g}) = (0.08206 \dfrac{L \cdot}{K \cdot mol})(300K)(1.271 \times 10^{-6} \dfrac{atm \cdot L}{g}) = {\color{red}3.13 \times 10^{-5} \dfrac{g}{mol}}$
Q5.65
Assume that you have made an ideal solution of an unknown compound made of two components. Why does the mole fraction 0.5 achieve the maximum $Δ_{mix}S$?
S5.65
Use the equation.
$\Delta_{mix} S = -nR ( x_1 \ln x_1+x_2 \ln x_2 )$
Find the partial derivative since we want to find out the max value for this equation for x1 . We find x1 since it will show us x2.
\begin{align} [\dfrac{\partial }{\partial x_1} \Delta_{mix} S]_{x_2,n,R} &= [\dfrac{\partial }{\partial x_1}-nR \left( x_1 \ln x_1+x_2 \ln x_2 \right) ]_{x_2,n,R} \ since & \ x_1+x_2 =1 \ &= - [\dfrac{\partial }{\partial x_1}nR \left( x_1 \ln x_1+(1-x_1) \ln (1-x_1) \right) ]_{x_2,n,R} \ derive & \ &= -\left [(x_1 \cdot \dfrac{1}{x_1})+(\ln x_1)+ \left((1-x_1) \cdot \dfrac{-1}{1-x_1} \right ) + \left( (-1) \cdot \ln (1-x_1) \right ) \right] \ &= -1- \ln x_1+1+ \ln (1-x_1) \ &= \ln (1-x_1) - \ln x_1 \ &= \ln \dfrac{1-x_1}{x_1} \end{align}
Now that we have the partial derivative, we set it to 0 and solve to attempt to find the maximum value.
\begin{align} & 0 = \ln \dfrac{1-x_1}{x_1} \ &e^{0} = e^{\ln \dfrac{1-x_1}{x_1}} \ &1 =\dfrac{1-x_1}{x_1} \ &x_1 = 1-x_1 \ &2x_1 = 1 \ &\underline{x_1 = 0.5} \end{align}
Now we know that x1 = 0.5 is either a min or max. If we plug in a number smaller then 0.5 and bigger than 0.5 into the partial derivative we found earlier, we can conclude that the slope of the equation is positive to the left of 0.5 and negative to the right of the equation. This is characteristic of maximum values. Thus x1 = 0.5 and x2 = 0.5 attain the maximum values of ΔmixS since
x1 = 0.5 is a max.
See here for more details about ΔmixS. See figure (11) on that page for more details about the max.
Q5.66
Producing a certain gas mixture involves mixing high-pressure nitrogen and low-pressure hydrogen together. In one such production, 3.22 moles of N2 at 4.2 atm are mixed with 1.25 moles of H2 at 0.8 atm. Determine the change in Gibbs’ free energy.
5.9: Colligative Properties of Electrolyte Solutions
Q5.72
You are trying to produce a passive reverse osmosis system to obtain pure water from the ocean by using a semi-permeable membrane on one end of the intake pipe. At what depth would the bottom of the pipe need to be in order for pure water to begin to fill the pipe? Assume the ocean temperature is 17ºC, and the seawater is a 0.70 M NaCl solution. The density of seawater is 1.03 g/cm3.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/05%3A_Solutions/5.E%3A_Solutions_%28Exercises%29.txt
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Q6.1
Imagine a catalytic reaction that changes a reagent between two states, i.e.:
$yY + cC \rightleftharpoons cC + zZ$
The equilibrium constants $K_c$, $K_{\chi}$, and $K_p$ with respect to concentration, mole fraction, and pressure are given by
$K_c = \dfrac{[Z]^z}{[Y]^y}$
$K_{\chi} = \dfrac{\chi_Z^z}{\chi_Y^y}$
$K_p = \dfrac{P_Z^z}{P_Y^y}$
Given this information, express $K_p$ in terms of $K_c$ and $K_{\chi}$.
Q6.2a
$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(s)}$
1. The partial pressure of CO2 is 0.1 atm, calculate $K_p$
2. If the reaction occurred in a 1L flask, what is the amount of CaO formed when 0.12 moles CaCO3 was set to react?
3. What is the minimum about of CaO needed to cause the formation of CaCO3?
Q6.2b
The decomposition of sodium chlorate is:
$2NaClO_{3(s)} \rightleftharpoons 2NaCl_{(s)} + 3O_{2(g)}$
Suppose 0.760 mole of sodium chlorate was to be decomposed in a 3.75 L container. At the temperature 336°C, the percent decomposition of NaClO3 is 3.5%. Find the
1. pressure of $O_2$ gas in the container in atm and
2. equilibrium constant $K_p$ of the reaction. Assume $O_2$ gas is an ideal gas.
Hint: Use partial pressure and look at chemical equilibrium of gaseous systems
S6.2b
a) number of moles of NaClO3 decomposed is:
$\dfrac{ \# moles \, decomposed}{.760 moles} = 0.035 \Rightarrow \# mole\,decomposed = 0.0266 mole \,NaClO_{3}$
number of moles of O2 formed is:
$0.0266 mol \,NaClO_{3} \times \dfrac{3 mol O_{3}}{2 mol \,NaClO_{3}} = 0.0399 mol \,O_{2}$
$P_{O_{2}} V = nRT \Rightarrow P_{O_{2}} = \dfrac{nRT}{V}$
$P_{O_{2}} = \dfrac{(0.0399 mol)(0.08206 \dfrac{L \cdot K}{K \cdot mol})(336+273)K}{3.75 L} = {\color{red} 0.532 atm}$
b) $K_{p} = \dfrac{P_{O_{2}}}{P^{\circ}} = \dfrac{0.532 atm \times \dfrac{1.013 bar}{1atm}}{1bar} = {\color{red} 0.539 }$
Q6.2c
Using the decomposition reaction of hydrogen peroxide (H2O2), assume that at 1048 °C and the pressure of the oxygen gas (O2) is 1.5 bar.
$\ce{2H2O2 (aq) \rightleftharpoons 2H_2O (l) + O2 (g) }$
1. Determine the $K_P$ for the reaction above.
2. If 0.92 mole of $H_2O_2$ is placed in a 3.0-L beaker at 1048°C, determine the fraction of $H_2O_2$ that will decompose.
3. Determine the same as (b) if 1.3 moles of $H_2O_2$ were placed inside the beaker instead.
4. What is the minimum amount of $H_2O_2$ (in moles) necessary in order to reach equilibrium?
S6.2c
a.
b. In order to calculate the number of moles of H2O2 that are decomposed, you first need to determine the moles of O2 formed by the reaction. Treat O2 as an ideal gas.
c. If 1.3 moles of H2O2 were used instead of 0.92 mole, the pressure of the O2 would not be affected and would remain at 1.5 bar. The number of moles of H2O2 decomposed would still be 0.0802 mole, therefore, the fraction of H2O2 decomposed would be as follows:
d. The pressure of O2 must be greater than or equal to 1.5 bar in order for equilibrium to take place, therefore, the number of moles of H2O2 cannot be less than 0.0802 mol.
Q6.3
Find the value of Kp for cellular respiration if $P_{O_{2}}=350torr$, with a 3:2 ratio to the pressure of CO2
S6.3
$C_6H_{12}O_6(s)+O_2(g)\rightleftharpoons CO_2(g)+H_2O(l)$
$P_{CO_{2}}=(350torr)\dfrac{2}{3}=233.3torr$
$K_p=\dfrac{\dfrac{233.3torr}{750torr}}{\dfrac{350torr}{750torr}}=0.666$
Q6.5
For the reaction of
$\ce{N2O4 → 2 NO2}$
with Kp = 0.167 at 300 K. If 1.0 g of N2O4 is placed into a 250.0 mL container:
1. What would be its pressure if none dissociates?
2. What is its partial pressure at equilibrium (with dissociation)?
3. What is the total pressure in the container?
Q6.9a
Calculate $\Delta_{vap}H$ for the evaporation of methanol when the temperature is raised from 20 °C to 100 °C and if the $K_2/K_1$ ratio is 22.14?
$CH_3OH_{(l)} \rightarrow CH_3OH_{(g)}$
Q6.9b
Using the decomposition reaction of coppr (II) oxide below:
$4CuO_{(s)} \rightleftharpoons 2Cu_2O_{(s)} + O_{2(g)}$
determine the standard enthalpy of the reaction. (Note: The equilibrium vapor pressures of O2 are 15.4 mmHg at 600°C and 927 mmHg at 850°C).
S6.9b
In this case, we must use the van't Hoff equation:
$\ln \dfrac{K_2}{K_1} = \dfrac{\Delta_ H^o}{R} \left( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right)$
in order to solve this particular problem. We can conclude that is proportional to due to the defined relationship of:
.
Therefore,
Q6.10
A container of water at 20ºC was placed in a freezer that was at a temperature of -5.0ºC. The vapor pressure of water in the container went from 0.60 bar to 0.38 bar. Calculate the enthalpy of fusion for the reaction that took place.
S6.10
• $k_2=0.38 \;bar$
• $k_1=0.60\; bar$
• $T_2= 268\;K$
• $T_1=293\;K$
$\ln\dfrac{k_2}{k_1}= \dfrac{\Delta_r H^\circ }{R}\dfrac{T_2-T_1}{T_2T_1}$
$\rightarrow \Delta_r H^\circ = \ln\dfrac{k_2}{k_1}(R)\dfrac{T_2T_1}{T_2-T_1}$
$\ln\dfrac{0.38 bar}{0.60 bar} 8.314 \dfrac{J}{mol\cdot K} \dfrac{(293\; K)(268\; K)}{268\; K-293\; K}(\dfrac{1\;kJ}{1000\; J})= 11.9 \dfrac{kJ}{mol}$
Q6.11
The chemical responsible for the brown air throughout the Los Angeles area is NO2(g). To learn more about NO2(g), you decide to study this pollutant spectroscopically (by light absorption). You fill a gas cell with N2O, equilibrate the temperature to 298.0 K and then open the stopcock on the cell to equilibrate the pressure to the barometric pressure (723.4 mm Hg) that day. You then reequilibrate the cell at 323.1 K, 348.0 K, and 372.9 K. The following data is obtained:
T/K 298.0 323.1 348.0 372.9
[NO2]/M 0.01262 0.02140 0.02756 0.02920
Find KP, Δ rxn G, Δ rxn H, and Δ rxn S at each temperature for the reaction
$\ce{N2O4(g) ⇌ 2 NO2(g)}.$
Q6.14
What is the equilibrium constant and standard Gibbs energy change for:
$CO_{(g)} \rightleftharpoons CO_{(g)} + ½O_{2(g)}$
Assume that when $CO_{(g)}$ dissociates into $CO_{(g)}$ and $O_{2(g)}$ at 300 K and 1.5 atm, the overall dissociation is one two fifths complete.
Q6.15
The standard Gibbs energies of formation of proban-1-ol and proban-2-ol are -171.3 kJ mol-1 and -180.3 kJ mol-1, respectively. Find the ratio of equilibrium vapor pressures of each isomer at 300 K.
S6.15
The ratio of equilibrium pressure is : 171.3/180.3 = 0.95
Q6.16
At what temperature does a particular reaction favor the formation of products at equilibrium if ΔrHº = 215.7 kJ mol-1 and ΔrSº = 348.8 J mol-1 K-1?
S6.16
The reaction favors the formation of products at equilibrium when
$\Delta_{r} G^{\circ} = \Delta_{r} H^{\circ}-T\Delta_{r} S^{\circ}<0$
$\Delta_{r} H^{\circ}-T\Delta_{r} S^{\circ} = 215.7 \ast 10^{3} J\ mol^{-1} - T(348.8 J\ mol^{-1}\ K^{-1})<0$
$T>\dfrac{215.7 \ast 10^{3} J\ mol^{-1}}{348.8 J\ mol^{-1}\ K^{-1}}$
$T > 616 K$
Q6.17
Using a table of thermodynamic data, find Ksp at 298.15K for:
$CO_{(g)} \rightarrow CO_{(g)} + ½O_{2(g)}$
(Hint, this problem can be solved using the van’t Hoff equation in its integrated form)
Q6.18
Consider the the Haber process:
$N_{2(g)} + 3H_{2 (g)} \rightleftharpoons 2NH_{3 (g)}$
How does the equilibrium shift under the following changes
1. the volume of nitrogen gas is increased
2. the temperature is decreased
3. the pressure of the system is decreased
S6.18
1. The reaction shifts to the right : more products formed.
2. The reaction shifts to the right : favor exothermic reaction
3. The reaction shifts to the left : more pressure applied.
Q6.19
Given the general reaction A(g) ⇔2B(g), calculate the degree of dissociation of A at 25ºC and 7.00 bar if ΔrGº = 6.76 kJ mol-1. According to Le Chatelier's Principal, in what direction should this reaction proceed?
S6.19
$\Delta_{r} G^{\circ} = -RTlnK_{P}$
$K_{P} = e^{\dfrac{\Delta_{r} G^{\circ}}{-RT}} = e^{\dfrac{6.76\ast 10^{3}J\ mol^{-1}}{-(8.314J\ mol^{-1}\ K^{-1})298K}}$
$K_{P} = 0.0653$
$K_{P} = \dfrac{4\alpha ^{2}}{1-\alpha ^{2}}P$
$\dfrac{4\alpha ^{2}}{1-\alpha ^{2}} = \dfrac{0.0653}{7}$
$\alpha = .048$
Degree of dissociation α = 0.048, the reaction proceeds 4.8%. Le Chatelier's Principal says that reactions will move towards the side with fewer moles of gas at high pressures.
Q6.19
How does Le Chatelier's Principle relate to the following equation?
S6.19
Le Chatelier's Principle basically states that a system will adjust itself in efforts to re-establish equilibrium when an outside stress is placed on it. Using the equation provided, we can conclude that raising the temperature causes the equilibrium to shift from left to right in an endothermic reaction. This suggests that it is favoring the formation of products. We can also conclude that the opposite is true. This conclusion reinforces Le Chatelier's Principle because the temperature acts as the external stress placed on the system in this case.
Q6.20
H2 and H molecules are at equilibrium pressures of .35 bar and .30 bar, respectively. If the size of the container they are in is reduced by a factor of two, what will be the new partial pressures?
S6.20
$H_2(g)\rightleftharpoons 2H(g)$
$K_p=\dfrac{P_H^2}{P_{H_{2}}}=\dfrac{.30^2}{.35}=.675$
At new volume: $P_H=.60$
$P_{H_{2}}=.70$
Pressures will increase with decreased volume. Less molecules of gas will be favored.
$P_{H}$ $P_{H_2}$
0.60 bar - 2x 0.70 bar + x
$.675=\dfrac{.60-(2x^2)}{.70+x}\rightarrow 4x^2 +.675x+.113\rightarrow x=.272$
$P_{H_{2}}=.70bar+.272bar=.972bar$
$P_H=.60bar-(2).272bar=.056bar$
Q6.21
With Le Chatlier's principle in mind, does raising the temperature favor the forward reaction or the reverse?
S6.21
This reaction favors the reverse reaction.
Q6.23a
When a gas was heated at atmospheric pressure and 25°C, its color deepened. Heating above 150°C caused the color to fade, and at 550°C the color was barely detectable. At 550°C, however, the color was partially restored by increasing the pressure of the system. Which of the following scenarios best fits the above description? Justify your answer.
1. A mixture of Gas A and Gas B
2. Pure Gas B
3. A mixture of Gas X and Gas Y
(Hint: Gas B is bluish, and Gas Y is yellow. The other gases are colorless. Gas A and B are in their natural state, Gas AB has a ΔfH° = -43.2 kJ / mol. The reaction of Gas X into Gas Y is endothermic.)
Q6.23b
Hydrogen gas and iodine react at equilibrium in a glass canister to form Hydrogen Iodide, a strong acid:
$H_{2(g)}+I_{2(g)}\rightleftharpoons HI_{(g)}$
Iodine gas is a deep purple color. Both Hydrogen Iodide and hydrogen gas are colorless. Assume that iodine sublimates readily at 37o C, and that the reaction is endothermic in the direction written. What color is the gas mixture in the canister in the following scenarios? Provide an explanation for each.
1. The canister is heated to 40o C
2. The canister is heated to 500o C
3. Negative pressure is applied to the cannister at 500o C
Hint: At room temperature, the gas is colorless. Solid iodine must be in gas phase to react.
S6.23b
a.) The color of the gas will be a deep purple. The iodine will have sublimated, but the temperature is not high enough to drive the reaction forward.
b.) The gas will be colorless, or almost completely so. This is because an increase in the temperature in an endothermic reaction drives the equilibrium constant higher. This can be justified using Le Chatelier's principle, which states that added stress to an equilibrium will be offset by the system.
$H_{2(g)}+I_{2(g)}+heat\rightleftharpoons HI_{(g)}$
Thus, when heat is added, the system will compensate by driving the reactants of the reaction forward into products. Another way to assess this is by the altered form of the van't Hoff equation:
$lnK=-\dfrac{\Delta _{r}H^{\circ}}{RT}+\dfrac{\Delta _{r}S^{\circ}}{R}$
Assuming that neither change in entropy nor enthalpy changes due to change in temperature, and that in an endothermic reaction the enthalpy is positive, as the temperature increases, so does K.
c.) The gas will become slightly purple. This is because a decrease in pressure will alter K and drive the reaction in the direction which produces more moles of gas. By this, Le Chatelier's principle once again holds.
Q6.24
$Mg + Pb^{2+} \rightarrow Mg^{2+}+Pb$
Write the two separate reactions happening here. Label which one is happening at the cathode end and which is happening at the anode end.
S6.24
Cathode:
$Pb^{2+}+2e^- \rightarrow Pb$
Anode:
$Mg \rightarrow Mg^{2+} + 2e^-$
Q6.24
The reaction of N2 (g) and H2 (g) gas produces NH3 (g). This reaction is exothermic explain what happens when you increase the temperature of the reaction. What happens when you increase the pressure?
S6.24
Since this reaction is exothermic heat is produced. Using Le Chatelier's principle we see that an increase in the temperature of the reaction will drive the reaction backward since heat is already in the product side. If we increase pressure we decrease the volume of the reaction therefore by Le Chatelier's principle an increase in the pressure should drive the reaction forward.
Q6.25
At places such as high mountain, the air pressure is lower than 1 atm, resulting lower partial pressure of Oxygen. What would you expect for the concentration of hemoglobin for the people living at such places?
Q6.26
Show the steps to get from
$Y = \dfrac{[PL]}{[L] + [P]}$
to
$\dfrac{1}{Y} = 1 + \dfrac{K_d}{L}$
Q6.27
After an experiment of protein-binding you find the data respectively
Total $\dfrac{Mg^{2+}}{\mu M}$: 60 120 180 240 480
$\dfrac{Mg^{2+} bound to protein}{\mu M}$: 33.8 120 180 240 480
Determine the dissociation constant of $Ca^{2+}$ graphically. The protein concentration was kept at 96 $\mu M$ for each run.
Solution:
$54.2\mu M\,pH=\,pKa$
$54.2\mu M(\dfrac{1\times10^{6}M}{1\mu M})=5.42\times10^{-5}$
$Ka=10^{-5.42E-5}$
$Ka=.9998$
Q6.28
The dissociation constant for the following reaction is 3.2x10-4. Dissolve 0.02M of C9H8O4 in water, calculate the molarity of reactants at equilibrium.
C9H8O4 + H2O <---> H3O+ + C9H7O4-
Q6.29
The reaction
\begin{align} &Glyceraldehyde \ 3-phosphate + NAD^{+} + HPO_{4}^{2-} \rightarrow 1,3-Biphosphoglycerate + NADH + H^{+} \ & \triangle_rG^{\circ'}=6.3 \ kJ \ mol^{-1} \end{align}
is catalyzed by GAPDH (Glyceraldehyde 3-phosphate Dehydrogenase). At 298 K, predict whether or not the reaction will be spontaneous with the following information:
[G3P] = 1.5x10-5 M; [BPG] = 3.0x10-3 M
[NAD+] = 1.2x10-5 M; [NADH]=1.0x10-4 M
[HPO42-]= 1.2x10-5 M; pH = 7.5
S6.29
Use the following equation.
Calculate [H+]
$[H^{+}] = 10^{-7.5} = 3.162 X 10^{-8} M \ since \ pH=-log([H^{+}])$
Plug values into equation and solve. Note: Since there is no coefficient in front of H+ in the reaction, x=1.
$\triangle_rG = 6.3 \ kJ mol^{-1} + (\dfrac{8.3145}{1000}kJ/mol^{-1} K^{-1}) (298K) \ln\dfrac{(3 \cdot 10^{-3} \ M)(3.162 \cdot 10^{-8} \ M/(1 \cdot 10^{-7})]^{1} \left(1.0 \cdot 10^{-4} \ M \right )}{ \left(1.5 \cdot 10^{-5} \ M \right ) \left(1.2 \cdot 10^{-5} \ M \right ) \left(1.2 \cdot 10^{-5} \ M \right )}$
$\underline {\triangle_rG = 49.90 \ kJ \ mol^{-1}}$
Click here for some more information on bioenergetics and free energy.
Q6.29
The reaction: $Glucose+ATP\rightleftharpoons Glucose 6-phosphate + ADP$ At 298K, the equilibrium constant for the reaction is $3.7 \times 10^{-3}$.Will the reaction occur spontaneously if the reaction is at the following concentrations:
[Glucose]=$3.2 \times 10^{-4}M$, [ATP]=$2.5 \times 10^{-3}M$, [G-6-P]=$1.2 \times 10^{-5}M$, [ADP]=$1.0 \times 10^{-5}M$.
S6.29
\begin{align*}\Delta G^\circ &=-RTln(K_{eq})=-(8.314J/K*mol)(298K)ln(3.7 \times 10^{-3})\&=13872.97J/mol=13.87297kJ/mol\end{align*}
\begin{align*} \Delta G &=\Delta G^\circ +RTln(\dfrac{[Product]}{[Reactant]})\&=13.87297kJ/mol + (8.314 \times 10^{-3}kJ/K*mol)(298K)ln(\dfrac{[1.2 \times 10^{-5}][1.0 \times 10^{-5}]}{[3.2 \times 10^{-4}][2.5 \times 10^{-3}]})\&=-7.942kJ/mol \end{align*}
$\Delta G$ is negative, so the reaction is spontaneous at the given concentrations.
Q6.30
The established standard Gibbs energy for hydrolysis of ATP to ADP at 310K is $-30.5kj\,mol^{-1}$. At $-4.6^oC$, determine the $\Delta rG^{o'}$ in the process of the muscle of a hippo. (Hint: $\Delta rH^{o'}=-20.1kjmol^{-1}$
S6.30
Step 1:
$\Delta rG^{o'}=\Delta rH^{o'}-T\Delta rS^{o'}$
Step 2:
$\Delta rS^{o'}=\dfrac{\Delta rH^{o'}-\Delta rG^{o'}}{T} =\dfrac{(-20.1kj\,mol^{-1})-(-3.0kj\,mol^{-1})}{310K}=3.355\times10^{-2}kj\,K^{-1}\,mol^{-1}$
Step 3:
$\Delta rG^{o'}=\Delta rH^{o'}-T\Delta rS^{o'}$
$=(-20.1kj\,mol^{-1})-(298.15K)(3.355\times10^{-2}kj\,K^{-1}\,mol^{-1})$
$=-29.1kj\,mol^{-1}$
Q6.31a
Which step of glycolysis would not occur spontaneously at standard-state conditions and why?
Q6.31b
Consider a hydrolysis of PEP, a phosphate compound.
PEP + H2O → pyruvate + Pi ΔrG°' = -61.9 kJ/mol
At the temperature of 288K, the following reaction took place and has a reaction Gibbs free energy of -49.5 kJ/mole. Find the concentration of PEP for the reaction if the other concentrations are: [Pi] = 3.54 x 10-1 M, [Pyruvate] = 1.85 x 10-2 M.
Hint: Use equation ΔrG° = -RTlnKp
S6.31b
$\Delta _{r} G = \Delta _{r} G^{\circ '} + RTln\dfrac{[pyruvate][P_i]}{[PEP]}$
$-49500 \dfrac{J}{mol} = -61900 \dfrac{J}{mol} + (8.314 \dfrac{J}{K \cdot mol})(288K)ln\dfrac{(1.85 \times 10^{-2}M)(3.54 \times 10^{-1}M)}{[PEP]}$
${\color{red} [PEP] = 3.69 \times 10^{-5} M}$
Q6.32
For the following reaction: $Fructose 1,6-bisphosphate\rightleftharpoons dihydroxyacetone phosphate + glyceraldehyde 3-phosphate$ The $\Delta G°'=5.7kcal/mol$. Calculate the equilibrium constant and determine if the reaction is spontaneous or not at 310K.
S6.32
$\Delta G°'=5.7kcal/mol=23.8488kJ/mol$
$\Delta G°'=-RTln(k_{eq}')\k_{eq}'=e^{\dfrac{\Delta G°'}{-RT}}=e^{\dfrac{23848.8J/mol}{(-8.314J/K*mol)(310K)}}=9.58 \times 10^{-5}$
Small $k_{eq}'$ suggest that this reaction is not a spontaneous process under the given conditions.
Q6.32a
Consider the following reaction:
Glucose + Fructose → Sucrose + H2O
Find:
1. the value of the standard free energy at 300 K.
2. the ratio between ∆rGo' at 300 K and ∆rGo' at 333 K.
S6.32a
a) The standard free energy = (-1544.3)- (-908.9 + -875.9) = 240.2 kJ/mol
b)delta Go at 300 K = delta G1 + RTln Q = -34.6 kJ/mol
delta GO at 333 K = delta G2 + RTln Q = -26.8 kJ/mol
Q6.32b
Consider the formation of the dipeptide glycylclycine. Using the following information, calculate the ΔrG.
\begin{align} & 2 \ Glycine \rightarrow Glycylglycine + H_2O \ & \triangle _{r}G^{\circ ' }=29.5 \ kJ \ mol^{-1} \ & \left [ Glycine \right ]= 1.4M\ & \left [ Glycylglycine \right ]= 0.7M \ & \left [ H_2O \right ]= 1.0M \end{align}
S6.32b
Use the following equation and plug in values.
\begin{align} & \triangle_rG^{\circ} = \triangle_rG^{\circ '} + RT\ln(K) \ & \triangle_rG^{\circ} = \triangle_rG^{\circ '} + RT\ln\dfrac{[Glycylglycine]\cdot [H_2O]}{[Glycine]^{2}} \ & \triangle_rG^{\circ} = 29.5 \ kJ \ mol^{-1} + (\dfrac{8.3145}{1000} \ kJ \ mol^{-1} \ K^{-1})(298K)\ln\dfrac{0.7M\cdot 1.0M}{1.4^{2} \ M} \ & \underline{\triangle_rG^{\circ} = 26.7 \ kJ \ mol^{-1}} \end{align}
Click here for some more information on bioenergetics and free energy.
Q6.32c
Consider the following reaction:
$2NH_{3(g)}+H_{2}O_{(l)}\rightleftharpoons 2NH^{+}_{4(aq)}$
$\Delta _{f}\bar{G}^{\circ}(NH_{3(g)})=-16.6 \dfrac{KJ}{mol}$
$\Delta _{f}\bar{G}^{\circ}(H_{2}O_{(l)})=-237.2 \dfrac{KJ}{mol}$
$\Delta _{f}\bar{G}^{\circ}(NH^{+}_{4(aq)})=-79.3 \dfrac{KJ}{mol}$
What is the equilibrium constant of this process? Is this process typically spontaneous process under standard conditions?
S6.32c
To calculate change in Gibbs free energy for the equation, we use the standard molar Gibbs energies of formation for the reactants and products:
$\Delta _{r}G = 2\Delta _{f}\bar{G}^{\circ}(NH^{+}_{4(aq)})-\Delta _{f}\bar{G}^{\circ}(H_{2}O_{(l)})-2\Delta _{f}\bar{G}^{\circ}(NH_{3(g)})$
$=2\left(-79.3\dfrac{kJ}{mol}\right)-\left(-267.2\dfrac{kJ}{mol}\right)-2\left(-16.6\dfrac{kJ}{mol}\right)$
$=141.8 \dfrac{kJ}{mol}$
To calculate the equilibrium constant:
$\ln K=-\dfrac{\Delta _{r}G }{RT}=-\dfrac{141800\dfrac{J}{mol}}{(8.3145\dfrac{J}{K\cdot mol})(298K)}=-57.23$
$K=e^{-57.23}=1.40 \times 10^{-25}$
Because the K value is so low for this process, the reaction does not occur spontaneously at 25o C
Q6.33a
Calculate the ΔrGº and equilibrium constant for the following reaction at 298K:
3-Phosphoglycerate → Phosphoenolpyruvate +H2O
Given that:
2-Phosphoglycerate → 3-Phosphoglycerate ΔrGº = -4.2 kJ mol-1
2-Phosphoglycerate → Phosphoenolpyruvate +H2O ΔrGº = -16.4 kJ mol-1
S6.33a
ΔrGº = 4.2 kJ mol-1 - 16.4 kJ mol-1 = 12.2 kJ mol-1
$\Delta _{r}G^{\circ} = -RTlnK$
$lnK = \dfrac{12.2 \ast 10^{3} J\ mol^{-1}}{-(8.314J\ mol^{-1}\ K^{-1})298K}$
$K = e^{.00492}$
$K = 1.00$
Q6.33b
Two acid dissociation reactions for carbonic and acetic acid and their corresponding pKa values are displayed below.
CH3CO2H CH3CO2- + H+ ; pKa = 4.75
2CO3 HCO3- + H+; pKa = 6.37
Assuming the reactions take place at 298K, what is the equilibrium constant and change in gibbs’ free energy associated with the following reaction?
CH3CO2- + H2CO3 HCO3- + CH3CO2H
Q6.34
Suppose the isomerization of DHAP to GAP in glycolysis has an enthalpy of -1.20 kJ/mol. At 25°C the Gibbs free energy of the reaction is 1.98 kJ/mol. Determine the equilibrium constant K of the isomerization at 25°C and at 35°C
Hint: Use van't Hoff equation
S6.34
For K at 25°C
$\Delta _{r} G^{\circ} = -RTln K_{25^{\circ}}$
$1980 \dfrac{J}{mol} = -(8.314 \dfrac{J}{K \cdot})(25+273)K (lnK_{25^{\circ}}) \Rightarrow {\color{red}K_{25^{\circ}} = 0.450}$
For K at 35°C
$ln\dfrac{K_{2}}{K_{1}} = \dfrac{ \Delta _{r} H^{\circ}}{R}(\dfrac{1}{T_1} - \dfrac{1}{T_2})$
$ln\dfrac{K_{35^{\circ}}}{0.45} = \dfrac{-1200 J/mol}{8.314 \dfrac{J}{K \cdot mol}}(\dfrac{1}{(25+273)K} - \dfrac{1}{(35+273)K})$
${\color{red} K_{35^{\circ}} = 0.152 }$
Q6.35a
Steady state and equilibrium state have an important role in understanding enzyme kinetics. What are some significant differences between the two?
S6.35a
In a steady state, there is no net change over time in concentrations of reactants and products of a reaction since they are being produced and consumed at constant rates. In this sense a steady state is a dynamic equilibrium. A steady state can be going in either the forward reaction or the backward reaction.
A chemical equilibrium on the other hand is when a reaction goes in the forward and backward reaction at the same rate so there is no net change in the system.
This is important because cells maintain steady states so that they are able to use particular reactions continuously. If a cell were at chemical equilibrium it would be dead because it would be at the point where all reactions are not going anymore, among other reasons.
Also, it is important to note that in a steady state, reactions are reversible compared to chemical equilibrium where the reaction rate is zero.
Click here to see more information about steady states.
Q6.35b
Is the following an example of a steady state, equilibrium state, or neither:
1. Glycolysis, ingestion, and respiration
2. NaCl_{(s)}\rightleftharpoons Na^{+}_{(aq)}+Cl^{-}_{(aq)}
3. Sodium potassium ATPase
4. Citric Acid Cycle
5. oxidation of gold
S6.35b
1. steady state
2. equilibrium state
3. steady state
4. steady state
5. neither
Q6.36
Diatomic hydrogen gas and diatomic iodine gas are in equilibrium with hydrogen Iodide gas in a closed environment of unknown volume at an unknown temperature.
Find the pressure equilibrium constant if the partial pressures of the gases are as follows: PH2 = 812 mmhg PI2= 587 mmhg PHI = 980 mmhg
Find the activity equilibrium constant if the activity constants of the gases are as follows γH2 = 1.45 γI2=0.844 γHI = 1.23
Use your results to find the thermodynamic equilibrium constant.
S6.36
The reaction in question is H2 + I2 ⇌ 2HI
The pressure equilibrium constant is given by PHI2/(PH2 PI2), the activity equilibrium constant is given by γHI2/(γH2 γI2), and the thermodynamic equilibrium constant is given by the product of those two.
Kp = (980 mmhg)2/(812 mmhg * 587 mmhg) = 2.01
Kγ = 1.232/(1.45 * 0.844) = 1.24
Kthermo = 1.24 * 2.01 = 2.49
Q6.38
$SO_2 + Cl_2 \rightarrow SO_2Cl_2$
This reaction happens at 273K. You are given a Kp of 0.683. The pressure for SO2 is 0.58 bar for Cl2 is 0.93 bar, and SO2Cl2 is 0.776 bar. From this information, determine (delta)rG.
=(-8.314Jmol-1K-1)(273K)ln(0.683)
=865.4 J mol-1
=1.69 kj mol-1
Q6.40
Assuming oxygen binding to hemoglobin can be represented by the following reaction:
$Hb_{(aq)} + O_{2(g)} \rightarrow HbO_{2(aq)}$
If the value of ΔrG° for the reaction is -11.2 kJ mol-1 at 37°C, calculate the value of ΔrG° for the reaction.
Q6.42
At T=300K, given the mole ratio between 2 isomers Cis-2-butene and Trans-2-butene in an equilibrium mixture is 1:4. Evaluate rG of the reversible reaction:
Cis-2-butene <------> Trans-2-butene
Q6.44
When discussing the reaction in biological cells, why would you use concentrations instead of activities?
S6.44
Concentrations are generally smaller and it's easier to compare the concentrations between two parts of a cell or between different molecules in a reaction whereas activity describes behavior.
Q6.46
The following data shows the oxygen binding concentration in snails. The protein concentration is 15mM. Find $n$ and $K_d$ by using the Scatchard plot
$[O_2]_{total}$ 10 mM 14 mM 18 mM 22 mM 26 mM 30 mM 34 mM 38 mM 42 mM 46 mM
$[O_2]_{bound}$ 9 mM 12 mM 16 mM 19 mM 22 mM 25 mM 28 mM 32 mM 35 mM 39 mM
S6.46
$y=mx+b\ \dfrac{Y}{[L]}=(-\dfrac{1}{K_d})Y+\dfrac{n}{K_d}\ Y=\dfrac{[O_2]_{bound}}{[P]}\.[L]=[O_2]_{free}$
[O2]tot
10 14 18 22 26 30 34 38 42 46
[O2]bound 9 12 16 19 22 25 28 32 35 39
[O2]free 1 2 2 3 4 5 6 6 7 7
Y 0.6 0.8 1.067 1.267 1.467 1.667 1.867 2.133 2.333 2.6
Y/[L] 0.6 0.4 0.833 0.422 0.367 0.333 0.311 0.356 0.333 0.371
$-\dfrac{1}{K_d}=-0.102\ K_d=9.804\ \dfrac{n}{K_d}=0.564\ n=K_d*0.564=5.529$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/06%3A_Chemical_Equilibrium/6.E%3A_Chemical_Equilibrium_%28Exercises%29.txt
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Electrochemistry is the study of chemical processes that cause electrons to move. This movement of electrons is called electricity, which can be generated by movements of electrons from one element to another in a reaction known as an oxidation-reduction ("redox") reaction.
07: Electrochemistry
Electrochemistry is the study of chemical processes that cause electrons to move. This movement of electrons is called electricity, which can be generated by movements of electrons from one element to another in a reaction known as an oxidation-reduction ("redox") reaction. A redox reaction is a reaction that involves a change in oxidation state of one or more elements. When a substance loses an electron, its oxidation state increases; thus, it is oxidized. When a substance gains an electron, its oxidation state decreases, thus being reduced. For example, for the redox reaction
$\ce{H_2 + F_2 → 2 HF} \nonumber$
can be rewritten as follows:
• Oxidation reaction
$\ce{H_2 \rightarrow 2H^+ + 2e^{-}} \nonumber$
• Reduction reaction:
$\ce{F_2 + 2e^{-} \rightarrow 2F^{-}} \nonumber$
• Overall Reaction
$\ce{H_2 + F_2 \rightarrow 2H^{+} + 2F^{-}} \nonumber$
Oxidation is the loss of electrons, whereas reduction refers to the acquisition of electrons, as illustrated in the respective reactions above. The species being oxidized is also known as the reducing agent or reductant, and the species being reduced is called the oxidizing agent or oxidant. In this case, H2 is being oxidized (and is the reducing agent), while F2 is being reduced (and is the oxidizing agent). The following acronym is useful in remembering this concept:
"OIL RIG"
"OIL RIG" is a useful mnemonic for remembering the definitions of oxidation nd reduction.
Oxidation Is Losing electrons; Reduction Is Gaining electrons
Example $1$: IRon-Vanadium Couple
Given the redox reaction
$\ce{Fe^{3+} + V^{2+}→ Fe^{2+} + V^{3+}} \nonumber$
which species is oxidized? Which is reduced? Identify the reducing agent and the oxidizing agent.
Solution
$\ce{Fe^{3+}}$ is reduced into $\ce{Fe^{2+}}$ and $\ce{V^{2+}}$ is oxidized into $\ce{V^{3+}}$. This is because the oxidized species loses electrons, and the reduced species gains electrons.
Iron gains an electron
$\ce{Fe^{3+}→ Fe^{2+}} \nonumber$
and vanadium loses an electron
$\ce{V^{2+}→ V^{3+}}. \nonumber$
Thus, $\ce{Fe^{3+}}$ is the oxidizing agent and $\ce{V^{2+}}$ is the reducing agent.
Oxidation States
Rules for Assigning Oxidation States
1. Free elements have an oxidation state of 0. (e.g., $\ce{He}$, $\ce{N2}$, $\ce{O2}$ has an oxidation state of 0)
2. The oxidation state of one ion must equal the net charge. (e.g., $\ce{F^{-}}$ oxidation state is -1, $K^{+}$ oxidation state is +1)
3. The sum of the oxidation state has to equal the total net charge for a compound. (e.g., $\ce{MnO4^{-}}$ has a net charge of -1 with Mn(+7)O4(-8)= -1)
4. The alkali metals (Group I elements) have an oxidation state of +1. (e.g., $\ce{Li2O}$, Li= +1)
5. The alkaline earth metals (Group II elements) always have an oxidation state of +2. (e.g., $\ce{CaO}$, Ca=+2)
6. Oxygen has an oxidation state of -2 in a compound
7. Fluorine has an oxidation state of -1 in a compound
8. Hydrogen has an oxidation state of +1 in a compound.
9. Transition metals and other metals may have more than one common ionic charge. (e.g., Chromium's common ionic charges are Cr+2 and Cr+3)
Exercise $1$
What is the oxidation state of magnesium in $\ce{MgF2}$?
Answer
Using rule 5 and 7.
$\ce{MgF2}$ total charge=0 Total Charge=(+2)+(-1*2)=0
Exercise $2$
What is the oxidation state of hydrogen in $\ce{H2O}$?
Answer
Using rule 6 and 8
$\ce{H2O}$ total charge=0 Total Charge=(+1*2)+(-2)=0
Balancing Redox Reactions
Method 1: Oxidation Number Method
• Step 1: Assign oxidation numbers to each atom.
• Step 2: Determine the net change in charge to determine the ratio of atoms
• Step 3: Use the ratio to eliminate the net charge change
• Step 4: Use the ratio as coefficients for the elements
• Step 5: Add H+ (under acidic conditions), OH- (under basic conditions), and H2O to balance charges.
Method 2: Half-Reaction Method
• Step 1: Determine oxidation numbers for each atom
• Step 2: Use oxidation numbers to determine what is oxidized and what is reduced.
• Step 3: Write a half-reaction for reduction
• Step 4: Write a half-reaction for oxidation
• Step 5: Balance all elements except H and O
• if have acid redox reaction: Balance the O using $H_2O$, balance the $H$ using protons
• if have base redox reaction: Balance O using $OH^-$
• Step 6: Add up the charge on each side
• Step 7: Balance the charges by adding electrons
• Step 8: Multiply the half-reactions by factors that cancel out electrons
• Step 9: Add the two half-reactions back together to eliminate out intermediates
Example $2$: Manganese
Balance the following reaction in an acidic aqueous Solution
$\ce{MnO_4^{-} + H_2C_2O_4 \rightarrow Mn^{2+} + CO_2} \nonumber$
Solution
• Reduction half-reaction:
$2 \times \left(\ce{5e^{-} + 8H^+ + MnO_4^{-} \rightarrow Mn^{2+} + 4H_2O}\right) \nonumber$
• Oxidation half-reaction:
$5 \times \left( \ce{H_2C_2O_4 \rightarrow 2CO_2 + 2H^{+} + 2e^{-}}\right) \nonumber$
Combining and canceling gives the following:
$\cancel{\ce{10e^{-}}} + \ce{16H^{+} + 2MnO_4^{1-} + 5H_2C_2O_4 \rightarrow 2Mn^{2+} + 8H_2O + 10 CO_2 + 10H^{+}} + \cancel{\ce{10e^{-}}} \nonumber$
Answer
$\ce{6H^{+} + 2MnO_4^{1-} + 5H_2C_2O_4 \rightarrow 2Mn^{2+} + 8H_2O + 10 CO_2} \nonumber$
Voltaic (Galvanic) Cells
In 1793, Alessandro Volta discovered that electricity could be produced by placing different metals on the opposite sides of a wet paper or cloth. He made his first battery by placing Ag and Zn on the opposite sides of a moistened cloth with salt or weak acid Solution. Therefore, these batteries acquired the name voltaic cells. Voltaic (galvanic) cells are electrochemical cells that contain a spontaneous reaction, and always have a positive voltage. The electrical energy released during the reaction can be used to do work. A voltaic cell consists of two compartments called half-cells. The half-cell where oxidation occurs is called the anode. The other half-cell, where reduction occurs, is called the cathode. The electrons in voltaic cells flow from the negative electrode to the positive electrode—from anode to cathode (see figure below). (Note: the electrodes are the sites of the oxidation and reduction reactions). The following acronym is useful in keeping this information straight:
Red Cat and An Ox
Reduction Cathode and Anode Oxidation
For an oxidation-reduction reaction to occur, the two substances in each respective half-cell are connected by a closed circuit such that electrons can flow from the reducing agent to the oxidizing agent. A salt bridge is also required to maintain electrical neutrality and allow the reaction to continue.
The figure above shows that $\ce{Zn(s)}$ is continuously oxidized, producing aqueous $\ce{Zn^{2+}}$:
$\ce{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-}} \nonumber$
Conversely, in the cathode, $\ce{Cu^{2+}}$ is reduced and continuously deposits onto the copper bar:
$\ce{Cu^{2+} (aq) + 2e^{-} \rightarrow Cu(s)} \nonumber$
As a result, the Solution containing $\ce{Zn(s)}$ becomes more positively charged as the Solution containing $\ce{Cu(s)}$ becomes more negatively charged. For the voltaic cell to work, the Solutions in the two half-cells must remain electrically neutral. Therefore, a salt bridge containing KNO3 is added to keep the Solutions neutral by adding NO3-, an anion, into the anode Solution and $\ce{K^{+}}$, a cation, into the cathode Solution. As oxidation and reduction proceed, ions from the salt bridge migrate to prevent charge buildup in the cell compartments.
The cell diagram (or cell notation) is a shorthand notation to represent the redox reactions of an electrical cell. For the cell described, the cell diagram is as follows:
$\ce{Zn(s) | Zn^{2+} (aq) || Cu^{2+} (aq) | Cu(s)} \nonumber$
• A double vertical line ($\ce{||}$) is used to separate the anode half reaction from the cathode half reaction. This represents the salt bridge.
• The anode (where oxidation occurs) is placed on the left side of the ($\ce{||}$).
• The cathode (where reduction occurs) is placed on the right side of the ($\ce{||}$).
• A single vertical line (|) is used to separate different states of matter on the same side, and a comma is used to separate like states of matter on the same side. For example:
$\ce{Fe^{2+} (aq), Fe^{3+} (aq) || Ag^{+} (aq) | Ag(s)} \nonumber$
Figure $2$: A voltaic cell works by the different reactivity of metal ions, and not require external battery source. Image taken at Hope College as part of their General Chemistry Lab curriculum.
The figure above shows a set of electrochemical half-cells that can be used to measure various voltages within galvanic cells. The cells shown are made of agar saturated with KCl Solution so as to act as a salt bridge. The zinc electrode in the middle can be used as a reference while the various concentrations of copper (labeled 1, 2, 3, 4 and 5) can be tested to form a calibration curve. The potential of the unknown can be used to determine the concentration of an unknown copper Solution. This application of the Nernst equation allows for rapid data collection without the need for a complicated salt bridge apparatus.
Example $3$: Copper-silver Reaction
Write the cell diagram for this reaction:
$\ce{Cu(s) + 2Ag{+}(aq) →Cu^{2+}(aq) + 2Ag(s)} \nonumber$
Solution
$\ce{Cu(s) | Cu^{2+}(aq) || Ag^{+}(aq) | Ag(s)} \nonumber$
Example $4$: Aluminum-Tin Reaction
Write cell reactions for this cell diagram:
$\ce{Al_(s) | Al^{3+}(aq) || Sn^{2+}(aq) | Sn(s)} \nonumber$
Answer
Oxidation: {Al(s) → Al3+(aq) +3e-} x 2
Reduction: {Sn2+(aq) +2e- → Sn(s)} x 3
Net: 2Al(s) + 3Sn2+(aq) → 2Al3+(aq) + 3Sn(s)
Cell Potentials
The oxidation of Zn(s) into Zn2+ and the reduction of Cu2+ to Cu(s) occur spontaneously. In other words, the redox reaction between Zn and Cu2+ is spontaneous. This is due to the difference in potential energy between the two substances. The difference in potential energy between the anode and cathode dictates the direction of electronic movement. Electrons move from areas of higher potential energy to areas of lower potential energy. In this case, the anode has a higher potential energy; electrons therefore move from anode to cathode. The potential difference between the two electrodes is measured in units of volts. One volt (V) is the potential difference necessary to generate a charge of 1 coulomb (C) from 1 joule (J) of energy.
For a voltaic cell, this potential difference is called the cell potential (or EMF for electromotive force, although it is not really a force), which is denoted Ecell. For a spontaneous reaction, Ecell is positive and ΔG (Gibbs free energy, used to determine if a reaction occurs spontaneously) is negative. Thus, when ΔG is negative the reaction is spontaneous. Merging electrochemistry with thermodynamics gives this formula:
$\Delta G = -n F E_{cell} \nonumber$
Cell potential is different for each voltaic cell; its value depends upon the concentrations of specific reactants and products as well as temperature of the reaction. For standard cell potential, temperature of the reaction is assumed to be 25o Celsius, the concentration of the reactants and products is 1 M, and reaction occurs at 1 atm pressure. The standard cell potential is denoted Eocell, and can be written as oxidation potential + reduction potential. For voltaic cells:
$E^o_{cell}=E^o_{cathode}-E^o_{Anode} \label{Celldef}$
Warning
To use Equation \ref{Celldef}, the cell potentials must be in reduction form.
Since standard potentials are given in the form of standard reduction potential for each half-reaction, to calculate the standard cell potential $E^o_{cell}$, the substance is being oxidized must be identified; then the standard reduction potential of the oxidation reaction is subtracted from the standard reduction potential of the reducing reaction.
Example $5$
What is the cell potential for the following reaction?
$\ce{Zn(s) + Cu^{2+} (aq) \rightarrow Zn^{2+}(aq) + Cu(s)} \nonumber$
Solution
$\ce{Zn(s)}$ is being oxidized, and $\ce{Cu(s)}$ is being reduced. The potentials for the two half reaction are given in the reduction form:
$\ce{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-}} \nonumber$
$\ce{Cu^{2+} (aq) + 2e^{-} \rightarrow Cu (s)} \nonumber$
The cell potentials indicate which reaction takes place at the anode and which at the cathode. The cathode has a more positive potential energy, and thus:
• $\ce{Cu(s)}$ is the cathode
• $\ce{Zn(s)}$ is the anode.
To calculate $E^o_{cell}$, subtract the $E^o$ of the oxidized half reaction from the $E^o_{cell}$ of the reduction half reaction, we use Equation \ref{Celldef}}:
$E^o_{cell}=E^o_{cathode}-E^o_{anode} \nonumber$
Oxidation half reaction: Eo= -0.763V
$\ce{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-}} \nonumber$
• Reduction half reaction: Eo= +0.342V
$\ce{Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)} \nonumber$
Net reaction:
$\ce{Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)} \nonumber$
Therefore:
$E^o_{cell}=E^o_{cathode}-E^o_{anode}=0.340\; V - (-0.763\; V)= +1.103 \; V \nonumber$
Exercise $6$
Calculate Eocell for the following redox reaction under standard conditions:
$\ce{2Al (s) + 3Sn^{2+} (aq) \rightarrow 2Al^{3+} (aq) + 3Sn(s)} \nonumber$
Answer
Oxidation: $\{\ce{Al(s) → Al^{3+} (aq) +3e^{-}}\} \times 2 \,\,\, -E^o = +1.676\,V \nonumber$
Reduction: $\{\ce{Sn^{2+}(aq) + 2e^{-} → Sn(s)}\} \times 3 \,\,\, E^o = -0.137\,V \nonumber$
Net reaction: $\ce{2Al(s) + 3Sn^{2+}(aq) → 2Al^{3+}(aq) + 3Sn(s)} \nonumber$
Then using Equation \ref{Celldef} \begin{align*} E^o_{cell} = -0.137\,V - (-1.676V) \[4pt] &= +1.539 \,V. \end{align*}
Voltage is an Intensive Property
Standard reduction potential is an intensive property, meaning that changing the stoichiometric coefficient in a half reaction does not affect the value of the standard potential. For example,
Oxidation:{Al(s) → Al3+(aq) +3e-} x 2 is still Eo= -1.676
Reduction:{Sn2+(aq) +2e- → Sn(s)} x 3 is still Eo= -0.137
If the stoichiometric coefficient is multiplied by 2, the standard potential does not change:
Example $7$: Iron/Vanadium Chemistry
Calculate the cell potential in the following redox reaction under standard conditions:
$\ce{ Fe^{3+} (aq) + V^{2+} (aq) \rightarrow Fe^{2+} (aq) + V^{3+}(aq)} \nonumber$
Solution
Consult the table of standard reduction potentials (Table P1) for each half reaction:
$Fe^{3+}_{(aq)}+e^- \rightarrow Fe^{2+}_{(aq)} \;\;\;\; \text{with } E^o=0.771\; V \nonumber$
$V^{2+}_{(aq)} \rightarrow V^{3+}_{(aq)} + e^- \;\;\;\; \text{with } E^o=-0.255\; V \nonumber$
The cell potential is
$E^o_{cell}=E^o_{cathode}-E^o_{anode}=0.771\; V -(-0.255\; V)=1.026 \; V \nonumber$
Glossary
• Anode: Electrode in an electrochemical cell on which the oxidation reaction occurs.
• Cathode: Electrode in an electrochemical cell on which the reduction reaction occurs
• Electrochemistry: A field of chemistry that focuses on the interchange between electrical and chemical energy
• Electricity: Flow of electrons over a wire that is affected by the presence and flow of electric charge.
• Electrolysis: The decomposition of a substance by means of electric current. This method pushes a redox reaction toward the non-spontaneous side.
• Electrolytic cell: Electrochemical cell that is being pushed toward the non-spontaneous direction by electrolysis.
• Electromotive force, EMF (or cell potential): Difference of potential energy of electrons between the two electrodes.
• Oxidation number: Charge on an atom if shared electrons where assigned to the more electronegative atom.
• Oxidation: Lose of electrons, can occur only in combination with reduction. [remember: Oxidation Is Loss, Reduction Is Gain = OIL RIG]
• Reduction: Gain of electrons, can occur only in combination with oxidation. [remember: OIL RIG]
• Redox reaction: Shorthand for reduction-oxidation reaction.
• Voltaic cell or galvanic cell: An electrochemical cell that uses redox reaction to produce electricity spontaneously.
Contributors and Attributions
• Matthew Bui (UCD), Wen Chung Chou (UCD)
• Justin Shorb (Hope College), Yong Chul Yoon (Hope College)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/07%3A_Electrochemistry/7.01%3A_Electrochemical_Cells.txt
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Concentration Cells
A concentration cell is an electrolytic cell that is comprised of two half-cells with the same electrodes, but differing in concentrations. A concentration cell acts to dilute the more concentrated solution and concentrate the more dilute solution, creating a voltage as the cell reaches an equilibrium. This is achieved by transferring the electrons from the cell with the lower concentration to the cell with the higher concentration.
The standard electrode potential, commonly written as Eocell, of a concentration cell is equal to zero because the electrodes are identical. But, because the ion concentrations are different, there is a potential difference between the two half-cells. One can find this potential difference via the Nernst Equation,
$E_{cell} = E^o_{cell} - \dfrac{0.0592}{n}logQ$
at 25oC. The E stands for the voltage that can be measured using a voltmeter (make sure if the voltmeter measures it in millivolts that you convert the number before using it in the equation). Note that the Nernst Equation indicates that cell potential is dependent on concentration, which results directly from the dependence of free energy on concentration. Remember that to find Q you use this equation:
$aA + bB \rightleftharpoons cC + dD$
$Q = \dfrac{(C)^c * (D)^d}{(A)^a * (B)^b}$
When Q=1, meaning that the concentrations for the products and reactants are the same, then taking the log of this equals zero. When this occurs, the Ecell is equal to the Eocell.
Another way to use the Eocell , or to find it, is using the equation below.
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$
A voltage can also be generated by constructing an electrochemical cell in which each compartment contains the same redox active solution but at different concentrations. The voltage is produced as the concentrations equilibrate. Suppose, for example, we have a cell with 0.010 M AgNO3 in one compartment and 1.0 M AgNO3 in the other. The cell diagram and corresponding half-reactions are as follows:
$Ag(s)\,|\,Ag^+(aq, 0.010 \;M)\,||\,Ag+(aq, 1.0 \;M)\,|\,Ag(s) \label{Eq8}$
cathode:
$Ag^+(aq, 1.0\; M) + e^− \rightarrow Ag(s) \label{Eq9}$
anode:
$Ag(s) \rightarrow Ag^+(aq, 0.010\; M) + e^− \label{Eq10}$
Overall
$Ag^+(aq, 1.0 \;M) \rightarrow Ag^+(aq, 0.010\; M) \label{Eq11}$
As the reaction progresses, the concentration of $Ag^+$ will increase in the left (oxidation) compartment as the silver electrode dissolves, while the $Ag^+$ concentration in the right (reduction) compartment decreases as the electrode in that compartment gains mass. The total mass of $Ag(s)$ in the cell will remain constant, however. We can calculate the potential of the cell using the Nernst equation, inserting 0 for E°cell because E°cathode = −E°anode:
$E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q=0-\left(\dfrac{\textrm{0.0591 V}}{1}\right)\log\left(\dfrac{0.010}{1.0}\right)=\textrm{0.12 V}$
An electrochemical cell of this type, in which the anode and cathode compartments are identical except for the concentration of a reactant, is called a concentration cell. As the reaction proceeds, the difference between the concentrations of Ag+ in the two compartments will decrease, as will Ecell. Finally, when the concentration of Ag+ is the same in both compartments, equilibrium will have been reached, and the measured potential difference between the two compartments will be zero (Ecell = 0).
Example $2$
Calculate the voltage in a galvanic cell that contains a manganese electrode immersed in a 2.0 M solution of MnCl2 as the cathode, and a manganese electrode immersed in a 5.2 × 10−2 M solution of MnSO4 as the anode (T = 25°C).
Given: galvanic cell, identities of the electrodes, and solution concentrations
Asked for: voltage
Strategy:
1. Write the overall reaction that occurs in the cell.
2. Determine the number of electrons transferred. Substitute this value into the Nernst equation to calculate the voltage.
Solution:
A This is a concentration cell, in which the electrode compartments contain the same redox active substance but at different concentrations. The anions (Cl and SO42−) do not participate in the reaction, so their identity is not important. The overall reaction is as follows:
$Mn^{2+}_{(aq, 2.0 M)} \rightarrow Mn^{2+}_{(aq, 5.2 \times 10^{−2} M)}$
B For the reduction of Mn2+(aq) to Mn(s), n = 2. We substitute this value and the given Mn2+ concentrations into Equation $\ref{Eq4}$:
$E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q=\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{5.2\times10^{-2}}{2.0}\right)=\textrm{0.047 V}$
Thus manganese will dissolve from the electrode in the compartment that contains the more dilute solution and will be deposited on the electrode in the compartment that contains the more concentrated solution.
Exercise $2$
Suppose we construct a galvanic cell by placing two identical platinum electrodes in two beakers that are connected by a salt bridge. One beaker contains 1.0 M HCl, and the other a 0.010 M solution of Na2SO4 at pH 7.00. Both cells are in contact with the atmosphere, with $P_\mathrm{O_2}$ = 0.20 atm. If the relevant electrochemical reaction in both compartments is the four-electron reduction of oxygen to water:
$O_{2(g)} + 4H^+_{(aq)} + 4e^− \rightarrow 2H_2O_{(l)}$
What will be the potential when the circuit is closed?
Answer 0.41 V
Contributors and Attributions
• SSReno, Rainna Lim
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/07%3A_Electrochemistry/7.04%3A_Types_of_Electrochemical_Cells.txt
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Before the concept of biological oxidation can be understood and explored, the fundamental chemical process through which oxidation and reduction take place should be first established.
Foundation
All reactions which involve electron flow are considered oxidation-reduction reactions. The basic definition can be defined as: One reactant is oxidized (loses electrons), while another is reduced (gains electrons). A couple of basic oxidation-reduction or "redox" example's are given here.
Example 1
The reaction of magnesium metal with oxygen, involves the oxidation of magnesium
$2Mg(s) + O_2(g)→ 2MgO(s) \label{1}$
Since the magnesium solid is oxidized, we expect to see a loss of electrons. Similarly, since oxygen must therefore be reduced, we should see a gain of electrons.
As the magnesium is oxidized there is a loss of 2 electrons while simultaneously, oxygen gains those two electrons. Another example of a redox reaction is with the two gasses CO2 and H2. This redox reaction also demonstrates the importance of implementing "oxidation numbers" in the methodology of redox reactions, allowing for the determination of which reactant is being reduced and which reactant is being oxidized.
Example 2
The reaction of carbon dioxide gas with hydrogen gas, involving the oxidation of hydrogen
$CO_2 (g) + H_2 (g) → 2CO (g) + H_2O (g) \label{3}$
Since the hydrogen gas is being oxidized (reductant), we expect to see an overall loss of electrons for the resulting molecule. Similarly, we expect to see a gain in the overall number of electrons for the resulting molecule of the oxidant (CO2).
Here it is possible to infer that the carbon of CO2 is being reduced by review of its unique oxidation number. Such that, C (of CO2) goes from an oxidation number of +4 to C (of CO) having an oxidation number of +2, representing a loss of two electrons. Similarly, H2 is noted as going from an oxidation number of 0 to +1, or gaining one electron in a reduction process. For more information on oxidation numbers, review the following link: Oxidation-Reduction Reactions
A Basic Biological Model
The flow of electrons is a vital process that provides the necessary energy for the survival of all organisms. The primary source of energy that drives the electron flow in nearly all of these organisms is the radiant energy of the sun, in the form of electromagnetic radiation or Light. Through a series of nuclear reactions, the sun is able to generate thermal energy (which we can feel as warmth) from electromagnetic radiation (which we perceive as light). However, the particular wavelength of the electromagnetic spectrum we are able to detect with the human eye is only between 400 and 700 nm in wavelength. It should therefore be noted that the visible part of the electromagnetic spectrum is actually a small percentage of the whole; where a much greater percentage remains undetectable for the human eye.
In physics, the use of the term "light" refers to electromagnetic radiation of any wavelength, independent of its detectability for the human eye. For plants, the upper and lower ends of the visible spectrum are the wavelengths that help drive the process of splitting water (H2O) during photosynthesis, to release its electrons for the biological reduction of carbon dioxide (CO2) and the release of diatomic oxygen (O2) to the atmosphere. It is through the process of photosynthesis that plants are able to use the energy from light to convert carbon dioxide and water into the chemical energy storage form called glucose.
Plants represent one of the most basic examples of biological oxidation and reduction. The chemical conversion of carbon dioxide and water into sugar (glucose) and oxygen is a light-driven reduction process:
$6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2 \label{5}$
The process by which non photosynthetic organisms and cells obtain energy, is through the consumption of the energy rich products of photosynthesis. By oxidizing these products, electrons are passed along to make the products carbon dioxide, and water, in an environmental recycling process. The process of oxidizing glucose and atmospheric oxygen allowed energy to be captured for use by the organism that consumes these products of the plant. The following reaction represents this process:
$C_6H_{12}O_6 + O_2 \rightarrow 6CO_2 + 6H_2O +Energy \label{6}$
It is therefore through this process that heterotrophs (most generally "animals" which consume other organisms obtain energy) and autotrophs (plants which are able to produce their own energy) participate in an environmental cycle of exchanging carbon dioxide and water to produce energy containing glucose for organismal oxidation and energy production, and subsequently allowing the regeneration of the byproducts carbon dioxide and water, to begin the cycle again. Therefore, these two groups of organisms have been allowed to diverge interdependently through this natural life cycle.
Physical Chemistry's Understanding
Biological oxidation-reduction reactions, or simply biological oxidations utilize multiple stages or processes of oxidation to produce large amounts of Gibbs energy, which is used to synthesize the energy unit called adenosine triphosphate or ATP. To efficiently produce ATP, the process of glycolysis must be near an abundance of oxygen. Since glycolysis by nature is not an efficient process, if it lacks sufficient amounts of oxygen the end product pyruvate, is reduced to lactate with NADH as the reducing agent. However, in a more favorable aerobic process, the degradation of glucose through glycolysis proceeds with two additional processes known as the citric acid cycle and the terminal respiratory chain; yielding the end products carbon dioxide and water, which we exhale with each breath.
The products NADH and FADH2 formed during glycolysis and the citric acid cycle are able to reduce molecular oxygen (O2) thereby releasing large amounts of Gibbs energy used to make ATP. The process by which electrons are transferred from NADH or FADH2 to O2 by a series of electron transfer carriers, is known as oxidative phosphorylation. It is through this process that ATP is able to form as a result of the transfer of electrons.
Thee specific examples of redox reactions that are used in biological processes, involving the transfer of electrons and hydrogen ions as follows. During some biological oxidation reactions, there is a simultaneous transfer of hydrogen ions with electrons (1). In other instances, hydrogen ions may be lost by the substance being oxidized while transferring only its electrons to the substance being reduced (2). A third type of biological oxidation might involve only a transfer of electrons (3). It should be noted that biological oxidation rarely proceeds in a direct manner, and generally involves complex mechanisms of several enzymes. The outline below recaps the three processes of biological oxidation stated above, in descending order.
Table 1: Transfer of hydrogen ions and electrons for the general reaction scheme of A + B with intermediate stage shown
Reactants Intermediate Stage Products
AH2 + B [A + 2H+ + 2e- + B] A + BH2
AH2 + B [A + 2H+ + 2e- + B] A + B2- + 2H+
A2- + B [A + 2e- + B] A + B2-
In the last stage of the metabolic process (the terminal respiratory chain), the sequence by which electrons are carried is determined by relative redox potentials. The carrier molecules used to transfer electrons in this stage are called cytochromes, which are an electron-carrying protein containing a heme group. The iron atom of each cytochrome molecule can exist either in the oxidized (Fe3+) or reduced (Fe2+) form. Within the terminal respiratory chain, each carrier molecule alternates between the reduced state and the oxidized state, with molecular oxygen as the final electron acceptor at the end.
It is through the knowledge of redox potentials, that the knowledge of biological processes can be further expanded. The standard reduction potential is denoted as Eo' and is often based on the hydrogen electrode scale of pH 7, rather than pH 0, a common reference point for listed values. Moreover, the superscript symbol ( o ) denotes standard-state conditions, while the adjacent superscript symbol ( ' ) denotes the pH scale of 7 for biochemical processes.
It therefore becomes possible to trace the energy transfer in cells back to the fundamental flow of electrons from one particular molecule to another. Where this electron flow occurs via the physics principle of higher potential to lower potential; similar to a ball rolling down a hill, as opposed to the opposite direction. All of these reactions involving electron flow can be attributed to the basic definition of the oxidation-reduction pathway stated above.
Contributors and Attributions
• Brent Younglove (Hope)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/07%3A_Electrochemistry/7.06%3A_Biological_Oxidation.txt
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Membrane potential is what we use to describe the difference in voltage (or electrical potential) between the inside and outside of a cell.
Introduction
Without membrane potentials human life would not be possible. All living cells maintain a potential difference across their membrane. Simply stated, membrane potential is due to disparities in concentration and permeability of important ions across a membrane. Because of the unequal concentrations of ions across a membrane, the membrane has an electrical charge. Changes in membrane potential elicit action potentials and give cells the ability to send messages around the body. More specifically, the action potentials are electrical signals; these signals carry efferent messages to the central nervous system for processing and afferent messages away from the brain to elicit a specific reaction or movement. Numerous active transports embedded within the cellular membrane contribute to the creation of membrane potentials, as well as the universal cellular structure of the lipid bilayer. The chemistry involved in membrane potentials reaches to many scientific disciplines. Chemically it involves molarity, concentration, electrochemistry and the Nernst equation. From a physiological standpoint, membrane potential is responsible for sending messages to and from the central nervous system. It is also very important in cellular biology and shows how cell biology is fundamentally connected with electrochemistry and physiology. The bottom line is that membrane potentials are at work in your body right now and always will be as long as you live.
History
The subject of membrane potential stretches across multiple scientific disciplines; Membrane Potential plays a role in the studies of Chemistry, Physiology and Biology. The culmination of the study of membrane potential came in the 19th and early 20th centuries. Early in the 20th century, a man named professor Bernstein hypothesized that there were three contributing factors to membrane potential; the permeability of the membrane and the fact that [K+] was higher inside and lower on the outside of the cell. He was very close to being correct, but his proposal had some flaws. Walther H. Nernst, notable for the development of the Nernst equation and winner of 1920 Nobel Prize in chemistry, was a major contributor to the study of membrane potential. He developed the Nernst equation to solve for the equilibrium potential for a specific ion. Goldman, Hodgkin and Katz furthered the study of membrane potential by developing the Goldman-Hodgkin-Katz equation to account for any ion that might permeate the membrane and affect its potential. The study of membrane potential utilizes electrochemistry and physiology to formulate a conclusive idea of how charges are separated across a membrane.
Figure 1. Differences in concentration of ions on opposite sides of a cellular membrane produce a voltage difference called the membrane potential. The largest contributions usually come from sodium (Na+) and chloride (Cl) ions which have high concentrations in the extracellular region, and potassium (K+) ions, which along with large protein anions have high concentrations in the intracellular region. Calcium ions, which sometimes play an important role, are not shown.
Membrane Potential and Cellular Biology
In discussing the concept of membrane potentials and how they function, the creation of a membrane potential is essential. The lipid bilayer structure of the cellular membrane, with its lipid-phosphorous head and fatty acid tail, provides a perfect building material that creates both a hydrophobic and hydrophilic side to the cellular membrane. The membrane is often referred to as a mosaic model because of its semi-permeability and its ability to keep certain substances from entering the cell. Molecules such as water can diffuse through the cell based on concentration gradients; however, larger molecules such as glucose or nucleotides require channels. The lipid bilayer also houses the Na+/K+ pump, ATPase pump, ion transporters, and voltage gated channels, and it is the site of vesicular transport. The structure regulates which ions enter and exit to determine the concentration of specific ions inside of the cell.
Why is membrane potential essential to the survival of all living creatures?
Animals and plants require the breakdown of organic substances through cellular respiration to generate energy. This process, which produces ATP, is dependent on the electron transport chain. Electrons travel down this path to be accepted by oxygen or other electron acceptors. The initial electrons are obtained from the breakdown of water molecules. The hydrogen build up in the extracellular fluid leaving a gradient. As per membrane potentials, when there a gradient, the molecules flow in the opposite direction. In this case, hydrogen flows back into the cell through a protein known as ATP synthase which creates ATP in the process. This action is essential to life because the number of ATP created from each glucose increases drastically. Chemical disequilibrium and membrane potentials allow bodily functions to take place.
Figure 2
Transport proteins, more specifically the 'active' transport proteins, can pump ions and molecules against their concentration gradient. This is the main source of charge difference across the cellular membrane.
Physiology of Membrane Potential
Understanding Membrane Potential
The following points should help you to understand how membrane potential works
• The difference between the electrical and chemical gradient is important.
• Electrical Gradient
• Opposes the chemical gradient.
• Represents the difference in electrical charge across the membrane
• Chemical Gradient
• Opposes the electrical gradient
• Represents the difference in the concentration of a specific ion across the membrane.
• A good example is K+. The membrane is very permeable to K+ and the [K+] inside the cell is great, therefore a positive charge is flowing out of the cell along with K+. The [K+] inside the cell decreases causing the concentration gradient to flow towards the outside of the cell. This also causes the inside of the cell to become more electronegative increasing its electrical gradient.
• The Nernst equation can help us relate the numerical values of concentration to the electrical gradient.
• Leak Channels
• Channels that are always open
• Permit unregulated flow of ions down an electrochemical gradient.
• Na+/K+ ATPase Pump
• Actively transports Na+ out of the cell and K+ into the cell.
• Helps to maintain the concentration gradient and to counteract the leak channels.
Membrane Potential and Physiology of Human Nerve Cells
Human nerve cells work mainly on the concept of membrane potentials. They transmit chemicals known as serotonin or dopamine through gradients. The brain receives these neurotransmitters and uses it to perform functions.
• Na+ has a much higher concentration outside of the cell and the cell membrane is very impermeable to Na+
• K+ has a high concentration inside the cell due to the fact that the cell membrane is very permeable to K+
• A- is used to refer to large ions that are found completely inside of the cell and cannot penetrate the cell membrane.
Concentration (in Millimoles/ Liter) and permeability of Ions Responsible for Membrane Potential in a Resting Nerve Cell
ION Extracellular Intracellular Relative Permeability
Na+ 150 15 1
K+ 5 150 25-30
A- 0 65 0
Check out this YouTube video if you want to know more about how the Na+/K+ pump and how the membrane potential works. www.youtube.com/watch?v=iA-Gdkje6pg
How To Calculate A Membrane Potential
The calculation for the charge of an ion across a membrane, The Nernst Potential, is relatively easy to calculate. The equation is as follows: (RT/zF) log([X]out/[X]in). RT/F is approximately 61, therefore the equation can be written as
(61/z) ln([X]out/[X]in)
• R is the universal gas constant (8.314 J.K-1.mol-1).
• T is the temperature in Kelvin (°K = °C + 273.15).
• z is the ionic charge for an ion. For example, z is +1 for K+, +2 for Mg2+, -1 for F-, -1 for Cl-, etc. Remember, z does not have a unit.
• F is the Faraday's constant (96485 C.mol-1).
• [X]out is the concentration of the ion outside of the species. For example the molarity outside of a neuron.
• [X]in is the concentration of the ion inside of the species. For example, the molarity inside of a neuron.
The only difference in the Goldman-Hodgkin-Katz equation is that is adds together the concentrations of all permeable ions as follows
(RT/zF) log([K+]o+[Na+]o+[Cl-]o /[K+]i+[Na+]i+[Cl-]i)
Figure 3. (Clockwise From Upper Left) 1) The charges are equal on both sides; therefore the membrane has no potential. 2)There is an unbalance of charges, giving the membrane a potential. 3) The charges line up on opposite sides of the membrane to give the membrane its potential. 4) A hypothetical neuron in the human body; a large concentration of potassium on the inside and sodium on the outside.
Problems
1. List the following in order from highest to lowest permeability. A-, K+, Na+
2. Which of the following statement is NOT true?
1. The membrane potential usually requires a minimal difference of electrocharges across the membrane
2. Membrane impermeability plays a role in membrane potentials.
3. Membrane potential exists in all cellular structures, except for neurons.
4. The active transports play a vital role in membrane potentials.
3. What would be the equilibrium potential for the ion K+ be if [K+]in= 5mM and [K+]in=150mM?
4. True or false: At resting membrane potential, the inside of the membrane is slightly negatively charged while the outside is slightly positively charged.
Answers:
1. K+ > Na+ > A-
2. Answer c.)is not true; membrane potential exists in neurons and is responsible for action potential propagation in neurons.
3. Ek+ = (61/z) log([K+]out/[K+]in) = (61/1) log([5mM]/[150mM]) = -90mV
z=1
4. True. The resting membrane potential is negative as a result of this disparity in concentration of charges.
Contributors
• Dan Chong, Matt Klingler (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/07%3A_Electrochemistry/7.07%3A_Membrane_Potential.txt
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Q7.1
For an electrochemical cell driven by the reaction:
$\ce{Cl2 + 2Tl → 2Cl- + 2Tl+}$
Determine the electromotive force of the cell under standard conditions. Standard reduction potentials for the half cells are given below
$\ce{Cl2 (g) + 2e- → 2Cl- }$
with Eo= +1.36 V
$\ce{Tl+ + e- → Tl}$
with Eo= -0.336 V
S7.1
The change in gibbs free energy is an extrinsic quality that can be used to determine the EMF of the cell based on the SRPs of the half-cells. Because the number of electrons exchanged in each reaction are known, we can cancel out the Faraday constants in the equation to relate Gibbs' free energy directly to the EMF of the cell.
ΔGTot = ΔG1 + ΔG2 → EototntotF = Eo1n1F + Eo2n2F → Eotot = (Eo1n1 + Eo2n2)/ntot = (1.36V*(2) -0.336V(1))/2 = 2.38 V
Q7.8
For the following electrochemical cells, the value of K is given. Determine the change in Gibbs free energy for the reaction in the cell described, and determine the standard potential of the cell.
Cu2+ + Sn -> Sn2+ + Cu; K = e38.8
2Li + Hg2+ -> 2Li+ + Hg; K = e381
Au3+ + 3Zn -> 3Zn+ + Au; K = e264
Q7.9
Refer to the reaction below for the following question.
$Zn + Sn^{4+} \rightleftharpoons Zn^{2+} +Sn^{2+}$
Calculate the for the specified reaction at 298 K.
S7.9
First, identify the anode and the cathode:
Anode:
Cathode:
With this information established, you can then calculate the using the half-reactions.
Q7.9
$Cu^{2+} + 2Tl \rightarrow Cu + 2Tl^{2+}$
If the equilibrium constant is $8.66 \times 10^{22}$, what is the electric potential of the cell?
Q7.10a
In an experiment using standard conditions, you find the cell emf to be 0.0428 V for the following cell:
$Ag_{(s)}|Ag^{+}||Ag^{+}|Ag_{(s)}$
If the concentration of Ag+ is 4.5 M at the cathode, calculate the concentration of Ag+ at the anode.
S7.10a
Firstly, note the half reaction.
$Ag^{+} + e^{-} \rightarrow Ag$
Then solve using the Nernst equation, noting that E is 0 for concentration cells (same electrodes with different concentrations).
\begin{align} E &= E^o - \dfrac{RT}{vF}\ln \dfrac{[Anode]}{[Cathode]} \ 0.0428&=0 - \frac{(8.3145 \ J \ mol^{-1})(298^{\circ}K)}{(1)(96,485 J)} \ \cdot \ \ln\frac{X}{4.5M} \ 0.0428&=-0.0257 \ \cdot \ \ln\frac{X}{4.5M} \ -1.665&= \ln(x) - \ln(4.5) \ \ln(4.5)+ \left (-1.665 \right) &= \ln (x) \ e^{-0.161} &= e^{\ln(x)} \ X&=0.85 M \rightarrow \underline{ [Ag^{+}]=0.85M } \end{align}
Click here for more information on concentration cells.
Q7.10b
A concentration cell contains $I$ and $I_2$. What is the pressure of $I$ if the cell emf is -.05 V and the pressure of $I_2$is 3.75 bar?
S7.10b
$2I_{(g)} \rightleftharpoons I_{2(g)}$
$E=\dfrac{-.0257V}{2}ln\dfrac{3.75bar}{x}\rightarrow x=13.0bar$
Q7.14
Cu(s)|Cu2+ (0.5M) || Cu2+(aq)|Cu(s)
Calculate the electric potential difference for the above concentration cell
Q7.14
Using the above concentration cell, determine the emf. Assume the temperature is at 25°C.
S7.14
The overall reaction is:
The emf of the cell is dependent upon the concentrations of Co2+ at the anode and at the cathode.
Q7.16
Referring to the reaction:
and taking into account that at 298 K, the E°' is -0.219 V. Use this information to determine the E' at pH = 3. Note: Both FAD and FADH2 have equal moles of concentration.
Q7.15
Calculate E for the following reaction:
[(Zn(s)+Cu^{2+}(aq) \rightleftharpoons Cu(s) + Zn^{2+}(aq)\]
S7.15
$Zn(s) \rightarrow 2e^- + Zn^{2+}(aq)$ -Anode [Zn2+]=.25M
$Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$ -Cathode [Cu2+]=.35M
$E^\circ = E_{cathode}-E_{anode}=.342V-(-.762V)=1.104 V$
$E=E^\circ -\dfrac{.0257V}{2}ln\dfrac{Cu^{2+}}{Zn^{2+}}=1.10V -\dfrac{0.0257V}{2}ln\dfrac{.35M}{.25M}= 1.08V$
Q7.14
Co(s)|Co2+ (0.05M) ||Co2+(aq) (0.1M)|Co(s)
Calculate the emf for the above concentration cell.
Q7.21
To breakdown consumed alcohol, our body converted ethanol into an acetyldehyde by forming a redox reaction with $NAD^+$.
$\text{ethanol} + NAD^+ → \text{acetaldehyde} + NADH + H^+$
Determine ΔrG°' of reaction at 298 K and pH = 7 (Hint: Set up half reactions and find E°' of the whole reaction.)
S7.21
$ethanol \rightarrow acetaldehyde + 2H^{+} + 2e^{-} \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}{'} = +0.197 V$
$NAD^{+} + H^{+} + 2e^{-} \rightarrow NADH \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}{'} = -0.320 V$
$Overall: \, ethanol + NAD^{+} \rightarrow acetaldehyde + NADH + H^{+} \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}{'} = -0.123$
$\Delta rG^{\circ}{'} = -vF\Delta E^{\circ}{'} = -(2)(96500 \frac{C}{mol})(-0.123V) = 23739 J = {\color{red}23.7 kJ/mol}$
Q7.21
The hydrolysis of ATP with the following concentrations:
ATP4- + H2O ADP3- + H+ + HPO4²-
10mM 0.5mM pH 6 5mM
Calculate the change in biochemical standard Gibbs energy of reaction.
Q7.22
Calculate the equilibrium constant and ∆rGo' for hydrolysis of ATP at 25.0oC:
$\text{Glucose} + \text{ ATP} \rightarrow \text{ADP} + \text{ glucose-6-phosphate}$
• [Glucose]initial = 0.05M
• [Glucose]equilibrium = 0.01M
S7.22
KC = (0.04M/1M)/(0.01M/1M)=4.0
delta GO = -8.3145J/Kmole *(25+273.15)K*ln 4.0=-3.44 *103kJ/mol
Q7.23
Calculate the Δr/ of a certain biochemical reaction that involves the exchange of 2 electrons and where Eº/ = 3.25 V
S7.23
$\Delta _{r}G^{\circ}{}'= -vFE^{\circ}{}'$
$\Delta _{r}G^{\circ}{}'=-(2)(96500 C\ mol^{-1})(3.25 V)$
$\Delta _{r}G^{\circ}{}'= -627 kJ\ mol^{-1}$
Q7.25
In intestinal epithelial cells, transport of lactose across a membrane is coupled with the transport of 2 H+ ions from a high to low concentration.
$\Delta _{r}G^{\circ'} = 16.2 kJ\, \, \, \, \, \, \, \, \, \psi=-0.15V$
How much will the pH change across the membrane at 298 K and 1 atm be if one lactose molecule is transported?
Hint: What is the change in Gibbs Free Energy for the movement of 2H+ across this membrane? Write this as an expression as a function of change in pH.
S7.25
When we consider the chemical potential of a single H+
$\mu _{H^{+}}=\mu^{\circ} _{H^{+}}+RTln[H^{+}]+Fz\psi$
• z = the ion charge
• F = Faraday constant
• \psi = electrical potential
$pH=-log[H^{+}]$
$ln[H^{+}]=-2.3log[H^{+}]=-2.3pH$
$\mu _{H^{+}}=\mu^{\circ} _{H^{+}}-2.3RTpH+F\psi$
Change in Gibbs free energy is defined as the initial and final chemical potentials, or:
$\Delta _{r}G = \mu _{H^{+}(low)}-\mu _{H^{+}(high)}$
$=-2.3RT\Delta pH+F\Delta \psi$
As a function of change in pH, the equation can be written as
$\Delta pH=\frac{\Delta_{r}G-F\Delta \psi }{-2.3RT}$
Since this equation describes chemical potential across a membrane for a H+ ions, the sign for change in Gibb's free energy should be opposite that of the transport lactose, or -16.2 KJ. Setting change in electric potential as -0.15 V and converting 12.7 KJ to J
$\Delta pH= \frac{-16200J-(96500\frac{C}{mol})(-0.15 V)}{(298K)(-2.3)(8.3145\frac{J}{K\cdot mol})}$
$=0.30$
Q7.28
You are studying muscle cells. You notice inside the muscle there is 255 mM of Li+ and 16 mM of Rb+ and there is 58 Mm of Li+ and 249 mM of Rb+ outside the cell. From this information, decide if the membrane is more permeable to Li+ or Rb+.
S7.28
=-0.075
=0.0407
So Rb+ has a greater magnitude so the membrane is more permeable to it
Q7.29
Calculate the membrane potential of a cell that has [K+]out = 8 and [K+]in = 200. Assume standard temperature of 298 K.
Q7.30
Find E° for:
Cu²+ + e- -> Cu+
K+ + e- -> K
How can E° be used to find Ksp?
Q7.30a
Given the values of E° for the following reactions:
Ag+ + e- → Ag E° = +0.851 V
AgCl + e- → Ag + Cl- E° = +0.222 V
Determine the solubility product (Ksp) of AgCl at 23.5°C.
Q7.30b
Given the half-reduction reaction:
$Fe^{3+} + 3e^- \rightarrow Fe$
$FeCl_3 + e^- \rightarrow Fe + 3Cl^-$
Calculate the Ksp of FeCl3 at 298 K
Q7.31
Given the following reaction at pH 7.0:
$CH_3COOH + CO_{2(g)} + 2H^+ + 2e^- \rightarrow CH_3COCOOH + H_2O$
1. Write down the reaction that produces $H^+$ ion in term of Eo
2. Calculate the concentration of $CH_3COCOOH$ at 25oC. Given the emf of the cell is -0.70 V.
Q7.31
A certain reaction in solution has provides H+ ions and has an Eº of .48V. If a calomel electrode is placed in solution (as part of a pH meter) and has an Eref of .11 V. Find the pH of the solution.
S7.31
$pH = \dfrac{E - E_{ref}}{0.0591V}$
$pH = \dfrac{.37V}{0.0591V}$
$pH = 6.26$
Q7.32a
If a since and iron block were to be exposed for an extended period of time, which would rust first?
S7.32a
Zinc is a more electropositive metal with a reduction potential of -0.762 V while iron has a reduction potential of -0.447 V. Since Zinc is more electropositive, it will be more likely to be oxidized so it will rust first.
Q7.32b
Aluminum is used widely to be a better protective metal than lead because it forms a corrosion resistance layer in most of environments. Explain this using Standard Reduction Potentials values?
Q7.34
The $\Delta rS^o$ of the Daniell cell is found to be $-26.8JK^{-1}mol^{-1}$. Given this value at $95^oC$ calculate the emf of the cell and the temperature coefficient.
S7.34
Step 1: (Hint: $\Delta rS^o$ is directly related to temperature coefficient.)
$\Delta rS^o=vF(\dfrac{ ∂ E^o}{\ ∂})_P$
$\Delta rS^o=vF(\dfrac{∂ E^o}{∂T})_P=\dfrac{\Delta rS^o}{vF}=\dfrac{-26.8JK^{-1}\,mol^{-1}}{2(96500C\,mol^{-1})}=-1.38860\times10^{-4}vK^{-1}$
Step 2:
$\dfrac{∂ E^o}{∂})_P=\dfrac{E^o_{369.15K}-E^o_{298.15K}}{368.15-298.15K}=\dfrac{E^o_{368.15K-1.104V}}{70K}=1.39\times10^{-4}VK^{-1}$
$E^o_{268.15K}=1.094V$
Q7.36
Calculate the equilibrium constant for the following reaction:
$FeCl_2\rightleftharpoons Fe^{2+} + 2Cl^-$
(Hint: look up the reduction potential for the half reactions)
S7.36
Half reactions are\begin{align*} Fe\rightarrow &Fe^{2+}+2e^-\ &\varepsilon ^\circ=0.44V (this-is-oxidation)\ Cl_2+ 2e^-\rightarrow &2Cl^- \ &\varepsilon ^\circ = 1.36V (this-is-reduction)\end{align*}
$\varepsilon ^\circ=\varepsilon ^\circ_{cathode}-\varepsilon ^\circ_{anode}=1.36V-0.44V=0.92V$
$K=e^\dfrac{vF\varepsilon ^\circ}{RT}=e^\dfrac{(96500C/mol)(0.92V)}{(8.314J/K*mol)(298.15K)}=3.58*10^{15}$
S7.36
\begin{align*}Anode:2K\rightarrow &2K^+ +2e^-\ &\varepsilon ^\circ =5.84V\ Cathode:2H^+2e^-\rightarrow &H_2\ &\varepsilon ^\circ=0V\end{align*}
\varepsilon ^\circ=\varepsilon ^\circ_{cathode}-\varepsilon ^\circ_{anode}=-5.84V\ \Delta _r\bar{G}^\circ=-vF\varepsilon ^\circ=2\Delta _f\bar{G}^\circ[K^+]+\Delta _f\bar{G}^\circ[H_2]-2\Delta _f\bar{G}^\circ[K]-2\Delta _f\bar{G}^\circ[H^+]\ \begin{align*}\Delta _f\bar{G}^\circ[K^+]&=-\dfrac{1}{2}vF\varepsilon ^\circ-\dfrac{1}{2}\Delta _f\bar{G}^\circ[H_2]+\Delta _f\bar{G}^\circ[K]+\Delta _f\bar{G}^\circ[H^+]\ &=-\dfrac{1}{2}(2)(96500C/mol)(-5.84V)-\dfrac{1}{2}(0kJ/mol)+(0kJ/mol)+(0kJ/mol)=5.6*10^5J/mol\end{align*}
Q7.36
Do you expect a high Kw value at 276 K or 373 K? Explain.
S7.36
We expect a higher Kw value for 373 K.
Q7.38a
What is the the reaction for a solution with Au3++3e- →Au and Ag++e- →Ag
S7.38a
The first thing we do is look at the standard reduction potentials table. We see that the reduction potential is higher for Au3+ so that will be the one that gets reduced. This means we reverse the reaction for Ag+.
$Au^{3+}+3e^- →Au$
3x(Ag→Ag++e-) we need the charges to add up to zero so we multiply it by 3 so there will be 3 electrons here as well
We end up with:
$Au^{3+}+3Ag → Au + 3Ag^+$
Q7.38b
The magnitudes of the standard electrode potentials of two metals, A and B, are
A2+ + 2e- → A |E°| = 0.63 V
B2+ + 2e- →B |E°| = 1.24 V
Where the | | notation denotes that only the magnitude (but not the sign) of the E° value is shown. When the half-cells of A and B are connected, electrons flow from A to B. When A is connected to a SHE, electrons flow from A to SHE. (a) What are the signs for each of the E°? (b) What is the standard emf of a cell made up of A and B?
Q7.38c
The electrochemical properties of a pair of newly discovered metals, Q and G, are being studied in a lab. Their associated reduction reactions are shown below. A lab technician collecting the data makes a mistake, and does not record the sign of the standard potential for the half-cells in their notes. However, he remembers that when he connected the cell containing Q to the standard hydrogen electrode, the Q electrode was oxidized, and when he connected G to the standard hydrogen electrode, it was reduced. What are the signs of the standard potentials for the half-cells? When the two half-cells are connected, what is the electromotive force of the complete Q/G cell?
$Q+ + e- -> Q= 0.38 V$
$G+ + e- -> G = 0.72 V$
Q7.39
Consider a galvanic (voltaic) cell with silver wire suspended in 0.5M Ag+ solution and a cobalt wire in 1.0M Co2+ solution. Write out the half cell reactions and overall reaction to find the cell's equilibrium constant and observed emf.
Hint: Look at examples of setting up half reactions
S7.39
Half cell reactions:
$Cathode: \, \, \, \, \, \, \, \, \,2[Ag^{+}(aq) + e^{-} \rightarrow Ag(s)]$
$Anode: \, \, \, \, \, \, \, \, \,Co(s) \rightarrow Co^{2+}(aq) + 2e^{-}$
$Overall: \, \, \, \, \, \, \, \,2Ag^{+}(aq) + Co(s) \rightarrow 2Ag(s) + Co^{2+}(aq)$
$E^{\circ} = 0.800 \,V - (-.277 \,V) = +1.077 \,V$
$E^{\circ} = \frac{RT}{vF}lnK$
$+1.077 \,V = \frac{8.314 \frac{J}{K \cdot mol}(298K)}{(2)(26485) \frac{C}{mol}}lnK$
${\color{red} K = 2.51 \times 10^{36}}$
$E = E^{\circ} - \frac{RT}{v} ln \frac{[Co^{2+}]}{[Ag^{+}]} = +1.077V - \frac{0.0257V}{2}ln\frac{1.0M}{0.5M}$
${\color{red} E = +1.068 \,V }$
Q7.40a
Answer the following questions about aluminum:
1. Given that the standard electrical potential of Al3+ is -1.662, why does aluminum tarnish in air?
2. Would the the following reaction be spontaneous? $2 Al + 3 Br_2 \rightarrow Al_2Br_6$
3. Predict the reaction between aluminum and chloride gas.
Hint: look up values for Eand appropriate half reactions.
S7.40a
(a) Silver tarnishes in air because the reduction potential for oxygen is high enough that it oxidizes aluminum.
\begin{align} & E_{cell}^{\circ}=E_{cathode}^{\circ}-E_{anode}^{\circ} \ & E_{cell}^{\circ}=E_{oxygen}^{\circ}-E_{aluminum}^{\circ} \ & E_{cell}^{\circ}=1.23 \ V \ - \left ( -1.662 \ V \right ) > 0 \end{align}
Since E > 0, then rG is less than zero (since rG = -vFE). Thus the reaction is spontaneous and aluminum will be oxidized (tarnish) in contact with air.
(b) Find the appropriate half reactions to fulfill the equation.
\begin{align} 2 Al + 3 Br_2 & \rightarrow Al_2Br_6\ \left ( 1 \right ) \ Al^{3^{+}} + 3 e^{-} & \rightarrow Al, \ where \ E_{Al}^{\circ}=-1.662 \ Br_2 + 2 e^{-}& \rightarrow 2Br^{-}, \ where \ E_{Br}^{\circ}=+1.087 \ \end{align}
Multiply both sides by the appropriate integer to fulfill equation (1).
\begin{align} -2 \ times \ Al^{3^{+}} + 3 e^{-} & \rightarrow Al, \ where \ E_{Al}^{\circ}=-1.662 \ 3 \ times \ Br_2 + 2 e^{-}& \rightarrow 2Br^{-}, \ where \ E_{Br}^{\circ}=+1.087 \end{align}
Switching around the half reactions you should achieve the following, which matches up to the equation (1).
$2Al + 3Br_2 + 6e^{-} \rightarrow 2Al^{3^{+}} + 6Br^{-} + 6e^{-}$
Use the equation to find the E. Keep in mind cathodes are reduced in the half reaction (you want the electrons to be on the left hand side of the half reaction after balancing) and anodes are oxidized in the half reaction (electrons on the right). Also note how standard reduction potential is an intrinsic property so there is no need to multiply by molar coefficients represented in the balanced equation.
\begin{align} E_{cell}^{\circ}&=E_{cathode}^{\circ}-E_{anode}^{\circ} \ E_{cell}^{\circ}&=E_{bromide}^{\circ}- E_{aluminum}^{\circ} \ E_{cell}^{\circ}&= \left ( 1.087 \ V \right ) - \left ( -1.662 \ V \right ) > 0 \ E_{cell}^{\circ}&=2.749 \ V \ \end{align}
Since E > 0, then rG is less than zero (since rG = -vFE). Thus the reaction is spontaneous.
(c)2Al + 3Cl2 → 2AlCl3 since Al3+ is aluminum's most stable form and Cl-. Then it's a matter of balancing the equation.
Check here for additional information on Redox Chemistry.
Q7.40b
Using the electrochemical principle to explain how the reduction potential for oxygen is not adequate to oxidize the gold metal.
Q7.40c
Answer the following questions for the half-reaction below and E°[Pt|O2,H2O] = 1.229 V.
Fe2+(aq) + e- → Fe(s) E° = -0.447 V
(a) Predict whether or not iron will rust in air.
(b) Will the following half-reaction have a positive E°? (E°[Pt|Fe3+,Fe2+] = 0.771 V)
Fe(s) → Fe3+(aq) + 3e-
(c) Predict the reaction between iron and bromine gas (E°[Br2] = 1.087 V)
Q7.40d
The standard reduction potential for Platinum, Pt, is as follows:
$Pt^{2+} + 2e^{-} \rightarrow Pt_{(s)}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}= +1.2 V$
1. Platinum is often used as a metal in fine jewelry because it does not tarnish noticeably. Why is this the case?
2. Based on the above standard reduction potential, will the following reaction be spontaneous,
$Pt^{2+}+Cl^{2-}\rightarrow PtCl_{4}^{2-}$
given that
$PtCl_{4}^{2-}+ 2e^{-}\rightarrow Pt_{(s)}+4Cl^{-}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}= +0.73 V$
c.) Gold is also used as a metal in fine jewelry. Based on your answer to part (a.) and the standard reduction potential for gold Au3+, Eo= 1.69, does gold tarnish?
S7.40d
a.) Platinum does not tarnish, or oxidize, very quickly in air since its reduction potential is only slightly lower than that of oxygen. Thus, any oxidation which does occur does so very slowly.
$O_{2}+4H^{+}+4e^{-}\rightarrow H_{2}O\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}= +1.229 V$
b.) If the reaction has a positive standard reduction potential, then it is spontaneous. The reaction can be broken up into two half reactions whose values are provided.
$Pt_{(s)}+4Cl^{-}\rightarrow PtCl_{4}^{2-}+ 2e^{-}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}= -0.73 V$
$\, \, \, \, \, \, Pt^{2+} + 2e^{-} \rightarrow Pt_{(s)}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}= +1.2 V$
$Pt^{2+}+Cl^{2-}\rightarrow PtCl_{4}^{2-}$
so the standard reduction potential is:
$E^{o}=-0.73V+1.2V=+0.47V$
Thus, the reaction is spontaneous.
c.) No. Since the standard reduction potential of gold is greater than that of oxygen, oxygen cannot spontaneously react with gold and oxidize it.
Q7.41
In figure 7.1 we observe a voltaic cell. Here the anode is oxidized (loss of electrons) and the cathode is reduced (gains electrons). The Zinc metal loses electron and Zn (s) becomes Zn 2+ ions. Explain how the Cu2+ ions become solid Copper.
Q7.42a
In the following reaction at 298K,
$Ca_{(s)}+2H^+ \rightarrow \leftarrow Ca_{(aq)}+H_{2(g)}$
Calculate the pressure of $H_2$ in bar to maintain equilibrium. The concentration of $[Ca^{2+}]=0.026M and the pH of the solution is buffered at 1.60 pH. S7.42a Step1: $Anode: Ca^{+2}\rightarrow Ca^{+2}+2e^-$ $Cathode: 2H^++2e^-\rightarrow H_2$ $E^o=0V-(-0.34)=0.34$ Step 2: EMF calculation $E=E^o-\dfrac{0.0257V}{v}ln\dfrac{[Ca^{+2}]P_{H_{2}}}{[H^+]^{2}}$ \(From\;the\;pH, [H^+]=10^{-1.60}=2.51\times10^{-2}M$
At E=0,
$0=0.34-\dfrac{0.0257V}{2}ln\dfrac{(0.026)P_{H{2}}}{(2.51\times10^{-2})^2}$
$P_{H_2}=7.5\times10^{9}bar$
Q7.42b
For the reaction at 300K:
$Mg(s) +2H^+(aq) \rightleftharpoons Mg^{2+} (aq) + H_2(g)$
Given pressure following the reaction is 296.07 atm, and molarity of acid used is 0.025M. Compute the molarity of the Mg2+.
Q7.44a
For the ion $Sn^{2+}$, use the electric potential table to determine the value of $\Delta f\bar G^o$.
S7.44a
$Sn^{2+}_{(aq)}+2e^{-}\rightarrow Sn_{(s)}$
$E^o=-0.137$
$Anode: Sn_{(s)}\rightarrow Sn^{2+}+2e^-$
$Cathode: 2H^{+}_{(aq)}+2e^{-}\rightarrow H_{2(g)}$
Overall Equation,
$Sn_{(s)}+2H^+_{(aq)}\rightarrow Sn^{2+}_{(aq)}+H_{2}\;\;\;E^o=0.137V$
$\Delta rG^o=-vFE^o=\Delta _f\bar{G}^o[Sn^{2+}_{(aq)}]+\Delta _f\bar{G}^o[H_{2(g)}]-\Delta _f\bar{G}^o[Sn_{(s)}]-2\Delta _f\bar{G}^o[H^+_{(aq)}]$
$-(2)(96500C\,mol^{-1})(0.137V)=\Delta _f\bar{G}^o[Sn^{2+}_{(aq)}]+0kj\,mol^{-1}-0kj\,mol^{-1}-2(0kj\,mol^{-1})$
$\Delta _f\bar{G}^o[Sn^{2+}_{(aq)}]=-2.6\times10^4J\,mol^{-1}$
Q7.44b
Calculate the $\Delta _f\bar{G}^\circ$ for $K^+$ in the following reaction: $2K+2H^+\rightarrow 2K^+ +H_2$
Q7.46a
Determine the standard emf Eo for the following reaction (at 298 K and pH 0)
I-(aq) + H+(aq) + Cr2O72- (aq) → I2(s) + Cr3+(aq) + H2O(l)
Hint: Use standard change in free energy of formation and use the relationship between Go and Eo
S7.46a
$oxidation: 6I^{-}(aq) \rightarrow 3I_{2} + 6e^{-}$
$reduction: 6e^{-} + 14H^{+}(aq) + Cr_{2}O{7}^{-2}(aq) \rightarrow 2Cr^{3+}(aq) +7H_{2}O (l)$
$Overall: 6I^{-}(aq) + 14H^{+}(aq) + Cr_{2}O{7}^{-2}(aq) \rightarrow 2Cr^{3+}(aq) +7H_{2}O +3I_{2}(s)$
$\Delta _{r}G^{\circ} = 2\Delta _{f}G^{\circ}(Cr^{3+})+3\Delta _{f}G^{\circ}(I_{2})+7\Delta _{f}G^{\circ}({H_{2}O}) - 6\Delta _{f}G^{\circ}(I^{-})-14\Delta _{f}G^{\circ}(H^{+})-\Delta _{f}G^{\circ}(Cr_{2}O_{7}\,^{2-})$
$\Delta _{r}G^{\circ} = [2(-214.2) + 3(0) + 7(-237.2) -6(51.57)-14(0)-(-1301.1)]\frac{kJ}{mol} = -1097.12 \frac{kJ}{mol}$
$E^{\circ} = -\frac{ \Delta _{r}G^{\circ}}{vF} = -\frac{-1097.12 \frac{kJ}{mol} \times \frac{1000J}{1kJ}}{(6)(96500 \frac{C}{mol})} = {\color{red} +1.89V}$
Q7.46b
Calculate the $\varepsilon ^\circ$ for the combustion of ethanol given that $\Delta _f\bar{G}^\circ_{ethanol} =-174.8kJ/mol$.
S7.46b
$Overall:C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O\ Anode: C_2H_5OH +3H_2O \rightarrow 2CO_2+12H^+ +12e^-\ Cathode: 3O_2+12H^+ +12e^- \rightarrow 6H_2O$
\begin{align*} \Delta _rG^\circ&=2\Delta _f\bar{G}^\circ[CO_2]+3\Delta _f\bar{G}^\circ[H_2O]-\Delta _f\bar{G}^\circ[C_2H_5OH]-3\Delta _f\bar{G}^\circ[O_2]\&=2(-394.4kJ/mol)+3(-237.14kJ/mol)-(-174.8kJ/mol)-3(0kJ/mol)\&=-1325.42kJ/mol\end{align*}
$\varepsilon ^\circ=-\dfrac{\Delta _rG^\circ}{vF}=-\dfrac{-1325.42*10^3J/mol}{(12)(96500C/mol)}=1.145V$
Q7.47a
Explain the biological significance of cytochromes by comparing the following half reaction in cytochrome c, a, or f vs. an electrode:
$Fe^{3^{+}} + e^{-}\rightarrow Fe^{2^{+}}$
S7.47a
For cytochrome c, E∘' = +0.254
For Pt|Fe3+ , Fe2+, E = +0.771
Since cytochrome c has a smaller reduction potential, it makes it easier for for Iron to exist in both the oxidized form Fe3+ or the reduced form Fe2+. This allows cytochromes to transfer electrons easily between other cytochromes in the electron transport chain. keep in mind that different cytochromes have slightly different reduction potentials.
See here more details about Electron Transfer Proteins.
Q7.47b
Practice
The standard reduction potential for the reduction of copper in aqueous solution is as follows:
$Cu^{2+} + e^{-} \rightarrow Cu^{+}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ}= +0.153 V$
In hemocyanin, the primary oxygen carrier in invertebrate blood, however, the standard reduction potential is:
$Cu^{2+} + e^{-} \rightarrow Cu^{+}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E^{\circ'}= +0.540 V$
Why do these reduction potentials differ?
S7.47b
The standard reduction potential is used to measure the readiness with which an atom may be reduced at standard conditions (25 oC, 1 atm, reactants at 1M). In biological half-reactions, standard reduction potentials differ because the process cannot occur at standard concentrations. Eo' occurs at pH 7, as opposed to pH 0, the reference point for electrode half reactions. To obtain a more intuitive understanding, we may use the Nernst equation:
$E^{o'}=E^{o}-\frac{RT}{\gamma F}\ln \frac{[Cu^{+}]}{[Cu^{2+}]}$
At a higher pH, more electrons are found in solution, due to the formation of Hydrogen ions. By le chatelier's principle, this pushes the above half reaction forward so that the concentration of Cu2+ is greater than that of Cu+. This makes the value produced by the natural log negative. Thus. Eo' is greater than Eo at higher pH.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/07%3A_Electrochemistry/7.E%3A_Electrochemistry_%28Exercises%29.txt
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Chemical kinetics is the study of the speed with which a chemical reaction occurs and the factors that affect this speed. This information is especially useful for determining how a reaction occurs.
09: Chemical Kinetics
In this Module, the quantitative determination of a reaction rate is demonstrated. Reaction rates can be determined over particular time intervals or at a given point in time. A rate law describes the relationship between reactant rates and reactant concentrations. Reaction rates are usually expressed as the concentration of reactant consumed or the concentration of product formed per unit time. The units are thus moles per liter per unit time, written as M/s, M/min, or M/h. To measure reaction rates, chemists initiate the reaction, measure the concentration of the reactant or product at different times as the reaction progresses, perhaps plot the concentration as a function of time on a graph, and then calculate the change in the concentration per unit time.
The progress of a simple reaction (A → B) is shown in Figure $1$; the beakers are snapshots of the composition of the solution at 10 s intervals. The number of molecules of reactant (A) and product (B) are plotted as a function of time in the graph. Each point in the graph corresponds to one beaker in Figure $1$. The reaction rate is the change in the concentration of either the reactant or the product over a period of time. The concentration of A decreases with time, while the concentration of B increases with time.
$\textrm{rate}=\dfrac{\Delta [\textrm B]}{\Delta t}=-\dfrac{\Delta [\textrm A]}{\Delta t} \label{Eq1}$
Square brackets indicate molar concentrations, and the capital Greek delta (Δ) means “change in.” Because chemists follow the convention of expressing all reaction rates as positive numbers, however, a negative sign is inserted in front of Δ[A]/Δt to convert that expression to a positive number. The reaction rate calculated for the reaction A → B using Equation $\ref{Eq1}$ is different for each interval (this is not true for every reaction, as shown below). A greater change occurs in [A] and [B] during the first 10 s interval, for example, than during the last, meaning that the reaction rate is greatest at first.
Reaction rates generally decrease with time as reactant concentrations decrease.
Study 1: The Hydrolysis of Aspirin
We can use Equation $\ref{Eq1}$ to determine the reaction rate of hydrolysis of aspirin, probably the most commonly used drug in the world (more than 25,000,000 kg are produced annually worldwide). Aspirin (acetylsalicylic acid) reacts with water (such as water in body fluids) to give salicylic acid and acetic acid, as shown in Figure $2$.
Because salicylic acid is the actual substance that relieves pain and reduces fever and inflammation, a great deal of research has focused on understanding this reaction and the factors that affect its rate. Data for the hydrolysis of a sample of aspirin are in Table $1$ and are shown in the graph in Figure $3$. These data were obtained by removing samples of the reaction mixture at the indicated times and analyzing them for the concentrations of the reactant (aspirin) and one of the products (salicylic acid).
Table $1$: Data for Aspirin Hydrolysis in Aqueous Solution at pH 7.0 and 37°C*
Time (h) [Aspirin] (M) [Salicylic Acid] (M)
*The reaction at pH 7.0 is very slow. It is much faster under acidic conditions, such as those found in the stomach.
0 5.55 × 10−3 0
2.0 5.51 × 10−3 0.040 × 10−3
5.0 5.45 × 10−3 0.10 × 10−3
10 5.35 × 10−3 0.20 × 10−3
20 5.15 × 10−3 0.40 × 10−3
30 4.96 × 10−3 0.59 × 10−3
40 4.78 × 10−3 0.77 × 10−3
50 4.61 × 10−3 0.94 × 10−3
100 3.83 × 10−3 1.72 × 10−3
200 2.64 × 10−3 2.91 × 10−3
300 1.82 × 10−3 3.73 × 10−3
The average reaction rate for a given time interval can be calculated from the concentrations of either the reactant or one of the products at the beginning of the interval (time = t0) and at the end of the interval (t1). Using salicylic acid, the reaction rate for the interval between t = 0 h and t = 2.0 h (recall that change is always calculated as final minus initial) is calculated as follows:
\begin{align}\textrm{rate}_{(t=0-2.0\textrm{ h})}&=\frac{[\textrm{salicyclic acid}]_2-[\textrm{salicyclic acid}]_0}{\textrm{2.0 h}-\textrm{0 h}} \&=\frac{0.040\times10^{-3}\textrm{ M}-0\textrm{ M}}{\textrm{2.0 h}}=2.0\times10^{-5}\textrm{ M/h} \end{align}
The reaction rate can also be calculated from the concentrations of aspirin at the beginning and the end of the same interval, remembering to insert a negative sign, because its concentration decreases:
\begin{align}\textrm{rate}_{(t=0-2.0\textrm{ h})}&=-\dfrac{[\textrm{aspirin}]_2-[\textrm{aspirin}]_0}{\mathrm{2.0\,h-0\,h}} \&=-\dfrac{(5.51\times10^{-3}\textrm{ M})-(5.55\times10^{-3}\textrm{ M})}{\textrm{2.0 h}} \&=2\times10^{-5}\textrm{ M/h}\end{align}
If the reaction rate is calculated during the last interval given in Table $1$(the interval between 200 h and 300 h after the start of the reaction), the reaction rate is significantly slower than it was during the first interval (t = 0–2.0 h):
\begin{align}\textrm{rate}_{(t=200-300\textrm{h})}&=\dfrac{[\textrm{salicyclic acid}]_{300}-[\textrm{salicyclic acid}]_{200}}{\mathrm{300\,h-200\,h}} \&=-\dfrac{(3.73\times10^{-3}\textrm{ M})-(2.91\times10^{-3}\textrm{ M})}{\textrm{100 h}} \&=8.2\times10^{-6}\textrm{ M/h}\end{align}
Study 2: The Fermentation of Sucrose
In the preceding example, the stoichiometric coefficients in the balanced chemical equation are the same for all reactants and products; that is, the reactants and products all have the coefficient 1. Consider a reaction in which the coefficients are not all the same, the fermentation of sucrose to ethanol and carbon dioxide:
$\underset{\textrm{sucrose}}{\mathrm{C_{12}H_{22}O_{11}(aq)}}+\mathrm{H_2O(l)}\rightarrow\mathrm{4C_2H_5OH(aq)}+4\mathrm{CO_2(g)} \label{Eq2}$
The coefficients indicate that the reaction produces four molecules of ethanol and four molecules of carbon dioxide for every one molecule of sucrose consumed. As before, the reaction rate can be found from the change in the concentration of any reactant or product. In this particular case, however, a chemist would probably use the concentration of either sucrose or ethanol because gases are usually measured as volumes and the volume of CO2 gas formed depends on the total volume of the solution being studied and the solubility of the gas in the solution, not just the concentration of sucrose. The coefficients in the balanced chemical equation tell us that the reaction rate at which ethanol is formed is always four times faster than the reaction rate at which sucrose is consumed:
$\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t}=-\dfrac{4\Delta[\textrm{sucrose}]}{\Delta t} \label{Eq3}$
The concentration of the reactant—in this case sucrose—decreases with time, so the value of Δ[sucrose] is negative. Consequently, a minus sign is inserted in front of Δ[sucrose] in Equation $\ref{Eq3}$ so the rate of change of the sucrose concentration is expressed as a positive value. Conversely, the ethanol concentration increases with time, so its rate of change is automatically expressed as a positive value.
Often the reaction rate is expressed in terms of the reactant or product with the smallest coefficient in the balanced chemical equation. The smallest coefficient in the sucrose fermentation reaction (Equation $\ref{Eq2}$) corresponds to sucrose, so the reaction rate is generally defined as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm{sucrose}]}{\Delta t}=\dfrac{1}{4}\left (\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t} \right ) \label{Eq4}$
Example $1$
Consider the thermal decomposition of gaseous N2O5 to NO2 and O2 via the following equation:
$\mathrm{2N_2O_5(g)}\xrightarrow{\,\Delta\,}\mathrm{4NO_2(g)}+\mathrm{O_2(g)}$
Write expressions for the reaction rate in terms of the rates of change in the concentrations of the reactant and each product with time.
Given: balanced chemical equation
Asked for: reaction rate expressions
Strategy:
1. Choose the species in the equation that has the smallest coefficient. Then write an expression for the rate of change of that species with time.
2. For the remaining species in the equation, use molar ratios to obtain equivalent expressions for the reaction rate.
Solution
A Because O2 has the smallest coefficient in the balanced chemical equation for the reaction, define the reaction rate as the rate of change in the concentration of O2 and write that expression.
B The balanced chemical equation shows that 2 mol of N2O5 must decompose for each 1 mol of O2 produced and that 4 mol of NO2 are produced for every 1 mol of O2 produced. The molar ratios of O2 to N2O5 and to NO2 are thus 1:2 and 1:4, respectively. This means that the rate of change of [N2O5] and [NO2] must be divided by its stoichiometric coefficient to obtain equivalent expressions for the reaction rate. For example, because NO2 is produced at four times the rate of O2, the rate of production of NO2 is divided by 4. The reaction rate expressions are as follows:
$\textrm{rate}=\dfrac{\Delta[\mathrm O_2]}{\Delta t}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t}$
Exercise $1$
The key step in the industrial production of sulfuric acid is the reaction of SO2 with O2 to produce SO3.
$2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)}$
Write expressions for the reaction rate in terms of the rate of change of the concentration of each species.
Answer
$\textrm{rate}=-\dfrac{\Delta[\mathrm O_2]}{\Delta t}=-\dfrac{\Delta[\mathrm{SO_2}]}{2\Delta t}=\dfrac{\Delta[\mathrm{SO_3}]}{2\Delta t}$
Example $2$
Using the reaction shown in Example $1$, calculate the reaction rate from the following data taken at 56°C:
$2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$
Time (s) [N2O5] (M) [NO2] (M) [O2] (M)
240 0.0388 0.0314 0.00792
600 0.0197 0.0699 0.0175
Given: balanced chemical equation and concentrations at specific times
Asked for: reaction rate
Strategy:
1. Using the equations in Example $1$, subtract the initial concentration of a species from its final concentration and substitute that value into the equation for that species.
2. Substitute the value for the time interval into the equation. Make sure your units are consistent.
Solution
A Calculate the reaction rate in the interval between t1 = 240 s and t2 = 600 s. From Example $1$, the reaction rate can be evaluated using any of three expressions:
$\textrm{rate}=\dfrac{\Delta[\mathrm O_2]}{\Delta t}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t}$
Subtracting the initial concentration from the final concentration of N2O5 and inserting the corresponding time interval into the rate expression for N2O5,
$\textrm{rate}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t}=-\dfrac{[\mathrm{N_2O_5}]_{600}-[\mathrm{N_2O_5}]_{240}}{2(600\textrm{ s}-240\textrm{ s})}$
B Substituting actual values into the expression,
$\textrm{rate}=-\dfrac{\mathrm{\mathrm{0.0197\;M-0.0388\;M}}}{2(360\textrm{ s})}=2.65\times10^{-5} \textrm{ M/s}$
Similarly, NO2 can be used to calculate the reaction rate:
$\textrm{rate}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=\dfrac{[\mathrm{NO_2}]_{600}-[\mathrm{NO_2}]_{240}}{4(\mathrm{600\;s-240\;s})}=\dfrac{\mathrm{0.0699\;M-0.0314\;M}}{4(\mathrm{360\;s})}=\mathrm{2.67\times10^{-5}\;M/s}$
Allowing for experimental error, this is the same rate obtained using the data for N2O5. The data for O2 can also be used:
$\textrm{rate}=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=\dfrac{[\mathrm{O_2}]_{600}-[\mathrm{O_2}]_{240}}{\mathrm{600\;s-240\;s}}=\dfrac{\mathrm{0.0175\;M-0.00792\;M}}{\mathrm{360\;s}}=\mathrm{2.66\times10^{-5}\;M/s}$
Again, this is the same value obtained from the N2O5 and NO2 data. Thus, the reaction rate does not depend on which reactant or product is used to measure it.
Exercise $2$
Using the data in the following table, calculate the reaction rate of SO2(g) with O2(g) to give SO3(g).
$2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)}$
Time (s) [SO2] (M) [O2] (M) [SO3] (M)
300 0.0270 0.0500 0.0072
720 0.0194 0.0462 0.0148
Answer 9.0 × 10−6 M/s
Instantaneous Rates of Reaction
The instantaneous rate of a reaction is the reaction rate at any given point in time. As the period of time used to calculate an average rate of a reaction becomes shorter and shorter, the average rate approaches the instantaneous rate. Comparing this to calculus, the instantaneous rate of a reaction at a given time corresponds to the slope of a line tangent to the concentration-versus-time curve at that point—that is, the derivative of concentration with respect to time.
$rate = \lim_{\Delta t \rightarrow 0} \dfrac{-\Delta [R]}{\Delta t} = - \dfrac{d[R]}{dt}$
The distinction between the instantaneous and average rates of a reaction is similar to the distinction between the actual speed of a car at any given time on a trip and the average speed of the car for the entire trip. Although the car may travel for an extended period at 65 mph on an interstate highway during a long trip, there may be times when it travels only 25 mph in construction zones or 0 mph if you stop for meals or gas. The average speed on the trip may be only only 50 mph, whereas the instantaneous speed on the interstate at a given moment may be 65 mph. Whether the car can be stopped in time to avoid an accident depends on its instantaneous speed, not its average speed. There are important differences between the speed of a car during a trip and the speed of a chemical reaction, however. The speed of a car may vary unpredictably over the length of a trip, and the initial part of a trip is often one of the slowest. In a chemical reaction, the initial interval typically has the fastest rate (though this is not always the case), and the reaction rate generally changes smoothly over time.
Chemical kinetics generally focuses on one particular instantaneous rate, which is the initial reaction rate, t = 0. Initial rates are determined by measuring the reaction rate at various times and then extrapolating a plot of rate versus time to t = 0.
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The kinetic theory of gases can be used to model the frequency of collisions between hard-sphere molecules, which is proportional to the reaction rate. Most systems undergoing a chemical reaction, however, are much more complex. The reaction rates may be dependent on specific interactions between reactant molecules, the phase(s) in which the reaction takes place, etc. The field of chemical kinetics is thus by-and-large based on empirical observations. From experimental observations, scientists have established that reaction rates almost always have a power-law dependence on the concentrations of one or more of the reactants. In the following sections, we will discuss different power laws that are commonly observed in chemical reactions.
$0^{th}$ Order Reaction Kinetics
Consider a closed container initially filled with chemical species $A$. At $t = 0$, a stimulus, such as a change in temperature, the addition of a catalyst, or irradiation, causes an irreversible chemical reaction to occur in which $A$ transforms into product $B$:
$a \text{A} \longrightarrow b \text{B}$
The rate that the reaction proceeds, $r$, can be described as the change in the concentrations of the chemical species with respect to time:
$r = -\dfrac{1}{a} \dfrac{d \left[ \text{A} \right]}{dt} = \dfrac{1}{b} \dfrac{d \left[ \text{B} \right]}{dt} \label{19.1}$
where $\left[ \text{X} \right]$ denotes the molar concentration of chemical species $\text{X}$ with units of $\frac{\text{mol}}{\text{L}^3}$.
Let us first examine a reaction $\text{A} \longrightarrow \text{B}$ in which the reaction rate, $r$, is constant with time:
$r = -\dfrac{d \left[ \text{A} \right]}{dt} = k \label{19.2}$
where $k$ is a constant, also known as the rate constant with units of $\dfrac{\text{mol}}{\text{m}^3 \text{s}}$. Such reactions are called zeroth order reactions because the reaction rate depends on the concentrations of species $\text{A}$ and $\text{B}$ to the $0^{th}$ power. Integrating $\left[ \text{A} \right]$ with respect to $T$, we find that
$\left[ \text{A} \right] = -kt + c_1 \label{19.3}$
At $t = 0$, $\left[ \text{A} \right] \left( 0 \right) = \left[ \text{A} \right]_0$. Plugging these values into the equation, we find that $c_1 = \left[ \text{A} \right]_0$. The final form of the equation is:
$\left[ \text{A} \right] = \left[ \text{A} \right]_0 - kt \label{19.4}$
A plot of the concentration of species $\text{A}$ with time for a $0^{th}$ order reaction is shown in Figure $1$, where the slope of the line is $-k$ and the $y$-intercept is $\left[ \text{A} \right]_0$. Such reactions in which the reaction rates are independent of the concentrations of products and reactants are rare in nature. An example of a system displaying $0^{th}$ order kinetics would be one in which a reaction is mediated by a catalyst present in small amounts.
$1^{st}$ Order Reaction Kinetics
Experimentally, it is observed than when a chemical reaction is of the form
$\sum_i \nu_i \text{A}_i = 0 \label{19.5}$
the reaction rate can be expressed as
$r = k \prod_\text{reactants} \left[ \text{A}_i \right]^{\nu_i} \label{19.6}$
where it is assumed that the stoichiometric coefficients $\nu_i$ of the reactants are all positive. Thus, or a reaction $\text{A} \longrightarrow \text{B}$, the reaction rate depends on $\left[ \text{A} \right]$ raised to the first power:
$r = \dfrac{d \left[ \text{A} \right]}{dt} = -k \left[ \text{A} \right] \label{19.7}$
For first order reactions, $k$ has the units of $\dfrac{1}{\text{s}}$. Integrating and applying the condition that at $t = 0 \: \text{s}$, $\left[ \text{A} \right] = \left[ \text{A} \right]_0$, we arrive at the following equation:
$\left[ \text{A} \right] = \left[ \text{A} \right] e^{-kt} \label{19.8}$
Figure $2$ displays the concentration profiles for species $\text{A}$ and $\text{B}$ for a first order reaction. To determine the value of $k$ from
experimental data, it is convenient to take the natural log of Equation $\ref{19.8}$:
$\text{ln} \left( \left[ \text{A} \right] \right) = \text{ln} \left( \left[ \text{A} \right]_0 \right) - kt \label{19.9}$
For a first order irreversible reaction, a plot of $\text{ln} \left( \left[ \text{A} \right] \right)$ vs. $t$ is straight line with a slope of $-k$ and a $y$-intercept of $\text{ln} \left( \left[ \text{A} \right]_0 \right)$.
$2^{nd}$ Order Reaction Kinetics
Another type of reaction depends on the square of the concentration of species $\text{A}$ - these are known as second order reactions. For a second order reaction in which $2 \text{A} \longrightarrow \text{B}$, we can write the reaction rate to be
$r = -\dfrac{1}{2} \dfrac{d \left[ \text{A} \right]}{dt} = k \left[ \text{A} \right]^2 \label{19.10}$
For second order reactions, $k$ has the units of $\dfrac{\text{m}^3}{\text{mol} \cdot \text{s}}$. Integrating and applying the condition that at $t = 0 \: \text{s}$, $\left[ \text{A} \right] = \left[ \text{A} \right]_0$, we arrive at the following equation for the concentration of $\text{A}$ over time:
$\left[ \text{A} \right] = \dfrac{1}{2kt + \dfrac{1}{\left[ \text{A} \right]_0}} \label{19.11}$
Figure $3$ shows concentration profiles of $\text{A}$ and $\text{B}$ for a second order reaction. To determine $k$ from experimental data for
second-order reactions, it is convenient to invert Equation $\ref{19.11}$:
$\dfrac{1}{\left[ \text{A} \right]} = \dfrac{1}{\left[ \text{A} \right]_0} + 2kt \label{19.12}$
A plot of $1/\left[ \text{A} \right]$ vs. $t$ will give rise to a straight line with slope $k$ and intercept $1/\left[ \text{A} \right]_0$.
Second order reaction rates can also apply to reactions in which two species react with each other to form a product:
$\text{A} + \text{B} \overset{k}{\longrightarrow} \text{C}$
In this scenario, the reaction rate will depend on the concentrations of both $\text{A}$ and $\text{B}$ to the first order:
$r = -\dfrac{d \left[ \text{A} \right]}{dt} = -\dfrac{d \left[ \text{B} \right]}{dt} = k \left[ \text{A} \right] \left[ \text{B} \right] \label{19.13}$
To integrate the above equation, we need to write it in terms of one variable. Since the concentrations of $\text{A}$ and $\text{B}$ are related to each other via the chemical reaction equation, we can write:
$\left[ \text{B} \right] = \left[ \text{B} \right]_0 - \left( \left[ \text{A} \right]_0 - \left[ \text{A} \right] \right) = \left[ \text{A} \right] + \left[ \text{B} \right]_0 - \left[ \text{A} \right]_0 \label{19.14}$
$\dfrac{d \left[ \text{A} \right]}{dt} = -k \left[ \text{A} \right] \left( \left[ \text{A} \right] + \left[ \text{B} \right]_0 - \left[ \text{A} \right]_0 \right) \label{19.15}$
We can then use partial fractions to integrate:
$kdt = \dfrac{d \left[ \text{A} \right]}{\left[ \text{A} \right] \left( \left[ \text{A} \right] + \left[ \text{B} \right]_0 - \left[ \text{A} \right]_0 \right)} = \dfrac{1}{\left[ \text{B} \right]_0 - \left[ \text{A} \right]_0} \left( \dfrac{d \left[ \text{A} \right]}{\left[ \text{A} \right]} - \dfrac{d \left[ \text{A} \right]}{\left[ \text{B} \right]_0 - \left[ \text{A} \right]_0 + \left[ \text{A} \right]} \right) \label{19.16}$
$kt = \dfrac{1}{\left[ \text{A} \right]_0 - \left[ \text{B} \right]_0} \text{ln} \dfrac{\left[ \text{A} \right] \left[ \text{B} \right]_0}{\left[ \text{B} \right] \left[ \text{A} \right]_0} \label{19.17}$
Figure $4$ displays the concentration profiles of species $\text{A}$, $\text{B}$, and $\text{C}$ for a second order reaction in which the initial concentrations of $\text{A}$ and $\text{B}$ are not equal.
9.03: Molecularity of a Reaction
In general, it is necessary to experimentally measure the concentrations of species over time in order to determine the apparent rate law governing the reaction. If the reactions are elementary reactions, (i.e. they cannot be expressed as a series of simpler reactions), then we can directly define the rate law based on the chemical equation. For example, an elementary reaction in which a single reactant transforms into a single product, is unimolecular reaction. These reactions follow $1^{st}$ order rate kinetics. An example of this type of reaction would be the isomerization of butane:
$nC_4 H_{10} \longrightarrow iC_4 H_{10}$
From the chemical reaction equation, we can directly write the rate law as
$\dfrac{d \left[ nC_4 H_{10} \right]}{dt} = -k \left[ nC_4 H_{10} \right] \label{19.18}$
without the need to carry out experiments.
Elementary bimolecular reactions that involve two molecules interacting to form one or more products follow second order rate kinetics. An example would be the following reaction between a nitrate molecule and carbon monoxide to form nitrogen dioxide and carbon dioxide:
$NO_3 + CO \longrightarrow NO_2 + CO_2$
For the above elementary reaction, we can directly write the rate law as:
$\dfrac{d \left[ NO_3 \right]}{dt} = -k \left[ NO_3 \right] \left[ CO \right] \label{19.19}$
Trimolecular elementary reactions involving three reactant molecules to form one or more products are rare due to the low probability of three molecules simultaneously colliding with one another.
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A major goal in chemical kinetics is to determine the sequence of elementary reactions, or the reaction mechanism, that comprise complex reactions. For example, Sherwood Rowland and Mario Molina won the Nobel Prize in Chemistry in 1995 for proposing the elementary reactions involving chlorine radicals that contribute to the overall reaction of $O_3 \rightarrow O_2$ in the troposphere. In the following sections, we will derive rate laws for complex reaction mechanisms, including reversible, parallel and consecutive reactions.
Parallel Reactions
Consider the reaction in which chemical species $\text{A}\;$ undergoes one of two irreversible first order reactions to form either species $\text{B}\;$ or species $\text{C}\;$:
\begin{align} \text{A} &\overset{k_1}{\rightarrow} \text{B} \ \text{A} &\overset{k_2}{\rightarrow} \text{C} \end{align}
The overall reaction rate for the consumption of $\text{A}$ can be written as:
$\dfrac{d \left[ \text{A} \right]}{dt} = -k_1 \left[ \text{A} \right] - k_2 \left[ \text{A} \right] = - \left( k_1 + k_2 \right) \left[ \text{A} \right] \label{21.1}$
Integrating $\left[ \text{A} \right]$ with respect to $t$, we obtain the following equation:
$\left[ \text{A} \right] = \left[ \text{A} \right]_0 e^{-\left( k_1 + k_2 \right) t} \label{21.2}$
Plugging this expression into the equation for $\dfrac{d \left[ \text{B} \right]}{dt}$, we obtain:
$\dfrac{d \left[ \text{B} \right]}{dt} = k_1 \left[ \text{A} \right] = k_1 \left[ \text{A} \right]_0 e^{- \left( k_1 + k_2 \right) t} \label{21.3}$
Integrating $\left[ \text{B} \right]$ with respect to $t$, we obtain:
$\left[ \text{B} \right] = -\dfrac{k_1 \left[ \text{A} \right]_0}{k_1 + k_2} \left( e^{-\left( k_1 + k_2 \right) t} \right) + c_1 \label{21.4}$
At $t = 0$, $\left[ \text{B} \right] = 0$. Therefore,
$c_1 = \dfrac{k_1 \left[ \text{A} \right]_0}{k_1 + k_2} \label{21.5}$
$\left[ \text{B} \right] = \dfrac{k_1 \left[ \text{A} \right]_0}{k_1 + k_2} \left( 1 - e^{-\left( k_1 + k_2 \right) t} \right) \label{21.6}$
Likewise,
$\left[ \text{C} \right] = \dfrac{k_2 \left[ \text{A} \right]_0}{k_1 + k_2} \left( 1 - e^{-\left( k_1 + k_2 \right) t} \right) \label{21.7}$
The ratio of $\left[ \text{B} \right]$ to $\left[ \text{C} \right]$ is simply:
$\dfrac{\left[ \text{B} \right]}{\left[ \text{C} \right]} = \dfrac{k_1}{k_2} \label{21.8}$
An important parallel reaction in industry occurs in the production of ethylene oxide, a reagent in many chemical processes and also a major component in explosives. Ethylene oxide is formed through the partial oxidation of ethylene:
$2 \: C_2 H_4 + O_2 \overset{k_1}{\longrightarrow} 2 \: C_2 H_4 O$
However, ethylene can also undergo a combustion reaction:
$C_2 H_4 + 3 \: O_2 \overset{k_2}{\longrightarrow} 2 \: CO_2 + 2 \: H_2 O$
To select for the first reaction, the oxidation of ethylene takes place in the presence of a silver catalyst, which significantly increases $k_1$ compared to $k_2$. Figure $1$ displays the concentration profiles for species $\text{A}$, $\text{B}$, and $\text{C}$ in a parallel reaction in which $k_1 > k_2$.
Consecutive Reactions
Consider the following series of first-order irreversible reactions, where species $\text{A}$ reacts to form an intermediate species, $\text{I}$, which then reacts to form the product, $\text{P}$:
$\text{A} \overset{k_1}{\longrightarrow} \text{I} \overset{k_2}{\longrightarrow} \text{P}$
We can write the reaction rates of species $\text{A}$, $\text{I}$ and $\text{P}$ as follows:
$\dfrac{d \left[ \text{A} \right]}{dt} = -k_1 \left[ \text{A} \right] \label{21.9}$
$\dfrac{d \left[ \text{I} \right]}{dt} = k_1 \left[ \text{A} \right] - k_2 \left[ \text{I} \right] \label{21.10}$
$\dfrac{d \left[ \text{P} \right]}{dt} = k_2 \left[ \text{I} \right] \label{21.11}$
As before, integrating $\left[ \text{A} \right]$ with respect to $t$ leads to:
$\left[ \text{A} \right] = \left[ \text{A} \right]_0 e^{-k_1 t} \label{21.12}$
The concentration of species $\text{I}$ can be written as
$\left[ \text{I} \right] = \dfrac{k_1 \left[ \text{A} \right]_0}{k_2 - k_1} \left( e^{-k_1 t} - e^{-k_2 t} \right) \label{21.13}$
Then, solving for $\left[ \text{P} \right]$, we find that:
$\left[ \text{P} \right] = \left[ \text{A} \right]_0 \left[ 1 + \dfrac{1}{k_1 - k_2} \left( k_2 e^{-k_1 t} - k_1 e^{-k_2 t} \right) \right] \label{21.14}$
Figure $2$ displays the concentration profiles for species $\text{A}$, $\text{I}$, and $\text{P}$ in a consecutive reaction in which $k_1 = k_2$. As can be seen from the figure, the concentration of species $\text{I}$ reaches a maximum at some time, $t_\text{max}$. Oftentimes, species $\text{I}$ is the desired product. Returning to the oxidation of ethylene into ethylene oxide, it is important to note another reaction in which ethylene oxide can decompose into carbon dioxide and water through the following reaction
$C_2 H_4 O + \dfrac{5}{2} \: O_2 \overset{k_3}{\longrightarrow} 2 \: CO_2 + 2 \: H_2 O$
Thus, to maximize the concentration of ethylene oxide, the oxidation of ethylene is only allowed proceed to partial completion before the reaction is stopped.
Finally, in the limiting case when $k_2 \gg k_1$, we can write the concentration of $\text{P}$ as
$\left[ \text{P} \right] \approx \left[ \text{A} \right]_0 \left\{ 1 + \dfrac{1}{-k_2} k_2 e^{-k_1 t} \right\} = \left[ \text{A} \right]_0 \left( 1 - e^{-k_1 t} \right) \label{21.15}$
Thus, when $k_2 \gg k_1$, the reaction can be approximated as $\text{A} \rightarrow \text{P}$ and the apparent rate law follows $1^{st}$ order kinetics.
Consecutive Reactions With an Equilibrium
Consider the reactions
$\text{A} \overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \text{I} \overset{k_2}{\rightarrow} \text{P}$
We can write the reaction rates as:
$\dfrac{d \left[ \text{A} \right]}{dt} = -k_1 \left[ \text{A} \right] + k_{-1} \left[ \text{I} \right] \label{21.16}$
$\dfrac{d \left[ \text{I} \right]}{dt} = k_1 \left[ \text{A} \right] - k_{-1} \left[ \text{I} \right] - k_2 \left[ \text{I} \right] \label{21.17}$
$\dfrac{ d \left[ \text{P} \right]}{dt} = k_2 \left[ \text{I} \right] \label{21.18}$
The exact solutions of these is straightforward, in principle, but rather involved, so we will just state the exact solutions, which are
\begin{align} \left[ \text{A} \right] \left( t \right) &= \dfrac{ \left[ \text{A} \right]_0}{2 \lambda} \left[ \left( \lambda - k_1 + K \right) e^{-\left( k_1 + K - \lambda \right) t/2} + \left( \lambda + k_1 - K \right) e^{-\left( k_1 + K + \lambda \right) t/2} \right] \ \left[ \text{I} \right] \left( t \right) &= \dfrac{ k_1 \left[ \text{A} \right]_0}{\lambda} \left[ e^{- \left( k_1 + K - \lambda \right) t/2} - e^{-\left( k_1 + K + \lambda \right) t/2} \right] \ \left[ \text{P} \right] \left( t \right) &= 2 k_1 k_2 \left[ \text{A} \right]_0 \left[ \dfrac{2}{\left( k_1 + K \right)^2 - \lambda^2} - \dfrac{1}{\lambda} \left( \dfrac{ e^{-\left( k_1 + K - \lambda \right) t/2}}{k_1 + K - \lambda} - \dfrac{e^{-\left( k_1 + K + \lambda \right) t/2}}{k_1 + K + \lambda} \right) \right] \end{align} \label{21.19}
where
\begin{align} K &= k_2 + k_{-1} \ \lambda &= \sqrt{\left( k_1 - K \right)^2 - 4k_1 k_{-1}} \end{align} \label{21.20}
Steady-State Approximations
Consider the following consecutive reaction in which the first step is reversible:
$\text{A} \overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \text{I} \overset{k_2}{\rightarrow} \text{P}$
We can write the reaction rates as:
$\dfrac{d \left[ \text{A} \right]}{dt} = -k_1 \left[ \text{A} \right] + k_{-1} \left[ \text{I} \right] \label{21.21}$
$\dfrac{ d \left[ \text{I} \right]}{dt} = k_1 \left[ \text{A} \right] - k_{-1} \left[ \text{I} \right] - k_2 \left[ \text{I} \right] \label{21.22}$
$\dfrac{d \left[ \text{P} \right]}{dt} = k_2 \left[ \text{I} \right] \label{21.23}$
These equations can be solved explicitly in terms of $\left[ \text{A} \right]$, $\left[ \text{I} \right]$, and $\left[ \text{P} \right]$, but the math becomes very complicated quickly. If, however, $k_2 + k_{-1} \gg k_1$ (in other words, the rate of consumption of $\text{I}$ is much faster than the rate of production of $\text{I}$), we can make the approximation that the concentration of the intermediate species, $\text{I}$, is small and constant with time:
$\dfrac{d \left[ \text{I} \right]}{dt} \approx 0 \label{21.24}$
Equation 21.22 can now be written as
$\dfrac{d \left[ \text{I} \right]}{dt} = k_1 \left[ \text{A} \right] - k_{-1} \left[ \text{I} \right]_{ss} - k_2 \left[ \text{I} \right]_{ss} \approx 0 \label{21.25}$
where $\left[ \text{I} \right]_{ss}$ is a constant represents the steady state concentration of intermediate species, $\left[ \text{I} \right]$. Solving for $\left[ \text{I} \right]_{ss}$,
$\left[ \text{I} \right]_{ss} = \dfrac{k_1}{k_{-1} + k_2} \left[ \text{A} \right] \label{21.26}$
We can then write the rate equation for species $\text{A}$ as
$\dfrac{d \left[ \text{A} \right]}{dt} = -k_1 \left[ \text{A} \right] + k_{-1} \left[ \text{I} \right]_{ss} = -k_1 \left[ \text{A} \right] + k_{-1} \dfrac{k_1}{k_{-1} + k_2} \left[ \text{A} \right] = -\dfrac{k_1 k_2}{k_{-1} + k_2} \left[ \text{A} \right] \label{21.27}$
Integrating,
$\left[ \text{A} \right] = \left[ \text{A} \right]_0 e^{-\dfrac{k_1 k_2}{k_{-1} + k_2} t} \label{21.28}$
Equation 21.28 is the same equation we would obtain for apparent $1^{st}$ order kinetics of the following reaction:
$\text{A} \overset{k'}{\longrightarrow} \text{P}$
where
$k' = \dfrac{k_1 k_2}{k_{-1} + k_2} \label{21.29}$
Figure $3$ displays the concentration profiles for species, $\text{A}$, $\text{I}$, and $\text{P}$ with the condition that $k_2 + k_{-1} \gg k_1$. These types of reaction kinetics appear when the intermediate species, $\text{I}$, is highly reactive.
Lindemann Mechanism
Consider the isomerization of methylisonitrile gas, $CH_3 NC$, to acetonitrile gas, $CH_3 CN$:
$CH_3 NC \overset{k}{\longrightarrow} CH_3 CN$
If the isomerization is a unimolecular elementary reaction, we should expect to see $1^{st}$ order rate kinetics. Experimentally, however, $1^{st}$ order rate kinetics are only observed at high pressures. At low pressures, the reaction kinetics follow a $2^{nd}$ order rate law:
$\dfrac{d \left[ CH_3 NC \right]}{dt} = -k \left[ CH_3 NC \right]^2 \label{21.30}$
To explain this observation, J.A. Christiansen and F.A. Lindemann proposed that gas molecules first need to be energized via intermolecular collisions before undergoing an isomerization reaction. The reaction mechanism can be expressed as the following two elementary reactions
\begin{align} \text{A} + \text{M} &\overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \text{A}^* + \text{M} \ \text{A}^* &\overset{k_2}{\rightarrow} \text{B} \end{align}
where $\text{M}$ can be a reactant molecule, a product molecule or another inert molecule present in the reactor. Assuming that the concentration of $\text{A}^*$ is small, or $k_1 \ll k_2 + k_{-1}$, we can use a steady-state approximation to solve for the concentration profile of species $\text{B}$ with time:
$\dfrac{d \left[ \text{A}^* \right]}{dt} = k_1 \left[ \text{A} \right] \left[ \text{M} \right] - k_{-1} \left[ \text{A}^* \right]_{ss} \left[ \text{M} \right] - k_2 \left[ \text{A}^* \right]_{ss} \approx 0 \label{21.31}$
Solving for $\left[ \text{A}^* \right]$,
$\left[ \text{A}^* \right] = \dfrac{k_1 \left[ \text{M} \right] \left[ \text{A} \right]}{k_2 + k_{-1} \left[ \text{M} \right]} \label{21.32}$
The reaction rates of species $\text{A}$ and $\text{B}$ can be written as
$-\dfrac{d \left[ \text{A} \right]}{dt} = \dfrac{d \left[ \text{B} \right]}{dt} = k_2 \left[ \text{A}^* \right] = \dfrac{k_1 k_2 \left[ \text{M} \right] \left[ \text{A} \right]}{k_2 + k_{-1} \left[ \text{M} \right]} = k_\text{obs} \left[ \text{A} \right] \label{21.33}$
where
$k_\text{obs} = \dfrac{k_1 k_2 \left[ \text{M} \right]}{k_2 + k_{-1} \left[ \text{M} \right]} \label{21.34}$
At high pressures, we can expect collisions to occur frequently, such that $k_{-1} \left[ \text{M} \right] \gg k_2$. Equation 21.33 then becomes
$-\dfrac{d \left[ \text{A} \right]}{dt} = \dfrac{k_1 k_2}{k_{-1}} \left[ \text{A} \right] \label{21.35}$
which follows $1^{st}$ order rate kinetics.
At low pressures, we can expect collisions to occurs infrequently, such that $k_{-1} \left[ \text{M} \right] \ll k_2$. In this scenario, equation 21.33 becomes
$-\dfrac{d \left[ \text{A} \right]}{dt} = k_1 \left[ \text{A} \right] \left[ \text{M} \right] \label{21.36}$
which follows second order rate kinetics, consistent with experimental observations.
Equilibrium Approximations
Consider again the following consecutive reaction in which the first step is reversible:
$\text{A} \overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \text{I} \overset{k_2}{\rightarrow} \text{P}$
Now let us consider the situation in which $k_2 \ll k_1$ and $k_{-1}$. In other words, the conversion of $\text{I}$ to $\text{P}$ is slow and is the rate-limiting step. In this situation, we can assume that $\left[ \text{A} \right]$ and $\left[ \text{I} \right]$ are in equilibrium with each other. As we derived before for a reversible reaction in equilibrium,
$K_\text{eq} = \dfrac{k_1}{k_{-1}} \approx \dfrac{ \left[ \text{I} \right]}{\left[ \text{A} \right]} \label{21.37}$
or, in terms of $\left[ \text{I} \right]$,
$\left[ \text{I} \right] = K_\text{eq} \left[ \text{A} \right] \label{21.38}$
These conditions also result from the exact solution when we set $k_2 \approx 0$. When this is done, we have the approximate expressions from the exact solution:
\begin{align} K &\approx k_{-1} \ \lambda &\approx \sqrt{ \left( k_1 - k_{-1} \right)^2 + 4k_1 k_{-1}} = \sqrt{k_1^2 + 2k_1 k_{-1} + k_{-1}^2} = k_1 + k_{-1} \ \lambda - k_1 + K &\approx k_1 + k_{-1} + k_1 - k_{-1} = 2k_1 \ \lambda + k_1 - K &\approx k_1 + k_{-1} + k_1 - k_{-1} = 2k_1 \ k_1 + K - \lambda &\approx k_1 + k_{-1} - k_1 - k_{-1} = 0 \ k_1 + K + \lambda &\approx k_1 + k_{-1} + k_1 + k_{-1} = 2 \left( k_1 + k_{-1} \right) \end{align} \label{21.39}
and the approximate solutions become
\begin{align} \left[ \text{A} \right] \left( t \right) &= \dfrac{ \left[ \text{A} \right]_0}{2 \left( k_1 + k_{-1} \right)} \left[ 2k_{-1} + 2k_1 e^{-\left( k_1 + k_{-1} \right) t} \right] \ \left[ \text{I} \right] \left( t \right) &= \dfrac{k_1 \left[ \text{A} \right]_0}{\left( k_1 + k_{-1} \right)} \left[ 1 - e^{-\left( k_1 + k_{-1} \right) t} \right] \end{align} \label{21.40}
In the long-time limit, when equilibrium is reached and transient behavior has decayed away, we find
$\dfrac{ \left[ \text{I} \right]}{\left[ \text{A} \right]} \equiv K_\text{eq} \rightarrow \dfrac{k_1}{k_{-1}} \label{21.41}$
Plugging the above equation into the expression for $d \left[ \text{P} \right]/dt$,
$\dfrac{d \left[ \text{P} \right]}{dt} = k_2 \left[ \text{I} \right] = k_2 K_\text{eq} \left[ \text{A} \right] = \dfrac{k_1 k_2}{k_{-1}} \left[ \text{A} \right] \label{21.42}$
The reaction can thus be approximated as a $1^{st}$ order reaction
$\text{A} \overset{k'}{\longrightarrow} \text{P}$
with
$k' = \dfrac{k_1 k_2}{k_{-1}} \label{21.43}$
Figure $4$ displays the concentration profiles for species, $\text{A}$, $\text{I}$, and $\text{P}$ with the condition that $k_2 \ll k_1 = k_{-1}$. When $k_1 = k_{-1}$, we expect $\left[ \text{A} \right] = \left[ \text{I} \right]$. As can be seen from the figure, after a short initial startup time, the concentrations of species $\text{A}$ and $\text{I}$ are approximately equal during the reaction.
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When molecules collide, the kinetic energy of the molecules can be used to stretch, bend, and ultimately break bonds, leading to chemical reactions. If molecules move too slowly with little kinetic energy, or collide with improper orientation, they do not react and simply bounce off each other. However, if the molecules are moving fast enough with a proper collision orientation, such that the kinetic energy upon collision is greater than the minimum energy barrier, then a reaction occurs. The minimum energy requirement that must be met for a chemical reaction to occur is called the activation energy, $E_a$.
The reaction pathway is similar to what happens in Figure $1$. To get to the other end of the road, an object must roll with enough speed to completely roll over the hill of a certain height. The faster the object moves, the more kinetic energy it has. If the object moves too slowly, it does not have enough kinetic energy necessary to overcome the barrier; as a result, it eventually rolls back down. In the same way, there is a minimum amount of energy needed in order for molecules to break existing bonds during a chemical reaction. If the kinetic energy of the molecules upon collision is greater than this minimum energy, then bond breaking and forming occur, forming a new product (provided that the molecules collide with the proper orientation).
The activation energy ($E_a$), labeled $\Delta{G^{\ddagger}}$ in Figure $2$, is the energy difference between the reactants and the activated complex, also known as transition state. In a chemical reaction, the transition state is defined as the highest-energy state of the system. If the molecules in the reactants collide with enough kinetic energy and this energy is higher than the transition state energy, then the reaction occurs and products form. In other words, the higher the activation energy, the harder it is for a reaction to occur and vice versa.
Overcoming the energy barrier from thermal energy involves addressing the fraction of the molecules that possess enough kinetic energy to react at a given temperature. According to kinetic molecular theory, a population of molecules at a given temperature is distributed over a variety of kinetic energies that is described by the Maxwell-Boltzman distribution law.
The two distribution plots shown here are for a lower temperature T1 and a higher temperature T2. The area under each curve represents the total number of molecules whose energies fall within particular range. The shaded regions indicate the number of molecules which are sufficiently energetic to meet the requirements dictated by the two values of Ea that are shown. It is clear from these plots that the fraction of molecules whose kinetic energy exceeds the activation energy increases quite rapidly as the temperature is raised. This the reason that virtually all chemical reactions (and all elementary reactions) proceed more rapidly at higher temperatures.
Arrhenius Equation
By 1890 it was common knowledge that higher temperatures speed up reactions, often doubling the rate for a 10-degree rise, but the reasons for this were not clear. Finally, in 1899, the Swedish chemist Svante Arrhenius (1859-1927) combined the concepts of activation energy and the Boltzmann distribution law into one of the most important relationships in physical chemistry:
Take a moment to focus on the meaning of this equation, neglecting the A factor for the time being. First, note that this is another form of the exponential decay law discussed in the previous section of this series. What is "decaying" here is not the concentration of a reactant as a function of time, but the magnitude of the rate constant as a function of the exponent –Ea /RT. And what is the significance of this quantity? Recalling that RT is the average kinetic energy, it becomes apparent that the exponent is just the ratio of the activation energy Ea to the average kinetic energy. The larger this ratio, the smaller the rate (hence the negative sign). This means that high temperature and low activation energy favor larger rate constants, and thus speed up the reaction. Because these terms occur in an exponent, their effects on the rate are quite substantial.
Svante August Arrhenius
Svante August Arrhenius (19 February 1859 – 2 October 1927) was a Swedish scientist, originally a physicist, but often referred to as a chemist, and one of the founders of the science of physical chemistry. He received the Nobel Prize for Chemistry in 1903, becoming the first Swedish Nobel laureate, and in 1905 became director of the Nobel Institute where he remained until his death. The Arrhenius equation, Arrhenius definition of an acid, lunar crater Arrhenius, the mountain of Arrheniusfjellet and the Arrhenius Labs at Stockholm University are named after him. Today, Arrhenius is best known for his study published in 1896, on the greenhouse effect.
The two plots in Figure $4$ show the effects of the activation energy (denoted here by E) on the rate constant. Even a modest activation energy of 50 kJ/mol reduces the rate by a factor of 108.
Looking at the role of temperature, a similar effect is observed. (If the x-axis were in "kilodegrees" the slopes would be more comparable in magnitude with those of the kilojoule plot at the above right.)
Determining the activation energy
The Arrhenius equation
$k = A e^{-E_a/RT} \label{1}$
can be written in a non-exponential form that is often more convenient to use and to interpret graphically (Figure $4$). Taking the logarithms of both sides and separating the exponential and pre-exponential terms yields
$\ln k = \ln \left(Ae^{-E_a/RT} \right) = \ln A + \ln \left(e^{-E_a/RT}\right) \label{2}$
$\ln k = \ln A + \dfrac{-E_a}{RT} = \left(\dfrac{-E_a}{R}\right) \left(\dfrac{1}{T}\right) + \ln A \label{3}$
which is the equation of a straight line whose slope is $–E_a /R$. This affords a simple way of determining the activation energy from values of k observed at different temperatures, by plotting $\ln k$ as a function of $1/T$.
Example $1$: Isomerization of Cyclopropane
For the isomerization of cyclopropane to propene,
the following data were obtained (calculated values shaded in pink):
T, °C 477 523 577 623
1/T, K–1 × 103 1.33 1.25 1.18 1.11
k, s–1 0.00018 0.0027 0.030 0.26
ln k –8.62 –5.92 –3.51 –1.35
From the calculated slope, we have
– (Ea/R) = –3.27 × 104 K
Ea=– (8.314 J mol–1 K–1) (–3.27 × 104 K) = 273 kJ mol–1
This activation energy is high, which is not surprising because a carbon-carbon bond must be broken in order to open the cyclopropane ring. (C–C bond energies are typically around 350 kJ/mol.) This is why the reaction must be carried out at high temperature.
Calculating $E_a$ without a plot
Because the ln k vs.-1/T plot yields a straight line, it is often convenient to estimate the activation energy from experiments at only two temperatures. The ln A term is eliminated by subtracting the expressions for the two ln-k terms via the following steps:
$\ln k_{1}=\ln A - \dfrac{E_{a}}{k_{B}T_1}$
at $T_1$ and
$\ln k_{2}=\ln A - \dfrac{E_{a}}{k_{B}T_2}$
at $T_2$. By rewriting the second equation:
$\ln A = \ln k_{2} + \dfrac{E_{a}}{k_{B}T_2}$
and substitute for $\ln A$ into the first equation:
$\ln k_{1}= \ln k_{2} + \dfrac{E_{a}}{k_{B}T_2} - \dfrac{E_{a}}{k_{B}T_1}$
This simplifies to:
$\ln k_{1} - \ln k_{2} = -\dfrac{E_{a}}{k_{B}T_1} + \dfrac{E_{a}}{k_{B}T_2}$
$\ln \dfrac{k_{1}}{k_{2}} = -\dfrac{E_{a}}{k_{B}} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$
Example $2$
A widely used rule-of-thumb for the temperature dependence of a reaction rate is that a ten degree rise in the temperature approximately doubles the rate. This is not generally true, especially when a strong covalent bond must be broken. For a reaction that does show this behavior, what would the activation energy be?
Solution
Center the ten degree interval at 300 K. Substituting into the above expression yields
$E_a = \dfrac{(8.314)(\ln 2/1)}{\dfrac{1}{295} – \dfrac{1}{305}} = \dfrac{(8.314)(0.693)}{0.00339 K^{-1} – 0.00328 \; K^{-1}}$
= (5.76 J mol–1 K–1) / (0.00011 K–1) = 52400 J mol–1 = 52.4 kJ mol–1
Example $3$
It takes about 3.0 minutes to cook a hard-boiled egg in Los Angeles, but at the higher altitude of Denver, where water boils at 92°C, the cooking time is 4.5 minutes. Use this information to estimate the activation energy for the coagulation of egg albumin protein.
Solution
The ratio of the rate constants at the elevations of Los Angeles and Denver is 4.5/3.0 = 1.5, and the respective temperatures are $373 \; \rm{K }$ and $365\; \rm{K}$. With the subscripts 2 and 1 referring to Los Angeles and Denver respectively:
$E_a = \dfrac{(8.314)(\ln 1.5)}{\dfrac{1}{365\; \rm{K}} – \dfrac{1}{373 \; \rm{K}}} = \dfrac{(8.314)(0.405)}{0.00274 \; \rm{K^{-1}} – 0.00268 \; \rm{K^{-1}}}$
$\ = \dfrac{(3.37\; \rm{J\; mol^{–1} K^{–1}})}{5.87 \times 10^{-5}\; \rm{K^{–1}}} = 5740\; \rm{ J\; mol^{–1}} = 5.73 \; \rm{kJ \;mol^{–1}}$
Comment: This low value seems reasonable because thermal denaturation of proteins primarily involves the disruption of relatively weak hydrogen bonds; no covalent bonds are broken (although disulfide bonds can interfere with this interpretation).
The pre-exponential factor
Up to this point, the pre-exponential term, A in the Arrhenius equation, has been ignored because it is not directly involved in relating temperature and activation energy, which is the main practical use of the equation.
However, because A multiplies the exponential term, its value clearly contributes to the value of the rate constant and thus of the rate. Recall that the exponential part of the Arrhenius equation expresses the fraction of reactant molecules that possess enough kinetic energy to react, as governed by the Maxwell-Boltzmann law. This fraction can run from zero to nearly unity, depending on the magnitudes of $E_a$ and of the temperature.
If this fraction were 0, the Arrhenius law would reduce to
$k = A$
In other words, $A$ is the fraction of molecules that would react if either the activation energy were zero, or if the kinetic energy of all molecules exceeded $E_a$ — admittedly, an uncommon scenario (although barrierless reactions have been characterized).
The Role of Collisions
What would limit the rate constant if there were no activation energy requirements? The most obvious factor would be the rate at which reactant molecules come into contact. This can be calculated from kinetic molecular theory and is known as the frequency- or collision factor, $Z$.
In some reactions, the relative orientation of the molecules at the point of collision is important, so a geometrical or steric factor (commonly denoted by $\rho$ (Greek lower case rho) can be defined. In general, we can express $A$ as the product of these two factors:
$A=Z\rho$
Values of ρ are generally very difficult to assess; they are sometime estimated by comparing the observed rate constant with the one in which $A$ is assumed to be the same as $Z$. Usually, the more complex the reactant molecules, the lower the steric factors. The deviation from unity can have different causes: the molecules are not spherical, so different geometries are possible; not all the kinetic energy is delivered into the right spot; the presence of a solvent (when applied to solutions) and other factors (Figure $4$).
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A potential energy surface (PES) describes the potential energy of a system, especially a collection of atoms, in terms of certain parameters, normally the positions of the atoms. The surface might define the energy as a function of one or more coordinates; if there is only one coordinate, the surface is called a potential energy curve or energy profile. It is helpful to use the analogy of a landscape: for a system with two degrees of freedom (e.g. two bond lengths), the value of the energy (analogy: the height of the land) is a function of two bond lengths (analogy: the coordinates of the position on the ground). The Potential Energy Surface represents the concepts that each geometry (both external and internal) of the atoms of the molecules in a chemical reaction is associated with it a unique potential energy. This creates a smooth energy “landscape” and chemistry can be viewed from a topology perspective (of particles evolving over "valleys""and passes").
Potential Energy Curves (1-D Potential Energy Surfaces)
The PES is the energy of a molecule as a function of the positions of its nuclei $r$. This energy of a system of two atoms depends on the distance between them. At large distances the energy is zero, meaning “no interaction”. At distances of several atomic diameters attractive forces dominate, whereas at very close approaches the force is repulsive, causing the energy to rise. The attractive and repulsive effects are balanced at the minimum point in the curve. Plots that illustrate this relationship are quite useful in defining certain properties of a chemical bond.
The internuclear distance at which the potential energy minimum occurs defines the bond length. This is more correctly known as the equilibrium bond length, because thermal motion causes the two atoms to vibrate about this distance. In general, the stronger the bond, the smaller will be the bond length.
Attractive forces operate between all atoms, but unless the potential energy minimum is at least of the order of RT, the two atoms will not be able to withstand the disruptive influence of thermal energy long enough to result in an identifiable molecule. Thus we can say that a chemical bond exists between the two atoms in H2. The weak attraction between argon atoms does not allow Ar2 to exist as a molecule, but it does give rise to the van Der Waals force that holds argon atoms together in its liquid and solid forms.
Potential, Kinetic, and Total Energy for a System
Potential energy and kinetic energy Quantum theory tells us that an electron in an atom possesses kinetic energy $K$ as well as potential energy $V$, so the total energy $E$ is always the sum of the two: $E = V + K$. The relation between them is surprisingly simple: $K = –0.5 V$. This means that when a chemical bond forms (an exothermic process with $ΔE < 0$), the decrease in potential energy is accompanied by an increase in the kinetic energy (embodied in the momentum of the bonding electrons), but the magnitude of the latter change is only half as much, so the change in potential energy always dominates. The bond energy $–ΔE$ has half the magnitude of the fall in potential energy.
Mathematical definition and computation
The geometry of a set of atoms can be described by a vector, r, whose elements represent the atom positions. The vector $r$ could be the set of the Cartesian coordinates of the atoms, or could also be a set of inter-atomic distances and angles. Given $r$, the energy as a function of the positions, $V(r)$, is the value of $V(r)$ for all values of $r$ of interest. Using the landscape analogy from the introduction, $V(r)$ gives the height on the "energy landscape" so that the concept of a potential energy surface arises. An example is the PES for water molecule (Figure $1$) that show the energy minimum corresponding to optimized molecular structure for water- O-H bond length of 0.0958 nm and H-O-H bond angle of 104.5°
Figure $2$: PES for water molecule: Shows the energy minimum corresponding to optimized molecular structure for water- O-H bond length of 0.0958nm and H-O-H bond angle of 104.5°. of Wikipedia (Credit: Aimnature).
The Dimensionality of a Potential Energy Surface
To define an atom’s location in 3-dimensional space requires three coordinates (e.g., $x$, $y$,and $z$ or $r$, $\theta$ and $phi$ in Cartesian and Spherical coordinates) or degrees of freedom. However, a reaction and hence the corresponding PESs do not depend of the absolute position of the reaction, only the relative positions (internal degrees). Hence both translation and rotation of the entire system can be removed (each with 3 degree of freedom, assuming non-linear geometries). So the dimensionality of a PES is
$3N-6$
where $N$ is the number of atoms involves in the reaction, i.e., the number of atoms in each reactants). The PES is a hypersurface with many degrees of freedom and typically only a few are plotted at any one time for understanding. See Calculate Number of Vibrational Modes to get a more details picture of how this applies to calculating the number of vibrations in a molecule
To study a chemical reaction using the PES as a function of atomic positions, it is necessary to calculate the energy for every atomic arrangement of interest. Methods of calculating the energy of a particular atomic arrangement of atoms are well described in the computational chemistry article, and the emphasis here will be on finding approximations of $(V(r)$ to yield fine-grained energy-position information.
For very simple chemical systems or when simplifying approximations are made about inter-atomic interactions, it is sometimes possible to use an analytically derived expression for the energy as a function of the atomic positions. An example is
$H + H_2 \rightarrow H_2 + H$
system as a function of the three H-H distances. For more complicated systems, calculation of the energy of a particular arrangement of atoms is often too computationally expensive for large scale representations of the surface to be feasible.
Application of Potential Energy Surfaces
A PES is a conceptual tool for aiding the analysis of molecular geometry and chemical reaction dynamics. Once the necessary points are evaluated on a PES, the points can be classified according to the first and second derivatives of the energy with respect to position, which respectively are the gradient and the curvature. Stationary points (or points with a zero gradient) have physical meaning: energy minima correspond to physically stable chemical species and saddle points correspond to transition states, the highest energy point on the reaction coordinate (which is the lowest energy pathway connecting a chemical reactant to a chemical product). Three
• PES do not show kinetic energy, only potential energy.
• At T = 0 K (no KE), species will want to be at the lowest possible potential energy, (i.e., at a minimum on the PES).
• Between any two minima (valley bottoms) the lowest energy path will pass through a maximum at a saddle point, which we call that saddle point a transition-state structure.
The PES concept finds application in fields such as chemistry and physics, especially in the theoretical sub-branches of these subjects. It can be used to theoretically explore properties of structures composed of atoms, for example, finding the minimum energy shape of a molecule or computing the rates of a chemical reaction.
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The macroscopic discussion of kinetics discussed in previous sections can be now expanded into a more microscopic picture in terms of molecular level properties (e..g, mass and velocities) involving two important theories: (1) collision theory and (2) transition-state theory.
Collision Theory
If two molecules need to collide in order for a reaction to take place, then factors that influence the ease of collisions will be important. The more energy there is available to the molecules, the faster they will move around, and the more likely they are to bump into each other. Higher temperatures ought to lead to more collisions and a greater frequency of reactions between molecules. In the drawing below, the cold, sluggish molecules on the left are not likely to collide, but the energetic molecules on the right are due to collide at any time.
The rate at which molecules collide which is the frequency of collisions is called the collision frequency, $Z$, which has units of collisions per unit of time. Given a container of molecules $A$ and $B$, the collision frequency between $A$ and $B$ is defined by:
$Z=N_{A}N_{B}\sigma_{AB}\sqrt{\dfrac{8k_{B}T}{\pi\mu_{AB}}}$
where:
• NA and NB are the numbers of molecules A and B, and is directly related to the concentrations of A and B.
• The mean speed of molecules obtained from the Maxwell-Boltzmann distribution for thermalized gases
$\sqrt{\dfrac{8k_{B}T}{\pi\mu_{AB}}}$
• $\sigma_{AB}$ is the averaged sum of the collision cross-sections of molecules A and B. The collision cross section represents the collision region presented by one molecule to another.
• $\mu$ is the reduced mass and is given by
$\mu = \dfrac{m_\text{A} m_\text{B}}{m_\text{A} + m_\text{B}}$
The concepts of collision frequency can be applied in the laboratory: (1) The temperature of the environment affects the average speed of molecules. Thus, reactions are heated to increase the reaction rate. (2) The initial concentration of reactants is directly proportional to the collision frequency; increasing the initial concentration will speed up the reaction.
For a successful collision to occur, the reactant molecules must collide with enough kinetic energy to break original bonds and form new bonds to become the product molecules. This energy is called the activation energy for the reaction; it is also often referred to as the energy barrier.
The fraction of collisions with enough energy to overcome the activation barrier is given by:
$f = e^{\frac{-E_a}{RT}}$
where:
• $f$ is the fraction of collisions with enough energy to react
• $E_a$ is the activation energy
The fraction of successful collisions is directly proportional to the temperature and inversely proportional to the activation energy.
The fraction of successful collisions is directly proportional to the temperature and inversely proportional to the activation energy.
The more complicated the structures of the reactants, the more likely that the value of the rate constant will depend on the trajectories at which the reactants approach each other. This kind of electrophilic addition reaction is well-known to all students of organic chemistry. Consider the addition of a hydrogen halide such as HCl to the double bond of an alkene, converting it to a chloroalkane.
Experiments have shown that the reaction only takes place when the HCl molecule approaches the alkene with its hydrogen-end, and in a direction that is approximately perpendicular to the double bond, as shown at below.
The reason for this becomes apparent when we recall that HCl is highly polar owing to the high electronegativity of chlorine, so that the hydrogen end of the molecule is slightly positive. The steric factor, $\rho$ is then introduced to represent is the probability of the reactant molecules colliding with the right orientation and positioning to achieve a product with the desirable geometry and stereospecificity. Values of $\rho$ are generally very difficult to assess and range from 0 to 1, but are sometime estimated by comparing the observed rate constant with the one in which the preexponential constant $A$ is assumed to be the same as $Z$.
The lesson you should take from this example is that once you start combining a variety of chemical principles, you gradually develop what might be called "chemical intuition" which you can apply to a wide variety of problems. This is far more important than memorizing specific examples.
All Three Factors Combined
The rate constant of the gas-phase reaction is proportional to the product of the collision frequency and the fraction of successful reactions. As stated above, sufficient kinetic energy is required for a successful reaction; however, they must also collide properly. Compare the following equation to the Arrhenius equation:
$k = Z\rho{e^{\frac{-E_a}{RT}}}$
where
• k is the rate constant for the reaction
• ρ is the steric factor.
• is the pre-exponential factor, A, of the Arrhenius equation. It is the frequency of total collisions that collide with the right orientation. In practice, it is the pre-exponential factor that is directly determined by experiment and then used to calculate the steric factor.
• Ea is activation energy
• T is absolute temperature
• R is gas constant.
Although the collision theory deals with gas-phase reactions, its concepts can also be applied to reactions that take place in solvents; however, the properties of the solvents (for example: solvent cage) will affect the rate of reactions. Ultimately, collision theory illustrates how reactions occur; it can be used to approximate the rate constants of reactions, and its concepts can be directly applied in the laboratory. Read this for a more detailed discussion of Collision Theory.
Transition-State Theory
Transition state theory (TST) provides a more accurate alternative to the previously used Arrhenius equation and the collision theory. The transition state theory attempts to provide a greater understanding of activation energy, $E_a$, and the thermodynamic properties involving the transition state. Collision theory of reaction rate, although intuitive, lacks an accurate method to predict the probability factor for the reaction. The theory assumes that reactants are hard spheres rather than molecules with specific structures. In 1935, Henry Eyring helped develop a new theory called the transition state theory to provide a more accurate alternative to the previously used Arrhenius equation and the collision theory. The Eyring equation involves the statistical frequency factory, v, which is fundamental to the theory.
According to TST, between the state where molecules are reactants and the state where molecules are products, there is a state known as the transition state. In the transition state, the reactants are combined in a species called the activated complex. The theory suggests that there are three major factors that determine whether a reaction will occur:
1. The concentration of the activated complex
2. The rate at which the activated complex breaks apart
3. The way in which the activated complex breaks apart: whether it breaks apart to reform the reactants or whether it breaks apart to form a new complex, the products.
Collision theory proposes that not all reactants that combine undergo a reaction. However, assuming the stipulations of the collision theory are met and a successful collision occurs between the molecules, transition state theory allows one of two outcomes: a return to the reactants, or a rearranging of bonds to form the products.
Consider a bimolecular reaction:
$A~+B~\rightarrow~C \label{2}$
$K = \dfrac{[C]}{[A][B]} \label{3}$
where $K$ is the equilibrium constant. In the transition state model, the activated complex AB is formed:
$A~+~B~\rightleftharpoons ~AB^\ddagger~\rightarrow ~C \label{4}$
$K^\ddagger=\dfrac{[AB]^\ddagger}{[A][B]} \label{5}$
There is an energy barrier, called activation energy, in the reaction pathway. A certain amount of energy is required for the reaction to occur. The transition state, $AB^\ddagger$, is formed at maximum energy. This high-energy complex represents an unstable intermediate. Once the energy barrier is overcome, the reaction is able to proceed and product formation occurs.
The rate of a reaction is equal to the number of activated complexes decomposing to form products. Hence, it is the concentration of the high-energy complex multiplied by the frequency of it surmounting the barrier.
$\begin{eqnarray} rate~&=&~v[AB^\ddagger] \label{6} \ &=&~v[A][B]K^\ddagger \label{7} \end{eqnarray}$
The rate can be rewritten:
$rate~=~k[A][B] \label{8}$
Combining Equations $\ref{8}$ and $\ref{7}$ gives:
$\begin{eqnarray} k[A][B]~&=&~v[A][B]K^\ddagger \label{9} \ k~&=&~vK^\ddagger \label{10} \end{eqnarray}$
where
• $v$ is the frequency of vibration,
• $k$ is the rate constant and
• $K ^\ddagger$ is the thermodynamic equilibrium constant.
Statistical mechanics (not shown) provides that the frequency, v, is equivalent to the thermal energy, kBT, divided by Planck's constant, h.
$v~=~\dfrac{k_BT}{h} \label{11}$
where
• $k_B$ is the Boltzmann's constant (1.381 x 10-23 J/K),
• $T$ is the absolute temperature in Kelvin (K) and
• $h$ is Planck's constant (6.626 x 10-34 Js).
Substituting Equation $\ref{11}$ into Equation $\ref{10}$ :
$k~=~\dfrac{k_BT}{h}K^\ddagger \label{12}$
Equation ${ref}$ is often tagged with another term $(M^{1-m})$ that makes the units equal with $M$ is the molarity and $m$ is the molecularly of the reaction.
$k~=~\dfrac{k_BT}{h}K^\ddagger (M^{1-m}) \label{E12}$
It is important to note here that the equilibrium constant $K^\ddagger$ can be calculated by absolute, fundamental properties such as bond length, atomic mass, and vibration frequency. This gives the transition rate theory the alternative name absolute rate theory, because the rate constant, k, can be calculated from fundamental properties.
Thermodynamics of Transition State Theory
To reveal the thermodynamics of the theory, $K^\ddagger$ must be expressed in terms of $\Delta {G^{\ddagger}}$. $\Delta {G^{\ddagger}}$ is simply,
$\Delta{G}^{o\ddagger} = G^o (transitionstate) - G^o (reactants)$
By definition, at equilibrium, $\Delta {G^{\ddagger}}$ can be expressed as:
$\Delta {G^{\ddagger}} = -RT \ln K^{\ddagger}$
Rearrangement gives:
$[K]^\ddagger = e^{ -\frac{\Delta{G}^\ddagger}{RT} }$
From Equation $\ref{E12}$
$\color{red} k = ve^{ -\frac{\Delta{G}^{\ddagger}}{RT}} (M^{1-m})$
It is also possible to obtain terms for the change in enthalpy and entropy for the transition state. Because
$\Delta{G^\ddagger} = \Delta{H^\ddagger} - T\Delta{S^\ddagger}$
it follows that the derived equation becomes,
$k = \dfrac{k_BT}{h} e^{\Delta{S}^\ddagger /R} e^{-\Delta H^\ddagger/RT} M^{1-m} \label{Eyring}$
Equation $\ref{Eyring}$ is known as the Eyring Equation and was developed by Henry Eyring in 1935, is based on transition state theory and is used to describe the relationship between reaction rate and temperature. It is similar to the Arrhenius Equation, which also describes the temperature dependence of reaction rates.
The linear form of the Eyring Equation is given below:
$\ln{\dfrac{k}{T}}~=~\dfrac{-\Delta H^\dagger}{R}\dfrac{1}{T}~+~\ln{\dfrac{k_B}{h}}~+~\dfrac{\Delta S^\ddagger}{R} \label{17}$
The values for $\Delta H^\ddagger$ and $\Delta S^\ddagger$ can be determined from kinetic data obtained from a $\ln{\dfrac{k}{T}}$ vs. $\dfrac{1}{T}$ plot. The Equation is a straight line with negative slope, $\dfrac{-\Delta H^\ddagger}{R}$, and a y-intercept, $\dfrac{\Delta S^\ddagger}{R}+\ln{\dfrac{k_B}{h}}$.
Conclusion
In this article, the complete thermodynamic formulation of the transition state theory was derived. This equation is more reliable than either the Arrhenius equation and the equation for the Collision Theory. However, it has its limitations, especially when considering the concepts of quantum mechanics. Quantum mechanics implies that tunneling can occur, such that particles can bypass the energy barrier created by the transition state. This can especially occur with low activation energies, because the probability of tunneling increases when the barrier height is lowered.
In addition, transition state theory assumes that an equilibrium exists between the reactants and the transition state phase. However, in solution non-equilibrium situations can arise, upsetting the theory. Several more complex theories have been presented to correct for these and other discrepancies. This theory still remains largely useful in calculating the thermodynamic properties of the transition state from the overall reaction rate. This presents immense usefulness in medicinal chemistry, in which the study of transition state analogs is widely implemented.
Contributors and Attributions
• Phong Dao, Bilal Latif, Lu Zhao (UC Davis)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/09%3A_Chemical_Kinetics/9.07%3A_Theories_of_Reaction_Rates.txt
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The kinetic isotope effect (KIE) is a phenomenon associated with isotopically substituted molecules exhibiting different reaction rates. Isotope effects such as KIEs are invaluable tools in both physical and biological sciences and are used to aid in the understanding of reaction kinetics, mechanisms, and solvent effects.
Introduction
Research was first introduced on this topic over 50 years ago and has grown into an enormous field. The scientists behind much of the understanding and development of kinetic isotope effects were Jacob Bigeleisen and Maria Goeppert Mayer who published the first paper on isotope effects [J. Chem. Phys., 15, 261 (1947)]. Kinetic isotope effects specifically explore the change in rate of a reaction due to isotopic substitution.
An element is identified by its symbol, mass number, and atomic number. The atomic number is the number of protons in the nucleus while the mass number is the total number of protons and neutrons in the nucleus. Isotopes are two atoms of the same element that have the same number of protons but different numbers of neutrons. Isotopes are specified by the mass number.
As an example consider the two isotopes of chlorine, you can see that their mass numbers vary, with 35Cl being the most abundant isotope, while their atomic numbers remain the same at 17.
$^{35}Cl \;\text{and}\; ^{37}Cl$
The most common isotope used in light atom isotope effects is hydrogen ($^{1}H$) commonly replaced by its isotope deuterium ($^{2}H$). Note: Hydrogen also has a third isotope, tritium ($^{2}H$). Isotopes commonly used in heavy atom isotope effects include carbon ($^{12}C$, $^{13}C$, nitrogen ($^{14}N$, $^{15}N$), oxygen, sulfur, and bromine. Not all elements exhibit reasonably stable isotopes (i.e. Fluorine, $^{19}F$), but those that due serve as powerful tools in isotope effects.
Potential Energy Surfaces
Understanding potential energy surfaces is important in order to be able to understand why and how isotope effects occur as they do. The harmonic oscillator approximation is used to explain the vibrations of a diatomic molecule. The energies resulting from the quantum mechanic solution for the harmonic oscillator help to define the internuclear potential energy of a diatomic molecule and are
$E_n = \left(n + \dfrac{1}{2}\right)h \nu \label{1}$
where
• n is a positive integer (n=1,2,3...),
• h is Planck's constant and
• $\nu$ is the frequency of vibration.
The Morse potential is an analytic expression that is used as an approximation to the intermolecular potential energy curves:
$V(l) = D_e{\left(1-e^{-\beta(l-l_o)}\right)}^2 \label{2}$
where
• $V(l)$ is the potential energy,
• $D_e$ is the dissociation energy of the molecule,
• $\beta$ is the measure of the curvature of the potential at its minimum,
• $l$ is displacement, and
• $l_o$ is the equilibrium bond length.
The $D_e$, $\beta$, and $l_o$ variables can be looked up in a textbook or CRC handbook.
Below is an example of a Morse potential curve with the zero point vibrational energies of two isotopic molecules (for example R-H and R-D where R is a group/atom that is much heavier than H or D). The y-axis is potential energy and the x axis is internuclear distance. In this figure EDo and EHo correspond to the zero point energies of deuterium and hydrogen. The zero point energy is the lowest possible energy of a system and equates to the ground state energy. Zero point energy is dependent upon the reduced mass of the molecule as will be shown in the next section. The heavier the molecule or atom, the lower the frequency of vibration and the smaller the zero point energy. Lighter molecules or atoms have a greater frequency of vibration and a higher zero point energy. We see this is the figure below where deuterium is heavier than hydrogen and therefore has the lower zero point energy.
This results in different bond dissociation energies for R-D and R-H. The bond dissociation energy for R-D (ED) is greater than the bond dissociation energy of R-H (EH). This difference in energy due to isotopic replacement results in differing rates of reaction, the effect that is measured in kinetic isotope effects. The reaction rate for the conversion of R-D is slower than the reaction rate for the conversion of R-H.
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It is important to note that isotope replacement does not change the electronic structure of the molecule or the potential energy surfaces of the reactions the molecule may undergo. Only the rate of the reaction is affected.
Activation Energies
The energy of the vibrational levels of a vibration (i.e., a bond) in a molecule is given by
$E_n = \left(n + \dfrac{1}{2}\right)h \nu \label{3}$
where we assume that the molecule is in its ground state and we can compare zero-point vibrational energies,
$E_o = \left(\dfrac{1}{2}\right)hv \label{4}$
Using the harmonic oscillator approximation the fundamental vibrational frequency is
$\nu = \dfrac{1}{2 \pi} \sqrt{ \dfrac {k}{\mu} } \label{5}$
where
• $k$ is the force constant of the bond and
• $\mu$ is the reduced mass
$\mu = \dfrac{m_1m_2}{m_1+m_2} \label{6}$
The Arrhenius equation is used to determine reaction rates and activation energies and since we are interested in the change in rate of reactions with different isotopes, this equation is very important,
$k = Ae^{-\frac{E_a}{kT}} \label{7}$
where
• $k$ is the reaction rate,
• $E_a$ is the activation energy, and
• $A$ is the Arrhenius constant.
The Arrhenius equation can be used to compare the rates of a reaction with R-H and R-D,
$k_H = A_He^{-\frac{E_a^H}{kT}} \label{8}$
$k_D = A_De^{-\frac{E_a^D}{kT}} \label{9}$
where kH and kD are the rates of reaction associated with R-H and the isotope substituted R-D. We will then assume the Arrhenius constants are equal ($A_H=A_D$). The ratio of the rates of reaction gives an approximation for the isotope effect resulting in:
$\dfrac{k_H}{k_D} = e^{-\frac{E_a^H - E_a^D}{kT}} \label{10}$
By using the relationship that for both R-H and R-D
$E_o = \left(\dfrac{1}{2}\right)h\nu \label{11}$
a substitution can be made resulting in
$\dfrac{k_H}{k_D} = e^{\frac{h(\nu_H - \nu_D)}{2kT}} \label{12}$
The vibrational frequency (Equation 5) can then be substituted for R-H and R-D and the value of the expected isotope effect can be calculated.
$\dfrac{k_H}{k_D} = e^{\dfrac {h \left( \dfrac{k_{RH}}{\mu_{RH}} - \dfrac{k_{RD}}{\mu_{RD}} \right)}{4\pi kT}} \label{13}$
The same general procedure can be followed for any isotope substitution.
In summary, the greater the mass the more energy is needed to break bonds. A heavier isotope forms a stronger bond. The resulting molecule has less of a tendency to dissociate. The increase in energy needed to break the bond results in a slower reaction rate and the observed isotope effect.
Kinetic Isotope Effects
Kinetic Isotope Effects (KIEs) are used to determine reaction mechanisms by determining rate limiting steps and transition states and are commonly measured using NMR to detect isotope location or GC/MS to detect mass changes. In a KIE experiment an atom is replaced by its isotope and the change in rate of the reaction is observed. A very common isotope substitution is when hydrogen is replaced by deuterium. This is known as a deuterium effect and is expressed by the ratio kH/kD (as explained above). Normal KIEs for the deuterium effect are around 1 to 7 or 8. Large effects are seen because the percentage mass change between hydrogen and deuterium is great. Heavy atom isotope effects involve the substitution of carbon, oxygen, nitrogen, sulfur, and bromine, with effects that are much smaller and are usually between 1.02 and 1.10. The difference in KIE magnitude is directly related to the percentage change in mass. Large effects are seen when hydrogen is replaced with deuterium because the percentage mass change is very large (mass is being doubled) while smaller percent mass changes are present when an atom like sulfur is replaced with its isotope (increased by two mass units).
Primary KIEs
Primary kinetic isotope effects are rate changes due to isotopic substitution at a site of bond breaking in the rate determining step of a reaction.
Example
Consider the bromination of acetone: kinetic studies have been performed that show the rate of this reaction is independent of the concentration of bromine. To determine the rate determining step and mechanism of this reaction the substitution of a deuterium for a hydrogen can be made.
When hydrogen was replaced with deuterium in this reaction a $k_H \over k_D$ of 7 was found. Therefore the rate determining step is the tautomerization of acetone and involves the breaking of a C-H bond. Since the breaking of a C-H bond is involved, a substantial isotope effect is expected.
Heavy Atom Isotope Effects
A rule of thumb for heavy atom isotope effects is that the maximum isotopic rate ratio is proportional to the square root of the inverse ratio of isotopic masses.
• Expected: $\dfrac{k_{32}}{k_{34}} = \sqrt{\dfrac{34}{32}}=1.031$
• Experimental: $\dfrac{k_{32}}{k_{34}} = 1.072$
Secondary KIEs
Secondary kinetic isotope effects are rate changes due to isotopic substitutions at a site other than the bond breaking site in the rate determining step of the reaction. These come in three forms: $\alpha$, $\beta$, and $\gamma$ effects.
$\beta$ secondary isotope effects occur when the isotope is substituted at a position next to the bond being broken.
$\ce{(CH3)2CHBr + H2O ->[k_H] (CH3)2CHOH}$
$\ce{(CD3)2CHBr + H2O ->[k_D] (CD3)2CHOH}$
This is thought to be due to hyperconjugation in the transition state. Hyperconjugation involves a transfer of electron density from a sigma bond to an empty p orbital (for more on hyperconjugation see outside links).
Solvent Effects in Reactions
Reactions may be affected by the type of solvent used (for example H2O to D2O or ROH to ROD). There are three main ways solvents effect reactions:
1. The solvent can act as a reactant resulting in a primary isotope effect.
2. Rapid hydrogen exchange can occur between substrate molecules labeled with deuterium and hydrogen atoms in the solvent. Deuterium may change positions in the molecule resulting in a new molecule that is then reacted in the rate determining step of the reaction.
3. The nature of solvent and solute interactions may also change with differing solvents. This could change the energy of the transition state and result in a secondary isotope effects.
References
1. Baldwin, J.E., Gallagher, S.S., Leber, P.A., Raghavan, A.S., Shukla, R.; J. Org. Chem. 2004, 69, 7212-7219 (This is a great paper using kinetic isotope effects to determine a reaction mechanism. It will interest the organic chemistry oriented reader.)
2. Bigeleisen, J., Goeppert, M., J. Chem. Phys. 1947, 15, 261.
3. Chang, R.; Physical Chemistry for the Chemical and Biological Sciences; University Science Books: Sausalito, CA, 2000, pp 480-483.
4. Isaacs, N.; Physical Organic Chemistry; John Wiley & Sons Inc.: New York, NY; 1995, 2nd ed, pp 287-313.
5. March, J., Smith, M.B.; March’s Advanced Organic Chemistry; John Wiley & Sons, Inc.: Hoboken, NJ, 2007; 6th ed.
6. McMurry, J.; Organic Chemistry; Brooks & Cole: Belmont, CA; 2004, 6th ed.
7. McQuarrie, D.; Quantum Chemistry; University Science Books: Sausalito, CA, 2008, 2nd ed.
8. Rouhi, A.; C&EN. 1997, 38-42.
Problems
1. Describe the difference between primary and secondary kinetic isotope effects.
2. Estimate the kN-H/kN-D for a deuterium substitution on nitrogen given that vH=9.3x1013 Hz and the activation energy is equal to 5.31 kJ/mol.
3. Using the 'rule of thumb' for heavy isotope effects, calculate the expected effect for a bromine isotope substitution, 79Br and 81Br.
4. Explain some of the main ways kinetic isotope effects are used.
5. As discussed, the rate-limiting step in the bromination of acetone is the breaking of a carbon-hydrogen bond. Estimate kC-H/KC-D for this reaction at 285 K. (Given: vtildeC-H=3000 cm-1 and vtildeC-D=2100 cm-1)
Solutions
1. Primary isotope effects involve isotopic substitution at the bond being broken in a reaction, while secondary isotope effects involve isotopic substituion on bonds adjacent to the bond being broken.
2. 8.5
3. 1.0126
4. To determine reaction mechanisms, to determine rate limiting steps in reactions, to determine transition states in reactions.
5. 9.685
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/09%3A_Chemical_Kinetics/9.08%3A_Isotope_Effects_in_Chemical_Reactions.txt
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Learning Objectives
Make sure you thoroughly understand the following essential ideas:
• Describe some of the major differences between the kinetics of reactions in the gas phase, compared with those in liquid solutions.
• What role do solvent cages play in solution kinetics?
• Explain the distinction between diffusion-control and activation-control of reaction rates in solutions.
• How can the polarity of a solvent affect the energetics of a reaction mechanism?
The kinetics fundamentals we covered in the earlier sections of this lesson group relate to processes that take place in the gas phase. But chemists and biochemists are generally much more concerned with solutions. This lesson will take you through some of the extensions of basic kinetics that you need in order to understand the major changes that occur when reactions take place in liquid solutions.
What's different about kinetics in liquid solutions?
Most of the added complications of kinetics and rate processes in liquid solutions arise from the much higher density of the liquid phase. In a typical gas at atmospheric pressure, the molecules occupy only about 0.2 per cent of the volume; the other 99.8 percent is empty space. In a liquid, molecules may take up more than half the volume, and the "empty" spaces are irregular and ever-changing as the solvent molecules undergo thermal motions of their own.
In a typical liquid solution, the solvent molecules massively outnumber the reactant solute molecules, which tend to find themselves momentarily (~10–11 sec) confined to a "hole" within the liquid. This trapping becomes especially important when the solvent is strongly hydrogen-bonded as is the case with water or alcohol.
When thermal motions occasionally release a solute molecule from this trap, it will jump to a new location. The jumps are very fast (10–12 - 10–13 sec) and short (usually a few solvent-molecule diameters), and follow an entirely random pattern, very much as in Brownian motion. Consider a simple bimolecular process A + B → products. The reactant molecules will generally be jumping from hole to hole in the solvent matrix, only occasionally finding themselves in the same solvent cage where thermal motions are likely to bring them into contact.
Table $1$: Solvent cages and encounter pairs
A pair of reactants end up in the same solvent cage, where they bounce around randomly and exchange kinetic energy with the solvent molecules. Eventually the two reactants form an encounter pair. If they fail to react the first time, they have many more opportunities during the lifetime of the cage. The products form and begin to move away from each other. Finally, after about 10–11 sec, the solvent cage breaks up and the products diffuse away.
The process can be represented as
$A + B \rightarrow \{AB\} → \text{products}$
in which the $\{AB\}$ term represents the caged reactants including the encounter pair and the activated complex.
Contrast this scenario with a similar reaction taking place in the gas phase; the molecules involved in the reaction will often be the only ones present, so a significant proportion of the collisions will be $A$-$B$ encounters. However, if the collision should fail to be energetically or geometrically viable, the reactant molecules fly apart and are unlikely to meet again anytime soon. In a liquid, however, the solute molecules are effectively in a constant state of collision — if not with other reactants, then with solvent molecules which can exchange kinetic energy with the reactants. So once an A-B encounter pair forms, the two reactants get multiple whacks at each other, greatly increasing the probability that they will obtain the kinetic energy needed to kick them over the activation hump before the encounter pair disintegrates.
Limiting Cases: Diffusion-Controlled and Activation-Controlled Reactions
The encounter pair model introduces some new rate parameters:
$\ce{A + B <=>[k1][k_{-1}] {AB} -> products}$
The first step is an equilibrium between the reactants outside and inside the solvent cage. The rate constants $k_1$ and $k_2$ reflect those relating to diffusion of molecules through the solvent; their values are strongly dependent on the viscosity (and thus the temperature) of the solvent. (Note that $k_1$ is a second-order rate constant, while $k_2$ is first-order.)
Diffusion is the transport of a substance through a concentration gradient; that is, from a region of higher concentration to one of lower concentration. Think of the way the color of tea spreads out when a tea bag is immersed in hot water. Diffusion occurs because random thermal motions are statistically more likely to move molecules out of a region of higher concentration than in the reverse direction, simply because in the latter case fewer molecules are available to make the reverse trip. Eventually the concentrations become uniform and equilibrium is attained.
As molecules diffuse through a liquid, they must nudge neighboring molecules out of the way. The work required to do this amounts to an activation energy, so diffusion can be thought of as a kinetic process with its own rate constant kd and activation energy. These parameters depend on the sizes of the solute and solvent molecules and on how strongly the latter interact with each other. This suggests two important limiting cases for reactions in solution.
For water at room temperature, k1 is typically 109-1010 dm–3 mol–1 s–1 and k2 is around 10–9-10–10 dm–3 mol–1 s–1. Given these values, k3 > 1012 s–1 implies diffusion control, while values < 109 s–1 are indicative of activation control.
• Diffusion Controlled ($k_3 \gg k_2$): If the activation energy of the A+B reaction is very small or if escape of molecules from the {AB} cage is difficult, the kinetics will be dominated by k1, and thus by the activation energy of diffusion. Such a process is said to be diffusion controlled. Reactions in aqueous solution in which Ea > 20 kJ/mol are likely to fall into this category.
• Activation Controlled ($k_3 \ll k_2$): Alternatively, if the activation energy of the A+B reaction dominates the kinetics, and the reaction is activation-controlled.
Several general kinds of reactions are consistently very "fast" and thus are commonly found to be diffusion-controlled in most solvents:
Units Matter
Gas-phase rate constants are normally expressed in units of mol s–1, but rate constants of reactions in solution are conventionally given in mol/L units, or dm3 mol–1 s–1. Conversion between them depends on a number of assumptions and is non-trivial.
• Recombination of atoms and radicals
For example the formation of I2 from I atoms in hexane at 298 K has k3 = 1.3×1012 dm3 mol–1 s–1.
• Acid-base reactions which involve the transport of H+ and OH ions tend to be very fast.
The most famous of these is one of the fastest reactions known:$H^+ + OH^– → H_2O \nonumber$ for which k3 = 1.4×1011 dm3 mol–1 s–1 at 298 K.
Polar solvents such as water and alcohols interact with ions and polar molecules through attractive dipole-dipole and ion-dipole interactions, leading to lower-energy solvated forms which stabilize these species. In this way, a polar solvent can alter both the thermodynamics and kinetics (rate) of a reaction.
Solvent Thermodynamic Effect
If the products of the reaction are markedly more or less polar than the reactants, solvent polarity can change the overal thermodynamics (equilibrium constant) of the reaction. Nowhere is this more apparent than when an ionic solid such as salt dissolves in water. The Na+ and Cl ions are bound together in the solid through strong coulombic forces; pulling the solid apart in a vacuum or in a nonpolar solvent is a highly endothermic process. In contrast, dissolution of NaCl in water is slightly exothermic and proceeds spontaneously.
The water facilitates this process in two important ways. First, its high dielectric constant of 80 reduces the force between the separated ions to 1/80 of its normal value. Secondly, the water molecules form a solvation shell around the ions (lower left), rendering them energetically (thermodynamically) more stable than they were in the NaCl solid.
Solvent Kinetic Effect
In the same way, a reaction whose mechanism involves the formation of an intermediate or activated complex having a polar or ionic character will have its activation energy, and thus its rate, subject to change as the solvent polarity is altered. As an example we will consider an important class of reactions that you will hear much about if you take a course in organic chemistry. When an aqueous solution of a strong base such as KOH is added to a solution of tertiary-butyl chloride in ethanol, the chlorine is replaced by a hydroxyl group, leaving t-butyl alcohol as a product:
This reaction is one of a large and important class known as SN1 nucleophilic substitution processes that are discussed in most organic chemistry courses. In these reactions, a species that possesses a pair of non-bonding electrons (also called a nucleophile or Lewis base) uses them to form a new bond with an electrophile — a compound in which a carbon atom has a partial positive charge owing to its bonds to electron-withdrawing groups. In the example here, other nucleophiles such as NH3 or even H2O would serve as well.
In order to reflect the generality of this process and to focus on the major changes that take place, we will represent this reaction as
Extensive studies of this class of reactions in the 1930's revealed that it proceeds in two activation energy-controlled steps, followed by a simple dissociation into the products:
In step , which is rate-determining, the chlorine leaves the alkyl chloride which becomes an intermediate known as a carbocation ("cat-ion"). These ions, in which the central carbon atom lacks a complete octet, are highly reactive, and in step the carbocation is attacked by the hydroxide ion which supplies the missing electron. The immediate product is another cation in which the positive charge is on the oxygen atom. This oxonium ion is unstable and rapidly dissociates ()into the alcohol and a hydrogen ion.
The reaction coordinate diagram helps us understand the effect of solvent polarity on this reaction. Polar solvent molecules interact most strongly with species in which the electric charge is concentrated in one spot. Thus the carbocation is stabilized to a greater degree than are the activated complexes in which the charge is spread out between the positive and negative ends. As the heavy green arrows indicate, a more polar solvent will stabilize the carbocation more than it will either of the activated complexes; the effect is to materially reduce the activation energy of the rate-determining step, and thus speed up the reaction. Because neither the alkyl chloride nor the alcohol is charged, the change in solvent polarity has no effect on the equilibrium constant of the reaction. This is dramatically illustrated by observing the rate of the reaction in solvents composed of ethanol and water in varying amounts:
Table $2$: Data
% water 10 20 30 40 50 60
$k_1 \times 10^6$
1.7 9.1 40.3 126 367 1294
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The traditional experimental methods described above all assume the possibility of following the reaction after its components have combined into a homogeneous mixture of known concentrations. But what can be done if the time required to complete the mixing process is comparable to or greater than the time needed for the reaction to run to completion?
Flow methods
Flow instruments are a rapid mixing devices used to study the chemical kinetics of fast reactions in solution. There are different flavors that can be implement depending on the nature of the reaction as discussed below.
Continuous Flow Approach
For reactions that take place in milliseconds, the standard approach since the 1950s has been to employ a flow technique of some kind. An early example was used to study fast gas-phase reactions in which one of the reactants is a free radical such as OH that can be produced by an intense microwave discharge acting on a suitable source gas mixture. This gas, along with the other reactant being investigated, is made to flow through a narrow tube at a known velocity.
If the distance between the point at which the reaction is initiated and the product detector is known, then the time interval can be found from the flow rate. By varying this distance, the time required to obtain the maximum yield can then be determined. Although this method is very simple in principle, it can be complicated in practice.
Stopped Flow Approach
Owing to the rather large volumes required, continuous flow method is more practical for the study of gas-phase reactions than for solutions, for which the stopped-flow method described below is generally preferred. These are by far the most common means of studying fast solution-phase reactions over time intervals of down to a fraction of a millisecond. The use of reasonably simple devices is now practical even in student laboratory experiments. These techniques make it possible to follow not only changes in the concentrations of reactants and products, but also the buildup and decay of reaction intermediates.
The basic stopped-flow apparatus consists of two or more coupled syringes that rapidly inject the reactants into a small mixing chamber and then through an observation cell that can be coupled to instruments that measure absorption, fluorescence, light scattering, or other optical or electrical properties of the solution. As the solution flows through the cell, it empties into a stopping syringe that, when filled, strikes a backstop that abruptly stops the flow. The volume that the stopping syringe can accept is adjusted so that the mixture in the cell has just become uniform and has reached a steady state; at this point, recording of the cell measurement begins and its change is followed.
Quenched Flow Approach
In a quenched-flow instrument, the reaction is stopped after a certain amount of time has passed after mixing. The stopping of the reaction is called quenching and it can be achieved by various means, for example by mixing with another solution, which stops the reaction (chemical quenching), quickly lowering the temperature (freeze quenching) or even by exposing the sample to light of a certain wavelength (optical quenching).
Of course, there are many reactions that cannot be followed by changes in light absorption or other physical properties that are conveniently monitored. In such cases, it is often practical to quench (stop) the reaction after a desired interval by adding an appropriate quenching agent. For example, an enzyme-catalyzed reaction can be stopped by adding an acid, base, or salt solution that denatures (destroys the activity of) the protein enzyme. Once the reaction has been stopped, the mixture is withdrawn and analyzed in an appropriate manner.
The quenched-flow technique works something like the stopped-flow method described above, with a slightly altered plumbing arrangement. The reactants A and B are mixed and fed directly through the diverter valve to the measuring cell, which is not shown in this diagram. After a set interval that can vary from a few milliseconds to 200 sec or more, the controller activates the quenching syringe and diverter valve, flooding the cell with the quenching solution.
Relaxation Methods
To investigate reactions that are complete in less than a millisecond, one can start with a pre-mixed sample in which one of active reactants is generated in situ. Alternatively, a rapid change in pressure or temperature can alter the composition of a reaction that has already achieved equilibrium.
Flash Photolysis
Many reactions are known which do not take place without light of wavelength sufficiently short to supply the activation energy needed to break a bond, often leading to the creation of a highly reactive radical. A good example is the combination of gaseous Cl2 with H2, which proceeds explosively when the system is illuminated with visible light. In flash photolysis, a short pulse of light is used to initiate a reaction whose progress can be observed by optical or other means.
Photolysis refers to the use of light to decompose a molecule into simpler units, often ions or free radicals. In contrast to thermolysis (decomposition induced by high temperature), photolysis is able to inject energy into a molecule almost instantaneously and can be much "cleaner," meaning that there are fewer side reactions that often lead to complex mixtures of products. Photolysis can also be highly specific; the wavelength of the light that triggers the reaction can often be adjusted to activate one particular kind of molecule without affecting others that might be present.
Norrish and Porter
All this had been known for a very long time, but until the mid-1940's there was no practical way of studying the kinetics of the reactions involving the highly reactive species produced by photolysis. In 1945, Ronald Norrish of Cambridge University and his graduate student George Porter conceived the idea of using a short-duration flash lamp to generate gas-phase CH2 radicals, and then following the progress of the reaction of these radicals with other species by means of absorption spectroscopy.
In a flash photolysis experiment, recording of the absorbance of the sample cell contents is timed to follow the flash by an interval that can be varied in order to capture the effects produced by the product or intermediate as it is formed or decays. Norrish and Porter shared the 1967 Nobel Prize in Chemistry for this work.
Many reactions, especially those that take place in solution, occur too rapidly to follow by flow techniques, and can therefore only be observed when they are already at equilibrium. The classical examples of such reactions are two of the fastest ones ever observed, the dissociation of water
$2 H_2O \rightarrow H_3O^+ + OH^-$
and the formation of the triiodide ion in aqueous solution
$I^– + I_2 \rightarrow I_3^–$
Reactions of these kinds could not be studied until the mid-1950s when techniques were developed to shift the equilibrium by imposing an abrupt physical change on the system.
Temperature Jumps
The rate constants of reversible reactions can be measured using a relaxation method. In this method, the concentrations of reactants and products are allowed to achieve equilibrium at a specific temperature. Once equilibrium has been achieved, the temperature is rapidly changed, and then the time needed to achieve the new equilibrium concentrations of reactants and products is measured. For example, if the reaction
$\text{A} \overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \text{B} \label{relax}$
is endothermic, then according to the Le Chatelier principle, subjecting the system to a rapid jump in temperature will shift the equilibrium state to one in which the product B has a higher concentration. The composition of the system will than begin to shift toward the new equilibrium composition at a rate determined by the kinetics of the process.
For the general case illustrated here, the quantity "x" being plotted is a measurable quantity such as light absorption or electrical conductivity that varies linearly with the composition of the system. In a first-order process, x will vary with time according to
$x_t = x_o e^{-kt}$
After the abrupt perturbation at time to, the relaxation time t* is defined as the half-time for the return to equilibrium — that is, as the time required for $x_o$ to decrease by $Δx/e = Δx/2.718$. The derivation of t* and the relations highlighted in yellow can be found in most standard kinetics textbooks. Temperature jumps are likely most commonly used.
The rate law for the reversible reaction in Equation $\ref{relax}$ can be written as
$\dfrac{d \left[ \text{B} \right]}{dt} = k_1 \left[ \text{A} \right] - k_{-1} \left[ \text{B} \right] \label{19.33}$
Consider a system comprising $\text{A}$ and $\text{B}$ that is allowed to achieve equilibrium concentrations at a temperature, $T_1$. After equilibrium is achieved, the temperature of the system is instantaneously lowered to $T_2$ and the system is allowed to achieve new equilibrium concentrations of $\text{A}$ and $\text{B}$, $\left[ \text{A} \right]_{\text{eq},2}$ and $\left[ \text{B} \right]_{\text{eq}, 2}$. During the transition time from the first equilibrium state to the second equilibrium state, we can write the instantaneous concentration of $\text{A}$ as
$\left[ \text{A} \right] = \left[ \text{B} \right]_{\text{eq}, 1} - \left[ \text{B} \right] \label{19.34}$
The rate of change of species $\text{B}$ can then be written as
$\dfrac{d \left[ \text{B} \right]}{dt} = k_1 \left( \left[ \text{B} \right]_{\text{eq}, 1} - \left[ \text{B} \right] \right) - k_{-1} \left[ \text{B} \right] = k_1 \left[ \text{B} \right]_{\text{eq}, 1} - \left( k_1 + k_{-1} \right) \left[ \text{B} \right] \label{19.35}$
At equilibrium, $d \left[ \text{B} \right]/dt = 0$ and $\left[ \text{B} \right] = \left[ \text{B} \right]_{\text{eq}, 2}$, allowing us to write
$k_1 \left[ \text{B} \right]_{\text{eq}, 1} = \left( k_1 + k_{-1} \right) \left[ \text{B} \right]_{\text{eq}, 2} \label{19.36}$
Using the above equation, we can rewrite the rate equation as
$\dfrac{ d \text{B}}{\left( \left[ \text{B} \right]_{\text{eq}, 2} - \left[ \text{B} \right] \right)} = \left( k_1 + k_{-1} \right) dt \label{19.37}$
Integrating yields
$-\text{ln} \left( \left[ \text{B} \right] - \left[ \text{B} \right]_{\text{eq}, 2} \right) = -\left( k_1 + k_{-1} \right) t + C \label{19.38}$
We can rearrange the above equation in terms of $\text{B}$
$\left[ \text{B} \right] = Ce^{- \left( k_1 + k_{-1} \right) t} + \left[ \text{B} \right]_{\text{eq}, 2} \label{19.39}$
At $t = 0$, $\left[ \text{B} \right] = \left[ \text{B} \right]_{\text{eq}, 1}$, so $C = \left[ \text{B} \right]_{\text{eq}, 1} - \left[ \text{B} \right]_{\text{eq}, 2}$. Plugging the the value of $C$, we arrive at
$\left[ \text{B} \right] - \left[ \text{B} \right]_{\text{eq}, 2} = \left( \left[ \text{B} \right]_{\text{eq}, 1} - \left[ \text{B} \right]_{\text{eq}, 2} \right) e^{-\left( k_1 + k_{-1} \right) t} \label{19.40}$
which can also be expressed as
$\Delta \left[ \text{B} \right] = \Delta \left[ \text{B} \right]_0 e^{-\left( k_1 + k_{-1} \right) t} = \Delta \left[ \text{B} \right]_0 e^{-t/\tau} \label{19.41}$
where $\Delta \left[ \text{B} \right]$ is the difference in the concentration of $\text{B}$ from the final equilibrium concentration after the perturbation, and $\tau$ is the relaxation time. A plot of $\text{ln} \left( \Delta \left[ \text{B} \right]/\Delta \left[ \text{B} \right]_0 \right)$ versus $t$ will be linear with a slope of $-\left( k_1 + k_{-1} \right)$, where $k_1$ and $k_{-1}$ are the rate constants at temperature, $T_2$.
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It should be clear by now that chemical kinetics is governed by the mathematics of systems of differential equations. Thus far, we have only looked at reaction systems that give rise to purely linear differential equations, however, in many instances the rate equations are nonlinear. When the differential equations are nonlinear, the behavior is considerably more complex. In particular, nonlinear equations can lead to oscillatory solutions and can also exhibit the phenomenon of chaos. Chaotic systems are systems that are highly sensitive to small changes in the parameters of the equations or in the initial conditions. Basically, this means that the behavior of a chaotic system can be unpredictable, since such small changes can occur in the form of small errors in determining the parameters (rounding to the nearest tenth or hundredth) or in specifying the initial conditions, and these small changes can cause the system to evolve in time in a very different way.
The Iodine Clock Reaction
The iodine clock reaction is a popular chemistry experiment in which one can visualize how different rate constants in consecutive reactions affect the concentration of species during the reaction. Iodine anions ($I^-$) are colorless. When $I^-$ is reacted with hydrogen peroxide and protons, triiodide is formed, which has a dark blue color. Consider the following series of irreversible reactions:
$H_2 O_2 + 3 \: I^- + 2 \: H^+ \overset{k_1}{\longrightarrow} I_3^- + 2 \: H_2O$
$I_3^- + 2 \: S_2 O_3^{2-} \overset{k_2}{\longrightarrow} 3 \: I^- + S_4 O_6^{2-}$
The rate laws for this system are
\begin{align} \dfrac{ d \left[ I_3^- \right]}{dt} &= k_1 \left[ I^- \right]^3 - k_2 \left[ I_3^- \right] \left[ S_2 O_3^{2-} \right]^2 \ \dfrac{d \left[ I^- \right]}{dt} &= -k_1 \left[ I^- \right]^3 + 3k_2 \left[ I_3^- \right] \left[ S_2 O_3^{2-} \right]^2 \ \dfrac{d \left[ S_2 O_3^{2-} \right]}{dt} &= -k_2 \left[ I_3^- \right] \left[ S_2 O_3^{2-} \right]^2 \end{align} \label{Eq2}
In order to make the equations look a little simpler, let us introduce the variables:
$x = \left[ I^- \right], \: \: \: \: \: \: \: y = \left[ I_3^- \right], \: \: \: \: \: \: \: z = \left[ S_2 O_3^{2-} \right] \label{Eq3}$
In terms of these, the rate equations are
\begin{align} \dfrac{dx}{dt} &= -k_1 x^3 + 3 k_2 yz^2 \ \dfrac{dy}{dt} &= k_1 x^3 - k_2 yz^2 \ \dfrac{dz}{dt} &= -k_2 yz^2 \end{align} \label{Eq4}
If we solve these numerically, we find the following time dependence of the three concentrations: This is a clear example of nonlinearity. Note how the concentration of $I_3^-$ remains close to $0$ for a period of time and then suddenly starts to increase. In a sense, think of the “straw that broke the camel’s back”. As we pile straws on the back of the camel, the camel remains upright until that last straw, which suddenly breaks the back of the camel, and the camel suddenly falls to the ground. This is also an illustration of nonlinearity.
Video $1$: Famous iodine clock reaction: oxidation of potassium iodide by hydrogen peroxide (https://www.youtube.com/watch?v=_qhYDuJt8fI).
Despite the complexity of the rate equations, we can still analyze the approximately and predict the behavior seen in Figure $1$. In this reaction mechanism, $k_2 \gg k_1$. Given the rate law for $I_3^-$,
$\dfrac{d \left[ I_3^- \right]}{dt} = k_1 \left[ I^- \right]^3 - k_2 \left[ I_3^- \right] \left[ S_2 O_3^{2-} \right]^2 \label{Eq5}$
if we use the steady-state approximation, we can set the $\dfrac{d \left[ I_3^- \right]}{dt}$ equal to $0$, yielding
$\left[ I_3^- \right] = \dfrac{k_1}{k_2} \dfrac{\left[ I^- \right]^3}{\left[ S_2 O_3^{2-} \right]^2} \label{Eq6}$
Since $k_2 \gg k_1$, the concentration of $\left[ I_3^- \right]$ is approximately $0$ as long as there are $S_2 O_3^{2-}$ ions present. As soon as all of the $S_2 O_3^{2-}$ is consumed, the concentration of $I_3^-$ can build up in the solution, changing the solution to a dark blue color. Figure $1$ displays the concentration profiles for $I^-$, $I_3^-$, and $S_3 O_3^{2-}$. As can be seen from the figure, the concentration of $I_3^-$ (red line) remains at approximately $0 \: \text{mol/L}$ until all of the $S_2 O_3^{2-}$ (blue line) has been depleted.
Oscillating Reactions
In all of the examples we have seen thus far, the concentration of intermediate species displays a single maximum during the course of the reaction. There is another class of reactions called oscillating reactions in which the concentration of intermediate species oscillates with time. Consider the following series of reactions
$\text{A} + \text{Y} \overset{k_1}{\longrightarrow} \text{X}$
$\text{X} + \text{Y} \overset{k_2}{\longrightarrow} \text{P}$
$\text{B} + \text{X} \overset{k_3}{\longrightarrow} 2 \: \text{X} + \text{Z}$
$2 \: \text{X} \overset{k_4}{\longrightarrow} \text{Q}$
$\text{Z} \overset{k_5}{\longrightarrow} \text{Y}$
In the above reaction mechanism, $\text{A}$ and $\text{B}$ are reactants; $\text{X}$, $\text{Y}$, and $\text{Z}$ are intermediates; and $\text{P}$ and $\text{Q}$ are products. The third reaction in which $\text{B}$ and $\text{X}$ react to form $\text{X}$ and $\text{Z}$ is known as an ”autocatalytic reaction” in which at least one of the reactants is also a product. Such reactions are a key feature of oscillating reactions, as will be discussed below.
Video $2$: The famous Belousov Zhabotinsky chemical reaction in a petri dish. The action is speeded up 8 x from real life. Pacemaker nucleation sites emit circular waves. Breaking the wavefront with a wire triggers pairs of spiral defects which emit more closely spaced waves which eventually fill the container.
Let us assume that the concentrations of $\text{A}$ and $\text{B}$ are large, such that we can approximate them to be constant with time. The rate equation for species $\text{X}$ can be written as
$\dfrac{d \left[ \text{X} \right]}{dt} = k_1 \left[ \text{A} \right] \left[ \text{Y} \right] - k_2 \left[ \text{X} \right] \left[ \text{Y} \right] + k_3 \left[ \text{B} \right] \left[ \text{X} \right] - 2k_4 \left[ \text{X} \right]^2 \label{Eq7}$
Using the steady-state approximation, we can set $d \text{X}/dt = 0$ and rewrite Equation $\ref{Eq7}$ as
$\left( -2 k_4 \right) \left[ \text{X} \right]^2 + \left( k_2 \left[ \text{Y} \right] - k_3 \left[ \text{B} \right] \right) \left[ \text{X} \right] + k_1 \left[ \text{A} \right] \left[ \text{Y} \right] = 0 \label{Eq8}$
We can then use the quadratic formula to solve for $\text{X}$:
$\left[ \text{X} \right] = -\dfrac{ \left( k_2 \left[ \text{Y} \right] - k_3 \left[ \text{B} \right] \right) \pm \sqrt{\left( k_2 \left[ \text{Y} \right] - k_3 \left[ \text{B} \right] \right)^2 - 4 \left( -2k_4 \right) \left( k_1 \left[ \text{A} \right] \left[ \text{Y} \right] \right)}}{2 \left( -2k_4 \right)} \label{Eq9}$
Thus, there are two solutions for the concentration of $\text{X}$ accessible to the reaction system. To examine solutions for $\left[ \text{X} \right]$, let us first assume that $\left[ \text{Y} \right]$ is large. Under these conditions, the first two reactions in the reaction mechanism largely determine the concentration of $\left[ \text{X} \right]$. We can thus approximate Equation $\ref{Eq7}$ as
$0 \approx k_1 \left[ \text{A} \right] \left[ \text{Y} \right] - k_2 \left[ \text{X} \right] \left[ \text{Y} \right] \label{Eq10}$
Solving for $\left[ \text{X} \right]$ yields
$\left[ \text{X} \right] \approx \dfrac{k_1 \left[ \text{A} \right]}{k_2} \label{Eq11}$
As the reaction continues, species $\text{Y}$ is depleted and the assumption that $\left[ \text{Y} \right]$ is large becomes invalid. Instead the $3^{rd}$ and $4^{th}$ steps of the reaction mechanism determine the concentration of $\text{X}$. In this limit, we can approximate Equation $\ref{Eq7}$ as
$0 \approx k_3 \left[ \text{B} \right] \left[ \text{X} \right] - 2 k_4 \left[ \text{X} \right]^2 \label{Eq12}$
Solving for $\left[ \text{X} \right]$ yields
$\left[ \text{X} \right] \approx \dfrac{k_3 \left[ \text{B} \right]}{2k_4} \label{Eq13}$
In the second mechanism, the autocatalytic reaction step leads to an increase in the concentration of $\text{X}$ and $\text{Z}$, which in turn leads to an increase in the concentration of $\text{Y}$. The feedback loop between the production of species $\text{X}$ and $\text{Y}$ leads to oscillatory behavior in the system. This reaction mechanism is known as the Belousov-Zhabotinksii reaction first discovered in the 1950s.
• The autocatalytic reaction of $\text{X}$ in which $\text{X}$ reacts with $\text{B}$ to form more $\text{X}$ in reaction $3$
• The regeneration of species $\text{Y}$ in reaction $5$
• The competition between reaction $2$ and $3$ for the consumption of $\text{X}$ and the involvement of $\text{Y}$ in reaction $2$
The actual Belousov-Zhabotinskii reaction is complex, involving many individual steps, and involves the oscillation between the concentration of $HBrO_2$ and $Br^-$. The reaction equations are
\begin{align} BrO_3^- + Br^- + 2 \: H^+ &\longrightarrow HBrO_2 + HOBr \ HBrO_2 + Br^- + H^+ &\longrightarrow 2 \: HOBr \ HOBr + Br^- + H^+ &\longrightarrow Br_2 + H_2 O \ 2 \: HBrO_2 &\longrightarrow BrO_3^- + HOBr + H^+ \ BrO_3^- + HBrO_2 + H^+ &\longrightarrow 2 \: BrO_2^. + H_2 O \ BrO_2^. + Ce^{3+} + H^+ &\longrightarrow HBrO_2 + Ce^{4+} \ BrO_2^. + Ce^{4+} + H_2 O &\longrightarrow BrO_3^- + Ce^{3+} + 2 \: H^+ \end{align} \label{Eq14}
The essential steps in this mechanism can be reduced to the following set of reactions. Note that we leave this unbalanced and only include the species whose concentrations as functions of time we seek.
\begin{align} BrO_3^- + Br^- &\overset{k_1}{\longrightarrow} HBrO_2 + HOBr \ HBrO_2 + Br^- &\overset{k_2}{\longrightarrow} 2 \: HOBr \ BrO_3^- + HBrO_2 &\overset{k_3}{\longrightarrow} 2 \: HBrO_2 + 2 \: Ce^{4+} \ 2 \: HBrO_2 &\overset{k_4}{\longrightarrow} BrO_3^- + HOBr \ Ce^{4+} &\overset{k_5}{\longrightarrow} fBr^- \end{align} \label{Eq15}
Setting the variables as follows:
$x = \left[ HBrO_2 \right], \: \: \: \: \: \: \: y = \left[ Br^- \right], \: \: \: \: \: \: \: z = \left[ Ce^{4+} \right] \label{Eq16}$
We make the approximation that $\left[ BrO_3^- \right]$ to be a constant $a$. In this case, the rate equations become
\begin{align} \dfrac{dx}{dt} &= k_1 ay - k_2 xy + k_3 ax - k_4 x^2 \ \dfrac{dy}{dt} &= -k_1 ay - k_2 xy + f k_5 z \ \dfrac{dz}{dt} &= 2k_3 ax - k_5 z \end{align} \label{Eq17}
Solving these equations numerically, we obtain the trajectory of two of the species show in the Figure $2$.
On the other hand, we can drive this system to become chaotic by changing the parameters a little. When this is done, we find the follow plot of the concentration of $x$:
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9.1: Reaction Rates
Q9.1
Write the rate of reaction in terms of the rate of disappearance of reactant and the rate of formation of products:
1. $NO_{(g)} + O_{3 (g)} \rightarrow NO_{2(g)} + O_{2(g)}$
2. $2C_2H_{6 (g)} + 7O_{2(g)} \rightarrow 4 CO_{2(g)} + 6 H_2O_{(aq)}$
3. $H_{2 (g)} + I_{2 (g)} \rightarrow 2HI_{(g)}$
4. $4OH_{(g)} + H_2S_{(g)} \rightarrow SO_{2(g)} + 2H_2O_{(aq)} + H_{2(g)}$
S9.1
1. $\text{rate of reaction} = \dfrac{-∆[NO]}{∆t} = \dfrac{-∆[O_3]}{∆t} = \dfrac{∆[NO_2]}{∆t} = \dfrac{∆[O_2]}{∆t}$
2. $\text{rate of reaction} = \dfrac{-∆[C_2H_6]}{2∆t} = \dfrac{-∆[O_2]}{ 7∆t} =\dfrac{∆[CO_2]}{4∆t} = \dfrac{∆[H_2O]}{6∆t}$
3. $\text{rate of reaction} = \dfrac{-∆[H_2]}{ ∆t} = \dfrac{-∆[I_2]}{∆t} = \dfrac{∆[HI]}{2∆t}$
4. $\text{rate of reaction} = \dfrac{-∆[OH]}{4∆t} = \dfrac{-∆[H_2S]}{∆t} = \dfrac{∆[SO_2]}{∆t} = \dfrac{∆[H_2O]}{∆t} = \dfrac{∆[H_2]}{∆t}$
9.2: Reaction Order
Q9.2
Determine the value of the rate constant for the elementary reaction:
$I_{2(g)} + H_{2 (g)} \rightarrow 2HI_{(aq)}$
given that when [Br2] is 0.15 M and [H2] is 0.2M, the rate of reaction is 0.005 M s-1 at 298 K.
S9.2
rate of reaction = k[Br2][H2]
0.005 Ms-1 =k ( 0.15M)2(0.2M)
k= 1.11 M-1s-1
Q9.3
Given the rate of the third order reaction:
$A + B + C \rightarrow P$
is 0.05 Ms-1
If the [A] = 0.05 M, [B] = 0.01M, and [C] = 0.25M. What is the third order rate constant?
S9.3
rate of reaction = k[A][B][C]
0.05Ms-1 =k (0.05M)(0,01M)(0.25M)
k= 0.05Ms-1/( 0.05) (0.01)( 0.25) M3
k= 400M-2s-1
Q9.3
What are the units of the rate constant for a second-order reaction?
S9.3
The reaction rate:
$Rate = \dfrac{d[A]}{dt} = -k[A]^2$
The rate constant of a second-order reaction, k, is M-1min-1 or M-1s-1
Q9.4
Calculate the fraction of the starting quantity of A that will be used up after 60 s. Given the reaction below which is found to be the first order in A and $t_{1/2} = 40\;s$
$A \rightarrow B + C$
Q9.5
Given the first order reaction is completed 90% in 30 mins at 298 K. Calculate the rate constant.
Q9.6
Assume the half life of the first order decay of radioactive isotope takes about 1 year (365 days). How long will it take the radioactivity of that isotope to decay by 60%?
Q9.7a
The decomposition of dinitrogen peroxide ($N_2O_5$) is a first-order reaction with a rate constant of 0.045 min-1 at 300 K.
$2 N_{2}O_{5(g)} \rightarrow 4NO_{2}(g)+O_{2}$
If there were initially 0.040 mol of $N_2O_5$, calculate the moles of $N_2O_5$ remaining after 5 minutes.
S9.7a
The integrated rate equation of a first-order reaction is:
$[A] = [A]_{0}e^{-kt}$
Substituting concentration for moles of reactant and plugging in the known values:
$n_{A} = 0.040_{0}e^{-0.045*5}$
$n_{A} = 0.032$
Q9.7b
You forgot to do your lab assignment the day before and realize you only have 5 hours to reduce your initial concentration of your reactant by 90%. You know the half life is 2 hours and before you get out of bed, you wonder if you can even finish the lab on time. Assuming you have a first order reactant, will 5 hours be enough time for your compound to reduce by 90%?
S9.7b
First, you need to figure the half life of your compound. We do this by solving for k:
$t_{1/2}=\dfrac{0.693}{k}$
$k=\dfrac{0.693}{t\dfrac{1}{2}}$
$k=\dfrac{0.693}{2 hrs} =0.3465$
with this half life, we can find the time it will take by solving for t:
$ln\dfrac{[A]}{[A]^{_{o}}}=-kt$
We do not have the initial and final concentration, but that is okay.
Assuming the initial concentration is 100 g, we can assume that the final concentration is 10 g because that would be reduced 90% as stated in the problem.
$ln\dfrac{[10]}{[100]}=-kt$
$\dfrac{ln[0.1]}{-0.3465}=t=6.65\; hours$
So 5 hours will be definitely not enough time to reduce the compound
Q9.8a
Assume you are dealing with a first order reactant. You know your rate constant is 1.5x10-4/min, but your employer wants to the half-life of the reactant?
S9.8a
$t_{1/2}=\dfrac{0.693}{k}$
$t_{1/2}=\dfrac{0.693}{1.5x10^{^{-4}}}$
$=4.62 \times 10^3\; mins$
Q9.8b
The half-life of a second order reaction $2A\rightarrow P$ is given by:
$t_{1/2}=\dfrac{1}{k[A]_{o}}$
Calculate the half-life of a reaction with initial reactant concentration [A] = 0.90 M and a rate constant of 0.20 M-1 min-1.
S9.8b
$t_{1/2} = \dfrac{1}{0.20\times 0.90} = 0.18 min^{-1}$
Q9.9a
A reaction $A\rightarrow B$ is observed. You do not know the stoichiometry of the reactants or products, but observe that when you increase the initial concentration of A from 0.4 M to 0.8 M, the half-life decreased from 10 minutes to 5 minutes. Determine the reaction order and the rate constant.
S9.9a
A reaction whose half-life changes when the reactant concentration is changed is a second-order reaction. $t_{1/2} = \dfrac{1}{k[A]_{o}}$ $k = \dfrac{1}{[A]_{0}t_{1/2}}$ $k = \dfrac{1}{(0.4M)(5min)} = 0.5 M^{-1}min^{-1}$
Q9.9b
You notice the initial concentration has no affect on the half-life. From this information alone, what can you determine about the the order of the reactant?
S9.9b
You can determine from this information that it would be a first order reactant. The concentration of first order reactants has not affect on the rate or half-life of the reactant. This can be seen in the formula. t_{\frac{1}{2}}=\frac{0.693}{k} This formula does not account for the initial concentration of the reactant.
Q9.10a
Calculate the order of the reaction and the rate constant of Cytobutane decompose to ethylene based on equation
$C_4H_{8 g)} \rightarrow 2C_2H_{4(g)}$
and temperature at 2150C, constant volume, the pressure 200, 158, 124, 98, 77.5, 61 mmHg.
Q9.10b
Given the following data of concentration [A] over a period of time, decide if the data represents first order or second order. Solve for K. Show graphs.
S9.10b
The data best fits a Second order graph.
The equation for a second order reaction is:
1/[A]= kt+ 1/[A0]
When solving for K, the equation is rearranged to:
1/[A]- 1/[A0] =k
t
Plug in the numbers given:
1/[.3]- 1/[.5]= k
(0-54)
.025=k
One, can also obtain the slop of the graph and k=.025
Q9.11a
If a compound’s ½ life is 15.6 days. What is the value of k? How long will it take to decompose to 10%. Use first order reaction.
S9.11a
Equation for half-life of a First-Order Reaction is:
$t_½ = \dfrac{\ln (2)}{k}$
If we plug in the information given:
$15.6 days= \dfrac{\ln(2)}{k}$
$k=4.4 \times 10^{-2}\, day^{-1}$
The second equation needed is the first order reaction, which is:
ln([A]/[A0])= -kt1/2
so, ln ([A]/[A0])=.1
t=-1/k x ln([A]/[A0])
t= -1/(4.4 x 10-2 day-1) x (.1)
t=52 days
12) In a second order reaction 2A--> products, the final concentration is .28M. What is the initial concentration if k=.32M-1s-1 and the times is 5 seconds.
Solution:
Second order reaction equation is:
1/[A]= kt+ 1/[A0]
Filling the information given:
1/[.28M]=.32M-1s-1(5)+ 1/[A0]
1/[A0]= 1.97
[A0]= .50M
Q9.11b
Calculate the half life of a compound if 90% of a given sample of the compound decomposed in 30min
S9.11b
[A]=[A]0 e^-kt
[A]0 = 90%
[A] = 10%
t = 30min * 60s/1min = 1800s
[10%]/[90%]=e^ -k*1800
ln[10%/90%]/1800 = k = 3.95E-4
t1/2= ln(2)/k
t1/2 = .693/3.95E-4= 1754s
1754s*1min/60s = 29 min
Q9.12
Given rate constant for second order reaction
$2 NO_{2(g)} \rightarrow 2NO_{(g)} + O_{2(g)}$
is 1.08 M-1s-1 at 600 0 C. Find the time that would take for the concentration of NO2 from 1.24 M to 0.56 M?
S9.12
1/ [A] – 1/[A]0 = kt
t= 1/k ( 1/[A] – 1/[A]0)
= (1/ 1.08 M-1s-1 ) (1/0.56 – 1/1.24) = 7.2 s
Q9.13a
Is which order of reaction half-life is independent of initial concentration?
S9.13a
First order because the half-life equation for first order is
t1/2= ln(2)/k, it does not have [A0]
Q9.13b
The decomposition of N2) is the first order. At 3650C, t1/2 is 1.79 x 103 min. given intial pressure of 1.05 atm.. Calculate total pressure.
S9.13b
\begin{align} P &= P_{N_2O} + P_{N_2} + P_{O_2} \ &= 0.525 + 0.525 +0.2625 = 1.31 \;atm \end{align}
Q9.14a
The integrated rate law for the zero-order reaction A → B is [A]t = [A]0 - kt
a) skektch the following plots:
(i) rate vs. [A]
(ii) [A] vs. t
S9.14a
rate = k
rate is independent of [A]
(ii) [A] vs. t
[A]t = [A]0 - kt
b) Derive an expression for the half-life of the reaction.
At t = t1/2 , [A] = [A]0/2 so, [A]0/2 = [A]0 - kt1/2
t1/2 = 1/2k [A}0
c) Calculate the time in half-lives when the integrated rate law is no longer valid (that is, when [A] = 0)
[A]t = 0 = [A]0 - kt
t1/2 = 1/2k [A}0 ⇒ k = 1/2t1/2 [A}0
Therefore, to consume all of the reactants it takes
t = [A}0/k = [A]0/2 so, [A]0 / (1/(2t1/2)) [A]0 = 2t1/2
integrated rate law is no longer valid after 2 half-lives
Q9.14b
Jack, Jill, and you are in a physical chemistry class. The professor writes the following equations on the black board.
$A\rightarrow B$
$[A]=[A_{0}]e^{-kt}$
1. The professor assigns you to derive the first-order reaction in front of the class.
2. Jack was assigned to find the rate constant if the reaction half-life is 10 hours. Since you are a good friend of Jack, you decide to help him double check his answer by solving for the rate constant.
S9.14b
a)
$\int ^{[A]}_{[A]_{0}} \dfrac{d[A]}{[A]}=\int^{t}_{0} -kdt$
$\ln[A] \mid ^{[A]}_{[A]_{0}} = -kt$
$\ln[A] -\ln[A]_{0}=-kt$
$\ln\dfrac{[A]}{[A]_{0}} =-kt$
$\dfrac{[A]}{[A]_{0}}=e^{-kt}$
$[A]=[A_{0}]e^{-kt}$
b)
$t_{1/2}=\dfrac{\ln 2}{k}$
$10 hours = \dfrac{\ln2}{k}$
$k=\dfrac{\ln2}{10 hours}\times\dfrac{1hour}{60min}\times\dfrac{1min}{60secs}$
$k=$
Q9.15a
In the nuclear industry, workers use a rule of thumb that the readioactivity from any sample will be relatively harmless after 10 half-lives. Calculate the fraction of a radioactive sample that remains after this time period (hint: Radioactive decay obeys first-order kinetics)
S9.15a
[A] = [A]oe-kt
t1/2 = ln2 / k
[A] = (100)e-(1)(10) =.00454
% remaining after 10 half-lives
Q9.15b
As a school field trip to apply what you learned in physical chemistry class, you and your class went to a nuclear power plant. The plant's manager, Mr. Setsaw gives the class a tour of the facilities. When the class reaches the simulation room, a meltdown simulation is offered to the class, but only if the class can answer his questions can the class experience it.
1. Radioactive decay occurs in what order?
2. The gas used in the simulation is harmless because the fraction of the original remaining is 0.004 and the time that passed 30 years. What is the rate constant?
S9.15b
a) Radioactive decay occurs in first order.
b)
$\dfrac{[A]}{[A]_{0}}=e^{-kt}$
$0.004=e^{-k\,30years}$
$\ln[0.004]=-k\,30years$
$k=\dfrac{-30}{\ln[0.004]}$
$k=$
Q9.16a
Many reactions involving heterogeneous catalysis are zero order; that is, rate = k. An example is the decomposition of phosphine ($\ce{PH3}$) over tungsten (W):
$\ce{4PH3(g) →P4(g) + 6H2(g)}$
The rate for this reaction is independent of [PH3] as long as phosphine's pressure is sufficiently high (>= 1 atm). Explain.
S9.16a
With sufficient PH3, all of the catalytic sites on the tungsten surface are occupied. Further increases in the amount of phosphine cannot affect the reaction, and the rate is independent of [PH3].
Q9.16b
Jill and you just went to an enzyme kinetic workshop. However, Jill is confused about why at high concentrations of substrates the reaction order is zero-order. Please explain to Jill in a way that anyone can understand.
S9.16b
At high concentrations of substrates, all the enzymes are working at their fastest rate. As soon as the enzyme is done converting one substrate to its product, another substrate comes into the active site. Therefore, the rate of the enzyme working is constant and this is the definition for zero-order reaction.
9.3: Molecularity of a Reaction
Q9.41a
A mixture of compounds M and N whose half-lives are 40 minutes and 17 minutes, respectively. They decompose by first-order kinetics. If their concentrations are equal initially, how long does concentration of N to be half that of M?
S9.41a
[I] is the initial concentration of M and N
-> t = 30.1 min
Q9.41b
Compounds A and B both decay by first-order kinetics. The half-life of A is 20 minutes and the half-life of B is 48 minutes. If a container initially contains equal concentrations of compounds A and B, after how long will the concentration of B be twice that of A?
S9.41b
1. Write, in mathematical terms, the information given in the problem and what the problem is asking for.
$t_{1/2, A} = 20.0 min$
$t_{1/2, B} = 48.0 min$
$[A]_0 = [B]_0$
A and B decay by first-order, so
$-\dfrac{d[A]}{dt} = k_A [A]$
$-\dfrac{d[B]}{dt} = k_B [B]$
(Note: the rate constants for A and B are not equal, so indicate which is which with subscripts.)
Want find t at which the following is true:
$[B] = 2[A]$
2. Substitute the integrated rate equations for [A] and [B]
$[B]_0 e^{-k_B t} = 2 [A]_0 e^{-k_A t}$
3. Write expressions for the rate constants in terms of half-lives, and substitute into the equation.
$t_{1/2, A} = \dfrac{ln2}{k_A} \Rightarrow k_A = \dfrac{ln2}{t_{1/2, A}}$
$t_{1/2, B} = \dfrac{ln2}{k_B} \Rightarrow k_B = \dfrac{ln2}{t_{1/2, B}}$
$[B]_0 e^{-\dfrac{ln2}{t_{1/2, B}} t} = 2 [A]_0 e^{-\dfrac{ln2}{t_{1/2, B}} t}$
4. Solve for t
Since initial concentrations of A and B are equal:
$e^{-\dfrac{ln2}{t_{1/2, B}} t} = 2 e^{-\dfrac{ln2}{t_{1/2, B}} t}$
Take natural log of both sides:
$-\dfrac{t}{t_{1/2, B}} ln2 = \left ( 1 - \dfrac{t}{t_{1/2, A}} \right ) ln2$
$t = \dfrac{1}{\dfrac{1}{t_{1/2, A}} - \dfrac{1}{t_{1/2, B}}}$
5. Plug in values for half-lives
$t = \dfrac{1}{\dfrac{1}{20.0 min} - \dfrac{1}{48.0 min}} = 34.2\ min$
Answer: 34.2 minutes
Q9.42a
In Q3 thermodynamics and in Q9 chemical kinetics, the term “reversible “ is used. How do you understand this term? (it has a same meaning in these two chapters)
S9.42a
Actually, this word is used to describe a “reversible” reaction in which both forward and backward reactions can happen in kinetics. In thermodynamics, “reversible” is used to describe a process that is in equilibrium along the pathway from the initial to final states.
Q9.42b
If a reaction has come to thermodynamic equilibrium, can we say anything in particular about the system's kinetics?
S9.42b
Equilibrium occurs when all reactants and products are being consumed at the same rate that they are created. Take the simple example:
$A\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}}B$
When the system is at equilibrium,
$\dfrac{d[A]}{dt} = \dfrac{d[B]}{dt} = 0$
$k_{-1} [A] = k_1 [B]$
$\dfrac{k_{-1}}{k_1} = \dfrac{[B]}{[A]}$
$K_{eq} = \dfrac{k_{-1}}{k_1}$
Q9.43a
The recombination of bromine atoms in an organic solvent, like carbon tetrafloride, is considered as a diffusion-controlled process
$Br + Br \rightarrow Br_2$
we have the viscosity of CF4 is 9.80 x 10-4 Nsm-2 at 30oC, what is the rate of recombination at 30oC?
Q9.43b
Calculate the rate constant of the diffusion-controlled reaction
$2 I \rightarrow I_2$
in dichloromethane at 15°C, which has a viscosity of 0.449 mPa·s at 15°C.
S9.43b
1. Use equation 9.50 to calculate the rate constant.
$k_D = \dfrac{8}{3}\dfrac{RT}{\eta}$
$k_D = \dfrac{8}{3} \dfrac{8.314 \dfrac{J}{mol \ K} \times 288\ K \times \dfrac{N\ m}{J}}{0.449\ mPa \ s \times \dfrac{Pa}{1000\ mPa} \times\dfrac{N/m^2}{Pa}} \times \dfrac{1000\ L}{m^3}$
Answer:
$k_D = 1.42 \times 10^{10} M^{-1} s^{-1}$
Q9.52
Japanese survivors have been exposed to the risk of radiation after the atomic bomb. One man was measured to have iodine-131 levels at 9.7 mC. Calculate the number of atoms of I-131 to which this radioactivity corresponds.
S9.52
First, convert the rate mCi s-1
1 mCi=1.10X10-3 Ci
1 Ci=3.7 X 1010 s-1
The rate ($r$ can be derived as such
$r= (9.7\; Ci) \left( \dfrac{1\;Ci}{1000\; mCi} \right) \left( \dfrac{3.7 \times 10^{10}\; s^{-1}}{1\; Ci} \right) = 3.59 \times 10^8 s^{-1}$
The accepted value for the half life of I-131 is 8.02 days. Using this information, the number of I-131 atoms can be calculated using the nuclear decay equation. Use the
Nuclear Decay Equation:
$\lambda N = \dfrac{\ln 2}{t_{1/2}} N$
or
$N = \dfrac{\ln 2}{t_{1/2}} r$
The half-life for the radioactive beta decay of iodine is 8.02 days
$\ce{^{131}_{53}I} \rightarrow \ce{^{131}_{54}Xe} + \beta + \bar{\nu_e}$
therefore, the rate is
$r= 3.59 \times 10^8 s^{-1}$
Plug into the equation and convert days to seconds.
(8.02 days/ln2)(24 hours/1 day)(3600 seconds/1 hour)(3.59X 10-8) = 2.49 X 1014 I-131 atoms.
N= (
Q9.54
Calculate the rate law for the following acid-catalyzed reaction:
$CH_3COCH_3 + Br_2 \overset{H^+}{\longrightarrow} CH_3COCH_2Br + H^+ + Br^-$
Expt. # $[CH_3COCH_3]_o$/ M $[Br_2]_o$ /M $[H^+]_o$ /M Rate of Disappearance Br2/ M*s-1
1
0.3
0.05 0.05 $5.7 \times 10^{-5}$
2 0.3 0.1 0.05 $5.7 \times 10^{-5}$
3 0.3 0.05 0.05 $1.2 \times 10^{-4}$
4 0.4 0.05 0.2 $3.1 \times 10^{-4}$
5 0.4 0.05 0.05 $7.6 \times 10^{-5}$
S9.54
Find the rate law.
rate = k[CH3COCH3]^x [Br2]^y [H+]^z
[exp 1] / [exp 5]:
5.7e-5 / 7.6e-5 = (0.30/0.40)^x (0.050/0.050)^y (0.050/0.050)^z
3/4 = (3/4)^x
x = 1
[exp 1] / [exp 2]:
5.7e-5 / 5.7e-5 = (0.30/0.30) (0.050/0.10)^y (0.050/0.050)^z
1 = (1/2)^y
y = 0
[exp 1] / [exp 3]:
5.7e-5 / 1.2e-4 = (0.30/0.30) (0.050/0.10)^z
19/40 = (1/2)^z
z ≈ 1
Use [exp 1] to find rate constant.
5.7e-5 M/s = k(0.30 M)(0.050 M)
k = 3.8e-3 M^(-1)-s^(-1)
"calculate the rate of disappearance of bromine if the initial concentration are .600mol/L, 0.200 mol/L, and .10 mol/L for propanone, bromine and H+."
rate = (3.8e-3 M^(-1)-s^(-1))(0.600 M)(0.10 M)
rate = 2.28e-4 M/s
Q9.55
Determine the rate law for the following reaction:
$N_2O_2 + H_2 \rightarrow H_2O + N_2O$
In addition, determine which of the following actions would alter the value of $k$?
1. Increase in pressure of $N_2O_2$
2. Increase in volume size of container
3. Increase in temperature
4. Addition of catalyst to the container.
5. None of the above ;a rate constant is always constant.
Q9.56
Consider the mechanism for the association of iodine atoms to create molecular iodine.
$2I_{(g)} \rightleftharpoons I_{2(g)}^*$
$I_2^*(g) + M(g) \rightarrow I_{2(g)} + M_{(g)}$
With the respect of the first step is at equilibrium, determine the expected rate law (d/dt)[I2(g)] in terms of k1, k-1, k2, [I], and [M].
Q9.57
Consider the following reaction:
$C_3H_{8(g)} + 5O_{2(g)} -> 3CO_{2(g)} + 4H_2O_{(g)}$
If propane (C3H8) is burning at a rate of 0.15 M/s^-1, calculate the rate of formation of CO2.
9.4: More Complex Reactions
S9.57
First, express the reaction with the differential rate equation for the reactants and products involved.
-(d/dt)[C3H8] = (1/3)(d/dt)[CO2]
Then, use the given burning rate of propane and plug it into the differential equation.
(d/dt)[CO2] = 3(0.15M/s^-1)
(d/dt)[CO2] = 0.45 M/s^-1
9.5: The Effect of Temperature on Reaction Rates
Q9.26
If kept in a refrigerator, fresh fish will last for 3 days. If kept in a freezer, it will last for 6 months. Assuming that the temperature in the refrigerator is 5°C, and the temperature in the freezer is -10°C, calculate the activation energy for the bacterial spoiling of fish. Assume that the spoiling time is the 1/e lifetime instead of the half-life.
Q9.27
Find the activation energy of a reaction whose rate constant is multiplied by 6.50 when T is increased from 300.0 K to 310.0 K. For a reaction with Ea = 19 kJ/mol, by what factor is k multiplied when T increases from 300.0 K to 310.0 K?
Q9.28
The kinetics of the browning of juice from Golden Delicious apples was studied; at 20°C k=7.87×10-3/week, and at 37°C k=0.139/week. What is the activation energy for the browning of Golden Delicious apple juice?
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/09%3A_Chemical_Kinetics/9.E%3A_Chemical_Kinetics_%28Exercises%29.txt
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Enzyme kinetics is the study of the chemical reactions that are catalyzed by enzymes. In enzyme kinetics, the reaction rate is measured and the effects of varying the conditions of the reaction are investigated.
• 10.1: General Principles of Catalysis
Catalysts provide a means of reducing the energy barrier and increasing the reaction rate. Catalysts are defined as substances that participate in a chemical reaction but are not changed or consumed. Instead they provide a new mechanism for a reaction to occur which has a lower activation energy than that of the reaction without the catalyst. Homogeneous catalysis refers to reactions in which the catalyst is in solution with at least one of the reactants whereas heterogeneous catalysis refers t
• 10.2: The Equations of Enzyme Kinetics
In biological systems, enzymes act as catalysts and play a critical role in accelerating reactions many times faster than the reaction would normally proceed. Enzymes are high-molecular weight proteins that act on a substrate, or reactant molecule, to form one or more products. Enzymes are highly specific catalysts for biochemical reactions, with each enzyme showing a selectivity for a single reactant, or substrate.
• 10.3: Chymotrypsin- A Case Study
Chymotrypsin is a digestive enzyme belonging to a super family of enzymes called serine proteases. It uses an active serine residue to perform hydrolysis on the C-terminus of the aromatic amino acids of other proteins. Chymotrypsin is a protease enzyme that cleaves on the C-terminal phenylalanine (F), tryptophan (W), and tyrosine (Y) on peptide chains. It shows specificity for aromatic amino acids because of its hydrophobic pocket.
• 10.4: Multisubstrate Systems
The Michaelis –Menten model of enzyme kinetics was derived for single substrate reactions. Enzymatic reactions requiring multiple substrates and yielding multiple products are more common and yielding multiple products are more common than single-substrate reaction. In these types of reactions, the all the substrates involved are bound to the enzyme before catalysis of the reaction takes place to release the products.
• 10.5: Enzyme Inhibition
Enzymes can be regulated in ways that either promote or reduce their activity. In some cases of enzyme inhibition, for example, an inhibitor molecule is similar enough to a substrate that it can bind to the active site and simply block the substrate from binding. When this happens, the enzyme is inhibited through competitive inhibition, because an inhibitor molecule competes with the substrate for active site binding
• 10.6: Allosteric Interactions
Allosteric regulation is the regulation of an enzyme or other protein by binding an effector molecule at the protein's allosteric site (that is, a site other than the protein's active site). Effectors that enhance the protein's activity are referred to as allosteric activators, whereas those that decrease the protein's activity are called allosteric inhibitors. The term allostery refers to the fact that the regulatory site of an allosteric protein is physically distinct from its active site.
• 10.7: The Effect of pH on Enzyme Kinetics
Enzymes are affected by changes in pH. The most favorable pH value - the point where the enzyme is most active - is known as the optimum pH.
• 10.8: The Effect of Temperature on Enzyme Kinetics
Enzyme structures unfold (denature) when heated or exposed to chemical denaturants and this disruption to the structure typically causes a loss of activity. Protein folding is key to whether a globular protein or a membrane protein can do its job correctly. It must be folded into the right shape to function.
• 10.E: Exercises
This are exercises that to accompany the TextMap organized around Raymond Chang's Physical Chemistry for the Biosciences textbook.
Thumbnail: Enzyme changes shape by induced fit upon substrate binding to form enzyme-substrate complex. Hexokinase has a large induced fit motion that closes over the substrates adenosine triphosphate and xylose. (PDB: 2E2N, 2E2Q). (CC BY 4.0; Thomas Shafee).
10: Enzyme Kinetics
As can be seen from the Arrhenius equation, the magnitude of the activation energy, $E_a$, determines the value of the rate constant, $k$, at a given temperature and thus the overall reaction rate. Catalysts provide a means of reducing $E_a$ and increasing the reaction rate. Catalysts are defined as substances that participate in a chemical reaction but are not changed or consumed. Instead they provide a new mechanism for a reaction to occur which has a lower activation energy than that of the reaction without the catalyst. Homogeneous catalysis refers to reactions in which the catalyst is in solution with at least one of the reactants whereas heterogeneous catalysis refers to reactions in which the catalyst is present in a different phase, usually as a solid, than the reactants. Figure $1$ shows a comparison of energy profiles of a reaction in the absence and presence of a catalyst.
Consider a non-catalyzed elementary reaction
$\text{A} \overset{k}{\longrightarrow} \text{P}$
which proceeds at rate $k$ at a certain temperature. The reaction rate can be expressed as
$\dfrac{d \left[ \text{A} \right]}{dt} = -k \left[ \text{A} \right] \label{Eq1}$
In the presence of a catalyst $\text{C}$, we can write the reaction as
$\text{A} + \text{C} \overset{k_\text{cat}}{\longrightarrow} \text{P} + \text{C}$
and the reaction rate as
$\dfrac{d \left[ \text{A} \right]}{dt} = -k \left[ \text{A} \right] - k_\text{cat} \left[ \text{A} \right] \left[ \text{C} \right] \label{Eq2}$
where the first term represents the uncatalyzed reaction and the second term represents the catalyzed reaction. Because the reaction rate of the catalyzed reaction is often magnitudes larger than that of the uncatalyzed reaction (i.e. $k_\text{cat} \gg k$), the first term can often be ignored.
Acid Catalysis
A common example of homogeneous catalysts are acids and bases. For example, given an overall reaction is $\text{S} \rightarrow \text{P}$. If $k$ is the rate, then
$\dfrac{d \left[ \text{P} \right]}{dt} = k \left[ \text{S} \right]$
The purpose of an enzyme is to enhance the rate of production of the product $\text{P}$. The equations of the acid-catalyzed reaction are
\begin{align} \text{S} + \text{A}H &\overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \text{S}H^+ + \text{A}^- \ \text{S}H^+ + H_2 O &\overset{k_2}{\rightarrow} \text{P} + H_3 O^+ \ H_3 O^+ + \text{A}^- &\overset{k_3}{\underset{k_{-3}}{\rightleftharpoons}} \text{A}H + H_2 O \end{align}
The full set of kinetic equations is
\begin{align} \dfrac{d \left[ \text{S} \right]}{dt} &= -k_1 \left[ \text{S} \right] \left[ \text{A} H \right] + k_{-1} \left[ \text{S} H^+ \right] \left[ \text{A}^- \right] \ \dfrac{ d \left[ \text{A} H \right]}{dt} &= -k_1 \left[ \text{S} \right] \left[ \text{A} H \right] + k_{-1} \left[ \text{S} H^+ \right] \left[ \text{A}^- \right] - k_{-3} \left[ \text{A} H \right] + k_3 \left[ H_3 O^+ \right] \left[ \text{A}^- \right] \ \dfrac{d \left[ \text{S} H^+ \right]}{dt} &= k_1 \left[ \text{S} \right] \left[ \text{A} H \right] - k_{-1} \left[ \text{S} H^+ \right] \left[ \text{A}^- \right] - k_2 \left[ \text{S} H^+ \right] \ \dfrac{d \left[ \text{A}^- \right]}{dt} &= k_1 \left[ \text{S} \right] \left[ \text{A} H \right] - k_{-1} \left[ \text{S} H^+ \right] \left[ \text{A}^- \right] -k_2 \left[ \text{A}^- \right] \left[ H_3 O^+ \right] + k_{-3} \left[ \text{A} H \right] \ \dfrac{d \left[ \text{P} \right]}{dt} &= k_2 \left[ \text{S} H^+ \right] \ \dfrac{d \left[ H_3 O^+ \right]}{dt} &= -k_2 \left[ \text{S} H^+ \right] - k_3 \left[ H_3 O^+ \right] \left[ \text{A}^- \right] + k_{-3} \left[ \text{A} H \right] \end{align} \label{Eq3}
We cannot easily solve these, as they are nonlinear. However, let us consider two cases $k_2 \gg k_{-1} \left[ \text{A}^- \right]$ and $k_2 \ll k_{-1} \left[ \text{A}^- \right]$. In both cases, $\text{S} H^+$ is consumed quickly, and we can apply a steady-state approximation:
$\dfrac{d \left[ \text{S} H^+ \right]}{dt} = k_1 \left[ \text{S} \right] \left[ \text{A} H \right] - k_{-1} \left[ \text{A}^- \right] \left[ \text{S} H^+ \right] - k_2 \left[ \text{S} H^+ \right] = 0 \label{Eq4}$
Rearranging in terms of $\text{S} H^+$ yields
$\left[ \text{S} H^+ \right] = \dfrac{k_1 \left[ \text{S} \right] \left[ \text{A} H \right]}{k_{-1} \left[ \text{A}^- \right] + k_2} \label{5}$
and the rate of production of $\text{P}$ can be written as
$\dfrac{d \left[ \text{P} \right]}{dt} = k_2 \left[ \text{S} H^+ \right] = \dfrac{k_1 k_2 \left[ \text{S} \right] \left[ \text{A} H \right]}{k_{-1} \left[ \text{A}^- \right] + k_2} \label{Eq6}$
In the case where $k_2 \gg k_{-1} \left[ \text{A}^- \right]$, Equation $\ref{Eq6}$ can be written as
$\dfrac{d \left[ \text{P} \right]}{dt} = k_1 \left[ \text{S} \right] \left[ \text{A} H \right] \label{Eq7}$
which is known as a general acid-catalyzed reaction. On the other hand, if $k_2 \ll k_{-1} \left[ \text{A}^- \right]$, we can use an equilibrium approximation to write the rate of production of $\text{P}$ as
$\dfrac{d \left[ \text{P} \right]}{dt} = \dfrac{k_1 k_2 \left[ \text{S} \right] \left[ \text{A} H \right]}{k_{-1} \left[ \text{A}^- \right]} = \dfrac{k_1 k_2}{k_{-1} K} \left[ \text{S} \right] \left[ H^+ \right] \label{Eq8}$
where $K$ is the acid dissociation constant:
$K = \dfrac{ \left[ \text{A}^- \right] \left[ H^+ \right]}{\left[ \text{A} H \right]} \label{Eq9}$
In this case, the reaction is hydrogen ion-catalyzed.
Enzyme Catalysis
To live, grow, and reproduce, microorganisms undergo a variety of chemical changes. They alter nutrients so they can enter the cell and they change them once they enter in order to synthesize cell parts and obtain energy.
Metabolism refers to all of the organized chemical reactions in a cell. Reactions in which chemical compounds are broken down are called catabolic reactions while reactions in which chemical compounds are synthesized are termed anabolic reactions. All of these reactions are under the control of enzymes.
Enzymes are substances present in the cell in small amounts that function to speed up or catalyze chemical reactions. On the surface of the enzyme is usually a small crevice that functions as an active site or catalytic site to which one or two specific substrates are able to bind. (Anything that an enzyme normally combines with is called a substrate.) The binding of the substrate to the enzyme causes the flexible enzyme to change its shape slightly through a process called induced fit to form a tempore intermediate called an enzyme-substrate complex (Figure 1).
Enzymes speed up the rate of chemical reactions because they lower the energy of activation, the energy that must be supplied in order for molecules to react with one another (Figure $3$). Like homogeneous catalysts discussed above, enzymes lower the energy of activation by forming an enzyme-substrate complex allowing products of the enzyme reaction to be formed and released (Figure $2$).
Enzyme-Substrate Reactions
Enzymes are substances present in the cell in small amounts which speed up or catalyze chemical reactions. Enzymes speed up the rate of chemical reactions because they lower the energy of activation, the energy that must be supplied in order for molecules to react with one another. Enzymes lower the energy of activation by forming an enzyme-substrate complex.
Many enzymes require a nonprotein cofactor to assist them in their reaction. In this case, the protein portion of the enzyme, called an apoenzyme, combines with the cofactor to form the whole enzyme or haloenzyme (Figure $4$). Some cofactors are ions such as Ca++, Mg++, and K+; other cofactors are organic molecules called coenzymes which serve as carriers for chemical groups or electrons. NAD+, NADP+, FAD, and coenzyme A (CoA) are examples of coenzymes.
Enzymes are generally globular proteins. (Some RNA molecules called ribozymes can also be enzymes. These are usually found in the nuclear region of cells and catalyze the splitting of RNA molecules.). Enzymes are catalysts that breakdown or synthesize more complex chemical compounds. They allow chemical reactions to occur fast enough to support life. Enzymes speed up the rate of chemical reactions because they lower the energy of activation, the energy that must be supplied in order for molecules to react with one another. Anything that an enzyme normally combines with is called a substrate. Enzymes are very efficient with a typically enzyme generally able to catalyze between 1 and 10,000 molecules of substrate per second. The means that enzymes are only have to be present in small amounts in the cell since. They are not altered during their reaction and are highly specific for their substrate, with generally one specific enzyme dedicated for each specific chemical reaction.
Factors that affect the rate of enzyme reactions
Enzyme activity is affected by a number of factors including:
1. The concentration of enzyme: Assuming a sufficient concentration of substrate is available, increasing enzyme concentration will increase the enzyme reaction rate.
2. The concentration of substrate: At a constant enzyme concentration and at lower concentrations of substrates, the substrate concentration is the limiting factor. As the substrate concentration increases, the enzyme reaction rate increases. However, at very high substrate concentrations, the enzymes become saturated with substrate and a higher concentration of substrate does not increase the reaction rate.
3. Inhibitors: inhibitors will inhibit the activity of enzyme and decrease the rate of reaction. Enzyme inhibitors will bind to enzyme active sites and could modify the chemistry of an active site which can stop a substrate from entering.
4. The temperature: Each enzyme has an optimum temperature at which it works best. A higher temperature generally results in an increase in enzyme activity (Arrhenius kinetics). As the temperature increases, molecular motion increases resulting in more molecular collisions. If, however, the temperature rises above a certain point, the heat will denature the enzyme, causing it to lose its three-dimensional functional shape by denaturing its hydrogen bonds. Cold temperature, on the other hand, slows down enzyme activity by decreasing molecular motion.
5. The pH: Each enzyme has an optimal pH that helps maintain its three-dimensional shape. Changes in pH may denature enzymes by altering the enzyme's charge. This alters the ionic bonds of the enzyme that contribute to its functional shape.
6. The salt concentration: Each enzyme has an optimal salt concentration. Changes in the salt concentration may also denature enzymes.
Applications of Enzymes
Enzymes are essential to maintain homeostasis because any malfunction of an enzyme could lead to diseases. Therefore, pharmaceutical companies study enzyme to manipulate and synthesis new medicine. Besides their medicinal applications, enzymes in industry are important because enzymes help breaking down cellulose, wastes, etc. Enzymes are essential in the process of making new products in many industries such as pharmaceutical, food, paper, wine, etc.
Contributors and Attributions
• Dr. Gary Kaiser (COMMUNITY COLLEGE OF BALTIMORE COUNTY, CATONSVILLE CAMPUS)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/10%3A_Enzyme_Kinetics/10.01%3A_General_Principles_of_Catalysis.txt
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In biological systems, enzymes act as catalysts and play a critical role in accelerating reactions, anywhere from $10^3$ to $10^{17}$ times faster than the reaction would normally proceed. Enzymes are high-molecular weight proteins that act on a substrate, or reactant molecule, to form one or more products.
Michaelis-Menten Enzyme Kinetics
Enzymes are highly specific catalysts for biochemical reactions, with each enzyme showing a selectivity for a single reactant, or substrate. For example, the enzyme acetylcholinesterase catalyzes the decomposition of the neurotransmitter acetylcholine to choline and acetic acid. Many enzyme–substrate reactions follow a simple mechanism that consists of the initial formation of an enzyme–substrate complex, $ES$, which subsequently decomposes to form product, releasing the enzyme to react again.
This is described within the following multi-step mechanism
$E + S \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} ES \underset{k_{-2}}{\overset{k_2}{\rightleftharpoons}} E + P \label{13.20}$
where $k_1$, $k_{–1}$, $k_2$, and $k_{–2}$ are rate constants. The reaction’s rate law for generating the product $[P]$ is
$rate = \dfrac{d[P]}{dt} = k_2[ES] - k_{-2}[E][P] \label{13.21A}$
However, if we make measurement early in the reaction, the concentration of products is negligible, i.e.,
$[P] \approx 0$
and we can ignore the back reaction (second term in right side of Equation $\ref{13.21A}$). Then under these conditions, the reaction’s rate is
$rate = \dfrac{d[P]}{dt} = k_2[ES] \label{13.21}$
To be analytically useful we need to write Equation $\ref{13.21}$ in terms of the reactants (e.g., the concentrations of enzyme and substrate). To do this we use the steady-state approximation, in which we assume that the concentration of $ES$ remains essentially constant. Following an initial period, during which the enzyme–substrate complex first forms, the rate at which $ES$ forms
$\dfrac{d[ES]}{dt} =k_1[E] [S] = k_1([E]_0 − [ES])[S] \label{13.22}$
is equal to the rate at which it disappears
$− \dfrac{d[ES]}{dt} = k_{−1}[ES] + k_2[ES] \label{13.23}$
where $[E]_0$ is the enzyme’s original concentration. Combining Equations $\ref{13.22}$ and $\ref{13.23}$ gives
$k_1([E]_0 − [ES]) [S] = k_{−1}[ES] + k_2[ES]$
which we solve for the concentration of the enzyme–substrate complex
$[ES] = \dfrac{ [E]_0[S] }{ \dfrac{k_{−1} + k_2}{k_1} + [S] } = \dfrac{[E]_0[S] }{K_m + [S]} \label{Eq13.24}$
where $K_m$ is the Michaelis constant. Substituting Equation $\ref{Eq13.24}$ into Equation $\ref{13.21}$ leaves us with our final rate equation.
$\dfrac{d[P]}{dt} = \dfrac{k_2[E]_0[S]}{K_m + [S]} \label{Eq13.25}$
A plot of Equation $\ref{Eq13.25}$, as shown in Figure $1$, is instructive for defining conditions where we can use the rate of an enzymatic reaction for the quantitative analysis of an enzyme or substrate.
For high substrate concentrations, where $[S] \gg K_m$, Equation $\ref{Eq13.25}$ simplifies to
$\dfrac{d[P]}{dt} = \dfrac{k_2[E]_0[S]}{ K_m + [S]} \approx \dfrac{k_2[E]_0[S]}{[S]} = k_2[E]_0 = V_{max} \label{Eq13.26}$
where $V_{max}$ is the maximum rate for the catalyzed reaction. Under these conditions the reaction is zero-order in substrate and we can use $V_{max}$ to calculate the enzyme’s concentration, typically using a variable-time method. At lower substrate concentrations, where $[S] \ll K_m$, Equation $\ref{Eq13.25}$ becomes
$\dfrac{d[P]}{dt} = \dfrac{k_2[E]_0[S]}{K_m + [S]} \approx \dfrac{k_2[E]_0[S]}{ K_m} =\dfrac{V_{max}[S]}{K_m} \label{13.27}$
The reaction is now first-order in substrate, and we can use the rate of the reaction to determine the substrate’s concentration by a fixed-time method.
The Michaelis constant $K_m$ is the substrate concentration at which the reaction rate is at half-maximum, and is an inverse measure of the substrate's affinity for the enzyme—as a small $K_m$ indicates high affinity, meaning that the rate will approach $V_{max}$ more quickly. The value of $K_m$ is dependent on both the enzyme and the substrate, as well as conditions such as temperature and pH.
The Michaelis constant $K_m$ is the substrate concentration at which the reaction rate is at half-maximum.
From the last two terms in Equation $\ref{13.27}$, we can express $V_{max}$ in terms of a turnover number ($k_{cat}$):
$V_{max} = k_{cat}[E]_o$
where $[E]_0$ is the enzyme concentration and $k_{cat}$ is the turnover number, defined as the maximum number of substrate molecules converted to product per enzyme molecule per second. Hence, the turnover number is defined as the maximum number of chemical conversions of substrate molecules per second that a single catalytic site will execute for a given enzyme concentration $[E]_o$.
Example $1$: Turnover number of acetylcholinesterase
Acetylcholinesterase (AChE) may be one of the fastest enzymes. It hydrolyzes acetylcholine to choline and an acetate group. One of the earliest values of the turnover number was $3 \times 10^7$ (molecules of acetylcholine) per minute per molecule of enzyme. A more recent value at 25°C, pH = 7.0, acetylcholine concentration of $2.5 \times 10^{−3}\; M$, was found to be $7.4 \times 10^5\; min^{−1}$ (J Biol Chem. 236 (8): 2292–5.).
There may be some 30 active centers per molecule. AChE is a serine hydrolase that reacts with acetylcholine at close to the diffusion-controlled rate. A diffusion-controlled reaction occurs so quickly that the reaction rate is the rate of transport of the reactants through the solution; as quickly as the reactants encounter each other, they react.
The Significance of $K_M$ and $V_{max}$
The Michaelis-Menten model is used in a variety of biochemical situations other than enzyme-substrate interaction, including antigen-antibody binding, DNA-DNA hybridization, and protein-protein interaction. It can be used to characterize a generic biochemical reaction, in the same way that the Langmuir equation can be used to model generic adsorption of biomolecular species. When an empirical equation of this form is applied to microbial growth. The experimentally determined parameters values vary wildly between enzymes (Table $1$):
Table $1$: Enzyme Kinetic parameters
Enzyme $K_m$ (M) $k_{cat}$ (1/s) $k_{cat}/K_m$ (1/M.s)
Chymotrypsin 1.5 × 10−2 0.14 9.3
Pepsin 3.0 × 10−4 0.50 1.7 × 103
Tyrosyl-tRNA synthetase 9.0 × 10−4 7.6 8.4 × 103
Ribonuclease 7.9 × 10−3 7.9 × 102 1.0 × 105
Carbonic anhydrase 2.6 × 10−2 4.0 × 105 1.5 × 107
Fumarase 5.0 × 10−6 8.0 × 102 1.6 × 108
While $K_m$ is equal to the substrate concentration at which the enzyme converts substrates into products at half its maximal rate and hence is related to the affinity of the substrate for the enzyme. The catalytic rate $k_{cat}$ is the rate of product formation when the enzyme is saturated with substrate and therefore reflects the enzyme's maximum rate. The rate of product formation is dependent on both how well the enzyme binds substrate and how fast the enzyme converts substrate into product once substrate is bound. For a kinetically perfect enzyme, every encounter between enzyme and substrate leads to product and hence the reaction velocity is only limited by the rate the enzyme encounters substrate in solution. From Equation $\ref{Eq13.24}$, the catalytic efficiency of a protein can be evaluated.
$\dfrac{k_{cat}}{K_m} = \dfrac{k_2}{K_m} = \dfrac{k_1k_2}{k_{-1} + k_2}$
This $k_{cat}/K_m$ ratio is called the specificity constant measure of how efficiently an enzyme converts a substrate into product. It has a theoretical upper limit of 108 – 1010 /M.s; enzymes working close to this, such as fumarase, are termed superefficient (Table $1$).
Determining $V_m$ and $K_m$ from experimental data can be difficult and the most common way is to determine initial rates, $v_0$, from experimental values of $[P]$ or $[S]$ as a function of time. Hyperbolic graphs of $v_0$ vs. $[S]$ can be fit or transformed as we explored with the different mathematical transformations of the hyperbolic binding equation to determine $K_d$. These included:
• nonlinear hyperbolic fit (e.g., Figure $1$)
• double reciprocal plot (e.g., Lineweaver–Burk plot discussed below
• Eadie-Hofstee plot
Lineweaver–Burk plot
Another commonly-used plot in examining enzyme kinetics is the Lineweaver-Burk plot, in with the inverse of the reaction rate, $1/r$, is plotted against the inverse of the substrate concentration $1/\left[ \text{S} \right]$. Rearranging Equation $\ref{Eq13.26}$,
$\dfrac{1}{r} = \dfrac{K_M + \left[ \text{S} \right]}{k_2 \left[ \text{E} \right]_0 \left[ \text{S} \right]} = \dfrac{K_M}{k_2 \left[ \text{E} \right]_0} \dfrac{1}{\left[ \text{S} \right]} + \dfrac{1}{k_2 \left[ \text{E} \right]_0} \label{Eq28}$
Tthe Lineweaver–Burk plot (or double reciprocal plot) is a graphical representation of the Lineweaver–Burk equation of enzyme kinetics, described by Hans Lineweaver and Dean Burk in 1934 (Figure $2$). The Lineweaver-Burk plot results in a straight line with the slope equal to $K_M/k_2 \left[ \text{E} \right]_0$ and $y$-intercept equal to $1/k_2 \left[ \text{E} \right]_0$ which is $1/V_{max}$ via Equation $\ref{Eq13.26}$.
The plot provides a useful graphical method for analysis of the Michaelis–Menten equation:
$V = \dfrac{V_{\max} [S]}{K_m + [S]}$
Taking the reciprocal gives
$\dfrac{1}{V} = \dfrac {K_m + [S]} {V_{max}[S]} = \dfrac{K_m}{V_{max}} \dfrac{1}{[S]} + \dfrac{1}{V_{max}}$
where
• $V$ is the reaction velocity (the reaction rate),
• $K_m$ is the Michaelis–Menten constant,
• $V_{max}$ is the maximum reaction velocity, and
• $[S]$ is the substrate concentration.
The Lineweaver–Burk plot was widely used to determine important terms in enzyme kinetics, such as $K_m$ and $V_{max}$, before the wide availability of powerful computers and non-linear regression software. The y-intercept of such a graph is equivalent to the inverse of $V_{max}$; the x-intercept of the graph represents $−1/K_m$. It also gives a quick, visual impression of the different forms of enzyme inhibition.
Example $2$
The reaction between nicotineamide mononucleotide and ATP to form nicotineamide–adenine dinucleotide and pyrophosphate is catalyzed by the enzyme nicotinamide mononucleotide adenylyltransferase. The following table provides typical data obtained at a pH of 4.95. The substrate, S, is nicotinamide mononucleotide and the initial rate, v, is the μmol of nicotinamide–adenine dinucleotide formed in a 3-min reaction period.
[S] (mM) v (μmol) [S] (mM) v (μmol)
0.138 0.148 0.560 0.324
0.220 0.171 0.766 0.390
0.291 0.234 1.460 0.493
Determine values for Vmax and Km.
Solution
Figure 13.12 shows the Lineweaver–Burk plot for this data and the resulting regression equation. Using the y-intercept, we calculate Vmax as
Vmax= 1 / y−intercept = 1 / 1.708 mol = 0.585 mol
and using the slope we find that Km is
Km = slope × Vmax = 0.7528 molimM × 0.585 mol = 0.440 mM
-diphenyl oxidase
The following data are for the oxidation of catechol (the substrate) to o-quinone by the enzyme o-diphenyl oxidase. The reaction is followed by monitoring the change in absorbance at 540 nm. The data in this exercise are adapted from jkimball.
[catechol] (mM)
0.3
0.6
1.2
4.8
rate (∆AU/min)
0.020
0.035
0.048
0.081
Determine values for Vmax and Km.
Click here to review your answer to this exercise.
The double reciprocal plot distorts the error structure of the data, and it is therefore unreliable for the determination of enzyme kinetic parameters. Although it is still used for representation of kinetic data, non-linear regression or alternative linear forms of the Michaelis–Menten equation such as the Hanes-Woolf plot or Eadie–Hofstee plot are generally used for the calculation of parameters.
Problems with the Method
The Lineweaver–Burk plot is classically used in older texts, but is prone to error, as the y-axis takes the reciprocal of the rate of reaction – in turn increasing any small errors in measurement. Also, most points on the plot are found far to the right of the y-axis (due to limiting solubility not allowing for large values of $[S]$ and hence no small values for $1/[S]$), calling for a large extrapolation back to obtain x- and y-intercepts.
When used for determining the type of enzyme inhibition, the Lineweaver–Burk plot can distinguish competitive, non-competitive and uncompetitive inhibitors. Competitive inhibitors have the same y-intercept as uninhibited enzyme (since $V_{max}$ is unaffected by competitive inhibitors the inverse of $V_{max}$ also doesn't change) but there are different slopes and x-intercepts between the two data sets. Non-competitive inhibition produces plots with the same x-intercept as uninhibited enzyme ($K_m$ is unaffected) but different slopes and y-intercepts. Uncompetitive inhibition causes different intercepts on both the y- and x-axes but the same slope.
Eadie–Hofstee Plot
The Eadie–Hofstee plot is a graphical representation of enzyme kinetics in which reaction rate is plotted as a function of the ratio between rate and substrate concentration and can be derived from the Michaelis–Menten equation ($\ref{Eq13.25}$) by inverting and multiplying with $V_{max}$:
$\dfrac{V_{max}}{v} = \dfrac{V_{max}(K_m+[S])}{V_{max}[S]} = \dfrac{K_m+[S]}{[S]}$
Rearrange:
$V_{max} = \dfrac{vK_m}{[S]} + \dfrac{v[S]}{[S]} = \dfrac{v K_m}{[S]} + v$
Isolate v:
$v = -K_m \dfrac{v}{[S]} + V_{max}$
A plot of v against $v/[S]$ will hence yield $V_{max}$ as the y-intercept, $V_{max}/K_m$ as the x-intercept, and $K_m$ as the negative slope (Figure $3$). Like other techniques that linearize the Michaelis–Menten equation, the Eadie-Hofstee plot was used historically for rapid identification of important kinetic terms like $K_m$ and $V_{max}$, but has been superseded by nonlinear regression methods that are significantly more accurate and no longer computationally inaccessible. It is also more robust against error-prone data than the Lineweaver–Burk plot, particularly because it gives equal weight to data points in any range of substrate concentration or reaction rate (the Lineweaver–Burk plot unevenly weights such points). Both Eadie-Hofstee and Lineweaver–Burk plots remain useful as a means to present data graphically.
Problems with the Method
One drawback from the Eadie–Hofstee approach is that neither ordinate nor abscissa represent independent variables: both are dependent on reaction rate. Thus any experimental error will be present in both axes. Also, experimental error or uncertainty will propagate unevenly and become larger over the abscissa thereby giving more weight to smaller values of v/[S]. Therefore, the typical measure of goodness of fit for linear regression, the correlation coefficient R, is not applicable.
Contributors and Attributions
• World Public Library
• Wikipedia
• Dr. S.K. Khare (IIT Delhi) via NPTEL
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/10%3A_Enzyme_Kinetics/10.02%3A_The_Equations_of_Enzyme_Kinetics.txt
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Chymotrypsin is a digestive enzyme belonging to a super family of enzymes called serine proteases. It uses an active serine residue to perform hydrolysis on the C-terminus of the aromatic amino acids of other proteins. Chymotrypsin is a protease enzyme that cleaves on the C-terminal phenylalanine (F), tryptophan (W), and tyrosine (Y) on peptide chains. It shows specificity for aromatic amino acids because of its hydrophobic pocket.
Introduction
Chymotrypsin is one of the most studied enzymes due to its two phase kinetics: pre-steady-state and steady state. The study of these two kinetic states gives evidence of the "Ping-Pong" mechanism, the formation of covalent complexes leading to covalent hydrolysis reactions, and the rate of the catalyzed reactions. Synthesis of chymotrypsin occurs primarily in the pancreas. Instead of the active form, however, it is produced as an inactive zymogen called chymotrypsinogen to prevent its protease activity from digesting the pancreas. Upon secretion into the lumen of the small intestine, it is converted to its active form by another enzyme called trypsin. This dependence of a different enzyme for the activation of a protease is a common way for the body to prevent the digestion of organs and other harmful enzymatic side-effects.
Chymotrypsin operates through a general mechanism known as the ping-pong mechanism (Figure $1$) whereby the enzyme reacts with a substrate to form an enzyme intermediate. This intermediate has different properties than the initial enzyme, so to regenerate the initial enzymatic activity, it must react with a secondary substrate. This process is illustrated below:
More specifically, chymotrypsin operates through a particular type of ping-pong mechanism called covalent hydrolysis. This means that the enzyme first forms a covalent bond with the target substrate, displacing the more stable moiety into solution. This enzyme-substrate complex is called the enzyme intermediate. The intermediate then reacts with water, which displaces the remaining part of the initial substrate and reforms the initial enzyme.
Chymotrypsin, like most enzymes, is specific in the types of substrates with which it reacts. As a protease, it cleaves polypeptides, and its inherent specificity allows it to act only on the carboxy-terminal of aromatic residues. It is a somewhat complicated mechanism, and is best explained in a series of steps.
Step 1: The target enters the active site of chymotrypsin, and it is held there by hydrophobic interactions between exposed non-polar groups of enzyme residues and the non-polar aromatic side-chain of the substrate. It is important to note the hydrogen bond between the Schiff nitrogen on histidine-57 and the oxygen side-chain of serine-195.
Step 2: Aided by the histidine-serine hydrogen bonding, the hydroxyl group on serine-195 performs a nucleophilic attack on the carbonyl carbon of an aromatic amino acid while simultaneously transferring the hydroxyl hydrogen to the histidine Schiff nitrogen. This attack pushes the pi carbonyl electrons onto the carbonyl oxygen, forming a short-lived intermediate consisting of a c-terminal carbon with four single bonds: an oxygen anion, the beta-carbon of the aromatic amino acid, the n-terminus of the subsequent amino acid of the substrate protein, and the serine-195 side-chain oxygen.
Step 3: This intermediate is short-lived, as the oxyanion electrons reform the pi bond with the c-terminus of the aromatic amino acid. The bond between the carboxy-terminus of the aromatic amino acid and the n-terminus of the subsequent residue is cleaved, and its electrons are used to extract the hydrogen of the protonated Schiff nitrogen on histidine-57. The bonds between the carbonyl carbon and the serine-195 oxygen remain in an ester configuration. This is called the acyl-enzyme intermediate. The c-terminal side of the polypeptide is now free to dissociate from the active site of the enzyme.
Step 4: Water molecules are now able to enter and bind to active site through hydrogen bonding between the hydrogen atoms of water and the histidine-57 Schiff nitrogen.
Step 5: The water oxygen now makes a nucleophilic attack on the carbonyl carbon of the acyl-enzyme intermediate, pushing the carbonyl’s pi electrons onto the carbonyl carbon as histidine-57 extracts one proton from water. This forms another quaternary carbon covalently bonded with serine, a hydroxyl, an oxyanion, and the aromatic amino acid. The proton on the recently protonated histidine-57 is now able to make a hydrogen bond with the serine oxygen.
Step 6: The oxyanion electrons reform the carbonyl pi bond, cleaving the bond between the carbonyl carbon and the serine hydroxyl. The electrons in this bond are used by the serine oxygen to deprotonate the histidine Schiff nitrogen and reform the original enzyme. The substrate no longer has affinity for the active site, and it soon dissociates from the complex.
Kinetics
Experiments were conducted in 1953 by B.S. Hartley and B.A. Kilby to investigate the kinetics of chymotrypsin-catalyzed hydrolysis. Instead of using a poly-peptide chain as a substrate, they used a nitro-phenyl ester, p-nitrophenyl acetate, that resembles an aromatic amino acid. Hydrolysis of this compound by chymotrypsin at the carbonyl group yields acetate and nitrophenolate, the latter of which absorbs near 400 nm light and its concentration can thus be measured by spectrophotometry (Figure $2$).
Spectrophotometric analysis of chymotrypsin acting on nitrophenylacetate showed that nitrophenolate was produced at a rate independent of substrate concentration, proving that the only factor contributing to the rate of product formation is the concentration of enzyme; this is typical for enzyme-substrate kinetics. However, when the slope of the 0-order absorbance plot was traced back to the starting point (time = 0), it was found that the initial concentration of nitrophenolate was not 0. In fact, it showed a 1:1 stoichiometric ratio with the amount of chymotrypsin used in the assay. This can only be explained by the fact that hydrolysis by chymotrypsin is biphasic in nature, meaning that it proceeds in two distinct steps.
1. The first step, which describes the initial burst of nitrophenolate seen in Hartley and Kilby’s absorbance plot, is the fastest. The attack of the nitrophenyl acetate substrate by chymotrypsin immediately cleaves the nitrophenolate moiety and leaves the acetate group attached to chymotrypsin, rendering the enzyme inactive.
2. The second step has been deduced to involve the hydrolysis of the acetate group from the inactivated chymotrypsin to regenerate the original enzyme.
To analytically determine the rate of catalysis, all substrates, products, and intermediates must be defined. Refer to the figure below:
Using these abbreviations, kinetic analysis becomes less cumbersome.
1. The initial amount of enzyme can be represented as the sum of the free enzyme, the bound enzyme, and the inactive intermediate.
$[E]_o = [ES] + [^*ES] + [E] \nonumber$
2. Assuming the initial step is the faster than the subsequent steps, the rate of nitrophenolate production can be described as:
$\dfrac{d[P_1]}{dt}=k_2[ES] \nonumber$
3. Likewise, the rate of acetate formation can be represented by the equation:
$\dfrac{d[P_2]}{dt}=k_3[^*ES] \nonumber$
4. Therefore, the net change in concentration of the inactive intermediate can be deduced:
$\dfrac{d[*ES]}{dt}=k_2[ES]-k_3[^*ES] \nonumber$
5. The last inference that can be made from analysis of the measured kinetics data (Figure $2$) is that the first step of the reaction equilibrates rapidly, and thus the change in bound substrate can be described in the following equation. This is a principal tenet in analyzing the kinetics of chymotrypsin and is a ubiquitous mechanism in biological enzyme catalysis.
$\dfrac{d[ES]}{dt}=k_1[E][S]-k_{-1}[ES]=0 \nonumber$
6. Where:
$\dfrac{k_{-1}}{k_1} = K_s = \dfrac{[E][S]}{[ES]} \nonumber$
7. Combining all of these quantities, we can deduce the catalytic rate constant as:
$k_{cat} = \dfrac{k_2k_3}{k_2+k_3} \nonumber$
8. In ester hydrolysis, $k_3 >> k_2$, so the resultant catalytic rate constant simplifies to:
$k_{cat}=k_2 \nonumber$
which is in agreement with the observed zeroth-order kinetics of Figure $2$.
Table $1$: Kinetic Constants of the Chymotrypsin-Catalyzed Hydrolysis of p-Nitrophenyl Trimethylacetate at pH 8.2. 0.01 M tris-HCL buffer, ionic strength 0.06, 25.6 ± 0.1 °C, 1.8% (v/v) acetonitrile-water. From M.L. Bender, F.J. Kezdy and F.C. Wedler, J. Chem. Educ. 44, 84 (1967
k2 0.37 ± 0.11 s-1
k3 (1.3 ± 0.02) × 10-4 s-1
Ks (1.6 ± 0.5) × 10-3
kcat 1.3 × 10-4 s-1
KM 5.6 × 10-7
Exercise $1$
Speculate on how the catalytic rate constant can be determined from the spectrophotogram.
Answer
The catalytic rate constant can be deduced from the graph by simply determining the slope of the line where the reaction demonstrates 0-order kinetics (the linear part)
Exercise $2$
How can product be consistently produced if the rate of change of the ES complex is 0?
Answer
This is pre-equilibrium kinetics in action. The ES complex is formed from E and S at a faster rate than any other step in the reaction. As soon as ES is converted to *ES, another mole of ES is produced from an infinite supply of E + S. This means that the amount of ES and E + S is constantly at equilibrium, and thus the change of either with respect to time is 0.
Exercise $3$
How would the rate of product formation change if:
1. the substrate concentration was doubled.
2. the enzyme concentration was doubled.
3. The reaction was carried out in mono-deuterated water instead of H2O (comment qualitatively).
Answer
1. No change.
2. Two fold increase.
3. Since water is involved in the final, slowest step of the mechanism, deuterating the water would decrease the rate of the overall reaction from 5 to 30-fold.
Exercise $4$
Explain the role of hydrogen bonding in protein hydrolysis catalyzed by chymotrypsin.
Answer
Initially, hydrogen bonding between the enzymes histidine and serine side chains weakens the bond of serine’s O-H. This allows for a facilitated nucleophilic attack of the hydroxyl Oxygen on the substrates carbonyl group. Conversely, in the final step of the reaction, the bound serine oxygen forms a hydrogen bond with a protonated histidine, which allows for easier cleavage from the substrate.
Exercise $5$
What would the spectrophotogram look like if the reaction proceeded via a steady-state mechanism instead of pre-equilibrium.
Answer
The graph would show similar 0-order kinetics, but the line would intercept the Y-axis at an absorbance of 0 instead of the 1:1 mole ratio of nitrophenolate to enzyme.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/10%3A_Enzyme_Kinetics/10.03%3A_Chymotrypsin-_A_Case_Study.txt
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The Michaelis –Menten model of enzyme kinetics was derived for single substrate reactions. Enzymatic reactions requiring multiple substrates and yielding multiple products are more common and yielding multiple products are more common than single-substrate reaction. In these types of reactions, the all the substrates involved are bound to the enzyme before catalysis of the reaction takes place to release the products. Sequential reactions can be either ordered or random. In contrast to the Michealis-Menton kinetics where a binary Enzyme-Substrate complex is generated in the mechanism (\([ES]\), in bisubstrate enzyme reactions, a ternary complex of the enzyme and two substrates is generated:
\[ A + B \overset{E}{\longrightleftharpoons} P+ Q\]
Bisubstrate reactions account for ~ 60% of the known enzymatic reactions. Multi-substrate reactions follow complex rate equations that describe how the substrates bind and in what sequence. The analysis of these reactions is much simpler if the concentration of substrate \(A\) is kept constant and substrate \(B\) varied. Under these conditions, the enzyme behaves just like a single-substrate enzyme and a plot of \(v\) by \([S]\) gives apparent \(K_M\) and \(V_{max}\) constants for substrate B. If a set of these measurements is performed at different fixed concentrations of A, these data can be used to work out what the mechanism of the reaction is. For an enzyme that takes two substrates A and B and turns them into two products P and Q, there are two types of mechanism: ternary complex and ping–pong.
How do you resolve the enzymes kinetics of these more complicated systems? The answer is fairly straightforward. You keep one of the substrates (B, for example) fixed, and vary the other substrate (A) and obtain a series of hyperbolic plots of \(v_o\) vs \(A\) at different fixed \(B\) concentrations. This would give a series of linear \(1/v\) vs. \(1/A\) double-reciprocal plots (Lineweaver-Burk plots) as well. The pattern of Lineweaver-Burk plots depends on how the reactants and products interact with the enzyme.
Sequential Mechanism
In this mechanism, both substrates must bind to the enzyme before any products are made and released.The substrates might bind to the enzyme in a random fashion (A first then B or vice-versa) or in an ordered fashion (A first followed by B). An abbreviated notation scheme developed by W.W. Cleland is shown in Figure \(1\) for the sequential random and sequential ordered mechanisms (Figure \(1\)). For both mechanisms, Lineweaver-Burk plots at varying A and different fixed values of B give a series of intersecting lines. Derivative curves can be solved to obtain appropriate kinetic constants. In ordered sequential reactions, all the substrates are first bound to the enzyme in a defined order or sequence. The products, too, are released after catalysis in a defined order or sequence.
An example is the lactate dehydrogenase enzyme, which is a protein that catalyzes glucose metabolism. In this ordered mechanism, the coenzyme, NADH, always binds first, with pyruvate binding afterward. During the reaction, the pyruvate is reduced to lactate while NADH is oxidized to NAD+ by the enzyme. Lactate is then released first, followed by the release of NAD+.
This is a characteristic of a ternary complex, which consists of three molecules that are bound together. Before catalysis, the substrates and coenzyme are bound to the enzyme. After catalysis, the complex consists of the enzyme and products, NAD+ and lactate.
In random sequential reactions, the substrates and products are bound and then released in no preferred order, or "random" order (Figure \(1\)). An example is the creatine kinase enzyme, which catalyzes creatine and ATP (the two substrates) to form phosphocreatine and ADP (teh Products) in Figure \(4\). In this case, either substrates may bind first and either products may be released first. A ternary complex is still observed for random sequential reactions. Before catalysis, the complex is generated that includes the enzyme, ATP, and creatine. After catalysis, the complex consists of the enzyme, ADP, and phosphocreatine.
Figure \(4\): The metabolism of creatine follows a random sequential mechanism.
Ping-Pong Mechanism
In this mechanism, one substrate bind first to the enzyme followed by product P release. Typically, product P is a fragment of the original substrate A.The rest of the substrate is covalently attached to the enzyme E, which is designated as E'. Now the second reactant, B, binds and reacts with the enzyme to form a covalent adduct with the A as it is covalentattached to the enzyme to form product Q. This is now released and the enzyme is restored to its initial form, E. This represents a ping-pong mechanism. An abbreviated notation scheme is shown below for the ping-pong mechanisms. For this mechanism, Lineweaver-Burk plots at varying A and different fixed values of B give a series of parallel lines. An example of this type of reaction might be low molecular weight protein tyrosine phosphatase against the small substrate p-initrophenylphosphate (A) which binds to the enzyme covalently with the expulsion of the product P, the p-nitrophenol leaving group. Water (B) then comes in and covalently attacks the enzyme, forming an adduct with the covalently bound phosphate releasing it as inorganic phosphate. In this particular example, however, you cannot vary the water concentration and it would be impossible to generate the parallel Lineweaver-Burk plots characteristic of ping-pong kinetics.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/10%3A_Enzyme_Kinetics/10.04%3A_Multisubstrate_Systems.txt
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Enzymes can be regulated in ways that either promote or reduce their activity. There are many different kinds of molecules that inhibit or promote enzyme function, and various mechanisms exist for doing so. In some cases of enzyme inhibition, for example, an inhibitor molecule is similar enough to a substrate that it can bind to the active site and simply block the substrate from binding. When this happens, the enzyme is inhibited through competitive inhibition, because an inhibitor molecule competes with the substrate for active site binding. On the other hand, in noncompetitive inhibition, an inhibitor molecule binds to the enzyme in a location other than an allosteric site and still manages to block substrate binding to the active site.
Elucidating Mechanisms for the Inhibition of Enzyme Catalysis
When an inhibitor interacts with an enzyme it decreases the enzyme’s catalytic efficiency. An irreversible inhibitor covalently binds to the enzyme’s active site, producing a permanent loss in catalytic efficiency even if we decrease the inhibitor’s concentration. A reversible inhibitor forms a noncovalent complex with the enzyme, resulting in a temporary decrease in catalytic efficiency. If we remove the inhibitor, the enzyme’s catalytic efficiency returns to its normal level.
There are several pathways for the reversible binding of an inhibitor to an enzyme, as shown in Figure \(1\). In competitive inhibition the substrate and the inhibitor compete for the same active site on the enzyme. Because the substrate cannot bind to an enzyme–inhibitor complex, EI, the enzyme’s catalytic efficiency for the substrate decreases. With noncompetitive inhibition the substrate and the inhibitor bind to different active sites on the enzyme, forming an enzyme–substrate–inhibitor, or ESI complex. The formation of an ESI complex decreases catalytic efficiency because only the enzyme–substrate complex reacts to form the product. Finally, in uncompetitive inhibition the inhibitor binds to the enzyme–substrate complex, forming an inactive ESI complex.
We can identify the type of reversible inhibition by observing how a change in the inhibitor’s concentration affects the relationship between the rate of reaction and the substrate’s concentration. As shown in Figure 13.14, when we display kinetic data using as a Lineweaver-Burk plot it is easy to determine which mechanism is in effect. For example, an increase in slope, a decrease in the x-intercept, and no change in the y-intercept indicates competitive inhibition. Because the inhibitor’s binding is reversible, we can still obtain the same maximum velocity—thus the constant value for the y-intercept—by adding enough substrate to completely displace the inhibitor. Because it takes more substrate, the value of Km increases, which explains the increase in the slope and the decrease in the x-intercept’s value.
Example \(1\)
Practice Exercise 13.3 provides kinetic data for the oxidation of catechol (the substrate) to o-quinone by the enzyme o-diphenyl oxidase in the absence of an inhibitor. The following additional data are available when the reaction is run in the presence of p-hydroxybenzoic acid, PBHA. Is PBHA an inhibitor for this reaction and, if so, what type of inhibitor is it?
[catechol] (mM) 0.3 0.6 1.2 4.8
rate (∆AU/min) 0.011 0.019 0.022 0.060
Solution
Figure \(3\) shows the resulting Lineweaver–Burk plot for the data in Practice Exercise 13.3 and Example 13.7. Although the y-intercepts are not identical in value—the result of uncertainty in measuring the rates—the plot suggests that PBHA is a competitive inhibitor for the enzyme’s reaction with catechol.
Exercise \(1\)
Practice Exercise 13.3 provides kinetic data for the oxidation of catechol (the substrate) to o-quinone by the enzyme o-diphenyl oxidase in the absence of an inhibitor. The following additional data are available when the reaction is run in the presence of phenylthiourea. Is phenylthiourea an inhibitor for this reaction and, if so, what type of inhibitor is it? The data in this exercise are adapted from jkimball.
[catechol] (mM) 0.3 0.6 1.2 4.8
rate (∆AU/min) 0.010 0.016 0.024 0.040
Click here to review your answer to this exercise.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/10%3A_Enzyme_Kinetics/10.05%3A_Enzyme_Inhibition.txt
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Learning Objectives
• When a substrate binds to one enzymatic subunit, the rest of the subunits are stimulated and become active. Ligands can either have non-cooperativity, positive cooperativity or negative cooperativity.
A significant portion of enzymes function such that their properties can be studied using the Michaelis-Menten equation. However, a particular class of enzymes exhibit kinetic properties that cannot be studied using the Michaelis-Menten equation. The rate equation of these unique enzymes is characterized by an “S-shaped” sigmoidal curve, which is different from the majority of enzymes whose rate equation exhibits hyberbolic curves. Allosteric regulation is the regulation of an enzyme or other protein by binding an effector molecule at the protein's allosteric site (that is, a site other than the protein's active site). Effectors that enhance the protein's activity are referred to as allosteric activators, whereas those that decrease the protein's activity are called allosteric inhibitors. The term allostery refers to the fact that the regulatory site of an allosteric protein is physically distinct from its active site. Allosteric regulations are a natural example of control loops, such as feedback from downstream products or feedforward from upstream substrates. Long-range allostery is especially important in cell signaling.
Allosteric Modulation (Cooperativity)
Cooperativity is a phenomenon displayed by enzymes or receptors that have multiple binding sites where the affinity of the binding sites for a ligand is increased, positive cooperativity, or decreased, negative cooperativity, upon the binding of a ligand to a binding site. We also see cooperativity in large chain molecules made of many identical (or nearly identical) subunits (such as DNA, proteins, and phospholipids), when such molecules undergo phase transitions such as melting, unfolding or unwinding. This is referred to as subunit cooperativity (discussed below).
An example of positive cooperativity is the binding of oxygen to hemoglobin. One oxygen molecule can bind to the ferrous iron of a heme molecule in each of the four chains of a hemoglobin molecule. Deoxy-hemoglobin has a relatively low affinity for oxygen, but when one molecule binds to a single heme, the oxygen affinity increases, allowing the second molecule to bind more easily, and the third and fourth even more easily. The oxygen affinity of 3-oxy-hemoglobin is ~300 times greater than that of deoxy-hemoglobin. This behavior leads the affinity curve of hemoglobin to be sigmoidal, rather than hyperbolic as with the monomeric myoglobin. By the same process, the ability for hemoglobin to lose oxygen increases as fewer oxygen molecules are bound.
Negative allosteric modulation (also known as allosteric inhibition) occurs when the binding of one ligand decreases the affinity for substrate at other active sites. For example, when 2,3-BPG binds to an allosteric site on hemoglobin, the affinity for oxygen of all subunits decreases. This is when a regulator is absent from the binding site.
Another instance in which negative allosteric modulation can be seen is between ATP and the enzyme Phosphofructokinase within the negative feedback loop that regulates glycolysis. Phosphofructokinase (generally referred to as PFK) is an enzyme that catalyses the third step of glycolysis: the phosphorylation of Fructose-6-phosphate into Fructose 1,6-bisphosphate. PFK can be allosterically inhibited by high levels of ATP within the cell. When ATP levels are high, ATP will bind to an allosteric site on phosphofructokinase, causing a change in the enzyme's three-dimensional shape. This change causes its affinity for substrate (fructose-6-phosphate and ATP) at the active site to decrease, and the enzyme is deemed inactive. This causes glycolysis to cease when ATP levels are high, thus conserving the body's glucose and maintaining balanced levels of cellular ATP. In this way, ATP serves as a negative allosteric modulator for PFK, despite the fact that it is also a substrate of the enzyme.
Sigmoidal kinetic profiles are the result of enzymes that demonstrate positive cooperative binding. cooperativity refers to the observation that binding of the substrate or ligand at one binding site affects the affinity of other sites for their substrates. For enzymatic reactions with multiple substrate binding sites, this increased affinity for the substrate causes a rapid and coordinated increase in the velocity of the reaction at higher $[S]$ until $V_{max}$ is achieved. Plotting the $V_0$ vs. $[S]$ for a cooperative enzyme, we observe the characteristic sigmoidal shape with low enzyme activity at low substrate concentration and a rapid and immediate increase in enzyme activity to $V_{max}$ as $[S]$ increases. The phenomenon of cooperativity was initially observed in the oxygen-hemoglobin interaction that functions in carrying oxygen in blood. Positive cooperativity implies allosteric binding – binding of the ligand at one site increases the enzyme’s affinity for another ligand at a site different from the other site. Enzymes that demonstrate cooperativity are defined as allosteric. There are several types of allosteric interactions: homotropic (positive) and heterotropic (negative).
Positive and negative allosteric interactions (as illustrated through the phenomenon of cooperativity) refer to the enzyme's binding affinity for other ligands at other sites, as a result of ligand binding at the initial binding site. When the ligands interacting are all the same compounds, the effect of the allosteric interaction is considered homotropic. When the ligands interacting are different, the effect of the allosteric interaction is considered heterotropic. It is also very important to remember that allosteric interactions tend to be driven by ATP hydrolysis.
The Hill Equation
The degree of cooperativity is determined by Hill equation (Equation \ref{Eq1}) for non-Michaelis-Menten kinetics. The Hill equation accounts for allosteric binding at sites other than the active site. $n$ is the "Hill coefficient."
$\theta = \dfrac{[L]^n}{K_d+[L]^n} = \dfrac{[L]^n}{K_a^n+[L]^n} \label{Eq1}$
where
• $\theta$ is the fraction of ligand binding sites filled
• $[L]$ is the ligand concentration
• $K_d$ is the apparent dissociation constant derived from the law of mass action (equilibrium constant for dissociation)
• $K_a$ is the ligand concentration producing half occupation (ligand concentration occupying half of the binding sides), that is also the microscopic dissociation constant
• $n$ is the Hill coefficient that describes the cooperativity
Taking the logarithm of both sides of the equation leads to an alternative formulation of the Hill Equation.
$\log \left( \dfrac{\theta}{1-\theta} \right) = n\log [L] - \log K_d \label{Eq2}$
• when $n < 1$, there is negative cooperativity
• when $n = 1$, there is no cooperativity
• when $n > 1$, there is positive cooperativity
Allosteric Models
Currently, there are 2 models for illustrating cooperativity: the concerted model and the sequential model. Most allosteric effects can be explained by the concerted MWC model put forth by Monod, Wyman, and Changeux, or by the sequential model described by Koshland, Nemethy, and Filmer. Both postulate that enzyme subunits exist in one of two conformations, tensed (T) or relaxed (R), and that relaxed subunits bind substrate more readily than those in the tense state. The two models differ most in their assumptions about subunit interaction and the preexistence of both states.
The Concerted model
The concerted model of allostery, also referred to as the symmetry model or MWC model, postulates that enzyme subunits are connected in such a way that a conformational change in one subunit is necessarily conferred to all other subunits. Thus, all subunits must exist in the same conformation. The model further holds that, in the absence of any ligand (substrate or otherwise), the equilibrium favours one of the conformational states, T or R. The equilibrium can be shifted to the R or T state through the binding of one ligand (the allosteric effector or ligand) to a site that is different from the active site (the allosteric site).
The Sequential model
The sequential model of allosteric regulation holds that subunits are not connected in such a way that a conformational change in one induces a similar change in the others. Thus, all enzyme subunits do not necessitate the same conformation. Moreover, the sequential model dictates that molecules of substrate bind via an induced fit protocol. In general, when a subunit randomly collides with a molecule of substrate, the active site, in essence, forms a glove around its substrate. While such an induced fit converts a subunit from the tensed state to relaxed state, it does not propagate the conformational change to adjacent subunits. Instead, substrate-binding at one subunit only slightly alters the structure of other subunits so that their binding sites are more receptive to substrate. To summarize:
• subunits need not exist in the same conformation
• molecules of substrate bind via induced-fit protocol
• conformational changes are not propagated to all subunits
Note: Allosteric database
Allostery is a direct and efficient means for regulation of biological macromolecule function, produced by the binding of a ligand at an allosteric site topographically distinct from the orthosteric site. Due to the often high receptor selectivity and lower target-based toxicity, allosteric regulation is also expected to play an increasing role in drug discovery and bioengineering. The AlloSteric Database (ASD, provides a central resource for the display, search and analysis of the structure, function and related annotation for allosteric molecules. Currently, ASD contains allosteric proteins from more than 100 species and modulators in three categories (activators, inhibitors, and regulators). Each protein is annotated with detailed description of allostery, biological process and related diseases, and each modulator with binding affinity, physicochemical properties and therapeutic area. Integrating the information of allosteric proteins in ASD should allow the prediction of allostery for unknown proteins, to be followed with experimental validation. In addition, modulators curated in ASD can be used to investigate potential allosteric targets for a query compound, and can help chemists to implement structure modifications for novel allosteric drug design.
Summary
Allosteric enzymes are an exception to the Michaelis-Menten model. Because they have more than two subunits and active sites, they do not obey the Michaelis-Menten kinetics, but instead have sigmoidal kinetics. Since allosteric enzymes are cooperative, a sigmoidal plot of $V_0$ versus $[S]$ results: There are distinct properties of Allosteric Enzymes that makes it different compared to other enzymes.
1. One is that allosteric enzymes do not follow the Michaelis-Menten Kinetics. This is because allosteric enzymes have multiple active sites. These multiple active sites exhibit the property of cooperativity, where the binding of one active site affects the affinity of other active sites on the enzyme. As mentioned earlier, it is these other affected active sites that result in a sigmoidal curve for allosteric enzymes.
2. Allosteric Enzymes are influenced by substrate concentration. For example, at high concentrations of substrate, more enzymes are found in the R state. The T state is favorite when there is an insufficient amount of substrate to bind to the enzyme. In other words, the T and R state equilibrium depends on the concentration of the substrate.
3. Allosteric Enzymes are regulated by other molecules. This is seen when the molecules 2,3-BPG, pH, and CO2 modulates the binding affinity of hemoglobin to oxygen. 2,3-BPG reduces binding affinity of O2 to hemoglobin by stabilizing the T- state. Lowering the pH from physiological pH=7.4 to 7.2 (pH in the muscles and tissues) favors the release of $O_2$. Hemoglobin is more likely to release oxygen in $CO_2$ rich areas in the body.
There are two primary models for illustrating cooperativity. The concerted model (also called the Monod-Wyman-Changeux model) illustrates cooperativity by assuming that proteins have two or more subunits, and that each part of the protein molecule is able to exist in either the relaxed (R) state or the tense (T) state - the tense state of a protein molecule is favored when it doesn't have any substrates bound. All aspects, including binding and dissociation constants are the same for each ligand at the respective binding sites. The sequential model aims to demonstrate cooperativity by assuming that the enzyme/protein molecule affinity is relative and changes as substrates bind. Unlike the concerted model, the sequential model accounts for different species of the protein molecule.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/10%3A_Enzyme_Kinetics/10.06%3A_Allosteric_Interactions.txt
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In the same way that every enzyme has an optimum temperature, so each enzyme also has an optimum pH at which it works best. For example, trypsin and pepsin are both enzymes in the digestive system which break protein chains in the food into smaller bits - either into smaller peptide chains or into individual amino acids. Pepsin works in the highly acidic conditions of the stomach. It has an optimum pH of about 1.5. On the other hand, trypsin works in the small intestine, parts of which have a pH of around 7.5. Trypsin's optimum pH is about 8.
Table $1$: pH for Optimum Activity
Enzyme Optimal pH Enzyme Optimal pH
Lipase (pancreas) 8.0 Invertase 4.5
Lipase (stomach) 4.0 - 5.0 Maltase 6.1 - 6.8
Lipase (castor oil) 4.7 Amylase (pancreas) 6.7 - 7.0
Pepsin 1.5 - 1.6 Amylase (malt) 4.6 - 5.2
Trypsin 7.8 - 8.7 Catalase 7.0
Urease 7.0
If you think about the structure of an enzyme molecule, and the sorts of bonds that it may form with its substrate, it isn't surprising that pH should matter. Suppose an enzyme has an optimum pH around 7. Imagine that at a pH of around 7, a substrate attaches itself to the enzyme via two ionic bonds. In the diagram below, the groups allowing ionic bonding are caused by the transfer of a hydrogen ion from a -COOH group in the side chain of one amino acid residue to an -NH2 group in the side chain of another.
In this simplified example, that is equally true in both the substrate and the enzyme.
Now think about what happens at a lower pH - in other words under acidic conditions. It won't affect the -NH3+ group, but the -COO- will pick up a hydrogen ion. What you will have will be this:
You no longer have the ability to form ionic bonds between the substrate and the enzyme. If those bonds were necessary to attach the substrate and activate it in some way, then at this lower pH, the enzyme won't work. What if you have a pH higher than 7 - in other words under alkaline conditions. This time, the -COO- group won't be affected, but the -NH3+ group will lose a hydrogen ion. That leaves . . .
Again, there is no possibility of forming ionic bonds, and so the enzyme probably won't work this time either. At extreme pH's, something more drastic can happen. Remember that the tertiary structure of the protein is in part held together by ionic bonds just like those we've looked at between the enzyme and its substrate. At very high or very low pH's, these bonds within the enzyme can be disrupted, and it can lose its shape. If it loses its shape, the active site will probably be lost completely. This is essentially the same as denaturing the protein by heating it too much.
Kinetics
The rates of enzyme-catalysed reactions vary with pH and often pass through a maximum as the pH is varied. If the enzyme obeys Michaelis-Menten kinetics the kinetic parameters k0 and kA often behave similarly. The pH at which the rate or a suitable parameter is a maximum is called the pH optimum and the plot of rate or parameter against pH is called a pH profile. Neither the pH optimum nor the pH profile of an enzyme has any absolute significance and both may vary according to which parameter is plotted and according to the conditions of the measurements.
If the pH is changed and then brought back to its original value, the behavior is said to be reversible if the original properties of the enzyme are restored; otherwise it is irreversible. Reversible pH behavior may occur over a narrow range of pH, but effects of large changes in pH are in most cases irreversible. The diminution in rate as the pH is taken to the acid side of the optimum can be regarded as inhibition by hydrogen ions. The diminution in rate on the alkaline side can be regarded as inhibition by hydroxide ions. The equations describing pH effects are therefore analogous to inhibition equations. For single-substrate reactions the pH behavior of the parameters k0 and kA can sometimes be represented by an equation of the form
$k = \dfrac{k_{opt}}{1 + \dfrac{[H^+]}{K_1} + \dfrac{K_2}{[H^+]}} \label{eq1}$
in which k represents k0 or kA, and $k_{opt}$ is the value of the same parameter that would be observed if the enzyme existed entirely in the optimal state of protonation; it may be called the pH-independent value of the parameter. The constants K1 and K2 can sometimes be identified as acid dissociation constants for the enzyme. substrates or other species in the reaction mixture. The identification is, however, never straight forward and has to be justified by independent evidence. The behavior is frequently much more complicated than represented by Equation $\ref{eq1}$.
It is not accidental that this section has referred exclusively to pH dependences of k0 and kA. The pH dependence of the initial rate or, worse, the extent of reaction after a given time is rarely meaningful; the pH dependence of the Michaelis constant is often too complex to be readily interpretable.
The pH dependence of the Michaelis constant is often too complex to be readily interpretable.
Quenching Enzyme Activity
In principle, we can use any of the range of possible analytical techniques to follow a reaction’s kinetics provided that the reaction does not proceed to any appreciable extent during the time it takes to make a measurement. As you might expect, this requirement places a serious limitation on kinetic methods of analysis. If the reaction’s kinetics are slow relative to the analysis time, then we can make our measurements without the analyte undergoing a significant change in concentration. When the reaction’s rate is too fast—which often is the case—then we introduce a significant error if our analysis time is too long.
One solution to this problem is to stop, or quench the reaction by adjusting experimental conditions. For example, many reactions show a strong pH dependency, and may be quenched by adding a strong acid or a strong base. Figure 13.7 shows a typical example for the enzymatic analysis of p-nitrophenylphosphate using the enzyme wheat germ acid phosphatase to hydrolyze the analyte to p-nitrophenol.
The reaction has a maximum rate at a pH of 5. Increasing the pH by adding NaOH quenches the reaction and converts the colorless p-nitrophenol to the yellow-colored p-nitrophenolate, which absorbs at 405 nm.
10.08: The Effect of Temperature on Enzyme Kinetics
Enzymes are generally globular proteins, acting alone or in larger complexes. Like all proteins, enzymes are linear chains of amino acids that fold to produce a three-dimensional structure. The sequence of the amino acids specifies the structure which in turn determines the catalytic activity of the enzyme. Although structure determines function, a novel enzyme's activity cannot yet be predicted from its structure alone. Enzyme structures unfold (denature) when heated or exposed to chemical denaturants and this disruption to the structure typically causes a loss of activity.
Protein folding is key to whether a globular protein or a membrane protein can do its job correctly. It must be folded into the right shape to function. But hydrogen bonds, which play a big part in folding, are rather weak, and it does not take much heat, acidity, or other stress to break some and form others, denaturing the protein. This is one reason why tight homeostasis is physiologically necessary in many life forms.
Denaturation
Denaturation is a process in which proteins or nucleic acids lose the quaternary structure, tertiary structure and secondary structure which is present in their native state, by application of some external stress or compound such as a strong acid or base, a concentrated inorganic salt, an organic solvent (e.g., alcohol or chloroform), radiation or heat. If proteins in a living cell are denatured, this results in disruption of cell activity and possibly cell death. Denatured proteins can exhibit a wide range of characteristics, from conformational change and loss of solubility to aggregation due to the exposure of hydrophobic groups.
Enzyme denaturation is normally linked to temperatures above a species' normal level; as a result, enzymes from bacteria living in volcanic environments such as hot springs are prized by industrial users for their ability to function at high temperatures, allowing enzyme-catalyzed reactions to be operated at a very high rate.
• Wikipedia
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/10%3A_Enzyme_Kinetics/10.07%3A_The_Effect_of_pH_on_Enzyme_Kinetics.txt
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10.1: General Principles of Catalysis
Q10.1a
One day in class about enzyme kinetics, Jack comes over to you and asks, "I know enzymes are biological catalysts, but I do not understand how it works. Can you explain how enzymes make reactions go faster? And is it only faster in one direction?"
S10.1a
Because the activation energy is the energy hill between reactants and products, enzymes decreasing the size of the hill also decreases the amount of energy needed for reactions to go in either direction. A smaller energy hill allows reactants and products to overcome the barrier quicker, resulting a faster reaction rate.
Q10.1b
If your student colleagues argues that a catalysts affects only the rate of only one direction of a reaction. Explain why he is correct or not.
S10.1b
False. Catalysts affect rate by providing an alternative mechanism which has a lower transition state energy. It's impossible to lower transition state energy for only one direction of a reaction. (It'd be like making a hill shorter from the north, but keeping it the same height from the south.)
Q10.1c
How does the enzyme catalysis affect both forward and reverse reaction?
S10.1c
The enzyme catalyst lowers the Gibb energy of transition state, which reduces the activation energy of both reactions. Therefore, it makes reactions occur faster.
Q10.2a
Given enzyme-catalyzed reaction k1 = 4x106 M-1 s-1 , k-1 =6x104 s-1 and k2= 2.0x103 s-1. Determine if the enzyme –substrate binding follow the equilibrium or not ?
S10.2a
The dissociation constant ks = k-1 / k1 =6x104 s-1/ 4x106 M-1 s-1 = 0.15 M
The Michaelis constant kM = k-1 +k2 / k1= 6x104 s-1+ 2.0x103 s-1/ 4x106 M-1 s-1 =0.155M
The two constant are not equal. Therefore, the binding does not follow the equilibrium scheme.
Q10.2b
Is it appropriate to use the rapid equilibrium scheme to model the kinetics of a catalyzed reaction with the following rate constants?
$k_1 = 7 \times 10^7\ M^{-1}\ s^{-1}$
$k_{-1} = 8 \times 10^5\ s^{-1}$
$k_2 = 5 \times 10^4\ s^{-1}$
S10.2b
Assuming rapid equilibrium is appropriate when $k_{-1} \gg k_2$, so that KM = KS. In this case, the values of k-1 and k2 are not dramatically different, so we can calculate KS and KM and compare.
1. Calculate $K_M$:
$K_M = \dfrac{k_{-1} + k_2}{k_1}$
$K_M = \dfrac{8 \times 10^5\ s^{-1} + 5 \times 10^4\ s^{-1}}{7 \times 10^7\ M^{-1}\ s^{-1}}$
$K_M = 0.01\ M$
2. Calculate $K_S$:
$K_S = \dfrac{k_{-1}}{k_1}$
$K_M = \dfrac{8 \times 10^5\ s^{-1}}{7 \times 10^7\ M^{-1}\ s^{-1}}$
$K_M = 0.01\ M$
Since $K_M = K_S$, it is appropriate to assume rapid equilibrium.
Q10.2c
Given k1= 7 x 10^6 M-1s-1, k-1=6 x 10^4 s-1, k3= 2 x 10^3 s-1, determine if the enzyme substance binding follow the equilibrium or steady state scheme?
\begin{align} K_s &= \dfrac{k_{-1}}{k_1} \ &= \dfrac{6 \times 10^4\, s^{-1}}{7 \times 10^6 M^{-1}s^{-1}} \ &=8.8 \times 10^{-3} M \end{align}
\begin{align} K_M &= k_{-1}+k_{2} \ &= \dfrac{6 \times 10^4 s^{-1} +2 \times 10^3s^{-1}}{ 7 \times 10^6 M ^{-1} s^{-1}} \ &= 8.9 \times 10^{-3} M \end{align}
Q10.3a
The substrate N-acetylglycine ethyl ester can be catalyzed by the enzyme carbonic anhydrase. This enzyme has a turnover rate of 30,000 s-1. Determine how long it will take carbonic anhydrase to cleave the substrate.
S10.3a
We already know the turnover number ($k_{cat}$). The amount of time necessary to cleave the substrate is the reciprocal of the turnover rate.
$t= \dfrac{1}{k} = \dfrac{1}{30,000 \;s^{-1}} = 3.33 \times 10^{-5}$
$(5 \;minutes)(7.1 \times 10^{-6}) = 3.5 \times 10^{-5}$
Q10.3b
RuBisCO is an enzyme in the Calvin cycle that fixes atmospheric carbon and has a turnover rate of 3.3 s-1. How long does it take RuBisCO to fix one molecule of carbon dioxide?
Spacefilling structure of RuBisCO created using Rasmol and the 8RUC file from the Protein Data Bank.
S10.3b
The turnover number is the number of molecules of substrate per unit time (when the enzyme is fully saturated). So simply take the reciprocal to find the time per molecule of substrate.
$\dfrac{1}{3.3\ s^{-1}} = 0.30\ s$
Note: RuBisCO is a notoriously slow enzyme.
Q10.3c
The catalyze of acetylcholine has a rate 50000 s-1 . Calculate the time for the enzyme to cleave one Ach molecule.
S10.3c
t= 1/k2 =1/ 50000 = 2.0 x10-5s
Q10.3d
Carbonic anhydrase, an enzyme that catalyzes the dehydration of carbonic acid to form carbonic acid, has the turnover rate of kcat 4.0x 105 s-1. Calculate how long does it take does it take for the enzyme to cleave one molecule carbonic acid?
S10.3d
The time required for the enzyme to cleave one molecule carbonic acid:
t=1/kcat = 1/ 4x105s-1 =2.5x10-6s = 2.5µs
Q10.3e
p-nitrophenyl acetate(PNPA) is catalyzed by chymotrypsin to yield p-nitrophenolate ion and acetate ion. The turnover rate of that enzyme is 40,000 s-1. How long will it take for the enzyme to produce 1 mole of Nitrophenyl acetate?
S10.3e
It takes 1 second to convert all 40,000 molecule substrate into the product, so: $t=\dfrac{40,000}{9.5x10^5}$ = 3.8 sec.
10.2: The Equations of Enzyme Kinetics
Q10.4
What is plotted on the x and y axes on a Lineweaver-Burk plot? Show how to derive the equation for the plot from the equation
$v_0 = \dfrac{V_{max}[S]}{K_M + [S]}$
and explain how Vmax and KM can be found from the graph's intercepts. Hint: A Lineweaver-Burk plot is also sometimes called a double reciprocal plot.
S10.4
The x-axis is 1/ν0, and the x-axis is 1/Vmax.
1. Take the reciprocal of both sides of the equation.
$\dfrac{1}{v_0} = \dfrac{K_M + [S]}{V_{max} [S]}$
$\dfrac{1}{v_0} = \dfrac{K_M}{V_{max}} \dfrac{1}{[S]} + \dfrac{1}{V_{max}}$
2. Set 1/[S] = 0 to find the y-intercept, and show that it relates to Vmax.
$Yint = \dfrac{1}{V_{max}}$
$V_{max} = \dfrac{1}{Yint}$
3. Set 1/ν0 = 0 to find the x-intercept, and show that it relates to KM.
$0 = \dfrac{K_M}{V_{max}} Xint + \dfrac{1}{V_{max}}$
$\dfrac{K_M}{V_{max}} Xint = - \dfrac{1}{V_{max}}$
$Xint = -\dfrac{1}{K_M}$
$K_M = -\dfrac{1}{Xint}$
Q10.4a
Derive the Michaelis-Menten equation by assuming rapid equilibrium.
$E + S \overset{K_{1}}{\rightleftharpoons} ES \overset{K_{2}}{\rightarrow} E + P$
$v_{0} = \dfrac{V_{max}[S]}{K_{M}+[S]}$
S10.4a
$\dfrac{[E][S]}{dt} = K_{-1}[ES]$ $\dfrac{[ES]}{dt} = K_{1}[E][S]$ $\dfrac{[E][S]}{dt}=\dfrac{[ES]}{dt}$ $K_{-1}[ES]=K_{1}[E][S]$ $K_{M}\dfrac{k_{-1}}{K_{1}}=\dfrac{[E][S]}{[ES]}$
given that $[E]_{0}=[E]+[ES]$
$K_{M}=\dfrac{[E]_{0}-[ES][S]}{[ES]}$ $K_{M}=\dfrac{[E]_{0}[S]-[ES][S]}{[ES]}$ $K_{M}\times[ES]=[E]_{0}[S]-[ES][S]$ $K_{M}\times[ES]+[ES][S]=[E]_{0}[S]$ $(K_{M}+[S])[ES]=[E_{0}][S]$ $[ES]=\dfrac{[E_{0}][S]}{K_{M}+[S]}$
since $\dfrac{d[P]}{dt}=k_{2}[ES]$
$v_{0}=\dfrac{K_{2}[E_{0}][S]}{K_{M}+[S]}$
since $v_{max}=k_{2}[E]_{0}$
$v_{0} = \dfrac{V_{max}[S]}{K_{M}+[S]}$
Q10.4b
We know that the Michaelis Menten derivation for the following reaction:
$\ce{E + S \rightleftharpoons ES -> E + P}$
However, what if the reaction took place in a different scenario whereby:
$\ce{E + S \rightleftharpoons ES1 -> ES2 -> E + P}$
What would be the corresponding Michaelis-Menten Ebe quation now?
S10.4b
This is the an outline for determining an expression for the rate of substrate conversion in the given case:
1. Set up the reaction with rate constants, assuming $k_{-2}\approx k_{-3}\approx0$: $\ce{E + S <=>[k_1][k_{-1}] ES_1->[k_2] ES_2 ->[k_3] E + P}$
2. Set up the differential equations describing the reaction, i.e. the rate of change for each component with time. The rate of substrate change, for example, will be $\frac{d[\ce{S}]}{dt}=-k_{1}[\ce{E}][\ce{S}] +k_{-1}[\ce{ES_1}]$
3. Choose initial conditions and set up two equations for conservation of mass. For example, the initial concentration of enzyme must equal the sum of the concentrations of E, ES1 and ES2.
4. Make the steady-state assumpt: assume that the concentrations of the intermediate complexes do not change on the time-scale of product formation, i.e.$\frac{d[\ce{ES_1}]}{dt}\approx \frac{d[\ce{ES_2}]}{dt}\approx0$
5. Solve for $-r_S$, the negative rate of substrate conversion, obtaining the Michaelis-Menten expression describing the kinetics of the given situtation.
Q10.4c
Prove that $K_s$ equals the concentration S when the initial rate is half its maximum value.
Q10.5a
An enzyme that has a $K_m$ value of $4.6 \times 10^{-5}\; M$ is studied at an initial substrate concentration of 0.041 M. After a minute, it is found that 7.3 uM of product has been produced. Calculate the value of Vmax and the amount of product formed after 4.5 minutes.
Q10.5b
An solution initially contains a catalytic amount of an enzyme with KM = 1.5 mM, 0.25 M of substrate, and no product. After 45 seconds, the solution contains 25 µM of product. Find Vmax and the concentration of product after 2.0 minutes.
Hint: [S] >> KM
S10.5b
1. Find the initial velocity.
$v_0 = \dfrac{25\ \mu M}{0.75\ min}$
$v_0 = 33.3\ \mu M/min$
2. Use v0, [S], and KM to solve for Vmax.
$v_0 = \dfrac{V_{max} [S]}{K_M + [S]}$
$V_{max} = v_0 \left ( \dfrac{K_M}{[S]} + 1 \right )$
$V_{max} = 33.3\ \mu M/min \left ( \dfrac{1.5\ mM}{0.25\ M} \times \dfrac{M}{1000\ mM} + 1 \right )$
$V_{max} = 33.3\ \mu M/min$
Notice that ν0 = Vmax. Since [S] >>KM, the reaction will continue with a velocity of Vmax for the remainder of the two minutes.
3. Predict [P] after 2.0 minutes at the rate Vmax.
$[P] = V_{max} \times t$
$[P] = 33.3\ \mu M/min \times 2.0\ min$
$[P] = 66.6\ \mu M$
Q10.5c
A particular enzyme at a research facility is being studied by a group of graduate students. This enzyme has a Km value of 5.0 X 10-6 M. The students study this enzyme with an initial substrate concentration of 0.055 M. At one minute, 7 µM of product was made. What is the amount of product produced after 5 minutes. What is the Vmax ?
S10.5c
$V_{o}=V_{max}[substrate]/K_{m}+ [substrate]$ 7.0 X 10-6 M = Vmax (0.055 M) / (5.0 X 10-6 M + .055 M) Vmax = 7.1 X 10-6 M/min At 5 minutes the amount of product formed is:
Q10.6a
Given the values, [S]/10^-4 M 3.0 4.6 10.5 16.5
vo/10^-6 M * min^-1. 2.64 3.5 6.2 7.8
construct a Lineweaver-Burk plot, and assuming Michaelis-Menten kinetics, calculate the values of Vmax, Km, and k2 using the constructed plot.
Q10.6b
The data below represents the data recorded after the hydrolysis of a substrate by an enzyme.
[S]/10-4 M 2.1 4.2 9.3 14.2
vo / 10-6 M · min-1 1.2 3.1 6.3 9.1
Calculate Vmax, Km and K2 using a Lineweaver-Burk plot. Assume Michaelis. Given [E}0 is 5.0 X 10-6 M
S10.6b
(1/[S])/10-3 M-1 (1/vo )/ 10-5 M-1 · min
4.8 8.3
2.4 3.2
1.1 1.6
0.7 1.1
Figure: Lineweaver-Burk Plot The linear equation for the graph above is: y=176x+4.23X104 To solve for Vmax use: Intercept= 1/Vmax Vmax = 1/(.4.23X104) = 2.36 X 10-5 To solve for Km use Slope= Km/Vmax (176)( 2.36 X 10-5) = 4.2 X 10-5 To solve for K2 use K2 = Vmax/ [E}0 = (2.36 X 10-5)(5.0 X 10-6 M)= 4.72 min-1
Q10.6c
Using the table below, calculate the KM, Vmax, and slope.
$\dfrac{1}{V_{0}} (10^{-3}\dfrac{sec}{M^{-1}})$ $\dfrac{1}{[S]}(10^{2}M^{-1})$
2.9 0.3
3.2 1.2
4.4 1.8
5.1 3.3
S10.6c
slope=$\dfrac{5.1\times10^{-3}-2.9\times10^{-3}}{3.3\times10^{2}-0.3\times10^{2}}=6.73\times10^{-6}$ $6.73\times10^{-6}\times0.3\times10^{2}+\dfrac{1}{V_{max}}=2.9\times10^{-3}$ $\dfrac{1}{V_{max}}=0.00270$ $V_{max}=370.62Ms^{-1}$ $6.73\times10^{-6}\times\dfrac{-1}{K_{M}}+0.00270=0$ $\dfrac{-1}{K_{M}}=-0.00943$ $K_{M}=106.044$
Q10.7a
From this graph determine the KM and Vmax?
S10.7b
From his graph we can see that the value KM is 2. Then we look to see where KM is half. At that point, we see that KM/2 is 1 and the x-value for that coordinate is 1. This means Vmax is 1.
Q10.7b
Neutral sphingomyelinase 2 converts sphingomyelin into ceramide and phosphocholine. Assume its Vmax is 35 µM min-1. When you provide 3 x 10-5 M of sphingomyelin, you observe an initial velocity of 6.0 µM min-1. Calculate the KM for this reaction, rounding to 3 significant figures.
S10.7b
$\dfrac{1}{V_{o}} = \dfrac{K_{M}}{V_{max}[S]}+\dfrac{1}{V_{max}}$
$\dfrac{1}{6.0} = \dfrac{K_{M}}{(35)(30)}+\dfrac{1}{35}$
$\dfrac{1}{6}-\dfrac{1}{35} = \dfrac{K_{M}}{1050}$
KM = 145 µM
Q10.8a
We are given a second order equation: r=k[A][B]. The concentration of A is 0.05g and the concentration of B is 2.5g. We decide this difference is great enough to treat this as a pseudo first reaction? What concentration is held constant and why? Write the new equation.
S10.8a
We hold B constant because the concentration is so much larger, so it should be close to constant for the reaction. The new equation would be r=k'[a]
Q10.9a
Name and briefly describe two types of reactions that do not follow Michaelis-Menten kinetics.
S10.9a
Irreversible inhibition - the inhibitor binds covalently and irreversibly to the enzyme Allosteric interactions - the binding of effectors at allosteric sites (away from the active site) influence substrate binding.
Q10.9b
For the reaction mechanism below, how does the concentration of $C$ affect the concentration of $B$?
$A\leftrightharpoons B\rightarrow C$
S10.9b
Because the second state is irreversible (i.e., since arrow), it does not matter if you have a large or little concentration of $C$, it would not affect $B$ and hence the kinetics of the reaction. However, the concentration of $A\0 would clearly affect the concentration of \(B$.
Q1
1. Speculate on how the catalytic rate constant can be determined from the spectrophotogram.
2. How can product be consistently produced if the rate of change of the ES complex is 0?
3. How would the rate of product formation change if:
1. the substrate concentration were doubled?
2. the enzyme concentration were doubled?
3. The reaction was carried out in mono-deuterated water instead of H2O (comment qualitatively)?
4. Explain the role of hydrogen bonding in protein hydrolysis catalyzed by chymotrypsin.
5. What would the spectrophotogram look like if the reaction proceeded via a steady-state mechanism instead of pre-equilibrium.
S2
1. The catalytic rate constant can be deduced from the graph by simply determining the slope of the line where the reaction demonstrates 0-order kinetics (the linear part).
2. This is pre-equilibrium kinetics in action. The ES complex is formed from E and S at a faster rate than any other step in the reaction. As soon as ES is converted to *ES, another mole of ES is produced from an infinite supply of E + S. This means that the amount of ES and E + S is constantly at equilibrium, and thus the change of either with respect to time is 0.
1. No change.
2. Two-fold increase.
3. Because water is involved in the final, slowest step of the mechanism, deuterating the water would decrease the rate of the overall reaction from 5- to 30-fold.
3. Initially, hydrogen bonding between the enzymes histidine and serine side chains weakens the bond of serine’s O-H. This allows a facilitated nucleophilic attack of the hydroxyl oxygen on the substrates carbonyl group. Conversely, in the final step of the reaction, the bound serine oxygen forms a hydrogen bond with a protonated histidine, which allows for easier cleavage from the substrate.
4. The graph would show similar 0-order kinetics, but the line would intercept the Y-axis at an absorbance of 0 instead of the 1:1 mole ratio of nitrophenolate to enzyme.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/10%3A_Enzyme_Kinetics/10.E%3A_Exercises.txt
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Quantum mechanics is a fundamental branch of physics concerned with processes involving small particles (e.g, atoms and photons). Such processes as said to be quantized with properties that are observed in only in integer multiples of the Planck constant. This is utterly inexplicable in classical physics which is a continuum approach.
• 11.1: The Wave Theory of Light
Water waves transmit energy through space by the periodic oscillation of matter (the water). In contrast, energy that is transmitted, or radiated, through space in the form of periodic oscillations of electric and magnetic fields is known as electromagnetic radiation, which is energy that is transmitted, or radiated, through space in the form of periodic oscillations of electric and magnetic fields.
• 11.2: Planck's Quantum Theory
One phenomenon that seemed to contradict the theories of classical physics was blackbody radiation, which is electromagnetic radiation whose wavelength and color that depends on the temperature of the object. The wavelength of energy emitted by an object depends on only its temperature, not its surface or composition. Hence an electric stove burner or the filament of a space heater glows dull red or orange when heated and much hotter tungsten wire in a light bulbs gives off a yellowish light.
• 11.3: The Photoelectric Effect
The photoelectric effect was first documented in 1887 by the German physicist Heinrich Hertz and is therefore sometimes referred to as the Hertz effect. While working with a spark-gap transmitter (a primitive radio-broadcasting device), Hertz discovered that upon absorption of certain frequencies of light, substances would give off a visible spark. In 1899, this spark was identified as light-excited electrons (also called photoelectrons) leaving the metal's surface by J.J. Thomson.
• 11.4: Bohr's Theory of the Hydrogen Emission Spectrum
• 11.5: de Broglie's Postulate
De Broglie suggested that if waves (photons) could behave as particles, then the converse, namely that particles could behave as waves, should be true. He associated a wavelength λ to a particle with momentum p using Planck's constant as the constant of proportionality, which is called the de Broglie wavelength. The fact that particles can behave as waves but also as particles, depending on which experiment you perform on them, is known as the particle-wave duality.
• 11.6: The Heisenberg Uncertainty Principle
In 1927 the German physicist Werner Heisenberg described such limitations as the Heisenberg Uncertainty Principle, or simply the Uncertainty Principle, stating that it is not possible to measure both the momentum and position of a particle simultaneously. The Heisenberg Uncertainty Principle is a fundamental theory in quantum mechanics that defines why a scientist cannot measure multiple quantum variables simultaneously.
• 11.7: The Schrödinger Wave Equation
Beginning in the early 20th century, physicists began to acknowledge that matter--much like electromagnetic radiation--possessed wave-like behaviors. While electromagnetic radiation were well understood to obey Maxwell's Equations, matter obeyed no known equations.
• 11.8: Particle in a One-Dimensional Box
A particle in a 1-dimensional box is a fundamental quantum mechanical approximation describing the translational motion of a single particle confined inside an infinitely deep well from which it cannot escape.
• 11.9: Quantum-Mechanical Tunneling
Tunneling is a quantum mechanical phenomenon when a particle is able to penetrate through a potential energy barrier that is higher in energy than the particle’s kinetic energy. This amazing property of microscopic particles play important roles in explaining several physical phenomena including radioactive decay. Additionally, the principle of tunneling leads to the development of Scanning Tunneling Microscope (STM) which had a profound impact on chemical, biological and material research.
• 11.10: The Schrödinger Wave Equation for the Hydrogen Atom
• 11.11: Many-Electron Atoms and the Periodic Table
Quantum mechanics can account for the periodic structure of the elements, by any measure a major conceptual accomplishment for any theory. Although accurate computations become increasingly more challenging as the number of electrons increases, the general patterns of atomic behavior can be predicted with remarkable accuracy.
• 11.E: Quantum Mechanics and Atomic Structure (Exercises)
This are exercises that to accompany the TextMap organized around Raymond Chang's Physical Chemistry for the Biosciences textbook.
11: Quantum Mechanics and Atomic Structure
Learning Objectives
• to learn about the characteristics of electromagnetic waves. Light, X-Rays, infrared and microwaves among the types of electromagnetic waves.
Scientists discovered much of what we know about the structure of the atom by observing the interaction of atoms with various forms of radiant, or transmitted, energy, such as the energy associated with the visible light we detect with our eyes, the infrared radiation we feel as heat, the ultraviolet light that causes sunburn, and the x-rays that produce images of our teeth or bones. All these forms of radiant energy should be familiar to you. We begin our discussion of the development of our current atomic model by describing the properties of waves and the various forms of electromagnetic radiation.
Properties of Waves
A wave is a periodic oscillation that transmits energy through space. Anyone who has visited a beach or dropped a stone into a puddle has observed waves traveling through water (Figure 1.1.1). These waves are produced when wind, a stone, or some other disturbance, such as a passing boat, transfers energy to the water, causing the surface to oscillate up and down as the energy travels outward from its point of origin. As a wave passes a particular point on the surface of the water, anything floating there moves up and down.
Waves have characteristic properties (Figure 1.1.2). As you may have noticed in Figure 1.1.1, waves are periodic, that is, they repeat regularly in both space and time. The distance between two corresponding points in a wave—between the midpoints of two peaks, for example, or two troughs—is the wavelength (λ), distance between two corresponding points in a wave—between the midpoints of two peaks or two troughs. $\lambda$ is the lowercase Greek lambda, and $\nu$ is the lowercase Greek nu. Wavelengths are described by a unit of distance, typically meters. The frequency (ν), the number of oscillations (i.e., of a wave) that pass a particular point in a given period of time. of a wave is the number of oscillations that pass a particular point in a given period of time. The usual units are oscillations persecond (1/s = s−1), which in the SI system is called the hertz (Hz). It is named after German physicist Heinrich Hertz (1857–1894),a pioneer in the field of electromagnetic radiation. The amplitude (the vertical height of a wave, which is defined as half the peak-to-trough height), or vertical height, of a wave is defined as half the peak-to-trough height; as the amplitude of a wave with a given frequency increases, so does its energy. As you can see in Figure 1.1.2 , two waves can have the same amplitude but different wavelengths and vice versa. The distance traveled by a wave per unit time is its speed (v), the distance traveled by a wave per unit time, which is typically measured in meters per second (m/s). The speed of a wave is equal to the product of its wavelength and frequency:
(wavelength)(frequency) = speed
$\lambda \nu =v \tag{1.1.1a}$
$\left ( \dfrac{meters}{\cancel{wave}} \right )\left ( \dfrac{\cancel{wave}}{second} \right )=\dfrac{meters}{second} \tag{1.1.1b}$
Be careful not to confuse the symbols for the speed, $v$, with the frequency, $\nu$. Water waves are slow compared to sound waves, which can travel through solids, liquids, and gases. Whereas water waves may travel a few meters per second, the speed of sound in dry air at 20°C is 343.5 m/s. Ultrasonic waves, which travel at an even higher speed (>1500 m/s) and have a greater frequency, are used in such diverse applications as locating underwater objects and the medical imaging of internal organs.
Electromagnetic Radiation
Water waves transmit energy through space by the periodic oscillation of matter (the water). In contrast, energy that is transmitted, or radiated, through space in the form of periodic oscillations of electric and magnetic fields is known as electromagnetic radiation, which is energy that is transmitted, or radiated, through space in the form of periodic oscillations of electric and magnetic fields. (Figure 1.1.3). Some forms of electromagnetic radiation are shown in Figure 1.1.4. In a vacuum, all forms of electromagnetic radiation—whether microwaves, visible light, or gamma rays—travel at the speed of light (c), which is the speed with which all forms of electromagnetic radiation travel in a vacuum, a fundamental physical constant with a value of 2.99792458 × 108 m/s (which is about 3.00 ×108 m/s or 1.86 × 105 mi/s). This is about a million times faster than the speed of sound.
Because the various kinds of electromagnetic radiation all have the same speed (c), they differ in only wavelength and frequency. As shown in Figure 1.1.4 and Table 1.1.1 , the wavelengths of familiar electromagnetic radiation range from 101 m for radio waves to 10−12 m for gamma rays, which are emitted by nuclear reactions. By replacing v with c in Equation 6.1.1, we can show that the frequency of electromagnetic radiation is inversely proportional to its wavelength:
$\begin{array}{cc} c=\lambda \nu \ \nu =\dfrac{c}{\lambda } \end{array} \tag{1.1.2}$
For example, the frequency of radio waves is about 108 Hz, whereas the frequency of gamma rays is about 1020 Hz. Visible light, which is electromagnetic radiation that can be detected by the human eye, has wavelengths between about 7 × 10−7 m (700 nm, or 4.3 × 1014 Hz) and 4 × 10−7 m (400 nm, or 7.5 × 1014 Hz). Note that when frequency increases, wavelength decreases; c being a constant stays the same. Similarly when frequency decreases, the wavelength increases.
Within this visible range our eyes perceive radiation of different wavelengths (or frequencies) as light of different colors, ranging from red to violet in order of decreasing wavelength. The components of white light—a mixture of all the frequencies of visible light—can be separated by a prism, as shown in part (b) in Figure 1.1.4. A similar phenomenon creates a rainbow, where water droplets suspended in the air act as tiny prisms.
Table 1.1.1: Common Wavelength Units for Electromagnetic Radiation
Unit Symbol Wavelength (m) Type of Radiation
picometer pm 10−12 gamma ray
angstrom Å 10−10 x-ray
nanometer nm 10−9 x-ray
micrometer μm 10−6 infrared
millimeter mm 10−3 infrared
centimeter cm 10−2 microwave
meter m 100 radio
As you will soon see, the energy of electromagnetic radiation is directly proportional to its frequency and inversely proportional to its wavelength:
$E\; \propto\; \nu \tag{1.1.3}$
$E\; \propto\; \dfrac{1}{\lambda } \tag{1.1.4}$
Whereas visible light is essentially harmless to our skin, ultraviolet light, with wavelengths of ≤ 400 nm, has enough energy to cause severe damage to our skin in the form of sunburn. Because the ozone layer absorbs sunlight with wavelengths less than 350 nm, it protects us from the damaging effects of highly energetic ultraviolet radiation.
The energy of electromagnetic radiation increases with increasing frequency and decreasing wavelength.
Example 1.1.1
Your favorite FM radio station, WXYZ, broadcasts at a frequency of 101.1 MHz. What is the wavelength of this radiation?
Given: frequency
Asked for: wavelength
Strategy:
Substitute the value for the speed of light in meters per second into Equation 1.1.2 to calculate the wavelength in meters.
Solution:
From Equation 1.1.2 , we know that the product of the wavelength and the frequency is the speed of the wave, which for electromagnetic radiation is 2.998 × 108 m/s:
$λν = c = 2.998 \times 10^8 m/s \nonumber$
Thus the wavelength λ is given by
$\lambda =\dfrac{c}{\nu }=\left ( \dfrac{2.988\times 10^{8}\; m/\cancel{s}}{101.1\; \cancel{MHz}} \right )\left ( \dfrac{1\; \cancel{MHz}}{10^{6}\; \cancel{s^{-1}}} \right )=2.965\; m \nonumber$
Exercise 1.1.1
As the police officer was writing up your speeding ticket, she mentioned that she was using a state-of-the-art radar gun operating at 35.5 GHz. What is the wavelength of the radiation emitted by the radar gun?
Click to Check Answer
8.45 mm
External Videos & Examples
Answers for these quizzes are included. There are also questions covering more topics in Chapter 6.
Summary
• Understanding the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation.
A basic knowledge of the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation. A wave is a periodic oscillation by which energy is transmitted through space. All waves are periodic, repeating regularly in both space and time. Waves are characterized by several interrelated properties: wavelength (λ), the distance between successive waves; frequency (ν), the number of waves that pass a fixed point per unit time; speed (v), the rate at which the wave propagates through space; and amplitude, the magnitude of the oscillation about the mean position. The speed of a wave is equal to the product of its wavelength and frequency. Electromagnetic radiation consists of two perpendicular waves, one electric and one magnetic, propagating at the speed of light (c). Electromagnetic radiation is radiant energy that includes radio waves, microwaves, visible light, x-rays, and gamma rays, which differ in their frequencies and wavelengths.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/11%3A_Quantum_Mechanics_and_Atomic_Structure/11.01%3A_The_Wave_Theory_of_Light.txt
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Learning Objectives
• To understand how energy is quantized in blackbody radiation
By the late 19th century, many physicists thought their discipline was well on the way to explaining most natural phenomena. They could calculate the motions of material objects using Newton’s laws of classical mechanics, and they could describe the properties of radiant energy using mathematical relationships known as Maxwell’s equations, developed in 1873 by James Clerk Maxwell, a Scottish physicist. The universe appeared to be a simple and orderly place, containing matter, which consisted of particles that had mass and whose location and motion could be accurately described, and electromagnetic radiation, which was viewed as having no mass and whose exact position in space could not be fixed. Thus matter and energy were considered distinct and unrelated phenomena. Soon, however, scientists began to look more closely at a few inconvenient phenomena that could not be explained by the theories available at the time.
One experimental phenomenon that could not be adequately explained by classical physics was blackbody radiation (Figure 1.2.1 ). Attempts to explain or calculate this spectral distribution from classical theory were complete failures. A theory developed by Rayleigh and Jeans predicted that the intensity should go to infinity at short wavelengths. Since the intensity actually drops to zero at short wavelengths, the Rayleigh-Jeans result was called the ultraviolet catastrophe (Figure 1.2.1 dashed line). There was no agreement between theory and experiment in the ultraviolet region of the blackbody spectrum.
Quantizing Electrons in the Radiator
In 1900, the German physicist Max Planck (1858–1947) explained the ultraviolet catastrophe by proposing that the energy of electromagnetic waves is quantized rather than continuous. This means that for each temperature, there is a maximum intensity of radiation that is emitted in a blackbody object, corresponding to the peaks in Figure 1.2.1 , so the intensity does not follow a smooth curve as the temperature increases, as predicted by classical physics. Thus energy could be gained or lost only in integral multiples of some smallest unit of energy, a quantum (the smallest possible unit of energy). Energy can be gained or lost only in integral multiples of a quantum.
Quantization
Although quantization may seem to be an unfamiliar concept, we encounter it frequently in quantum mechanics (hence the name). For example, US money is integral multiples of pennies. Similarly, musical instruments like a piano or a trumpet can produce only certain musical notes, such as C or F sharp. Because these instruments cannot produce a continuous range of frequencies, their frequencies are quantized. It is also similar to going up and down a hill using discrete stair steps rather than being able to move up and down a continuous slope. Your potential energy takes on discrete values as you move from step to step. Even electrical charge is quantized: an ion may have a charge of −1 or −2, but not −1.33 electron charges.
Planck's quantization of energy is described by the his famous equation:
$E=h \nu \label{Eq1.2.1}$
where the proportionality constant $h$ is called Planck’s constant, one of the most accurately known fundamental constants in science
$h=6.626070040(81) \times 10^{−34}\, J\cdot s \nonumber$
However, for our purposes, its value to four significant figures is sufficient:
$h = 6.626 \times 10^{−34} \,J\cdot s \nonumber$
As the frequency of electromagnetic radiation increases, the magnitude of the associated quantum of radiant energy increases. By assuming that energy can be emitted by an object only in integral multiples of $hν$, Planck devised an equation that fit the experimental data shown in Figure 1.2.1 . We can understand Planck’s explanation of the ultraviolet catastrophe qualitatively as follows: At low temperatures, radiation with only relatively low frequencies is emitted, corresponding to low-energy quanta. As the temperature of an object increases, there is an increased probability of emitting radiation with higher frequencies, corresponding to higher-energy quanta. At any temperature, however, it is simply more probable for an object to lose energy by emitting a large number of lower-energy quanta than a single very high-energy quantum that corresponds to ultraviolet radiation. The result is a maximum in the plot of intensity of emitted radiation versus wavelength, as shown in Figure 1.2.1 , and a shift in the position of the maximum to lower wavelength (higher frequency) with increasing temperature.
At the time he proposed his radical hypothesis, Planck could not explain why energies should be quantized. Initially, his hypothesis explained only one set of experimental data—blackbody radiation. If quantization were observed for a large number of different phenomena, then quantization would become a law. In time, a theory might be developed to explain that law. As things turned out, Planck’s hypothesis was the seed from which modern physics grew.
Max Planck explain the spectral distribution of blackbody radiation as result from oscillations of electrons. Similarly, oscillations of electrons in an antenna produce radio waves. Max Planck concentrated on modeling the oscillating charges that must exist in the oven walls, radiating heat inwards and—in thermodynamic equilibrium—themselves being driven by the radiation field. He found he could account for the observed curve if he required these oscillators not to radiate energy continuously, as the classical theory would demand, but they could only lose or gain energy in chunks, called quanta, of size $h\nu$, for an oscillator of frequency $\nu$ (Equation $\ref{Eq1.2.1}$).
With that assumption, Planck calculated the following formula for the radiation energy density inside the oven:
\begin{align} d\rho(\nu,T) &= \rho_\nu (T) d\nu \[4pt] &= \dfrac {2 h \nu^3}{c^2} \cdot \dfrac {1 }{\exp \left( \dfrac {h\nu}{k_B T}\right)-1} d\nu \label{Eq2a} \end{align}
with
• $\pi = 3.14159$
• $h$ = $6.626 \times 10^{-34} J\cdot s$
• $c$ = $3.00 \times 10^{8}\, m/s$
• $\nu$ = $1/s$
• $k_B$ = $1.38 \times 10^{-23} J/K$
• $T$ is absolute temperature (in Kelvin)
Planck's radiation energy density (Equation $\ref{Eq2a}$) can also be expressed in terms of wavelength $\lambda$.
$\rho (\lambda, T) = \dfrac {2 hc^2}{\lambda ^5} \left(\dfrac {1}{ e^{\dfrac {hc}{\lambda k_B T}} - 1}\right) \label{Eq2b}$
With a wavelength of maximum energy density at:
$\lambda_{max}=\frac{hc}{4.965kT} \nonumber$
Planck's equation (Equation $\ref{Eq2b}$) gave an excellent agreement with the experimental observations for all temperatures (Figure 1.2.2 ).
Max Planck (1858–1947)
Planck made many substantial contributions to theoretical physics, but his fame as a physicist rests primarily on his role as the originator of quantum theory. In addition to being a physicist, Planck was a gifted pianist, who at one time considered music as a career. During the 1930s, Planck felt it was his duty to remain in Germany, despite his open opposition to the policies of the Nazi government.
One of his sons was executed in 1944 for his part in an unsuccessful attempt to assassinate Hitler and bombing during the last weeks of World War II destroyed Planck’s home. After WWII, the major German scientific research organization was renamed the Max Planck Society.
Exercise 1.2.1
Use Equation $\ref{Eq2b}$ to show that the units of $ρ(λ,T)\,dλ$ are $J/m^3$ as expected for an energy density.
The near perfect agreement of this formula with precise experiments (e.g., Figure 1.2.3 ), and the consequent necessity of energy quantization, was the most important advance in physics in the century. His blackbody curve was completely accepted as the correct one: more and more accurate experiments confirmed it time and again, yet the radical nature of the quantum assumption did not sink in. Planck was not too upset—he didn’t believe it either, he saw it as a technical fix that (he hoped) would eventually prove unnecessary.
Part of the problem was that Planck’s route to the formula was long, difficult and implausible—he even made contradictory assumptions at different stages, as Einstein pointed out later. However, the result was correct anyway!
The mathematics implied that the energy given off by a blackbody was not continuous, but given off at certain specific wavelengths, in regular increments. If Planck assumed that the energy of blackbody radiation was in the form
$E = nh \nu \nonumber$
where $n$ is an integer, then he could explain what the mathematics represented. This was indeed difficult for Planck to accept, because at the time, there was no reason to presume that the energy should only be radiated at specific frequencies. Nothing in Maxwell’s laws suggested such a thing. It was as if the vibrations of a mass on the end of a spring could only occur at specific energies. Imagine the mass slowly coming to rest due to friction, but not in a continuous manner. Instead, the mass jumps from one fixed quantity of energy to another without passing through the intermediate energies.
To use a different analogy, it is as if what we had always imagined as smooth inclined planes were, in fact, a series of closely spaced steps that only presented the illusion of continuity.
Summary
The agreement between Planck’s theory and the experimental observation provided strong evidence that the energy of electron motion in matter is quantized. In the next two sections, we will see that the energy carried by light also is quantized in units of $h \bar {\nu}$. These packets of energy are called “photons.”
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/11%3A_Quantum_Mechanics_and_Atomic_Structure/11.02%3A_Planck%27s_Quantum_Theory.txt
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Learning Objectives
• To be familiar with the photoelectron effect for bulk materials
• Understand how the photoelectron kinetic energy and intensity vary as a function of incident light wavelength
• Understand how the photoelectron kinetic energy and intensity vary as a function of incident light intensity
• Describe what a workfunction is and relate it to ionization energy
• Describe the photoelectric effect with Einstein's quantized photon model of light
Nature, it seemed, was quantized (non-continuous, or discrete). If this was so, how could Maxwell’s equations correctly predict the result of the blackbody radiator? Planck spent a good deal of time attempting to reconcile the behavior of electromagnetic waves with the discrete nature of the blackbody radiation, to no avail. It was not until 1905, with yet another paper published by Albert Einstein, that the wave nature of light was expanded to include the particle interpretation of light which adequately explained Planck’s equation.
The photoelectric effect was first documented in 1887 by the German physicist Heinrich Hertz and is therefore sometimes referred to as the Hertz effect. While working with a spark-gap transmitter (a primitive radio-broadcasting device), Hertz discovered that upon absorption of certain frequencies of light, substances would give off a visible spark. In 1899, this spark was identified as light-excited electrons (called photoelectrons) leaving the metal's surface by J.J. Thomson (Figure 1.3.1 ).
The classical picture underlying the photoelectron effect was that the atoms in the metal contained electrons, that were shaken and caused to vibrate by the oscillating electric field of the incident radiation. Eventually some of them would be shaken loose, and would be ejected from the cathode. It is worthwhile considering carefully how the number and speed of electrons emitted would be expected to vary with the intensity and color of the incident radiation along with the time needed to observe the photoelectrons.
• Increasing the intensity of radiation would shake the electrons more violently, so one would expect more to be emitted, and they would shoot out at greater speed, on average.
• Increasing the frequency of the radiation would shake the electrons faster, so it might cause the electrons to come out faster. For very dim light, it would take some time for an electron to work up to a sufficient amplitude of vibration to shake loose.
Lenard's Experimental Results (Intensity Dependence)
In 1902, Hertz's student, Philipp Lenard, studied how the energy of the emitted photoelectrons varied with the intensity of the light. He used a carbon arc light and could increase the intensity a thousand-fold. The ejected electrons hit another metal plate, the collector, which was connected to the cathode by a wire with a sensitive ammeter, to measure the current produced by the illumination (Figure 1.3.2 ). To measure the energy of the ejected electrons, Lenard charged the collector plate negatively, to repel the electrons coming towards it. Thus, only electrons ejected with enough kinetic energy to get up this potential hill would contribute to the current.
Lenard discovered that there was a well defined minimum voltage that stopped any electrons getting through ($V_{stop}$). To Lenard's surprise, he found that $V_{stop}$ did not depend at all on the intensity of the light! Doubling the light intensity doubled the number of electrons emitted, but did not affect the kinetic energies of the emitted electrons. The more powerful oscillating field ejected more electrons, but the maximum individual energy of the ejected electrons was the same as for the weaker field (Figure 1.3.2 ).
Millikan's Experimental Results (Wavelength Dependence)
The American experimental physicist Robert Millikan followed up on Lenard's experiments and using a powerful arc lamp, he was able to generate sufficient light intensity to separate out the colors and check the photoelectric effect using light of different colors. He found that the maximum energy of the ejected electrons did depend on the color - the shorter wavelength, higher frequency light eject photoelectrons with greater kinetic energy (Figures 1.3.3 ).
As shown in Figure 1.3.4 , just the opposite behavior from classical is observed from Lenard's and Millikan's experiments. The intensity affects the number of electrons, and the frequency affects the kinetic energy of the emitted electrons. From these sketches, we see that
• the kinetic energy of the electrons is linearly proportional to the frequency of the incident radiation above a threshold value of $ν_0$ (no current is observed below $ν_0$), and the kinetic energy is independent of the intensity of the radiation, and
• the number of electrons (i.e. the electric current) is proportional to the intensity and independent of the frequency of the incident radiation above the threshold value of $ν_0$ (i.e., no current is observed below $ν_0$).
Classical Theory does not Describe Experiment
Classical theory predicts that energy carried by light is proportional to its amplitude independent of its frequency, and this fails to correctly explain the observed wavelength dependence in Lenard's and Millikan's observations.
As with most of the experimental results we discuss in this text, the behavior described above is a simplification of the true experimental results observed in the laboratory. A more complex description involves a greater introduction of more complex physics and instrumentation, which will be ignored for now.
Einstein's Quantum Picture
In 1905 Einstein gave a very simple interpretation of Lenard's results and borrowed Planck's hypothesis about the quantized energy from his blackbody research and assumed that the incoming radiation should be thought of as quanta of energy $h\nu$, with $\nu$ the frequency. In photoemission, one such quantum is absorbed by one electron. If the electron is some distance into the material of the cathode, some energy will be lost as it moves towards the surface. There will always be some electrostatic cost as the electron leaves the surface, which is the workfunction, $\Phi$. The most energetic electrons emitted will be those very close to the surface, and they will leave the cathode with kinetic energy
$KE = h\nu - \Phi \label{Eq1}$
On cranking up the negative voltage on the collector plate until the current just stops, that is, to $V_{stop}$, the highest kinetic energy electrons ($KE_e$) must have had energy $eV_{stop}$ upon leaving the cathode. Thus,
$eV_{stop} = h\nu - \Phi \label{Eq2}$
Thus, Einstein's theory makes a very definite quantitative prediction: if the frequency of the incident light is varied, and $V_{stop}$ plotted as a function of frequency, the slope of the line should be $\frac{h}{e}$ (Figure $\PageIndex{4A}$). It is also clear that there is a minimum light frequency for a given metal $\nu_o$, that for which the quantum of energy is equal to $\Phi$ (Equation \ref{Eq1}). Light below that frequency, no matter how bright, will not eject electrons.
According to both Planck and Einstein, the energy of light is proportional to its frequency rather than its amplitude, there will be a minimum frequency $\nu_0$ needed to eject an electron with no residual energy.
Since every photon of sufficient energy excites only one electron, increasing the light's intensity (i.e. the number of photons/sec) only increases the number of released electrons and not their kinetic energy. In addition, no time is necessary for the atom to be heated to a critical temperature and therefore the release of the electron is nearly instantaneous upon absorption of the light. Finally, because the photons must be above a certain energy to satisfy the workfunction, a threshold frequency exists below which no photoelectrons are observed. This frequency is measured in units of Hertz (1/second) in honor of the discoverer of the photoelectric effect.
Einstein's Equation $\ref{Eq1}$ explains the properties of the photoelectric effect quantitatively. A strange implication of this experiment is that light can behave as a kind of massless "particle" now known as a photon whose energy $E=h\nu$ can be transferred to an actual particle (an electron), imparting kinetic energy to it, just as in an elastic collision between to massive particles such as billiard balls.
Robert Millikan initially did not accept Einstein's theory, which he saw as an attack on the wave theory of light, and worked for ten years until 1916, on the photoelectric effect. He even devised techniques for scraping clean the metal surfaces inside the vacuum tube. For all his efforts he found disappointing results: he confirmed Einstein's theory after ten years. In what he writes in his paper, Millikan is still desperately struggling to avoid this conclusion. However, by the time of his Nobel Prize acceptance speech, he has changed his mind rather drastically!
Einstein's simple explanation (Equation \ref{Eq1}) completely accounted for the observed phenomena in Lenard's and Millikan's experiments (Figure 1.3.4 ) and began an investigation into the field we now call quantum mechanics. This new field seeks to provide a quantum explanation for classical mechanics and create a more unified theory of physics and thermodynamics. The study of the photoelectric effect has also lead to the creation of new field of photoelectron spectroscopy. Einstein's theory of the photoelectron presented a completely different way to measure Planck's constant than from black-body radiation.
The Workfunction (Φ)
The workfunction is an intrinsic property of the metal. While the workfunctions and ionization energies appear as similar concepts, they are independent. The workfunction of a metal is the minimum amount of energy ($\ce{E}$) necessary to remove an electron from the surface of the bulk (solid) metal (sometimes referred to as binding energy).
$\ce{M (s) + \Phi \rightarrow M^{+}(s) + e^{-}}(\text{free with no kinetic energy}) \nonumber$
The workfunction is qualitatively similar to ionization energy ($\ce{IE}$), which is the amount of energy required to remove an electron from an atom or molecule in the gaseous state.
$\ce{M (g) + IE \rightarrow M^{+}(g) + e^{-}} (\text{free with no kinetic energy}) \nonumber$
However, these two energies differ in magnitude (Table 1.3.1 ). For instance, copper has a workfunction of about 4.7 eV, but has a higher ionization energy of 7.7 eV. Generally, the ionization energies for metals are greater than the corresponding workfunctions (i.e., the electrons are less tightly bound in bulk metal).
Element Workfunction $\Phi$ (eV) Ionization Energy (eV)
Table 1.3.1 : Workfunctions and Ionization Energies of Select Elements
Lithium (Li) 2.93 5.39
Beryllium (Be) 4.98 9.32
Boron (B) 4.45 8.298
Carbon (C) 5.0 11.26
Sodium (Na) 2.36 5.13
Aluminum (Al) 4.20 5.98
Silicon (Si) 4.85 8.15
Potassium (K) 2.3 4.34
Iron (Fe) 4.67 7.87
Cobalt (Co) 4.89 7.88
Copper (Cu) 4.7 7.7
Gallium (Ga) 4.32 5.99
Germanium (Ge) 5.0 7.89
Arsenic (As) 3.75 9.81
Selenium (Se) 5.9 9.75
Silver (Ag) 4.72 7.57
Tin (Sn) 4.42 7.34
Cesium (Cs) 1.95 3.89
Gold (Au) 5.17 9.22
Mercury (Hg)liquid 4.47 10.43
Bismuth (Bi) 4.34 7.29
Example 1.3.1 : Calcium
1. What is the energy in joules and electron volts of a photon of 420-nm violet light?
2. What is the maximum kinetic energy of electrons ejected from calcium by 420-nm violet light, given that the workfunction for calcium metal is 2.71 eV?
Strategy
To solve part (a), note that the energy of a photon is given by $E=h\nu$. For part (b), once the energy of the photon is calculated, it is a straightforward application of Equation \ref{Eq1} to find the ejected electron’s maximum kinetic energy, since $\Phi$ is given.
Solution for (a)
Photon energy is given by
$E = h\nu \nonumber$
Since we are given the wavelength rather than the frequency, we solve the familiar relationship $c=\nu\lambda$ for the frequency, yielding
$\nu=\dfrac{c}{\lambda} \nonumber$
Combining these two equations gives the useful relationship
$E=\dfrac{hc}{\lambda} \nonumber$
Now substituting known values yields
\begin{align*} E &= \dfrac{(6.63 \times 10^{-34}\; J \cdot s)(3.00 \times 10^8 m/s)}{420 \times 10^{-9}\; m} \[4pt] &= 4.74 \times 10^{-19}\; J \end{align*} \nonumber
Converting to eV, the energy of the photon is
\begin{align*} E&=(4.74 \times 10^{-19}\; J) \left( \dfrac{1 \;eV}{1.6 \times 10^{-19}\;J} \right) \[4pt] &= 2.96\; eV. \nonumber \end{align*} \nonumber
Solution for (b)
Finding the kinetic energy of the ejected electron is now a simple application of Equation \ref{Eq1}. Substituting the photon energy and binding energy yields
\begin{align*} KE_e &=h\nu – \Phi \[4pt] &= 2.96 \;eV – 2.71 \;eV \[4pt] &= 0.246\; eV.\nonumber \end{align*} \nonumber
Discussion
The energy of this 420-nm photon of violet light is a tiny fraction of a joule, and so it is no wonder that a single photon would be difficult for us to sense directly—humans are more attuned to energies on the order of joules. But looking at the energy in electron volts, we can see that this photon has enough energy to affect atoms and molecules. A DNA molecule can be broken with about 1 eV of energy, for example, and typical atomic and molecular energies are on the order of eV, so that the UV photon in this example could have biological effects.
The ejected electron (called a photoelectron) has a rather low energy, and it would not travel far, except in a vacuum. The electron would be stopped by a retarding potential of 0.26 eV. In fact, if the photon wavelength were longer and its energy less than 2.71 eV, then the formula would give a negative kinetic energy, an impossibility. This simply means that the 420-nm photons with their 2.96-eV energy are not much above the frequency threshold. You can show for yourself that the threshold wavelength is 459 nm (blue light). This means that if calcium metal is used in a light meter, the meter will be insensitive to wavelengths longer than those of blue light. Such a light meter would be insensitive to red light, for example.
Exercise 1.3.1 : Silver
What is the longest-wavelength electromagnetic radiation that can eject a photoelectron from silver? Is this in the visible range?
Answer
Given that the workfunction is 4.72 eV from Table 1.3.1 , then only photons with wavelengths lower than 263 nm will induce photoelectrons (calculated via $E=h \nu$). This is ultraviolet and not in the visible range.
Exercise 1.3.2
Why is the workfunction of an element generally lower than the ionization energy of that element?
Answer
The workfunction of a metal refers to the minimum energy required to extract an electron from the surface of a (bulk) metal by the absorption a photon of light. The workfunction will vary from metal to metal. In contrast, ionization energy is the energy needed to detach electrons from atoms and also varies with each particular atom, with the valence electrons require less energy to extract than core electrons (i.e., from lower shells) that are more closely bound to the nuclei. The electrons in the metal lattice there less bound (i.e., free to move within the metal) and removing one of these electrons is much easier than removing an electron from an atom because the metallic bonds of the bulk metal reduces their binding energy. As we will show in subsequent chapters, the more delocalized a particle is, the lower its energy.
Summary
Although Hertz discovered the photoelectron in 1887, it was not until 1905 that a theory was proposed that explained the effect completely. The theory was proposed by Einstein and it made the claim that electromagnetic radiation had to be thought of as a series of particles, called photons, which collide with the electrons on the surface and emit them. This theory ran contrary to the belief that electromagnetic radiation was a wave and thus it was not recognized as correct until 1916 when Robert Millikan experimentally confirmed the theory
The photoelectric effect is the process in which electromagnetic radiation ejects electrons from a material. Einstein proposed photons to be quanta of electromagnetic radiation having energy $E=h\nu$ is the frequency of the radiation. All electromagnetic radiation is composed of photons. As Einstein explained, all characteristics of the photoelectric effect are due to the interaction of individual photons with individual electrons. The maximum kinetic energy $KE_e$ of ejected electrons (photoelectrons) is given by $KE_e=h\nu – \Phi$, where $h\nu$ is the photon energy and $\Phi$ is the workfunction (or binding energy) of the electron to the particular material.
Conceptual Questions
1. Is visible light the only type of electromagnetic radiation that can cause the photoelectric effect?
2. Which aspects of the photoelectric effect cannot be explained without photons? Which can be explained without photons? Are the latter inconsistent with the existence of photons?
3. Is the photoelectric effect a direct consequence of the wave character of electromagnetic radiation or of the particle character of electromagnetic radiation? Explain briefly.
4. Insulators (nonmetals) have a higher $\Phi$ than metals, and it is more difficult for photons to eject electrons from insulators. Discuss how this relates to the free charges in metals that make them good conductors.
5. If you pick up and shake a piece of metal that has electrons in it free to move as a current, no electrons fall out. Yet if you heat the metal, electrons can be boiled off. Explain both of these facts as they relate to the amount and distribution of energy involved with shaking the object as compared with heating it.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/11%3A_Quantum_Mechanics_and_Atomic_Structure/11.03%3A_The_Photoelectric_Effect.txt
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Overview
• To introduce the concept of absorption and emission line spectra and describe the Balmer equation to describe the visible lines of atomic hydrogen.
• Describe Rydberg's theory for the hydrogen spectra.
• Interpret the hydrogen spectrum in terms of the energy states of electrons.
The first person to realize that white light was made up of the colors of the rainbow was Isaac Newton, who in 1666 passed sunlight through a narrow slit, then a prism, to project the colored spectrum on to a wall. This effect had been noticed previously, of course, not least in the sky, but previous attempts to explain it, by Descartes and others, had suggested that the white light became colored when it was refracted, the color depending on the angle of refraction. Newton clarified the situation by using a second prism to reconstitute the white light, making much more plausible the idea that the white light was composed of the separate colors. He then took a monochromatic component from the spectrum generated by one prism and passed it through a second prism, establishing that no further colors were generated. That is, light of a single color did not change color on refraction. He concluded that white light was made up of all the colors of the rainbow, and that on passing through a prism, these different colors were refracted through slightly different angles, thus separating them into the observed spectrum.
Atomic Line Spectrum
The spectrum of hydrogen, which turned out to be crucial in providing the first insight into atomic structure over half a century later, was first observed by Anders Angstrom in Uppsala, Sweden, in 1853. His communication was translated into English in 1855. Angstrom, the son of a country minister, was a reserved person, not interested in the social life that centered around the court. Consequently, it was many years before his achievements were recognized, at home or abroad (most of his results were published in Swedish).
Most of what is known about atomic (and molecular) structure and mechanics has been deduced from spectroscopy. Figure 1.4.1 shows two different types of spectra. A continuous spectrum can be produced by an incandescent solid or gas at high pressure (blackbody radiation, for example, is a continuum). An emission spectrum can be produced by a gas at low pressure excited by heat or by collisions with electrons. An absorption spectrum results when light from a continuous source passes through a cooler gas, consisting of a series of dark lines characteristic of the composition of the gas.
In 1802, William Wollaston in England had discovered (perhaps by using a thinner slit or a better prism) that in fact the solar spectrum itself had tiny gaps - there were many thin dark lines in the rainbow of colors. These were investigated much more systematically by Joseph von Fraunhofer, beginning in 1814. He increased the dispersion by using more than one prism. He found an "almost countless number" of lines. He labeled the strongest dark lines A, B, C, D, etc. Frauenhofer between 1814 and 1823 discovered nearly 600 dark lines in the solar spectrum viewed at high resolution and designated the principal features with the letters A through K, and weaker lines with other letters (Table $1$). Modern observations of sunlight can detect many thousands of lines. It is now understood that these lines are caused by absorption by the outer layers of the Sun.
Table $1$ : Major Fraunhofer lines, and the elements they are associated with
Designation Element Wavelength (nm)
y O2 898.765
Z O2 822.696
A O2 759.370
B O2 686.719
C H 656.281
a O2 627.661
D1 Na 589.592
D2 Na 588.995
D3 or d He 587.5618
Fraunhofer Absorption Lines
The Fraunhofer lines are typical spectral absorption lines. These dark lines are produced whenever a cold gas is between a broad spectrum photon source and the detector. In this case, a decrease in the intensity of light in the frequency of the incident photon is seen as the photons are absorbed, then re-emitted in random directions, which are mostly in directions different from the original one. This results in an absorption line, since the narrow frequency band of light initially traveling toward the detector, has been turned into heat or re-emitted in other directions.
By contrast, if the detector sees photons emitted directly from a glowing gas, then the detector often sees photons emitted in a narrow frequency range by quantum emission processes in atoms in the hot gas, resulting in an emission line. In the Sun, Fraunhofer lines are seen from gas in the outer regions of the Sun, which are too cold to directly produce emission lines of the elements they represent.
Gases heated to incandescence were found by Bunsen, Kirkhoff and others to emit light with a series of sharp wavelengths. The emitted light analyzed by a spectrometer (or even a simple prism) appears as a multitude of narrow bands of color. These so called line spectra are characteristic of the atomic composition of the gas. The line spectra of several elements are shown in Figure $3$.
The Balmer Series of Hydrogen
Obviously, if any pattern could be discerned in the spectral lines for a specifc atom (in contract to the mixture that Fraunhofer lines represent), that might be a clue as to the internal structure of the atom. One might be able to build a model. A great deal of effort went into analyzing the spectral data from the 1860's on. The big breakthrough was made by Johann Balmer, a math and Latin teacher at a girls' school in Basel, Switzerland. Balmer had done no physics before, and made his great discovery when he was almost sixty.
Balmer decided that the most likely atom to show simple spectral patterns was the lightest atom, hydrogen. Angstrom had measured the four visible spectral lines to have wavelengths 656.21, 486.07, 434.01 and 410.12 nm (Figure $4$). Balmer concentrated on just these four numbers, and found they were represented by the formula:
$\lambda = b \left( \dfrac{n_2^2}{n_2^2 -4} \right) \label{1.4.1}$
where
• $b$ = 364.56 nm and
• $n_2 = 3, 4, 5, 6$.
The first four wavelengths of Equation $\ref{1.4.1}$ (with n_2=3,4,5,6) were in excellent agreement with the experimental lines from Angstrom (Table $2$). Balmer predicted that other lines exist in the infrared that correspond to n = 7, 8, etc., and in fact some of them had already been observed, unbeknown to Balmer.
Table $2$ : The Balmer Series of Hydrogen Emission Lines
$n_2$ 3 4 5 6 7 8 9 10
$\lambda$ 656 486 434 410 397 389 383 380
color red teal blue indigo violet not visible not visible not visible
The $n_2$ integer in the Balmer series extends theoretically to infinity and the series resents a monotonically increasing energy (and frequency) of the absorption lines with increasing $n_2$ values. Moreover, the energy difference between successive lines decreased as $n_2$ increases (Figure 1.4.4). This behavior converges to a highest possible energy as Example $1$ demonstrates. If the lines are plot according to their $\lambda$ on a linear scale, you will get the appearance of the spectrum in Figure $4$; these lines are called the Balmer series.
Balmer's general formula (Equation $\ref{1.4.1}$) can be rewritten in terms of the inverse wavelength typically called the wavenumber ($\widetilde{\nu}$).
$\widetilde{\nu}= \dfrac{1}{ \lambda} =R_H \left( \dfrac{1}{4} -\dfrac{1}{n_2^2}\right) \label{1.4.2}$
where
• $n_2 = 3, 4, 5, 6$ and
• $R_H$ is the Rydberg constant (discussed below) and is equal to 109,737 cm-1.
He further conjectured that the 4 could be replaced by 9, 16, 25, … and this also turned out to be true - but these lines, further into the infrared, were not detected until the early twentieth century, along with the ultraviolet lines.
Wavenumbers
The relation between wavelength and frequency for electromagnetic radiation is
$\lambda \nu= c$
In the SI system of units the wavelength, \lambda\) is measured in meters (m) and since wavelengths are usually very small one often uses the nanometer (nm) which is $10^{-9}\; m$. The frequency ($\nu$) in the SI system is measured in reciprocal seconds 1/s − which is called a Hertz (after the discover of the photoelectron effect) and is represented by Hz.
It is common to use the reciprocal of the wavelength in centimeters as a measure of the frequency of radiation. This unit is called a wavenumber and is represented by ($\widetilde{\nu}$) and is defined by
$\widetilde{\nu}= \dfrac{1}{ \lambda} = \dfrac{\nu}{c}$
Wavenumbers is a convenient unit in spectroscopy because it is directly proportional to energy.
$E = \dfrac{hc}{\lambda}$
$E = hc \times \dfrac{1}{\lambda}$
$E = hc\widetilde{\nu}$
$E \propto \widetilde{\nu}$
Balmer Series
Calculate the longest and shortest wavelengths (in nm) emitted in the Balmer series of the hydrogen atom emission spectrum.
Solution
From the behavior of the Balmer equation (Equation $\ref{1.4.1}$ and Table $2$), the value of $n_2$ that gives the longest (i.e., greatest) wavelength ($\lambda$) is the smallest value possible of $n_2$, which is $n_2=3$ for this series. This results in
$\lambda_{longest} = (364.56 \;nm) \left( \dfrac{9}{9 -4} \right)$
$\lambda_{longest} = (364.56 \;nm) \left( 1.8 \right) = 656.2\; nm$
This is also known as the $H_{\alpha}$ line of atomic hydrogen and is bight red (Figure 1.4.4).
For the shortest wavelength, it should be recognized that the shortest wavelength (greatest energy) is obtained at the limit of greatest (n_2\):
$\lambda_{shortest} = \lim_{n_2 \rightarrow \infty} (364.56 \;nm) \left( \dfrac{n_2^2}{n_2^2 -4} \right)$
This can be solved via L'Hôpital's Rule, or alternatively the limit can be expressed via the equally useful energy expression (Equation 1.4.2) and simply solved:
$\widetilde{\nu}_{greatest}= \lim_{n_2 \rightarrow \infty} R_H \left( \dfrac{1}{4} -\dfrac{1}{n_2^2}\right)$
$\widetilde{\nu}_{greatest}= \lim_{n_2 \rightarrow \infty} R_H \left( \dfrac{1}{4}\right) = 27,434 \;cm^{-1}$
Since $\dfrac{1}{\widetilde{\nu}}= \lambda$ in units of cm, this converts to 364 nm as the shortest wavelength possible for the Balmer series.
The Balmer series is particularly useful in astronomy because the Balmer lines appear in numerous stellar objects due to the abundance of hydrogen in the universe, and therefore are commonly seen and relatively strong compared to lines from other elements.
The Generalized Rydberg Equation
In an amazing demonstration of mathematical insight, in 1885 Balmer came up with a simple formula for predicting the wavelength of any of the lines in atomic Hydrogen in what we now know as the Balmer series (Equation $\ref{1.4.2}$). Three years later, Rydberg generalized this so that it was possible to determine the wavelengths of any of the lines in the hydrogen emission spectrum. Rydberg suggested that all atomic spectra formed families with this pattern (he was unaware of Balmer's work). It turns out that there are families of spectra following Rydberg's pattern, notably in the alkali metals, sodium, potassium, etc., but not with the precision the hydrogen atom lines fit the Balmer formula, and low values of $n_2$ predicted wavelengths that deviate considerably.
Rydberg's general equation is as follows:
$\color{ref} \widetilde{\nu}= \dfrac{1}{ \lambda} =R_H \left( \dfrac{1}{n_1^2} -\dfrac{1}{n_2^2}\right) \label{1.5.1}$
where
• $R_H$ is the Rydberg constant and is equal to 109,737 cm-1.
• $n_1$ and $n_2$ are integers (whole numbers) with $n_2 > n_1$.
For the Balmer lines, $n_1 =2$ and $n_2$ can be any whole number between 3 and infinity. The various combinations of numbers that can be substituted into this formula allow the calculation the wavelength of any of the lines in the hydrogen emission spectrum; there is close agreement between the wavelengths generated by this formula and those observed in a real spectrum.
Other Series
The results given by Balmer and Rydberg for the spectrum in the visible region of the electromagnetic radiation start with $n_2 = 3$, and $n_1^2=2$. Is there a different series with the following formula (e.g., $n_1=1$?
$\dfrac{1}{\lambda} = R_{\textrm H} \left(\dfrac{1}{1^2} - \dfrac{1}{n^2} \right ) \label{1.5.2}$
The values for $n_2$ and wavenumber $\widetilde{\nu}$ for this series would be:
Table $3$: The Lyman Series of Hydrogen Emission Lines ($n_1=1$)
$n_2$ 2 3 4 5 ...
$\lambda$ (nm) 121 102 97 94 ...
$\widetilde{\nu}$ (cm-1) 82,2291 97,530 102,864 105,332 ...
These lines are in the UV region, and they are not visible, but they are detected by instruments; these lines form a Lyman series. The existences of the Lyman series and Balmer's series suggest the existence of more series. For example, the series with $n_2^2 = 3$ and $n_1^2$ = 4, 5, 6, 7, ... is called Pashen series.
Note
The spectral lines are grouped into series according to $n_1$ values. Lines are named sequentially starting from the longest wavelength/lowest frequency of the series, using Greek letters within each series. For example, the ($n_1=1/n_2=2$) line is called "Lyman-alpha" (Ly-α), while the ($n_1=3/n_2=7$) line is called "Paschen-delta" (Pa-δ). The first six series have specific names:
• Lyman series with $n_1 = 1$
• Balmer series with $n_1 = 2$
• Paschen series (or Bohr series) with $n_1 = 3$
• Brackett series with $n_1 = 4$
• Pfund series with $n_1 = 5$
• Humphreys series with $n_1 = 6$
The spectral series of hydrogen based of the Rydberg Equation (on a logarithmic scale).
Example $2$: The Lyman Series
The so-called Lyman series of lines in the emission spectrum of hydrogen corresponds to transitions from various excited states to the n = 1 orbit. Calculate the wavelength of the lowest-energy line in the Lyman series to three significant figures. In what region of the electromagnetic spectrum does it occur?
Given: lowest-energy orbit in the Lyman series
Asked for: wavelength of the lowest-energy Lyman line and corresponding region of the spectrum
Strategy:
1. Substitute the appropriate values into Equation 1.5.1 (the Rydberg equation) and solve for $\lambda$.
2. Locate the region of the electromagnetic spectrum corresponding to the calculated wavelength.
Solution:
We can use the Rydberg equation to calculate the wavelength:
$\dfrac{1}{\lambda }=\Re \left ( \dfrac{1}{n_{1}^{2}} - \dfrac{1}{n_{2}^{2}}\right )$
A For the Lyman series, $n_1 = 1$.
$\dfrac{1}{\lambda }=\Re \left ( \dfrac{1}{n_{1}^{2}} - \dfrac{1}{n_{2}^{2}}\right )=1.097 \times m^{-1}\left ( \dfrac{1}{1}-\dfrac{1}{4} \right )= 8.228 \times 10^{6}\; m^{-1}$
Spectroscopists often talk about energy and frequency as equivalent. The cm-1 unit is particularly convenient. The infrared range is roughly 200 - 5,000 cm-1, the visible from 11,000 to 25.000 cm-1 and the UV between 25,000 and 100,000 cm-1. The units of cm-1 are called wavenumbers, although people often verbalize it as inverse centimeters. We can convert the answer in part A to cm-1.
$\widetilde{\nu} =\dfrac{1}{\lambda }=8.228\times 10^{6}\cancel{m^{-1}}\left (\dfrac{\cancel{m}}{100\;cm} \right )=82,280\: cm^{-1}$
and
$\lambda = 1.215 \times 10^{−7}\; m = 122\; nm$
This emission line is called Lyman alpha. It is the strongest atomic emission line from the sun and drives the chemistry of the upper atmosphere of all the planets producing ions by stripping electrons from atoms and molecules. It is completely absorbed by oxygen in the upper stratosphere, dissociating O2 molecules to O atoms which react with other O2 molecules to form stratospheric ozone
B This wavelength is in the ultraviolet region of the spectrum.
Exercise $2$: The Pfund Series
The Pfund series of lines in the emission spectrum of hydrogen corresponds to transitions from higher excited states to the $n_1 = 5$ orbit. Calculate the wavelength of the second line in the Pfund series to three significant figures. In which region of the spectrum does it lie?
Answer: 4.65 × 103 nm; infrared
The above discussion presents only a phenomenological description of hydrogen emission lines and fails to provide a probe of the nature of the atom itself. Clearly a continuum model based on classical mechanics is not applicable.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/11%3A_Quantum_Mechanics_and_Atomic_Structure/11.04%3A_Bohr%27s_Theory_of_the_Hydrogen_Emission_Spectrum.txt
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Learning Objectives
• To introduce the wave-particle duality of light extends to matter
• To describe how matter (e.g., electrons and protons) can exhibit wavelike properties, e.g., interference and diffraction patterns
• To use algebra to find the de Broglie wavelength or momentum of a particle when either one of these quantities is given
The next real advance in understanding the atom came from an unlikely quarter - a student prince in Paris. Prince Louis de Broglie was a member of an illustrious family, prominent in politics and the military since the 1600's. Louis began his university studies with history, but his elder brother Maurice studied x-rays in his own laboratory, and Louis became interested in physics. After World War I, de Broglie focused his attention on Einstein's two major achievements, the theory of special relativity and the quantization of light waves. He wondered if there could be some connection between them. Perhaps the quantum of radiation really should be thought of as a particle. De Broglie suggested that if waves (photons) could behave as particles, as demonstrated by the photoelectric effect, then the converse, namely that particles could behave as waves, should be true. He associated a wavelength $\lambda$ to a particle with momentum $p$ using Planck's constant as the constant of proportionality:
$\lambda =\dfrac{h}{p} \label{1.6.1}$
which is called the de Broglie wavelength. The fact that particles can behave as waves but also as particles, depending on which experiment you perform on them, is known as the wave-particle duality.
Deriving the de Broglie Wavelength
From the discussion of the photoelectric effect, we have the first part of the particle-wave duality, namely, that electromagnetic waves can behave like particles. These particles are known as photons, and they move at the speed of light. Any particle that moves at or near the speed of light has kinetic energy given by Einstein's special theory of relatively. In general, a particle of mass $m$ and momentum $p$ has an energy
$E=\sqrt{p^2 c^2+m^2 c^4} \label{1.6.2}$
Note that if $p=0$, this reduces to the famous rest-energy expression $E=mc^2$. However, photons are massless particles (technically rest-massless) that always have a finite momentum $p$. In this case, Equation \ref{1.6.2} becomes
$E=pc. \nonumber$
From Planck's hypothesis, one quantum of electromagnetic radiation has energy $E=h\nu$. Thus, equating these two expressions for the kinetic energy of a photon, we have
$h\nu =\dfrac{hc}{\lambda}=pc \label{1.6.4}$
Solving for the wavelength $\lambda$ gives Equation \ref{1.6.1}:
$\lambda=\dfrac{h}{p}= \dfrac{h}{mv} \nonumber$
where $v$ is the velocity of the particle. Hence, de Broglie argued that if particles can behave as waves, then a relationship like this, which pertains particularly to waves, should also apply to particles.
Equation \ref{1.6.1} allows us to associate a wavelength $\lambda$ to a particle with momentum $p$. As the momentum increases, the wavelength decreases. In both cases, this means the energy becomes larger. i.e., short wavelengths and high momenta correspond to high energies.
It is a common feature of quantum mechanics that particles and waves with short wavelengths correspond to high energies and vice versa.
Having decided that the photon might well be a particle with a rest mass, even if very small, it dawned on de Broglie that in other respects it might not be too different from other particles, especially the very light electron. In particular, maybe the electron also had an associated wave. The obvious objection was that if the electron was wavelike, why had no diffraction or interference effects been observed? But there was an answer. If de Broglie's relation between momentum and wavelength also held for electrons, the wavelength was sufficiently short that these effects would be easy to miss. As de Broglie himself pointed out, the wave nature of light is not very evident in everyday life. As the next section will demonstrate, the validity of de Broglie’s proposal was confirmed by electron diffraction experiments of G.P. Thomson in 1926 and of C. Davisson and L. H. Germer in 1927. In these experiments it was found that electrons were scattered from atoms in a crystal and that these scattered electrons produced an interference pattern. These diffraction patterns are characteristic of wave-like behavior and are exhibited by both electrons (i.e., matter) and electromagnetic radiation (i.e., light).
Example 1.6.1 : Electron Waves
Calculate the de Broglie wavelength for an electron with a kinetic energy of 1000 eV.
Solution
To calculate the de Broglie wavelength (Equation \ref{1.6.1}), the momentum of the particle must be established and requires knowledge of both the mass and velocity of the particle. The mass of an electron is $9.109383 \times 10^{−28}\; g$ and the velocity is obtained from the given kinetic energy of 1000 eV:
\begin{align*} KE &= \dfrac{mv^2}{2} \[4pt] &= \dfrac{p^2}{2m} = 1000 \;eV \end{align*} \nonumber
Solve for momentum
$p = \sqrt{2 m KE} \nonumber$
convert to SI units
$p = \sqrt{(1000 \; \cancel{eV}) \left( \dfrac{1.6 \times 10^{-19} \; J}{1\; \cancel{ eV}} \right) (2) (9.109383 \times 10^{-31}\; kg)} \nonumber$
expanding definition of joule into base SI units and cancel
\begin{align*} p &= \sqrt{(3.1 \times 10^{-16} \;kg \cdot m^2/s^2 ) (9.109383 \times 10^{-31}\; kg)} \[4pt] &= \sqrt{ 2.9 \times 10^{-40 }\, kg^2 \;m^2/s^2 } \[4pt] &= 1.7 \times 10^{-23} kg \cdot m/s \end{align*} \nonumber
Now substitute the momentum into the equation for de Broglie's wavelength (Equation $\ref{1.6.1}$) with Planck's constant ($h = 6.626069 \times 10^{−34}\;J \cdot s$). After expanding units in Plank's constant
\begin{align*} \lambda &=\dfrac{h}{p} \[4pt] &= \dfrac{6.626069 \times 10^{−34}\;kg \cdot m^2/s}{1.7 \times 10^{-23} kg \cdot m/s} \[4pt] &= 3.87 \times 10^{-11}\; m \[4pt] &=38.9\; pm \end{align*} \nonumber
Exercise 1.6.1 : Baseball Waves
Calculate the de Broglie wavelength for a fast ball thrown at 100 miles per hour and weighing 4 ounces. Comment on whether the wave properties of baseballs could be experimentally observed.
Answer
Following the unit conversions below, a 4 oz baseball has a mass of 0.11 kg. The velocity of a fast ball thrown at 100 miles per hour in m/s is 44.7 m/s.
$m = \left(4 \; \cancel{oz}\right)\left(\frac{0.0283 \; kg}{1 \; \cancel{oz}}\right) = 0.11 kg \nonumber$
$v = \left(\frac{100 \; \cancel{mi}}{\cancel{hr}}\right) \left(\frac{1609.34 \; m}{\cancel{mi}}\right) \left( \frac{1 \; \cancel{hr}}{3600 \; s}\right) = 44.7 \; m/s \nonumber$
The de Broglie wavelength of this fast ball is:
$\lambda = \frac{h}{mv} = \frac{6.626069 \times 10^{-34}\;kg \cdot m^2/s}{(0.11 \; kg)(44.7 \;m/s)} = 1.3 \times 10^{-34} m \nonumber$
Exercise 1.6.2 : Electrons vs. Protons
If an electron and a proton have the same velocity, which would have the longer de Broglie wavelength?
1. The electron
2. The proton
3. They would have the same wavelength
Answer
Equation \ref{1.6.1} shows that the de Broglie wavelength of a particle's matter wave is inversely proportional to its momentum (mass times velocity). Therefore the smaller mass particle will have a smaller momentum and longer wavelength. The electron is the lightest and will have the longest wavelength.
This was the prince's Ph.D. thesis, presented in 1924. His thesis advisor was somewhat taken aback, and was not sure if this was sound work. He asked de Broglie for an extra copy of the thesis, which he sent to Einstein. Einstein wrote shortly afterwards: "I believe it is a first feeble ray of light on this worst of our physics enigmas" and the prince got his Ph.D.
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Learning Objectives
• To understand that sometime you cannot know everything about a quantum system as demonstrated by the Heisenberg uncertainly principle.
In classical physics, studying the behavior of a physical system is often a simple task due to the fact that several physical qualities can be measured simultaneously. However, this possibility is absent in the quantum world. In 1927 the German physicist Werner Heisenberg described such limitations as the Heisenberg Uncertainty Principle, or simply the Uncertainty Principle, stating that it is not possible to measure both the momentum and position of a particle simultaneously.
The Heisenberg Uncertainty Principle is a fundamental theory in quantum mechanics that defines why a scientist cannot measure multiple quantum variables simultaneously. Until the dawn of quantum mechanics, it was held as a fact that all variables of an object could be known to exact precision simultaneously for a given moment. Newtonian physics placed no limits on how better procedures and techniques could reduce measurement uncertainty so that it was conceivable that with proper care and accuracy all information could be defined. Heisenberg made the bold proposition that there is a lower limit to this precision making our knowledge of a particle inherently uncertain.
Probability
Matter and photons are waves, implying they are spread out over some distance. What is the position of a particle, such as an electron? Is it at the center of the wave? The answer lies in how you measure the position of an electron. Experiments show that you will find the electron at some definite location, unlike a wave. But if you set up exactly the same situation and measure it again, you will find the electron in a different location, often far outside any experimental uncertainty in your measurement. Repeated measurements will display a statistical distribution of locations that appears wavelike (Figure 1.9.1 ).
After de Broglie proposed the wave nature of matter, many physicists, including Schrödinger and Heisenberg, explored the consequences. The idea quickly emerged that, because of its wave character, a particle’s trajectory and destination cannot be precisely predicted for each particle individually. However, each particle goes to a definite place (Figure 1.9.1 ). After compiling enough data, you get a distribution related to the particle’s wavelength and diffraction pattern. There is a certain probability of finding the particle at a given location, and the overall pattern is called a probability distribution. Those who developed quantum mechanics devised equations that predicted the probability distribution in various circumstances.
It is somewhat disquieting to think that you cannot predict exactly where an individual particle will go, or even follow it to its destination. Let us explore what happens if we try to follow a particle. Consider the double-slit patterns obtained for electrons and photons in Figure 1.9.2 . The interferrence patterns build up statistically as individual particles fall on the detector. This can be observed for photons or electrons—for now, let us concentrate on electrons. You might imagine that the electrons are interfering with one another as any waves do. To test this, you can lower the intensity until there is never more than one electron between the slits and the screen. The same interference pattern builds up!
This implies that a particle’s probability distribution spans both slits, and the particles actually interfere with themselves. Does this also mean that the electron goes through both slits? An electron is a basic unit of matter that is not divisible. But it is a fair question, and so we should look to see if the electron traverses one slit or the other, or both. One possibility is to have coils around the slits that detect charges moving through them. What is observed is that an electron always goes through one slit or the other; it does not split to go through both.
But there is a catch. If you determine that the electron went through one of the slits, you no longer get a double slit pattern—instead, you get single slit interference. There is no escape by using another method of determining which slit the electron went through. Knowing the particle went through one slit forces a single-slit pattern. If you do not observe which slit the electron goes through, you obtain a double-slit pattern. How does knowing which slit the electron passed through change the pattern? The answer is fundamentally important—measurement affects the system being observed. Information can be lost, and in some cases it is impossible to measure two physical quantities simultaneously to exact precision. For example, you can measure the position of a moving electron by scattering light or other electrons from it. Those probes have momentum themselves, and by scattering from the electron, they change its momentum in a manner that loses information. There is a limit to absolute knowledge, even in principle.
Heisenberg’s Uncertainty Principle
It is mathematically possible to express the uncertainty that, Heisenberg concluded, always exists if one attempts to measure the momentum and position of particles. First, we must define the variable “x” as the position of the particle, and define “p” as the momentum of the particle. The momentum of a photon of light is known to simply be its frequency, expressed by the ratio $h/λ$, where h represents Planck’s constant and $\lambda$ represents the wavelength of the photon. The position of a photon of light is simply its wavelength ($\lambda$). To represent finite change in quantities, the Greek uppercase letter delta, or Δ, is placed in front of the quantity. Therefore,
$\Delta{p}=\dfrac{h}{\lambda} \label{1.9.1}$
$\Delta{x}= \lambda \label{1.9.2}$
By substituting $\Delta{x}$ for $\lambda$ into Equation $\ref{1.9.1}$, we derive
$\Delta{p}=\dfrac{h}{\Delta{x}} \label{1.9.3}$
or,
$\underset{\text{early form of uncertainty principle }}{\Delta{p}\Delta{x}=h} \label{1.9.4}$
A Common Trend in Quantum Systems
Equation $\ref{1.9.4}$ can be derived by assuming the particle of interest is behaving as a particle, and not as a wave. Simply let $\Delta p=mv$, and $Δx=h/(m v)$ (from De Broglie’s expression for the wavelength of a particle). Substituting in $Δp$ for $mv$ in the second equation leads to Equation $\ref{1.9.4}$.
Equation \ref{1.9.4} was further refined by Heisenberg and his colleague Niels Bohr, and was eventually rewritten as
$\Delta{p_x}\Delta{x} \ge \dfrac{h}{4\pi} = \dfrac{\hbar}{2} \label{1.9.5}$
with $\hbar = \dfrac{h}{2\pi}= 1.0545718 \times 10^{-34}\; m^2 \cdot kg / s$.
Equation $\ref{1.9.5}$ reveals that the more accurately a particle’s position is known (the smaller $Δx$ is), the less accurately the momentum of the particle in the x direction ($Δp_x$) is known. Mathematically, this occurs because the smaller $Δx$ becomes, the larger $Δp_x$ must become in order to satisfy the inequality. However, the more accurately momentum is known the less accurately position is known (Figure 1.9.2 ).
What is the Proper Definition of Uncertainty?
Equation $\ref{1.9.5}$ relates the uncertainty of momentum and position. An immediate questions that arise is if $\Delta x$ represents the full range of possible $x$ values or if it is half (e.g., $\langle x \rangle \pm \Delta x$). $\Delta x$ is the standard deviation and is a statistic measure of the spread of $x$ values. The use of half the possible range is more accurate estimate of $\Delta x$. As we will demonstrated later, once we construct a wavefunction to describe the system, then both $x$ and $\Delta x$ can be explicitly derived. However for now, Equation \ref{1.9.5} will work.
For example: If a problem argues a particle is trapped in a box of length, $L$, then the uncertainly of it position is $\pm L/2$. So the value of $\Delta x$ used in Equation $\ref{1.9.5}$ should be $L/2$, not $L$.
Example 1.9.1
An electron is confined to the size of a magnesium atom with a 150 pm radius. What is the minimum uncertainty in its velocity?
Solution
The uncertainty principle (Equation $\ref{1.9.5}$):
$\Delta{p}\Delta{x} \ge \dfrac{\hbar}{2} \nonumber$
can be written
$\Delta{p} \ge \dfrac{\hbar}{2 \Delta{x}} \nonumber$
and substituting $\Delta p=m \Delta v$ since the mass is not uncertain.
$\Delta{v} \ge \dfrac{\hbar}{2\; m\; \Delta{x}} \nonumber$
the relevant parameters are
• mass of electron $m=m_e= 9.109383 \times 10^{-31}\; kg$
• uncertainty in position: $\Delta x=150 \times 10^{-12} m$
\begin{align*} \Delta{v} &\ge \dfrac{1.0545718 \times 10^{-34} \cancel{kg} m^{\cancel{2}} / s}{(2)\;( 9.109383 \times 10^{-31} \; \cancel{kg}) \; (150 \times 10^{-12} \; \cancel{m}) } \[4pt] &= 3.9 \times 10^5\; m/s \end{align*} \nonumber
Exercise 1.9.1
What is the maximum uncertainty of velocity the electron described in Example 1.9.1 ?
Answer
Infinity. There is no limit in the maximum uncertainty, just the minimum uncertainty.
Understanding the Uncertainty Principle through Wave Packets and the Slit Experiment
It is hard for most people to accept the uncertainty principle, because in classical physics the velocity and position of an object can be calculated with certainty and accuracy. However, in quantum mechanics, the wave-particle duality of electrons does not allow us to accurately calculate both the momentum and position because the wave is not in one exact location but is spread out over space. A "wave packet" can be used to demonstrate how either the momentum or position of a particle can be precisely calculated, but not both of them simultaneously. An accumulation of waves of varying wavelengths can be combined to create an average wavelength through an interference pattern: this average wavelength is called the "wave packet". The more waves that are combined in the "wave packet", the more precise the position of the particle becomes and the more uncertain the momentum becomes because more wavelengths of varying momenta are added. Conversely, if we want a more precise momentum, we would add less wavelengths to the "wave packet" and then the position would become more uncertain. Therefore, there is no way to find both the position and momentum of a particle simultaneously.
Several scientists have debated the Uncertainty Principle, including Einstein. Einstein created a slit experiment to try and disprove the Uncertainty Principle. He had light passing through a slit, which causes an uncertainty of momentum because the light behaves like a particle and a wave as it passes through the slit. Therefore, the momentum is unknown, but the initial position of the particle is known. Here is a video that demonstrates particles of light passing through a slit and as the slit becomes smaller, the final possible array of directions of the particles becomes wider. As the position of the particle becomes more precise when the slit is narrowed, the direction, or therefore the momentum, of the particle becomes less known as seen by a wider horizontal distribution of the light.
Example 1.9.2
The speed of a 1.0 g projectile is known to within $10^{-6}\;m/s$.
1. Calculate the minimum uncertainty in its position.
2. What is the maximum uncertainty of its position?
Solution
a
From Equation $\ref{1.9.5}$, the $\Delta{p_x} = m \Delta v_x$ with $m=1.0\;g$. Solving for $\Delta{x}$ to get
\begin{align*} \Delta{x} &= \dfrac{\hbar}{2m\Delta v} \[4pt] &= \dfrac{1.0545718 \times 10^{-34} \; m^2 \cdot kg / s}{(2)(0.001 \; kg)(10^{-6} \;m/s)} \[4pt] &= 5.3 \times 10^{-26} \,m \end{align*} \nonumber
This negligible for all intents and purpose as expected for any macroscopic object.
b
Unlimited (or the size of the universe). The Heisenberg uncertainty principles does not quantify the maximum uncertainty.
Exercise 1.9.2
Estimate the minimum uncertainty in the speed of an electron confined to a hydrogen atom within a diameter of $1 \times 10^{-10} m$?
Answer
We need to quantify the uncertainty of the electron in position. We can estimate that as $\pm 5 \times 10^{-10} m$. Hence, substituting the relavant numbers into Equation \ref{1.9.5} and solving for $\Delta v$ we get
$\Delta v= 1.15 \times 10^6\, km/s \nonumber$
Notice that the uncertainty is significantly greater for the electron in a hydrogen atom than in the magnesium atom (Example 1.9.1 ) as expected since the magnesium atom is appreciably bigger.
Heisenberg’s Uncertainty Principle not only helped shape the new school of thought known today as quantum mechanics, but it also helped discredit older theories. Most importantly, the Heisenberg Uncertainty Principle made it obvious that there was a fundamental error in the Bohr model of the atom. Since the position and momentum of a particle cannot be known simultaneously, Bohr’s theory that the electron traveled in a circular path of a fixed radius orbiting the nucleus was obsolete. Furthermore, Heisenberg’s uncertainty principle, when combined with other revolutionary theories in quantum mechanics, helped shape wave mechanics and the current scientific understanding of the atom.
Humor: Heisenberg and the Police
• Heisenberg get pulled over for speeding by the police. The officer asks him "Do you know how fast you were going?"
• Heisenberg replies, "No, but we know exactly where we are!"
• The officer looks at him confused and says "you were going 108 miles per hour!"
• Heisenberg throws his arms up and cries, "Great! Now we're lost!"
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Beginning in the early 20th century, physicists began to acknowledge that matter--much like electromagnetic radiation--possessed wave-like behaviors. While electromagnetic radiation were well understood to obey Maxwell's Equations, matter obeyed no known equations.
Introduction
In 1926, the Austrian physicist Erwin Schrödinger formulated what came to be known as the (time-dependent) Schrödinger Equation:
$i\hbar\dfrac{\partial}{\partial t}\Psi(x,t)=\dfrac{-\hbar}{2m}\nabla^2\Psi(x,t) +V(x)\Psi(x,t) \label{1.1}$
Equation $\ref{1.1}$ effectively describes matter as a wave that fluctuates with both displacement and time. Since the imaginary portion of the equation dictates its time dependence, it is sufficed to say that for most purposes it can be treated as time-independent. The result is seen in Equation $\ref{1.3}$:
$\dfrac{-\hbar^2}{2m}\dfrac{d^2\psi(x)}{dx^2}+V(x)\psi(x) = E\psi(x) \label{1.3}$
Although this time-independent Schrödinger Equation can be useful to describe a matter wave in free space, we are most interested in waves when confined to a small region, such as an electron confined in a small region around the nucleus of an atom. Several different models have been developed that provide a means by which to study a matter-wave when confined to a small region: the particle in a box (infinite well), finite well, and the Hydrogen atom. We will discuss each of these in order to develop a greater understanding for how a wave behaves when it is in a bound state.
There are four general aspects that are applicable to an acceptable wavefunction
1. An acceptable wavefunction will be the solution of the Schrödinger equation (either Equations $\ref{1.1}$ or $\ref{1.3}$).
2. An acceptable wavefunction must be normalizable so will approaches zero as position approaches infinity.
3. An acceptable wavefunction must be a continuous function of position.
4. An acceptable wavefunction will have a continuous slope (Figure $1$).
Interpretation of the Wavefunction
Since wavefunctions can in general be complex functions, the physical significance cannot be found from the function itself because the $\sqrt {-1}$ is not a property of the physical world. Rather, the physical significance is found in the product of the wavefunction and its complex conjugate, i.e. the absolute square of the wavefunction, which also is called the square of the modulus.
$\psi ^* (r , t ) \psi (r , t ) = {|\psi (r , t ) |}^2 \label{3-34}$
where $r$ is a vector (x, y, z) specifying a point in three-dimensional space. The square is used, rather than the modulus itself, just like the intensity of a light wave depends on the square of the electric field. At one time it was thought that for an electron described by the wavefunction $\psi(r)$, the quantity $e\psi^*(r_i)\psi(r_i)d\tau$ was the amount of charge to be found in the volume $d\tau$ located at $r_i$. However, Max Born found this interpretation to be inconsistent with the results of experiments.
The Born interpretation, which generally is accepted today, is that $\psi^*(r_i)\psi(r_i)\, d\tau$ is the probability that the electron is in the volume dτ located at $r_i$. The Born interpretation therefore calls the wavefunction the probability amplitude, the absolute square of the wavefunction is called the probability density, and the probability density times a volume element in three-dimensional space ($d\tau$) is the probability. The idea that we can understand the world of atoms and molecules only in terms of probabilities is disturbing to some, who are seeking more satisfying descriptions through ongoing research.
Normalization of the Wavefunction
A probability is a real number between 0 and 1. An outcome of a measurement which has a probability 0 is an impossible outcome, whereas an outcome which has a probability 1 is a certain outcome. The probability of a measurement of $x$ yielding a result between $-\infty$ and $+\infty$ is
$P_{x \in -\infty:\infty}(t) = \int_{-\infty}^{\infty}\vert\psi(x,t)\vert^{ 2} dx. \label{3.6.2}$
However, a measurement of $x$ must yield a value between $-\infty$ and $+\infty$, since the particle has to be located somewhere. It follows that $P_{x \in -\infty:\infty}(t) =1$, or
$\int_{-\infty}^{\infty}\vert\psi(x,t)\vert^{ 2} dx = 1, \label{3.6.3}$
which is generally known as the normalization condition for the wavefunction.
Example $1$: Normalizing a Gaussian wavefunction
Normalize the wavefunction of a Gaussian wave packet, centered on $x=x_o$, and of characteristic width $\sigma$: i.e.,
$\psi(x) = \psi_0 {\rm e}^{-(x-x_0)^{ 2}/(4 \sigma^2)}. \label{3.6.4}$
Solution
To determine the normalization constant $\psi_0$, we simply substitute Equation $\ref{3.6.4}$ into Equation $\ref{3.6.3}$, to obtain
$\vert\psi_0\vert^{ 2}\int_{-\infty}^{\infty}{\rm e}^{-(x-x_0)^{ 2}/(2 \sigma^2)} dx = 1. \label{3.6.5}$
Changing the variable of integration to $y=(x-x_0)/(\sqrt{2} \sigma)$, we get
$\vert\psi_0\vert^{ 2}\sqrt{2} \sigma \int_{-\infty}^{\infty}{\rm e}^{-y^2} dy=1. \label{3.6.6}$
However,
$\int_{-\infty}^{\infty}{\rm e}^{-y^2} dy = \sqrt{\pi}, \label{3.6.7}$
which implies that
$\vert\psi_0\vert^{ 2} = \dfrac{1}{(2\pi \sigma^2)^{1/2}}. \label{3.6.8}$
Hence, a general normalized Gaussian wavefunction takes the form
$\psi(x) = \dfrac{e^{\rm{i} \phi}}{(2\pi \sigma^2)^{1/4}}e^{-(x-x_0)^2/(4 \sigma^2)} \label{3.6.9}$
where $\phi$ is an arbitrary real phase-angle.
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A particle in a 1-dimensional box is a fundamental quantum mechanical approximation describing the translational motion of a single particle confined inside an infinitely deep well from which it cannot escape.
Introduction
The particle in a box problem is a common application of a quantum mechanical model to a simplified system consisting of a particle moving horizontally within an infinitely deep well from which it cannot escape. The solutions to the problem give possible values of E and $\psi$ that the particle can possess. E represents allowed energy values and $\psi(x)$ is a wavefunction, which when squared gives us the probability of locating the particle at a certain position within the box at a given energy level.
recipe for Quantum Mechanics
To solve the problem for a particle in a 1-dimensional box, we must follow our Big, Big recipe for Quantum Mechanics:
1. Define the Potential Energy, $V$
2. Solve the Schrödinger Equation
3. Solve for the wavefunctions
4. Solve for the allowed energies
Step 1: Define the Potential Energy V
The potential energy is 0 inside the box (V=0 for 0<x<L) and goes to infinity at the walls of the box (V=∞ for x<0 or x>L). We assume the walls have infinite potential energy to ensure that the particle has zero probability of being at the walls or outside the box. Doing so significantly simplifies our later mathematical calculations as we employ these boundary conditions when solving the Schrödinger Equation.
Step 2: Solve the Schrödinger Equation
The time-independent Schrödinger equation for a particle of mass $m$ moving in one direction with energy $E$ is
$-\dfrac{\hbar^2}{2m} \dfrac{d^2 \psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x) \label{5.5.1}$
with
• $\hbar$ is the reduced Planck constant where $\hbar = \frac{h}{2\pi}$
• $m$ is the mass of the particle
• $\psi(x)$ is the stationary time-independent wavefunction
• $V(x)$ is the potential energy as a function of position
• $E$ is the energy, a real number
This equation can be modified for a particle of mass $m$ free to move parallel to the x-axis with zero potential energy (V = 0 everywhere) resulting in the quantum mechanical description of free motion in one dimension:
$-\dfrac{\hbar^2}{2m} \dfrac{d^2\psi(x)}{dx^2} = E\psi(x) \label{5.5.2}$
This equation has been well studied and gives a general solution of:
$\psi(x) = A\sin(kx) + B\cos(kx) \label{5.5.3}$
where A, B, and k are constants.
Step 3: Define the Wavefunction
The solution to the Schrödinger equation we found above is the general solution for a 1-dimensional system. We now need to apply our boundary conditions to find the solution to our particular system. According to our boundary conditions, the probability of finding the particle at $x=0$ or $x=L$ is zero. When $x=0$, then $\sin(0)=0$ and $\cos(0)=1$; therefore, $B$ must equal 0 to fulfill this boundary condition giving:
$\psi(x) = A\sin(kx) \label{5.5.4}$
We can now solve for our constants ($A$ and $k$) systematically to define the wavefunction.
Solving for $k$
Differentiate the wavefunction with respect to $x$:
$\dfrac{d\psi}{dx} = kA\cos(kx) \label{5.5.5}$
Differentiate the wavefunction algain with respect to $x$:
$\dfrac{d^{2}\psi}{dx^{2}} = -k^{2}A\sin(kx) \label{5.5.6}$
Since $\psi(x) = A\sin(kx)$, then
$\dfrac{d^{2}\psi}{dx^{2}} = -k^{2}\psi \label{5.5.7}$
If we then solve for k by comparing with the Schrödinger equation above, we find:
$k = \left( \dfrac{8\pi^2mE}{h^2} \right)^{1/2} \label{5.5.8}$
Now we plug $k$ into our wavefunction (Equation \ref{5.5.4}):
$\psi = A\sin\left(\dfrac{8\pi^{2}mE}{h^{2}}\right)^{1/2}x \label{5.5.9}$
Solving for $A$
To determine A, we have to apply the boundary conditions again. Recall that the probability of finding a particle at $x = 0$ or $x = L$ is zero.
When x = L:
$0 = A\sin\left(\dfrac{8\pi^{2}mE}{h^{2}}\right)^{1/2}L \label{5.5.10}$
This is only true when
$\left(\dfrac{8\pi^{2}mE}{h^{2}}\right)^{1/2}L = n\pi \label{5.5.11}$
where $n = 1,2,3, …$
Plugging this back in gives us:
$\psi = A\sin{\dfrac{n\pi}{L}}x \label{5.5.12}$
To determine $A$, recall that the total probability of finding the particle inside the box is 1, meaning there is no probability of it being outside the box. When we find the probability and set it equal to 1, we are normalizing the wavefunction.
$\int^{L}_{0}\psi^{2}dx = 1 \label{5.5.13}$
For our system, the normalization looks like:
$A^2 \int^{L}_{0}\sin^2\left(\dfrac{n\pi}{L}\right)x\,dx = 1 \label{5.5.14}$
Using the solution for this integral from an integral table, we find our normalization constant, $A$:
$A = \sqrt{\dfrac{2}{L}} \label{5.5.15}$
Which results in the normalized wavefunctions for a particle in a 1-dimensional box:
$\psi_n = \sqrt{\dfrac{2}{L}}\sin{\dfrac{n\pi}{L}}x \label{5.5.16}$
where $n = 1,2,3, …$
Step 4: Determine the Allowed Energies
Solving for the energy of each $\psi$ requires substituting Equation $\ref{5.5.16}$ into Equation $\ref{5.5.2}$ to get the allowed energies for a particle in a box:
$E_n = \dfrac{n^{2}h^{2}}{8mL^{2}} \label{5.5.17}$
Equation $\ref{5.5.17}$ is a very important result and tells us that:
1. The energy of a particle is quantized.
2. The lowest possible energy of a particle is NOT zero. This is called the zero-point energy and means the particle can never be at rest because it always has some kinetic energy.
This is also consistent with the Heisenberg Uncertainty Principle: if the particle had zero energy, we would know where it was in both space and time.
What does all this mean?
The wavefunction for a particle in a box at the $n=1$ and $n=2$ energy levels look like this:
The probability of finding a particle a certain spot in the box is determined by squaring $\psi$. The probability distribution for a particle in a box at the $n=1$ and $n=2$ energy levels looks like this:
Notice that the number of nodes (places where the particle has zero probability of being located) increases with increasing energy n. Also note that as the energy of the particle becomes greater, the quantum mechanical model breaks down as the energy levels get closer together and overlap, forming a continuum. This continuum means the particle is free and can have any energy value. At such high energies, the classical mechanical model is applied as the particle behaves more like a continuous wave. Therefore, the particle in a box problem is an example of Wave-Particle Duality.
Example $1$
What is the $\Delta E$ between the $n = 4$ and $n = 5$ states for an $F_2$ molecule trapped within in a one-dimension well of length 3.0 cm? At what value of $n$ does the energy of the molecule reach $¼k_BT$ at 450 K, and what is the separation between this energy level and the one immediately above it?
Solution
Since this is a one-dimensional particle in a box problem, the particle has only kinetic energy (V = 0), so the permitted energies are:
$E_n = \dfrac{n^2 h^2}{8 m L^2} \nonumber$
with $n=1,2,...$
The energy difference between $n = 4$ and $n=5$ is then
$\Delta E = E_5 - E_4=\dfrac{5^2 h^2}{8 m L^2} - \dfrac{4^2 h^2}{8 m L^2}\nonumber$
Using Equation $\ref{5.5.17}$ with the mass of $F_2$ (37.93 amu = $6.3 \times 10^{-26}\;kg$) and the length of the box ($L= 3 \times 3.0\times10^{-2}\;\text{m}^2$:
$\Delta E=\dfrac{9 h^2}{8 m L^2} = \dfrac{9 (6.626\times10^{-34}\;\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-1})^2 }{8(6.30938414\times10^{-26}\;\text{kg})(3.0\times10^{-2}\;\text{m}^2)^2 }\nonumber$
$\Delta E=8.70\times10^{-39}\; J \nonumber$
The $n$ value for which the energy reaches $\frac{1}{4} k_B T$:
$\dfrac{n^2 h^2}{8 m L^2} = \dfrac{1}{4} k_B T \nonumber$
$n=1.79 \times 10^9 \nonumber$
The separation between $n+1$ and $n$:
$\Delta E = E_{n+1} - E_{n} = \dfrac{(n+1)^2h^2 - (n)^2h^2}{8mL^2} \nonumber$
$\Delta E = \dfrac{(2n+1)h^2}{8mL^2} \nonumber$
$\Delta E = 3.47\times10^{-30}\;J \nonumber$
Important Facts to Learn from the Particle in the Box
• The energy of a particle is quantized. This means it can only take on discreet energy values.
• The lowest possible energy for a particle is NOT zero (even at 0 K). This means the particle always has some kinetic energy.
• The square of the wavefunction is related to the probability of finding the particle in a specific position for a given energy level.
• The probability changes with increasing energy of the particle and depends on the position in the box you are attempting to define the energy for
• In classical physics, the probability of finding the particle is independent of the energy and the same at all points in the box
Helpful Links
• Provides a live quantum mechanical simulation of the particle in a box model and allows you to visualize the solutions to the Schrödinger Equation: www.falstad.com/qm1d/
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Tunneling is a quantum mechanical phenomenon when a particle is able to penetrate through a potential energy barrier that is higher in energy than the particle’s kinetic energy. This amazing property of microscopic particles play important roles in explaining several physical phenomena including radioactive decay. Additionally, the principle of tunneling leads to the development of Scanning Tunneling Microscope (STM) which had a profound impact on chemical, biological and material science research.
Violating Classical Mechanics
Consider a ball rolling from one valley to another over a hill (Figure $1$). If the ball has enough energy ($E$) to overcome the potential energy ($V$) at the top of the barrier between each valley, then it can roll from one valley to the other. This is the classical picture and is controlled by the simple Law of Conservation of Energy approach taught in beginning physics courses. However, If the ball does not have enough kinetic energy ($E<V$), to overcome the barrier it will never roll from one valley to the other. In contrast, when quantum effects are taken into effect, the ball can "tunnel" through the barrier to the other valley, even if its kinetic energy is less than the potential energy of the barrier to the top of one of the hills.
The Heisenberg Uncertainty Principle in Action
The reason for the difference between classical and quantum motion comes from wave-particle nature of matter. One interpretation of this duality involves the Heisenberg uncertainty principle, which defines a limit on how precisely the position and the momentum of a particle can be known at the same time. This implies that there are no solutions with a probability of exactly zero (or one), though a solution may approach infinity if, for example, the calculation for its position was taken as a probability of 1, the other, i.e. its speed, would have to be infinity. Hence, the probability of a given particle's existence on the opposite side of an intervening barrier is non-zero, and such particles will appear on the 'other' (a semantically difficult word in this instance) side with a relative frequency proportional to this probability.
Microscopic particles such as protons, or electrons would behave differently as a consequence of wave-particle duality. Consider a particle with energy $E$ that is confined in a box which has a barrier of height $V$. Classically, the box will prevent these particles from escaping due to the insufficiency in kinetic energy of these particles to get over the barrier. However, if the thickness of the barrier is thin, the particles have some probability of penetrating through the barrier without sufficient energy and appear on the other side of the box (Figure $2$).
When it reaches a barrier it cannot overcome, a particle's wave function changes from sinusoidal to exponentially diminishing in form. The solution for the Schrödinger equation in such a medium (Figure $2$; blue region) is:
$\psi = N e^{-\beta x}$
where
• $N$ is a normalization constant and
• $\beta= \sqrt{\dfrac{2m(V-E)}{\hbar^2}}$
For a quantum particle to appreciably tunnel through a barrier three conditions must be met (Figure $2$):
1. The height of the barrier must be finite and the thickness of the barrier should be thin.
2. The potential energy of the barrier exceeds the kinetic energy of the particle ($E<V$).
3. The particle has wave properties because the wavefunction is able to penetrate through the barrier. This suggests that quantum tunneling only apply to microscopic objects such protons or electrons and does not apply to macroscopic objects.
If these conditions are met, there would be some probability of finding the particles on the other side of the barrier. Beginning as a sinusoidal wave, a particle begins tunneling through the barrier and goes into exponential decay until it exits the barrier and gets transmitted out the other side as a final sinusoidal wave with a smaller amplitude. The act of tunneling decreases the wave amplitude due the reflection of the incident wave when it comes into the contact with the barrier but does not affect the wave equation.
The probability, $P$, of a particle tunneling through the potential energy barrier is derived from the Schrödinger Equation and is described as,
$P = \exp\left(\dfrac{-4a\pi}{h} \sqrt{2m(V-E)} \right) \label{prob}$
with $E<V$
where
• $V$ is the potential barrier,
• $E$ is the kinetic energy possessed by the particle, and
• $a$ is the thickness of the barrier.
• $m$ is mass of the particle
• $h$ is Planks Constant ($6.6260 \times 10^{-34} m^2 kg / s$)
Therefore, the probability of an object tunneling through a barrier decreases with the object's increasing mass and with the increasing gap between the energy of the object and the energy of the barrier. And although the wave function never quite reaches 0 (as can be determined from the functionality), this explains how tunneling is frequent on nanoscale, but negligible at the macroscopic level. Equation $\ref{prob}$ argues that the probability of tunneling decreases exponentially with the square root of the particles mass (particles with a small mass can effectively tunnel through barriers more easily than those with larger mass) and the thickness of the barrier ($a$)
Example $1$
An electron having total kinetic energy $E$ of 4.50 eV approaches a rectangular energy barrier with $V= 5.00\, eV$ and $L= 950\, pm$. Classically, the electron cannot pass through the barrier because $E<V$. Calculate probability of tunneling of this electron through the barrier.
Solution
This is a straightforward application of Equation $\ref{prob}$.
$P = \exp\left(\dfrac{-4a\pi}{h} \sqrt{2m(V-E)} \right) \nonumber$
The electronvolt (eV) is a unit of energy that is equal to approximately $1.6 \times 10^{−19}\; J$, which is the conversion used below.
The mass of an electron is $9.10 \times 10^{-31}\; kg$
\begin{align*} P &= \exp \left[\left(- \dfrac{ (4) (950 \times 10^{-12}\, m) (\pi)}{ 6.6260 \times 10^{-34} m^2 kg / s} \right) \sqrt{(2)(9.10 \times 10^{-31}\; kg) (5.00 - 4.50 \; \cancel{eV})( 1.60 \times 10^{-19} J/\cancel{eV}) } \right] \[4pt] &= \exp^{-6.88} = 1.03 \times 10^{-3} \end{align*}
There is a ~0.1% probability of the electrons tunneling though the barrier.
Alpha Decay Radioactivity
Protons and neutrons in a nucleus have kinetic energy, but it is about 8 MeV less than that needed to get out from attractive nuclear potential (Figure $4$). Hence, they are bound by an average of 8 MeV per nucleon. The slope of the hill outside the bowl is analogous to the repulsive Coulomb potential for a nucleus, such as for an α particle outside a positive nucleus. In $\alpha$ decay, two protons and two neutrons spontaneously break away as a 4He unit. Yet the protons and neutrons do not have enough kinetic energy to classically get over the rim.
The $\alpha$ article tunnels through a region of space it is forbidden to be in, and it comes out of the side of the nucleus. Like an electron making a transition between orbits around an atom, it travels from one point to another without ever having been in between (Figure $5$). The wave function of a quantum mechanical particle varies smoothly, going from within an atomic nucleus (on one side of a potential energy barrier) to outside the nucleus (on the other side of the potential energy barrier). Inside the barrier, the wave function does not become zero but decreases exponentially, and we do not observe the particle inside the barrier. The probability of finding a particle is related to the square of its wave function, and so there is a small probability of finding the particle outside the barrier, which implies that the particle can tunnel through the barrier.
Scanning Tunneling Microscopy (STM)
A metal tip usually made out of tungsten is placed between a very small distance above a conducting or semiconducting surface. This distance acts as a potential barrier for tunneling. The space between the tip and the surface normally is vacuum. When electrons tunnel from the metal tip to the surface, a current is created and monitored by a computer (Figure $6$). The current depends on the distance between the tip and the surface, which is controlled by a piezoelectric cylinder. If there is a strong current, the tip will move away from the surface. The increase of the potential barrier will decrease the probability of tunneling and decrease the current. If the current becomes too weak, the tip moves closer to the surface. The potential barrier will be reduced and the current will increase. The variations in the current as the tip moves over the sample are reconstructed by the computer to produce topological image of the scanned surface.
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The hydrogen atom, consisting of an electron and a proton, is a two-particle system, and the internal motion of two particles around their center of mass is equivalent to the motion of a single particle with a reduced mass. This reduced particle is located at $r$, where $r$ is the vector specifying the position of the electron relative to the position of the proton. The length of $r$ is the distance between the proton and the electron, and the direction of $r$ and the direction of $r$ is given by the orientation of the vector pointing from the proton to the electron. Since the proton is much more massive than the electron, we will assume throughout this chapter that the reduced mass equals the electron mass and the proton is located at the center of mass.
Since the internal motion of any two-particle system can be represented by the motion of a single particle with a reduced mass, the description of the hydrogen atom has much in common with the description of a diatomic molecule discussed previously. The Schrödinger Equation for the hydrogen atom
$\hat {H} (r , \theta , \varphi ) \psi (r , \theta , \varphi ) = E \psi ( r , \theta , \varphi) \label {6.1.1}$
employs the same kinetic energy operator, $\hat {T}$, written in spherical coordinates. For the hydrogen atom, however, the distance, r, between the two particles can vary, unlike the diatomic molecule where the bond length was fixed, and the rigid rotor model was used. The hydrogen atom Hamiltonian also contains a potential energy term, $\hat {V}$, to describe the attraction between the proton and the electron. This term is the Coulomb potential energy,
$\hat {V} (r) = - \dfrac {e^2}{4 \pi \epsilon _0 r } \label {6.1.2}$
where $r$ is the distance between the electron and the proton. The Coulomb potential energy depends inversely on the distance between the electron and the nucleus and does not depend on any angles. Such a potential is called a central potential.
It is convenient to switch from Cartesian coordinates $x, y, z$ to spherical coordinates in terms of a radius $r$, as well as angles $\phi$, which is measured from the positive x axis in the xy plane and may be between 0 and $2\pi$, and $\theta$, which is measured from the positive z axis towards the xy plane and may be between 0 and $\pi$.
The time-indepdent Schrödinger equation (in spherical coordinates) for a electron around a positively charged nucleus is then
$\left \{ -\dfrac {\hbar ^2}{2 \mu r^2} \left [ \dfrac {\partial}{\partial r} \left (r^2 \dfrac {\partial}{\partial r} \right ) + \dfrac {1}{\sin \theta } \dfrac {\partial}{\partial \theta } \left ( \sin \theta \dfrac {\partial}{\partial \theta} \right ) + \dfrac {1}{\sin ^2 \theta} \dfrac {\partial ^2}{\partial \varphi ^2} \right ] - \dfrac {e^2}{4 \pi \epsilon _0 r } \right \} \psi (r , \theta , \varphi ) = E \psi (r , \theta , \varphi ) \label {6.1.4}$
Since the angular momentum operator does not involve the radial variable, $r$, we can separate variables in Equation $\ref{6.1.4}$ by using a product wavefunction. We know that the eigenfunctions of the angular momentum operator are the Spherical Harmonic functions (Table M4), $Y (\theta , \varphi )$, so a good choice for a product function is
$\color{red} \psi (r , \theta , \varphi ) = R (r) Y (\theta , \varphi ) \label {6.1.8}$
The Spherical Harmonic $Y (\theta , \varphi )$ functions provide information about where the electron is around the proton, and the radial function $R(r)$ describes how far the electron is away from the proton. A solution for both $R(r)$ and $Y (\theta , \varphi )$ with $E_n$ that depends on only one quantum number $n$, although others are required for the proper description of the wavefunction:
$\color{red} E_n = -\dfrac {m_e e^4}{8\epsilon_0^2 h^2 n^2}\label{6}$
with $n=1,2,3 ...\infty$
The hydrogen atom wavefunctions, $\psi (r, \theta , \phi )$, are called atomic orbitals. An atomic orbital is a function that describes one electron in an atom. The wavefunction with n = 1, $l$ $l$ = 0 is called the 1s orbital, and an electron that is described by this function is said to be “in” the ls orbital, i.e. have a 1s orbital state. The constraints on n, $l$ $l)$, and $m_l$ that are imposed during the solution of the hydrogen atom Schrödinger equation explain why there is a single 1s orbital, why there are three 2p orbitals, five 3d orbitals, etc. We will see when we consider multi-electron atoms, these constraints explain the features of the Periodic Table. In other words, the Periodic Table is a manifestation of the Schrödinger model and the physical constraints imposed to obtain the solutions to the Schrödinger equation for the hydrogen atom.
The Three Quantum Numbers
Schrödinger’s approach requires three quantum numbers ($n$, $l$, and $m_l$) to specify a wavefunction for the electron. The quantum numbers provide information about the spatial distribution of an electron. Although $n$ can be any positive integer (NOT zero), only certain values of $l$ and $m_l$ are allowed for a given value of $n). The principal quantum number (n): One of three quantum numbers that tells the average relative distance of an electron from the nucleus. indicates the energy of the electron and the average distance of an electron from the nucleus $n = 1,\;2,\;3,\;4,\;.\;.\;.\; \label{2.2.2}$ As\($n increases for a given atom, so does the average distance of an electron from the nucleus. A negatively charged electron that is, on average, closer to the positively charged nucleus is attracted to the nucleus more strongly than an electron that is farther out in space. This means that electrons with higher values of n are easier to remove from an atom. All wave functions that have the same value of n are said to constitute a principal shell. All the wave functions that have the same value of n because those electrons have similar average distances from the nucleus. because those electrons have similar average distances from the nucleus. As you will see, the principal quantum number n corresponds to the n used by Bohr to describe electron orbits and by Rydberg to describe atomic energy levels.
The Azimuthal Quantum Number: The second quantum number is often called the azimuthal quantum number (l). One of three quantum numbers that describes the shape of the region of space occupied by an electron.. The value of l describes the shape of the region of space occupied by the electron. The allowed values of l depend on the value of n and can range from 0 to n − 1:
$l = 0,\;1,.,2,\;3,\;.\;.\;\left (n-1 \right ) \label{2.2.3}$
For example, if n = 1, l can be only 0; if n = 2, l can be 0 or 1; and so forth. For a given atom, all wave functions that have the same values of both n and l form a subshell. A group of wave functions that have the same values of n and l. The regions of space occupied by electrons in the same subshell usually have the same shape, but they are oriented differently in space.
The Magnetic Quantum Number: The third quantum number is the magnetic quantum number ($m_l$). One of three quantum numbers that describes the orientation of the region of space occupied by an electron with respect to an applied magnetic field.. The value of $m_l$ describes the orientation of the region in space occupied by an electron with respect to an applied magnetic field. The allowed values of $m_l$ depend on the value of l: $m_l$ can range from −l to l in integral steps:
$m = -l,\;-l+1,\,.\;.\;.0,\;\,.\;.\;.l-1,l \label{2.2.4}$
For example, if $l = 0$, ml can be only 0; if l = 1, $m_l$ can be −1, 0, or +1; and if l = 2, $m_l$ can be −2, −1, 0, +1, or +2.
Each wave function with an allowed combination of n, l, and ml values describes an atomic orbital A wave function with an allowed combination of n, l and ml quantum numbers., a particular spatial distribution for an electron. For a given set of quantum numbers, each principal shell has a fixed number of subshells, and each subshell has a fixed number of orbitals.
Example $1$: Number of subshells
How many subshells and orbitals are contained within the principal shell with n = 4?
Given: value of n
Asked for: number of subshells and orbitals in the principal shell
Strategy:
A Given n = 4, calculate the allowed values of l. From these allowed values, count the number of subshells.
B For each allowed value of l, calculate the allowed values of ml. The sum of the number of orbitals in each subshell is the number of orbitals in the principal shell.
Solution
A We know that l can have all integral values from 0 to n − 1. If n = 4, then l can equal 0, 1, 2, or 3. Because the shell has four values of l, it has four subshells, each of which will contain a different number of orbitals, depending on the allowed values of ml.
B For l = 0, ml can be only 0, and thus the l = 0 subshell has only one orbital. For l = 1, ml can be 0 or ±1; thus the l = 1 subshell has three orbitals. For l = 2, ml can be 0, ±1, or ±2, so there are five orbitals in the l = 2 subshell. The last allowed value of l is l = 3, for which ml can be 0, ±1, ±2, or ±3, resulting in seven orbitals in the l = 3 subshell. The total number of orbitals in the n = 4 principal shell is the sum of the number of orbitals in each subshell and is equal to n2:
$\mathop 1\limits_{(l = 0)} + \mathop 3\limits_{(l = 1)} + \mathop 5\limits_{(l = 2)} + \mathop 7\limits_{(l = 3)} = 16\; {\rm{orbitals}} = {(4\; {\rm{principal\: shells}})^2}$
Exercise $1$
How many subshells and orbitals are in the principal shell with n = 3?
Answer: three subshells; nine orbitals
Rather than specifying all the values of n and l every time we refer to a subshell or an orbital, chemists use an abbreviated system with lowercase letters to denote the value of l for a particular subshell or orbital:
l = 0 1 2 3
Designation s p d f
The principal quantum number is named first, followed by the letter s, p, d, or f as appropriate. These orbital designations are derived from corresponding spectroscopic characteristics of lines involving them: sharp, principle, diffuse, and fundamental. A 1s orbital has n = 1 and l = 0; a 2p subshell has n = 2 and l = 1 (and has three 2p orbitals, corresponding to ml = −1, 0, and +1); a 3d subshell has n = 3 and l = 2 (and has five 3d orbitals, corresponding to ml = −2, −1, 0, +1, and +2); and so forth.
We can summarize the relationships between the quantum numbers and the number of subshells and orbitals as follows (Table $1$):
• Each principal shell has n subshells. For n = 1, only a single subshell is possible (1s); for n = 2, there are two subshells (2s and 2p); for n = 3, there are three subshells (3s, 3p, and 3d); and so forth. Every shell has an ns subshell, any shell with n ≥ 2 also has an np subshell, and any shell with n ≥ 3 also has an nd subshell. Because a 2d subshell would require both n = 2 and l = 2, which is not an allowed value of l for n = 2, a 2d subshell does not exist.
• Each subshell has 2l + 1 orbitals. This means that all ns subshells contain a single s orbital, all np subshells contain three p orbitals, all nd subshells contain five d orbitals, and all nf subshells contain seven f orbitals.
Note
Each principal shell has n subshells, and each subshell has 2l + 1 orbitals.
Table $1$: Allowed values of $n$, $l$, and $m_l$ through n = 4
n l Subshell Designation ml Number of Orbitals in Subshell Number of Orbitals in Shell
1 0 1s 0 1 1
2 0 2s 0 1 4
1 2p −1, 0, 1 3
3 0 3s 0 1 9
1 3p −1, 0, 1 3
2 3d −2, −1, 0, 1, 2 5
4 0 4s 0 1 16
1 4p −1, 0, 1 3
2 4d −2, −1, 0, 1, 2 5
3 4f −3, −2, −1, 0, 1, 2, 3 7
The Radial Component
The first six radial functions are provided in Table $2$. Note that the functions in the table exhibit a dependence on $Z$, the atomic number of the nucleus. Other one electron systems have electronic states analogous to those for the hydrogen atom, and inclusion of the charge on the nucleus allows the same wavefunctions to be used for all one-electron systems. For hydrogen, $Z = 1$ and for helium, $Z=2$.
Table $2$: Hydrogen-like atomic wavefunctions for $n$ values $1,2,3$: $Z$ is the atomic number of the nucleus, and $\rho = \dfrac {Zr}{a_0}$, where $a_0$ is the Bohr radius and r is the radial variable.
$n$
$\ell$
$m$ Radial Component
$n=1$ $\ell=0$ $m=0$ $\psi_{100} = \dfrac {1}{\sqrt {\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} e^{-\rho}$
$n=2$ $\ell=0$ $m=0$ $\psi_{200} = \dfrac {1}{\sqrt {32\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} (2-\rho)e^{\dfrac {-\rho}{2}}$
$\ell=1$ $m=0$ $\psi_{210} = \dfrac {1}{\sqrt {32\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} \rho e^{-\rho/2} \cos(\theta)$
$\ell=1$ $m=\pm 1$ $\psi_{21\pm\ 1} = \dfrac {1}{\sqrt {64\pi}} \left(\dfrac {Z}{a_0}\right)^{\dfrac {3}{2}} \rho e^{-\rho/2} \sin(\theta) e^{\pm\ i\phi}$
$n=3$ $\ell=0$ $m=0$ $\psi_{300} = \dfrac {1}{81\sqrt {3\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} (27-18\rho +2\rho^2)e^{-\rho/3}$
$\ell=1$ $m=0$ $\psi_{310} = \dfrac{1}{81} \sqrt {\dfrac {2}{\pi}} \left(\dfrac {Z}{a_0}\right)^{\dfrac {3}{2}} (6r - \rho^2)e^{-\rho/3} \cos(\theta)$
$\ell=1$ $m=\pm 1$ $\psi_{31\pm\ 1} = \dfrac {1}{81\sqrt {\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} (6\rho - \rho^2)e^{-r/3} \sin(\theta)e^{\pm\ i \phi}$
$\ell=2$ $m=0$ $\psi_{320} = \dfrac {1}{81\sqrt {6\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} \rho^2 e^{-\rho/3}(3cos^2(\theta) -1)$
$\ell=2$ $m=\pm 1$ $\psi_{32\pm\ 1} = \dfrac {1}{81\sqrt {\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} \rho^2 e^{-\rho/3} \sin(\theta)\cos(\theta)e^{\pm\ i \phi}$
$\ell=2$ $m=\pm 2$
$\psi_{32\pm\ 2} = \dfrac {1}{162\sqrt {\pi}} \left(\dfrac {Z}{a_0}\right)^{\frac {3}{2}} \rho^2 e^{-\rho/3}{\sin}^2(\theta)e^{\pm\ 2i\phi}$
Visualizing the variation of an electronic wavefunction with r,$\theta$, and $\phi$ is important because the absolute square of the wavefunction depicts the charge distribution (electron probability density) in an atom or molecule. The charge distribution is central to chemistry because it is related to chemical reactivity. For example, an electron deficient part of one molecule is attracted to an electron rich region of another molecule, and such interactions play a major role in chemical interactions ranging from substitution and addition reactions to protein folding and the interaction of substrates with enzymes.
Methods for separately examining the radial portions of atomic orbitals provide useful information about the distribution of charge density within the orbitals. Graphs of the radial functions, R(r), for the 1s, 2s, and 2p orbitals plotted in Figure $2$ left). The quantity $R(r)^* R(r)$ gives the radial probability density; i.e., the probability density for the electron to be at a point located the distance r from the proton. Radial probability densities for three types of atomic orbitals are plotted in Figure $2$ (right).
For the hydrogen atom, the peak in the radial probability plot occurs at r = 0.529 Å (52.9 pm), which is exactly the radius calculated by Bohr for the n = 1 orbit. Thus the most probable radius obtained from quantum mechanics is identical to the radius calculated by classical mechanics. In Bohr’s model, however, the electron was assumed to be at this distance 100% of the time, whereas in the Schrödinger model, it is at this distance only some of the time. The difference between the two models is attributable to the wavelike behavior of the electron and the Heisenberg uncertainty principle.
Figure $3$ compares the electron probability densities for the hydrogen 1s, 2s, and 3s orbitals. Note that all three are spherically symmetrical. For the 2s and 3s orbitals, however (and for all other s orbitals as well), the electron probability density does not fall off smoothly with increasing r. Instead, a series of minima and maxima are observed in the radial probability plots (part (c) in Figure $3$). The minima correspond to spherical nodes (regions of zero electron probability), which alternate with spherical regions of nonzero electron probability.
The Angular Component
The angular component of the wavefunction $Y(\theta,\phi)$ in Equation $\ref{6.1.8}$ does much to give an orbital its distinctive shape. $Y(\theta,\phi)$ is typically normalized so the the integral of $Y^2(\theta,\phi)$ over the unit sphere is equal to one. In this case, $Y^2(\theta,\phi)$ serves as a probability function. The probability function can be interpreted as the probability that the electron will be found on the ray emitting from the origin that is at angles $(\theta,\phi)$ from the axes. The probability function can also be interpreted as the probability distribution of the electron being at position $(\theta,\phi)$ on a sphere of radius r, given that it is r distance from the nucleus. $Y_{l,m_l}(\theta,\phi)$ are also the wavefunction solutions to Schrödinger’s equation for a rigid rotor consisting of rotating bodies, for example a diatomic molecule. These are called Spherical Harmonic functions (Table M4).
s Orbitals (l=0)
Three things happen to s orbitals as n increases (Figure 6.6.2):
1. They become larger, extending farther from the nucleus.
2. They contain more nodes. This is similar to a standing wave that has regions of significant amplitude separated by nodes, points with zero amplitude.
3. For a given atom, the s orbitals also become higher in energy as n increases because of their increased distance from the nucleus.
Orbitals are generally drawn as three-dimensional surfaces that enclose 90% of the electron density. Although such drawings show the relative sizes of the orbitals, they do not normally show the spherical nodes in the 2s and 3s orbitals because the spherical nodes lie inside the 90% surface. Fortunately, the positions of the spherical nodes are not important for chemical bonding.
p Orbitals (l=1)
Only s orbitals are spherically symmetrical. As the value of l increases, the number of orbitals in a given subshell increases, and the shapes of the orbitals become more complex. Because the 2p subshell has l = 1, with three values of ml (−1, 0, and +1), there are three 2p orbitals).
The electron probability distribution for one of the hydrogen 2p orbitals is shown in Figure $4$. Because this orbital has two lobes of electron density arranged along the z axis, with an electron density of zero in the xy plane (i.e., the xy plane is a nodal plane), it is a 2pz orbital. As shown in Figure $5$, the other two 2p orbitals have identical shapes, but they lie along the x axis (2px) and y axis (2py), respectively. Note that each p orbital has just one nodal plane. In each case, the phase of the wave function for each of the 2p orbitals is positive for the lobe that points along the positive axis and negative for the lobe that points along the negative axis. It is important to emphasize that these signs correspond to the phase of the wave that describes the electron motion, not to positive or negative charges.
The surfaces shown enclose 90% of the total electron probability for the 2px, 2py, and 2pz orbitals. Each orbital is oriented along the axis indicated by the subscript and a nodal plane that is perpendicular to that axis bisects each 2p orbital. The phase of the wave function is positive (orange) in the region of space where x, y, or z is positive and negative (blue) where x, y, or z is negative. Just as with the s orbitals, the size and complexity of the p orbitals for any atom increase as the principal quantum number n increases. The shapes of the 90% probability surfaces of the 3p, 4p, and higher-energy p orbitals are, however, essentially the same as those shown in Figure Figure $5$.
d Orbitals (l=2)
Subshells with l = 2 have five d orbitals; the first principal shell to have a d subshell corresponds to n = 3. The five d orbitals have ml values of −2, −1, 0, +1, and +2 (Figure $6$).
The hydrogen 3d orbitals have more complex shapes than the 2p orbitals. All five 3d orbitals contain two nodal surfaces, as compared to one for each p orbital and zero for each s orbital. In three of the d orbitals, the lobes of electron density are oriented between the x and y, x and z, and y and z planes; these orbitals are referred to as the $3d_{xy}$, \)3d_{xz}\_, and \)3d_{yz}\) orbitals, respectively. A fourth d orbital has lobes lying along the x and y axes; this is the 3dx2−y2 orbital. The fifth 3d orbital, called the $3d_{z^2}$ orbital, has a unique shape: it looks like a $2p_z$ orbital combined with an additional doughnut of electron probability lying in the xy plane. Despite its peculiar shape, the $3d_{z^2}$ orbital is mathematically equivalent to the other four and has the same energy. In contrast to p orbitals, the phase of the wave function for d orbitals is the same for opposite pairs of lobes. As shown in Figure $6$, the phase of the wave function is positive for the two lobes of the $dz^2$ orbital that lie along the z axis, whereas the phase of the wave function is negative for the doughnut of electron density in the xy plane. Like the s and p orbitals, as n increases, the size of the d orbitals increases, but the overall shapes remain similar to those depicted in Figure 6.6.5.
f Orbitals (l=3)
Principal shells with n = 4 can have subshells with l = 3 and ml values of −3, −2, −1, 0, +1, +2, and +3. These subshells consist of seven f orbitals. Each f orbital has three nodal surfaces, so their shapes are complex (not shown).
Energies
The constraint that $n$ be greater than or equal to $l +1$ also turns out to quantize the energy, producing the same quantized expression for hydrogen atom energy levels that was obtained from the Bohr model of the hydrogen atom.
$E=−\dfrac{Z^2}{n^2}Rhc \label{6.6.1}$
or
$E_n = - \dfrac {Z^2 \mu e^4}{8 \epsilon ^2_0 h^2 n^2}$
The relative energies of the atomic orbitals with n ≤ 4 for a hydrogen atom are plotted in Figure $7$ ; note that the orbital energies depend on only the principal quantum number n. Consequently, the energies of the 2s and 2p orbitals of hydrogen are the same; the energies of the 3s, 3p, and 3d orbitals are the same; and so forth. The orbital energies obtained for hydrogen using quantum mechanics are exactly the same as the allowed energies calculated by Bohr. In contrast to Bohr’s model, however, which allowed only one orbit for each energy level, quantum mechanics predicts that there are 4 orbitals with different electron density distributions in the n = 2 principal shell (one 2s and three 2p orbitals), 9 in the n = 3 principal shell, and 16 in the n = 4 principal shell.
Note
For a single electron system, the energy of that electron is a function of only the principal quantum number (Equation $\ref{6.6.1}$).
The different values of l and ml for the individual orbitals within a given principal shell are not important for understanding the emission or absorption spectra of the hydrogen atom under most conditions, but they do explain the splittings of the main lines that are observed when hydrogen atoms are placed in a magnetic field. As we have just seen, however, quantum mechanics also predicts that in the hydrogen atom, all orbitals with the same value of n (e.g., the three 2p orbitals) are degenerate, meaning that they have the same energy. Figure $7$ shows that the energy levels become closer and closer together as the value of n increases, as expected because of the 1/n2 dependence of orbital energies.
In general, both energy and radius decrease as the nuclear charge increases. Thus the most stable orbitals (those with the lowest energy) are those closest to the nucleus. For example, in the ground state of the hydrogen atom, the single electron is in the 1s orbital, whereas in the first excited state, the atom has absorbed energy and the electron has been promoted to one of the n = 2 orbitals. In ions with only a single electron, the energy of a given orbital depends on only n, and all subshells within a principal shell, such as the px, py, and pz orbitals, are degenerate.
Note: Bohr vs. Schrödinger
It is interesting to compare the results obtained by solving the Schrödinger equation with Bohr’s model of the hydrogen atom. There are several ways in which the Schrödinger model and Bohr model differ.
1. First, and perhaps most strikingly, the Schrödinger model does not produce well-defined orbits for the electron. The wavefunctions only give us the probability for the electron to be at various directions and distances from the proton.
2. Second, the quantization of angular momentum is different from that proposed by Bohr. Bohr proposed that the angular momentum is quantized in integer units of $\hbar$, while the Schrödinger model leads to an angular momentum of $(l (l +1) \hbar ^2)^{\dfrac {1}{2}}$.
3. Third, the quantum numbers appear naturally during solution of the Schrödinger equation while Bohr had to postulate the existence of quantized energy states. Although more complex, the Schrödinger model leads to a better correspondence between theory and experiment over a range of applications that was not possible for the Bohr model.
Electron Spin: The Fourth Quantum Number
The quantum numbers $n, \ l, \ m$ are not sufficient to fully characterize the physical state of the electrons in an atom. In 1926, Otto Stern and Walther Gerlach carried out an experiment that could not be explained in terms of the three quantum numbers $n, \ l, \ m$ and showed that there is, in fact, another quantum-mechanical degree of freedom that needs to be included in the theory. The experiment is illustrated in the Figure $8$. A beam of atoms (e.g. hydrogen or silver atoms) is sent through a spatially inhomogeneous magnetic field with a definite field gradient toward one of the poles. It is observed that the beam splits into two beams as it passes through the field region.
The fact that the beam splits into 2 beams suggests that the electrons in the atoms have a degree of freedom capable of coupling to the magnetic field. That is, an electron has an intrinsic magnetic moment $M$ arising from a degree of freedom that has no classical analog. The magnetic moment must take on only 2 values according to the Stern-Gerlach experiment. The intrinsic property that gives rise to the magnetic moment must have some analog to a spin, $S$; unlike position and momentum, which have clear classical analogs, spin does not. The implication of the Stern-Gerlach experiment is that we need to include a fourth quantum number, $m_s$ in our description of the physical state of the electron. That is, in addition to give its principle, angular, and magnetic quantum numbers, we also need to say if it is a spin-up electron or a spin-down electron.
Note: Spin is a Quantum Phenomenon
Unlike position and momentum, which have clear classical analogs, spin does not.
George Uhlenbeck (1900–1988) and Samuel Goudsmit (1902–1978), proposed that the splittings were caused by an electron spinning about its axis, much as Earth spins about its axis. When an electrically charged object spins, it produces a magnetic moment parallel to the axis of rotation, making it behave like a magnet. Although the electron cannot be viewed solely as a particle, spinning or otherwise, it is indisputable that it does have a magnetic moment. This magnetic moment is called electron spin.
In an external magnetic field, the electron has two possible orientations (Figure $9$). These are described by a fourth quantum number (ms), which for any electron can have only two possible values, designated +½ (up) and −½ (down) to indicate that the two orientations are opposites; the subscript s is for spin. An electron behaves like a magnet that has one of two possible orientations, aligned either with the magnetic field or against it. The implications of electron spin for chemistry were recognized almost immediately by an Austrian physicist, Wolfgang Pauli (1900–1958; Nobel Prize in Physics, 1945), who determined that each orbital can contain no more than two electrons whoe developed the Pauli exclusion principle.
Note: Pauli Exclusion Principle
No two electrons in an atom can have the same values of all four quantum numbers ($n$, $l$, $m_l$, $m_s$).
By giving the values of n, l, and ml, we also specify a particular orbital (e.g., 1s with n = 1, l = 0, ml = 0). Because ms has only two possible values (+½ or −½), two electrons, and only two electrons, can occupy any given orbital, one with spin up and one with spin down. With this information, we can proceed to construct the entire periodic table, which was originally based on the physical and chemical properties of the known elements.
Example $2$
List all the allowed combinations of the four quantum numbers (n, l, ml, ms) for electrons in a 2p orbital and predict the maximum number of electrons the 2p subshell can accommodate.
Given: orbital
Asked for: allowed quantum numbers and maximum number of electrons in orbital
Strategy:
1. List the quantum numbers (n, l, ml) that correspond to an n = 2p orbital. List all allowed combinations of (n, l, ml).
2. Build on these combinations to list all the allowed combinations of (n, l, ml, ms).
3. Add together the number of combinations to predict the maximum number of electrons the 2p subshell can accommodate.
Solution:
A For a 2p orbital, we know that n = 2, l = n − 1 = 1, and ml = −l, (−l +1),…, (l − 1), l. There are only three possible combinations of (n, l, ml): (2, 1, 1), (2, 1, 0), and (2, 1, −1).
B Because ms is independent of the other quantum numbers and can have values of only +½ and −½, there are six possible combinations of (n, l, ml, ms): (2, 1, 1, +½), (2, 1, 1, −½), (2, 1, 0, +½), (2, 1, 0, −½), (2, 1, −1, +½), and (2, 1, −1, −½).
C Hence the 2p subshell, which consists of three 2p orbitals (2px, 2py, and 2pz), can contain a total of six electrons, two in each orbital.
Exercise $2$
List all the allowed combinations of the four quantum numbers (n, l, ml, ms) for a 6s orbital, and predict the total number of electrons it can contain.
Answer: (6, 0, 0, +½), (6, 0, 0, −½); two electrons
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/11%3A_Quantum_Mechanics_and_Atomic_Structure/11.10%3A_The_Schrodinger_Wave_Equation_for_the_Hydrogen_Atom.txt
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Quantum mechanics can account for the periodic structure of the elements, by any measure a major conceptual accomplishment for any theory. Although accurate computations become increasingly more challenging as the number of electrons increases, the general patterns of atomic behavior can be predicted with remarkable accuracy.
Figure $1$ shows a schematic representation of a helium atom with two electrons whose coordinates are given by the vectors $r_1$ and $r_2$. The electrons are separated by a distance $r_{12} = |r_1-r_2|$. The origin of the coordinate system is fixed at the nucleus. As with the hydrogen atom, the nuclei for multi-electron atoms are so much heavier than an electron that the nucleus is assumed to be the center of mass. Fixing the origin of the coordinate system at the nucleus allows us to exclude translational motion of the center of mass from our quantum mechanical treatment.
The Schrödinger equation operator for the hydrogen atom serves as a reference point for writing the Schrödinger equation for atoms with more than one electron. Start with the same general form we used for the hydrogen atom Hamiltonian
$(\hat {T} + \hat {V}) \psi = E \psi \label {9-1}$
Include a kinetic energy term for each electron and a potential energy term for the attraction of each negatively charged electron for the positively charged nucleus and a potential energy term for the mutual repulsion of each pair of negatively charged electrons. The He atom Schrödinger equation is
$\left( -\dfrac {\hbar ^2}{2m_e} (\nabla ^2_1 + \nabla ^2_2) + V (r_1) + V (r_2) + V (r_{12}) \right) \psi = E \psi \label {9-2}$
where
$V(r_1) = -\dfrac {2e^2}{4 \pi \epsilon _0 r_1} \label {9-3}$
$V(r_2) = -\dfrac {2e^2}{4 \pi \epsilon _0 r_2} \label {9-4}$
$V(r_{12}) = -\dfrac {e^2}{4 \pi \epsilon _0 r_{12}} \label {9-5}$
Equation $\ref{9-2}$ can be extended to any atom or ion by including terms for the additional electrons and replacing the He nuclear charge +2 with a general charge Z; e.g.
$V(r_1) = -\dfrac {Ze^2}{4 \pi \epsilon _0 r_1} \label {9-6}$
Equation $\ref{9-2}$ then becomes
$\left( -\dfrac {\hbar ^2}{2m_e} \sum _i \nabla ^2_i + \sum _i V (r_i) + \sum _{i \ne j} V (r_{ij}) \right) \psi = E \psi \label {9-7}$
Given what we have learned from the previous quantum mechanical systems we’ve studied, we predict that exact solutions to the multi-electron Schrödinger equation would consist of a family of multi-electron wavefunctions, each with an associated energy eigenvalue. These wavefunctions and energies would describe the ground and excited states of the multi-electron atom, just as the hydrogen wavefunctions and their associated energies describe the ground and excited states of the hydrogen atom. We would predict quantum numbers to be involved, as well.
The fact that electrons interact through their electron-electron repulsion means that an exact wavefunction for a multi-electron system would be a single function that depends simultaneously upon the coordinates of all the electrons; i.e., a multi-electron wavefunction:
$\Psi (r_1, r_2, \cdots r_i) \label{8.3.4}$
Unfortunately, the electron-electron repulsion terms make it impossible to find an exact solution to the Schrödinger equation for many-electron atoms. The most basic ansatz to the exact solutions involve writing a multi-electron wavefunction as a simple product of single-electron wavefunctions
$\psi (r_1, r_2, \cdots , r_i) = \varphi _1 (r_1) \varphi _2 (r_2) \cdots \varphi _i(r_i) \label{8.3.5}$
Obtaining the energy of the atom in the state described by that wavefunction as the sum of the energies of the one-electron components.
By writing the multi-electron wavefunction as a product of single-electron functions in Equation $\ref{8.3.5}$, we conceptually transform a multi-electron atom into a collection of individual electrons located in individual orbitals whose spatial characteristics and energies can be separately identified. For atoms these single-electron wavefunctions are called atomic orbitals. For molecules, as we will see in the next chapter, they are called molecular orbitals. While a great deal can be learned from such an analysis, it is important to keep in mind that such a discrete, compartmentalized picture of the electrons is an approximation, albeit a powerful one.
Electron Configurations
The specific arrangement of electrons in orbitals of an atom determines many of the chemical properties of that atom and is formulated via the Aufbau principle, which means "building-up" in German. Aufbau principles determine the order in which atomic orbitals are filled as the atomic number is increased. For the hydrogen atom, the order of increasing orbital energy is given by 1s < 2s = 2p < 3s = 3p = 3d, etc. The dependence of energy on n alone leads to extensive degeneracy, which is however removed for orbitals in many-electron atoms. Thus 2s lies below 2p, as already observed in helium. Similarly, 3s, 3p and 3d increase energy in that order, and so on. The 4s is lowered sufficiently that it becomes comparable to 3d. The general ordering of atomic orbitals is summarized in the following scheme:
$1s < 2s < 2p < 3s < 3p < 4s \sim 3d < 4p < 5s \sim 4d\< 5p < 6s \sim 5d \sim 4f < 6p < 7s \sim 6d \sim 5f \label{4}$
and illustrated in Figure $2$. This provides enough orbitals to fill the ground states of all the atoms in the periodic table. For orbitals designated as comparable in energy, e.g., 4s $\sim$ 3d, the actual order depends which other orbitals are occupied. The energy of atomic orbitals increases as the principal quantum number, $n$, increases. In any atom with two or more electrons, the repulsion between the electrons makes energies of subshells with different values of $l$ differ so that the energy of the orbitals increases within a shell in the order s < p < d < f. Figure $2$ depicts how these two trends in increasing energy relate. The 1s orbital at the bottom of the diagram is the orbital with electrons of lowest energy. The energy increases as we move up to the 2s and then 2p, 3s, and 3p orbitals, showing that the increasing n value has more influence on energy than the increasing l value for small atoms. However, this pattern does not hold for larger atoms. The 3d orbital is higher in energy than the 4s orbital. Such overlaps continue to occur frequently as we move up the chart.
Electrons in successive atoms on the periodic table tend to fill low-energy orbitals first. The arrangement of electrons in the orbitals of an atom is called the electron configuration of the atom. We describe an electron configuration with a symbol that contains three pieces of information ( Figure $3$):
1. The number of the principal quantum shell, n,
2. The letter that designates the orbital type (the subshell, l), and
3. A superscript number that designates the number of electrons in that particular subshell.
For example, the notation 2p4 (read "two–p–four") indicates four electrons in a p subshell (l = 1) with a principal quantum number (n) of 2. The notation 3d8 (read "three–d–eight") indicates eight electrons in the d subshell (i.e., l = 2) of the principal shell for which n = 3.
To determine the electron configuration for any particular atom, we can “build” the structures in the order of atomic numbers. Beginning with hydrogen, and continuing across the periods of the periodic table, we add one proton at a time to the nucleus and one electron to the proper subshell until we have described the electron configurations of all the elements. This procedure is called the Aufbau principle, from the German word Aufbau (“to build up”). Each added electron occupies the subshell of lowest energy available (in the order shown in Figure $4$), subject to the limitations imposed by the allowed quantum numbers according to the Pauli exclusion principle. Electrons enter higher-energy subshells only after lower-energy subshells have been filled to capacity. Figure $3$ illustrates the traditional way to remember the filling order for atomic orbitals.
We will now construct the ground-state electron configuration and orbital diagram for a selection of atoms in the first and second periods of the periodic table. Orbital diagrams are pictorial representations of the electron configuration, showing the individual orbitals and the pairing arrangement of electrons. We start with a single hydrogen atom (atomic number 1), which consists of one proton and one electron. Referring to either Figure $4$, we would expect to find the electron in the 1s orbital. By convention, the $m_s=+\dfrac{1}{2}$ value is usually filled first. The electron configuration and the orbital diagram are:
Following hydrogen is the noble gas helium, which has an atomic number of 2. The helium atom contains two protons and two electrons. The first electron has the same four quantum numbers as the hydrogen atom electron (n = 1, l = 0, ml = 0, $m_s=+\dfrac{1}{2}$). The second electron also goes into the 1s orbital and fills that orbital. The second electron has the same n, l, and ml quantum numbers, but must have the opposite spin quantum number, $m_s=−\dfrac{1}{2}$. This is in accord with the Pauli exclusion principle: No two electrons in the same atom can have the same set of four quantum numbers. For orbital diagrams, this means two arrows go in each box (representing two electrons in each orbital) and the arrows must point in opposite directions (representing paired spins). The electron configuration and orbital diagram of helium are:
The n = 1 shell is completely filled in a helium atom.
The next atom is the alkali metal lithium with an atomic number of 3. The first two electrons in lithium fill the 1s orbital and have the same sets of four quantum numbers as the two electrons in helium. The remaining electron must occupy the orbital of next lowest energy, the 2s orbital (Figure $4$ ). Thus, the electron configuration and orbital diagram of lithium are:
An atom of the alkaline earth metal beryllium, with an atomic number of 4, contains four protons in the nucleus and four electrons surrounding the nucleus. The fourth electron fills the remaining space in the 2s orbital.
An atom of boron (atomic number 5) contains five electrons. The n = 1 shell is filled with two electrons and three electrons will occupy the n = 2 shell. Because any s subshell can contain only two electrons, the fifth electron must occupy the next energy level, which will be a 2p orbital. There are three degenerate 2p orbitals (ml = −1, 0, +1) and the electron can occupy any one of these p orbitals. When drawing orbital diagrams, we include empty boxes to depict any empty orbitals in the same subshell that we are filling.
Carbon (atomic number 6) has six electrons. Four of them fill the 1s and 2s orbitals. The remaining two electrons occupy the 2p subshell. We now have a choice of filling one of the 2p orbitals and pairing the electrons or of leaving the electrons unpaired in two different, but degenerate, p orbitals. The orbitals are filled as described by Hund’s rule: the lowest-energy configuration for an atom with electrons within a set of degenerate orbitals is that having the maximum number of unpaired electrons. Thus, the two electrons in the carbon 2p orbitals have identical n, l, and ms quantum numbers and differ in their ml quantum number (in accord with the Pauli exclusion principle). The electron configuration and orbital diagram for carbon are:
Nitrogen (atomic number 7) fills the 1s and 2s subshells and has one electron in each of the three 2p orbitals, in accordance with Hund’s rule. These three electrons have unpaired spins. Oxygen (atomic number 8) has a pair of electrons in any one of the 2p orbitals (the electrons have opposite spins) and a single electron in each of the other two. Fluorine (atomic number 9) has only one 2p orbital containing an unpaired electron. All of the electrons in the noble gas neon (atomic number 10) are paired, and all of the orbitals in the n = 1 and the n = 2 shells are filled. The electron configurations and orbital diagrams of these four elements are:
Similarly, the abbreviated configuration of lithium can be represented as [He]2s1, where [He] represents the configuration of the helium atom, which is identical to that of the filled inner shell of lithium. Writing the configurations in this way emphasizes the similarity of the configurations of lithium and sodium. Both atoms, which are in the alkali metal family, have only one electron in a valence s subshell outside a filled set of inner shells.
$\ce{Li:[He]}\,2s^1\ \ce{Na:[Ne]}\,3s^1$
The alkaline earth metal magnesium (atomic number 12), with its 12 electrons in a [Ne]3s2 configuration, is analogous to its family member beryllium, [He]2s2. Both atoms have a filled s subshell outside their filled inner shells. Aluminum (atomic number 13), with 13 electrons and the electron configuration [Ne]3s23p1, is analogous to its family member boron, [He]2s22p1.
The electron configurations of silicon (14 electrons), phosphorus (15 electrons), sulfur (16 electrons), chlorine (17 electrons), and argon (18 electrons) are analogous in the electron configurations of their outer shells to their corresponding family members carbon, nitrogen, oxygen, fluorine, and neon, respectively, except that the principal quantum number of the outer shell of the heavier elements has increased by one to n = 3. Figure
Beginning with the transition metal scandium (atomic number 21), additional electrons are added successively to the 3d subshell. This subshell is filled to its capacity with 10 electrons (remember that for l = 2 [d orbitals], there are 2l + 1 = 5 values of ml, meaning that there are five d orbitals that have a combined capacity of 10 electrons). The 4p subshell fills next. Note that for three series of elements, scandium (Sc) through copper (Cu), yttrium (Y) through silver (Ag), and lutetium (Lu) through gold (Au), a total of 10 d electrons are successively added to the (n – 1) shell next to the n shell to bring that (n – 1) shell from 8 to 18 electrons. For two series, lanthanum (La) through lutetium (Lu) and actinium (Ac) through lawrencium (Lr), 14 f electrons (l = 3, 2l + 1 = 7 ml values; thus, seven orbitals with a combined capacity of 14 electrons) are successively added to the (n – 2) shell to bring that shell from 18 electrons to a total of 32 electrons.
Example $1$
Quantum Numbers and Electron Configurations What is the electron configuration and orbital diagram for a phosphorus atom? What are the four quantum numbers for the last electron added?
Solution
The atomic number of phosphorus is 15. Thus, a phosphorus atom contains 15 electrons. The order of filling of the energy levels is 1s, 2s, 2p, 3s, 3p, 4s, . . . The 15 electrons of the phosphorus atom will fill up to the 3p orbital, which will contain three electrons:
The last electron added is a 3p electron. Therefore, n = 3 and, for a p-type orbital, l = 1. The ml value could be –1, 0, or +1. The three p orbitals are degenerate, so any of these ml values is correct. For unpaired electrons, convention assigns the value of $+\dfrac{1}{2}$ for the spin quantum number; thus, $m_s=+\dfrac{1}{2}$.
Exercise $1$
Identify the atoms from the electron configurations given:
1. [Ar]4s23d5
2. [Kr]5s24d105p6
Answer
(a) Mn (b) Xe
Effective Charge, Shielding and Penetration
For an atom or an ion with only a single electron, we can calculate the potential energy by considering only the electrostatic attraction between the positively charged nucleus and the negatively charged electron. When more than one electron is present, however, the total energy of the atom or the ion depends not only on attractive electron-nucleus interactions but also on repulsive electron-electron interactions. When there are two electrons, the repulsive interactions depend on the positions of both electrons at a given instant, but because we cannot specify the exact positions of the electrons, it is impossible to exactly calculate the repulsive interactions. Consequently, we must use approximate methods to deal with the effect of electron-electron repulsions on orbital energies.
If an electron is far from the nucleus (i.e., if the distance r between the nucleus and the electron is large), then at any given moment, most of the other electrons will be between that electron and the nucleus. Hence the electrons will cancel a portion of the positive charge of the nucleus and thereby decrease the attractive interaction between it and the electron farther away. As a result, the electron farther away experiences an effective nuclear charge (Zeff) that is less than the actual nuclear charge Z (Figure $6$). This effect is called electron shielding.
As the distance between an electron and the nucleus approaches infinity, Zeff approaches a value of 1 because all the other (Z − 1) electrons in the neutral atom are, on the average, between it and the nucleus. If, on the other hand, an electron is very close to the nucleus, then at any given moment most of the other electrons are farther from the nucleus and do not shield the nuclear charge. At r ≈ 0, the positive charge experienced by an electron is approximately the full nuclear charge, or ZeffZ. At intermediate values of r, the effective nuclear charge is somewhere between 1 and Z: 1 ≤ ZeffZ. Thus the actual Zeff experienced by an electron in a given orbital depends not only on the spatial distribution of the electron in that orbital but also on the distribution of all the other electrons present. This leads to large differences in Zeff for different elements, as shown in Figure 2.5.1 for the elements of the first three rows of the periodic table. Notice that only for hydrogen does Zeff = Z, and only for helium are Zeff and Z comparable in magnitude.
Because of the effects of shielding and the different radial distributions of orbitals with the same value of n but different values of l, the different subshells are not degenerate in a multielectron atom. For a given value of n, the ns orbital is always lower in energy than the np orbitals, which are lower in energy than the nd orbitals, and so forth. As a result, some subshells with higher principal quantum numbers are actually lower in energy than subshells with a lower value of n; for example, the 4s orbital is lower in energy than the 3d orbitals for most atoms.
Note
Except for the single electron containing hydrogen atom, in every other element $Z_{eff}$ is always less than $Z$.
Ionization Energy
Ionization energy is the energy required to remove an electron from a neutral atom in its gaseous phase. Conceptually, ionization energy is the opposite of electronegativity. The lower this energy is, the more readily the atom becomes a cation. Therefore, the higher this energy is, the more unlikely it is the atom becomes a cation. Generally, elements on the right side of the periodic table have a higher ionization energy because their valence shell is nearly filled. Elements on the left side of the periodic table have low ionization energies because of their willingness to lose electrons and become cations. Thus, ionization energy increases from left to right on the periodic table.
Another factor that affects ionization energy is electron shielding. Electron shielding describes the ability of an atom's inner electrons to shield its positively-charged nucleus from its valence electrons. When moving to the right of a period, the number of electrons increases and the strength of shielding increases. As a result, it is easier for valence shell electrons to ionize, and thus the ionization energy decreases down a group. Electron shielding is also known as screening.
Note
• The ionization energy of the elements within a period generally increases from left to right. This is due to valence shell stability.
• The ionization energy of the elements within a group generally decreases from top to bottom. This is due to electron shielding.
• The noble gases possess very high ionization energies because of their full valence shells as indicated in the graph. Note that helium has the highest ionization energy of all the elements.
Some elements have several ionization energies; these varying energies are referred to as the first ionization energy, the second ionization energy, third ionization energy, etc. The first ionization energy is the energy requiredto remove the outermost, or highest, energy electron, the second ionization energy is the energy required to remove any subsequent high-energy electron from a gaseous cation, etc. Below are the chemical equations describing the first and second ionization energies:
First Ionization Energy:
$X_{(g)} \rightarrow X^+_{(g)} + e^-$
Second Ionization Energy:
$X^+_{(g)} \rightarrow X^{2+}_{(g)} + e^-$
Generally, any subsequent ionization energies (2nd, 3rd, etc.) follow the same periodic trend as the first ionization energy.
Ionization energies decrease as atomic radii increase. This observation is affected by $n$ (the principal quantum number) and $Z_{eff}$ (based on the atomic number and shows how many protons are seen in the atom) on the ionization energy (I). The relationship is given by the following equation:
$I = \dfrac{R_H Z^2_{eff}}{n^2}$
• Across a period, $Z_{eff}$ increases and n (principal quantum number) remains the same, so the ionization energy increases.
• Down a group, $n$ increases and $Z_{eff}$ increases slightly; the ionization energy decreases.
The periodic structure of the elements is evident for many physical and chemical properties, including chemical valence, atomic radius, electronegativity, melting point, density, and hardness. The classic prototype for periodic behavior is the variation of the first ionization energy with atomic number, which is plotted in Figure $8$.
Electron Affinity
The electron affinity (EA) of an element E is defined as minus the internal energy change associated with the gain of an electron by a gaseous atom, at 0 K :
$E_{(g)} + e^- → E^-_{(g)}$
Unlike ionization energies, which are always positive for a neutral atom because energy is required to remove an electron, electron affinities can be positive (energy is released when an electron is added), negative (energy must be added to the system to produce an anion), or zero (the process is energetically neutral).
Figure $9$: Periodic trends in electron affinities.
The periodic trends of electron affinity (Figure $9$) shows that chlorine has the most positive electron affinity of any element, which means that more energy is released when an electron is added to a gaseous chlorine atom than to an atom of any other element, EA= 348.6 kJmol-1 and the Group 17 elements have the largest values overall. The addition of a second electron to an element is expected to be much less favored since there will be repulsion between the negatively charged electron and the overall negatively charged anion. For example, for O the values are:
$O_{(g)} + e^ \rightarrow O^-_{(g)} \;\;\;\; EA = +141\; kJmol^{-1}$
$O^-_{(g)} + e^- → O^{2-}){(g)} \;\;\;\; EA = -798\; kJmol^{-1}$
David M. Hanson, Erica Harvey, Robert Sweeney, Theresa Julia Zielinski ("Quantum States of Atoms and Molecules")
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/11%3A_Quantum_Mechanics_and_Atomic_Structure/11.11%3A_Many-Electron_Atoms_and_the_Periodic_Table.txt
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Q11.1
What is the wavelength associated with a photon of a light with the energy is $3.6 \times 10^{-19}\;J$?
S11.1
$E= hv = \dfrac{hc}{λ} \nonumber$
$3.6 \times 10^{-19}\; J = \dfrac{(6.626 \times 10^{-34}\; J\;s) ( 3 \times 10^8\; ms^{-1})}{\lambda}\nonumber$
$λ = 0.55 \times 10^{-7}\; m = 550\; nm\nonumber$
Q11.2
Calculate the energy of a photon of a light with the frequency is $6.5 \times 10^{-14}\; s^{-1}$ ?
S11.2
$E= hv\nonumber$
$E=( 6.626 \times 10^{-34}J\;s) ( 6.5 \times 10^{-14} s^{-1})\nonumber$
$E= 4.3 \times 10^{-19}\;J\nonumber$
Q11.3a
What are the wavelength of the light emitted by atomic hydrogen when the electron move from $n=5$ to $n=3$ orbit in the Bohr model?
S11.3a
Bohr model:
$\dfrac{1}{λ} = R_H \left(\dfrac{1}{n_a^2} -\dfrac{1}{n_b^2} \right)\nonumber$
$\dfrac{1}{λ} = ( 1.09678 \times 10^{-2} nm^{-1} ) \left(\dfrac{1}{9}- \dfrac{1}{25}\right) \nonumber$
$λ = 1,282\; nm\nonumber$
Q11.3b
Calculate the different energy between $n=3$ and $n=4$ of atomic hydrogen based on Bohr orbits
S11.3b
$\Delta E=-hcR_H \left(\dfrac{1}{3^2}\ - \dfrac{1}{4^2} \right) \nonumber$
$\Delta E=-6.626 \times 10^{-34} J.s)(2.9979 \times 10^8\; m/s)(109,737\; cm^{-1}) \left(\dfrac{1}{3^2} - \dfrac{1}{4^2}\right))\nonumber$
$\Delta E= 1.8 \times 10^{-21}\; J\nonumber$
Q11.4
Calculate the wavenumber of the wavelength of the light emitted from the $n=8$ to $n=6$ transition.
S11.4
This is a simple application of the Rydberg equation
$\tilde{\nu} R_H \left (\dfrac{1}{n_{f}^{2}} -\dfrac{1}{n_{i}^{2}} \right)\nonumber$
with $n_i = 6$ and $n_f = 8$.
$\tilde{\nu}=109,737cm^{-1}\left (\dfrac{1}{6^{2}}-\dfrac{1}{8^{2}} \right)=1333.61 \; cm^{-1}\nonumber$
Q11.5
Calculate the wavelength associated with a 42 g baseball with speed of 80 m/s.
S11.5
$\lambda =\dfrac{h}{p} = \dfrac{h}{mv}\nonumber$
$\lambda = \dfrac{6.626 \times 10^{34} J s}{(0.042 \; kg)( 80 m/s)} = 1.97 \times 10^{-34} \, m\nonumber$
This is a very small wavelength as expected since a ball behaves rather classically (i.e., non-quantum).
Q11.6
Calculate the energy of a 530-nm photon.
S11.6
$p =\dfrac{E}{c}\nonumber$
$\lambda = \dfrac{h}{p}\nonumber$
$\lambda = \dfrac{hc}{E}\nonumber$
$E= \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 1-^{-34})( 3 \times 10^{8})}{530 \times 10^{-9}} = 3.75 \times 10^{-9} \; J\nonumber$
11.3: The Photoelectric Effect
Q11.7
Describe Planck's three experimental observations that explain the photoelectric effect.
S11.7
"When light of a certain frequency shines on a clean metal surface, electrons are ejectedd from the metal. Experimentally, it is found that:
1. The number of electrons ejected is proportional to the intensity of light
2. The kinetic energy of the ejected electrons is proportional to the frequency of the ejected light
3. No electrons can be ejected if the frequency of the light is lower than a certain value, called the threshold frequency ($nu_o$).
Q11.8a
What is the difference between classical and quantum mechanics? Provide the equation relating the energy of emitted radiation to frequency.
S11.8a
Classical mechanics predicts that the radiant energy produced by oscillating objects is continuous. Quantum mechanics predicts that their energy can be viewed as existing in discrete levels. The equation is $E = h\nu\nonumber$. Energy is emitted in discrete multiples of $h\nu$, where h = Planck's constant.
Q11.8b
Do you expect the first ionization of an atom to be an endothermic process or an exothermic process?
S11.8b
It will be endothermic because energy is needed to ionize the atom.
Q11.9a
Calculate the wavelength of light with an energy of $5.22 \times 10^{-19}\; J$.
S11.9a
Rearrange the equation $E = \dfrac{hc}{\lambda }\nonumber$ so that the unknown value is the wavelength of light $\lambda = \dfrac{hc}{E}\nonumber$ $\lambda =\dfrac{(6.626\times 10^{-34}Js)(3.0\times 10^{8}m/s)}{5.22\times 10^{-19}J} = 3.81\times 10^{-7}m = 381 nm\nonumber$
Q11.9b
You realize after class that you forgot to write down the final wavelength of your proton. You do, however, remember that the potential difference was 114 V. From this information, figure out the final wavelength:
S11.9b
$\frac{1}{2}mv^{2}=eV$
$e=1.6x10^{-19} C$ This is the charge of the electron
m=9.11x10-31kg
$v=\sqrt{\frac{2eV}{m}}$
$\sqrt{\frac{2(1.6\cdot 10^{-19})(114)}{1.67\cdot 10^{-31}}}$
=2.0x109m/s
$\lambda =\frac{h}{mv}$
$\lambda =\frac{6.626\cdot 10^{-34}}{(9.11\cdot 10^{-31})(2\cdot 10^{9})}$
$\lambda =3.64\cdot 10^{-13}m$
Q11.10a
given the uncertainty for position of an electron circling an atom is 0.5A0. Find uncertainty in its velocity?
S11.10a
Δp ≥ h/ 4πΔx = 6.626 x 10-34 Js/ 4π(0.5 x 10-10m) = 1.311.05 x 10-24kg m s-1
Δ v= Δp/ m ≥ 1.3 x 10-24 kg m s-1/ 9.109 x 10-31 kg = 1 x 106 m s-1
Q11.10
If the uncertainty of measuring the position of an electron is 2.0 Å, what is the uncertainty of simultaneously measuring its velocity? Hint: What formula deals with uncertainty of measurements?
11.7: The Schrödinger Wave Equation
Q11.11
A typical mass for a horse is 510 kg, and a typical galloping speed is 22 kilometers per hour. Use these values to answer the following questions.
1. What is the momentum of a galloping horse? What is its wavelength?
2. If a galloping horse's velocity and position are simultaneously measured, and the velocity is measured to within ± 1.0%, what is the uncertainty of its position?
3. Suppose Planck's constant was actually 0.01 J s. How would that change your answers to (a) and (b)? Which values would be unchanged?
Hints:
• de Broglie's postulate deals with the wave-like properties of particles.
• Heisenberg's uncertainty principle deals with uncertainty of simultaneous measurements.
S11.11
(a)
1. Use the relationship between velocity and momentum to find the momentum.
$p = m v \nonumber$
$p = 510\ kg \times 22\ \dfrac{km}{hr} \times \dfrac{1000\ m}{km} \times \dfrac{hr}{3600\ s} \nonumber$
$p = 3.1 \times 10^3\ \dfrac{kg\ m}{s} \nonumber$
2. Find the de Broglie wavelength.
$\lambda = \dfrac{h}{p} \nonumber$
$\lambda = \dfrac{6.626 \times 10^{-34}\ J\ s}{3.1 \times 10^3\ \dfrac{kg\ m}{s}} \times \dfrac{\dfrac{kg\ m^2}{s^2}}{J} \nonumber$
$\lambda = 2.1 \times 10^{-37}\ m \nonumber$
(b)
1. Find the uncertainty of momentum from the uncertainty of velocity.
$\Delta p = m \Delta v \nonumber$
$\Delta v = 0.01 \times v \nonumber$
$\Delta p = 510\ kg \times 0.01 \times 22\ \dfrac{km}{hr} \times \dfrac{1000\ m}{km} \times \dfrac{hr}{3600\ s} \nonumber$
$\Delta p = 31 \dfrac{kg\ m}{s} \nonumber$
2. Use Heisenberg's uncertainty principle to find the uncertainty of position.
$\Delta x \Delta p \geq \dfrac{h}{4 \pi} \nonumber$
$\Delta x \geq \dfrac{h}{4 \pi \Delta p} \nonumber$
$\Delta x \geq \dfrac{6.626 \times 10^{-34}\ J\ s}{4 \pi \times 3.1 \times 10^3 \dfrac{kg\ m}{s}} \times \dfrac{\dfrac{kg\ m^2}{s^2}}{J} \nonumber$
$\Delta x \geq 1.7 \times 10^{-36}\ m \nonumber$
(c)
The calculations that involve Planck's constant (h) will change. This much larger value will make the wave-like properties of the horse more important; its wavelength and uncertainty of position will increase dramatically:
$\lambda ' = \dfrac{h'}{p} = \dfrac{0.010\ J\ s}{3.1 \times 10^3\ \dfrac{kg\ m}{s} \times \dfrac{\dfrac{kg\ m^2}{s^2}}{J}} \nonumber$
$\lambda ' = 3.2 \times 10^{-6}\ m = 3200\ nm \nonumber$
$\Delta x' \geq \dfrac{h'}{4 \pi \Delta p} = \dfrac{0.010\ J\ s}{4 \pi \times 3.1 \times 10^3 \dfrac{kg\ m}{s}} \times \dfrac{\dfrac{kg\ m^2}{s^2}}{J} \nonumber$
$\Delta x' \geq 2.6 \times 10^{-5} m = 26\ \mu m \nonumber$
The momentum would not change.
Q11.12a
how fast must a person weighing 42 kg move through a door 0.5 m wide in order to be diffracted?
S11.12a
p= h/ λ = 6.626 x 10 -34 J s/ 0.5
= 3.313 x 10-34 kg m s-1
V= p/m = 3.313 x 10-34 / 42 kg = 7.88 x 10-36 ms-1
Q11.12b
When a particle passes through a slit, diffraction occurs if the particle's wavelength is on the same order of magnitude as the width of the slit. At approximately what velocity would a 1670 kg car have to move through the Lincoln Tunnel (6.6 m width) for diffraction to occur? Hint: How can you find the wavelength of the car?
S11.12b
1. Use de Broglie's equation to relate velocity to wavelength; solve for velocity.
$\lambda = \dfrac{h}{m v} \nonumber$
$v = \dfrac{h}{m \lambda} \nonumber$
2. Set the wavelength equal to the tunnel width and solve.
$v = \dfrac{6.626 \times 10^{-34}\ J\ s}{1670\ kg \times 6.6\ m} \times \dfrac{\dfrac{kg\ m^2}{s^2}}{J} \nonumber$
Answer: the car would have to be driving very slowly.
$v = 6.0 \times 10^{-38}\ m/s \nonumber$
Q11.13a
The Paschen emission spectrum is the collection of spectral lines emitted by H atoms, where the final state is n = 3. What are the shortest and longest wavelengths (in nm) of Paschen spectral lines? Hints: The minimum value for ni is 4, the maximum value for ni is infinity and the Rydberg formula describes the emission spectra of hydrogen
Q11.13b
Spectral lines of Lyman and Balmer series do not overlap. Verify this statement.
S11.13b
A photon has a specific energy according to the energy between the excited and ground state.
E = (1/nf2 - 1/ni2), A = hv
The lyman series is when nf is the ground state or nf =1 and Balmer series is when nf =2
nf = 1 the value , ni from 2 to infinity with the values between 1 and ½. For the balmer series ni is from 3 to infinity the range between 1/4 and 5/36. Thus, these intervals don not overlap --> the energies do not overlap --> the lines in the series cant overlap
Q11.14
Assume the Rydberg constant for He+ is 9.72x10-18nm, calculate the wavelength of He+ ions from n=3 to n=2.
S11.14
$\dfrac{1}{\lambda}=R_{H} \left | \dfrac{1}{n^{2}_{i}}-\dfrac{1}{n^{2}_{f}} \right | \nonumber$
$\dfrac{1}{\lambda}=9.72\times10^{-18} \left | \dfrac{1}{3^{2}}-\dfrac{1}{2^{2}} \right | \nonumber$
$\lambda=7.41\times10^{17}nm\nonumber$
Q11.15
Derive from the following equation to solve for wavelength.
$\Delta E=hcR_{H}(\dfrac{1}{n_{i}^{2}}-\dfrac{1}{n_{f}^{2}})$
S11.15
$\Delta E=R_{H} \left | \dfrac{1}{n^{2}_{i}}-\dfrac{1}{n^{2}_{f}} \right | \nonumber$
$\dfrac{v}{c}=\Delta E=R_{H} \left | \dfrac{1}{n^{2}_{i}}-\dfrac{1}{n^{2}_{f}} \right | \nonumber$
$\dfrac{1}{\lambda}=\dfrac{v}{c}=\Delta E=R_{H} \left | \dfrac{1}{n^{2}_{i}}-\dfrac{1}{n^{2}_{f}} \right | \nonumber$
$\lambda=\dfrac{1}{R_{H} \left | \dfrac{1}{n^{2}_{i}}-\dfrac{1}{n^{2}_{f}} \right |}\nonumber$
Q11.16
Photosynthesis occurs when light excites the electrons in the chloroplasts of leaves. If you flash a light of 500 nm at the plant and assuming your flashlight emits 4.5x1020 numbers of photons, how much energy in joules was that one flash?
Q11.50
What is an antiatom? How is it different from a regular hydrogen? What will happen if these two collided with one another?
S11.50
An antiatom is a hydrogen atom with reversed electrical charges compared to that of a hydrogen atom. The proton is called an antiproton and it bears a negative charge rather than a positive charge. The electron, or antielectron, has a positive charge rather than a negative charge. As a result, an antiatom and regular hydrogen would obliterate one another should they collide. Subsequent energy would then be released.
Q11.51
A scientist determined the kinetic energy released electrons of cesium metal through a photoelectric experiment. Determine h and the \work function for cesium graphically from the following results:
λ/nm 410 420 475 500 550 630
V/volt 1.6 1.4 1.1 0.91 0.72
0.37
S11.51
First, convert wavelength to frequency and volt to energy in J.
f (1/sec) Energy (J)
$7.31707 \times 10^{15}$ 1.6
$7.14286 \times 10^{15}$ 1.4
$6.3157 \times 10^{15}$ 1.1
$6 \times 10^{15}$ 0.91
$5.45455 \times 10^{15}$ 0.72
$4.7619 \times 10^{15}$ 0.32
Then graph the values and determine the linear equation.
The linear equation is
$y=4.7 \times 10^{-16} f - 1.89\nonumber$
Use the equation below:
hf-$\phi$=Ek (max kinetic energy)
h=slope
$\phi$= y-intercept
h= 4.7 X 10-16
$\phi$= 1.89
Q11.52
Calculate the de Broglie wavelength of a Cl2 molecule at 300 K.
S11.52
$v_{rms} = \sqrt{\dfrac{3RT}{M}}\nonumber$
=((3)(8.314)(300)/(70.8X10-3 kg) = 325 m/s
$\lambda = \dfrac{h}{mv_{rms}}\nonumber$
=(6.626X10-34 J*s)/[(70.8 amu)(1.66X10-27 kg/amu)(325 m/s)] = 1.73 X 10-11 m
Q11.53
Consider a balloon with a diameter of $2.5 \times 10^{-5}\; m$. What is the uncertainty of the velocity of an oxygen molecule that is trapped inside.
S11.53
Use the direct application of the uncertainty principle:
$\Delta x \Delta p \ge \dfrac{h}{4 \pi}\nonumber$
Let's consider the molecule has an uncertainly that is $\pm\, radius$ of the balloon.
$\Delta x = 1.3 \times 10^{-5}\; m\nonumber$
The uncertainty of the momentum of the molecule can be estimated via the uncertainty principle.
$\Delta p =\dfrac{h}{4 \pi \Delta x} = \dfrac{(6.626 \times 10^{-34} J*s)}{(4\pi) (1.3 \times 10^{-5}\; m)} = 4.05 \times 10^{-30} kg\,m\,s^{-1}\nonumber$
This can be converted to uncertainty in velocity via
$p=mv\nonumber$
or
$\Delta p =m_e \Delta v\nonumber$
with the electron mass $m_e$ equal to $9.109 \times 10^{-31}\; kg$. since the mass of the molecule is not uncertain.
$\Delta v = \dfrac{ \Delta p }{ m_e } = \dfrac{4.05 \times 10^{-30}}{9.109 \times 10^{-31}\; kg} = 4.5\; m/s\nonumber$
Q11.55
What conditions must be satisfied for the Schrodinger's equation for $\psi$ to be an acceptable wavefunction. What are some examples of unacceptable wave functions?
Q11.56
The equation for calculating the energies of the electron in a hydrogen atom or a hydrogenlike ion is given by En = -(2.18 x10^-18 J)Z^2(1/n^2), where Z is the atomic number of the element. One way to modify this equation for many-electron atoms is to replace Z with (Z- [sigma]), where [sigma] is a positive dimensionless quantity called the shielding constant. Consider the lithium atom as an example. The physical significance of sigma that it represents the extent of shielding that the two 1s electrons exert on each other. Thus the quantity (Z - [sigma]) is appropriated called the "effective nuclear change." Use the first ionization energy of lithium to calculate the value of [sigma].
Q11.57
Calculate the wavelength and frequency of an emitted gamma particle that has energy of 3.11 x 10^12 J mol^-1.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/11%3A_Quantum_Mechanics_and_Atomic_Structure/11.E%3A_Quantum_Mechanics_and_Atomic_Structure_%28Exercises%29.txt
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Chemical bond refers to the forces holding atoms together to form molecules and solids. This force is of an electric nature, and the attraction between electrons of one atom to the nucleus of another atom contributes to what is known as chemical bonds. A chemical bond is a lasting attraction between atoms that enables the formation of chemical compounds and may result from the electrostatic force of attraction between atoms with opposite charges, or through the sharing of electrons as in the covalent bonds. The strength of chemical bonds varies considerably.
• 12.1: Lewis Structures
hemical bond refers to the forces holding atoms together to form molecules and solids. This force is of an electric nature, and the attraction between electrons of one atom to the nucleus of another atom contributes to what is known as chemical bonds. Various theories regarding chemical bonds have been proposed over the past 300 years, during which our interpretation of the world has also changed.
• 12.2: Valence Bond Theory
Valence bond theory describes bonding as a consequence of the overlap of two separate atomic orbitals on different atoms that creates a region with one pair of electrons shared between the two atoms. When the orbitals overlap along an axis containing the nuclei, they form a σ bond. When they overlap in a fashion that creates a node along this axis, they form a π bond.
• 12.3: Hybridization of Atomic Orbitals
• 12.4: Electronegativity and Dipole Moment
• 12.5: Molecular Orbital Theory
The positions and energies of electrons in molecules can be described in terms of molecular orbitals. A particular spatial distribution of electrons in a molecule that is associated with a particular orbital energy.—a spatial distribution of electrons in a molecule that is associated with a particular orbital energy. As the name suggests, molecular orbitals are not localized on a single atom but extend over the entire molecule. Hence, the molecular orbital approach is a delocalized approach.
• 12.6: Diatomic Molecules
• 12.7: Resonance and Electron Delocalization
Resonance is a mental exercise and method within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. It compares and contrasts two or more possible Lewis structures that can represent a particular molecule. Resonance structures are used when one Lewis structure for a single molecule cannot fully describe the bonding that takes place between neighboring atoms relative to the empirical data for the actual bond lengths between those atoms.
• 12.8: Coordination Compounds
• 12.9: Coordination Compounds in Biological Systems
• 12.E: The Chemical Bond (Exercises)
This are exercises that to accompany the TextMap organized around Raymond Chang's Physical Chemistry for the Biosciences textbook.
Thumbnail: Boundary surface of a delta chemical bond. (Public Domain; Ben Mills via Wikipedia)
12: The Chemical Bond
Chemical bond refers to the forces holding atoms together to form molecules and solids. This force is of an electric nature, and the attraction between electrons of one atom to the nucleus of another atom contributes to what is known as chemical bonds. Various theories regarding chemical bonds have been proposed over the past 300 years, during which our interpretation of the world has also changed. Some old concepts such as Lewis dot structure and valency are still rather useful in our understanding of the chemical properties of atoms and molecules, and new concepts involving quantum mechanics of chemical bonding interpret modern observations very well. You learn new concepts such as bond length, bond energy, bond order, covalent bond, ionic bond, polar and non-polar bond etc. These concepts help you understand the material world at the molecular level.
Of the three primary theories of bonding discussed in this class (Lewis Theory, Valance Bond Theory and Molecular orbital theory), Lewis theory does not take full advantage of the quantum mechanical concepts discussed previously. Key to this theory is the Lewis Structure, which is a very simplified representation of the electrons in a molecule and is use to show how the electrons are arranged around individual atoms in a molecule. Between 1916 and 1919, Gilbert Newton Lewis, Walther Kossel, and Irving Langmuir formulated a theory to explain chemical bonding. This theory is now called Lewis Theory and it is based on the following five principles:
1. Valence electrons, or the electrons in the outermost electron shell, have an essential role in chemical bonding.
2. Ionic bonds are formed between atoms when electrons are transferred from one atom to another. Ionic bond is a bond between nonmetals and metals .
3. Covalent bonds are formed between atoms when pairs of electrons are shared between atoms. A covalent bond is between two nonmetals.
4. Electrons are transferred/shared so that each atom may reach a more stable electron configuration, i.e., the noble gas configuration which contains 8 valence electrons; this is called octet rule.
5. Although many different Lewis structures can be constructed, those that are most stable have the lowest formal charge on the atoms.
Writing out Lewis structures can be at times, tricky and somewhat difficult. Moreover, a compound can have multiple Lewis Structures that contribute to the shape of the overall compound, so one Lewis structure of a compound may not necessarily be exactly what the compound looks like. If you need a refresher on how to constructing Lewis Structures consult the following Modules: Writing Lewis Structures, Lewis Structures I and Lewis Structures II.
In Lewis Structures, valence electrons are represented as dots and only outer-shell electrons are considered (e.g., the valence electrons and not the core electrons). Single (covalent) bonds are made up of two shared electrons and are represented by two dots between the bonded atoms or a single line. Double bonds and triple bonds are represented as two and three lines/(pairs of electrons), respectively. Lone pairs on the outer rims of an atom are represented as two dots. This is because they are the ones involved in chemical reactions. For the 2nd and 3rd periods elements, the number of valence electrons range from 1 to 8. Lewis dot structure for them are as indicated:
$\mathrm{ \overset{\Large{.}}Li \hspace{20px} \underset{\Large{.}}{\overset{\Large{.}}Be}\hspace{20px} \cdot \overset{\Large{.}}{B} \cdot\hspace{20px} \cdot \underset{\Large{.}}{\overset{\Large{.}}{C}} \cdot\hspace{20px} \cdot \underset{\Large{.}}{\overset{\Large{.}}{N}} :\hspace{20px} : \underset{\Large{.}}{\overset{\Large{.}}{O}} :\hspace{20px} : \underset{\Large{.}}{\overset{\Large{..}}{F}} :\hspace{20px} : \underset{\Large{..}}{\overset{\Large{..}}{Ne}} :}$
Lewis made the valence electron visible with dots. Other examples include the following molecules:
$\mathrm{H : H}$ $\mathrm{H : \underset{\Large{..}}{\overset{\Large{..}}{O}} : H}$ $\mathrm{H : \underset{\Large{..}}{\overset{\Large{..}}{F}} :}$ $\mathrm{H : \underset{\Large{..}}{\overset{\Large{\underset{\huge{..}}H}}{N}} : H}$ $\mathrm{H : \underset{\Large{\overset{\huge{..}}H}}{\overset{\Large{\underset{\huge{..}}H}}{C}} : H}$ $\mathrm{: N ::: N :}$ $\mathrm{: \overset{\Large{..}}O :: \overset{\Large{..}}O :}$ $\mathrm{: \overset{\Large{..}}F : \overset{\Large{..}}F :}$ $\mathrm{: \overset{\Large{..}}O :: C :: \overset{\Large{..}}O : }$
Note
Lewis structure does NOT attempt to explain the geometry of molecules, how the bonds form, or how the electrons are shared between the atoms. It is the simplest and most limited theory on electronic structure.
Principle 1: Valence Electrons
The electrons of an atom can divided into two categories: valence and core electrons. Valence electrons are those occupying the outermost shell or highest energy level of an atom while core electrons are those occupying the innermost shell or lowest energy level. This difference greatly influences the role of the two types of electrons in a chemical reaction. Generally, valence electrons can participate in the formation of chemical bonding, but core electrons cannot. While core electrons are not involved in bonding, they do have an influence on the chemical reactivity of an atom.
The electron configuration of the oxygen atoms is
$O: \,1s^22s^22p^4 \label{1}$
As discussed before, this configuration may be shorted
$O:\, [He]2s^22p^4 \label{2}$
where the $[He]$ stands for the configuration of helium ($1s^22s^2$). The $1s$ electrons in oxygen (Equation $\ref{1}$) do not participate in bonding, i.e., chemistry, and are called core electrons. The other electrons (i.e., the $2s^22p^4$ part) are call valence electrons and are responsible for the making and breaking of chemical bonds.
The configuration of calcium with 20 electrons can be written
$Ca:\, [Ar]4s^2 \label{3}$
where the $[Ar]$ stands for the configuration of argon ($1s^22s^22p^63s^23p^6$). Electronic configurations that are the same as noble gases are very stable since they have a full octet (except helium with a full 1s orbital). The electrons in the argon-like closed shell of Equation $\ref{3}$ are the core electrons and the two electrons in the 4s orbital are valence electrons.
Example $1$: Cobalt
What are the core and valence electrons in cobalt?
Solution
Start by writing the electron configuration of cobalt (i.e., 27 electrons):
$1s^22s^22p^63s^23p^64s^23d^7 \nonumber$
However, argon has the electronic structure $1s^22s^22p^63s^23p^6$, so we can write the configuration of cobalt as
$[Ar]4s^23d^7 \nonumber$
The two electrons in the $4s$ orbital and the seven electrons in the 3d are the valence electrons: all others are core electrons.
Principle 2: Ionic Bonding
In ionic bonding, electrons are transferred from one atom to another resulting in the formation of positive and negative ions. The electrostatic attractions between the positive and negative ions hold the compound together. The predicted overall energy of the ionic bonding process, which includes the ionization energy of the metal and electron affinity of the nonmetal, is usually positive, indicating that the reaction is endothermic and unfavorable. However, this reaction is highly favorable because of their electrostatic attraction.
The energy of the electrostatic attraction ($E$), a measure of the force’s strength, is inversely proportional to the internuclear distance between the charged particles ($r$):
$E = k\dfrac{Q_{1}Q_{2}}{r} \label{8.2.1b}$
where each ion’s charge is represented by the symbol Q. The proportionality constant k is equal to 2.31 × 10−28 J·m.
Example $2$: Ionic Bonding in NaCl
For example, in the reaction of Na (sodium) and Cl (chlorine), each Cl atom takes one electron from a Na atom. Therefore each Na becomes a Na+ cation and each Cl atom becomes a Cl- anion. Due to their opposite charges, they attract each other to form an ionic lattice. The formula (ratio of positive to negative ions) in the lattice is NaCl.
$2Na_{(s)} + Cl_{2(g)} \rightarrow 2NaCl_{(s)} \nonumber$
Sodium has one valence electron and chlorine has seven valence electrons; the two elements react such that the chlorine atom takes the valence electron from the sodium atom leaving the chlorine atom with one extra electron and thus negatively charged and the Sodium atom without an electron and thus positively charged. The two atoms then become ions and because of their opposite charges the ions are held together in an ionic bond (Equation $\ref{8.2.1b}$).
The chlorine has a high affinity for electrons, and the sodium has a low ionization potential. Thus the chlorine gains an electron from the sodium atom. The Lewis Structure of this reaction is (here we will consider one chlorine atom, rather than Cl2) is:
The arrow indicates the transfer of the electron from sodium to chlorine to form the Na+ metal ion and the Cl- chloride ion
Principle 3: Covalent Bonding
Covalent bonding occurs when pairs of electrons are shared by atoms. Atoms will covalently bond with other atoms in order to gain more stability, which is gained by forming a full electron shell. By sharing their outer most (valence) electrons, atoms can fill up their outer electron shell and gain stability. Nonmetals will readily form covalent bonds with other nonmetals in order to obtain stability, and can form anywhere between one to three covalent bonds with other nonmetals depending on how many valence electrons they posses. Although it is said that atoms share electrons when they form covalent bonds, they do not usually share the electrons equally.
The diatomic hydrogen molecule (H2) is the simplest model of a covalent bond, and is represented in Lewis structures as:
The shared pair of electrons provides each hydrogen atom with two electrons in its valence shell (the 1s) orbital. In a sense, it has the electron configuration of the noble gas helium. When two chlorine atoms covalently bond to form $Cl_2Cl_2$, the following sharing of electrons occurs:
Each chlorine atom shared the bonding pair of electrons and achieves the electron configuration of the noble gas argon. In Lewis structures the bonding pair of electrons is usually displayed as a line, and the unshared electrons as dots:
The shared electrons are not located in a fixed position between the nuclei. In the case of the $H_2H_2$ compound, the electron density is concentrated between the two nuclei:
The two atoms are bound into the $H_2H_2$ molecule mainly due to the attraction of the positively charged nuclei for the negatively charged electron cloud located between them.
Example $3$: The Chlorate Ion
Draw the Lewis Structure for the chlorate ion (ClO3-).
Solution
First, lets find the how many valence electrons chlorate has:
ClO3- : 7 e-(from Cl) + 3(6) e-(from 3 O atoms) + 1 (from the total charge of -1) = 26
There are 26 valence electrons. Next lets draw the basic framework of the molecule:
The molecule uses covalent bonds to hold together the atoms to the central Chlorine. The remaining electrons become non-bonding electrons. Since 6 electrons were used for the bonds, the 20 others become those un-bonding electrons to complete the octet:
The oxygen atom's shells fill up with 18 electrons, and the other 2 complete Chlorine's octet
Example $4$: Formaldehyde
Constructing the Lewis Structure of the formaldehyde (H2CO) molecule.
Solution
First find number of valence electrons:
H2CO: 2(1) e- (from the H atoms) + 4 e- (from the C atom) + 6 e- (from the O atom)
There are 12 valence electrons. Next draw out the framework of the molecule:
To satisfy the octet of Carbon, one of the pairs of electrons on Oxygen must be moved to create a double bond with Carbon. Therefore our Lewis Structure would look as it does below:
The Hydrogen atoms are each filled up with their two electrons and both the Carbon and the Oxygen atoms' octets are filled.
Principle 4: The Octet Rule
The octet rule is a chemical rule of thumb that reflects observation that atoms of main-group elements tend to combine in such a way that each atom has eight electrons in its valence shell, giving it the same electronic configuration as a noble gas. The Octet Rule requires all atoms in a molecule to have 8 valence electrons--either by sharing, losing or gaining electrons--to become stable. For covalent bonds, atoms to share their electrons with each other to satisfy the octet rule. It requires eight electrons because that is the amount of electrons needed to fill a s- and p- orbital (electron configuration); also known as a noble gas configuration. Although it is important to remember the "magic number" is 8, but there are many octet rule exceptions.
Octet Rule
A stable arrangement is attended when the atom is surrounded by eight electrons. This octet can be made up by own electrons and some electrons which are donated (e.g, ionic bonding) or shared (covalent bonding). An atom continues to form bonds until an octet of electrons is reached.
For the first rows in the periodic table, the magic number 8 can easily be explained from quantum mechanics of the multiple electron atoms discussed previously. In particular, electrons possess four quantum numbers (usually $n$, $l$, $m_l$, and $m_s$), which are restricted as such:
$n = 1, 2, 3, ...$
$l = 0, 1, 2, ..., n - 1$
$m_l = -l, -l + 1, ..., 0, ..., l - 1, l$
$m_s = -\frac{1}{2}, \frac{1}{2}$
where
• $n$ is the principal quantum number (dealing with energy states),
• $l$ is the orbital quantum number (dealing with subshells),
• $m_l$ is the orbital magnetic quantum number (direction of $l$) and
• $m_s$ is the spin quantum number (direction of spin).
By the Pauli's Exclusion Principle all of these numbers cannot be the same for any two electrons in an atom. So in the $n = 2$ shell you can have two possible values for $l$, one possible value for $m_l$ when $l = 0$, three possible values for $m_l$ when $l = 1$, and two possible values for $m_s$. This sums to a total of eight possible value vectors:
$(1 + 3) \cdot 2 = 8$
In the $n = 3$ shell, the Octet rule also holds for $l < 2$, which gives you sort of the same combinatorics all over again.
Since the quantum numbers cannot be the same for any electrons in an atom, adding an extra electron to an atom with eight electrons in its outermost shell forces the electron to occupy an orbital with a higher principal quantum number ($n$). Since the binding energy of the electron increases with $n$, this is energetically unfavorable to binding with an atom that has more space in its outermost shell. For example, for the nonmetals (and the 's' block metals) to achieve an octet, the number of valence electrons is equal to the group number (Table $1$).
Table $1$: Select electronic properties of p-block elements of the 2nd row.
Element
Group
Valence electrons
Number of Covalent Bonds needed to form valence octet
F
17
7
1
O
16
6
2
N
15
5
3
C
14
4
4
Note: The 18 Electron Rule
The octet rule is not a law and it only works for a handful of elements, mainly the second-row p-block ones; there are many exceptions. Transition metals compounds often have 18 electrons on the metal atom because eighteen is the number of electrons in the nearest noble gas configuration in transition metals, which includes s + p + d electrons. The second and third row of transition metals have f electrons, too, but we usually simplify and don't worry about those electrons for electron counting purposes; we treat them like core electrons, not valence electrons.
Lewis theory successfully describe the covalent interactions between various nonmetal elements. Examples of hydride compounds of the elements in Table 12.2.1 (covalent compounds with hydrogen):
The sharing of a pair of electrons represents a single covalent bond, usually just referred to as a single bond. However, in many molecules atoms attain complete octets by sharing more than one pair of electrons between them.
• Two electron pairs shared a double bond
• Three electron pairs shared a triple bond
Because each nitrogen contains five valence electrons, they need to share three pairs to each achieve a valence octet. The strong nitrogen triple bond makes the molecule fairly inert. Moreover, the N - N bond distance in N2 is 1.10 Å, which is appreciably shorter than the average N-N single bonds (14.7Å).
Principle 5: Formal Charges
It is sometimes possible to write more than one Lewis structure for a substance that does not violate the octet rule, as we saw for CH2O, but not every Lewis structure may be equally reasonable. In these situations, we can choose the most stable Lewis structure by considering the formal charge on the atoms, which is the difference between the number of valence electrons in the free atom and the number assigned to it in the Lewis electron structure. The formal charge is a way of computing the charge distribution within a Lewis structure; the sum of the formal charges on the atoms within a molecule or an ion must equal the overall charge on the molecule or ion. A formal charge does not represent a true charge on an atom in a covalent bond but is simply used to predict the most likely structure when a compound has more than one valid Lewis structure. To calculate formal charges, we assign electrons in the molecule to individual atoms according to these rules:
• Nonbonding electrons are assigned to the atom on which they are located.
• Bonding electrons are divided equally between the bonded atoms.
For each atom, we then compute a formal charge:
$\begin{matrix} \text{formal charge}= & valence\; e^{-}- & \left ( nonbonding\; e^{-}+\dfrac{bonding\;e^{-}}{2} \right )\ & ^{\left ( free\; atom \right )} & ^{\left ( atom\; in\; Lewis\; structure \right )} \end{matrix} \label{8.5.1}$
The most favorable Lewis Structure has the smallest formal charge for the atoms, and negative formal charges tend to come from more electronegative atoms. An example of determining formal charge can be seen below with the nitrate ion, NO3-:
• The double bonded O atom has 6 electrons: 4 non-bonding and 2 bonding (one electron for each bond). Since O should have 6 electrons, the formal charge is 0.
• The two singly bonded O atoms each have 7 electrons: 6 non-bonding and 1 bonding electron. Since O should have 6 electrons, and there is one extra electron, those O atoms each have formal charges of -1.
• The N atom has 4 electrons: 4 bonding and 0 non-bonding electrons. Since N should have 5 electrons and there are only 4 electrons for this N, the N atom has a formal charge of +1.
• The charges add up to the overall charge of the ion. 0 + (-1) + (-1) + 1 = -1. Thus, these charges are correct, as the overall charge of nitrate is -1.
In general you want the fewest number of formal charges possible, i.e. formal charges of 0 for as many of the atoms in a structure as possible. Also the formal charges should match the electronegativity of the atom, that is negative charges should be on the more electronegative atoms and positive charges on the least electronegative atoms if possible. Charges of -1 and +1 on adjacent atoms can usually be removed by using a lone pair of electrons from the -1 atom to form a double (or triple) bond to the atom with the +1 charge.
Summary
Lewis Structures are visual representations of the bonds between atoms and illustrate the lone pairs of electrons in molecules. They can also be called Lewis dot diagrams and are used as a simple way to show the configuration of atoms within a molecule. Electrons are shown as "dots" or for bonding electrons as a line between the two atoms. The goal is to obtain the "best" electron configuration, i.e. the octet rule and formal charges need to be satisfied. Lewis structures can also be useful in predicting molecular geometry in conjunction with hybrid orbitals. A compound may have multiple resonance forms that are also all correct Lewis structures. This section will discuss the rules for writing out Lewis structures correctly.
Contributors and Attributions
• Andreas Hagen (via StackExchange)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/12%3A_The_Chemical_Bond/12.01%3A_Lewis_Structures.txt
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Valence bond (VB) theory is one of two basic theories, along with molecular orbital (MO) theory, that were developed to use the methods of quantum mechanics to explain chemical bonding. It focuses on how the atomic orbitals of the dissociated atoms combine to give individual chemical bonds when a molecule is formed. In contrast, molecular orbital theory, which will be discussed elsewhere, predict wavefunctions that cover the entire molecule.
The simplest case to consider is the hydrogen molecule, $H_2$. When we say that the two hydrogen nuclei share their electrons to form a covalent bond, what we mean in VB theory terms is that the two spherical 1s orbitals (the grey spheres in Figure $1$) overlap and contain two electrons.
How far apart are the two nuclei? If they are too far apart, their respective 1s orbitals cannot overlap, and thus no covalent bond can form - they are still just two separate hydrogen atoms. As they move closer and closer together, orbital overlap begins to occur, and a bond begins to form. This lowers the potential energy of the system, as new, attractive positive-negative electrostatic interactions become possible between the nucleus of one atom and the electron of the second (Figure $2$). However, something else is happening at the same time: as the atoms get closer, the repulsive positive-positive interaction between the two nuclei also begins to increase.
At first this repulsion is more than offset by the attraction between nuclei and electrons, but at a certain point, as the nuclei get even closer, the repulsive forces begin to overcome the attractive forces, and the potential energy of the system rises quickly. When the two nuclei are ‘too close’, we have an unstable, high-energy situation. There is a defined optimal distance between the nuclei in which the potential energy is at a minimum, meaning that the combined attractive and repulsive forces add up to the greatest overall attractive force. This optimal internuclear distance is the bond length; the distance is 74 pm for $H_2$. Likewise, the difference in potential energy between the lowest energy state (at the optimal internuclear distance) and the state where the two atoms are completely separated is called the bond dissociation energy, or, more simply, bond strength; the $H_2$ bond strength is 435 kJ/mol.
A comparison of some bond lengths and energies is shown in Table $1$. We can find many of these bonds in a variety of molecules, and this table provides average values. For example, breaking the first C–H bond in CH4 requires 439.3 kJ/mol, while breaking the first C–H bond in $\ce{H–CH2C6H5}$ (a common paint thinner) requires 375.5 kJ/mol.
Table $1$: Representative Bond Energies and Lengths
Bond Length (pm) Energy (kJ/mol) Bond Length (pm) Energy (kJ/mol)
H–H 74 436 C–O 140.1 358
H–C 106.8 413 $\mathrm{C=O}$ 119.7 745
H–N 101.5 391 $\mathrm{C≡O}$ 113.7 1072
H–O 97.5 467 H–Cl 127.5 431
C–C 150.6 347 H–Br 141.4 366
$\mathrm{C=C}$ 133.5 614 H–I 160.9 298
$\mathrm{C≡C}$ 120.8 839 O–O 148 146
C–N 142.1 305 $\mathrm{O=O}$ 120.8 498
$\mathrm{C=N}$ 130.0 615 F–F 141.2 159
$\mathrm{C≡N}$ 116.1 891 Cl–Cl 198.8 243
Example $1$: A Morse Potential Approximation
The Morse potential, named after physicist Philip M. Morse, is a convenient interatomic interaction approximation for the potential energy of a diatomic molecule.
$V(r) = D_e \left[ 1-e^{-\beta(r-r_e)} \right]^2 \nonumber$
Here $r$ is the distance between the atoms, $r_e$ is the equilibrium bond distance, $D_e$ is the well depth (defined relative to the dissociated atoms), and $\beta$ controls the 'width' of the potential (the smaller $\beta$ is, the larger the well). For $H_2$, the equilibrium bond length $r_e = 74.1 \;pm$, the bond energy $D_e = 7.61 \times 10^{-19} J$, and the Morse potential parameter $\beta= 0.0193\; pm^{-1}$.
The force on the system is given by
$F(r) = -\dfrac{dV(r)}{dr} \nonumber$
Calculate the force for $H_2$ and explain the status of the system (e.g., compressing, expanding, or stationary) at:
1. 45 pm,
2. 74 pm,
3. 150 pm, and
4. 500 pm.
Overlap
Valence bond theory describes a covalent bond as the overlap of half-filled atomic orbitals (each containing a single electron) that yield a pair of electrons shared between the two bonded atoms. We say that orbitals on two different atoms overlap when a portion of one orbital and a portion of a second orbital occupy the same region of space. According to valence bond theory, a covalent bond results when two conditions are met:
1. an orbital on one atom overlaps an orbital on a second atom and
2. the single electrons in each orbital combine to form an electron pair.
The mutual attraction between this negatively charged electron pair and the two atoms’ positively charged nuclei serves to physically link the two atoms through a force we define as a covalent bond. The strength of a covalent bond depends on the extent of overlap of the orbitals involved. Orbitals that overlap extensively form bonds that are stronger than those that have less overlap.
In addition to the distance between two orbitals, the orientation of orbitals also affects their overlap (other than for two s orbitals, which are spherically symmetric). Greater overlap is possible when orbitals are oriented such that they overlap on a direct line between the two nuclei. Figure $3$ illustrates this for two p orbitals from different atoms; the overlap is greater when the orbitals overlap end to end rather than at an angle.
The overlap of two s orbitals (as in H2), the overlap of an s orbital and a p orbital (as in HCl), and the end-to-end overlap of two p orbitals (as in Cl2) all produce sigma bonds (σ bonds), as illustrated in Figure $4$. A σ bond is a covalent bond in which the electron density is concentrated in the region along the internuclear axis; that is, a line between the nuclei would pass through the center of the overlap region. Single bonds in Lewis structures are described as σ bonds in valence bond theory.
A pi bond (π bond) is a type of covalent bond that results from the side-by-side overlap of two p orbitals, as illustrated in Figure $5$. In a π bond, the regions of orbital overlap lie on opposite sides of the internuclear axis. Along the axis itself, there is a node, that is, a plane with no probability of finding an electron.
While all single bonds are σ bonds, multiple bonds consist of both σ and π bonds. As the Lewis structures in suggest, O2 contains a double bond, and N2 contains a triple bond. The double bond consists of one σ bond and one π bond, and the triple bond consists of one σ bond and two π bonds. Between any two atoms, the first bond formed will always be a σ bond, but there can only be one σ bond in any one location. In any multiple bond, there will be one σ bond, and the remaining one or two bonds will be π bonds. These bonds are described in more detail later in this chapter.
As seen in Table $2$, an average carbon-carbon single bond is 347 kJ/mol, while in a carbon-carbon double bond, the π bond increases the bond strength by 267 kJ/mol. Adding an additional π bond causes a further increase of 225 kJ/mol. We can see a similar pattern when we compare other σ and π bonds. Thus, each individual π bond is generally weaker than a corresponding σ bond between the same two atoms. In a σ bond, there is a greater degree of orbital overlap than in a π bond.
Counting σ and π Bonds
Butadiene, C6H6, is used to make synthetic rubber. Identify the number of σ and π bonds contained in this molecule.
Butadiene
Solution
There are six σ C–H bonds and one σ C–C bond, for a total of seven from the single bonds. There are two double bonds that each have a π bond in addition to the σ bond. This gives a total nine σ and two π bonds overall.
Exercise $2$
Identify each illustration as depicting a σ or π bond:
1. side-by-side overlap of a 4p and a 2p orbital
2. end-to-end overlap of a 4p and 4p orbital
3. end-to-end overlap of a 4p and a 2p orbital
Answer
(a) is a $π$ bond with a node along the axis connecting the nuclei while (b) and (c) are $σ$ bonds that overlap along the axis.
The Valence Bond Wavefunction
In the hydrogen molecule, VB theory uses two electrons (electron 1 and electron 2) in two 1s atomic orbitals ($\psi_a$ and $\psi_b$) on the hydrogen atoms(atom $a$ and atom $b$). The simplest description of a $H_2$ molecule with two electrons and two nuclei is
$\psi_{simple} = \psi_a(1)\psi_b(2) \label{VBA}$
Equation $\ref{VBA}$ means electron $1$ is in $\psi_a$ of atom $a$ and electron $2$ is in $\psi_a$ on the atom $b$. Where $\psi_a$ and $\psi_b$ may be the $1s$ orbitals as the simplest wavefunction ($n=1$, $l=0$, $m_l$). The $\psi_{simple}$ wavefunction predicts an energy that is about 6% of the experimental value at a bond length of 0.90 pm, which is considerably longer than the experimental value of 0.74 pm (Table $1$). However, Equation $\ref{VBA}$ does not address the indistinguishability of the problem, i.e, we cannot distinguished each electron from the other.
Distinguishable vs. Indistinguishable Particles
Assume you have two particle A and B in states 1 and 2. If the two particle are distinguishable, then by exchanging the particles A and B, you will obtain a new state that will have the same properties as the old state. However, for indistinguishable particles, exchanging A and B is a transformation that does nothing and you have the same physical state. This means that for indistinguishable particles, particle labels are unphysical and they represent a redundancy in describing the physical state.
To account for the indistinguishable nature of the two electrons in the system, a covalent wavefunction that is better than Equation $\ref{VBA}$ can be constructed (ignoring spin effect):
$\psi_{covalent} = \psi_a(1)\psi_b(2) + \psi_a(2)\psi_b(1) \label{VBC}$
Equation $\ref{VBC}$ means electron $1$ is sometimes in $\psi_a$ on atom $a$ and electron $2$ is on the atom $b$ (left term of right hand side of Equation) and sometime electron $1$ is sometimes in $\psi_b$ on atom $b$ and electron $2$ is in $\psi_a$ on the atom $a$ (right term of right hand side of Equation). Notice that in contrast to Equation $\ref{VBA}$, if the two electron labels are switched, the wavefunction in Equation $\ref{VBC}$ does not change. The potential energy curve for the $\psi_{covalent}$ wavefunction predicts a bond energy that is about 72% at a bond length of 0.87 pm (Table $2$), so the $\psi_{covalent}$ wavefunction is more accurate than $\psi_{simple}$.
To be more accurate within VB theory, the possibility that both electrons may be on atom a ($\psi_a(1)\psi_a(2)$) or on atom b ($\psi_b(1)\psi_b(2)$) should also be taken into account. This introduces an ionic contribution:
$\psi_{ionic} = \psi_a(1)\psi_a(2) + \psi_b(2)\psi_b(1) \label{VBD}$
Notice that Equation $\ref{VBD}$ addresses indistinguishability since the wavefunction is unchanged if the two electron labels are switched. The more complete VB wavefunction is a linear combination of both $\psi_{covalent}$ and $\psi_{ionic}$:
$\psi_{AB} = Ψ_{covalent} + λ\psi_{ionic} \label{VBE}$
where $\lambda$ quantifies the amount of ionic character to the bond, e.g., if $\lambda=0$, then the bond is purely covalent. The potential energy curve for the $\psi_{AB}$ wavefunction predicts a bond energy that is ~80% of the experimental value and a reasonably accurate bond length of 0.77 pm (Table $1$). Additional tinkering with the wavefunction leads to even better results, but is beyond the scope of this discussion. It should be noted that the wavefunctions in Equations $\ref{VBC}$, $\ref{VBD}$, and $\ref{VBE}$ are unnormalized.
Table $2$: Valence Bond Theory Results for $H_2$
Wavefunction Bond Energy (kJ mol-1) Bond Length (pm)
Experiment
435 0.74
$\psi_{simple}$
25 0.90
$\psi_{covalent}$
301 0.87
$\psi_{AB}$
335 0.77
Two-Atom Bonds
The central gist of VB theory is that a covalent bond is formed between the two atoms by the overlap of half filled valence atomic orbitals of each atom containing one unpaired electron.
Example $3$: Percent Ionic Character
Calculate the relative fraction of the contribution of the ionic character to the VB wavefunction. Assuming that the ionic character in $HBr$ bond is 20%.
Solution
From Equation $\ref{VBE}$, the percent ionic character is given by:
$\dfrac{λ^2}{1 + λ^2} \times 100\% \nonumber$
Therefore,
$20 \% = 100\% \dfrac{λ^2}{1 + λ^2} \nonumber$
$20 \% + 20 \% λ^2 = 100 \% λ^2 \nonumber$
$20 \% = 100 \%λ^2 – 20λ^2 \nonumber$
$20 \% = 80 \%λ^2 \nonumber$
Therefore,
$λ^2 = \dfrac{20}{80} \nonumber$
Hence, $λ=\sqrt{20/80}= 0.5$ and the fraction of contribution of ionic character to $\psi_{AB}$ is 0.5 or 50%.
Linus Pauling used the pair bonding ideas of Lewis together with Heitler–London theory to develop two key concepts in VB theory: resonance and orbital hybridization. These will be addressed in later Modules.
Summary
Valence bond theory describes bonding as a consequence of the overlap of two separate atomic orbitals on different atoms that creates a region with one pair of electrons shared between the two atoms. When the orbitals overlap along an axis containing the nuclei, they form a σ bond. When they overlap in a fashion that creates a node along this axis, they form a π bond.
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Hybridization was introduced to explain molecular structure when the valence bond theory failed to correctly predict them. It is experimentally observed that bond angles in organic compounds are close to 109°, 120°, or 180°. According to Valence Shell Electron Pair Repulsion (VSEPR) theory, electron pairs repel each other and the bonds and lone pairs around a central atom are generally separated by the largest possible angles.
In this section, we will use a model called valence bond theory to describe bonding in molecules. In this model, bonds are considered to form from the overlapping of two atomic orbitals on different atoms, each orbital containing a single electron. In looking at simple inorganic molecules such as H2 or HF, our present understanding of s and p atomic orbitals will suffice. In order to explain the bonding in organic molecules, however, we will need to introduce the concept of hybrid orbitals.
Hybrid orbitals: sp3 hybridization and tetrahedral bonding
Now let’s turn to methane, the simplest organic molecule. Recall the valence electron configuration of the central carbon:
This picture, however, is problematic. How does the carbon form four bonds if it has only two half-filled p orbitals available for bonding? A hint comes from the experimental observation that the four C-H bonds in methane are arranged with tetrahedral geometry about the central carbon, and that each bond has the same length and strength. In order to explain this observation, valence bond theory relies on a concept called orbital hybridization. In this picture, the four valence orbitals of the carbon (one 2s and three 2p orbitals) combine mathematically (remember: orbitals are described by equations) to form four equivalent hybrid orbitals, which are named sp3 orbitals because they are formed from mixing one s and three p orbitals. In the new electron configuration, each of the four valence electrons on the carbon occupies a single sp3 orbital.
The sp3 hybrid orbitals, like the p orbitals of which they are partially composed, are oblong in shape, and have two lobes of opposite sign. Unlike the p orbitals, however, the two lobes are of very different size. The larger lobes of the sp3 hybrids are directed towards the four corners of a tetrahedron, meaning that the angle between any two orbitals is 109.5o.
This geometric arrangement makes perfect sense if you consider that it is precisely this angle that allows the four orbitals (and the electrons in them) to be as far apart from each other as possible. This is simply a restatement of the Valence Shell Electron Pair Repulsion (VSEPR) theory that you learned in General Chemistry: electron pairs (in orbitals) will arrange themselves in such a way as to remain as far apart as possible, due to negative-negative electrostatic repulsion.
Each C-H bond in methane, then, can be described as an overlap between a half-filled 1s orbital in a hydrogen atom and the larger lobe of one of the four half-filled sp3 hybrid orbitals in the central carbon. The length of the carbon-hydrogen bonds in methane is 1.09 Å (1.09 x 10-10 m).
While previously we drew a Lewis structure of methane in two dimensions using lines to denote each covalent bond, we can now draw a more accurate structure in three dimensions, showing the tetrahedral bonding geometry. To do this on a two-dimensional page, though, we need to introduce a new drawing convention: the solid / dashed wedge system. In this convention, a solid wedge simply represents a bond that is meant to be pictured emerging from the plane of the page. A dashed wedge represents a bond that is meant to be pictured pointing into, or behind, the plane of the page. Normal lines imply bonds that lie in the plane of the page.
This system takes a little bit of getting used to, but with practice your eye will learn to immediately ‘see’ the third dimension being depicted.
interactive 3D model of the bonding in methane
Exercise
Imagine that you could distinguish between the four hydrogens in a methane molecule, and labeled them Ha through Hd. In the images below, the exact same methane molecule is rotated and flipped in various positions. Draw the missing hydrogen atom labels. (It will be much easier to do this if you make a model.)
In the ethane molecule, the bonding picture according to valence orbital theory is very similar to that of methane. Both carbons are sp3-hybridized, meaning that both have four bonds arranged with tetrahedral geometry. The carbon-carbon bond, with a bond length of 1.54 Å, is formed by overlap of one sp3 orbital from each of the carbons, while the six carbon-hydrogen bonds are formed from overlaps between the remaining sp3 orbitals on the two carbons and the 1s orbitals of hydrogen atoms. All of these are sigma bonds.
Because they are formed from the end-on-end overlap of two orbitals, sigma bonds are free to rotate. This means, in the case of ethane molecule, that the two methyl (CH3) groups can be pictured as two wheels on a hub, each one able to rotate freely with respect to the other.
In another module we will learn more about the implications of rotational freedom in sigma bonds, when we discuss the ‘conformation’ of organic molecules.
The sp3 bonding picture is also used to described the bonding in amines, including ammonia, the simplest amine. Just like the carbon atom in methane, the central nitrogen in ammonia is sp3-hybridized. With nitrogen, however, there are five rather than four valence electrons to account for, meaning that three of the four hybrid orbitals are half-filled and available for bonding, while the fourth is fully occupied by a (non-bonding) pair of electrons.
The bonding arrangement here is also tetrahedral: the three N-H bonds of ammonia can be pictured as forming the base of a trigonal pyramid, with the fourth orbital, containing the lone pair, forming the top of the pyramid.
Recall from your study of VSEPR theory in General Chemistry that the lone pair, with its slightly greater repulsive effect, ‘pushes’ the three N-H sigma bonds away from the top of the pyramid, meaning that the H-N-H bond angles are slightly less than tetrahedral, at 107.3˚ rather than 109.5˚.
VSEPR theory also predicts, accurately, that a water molecule is ‘bent’ at an angle of approximately 104.5˚. It would seem logical, then, to describe the bonding in water as occurring through the overlap of sp3-hybrid orbitals on oxygen with 1sorbitals on the two hydrogen atoms. In this model, the two nonbonding lone pairs on oxygen would be located in sp3 orbitals.
Some experimental evidence, however, suggests that the bonding orbitals on the oxygen are actually unhybridized 2p orbitals rather than sp3 hybrids. Although this would seem to imply that the H-O-H bond angle should be 90˚ (remember that p orbitals are oriented perpendicular to one another), it appears that electrostatic repulsion has the effect of distorting this p-orbital angle to 104.5˚. Both the hybrid orbital and the nonhybrid orbital models present reasonable explanations for the observed bonding arrangement in water, so we will not concern ourselves any further with the distinction.
Exercise
Draw, in the same style as the figures above, an orbital picture for the bonding in methylamine.
Solution
Formation of pi bonds - sp2 and sp hybridization
The valence bond theory, along with the hybrid orbital concept, does a very good job of describing double-bonded compounds such as ethene. Three experimentally observable characteristics of the ethene molecule need to be accounted for by a bonding model:
1. Ethene is a planar (flat) molecule.
2. Bond angles in ethene are approximately 120o, and the carbon-carbon bond length is 1.34 Å, significantly shorter than the 1.54 Å single carbon-carbon bond in ethane.
3. There is a significant barrier to rotation about the carbon-carbon double bond.
Clearly, these characteristics are not consistent with an sp3 hybrid bonding picture for the two carbon atoms. Instead, the bonding in ethene is described by a model involving the participation of a different kind of hybrid orbital. Three atomic orbitals on each carbon – the 2s, 2px and 2py orbitals – combine to form three sp2 hybrids, leaving the 2pz orbital unhybridized.
The three sp2 hybrids are arranged with trigonal planar geometry, pointing to the three corners of an equilateral triangle, with angles of 120°between them. The unhybridized 2pz orbital is perpendicular to this plane (in the next several figures, sp2 orbitals and the sigma bonds to which they contribute are represented by lines and wedges; only the 2pz orbitals are shown in the 'space-filling' mode).
another view
The carbon-carbon double bond in ethene consists of one sigma bond, formed by the overlap of two sp2 orbitals, and a second bond, called a π (pi) bond, which is formed by the side-by-side overlap of the two unhybridized 2pz orbitals from each carbon.
another view
interactive model
The pi bond does not have symmetrical symmetry. Because they are the result of side-by-side overlap (rather then end-to-end overlap like a sigma bond), pi bonds are not free to rotate. If rotation about this bond were to occur, it would involve disrupting the side-by-side overlap between the two 2pz orbitals that make up the pi bond. The presence of the pi bond thus ‘locks’ the six atoms of ethene into the same plane. This argument extends to larger alkene groups: in each case, the six atoms of the group form a single plane.
Conversely, sigma bonds such as the carbon-carbon single bond in ethane (CH3CH3) exhibit free rotation, and can assume many different conformations, or shapes.
Example
Circle the six atoms in the molecule below that are ‘locked’ into the same plane.
Exercise
What kinds of orbitals are overlapping in bonds a-d indicated below?
A similar picture can be drawn for the bonding in carbonyl groups, such as formaldehyde. In this molecule, the carbon is sp2-hybridized, and we will assume that the oxygen atom is also sp2 hybridized. The carbon has three sigma bonds: two are formed by overlap between two of its sp2 orbitals with the 1sorbital from each of the hydrogens, and the third sigma bond is formed by overlap between the remaining carbon sp2 orbital and an sp2 orbital on the oxygen. The two lone pairs on oxygen occupy its other two sp2 orbitals.
interactive 3D model
The pi bond is formed by side-by-side overlap of the unhybridized 2pz orbitals on the carbon and the oxygen. Just like in alkenes, the 2pz orbitals that form the pi bond are perpendicular to the plane formed by the sigma bonds.
Finally, the hybrid orbital concept applies well to triple-bonded groups, such as alkynes and nitriles. Consider, for example, the structure of ethyne (common name acetylene), the simplest alkyne.
This molecule is linear: all four atoms lie in a straight line. The carbon-carbon triple bond is only 1.20Å long. In the hybrid orbital picture of acetylene, both carbons are sp-hybridized. In an sp-hybridized carbon, the 2s orbital combines with the 2px orbital to form two sp hybrid orbitals that are oriented at an angle of 180°with respect to each other (eg. along the x axis). The 2py and 2pz orbitals remain unhybridized, and are oriented perpendicularly along the y and z axes, respectively.
another view
The C-C sigma bond, then, is formed by the overlap of one sp orbital from each of the carbons, while the two C-H sigma bonds are formed by the overlap of the second sp orbital on each carbon with a 1s orbital on a hydrogen. Each carbon atom still has two half-filled 2py and 2pz orbitals, which are perpendicular both to each other and to the line formed by the sigma bonds. These two perpendicular pairs of p orbitals form two pi bonds between the carbons, resulting in a triple bond overall (one sigma bond plus two pi bonds).
another view
The hybrid orbital concept nicely explains another experimental observation: single bonds adjacent to double and triple bonds are progressively shorter and stronger than ‘normal’ single bonds, such as the one in a simple alkane. The carbon-carbon bond in ethane (structure A below) results from the overlap of two sp3 orbitals.
In alkene B, however, the carbon-carbon single bond is the result of overlap between an sp2 orbital and an sp3 orbital, while in alkyne C the carbon-carbon single bond is the result of overlap between an sp orbital and an sp3 orbital. These are all single bonds, but the bond in molecule C is shorter and stronger than the one in B, which is in turn shorter and stronger than the one in A.
The explanation here is relatively straightforward. An sp orbital is composed of one s orbital and one p orbital, and thus it has 50% s character and 50% p character. sp2 orbitals, by comparison, have 33% s character and 67% p character, while sp3 orbitals have 25% s character and 75% p character. Because of their spherical shape, 2s orbitals are smaller, and hold electrons closer and ‘tighter’ to the nucleus, compared to 2p orbitals. Consequently, bonds involving sp + sp3 overlap (as in alkyne C) are shorter and stronger than bonds involving sp2 + sp3 overlap (as in alkene B). Bonds involving sp3-sp3overlap (as in alkane A) are the longest and weakest of the group, because of the 75% ‘p’ character of the hybrids.
Example
For each of the bonds indicated by arrows a-e in the figures below, describe the bonding picture by completing this sentence:
"The sigma bond indicated by this arrow is formed by the overlap of an ________ orbital of a _________atom and an ________orbital of a _______atom."
Solution
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Within a group of the periodic table, bond lengths tend to increase with increasing atomic number $Z$. Consider the Group 17 elements:
\begin{align*}& F_2 \;\;\;\; d=141.7 \;pm\ & Cl_2 \;\;\;\; d=199.1 \, pm \ & Br_2 \;\;\;\; d=228.6 \, pm\ & I_2 \;\;\;\; d=266.9 \, pm\end{align*}
which corresponds to an increased valence shell size, hence increased electron-electron repulsion. An important result from experiment, which has been corroborated by theory, is that bond lengths tend not to vary much from molecule to molecule. Thus, a $CH$ bond will have roughly the same value in methane, $CH_4$ as it will in aspirin, $C_9 H_8 O_4$.
Bond dissociation energies. $\Delta E_d$ measured in $kJ/mol$, measure the energy required to break a mole of a particular kind of bond. A similar periodic trend exists for bond dissociation energies. Consider the hydrogen halides:
\begin{align*} & HF \;\;\;\; \Delta E_d =565 \ kJ/mol \;\;\;\; d= 0.926 \ \, pm\ & HCl \;\;\;\; \Delta E_d =429 \ kJ/mol \;\;\;\; d= 128.4 \ \, pm\ & HBr \;\;\;\; \Delta E_d =363 \ kJ/mol \;\;\;\; d= 142.4 \ \, pm\ & HI \;\;\;\; \Delta E_d =295 \ kJ/mol \;\;\;\; d= 162.0 \ \, pm \end{align*}
Thus, as bond lengths increase with increasing $Z$, there is a corresponding decrease in the bond dissociation energy.
$CC$ bonds are an exception to the the rule of constancy of bond lengths across different molecules. Because $CC$ bonds can be single, double, or triple bonds, some differences can occur. For example, consider the $CC$ bond in the molecules ethane $(C_2 H_6)$, ethylene $(C_2 H_4)$ and acetylene $(C_2 H_2)$:
\begin{align*} & C_2 H_6 \;\;\;\; (single)\;\;\;\; d=1.536 \ \stackrel{\circ}{A}\;\;\;\; \Delta E_d=345 \ kJ/mol\ & C_2 H_4 \;\;\;\; (double)\;\;\;\; d=133.7 \, pm\;\;\;\; \Delta E_d=612 \ kJ/mol\ & C_2 H_2 \;\;\;\; (triple)\;\;\;\; d=126.4 \, pm\;\;\;\; \Delta E_d=809 \ kJ/mol\end{align*}
The greater the bond order, i.e., number of shared electron pairs, the greater the dissociation energy. The same will be true for any kind of bond that can come in such different flavors'', e.g., $NN$ bonds, $OO$ bonds, $NO$ bonds, $CO$ bonds, etc.
The two idealized extremes of chemical bonding:
1. Ionic bonding—in which one or more electrons are transferred completely from one atom to another, and the resulting ions are held together by purely electrostatic forces—and
2. Covalent bonding, in which electrons are shared equally between two atoms.
Most compounds, however, have polar covalent bonds, which means that electrons are shared unequally between the bonded atoms. Figure $1$ compares the electron distribution in a polar covalent bond with those in an ideally covalent and an ideally ionic bond. Recall that a lowercase Greek delta ($δ$) is used to indicate that a bonded atom possesses a partial positive charge, indicated by $δ^+$, or a partial negative charge, indicated by $δ^−$, and a bond between two atoms that possess partial charges is a polar bond.
Most real chemical bonds in nature are neither truly covalent nor truly ionic. Only homonuclear bonds are truly covalent, and nearly perfect ionic bonds can form between group I and group VII elements, for example, KF. Generally, however, bonds are partially covalent and partially ionic, meaning that there is partial transfer of electrons between atoms and partial sharing of electrons. To quantify how much ionic character (and how much covalent character) a bond possesses, electronegativity differences between the atoms in the bond can be used. In 1936, Linus Pauling came up a method for estimating atomic electronegativities forms the basis of our understanding of electronegativity today.
Only homonuclear bonds are truly covalent, or as covalent as a bond can get. As discussed in Section 12.2, a quantum-mechanical treatment has shown that the two ionic structures (e.g., $H^+H^−$ and $H^−H^+$ for $H_2$) also contribute via a resonance with the covalent structure $H−H$. However, this is often to a small extent, for example in $H_2$, each ionic structure contributes only ~2% to the bonding of the molecule.
Pauling Electronegativity
Linus Pauling described electronegativity as “the power of an atom in a molecule to attract electrons to itself.” Basically, the electronegativity of an atom is a relative value of that atom's ability to attract election density toward itself when it bonds to another atom. The higher the electronegative of an element, the more that atom will attempt to pull electrons towards itself and away from any atom it bonds to. The main properties of an atom dictate it's electronegativity are it's atomic number as well as its atomic radius. The trend for electronegativity is to increase as you move from left to right and bottom to top across the periodic table. This means that the most electronegative atom is Fluorine and the least electronegative is Francium.
Recall the Mulliken's method was based on the arithmetic average of the first ionization energy $IE_1$ and the electron affinity $EA$. Both of these energies are properties of individual atoms, hence this method is appealing in its simplicity. However, there is no information about bonding in the Mulliken method. Pauling's method includes such information, and hence is a more effective approach.
To see how the Pauling method works, consider a diatomic $AB$, which is polar covalent. Let $\Delta E_{AA}$ and $\Delta E_{BB}$ be the dissociation energies of the diatomics $A_2$ and $B_2$, respectively. Since $A_2$ and $B_2$ are purely covalent bonds, these two dissociation energies can be used to estimate the pure covalent contribution to the bond $AB$. Pauling proposed the geometric mean of $\Delta E_{AA}$ and $\Delta E_{BB}$, this being more sensitive to large differences between these energies than the arithmetic average:
$pure \ covalent \ contribution=\sqrt{\Delta E_{AA} \Delta E_{BB}}$
If $\Delta E_{AB}$ is the true bond dissociation energy, then the difference
$\Delta E_{AB}-\sqrt{\Delta E_{AA} \Delta E_{BB}}$
is a measure of the ionic contribution. Let us define this difference to be $\Delta$:
$\Delta =\Delta E_{AB}-\sqrt{\Delta E_{AA} \Delta E_{BB}}$
Then Pauling defined the electronegativity difference $\chi_A -\chi_B$ between atoms $A$ and $B$ to be
$\chi_A -\chi_B = 0.102 \sqrt{\Delta}$
where $\Delta$ is measured in $kJ/mol$, and the constant $0.102$ has units $mol^{1/2} /kJ^{1/2}$, so that the electronegativity difference is dimensionless. Thus, with some extra input information, he was able to generate a table of atomic electronegativities that are still used today and is Tablated in Table A2 (Figure $2$).
To use the electronegativities to estimate degree of ionic character, simply compute the absolute value of the difference for the two atoms in the bond. As an example, consider again the hydrogen halides:
\begin{align*} & HF \;\;\;\; |\chi_F -\chi_H|=1.78\ & HCl \;\;\;\; |\chi_{Cl} -\chi_H|=0.96\ & HBr \;\;\;\; |\chi_{Br} -\chi_H|=0.76\ & HI \;\;\;\; |\chi_I -\chi_H|=0.46\end{align*}
As the electronegativity difference decreases, so does the ionic character of the bond. Hence its covalent character increases. It is possible to predict whether a given bond will be non-polar, polar covalent, or ionic based on the electronegativity difference, since the greater the difference, the more polar the bond (Figure $3$)
Electronegativity difference, ΔχP Bond
Δχ < 0.4 covalent
0.4 < Δχ < 1.7 polar covalent
Δχ > 1.7 ionic
As the electronegativity difference decreases, so does the ionic character of the bond. Hence its covalent character increases.
Dipole moment
In a nearly perfect ionic bond, such as $KF$, where electron transfer is almost complete, representing the molecule as
$K^+ F^-$
is a very good approximation, since the charge on the potassium will be approximately $1e$ and the charge on the fluorine will be approximately $-1e$. For a polar covalent bond, such as $HF$, in which only partial charge transfer occurs, a more accurate representation would be
$H^{+\delta}F^{-\delta}$
where $\delta$, expressed in units of $e$, is known as a partial charge. It suggests that a fraction of an electron is transferred, although the reality is that there is simply a little more electron density on the more electronegative atom and a little less on the electropositive atom.
How much charge is actually transferred can be quantified by studying the electric dipole moment of the bond, which is a quantity that can be measured experimentally. The electric dipole moment for a diatomic with charges $Q_1 =Q=\delta e$ and $Q_2 =-Q =-\delta e$ on atoms 1 and 2, respectively, is
\begin{align*}\mu &= Q_1 r_1 +Q_2 r_2\ &= Qr_1 -Qr_2\ &=Q(r_1 -r_2)\end{align*}
Hence, the magnitude of the dipole moment is
$\mu = |\mu|=Q|r_1 -r_2|=QR \label{Dipole}$
where $R$ is the bond length. As an example, consider $HF$, which has a partial charge on $H$ of $0.41 \;e$, which means $\delta =0.41$, and a bond length of $0.926 \ \stackrel{\circ}{A}$. Thus, the magnitude of the dipole moment is
$|\mu|=0.41(1.602 \times 10^{-19}C)(0.926 \times 10^{-10}m)=6.08\times 10^{-30}C\cdot m$
Thus, the units of the dipole moment are Coulomb-meters. However, as this example makes clear, this is a very large unit and awkward to work with for molecules. A more convenient unit is the Debye $(D)$, defined to be
$1\;D=3.336\times 10^{-30}\; \text{Coulomb} \cdot \text{meters}$
Historically, the Debye was defined in terms of the dipole moment resulting from two equal charges of opposite sign and separated by 1 Ångstrom ($10^{-10}\; m$) as 4.801 D from Equation $\ref{Dipole}$. This value arises from
$\dfrac{ (1.602 \times 10^{-19} ) (1 \times 10^{-10}) }{3.336 \times 10^{-30}} \nonumber$
where
$D = 3.336 \times 10^{-30}\; C\, m \nonumber$
or
$1\; C\, m = 2.9979 \times 10^{29}\; D \nonumber$
Thus, for a diatomic with partial charges $+\delta$ and $-\delta$, the dipole moment in $D$ is given by
$\mu (D)=\dfrac{\delta *R(\stackrel{\circ}{A})}{0.2082 \ \stackrel{\circ}{A}D^{-1}}$
and the percent ionic character is defined in terms of the partial charge $\delta$ by
$percent \ ionic \ character=100\% *\delta \label{Ea1}$
Typical dipole moments for simple diatomic molecules are in the range of 0 to 11 D (Table $1$). Equation $\ref{Ea1}$ can be expressed differently in terms of the expected dipole assuming a full charge separation ($\mu_{ionic}$) compared to the experimental dipole moment ($\mu_{exp}$)
$percent \ ionic \ character=100\% *\dfrac{\mu_{exp}}{\mu_{ionic}} \label{Ea2}$
Example $1$: KBr
Calculate a theoretical dipole moment for the KBr molecule, assuming opposite charges of one fundamental unit located at each nucleus, and hence the percentage ionic character of KBr.
Solution
From Table $1$, the observed dipole moment of KBr is given as 10.41 D, (3.473 x 10-29 Coulomb-meters), which being close to the upper level of 11 indicates that it is a highly polar molecule. The interatomic distance between K+ and Br- is 282 pm. From this it is possible to calculate a theoretical dipole moment for the KBr molecule, assuming opposite charges of one fundamental unit located at each nucleus, and hence the percentage ionic character of KBr.
Dipole moment
$μ = q \times e \times d \, (\text{in Coulomb-meters})$
• q = 1 for complete separation of unit charge
• e = 1.602 x 10-19 C
• d = 2.82 x 10-10 m for KBr (282 pm)
Hence calculated ionic dipole moment is
$μ_{KBr}= (1) (1.602 \times 10^{-19})( 2.82 \times 10^{-10}) = 4.518 \times 10^{-29}\; Cm = 13.54\; D \nonumber$
The observed dipole moment is
$μ_{KBr} = 3.473 \times 10^{-29}\; Cm = 10.41\; D \nonumber$
the % ionic character from Equation $\ref{Ea2}$ is
$KBr = \dfrac{3.473 \times 10^{-29}}{4.518 \times 10^{-29}} \times 100\%= \dfrac{10.41\, D}{13.54\;D} \times 100\% = 76.87\% \nonumber$
and the % covalent character is therefore about 23% (100% - 77%).
Given the observed dipole moment is 10.41 D (3.473 x 10-29) it is possible to estimate the charge distribution from the same equation by now solving for q.
Dipole moment μ = q * e * d Coulomb metre, but since q is no longer 1 we can substitute in values for μ and d to obtain an estimate for it.
q = μ /(e * d) = 3.473 x 10-29 / (1.602 x 10-19 * 2.82 x 10-10)
thus q = 3.473 x 10-29 / (4.518 x 10-29) = 0.77 and the δ- and δ+ are -0.8 and +0.8 respectively.
Example $2$: NaCl
In the gas phase, NaCl has a dipole moment of 9.001 D and an Na–Cl distance of 236.1 pm. Calculate the percent ionic character in NaCl.
Given: chemical species, dipole moment, and internuclear distance
Asked for: percent ionic character
Strategy:
A Compute the charge on each atom using the information given and Equation 8.4.2.
B Find the percent ionic character from the ratio of the actual charge to the charge of a single electron.
Solution:
A The charge on each atom is given by
$Q=\dfrac{\mu }{r} =9.001\;\cancel{D}\left ( \dfrac{3.3356\times 10^{-30}\; C\cdot \cancel{m}}{1\; \cancel{D}} \right )\left ( \dfrac{1}{236.1\; \cancel{pm}} \right )\left ( \dfrac{1\; \cancel{pm}}{10^{-12\;} \cancel{m}} \right )=1.272\times 10^{-19}\;C$
Thus NaCl behaves as if it had charges of 1.272 × 10−19 C on each atom separated by 236.1 pm.
B The percent ionic character is given by the ratio of the actual charge to the charge of a single electron (the charge expected for the complete transfer of one electron):
$\% \; ionic\; character=\left ( \dfrac{1.272\times 10^{-19}\; \cancel{C}}{1.6022\times 10^{-19}\; \cancel{C}} \right )\left ( 100 \right )=79.39\%\simeq 79\%$
Exercise $2$: AgCl
In the gas phase, silver chloride (AgCl) has a dipole moment of 6.08 D and an Ag–Cl distance of 228.1 pm. What is the percent ionic character in silver chloride?
Answer
55.5%
Table $1$: Bond characteristics of select diatomics
diatomic Δ χ %ionic bond dist (pm) $μ_{exp}$ (D) $μ_{ionic}$ (D)
Cl2 0.0 0.0 200 0.00 9.60
IBr 0.3 5.9 247 0.70 11.86
HI 0.4 5.7 161 0.44 7.73
ICl 0.5 5.4 232 0.60 11.14
HBr 0.7 12.1 141 0.82 6.77
HCl 0.9 17.7 127 1.08 6.10
ClF 1.0 11.2 163 0.88 7.83
BrF 1.2 15.1 178 1.29 8.55
LiI 1.5 65.0 238 7.43 11.43
HF 1.9 41.2 92 1.82 4.42
LiBr 1.8 69.8 217 7.27 10.42
KI 1.7 73.7 305 10.80 14.65
LiCl 2.0 73.5 202 7.13 9.70
KBr 2.0 76.9 282 10.41 13.54
NaCl 2.1 79.4 236 9.00 11.33
KCl 2.2 80.1 267 10.27 12.82
CsCl 2.3 74.6 291 10.42 13.97
LiF 3.0 86.7 152 6.33 7.30
KF 3.2 82.5 217 8.60 10.42
CsF 3.3 64.4 255 7.88 12.25
Pauling proposed an empirical relationship (instead of the defintion in Equation $\ref{Ea2}$) which relates the percent ionic character in a bond to the electronegativity difference.
$percent \ ionic \ character= 100\% \left( 1 - e^{(\Delta χ/2)^2} \right)$
This is shown as the curve in Figure $4$ and is compared to the values for some diatomic molecules calculated from observed and calculated dipole moments.
As an example, consider $HF$ again, for which $\delta = 0.41$. The bond length is $R=0.926 \ \stackrel{\circ}{A}$. Thus, its dipole moment will be
$\mu (D)=\dfrac{0.41*0.926 \stackrel{\circ}{A}}{0.2082 \ \stackrel{\circ}{A}D^{-1}}=1.82D$
and its percent ionic character is $41\%$.
Example $3$: Bond Polarity
Without consulting the table of electronegativities (use the periodic table), arrange the following bonds in order of decreasing polarity:
B—Cl, Ba—Cl, Be—Cl, Br—Cl, Cl—Cl.
Solution
We first need to arrange the elements in order of increasing electronegativity. Since the electronegativity increases in going up a column of the periodic table, we have the following relationships:
Ba < Be and Br < Cl
Also since the electronegativity increases across the periodic table, we have
Be < B
Since B is a group III element on the borderline between metals and non-metals, we easily guess that
B < Br
which gives us the complete order
Ba < Be < B < Br < Cl
Among the bonds listed, therefore, the Ba—Cl bond corresponds to the largest difference in electronegativity, i.e., to the most nearly ionic bond. The order of bond polarity is thus
Ba—Cl > Br—Cl > B—Cl > Br—Cl > Cl—Cl
where the final bond, Cl—Cl,is, of course, purely covalent.
Contributors
• Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.
• The Department of Chemistry, University of the West Indies)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/12%3A_The_Chemical_Bond/12.04%3A_Electronegativity_and_Dipole_Moment.txt
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None of the approaches we have described so far can adequately explain why some compounds are colored and others are not, why some substances with unpaired electrons are stable, and why others are effective semiconductors. These approaches also cannot describe the nature of resonance. Such limitations led to the development of a new approach to bonding in which electrons are not viewed as being localized between the nuclei of bonded atoms but are instead delocalized throughout the entire molecule. Just as with the valence bond theory, the approach we are about to discuss is based on a quantum mechanical model.
Previously, we described the electrons in isolated atoms as having certain spatial distributions, called orbitals, each with a particular orbital energy. Just as the positions and energies of electrons in atoms can be described in terms of atomic orbitals (AOs), the positions and energies of electrons in molecules can be described in terms of molecular orbitals (MOs) A particular spatial distribution of electrons in a molecule that is associated with a particular orbital energy.—a spatial distribution of electrons in a molecule that is associated with a particular orbital energy. As the name suggests, molecular orbitals are not localized on a single atom but extend over the entire molecule. Consequently, the molecular orbital approach, called molecular orbital theory is a delocalized approach to bonding.
Although the molecular orbital theory is computationally demanding, the principles on which it is based are similar to those we used to determine electron configurations for atoms. The key difference is that in molecular orbitals, the electrons are allowed to interact with more than one atomic nucleus at a time. Just as with atomic orbitals, we create an energy-level diagram by listing the molecular orbitals in order of increasing energy. We then fill the orbitals with the required number of valence electrons according to the Pauli principle. This means that each molecular orbital can accommodate a maximum of two electrons with opposite spins.
The Hydrogen Molecule-Ion
Molecular orbital theory is a conceptual extension of the orbital model, which was so successfully applied to atomic structure. As was once playfully remarked, "a molecule is nothing more than an atom with more nuclei." This may be overly simplistic, but we do attempt, as far as possible, to exploit analogies with atomic structure. Our understanding of atomic orbitals began with the exact solutions of a prototype problem – the hydrogen atom. We will begin our study of homonuclear diatomic molecules beginning with another exactly solvable prototype, the hydrogen molecule-ion $H_{2}^{+}$. This species actually has a transient existence in electrical discharges through hydrogen gas and has been detected by mass spectrometry. It also has been detected in outer space. The Schrödinger equation for H$_{2}^{+}$ can be solved exactly within the Born-Oppenheimer approximation. For fixed internuclear distance R, this reduces to a problem of one electron in the field of two protons, designated A and B. We can write$\left\{-\frac{1}{2}\nabla^2\frac{1}{r_A}-\frac{1}{r_B}+\frac{1}{R} \right\} \psi(r)=E\psi(r) \tag{1}$where rA and rB are the distances from the electron to protons A and B, respectively. This equation was solved by Burrau (1927), after separating the variables in prolate spheroidal coordinates.
Molecular Orbitals Involving Only ns Atomic Orbitals
We begin our discussion of molecular orbitals with the simplest molecule, H2, formed from two isolated hydrogen atoms, each with a $1s^1$ electron configuration. As discussed before, electrons can behave like waves. In the molecular orbital approach, the overlapping atomic orbitals are described by mathematical equations called wave functions. The 1s atomic orbitals on the two hydrogen atoms interact to form two new molecular orbitals, one produced by taking the sum of the two H 1s wave functions, and the other produced by taking their difference:
$\begin{matrix} MO(1)= & AO(atom\; A) & +& AO(atomB) \ MO(1)= & AO(atom\; A) & -&AO(atomB) \end{matrix} \label{Eq1}$
The molecular orbitals created from the sum and the difference of two wavefunctions (atomic orbitals) from Equation $\ref{Eq1}$ are called Linear Combinations of Atomic Orbitals (LCAOs). A molecule must have as many molecular orbitals as there are atomic orbitals.
Adding two atomic orbitals corresponds to constructive interference between two waves, thus reinforcing their intensity; the internuclear electron probability density is increased. The molecular orbital corresponding to the sum of the two H 1s orbitals is called a σ1s combination (pronounced “sigma one ess”) (part (a) and part (b) in Figure $1$). In a sigma (σ) orbital, A bonding molecular orbital in which the electron density along the internuclear axis and between the nuclei has cylindrical symmetry, the electron density along the internuclear axis and between the nuclei has cylindrical symmetry; that is, all cross-sections perpendicular to the internuclear axis are circles. The subscript 1s denotes the atomic orbitals from which the molecular orbital was derived: The ≈ sign is used rather than an = sign because we are ignoring certain constants that are not important to our argument.
$\sigma _{1s} \approx 1s\left ( A \right ) + 1s\left ( B \right ) \label{Eq2}$
Conversely, subtracting one atomic orbital from another corresponds to destructive interference between two waves, which reduces their intensity and causes a decrease in the internuclear electron probability density (part (c) and part (d) in Figure $1$ ). The resulting pattern contains a node where the electron density is zero. The molecular orbital corresponding to the difference is called $\sigma _{1s}^{*}$ (“sigma one ess star”). In a sigma star (σ*) orbital An antibonding molecular orbital in which there is a region of zero electron probability (a nodal plane) perpendicular to the internuclear axis., there is a region of zero electron probability, a nodal plane, perpendicular to the internuclear axis:
$\sigma _{1s}^{\star } \approx 1s\left ( A \right ) - 1s\left ( B \right ) \label{Eq3}$
A molecule must have as many molecular orbitals as there are atomic orbitals.
The electron density in the σ1s molecular orbital is greatest between the two positively charged nuclei, and the resulting electron–nucleus electrostatic attractions reduce repulsions between the nuclei. Thus the σ1s orbital represents a bonding molecular orbital. A molecular orbital that forms when atomic orbitals or orbital lobes with the same sign interact to give increased electron probability between the nuclei due to constructive reinforcement of the wave functions. In contrast, electrons in the $\sigma _{1s}^{\star }$ orbital are generally found in the space outside the internuclear region. Because this allows the positively charged nuclei to repel one another, the $\sigma _{1s}^{\star }$ orbital is an antibonding molecular orbital (a molecular orbital that forms when atomic orbitals or orbital lobes of opposite sign interact to give decreased electron probability between the nuclei due to destructive reinforcement of the wave functions).
Antibonding orbitals contain a node perpendicular to the internuclear axis; bonding orbitals do not.
Energy-Level Diagrams
Because electrons in the σ1s orbital interact simultaneously with both nuclei, they have a lower energy than electrons that interact with only one nucleus. This means that the σ1s molecular orbital has a lower energy than either of the hydrogen 1s atomic orbitals. Conversely, electrons in the $\sigma _{1s}^{\star }$ orbital interact with only one hydrogen nucleus at a time. In addition, they are farther away from the nucleus than they were in the parent hydrogen 1s atomic orbitals. Consequently, the $\sigma _{1s}^{\star }$ molecular orbital has a higher energy than either of the hydrogen 1s atomic orbitals. The σ1s (bonding) molecular orbital is stabilized relative to the 1s atomic orbitals, and the $\sigma _{1s}^{\star }$ (antibonding) molecular orbital is destabilized. The relative energy levels of these orbitals are shown in the energy-level diagram (a schematic drawing that compares the energies of the molecular orbitals (bonding, antibonding, and nonbonding) with the energies of the parent atomic orbitals) in Figure $2$
A bonding molecular orbital is always lower in energy (more stable) than the component atomic orbitals, whereas an antibonding molecular orbital is always higher in energy (less stable).
To describe the bonding in a homonuclear diatomic molecule (a molecule that consists of two atoms of the same element) such as H2, we use molecular orbitals; that is, for a molecule in which two identical atoms interact, we insert the total number of valence electrons into the energy-level diagram (Figure $2$). We fill the orbitals according to the Pauli principle and Hund’s rule: each orbital can accommodate a maximum of two electrons with opposite spins, and the orbitals are filled in order of increasing energy. Because each H atom contributes one valence electron, the resulting two electrons are exactly enough to fill the σ1s bonding molecular orbital. The two electrons enter an orbital whose energy is lower than that of the parent atomic orbitals, so the H2 molecule is more stable than the two isolated hydrogen atoms. Thus molecular orbital theory correctly predicts that H2 is a stable molecule. Because bonds form when electrons are concentrated in the space between nuclei, this approach is also consistent with our earlier discussion of electron-pair bonds.
Bond Order in Molecular Orbital Theory
In the Lewis electron structures, the number of electron pairs holding two atoms together was called the bond order. In the molecular orbital approach, bond order One-half the net number of bonding electrons in a molecule. is defined as one-half the net number of bonding electrons:
$bond\; order=\dfrac{number\; of \; bonding\; electrons-number\; of \; antibonding\; electrons}{2} \label{Eq4}$
To calculate the bond order of H2, we see from Figure $2$ that the σ1s (bonding) molecular orbital contains two electrons, while the $\sigma _{1s}^{\star }$ (antibonding) molecular orbital is empty. The bond order of H2 is therefore
$\dfrac{2-0}{2}=1 \label{Eq5}$
This result corresponds to the single covalent bond predicted by Lewis dot symbols. Thus molecular orbital theory and the Lewis electron-pair approach agree that a single bond containing two electrons has a bond order of 1. Double and triple bonds contain four or six electrons, respectively, and correspond to bond orders of 2 and 3.
We can use energy-level diagrams such as the one in Figure $2$ to describe the bonding in other pairs of atoms and ions where n = 1, such as the H2+ ion, the He2+ ion, and the He2 molecule. Again, we fill the lowest-energy molecular orbitals first while being sure not to violate the Pauli principle or Hund’s rule.
Part (a) in Figure $3$ shows the energy-level diagram for the H2+ ion, which contains two protons and only one electron. The single electron occupies the σ1s bonding molecular orbital, giving a (σ1s)1 electron configuration. The number of electrons in an orbital is indicated by a superscript. In this case, the bond order is (1-0)/2=1/2 Because the bond order is greater than zero, the H2+ ion should be more stable than an isolated H atom and a proton. We can therefore use a molecular orbital energy-level diagram and the calculated bond order to predict the relative stability of species such as H2+. With a bond order of only 1/2 the bond in H2+ should be weaker than in the H2 molecule, and the H–H bond should be longer. As shown in Table $1$, these predictions agree with the experimental data.
Part (b) in Figure $3$ is the molecular orbital energy-level diagram for He2+. This ion has a total of three valence electrons. Because the first two electrons completely fill the σ1s molecular orbital, the Pauli principle states that the third electron must be in the $\sigma _{1s}^{\star}$ antibonding orbital, giving a $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{1}$ electron configuration. This electron configuration gives a bond order of (2-1)/2=1/2. As with H2+, the He2+ ion should be stable, but the He–He bond should be weaker and longer than in H2. In fact, the He2+ ion can be prepared, and its properties are consistent with our predictions (Table $1$).
Table $1$: Molecular Orbital Electron Configurations, Bond Orders, Bond Lengths, and Bond Energies for some Simple Homonuclear Diatomic Molecules and Ions
Molecule or Ion Electron Configuration Bond Order Bond Length (pm) Bond Energy (kJ/mol)
H2+ 1s)1 1/2 106 269
H2 1s)2 1 74 436
He2+ $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{1}$ 1/2 108 251
He2 $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{2}$ 0 not observed not observed
Finally, we examine the He2 molecule, formed from two He atoms with 1s2 electron configurations. Part (c) in Figure $3$ is the molecular orbital energy-level diagram for He2. With a total of four valence electrons, both the σ1s bonding and $\sigma _{1s}^{\star }$ antibonding orbitals must contain two electrons. This gives a $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{1}$ electron configuration, with a predicted bond order of (2 − 2) ÷ 2 = 0, which indicates that the He2 molecule has no net bond and is not a stable species. Experiments show that the He2 molecule is actually less stable than two isolated He atoms due to unfavorable electron–electron and nucleus–nucleus interactions.
In molecular orbital theory, electrons in antibonding orbitals effectively cancel the stabilization resulting from electrons in bonding orbitals. Consequently, any system that has equal numbers of bonding and antibonding electrons will have a bond order of 0, and it is predicted to be unstable and therefore not to exist in nature. In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory is able to accommodate systems with an odd number of electrons, such as the H2+ ion.
Note
In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory can accommodate systems with an odd number of electrons.
Example $1$
Use a molecular orbital energy-level diagram, such as those in Figure $3$, to predict the bond order in the He22+ ion. Is this a stable species?
Given: chemical species
Asked for: molecular orbital energy-level diagram, bond order, and stability
Strategy:
1. Combine the two He valence atomic orbitals to produce bonding and antibonding molecular orbitals. Draw the molecular orbital energy-level diagram for the system.
2. Determine the total number of valence electrons in the He22+ ion. Fill the molecular orbitals in the energy-level diagram beginning with the orbital with the lowest energy. Be sure to obey the Pauli principle and Hund’s rule while doing so.
3. Calculate the bond order and predict whether the species is stable.
Solution:
A Two He 1s atomic orbitals combine to give two molecular orbitals: a σ1s bonding orbital at lower energy than the atomic orbitals and a $\sigma _{1s}^{\star }$ antibonding orbital at higher energy. The bonding in any diatomic molecule with two He atoms can be described using the following molecular orbital diagram:
B The He22+ ion has only two valence electrons (two from each He atom minus two for the +2 charge). We can also view He22+ as being formed from two He+ ions, each of which has a single valence electron in the 1s atomic orbital. We can now fill the molecular orbital diagram:
The two electrons occupy the lowest-energy molecular orbital, which is the bonding (σ1s) orbital, giving a (σ1s)2 electron configuration. To avoid violating the Pauli principle, the electron spins must be paired. C So the bond order is
$\dfrac{2-0}{2} =1$
He22+ is therefore predicted to contain a single He–He bond. Thus it should be a stable species.
Exercise $1$
Use a molecular orbital energy-level diagram to predict the valence-electron configuration and bond order of the H22− ion. Is this a stable species?
Answer:
H22− has a valence electron configuration of $\left (\sigma _{1s} \right )^{2}\left (\sigma _{1s}^{\star } \right )^{2}$ with a bond order of 0. It is therefore predicted to be unstable.
So far, our discussion of molecular orbitals has been confined to the interaction of valence orbitals, which tend to lie farthest from the nucleus. When two atoms are close enough for their valence orbitals to overlap significantly, the filled inner electron shells are largely unperturbed; hence they do not need to be considered in a molecular orbital scheme. Also, when the inner orbitals are completely filled, they contain exactly enough electrons to completely fill both the bonding and antibonding molecular orbitals that arise from their interaction. Thus the interaction of filled shells always gives a bond order of 0, so filled shells are not a factor when predicting the stability of a species. This means that we can focus our attention on the molecular orbitals derived from valence atomic orbitals.
A molecular orbital diagram that can be applied to any homonuclear diatomic molecule with two identical alkali metal atoms (Li2 and Cs2, for example) is shown in part (a) in Figure $4$, where M represents the metal atom. Only two energy levels are important for describing the valence electron molecular orbitals of these species: a σns bonding molecular orbital and a σ*ns antibonding molecular orbital. Because each alkali metal (M) has an ns1 valence electron configuration, the M2 molecule has two valence electrons that fill the σns bonding orbital. As a result, a bond order of 1 is predicted for all homonuclear diatomic species formed from the alkali metals (Li2, Na2, K2, Rb2, and Cs2). The general features of these M2 diagrams are identical to the diagram for the H2 molecule in Figure $2$. Experimentally, all are found to be stable in the gas phase, and some are even stable in solution.
Similarly, the molecular orbital diagrams for homonuclear diatomic compounds of the alkaline earth metals (such as Be2), in which each metal atom has an ns2 valence electron configuration, resemble the diagram for the He2 molecule in part (c) in Figure $3$ As shown in part (b) in Figure $4$, this is indeed the case. All the homonuclear alkaline earth diatomic molecules have four valence electrons, which fill both the σns bonding orbital and the σns* antibonding orbital and give a bond order of 0. Thus Be2, Mg2, Ca2, Sr2, and Ba2 are all expected to be unstable, in agreement with experimental data.In the solid state, however, all the alkali metals and the alkaline earth metals exist as extended lattices held together by metallic bonding. At low temperatures, $Be_2$ is stable.
Example $2$
Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Na2 ion.
Given: chemical species
Asked for: molecular orbital energy-level diagram, valence electron configuration, bond order, and stability
Strategy:
1. Combine the two sodium valence atomic orbitals to produce bonding and antibonding molecular orbitals. Draw the molecular orbital energy-level diagram for this system.
2. Determine the total number of valence electrons in the Na2 ion. Fill the molecular orbitals in the energy-level diagram beginning with the orbital with the lowest energy. Be sure to obey the Pauli principle and Hund’s rule while doing so.
3. Calculate the bond order and predict whether the species is stable.
Solution:
A Because sodium has a [Ne]3s1 electron configuration, the molecular orbital energy-level diagram is qualitatively identical to the diagram for the interaction of two 1s atomic orbitals.
B The Na2 ion has a total of three valence electrons (one from each Na atom and one for the negative charge), resulting in a filled σ3s molecular orbital, a half-filled σ3s* and a $\left ( \sigma _{3s} \right )^{2}\left ( \sigma _{3s}^{\star } \right )^{1}$ electron configuration.
C The bond order is (2-1)÷2=1/2 With a fractional bond order, we predict that the Na2 ion exists but is highly reactive.
Exercise $2$
Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Ca2+ ion.
Answer: Ca2+ has a $\left ( \sigma _{4s} \right )^{2}\left ( \sigma _{4s}^{\star } \right )^{1}$ electron configurations and a bond order of 1/2 and should exist.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/12%3A_The_Chemical_Bond/12.05%3A_Molecular_Orbital_Theory.txt
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For almost every covalent molecule that exists, we can now draw the Lewis structure, predict the electron-pair geometry, predict the molecular geometry, and come close to predicting bond angles. However, one of the most important molecules we know, the oxygen molecule O2, presents a problem with respect to its Lewis structure. We would write the following Lewis structure for O2:
This electronic structure adheres to all the rules governing Lewis theory. There is an O=O double bond, and each oxygen atom has eight electrons around it. However, this picture is at odds with the magnetic behavior of oxygen. By itself, O2 is not magnetic, but it is attracted to magnetic fields. Thus, when we pour liquid oxygen past a strong magnet, it collects between the poles of the magnet and defies gravity. Such attraction to a magnetic field is called paramagnetism, and it arises in molecules that have unpaired electrons. And yet, the Lewis structure of O2 indicates that all electrons are paired. How do we account for this discrepancy?
Magnetic susceptibility measures the force experienced by a substance in a magnetic field. When we compare the weight of a sample to the weight measured in a magnetic field (Figure $1$), paramagnetic samples that are attracted to the magnet will appear heavier because of the force exerted by the magnetic field. We can calculate the number of unpaired electrons based on the increase in weight.
Experiments show that each O2 molecule has two unpaired electrons. The Lewis-structure model does not predict the presence of these two unpaired electrons. Unlike oxygen, the apparent weight of most molecules decreases slightly in the presence of an inhomogeneous magnetic field. Materials in which all of the electrons are paired are diamagnetic and weakly repel a magnetic field. Paramagnetic and diamagnetic materials do not act as permanent magnets. Only in the presence of an applied magnetic field do they demonstrate attraction or repulsion.
Video $1$: Water, like most molecules, contains all paired electrons. Living things contain a large percentage of water, so they demonstrate diamagnetic behavior. If you place a frog near a sufficiently large magnet, it will levitate. You can see videos of diamagnetic floating frogs, strawberries, and more (https://www.youtube.com/watch?v=A1vyB-O5i6E)
Molecular orbital theory (MO theory) provides an explanation of chemical bonding that accounts for the paramagnetism of the oxygen molecule. It also explains the bonding in a number of other molecules, such as violations of the octet rule and more molecules with more complicated bonding (beyond the scope of this text) that are difficult to describe with Lewis structures. Additionally, it provides a model for describing the energies of electrons in a molecule and the probable location of these electrons. Unlike valence bond theory, which uses hybrid orbitals that are assigned to one specific atom, MO theory uses the combination of atomic orbitals to yield molecular orbitals that are delocalized over the entire molecule rather than being localized on its constituent atoms. MO theory also helps us understand why some substances are electrical conductors, others are semiconductors, and still others are insulators. Table $1$ summarizes the main points of the two complementary bonding theories. Both theories provide different, useful ways of describing molecular structure.
Table $1$ : Comparison of Bonding Theories
Valence Bond Theory Molecular Orbital Theory
considers bonds as localized between one pair of atoms considers electrons delocalized throughout the entire molecule
creates bonds from overlap of atomic orbitals (s, p, d…) and hybrid orbitals (sp, sp2, sp3…) combines atomic orbitals to form molecular orbitals (σ, σ*, π, π*)
forms σ or π bonds creates bonding and antibonding interactions based on which orbitals are filled
predicts molecular shape based on the number of regions of electron density predicts the arrangement of electrons in molecules
needs multiple structures to describe resonance
Molecular Orbitals Formed from ns and np Atomic Orbitals
Atomic orbitals other than ns orbitals can also interact to form molecular orbitals. Because individual p, d, and f orbitals are not spherically symmetrical, however, we need to define a coordinate system so we know which lobes are interacting in three-dimensional space. Recall that for each np subshell, for example, there are npx, npy, and npz orbitals. All have the same energy and are therefore degenerate, but they have different spatial orientations.
$\sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right ) \label{Eq6}$
Just as with ns orbitals, we can form molecular orbitals from np orbitals by taking their mathematical sum and difference. When two positive lobes with the appropriate spatial orientation overlap, as illustrated for two npz atomic orbitals in part (a) in Figure $2$, it is the mathematical difference of their wave functions that results in constructive interference, which in turn increases the electron probability density between the two atoms. The difference therefore corresponds to a molecular orbital called a $\sigma _{np_{z}}$ bonding molecular orbital because, just as with the σ orbitals discussed previously, it is symmetrical about the internuclear axis (in this case, the z-axis):
$\sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right ) \label{Eq6a}$
The other possible combination of the two npz orbitals is the mathematical sum:
$\sigma _{np_{z}}=np_{z}\left ( A \right )+np_{z}\left ( B \right ) \label{Eq7}$
In this combination, shown in part (b) in Figure $2$, the positive lobe of one npz atomic orbital overlaps the negative lobe of the other, leading to destructive interference of the two waves and creating a node between the two atoms. Hence this is an antibonding molecular orbital. Because it, too, is symmetrical about the internuclear axis, this molecular orbital is called a $\sigma _{np_{z}}=np_{z}\left ( A \right )-np_{z}\left ( B \right )$ antibonding molecular orbital. Whenever orbitals combine, the bonding combination is always lower in energy (more stable) than the atomic orbitals from which it was derived, and the antibonding combination is higher in energy (less stable).
Overlap of atomic orbital lobes with the same sign produces a bonding molecular orbital, regardless of whether it corresponds to the sum or the difference of the atomic orbitals.
The remaining p orbitals on each of the two atoms, npx and npy, do not point directly toward each other. Instead, they are perpendicular to the internuclear axis. If we arbitrarily label the axes as shown in Figure $3$, we see that we have two pairs of np orbitals: the two npx orbitals lying in the plane of the page, and two npy orbitals perpendicular to the plane. Although these two pairs are equivalent in energy, the npx orbital on one atom can interact with only the npx orbital on the other, and the npy orbital on one atom can interact with only the npy on the other. These interactions are side-to-side rather than the head-to-head interactions characteristic of σ orbitals. Each pair of overlapping atomic orbitals again forms two molecular orbitals: one corresponds to the arithmetic sum of the two atomic orbitals and one to the difference. The sum of these side-to-side interactions increases the electron probability in the region above and below a line connecting the nuclei, so it is a bonding molecular orbital that is called a pi (π) orbital (a bonding molecular orbital formed from the side-to-side interactions of two or more parallel np atomic orbitals). The difference results in the overlap of orbital lobes with opposite signs, which produces a nodal plane perpendicular to the internuclear axis; hence it is an antibonding molecular orbital, called a pi star (π*) orbital An antibonding molecular orbital formed from the difference of the side-to-side interactions of two or more parallel np atomic orbitals, creating a nodal plane perpendicular to the internuclear axis..
$\pi _{np_{x}}=np_{x}\left ( A \right )+np_{x}\left ( B \right ) \label{Eq8}$
$\pi ^{\star }_{np_{x}}=np_{x}\left ( A \right )-np_{x}\left ( B \right ) \label{Eq9}$
The two npy orbitals can also combine using side-to-side interactions to produce a bonding $\pi _{np_{y}}$ molecular orbital and an antibonding $\pi _{np_{y}}^{\star }$ molecular orbital. Because the npx and npy atomic orbitals interact in the same way (side-to-side) and have the same energy, the $\pi _{np_{x}}$ and $\pi _{np_{y}}$molecular orbitals are a degenerate pair, as are the $\pi _{np_{x}}^{\star }$ and $\pi _{np_{y}}^{\star }$ molecular orbitals.
Figure $4$ is an energy-level diagram that can be applied to two identical interacting atoms that have three np atomic orbitals each. There are six degenerate p atomic orbitals (three from each atom) that combine to form six molecular orbitals, three bonding and three antibonding. The bonding molecular orbitals are lower in energy than the atomic orbitals because of the increased stability associated with the formation of a bond. Conversely, the antibonding molecular orbitals are higher in energy, as shown. The energy difference between the σ and σ* molecular orbitals is significantly greater than the difference between the two π and π* sets. The reason for this is that the atomic orbital overlap and thus the strength of the interaction are greater for a σ bond than a π bond, which means that the σ molecular orbital is more stable (lower in energy) than the π molecular orbitals.
Although many combinations of atomic orbitals form molecular orbitals, we will discuss only one other interaction: an ns atomic orbital on one atom with an npz atomic orbital on another. As shown in Figure $5$, the sum of the two atomic wave functions (ns + npz) produces a σ bonding molecular orbital. Their difference (nsnpz) produces a σ* antibonding molecular orbital, which has a nodal plane of zero probability density perpendicular to the internuclear axis.
Second Row Diatomic Molecules
If we combine the splitting schemes for the 2s and 2p orbitals, we can predict bond order in all of the diatomic molecules and ions composed of elements in the first complete row of the periodic table. Remember that only the valence orbitals of the atoms need be considered; as we saw in the cases of lithium hydride and dilithium, the inner orbitals remain tightly bound and retain their localized atomic character.
We now describe examples of systems involving period 2 homonuclear diatomic molecules, such as N2, O2, and F2. When we draw a molecular orbital diagram for a molecule, there are four key points to remember:
1. The number of molecular orbitals produced is the same as the number of atomic orbitals used to create them (the law of conservation of orbitals).
2. As the overlap between two atomic orbitals increases, the difference in energy between the resulting bonding and antibonding molecular orbitals increases.
3. When two atomic orbitals combine to form a pair of molecular orbitals, the bonding molecular orbital is stabilized about as much as the antibonding molecular orbital is destabilized.
4. The interaction between atomic orbitals is greatest when they have the same energy.
The number of molecular orbitals is always equal to the total number of atomic orbitals we started with.
We illustrate how to use these points by constructing a molecular orbital energy-level diagram for F2. We use the diagram in part (a) in Figure $6$; the n = 1 orbitals (σ1s and σ1s*) are located well below those of the n = 2 level and are not shown. As illustrated in the diagram, the σ2s and σ2s* molecular orbitals are much lower in energy than the molecular orbitals derived from the 2p atomic orbitals because of the large difference in energy between the 2s and 2p atomic orbitals of fluorine. The lowest-energy molecular orbital derived from the three 2p orbitals on each F is $\sigma _{2p_{z}}$ and the next most stable are the two degenerate orbitals, $\pi _{2p_{x}}$ and $\pi _{2p_{y}}$. For each bonding orbital in the diagram, there is an antibonding orbital, and the antibonding orbital is destabilized by about as much as the corresponding bonding orbital is stabilized. As a result, the $\sigma ^{\star }_{2p_{z}}$ orbital is higher in energy than either of the degenerate $\pi _{2p_{x}}^{\star }$ and $\pi _{2p_{y}}^{\star }$ orbitals. We can now fill the orbitals, beginning with the one that is lowest in energy.
Each fluorine has 7 valence electrons, so there are a total of 14 valence electrons in the F2 molecule. Starting at the lowest energy level, the electrons are placed in the orbitals according to the Pauli principle and Hund’s rule. Two electrons each fill the σ2s and σ2s* orbitals, 2 fill the $\sigma _{2p_{z}}$ orbital, 4 fill the two degenerate π orbitals, and 4 fill the two degenerate π* orbitals, for a total of 14 electrons. To determine what type of bonding the molecular orbital approach predicts F2 to have, we must calculate the bond order. According to our diagram, there are 8 bonding electrons and 6 antibonding electrons, giving a bond order of (8 − 6) ÷ 2 = 1. Thus F2 is predicted to have a stable F–F single bond, in agreement with experimental data.
We now turn to a molecular orbital description of the bonding in O2. It so happens that the molecular orbital description of this molecule provided an explanation for a long-standing puzzle that could not be explained using other bonding models. To obtain the molecular orbital energy-level diagram for O2, we need to place 12 valence electrons (6 from each O atom) in the energy-level diagram shown in part (b) in Figure $6$. We again fill the orbitals according to Hund’s rule and the Pauli principle, beginning with the orbital that is lowest in energy. Two electrons each are needed to fill the σ2s and σ2s* orbitals, 2 more to fill the $\sigma _{2p_{z}}$ orbital, and 4 to fill the degenerate $\pi _{2p_{x}}^{\star }$ and $\pi _{2p_{y}}^{\star }$ orbitals. According to Hund’s rule, the last 2 electrons must be placed in separate π* orbitals with their spins parallel, giving two unpaired electrons. This leads to a predicted bond order of (8 − 4) ÷ 2 = 2, which corresponds to a double bond, in agreement with experimental data (Table 4.5): the O–O bond length is 120.7 pm, and the bond energy is 498.4 kJ/mol at 298 K.
None of the other bonding models can predict the presence of two unpaired electrons in O2. Chemists had long wondered why, unlike most other substances, liquid O2 is attracted into a magnetic field. As shown in Video $2$, it actually remains suspended between the poles of a magnet until the liquid boils away. The only way to explain this behavior was for O2 to have unpaired electrons, making it paramagnetic, exactly as predicted by molecular orbital theory. This result was one of the earliest triumphs of molecular orbital theory over the other bonding approaches we have discussed.
Video $2$: Liquid O2 Suspended between the Poles of a Magnet.Because the O2 molecule has two unpaired electrons, it is paramagnetic. Consequently, it is attracted into a magnetic field, which allows it to remain suspended between the poles of a powerful magnet until it evaporates. Full video can be found at https://www.youtube.com/watch?featur...&v=Lt4P6ctf06Q.
The magnetic properties of O2 are not just a laboratory curiosity; they are absolutely crucial to the existence of life. Because Earth’s atmosphere contains 20% oxygen, all organic compounds, including those that compose our body tissues, should react rapidly with air to form H2O, CO2, and N2 in an exothermic reaction. Fortunately for us, however, this reaction is very, very slow. The reason for the unexpected stability of organic compounds in an oxygen atmosphere is that virtually all organic compounds, as well as H2O, CO2, and N2, have only paired electrons, whereas oxygen has two unpaired electrons. Thus the reaction of O2 with organic compounds to give H2O, CO2, and N2 would require that at least one of the electrons on O2 change its spin during the reaction. This would require a large input of energy, an obstacle that chemists call a spin barrier. Consequently, reactions of this type are usually exceedingly slow. If they were not so slow, all organic substances, including this book and you, would disappear in a puff of smoke!
For period 2 diatomic molecules to the left of N2 in the periodic table, a slightly different molecular orbital energy-level diagram is needed because the $\sigma _{2p_{z}}$ molecular orbital is slightly higher in energy than the degenerate $\pi ^{\star }_{np_{x}}$ and $\pi ^{\star }_{np_{y}}$ orbitals. The difference in energy between the 2s and 2p atomic orbitals increases from Li2 to F2 due to increasing nuclear charge and poor screening of the 2s electrons by electrons in the 2p subshell. The bonding interaction between the 2s orbital on one atom and the 2pz orbital on the other is most important when the two orbitals have similar energies. This interaction decreases the energy of the σ2s orbital and increases the energy of the $\sigma _{2p_{z}}$ orbital. Thus for Li2, Be2, B2, C2, and N2, the $\sigma _{2p_{z}}$ orbital is higher in energy than the $\sigma _{3p_{z}}$ orbitals, as shown in Figure $7$ Experimentally, it is found that the energy gap between the ns and np atomic orbitals increases as the nuclear charge increases (Figure $7$ ). Thus for example, the $\sigma _{2p_{z}}$ molecular orbital is at a lower energy than the $\pi _{2p_{x,y}}$ pair.
Completing the diagram for N2 in the same manner as demonstrated previously, we find that the 10 valence electrons result in 8 bonding electrons and 2 antibonding electrons, for a predicted bond order of 3, a triple bond. Experimental data show that the N–N bond is significantly shorter than the F–F bond (109.8 pm in N2 versus 141.2 pm in F2), and the bond energy is much greater for N2 than for F2 (945.3 kJ/mol versus 158.8 kJ/mol, respectively). Thus the N2 bond is much shorter and stronger than the F2 bond, consistent with what we would expect when comparing a triple bond with a single bond.
Example $1$
Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in S2, a bright blue gas at high temperatures.
Given: chemical species
Asked for: molecular orbital energy-level diagram, bond order, and number of unpaired electrons
Strategy:
1. Write the valence electron configuration of sulfur and determine the type of molecular orbitals formed in S2. Predict the relative energies of the molecular orbitals based on how close in energy the valence atomic orbitals are to one another.
2. Draw the molecular orbital energy-level diagram for this system and determine the total number of valence electrons in S2.
3. Fill the molecular orbitals in order of increasing energy, being sure to obey the Pauli principle and Hund’s rule.
4. Calculate the bond order and describe the bonding.
Solution:
A Sulfur has a [Ne]3s23p4 valence electron configuration. To create a molecular orbital energy-level diagram similar to those in Figure $6$ and Figure $7$, we need to know how close in energy the 3s and 3p atomic orbitals are because their energy separation will determine whether the $\pi _{3p_{x,y}}$ or the $\sigma _{3p_{z}}$> molecular orbital is higher in energy. Because the nsnp energy gap increases as the nuclear charge increases (Figure $7$), the $\sigma _{3p_{z}}$ molecular orbital will be lower in energy than the $\pi _{3p_{x,y}}$ pair.
B The molecular orbital energy-level diagram is as follows:
Each sulfur atom contributes 6 valence electrons, for a total of 12 valence electrons.
C Ten valence electrons are used to fill the orbitals through $\pi _{3p_{x}}$ and $\pi _{3p_{y}}$, leaving 2 electrons to occupy the degenerate $\pi ^{\star }_{3p_{x}}$ and $\pi ^{\star }_{3p_{y}}$ pair. From Hund’s rule, the remaining 2 electrons must occupy these orbitals separately with their spins aligned. With the numbers of electrons written as superscripts, the electron configuration of S2 is $\left ( \sigma _{3s} \right )^{2}\left ( \sigma ^{\star }_{3s} \right )^{2}\left ( \sigma _{3p_{z}} \right )^{2}\left ( \pi _{3p_{x,y}} \right )^{4}\left ( \pi _{3p ^{\star }_{x,y}} \right )^{2}$ with 2 unpaired electrons. The bond order is (8 − 4) ÷ 2 = 2, so we predict an S=S double bond.
Exercise $1$
Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in the peroxide ion (O22−).
Answer
$\left ( \sigma _{2s} \right )^{2}\left ( \sigma ^{\star }_{2s} \right )^{2}\left ( \sigma _{2p_{z}} \right )^{2}\left ( \pi _{2p_{x,y}} \right )^{4}\left ( \pi _{2p ^{\star }_{x,y}} \right )^{4}$ bond order of 1; no unpaired electrons
Molecular Orbitals for Heteronuclear Diatomic Molecules
Diatomic molecules with two different atoms are called heteronuclear diatomic molecules. When two nonidentical atoms interact to form a chemical bond, the interacting atomic orbitals do not have the same energy. If, for example, element B is more electronegative than element A (χB > χA), the net result is a “skewed” molecular orbital energy-level diagram, such as the one shown for a hypothetical A–B molecule in Figure $8$. The atomic orbitals of element B are uniformly lower in energy than the corresponding atomic orbitals of element A because of the enhanced stability of the electrons in element B. The molecular orbitals are no longer symmetrical, and the energies of the bonding molecular orbitals are more similar to those of the atomic orbitals of B. Hence the electron density of bonding electrons is likely to be closer to the more electronegative atom. In this way, molecular orbital theory can describe a polar covalent bond.
A molecular orbital energy-level diagram is always skewed toward the more electronegative atom.
An Odd Number of Valence Electrons: NO
Nitric oxide (NO) is an example of a heteronuclear diatomic molecule. The reaction of O2 with N2 at high temperatures in internal combustion engines forms nitric oxide, which undergoes a complex reaction with O2 to produce NO2, which in turn is responsible for the brown color we associate with air pollution. Recently, however, nitric oxide has also been recognized to be a vital biological messenger involved in regulating blood pressure and long-term memory in mammals.
Because NO has an odd number of valence electrons (5 from nitrogen and 6 from oxygen, for a total of 11), its bonding and properties cannot be successfully explained by either the Lewis electron-pair approach or valence bond theory. The molecular orbital energy-level diagram for NO (Figure $9$) shows that the general pattern is similar to that for the O2 molecule (Figure $7$). Because 10 electrons are sufficient to fill all the bonding molecular orbitals derived from 2p atomic orbitals, the 11th electron must occupy one of the degenerate π* orbitals. The predicted bond order for NO is therefore (8-3) ÷ 2 = 2 1/2 . Experimental data, showing an N–O bond length of 115 pm and N–O bond energy of 631 kJ/mol, are consistent with this description. These values lie between those of the N2 and O2 molecules, which have triple and double bonds, respectively. As we stated earlier, molecular orbital theory can therefore explain the bonding in molecules with an odd number of electrons, such as NO, whereas Lewis electron structures cannot.
Molecular orbital theory can also tell us something about the chemistry of $NO$. As indicated in the energy-level diagram in Figure $9$, NO has a single electron in a relatively high-energy molecular orbital. We might therefore expect it to have similar reactivity as alkali metals such as Li and Na with their single valence electrons. In fact, $NO$ is easily oxidized to the $NO^+$ cation, which is isoelectronic with $N_2$ and has a bond order of 3, corresponding to an N≡O triple bond.
Nonbonding Molecular Orbitals
Molecular orbital theory is also able to explain the presence of lone pairs of electrons. Consider, for example, the HCl molecule, whose Lewis electron structure has three lone pairs of electrons on the chlorine atom. Using the molecular orbital approach to describe the bonding in HCl, we can see from Figure $10$ that the 1s orbital of atomic hydrogen is closest in energy to the 3p orbitals of chlorine. Consequently, the filled Cl 3s atomic orbital is not involved in bonding to any appreciable extent, and the only important interactions are those between the H 1s and Cl 3p orbitals. Of the three p orbitals, only one, designated as 3pz, can interact with the H 1s orbital. The 3px and 3py atomic orbitals have no net overlap with the 1s orbital on hydrogen, so they are not involved in bonding. Because the energies of the Cl 3s, 3px, and 3py orbitals do not change when HCl forms, they are called nonbonding molecular orbitals. A nonbonding molecular orbital occupied by a pair of electrons is the molecular orbital equivalent of a lone pair of electrons. By definition, electrons in nonbonding orbitals have no effect on bond order, so they are not counted in the calculation of bond order. Thus the predicted bond order of HCl is (2 − 0) ÷ 2 = 1. Because the σ bonding molecular orbital is closer in energy to the Cl 3pz than to the H 1s atomic orbital, the electrons in the σ orbital are concentrated closer to the chlorine atom than to hydrogen. A molecular orbital approach to bonding can therefore be used to describe the polarization of the H–Cl bond to give $H^{\delta +} - Cl^{\delta -}$.
Electrons in nonbonding molecular orbitals have no effect on bond order.
Example $2$
Use a “skewed” molecular orbital energy-level diagram like the one in Figure $8$ to describe the bonding in the cyanide ion (CN). What is the bond order?
Given: chemical species
Asked for: “skewed” molecular orbital energy-level diagram, bonding description, and bond order
Strategy:
1. Calculate the total number of valence electrons in CN. Then place these electrons in a molecular orbital energy-level diagram like Figure $8$ in order of increasing energy. Be sure to obey the Pauli principle and Hund’s rule while doing so.
2. Calculate the bond order and describe the bonding in CN.
Solution:
A The CN ion has a total of 10 valence electrons: 4 from C, 5 from N, and 1 for the −1 charge. Placing these electrons in an energy-level diagram like Figure $8$ fills the five lowest-energy orbitals, as shown here:
Because χN > χC, the atomic orbitals of N (on the right) are lower in energy than those of C. B The resulting valence electron configuration gives a predicted bond order of (8 − 2) ÷ 2 = 3, indicating that the CN ion has a triple bond, analogous to that in N2.
Exercise $2$
Use a qualitative molecular orbital energy-level diagram to describe the bonding in the hypochlorite ion (OCl). What is the bond order?
Answer
All molecular orbitals except the highest-energy σ* are filled, giving a bond order of 1.
Although the molecular orbital approach reveals a great deal about the bonding in a given molecule, the procedure quickly becomes computationally intensive for molecules of even moderate complexity. Furthermore, because the computed molecular orbitals extend over the entire molecule, they are often difficult to represent in a way that is easy to visualize. Therefore we do not use a pure molecular orbital approach to describe the bonding in molecules or ions with more than two atoms. Instead, we use a valence bond approach and a molecular orbital approach to explain, among other things, the concept of resonance, which cannot adequately be explained using other methods.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/12%3A_The_Chemical_Bond/12.06%3A_Diatomic_Molecules.txt
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Resonance is a mental exercise and method within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. It compares and contrasts two or more possible Lewis structures that can represent a particular molecule. Resonance structures are used when one Lewis structure for a single molecule cannot fully describe the bonding that takes place between neighboring atoms relative to the empirical data for the actual bond lengths between those atoms. The net sum of valid resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. A molecule that has several resonance structures is more stable than one with fewer. Some resonance structures are more favorable than others.
Introduction
Electrons have no fixed position in atoms, compounds and molecules (see image below) but have probabilities of being found in certain spaces (orbitals). Resonance forms illustrate areas of higher probabilities (electron densities). This is like holding your hat in either your right hand or your left. The term Resonance is applied when there are two or more possibilities available. Chemists use Lewis diagrams to depict structure and bonding of covalent entities, such as molecules and polyatomic ions, henceforth, molecules. The Lewis diagram of many a molecule, however, is not consistent with the observed properties of the molecule.
Example $1$: Nitrate Ion
The nitrate ion, according to its Lewis diagram, has two types of nitrogen-oxygen bonds, one double bond and two single bonds, suggesting that one nitrogen-oxygen bond in the nitrate ion is shorter and stronger than each of the other two. Also, the Lewis structure implies, with respect to formal charge, that there are two types of oxygen atoms in the nitrate ion, one formally neutral and each of the other two bearing a formal charge of –1. Experimentally, however, the three nitrogen-oxygen bonds in the nitrate ion have the same bond length and the same bond energy, and the three oxygen atoms are indistinguishable. The Lewis diagram fails to explain the structure and bonding of the nitrate ion satisfactorily.
Two additional Lewis diagrams can be drawn for the nitrate ion.
However, none of them are consistent with the observed properties of the nitrate ion and, therefore, does not correctly depict the nitrate ion.
Example $2$: Benzene
Benzene, according to its Lewis diagram, has two types of carbon-carbon bonds, three double bonds and three single bonds, suggesting that each of three carbon-carbon bonds in benzene is shorter and stronger than each of the other three. Experimentally, however, the six carbon-carbon bonds in benzene have the same bond length and the same bond energy. The Lewis diagram fails to explain the structure and bonding of benzene satisfactorily.
An attribute of molecules of which the classical Lewis diagram is not consistent with the observed properties is that other valid Lewis diagrams can be generated for them. One additional Lewis diagram can be drawn for benzene.
However, none of them are consistent with the observed properties of benzene and, therefore, does not correctly depict benzene.
Resonance theory is an attempt to explain the structure of a species, like the nitrate ion or benzene, no Lewis diagram of which is consistent with the observed properties of the species. The major advantage of resonance theory is that, although based on rigorous mathematical analysis, resonance theory can be applied successfully invoking little or no math. Resonance theory is explained below using the nitrate ion as the example.
According to resonance theory, the structure of the nitrate ion is not 1 nor 2 nor 3 but the average of all three of them, weighted by stability. Lewis diagrams 1, 2, and 3 are called resonance forms, resonance structures, or resonance contributors of the nitrate ion. The structure of the nitrate ion is said to be a resonance hybrid or, simply, hybrid of resonance forms 1, 2, and 3. Whenever it is necessary to show the structure of the nitrate ion, resonance forms 1, 2, and 3 are drawn, connected by a double-headed arrows.
The three resonance forms of the nitrate ion, 1, 2, and 3, are identical, so they have the same stability and, therefore, contribute equally to the hybrid. Since the exact extent to which each resonance form of the nitrate ion contributes to the hybrid is known, the bond order of each nitrogen-oxygen bond as well as the formal charge on each oxygen atom in the hybrid can be easily determined:
According to resonance theory, each bond in the nitrate ion is one and one-third of a bond, which is consistent with the observation that the three bonds in the nitrate ion have the same bond length and the same bond energy.
According to resonance theory, each oxygen atom in the nitrate ion has a formal charge of –2/3, which, in conjunction with the fact that the three nitrogen-oxygen bonds are identical, is consistent with the observation that the three oxygen atoms in the nitrate ion are indistinguishable. In each resonance form of the nitrate ion, there are two $\pi$ electrons, and they are shared only by two atoms. An electron shared only by two atoms is said to be localized. Thus, the two $\pi$ electrons in each resonance form of the nitrate ion are localized. The nitrate ion, as represented by the hybrid, has two $\pi$ electrons:
• # electrons in one $\pi$ bond = 2
• # electrons in one-third of a $\pi$ bond = 2/3
• # electrons in three of them = 3 x (2/3) = 2
The two $\pi$ electrons in the nitrate ion are shared by a total of four atoms, one nitrogen atom and three oxygen atoms. An electron shared by more than two atoms is said to be delocalized. Thus, the two $\pi$ electrons in the nitrate ion are delocalized. Delocalization of $\pi$ electrons in the nitrate ion requires that the four atoms be on the same plane, allowing lateral overlap of the p orbitals on them.
If the energy of the nitrate ion were the weighted average of the energies of its three resonance forms, just as the structure of the nitrate ion is the weighted average of the structures of its three resonance forms, it should be equal to the energy of one of the three identical resonance forms:
If the energy of the hybrid were equal to that of a resonance form, given that all chemical entities (elementary particles, atoms, molecules, etc.) naturally tend to be in the lowest possible energy state, there would be no advantage for the nitrate ion to exist as the hybrid; it could simply exist as a resonance form. Since the nitrate ion exists as the hybrid, not as a resonance form, it can be inferred that the energy of the hybrid is lower than that of any of the resonance forms.
According to resonance theory then, the energy of a molecule is lower than that of the lowest-energy resonance form. Since the nitrate ion has lower energy and, therefore, is more stable than any of its resonance forms, the nitrate ion is said to be resonance stabilized.
Common Misconceptions
There are two misconceptions about resonance theory among beginning students, likely due to literal interpretation of the word resonance. They are described below, using the nitrate ion as the example.
Misconception 1: The nitrate ion exists as resonance form 1 for a moment and then changes either to resonance form 2 or to resonance form 3, which interconvert, or revert to 1.
The structure of the nitrate ion is not 1 nor 2 nor 3 but the hybrid and does not change with time unless undergoing a reaction.
Misconception 2: In a sample of nitrate ions, at a given moment, one-third of the ions exist as resonance form 1, another one-third as resonance form 2, and the remaining one-third as resonance form 3.
In a sample of nitrate ions, at a given moment, all ions have the same structure, which is the hybrid.
Mules and Rhinoceros
The classic analogy used to clarify these two misconceptions is the mule (Morrison, R. T.; Boyd, R.N. Organic Chemistry, fifth edition; Allyn and Bacon: Boston, 1987, pg. 373). Biologically, a mule is a hybrid of a horse and a donkey. This does not mean that a mule resembles a horse for a moment and then changes to resemble a donkey. The appearance of a mule is a combination of that of a horse and that of a donkey and does not change with time. Nor does it mean that, in a herd, some mules resemble a horse and the others a donkey. In a herd, all mules have the same appearance, which is a combination of a horse and a donkey. The weakness of this analogy is that horses and donkeys do exist, whereas resonance forms are strictly hypothetical. A better analogy, cited in Morrison and Boyd, is the rhinoceros. Upon seeing a rhinoceros, one could describe it as the hybrid of a dragon and a unicorn, two creatures that do not exist.
Peptide Bonds
A consideration of resonance contributors is crucial to any discussion of the amide functional group. One of the most important examples of amide groups in nature is the ‘peptide bond’ that links amino acids to form polypeptides and proteins.
Critical to the structure of proteins is the fact that, although it is conventionally drawn as a single bond, the C-N bond in a peptide linkage has a significant barrier to rotation, almost as if it were a double bond.
This, along with the observation that the bonding around the peptide nitrogen has trigonal planar geometry, strongly suggests that the nitrogen is sp2-hybridized. An important resonance contributor has a C=N double bond and a C-O single bond, with a separation of charge between the oxygen and the nitrogen.
Although B is a minor contributor due to the separation of charges, it is still very relevant in terms of peptide and protein structure – our proteins would simply not fold up properly if there was free rotation about the peptide C-N bond.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/12%3A_The_Chemical_Bond/12.07%3A_Resonance_and_Electron_Delocalization.txt
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Learning Objectives
• To know the most common structures observed for metal complexes.
• To predict the relative stabilities of metal complexes with different ligands
One of the most important properties of metallic elements is their ability to act as Lewis acids that form complexes with a variety of Lewis bases. A metal complex consists of a central metal atom or ion that is bonded to one or more ligands (from the Latin ligare, meaning “to bind”), which are ions or molecules that contain one or more pairs of electrons that can be shared with the metal. Metal complexes can be neutral, such as Co(NH3)3Cl3; positively charged, such as [Nd(H2O)9]3+; or negatively charged, such as [UF8]4−. Electrically charged metal complexes are sometimes called complex ions. A coordination compound contains one or more metal complexes.
Coordination compounds are important for at least three reasons. First, most of the elements in the periodic table are metals, and almost all metals form complexes, so metal complexes are a feature of the chemistry of more than half the elements. Second, many industrial catalysts are metal complexes, and such catalysts are steadily becoming more important as a way to control reactivity. For example, a mixture of a titanium complex and an organometallic compound of aluminum is the catalyst used to produce most of the polyethylene and polypropylene “plastic” items we use every day. Finally, transition-metal complexes are essential in biochemistry. Examples include hemoglobin, an iron complex that transports oxygen in our blood; cytochromes, iron complexes that transfer electrons in our cells; and complexes of Fe, Zn, Cu, and Mo that are crucial components of certain enzymes, the catalysts for all biological reactions.
History of the Coordination Compounds
Coordination compounds have been known and used since antiquity; probably the oldest is the deep blue pigment called Prussian blue: $\ce{KFe2(CN)6}$. The chemical nature of these substances, however, was unclear for a number of reasons. For example, many compounds called “double salts” were known, such as $\ce{AlF3·3KF}$, $\ce{Fe(CN)2·4KCN}$, and $\ce{ZnCl2·2CsCl}$, which were combinations of simple salts in fixed and apparently arbitrary ratios. Why should $\ce{AlF3·3KF}$ exist but not $\ce{AlF3·4KF}$ or $\ce{AlF3·2KF}$? And why should a 3:1 KF:AlF3 mixture have different chemical and physical properties than either of its components? Similarly, adducts of metal salts with neutral molecules such as ammonia were also known—for example, $\ce{CoCl3·6NH3}$, which was first prepared sometime before 1798. Like the double salts, the compositions of these adducts exhibited fixed and apparently arbitrary ratios of the components. For example, $\ce{CoCl3·6NH3}$, $\ce{CoCl3·5NH3}$, $\ce{CoCl3·4NH3}$, and $\ce{CoCl3·3NH3}$ were all known and had very different properties, but despite all attempts, chemists could not prepare $\ce{CoCl3·2NH3}$ or $\ce{CoCl3·NH3}$.
Although the chemical composition of such compounds was readily established by existing analytical methods, their chemical nature was puzzling and highly controversial. The major problem was that what we now call valence (i.e., the oxidation state) and coordination number were thought to be identical. As a result, highly implausible (to modern eyes at least) structures were proposed for such compounds, including the “Chattanooga choo-choo” model for CoCl3·4NH3 shown here.
The modern theory of coordination chemistry is based largely on the work of Alfred Werner (1866–1919; Nobel Prize in Chemistry in 1913). In a series of careful experiments carried out in the late 1880s and early 1890s, he examined the properties of several series of metal halide complexes with ammonia. For example, five different “adducts” of ammonia with PtCl4 were known at the time: PtCl4·nNH3 (n = 2–6). Some of Werner’s original data on these compounds are shown in Table $1$. The electrical conductivity of aqueous solutions of these compounds was roughly proportional to the number of ions formed per mole, while the number of chloride ions that could be precipitated as AgCl after adding Ag+(aq) was a measure of the number of “free” chloride ions present. For example, Werner’s data on PtCl4·6NH3 in Table $1$ showed that all the chloride ions were present as free chloride. In contrast, PtCl4·2NH3 was a neutral molecule that contained no free chloride ions.
Alfred Werner (1866–1919)
Werner, the son of a factory worker, was born in Alsace. He developed an interest in chemistry at an early age, and he did his first independent research experiments at age 18. While doing his military service in southern Germany, he attended a series of chemistry lectures, and he subsequently received his PhD at the University of Zurich in Switzerland, where he was appointed professor of chemistry at age 29. He won the Nobel Prize in Chemistry in 1913 for his work on coordination compounds, which he performed as a graduate student and first presented at age 26. Apparently, Werner was so obsessed with solving the riddle of the structure of coordination compounds that his brain continued to work on the problem even while he was asleep. In 1891, when he was only 25, he woke up in the middle of the night and, in only a few hours, had laid the foundation for modern coordination chemistry.
Table $1$: Werner’s Data on Complexes of Ammonia with $PtCl_4$
Complex Conductivity (ohm−1) Number of Ions per Formula Unit Number of Cl Ions Precipitated by Ag+
PtCl4·6NH3 523 5 4
PtCl4·5NH3 404 4 3
PtCl4·4NH3 299 3 2
PtCl4·3NH3 97 2 1
PtCl4·2NH3 0 0 0
These data led Werner to postulate that metal ions have two different kinds of valence: (1) a primary valence (oxidation state) that corresponds to the positive charge on the metal ion and (2) a secondary valence (coordination number) that is the total number of ligand-metal bonds bound to the metal ion. If $\ce{Pt}$ had a primary valence of 4 and a secondary valence of 6, Werner could explain the properties of the $\ce{PtCl4·NH3}$ adducts by the following reactions, where the metal complex is enclosed in square brackets:
\begin{align*} \mathrm{[Pt(NH_3)_6]Cl_4} &\rightarrow \mathrm{[Pt(NH_3)_6]^{4+}(aq)+4Cl^-(aq)} \[4pt] \mathrm{[Pt(NH_3)_5Cl]Cl_3} &\rightarrow \mathrm{[Pt(NH_3)_5Cl]^{3+}(aq) +3Cl^-(aq)}\[4pt] \mathrm{[Pt(NH_3)_4Cl_2]Cl_2} &\rightarrow \mathrm{[Pt(NH_3)_4Cl_2]^{2+}(aq) +2Cl^-(aq)}\[4pt] \mathrm{[Pt(NH_3)_3Cl_3]Cl} &\rightarrow \mathrm{[Pt(NH_3)_3Cl_3]^+(aq) + Cl^-(aq)}\[4pt] \mathrm{[Pt(NH_3)_2Cl_4]} &\rightarrow \mathrm{[Pt(NH_3)_2Cl_4]^0(aq)} \end{align*} \label{23.9}
Further work showed that the two missing members of the series—[Pt(NH3)Cl5] and [PtCl6]2−—could be prepared as their mono- and dipotassium salts, respectively. Similar studies established coordination numbers of 6 for Co3+ and Cr3+ and 4 for Pt2+ and Pd2+.
Werner’s studies on the analogous Co3+ complexes also allowed him to propose a structural model for metal complexes with a coordination number of 6. Thus he found that [Co(NH3)6]Cl3 (yellow) and [Co(NH3)5Cl]Cl2 (purple) were 1:3 and 1:2 electrolytes. Unexpectedly, however, two different [Co(NH3)4Cl2]Cl compounds were known: one was red, and the other was green (Figure $\PageIndex{1a}$). Because both compounds had the same chemical composition and the same number of groups of the same kind attached to the same metal, there had to be something different about the arrangement of the ligands around the metal ion. Werner’s key insight was that the six ligands in [Co(NH3)4Cl2]Cl had to be arranged at the vertices of an octahedron because that was the only structure consistent with the existence of two, and only two, arrangements of ligands (Figure $\PageIndex{1b}$. His conclusion was corroborated by the existence of only two different forms of the next compound in the series: Co(NH3)3Cl3.
Example $1$
In Werner’s time, many complexes of the general formula MA4B2 were known, but no more than two different compounds with the same composition had been prepared for any metal. To confirm Werner’s reasoning, calculate the maximum number of different structures that are possible for six-coordinate MA4B2 complexes with each of the three most symmetrical possible structures: a hexagon, a trigonal prism, and an octahedron. What does the fact that no more than two forms of any MA4B2 complex were known tell you about the three-dimensional structures of these complexes?
Given: three possible structures and the number of different forms known for MA4B2 complexes
Asked for: number of different arrangements of ligands for MA4B2 complex for each structure
Strategy:
Sketch each structure, place a B ligand at one vertex, and see how many different positions are available for the second B ligand.
Solution
The three regular six-coordinate structures are shown here, with each coordination position numbered so that we can keep track of the different arrangements of ligands. For each structure, all vertices are equivalent. We begin with a symmetrical MA6 complex and simply replace two of the A ligands in each structure to give an MA4B2 complex:
For the hexagon, we place the first B ligand at position 1. There are now three possible places for the second B ligand: at position 2 (or 6), position 3 (or 5), or position 4. These are the only possible arrangements. The (1, 2) and (1, 6) arrangements are chemically identical because the two B ligands are adjacent to each other. The (1, 3) and (1, 5) arrangements are also identical because in both cases the two B ligands are separated by an A ligand.
Turning to the trigonal prism, we place the first B ligand at position 1. Again, there are three possible choices for the second B ligand: at position 2 or 3 on the same triangular face, position 4 (on the other triangular face but adjacent to 1), or position 5 or 6 (on the other triangular face but not adjacent to 1). The (1, 2) and (1, 3) arrangements are chemically identical, as are the (1, 5) and (1, 6) arrangements.
In the octahedron, however, if we place the first B ligand at position 1, then we have only two choices for the second B ligand: at position 2 (or 3 or 4 or 5) or position 6. In the latter, the two B ligands are at opposite vertices of the octahedron, with the metal lying directly between them. Although there are four possible arrangements for the former, they are chemically identical because in all cases the two B ligands are adjacent to each other.
The number of possible MA4B2 arrangements for the three geometries is thus: hexagon, 3; trigonal prism, 3; and octahedron, 2. The fact that only two different forms were known for all MA4B2 complexes that had been prepared suggested that the correct structure was the octahedron but did not prove it. For some reason one of the three arrangements possible for the other two structures could have been less stable or harder to prepare and had simply not yet been synthesized. When combined with analogous results for other types of complexes (e.g., MA3B3), however, the data were best explained by an octahedral structure for six-coordinate metal complexes.
Exercise $1$
Determine the maximum number of structures that are possible for a four-coordinate MA2B2 complex with either a square planar or a tetrahedral symmetrical structure.
Answer
square planar, 2; tetrahedral, 1
Structures of Metal Complexes
The coordination numbers of metal ions in metal complexes can range from 2 to at least 9. In general, the differences in energy between different arrangements of ligands are greatest for complexes with low coordination numbers and decrease as the coordination number increases. Usually only one or two structures are possible for complexes with low coordination numbers, whereas several different energetically equivalent structures are possible for complexes with high coordination numbers (n > 6). The following presents the most commonly encountered structures for coordination numbers 2–9. Many of these structures should be familiar to you from our discussion of the valence-shell electron-pair repulsion (VSEPR) model because they correspond to the lowest-energy arrangements of n electron pairs around a central atom.
Compounds with low coordination numbers exhibit the greatest differences in energy between different arrangements of ligands.
Coordination Number 2
Although it is rare for most metals, this coordination number is surprisingly common for d10 metal ions, especially Cu+, Ag+, Au+, and Hg2+. An example is the [Au(CN)2] ion, which is used to extract gold from its ores. As expected based on VSEPR considerations, these complexes have the linear L–M–L structure shown here.
Coordination Number 3
Although it is also rare, this coordination number is encountered with d10 metal ions such as Cu+ and Hg2+. Among the few known examples is the HgI3 ion. Three-coordinate complexes almost always have the trigonal planar structure expected from the VSEPR model.
Coordination Number 4
Two common structures are observed for four-coordinate metal complexes: tetrahedral and square planar. The tetrahedral structure is observed for all four-coordinate complexes of nontransition metals, such as [BeF4]2−, and d10 ions, such as [ZnCl4]2−. It is also found for four-coordinate complexes of the first-row transition metals, especially those with halide ligands (e.g., [FeCl4] and [FeCl4]2−). In contrast, square planar structures are routinely observed for four-coordinate complexes of second- and third-row transition metals with d8 electron configurations, such as Rh+ and Pd2+, and they are also encountered in some complexes of Ni2+ and Cu2+.
Coordination Number 5
This coordination number is less common than 4 and 6, but it is still found frequently in two different structures: trigonal bipyramidal and square pyramidal. Because the energies of these structures are usually rather similar for most ligands, many five-coordinate complexes have distorted structures that lie somewhere between the two extremes.
Coordination Number 6
This coordination number is by far the most common. The six ligands are almost always at the vertices of an octahedron or a distorted octahedron. The only other six-coordinate structure is the trigonal prism, which is very uncommon in simple metal complexes.
Coordination Number 7
This relatively uncommon coordination number is generally encountered for only large metals (such as the second- and third-row transition metals, lanthanides, and actinides). At least three different structures are known, two of which are derived from an octahedron or a trigonal prism by adding a ligand to one face of the polyhedron to give a “capped” octahedron or trigonal prism. By far the most common, however, is the pentagonal bipyramid.
Coordination Number 8
This coordination number is relatively common for larger metal ions. The simplest structure is the cube, which is rare because it does not minimize interligand repulsive interactions. Common structures are the square antiprism and the dodecahedron, both of which can be generated from the cube.
Coordination Number 9
This coordination number is found in larger metal ions, and the most common structure is the tricapped trigonal prism, as in [Nd(H2O)9]3+.
Key Takeaways
• Coordination compounds are a major feature of the chemistry of over half the elements.
• Coordination compounds have important roles as industrial catalysts in controlling reactivity, and they are essential in biochemical processes.
Summary
Transition metals form metal complexes, polyatomic species in which a metal ion is bound to one or more ligands, which are groups bound to a metal ion. Complex ions are electrically charged metal complexes, and a coordination compound contains one or more metal complexes. Metal complexes with low coordination numbers generally have only one or two possible structures, whereas those with coordination numbers greater than six can have several different structures. Coordination numbers of two and three are common for d10 metal ions. Tetrahedral and square planar complexes have a coordination number of four; trigonal bipyramidal and square pyramidal complexes have a coordination number of five; and octahedral complexes have a coordination number of six. At least three structures are known for a coordination number of seven, which is generally found for only large metal ions. Coordination numbers of eight and nine are also found for larger metal ions.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/12%3A_The_Chemical_Bond/12.08%3A_Coordination_Compounds.txt
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12.1: Lewis Structures
Q12.1
Which of the following molecules has the shortest bond length?
• $O_2$
• $Cl_2$
• $Br_2$
• $I_2$
Q12.2
Draw a plausible Lewis structure for 1,3-Bisphosphoglyerate (1,3-BPG), (C3H8O10P2).
Q12.3
Draw resonance structures for NO3- and ClO3-, please show formal charges in drawings.
Q12.4
Calculate the dipole moment ($\mu$) of a two ions of +2e and -2e that are separated by 100 pm. Express dipole moment in both units of $C\,m$ and $D$.
S12.4
$\mu = Q \times r = 2 ( 1.602 \times 10^{-19})( 8 \times 10^{-10}) = 2.56 \times 10^{-28} C\,m$
$(2.56 \times 10^{-28} \cancel{C\,m}) \left( \dfrac{1\,D}{3.336 \times 10^{-30} \cancel{C\,m}} \right) = 76.8 \;D$
12.2: Valence Bond Theory
Q12.5
What are the resonance of $NO_3^0$?
S12.5
Resonance structure in basic means that there are multiple way to draw a Lewis structure. For example:
*Source from Michael Blaber
Q12.6
Draw resonance structures for the carbonate ion ($CO_3^{-2}$).
Q12.7a
How is it that this structure, PF5, have P have 5 bonds to it?
S12.7a
The d orbitals are available to make extra bonds allowing the P to bond to 5 Flourines instead of just Four.
Q12.7b
Draw the Lewis structure for nitric acid ($HNO_3$). This may or may not include resonance structures.
12.3: Hybridization of Atomic Orbitals
Q12.8a
What is the hybridization of carbon in CH4?
S12.8a
Carbon has the electron configuration 1s22s2sp2. In the excited state, because the 2s and 2p orbitals are so close one of the s electrons is able to fill the empty 2p orbital. All the carbon bonds are identical so the four unpaired electrons will be in a sp3 hybridization. The sp3 hybridization is the only hybridization that accounts for four identical orbitals.
Q12.8b
Describe the pattern of valence electrons across the periodic table (left to right and top to bottom), excluding the transition metals.
S12.8b
The number of valence electrons increases going left to right along the same row and stays constant going from top to bottom along the same column.
Q12.9a
What is the difference between covalent bond and ionic bond wave functions?
S12.9a
An ionic bond wave function takes into account the probability that the electrons of a multiatomic molecule might exist on the same atom; the covalent bond wave function assumes they exist on separate atoms.
Q12.9b
What is the hybridization of HCl?
S12.9b
There is no orbital hybridization in HCl. The $3p_z$ of Cl can overlap with the 1s of H just fine; they are only 0.06 eV apart in energy However, the 3s of Cl and the 1s of H are too far away in energy to interact, so there is no reason why the 3p of Cl can interact with the 3s of Cl anyway.
Q13.7
Calculate the potential energy of interaction between an electron situated 5.5 Å away from an HF molecule with a dipole moment of 2.5 D.
S13.7
$\mu = 2.5 D \times \dfrac{3.3356\times 10^{30}C m}{D} = 8.39 \times 10^{-30}Cm$
$r = 5.5 \; angstroms= 5.5 \times 10^{-10}m$
$V = \dfrac{q\mu}{4\pi \varepsilon _{o}r^{2}} = \dfrac{(+1.602\times10^{-19}C)(8.39\times 10^{-30}Cm)}{4\pi (8.851\times 10^{-12}C^{-2}N^{-1}m^{-2})(5.5\times 10^{-10})} = 2.196\times 10^{-11} J$
Q12.10
Give two types of bonding that present in the molecule CH2. how do you do this?
S12.10
1 is C atom to be not hydrid, no paired electrons and the molecule will be diamagnetic.
2 is C atom is sp3 hybbridized with 2 of the hybrid orbitals used for bonding to H's , unpaired electron. this is paramagnetic.
Q12.11
Describe bonding CO2 and C3H4 as hybridization. Draw diagram of forming of sigma and pi bond of C3H4 there is a double bond CO2 and its hydridization is sp and C3H4 is sp2 belong to CH2 and 1 sp of C=C=C.
Q12.12
Give the bonding scheme in term of molecular orbital H2+, H2, He2+, He. Besides, give the species in the following decreasing stability.
Q12.13
Between F2 and F2+ which one has longer bond?
S12.13
$F_2^+$ has bond order 1.5, $F_2$ has bond order 1.0. The bond lengths are inverse to the bond order, so the order is F2+ < F2 .There is 1 unpaired electron in F2+, 0 unpaired electrons in F2
Q12.14
Which of the following molecules have the shortest bond: F2, N2, O2? Why?
S12.14
N2 has the shortest bond because the bond order is 3 while O2 is 2 and F2 is 1.
12.5: Molecular Orbital Theory
Q12.15
Draw the molecular orbital diagram for F2 with the atomic orbitals labeled and find the bond order.
Q12.16
Comparing benezene and cyclopentadiene, which one has a lesser degree of $\pi$-electron delocalization?
Q12.19
Draw the molecular orbital energy diagram for the F2 Then compare the relative stability of the molecule to F2+.
Answer
BOND ORDER OF F2 = (8-6)/2 =1
BOND ORDER OF F2+ (8-5)/2 =1.5
Based on bond order F2+ is stable than F2
Q12.21
Diatomic carbon (C2) exists at very high temperatures. What type or types of bonds do you expect a molecule of diatomic carbon to have?
S12.21
1. Draw a molecular orbital diagram.
Answer: The molecular orbital diagram predicts that there will be a double bond consisting of two π bonds.
Q12.21
Determine the second period for molecule O2 by using molecular orbital?
S12.21
Based on general diagram for molecular orbital of second row of O2 molecule, the 2p MO is below the 2p molecular orbitals.
Q12.30
Though there are some exceptions, the double bond rule states that atoms of period 3 and greater do not form double or triple bonds. Explain why this is generally true. Hint: What properties of elements change with increasing period?
S12.30
Pi bonds are usually weaker than sigma bonds because there is less orbital overlap. The larger atomic radius of period 3 and greater elements means that there is even less overlap between the orbitals that would form pi bonds. There is not enough overlap to form the pi bonds that would make for double or triple bonds.
Q12.30
Explain why the Si=Si double bond is not stable than the C=C. Given the covalent radius of carbon is 77 pm and Silicon is 111pm?
S12.30
Silicon is significantly bigger than carbon. Thus the Si-Si bond is longer than C-C bond. However, the longer bond result in lowering the effective overlap of 3p orbital to form a π bonds. In contrast, C atoms can form the π bond from 2p orbital. Thus the Si=Si is less stable despite its larger size.
Q12.33
Myohemerythrin is an iron-containing protein that binds oxygen in marine invertebrates. The protein is monomeric (has one subunit) and contains 0.8107% iron by mass. Estimate the molar mass of myohemerythrin. Is your answer the same as the known molar mass of myohemerythrin, 13,780 grams per mole? If it is not, account for the difference between your estimate and the known value.
Myohemerythrin ribbon schematic, a 4-helix bundle protein structure with a bound iron. Hand drawn by Jane Richardson in 1980, from PDB file 2MHR.
S12.33
1. Find the mass of the protein if there is one atom of iron per protein.
$55.85\ g\ Fe/mol \times \dfrac{100.0\ g\ protein}{0.8107\ g\ Fe}$
$6889 g/mol$
2. This is much less than the known molar mass. In fact, it is half.
$\dfrac{13780\ g/mol}{6889\ g/mol} = 2.000$
We know the protein is monomeric, so there must be two atoms of iron in each protein.
Q12.19
Neon is a noble gas. Draw the molecular orbital diagram of $Ne_2$ and $Ne_2$. From that, describe the bonding scheme of those two molecules based on molecular orbital theory.
S12.19
Based on the molecular orbital diagram, there are more anti-bonding than bonding state. The bonding order of Ne2 is:
BO = 1/2 (8-8) = 0
Therefore, $Ne_2$ does not exist in nature.
Q12.21
What is the difference between the last three group, including O2, F2 and the other diatomic molecules in the second row of Periodic table in term of Molecular orbital theory?
S12.21
For O2 and F2, the pi bonds(2p) have higher energy than the sigma bonds (2p). Whereas, a group of B2, C2, and N2, the sigma bonds(2p) have higher energy than the pi bonds(2p)
Q12.30
Be is in the second period, and very rare Be2 exist in nature. Explain the instability of Be2 , and calculate the bond order.
S12.30
Be2 molecule has 2 valence electrons. The bonding order, BO= 1/2 (2-2)= 0 Therefore, diatomic Be2 is not stable! Not exist in nature
12.4: Electronegativity and Dipole Moment
Q12.36
Will the dipole moment of a cis-dichloroethylene molecule increase or decrease upon heating? Why?
S12.36
The dipole moment will decrease because cis-trans isomerization will occur when heated. This is because the trans isomer of the molecule does not have a dipole moment present.
Q12.37
Draw the $\sigma$ and $\sigma^*$ molecular orbital of $CO$. Draw the MO energy level diagram and write the electron coefficient. Calculate the Bond order.
S12.37
Electron Configuration:
($\sigma$1s)2($\sigma$*1s)2($\sigma$2s)2($\sigma$*2s)2($\pi$x)2($\pi$y)2($\sigma$2sp)2
Bond Order: 1/2(Bonding Electrons-Antibonding Electrons)
=1/2(8-2) = 3
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Intermolecular forces are forces of attraction or repulsion which act between neighboring particles (atoms, molecules, or ions). They are weak compared to the intramolecular forces, the forces which keep a molecule together.
13: Intermolecular Forces
Learning Objectives
• Classify intermolecular forces as ionic, covalent, London dispersion, dipole-dipole, or hydrogen bonding.
• Explain properties of material in terms of type of intermolecular forces.
• Predict the properties of a substance based on the dominant intermolecular force.
Forces binding atoms in a molecule are due to chemical bonding. The energy required to break a bond is called the bond-energy. For example, the average bond-energy for $\ce{O-H}$ bonds in water is 463 kJ/mol. On average, 463 kJ is required to break 6.023x1023 $\ce{O-H}$ bonds, or 926 kJ to convert 1.0 mole of water into 1.0 mol of $\ce{O}$ and 2.0 mol of $\ce{H}$ atoms. The forces holding molecules together are generally called intermolecular forces. The energy required to break molecules apart is much smaller than a typical bond-energy, but intermolecular forces play important roles in determining the properties of a substance. Intermolecular forces are particularly important in terms of how molecules interact and form biological organisms or even life. This link gives an excellent introduction to the interactions between molecules.
Classifying Intermolecular Forces
In general, intermolecular forces can be divided into several categories. The four prominent types are:
• Ion-Ion Interactions: Recall lattice energy and its relation to properties of solids. The more ionic, the higher the lattice energy. Examine the following list and see if you can explain the observed values by way of ionic attraction: $\mathrm{LiF = 1036}$; $\mathrm{LiI = 737}$ ; $\mathrm{KF = 821}$; $\mathrm{MgF_2 = 2957\: kJ/mol}$
• Dipole-dipole Interactions: Substances whose molecules have dipole moment have a higher melting point or boiling point than those of similar molecular mass, but whose molecules have no dipole moment.
• London dispersion forces or van der Waal's force: These forces always operate in any substance. The force arisen from induced dipole and the interaction is weaker than the dipole-dipole interaction. In general, the heavier the molecule, the stronger the van der Waal's force of interaction. For example, the boiling points of inert gases increase as their atomic masses increase due to stronger London dispersion interactions.
• Hydrogen bonds: Certain substances such as $\ce{H2O}$, $\ce{HF}$, and $\ce{NH3}$ form hydrogen bonds, which affects properties (mp, bp, solubility) of the substance. Other compounds containing $\ce{OH}$ and $\ce{NH2}$ groups also form hydrogen bonds. Molecules of many organic compounds such as alcohols, acids, amines, and amino acids contain these groups, and thus hydrogen bonding plays a important role in biological science.
• Covalent bonding: Covalent bonding is really intramolecular force rather than intermolecular force. It is mentioned here, because some solids are formed due to covalent bonding. For example, in diamond, silicon, quartz etc., all the atoms in the entire crystal are linked together by covalent bonding. These solids are hard, brittle, and have high melting points. Covalent bonding holds atoms tighter than ionic attraction.
• Metallic bonding: Forces between atoms in metallic solids belong to another category. Valence electrons in metals are rampant. They are not restricted to certain atoms or bonds. Rather they run freely in the entire solid, providing good conductivity for heat and electric energy. This behavior of electrons gives special properties such as ductility and mechanical strength to metals.
The division into types is for convenience in their discussion. Of course all types can be present simultaneously for many substances. Usually, intermolecular forces are discussed together with The States of Matter. Intermolecular forces also play important roles in solutions, a discussion of which is given in Hydration, solvation in water. A summary of the interactions is illustrated in the following diagram:
See if you can answer the following questions.
• What are dipoles?
• What are dipole moments?
• How do dipoles interact?
• Why do molecules attract one another?
• How do London dispersion forces come about?
• What parameters cause an increase of the London dispersion forces?
• What is a hydrogen bond?
• What type of hydrogen bonds are strong?
• What chemical groups are hydrogen acceptors for hydrogen bonds?
If you are looking for specific information, your study will be efficient. Some answers can be found in the Confidence Building Questions. Consider carefully the purpose of each question, and figure out what there is to be learned in it.
Confidence Building Questions
1. Which of the following molecules have a permanent dipole moment?
1. $\ce{H2O}$
2. $\ce{CO2}$
3. $\ce{CH4}$
4. $\ce{N2}$
5. $\ce{CO}$
6. $\ce{NH3}$
Hint: a e f
Discussion -
$\ce{CO2}$, $\ce{CH4}$, and $\ce{N2}$ are symmetric, and hence they have no permanent dipole moments. A molecule with polar bonds unsymmetrically arranged will possess a permanent dipole.
2. Which has the higher boiling point, $\ce{Br2}$ or $\ce{ICl}$?
Hint: $\ce{ICl}$
Discussion -
They have similar molecular weights: $\mathrm{Br_2 = 160}$; $\mathrm{ICl = 162}$. Their boiling points are 332 K and 370 K respectively.
3. An atom or molecule can be temporarily polarized by a nearby species. Polarization separates centers of charge giving
1. permanent dipole
2. temporary charges
3. hydrogen bonding
4. induced dipole
5. induced ions
6. radicals
Hint: d.
Discussion -
Induced dipoles are responsible for the London dispersion forces. The heavier the molecule, the larger the induced dipole will be. Thus, London dispersion forces are strong for heavy molecules.
4. Which has a higher boiling point, $\ce{I2}$ or $\ce{Br2}$?
Hint: iodine, $\ce{I2}$.
Discussion -
Atomic weights for $\ce{Br}$ and $\ce{I}$ are 80 and 127 respectively. The higher the molecular weight, the stronger the London dispersion forces.
5. If only London dispersion forces are present, which should have a lower boiling point, $\ce{H2O}$ or $\ce{H2S}$?
Hint: water, $\ce{H2O}$
Discussion -
The b.p. for $\ce{H2O}$ is 100 deg C, and that of $\ce{H2S}$ is -70 deg C. Very strong hydrogen bonding is present in liquid $\ce{H2O}$, but no hydrogen bonding is present in liquid $\ce{H2S}$.
6. Contrary to most other substances, the density of water decreases as temperature decreases between 4 and 0 deg C. This is due to
1. dipole-dipole interaction
2. London dispersion
3. decreasing number of hydrogen bonds
4. increasing number of hydrogen bonds formed
Hint: d.
Discussion -
As more hydrogen bonds form when the temperature decreases, the volume expands, causing a decrease in density. Above 4 deg C, the thermal expansion is more prominent than the effect of hydrogen bonds.
7. Ethanol ($\ce{C2H5OH}$, molar mass 46) boils at 351 K, but water ($\ce{H2O}$, molar mass 18) boils at higher temperature, 373 K. This is because:
1. water is denser
2. water has stronger London dispersion forces
3. water has stronger hydrogen bonds
4. water molecules contain no carbon
Hint: c.
Discussion -
A hydrogen atom between two small, electronegative atoms (such as $\ce{F}$, $\ce{O}$, $\ce{N}$) causes a strong intermolecular interaction known as the hydrogen bond. The strength of a hydrogen bond depends upon the electronegativities and sizes of the two atoms.
8. Ethanol ($\ce{C2H5OH}$) and methyl ether ($\ce{CH3OCH3}$) have the same molar mass. Which has a higher boiling point?
Hint: Ethanol has a higher boiling point.
Discussion -
$\ce{R-OH}$ group is both proton donor and acceptor for hydrogen bonding. Methyl groups have very weak hydrogen bonding, if any.
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The interactions between ions (ion - ion interactions) are the easiest to understand: like charges repel each other and opposite charges attract. These Coulombic forces operate over relatively long distances in the gas phase. The force depends on the product of the charges (\(Z_1\), \(Z_2\)) divided by the square of the distance of separation (\(d^2\)):
\[F \propto \dfrac{- Z_1Z_2}{d^2} \tag{1}\]
Two oppositely-charged particles flying about in a vacuum will be attracted toward each other, and the force becomes stronger and stronger as they approach until eventually they will stick together and a considerable amount of energy will be required to separate them.
Figure 1: Ion - Ion Interactions in the Gas Phase
They form an ion-pair, a new particle which has a positively-charged area and a negatively-charged area. There are fairly strong interactions between these ion pairs and free ions, so that these the clusters tend to grow, and they will eventually fall out of the gas phase as a liquid or solid (depending on the temperature).
13.04: Hydrogen Bonding
This page explains the origin of hydrogen bonding - a relatively strong form of intermolecular attraction.
The evidence for hydrogen bonding
Many elements form compounds with hydrogen. If you plot the boiling points of the compounds of the Group 4 elements with hydrogen, you find that the boiling points increase as you go down the group.
The increase in boiling point happens because the molecules are getting larger with more electrons, and so van der Waals dispersion forces become greater. If you repeat this exercise with the compounds of the elements in Groups 5, 6 and 7 with hydrogen, something odd happens.
Although for the most part the trend is exactly the same as in group 4 (for exactly the same reasons), the boiling point of the compound of hydrogen with the first element in each group is abnormally high. In the cases of NH3, H2O and HF there must be some additional intermolecular forces of attraction, requiring significantly more heat energy to break. These relatively powerful intermolecular forces are described as hydrogen bonds.
The origin of hydrogen bonding
The molecules which have this extra bonding are:
The solid line represents a bond in the plane of the screen or paper. Dotted bonds are going back into the screen or paper away from you, and wedge-shaped ones are coming out towards you. Notice that in each of these molecules:
• The hydrogen is attached directly to one of the most electronegative elements, causing the hydrogen to acquire a significant amount of positive charge.
• Each of the elements to which the hydrogen is attached is not only significantly negative, but also has at least one "active" lone pair.
• Lone pairs at the 2-level have the electrons contained in a relatively small volume of space which therefore has a high density of negative charge. Lone pairs at higher levels are more diffuse and not so attractive to positive things.
Consider two water molecules coming close together.
The + hydrogen is so strongly attracted to the lone pair that it is almost as if you were beginning to form a co-ordinate (dative covalent) bond. It doesn't go that far, but the attraction is significantly stronger than an ordinary dipole-dipole interaction. Hydrogen bonds have about a tenth of the strength of an average covalent bond, and are being constantly broken and reformed in liquid water. If you liken the covalent bond between the oxygen and hydrogen to a stable marriage, the hydrogen bond has "just good friends" status.
Water as a "perfect" example of hydrogen bonding
Notice that each water molecule can potentially form four hydrogen bonds with surrounding water molecules. There are exactly the right numbers of + hydrogens and lone pairs so that every one of them can be involved in hydrogen bonding. This is why the boiling point of water is higher than that of ammonia or hydrogen fluoride. In the case of ammonia, the amount of hydrogen bonding is limited by the fact that each nitrogen only has one lone pair. In a group of ammonia molecules, there aren't enough lone pairs to go around to satisfy all the hydrogens. In hydrogen fluoride, the problem is a shortage of hydrogens. In water, there are exactly the right number of each. Water could be considered as the "perfect" hydrogen bonded system.
The diagram shows the potential hydrogen bonds formed to a chloride ion, Cl-. Although the lone pairs in the chloride ion are at the 3-level and wouldn't normally be active enough to form hydrogen bonds, in this case they are made more attractive by the full negative charge on the chlorine.
However complicated the negative ion, there will always be lone pairs that the hydrogen atoms from the water molecules can hydrogen bond to.
Hydrogen bonding in alcohols
An alcohol is an organic molecule containing an -O-H group. Any molecule which has a hydrogen atom attached directly to an oxygen or a nitrogen is capable of hydrogen bonding. Such molecules will always have higher boiling points than similarly sized molecules which don't have an -O-H or an -N-H group. The hydrogen bonding makes the molecules "stickier", and more heat is necessary to separate them.
Ethanol, CH3CH2-O-H, and methoxymethane, CH3-O-CH3, both have the same molecular formula, C2H6O.
They have the same number of electrons, and a similar length to the molecule. The van der Waals attractions (both dispersion forces and dipole-dipole attractions) in each will be much the same. However, ethanol has a hydrogen atom attached directly to an oxygen - and that oxygen still has exactly the same two lone pairs as in a water molecule. Hydrogen bonding can occur between ethanol molecules, although not as effectively as in water. The hydrogen bonding is limited by the fact that there is only one hydrogen in each ethanol molecule with sufficient + charge.
In methoxymethane, the lone pairs on the oxygen are still there, but the hydrogens aren't sufficiently + for hydrogen bonds to form. Except in some rather unusual cases, the hydrogen atom has to be attached directly to the very electronegative element for hydrogen bonding to occur. The boiling points of ethanol and methoxymethane show the dramatic effect that the hydrogen bonding has on the stickiness of the ethanol molecules:
ethanol (with hydrogen bonding) 78.5°C
methoxymethane (without hydrogen bonding) -24.8°C
The hydrogen bonding in the ethanol has lifted its boiling point about 100°C.
It is important to realize that hydrogen bonding exists in addition to van der Waals attractions. For example, all the following molecules contain the same number of electrons, and the first two are much the same length. The higher boiling point of the butan-1-ol is due to the additional hydrogen bonding.
Comparing the two alcohols (containing -OH groups), both boiling points are high because of the additional hydrogen bonding due to the hydrogen attached directly to the oxygen - but they are not the same. The boiling point of the 2-methylpropan-1-ol isn't as high as the butan-1-ol because the branching in the molecule makes the van der Waals attractions less effective than in the longer butan-1-ol.
Hydrogen bonding in nitrogen containing organic molecules
Hydrogen bonding also occurs in organic molecules containing N-H groups - in the same sort of way that it occurs in ammonia. Examples range from simple molecules like CH3NH2 (methylamine) to large molecules like proteins and DNA. The two strands of the famous double helix in DNA are held together by hydrogen bonds between hydrogen atoms attached to nitrogen on one strand, and lone pairs on another nitrogen or an oxygen on the other one.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/13%3A_Intermolecular_Forces/13.02%3A_The_Ionic_Bond.txt
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With 70% of our earth being ocean water and 65% of our bodies being water, it is hard to not be aware of how important it is in our lives. There are 3 different forms of water, or H2O: solid (ice), liquid (water), and gas (steam). Because water seems so ubiquitous, many people are unaware of the unusual and unique properties of water, including:
• Boiling Point and Freezing Point
• Surface Tension, Heat of Vaporization, and Vapor Pressure
• Viscosity and Cohesion
• Solid State
• Liquid State
• Gas State
Boiling Point and Freezing Point
If you look at the periodic table and locate tellurium (atomic number: 52), you find that the boiling points of hydrides decrease as molecule size decreases. So the hydride for tellurium: H2Te (hydrogen telluride) has a boiling point of -4°C. Moving up, the next hydride would be H2Se (hydrogen selenide) with a boiling point of -42°C. One more up and you find that H2S (hydrogen sulfide) has a boiling point at -62°C. The next hydride would be H2O (WATER!). And we all know that the boiling point of water is 100°C. So despite its small molecular weight, water has an incredibly big boiling point. This is because water requires more energy to break its hydrogen bonds before it can then begin to boil. The same concept is applied to freezing point as well, as seen in the table below. The boiling and freezing points of water enable the molecules to be very slow to boil or freeze, this is important to the ecosystems living in water. If water was very easy to freeze or boil, drastic changes in the environment and so in oceans or lakes would cause all the organisms living in water to die. This is also why sweat is able to cool our bodies.
COMPOUND BOILING POINT FREEZING POINT
Hydrogen Telluride -4°C -49°C
Hydrogen Selenide -42°C -64°C
Hydrogen Sulfide -62°C -84°C
Water 100°C 0 °C
Surface Tension, Heat of Vaporization, and Vapor Pressure
Besides mercury, water has the highest surface tension for all liquids. Water's high surface tension is due to the hydrogen bonding in water molecules. Water also has an exceptionally high heat of vaporization. Vaporization occurs when a liquid changes to a gas, which makes it an endothermic reaction. Water's heat of vaporization is 41 kJ/mol. Vapor pressure is inversely related to intermolecular forces, so those with stronger intermolecular forces have a lower vapor pressure. Water has very strong intermolecular forces, hence the low vapor pressure, but it's even lower compared to larger molecules with low vapor pressures.
• Viscosity is the property of fluid having high resistance to flow. We normally think of liquids like honey or motor oil being viscous, but when compared to other substances with like structures, water is viscous. Liquids with stronger intermolecular interactions are usually more viscous than liquids with weak intermolecular interactions.
• Cohesion is intermolecular forces between like molecules; this is why water molecules are able to hold themselves together in a drop. Water molecules are very cohesive because of the molecule's polarity. This is why you can fill a glass of water just barely above the rim without it spilling.
Solid State (Ice)
All substances, including water, become less dense when they are heated and more dense when they are cooled. So if water is cooled, it becomes more dense and forms ice. Water is one of the few substances whose solid state can float on its liquid state! Why? Water continues to become more dense until it reaches 4°C. After it reaches 4°C, it becomes LESS dense. When freezing, molecules within water begin to move around more slowly, making it easier for them to form hydrogen bonds and eventually arrange themselves into an open crystalline, hexagonal structure. Because of this open structure as the water molecules are being held further apart, the volume of water increases about 9%. So molecules are more tightly packed in water's liquid state than its solid state. This is why a can of soda can explode in the freezer.
Liquid State (Liquid Water)
It is very rare to find a compound that lacks carbon to be a liquid at standard temperatures and pressures. So it is unusual for water to be a liquid at room temperature! Water is liquid at room temperature so it's able to move around quicker than it is as solid, enabling the molecules to form fewer hydrogen bonds resulting in the molecules being packed more closely together. Each water molecule links to four others creating a tetrahedral arrangement, however they are able to move freely and slide past each other, while ice forms a solid, larger hexagonal structure.
Gas State (Steam)
As water boils, its hydrogen bonds are broken. Steam particles move very far apart and fast, so barely any hydrogen bonds have the time to form. So, less and less hydrogen bonds are present as the particles reach the critical point above steam. The lack of hydrogen bonds explains why steam causes much worse burns that water. Steam contains all the energy used to break the hydrogen bonds in water, so when steam hits your face you first absorb the energy the steam has taken up from breaking the hydrogen bonds it its liquid state. Then, in an exothermic reaction, steam is converted into liquid water and heat is released. This heat adds to the heat of boiling water as the steam condenses on your skin.
Water as the "Universal Solvent"
Because of water's polarity, it is able to dissolve or dissociate many particles. Oxygen has a slightly negative charge, while the two hydrogens have a slightly positive charge. The slightly negative particles of a compound will be attracted to water's hydrogen atoms, while the slightly positive particles will be attracted to water's oxygen molecule; this causes the compound to dissociate. Besides the explanations above, we can look to some attributes of a water molecule to provide some more reasons of water's uniqueness:
• Forgetting fluorine, oxygen is the most electronegative non-noble gas element, so while forming a bond, the electrons are pulled towards the oxygen atom rather than the hydrogen. This creates 2 polar bonds, which make the water molecule more polar than the bonds in the other hydrides in the group.
• A 104.5° bond angle creates a very strong dipole.
• Water has hydrogen bonding which probably is a vital aspect in waters strong intermolecular interaction
Why is this important for the real world?
The properties of water make it suitable for organisms to survive in during differing weather conditions. Ice freezes as it expands, which explains why ice is able to float on liquid water. During the winter when lakes begin to freeze, the surface of the water freezes and then moves down toward deeper water; this explains why people can ice skate on or fall through a frozen lake. If ice was not able to float, the lake would freeze from the bottom up killing all ecosystems living in the lake. However ice floats, so the fish are able to survive under the surface of the ice during the winter. The surface of ice above a lake also shields lakes from the cold temperature outside and insulates the water beneath it, allowing the lake under the frozen ice to stay liquid and maintain a temperature adequate for the ecosystems living in the lake to survive.
Resources
1. Cracolice, Mark S. and Edward Peters I. Basics of Introductory Chemistry. Thompson, Brooks/Cole Publishing Company. 2006
2. Petrucci, et al. General Chemistry: Principles & Modern Applications: AIE (Hardcover). Upper Saddle River: Pearson/Prentice Hall, 2007.
Contributors and Attributions
• Corinne Yee (UCD), Desiree Rozzi (UCD)
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Hydrophobic interactions describe the relations between water and hydrophobes (low water-soluble molecules). Hydrophobes are nonpolar molecules and usually have a long chain of carbons that do not interact with water molecules. The mixing of fat and water is a good example of this particular interaction. The common misconception is that water and fat doesn’t mix because the Van der Waals forces that are acting upon both water and fat molecules are too weak. However, this is not the case. The behavior of a fat droplet in water has more to do with the enthalpy and entropy of the reaction than its intermolecular forces.
Causes of Hydrophobic Interactions
American chemist Walter Kauzmann discovered that nonpolar substances like fat molecules tend to clump up together rather than distributing itself in a water medium, because this allow the fat molecules to have minimal contact with water.
The image above indicates that when the hydrophobes come together, they will have less contact with water. They interact with a total of 16 water molecules before they come together and only 10 atoms after they interact.
Thermodynamics of Hydrophobic Interactions
When a hydrophobe is dropped in an aqueous medium, hydrogen bonds between water molecules will be broken to make room for the hydrophobe; however, water molecules do not react with hydrophobe. This is considered an endothermic reaction, because when bonds are broken heat is put into the system. Water molecules that are distorted by the presence of the hydrophobe will make new hydrogen bonds and form an ice-like cage structure called a clathrate cage around the hydrophobe. This orientation makes the system (hydrophobe) more structured with an decrease of the total entropy of the system; therefore $\Delta S < 0$.
The change in enthalpy ($\Delta H$) of the system can be negative, zero, or positive because the new hydrogen bonds can partially, completely, or over compensate for the hydrogen bonds broken by the entrance of the hydrophobe. The change in enthalpy, however, is insignificant in determining the spontaneity of the reaction (mixing of hydrophobic molecules and water) because the change in entropy ($\Delta S$) is large.
According to the Gibbs Energy formula
$\Delta G = \Delta H - T\Delta S \label{eq1}$
with a small unknown value of $\Delta H$ and a large negative value of $\Delta{S}$, the value of $\Delta G$ will turn out to be positive. A positive $\Delta G$ indicates that the mixing of the hydrophobe and water molecules is not spontaneous.
Formation of Hydrophobic Interactions
The mixing hydrophobes and water molecules is not spontaneous; however, hydrophobic interactions between hydrophobes are spontaneous. When hydropobes come together and interact with each other, enthalpy increases ( $\Delta H$ is positive) because some of hydrogen bonds that form the clathrate cage will be broken. Tearing down a portion of the clathrate cage will cause the entropy to increase ( $\Delta S$ is positive), since forming it decreases the entropy.
According to the Equation $\ref{eq1}$
• $\Delta{H}$ = small positive value
• $\Delta{S}$ = large positive value
Result: $\Delta{G}$ is negative and hence hydrophobic interactions are spontaneous.
Strength of Hydrophobic Interactions
Hydrophobic interactions are relatively stronger than other weak intermolecular forces (i.e., Van der Waals interactions or Hydrogen bonds). The strength of Hydrophobic Interactions depend on several factors including (in order of strength of influence):
1. Temperature: As temperature increases, the strength of hydrophobic interactions increases also. However, at an extreme temperature, hydrophobic interactions will denature.
2. Number of carbons on the hydrophobes: Molecules with the greatest number of carbons will have the strongest hydrophobic interactions.
3. The shape of the hydrophobes: Aliphatic organic molecules have stronger interactions than aromatic compounds. Branches on a carbon chain will reduce the hydrophobic effect of that molecule and linear carbon chain can produce the largest hydrophobic interaction. This is so because carbon branches produce steric hindrance, so it is harder for two hydrophobes to have very close interactions with each other to minimize their contact to water.
Biological Importance of Hydrophobic Interactions
Hydrophobic Interactions are important for the folding of proteins. This is important in keeping a protein stable and biologically active, because it allow to the protein to decrease in surface are and reduce the undesirable interactions with water. Besides from proteins, there are many other biological substances that rely on hydrophobic interactions for its survival and functions, like the phospholipid bilayer membranes in every cell of your body!
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13.1: Intermolecular Interactions
Q13.1a
List all the intermolecular interactions that take place in each of the follow kings of molecules: $CCl_3F$, $CCl_2F_2$, $CClF_3$, and $CF_4$.
Q13.1b
Determine what type of intermolecular forces exist in the following molecules: LiF, MgF2, H2O, and HF.
S13.1b
• H2O: London Force, Dipole-Dipole interaction, Hydrogen bonds.
• HF: Dipole-Dipole intermolecular forces, Hydrogen bonds.
• MgF2 and LiF: strong ionic attraction.
Q13.2a
Arrange the follow species in order of decreasing melting points: CsBr, KI, KCL, MgF2.
Q13.2b
Which has the highest boiling point I2, Br2, and Cl2. Explain why?
S13.2b
The atomic weigh of Iodine = 127, Bromine = 80, and Chlorine = 35.5. The weigh is proportion to the London dispersion force, and the higher molecular weigh, the larger the force. Thus, I2 has a highest boiling point.
Q13.3
1-Propanol C3H7OH and methoxyethane CH3O C2H5 have the same molecular weigh. Which has the higher boiling point?
S13.3
The 1-Propanol can form London Force, Dipole- Dipole, and H- bonding due to the H bonded to O atom of OH group, whereas the methoxyethane can not form the H-bonding. Therefore, the 1-Propanol has higher intermolecular attractive force and thus a higher boiling point.
Q13.3
Why do the lightest compounds such as NH3, H2O, and HF have the highest boiling points?
Q13.4a
What kind of attractive interaction exists between atoms and between nonpolar molecules?
Q13.4b
Why nature gas CH4 is a good choice to storage tank in winter?
S13.4b
It's about boiling point. The methane has the boiling point at -161 °C, making it to be a good choice for winter season.
Q13.4
Explain why methane (CH_4\) is used as the primary heating gas in Alaska during wintertime instead of the more commonly used butant or propane gases use in the lower 48 states.
S13.4
Methane ($CH_4$) remains gas because its boiling point is about -160°C. Other gases, such as propane or butane, would liquefy under freezing condition. Therefore, methane is more likely to be used during wintertime at Alaska
Q13.5
Define types of intermolecular forces and give example for each.
S13.5
There are 3 types of intermolecular force: London Dispersion, Dipole-Dipole (Example: Two $NaCl$) and Ion-Dipole (Example: $Mg^+$ and $HCl$)
• Dipole- Dipole occurs between polar molecules
• Ion- Dipole occurs between an ion and polar molecules
• London Dispersion occurs between the nonpolar molecules.
Q13.6
How does the intermolecular determine the boiling point?
S13.6
The weakest intermolecular, the lowest boiling point.
Q13.7a
What is their dipole-dipole interaction of wo HCl molecules are co-linear head-to tail.
Given: The dipole moment of HF is 1.86 D. The dipole moment of HCl is 1.05 D. The distance between the two is 1.78
H+-F- - - - - H+-Cl-
S13.7a
$V=-\dfrac{2\mu _{A}\mu _{B}}{4\pi \varepsilon _{0}r^{3}}$
$V=-\dfrac{2(1.05)(1.86)}{4\pi (8.854187817\cdot 10^{-12})(1.78)^{3}}$
13.2: The Ionic Bond
Q13.7b
Calculate an ion-ion interaction energy between $K^+$ and $Cl^-$ at a distance of 600 pm.
S13.7b
The ion-ion interaction energy is given by Coulomb's law.
$V = \dfrac{q_1q_2}{4 \pi \epsilon_o r}$
$V = \dfrac{- (1.602 \times 10^{-19}\;\cancel{C})(1.602 \times 10^{-19} \cancel{C})}{4 \pi (8.853 \times 10^{-12} \cancel{C^2} \cdot N^{−1} \cdot m \cancel{^{−2}})(6 \times 10^{-10}\; \cancel{m})} = -3.84 \times 10^{-19} \; J$
Q13.7
Calculate an ion-dipole interaction energy between $K^+$ and $HCl$ at a distance of 600 pm. $HCl$ has a dipole moment of $1.08\;D$.
S13.7
$\mu = 1.08 \cancel{D} \times \dfrac{3.3356 \times 10^{30} \; C \cdot m}{1\;\cancel{D}} = 3.6 \times 10^{-30}\; C \cdot m$
$V = \dfrac{-q\;\mu}{4 \pi \epsilon_o r^2} = \dfrac{- (1.602 \times 10^{-19}\;\cancel{C})(3.6 \times 10^{-30} \cancel{C} \cdot \cancel{m})}{4 \pi (8.853 \times 10^{-12} \cancel{C^2} \cdot N^{−1} \cdot m \cancel{^{−2}})(6 \times 10^{-10}\; \cancel{m})^2} = -1.44 \times 10^{-20} \; J$
As expected this is appreciably smaller in energy than covalent bonds (e..g, $HCl$ has a bond enthalpy of $7.0 \times 10^{-19}\;J$).
13.3: Types of Intermolecular Forces
Q13.8a
Rank the interactions from weakest to strongest:
1. H2O - - OH2
2. Li+ - - F-
3. Li+ - - OH2
S13.8a
1. ion-ion interaction: Li+ - - F-
2. ion-dipole interaction: Li+ - - OH2
3. dipole-dipole interaction: H2O - - OH2
Q13.8b
A low concentration electrolytic solution behaves non-ideally while a high concentration of the same solution behaves ideally. Explain this phenomenon in terms of forces, noting that Coulomb forces depend on 1/r2 while van der Waals forces depend on 1/r7. Which of these forces are low concentration electrolytic solutions likely to follow? High concentration?
S13.8b
The interatomic distances in a low concentration electrolytic solution are greater than those in a high concentration solution. They follow van der Waals forces and thus behave less ideally. High concentration electrolytic solutions follow Coulomb forces.
Q13.9a
Calculate the ion-dipole interaction between H2O and Li+. You are given the dipole moment of H2O is 1.82 D. The distance between these two is 2 Å.
S13.9a
$V=-\frac{q\mu }{4\pi \varepsilon _{0}r^{2}}$
$=\frac{1.82D\cdot(\frac{3.3356\cdot 10^{-30}Cm)}{1D}}{4\pi (8.85\cdot 10^{-12})(2\cdot 10^{-10}m)^{^{2}}} =1.36\; kJ/mol$
Q13.9b
Calculate the potential energy of interaction between a Cl- ion situated 120 pm away from an $H_2O$ molecule with a dipole moment of 1.85 D.
S13.9b
$\mu = 1.85 \cancel{D} \times \dfrac{3.3356 \times 10^{30} \; C \cdot m}{1\;\cancel{D}} = 6.18 \times 10^{-30}\; C \cdot m$
$r = 1.2 \times 10^{-10}\; m$
$V = \dfrac{q\mu}{4\pi \varepsilon _{o}r^{2}} = \dfrac{(-1.602\times10^{-19}\;C)(6.18 \times 10^{-30}\; C \cdot m)}{4\pi (8.851 \times 10^{-12}\; C^{-2}N^{-1}m^{-2})(1.2 \times 10^{-10} \; m) ^2}$
Q13.10
Do you expect a greater dipole-dipole interaction between two molecules that are antiparallel or between two molecules that are co-linear head-to-tail?
S13.10
You expect a stronger interaction when the two are co-linear head-to tail. This can be seen by looking at the formula or in the images of the two.
Q13.10
Express the equilibrium distance re in term ð and show V = - €
Q13.12
1. Determine Vander Waals radius of Argon.
2. Use this radius to find fraction of volume by 2 mole of argon at room temperature at 1 atm.
S13.12
a. r= σ/2 = 3.40 A0 /2 = 1.70 A0
b. Volume of 2 mole of Ar
4/3 πr^3 ((6.022 x 10^23)/(2 mol))= 4/3 π (1.70 x 10^(-10) m)^3 ((6.022 x 10^23)/(2 mol))
= 6.19 x 10-6 L mol-1
V/n=RT/P= ((0.08206 L atm K^(-1) mol^(-1) (298.2 K))/1atm
= 24.5 L mol-1
The fraction of this volume occupied by 2 mole of Ar
(1.239 x 10^-2 L mol-1)/ 24.47 L mol-1 = 2.5 x10 -7
Q13.15
a) What is the original of polarity in a molecule?
b) Is CO2 polar? Explain.
Q13.16
Of the following compound, which one(s) is/are soluble?
1. CH4
2. NCl3
3. C6H6
4. CO(NH2)2
Q13.17
What makes a compound soluble in water? Explain using examples.
Q13.18
Explain why does water have a high specific heat.
Q13.20
The energy of a hydrogen bond for each base pair in DNA is 15 kJ/mol. Two complimentary strands has 50 base pairs each. What is the ratio of the 2 different strands to hydrogen double helix in a solution given a temperature of 300 K.
S13.20
First calculate the ratio of the two different strands for just one pair.
$e^{\Delta E/RT}=exp[(15 \times 10^3\; J/mol)/(8.314\; J/K*mol)(300\; K) = 2.4 \times 10^{-3}$
Since there are 50 base pairs, we need to multiply by 50 to account for all the base pairs.
exp[100X(15X103 J/mol)/(8.314 J/K*mol)(300K) = 0
Q13.24
Consider two pure liquids. One has strong intermolecular interactions, and the other has relatively weak intermolecular interactions. For the following properties, indicate which of the liquids you would expect to have a higher value (answer with "strong" or "weak").
1. viscosity
2. vapor pressure
3. freezing point
4. surface tension
S13.24
1. Strong. Higher viscosity results from stronger interactions between the liquid molecules.
2. Weak. The liquid with weaker bonds takes less energy to turn into vapor, so it will exert a higher vapor pressure.
3. Strong. The freezing point is the same as the melting point; it takes more energy to melt a solid with stronger intermolecular interactions.
4. Strong. Surface tension is a result of intermolecular interactions. The stronger these interactions, the greater the surface tension.
Q13.25
Fun fact: if the DNA in a single human cell were stretched out (but still in its familiar double helix conformation), it would be approximately 2 meters long. The distance, along the helix, between nucleotides is 3.4 Å.
1. Estimate the number of basepairs in the haploid human genome, from the 2 meter fun fact.
2. The human body contains about 100 trillion cells. About a quarter of these are erythrocytes (red blood cells) and contain no genomic DNA. Use the average molar mass for a basepair, 650 grams per mole, to estimate how much of a human's mass is human genomic DNA.
3. At its closest, Pluto is 4.28 billion km from Earth. Do you have enough DNA to reach Pluto?
Hint: Humans are diploid.
S13.25
(a)
$2\ m/cell \times \dfrac{bp}{3.4\ Å} \times \dfrac{10^{10}\ Å}{m} \times \dfrac{cell}{2\ haploid\ genomes} = 3 \times 10^{9} \dfrac{bp}{haploid\ genome} \] Three billion basepairs. (b) 75 trillion of the human cells in your body have genomic DNA.$ 75 \times 10^{20}\ cells \times \dfrac{haploid\ genomes}{cell} \times \dfrac{3 \times 10^9\ bp}{haploid\ genome} \times \dfrac{mol}{6.022 \times 10^{23}} \times 650 \dfrac{g}{mol\ bp} = 200\ g \]
That's about half a pound.
(c)
\dfrac{2\ m}{cell} \times 75 \times 10^{12}\ cells \times \dfrac{km}{1000\ m} = 2 \times 10^{11}\ km \]
Yes, you have way more DNA than you need to stretch it from Earth to Pluto.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/13%3A_Intermolecular_Forces/13.E%3A_Intermolecular_Forces_%28Exercises%29.txt
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