# 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} = N_{j}/V we have

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

where c^{o} is in units of molarity (moles/liter). In reality, c^{o} = 1 molar and thus K_{c} 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 P^{o} = 1 bar and the factors of P^{o} guarantee that K_{p} will be unitless. Ignoring the constants P^{o} we can compare K_{p} and K_{c}.

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 H_{2}

and for I_{2}

Note that the symmetry number s = 2 for H_{2} and I_{2}, 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_{0 }(kJ/mol) |
---|---|---|---|

H_{2} |
6215 | 85.3 | 432.1 |

I_{2} |
308 | 0.0537 | 148.8 |

HI | 3266 | 9.25 | 294.7 |

we find the following temperature dependence for the equilibrium constant:

T/K |
K_{p}(T) |
ln K_{p}(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 k_{B} = 0.00831 kJ/mol/K so that we may also define an electronic temperature. For example, for H_{2} we see that D_{0} = 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.