# Deriving chemical equilibrium constant from non-reacting or reacting ideal gas mixture partition function?

Given a chemical reaction of ideal gasses:

A <-> B

From my understanding, there are two partition functions that you can write depending on whether A and B can react to form each other.

Non-reacting:

$$Q_{nr} = \frac{q_A^{N_A}}{N_A!} \frac{q_B^{N_B}}{N_B!}$$

Reacting:

$$Q_r = \frac{(q_A+q_B)^{N_A+N_B}}{(N_A+N_B)!}$$

Where $$q_i$$ is the partition function for the ith subsystem and $$N_i$$ is the number of particles of i.

I have read that to derive the chemical equilibrium constant ($$K$$), you maximize the partition function with respect to $$N_A$$.

$$\frac{\partial\ln(Q)}{∂N_A} = 0$$

If you use $$Q_{nr}$$, then the math works out this way ($$N = N_A + N_B$$):

$$\frac{\partial\ln(Q)}{∂N_A} = \ln(q_A) - \ln(N_A) - \ln(q_B) + \ln(N-N_A) = 0$$

$$= \frac{q_B}{q_A} = \frac{N_B}{N_A} = K$$

Why do you use $$Q_{nr}$$ for the calculation? Wouldn't you use $$Q_{r}$$?

• Where did you get your $Q_r$ from? Do you have a reference? Apr 20, 2023 at 18:58
• I was looking for a rigorous statistical mechanics derivation of the equilibrium constant from a partition function but all I could find were these slides. The math seemed to make sense to me. I got $Q_r$ from the slides: slideplayer.com/slide/8364265 If you have a reference that I could look at I'd appreciate it as well. Apr 21, 2023 at 19:13

Here is what is going on. The expression $$\tag{1} q_R = \frac{(q_A+q_B)^{N_A+N_B}}{(N_A+N_B)!}$$ with $$N_A+N_B=N_\text{tot}$$ fixed considers every possibility for a mixture of A and B with constant number of particles, from all particles being A, to all particles being B and everything in between. The reason is that A and B can be converted into each other via the reaction. This is the grand canonical partition function that allows the mixture to have any composition under the conditions that components convert into each other.
Expanding $$q_r$$ we obtain $$\tag{2} q_r = \frac{1}{(N_A+N_B)!} \sum_{N_A=0}^{N_\text{tot}} \frac{(N_A+N_B)!}{N_A! N_B!} q_A^{N_A} q_B^{N_B} = \sum_{N_A=0}^{N_\text{tot}} \frac{q_A^{N_A} q_B^{N_B}}{N_A! N_B!}$$ The result on the right-hand side is a summation of the terms you called $$q_{nr}$$ over all combinations of $$N_A$$ and $$N_B$$ under fixed $$N_A+N_B$$. Each term is a canonical partition function for a system with exactly $$N_A$$ A's and $$N_B=N_\text{tot}-N_A$$ B's.
We can calculate the equilibrium composition from either partition function. To work with $$q_r$$ we calculate $$N_A^*$$ as the average number of A's in the grand canonical partition function, each weighted by the grand canonical probability. This is done on p23 of the reference you cited.
Alternatively, we can obtain it from $$q_{nr}$$, which amounts to identifying the maximum tern in Eq. 1. This calculation is done on p26 of the same reference.