# Statistical Thermodynamics and Rate Theories/Printable version

Statistical Thermodynamics and Rate Theories

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# Molecular Energy Levels

To predict the properties of gases from their molecular properties, we must first know the different states the molecule can occupy and their energy levels.

# Degrees of freedom

Molecular degrees of freedom refer to the number of ways a molecule in the gas phase may move, rotate, or vibrate in space. Three types of degrees of freedom exist, those being translational, rotational, and vibrational. The number of degrees of freedom of each type possessed by a molecule depends on both the number of atoms in the molecule and the geometry of the molecule, with geometry referring to the way in which the atoms are arranged in space. The number of degrees of freedom a molecule possesses plays a role in estimating the values of various thermodynamic variables using the equipartition theorem. These molecular degrees of freedom essentially describe how a molecule is able to contain and distribute its energy.

Degree of freedom Monatomic Linear molecules Non-linear molecules
Translational 3 3 3
Rotational 0 2 3
Vibrational 0 3N - 5 3N - 6
Total 3 3N 3N

## Translational degrees of freedom

Translational degrees of freedom arise from a gas molecule's ability to move freely in space. A molecule may move in the x, y, and z directions of a Cartesian coordinate system, appearing at a new position in space (relative to a starting position) via translation. A gas molecule is not restricted in which direction it may move, thus it has three translational degrees of freedom. This holds true for all gas molecules, whether they are monatomic, diatomic, or polyatomic, as any molecule may move freely in all directions in three-dimensional space.

## Rotational degrees of freedom

A molecule's rotational degrees of freedom represent the number of unique ways the molecule may rotate in space about its center of mass which a change in the molecule's orientation. These axes may pass through atoms or bonds. By examining the molecule's symmetry (and, by extension, its geometry), the number of rotational degrees of freedom can quickly and easily be determined. A monatomic gaseous molecule such as a noble gas possesses no rotational degrees of freedom, as the center of mass sits directly on the atom and no rotation which creates change is possible.

A diatomic molecule, like H2 or HCl, has two rotational degrees of freedom. The center of mass of a linear molecule rests somewhere between the two terminal atoms. In the case of HCl it exists somewhere along the bond. The center of mass can be taken as the origin of a three-dimensional Cartesian grid, the z axis of which runs along the bond and through the two atoms. Rotation about the x and y axes generates a noticeable change in the molecule's orientation, while rotation about the z axis (analogous to the monatomic case) produces no change in the molecule and is considered a 'lost' rotational degree of freedom.

A polyatomic molecule may have either two or three rotational degrees of freedom, depending on the geometry of the molecule. For a linear polyatomic, such as CO2 or C2H2, the molecule has only two rotational degrees of freedom. The reason for this is discussed in the previous paragraph.

## Vibrational degrees of freedom

The number of vibrational degrees of freedom (or vibrational modes) of a molecule is determined by examining the number of unique ways the atoms within the molecule may move relative to one another, such as in bond stretches or bends. This can be determined mathematically using the ${\displaystyle 3N-6=n_{DOF}}$ rule, where ${\displaystyle N}$ is the number of atoms in the molecule and ${\displaystyle n_{DOF}}$ is the number of vibrational degrees of freedom. Note that for linear molecules, the rule instead becomes ${\displaystyle 3N-5=n_{DOF}}$, meaning a linear polyatomic made up of ${\displaystyle N}$ atoms will have one more vibrational degree of freedom than a non-linear polyatomic with ${\displaystyle N}$ atoms. Application of the rule for linear molecules shows that a diatomic molecule has only one vibrational degree of freedom. This is a logical conclusion, as there is only one bond vibration possible, a stretch of the bond between the two atoms.

As an example, water, a non-linear triatomic, should have three vibrational degrees of freedom, since ${\displaystyle 3(3)-6=3}$. The three vibrational modes corresponding to these three vibrational degrees of freedom can be seen above. These vibrational modes were determined computationally.

# Translational energy

A gas particle can move through space in three independent directions: x, y, and z. Each of these directions is a distinct translational degree of freedom. For ideal gases, these particles can move freely through space in any of these directions until it collides with the wall of its container. For isolated systems, collisions with the wall of the container are assumed to be elastic, meaning no energy is lost upon impact.

## Particle in a 1D Box

The first quantum mechanical model used for describing translation is a particle in a simple 1D box. It is free to move anywhere along one axis (usually assigned to be the x axis) between the arbitrarily assigned boundary limits 0 and a. At distances smaller than 0 and greater than “a” the potential energy function immediately rises to infinity. The particle cannot move past these points. 0 and a represent the walls of the 1D box. We can mathematically assign boundary conditions using this information which allows the derivation of the wave function for a particle in a 1D box. The resulting piece-wise function, wave function and energy level equation are given as follows,

${\displaystyle {\mathcal {V}}(x)={\begin{cases}\infty ,x<0\\0,0{\leq }x{\leq }a\\\infty ,x>a\end{cases}}}$
${\displaystyle \Psi (x)={\sqrt {\frac {2}{a}}}\sin {\left({\frac {n\pi x}{a}}\right)}}$
${\displaystyle n=1,2,3,...}$

As well as the corresponding energy levels of the system.

${\displaystyle \epsilon _{n}={\frac {h^{2}n^{2}}{8ma^{2}}}}$

Note: In the equations above m represents the mass of the gas particle, and h is Planck's constant.

## Zero-Point Energy

The energy levels for translation within the box are quantized, only discrete energy levels can be occupied by the particle at any given time. Each of these energy levels are defined by a single quantum number n. n can take on any integer values starting from 1 and up to a hypothetical infinity. The n = 0 state for a particle in a box does not exist. The particle, in accordance with the Heisenberg uncertainty principle cannot be motionless. If this were to be the case then both the momentum and position of the particle could be determined simultaneously, which is a violation of the principle. The energy of a particle in a 1D box at the lowest translational energy level is therefore non-zero (i.e., n = 1).

## Probability Density Plots and the Correspondence Principle

It is possible to construct a probability density plot of the particle in a 1D box wave function. The plot is characterized by sizable “humps” at low values of n. For example, at n=1 there is one large hump which spans from a minimum at x = 0, a maximum somewhere in the center and a second minimum at x = a. These minimums represent regions of zero probability density. In other words, the particle will not be found in these regions.

As n increases the spacing between humps of the probability density plot becomes smaller and smaller until a near continuum is achieved at sufficiently high values of n. It must also be noted that the magnitude of the energy spacing’s between translational energy levels is extremely small relative to the amount of energy available to a particle under normal conditions. At room temperature ideal gas particles will occupy very high energy levels. As we increase the quantum number to a large enough value the behavior of the system will begin to reproduce that of classical mechanics in that the particle is essentially equally likely to be found anywhere in the box rather than within discrete hump regions seen at lower values of n. This is what is known as the correspondence principle.

## Particle in a 3D Box

The correspondence principle applies to a 3D box as well. For a particle in a 3D box, the particle is able to move in any direction along the x, y and z axes. Since its motion now incorporates a combination of three possible directions the system must include two additional quantum numbers to compensate for the difference. nx, ny, and nz for the x, y, and z dimensions, respectively. Both the wave function and the equation for the energy levels must be adjusted to compensate for the new quantum numbers. These are given as follows,

${\displaystyle \Psi _{n_{x},n_{y},n_{z}}={\sqrt {\frac {8}{abc}}}\sin \left({\frac {n_{x}{\pi }x}{a}}\right)\sin \left({\frac {n_{y}{\pi }y}{b}}\right)\sin \left({\frac {n_{z}{\pi }z}{c}}\right)}$
${\displaystyle \epsilon _{n_{x},n_{y},n_{z}}={\frac {h^{2}}{8m}}\left({\frac {{n_{x}}^{2}}{a^{2}}}+{\frac {{n_{y}}^{2}}{b^{2}}}+{\frac {{n_{z}}^{2}}{c^{2}}}\right)}$

a, b, and c represent the length of each corresponding side of the box. If the box were a cube then the energy level equation can be simplified because sides a, b, and c of the cube are all equal. The resulting equation would take on the form,

${\displaystyle \epsilon _{n_{x},n_{y},n_{z}}={\frac {h^{2}}{8m{a^{2}}}}({{n_{x}}^{2}}+{{n_{y}}^{2}}+{{n_{z}}^{2}})}$

For a particle in a cube some combinations of quantum numbers will give the same energy level. Energy levels with a different set of quantum numbers but having the same energy are said to be degenerate and unless all three of the quantum numbers are identical there is always another combination of the three quantum numbers that will give rise to a degenerate state.

## Example

Calculate the energy difference, ${\displaystyle \Delta E}$, for the translation of N2 from the ground state to the first excited state. Assume the box is a cube with an edge length of 10 cm.

### Solution

The ground state, where ${\displaystyle n_{x}=n_{y}=n_{z}=1}$, has a degeneracy of 1. The first excited state has a degeneracy of 3, involving the different combinations of quantum numbers ${\displaystyle n=1,1,}$ and 2. Can be determined by using the following equation:

${\displaystyle \epsilon _{n_{x},n_{y},n_{z}}={\frac {h^{2}}{8m}}\left({\frac {{n_{x}}^{2}}{a^{2}}}+{\frac {{n_{y}}^{2}}{b^{2}}}+{\frac {{n_{z}}^{2}}{c^{2}}}\right)}$

However because all edges of the cube is 10 cm, then a=b=c=10 cm = 0.1 m. Then the equation becomes:

${\displaystyle \epsilon _{n_{x},n_{y},n_{z}}={\frac {h^{2}}{8m{a^{2}}}}\left({{n_{x}}^{2}}+{{n_{y}}^{2}}+{{n_{z}}^{2}}\right)}$

The mass of N2 can be determined with the following equation:

${\displaystyle \mu ={\cfrac {m_{N}m_{N}}{m_{N}+m_{N}}},\!\,}$
${\displaystyle \mu ={\cfrac {m_{N}^{2}}{2m_{N}}},\!\,}$
${\displaystyle \mu ={\cfrac {m_{N}}{2}},\!\,}$
${\displaystyle \mu ={\cfrac {14.0067u}{2}},\!\,}$
${\displaystyle \mu =(7.00335u)(1.660549\times 10^{-27}kg/u),\!\,}$
${\displaystyle \mu =1.16294\times 10^{-26}kg,\!\,}$

Using the mass the ${\displaystyle \Delta E}$ can now be determined.

${\displaystyle \epsilon _{n_{x},n_{y},n_{z}}={\frac {h^{2}}{8m{a^{2}}}}({{n_{x}}^{2}}+{{n_{y}}^{2}}+{{n_{z}}^{2}})-({{n_{x}}^{2}}+{{n_{y}}^{2}}+{{n_{z}}^{2}})}$
${\displaystyle \epsilon _{n_{x},n_{y},n_{z}}={\frac {({6.626\times 10^{-34}})^{2}Js}{8(1.16294\times 10^{-26}kg){0.1m^{2}}}}({{2}^{2}}+{{1}^{2}}+{{1}^{2}})-({{1}^{2}}+{{1}^{2}}+{{1}^{2}})}$
${\displaystyle \epsilon _{n_{x},n_{y},n_{z}}={\frac {{4.3904\times 10^{-67}}J^{2}s^{2}}{9.30352\times 10^{-28}kgm^{2}}}(6-3)}$
${\displaystyle \epsilon _{n_{x},n_{y},n_{z}}=1.4157\times 10^{-39}J}$

# Rotational Energy

Molecules that consist of more than one atom have rotational angular momentum. The rotational motion which is generated by these molecules is purely kinetic as each molecule rotates around the molecules centre of mass. A good approximation which makes interpreting the rotational energy levels easier is the rigid rotor approximation which is mentioned below. Essentially this approximation assumes that the molecule has a fixed bond length and it does not vibrate while it undergoes rotation about its centre of mass. This approximation makes it easy to calculate the bond length by simply knowing the rotational energy spacings of the molecule observed in a spectroscopic experiment.

The rotational kinetic energy of an object depends on its moment of inertia, I,

${\displaystyle I=\Sigma _{i}m_{i}r_{i}^{2}}$

where ${\displaystyle m_{i}}$ is the mass of the atom ${\displaystyle i}$, ${\displaystyle r_{i}}$ is the distance of that atom from the centre of mass of the molecule. Any molecule has three perpendicular axes of rotation each of which intersect at the centre of mass for that molecule, the moment of inertia is the rotational motion found at the center of mass of a molecule. Each of these axes of rotation could be distant from one another, or they could all be equal which leads to different types of rigid rotors. A large moment of inertia is basically based on either larger atomic masses and/or longer bonds.

## Types of Rigid Rotor

Type Moments of Inertia Examples
Linear Rotors ${\displaystyle I_{1}=I_{2}=I_{3}=0}$ N2, CO2, C2H2
Spherical Rotors ${\displaystyle I_{1}=I_{2}=I_{3}\neq 0}$ CH4, SF6
Symmetric Rotors ${\displaystyle I_{1}=I_{2}\neq I_{3}}$ NH3, CH3Cl
Asymmetric Rotors ${\displaystyle I_{1}\neq I_{2}\neq I_{3}}$ H2O, H2CO, Most Molecules

## Linear Rotors

In this course we are mostly concerned with linear rotors, using these systems simplifies the math. If we are using a linear rigid rotor, the bond length is constant due to the rigid rotor approximation, and the centre of mass is constant, therefore the moment of inertia can be simplifies using the reduced mass as follows:

${\displaystyle I=\Sigma \mu r_{e}^{2}}$

here, ${\displaystyle \mu }$ is the reduced mass of the linear rotor where ${\displaystyle {\frac {m_{1}m_{2}}{m_{1}+m_{2}}}}$, and ${\displaystyle r_{e}}$ is the bond length of the diatomic linear rotor. The reduced mass the inertial mass of two atom system following Newtonian Mechanics. This can then be used to solve the Schrödinger equation which is given in example 2 below. By solving this equation we find that there are two quantum numbers which determine the energy levels of a rotating rigid rotor, M and J both of which are described below in example 2.

### Zero Point Energy of Linear Rotors

Unlike the particles in a box model, which has a non zero energy minima due to the fact that the lowest energy state corresponds to the n=1 state, the linear rotor can have an energy of zero in the ground state corresponding to a molecule which is not rotating in space. This is the case since ${\displaystyle J=0,1,2...}$ when ${\displaystyle J=0}$ the energy of the system is equal to zero. Since the orientation in space is not known for a linear rotor it obeys the Heisenberg uncertainty principle which states that we cannot simultaneously determine the momentum and the position of the particle. Where each energy level can exist using following equation

${\displaystyle E_{J}={\frac {\hbar }{\mu r_{e}^{2}}}\times J(J+1)}$

where J is the quantum number, ${\displaystyle \mu }$ is the reduced mass, re2 is the length of the bond from the center of mass and ħ is reduced Planck's constant where Planck's constant is divided by 2π. From inserting J=0 in this equation above, we see ${\displaystyle \epsilon _{0}=0}$.

## Degeneracy of Rotational States

The rotational levels of a diatomic only depend on the quantum number J (orbital angular momentum quantum number) and does not have any dependence on the quantum number M. However, as the quantum number J increases, the degeneracy of the states increases since there are more possible values of M, which is the magnetic quantum number which are as follows

${\displaystyle J=0,1,2...}$
${\displaystyle M=-J,...,0,...,+J}$

Therefore, as the quantum number J increases, the degeneracy of the states also increases. The degeneracy of the states can be calculated using the following equation:

${\displaystyle g_{J}=2J+1}$

Degeneracy occurs when there are accessible states present at the same energy level, for example, when J=1, MJ= -1,0,+1 meaning there are 3 different degeneracies at quantum number J=1

## Rotational Spectroscopy of Linear Rotors

Linear rotors which contain a permanent dipole are able to be observed through rotational-vibrational spectroscopy. These molecules exhibit fine structure of lines which are all separated by a constant value. This value can be used to determine the bond length of the diatomic in question but using the following equations.

${\displaystyle v=2B(J+1)}$ and ${\displaystyle B={\frac {h}{8\pi ^{2}cI}}}$

where ${\displaystyle v}$ is the spacing of the rotational lines, ${\displaystyle B}$ is the rotational constant, ${\displaystyle h}$ is planks constant which has a value of ${\displaystyle 6.626\times 10^{-34}Js}$, ${\displaystyle c}$ is the speed of light and ${\displaystyle I}$ is the moment of inertia. Once we know the value of ${\displaystyle B}$ we can then use this value to calculate the bond length of the diatomic which is being observed.

## Example 1

Calculation of the bond length of HCl from the rotational constant.

Rotational Angular Momentum and the Rigid Rotor Approximation

Rotational energy is purely kinetic, which is a result of angular momentum in the molecule (How quickly the molecule is spinning in space). It can be assumed that the molecule is rigid for an ideal gas, this is because the change in the bond length for a vibration is small in comparison to the total length of the bond, this is called the Rigid Rotor Approximation.

Linear Rotors

The rotation of an object depends on its moment of inertia ${\displaystyle I=\Sigma _{i}m_{i}r_{i}^{2}}$ where ${\displaystyle m_{i}}$ is the mass of atom i, and ${\displaystyle r_{i}}$ is the distance of atom i to the axis of rotation. This expression works well if the diatomic molecule consists of only the same element (e.g. ${\displaystyle O_{2},N_{2}}$). For diatomic molecules which do not consist of the same elements (e.g. ${\displaystyle HCl,NO,NaF}$) this expression can be more work intensive then needed. For diatomic molecules which do not comprise of the same elements the expression ${\displaystyle I=\mu r_{e}^{2}}$ is significantly less work intensive to use in comparison to the previously stated equation. Where ${\displaystyle r_{e}^{2}}$ is the bond length of the diatomic molecule and ${\displaystyle \mu }$ is the reduced mass of the molecule, ${\displaystyle \mu }$ is given by the equation ${\displaystyle \mu ={\frac {m_{1}m_{2}}{m_{1}+m_{2}}}}$ where ${\displaystyle m_{1}}$ is the atomic mass of atom 1, and ${\displaystyle m_{2}}$ is the atomic mass of atom 2.

Rotational Spacings

Example of a pure rotational absorption spectrum

The spacings of rotational transitions are constant, and are commonly referred to as a rotational constants ${\displaystyle {\bar {B}}}$. For linear rotors the spacing is ${\displaystyle 2{\bar {B}}}$ between each rotational transition as can be seen in the image to the right.

Utilizing the equation for the rotational constant ${\displaystyle {\bar {v}}=2{\bar {B}}(J+1)}$ we can find the value for ${\displaystyle {\bar {B}}}$ and then use it in the equation ${\displaystyle {\bar {B}}={\frac {h}{8\pi ^{2}cI}}}$ and solve for the moment of inertia ${\displaystyle I}$, where ${\displaystyle h}$ is Planck's constant, and ${\displaystyle c}$ is the speed of light. Once the moment of inertia is calculated it is simply the matter of solving for the diatomic bond length ${\displaystyle r_{e}}$ which can be found by ${\displaystyle r_{e}={\sqrt {\frac {I}{\mu }}}}$.

Sample Problem

Calculate the Bond Length of HCl from the Rotational Constant.

Knowing HCl has a rotational constant value of 10.59341 cm−1, the Planck's constant is 6.626 × 10−34J s, and the speed of light being 2.998 × 1010 cm s−1. We then solve for the moment of inertia such that:

${\displaystyle {\bar {B}}={\frac {h}{8\pi ^{2}cI}}}$

${\displaystyle I={\frac {h}{8\pi ^{2}c{\bar {B}}}}}$

${\displaystyle I={\frac {6.626\times 10^{-34}Js}{8\pi ^{2}(2.998\times 10^{10}cm/s)(10.59341cm^{-1})}}}$

${\displaystyle I=2.642401\times 10^{-47}Js^{2}}$

${\displaystyle I=2.642401\times 10^{-47}(kgm^{2}/s^{2})s^{2}}$

${\displaystyle I=2.642401\times 10^{-47}kgm^{2}}$

The next step is to calculate the reduced mass ${\displaystyle \mu }$. Knowing that the mass of hydrogen is 1.00794 u, and the mass of chlorine is 35.4527 u.

${\displaystyle \mu ={\frac {m_{1}m_{2}}{m_{1}+m_{2}}}}$

${\displaystyle \mu ={\frac {(1.00794u)\times (35.4527u)}{(1.00794u)+(35.4527u)}}}$

${\displaystyle \mu ={\frac {35.73419u^{2}}{36.46064u}}}$

${\displaystyle \mu =0.9800758u}$

${\displaystyle \mu =0.9800758u\times {\frac {1.6605\times 10^{-27}kg}{1u}}}$

${\displaystyle \mu =1.627454\times 10^{-27}kg}$

Now knowing all of the variable it is just a matter of solving for the bond length ${\displaystyle r_{e}}$.

${\displaystyle r_{e}={\sqrt {\frac {I}{\mu }}}}$

${\displaystyle r_{e}={\sqrt {\frac {2.642401\times 10^{-47}kgm^{2}}{1.627454\times 10^{-27}kg}}}}$

${\displaystyle r_{e}={\sqrt {1.623641\times 10^{-20}m^{2}}}}$

${\displaystyle r_{e}=1.274222\times 10^{-10}m}$

${\displaystyle r_{e}=1.274222}$ Å

${\displaystyle r_{e}=1.274}$ Å

## Example 2

Calculation of the energy for the transition of N2 from the ground to first excited rotational state.

### Solution

For N2, the reduced mass is calculated using an atomic mass of 14.0067 u for nitrogen. Since it is a diatomic molecule, the equation can be simplified to the following where ${\displaystyle m}$ represents the mass of a nitrogen atom:

${\displaystyle \mu ={\frac {m^{2}}{2m}}}$
${\displaystyle \mu ={\frac {m}{2}}}$
${\displaystyle \mu ={\frac {14.0067u}{2}}}$
${\displaystyle \mu =7.0033u}$
${\displaystyle \mu =(7.0033u)\left({\frac {1.660539040\times 10^{-27}kg}{u}}\right)}$
${\displaystyle \mu =1.1629\times 10^{-26}kg}$

For the rigid rotor, the lowest possible energy state is J=0 so the ground state rotational energy is always zero. The rotational energy levels are only dependent on quantum number J and only the ∆𝐽 = ±1 transitions can occur. In the case of the transition from the ground to the first excited state, the following is a diagram that represents the transition.

The energy for the transition J=1 ← J=0 is calculated as follows:

${\displaystyle \Delta \epsilon ={\frac {\hbar ^{2}}{2\mu r_{e}^{2}}}J(J+1)-{\frac {\hbar ^{2}}{2\mu r_{e}^{2}}}J(J+1)}$
${\displaystyle \Delta \epsilon ={\frac {\hbar ^{2}}{2\mu r_{e}^{2}}}1(1+1)-{\frac {\hbar ^{2}}{2\mu r_{e}^{2}}}0(0+1)}$
${\displaystyle \Delta \epsilon ={\frac {\hbar ^{2}}{\mu r_{e}^{2}}}}$
${\displaystyle \Delta \epsilon ={\frac {(1.0545718Js)^{2}}{(1.1629\times 10^{-26}kg)(1.098\times 10^{-10}m)^{2}}}}$
${\displaystyle \Delta \epsilon =7.9320\times 10^{-23}J}$

# Vibrational energy

Molecular vibrations occur when atoms in a molecule are in periodic motion while the molecule as a whole has constant translational and vibrational motion. Vibrational energy is different from translational energy (how a molecule translates through 3-dimensional space along the x,y, and z Cartesian coordinates) and rotational energy (how fast a molecule spins) in that the focus is placed on how a chemical bond behaves with respect to the quantum harmonic oscillator approximation.

## Vibrational Energy Levels

Rotational-vibrational spectrum of HCl

### The Potential Energy Surface of Molecules

Chemical bonds are formed due to electrostatic interactions between protons and electrons, and can occur as nuclear-electron attractions, electron-electron repulsion, and inter-nuclear repulsion. As a bond forms due to the nuclear electron attractions between a proton and electron, the electron density increases in the space between two nuclei. For a diatomic, the potential energy is only a function of the distance between the two atoms, and the strength of these components changes depending on the distance between the atoms. Bonding is generally stronger when two nuclei are close, but this close proximity also increases internuclear and core-core repulsions. Hence, the potential energy is lowest at the equilibrium bond length ${\displaystyle {r_{e}}}$. Short distances lead to high repulsion, long distances result in no interactions, and middle distances result in stabilized bonding.

### Application of Hooke's Law

Molecules will have some amount of kinetic energy along this degree of freedom and harmonic oscillation will occur along the bond length. A bond between molecules can be approximated as a parabola centred around the equilibrium bond length ${\displaystyle {r_{e}}}$. Hooke's Law treats the bond as a spring. When a bond is stretched or compressed its potential energy increases, just like the behaviour of a spring under the harmonic oscillation approximation. ${\displaystyle {k}}$ is the constant corresponding to the stiffness of the spring, called the spring constant, and can be correlated to the stiffness of a molecular bond. Higher spring constants correlate to stiffer (stronger) bonds. For example, the value of ${\displaystyle {k}}$ will be larger for a double bond than in a single bond. Hence, a double bond is more stiff.

We can approximate the bond as a parabola centred around the equilibrium bond length ${\displaystyle {r_{e}}}$:

${\displaystyle {\mathcal {V}}(r)={\frac {1}{2}}k_{bond}\left(r-r_{eq}\right)^{2}}$

### Quantum Harmonic Oscillator Approximation

Internuclear separation with respect to bond length.

There are several significant approximations associated with describing chemical bonds as harmonic oscillators:

1. The bond can never dissociate. In reality if we keep pulling two atoms apart the bond will eventually break, but according to the quantum harmonic oscillator approximation the potential energy just continues to increase and bonds are not allowed to break. This can be referred to as anharmonic effects; if bonds were really harmonic no chemical reaction would ever take place.
2. The repulsive wall isn't repulsive enough. In reality, short distances are dominated by repulsive interactions. The low potential energy structures of most important molecules are near the minimum energy geometry.
3. The spacing of the vibrational energy levels are exactly equal. However, in reality they become slightly smaller as quantum number ${\displaystyle {n}}$ increases.

Energy levels of a quantum harmonic oscillator follow:

${\displaystyle E_{n}=h{\nu }\left(n+{1 \over 2}\right)}$

where ${\displaystyle {h}}$ is Planck's Constant (6.626×10−34 J s), ${\displaystyle \nu }$ is the vibrational frequency of the oscillator in s-1 and quantum number ${\displaystyle {n}}$= 0,1,2...

${\displaystyle \mu ={\cfrac {m_{1}m_{2}}{m_{1}+m_{2}}},\!\,}$

${\displaystyle {v}}$ is related to the spring constant, ${\displaystyle {k}}$, and reduced mass of a diatomic,${\displaystyle {\mu }}$.

${\displaystyle v={1 \over {2\pi }}{\sqrt {k \over {\mu }}}\!}$

### Other Important Formulae

Since Planck's constant, ${\displaystyle h}$, and vibrational frequency, ${\displaystyle \nu }$, are related by ${\displaystyle E=h\nu }$, the relation can also be expressed as: ${\displaystyle E={\frac {hc}{\lambda }}.}$

Wavenumbers, also called vibrational frequency, are reported in reciprocal distance and chemists typically use reciprocal centimetres.

${\displaystyle v={\frac {1}{\lambda }}={\frac {E}{hc}}}$

where ${\displaystyle {E}}$ is Energy in Joules, ${\displaystyle {c}}$ is the speed of light (2.9979×108 m/s = 2.9979×1010 cm/s) and, ${\displaystyle {\lambda }}$ is wavelength. This molecular transition energy is usually extremely small and typically follows the order of 1×10−20 joules, J.

### Vibrational Zero Point Energy

The ground state does not have zero energy. This is the vibrational zero-point energy, and is consistent with the Heisenberg Uncertainty principle. If a molecule had zero vibrational energy it would be at rest at the minimum, and therefore we would know its position and momentum completely. Even in the lowest energy state, molecules still vibrate.

${\displaystyle E_{n=0}=h\nu \left(n+{1 \over 2}\right)=h\nu \left(0+{1 \over 2}\right)={1 \over 2}h\nu }$

Vibrational transitions follow the selection rule ${\displaystyle \Delta n=\pm 1$,andthereforeallvibrationaltransitionshaveexactlythesameenergy.[itex]\Delta E=h{\nu }\left((n'+1)+{1 \over 2}\right)-h{\nu }\left(n'+{1 \over 2}\right)}$ ${\displaystyle \Delta E=h{\nu }\left((n'+1)+{1 \over 2}-(n'+{1 \over 2})\right)}$ ${\displaystyle \Delta E=h{\nu }(1)=h{\nu }}$ The vibrational spectrum of diatomics is coupled with rotational transitions. The total selection rule is${\displaystyle \Delta n=\pm 1[itex]and[itex]\Delta J=\pm 1[itex].===VibrationalStatesofPolyatomics===Whilemonatomicgaseousatomshaveavibrationaldegreeoffreedomofzero,polyatomicmoleculeswillhaveseveralvibrationaldegreesoffreedom.Thesevibrationsaretreatedasindependentharmonicoscillators.Polyatomicmoleculesaredividedintotwocategories:Linearandnon-linearmolecules.Forlinearmolecules:[itex]n_{vib}=3N_{atom}-5}$ For non-linear molecules: ${\displaystyle n_{vib}=3N_{atom}-6}$ ### Trends in Vibrational Frequencies • Strong bonds are stiff bonds, which have a large spring constant (${\displaystyle k}$). • Stiff bonds have large vibrational energy level spacings. • Light atoms give small reduced masses (${\displaystyle {\mu }}$). • Molecules with small reduced masses have large vibrational energy level spacings. ## Vibrational Energy Example 1 The harmonic vibrational frequency of HF in wavenumbers is 4141.3 cm-1. [1] Using this information, calculate the energy for the transition of HF from the ground to first excited vibrational state. ### Solution Vibrational energy levels of a quantum harmonic oscillator follow: ${\displaystyle E_{n}=h{v}\left(n+{1 \over 2}\right)}$ However, this relationship requires the vibrational frequency to be in units of s-1. To do this, we multiply the given value of 4141.3 cm-1 by the speed of light, in centimetres. ${\displaystyle v={\widetilde {v}}c}$${\displaystyle v=(4141.3cm^{-1})(2.9979\times 10^{10}cm/s)}$ ${\displaystyle v=1.2415\times 10^{14}s^{-1}}$ Now that we have the correct units, the ground state vibrational energy can be calculated as follows: ${\displaystyle E_{0}=h{v}\left(n+{1 \over 2}\right)=h{v}\left(0+{1 \over 2}\right)={1 \over 2}h{v}}$ ${\displaystyle E_{0}=\left({1 \over 2}\right)(6.626\times 10^{-34}Js)(1.2415\times 10^{14}s^{-1})}$ ${\displaystyle E_{0}=4.1131\times 10^{-20}J}$ Next, the first excited state vibrational energy is: ${\displaystyle E_{1}=h{v}\left(n+{1 \over 2}\right)=h{v}\left(1+{1 \over 2}\right)={3 \over 2}h{v}}$ ${\displaystyle E_{1}=\left({3 \over 2}\right)(6.626\times 10^{-34}Js)(1.2415\times 10^{14}s^{-1})}$ ${\displaystyle E_{1}=1.2339\times 10^{-19}J}$ Finally, the energy for the transition from the ground state (n=0) to the first excited state (n=1) level is: ${\displaystyle \Delta E=E_{1}-E_{0}}$ ${\displaystyle \Delta E=1.2339\times 10^{-19}J-4.1131\times 10^{-20}J}$ ${\displaystyle \Delta E=8.2259\times 10^{-20}J}$ ## Vibrational Energy Example 2 Calculation of the vibrational wavenumber of CO from the force constant and reduced mass. ### Harmonic Oscillator Approximation As there is some kinetic energy along the vibrational degree of freedom, the bond length undergoes oscillation. The Schrӧdinger equation for the movement of quantum mechanical particles given the potential energy function V(r) is complicated to solve. Instead, the Harmonic Oscillator Approximation is used which proposes that the bond behaves like a spring governed by Hooke's law. Hooke's law: ${\displaystyle {\mathcal {V}}(r)={\frac {1}{2}}k(r-r_{eq})^{2}}$ where r is the internuclear distance and k is a constant corresponding to the stiffness of the spring Given this approximation, the following relationship arises ${\displaystyle \nu ={1 \over {2\pi }}{\sqrt {k \over {\mu }}}\!}$ where, • ${\displaystyle k}$ is the force constant of the bond in ${\displaystyle {\textrm {kg}}\ {\textrm {s}}^{-1}}$ • ${\displaystyle \mu }$ is the reduced mass in kilograms • ${\displaystyle v}$ is the vibrational frequency of the oscillator in ${\displaystyle s^{-1}}$ The reduced mass ${\displaystyle {\mu }}$ for a heteronuclear diatomic can be calculated as follows ${\displaystyle {\mu }}$ ${\displaystyle ={\cfrac {m_{1}m_{2}}{m_{1}+m_{2}}}}$ where ${\displaystyle m_{1}}$ and ${\displaystyle m_{2}}$ corresponds to the masses of each atom. ### Solution Firstly, the reduced mass of the carbon monoxide molecule must be calculated given that the atomic masses of carbon and oxygen is ${\displaystyle 12.0107u}$ and ${\displaystyle 15.9994u}$ respectively. ${\displaystyle \mu ={\cfrac {m_{1}m_{2}}{m_{1}+m_{2}}}}$ ${\displaystyle \mu ={\cfrac {(12.0107\ u)(15.9994\ u)}{12.0107\ u+15.9994\ u}}}$ ${\displaystyle \mu ={\cfrac {192.164\ u^{2}}{28.0101\ u}}}$ ${\displaystyle \mu =6.8605\ u}$ Convert into units of kilogram as follows ${\displaystyle \mu =(6.8605\ u)\left({\cfrac {1\ kg}{1000\ g}}\right)\left({\cfrac {1}{6.02214129\times 10^{23}\ mol^{-1}}}\right)}$ ${\displaystyle \mu =1.1392\times 10^{-26}\ kg}$ The next step is to derive the vibrational frequency in ${\displaystyle s^{-1}}$ given the value of the reduced mass,${\displaystyle \mu =(1.1392)(10^{-26})kg}$ , and the force constant ${\displaystyle k=1860Nm^{-1}}$ [2]. Note that ${\displaystyle 1\ N\ m^{-1}=1\ kg\ s^{-2}}$ ${\displaystyle \nu ={1 \over {2\pi }}{\sqrt {k \over {\mu }}}\!}$ ${\displaystyle \nu ={1 \over {2\pi }}{\sqrt {1860kg/s^{2} \over {1.1392\times 10^{-26}\ kg}}}\!}$ ${\displaystyle \nu =(0.15915)({\sqrt {(1.6327\times 10^{29})\ s^{-2}}}\!}$ ${\displaystyle \nu =(0.15915)(4.0407\times 10^{14})\ s^{-1}\!}$ ${\displaystyle \nu =6.4308\times 10^{13}\ s^{-1}\!}$ Convert the vibrational frequency into wavenumbers given the following relationship ${\displaystyle {\tilde {v}}={\cfrac {vibrationalfrequency,v\ (s^{-1})}{\textrm {speedoflight}}}}$ ${\displaystyle {\tilde {v}}={\cfrac {6.4308\times 10^{13})\ s^{-1}}{\textrm {speedoflight}}}\!}$ ${\displaystyle {\tilde {v}}={\cfrac {6.4308\times 10^{13}s^{-1}}{2.99792458\times \ 10^{10}cm\ s^{-1}}}\!}$ ${\displaystyle {\tilde {v}}=2145.1\ cm^{-1}\!}$ ## References 1. Atkins, Peter and DePaula, Julio. "Elements of Physical Chemistry". Oxford University Press, 2013, p. 501. 2. HyperPhysics.Quantum Physics: Vibrational Spectra of Diatomic Molecules. http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/vibspe.html (accessed Feb. 2, 2016) # Electronic energy Atoms and molecules can hold different energy levels that correspond to different distributions in how the electron density is distributed around them. These are counted as electronic degrees of freedom. For example, the simplest atom is the hydrogen atom, with 1 proton and 1 electron. The attraction between the proton and electron in the hydrogen atom is governed by Coulomb's Law,  principal quantum number n=1,2,3,... angular momentum quantum number ${\displaystyle l=0,1,...,n-1}$ magnetic quantum number ${\displaystyle m_{l}=-l,-l+1,...,0,l-1,l}$ spin quantum number ${\displaystyle m_{s}=+{\frac {1}{2}},-{\frac {1}{2}}}$ ${\displaystyle {\mathcal {V}}(r)=-{\frac {e^{2}}{4\pi \epsilon _{0}r}}}$ Where ${\displaystyle e}$ is the elementary charge of an electron, ${\displaystyle r}$ is the distance between the proton and electron in the hydrogen atom, and ${\displaystyle \epsilon _{o}}$ is the vacuum permittivity constant. When the Schrödinger equation is solved for this potential, four quantum numbers result, n, l, ml, and ms. This means that the particles can only take on certain discrete values of energy. The electronic energy of the hydrogen atom depends only on the principal quantum number ${\displaystyle n}$ via the following equation, ${\displaystyle \epsilon _{n}=-{\frac {m_{e}e^{4}}{32\pi ^{2}\epsilon _{o}^{2}\hbar ^{2}}}{\frac {1}{n^{2}}}}$ Where ${\displaystyle m_{e}}$ is the mass of an electron. This means that all sets of angular momentum, l, and magnetic quantum numbers, ml, and ms, with the same value of n are degenerate, as they do not appear in this solution to the Schrödinger equation. This quantization of allowed energies give rise to the electronic energy levels of the hydrogen atom. This equation can also be used to calculate the energy levels of hydrogen like species such as He+. An atomic factor, ${\displaystyle Z^{2}}$, would be added in the numerator to account for the extra protons in the atom, which would pull the electrons in more than the hydrogen atom, and thereby increase the energy. ## Example Calculation of the energy for the excitation of O2 from the triplet to singlet electronic state. Singlet oxygen comes from the electronic excitation of triple oxygen, the energy is experimentally 0.98 electron volts per mole of electrons. The figure below illustrates the different states of oxygen. 1Δg and 1Σg+ represents singlet oxygen as it has a spin state equal to one, the triplet degenerate state is represented by 3Σg-, the transition is represented by 1Δg3Σg- The degeneracy is calculated for this state is calculated from the formula gel= 2S+1 where S is the sum of the spin of the electrons. For singlet oxygen it has one spin state of -1/2 and a spin state of +1/2 giving the sum to be 0 and the degeneracy to be calculated by gel= 2(0)+1 to give a degeneracy of 1. Triple oxygen has two electrons in the spin up state to give it a value of 1. Which gives the degeneracy of 3 for this state. The experimental value for this transition is experimentally calculated to be 0.98 eV found on Singlet Oxygen. This is for a wavelength of 1270 nm. the conversion factor from eV (electron volts) to kJ/mol is given by multiplying the number of electron volts by 96 kJ/mol, for the transition of triplet degenerate state to the singlet energy state it gives a value of 94.56 kJ/mol. # Raman spectroscopy ## The Raman Effect Examples of energy transitions for each type of Raman line The Raman effect is a form of light scattering by a material. It can be observed when a sample is exposed to an intense, high energy, monochromatic light source, such as a laser. Most photons of this light that are scattered by the material will be scattered elastically, meaning that there will be no change in the frequency of light. This results in an intense line at the same frequency of the source, known as the Rayleigh line. Photons can also undergo Raman scattering, where they are inelastically scattered. This means that they will emerge at a higher or lower frequency. This type of scattering is less common than elastic scattering, with only 1 in 107 photons experiencing Raman scattering. Stokes lines occur when the scattered light emerges at a lower frequency. Anti-stokes lines occur when the scattered light emerges at a higher frequency. The Raman effect is applicable even when the frequency of radiation does not correspond to a transition between molecular energy levels. This cannot be explained by standard absorption or emission and therefore must be the Raman effect. ## Raman Spectroscopy Raman spectroscopy is a spectroscopic technique which measures inelastic scattering. The gain or loss of energy by incident light corresponds to transitions between the energy levels of the molecules. These transitions are both rotational and vibrational. A high energy photon excites a molecule to a "virtual state" (v1), following this the molecule returns to a different state (v2) on emission. A different frequency of light is emitted due to the difference in initial and final energy levels. ## The Gross Selection Rule for Raman Spectroscopy The gross selection rule for Raman spectroscopy states that a molecule must be anisotropically polarizable to have a Raman spectrum. Anisotropic indicates that something is different in some directions, in this case the electron density of the molecule must be polarized heterogeneously allowing for it to be observed via Raman spectroscopy. Examples of some molecules which are spherical tops This selection rule explains why Raman spectroscopy can successfully analyze homonuclear diatomics, such as H2 and N2, when rotational and vibrational spectroscopic techniques cannot. The Raman effect is present for any molecule that can be polarized anisotropically. Molecules that are spherical tops cannot be anisotropically polarized, this is because they have three equal moments of inertia. Therefore these molecules cannot be observed using Raman spectroscopy. ## The Specific Selection Rule of Rotational Raman Spectroscopy The specific selection rule for Raman spectroscopy of linear molecules is ${\displaystyle \Delta J=0,\pm 2}$. The Raman spectrum has regular spacing of lines, as seen previously in absorption spectra, but separation between the lines is doubled. The evenly spaced lines have a separation of: ${\displaystyle {\widetilde {v}}=4{\widetilde {B}}(J+1)}$ where: ${\displaystyle {\widetilde {B}}={\frac {h}{8\pi ^{2}cI}}}$ The rotational energy of the molecule can be calculated using the following equation: ${\displaystyle E_{rot}={\frac {h^{2}}{4\pi ^{2}\mu r_{e}^{2}}}J(J+1)}$ ## The Specific Selection Rule of Vibrational Raman Spectroscopy The specific selection rule for vibrational Raman spectroscopy states that only Δv = ±1 transitions are allowed. This is the same as for vibrational absorption spectroscopy. Vibrational Raman transitions occur simultaneously with rotational Raman transitions, this results in branching caused by rotational transitions in the Δv = ±1 peaks. The vibrational transitions result in 3 branches with fine structure from rotational transitions. In the spectrum each line corresponds to a change in quantum numbers v, J, or both. Branch Energy Level Transition O ΔJ=-2 Q ΔJ=0 S ΔJ=2 ## Raman Spectroscopy Example 1 Calculation of rotational energy levels of N2 from pure rotational Raman spectrum. For a molecule to have a Raman spectrum it must be anisotropically polarizable and thus have a different electron cloud distortion along different axis. Given the Pure Rotational Raman Spectrum of a molecule it is possible to calculate the rotational energy levels based on the spectrum itself. Transitions of ${\displaystyle \Delta J=\pm 1}$ can be observed in the spectrum and used to find the energy levels. A sample calculation to find the rotational energy levels of Nitrogen can be computed using the spacings and J values of the Raman spectrum below. This calculation will be done in three steps: 1. Finding the rotational constant, ${\displaystyle {\widetilde {B}}}$. 2. Finding the moment of Inertia, ${\displaystyle I}$. 3. Finding the rotational energy, ${\displaystyle E_{rot}}$. Step 1: Computing the rotational constant B: As seen from the figure above, the spacing, or v, is approximately 8.00 cm-1 which corresponds to the quantum number J = 1. Using these values it is possible to compute B by re-arranging the formula below: ${\displaystyle {\widetilde {\nu }}=4{\widetilde {B}}(J+1)}$ ${\displaystyle {\widetilde {B}}={\frac {\widetilde {\nu }}{4(J+1)}}}$ ${\displaystyle {\widetilde {B}}={\frac {8.00cm^{-1}}{4(1+1)}}}$ ${\displaystyle {\widetilde {B}}=1.00cm^{-1}}$ Step 2: Compute the moment of Inertia I using the formula below and the constant B determined above: ${\displaystyle {\widetilde {B}}={\frac {h}{8\pi ^{2}cI}}}$ ${\displaystyle I={\frac {h}{8\pi ^{2}c{\widetilde {B}}}}}$ Note: h is Planck's constant ${\displaystyle 6.62607004\times 10^{-34}J/s}$ and c is the speed of light in cm/s: ${\displaystyle 2.99792458\times 10^{10}cm/s}$. When using frequencies in units of wavenumber (${\displaystyle cm^{-1}}$), the speed of light in units of cm/s should be used in order for them to cancel out during the calculation.  ${\displaystyle I={\frac {6.626\times 10^{-34}J/s}{8\pi ^{2}(2.998\times 10^{10}cm/s)(1.00cm^{-1})}}}$ ${\displaystyle I=2.799\times 10^{-46}kg/m^{2}}$ Step 3: Compute the rotational energy level using the value of I from above and the following two equations: ${\displaystyle I=\mu r_{e}^{2}}$ ${\displaystyle E_{rot}={\frac {h}{4\pi \mu r_{e}^{2}}}J(J+1)}$ Note: Substituting the equation for I into the energy equation will give: ${\displaystyle E_{rot}={\frac {h^{2}}{4\pi ^{2}I}}J(J+1)}$ ${\displaystyle E_{rot}={\frac {(6.626\times 10^{-34}J/s)^{2}}{4\pi ^{2}(2.799\times 10^{-46}kg/m^{2})}}1(1+1)}$ ${\displaystyle E_{rot}=3.94\times 10^{-23}J}$ Note: Since the calculated value of ${\displaystyle E_{rot}}$ was done using ${\displaystyle J=1}$, this corresponds to the energy of the ${\displaystyle J=1}$ energy level. This calculation can be repeated using different values of ${\displaystyle J}$ to determine the different energies of each rotational energy level. # Postulates of Statistical Thermodynamics ## Postulates of Statistical Thermodynamics Statistical thermodynamics is a branch of science which utilizes statistics in order to relate the microscopic properties of a system to the macroscopic properties. Classical thermodynamics describes macroscopic properties of a system composed of atoms or molecules such as pressure, enthalpy and internal energy. On the other hand, quantum mechanics utilizes quantized values of microscopic properties of a system, such as rotational and vibrational movement, to calculate the energy of the system. Using two different methodologies, both thermodynamics and quantum mechanics can be utilized to evaluate similar properties of a system without any clear connection between the two fields. Statistical thermodynamics provides this connection by averaging the microscopic values across a large set of microcanonical ensembles at an instant in time to arrive at macroscopic values. These microcanonical ensembles are physical systems in which the volume, energy and total number of particles are held constant. The total number of microcanonical ensembles (𝒜), can be averaged to obtain a mean result, known as the mechanical value. ## First Postulate of Statistical Thermodynamics The First Postulate of Statistical Thermodynamics states that the time average of a mechanical variable M in the thermodynamic system of interest is equal to the ensemble average of M as the limit of 𝒜 → ∞. This law states that the average value of a mechanical variable taken from the ensembles of microstates matches the mechanical value predicted by classical thermodynamics, so long as the number of microstates 𝒜 is an exceedingly high number. The First Postulate of Statistical Thermodynamics can be extended to arrive at the Gibbs Postulate, which relates the energy of said microstates to internal energy of a system as calculated by classical thermodynamics. ## Gibbs's Postulate Gibbs's Postulate relates the internal energy (U) of a system, determined by thermodynamics, to the average ensemble energy (E), determined by statistical mechanics. ${\displaystyle U=\langle E\rangle }$ The average energy over the 𝒜 copies in the ensemble is given by the following equation. ${\displaystyle \langle E\rangle ={\frac {1}{A}}\sum _{i}^{A}E^{i}}$ ${\displaystyle E^{i}=hv(n_{1}^{i}+n_{2}^{i}+n_{3}^{i}+...)}$ Through the Gibbs Postulate, the average ensemble energy can be used to define the thermodynamic values of Helmholtz energy (${\displaystyle A}$), entropy (${\displaystyle S}$), pressure (${\displaystyle p}$), and thermodynamic potential (${\displaystyle \mu }$) through the following relationships. ${\displaystyle {A=U-TS}}$ ${\displaystyle {S=-\left({\frac {\partial A}{\partial T}}\right)_{N,V,}}}$ ${\displaystyle {p=-\left({\frac {\partial A}{\partial V}}\right)_{N,T,}}}$ ${\displaystyle {\mu =\left({\frac {\partial A}{\partial N}}\right)_{T,V,}}}$ ## Second Postulate of Statistical Thermodynamics The Second Postulate of Statistical Thermodynamics states that for an ensemble representative of an isolated system, the systems of the ensemble are distributed uniformly. All states consistent with the specified microcanonical system will occur with equal probability. This is otherwise known as the principle of equal 'a priori' probabilities. For example, in two microcanonical systems with three particles capable of occupying a quantum level ${\displaystyle n}$, with ${\displaystyle n=0,1,2,3...}$, there is an equal probability of the first system occupying the states ${\displaystyle n_{1}=1,n_{2}=0,n_{3}=2}$, as there is the second system occupying the states ${\displaystyle n_{1}=6,n_{2}=4,n_{3}=9}$. Across the distribution for each system there is an indiscriminate occupation of each quantum state which is just as likely as any other. # Microcanonical ensemble A microcanonical ensemble is a statistical ensemble which is used to represent the possible states of a mechanical system in which the total energy of the system is exactly specified. A micro-canonical ensemble is assumed to be isolated, thus the system cannot exchange energy or particles with the environment. The system, which can be composed of a solid, liquid or a gas is completely isolated from its surroundings and has constant energy (E), number of particles (N), and volume (V). This ensemble composes of an assembly of “copies” of this isolated system; the energy of each copy is the same, thus E=U, considering one copy of this system is sufficient. ## Combinatorial Mathematics Even though the energy of each copy of the system in a microcanonical ensemble is the same, these copies can exist in different quantum states. For example, consider a system composed of 2 hydrogen molecules, with the total vibrational energy of 2hυ. ${\displaystyle E_{vib}=\sum _{i}n_{i}h\nu =2h\nu }$ ${\displaystyle \sum _{i}n_{i}=2}$ The vibrational quantum numbers must add up to 2, however, there are, in fact, three ways of achieving this. Two possible configurations are where one molecule has n = 2, and the other has n = 0; and the third configuration is where both molecules have n = 1. It can be noted that in the example above that the three configurations (also called microstates) form two macrostates (states with unique sets of quantum numbers) - {0,2} with a weight of 2 (because there are two ways of arriving at this configuration), and {1,1} with a weight of one. The weights of these states play an important role in Statistical Mechanics, so it is important to know a way to quickly determine them, rather than just counting them directly. Let's consider the weights of configurations {1,2}, {1,2,3} and {1,2,3,4}. Counting the weights gives 2, 6 and 24 or 2!, 3! and 4! respectively for these three states where ${\displaystyle n!=1\cdot 2\cdot 3\cdot ...\cdot n}$ represents factorial of the number n. The weights of the configurations are just permutations or ways a given set of the quantum numbers can be arranged. Using combinatorial mathematics, the weight of a certain vibrational state in the example above can be found to be ${\displaystyle W(n_{1},n_{2},n_{3},...)={\frac {N!}{n_{1}\cdot n_{2}\cdot n_{3}\cdot ...}}={\frac {N!}{\prod _{i}n_{i}!}}}$ where N is the total number of molecules and ni is the number of molecules in the state i. Generalizing, the weight of a system with occupancies ${\displaystyle a_{1},a_{2},a_{3},...}$ in an ensemble containing ${\displaystyle {\mathcal {A}}}$ systems is ${\displaystyle W(a_{1},a_{2},a_{3},...)={\frac {{\mathcal {A}}!}{a_{1}\cdot a_{2}\cdot a_{3}\cdot ...}}={\frac {{\mathcal {A}}!}{\prod _{i}a_{i}!}}}$ ## Example If there are three molecules and four units of vibrational energy, what are the possible macrostates of this system and what is the weight of each macrostate? A number of microstates in the same system can have the same composition. These systems can be weighted depending on the occupancies of the systems. The weight of a macrostate can be defined by the equation: ${\displaystyle W(a_{1},a_{2},a_{3},..)={\frac {{\mathcal {A}}!}{\prod _{i}a_{i}!}}}$ Where, ${\displaystyle {\mathcal {A}}!}$ represents the total number of systems in the ensemble and ${\displaystyle \prod _{i}a_{i}!}$ represents the product of the factorial of the occupancies of each level. For three molecules with four units of vibrational energy, we have four unique ways to distribute the four units of vibrational energy between the three molecules, {0,0,4}, {0,1,3}, {1,1,2} and {0,2,2,}. So for this example, ${\displaystyle {\mathcal {A}}}$ = 4 and the weights of each variation of the system would then be: ${\displaystyle W(0,0,4)={\frac {{\mathcal {A}}!}{\prod _{i}a_{i}!}}={\frac {4!}{4!0!0!}}=1}$ ${\displaystyle W(0,1,3)={\frac {{\mathcal {A}}!}{\prod _{i}a_{i}!}}={\frac {4!}{0!1!3!}}=4}$ ${\displaystyle W(1,1,2)={\frac {{\mathcal {A}}!}{\prod _{i}a_{i}!}}={\frac {4!}{1!1!2!}}=12}$ ${\displaystyle W(0,2,2)={\frac {{\mathcal {A}}!}{\prod _{i}a_{i}!}}={\frac {4!}{0!2!2!}}=6}$ ## References University of Delaware.http://www.physics.udel.edu/~glyde/PHYS813/Lectures/chapter_6.pdf (accessed Feb 24, 2017). Balakrishnan, V. The Microcanonical Ensemble, 2017. # Boltzmann Distribution ## Developing the Canonical Ensemble The first step is to develop a model of how the energy is distributed; the model which will be used is a canonical ensemble which is a system at constant number of particles, volume and temperature. A canonical ensemble can be created by first taking a microcanonical ensemble of a large number of systems, ${\displaystyle {\mathcal {A}}}$, all at the same number of particles, volume and energy. The entire microcanonical ensemble is then immersed into a heat bath. The systems are then allowed to exchange energy with the heat bath until all the systems come to a thermal equilibrium. The microcanonical ensemble is the removed from the heat bath such that the systems can only exchange energy with the other systems around it. The systems will now have a distribution of all possible total energies. These energy states can then be numbered in increasing energy (i.e., lowest energy is ${\displaystyle E_{1}}$). Since each of these energy states can have any number of systems with that energy, then the number of systems with total energy ${\displaystyle E_{i}}$ can be defined as the occupation of the state ${\displaystyle i}$ with variable ${\displaystyle a_{i}}$. The entire ensemble can then be defined as the occupancies of the energy states, where the total number of systems, ${\displaystyle {\mathcal {A}}}$, is: ${\displaystyle \Sigma _{i}a_{i}={\mathcal {A}}}$ and the total energy of the entire ensemble, or energy level, ${\displaystyle \varepsilon }$, is: ${\displaystyle \Sigma _{i}a_{i}E_{i}=\varepsilon }$ ## The Weight of a Configuration The occupancy of any of these energy levels depends on how many possible ways the total energy of the ensemble can be distributed amongst the systems or can be described by the occupancies of the energy states. Some of these configurations of the occupancies are mathematically more likely than others. The weight of a configuration, or the number of ways to distribute the energy of the ensemble, is important. By combinatorial mathematics the weight of a system with occupancies ${\displaystyle a_{1},a_{2},a_{3},...}$ in an ensemble containing ${\displaystyle {\mathcal {A}}}$ systems is: ${\displaystyle W(a_{1},a_{2},a_{3},...)={\frac {{\mathcal {A}}!}{a_{1}!a_{2}!a_{3}!...}}={\frac {{\mathcal {A}}!}{\Pi _{i}a_{i}!}}}$ ## System Weighting Consider an ensemble of ${\displaystyle {\mathcal {A}}}$ systems divided into two states. The occupancy of the first state is N and therefore the occupancy of the second state is A-N. The weight of this configuration as a function of N is: ${\displaystyle W(N)={\frac {{\mathcal {A}}!}{N!({\mathcal {A}}-N)!}}}$ It can be shown that as ${\displaystyle {\mathcal {A}}}$ increases the most probable distribution (N=${\displaystyle {\mathcal {A}}}$ /2) becomes much more heavily weighted than any other configuration. ## The Most Probable Distribution To describe a canonical ensemble, only the most probable set of occupancies need to be determined, denoted as a*. For a very large ${\displaystyle {\mathcal {A}}}$, the configuration with the largest weight will be much more heavily weighted than the other configurations. Therefore, this set of occupancies will be the most probable, and any configuration with a significantly large weight will be close to a*. However, for any given ensemble, not all occupancies are possible. This is due to the imposition of constraints upon the occupancies by the specific ensemble. In a canonical ensemble there are two constraints which were previously used to define a canonical system: 1. The sum over all energy levels has to add up to the total energy of the ensemble, denoted as ${\displaystyle \varepsilon }$ :${\displaystyle \Sigma _{i}a_{i}E_{i}=\varepsilon }$ 2. The sum of all of the occupancies must be equal to ${\displaystyle {\mathcal {A}}}$ :${\displaystyle \Sigma _{i}a_{i}={\mathcal {A}}}$ Therefore, given these two constraints, we must find a* that agrees with the above imposed constraints of the canonical ensemble. ## Calculating the Most Probable Occupancy To calculate the most probable occupancy, we must calculate the occupancies that would give the maximum weight. Lagrange multipliers are used to maximize this function. ## Mathematical Manipulation of W It was previously given that ${\displaystyle W(a_{1},a_{2},a_{3},...)={\frac {{\mathcal {A}}!}{a_{1}!a_{2}!a_{3}!...}}={\frac {{\mathcal {A}}!}{\Pi _{i}a_{i}!}}}$ However, this equation is difficult to manipulate. But, for a positive function, the maximum occurs at the same location as it does in the natural logarithm of the function. Since the maximum is what we’re really interested in finding, we can use ${\displaystyle \ln {W}}$ to find the maximum, which makes the mathematics easier to manipulate. Therefore the equation becomes: ${\displaystyle \ln {W}(a_{1},a_{2},a_{3},...)=\ln {\left({\frac {{\mathcal {A}}!}{\Pi _{i}a_{i}!}}\right)}}$ ${\displaystyle =\ln {{\mathcal {A}}!}-\ln {\Pi _{i}a_{i}!}}$ ${\displaystyle =\ln {{\mathcal {A}}!}-\Sigma _{i}\ln {a_{i}!}}$ ## Applying the Math to a Canonical Ensemble Using the Lagrange multipliers as previously described, we can determine the most probable occupancy of the canonical system. When Lagrange multipliers are applied to the ${\displaystyle \ln {W}}$ equation, following the aforementioned constraints, the new equation becomes: ${\displaystyle {\frac {\partial }{\partial a_{j}}}\left[\ln W(a_{1},a_{2},a_{3}...)-{\alpha }[({\Sigma _{i}a_{i}})-{\mathcal {A}}]-{\beta }[({\Sigma _{i}a_{i}E_{i}})-{\varepsilon }]\right]=0}$ Next, we must simplify each of the three terms in the equation and solve for the unknown constants \alpha and \beta. Term 1 ${\displaystyle {\frac {\partial }{\partial a_{j}}}\ln W(a_{1},a_{2},a_{3}...)={\frac {\partial }{\partial a_{j}}}\ln {{\mathcal {A}}!}-\ln {\Pi _{i}a_{i}!}=-{\frac {\partial }{\partial a_{j}}}\ln {a_{j}!}}$ Using Stirling’s Approximation: ${\displaystyle -{\frac {\partial }{\partial a_{j}}}\ln {(a_{j}!)}=-{\frac {\partial }{\partial a_{j}}}[{a_{j}}\ln {(a_{j})}-{a_{j}}]={-a_{j}}{\frac {1}{a_{j}}}-\ln {(a_{j})}+{1}=-\ln {(a_{j})}}$ Term 2 ${\displaystyle {\frac {\partial }{\partial a_{j}}}\alpha [(\Sigma _{i}a_{i})-{\mathcal {A}}]}$ For ${\displaystyle {\Sigma _{i}a_{i}}}$ , there will be one term of the sum where i = j, everything else is equal to zero. Therefore, this term becomes: ${\displaystyle {\frac {\partial }{\partial a_{j}}}{\alpha }{a_{j}}={\alpha }}$ Term 3 ${\displaystyle {\frac {\partial }{\partial a_{j}}}{\beta }[({\Sigma _{i}a_{i}E_{i}})-{\varepsilon }]}$ Like term 2, for ${\displaystyle {\Sigma _{i}E_{i}}}$ , there will be one term of the sum where i = j, everything else is equal to zero. Therefore, this term becomes: ${\displaystyle {\frac {\partial }{\partial a_{j}}}\beta {E_{j}a_{j}}=\beta {E_{j}}}$ Now that the three terms have been simplified, we can combine them into one equation to solve. This gives us: ${\displaystyle -\ln {(a_{j})}-{\alpha }-{\beta }{E_{j}}=0}$ Solving for ${\displaystyle {a_{j}}}$ gives: ${\displaystyle {a_{j}}={e^{-\alpha }}{e^{-{\beta }{E_{j}}}}}$ ## Determining ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ The next step of the derivation is to determine the constants ${\displaystyle \alpha }$ and ${\displaystyle \beta }$. We can use one of the constraints to determine ${\displaystyle \alpha }$. ${\displaystyle a_{i}={e^{-\alpha }}{e^{-\beta {E_{i}}}}}$ ${\displaystyle \Sigma _{i}a_{i}={\mathcal {A}}}$ ${\displaystyle {\Sigma _{i}}{e^{-\alpha }}{e^{{-\beta }{E_{i}}}}={\mathcal {A}}}$ ${\displaystyle {e^{-\alpha }}{\Sigma _{i}}{e^{{-\beta }{E_{i}}}}={\mathcal {A}}}$ ${\displaystyle {e^{\alpha }}={\frac {1}{\mathcal {A}}}{\Sigma _{i}}{e^{{-\beta }{E_{i}}}}}$ Using this equation and the constraint, we can define the probability of a system in the ensemble as the occupation of state i divided by the total number of systems in the ensemble, which can be represented by the following equation. ${\displaystyle {P_{i}}={\frac {a_{i}}{\mathcal {A}}}={\frac {e^{{-\beta }{E_{i}}}}{{\Sigma _{i}}{e^{{-\beta }{E_{j}}}}}}}$ Determining ${\displaystyle \beta }$ requires connecting it to classical thermodynamics. We can determine the average of a variable (<M>) over the states of a system by employing the equation ${\displaystyle \langle M\rangle ={M_{i}}{P_{i}}}$ The average energy is ${\displaystyle \langle E\rangle ={\frac {{\Sigma _{j}}{E_{j}}{e^{{-\beta }{E_{j}}}}}{{\Sigma _{i}}{e^{{-\beta }{E_{j}}}}}}}$ The average pressure is ${\displaystyle \langle p\rangle ={\frac {{\Sigma _{j}}-\left({\frac {\partial E_{j}}{\partial V}}\right)_{N}{e^{{-\beta }{E_{i}}}}}{{\Sigma _{i}}{e^{{-\beta }{E_{j}}}}}}}$ Knowing these equations and combining two other equations we can come up with the following equation, which can be compared to a classical thermodynamic equation shown below. ${\displaystyle \left({\frac {\partial \langle E\rangle }{\partial V}}\right)_{N,{\beta }}+{\beta }\left({\frac {\partial \langle p\rangle }{\partial {\beta }}}\right)_{N,V}=-\langle p\rangle }$ ${\displaystyle \left({\frac {\partial U}{\partial V}}\right)_{T,N}-{T}\left({\frac {\partial p}{\partial T}}\right)_{N,V}=-{p}}$ In order to better compare the equations, we can perform a calculation such that the signs in front of ${\displaystyle {\beta }}$ and T are the same, giving the following equations: ${\displaystyle \left({\frac {\partial \langle E\rangle }{\partial V}}\right)_{N,{\beta }}+{\beta }\left({\frac {\partial \langle p\rangle }{\partial {\beta }}}\right)_{N,V}=-{\langle p\rangle }}$ ${\displaystyle \left({\frac {\partial U}{\partial V}}\right)_{T,N}+{\frac {1}{T}}\left({\frac {\partial p}{\partial {\frac {1}{T}}}}\right)_{N,V}=-p}$ Since multiplying through by a constant would not change the equations, we can determine that ${\displaystyle {\beta }}$ is only proportional to ${\displaystyle {\frac {1}{T}}}$. ${\displaystyle \beta \propto {\frac {1}{T}}}$ We can introduce ${\displaystyle {k_{B}}}$ as a proportionality constant to convert the proportionality to an equation as follows: ${\displaystyle \beta ={\frac {1}{k_{B}T}}}$ ## The Boltzmann Distribution Earlier we determined the probability of a state within a system. Now that we know ${\displaystyle {\alpha }}$ and ${\displaystyle {\beta }}$, we can complete the equation to determine the Boltzmann Distribution. ${\displaystyle {P_{i}}={\frac {a_{i}}{\mathcal {A}}}={\frac {e^{-\beta E_{i}}}{\sum _{i}e^{-\beta E_{j}}}}={\frac {e^{-E_{i}/{k_{B}T}}}{\sum _{i}e^{-E_{i}/{k_{B}T}}}}}$ ## Example Consider a system with two singly-degenerate energy levels separated by ${\displaystyle 10^{-21}}$ J. Derive an equation for the probability of the system being in the ground state. Plot the probability of the system being in the ground state between temperatures 1 K and 1000 K. ${\displaystyle {P_{i}}={\frac {a_{i}}{\mathcal {A}}}={\frac {e^{-\beta E_{i}}}{\sum _{i}e^{-\beta E_{j}}}}={\frac {e^{-E_{i} \over {k_{B}T}}}{\sum _{i}e^{-E_{i} \over {k_{B}T}}}}}$ ${\displaystyle k_{B}=1.380\times 10^{-23}{J \over k}}$ is the Boltzmann constant in joules per Kelvin. The ground state has an energy level of zero (${\displaystyle {E_{o}=0}}$). Since the ground energy level is known and the difference between the ground energy level and the first high energy level is given (∆E = ${\displaystyle 10^{-21}}$J), the high energy level can be determined using the following equation: ${\displaystyle {\Delta E=E_{1}-E_{0}}}$ After using the actual values, the equation should look like this ${\displaystyle {10^{-21}=E_{1}-0}}$ ${\displaystyle {E_{1}=10^{-21}}}$ The probability equation can then be simplified to the following: ${\displaystyle {P_{i}}={\frac {e^{-E_{i} \over {k_{B}T}}}{e^{-E_{i} \over {k_{B}T}}+{e^{-E_{i} \over {k_{B}T}}}}}}$ ${\displaystyle {P_{0}}={\frac {e^{-E_{0} \over {k_{B}T}}}{e^{-E_{0} \over {k_{B}T}}+{e^{-E_{1} \over {k_{B}T}}}}}}$ After substituting the values determined into the equation, the probability equation should be the following: ${\displaystyle {P_{0}}={\frac {e^{0 \over {k_{B}T}}}{e^{0 \over {k_{B}T}}+{e^{{\Biggl (}{-(10^{-21}) \over {1.380\times 10^{-}23{\bigl (}{J \over K}{\bigr )}\times T}}{\Biggr )}}}}}}$ Since the ${\displaystyle e^{0}=1}$, the equation can be simplified to ${\displaystyle P_{0}={\frac {1}{1+{e^{{\Biggl (}{-(10^{-21}) \over {{1.380\times 10^{-23}{\bigl (}{J \over K}{\bigr )}}\times T}}{\Biggr )}}}}}}$ Note: The probability does not have units. This equation can be used to determine the probability of the system as the temperature changes in Kelvin. Zoomed in graph of the probability of the Boltzmann distribution over the change of temperature in Kelvin The probability of the system, which starts from one, gradually decreases as the temperature increases. the reason why can be observed from the equation. Since the temperature is in the denominator, the probability should be smaller as the denominator increases. After about 300 k, the line pattern starts to have a slight decrease toward 0.5 of probability. In other words, the probability of Boltzmann distribution at this system becomes less probable as the temperature exceeds 1000 K. # Molecular partition functions The partition function of a system, Q, provides the tools to calculate the probability of a system occupying state i .Partition function depends on composition,volume and number of particle. Larger the partition function allows to have more accessible energy states at that temperature.The general form of a partition function is a sum over the states of the system, ${\displaystyle Q=\sum _{i}\exp \left({\frac {-E_{i}}{k_{B}T}}\right)}$ Two equivalent ways could be used to write the the partition function. Sum over states approach allows to give different indices to the states with the same energy. Energy levels approach suggests that only energy levels with distinct energies have their own index. This requires that the energies levels of the entire system must be known and the calculations have to be calculated from sums over states. This limits the types of the systems that we can derive properties for. For ideal gases, we assume that the energy states of molecules are independent of those in other molecules. The molecular partition function, q, is defined as the sum over the states of an individual molecule. ${\displaystyle q=\sum _{i}\exp \left({\frac {-\epsilon _{i}}{k_{B}T}}\right)}$ The particles of an ideal gas could be considered distinguishable if they are different from each other and, therefore, the unique label could be assigned to each.On the contrary, the indistinguishable particles are impossible to assign a unique label, as they are the identical to each other.This article considers the indistinguishable particles of an ideal gas for which the partition function of the system (Q) could be expressed in terms of the molecular partition function (q),and the number of particles in the system (N).As the partition function allows to calculate the probability of the system occupying the state j.Such system could be assumed as isolated, as a microcanonical ensemble of particles, where the total volume, the total energy and number of particles are constant. However, the range of energies appropriate for composition, volume and temperature of canonical ensembles contributing to the system must be still considered for composition, volume and temperature.  Partition function of an ideal gas of indistinguishable particles ${\displaystyle Q={\frac {q^{N}}{N!}}}$ ## Molecular Partition Functions The energy levels of a molecule can be approximated as the sum of energies in the various degrees of freedom of the molecule, ${\displaystyle \epsilon =\epsilon _{trans}+\epsilon _{rot}+\epsilon _{vib}+\epsilon _{elec}}$ Correspondingly, we can divide molecular partition function (q), ${\displaystyle q=\sum _{i}\exp \left({\frac {-(\epsilon _{trans}+\epsilon _{rot}+\epsilon _{vib}+\epsilon _{elec})}{k_{B}T}}\right)}$ ${\displaystyle q=\sum _{i}\exp \left({\frac {-\epsilon _{trans,i}}{k_{B}T}}\right)\sum _{i}\exp \left({\frac {-\epsilon _{rot,i}}{k_{B}T}}\right)\sum _{i}\exp \left({\frac {-\epsilon _{vib,i}}{k_{B}T}}\right)\sum _{i}\exp \left({\frac {-\epsilon _{elec,i}}{k_{B}T}}\right)}$ ${\displaystyle q=\sum _{i}\exp \left({\frac {-\epsilon _{trans,i}}{k_{B}T}}\right)\times \sum _{i}\exp \left({\frac {-\epsilon _{rot,i}}{k_{B}T}}\right)\times \sum _{i}\exp \left({\frac {-\epsilon _{vib,i}}{k_{B}T}}\right)\times \sum _{i}\exp \left({\frac {-\epsilon _{elec,i}}{k_{B}T}}\right)}$ ${\displaystyle q=q_{trans}\times q_{rot}\times q_{vib}\times q_{elec}}$ ## Translational Partition Function The translational partition function, qtrans, is the sum of all possible translational energy states, which could be represented using one,two and three dimensional models for a particle in the box equation, depending on the system of the coordinates .The one and two dimensionsal spaces for a particle in the box equation forms are less commonly used than the three dimensional form as those do not account for the force acting on the particles inside of a box. For a molecule in the three dimensional space, the energy term in the general partition function equation is replaced with the particle in a 3D box equation. All molecules have three translational degrees of freedom, one for each axis the molecule can move along in three dimensional space.  Particle in a 3D Box Equation ${\displaystyle E_{n_{x},n_{y},n_{z}}={\frac {h^{2}}{8mL^{2}}}(n_{x}^{2}+n_{y}^{2}+n_{z}^{2})}$ The assumption that energy levels are continuous is valid because the space between energy levels is extremely small, resulting in minimal error. This form is convenient as it does not include a sum to infinity and can therefore be solved with relative ease. The translational partition function can be simplified further by defining a DeBroglie wavelength, ${\displaystyle \Lambda }$, of a molecule at a given temperature. ${\displaystyle q_{trans}={\frac {V}{\Lambda ^{3}}}}$ ${\displaystyle \Lambda =\left({\frac {2\pi mk_{B}T}{h^{2}}}\right)^{-1/2}}$ ### Thermal de Broglie Wavelength ## Rotational Partition Function The rotational partition function, ${\displaystyle q_{rot}}$, is the sum of all possible rotational energy levels. This sum is found by substituting the equation for the energy levels of a linear rigid rotor: ${\displaystyle E_{j}={\frac {\hbar ^{2}}{2\mu r^{2}}}J(J+1),J=0,1,2,...}$ Into the partition function, to produce an open form of the rotational partition function: ${\displaystyle q_{rot}=\sum _{j=1}^{\infty }\ g_{i}\exp \left({\frac {-E_{j}}{k_{B}T}}\right)}$ By solving this sum using a definite integral from zero to infinity, a closed form of this function can be found, making numerical evaluation much easier: ${\displaystyle q_{rot}={\frac {2k_{B}T\mu r^{2}}{{\text{ħ}}^{2}}}}$ The full derivation of the closed form of the rotational partition function of a linear rotor is given here. The special case of rotational partition function for homonuclear diatomics could cause the lower weight for alternating states,and result in the change of the rotational partition function. This case is based on the assumption that the total wavefunction of the two exchanging nuclei must be either asymmetric for the nuclei with the even spins (integer spins),or symmetric for the nuclei with odd spins. This function applies only to heteronuclear diatomic molecules. However, this equation can be altered by adding a variable to alter the equation based on the nature of the diatomic molecule being studied: ${\displaystyle q_{rot}={\frac {2k_{B}T\mu r^{2}}{\sigma \hbar ^{2}}}}$ where ${\displaystyle \sigma }$ is 1 for heteronuclear diatomics, and 2 for homonuclear diatomics. The characteristic temperature is the constant that combines many constants calculated for partition functions of rotational and vibrational levels,introducing the fact of partition function dependence on the temperature change. The characteristic temperature is now defined as , ${\displaystyle \Theta _{rot}}$ where, ${\displaystyle \Theta _{rot}={\frac {\hbar ^{2}}{2k_{B}\mu r^{2}}}}$ determining the rotational partition function can be calculated much easier using, ${\displaystyle q_{rot}={\frac {T}{\sigma \Theta _{rot}}}}$ ## Vibrational Partition Function A Partition Function (Q) is the denominator of the probability equation. It corresponds to the number of accessible states in a given molecule. Q represents the partition function for the entire system, which is broken down and calculated from each individual partition function of each molecule in the system. These individual partition functions are denoted by q. All molecules have four different types of partition functions: translational, rotational, vibrational, and electronic. Looking only at the vibrational aspect of the system, there is a specific unique equation used to calculate its partition function: The vibrational partition function of a linear molecule is, ${\displaystyle q_{vib}={\frac {1}{1-\exp \left({\frac {-h\nu }{k_{B}T}}\right)}}}$ In general, the molecule partition function can be written as an infinite sum. This is called the open form of the equation: ${\displaystyle q=\sum _{j}g_{i}\exp \left({\frac {-E_{j}}{k_{B}T}}\right)}$ It is much easier and more convenient to write this as a closed sum. This turns the equation into an approximative algebraic expression, with the following parameters in variables: • degeneracy ${\displaystyle g_{j}=1}$ • quantum number ${\displaystyle n=1,2,3...}$ • energy levels ${\displaystyle \varepsilon _{n}=h\nu (n+({\frac {1}{2}}))}$ The energy levels, ${\displaystyle \Delta E_{J}}$ are defined relative to the ground state of the system (i.e., the zero point energy is subtracted from each level), ${\displaystyle \Delta E_{J}=\varepsilon _{j}-\varepsilon _{0}}$ ${\displaystyle =h\nu (j+{\frac {1}{2}})-h\nu (0+{\frac {1}{2}})=h{\nu }j+{\frac {1}{2}}h{\nu }-{\frac {1}{2}}h{\nu }=h{\nu }j}$ By exploring some substitutions and derivations, the equation listed above for a linear molecule is achieved. The substitutions made include: • ${\displaystyle g_{j}=1}$ • ${\displaystyle E_{j}=h{\nu }j}$ • ${\displaystyle j=0,1,2...}$ At the same time, it is important to note that ${\displaystyle {\nu }}$ represents the vibrational frequency of the molecule. It can be calculated on its own prior by the following relation: ${\displaystyle v={\frac {1}{2\pi }}\left({\frac {k}{\mu }}\right)^{\frac {1}{2}}}$ where k represents the spring constant of the molecule and μ represents the reduced mass of the same molecule. ## Electronic Partition Function The electronic partition function (qel) of a system describes the electronic states of the system at thermodynamic equilibrium. This can be written as a sum over states, ${\displaystyle q_{el}=\sum g_{j}\exp(-\beta \epsilon _{j})}$ however due to high energy levels being present under most circumstances, the electronic partition function can be reduced to: ${\displaystyle q_{el}=g_{0}}$ Thus, the electronic partition function can usually be approximated as the ground state degeneracy of the atom or molecule. ## Molecular Partition Function The molecular partition function, q, is the total number of states accessible to the atom or molecule. It is the product of the vibrational, rotational and translational partition functions: ${\displaystyle q=q_{trans}\times q_{rot}\times q_{vib}\times q_{el}}$ # Sample problems ## Problem 1 Calculate the probability of a molecule of N2 being in the ground vibrational state at 298 K. The probability that a system occupies a given state at an instant of time and a specific temperature is given by the Boltzmann distribution. ${\displaystyle P_{i}={\frac {\exp \left({\frac {-E_{i}}{k_{B}T}}\right)}{\sum _{j}\exp \left({\frac {-E_{j}}{k_{B}T}}\right)}}}$ ${\displaystyle P_{i}={\frac {\exp \left({\frac {-E_{i}}{k_{B}T}}\right)}{Q}}}$ where: • i is the energy of the specific state, i, of interest • kB is Boltzmann's constant, which equals ${\displaystyle 1.3806\times 10^{-34}}$ JK-1 • T is the temperature in Kelvin The denominator of this function is known as the partition function, Q, which corresponds to the total number of accessible states of the molecule. The closed form of the molecular vibrational partition function is given by: ${\displaystyle q_{vib}={\frac {1}{1-e^{-h\nu /k_{B}T}}}}$ where: • ${\displaystyle \nu }$ is the fundamental vibrational frequency of N2 in s-1 • h is Planck's constant, which is ${\displaystyle 6.62607\times 10^{-34}}$ Js This is equivalent to Q since only the vibrational energy states are of interest and there is only one molecule of N2. The equation for determining the partition function Q, from molecular partition functions, q, is given by: ${\displaystyle Q={\frac {q^{N}}{N!}}}$ where: • N is the number of molecules The fundamental vibrational frequency of N2 in wavenumbers, ${\displaystyle {\tilde {\nu }}}$, is 2358.6cm-1 [1] The fundamental vibrational frequency in s-1 is given by: ${\displaystyle \nu ={\tilde {\nu }}\times c}$ where • c is the speed of light, which is ${\displaystyle 2.9979\times 10^{10}}$ cm/s For N2, ${\displaystyle \nu }$ = (2358.6cm-1) \times (2.9979 \times 10^{10| cm/s) = 7.0708 \times 10^{13}$

For N2 at 298 K,

${\displaystyle q_{v}=\left({\frac {1}{1-e^{-(6.62607\times 10^{-34}Js\times 7.0708\times 10^{13}s^{-1})/(1.3806\times 10^{-23}JK^{-1}\times 298K)}}}\right)=1.000011333}$

The vibrational energy levels follow that of a quantum mechanical harmonic oscillator. The energy levels are represented by:

${\displaystyle E_{n}=h\nu (n+{\frac {1}{2}})}$

where:

• n is the quantum vibrational number, which equals 0, 1, 2,...

For the ground state (n=0), the energy becomes:

${\displaystyle E_{0}={\frac {1}{2}}h\nu }$

Since the vibrational zero point energy is not zero, the energy levels are defined relative to the n=0 level. This is used in the molecular partition function above and therefore, the ground state is regarded as having zero energy.

For N2 the probability of being in the ground state at 298K is:

${\displaystyle P_{0}={\frac {e^{-E_{0}/k_{B}T}}{q_{v}}}}$
${\displaystyle P_{0}={\frac {e^{(0J)/(1.3806\times 10^{-23}\times 298K)}}{1.000011333}}}$
${\displaystyle P_{0}=0.999988667}$

This means that at room temperature, the probability of a molecule of N2 being in the ground vibrational state is 99.9988667%.

### References

1. Lide, D. R., (84th ed.). (2003-2004). Handbook of Chemistry and Physics. pg.9-85.

## Problem 2

Derive an equation for the population of rotational state i of a linear diatomic. Make a bar graph of the distribution of rotational states of N2 at 298 K.

1. Equation for the population of rotational state i

For a diatomic molecule, it can be approximated as a rigid rotor. Solving Schrödinger equation of rigid rotors gives the energy levels of the molecule at state J:

${\displaystyle E(J)=BJ(J+1)}$

where ${\displaystyle J}$ is the quantum number for total rotational angular momentum; ${\displaystyle B}$ is the rotational constant in cm-1.

• ${\displaystyle B={\frac {h}{8\pi ^{2}cI}}}$,

where ${\displaystyle h}$ is Planck constant; ${\displaystyle c}$ is the vacuum light speed in cm/s; and ${\displaystyle I}$ is the moment of inertia.

• ${\displaystyle I=\mu r^{2}}$,

where ${\displaystyle \mu }$ is the reduced mass and ${\displaystyle r}$ is the bond length.

By the Maxwell–Boltzmann_distribution, the population of rotational state i comparing to ground state is:

${\displaystyle {\frac {N_{i}}{N_{0}}}={\frac {g_{i}}{g_{0}}}\left({\frac {e^{-E_{i}/k_{B}T}}{e^{-E_{0}/k_{B}T}}}\right)}$

where ${\displaystyle g}$ is the degeneracy of the state; ${\displaystyle E}$ is the energy of the state; ${\displaystyle k_{B}}$ is Boltzmann constant and ${\displaystyle T}$ is the temperature.

Substitute ${\displaystyle E_{i}=hcBi(i+1)}$ ,${\displaystyle E_{0}=hcB0(0+1)=0}$, and the degeneracy of the state i ${\displaystyle g_{i}=2i+1}$ to the equation, get:

${\displaystyle {\frac {N_{i}}{N_{0}}}=(2i+1)\left({\frac {e^{-hcBi(i+1)/k_{B}T}}{e^{0/k_{B}T}}}\right)=(2i+1)e^{-hcBi(i+1)/k_{B}T}}$

### 2. Distribution of rotational states of N2

For N2, the population of state i is:

${\displaystyle {\frac {N_{i}}{N_{0}}}=(2i+1)e^{-hcBi(i+1)/k_{B}T}}$

Combine the constant part, define constant ${\displaystyle a=hcB/k_{B}T={\frac {h^{2}}{8\pi ^{2}k_{B}T\mu r^{2}}}}$.

The reduced mass μ of Nitrogen is 7.00D a=1.16×10-26 kg and the bond length r of N2 is 110 pm = 1.10×10-10m.

Substitute T = 298 K, ${\displaystyle a={\frac {h^{2}}{8\pi ^{2}k_{B}T\mu r^{2}}}=9.60\times 10^{-3}}$

By sum over states, qrot=1/a = 104, so the probability of N2 occupying the ground vibrational state at 298 K is 1/104 = 9.60×10-3

The bar graph of the distribution of rotational states of N2 at 298 K is:

Distribution of rotational states of nitrogen gas at 298 K.

## Problem 3

Estimate the number of translational states that are available to a molecule of N2 in 1 m3 container at 298 K.

The equation that is used to determine translational states of the molecule of N2 at 298 K is shown below.

${\displaystyle q_{trans}(V,T)=\left({\frac {2\pi mk_{B}T}{h^{2}}}\right)^{\frac {3}{2}}}$${\displaystyle V}$

Where ${\displaystyle q_{trans}}$(V,T) represents the partition function for translation, ${\displaystyle m}$ represents the mass of the particle in kilograms (kg), ${\displaystyle k_{B}}$ represents Boltzmann's constant ${\displaystyle (1.38064852\times 10^{-23}J\times K^{-1})}$, ${\displaystyle T}$ represents the temperature in kelvins ${\displaystyle (K)}$, ${\displaystyle h^{2}}$ represents Planck's constant ${\displaystyle (6.626\times 10^{-34}J\times s)}$ and ${\displaystyle V}$ represents the volume in 3 dimensions in ${\displaystyle m^{3}}$.

The steps for solving this problem is shown below:

m = ${\displaystyle 28.0102}$ amu

m = ${\displaystyle (28.0102amu)(1.6605\times 10^{-27}kg)}$

m = ${\displaystyle 4.6512\times 10^{-26}kg}$

${\displaystyle q_{trans}(V,T)=\left({\frac {2\pi mk_{B}T}{h^{2}}}\right)^{3/2}V}$

${\displaystyle q_{trans}(V,T)=\left({\frac {2\times 3.14159\times (4.6512\times 10^{-26}kg)\times (1.38064852\times 10^{-23}J/K)\times 298K}{{(6.626\times 10^{-34}J\times s})^{2}}}\right)^{3/2}(1.000m^{3})}$

${\displaystyle q_{trans}(V,T)=1.4322\times 10^{32}}$

Therefore, there should be ${\displaystyle 1.4322\times 10^{32}}$ translational states for N2 at 298 K.

## Problem 4

Calculate the DeBroglie wavelength, rotational temperature, and vibrational temperature of N2 and Cl2 at 298 K.

### The DeBroglie Wavelength

${\displaystyle \Lambda =\left({\frac {2{\pi }mk_{B}T}{h^{2}}}\right)^{-1/2}}$

where:

• m is the mass of the molecule.
• Boltzmann's constant kB = 1.3806488×10-23 J K-1
• Planck's constant h = 6.62606957×10-34 J s

For N2 at 298K,

${\displaystyle m=2\times 14.0067u}$
${\displaystyle m=28.0134u\times {\frac {1.66054\times 10^{-27}kg}{u}}}$
${\displaystyle m=4.65173712\times 10^{-26}kg}$
${\displaystyle \Lambda =\left({\frac {2{\pi }(4.65173712\times 10^{-26}kg)(1.3806488\times 10^{-23}JK^{-1})(298K)}{(6.62606957\times 10^{-34}Js)^{2}}}\right)^{-1/2}}$
${\displaystyle \Lambda =\left({\frac {1.20252611\times 10^{-45}kgJ}{4.39047979\times 10^{-67}J^{2}s^{2}}}\right)^{-1/2}}$
${\displaystyle \Lambda =(2.738940097\times 10^{-21}kgJ^{-1}s^{-2})^{-1/2}}$
${\displaystyle \Lambda =1.9107714\times 10^{-11}m}$

For Cl2 at 298K,

${\displaystyle m=2\times 35.453u}$
${\displaystyle m=70.906u\times {\frac {1.66054\times 10^{-27}kg}{u}}}$
${\displaystyle m=1.17742249\times 10^{-25}kg}$
${\displaystyle \Lambda =\left({\frac {2{\pi }(1.17742249\times 10^{-25}kg)(1.3806488\times 10^{-23}JK^{-1})(298K)}{(6.62606957\times 10^{-34}Js)^{2}}}\right)^{-1/2}}$
${\displaystyle \Lambda =\left({\frac {3.04376893\times 10^{-45}kgJ}{4.39047979\times 10^{-67}J^{2}s^{2}}}\right)^{-1/2}}$
${\displaystyle \Lambda =(6.932656726\times 10^{21}kgJ^{-1}s^{-2})^{-1/2}}$
${\displaystyle \Lambda =1.20101976\times 10^{-11}m}$

Unit Conversion for the DeBroglie Wavelength:

${\displaystyle (kgJ^{-1}s^{-2})^{-1/2}=\left({\frac {kg}{kgm^{2}s^{-2}s^{2}}}\right)^{-1/2}}$
${\displaystyle (kgJ^{-1}s^{-2})^{-1/2}=(m^{-2})^{-1/2}}$
${\displaystyle (kgJ^{-1}s^{-2})^{-1/2}=m}$

### The Rotational Temperature

${\displaystyle \Theta _{r}={\frac {\hbar ^{2}}{2k_{B}\mu r_{e}^{2}}}}$

where:

• Planck's constant ${\displaystyle \hbar ={\frac {h}{2\pi }}={\frac {6.62606957\times 10^{-34}Js}{2\pi }}=1.054571726\times 10^{-34}Js}$
• μ is the reduced mass.
• re is the bond length between two atoms in a molecule

For N2,

re = 1.09769Å = 1.09769×10-10m [1]
${\displaystyle \mu ={\frac {m_{N}m_{N}}{m_{N}+m_{N}}}={\frac {m_{N}^{2}}{2m_{N}}}={\frac {m_{N}}{2}}={\frac {14.00307400529u}{2}}=7.001537003u}$
${\displaystyle \mu =7.001537003u\times {\frac {1.66054\times 10^{-27}kg}{u}}}$
${\displaystyle \mu =1.16263323\times 10^{-26}kg}$
${\displaystyle \Theta _{r}={\frac {(1.054571726\times 10^{-34}Js)^{2}}{2(1.3806488\times 10^{-23}JK^{-1})(1.16263323\times 10^{-26}kg)(1.09769\times 10^{-10}m)^{2}}}}$
${\displaystyle \Theta _{r}={\frac {1.11212153\times 10^{-68}J^{2}s^{2}}{3.86825738\times 10^{-69}JK^{-1}kgm^{2}}}}$
${\displaystyle \Theta _{r}=2.874993624Js^{2}Kkg^{-1}m^{-2}}$
${\displaystyle \Theta _{r}=2.874993624K}$

For Cl2,

re = 1.988Å = 1.988×10-10m [2]
${\displaystyle \mu ={\frac {m_{Cl}m_{Cl}}{m_{Cl}+m_{Cl}}}={\frac {m_{Cl}^{2}}{2m_{Cl}}}={\frac {m_{Cl}}{2}}={\frac {34.968852714u}{2}}=17.48442636u}$
${\displaystyle \mu =17.4842636u\times {\frac {1.66054\times 10^{-27}kg}{u}}}$
${\displaystyle \mu =2.90335893\times 10^{-26}kg}$
${\displaystyle \Theta _{r}={\frac {(1.054571726\times 10^{-34}Js)^{2}}{2(1.3806488\times 10^{-23}JK^{-1})(2.90335893\times 10^{-26}kg)(1.988\times 10^{-10}m)^{2}}}}$
${\displaystyle \Theta _{r}={\frac {1.11212153\times 10^{-68}J^{2}s^{2}}{3.16844888\times 10^{-68}JK^{-1}kgm^{2}}}}$
${\displaystyle \Theta _{r}=0.350998729Js^{2}Kkg^{-1}m^{-2}}$
${\displaystyle \Theta _{r}=0.350998729K}$

Unit Conversion for the Rotational Temperature:

${\displaystyle Js^{2}Kkg^{-1}m^{-2}=J(kg^{-1}m^{-2}s^{2})K}$
${\displaystyle Js^{2}Kkg^{-1}m^{-2}=JJ^{-1}K}$
${\displaystyle Js^{2}Kkg^{-1}m^{-2}=K}$

### The Vibrational Temperature

${\displaystyle \Theta _{v}={\frac {h\nu }{k_{B}}}}$
${\displaystyle \Theta _{v}={\frac {h{\tilde {\nu }}c}{k_{B}}}}$

where:

• ${\displaystyle {\tilde {\nu }}}$ the vibrational frequency of the molecule in wavenumbers.
• The speed of light c = 2.99792458×1010cm s-1

For N2,

${\displaystyle {\tilde {\nu }}=2358.57cm^{-1}}$ [1]
${\displaystyle \Theta _{v}={\frac {(6.62606957\times 10^{-34}Js)(2358.57cm^{-1})(2.99792458\times 10^{10}cms^{-1})}{1.3806488\times 10^{-23}JK^{-1}}}}$
${\displaystyle \Theta _{v}={\frac {4.6851712\times 10^{-20}J}{1.3806488\times 10^{-23}JK^{-1}}}}$
${\displaystyle \Theta _{v}=3393.456175K}$

For Cl2,

${\displaystyle {\tilde {\nu }}=559.7cm^{-1}}$ [2]
${\displaystyle \Theta _{v}={\frac {(6.62606957\times 10^{-34}Js)(559.7cm^{-1})(2.99792458\times 10^{10}cms^{-1})}{1.3806488\times 10^{-23}JK^{-1}}}}$
${\displaystyle \Theta _{v}={\frac {1.11181365\times 10^{-20}J}{1.3806488\times 10^{-23}JK^{-1}}}}$
${\displaystyle \Theta _{v}=805.2834645K}$

### References

1. a b Lide, D. R., (84th ed.). (2003-2004). Handbook of Chemistry and Physics. CRC Press. pg.9-85.
2. a b Lide, D. R., (84th ed.). (2003-2004). Handbook of Chemistry and Physics. CRC Press. pg.9-83.
Molecule N2 Cl2
${\displaystyle \Lambda }$ 1.91×10-11 m 1.20×10-11m
${\displaystyle \Theta _{r}}$ 2.87 K 0.351 K
${\displaystyle \Theta _{v}}$ 3.39×103 K 805 K

## Problem 5

At what temperature would the probability for N2 being the ground vibrational state be reduced to 50%?

50% of the N2 molecules are in the ground state, and 50% are in the excited state. So, the population can be expressed in the following equation:

${\displaystyle P_{0}=0.5}$

The value for the ${\displaystyle {\tilde {\nu }}}$ wavenumber value for the N2 molecule, taken from the NIST Webbook, is 2358.57 cm-1

This can be converted into the fundamental vibrational frequency by the relation,

${\displaystyle \nu ={\tilde {\nu }}\times c}$
where:
• ${\displaystyle {\tilde {\nu }}}$ is fundamental wavenumber for the molecule in cm-1.
• ${\displaystyle c}$ is the speed of light, 2.998x1010 cm/s By knowing this relation, the fundamental vibrational frequency can be calculated.
${\displaystyle \nu ={\tilde {\nu }}\times c=2358.57cm^{-1}\times 2.998\times 10^{10}cm/s=7.07099\times 10^{13}s^{-1}}$
The population of a system can be expressed by the following equation:
${\displaystyle P_{0}={\frac {\exp \left({\frac {-E_{0}}{k_{B}T}}\right)}{q_{vib}}}}$
where:
• ${\displaystyle P_{0}}$ is the population of the ground state molecules
• ${\displaystyle E_{0}}$ is the energy at the ground state
• ${\displaystyle k_{B}}$is the Boltzmann constant, 1.38064853 \times 10-23 J/K
• ${\displaystyle T}$ is the temperature of the system in K
• ${\displaystyle q_{vib}}$ is the molecular vibrational partition function

${\displaystyle q_{vib}}$ is related by the following equation:

${\displaystyle q_{vib}={\frac {1}{1-\exp \left({\frac {-h\nu }{k_{B}T}}\right)}}}$
where:
• h is the Planck constant, 6.626*10-34 J*s
Knowing these relations, we can solve for the temperature of the system when the molecule is reduced to 50% ground state, which is when 50% of the molecules are in the ground state and 50% of the molecules are excited.
${\displaystyle P_{0}={\frac {\exp {\frac {-E_{0}}{k_{B}T}}}{q_{vib}}}=50\%=0.5}$
${\displaystyle q_{vib}={\frac {\exp \left({\frac {-E_{0}}{k_{B}T}}\right)}{P_{0}}}}$
${\displaystyle q_{vib}={\frac {\exp(0)}{0.5}}=2}$
Knowing the value of ${\displaystyle q_{vib}}$, we can solve the molecular vibrational partition function to obtain the exact value of T for when 50% of the molecules are in the ground state.
${\displaystyle q_{vib}={\frac {1}{1-e^{\frac {-h\nu }{k_{B}T}}}}}$${\displaystyle =2}$
${\displaystyle 2={\frac {1}{1-\exp ^{\frac {-h\nu }{k_{B}T}}}}}$
${\displaystyle 2(1-\exp ^{\frac {-h\nu }{k_{B}T}})=1}$
${\displaystyle 2-2\exp ^{\frac {-h\nu }{k_{B}T}}=1}$
${\displaystyle -2\exp \left({\frac {-h\nu }{k_{B}T}}\right)=-1}$
${\displaystyle \exp \left({\frac {-h\nu }{k_{B}T}}\right)=0.5}$
${\displaystyle \ln(\exp \left({\frac {-h\nu }{k_{B}T}}\right)=\ln(0.5)}$
${\displaystyle {\frac {-h\nu }{k_{B}T}}=-0.6931472}$
${\displaystyle T={\frac {h\nu }{0.6931472k_{B}}}}$
Therefore, by substituting the Planck constant, fundamental vibrational frequency and the Boltzmann constant into the equation, we can obtain the temperature for when half of the molecules are in the ground state.
${\displaystyle T=6.626\times 10^{-34}J\cdot s\times 7.07099\times 10^{13}s^{-}1/(0.6931472\times 1.38064853\times 10^{-23}J/K)=4895.792994K}$
The temperature at which 50% of the N2 molecules are in the ground vibrational state is 4895 K.

# Variables of the Canonical Ensemble

The Canonical Ensemble partition function depends on variables including the composition (N), volume (V) and temperature (T) of a given system, where the above partition function equation is still valid with ${\displaystyle Q_{N,V,T}=Q(N,V,T)}$.

The partition function of the canonical ensemble is defined as the sum over all states of a particular system involving each states respective energies, and represented by the following equation,

${\displaystyle Q=\sum _{j}\exp \left({\frac {-E_{j}}{k_{B}T}}\right)}$

In this equation, ${\displaystyle k_{B}}$ represents the Boltzmann Constant with a value of ${\displaystyle k_{B}=1.3806503\times 10^{-23}JK^{-1}}$, T represents the temperature in kelvin and ${\displaystyle E_{j}}$ is the energy at state j.

## Internal Energy

The partition function can also be related to all state functions from classical thermodynamics, such as U, A, G and S. The ensemble average of the internal energy in a given system is the thermodynamic equivalent to internal energy, as stated by the Gibbs postulate, and defined by,

${\displaystyle U=\langle E\rangle ={\frac {\sum _{j}E_{j}\exp \left({\frac {-E_{j}}{k_{B}T}}\right)}{Q}}}$

Where the variable Q here represents the partition function term. For a canonical ensemble, the internal energy can be derived from the above equation by considering the derivative of Q with respect to T,

{\displaystyle {\begin{alignedat}{7}{\frac {\partial {}}{\partial {T}}}Q={\frac {\partial {}}{\partial {T}}}\sum _{j}\exp \left({\frac {-E_{j}}{k_{B}T}}\right)&\\=\sum _{j}{\frac {\partial {}}{\partial {T}}}\exp \left({\frac {-E_{j}}{k_{B}T}}\right)&\\=\sum _{j}\exp \left({\frac {-E_{j}}{k_{B}T}}\right){\frac {\partial {}}{\partial {T}}}\left({\frac {-E_{j}}{k_{B}T}}\right)&\\={\frac {E_{j}}{k_{B}}}{\frac {1}{T^{2}}}\exp \left({\frac {-E_{j}}{k_{B}T}}\right)&\\\end{alignedat}}}
{\displaystyle {\begin{alignedat}{7}{\frac {\partial {}}{\partial {T}}}Q=\sum _{j}{\frac {E_{j}}{k_{B}}}{\frac {1}{T^{2}}}\exp \left({\frac {-E_{j}}{k_{B}T}}\right)={\frac {1}{k_{B}T^{2}}}\sum _{j}E_{j}\exp \left({\frac {-E_{j}}{k_{B}T}}\right)&\\={\frac {Q}{Q}}{\frac {1}{k_{B}T^{2}}}\sum _{j}E_{j}\exp \left({\frac {-E_{j}}{k_{B}T}}\right)&\\=Q{\frac {1}{k_{B}T^{2}}}{\frac {\sum _{j}E_{j}\exp \left({\frac {-E_{j}}{k_{B}T}}\right)}{Q}}&\\={\frac {Q}{k_{B}T^{2}}}{\frac {\sum _{j}E_{j}\exp \left({\frac {-E_{j}}{k_{B}T}}\right)}{Q}}={\frac {Q}{k_{B}T^{2}}}\langle E\rangle \end{alignedat}}}

Rearranging this resulting equation for ${\displaystyle \langle E\rangle }$ yields,

${\displaystyle \langle E\rangle =k_{B}T^{2}{\frac {1}{Q}}\left({\frac {\partial {Q}}{\partial {T}}}\right)_{N,V}}$

and from the Gibbs Postulate,

${\displaystyle U=k_{B}T^{2}\left({\frac {\partial {\textrm {ln}}Q}{\partial T}}\right)_{N,V}}$

## Helmholtz Energy

The Helmholtz Energy, or Helmholtz Free Energy, of a Canonical Ensemble represents the amount of work and energy obtainable by a certain closed system under constant concentration, volume and temperature. The expression for A utilizes both the internal energy and entropy of the system, and is derived by,

${\displaystyle A=U-TS}$

Where A is the Helmholtz Energy term, U represents the internal energy and S represents the entropy of the system being studied. Substituting the per-determined values for the canonical ensemble,

${\displaystyle A=\langle E\rangle -T\left({\frac {\langle E\rangle }{T}}+k_{B}\ln {Q}\right)}$

${\displaystyle A=\langle E\rangle -\langle E\rangle -k_{B}T\ln {Q}}$

${\displaystyle A=-k_{B}T\ln \left(Q\right)}$

Q represents the partition function for this particular system, and is proportional to the absolute free energy. Thus, an increase in ${\displaystyle \ln(Q)}$ will occur with an increase in the number of total accessible states, and a more negative free energy.

## Entropy

The Gibbs definition of Entropy is described by,

${\displaystyle S=-k_{B}\sum _{j}P_{j}\ln {P_{j}}}$

where ${\displaystyle P_{j}}$ represents the Probability of being in state i and described by the weight of the state divided by the sum of all possible weights, such as,

${\displaystyle P_{i}={\frac {\exp \left({\frac {-E_{i}}{k_{B}T}}\right)}{\sum _{j}\exp \left({\frac {-E_{j}}{k_{B}T}}\right)}}}$

and thus large probabilities of multiple states correlate with large values of S. By using the above Gibbs definition, the entropy of a Canonical Ensemble (NVT) can be derived, in terms of Q, to yield the following expression,

${\displaystyle S={\frac {\langle E\rangle }{T}}+k_{B}\ln {Q}}$ or ${\displaystyle S=k_{B}T\left({\frac {\partial {\ln {Q}}}{\partial {T}}}\right)_{N,V}+k_{B}\ln {Q}}$

The value of Q in this equation can also be represented as ${\displaystyle Q={\frac {q^{n}}{N!}}}$, where q is equivalent to the partition function of the molecules, with N representing the number of molecules, in the system.

The partition function is also represented by the denominator of the probability term for a certain state, given by the following,

${\displaystyle P_{i}={\frac {\exp \left({\frac {-E_{j}}{k_{B}T}}\right)}{Q}}}$

## Chemical potential

The Chemical Potential, μ, represents the change in the Helmholtz free energy when an additional particle is added to the canonical ensemble system. A heterogeneous system represents a system containing more than one type of gas and the expression for the chemical potential here is μi, where i represents each differing species. This term is derived from Helmholtz free energy, A, with respect to number of molecules for a certain species i. Thus,

${\displaystyle \mu _{i}=\left({\frac {\partial {A}}{\partial {N_{i}}}}\right)_{T,V,N}}$

And utilizing the above Helmholtz expression, the following can be derived,

${\displaystyle =\left({\frac {\partial {}-k_{B}T\ln {Q}}{\partial {N_{i}}}}\right)_{T,V,N}}$

${\displaystyle =-k_{B}T\left({\frac {\partial {\ln {\frac {q_{i}\left(V,T\right)^{N_{i}}}{N_{i}!}}}}{\partial {N_{i}}}}\right)_{T,V,N}}$

${\displaystyle =-k_{B}T\left({\frac {\partial {\ln {q_{i}\left(V,T\right)^{N_{i}}}}-\ln {N_{i}!}}{\partial {N_{i}}}}\right)_{T,V,N}}$

${\displaystyle =-k_{B}T\left({\frac {\partial {N_{i}\ln {q_{i}\left(V,T\right)}}-N_{i}\ln {N_{i}}+N_{i}}{\partial {N_{i}}}}\right)_{T,V,N}}$

${\displaystyle =-k_{B}T\left(\ln {q_{i}}\left(V,T\right)-\ln {N_{i}}-1+1\right)_{T,V,N}}$

${\displaystyle \mu =-k_{B}T\ln {}\left({\frac {q_{i}\left(V,T\right)}{N_{i}}}\right)}$

This definition helps describe the change in the Helmholtz free energy with a changing system composition, such as a reaction forming and/or consuming certain molecules in a system.

## Heat Capacity

In this section, we will complete the deviation of heat capacity in terms of the partition function Q.

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