# 26.8: Equilibrium Constants in Terms of Partition Functions

We can use statistical mechanics to calculate equilibrium constants in terms of molecular parameters. Consider the general gas phase reaction

in a reaction vessel at fixed volume and temperature.

The Helmholtz energy is

where m  is the chemical potential

Under the conditions of fixed volume and temperature dV = 0 and dT = 0 so

where n  is the stoichiometric coefficient for the reaction and l  is a measure of the extent of reaction.

Many physical chemistry texts derive expressions like this one using the number of moles, n instead of the number of molecules, N. We can write the above reaction as a thermochemical reaction at chemical equilibrium

In a mixture of ideal gases, the species are independent, and so the partition function of the mixture is a product of the partition functions of the individual components.

We have seen that

so the chemical potential is

Note that we used Stirling's approximation to make the last step. Substituting the above expression for the chemical potential into the equilibrium condition we have

and thus

For an ideal gas, the molecular partition function is a function of the form f(T)V so that q/V is a function of the temperature only. If we divide all of the terms on both sides by V and insert the definition of the number density r j  = Nj/V we have

where Kc is a function of temperature only. Kc is known as the equilibrium constant and can be expressed in terms of the concentrations (or activities) of the products and reactions.

where co is in units of molarity (moles/liter). In reality, co = 1 molar and thus Kc is unitless. In gas phase reactions the equilibrium constant can be expressed in terms of the partial pressures of the reactants and products.

Likewise, the standard state requires that Po = 1 bar and the factors of Po guarantee that Kp will be unitless. Ignoring the constants Po we can compare Kp and Kc.

Numerical tabulations of partition functions can be used to calculate the equilibrium constants of gas phase reactions.

Example $$\PageIndex{1}$$: Hydrogen Iodide

We can apply the above idea to the calculation of an equilibrium constant for a reaction involving diatomic molecules.

$H_2 + I_2 \rightleftharpoons 2HI$

The equilibrium constant is given by

For a diatomic we have in the general case

where M is the total mass of the diatomic molecule and the rotational and vibrational temperature appear in their respective partition functions.

For HI

where we have ignored the electronic partition function since it is approximately 1. Similarly for H2

and for I2

Note that the symmetry number s  = 2 for H2 and I2, but s  = 1 for HI.

The equilibrium constant is

Substituting the data obtained from tables into the above expression

Molecule Q vib /K Q rot /K D(kJ/mol)
H2 6215 85.3 432.1
I2 308 0.0537 148.8
HI 3266 9.25 294.7

we find the following temperature dependence for the equilibrium constant:

T/K Kp(T) ln Kp(T)
500 138 4.92
750 51.1 3.93
1000 28.5 3.35
1250 19.1 2.95
1500 14.2 2.65

Note that we did include the electronic partition function for this calculation since the difference in the values for the electronic transitions turns out to be a reasonably small energy and therefore does contribute to the equilibrium. Keep in mind that kB = 0.00831 kJ/mol/K so that we may also define an electronic temperature. For example, for H2 we see that D0 = 432.1 kJ/mole and Q elec  = 51,970 K. Compare this to the rotational and vibrational temperatures in the table above. In spite of this large value, the value of the difference is not so large

Corresponding approximately to a temperature of 1021.7 K. Compare this value to the rotational temperatures of the system and you can see that we cannot neglect electronic states when calculating the equilibrium constant. Note that since there are an equal number of diatomic molecules in the reactants and products there is no contribution to the temperature dependence from the ratio of either the translation or rotation partition functions.