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Nuclear Stability is a concept that helps to identify the stability of an isotope. The two main factors that determine nuclear stability are the neutron/proton ratio and the total number of nucleons in the nucleus.
Introduction
A isotope is an element that has same atomic number but different atomic mass compared to the periodic table. Every element has a proton, neutron, and electron. The number of protons is equal to the atomic number, and the number of electrons is equal the protons, unless it is an ion. To determine the number of neutrons in an element you subtract the atomic number from the atomic mass of the element. Atomic mass is represented as ($A$) and atomic number is represented as ($Z$) and neutrons are represented as ($N$).
$A=N + Z \label{1}$
The principal factor for determining whether a nucleus is stable is the neutron to proton ratio. Elements with ($Z<20$) are lighter and these elements' nuclei and have a ratio of 1:1 and prefer to have the same amount of protons and neutrons.
Example $1$: Carbon Isotopes
Carbon has three isotopes that scientists commonly used: $\ce{^12C}$, $\ce{^13C}$, $\ce{^14C}$. What is the the number of neutron, protons, total nucleons and $N:Z$ ratio for the $\ce{^12C}$ nuclide?
Solution
For this specific isotope, there are 12 total nucleons ($A$). From the periodic table, we can see that $Z$ for carbon (any of the isotopes) is 6, therefore $N=A-Z$ (from Equation \ref{1}):
$12-6=6 \nonumber$
The N:P ratio therefore is 6:6 or a 1:1. In fact 99% of all carbon in the earth is this isotope.
Exercise $1$: Oxygen
Identify the number of neutron, protons, total nucleons and N:Z ratio in the $\ce{^12_8O}$ nuclide?
Elements that have atomic numbers from 20 to 83 are heavy elements, therefore the ratio is different. The ratio is 1.5:1, the reason for this difference is because of the repulsive force between protons: the stronger the repulsion force, the more neutrons are needed to stabilize the nuclei.
Neutrons help to separate the protons from each other in a nucleus so that they do not feel as strong a repulsive force from other.
Isotope Stability
The graph of stable elements is commonly referred to as the Band (or Belt) of Stability. The graph consists of a y-axis labeled neutrons, an x-axis labeled protons, and a nuclei. At the higher end (upper right) of the band of stability lies the radionuclides that decay via alpha decay, below is positron emission or electron capture, above is beta emissions and elements beyond the atomic number of 83 are only unstable radioactive elements. Stable nuclei with atomic numbers up to about 20 have an neutron:proton ratio of about 1:1 (solid line).
The deviation from the $N:Z=1$ line on the belt of stability originates from a non-unity $N:Z$ ratio necessary for total stability of nuclei. That is, more neutrons are required to stabilize the repulsive forces from a fewer number of protons within a nucleus (i.e., $N>Z$).
The belt of stability makes it is easy to determine where the alpha decay, beta decay, and positron emission or electron capture occurs.
• Alpha $\alpha$ Decay: Alpha decay is located at the top of the plotted line, because the alpha decay decreases the mass number of the element to keep the isotope stable. This is accomplished by emitting a alpha particle, which is just a helium ($\ce{He}$) nucleus. In this decay pathway, the unstable isotope's proton number $P$ is decreased by 2 and its neutron ($N$) number is decreased by 2. The means that the nucleon number $A$ decreases by 4 (Equation \ref{1}).
• Beta $\beta^-$ Decay: Beta $\beta^-$ decay accepts protons so it changes the amount of protons and neutrons. the number of protons increase by 1 and the neutron number decreases by 1. This pathway occurs in unstable nuclides that have too many neutrons lie above the band of stability (blue isotopes in Figure $1$).
• Positron $\beta^+$ Decay: Positron $\beta^+$ emission and electron capture is when the isotope gains more neutrons. Positron emission and electron capture are below the band of stability because the ratio of the isotope has more protons than neutrons, think of it as there are too few protons for the amount of neutrons and that is why it is below the band of stability (yellow isotopes in Figure $1$).
As with all decay pathways, if the daughter nuclides are not on the Belt, then subsequent decay pathways will occur until the daughter nuclei are on the Belt.
Magic Numbers
The Octet Rule was formulated from the observation that atoms with eight valence electrons were especially stable (and common). A similar situation applies to nuclei regarding the number of neutron and proton numbers that generate stable (non-radioactive) isotopes. These "magic numbers" are natural occurrences in isotopes that are particularly stable. Table 1 list of numbers of protons and neutrons; isotopes that have these numbers occurring in either the proton or neutron are stable. In some cases there the isotopes can consist of magic numbers for both protons and neutrons; these would be called double magic numbers. The double numbers only occur for isotopes that are heavier, because the repulsion of the forces between the protons. The magic numbers are:
• proton: 2, 8, 20, 28, 50, 82, 114
• neutron: 2, 8, 20, 28, 50, 82, 126, 184
Also, there is the concept that isotopes consisting a combination of even-even, even-odd, odd-even, and odd-odd are all stable. There are more nuclides that have a combination of even-even than odd-odd. Just like there exist violations to the octet rule, many isotopes with no magic numbers of nucleons are stable.
Table $1$: Distribution of Stable and Unstable Isotopes based on Neutron and Proton Numbers
Proton number (Z) Neutron Number # of stable Isotopes
Even Even 163
Even Odd 53
Odd Even 50
Odd Odd 4
Note
Although rare, four stable odd-odd nuclides exist: $\ce{^2_1H}$, $\ce{^{6}_3Li}$, $\ce{^{10}_5B}$, $\ce{^{14}_7N}$
Unstable or Stable
Here is a simple chart that can help you decide is an element is likely stable.
• Calculate the total number of nucleons (protons and neutrons) in the nuclide. If the number of nucleons is even, there is a good chance it is stable.
• Are there a magic number of protons or neutrons? 2,8,20,28,50,82,114 (protons), 126 (neutrons), 184 (neutrons) are particularly stable in nuclei.
• Calculate the N/Z ratio and use the belt of stability (Figure $1$:) to determine the best way to get from an unstable nucleus to a stable nucleus
Exercise $1$
Using the above chart state if this isotope is alpha-emitter, stable, or unstable:
1. $\ce{^{40}_{20}Ca}$
2. $\ce{^{54}_{25}Mn}$
3. $\ce{^{210}_{84}Po}$
Answer
Add texts here. Do not delete this text first.
Exercise $2$
If the isotope is located above the band of stability what type of radioactivity is it? what if it was below?
Answer
Based off the belt of stability:
1. Stable, because this Ca isotope has 20 neutrons, which is on of the magic numbers
2. Unstable, because there is an odd number (25 and 29) of protons and neutrons
3. Alpha-emitter, because Z=84, which follows rule/step one on the chart
Carbon is stable
Answer
Carbon is stable
Exercise $4$
Name one of the isotopes that consist of odd-odd combination in the nuclei?
Answer
Hydrogen-2, Lithium-6, Boron-10, nitrogen-14
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As with chemical reactions, the nuclear reactions are not instantaneous and evolve on differing times (ranging from billions of years to microseconds). Also, as with chemical reactions, nuclear reactions follow comparable rate laws.
• Half-Lives and Radioactive Decay Kinetics
Another approach to describing reaction rates is based on the time required for the concentration of a reactant to decrease to one-half its initial value. This period of time is called the half-life of the reaction, written as t1/2. Thus the half-life of a reaction is the time required for the reactant concentration to decrease from [A]0 to [A]0/2. If two reactions have the same order, the faster reaction will have a shorter half-life, and the slower reaction will have a longer half-life.
• How to Change Nuclear Decay Rates
• Radioactive Decay Rates
Nuclear Kinetics
Learning Objectives
• To know how to use half-lives to describe the rates of first-order reactions
Another approach to describing reaction rates is based on the time required for the concentration of a reactant to decrease to one-half its initial value. This period of time is called the half-life of the reaction, written as t1/2. Thus the half-life of a reaction is the time required for the reactant concentration to decrease from [A]0 to [A]0/2. If two reactions have the same order, the faster reaction will have a shorter half-life, and the slower reaction will have a longer half-life.
The half-life of a first-order reaction under a given set of reaction conditions is a constant. This is not true for zeroth- and second-order reactions. The half-life of a first-order reaction is independent of the concentration of the reactants. This becomes evident when we rearrange the integrated rate law for a first-order reaction to produce the following equation:
$\ln\dfrac{[\textrm A]_0}{[\textrm A]}=kt \label{21.4.1}$
Substituting [A]0/2 for [A] and t1/2 for t (to indicate a half-life) into Equation $\ref{21.4.1}$ gives
$\ln\dfrac{[\textrm A]_0}{[\textrm A]_0/2}=\ln 2=kt_{1/2}$
Substituting $\ln{2} \approx 0.693$ into the equation results in the expression for the half-life of a first-order reaction:
$t_{1/2}=\dfrac{0.693}{k} \label{21.4.2}$
Thus, for a first-order reaction, each successive half-life is the same length of time, as shown in Figure $1$, and is independent of [A].
If we know the rate constant for a first-order reaction, then we can use half-lives to predict how much time is needed for the reaction to reach a certain percent completion.
Number of Half-Lives Percentage of Reactant Remaining
1 $\dfrac{100\%}{2}=50\%$ $\dfrac{1}{2}(100\%)=50\%$
2 $\dfrac{50\%}{2}=25\%$ $\dfrac{1}{2}\left(\dfrac{1}{2}\right)(100\%)=25\%$
3 $\dfrac{25\%}{2}=12.5\%$ $\dfrac{1}{2}\left(\dfrac{1}{2}\right )\left (\dfrac{1}{2}\right)(100\%)=12.5\%$
n $\dfrac{100\%}{2^n}$ $\left(\dfrac{1}{2}\right)^n(100\%)=\left(\dfrac{1}{2}\right)^n\%$
As you can see from this table, the amount of reactant left after n half-lives of a first-order reaction is (1/2)n times the initial concentration.
For a first-order reaction, the concentration of the reactant decreases by a constant with each half-life and is independent of [A].
Example $1$
The anticancer drug cis-platin hydrolyzes in water with a rate constant of 1.5 × 10−3 min−1 at pH 7.0 and 25°C. Calculate the half-life for the hydrolysis reaction under these conditions. If a freshly prepared solution of cis-platin has a concentration of 0.053 M, what will be the concentration of cis-platin after 5 half-lives? after 10 half-lives? What is the percent completion of the reaction after 5 half-lives? after 10 half-lives?
Given: rate constant, initial concentration, and number of half-lives
Asked for: half-life, final concentrations, and percent completion
Strategy:
1. Use Equation $\ref{21.4.2}$ to calculate the half-life of the reaction.
2. Multiply the initial concentration by 1/2 to the power corresponding to the number of half-lives to obtain the remaining concentrations after those half-lives.
3. Subtract the remaining concentration from the initial concentration. Then divide by the initial concentration, multiplying the fraction by 100 to obtain the percent completion.
Solution
A We can calculate the half-life of the reaction using Equation $\ref{21.4.2}$:
$t_{1/2}=\dfrac{0.693}{k}=\dfrac{0.693}{1.5\times10^{-3}\textrm{ min}^{-1}}=4.6\times10^2\textrm{ min}$
Thus it takes almost 8 h for half of the cis-platin to hydrolyze.
B After 5 half-lives (about 38 h), the remaining concentration of cis-platin will be as follows:
$\dfrac{0.053\textrm{ M}}{2^5}=\dfrac{0.053\textrm{ M}}{32}=0.0017\textrm{ M}$
After 10 half-lives (77 h), the remaining concentration of cis-platin will be as follows:
$\dfrac{0.053\textrm{ M}}{2^{10}}=\dfrac{0.053\textrm{ M}}{1024}=5.2\times10^{-5}\textrm{ M}$
C The percent completion after 5 half-lives will be as follows:
$\textrm{percent completion}=\dfrac{(0.053\textrm{ M}-0.0017\textrm{ M})(100)}{0.053}=97\%$
The percent completion after 10 half-lives will be as follows:
$\textrm{percent completion}=\dfrac{(0.053\textrm{ M}-5.2\times10^{-5}\textrm{ M})(100)}{0.053\textrm{ M}}=100\%$
Thus a first-order chemical reaction is 97% complete after 5 half-lives and 100% complete after 10 half-lives.
Exercise $1$
Ethyl chloride decomposes to ethylene and HCl in a first-order reaction that has a rate constant of 1.6 × 10−6 s−1 at 650°C.
1. What is the half-life for the reaction under these conditions?
2. If a flask that originally contains 0.077 M ethyl chloride is heated at 650°C, what is the concentration of ethyl chloride after 4 half-lives?
Answer a
4.3 × 105 s = 120 h = 5.0 days;
Answer b
4.8 × 10−3 M
Radioactive Decay Rates
Radioactivity, or radioactive decay, is the emission of a particle or a photon that results from the spontaneous decomposition of the unstable nucleus of an atom. The rate of radioactive decay is an intrinsic property of each radioactive isotope that is independent of the chemical and physical form of the radioactive isotope. The rate is also independent of temperature. In this section, we will describe radioactive decay rates and how half-lives can be used to monitor radioactive decay processes.
In any sample of a given radioactive substance, the number of atoms of the radioactive isotope must decrease with time as their nuclei decay to nuclei of a more stable isotope. Using N to represent the number of atoms of the radioactive isotope, we can define the rate of decay of the sample, which is also called its activity (A) as the decrease in the number of the radioisotope’s nuclei per unit time:
$A=-\dfrac{\Delta N}{\Delta t} \label{21.4.3}$
Activity is usually measured in disintegrations per second (dps) or disintegrations per minute (dpm).
The activity of a sample is directly proportional to the number of atoms of the radioactive isotope in the sample:
$A = kN \label{21.4.4}$
Here, the symbol k is the radioactive decay constant, which has units of inverse time (e.g., s−1, yr−1) and a characteristic value for each radioactive isotope. If we combine Equation $\ref{21.4.3}$ and Equation $\ref{21.4.4}$, we obtain the relationship between the number of decays per unit time and the number of atoms of the isotope in a sample:
$-\dfrac{\Delta N}{\Delta t}=kN \label{21.4.5}$
Equation $\ref{21.4.5}$ is the same as the equation for the reaction rate of a first-order reaction, except that it uses numbers of atoms instead of concentrations. In fact, radioactive decay is a first-order process and can be described in terms of either the differential rate law (Equation $\ref{21.4.5}$) or the integrated rate law:
$N = N_0e^{−kt}$
$\ln \dfrac{N}{N_0}=-kt \label{21.4.6}$
Because radioactive decay is a first-order process, the time required for half of the nuclei in any sample of a radioactive isotope to decay is a constant, called the half-life of the isotope. The half-life tells us how radioactive an isotope is (the number of decays per unit time); thus it is the most commonly cited property of any radioisotope. For a given number of atoms, isotopes with shorter half-lives decay more rapidly, undergoing a greater number of radioactive decays per unit time than do isotopes with longer half-lives. The half-lives of several isotopes are listed in Table 14.6, along with some of their applications.
Table $2$: Half-Lives and Applications of Some Radioactive Isotopes
Radioactive Isotope Half-Life Typical Uses
*The m denotes metastable, where an excited state nucleus decays to the ground state of the same isotope.
hydrogen-3 (tritium) 12.32 yr biochemical tracer
carbon-11 20.33 min positron emission tomography (biomedical imaging)
carbon-14 5.70 × 103 yr dating of artifacts
sodium-24 14.951 h cardiovascular system tracer
phosphorus-32 14.26 days biochemical tracer
potassium-40 1.248 × 109 yr dating of rocks
iron-59 44.495 days red blood cell lifetime tracer
cobalt-60 5.2712 yr radiation therapy for cancer
technetium-99m* 6.006 h biomedical imaging
iodine-131 8.0207 days thyroid studies tracer
radium-226 1.600 × 103 yr radiation therapy for cancer
uranium-238 4.468 × 109 yr dating of rocks and Earth’s crust
americium-241 432.2 yr smoke detectors
Note
Radioactive decay is a first-order process.
Radioisotope Dating Techniques
In our earlier discussion, we used the half-life of a first-order reaction to calculate how long the reaction had been occurring. Because nuclear decay reactions follow first-order kinetics and have a rate constant that is independent of temperature and the chemical or physical environment, we can perform similar calculations using the half-lives of isotopes to estimate the ages of geological and archaeological artifacts. The techniques that have been developed for this application are known as radioisotope dating techniques.
The most common method for measuring the age of ancient objects is carbon-14 dating. The carbon-14 isotope, created continuously in the upper regions of Earth’s atmosphere, reacts with atmospheric oxygen or ozone to form 14CO2. As a result, the CO2 that plants use as a carbon source for synthesizing organic compounds always includes a certain proportion of 14CO2 molecules as well as nonradioactive 12CO2 and 13CO2. Any animal that eats a plant ingests a mixture of organic compounds that contains approximately the same proportions of carbon isotopes as those in the atmosphere. When the animal or plant dies, the carbon-14 nuclei in its tissues decay to nitrogen-14 nuclei by a radioactive process known as beta decay, which releases low-energy electrons (β particles) that can be detected and measured:
$\ce{^{14}C \rightarrow ^{14}N + \beta^{−}} \label{21.4.7}$
The half-life for this reaction is 5700 ± 30 yr.
The 14C/12C ratio in living organisms is 1.3 × 10−12, with a decay rate of 15 dpm/g of carbon (Figure $2$). Comparing the disintegrations per minute per gram of carbon from an archaeological sample with those from a recently living sample enables scientists to estimate the age of the artifact, as illustrated in Example 11.Using this method implicitly assumes that the 14CO2/12CO2 ratio in the atmosphere is constant, which is not strictly correct. Other methods, such as tree-ring dating, have been used to calibrate the dates obtained by radiocarbon dating, and all radiocarbon dates reported are now corrected for minor changes in the 14CO2/12CO2 ratio over time.
Example $2$
In 1990, the remains of an apparently prehistoric man were found in a melting glacier in the Italian Alps. Analysis of the 14C content of samples of wood from his tools gave a decay rate of 8.0 dpm/g carbon. How long ago did the man die?
Given: isotope and final activity
Asked for: elapsed time
Strategy:
A Use Equation $\ref{21.4.4}$ to calculate N0/N. Then substitute the value for the half-life of 14C into Equation $\ref{21.4.2}$ to find the rate constant for the reaction.
B Using the values obtained for N0/N and the rate constant, solve Equation $\ref{21.4.6}$ to obtain the elapsed time.
Solution
We know the initial activity from the isotope’s identity (15 dpm/g), the final activity (8.0 dpm/g), and the half-life, so we can use the integrated rate law for a first-order nuclear reaction (Equation $\ref{21.4.6}$) to calculate the elapsed time (the amount of time elapsed since the wood for the tools was cut and began to decay).
\begin{align}\ln\dfrac{N}{N_0}&=-kt \ \dfrac{\ln(N/N_0)}{k}&=t\end{align}
A From Equation $\ref{21.4.4}$, we know that A = kN. We can therefore use the initial and final activities (A0 = 15 dpm and A = 8.0 dpm) to calculate N0/N:
$\dfrac{A_0}{A}=\dfrac{kN_0}{kN}=\dfrac{N_0}{N}=\dfrac{15}{8.0}$
Now we need only calculate the rate constant for the reaction from its half-life (5730 yr) using Equation $\ref{21.4.2}$:
$t_{1/2}=\dfrac{0.693}{k}$
This equation can be rearranged as follows:
$k=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{5730\textrm{ yr}}=1.22\times10^{-4}\textrm{ yr}^{-1}$
B Substituting into the equation for t,
$t=\dfrac{\ln(N_0/N)}{k}=\dfrac{\ln(15/8.0)}{1.22\times10^{-4}\textrm{ yr}^{-1}}=5.2\times10^3\textrm{ yr}$
From our calculations, the man died 5200 yr ago.
Exercise $2$
It is believed that humans first arrived in the Western Hemisphere during the last Ice Age, presumably by traveling over an exposed land bridge between Siberia and Alaska. Archaeologists have estimated that this occurred about 11,000 yr ago, but some argue that recent discoveries in several sites in North and South America suggest a much earlier arrival. Analysis of a sample of charcoal from a fire in one such site gave a 14C decay rate of 0.4 dpm/g of carbon. What is the approximate age of the sample?
Answer
30,000 yr
Summary
• The half-life of a first-order reaction is independent of the concentration of the reactants.
• The half-lives of radioactive isotopes can be used to date objects.
The half-life of a reaction is the time required for the reactant concentration to decrease to one-half its initial value. The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction: t1/2 = 0.693/k. Radioactive decay reactions are first-order reactions. The rate of decay, or activity, of a sample of a radioactive substance is the decrease in the number of radioactive nuclei per unit time.
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"One of the paradigms of nuclear science since the very early days of its study has been the general understanding that the half-life, or decay constant, of a radioactive substance is independent of extranuclear considerations." (Emery) Like all paradigms, this one is subject to some interpretation. Normal decay of radioactive stuff proceeds via one of four mechanisms:
• Alpha decay: the emission of an alpha particle (a helium-4 nucleus) which reduces the number of protons and neutrons present in the parent nucleus by two each;
• Beta decay: encompassing several related phenomena in which a neutron in the nucleus is replaced by a proton, or a proton is replaced by a neutron, along with some other things involving electrons, positrons, neutrinos, and antineutrinos. These other things are, as we shall see, at the bottom of several questions involving perturbation of decay rates;
• Gamma decay: the emission of one or more gamma rays — very energetic photons — that take a nucleus from an excited state to some other (typically ground) state; some of these photons may be replaced by "conversion electrons", of which more shortly;
• Spontaneous fission: in which a sufficiently heavy nucleus simply breaks in half. Most of the discussion about alpha particles will also apply to spontaneous fission.
$\gamma$ decay often occurs from the daughter nucleus of one of the other decay modes. We neglect very exotic processes like C-14 emission or double beta decay in this analysis.
$\beta$ decay happens most often to a nucleus with a neutron excess, which decays by converting a neutron into a proton:
$\underset{\beta \text{ decay}}{n \rightarrow p^+ + e^- + \bar{\nu_e}} \label{1}$
where
• $n$ is a neutron
• $p$ is a proton
• $e^-$ is an electrons and
• $\bar{\nu_e}$ is an antineutrono of the electron type.
The type of beta decay that involves destruction of a proton is not familiar to many people, so deserves a little elaboration. Either of two processes may occur when this kind of decay happens:
$\underset{\text{positron emission}}{p^+ \rightarrow n + e^+ + \nu_e} \label{2}$
where $e^+$ is a positron (anti-electron) and $\nu_e$ is the electron neutrino; or
$\underset{\text{electron capture}}{p^+ + e^- \rightarrow n + \nu_e} \label{3}$
where the electron is captured from the neighborhood of the nucleus undergoing decay. These processes are called "positron emission" and "electron capture" respectively. A given nucleus that has too many protons for stability may undergo beta decay through either, and typically both, of these reactions.
"Conversion electrons" are produced by the process of "internal conversion", whereby the photon that would normally be emitted in gamma decay is virtual and its energy is absorbed by an atomic electron. The absorbed energy is sufficient to unbind the electron from the nucleus (ignoring a few exceptional cases), and it is ejected from the atom as a result.
Now for the tie-in to decay rates. Both the electron-capture and internal conversion phenomena require an electron somewhere close to the decaying nucleus. In any normal atom, this requirement is satisfied in spades: the innermost electrons are in states such that their probability of being close to the nucleus is both large and insensitive to things in the environment. The decay rate depends only very weakly on the electron wave functions, i.e., on how much of their time the inner electrons spend very near the nucleus. For most nuclides that decay by electron capture or internal conversion, the probability of grabbing or converting an electron is usually also insensitive to the environment, as the innermost electrons are the ones most likely to get grabbed/converted.
All told, the existence of changes in radioactive decay rates due to the environment of the decaying nuclei is on solid grounds both experimentally and theoretically. However, the magnitude of the changes is nothing to get very excited about.
Contributors and Attributions
• Bill Johnson.
This page is part of the Physics and Relativity FAQ, as a collection, is copyright 1992—2009 by Scott Chase, Michael Weiss, Philip Gibbs, Chris Hillman, and Nathan Urban. The individual articles are copyright 1992—2009 by the individual authors. All rights are reserved. Permission to use, copy and distribute this unmodified document, as a whole, by any means and for any purpose EXCEPT PROFIT PURPOSES is hereby granted, provided that all author attributions, the above Copyright notice, and this permission notice appear in all copies of the FAQ itself.
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Radioactive decay is the loss of elementary particles from an unstable nucleus, ultimately changing the unstable element into another more stable element. There are five types of radioactive decay: alpha emission, beta emission, positron emission, electron capture, and gamma emission. Each type of decay emits a specific particle which changes the type of product produced. The number of protons and neutrons found in the daughter nuclei (the nuclei produced from the decay) are determined by the type of decay or emission that the original element goes through.
Introduction
In terms of entropy, radioactive decay can be defined as the tendency for matter and energy to gain inert uniformity or stability. For elements, uniformity is produced by having an equal number of neutrons and protons which in turn dictates the desired nuclear forces to keep the nuclear particles inside the nucleus. However, any instance where one particle becomes more frequent than another creates a nucleus that becomes unstable. The unstable nucleus then releases radiation in order to gain stability. For example, the stable element Beryllium usually contains 4 protons and 5 neutrons in its nucleus (this is not considered a very large difference). However, there exists a lighter isotope of Beryllium which contains 4 protons and only 3 neutrons, which gives a total mass of 7 amu. This lighter isotope decays into Lithium-7 through electron capture. A proton from Beryllium-7 captures a single electron and becomes a neutron. This reaction produces a new isotope (Lithium-7) that has the same atomic mass unit as Beryllium-7 but one less proton which stabilizes the element.
Another example is the element Uranium-238 which has 54 more neutrons than its protons (Atomic umber =92). This element gains stability by passing through various types of decays (19 steps-- also known as the Uranium series) and is converted into Pb-206 (atomic number 82).For further information about different types of decay that Uranium goes through, refer to Decay Pathways).
Decay Rates
Due to the smaller size of the nucleus compared to the atom and the enormity of electromagnetic forces, it is impossible to predict radioactive decay. The atomic nucleus which is in the center of the atom is buffered by surrounding electrons and external conditions. Because of this, the study of decay is independent of the element's environment. In other words, the decay rate is independent of an element's physical state such as surrounding temperature and pressure. For a given element, the decay or disintegration rate is proportional to the number of atoms and the activity measured in terms of atoms per unit time. If "A" represents the disintegration rate and "N" is number of radioactive atoms, then the direct relationship between them can be shown as below:
$A \propto N \label{1A}$
or mathematically speaking
$A= \lambda N \label{1B}$
where
• $A$ is the Total activity and is the number of decays per unit time of a radioactive sample.
• $N$ is the total number of particles in the sample.
• $\lambda$ is the constant of proportionality or decay constant.
Decay Rate & Chemical Kinetics
Since the decay rate is dependent upon the number of radioactive atoms, in terms of chemical kinetics, one can say that radioactive decay is a first order reaction process. Even though radioactive decay is a first order reaction, where the rate of the reaction depends upon the concentration of one reactant (r = k [A][B] = k [A}) , it is not affected by factors that alter a typical chemical reactions. In other words, the reaction rate does not depend upon the temperature, pressure, and other physical determinants. However, like a typical rate law equation, radioactive decay rate can be integrated to link the concentration of a reactant with time. Also, radioactive decay is an exponential decay function which means the larger the quantity of atoms, the more rapidly the element will decay. Mathematically speaking, the relationship between quantity and time for radioactive decay can be expressed in following way:
$\dfrac{dN}{dt} = - \lambda N \label{2A}$
or more specifically
$\dfrac{dN(t)}{dt} = - \lambda N \label{2B}$
or via rearranging the separable differential equation
$\dfrac{dN(t)}{N (t)} = - \lambda dt \label{3}$
by Integrating the equation
$\ln N(t) = - \lambda t + C \label{4}$
with
• $C$ is the constant of integration
• $N(t)$ is the amplitude of $N$ after lapse of time $t$
• $\lambda$ is the decay rate constant.
One could derive equation 4 in following manner, too. The decay rate constant, $\lambda$, is in the units time-1. For further information about first-order reactions, refer to First-Order Reactions.
Ways to Characterize Decay Constant
There are two ways to characterize the decay constant: mean-life and half-life. In both cases the unit of measurement is seconds. As indicated by the name, mean-life is the average of an element's lifetime and can be shown in terms of following expression
$N_t=N_o e^{-\lambda t} \label{5}$
$1 = \int^{\infty}_ 0 c \cdot N_0 e^{-\lambda t} dt = c \cdot \dfrac{N_0}{\lambda} \label{6}$
Rearranging the equation:
$c= \dfrac{\lambda}{N_o}$
Decay Rate Half-Life
Half-life is the time period that is characterized by the time it takes for half of the substance to decay (both radioactive and non-radioactive elements).The rate of decay remains constant throughout the decay process. There are three ways to show the exponential nature of half-life.
$N_t=N_o\left( \dfrac{1}{2} \right)^{t/t_{1/2}} \label{7}$
$N_t=N_o e^{-t/\tau} \label{8}$
By comparing Equations 1, 3 and 4, one will get following expressions
$\ln {\left( \dfrac{1}{2} \right)^{t/t_{1/2}}}= \ln(e^{-t/\tau}) = \ln (e^{-\lambda t} ) \label{9}$
or with $\ln(e) = 1$, then
$\dfrac{t}{t_{1/2}} \ln \left( \frac{1}{2} \right) = \dfrac{-t}{\tau} = -\lambda t \label{10}$
By canceling $t$ on both sides, one will get following equation (for half-life)
$t_{1/2}= \dfrac{\ln(2)}{\lambda} \approx \dfrac{0.693}{\lambda} \label{11}$
or combining equations 1B and 11
$A = \dfrac{0.693}{t_{1/2}}N \label{12}$
Equation 11 is a constant, meaning the half-life of radioactive decay is constant. Half-life and the radioactive decay rate constant λ are inversely proportional which means the shorter the half-life, the larger $\lambda$ and the faster the decay. This is a hypothetical radioactive decay graph. If the half-life were shorter, then the exponential decay graph would be steeper and the line would be decreasing at a faster rate; therefore, the amount of the radioactive nuclei would decrease as well.
Radioactive decay is not always a one step phenomenon. Often times the parent nuclei changes into a radioactive daughter nuclei which also decays. In such cases, it is possible that the half-life of the parent nuclei is longer or shorter than the half-life of the daughter nuclei. Depending upon the substance, it is possible that both parent and daughter nuclei have similar half lives.
Ba-140 Parent has a longer half-life than the daughter nuclei (La and Ce)
Po-218 has a smaller half-life than its daughter nuclei (different species of Pb and Bi).
Example $3$: Iodine Decay
Almost equal half-life for both parents ($^{135}I$) and daughter nuclei ($^{135}Xe$ and $^{135}Cs$)
Constant Decay Rate
Since the decay rate is constant, one can use the radioactive decay law and the half-life formula to find the age of organic material, which is known as radioactive dating. One of the forms of radioactive dating is radiocarbon dating. Carbon 14 (C-14) is produced in the upper atmosphere through the collision of cosmic rays with atmospheric 14N. This radioactive carbon is incorporated in plants and respiration and eventually with animals that feed upon plants. The ratio of C-14 to C-12 is 1:10^12 within plants as well as in the atmosphere. This ratio, however, increases upon the death of an animal or when a plant decays because there is no new income of carbon 14. By knowing the half-life of carbon-14 (which is 5730 years) one can calculate the rate of disintegration of the nuclei within the organism or substance and thereby determine its age. It is possible to use other radioactive elements in order to determine the age of nonliving substances as well.
Problems
1. Determine the decay constant for Carbon-14, if it has a half-life of 5730 years?
2. If Radium-223 has a half life of 10.33 days, what time duration would it require for the activity associated with this sample to decrease 1.5% of its present value?
3. Determine the number of atoms in a 1.00 mg sample of Carbon-14?
4. What mass of Carbon-14 must be in a sample to have an activity of 2.00 mCi?
5. The disintegration rate for a sample Co-60 is 6800 dis/h. If the half-life is 5.3 years, determine the quantity of atoms in this sample?
6. Isotope A requires 6.0 days for its decay rate to fall to 1/20 its initial value. Isotope B has a half-life that is 1.5 times that of A. How long will it take for isotope B to decrease to 1/16 of its initial value?
7. A sample disintegrates at the following rate of counts per minute (cpm): t=0, 2000 cpm; t=5 hr, 1984 cpm; t=50 hr, 1848 cpm; t=500 hr, 904 cpm; t=1250, 276 cpm. What is the half-life of this nuclide?
Solutions
Odd Problem Solutions:
1. Using Equation 11, we can set $t_{1/2} = 573\, yrs$ and solve for $\lambda$.
$\lambda=1.209 \times 10^{-4}\; yr^{-1}$
3. To find the number of atoms in a Carbon-14 sample, we will use dimensional analysis. First we convert 1.00mg to 0.001 grams. From the name, we know the atomic mass of Carbon-14 to be 14 g/mol. We know Avogadro's number expresses $6.022 \times 10^{23}$ atoms. Now we have the equation:
$N= \dfrac{0.001\; g}{(1\; mol/14\;g)(6.022 \times 10^{23})} \;atoms/ 1\;mol$
$N=4.301 \times 10^{19} \, \text{atoms}$
5. Using Equation 1B and Equation 12, we can combine them and solve for $N$. By rearranging Equation 11, $\lambda=\ln\; 2/t_{1/2}$ we can insert that into Equation 1B. Now we have the formula $A=\ln 2/t_{1/2} N$. Now we have to convert 5.3 years to hours because the activity is measured in disintegration (atoms) per hour.
$t_{1/2}=5.3\; \cancel{years} \times \left(\dfrac{365\; \cancel{days}}{1\; \cancel{year}}\right) \times \left(\dfrac{24\;hr}{1\;\cancel{day}} \right)=46,428\; hours$
From equation 12, $N$ can be calculated
$N = A\; \dfrac{t_{1/2}}{\ln 2}$
or
$N = (6,800\; dis/hr)\; \dfrac{46,428\; hr}{\ln 2} \approx 4.56 \times 10^8\; \text{atoms}$
The Cobalt-60 sample has 456,000,000 atoms.
7. We want to determine the decay constant. By looking at the first and last given values, we can use Equation 2 to solve for λ.
$\ln(276\;cpm / 2000\;cpm)=-\lambda \times 1250\;hr$
$\lambda = 0.00158 \;hr^{-1}$
Then using Equation 11, we can solve for half-life.
$t_{1/2}=\dfrac{\ln 2}{0.00158}$
The half-life of the sample is 438 hours.
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Radioactivity is the ascribed to the emitting or relating to the emission of ionizing radiation or particles.
• Artificially Induced Radioactivity
Radioactivity is the process by which the nucleus of an unstable atom loses energy by emitting radiation, including alpha particles, beta particles, gamma rays and conversion electrons Although radioactivity is observed as a natural occurring process, it can also be artificially induced typically via the bombarding atoms of a specific element by radiating particles, thus creating new atoms.
• Discovery of Radioactivity
The discovery of radioactivity took place over several years beginning with the discovery of x-rays in 1895 by Wilhelm Conrad Roentgen and continuing with such people as Henri Becquerel and the Curie family. The application of x-rays and radioactive materials is far reaching in medicine and industry.
• Nuclear Decay Pathways
Nuclear reactions that transform atomic nuclei alter their identity and spontaneously emit radiation via processes of radioactive decay.
• Purifying Radioactive Materials
The production of everyday used materials such as oil and gas results in the buildup of radioactive materials in high concentrations. As a result of the commonality of this occurrence, a solution has been presented as to rid the world of this toxic waste. Purifying these radioactive materials with the use of various methods allows the pure substance to be reused and prevents the depletion of resources.
• The Effects of Radiation on Matter
All radioactive particles and waves, from the entire electromagnetic spectrum, to alpha, beta, and gamma particles, possess the ability to eject electrons from atoms and molecules to create ions.
• Transmutation of the Elements
Although the conversion of one element to another is the basis of natural radioactive decay, it is also possible to convert one element to another experimentally. The conversion of one element to another is the process of transmutation.
Radioactivity
Radioactivity is the process by which the nucleus of an unstable atom loses energy by emitting radiation, including alpha particles, beta particles, gamma rays and conversion electrons Although radioactivity is observed as a natural occurring process, it can also be artificially induced typically via the bombarding atoms of a specific element by radiating particles, thus creating new atoms.
Introduction
Ernest Rutherford was a prominent New Zealand scientist, and a winner of the Nobel Prize in chemistry in 1908. Amongst his vast list of discoveries, Rutherford was also the first to discover artificially induced radioactivity. Through the bombardment of alpha particles against nuclei of $\ce{^{14}N}$ with 7 protons/electrons, Rutherford produced $\ce{^{17}O}$ (8 protons/electrons) and protons (Figure $1$). Through this observation, Rutherford concluded that atoms of one specific element can be made into atoms of another element. If the resulting element is radioactive, then this process is called artificially induced radioactivity
Rutherford was the first researcher to create protons outside of the atomic nuclei and the $\ce{^{17}O}$ isotope of oxygen, which is nonradioactive. Similarly, other nuclei when bombarded with alpha particles will generate new elements (Figure $2$) that may be radioactive and decay naturally or that may be stable and persist like $\ce{^{17}O}$.
Before this discovery of artificial induction of radioactivity, it was a common belief that atoms of matter are unchangeable and indivisible. After the very first discoveries made by Ernest Rutherford, Irene Joliot-Curie and her husband, Frederic Joliot, a new point of view was developed. The point of view that although atoms appear to be stable, they can be transformed into new atoms with different chemical properties. Today over one thousand artificially created radioactive nuclides exist, which considerably outnumber the nonradioactive ones created.
Note: Irene Joliet-Curie and Frederic Joliot
Irene Joliet-Curie and her husband Frédéric both were French scientists who shared winning the Nobel Prize award in chemistry in 1935 for artificially synthesizing a radioactive isotope of phosphorus by bombarding aluminum with alpha particles. $\ce{^{30}P}$ with 15 protons was the first radioactive nuclide obtained through this method of artificially inducing radioactivity.
$\ce{^27_13Al + ^4_2He \rightarrow ^30_15P + ^1_0n}$
$\ce{^30_15P \rightarrow ^30_14Si + ^0_{-1}\beta}$
Activation (or radioactivation) involves making a radioactive isotope by neutron capture, e.g. the addition of a neutron to a nuclide resulting in an increase in isotope number by 1 while retaining the same atomic number (Figure $3$). Activation is often an inadvertent process occurring inside or near a nuclear reactor, where there are many neutrons flying around. For example, Coba in or near a nuclear reactor will capture a neutron forming the radioactive isotope Co-60.
$\ce{ ^1_0n + ^{59}Co \rightarrow ^{60}Co }$
The $\ce{ ^{60}Co}$ isotope is unstable (half life of 5.272 years) and disintegrates into $\ce{ ^{60}Ni }$ via the emission of $\beta$ particle and $\gamma$ radiation Figure $4$.
Example $1$: Neutron Bombardment
Write a nuclear equation for the creation of 56Mn through the bombardment of 59Co with neutrons.
Solution
A unknown particle is produced with 56Mn, in order to find the mass number (A) of the unknown we must subtract the mass number of the Manganese atom from the mass number of the Cobalt atom plus the neutron being thrown. In simpler terms,
Now, by referring to a periodic table to find the atomic numbers of Mn and Co, and then subtracting the atomic number of Mn from Co, we will receive the atomic number of the unknown particle
Thus, the unknown particle has A = 4, and Z = 2, which would make it a Helium particle, and the nuclear formula would be as follows:
$\ce{^{50}_{27}Co + ^1_0n \rightarrow ^{56}_{25}Mn + ^{4}_{2}\alpha } \nonumber$
Example $2$: Calcium Bombardment
Write a nuclear equation for the production of $\ce{^{147}Eu}$ by bombardment of $\ce{^{139}La}$ with $\ce{^{12}Ca}$.
Solution
Like the above example, you must first find the mass number of the unknown particle.
Thus, the mass number of the unknown particle is 4. Again by referring to a periodic table and finding the atomic numbers of Lanthanum, Carbon and Europium, we are able to calculate the atomic number of the unknown particle,
The atomic number for the unknown particle equals to zero, therefore 4 neutrons are emitted, and the nuclear equation is written as follows:
$\ce{^{139}_{57}La + ^{12}_6C \rightarrow ^{147}_{63}Eu + 4 ^{0}_{1}n } \nonumber$
Summary
Induced radioactivity occurs when a previously stable material has been made radioactive by exposure to specific radiation. Most radioactivity does not induce other material to become radioactive. This Induced radioactivity was discovered by Irène Curie and F. Joliot in 1934. This is also known as man-made radioactivity. The phenomenon by which even light elements are made radioactive by artificial or induced methods is called artificial radioactivity.
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The discovery of radioactivity took place over several years beginning with the discovery of x-rays in 1895 by Wilhelm Conrad Roentgen and continuing with such people as Henri Becquerel and the Curie family. The application of x-rays and radioactive materials is far reaching in medicine and industry. Radioactive material is used in everything from nuclear reactors to isotope infused saline solutions. These technologies allow us to utilize great amounts of energy and observe biological systems in ways which were unthinkable less than a century ago.
Introduction
What is the definition of radioactive? If you look up the meaning in the dictionary the convoluted answer that you will receive is: Radioactive- adjective: emitting or relating to the emission of ionizing radiation or particles. This definition begs the questions: What are ionizing radiation or particles? What exactly is meant by emission? Can you see or feel these particles? What makes something radioactive?
The Discovery Of Radioactivity
Wilhelm Conrad Roentgen (1845-1923)
Contribution: Received the first Noble Prize in physics for his discovery of x-rays in 1901.
On November 8, 1895, at the University of Wurzburg, Roentgen was working in the lab when he noticed a strange fluorescence coming from a nearby table. Upon further observation he found that it originated from a partially evacuated Hittof-Crookes tube, covered in opaque black paper which he was using to study cathode rays. He concluded that the fluorescence, which penetrated the opaque black paper, must have been caused by rays. This phenomenon was later coined x-rays and though the phenomenon of x-rays is not the same as radioactivity, Roentgen opened the door for radioactive discovery.
Antoine Henri Becquerel (1852-1908)
Contributions: Received the Noble Prize in physics for being the first to discover radioactivity as a phenomenon separate from that of x-rays and document the differences between the two.
Henri Becquerel learned of Roentgen's discovery of x-rays through the fluorescence that some materials produce. Using a method similar to that of Roentgen, Becquerel surrounded several photographic plates with black paper and florescent salts. With the intention of further advancing the study of x-rays, Becquerel intended to place the concealed photographic paper in the sunlight and observe what transpired. Unfortunately, he had to delay his experiment because the skies over Paris were overcast. He placed the wrapped plates into a dark desk drawer. After a few days Becquerel returned to his experiment unwrapping the photographic paper and developing it, expecting only a light imprint from the salts. Instead, the salts left very distinct outlines in the photographic paper suggesting that the salts, regardless of lacking an energy source, continually fluoresced. What Becquerel had discovered was radioactivity.
Pierre (1859-1906) and Marie (1867-1934) Curie
Contributions: Pierre and Marie were award the Noble Prize in Physics in 1903 for their work on radioactivity. Marie Curie became the first woman to be awarded the nobel prize and the first person to obtain two nobel prizes when she won the prize for the discovery of Polonium and Radium in 1911.
Though it was Henri Becquerel that discovered radioactivity, it was Marie Curie who coined the term. Using a device invented by her husband and his brother, that measured extremely low electrical currents, Curie was able to note that uranium electrified the air around it. Further investigation showed that the activity of uranium compounds depended upon the amount of uranium present and that radioactivity was not a result of the interactions between molecules, but rather came from the atom itself. Using Pitchblende and chalcolite Curie found that Thorium was radioactive as well. She later discovered two new radioactive elements: Radium and Polonium which took her several years since these elements are difficult to extract and extremely rare. Unfortunately, the Curies died young. Pierre Curie was killed in a street accident and Marie died of aplastic anemia, almost certainly a result of radiation exposure.
Ernest Rutherford (1871-1937)
Contributions: Ernest Rutherford is considered the father of nuclear physics. With his gold foil experiment he was able to unlock the mysteries of the atomic structure. He received the noble prize in chemistry in 1908.
In 1909 at the University of Manchester, Rutherford was bombarding a piece of gold foil with Alpha particles. Rutherford noted that although most of the particles went straight through the foil, one in every eight thousand was deflected back. "It was as if you fired a fifteen inch naval shell at a piece of tissue paper and the shell came right back and hit you," Rutherford said. He concluded that though an atom consists of mostly empty space, most of its mass is concentrated in a very small positively charged region known as the nucleus, while electrons buzz around on the outside.
Rutherford was also able to observe that radioactive elements underwent a process of decay over time which varied from element to element. In 1919, Rutherford used alpha particles to transmutate one element (Oxygen) into another element (Nitrogen). Papers at the timed called it "splitting the atom."
What They Had Discovered:
We now have the essentials to utilize radioactive elements. Roentgen gave us x-rays, Becquerel discovered radioactivity, the Curies were able to discover which elements were radioactive, and Rutherford brought about transmutation and the "splitting of the atom." All of these discoveries and curiosity came with a price. Time showed the damaging effects of radiation exposure and the incredible destruction that could be harnessed from these elements.
Applications
Radioactive isotopes are presently used in many aspects of human life today. Most people recognize radioactivity's contributions to industry, research and war, but it is even used within many peoples homes. Here are a few examples of how radioactive isotopes are utilized today.
At Home
Most people have radioactive material in their very own homes, or at least we would hope so. Why? Because in most every smoke detector unit today there is a very small amount of Americium-241. How does it work? Well Americium-241 is present in the detector in oxide form and it emits alpha particles and very low energy gamma rays. The alpha rays are absorbed in the detector, while the non-harmful gamma rays are able to escape. The alpha particles collide with oxygen and nitrogen in the air of the detector's ionization chamber producing charged particles, or ions. A small electric voltage runs across the chamber which is used to collect these ions and operate a small electric current between two electrodes. When smoke enters the chamber it absorbs the alpha particles disrupting the rate of ionization in the chamber, thereby turning off the electrical current, which sets off the alarm.
For more information go to: http://home.howstuffworks.com/smoke2.htm
Nuclear Power
On June 7th 1954 the the USSR produced the world's very first nuclear power plant. These plants, though clean burning, produce a great deal of toxic nuclear waste which is difficult to eliminate. To date, approximately 15% of the worlds electricity and 6% of the worlds power is produced in nuclear power plants. With the rise in gas prices many countries around the world considered increasing their use nuclear energy.
The problem with nuclear energy is that although it is "clean" in the sense that only water vapor is emitted into the atmosphere, it has its share of problems. It must be kept constantly regulated, and is extremely hard to dispose of. In the past, poor regulation of nuclear power has caused major problems, such as the Chernobyl incident in 1986. Even when regulated properly, the waste can cause contamination which lasts for many years and destroys natural resources.
For more information and a specific example go to: www.world-nuclear.org/info/ch...byl/inf07.html
Industry
Gamma Sterilization
Large scale gamma irradiation is used to sterilize disposable medical supplies such as syringes, gloves and other instruments that would be damaged by heat sterilization. Large scale gamma irradiation is also used for killing parasites found in wool, wood and other widely distributed products. In the 1960's the irradiation of meat was allowed by the US, and it is now a commonly used food sterilization method. Small scale irradiates are also used for blood transfusions and other medical sterilization procedures.
Gamma Ray Analysis
Gamma Rays can be used to determine the ash content of coal. By bombarding stable elements with radioactive rays one can cause a fluorescence, the energy of fluorescent x-rays can help identify if any elements are represented in a material. The intensity of the rays can indicate the quantity of that material. This process is commonly used in element processing plants.
Medicine
Radioisotopes are used as tracers in medical research. People ingest these isotopes which allow researchers to study processes like digestion and locate medical problems like cancers and obstructions within an individual's digestive tract.
Radioactive elements are also used in clearing angioplasty obstructions and eliminating cancer.
War
To date the only country to utilize nuclear weapons and actually use them is the United States. On August 6th and 9th 1945, the US dropped nuclear weapons on Nagasaki and Hiroshima, Japan. These weapons were a part of a top secret project known today as the Manhattan project. Though those within the blast zone were instantly killed, the effects of these weapons would be felt for many years to come. Many more people died in the months following the bombing due to radiation poisoning, and years later, birth defects would prove the effects of radioactive bombardment upon DNA.
A good resource on the industrial and medical uses of radioactive isotopes: www.world-nuclear.org/info/inf56.htm
List of Radioactive Elements
All of the naturally occuring radioactive elements are concentrated between atomic numbers 84 and 118 on the periodic table, though Tc and Pm are an exception. Also note that there is a break between 110 and 118 on the table, which are suspected radioactive elements that have yet to be discovered. 29 radioactive elements have been identified by scientists to date:
• Technetium (TC)- Transition metal
• Promethium (Pm)- Rare earth metal
• Polonium (Po)- Metalliod
• Astatine (At)- Halogen
• Radon (Rn)- Noble gas
• Francium (Fr)- Alkali Metal
• Radium (Ra)- Alkali Earth Metal
• Actinium (Ac)- Rare Earth metal
• Thorium (Th)- Rare Earth Metal
• Protactinium (Pa)- Rare Earth Metal
• Uranium (U)- Rare Earth Metal
• Neptunium (Np)- Rare Earth Metal
• Plutonium (Pu)- Rare Earth Metal
• Americium (Am)- Rare Earth Metal
• Curium (Cm)- Rare Earth Metal
• Berkelium (Bk)- Rare Earth Metal
• Californium (Cf)- Rare Earth Metal
• Einsteinium (Es)- Rare Earth Metal
• Fermium (Fm)- Rare Earth Metal
• Mendelevium (Md)- Rare Earth Metal
• Nobelium (No)- Rare Earth Metal
• Lawrencium (Lr)- Rare Earth Metal
• Rutherfordium (Rt) or Kurchatovium (Ku)- Transition Metal
• Dubnium (Db) or Nilsborium (Ns)- Transition Metal
• Seaborgium (Sg)- Transition Metal
• Bohrium (Bh)- Transition Metal
• Hassium (Hs)- Transition Metal
• Meitnerium (Mt)- Transition Metal
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Nuclear reactions that transform atomic nuclei alter their identity and spontaneously emit radiation via processes of radioactive decay.
Types of Nuclear Decay
In 1889, Ernest Rutherford recognized and named two modes of radioactive decay, showing the occurrence of both processes in a decaying sample of natural uranium and its daughters. Rutherford named these types of radiation based on their penetrating power: heavier alpha and lighter beta radiation. Gamma rays, a third type of radiation, were discovered by P. Villard in 1900 but weren't recognized as electromagnetic radiation until 1914. Since gamma radiation is only the discharge of a high-energy photon from an over-excited nucleus, it does not change the identity of the atom from which it originates and therefore will not be discussed in depth here.
Because nuclear reactions involve the breaking of very powerful intra nuclear bonds, massive amounts of energy can be released. At such high energy levels, the matter can be converted directly to energy according to Einstein's famous Mass-Energy relationship E = mc2. The sum of mass and energy are conserved in nuclear decay. The free energy of any spontaneous reaction must be negative according to thermodynamics (ΔG < 0), and ΔG is essentially equal to the energy change ΔE of nuclear reactions because ΔE is so massive. Therefore, a nuclear reaction will occur spontaneously when:
$ΔE = Δmc^2 < 0$
$ΔE < 0$ or $Δm < 0$
When the mass of the products of a nuclear reaction weigh less than the reactants, the difference in mass has been converted to energy.
There are three types of nuclear reactions that are classified as beta decay processes. Beta decay processes have been observed in 97% of all known unstable nuclides and are thus the most common mechanism for radioactive decay by far. The first type (here referred to as beta decay) is also called Negatron Emission because a negatively charged beta particle is emitted, whereas the second type (positron emission) emits a positively charged beta particle. In electron capture, an orbital electron is captured by the nucleus and absorbed in the reaction. All these modes of decay represent changes of one in the atomic number Z of the parent nucleus but no change in the mass number A. Alpha decay is different because both the atomic and mass number of the parent nucleus decrease. In this article, the term beta decay will refer to the first process described in which a true beta particle is the product of the nuclear reaction.
Beta Decay / Negatron Emission
Nuclides can be radioactive and undergo nuclear decay for many reasons. Beta decay can occur in nuclei that are rich in neutrons - that is - the nuclide contains more neutrons than stable isotopes of the same element. These "proton deficient" nuclides can sometimes be identified simply by noticing that their mass number A (the sum of neutrons and protons in the nucleus) is significantly more than twice that of the atomic number Z (number of protons in nucleus). In order to regain some stability, such a nucleus can decay by converting one of its extra neutrons into a proton, emitting an electron and an antineutrino(ν). The high energy electron emitted in this reaction is called a beta particle and is represented by $_{-1}^{0}\textrm{e}^{-}$ in nuclear equations. Lighter atoms (Z < 60) are the most likely to undergo beta decay. The decay of a neutron to a proton, a beta particle, and an antineutrino ($\bar{\nu}$) is
$\ce{_{0}^{1}n^0 \rightarrow _{0}^{1}p^+ + _{-1}^{0}e^-}+ \bar{\nu}$
Some examples of beta decay are
$\ce{_{2}^{6}He \rightarrow _{3}^{6}Li + _{-1}^{0}e^-} +\bar{\nu}$
$\ce{_{11}^{24}Na \rightarrow _{12}^{24}Mg + _{-1}^{0}e^-} + \bar{\nu}$
In order for beta decay to occur spontaneously according to Δm < 0, the mass of the parent nucleus (not atom) must have a mass greater than the sum of the masses of the daughter nucleus and the beta particle:
m[AZ] > m[A(Z+1)] + m[0-1e-]
(Parent nucleus) > (Daughter nucleus) + (electron)
The mass of the antineutrino is almost zero and can therefore be neglected. The equation above can be reached easily from any beta decay reaction, however, it is not useful because mass spectrometers measure the mass of atoms rather than just their nuclei. To make the equation useful, we must make these nuclei into neutral atoms by adding the mass of Z + 1 electrons to each side of the equation. The parent nucleus then becomes the neutral atom [AZ] plus the mass of one electron, while the daughter nucleus and the beta particle on the right side of the equation become the neutral atom [A(Z+1)] plus the mass of the beta particle. The extra electron on the left cancels the mass of the beta particle on the right, leaving the inequality
m[AZ] > m[A(Z+1)]
(Parent atom) > (Daughter atom)
The change in mass then equals
Δm = m[A(Z+1)] - m[AZ]
The energy released in this reaction is carried away as kinetic energy by the beta particle and antineutrino, with an insignificant of energy causing recoil in the daughter nucleus. The beta particle can carry anywhere from all to none of this energy, therefore the maximum kinetic energy of a beta particle in any instance of beta decay is -ΔE.
Positron Emission
Nuclides that are imbalanced in their ratio of protons to neutrons undergo decay to correct the imbalance. Nuclei that are rich in protons relative to their number of neutrons can decay by conversion of a proton to a neutron, emitting a positron ($^0_1e^+$) and a neutrino (ν). Positrons are the antiparticles of electrons, therefore a positron has the same mass as an electron but with the opposite (positive) charge. In positron emission, the atomic number Z decreases by 1 while the mass number A remains the same.
Some examples of positron emission are
$\ce{^8_5B \rightarrow ^8_4Be + ^0_1e^{+}} + \nu_e$
$\ce{^{50}_{25}Mg \rightarrow ^{50}_{24}Cr + ^0_1e^+} + \nu_e$
Positron emission is only one of the two types of decay that tends to happen in "neutron deficient" nuclides, therefore it is very important to establish the correct mass change criterion. Positron emission occurs spontaneously when
m[AZ] > m[A(Z-1)] + m[0+1e+]
(Parent nucleus) > (Daughter nucleus) + (positron)
In order to rewrite this inequality in terms of the masses of neutral atoms, we add the mass of Z electrons to both sides of the equation, giving the mass of a neutral [AZ] atom on the left and the mass of a neutral [A(Z-1)] atom, plus an extra electron, (since only Z-1 electrons are needed to make the neutral atom), and a positron on the right. Because positrons and electrons have equal mass, the inequality can be written as
m[AZ] > m[A(Z-1)] + 2m[0-1e-]
(Parent atom) > (Daughter atom) + (2 electrons)
The change in mass for positron emission decay is
Δm = m[A(Z)] - m[AZ] - 2m[0-1e-]
As with beta decay, the kinetic energy -ΔE is split between the emitted particles - in this case the positron and neutrino.
Electron Capture
As mentioned before, there are two ways in which neutron-deficient / proton-rich nuclei can decay. When the mass change $Δm < 0$ yet is insufficient to cause spontaneous positron emission, a neutron can form by an alternate process known as electron capture. An outside electron is pulled inside the nucleus and combined with a proton to make a neutron, emitting only a neutrino.
$\ce{^1_1p + ^0_{-1}e^{-} → ^1_0n + \nu }$
Some examples of electron capture are
$\ce{^{231}_{92}U + ^0_{-1}e^{-} → ^{231}_{91}Pa + \nu }$
$\ce{ ^{81}{36}Kr + ^0_{-1}e^- → ^{81}_{35}Br + \nu }$
Electron capture happens most often in the heavier neutron-deficient elements where the mass change is smallest and positron emission isn't always possible. For $Δm < 0$, the following inequality applies:
m[AZ] + m[0-1e-] > m[A(Z-1)]
(Parent nucleus) + (electron) > (Daughter nucleus)
Adding $Z$ electrons to each side of the inequality changes it to its useful form in which the captured electron on the left cancels out the extra electron on the right
m[AZ] > m[A(Z-1)]
(Parent atom) > (Daughter atom)
The change in mass then equals
Δm = m[A(Z-1)] - m[AZ]
When the loss of mass in a nuclear reaction is greater than zero, but less than 2m[0-1e-], the process cannot occur by positron emission and is spontaneous for electron capture.
Alpha Decay
The other three processes of nuclear decay involve the formation of a neutron or a proton inside the nucleus to correct an existing imbalance. In alpha decay, unstable, heavy nuclei (typically $Z > 83$) reduce their mass number $A$ by 4 and their atomic number $Z$ by 2 with the emission of a helium nuclei ($\ce{^4_2He^{2+}}$), known as an alpha particle.
Some examples of alpha decay are
$\ce{^{222}_{88}Ra} \rightarrow \ce{^{218}_{86}Rn + ^4_2He^{2+}}$
$\ce{^{233}_{92}U} \rightarrow \ce{^{229}_{90}Th + ^4_2He^{2+}}$
As with beta decay and electron capture, Δm must only be less than zero for spontaneous alpha decay to occur. Since the number of total protons on each side of the reaction does not change, equal numbers of electrons are added to each side to make neutral atoms. Therefore, the mass of the parent atom must simply be greater than the sum of the masses of its daughter atom and the helium atom.
m[AZ] > m[A-4(Z-2)] + m[42He2+]
The change in mass then equals
Δm = m[A(Z)] - m[A-4(Z-2)] - m[42He2+]
The energy released in an alpha decay reaction is mostly carried away by the lighter helium, with a small amount of energy manifesting itself in the recoil of the much heavier daughter nucleus. Alpha decay is a form of spontaneous fission, a reaction in which a massive nuclei can lower its mass and atomic number by splitting. Other heavy unstable elements undergo fission reactions in which they split into nuclei of about equal size.
Summary
Proton-deficient or neutron-deficient nuclei undergo nuclear decay reactions that serve to correct unbalanced neutron/proton ratios. Proton-deficient nuclei undergo beta decay - emitting a beta particle (electron) and an antineutrino to convert a neutron to a proton - thus raising the elements atomic number Z by one. Neutron-deficient nuclei can undergo positron emission or electron capture (depending on the mass change), either of which synthesizes a neutron - emitting a positron and a neutrino or absorbing an electron and emitting a neutrino respectively - thus lowering Z by one. Nuclei with Z > 83 which are unstable and too massive will correct by alpha decay, emitting an alpha particle (helium nucleus) and decreasing both mass and atomic number. Very proton-deficient or neutron-deficient nuclei can also simply eject an excess particle directly from the nucleus. These types of decay are called proton and neutron emission. These processes are summarized in the table below.
Table: Characteristics of Radioactive Decay
Decay Type Emitted Particle ΔZ ΔA Occurrence
Alpha 42He2+ -2 -4 Z > 83
Beta Energetic e-, γ +1 0 A/Z > (A/Z)stable
PE Energetic e+, γ -1 0 A/Z < (A/Z)stable, light nuclei
EC ν -1 0 A/Z < (A/Z)stable, heavy nuclei
γ Photon 0 0 Any excited nucleus
Problems
1. Write the balanced equation for the beta decay of 14C.
2. Write the balanced equation for the positron emission decay of 22Na.
3. Write the balanced equation for electron capture in 207Bi.
4. Write the balanced equation for the alpha decay of 238U.
5. Calculate the maximum kinetic energy of the emitted beta particle in the decay
2411Na → 2412Mg + 0-1e- + ν
Use Table A4 of particle masses to do this calculation.
6. Calculate the maximum kinetic energy of the positron emitted in the decay
85B → 84Be + 01e+ + ν
Use Table A4 of particle masses to do this calculation.
7. Will 23192U likely decay to 23191Pa by positron emission or by electron capture? Use the mass criterion equations.
Solutions
1. 146C → 147N + 0-1e- + ν
2. 2211Na → 2210Ne + 0+1e+ + ν
3. 20783Bi + 0-1e-20782Pb + ν
4. 23892U → 23490Th + 42He
5. Given the masses of relevant atoms and the mass change criteria for beta decay, we calculate:
Δm = m[2412Mg] - m[2411Na]
Δm = 23.9850419 u - 23.990963 u = -0.0059211 u
ΔE = (-0.0059211 u)(931.494 MeV / u) = -5.515 MeV
The maximum kinetic energy of the beta particle is 5.515 MeV.
6. The change in mass is:
Δm = m[84Be] + 2m[0-1e-] - m[85B]
Δm = 8.005305 u + 2(0.000548579911 u) - 8.024606 u = -0.01820384 u
ΔE = (-0.01820384 u)(931.494 MeV / u) = -16.9568 MeV
The maximum kinetic energy of the positron is 16.9568 MeV.
7. The difference in mass between the daughter and parent atom is:
Δm = m[23191Pa] - m[23192U]
Δm = 231.035879 u - 231.03689 u = -0.00041 u
2m[0-1e-] = 0.0010971598 u
Since 0.00041 u is less than 2m[0-1e-], the process cannot occur by positron emission. The mass criterion Δm < 0 for electron capture is met, therefore 231U decays by electron capture.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Nuclear_Chemistry/Radioactivity/Nuclear_Decay_Pathways.txt
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The production of everyday used materials such as oil and gas results in the buildup of radioactive materials in high concentrations. As a result of the commonality of this occurrence, a solution has been presented as to rid the world of this toxic waste. Purifying these radioactive materials with the use of various methods allows the pure substance to be reused and prevents the depletion of resources.
Introduction
Before we can understand how to purify these substances, it is important to understand the chemistry involved within to get a deeper understanding. Radioactivity was first discovered by Henri Becquerel in 1896 when he considered that the phosphorescent materials could be related to the glow emitted by X-rays. Rutherford further enhanced this discovery with his gold foil experiment to class the particles emitted by radioactive materials into classes based on their ability to penetrate through materials. Alpha particles are larger in size and therefore the least harmful because they cannot pass through something as thin as a sheet of paper. Through analyzing radioactive decay, it can be determined that alpha particles are positive. Beta particles are larger, carry a negative charge, and require a more dense substance to hinder their path. Gamma rays are neutral and by far the most dangerous of the three. These rays cannot be easily deflected and can even go through concrete. The term radioactive is defined as an unstable particle that releases subatomic particles. Examples include carbon-14, radium, uranium. Usually isotopes of elements have enough instability to fit this definition. Although these rays have a positive effect on some fields such as medicine in terms of x-rays, caution is used to prevent bodily damage. Scientists working with these substances wear protective gear as well as gadgets that records their exposure to the substance.
Methods of Purification
The most common method of purification of radioactive materials in very minute quantities is distillation. However, radioactive materials occur in large quantities and pose a bigger question. Rather than mass purify radioactive materials, they are currently placed deep underground until radioactive decay keeps it from being harmful.
Electro filtration method: Separation of liquid and solid phases to extract the pure substance with the use of electrodes
Filtration through a substance that reduces the radioactive material in question and then using a substance that will bind this reduced radioactive material will allow it to be separated from the remaining solution. This method has recently acquired a patent and is still undergoing experimental procedures but remains effective.
Purification of Radioactive Water
In nature water generally contains a plethora of impurities. These impurities can include small microbes to something as dangerous as radioactive substances. Methods such as boiling, Chlorination (use of household chlorine bleach), and purification tablets remove microorganisms. More rigorous modes of purification are used to rid the water of other wastes including radioactive materials. Groundwater is a common example in which radium, a radioactive element, is mixed with the water. This way produces a black sludge of radioactive water which is unhealthy for consumer usage. The radium can be removed through ion exchange or the conditioning of water. Other unnatural occurrences of radioactive materials require more meticulous methods. Distillation removed salts, heavy metals, and radioactive fallout (since water itself cannot become radioactive, the radioactive components are referred to as radioactive fallout). Filtering the water will also remove the radioactive fallout.
Problems
1. What are way in which to purify water that contains radium?
2. What does the term radioactive mean?
3. Define the different kinds of emissions and their range of harmfulness.
4. What does distillation do that chlorination cannot achieve?
5. Why is it important to purify radioactive materials?
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All radioactive particles and waves, from the entire electromagnetic spectrum, to alpha, beta, and gamma particles, possess the ability to eject electrons from atoms and molecules to create ions.
Introduction
There are many types of radiation, but the two most common are electromagnetic radiation and ionizing radiation. Ionizing radiation refers to radioactive particles, such as alpha and beta particles, or electromagnetic waves, such as gamma or ultraviolet rays, which have sufficient energy to detach electrons off of atoms to create ions, hence the name “ionizing radiation.” Electromagnetic radiation, which sometimes can be placed as a subcategory of ionizing radiation, deals with waves or photons from the electromagnetic spectrum. Unlike ionizing radiation, electromagnetic radiation deals with electric and magnetic field oscillations such as with X-rays, radio waves, or gamma rays.
Radioactive decay of atoms creates three radioactive particles, alpha, beta, and gamma. Of the three, alpha particles are known to have the most “ionizing power,” a term describing the number of ion pairs produced per centimeter through a material, followed by beta, then gamma. However, a common misconception is that the higher ionizing power a particle has, the more damaging it is to matter. Electromagnetic waves can also ionize, hence the reason electromagnetic radiation is often placed as part of ionizing radiation.
Primary Electrons and Secondary Ionization
The main effect radiation has on matter is its ability to ionize atoms to become ions, a phenomenon known as ionization, which is very similar to the photoelectric effect. Radioactive particles or electromagnetic waves with sufficient energy collide with electrons on the atom to knock electrons off the atom. The electron ejected off the atom is called the primary electron. When the primary electrons hold energy, a particle ejecting the primary electron may cause it to eject another electron, either on their own atom or on another atom. This is known as secondary ionization.
However, ionization does not have to completely eject an electron off the atom. It can raise the energy of the electron instead, raising the electron energy to a higher energy state. When the electron reverts to its normal energy level, it emits energy in the form of radiation, usually in the forms of ultraviolet rays or radio waves.
Production of X-Rays and Electromagnetic Radiation
Radiation can be both natural and synthetic. Artificially induced radioactivity utilizes primary and secondary ionizations in order to emit X-rays. Most X-ray emission is due to the bombardment of electrons on a metal target. If the electrons have sufficient energy, the inner shell electrons of the atom fall out, and higher-leveled electrons fill in the hole left by the previous electrons. By doing so, packets of energy are released in the forms of X-ray photons. Other forms of ionizing radiation can produce UV and gamma rays in a similar manner. This type of radiation is known as “ionizing radiation.”
All charged particles and rays have the ability to be radioactive; however, not all rays and particles have the energy per photon to ionize atoms. This is known as “non-ionizing radiation.” Non-ionizing radiation has enough energy to excite electrons to move to a higher state, releasing photons of electromagnetic radiation such as visible light, near ultraviolet, and microwaves. Radio waves, microwaves, and neutron radiation (an important application in fission and fusion) all fall under non-ionizing radiation, as their respective energies are too low to ionize atoms.
(Courtesy of iforms.osha-slc.gov/SLTC/radiation/index.html)
Effects of Radiation on Living Matter
Prolonged exposure to radiation often has detrimental effects on living matter. This is due to radiation’s ionizing ability, which can damage the internal functioning of cells. Radiation either ionizes or excites atoms or molecules in living cells, leading to the dissociation of molecules within an organism. The most destructive effect radiation has on living matter is ionizing radiation on DNA. Damage to DNA can cause cellular death, mutagenesis (the process by which genetic information is modified by radiation or chemicals), and genetic transformation. Effects from exposure to radiation include leukemia, birth defects, and many forms of cancer.
Most external radiation is absorbed by the environment; for example, most ultraviolet radiation is absorbed by the ozone layer, preventing deadly levels of ultraviolet radiation to come in contact with the surface of the earth. Sunburn is an effect of UV radiation damaging skin cells, and prolonged exposure to UV radiation can cause genetic information in skin cells to mutate, leading to skin cancer.
Alpha, beta, and gamma rays also cause damage to living matter, in varying degrees. Alpha particles have a very small absorption range, and thus are usually not harmful to life, unless ingested, due to its high ionizing power. Beta particles are also damaging to DNA, and therefore are often used in radiation therapy to mutate and kill cancer cells. Gamma rays are often considered the most dangerous type of radiation to living matter. Unlike alpha and beta particles, which are charged particles, gamma rays are instead forms of energy. They have large penetrating range and can diffuse through many cells before dissipating, causing widespread damage such as radiation sickness. Because gamma rays have such high penetrating power and can damage living cells to a great extent, they are often used in irradiation, a process used to kill living organisms.
Radiation Dosage and Decay
There are several methods to measure radiation; hence, there are several radiation units based on different radiation factors. Radiation units can measure radioactive decay, absorbed dosage, and human absorbed doses. Bq and Ci measure radioactive decay, while Gy and Rad measures absorbed doses. Sv and Rem measure absorbed doses in Gy and Rad equivalents. Rem takes into account different radiation types and the speed of particles. Below is a chart to help organize the different units:
Units for Radioactive Decay
Becquerel, Bq Measured in s-1, as disintegration per second
Curie, Ci
Measured as amount of decay at the same rate as 1 gram of radium
1 Ci = 3.70∙1010 Bq
Units for Absorbed Dose
Gray, Gy 1 Gy deposits 1 Joule of energy per kilogram of matter
Rad 1 rad = 0.01 Gy
Equivalent Doses
Sievert, Sv 1Sv= 100 rem
Rem
1 rem = 1 rad∙Q
Q = 1 for X-rays, gamma rays, and beta particles
Q = 3 for slow neutrons
Q = 10 for protons and fast neutrons
Q = 20 for alpha particles
The most commonly used unit is the "rad," which stands for "radiation absorbed dose," and the "rem," which stands for "radiation equivalent for man." One rad corresponds to the absorption of 0.01 Joules of energy per kilogram of matter. Rem is the rad multiplied by the relative biological effectiveness, which is most often expressed as the variable "Q." The factor Q is used to take into account the different effects caused by different radiation.
Concept Review Questions
1. Classify the following interactions that occur as either primary ionization, secondary ionization, or electron excitement.
1. Photons are ejected from the atom.
2. An electron from a nearby atom is ejected, knocking out an electron from a neighboring atom.
3. Electrons are ejected from the atom.
2. Describe the difference between ionizing and non-ionizing radiation.
3. Explain why radiation has such a harmful effect on living matter.
4. Consider modern microwave oven used in kitchens. Are the microwaves ejected to heat water and food harmful to the human body?
5. What is the Q in the calculation or REM?
Answers
1. Electron excitement. An electron is excited to a higher energy level. When it falls, it releases a packet of energy in the form of a photon.
2. Secondary ionization. The ejection of the second electron was caused by another electron, as opposed to another charged particle or radiating ray.
3. Primary ionization. The electron was ejected by a charged particle or ray.
1. Ionizing radiation describes atoms becoming ionized to ions. During ionizing radiation, an electron is ejected off the atom, causing the atom to lose an electron and become ionized. Non-ionizing radiation is generally caused by excitation of electrons. When a particle or electromagnetic ray does not have sufficient energy to completely knock an electron off an atom, it can instead excite the electron to go to a higher energy level. When the electron falls, it emits photons of energy.
2. Radiation can ionize atoms, but it can also mutant molecules by ionizing atoms. It can affect the structure of a cell by debilitating organelles or other cellular functions, but its most damaging effect is on DNA. Radiation mutates DNA by ionizing base sequences, or by altering the backbone of DNA. DNA mutations arising from irradiation can cause cancer, or otherwise kill the cell.
3. Although there are many myths concerning microwave radiation, microwaves fall under "non-ionizing" radiation and hence does not cause any of the effects that ionizing radiation causes, such as cancer. A microwave oven is also designed to minimise microwaves escaping outside of the oven by using metal to absorb the microwaves. This is why the door of the microwave oven is not transparent; it is lined with strategically placed metal atoms for maximum absorbing efficiency.
4. Q is the constant that is used depending on what radioactive particle you are calculating. It is based on the type of particle and its effect on matter.
Transmutation of the Elements
Although the conversion of one element to another is the basis of natural radioactive decay, it is also possible to convert one element to another artificially. The conversion of one element to another is the process of transmutation. Between 1921 and 1924, Patrick Blackett conducted experiments in which he converted a stable isotope of nitrogen to a stable isotope of oxygen. By bombarding $\ce{^{14}N}$ with $\alpha$ particles he created $\ce{^{17}O}$. Transmutation may also be accomplished by bombardment with neutrons.
$\ce{^{14}_7N + ^4_2He \rightarrow ^{17}_8O + ^1_1H}$
Historically, part of Alchemy was the study of methods of creating gold from base metals, such lead. Where the Alchemists failed in this quest, we can now succeed. Thus, bombardment of platinum-198 with a neutron creates an unstable isotope of platinum that undergoes beta decay to gold-199. Unfortunately, while we may succeed in making gold, the platinum we make it from is actually worth more than the gold making this particular transmutation economically non-viable!
$\ce{^{198}_{78}Pt + ^1_0n \rightarrow ^{199}_{78}Pt \rightarrow ^{199}_{79}Au + ^0_{-1}\beta}$
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The atomic and molecular properties are the intrinsic features associated with the system at the atomic and sub-atomic level.
• Atomic and Ionic Radius
This page explains the various measures of atomic radius, and then looks at the way it varies around the Periodic Table - across periods and down groups. It assumes that you understand electronic structures for simple atoms written in s, p, d notation.
• Atomic Radii
Atomic radii is useful for determining many aspects of chemistry such as various physical and chemical properties. The periodic table greatly assists in determining atomic radius and presents a number of trends.
• Dipole Moments
Dipole moments occur when there is a separation of charge. They can occur between two ions in an ionic bond or between atoms in a covalent bond; dipole moments arise from differences in electronegativity. The larger the difference in electronegativity, the larger the dipole moment. The distance between the charge separation is also a deciding factor into the size of the dipole moment. The dipole moment is a measure of the polarity of the molecule.
• Electronegativity
Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The Pauling scale is the most commonly used. Fluorine (the most electronegative element) is assigned a value of 4.0, and values range down to cesium and francium which are the least electronegative at 0.7.
• Electron Affinity
Electron affinity is defined as the change in energy (in kJ/mole) of a neutral atom (in the gaseous phase) when an electron is added to the atom to form a negative ion. In other words, the neutral atom's likelihood of gaining an electron.
• Formal Charges
Knowing the formal charges on specific atoms in a molecule is an important step in keeping tract of the electrons and determine the chemical reactivity of the molecule. Formal charges can be calculated mathematically, but they can also be determined by intuition. The instinctive method is faster but requires more skill and knowledge of common structures.
• Intermolecular Forces
Intermolecular forces are the attractive or repulsive forces between molecules. They are separated into two groups; short range and long range forces. Short range forces happen when the centers of the molecules are separated by three angstroms (10-8 cm) or less. Short range forces tend to be repulsive, where the long range forces that act outside the three angstroms range are attractive. Long range forces are also known as Van der Waals forces. They are responsible for surface tension, friction,
• Ionization Energy
Ionization energy is the quantity of energy that an isolated, gaseous atom in the ground electronic state must absorb to discharge an electron, resulting in a cation.
• Lewis Structures
A Lewis Structure is a very simplified representation of the valence shell electrons in a molecule. It is used to show how the electrons are arranged around individual atoms in 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.
• Magnetic Properties
Anything that is magnetic, like a bar magnet or a loop of electric current, has a magnetic moment. A magnetic moment is a vector quantity, with a magnitude and a direction. An electron has an electron magnetic dipole moment, generated by the electron's intrinsic spin property, making it an electric charge in motion. There are many different magnetic forms: including paramagnetism, and diamagnetism, ferromagnetism, and anti-ferromagnetism. Only paramagnetism, and diamagnetism are discussed here.
• Molecular Polarity
Polarity is a physical property of compounds which relates other physical properties such as melting and boiling points, solubility, and intermolecular interactions between molecules. For the most part, there is a direct correlation between the polarity of a molecule and number and types of polar or non-polar covalent bonds which are present. In a few cases, a molecule may have polar bonds, but in a symmetrical arrangement which then gives rise to a non-polar molecule such as carbon dioxide.
• Polarizability
Having now revised the basics of trends across and down the Periodic Table, we can use the concepts of Effective Nuclear Charge and Electronegativity to discuss the factors that contribute to the types of bonds formed between elements.
Thumbnail: pixabay.com/illustrations/mo...ry-3d-1818492/
Atomic and Molecular Properties
Atomic radii is useful for determining many aspects of chemistry such as various physical and chemical properties. The periodic table greatly assists in determining atomic radius and presents a number of trends.
Definition
Atomic radius is generally stated as being the total distance from an atom’s nucleus to the outermost orbital of electron. In simpler terms, it can be defined as something similar to the radius of a circle, where the center of the circle is the nucleus and the outer edge of the circle is the outermost orbital of electron. As you begin to move across or down the periodic table, trends emerge that help explain how atomic radii change.
The effective nuclear charge (\(Z_{eff}\)) of an atom is the net positive charge felt by the valence electron. Some positive charge is shielded by the core electrons therefore the total positive charge is not felt by the valence electron. A detailed description of shielding and effective nuclear charge can be found here. \(Z_{eff}\) greatly affects the atomic size of an atom. So as the \(Z_{eff}\) decreases, the atomic radius will grow as a result because there is more screening of the electrons from the nucleus, which decreases the attraction between the nucleus and the electron. Since \(Z_{eff}\)decreases going down a group and right to left across the periodic table, the atomic radius will increase going down a group and right to left across the periodic table.
Types of Radius with Respect to Types of Bonds
Determining the atomic radii is rather difficult because there is an uncertainty in the position of the outermost electron – we do not know exactly where the electron is. This phenomenon can be explained by the Heisenberg Uncertainty Principle. To get a precise measurement of the radius, but still not an entirely correct measurement, we determine the radius based on the distance between the nuclei of two bonded atoms. The radii of atoms are therefore determined by the bonds they form. An atom will have different radii depending on the bond it forms; so there is no fixed radius of an atom.
Covalent Radius
When a covalent bond is present between two atoms, the covalent radius can be determined. When two atoms of the same element are covalently bonded, the radius of each atom will be half the distance between the two nuclei because they equally attract the electrons. The distance between two nuclei will give the diameter of an atom, but you want the radius which is half the diameter.
Covalent radii will increase in the same pattern as atomic radii. The reason for this trend is that the bigger the radii, the further the distance between the two nuclei. See explanation for \(Z_{eff}\) for more details.
The covalent radius depicted below in Figure 1 will be the same for both atoms because they are of the same element as shown by X.
Ionic Radius
The ionic radius is the radius of an atom forming ionic bond or an ion. The radius of each atom in an ionic bond will be different than that in a covalent bond. This is an important concept. The reason for the variability in radius is due to the fact that the atoms in an ionic bond are of greatly different size. One of the atoms is a cation, which is smaller in size, and the other atom is an anion which is a lot larger in size. So in order to account for this difference, one most get the total distance between the two nuclei and divide the distance according to atomic size. The bigger the atomic size, the larger radius it will have. This is depicted in Figure 2 as shown below where the cation is displayed on the left as X+, and clearly has a smaller radius than the anion, which is depicted as Y- on the right.
Example 1: Cadmium Sulfide
If we were able to determine the atomic radius of an atom from experimentation, say Se, which had an atomic radius of 178 pm, then we could determine the atomic radius of any other atom bonded to Se by subtracting the size of the atomic radius of Se from the total distance between the two nuclei. So, if we had the compound CaSe, which had a total distance of 278 pm between the nucleus of the Ca atom and Se atom, then the atomic radius of the Ca atom will be 278 pm (total distance) - 178 pm (distance of Se), or 100 pm. This process can be applied to other examples of ionic radius.
Cations have smaller ionic radii than their neutral atoms. In contrast, anions have bigger ionic radii than their corresponding neutral atoms.
A detailed explanation is given below:
• The cation, which is an ion with a positive charge, by definition has fewer electrons than protons. The loss in an electron will consequently result in a change in atomic radii in comparison to the neutral atom of interest (no charge).
• The loss of an electron means that there are now more protons than electrons in the atom, which is stated above. This will cause a decrease in atomic size because there are now fewer electrons for the protons to pull towards the nucleus and will result in a stronger pull of the electrons towards the nucleus. It will also decrease because there are now less electrons in the outer shell, which will decrease the radius size.
• An analogy to this can be of a magnet and a metallic object. If ten magnets and ten metallic objects represent a neutral atom where the magnets are protons and the metallic objects are electrons, then removing one metallic object, which is like removing an electron, will cause the magnet to pull the metallic objects closer because of a decrease in number of the metallic objects. This can similarly be said about the protons pulling the electrons closer to the nucleus, which as a result decreases atomic size.
Figure 3 below depicts this process. A neutral atom X is shown here to have a bond length of 180 pm and then the cation X+ is smaller with a bond length of 100 pm.
An anion, on the other hand, will be bigger in size than that of the atom it was made from because of a gain of an electron. This can be seen in the Figure 4 below. The gain of an electron adds more electrons to the outermost shell which increases the radius because there are now more electrons further away from the nucleus and there are more electrons to pull towards the nucleus so the pull becomes slightly weaker than of the neutral atom and causes an increase in atomic radius.
Metallic Radius
The metallic radius is the radius of an atom joined by metallic bond. The metallic radius is half of the total distance between the nuclei of two adjacent atoms in a metallic cluster. Since a metal will be a group of atoms of the same element, the distance of each atom will be the same (Figure 5).
Periodic Trends of Atomic Radius
• An atom gets larger as the number of electronic shells increase; therefore the radius of atoms increases as you go down a certain group in the periodic table of elements.
• In general, the size of an atom will decrease as you move from left to the right of a certain period.
Vertical Trend
The radius of atoms increases as you go down a certain group.
Horizontal Trend
The size of an atom will decrease as you move from left to the right of a period.
EXCEPTIONS: Because the electrons added in the transition elements are added in the inner electron shell and at the same time, the outer shell remains constant, the nucleus attracts the electrons inward. The electron configuration of the transition metals explains this phenomenon. This is why Ga is the same size as its preceding atom and why Sb is slightly bigger than Sn.
Problems
1. Which atom is larger: K or Br?
2. Which atom is larger: Na or Cl?
3. Which atom is smaller: Be or Ba?
4. Which atom is larger: K+ or K?
5. Put in order of largest to smallest: F, Ar, Sr, Cs.
6. Which has a bigger atomic radius: Sr2+ or Se2-?
7. If Br has an ionic radius of 100 pm and the total distance between K and Br in KBr is 150 pm, then what is the ionic radius of K?
8. Which has a smaller atomic radius: Cs+ or Xe?
9. If the distance between the nuclei of two atoms in a metallic bond is 180 pm, what is the atomic radius of one atom?
10. If Z effective is increasing, is the atomic radius also increasing?
*Hint* When solving a radius-bond problem, identify the bond first and then use the standard method of finding the radius for that particular bond. Also remember the trend for the atomic radii.
Answers
1. K
2. Na
3. Be
4. K
5. Cs, Sr, Ar, F
6. Se2-
7. 50 pm
8. Cs+
9. 90 pm
10. No
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This page explains the various measures of atomic radius, and then looks at the way it varies around the Periodic Table - across periods and down groups. It assumes that you understand electronic structures for simple atoms written in s, p, d notation.
Atomic Radius
Unlike a ball, an atom does not have a fixed radius. The radius of an atom can only be found by measuring the distance between the nuclei of two touching atoms, and then halving that distance.
As you can see from the diagrams, the same atom could be found to have a different radius depending on what was around it. The left hand diagram shows bonded atoms. The atoms are pulled closely together and so the measured radius is less than if they are just touching. This is what you would get if you had metal atoms in a metallic structure, or atoms covalently bonded to each other. The type of atomic radius being measured here is called the metallic radius or the covalent radius depending on the bonding.
The right hand diagram shows what happens if the atoms are just touching. The attractive forces are much less, and the atoms are essentially "unsquashed". This measure of atomic radius is called the van der Waals radius after the weak attractions present in this situation.
Trends in atomic radius in the Periodic Table
The exact pattern you get depends on which measure of atomic radius you use - but the trends are still valid. The following diagram uses metallic radii for metallic elements, covalent radii for elements that form covalent bonds, and van der Waals radii for those (like the noble gases) which don't form bonds.
Trends in atomic radius down a group
It is fairly obvious that the atoms get bigger as you go down groups. The reason is equally obvious - you are adding extra layers of electrons.
Trends in atomic radius across periods
You have to ignore the noble gas at the end of each period. Because neon and argon don't form bonds, you can only measure their van der Waals radius - a case where the atom is pretty well "unsquashed". All the other atoms are being measured where their atomic radius is being lessened by strong attractions. You aren't comparing like with like if you include the noble gases.
Leaving the noble gases out, atoms get smaller as you go across a period. If you think about it, the metallic or covalent radius is going to be a measure of the distance from the nucleus to the electrons which make up the bond. (Look back to the left-hand side of the first diagram on this page if you aren't sure, and picture the bonding electrons as being half way between the two nuclei.)
From lithium to fluorine, those electrons are all in the 2-level, being screened by the 1s2 electrons. The increasing number of protons in the nucleus as you go across the period pulls the electrons in more tightly. The amount of screening is constant for all of these elements.
In the period from sodium to chlorine, the same thing happens. The size of the atom is controlled by the 3-level bonding electrons being pulled closer to the nucleus by increasing numbers of protons - in each case, screened by the 1- and 2-level electrons.
Trends in the transition elements
Although there is a slight contraction at the beginning of the series, the atoms are all much the same size. The size is determined by the 4s electrons. The pull of the increasing number of protons in the nucleus is more or less offset by the extra screening due to the increasing number of 3d electrons.
Ionic Radius
Ionic radii are difficult to measure with any degree of certainty, and vary according to the environment of the ion. For example, it matters what the co-ordination of the ion is (how many oppositely charged ions are touching it), and what those ions are. There are several different measures of ionic radii in use, and these all differ from each other by varying amounts. It means that if you are going to make reliable comparisons using ionic radii, they have to come from the same source.
What you have to remember is that there are quite big uncertainties in the use of ionic radii, and that trying to explain things in fine detail is made difficult by those uncertainties. What follows will be adequate for UK A level (and its various equivalents), but detailed explanations are too complicated for this level.
Trends in ionic radius in the Periodic Table
Trends in ionic radius down a group: This is the easy bit! As you add extra layers of electrons as you go down a group, the ions are bound to get bigger. The two tables below show this effect in Groups 1 and 7.
electronic structure of ion ionic radius (nm)
Li+ 2 0.076
Na+ 2, 8 0.102
K+ 2, 8, 8 0.138
Rb+ 2, 8, 18, 8 0.152
Cs+ 2, 8, 18, 18, 8 0.167
electronic structure of ion ionic radius (nm)
F- 2, 8 0.133
Cl- 2, 8, 8 0.181
Br- 2, 8, 18, 8 0.196
I- 2, 8, 18, 18, 8 0.220
Trends in ionic radius across a period
Let's look at the radii of the simple ions formed by elements as you go across Period 3 of the Periodic Table - the elements from Na to Cl.
Na+ Mg2+ Al3+ P3- S2- Cl-
no of protons 11 12 13 15 16 17
electronic structure of ion 2,8 2,8 2,8 2,8,8 2,8,8 2,8,8
ionic radius (nm) 0.102 0.072 0.054 (0.212) 0.184 0.181
The table misses out silicon which does not form a simple ion. The phosphide ion radius is in brackets because it comes from a different data source, and I am not sure whether it is safe to compare it. The values for the oxide and chloride ions agree in the different source, so it is probably OK. The values are again for 6-co-ordination, although I can't guarantee that for the phosphide figure.
First of all, notice the big jump in ionic radius as soon as you get into the negative ions. Is this surprising? Not at all - you have just added a whole extra layer of electrons. Notice that, within the series of positive ions, and the series of negative ions, that the ionic radii fall as you go across the period. We need to look at the positive and negative ions separately.
• The positive ions: In each case, the ions have exactly the same electronic structure - they are said to be isoelectronic. However, the number of protons in the nucleus of the ions is increasing. That will tend to pull the electrons more and more towards the center of the ion - causing the ionic radii to fall. That is pretty obvious!
• The negative ions: Exactly the same thing is happening here, except that you have an extra layer of electrons. What needs commenting on, though is how similar in size the sulphide ion and the chloride ion are. The additional proton here is making hardly any difference.
The difference between the size of similar pairs of ions actually gets even smaller as you go down Groups 6 and 7. For example, the Te2- ion is only 0.001 nm bigger than the I- ion.
As far as I am aware there is no simple explanation for this - certainly not one which can be used at this level. This is a good illustration of what I said earlier - explaining things involving ionic radii in detail is sometimes very difficult.
Trends in ionic radius for some more isoelectronic ions
This is only really a variation on what we have just been talking about, but fits negative and positive isoelectronic ions into the same series of results. Remember that isoelectronic ions all have exactly the same electron arrangement.
N3- O2- F- Na+ Mg2+ Al3+
no of protons 7 8 9 11 12 13
electronic structure of ion 2, 8 2, 8 2, 8 2, 8 2, 8 2, 8
ionic radius (nm) (0.171) 0.140 0.133 0.102 0.072 0.054
Note: The nitride ion value is in brackets because it came from a different source, and I don't know for certain whether it relates to the same 6-co-ordination as the rest of the ions. This matters. My main source only gave a 4-coordinated value for the nitride ion, and that was 0.146 nm.
You might also be curious as to how the neutral neon atom fits into this sequence. Its van der Waals radius is 0.154 or 0.160 nm (depending on which source you look the value up in) - bigger than the fluoride ion. You can't really sensibly compare a van der Waals radius with the radius of a bonded atom or ion.
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Dipole moments occur when there is a separation of charge. They can occur between two ions in an ionic bond or between atoms in a covalent bond; dipole moments arise from differences in electronegativity. The larger the difference in electronegativity, the larger the dipole moment. The distance between the charge separation is also a deciding factor in the size of the dipole moment. The dipole moment is a measure of the polarity of the molecule.
Introduction
When atoms in a molecule share electrons unequally, they create what is called a dipole moment. This occurs when one atom is more electronegative than another, resulting in that atom pulling more tightly on the shared pair of electrons, or when one atom has a lone pair of electrons and the difference of electronegativity vector points in the same way. One of the most common examples is the water molecule, made up of one oxygen atom and two hydrogen atoms. The differences in electronegativity and lone electrons give oxygen a partial negative charge and each hydrogen a partial positive charge.
Dipole Moment
When two electrical charges, of opposite sign and equal magnitude, are separated by a distance, an electric dipole is established. The size of a dipole is measured by its dipole moment ($\mu$). Dipole moment is measured in Debye units, which is equal to the distance between the charges multiplied by the charge (1 Debye equals $3.34 \times 10^{-30}\; C\, m$). The dipole moment of a molecule can be calculated by Equation $\ref{1}$:
$\vec{\mu} = \sum_i q_i \, \vec{r}_i \label{1}$
where
• $\vec{\mu}$ is the dipole moment vector
• $q_i$ is the magnitude of the $i^{th}$ charge, and
• $\vec{r}_i$ is the vector representing the position of $i^{th}$ charge.
The dipole moment acts in the direction of the vector quantity. An example of a polar molecule is $\ce{H_2O}$. Because of the lone pair on oxygen, the structure of $\ce{H_2O}$ is bent (via VSEPR theory), which means that the vectors representing the dipole moment of each bond do not cancel each other out. Hence, water is polar.
The vector points from positive to negative, on both the molecular (net) dipole moment and the individual bond dipoles. Table A2 shows the electronegativity of some of the common elements. The larger the difference in electronegativity between the two atoms, the more electronegative that bond is. To be considered a polar bond, the difference in electronegativity must be large. The dipole moment points in the direction of the vector quantity of each of the bond electronegativities added together.
It is relatively easy to measure dipole moments: just place a substance between charged plates (Figure $2$); polar molecules increase the charge stored on the plates, and the dipole moment can be obtained (i.e., via the capacitance of the system). Nonpolar $\ce{CCl_4}$ is not deflected; moderately polar acetone deflects slightly; highly polar water deflects strongly. In general, polar molecules will align themselves: (1) in an electric field, (2) with respect to one another, or (3) with respect to ions (Figure $2$).
Equation $\ref{1}$ can be simplified for a simple separated two-charge system like diatomic molecules or when considering a bond dipole within a molecule
$\mu_{diatomic} = Q \times r \label{1a}$
This bond dipole is interpreted as the dipole from a charge separation over a distance $r$ between the partial charges $Q^+$ and $Q^-$ (or the more commonly used terms $δ^+$ - $δ^-$); the orientation of the dipole is along the axis of the bond. Consider a simple system of a single electron and proton separated by a fixed distance. When the proton and electron are close together, the dipole moment (degree of polarity) decreases. However, as the proton and electron get farther apart, the dipole moment increases. In this case, the dipole moment is calculated as (via Equation $\ref{1a}$):
\begin{align*} \mu &= Qr \nonumber \[4pt] &= (1.60 \times 10^{-19}\, C)(1.00 \times 10^{-10} \,m) \nonumber \[4pt] &= 1.60 \times 10^{-29} \,C \cdot m \label{2} \end{align*}
The Debye characterizes the size of the dipole moment. When a proton and electron are 100 pm apart, the dipole moment is $4.80\; D$:
\begin{align*} \mu &= (1.60 \times 10^{-29}\, C \cdot m) \left(\dfrac{1 \;D}{3.336 \times 10^{-30} \, C \cdot m} \right) \nonumber \[4pt] &= 4.80\; D \label{3} \end{align*}
$4.80\; D$ is a key reference value and represents a pure charge of +1 and -1 separated by 100 pm. If the charge separation is increased then the dipole moment increases (linearly):
• If the proton and electron are separated by 120 pm: $\mu = \dfrac{120}{100}(4.80\;D) = 5.76\, D \label{4a}$
• If the proton and electron are separated by 150 pm: $\mu = \dfrac{150}{100}(4.80 \; D) = 7.20\, D \label{4b}$
• If the proton and electron are separated by 200 pm: $\mu = \dfrac{200}{100}(4.80 \; D) = 9.60 \,D \label{4c}$
Example $1$: Water
The water molecule in Figure $1$ can be used to determine the direction and magnitude of the dipole moment. From the electronegativities of oxygen and hydrogen, the difference in electronegativity is 1.2e for each of the hydrogen-oxygen bonds. Next, because the oxygen is the more electronegative atom, it exerts a greater pull on the shared electrons; it also has two lone pairs of electrons. From this, it can be concluded that the dipole moment points from between the two hydrogen atoms toward the oxygen atom. Using the equation above, the dipole moment is calculated to be 1.85 D by multiplying the distance between the oxygen and hydrogen atoms by the charge difference between them and then finding the components of each that point in the direction of the net dipole moment (the angle of the molecule is 104.5˚).
The bond moment of the O-H bond =1.5 D, so the net dipole moment is
$\mu=2(1.5) \cos \left(\dfrac{104.5˚}{2}\right)=1.84\; D \nonumber$
Polarity and Structure of Molecules
The shape of a molecule and the polarity of its bonds determine the OVERALL POLARITY of that molecule. A molecule that contains polar bonds might not have any overall polarity, depending upon its shape. The simple definition of whether a complex molecule is polar or not depends upon whether its overall centers of positive and negative charges overlap. If these centers lie at the same point in space, then the molecule has no overall polarity (and is non polar). If a molecule is completely symmetric, then the dipole moment vectors on each molecule will cancel each other out, making the molecule nonpolar. A molecule can only be polar if the structure of that molecule is not symmetric.
A good example of a nonpolar molecule that contains polar bonds is carbon dioxide (Figure $\PageIndex{3a}$). This is a linear molecule and each C=O bond is, in fact, polar. The central carbon will have a net positive charge, and the two outer oxygen atoms a net negative charge. However, since the molecule is linear, these two bond dipoles cancel each other out (i.e. the vector addition of the dipoles equals zero) and the overall molecule has a zero dipole moment ($\mu=0$).
Although a polar bond is a prerequisite for a molecule to have a dipole, not all molecules with polar bonds exhibit dipoles
For $AB_n$ molecules, where $A$ is the central atom and $B$ are all the same types of atoms, there are certain molecular geometries which are symmetric. Therefore, they will have no dipole even if the bonds are polar. These geometries include linear, trigonal planar, tetrahedral, octahedral and trigonal bipyramid.
Example $3$: $\ce{C_2Cl_4}$
Although the C–Cl bonds are rather polar, the individual bond dipoles cancel one another in this symmetrical structure, and $\ce{Cl_2C=CCl_2}$ does not have a net dipole moment.
Example $3$: $\ce{CH_3Cl}$
C-Cl, the key polar bond, is 178 pm. Measurement reveals 1.87 D. From this data, % ionic character can be computed. If this bond were 100% ionic (based on proton & electron),
\begin{align*} \mu &= \dfrac{178}{100}(4.80\; D) \nonumber \[4pt] &= 8.54\; D \nonumber \end{align*}
Although the bond length is increasing, the dipole is decreasing as you move down the halogen group. The electronegativity decreases as we move down the group. Thus, the greater influence is the electronegativity of the two atoms (which influences the charge at the ends of the dipole).
Table $1$: Relationship between Bond length, Electronegativity and Dipole moments in simple Diatomics
Compound Bond Length (Å) Electronegativity Difference Dipole Moment (D)
HF 0.92 1.9 1.82
HCl 1.27 0.9 1.08
HBr 1.41 0.7 0.82
HI 1.61 0.4 0.44
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Electron affinity is defined as the change in energy (in kJ/mole) of a neutral atom (in the gaseous phase) when an electron is added to the atom to form a negative ion. In other words, the neutral atom's likelihood of gaining an electron.
Introduction
Energy of an atom is defined when the atom loses or gains energy through chemical reactions that cause the loss or gain of electrons. A chemical reaction that releases energy is called an exothermic reaction and a chemical reaction that absorbs energy is called an endothermic reaction. Energy from an exothermic reaction is negative, thus energy is given a negative sign; whereas, energy from an endothermic reaction is positive and energy is given a positive sign. An example that demonstrates both processes is when a person drops a book. When he or she lifts a book, he or she gives potential energy to the book (energy absorbed). However, once the he or she drops the book, the potential energy converts itself to kinetic energy and comes in the form of sound once it hits the ground (energy released).
When an electron is added to a neutral atom (i.e., first electron affinity) energy is released; thus, the first electron affinities are negative. However, more energy is required to add an electron to a negative ion (i.e., second electron affinity) which overwhelms any the release of energy from the electron attachment process and hence, second electron affinities are positive.
• First Electron Affinity (negative energy because energy released):
$\ce{X (g) + e^- \rightarrow X^{-} (g)} \label{1}$
• Second Electron Affinity (positive energy because energy needed is more than gained):
$\ce{X^- (g) + e^- \rightarrow X^{2-} (g)} \label{2}$
First Electron Affinity
Ionization energies are always concerned with the formation of positive ions. Electron affinities are the negative ion equivalent, and their use is almost always confined to elements in groups 16 and 17 of the Periodic Table. The first electron affinity is the energy released when 1 mole of gaseous atoms each acquire an electron to form 1 mole of gaseous -1 ions. It is the energy released (per mole of X) when this change happens. First electron affinities have negative values. For example, the first electron affinity of chlorine is -349 kJ mol-1. By convention, the negative sign shows a release of energy.
When an electron is added to a metal element, energy is needed to gain that electron (endothermic reaction). Metals have a less likely chance to gain electrons because it is easier to lose their valance electrons and form cations. It is easier to lose their valence electrons because metals' nuclei do not have a strong pull on their valence electrons. Thus, metals are known to have lower electron affinities.
Example $1$: Group 1 Electron Affinities
This trend of lower electron affinities for metals is described by the Group 1 metals:
• Lithium (Li): -60 KJ mol-1
• Sodium (Na): -53 KJ mol-1
• Potassium (K): -48 KJ mol-1
• Rubidium (Rb): -47 KJ mol-1
• Cesium (Cs): -46 KJ mol-1
Notice that electron affinity decreases down the group.
When nonmetals gain electrons, the energy change is usually negative because they give off energy to form an anion (exothermic process); thus, the electron affinity will be negative. Nonmetals have a greater electron affinity than metals because of their atomic structures: first, nonmetals have more valence electrons than metals do, thus it is easier for the nonmetals to gain electrons to fulfill a stable octet and secondly, the valence electron shell is closer to the nucleus, thus it is harder to remove an electron and it easier to attract electrons from other elements (especially metals). Thus, nonmetals have a higher electron affinity than metals, meaning they are more likely to gain electrons than atoms with a lower electron affinity.
Example $2$: Group 17 Electron Affinities
For example, nonmetals like the elements in the halogens series in Group 17 have a higher electron affinity than the metals. This trend is described as below. Notice the negative sign for the electron affinity which shows that energy is released.
• Fluorine (F) -328 kJ mol-1
• Chlorine (Cl) -349 kJ mol-1
• Bromine (Br) -324 kJ mol-1
• Iodine (I) -295 kJ mol-1
Notice that electron affinity decreases down the group, but increases up with the period.
As the name suggests, electron affinity is the ability of an atom to accept an electron. Unlike electronegativity, electron affinity is a quantitative measurement of the energy change that occurs when an electron is added to a neutral gas atom. The more negative the electron affinity value, the higher an atom's affinity for electrons.
Periodic Table showing Electron Affinity Trend
Nonmetals vs. Metals
To summarize the difference between the electron affinity of metals and nonmetals (Figure $1$):
• Metals: Metals like to lose valence electrons to form cations to have a fully stable octet. They absorb energy (endothermic) to lose electrons. The electron affinity of metals is lower than that of nonmetals.
• Nonmetals: Nonmetals like to gain electrons to form anions to have a fully stable octet. They release energy (exothermic) to gain electrons to form an anion; thus, electron affinity of nonmetals is higher than that of metals.
Patterns in Electron Affinity
Electron affinity increases upward for the groups and from left to right across periods of a periodic table because the electrons added to energy levels become closer to the nucleus, thus a stronger attraction between the nucleus and its electrons. Remember that greater the distance, the less of an attraction; thus, less energy is released when an electron is added to the outside orbital. In addition, the more valence electrons an element has, the more likely it is to gain electrons to form a stable octet. The less valence electrons an atom has, the least likely it will gain electrons.
Electron affinity decreases down the groups and from right to left across the periods on the periodic table because the electrons are placed in a higher energy level far from the nucleus, thus a decrease from its pull. However, one might think that since the number of valence electrons increase going down the group, the element should be more stable and have higher electron affinity. One fails to account for the shielding affect. As one goes down the period, the shielding effect increases, thus repulsion occurs between the electrons. This is why the attraction between the electron and the nucleus decreases as one goes down the group in the periodic table.
As you go down the group, first electron affinities become less (in the sense that less energy is evolved when the negative ions are formed). Fluorine breaks that pattern, and will have to be accounted for separately. The electron affinity is a measure of the attraction between the incoming electron and the nucleus - the stronger the attraction, the more energy is released. The factors which affect this attraction are exactly the same as those relating to ionization energies - nuclear charge, distance and screening. The increased nuclear charge as you go down the group is offset by extra screening electrons. Each outer electron in effect feels a pull of 7+ from the center of the atom, irrespective of which element you are talking about.
Example $3$: Fluorine vs. Chlorine
A fluorine atom has an electronic structure of 1s22s22px22py22pz1. It has 9 protons in the nucleus.The incoming electron enters the 2-level, and is screened from the nucleus by the two 1s2 electrons. It therefore feels a net attraction from the nucleus of 7+ (9 protons less the 2 screening electrons).
In contrast, chlorine has the electronic structure 1s22s22p63s23px23py23pz1 with 17 protons in the nucleus. But again the incoming electron feels a net attraction from the nucleus of 7+ (17 protons less the 10 screening electrons in the first and second levels). There is also a small amount of screening by the 2s electrons in fluorine and by the 3s electrons in chlorine. This will be approximately the same in both these cases and so does not affect the argument in any way (apart from complicating it!).
The over-riding factor is therefore the increased distance that the incoming electron finds itself from the nucleus as you go down the group. The greater the distance, the less the attraction and so the less energy is released as electron affinity.
Comparing fluorine and chlorine is not ideal, because fluorine breaks the trend in the group. However, comparing chlorine and bromine, say, makes things seem more difficult because of the more complicated electronic structures involved. What we have said so far is perfectly true and applies to the fluorine-chlorine case as much as to anything else in the group, but there's another factor which operates as well which we haven't considered yet - and that over-rides the effect of distance in the case of fluorine.
Why is Fluorine an Anomaly?
The incoming electron is going to be closer to the nucleus in fluorine than in any other of these elements, so you would expect a high value of electron affinity. However, because fluorine is such a small atom, you are putting the new electron into a region of space already crowded with electrons and there is a significant amount of repulsion. This repulsion lessens the attraction the incoming electron feels and so lessens the electron affinity. A similar reversal of the expected trend happens between oxygen and sulfur in Group 16. The first electron affinity of oxygen (-142 kJ mol-1) is smaller than that of sulfur (-200 kJ mol-1) for exactly the same reason that fluorine's is smaller than chlorine's.
Comparing Group 16 and Group 17 values
As you might have noticed, the first electron affinity of oxygen ($-142\; kJ\; mol^{-1}$) is less than that of fluorine ($-328\; kJ\; mol^{-1}$). Similarly sulfur's ($-200\; kJ\; mol^{-1}$) is less than chlorine's ($-349\; kJ\; mol^{-1}$). Why? It's simply that the Group 16 element has 1 less proton in the nucleus than its next door neighbor in Group 17. The amount of screening is the same in both. That means that the net pull from the nucleus is less in Group 16 than in Group 17, and so the electron affinities are less.
The reactivity of the elements in group 17 falls as you go down the group - fluorine is the most reactive and iodine the least. Often in their reactions these elements form their negative ions. The first impression that is sometimes given that the fall in reactivity is because the incoming electron is held less strongly as you go down the group and so the negative ion is less likely to form. That explanation looks reasonable until you include fluorine!
An overall reaction will be made up of lots of different steps all involving energy changes, and you cannot safely try to explain a trend in terms of just one of those steps. Fluorine is much more reactive than chlorine (despite the lower electron affinity) because the energy released in other steps in its reactions more than makes up for the lower amount of energy released as electron affinity.
Second Electron Affinity
You are only ever likely to meet this with respect to the group 16 elements oxygen and sulfur which both form -2 ions. The second electron affinity is the energy required to add an electron to each ion in 1 mole of gaseous 1- ions to produce 1 mole of gaseous 2- ions. This is more easily seen in symbol terms.
$X^- (g) + e^- \rightarrow X^{-2} (g) \label{3}$
It is the energy needed to carry out this change per mole of $X^-$.
Why is energy needed to do this? You are forcing an electron into an already negative ion. It's not going to go in willingly!
$O_{g} + e^- \rightarrow O^- (g) \;\;\; \text{1st EA = -142 kJ mol}^{-1} \label{4}$
$O^-_{g} + e^- \rightarrow O^{2-} (g) \;\;\; \text{2nd EA = +844 kJ mol}^{-1} \label{5}$
The positive sign shows that you have to put in energy to perform this change. The second electron affinity of oxygen is particularly high because the electron is being forced into a small, very electron-dense space.
Practice Problems
1. When an electron is added to a nonmetal atom, is energy released or absorbed?
2. Why do nonmetal atoms have a greater electron affinity than metal atoms?
3. Why are atoms with a low electron affinity more likely to lose electrons than gain electrons?
4. As you move down a group of the periodic table, does electron affinity increase or decrease, if so, why?
5. Why do nonmetals want to gain electrons?
6. Why do metals have a low electron affinity?
Answers
1. Energy is released when a electron is added to a nonmetal.
2. Nonmetals have a greater electron affinity than metals because their atomic structure allows them to gain electrons rather than lose them.
3. Atoms with a low electron affinity want to give up their valence electrons because they are further from the nucleus; as a result, they do not have a strong pull on the valence electrons.
4. As you move down a group on the periodic table, electron affinity decreases. First, the electrons are placed in energy levels further away from the nucleus, which results in electrons not having a strong attraction to the nucleus; secondly, the atom does not want gain electrons because there is minimal charge on the outer energy levels from the nucleus; and lastly, the shielding effect increases, causing repulsion between the electrons, thus they move further from each other and the nucleus itself.
5. Nonmetals want to gain electrons because they have more valence electrons than metals, so it is easier for them to gain electrons than lose the valance electrons to fulfill a stable octet. In addition, nonmetals' valance electrons are closer to the nucleus, thus allowing more attraction between the two.
6. Metals have a low electron affinity (a less likely chance to gain electrons) because they want to give up their valence electrons rather than gain electrons, which require more energy than necessary. In addition, they do not have a strong pull on the valance electrons because they are far away from the nucleus, thus they have less energy for an attraction.
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Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The Pauling scale is the most commonly used. Fluorine (the most electronegative element) is assigned a value of 4.0, and values range down to cesium and francium which are the least electronegative at 0.7.
What if two atoms of equal electronegativity bond together?
Consider a bond between two atoms, A and B. If the atoms are equally electronegative, both have the same tendency to attract the bonding pair of electrons, and so it will be found on average half way between the two atoms:
To get a bond like this, A and B would usually have to be the same atom. You will find this sort of bond in, for example, H2 or Cl2 molecules. Note: It's important to realize that this is an average picture. The electrons are actually in a molecular orbital, and are moving around all the time within that orbital. This sort of bond could be thought of as being a "pure" covalent bond - where the electrons are shared evenly between the two atoms.
What if B is slightly more electronegative than A?
B will attract the electron pair rather more than A does.
That means that the B end of the bond has more than its fair share of electron density and so becomes slightly negative. At the same time, the A end (rather short of electrons) becomes slightly positive. In the diagram, "$\delta$" (read as "delta") means "slightly" - so $\delta+$ means "slightly positive".
A polar bond is a covalent bond in which there is a separation of charge between one end and the other - in other words in which one end is slightly positive and the other slightly negative. Examples include most covalent bonds. The hydrogen-chlorine bond in HCl or the hydrogen-oxygen bonds in water are typical.
If B is a lot more electronegative than A, then the electron pair is dragged right over to B's end of the bond. To all intents and purposes, A has lost control of its electron, and B has complete control over both electrons. Ions have been formed. The bond is then an ionic bond rather than a covalent bond.
A "spectrum" of bonds
The implication of all this is that there is no clear-cut division between covalent and ionic bonds. In a pure covalent bond, the electrons are held on average exactly half way between the atoms. In a polar bond, the electrons have been dragged slightly towards one end. How far does this dragging have to go before the bond counts as ionic? There is no real answer to that. Sodium chloride is typically considered an ionic solid, but even here the sodium has not completely lost control of its electron. Because of the properties of sodium chloride, however, we tend to count it as if it were purely ionic. Lithium iodide, on the other hand, would be described as being "ionic with some covalent character". In this case, the pair of electrons has not moved entirely over to the iodine end of the bond. Lithium iodide, for example, dissolves in organic solvents like ethanol - not something which ionic substances normally do.
Summary
• No electronegativity difference between two atoms leads to a pure non-polar covalent bond.
• A small electronegativity difference leads to a polar covalent bond.
• A large electronegativity difference leads to an ionic bond.
Example 1: Polar Bonds vs. Polar Molecules
In a simple diatomic molecule like HCl, if the bond is polar, then the whole molecule is polar. What about more complicated molecules?
Consider CCl4, (left panel in figure above), which as a molecule is not polar - in the sense that it doesn't have an end (or a side) which is slightly negative and one which is slightly positive. The whole of the outside of the molecule is somewhat negative, but there is no overall separation of charge from top to bottom, or from left to right.
In contrast, CHCl3 is a polar molecule (right panel in figure above). The hydrogen at the top of the molecule is less electronegative than carbon and so is slightly positive. This means that the molecule now has a slightly positive "top" and a slightly negative "bottom", and so is overall a polar molecule.
A polar molecule will need to be "lop-sided" in some way.
Patterns of electronegativity in the Periodic Table
The distance of the electrons from the nucleus remains relatively constant in a periodic table row, but not in a periodic table column. The force between two charges is given by Coulomb’s law.
$F=k\dfrac{Q_1Q_2}{r^2}$
In this expression, Q represents a charge, k represents a constant and r is the distance between the charges. When r = 2, then r2= 4. When r = 3, then r2 = 9. When r = 4, then r2 = 16. It is readily seen from these numbers that, as the distance between the charges increases, the force decreases very rapidly. This is called a quadratic change.
The result of this change is that electronegativity increases from bottom to top in a column in the periodic table even though there are more protons in the elements at the bottom of the column. Elements at the top of a column have greater electronegativities than elements at the bottom of a given column.
The overall trend for electronegativity in the periodic table is diagonal from the lower left corner to the upper right corner. Since the electronegativity of some of the important elements cannot be determined by these trends (they lie in the wrong diagonal), we have to memorize the following order of electronegativity for some of these common elements.
F > O > Cl > N > Br > I > S > C > H > metals
The most electronegative element is fluorine. If you remember that fact, everything becomes easy, because electronegativity must always increase towards fluorine in the Periodic Table.
Note
This simplification ignores the noble gases. Historically this is because they were believed not to form bonds - and if they do not form bonds, they cannot have an electronegativity value. Even now that we know that some of them do form bonds, data sources still do not quote electronegativity values for them.
Trends in electronegativity across a period
The positively charged protons in the nucleus attract the negatively charged electrons. As the number of protons in the nucleus increases, the electronegativity or attraction will increase. Therefore electronegativity increases from left to right in a row in the periodic table. This effect only holds true for a row in the periodic table because the attraction between charges falls off rapidly with distance. The chart shows electronegativities from sodium to chlorine (ignoring argon since it does not does not form bonds).
Trends in electronegativity down a group
As you go down a group, electronegativity decreases. (If it increases up to fluorine, it must decrease as you go down.) The chart shows the patterns of electronegativity in Groups 1 and 7.
Explaining the patterns in electronegativity
The attraction that a bonding pair of electrons feels for a particular nucleus depends on:
• the number of protons in the nucleus;
• the distance from the nucleus;
• the amount of screening by inner electrons.
Why does electronegativity increase across a period?
Consider sodium at the beginning of period 3 and chlorine at the end (ignoring the noble gas, argon). Think of sodium chloride as if it were covalently bonded.
Both sodium and chlorine have their bonding electrons in the 3-level. The electron pair is screened from both nuclei by the 1s, 2s and 2p electrons, but the chlorine nucleus has 6 more protons in it. It is no wonder the electron pair gets dragged so far towards the chlorine that ions are formed. Electronegativity increases across a period because the number of charges on the nucleus increases. That attracts the bonding pair of electrons more strongly.
Why does electronegativity fall as you go down a group?
As you go down a group, electronegativity decreases because the bonding pair of electrons is increasingly distant from the attraction of the nucleus. Consider the hydrogen fluoride and hydrogen chloride molecules:
The bonding pair is shielded from the fluorine's nucleus only by the 1s2 electrons. In the chlorine case it is shielded by all the 1s22s22p6 electrons. In each case there is a net pull from the center of the fluorine or chlorine of +7. But fluorine has the bonding pair in the 2-level rather than the 3-level as it is in chlorine. If it is closer to the nucleus, the attraction is greater.
Diagonal relationships in the Periodic Table
At the beginning of periods 2 and 3 of the Periodic Table, there are several cases where an element at the top of one group has some similarities with an element in the next group. Three examples are shown in the diagram below. Notice that the similarities occur in elements which are diagonal to each other - not side-by-side.
For example, boron is a non-metal with some properties rather like silicon. Unlike the rest of Group 2, beryllium has some properties resembling aluminum. And lithium has some properties which differ from the other elements in Group 1, and in some ways resembles magnesium. There is said to be a diagonal relationship between these elements. There are several reasons for this, but each depends on the way atomic properties like electronegativity vary around the Periodic Table. So we will have a quick look at this with regard to electronegativity - which is probably the simplest to explain.
Explaining the diagonal relationship with regard to electronegativity
Electronegativity increases across the Periodic Table. So, for example, the electronegativities of beryllium and boron are:
Be 1.5
B 2.0
Electronegativity falls as you go down the Periodic Table. So, for example, the electronegativities of boron and aluminum are:
B 2.0
Al 1.5
So, comparing Be and Al, you find the values are (by chance) exactly the same. The increase from Group 2 to Group 3 is offset by the fall as you go down Group 3 from boron to aluminum. Something similar happens from lithium (1.0) to magnesium (1.2), and from boron (2.0) to silicon (1.8). In these cases, the electronegativities are not exactly the same, but are very close.
Similar electronegativities between the members of these diagonal pairs means that they are likely to form similar types of bonds, and that will affect their chemistry. You may well come across examples of this later on in your course.
Electronegativity
Allred-Rochow Electronegativity is a measure that determines the values of the electrostatic force exerted by the effective nuclear charge on the valence electrons. The value of the effective nuclear charges is estimated from Slater's rules. The higher charge, the more likely it will attract electrons. Although, Slater's rule are partly empirical. So the Allred-Rochow electronegativity is no more rigid than the Pauling Electronegativity.
Electronegativity
Pauling established Electronegativity as the "power" of an atom in a molecule to attract electron to itself. It is a measure of the atom's ability to attract electron to itself while the electron is still attached to another atom. The higher the values, the more likely that atom can pull electron from another atom and into itself. Electronegativity correlates with bond polarity, ionization energy, electron affinity, effective nuclear charge, and atomic size.
Table 1: Pauling Electronegativity Values
H
2.1
Li
1.0
Be
1.6
B
2.0
C
2.50
N
3.0
O
3.5
F
4.0
Na
0.9
Mg
1.3
Al
1.6
Si
1.9
P
2.2
S
2.5
Cl
3.0
K
0.8
Ca
1.3
Sc
1.4
Ti
1.5
V
1.6
Cr
1.7
Mn
1.6
Fe
1.8
Co
1.9
Ni
1.9
Cu
1.9
Zn
1.7
Ga
1.6
Ge
2.0
As
2.2
Se
2.6
Br
2.8
Rb
0.8
Sr
1.0
Y
1.2
Zr
1.3
Nb
1.6
Mo
2.2
Te
2.1
Ru
2.2
Rh
2.3
Pd
2.2
Ag
1.9
Cd
1.7
In
1.8
Sn
2.0
Sb
2.1
Te
2.1
I
2.7
Cs
0.8
Ba
0.9
La
1.1
Hf
1.3
Ta
1.5
W
1.7
Re
1.9
Os
2.2
Ir
2.2
Pt
2.2
Au
2.4
Hg
1.9
Tl
2.0
Pb
2.3
Bi
2.0
Po
2.0
At
2.2
The periodic trend for electronegativity generally increases from left to right and decreases as it go down the group. The exception are Hydrogen and the noble gases because the noble gases are content with their filled outermost shells, and hydrogen cannot bear to lose a valence electron unlike the rest of the group 1 metals. The elements in the halogen group usually have the highest electronegativity values because they only need to attract one valence electron to complete the octet in their outer shell. Whereas the group 1 elements except for Hydrogen, are willing to give up their only valence electron so they can fulfill having a complete, filled outer shell.
Slater's rules
Slater's rules are rules that provides the values for the effective nuclear charge concept, or $Z_{eff}$. These rules are based on experimental data for electron promotion and ionization energies, and $Z_{eff}$ is determined from this equation:
$Z_{eff} = Z - S \label{A}$
where
• $Z$ is the nuclear charge,
• $Z_{eff}$ is the effective nuclear charge, and
• $S$ is the shielding constant
Through this equation, this tells us that electron may get reduced nuclear charge due to high shielding. Allred and Rochow used $Z_{eff}$ because it is accurate due to the involvement of shielding that prevents electron to reach its true nuclear charge: $Z$. When an atom with filled s-shell attracts electrons, those electrons will go to the unfilled p-orbital. Since the electrons have the same negative charge, they will not only repel each other, but also repel the electrons from the filled s-shell. This creates a shielding effect where the inner core electrons will shield the outer core electrons from the nucleus. Not only would the outer core electrons experience effective nuclear charge, but it will make them easily removed from the outer shell. Thus, It is easier for outer electrons to penetrate the p shell, which has little likelihood of being near the nuclear, rather than the s shell. Consider this, each of the outer electron in the (ns, np) group contributes S = 0.35, S = 0.85 in the (n - 1) shell, and S = 1.00 in the (n - 2) or lower shells.
Example 1: Slater's Rules
What is the $Z_{eff}$ for the 4s electrons in Ca.
Solution
Since $\ce{Ca}$ has atomic number of 20, $Z = 20$.
Then, we find the electron configuration for Ca, which is 1s22s22p63s23p64s2.
Now we got that, we can use Slater's rules:
\begin{align*} Z_{eff} &= Z - S \[4pt] &= 20 - ((8\times 0.85) + (10 \times 1.00)) \[4pt] &= 3.2 \end{align*}
So, Ca has a $Z_{eff}$ of 3.2.
Allred-Rochow Electronegativity
Allred and Rochow were two chemists who came up with the Allred-Rochow Electronegativity values by taking the electrostatic force exerted by effective nuclear charge, Zeff, on the valence electron. To do so, they came up with an equation:
$\chi^{AR} = \left(\dfrac{3590 \times Z_{eff}}{r^2_{cov}}\right) + 0.744 \label{1}$
At the time, the values for the covalent radius, $r_{cov}$, were inaccurate. Allred and Rochow added certain perimeters so that it would more closely correspond to Pauling's electronegativity scale.
Table 2: Allred-Rochow Electronegativity Values
H
2.20
Li
0.97
Be
1.47
B
2.01
C
2.50
N
3.07
O
3.50
F
4.10
Na
1.01
Mg
1.23
Al
1.47
Si
1.74
P
2.06
S
2.44
Cl
2.83
K
0.91
Ca
1.04
Sc
1.20
Ti
1.32
V
1.45
Cr
1.56
Mn
1.60
Fe
1.64
Co
1.70
Ni
1.75
Cu
1.75
Zn
1.66
Ga
1.82
Ge
2.02
As
2.20
Se
2.48
Br
2.74
Rb
0.89
Sr
0.99
Y
1.11
Zr
1.22
Nb
1.23
Mo
1.30
Te
1.36
Ru
1.42
Rh
1.45
Pd
1.35
Ag
1.42
Cd
1.46
In
1.49
Sn
1.72
Sb
1.82
Te
2.01
I
2.21
Cs
0.86
Ba
0.97
La
1.08
Hf
1.23
Ta
1.33
W
1.40
Re
1.46
Os
1.52
Ir
1.55
Pt
1.44
Au
1.42
Hg
1.44
Tl
1.44
Pb
1.55
Bi
1.67
Po
1.76
At
1.90
In this table, the electronegativities increases from left to right just like Pauling's scale because the $Z$ is increasing. As we go down the group, it decreases because of the larger atomic size that increases the distance between the electrons and nucleus.
Problems
1. From lowest to highest, order the elements in terms of Zeff: Ni, Cu, Zn, Ga, Ge
2. Using the equations above, find the Zeff for the Br by using Slater's rules.
3. Using the equations above, Find the Xar for Br.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Electronegativity/Allred-Rochow_Electronegativity.txt
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A method for estimating electronegativity was developed by Robert Mulliken (1896–1986; Nobel Prize in Chemistry 1966) who noticed that elements with large first ionization energies tend to have very negative electron affinities and gain electrons in chemical reactions. Conversely, elements with small first ionization energies tend to have slightly negative (or even positive) electron affinities and lose electrons in chemical reactions. Mulliken recognized that an atom’s tendency to gain or lose electrons could therefore be described quantitatively by the average of the values of its first ionization energy and the absolute value of its electron affinity.
Robert S. Mulliken proposed that the arithmetic mean of the first ionization energy ($E_{I_1}$) and the electron affinity ($E_{ea}$) should be a measure of the tendency of an atom to attract electrons. As this definition is not dependent on an arbitrary relative scale, it has also been termed absolute electronegativity. Using our definition of electron affinity, we can write Mulliken’s original expression for electronegativity as follows:Mulliken’s definition used the magnitude of the ionization energy and the electron affinity. By definition, the magnitude of a quantity is a positive number. Our definition of electron affinity produces negative values for the electron affinity for most elements, so vertical lines indicating absolute value are needed in Equation $\ref{1}$ to make sure that we are adding two positive numbers in the numerator.
$\chi = \dfrac{|E_{I_1} + E_{ea}|}{2} \label{1}$
Elements with a large first ionization energy and a very negative electron affinity have a large positive value in the numerator of Equation $\ref{1}$, so their electronegativity is high. Elements with a small first ionization energy and a small electron affinity have a small positive value for the numerator in Equation $\ref{1}$, so they have a low electronegativity. Inserting the appropriate data into Equation $\ref{1}$ gives a Mulliken electronegativity value for fluorine of 1004.6 kJ/mol. To compare Mulliken’s electronegativity values with those obtained by Pauling, Mulliken’s values are divided by 252.4 kJ/mol, which gives Pauling’s value (3.98).
However, it is more usual to use a linear transformation to transform these absolute values into values that resemble the more familiar Pauling values. For ionization energies and electron affinities in electronvolts:
$\chi_{Mulliken} = 0.187 (E_{I_1}+E_{ea})+0.17 \label{2}$
and for energies in kJ/mol,
$\chi_{Mulliken} = (1.97 \times 10^{-3})(E_{I_1}+E_{ea})+0.19 \label{3}$
The Mulliken electronegativity can only be calculated for an element for which the electron affinity is known, fifty-seven elements as of 2006. The Mulliken electronegativity of an atom is sometimes said to be the negative of the chemical potential. By inserting the energetic definitions of the ionization potential and electron affinity into the Mulliken electronegativity, it is possible to show that the Mulliken chemical potential is a finite difference approximation of the electronic energy with respect to the number of electrons., i.e.,
$\mu_{Mulliken}= -\chi_{Mulliken} = -\dfrac{E_{I_1} + E_{ea}}{2} \label{4}$
All electronegativity scales give essentially the same results for one element relative to another. Even though the Mulliken scale is based on the properties of individual atoms and the Pauling scale is based on the properties of atoms in molecules, they both apparently measure the same basic property of an element. In the following discussion, we will focus on the relationship between electronegativity and the tendency of atoms to form positive or negative ions. We will therefore be implicitly using the Mulliken definition of electronegativity. Because of the parallels between the Mulliken and Pauling definitions, however, the conclusions are likely to apply to atoms in molecules as well.
Significance
Despite being developed from a very different set of principles than Pauling Electronegativity, which is based on bond dissociation energies, there is a good correlation between Mullikin and Pauling Electronegativities for the atoms, as shown in the plot below.
Although Pauling electronegativities are usually what are found in textbooks, the Mullikin electronegativity more intuitively corresponds to the "ability of an atom to draw electrons toward itself in bonding," and is probably a better indicator of that property. However, because of the good correlation between the two scales, using the Pauling scale is sufficient for most purposes.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Electronegativity/Mulliken_Electronegativity.txt
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Linus Pauling described electronegativity as “the power of an atom in a molecule to attract electrons to itself.”1 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 electronegativity 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.
There are a few different 'types' of electronegativity which differ only in their definitions and the system by which they assign values for electronegativity. For example, Mulliken electronegativity defines electronegativity as the "the average of the ionization energy and electron affinity of an atom."3 As we will see, this definition differs slightly from Pauling's definition of electronegativity.
Pauling Electronegativity
Linus Pauling was the original scientist to describe the phenomena of electronegativity. The best way to describe his method is to look at a hypothetical molecule that we will call XY. By comparing the measured X-Y bond energy with the theoretical X-Y bond energy (computed as the average of the X-X bond energy and the Y-Y bond energy), we can describe the relative affinities of these two atoms with respect to each other.
$Δ\text{Bond Energies} = (X-Y)_{measured} – (X-Y)_{expected} \nonumber$
If the electonegativities of X and Y are the same, then we would expect the measured bond energy to equal the theoretical (expected) bond energy and therefore the Δ bond energies would be zero. If the electronegativities of these atoms are not the same, we would see a polar molecule where one atom would start to pull electron density toward itself, causing it to become partially negative.
By doing some careful experiments and calculations, Pauling came up with a slightly more sophisticated equation for the relative electronegativities of two atoms in a molecule:
$EN(X) - EN(Y) = 0.102 \sqrt{Δ}.\nonumber$
In that equation, the factor 0.102 is simply a conversion factor between kJ and eV to keep the units consistent with bond energies.
By assigning a value of 4.0 to Fluorine (the most electronegative element), Pauling was able to set up relative values for all of the elements. This was when he first noticed the trend that the electronegativity of an atom was determined by it's position on the periodic table and that the electronegativity tended to increase as you moved left to right and bottom to top along the table. The range of values for Pauling's scale of electronegativity ranges from Fluorine (most electronegative = 4.0) to Francium (least electronegative = 0.7). 2 Furthermore, if the electronegativity difference between two atoms is very large, then the bond type tends to be more ionic, however if the difference in electronegativity is small then it is a nonpolar covalent bond.
Exercise $1$
Explain the difference between Electronegativity and Electron Affinity
Exercise $2$
Predict the order or increasing electronegativity from the following elements
1. F, Li, C, O
2. Te, Cl, S, Se
3. Cs, At, Tl, I
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Electronegativity/Pauling_Electronegativity.txt
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Key Points
• Not all atoms within a neutral molecule have to be neutral.
• Atoms within a molecule with formal charges are often the site of chemical reactions.
• You should be able to determine the formal charge for any atom in any molecule.
Knowing the formal charges on specific atoms in a molecule is an important step in keeping tract of the electrons and determine the chemical reactivity of the molecule. Formal charges can be calculated mathematically, but they can also be determined by intuition. The instinctive method is faster but requires more skill and knowledge of common structures.
Mathematical Method
Formal charge equation is based on the comparing the number of electrons in the individual atom with that in the structure. For each atom, we then compute a formal charge:
$\text{formal charge}= \underbrace{valence\; e^-}_{free\; atom } - \underbrace{ \left ( nonbonding\; e^{-}+\dfrac{bonding\;e^{-}}{2} \right ) }_{atom\; in\; Lewis\; structure } \label{8.5.1}$
where the group number is equal to the number of electrons of the neutral atom.
Intuitive Method
The instinctive method relies on experience and the understanding of common, known neutral structures. To do this you need to know the common bonding patterns for C, N, O, and the halides:
• C: 4 bonds;
• N: 3 bonds, 1 lone pair;
• O: 2 bonds, 2 lone pairs;
• F: 1 bond, 3 lone pairs (or any other halide).
Once mastered, this is much easier and quicker. The diagram below shows the common bonding situations for C, N, O, and F. The central column is the neutral atom, in the left most column the atom has gained an electron by having a bonding pair converted to a lone pair of electrons. The right most column has a formal charge of +1 due to either: a) using a lone pair of electrons as a bonding pair (N and O), b) losing a bonding pair of electrons (C), or c) losing a lone pair of electrons (F).
Formal Charges
A formal charge (FC) is the charge assigned to an atom in a molecule, assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity. When determining the best Lewis structure (or predominant resonance structure) for a molecule, the structure is chosen such that the formal charge on each of the atoms is as close to zero as possible.
The formal charge of any atom in a molecule can be calculated by the following equation:
${\displaystyle FC=V-N-{\frac {B}{2}}\ }$
where V is the number of valence electrons of the neutral atom in isolation (in its ground state); N is the number of non-bonding valence electrons on this atom in the molecule; and B is the total number of electrons shared in bonds with other atoms in the molecule.
Examples
• Example: CO2 is a neutral molecule with 16 total valence electrons. There are three different ways to draw the Lewis structure
• Carbon single bonded to both oxygen atoms (carbon = +2, oxygens = −1 each, total formal charge = 0)
• Carbon single bonded to one oxygen and double bonded to another (carbon = +1, oxygendouble = 0, oxygensingle = −1, total formal charge = 0)
• Carbon double bonded to both oxygen atoms (carbon = 0, oxygens = 0, total formal charge = 0)
Even though all three structures gave us a total charge of zero, the final structure is the superior one because there are no charges in the molecule at all.
Alternative Method
The following is equivalent:
• Draw a circle around the atom for which the formal charge is requested (as with carbon dioxide, below)
• Count up the number of electrons in the atom's "circle." Since the circle cuts the covalent bond "in half," each covalent bond counts as one electron instead of two.
• Subtract the number of electrons in the circle from the group number of the element (the Roman numeral from the older system of group numbering, NOT the IUPAC 1-18 system) to determine the formal charge.
• The formal charges computed for the remaining atoms in this Lewis structure of carbon dioxide are shown below.
It is important to keep in mind that formal charges are just that – formal, in the sense that this system is a formalism. The formal charge system is just a method to keep track of all of the valence electrons that each atom brings with it when the molecule is formed.
Formal charge compared to oxidation state
Formal charge is a tool for estimating the distribution of electric charge within a molecule. The concept of oxidation states constitutes a competing method to assess the distribution of electrons in molecules. If the formal charges and oxidation states of the atoms in carbon dioxide are compared, the following values are arrived at:
The reason for the difference between these values is that formal charges and oxidation states represent fundamentally different ways of looking at the distribution of electrons amongst the atoms in the molecule. With formal charge, the electrons in each covalent bond are assumed to be split exactly evenly between the two atoms in the bond (hence the dividing by two in the method described above). The formal charge view of the CO2 molecule is essentially shown below:
The covalent (sharing) aspect of the bonding is overemphasized in the use of formal charges, since in reality there is a higher electron density around the oxygen atoms due to their higher electronegativity compared to the carbon atom. This can be most effectively visualized in an electrostatic potential map.
With the oxidation state formalism, the electrons in the bonds are "awarded" to the atom with the greater electronegativity. The oxidation state view of the CO2 molecule is shown below:
Oxidation states overemphasize the ionic nature of the bonding; the difference in electronegativity between carbon and oxygen is insufficient to regard the bonds as being ionic in nature.
In reality, the distribution of electrons in the molecule lies somewhere between these two extremes. The inadequacy of the simple Lewis structure view of molecules led to the development of the more generally applicable and accurate valence bond theory of Slater, Pauling, et al., and henceforth the molecular orbital theory developed by Mulliken and Hund.
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Intermolecular forces are the attractive or repulsive forces between molecules. They are separated into two groups; short range and long range forces. Short range forces happen when the centers of the molecules are separated by three angstroms (10-8 cm) or less. Short range forces tend to be repulsive, where the long range forces that act outside the three angstroms range are attractive. Long range forces are also known as Van der Waals forces. They are responsible for surface tension, friction, viscosity and differences between actual behavior of gases and that predicted by the ideal gas law. Intermolecular forces are responsible for most properties of all the phases. The viscosity, diffusion, and surface tension are examples of physical properties of liquids that depend on intermolecular forces. Vapor pressure, critical point, and boiling point are examples of properties of gases. Melting and sublimation are examples of properties of solids that depend on intermolecular forces.
• Hydrogen Bonding
A hydrogen bond is a special type of dipole-dipole attraction which occurs when a hydrogen atom bonded to a strongly electronegative atom exists in the vicinity of another electronegative atom with a lone pair of electrons. These bonds are generally stronger than ordinary dipole-dipole and dispersion forces, but weaker than true covalent and ionic bonds.
• Hydrophobic Interactions
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.
• Multipole Expansion
A multipole expansion is a series expansion of the effect produced by a given system in terms of an expansion parameter which becomes small as the distance away from the system increases. Therefore, the leading one or terms in a multipole expansion are generally the strongest. The first-order behavior of the system at large distances can therefore be obtained from the first terms of this series, which is generally much easier to compute than the general solution.
• Overview of Intermolecular Forces
Intermolecular forces are forces between molecules. Depending on its strength, intermolecular forces cause the forming of three physical states: solid, liquid and gas. The physical properties of melting point, boiling point, vapor pressure, evaporation, viscosity, surface tension, and solubility are related to the strength of attractive forces between molecules. These attractive forces are called Intermolecular Forces.
• Specific Interactions
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, which keep a molecule together (e.g., covalent and ionic bonding).
• Van der Waals Forces
Van der Waals forces' is a general term used to define the attraction of intermolecular forces between molecules. There are two kinds of Van der Waals forces: weak London Dispersion Forces and stronger dipole-dipole forces.
Intermolecular Forces
A hydrogen bond is a special type of dipole-dipole attraction which occurs when a hydrogen atom bonded to a strongly electronegative atom exists in the vicinity of another electronegative atom with a lone pair of electrons. These bonds are generally stronger than ordinary dipole-dipole and dispersion forces, but weaker than true covalent and ionic bonds.
Introduction
For a hydrogen bond to occur there must be both a hydrogen donor and an acceptor present. The donor in a hydrogen bond is the atom to which the hydrogen atom participating in the hydrogen bond is covalently bonded, and is usually a strongly electronegative atom such as N, O, or F. The hydrogen acceptor is the neighboring electronegative ion or molecule, and must posses a lone electron pair in order to form a hydrogen bond.
Since the hydrogen donor is strongly electronegative, it pulls the covalently bonded electron pair closer to its nucleus, and away from the hydrogen atom. The hydrogen atom is then left with a partial positive charge, creating a dipole-dipole attraction between the hydrogen atom bonded to the donor, and the lone electron pair on the acceptor.
Types of hydrogen bonds
Hydrogen bonds can occur within one single molecule, between two like molecules, or between two unlike molecules.
• Intramolecular hydrogen bonds: Intramolecular hydrogen bonds are those which occur within one single molecule. This occurs when two functional groups of a molecule can form hydrogen bonds with each other. In order for this to happen, both a hydrogen donor an acceptor must be present within one molecule, and they must be within close proximity of each other in the molecule. For example, intramolecular hydrogen bonding occurs in ethylene glycol (\(C_2H_4(OH)_2\)) between its two hydroxyl groups due to the molecular geometry.
• Intermolecular hydrogen bonds: Intermolecular hydrogen bonds occur between separate molecules in a substance. They can occur between any number of like or unlike molecules as long as hydrogen donors and acceptors are present an in positions in which they can interact. For example, intermolecular hydrogen bonds can occur between \(NH_3\) molecules alone, between \(H_2O\) molecules alone, or between \(NH_3\) and \(H_2O\) molecules.
Properties and effects of hydrogen bonds
When we consider the boiling points of molecules, we usually expect molecules with larger molar masses to have higher normal boiling points than molecules with smaller molar masses. This, without taking hydrogen bonds into account, is due to greater dispersion forces. Larger molecules have more space for electron distribution and thus more possibilities for an instantaneous dipole moment. However, when we consider the table below, we see that this is not always the case.
We see that \(H_2O\), \(HF\), and \(NH_3\) each have higher boiling points than the same compound formed between hydrogen and the next element moving down its respective group, indicating that the former have greater intermolecular forces. This is because \(H_2O\), \(HF\), and \(NH_3\) all exhibit hydrogen bonding, whereas the others do not. Furthermore, \(H_2O\) has a smaller molar mass than \(HF\), but partakes in more hydrogen bonds per molecule, so its boiling point is consequently higher.
Viscosity
The same effect that is seen on boiling point as a result of hydrogen bonding can also be observed in the viscosity of certain substances. Those substances which are capable of forming hydrogen bonds tend to have a higher viscosity than those that do not. Substances which have the possibility for multiple hydrogen bonds exhibit even higher viscosities.
Factors preventing Hydrogen bonding
• Electronegativity: Hydrogen bonding cannot occur without significant electronegativity differences between hydrogen and the atom it is bonded to. Thus, we see molecules such as \(PH_3\), which no not partake in hydrogen bonding. \(PH_3\) exhibits a trigonal pyramidal molecular geometry like that of ammmonia, but unlike \(NH_3\) it cannot hydrogen bond. This is due to the similarity in the electronegativities of phosphorous and hydrogen. Both atoms have an electronegativity of 2.1, and thus, no dipole moment occurs. This prevents the hydrogen bonding from acquiring the partial positive charge needed to hydrogen bond with the lone electron pair in another molecule (see Polarizability)
• Atom Size: The size of donors and acceptors can also effect the ability to hydrogen bond. This can account for the relatively low ability of Cl to form hydrogen bonds. When the radii of two atoms differ greatly or are large, their nuclei cannot achieve close proximity when they interact, resulting in a weak interaction.
Hydrogen Bonding in Nature
Hydrogen bonding plays a crucial role in many biological processes and can account for many natural phenomena such as the unusual properties of water. In addition to being present in water, hydrogen bonding is also important in the water transport system of plants, secondary and tertiary protein structure, and DNA base pairing.
Plants
The cohesion-adhesion theory of transport in vascular plants uses hydrogen bonding to explain many key components of water movement through the plant's xylem and other vessels. Within a vessel, water molecules hydrogen bond not only to each other, but also to the cellulose chain which comprises the wall of plant cells. This creates a sort of capillary tube which allows for capillary action to occur since the vessel is relatively small. This mechanism allows plants to pull water up into their roots. Furthermore,hydrogen bonding can create a long chain of water molecules which can overcome the force of gravity and travel up to the high altitudes of leaves.
Proteins
Hydrogen bonding is present abundantly in the secondary structure of proteins, and also sparingly in tertiary conformation. The secondary structure of a protein involves interactions (mainly hydrogen bonds) between neighboring polypeptide backbones which contain Nitrogen-Hydrogen bonded pairs and oxygen atoms. Since both N and O are strongly electronegative, the hydrogen atoms bonded to nitrogen in one polypeptide backbone can hydrogen bond to the oxygen atoms in another chain and visa-versa. Though they are relatively weak,these bonds offer great stability to secondary protein structure because they repeat a great number of times.
In tertiary protein structure, interactions are primarily between functional R groups of a polypeptide chain; one such interaction is called a hydrophobic interaction. These interactions occur because of hydrogen bonding between water molecules around the hydrophobe and further reinforce conformation.
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/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Hydrogen_Bonding/Hydrogen_Bondi.txt
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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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A multipole expansion is a mathematical series representing a function that depends on angles—usually the two angles on a sphere. These series are useful because they can often be truncated, meaning that only the first few terms need to be retained for a good approximation to the original function. Multipole expansions are very frequently used in the study of electromagnetic and gravitational fields, where the fields at distant points are given in terms of sources in a small region. The multipole expansion with angles is often combined with an expansion in radius. Such a combination gives an expansion describing a function throughout three-dimensional space.
The multipole expansion is expressed as a sum of terms with progressively finer angular features. For example, the initial term—called the zeroth, or monopole, moment—is a constant, independent of angle. The following term—the first, or dipole, moment—varies once from positive to negative around the sphere. Higher-order terms (like the quadrupole and octupole) vary more quickly with angles. A multipole moment usually involves powers (or inverse powers) of the distance to the origin, as well as some angular dependence.
Setting up the System
Consider an arbitrary charge distribution $\rho (\mathbf {r} ')$. We wish to find the electrostatic potential due to this charge distribution at a given point $\mathbf {r}$. We assume that this point is at a large distance from the charge distribution, that is if $\mathbf {r} '$ varies over the charge distribution, then $\mathbf {r} \gg \mathbf {r} '$.
Now, the Coulomb potential for an arbitrary charge distribution is given by
$V(\mathbf {r} ) ={\dfrac {1}{4\pi \epsilon _{0}}}\int _{V'}{\dfrac {\rho (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r'} |}}dV' \label{eq2}$
Here,
\begin{align} | \mathbf {r} -\mathbf {r'} |&= \sqrt{ |r^{2}-2\mathbf {r} \cdot \mathbf {r} '+r'^{2}|} \[5pt] &=r \sqrt{ \left|1-2 \dfrac {\hat {\mathbf {r}} \cdot \mathbf {r} '}{r} +\left( \dfrac {r'}{r} \right)^2 \right|} \end{align}
where $\hat {\mathbf {r} } = \mathbf {r} /r$
Thus, using the fact that ${ \mathbf {r} }$ is much larger than $\mathbf {r} '$, we can write
$\dfrac {1}{|\mathbf {r} -\mathbf {r'} |} = \dfrac {1}{r} \dfrac {1}{ \sqrt{\left|1-2 \dfrac {\hat {\mathbf {r} }\cdot \mathbf {r} '}{r} +\left(\dfrac {r'}{r}\right)^2 \right|}} \label{eq6}$
and using the binomial expansion (see below),
$\dfrac {1} {\sqrt{ \left|1-2 \dfrac {\hat {\mathbf {r}} \cdot \mathbf {r} '}{r} + \left( \dfrac {r'}{r} \right)^{2} \right|} } =1-{\dfrac {\hat {\mathbf {r} }\cdot \mathbf {r} '}{r}} + {\dfrac {1}{2r^{2}}}\left(r'^{2}-3({\hat {\mathbf {r} }}\cdot \mathbf {r} '\right)^{2})+O\left({\dfrac {r'}{r}}\right)^{3} \label{eq10}$
(we neglect the third and higher order terms).
Binomial Theorem
The binomial theory can be used to expand specific functions into an infinite series:
\begin{align} (1+ x )^s &= \sum_{n=0}^{\infty} \dfrac{s!}{n! (s-n)!} x^n \[5pt] &= 1 + \dfrac{s}{1!} x + \dfrac{s(s-1)}{2!} x^2 + \dfrac{s(s-1)(s-2)}{3!} x^3 + \ldots \end{align}
Equation \ref{eq6} can be rewritten as
$\dfrac {1}{|\mathbf {r} -\mathbf {r'} |} = \dfrac {1}{r} \dfrac {1}{ \sqrt{1 + \epsilon } } \label{eq6B}$
where
$\epsilon = -2 \dfrac {\hat {\mathbf {r} }\cdot \mathbf {r} '}{r} +\left(\dfrac {r'}{r}\right)^2 \label{eq6c}$
Applying the Binomial Theorem to Equation \ref{eq6B} (with $s = -1/2$) results in
$\dfrac {1}{|\mathbf {r} -\mathbf {r'} |} = \dfrac {1}{r} \left( 1 - \dfrac{1}{2} \epsilon + \dfrac{3}{8} \epsilon^2 - \dfrac{5}{16} \epsilon^3 + \ldots \right) \label{eq6d}$
Equation \ref{eq10} originates from substituting Equation \ref{eq6c} into Equation \ref{eq6d}.
The Expansion
Inserting Equation \ref{eq10} into Equation \ref{eq2} shows that the potential can be written as
$V(\mathbf {r} )={\dfrac {1}{4\pi \epsilon _{0}r}}\int _{V'}\rho (\mathbf {r} ')\left(1-{\dfrac {\hat {\mathbf {r} }\cdot \mathbf {r} '}{r}}+{\dfrac {1}{2r^{2}}}\left(3({\hat {\mathbf {r} }}\cdot \mathbf {r} '\right)^{2}-r'^{2})+O\left({\dfrac {r'}{r}}\right)^{3}\right)dV'$
We write this as
$V(\mathbf {r} )=V_{\text{mon}}(\mathbf {r} )+V_{\text{dip}}(\mathbf {r} )+V_{\text{quad}}(\mathbf {r} )+\ldots \label{expand}$
The first (the zeroth-order) term in the expansion is called the monopole moment, the second (the first-order) term is called the dipole moment, the third (the second-order) is called the quadrupole moment, the fourth (third-order) term is called the octupole moment, and the fifth (fourth-order) term is called the hexadecapole moment. Given the limitation of Greek numeral prefixes, terms of higher order are conventionally named by adding "-pole" to the number of poles—e.g., 32-pole (i.e., dotriacontapole) and 64-pole (hexacontatetrapole).
These moments can be expanded thusly
\begin{align} V_{\text{mon}}(\mathbf {r} ) &={\dfrac {1}{4\pi \epsilon _{0}r}}\int _{V'}\rho (\mathbf {r} ')dV' \[5pt] V_{\text{dip}}(\mathbf {r} ) &=-{\dfrac {1}{4\pi \epsilon _{0}r^{2}}}\int _{V'}\rho (\mathbf {r} ')\left({\hat {\mathbf {r} }}\cdot \mathbf {r} '\right)dV' \[5pt] V_{\text{quad}}(\mathbf {r} ) &={\dfrac {1}{8\pi \epsilon _{0}r^{3}}}\int _{V'}\rho (\mathbf {r} ')\left(3\left({\hat {\mathbf {r} }}\cdot \mathbf {r} '\right)^{2}-r'^{2}\right)dV' \end{align}
and so on.
In principle, a multipole expansion provides an exact description of the potential and generally converges under two conditions:
1. if the sources (e.g. charges) are localized close to the origin and the point at which the potential is observed is far from the origin; or
2. the reverse, i.e., if the sources are located far from the origin and the potential is observed close to the origin.
In the first (more common) case, the coefficients of the series expansion are called exterior multipole moments or simply multipole moments whereas, in the second case, they are called interior multipole moments.
The Monopole Scalar
Observe that
$V_{mon}(\mathbf {r}) =\dfrac {1}{4\pi \epsilon _0r}\int _{V'}\rho (\mathbf {r} ')dV' = \dfrac{q}{ 4\pi \epsilon _0 r}$
is a scalar, (actually the total charge in the distribution) and is called the electric monopole. This term indicates point charge electrical potential with charge $q$.
The Dipole Vector
If a charge distribution has a net total charge, it will tend to look like a monopole (point charge) from large distances. We can write
$V_{\text{dip}}(\mathbf {r} )=-{\dfrac {\hat {\mathbf {r} }}{4\pi \epsilon _{0}r^{2}}}\cdot \int _{V'}\rho (\mathbf {r} ')\mathbf {r} 'dV'$
The vector
$\mathbf {p} =\int _{V'}\rho (\mathbf {r} ')\mathbf {r} 'dV'$
is called the electric dipole. And its magnitude is called the dipole moment of the charge distribution. This terms indicates the linear charge distribution geometry of a dipole electrical potential.
The Quadrupole Tensor
Let $\hat {\mathbf {r} }$ and $\mathbf {r} '$ be expressed in Cartesian coordinates as $(r_{1},r_{2},r_{3})$ and $(x_{1},x_{2},x_{3})$. Then, $({\hat {\mathbf {r} }}\cdot \mathbf {r} ')^{2}=(r_{i}x_{i})^{2}=r_{i}r_{j}x_{i}x_{j}$.
We define a dyad to be the tensor ${\hat {\mathbf {r} }}{\hat {\mathbf {r} }}$ given by
$\left({\hat {\mathbf {r} }}{\hat {\mathbf {r} }}\right)_{ij}=r_{i}r_{j}$
Define the Quadrupole tensor as
$T=\int _{V'} \rho (\mathbf {r} ') \left(3 (\mathbf {r} '\mathbf {r} ')-\mathbf {I} r'^{2}\right)dV'$
Then, we can write $V_{\text{qua}}$ as the tensor contraction
$V_{\text{qua}}(\mathbf {r} )=- \dfrac {\hat {\mathbf {r}} \hat {\mathbf {r}}} {4\pi \epsilon _{0}r^{3}} ::T$
this term indicates the three dimensional distribution of a quadruple electrical potential.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Multipole_Expansion.txt
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Intermolecular forces are forces between molecules. Depending on its strength, intermolecular forces cause the forming of three physical states: solid, liquid and gas. The physical properties of melting point, boiling point, vapor pressure, evaporation, viscosity, surface tension, and solubility are related to the strength of attractive forces between molecules. These attractive forces are called Intermolecular Forces. The amplitude of interaction is important in the interpretation of the various properties listed above.
Introduction
There are four types of intermolecular forces. Most of the intermolecular forces are identical to bonding between atoms in a single molecule. Intermolecular forces just extend the thinking to forces between molecules and follows the patterns already set by the bonding within molecules.
Coulombic Forces
The forces holding ions together in ionic solids are electrostatic forces. Opposite charges attract each other. These are the strongest intermolecular forces. Ionic forces hold many ions in a crystal lattice structure. According to Coulomb’s law:
$V=-\dfrac{q_1 q_2}{4 \pi \epsilon r} \tag{1}$
• $q$ is the charges.
• $r$ is the distance of separation.
Based on Coulomb’s law, we can find the potential energy between different types of molecules.
Dipole Forces
Polar covalent molecules are sometimes described as "dipoles", meaning that the molecule has two "poles". One end (pole) of the molecule has a partial positive charge while the other end has a partial negative charge. The molecules will orientate themselves so that the opposite charges attract principle operates effectively.
In the figure below, hydrochloric acid is a polar molecule with the partial positive charge on the hydrogen and the partial negative charge on the chlorine. A network of partial + and - charges attract molecules to each other.
$V=-\dfrac{3}{2}\dfrac{I_A I_B}{I_A+I_B}\dfrac{\alpha_A \alpha_B}{r^6} \tag{3}$
Dipole-Dipole Interactions
When a polar molecule encounters another polar molecule, the positive end of one molecule is attracted to the negative end of the other polar molecule. Many molecules have dipoles, and their interaction occur by dipole-dipole interaction. For example: SO2 ↔SO2. (approximate energy: 15 kJ/mol). Polar molecules have permanent dipole moments, so in this case, we consider the electrostatic interaction between the two dipoles:
$V=-\dfrac{2}{3}\dfrac{\mu_1^2 \mu_2^2}{(4\pi\epsilon_0)^2 r^6} \tag{4}$
µ is the permanent dipole moment of the molecule 1 and 2.
Ion-Dipole Interactions
Ion-Dipole interaction is the interaction between an ion and polar molecules. For example, the sodium ion/water cluster interaction is approximately 50 KJ/mol.
$Na^+ ↔ (OH_2)_n \tag{5}$
Because the interaction involves in the charge of the ion and the dipole moment of the polar molecules, we can calculate the potential energy of interaction between them using the following formula:
$V=-\dfrac{q\mu}{(4\pi\epsilon_0) r^2} \tag{6}$
• r is the distance of separation.
• q is the charge of the ion ( only the magnitude of the charge is shown here.)
• $\mu$ is the permanent dipole moment of the polar molecule.
Hydrogen Bonding
The hydrogen bond is really a special case of dipole forces. A hydrogen bond is the attractive force between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule. Usually the electronegative atom is oxygen, nitrogen, or fluorine. In other words - The hydrogen on one molecule attached to O or N that is attracted to an O or N of a different molecule.
In the graphic below, the hydrogen is partially positive and attracted to the partially negative charge on the oxygen or nitrogen. Because oxygen has two lone pairs, two different hydrogen bonds can be made to each oxygen. This is a very specific bond as indicated. Some combinations that are not hydrogen bonds include: hydrogen to another hydrogen or hydrogen to a carbon.
Induced Dipole Forces
Forces between essentially non-polar molecules are the weakest of all intermolecular forces. "Temporary dipoles" are formed by the shifting of electron clouds within molecules. These temporary dipoles attract or repel the electron clouds of nearby non-polar molecules.
The temporary dipoles may exist for only a fraction of a second but a force of attraction also exist for that fraction of time. The strength of induced dipole forces depends on how easily electron clouds can be distorted. Large atoms or molecules with many electrons far removed from the nucleus are more easily distorted. Review - Non-Polar Bonds
London Dispersion force
London Dispersion Force, also called induced dipole-induced dipole, is the weakest intermolecular force. It is the interaction between two nonpolar molecules. For example: interaction between methyl and methyl group: (-CH3 ↔ CH3-). (approximate energy: 5 KJ/mol)
When two non polar molecules approach each other, attractive and repulsive forces between their electrons and nuclei will lead to distortions in these electron clouds, which mean they are induced by the neighboring molecules, and this leads to the interaction. We can calculate the potential energy between the two identical nonpolar molecule as followed formula:
$V=-\dfrac{3}{4} \dfrac{\alpha^2 I}{r^6} \tag{2}$
• α is the polarizability of nonpolar molecule.
• r is the distance between the two molecule.
• I = the first ionization energy of the molecule.
Negative sign indicates the attractive interaction. For two nonidentical nonpolar molecules A and B, we have the formula:
Problems
1. Two HBr molecules (u= 0.78 D) are separate by 300pm in air. Calculate the dipole-dipole interaction energy in KJ/mol if they are oriented from end to end.
2. Calculate the potential energy of interaction of a sodium ion that is at 3 Amstrong from HCl molecule with a dipole moment of 1.08 D.
3. Calculate the potential energy of interaction between two He atoms separated by 5 Amstrong in the air. (polarizability = 0.20x 10-30 m3 . I = 2373 KJ/mol)
Solution
1. 1D=3.3356x10-30Cm; 300pm=3 Amstrong= 3x10-10m. V=-2.7 KJ/mol
2. -35 KJ/mol
3. -0.0045 KJ/mol
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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, which keep a molecule together (e.g., covalent and ionic bonding).
Specific Interactions
Dipole-Dipole interactions result when two dipolar molecules interact with each other through space. When this occurs, the partially negative portion of one of the polar molecules is attracted to the partially positive portion of the second polar molecule. This type of interaction between molecules accounts for many physically and biologically significant phenomena such as the elevated boiling point of water.
Definition of a Dipole
Molecular dipoles occur due to the unequal sharing of electrons between atoms in a molecule. Those atoms that are more electronegative pull the bonded electrons closer to themselves. The buildup of electron density around an atom or discreet region of a molecule can result in a molecular dipole in which one side of the molecule possesses a partially negative charge and the other side a partially positive charge. Molecules with dipoles that are not canceled by their molecular geometry are said to be polar.
Example $1$: Carbon Dioxide and Hydrogen Flouride
In Figure 1 above, the more electronegative Oxygen atoms pull electron density towards themselves as demonstrated by the arrows. Carbon Dioxide is not polar however because of its linear geometry. A molecule's overall dipole is directional, and is given by the vector sum of the dipoles between the atoms. If we imagined the Carbon Dioxide molecule centered at 0 in the XY coordinate plane, the molecule's overall dipole would be given by the following equation:
$\mu \cos(0) + -\mu \cos(0) = 0.$
Where $μ$ is the dipole moment of the bond (given by μ=Q x r where Q is the charge and r is the distance of separation). Therefore, the two dipoles cancel each other out to yield a molecule with no net dipole.
In contrast, figure 2 demonstrates a situation where a molecular dipole does result. There is no opposing dipole moment to cancel out the one that is shown above. If we were to imagine the hydrogen flouride molecule placed so that the Hydrogen sat at the origin in the XY coordinate plane, the dipole would be given by $\mu \cos(0)=\mu$.
Potential Energy of Dipole Interaction
Potential energy is the maximum energy that is available for an object to do work. In physics, work is a quantity that describes the energy expended as a force operates over a distance. Potential energy is positional because it depends on the forces acting on an object at its position in space. For instance, we could say that an object held above the ground has a potential energy equal to its mass x acceleration due to gravity x its height above the ground (i.e., $mgh$). This potential energy that an object has as a result of its position can be used to do work. For instance we could use a pulley system with a large weight held above the ground to hoist a smaller weight into the air. As we drop the large weight it converts its potential energy to kinetic energy and does work on the rope which lifts the smaller weight into the air. It is important to remember that due to the second law of thermodynamics, the amount of work done by an object can never exceed (and is often considerably less) than the objects potential energy.
On a subatomic level, charged atoms have an electric potential which allows them to interact with each other. Electric potential refers to the energy held by a charged particle as a result of it's position relative to a second charged particle. Electric potential depends on charge polarity, charge strength and distance. Molecules with the same charge will repel each other as they come closer together while molecules with opposite charges will attract.
For two positively charged particles interacting at a distance r, the potential energy possessed by the system can be defined using Coulomb's Law:
$V = \dfrac{kQq}{r} \label{1}$
where
• $k$ is the Coulomb constant and
• $Q$ and $q$ refer to the magnitude of the charge for each particle in Coulombs.
The above equation can also be used to calculate the distance between two charged particles ($r$) if we know the potential energy of the system. While Coulomb's law is important, it only gives the potential energy between two point particles. Since molecules are much larger than point particles and have charge concentrated over a larger area, we have to come up with a new equation.
The potential energy possessed by two polar atoms interacting with each other depends on the dipole moment, μ, of each molecule, the distance apart, r, and the orientation in which the two molecules interact. For the case in which the partially positive area of one molecule interacts only with the partially negative area of the other molecule, the potential energy is given by:
$V(r) = -\dfrac{2\mu_{1}\mu_{2}}{4\pi\epsilon_{o}r^{3}} \label{2}$
where $\epsilon_o$ is the permeability of space. If it is not the case that the molecular dipoles interact in this straight end to end manor, we have to account mathematically for the change in potential energy due to the angle between the dipoles. We can add an angular term to the above equation to account for this new parameter of the system:
$V (r) =-\dfrac{\mu_{1}\mu_{2}}{4\pi\epsilon_{0}r_{12}^{3}}{(\cos\theta_{12}- 3\cos\theta_{1}\cos\theta_{2})} \label{3}$
In this formula $\theta_{12}$ is the angle made by the two oppositely charged dipoles, and $r_{12}$ is the distance between the two molecules. Also, $\theta_{1}$ and $\theta_{2}$ are the angles formed by the two dipoles with respect to the line connecting their centers.
It is also important to find the potential energy of the dipole moment for more than two interacting molecules. An important concept to keep in mind when dealing with multiple charged molecules interacting is that like charges repel and opposite charges attract. So for a system in which three charged molecules (2 positively charged molecules and 1 negatively charged molecule) are interacting, we need to consider the angle between the attractive and repellant forces.
The potential energy for the dipole interaction between multiple charged molecules is:
$V = \dfrac{kp \cos\theta}{r^{2}} \label{4}$
where
• $k$ is the Coulomb constant, and
• $r$ is the distance between the molecules.
Example $2$
Calculate the potential energy of the dipole-dipole interaction between 2 $\ce{HF}$ molecules oriented along the x axis in an XY coordinate plane whose area of positive charge is separated by 5.00 Angstroms from the area of negative charge of the adjacent molecule:
Solution
The Dipole moment of the HF molecules can be found in many tables, μ=1.92 D. Assume the molecules exist in a vacuum such that $\epsilon_{0}=8.8541878\times10^{-12}C^2N^{-1}m^{-2}$
$\mu=1.92D\times3.3356\times10^{-30}\dfrac{Cm}{D}=6.4044\times10^{-30}Cm$
Now use Equation \ref{2} to calculate the interaction energy
\begin{align*}V&=-\dfrac{2(6.4044\times10^{-30}Cm)^2}{4(8.8541878\times10^{-12}C^2N^{-1}m^{-2})(5.0\times10^{-10})^3} \[4pt] &=1.4745\times10^{-19}Nm \end{align*}
Example $3$
Now imagine the same two HF molecules in the following orientation:
Given: $\theta_{1}=\dfrac{3\pi}{4}$, $\theta_{2}=\dfrac{\pi}{3}$ and $\theta_{12}=\dfrac{5\pi}{12}$
Solution
\begin{align*} V&=-\dfrac{(6.4044\times10^{-30}Cm)^2}{4\pi(8.8541878\times10^{-12}C^2N^{-1}m^{-2}(5.00\times10^{-10}m)^3}(\cos\dfrac{5\pi}{12}-3\cos\dfrac{3\pi}{4}\cos\dfrac{\pi}{3}) \[4pt] &=-9.73\times10^{-20}Nm=9.73\times10^{-20}J\end{align*}
Dipole-Dipole Interactions in Macroscopic Systems
It would seem, based on the above discussion, that in a system composed of a large number of dipolar molecules randomly interacting with one another, V should go to zero because the molecules adopt all possible orientations. Thus the negative potential energy of two molecular dipoles participating in a favorable interaction would be cancelled out by the positive energy of two molecular dipoles participating in a high potential energy interaction. Contrary to our assumption, in bulk systems, it is more probable for dipolar molecules to interact in such a way as to minimize their potential energy (i.e., dipoles form less energetic, more probable configurations in accordance with the Boltzmann's Distribution). For instance, the partially positive area of a molecular dipole being held next to the partially positive area of a second molecular dipole is a high potential energy configuration and few molecules in the system will have sufficient energy to adopt it at room temperature. Generally, the higher potential energy configurations are only able to be populated at elevated temperatures. Therefore, the interactions of dipoles in a bulk Solution are not random, and instead adopt more probable, lower energy configurations. The following equation takes this into account:
$V=-\dfrac{2\mu_{A}^2\mu_{B}^2}{3(4\pi\epsilon_{0})^2r^6}\dfrac{1}{k_{B}T} \label{5}$
Example $4$
Looking at Equation \ref{5}, what happens to the potential energy of the interaction as temperature increases.
Solution
The potential energy of the dipole-dipole interaction decreases as T increases. This can be seen from the form of the above equation, but an explanation for this observation is relatively simple to come by. As the temperature of the system increases, more molecules have sufficient energy to occupy the less favorable configurations. The higher, less favorable, configurations are those that give less favorable interactions between the dipoles (i.e., higher potential energy configurations).
Example $5$
Calculate the average energy of HF molecules interacting with one another in a bulk Solution assuming that the molecules are 4.00 Angstroms apart in room temperature Solution.
Solution
Using Equation \ref{5} to calculate the bulk potential energy:
\begin{align*} V&=-\dfrac{2}{3}\dfrac{(6.4044\times10^{-30}Cm)^4}{(4\pi(8.8541878\times10^{-12}C^2N^{-1}m^{-2})^2(4.00\times10^{-10}m)^6}\dfrac{1}{(1.381\times10^{-23}Jk^{-1})(298k)} \[4pt] &=-5.46\times10^{-21}J\end{align*}
Example $6$
What is the amount of energy stabilization that is provided to the system when 1 mole of HF atoms interact through dipole-dipole interactions.
Solution
Since we have already calculated above the average potential energy of the HF dipole-dipole interaction this problem can be easily solved.
\begin{align*} V &=-5.46\times10^{-21}J\times(6.022\times10^{23}mol^{-1}) \[4pt] &=-3288\dfrac{J}{mol}=3.29\dfrac{kJ}{mol} \end{align*}
Biological Importance of Dipole Interactions
The potential energy from dipole interactions is important for living organisms. The biggest impact dipole interactions have on living organisms is seen with protein folding. Every process of protein formation, from the binding of individual amino acids to secondary structures to tertiary structures and even the formation of quaternary structures is dependent on dipole-dipole interactions.
A prime example of quaternary dipole interaction that is vital to human health is the formation of erythrocytes. Erythrocytes, commonly known as red blood cells are the cell type responsible for the gas exchange (i.e. respiration). Inside the erythrocytes, the molecule involved in this crucial process, is 'hemoglobin', formed by four protein subunits and a heme group'. For an heme to form properly, multiple steps must occur, all of which involve dipole interactions. The four protein subunits—two alpha chains, two beta chains—and the heme group, interact with each other through a series of dipole-dipole interactions which allow the erythrocyte to take its final shape. Any mutation that destroys these dipole-dipole interactions prevents the erythrocyte from forming properly, and impairs their ability to carry oxygen to the tissues of the body. So we can see that without the dipole-dipole interactions, proteins would not be able to fold properly and all life as we know it would cease to exist.
Dipole Mo
Let's go back to that first ion pair which was formed when the positive ion and the negative ion came together. If the electronegativities of the elements are sufficiently different (like an alkali metal and a halide), the charges on the paired ions will not change appreciably - there will be a full electron charge on the blue ion and a full positive charge on the red ion. The bond formed by the attraction of these opposite charges is called an ionic bond. If the difference in electronegativity is not so great, however, there will be some degree of sharing of the electrons between the two atoms. The result is the same whether two ions come together or two atoms come together:
The combination of atoms or ions is no longer a pair of ions, but rather a polar molecule which has a measureable dipole moment. The dipole moment (D) is defined as if there were a positive (+q) and a negative (-q) charge separated by a distance (r):
\[D = qr\]
If there is no difference in electronegativity between the atoms (as in a diatomic molecule such as \(O_2\) or \(F_2\)) there is no difference in charge and no dipole moment. The bond is called a covalent bond, the molecule has no dipole moment, and the molecule is said to be non-polar. Bonds between different atoms have different degrees of ionicity depending on the difference in the electronegativities of the atoms. The degree of ionicity may range from zero (for a covalent bond between two atoms with the same electronegativity) to one (for an ionic bond in which one atom has the full charge of an electron and the other atom has the opposite charge).
In some cases, two or more partially ionic bonds arranged symmetrically around a central atom may mutually cancel each other's polarity, resulting in a non-polar molecule. An example of this is seen in the carbon tetrachloride (\(CCl_4\)) molecule. There is a substantial difference between the electronegativities of carbon (2.55) and chlorine (3.16), but the four chlorine atoms are arranged symmetrically about the carbon atom in a tetrahedral configuration, and the molecule has zero dipole moment. Saturated hydrocarbons (\(C_nH_{n+2}\)) are non-polar molecules because of the small difference in the electronegativities of carbon and hydrogen plus the near symmetry about each carbon atom.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Specific_Interactions/Dipole-Di.txt
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A hydrogen bond is an intermolecular force (IMF) that forms a special type of dipole-dipole attraction when a hydrogen atom bonded to a strongly electronegative atom exists in the vicinity of another electronegative atom with a lone pair of electrons. Intermolecular forces (IMFs) occur between molecules. Other examples include ordinary dipole-dipole interactions and dispersion forces. Hydrogen bonds are are generally stronger than ordinary dipole-dipole and dispersion forces, but weaker than true covalent and ionic bonds.
The evidence for hydrogen bonding
Many elements form compounds with hydrogen. If you plot the boiling points of the compounds of the group 14 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 15, 16, and 17 with hydrogen, something odd happens.
Although the same reasoning applies for group 4 of the periodic table, the boiling point of the compound of hydrogen with the first element in each group is abnormally high. In the cases of $NH_3$, $H_2O$ and $HF$ there must be some additional intermolecular forces of attraction, requiring significantly more heat energy to break the IMFs. These relatively powerful intermolecular forces are described as hydrogen bonds.
Origin of Hydrogen Bonding
The molecules capable of hydrogen bonding include the following:
Notice that in each of these molecules:
• The hydrogen is attached directly to a highly electronegative atoms, causing the hydrogen to acquire a highly positive charge.
• Each of the highly electronegative atoms attains a high negative charge and has at least one "active" lone pair. Lone pairs at the 2-level have electrons contained in a relatively small volume of space, resulting in a high negative charge density. Lone pairs at higher levels are more diffuse and, resulting in a lower charge density and lower affinity for positive charge.
If you are not familiar with electronegativity, you should follow this link before you go on.
Consider two water molecules coming close together.
The $\delta^+$ 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 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 is an ideal example of hydrogen bonding. Notice that each water molecule can potentially form four hydrogen bonds with surrounding water molecules: two with the hydrogen atoms and two with the with the oxygen atoms. There are exactly the right numbers of $\delta^+$ hydrogens and lone pairs for every one of them to 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 are not enough lone pairs to go around to satisfy all the hydrogens. In hydrogen fluoride, the problem is a shortage of hydrogens. In water, two hydrogen bonds and two lone pairs allow formation of hydrogen bond interactions in a lattice of water molecules. Water is thus considered an ideal hydrogen bonded system.
More complex examples of hydrogen bonding
The hydration of negative ions
When an ionic substance dissolves in water, water molecules cluster around the separated ions. This process is called hydration. Water frequently attaches to positive ions by co-ordinate (dative covalent) bonds. It bonds to negative ions using hydrogen bonds.
If you are interested in the bonding in hydrated positive ions, you could follow this link to co-ordinate (dative covalent) bonding.
The diagram shows the potential hydrogen bonds formed with a chloride ion, Cl-. Although the lone pairs in the chloride ion are at the 3-level and would not normally be active enough to form hydrogen bonds, they are made more attractive by the full negative charge on the chlorine in this case.
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 -OH group. Any molecule which has a hydrogen atom attached directly to an oxygen or a nitrogen is capable of hydrogen bonding. Hydrogen bonds also occur when hydrogen is bonded to fluorine, but the HF group does not appear in other molecules. Molecules with hydrogen bonds 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," such that more heat (energy) is required to separate them. This phenomenon can be used to analyze boiling point of different molecules, defined as the temperature at which a phase change from liquid to gas occurs.
Ethanol, $\ce{CH3CH2-O-H}$, and methoxymethane, $\ce{CH3-O-CH3}$, both have the same molecular formula, $\ce{C2H6O}$.
They have the same number of electrons, and a similar length. The van der Waals attractions (both dispersion forces and dipole-dipole attractions) in each will be similar. However, ethanol has a hydrogen atom attached directly to an oxygen; here the oxygen still has two lone pairs like 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 are not 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 have similar chain lengths. 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; however, the values 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 organic molecules containing nitrogen
Hydrogen bonding also occurs in organic molecules containing N-H groups; recall the hydrogen bonds that occur with 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.
Donors and Acceptors
In order for a hydrogen bond to occur there must be both a hydrogen donor and an acceptor present. The donor in a hydrogen bond is usually a strongly electronegative atom such as N, O, or F that is covalently bonded to a hydrogen bond.
The hydrogen acceptor is an electronegative atom of a neighboring molecule or ion that contains a lone pair that participates in the hydrogen bond.
Why does a hydrogen bond occur?
Since the hydrogen donor (N, O, or F) is strongly electronegative, it pulls the covalently bonded electron pair closer to its nucleus, and away from the hydrogen atom. The hydrogen atom is then left with a partial positive charge, creating a dipole-dipole attraction between the hydrogen atom bonded to the donor and the lone electron pair of the acceptor. This results in a hydrogen bond.(see Interactions Between Molecules With Permanent Dipoles)
Types of hydrogen bonds
Although hydrogen bonds are well-known as a type of IMF, these bonds can also occur within a single molecule, between two identical molecules, or between two dissimilar molecules.
Intramolecular hydrogen bonds
Intramolecular hydrogen bonds are those which occur within one single molecule. This occurs when two functional groups of a molecule can form hydrogen bonds with each other. In order for this to happen, both a hydrogen donor a hydrogen acceptor must be present within one molecule, and they must be within close proximity of each other in the molecule. For example, intramolecular hydrogen bonding occurs in ethylene glycol (C2H4(OH)2) between its two hydroxyl groups due to the molecular geometry.
Intermolecular hydrogen bonds
Intermolecular hydrogen bonds occur between separate molecules in a substance. They can occur between any number of like or unlike molecules as long as hydrogen donors and acceptors are present in positions where they can interact with one another. For example, intermolecular hydrogen bonds can occur between NH3 molecules alone, between H2O molecules alone, or between NH3 and H2O molecules.
Properties and effects of hydrogen bonds
On Boiling Point
When we consider the boiling points of molecules, we usually expect molecules with larger molar masses to have higher normal boiling points than molecules with smaller molar masses. This, without taking hydrogen bonds into account, is due to greater dispersion forces (see Interactions Between Nonpolar Molecules). Larger molecules have more space for electron distribution and thus more possibilities for an instantaneous dipole moment. However, when we consider the table below, we see that this is not always the case.
Compound Molar Mass Normal Boiling Point
$H_2O$ 18 g/mol 373 K
$HF$ 20 g/mol 292.5 K
$NH_3$ 17 g/mol 239.8 K
$H_2S$ 34 g/mol 212.9 K
$HCl$ 36.4 g/mol 197.9 K
$PH_3$ 34 g/mol 185.2 K
We see that H2O, HF, and NH3 each have higher boiling points than the same compound formed between hydrogen and the next element moving down its respective group, indicating that the former have greater intermolecular forces. This is because H2O, HF, and NH3 all exhibit hydrogen bonding, whereas the others do not. Furthermore, $H_2O$ has a smaller molar mass than HF but partakes in more hydrogen bonds per molecule, so its boiling point is higher.
On Viscosity
The same effect that is seen on boiling point as a result of hydrogen bonding can also be observed in the viscosity of certain substances. Substances capable of forming hydrogen bonds tend to have a higher viscosity than those that do not form hydrogen bonds. Generally, substances that have the possibility for multiple hydrogen bonds exhibit even higher viscosities.
Factors preventing Hydrogen bonding
Electronegativity
Hydrogen bonding cannot occur without significant electronegativity differences between hydrogen and the atom it is bonded to. Thus, we see molecules such as PH3, which do not participate in hydrogen bonding. PH3 exhibits a trigonal pyramidal molecular geometry like that of ammonia, but unlike NH3 it cannot hydrogen bond. This is due to the similarity in the electronegativities of phosphorous and hydrogen. Both atoms have an electronegativity of 2.1, and thus, there is no dipole moment. This prevents the hydrogen atom from acquiring the partial positive charge needed to hydrogen bond with the lone electron pair in another molecule. (see Polarizability)
Atom Size
The size of donors and acceptors can also affect the ability to hydrogen bond. This can account for the relatively low ability of Cl to form hydrogen bonds. When the radii of two atoms differ greatly or are large, their nuclei cannot achieve close proximity when they interact, resulting in a weak interaction.
Hydrogen Bonding in Nature
Hydrogen bonding plays a crucial role in many biological processes and can account for many natural phenomena such as the Unusual properties of Water. In addition to being present in water, hydrogen bonding is also important in the water transport system of plants, secondary and tertiary protein structure, and DNA base pairing.
Plants
The cohesion-adhesion theory of transport in vascular plants uses hydrogen bonding to explain many key components of water movement through the plant's xylem and other vessels. Within a vessel, water molecules hydrogen bond not only to each other, but also to the cellulose chain that comprises the wall of plant cells. Since the vessel is relatively small, the attraction of the water to the cellulose wall creates a sort of capillary tube that allows for capillary action. This mechanism allows plants to pull water up into their roots. Furthermore, hydrogen bonding can create a long chain of water molecules, which can overcome the force of gravity and travel up to the high altitudes of leaves.
Proteins
Hydrogen bonding is present abundantly in the secondary structure of proteins, and also sparingly in tertiary conformation. The secondary structure of a protein involves interactions (mainly hydrogen bonds) between neighboring polypeptide backbones which contain nitrogen-hydrogen bonded pairs and oxygen atoms. Since both N and O are strongly electronegative, the hydrogen atoms bonded to nitrogen in one polypeptide backbone can hydrogen bond to the oxygen atoms in another chain and vice-versa. Though they are relatively weak, these bonds offer substantial stability to secondary protein structure because they repeat many times and work collectively.
In tertiary protein structure, interactions are primarily between functional R groups of a polypeptide chain; one such interaction is called a hydrophobic interaction. These interactions occur because of hydrogen bonding between water molecules around the hydrophobe that further reinforces protein conformation.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Specific_Interactions/Hydrogen_.txt
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Polar molecules can interact with ions:
Figure 1: Ion-Dipole Interactions (Gary L Bertrand)
Ion - Ind
The charges on ions and the charge separation in polar molecules explain the fairly strong interactions between them, with very strong ion - ion interactions, weaker ion - dipole interactions, and considerably weaker dipole-dipole interactions. Even in a non-polar molecule, however, the valence electrons are moving around and there will occasionally be instances when more are on one side of the molecule than on the other. This gives rise to fluctuating or instantaneous dipoles:
Figure 1: Fluctuating Dipole in a Non-polar Molecule
These instantaneous dipoles may be induced and stabilized as an ion or a polar molecule approaches the non-polar molecule.
Figure 2: (Left) Ion - Induced Dipole Interaction. (Right) Dipole - Induced Dipole Interaction
Ion - Ion
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).
Lennard-J
Proposed by Sir John Edward Lennard-Jones, the Lennard-Jones potential describes the potential energy of interaction between two non-bonding atoms or molecules based on their distance of separation. The potential equation accounts for the difference between attractive forces (dipole-dipole, dipole-induced dipole, and London interactions) and repulsive forces.
Introduction
Imagine two rubber balls separated by a large distance. Both objects are far enough apart that they are not interacting. The two balls can be brought closer together with minimal energy, allowing interaction. The balls can continuously be brought closer together until they are touching. At this point, it becomes difficult to further decrease the distance between the two balls. To bring the balls any closer together, increasing amounts of energy must be added. This is because eventually, as the balls begin to invade each other’s space, they repel each other; the force of repulsion is far greater than the force of attraction.
This scenario is similar to that which takes place in neutral atoms and molecules and is often described by the Lennard-Jones potential.
The Lennard-Jones Potential
The Lennard-Jones model consists of two 'parts'; a steep repulsive term, and smoother attractive term, representing the London dispersion forces. Apart from being an important model in itself, the Lennard-Jones potential frequently forms one of 'building blocks' of many force fields. It is worth mentioning that the 12-6 Lennard-Jones model is not the most faithful representation of the potential energy surface, but rather its use is widespread due to its computational expediency. The Lennard-Jones Potential is given by the following equation:
$V(r)= 4 \epsilon \left [ {\left (\dfrac{\sigma}{r} \right )}^{12}-{\left (\dfrac{\sigma}{r} \right )}^{6} \right] \label{1}$
or is sometimes expressed as:
$V(r) = \frac{A}{r^{12}}- \dfrac{B}{r^6} \label{2}$
where
• $V$ is the intermolecular potential between the two atoms or molecules.
• $\epsilon$ is the well depth and a measure of how strongly the two particles attract each other.
• $\sigma$ is the distance at which the intermolecular potential between the two particles is zero (Figure $1$). $\sigma$ gives a measurement of how close two nonbonding particles can get and is thus referred to as the van der Waals radius. It is equal to one-half of the internuclear distance between nonbonding particles.
• $r$ is the distance of separation between both particles (measured from the center of one particle to the center of the other particle).
• $A= 4\epsilon \sigma^{12}$, $B= 4\epsilon \sigma^{6}$
• Minimum value of $\Phi_{12}(r)$ at $r = r_{min}$.
Example $1$
The $\epsilon$ and $\sigma$ values for Xenon (Xe) are found to be 1.77 kJ/mol and 4.10 Angstroms, respectively. Determine the van der Waals radius for the Xenon atom.
Solution
Recall that the van der Waals radius is equal to one-half of the internuclear distance between nonbonding particles. Because $\sigma$ gives a measure of how close two non-bonding particles can be, the van der Waals radius for Xenon (Xe) is given by:
r = $\sigma$/2 = 4.10 Angstroms/2 = 2.05 Angstroms
Bonding Potential
The Lennard-Jones potential is a function of the distance between the centers of two particles. When two non-bonding particles are an infinite distance apart, the possibility of them coming together and interacting is minimal. For simplicity's sake, their bonding potential energy is considered zero. However, as the distance of separation decreases, the probability of interaction increases. The particles come closer together until they reach a region of separation where the two particles become bound; their bonding potential energy decreases from zero to a negative quantity. While the particles are bound, the distance between their centers continue to decrease until the particles reach an equilibrium, specified by the separation distance at which the minimum potential energy is reached.
If the two bound particles are further pressed together, past their equilibrium distance, repulsion begins to occur: the particles are so close together that their electrons are forced to occupy each other’s orbitals. Repulsion occurs as each particle attempts to retain the space in their respective orbitals. Despite the repulsive force between both particles, their bonding potential energy increases rapidly as the distance of separation decreases.
Example $2$
Calculate the intermolecular potential between two Argon (Ar) atoms separated by a distance of 4.0 Angstroms (use $\epsilon$=0.997 kJ/mol and $\sigma$=3.40 Angstroms).
Solution
To solve for the intermolecular potential between the two Argon atoms, we use equation 2.1 where V is the intermolecular potential between two non-bonding particles.
$V= 4 \epsilon \left [ {\left (\dfrac{\sigma}{r} \right )}^{12}-{\left (\dfrac{\sigma}{r} \right )}^{6} \right] \nonumber$
The data given are $\epsilon$=0.997 kJ/mol, $\sigma$=3.40 Angstroms, and the distance of separation, r=4.0 Angstroms. We plug these values into equation 2.1 and solve as follows:
\begin{align*} V &= 4(0.997\;\text{kJ/mol}) \left[\left(\dfrac{3.40\;\text{Angstroms}}{4.0\;\text{Angstroms}}\right)^{12}-\left(\dfrac{3.40 \;\text{Angstroms}}{4.0 \;\text{Angstroms}}\right)^6\right] \[4pt] &= 3.988(0.14-0.38) \[4pt] & = 3.988(-0.24) \[4pt] & = -0.96 kJ/mol \end{align*}
Stability and Force of Interactions
Like the bonding potential energy, the stability of an arrangement of atoms is a function of the Lennard-Jones separation distance. As the separation distance decreases below equilibrium, the potential energy becomes increasingly positive (indicating a repulsive force). Such a large potential energy is energetically unfavorable, as it indicates an overlapping of atomic orbitals. However, at long separation distances, the potential energy is negative and approaches zero as the separation distance increases to infinity (indicating an attractive force). This indicates that at long-range distances, the pair of atoms or molecules experiences a small stabilizing force. Lastly, as the separation between the two particles reaches a distance slightly greater than σ, the potential energy reaches a minimum value (indicating zero force). At this point, the pair of particles is most stable and will remain in that orientation until an external force is exerted upon it.
Example $3$
Two molecules, separated by a distance of 3.0 angstroms, are found to have a $\sigma$ value of 4.10 angstroms. By decreasing the separation distance between both molecules to 2.0 angstroms, the intermolecular potential between the molecules becomes more negative. Do these molecules follow the Lennard-Jones potential? Why or why not?
Solution
Recall that $\sigma$ is the distance at which the bonding potential between two particles is zero. On a graph of the Lennard-Jones potential, then, this value gives the x-intersection of the graph. According to the Lennard-Jones potential, any value of $r$ greater than $\sigma$ should yield a negative bonding potential and any value of r smaller than $\sigma$ should yield a positive bonding potential. In this scenario, as the separation between the two molecules decreases from 3.0 angstroms to 2.0 angstroms, the bonding potential is becomes more negative. In essence however, because the starting separation (3.0 angstroms) is already less than $\sigma$ (4.0 angstroms), decreasing the separation even further (2.0 angstroms) should result in a more positive bonding potential. Therefore, these molecules do not follow the Lennard-Jones potential.
Problems
1. The second part of the Lennard-Jones equation is ($\sigma$/r)6 and denotes attraction. Name at least three types of intermolecular interactions that represent attraction.
Answer
From section 2.1, dipole-dipole, dipole-induced dipole, and London interactions are all attractive forces.
1. At what separation distance in the Lennard-Jones potential does a species have a repulsive force acting on it? An attractive force? No force?
Answer
See Figure C. A species will have a repulsive force acting on it when r is less than the equilibrium distance between the particles. A species will have an attractive force acting on it when r is greater than the equilibrium distance between the particles. Lastly, when $r$ is equal to the equilibrium distance between both particles, the species will have no force acting upon it.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Specific_Interactions/Ion_-_Dip.txt
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The term polarizability refers to the tendency of molecules to generate induced electric dipole moments when subjected to an electric field. It originates from the fact that nuclei and electrons are generally not fixed in space and that when molecules are subject to an electric field, the negatively charged electrons and positively charged atomic nuclei are subject to opposite forces and undergo charge separation. Polarizability allows us to better understand the interactions between nonpolar atoms and molecules and other electrically charged species, such as ions or polar molecules with dipole moments.
Introduction
Neutral nonpolar species have spherically symmetric arrangements of electrons in their electron clouds. When in the presence of an electric field, their electron clouds can be distorted (Figure $1$). The ease of this distortion is the polarizability of the atom or molecule. The created distortion of the electron cloud causes the originally nonpolar molecule or atom to acquire a dipole moment. This induced dipole moment is related to the polarizability of the molecule or atom and the strength of the electric field by the following equation:
$μ_{ind} = \alpha E \label{1}$
where $E$ denotes the strength of the electric field and $\alpha$ is the polarizability of the atom or molecule with units of C m2V-1.
In general, polarizability correlates with the interaction between electrons and the nucleus. The amount of electrons in a molecule affects how tight the nuclear charge can control the overall charge distribution. Atoms with fewer electrons will have smaller, denser electron clouds, as there is a strong interaction between the few electrons in the atoms’ orbitals and the positively charged nucleus. There is also less shielding in atoms with fewer electrons contributing to the stronger interaction of the outer electrons and the nucleus. With the electrons held tightly in place in these smaller atoms, these atoms are typically not easily polarized by external electric fields. In contrast, large atoms with many electrons, such as negative ions with excess electrons, are easily polarized. These atoms typically have very diffuse electron clouds and large atomic radii that limit the interaction of their external electrons and the nucleus.
Factors that Influence Polarizability
The relationship between polarizability and the factors of electron density, atomic radii, and molecular orientation is as follows:
1. The greater the number of electrons, the less control the nuclear charge has on charge distribution, and thus the increased polarizability of the atom.
2. The greater the distance of electrons from nuclear charge, the less control the nuclear charge has on the charge distribution, and thus the increased polarizability of the atom.
3. Molecular orientation with respect to an electric field can affect polarizability (labeled Orientation-dependent), except for molecules that are: tetrahedral, octahedral or icosahedral (labeled Orientation-independent). This factor is more important for unsaturated molecules that contain areas of electron-dense regions, such as 2,4-hexadiene. Greatest polarizability in these molecules is achieved when the electric field is applied parallel to the molecule rather than perpendicular to the molecule.
Polarizability Influences Dispersion Forces
The dispersion force is the weakest intermolecular force. It is an attractive force that arises from surrounding temporary dipole moments in nonpolar molecules or species. These temporary dipole moments arise when there are instantaneous deviations in the electron clouds of the nonpolar species. Surrounding molecules are influenced by these temporary dipole moments and a sort of chain reaction results in which subsequent weak, dipole-induced dipole interactions are created (Figure $2$).
These cumulative dipole-induced dipole interactions create attractive dispersion forces. Dispersion forces are the forces that make nonpolar substances condense to liquids and freeze into solids when the temperature is low enough.
Polarizability affects dispersion forces in the following ways:
• As polarizability increases, the dispersion forces also become stronger. Thus, molecules attract one another more strongly and melting and boiling points of covalent substances increase with larger molecular mass.
• Polarizability also affects dispersion forces through the molecular shape of the affected molecules. Elongated molecules with electrons that are delocalized (e.g., with double or triple bonds) will have some of those electrons easily shifted over a greater distance. This is manifested in an greater polarizability and thus greater dispersion forces.
Structure Dependence to Dispersion Forces
The influence of polarizability on the strength of the dispersion forces is also modulated by the 3-D structure of the molecules. For example, neo-pentane and n-pentane (Figure $3$) exhibit comparable polarizabilities, but have different boiling points. The lack of polar bonding in both indicates that dispersion is the dominant interaction responsible in both substances. The higher boiling point of n-pentane indicates there are stronger intermolecular interactions.
Table $3$: Properties of neopentane and n-pentane
molecule Boiling Points (ºC) $\alpha$ (Å3)
neopentane 9.5 10.24
n-pentane 36.0 9.879
Both neopentane and n-pentane would be expected to exhibits comparable dispersion forces since they have comparable polarizabilities. However, the structure of the n-pentane molecule facilitates greater dispersion forces due to more contacts between n-pentane molecules. In contrast, neopentane has a smaller dispersion force since the interactions between neopentane are reduced since neopentane is more compact and symmetrical.
The relationship between polarizability and dispersion forces can be seen in the following equation, which can be used to quantify the interaction between two like nonpolar atoms or molecules (e.g., $\ce{A}$ with $\ce{A}$):
$V(r) = \dfrac{-3}{4} \dfrac{\alpha^2 I}{r^6} \label{2}$
where $r$ is the distance between the atoms or molecules, $I$ is the first ionization energy of the atom or molecule, and $\alpha$ is the polarizability constant expressed in units of m3.
This expression of $\alpha$ is related to $\alpha'$ by the following equation:
$\alpha' = \dfrac{\alpha}{4 \pi \epsilon_o} \label{3}$
To quantify the interaction between unlike atoms or molecules ($\ce{A}$ and $\ce{B}$), Equation $\ref{2}$ becomes:
$V(r) = \dfrac{-3}{2}\dfrac{I_AI_B}{I_A+I_B} \dfrac{\alpha_A \alpha_B}{r^6} \label{4}$
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Van der Waals forces are driven by induced electrical interactions between two or more atoms or molecules that are very close to each other. Van der Waals interaction is the weakest of all intermolecular attractions between molecules. However, with a lot of Van der Waals forces interacting between two objects, the interaction can be very strong.
Introduction
Here is a chart to compare the relative weakness of Van der Waals forces to other intermolecular attractions.
Weak Intermolecular Interactions
Force Strength (kJ/mol) Distance (nm)
Van der Waals 0.4-4.0 0.3-0.6
Hydrogen Bonds 12-30 0.3
Ionic Interactions 20 0.25
Hydrophobic Interactions <40 varies
Causes of Van der Waals Forces
Quantum Mechanics strongly emphasizes the constant movement of electrons in an atom through the Schrödinger Equation and the Heisenberg’s Uncertainty Principle. The Heisenberg’s Uncertainty Principle proposes that the energy of the electron is never zero; therefore, it is constantly moving around its orbital. The square of the Schrödinger Equation for a particle in a box suggests that it is probable of finding the electron (particle) anywhere in the orbital of the atom (box).
These two important aspects of Quantum Mechanics strongly suggest that the electrons are constantly are moving in an atom, so dipoles are probable of occurring. A dipole is defined as molecules or atoms with equal and opposite electrical charges separated by a small distance.
It is probable to find the electrons in this state:
This is how spontaneous (or instantaneous) dipoles occur. When groups of electrons move to one end of the atom, it creates a dipole. These groups of electrons are constantly moving so they move from one end of the atom to the other and back again continuously. Therefore, the opposite state is as probable of occurring.
Opposite state due to fluctuation of dipoles:
Dipole-Dipole Interaction
Dipole-Dipole interactions occur between molecules that have permanent dipoles; these molecules are also referred to as polar molecules. The figure below shows the electrostatic interaction between two dipoles.
The potential energy of the interaction for the top pair of the image above is represented by the equation:
$V = -\dfrac{2\mu_A\mu_B}{4\pi\epsilon_o r^3} \tag{1}$
The potential energy of the interaction for the bottom pair is represented by the equation:
$V = -\dfrac{\mu_A\mu_B}{4\pi\epsilon_o r^3} \tag{2}$
with
• $V$ is the potential energy
• $\mu$ is the dipole moment
• $\epsilon_o$ is the vacuum permittivity
• $r$ is the length between the two nuclei
The negative sign indicates that energy is released out of the system, because energy is released when bonds are formed, even weak bonds. The negative sign also suggests that the interaction is driven by an attractive force (a positive sign would indicate a repulsion force between the two molecules). If the conditions of these two samples are the same except for their orientation, the second pair of the electron will always have a larger potential energy, because both the negative and positive ends are involve in the interactions.
Induced Dipoles
An induced dipole moment is a temporary condition during which a neutral nonpolar atom (i.e. Helium) undergo a separation of charges due to the environment. When an instantaneous dipole atom approaches a neighboring atom, it can cause that atom to also produce dipoles. The neighboring atom is then considered to have an induced dipole moment.
Even though these two atoms are interacting with each other, their dipoles may still fluctuate. However, they must fluctuate in synchrony in order to maintain their dipoles and stay interacted with each other. Result of synchronizing fluctuation of dipoles:
The potential energy representing the dipole-induced dipole interaction is:
$V = -\dfrac{\alpha\mu^2}{4\pi\epsilon_o r^6} \tag{4}$
• $\alpha$ = polarizability of the nonpolar molecule
Polarizability defines how easy the electron density of an atom or a molecule can be distorted by an external electric field.
Spontaneous Dipole-Induced Dipole Interaction
Spontaneous dipole-induced dipole interactions are also known as dispersion or London forces (name after the German physicist Fritz London). They are large networks of intermolecular forces between nonpolar and non-charged molecules and atoms (i.e. alkanes, noble gases, and halogens). Molecules that have induced dipoles may also induce neighboring molecules to have dipole moments, so a large network of induced dipole-induced dipole interactions may exist. The image below illustrates a network of induced dipole-induced dipole interactions.
The potential energy of an induced dipole-induced dipole interaction is represented by this equation:
$V = -\dfrac{3}{2}\dfrac{I_aI_b}{I_a + I_b}\dfrac{\alpha_a\alpha_b}{r^6} \tag{5}$
• $I$ = The first ionization energy of the molecule
The radius is a huge determinant of how large the potential energy is since the potential energy is inversely proportional to $r^6$. A small increase in the radius, would greatly decrease the potential energy of the interaction.
Contributors and Attributions
• Justin Than (UCD)
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Van der Waals forces' is a general term used to define the attraction of intermolecular forces between molecules. There are two kinds of Van der Waals forces: weak London Dispersion Forces and stronger dipole-dipole forces.
Introduction
The chance that an electron of an atom is in a certain area in the electron cloud at a specific time is called the "electron charge density." Since there is no way of knowing exactly where the electron is located and since they do not all stay in the same area 100 percent of the time, if the electrons all go to the same area at once, a dipole is formed momentarily. Even if a molecule is nonpolar, this displacement of electrons causes a nonpolar molecule to become polar for a moment.
Since the molecule is polar, this means that all the electrons are concentrated at one end and the molecule is partially negatively charged on that end. This negative end makes the surrounding molecules have an instantaneous dipole also, attracting the surrounding molecules' positive ends. This process is known as the London Dispersion Force of attraction.
The ability of a molecule to become polar and displace its electrons is known as the molecule's "polarizability." The more electrons a molecule contains, the higher its ability to become polar. Polarizability increases in the periodic table from the top of a group to the bottom and from right to left within periods. This is because the higher the molecular mass, the more electrons an atom has. With more electrons, the outer electrons are easily displaced because the inner electrons shield the nucleus' positive charge from the outer electrons which would normally keep them close to the nucleus.
When the molecules become polar, the melting and boiling points are raised because it takes more heat and energy to break these bonds. Therefore, the greater the mass, the more electrons present, and the more electrons present, the higher the melting and boiling points of these substances.
London dispersion forces are stronger in those molecules that are not compact, but long chains of elements. This is because it is easier to displace the electrons because the forces of attraction between the electrons and protons in the nucleus are weaker. The more readily displacement of electrons means the molecule is also more “polarizable.”
Dipole-Dipole Forces
These forces are similar to London Dispersion forces, but they occur in molecules that are permanently polar versus momentarily polar. In this type of intermolecular interaction, a polar molecule such as water or H2O attracts the positive end of another polar molecule with its negative end of its dipole. The attraction between these two molecules is the dipole-dipole force.
Van der Waals Equation
Van der Waals equation is required for special cases, such as non-ideal (real) gases, which is used to calculate an actual value. The equation consist of:
$\left (P+\frac{n^2a}{V^2} \right ) \left (V-nb \right )=nRT \tag{1}$
The V in the formula refers to the volume of gas, in moles n. The intermolecular forces of attraction is incorporated into the equation with the $\frac{n^2a}{V^2}$ term where a is a specific value of a particular gas. P represents the pressure measured, which is expected to be lower than in usual cases. The variable b expresses the eliminated volume per mole, which accounts for the volume of gas molecules and is also a value of a particular gas. R is a known constant, 0.08206 L atm mol-1 K-1, and T stands for temperature.
Unlike most equations used for the calculation of real, or ideal, gases, van der Waals equation takes into account, and corrects for, the volume of participating molecules and the intermolecular forces of attraction.
Contributors and Attributions
• Kathryn Rashe, Lisa Peterson, Seila Buth, Irene Ly
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The ionization energy is the quantity of energy that an isolated, gaseous atom in the ground electronic state must absorb to discharge an electron, resulting in a cation.
$\ce{ H(g) \rightarrow H^{+}(g) + e^{-}}$
This energy is usually expressed in kJ/mol, or the amount of energy it takes for all the atoms in a mole to lose one electron each.
When considering an initially neutral atom, expelling the first electron will require less energy than expelling the second, the second will require less energy than the third, and so on. Each successive electron requires more energy to be released. This is because after the first electron is lost, the overall charge of the atom becomes positive, and the negative forces of the electron will be attracted to the positive charge of the newly formed ion. The more electrons that are lost, the more positive this ion will be, the harder it is to separate the electrons from the atom.
In general, the further away an electron is from the nucleus, the easier it is for it to be expelled. In other words, ionization energy is a function of atomic radius; the larger the radius, the smaller the amount of energy required to remove the electron from the outer most orbital. For example, it would be far easier to take electrons away from the larger element of Ca (Calcium) than it would be from one where the electrons are held tighter to the nucleus, like Cl (Chlorine).
In a chemical reaction, understanding ionization energy is important in order to understand the behavior of whether various atoms make covalent or ionic bonds with each other. For instance, the ionization energy of Sodium (alkali metal) is 496KJ/mol (1) whereas Chlorine's first ionization energy is 1251.1 KJ/mol (2). Due to this difference in their ionization energy, when they chemically combine they make an ionic bond. Elements that reside close to each other in the periodic table or elements that do not have much of a difference in ionization energy make polar covalent or covalent bonds. For example, carbon and oxygen make CO2 (Carbon dioxide) reside close to each other on a periodic table; they, therefore, form a covalent bond. Carbon and chlorine make CCl4 (Carbon tetrachloride) another molecule that is covalently bonded.
Periodic Table and Trend of Ionization Energies
As described above, ionization energies are dependent upon the atomic radius. Since going from right to left on the periodic table, the atomic radius increases, and the ionization energy increases from left to right in the periods and up the groups. Exceptions to this trend is observed for alkaline earth metals (group 2) and nitrogen group elements (group 15). Typically, group 2 elements have ionization energy greater than group 13 elements and group 15 elements have greater ionization energy than group 16 elements. Groups 2 and 15 have completely and half-filled electronic configuration respectively, thus, it requires more energy to remove an electron from completely filled orbitals than incompletely filled orbitals.
Alkali metals (IA group) have small ionization energies, especially when compared to halogens or VII A group (see diagram 1). In addition to the radius (distance between nucleus and the electrons in outermost orbital), the number of electrons between the nucleus and the electron(s) you're looking at in the outermost shell have an effect on the ionization energy as well. This effect, where the full positive charge of the nucleus is not felt by outer electrons due to the negative charges of inner electrons partially canceling out the positive charge, is called shielding. The more electrons shielding the outer electron shell from the nucleus, the less energy required to expel an electron from said atom. The higher the shielding effect the lower the ionization energy (see diagram 2). It is because of the shielding effect that the ionization energy decreases from top to bottom within a group. From this trend, Cesium is said to have the lowest ionization energy and Fluorine is said to have the highest ionization energy (with the exception of Helium and Neon).
Table 1: showing the increasing trend of ionization energy in KJ/mol (exception in case of Boron) from left to right in the periodic table(8)
Li 520 Be 899 B 800 C 1086 N 1402 O 1314 F 1680
Table 2: showing decreasing trend of ionization energies (Kj/mol) from top to bottom (Cs is the exception in the first group) (8)
Li 520
Na 496
K 419
Rb 408
Cs 376
Fr 398
1st, 2nd, and 3rd Ionization Energies
The symbol $I_1$ stands for the first ionization energy (energy required to take away an electron from a neutral atom) and the symbol $I_2$ stands for the second ionization energy (energy required to take away an electron from an atom with a +1 charge. Each succeeding ionization energy is larger than the preceding energy. This means that $I_1 < I_2 < I_3 < ... < I_n$ will always be true.
Example of how ionization energy increases as succeeding electrons are taken away.
$Mg \,(g) \rightarrow Mg^+\,(g) + e^- \;\;\; I_1= 738\, kJ/mol$
$Mg^+ \,(g) \rightarrow Mg^{2+}\, (g) + e^- \;\;\; I_2= 1451\, kJ/mol$
See first, second, and third ionization energies of elements/ions in Table 3.
1
2
3
4
5
6
7
8
Table 3: Ionization Energies (kJ/mol)
H 1312
He 2372 5250
Li 520 7297 11810
Be 899 1757 14845 21000
B 800 2426 3659 25020 32820
C 1086 2352 4619 6221 37820 47260
N 1402 2855 4576 7473 9442 53250 64340
O 1314 3388 5296 7467 10987 13320 71320 84070
F 1680 3375 6045 8408 11020 15160 17860 92010
Ne 2080 3963 6130 9361 12180 15240
Na 496 4563 6913 9541 13350 16600 20113 25666
Mg 737 1450 7731 10545 13627 17995 21700 25662
The Effects of Electron Shells on Ionization Energy
Electron orbitals are separated into various shells which have strong impacts on the ionization energies of the various electrons. For instance, let us look at aluminum. Aluminum is the first element of its period with electrons in the 3p shell. This makes the first ionization energy comparably low to the other elements in the same period, because it only has to get rid of one electron to make a stable 3s shell, the new valence electron shell. However, once you've moved past the first ionization energy into the second ionization energy, there is a large jump in the amount of energy required to expel another electron. This is because you now are trying to take an electron from a fairly stable and full 3s electron shell. Electron shells are also responsible for the shielding that was explained above.
Ionization Energy and Electron Affinity--Similar Trend
Both ionization energy and electron affinity have similar trend in the periodic table. For example, just as ionization energy increases along the periods, electron affinity also increases. Likewise, electron affinity decreases from top to bottom due to the same factor, i.e., shielding effect. Halogens can capture an electron easily as compared to elements in the first and second group. This tendency to capture an electron in a gaseous state is termed as electronegativity. This tendency also determines one of the chemical differences between Non metallic and metallic elements.
Diagram 3: showing increasing trend of electron affinity from left to right (9).
B 27 C 123.4 N -7 O 142.5 F 331.4
Diagram 4: showing decreasing pattern of electron affinities of elements from top to bottom (9)
H 73.5
Li 60.4
Na 53.2
K 48.9
Rb 47.4
Cs 46.0
Fr 44.5
As indicated above, the elements to the right side of periodic table (diagram 3) have tendency to receive the electron while the one at the left are more electropositive. Also, from left to right, the metallic characteristics of elements decrease (4).
Prediction of Covalent and Ionic Bonds
The difference of electronegativity or ionization energies between two reacting elements determine the fate of the type of bond. For example, there is a big difference of ionization energies and electronegativity between Na and. Cl. Therefore, sodium completely removes the electron from its outermost orbital and chlorine completely accepts the electron, and as a result we have an ionic bond (4). However, in cases where there is no difference in electronegativity, the sharing of electrons produces a covalent bond. For example, electronegativity of Hydrogen is 2.1 and the combination of two Hydrogen atoms will definitely make a covalent bond (by sharing of electrons). The combination of Hydrogen and Fluorine (electronegativity=3.96) will produce a polar covalent bond because they have small differences between electronegativity (5).
Questions
1) By looking at following electronic configuration of elements, can you predict which element has the lowest first ionization energy?
1. 1s2 2s2 2p6
2. 1s2 2s2 2p4
3. 1s2 2s2 2p6 3s2
4. 1s2 2s2 2p6 3s1
5. 1s2 2s2 2p5.
2) The ionization energy of $Na^{+3}$ ion is one of the following (7)
1. More than first ionization only
2. More than second ionization only
3. Sum of first and second ionization energies
4. Sum of first, second, and third ionization energies
3) Ionization energies and electron affinities are
1. Dependent upon each other,
2. Similar trend of increasing/decreasing along the periods and within the group of periodic table,
3. Inversely related with each other,
4. Indirectly related with each other
4) Ionization energy is the ability to capture an electron:
1. False,
2. True
5) The second ionization energy of Mg is greater than second ionization energy of Al:
1. False,
2. True
6) Which group would generally have the lowest first ionization energy?
1. Transition Metals
2. Alkali Metals
3. Noble Gases
4. Alkaline Earth Metals
5. Halogens
7) Sulfur has a first ionization energy of 999.6 kJ/mol. Rubidium has a first ionization energy of 403 kJ/mol. What bond do they form when chemically combined?
1. Covalent
2. Polar Covalent
3. Ionic
8) Ionization energy, when supplied to an atom, results in a(n)
1. Anion and a proton
2. Cation and a proton
3. Cation and an Electron
4. Anion and an electron
9) Low first ionization energy is considered a property of
1. Metals
2. Nonmetals
10) Gallium has a first ionization energy of 578.8 kJ/mol, and calcium has a first ionization energy of 589.8 kJ/mol . According to periodic trends, one would assume that calcium, being to the left of gallium, would have the lower ionization energy. Explain, in terms of orbitals, why these numbers make sense.
Answers
1) Element D, 2) D, 3) B, 4) A, 5) B, 6) B, 7) C, 8) C, 9) A
10) Gallium has one electron in the 4p orbital, which can be expelled to reveal a more stable and full 4s orbital. Calcium, however, has a fully stable 4s orbital as its valence orbital, which you would have to disrupt to take an electron away from.
Ionization Energy
Ionization energy is the amount of energy required to remove an electron from a given chemical species. Reduction Potential is a measurement of the amount of force required for a chemical species to gain electrons. The ionization energy is a single step process and follows a constant trend by decreasing down a period within a group. Standard reduction potential is more closely related to a multistep process known as solvation. Solvation upon how easily a solute can break its own bonds, how easily the solute can break its bonds, and how much easily the solute can attract the solvent toward it self in order to form and ionic compound. is depend and upon the polarity of the molecules within the solution. As we will see, some species will have lower ionization energies as well as having higher reduction potential when compared to species of the same period.
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This page explains what first ionization energy is, and then looks at the way it varies around the Periodic Table - across periods and down groups. It assumes that you know about simple atomic orbitals, and can write electronic structures for simple atoms.
Definition: First Ionization Energy
The first ionization energy is the energy required to remove the most loosely held electron from one mole of neutral gaseous atoms to produce 1 mole of gaseous ions each with a charge of 1+.
This is more easily seen in symbol terms.
$X(g) \rightarrow X^+(g) + e^-$
It is the energy needed to carry out this change per mole of $X$. The state symbols - (g) - are essential. When you are talking about ionization energies, everything must be present in the gas state. i\Ionization energies are measured in kJ mol-1 (kilojoules per mole). They vary in size from 381 (which you would consider very low) up to 2370 (which is very high).
All elements have a first ionization energy - even atoms which do not form positive ions in test tubes. The reason that helium (1st I.E. = 2370 kJ mol-1) does not normally form a positive ion is because of the huge amount of energy that would be needed to remove one of its electrons.
Trends of first ionization energies
First ionization energy shows periodicity. That means that it varies in a repetitive way as you move through the Periodic Table. For example, look at the pattern from Li to Ne, and then compare it with the identical pattern from Na to Ar. These variations in first ionization energy can all be explained in terms of the structures of the atoms involved.
Factors affecting the size of ionization energy
ionization energy is a measure of the energy needed to pull a particular electron away from the attraction of the nucleus. A high value of ionization energy shows a high attraction between the electron and the nucleus. The size of that attraction will be governed by:
• The charge on the nucleus: The more protons there are in the nucleus, the more positively charged the nucleus is, and the more strongly electrons are attracted to it.
• The distance of the electron from the nucleus: Attraction falls off very rapidly with distance. An electron close to the nucleus will be much more strongly attracted than one further away.
• The number of electrons between the outer electrons and the nucleus: Consider a sodium atom, with the electronic structure 2,8,1. (There's no reason why you can't use this notation if it's useful!)
If the outer electron looks in towards the nucleus, it does not see the nucleus sharply. Between it and the nucleus there are the two layers of electrons in the first and second levels. The 11 protons in the sodium's nucleus have their effect cut down by the 10 inner electrons. The outer electron therefore only feels a net pull of approximately 1+ from the center. This lessening of the pull of the nucleus by inner electrons is known as screening or shielding.
• Whether the electron is on its own in an orbital or paired with another electron: Two electrons in the same orbital experience a bit of repulsion from each other. This offsets the attraction of the nucleus, so that paired electrons are removed rather more easily than you might expect.
Example 1: Helium vs. Lithium
Hydrogen has an electronic structure of 1s1. It is a very small atom, and the single electron is close to the nucleus and therefore strongly attracted. There are no electrons screening it from the nucleus and so the ionization energy is high (1310 kJ mol-1).
• Helium has a structure 1s2. The electron is being removed from the same orbital as in hydrogen's case. It is close to the nucleus and unscreened. The value of the ionization energy (2370 kJ mol-1) is much higher than hydrogen, because the nucleus now has 2 protons attracting the electrons instead of 1.
• Lithium is 1s22s1. Its outer electron is in the second energy level, much more distant from the nucleus. You might argue that that would be offset by the additional proton in the nucleus, but the electron does not feel the full pull of the nucleus - it is screened by the 1s2 electrons.
One can think of the electron as feeling a net 1+ pull from the center (3 protons offset by the two 1s2 electrons). If you compare lithium with hydrogen (instead of with helium), the hydrogen's electron also feels a 1+ pull from the nucleus, but the distance is much greater with lithium. Lithium's first ionization energy drops to 519 kJ mol-1 whereas hydrogen's is 1310 kJ mol-1.
The patterns in periods 2 and 3
Talking through the next 17 atoms one at a time would take ages. We can do it much more neatly by explaining the main trends in these periods, and then accounting for the exceptions to these trends. The first thing to realize is that the patterns in the two periods are identical - the difference being that the ionization energies in period 3 are all lower than those in period 2.
Explaining the general trend across periods 2 and 3
The general trend is for ionization energies to increase across a period. In the whole of period 2, the outer electrons are in 2-level orbitals - 2s or 2p. These are all the same sort of distances from the nucleus, and are screened by the same 1s2 electrons.
The major difference is the increasing number of protons in the nucleus as you go from lithium to neon. That causes greater attraction between the nucleus and the electrons and so increases the ionization energies. In fact the increasing nuclear charge also drags the outer electrons in closer to the nucleus. That increases ionization energies still more as you go across the period.
In period 3, the trend is exactly the same. This time, all the electrons being removed are in the third level and are screened by the 1s22s22p6 electrons. They all have the same sort of environment, but there is an increasing nuclear charge.
Why the drop between groups 2 and 3 (Be-B and Mg-Al)?
The explanation lies with the structures of boron and aluminum. The outer electron is removed more easily from these atoms than the general trend in their period would suggest.
Be 1s22s2 1st I.E. = 900 kJ mol-1
B 1s22s22px1 1st I.E. = 799 kJ mol-1
You might expect the boron value to be more than the beryllium value because of the extra proton. Offsetting that is the fact that boron's outer electron is in a 2p orbital rather than a 2s. 2p orbitals have a slightly higher energy than the 2s orbital, and the electron is, on average, to be found further from the nucleus. This has two effects.
• The increased distance results in a reduced attraction and so a reduced ionization energy.
• The 2p orbital is screened not only by the 1s2 electrons but, to some extent, by the 2s2 electrons as well. That also reduces the pull from the nucleus and so lowers the ionization energy.
The explanation for the drop between magnesium and aluminum is the same, except that everything is happening at the 3-level rather than the 2-level.
Mg 1s22s22p63s2 1st I.E. = 736 kJ mol-1
Al 1s22s22p63s23px1 1st I.E. = 577 kJ mol-1
The 3p electron in aluminum is slightly more distant from the nucleus than the 3s, and partially screened by the 3s2 electrons as well as the inner electrons. Both of these factors offset the effect of the extra proton.
Why the drop between groups 5 and 6 (N-O and P-S)?
Once again, you might expect the ionization energy of the group 6 element to be higher than that of group 5 because of the extra proton. What is offsetting it this time?
N 1s22s22px12py12pz1 1st I.E. = 1400 kJ mol-1
O 1s22s22px22py12pz1 1st I.E. = 1310 kJ mol-1
The screening is identical (from the 1s2 and, to some extent, from the 2s2 electrons), and the electron is being removed from an identical orbital. The difference is that in the oxygen case the electron being removed is one of the 2px2 pair. The repulsion between the two electrons in the same orbital means that the electron is easier to remove than it would otherwise be. The drop in ionization energy at sulfur is accounted for in the same way.
Trends in ionization energy down a group
As you go down a group in the Periodic Table ionization energies generally fall. You have already seen evidence of this in the fact that the ionization energies in period 3 are all less than those in period 2. Taking Group 1 as a typical example:
Why is the sodium value less than that of lithium? There are 11 protons in a sodium atom but only 3 in a lithium atom, so the nuclear charge is much greater. You might have expected a much larger ionization energy in sodium, but offsetting the nuclear charge is a greater distance from the nucleus and more screening.
Li 1s22s1 1st I.E. = 519 kJ mol-1
Na 1s22s22p63s1 1st I.E. = 494 kJ mol-1
Lithium's outer electron is in the second level, and only has the 1s2 electrons to screen it. The 2s1 electron feels the pull of 3 protons screened by 2 electrons - a net pull from the center of 1+. The sodium's outer electron is in the third level, and is screened from the 11 protons in the nucleus by a total of 10 inner electrons. The 3s1 electron also feels a net pull of 1+ from the center of the atom. In other words, the effect of the extra protons is compensated for by the effect of the extra screening electrons. The only factor left is the extra distance between the outer electron and the nucleus in sodium's case. That lowers the ionization energy.
Similar explanations hold as you go down the rest of this group - or, indeed, any other group.
Trends in ionization energy in a transition series
Apart from zinc at the end, the other ionization energies are all much the same. All of these elements have an electronic structure [Ar]3dn4s2 (or 4s1 in the cases of chromium and copper). The electron being lost always comes from the 4s orbital.
As you go from one atom to the next in the series, the number of protons in the nucleus increases, but so also does the number of 3d electrons. The 3d electrons have some screening effect, and the extra proton and the extra 3d electron more or less cancel each other out as far as attraction from the center of the atom is concerned.
The rise at zinc is easy to explain.
Cu [Ar]3d104s1 1st I.E. = 745 kJ mol-1
Zn [Ar]3d104s2 1st I.E. = 908 kJ mol-1
In each case, the electron is coming from the same orbital, with identical screening, but the zinc has one extra proton in the nucleus and so the attraction is greater. There will be a degree of repulsion between the paired up electrons in the 4s orbital, but in this case it obviously isn't enough to outweigh the effect of the extra proton.
Ionization Energies and Chemical reactivity
The lower the ionization energy, the more easily this change happens:
$X(g) \rightarrow X^+(g) + e^-$
You can explain the increase in reactivity of the Group 1 metals (Li, Na, K, Rb, Cs) as you go down the group in terms of the fall in ionization energy. Whatever these metals react with, they have to form positive ions in the process, and so the lower the ionization energy, the more easily those ions will form. The danger with this approach is that the formation of the positive ion is only one stage in a multi-step process.
For example, you would not be starting with gaseous atoms; nor would you end up with gaseous positive ions - you would end up with ions in a solid or in solution. The energy changes in these processes also vary from element to element. Ideally you need to consider the whole picture and not just one small part of it.
However, the ionization energies of the elements are going to be major contributing factors towards the activation energy of the reactions. Remember that activation energy is the minimum energy needed before a reaction will take place. The lower the activation energy, the faster the reaction will be - irrespective of what the overall energy changes in the reaction are.
The fall in ionization energy as you go down a group will lead to lower activation energies and therefore faster reactions.
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A Lewis Structure is a very simplified representation of the valence shell electrons in a molecule. It is used to show how the electrons are arranged around individual atoms in 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.
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.
How to draw Lewis Diagrams
The following is an example of how to draw the "best" Lewis structure for NO3- (learning by example).
1. First determine the total number of valence electrons in the molecule. This will be the sum of the group number a of all atoms plus the charge.
N 5
O (x 3) 18
charge 1
24
1. Draw a skeletal structure for the molecule which connects all atoms using only single bonds. The central atom will be the one that can form the greatest number of bonds and/or expand its octet. This usually means the atom lower and/or to the right in the Periodic Table, N in this case.
1. Now we need to add lone pairs of electrons. Of the 24 valence electrons available in NO3-, 6 were used to make the skeletal structure. Add lone pairs of electrons on the terminal atoms until their octet is complete or you run out of electrons.
1. If there are remaining electrons they can be used to complete the octet of the central atom. If you have run out of electrons you are required to use lone pairs of electrons from a terminal atom to complete the octet on the central atom by forming multiple bond(s). In this case the N is short 2 electrons so we can use a lone pair from the left most O atom to form a double bond and complete the octet on the N atom.
1. Now you need to determine the FORMAL CHARGES for all of the atoms. The formal charge is calculated by: (group number of atom) - (½ number of bonding electrons) - (number of lone pair electrons), i.e. see the figure below.
No Lewis structure is complete without the formal charges. 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.
• 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. Note: the octet can be expanded beyond 8 electrons but only for atoms in period 3 or below in the periodic table. In our present example N can not expand beyond 8 electrons so retains a formal charge of +1, but the S atom below can expand its octet.
1. You have determined the "best" Lewis structure (octets completed and lowest formal charges) for NO3-, but there are a number of ways to show this structure. Although it is most common to use a line to indicate a bonding pair of electrons they can be shown as electrons, see the left most image below. It is also common to show only the net charge on the ion rather than all of the formal charges, i.e. see the right most figure below.
Why are there different ways for the "same" Lewis structure? It depends what you want to show. While the most complete structure is more useful for the novice chemist, the simplest is quicker to draw and still conveys the same information for the experienced chemist. You should learn to recognize any of the possible Lewis structures.
Lewis Structures
A Lewis Structure is a very simplified representation of the valence shell electrons in a molecule. It is used to show how the electrons are arranged around individual atoms in 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.
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Learning Objectives
• To understand the difference between Ferromagnetism, paramagnetism and diamagnetism
• To identify if a chemical will be paramagnetic or diamagnetic when exposed to an external magnetic field
The magnetic moment of a system measures the strength and the direction of its magnetism. The term itself usually refers to the magnetic dipole moment. Anything that is magnetic, like a bar magnet or a loop of electric current, has a magnetic moment. A magnetic moment is a vector quantity, with a magnitude and a direction. An electron has an electron magnetic dipole moment, generated by the electron's intrinsic spin property, making it an electric charge in motion. There are many different magnetic behavior including paramagnetism, diamagnetism, and ferromagnetism.
An interesting characteristic of transition metals is their ability to form magnets. Metal complexes that have unpaired electrons are magnetic. Since the last electrons reside in the d orbitals, this magnetism must be due to having unpaired d electrons. The spin of a single electron is denoted by the quantum number \(m_s\) as +(1/2) or –(1/2). This spin is negated when the electron is paired with another, but creates a weak magnetic field when the electron is unpaired. More unpaired electrons increase the paramagnetic effects. The electron configuration of a transition metal (d-block) changes in a coordination compound; this is due to the repulsive forces between electrons in the ligands and electrons in the compound. Depending on the strength of the ligand, the compound may be paramagnetic or diamagnetic.
Ferromagnetism (Permanent Magnet)
Ferromagnetism is the basic mechanism by which certain materials (such as iron) form permanent magnets. This means the compound shows permanent magnetic properties rather than exhibiting them only in the presence of an external magnetic field (Figure \(1\)). In a ferromagnetic element, electrons of atoms are grouped into domains in which each domain has the same charge. In the presence of a magnetic field, these domains line up so that charges are parallel throughout the entire compound. Whether a compound can be ferromagnetic or not depends on its number of unpaired electrons and on its atomic size.
Ferromagnetism, the permanent magnetism associated with nickel, cobalt, and iron, is a common occurrence in everyday life. Examples of the knowledge and application of ferromagnetism include Aristotle's discussion in 625 BC, the use of the compass in 1187, and the modern-day refrigerator. Einstein demonstrated that electricity and magnetism are inextricably linked in his theory of special relativity.
Paramagnetism (Attracted to Magnetic Field)
Paramagnetism refers to the magnetic state of an atom with one or more unpaired electrons. The unpaired electrons are attracted by a magnetic field due to the electrons' magnetic dipole moments. Hund's Rule states that electrons must occupy every orbital singly before any orbital is doubly occupied. This may leave the atom with many unpaired electrons. Because unpaired electrons can spin in either direction, they display magnetic moments in any direction. This capability allows paramagnetic atoms to be attracted to magnetic fields. Diatomic oxygen, \(O_2\) is a good example of paramagnetism (described via molecular orbital theory). The following video shows liquid oxygen attracted into a magnetic field created by a strong magnet:
Diamagnetism (Repelled by Magnetic Field)
As shown in the video, molecular oxygen (\(\ce{O2}\)) is paramagnetic and is attracted to the magnet. In contrast, molecular nitrogen (\(\ce{N_2}\)) has no unpaired electrons and is diamagnetic; it is unaffected by the magnet. Diamagnetic substances are characterized by paired electrons, e.g., no unpaired electrons. According to the Pauli Exclusion Principle which states that no two electrons may occupy the same quantum state at the same time, the electron spins are oriented in opposite directions. This causes the magnetic fields of the electrons to cancel out; thus there is no net magnetic moment, and the atom cannot be attracted into a magnetic field. In fact, diamagnetic substances are weakly repelled by a magnetic field as demonstrated with the pyrolytic carbon sheet in Figure \(2\).
How to Tell if a Substance is Paramagnetic or Diamagnetic
The magnetic properties of a substance can be determined by examining its electron configuration: If it has unpaired electrons, then the substance is paramagnetic and if all electrons are paired, the substance is then diamagnetic. This process can be broken into three steps:
1. Write down the electron configuration
2. Draw the valence orbitals
3. Identify if unpaired electrons exist
4. Determine whether the substance is paramagnetic or diamagnetic
Example \(1\): Chlorine Atoms
Step 1: Find the electron configuration
For Cl atoms, the electron configuration is 3s23p5
Step 2: Draw the valence orbitals
Ignore the core electrons and focus on the valence electrons only.
Step 3: Look for unpaired electrons
There is one unpaired electron.
Step 4: Determine whether the substance is paramagnetic or diamagnetic
Since there is an unpaired electron, \(\ce{Cl}\) atoms are paramagnetic (albeit, weakly).
Example \(2\): Zinc Atoms
Step 1: Find the electron configuration
For Zn atoms, the electron configuration is 4s23d10
Step 2: Draw the valence orbitals
Step 3: Look for unpaired electrons
There are no unpaired electrons.
Step 4: Determine whether the substance is paramagnetic or diamagnetic
Because there are no unpaired electrons, \(\ce{Zn}\) atoms are diamagnetic.
Exercise \(1\)
1. How many unpaired electrons are found in oxygen atoms ?
2. How many unpaired electrons are found in bromine atoms?
3. Indicate whether boron atoms are paramagnetic or diamagnetic.
4. Indicate whether \(\ce{F^{-}}\) ions are paramagnetic or diamagnetic.
5. Indicate whether \(\ce{Fe^{2+}}\) ions are paramagnetic or diamagnetic.
Answer a
The O atom has 2s22p4 as the electron configuration. Therefore, O has 2 unpaired electrons.
Answer b
The Br atom has 4s23d104p5 as the electron configuration. Therefore, Br has 1 unpaired electron.
Answer c
The B atom has 2s22p1 as the electron configuration. Because it has one unpaired electron, it is paramagnetic.
Answer d
The \(\ce{F^{-}}\) ion has 2s22p6 as the electron configuration. Because it has no unpaired electrons, it is diamagnetic.
Answer e
The \(\ce{Fe^{2+}}\) ion has 3d6 as the electron configuration. Because it has 4 unpaired electrons, it is paramagnetic.
Molecular Polarity
Polarity is a physical property of compounds which relates other physical properties such as melting and boiling points, solubility, and intermolecular interactions between molecules. For the most part, there is a direct correlation between the polarity of a molecule and number and types of polar or non-polar covalent bonds which are present. In a few cases, a molecule may have polar bonds, but in a symmetrical arrangement which then gives rise to a non-polar molecule such as carbon dioxide.
Polarity - Dipole
Polarity results from the uneven partial charge distribution between various atoms in a compound. Atoms, such as nitrogen, oxygen, and halogens, that are more electronegative have a tendency to have partial negative charges. Atoms, such as carbon and hydrogen, have a tendency to be more neutral or have partial positive charges. Electrons in a polar covalent bond are unequally shared between the two bonded atoms, which results in partial positive and negative charges. The separation of the partial charges creates a dipole. The word dipole means two poles: the separated partial positive and negative charges. A polar molecule results when a molecule contains polar bonds in an unsymmetrical arrangement.
Nonpolar molecules are of two types. Molecules whose atoms have equal or nearly equal electronegativities have zero or very small dipole moments. A second type of nonpolar molecule has polar bonds, but the molecular geometry is symmetrical allowing the bond dipoles to cancel each other out.
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Having now revised the basics of trends across and down the Periodic Table, we can use the concepts of Effective Nuclear Charge and Electronegativity to discuss the factors that contribute to the types of bonds formed between elements.
Fajans' Rules
Rules formulated by Kazimierz Fajans in 1923, can be used to predict whether a chemical bond is expected to be predominantly ionic or covalent, and depend on the relative charges and sizes of the cation and anion. If two oppositely charged ions are brought together, the nature of the bond between them depends upon the effect of one ion on the other.
non-polar covalent polar covalent ionic
Fajan's rules for predicting whether a bond is predominantly Covalent or Ionic
Covalent Ionic
Small cation (< ~100 pm) Large cation (> ~100 pm)
Large anion Small anion
High charges Low charges
Although the bond in a compound like X+Y- may be considered to be 100% ionic, it will always have some degree of covalent character. When two oppositely charged ions (X+ and Y-) approach each other, the cation attracts electrons in the outermost shell of the anion but repels the positively charged nucleus. This results in a distortion, deformation or polarization of the anion. If the degree of polarization is quite small, an ionic bond is formed, while if the degree of polarization is large, a covalent bond results.
The ability of a cation to distort an anion is known as its polarization power and the tendency of the anion to become polarized by the cation is known as its polarizability. The polarizing power and polarizability that enhances the formation of covalent bonds is favoured by the following factors:
• Small cation: the high polarizing power stems from the greater concentration of positive charge on a small area. This explains why LiBr is more covalent than KBr (Li+ 90 pm cf. K+ 152 pm).
• Large anion: the high polarizability stems from the larger size where the outer electrons are more loosely held and can be more easily distorted by the cation. This explains why for the common halides, iodides, are the most covalent in nature (I- 206 pm).
• Large charges: as the charge on an ion increases, the electrostatic attractions of the cation for the outer electrons of the anion increases, resulting in the degree of covalent bond formation increasing.
Reminder. Large cations are to be found on the bottom left of the periodic table and small anions on the top right. The greater the positive charge, the smaller the cation becomes and the ionic potential is a measure of the charge to radius ratio.
The cation charge increases (size decreases) and on the right, the anion size increases, both variations leading to an increase in the covalency. Thus covalency increases in the order:
[Na+ Cl-, NaCl] < [Mg2+ 2(Cl)-, MgCl2] < [Al3+ 3(Cl)-, AlCl3] and [Al3+ 3(F)-, AlF3] < [Al3+ 3(Cl)-, AlCl3] < [Al3+ 3(Br)-, AlBr3]
Electronic configuration of the cation: for two cations of the same size and charge, the one with a pseudo noble-gas configuration (with 18 electrons in the outer-most shell) will be more polarizing than that with a noble gas configuration (with 8 electrons in the outermost shell). Thus zinc (II) chloride ( Zn(II) 1s2 2s2 2p6 3s2 3p6 3d10 and Cl- 1s2 2s2 2p6 3s2 3p6 ) is more covalent than magnesium chloride ( Mg(II) 1s2 2s2 2p6) despite the Zn2+ ion (74 pm) and Mg2+ ion (72 pm) having similar sizes and charges.
From an MO perspective, the orbital overlap disperses the charge on each ion and so weakens the electrovalent forces throughout the solid, this can be used to explain the trend seen for the melting points of lithium halides.
LiF = 870 °C, LiCl = 613 °C, LiBr = 547 °C, LiI = 446 °C
It is found that the greater the possibility of polarization, the lower is the melting point and heat of sublimation and the greater is the solubility in non-polar solvents.
Example \(1\):
The melting point of KCl is higher than that of AgCl though the crystal radii of Ag+ and K+ ions are almost the same.
Solution:
When the melting points of two compounds are compared, the one having the lower melting point is assumed to have the smaller degree of ionic character. In this case, both are chlorides, so the anion remains the same. The deciding factor must be the cation. (If the anions were different, then the answer could be affected by the variation of the anion.) Here the significant difference between the cations is in their electronic configurations. K+= [Ar] and Ag+ =[Kr] 4d10. This means a comparison needs to be made between a noble gas core and pseudo noble gas core, which as noted above holds that the pseudo noble gas would be the more polarizing.
Percentage of ionic character and charge distribution
Based on Fajan's rules, it is expected that every ionic compound will have at least some amount of covalent character. The percentage of ionic character in a compound can be estimated from dipole moments. The bond dipole moment uses the idea of electric dipole moment to measure the polarity of a chemical bond within a molecule. It occurs whenever there is a separation of positive and negative charges. The bond dipole μ is given by:
μ = δ d
A bond dipole is modeled as +δ - δ- with a distance d between the partial charges. It is a vector, parallel to the bond axis and by convention points from minus to plus (note that many texts appear to ignore the convention and point from plus to minus). The SI unit for an electric dipole moment is the coulomb-meter, (C m). This is thought to produce values too large to be practical on the molecular scale so bond dipole moments are commonly measured in Debye, represented by the symbol, D.
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. This value arises from (1.602 x 10-19 * 1 x 10-10) / 3.336 x 10-30 where D = 3.336 x 10-30 C m (or 1 C m = 2.9979 x 1029 D).
Typical dipole moments for simple diatomic molecules are in the range of 0 to 11 D (see Table below).
The % ionic character = μobserved / μcalculated (assuming 100% ionic bond) * 100 %
Example \(12\):
From the Table below the observed dipole moment of KBr is given as 10.41 D, (3.473 x 10-29 coulomb metre), 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.
Solution
dipole moment μ = q * e * d coulomb metre
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 μKBr = 1 * 1.602 x 10-19 * 2.82 x 10-10 = 4.518 x 10-29 Cm (13.54 D)
The observed μKBr = 3.473 x 10-29 Cm (10.41 D)
the % ionic character of KBr = 3.473 x 10-29/ 4.518 x 10-29 or 10.41 / 13.54 = 76.87% 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 meters, 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 \(3\):
For HI, calculate the % of ionic character given a bond length = 161 pm and an observed dipole moment 0.44 D.
Solution
To calculate μ considering it as a 100% ionic bond
μ = 1 * 1.602 x 10-19 * 1.61 x 10-10 / (3.336 x 10-30) = 7.73 D
the % ionic character = 0.44/7.73 * 100 = 5.7%
The calculated % ionic character is only 5.7% and the % covalent character is (100 - 5.7) = 94.3%. The ionic character arises from the polarizability and polarizing effects of H and I. Similarly, knowing the bond length and observed dipole moment of HCl, the % ionic character can be found to be 18%. Thus it can be seen that while HI is essentially covalent, HCl has significant ionic character.
Note that by this simplistic definition, to achieve 100 % covalent character a compound must have an observed dipole moment of zero. Whilst not strictly true for heteronuclear molecules it does provide a simple qualitative method for predicting the bond character.
Bond character based on electronegativity differences
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.
Electronegativity difference, ΔχP Bond
Δχ < 0.4 covalent
0.4 < Δχ < 1.7 polar covalent
Δχ > 1.7 ionic
Linus Pauling proposed an empirical relationship which relates the percent ionic character in a bond to the electronegativity difference.
percent ionic character= (1-e-(Δχ/2)^2 )* 100
This is shown as the curve in red below and is compared to the values for some diatomic molecules calculated from observed and calculated dipole moments.
diatomic Δ χ %ionic bond dist /pm Obs μ /D Calc μ /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
London dispersion forces
Intermolecular forces are the attractive forces between molecules without which all substances would be gases. The various types of these interactions span large differences in energy and for the halogens and interhalogens are generally quite small. The forces involved in these cases are called London dispersion forces (after Fritz Wolfgang London, 1900-1954). They are derived from momentary oscillations of electron charge in atoms and hence are present between all particles (atoms, ions and molecules).
The ease with which the electron cloud of an atom can be distorted to become asymmetric is termed the molecule's polarizability. The greater the number of electrons an atom has, the farther the outer electrons will be from the nucleus, and the greater the chance for them to shift positions within the molecule. This means that larger nonpolar molecules tend to have stronger London dispersion forces. This is evident when considering the diatomic elements in Group 17, the Halogens. All of these homo-nuclear diatomic elements are nonpolar, covalently bonded molecules. Descending the group, fluorine and chlorine are gases, bromine is a liquid, and iodine is a solid. For nonpolar molecules, the farther you go down the group, the stronger the London dispersion forces.
To picture how this occurs, compare the situation 1) where the electrons are evenly distributed and then consider 2) an instantaneous dipole that would arise from an uneven distribution of electrons on one side of the nucleus. When two molecules are close together, the instantaneous dipole of one molecule can induce a dipole in the second molecule. This results in synchronised motion of the electrons and an attraction between them. 3) when this effect is multiplied over numerous molecules the overall result is that the attraction keeps these molecules together, and for diiodine is sufficient to make this a solid.
On average the electron cloud for molecules can be considered to be spherical in shape. When two non-polar molecules approach, attractions or repulsions between the electrons and nuclei can lead to distortions in their electron clouds (i.e. dipoles are induced). When more molecules interact these induced dipoles lead to intermolecular attraction.
The changes seen in the variation of MP and BP for the dihalogens and binary interhalogens can be attributed to the increase in the London dispersion forces of attraction between the molecules. In general they increase with increasing atomic number.
Anomalous behavior of the 2nd row elements: Li, Be, B, C, N, O, F
For the elements in the 2nd row, as the atomic number increases, the atomic radius of the elements decreases, the electronegativity increases, and the ionization energy increases.
The 2nd row has two metals (lithium and beryllium), making it the least metallic period and it has the most nonmetals, with four. The elements in the 2nd row often have the most extreme properties in their respective groups; for example, fluorine is the most reactive halogen, neon is the most inert noble gas, and lithium is the least reactive alkali metal.
These differences in properties with the subsequent rows are a result of:
• the smaller size of the atoms
• an outer shell with a maximum of 8 electrons (2s and 2p) and an underlying shell with just 2 electrons
• no acessible d-orbitals - energy too high for use in bonding
Apart from the 2nd row (ignoring H/He 1st row) the later rows all end with inert gases but these do not have completed quantum levels. The 2nd row elements in general can only use the 2s and 2p electrons for bonding restricting the total number of bonds to 4.
So N is not expected to have more than 4 bonds and 3 is common, while for P 5 and 6 bonded species are quite common.
Reactivity of metals and metalloids
For Lithium, compared to other alkali metals
Reaction with water:
Li reacts slowly with water at 25 °C
Na reacts violently and K in flames
2M(s) + 2H2O(l) → 2M+(aq) + 2OH- + H2(g)
In general Li, Be, B, C, N, O, F are less reactive towards water than their heavier congeners.
Reaction with oxygen:
In conditions of excess oxygen, only Li forms a simple oxide, Li2O. Other metals form peroxides and superoxides
Reaction with nitrogen:
Li reacts directly with N2 to form Li3N
6Li(s) + N2 (g) → 2Li3N(s).
No other alkali metal reacts with N2
Solubility:
LiF, LiOH and Li2CO3 are less soluble than the corresponding Na and K compounds
For Beryllium compared to the other alkaline earth metals:
With water:
All Group 2 metals except Be, react with water
M(s) + 2H2O(l) → M2+(aq)+ 2OH-(aq) + H2 (g)
With oxygen (air):
Be only reacts with air above 600 °C if it is finely powdered. The BeO that is formed is amphoteric (other Group 2 oxides are basic).
Of the Group 2 elements only Be reacts with NaOH or KOH to liberate H2 and form [Be(OH)4]2-.
Li and Be are metals but are less conducting than the higher members of Group 1 and 2 elements due to their high IEs (electrons are close to nucleus).
Ionization of Boron to B3+ requires a large input of energy and B adopts a covalent polymeric structure with semi-metallic properties.
The other elements of Group 14 become increasingly metallic as the group is descended due to the decrease in ionization energies.
Crystalline Boron is chemically inert - unaffected by boiling HCl and only slowly oxidized by hot concentrated HNO3 when finely powdered.
Covalent character
Li+ and Be2+ are small and have strong polarizing abilities. Their compounds are more covalent than those of the heavier elements in their groups.
BeCl2 is covalent while MCl2 (M = Mg-Ba) are ionic. The conductivity of fused beryllium chloride is only 1/1000 that of sodium chloride under similar conditions.
Catenation
Catenation is the linkage of atoms of the same element into longer chains. Catenation occurs most readily in carbon, which forms covalent bonds with other carbon atoms to form longer chains and structures. This is the reason for the presence of the vast number of organic compounds in nature.
The ability of an element to catenate is primarily based on the bond energy of the element to itself, which decreases with more diffuse orbitals (those with higher azimuthal quantum number) overlapping to form the bond. Hence, carbon, with the least diffuse valence shell 2p orbital is capable of forming longer p-p sigma bonded chains of atoms than heavier elements which bond via higher valence shell orbitals.
Hetero-catenation is quite common in Inorganic Chemistry. Phosphates and silicates with P-O-P-O and Si-O-Si-O linkages are examples of this.
Multiple Bonds
C, N and O are able to form multiple bonds (double and/or triple). In Group 14, C=C double bonds are stable (134 pm) but Si=Si double bonds (227 pm) are uncommon. The diagram below shows how multiple bonds are formed involving π overlap of 2p orbitals. By comparison the 3p orbitals of the corresponding third row elements Si, P, and S are more diffuse and the longer bond distances expected for these larger atoms would result in poor π overlap.
C=C bond length = 134 pm and Si=Si bond length = 227 pm
Oxidizing ability of oxygen and fluorine
Due to the high electron affinities and electronegativities of oxygen and fluorine, they tend to form strong ionic bonds with other elements. They even react with noble gases to form compounds such as XeO3, XeO4, XeF4 and XeF6. In 1962 Neil Bartlett at the University of British Columbia reacted platinum hexafluoride and xenon, in an experiment that demonstrated the chemical reactivity of the noble gases. He discovered the mustard yellow compound, xenon hexafluoroplatinate, which is perhaps now best formulated as a mixture of species, [XeF+][PtF5]-, [XeF+][Pt2F11]-, and [Xe2F3]+[PtF6]-.
A few hundred compounds of other noble gases have subsequently been discovered: in 1962 for radon, radon difluoride (RnF2), and in 1963 for krypton, krypton difluoride (KrF2). The first stable compound of argon was reported in 2000 when argon fluorohydride (HArF) was formed at a temperature of 40 K (-233.2 °C). Neutral compounds in which helium and neon are involved in chemical bonds have still not been formed.
Noble gas compounds have already made an impact on our daily lives. XeF2 is a strong fluorinating agent and has been used to convert uracil to 5-fluorouracil, one of the first anti-tumor agents.
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• Dialysis
Dialysis is the separation of colloids from dissolved ions or molecules of small dimensions, or crystalloid, in a solution. Dialysis is a process that is like osmosis. Osmosis is the process in which there is a diffusion of a solvent through a semipermeable membrane.
• Recrystallization
Recrystallization, also known as fractional crystallization, is a procedure for purifying an impure compound in a solvent. The method of purification is based on the principle that the solubility of most solids increases with increased temperature. This means that as temperature increases, the amount of solute that can be dissolved in a solvent increases.
Case Studies
Dialysis is the separation of colloids from dissolved ions or molecules of small dimensions, or crystalloid, in a solution. A colloid is any substance that is made of particles that are of an extremely small size: larger than atoms but generally have the size of 10-7 cm ranging to 10-3 cm. A crystalloid is a substance that has some or all of the properties of a crystal or a substance that forms a true solution and diffuses through a membrane by dialysis. Dialysis is a process that is like osmosis. Osmosis is the process in which there is a diffusion of a solvent through a semipermeable membrane.
Introduction
In 1861, chemist Thomas Graham (how developed Graham's Law) used the process of dialysis, a process used to separate colloidal particles from dissolved ions or molecules. Dialysis is possible because of the unequal rates of diffusion through a semipermeable membrane. A semipermeable membrane is a membrane that lets some molecules to pass through it while not letting others (Figure \(1\)). Examples of semipermeable membranes include parchment and cellophane.
Figure \(1\): Graphic showing the diffusion of solutes across a membrane during dialysis. from Wikipedia (Potcherboy)
Another way to think of a semipermeable membrane is to think of a net like object that traps larger objects, but lets smaller object pass through because they can pass through the holes in the net.
Video \(1\): Osmotic Pressure Dialysis Tubing
When a colloidal mixture is places in a semipermeable membrane, which is then placed in an aqueous solution or pure water, dissolved ions and small molecules are allowed to pass through this membrane. This causes colloidal particles to stay in the membrane, because these particles are unable to pass through the small pores of the membrane.
The Rate of Dialysis
Dialysis is not a quick process; the rate of dialysis depends on the speed of the unequal diffusion rates between the crystalloids and the colloids and the differences in particle size. The rate of dialysis can be changed through heating, or if the crystalloids are charged, then applying an electric field, called electrodialysis. Electrodialysis is the type of dialysis in which electrodes are placed on the sides of the membrane. In this way, positive ions can pass through one side of this membrane while the negatively charged ions can pass through the other side of the membrane. This causes acceleration in the process of dialysis.
Hemodialysis
Hemodialysis is a method in which kidney failure is treated with the process of dialysis. In hemodialysis, blood is removed, purified through dialysis, and returned to the bloodstream. In kidney failure, there is a retention of salts and water, urea, and metabolic acids. The patient is then connected to a dialysis machine, which is also called a hemodialyzer. The blood flows through small channels made of semipermeable membranes (Figure \(2\)). The dissolved substances like urea and salts pass through a sterile solution. Compounds like sugar and amino acids are added to the sterile solution. The dialysis solution is on the other side of the membranes, and the molecules flow through the membranes. The molecules diffuse from a higher concentration to low concentration area. The concentrations of molecules needed to be removed from the blood are zero in the dialysis fluid.
The process of hemodialysis helps many patients who have kidney failure because a person who suffers from kidney failure are at great risk, because someone who has complete kidney failure will need a kidney transplant within two weeks, or else he/she will face death. Between the time that the person finds a suitable kidney to be transplanted, the hemodialyzer comes into great help in facing the fight against death.
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Recrystallization, also known as fractional crystallization, is a procedure for purifying an impure compound in a solvent. The method of purification is based on the principle that the solubility of most solids increases with increased temperature. This means that as temperature increases, the amount of solute that can be dissolved in a solvent increases.
Introduction
An impure compound is dissolved (the impurities must also be soluble in the solvent), to prepare a highly concentrated solution at a high temperature. The solution is cooled. Decreasing the temperature causes the solubility of the impurities in the solution and the substance being purified to decrease. The impure substance then crystallizes before the impurities- assuming that there was more impure substance than there were impurities. The impure substance will crystallize in a purer form because the impurities won't crystallize yet, therefore leaving the impurities behind in the solution. A filtration process must be used to separate the more pure crystals at this point. The procedure can be repeated. Solubility curves can be used to predict the outcome of a recrystallization procedure.
Note
Recrystallization works best when
1. the quantity of impurities is small
2. the solubility curve of the desired solute rises rapidly with temperature
The slower the rate of cooling, the larger the crystals are that form. The disadvantage of recrystallization is that it takes a long time. Also, it is very important that the proper solvent is used. This can only be determined by trial and error, based on predictions and observations. The solution must be soluble at high tempratures and insoluble at low temperatures. The advantage or recrystallization is that, when carried out correctly, it is a very effective way of obtaining a pure sample of some product, or precipitate.
Procedure
These are the important steps to the recrsytallization process.
1. Dissolve the solute in the solvent: Add boiling solvent to a beaker containing the impure compound. Heat the beaker and keep adding solvent until the solute is completely dissolved. See Figure 1
2. Cool the Solution: The solution is cooled in open air first, and then cooled in an ice bath. Slow cooling often leads to purer crystals. Crystals should form on the bottom of the beaker. The process of "seeding" can be used to aid the formation of crystals- this means adding a pure crystal of the compound. The pure crystal forms a surface for the solute to crystallize upon. See Figure 2
3. Obtain the crystals from the solute: The purer crystals precipitated from the solute are the desirable part of the mixture, and so they must be removed from the solvent. The process used for isolating the crystals that remain in the beaker still is called vacuum filtration. Suction is created using an aspirator, and whatever remains in the beaker is poured though a Buchner funnel. If for some reason there are no crystals visible, a gravity filtration can be performed. Activated carbon is added to the solution, the mixture is boiled, and a funnel system is used to transfer the new mixture to a new beaker of boiling solvent. Filter paper is used in the funnel to remove excess carbon. After this mixture cools slowly there should be large crystals present.
4. Dry the resulting crystals: The crystals are dried by leaving them in the aspirator and then by removing them to a glass dish to wait a while longer. The purity of the crystals can be tested by performing a "melting point determination".
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Colligative properties are the physical changes that result from adding solute to a solvent. Colligative Properties depend on how many solute particles are present as well as the solvent amount, but they do NOT depend on the type of solute particles, although do depend on the type of solvent.
• Anomalous Colligative Properties (Real Solutions)
Anomalous colligative properties are colligative properties that deviate from the ideal colligative behavior. It is quantified by the introduction of the Van't Hoff factor.
• Boiling Point Elevation
The boiling points of solutions are higher than that of the pure solvent. This effect is directly proportional to the molality of the solute.
• Freezing Point Depression
The freezing points of solutions are all lower than that of the pure solvent. The freezing point depression is directly proportional to the molality of the solute.
• Osmotic Pressure
The osmotic pressure of a solution is the pressure difference needed to stop the flow of solvent across a semipermeable membrane. The osmotic pressure of a solution is proportional to the molar concentration of the solute particles in solution.
• Vapor Pressure Lowering
The vapor pressure of a solvent in a solution is always lower than the vapor pressure of the pure solvent. The vapor pressure lowering is directly proportional to the mole fraction of the solute.
Colligative Properties
Anomalous colligative properties are colligative properties that deviate from the norm. Chemist Jacobus van 't Hoff was the first to describe anomalous colligative properties, but it was Svante Arrhenius who succeeded in explaining anomalous values of colligative properties.
Introduction
Colligative Properties are the properties of solutions that rely only on the number (concentration) of the solute particles, and not on the identity/type of solute particles, in an ideal solution (e.g., vapor pressure lowering, freezing point depression, boiling point elevation and osmotic pressure). There is a direct relationship between the concentration and the effect that is recorded. Therefore, the colligative properties are helpful when characterizing the nature of a solute after it is dissolved in a solvent.
There are some solutes that produce a greater effect on colligative properties than what is expected. Arrhenius explained this by using the following equation:
$\Delta{T_f} = -K_f \times m = -1.86 \; ^{\circ}C \; m^{-1} \times 0.0100 \; m = -0.0186 \; ^{\circ}C \tag{1}$
The expected freezing point for this solution would be: - 0.0186 $^{\circ}C$. Lets say that this solution was that of urea, the measured freezing point is close to -0.0186 $^{\circ}C$. If it were to be a solution of NaCl, then the measured freezing point would then be -0.0361 $^{\circ}C$. According to Van't Hoff the factor, $i$ is the ratio of the measured value of a colligative property to that of the expected value if the solute is a nonelectrolyte. Now, for 0.0100 m NaCl, it would be:
$i = \dfrac{\text{measured} \; \Delta{T_f}}{\text{expected} \; \Delta{T_f}} \tag{2}$
For the solute urea $i$ = 1. For a strong electrolyte like NaCl that produces 2 moles of ions in a solution/ mole of solute dissolved, the effect on the freezing point depression would be expected to be twice as much as that for a nonelectrolyte. The expected $i$ = 2. This leads the colligative properties to be rewritten as demonstrated in the table below.
Original Rewritten
$\Pi = M RT$ $\pi = i M RT$
$\Delta{T_f} = -K_f m$ $\Delta{T_f} = -i K_f m$
$\Delta{T_f} = -K_b m$ $\Delta{T_f} = i K_b m$
We can just substitute 1 for $i$ for nonelectrolytes and for strong electrolytes, find the value of $i$.
Resources
1. Petrucci, Harwood, Herring, Madura. General Chemistry, Principles & Modern Applications 9th Edition. Upper Saddle River, NJ: Pearson Education, Inc., 2007.
2. Zumdahl, Steven S. Chemistry 4th Edition. New York: Houghton Mifflin,1997.
Boiling Point Elevation
The colligative properties of a solution depend on the relative numbers (concentration) of solute and solvent particles, they do not depend on the nature of the particles. Colligative properties change in proportion to the concentration of the solute particles. We distinguish between four colligative properties: vapor pressure lowering, freezing point depression, boiling point elevation, and osmotic pressure. All four colligative properties fit the relationship
property = solute concentration x constant
Property Symbol Solute Concentration Proportionality Constant
Vapor pressure $\Delta P$ mole fraction Po (vapor pressure of pure solvent)
Boiling Point $\Delta{T_b}$ molal Kb (boiling point constant)
Freezing Point $\Delta{T_f}$ molal Kf (freezing point constant)
Osmotic Pressure $P$ molar RT
The determination of colligative properties allows us to determine the concentration of a solution and calculate molar masses of solutes
Boiling Point Elevation
The boiling points of solutions are all higher than that of the pure solvent. Difference between the boiling points of the pure solvent and the solution is proportional to the concentration of the solute particles:
$\Delta{T_b} = T_b (solution) - T_b (solvent) = K_b \times m$
where $\Delta{T_b}$ is the boiling point elevation, $K_b$ is the boiling point elevation constant, and m is the molality (mol/kg solvent) of the solute.
Exercise
A solution is prepared when 1.20 g of a compound is dissolved in 20.0 g of benzene. The boiling point of the solution is 80.94 oC.
• What is the boiling point of pure benzene?
• What is the molality of the solution?
• What is the molar mass of the compound?
Answer
• 1.8 x 102 g/mol)
Questions
• Explain how a non-volatile solute lowers the vapor pressure, raises the boiling point and lowers the freezing point of a solvent.
• Describe the process of osmosis and reverse osmosis
• Explain why ocean water is not fit for consumption by humans
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Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all lower than that of the pure solvent and is directly proportional to the molality of the solute.
\begin{align*}\Delta{T_f} &= T_f(solvent) - T_f (solution) \[4pt] &= K_f \times m \end{align*}
where $\Delta{T_f}$ is the freezing point depression, $T_f$ (solution) is the freezing point of the solution, $T_f$ (solvent) is the freezing point of the solvent, $K_f$ is the freezing point depression constant, and m is the molality.
Introduction
Nonelectrolytes are substances with no ions, only molecules. Strong electrolytes, on the other hand, are composed mostly of ionic compounds, and essentially all soluble ionic compounds form electrolytes. Therefore, if we can establish that the substance that we are working with is uniform and is not ionic, it is safe to assume that we are working with a nonelectrolyte, and we may attempt to solve this problem using our formulas. This will most likely be the case for all problems you encounter related to freezing point depression and boiling point elevation in this course, but it is a good idea to keep an eye out for ions. It is worth mentioning that these equations work for both volatile and nonvolatile solutions. This means that for the sake of determining freezing point depression or boiling point elevation, the vapor pressure does not effect the change in temperature. Also, remember that a pure solvent is a solution that has had nothing extra added to it or dissolved in it. We will be comparing the properties of that pure solvent with its new properties when added to a solution.
Adding solutes to an ideal solution results in a positive $ΔS$, an increase in entropy. Because of this, the newly altered solution's chemical and physical properties will also change. The properties that undergo changes due to the addition of solutes to a solvent are known as colligative properties. These properties are dependent on the number of solutes added, not on their identity. Two examples of colligative properties are boiling point and freezing point: due to the addition of solutes, the boiling point tends to increase, and freezing point tends to decrease.
The freezing point and boiling point of a pure solvent can be changed when added to a solution. When this occurs, the freezing point of the pure solvent may become lower, and the boiling point may become higher. The extent to which these changes occur can be found using the formulas:
$\Delta{T}_f = -K_f \times m$
$\Delta{T}_b = K_b \times m$
where $m$ is the solute molality and $K$ values are proportionality constants; ($K_f$ and $K_b$ for freezing and boiling, respectively).
Molality
Molality is defined as the number of moles of solute per kilogram solvent. Be careful not to use the mass of the entire solution. Often, the problem will give you the change in temperature and the proportionality constant, and you must find the molality first in order to get your final answer.
If solving for the proportionality constant is not the ultimate goal of the problem, these values will most likely be given. Some common values for $K_f$ and $K_b$ respectively, are in Table $1$:
Table $1$: Ebullioscopic and cryoscopic constants for select solvents. Note that the nature of the solute does not affect colligative property relations.
Solvent $K_f$ $K_b$
Water 1.86 .512
Acetic acid 3.90 3.07
Benzene 5.12 2.53
Phenol 7.27 3.56
The solute, in order for it to exert any change on colligative properties, must fulfill two conditions. First, it must not contribute to the vapor pressure of the solution, and second, it must remain suspended in the solution even during phase changes. Because the solvent is no longer pure with the addition of solutes, we can say that the chemical potential of the solvent is lower. Chemical potential is the molar Gibb's energy that one mole of solvent is able to contribute to a mixture. The higher the chemical potential of a solvent is, the more it is able to drive the reaction forward. Consequently, solvents with higher chemical potentials will also have higher vapor pressures.
The boiling point is reached when the chemical potential of the pure solvent, a liquid, reaches that of the chemical potential of pure vapor. Because of the decrease in the chemical potential of mixed solvents and solutes, we observe this intersection at higher temperatures. In other words, the boiling point of the impure solvent will be at a higher temperature than that of the pure liquid solvent. Thus, boiling point elevation occurs with a temperature increase that is quantified using
$\Delta{T_b} = K_b m$
where $K_b$ is known as the ebullioscopic constant and $m$ is the molality of the solute.
Freezing point is reached when the chemical potential of the pure liquid solvent reaches that of the pure solid solvent. Again, since we are dealing with mixtures with decreased chemical potential, we expect the freezing point to change. Unlike the boiling point, the chemical potential of the impure solvent requires a colder temperature for it to reach the chemical potential of the pure solid solvent. Therefore, a freezing point depression is observed.
Example $1$
2.00 g of some unknown compound reduces the freezing point of 75.00 g of benzene from 5.53 to 4.90 $^{\circ}C$. What is the molar mass of the compound?
Solution
First we must compute the molality of the benzene solution, which will allow us to find the number of moles of solute dissolved.
\begin{align*} m &= \dfrac{\Delta{T}_f}{-K_f} \[4pt] &= \dfrac{(4.90 - 5.53)^{\circ}C}{-5.12^{\circ}C / m} \[4pt] &= 0.123 m \end{align*}
\begin{align*} \text{Amount Solute} &= 0.07500 \; kg \; benzene \times \dfrac{0.123 \; m}{1 \; kg \; benzene} \[4pt] &= 0.00923 \; m \; solute \end{align*}
We can now find the molecular weight of the unknown compound:
\begin{align*} \text{Molecular Weight} =& \dfrac{2.00 \; g \; unknown}{0.00923 \; mol} \[4pt] &= 216.80 \; g/mol \end{align*}
The freezing point depression is especially vital to aquatic life. Since saltwater will freeze at colder temperatures, organisms can survive in these bodies of water.
Exercise $1$
Benzophenone has a freezing point of 49.00oC. A 0.450 molal solution of urea in this solvent has a freezing point of 44.59oC. Find the freezing point depression constant for the solvent.
Answer
$9.80\,^oC/m$
Applications
Road salting takes advantage of this effect to lower the freezing point of the ice it is placed on. Lowering the freezing point allows the street ice to melt at lower temperatures. The maximum depression of the freezing point is about −18 °C (0 °F), so if the ambient temperature is lower, $\ce{NaCl}$ will be ineffective. Under these conditions, $\ce{CaCl_2}$ can be used since it dissolves to make three ions instead of two for $\ce{NaCl}$.
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Introduction
Semipermiable membranes do not let the solute pass through (Think of the sugar example). A solvent will move to the side that is more concentrated to try to make each side more similar! Since there is a flow of solvents, the height of each side changes, which is osmotic pressure. When we work with aqueous solutions, we use mm of H2O to describe the difference.
Osmosis is the diffusion of a fluid through a semipermeable membrane. When a semipermeable membrane (animal bladders, skins of fruits and vegetables) separates a solution from a solvent, then only solvent molecules are able to pass through the membrane. The osmotic pressure of a solution is the pressure difference needed to stop the flow of solvent across a semipermeable membrane. The osmotic pressure of a solution is proportional to the molar concentration of the solute particles in solution.
$\Pi = i \dfrac{n}{V}RT = i M RT \label{eq1}$
where
• $\Pi$ is the osmotic pressure,
• $R$ is the ideal gas constant (0.0821 L atm / mol K),
• $T$ is the temperature in Kelvin,
• $i$ is the van 't Hoff factor
• $n$ is the number of moles of solute present,
• $V$ is the volume of the solution, and
• $M$ is the molar concentration of added solute (the $i$ factor accounts for how many species in solution are generated)
Exercise $1$
Calculate molarity of a sugar solution in water (300 K) has osmotic pressure of 3.00 atm.
Answer
Since it is sugar, we know it doesn't dissociate in water, so $i$ is 1. Then we use Equation \ref{eq1} directly
$M = \dfrac{\Pi}{RT} = \dfrac{3.00\, atm}{(0.0821\, atm.L/mol.K)(300\,K)} = 0.122\,M \nonumber$
Exercise $2$
Calculate osmotic pressure for 0.10 M $\ce{Na3PO4}$ aqueous solution at 20°C.
Answer
Since $\ce{Na3PO4}$ ionizes into four particles (3 Na+1 + $PO_4^{-3}$), then $i = 4$. We can then calculate the osmotic pressure via Equation \ref{eq1}
$\Pi = iMRT = (0.40)(0.0821)(293) = 9.6\, atm \nonumber$
Exercise $3$
Hemoglobin is a large molecule that carries oxygen in human blood. A water solution that contains 0.263 g of hemoglobin (Hb) in 10.0 mL of solution has an osmotic pressure of 7.51 torr at $25 ^oC$. What is the molar mass of the hemoglobin?
Answer
$6.51 \times 10^4 \; g/mol$
Vapor Pressure Lowering
The vapor pressure of a solvent in a solution is always lower than the vapor pressure of the pure solvent. The vapor pressure lowering is directly proportional to the mole fraction of the solute. This is Raoult's Law
$P_{solution} = \chi_{solvent}P^o_{solvent} \label{RLaw}$
where $P^o_{solvent}$ is the vapor pressure of the pure solvent and $\chi_{solvent}$ is the mole fraction of the solvent. Since this is a two-component system (solvent and solute), then
$\chi_{solvent} + \chi_{solute} = 1$
where $\chi_{solute}$ is the mole fraction of the solvent or solute. The change in vapor pressure ($\Delta P$) can be expressed
$\Delta P = P_{solution} - P^o_{solvent} = \chi_{solvent}P^o_{solvent} - P^o_{solvent}$
or
$\Delta P = ( \chi_{solvent} - 1) P^o_{solvent} = \chi_{solute} P^o_{solvent} \label{DP}$
Example $\PageIndex{1A}$: Non-electrolyte Solutions
Calculate the vapor pressure of a solution made by dissolving 50.0 g glucose, $C_6H_{12}O_6$, in 500 g of water. The vapor pressure of pure water is 47.1 torr at 37°C
Solution
To use Raoult's Law (Equation $\ref{RLaw}$), we need to calculate the mole fraction of water (the solvent) in this sugar-water solution.
$\chi_{solvent} = \dfrac{ \text{moles of water}}{\text{moles of solute} + \text{moles of solvent}} \nonumber$
$\chi_{solvent} = \dfrac{ n_{water}}{ n_{glucose} + n_{water} } \nonumber$
The molar mass of glucose is 180.2 g/mol and of water is 18 g/mol. So
$n_{water} = \dfrac{500\,g}{18\,g /mol} = 27.7 \,mol \nonumber$
and
$n_{glucose} = \dfrac{50\,g}{180.2\,g /mol} = 0.277 \,mol \nonumber$
and
$\chi_{solvent} = \dfrac{ 27.7 \,mol}{ 0.277 \,mol + 27.7 \,mol } = 0.99 \nonumber$
Note that this still relatively dilute. The pressure of the solution is then calculated via Raoult's Law (Equation $\ref{RLaw}$):
$P_{solution} = 0.99 \times 47.1 = 46.63 \, torr \nonumber$
not much of a change at all.
Example $\PageIndex{1B}$: Electrolyte Solutions
Calculate the vapor pressure of a solution made by dissolving 50.0 g CaCl2, $C_6H_{12}O_6$, in 500 g of water. The vapor pressure of pure water is 47.1 torr at 37°C
Solution
To use Raoult's Law (Equation $\ref{RLaw}$), we need to calculate the mole fraction of water (the solvent) in this salt-water solution.
$\chi_{solvent} = \dfrac{ \text{moles of water}}{\text{moles of solute} + \text{moles of solvent}} \nonumber$
$\chi_{solvent} = \dfrac{ n_{water}}{ n_{solutes} + n_{water} } \nonumber$
The molar mass of $CaCl_2$ is 111 g/mol and of water is 18 g/mol. So
$n_{water} = \dfrac{500\,g}{18\,g /mol} = 27.7 \,mol \nonumber$
and
$n_{solutes} = \dfrac{50\,g}{111 \,g /mol} = 0.45 \,mol \nonumber$
but this is really:
• $n_{Ca^+} = 0.45 \,mol$
• $n_{Cl^-} = 0.9 \,mol$
and
$\chi_{solvent} = \dfrac{ 27.7 \,mol}{ 0.45 \,mol + 0.9 \,mol + 27.7 \,mol } = 0.953 \nonumber$
Note that this still relatively dilute. The pressure of the solution is then calculated via Raoult's Law (Equation $\ref{RLaw}$):
$P_{solution} = 0.953 \times 47.1 = 44.88\, torr \nonumber$
A bigger change that the glucose in Example $\PageIndex{1A}$.
Exercise $1$: Electrolyte Solutions
At 25oC the vapor pressure of pure benzene is 93.9 torr. When a non-volatile solvent is dissolved in benzene, the vapor pressure of benzene is lowered to 91.5 torr. What is the concentration of the solute and the solvent, expressed in mole fraction?
Answer
Vapor pressure lowering $\Delta P= 2.4\; torr$ with $\chi_{solute} = 0.026$.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Solutions_and_Mixtures/Colligative_Properties/Osmotic_Pressure.txt
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A colloid is one of the three primary types of mixtures, with the other two being a solution and suspension. A colloid is a mixture that has particles ranging between 1 and 1000 nanometers in diameter, yet are still able to remain evenly distributed throughout the solution. These are also known as colloidal dispersions because the substances remain dispersed and do not settle to the bottom of the container. In colloids, one substance is evenly dispersed in another. The substance being dispersed is referred to as being in the dispersed phase, while the substance in which it is dispersed is in the continuous phase.
To be classified as a colloid, the substance in the dispersed phase must be larger than the size of a molecule but smaller than what can be seen with the naked eye. This can be more precisely quantified as one or more of the substance's dimensions must be between 1 and 1000 nanometers. If the dimensions are smaller than this the substance is considered a solution and if they are larger than the substance is a suspension.
Classifying Colloids
A common method of classifying colloids is based on the phase of the dispersed substance and what phase it is dispersed in. The types of colloids includes sol, emulsion, foam, and aerosol.
1. Sol is a colloidal suspension with solid particles in a liquid.
2. Emulsion is between two liquids.
3. Foam is formed when many gas particles are trapped in a liquid or solid.
4. Aerosol contains small particles of liquid or solid dispersed in a gas.
When the dispersion medium is water, the collodial system is often referred to as a hydrocolloid. The particles in the dispersed phase can take place in different phases depending on how much water is available. For example, Jello powder mixed in with water creates a hydrocolloid. A common use of hydrocolloids is in the creation of medical dressings.
Table 1: Examples of Colloids
Dispersion Medium Dispersed Phase Type of Colloid Example
Solid Solid Solid sol Ruby glass
Solid Liquid Solid emulsion/gel Pearl, cheese
Solid Gas Solid foam Lava, pumice
Liquid Solid Sol Paints, cell fluids
Liquid Liquid Emulsion Milk, oil in water
Liquid Gas Foam Soap suds, whipped cream
Gas Solid Aerosol Smoke
Gas Liquid Aerosol Fog, mist
An easy way of determining whether a mixture is colloidal or not is through use of the Tyndall Effect. When light is shined through a true solution, the light passes cleanly through the solution, however when light is passed through a colloidal solution, the substance in the dispersed phases scatters the light in all directions, making it readily seen. An example of this is shining a flashlight into fog. The beam of light can be easily seen because the fog is a colloid.
Another method of determining whether a mixture is a colloid is by passing it through a semipermeable membrane. The larger dispersed particles in a colloid would be unable to pass through the membrane, while the surrounding liquid molecules can. Dialysis takes advantage of the fact that colloids cannot diffuse through semipermeable membranes to filter them out of a medium.
Problems
1. Is dust a colloid? If so, what type is it?
2. Is whipped cream a colloid? if so, what type is it?
3. What does Sol mean?
4. When hit by light what happens to a colloidal mixture?
5. What is the mixture considered if the particles are larger than the particles of a colloidal substance
Answers
1. Dust is a colloid if suspended in air. It consists of a solid in a gas, so it is a aerosol.
2. Whipped cream is a colloid. It consists of a gas in a liquid, so it is a foam.
3. Sol is a colloidal suspension with solid particles in a liquid.
4. The light is reflected off the large particles and spread out.
5. It's considered a suspension if the particles are larger than 1000 nanometers.
Contributors and Attributions
• Jimmy Law (UCD), Abheetinder Brar (UCD)
• Thumbnail: pixabay.com/photos/milk-spra...-food-4755234/
Colloid
The Tyndall Effect is the effect of light scattering in colloidal dispersion, while showing no light in a true solution. This effect is used to determine whether a mixture is a true solution or a colloid.
Introduction
"To be classified colloidal, a material must have one or more of its dimensions (length, width, or thickness) in the approximate range of 1-1000 nm." Because a colloidal solution or substance (like fog) is made up of scattered particles (like dust and water in air), light cannot travel straight through. Rather, it collides with these micro-particles and scatters causing the effect of a visible light beam. This effect was observed and described by John Tyndall as the Tyndall Effect.
The Tyndall effect is an easy way of determining whether a mixture is colloidal or not. When light is shined through a true solution, the light passes cleanly through the solution, however when light is passed through a colloidal solution, the substance in the dispersed phases scatters the light in all directions, making it readily seen.
For example, light being shined through water and milk. The light is not reflected when passing through the water because it is not a colloid. It is however reflected in all directions when it passes through the milk, which is colloidal. A second example is shining a flashlight into fog or smog; the beam of light can be easily seen because the fog is a colloid.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Solutions_and_Mixtures/Colloid/Tyndall_Effect.txt
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The concept of an ideal solution is fundamental to chemical thermodynamics and its applications, such as the use of colligative properties. An ideal solution or ideal mixture is a solution in which the enthalpy of solution (\(\Delta{H_{solution}} = 0\)) is zero; with the closer to zero the enthalpy of solution, the more "ideal" the behavior of the solution becomes. Since the enthalpy of mixing (solution) is zero, the change in Gibbs energy on mixing is determined solely by the entropy of mixing (\(\Delta{S_{solution}}\)).
• Raoult's Law
Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present. At any given temperature for a particular solid or liquid, there is a pressure at which the vapor formed above the substance is in dynamic equilibrium with its liquid or solid form. At equilibrium, the rate at which the solid or liquid evaporates is equal to the rate that the gas is condensing.
• Henry's Law
Henry's law is one of the gas laws formulated by William Henry in 1803 and states: "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
Ideal Solutions
Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present:
$P_{solution} = \chi_{solvent}P^o_{solvent} \label{RLaw}$
Introduction
In the 1880s, French chemist François-Marie Raoult discovered that when a substance is dissolved in a solution, the vapor pressure of the solution will generally decrease. This observation depends on two variables:
1. the mole fraction of the amount of dissolved solute present and
2. the original vapor pressure (pure solvent).
At any given temperature for a particular solid or liquid, there is a pressure at which the vapor formed above the substance is in dynamic equilibrium with its liquid or solid form. This is the vapor pressure of the substance at that temperature. At equilibrium, the rate at which the solid or liquid evaporates is equal to the rate that the gas is condensing back to its original form. All solids and liquids have a vapor pressure, and this pressure is constant regardless of how much of the substance is present.
Ideal vs. Nonideal Solutions
Raoult's Law only works for ideal solutions. "An ideal solution shows thermodynamic mixing characteristics identical to those of ideal gas mixtures [except] ideal solutions have intermolecular interactions equal to those of the pure components."2 Like many other concepts explored in Chemistry, Raoult's Law only applies under ideal conditions in an ideal solution. However, it still works fairly well for the solvent in dilute solutions. In reality though, the decrease in vapor pressure will be greater than that calculated by Raoult's Law for extremely dilute solutions.3
Why Raoult's Law works
If you look review the concepts of colligative properties, you will find that adding a solute lowers vapor pressure because the additional solute particles will fill the gaps between the solvent particles and take up space. This means less of the solvent will be on the surface and less will be able to break free to enter the gas phase, resulting in a lower vapor pressure. There are two ways of explaining why Raoult's Law works - a simple visual way, and a more sophisticated way based on entropy. Below is the simple approach.
Remember that saturated vapor pressure is what you get when a liquid is in a sealed container. An equilibrium is set up where the number of particles breaking away from the surface is exactly the same as the number sticking on to the surface again.
Now suppose solute molecules were added so that the solvent molecules occupied only 50% of the surface of the solution.
A certain fraction of the solvent molecules will have sufficient energy to escape from the surface (e.g., 1 in 1000 or 1 in a million). If you reduce the number of solvent molecules on the surface, you are going to reduce the number which can escape in any given time. But it will not make any difference to the ability of molecules in the vapor to stick to the surface again. If a solvent molecule in the vapor hits a bit of surface occupied by the solute particles, it may well stick. There are obviously attractions between solvent and solute otherwise you would not have a solution in the first place.
The net effect of this is that when equilibrium is established, there will be fewer solvent molecules in the vapor phase - it is less likely that they are going to break away, but there is not any problem about them returning. However, if there are fewer particles in the vapor at equilibrium, the saturated vapor pressure is lower.
Limitations on Raoult's Law
In practice, there's no such thing as an ideal solution! However, features of one include:
• Ideal solutions satisfy Raoult's Law. The solution in the last diagram of Figure $3$ above would not actually obey Raoult's Law - it is far too concentrated, but was drawn so concentrated to emphasized the point.
• In an ideal solution, it takes exactly the same amount of energy for a solvent molecule to break away from the surface of the solution as it did in the pure solvent. The forces of attraction between solvent and solute are exactly the same as between the original solvent molecules - not a very likely event!
Suppose that in the pure solvent, 1 in 1000 molecules had enough energy to overcome the intermolecular forces and break away from the surface in any given time. In an ideal solution, that would still be exactly the same proportion. Fewer would, of course, break away because there are now fewer solvent molecules on the surface - but of those that are on the surface, the same proportion still break away. If there were strong solvent-solute attractions, this proportion may be reduced to 1 in 2000, or 1 in 5000 or whatever.
In any real solution of, say, a salt in water, there are strong attractions between the water molecules and the ions. That would tend to slow down the loss of water molecules from the surface. However, if the solution is sufficiently dilute, there will be good-sized regions on the surface where you still have water molecules on their own. The solution will then approach ideal behavior.
The nature of the solute
There is another thing that you have to be careful of if you are going to do any calculations on Raoult's Law. You may have noticed in the little calculation about mole fraction further up the page, that sugar was as a solute rather than salt. What matters is not actually the number of moles of substance that you put into the solution, but the number of moles of particles formed. For each mole of sodium chloride dissolved, you get 1 mole of sodium ions and 1 mole of chloride ions - in other words, you get twice the number of moles of particles as of original salt.
So, if you added 0.1 moles of sodium chloride, there would actually be 0.2 moles of particles in the solution - and that's the figure you would have to use in the mole fraction calculation. Unless you think carefully about it, Raoult's Law only works for solutes which do not change their nature when they dissolve. For example, they must not ionize or associate (e.g., if you put in substance A, it must not form A2 in solution). If it does either of these things, you have to treat Raoult's law with great care.
What matters is not actually the number of moles of substance that you put into the solution, but the number of moles of particles formed.
Raoult's Law and Colligative Properties
The effect of Raoult's Law is that the saturated vapor pressure of a solution is going to be lower than that of the pure solvent at any particular temperature. That has important effects on the phase diagram of the solvent. The next diagram shows the phase diagram for pure water in the region around its normal melting and boiling points. The 1 atmosphere line shows the conditions for measuring the normal melting and boiling points.
The line separating the liquid and vapor regions is the set of conditions where liquid and vapor are in equilibrium. It can be thought of as the effect of pressure on the boiling point of the water, but it is also the curve showing the effect of temperature on the saturated vapor pressure of the water. These two ways of looking at the same line are discussed briefly in a note about half-way down the page about phase diagrams (follow the last link above).
If you draw the saturated vapor pressure curve for a solution of a non-volatile solute in water, it will always be lower than the curve for the pure water.
If you look closely at the last diagram, you will see that the point at which the liquid-vapor equilibrium curve meets the solid-vapor curve has moved. That point is the triple point of the system - a unique set of temperature and pressure conditions at which it is possible to get solid, liquid and vapor all in equilibrium with each other at the same time. Since the triple point has solid-liquid equilibrium present (amongst other equilibria), it is also a melting point of the system - although not the normal melting point because the pressure is not one atmosphere.
The curves for the pure water and for the solution are often drawn parallel to each other. That has got to be wrong! Suppose you have a solution where the mole fraction of the water is 0.99 and the vapor pressure of the pure water at that temperature is 100 kPa. The vapor pressure of the solution will be 99 kPa - a fall of 1 kPa. At a lower temperature, where the vapor pressure of the pure water is 10 kPa, the fall will only be 0.1 kPa. For the curves to be parallel the falls would have to be the same over the whole temperature range. They aren't!
That must mean that the phase diagram needs a new melting point line (a solid-liquid equilibrium line) passing through the new triple point. That is shown in the next diagram.
Now we are finally in a position to see what effect a non-volatile solute has on the melting and freezing points of the solution. Look at what happens when you draw in the 1 atmosphere pressure line which lets you measure the melting and boiling points. The diagram also includes the melting and boiling points of the pure water from the original phase diagram for pure water (black lines).
Because of the changes to the phase diagram, you can see that:
• the boiling point of the solvent in a solution is higher than that of the pure solvent;
• the freezing point (melting point) of the solvent in a solution is lower than that of the pure solvent.
We have looked at this with water as the solvent, but using a different solvent would make no difference to the argument or the conclusions. The only difference is in the slope of the solid-liquid equilibrium lines. For most solvents, these slope forwards whereas the water line slopes backwards. You could prove to yourself that that does not affect what we have been looking at by re-drawing all these diagrams with the slope of that particular line changed. You will find it makes no difference whatsoever.
How to Calculate the Vapor Pressure of a Solution
We can calculate the vapor pressure of the solution in two ways, depending on the volatility of the solute. If the solute is volatile, it will exert its own vapor pressure and this amount is a significant contribution to the overall vapor pressure of the solution, and thus needs to be included in the calculations. On the other hand, if it is nonvolatile, the solute will not produce vapor pressure in solution at that temperature.
Nonvolatile solutes
These calculations are fairly straightforward if you are comfortable with stoichiometric conversions. Because the solute is nonvolatile, you need only determine the change in vapor pressure for the solvent. Using the equation for Raoult's Law, you will need to find the mole fraction of the solvent and the vapor pressure of the pure solvent is typically given.
Example $1$: Kool-Aid
1.5 moles of cherry Kool-Aid are added to a pitcher containing 2 liters of water on a nice day at 25o C. The vapor pressure of water alone is 23.8 mm Hg at 25o C. What is the new vapor pressure of Kool-Aid?
Solution
$P_{H_2O}$ = 23.8 mm Hg
To solve for the mole fraction, you must first convert the 2 L of water into moles:
1 L = 1000 mL = 1000 g
Knowing this, you can convert the mass of water (2000 g) into moles:
2000 g / 18.02 g (molar mass of water) = 110.9 moles H2O
Solve for the mole fraction, $\chi_{H_2O}$:
$\chi_{H_2O}$ = moles H2O / total moles
= 110.9 moles / 110.9 + 1.5 moles = 0.979
Finally, apply Raoult's Law
$P_{Kool-Aid} = \chi_{H_2O} \, P_{H_2O}$ = (0.979)(23.8 mm Hg) = 23.3 mm Hg
Example $\PageIndex{2A}$: Non-electrolyte
Calculate the vapor pressure of a solution made by dissolving 50.0 g glucose, $C_6H_{12}O_6$, in 500 g of water. The vapor pressure of pure water is 47.1 torr at 37°C
Solution
To use Raoult's Law (Equation $\ref{RLaw}$), we need to calculate the mole fraction of water (the solvent) in this sugar-water solution.
$\chi_{solvent} = \dfrac{ \text{moles of water}}{\text{moles of solute} + \text{moles of solvent}} \nonumber$
$\chi_{solvent} = \dfrac{ n_{water}}{ n_{glucose} + n_{water} } \nonumber$
The molar mass of glucose if 180.2 g/mol and of water is 18 g/mol. So
$n_{water} = \dfrac{500\,g}{18\,g /mol} = 27.7 \,mol \nonumber$
and
$n_{glucose} = \dfrac{50\,g}{180.2\,g /mol} = 0.277 \,mol \nonumber$
and
$\chi_{solvent} = \dfrac{ 27.7 \,mol}{ 0.277 \,mol + 27.7 \,mol } = 0.99 \nonumber$
Note that this still relatively dilute.
$P_{solution} = 0.99 \times 47.1 = 46.63 \, torr \nonumber$
not much of a change at all.
Example $\PageIndex{2B}$: Electrolyte
Calculate the vapor pressure of a solution made by dissolving 50.0 g CaCl2, $C_6H_{12}O_6$, in 500 g of water. The vapor pressure of pure water is 47.1 torr at 37°C
Solution
To use Raoult's Law (Equation $\ref{RLaw}$), we need to calculate the mole fraction of water (the solvent) in this salt-water solution.
$\chi_{solvent} = \dfrac{ \text{moles of water}}{\text{moles of solute} + \text{moles of solvent}} \nonumber$
$\chi_{solvent} = \dfrac{ n_{water}}{ n_{solutes} + n_{water} } \nonumber$
The molar mass of $\ce{CaCl_2}$ if 111 g/mol and of water is 18 g/mol. So
$n_{water} = \dfrac{500\,g}{18\,g /mol} = 27.7 \,mol \nonumber$
and
$n_{solutes} = \dfrac{50\,g}{111 \,g /mol} = 0.45 \,mol \nonumber$
but this is really:
• $n_{Ca^+} = 0.45 \,mol$
• $n_{Cl^-} = 0.9 \,mol$
and
$\chi_{solvent} = \dfrac{ 27.7 \,mol}{ 0.45 \,mol + 0.9 \,mol + 27.7 \,mol } = 0.953 \nonumber$
Note that this still relatively dilute.
$P_{solution} = 0.953 \times 47.1 = 44.88\, torr \nonumber$
A bigger change that the glucose example above.
Volatile Solutes
The only difference between volatile and nonvolatile solutes, is that the partial pressure exerted by the vapor pressure of the volatile solute and the vapor pressure of the solvent must be accounted for. The sum of the two will give you the total vapor pressure of the solution.
Example $3$
What are the partial pressures of benzene and toluene in a solution in which the mole fraction of benzene is 0.6? What is the total vapor pressure? The vapor pressure of pure benzene is 95.1 mm Hg and the vapor pressure of pure toluene 28.4 mm Hg at 25oC.
Solution
If $\chi_{benzene} = 0.6$, than $\chi_{toluene} = 0.4$ because $1 - 0.6 = 0.4$.
Now that we know the mole fractions and vapor pressures, this problem is a cinch.
Pbenzene = xbenzenePbenzene = (0.6)(95.1 mm Hg) = 57.1 mm Hg
Ptoluene = xtoluenePtoluene = (0.4)(28.4 mm Hg) = 11.4 mm Hg
The total vapor pressure is simply the sum of the partial pressures:
Ptotal = Pbenzene + Ptuolene = 57.1 mm Hg + 11.4 mm Hg = 68.5 mm Hg
Exercises
*MM = molar mass
1. What is the vapor pressure of a solution at 25oC containing 78.0 grams of glucose (MM = 180.16 g/mol) in 500 grams of water? The vapor pressure of pure water at this temperature is 23.8 mm Hg.
2. 25 grams of cyclohexane (Po = 80.5 torr, MM = 84.16 g/mol) and 30 grams of ethanol (Po = 52.3 torr, MM = 92.14 g/mol) are both volatile components present in a solution. What is the partial pressure of ethanol?
3. What is the vapor pressure of pure butane at 20o C if its partial pressure is 698 mm Hg in a butane-acetone mixture where the mole fraction of acetone is x = 0.577?
Solution
1.
Solve for xH2O.
Moles H2O = 500 g / 18.02 g/mol = 27.7 moles H2O
Moles glucose = 78 g / 180.16 g/mol = 0.433 moles glucose
xH2O = 27.7 moles / (27.7 + 0.433) moles = 0.985
Now we can apply Raoult's Law.
$P = XP^o = (0.985)(23.8\; mmHg) = 23.4\; mmHg$
2.
Calculate the moles of each component.
• Moles cyclohexane: $\dfrac{25\; g}{84.16\; g/mol} = 0.297 \; \text{moles of cyclohexane}$
• Moles ethanol: $\dfrac{30\; g}{92.14 \;g/mol} = 0.326 \; \text{moles ethanol}$
Determine the mole fraction of ethanol and apply Raoult's Law.
Xethanol = 0.326 moles / (0.326 + 0.297) moles = 0.523
P = xPo = (0.523)(52.3 torr) = 27.4 torr
3.
If xacetone = 0.577 then xmethanol = 0.423 because 1 - 0.577 = 0.423
If you rearrange the Raoult's Law equation, you can solve for Po.
$P = XP^o \rightarrow$
$P^o = \dfrac{P}{X} = \dfrac{698\; mmHg}{0.423} = 1650\; mmHg$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Solutions_and_Mixtures/Ideal_Solutions/Changes_In_Vapor_Pressure%2C_Raoult%27s_Law.txt
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Henry's law is one of the gas laws formulated by William Henry in 1803 and states: "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid:
$C =k P_{gas}$
where
• $C$ is the solubility of a gas at a fixed temperature in a particular solvent (in units of M or mL gas/L)
• $k$ is Henry's law constant (often in units of M/atm)
• $P_{gas}$ is the partial pressure of the gas (often in units of atm)
Key differences between Raoult’s Law and Henry’s Law
Raoult's and Henry's laws are limiting laws, generally applicable when the solute concentration goes to zero. In this limit the vapor pressure of any component in the solution depends linearly on its mole fraction, implying the absence of solute-solute interactions.
Raoult's law describes the dependence of the vapor pressure of a solvent as a function of its mole fraction $\chi_1$:
$\lim_{\chi_1\rightarrow1}\left( \frac{p}{\chi_1}\right) =p^\ast \nonumber$
where $p^{\ast}$ is the vapor pressure of the pure solvent.
Henry's law describes the dependence of the vapor pressure of a solute as a function of its concentration. In terms of mole fraction $\chi_2$:
$$\lim_{\chi_2\rightarrow0}\left( \frac{p}{\chi_2}\right) =K \nonumber$$
For a binary mixture of pure substances it can be shown that the laws are complementary: if one law holds for one component then the other law holds for the other component. It can also be shown that the laws imply that the solution satisfies other criteria for ideality, including zero enthalpy and volume of mixing.
Example $1$
What is Henry's constant for neon dissolved in water given: $C_{Ne}=23.5 \,mL/L$ solution and STP (22,414 mL/mole gas) and pressure (1 atm)?
Solution
Now we can rearrange our equation from above to solve for the constant:
$C =k P_{Ne} \nonumber$
To use C we must convert 23.5 mL/L solution to Molarity. Since Ne is a gas, we can use our standard molar volume. Thus giving us:
$\dfrac{\text{23.5 mL/L soln}}{\text{1 mole Ne/22,414 mL}}= 0.00105\,M.$
Now we have solved for the solubility of Ne in the solution. C= 0.00105M and we know the pressure at STP is 1 atm, so we can now use our rearranged equation:
$k= \dfrac{C}{P_{Ne}} \nonumber$
Where C= 0.00105M, PNe = 1 atm, thus giving us k=0.00105 M/atm
Example $2$
Compute the molar solubility in water that is saturated with air.
Solution
This time, we need to use constant (k) that we just calculated and our $P_{Ne}$ in air.
\begin{align*} C &=k P_{gas} \[4pt] &=(0.00105\; M/atm)(0.0341\; atm) \[4pt] &= 3.58 \times 10^{-5}\; M \end{align*}
Applicability of Henry's Law
• Henry's law only works if the molecules are at equilibrium.
• Henry's law does not work for gases at high pressures (e.g., $N_{2\;(g)}$ at high pressure becomes very soluble and harmful when in the blood supply).
• Henry's law does not work if there is a chemical reaction between the solute and solvent (e.g., $HCl_{(g)}$ reacts with water by a dissociation reaction to generate $H_3O^+$ and $Cl^-$ ions).
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Solutions_and_Mixtures/Ideal_Solutions/Dissolving_Gases_In_Liquids%2C_Henry%27s_Law.txt
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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.
• Activities and their Effects on Equilibria
Activity is a measure of the “effective concentration” of a species in a mixture, in the sense that the species' chemical potential depends on the activity of a real solution in the same way that it would depend on concentration for an ideal solution. By convention, activity is treated as a dimensionless quantity, although its value depends on customary choices of standard state for the species. The activity of pure substances in condensed phases (solid or liquids) is normally taken as 1.
• Activity
Activity is a measure of the effective concentration of a species under non-ideal (e.g., concentrated) conditions. This determines the real chemical potential for a real solution rather than an ideal one.
• Azeotropes
An azeotrope is a mixture that exhibits the same concentration in the vapor phase and the liquid phase. This is in contrast to ideal solutions with one component typically more volatile than the other; this is how we use distillation to separate materials. If the mixture forms an azeotrope, the vapor and the liquid concentrations are the same, which preventing separation via this approach.
• Debye-Hückel Theory
The Debye-Hückel theory of electrolytes 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 and (3) Each ion is surrounded by ions of the opposite charge, on average.
• Introduction to Non-ideal 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.
Nonideal Solutions
Effective Concentrations in Gases
In chemical thermodynamics, activity (symbol a) is a measure of the “effective concentration” of a species in a mixture, in the sense that the species' chemical potential depends on the activity of a real solution in the same way that it would depend on concentration for an ideal solution. By convention, activity is treated as a dimensionless quantity, although its value depends on customary choices of standard state for the species. The activity of pure substances in condensed phases (solid or liquids) is normally taken as unity (the number 1). Activity depends on temperature, pressure and composition of the mixture, among other things. For gases, the activity is the effective partial pressure, and is usually referred to as fugacity.
The difference between activity and other measures of composition arises because molecules in non-ideal gases or solutions interact with each other, either to attract or to repel each other. The activity of an ion is particularly influenced by its surroundings. The use of activities allows chemists to explain various discrepancies between ideal solutions and real solutions. The mathematical description of an ideal solution must be modified to describe a real solution, just as the law for ideal gases (PV = nRT) must be modified to describe real gases.
Example $1$: Ideal vs. Real Gases
Compare the pressures predicted for one mole of ethane at 298.15 K under the following equations of states:
1. ethane is an ideal gas or
2. ethane is a van der Waal gas
What is the deviation of the two?
Solution
a) Ideal gas law equation of state: Calculate the pressure of 1.000 mole of ethane at 298.15 K in a 1.000 L flask using the ideal gas law.
\begin{align} P_{ideal}V &=nRT \label{ideal gas} \[4pt] P_{ideal} &=\dfrac{nRT}{V} \nonumber \[4pt] &= \dfrac{ (1\;mole)(0.0821 L \; atm \; K^{-1} \; mol^{-1}) ( 298.15 \;K)}{1 \;L} \nonumber \[4pt] &= 24.47\; atm \nonumber \end{align}
b) Van der Waal's equation of state: Calculate the pressure of 1.00 mole of ethane at 298 K in a 1.00 L flask using the van der Waals equation. The van der Waals constants for ethane can found in Table A8.
• $a = 5.492 \;atm \; \dfrac{L^2}{M^2}$ and
• $b = 0.06499 \;L/m$
\begin{align} \left(P_{vdW} + \dfrac{an^2}{V^2}\right) (V − nb) &=nRT \label{vdW equation} \[4pt] P_{vdW} &= \dfrac{nRT}{V − nb} - \dfrac{an^2}{V^2} \nonumber \[4pt] &= 20.67\; atm \nonumber \end{align}
Calculate the percent error between the $P_{ideal}$ and the $P_{vdW}$
$Error = \dfrac{P_{vdW} - P_{ideal}}{P_{vdW}} = 18.36\% \nonumber$
An error this large is often too big to ignore when carrying out a gas-phase reaction or designing a vessel in which to carry out such a reaction.
There are two ways to deal with real systems that deviate appreciably from ideal conditions:
1. Use a more accurate phenomenologically (i.e., real) Equation of State like the van der Waal' (Equation $\ref{vdW equation}$) that models the system more accurately. This must explicitly address intermolecular forces and other effects that exist in real system and the concentration used is 1 mol/L (i.e., the real concentration).
2. Use an ideal equation pf state like the ideal gas Equation of State in Equation $\ref{ideal gas}$), but use an "effective concentration" of 0.816 mol/L to generate the observed pressure (that is, the gas behaves as if it has a reduced concentration of 100%-18.36% = 81.6% of the real concentration).
Effective Concentrations in Solutions
In a similar fashion, the difference between the calculated solute concentrations in an ideal solution and in a real solution can lead to wide variations in experimental results. The following three examples compare the results obtained when formal concentrations are used (assuming ideality) and when activities are used (assuming non-ideality). Just like gases, "ideal solutions" have certain predictable physical properties (e.g. colligative properties) that real solutions often deviate from. As with the van der Waal equation in Example 1, this deviation originates from solute-solvent, solvent-solvent and solute-solute interactions. The magnitude of this non-ideality naturally greater with higher with solute concentrations and with greater intermolecular interaction (e.g., ions vs. non-charged species). Lewis introduced idea of 'effective concentration' or 'activity' to deal with this problem by allowing an "ideal solution" description for non-ideal solutions.
Since the Van der Waals equation describes real gases instead of the ideal gases law, activity can be used in place of concentration to describe the behavior of real solutions vs. ideal solutions.
The activity of a substance (abbreviated as a) describes the effective concentration of that substance in the reaction mixture. Activity takes into account the non-ideality of the reaction mixture, including solvent-solvent, solvent-solute, and solute-solute interactions. Thus, activity provides a more accurate description of how all of the particles act in solution. For very dilute solutions, the activities of the substances in the solution closely approach the formal concentration (what the calculated concentration should be based on how much substance was measured out.) As solutions get more concentrated, the activities of all of the species tend to be smaller than the formal concentration. The decrease in activity as concentration increases is much more pronounced for ions than it is for neutral solutes.
Activities are actually unitless ratios that compare an effective pressure or an effective concentration to a standard state pressure or concentration (the correct term for the effective pressure is fugacity). There are several ways to define standard states for the different components of a solution, but a common system is
• the standard state for gas pressure, P°, is 1 bar (often approximated with 1 atm)
• the standard state for solute concentration, C°, is 1 molal (moles solute/kg solvent) for dilute solutions. Often molality is approximated with molarity (moles solute/Liter solution).
• the standard state for a liquid is the pure liquid
• the standard state for a solid is the pure solid
Thus, when we discuss the activity of a gas, we actually are discussing the ratio of the effective pressure to the standard state pressure:
$a_{gas} = \dfrac{P}{P^º} \label{1}$
• $a_{gas}$ is a ratio with no units.
Likewise, the activity of a solute in solution would be:
$a_{solute} = \dfrac{C}{C^º} \label{2}$
• $a_{solute}$ is a ratio with no units.
For all solids, the activity is a ratio of the concentration of a pure solid to the concentration of that same pure solid
$a_{solid} = \dfrac{C_{\text{effective solid}}}{C^º_{\text{effective solid}}} = 1 \label{3}$
• $a_{solid}$ always has a value of 1 with no units.
For all liquids, the activity is a ratio of the concentration of a pure liquid to the concentration of that same pure liquid:
$a_{liquid} =\dfrac{C_{\text{effective liquid}}}{C^º_{\text{effective liquid}}} = 1 \label{4}$
• $a_{liquid}$ always has a value of 1 with no units.
For most experimental situations, solutions are assumed to be dilute with respect to the solvent. This assumption implies the solvent can be approximated with pure liquid. According to Raoult's Law, the vapor pressure of the solvent in a solution is equal to the mole fraction of the solvent in the solution times the vapor pressure of the pure solvent:
$\chi= \dfrac{P}{P^º} \label{5}$
The mole fraction of solvent in a dilute solution is approximately 1, so the vapor pressure of the solution is essentially identical to the vapor pressure of the pure solvent. This means that the activity of a solvent in dilute solution will always has a value of 1, with no units.
Activity indicates how many particles "appear" to be present in the solution, which is different from how many actually are present. Hence, activity is a "fudge factor" to ideal solutions that correct the true concentration.
• $a_{gas}$ is a ratio with no units.
• $a_{solute}$ is a ratio with no units.
• $a_{solid}$ is always 1 with no units.
• $a_{liquid}$ is always 1 with no units.
Estimating Activities
The activity of a substance can be estimated from the nominal concentration of that substance (C) by using an activity coefficient, $\gamma$:
$a = \gamma \cdot [C] \label{6}$
The value of $\gamma$ depends upon the substance, the temperature, and the concentration of all solute particles in the solution. The lower the concentration of all solute particles in the solution, the closer the value of $\gamma$ for each solute approaches 1:
$\lim_{[C] \rightarrow 0} \gamma \rightarrow 1 \label{6a}$
Therefore, as $\gamma$ approaches 1, the value of $a$ for the solute approaches C.
$\lim_{\gamma \rightarrow 1} a \rightarrow [C] \label{6b}$
The activity coefficient for a nonvolatile, neutral solute is often estimated by non-linear curve fitting, taking into account the molality of the solute and the activity of the solvent (usually its vapor pressure). In most situations, it is more practical to look up the values of the activity coefficient for a given solute than it is to carry out the curve fitting.
Table $1$: Activity Coefficients
m/(mol kg-1) HCl LiCl NaCl LiNO3 NaNO3
0.01 0.904 0.903 0.902 0.903 0.900
0.02 0.875 0.873 0.870 0.872 0.866
0.05 0.830 0.825 0.820 0.825 0.811
0.10 0.796 0.790 0.778 0.788 0.762
0.2 0.767 0.757 0.735 0.752 0.703
0.4 0.755 0.740 0.693 0.728 0.638
0.6 0.763 0.743 0.673 0.727 0.599
0.8 0.783 0.755 0.662 0.733 0.570
1.0 0.809 0.774 0.657 0.743 0.548
1.2 0.840 0.796 0.654 0.757 0.530
1.4 0.876 0.823 0.655 0.774 0.514
1.6 0.916 0.853 0.657 0.792 0.501
1.8 0.960 0.885 0.662 0.812 0.489
2.0 1.009 0.921 0.668 0.835 0.478
2.5 1.147 1.026 0.688 0.896 0.455
3.0 1.316 1.156 0.714 0.966 0.437
3.5 1.518 1.317 0.746 1.044 0.422
4.0 1.762 1.510 0.783 1.125 0.408
4.5 2.04 1.741 0.826 1.215 0.396
5.0 2.38 2.02 0.874 1.310 0.386
Estimating the activity coefficient of electrolytes (solutes that dissolve or react with the solvent to form ions) depends upon the number of ions formed by the dissociation of the solute in solution or the reaction of the solute with the solution, because each ion formed is dealt with individually. In a theoretical, infinitely dilute ideal solution, an electrolyte would dissociate or react completely to form an integer number of independent ions. For example, 1 mole of NaCl would dissociate to form 2 moles of ions (1 mole of Na+ ions and 1 mole of Cl- ions). In reality, it is found that electrolytes almost always act as if they contain fewer moles of ions than expected based on the formal concentration. This non-ideality is attributed to the degree of dissociation/reaction of the solute, to the solute-solvent interactions such as complex ion formation, and to the solute-solute interactions such as ion pairing. An activity coefficient incorporates the particle interactions into a single term that modifies the formal concentration to give an estimate of the effective concentration, or activity, of each ion.
At infinite dilution, $gamma$ is solely determined by the Debye-Hückel limiting law (Equation $\ref{Debye Equation}$) and depends only on the number and charges of the cations and anions. This means that the same limiting mean ionic activity coefficient is found for sodium chloride and potassium chloride and that also the values for the 2-1 and 1-2 salts sodium sulfate and calcium chloride are identical. At higher electrolyte concentrations though, these values change very strongly and are usually modeled using empirical parameters regressed to the experimental data.
Electrolytes almost always act as if they contain fewer moles of ions than expected based on the formal concentration.
Single ion activity coefficients are calculated using various forms of the Debye-Hückel equation:
$\log \gamma = \dfrac{-0.51 z^2 \sqrt{\mu}}{1+ \dfrac{\alpha\sqrt{\mu}}{305}} \label{Debye Equation}$
This equation takes into account the solution environment as well as the individual characteristics of the specific ion of interest. It is not difficult to calculate single ion activity coefficients, but tables of these activity coefficients for many common ions in solutions of various concentrations are available (e.g., Table $1$).
Applications of Activities
The law of mass action states that a reaction at a constant temperature will proceed spontaneously and predominantly in one direction until a constant ratio of concentrations of products and reactants is obtained. For the generic reaction
$aA + bB \rightleftharpoons cC + dD \label{8}$
the ratio of concentrations (called the mass action expression or equilibrium constant expression) is
$\dfrac{[C]^c[D]^d}{[A]^a[B]^b} = Q \label{9}$
where [ ] represents concentration in
$\dfrac{\text{moles of solute}}{\text{Liter of solution}} \label{10}$\\\\\\\hhhufd
This ratio can take on any value greater than zero, depending on the reaction conditions. Thus, it is often called the instantaneous reaction quotient, Q. The term “instantaneous” signifies that the reaction will (and is) proceeding spontaneously to reach a constant ratio of products and reactants. When the reaction attains that constant ratio of products and reactants, it has reached a state of dynamic equilibrium, and the ratio of concentrations can be represented by the symbol K, the equilibrium constant:
$\dfrac{[C]_{eq}^c[D]_{eq}^d}{[A]_{eq}^a[B]_{eq}^b} = K \label{11}$
Laws of mass action and equilibrium constants are discussed in most general chemistry textbooks, but they are often discussed as if they were describing ideal systems. For instance, if all of the substances are gases, partial pressures are used in the mass-action expression. If the substances are in solution, molarities are used in the mass-action expression. To be thermodynamically correct, however, the activities of the substances must be compared in the mass-action expression.
Activities are needed for precise work because, unlike concentrations, activities contain information about the effects of the solvent and other surrounding particles on the behavior of the particles of interest. Using any unit of comparison other than activities will give an incorrect value for K, but it is assumed that the approximate value is close enough to the true value for most situations. Many tables list K values to 2-3 significant digits, but this degree of precision is valid only under the exact experimental conditions used to obtain those values.
The Derivation of Mass Action Expressions
Given all of the above information on activities, it is now possible to show how a true mass-action equation involving activities can be approximated by a mass-action equation involving molarites. It should be noted that just as the activities were unitless ratios, the molalities and molarities that appear in the following approximations should also be thought of as unitless ratios of concentrations divided by the standard state concentration.
Example $2$: A Solution of Ammonia in Water
Starting with the mass action equation in terms of activities, show all the approximations needed to obtain the mass action equation in terms of molar concentrations for the reaction:
$NH_{3(aq)} + H_2O_{(l)} \rightleftharpoons NH^+_{4(aq)} + OH^-_{(aq)}$
Solution
$Q = \dfrac{a_{NH_4^+}a_{OH^-}}{a_{NH_3}a_{H_2O}} = \dfrac{a_{NH_4^+}a_{OH^-}}{a_{NH_3}(1)}$
• assume $a_{H_2O} \approx 1$ because water is the solvent in a dilute solution
$Q=\dfrac{a_{NH_4^+}a_{OH^-}}{a_{NH_3}}=\dfrac{\gamma_{NH_4^+}\gamma_{OH^-}}{\gamma_{NH_3}} \dfrac{m_{NH_4^+}m_{OH^-}}{m_{NH_3}}$
• assume activity = (activity coefficient)(molality)
$Q =\dfrac{\gamma_{NH_4^+}\gamma_{OH^-}}{\gamma_{NH_3}} \dfrac{m_{NH_4^+}m_{OH^-}}{m_{NH_3}} \approx \dfrac{\gamma_{NH_4^+}\gamma_{OH^-}}{\gamma_{NH_3}} \dfrac{[NH_4^+][OH^-]}{[NH_3]}$
• assume molarity [ ] ≈ molality in dilute solutions
$Q \approx \dfrac{\gamma_{NH_4^+}\gamma_{OH^-}}{\gamma_{NH_3}} \dfrac{[NH_4^+][OH^-]}{[NH_3]}\approx \dfrac{[NH_4^+][OH^-]}{[NH_3]}$
• assume $\dfrac{\gamma_{NH_4^+}\gamma_{OH^-}}{\gamma_{NH_3}} \approx 1$ because the solution is dilute
Therefore the final mass action equation typically used for this reaction is
$Q \approx \dfrac{[NH_4^+][OH^-]}{[NH_3]}$
Example $3$: A Reaction of Two Solids that Produces a Solution
Starting with the mass action equation in terms of activities, show all the approximations needed to obtain the mass action equation in terms of molar concentrations for the reaction:
$Ba(OH)_2 \cdot 8H_2O(s) + 2NH_4NO_3(s) \rightleftharpoons Ba^{2+}(aq) + 2NO_3^-(aq) + 2NH_3(aq) + 10H_2O(l)$
Solution
$Q = \dfrac{a_{Ba^{2+}}a_{NO_3^-}^2a_{NH_3}^2a_{H_2O}^{10}}{a_{Ba(OH)_2·8H_2O}a_{NH_4NO_3}} = \dfrac{a_{Ba^{2+}}a_{NO_3^-}^2a_{NH_3}^2·(1)}{(1)·(1)} = a_{Ba^{2+}}a_{NO_3^-}^2a_{NH_3}^2$
• a =1 for solids and for solvent
$Q = a_{Ba^{2+}}a_{NO_3^-}^2a_{NH_3}^2 = \gamma_{Ba^{2+}} \gamma_{NO_3^-}^2\gamma_{NH_3}^2m_{Ba^{2+}}m_{NO_3^-}^2m_{NH_3}^2$
• activity = (activity coefficient)(molality)
$Q = \gamma_{Ba^{2+}} \gamma_{NO_3^-}^2 \gamma_{NH_3}^2m_{Ba^{2+}} {m_{NO_3^-}}^2 m_{NH_3}^2 \approx \gamma_{Ba^{2+}}\gamma_{NO_3^-}^2 \gamma_{NH_3}^2 [Ba^{2+}] [NO_3^-]^2 [NH_3]^2$
• molarity [ ] ≈ molality in dilute solutions
$Q \approx \gamma_{Ba^{2+}}\gamma_{NO_3^-}^2 \gamma_{NH_3}^2[Ba^{2+}][NO_3^-]^2[NH_3]^2 \approx [Ba^{2+}][NO_3^-]^2[NH_3]^2$
• $\gamma_{Ba^{2+}}\gamma_{NO_3^-}^2\gamma_{NH_3}^2 \approx 1$ because the solution is dilute
$Q \approx [Ba^{2+}][NO_3^-]^2[NH_3]^2$
Salt
Determine the molar solubility $\dfrac {moles}{Liter}$ of the slightly soluble solid, BaSO4, inPure Water and An Aqueous 0.1 M NaCl Solution?
Solution
Pure Water
Barium sulfate is a solid that is slightly soluble in water, with a Ksp value of 1.1 x 10-10:
$BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$
This solution is dilute enough that the Ba2+, SO42-, OH-, and H+ ions will not affect each other greatly, thus the activity of the ions closely approaches their formal concentration. In this nearly ideal aqueous solution, the mass action expression would be
$1.1 \times 10^{-10} = a_{Ba} · a_{SO_4} \approx [Ba^{2+}][SO_4^{2-}]$
Remember that BaSO4 is a solid, with an activity equal to 1. At equilibrium, the [Ba2+] = [SO42-] = 1.05 x 10-5 M.
An Aqueous 0.1 M NaCl Solution
A saturated aqueous solution of BaSO4 that also is 0.1 M in NaCl is no longer near to ideality. The Na+ and Cl- ions surround the Ba2+ and SO42- ions and prevent these ions from being able to reform solid BaSO4 readily as they did in pure water. The activities of the Ba2+ and the SO42- ions will be lower than their formal concentrations. However, the product of the activities must still be equal to the true (thermodynamic) equilibrium constant.
$1.1 x 10^{-10} = a_{Ba}·a_{SO_4} = \gamma_{Ba}[Ba^{2+}]·\gamma_{SO_4}[SO_4^{2-}]$
With the total amount of Na+, Cl-, Ba2+, SO42-, OH-, and H+ ions in the solution, $\gamma_{Ba}$ = 0.38 and $\gamma_{SO_4}$ = 0.355.
$1.1 x 10^{-10} = a_{Ba}·a_{SO_4} = (0.38)[Ba^{2+}](0.355)[SO_4^{2-}]$
$8.2 x 10^{-10} =[Ba^{2+}][SO_4^{2-}]$
$[Ba^{2+}] = [SO_4^{2-}] = 2.9 x 10^{-5}$
The net result is that more solid BaSO4 will dissolve in the 0.1M NaCl solution than in water, and the experimental equilibrium constant will seem to be larger than the thermodynamic equilibrium constant.
Colligative Properties
The van 't Hoff factor, i, is a term that often appears in colligative property calculations to account for the fact that electrolytes will form two or more moles of ions per every mole of electrolyte. In most cases, the solutions are treated as if they are ideal, in which case i will equal an integer representing the total number of independent ions per one formula unit of the solute (Table 2).
Table $2$: Integer van 't Hoff factors for Colligative Properties
Compound
i
Sucrose 1
NaCl 2
MgBr2 3
CaCl2 3
Na3PO4 4
Al2(SO4)3 5
The van 't Hoff factor is actually rarely an integer, and was, in fact, developed to take into account the non-ideality of solutes. Tables listing the i values for specific compounds in specific solutions are available, but it is also possible to use activities to estimate to effective concentrations of ions in solution for use in colligative property calculations.
Example $5$
What is the freezing point of a 0.1 m BaCl2 aqueous solution?
Solution
The calculation for an ideal solution would be $\Delta T = mki$, where $m = 0.1 molal$, $k = 1.86\dfrac{ºC}{molal}$, and $i = 3$. The resulting $\Delta T$ is
$\Delta T = (0.1 molal)(1.86\dfrac{ºC}{molal})(3) = 0.558ºC$
if non-ideality is assumed, the calculation becomes
$\Delta T = \gamma_{Ba} \cdot (\gamma_{Cl})^2•(m_{Ba})(m_{Cl})^2$
Substituting in the estimated $\gamma$ values of $\gamma_{Ba} = 0.38$ and$\gamma_{Cl} = 0.755$, the ion activities are
• $a_{Ba} = (0.38)(0.1) = 0.038$
• $a_{Cl} =(0.755)(0.2) = 0.151$
and the $\Delta T$ is
$\Delta T = (0.038 + 0.151)(1.86 ºC) = 0.351 ºC$
The $\Delta T$ obtained using activities is lower than the $\Delta T$ obtained when using an integer value for i because the activity values take into account the fact that the ions in the solution are not able to act as free and independent particles because of their interactions with each other and with the solvent,
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Solutions_and_Mixtures/Nonideal_Solutions/Activities_and_their_Effects_on_Equilibria.txt
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Activity is a measure of the effective concentration of a species under non-ideal (e.g., concentrated) conditions. This determines the real chemical potential for a real solution rather than an ideal one.
Introduction
Activities and concentrations can both be used to calculate equilibrium constants and reaction rates. However, most of the time we use concentration even though activity is also a measure of composition, similar to concentration. It is satisfactory to use concentration for diluted solutions, but when you are dealing with more concentrated solutions, the difference in the observed concentration and the calculated concentration in equilibrium increases. This is the reason that the activity was initially created.
$\large a=e^{\frac {\mu -\mu_o}{RT}} \tag{1}$
where
• $a$=Activity
• $\mu$ is chemical potential (dependent on standard state) which is Gibbs Energy per mole
• $\mu_0$ is the standard chemical potential
• $R$ is the gas constant
• $T$ is the absolute Temperature
Non-ideality in Gases (Fugacity)
Fugacity is the effective pressure for a non-ideal gas. The pressures of an ideal gas and a real gas are equivalent when the chemical potential is the same. The equation that relates the non-ideal to the ideal gas pressure is:
$f = \phi P \tag{2}$
with
• $f$ represents fugacity,
• $P$ is the pressure for an ideal gas, and
• $\phi$ is the fugacity coefficient.
For an ideal gas, the fugacity coefficient is 1.
Non-ideality in Solutions
pH
We have become accustomed to using the equation $pH= -\log[H^+]$, but this equation not accurate at all concentrations. A better expression for pH is $pH= -\log[a_H^+]$ which accounts for the activity, a. The only reason that other indicators may correctly seem to measure the acidity, which was equivalent to $–\log[H^+]$, is because of the use of Beer’s law, which uses concentration rather than activity.1
Problems
1. What is the purpose of using activity rather than concentration?
2. If the ratio of fugacity to the pressure of the ideal gas is 1 then what is the activity coefficient?
3. If pH does not= -log[H+] then why do the calculations seem to show the correct acidity?
4. If concentration is not accurate does that mean we should start only using activity instead?
Solutions
1. Activity is more accurate in more concentrated solutions
2. The coefficient is one because the gas is in a similar state as an ideal gas.
3. The use of Beer’s law which uses also concentration rather than activity results in the seemingly correct results.
4. Activity is only required highly concentrated solutions.
Azeotropes
An azeotrope is a mixture that exhibits the same concentration in the vapor phase and the liquid phase. This is in contrast to ideal solutions with one component typically more volatile than the other; this is how we use distillation to separate materials. If the mixture forms an azeotrope, the vapor and the liquid concentrations are the same, which preventing separation via this approach.
Introduction
Azeotropes are a mixture of at least two different liquids. Their mixture can either have a higher boiling point than either of the components or they can have a lower boiling point. Azeotropes occur when fraction of the liquids cannot be altered by distillation. Typically when dealing with mixtures, components can be extracted out of solutions by means of Fractional Distillation, or essentially repeated distillation in stages (hence the idea of 'fractional'). The more volatile component tends to vaporize and is collected separately while the least volatile component remains in the distillation container and ultimately, the result is two pure, separate solutions.
Ideal Solutions vs. Azeotropes
Ideal solutions are uniform mixtures of components that have physical properties connected to their pure components. These solutions are supported by Raoult’s law stating that interactions between molecules of solute and molecules of solvent are the same as those molecules each are by themselves. An example of ideal solutions would be benzene and toluene. Azeotropes fail to conform to this idea because, when boiling, the component ratio of unvaporized solution is equal to that of the vaporized solution. So an azeotrope can be defined as a solution whose vapor has the same composition its liquid. As you can imagine, it is extremely difficult to distil this type of substance. In fact, the most concentrated form of ethanol, an azeotrope, is around 95.6% ethanol by weight because pure ethanol is basically nonexistent.
Azeotropes exist in solution at a boiling point specific for that component. This is best represented graphically and the phase diagram of a maximum-boiling point azeotrope can be seen in the following figure. The point 3 represents where the azeotrope exists at a certain boiling point. Imagine that at point 3, the A-B solution is 64% B by mass while component A is water. If that same solution contained any less than 64%, the solution would then be water + the azeotrope. Conversely, if it were to be greater than 64% then the solution would be component B + the azeotrope. This demonstrates that an azeotrope can only exist at one temperature because any higher or lower temperature would result in a different concentration of component A or B.
Also, a maximum-boiling point azeotrope is said to be a negative azeotrope because the boiling point of the azeotrope itself is higher than the boiling point of its components. As you can imagine, a positive azeotrope would have a lower boiling point than any of its components.
Example 1
If pure ethanol has a boiling point of 78.3 °C and its azeotrope has a boiling point of 78.174 °C, what would its graph look like?
Solution
Since the azeotrope BP < pure ethanol BP, the azeotrope is a positive azeotrope and would have a graph that looks like the above figure upside-down (U shaped).
Debye-Hückel
A solution is defined as a homogeneous mixture of two or more components existing in a single phase. In this description, the focus will be on liquid solutions because within the realm of biology and chemistry, liquid solutions play an important role in multiple processes. Without the existence of solutions, a cell would not be able to carry out glycolysis and other signaling cascades necessary for cell growth and development. Chemists, therefore, have studied the processes involved in solution chemistry in order to further the understanding of the solution chemistry in nature.
Introduction
The mixing of solutions is driven by entropy, opposed to being driven by enthalpy. While an ideal gas by definition does not have interactions between particles, an ideal solution assumes there are interactions. Without the interactions, the solution would not be in a liquid phase. Rather, ideal solutions are defined as having an enthalpy of mixing or enthalpy of solution equal to zero (ΔHmixing or ΔHsolution = 0). This is because the interactions between two liquids, A-B, is the average of the A-A interactions and the B-B interactions. In an ideal solution the average A-A and B-B interactions are identical so there is no difference between the average A-B interactions and the A-A/B-B interactions.
Since in biology and chemistry the average interactions between A and B are not always equivalent to the interactions of A or B alone, the enthalpy of mixing is not zero. Consequently, a new term is used to describe the concentration of molecules in solution. Activity, $a_1$, is the effective concentration that takes into account the deviation from ideal behavior, with the activity of an ideal solution equal to one.
An activity coefficient, $\gamma_1$, is utilized to convert from the solute’s mole fraction, $x_1$, (as a unit of concentration, mole fraction can be calculated from other concentration units like molarity, molality, or percent by weight) to activity, $a_1$.
$a_1=\gamma_1x_1 \tag{1}$
Debye-Hückel Formula
The Debye-Hückel formula is used to calculate the activity coefficient.
$\log \gamma_\pm = - \dfrac{1.824 \times 10^6} { \left( \epsilon T \right)^{3/2}} | z_+ z_- | \sqrt I \tag{2}$
This form of the Debye-Hückel equation is used if the solvent is water at 298 K.
$\log \gamma_\pm = - 0.509 | z_+ z_- | \sqrt I \tag{3}$
$\gamma_\pm$ mean ionic activity coefficent
$z_+$ catonic charge of the electrolyte for $\gamma_\pm$
$z_-$ anionic charge of the electrolyte for $\gamma_\pm$
$I$ ionic strength
$\epsilon$ relative dielectric constant for the solution
$T$ temperature of the electrolyte solution
Example 1
Consider a solution of 0.01 M MgCl2 (aq) with an ionic strength of 0.030 M. What is the mean activity coefficient?
Solution
$\log \gamma_\pm = - \displaystyle \frac{1.824 \times 10^6} { \left( \epsilon T \right)^{3/2}} | z_+ z_- | \sqrt I$
$\log \gamma_\pm = - 0.509 | z_+ z_- | \sqrt I$
$\log \gamma_\pm = - \displaystyle \frac{1.824 \times 10^6} { \left( 78.54 \cdot 298 \mathrm {K} \right)^{3/2}} | 2 \cdot 1 | \sqrt {0.0030}$
$\log \gamma_\pm = - 0.509 | 2 \cdot 1 | \sqrt {0.0030} \; m$
$\gamma_\pm = 0.67$
$\gamma_\pm = 0.67$
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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.
Introduction
We use the concept of non-ideal solutions for concentrated solutions. A variety of forces act on real mixtures, making it difficult to predict the properties of such solutions. Non-ideal solutions are identified by determining the strength and specifics of the intermolecular forces between the different molecules in that particular solution.
Non-ideal solutions can occur two ways:
• When intermolecular forces between solute and solvent molecules are less strong than between molecules of similar (of the same type) molecules.
• When intermolecular forces between dissimilar molecules are greater than those between similar molecules.
Reminder: A solvent is the major component of a mixture (i.e. water, air) while a solute is the minor component (sugar, carbon dioxide, etc...). A concrete example would be your daily cup of coffee: the coffee itself is the solvent, and anything you add (may it be sugar or cream) will be the solute.
As mentioned above, non-ideal solutions are under study because their properties are not easily predictable, as forces between molecules can fluctuate over time. Non-ideal solutions cannot be defined by Raoult's law or by Henry's law, which are properties specifically unique to ideal mixtures:
• Raoult's Law: The vapor pressure of a solvent is proportional to its mole fraction. (for solutions)
• Henry's Law: The partial pressure of a gas is proportional to its mole fraction. (for gases)
Since these laws assume that there are no intermolecular interactions, it is evident that they cannot be used for real mixtures, since the mathematical formulas will not hold true anymore due to the fact that the forces will have to be taken into account. However, non-ideal solutions are limited on both sides by these two laws.
There are two main situations that can cause non-ideal solutions to form:
Situation 1: Non-ideal solutions can form when forces of attraction between dissimilar molecules are weaker than between similar molecules. At this point, a heterogeneous (non-mixing) solution may still occur, but it is not always the case.The resulting solution has a larger enthalpy of solution than pure components of the solution, causing the process to be endothermic (heat is absorbed to move the reaction forward).
Example 1
A common example of a type of solution where this behavior is seen is in mixtures of carbon disulfide and acetone. Carbon disulfide is non-polar and acetone is polar. Since carbon disulfide is non-polar, the intermolecular attractions are London dispersion forces, which are known to be weak compared to other types of intermolecular forces. However, since acetone is polar, it has dipole-dipole forces, which are known to be very strong.
Putting these two components together in a mixture results in dipole-induced dipole interactions. Since dipole-dipole induced forces are not nearly as strong as the dipole-dipole interactions between acetone molecules in a pure substance, carbon disulfide-acetone mixtures are non-ideal solutions.
Situation 2: Non-ideal solutions can also form when intermolecular forces between dissimilar molecules are larger than those between similar molecules. In this case, interactions between these two types of molecules release more energy than is taken in to separate the two types of molecules. This energy is released in the form of heat, making the solution process exothermic.
Example 2
An example of this kind of non-ideal solution is a mixture of acetone and chloroform.
Activity Coefficients
The chemical activity of a compound corresponds to the active concentration of that particular compound. However, due to intermolecular forces we known is not the case; therefore, we introduce an activity coefficient, labeled $\gamma$, as a unitless correctional factor. This coefficient takes into account the non-ideal characteristics of a mixture and it is between 0 and 1.
For example, the relationship between the activity of a component and its concentration for ideal mixtures is defined by:
$a_1=\dfrac{C}{C_{pure}}$
While the same relationship for real (non-ideal) mixtures is defined as follows:
$a_1=\gamma \dfrac{C}{C_{pure}}$
• If the interactions attract each other: $\gamma <1$
• If the interactions repel one another: $\gamma >1$
• If there are no interactions: $\gamma =1$
Problems
1. What is the definition of a nonideal solution?
2. hat kinds of non-ideal solutions can be represented through Raoult's Law?
3. What kind of solution would a carbon disulfide (CS2) and acetone ((CH3)2CO) mixture form? Why?
4. What kind of solution would a mixture of acetone ((CH3)2CO) and chloroform (CHCl3) form?
5. Is the volume of a non-ideal solution the sum of the volumes of its components?
Answers
1. A non-ideal solution is a solution whose properties are generally not very predictable on account of the intermolecular forces between the molecules.
2. None. Non-ideal solutions by definition cannot be dealt with through Raoult's Law. Raoult's Law is strictly for ideal solutions only.
3. A non-ideal solution. The interactions between the two types of molecules are weaker than interactions between acetone molecules (pure substance). An explanation is given in further detail in the text above.
4. A non-ideal solution. Hydrogen bonding between the two molecules would produce forces of attraction between unlike molecules that exceed those between like molecules.
5. No. If the solution were ideal, this statement would be true. However, for non-ideal solutions, this is usually not the case.
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Solutions are homogeneous mixtures containing one or more solutes in a solvent. The solvent that makes up most of the solution, whereas a solute is the substance that is dissolved inside the solvent.
• Electrolyte Solutions
An electrolyte solution is a solution that 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.
• Enthalpy of Solution
A solution is a homogeneous mixture of two or more substances and can either be in the gas phase, the liquid phase, the solid phase. The enthalpy change of solution refers to the amount of heat that is released or absorbed during the dissolving process (at constant pressure). This enthalpy of solution can either be positive (endothermic) or negative (exothermic). When understanding the enthalpy of solution, it is easiest to think of a hypothetical three-step process.
• Formation of Ionic Solutions
• Interionic Attractions
• Intermolecular Forces in Mixtures And Solutions
Some forces that interact within pure liquids are also present during mixtures and solutions. Forces such as Cohesive as well as Adhesive forces still apply to mixtures; however, more importantly we focus on the interaction between different molecules. Why is oil only soluble in benzene and not water? Why do only "like" molecules dissolve in "like" molecules?
• Units of Concentration
Solutions are homogeneous mixtures containing one or more solutes in a solvent. The solvent that makes up most of the solution, whereas a solute is the substance that is dissolved inside the solvent.
Thumbnail: pixabay.com/photos/chemistry...emist-3533039/
Solution Basics
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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A solution is a homogeneous mixture of two or more substances and can either be in the gas phase, the liquid phase, the solid phase. The enthalpy change of solution refers to the amount of heat that is released or absorbed during the dissolving process (at constant pressure). This enthalpy of solution ($ΔH_{solution}$) can either be positive (endothermic) or negative (exothermic). When understanding the enthalpy of solution, it is easiest to think of a hypothetical three-step process happening between two substances. One substance is the solute, let’s call that A. The other substance is the solvent, let’s call that B.
Step 1: Breaking up the Solute
The first process that happens deals only with the solute, A, which requires breaking all intramolecular forces holding it together. This means the solute molecules separate from each other. The enthalpy of this process is called $ΔH_1$. This since this is always an endothermic process (requiring energy to break interactions), then $ΔH_1 > 0$.
$\ce{A (s) ->[\text{energy in}] A (g)} \nonumber$
Step 2: Breaking up the Solvent
The second process is very similar to the first step. Much like how the solute, A, needed to break apart from itself, the solvent, B, also needs to overcome the intermolecular forces holding it together. This causes the solvent molecules separate from each other. The enthalpy of this process is called $ΔH_2$. Like the first step, this reaction is always endothermic ($ΔH_2 > 0$) because energy is required to break the interaction between the B molecules.
$\ce{B (l) ->[\text{energy in}] B (g)} \nonumber$
At this point, let us visualize what has happened so far. The solute, A, has broken from the intermolecular forces holding it together and the solvent, B, has broken from the intermolecular forces holding it together as well. It is at this time that the third process happens. We also have two values $ΔH_1$ and $ΔH_2$. that are both greater than zero (endothermic).
Step 3: Combining the Two Together
The third process is when substance A and substance B mix to for a solution. The separated solute molecules and the separated solvent molecules join together to form a solution. This solution will contain one mole of the solute A in an infinite amount of the solvent B.The enthalpy of combining these two substances to form the solution is $ΔH_3$ and is an exothermic reaction (releasing heat since interactions are formed) with $ΔH_3 < 0$.
$\ce{A (g) + B (g) ->[\text{energy out}] A(sol)} \nonumber$
The enthalpy of solution can expressed as the sum of enthalpy changes for each step:
$ΔH_{solution} = ΔH_1 + ΔH_2 + ΔH_3. \label{eq1}$
So the enthalpy of solution can either be endothermic, exothermic or neither $ΔH_{solution} = 0$), depending on how much heat is required or release in each step. If $ΔH_{solution} = 0$, then the solution is called an ideal solution and if $ΔH_{solution} > 0$ or $ΔH_{solution} < 0$, then these solutions are called non-ideal solutions.
The diagrams below can be used as visuals to help facilitate the understanding of this concept. Figure $1$ is for an endothermic reaction, where $ΔH_{solution} > 0.$ Figure $2$ is for an exothermic reaction, where $ΔH_{solution} < 0$. Figure $3$ is for an ideal solution, where $ΔH_{solution} = 0$.
Ideal Solutions
The enthalpy of solution depends on the strengths of intermolecular forces of the solute and solvent and solvent (Equation \ref{eq1}). If the solution is ideal, and $ΔH_{solution} = 0$, then
\begin{align*} ΔH_{solution} = ΔH_1 + ΔH_2 + ΔH_3 &= 0. \label{eq2} \[4pt] ΔH_1 + ΔH_2 &= - ΔH_3 \end{align*}
This means the forces of attraction between like (the solute-solute and the solvent-solvent) and unlike (solute-solvent) molecules are the same (Figure $3$). If the solution is non-ideal, then either $ΔH_1$ added to $ΔH_2$ is greater than $ΔH_3$ or $ΔH_3$ is greater than the sum of $ΔH_1$ and $ΔH_2$. The first case means the forces of attraction of unlike molecules is greater than the forces of attraction between like molecules. The second case means the forces of attraction between like molecules is greater than the forces of attraction between unlike molecules (Figure $2$).
Example $1$: Table salt
Table salt ($\ce{NaCl}$) dissolves readily in water. In solid ($\ce{NaCl}$), the positive sodium ions are attracted to the negative chloride ions. The same is true of the solvent, water; the partially positive hydrogen atoms are attracted to the partially negative oxygen atoms. While ($\ce{NaCl}$) dissolves in water, the positive sodium cations and chloride anions are stabilized by the water molecule electric dipoles. Thus, the intermolecular interactions (i.e., ionic bonds) between ($\ce{NaCl}$) are broken and the salt is dissolved. The overall chemical equation for this reaction is as follows:
$\ce{NaCl (s) ->[H_2O] Na^+ (aq) + Cl^- (aq)}$
Enthalpy of solution is only one part of the driving force in the formation of solutions; the other part is the entropy of solution.
Contributors and Attributions
• Zafir Javeed, Mark Tye (DVC)
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Using the rule "like dissolves like" with the formation of ionic solutions, we must assess first assess two things: 1) the strength of the ion-dipole forces of attraction between water and the ionic compound and 2) the strength of the interionic bond of the ionic compound.
For an ionic compound to form a solution, the ion-dipole forces between water and ionic compound must be greater than the interionic bonds. Therefore, to form a compound: ion-dipole forces > interionic bonds
When the ionic compound is surrounded by water, the water dipoles surround the crystal's clustered structure. The water's negative ends of the dipole will be attracted to the positive dipoles of the ion and the positive ends of the water's dipole will be attracted to the negative dipoles of the ion. If the force of this attraction is stronger than the interionic bonds, the crystal's interionic bonds will be broken, then surrounded by the water molecules or hydrated .
There is a 3-step process that we can use to approach the energy involved in ionic solution formation. 1) Breaking apart the ionic compound is endothermic and requires energy. 2) Hydrating cation is exothermic and therefore releases energy. 3) Hydrating the anion is exothermic and also releases energy. The sum of these 3 steps will then give us the enthalpy of the solution.
Example \(1\): CaCl2
1) CaCl2 (s) -> Ca2+ (g) + Cl 2(g) energy > 0
2) Ca2+(g) H2O > Ca2+(aq) energy < 0
3) Cl2(g) H2O > Cl2(aq) energy < 0
CaCl2 (s) H2O> Ca2+(aq) + Cl2 (aq) energy > 0 The dissolution is endothermic because in the formation of ionic solutions, you must take into account entropy in addition to the enthalpy of the solution to determine whether it will occur spontaneously.
Interionic Attractions
This theory was discovered due to Arrhenius's theory having deficiencies. Arrhenius's theory states that ions exist in a solid substance and dissociated from each other once the solid dissolves. Arhennius's theory did not take into account the fact that strong electrolytes are not as great as he originally thought and the values of the van 't Hoff factor i relied on the concentration of the solution. The theory of electrolyte solution was brought about by Peter Debye and Erich Huckel in 1923.
Interionic Attractions are when an ion is surrounded by an ionic atmosphere which has a net charge opposite for its own. For example an anion would be completely surrounded by ions mostly composed of cations and a cation would mostly be surrounded by ions of anions. The ionic atmosphere decreases the mobility of each ion by exerting a drag on it, which in turn also decreases the magnitude of colligative properties. The ionic atmosphere cannot created nor destroyed.
In solutions with weak electrolytes the number of ions is not large, therefore the effect of the interionic attraction is small. In a concentrated solution of strong electrolytes the ion count is large, and therefore the interionic attraction will be apparent. The reason behind the differences in the interionic attraction is that in concentrated solutions ions are closer together due to the large ion count, while in less concentrated solutions they are further apart.
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Some forces that interact within pure liquids are also present during mixtures and solutions. Forces such as Cohesive as well as Adhesive forces still apply to mixtures; however, more importantly we focus on the interaction between different molecules. Why is oil only soluble in benzene and not water? Why do only "like" molecules dissolve in "like" molecules?
The process of Mixing
Before we go on to the more specific mechanisms of mixing, let's discuss its process. Mixing is a spontaneous process that increases the entropy of the solution. In order to form a mixture of homogenous solutions by distributing the solute molecules evenly within the solvent molecules, heat transfers are inevitable. This heat transfer is denoted ΔHsoln for our general comprehension. ΔH is the change in heat energy found by subtracting the enthalpy of the reactant from that of the product:
\[H_{products} - H_{reactants}= ΔH_{soln}.\]
Enthalpy of Solution
What then is the significance of \(ΔH_{soln}\)? It presents a clear indication of the magnitude as well as direction of the heat transfer so that when:
• ΔH>0 : Endothermic Reaction (positive), because the products encompass more energy than the reactants
• ΔH<0 : Exothermic Reaction (negative), because the reactants consist of more energy than the products.
What we have to supply for our understanding for this equation is that the extra energy is seized either from or give to the surrounding. And to ascertain the enthalpy of solution, we take the three step approach in enthalpy when a solute is mixed with the solvent.
Three Step Approach to Finding the Enthalpy of Solution: ΔHsoln = ΔH1 + ΔH2+ ΔH3
1. Each molecule of solute is Separated from each other (expand the solute), endothermic reaction. (ΔH1)
2. Each molecule of solvent is separated from each other (expand solvent), endothermic reaction. (ΔH2) Now the molecules of solute and molecules of solvent can be permitted to attract one another in solution.
3. The molecules of solute and solvent react with each other and a solution will result. exothermic reaction (ΔH3)
Note
note that usually \(ΔH_1\) and \(ΔH_2\) are opposite in sign as \(ΔH_3\). Separating the solute and the solvent solutions alone are usually endothermic reactions in that their cohesive forces are broken while letting the molecules to react freely is an exothermic reaction. The figure below explains pictorially how positive and negative \(ΔH\) can be obtained through the three step process of mixing solution.
ΔH's Relationship to the Behavior of the Solution
Ideal solution is the mixture that has little to no net intermolecular interactions that differentiates it from its ideal behavior. Thus if the intermolecular forces of attraction are the same and have the same strength, both the solvent and solute will mix at random. This solution is called an ideal solution, which means that \(ΔH_{soln} = 0\). If the intermolecular forces of attraction of different molecules are greater than the forces of attraction of like molecules, then it is called a nonideal solution. This will result in an exothermic process (\(ΔH_{soln}<0\)). If the intermolecular forces of attraction of different molecules are a bit weaker than the forces of attraction of like molecules. This solution is a nonideal solution, has bigger enthalpy value than pure components, and it goes through an endothermic process. Lastly, if the intermolecular forces of attraction of different molecules is a lot weaker than the forces of attraction of like molecules, the solution becomes a heterogeneous mixture (e.g., water and olive oil).
The Effects of Intermolecular Forces in Solution.
The epitome of intermolecular forces in solution is the miracle of solubility, because when a matter precipitates it no longer interacts with the solvent. So what is the attraction between "like" molecules that makes them attract to each other? Let's take a phospholipid, the building block of a cell's membrane, as an example. This molecule is amphipathic, meaning that it is both hydrophilic and hydrophobic. Beginning with the structure of a phospholipid, it has a polar head which is hydrophilic and a nonpolar tail which is hydrophobic as the picture below.
How can a single molecule be both polar molecule loving and polar molecule disliking at the same time? This is because at the polar head, the phosphate has a net negative charge thus attracting the partial positive charge of the hydrogen molecules of water. Its nonpolar tails on the other hand, is a very organized form of hydrocarbon, consisting of no net charges. The tail is then repelled by water as it struggles to fit between the partial positive and partial negative of the water molecule.
Another side effect of the interactions of molecules is reflected by the use of the activity coefficient during thermodynamic equilibrium constant calculations. This constant differentiates ideal and nonideal solutions so that interactions for solution equilibrium can be more accurately estimated. Most versions of the equilibrium constant K utilizes activity instead of concentration so that the units would disappear more fluently. For an ideal solution, the activity coefficient is 1 [x]/ oCelcius, thus when the concentration is dived by it to yield activity, it is unaltered.
Example \(1\)
Based on the concept of intermolecular interactions, ascertain the reason behind freezing-point depression and boiling-point elevation.
Solution
When an ion is added into solution, it exerts an intermolecular force which binds loosely to the water molecules in solution. This weak force then increases the energy necessary to break each molecule loose, thus increasing temperature in relationship to vapor pressure. It now takes more energy input to obtain the same vapor pressure, thus elevating the boiling point. For freezing-point depression, the same force that is holding the water molecules from evaporating is holding them against being placed into an organized solid form. It now takes more energy to form the weak bonds between each water molecule because the intermolecular forces between water and the ions first have to be overcome, hence reducing the freezing point.
Example \(2\)
Give examples that present the involvement of intermolecular forces thus differentiating ideal from nonideal solutions.
Solution
Gibbs free energy (relating ΔG with ΔGo), calculating thermodynamic equilibrium constants, boiling-point elevation, and freezing-point depression
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Solutions are homogeneous mixtures containing one or more solutes in a solvent. The solvent that makes up most of the solution, whereas a solute is the substance that is dissolved inside the solvent.
Relative Concentration Units
Concentrations are often expressed in terms of relative unites (e.g. percentages) with three different types of percentage concentrations commonly used:
1. Mass Percent: The mass percent is used to express the concentration of a solution when the mass of a solute and the mass of a solution is given: $\text{Mass Percent}=\dfrac{\text{Mass of Solute}}{\text{Mass of Solution}} \times 100\% \label{1}$
2. Volume Percent: The volume percent is used to express the concentration of a solution when the volume of a solute and the volume of a solution is given: $\text{Volume Percent}= \dfrac{\text{Volume of Solute}}{\text{Volume of Solution}} \times 100\% \label{2}$
3. Mass/Volume Percent: Another version of a percentage concentration is mass/volume percent, which measures the mass or weight of solute in grams (e.g., in grams) vs. the volume of solution (e.g., in mL). An example would be a 0.9%( w/v) $NaCl$ solution in medical saline solutions that contains 0.9 g of $NaCl$ for every 100 mL of solution (see figure below). The mass/volume percent is used to express the concentration of a solution when the mass of the solute and volume of the solution is given. Since the numerator and denominator have different units, this concentration unit is not a true relative unit (e.g. percentage), however it is often used as an easy concentration unit since volumes of solvent and solutions are easier to measure than weights. Moreover, since the density of dilute aqueous solutions are close to 1 g/mL, if the volume of a solution in measured in mL (as per definition), then this well approximates the mass of the solution in grams (making a true reletive unit (m/m)).
$\text{Mass/Volume Percent}= \dfrac{\text{Mass of Solute (g)}}{\text{Volume of Solution (mL)}} \times 100\% \label{3}$
Example $1$: Alcohol "Proof" as a Unit of Concentration
For example, In the United States, alcohol content in spirits is defined as twice the percentage of alcohol by volume (v/v) called proof. What is the concentration of alcohol in Bacardi 151 spirits that is sold at 151 proof (hence the name)?
Solution
It will have an alcohol content of 75.5% (w/w) as per definition of "proof".
When calculating these percentages, that the units of the solute and solution should be equivalent units (and weight/volume percent (w/v %) is defined in terms of grams and mililiters).
You CANNOT plug in… You CANNOT plug in…
(2 g Solute) / (1 kg Solution) (2 g Solute) / (1000 g Solution)
or (0.002 kg Solute) / (1 kg Solution)
(5 mL Solute) / (1 L Solution) (5 mL Solute) / (1000 mL Solution)
or (0.005 L Solute) / (1 L Solution)
(8 g Solute) / (1 L Solution) (8 g Solute) / (1000 mL Solution)
or (0.008 kg Solute) / (1 L Solution)
Dilute Concentrations Units
Sometimes when solutions are too dilute, their percentage concentrations are too low. So, instead of using really low percentage concentrations such as 0.00001% or 0.000000001%, we choose another way to express the concentrations. This next way of expressing concentrations is similar to cooking recipes. For example, a recipe may tell you to use 1 part sugar, 10 parts water. As you know, this allows you to use amounts such as 1 cup sugar + 10 cups water in your equation. However, instead of using the recipe's "1 part per ten" amount, chemists often use parts per million, parts per billion or parts per trillion to describe dilute concentrations.
• Parts per Million: A concentration of a solution that contained 1 g solute and 1000000 mL solution (same as 1 mg solute and 1 L solution) would create a very small percentage concentration. Because a solution like this would be so dilute, the density of the solution is well approximated by the density of the solvent; for water that is 1 g/mL (but would be different for different solvents). So, after doing the math and converting the milliliters of solution into grams of solution (assuming water is the solvent): $\dfrac{\text{1 g solute}}{\text{1000000 mL solution}} \times \dfrac{\text{1 mL}}{\text{1 g}} = \dfrac{\text{1 g solute}}{\text{1000000 g solution}}$ We get (1 g solute)/(1000000 g solution). Because both the solute and the solution are both now expressed in terms of grams, it could now be said that the solute concentration is 1 part per million (ppm). $\text{1 ppm}= \dfrac{\text{1 mg Solute}}{\text{1 L Solution}}$ The ppm unit can also be used in terms of volume/volume (v/v) instead (see example below).
• Parts per Billion: Parts per billion (ppb) is almost like ppm, except 1 ppb is 1000-fold more dilute than 1 ppm. $\text{1 ppb} = \dfrac{1\; \mu \text{g Solute}}{\text{1 L Solution}}$
• Parts per Trillion: Just like ppb, the idea behind parts per trillion (ppt) is similar to that of ppm. However, 1 ppt is 1000-fold more dilute than 1 ppb and 1000000-fold more dilute than 1 ppm. $\text{1 ppt} = \dfrac{ \text{1 ng Solute}}{\text{1 L Solution}}$
Example $2$: ppm in the Atmosphere
Here is a table with the volume percent of different gases found in air. Volume percent means that for 100 L of air, there are 78.084 L Nitrogen, 20.946 L Oxygen, 0.934 L Argon and so on; Volume percent mass is different from the composition by mass or composition by amount of moles.
Elements Name Volume Percent (v/v) ppm (v/v)
Nitrogen 78.084 780,840
Oxygen 20.946 209,460
Water Vapor 4.0% 40,000
Argon 0.934 9,340
Carbon Dioxide 0.0379 379* (but growing rapidly)
Neon 0.008 8.0
Helium 0.000524 5.24
Methane 0.00017 1.7
Krypton 0.000114 1.14
Ozone 0.000001 0.1
Dinitrogen Monoxide 0.00003 0.305
Concentration Units based on moles
• Mole Fraction: The mole fraction of a substance is the fraction of all of its molecules (or atoms) out of the total number of molecules (or atoms). It can also come in handy sometimes when dealing with the $PV=nRT$ equation. $\chi_A= \dfrac{\text{number of moles of substance A}}{\text{total number of moles in solution}}$ Also, keep in mind that the sum of each of the solution's substances' mole fractions equals 1. $\chi_A + \chi_B + \chi_C \;+\; ... \;=1$
• Mole Percent: The mole percent (of substance A) is $\chi_A$ in percent form. $\text{Mole percent (of substance A)}= \chi_A \times 100\%$
• Molarity: The molarity (M) of a solution is used to represent the amount of moles of solute per liter of the solution. $M= \dfrac{\text{Moles of Solute}}{\text{Liters of Solution}}$
• Molality: The molality (m) of a solution is used to represent the amount of moles of solute per kilogram of the solvent. $m= \dfrac{\text{Moles of Solute}}{\text{Kilograms of Solvent}}$
The molarity and molality equations differ only from their denominators. However, this is a huge difference. As you may remember, volume varies with different temperatures. At higher temperatures, volumes of liquids increase, while at lower temperatures, volumes of liquids decrease. Therefore, molarities of solutions also vary at different temperatures. This creates an advantage for using molality over molarity. Using molalities rather than molarities for lab experiments would best keep the results within a closer range. Because volume is not a part of its equation, it makes molality independent of temperature.
Example $1$
In a solution, there is 111.0 mL (110.605 g) solvent and 5.24 mL (6.0508 g) solute present in a solution. Find the mass percent, volume percent and mass/volume percent of the solute.
Solution
Mass Percent
=(Mass of Solute) / (Mass of Solution) x 100%|
=(6.0508g) / (110.605g + 6.0508g) x 100%
=(0.0518688312) x 100%
=5.186883121%
Mass Percent= 5.186%
Volume Percent
=(Volume of Solute) / (Volume of Solution) x 100%
=(5.24mL) / (111.0mL + 5.24mL) x 100%
=(0.0450791466) x 100%
=4.507914659%
Volume Percent= 4.51%
Mass/Volume Percent
=(Mass of Solute) / (Volume of Solution) x 100%
=(6.0508g) / (111.0mL + 5.24mL) x 100%
=(0.0520) x 100%
=5.205%
Mass/Volume Percent= 5.2054%
Example $2$
With the solution shown in the picture below, find the mole percent of substance C.
Solution
Moles of C= (5 C molecules) x (1mol C / 6.022x1023 C molecules) = 8.30288941x10-24mol C
Total Moles= (24 molecules) x (1mol / 6.022x1023 molecules)= 3.98538691x10-23mol total
XC= (8.30288941x10-24mol C) / (3.98538691x10-23mol) = .2083333333
Mole Percent of C
= XC x 100%
=(o.2083333333) x 100%
=20.83333333
Mole Percent of C = 20%
Example $3$
A 1.5L solution is composed of 0.25g NaCl dissolved in water. Find its molarity.
Solution
Moles of NaCl= (0.25g) / (22.99g + 35.45g) = 0.004277 mol NaCl
Molarity
=(Moles of Solute) / (Liters of Solution)
=(0.004277mol NaCl) / (1.5L)
=0.002851 M
Molarity= 0.0029M
Example $4$
0.88g NaCl is dissolved in 2.0L water. Find its molality.
Solution
Moles of NaCl= (0.88g) / (22.99g + 35.45g) = 0.01506 mol NaCl
Mass of water= (2.0L) x (1000mL / 1L) x (1g / 1mL) x (1kg / 1000g) = 2.0kg water
Molality
=(Moles of Solute) / (kg of Solvent)
=(0.01506 mol NaCl) / (2.0kg)
=0.0075290897m
Molality= 0.0075m
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A true liquid is isotropic, meaning that its properties are uniform in all directions— the result of its molecules being in constant random motion. Crystalline solids, in contrast, are anisotropic; optical- and other properties such as thermal and electrical conductivity vary with direction. A liquid crystal phase has many of the physical attributes of a liquid, but its molecular units are sufficiently ordered to give rise to some anisotropy, most notably in their optical properties.
As with so many scientific discoveries, it all started with an unexpected observation. In 1888, an Austrian plant physiologist, working in Prague, attempted to measure the melting point of a cholesterol derivative that he had extracted from a plant. To his surprise, he found that this substance appeared to have two melting points. At 145° C the crystalline solid first melted into a cloudy liquid, and then at 178° the cloudiness suddenly disappeared, leaving the clear, transparent liquid that one ordinarily expects after melting.
A physicist who examined this material recognized that the cloudy liquid had a certain degree of order; he proposed that it was a hitherto unknown state of matter, and suggested the name "liquid crystal". But science was not quite ready to accept this concept, and despite a number of confirming experiments between 1910 and 1930, the field remained largely dormant until the mid-1960s when the French physicist Pierre-Gilles de Gennes (1932-2007) developed a thorough theoretical model for the properties of liquid crystals, particularly their ability to scatter light. For this and for related studies on polymers, de Gennes was awarded the 1991 Nobel Prize in Physics.
The structural units capable of forming liquid crystals are always molecules, usually rather elongated organic ones that possess dissimilar local structural regions that can interact in an organized way with their neighbors. Over a certain range of temperatures, these attractive forces can lead to a degree of self-organization in which crystal-like order persists in some directions even though it is lost in other directions. Although a large variety of molecules are known to form liquid crystals, the simplest and most common structures can be represented by the following generic scheme:
1. The two benzene rings confer a degree of planarity on the molecule that promotes attractions between neighboring molecules. This planarity is enhanced when the linkage group contains a double bond such as -(HC=N)- which keeps the rings in the same plane.
2. The terminal group is often (but not always) one that is somewhat polar, giving rise to intermolecular attractions along the long axis.
3. The side chain is commonly a hydrocarbon chain that serves to elongate the molecule.
Properties of Liquid Crystals
Liquid crystal phases are generally cloudy in appearance, which means that they scatter light in much the same way as colloids such as milk. This light scattering is a consequence of fluctuating regions of non-uniformity as small groups of molecules form and disperse.
The anisotropy of liquid crystals causes them to exhibit birefringence. That is, light that enters the crystal is broken up into two oppositely-polarized rays that travel at different velocities. Observation of a birefringent material between crossed polarizing filters reveals striking patterns and color effects.
The colors arise from interference between the ordinary ray and the extraordinary ray; the latter traverses a slightly longer path through the material, and thus emerges later (and out-of-phase) with the former.
Liquid crystals, like all other kinds of matter, are subject to thermal expansion. As the temperature rises, the average spacing between the aligned molecules of a nematic phase (see below) increases, thus causing the e-ray to be increasingly retarded with respect to the o-ray. If a suitable liquid crystal mixture is painted onto the surface of a patient's body, it can often reveal the sites of infection or tumors, which cause increases or reductions in local blood flow giving rise to temperature anomalies. Inexpensive thermometers can be made by printing a succession of suitably formulated LC mixtures on a paper or plastic strip which is held in contact with the surface whose temperature is to be measured.
Liquid Crystal Displays (LCDs)
The first working liquid crystal display ("LCD") was demonstrated at RCA in 1968. In the following year, James Fergason of Kent State University (OH) discovered the twisted nematic field effect which allowed a much higher-quality display and led to the first commercial LCD wristwatch in 1979. LCDs were first used in calculators in the late 1970s, but they are now widely encountered in computer- and television displays. The thickness of the chiral phase is such that polarized light passing through it is rotated by 90°, which corresponds to the orientation of the right-hand polarizing filter. So in this state, the light passes through and illuminates the pixel surface.
When an electric field is imposed on the liquid crystal phase, the component molecules line up in the field and the chirality is lost. Light passing through the cell does not undergo rotation of its polarization plane, and is therefore stopped by the right polarizing filter, turning the display off.
Nematic and Smectic Phases
There are many classes and sub-classes of liquid crystals, but for the purposes we will divide them into the two kinds depicted on the right side of this Figure below which also compares them with the two conventional condensed phases of matter:
In a nematic phase (the term means "thread-like") the molecules are aligned in the same direction but are free to drift around randomly, very much as in an ordinary liquid. Owing to their polarity, the alignment of the rod-like molecules can be controlled by applying an electric field; this is the physical basis for liquid crystal displays and certain other electrooptic devices.
In smectic ("soap-like") phases the molecules are arranged in layers, with the long molecular axes approximately perpendicular to the laminar planes. The only long-range order extends along this axis, with the result that individual layers can slip over each other (hence the "soap-like" nature) in a manner similar to that observed in graphite. Within a layer there is a certain amount of short-range order. There are a large number of sub-categories of smectic phases which we will not go into here.
Chiral phases
Special cases of nematic and smectic phases are sometimes formed by molecules that display chirality — that is, they can exist in either left- or right-handed forms that cannot be superposed on each other. In the resulting chiral phase, successive molecules positioned along the long axis are rotated around this axis, giving rise to a periodicity that repeats itself at distances corresponding to a complete rotation. These twisted phases are able to rotate the plane of polarized light that passes along the axis. If the molecules are polar, this twisting can be turned off by imposing an external electric field at either end of the long axis. Besides the very important application of this property (known as ferroelectricity) to liquid crystal displays, these materials can be used to make electrooptic shutters which can be switched open and closed in microseconds.
A typical chiral molecule capable of exhibiting ferroelectric behavior is shown below. The chiral part of the molecule is indicated by the asterisk. The chirality arises because this carbon atom is joined to four different groups.
Contributors and Attributions
Thumnbnail: Schematic of mesogen ordering in the smectic liquid crystal phases: smectic-A (layered) and smectic-C (layered and tilted). (CC -SA-BY 3.0; Kebes).
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• Boiling
Boiling is the process by which a liquid turns into a vapor when it is heated to its boiling point. The change from a liquid phase to a gaseous phase occurs when the vapor pressure of the liquid is equal to the atmospheric pressure exerted on the liquid. Boiling is a physical change and molecules are not chemically altered during the process.
• Clausius-Clapeyron Equation
The Clausius-Clapeyron equation allows us to estimate the vapor pressure at another temperature, if the vapor pressure is known at some temperature, and if the enthalpy of vaporization is known.
• Fundamentals of Phase Transitions
Phase transition is when a substance changes from a solid, liquid, or gas state to a different state. Every element and substance can transition from one phase to another at a specific combination of temperature and pressure.
• Phase Diagrams
Phase diagram is a graphical representation of the physical states of a substance under different conditions of temperature and pressure. A typical phase diagram has pressure on the y-axis and temperature on the x-axis. As we cross the lines or curves on the phase diagram, a phase change occurs. In addition, two states of the substance coexist in equilibrium on the lines or curves.
• Simple Kinetic Theory
This page takes a simple look at solids, liquids and gases, and changes of state such as melting and boiling, in terms of the behavior of the particles present.
• Vapor Pressure
This page looks at how the equilibrium between a liquid (or a solid) and its vapor leads to the idea of a saturated vapor pressure. It also looks at how saturated vapor pressure varies with temperature, and the relationship between saturated vapor pressure and boiling point.
Thumbnail: Structural details of melting icicles against gray sky. (CC SA-BY 3.0; Wing-Chi Poon).
Phase Transitions
Boiling is the process by which a liquid turns into a vapor when it is heated to its boiling point. The change from a liquid phase to a gaseous phase occurs when the vapor pressure of the liquid is equal to the atmospheric pressure exerted on the liquid. Boiling is a physical change and molecules are not chemically altered during the process.
How Does boiling Occur?
When atoms or molecules of a liquid are able to spread out enough to change from a liquid phase to a gaseous phase, bubbles form and boiling occurs.
Video: Boiling basics (https://www.youtube.com/embed/Py0GEByCke4).
The boiling point is the temperature at which boiling occurs for a specific liquid. For example, for water, the boiling point is 100ºC at a pressure of 1 atm. The boiling point of a liquid depends on temperature, atmospheric pressure, and the vapor pressure of the liquid. When the atmospheric pressure is equal to the vapor pressure of the liquid, boiling will begin.
A liquid will begin to boil when Atmospheric Pressure = Vapor Pressure of Liquid
Exercise 1: Boiling Basics
When a liquid boils, what is inside the bubbles?
Answer
The bubbles in a boiling liquid are made up of molecules of the liquid which have gained enough energy to change to the gaseous phase.
Exercise 2
Describe the formation of bubbles in a boiling liquid (see video for answer).
Temperature and Boiling
When boiling occurs, the more energetic molecules change to a gas, spread out, and form bubbles. These rise to the surface and enter the atmosphere. It requires energy to change from a liquid to a gas (see enthalpy of vaporization). In addition, gas molecules leaving the liquid remove thermal energy from the liquid. Therefore the temperature of the liquid remains constant during boiling. For example, water will remain at 100ºC (at a pressure of 1 atm or 101.3 kPa) while boiling. A graph of temperature vs. time for water changing from a liquid to a gas, called a heating curve, shows a constant temperature as long as water is boiling.
Exercise 3: Heating Curve for Water
Based on the heating curve below, when will the temperature of \(H_2O\) exceed 100ºC (in an open system)?
Answer
The temperature of \(H_2O\) will only exceed 100 ºC once it has entirely changed to the gaseous phase. As long as there is liquid the temperature will remain constant.
Atmospheric Pressure and Boiling
The pressure of gas above a liquid affects the boiling point. In an open system this is called atmospheric pressure. The greater the pressure, the more energy required for liquids to boil, and the higher the boiling point.
Higher Atmospheric Pressure = More Energy Required to Boil = Higher Boiling Point
In an open system this can be visualized as air molecules colliding with the surface of the liquid and creating pressure. This pressure is transmitted throughout the liquid and makes it more difficult for bubbles to form and for boiling to take place. If the pressure is reduced, the liquid requires less energy to change to a gaseous phase, and boiling occurs at a lower temperature.
Video: Atmospheric Pressure and Boiling (www.youtube.com/watch?v=aiwy...ature=youtu.be).
Exercise 4
Based on the atmospheric pressure, predict the boiling point for water at the following locations. Remember that water boils at 100ºC at sea level on earth. Assume constant temperature.
• Earth at Sea Level: 101.3 kPa
• Mount Everest Summit: 33.7 kPa
• Mars (average): 0.6 kPa
• Venus (surface): 9200 kPa
Answer
Since water boils at 100ºC, water would boil quickly on Mars (actual value us about 10ºC). The boiling point on Mt. Everest would be closer to water (actual value about 70ºC). On Venus water would boil well over 100ºC.
Vapor Pressure and Boiling
The molecules leaving a liquid through evaporation create an upward pressure as they collide with air molecules. This upward push is called the vapor pressure. Different substances have different vapor pressures and therefore different boiling points. This is due to differing intermolecular forces between molecules.
Video: Vapor Pressure and Bioling (youtu.be/ffBusZO-TO0)
The vapor pressure of a liquid lowers the amount of pressure exerted on the liquid by the atmosphere. As a result, liquids with high vapor pressures have lower boiling points. Vapor pressure can be increased by heating a liquid and causing more molecules to enter the atmosphere. At the point where the vapor pressure is equal to the atmospheric pressure boiling will begin. In effect, without any external pressure the liquid molecules will be able to spread out and change from a liquid to a gaseous phase. The gas, as bubbles in the liquid, will rise to the surface and be released into the atmosphere.
Contributors and Attributions
• Wayne Breslyn, NBCT, Ph.D. (Gaithersburg High School)
• Chadwick Wyler
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Learning Objectives
• Apply the Clausius-Clapeyron equation to estimate the vapor pressure at any temperature.
• Estimate the heat of phase transition from the vapor pressures measured at two temperatures.
The vaporization curves of most liquids have similar shapes with the vapor pressure steadily increasing as the temperature increases (Figure $1$).
A good approach is to find a mathematical model for the pressure increase as a function of temperature. Experiments showed that the vapor pressure $P$ and temperature $T$ are related,
$P \propto \exp \left(- \dfrac{\Delta H_{vap}}{RT}\right) \ \label{1}$
where $\Delta{H_{vap}}$ is the Enthalpy (heat) of Vaporization and $R$ is the gas constant (8.3145 J mol-1 K-1).
A simple relationship can be found by integrating Equation \ref{1} between two pressure-temperature endpoints:
$\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right) \label{2}$
where $P_1$ and $P_2$ are the vapor pressures at two temperatures $T_1$ and $T_2$. Equation \ref{2} is known as the Clausius-Clapeyron Equation and allows us to estimate the vapor pressure at another temperature, if the vapor pressure is known at some temperature, and if the enthalpy of vaporization is known.
Alternative Formulation
The order of the temperatures in Equation \ref{2} matters as the Clausius-Clapeyron Equation is sometimes written with a negative sign (and switched order of temperatures):
$\ln \left( \dfrac{P_1}{P_2} \right) = - \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_1}- \dfrac{1}{T_2} \right) \label{2B}$
Example $1$: Vapor Pressure of Water
The vapor pressure of water is 1.0 atm at 373 K, and the enthalpy of vaporization is 40.7 kJ mol-1. Estimate the vapor pressure at temperature 363 and 383 K respectively.
Solution
Using the Clausius-Clapeyron equation (Equation $\ref{2B}$), we have:
\begin{align} P_{363} &= 1.0 \exp \left[- \left(\dfrac{40,700}{8.3145}\right) \left(\dfrac{1}{363\;K} -\dfrac{1}{373\; K}\right) \right] \nonumber \[4pt] &= 0.697\; atm \nonumber \end{align} \nonumber
\begin{align} P_{383} &= 1.0 \exp \left[- \left( \dfrac{40,700}{8.3145} \right)\left(\dfrac{1}{383\;K} - \dfrac{1}{373\;K} \right) \right] \nonumber \[4pt] &= 1.409\; atm \nonumber \end{align} \nonumber
Note that the increase in vapor pressure from 363 K to 373 K is 0.303 atm, but the increase from 373 to 383 K is 0.409 atm. The increase in vapor pressure is not a linear process.
Discussion
We can use the Clausius-Clapeyron equation to construct the entire vaporization curve. There is a deviation from experimental value, that is because the enthalpy of vaporization varies slightly with temperature.
The Clausius-Clapeyron equation can be also applied to sublimation; the following example shows its application in estimating the heat of sublimation.
Example $2$: Sublimation of Ice
The vapor pressures of ice at 268 K and 273 K are 2.965 and 4.560 torr respectively. Estimate the heat of sublimation of ice.
Solution
The enthalpy of sublimation is $\Delta{H}_{sub}$. Use a piece of paper and derive the Clausius-Clapeyron equation so that you can get the form:
\begin{align} \Delta H_{sub} &= \dfrac{ R \ln \left(\dfrac{P_{273}}{P_{268}}\right)}{\dfrac{1}{268 \;K} - \dfrac{1}{273\;K}} \nonumber \[4pt] &= \dfrac{8.3145 \ln \left(\dfrac{4.560}{2.965} \right)}{ \dfrac{1}{268\;K} - \dfrac{1}{273\;K} } \nonumber \[4pt] &= 52,370\; J\; mol^{-1}\nonumber \end{align} \nonumber
Note that the heat of sublimation is the sum of heat of melting (6,006 J/mol at 0°C and 101 kPa) and the heat of vaporization (45,051 J/mol at 0 °C).
Exercise $2$
Show that the vapor pressure of ice at 274 K is higher than that of water at the same temperature. Note the curve of vaporization is also called the curve of evaporization.
Example $3$: Vaporization of Ethanol
Calculate $\Delta{H_{vap}}$ for ethanol, given vapor pressure at 40 oC = 150 torr. The normal boiling point for ethanol is 78 oC.
Solution
Recognize that we have TWO sets of $(P,T)$ data:
• Set 1: (150 torr at 40+273K)
• Set 2: (760 torr at 78+273K)
We then directly use these data in Equation \ref{2B}
\begin{align*} \ln \left(\dfrac{150}{760} \right) &= \dfrac{-\Delta{H_{vap}}}{8.314} \left[ \dfrac{1}{313} - \dfrac{1}{351}\right] \[4pt] \ln 150 -\ln 760 &= \dfrac{-\Delta{H_{vap}}}{8.314} \left[ \dfrac{1}{313} - \dfrac{1}{351}\right] \[4pt] -1.623 &= \dfrac{-\Delta{H_{vap}}}{8.314} \left[ 0.0032 - 0.0028 \right] \end{align*}
Then solving for $\Delta{H_{vap}}$
\begin{align*} \Delta{H_{vap}} &= 3.90 \times 10^4 \text{ joule/mole} \[4pt] &= 39.0 \text{ kJ/mole} \end{align*}
Advanced Note
It is important to not use the Clausius-Clapeyron equation for the solid to liquid transition. That requires the use of the more general Clapeyron equation
$\dfrac{dP}{dT} = \dfrac{\Delta \bar{H}}{T \Delta \bar{V}} \nonumber$
where $\Delta \bar{H}$ and $\Delta \bar{V}$ is the molar change in enthalpy (the enthalpy of fusion in this case) and volume respectively between the two phases in the transition.
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Phase transition is when a substance changes from a solid, liquid, or gas state to a different state. Every element and substance can transition from one phase to another at a specific combination of temperature and pressure.
Phase Changes
Each substance has three phases it can change into; solid, liquid, or gas(1). Every substance is in one of these three phases at certain temperatures. The temperature and pressure at which the substance will change is very dependent on the intermolecular forces that are acting on the molecules and atoms of the substance(2). There can be two phases coexisting in a single container at the same time. This typically happens when the substance is transitioning from one phase to another. This is called a two-phase state(4). In the example of ice melting, while the ice is melting, there is both solid water and liquid water in the cup.
There are six ways a substance can change between these three phases; melting, freezing, evaporating, condensing, sublimination, and deposition(2). These processes are reversible and each transfers between phases differently:
• Melting: The transition from the solid to the liquid phase
• Freezing: The transition from the liquid phase to the solid phase
• Evaporating: The transition from the liquid phase to the gas phase
• Condensing:The transition from the gas phase to the liquid phase
• Sublimination: The transition from the solid phase to the gas phase
• Deposition: The transition from the gas phase to the solid phase
How Phase Transition works
There are two variables to consider when looking at phase transition, pressure (P) and temperature (T). For the gas state, The relationship between temperature and pressure is defined by the equations below:
Ideal Gas Law:
$PV=nRT$
van der Waals Equation of State:
$\left(P+a*\frac{n^2}{V^2}\right)\left(V-nb\right)=nRT$
Where V is volume, R is the gas constant, and n is the number of moles of gas.
The ideal gas law assumes that no intermolecular forces are affecting the gas in any way, while the van der Waals equation includes two constants, a and b, that account for any intermolecular forces acting on the molecules of the gas.
Temperature
Temperature can change the phase of a substance. One common example is putting water in a freezer to change it into ice. In the picture above, we have a solid substance in a container. When we put it on a heat source, like a burner, heat is transferred to the substance increasing the kinetic energy of the molecules in the substance. The temperature increases until the substance reaches its melting point(2). As more and more heat is transferred beyond the melting point, the substance begins to melt and become a liquid(3). This type of phase change is called an isobaric process because the pressure of the system stays at a constant level.
Melting point (Tf)
Each substance has a melting point. The melting point is the temperature that a solid will become a liquid. At different pressures, different temperatures are required to melt a substance. Each pure element on the periodic table has a normal melting point, the temperature that the element will become liquid when the pressure is 1 atmosphere(2).
Boiling Point (Tb)
Each substance also has a boiling point. The boiling point is the temperature that a liquid will evaporate into a gas. The boiling point will change based on the temperature and pressure. Just like the melting point, each pure element has a normal boiling point at 1 atmosphere(2).
Pressure
Pressure can also be used to change the phase of the substance. In the picture above, we have a container fitted with a piston that seals in a gas. As the piston compresses the gas, the pressure increases. Once the boiling point has been reached, the gas will condense into a liquid. As the piston continues to compress the liquid, the pressure will increase until the melting point has been reached. The liquid will then freeze into a solid. This example is for an isothermal process where the temperature is constant and only the pressure is changing.
A Brief Explanation of a Phase Diagram
Phase transition can be represented with a phase diagram. A phase diagram is a visual representation of how a substance changes phases.
This is an example of a phase diagram. Often, when you are asked about a phase transition, you will need to refer to a phase diagram to answer it. These diagrams usually have the normal boiling point and normal melting point marked on them, and have the pressures on the y-axis and temperatures on the x-axis. The bottom curve marks the temperature and pressure combinations in which the substance will subliminate (1). The left left marks the temperature and pressure combinations in which the substance will melt (1). Finally, the right line marks the conditions under which the substance will evaporate (1).
Problems
1. Using the phase diagram for carbon dioxide below, explain what phase carbon dioxide is normally in at standard temperature and pressure, 1 atm and 273.15 K.
Phase diagram for CO2.from Wikipedia.
2: Looking at the same diagram, we see that carbon dioxide does not have a normal melting point or a normal boiling point. Explain what kind of a change carbon dioxide makes at 1 atm and estimate the temperature of this point.
Solutions
1: Before we can completely answer the question, we need to convert the given information to match the units in the diagram. First we convert 25 degrees Kelvin into Celsius: $K=273.15+C$ $298.15-273.25C$ Now we can look at the diagram and determine its phase. At 25 degrees Celsius and 1 atm carbon dioxide is in the gas phase.
2: Carbon dioxide sublimes at 1 atm because it transitions from the solid phase directly to the gas phase. The temperature of sublimation at 1 atm is about -80 degrees Celsius.
Contributors and Attributions
• Kirsten Amdahl (UC Davis)
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Phase diagram is a graphical representation of the physical states of a substance under different conditions of temperature and pressure. A typical phase diagram has pressure on the y-axis and temperature on the x-axis. As we cross the lines or curves on the phase diagram, a phase change occurs. In addition, two states of the substance coexist in equilibrium on the lines or curves.
Introduction
A phase transition is the transition from one state of matter to another. There are three states of matter: liquid, solid, and gas.
• Liquid: A state of matter that consists of loose, free moving particles which form the shape set by the boundaries of the container in which the liquid is in. This happens because the motion of the individual particles within a liquid is much less restricted than in a solid. One may notice that some liquids flow readily whereas some liquids flow slowly. A liquid's relative resistance to flow is viscosity.
• Solid: A state of matter with tightly packed particles which do not change the shape or volume of the container that it is in. However, this does not mean that the volume of a solid is a constant. Solids can expand and contract when temperatures change. This is why when you look up the density of a solid, it will indicate the temperature at which the value for density is listed. Solids have strong intermolecular forces that keep particles in close proximity to one another. Another interesting thing to think about is that all true solids have crystalline structures. This means that their particles are arranged in a three-dimensional, orderly pattern. Solids will undergo phase changes when they come across energy changes.
• Gas: A state of matter where particles are spread out with no definite shape or volume. The particles of a gas will take the shape and fill the volume of the container that it is placed in. In a gas, there are no intermolecular forces holding the particles of a gas together since each particle travels at its own speed in its own direction. The particles of a gas are often separated by great distances.
Phase diagrams illustrate the variations between the states of matter of elements or compounds as they relate to pressure and temperatures. The following is an example of a phase diagram for a generic single-component system:
• Triple point – the point on a phase diagram at which the three states of matter: gas, liquid, and solid coexist
• Critical point – the point on a phase diagram at which the substance is indistinguishable between liquid and gaseous states
• Fusion(melting) (or freezing) curve – the curve on a phase diagram which represents the transition between liquid and solid states
• Vaporization (or condensation) curve – the curve on a phase diagram which represents the transition between gaseous and liquid states
• Sublimation (or deposition) curve – the curve on a phase diagram which represents the transition between gaseous and solid states
Phase diagrams plot pressure (typically in atmospheres) versus temperature (typically in degrees Celsius or Kelvin). The labels on the graph represent the stable states of a system in equilibrium. The lines represent the combinations of pressures and temperatures at which two phases can exist in equilibrium. In other words, these lines define phase change points. The red line divides the solid and gas phases, represents sublimation (solid to gas) and deposition (gas to solid). The green line divides the solid and liquid phases and represents melting (solid to liquid) and freezing (liquid to solid). The blue divides the liquid and gas phases, represents vaporization (liquid to gas) and condensation (gas to liquid). There are also two important points on the diagram, the triple point and the critical point. The triple point represents the combination of pressure and temperature that facilitates all phases of matter at equilibrium. The critical point terminates the liquid/gas phase line and relates to the critical pressure, the pressure above which a supercritical fluid forms.
With most substances, the temperature and pressure related to the triple point lie below standard temperature and pressure and the pressure for the critical point lies above standard pressure. Therefore at standard pressure as temperature increases, most substances change from solid to liquid to gas, and at standard temperature as pressure increases, most substances change from gas to liquid to solid.
Exception: Water
Normally the solid/liquid phase line slopes positively to the right (as in the diagram for carbon dioxide below). However for other substances, notably water, the line slopes to the left as the diagram for water shows. This indicates that the liquid phase is more dense than the solid phase. This phenomenon is caused by the crystal structure of the solid phase. In the solid forms of water and some other substances, the molecules crystalize in a lattice with greater average space between molecules, thus resulting in a solid with a lower density than the liquid. Because of this phenomenon, one is able to melt ice simply by applying pressure and not by adding heat.
Moving About the Diagram
Moving about the phase diagram reveals information about the phases of matter. Moving along a constant temperature line reveals relative densities of the phases. When moving from the bottom of the diagram to the top, the relative density increases. Moving along a constant pressure line reveals relative energies of the phases. When moving from the left of the diagram to the right, the relative energies increases.
Important Definitions
• Sublimation is when the substance goes directly from solid to the gas state.
• Deposition occurs when a substance goes from a gas state to a solid state; it is the reverse process of sublimation.
• Melting occurs when a substance goes from a solid to a liquid state.
• Fusion is when a substance goes from a liquid to a solid state, the reverse of melting.
• Vaporization (or evaporation) is when a substance goes from a liquid to a gaseous state.
• Condensation occurs when a substance goes from a gaseous to a liquid state, the reverse of vaporization.
• Critical Point – the point in temperature and pressure on a phase diagram where the liquid and gaseous phases of a substance merge together into a single phase. Beyond the temperature of the critical point, the merged single phase is known as a supercritical fluid.
• Triple Point occurs when both the temperature and pressure of the three phases of the substance coexist in equilibrium.
Problems
Imagine a substance with the following points on the phase diagram: a triple point at .5 atm and -5°C; a normal melting point at 20°C; a normal boiling point at 150°C; and a critical point at 5 atm and 1000°C. The solid liquid line is "normal" (meaning positive sloping). For this, complete the following:
1. Roughly sketch the phase diagram, using units of atmosphere and Kelvin.
Answer
1-solid, 2-liquid, 3-gas, 4-supercritical fluid, point O-triple point, C-critical point -78.5 °C (The phase of dry ice changes from solid to gas at -78.5 °C)
2. Rank the states with respect to increasing density and increasing energy.
3. Describe what one would see at pressures and temperatures above 5 atm and 1000°C.
Answer
One would see a super-critical fluid, when approaching the point, one would see the meniscus between the liquid and gas disappear.
4. Describe what will happen to the substance when it begins in a vaccum at -15°C and is slowly pressurized.
Answer
The substance would begin as a gas and as the pressure increases, it would compress and eventually solidify without liquefying as the temperature is below the triple point temperature.
5. Describe the phase changes from -80°C to 500°C at 2 atm.
Answer
The substance would melt at somewhere around, but above 20°C and then boil at somewhere around, but above 150°C. It would not form a super-critical fluid as the neither the pressure nor temperature reach the critical pressure or temperature.
6. What exists in a system that is at 1 atm and 150°?
Answer
Depending on how much energy is in the system, there will be different amounts of liquid and gas at equilibrium. If just enough energy was added to raise the temperature of the liquid to 150°C, there will just be liquid. If more was added, there will be some liquid and some gas. If just enough energy was added to change the state of all of the liquid without raising the temperature of the gas, there will just be gas.
7. Label the area 1, 2, 3, and 4 and points O and C on the diagram.
8. A sample of dry ice (solid CO2) is cooled to -100 °C, and is set on a table at room temperature (25 °C). At what temperature is the rate of sublimation and deposition the same? (Assume pressure is held constant at 1 atm).
Contributors and Attributions
• Matthew McKinnell (UCD), Jessie Verhein (UCD), Pei Yu (UCD), Lok Ka Chan (UCD), Jessica Dhaliwal (UCD), Shyall Bhela (UCD), Candace Wong-Sing (UCD)
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This page takes a simple look at solids, liquids and gases, and changes of state such as melting and boiling, in terms of the behavior of the particles present.
The arrangements of particles in solids, liquids and gases
A simple view of the arrangement of the particles in solids, liquids and gases looks like this:
Solids
In the solid, the particles are touching, and the only motion allowed to them is vibration. The particles may be arranged regularly (in which case, the solid is crystalline), or at random (giving waxy solids like candles or some forms of polythene, for example).
The particles are held in the solid by forces which depend on the actual substance - ionic bonds, covalent bonds, metallic bonds, hydrogen bonds or van der Waals attractions.
Liquids
In a liquid, the particles are mainly touching, but some gaps have appeared in the structure. These gaps allow the particles to move, and so the particles are arranged randomly.The forces that held the solid particles together are also present in the liquid (unless melting has broken up a substance consisting only of covalent bonds - a giant covalent structure). However, the particles in the liquid have enough energy to prevent the forces holding them in a fixed arrangement.
For most liquids, the density of the liquid is slightly less than that of the solid, but there isn't much difference. That means that the particles in the liquid are almost as close together as they are in a solid. If you draw diagrams of liquids, make sure that most of the particles are touching, but at random, with a few gaps.
Gases
In a gas, the particles are entirely free to move. At ordinary pressures, the distance between individual particles is of the order of ten times the diameter of the particles. At that distance, any attractions between the particles are fairly negligible at ordinary temperatures and pressures.
Changes of state
Melting and freezing
If energy is supplied by heating a solid, the heat energy causes stronger vibrations until the particles eventually have enough energy to break away from the solid arrangement to form a liquid. The heat energy required to convert 1 mole of solid into a liquid at its melting point is called the enthalpy of fusion.
When a liquid freezes, the reverse happens. At some temperature, the motion of the particles is slow enough for the forces of attraction to be able to hold the particles as a solid. As the new bonds are formed, heat energy is evolved.
Boiling and condensing
If more heat energy is supplied, the particles eventually move fast enough to break all the attractions between them, and the liquid boils. The heat energy required to convert 1 mole of liquid into a gas at its boiling point is called the enthalpy of vaporization. If the gas is cooled, at some temperature the gas particles will slow down enough for the attractions to become effective enough to condense it back into a liquid. Again, as those forces are re-established, heat energy is released.
The evaporation of a liquid
The average energy of the particles in a liquid is governed by the temperature. The higher the temperature, the higher the average energy. But within that average, some particles have energies higher than the average, and others have energies lower than the average. Some of the more energetic particles on the surface of the liquid can be moving fast enough to escape from the attractive forces holding the liquid together. They evaporate.
Fig 1: The diagram shows a small region of a liquid near its surface.
Notice that evaporation only takes place on the surface of the liquid. That's quite different from boiling which happens when there is enough energy to disrupt the attractive forces throughout the liquid. That's why, if you look at boiling water, you see bubbles of gas being formed all the way through the liquid. If you look at water which is just evaporating in the sun, you don't see any bubbles. Water molecules are simply breaking away from the surface layer.
Eventually, the water will all evaporate in this way. The energy which is lost as the particles evaporate is replaced from the surroundings. As the molecules in the water jostle with each other, new molecules will gain enough energy to escape from the surface.
The evaporation of a liquid in a closed container
Now imagine what happens if the liquid is in a closed container. Common sense tells you that water in a sealed bottle doesn't seem to evaporate - or at least, it doesn't disappear over time. But there is constant evaporation from the surface. Particles continue to break away from the surface of the liquid - but this time they are trapped in the space above the liquid.
As the gaseous particles bounce around, some of them will hit the surface of the liquid again, and be trapped there. There will rapidly be an equilibrium set up in which the number of particles leaving the surface is exactly balanced by the number rejoining it.
In this equilibrium, there will be a fixed number of the gaseous particles in the space above the liquid. When these particles hit the walls of the container, they exert a pressure. This pressure is called the saturated vapor pressure (also known as saturation vapor pressure) of the liquid.
Sublimation
Solids can also lose particles from their surface to form a vapor, except that in this case we call the effect sublimation rather than evaporation. Sublimation is the direct change from solid to vapor (or vice versa) without going through the liquid stage. In most cases, at ordinary temperatures, the saturated vapor pressures of solids range from low to very, very, very low. The forces of attraction in many solids are too high to allow much loss of particles from the surface. However, there are some which do easily form vapors. For example, naphthalene (used in old-fashioned "moth balls" to deter clothes moths) has quite a strong smell. Molecules must be breaking away from the surface as a vapor, because otherwise you wouldn't be able to smell it.
Another fairly common example (discussed in detail elsewhere on the site) is solid carbon dioxide - "dry ice". This never forms a liquid at atmospheric pressure and always converts directly from solid to vapor. That's why it is known as dry ice.
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This page looks at how the equilibrium between a liquid (or a solid) and its vapor leads to the idea of a saturated vapor pressure. It also looks at how saturated vapor pressure varies with temperature, and the relationship between saturated vapor pressure and boiling point.
Evaporation: Liquid/Vapor Equilibrium
The average energy of the particles in a liquid is governed by the temperature. The higher the temperature, the higher the average energy. But within that average, some particles have energies higher than the average, and others have energies lower than the average. Some of the more energetic particles on the surface of the liquid can be moving fast enough to escape from the attractive forces holding the liquid together. They evaporate. The diagram shows a small region of a liquid near its surface.
Notice that evaporation only takes place on the surface of the liquid. That's quite different from boiling which happens when there is enough energy to disrupt the attractive forces throughout the liquid. That's why, if you look at boiling water, you see bubbles of gas being formed all the way through the liquid.
If you look at water which is just evaporating in the sun, you don't see any bubbles. Water molecules are simply breaking away from the surface layer. Eventually, the water will all evaporate in this way. The energy which is lost as the particles evaporate is replaced from the surroundings. As the molecules in the water jostle with each other, new molecules will gain enough energy to escape from the surface.
The evaporation of a liquid in a closed container
Now imagine what happens if the liquid is in a closed container. Common sense tells you that water in a sealed bottle does not seem to evaporate - or at least, it does not disappear over time. But there is constant evaporation from the surface. Particles continue to break away from the surface of the liquid - but this time they are trapped in the space above the liquid.
As the gaseous particles bounce around, some of them will hit the surface of the liquid again, and be trapped there. There will rapidly be an equilibrium set up in which the number of particles leaving the surface is exactly balanced by the number rejoining it.
In this equilibrium, there will be a fixed number of the gaseous particles in the space above the liquid. When these particles hit the walls of the container, they exert a pressure. This pressure is called the saturated vapor pressure (also known as saturation vapor pressure) of the liquid.
Measuring the saturated vapor pressure
It is not difficult to show the existence of this saturated vapor pressure (and to measure it) using a simple piece of apparatus. If you have a mercury barometer tube in a trough of mercury, at 1 atmosphere pressure the column will be 760 mm tall. 1 atmosphere is sometimes quoted as 760 mmHg ("millimetres of mercury").
If you squirt a few drops of liquid into the tube, it will rise to form a thin layer floating on top of the mercury. Some of the liquid will evaporate and you will get the equilibrium we've just been talking about - provided there is still some liquid on top of the mercury. It is only an equilibrium if both liquid and vapor are present.
The saturated vapor pressure of the liquid will force the mercury level down a bit. You can measure the drop - and this gives a value for the saturated vapor pressure of the liquid at this temperature. In this case, the mercury has been forced down by a distance of 760 - 630 mm. The saturated vapor pressure of this liquid at the temperature of the experiment is 130 mmHg. You could convert this into proper SI units (pascals) if you wanted to. 760 mmHg is equivalent to 101,325 Pa.
A value of 130 mmHg is quite a high vapor pressure if we are talking about room temperature. Water's saturated vapor pressure is about 20 mmHg at this temperature. A high vapor pressure means that the liquid must be volatile - molecules escape from its surface relatively easily, and aren't very good at sticking back on again either.
That will result in larger numbers of them in the gas state once equilibrium is reached. The liquid in the example must have significantly weaker intermolecular forces than water.
The variation of saturated vapor pressure with temperature
The effect of temperature liquid/ vapor equilibrium
You can look at this in two ways. (1) There is a common sense way. If you increase the temperature, you are increasing the average energy of the particles present. That means that more of them are likely to have enough energy to escape from the surface of the liquid. That will tend to increase the saturated vapor pressure. (2) Or you can look at it in terms of Le Chatelier's Principle - which works just as well in this kind of physical situation as it does in the more familiar chemical examples.
When the space above the liquid is saturated with vapor particles, you have this equilibrium occurring on the surface of the liquid:
The forward change (liquid to vapor) is endothermic. It needs heat to convert the liquid into the vapor. According to Le Chatelier, increasing the temperature of a system in a dynamic equilibrium favors the endothermic change. That means that increasing the temperature increases the amount of vapor present, and so increases the saturated vapor pressure.
The effect of temperature on the saturated vapor pressure of water
The graph shows how the saturated vapor pressure (svp) of water varies from 0°C to 100 °C. The pressure scale (the vertical one) is measured in kilopascals (kPa). 1 atmosphere pressure is 101.325 kPa.
Saturated vapor pressure and boiling point
A liquid boils when its saturated vapor pressure becomes equal to the external pressure on the liquid. When that happens, it enables bubbles of vapor to form throughout the liquid - those are the bubbles you see when a liquid boils.
If the external pressure is higher than the saturated vapor pressure, these bubbles are prevented from forming, and you just get evaporation at the surface of the liquid. If the liquid is in an open container and exposed to normal atmospheric pressure, the liquid boils when its saturated vapor pressure becomes equal to 1 atmosphere (or 101325 Pa or 101.325 kPa or 760 mmHg). This happens with water when the temperature reaches 100°C.
But at different pressures, water will boil at different temperatures. For example, at the top of Mount Everest the pressure is so low that water will boil at about 70°C. Whenever we just talk about "the boiling point" of a liquid, we always assume that it is being measured at exactly 1 atmosphere pressure. In practice, of course, that is rarely exactly true.
Sublimation: solid/vapor Equilibrium
Solids can also lose particles from their surface to form a vapor, except that in this case we call the effect sublimation rather than evaporation. Sublimation is the direct change from solid to vapor (or vice versa) without going through the liquid stage.
In most cases, at ordinary temperatures, the saturated vapor pressures of solids range from low to very, very, very low. The forces of attraction in many solids are too high to allow much loss of particles from the surface. However, there are some which do easily form vapors. For example, naphthalene (used in old-fashioned "moth balls" to deter clothes moths) has quite a strong smell. Molecules must be breaking away from the surface as a vapor, because otherwise you would not be able to smell it. Another fairly common example (discussed in detail on another page) is solid carbon dioxide - "dry ice". This never forms a liquid at atmospheric pressure and always converts directly from solid to vapor. That's why it is known as dry ice.
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A gas has NEITHER a definite volume NOR shape. At the microscopic level, gases particles are very far apart. These particles move fast and independently of each other. Such properties allow a gas to expand inside its container.
Thumbnail: Drifting smoke particles provide clues to the movement of the surrounding gas. (Public Domain; Macluskie).
Properties of Gases
What you Should Know
Before beginning this section, you should know and understand:
Chemical reactions between gaseous materials are quite similar to reactions between solids and liquids, except the Ideal Gas Law ($PV=nRT$) can now be included in the calculations. If a chemical reaction is reversible (such as the decay and formation of dinitrogen tetraoxide), then Dalton's Law of Partial Pressure may be used to determine the moles of reactants and products at which the reaction ceases (and subsequently, the temperature, pressure and volume of each gas can be determined as well).
Non-Reversible Reactions
A non-reversible reaction uses the reactants to form the products. The reaction goes in one direction; that is, using the product to recreate the reactants has greatly different requirements. One of the most common forms of non-reversible reactions is combustion (once an organic molecule has been converted to water and hydrogen gas, it is extremely difficult to reform). Other non-reversible reactions produce a state change, such as Hydrogen Peroxide (the gaseous material produces water, a liquid). To understand how the Ideal Gas Law applies to reactions, we shall use a nonreversible reaction as an example.
Example $1$: the decomposition of Hydrogen Peroxide to water and oxygen gas
If 4.000 grams of hydrogen peroxide is placed within a sealed 250 milliliter container at 500 K. What is the pressure of the oxygen gas produced in atmospheres?
$\ce{2H_2O_2 \rightarrow 2H_2O + O_2} \nonumber$
Solution
First, we need to determine the moles of $\ce{O_2}$ produced, just like any other stoichiometric problem.
$(4g\; \cancel{\ce{H_2O_2}}) \times \left(\dfrac{1\; mol\; \cancel{\ce{H_2O_2}}}{34.016\;g\; \cancel{ \ce{H_2O_2}}} \right) \left(\dfrac{1\; mol\; \ce{O_2}}{2\;mol\; \cancel{\ce{H_2O_2}}} \right) = 0.0588 \;mol\; \ce{O_2} \nonumber$
With the moles of oxygen determined, we can now use the Ideal Gas Law to determine the pressure.
$PV=nRT \nonumber$
The volume (250 mL = 0.25 L) and temperature (500 K) are already given to us, and R (0.0820574 Latm mol-1K-1) is a constant.
\begin{align*} P &=\dfrac{nRT}{V} \[4pt] &= \dfrac{(0.0588\; mol\; O_2) \times (0.0820\; L \;atm \;mol^{-1}\;K^{-1}) \times (500 \;K)}{0.25\;L} \[4pt] &= 9.65\; atm \end{align*}
By using the Ideal Gas Law for unit conversions, properties such as the pressure, volume, moles, and temperature of a gas involved in a reaction can be determined. However, a different approach is needed to solve reversible reactions.
For further clarification, when solving equations with gases, we must remember that gases behave differently under different conditions. For example, if we have a certain temperature or pressure, this can change the number of moles produced or the volume. This is unlike regular solids where we only had to account for the mass of the solids and solve for the mass of the product by stoichiometry. In order to solve for the temperature, pressure, or volume of a gas using chemical reactions, we often need to have information on two out of three of these variables. So we need either the temperature and volume, temperature and pressure, or pressure and volume. The mass we can find using stoichiometric conversions we have learned before.
The reason why gases require additional information is because gases behave as ideal gases and ideal gases behave differently under different conditions. To account for these conditions, we use the ideal gas equation PV=nRT where P is the pressure measured in atmosphere(atm), V is the volume measured in liters (L), n is the number of moles, R is the gas constant with a value of .08206 L atm mol-1 K-1, and T is the temperature measured in kelvin (K).
Example $2$
Suppose we have the following combustion reaction (below). If we are given 2 moles of ethane at STP, how many liters of CO2 are produced?
$\ce{2C2H6(s) + 7O2(g) -> 6H2O(l) + 4CO2(g)}$
Solution
Step 1
First use stoichiometry to solve for the number of moles of CO2 produced.
$(2\, mol\, \ce{C2H6} )\left(\dfrac{4 \,mol \ce{CO2}}{2\, mol\, \ce{ C2H6}}\right) = 4 mol \, \ce{CO2} \nonumber$
So 4 moles of Carbon Dioxide are produced if we react 2 moles of ethane gas.
Step 2
Now we simply need to manipulate the ideal gas equation to solve for the variable of interest. In this case we are solving for the number of liters.
Since we are told ethane is at STP, we know that the temperature is 273 K and the pressure is 1 atm.
So the variables we have are:
• V = ?
• T = 273K
• P = 1 atm
• n = 4 moles CO2
• R (gas constant) = 0.08206 L atm mol-1 K-1
Isolating the variable of interest from $PV=nRT$, we get
\begin{align*} V &=\dfrac{nRT}{P} \[4pt] &= \dfrac{4\, mol \times 0.08206 \,L \,atm \,mol^{-1} K^{-1} \times 273\,K}{1\, atm} \[4pt] &= 89.61\, L \end{align*}
So we have a volume of 89.61 liters.
Reversible Reactions in Gases
A reversible reaction is a chemical reaction in which reactants produces a product, which then decays back to the reactants. This continues until the products and reactants are in equilibrium. In other words, the final state of the gas includes both the reactants and the products. For example, Reactant A combines with Reactant B to form Product AB, which then breaks apart into A and B, until an equilibrium of the three is reached. In a reaction between gases, determining gas properties such as partial pressure and moles can be quite difficult. For this example, we consider the theral decomposition of Dinitrogen Tetraoxide into Nitrogen Dioxide.
Example $3$
For this example, we shall use Dinitrogen Tetraoxide, which decomposes to form Nitrogen Dioxide.
$\ce{N_2O_4 <=> 2NO_2} \nonumber$
2 atm of dinitrogen tetraoxide is added to a 500 mL container at 273 K. After several minutes, the total pressure of N2O4 and 2NO2 at equilibrium is found to be 3.2 atm. Find the partial pressures of both gases.
Solution
The simplest way of solving this problem is to begin with an ICE table.
$\ce{N2O4}$
$\ce{2NO2}$ Description of Each Letter
Initial 2atm 0atm The initial amounts of reactants and products
Change -X +2X The unknown change, X, each one multiplied by their stoichiometric factor
Equilibrium 2-Xatm 2x atm The initial + the change
With this data, a simple equation can be derived to determine the value of X.
$P_{total} = (2-X) + 2X = 3.2\;atm$
$X = 1.2\; atm$
$P_{NO_2}= 2x = 2.4\; atm$
$P_{N_20_4} = 2-x = 0.8\; atm$
Law of Combining Volumes
This law of combining volumes was first discovered by the famous scientist Gay-Lussac who noticed this relationship. He determined that if certain gases that are products and reactions in a chemical reaction are measured at the same conditions, temperature and pressure, then the volume of gas consumed/produced is equal to the ratio between the gases or the ratio of the coefficients.
Example 5.4.4
If ozone, hydrogen, and oxygen were all measured at 35oC and at 753 mmHg, then how many liters of ozone was consumed if you had 5 liters of oxygen gas?
$\ce{O3(g) + H2O(l) -> H2(g) + 2O2(g)} \nonumber$
Solution
Step 1
Identify what we are looking for and if any relationships can be spotted. In this case, we can see that there are three gases all at the same temperature and pressure, which follows Gay-Lussac’s Law of Combining Volumes. We can now proceed to use his law.
Step 2
We simply change the coefficients to volumetric ratios. So for every 1 Liter of Ozone gas we have, we produce 1 Liter of H2gas and 2 Liter of $O_2$ gas.
We are given 5 liters of Oxygen gas and want to solve for the amount of liters of ozone consumed. We simply use the 2:1 stoichiometry of the reaction.
$5 L O_2 \left(\dfrac{1\; L\; O_3}{2\; L\; O_2}\right) = 2.5\; L\; O_3$
Problems
1. A 450 mL container of oxygen gas is at STP. Hydrogen gas is pumped into the container, producing water. What is the least amount of mL of Hydrogen gas needed in order to react the oxygen to completion? $2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)}$
2. This reaction occurred at 427 Kelvin, with 37 g of $CH_4$ and an excess of oxygen. The carbon dioxide produced was captured in a 30L sealed container. What is the pressure of the carbon dioxide within the container? $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
Solution
1. $900\; mL\; H_2$: To solve this question you simply use Gay-Lussac’s law of combining volume because both gases are at the same temperature at pressure. $450\; mL\; O_2 \left(\dfrac{2\; mL\; H_2}{1\; mL\; O_2}\right) = 900\; mL\; H_2$
2. $2.7\; atm$: For a question like this, you want to first determine the number of moles of the compound in interest first using stoichiometry and then using the ideal gas law to solve for the variable of interest. $37 \;g \;CH_4 \left(\dfrac{1\; mol\; CH_4}{16\; g\; CH_4}\right)\left(\dfrac{1\; mol\; CO_2}{1\; mol\; CH_4}\right)= 2.3 \;mol\; CO_2$
Now use the ideal gas law to solve for the pressure of CO2.
• PV=nRT
• P = ?
• V = 30L
• n= 2.3 mol CO2
• T = 427 K
• P = nRT/V
P = (2.3 mol * .08206 L atm mol-1 K-1 * 427 K)/(30 L)
P=2.7 atm
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Chemical_Reactions_in_Gas_Phase.txt
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The gas laws are empirical laws that describe the properties of gases and typically include Avogadro's Law, Boyle's Law, and Charles' laws and Dalton's Laws. These are sometimes call the ABCD laws for convenience and can be derived from the Ideal Gas Law. Deviations of these laws occur when considering real gases under high pressures or low temperatures.
Gas Laws
Discovering that the volume of a gas was directly proportional to the number of particles it contained was crucial in establishing the formulas for simple molecules at a time (around 1811) when the distinction between atoms and molecules was not clearly understood. In particular, the existence of diatomic molecules of elements such as $H_2$, $O_2$, and $Cl_2$ was not recognized until the results of experiments involving gas volumes was interpreted.
Early chemists mistakenly assumed that the formula for water was $HO$, leading them to miscalculate the molecular weight of oxygen as 8 instead of 16. However, when chemists found that an assumed reaction of
$H+Cl→HCl$
yielded twice the volume of $HCl$, they realized hydrogen and chlorine were diatomic molecules. The chemists revised their reaction equation to be
$H_2+Cl_2→2HCl.$
When chemists revisited their water experiment and their hypothesis that
$HO \rightarrow H + O$
they discovered that the volume of hydrogen gas consumed was twice that of oxygen. By Avogadro's Law, this meant that hydrogen and oxygen were combining in a 2:1 ratio. This discovery led to the correct molecular formula for water ($H_2O$) and the correct reaction
$2H_2O→2H_2+O_2.$
Summary
• The number of molecules or atoms in a specific volume of ideal gas is independent of size or the gas' molar mass.
• Avogadro's Law is stated mathematically as follows: $\frac{V}{n} = k$, where $V$ is the volume of the gas, n is the number of moles of the gas, and k is a proportionality constant.
• Volume ratios must be related to the relative numbers of molecules that react; this relationship was crucial in establishing the formulas of simple molecules at a time when the distinction between atoms and molecules was not clearly understood.
Contributors and Attributions
• Boundless (www.boundless.com)
Boyle's Law
For a fixed mass of gas at constant temperature, the volume is inversely proportional to the pressure. This is mathematically:
\[pV = constant\]
That means that, for example, if you double the pressure, you will halve the volume. If you increase the pressure 10 times, the volume will decrease 10 times.
Is this consistent with pV = nRT ?
• You have a fixed mass of gas, so n (the number of moles) is constant.
• R is always constant - it is called the gas constant.
• Boyle's Law demands that temperature is constant as well.
That means that everything on the right-hand side of pV = nRT is constant, and so pV is constant - which is what we have just said is a result of Boyle's Law.
Simple Explanation
This is easiest to see if you think about the effect of decreasing the volume of a fixed mass of gas at constant temperature. Pressure is caused by gas molecules hitting the walls of the container. With a smaller volume, the gas molecules will hit the walls more frequently, and so the pressure increases.
You might argue that this isn't actually what Boyle's Law says - it wants you to increase the pressure first and see what effect that has on the volume. But, in fact, it amounts to the same thing. If you want to increase the pressure of a fixed mass of gas without changing the temperature, the only way you can do it is to squeeze it into a smaller volume. That causes the molecules to hit the walls more often, and so the pressure increases.
But everything in the nR/p part of this is constant. That means that V = constant x T, which is Charles's Law.
Charles's Law (Law of Volumes)
Charles' Law
For a fixed mass of gas at constant pressure, the volume is directly proportional to temperature (in Kelvin). This is mathematically
\[ V = constant x T \]
That means, for example, that if you double the temperature from, say to 300 K to 600 K, at constant pressure, the volume of a fixed mass of the gas will double as well. You can express this mathematically as
Is this consistent with pV = nRT ?
• You have a fixed mass of gas, so n (the number of moles) is constant.
• R is the gas constant.
• Charles' Law demands that pressure is constant as well.
If you rearrange the pV = nRT equation by dividing both sides by p, you will get
V = nR/p x T
But everything in the nR/p part of this is constant. That means that V = constant x T, which is Charles' Law.
Contributors and Attributions
Jim Clark (Chemguide.co.uk)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Laws/Avogadro%27s_Law.txt
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Dalton’s Law, or the Law of Partial Pressures, states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the gases in the mixture.
Explanation
Based on the kinetic theory of gases, a gas will diffuse in a container to fill up the space it is in and does not have any forces of attraction between the molecules. In other words, the different molecules in a mixture of gases are so far apart that they act independently; they do not react with each other. The pressure of an ideal gas is determined by its collisions with the container, not collisions with molecules of other substances since there are no other collisions. A gas will expand to fill the container it is in without affecting the pressure of another gas. So it can be concluded that the pressure of a certain gas is based on the number of moles of that gas and the volume and temperature of the system. Since the gases in a mixture of gases are in one container, the Volume (V) and Temperature (T) for the different gases are the same as well. Each gas exerts its own pressure on the system, which can be added up to find the total pressure of the mixture of gases in a container. This is shown by the equation
\[P_{total} = P_A + P_B + ... \tag{1}\]
Derivation
We have, from the Ideal Gas Law
\[PV = nRT \tag{2}\]
If we know the molar composition of the gas, we can write
\[n_{total} = n_a + n_b + ... \tag{3}\]
Again, based on the kinetic theory of gases and the ideal gas law, Dalton’s law can also be applied to the number of moles so that the total number of moles equals the sum of the number of moles of the individual gases. Here, the pressure, temperature and volume are held constant in the system. The total volume of a gas can be found the same way, although this is not used as much. This yields the equation,
\[P_{total} V=n_{total} RT \tag{4}\]
We can rearrange this equation to find the total number of moles. Sometimes masses of each sample of gas are given and students are asked to find the total pressure. This can be done by converting grams to moles and using Dalton's law to find the pressure.
Mole Ratio
From the partial pressure of a certain gas and the total pressure of a certain mixture, the mole ratio, called Xi, of a gas can be found. The mole ratio describes what fraction of the mixture is a specific gas. For example, if oxygen exerts 4 atm of pressure in a mixture and the total pressure of the system is 10 atm, the mole ratio would be 4/10 or 0.4. The mole ratio applies to pressure, volume, and moles as seen by the equation below. This also means that 0.4 moles of the mixture is made up of gas i.
\[X_i=\dfrac{P_i}{P_{tot}}=\dfrac{n_i}{n_{tot}}=\dfrac{V_i}{V_{tot}} \tag{5}\]
The mole ratio, (\(X_i\)) is often used to determine the composition of gases in a mixture. The sum of the mole ratios of each gas in a mixture should always equal one since they represent the proportion of each gas in the mixture.
Collection of a Gas Over Water
The Law of Partial Pressures is commonly applied in looking at the pressure of a closed container of gas and water. The total pressure of this system is the pressure that the gas exerts on the liquid. The gas is made up of whatever sample of gas there is plus the evaporated water. The pressure of the gas on the liquid consists of the pressure of the evaporated water and the pressure of the gas collected. Based on Dalton’s law, the pressure of the gas collected can be calculated by subtracting the pressure of the water vapor from the total pressure.
Real Gases
Real gases are gases that do not behave ideally. That is, they violate one or more of the rules of the kinetic theory of gases. Real gases behave ideally when the gases are at low pressure and high temperature. Therefore at high pressures and low temperatures, Dalton’s law is not applicable since the gases are more likely to react and change the pressure of the system. For example, if there are forces of attraction between the molecules, the molecules would get closer together and the pressure would be adjusted because the molecules are interacting with each other.
Problems
1. A sample of gas A evaporates over water in a closed system. What is the pressure of gas A if the total pressure is 780 torr and water vapor pressure is 1 atm?
2. There is a mixture of 4 moles of hydrogen gas, 8 moles of oxygen, 12 moles of helium, and 6 moles of nitrogen in a closed container. What is the total number of moles in this system?
3. If there is a mixture of hydrogen and oxygen gas in a container with 10 moles total and the mole ratio of hydrogen is .67 molHydrogen to 1 moltotal, how many moles of each gas are there?
4. 24.0 L of nitrogen gas at 2 atm and 12.0 L of oxygen gas at 2 atm are added to a 10 L container at 273 K. Find the partial pressure of nitrogen and oxygen and then find the total pressure.
5. Flourine gas is in a 5.0 L container that is 25 C and 2 atm. A certain amount of hydrogen with a partial pressure of .5 is added to the container. What is the mole ratio of hydrogen?
Solutions
1:
• Convert pressure to same units so 780 torr=1.03 atm
• Subtract water vapor pressure from total pressure to get partial pressure of gas A: PA=1.03 atm- 1 atm=0.03 atm
2. The law of partial pressures also applies to the total number of moles if the other values are constant, so
4 mol Hydrogen+8 mol Oxygen+12 mol Helium+6 mol Nitrogen=30 moles total
3.
• 10 moltotal x .67 molHydrogen/ 1 moltotal= 6.7 moles H2 gas
• 10 moltotal-6.7 molHydrogen=3.3 moles O gas
4.
• Find the number of moles of oxygen and nitrogen using PV=nRT which is n=PV/RT
1. oxygen: ((1 atm)(12L))/(0.08206 atm L mol-1 K-1)(273 K)=0.536 moles oxygen
2. nitrogen: ((1 atm)(24.0L))/(.08206 atm L mol-1 K-1)(273 K)=1.07 moles of Nitrogen
3. add to get ntot: .536 molOxygen+1.07 molNitrogen=1.61 moles total
• Use PV=nRT or P=(nRT)/V to find the total pressure
1. Ptot=((1.61 moltotal)(0.08206 atm L mol-1 K-1)(273 K))/(10.0 L)=3.61 atm
• PA/Ptot=nA/Ntot can be rearranged to PA=(Ptot)(nA/Ntot) to find the partial pressures
1. Poxygen=(3.61 atmtotal)(.536 molOxygen/1.61 moltotal)=1.20 atmOxygen
2. Pnitrogen=3.61 atmtotal-1.20 atmOxygen=2.41 atmNitrogen
5.
• total pressure is 2 atmFlourine+.5 atmHydrogen=2.5 atmtotal
• remember that pH/ptot=nH/ntot=vh/vtot so the mole ratio of hydrogen to the mixture is is .5/2.5=.2 molHydrogen to 1 moltotal
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Laws/Dalton%27s_Law_%28Law_of_Partial_Pressu.txt
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Created in the early 17th century, the gas laws have been around to assist scientists in finding volumes, amount, pressures and temperature when coming to matters of gas. The gas laws consist of three primary laws: Charles' Law, Boyle's Law and Avogadro's Law (all of which will later combine into the General Gas Equation and Ideal Gas Law).
Introduction
The three fundamental gas laws discover the relationship of pressure, temperature, volume and amount of gas. Boyle's Law tells us that the volume of gas increases as the pressure decreases. Charles' Law tells us that the volume of gas increases as the temperature increases. And Avogadro's Law tell us that the volume of gas increases as the amount of gas increases. The ideal gas law is the combination of the three simple gas laws.
Ideal Gases
Ideal gas, or perfect gas, is the theoretical substance that helps establish the relationship of four gas variables, pressure (P), volume(V), the amount of gas(n)and temperature(T). It has characters described as follow:
1. The particles in the gas are extremely small, so the gas does not occupy any spaces.
2. The ideal gas has constant, random and straight-line motion.
3. No forces between the particles of the gas. Particles only collide elastically with each other and with the walls of container.
Real Gases
Real gas, in contrast, has real volume and the collision of the particles is not elastic, because there are attractive forces between particles. As a result, the volume of real gas is much larger than of the ideal gas, and the pressure of real gas is lower than of ideal gas. All real gases tend to perform ideal gas behavior at low pressure and relatively high temperature.
The compressiblity factor (Z) tells us how much the real gases differ from ideal gas behavior.
$Z = \dfrac{PV}{nRT}$
For ideal gases, $Z = 1$. For real gases, $Z\neq 1$.
Boyle's Law
In 1662, Robert Boyle discovered the correlation between Pressure (P)and Volume (V) (assuming Temperature(T) and Amount of Gas(n) remain constant):
$P\propto \dfrac{1}{V} \rightarrow PV=x$
where x is a constant depending on amount of gas at a given temperature.
• Pressure is inversely proportional to Volume
Another form of the equation (assuming there are 2 sets of conditions, and setting both constants to eachother) that might help solve problems is:
$P_1V_1 = x = P_2V_2$
Example 1.1
A 17.50mL sample of gas is at 4.500 atm. What will be the volume if the pressure becomes 1.500 atm, with a fixed amount of gas and temperature?
Solution
$V_2= \dfrac {P_1 \centerdot V_1}{P_2}$ $=\dfrac{4.500 atm \centerdot 17.50mL}{1.500 atm}$ $= 52.50mL$
Charles' Law
In 1787, French physicists Jacques Charles, discovered the correlation between Temperature(T) and Volume(V) (assuming Pressure (P) and Amount of Gas(n) remain constant):
$V \propto T \rightarrow V=yT$
where y is a constant depending on amount of gas and pressure. Volume is directly proportional to Temperature
Another form of the equation (assuming there are 2 sets of conditions, and setting both constants to eachother) that might help solve problems is:
$\dfrac{V_1}{T_1} = y = \dfrac{V_2}{T_2}$
Example 1.2
A sample of Carbon dioxide in a pump has volume of 20.5 mL and it is at 40.0 oC. When the amount of gas and pressure remain constant, find the new volume of Carbon dioxide in the pump if temperature is increased to 65.0 oC.
SOUTION
$V_2=\dfrac{V_1 \centerdot T_2}{T_1}$
$=\dfrac{20.5mL \centerdot (60+273.15K)}{40+273.15K}$
$= 22.1mL$
Avogadro's Law
In 1811, Amedeo Avogadro fixed Gay-Lussac's issue in finding the correlation between the Amount of gas(n) and Volume(V) (assuming Temperature(T) and Pressure(P) remain constant):
$V \propto n \rightarrow V = zn$
where z is a constant depending on Pressure and Temperature.
• Volume(V) is directly proportional to the Amount of gas(n)
Another form of the equation (assuming there are 2 sets of conditions, and setting both constants to eachother) that might help solve problems is:
$\dfrac{P_1}{n_1} = z= \dfrac{P_2}{n_2}$
Example 1.3
A 3.80 g of oxygen gas in a pump has volume of 150 mL. constant temperature and pressure. If 1.20g of oxygen gas is added into the pump. What will be the new volume of oxygen gas in the pump if temperature and pressure held constant?
Solution
V1=150 mL
$n_1= \dfrac{m_1}{M_oxygen gas}$
$n_2= \dfrac{m_2}{M_oxygen gas}$
$V_2=\dfrac{V_1 \centerdot n_2}{n_1}$
$= \dfrac{150mL\centerdot \dfrac{5.00g}{32.0g \centerdot mol^-1} \dfrac{3.80g}{32.0g\centerdot mol^-1}$
$= 197ml$
Ideal Gas Law
The ideal gas law is the combination of the three simple gas laws. By setting all three laws directly or inversely proportional to Volume, you get:
$V \propto \dfrac{nT}{P}$
Next replacing the directly proportional to sign with a constant(R) you get:
$V = \dfrac{RnT}{P}$
And finally get the equation:
$PV = nRT$
where P= the absolute pressure of ideal gas
• V= the volume of ideal gas
• n = the amount of gas
• T = the absolute temperature
• R = the gas constant
Here, R is the called the gas constant. The value of R is determined by experimental results. Its numerical value changes with units.
R = gas constant = 8.3145 Joules · mol-1 · K-1 (SI Unit)
= 0.082057 L · atm·K-1 · mol-1
Example 1.4
At 655mm Hg and 25.0oC, a sample of Chlorine gas has volume of 750mL. How many moles of Chlorine gas at this condition?
• P=655mm Hg
• T=25+273.15K
• V=750mL=0.75L
n=?
Solution
$n=\frac{PV}{RT}$
$=\frac{655mm Hg \centerdot \frac{1 atm}{760mm Hg} \centerdot 0.75L}{0.082057L \centerdot atm \centerdot mol^-1 \centerdot K^-1 \centerdot (25+273.15K) }$
$=0.026 mol$
Evaluation of the Gas Constant, R
You can get the numerical value of gas constant, R, from the ideal gas equation, PV=nRT. At standard temperature and pressure, where temperature is 0 oC, or 273.15 K, pressure is at 1 atm, and with a volume of 22.4140L,
$R= \frac{PV}{RT}$
$\frac{1 atm \centerdot 22.4140L}{1 mol \centerdot 273.15K}$
$=0.082057 \; L \centerdot atm \centerdot mol^{-1} K^{-1}$
$R= \frac{PV}{RT}$
$= \frac{1 atm \centerdot 2.24140 \centerdot 10^{-2}m^3}{1 mol \centerdot 273.15K}$
$= 8.3145\; m^3\; Pa \centerdot mol^{-1} \centerdot K^{-1}$
General Gas Equation
In an Ideal Gas situation, $\frac{PV}{nRT} = 1$ (assuming all gases are "ideal" or perfect). In cases where $\frac{PV}{nRT} \neq 1$ or if there are multiple sets of conditions (Pressure(P), Volume(V), number of gas(n), and Temperature(T)), use the General Gas Equation:
Assuming 2 set of conditions:
Initial Case: Final Case:
$P_iV_i = n_iRT_i \; \; \; \; \; \; P_fV_f = n_fRT_f$
Setting both sides to R (which is a constant with the same value in each case), one gets:
$R= \dfrac{P_iV_i}{n_iT_i} \; \; \; \; \; \; R= \dfrac{P_fV_f}{n_fT_f}$
If one substitutes one R for the other, one will get the final equation and the General Gas Equation:
$\dfrac{P_iV_i}{n_iT_i} = \dfrac{P_fV_f}{n_fT_f}$
Standard Conditions
If in any of the laws, a variable is not give, assume that it is given. For constant temperature, pressure and amount:
1. Absolute Zero (Kelvin): 0 K = -273.15 oC
T(K) = T(oC) + 273.15 (unit of the temperature must be Kelvin)
2. Pressure: 1 Atmosphere (760 mmHg)
3. Amount: 1 mol = 22.4 Liter of gas
4. In the Ideal Gas Law, the gas constant R = 8.3145 Joules · mol-1 · K-1
= 0.082057 L · atm·K-1 · mol-1
The Van der Waals Equation For Real Gases
Dutch physicist Johannes Van Der Waals developed an equation for describing the deviation of real gases from the ideal gas. There are two correction terms added into the ideal gas equation. They are $1 +a\frac{n^2}{V^2}$, and $1/(V-nb)$.
Since the attractive forces between molecules do exist in real gases, the pressure of real gases is actually lower than of the ideal gas equation. This condition is considered in the van der waals equation. Therefore, the correction term $1 +a\frac{n^2}{V^2}$ corrects the pressure of real gas for the effect of attractive forces between gas molecules.
Similarly, because gas molecules have volume, the volume of real gas is much larger than of the ideal gas, the correction term $1 -nb$ is used for correcting the volume filled by gas molecules.
Practice Problems
1. If 4L of H2 gas at 1.43 atm is at standard temperature, and the pressure were to increase by a factor of 2/3, what is the final volume of the H2 gas? (Hint: Boyle's Law)
2. If 1.25L of gas exists at 35oC with a constant pressure of .70 atm in a cylindrical block and the volume were to be multiplied by a factor of 3/5, what is the new temperature of the gas? (Hint: Charles's Law)
3. A ballon with 4.00g of Helium gas has a volume of 500mL. When the temperature and pressure remain constant. What will be the new volume of Helium in the ballon if another 4.00g of Helium is added into the ballon? (Hint: Avogadro's Law)
Solutions
1. 2.40L
To solve this question you need to use Boyle's Law:
$P_1V_1 = P_2V_2$
Keeping the key variables in mind, temperature and the amount of gas is constant and therefore can be put aside, the only ones necessary are:
1. Initial Pressure: 1.43 atm
2. Initial Volume: 4 L
3. Final Pressure: 1.43x1.67 = 2.39
4. Final Volume(unknown): V2
Plugging these values into the equation you get:
V2=(1.43atm x 4 L)/(2.39atm) = 2.38 L
2. 184.89 K
To solve this question you need to use Charles's Law:
Once again keep the key variables in mind. The pressure remained constant and since the amount of gas is not mentioned, we assume it remains constant. Otherwise the key variables are:
1. Initial Volume: 1.25 L
2. Initial Temperature: 35oC + 273.15 = 308.15K
3. Final Volume: 1.25L*3/5 = .75 L
4. Final Temperature: T2
Since we need to solve for the final temperature you can rearrange Charles's:
Once you plug in the numbers, you get: T2=(308.15 K x .75 L)/(1.25 L) = 184.89 K
3. 1000 mL or 1L
Using Avogadro's Law to solve this problem, you can switch the equation into $V_2=\frac{n_1\centerdot V_2}{n_2}$. However, you need to convert grams of Helium gas into moles.
$n_1 = \frac{4.00g}{4.00g/mol} = \text{1 mol}$
Similarily, n2=2 mol
$V_2=\frac{n_2 \centerdot V_2}{n_1}$
$=\frac{2 mol \centerdot 500mL}{1 mol}$
$= \text{1000 mL or 1L }$
References
1. Petrucci, Ralph H. General Chemistry: Principles and Modern Applications. 9th Ed. Upper Saddle River, NJ: Pearson Prentice Hall, 2007.
2. Staley, Dennis. Prentice Hall Chemistry. Boston, MA: Pearson Prentice Hall, 2007.
3. Olander, Donald R. "Chapter2 Equation of State." General Thermodynamics. Boca Raton, NW: CRC, 2008. Print
4. O'Connell, John P., and J. M. Haile. "Properties Relative to Ideal Gases." Thermodynamics: Fundamentals for Applications. Cambridge: Cambridge UP, 2005. Print.
5. Ghare, Shakuntala. "Ideal Gas Laws for One Component." Ideal Gas Law, Enthalpy, Heat Capacity, Heats of Solution and Mixing. Vol. 4. New York, NY, 1984. Print. F.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Laws/Gas_Laws%3A_Overview.txt
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The Ideal Gas Law is very simply expressed:
$PV=nRT$
from which simpler gas laws such as Boyle's, Charles's, Avogadro's and Amonton's law be derived.
Introduction
Many chemists had dreamed of having an equation that describes relation of a gas molecule to its environment such as pressure or temperature. However, they had encountered many difficulties because of the fact that there always are other affecting factors such as intermolecular forces. Despite this fact, chemists came up with a simple gas equation to study gas behavior while putting a blind eye to minor factors.
When dealing with gas, a famous equation was used to relate all of the factors needed in order to solve a gas problem. This equation is known as the Ideal Gas Equation. As we have always known, anything ideal does not exist. In this issue, two well-known assumptions should have been made beforehand:
1. the particles have no forces acting among them, and
2. these particles do not take up any space, meaning their atomic volume is completely ignored.
An ideal gas is a hypothetical gas dreamed by chemists and students because it would be much easier if things like intermolecular forces do not exist to complicate the simple Ideal Gas Law. Ideal gases are essentially point masses moving in constant, random, straight-line motion. Its behavior is described by the assumptions listed in the Kinetic-Molecular Theory of Gases. This definition of an ideal gas contrasts with the Non-Ideal Gas definition, because this equation represents how gas actually behaves in reality. For now, let us focus on the Ideal Gas.
We must emphasize that this gas law is ideal. As students, professors, and chemists, we sometimes need to understand the concepts before we can apply it, and assuming the gases are in an ideal state where it is unaffected by real world conditions will help us better understand the behavior the gases. In order for a gas to be ideal, its behavior must follow the Kinetic-Molecular Theory whereas the Non-Ideal Gases will deviate from this theory due to real world conditions.
The Ideal Gas Equation
Before we look at the Ideal Gas Equation, let us state the four gas variables and one constant for a better understanding. The four gas variables are: pressure (P), volume (V), number of mole of gas (n), and temperature (T). Lastly, the constant in the equation shown below is R, known as the the gas constant, which will be discussed in depth further later:
$PV=nRT$
Another way to describe an ideal gas is to describe it in mathematically. Consider the following equation:
$\dfrac{PV}{nRT}=1$
The term $\frac{pV}{nRT}$ is also called the compression factor and is a measure of the ideality of the gas. An ideal gas will always equal 1 when plugged into this equation. The greater it deviates from the number 1, the more it will behave like a real gas rather than an ideal. A few things should always be kept in mind when working with this equation, as you may find it extremely helpful when checking your answer after working out a gas problem.
• Pressure is directly proportional to number of molecule and temperature. (Since P is on the opposite side of the equation to n and T)
• Pressure, however, is indirectly proportional to volume. (Since P is on the same side of the equation with V)
Simple Gas Laws
The Ideal Gas Law is simply the combination of all Simple Gas Laws (Boyle's Law, Charles' Law, and Avogadro's Law), and so learning this one means that you have learned them all. The Simple Gas Laws can always be derived from the Ideal Gas equation.
Boyle's Law
Boyle’s Law describes the inverse proportional relationship between pressure and volume at a constant temperature and a fixed amount of gas. This law came from a manipulation of the Ideal Gas Law.
$P \propto \dfrac{1}{V}$
or expressed from two pressure/volume points:
$P_1V_1=P_2V_2$
This equation would be ideal when working with problem asking for the initial or final value of pressure or volume of a certain gas when one of the two factor is missing.
Charles's Law
Charles's Law describes the directly proportional relationship between the volume and temperature (in Kelvin) of a fixed amount of gas, when the pressure is held constant.
$V\propto \; T$
or express from two volume/temperature points:
$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$
This equation can be used to solve for initial or final value of volume or temperature under the given condition that pressure and the number of mole of the gas stay the same.
Avogadro's Law
Volume of a gas is directly proportional to the amount of gas at a constant temperature and pressure.
$V \propto \; n$
or expressed as a two volume/number points:
$\dfrac{V_1}{n_1}=\dfrac{V_2}{n_2}$
Avogadro's Law can apply well to problems using Standard Temperature and Pressure (see below), because of a set amount of pressure and temperature.
Amontons's Law
Given a constant number of mole of a gas and an unchanged volume, pressure is directly proportional to temperature.
$P \propto \; T$
or expressed as two pressure/temperature points:
$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$
Boyle's Law, Charles' Law, and Avogradro's Law and Amontons's Law are given under certain conditions so directly combining them will not work. Through advanced mathematics (provided in outside link if you are interested), the properties of the three simple gas laws will give you the Ideal Gas Equation.
Standard Temperature and Pressure (STP)
Standard condition of temperature and pressure is known as STP. Two things you should know about this is listed below.
• The universal value of STP is 1 atm (pressure) and 0o C. Note that this form specifically stated 0o C degree, not 273 Kelvin, even thought you will have to convert into Kelvin when plugging this value into the Ideal Gas equation or any of the simple gas equations.
• In STP, 1 mole of gas will take up 22.4 L of the volume of the container.
Units of P, V and T
The table below lists the different units for each property.
Factor
Variable
Units
Pressure
P
atm
Torr
Pa
mmHg
Volume
V
L
Moles
n
mol
Temperature
T
K
Gas Constant
R*
see Values of R table below
Take note of certain things such as temperature is always in its SI units of Kelvin (K) rather than Celsius (C), and the amount of gas is always measured in moles. Gas pressure and volume, on the other hand, may have various different units, so be sure to know how to convert to the appropriate units if necessary.
Pressure Units
Use the following table as a reference for pressure.
Unit Symbol Equivalent to 1 atm
Common Units of Pressure
Atmosphere atm 1 atm
Millimeter of Mercury mmHg 760 mmHg
Torr Torr 760 Torr
Pascal Pa 101326 Pa
Kilopascal kPa* 101.326 kPa
Bar bar 1.01325 bar
Millibar mb 1013.25 mb
*note: This is the SI unit for pressure
The Gas Constant (R)
Here comes the tricky part when it comes to the gas constant, R. Value of R WILL change when dealing with different unit of pressure and volume (Temperature factor is overlooked because temperature will always be in Kelvin instead of Celsius when using the Ideal Gas equation). Only through appropriate value of R will you get the correct answer of the problem. It is simply a constant, and the different values of R correlates accordingly with the units given. When choosing a value of R, choose the one with the appropriate units of the given information (sometimes given units must be converted accordingly). Here are some commonly used values of R:
Values of R
0.082057 L atm mol-1 K-1
62.364 L Torr mol-1 K-1
8.3145 m3 Pa mol-1 K-1
8.3145 J mol-1 K-1*
*note: This is the SI unit for the gas constant
Example 1
So, which value of R should I use?
Solution
Because of the various value of R you can use to solve a problem. It is crucial to match your units of Pressure, Volume, number of mole, and Temperature with the units of R.
• If you use the first value of R, which is 0.082057 L atm mol-1K-1, your unit for pressure must be atm, for volume must be liter, for temperature must be Kelvin.
• If you use the second value of R, which is 62.364 L Torr mol-1K-1, your unit for pressure must be Torr, for volume must be liter, and for temperature must be Kelvin.
Ideal Gas Law Applications
How do you know the Ideal Gas Equation is the correct equation to use? Use the Ideal Gas Equation to solve a problem when the amount of gas is given and the mass of the gas is constant. There are various type of problems that will require the use of the Ideal Gas Equation.
• Solving for the unknown variable
• Initial and Final
• Partial Pressure
Other things to keep in mind: Know what Standard Temperature and Pressure (STP) values are. Know how to do Stoichiometry. Know your basic equations. Take a look at the problems below for examples of each different type of problem. Attempt them initially, and if help is needed, the solutions are right below them. Remark: The units must cancel out to get the appropriate unit; knowing this will help you double check your answer.
Example 2
5.0 g of neon is at 256 mm Hg and at a temperature of 35º C. What is the volume?
Solution
Step 1: Write down your given information:
• P = 256 mmHg
• V = ?
• m = 5.0 g
• R = 0.0820574 L•atm•mol-1K-1
• T = 35º C
Step 2: Convert as necessary:
Pressure: $256 \; \rm{mmHg} \times (1 \; \rm{atm/} 760 \; \rm{mmHg}) = 0.3368 \; \rm{atm}$
Moles: $5.0 \; \rm{g}\; Ne \times (1 \; \rm{mol} / 20.1797\; \rm{g}) = 0.25 \; \rm{mol}\; \rm{Ne}$
Temperature: $35º C + 273 = 308 \; \rm{K}$
Step 3: Plug in the variables into the appropriate equation.
$V = (nRT/P)$
$V = \dfrac{(0.25\; \rm{mol})(0.08206\; \rm{L atm}/\rm{K mol})(308\; \rm{K})}{(0.3368\; \rm{atm})}]$
$V = 19\; \rm{L}$
Example 3
What is a gas’s temperature in Celsius when it has a volume of 25 L, 203 mol, 143.5 atm?
Solution
Step 1: Write down your given information:
• P = 143.5 atm
• V= 25 L
• n = 203 mol
• R = 0.0820574 L•atm•mol-1 K-1
• T = ?
Step 2: Skip because all units are the appropriate units.
Step 3: Plug in the variables into the appropriate equation.
$T = \dfrac{PV}{nR}$
$T = \dfrac{(143.5\; \rm{atm})(25\; \rm{L})}{(203 \; \rm{mol})(0.08206 L•atm/K mol)}$
$T = 215.4\; \rm{K}$
Step 4: You are not done. Be sure to read the problem carefully, and answer what they are asking for. In this case, they are asking for temperature in Celsius, so you will need to convert it from K, the units you have.
$215.4 K - 273 = -57.4º C$
Example 4
What is the density of nitrogen gas ($N_2$) at 248.0 Torr and 18º C?
Solution
Step 1: Write down your given information
• P = 248.0 Torr
• V = ?
• n = ?
• R = 0.0820574 L•atm•mol-1 K-1
• T = 18º C
Step 2: Convert as necessary.
$(248 \; \rm{Torr}) \times \dfrac{1 \; \rm{atm}}{760 \; \rm{Torr}} = 0.3263 \; \rm{atm}$
$18ºC + 273 = 291 K$
Step 3: This one is tricky. We need to manipulate the Ideal Gas Equation to incorporate density into the equation. *Write down all known equations:
$PV = nRT$
$\rho=\dfrac{m}{V}$
where $\rho$=density, m=mass, V=Volume
$m=M \times n$
where m=mass, M=molar mass, n=moles
*Now take the density equation.
$\rho=\dfrac{m}{V}$
*Keeping in mind $m=M \times n$...replace $(M \times n)$ for $mass$ within the density formula.
$\rho=\dfrac{M \times n}{V}$
$\dfrac{\rho}{M} = \dfrac{n}{V}$
*Now manipulate the Ideal Gas Equation
$PV = nRT$
$\dfrac{n}{V} = \dfrac{P}{RT}$
*$(n/V)$ is in both equations.
$\dfrac{n}{V} = \dfrac{\rho}{M}$
$\dfrac{n}{V} = \dfrac{P}{RT}$
*Now combine them please.
$\dfrac{\rho}{M} = \dfrac{P}{RT}$
*Isolate density.
$\rho = \dfrac{PM}{RT}$
Step 4: Now plug in the information you have.
$\rho = \dfrac{PM}{RT}$
$\rho = \dfrac{(0.3263\; \rm{atm})(2*14.01 \; \rm{g/mol})}{(0.08206 L atm/K mol)(291 \; \rm{K})}$
$\rho = 0.3828 \; g/L$
Example 5
Find the volume, in mL, when 7.00 g of $O_2$ and 1.50 g of $Cl_2$ are mixed in a container with a pressure of 482 atm and at a temperature of 22º C.
Solution
Step 1: Write down your given information
• P = 482 atm
• V = ?
• n = ?
• R = 0.0820574 L•atm•mol-1 K-1
• T = 22º C + 273 = 295K
• 1.50g Cl2
• 7.00g O2
Step 2: Find the total moles of the mixed gases in order to use the Ideal Gas Equation.
$n_{total} = n_{O_2}+ n_{Cl_2}$
$= \left[7.0 \; \rm{g} \; O_2 \times \dfrac{1 \; \rm{mol} \; O_2}{32.00 \; \rm{g} \; O_2}\right] + \left[1.5 \; \rm{g}\; Cl_2 \times \dfrac{1 \; \rm{mol} \; Cl_2}{70.905 \; \rm{g} \; Cl_2}\right]$
$= 0.2188 \; \rm{mol} \; O_2 + 0.0212 \; \rm{mol} \; Cl_2$
$= 0.24 \; \rm{mol}$
Step 3: Now that you have moles, plug in your information in the Ideal Gas Equation.
$V= \dfrac{nRT}{P}$
$V= \dfrac{(0.24\; \rm{mol})(0.08206 L atm/K mol)(295\; \rm{K})}{(482\; \rm{atm})}$
$V= 0.0121\; \rm{L}$
Step 4: Almost done! Now just convert the liters to milliliters.
$0.0121\; \rm{L} \times \dfrac{1000\; \rm{ml}}{1\; \rm{L}} = 12.1\; \rm{mL}$
Example 6
A 3.00 L container is filled with $Ne_{(g)}$ at 770 mmHg at 27oC. A $0.633\;\rm{g}$ sample of $CO_2$ vapor is then added. What is the partial pressure of $CO_2$ and $Ne$ in atm? What is the total pressure in the container in atm?
Solution
Step 1: Write down all given information, and convert as necessary.
Before:
• P = 770mmHg --> 1.01 atm
• V = 3.00L
• nNe=?
• T = 27oC --> $300\; K$
Other Unknowns: $n_{CO_2}$= ?
$n_{CO_2} = 0.633\; \rm{g} \;CO_2 \times \dfrac{1 \; \rm{mol}}{44\; \rm{g}} = 0.0144\; \rm{mol} \; CO_2$
Step 2: After writing down all your given information, find the unknown moles of Ne.
$n_{Ne} = \dfrac{PV}{RT}$
$n_{Ne} = \dfrac{(1.01\; \rm{atm})(3.00\; \rm{L})}{(0.08206\;atm\;L/mol\;K)(300\; \rm{K})}$
$n_{Ne} = 0.123 \; \rm{mol}$
Because the pressure of the container before the $CO_2$ was added contained only $Ne$, that is your partial pressure of $Ne$. After converting it to atm, you have already answered part of the question!
$P_{Ne} = 1.01\; \rm{atm}$
Step 3: Now that have pressure for Ne, you must find the partial pressure for $CO_2$. Use the ideal gas equation.
$\dfrac{P_{Ne}V}{n_{Ne}RT} = \dfrac{P_{CO_2}V}{n_{CO_2}RT}$
but because both gases share the same Volume ($V$) and Temperature ($T$) and since the Gas Constant ($R$) is constants, all three terms cancel and can be removed them from the equation.
$\dfrac{P}{n_{Ne}} = \dfrac{P}{n_{CO_2}}$
$\dfrac{1.01 \; \rm{atm}}{0.123\; \rm{mol} \;Ne} = \dfrac{P_{CO_2}}{0.0144\; \rm{mol} \;CO_2}$
$P_{CO_2} = 0.118 \; \rm{atm}$
This is the partial pressure $CO_2$.
Step 4: Now find total pressure.
$P_{total}= P_{Ne} + P_{CO_2}$
$P_{total}= 1.01 \; \rm{atm} + 0.118\; \rm{atm}$
$P_{total}= 1.128\; \rm{atm} \approx 1.13\; \rm{atm} \; \text{(with appropriate significant figures)}$
Contributors and Attributions
• Duke LeTran (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Laws/The_Ideal_Gas_Law.txt
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Pressure is determined by the flow of a mass from a high pressure region to a low pressure region. Pressure measurements are made on the fluid states--liquids and gases. Air exerts a pressure which we are so accustomed to that we ignore it, however the pressure of water on a swimmer is more noticeable. You may be aware of pressure measurements in relations to the weather, your car, or bicycle tires.
Contributors and Attributions
• Michele Salles
• Charles Ophardt, Professor Emeritus, Elmhurst College & The Virtual Chembook
Gases
Learning Objectives
• Explain the characteristics of a gas.
• Identify the variables or parameters associated with a gas (pressure, volume, temperature, and amounts in mass or in mole).
• Evaluate volume, pressure, temperature, and amount of a gas in problem solving.
Gases - A Review
Skills Tested by This Quiz
• Identify the theory applicable to each problem, and calculate the desirable quantity from a given set of conditions.
Kinetic Theory of Gases
Learning Objectives
• Correlate energy to motion of gas molecules.
• Correlate temperature to kinetic energy of gas molecules.
• Interpret pressure in terms of gas molecule motion.
• Describe effusion rate in terms of molecular motion.
• Estimate effusion rate and time by comparison.
Temperature and pressure are macroscopic properties of gases. These properties are related to molecular motion, which is a microscopic phenomenon. The kinetic theory of gases correlates between macroscopic properties and microscopic phenomena. Kinetics means the study of motion, and in this case motions of gas molecules.
At the same temperature and volume, the same numbers of moles of all gases exert the same pressure on the walls of their containers. This is known as Avogadro's principle. His theory implies that the same numbers of moles of gas have the same number of molecules. Common sense tells us that the pressure is proportional to the average kinetic energy of all the gas molecules. Avogadro's principle also implies that the kinetic energies of various gases are the same at the same temperature. The molecular masses are different from gas to gas, and if all gases have the same average kinetic energy, the average speed of a gas is unique.
Based on the above assumption or theory, Boltzmann (1844-1906) and Maxwell (1831-1879) extended the theory to imply that the average kinetic energy of a gas depends on its temperature.
They let $u$ be the average or root-mean-square speed of a gas whose molar mass is M. Since N is the Avogadro's number, the average kinetic energy is (1/2) (M/N) u2 or
$\mathrm{K.E.} = \dfrac{M}{2 N}u^2 =\dfrac{3R T}{2 N}=\dfrac{3}{2}\,k T$
Note that M / N is the mass of a single molecule. Thus,
\begin{align} u &= \left(\dfrac{3k N T}{M}\right)^{1/2}\ &= \left(\dfrac{3 R T}{M}\right)^{1/2} \end{align}
where k (= R/N) is the Boltzmann constant. Note that u so evaluated is based on the average energy of gas molecules being the same, and it is called the root-mean-square speed; $u$ is not the average speed of gas molecules.
These formulas correlate temperature, pressure and kinetic energy of molecules. The distribution of gas speed has been studied by Boltzmann and Maxwell as well, but this is beyond the scope of this course. However, you may notice that at the same temperature, the average speed of hydrogen gas, $\ce{H2}$, is 4 times more than that of oxygen, $\ce{O2}$ in order to have the same average kinetic energy.
Calculation of Effusion Rate By Comparison
For two gases, at the same temperature, with molecular masses M1 and M2, and average speeds u1 and u2, Boltzmann and Maxwell theory implies the following relationship:
$M_1 u_1^2 = M_2 u_2^2$
Thus,
$\dfrac{M_1}{M_2}=\left(\dfrac{u_2}{u_1}\right)^2$
The consequence of the above property is that the effusion rate, the root mean square speed, and the most probable speed, are all inversely proportional to the square root (SQRT) of the molar mass. Simply formulated, the Graham's law of effusion is
$\textrm{rate of effusion} = \dfrac{k}{\mathrm d^{1/2}} = \dfrac{k'}{\mathrm M^{1/2}}$
Have you noticed that helium balloons were usually deflated the next day while sizes of normal air balloons will keep at least for a few days? Small helium molecules not only effuse through the tiny holes of the balloons, they also effuse much faster through them.
The theories covered here enable you to make many predictions. Apply these theories to solve the following problems.
Example 1
Calculate the kinetic energy of 1 mole of nitrogen molecules at 300 K.
Solution
Assume nitrogen behaves as an ideal gas, then
\begin{align*} E_{\textrm k} &= \dfrac{3}{2} R T\ &= \mathrm{\dfrac{3}{2}\times 8.3145\: \dfrac{J}{mol\: K} \times 300\: K}\ &= \mathrm{3742\: J / mol\: (or\: 3.74\: kJ/mol)} \end{align*}
Discussion
At 300 K, any gas that behaves like an ideal gas has the same energy per mol.
Example 2
Evaluate the root-mean-square speed of $\ce{H2}$, $\ce{He}$, $\ce{N2}$, $\ce{O2}$ and $\ce{CO2}$ at 310 K (the human body temperature).
Solution
Recall that
\begin{align*} u &= \left(\dfrac{3k N T}{M}\right)^{1/2}\ &= \left(\dfrac{3 R T}{M}\right)^{1/2}\ &= \left(\dfrac{3 \times 8.3145 \times 310}{0.002}\right)^{1/2}\ &= 1966\: \mathrm{m/s} \end{align*}
Note that the molecular mass of hydrogen is 0.002 kg/mol. These units are used because the constant R has been calculated using the SI units. The calculation for other gases is accomplished using their molar mass in kg.
\begin{align*} u &= \left(\dfrac{3k N T}{M}\right)^{1/2}\ &= \left(\dfrac{3 R T}{M}\right)^{1/2}\ &= \dfrac{(3\times 8.3145\times 310)^{1/2}}{M^{1/2}}\ &= \dfrac{87.9345}{M^{1/2}}\: \mathrm{m/s} \end{align*}
The root-mean-square speeds are:
Gas Molar Mass $u$ (root-mean-square speed in m/s)
$\ce{H2}$ 2 1966
$\ce{He}$ 4 1390
$\ce{N2}$ 28 525
$\ce{O2}$ 32 492
$\ce{CO2}$ 44 419
Discussion
Molar masses are 349 and 352 for $\ce{^235UF6}$ and $\ce{^238UF6}$ respectively. Using the method above, their root-mean-square speeds are 149 and 148 m/s respectively.
The separation of these two isotopes of uranium was a necessity during the time of war for the US scientists. Gas diffusion was one of the methods employed for their separation.
Example 3
Assume air and helium molecules pass through the undetected holes in balloons with equal opportunities. If a helium balloon takes 10.0 hours to reduce its size by 5.0 %, how long will it take a nitrogen balloon to reduce its size by 5.0 %?
Solution
The effusion rates are
$\textit{rate of effusion} =\dfrac{k}{\mathrm d^{1/2}}=\dfrac{k '}{\mathrm M^{1/2}} \nonumber$
Let's assume an average rate of effusion of helium to be 5/10 = 0.5, then the effusion rate of nitrogen is 0.5 * (4/28)1/2 = 0.189. The time required to effuse the same amount is thus 10*0.5/0.189 = 26.5 hr.
Discussion
The time required can be evaluated by
\begin{align*} time &= \mathrm{10\times(28/4)^{1/2}\: hr}\ &= \mathrm{26.5\: hr} \end{align*}
Exercise $1$
Calculate the root mean square speed of $\ce{N2}$ (molar mass = 28) at 37 °C (310 K, body temperature). R = 8.314 kg m2/(s2 mol K).
Answer
Hint: 525 m/s
Skill:
Calculate the root mean square speed u of gas molecules.
Exercise $2$
Which gas has a lowest root mean square speed: $\ce{H2}$, $\ce{N2}$, $\ce{O2}$, $\ce{CH4}$, or $\ce{CO2}$?
Answer
carbon dioxide
Skill:
Know that $\ce{CO2}$ has the highest molecular mass.
Exercise $3$
Which gas has a highest root mean square speed: $\ce{H2}$, $\ce{N2}$, $\ce{O2}$, $\ce{CH4}$, or $\ce{CO2}$?
Answer
Hint: hydrogen gas
Skill:
Knowing that $\ce{H2}$ has the lowest molecular mass makes your decision easy.
Exercise $4$
Assuming ideal gas behavior, calculate the kinetic energy of 1.00 mole of $\ce{N2}$ at 37 °C (= 310 K). R = 8.314 J/(mol K); $\ce{N2}$ molar mass = 28.0
Answer
3.87e3 J/mol
Calculate the kinetic energy of any amount of any gas at any temperature.
Exercise $5$
Which of the following gases has the highest effusion rate through a pinhole opening of its container (T = 300K): $\ce{He}$ (molar mass 4), $\ce{N2}$ (28), $\ce{O2}$ (32), $\ce{CO2}$ (44), $\ce{SO2}$ (64), or $\ce{Ar}$ (40)?
Answer
helium
Associate effusion rate with root mean square speed, and determine the effusion rates.
Exercise $6$
For which gas in the accompanying list is the effusion rate the smallest (T = 400 K): $\ce{NH3}$ (17), $\ce{CO2}$ (44), $\ce{Cl2}$ (71), $\ce{CH4}$ (16)?
Answer
Hint: chlorine gas
Calculate the root mean speed for these gases. The average speed is 324 m/s for $\ce{Cl2}$, 681 m/s for $\ce{CH4}$.
Exercise $7$
The most probable speed of $\ce{CH4}$ (molar mass 16.0) at a given temperature is 411 m/s. What is the most probable speed of $\ce{He}$ (molar mass 4.00) at the same temperature?
Answer
822 m/s
Discussion:
At the same T, the most probable speed of a gas is inversely proportional to the square-root of its molar mass.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Pressure.txt
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Learning Objectives
• Explain all the quantities involved in the ideal gas law.
• Evaluate the gas constant $R$ from experimental results.
• Calculate $T$, $V$, $P$, or $n$ of the ideal gas law: $P V = n R T$.
• Describe the ideal gas law using graphics.
The Ideal Gas Law
The volume (V) occupied by n moles of any gas has a pressure (P) at temperature (T) in Kelvin. The relationship for these variables,
$P V = n R T$
where R is known as the gas constant, is called the ideal gas law or equation of state. Properties of the gaseous state predicted by the ideal gas law are within 5% for gases under ordinary conditions. In other words, given a set of conditions, we can predict or calculate the properties of a gas to be within 5% by applying the ideal gas law. How to apply such a law for a given set of conditions is the focus of general chemistry.
At a temperature much higher than the critical temperature and at low pressures, however, the ideal gas law is a very good model for gas behavior. When dealing with gases at low temperature and at high pressure, correction has to be made in order to calculate the properties of a gas in industrial and technological applications. One of the common corrections made to the ideal gas law is the van der Waal's equation, but there are also other methods dealing with the deviation of gas from ideality.
The Gas Constant R
Repeated experiments show that at standard temperature (273 K) and pressure (1 atm or 101325 N/m2), one mole (n = 1) of gas occupies 22.4 L volume. Using this experimental value, you can evaluate the gas constant R,
\begin{align} R &= \dfrac{P V}{n T} = \mathrm{\dfrac{1\: atm\:\: 22.4\: L}{1\: mol\:\: 273\: K}}\ \ &= \mathrm{0.08205\: \dfrac{L\: atm}{mol\cdot K}} \end{align}
When SI units are desirable, P = 101325 N/m2 (Pa for pascal) instead of 1 atm. The volume is 0.0224 m3. The numerical value and units for R are
\begin{align} R &= \mathrm{\dfrac{101325\: \dfrac{N}{m^2}\: 0.0224\: m^3}{1\: mol\: 273\: K}}\ \ &= \mathrm{8.314\: \dfrac{J}{mol\cdot K}} \end{align}
Note that $\mathrm{1\: L\: atm = 0.001\: m^3 \times 101325\: \dfrac{N}{m^2} = 101.325\: J\: (or\: N\: m)}$. Since energy can be expressed in many units, other numerical values and units for R are frequently in use.
The Gas Constant
For your information, the gas constant can be expressed in the following values and units.
\begin{align} R &= \mathrm{0.08205\: \dfrac{L\: atm}{mol \cdot K}} &&\textrm{Notes:} \ &= \mathrm{8.3145\: \dfrac{L\: kPa}{mol\cdot K}} &&\mathrm{1\: atm = 101.32\: kPa} \ &= \mathrm{8.3145\: \dfrac{J}{mol\cdot K}} &&\mathrm{1\: J = 1\: L\: kPa} \ &= \mathrm{1.987\: \dfrac{cal}{mol\cdot K}} &&\mathrm{1\: cal = 4.182\: J} \ &= \mathrm{62.364\: \dfrac{L\: torr}{mol\cdot K}} &&\mathrm{1\: atm = 760\: torr}\ \end{align}
The gas constant R is such a universal constant for all gases that its values are usually listed in the "Physical Constants" of textbooks and handbooks. It is also listed in Constants of our HandbookMenu at the left bottom. Although we try to use SI units all the time, the use of atm for pressure is still common. Thus, we often use R = 8.314 J / (mol·K) or 8.3145 J / mol·K.
The volume occupied by one mole, n = 1, of substance is called the molar volume, $V_{\textrm{molar}} = \dfrac{V}{n}$. Using the molar volume notation, the ideal gas law is:
$P V_{\textrm{molar}} = R T$
Applications of the Ideal Gas Law
The ideal gas law has four parameters and a constant, R,
$P V = n R T$,
and it can be rearranged to give an expression for each of P, V, n or T. For example,
$P = \dfrac{n R T}{V}$ (Boyle's law)
$P = \left(\dfrac{n R}{V}\right) T$ (Charles's law)
These equations are Boyle's law and Charles's law respectively. Similar expressions can be derived for V, n and T in terms of other variables. Thus, there are many applications. However, you must make sure that you use the proper numerical value for the gas constant R according to the units you have for the parameters.
Furthermore, $\dfrac{n}{V}$ is number of moles per unit volume, and this quantity has the same units as the concentration (C). Thus, the concentration is a function of pressure and temperature,
$C = \dfrac{P}{R T}$
At 1.0 atm pressure and room temperature of 298 K, the concentration of an ideal gas is 0.041 mol/L.
Avogadro's law can be further applied to correlate gas density $\rho$ (weight per unit volume or n M / V) and molecular mass M of a gas. The following equation is easily derived from the ideal gas law:
$P M =\dfrac{n M}{V}R T$
Thus, we have
\begin{align} P M &= \dfrac{d R T}{M}\ \rho &= \dfrac{n M}{V} \leftarrow \textrm{definition, and}\ \rho &= \dfrac{P M}{R T}\ M &= \dfrac{d R T}{P} \end{align}
Example 1
An air sample containing only nitrogen and oxygen gases has a density of 1.3393 g / L at STP. Find the weight and mole percentages of nitrogen and oxygen in the sample.
Solution
From the density $\rho$, we can evaluate an average molecular weight (also called molar mass).
\begin{align} P M &= d R T\ M &= 22.4 \times d\ &= \mathrm{22.4\: L/mol \times 1.3393\: g/L}\ &= \mathrm{30.0\: g / mol} \end{align}
Assume that we have 1.0 mol of gas, and x mol of which is nitrogen, then (1 - x) is the amount of oxygen. The average molar mass is the mole weighted average, and thus,
$\mathrm{28.0\, x + 32.0 (1 - x) = 30.0}$
$\mathrm{- 4\, x = - 2}$
$\mathrm{x = 0.50\: mol\: of\: N_2,\: and\: 1.0 - 0.50 = 0.50\: mol\: O_2}$
Now, to find the weight percentage, find the amounts of nitrogen and oxygen in 1.0 mol (30.0 g) of the mixture.
$\mathrm{Mass\: of\: 0.5\: mol\: nitrogen = 0.5 \times 28.0 = 14.0\: g}$
$\mathrm{Mass\: of\: 0.5\: mol\: oxygen = 0.5 \times 32.0 = 16.0\: g}$
$\mathrm{Percentage\: of\: nitrogen = 100 \times \dfrac{14.0}{30.0} = 46.7 \% }$
$\mathrm{Percentage\: of\: oxygen = 100 \times \dfrac{16.0}{30.0} = 100 - 46.7 = 53.3 \%}$
DISCUSSION
We can find the density of pure nitrogen and oxygen first and evaluate the fraction from the density.
$\rho \mathrm{\: of\: N_2 = \dfrac{28.0}{22.4} = 1.2500\: g/L}$
$\rho \mathrm{\: of\: O_2 = \dfrac{32.0}{22.4} = 1.4286\: g/L}$
$\mathrm{1.2500\, x + 1.4286 (1 - x) = 1.3393}$
Solving for x gives
$\mathrm{x = 0.50}$ (same result as above)
Exercise 1
Now, repeat the calculations for a mixture whose density is 1.400 g/L.
Example 2
What is the density of acetone ($\ce{C3H6O}$) vapor at 1.0 atm and 400 K?
Solution
The molar mass of acetone = 3*12.0 + 6*1.0 + 16.0 = 58.0. Thus,
\begin{align} \rho &= \dfrac{P M}{R T}\ \ &= \mathrm{\dfrac{1.0 \times 58.0\: atm\: \dfrac{g}{mol}} {0.08205\: \dfrac{L\: atm}{mol\: K} \times 400\: K}}\ \ &= \mathrm{1.767\: g / L} \end{align}
Exercise 2
The density of acetone is 1.767 g/L; calculate its molar mass.
Confidence Building Questions
1. What does the variable n stand for in the ideal gas law,
$P V = n R T$?
Hint: number of moles of gas in a closed system.
Skill:
Describe the ideal gas law.
2. A closed system means no energy or mass flow into or out of a system. In a closed system, how many independent variables are there among n, T, V and P for a gas? Note: an independent variable can be of any arbitrary values.
Hint: one
Skill:
The ideal gas equation shows the interdependence of the variables. Only one of them can be varied independently.
3. What is the molar volume of an ideal gas at 2 atm and 1000 K?
Hint: 41.0
Skill:
Evaluate molar volume at any condition.
4. A certain amount of a gas is enclosed in a container of fixed volume. If you let heat (energy) flow into it, what will increase?
(In a multiple choice, you may have volume, pressure, temperature, and any combination of these to choose from.)
Hint: Both pressure and temperature will increase.
Skill:
Explain a closed system and apply ideal gas law.
5. For a certain amount (n = constant) of gas in a closed system, how does volume V vary with the temperature? In the following, k is a constant depending on n and P.
1. $V = k T$
2. $V = \dfrac{k}{T}$
3. $T V = k$
4. $V = k T^2$
5. $V = k$
Hint: a
Skill:
Explain Charles's law.
6. Boyle's law is P V = constant. A sketch of P vs V on graph paper is similar to a sketch of the equation x y = 5. What curve(s) does this equation represent?
1. a parabola
2. an ellipse
3. a hyperbola
4. a pair of hyperbolas
5. a straight line
6. a surface
Hint: d
Skill:
Apply the skills acquired in math courses to chemical problem solving.
7. For a certain amount of gas in a closed system, which one of the following equations is valid? Subscripts 1 and 2 refer to specific conditions 1 and 2 respectively.
1. $P_1 V_1 T_1 = P_2 V_2 T_2$
2. $P_1 V_1 T_2 = P_2 V_2 T_1$
3. $P_1 V_2 T_1 = P_2 V_1 T_2$
4. $P_2 V_1 T_1 = P_1 V_2 T_2$
5. $\dfrac{P_1 V_2}{T_1} = \dfrac{P_2 V_1}{T_2}$
Hint: b
Skill:
Rearrange a mathematical equation.
8. The gas constant R is 8.314 J / mol·K. Convert the numerical value of R so that its units are cal / (mol·K). A unit conversion table will tell you that 1 cal = 4.184 J. Make sure you know where to find it. During the exam, the conversion factor is given, but you should know how to use it.
Hint: 1.987 cal/(mol K).
Skill:
Use conversion factors, for example:
$\mathrm{8.314\: J\times\dfrac{1\: cal}{4.184\: J}=\: ?\: cal}$
9. At standard temperature and pressure, how many moles of $\ce{H2}$ are contained in a 1.0 L container?
Hint: 0.045 mol/L
Discussion:
There are many methods for calculating this value.
10. At standard temperature and pressure, how many grams of $\ce{CO2}$ are contained in a 3.0 L container? Molar mass of $\mathrm{CO_2 = 44}$.
Hint: 5.89 g in 3 L
One method:
It contains $n \mathrm{= \dfrac{1\: atm \times 3\: L}{0.08205\: \dfrac{L\: atm}{mol\cdot K}\times 273\: K}}$
11. What is the pressure if 1 mole of $\ce{N2}$ occupies 1 L of volume at 1000 K?
Hint: 82.1 atm
Discussion:
Depending on the numerical value and units of R you use, you will get the pressure in various units.
At 1000 K, some of the $\ce{N2}$ molecules may dissociate. If that is true, the pressure will be higher!
12. What is the temperature if 1 mole of $\ce{N2}$ occupying 100 L of volume has a pressure of 20 Pa (1 Pa = 1 Nm-2 )?
Hint: 240 K
Discussion:
At T = 240 K, ideal gas law may not apply to $\ce{CO2}$, because this gas liquifies at a rather high temperature. The ideal gas law is still good for $\ce{N2}$, $\ce{H2}$, $\ce{O2}$ etc, because these gases liquify at much lower temperatures.
The Simple Gas Laws
Learning Objectives
• Explain the following laws within the Ideal Gas Law:
• Avogadro's law of gases
• Boyle's law of gases
• Charles's law of gases
• Dalton's law of partial pressure
• Apply gas laws to solve problems involving gases.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gases_%28Waterloo%29/The_Ideal_Gas_Law.txt
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Basic kinetic theory ideas about solids, liquids and gases, and changes of state. Ideal and real gases. The ideal gas equation. Boyle's Law and Charles' Law.
Kinetic Theory of Gases
Learning Objectives
• To understand the five fundamentals of Kinetic Molecular Theory.
• To use Kinetic Molecular Theory to describe the behavior of the macroscopic gas laws.
The ideal gas equation
$PV = nRT$
describes how gases behave, e.g.:
• A gas expands when heated at constant pressure
• The pressure increases when a gas is compressed at constant temperature
However, the ideal gas law (nor any of the constituent gas laws) does not explain why gases behave this way? What happens to gas particles when conditions such as pressure and temperature change? This is addressed via Kinetic Molecule Theory.
The Fundamentals of Kinetic Molecular Theory (KMT)
The molecules of a gas are in a state of perpetual motion in which the velocity (that is, the speed and direction) of each molecule is completely random and independent of that of the other molecules. This fundamental assumption of the kinetic-molecular model helps us understand a wide range of commonly-observed phenomena. 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.
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.
Note
The Kinetic-Molecular Theory is "the theory of moving molecules." -Rudolf Clausius, 1857
Pressure
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,
$F = ma$
with a force, $F$, that is exerted on the surface of area $A$ exerting a pressure
$P = \dfrac{f}{A}$
• The pressure of a gas is causes by collisions of the molecules with the walls of the container.
• The magnitude of the pressure is related to how hard and how often the molecules strike the wall
• The "hardness" of the impact of the molecules with the wall will be related to the velocity of the molecules times the mass of the molecules
Kinetic Interpretation of Absolute 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:
$KE = \dfrac{1}{2}m v^2$
with
• $KE$ is the kinetic energy of a molecule,
• $m$ is the mass of the molecule, and
• $v$ is the magnitude of the velocity of a molecule.
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.
Molecular Speed
Although the molecules in a sample of gas have an average kinetic energy (and therefore an average speed) the individual molecules move at various speeds. Some are moving fast, others relatively slowly
At higher temperatures at greater fraction of the molecules are moving at higher speeds (Figure 3). What is the speed (velocity) of a molecule possessing average kinetic energy? KMT theory shows the the average kinetic energy (KE) is related to the root mean square (rms) speed $u$
$KE = \dfrac{1}{2} m u^2$
This is different from the typical definition of an average speed $\langle v \rangle$ as demonstrated in Example 5.6.1.
Example 5.6.1
Suppose a gas consists of four molecules with speeds of 3.0, 4.5, 5.2 and 8.3 m/s. What is the difference between the average speed and root mean square speed of this gas?
Solution
The average speed is:
$\langle v \rangle = \dfrac{3.0 + 4.5 + 5.2 + 8.3}{4}=5.25\; m/s$
The root mean square speed is:
$u= \sqrt{\dfrac{3.0^2+ 4.5^2+5.2^2+8.3^2}{4}}= 5.59 \; m/s$
The Gas Laws Explained from a KMT Perspective
• 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 first fundamental tenet of KMT theory is 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
Note
• The absolute temperature is a measure of the average kinetic energy of its molecules
• If two different gases are at the same temperature, their molecules have the same average kinetic energy
• If the temperature of a gas is doubled, the average kinetic energy of its molecules is doubled
Contributors and Attributions
Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Kinetic_Theory_of_Gases/Basics_of_Kinetic_Molecu.txt
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This page looks at the assumptions which are made in the Kinetic Theory about ideal gases, and takes an introductory look at the Ideal Gas Law: pV = nRT.
Kinetic Theory assumptions about ideal gases
There is no such thing as an ideal gas, of course, but many gases behave approximately as if they were ideal at ordinary working temperatures and pressures. Real gases are dealt with in more detail on another page. The assumptions are:
• Gases are made up of molecules which are in constant random motion in straight lines.
• The molecules behave as rigid spheres.
• Pressure is due to collisions between the molecules and the walls of the container.
• All collisions, both between the molecules themselves, and between the molecules and the walls of the container, are perfectly elastic. (That means that there is no loss of kinetic energy during the collision.)
• The temperature of the gas is proportional to the average kinetic energy of the molecules.
And then two absolutely key assumptions, because these are the two most important ways in which real gases differ from ideal gases:
• There are no (or entirely negligible) intermolecular forces between the gas molecules.
• The volume occupied by the molecules themselves is entirely negligible relative to the volume of the container.
The Ideal Gas Equation
The ideal gas equation is:
\[ pV = nRT \]
On the whole, this is an easy equation to remember and use. The problems lie almost entirely in the units, which should be in strict SI units
Pressure, p
Pressure is measured in Pascals, Pa - sometimes expressed as newtons per square meter, N m-2. These mean exactly the same thing. Be careful if you are given pressures in kPa (kilopascals). For example, 150 kPa is 150,000 Pa. You must make that conversion before you use the ideal gas equation.
Should you want to convert from other pressure measurements:
• 1 atmosphere = 101,325 Pa
• 1 bar = 100 kPa = 100,000 Pa
Volume, V
This is the most likely place for you to go wrong when you use this equation. That's because the SI unit of volume is the cubic metre, m3 - not cm3 or dm3 with 1 m3 = 1000 dm3 = 1,000,000 cm3. So if you are inserting values of volume into the equation, you first have to convert them into cubic metres.
You would have to divide a volume in dm3 by 1000, or in cm3 by a million.
Similarly, if you are working out a volume using the equation, remember to covert the answer in cubic metres into dm3 or cm3 if you need to - this time by multiplying by a 1000 or a million. If you get this wrong, you are going to end up with a silly answer, out by a factor of a thousand or a million. So it is usually fairly obvious if you have done something wrong, and you can check back again.
Number of moles, n
This is easy, of course - it is just a number. You already know that you work it out by dividing the mass in grams by the mass of one mole in grams. You will most often use the ideal gas equation by first making the substitution to give:
The gas constant, R
A value for R will be given you if you need it, or you can look it up in a data source. The SI value for R is 8.31441 J K-1 mol-1.
The temperature, T
The temperature has to be in Kelvin. Don't forget to add 273 if you are given a temperature in degrees Celsius.
Using the ideal gas equation
Calculations using the ideal gas equation are included in my calculations book (see the link at the very bottom of the page), and I can't repeat them here. There are, however, a couple of calculations that I haven't done in the book which give a reasonable idea of how the ideal gas equation works.
The molar volume at stp
If you have done simple calculations from equations, you have probably used the molar volume of a gas.
1 mole of any gas occupies 22.4 dm3 at stp (standard temperature and pressure, taken as 0°C and 1 atmosphere pressure). You may also have used a value of 24.0 dm3 at room temperature and pressure (taken as about 20°C and 1 atmosphere).
These figures are actually only true for an ideal gas, and we'll have a look at where they come from.
We can use the ideal gas equation to calculate the volume of 1 mole of an ideal gas at 0°C and 1 atmosphere pressure.
First, we have to get the units right.
0°C is 273 K. T = 273 K
1 atmosphere = 101325 Pa. p = 101325 Pa
We know that n = 1, because we are trying to calculate the volume of 1 mole of gas.
And, finally, R = 8.31441 J K-1 mol-1.
Slotting all of this into the ideal gas equation and then rearranging it gives:
And finally, because we are interested in the volume in cubic decimetres, you have to remember to multiply this by 1000 to convert from cubic metres into cubic decimetres.
The molar volume of an ideal gas is therefore 22.4 dm3 at stp.
And, of course, you could redo this calculation to find the volume of 1 mole of an ideal gas at room temperature and pressure - or any other temperature and pressure.
Finding the relative formula mass of a gas from its density
This is about as tricky as it gets using the ideal gas equation.
The density of ethane is 1.264 g dm-3 at 20°C and 1 atmosphere. Calculate the relative formula mass of ethane.
The density value means that 1 dm3 of ethane weighs 1.264 g.
Again, before we do anything else, get the awkward units sorted out.
A pressure of 1 atmosphere is 101325 Pa.
The volume of 1 dm3 has to be converted to cubic metres, by dividing by 1000. We have a volume of 0.001 m3.
The temperature is 293 K.
Now put all the numbers into the form of the ideal gas equation which lets you work with masses, and rearrange it to work out the mass of 1 mole.
The mass of 1 mole of anything is simply the relative formula mass in grams.So the relative formula mass of ethane is 30.4, to 3 sig figs. Now, if you add up the relative formula mass of ethane, C2H6 using accurate values of relative atomic masses, you get an answer of 30.07 to 4 significant figures. Which is different from our answer - so what's wrong?
There are two possibilities.
• The density value I have used may not be correct. I did the sum again using a slightly different value quoted at a different temperature from another source. This time I got an answer of 30.3. So the density values may not be entirely accurate, but they are both giving much the same sort of answer.
• Ethane isn't an ideal gas. Well, of course it isn't an ideal gas - there's no such thing! However, assuming that the density values are close to correct, the error is within 1% of what you would expect. So although ethane isn't exactly behaving like an ideal gas, it isn't far off.
If you need to know about real gases, now is a good time to read about them.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Kinetic_Theory_of_Gases/Common_Ideal_Gas_Difficu.txt
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When examining the ideal gas laws in conjunction with the kinetic theory of gases, we gain insights into the behavior of ideal gas. We can then predict how gas particles behaviors such as gas molecular speed, effusion rates, distances traveled by gas molecules. Graham's Law, which was formulated by the Scottish physical chemist Thomas Graham, is an important law that connects gas properties to the kinetic theory of gases.
Introduction
The Kinetic Molecular Theory states that the average energy of molecules is proportional to absolute temperature as illustrated by the following equation:
$e_K=\dfrac{3}{2}\dfrac{R}{N_A}T$
where
• ek is the average translation kinetic energy,
• R is the gas constant,
• NA is Avogadro's number, and
• T is temperature in Kelvins.
Since R and NA are constants, this means that the Kelvin temperature (T) of a gas is directly proportional to the average kinetic energy of its molecules. This means that at a given temperature, different gases (for example He or O2) will the same average kinetic energy.
Graham's Law
Gas molecules move constantly and randomly throughout the volume of the container they occupy. When examining the gas molecules individually, we see that not all of the molecules of a particular gas at a given temperature move at exactly the same speed. This means that each molecule of a gas have slightly different kinetic energy. To calculate the average kinetic energy (eK) of a sample of a gas, we use an average speed of the gas, called the root mean square speed (urms).
$e_K=\dfrac{1}{2}m{u_{rms}^2}$
with
• eK is the kinetic energy measures in Joules
• m is mass of a molecule of gas (kg)
• urms is the root mean square speed (m/s)
The root mean square speed, urms, can be determined from the temperature and molar mass of a gas.
$u_{rms}=\sqrt{\dfrac{3RT}{M}}$
with
• R the ideal gas constant (8.314 kg*m2/s2*mol*K)
• T temperature (Kelvin)
• M molar mass (kg/mol)
When examining the root mean square speed equation, we can see that the changes in temperature (T) and molar mass (M) affect the speed of the gas molecules. The speed of the molecules in a gas is proportional to the temperature and is inversely proportional to molar mass of the gas. In other words, as the temperature of a sample of gas is increased, the molecules speed up and the root mean square molecular speed increases as a result.
Graham's Law states that the rate of effusion of two different gases at the same conditions are inversely proportional to the square roots of their molar masses as given by the following equation:
$\dfrac {\it{Rate\;of\;effusion\; of \;A} }{ \it{Rate \;of \;effusion\; of \;B}}=\dfrac{(u_rms)_A}{(u_rms)_B}=\dfrac{\sqrt {3RT/M_A}}{\sqrt{3RT/M_B}}=\dfrac{\sqrt{M_B}}{\sqrt{M_A}}$
In according with the Kinetic Molecular Theory, each gas molecule moves independently. However, the net rate at which gas molecules move depend on their average speed. By examining the equation above, we can conclude that the heavier the molar mass of the gas molecules slower the gas molecules move. And conversely, lighter the molar mass of the gas molecules the faster the gas molecules move.
Limitations of Graham's Law
Graham's Law can only be applied to gases at low pressures so that gas molecules escape through the tiny pinhole slowly. In addition, the pinhole must be tiny so that no collisions occur as the gas molecules pass through. Since Graham's Law is an extension of the Ideal Gas Law, gases that follows Graham's Law also follows the Ideal Gas Law.
Molecular Effusion
The random and rapid motion of tiny gas molecules results in effusion. Effusion is the escape of gas molecules through a tiny hole or pinhole.
The behavior of helium gas in balloons is an example of effusion. The balloons are made of latex which is porous material that the small helium atom can effuse through. The helium inside a newly inflated balloon will eventually effuse out. This is the reason why balloons will deflate after a period of time. Molecular speeds are also used to explain why small molecules (such as He) diffuse more rapidly than larger molecules (O2). That is the reason why a balloon filled with helium gas will deflate faster than a balloon filled with oxygen gas.
The effusion rate, r, is inversely proportional to the square root of its molar mass, M.
$r\propto\sqrt{\dfrac{1}{M}}$
When there are two different gases the equation for effusion becomes
$\dfrac {\it{Rate\;of\;effusion\; of \;A} }{ \it{Rate \;of \;effusion\; of \;B}}=\dfrac{\sqrt {3RT/M_A}}{\sqrt{3RT/M_B}}$
$M_A$ is the molar mass of gas A, $M_B$ is the molar mass of gas B, $T$ Temperature in Kelvin, $R$ is the ideal gas constant.
From the equation above,
Rates of effusion of two gases
The relative rates of effusion of two gases at the same temperature is given as:
$\dfrac{\it{Rate\;of\;effusion\;of\;gas\;}\mathrm 1}{\it{Rate\;of \;effusion\;of\;gas\;}\mathrm 2}=\dfrac{\sqrt{M_2}}{\sqrt{M_1}}$
Units used to express rate of effusion includes: moles/seconds, moles/minutes, grams/seconds, grams/minutes.
Relative distances of two gases
The relative distances traveled by the two gases is given as:
$\dfrac{Distance\;traveled\;by\;gas\;1}{Distance\;traveled\;by\;gas\;2}=\dfrac{\sqrt{M_2}}{\sqrt{M_1}}$
By examine the Graham's law as stated above, we can conclude that a lighter gas will effuse or travel more rapidly than a heavier gas. Mathematically speaking, a gas with smaller molar mass will effuse faster than a gas with larger molar mass under the same condition.
Molecular Diffusion
Similar to effusion, the process of diffusion is the spread of gas molecules through space or through a second substance such as the atmosphere.
Diffusion has many useful applications. Here is an example of diffusion that is use in everyday households. Natural gas is odorless and used commercially daily. An undetected leakage can be very dangerous as it is highly flammable and can cause an explosion when it comes in contact with an ignition source. In addition, the long term breathing of natural gas can lead to asphyxiation. Fortunately, chemists have discovered a way to easily detect natural gas occur leak by adding a small quantity of a gaseous organic sulfur compound named methyl mercaptan, CH3SH, to the natural gas. When a leak happens, the diffusion of the odorous methyl mercaptan in the natural gas will serve as a sign of warning.
General guideline in solving effusion problems
The following flowchart outlines the steps in solving quantitative problems involving Graham's Law. It can be used as a general guideline.
Example $1$
Calculate the root mean square speed, $u_{rms}$, in m/s of helium at $30 ^o C$.
Solution
Start by converting the molar mass for helium from g/mol to kg/mol.
$M=(4.00\;g/mol)\times\dfrac{1\;kg}{1000\;g}$
$M=4.00\times10^{-3}\;kg/mol$
Now, using the equation for $u_{rms}$ substitute in the proper values for each variable and perform the calculation.
$u_{rms}=\sqrt{\dfrac{3RT}{M}}$
$u_{rms}=\sqrt{\dfrac{3 \times (8.314\; kg\;m^2/s^2*mol*K)(303K)}{4.00 \times 10^{-3}kg/mol}}$
$u_{rms}=1.37 \times 10^3 \;m/s$
Example $2$
What is the ratio of $u_{rms}$ values for helium vs. xenon at $30^oC$. Which is higher and why?
Solution
There are two approaches to solve this problem: the hard way and the easy way
Hard way:
The $u_{rms}$ speed of helium is calculated from the above example.
First convert the molar mass of xenon from g/mol to kg/mol as we did for helium in example 1
$M_{Xe}=(131.3\;g/mol)\times\dfrac{1\;kg}{1000\;g}$
$M=0.1313\;kg/mol$
Now, using the equation for the urms, insert the given and known values and solve for the variable of interest.
$u_{rms}=\sqrt{\dfrac{3(8.314\; kg\;m^2/s^2*mol*K)(303\;K)}{0.1313\;kg/mol}}$
$u_{rms}=2.40 \times 10^2\;m/s$
Compare the two values for xenon and helium and decide which is greater.
• $u_{Xe}=2.4 \times 10^2 \;m/s$
• $u_{He}=1.37 \times10^3 \;m/s$
So the ratio of RMS speeds is
$\dfrac{u_{Xe}}{u_{He}} \approx 0.18$
Helium has the higher $u_{rms}$ speed. This is in according with Graham's Law, because helium atoms are much lighter than xenon atoms.
Easy Way:
Since the temperature is the same for both gases, only the square root of the ratio of molar mass is needed to be calculated.
$\sqrt{\dfrac{M_{He}}{M_{Xe}}} = \sqrt{\dfrac{4.00 \; g/mol}{131.3\;g/mol }} \approx 0.18$
In either approach, helium has a faster RMS speed than xenon and this is due exclusively to its smaller mass.
Example $3$
If oxygen effuses from a container in 5.00 minutes, what is the molecular weight of a gas with the same given quantity of molecules effusing from the same container in 4.00 min?
Solution
Let oxygen be gas A and its rate is 1/5.00 minutes because it takes that much time for a certain quantity of oxygen to effuse and its molecular weight is 32 grams/mole (O2 (g)). Let the unknown gas be B and its rate is 1/4.00 minutes and Mb be the molecular weight of the unknown gas.
First, choose the appropriate equation,
$\dfrac{\it{Rate\;effusion\;of\;gas\;A}}{\it{Rate\;effusion\;of\;gas\;B}}=\dfrac{\sqrt{M_B}}{\sqrt{M_A}}$
Second, using algebra, solve for the variable of interest on one side of the equation and plug in known and given values from the problem.
$\sqrt{M_B}=\dfrac{(\mathit{Rate\;effusion\;of\;gas\;A} )(\sqrt{M_A})}{\mathit{Rate\;effusion\;of\;gas\;B}}$
$\sqrt{M_B}=\dfrac{(1/5.00\;minutes)(\sqrt{32\;grams/mole})}{1/4.00\;minutes}$
$\sqrt{M_B}=4.525$
${M_B}=(4.525)^2=20.5\;grams/mole$
Example $4$
Oxygen, O2 (g), effuses from a container at the rate of 3.64 mL/sec, what is the molecular weight of a gas effusing from the same container under identical conditions at 4.48 mL/sec?
Solution
Let label oxygen as gas A. Therefore, rate of effusion of gas A is equaled to 3.64 mL/sec and the molecular weight is 32 grams/mole. The unknown gas is B and its rate of effusion is 4.48 mL/sec. We are solving for the molecular weight of gas B which is labeled as MB.
First, choose the appropriate equation,
$\dfrac{\it{Rate\;effusion\;of\;gas\;A}}{\it{Rate\;effusion\;of\;gas\;B}}=\dfrac{\sqrt{M_B}}{\sqrt{M_A}}$
$\sqrt{M_B}=\dfrac{(\mathit{Rate\;effusion\;of\;gas\;A} )(\sqrt{M_A})}{\mathit{Rate\;effusion\;of\;gas\;B}}$
Second, using algebra, solve for the variable of interest on one side of the equation and plug in known and given values from the problem.
$\sqrt{M_B}=\dfrac{(\mathit{Rate\;effusion\;of\;gas\;A})(\sqrt{M_A})}{\mathit{Rate\;effusion\;of\;gas\;B}}$
$\sqrt{M_B}=\dfrac{(3.64\;mL/sec)(\sqrt{32\;grams/mole})}{4.48\;mL/sec}$
$\sqrt{M_B}=4.50$
${M_B}=(4.50)^2=20.2\;grams/mole$
Example $5$
Which answer(s) are true when comparing 1.0 mol O2 (g) at STP (Standard Temperature and Pressure) and 0.50 mole of S2 (g) at STP?
The two gases have equal through the same orifice, a tiny opening:
1. average molecular kinetic energies
2. root-mean-square speeds
3. masses
4. volumes
5. densities
6. effusion rates
Solution
The average kinetic energy of gas molecules depends on only the Temperature as exemplifies in this equation:
$e_K=\dfrac{3}{2}\dfrac{R}{N_A}T$
In addition, the mass of 0.50 mole of S2 is the same as that of 1.0 mole of O2. The rest of the choices are false.
Therefore, the correct answers are (a) and (c).
• Tram Anh Dao
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Kinetic_Theory_of_Gases/Connecting_Gas_Propertie.txt
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To better understand the molecular origins of the ideal gas law,
$PV=nRT$
the basics of the Kinetic Molecular Theory of Gases (KMT) should be understood. This model is used to describe the behavior of gases. More specifically, it is used to explain macroscopic properties of a gas, such as pressure and temperature, in terms of its microscopic components, such as atoms. Like the ideal gas law, this theory was developed in reference to ideal gases, although it can be applied reasonably well to real gases.
In order to apply the kinetic model of gases, five assumptions are made:
1. Gases are made up of particles with no defined volume but with a defined mass. In other words their volume is miniscule compared to the distance between themselves and other molecules.
2. Gas particles undergo no intermolecular attractions or repulsions. This assumption implies that the particles possess no potential energy and thus their total energy is simply equal to their kinetic energies.
3. Gas particles are in continuous, random motion.
4. Collisions between gas particles are completely elastic. In other words, there is no net loss or gain of kinetic energy when particles collide.
5. The average kinetic energy is the same for all gases at a given temperature, regardless of the identity of the gas. Furthermore, this kinetic energy is proportional to the absolute temperature of the gas.
Temperature and KMT
The last assumption can be written in equation form as:
$KE = \dfrac{1}{2}mv^2 = \dfrac{3}{2}k_BT$
where
• $k_B$ is Boltzmann's constant (kB = 1.381×10-23 m2 kg s-2 K-1) and
• $T$ is the absolution temperature (in Kelvin)
This equation says that the speed of gas particles is related to their absolute temperature. In other words, as their temperature increases, their speed increases, and finally their total energy increases as well. However, it is impossible to define the speed of any one gas particle. As such, the speeds of gases are defined in terms of their root-mean-square speed.
Pressure and KMT
The macroscopic phenomena of pressure can be explained in terms of the kinetic molecular theory of gases. Assume the case in which a gas molecule (represented by a sphere) is in a box, length L (Figure 1). Through using the assumptions laid out above, and considering the sphere is only moving in the x-direction, we can examine the instance of the sphere colliding elastically with one of the walls of the box.
The momentum of this collision is given by p=mv, in this case p=mvx, since we are only considering the x dimension. The total momentum change for this collision is then given by
$mv_x - m(-v_x) = 2mv_x$
Given that the amount of time it takes between collisions of the molecule with the wall is L/vx we can give the frequency of collisions of the molecule against a given wall of the box per unit time as vx/2L. One can now solve for the change in momentum per unit of time:
$(2mv_x)(v_x/2L) = mv_x^2/L$
Solving for momentum per unit of time gives the force exerted by an object (F=ma=p/time). With the expression that F=mvx2/L one can now solve for the pressure exerted by the molecular collision, where area is given as the area of one wall of the box, A=L2:
$P=\dfrac{F}{A}$
$P=\dfrac{mv_x^2}{[L(L^2)}$
The expression can now be written in terms of the pressure associated with collisions from N number of molecules:
$P=\dfrac{Nmv_x^2}{V}$
This expression can now be adjusted to account for movement in the x, y and z directions by using mean-square velocity for three dimensions and a large value of N. The expression now is written as:
$P={\dfrac{Nm\overline{v}^2}{3V}}$
This expression now gives pressure, a macroscopic quality, in terms of atomic motion. The significance of the above relationship is that pressure is proportional to the mean-square velocity of molecules in a given container. Therefore, as molecular velocity increases so does the pressure exerted on the container.
Contributors and Attributions
• Sevini Shahbaz, Andrew Cooley
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Kinetic_Theory_of_Gases/Kinetic_Molecular_Theory.txt
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Learning Objectives
Make sure you thoroughly understand the following essential ideas:
• State the three major properties of gases that distinguish them from condensed phases of matter.
• Define pressure, and explain why a gas exerts pressure on the walls of a container.
• Explain the operation of a simple barometer, and why its invention revolutionized our understanding of gases.
• Explain why a barometer that uses water as the barometric fluid is usually less practical than one which employs mercury.
• How are the Celsius and Fahrenheit temperature scales defined? How do the magnitudes of the "degree" on each scale related?
• Why must the temperature and pressure be specified when reporting the volume of a gas?
The invention of the sensitive balance in the early seventeenth century showed once and for all that gases have weight and are therefore matter. Guericke's invention of air pump (which led directly to his discovery of the vacuum) launched the “pneumatic era" of chemistry long before the existence of atoms and molecules had been accepted. Indeed, the behavior of gases was soon to prove an invaluable tool in the development of the atomic theory of matter.
Introduction
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.
First, we know that a gas has no definite volume or shape; a gas will fill whatever volume is available to it. Contrast this to the behavior of a liquid, which always has a distinct upper surface when its volume is less than that of the space it occupies. The other outstanding characteristic of gases is their low densities, compared with those of liquids and solids. One mole of liquid water at 298 K and 1 atm pressure occupies a volume of 18.8 cm3, whereas the same quantity of water vapor at the same temperature and pressure has a volume of 30200 cm3, more than 1000 times greater.
The most remarkable property of gases, however, is that to a very good approximation, they all behave the same way in response to changes in temperature and pressure, expanding or contracting by predictable amounts. This is very different from the behavior of liquids or solids, in which the properties of each particular substance must be determined individually. We will see later that each of these three macroscopic characteristics of gases follows directly from the microscopic view— that is, from the atomic nature of matter.
The Pressure of a Gas
The molecules of a gas, being in continuous motion, frequently strike the inner walls of their container. As they do so, they immediately bounce off without loss of kinetic energy, but the reversal of direction (acceleration) imparts a force to the container walls. This force, divided by the total surface area on which it acts, is the pressure of the gas.
The pressure of a gas is observed by measuring the pressure that must be applied externally in order to keep the gas from expanding or contracting. To visualize this, imagine some gas trapped in a cylinder having one end enclosed by a freely moving piston. In order to keep the gas in the container, a certain amount of weight (more precisely, a force, f ) must be placed on the piston so as to exactly balance the force exerted by the gas on the bottom of the piston, and tending to push it up. The pressure of the gas is simply the quotient f/A, where A is the cross-section area of the piston.
Pressure Units
The unit of pressure in the SI system is the pascal (Pa), defined as a force of one newton per square meter (1 Nm–2 = 1 kg m–1 s–2.) At the Earth's surface, the force of gravity acting on a 1 kg mass is 9.81 N. Thus if the weight is 1 kg and the surface area of the piston is 1 M2, the pressure of the gas would be 9.81 Pa. A 1-gram weight acting on a piston of 1 cm2 cross-section would exert a pressure of 98.1 pA. (If you wonder why the pressure is higher in the second example, consider the number of cm2 contained in 1 m2.)
In chemistry, it is often common to express pressures in units of atmospheres or torr: 1 atm = 101325 Pa = 760 torr. The older unit millimeter of mercury (mm Hg) is almost the same as the torr and is defined as one mm of level difference in a mercury barometer at 0°C. In meteorology, the pressure unit most commonly used is the bar:
1 bar = 105 N m–2 = 750.06 torr = 0.987 atm.
The pressures of gases encountered in nature span an exceptionally wide range, only part of which is ordinarily encountered in chemistry. Note that in the chart below, the pressure scales are logarithmic; thus 0 on the atm scale means 100 = 1 atm.
Atmospheric Pressure
The column of air above us exerts a force on each 1-cm2 of surface equivalent to a weight of about 1034 g. The higher into the air you go, the lower the mass of air above you, hence the lower the pressure (right).
This figure is obtained by solving Newton's law \(\textbf{F} = m\textbf{a}\) for \(m\), using the acceleration of gravity for \(\textbf{a}\):
\[ m = \dfrac{101375\; kg\, m^{-1} \, s^{-2}}{9.8 \, m \, s^{-2}} = 10340 \, kg\, m^{-1} =1034\; g \; cm^{-2}\]
Example \(1\)
If several kilos of air are constantly pressing down on your body, why do you not feel it?
Solution
Because every other part of your body (including within your lungs and insides) also experiences the same pressure, so there is no net force (other than gravity) acting on you.
This was the crucial first step that led eventually to the concept of gases and their essential role in the early development of Chemistry. In the early 17th century the Italian Evangelista Torricelli invented a device — the barometer — to measure the pressure of the atmosphere. A few years later, the German scientist and some-time mayor of Magdeburg Otto von Guericke devised a method of pumping the air out of a container, thus creating which might be considered the opposite of air: the vacuum.
As with so many advances in science, idea of a vacuum — a region of nothingness — was not immediately accepted. Torricelli's invention overturned the then-common belief that air (and by extension, all gases) are weightless. The fact that we live at the bottom of a sea of air was most spectacularly demonstrated in 1654, when two teams of eight horses were unable to pull apart two 14-inch copper hemispheres (the "Magdeburg hemispheres") which had been joined together and then evacuated with Guericke's newly-invented vacuum pump.
The classical barometer, still used for the most accurate work, measures the height of a column of liquid that can be supported by the atmosphere. As indicated below, this pressure is exerted directly on the liquid in the reservoir, and is transmitted hydrostatically to the liquid in the column.
Metallic mercury, being a liquid of exceptionally high density and low vapor pressure, is the ideal barometric fluid. Its widespread use gave rise to the "millimeter of mercury" (now usually referred to as the "torr") as a measure of pressure.
Example \(2\): Water Barometer
How is the air pressure of 1034 g cm–3 related to the 760-mm height of the mercury column in the barometer? What if water were used in place of mercury?
Solution
The density of Hg is 13.6 g cm–3, so in a column of 1-cm2 cross-section, the height needed to counter the atmospheric pressure would be (1034 g × 1 cm2) / (13.6 g cm–3) = 76 cm.
The density of water is only 1/13.6 that of mercury, so standard atmospheric pressure would support a water column whose height is 13.6 x 76 cm = 1034 cm, or 10.3 m. You would have to read a water barometer from a fourth-story window!
Water barometers were once employed to measure the height of the ground and the heights of buildings before more modern methods were adopted.
A modification of the barometer, the U-tube manometer, provides a simple device for measuring the pressure of any gas in a container. The U-tube is partially filled with mercury, one end is connected to container, while the other end can either be opened to the atmosphere. The pressure inside the container is found from the difference in height between the mercury in the two sides of the U-tube. The illustration below shows how the two kinds of manometer work.
The manometers ordinarily seen in the laboratory come in two flavors: closed-tube and open-tube. In the closed-tube unit shown at the left, the longer limb of the J-tube is evacuated by filling it with mercury and then inverting it. If the sample container is also evacuated, the mercury level will be the same in both limbs. When gas is let into the container, its pressure pushes the mercury down on one side and up on the other; the difference in levels is the pressure in torr. For practical applications in engineering and industry, especially where higher pressures must be monitored, many types of mechanical and electrical pressure gauges are available.
The Temperature of a Gas
If two bodies are at different temperatures, heat will flow from the warmer to the cooler one until their temperatures are the same. This is the principle on which thermometry is based; the temperature of an object is measured indirectly by placing a calibrated device known as a thermometer in contact with it. When thermal equilibrium is obtained, the temperature of the thermometer is the same as the temperature of the object.
A thermometer makes use of some temperature-dependent quantity, such as the density of a liquid, to allow the temperature to be found indirectly through some easily measured quantity such as the length of a mercury column. The resulting scale of temperature is entirely arbitrary; it is defined by locating its zero point, and the size of the degree unit. At one point in the 18th century, 35 different temperature scales were in use! The Celsius temperature scale (formally called centigrade) locates the zero point at the freezing temperature of water; the Celsius degree is defined as 1/100 of the difference between the freezing and boiling temperatures of water at 1 atm pressure.
The older Fahrenheit scale placed the zero point at the coldest temperature it was possible to obtain at the time (by mixing salt and ice.) The 100° point was set with body temperature (later found to be 98.6°F.) On this scale, water freezes at 32°F and boils at 212°F. The Fahrenheit scale is a finer one than the Celsius scale; there are 180 Fahrenheit degrees in the same temperature interval that contains 100 Celsius degrees, so 1 F° = 5/9 C°. Since the zero points are also different by 32F°, conversion between temperatures expressed on the two scales requires the addition or subtraction of this offset, as well as multiplication by the ratio of the degree size.
You should be able to derive the formulas for this conversion.
Absolute Temperature
In 1787 the French mathematician and physicist Jacques Charles discovered that for each Celsius degree that the temperature of a gas is lowered, the volume of the gas will diminish by 1/273 of its volume at 0°C. The obvious implication of this is that if the temperature could be reduced to –273°C, the volume of the gas would contract to zero. Of course, all real gases condense to liquids before this happens, but at sufficiently low pressures their volumes are linear functions of the temperature (Charles' Law), and extrapolation of a plot of volume as a function of temperature predicts zero volume at -273°C. This temperature, known as absolute zero, corresponds to the total absence of thermal energy.
The temperature scale on which the zero point is –273.15°C was suggested by Lord Kelvin, and is usually known as the Kelvin scale. Since the size of the Kelvin and Celsius degrees are the same, conversion between the two scales is a simple matter of adding or subtracting 273.15; thus room temperature, 20°, is about 293 K.
Because the Kelvin scale is based on an absolute, rather than on an arbitrary zero of temperature, it plays a special significance in scientific calculations; most fundamental physical relations involving temperature are expressed mathematically in terms of absolute temperature. In engineering work, an absolute scale based on the Fahrenheit degree is sometimes used; this is known as the Rankine scale.
The volume occupied by a gas
The volume of a gas is simply the space in which the molecules of the gas are free to move. If we have a mixture of gases, such as air, the various gases will coexist within the same volume. In these respects, gases are very different from liquids and solids, the two condensed states of matter. The volume of a gas can be measured by trapping it above mercury in a calibrated tube known as a gas burette. The SI unit of volume is the cubic meter, but in chemistry we more commonly use the liter and the milliliter (mL). The cubic centimeter (cc) is also frequently used; it is very close to 1 milliliter (mL).
It's important to bear in mind, however, that the volume of a gas varies with both the temperature and the pressure, so reporting the volume alone is not very useful. A common practice is to measure the volume of the gas under the ambient temperature and atmospheric pressure, and then to correct the observed volume to what it would be at standard atmospheric pressure and some fixed temperature, usually 0° C or 25°C.
Real Gases
Gases that deviate from ideality are known as Real Gases, which originate from two factors: (1) First, the theory assumes that as pressure increases, the volume of a gas becomes very small and approaches zero. While it does approach a small number, it will not be zero because molecules do occupy space (i.e. have volume) and cannot be compressed. (2) Intermolecular forces do exist in gases. These become increasingly important in low temperatures, when translational (definition of translational, please) molecular motion slows down, almost to a halt. However, at high temperatures, or even normal, every day temperatures, the intermolecular forces are very small and tend to be considered negligible.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Properties_of_Gas.txt
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Ensemble properties result from being or relating to the greater in number of atoms in a sample. This are in contrast to atomic or molecular properties.
• Capillary Action
Capillary action can be defined as the ascension of liquids through slim tube, cylinder or permeable substance due to adhesive and cohesive forces interacting between the liquid and the surface. When intermolecular bonding of a liquid itself is substantially inferior to a substances’ surface it is interacting, capillarity occurs. Also, the diameter of the container as well as the gravitational forces will determine amount of liquid raised.
• Cohesive and Adhesive Forces
Cohesive and adhesive forces are associated with bulk (or macroscopic) properties and hence the terms are not applicable to discussion of atomic and molecular properties. When a liquid comes into contact with a surface (such as the walls of a graduated cylinder or a tabletop), both cohesive and adhesive forces will act on it. These forces govern the shape which the liquid takes on.
• Contact Angles
Contact angle is one of the common ways to measure the wettability of a surface or material. Wetting refers to the study of how a liquid deposited on a solid (or liquid) substrate spreads out or the ability of liquids to form boundary surfaces with solid states. The wetting, as mentioned before is determined by measuring the contact angle, which the liquid forms in contact with the solids or liquids. The wetting tendency is larger, the smaller the contact angle or the surface tension is.
• Surface Tension
Surface tension is the energy, or work, required to increase the surface area of a liquid due to intermolecular forces. Since these intermolecular forces vary depending on the nature of the liquid (e.g. water vs. gasoline) or solutes in the liquid (e.g. surfactants like detergent), each solution exhibits differing surface tension properties.
• Unusual Properties of Water
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:
• Vapor Pressure
Pressure is the average force that material (gas, liquid or solid) exert upon the surface, e.g. walls of a container or other confining boundary. Vapor pressure or equilibrium vapor pressure is the pressure of a vapor in thermodynamic equilibrium with its condensed phases in a closed container. All liquids and solids have a tendency to evaporate or sublime into a gaseous form and all gases have a tendency to condense back to their liquid or solid form.
• Viscosity
Viscosity is another type of bulk property defined as a liquid’s resistance to flow. When the intermolecular forces of attraction are strong within a liquid, there is a larger viscosity. An example of this phenomenon is imagining a race between two liquids down a windshield. Which would you expect to roll down the windshield faster honey or water? Obviously from experience one would expect water to easily speed right past the honey, a fact that reveals honey has a much higher viscocity than wate
• Wetting Agents
A substance is referred to as a wetting agent if it lowers the surface tension of a liquid and thus allows it to spread more easily.
Thumbnail: A water drop on a lotus leaf surface showing contact angles of approximately 147°. (Public Domain; Na2jojon).
Properties of Liquids
Capillary action can be defined as the ascension of liquids through slim tube, cylinder or permeable substance due to adhesive and cohesive forces interacting between the liquid and the surface. When intermolecular bonding of a liquid itself is substantially inferior to a substances’ surface it is interacting, capillarity occurs. Also, the diameter of the container as well as the gravitational forces will determine amount of liquid raised. While, water possesses this unique property, a liquid like mercury will not display the same attributes due to the fact that it has higher cohesive force than adhesive force.
Forces in Capillary Action
Three main variables that determine whether a liquid possesses capillary action are:
• Cohesive force: It is the intermolecular bonding of a substance where its mutual attractiveness forces them to maintain a certain shape of the liquid.
• Surface tension: This occurs as a result of like molecules, cohesive forces, banding together to form a somewhat impenetrable surface on the body of water. The surface tension is measured in Newton/meter.
• Adhesive force: When forces of attraction between unlike molecules occur, it is called adhesive forces.
Capillary action only occurs when the adhesive forces are stronger than the cohesive forces, which invariably becomes surface tension, in the liquid.
A good way to remember the difference between adhesive and cohesive forces is that with adhesive forces you add another set of molecules, the molecules of the surface, for the liquid to bond with. With cohesive forces, the molecules of the liquid will only cooperate with their own kind. Decreased surface tension also increases capillary action. This is because decreased surface tension means that the intermolecular forces are decreased, thus decreasing cohesive forces. As a result, capillary action will be even greater.
Applications
Practical use of capillary action is evident in all forms of our daily lives. It makes performing our tasks efficiently and effectively. Some applications of this unique property include:
• The fundamental properties are used to absorb water by using paper towels. The cohesive and adhesive properties draw the fluid into the paper towel. The liquid flows into the paper towel at a certain rate.
• A technique called thin layer chromatography uses capillary action in which a layer of liquid is used to separate mixtures from substances.
• Capillary action helps us naturally by pumping out tear fluid in the eye. This process cleanses the eye and clears all of the dust and particles that are around the ducts of the eye.
• To generate energy: A possible use for capillary action is as a source of renewable energy. By allowing water to climb through capillaries, evaporate once it reaches the top, the condensate and drop back down to the bottom spinning a turbine on its way to create the energy, capillary action can make electricity! Although this idea is still in the works, it goes to show the potential that capillary action holds and how important it is.
When measuring the level of liquid of a test tube or buret, it is imperative to measure at the meniscus line for an accurate reading. It is possible to measure the height (represented by h) of a test tube, buret, or other liquid column using the formula:
$h = \dfrac{2\gamma \cos\theta}{\rho\;g\;r}$
In this formula,
• γ represents the surface tension in a liquid-air environment,
• θ is the angle of contact or the degree of contact,
• ρ is the density of the liquid in the representative column,
• g is the acceleration due to the force of gravity and
• r is the radius of the tube in which the liquid is presented in.
At optimum level, in which a glass tube filled with water is present in air, this formula can determine the height of a specific column of water in meters (m):
$h\approx\dfrac{1.4 \times 10^{-5}}{r}$
However, the following conditions must be met for this formula to occur.
• γ= 0.0728 N/m (when water is at a temperature of 20°C)
• θ= 20°
• ρis 1000 kg/m3
• g= 9.8 m/s2
Formula for Volume of Liquid Transport in Medium:
When certain objects that are porous encounter a liquid medium, it will begin to absorb the liquid at a rate which actually decreases over a period of time. This formula is written as:
$V = S*A\sqrt{t}$
In this specific formula,
• A is the wet area (cross-section),
• S is the sorptivity (capacity of medium to absorb using the process of capillary action),
• V is the volume of liquid absorbed in time, t.
Questions
1. Name one way to increase capillary action, and one way to decrease it.
2. If cohesion is greater than adhesion, will the meniscus be convex or concave?
3. What would be the height of a liquid in a column, on earth, with a liquid-air surface tension 0f .0973 N/m, contact angle of 30 degrees, density of 1200 kg/m3? Note that the radius of the tube is 0.2 meters.
4. What would be the height of water in a glass tube with a radius of .6mm?
Solutions
1. Increase capillary action: Increase temperature, decrease capillary tube diameter, perform any number of actions to decrease surface tension, etc…! Decrease capillary action: The opposite of the steps you would take to increase, also, increasing the density of the liquid you're working with.
2. The meniscus will result in a convex formation.
3. Using the formula above, the height of the liquid will be 7.165* 10-5m high.
4. Using the formula above, the height of the water in the glass tube would be .014m high.
Contributors and Attributions
• Achille Peiris, Becky Stein
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Liquids/Capillary_Action.txt
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Cohesive and adhesive forces are associated with bulk (or macroscopic) properties, and hence the terms are not applicable to the discussion of atomic and molecular properties. When a liquid comes into contact with a surface (such as the walls of a graduated cylinder or a tabletop), both cohesive and adhesive forces will act on it. These forces govern the shape which the liquid takes on. Due to the effects of adhesive forces, liquid on a surface can spread out to form a thin, relatively uniform film over the surface, a process known as wetting. Alternatively, in the presence of strong cohesive forces, the liquid can divide into a number of small, roughly spherical beads that stand on the surface, maintaining minimal contact with the surface.
Adhesive and Cohesive Forces
The term "cohesive forces" is a generic term for the collective intermolecular forces (e.g., hydrogen bonding and van der Waals forces) responsible for the bulk property of liquids resisting separation. Specifically, these attractive forces exist between molecules of the same substance. For instance, rain falls in droplets, rather than a fine mist, because water has strong cohesion, which pulls its molecules tightly together, forming droplets. This force tends to unite molecules of a liquid, gathering them into relatively large clusters due to the molecules' dislike for its surroundings.
Similarly, the term "adhesive forces" refers to the attractive forces between unlike substance, such as mechanical forces (sticking together) and electrostatic forces (attraction due to opposing charges). In the case of a liquid wetting agent, adhesion causes the liquid to cling to the surface on which it rests. When water is poured on clean glass, it tends to spread, forming a thin, uniform film over the glasses surface. This is because the adhesive forces between water and glass are strong enough to pull the water molecules out of their spherical formation and hold them against the surface of the glass, thus avoiding the repulsion between like molecules.
Macroscopic Effects of Cohesive and Adhesive Forces
When a liquid is placed on a smooth surface, the relative strengths of the cohesive and adhesive forces acting on that liquid determine the shape it will take (and whether or not it will wet the surface). If the adhesive forces between a liquid and a surface are stronger, they will pull the liquid down, causing it to wet the surface. However, if the cohesive forces among the liquid itself are stronger, they will resist such adhesion and cause the liquid to retain a spherical shape and bead the surface.
Case I: The Meniscus
The meniscus is the curvature of a liquid's surface within a container, such as a graduated cylinder. However, before we explain why some liquid have a concave up meniscus while others share a concave down meniscus, we have to understand the adhesive forces at work of surface tension. Water, for example, is a polar molecule that consists of a partial positive charge on the hydrogens and a partial negative charge on the oxygen. Thus, within liquid water, each molecule's partial positive charge is attracted to its neighbor's partial negative charge. This is the origin of the cohesive forces within the water. Water molecules buried inside the liquid is then being pulled and pushed evenly in every direction, producing no net pull. Meanwhile, the molecules on the surface of the liquid, lacking pulling forces in the upward direction, thus encompass a net downward pull.
How does this cohesive force create both a concave up and concave down surface then? The answer is in its relationship to the adhesive forces between the water molecules and the container's surface. When the cohesive force of the liquid is stronger than the adhesive force of the liquid to the wall, the liquid concaves down in order to reduce contact with the surface of the wall. When the adhesive force of the liquid to the wall is stronger than the cohesive force of the liquid, the liquid is more attracted to the wall than its neighbors, causing the upward concavity.
Case II: Tears of Wine
In agitated glasses of wine, droplets of wine seemingly "float" above the meniscus of the liquid and form "tears." This age-old phenomenon is the result of surface tension and cohesive and adhesive forces. Alcohol is more volatile than water. As a result, "evaporation of alcohol produces a surface tension gradient driving a thin film up along the side of a wine glass" (Adamson). This process is called the "solutal Marangoni effect."2 Due to adhesive forces, some waters clings to the walls of the glass. The "tears" form from the cohesive forces within the water holding it together. It is important to note that the surface tension gradient is "the driving force for the motion of the liquid" (Gugliotti), but the actual formation of the tears is a result of cohesive and adhesive forces.
Problems
1. Name two examples where the cohesive force dominates over the adhesive force and vice versa.
2. When in a glass graduated cylinder, water presents an upwardly concave meniscus. However, when water is filled to the tip of the cylinder, the water level could maintain higher than the wall of the cylinder without pouring out resembling a concave down meniscus. Use the principles of cohesive and adhesive forces to explain this situation.
3. Explain why a water strider can glide on the water with the knowledge of cohesion in water.
4. Propose different types of forces that adhesive forces can build on.
Answers
1. When cohesive force is stronger than the adhesive force: concave up meniscus, water forms droplets on the surface. When adhesive force is stronger than the cohesive force: concave down meniscus, the surfaces are covered by the wetting agent, the last drops of liquid in the bottle always refuse to come out.
2. Since water forms a concave up meniscus, the adhesion of the molecules to the glass is stronger than the cohesion among the molecules. However, in the absence of the adhesive force (when water reaches the tip of the glass), the cohesive force remains present. Thus cohesive force alone proves that it can still hold itself in place without pouring out of the cylinder. This example emphasizes the importance of that cohesive force and adhesive forces do not simply cancel each other out, yet it is the difference between the two that determines the characteristic of the liquid.
3. This problem addresses once again the concept of surface tension. Because the cohesion of the water is built on the weak intermolecular forces of the water, when a water strider stepson to the surface, an extra energy will be necessary to overcome to break those bonds to increase surface area. Moreover, since the gravitational pull down on the water strider cannot overcome the activation energy to break these intermolecular forces, the water strider is able to glide freely on the water.
4. Complementary shape, chemical bonds form, weak intermolecular forces such as H-bonding or Van der Waals forces.
Contributors and Attributions
• Cameron Tracy, Ling Xie, Irene Lym, Henry Li
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Liquids/Cohesive_and_Adhesive_Forces.txt
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Contact angle is one of the common ways to measure the wettability of a surface or material. Wetting refers to the study of how a liquid deposited on a solid (or liquid) substrate spreads out or the ability of liquids to form boundary surfaces with solid states. The wetting, as mentioned before is determined by measuring the contact angle, which the liquid forms in contact with the solids or liquids. The wetting tendency is larger, the smaller the contact angle or the surface tension is. A wetting liquid is a liquid that forms a contact angle with the solid which is smaller then 90º. A non-wetting liquid creates a contact angle between 90º and 180º with the solid.
Introduction
The contact angle is an angle that a liquid creates with a solid surface or capillary walls of a porous material when both materials come in contact together. This angle is determined by both properties of the solid and the liquid and the interaction and repulsion forces between liquid and solid and by the three phase interface properties (gas, liquid and solid). Those interactions are described by cohesion and adhesion forces which are intermolecular forces. The balance between the cohesive forces of similar molecules such as between the liquid molecules (i.e. hydrogen bonds and Van der Waals forces) and the adhesive forces between dissimilar molecules such as between the liquid and solid molecules (i.e. mechanical and electrostatic forces) will determine the contact angle created in the solid and liquid interface. The traditional definition of a contact angle is the angle a liquid creates with the solid or liquid when it is deposited on it. A less traditional definition is the angle a liquid creates with the sides of a capillary when it rises in it to create a meniscus.
• The smaller the contact angle - cohesive forces are weaker then adhesive forces and molecules of the liquid tend to interact more with solid molecules then liquid molecules.
• The larger the contact angle - cohesive forces are stronger then adhesive forces and the molecules of the liquid tend to interact more with each other then with the solid molecules.
Thermodynamics
Thomas Young (13 June 1773 – 10 May 1829) proposed treating the contact angle of a liquid with a surface as the mechanical equilibrium of a drop resting on a plane solid surface under the restrains of three surface tensions:
• $\gamma_{lv}$ (at the interface of the liquid and vapor phases),
• $\gamma_{sl}$ (at the interface of the solid and liquid phases) and
• $\gamma_{sv}$ (at the interface of the solid and vapor phase). This lead to Young’s equation:
$\gamma_{sv} - \gamma_{sl} = \gamma_{lv} \cos \theta \label{1}$
This equation is deceptively simple, but there are a few difficulties with it because of the definitions of the surface tension of the solid-vapor and solid liquid phases.
Another approach avoids specifying the field of intermolecular forces between solid and liquid and instead offers a thermodynamic solution. This lead to Young and Dupre’ equation which introduces the reversible work of adhesion of liquid and solid and it’s relation to the surface tension between liquid and vapor phases and the contact angle:
$W_{sl}^{adh} = \gamma_{lv} (1 + \cos \theta) \label{2}$
Where $WA$ is the reversible work of adhesion of the liquid to the solid when coated with an adsorbed film of the saturated vapor.
Less Traditional Definition
Capillarity is the ability of a substance to draw another substance into it and It occurs when the adhesive intermolecular forces between the liquid and a substance are stronger than the cohesive intermolecular forces inside the liquid. The effect causes a concave meniscus to form where the substance is touching a vertical surface (Figure $3$). The same effect is what causes porous materials to soak up liquids. Capillary forces pull a wetting liquid has a low contact angle with the surface, it wets the surface. Assuming that there are no other factors involved (e.g. roughness), a low contact angle with water means that the surface is hydrophilic. A completely wetting liquid forms a zero contact angle into a capillary by creating a curved meniscus at the rising liquid front. This phenomenon can be described with the Young-Laplace equation and the Laplace pressure inside a capillary.
Measuring methods
Many measuring methods have been described in the literature but only a few of them have been found to be widely applicable.
The Sessile Drop Method
The most frequently used is the goniometer-telescope measurement of sessile-drop contact angles. Commercial contact angle goniometers employ a microscope objective to view the angle directly. In the static method a drop is deposited on a surface and the contact angle can be measured by looking at the drop through a goniometer (an instrument that measures contact angles). The dynamic method is similar to the static one but the drop of liquid which is deposited on a surface is modified. The droplet is being deposited via a syringe and the droplet’s volume is changed dynamically without increasing its solid-liquid interface area and this maximum angle is the advancing angle. Volume is then removed to produce the smallest possible angle, which is called the receding angle. The difference between those two measured angles is called contact angle hysteresis.
Wilhelmy Plate Method
The capillary rise on a Wilhelmy plate is a good way to obtain the contact angle by measuring the height of the meniscus created on a partially immersed plate. The contact angle is obtained directly form the height trough:
$\sin ( \theta) = 1 - \rho g h^2 / 2 \gamma \label{3}$
where:
• h – height of the meniscus created
• γ - surface tension of the tested liquid (N/m)
• ρ - density of the tested liquid (kg/m3)
• θ - contact angle of the liquid on the solid (º)
• g – gravitational acceleration = 9.8 m/s2
Capillary Rise
The common direct measurement of contact angles of a liquid drop on a flat and smooth solid is not applicable to powders and dry porous foods. The common method of measuring the contact angle in capillary rise in porous media is by using the Lucas-Washburn’s equation which is derived from Poiseulle's law of liquid flow in capillary rise:
$h^2 = r c(\gamma \cos \theta / 2 \eta) t \label{4}$
$w^2 = c (\rho^2 \gamma \cos \theta / \eta) t \label{5}$
Where:
• h – height of the rising liquid
• w – weight gain of the sample (caused by the water absorbed into it)
• γ - surface tension of the tested liquid (N/m)
• ρ - density of the tested liquid (kg/m3)
• θ - contact angle of the liquid on the solid (º)
• r - mean static radius of the pores (m)
• η - liquid viscosity (Pa s)
• c - geometric factor (m5)
The common method to measure the contact angles in this case is to conduct capillary rise experiments where your sample is hang bellow a balance and you dip it in the tested liquid. The main deficiency of this approach (using Washburn’s equation) is its inability to separate between those two variables (r and cos θ or c and cos θ). The common method to overcome the problem of the unknown term r•cos(θ) is by using a reference liquid that completely wets the sample (θ=0, cos(θ)=1). It was discovered however that dynamic advancing contact angle is generally larger than the static one, even for a total wetting liquid and as a result this method can be erroneous.
Seibold et al, (2000) suggested a way to overcome this problem. It was observed that this constant term varies as a function of the liquid used, in contradiction with the Washburn approach which uses Hexane as a completely wetting liquid in order to find the radius of the pores. In this study, it was proposed to get the actual constant term r in Washburn equation by plotting the measured value r cos θ versus the rate of rise of the alkanes (the slope in Equations \ref{4} and \ref{5}). The intercept at zero velocity gives the value of r. Siebold el al. (2000) carried out capillary rise experiments with different n-alkanes, which are all considered completely wetting, due to their low surface tension. In each case, a linear relationship was obtained between squared height of the rising liquid and time, their results of the term r•cos(θ) were calculated from the slopes of those curves. This finding was transformed into a method to calculate or measure the contact angle at zero velocity. Siebold et al. (2000) suggested that for porous media the term r•cos(θ) for each alkane (the term was calculated from Washburn equation) can be ploted against the initial front rate of each liquid. The curve that was created can be extrapolated to velocity zero (and therefore, cos(θ)=1) and this enables the determination of the representative radius r. After finding r, which is a property of the solid sample and doesn’t change as a function of the liquids used, we can conduct capillary rise experiment with our test liquid and calculate the contact angles this liquid creates with the solid porous sample.
Applications
The interest in contact angles is because it plays a significant role in a number of technological, environmental and biological phenomena. Water imbibition into porous media theory has been shown to have a multidisciplinary validity in food, soil physics, geology, printings and more. Imbibition of a liquid by a porous solid is a phenomenon highly dependent on wetting. Capillary imbibition is a mechanism that plays a significant role during rehydration of dry food particles that are considered as porous media. Imbibition is highly dependent on the wettability of the porous media, which is usually determined by measuring contact angles which liquids form with the solid.
Outside links
1. http://en.Wikipedia.org/wiki/Contact_angle
2. Contact angle measurements – The static and dynamic sessile drop method:
1. www.youtube.com/watch?v=u265qlIUNrw
2. www.youtube.com/watch?v=5q9qOMesc88&feature=related
3. Equilibrium of water droplet on a hydrophobic surface:
1. www.youtube.com/watch?v=HQAyyJsg17w&NR=1
4. Super hydrophobic materials:
1. www.youtube.com/watch?v=EaE9k-xUtrQ&feature=related
Contributors and Attributions
• Keren Kles (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Liquids/Contact_Angles.txt
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Surface tension is the energy, or work, required to increase the surface area of a liquid due to intermolecular forces. Since these intermolecular forces vary depending on the nature of the liquid (e.g. water vs. gasoline) or solutes in the liquid (e.g. surfactants like detergent), each solution exhibits differing surface tension properties. Whether you know it or not, you already have seen surface tension at work. Whenever you fill a glass of water too far, you may notice afterward that the level of the water in the glass is actually higher than the height of the glass. You may have also noticed that the water that you spilled has formed into pools that rise up off the counter. Both of these phenomena are due to surface tension.
Molecular Perspective
In a sample of water, there are two types of molecules. Those that are on the outside, exterior, and those that are on the inside, interior. The interior molecules are attracted to all the molecules around them, while the exterior molecules are attracted to only the other surface molecules and to those below the surface. This makes it so that the energy state of the molecules on the interior is much lower than that of the molecules on the exterior. Because of this, the molecules try to maintain a minimum surface area, thus allowing more molecules to have a lower energy state. This is what creates what is referred to as surface tension. An illustration of this can be seen in Figure \(1\).
The water molecules attract one another due to the water's polar property. The hydrogen ends, which are positive in comparison to the negative ends of the oxygen cause water to "stick" together. This is why there is surface tension and takes a certain amount of energy to break these intermolecular bonds. Same goes for other liquids, even hydrophobic liquids such as oil. There are forces between the liquid such as Van der Waals forces that are responsible for the intermolecular forces found within the liquid. It will then take a certain amount of energy to break these forces, and the surface tension. Water is one liquid known to have a very high surface tension value and is difficult to overcome.
Surface tension of water can cause things to float which are denser than water, allowing organisms to literally walk on water (Figure \(2\)). An example of such an organism is the water strider, which can run across the surface of water, due to the intermolecular forces of the molecules, and the force of the strider which is distributed to its legs. Surface tension also allows for the formation of droplets that we see in nature.
Cohesive and Adhesive Forces
There are several other important concepts that are related to surface tension. The first of these is the idea of cohesive and adhesive Forces. Cohesive forces are those that hold the body of a liquid together with minimum surface area and adhesive forces are those that try to make a body of a liquid spread out. So if the cohesive forces are stronger than the adhesive forces, the body of water will maintain its shape, but if the opposite is true than the liquid will be spread out, maximizing its surface area. Any substance that you can add to a liquid that allows a liquid to increase its surface area is called a wetting agent.
In the lab there are also several important points to remember about surface tension. The first you've probably noticed before. This is the idea of a meniscus (Figure \(\PageIndex{3a}\)). This is the concave (curved in) or convex (curved out) look that water or other liquids have when they are in test tubes. This is caused by the attraction between the glass and the liquid. With water, this causes it to climb up the sides of a test tube. This attraction is amplified as the diameter of the tubes increases; this is called capillary action. This can be seen if you take a tube with a very small diameter (a capillary tube) and lower it into a body of water. The liquid will climb up into the tube, even though there is no outside force. You may have also seen this when you put a straw into a drink and notice that the liquid level inside the straw is higher than it is in your drink. All of this however, requires that the adhesive forces (between the liquid and the capillary surface) be higher than the cohesive forces (between the liquid and itself), otherwise there will be no capillary action or the opposite can even happen. Mercury has higher cohesive forces than adhesive forces, so the level of the liquid will actually be lower in the capillary tubes than compared to the rest of the mercury (Figure \(\PageIndex{3b}\)).
Contributors and Attributions
• Benjamin Aldridge, Nivaz Brar
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Liquids/Surface_Tension.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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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Liquids/Unusual_Properties_of_Water.txt
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Pressure is the average force that material (gas, liquid or solid) exert upon the surface, e.g. walls of a container or other confining boundary. Vapor pressure or equilibrium vapor pressure is the pressure of a vapor in thermodynamic equilibrium with its condensed phases in a closed container. All liquids and solids have a tendency to evaporate or sublime into a gaseous form and all gases have a tendency to condense back to their liquid or solid form.
Introduction
When a liquid is in a confined, closed, container, an equilibrium exists between the liquid and its gaseous phase. This equilibrium exists regardless of the temperature inside the container and the temperature of the liquid. The equilibrium exists due to the fact that some of the particles in the liquid, essentially at any temperature, will always have enough energy to escape the intrinsic cohesive forces and enter the gaseous phase (Figure $1$).
The vapor pressure of a substance is the pressure that the gaseous part of the substance exerts on the container of said substance.
Boiling a elevated altitudes
The fact that the vapor pressure is equal to the external pressure can become important when talking about boiling temperatures at various altitudes. At higher altitudes the pressure is lower than at sea level and therefore liquids such as water boil at lower temperatures. This would translate it into a longer time to cook something in the water as compared to cooking the same object at sea level. The opposite is also true if one boils water at an altitude lower than sea level.
Characteristics of Vapor Pressure
Vapor pressures are dependent only on temperature and nothing else. The vapor pressure of a liquid does not depend on the amount on the liquid in the container, be it one liter or thirty liters; at the same temperature, both samples will have the same vapor pressure. Vapor pressures have an exponential relationship with temperature and always increase as temperature increases (Figure 2: Vapor Pressure Curves). It is important to note that when a liquid is boiling, its vapor pressure is equal to the external pressure. For example, as water boils at sea level, its vapor pressure is 1 atmosphere because the external pressure is also 1 atmosphere.
Figure $2$: This chart shows the general relationship between a substance's vapor pressure and temperature change. Generally a substance's vapor pressure increases as temperature increases and decreases as temperature decreases (i.e. vapor pressure is directly proportional to temperature). This chart shows that this trend is true for various substances with differing chemical properties.
The change in vapor pressure of a pure substance as temperature changes can be described using the equation known as the Clausius-Clapeyron Equation:
$ln \dfrac{P_2}{P_1} = \dfrac{\Delta H_{vap}}{R} \left ( \dfrac{1}{T_1}-\dfrac{1}{T_2} \right) \label{CC}$
Where:
• P1 is the partial pressure of the liquid at T1
• P2 is the partial pressure of the liquid at T2
Example $1$
If the vapor pressure of water at 293 K is 17.5 mmHg, what is the vapor pressure of water at 300 K?
Solution
Step 1: Use the Clausius Clapeyron equation (Equation \ref{CC}). Assume 293 K to be T1 and 17.5 mmHg to be P1 and 300 K to be T2. We know the enthalpy of vaporization of water is 44000 J mol-1. Therefore we plug in everything we are given into the equation.
$ln\dfrac{P_2}{17.5mmHg} = \dfrac{44000 J mol^{-1}}{8.3145 Jmol^{-1}K^{-1}}\ \left(\dfrac{1}{293K}-\dfrac{1}{300K} \right)$
Step 2: Calculate everything possible, which is everything on the right side of the equation.
$ln \dfrac{P_2}{17.5 mmHg} = (5291.96 K)(0.0000796 K^{-1})$
Step 3: To isolate the variable, we need to get rid of the natural log function on the left side. To do so, we must exponentiate both sides of the equation after calculating the numerical value of the right side of the equation:
$\dfrac{P_2}{17.5 mmHg} = e^{0.421}$
Step 4: To solve for P2, we multiply both sides of the equation by 17.5mmHg.
$P_2 = 26.7 mmHg$
Example $2$
The vapor pressure of water at 298K is 23.8 mmHg. At what temperature is the vapor pressure of water 1075 mmHg?
Solution
Step 1: Once again this can be solved using the Clausius-Clapeyron equation. If we set P1 = 23.8 mmHg, T1 = 298K, and P2 = 1075 mmHg, all we need to do is to solve for T2. So plug everything we know into the Clausius-Clapeyron Equation and we get:
$\ln\dfrac{1075 mmHg}{23.8 mmHg} = \dfrac{44000 Jmol^{-1}}{8.3145 Jmol^{-1}K^{-1}}\ \left (\dfrac{1}{298 K}-\dfrac{1}{T_2} \right)$
Step 2: We can solve the right side of the equation for a numerical answer and we can simplify the right side of the equation to:
$3.81 = (5291.96 K)(\dfrac{1}{298K}-\dfrac{1}{T_2})$
Step 3: Further simplifying the equation by distributing the 5291.96 K, we get:
$3.81 = (17.76) - \dfrac{5291.96K}{T_{2}}$
Step 4: After subtracting 17.76 from both sides and multiplying both sides by T2 to isolate the T2 term, we get:
$-13.95T_2 = -5291.96K$
Step 5: Solving for T2 we get:
$T_2 = 383K$
*Note: There may be some slight variations in this answer if you try working this problem out. The rounding in this problem will make a relatively large difference in mmHg (10-20 mmHg).
Vapor Pressure of Solutions: Raoult's Law
While the Clausius-Clapeyron equation is useful for describing the vapor pressure behavior of a pure substance, it does not quite help us when we need to describe the vapor pressure of a solution comprised of two ore more different liquids with different vapor pressures, that is where Raoult's Law comes in. Raoult's Law is:
$\displaystyle P_{total} = \sum_{i} P_{i}X_{i} \label{Rlaw}$
Where Pi is the vapor pressure of that particular substance and Xi is the corresponding mole fraction of that substance. "i" is an indexing component that keeps track of each substance in the solution.
Figure $3$: Raoult's Law is used for calculating the vapor pressure of solutions with two or more substances in it. The total vapor pressure is a function of the vapor pressure of the individual vapor pressures of the components and their respective mole fractions.
Essentially what Raoult's Law states is that the vapor pressure of a solution with two or more components is directly proportional to the vapor pressures of each component and their respective amounts in the solution (Figure 3). However, Raoult's Law is used to describe solutions that are essentially ideal solutions, meaning it is assumed that there are no interactions between the components of the solution. If the solution is non-ideal, it will deviate from the relationship described by Raoult's Law.
Example $3$
A solution is comprised of 0.5000 mole H2O, 1.000 mole ethanol, and 2.000 moles acetaldehyde at 293 K. What is the total vapor pressure of this solution? The partial pressures of these substances at 293 K are, respectively, 18 mmHg, 67.5 mmHg, and 740 mmHg (assume the liquids mix homogeneously).
Solution
Step 1: Calculate the mole fractions (moles of each substance divided by total moles) of each substance in the solution.
$X_{water} = 0.5000/3.500 = 0.143$
$X_{ethanol} = 1.000/3.500 = 0.286$
$X_{acetaldehyde} = 2.000/3.500 = 0.571$
Step 2: Plug in all the values and solve for Ptotal
$P_{total} = (18mmHg)(0.143) + (67.5 mmHg)(0.286) + (740mmHg)(0.571mmHg)$
$P_{total} = 444.4 mmHg$
Example $4$
A solution is comprised of only water and ethylene glycol and is at 293 K. Water's mole fraction in this solution is 0.379 and water's vapor pressure at this temperature is 18 mmHg (assume the water and ethylene glycol mix homogeneously). The total vapor pressure of this solution is 9.15 mmHg. Calculate the vapor pressure of pure ethylene glycol at this temperature.
Solution
Step 1: Since this solution is only comprised of water and ethylene glycol, we can easily calculate the mole fraction of ethylene glycol in this solution by subtracting water's mole fraction from 1.
$X_{ethylene glycol} = 1 - 0.379 = 0.621$
Step 2: Once we have the mole fraction of ethylene glycol, we have everything we need to solve for the partial pressure of pure ethylene glycol at 293 K using Raoult's Law:
$9.15 = (18mmHg)(0.379) + (P_{ethylene glycol})(0.621)$
Solving for $P_{ethylene glycol}$ we get:
$P_{ethylene glycol} = 3.75mmHg$
Henry's Law
Both the Clausius-Clapeyron Equation (Equation \ref{CC}) and Raoult's Law (Equation \ref{Rlaw}) describe liquids without any significant solutes in them. So what happens if the liquid contains a nonvolatile solute, in other words, a solute that does not evaporate and merely stays in the solution? Will the vapor pressure of the liquid increase or decrease? It turns out that having a nonvolatile solute in the liquid will decrease the vapor pressure of the liquid because the solutes will interfere with the high-energy liquid molecules' path to the surface to break free into the gaseous phase. An example of a nonvolatile solute in a liquid would be glucose in water (Figure $4$: Henry's Law).
Henry's Law can also be used the describe the partial pressure of a volatile solute in a liquid as a function of its concentration in the liquid. An example of this would be oxygen or carbon dioxide in a soda. Henry's Law is as follows:
$C = (K_{H})P_{gas}$
Where P is the partial pressure of either the volatile solute or of the solvent with a nonvolatile solute in it and C is the solubility of the volatile solute or the concentration of the solvent with a nonvolatile solute in it. KH is Henry's Constant and is different for various substances and differs with temperature and comes in many different units.
Example $5$: Henry's Law Application
Assume at 273 K and a CO2 pressure of 3.6 atm, the aqueous solubility of CO2 is 24.8 ml CO2 per liter. What is the molarity of a saturated water solution when the CO2 is under its normal partial pressure in air of 0.000395 atm.
Step 1: We need to calculate Henry's Constant for CO2 when the partial pressure of CO2 is 3.6 atm. To do this, we must first calculate the molarity of CO2 under these conditions:
$Molarity = \dfrac{0.0248L CO_{2} * \dfrac{1 mol CO_{2}}{22.4L CO_{2}}}{1 L solution} = 0.00111 M CO_{2}$
Step 2: Next we must solve for Henry's Constant under these conditions:
$K_H = \dfrac{C}{P_{CO_2}} = 0.000308 \dfrac{M CO_2}{atm}$
Step 3: Solve for the concentration (molarity) using Henry's Law:
$C = K_H * P_{CO_2} = 0.000308M CO_{2}atm^{-1} * 0.000395atm = 1.22 M CO_2$
"The bends"
One great use of Henry's Law is to calculate the various solubilities of gases in solutions under different pressures. One example of this is what is colloquially referred to as "the bends" (scientifically called decompression sickness). When divers go underwater, they are subjected to much higher pressures than they would be at sea level therefore their blood is much more soluble to gases and as such much larger quantities of gases such as nitrogen dissolve in the blood. If the divers then rise up back to the surface too quickly, these gases will be forced out of their dissolved form and form bubbles in the blood vessels. These gas bubbles can then cause the blood vessels to pop, causing much physical pain and serious medical issues. To avoid this, divers rise from deep depths slowly so their bodies can expel the extraneous gas through the lungs and allow the amounts of gas in the blood to equilibrate with the external pressure. This way the divers can avoid gas bubbles forming in blood vessels and therefore any personal injury.
Problems
The vapor pressure of water at 283 K is 9.2 mmHg, at what temperature is the vapor pressure of water 546 mmHg?
1. At 393 K the vapor pressure of water is 1489 mmHg; what is the vapor pressure of water at 343 K?
2. A solution's partial pressure is 34.93 mmHg. This solution is comprised of Chemical A and Chemical B. The partial pressure of Chemical A is 24.5 mmHg and its mole fraction is .145. What is the partial pressure of Chemical B?
3. Chemical A's partial pressure is 35.5 mmHg and Chemical B's partial pressure is 895.5 mmHg. Their respective mole fractions are 0.79 and 0.21. What is the the partial pressure of a homogenous solution of Chemical A and Chemical B?
4. Assume at 273 K and an O2 pressure of 4.7 atm, the aqueous solubility of CO2 is 79.6 ml CO2 per liter. What is the molarity of a saturated water solution when the CO2 is under its normal partial pressure in air of 0.21 atm.
Answers
1. 364K
2. 233.7mmHg
3. 36.7mmHg
4. 216.1mmHg
5. 0.000159 M O2
Contributors and Attributions
• Shuyang (Scott), Liu (UCD), Caroline Tran
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Liquids/Vapor_Pressure.txt
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Viscosity is another type of bulk property defined as a liquid’s resistance to flow. When the intermolecular forces of attraction are strong within a liquid, there is a larger viscosity. An example of this phenomenon is imagining a race between two liquids down a windshield. Which would you expect to roll down the windshield faster honey or water? Obviously from experience one would expect water to easily speed right past the honey, a fact that reveals honey has a much higher viscocity than water.
Introduction
Viscosity can be not only a fluid’s resistance to flow but also a gas’ resistance to flow, change shape or movement. The opposite of viscosity is fluidity which measures the ease of flow while liquids such as motor oil or honey which are “sluggish” and high in viscosity are known as viscous. One may ask the question of what is actually going on in the liquids to make one type flow faster and the other more resistant to flow such as the comparison between honey and water earlier. Because part of a fluid moves, it forces other adjacent parts of the liquid to move along with it causing an internal friction between the molecules which ultimately leads to a reduced rate of flow.
It is also important to note that the viscosity of liquids and gases are affected by temperature but in opposite ways meaning that upon heating, the viscosity of a liquid decreases rapidly, whereas gases flow more sluggishly. Why is this the case? As temperature increases, the average speed of molecules in a liquid also increases and as a result, they spend less time with their "neighbors." Therefore, as temperature increases, the average intermolecular forces decrease and the molecules are able to interact without being "weighed down" by one another. The viscosity of a gas, however, increases as temperature increases because there is an increase in frequency of intermolecular collisions at higher temperatures. Since the molecules are flying around in the void most of the time, any increase in the contact they have with one another will increase the intermolecular force which will ultimately lead to a disability for the whole substance to move.
Measuring Viscosity
There are numerous ways to measure viscosity. One of the most elementary ways is to allow a sphere, such as a metal ball, to drop through a fluid and time the fall of the metal ball: the slower the sphere falls, the greater the viscosity that is measured. Another more advanced design of measuring viscosity known as the Ostwald Viscometer that is much more accurate than dropping a metal ball. An Ostwald Viscometer consists of two reservoir bulbs and a capillary tube. The viscometer is filled with liquid until the liquid reaches the mark A with the aid of a pipette to accurately measure out the volume of needed liquid. The viscometer is then put into a water bath which equilibrates the temperature of the test liquid. As noted before, the equilibration is important to maintain a constant temperature as to not affect the viscosity otherwise. The liquid is then drawn through the side 2 of the U-tube by use of suction and lastly, the flow is time between marks C and B. The viscosity is calculated with Equation $\ref{1}$
$\eta =K t \label{1}$
where $K$ is the value of a liquid with known viscosity and density such as water. Once the value of K is known, the viscosity can be determined by measuring the amount of time the test liquid flows between the two graduated marks.
Units of Measure:
1 Pascal-second (Pa•s) = 1 kg•m?1•s?1 1 Poise = 1 g•cm?1•s?1
Relationship between pascal-second to poise:
10 P = 1 kg•m?1•s?1 = 1 Pa•s 1 cP = 0.001 Pa•s = 1 mPa•s
When measuring viscosity with any type of viscometer, accurate temperature is so important that viscosity can double with a change of only 5 Celsius.
Table 1: Compound: Olive Oil
Temperature Viscosity
290.00 96.102
295.00 75.392
300.00 59.906
305.00 48.167
310.00 39.153
315.00 32.150
320.00 26.649
325.00 22.283
330.00 18.785
335.00 15.956
340.00 13.651
notice how the increase in temperature lowers the viscosity of the fluid which in this case is olive oil
• Pitch (any highly viscous liquid which appear solid) has a viscosity of nearly 100 billion times the viscosity of water
Contributors and Attributions
• Christopher Wen
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Liquids/Viscosity.txt
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There are two general types of bulk intermolecular forces:
• Cohesive Forces: The forces exerted between molecules holding them together. If cohesive forces are strong, a liquid tends to form droplets on a surface.
• Adhesive Forces: The forces between liquid molecules and a surface. If adhesive forces are strong, a liquid tends to spread across a surface.
Wetting agents are substances that reduce the surface tension of water to allow it to spread drops onto a surface, increasing the spreading abilities of a liquid. Lowering the surface tension lowers the energy required to spread drops onto a film, thus weakening the cohesive properties of the liquid and strengthening its adhesive properties. One example of how wetting agents work is in the formation of micelles. Micelles consist of hydrophilic heads forming an outer layer around lipophilic tails. When in water, the micelles' tails can surround an oil droplet while the heads are attracted to the water.
Dish soap is a great example of a wetting agent. With all the food oils and such on the plate cohesive forces make it difficult for the water to spread and clean the plate. The soap dissolves all theses unwanted particles, exposing a clean surface. The soap also lowers the surface tension of water, allowing it to spread evenly across the entire surface.
There are four main types of wetting agents: anionic, cationic, amphoteric, and nonionic.
• Anionic, cationic, and amphoteric wetting agents ionize when mixed with water.
• Anions have a negative charge, while cations have a positive charge.
• Amphoteric wetting agents can act as either anions or cations, depending on the acidity of the solution.
• Nonionic wetting agents do not ionize in water. A possible advantage for using a nonionic wetting agent is that it does not react with other ions in the water, which could lead to formation of a precipitate.
How to Tell if a Liquid Contains a Wetting Agent
One method of knowing whether or not a liquid has a wetting agent in it is to spread the liquid on a surface that is coated in grease. If the liquid does not contain a wetting agent, the its cohesive forces would overpower adhesive forces, causing the liquid to for droplets on the surface (Figure \(2\), left). If the liquid does contain a wetting agent, the grease would be dissolved and the surface tension of the liquid would be lowered, causing the adhesive forces to overpower the cohesive forces. This would result in the liquid spreading evenly along the surface (Figure \(2\), right).
Another method is to place the liquid in a test tube and observe the liquid's meniscus (Figure \(3\)). If the liquid contains a wetting agent, its adhesive forces are stronger than cohesive forces, which means the liquid molecules are more inclined to stick to the surface than other liquid molecules. This results in a concave meniscus. If the liquid does not contain a wetting agent and is naturally very cohesive, like mercury, it forms a convex meniscus. This is caused by the fact the the molecules of the liquid have a stronger attraction to each other than to the surface of the test tube.
Intermolecular forces also cause a phenomenon called capillary action, which is the tendency of a polar liquid to rise against gravity into a small-diameter tube (a capillary), as shown in Figure \(3\). When a glass capillary is put into a dish of water, water is drawn up into the tube. The height to which the water rises depends on the diameter of the tube and the temperature of the water but not on the angle at which the tube enters the water. The smaller the diameter, the higher the liquid rises.
Problems
1. Would it be beneficial to use a wetting agent when waxing a car?
2. An unknown liquid forms a convex meniscus when poured into a test tube. Does the liquid wet the test tube?
3. Do wetting agents increase or decrease the adhesive properties of a liquid?
4. Soap can form a precipitate when used as a wetting agent. Is it a nonionic or ionic wetting agent?
5. A liquid's cohesive forces overwhelm its adhesive forces. Do you think it contains a wetting agent?
Answers
1. No, when waxing a car, you do not want water to wet the car's surface.
2. No, if it forms a convex meniscus, its cohesive forces overpower its adhesive forces, causing the liquid's molecules to want to stick to each other as much as possible.
3. They increase the adhesive properties of a liquid.
4. It must be an ionic wetting agent, since nonionic wetting agents do not form precipitates.
5. The liquid most likely does not contain a wetting agent, since it is more inclined to stick to itself than to wet the surface.
Contributors and Attributions
• Ulysses Morazan (UCD), Abheetinder Brar (UCD)
Properties of Plasma
A plasma is an ionized gas, a gas into which sufficient energy is provided to free electrons from atoms or molecules and to allow both species, ions and electrons, to coexist.
Plasma is the fourth state of matter. Many places teach that there are three states of matter; solid, liquid and gas, but there are actually four. The funny thing about that is, that as far as we know, plasmas are the most common state of matter in the universe. They are even common here on earth. A plasma is a gas that has been energized to the point that some of the electrons break free from, but travel with, their nucleus. Gases can become plasmas in several ways, but all include pumping the gas with energy. A spark in a gas will create a plasma. A hot gas passing through a big spark will turn the gas stream into a plasma that can be useful. Plasma torches like that are used in industry to cut metals. The biggest chunk of plasma you will see is the sun. The sun's enormous heat rips electrons off the hydrogen and helium molecules that make up the sun. Essentially, the sun, like most stars, is a great big ball of plasma.
Contributors and Attributions
• Brian Kross, Chief Detector Engineer (Thomas Jefferson National Accelerator Facility - Office of Science Education)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Liquids/Wetting_Agents.txt
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Solid are characterized by structural rigidity and resistance to changes of shape or volume. Unlike a liquid, a solid object does not flow to take on the shape of its container, nor does expands to fill the entire volume available to it like a gas . The atoms in a solid are tightly bound to each other, either in a regular geometric lattice (crystalline solids, which include metals and ordinary water ice) or irregularly (an amorphous solid such as common window glass).
Properties of Solids
Crystals are composed of three-dimensional patterns. These patterns consist of atoms or groups of atoms in ordered and symmetrical arrangements which are repeated at regular intervals keeping the same orientation to one another. By replacing each group of atoms by a representative point a crystal lattice is obtained. Keep in mind; lattice sites are not necessarily associated with the position of atoms. Thus, a crystal lattice is a set of infinite, arranged points related to each other by transitional symmetry. The outlines for such patterns are called lattices. Lattices are comprised of the intersections of three parallel planes. The planes intersect producing three-dimensional figures which have six faces (like a cube) these are set in three sets of parallel planes, thus making a figure known as a parallelepiped.
Contributors and Attributions
• Cassandra Patterson (UCD)
Crystal Lattice
The term "closest packed structures" refers to the most tightly packed or space-efficient composition of crystal structures (lattices). Imagine an atom in a crystal lattice as a sphere. While cubes may easily be stacked to fill up all empty space, unfilled space will always exist in the packing of spheres. To maximize the efficiency of packing and minimize the volume of unfilled space, the spheres must be arranged as close as possible to each other. These arrangements are called closest packed structures.
The packing of spheres can describe the solid structures of crystals. In a crystal structure, the centers of atoms, ions, or molecules lie on the lattice points. Atoms are assumed to be spherical to explain the bonding and structures of metallic crystals. These spherical particles can be packed into different arrangements. In closest packed structures, the arrangement of the spheres are densely packed in order to take up the greatest amount of space possible.
Types of Holes From Close-Packing of Spheres
When a single layer of spheres is arranged into the shape of a hexagon, gaps are left uncovered. The hole formed between three spheres is called a trigonal hole because it resembles a triangle. In the example below, two out of the the six trigonal holes have been highlighted green.
Once the first layer of spheres is laid down, a second layer may be placed on top of it. The second layer of spheres may be placed to cover the trigonal holes from the first layer. Holes now exist between the first layer (the orange spheres) and the second (the lime spheres), but this time the holes are different. The triangular-shaped hole created over a orange sphere from the first layer is known as a tetrahedral hole. A hole from the second layer that also falls directly over a hole in the first layer is called an octahedral hole.
Closest Pack Crystal Structures
Hexagonal Closest Packed (HCP)
In a hexagonal closest packed structure, the third layer has the same arrangement of spheres as the first layer and covers all the tetrahedral holes. Since the structure repeats itself after every two layers, the stacking for hcp may be described as "a-b-a-b-a-b." The atoms in a hexagonal closest packed structure efficiently occupy 74% of space while 26% is empty space.
Cubic Closest Packed (CCP)
The arrangement in a cubic closest packing also efficiently fills up 74% of space. Similar to hexagonal closest packing, the second layer of spheres is placed on to of half of the depressions of the first layer. The third layer is completely different than that first two layers and is stacked in the depressions of the second layer, thus covering all of the octahedral holes. The spheres in the third layer are not in line with those in layer A, and the structure does not repeat until a fourth layer is added. The fourth layer is the same as the first layer, so the arrangement of layers is "a-b-c-a-b-c."
Coordination Number and Number of Atoms Per Unit Cell
A unit cell is the smallest representation of an entire crystal. All crystal lattices are built of repeating unit cells. In a unit cell, an atom's coordination number is the number of atoms it is touching.
• The hexagonal closest packed (hcp) has a coordination number of 12 and contains 6 atoms per unit cell.
• The face-centered cubic (fcc) has a coordination number of 12 and contains 4 atoms per unit cell.
• The body-centered cubic (bcc) has a coordination number of 8 and contains 2 atoms per unit cell.
• The simple cubic has a coordination number of 6 and contains 1 atom per unit cell.
Unit Cell Coordination Number # of Atoms Per Unit Cell % space
Simple Unit Cell
6
1 52%
Body-Centered Cubic
8 2 68%
Face-centered Cubic
12 4 74.04%
Cubic Closest Packed
12 4 74.04%
Hexagonal Closest Packed
12 6* (2) see note below 74.04%
Note
*For the hexagonal close-packed structure the derivation is similar. Here the unit cell consist of three primitive unit cells is a hexagonal prism containing six atoms (if the particles in the crystal are atoms). Indeed, three are the atoms in the middle layer (inside the prism); in addition, for the top and bottom layers (on the bases of the prism), the central atom is shared with the adjacent cell, and each of the six atoms at the vertices is shared with other five adjacent cells. So the total number of atoms in the cell is 3 + (1/2)×2 + (1/6)×6×2 = 6, however this results in 2 per primitive unit cell.
www.quora.com/What-is-the-nu...it-cell-of-HCP
Contributors and Attributions
• Brittanie Harbick, Laura Suh, Jenny Fong
Crystal Planes and Miller Indic
Crystal planes come from the structures known as crystal lattices. These lattices are three dimensional patterns that consist of symmetrically organized atoms intersecting three sets of parallel planes. These parallel planes are "crystal planes" and are used to determine the shape and structure of the unit cell and crystal lattice. The planes intersect with each other and make 3D shapes that have six faces. These crystal planes define the crystal structure by making axes visible and are the means by which we can calculate the Miller Indices.
Contributors and Attributions
• Timothy Ulleseit (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Solids/Crystal_Lattice/Closest_Pack_Structures.txt
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Contributors and Attributions
• Nathalie Interiano
Network Covalent Solids
Covalent Network Solids are giant covalent substances like diamond, graphite and silicon dioxide (silicon(IV) oxide). This page relates the structures of covalent network solids to the physical properties of the substances.
Diamond
Carbon has an electronic arrangement of 2,4. In diamond, each carbon shares electrons with four other carbon atoms - forming four single bonds.
In the diagram some carbon atoms only seem to be forming two bonds (or even one bond), but that's not really the case. We are only showing a small bit of the whole structure. This is a giant covalent structure - it continues on and on in three dimensions. It is not a molecule, because the number of atoms joined up in a real diamond is completely variable - depending on the size of the crystal.
How to draw the structure of diamond
Don't try to be too clever by trying to draw too much of the structure! Learn to draw the diagram given above. Do it in the following stages:
Practice until you can do a reasonable free-hand sketch in about 30 seconds.
Physical Properties of Diamond
• has a very high melting point (almost 4000°C). Very strong carbon-carbon covalent bonds have to be broken throughout the structure before melting occurs.
• is very hard. This is again due to the need to break very strong covalent bonds operating in 3-dimensions.
• doesn't conduct electricity. All the electrons are held tightly between the atoms, and aren't free to move.
• is insoluble in water and organic solvents. There are no possible attractions which could occur between solvent molecules and carbon atoms which could outweigh the attractions between the covalently bound carbon atoms.
Graphite
Graphite has a layer structure which is quite difficult to draw convincingly in three dimensions. The diagram below shows the arrangement of the atoms in each layer, and the way the layers are spaced.
Notice that you cannot really draw the side view of the layers to the same scale as the atoms in the layer without one or other part of the diagram being either very spread out or very squashed. In that case, it is important to give some idea of the distances involved. The distance between the layers is about 2.5 times the distance between the atoms within each layer. The layers, of course, extend over huge numbers of atoms - not just the few shown above.
You might argue that carbon has to form 4 bonds because of its 4 unpaired electrons, whereas in this diagram it only seems to be forming 3 bonds to the neighboring carbons. This diagram is something of a simplification, and shows the arrangement of atoms rather than the bonding.
The Bonding in Graphite
Each carbon atom uses three of its electrons to form simple bonds to its three close neighbors. That leaves a fourth electron in the bonding level. These "spare" electrons in each carbon atom become delocalized over the whole of the sheet of atoms in one layer. They are no longer associated directly with any particular atom or pair of atoms, but are free to wander throughout the whole sheet. The important thing is that the delocalized electrons are free to move anywhere within the sheet - each electron is no longer fixed to a particular carbon atom. There is, however, no direct contact between the delocalized electrons in one sheet and those in the neighboring sheets. The atoms within a sheet are held together by strong covalent bonds - stronger, in fact, than in diamond because of the additional bonding caused by the delocalized electrons.
So what holds the sheets together? In graphite you have the ultimate example of van der Waals dispersion forces. As the delocalized electrons move around in the sheet, very large temporary dipoles can be set up which will induce opposite dipoles in the sheets above and below - and so on throughout the whole graphite crystal.
Graphite has a high melting point, similar to that of diamond. In order to melt graphite, it isn't enough to loosen one sheet from another. You have to break the covalent bonding throughout the whole structure. It has a soft, slippery feel, and is used in pencils and as a dry lubricant for things like locks. You can think of graphite rather like a pack of cards - each card is strong, but the cards will slide over each other, or even fall off the pack altogether. When you use a pencil, sheets are rubbed off and stick to the paper. Graphite has a lower density than diamond. This is because of the relatively large amount of space that is "wasted" between the sheets.
Graphite is insoluble in water and organic solvents - for the same reason that diamond is insoluble. Attractions between solvent molecules and carbon atoms will never be strong enough to overcome the strong covalent bonds in graphite. conducts electricity. The delocalized electrons are free to move throughout the sheets. If a piece of graphite is connected into a circuit, electrons can fall off one end of the sheet and be replaced with new ones at the other end.
Silicon dioxide: SiO2
Silicon dioxide is also known as silica or silicon(IV) oxide has three different crystal forms. The easiest one to remember and draw is based on the diamond structure. Crystalline silicon has the same structure as diamond. To turn it into silicon dioxide, all you need to do is to modify the silicon structure by including some oxygen atoms.
Notice that each silicon atom is bridged to its neighbors by an oxygen atom. Don't forget that this is just a tiny part of a giant structure extending on all 3 dimensions.
Silicon Dioxide has a high melting point - varying depending on what the particular structure is (remember that the structure given is only one of three possible structures), but around 1700°C. Very strong silicon-oxygen covalent bonds have to be broken throughout the structure before melting occurs. Morevoer, it hard due to the need to break the very strong covalent bonds.Silicon Dioxide does not conduct electricity since there aren't any delocalized electrons with all the electrons are held tightly between the atoms, and are not free to move.Silicon Dioxide is insoluble in water and organic solvents. There are no possible attractions which could occur between solvent molecules and the silicon or oxygen atoms which could overcome the covalent bonds in the giant structure.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Solids/Ionic_Solids.txt
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A unit cell is the most basic and least volume consuming repeating structure of any solid. It is used to visually simplify the crystalline patterns solids arrange themselves in. When the unit cell repeats itself, the network is called a lattice.
Introduction
The work of Auguste Bravais in the early 19th century revealed that there are only fourteen different lattice structures (often referred to as Bravais lattices). These fourteen different structures are derived from seven crystal systems, which indicate the different shapes a unit cell take and four types of lattices, which tells how the atoms are arranged within the unit. The kind of cell one would pick for any solid would be dependent upon how the latices are arranged on top of one another. A method called X-ray Diffraction is used to determine how the crystal is arranged. X-ray Diffraction consists of a X-ray beam being fired at a solid, and from the diffraction of the beams calculated by Bragg's Law the configuration can be determined.
The unit cell has a number of shapes, depending on the angles between the cell edges and the relative lengths of the edges. It is the basic building block of a crystal with a special arrangement of atoms. The unit cell of a crystal can be completely specified by three vectors, a, b, c that form the edges of a parallelepiped. A crystal structure and symmetry is also considered very important because it takes a role in finding a cleavage, an electronic band structure, and an optical property. There are seven crystal systems that atoms can pack together to produce 3D space lattice.
The unit cell is generally chosen so as to contain only complete complement of the asymmetric units. In a geometrical arrangement, the crystal systems are made of three set of ($a$, $b$, $c$), which are coincident with unit cell edges and lengths. The lengths of a, b, c are called the unit cell dimensions, and their directions define the major crystallographic axes. A unit cell can be defined by specifying a, b, c, or alternately by specifying the lengths |a|, |b|, |c| and the angles between the vectors, $\alpha$, $\beta$, and $\gamma$ as shown in Fig. 1.1. Unit cell cannot have lower or higher symmetry than the aggregate of asymmetric units. There are seven crystal systems and particular kind of unit cell chosen will be determined by the type of symmetry elements relating the asymmetric units in the cell.
The unit cell is chosen to contain only one complete complement of the asymmetric units, which is called primitive (P). Unit cells that contain an asymmetric unit greater than one set are called centered or nonprimitive unit cells. The additional asymmetric unit sets are related to the first simple fractions of unit cells edges. For example, (1/2, 1/2, 1/2) for the body centered cell $I$ and (1/2,1/2, 0) for the single-face-centered cell $C$. The units can be completely specified by three vectors (a, b, c) and the lengths of vectors in angstroms are called the unit cell dimensions. Vectors directions are defined the major crystallographic axes. Unit cell can also be defied by specifying the lengths (|a|, |b|, |c|) and the angles between the vectors ($\alpha$, $\beta$, and $\gamma$) as shown in Fig.1.1.
Table 1: includes the allowable unit cell types found in crystals and their distinguishing characteristics.
Crystal System Types of Lattices (number of particles) Description of Cell
Cubic Simple or Primitive (1) Face centered (4) Body centered (2) Quadratic prism that is equilateral and equiangular (i.e. a cube)
Tetragonal Simple or Primitive (1) Body centered (2) Quadratic prism that has two equal edges and one different sized edge, and is equiangular. (i.e. a rectangular prism with a square base)
Orthorhombic Simple or Primitive (1) Face centered (4) Body centered (2) End or Base centered (2) Quadratic prism that has no equal edges, and is equiangular. (i.e. a rectangular prism without any square base faces)
Monoclinic Simple or Primitive (1) End or Base centered (2) Quadratic prism that has no equal edges, two edges at 90 degrees of each other, and one thats not at 90. (i.e. a parallelogram extended to some distant not equal to the width of the base)
Rhombohedral Simple or Primitive (1) Quadratic prism that has equal length edges, but one slanted side (i.e. a slanted cube)
Triclinic Simple or Primitive (1) Quadratic prism with no equal length edges, two unequal angles and one angle at 90 degrees.
Hexagonal Simple or Primitive (1) Hexagonal prism
Note: Edges refers to all parallel edges. And all parallel edges are equal to each other.
This table describes the fourteen different kind of unit cells available. As you can see not every crystal systems can have all the different types of lattices.
Volumes
Calculating the volume for a unit cell is the same as calculating the volume for any prism - base area multiplied by height. The equations for each different crystal system are as follow:
Crystal System Volume =
Cubic abc
Tetragonal abc
Orthorhombic abc
Monoclinic abc sin(?), where ? is the acute non 90 degree angle.
Rhombohedral abc sin(60°)
Triclinic abc ((1- cos²? - cos²? - cos²?) + 2(cos(?) cos(?) cos(?))½
Hexagonal abc sin(60°)
Note: a, b, & c are represent the edges. ?, ?, & ? are the angles.
Most calculations involving unit cells can be solved with the formula: density = Mass/Volume. Then in addition to the obvious three the number of particles per cell can also be calculated by the density/molar mass.
Density - Particles
It can easily be seen that not all the particles are complete in the unit cell form. For the fractional particles in unit cell, its corner particles will always sum to one whole particle, its face particles (for face centered lattices) will sum to three whole particles, and for the base particles will sum to one.
Contributors and Attributions
• Minh Nguyen (UCD), Mandeep Singh (UCD)
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Matter can be pushed to temperatures and pressures beyond those of its critical point. This stage is characterized by the inability to distinguish whether the matter is a liquid or a gas, as a result, Supercritical fluids (SCF) do not have a definite phase. In 1822 Baron Charles Cagniard de la Tour discovered supercritical fluids while conducting experiments involving the discontinuities of the sound of a flint ball in a sealed cannon barrel filled with various fluids at various temperatures ("Charles Cagniard de la Tour").
Supercritical fluids have the low viscosity of a gas and the high density of a liquid, making it impossible to liquefy the matter using any amount of pressure. However, it is possible to go from a gas to a liquid without crossing the boundary between the vapor and liquid phase using a supercritical fluid just by lowering the temperature of the liquid (Observe phase diagram below).
Volatile liquids and solids, or liquids and solids with a high vapor pressure or low boiling point, are soluble in gas. It becomes especially easy to dissolve liquids and solids such as these in a supercritical fluid because of the high density. As can be noted from the Mole Fraction Solubility. The Mole Fraction Solubility is simply the ratio of the sublimation (or vapor pressure) to the total gas pressure.
Supercritical fluids have no surface tension because they are not subject to the vapor-liquid boundary so no molecules have the attraction to the interior of the liquid. The densities and viscosity of a supercritical fluid are subject to change when pressure or temperature are tampered with, and the supercritical fluid of a substance can have very different properties than the regular fluids. For instance, water that is supercritical differs from regular water in the fact that it is non-polar and acidic (Benner 680).
Examples: A Phase Diagram:
Notice the yellow and blue mix to create green area that follows the Coordinates of the critical point, that is where the supercritical fluids occur on the graph. Each element and molecule have unique critical points. The arrow shows how it is possible to go from a vapor to a liquid by using supercritical fluids, pressure and temperature. Notice how when pressure and temperature on a gas are increased into the supercritical range, and the temperature is lowered, the substance moves into the liquid phase.
Table 1 : Critical Points of Common SCFs
Liquid Critical Temperature (K) Critical Pressure (atm)
Hydrogen (H) 33.3 12.8
Neon (Ne) 44.4 26.3
Nitrogen (N) 126 33.5
Argon (Ar) 151 48.5
Methane (CH4) 191 45.8
Ethane (C2H6) 305 48.2
Carbon Dioxide (CO2) 305 72.9
Ammonia (NH3) 406 112
Water (H2O) 647 218
(Benner 680)
Natural Occurences
Supercritical fluids can occur in nature. For example, in places like underwater volcanoes, specifically those located deep beneath the ocean's surface, supercritical water is formed because of the immense pressure due to the depth and the intense heat from the vents of the volcano. This water can lead to the formation of crystals used in some jewelry (Benner 680).
Venus
The atmospheric pressure of Venus is approximately 90 times greater than that of the Earth, with an average temperature of 467 °C, and about 97% of its atmosphere is Carbon Dioxide. Therefore it would be reasonable to consider the atmosphere of Venus a supercritical fluid because both the pressure and temperature exceed that of Carbon Dioxide's critical point however this theory has not been proven. Examples similar to this one can be found throughout the solar system, particularly in the Gas Giants (Benner 680).
Supercritical fluids are useful in science today for purposes ranging from the extraction of floral fragrance from flowers and the process of creating decaffeinated coffee, to applications in food science and functional food ingredients, pharmaceuticals, cosmetics, polymers, powders, bio- and functional materials, nano-systems, natural products, biotechnology, fossil and biofuels, microelectronics and environment (Bottini 133).
Extraction using Supercritical Fluids is a fairly simple concept, and much more efficient than normal extraction methods, which require both heating and ventilation of the solution to the atmosphere. Supercritical fluids allow continuous extraction using common, inexpensive, and more importantly non-toxic materials, and only requires venting to separate the solvent from the material being removed. The extraction involves applying the supercritical solvent to whatever material is being eradicated, for example, coffee beans which are being decaffeinated, and allowing the solvent to remove the substance being extracted. Returning to the coffee bean example, supercritical CO2 would be applied to the beans, then when it had extracted the caffeine, the CO2 would be put through carbon filters, which would separate the caffeine from the solvent which can then be removed simply by venting because of its unique properties and similarities to vapor. Likewise, supercritical fluids can be used as solvents to apply substances like dyes to clothing, the process for this is more or less the reverse of extraction. The most commonly used solvents are supercritical Carbon Dioxide and Water because of their availability and low critical temperatures (Hardy).
Contributors and Attributions
• Kiana Samadzadeh
Supercritical Fluids
Caffeine molecules are naturally found in coffee beans, tea leaves, cocoa and a variety of exotic berries. When ingested, caffeine can act as a stimulant in humans or a toxin in small animals and insects. A certain portion of the human population can’t tolerate increased levels of caffeine in their body. They can experience extreme side effects including, but not limited to irritability, muscle twitching, dehydration, headaches, increased heart rate, and frequent urination. These side effects can be quite unpleasant, which is why many coffee manufacturers decaffeinate coffee.
Introduction
Decaffeination is a fairly easy process since caffeine is polar and water-soluble. The most popular methods of decaffeinating coffee today are, Swiss Water Processing, Ethyl Acetate Processing, Methylene Chloride Processing (Direct and Indirect), and Supercritical Carbon Dioxide Processing.
Swiss Water Processing
The Swiss Water Processing method removes caffeine without using any chemicals, but instead applies the law of simple diffusion. First, unroasted (green) coffee beans are soaked in water until caffeine is dissolved in water. The beans are then discarded, and the solution of water, caffeine, and coffee solids is passed through a carbon filter. The carbon filter is made out of activated carbon, carbon that has been made porous through the process of carbonization (reacting carbon in anaerobic conditions until the gaps between carbon atoms are large enough to allow molecules to pass through). The activated carbon filter has holes large enough to allow water and coffee solids (smaller molecules) to pass through, but not caffeine (relatively larger molecule). After filtration, the mixture that is left is water saturated with coffee flavor molecules – referred as “coffee solids” by the manufacturers. The mixture creates a concentration gradient when added to a fresh batch of coffee beans. Concentration gradients take advantage of the law of simple diffusion- the movement of molecules from an area of high solute concentration to an area of low solute concentration in order to 'even out' the uneven distribution of molucules. Since the only difference between the mixture and the fresh coffee beans is the caffeine concentration, caffeine molecules will diffuse out of the beans into the mixture of coffee solids, leaving the coffee beans caffeine free. This method is repeated until the coffee beans are 99.9% decaffeinated, and the flavor is left intact.
Ethyl Acetate Processing
Ethyl Acetate occurs naturally in many fruits, which is why this method is often referred to as natural decaffeination. It is however much cheaper commercially to use synthetic ethyl acetate. This method requires a thorough steaming of the beans until swell. An ethyl acetate aqueous solution is used to wash the swollen beans repeatedly. Ethyl acetate is a polar molecule, which makes it a good solvent for capturing the polar caffeine molecules from the coffee beans (since 'like dissolves like'). The caffeine molecules bind to the ethyl acetate molecules, and migrate through the cell membranes of cells of the beans. The beans are once again steamed in order to eliminate any ethyl acetate that remains. This method decaffeinates the coffee beans by approximately 97%.
Methylene Chloride Processing
• Direct Method- Steamed coffee beans are rinsed directly with methylene chloride which is a polar molecule and is good solvent to organic molecules. The caffeine molecules hydrogen bond to the methylene chloride molecules, and are removed from the coffee beans, leaving the coffee solids (flavor) intact. The resulting coffee beans are 97% caffeine free.
• Indirect Method- Coffee beans are rinsed with water, removing the caffeine molecules and coffee solids (similar to the first part of the Swiss Water Process). This solution is then treated with methylene chloride. The caffeine forms hydrogen bonds with the methylene chloride, leaving a coffee flavor aqueous solution. The original coffee beans are soaked on this solution, allowing for the reabsorption of the coffee solids (flavor). The original flavor of the coffee beans is preserved, but are 97% decaffeinated.
Supercritical Carbon Dioxide Processing
Carbon dioxide supercritical fluid (temperature above 31.1 °C and pressure above 73 atm) exhibits both liquid and gas-like behavior. It behaves like gas, and permeates a porous substance, while also exhibiting liquid properties to dissolve substances. Although supercritical carbon dioxide is non-polar, and should only be able to dissolve non-polar substances, certain co-solvents, like water, can be added so that supercritical carbon dioxide can actually dissolve polar molecules like caffeine. Water is more polar than caffeine is, so supercritical carbon dioxide, in the presence of a co-solvent like water, will dissolve the more non-polar substance, in this case, caffeine. In order to use supercritical carbon dioxide to decaffeinate coffee beans, the beans are first steamed until they swell (this is where the co-solvent, water, comes into play). After this, they are immersed in supercritical carbon dioxide which binds to the caffeine molecules and draws them out of the beans, leaving the coffee solids (flavor) embedded in the bean. The resulting coffee beans are about 97% caffeine free. The carbon dioxide is then passed through a charcoal membrane that is selective toward carbon dioxide molecules. Caffeine is stopped by the membrane, because of its larger size relative to carbon dioxide, and collected.
Usage of Removed Caffeine
Once coffee beans have been decaffeinated, all of the extracted caffeine is made into a white powder and sold to the pharmaceutical or food industries. The pharmaceutical industry adds caffeine into certain drugs, including many pain killers. Food industries add caffeine to certain foods like soda, because of its stimulating effect.
Contributors
• Mavish Mahomed (UC Davis)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Supercritical_Fluids/Case_Study%3A_Removing_caffeine_from_Coffee.txt
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This module refers to a finite amount of particles placed in a closed container (i.e. no volume change) in which boiling cannot occur. The inability for boiling to occur- because the particles in the container are not exposed to the atmosphere, results in the incessant increase of temperature and pressure. The critical point is the temperature and pressure at which the distinction between liquid and gas can no longer be made.
Introduction
At the critical point, the particles in a closed container are thought to be vaporizing at such a rapid rate that the density of liquid and vapor are equal, and thus form a supercritical fluid. As a result of the high rates of change, the surface tension of the liquid eventually disappears. You will have noticed that this liquid-vapor equilibrium curve has a top limit (labeled as C in the phase diagram in Figure 1), which is known as the critical point. The temperature and pressure corresponding to this are known as the critical temperature and critical pressure. If you increase the pressure on a gas (vapor) at a temperature lower than the critical temperature, you will eventually cross the liquid-vapor equilibrium line and the vapor will condense to give a liquid.
Figure 1: Phase diagram for a single component system with critical point emphasized
This works fine as long as the gas is below the critical temperature. What, though, if your temperature was above the critical temperature? There wouldn't be any line to cross! That is because, above the critical temperature, it is impossible to condense a gas into a liquid just by increasing the pressure. All you get is a highly compressed gas. The particles have too much energy for the intermolecular attractions to hold them together as a liquid. The critical temperature obviously varies from substance to substance and depends on the strength of the attractions between the particles. The stronger the intermolecular attractions, the higher the critical temperature.
Why the Critical Point is Important
The condensation of a gas will never occur above the critical point. A massive amount of pressure can be applied to a gas in a closed container, and it may become highly dense, but will not exhibit a meniscus. Molecules at critical temperatures possess high kinetic energy, and as a result the intermolecular forces in the molecules are weakened.
The Declined Critical Points of Polymer Solutions
A novel discovery made by the University of Manchester, identified that lower critical temperatures are existent in polymer solutions. It has been manifested that hydrocarbon polymers integrated with a hydrocarbon solution portrays what the university terms a "L.C.S.T." or a lower critical solution temperature. This lower critical solution temperature of polymers has been proclaimed to be in a range near the gas- liquid critical point of the polymer's solvent, and can reach up to 170 degrees Celsius. Such a lower critical solution temperature can be contributed to the assimilation of the heat and volume of the substance n-pentane with most hydrocarbon polymers at room temperature (Freeman, P.I., Rowlinson, J.S.).
The Effects of Wetting on the Critical Point
When a fluid is present in two phases, in a container, and a critical point is near establishment, contact with the imminently forming third phase does not occur. This phenomena can be accounted for by examining the other two existing phases; the third phase does not immediately form because one of the other two phases wets the third phase, causing it to be eliminated. This wetting phase will continually occur when a phase is not entirely stable as a whole.
Problems
Temperature and vapor pressure are essential to the stimulation of a critical point; the following problems interconnect the two concepts.
1. Determine the vapor pressure of liquid gold at 1936 °C if 200 L of Ne gas is permitted to assimilate with the metal causing it to decline in mass by 0.213 g.
2. Approximately 0.423 g H2O is preserved at a temperature of 62 °C in a sealed flask at a volume of 0.726 L. When in equilibrium, will the water be present solely as a liquid?
3. Taking into account the previous problem, would it be probable for the sample of water in the flask to be solitarily present as a vapor?
4. Is it possible for the the sample of water in problem number two to exist in both the liquid and vapor form at equilibrium?
5. The vapor pressure of H2O at 30 °C is 46.2 mmHg. What is the vapor pressure of water at 28 °C, if the enthalpy of fusion is 44 kJ/mol?
Solutions
1. In correlation with Dalton's Law of partial pressure, the gold vapor can be interpreted as a solitary gas occurring at the given volume of Neon. The ideal-gas law will be utilized to determine the vapor pressure.
$PV=nRT$
• P=?
• V=200L
• T=1936.C (to convert to K add 273)=2209 K
• R=00820L atm/mol K
Convert to mol the molar mass of Au
$n= 0.213\; \cancel{g} \left( \dfrac{1 \;mol\; Au}{196.9\; \cancel{g}\; Au}\right) = 0.00108\; mol\; Au$
$PV=nRT$ can then be algebraically converted into $P=\dfrac{nRT}{V}$. Substitute you attained values.
$P= \dfrac{ (0.00108 \;mol) \times (0.08206\; L \; atm / mol K) \times (2209 \; K)}{ 200 \; L}$
$P= 9.78\times 10^{-4}\; atm$
2. Taking into account the density of water, 1 g/ml, a 0.423 g sample of $H_2O$ would amount to 0.42 ml. Considering this, it can be concluded that the water in the flask could not solely exist as a liquid at equilibrium since a 0.42 ml sample of water is incapable of accounting for the overall volume present in the flask (0.726L).
3. The ideal gas law can be applied to determine whether or not the water in the flask could solely exist in the vapor form. Rearrange the Ideal Gas Law:
$P=\dfrac{nRT}{V}$
• n= (0.423g)*(1 mol/18 g H2O)=0.0235mol
• R=0.08206 L atm/ mol K
• T=62°C + 273 = 335K
• V=0.726 L
$P= \dfrac{(0.0235 \;mol) \times (0.08206 \;L\; atm/mol\; K) \times (335\; K)}{0.726\; L}$
$P=0.889 atm$
The actual Vapor pressure of water at 60 °C is $149.4\; mmHg$.
To make a comparison to the vapor pressure attained at 62°C (which is within a close range of the temperature of 60 degrees Celsius), convert the vapor pressure calculated in atm to mmHg.
$(0.889\;\cancel{atm})* \left(\dfrac{760\;mmHg}{1\; \cancel{atm}}\right)=676.27 \;mmHg$
Comparing this vapor pressure with the actual vapor pressure of water at 60 °C, it can be concluded that it is improbable that water solely exists in the flask as vapor since this attained vapor pressure exceeds that of the actual vapor pressure that occurs naturally.
4. As the prior explications and calculations have proven, the sample of water is incapable of existing in the flask as either a liquid or vapor alone. Therefore vapor and liquid must exist together at 60 °C and 149.4 mmHg.
5. To evaluate the pressure of water at 28 °C, the Clausius-Clapeyron equation must be used.
$\ln \dfrac{P_1}{P_2}= \dfrac{H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)$
Begin by designating the variables in the equation to the given values in the problem.
If $T_2 =30 °C$ and $P_2= 46.2\; mmHg$, what is $P_1$ when $T_1=28 °C$
The units for temperature must be in Kelvin.
• $T_2= 30 °C + 273.15 K = 303.15\; K$
• $T_1 =28 °C\; +273.15 =301.15\; K$
The enthalpy of vaporization (44 kJ/ mol) should be converted to J/ mol to allow for unit cancelation.
$44 \;\cancel{kJ}/mol \left(\dfrac{1000\; J}{1\; \cancel{kJ}}\right)= 44.0 \times 10^{3} J/mol.$
R= 8.3145 J/mol K
Substituting the values in the equation one you obtain:
$\ln \left(\dfrac{46.2\; mmHg}{P_1}\right) = \dfrac{44,000\; J \;mol^{-1}}{8.3145 \;J/mol\; K} \left(\dfrac{1}{ 301.15 \;K} - \dfrac{1}{ 303.15 \;K}\right)$
To eliminate the natural logarithm, take the exponentl of both sides:
$e^{\ln \left(\dfrac{46.2\; mmHg}{P_1}\right) }= e^{ \dfrac{44,000\; J \;mol^{-1}}{8.3145 \;J/mol\; K} \left(\dfrac{1}{ 301.15 \;K} - \dfrac{1}{ 303.15 \;K}\right)}$
$\dfrac{46.2\; mmHg}{P_1} = e^{ \dfrac{44,000\; J \;mol^{-1}}{8.3145 \;J/mol\; K} \left(\dfrac{1}{ 301.15 \;K} - \dfrac{1}{ 303.15 \;K}\right)}$
$\dfrac{46.2\; mmHg}{P_1} = e^{0.11593}$
$\dfrac{46.2\; mmHg}{P_1} = 1.1229$
$\dfrac{1}{ P_1} =\dfrac{1.1229 }{46.2\; mmHg}$
$P_1 = \dfrac{46.2\; mmHg}{1.1229}$
$P_1 = 41.143\; mmHg$, which gets rounded to two significant digits (from the two significant digits in 44 kJ/mol).
$P_1$ vapor pressure is 41 mmHg.
Outside Links
• Brown, R. J. C.; Brown, R. F. C. "Melting Point and Molecular Symmetry." J. Chem. Educ. 2000 77 724.
Contributors
• Minh Nguyen, Cynthia Dvorsky Garcia
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Blackbody radiation is a cornerstone in the study of quantum mechanics.This experiment is what led to the discovery of a field that would revolutionize physics and chemistry. Quantum mechanics gives a more complete understanding of the fundamental mechanisms at the sub-atomic level.
Introduction
The work done at the turn of the 20th century on blackbody radiation was the beginning of a totally new field of science. Blackbody radiation is a theoretical concept in quantum mechanics in which a material or substance completely absorbs all frequencies of light. Because of the laws of thermodynamics, this ideal body must also re-emit as much light as it absorbs.Although there is no material that can truly be a blackbody, some have come close.Carbon in its graphite form is about 96% efficient in its absorption of light.
The concept of blackbody radiation is seen in many different places.The intensity of the energy coming from the radiator is a function only of temperature. A good example of this temperature dependence is a flame. The flame starts out with a low frequency emitting red light in the visible range, as the temperature increases the flame turns white and then blue as is moves across the visible spectrum with an increasing temperature. Also, with each temperature corresponds a new maximum radiance which can be emitted. As the temperature increases, the total radiation emitted also increases due to an increase in the area under the curve.
Electromagnetic spectrum:click for citation
Lord Rayleigh and J. H. Jeans developed an equation which explained blackbody radiation at low frequencies.The equation which seemed to express blackbody radiation was built upon all the known assumptions of physics at the time. The big assumption which Rayleigh and Jean implied was that infinitesimal amounts of energy were continuously added to the system when the frequency was increased. Classical physics assumed that energy emitted by atomic oscillations could have any continuous value. This was true for anything that had been studied up until that point, including things like acceleration, position, or energy.They came up with Rayleigh-Jeans law and the equation they derived was
$d\rho \left( \nu ,T \right) = \rho_{\nu} \left( T \right) d \nu = \frac{8 \pi k_B T}{c^3} \nu^2 d\nu$
Experimental data performed on the black box showed slightly different results than what was expected by the Rayleigh-Jeans law. The law had been studied and widely accepted by many physicists of the day, but the experimental results did not lie, something was different between what was theorized and what actually happens.The experimental results showed a bell type of curve but according to the Rayleigh-Jeans law the frequency diverged as it neared the ultraviolet region.This inconsistency was termed the ultraviolet catastrophe.
Ultraviolet Catastrophe
During the 19th century much attention was given to the study of heat properties of various objects. An idealised model that was considered was the Black Body, an object which absorbs all incident radiation and then re-emits all this energy again. We can think of the radiating energy as standing waves inside our blackbody cavity. The energy of the radiating waves at a given frequency ν, should be proportional to the number of modes at this frequency. Classical physics states that all these modes have the same energy kT (a result derived from classical thermodynamics) and as the number of modes is proportional to $\nu^2$:
$E \propto \nu^2 kT \label{1.1.1}$
This implies that we would expect most of the energy at higher frequency, and this energy diverges with frequency. If we try and sum the energies at each frequency we find that there is an infinite energy in ths system! This paradox was called the ULTRAVIOLET CATASTROPHE.
It was left to Planck to resolve this gaping paradox, but postulated that the energy of the modes could only come in discrete packets – quanta – of energy:
$E = h\nu, 2h\nu, 3h\nu, \ldots \qquad \Delta E = h\nu \label{1.1.2}$
Using statistical mechanics Planck found that the modes at higher frequency were less likely excited so the average energy of these modes would decrease with the frequency. The exact expression for the average energy of each mode is given by the Planck distribution:
$\langle E \rangle = \frac{h\nu}{\exp(\frac{h\nu}{KT}) - 1} \label{1.1.3}$
You can see that if the frequency is low then the average energy tends towards the classical result, and as frequency goes to infinity we get that the average energy goes to zero as expected.
Max Planck was the first person to properly explain this experimental data. Rayleigh and Jean made the assumption that energy is continuous, but Planck took a slightly different approach. He said energy must come in certain unit intervals instead of being any random unit or number. He instead “quantized” energy in the form of $E= nh\nu$ where $n$ is an integer, $h$ is a constant, and $\nu$ is the frequency. This assumption proved to be the missing piece of the puzzle and Planck derived an expression which could explain the experimental data.
$d\rho \left( \nu ,T \right) = \rho_{\nu} \left( T \right) d\nu = \dfrac{8 \pi k_B T}{c^3} \dfrac{nu^2}{e^{hv/K_bT}-1} d\nu$
This now famous equation is known as the Planck Distribution Law for Blackbody Radiation.The h in this equation is the famous Planck’s constant, which has a value of $6.626 \times 10^-34\; J \;s$.
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Atomic and Subatomic Particles
The notion that the building blocks of matter are invisibly tiny particles called atoms is usually traced back to the Greek philosophers Leucippus of Miletus and Democritus of Abdera in the 5th Century BC. The English chemist John Dalton developed the atomic philosophy of the Greeks into a true scientific theory in the early years of the 19th Century. His treatise New System of Chemical Philosophy gave cogent phenomenological evidence for the existence of atoms and applied the atomic theory to chemistry, providing a physical picture of how elements combine to form compounds consistent with the laws of definite and multiple proportions. Table $1$ summarizes some very early measurements (by Sir Humphrey Davy) on the relative proportions of nitrogen and oxygen in three gaseous compounds.
Compound Percent N Percent O Ratio
Table $1$: Oxides of Nitrogen
I 29.50 70.50 0.418
II 44.05 55.95 0.787
III 63.30 36.70 1.725
We would now identify these compounds as NO2, NO and N2O, respectively. We see in data such as these a confirmation of Dalton's atomic theory: that compounds consist of atoms of their constituent elements combined in small whole number ratios. The mass ratios in Table $1$ are, with modern accuracy, 0.438, 0.875 and 1.750.
After over 2000 years of speculation and reasoning from indirect evidence, it is now possible in a sense to actually see individual atoms, as shown for example in Figure $1$. The word "atom" comes from the Greek atomos, meaning literally "indivisible." It became evident in the late 19th Century, however, that the atom was not truly the ultimate particle of matter. Michael Faraday's work had suggested the electrical nature of matter and the existence of subatomic particles. This became manifest with the discovery of radioactive decay by Henri Becquerel in 1896 the emission of alpha, beta and gamma particles from atoms. In 1897, J. J. Thompson identified the electron as a universal constituent of all atoms and showed that it carried a negative electrical charge, now designated -e.
To probe the interior of the atom, Ernest Rutherford in 1911 bombarded a thin sheet of gold with a stream of positively-charged alpha particles emitted by a radioactive source. Most of the high-energy alpha particles passed right through the gold foil, but a small number were strongly detected in a way that indicated the presence a small but massive positive charge in the center of the atom (Figure $2$). Rutherford proposed the nuclear model of the atom. As we now understand it, an electrically-neutral atom of atomic number Z consists of a nucleus of positive charge +Ze, containing almost the entire the mass of the atom, surrounded by Z electrons of very small mass, each carrying a charge -e. The simplest atom is hydrogen, with Z = 1, consisting of a single electron outside a single proton of charge +e.
With the discovery of the neutron by Chadwick in 1932, the structure of the atomic nucleus was clarified. A nucleus of atomic number Z and mass number A was composed of Z protons and A-Z neutrons. Nuclei diameters are of the order of several times 10-15m. From the perspective of an atom, which is 105 times larger, a nucleus behaves, for most purposes, like a point charge +Ze.
During the 1960's, compelling evidence began to emerge that protons and neutrons themselves had composite structures, with major contributions by Murray Gell-Mann. According to the currently accepted "Standard Model," the protons and neutron are each made of three quarks, with compositions uud and udd, respectively. The up quark u has a charge of $+ \frac{2}{3}e$, while the down quark d has a charge of $-\frac{1}{3}e$. Despite heroic experimental efforts, individual quarks have never been isolated, evidently placing them in the same category with magnetic monopoles. By contrast, the electron maintains its status as an indivisible elementary particle.
Electromagnetic Waves
Perhaps the greatest achievement of physics in the 19th century was James Clerk Maxwell's unification in 1864 of the phenomena of electricity, magnetism and optics. An (optional) summary of Maxwell's equations is given in Supplement 1A. Heinrich Hertz in 1887 was the first to demonstrate experimentally the production and detection of the electromagnetic waves predicted by Maxwell-specifically radio waves-by acceleration of electrical charges. As shown in Figure $3$, electromagnetic waves consist of mutually perpendicular electric and magnetic fields, E and B respectively, oscillating in synchrony at high frequency and propagating in the direction of E x B.
The wavelength $\lambda$ is the distance between successive maxima of the electric (or magnetic) field. The frequency $\nu$ represents the number of oscillations per second observed at a fixed point in space. The reciprocal of frequency $\tau = \frac{1}{\nu}$ represents period of oscillation-the time it takes for one wavelength to pass a fixed point. The speed of propagation of the wave is therefore determined by $\lambda = c \tau$ or in more familiar form
$\lambda \nu = c \label{1}$
where $c = 2.9979 \times 10^8m/sec$, usually called the speed of light, applies to all electromagnetic waves in vacuum. Frequencies are expressed in hertz (Hz), defined as the number of oscillations per second.
Electromagnetic radiation is now known to exist in an immense range of wavelengths including gamma rays, X-rays, ultraviolet, visible light, infrared, microwaves and radio waves, as shown in Figure $4$.
Three Failures of Classical Physics
Isaac Newton's masterwork, Pincipia, published in 1687, can be considered to mark the beginning of modern physical science. Not only did Newton delineate the fundamental laws governing motion and gravitation but he established a general philosophical worldview which pervaded all scientific theories for two centuries afterwards. This system of thinking about the physical world is known as "Classical Physics." Its most notable feature is the primacy of cause and effect relationships. Given sufficient information about the present state of part of the Universe, it should be possible, at least in principle, to predict its future behavior (as well as its complete history.) This capability is known as determinism. For example, solar and lunar eclipses can be predicted centuries ahead, within an accuracy of several seconds. (But interestingly, we can't predict even a couple of days in advance if the weather will be clear enough to view the eclipse!) The other great pillar of classical physics is Maxwell's theory of electromagnetism.
The origin of quantum theory can be marked by three diverse phenomena involving electromagnetic radiation, which could not be adequately explained by the methods of classical physics. First among these was blackbody radiation, which led to the contribution of Max Planck in 1900. Next was the photoelectric effect, treated by Albert Einstein in 1905. Third was the origin of line spectra, the hero being Neils Bohr in 1913. A coherent formulation of quantum mechanics was eventually developed in 1925 and 1926, principally the work of Schrödinger, Heisenberg and Dirac. The remainder of this Chapter will describe the early contributions to the quantum theory by Planck, Einstein and Bohr.
Blackbody Radiation
It is a matter of experience that a hot object can emit radiation. A piece of metal stuck into a flame can become "red hot." At higher temperatures, its glow can be described as "white hot." Under even more extreme thermal excitation it can emit predominantly blue light (completing a very patriotic sequence of colors!). Josiah Wedgwood, the famous pottery designer, noted as far back as 1782 that different materials become red hot at the same temperature. The quantitative relation between color and temperature is described by the blackbody radiation law. A blackbody is an idealized perfect absorber and emitter of all possible wavelengths $\lambda$ of the radiation. Figure $5$ shows experimental wavelength distributions of thermal radiation at several temperatures. Consistent with our experience, the maximum in the distribution, which determines the predominant color, increases with temperature. This relation is given by Wien's displacement law, which can be expressed
$T \lambda_{max} = 2.898 \times 10^6\, nm K$
where the wavelength is expressed in nanometers (nm). At room temperature (300 K), the maximum occurs around 10$\mu m$, in the infrared region. In Figure $5$, the approximate values of $\lambda_{max}$ are 2900 nm at 1000 K, 1450 nm at 2000 K and 500n m at 5800 K, the approximate surface temperature of the Sun. The Sun's $\lambda_{max}$ is near the middle of the visible range (380-750nm) and is perceived by our eyes as white light.
The origin of blackbody radiation was a major challenge to 19th Century physics. Lord Rayleigh proposed that the electromagnetic field could be represented by a collection of oscillators of all possible frequencies. By simple geometry, the higher-frequency (lower wavelength) modes of oscillation are increasingly numerous since it it possible to fit their waves into an enclosure in a larger number of arrangements. In fact, the number of oscillators increases very rapidly as $\lambda^{-4}$. Rayleigh assumed that every oscillator contributed equally to the radiation (the equipartition principle). This agrees fairly well with experiment at low frequencies. But if ultraviolet rays and higher frequencies were really produced in increasing number, we would get roasted like marshmallows by sitting in front of a fireplace! Fortunately, this doesn't happen, and the incorrect theory is said to suffer from an "ultraviolet catastrophe."
Max Planck in 1900 derived the correct form of the blackbody radiation law by introducing a bold postulate. He proposed that energies involved in absorption and emission of electromagnetic radiation did not belong to a continuum, as implied by Maxwell's theory, but were actually made up of discrete bundles which he called "quanta." Planck's idea is traditionally regarded as marking the birth of the quantum theory. A quantum associated with radiation of frequency $\nu$ has the energy
$E = h \nu \label2$
where the proportionality factor h = 6.626 x 10-34 J sec is known as Planck's constant. For our development of the quantum theory of atoms and molecules, we need only this simple result and do not have to follow the remainder of Planck's derivation. If you insist, however, the details are given in Supplement 1B.
The Photoelectric Effect
A familiar device in modern technology is the photocell or "electric eye," which runs a variety of useful gadgets, including automatic door openers. The principle involved in these devices is the photoelectric effect, which was first observed by Heinrich Hertz in the same laboratory in which he discovered electromagnetic waves. Visible or ultraviolet radiation impinging on clean metal surfaces can cause electrons to be ejected from the metal. Such an effect is not, in itself, inconsistent with classical theory since electromagnetic waves are known to carry energy and momentum. But the detailed behavior as a function of radiation frequency and intensity can not be explained classically.
The energy required to eject an electron from a metal is determined by its work function $\Phi$. For example, sodium has $\Phi = 1.82 eV$. The electron-volt is a convenient unit of energy on the atomic scale: 1 eV = 1.602 x 10-19J. This corresponds to the energy which an electron picks up when accelerated across a potential difference of 1 volt. The classical expectation would be that radiation of sufficient intensity should cause ejection of electrons from a metal surface, with their kinetic energies increasing with the radiation intensity. Moreover, a time delay would be expected between the absorption of radiation and the ejection of electrons. The experimental facts are quite different. It is found that no electrons are ejected, no matter how high the radiation intensity, unless the radiation frequency exceeds some threshold value $\nu_{0}$ for each metal. For sodium $\nu_{0}$ = 4.39 x 1014 Hz (corresponding to a wavelength of 683 nm), as shown in Figure $6$. For frequencies $\nu$ above the threshold, the ejected electrons acquire a kinetic energy given by
$\frac{1}{2}mv^{2} =h( \nu - \nu_{0}) =h \nu - \Phi \label3$
Evidently, the work function $\Phi$ can be identified with $h \nu_{0}$, equal to 3.65 x 10-19J=1.82 eV for sodium. The kinetic energy increases linearly with frequency above the threshold but is independent of the radiation intensity. Increased intensity does, however, increase the number of photoelectrons.
Einstein's explanation of the photoelectric effect in 1905 appears trivially simple once stated. He accepted Planck's hypothesis that a quantum of radiation carries an energy $h \nu$. Thus, if an electron is bound in a metal with an energy $\Phi$, a quantum of energy $h \nu_{0}$ = $\Phi$ will be sufficient to dislodge it. And any excess energy $h( \nu - \nu_{0})$ will appear as kinetic energy of the ejected electron. Einstein believed that the radiation field actually did consist of quantized particles, which he named photons. Although Planck himself never believed that quanta were real, Einstein's success with the photoelectric effect greatly advanced the concept of energy quantization.
Line Spectra
Most of what is known about atomic (and molecular) structure and mechanics has been deduced from spectroscopy. Figure $7$ 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. Frauenhofer between 1814 and 1823 discovered nearly 600 dark lines in the solar spectrum viewed at high resolution. It is now understood that these lines are caused by absorption by the outer layers of the Sun.
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 $8$.
It is consistent with classical electromagnetic theory that motions of electrical charges within atoms can be associated with the absorption and emission of radiation. What is completely mysterious is how such radiation can occur for discrete frequencies, rather than as a continuum. The breakthrough that explained line spectra is credited to Neils Bohr in 1913. Building on the ideas of Planck and Einstein, Bohr postulated that the energy levels of atoms belong to a discrete set of values $E_{n}$, rather than a continuum as in classical mechanics. When an atom makes a downward energy transition from a higher energy level $E_{m}$ to a lower energy level $E_{n}$, it caused the emission of a photon of energy
$h \nu =E_{m} - E_{n} \label{4}$
This is what accounts for the discrete values of frequency $\nu$ in emission spectra of atoms. Absorption spectra are correspondingly associated with the annihilation of a photon of the same energy and concomitant excitation of the atom from $E_{n}$ to $E_{m}$. Figure $9$ is a schematic representation of the processes of absorption and emission of photons by atoms. Absorption and emission processes occur at the same set frequencies, as is shown by the two line spectra in Figure $7$.
Rydberg (1890) found that all the lines of the atomic hydrogen spectrum could be fitted to a simple empirical formula
$\dfrac{1}{ \lambda} =R\left( \dfrac{1}{n_1^2} -\dfrac{1}{n_2^2}\right), n = 1,2,3..., n_2>n_1 \label{5}$
where R, known as the Rydberg constant, has the value 109,677 cm-1. This formula was found to be valid for hydrogen spectral lines in the infrared and ultraviolet regions, in addition to the four lines in the visible region. No analogously simple formula has been found for any atom other than hydrogen. Bohr proposed a model for the energy levels of a hydrogen atom which agreed with Rydberg's formula for radiative transition frequencies. Inspired by Rutherford's nuclear atom, Bohr suggested a planetary model for the hydrogen atom in which the electron goes around the proton in one of a set of allowed circular orbits, as shown in Fig 8. A more fundamental understanding of the discrete nature of orbits and energy levels had to await the discoveries of 1925-26, but Bohr's model provided an invaluable stepping-stone to the development of quantum mechanics. We will consider the hydrogen atom in greater detail in Chap. 7.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/01._Waves_and_Particles/Chapter_1%3A_Atoms_and_Photons%3A_Origin_of_Quantum_Theory.txt
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Quantum mechanics is the theoretical framework which describes the behavior of matter on the atomic scale. It is the most successful quantitative theory in the history of science, having withstood thousands of experimental tests without a single verifiable exception. It has correctly predicted or explained phenomena in fields as diverse as chemistry, elementary-particle physics, solid-state electronics, molecular biology and cosmology. A host of modern technological marvels, including transistors, lasers, computers and nuclear reactors are offspring of the quantum theory. Possibly 30% of the US gross national product involves technology which is based on quantum mechanics. For all its relevance, the quantum world differs quite dramatically from the world of everyday experience. To understand the modern theory of matter, conceptual hurdles of both psychological and mathematical variety must be overcome. A paradox which stimulated the early development of the quantum theory concerned the indeterminate nature of light. Light usually behaves as a wave phenomenon but occasionally it betrays a particle-like aspect, a schizoid tendency known as the wave-particle duality. We consider first the wave theory of light.
The Double-Slit Experiment
Figure $1$ shows a modernized version of the famous double-slit diffraction experiment first performed by Thomas Young in 1801. Light from a monochromatic (single wavelength) source passes through two narrow slits and is projected onto a screen. Each slit by itself would allow just a narrow band of light to illuminate the screen. But with both slits open, a beautiful interference pattern of alternating light and dark bands appears, with maximum intensity in the center. To understand what is happening, we review some key results about electromagnetic waves.
Maxwell's theory of electromagnetism was an elegant unification of the diverse phenomena of electricity, magnetism and radiation, including light. Electromagnetic radiation is carried by transverse waves of electric and magnetic fields, propagating in vacuum at a speed $c≈3\times 10^8m/sec$, known as the "speed of light." As shown in Figure 2, the E and B fields oscillate sinusoidally, in synchrony with one another. The magnitudes of E and B are proportional ($B=E/c$ in SI units). The distance between successive maxima (or minima) at a given instant of time is called the wavelength $\lambda$. At every point in space, the fields also oscillate sinusoidally as functions of time. The number of oscillations per unit time is called the frequency $\nu$. Since the field moves one wavelength in the time $\lambda/c$, the wavelength, frequency and speed for any wave phenomenon are related by $\lambda\nu=c\label{1}$
The energy density contained in an electromagnetic field, even a static one, is given by
$\rho=\dfrac{1}{2}\left(\epsilon_{0}E^{2}+\dfrac{B^{2}}{\mu_{0}}\right)\label{3}$
Note that both of the above energy quantities depend quadratically on the fields E and B. To discuss the diffraction experiments described above, it is useful to define the amplitude of an electromagnetic wave at each point in space and time r, t by the function
$\Psi\left(\vec{r},t\right)=\sqrt{\epsilon_{0}}E\left(\vec{r},t\right)=\dfrac{B\left(\vec{r},t\right)}{\sqrt{\mu_{0}}}\label{4}$
such that the intensity is given by
$\rho\left(\vec{r},t\right)=[\Psi\left(\vec{r},t\right)]^2\label{5}$
The function $\Psi(\vec{r},t)$ will, in some later applications, have complex values. In such cases we generalize the definition of intensity to
$\rho\left(\vec{r},t\right)=|\Psi\left(\vec{r},t\right)|^2=\Psi\left(\vec{r},t\right)^\ast\Psi\left(\vec{r},t\right)\label{6}$
where $\Psi\left(\vec{r},t\right)^\ast$ represents the complex conjugate of $\Psi\left(\vec{r},t\right)$. In quantum mechanical applications, the function $\Psi$ is known as the wavefunction.
Figure $1$: Interference of two equal sinusoidal waves. Top: constructive interference. Bottom: destructive interference. Center: intermediate case. The resulting intensities $\rho=\Psi^2$ is shown on the right.
The electric and magnetic fields, hence the amplitude $\Psi$, can have either positive and negative values at different points in space. In fact constructive and destructive interference arises from the superposition of waves, as illustrated in Figure $3$. By Equation $\ref{5}$, the intensity $\rho\ge0$ everywhere. The light and dark bands on the screen are explained by constructive and destructive interference, respectively. The wavelike nature of light is convincingly demonstrated by the fact that the intensity with both slits open is not the sum of the individual intensities, ie, $\rho\neq\rho_{1}+\rho_{2}$. Rather it is the wave amplitudes which add:
$\Psi=\Psi_{1}+\Psi_{2}\label{7}$
with the intensity given by the square of the amplitude:
$\rho=\Psi^2=\Psi_{1}^2+\Psi_{2}^2+2\Psi_{1}\Psi_{2}\label{8}$
The cross term $2\Psi_{1}\Psi_{2}$ is responsible for the constructive and destructive interference. Where $\Psi_{1}$ and $\Psi_{2}$ have the same sign, constructive interference makes the total intensity greater than the the sum of $\rho_{1}$ and $\rho_{2}$. Where $\Psi_{1}$ and $\Psi_{2}$ have opposite signs, there is destructive interference. If, in fact, $\Psi_{1}$ = $-\Psi_{2}$ then the two waves cancel exactly, giving a dark fringe on the screen.
Wave-Particle Duality
The interference phenomena demonstrated by the work of Young, Fresnel and others in the early 19th Century, apparently settled the matter that light was a wave phenomenon, contrary to the views of Newton a century earlier--case closed! But nearly a century later, phenomena were discovered which could not be satisfactorily accounted for by the wave theory, specifically blackbody radiation and the photoelectric effect.
Deviating from the historical development, we will illustrate these effects by a modification of the double slit experiment. Let us equip the laser source with a dimmer switch capable of reducing the light intensity by several orders of magnitude, as shown in Figure $4$. With each successive filter the diffraction pattern becomes dimmer and dimmer. Eventually we will begin to see localized scintillations at random positions on an otherwise dark screen. It is an almost inescapable conclusion that these scintillations are caused by photons, the bundles of light postulated by Planck and Einstein to explain blackbody radiation and the photoelectric effect.
But wonders do not cease even here. Even though the individual scintillations appear at random positions on the screen, their statistical behavior reproduces the original high-intensity diffraction pattern. Evidently the statistical behavior of the photons follows a predictable pattern, even though the behavior of individual photons is unpredictable. This implies that each individual photon, even though it behaves mostly like a particle, somehow carry with it a "knowledge" of the entire wavelike diffraction pattern. In some sense, a single photon must be able to go through both slits at the same time. This is what is known as the wave-particle duality for light: under appropriate circumstances light can behave as a wave or as a particle.
Planck's resolution of the problem of blackbody radiation and Einstein's explanation of the photoelectric effect can be summarized by a relation between the energy of a photon to its frequency:
$E=h \nu \label{8b}$
where $h = 6.626\times 10^{-34} J sec$, known as Planck's constant. Much later, the Compton effect was discovered, wherein an x-ray or gamma ray photon ejects an electron from an atom, as shown in Figure $5$. Assuming conservation of momentum in a photon-electron collision, the photon is found to carry a momentum p, given by
$p=\dfrac{h}{\lambda} \label{9}$
Equation $\ref{8b}$ and $\ref{9}$ constitute quantitative realizations of the wave-particle duality, each relating a particle-like property--energy or momentum--to a wavelike property--frequency or wavelength.
According to the special theory of relativity, the last two formulas are actually different facets of the same fundamental relationship. By Einstein's famous formula, the equivalence of mass and energy is given by
$E=mc^2\label{10}$
The photon's rest mass is zero, but in travelling at speed c, it acquires a finite mass. Equating Equation $\ref{8}$) and $\ref{10}$ for the photon energy and taking the photon momentum to be $p = mc$, we obtain
$p = \dfrac{E}{c} = \dfrac{h\nu}{c} = \dfrac{h}{\lambda} \label{11}$
Thus, the wavelength-frequency relation (Equation $\ref{1}$), implies the Compton-effect formula (Equation $\ref{9}$).
The best we can do is to describe the phenomena constituting the wave-particle duality. There is no widely accepted explanation in terms of everyday experience and common sense. Feynman referred to the "experiment with two holes" as the "central mystery of quantum mechanics." It should be mentioned that a number of models have been proposed over the years to rationalize these quantum mysteries. Bohm proposed that there might exist hidden variables which would make the behavior of each photon deterministic, ie, particle-like. Everett and Wheeler proposed the "many worlds interpretation of quantum mechanics" in which each random event causes the splitting of the entire universe into disconnected parallel universes in which each possibility becomes the reality. Needless to say, not many people are willing to accept such a metaphysically unwieldy view of reality. Most scientists are content to apply the highly successful computational mechanisms of quantum theory to their work, without worrying unduly about its philosophical underpinnings. Sort of like people who enjoy eating roast beef but would rather not think about where it comes from.
There was never any drawn-out controversy about whether electrons or any other constituents of matter were other than particle-like. Yet a variant of the double-slit experiment using electrons instead of light proves otherwise. The experiment is technically difficult but has been done. An electron gun, instead of a light source, produces a beam of electrons at a selected velocity, which is focused and guided by electric and magnetic fields. Then, everything that happens for photons has its analog for electrons. Individual electrons produce scintillations on a phosphor screen-this is how TV works. But electrons also exhibit diffraction effects, which indicates that they too have wavelike attributes. Diffraction experiments have been more recently carried out for particles as large as atoms and molecules, even for the C60 fullerene molecule.
De Broglie in 1924 first conjectured that matter might also exhibit a wave-particle duality. A wavelike aspect of the electron might, for example, be responsible for the discrete nature of Bohr orbits in the hydrogen atom. According to de Broglie's hypothesis, the "matter waves" associated with a particle have a wavelength given by
$\lambda=h/p\label{12}$
which is identical in form to Compton's result (Equation \ref{9}) (which, in fact, was discovered later). The correctness of de Broglie's conjecture was most dramatically confirmed by the observations of Davisson and Germer in 1927 of diffraction of monoenergetic beams of electrons by metal crystals, much like the diffraction of x-rays. And measurements showed that de Broglie's formula (Equation $\ref{12}$) did indeed give the correct wavelength (see Figure $6$).
The Schrödinger Equation
Schrödinger in 1926 first proposed an equation for de Broglie's matter waves. This equation cannot be derived from some other principle since it constitutes a fundamental law of nature. Its correctness can be judged only by its subsequent agreement with observed phenomena (a posteriori proof). Nonetheless, we will attempt a heuristic argument to make the result at least plausible. In classical electromagnetic theory, it follows from Maxwell's equations that each component of the electric and magnetic fields in vacuum is a solution of the wave equation
$\nabla^2\Psi-\dfrac{1}{c^2}\dfrac{\partial ^2\Psi}{\partial t^2}=0\label{13}$
where the Laplacian or "del-squared" operator is defined by
$\nabla^2=\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}+\dfrac{\partial^2}{\partial z^2}\label{14}$
We will attempt now to create an analogous equation for de Broglie's matter waves. Accordingly, let us consider a very general instance of wave motion propagating in the x-direction. At a given instant of time, the form of a wave might be represented by a function such as
$\psi(x)=f\left(\dfrac {2\pi x}{ \lambda}\right)\label{15}$
where $f(\theta)$ represents a sinusoidal function such as $\sin\theta$, $\cos\theta$, $e^{i\theta}$, $e^{-i\theta}$ or some linear combination of these. The most suggestive form will turn out to be the complex exponential, which is related to the sine and cosine by Euler's formula
$e^{i\theta}=\cos\theta + i \sin\theta\label{16}$
Each of the above is a periodic function, its value repeating every time its argument increases by $2\pi$. This happens whenever x increases by one wavelength $\lambda$. At a fixed point in space, the time-dependence of the wave has an analogous structure:
$T(t)=f(2\pi\nu t)\label{17}$
where $\nu$ gives the number of cycles of the wave per unit time. Taking into account both x- and t-dependence, we consider a wavefunction of the form
$\Psi(x,t)=exp\left[2\pi i\left(\dfrac{x}{\lambda}-\nu t\right)\right]\label{18}$
representing waves travelling from left to right. Now we make use of the Planck and de Broglie formulas (Equation $\ref{8}$ and $\ref{12}$, respectively) to replace $\nu$ and $\lambda$ by their particle analogs. This gives
$\Psi(x,t)=exp[i(px-Et)/\hbar]\label{19}$
where
$\hbar\equiv\dfrac{h}{2\pi}\label{20}$
Since Planck's constant occurs in most formulas with the denominator $2\pi$, this symbol was introduced by Dirac. Now Equation $\ref{17}$ represents in some way the wavelike nature of a particle with energy E and momentum p. The time derivative of Equation \ref{19} gives
$\dfrac{\partial\Psi}{\partial t} = -(iE/\hbar)\times \exp \left[\dfrac{i(px-Et)}{\hbar} \right]\label{21}$
Thus
$i\hbar\dfrac{\partial\Psi}{\partial t} = E\Psi\label{22}$
Analogously
$-i\hbar\dfrac{\partial\Psi}{\partial x} = p\Psi\label{23}$
and
$-\hbar^2\dfrac{\partial^2\Psi}{\partial x^2} = p^2\Psi\label{24}$
The energy and momentum for a nonrelativistic free particle are related by
$E=\dfrac{1}{2}mv^2=\dfrac{p^2}{2m}\label{25}$
Thus $\Psi(x,t)$ satisfies the partial differential equation
$i\hbar\dfrac{\partial\Psi}{\partial t}=-\dfrac{\hbar^2}{2m}\dfrac{\partial^2\Psi}{\partial x^2}\label{26}$
For a particle with a potential energy $V(x)$,
$E=\dfrac{p^2}{2m}+V(x)\label{27}$
we postulate that the equation for matter waves generalizes to
$i\hbar\dfrac{\partial\Psi}{\partial t}=\left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}+V(x)\right]\Psi\label{28}$
For waves in three dimensions should then have
$i\hbar\dfrac{\partial}{\partial t}\Psi(\vec{r},t)=\left[-\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\right]\Psi(\vec{r},t)\label{29}$
Here the potential energy and the wavefunction depend on the three space coordinates x, y, z, which we write for brevity as r. This is the time-dependent Schrödinger equation for the amplitude $\Psi(\vec{r}, t)$ of the matter waves associated with the particle. Its formulation in 1926 represents the starting point of modern quantum mechanics. (Heisenberg in 1925 proposed another version known as matrix mechanics.)
For conservative systems, in which the energy is a constant, we can separate out the time-dependent factor from (19) and write
$\Psi(\vec{r},t)=\psi(\vec{r})e^{-iEt\div\hbar}\label{30}$
where $\psi(\vec{r})$ is a wavefunction dependent only on space coordinates. Putting Equation \ref{30} into Equation \ref{29} and cancelling the exponential factors, we obtain the time-independent Schrödinger equation:
$\left[-\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\right]\psi(\vec{r})=E\psi(\vec{r})\label{31}$
Most of our applications of quantum mechanics to chemistry will be based on this equation.
The bracketed object in Equation $\ref{31}$ is called an operator. An operator is a generalization of the concept of a function. Whereas a function is a rule for turning one number into another, an operator is a rule for turning one function into another. The Laplacian ($\nabla$) is an example of an operator. We usually indicate that an object is an operator by placing a hat' over it, eg, $\hat{A}$. The action of an operator that turns the function f into the function g is represented by
$\hat{A}f=g\label{32}$
Equation $\ref{23}$ implies that the operator for the x-component of momentum can be written
$\hat{p}_{x}=-i\hbar\dfrac{\partial}{\partial x}\label{33}$
and by analogy, we must have
$\hat{p}_{y}=-i\hbar\dfrac{\partial}{\partial y}$
and
$\hat{p}_{z}=-i\hbar\dfrac{\partial}{\partial z}\label{34}$
The energy, as in Equation $\ref{27}$, expressed as a function of position and momentum is known in classical mechanics as the Hamiltonian. Generalizing to three dimensions,
$\hat{H}=\dfrac{p^2}{2m}+V(\vec{r})=\dfrac{1}{2m}(p_{x}^2+p_{y}^2+p_{z}^2)+V(x,y,z)\label{35}$
We construct thus the corresponding quantum-mechanical operator
$\hat{H}=-\dfrac{\hbar^2}{2m}\left(\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}+\dfrac{\partial^2}{\partial z^2}\right)+V(x,y,z)=-\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\label{36}$
The time-independent Schrödinger equation (Equation $\ref{31}$) can then be written symbolically as
$\hat{H}\Psi=E\Psi\label{37}$
This form is actually more generally to any quantum-mechanical problem, given the appropriate Hamiltonian and wavefunction. Most applications to chemistry involve systems containing many particles--electrons and nuclei. An operator equation of the form
$\hat{A}\psi=const \psi\label{38}$
is called an eigenvalue equation. Recall that, in general, an operator acting on a function gives another function (e.g., Equation $\ref{32}$). The special case (Equation $\ref{38}$) occurs when the second function is a multiple of the first. In this case, $\psi$ is known as an eigenfunction and the constant is called an eigenvalue. (These terms are hybrids with German, the purely English equivalents being characteristic function' and `characteristic value.') To every dynamical variable $A$ in quantum mechanics, there corresponds an eigenvalue equation, usually written
$\hat{A}\psi=a\psi \label{39}$
The eigenvalues a represent the possible measured values of the variable $A$. The Schrödinger Equation ($\ref{37}$) is the best known instance of an eigenvalue equation, with its eigenvalues corresponding to the allowed energy levels of the quantum system.
The Wavefunction
For a single-particle system, the wavefunction $\Psi(\vec{r},t)$, or $\psi(\vec{r})$ for the time-independent case, represents the amplitude of the still vaguely defined matter waves. The relationship between amplitude and intensity of electromagnetic waves we developed for Equation $\ref{6}$ can be extended to matter waves.
The most commonly accepted interpretation of the wavefunction is due to Max Born (1926), according to which $\rho(r)$, the square of the absolute value of $\psi(r)$ is proportional to the probability density (probability per unit volume) that the particle will be found at the position r. Probability density is the three-dimensional analog of the diffraction pattern that appears on the two-dimensional screen in the double-slit diffraction experiment for electrons described in the preceding Section. In the latter case we had the relative probability a scintillation would appear at a given point on the screen. The function $\rho(r)$ becomes equal, rather than just proportional to, the probability density when the wavefunction is normalized, that is,
$\int|\psi(\vec{r})|^2d\tau=1\label{40}$
This simply accounts for the fact that the total probability of finding the particle somewhere adds up to unity. The integration in Equation $\ref{40}$ extends over all space and the symbol $d\tau$ designates the appropriate volume element. For example, the volume differential in Cartesian coordinates, $d\tau=dx\,dy\,dz$ is changed in spherical coordinates to $d\tau=r^2sin\theta\, dr \,d\theta \, d\phi$. The physical significance of the wavefunctions makes certain demands on its mathematical behavior. The wavefunction must be a single-valued function of all its coordinates, since the probability density ought to be uniquely determined at each point in space.
Moreover, the wavefunction should be finite and continuous everywhere, since a physically-meaningful probability density must have the same attributes. The conditions that the wavefunction be single-valued, finite and continuous--in short, "well behaved"-- lead to restrictions on solutions of the Schrödinger equation such that only certain values of the energy and other dynamical variables are allowed. This is called quantization and is in the feature that gives quantum mechanics its name.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/01._Waves_and_Particles/Chapter_2%3A_Waves_and_Particles.txt
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Quantum Mechanics is the cornerstone of physical theories dealing with the most fundamental issues of nature. Its principles appear to be different from classical laws of nature
02. Fundamental Concepts of Quantum Mechanics
The Rayleigh-Jeans Radiation Law was a useful, but not completely successful attempt at establishing the functional form of the spectra of thermal radiation. The energy density $u_ν$ per unit frequency interval at a frequency $ν$ is, according to the The Rayleigh-Jeans Radiation,
$u_ν = \dfrac{8πν^2kT}{c^2} \nonumber$
where $k$ is Boltzmann's constant, $T$ is the absolute temperature of the radiating body, and $c$ is the speed of light in a vacuum.
This formula fits the empirical measurements for low frequencies, but fails increasingly for higher frequencies. The failure of the formula to match the new data was called the ultraviolet catastrophe. The significance of this inadequate so-called law is that it provides an asymptotic condition which other proposed formulas, such as Planck's, need to satisfy. It gives a value to an otherwise arbitrary constant in Planck's thermal radiation formula.
The Derivation
Consider a cube of edge length $L$ in which radiation is being reflected and re-reflected off its walls. Standing waves occur for radiation of a wavelength $λ$ only if an integral number of half-wave cycles fit into an interval in the cube. For radiation parallel to an edge of the cube this requires
$\dfrac{L}{λ/2} = m \nonumber$
where $m$ is an integer or, equivalently
$λ = \dfrac{2L}{m} \nonumber$
Between two end points there can be two standing waves, one for each polarization. In the following the matter of polarization will be ignored until the end of the analysis and there the number of waves will be doubled to take into account the matter of polarization.
Since the frequency $\nu$ is equal to $c/λ$, where $c$ is the speed of light
$ν = \dfrac{cm}{2L} \nonumber$
It is convenient to work with the quantity $q$, known as the wavenumber, which is defined as
$q = \dfrac{2π}{λ} \nonumber$
and hence
$q = \dfrac{2πν}{c} \nonumber$
In terms of the relationship for the cube,
$q = \dfrac{2πm}{2L} = π\left(\dfrac{m}{L}\right) \nonumber$
and hence
$q^2 = π^2\left(\dfrac{m}{L}\right)^2 \nonumber$
Another convenient term is the radian frequency $ω=2πν$. From this it follows that $q=ω/c$.
If $m_X$, $m_Y$, $m_Z$ denote the integers for the three different directions in the cube then the condition for a standing wave in the cube is that
$q^2 = π^2\left[ \left(\dfrac{m_X}{L}\right)^2 + \left(\dfrac{m_Y}{L}\right)^2 + \left(\dfrac{m_Z}{L}\right)^2\right] \nonumber$
which reduces to
$m_X^2 + m_Y^2 + m_Z^2 = \dfrac{4L^2ν^2}{c^2} \nonumber$
Now the problem is to find the number of nonnegative combinations of ($m_X$, $m_Y$, $m_Z$) that fit between a sphere of radius $R$ and and one of radius $R+dR$. First the number of combinations ignoring the nonnegativity requirement can be determined.
The volume of a spherical shell of inner radius $R$ and outer radius $R+dR$ is given by
$dV = 4πR^2\,dR \nonumber$
If
$R = \sqrt{m_X^2+m_Y^2+m_Z^2} \nonumber$
then
$R = \sqrt{\dfrac{4L^2ν^2}{c^2}} = \dfrac{2Lν}{c} \nonumber$
and hence
$dR=\dfrac{2L\,dν}{c}. \nonumber$
This means that
\begin{align*} dV &= 4π\left(\dfrac{2Lν}{c}\right)^2 \left(\dfrac{2L}{c}\right)\,dν \[4pt] &= 32π \left(\dfrac{L^3ν^2}{c^3}\right) dν \end{align*}
Now the nonnegativity require for the combinations ($m_X$, $m_Y$, $m_Z$) must be taken into account. For the two dimensional case the nonnegative combinations are approximately those in one quadrant of circle. The approximation arises from the matter of the combinations on the boundaries of the nonnegative quadrant. For the three dimensional case the nonnegative combinations constitute approximately one octant of the total. Thus the number $dN$ for the nonnegative combinations of ($m_X$, $m_Y$, $m_Z$) in this volume is equal to $(1/8)dV$ and hence
$dN = 4πν^2\left(\dfrac{L^3}{c^3}\right)\,dν \nonumber$
The average kinetic energy per degree of freedom is $½kT$, where $k$ is Boltzmann's constant. For harmonic oscillators there is an equality between kinetic and potential energy so the average energy per degree of freedom is $kT$. This means that the average radiation energy $E$ per unit frequency is given by
$\dfrac{dE}{dν} = kT\left(\dfrac{dN}{dν}\right) = 4πkT\left(\dfrac{L^3}{c^3}\right)ν^2 \nonumber$
and the average energy density, $u_ν$, is given by
$\dfrac{du_ν}{dν} = \left(\dfrac{1}{L^3}\right)\left(\dfrac{dE}{dν}\right) = \dfrac{4πkTν^2}{c^3} \nonumber$
The previous only considered one direction of polarization for the radiation. If the two directions of polarization are taken into account a factor of 2 must be included in the above formula; i.e.,
$\dfrac{du_ν}{dν} = \dfrac{8πkTν^2}{c^3} \nonumber$
This is the Raleigh-Jeans Law of Radiation and holds empirically as the frequency goes to zero.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/02._Fundamental_Concepts_of_Quantum_Mechanics/Deriving_the_Rayleigh-Jeans_Radiation_Law.txt
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Wien's displacement law states that the blackbody radiation curve for different temperatures peaks at a wavelength inversely proportional to the temperature. The shift of that peak is a direct consequence of the Planck radiation law which describes the spectral brightness of black body radiation as a function of wavelength at any given temperature. However it had been discovered by Wilhelm Wien several years before Max Planck developed that more general equation, and describes the entire shift of the spectrum of black body radiation toward shorter wavelengths as temperature increases.
Derive Wien's displacement law from Planck's law. Proceed as follows:
$\rho (\nu, T) = \dfrac {2 h \nu^3 }{c^3\left(e^{\frac {h\nu}{k_B T}}-1\right)} \label{Planck2}$
We need to evaluate the derivative of Equation \ref{Planck2} with respect to $\nu$ and set it equal to zero to find the peak wavelength.
$\dfrac{d}{d\nu} \left \{ \rho (\nu, T) \right \} = \dfrac{d}{d\nu} \left \{ \dfrac {2 h \nu^3 }{c^3\left(e^{\frac {h\nu}{k_B T}}-1\right)} \right \} =0 \label{eq2}$
This can be solved via the quotient rule or product rule for differentiation. Selecting the latter for convenience requires rewriting Equation \ref{eq2} as a product:
$\dfrac{d}{d\nu} \left \{ \rho (\nu, T) \right \} = \dfrac{2h}{c^3} \dfrac{d}{d\nu} \left \{ ( \nu^3) \left( e^{\frac {h\nu}{k_B T}}-1 \right)^{-1} \right \} = 0$
applying the product rule (and power rule and chain rule)
$= \dfrac{2h}{c^3} \left [ (3 \nu^2) \left( e^{\frac {h\nu}{k_B T}}-1 \right)^{-1} - ( \nu^3) \left( e^{\frac {h\nu}{k_B T}}-1 \right)^{-2} \left(\dfrac{h}{k_BT}\right) e^{\frac {h\nu}{k_B T}} \right] = 0$
so this expression is zero when
$(3 {\nu^2}) {\left( e^{\frac {h\nu}{k_B T}}-1 \right)^{-1}} = ( \nu^{3}) \left( e^{\frac {h\nu}{k_B T}}-1 \right)^{-2} \left(\dfrac{h}{k_BT}\right) e^{\frac {h\nu}{k_B T}}$
or when simplified
$3 \left( e^{\frac {h\nu}{k_B T}}-1 \right) - \left(\dfrac{hv}{k_BT}\right) e^{\frac {h\nu}{k_B T}} =0 \label{eq10}$
We can do a substitution $u=\frac {h\nu}{k_B T}$ and Equation \ref{eq10} becomes
$3 (e^u - 1) - u e^u = 0$
Finding the solutions to this equation requires using Lambert's W-functions and results numerically in
$u = 3 +W(-3e^{-3}) \approx 2.8214$
so unsubstituting the $u$ variable
$u = \dfrac {h\nu}{k_B T} \approx 2.8214 \label{eq20}$
or
\begin{align} \nu &\approx \dfrac{2.8214\, k_B}{h} T \[4pt] &\approx \dfrac{(2.8214 )(1.38 \times 10^{-23} J/K) }{6.63 \times 10^{-34} J\,s} T \[4pt] &\approx (5.8 \times 10^{10} Hz/K)\, T \end{align}
The consequence is that the shape of the blackbody radiation function would shift proportionally in frequency with temperature. When Max Planck later formulated the correct blackbody radiation function it did not include Wien's constant explicitly. Rather, Planck's constant h was created and introduced into his new formula. From Planck's constant h and the Boltzmann constant k, Wien's constant (Equation \ref{eq20}) can be obtained.
• Wikipedia
Deriving the de Broglie Wavelength
The de Broglie wavelength is the wavelength, $\lambda$, associated with a object and is related to its momentum and mass.
Introduction
In 1923, Louis de Broglie, a French physicist, proposed a hypothesis to explain the theory of the atomic structure. By using a series of substitution de Broglie hypothesizes particles to hold properties of waves. Within a few years, de Broglie's hypothesis was tested by scientists shooting electrons and rays of lights through slits. What scientists discovered was the electron stream acted the same was as light proving de Broglie correct.
Deriving the de Broglie Wavelength
De Broglie derived his equation using well established theories through the following series of substitutions:
De Broglie first used Einstein's famous equation relating matter and energy:
$E = mc^2 \label{0}$
with
• $E$ = energy,
• $m$ = mass,
• $c$ = speed of light
Using Planck's theory which states every quantum of a wave has a discrete amount of energy given by Planck's equation:
$E= h \nu \label{1}$
with
• $E$ = energy,
• $h$ = Plank's constant (6.62607 x 10-34 J s),
• $\nu$= frequency
Since de Broglie believed particles and wave have the same traits, he hypothesized that the two energies would be equal:
$mc^2 = h\nu \label{2}$
Because real particles do not travel at the speed of light, De Broglie submitted velocity ($v$) for the speed of light ($c$).
$mv^2 = h\nu \label{3}$
Through the equation $\lambda$, de Broglie substituted $v/\lambda$ for $\nu$ and arrived at the final expression that relates wavelength and particle with speed.
$mv^2 = \dfrac{hv}{\lambda} \label{4}$
Hence
$\lambda = \dfrac{hv}{mv^2} = \dfrac{h}{mv} \label{5}$
A majority of Wave-Particle Duality problems are simple plug and chug via Equation \ref{5} with some variation of canceling out units
Example $1$
Find the de Broglie wavelength for an electron moving at the speed of $5.0 \times 10^6\; m/s$ (mass of an electron is $9.1 \times 10^{-31}\; kg$).
Solution
$\lambda = \dfrac{h}{p}= \dfrac{h}{mv} =\dfrac{6.63 \times 10^{-34}\; J \cdot s}{(9.1 \times 10^{-31} \; kg)(5.0 \times 10^6\, m/s)}= 1.46 \times 10^{-10}\;m$
Although de Broglie was credited for his hypothesis, he had no actual experimental evidence for his conjecture. In 1927, Clinton J. Davisson and Lester H. Germer shot electron particles onto onto a nickel crystal. What they saw was the diffraction of the electron similar to waves diffraction against crystals (x-rays). In the same year, an English physicist, George P. Thomson fired electrons towards thin metal foil providing him with the same results as Davisson and Germer.
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Heisenberg’s Uncertainty Principle is one of the most celebrated results of quantum mechanics and states that one (often, but not always) cannot know all things about a particle (as it is defined by it’s wave function) at the same time. This principle is mathematically manifested as non-commuting operators.
Introduction
Heisenberg's Uncertainty Principle states that there is inherent uncertainty in the act of measuring a variable of a particle. Commonly applied to the position and momentum of a particle, the principle states that the more precisely the position is known the more uncertain the momentum is and vice versa. This is contrary to classical Newtonian physics which holds all variables of particles to be measurable to an arbitrary uncertainty given good enough equipment. 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.
More specifically, if one knows the precise momentum of the particle, it is impossible to know the precise position, and vice versa. This relationship also applies to energy and time, in that one cannot measure the precise energy of a system in a finite amount of time. Uncertainties in the products of “conjugate pairs” (momentum/position) and (energy/time) were defined by Heisenberg as having a minimum value corresponding to Planck’s constant divided by $4\pi$. More clearly:
$\Delta{p}\Delta{x} \ge \dfrac{h}{4\pi}$
$\Delta{t}\Delta{E} \ge \dfrac{h}{4\pi}$
Where $\Delta$ refers to the uncertainty in that variable and h is Planck's constant.
Aside from the mathematical definitions, one can make sense of this by imagining that the more carefully one tries to measure position, the more disruption there is to the system, resulting in changes in momentum. For example compare the effect that measuring the position has on the momentum of an electron versus a tennis ball. Let’s say to measure these objects, light is required in the form of photon particles. These photon particles have a measurable mass and velocity, and come into contact with the electron and tennis ball in order to achieve a value in their position. As two objects collide with their respective momenta (p=m*v), they impart theses momenta onto each other. When the photon contacts the electron, a portion of its momentum is transferred and the electron will now move relative to this value depending on the ratio of their mass. The larger tennis ball when measured will have a transfer of momentum from the photons as well, but the effect will be lessened because its mass is several orders of magnitude larger than the photon. To give a more practical description, picture a tank and a bicycle colliding with one another, the tank portraying the tennis ball and the bicycle that of the photon. The sheer mass of the tank although it may be traveling at a much slower speed will increase its momentum much higher than that of the bicycle in effect forcing the bicycle in the opposite direction. The final result of measuring an object’s position leads to a change in its momentum and vice versa.
All Quantum behavior follows this principle and it is important in determining spectral line widths, as the uncertainty in energy of a system corresponds to a line width seen in regions of the light spectrum explored in Spectroscopy.
What does it mean?
It is hard to imagine not being able to know exactly where a particle is at a given moment. It seems intuitive that if a particle exists in space, then we can point to where it is; however, the Heisenberg Uncertainty Principle clearly shows otherwise. This is because of the wave-like nature of a particle. A particle is spread out over space so that there simply is not a precise location that it occupies, but instead occupies a range of positions. Similarly, the momentum cannot be precisely known since a particle consists of a packet of waves, each of which have their own momentum so that at best it can be said that a particle has a range of momentum.
Let's consider if quantum variables could be measured exactly. A wave that has a perfectly measurable position is collapsed onto a single point with an indefinite wavelength and therefore indefinite momentum according to de Broglie's equation. Similarly, a wave with a perfectly measurable momentum has a wavelength that oscillates over all space infinitely and therefore has an indefinite position.
You could do the same thought experiment with energy and time. To precisely measure a wave's energy would take an infinite amount of time while measuring a wave's exact instance in space would require to be collapsed onto a single moment which would have indefinite energy.
Consequences
The Heisenberg Principle has large bearing on practiced science and how experiments are designed. Consider measuring the momentum or position of a particle. To create a measurement, an interaction with the particle must occur that will alter it's other variables. For example, in order to measure the position of an electron there must be a collision between the electron and another particle such as a photon. This will impart some of the second particle's momentum onto the electron being measured and thereby altering it. A more accurate measurement of the electron's position would require a particle with a smaller wavelength, and therefore be more energetic, but then this would alter the momentum even more during collision. An experiment designed to determine momentum would have a similar effect on position. Consequently, experiments can only gather information about a single variable at a time with any amount of accuracy.
Problems
1. The uncertainty in the momentum $\Delta{p}$ of a football thrown by Tom Brady during the superbowl traveling at $40\; m/s$ is $\ 1 \times 10^{-6}$ of its momentum. What is its uncertainty in position $\Delta{x}$? Mass=0.40kg
2. You notice there is 2 mL of water traveling on the football at the same speed and $\Delta{p}$. Calculate its $\Delta{x}$.
3. An electron in that molecule of water traveling at the same speed has the same $\Delta{p}$. Calculate its $\Delta{x}$ if the mass of an electron is $\ 9.1 \times 10^{-31} kg$.
4. Comment on the differences in the uncertainty of momentum between the ball, water, and electron. How does the mass effect this value?
5. Taking into account all of the information presented above, can you state a situation in which the Heisenberg Uncertainty Principle has little effect on measuring the momentum and position of one object, but dominates for that of another when both objects are part of the same system?
Answers
1.
\begin{align} p&=mv\nonumber \[4pt] &= (0.40\, kg)(40\,m/s)\nonumber \[4pt] &= 16\,\dfrac{kg \, m}{s}\nonumber \end{align} \nonumber
\begin{align} \Delta{p} &= p (1 \times 10^{-6})\nonumber \[4pt] &=16\dfrac{kg \, m}{s} (1 \times 10^{-6})\nonumber \[4pt] &=16 \times 10^{-6}\dfrac{kg \, m}{s}\nonumber \end{align} \nonumber
\begin{align} \Delta{p}\Delta{x} & \ge \dfrac{h}{4\pi} \nonumber \[4pt] \Delta{x} &\ge \dfrac{h}{4\pi\Delta{p}} \nonumber \[4pt] &\ge \dfrac{6.626 \times 10^{-34}J \,s}{4\pi \left(16 \times 10^{-6}\dfrac{kg \, m}{s}\right)}\nonumber \[4pt] &\ge 3.3 \times 10^{-30}m\nonumber \end{align} \nonumber
Note that $1\,J = 1 \dfrac{kg \, m^2}{s^2}$.
2.
The volume is not the property that matters, but the mass. So convert to mass with density.
$(2\, mL ) \underbrace{\left(\dfrac{1\,g}{1\,mL}\right)}_{\text{density of water}} \left(\dfrac{1\,kg}{1,000\,g}\right) =2 \times 10^{-3} kg \nonumber$
\begin{align} p&=m v\nonumber \[4pt] &= (2 \times 10^{-3} kg)(40\,m/s)\nonumber \[4pt] &= 8 \times 10^{-2}\dfrac{kg \, m}{s}\nonumber \end{align} \nonumber
\begin{align} \Delta{p} &=p (1 \times 10^{-6}) = (8 \times 10^{-2} \dfrac{kg \, m}{s})(1 \times 10^{-6}) = 8 \times 10^{-8} \dfrac{kg \, m}{s} \nonumber \end{align} \nonumber
\begin{align} \Delta{p}\Delta{x} &\ge \dfrac{h}{4\pi} \ \Delta{x} &\ge \dfrac{h}{4\pi\Delta{p}} \nonumber \[4pt] &\ge \dfrac{6.626 \times 10^{-34} J \, s}{4\pi \left(8 \times 10^{-8} \dfrac{kg \, m}{s}\right)}\nonumber \[4pt] &\ge 6.6 \times 10^{-28}m \nonumber \end{align} \nonumber
3
\begin{align} p&=m v \nonumber \[4pt]&= (9.1 \times 10^{-31}kg)(40\,m/s) \nonumber \[4pt] &= 3.6 \times 10^{-29}\dfrac{kg \, m}{s} \nonumber \end{align} \nonumber
\begin{align}\Delta{p} &= p \times 10^{-6} \nonumber \[4pt] &= 3.6 \times 10^{-29}\dfrac{kg \, m}{s} \times 1 \times 10^{-6} \nonumber \[4pt] &= 3.6 \times 10^{-35} \dfrac{kg \, m}{s} \nonumber \end{align} \nonumber
\begin{align} \Delta{p}\Delta{x} &\ge \dfrac{h}{4\pi} \nonumber \[4pt] \Delta{x} &\ge \dfrac{h}{4\pi\Delta{p}} \nonumber \[4pt] &\ge \dfrac{6.626 \times 10^{-34}J \times s}{4\pi 3.6 \times 10^{-35} \dfrac{kg \, m}{s}} \nonumber \[4pt] &\ge1.5\,m \nonumber \end{align} \nonumber
4
The mass of the football is $\ 4 \times 10^{-1} kg$, the water is $\ 2 \times 10^{-3} kg$, and the electron is $\ 9.1 \times 10^{-31} kg$. The mass of the water is 2 orders of magnitude smaller than that of the football, and the resulting position uncertainty is 2 orders of magnitude larger. Between the electron and water there is a difference of 28 orders of magnitude for both mass and $\Delta{x}$. There is a direct correlation of inverse proportionality between the $\Delta{x}$ and $\Delta{p}$ as described by the Heisenberg Uncertainty Principle, and the much smaller electron has a larger position of uncertainty of 1.5 m compared to the larger football's of $\ 3.3 \times 10^{-30} m$
5
One example that can be used is a glass of water in a cup holder inside a moving car. This glass of water has multiple water molecules each consisting of electrons. The water in the glass is a macroscopic object and can be viewed with the naked eye. The electrons however occupy the same space as the water, but cannot be seen and therefore must be measured microscopically. As stated above in the introduction, the effect of measuring a tiny particle causes a change in its momentum and time in space, but this is not the case for the larger object. Thus, the uncertainty principle has much greater bearing on the electrons rather than the macroscopic water.
Contributors and Attributions
• Sarah Woods, Kris Baumgartner (UC Davis)
Heisenberg's Uncertainty Principle
If two operators commute then both quantities can be measured at the same time, if not then there is a tradeoff in the accuracy in the measurement for one quantity vs. the other.
Introduction
Operators are commonly used to perform a specific mathematical operation on another function. The operation can be to take the derivative or integrate with respect to a particular term, or to multiply, divide, add or subtract a number or term with regards to the initial function. Operators are commonly used in physics, mathematics and chemistry, often to simplifiy complicated equations such as the Hamiltonian operator, used to solve the Schrödinger equation.
Operators
Operators are generally characterized by a hat. Thus they generally appear like the following equation with $\hat{E}$ being the operator operating on $f(x)$
$\hat{E} f(x) = g(x)$
For example if $\hat{E} = d/dx$ then:
$g(x) = f'(x)$
Commuting Operators
One property of operators is that the order of operation matters. Thus:
$\hat{A}\hat{E}f(x) \not= \hat{E}\hat{A}f(x)$
unless the two operators commute. Two operators commute if the following equation is true:
$\left[\hat{A},\hat{E}\right] = \hat{A}\hat{E} - \hat{E}\hat{A} = 0$
To determine whether two operators commute first operate $\hat{A}\hat{E}$ on a function $f(x)$. Then operate$\hat{E}\hat{A}$ the same function $f(x)$. If the same answer is obtained subtracting the two functions will equal zero and the two operators will commute.on
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The uncertainty principle comes about fundamentally from the commuting properties of any two quantum operators. So for any two observables, $A$ and $B$, then the generalized uncertainty principle states that $\langle(\Delta_A)^2\rangle \langle (\Delta_B)^2 \rangle \geq (1/4)|\langle[A,B]\rangle|^2$ where the $\langle \rangle$ denote expectation values, the $\Delta_Q$ denotes the variance in the $Q$ operator, and $[A,B] = AB - BA$ is the commutator. Now this definition shows you that there are really only two choices, either the observables commute or they don't. If they commute $([A,B] = 0)$ then there are no restrictions on how accurately we may determine them, if they do not commute then the generalized uncertainty principle holds. This definition shows that it is possible (and relatively trivial) to create operators that are not conjugate, but where they do not commute and below are three examples.
Position and Momentum
If we examine a particle traveling freely described by a quantum-mechanical plane wave, $\Psi = Ae^{ikx}$ if the particle is traveling in the $+x$ direction. We have seen that the momentum of this particle can be calculated exactly using the momentum operator
$i\hbar \dfrac{\partial}{\partial x}e^{ikx} = \hbar k e^{ikx} = Pe^{ikx}$
If we now ask where this particle is we examine the probability density of the particle $\Psi^*Psi = Ae^{-ikx}Ae^{ikx} = A^2$. Since the square of A has no x-dependence we have no information regarding the position of the particle. In other words, if we know the momentum of the particle exactly, the position is completely unknown.
Suppose we assume that the particle can be located in space and has a Gaussian probability distribution. Then instead of a constant prefactor $A$ we would have a Gaussian prefactor and the wavefunction becomes
$\Psi = B e^{-(x-x_o)^2/2}e^{ikx}$
Note that the Gaussian function expressed in terms of $(x – x_0)$ signifies a Gaussian centered at $x_0$. This distribution in space implies a distribution of momenta since
$\Psi(p) \int_{-\infty}^{\infty} e^{-(x-x_o)^2/2} e^{ipx/h}dx$
Which is a Fourier transform of the two conjugate variables position and momentum. Note how the relationship between these two conjugate variables implies a quantitative relationship between their distributions. To calculate $\Psi(p)$ we need to complete the square. In order to demonstrate this we will calculate a simpler distribution. Imagine that our particle is found centered at the origin so $x_0 = 0$. We can then solve the integral
$\Psi(k) = \int_{-\infty}^{\infty} e^{-x^2/2} e^{ikx} dx$
By multiplying by a term equal to $e^{-k^2}$ inside and $e^{k^2}$ outside the integral.
$\Psi(k)= e^{-k^2/2} \int_{-\infty}^{\infty} e^{-x^2/2}e^{ikx}e^{k^2/2} dx$
Thus, we have
$\Psi(k) = e^{-k^2/2} \int_{-\infty}^{\infty} e^{-(x-ik)^2/2}dx = \dfrac{1}{\sqrt{\pi}} e^{-k^2/2}$
Note that in fact this gives a moment distribution since $k = p/\hbar$. Note also that substituting $x – x_0$ for $x$ will not change this result. This result shows that the Fourier transform of a Gaussian is also a Gaussian (in the space of the conjugate variable). This result is quite useful for quantifying the relationships between conjugate variables.
How does this result give us an uncertainty principle? We now include a Gaussian standard deviation that we will call $\Delta{x}$ and one for momentum that we will call $\Delta{k}$.
$\Psi(k) = e^{-k^2/2\Delta_k^2} \int_{-\infty}^{\infty} e^{-x^2/2\Delta_x^2}e^{ikx}e^{k^2/2\Delta_k^2}dx$
The Fourier transform then has the form
$\Psi(x) = \dfrac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \Psi(k) e^{-ikx} dk$
which implies that $\Delta{x} \Delta{k} = 1$ in order for $e^{ikx}$ to conform to the definition of a Fourier transform.
This in turn implies that $\Delta{x} \Delta{p} = \hbar$. The relationship described here implies that the linear momentum and position along the x-axis do not commute, $[x,p_x] \neq 0$. In fact, $[x,p_x] = -i\hbar$.
Angular Momenta
The argument that we have made for linear momentum also applies to the angular momentum of a particle. For example, consider a particle traversing a circular trajectory. In this case we have seen that the wave function is $\Psi = e^{im}\phi$. This implies azimuthal angular momenta of $m_h$. For each of these momentum states if we now ask where the particle is on the circle we find.
$\int _o^{2\pi} \Psi^*\Psi d\phi$
for
$\Psi=\dfrac{1}{\sqrt{2\pi}} e^{im\phi}$
So that the probability of finding the particle in any region of space is $1/2\pi$. There is no $\phi$ dependence and hence the location of the particle on the circle is completely undetermined. We can therefore understand that the same type of relationship exists between angular momenta and angular position as exists for linear momenta and linear position. The angular momentum relationships can also be expressed in terms of the total angular momentum and the azimuthal angular momentum. That is that we can know both $L_z$ and $L$ simultaneously, but not $L_x$ or $L_y$ and $L_z$ simultaneously.
$[L_z,L] = 0$
although the individial components do not commute:
• $[L_z,L_x] = i\hbar L_y$
• $[L_x,L_y] = i \hbar L_z$
• $[L_y,L_z] = i\hbar L_x$
Energy and Time
It is important that the same type of conjugate variable relationship exists between the energy e and time t such that the uncertainty in the energy times the uncertainty in the time is of the order of Planck’s constant as well. $\Delta_e \Delta_t = \hbar$ exactly analogous to the relationship between momentum and position given above. The mathematical form of the of the Fourier transform is identical if
$k \rightarrow w$
and
$x \rightarrow t.$
The latter relationship is important since it implies that the lifetime of a state results in an energy width or the duration of a laser pulse implies a certain spectral width. If we use the properties of Gaussians shown above for the position/momentum conjugate variables we can calculate a spectral width for a given time duration. We do this in terms of Joules and wave numbers. The wave number unit is particularly valuable since spectra are reported in cm-1.
Time
(sec)
Joules
cm-1
1 ns
10-9
10-25
0.005
10 ns
10-10
10-24
0.05
100 ns
10-11
10-23
0.5
1 ps
10-12
10-22
5
10 ps
10-13
10-21
50
100 ps
10-14
10-20
500
1 fs
10-15
10-19
5000
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A photon is a tiny particle that comprises waves of electromagnetic radiation. As shown by Maxwell, photons are just electric fields traveling through space. Photons have no charge, no resting mass, and travel at the speed of light. Photons are emitted by the action of charged particles, although they can be emitted by other methods including radioactive decay. Since they are extremely small particles, the contribution of wavelike characteristics to the behavior of photons is significant. In diagrams, individual photons are represented by a squiggly arrow.
Description
Photons are often described as energy packets. This is a very fitting analogy, as a photon contains energy that cannot be divided. This energy is stored as an oscillating electric field. These fields may oscillate at almost any frequency. Although they have never been observed, the longest theoretical wavelength of light is the size of the universe, and some theories predict the shortest possible at the Planck length. These packets of energy can be transmitted over vast distances with no decay in energy or speed. Photons travel at the speed of light, 2.997x108 m/s in empty space. The speed of a photon through space can be directly derived from the speed of an electric field through free space. Maxwell unveiled this proof in 1864. Even though photons have no mass, they have an observable momentum which follows the de Broglie equation. The momentum of photons leads to interesting practical applications such as optical tweezers.
Generally speaking, photons have similar properties to electromagnetic waves. Each photon has a wavelength and a frequency. The wavelength is defined as the distance between two peaks of the electric field with the same vector. The frequency of a photon is defined as how many wavelengths a photon propagates each second.
Unlike an electromagnetic wave, a photon cannot actually be of a color. Instead, a photon will correspond to light of a given color. As color is defined by the capabilities of the human eye, a single photon cannot have color because it cannot be detected by the human eye. In order for the retina to detect and register light of a given color, several photons must act on it. Only when many photons act in unison on the retina, as an electromagnetic wave, can color be perceived.
As Described by Maxwell's Equations
The most accurate descriptions we have about the nature of photons are given by Maxwell's equations. Maxwell's equations mathematically predict how photons move through space. Fundamentally, an electric field undergoing flux will create an orthogonal magnetic field. The flux of the magnetic field then recreates the electric field. The creation and destruction of each corresponding wave allows the wave pair to move through space at the speed of light. Maxwell's equations correctly describe the nature of individual photons within the framework of quantum dynamics.
Creation of Photons
Photons can be generated in many different ways. This section will discuss some of the ways photons may be emitted. As photons are electric field propagating through space, the emission of photons requires the movement of charged particles.
Blackbody Radiation
As a substance is heated, the atoms within it vibrate at higher energies. These vibrations rapidly change the shape and energies of electron orbitals. As the energy of the electrons changes, photons emitted and absorbed at energies corresponding to the energy of the change. Blackbody radiation is what causes light bulbs to glow, and the heat of an object to be felt from a great distance. The simplification of objects as blackbodies allows indirect temperature calculation of distant objects. Astronomers and kitchen infrared thermometers use this principle every day.
Spontaneous Emission
Photons may be spontaneously emitted when electons fall from an excited state to a lower energy state(usually the ground state). The technical term for this drop in energy is a relaxation. Electrons undergoing this type of emission will produce a very distinctive set of photons based on the available energy levels of their environment. This set of possible photons is the basis for an emission spectrum.
Flourescence
Florescence is special case of spontaneous emission. In florescence, the energy of a photon emitted does not match the energy used to excite the electron. An electron will fluoresce when it loses a considerable amount of energy to its surroundings before undergoing a relaxation. Generally florescence is employed in a laboratory setting to visualize the presence of target molecules. UV light is used to excite electrons, which then emit light at visible wavelengths that researchers can see.
Stimulated Emission
An excited electron can be artificially caused to relax to a lower energy state by a photon matching the difference between these energy states. The electric field's phase and orientation of the resultant photon, as well as its energy and direction will be identical to that of the incident photon. The light produced by stimulated emission is said to be coherent as it is similar in every way to the photon that caused it. Lasers produce coherent electromagnetic radiation by stimulated emission.
Synchrotrons (electron bending)
Electrons with extremely high kinetic energy, such as those in particle accelerators, will produce high energy photons when their path is altered. This alteration is accomplished by a strong magnetic field. All free electrons will emit light in this manner, but synchrotron radiation has special practical implications. Synchrotron radiation is currently the best technology available for producing directional x-ray radiation at precise frequencies. Synchrotrons, such as the Advanced Light Source (ALS) at Lawrence Berkeley Labs and Stanford Synchrotron Radiation Light Source (SSRL) are hotbeds of x-ray spectroscopy due to the excellent quality of x-rays produced.
Nuclear Decay
Certain types of radioactive decay can involve the release of high energy photons. One such type of decay is a nuclear isomerization. In an isomerization, a nucleus rearranges itself to a more stable configuration and emits a gamma ray. While it is only theorized to occur, proton decay will also emit extremely high energy photons.
The Photoelectric Effect
Light incident on a metal plate may cause electrons to break loose from the plate surface (Fig. 1). This interaction between light and electrons is called the photoelectric effect. The photoelectric effect provided the first conclusive evidence that beams of light was made of quantized particles. The energy required to eject an electron from the surface of the metal is usually on the same order of magnitude as the ionization energy. As metals generally have ionization energies of several electron-volts, the photoelectric effect is generally observed using visible light or light of even higher energy.
Fig. 1, The photoelectric effect.
At the time this phenomenon was studied, light was thought to travel in waves. Contrary to what the wave model of light predicted, an increase in the intensity of light resulted in an increase in current, not an increase in the kinetic energy of the emitted electron. Einstein later explained this difference by showing that light was comprised of quantized packets of energy called photons. His work on the photoelectric effect earned him the Nobel Prize.
The photoelectric effect has many practical applications, as current may be generated from a light source. Generally, the photoelectric effect is used as a component in switches that respond to light. Some examples are nightlights and photomultipliers. Usually the current is so small that it must be amplified in order to be an effective switch
Energy of a Photon
The energy of a photon is a discrete quantity determined by its frequency. This result can be determined experimentally by studying the photoelectric effect. The kinetic energy of an emitted electron varies directly with the frequency of the incident light. If the experimental values of these energies are fitted to a line, the slope of that line is Planck's constant. The point at which electrons begin to be emitted from the surface is called the threshold frequency, and is denoted by $\nu_0$. The principle of conservation of energy dictates that the energy of a photon must all go somewhere. Assuming that the energy $h\nu_0$ is the initial energy requirement to pry an electron from its orbital, the kinetic energy of a photon is equal to the kinetic energy of the emitted electron plus the ionization energy. Therefore the energy of a free photon becomes $E = h\nu$ where nu is the frequency of the photon and h is Planck's constant.
Fig. 2, Photoelectric effect results
The results from a photoelectric experiment are shown in Figure 2. $\nu_0$ is the minimum frequency at which electrons start to be detected. The solid lines represent the actual observed kinetic energies of released electrons. The dotted red line shows how a linear result can be obtained by tracing back to the y axis. Electrons cannot actually have negative kinetic energies.
Photon Interference
Whereas the double slit experiment initially indicated that a beam of light was a wave, more advanced experiments confirm the electron as a particle with wavelike properties. The diffraction of a beam of light though a double slit is observed to diffract producing constructive and destructive interference. Modern technology allows the emission and detection of single photons. In an experiment conducted by Philippe Grangier, a single photon is passed through a double slit. The photon then is detected on the other side of the slits. Across a large sample size, a trend in the final position of the photons can be determined. Under the wave model of light, an interference pattern will be observed as the photon splits over and over to produce a pattern. However, the results disagree with the wave model of light. Each photon emitted corresponds with a single detection on the other side of the slits(Fig. 3). With a certain probability, each photon is be detected at 100% strength. Over a series of measurements, photons produce the same interference pattern expected of a beam of photons. When one slit is closed, no interference pattern is observed and each photon travels in a linear path through the open slit.
Fig 3, Proof for the particle-nature of photons. One possible result is shown.
This interference has a profound implication which is that photons do not necessarily interact with each other to produce an interference pattern. Instead, they interact and interfere with themselves. Furthermore, this shows that the electron does not pass through one slit or the other, but rather passes through both slits simultaneously. Richard Feynman's theory of quantum electrodynamics explains this phenomenon by asserting that a photon will travel not in a single path, but all possible paths in the universe. The interference between these paths will give the probability of the photon taking any given path, as the majority of the paths cancel with each other. He has used this theory to explain the nature of wide ranges of the actions of photons, such as reflection and refraction, with absolute precision.
Problems
1) The peak wavelength of a light bulb is 500 nm. Calculate the energy of a single photon at this wavelength.
Solution
$E = h\nu$
$\nu = \dfrac{c}{\lambda}$
$E = h*\dfrac{c}{\lambda}$
$= 6.626x10^{-34}m^2kg/s^2*\dfrac{3.00x10^{8}m/s}{500x10^{-9}m}$
$= 3.97x10^{-19}J$
2) The work function of an metal surface is 9.4eV. What is the frequency of a photon which ejects an electron from this surface at 420km/s?
Solution
$h\nu_0 = 9.4eV x 1.6x10^{-19}J/eV = 1.51x10^{-18} J$
$KE = \dfrac{1}{2}mv^2 = h\nu-h\nu_0$
$\dfrac{1}{2} ( 9.11x10^{-31}kg)/{(420,000km/s)}^2 = 6.626x10^{-34}m^2kg/s*\nu - 1.51x10^{-18}J$
$\nu = 2.28x10^{15}hz$
3) A single photon passes through a double slit 20nm apart. A photomultiplier detects at least one particle in the 20 nm directly behind the slit. What fraction of the photon is detected here?
Solution
The entire photon is detected. Protons are quantized particles. Although they can pass through both slits, it is still a single particle and will be detected accordingly.
4) A photon removes an electron from an atom. The kinetic energy of the exiting electron is found to be less than that of the photon that removed it. Why isn't the energy the same?
Solution
Recall the photoelectric equation : $KE = h\nu-h\nu_0$. This equation relates the energies of photons and electrons from an ejection. The second term of the equation, $-h\nu_0$ is the amount of energy required to remove an electron from its orbital. The extra energy goes into breaking the association of an electron with a nucleus. Keep in mind that for a metal this is not the ionization energy due to the delocalization of electrons involved in metallic bonding.
5) Keeping in mind the relationship between the energy and frequency of light, design an experiment to determine if photons lose energy as they travel through space.
Solution
One possible experiment utilizes the photoelectric effect. A light source is shone on a piece of metal, and the kinetic energy of ejected electrons is calculated. By shining the light at different distances from the metal plate, individual photons may be shown to undergo lossless transmission. The experiment will show that while the number of electrons ejected may decrease as a function of distance, their kinetic energy will remain the same.
Contributors and Attributions
• Michael Kennedy
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How many of the above questions can you answer? You can either click on any you are unsure about, or browse through them by simply scrolling down this page.
Contributors and Attributions
Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook
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 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 Wave-Particle Duality theory states that waves can exhibit particle-like properties while particles can exhibit wave-like properties. This definition opposes classical mechanics or Newtonian Physics.
Double Slit Experiment
In the 17thcentury, Newton demonstrated that, similar to wave, beams of light can also diffract and interfere with one another by shining white light into a prism to collect seven different colors and recombining them with a second prism to produce white light. This wave theory of light (classical physics) was confirmed by Young's double slit experiment in 1801 (figure 1).
This classical theory was also proven by Davisson and Germer in 1925, when they aimed a beam of electrons at nickel, and the diffraction of the electrons produced fringes (Figure 2A). Fringes are properties of waves, and the diffraction is explained using the interference properties of waves. The dark fringes are produced when the waves are in phase, and light fringes are produced when the waves are out of phase (Electromagnetic Radiation).
Black Body Radiation
Based on the classical theory, light's energy will follow Rayleigh-Jeans Law:
$\rho = \dfrac{8 \pi T}{\lambda^4}$
where
• ρ is the radiant energy,
• λ is the wavelength,
• k is Boltzmann constant, and
• T is temperature
According to this equation, radiance energy is continous and will increase to infinity if the wavelength gets very small. However, in 1899, Otto R. Lummer and Ernst Pringsheim discovered blackbody radiation which showed that radiant energy is discreet and has a max value. The energy doesn't go to infinity as the classical physics had predicted but declines after reaching a max value.
Photoelectron Effects
The first experiments towards Wave-Particle duality were done by German Physicist Max Planck (1858-1947). Using blackbody radiator (equal emitter and absorber of radiation at all wavelengths), Planck derived the equation for the smallest amount of energy that can be changed into light
$E=h \nu$
where h is Planck’s constant 6.626x10-34 J.S and v is the frequency.
He also formulated the quantum theory by saying that light that was emitted had discrete levels of energy, and that energy that was radiated was quantized;
E=nhv
(where n is an integer, and can be zero or a positive number).
Quantization of energy states that there are discrete values or states, and energies in between the values of n are forbidden. Hence, he stated that if x number of particles were present with a certain frequency value, than energy would be
E=xhv
Frequency is related to the wavelength where c=vλ or v=c/λ
Replace v=c/λ into the above equation, we have
E=xhc/λ
In 1905, Einstein assumed that Planck’s discrete energies are packets of energy called photons. The total energy of a system is equal to the kinetic energy plus the potential energy, and as always the Law of Conservation of energy applies. Einstein explained that in the photoelectric effect energy each photon's energy is absorbed by one electrons in a given metal, and as a result the electron was able to eject if the photon's energy is equal or greater than the threshold energy (Figure 2). The threshold energy is the amount of energy needed to eject an electron, and is called work function Φ.
Since E=hv
we can rewrite the equation to show that the total energy is equal to Φ plus the kinetic energy
E = Φ + KE = hv
The photoelectric effect shows that light behaves like a photon or a particle packed with energy, in other words light waves behave like particles.
According to Particle theory of light, light energy will increase to a discreet and finite value unless λ goes to zero, which will never happen according to the Particle in a One-Dimensional Box theory. This helps explain the blackbody radiation observation.
Photoelectric Checklist
The number of electrons ejected from the metal increases as the intensity of the light increases. An electron that is not held strongly will have more kinetic energy. The threshold energy must be absorbed in order for an electron to be ejected. Due to conservation of energy, the kinetic energy (T) of the electron is dependent on the frequency of the wavelength of incident light. Remember, high frequencies have short wavelengths therefore photons with short wavelengths will be higher in energy. There is a linear relationship between kinetic energy of the ejected electron and frequency. After the work function energy is absorbed by the electron, the rest of the energy that was provided by the photon changes into kinetic energy T=1/2mv2, and hence the equation E=Φ+T=hv. An electron that only absorbs the threshold energy has no kinetic once it has traveled outside the metal.
Wave-Particle Duality
Since both particle and wave theories of light seem to explain a portion of light properties correctly, which is the correct one? In 1924, de Broglie (1892-1987) proposed an answer to this question. He assumed that all moving objects have wave-like properties. He combined Planck's constant and linear momentum
E = hv = hc/λ
so
$\lambda = \dfrac{hc}{E} \tag{1}$
and
p = E/c = mv
(p is the momentum of the object, m is the object's mass, and v is the velocity of the object) so
E = mvc
Plug this into (1), we have
λ = hc/mvc = h/mv
This equation postulates that all moving object with a mass will have a wavelength which is called de Broglie wavelength, but these wavelengths are only seen with objects that have a very small mass. Since h is very small (6.626 x 10-34Js), any object that has a large mass will have its wavelength close to zero. That is why we cannot see a walking human's wavelength. This relationship was confirmed by the Davisson and Germer diffraction experiments, where the wavelengths of the electrons, that gave diffraction patterns were same as the predicted wavelength using de Broglie relationship.
Prolems
1. Find the energy of an electron with wavelength of 30 nm using Rayleigh-Jeans Law at T = 280K.
2. Find the energy of the same electron using Planck law.
3. Find the wavelength of an electron that travels at 10m/s (me = 9.10939 x 10-31kg).
4. Find the wavelenth of a person who has the same speed as the electron in question 2 with a mass of 60 kg.
5. How long does it take to melt 1g of ice if there are 3 x 105 photons stricking the ice from the sun light per second with a wavelength of 6 nm (energy to melt 1g of ice is 334 J).
Solutions
1. ρ = 8πkT/λ^4 = 8(3.14)(1.38 x 10-23mkg/s2K)(280K)/(30 X 10-9m)4 = 1.2 x 1011 J
2. E = hc/λ = 6.626 x 10-34Js(3 x 108m/s)/(30 x 10-9m) = 6.626 x 10-17 J
3. λ = h/(mv) = 6.626 x 10-34Js/[(9.109 x 10-31kg)(10 m/s)] = 7.27 x 10-5 m
4. λ = h/(mv) = 6.626 x 10-34Js/[(60kg)(10 m/s)] = 1.1 x 10-36 m
5. E = Xhc/λ
or we can write n = λE/hc = 334J(6 x 10-9m)/[(6.626 x 10-34Js)(3 x 108)] = 1.01 x 1019 photons
t = 1.01 x 1019photons/(3 x 105photons/s) = 3.37 x 1013 s
Outside Links
• Professor Guo from UCD Chemistry sings the Duality blues.
• en.Wikipedia.org/wiki/Wave%E2...rticle_duality
• video about de Broglie's wavelength www.youtube.com/watch?v=lDYMu...eature=related
Contributors and Attributions
• Artika Singh, Tu Quach
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• Bra-Ket Notation
Bra-ket notation is a standard notation for describing quantum states in the theory of quantum mechanics composed of angle brackets and vertical bars. It can also be used to denote abstract vectors and linear functionals in mathematics. It is so called because the inner product (or dot product) of two states is denoted by a bracket, <Φ|Ψ>, consisting of a left part, <Φ|, called the bra, and a right part, |Ψ>, called the ket.
• Chapter 4. Principles of Quantum Mechanics
Here we will continue to develop the mathematical formalism of quantum mechanics, using heuristic arguments as necessary. This will lead to a system of postulates which will be the basis of our subsequent applications of quantum mechanics.
• Collapsing Wavefunctions
One of the postulates is that all measurable quantities in a quantum system are represented mathematically by so called observables. An observable is thus a mathematical object, more specifically a real linear operator whose 'eigenstates' form a complete set. This essentially means that any quantum state can be expressed as a linear combination of these eigenstates of the observable.
• Expectation Values (Averages)
The expected value (or expectation, mathematical expectation, mean, or first moment) refers to the value of a variable one would "expect" to find if one could repeat the random variable process an infinite number of times and take the average of the values obtained. More formally, the expected value is a weighted average of all possible values.
• Matrix Representation of Operators and Wavefunctions
• Properties of the solutions to the Schrödinger 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.
03. The Tools of Quantum Mechanics
Bra-ket notation is a standard notation for describing quantum states in the theory of quantum mechanics composed of angle brackets and vertical bars. It can also be used to denote abstract vectors and linear functionals in mathematics. It is so called because the inner product (or dot product) of two states is denoted by a bracket, $\langle Φ|Ψ \rangle$, consisting of a left part, $\langle Φ |$, called the bra, and a right part, $|Ψ \rangle$, called the ket. The notation was introduced in 1939 by Paul Dirac, and is also known as Dirac notation. Bra-ket notation is widespread in quantum mechanics: almost every phenomenon that is explained using quantum mechanics—including a large proportion of modern physics—is usually explained with the help of bra-ket notation.
Bra-Ket Notation
A snack machine inputs coins plus some code entered on a key pad, and (hopefully) outputs a snack. It also does so in a deterministic manner: i.e., the same money plus the same code produces the same snack (or the same error message) time after time. Note that the input and output of the machine have completely different natures. We can imagine building a rather abstract snack machine which inputs ket vectors and outputs complex numbers in a deterministic fashion. Mathematicians call such a machine a functional. Imagine a general functional, labeled , acting on a general ket vector, labeled , and spitting out a general complex number . This process is represented mathematically by writing
(9)
Let us narrow our focus to those functionals which preserve the linear dependencies of the ket vectors upon which they operate. Not surprisingly, such functionals are termed linear functionals. A general linear functional, labeled , satisfies
(10)
where and are any two kets in a given ket space.
Consider an -dimensional ket space [i.e., a finite-dimensional, or denumerably infinite dimensional (i.e., ), space]. Let the (where runs from 1 to ) represent independent ket vectors in this space. A general ket vector can be written1
(11)
where the are an arbitrary set of complex numbers. The only way the functional can satisfy Eq. (10) for all vectors in the ket space is if
(12)
where the are a set of complex numbers relating to the functional.
Let us define basis functionals which satisfy
(13)
It follows from the previous three equations that
(14)
But, this implies that the set of all possible linear functionals acting on an -dimensional ket space is itself an -dimensional vector space. This type of vector space is called a bra space (after Dirac), and its constituent vectors (which are actually functionals of the ket space) are called bra vectors. Note that bra vectors are quite different in nature to ket vectors (hence, these vectors are written in mirror image notation, and , so that they can never be confused). Bra space is an example of what mathematicians call a dual vector space (i.e., it is dual to the original ket space). There is a one to one correspondence between the elements of the ket space and those of the related bra space. So, for every element of the ket space, there is a corresponding element, which it is also convenient to label , in the bra space. That is,
(15)
where DC stands for dual correspondence.
There are an infinite number of ways of setting up the correspondence between vectors in a ket space and those in the related bra space. However, only one of these has any physical significance. For a general ket vector , specified by Eq. (11), the corresponding bra vector is written
(16)
where the are the complex conjugates of the . is termed the dual vector to . It follows, from the above, that the dual to is , where is a complex number. More generally,
(17)
Recall that a bra vector is a functional which acts on a general ket vector, and spits out a complex number. Consider the functional which is dual to the ket vector
(18)
acting on the ket vector . This operation is denoted . Note, however, that we can omit the round brackets without causing any ambiguity, so the operation can also be written . This expression can be further simplified to give . According to Eqs. (11), (12), (16), and (18),
(19)
Mathematicians term the inner product of a bra and a ket.2 An inner product is (almost) analogous to a scalar product between a covariant and contravariant vector in some curvilinear space. It is easily demonstrated that
(20)
Consider the special case where . It follows from Eqs. (12) and (20) that is a real number, and that
(21)
The equality sign only holds if is the null ket [i.e., if all of the are zero in Eq. (11)]. This property of bra and ket vectors is essential for the probabilistic interpretation of quantum mechanics, as will become apparent later.
Two kets and are said to be orthogonal if
(22)
which also implies that .
Given a ket which is not the null ket, we can define a normalized ket , where
(23)
with the property
(24)
Here, is known as the norm or length'' of , and is analogous to the length, or magnitude, of a conventional vector. Since and represent the same physical state, it makes sense to require that all kets corresponding to physical states have unit norms.
It is possible to define a dual bra space for a ket space of nondenumerably infinite dimensions in much the same manner as that described above. The main differences are that summations over discrete labels become integrations over continuous labels, Kronecker delta-functions become Dirac delta-functions, completeness must be assumed (it cannot be proved), and the normalization convention is somewhat different. More of this later.
Eigenvalues and eigenvectors
In general, the ket $X\vert A\rangle$ is not a constant multiple of $\vert A\rangle$. However, there are some special kets known as the eigenkets of operator $X$. These are denoted
$\vert x'\rangle, \vert x''\rangle, \vert x'''\rangle \ldots,$
and have the property
$X\vert x'\rangle = x'\vert x'\rangle, X\vert x''\rangle = x''\vert x''\rangle \dots,$
where $x'$, $x''$, $\ldots$ are numbers called eigenvalues. Clearly, applying $X$ to one of its eigenkets yields the same eigenket multiplied by the associated eigenvalue.
Consider the eigenkets and eigenvalues of a Hermitian operator $\xi$. These are denoted
$\xi \vert\xi'\rangle = \xi' \vert\xi' \rangle, \label{44}$
where $\vert\xi'\rangle$ is the eigenket associated with the eigenvalue $\xi'$. Three important results are readily deduced:
(i) The eigenvalues are all real numbers, and the eigenkets corresponding to different eigenvalues are orthogonal. Since $\xi$ is Hermitian, the dual equation to Equation \ref{44} (for the eigenvalue $\xi''$ reads
$\langle \xi''\vert\xi = \xi''^\ast \langle \xi''\vert. \label{45}$
If we left-multiply Equation \ref{44} by $\langle \xi''\vert$, right-multiply the above equation by $\vert\xi'\rangle$, and take the difference, we obtain
$(\xi' - \xi''^\ast) \langle \xi''\vert\xi'\rangle = 0. \label{46}$
Suppose that the eigenvalues $\xi'$ and $\xi''$ are the same. It follows from the above that
$\xi' = \xi'^\ast, \label{47}$
where we have used the fact that $\vert\xi'\rangle$ is not the null ket. This proves that the eigenvalues are real numbers. Suppose that the eigenvalues $\xi'$ and $\xi''$ are different. It follows that
$\langle \xi''\vert\xi'\rangle = 0, \label{48}$
which demonstrates that eigenkets corresponding to different eigenvalues are orthogonal.
(ii) The eigenvalues associated with eigenkets are the same as the eigenvalues associated with eigenbras. An eigenbra of $\xi$ corresponding to an eigenvalue $\xi'$ is defined
$\langle \xi'\vert\xi = \langle \xi'\vert\xi'. \label{49}$
(iii) The dual of any eigenket is an eigenbra belonging to the same eigenvalue, and conversely.
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Expectation Values
Consider an ensemble of microscopic systems prepared in the same initial state $\mid A \rangle$. Suppose a measurement of the observable $\xi'$ is made on each system. We know that each measurement yields the value $\xi'$ with probability $P(\xi')$. What is the mean value of the measurement? This quantity, which is generally referred to as the expectation value of $\xi'$, is given by
$\langle\xi\rangle = \displaystyle \sum^{}_{\xi'} \xi'P(\xi') = \displaystyle \sum^{}_{\xi'} \xi' \mid\langle A\mid\xi\rangle\mid^{2} \label{58}$
$= \displaystyle \sum^{}_{\xi'} \xi' \langle A\mid\xi\rangle\langle \xi'\mid A\rangle = \displaystyle \sum^{}_{\xi'} \xi' \langle A\mid\xi\mid\xi'\rangle\langle \xi'\mid A\rangle$
which reduces to
$\langle\xi\rangle = \langle A\mid\xi\mid A\rangle \label{59}$
with the aid of Eq. (54).
Consider the identity operator, 1. All states are eigenstates of this operator with the eigenvalue unity. Thus, the expectation value of this operator is always unity: i.e.,
$\langle A\mid 1\mid A\rangle = \langle A\mid A\rangle = 1 \label{60}$
for all $\mid A \rangle$. Note that it is only possible to normalize a given ket $\mid A \rangle$ such that Equation $\ref{60}$ is satisfied because of the more general property (21) of the norm. This property depends on the particular correspondence (16), that we adopted earlier, between the elements of a ket space and those of its dual bra space.
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Consider a microscopic system composed of particles or bodies with specific properties (mass, moment of inertia, etc.) interacting according to specific laws of force. There will be various possible motions of the particles or bodies consistent with the laws of force. Let us term each such motion a state of the system. According to the principle of superposition of states, any given state can be regarded as a superposition of two or more other states. Thus, states must be related to mathematical quantities of a kind which can be added together to give other quantities of the same kind. The most obvious examples of such quantities are vectors.
Let us consider a particular microscopic system in a particular state, which we label $A$: e.g., a photon with a particular energy, momentum, and polarization. We can represent this state as a particular vector, which we also label $A$, residing in some vector space, where the other elements of the space represent all of the other possible states of the system. Such a space is called a ket space (after Dirac). The state vector $A$ is conventionally written
$\mid A\rangle \tag{1}$
Suppose that state $A$ is, in fact, the superposition of two different states, $B$ and $C$. This interrelation is represented in ket space by writing
$\mid A\rangle = \mid B\rangle + \mid C\rangle \tag{2}$
where $\mid B \rangle$ is the vector relating to the state $B$, etc. For instance, state $\mid B \rangle$ might represent a photon propagating in the -direction, and plane polarized in the $x$,-direction, and state $\mid B \rangle$ might represent a similar photon plane polarized in the $y$-direction. In this case, the sum of these two states represents a photon whose plane of polarization makes an angle of with both the $x$- and $"y$-directions (by analogy with classical physics). This latter state is represented by $\mid B \rangle + \mid C \rangle$ in ket space.
Suppose that we want to construct a state whose plane of polarization makes an arbitrary angle $\alpha$, with the $x$,-direction. We can do this via a suitably weighted superposition of states $B$, and $C$. By analogy with classical physics, we require $\cos(\alpha)$ of state $B$, and $\sin(\alpha)$ of state $C$. This new state is represented by
$\cos(\alpha)\mid B\rangle + sin(\alpha)\mid C\rangle \tag{3}$
in ket space. Note that we cannot form a new state by superposing a state with itself. For instance, a photon polarized in the $y$,-direction superposed with another photon polarized in the $y$,-direction (with the same energy and momentum) gives the same photon. This implies that the ket vector
$c_1\mid A\rangle + c_2\mid A\rangle = (c_1 + c_2)\mid A\rangle \tag{4}$
corresponds to the same state that $\mid A \rangle$ does. Thus, ket vectors differ from conventional vectors in that their magnitudes, or lengths, are physically irrelevant. All the states of the system are in one to one correspondence with all the possible directions of vectors in the ket space, no distinction being made between the directions of the ket vectors $\mid A \rangle$ and $-\mid A \rangle$. There is, however, one caveat to the above statements. If then the superposition process yields nothing at all: i.e., no state. The absence of a state is represented by the null vector $\mid 0 \rangle$ in ket space. The null vector has the fairly obvious property that
$\mid A\rangle + \mid 0\rangle = \mid A\rangle \tag{5}$
for any vector $\mid A \rangle$. The fact that ket vectors pointing in the same direction represent the same state relates ultimately to the quantization of matter: i.e., the fact that it comes in irreducible packets called photons, electrons, atoms, etc. If we observe a microscopic system then we either see a state (i.e., a photon, or an atom, or a molecule, etc.) or we see nothing--we can never see a fraction or a multiple of a state. In classical physics, if we observe a wave then the amplitude of the wave can take any value between zero and infinity. Thus, if we were to represent a classical wave by a vector, then the magnitude, or length, of the vector would correspond to the amplitude of the wave, and the direction would correspond to the frequency and wave-length, so that two vectors of different lengths pointing in the same direction would represent different wave states.
We have seen, in Eq. (3), that any plane polarized state of a photon can be represented as a linear superposition of two orthogonal polarization states in which the weights are real numbers. Suppose that we want to construct a circularly polarized photon state. Well, we know from classical physics that a circularly polarized wave is a superposition of two waves of equal amplitude, plane polarized in orthogonal directions, which are in phase quadrature. This suggests that a circularly polarized photon is the superposition of a photon polarized in the $x$-direction (state $B$) and a photon polarized in the $y$-direction (state $C$, with equal weights given to the two states, but with the proviso that state $C$ is out of phase with state . By analogy with classical physics, we can use complex numbers to simultaneously represent the weighting and relative phase in a linear superposition. Thus, a circularly polarized photon is represented by
$\mid B\rangle + i\mid C\rangle \tag{6}$
in ket space. A general elliptically polarized photon is represented by
$c_1\mid B\rangle + c_2\mid C\rangle \tag{7}$
where $c_1$ and $c_2$ are complex numbers. We conclude that a ket space must be a complex vector space if it is to properly represent the mutual interrelations between the possible states of a microscopic system.
Suppose that the ket $\mid R \rangle$ is expressible linearly in terms of the kets $\mid A \rangle$ and $\mid B \rangle$, so that
$\mid R\rangle=c_1\mid A\rangle + c_2\mid B\rangle \tag{8}$
We say that $\mid R \rangle$ is dependent on $\mid A \rangle$ and $\mid B \rangle$. It follows that the state $R$ can be regarded as a linear superposition of the states $A$ and $B$. So, we can also say that state $R$ is dependent on states $\mid A \rangle$ and $\mid B \rangle$. In fact, any ket vector (or state) which is expressible linearly in terms of certain others is said to be dependent on them. Likewise, a set of ket vectors (or states) are termed independent if none of them are expressible linearly in terms of the others.
The dimensionality of a conventional vector space is defined as the number of independent vectors contained in the space. Likewise, the dimensionality of a ket space is equivalent to the number of independent ket vectors it contains. Thus, the ket space which represents the possible polarization states of a photon propagating in the -direction is two-dimensional (the two independent vectors correspond to photons plane polarized in the $x$- and $y$-directions, respectively). Some microscopic systems have a finite number of independent states (e.g., the spin states of an electron in a magnetic field). If there are $N$ independent states, then the possible states of the system are represented as an $N$-dimensional ket space. Some microscopic systems have a denumerably infinite number of independent states (e.g., a particle in an infinitely deep, one-dimensional potential well). The possible states of such a system are represented as a ket space whose dimensions are denumerably infinite. Such a space can be treated in more or less the same manner as a finite-dimensional space. Unfortunately, some microscopic systems have a nondenumerably infinite number of independent states (e.g., a free particle). The possible states of such a system are represented as a ket space whose dimensions are nondenumerably infinite. This type of space requires a slightly different treatment to spaces of finite, or denumerably infinite, dimensions.
In conclusion, the states of a general microscopic system can be represented as a complex vector space of (possibly) infinite dimensions. Such a space is termed a Hilbert space by mathematicians.
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We have seen that a measurement of some observable $\xi$ of a microscopic system causes the system to jump into one of the eigenstates of $\xi$. The result of the measurement is the associated eigenvalue (or some function of this quantity). It is impossible to determine into which eigenstate a given system will jump, but it is possible to predict the probability of such a transition. So, what is the probability that a system in some initial state $\mid B \rangle$ makes a transition to an eigenstate $\mid \xi' \rangle$ of an observable $\xi$, as a result of a measurement made on the system? Let us start with the simplest case. If the system is initially in an eigenstate $\mid \xi' \rangle$ then the transition probability to a eigenstate $\mid \xi'' \rangle$ corresponding to a different eigenvalue is zero, and the transition probability to the same eigenstate $\mid \xi' \rangle$ is unity. It is convenient to normalize our eigenkets such that they all have unit norms. It follows from the orthogonality property of the eigenkets that
(50)
where is unity if , and zero otherwise. For the moment, we are assuming that the eigenvalues of $\xi$ are all different.
Note that the probability of a transition from an initial eigenstate $\mid \xi' \rangle$ to a final eigenstate is the same as the value of the inner product . Can we use this correspondence to obtain a general rule for calculating transition probabilities? Well, suppose that the system is initially in a state $\mid B \rangle$ which is not an eigenstate of $\xi$. Can we identify the transition probability to a final eigenstate $\mid \xi' \rangle$ with the inner product ? The straight answer is no'', since is, in general, a complex number, and complex probabilities do not make much sense. Let us try again. How about if we identify the transition probability with the modulus squared of the inner product, ? This quantity is definitely a positive number (so it could be a probability). This guess also gives the right answer for the transition probabilities between eigenstates. In fact, it is the correct guess.
Since the eigenstates of an observable $\xi$ form a complete set, we can express any given state $\mid B \rangle$ as a linear combination of them. It is easily demonstrated that
(51)
(52)
(53)
where the summation is over all the different eigenvalues of $\xi$, and use has been made of Eq. (20), and the fact that the eigenstates are mutually orthogonal. Note that all of the above results follow from the extremely useful (and easily proved) result
(54)
where 1 denotes the identity operator. The relative probability of a transition to an eigenstate $\mid \xi' \rangle$, which is equivalent to the relative probability of a measurement of $\xi$ yielding the result $\xi'$, is
(55)
The absolute probability is clearly
(56)
If the ket $\mid B \rangle$ is normalized such that its norm is unity, then this probability simply reduces to
Observables
We have developed a mathematical formalism which comprises three types of objects--bras, kets, and linear operators. We have already seen that kets can be used to represent the possible states of a microscopic system. However, there is a one to one correspondence between the elements of a ket space and its dual bra space, so we must conclude that bras could just as well be used to represent the states of a microscopic system. What about the dynamical variables of the system (e.g., its position, momentum, energy, spin, etc.)? How can these be represented in our formalism? Well, the only objects we have left over are operators. We, therefore, assume that the dynamical variables of a microscopic system are represented as linear operators acting on the bras and kets which correspond to the various possible states of the system. Note that the operators have to be linear, otherwise they would, in general, spit out bras/kets pointing in different directions when fed bras/kets pointing in the same direction but differing in length. Since the lengths of bras and kets have no physical significance, it is reasonable to suppose that non-linear operators are also without physical significance.
We have seen that if we observe the polarization state of a photon, by placing a polaroid film in its path, the result is to cause the photon to jump into a state of polarization parallel or perpendicular to the optic axis of the film. The former state is absorbed, and the latter state is transmitted (which is how we tell them apart). In general, we cannot predict into which state a given photon will jump (except in a statistical sense). However, we do know that if the photon is initially polarized parallel to the optic axis then it will definitely be absorbed, and if it is initially polarized perpendicular to the axis then it will definitely be transmitted. We also known that after passing though the film a photon must be in a state of polarization perpendicular to the optic axis (otherwise it would not have been transmitted). We can make a second observation of the polarization state of such a photon by placing an identical polaroid film (with the same orientation of the optic axis) immediately behind the first film. It is clear that the photon will definitely be transmitted through the second film.
There is nothing special about the polarization states of a photon. So, more generally, we can say that when a dynamical variable of a microscopic system is measured the system is caused to jump into one of a number of independent states (note that the perpendicular and parallel polarization states of our photon are linearly independent). In general, each of these final states is associated with a different result of the measurement: i.e., a different value of the dynamical variable. Note that the result of the measurement must be a real number (there are no measurement machines which output complex numbers). Finally, if an observation is made, and the system is found to be a one particular final state, with one particular value for the dynamical variable, then a second observation, made immediately after the first one, will definitely find the system in the same state, and yield the same value for the dynamical variable.
How can we represent all of these facts in our mathematical formalism? Well, by a fairly non-obvious leap of intuition, we are going to assert that a measurement of a dynamical variable corresponding to an operator $X$ in ket space causes the system to jump into a state corresponding to one of the eigenkets of $X$. Not surprisingly, such a state is termed an eigenstate. Furthermore, the result of the measurement is the eigenvalue associated with the eigenket into which the system jumps. The fact that the result of the measurement must be a real number implies that dynamical variables can only be represented by Hermitian operators (since only Hermitian operators are guaranteed to have real eigenvalues). The fact that the eigenkets of a Hermitian operator corresponding to different eigenvalues (i.e., different results of the measurement) are orthogonal is in accordance with our earlier requirement that the states into which the system jumps should be mutually independent. We can conclude that the result of a measurement of a dynamical variable represented by a Hermitian operator $\xi$ must be one of the eigenvalues of $\xi$. Conversely, every eigenvalue of $\xi$ is a possible result of a measurement made on the corresponding dynamical variable. This gives us the physical significance of the eigenvalues. (From now on, the distinction between a state and its representative ket vector, and a dynamical variable and its representative operator, will be dropped, for the sake of simplicity.)
It is reasonable to suppose that if a certain dynamical variable $\xi$ is measured with the system in a particular state, then the states into which the system may jump on account of the measurement are such that the original state is dependent on them. This fairly innocuous statement has two very important corollaries. First, immediately after an observation whose result is a particular eigenvalue $\xi'$, the system is left in the associated eigenstate. However, this eigenstate is orthogonal to (i.e., independent of) any other eigenstate corresponding to a different eigenvalue. It follows that a second measurement made immediately after the first one must leave the system in an eigenstate corresponding to the eigenvalue $\xi'$. In other words, the second measurement is bound to give the same result as the first. Furthermore, if the system is in an eigenstate of $\xi$, corresponding to an eigenvalue $\xi'$, then a measurement of $\xi$ is bound to give the result $\xi'$. This follows because the system cannot jump into an eigenstate corresponding to a different eigenvalue of $\xi$, since such a state is not dependent on the original state. Second, it stands to reason that a measurement of $\xi$ must always yield some result. It follows that no matter what the initial state of the system, it must always be able to jump into one of the eigenstates of $\xi$. In other words, a general ket must always be dependent on the eigenkets of $\xi$. This can only be the case if the eigenkets form a complete set (i.e., they span ket space). Thus, in order for a Hermitian operator $\xi$ to be observable its eigenkets must form a complete set. A Hermitian operator which satisfies this condition is termed an observable. Conversely, any observable quantity must be a Hermitian operator with a complete set of eigenstates.
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Here we will continue to develop the mathematical formalism of quantum mechanics, using heuristic arguments as necessary. This will lead to a system of postulates which will be the basis of our subsequent applications of quantum mechanics.
Hermitian Operators
An important property of operators is suggested by considering the Hamiltonian for the particle in a box:
$\hat{H}=-\frac{h^2}{2m}\frac{d^2}{dx^2} \label{1}$
Let $f(x)$ and $g(x)$ be arbitrary functions which obey the same boundary values as the eigenfunctions of $\hat{H}$, namely that they vanish at $x = 0$ and $x = a$. Consider the integral
$\int_0^a \! f(x) \, \hat{H} \, g(x) \, \mathrm{d}x =-\frac{\hbar^2}{2m} \int_0^a \! f(x) \, g''(x) \, \mathrm{d}x \label{2}$
Now, using integration by parts,
$\int_0^a \! f(x) \, g''(x) \, \mathrm{d}x = - \int_0^a \! f'(x) \, g'(x) \, \mathrm{d}x + \, \Biggl[f(x) \, g'(x) \Biggr]_0^a \label{3}$
The boundary terms vanish by the assumed conditions on $f$ and $g$. A second integration by parts transforms Equation $\ref{3}$ to
$\int_0^a \! f''(x) \, g(x) \, \mathrm{d}x \, - \, \Biggl[f'(x) \, g(x) \Biggr]_0^a$
It follows therefore that
$\int_0^a \! f(x) \, \hat{H} \, g(x) \, \mathrm{d}x=\int_0^a g(x) \, \hat{H} \, f(x) \, \mathrm{d}x \label{4}$
An obvious generalization for complex functions will read
$\int_0^a \! f^*(x) \, \hat{H} \, g(x) \, \mathrm{d}x=\Biggl(\int_0^a g^*(x) \, \hat{H} \, f(x) \, \mathrm{d}x\Biggr)^* \label{5}$
In mathematical terminology, an operator $\hat{A}$ for which
$\int \! f^* \, \hat{A} \, g \, \mathrm{d}\tau=\Biggl(\int \! g^* \, \hat{A} \, f \, \mathrm{d}\tau\Biggr)^* \label{6}$
for all functions $f$ and $g$ which obey specified boundary conditions is classified as hermitian or self-adjoint. Evidently, the Hamiltonian is a hermitian operator. It is postulated that all quantum-mechanical operators that represent dynamical variables are hermitian.
Properties of Eigenvalues and Eigenfunctions
The sets of energies and wavefunctions obtained by solving any quantum-mechanical problem can be summarized symbolically as solutions of the eigenvalue equation
$\hat{H} \, \psi_n=E_n \, \psi_n \label{7}$
For another value of the quantum number, we can write
$\hat{H} \, \psi_m=E_m \, \psi_m \label{8}$
Let us multiply Equation $\ref{7}$ by $\psi_m^*$ and the complex conjugate of Equation $\ref{8}$ by $\psi_n$. Then we subtract the two expressions and integrate over $\mathrm{d}\tau$. The result is
$\int \! \psi_m^* \, \hat{H} \, \psi_n \, \mathrm{d}\tau \, - \, \Biggl(\int \! \psi_n^* \, \hat{H} \, \psi_m \, \mathrm{d}\tau\Biggr)^*=(E_n-E_m^*)\int \! \psi_m^* \, \psi_n \, \mathrm{d}\tau \label{9}$
But by the hermitian property (Equation $\ref{5}$), the left-hand side of Equation $\ref{9}$ equals zero. Thus
$(E_n-E_m^*)\int \! \psi_m^* \, \psi_n \, \mathrm{d}\tau=0 \label{10}$
Consider first the case $m = n$. The second factor in Equation $\ref{10}$ then becomes the normalization integral $\int \! \psi_n^* \, \psi_n \, \mathrm{d}\tau$, which equals 1 (or at least a nonzero constant). Therefore the first factor in Equation $\ref{10}$ must equal zero, so that
$E_n^*=E_n \label{11}$
implying that the energy eigenvalues must be real numbers. This is quite reasonable from a physical point of view since eigenvalues represent possible results of measurement. Consider next the case when $E_m \not= E_n$. Then it is the second factor in Equation $\ref{10}$ that must vanish and
$\int \! \psi_m^* \, \psi_n \, \mathrm{d}\tau=0 \,\,\,\, when \,\, E_m \not= E_n \label{12}$
Thus eigenfunctions belonging to different eigenvalues are orthogonal. In the case that $\psi_m$ and $\psi_n$ are degenerate eigenfunctions, so $m \not= n$ but $E_m = E_n$, the above proof of orthogonality does not apply. But it is always possible to construct degenerate functions that are mutually orthogonal. A general result is therefore the orthonormalization condition
$\int \! \psi_m^* \, \psi_n \, \mathrm{d}\tau=\delta_{mn} \label{13}$
It is easy to prove that a linear combination of degenerate eigenfunctions is itself an eigenfunction of the same energy. Let
$\hat{H} \, \psi_{nk}= E_n \, \psi_{nk}, \,\,\,\,\,\,\, k=1,2,...d \label{14}$
where the $\psi_{nk}$ represent a d-fold degenerate set of eigenfunctions with the same eigenvalue $E_n$. Consider now the linear combination
$\psi = c_1\psi_{n,1} + c_2\psi_{n,2} + ... + c_d\psi_{n,d} \label{15}$
Operating on $\psi$ with the Hamiltonian and using (14), we find
$\hat{H} \, \psi = c_1\hat{H} \,\psi_{n,1} + c_2\hat{H} \,\psi_{n,2} + ... =E_n (c_1\psi_{n,1} + c_2\psi_{n,2} + ... )=E_n \, \psi \label{16}$
which shows that the linear combination $\psi$ is also an eigenfunction of the same energy. There is evidently a limitless number of possible eigenfunctions for a degenerate eigenvalue. However, only d of these will be linearly independent.
Dirac Notation
The term orthogonal has been used both for perpendicular vectors and for functions whose product integrates to zero. This actually connotes a deep connection between vectors and functions. Consider two orthogonal vectors a and b. Then, in terms of their x, y, z components, labeled by 1, 2, 3, respectively, the scalar product can be written
$\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 = 0 \label{17}$
Suppose now that we consider an analogous relationship involving vectors in n-dimensional space (which you need not visualize!). We could then write
${a} \cdot {b}= \sum_{k=0}^{n} a_kb_k = 0 \label{18}$
Finally let the dimension of the space become non-denumerably infinite, turning into a continuum. The sum in Equation $\ref{18}$ would then be replaced by an integral such as
$\int \! a(x) \, b(x) dx = 0 \label{19}$
But this is just the relation for orthogonal functions. A function can therefore be regarded as an abstract vector in a higher-dimensional continuum, known as Hilbert space. This is true for eigenfunctions as well. Dirac denoted the vector in Hilbert space corresponding to the eigenfunction $\psi_n$ by the symbol $|n \rangle$. Correspondingly, the complex conjugate $\psi_m^*$ is denoted by $\langle m|$. The integral over the product of the two functions is then analogous to a scalar product (or inner product in linear algebra) of the abstract vectors, written
$\int \! \psi_m^* \, \psi_n \, \mathrm{d}\tau= \langle m| \cdot |n \rangle\equiv \langle m|n\rangle \label{20}$
The last quantity is known as a bracket, which led Dirac to designate the vectors $\langle m|$ and $|n \rangle$ as a "bra" and a "ket," respectively. The orthonormality conditions (Equation $\ref{13}$) can be written
$\langle m|n\rangle = \delta_{mn} \label{21}$
The integral of a "sandwich" containing an operator $\hat{A}$ can be written very compactly in the form
$\int \! \psi_m^* \, \hat{A} \, \psi_n \, \mathrm{d}\tau=\langle m| A |n \rangle \label{22}$
The hermitian condition on $\hat{A}$ [cf. Eq (6)] is therefore expressed as
$\langle m| A |n \rangle=\langle n| A |m \rangle^* \label{23}$
Expectation Values
One of the extraordinary features of quantum mechanics is the possibility for superpositions of states. The state of a system can sometimes exist as a linear combination of other states, for example,
$\psi = c_1\psi_{1} + c_2\psi_{2} \label{24}$
Assuming that all three functions are normalized and that $\psi_1$ and $\psi_2$ are orthogonal, we find
$\int \! \psi^* \, \psi \, \mathrm{d}\tau=|c_1|^2 + |c_2|^2=1 \label{25}$
We can interpret $|c_1|^2$ and $|c_2|^2$ as the probabilities that a system in a state described by $\psi$ can have the attributes of the states $\psi_1$ and $\psi_2$, respectively. Suppose $\psi_1$ and $\psi_2$ represent eigenstates of an observable $A$, satisfying the respective eigenvalue equations
$\hat{A} \psi_1=a_1\psi_1 \,\,\,\,\,\, and \,\,\,\,\,\, \hat{A} \psi_2=a_2\psi_2 \label{26}$
Then a large number of measurements of the variable $A$ in the state $\psi$ will register the value $a_1$ with a probability $|c_1|^2$ and the value $a_2$ with a probability $|c_2|^2$. The average value or expectation value of $A$ will be given by
$\langle{A}\rangle =|c_1|^2 a_1+|c_2|^2 a_2 \label{27}$
This can be obtained directly from $\psi$ by the "sandwich construction"
$\langle{A}\rangle=\int \! \psi^* \hat{A} \, \psi \, \mathrm{d}\tau \label{28}$
or, if $\psi$ is not normalized,
$\langle{A}\rangle=\frac{\int \! \psi^* \hat{A} \, \psi \, \mathrm{d}\tau}{\int \! \psi^* \, \psi \, \mathrm{d}\tau} \label{29}$
Note that the expectation value need not itself be a possible result of a single measurement (like the centroid of a donut, which is located in the hole!). When the operator $\hat{A}$ is a simple function, not containing differential operators or the like, then Equation $\ref{28}$ reduces to the classical formula for an average value:
$\langle{A}\rangle=\int \, A \, \rho \,\mathrm{d}\tau \label{30}$
More on Operators
An operator represents a prescription for turning one function into another: in symbols, $\hat{A}\psi=\phi$. From a physical point of view, the action of an operator on a wavefunction can be pictured as the process of measuring the observable $A$ on the state $\psi$. The transformed wavefunction $\phi$ then represents the state of the system after the measurement is performed. In general, $\phi$ is different from $\psi$, consistent with the fact that the process of measurement on a quantum system produces an irreducible perturbation of its state. Only in the special case that $\psi$ is an eigenstate of $A$, does a measurement preserve the original state. The function $\phi$ is then equal to an eigenvalue $a$ times $\psi$.
The product of two operators, say $\hat{A}\hat{B}$, represents the successive action of the operators, reading from right to left---i.e., first $\hat{B}$ then $\hat{A}$. In general, the action of two operators in the reversed order, say $\hat{B}\hat{A}$, gives a different result, which can be written
$\hat{A}\hat{B}\not=\hat{B}\hat{A}.$
We say that the operators do not commute. This can be attributed to the perturbing effect one measurement on a quantum system can have on subsequent measurements. An example of non-commuting operators from everyday life. In our usual routine each morning, we shower and we get dressed. But the result of carrying out these operations in reversed order will be dramatically different!
The commutator of two operators is defined by
$\left[ \hat{A}, \, \hat{B} \, \right] \equiv \hat{A}\hat{B}-\hat{B}\hat{A} \label{31}$
When $\left[ \hat{A}, \, \hat{B}\, \right]=0$, the two operators are said to commute. This means their combined effect will be the same whatever order they are applied (like brushing your teeth and showering).
The uncertainty principle for simultaneous measurement of two observables $A$ and $B$ is closely related to their commutator. The uncertainty $\Delta a$ in the observable $A$ is defined in terms of the mean square deviation from the average:
$(\Delta a)^2 = \langle{(\hat{A}-\langle{A}\rangle)^2}\rangle=\langle{A^2}\rangle-\langle{A}\rangle^2 \label{32}$
It corresponds to the standard deviation in statistics. The following inequality can be proven for the product of two uncertainties:
$\Delta{a}\Delta{b} \ge \frac{1}{2}|\langle{\left[ \hat{A}, \, \hat{B}\, \right]}\rangle| \label{33}$
The best known application of Equation $\ref{33}$ is to the position and momentum operators, say $\hat{x}$ and $\hat{p_x}$. Their commutator is given by
$[ \hat{x}, \, \hat{p_x} \, ] = i\hbar \label{34}$
so that
$\Delta{x}\Delta{p} \ge \frac{\hbar}{2} \label{35}$
which is known as the Heisenberg uncertainty principle. This fundamental consequence of quantum theory implies that the position and momentum of a particle cannot be determined with arbitrary precision--the more accurately one is known, the more uncertain is the other. For example, if the momentum is known exactly, as in a momentum eigenstate, then the position is completely undetermined.
If two operators commute, there is no restriction on the accuracy of their simultaneous measurement. For example, the $x$ and $y$ coordinates of a particle can be known at the same time. An important theorem states that two commuting observables can have simultaneous eigenfunctions. To prove this, write the eigenvalue equation for an operator $\hat{A}$
$\hat{A} \, \psi_n=a_n \, \psi_n \label{36}$
then operate with $\hat{B}$ and use the commutativity of $\hat{A}$ and $\hat{B}$ to obtain
$\hat{B} \, \hat{A} \, \psi_n=\hat{A} \, \hat{B} \, \psi_n=a_n \, \hat{B} \, \psi_n \label{37}$
This shows that $\hat{B} \, \psi_n$ is also an eigenfunction of $\hat{A}$ with the same eigenvalue $a_n$. This implies that
$\hat{B} \, \psi_n=const \, \psi_n=b_n \, \psi_n \label{38}$
showing that $\psi_n$ is a simultaneous eigenfunction of $\hat{A}$ and $\hat{B}$ with eigenvalues $a_n$ and $b_n$, respectively. The derivation becomes slightly more complicated in the case of degenerate eigenfunctions, but the same conclusion follows.
After the Hamiltonian, the operators for angular momenta are probably the most important in quantum mechanics. The definition of angular momentum in classical mechanics is $\mathbf{L} = \mathbf{r} \times \mathbf{p}$. In terms of its Cartesian components,
$L_x = yp_z - zp_y$L_y = zp_x - xp_z$L_z = xp_y - yp_x \label{39}$
In future, we will write such sets of equation as "$L_x = yp_z - zp_y, \, et \, cyc$," meaning that we add to one explicitly stated relation, the versions formed by successive cyclic permutation $x \rightarrow y \rightarrow z \rightarrow x$. The general prescription for turning a classical dynamical variable into a quantum-mechanical operator was developed in Chap 2. The key relations were the momentum components
$\hat{p_x}=-i \hbar \frac{\partial}{\partial x}, \,\,\, \hat{p_y}=-i \hbar \frac{\partial}{\partial y}, \,\,\, \hat{p_z}=-i \hbar \frac{\partial}{\partial z} \label{40}$
with the coordinates $x, y, z$ simply carried over into multiplicative operators. Applying Equation $\ref{40}$ to Equation $\ref{39}$, we construct the three angular momentum operators
$\hat{L_x}=-i \hbar \, \left( y \frac{\partial}{\partial z}-z \frac{\partial}{\partial y}\right) \,\,\,\,\,\, et \, cyc \label{41}$
while the total angular momentum is given by
$\hat{L}^2=\hat{L}_x^2+\hat{L}_y^2+\hat{L}_z^2 \label{42}$
The angular momentum operators obey the following commutation relations:
$\left[ \hat{L_x}, \, \hat{L_y}\right]=i \hbar \hat{L_z} \,\,\,\, et \, cyc \label{43}$
but
$\left[ \hat{L}^2, \, \hat{L_z}\right]=0 \label{44}$
and analogously for $\hat{L_x}$ and $\hat{L_y}$. This is consistent with the existence of simultaneous eigenfunctions of $\hat{L}^2$ and any one component, conventionally designated $\hat{L_z}$. But then these states cannot be eigenfunctions of either $\hat{L_x}$ or $\hat{L_y}$.
Postulates of Quantum Mechanics
Our development of quantum mechanics is now sufficiently complete that we can reduce the theory to a set of five postulates.
Postulate 1: Wavefunctions
The state of a quantum-mechanical system is completely specified by a wavefunction $\Psi$ that depends on the coordinates and time. The square of this function $\Psi^* \Psi$ gives the probability density for finding the system with a specified set of coordinate values.
The wavefunction must fulfill certain mathematical requirements because of its physical interpretation. It must be single-valued, finite and continuous. It must also satisfy a normalization condition
$\int \! \Psi^* \, \Psi \, \mathrm{d}\tau=1 \label{45}$
Postulate 2: Observables
Every observable in quantum mechanics is represented by a linear, hermitian operator.
The hermitian property was defined in Equation \ref{6}. A linear operator is one which satisfies the identity
$\hat{A} (c_1\psi_{1} + c_2\psi_{2})=c_1 \, \hat{A} \psi_{1} + c_2 \, \hat{A} \psi_{2} \label{46}$
which is required in order to have a superposition property for quantum states. The form of an operator which has an analog in classical mechanics is derived by the prescriptions
$\mathbf{\hat{r}}=\mathbf{r}, \,\,\,\,\,\, \mathbf{\hat{p}}=-i \hbar \nabla \label{47}$
which we have previously expressed in terms of Cartesian components [cf. Equation $\ref{40}$].
Postulate 3: Eigenstates
In any measurement of an observable $A$, associated with an operator $\hat{A}$, the only possible results are the eigenvalues $a_n$, which satisfy an eigenvalue equation
$\hat{A} \psi_n=a_n \, \psi_n \label{48}$
This postulate captures the essence of quantum mechanics--the quantization of dynamical variables. A continuum of eigenvalues is not forbidden, however, as in the case of an unbound particle.
Every measurement of $A$ invariably gives one of the eigenvalues. For an arbitrary state (not an eigenstate of $A$), these measurements will be individually unpredictable but follow a definite statistical law, which is the subject of the fourth postulate:
Postulate 4: Expectation Values
For a system in a state described by a normalized wave function $\Psi$, the average or expectation value of the observable corresponding to $A$ is given by
$\langle{A}\rangle=\int \! \Psi^* \, \hat{A} \, \Psi \, \mathrm{d}\tau \label{49}$
Finally,
Postulate 5: Time-dependent Evolution
The wavefunction of a system evolves in time in accordance with the time-dependent Schrödinger equation
$i\hbar \frac{\partial \Psi}{\partial t}=\hat{H} \, \Psi \label{50}$
For time-independent problems this reduces to the time-independent Schrödinger equation
$\hat{H} \, \psi=E \, \psi \label{51}$
which is the eigenvalue equation for the Hamiltonian operator.
The Variational Principle
Except for a small number of intensively-studied examples, the Schrödinger equation for most problems of chemical interest cannot be solved exactly. The variational principle provides a guide for constructing the best possible approximate solutions of a specified functional form. Suppose that we seek an approximate solution for the ground state of a quantum system described by a Hamiltonian $\hat{H}$. We presume that the Schrödinger equation
$\hat{H} \, \psi_0=E \, \psi_0 \label{52}$
is too difficult to solve exactly. Suppose, however, that we have a function $\tilde{\psi}$ which we think is an approximation to the true ground-state wavefunction. According to the variational principle (or variational theorem), the following formula provides an upper bound to the exact ground-state energy $E_0$:
$\tilde{E} \equiv \frac{\int \! \tilde{\psi}^* \hat{H} \, \tilde{\psi} \, \mathrm{d}\tau}{\int \! \tilde{\psi}^* \, \tilde{\psi} \, \mathrm{d}\tau} \ge E_0 \label{53}$
Note that this ratio of integrals has the same form as the expectation value $\langle{H}\rangle$ defined by Equation $\ref{29}$. The better the approximation $\tilde{\psi}$, the lower will be the computed energy $\tilde{E}$, though it will still be greater than the exact value. To prove Equation $\ref{53}$, we suppose that the approximate function can, in concept, be represented as a superposition of the actual eigenstates of the Hamiltonian, analogous to Equation $\ref{24}$,
$\tilde{\psi}=c_0\psi_0+c_1\psi_1+... \label{54}$
This means that $\tilde{\psi}$, the approximate ground state, might be close to the actual ground state $\psi_0$ but is "contaminated" by contributions from excited states $\psi_1$, ... Of course, none of the states or coefficients on the right-hand side is actually known, otherwise there would be no need to worry about approximate computations. By Equation $\ref{25}$, the expectation value of the Hamiltonian in the state Equation $\ref{54}$ is given by
$\tilde{E}=|c_0|^2E_0+|c_1|^2E_1+... \label{55}$
Since all the excited states have higher energy than the ground state, $E_1, \, E_2... \ge E_0$, we find
$\tilde{E} \ge (|c_0|^2+|c_1|^2+...) \, E_0=E_0 \label{56}$
assuming $\tilde{\psi}$ has been normalized. Thus $\tilde{E}$ must be greater than the true ground-state energy $E_0$, as implied by Equation $\ref{53}$.
As a very simple, although artificial, illustration of the variational principle, consider the ground state of the particle in a box. Suppose we had never studied trigonometry and knew nothing about sines or cosines. Then a reasonable approximation to the ground state might be an inverted parabola such as the normalized function
$\tilde{\psi}(x)=\left( \frac{30}{a^5} \right)^\frac{1}{2} \, x(a-x) \label{57}$
Fig. 1 shows this function along with the exact ground-state eigenfunction
$\psi_1 (x)=\left( \frac{2}{a} \right)^\frac{1}{2} \, sin \frac{\pi x}{a} \label{58}$
$\frac{x}{a}$
Figure $1$: Variational approximation for particle in a box. Red line represents $\tilde{\psi}$ and black line represents $\psi_1$
A variational calculation gives
$\tilde{E}=\int^a_0 \tilde{\psi} (x) \, \left( -\frac{\hbar^2}{2m} \right) \, \tilde{\psi}''(x) \, dx = \frac{5}{4\pi^2}\frac{h^2}{ma^2}=\frac{10}{\pi^2} \, E_1 \approx 1.01321E_1 \label{59}$
in terms of the exact ground state energy $E_1 = \frac{h^2}{8ma^2}$. In accord with the variational theorem, $\tilde{E} > E_1$. The computation is in error by about 1%.
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One of the postulates is that all measurable quantities in a quantum system are represented mathematically by so called observables. An observable is thus a mathematical object, more specifically a real linear operator whose 'eigenstates' form a complete set. This essentially means that any quantum state can be expressed as a linear combination of these eigenstates of the observable.
A simple example of an observable is the spin operator. If we apply the postulate to this case it simply means that any spin state can be expressed as a combination of the eigenstates of the spin operator. If we are talking about the spin of an electron, for example, the eigenstates are 'spin up' and 'spin down' (naively one could think of an electron spinning counterclockwise or clockwise, respectively). So any spin state can be seen as a linear combination of these spin up and spin down states.
Now, when we do a measurement of the spin of a particular electron, we find out what the spin of the electron is at that moment. Another postulate states that the only possible outcomes of such a measurement is an eigenstate. So the only possible results of measuring the spin of an electron is either spin up, or spin down. After this measurement we thus know that the electron has one of these spins, it's previous spin state has 'collapsed' onto one of these states.
Now there are other postulates which explicitly tell us exactly how the state of a quantum system evolves with time. So if we wait a while after we measured the spin state of the electron, it's spin state might have changed if for example it interacts with some other particle. Using the laws of quantum mechanics, we can thus calculate the probabilities of measuring spin up or spin down at a later time. So quantum mechanics really does not state anything about quantum states being constantly observed, or about observation apart from measurement at all for that matter. It is only concerned with measurements of states and evolution of states over time.
Say that we have a complex quantum system consisting of many parts (particles, fields, etc). We can measure some properties of this system at the outset, providing us with a specific initial state of the system. These different parts of the system then might go on to interact with each other and evolve by the laws of quantum mechanics into some new state (i.e. by the Schrödinger equation). After this, we can do new measurements, and we can in principle calculate, exactly, the probabilities of the different possible outcomes of each of these measurements. When we do these new measurements, the probabilities stop being probabilities however and we get a new definite state, the previous 'probabilistic state' has 'collapsed' (the probabilistic state being a linear combination of eigenstates, and the collapsed state a specific eigenstate).
Quantum mechanics really does not state anything about quantum states being constantly observed, or about observation apart from measurement at all for that matter. It is only concerned with measurements of states and evolution of states over time.
How to Collapse the Wavefunction
From the wavefunction, one can determine the probability that a measurement performed on the system will yield a particular result. For instance, if we know the wavefunction of an electron in a hydrogen atom, we can find the probability that a measurement of the electron's position will find it at 1 angstrom away from the nucleus. We can also find the probability that the electron will be found 1 meter away from the nucleus (it's very low), or half an angstrom away from the nucleus (probably higher).
A counter-intuitive property of quantum systems is that each state can be expressed as a linear combination of other eigenstates (this is known as a "superposition"). Given a sufficiently large number of states, every other state can be expressed as a superposition of the original states. When a measurement of a property is carried out, the wavefunction "collapses" to one of the state with a defined value for that property, and the measurement corresponding to that particular state is observed. A system in a superposition of states 1, 3, 5, and 6 might collapse to state 3. The probability of collapsing to a given state is determined by the wavefunction of the system before the collapse.
Example $1$: A Three-state Wavefunction
Suppose that we have a hypothetical quantum system with several eigenstates with well-defined momentum represented by $|1\rangle$, $|2\rangle$, $|3\rangle$, $|4\rangle$, .... Now suppose that the wavefunction of the system $| \Psi \rangle$ is in a superposition of states $|1\rangle$, $|2\rangle$, and $|6\rangle$, i.e.,
$| \Psi \rangle = c_1 | 1 \rangle + c_2 | 2 \rangle + c_6 | 6 \rangle$
Now, say that we want to measure the momentum of the system, which involves applying the momentum operator on the wavefunction. The instant we make the measurement, the wavefunction collapses to one of the three basis eigenstates. If it collapses to $| 1 \rangle$, then the value for momentum associated with the state is measured. If it collapses to $|2\rangle$ instead, we measure a value for momentum associated with that state and the same is true if the system collapses to state $|(6\rangle$.
Note that a superposition of states is never actually observed, since the system collapses to a single state at the instant that a measurement takes place. The superposition can be interpreted as the description of potential measurement outcomes, while the state of the system after the collapse takes place is the actual realized outcome. The collapse can thus be defined as the transition between the potential and the actual. However, the situation is a bit more complicated than that, since whether something is a "superposition" of states, or a "pure" state, depends on the property being measured. A state with a well-defined position will be a superposition of states with well-defined momentum, and a state with well-defined momentum will be a superposition of states with well-defined position (the fact that no state is both well-defined is related to the Heisenberg uncertainty principle).
Example $2$: Schrödinger's Cat
A cat, a flask of poison, and a radioactive source are placed in a sealed box. If an internal monitor detects radioactivity (i.e., a single atom decaying), the flask is shattered, releasing the poison that kills the cat. The Copenhagen interpretation of quantum mechanics implies that after a while, the cat is simultaneously alive and dead. Yet, when one looks in the box, one sees the cat either alive or dead, not both alive and dead. This poses the question of when exactly quantum superposition ends and reality collapses into one possibility or the other.
Schrödinger's cat poses the question, "when does a quantum system stop existing as a superposition of states and become one or the other?" (More technically, when does the actual quantum state stop being a linear combination of states, each of which resembles different classical states, and instead begin to have a unique classical description?)
Collapse is fancy terminology for measurement
Five Interpretations of Collapsing Wavefunctions
So far, the discussion of quantm mechanics has focused on the numerics and mathematics, but there exists a rich effort in discussing the nature of the quantum mechanical world which often more philosophical than mathematical. Below are the five most popular interpretations of quantum mechanics, which the definition of "measurement" differ and hence the interpretation of the wavefunction collapse also differ.
1. The Copenhagen/von Neuman interpretations argues the collapse of the wave function is triggered by the observer. This person has the special property which no other object in universe is capable of. In the Copenhagen interpretation, the collapse can be triggered by any system which is connected to the observer, including the measurement apparatus and external medium (if the observer is not isolated from it). the Copenhagen interpretation his is the most popular interpretation of quantum mechanics.
2. The the von Neuman interpretation argues the collapse of the wave function happens when the observer feels any feeling depended on the measured value.
3. The Bohm interpretation argues collapse of the wavefunction happens when the observer introduces into the measured system some perturbation, which is inevitable when performing the measurement. The difference between the measurement and any other interaction is in that the perturbation introduced by measurement is unknown beforehand.
4. The Relational interpretation argue the collapse happens when the interaction affects the ultimate measurement performed by ultimate observer on the universal wavefunction at infinite future. As such, for the collapse to happen, the result of interaction should somehow affect the external medium, the stars, etc, either now or in the future, rather than being lost.
5. The Many-worlds interpretation argues the wavefunction collapse never happens. Instead what the observer perceives as the collapse is just the event of entanglement of the observer with the observed system.
Figure $2$: The quantum-mechanical "Schrödinger's cat" paradox according to the many-worlds interpretation. In this interpretation, every event is a branch point. The cat is both alive and dead—regardless of whether the box is opened—but the "alive" and "dead" cats are in different branches of the universe that are equally real but cannot interact with each other. Imge used with permission from Wikipedia.
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The expected value (or expectation, mathematical expectation, mean, or first moment) refers to the value of a variable one would "expect" to find if one could repeat the random variable process an infinite number of times and take the average of the values obtained. More formally, the expected value is a weighted average of all possible values.
Expectation value of momentum
$\langle p_z \rangle = \int \Psi^* p_z \Psi\;dz \nonumber$
where
$p_z = -i \hbar \dfrac{\partial}{\partial z} \nonumber$
Example $1$: Free Particles
For a free particle $\Psi= e^{ikz}$.
$-i\hbar\dfrac{\partial}{\partial z}\ e^{ikz}=\hbar k\ e^{ikz} \nonumber$
Example $2$: Bound in a Box
For a bound particle in a box of length $L$, $\Psi = \sqrt{(2/L)}\sin(npz/L)$
$\langle p_z\rangle=\dfrac{2}{L} \left(-i \dfrac{n\pi\hbar}{L}\right)\int_{0}^{L}\sin\left(\dfrac{n\pi\,z}{L}\right)\cos \left(\dfrac{n\pi\,z}{L}\right)dz \nonumber$
$-i\hbar \dfrac{\partial }{\partial z}\sin \left(\frac{n\pi\,z}{L}\right)=-i \dfrac{\hbar\ n\pi}{L}\cos \left(\frac{n\pi\,z}{L}\right) \nonumber$
Following a simple substitution approach to solving this integral, let $u = npz/L$ and $du = np/L dz$, then $dz = L/np du$
$\langle p_z\rangle=\frac{2}{L}(-i\hbar)\int_{0}^{n\pi}\sin(u)\cos(u) du=0 \nonumber$
Note that this makes sense since the particles spends an equal amount of time traveling in the $+x$ and $–x$ direction.
Expectation value of energy
$\langle E_z \rangle=\int\Psi^* T_z\Psi\,dz \nonumber$
where the translation energy operation in the z-direction:
$T_z=-\dfrac{\hbar^2}{2m}\dfrac{\partial^2 }{\partial z^2} \nonumber$
For a bound particle
$-\dfrac{\hbar^2}{2m}\dfrac{\partial^2 }{\partial z^2}\sin\left(\dfrac{n\pi\,z}{L}\right)=\dfrac{\hbar^2n^2\pi^2}{2mL^2}\sin \left(\dfrac{n\pi\,z}{L}\right) \nonumber$
and using the normalized wavefunction
$\langle E_z\rangle=\dfrac{2}{L}\left(\dfrac{\hbar^2n^2\pi^2}{2mL^2}\right)\int_{0}^{L}\sin^2\left(\dfrac{n\pi\,z}{L}\right)\,dz \nonumber$
Let $u = npz/L$ and $du = np/L\, dz$, then $dz = L/np\, du$
\begin{align*} \langle E_z\rangle &=\dfrac{2}{L}\left(\dfrac{\hbar^2n^2\pi^2}{2mL^2}\right)\left(\dfrac{L}{n\pi}\right)\int_{0}^{n\pi}\sin^2(u)\,du \[4pt] &=\dfrac{2}{L}\left(\dfrac{\hbar^2n^2\pi^2}{2mL^2}\right)\left(\dfrac{L}{n\pi}\right)\dfrac{u}{2}\Biggr\rvert_{0}^{n\pi}\&=\dfrac{\hbar^2n^2\pi^2}{2mL^2} \[4pt] &=\dfrac{h^2n^2}{8mL^2} \end{align*}
Expectation value of position
$\langle z\rangle=\int\Psi^* z\Psi\,dz \nonumber$
For the bound particle in a box.
$\langle z\rangle=\frac{2}{L}\int_{0}^{L}\sin^2\left(\frac{n\pi\,z}{L}\right)\,z\,dz \nonumber$
Let $u = npz/L$ and $du = np/L\; dz$, then $dz = L/np\; du$
$\langle z\rangle=\frac{2}{L}\left(\frac{L^2}{n^2\pi^2}\right)\int_{0}^{n\pi}\sin ^2(u)u\,du \nonumber$
$\int\sin^2(x) x\,dx=\frac{x^2}{4}+\frac{\sin^2(x)}{4}-\frac{\cos(x)\sin (x)}{2} \nonumber$
Therefore, $\langle z \rangle = L/2$. This is logical since the average position of the particle is in the middle of the box.
Matrix Representation of Operators and Wavefunctions
Vector Representation of an eigenstate
For a set of vectors $\{|1 \rangle, |2 \rangle, ... | \infty \rangle \}$ that spans the space we are interested in, the arbitrary eigenstate can be decomposed
$| \psi \rangle = \sum_i^n c_i | i \rangle = \begin{pmatrix} c_1 \ c_2 \ ... \ c_n \end{pmatrix} \label{1A}$
The $\{|1 \rangle, |2 \rangle, ... | \infty \rangle \}$ constitutes a basis (one of many possible) for the space.
Most of the operators we are discussed are linear so
$\hat{A} | \psi \rangle = \hat{A} \left( \sum_i^n c_i | i \rangle \right) = \sum_i^n c_i \hat{A} | i \rangle \label{2A}$
Matrix Representation of an Operator
Operators can be expressed as matrices that "operator" on the eigenvector discussed above
$\hat{A} | i \rangle = \sum_i^n A_{ij} | i \rangle \label{3A}$
The number $A_{ij}$ is the $ij^{th}$ matrix element of $A$ in the basis select.
Hermitian Operators
Hermitian operators are operators that satisfy the general formula
$\langle \phi_i | \hat{A} | \phi_j \rangle = \langle \phi_j | \hat{A} | \phi_i \rangle \label{Herm1}$
If that condition is met, then $\hat{A}$ is a Hermitian operator. For any operator that generates a real eigenvalue (e.g., observables), then that operator is Hermitian. The Hamiltonian $\hat{H}$ meets the condition and a Hermitian operator. Equation \ref{Herm1} can be rewriten as
$A_{ij} =A_{ji}$
where
$A_{ij} = \langle \phi_i | \hat{A} | \phi_j \rangle$
and
$A_{ji} = \langle \phi_j | \hat{A} | \phi_i \rangle$
Therefore, when applied to the Hamiltonian operator
$\boxed{H_{ij}^* =H_{ji}.}$
Multiplication
We can define an inner product (dot product) of two eigenstates $| \phi_1 \rangle$ and $| \phi_2\ \rangle$
$\langle \phi_1 | \phi_2 \rangle = \sum_{i=1}^n c_i^* c_j \label{4A}$
which looks like this in vector representations
$\langle \phi_1 | \phi_2 \rangle = ( c_1^* \; c_2^* \; ... c_n^* ) \begin{pmatrix} c_1 \ c_2 \ ... \ c_n \end{pmatrix} \label{5A}$
This form emphasizes the dot product nature of the inner product multiplication.
From equation $\ref{5A}$, we can express the bra form of the eigenstate as
$\langle \phi | = \sum_i^n c_i^* \langle i | \label{6A}$
Completeness Relation
For vectors $|i \rangle$ forming an orthonormal basis $\langle i | j \rangle = \delta_{ij}$ for all space then
$\sum_i^n | i \rangle \langle j| = 1 \label{7A}$
Those terms in this sum are outer products and are matrices (remember inner products are scalars).
Diagonal Representation of an operator
$\hat{A} = \sum_i ^n \lambda_i | i \rangle \langle i| \label{8A}$
This is a matrix that has non-zero element everywhere except the diagonal. E.g.,
$\begin{pmatrix} 4 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -4 \end{pmatrix}$
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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 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. However, in most applications of the wave behavior of matter, we are only interested in the probability of finding the particle in a particular region. This probability in a one dimensional system is defined as:
$\int\psi(x,t)^*\psi(x,t)dxdt\label{1.2}$
Equation \ref{1.2} describes the probability of finding a particle as the integral of the wavefunction times its complex conjugate. The complex conjugate is effective in that it negates all imaginary components of the wavefunction. 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.
The Particle in a One Dimensional Box (The Infinite Well)
The particle in a box is a very simple model developed in order to understand the behavior of a particle (i.e. an electron) when confined to a small region of length, L, with infinite potential barriers, V, at x=0 and x=L, as seen in Figure 1.
When a particle of wavelength λ is placed in a box with length L ≈ λ, the particle begins to exhibit significant wave-like characteristics, as seen in Figure 1.
Now, in this model, we first assume that the particle has zero potential energy and is bound in a well with barriers of infinite potential energy. Therefore, we may negate the potential energy, V, component of Equation \ref{1.3}:
$\dfrac{-\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} = E\psi(x) \label{1.4}$
Now, if we rearrange Equation \ref{1.4}, we obtain:
$\frac{d^2\psi(x)}{dx^2} = \dfrac{-2mE}{\hbar^2}\psi(x) \label{1.5}$
Equation \ref{1.5} is a second order differential equation, in which we are interested in solving for Ψ and E. Looking at Equation \ref{1.4}, we see that we have a wavefunction, Ψ, whose second derivative is equal to the negative of itself times some constants. This means that Ψ must be a trigonometric function, represented as:
$\psi(x) = A\sin(kx) + B\cos(kx) \label{1.6}$
Next, we mandate that the wavefunction be continuous. Since we have infinite potential energy barriers, there is zero probability that the particle would be found at x<0 or x>L. Therefore, we must have Ψ(0)=0 and Ψ(L)=0. Applying these conditions to Equation \ref{1.6}, we have
$0 = \psi(0) = A\sin(0) + B\cos(0)$
Because cos(0)≠ 0, we must have B=0, so this means that Ψ cannot involve a cosine function. Thus, Equation \ref{1.6} may be reduced to
$\psi(x) = Asin(kx) \label{1.7}$
We now differentiate $\psi(x)$ twice to solve for k; we obtain
$\frac{d\psi(x)}{dx} = kAcos(kx)$
$\frac{d^2\psi(x)}{dx^2} = -k^2Asin(kx)$
$\frac{d^2\psi(x)}{dx^2} = -k^2\psi(x) \label{1.8}$
Now if we relate Equation \ref{1.5} to Equation \ref{1.8}, we have
$k^2 = \frac{2mE}{\hbar^2}$
$k = \frac{\sqrt{2mE}}{\hbar} \label{1.9}$
Because the wavefunction is limited in the confines of the infinite well, the wavenumber, k, can only take on certain discrete values that would allow Ψ(0)=0 and Ψ(L)=0. Therefore, it is found that
$k=\frac{n\pi}{L} \label{1.10}$
where L is the length of the well and n is the energy level or fundamental quantum number of the particle in the well. If we now set Equation 1.9 equal to Equation 1.10 and solve for the energy we obtain
$E=\frac{n^2\pi^2\hbar^2}{2mL^2} \label{1.11}$
Equation \ref{1.11} is a very useful equation because it relates the energy of a particle in a system to the size of its confines, L, its mass, m, and its energy level, n.
Now that we have solved for the Energy of a particle in an infinite well, we can return to solving for the wavefunction Ψ(x). By substituting Equation 1.10 into Equation 1.7, we will have
$\psi(x) = Asin(\frac{n\pi x}{L}) \label{1.12}$
Finally in order to solve for A, we can apply the condition of normalizability; that is, the probability of finding the wavefunction between x=0 and x=L must be equal to unity. So, we have
$1=\int \psi^2(x)dx=\int (Asin(\frac{n\pi x}{L}))^2dx \label{1.13}] Equation \ref{1.13} can be used to solve for the normalization constant, A, while making use of the identity wherein $sin^2(ax)= \frac{x}{2} - \frac{sin(2ax)}{4a}$. We will get \[|A|=\sqrt{\frac{2}{L}}$
Therefore, we can now rewrite Equation \ref{1.10} as
$\psi(x) = \sqrt{\frac{2}{L}}sin(\frac{n\pi x}{L}) \label{1.14}$
We now have solved the Schrödinger Equation for the Particle in a Box, and can apply Equations \ref{1.11} and \ref{1.14}, which describe the wavefunction, Ψ, and the energy, E, respectively, to many applications, especially when trying to understand the behaviors of quantum mechanical particles in small confines (i.e. an electron orbiting a nucleus).
Particle in a Finite Well
What distinguishes the Finite Well from the Particle in a Box scenarios are that for the finite well, the potential energy, V, of the barriers do not approach infinity. As a result, we find that the particle is not totally restricted to the region between the barrier, as shown in Figure 2 below.
In solving the Schrödinger Equation for the Finite Well, we cannot apply some of the same conditions that were imposed on the Particle in a Box; in fact, the Schrödinger Equation for a Finite Well cannot be solved exactly, but must employ numeric approximations. However, it is still useful to show the basic functions that describe the wave-like particle. So, we must reapply the same methodology that we used for the Particle in a Box. For the region between x=0 and x=L, we have a potential energy, V, equal to zero. Thus, the time-independent Schrödinger Equation can be reduced to
$\frac{-\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} = E\psi(x)$ (2.1)
Moreover for this region inside the box, we again find that the wavefunction is defined as
$\psi(x) = Asin(kx) + Bcos(kx)$ (2.2)
However, we can no longer utilize the same barrier conditions to remove the cosine portion of the wavefunction. So, we will not be able to solve it further without numerical approximations. We can, however, make use of our result in the derivations for the Particle in a Box to obtain a formula for the Energy of a particle in a Finite Well. From Equation , we have
$k = \frac{\sqrt{2mE}}{\hbar}$ (2.3)
Therefore, we obtain
$E=\frac{k^2\hbar^2}{2m}$ (2.4)
For the regions outside the well, x<0 and x>L, we cannot employ the same analysis we used earlier because the potential energy,V, is not infinite but equal to a constant, P. Therefore, we will again solve the regions outside the potential well beginning with the time-independent Schrödinger Equation
$\dfrac{-\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x) = E\psi(x)$ (2.5)
For these regions outside the well, we have the potential energy V(x) = P. Therefore, Equation 2.5 may be rewritten as
$\dfrac{-\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+P\psi(x) = E\psi(x)$ (2.6)
Rearranging we have
$\dfrac{d^2\psi(x)}{dx^2} = \frac{2m(P-E)\psi(x)}{\hbar^2}$ (2.7)
Now, for the regions outside the well, the particle may have energy that is greater than or less than the potential energy of the well. Where, E>P, the particle is said to be in a free state and we again find that
$k = \dfrac{\sqrt{2m(E-P)}}{\hbar}$ (2.8)
and where E<P, the particle is said to be in a bound state and we define
$\alpha = \dfrac{\sqrt{2m(P-E)}}{\hbar}$ (2.9)
In order to describe the wavefunction in the regions outside the well, a closer examination of Equation 2.7 is needed. We have a second order differential equation where the second derivative of the wavefunction, Ψ(x), is equal the positive of the wavefunction times some constants. Trigonometric functions do not satisfy this condition, but exponentials do. Thus, we have
$\psi(x) = Ce^{-\alpha x} + De^{\alpha x}$ (2.10)
Now, referring back to Figure 2, we see that in the region, where 0<x<L, the function takes on a trigonometric form. In the region where x<0, we see that as x goes to -∞, the C term goes to infinity. Since, the wavefunction must have a finite total integral, the C term must be negated; so we have
$\psi(x) = De^{\alpha x}$ where x<0. (2.11)
By the same condition, we see that for the region where x>L, as x goes to ∞, the D term goes to infinity. As a result, this D term must be negated. Thus, we have
$\psi(x) = Ce^{-\alpha x}$ where x>L (2.12)
Solving for the constants, C and D, is not possible without numerical approximations, so we will be sufficed with Equations 2.11 and 2.12 as descriptions of the wavefunction, Ψ, in the regions outside the Finite Well.
The Hydrogen Atom
The solutions to the Infinite Well and Finite Well are useful for describing the behavior of a particle when confined to a small region of space with large and small potential barriers, respectively. However, this one dimensional treatment of the displacement of a particle in space does not provide a comprehensive understanding of the behavior of an electron orbiting a nucleus (i.e. a hydrogen atom). Beginning with Equation 1.1, we can describe the wavefunction in terms of the three dimensions, x, y, and z, as follows
\]\frac{-\hbar^2}{2m_e}[\frac{\partial^2\psi}{\partial x^2} + \frac{\partial^2\psi}{\partial y^2} +\frac{\partial^2\psi}{\partial z^2}] + V(x,y,z)\psi(x,y,z) = E\psi(x,y,z) \label{3.1\]
Treatment of Equation 3.1 in Spherical Polar Coordinates significantly facilitates the problem. So, after some rearranging, we have
$\frac{\partial^2\psi}{\partial r^2} + \frac{2}{r}\frac{\partial\psi}{\partial r} + \frac{1}{r^2sin\theta}\frac{\partial[sin\theta(\partial\psi/\partial\theta)]}{\partial\theta} + \frac{1}{r^2 sin\theta}\frac{\partial^2\psi}{\partial\phi^2} + \frac{2m_e}{\hbar^2}(E+\frac{e^2}{4\pi\epsilon_0 r})\psi = 0 \label{3.2}$
Equation 3.2 provides a very complex description for the wavefunction of an electron orbiting in a Hydrogen Atom. Fortunately, it has already been solved, and so we will simply examine the solutions. But first, it is important to note that the wavefunction can be described as the product of three distinct functions
$\psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)\label{3.3}] where $R(r)$ is the radial part and $\Theta(\theta)\Phi(\phi)$ is the angular part. A deeper analysis of Equation \ref{3.2} reveals three quantum numbers, $n$, $l$, and $m_l$. The fundamental quantum number,$n$, describes the size of the orbital in which the electron orbits. The angular momentum quantum number, $l$, describes the shape of the orbital. Finally, the magnetic quantum number, $m_l$, describes the orientation of the orbital in space. Now, for any given value of n (n = 1,2,3,...), there are n values of l, which are given by l = 0,1,2,...,(n-1). While, the magnetic quantum number, $m_l$, takes on values of $m_l$=0, $\pm$1, $\pm$2,...$\pm l$. A more in-depth analysis of Equation 3.2, which is beyond the scope of this module, allows for a derivation of the energy of an electron in a Hydrogen atom. We will just make use of the result where \[E_n = -hcR_H (\frac{1}{n^2}); n =1,2,3,... \label{3.4}$
or
$E_n = -13.6eV (\frac{1}{n^2}) ; n =1,2,3,...\label{3.5}$
As Equations \ref{3.4} and \ref{3.5} show, the energy of an electron in a Hydrogen atom is only dependent on the fundamental quantum number, n. This is not the case for more complex atoms, whose energy also depends on the angular momentum quantum number, $l$.
By solving the Schrödinger Equation for the Infinite Well, Finite Well, and the Hydrogen Atom, we are able to establish models that allow for a good understanding of the behavior of a particle when confined to a small region of space, whose length is proportional to the Planck wavelength of the particle. By restricting the particle to portential wells, we are able to derive the particle's wavefunction--both inside and outside of the well--and the quantum mechanical energy that the particle possesses due to its wave-like behavior.
Problems
1. Calculate the ground state energy of a baseball that weighs 145 grams that is in a football field 100 yards long.
Solution:
We have m = 0.145 kg and L = $100yards(\frac{0.9144meter}{1 yard} = 91.44meters$
Plugging these values into Equation 1.11, we have
$E_{baseball}=\dfrac{1^2\pi^2(1.05x10^{-34})^2}{2(0.145)(91.44)^2}= 4.49x10^{-71}J$
2. Calculate the energy of an electron in a well that is 1 Å long.
Solution:
We apply the same technique as in Problem 1
$E_{electron}=\dfrac{1^2\pi^2(1.05x10^{-34})^2}{2(9.11x10^{-31})(1x10^{-10})^2}= 5.69x10^{16}J$
Note: It is important to realize that in the case of the baseball in a football field, the quantum mechanical energies associated with its wavelike behavior are so small that they are essetially negligible. This is why we see a continuum of energy associated with its motion, whereas the electron has defined quanta of energy in which it can orbit.
3. Derive the wavefunction, Ψ, and energy, E, for the Particle in a 1 dimensional box (infinite well), while describing any conditions that must be imposed on the wavefunction.
Solution:
Refer to Section 1 above.
4. Find the probability that an electron in the n=2 state of an infinite well (1 Å long) is found between x=L/4 and x=3L/4.
Solution:
Although this problem could be solved by integration as follows
$P = \int^{3L/4}_{L/4}\psi(x)^2dx$
where
$\psi(x) = \sqrt\frac{2}{L}sin(\frac{n\pi}{L}x) \label{P.1}$
it can be more readily solved by taking a look at the graph of the probability of the wavefunction in an infinite well as seen in Figure P.4 below
As Figure P.4 shows, we can infer the probability of finding the particular between x=L/4 and x=3L/4 to be P=1/2, simply because of symmetry. This assessment only works for symmetrical wavefunction distributions. For all others, Equation P.1 must be used.
5. For an electron in the n=4 energy level, what values of $l$ and $m_l$ are allowed?
Solution:
$l = 0,1,2, and 3$
$m_l = -3,-2,-1,0,1,2,3$
Contributors and Attributions
• John Paul Aboubechara, UC Davis 2012
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The following summary of the mathematical framework of quantum mechanics.
04. The 7 Postulates of Quantum Mechanics
Every physically-realizable state of the system is described in quantum mechanics by a state function $\psi$ that contains all accessible physical information about the system in that state.
Discussion
The properties of a quantum mechanical system are determined by a wavefunction Ψ(r,t) that depends upon the spatial coordinates of the system and time, $r$ and $t$. For a single particle system, r is the set of coordinates of that particle $r = (x_1, y_1, z_1)$. For more than one particle, $r$ is used to represent the complete set of coordinates $r = (x_1, y_1, z_1, x_2, y_2, z_2,\dots x_n, y_n, z_n)$. Since the state of a system is defined by its properties, $\Psi$ specifies or identifies the state and sometimes is called the state function rather than the wavefunction.
Postulate 2: Quantum Mechanics
If a system is in a quantum state represented by a wavefunction $\psi$, then
$P = \int \Psi^2 \;dV$
is the probability that in a position measurement at time $t$ the particle will be detected in the infinitesimal volume $dV$.
Discussion
The wavefunction is interpreted to be the probability amplitude and the absolute square of the wavefunction, $Ψ^*(r,t)Ψ(r,t)$, is interpreted to be the probability density at time t. A probability density times a volume is a probability, so for one particle
$\Psi^*(x_1,y_1,z_1,t)\Psi(x_1,y_1,z_1,t)dx_1dy_1dz_1$
is the probability that the particle is in the volume $dx\;dy\;dz$ located at $x_l, y_l, z_l$ at time $t$. For a many particle system, we write the volume element as $dτ = dx_1dy_1dz_1\dots dx_ndy_ndz_n$; and $Ψ^*(r,t)Ψ(r,t)dτ$ is the probability that particle 1 is in the volume $dx_ldy_ldz_1$ at $x_ly_lz_l$ and particle 2 is in the volume $dx_2dy_2dz_2$ at $x_2y_2z_2$, etc. Because of this probabilistic interpretation, the wavefunction must be normalized.
$\int \limits _{all space} \Psi ^* (r, t) \psi (r , t) d \tau = 1 \tag {3-38}$
The integral sign here represents a multi-dimensional integral involving all coordinates: $x_l \dots z_n$.
.
Postulate 3: Quantum Mechanics
Every observable in quantum mechanics is represented by an operator which is used to obtain physical information about the observable from the state function. For an observable that is represented in classical physics by a function $Q(x,p)$, the corresponding operator is $Q(\hat{x},\hat{p})$.
Discussion
For every observable property of a system there is a quantum mechanical operator. The operator for position of a particle in three dimensions is just the set of coordinates x, y, and z, which is written as a vector
$r = (x , y , z ) = x \vec {x} + y \vec {y} + z \vec {z} \tag {3.1}$
The operator for a component of momentum is
$\hat {P} _x = -i \hbar \dfrac {\partial}{\partial x} \tag {3.2}$
and the operator for kinetic energy in one dimension is
$\hat {T} _x = \left ( -\dfrac {\hbar ^2}{2m} \right ) \dfrac {\partial ^2}{\partial x^2} \tag {3.3}$
and in three dimensions
$\hat {p} = -i \hbar \nabla \tag {3.4}$
and
$\hat {T} = \left ( -\dfrac {\hbar ^2}{2m} \right ) \nabla ^2 \tag {3.5}$
The Hamiltonian operator $\hat{H}$ is the operator for the total energy. In many cases only the kinetic energy of the particles and the electrostatic or Coulomb potential energy due to their charges are considered, but in general all terms that contribute to the energy appear in the Hamiltonian. These additional terms account for such things as external electric and magnetic fields and magnetic interactions due to magnetic moments of the particles and their motion.
Postulate 4: Quantum Mechanics
The time development of the state functions of an isolated quantum system is governed by the time-dependent Schrödinger equation
$\hat {H} (r , t) \psi (r , t) = i \hbar \dfrac {\partial}{\partial t} \Psi (r , t ) \tag {4.1}$
where $H=T+V$ is the Hamiltonian of the system.
Discussion
The time evolution or time dependence of a state is found by solving the time-dependent Schrödinger equation (Eq. 4.1). For the case where $\hat{H}$ is independent of time, the time dependent part of the wavefunction is $e^{-i\omega t}$ where $\omega = \frac {E}{ħ}$ or equivalently $\nu = \frac {E}{h}$, which shows that the energy-frequency relation used by Planck, Einstein, and Bohr results from the time-dependent Schrödinger equation. This oscillatory time dependence of the probability amplitude does not affect the probability density or the observable properties because in the calculation of these quantities, the imaginary part cancels in multiplication by the complex conjugate.
Postulate 5: Quantum Mechanics
If a system is in a state described by a wave function $\psi$, then the average value of the observable corresponding to the $\hat{A}$ operator is given by
$\langle A \rangle = \dfrac{\int_{\infty}^{\infty} \psi^* \hat{A} \psi \;d\tau}{\int_{\infty}^{\infty} \psi^* \psi \;d\tau} \tag{5.1}$
If the wavefunction is normalizedt, then this expression simplifies to
$\langle A \rangle = \int_{\infty}^{\infty} \psi^* \hat{A} \psi \;d\tau \tag{5.2}$
since
$\int_{\infty}^{\infty} \psi^* \psi \;d\tau = 1 \tag{5.3}$
Postulate 6: Quantum Mechanics
If a system is described by the eigenfunction $\Psi$ of an operator $\hat{A}$ then the value measured for the observable property corresponding to $\hat{A}$ will always be the eigenvalue $a$, which can be calculated from the eigenvalue equation.
$\hat {A} \Psi = a \Psi \tag {6.1}$
Postulate 7: Quantum Mechanics
If a system is described by a wavefunction $\Psi$, which is not an eigenfunction of an operator $\hat{A}$, then a distribution of measured values will be obtained, and the average value of the observable property is given by the expectation value integral:
$\left \langle A \right \rangle = \dfrac {\int \Psi ^* \hat {A} \Psi d \tau}{\int \Psi ^* \Psi d \tau} \tag {7.1}$
where the integration is over all coordinates involved in the problem. The average value $\left \langle A \right \rangle$, also called the expectation value, is the average of many measurements. If the wavefunction is normalized, then the normalization integral in the denominator of Eq. 4.1 equals 1, which then converts to
$\left \langle A \right \rangle = \int \Psi ^* \hat {A} \Psi d \tau \tag {7.2}$
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The complexities of quantum mechanics are demonstrated mostly in trapped particles.
• Free-Electron Model
The particle in a box quantum-mechanical problem can provide an instructive application to chemistry: the free-electron model (FEM) for delocalized π -electrons.
• Particle in a 1-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.
• Particle in a 2-Dimensional Box
A particle in a 2-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.
• Particle in a 3-Dimensional box
A particle in a 3-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.
• Particle in a Sphere
Particle on a sphere is one out of the two models that describe rotational motion. A single particle travels on the surface of the sphere. Unlike particle in a box, the particle on a sphere requires angular momentum, J.
• Particle on a Ring
The case of a particle in a one-dimensional ring is similar to the particle in a box.
05.5: Particle in Boxes
The simple quantum-mechanical problem we have just solved can provide an instructive application to chemistry: the free-electron model (FEM) for delocalized $\pi$-electrons. The simplest case is the 1,3-butadiene molecule
$\rho =2\psi _{1}^2+2\psi _{2}^2\label{28}$
A chemical interpretation of this picture might be that, since the $\pi$-electron density is concentrated between carbon atoms 1 and 2, and between 3 and 4, the predominant structure of butadiene has double bonds between these two pairs of atoms. Each double bond consists of a $\pi$-bond, in addition to the underlying $\sigma$-bond. However, this is not the complete story, because we must also take account of the residual $\pi$-electron density between carbons 2 and 3. In the terminology of valence-bond theory, butadiene would be described as a resonance hybrid with the contributing structures CH2=CH-CH=CH2 (the predominant structure) and ºCH2-CH=CH-CH2º (a secondary contribution). The reality of the latter structure is suggested by the ability of butadiene to undergo 1,4-addition reactions.
The free-electron model can also be applied to the electronic spectrum of butadiene and other linear polyenes. The lowest unoccupied molecular orbital (LUMO) in butadiene corresponds to the n=3 particle-in-a-box state. Neglecting electron-electron interaction, the longest-wavelength (lowest-energy) electronic transition should occur from n=2, the highest occupied molecular orbital (HOMO).
The energy difference is given by
$\Delta E=E_{3}-E_{2}=(3^2-2^2)\dfrac{h^2}{8mL^2}\label{29}$
Here m represents the mass of an electron (not a butadiene molecule!), 9.1x10-31 Kg, and L is the effective length of the box, 4x1.40x10-10 m. By the Bohr frequency condition
$\Delta E=h\upsilon =\dfrac{hc}{\lambda }\label{30}$
The wavelength is predicted to be 207 nm. This compares well with the experimental maximum of the first electronic absorption band, $\lambda_{max} \approx$ 210 nm, in the ultraviolet region.
We might therefore be emboldened to apply the model to predict absorption spectra in higher polyenes CH2=(CH-CH=)n-1CH2. For the molecule with 2n carbon atoms (n double bonds), the HOMO → LUMO transition corresponds to n → n + 1, thus
$\dfrac{hc}{\lambda} \approx \begin{bmatrix}(n+1)^2-n^2\end{bmatrix}\dfrac{h^2}{8m(2nL_{CC})^2}\label{31}$
A useful constant in this computation is the Compton wavelength
$\dfrac{h}{mc}= 2.426 \times 10^{-12} m.$
For n=3, hexatriene, the predicted wavelength is 332 nm, while experiment gives $\lambda _{max}\approx$ 250 nm. For n=4, octatetraene, FEM predicts 460 nm, while $\lambda _{max}\approx$ 300 nm. Clearly the model has been pushed beyond range of quantitate validity, although the trend of increasing absorption band wavelength with increasing n is correctly predicted. Incidentally, a compound should be colored if its absorption includes any part of the visible range 400-700 nm. Retinol (vitamin A), which contains a polyene chain with n=5, has a pale yellow color. This is its structure:
Contributors and Attributions
Seymour Blinder (Professor Emeritus of Chemistry and Physics at the University of Michigan, Ann Arbor)
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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.
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. Define the wavefunction
4. Define the allowed energies
Step 1: Define the Potential Energy V
A particle in a 1D infinite potential well of dimension $L$.
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 $x=0$, $\sin(0)=0$, 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:
$\dfrac{d\psi}{dx} = kA\cos(kx) \label{5.5.5}$
$\dfrac{d^{2}\psi}{dx^{2}} = -k^{2}A\sin(kx) \label{5.5.6}$
Since $\psi(x) = Asin(kx)$, 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:
$\psi = A\sin\left(\dfrac{8\pi^{2}mE}{h^{2}}\right)^{1/2}x \label{5.5.9}$
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:
$A^2 \int^{L}_{0}\sin^2\left(\dfrac{n\pi x}{L}\right) dx = 1 \label{5.5.14}$
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 wavefunction for a particle in a 1-dimensional box:
$\psi = \sqrt{\dfrac{2}{L}}\sin{\dfrac{n\pi}{L}}x \label{5.5.16}$
Step 4: Determine the Allowed Energies
Solving for E results in the allowed energies for a particle in a box:
$E_n = \dfrac{n^{2}h^{2}}{8mL^{2}} \label{5.5.17}$
This is an important result that tells us:
1. The energy of a particle is quantized and
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.
Important Facts to Learn from the Particle in the Box
• The energy of a particle is quantized. This means it can only take on discrete 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
Questions
1. Draw the wave function for a particle in a box at the $n = 4$ energy level.
2. Draw the probability distribution for a particle in a box at the $n = 3$ energy level.
3. What is the probability of locating a particle of mass m between $x = L/4$ and $x = L/2$ in a 1-D box of length $L$? Assume the particle is in the $n=1$ energy state.
4. Calculate the electronic transition energy of acetylaldehyde (the stuff that gives you a hangover) using the particle in a box model. Assume that aspirin is a box of length $300 pm$ that contains 4 electrons.
5. Suggest where along the box the $n=1$ to $n=2$ electronic transition would most likely take place.
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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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/05.5%3A_Particle_in_Boxes/Particle_in_a_1-Dimensional_box.txt
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