# Statistical Thermodynamics and Rate Theories/Chemical Equilibrium

## Chemical Equilibrium from Statistical Thermodynamics

Consider the general gas phase chemical reaction represented by

${\displaystyle {\ce {\nu_{A}A + \nu_{B}B <=>\nu_{C}C + \nu_{D}D}}}$

where A, B, C and D are the reactants and products of the reaction, and ${\displaystyle \nu _{A}}$ is the stoichiometric coefficient of chemical A, ${\displaystyle \nu _{B}}$ is the stoichiometric coefficient of chemical B, and so on. Each of the gases involved in the reaction will eventually reach an equilibrium concentration according to their equilibrium constant(${\displaystyle K_{c}}$):

${\displaystyle K_{c}={\frac {[C]^{\nu _{C}}[D]^{\nu _{D}}}{[A]^{\nu _{A}}[B]^{\nu _{B}}}}={\frac {\rho _{C}^{\nu _{C}}\rho _{D}^{\nu _{D}}}{\rho _{A}^{\nu _{A}}\rho _{B}^{\nu _{B}}}}}$

The chemical potential of species i is given by the equation

${\displaystyle \mu _{i}={\left({\frac {\partial A}{\partial N_{i}}}\right)}_{T,V,N_{j\neq 1}}}$

where A is the Helmholtz energy, and ${\displaystyle N_{i}}$ is the number of molecules of species i. The Helmholtz energy can be determined as a function of the total partition function, Q:

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

where ${\displaystyle k_{B}}$ is the Boltzmann constant and T is the temperature of the system. The total partition function is given by

${\displaystyle Q={\frac {q_{i}(V,T)^{Ni}}{N_{i}!}}}$

where ${\displaystyle q_{i}}$ is the molecular partition function of chemical species i. Substituting these definitions into the equation for chemical potential yields:

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

A variable, ${\displaystyle \lambda }$, is then defined such that ${\displaystyle dN_{j}=\nu _{j}d\lambda }$, where j = A, B, C or D and ${\displaystyle \nu _{j}}$ is taken to be positive for products and negative for reactants. A change in ${\displaystyle \lambda }$ therefore corresponds to a change in the concentrations of the reactants and products. Thus, at equilibrium,

${\displaystyle \left({\frac {\partial A}{\partial \lambda }}\right)_{T,V}=0}$

From Classical thermodynamics, the total differential of A is:

${\displaystyle dA=-SdT-pdV+\sum _{j}\mu _{j}dN_{j}}$

For a reaction at a fixed volume and temperature (such as in the canonical ensemble), ${\displaystyle dT}$ and ${\displaystyle dV}$ equal 0. Therefore,

${\displaystyle dA=\sum _{j}\mu _{j}dN_{j}}$
${\displaystyle dA=\sum _{j}\mu _{j}\nu _{j}d\lambda }$
${\displaystyle dA=d\lambda \sum _{j}\mu _{j}\nu _{j}}$
${\displaystyle \sum _{j}\mu _{j}\nu _{j}=0}$

Substituting the expanded form of chemical potential:

${\displaystyle -k_{B}T\sum _{j}\ln \left({\frac {q_{i}}{N_{i}}}\right)\nu _{j}=0}$
${\displaystyle \sum _{j}\ln \left({\frac {q_{i}}{N_{i}}}\right)\nu _{j}=0}$
${\displaystyle \sum _{j}\nu _{j}[\ln(q_{j})-\ln(N_{j})]=0}$

For the reaction ${\displaystyle {\ce {\nu_{a}A + \nu_{b}B <=>\nu_{c}C + \nu_{d}D}}}$:

${\displaystyle [\nu _{C}\ln(q_{C})-\nu _{C}\ln(N_{C})]+[\nu _{D}\ln(q_{D})-\nu _{D}\ln(N_{D})]-[\nu _{A}\ln(q_{A})-\nu _{A}\ln(N_{A})]-[\nu _{B}\ln(q_{B})-\nu _{B}\ln(N_{B})]=0}$

This equation simplifies to

${\displaystyle {\frac {(q_{C})^{\nu _{C}}(q_{D})^{\nu _{D}}}{(q_{A})^{\nu _{A}}(q_{B})^{\nu _{B}}}}={\frac {(N_{C})^{\nu _{C}}(N_{D})^{\nu _{D}}}{(N_{A})^{\nu _{A}}(N_{B})^{\nu _{B}}}}}$

By dividing all terms by volume, and noting the relationship ${\displaystyle {\frac {N_{A}}{V}}={\frac {\#molecules}{volume}}=\rho _{A}=[A]}$, the following equation is obtained:

${\displaystyle K_{c}={\frac {\rho _{C}^{\nu _{C}}\rho _{D}^{\nu _{D}}}{\rho _{A}^{\nu _{A}}\rho _{B}^{\nu _{B}}}}={\frac {(q_{C}/V)^{\nu _{C}}(q_{D}/V)^{\nu _{D}}}{(q_{A}/V)^{\nu _{A}}(q_{B}/V)^{\nu _{B}}}}}$

## Partition Functions

The molecular partition function, ${\displaystyle q_{i}}$ is defined as the product of the transnational, rotational, vibration and electronic partition functions:

${\displaystyle q=q_{trans}q_{rot}q_{vib}q_{elec}}$

These components of the molecular partition function may be defined as follows:

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

where m is the mass of a single particle and h is Planck's constant, and T is temperature.

${\displaystyle q_{rot}=\left({\frac {2k_{B}T\mu {r_{e}}^{2}}{\sigma \hbar ^{2}}}\right)}$

where ${\displaystyle \mu }$ is the reduced mass of the molecule, ${\displaystyle r_{e}}$ is the bond length between the atoms in the molecule, and ${\displaystyle \hbar }$ is the reduced Planck's constant.

${\displaystyle q_{vib}={\frac {1}{1-\exp \left({\frac {-h\nu }{k_{B}T}}\right)}}}$

where ${\displaystyle \nu }$ is the vibrational frequency of the molecule.

${\displaystyle q_{elec}=g_{1}e^{D_{0}/k_{b}T}}$

where ${\displaystyle g_{1}}$ is the degeneracy of the ground state, and ${\displaystyle D_{0}}$ is the bond energy of the molecule.

Thus, the equilibrium constant of a chemical reaction can be expressed in terms of the molecular partition functions and the difference in atomization energies of the products and reactants ${\displaystyle \Delta D_{0}}$.

${\displaystyle K_{eq}={\frac {\prod _{i}^{products}\left(q_{i}/V\right)^{v_{i}}}{\prod _{i}^{reactants}\left(q_{i}/V\right)^{v_{i}}}}\exp \left({\frac {\Delta D_{0}}{k_{B}T}}\right)}$
${\displaystyle \Delta D_{0}=\sum _{products}^{i}D_{0,i}-\sum _{reactants}^{i}D_{0,i}}$

## Example

Calculate the equilibrium constant for the reaction of ${\displaystyle {\ce {H2 (g)}}}$ and ${\displaystyle {\ce {Cl2 (g)}}}$ at 650 K.

${\displaystyle {\ce {H2 (g) + Cl2 (g) <=> 2HCl (g)}}}$

 Equilibrium Constant Equation (From Molecular Partition Functions) ${\displaystyle K_{c}(T)={\left({\frac {{\Bigl (}{\frac {q_{\text{HCl}}}{V}}{\Bigr )}^{2}}{{\Bigl (}{\frac {q_{{\text{H}}_{2}}}{V}}{\Bigr )}{\Bigl (}{\frac {q_{{\text{Cl}}_{2}}}{V}}{\Bigr )}}}\right)}}$

${\displaystyle K_{c}(T)={\left({\frac {{\Bigl (}{\frac {q_{trans{\text{HCl}}}q_{rot{\text{HCl}}}q_{vib{\text{HCl}}}q_{elec{\text{HCl}}}}{V}}{\Bigr )}^{2}}{{\Bigl (}{\frac {q_{trans{\text{H}}_{2}}q_{rot{\text{H}}_{2}}q_{vib{\text{H}}_{2}}q_{elec{\text{H}}_{2}}}{V}}{\Bigr )}({\frac {q_{trans{\text{Cl}}_{2}}q_{rot{\text{Cl}}_{2}}q_{vib{\text{Cl}}_{2}}q_{elec{\text{Cl}}_{2}}}{V}}{\Bigr )}}}\right)}}$
A simple problem solving strategy for finding equilibrium constants via statistical mechanics is to separate the equation into the molecular partition functions of each of the reactant and product species, solve for each one, and recombine them to arrive at a final answer.

${\displaystyle \left({\frac {q_{\text{HCl}}}{V}}\right)=\left({\frac {2\pi mk_{B}T}{h^{2}}}\right)^{(3/2)}\times {\frac {2k_{B}T\mu r_{e}^{2}}{\sigma \hbar ^{2}}}\times {\frac {1}{1-\exp {{\Bigl (}{\frac {-h\nu }{k_{B}T}}{\Bigr )}}}}\times g_{1}\exp(D_{0}/K_{B}T)}$

In order to simplify the calculations of molecular partition functions, the characteristic temperature of rotation (${\displaystyle \Theta _{r}}$) and vibration (${\displaystyle \Theta _{\nu }}$) are used. These values are constants that incorporate the physical constants found in the rotational and vibrational partition functions of the molecules. Tabulated values of ${\displaystyle \Theta _{r}}$ and ${\displaystyle \Theta _{\nu }}$ for select molecules can be found here.

Species ${\displaystyle \Theta _{/nu}}$ (K) ${\displaystyle \Theta _{r}}$ (K) ${\displaystyle D_{0}}$ (kJ mol-1)
${\displaystyle {\ce {Cl2 (g)}}}$ 6125 0.351 239.0
${\displaystyle {\ce {H2 (g)}}}$ 808 87.6 431.9
${\displaystyle {\ce {HCl (g)}}}$ 4303 15.2 427.7

 ${\displaystyle \left({\frac {q_{\text{HCl}}}{V}}\right)=\left({\frac {2\pi mk_{B}T}{h^{2}}}\right)^{(3/2)}\times {\frac {T}{\sigma \Theta _{r}}}\times {\frac {1}{1-\exp {{\Bigl (}{\frac {-\Theta _{\nu }}{T}}{\Bigr )}}}}\times g_{1}\exp(D_{0}/RT)}$ ${\displaystyle \left({\frac {q_{\text{HCl}}}{V}}\right)=\left({\frac {2\pi (2.1957\times 10^{-24}{\text{kg}})(1.38065\times 10^{-23}{\text{JK}}^{-1})(650K)}{(6.62607\times 10^{-34}{\text{Js}})^{2}}}\right)^{(3/2)}\times {\frac {650{\text{K}}}{(2)(15.2{\text{K}})}}\times {\frac {1}{1-\exp {{\Bigl (}{\frac {-4303{\text{K}}}{650{\text{K}}}}{\Bigr )}}}}\times (1)\exp((427700{\text{Jmol}}^{-1})/(8.3145{\text{JKmol}}^{-1})(650{\text{K}}))}$ ${\displaystyle \left({\frac {q_{\text{HCl}}}{V}}\right)=(1.4975\times 10^{35}{\text{m}}^{-3})(21.4)(1.0013)(2.3419\times 10^{34})}$

 ${\displaystyle \left({\frac {q_{{\text{H}}_{2}}}{V}}\right)=\left({\frac {2\pi mk_{B}T}{h^{2}}}\right)^{(3/2)}\times {\frac {T}{\sigma \Theta _{r}}}\times {\frac {1}{1-\exp {{\Bigl (}{\frac {-\Theta _{\nu }}{T}}{\Bigr )}}}}\times g_{1}\exp(D_{0}/RT)}$ ${\displaystyle \left({\frac {q_{{\text{H}}_{2}}}{V}}\right)=\left({\frac {2\pi (1.2140\times 10^{-25}{\text{kg}})(1.38065\times 10^{-23}{\text{JK}}^{-1})(650K)}{(6.62607\times 10^{-34}{\text{Js}})^{2}}}\right)^{(3/2)}\times {\frac {650{\text{K}}}{(2)(87.6{\text{K}})}}\times {\frac {1}{1-\exp {{\Bigl (}{\frac {-808{\text{K}}}{650{\text{K}}}}{\Bigr )}}}}\times (1)\exp((431900{\text{Jmol}}^{-1})/(8.3145{\text{JKmol}}^{-1})(650{\text{K}}))}$ ${\displaystyle \left({\frac {q_{{\text{H}}_{2}}}{V}}\right)=(1.9468\times 10^{33}{\text{m}}^{-3})(3.71)(1.41)(5.0942\times 10^{34})}$

 ${\displaystyle \left({\frac {q_{{\text{Cl}}_{2}}}{V}}\right)=\left({\frac {2\pi mk_{B}T}{h^{2}}}\right)^{(3/2)}\times {\frac {T}{\sigma \Theta _{r}}}\times {\frac {1}{1-\exp {{\Bigl (}{\frac {-\Theta _{\nu }}{T}}{\Bigr )}}}}\times g_{1}\exp(D_{0}/RT)}$ ${\displaystyle \left({\frac {q_{{\text{Cl}}_{2}}}{V}}\right)=\left({\frac {2\pi (4.2700\times 10^{-24}{\text{kg}})(1.38065\times 10^{-23}{\text{JK}}^{-1})(650K)}{(6.62607\times 10^{-34}{\text{Js}})^{2}}}\right)^{(3/2)}\times {\frac {650{\text{K}}}{(2)(0.351{\text{K}})}}\times {\frac {1}{1-\exp {{\Bigl (}{\frac {-6125{\text{K}}}{650{\text{K}}}}{\Bigr )}}}}\times (1)\exp((239000{\text{Jmol}}^{-1})/(8.3145{\text{JKmol}}^{-1})(650{\text{K}}))}$ ${\displaystyle \left({\frac {q_{{\text{Cl}}_{2}}}{V}}\right)=(5.4839\times 10^{35}{\text{m}}^{-3})(925.9)(1.00)(1.606\times 10^{19})}$

Combining the terms from each species, the following expression is obtained:

 ${\displaystyle K_{c}={\frac {(1.9468\times 10^{33}{\text{m}}^{-3})^{2}}{(1.9468\times 10^{33}{\text{m}}^{-3})(5.4839\times 10^{35}{\text{m}}^{-3})}}\times {\frac {(21.4)^{2}}{(3.71)(925.9)}}\times {\frac {(1.0013)^{2}}{(1.41)(1.00)}}\times {\frac {(2.3419\times 10^{34})^{2}}{(1.606\times 10^{19})(5.0942\times 10^{34})}}}$ ${\displaystyle K_{c}=(0.003550)(0.1333)(0.711)(6.7037\times 10^{14})}$ ${\displaystyle K_{c}=(2.26\times 10^{11})}$

At 650 K, the reaction between ${\displaystyle {\ce {H2 (g)}}}$ and ${\displaystyle {\ce {Cl2 (g)}}}$ proceeds spontaneously towards the products. From a statistical mechanics point of view, the product ${\displaystyle {\ce {HCl (g)}}}$ molecule has more states accessible to it than the reactant species. The sponteneity of this reaction is largely due to the electronic partition function: two very strong H—Cl bonds are formed at the expense of a very strong H—H bond and a relatively weak Cl—Cl bond.