# Statistical Mechanics Reaction problem

I am trying to do a problem that I encountered on a test, that I couldn't do. It reads as follows: "Consider a system comprised of two types of molecules, A and B, trapped in a volume V. They can react to form a third type of molecule, C, which has an internal energy of $$-\epsilon$$, while A and B have $$0$$ internal energy. If initially $$N_A=N_B$$ and $$N_C=0$$, find the equilibrium amount of C molecules if the system is in contact with a reservoir with temperature T. Consider ideal gases."

I know that equilibrium means that the chemical potentials $$\mu_i$$ have to be equal, that is, $$\mu_A+\mu_B=\mu_C$$

Additionally,the partition function for the molecules are $$Z_A=\frac{e^{-\beta HN_A}}{N_A!}=\frac{e^{0}}{N_A!}=\frac{1}{N_A!}$$ $$Z_B=\frac{e^{-\beta HN_B}}{N_B!}=\frac{1}{N_B!}$$ $$Z_C=\frac{e^{-\beta HN_C}}{N_C!}=\frac{e^{\beta \epsilon N_C}}{N_C!}$$

from here, I use that the Helmholtz Free energy $$F=-k_BTln(Z)$$ and the chemical potential $$\mu =\frac{dF}{dN}$$ to obtain $$\mu_A=k_BTln(N_A)$$, $$\mu_B=k_BTln(N_B)$$ and $$\mu_C=k_BTln(\frac{N_C}{e^{\beta \epsilon}})$$

Now I plug this into the initial relationship between chemical potentials to obtain

$$k_BTln(N_A)+k_BTln(N_B)=k_BTln(\frac{N_C}{e^{\beta \epsilon}})$$ $$ln(\frac{N_AN_Be^{\beta \epsilon}}{N_C})=0$$ $$N_C=N_AN_Be^{\beta \epsilon}$$

And I think that I now have to plug in the relationship that $$N_A=N_B$$ $$N_C=0$$, but the expression blows up. What can I do from here?

It looks all good, just I guess that you are looking at the final concentration at equilibrium, so you don't care about initial conditions! I will discard $$N_c=0$$ and set instead conservation of the number of molecules

$$N_{c \, final} + N_{a \, final} + N_{b \, final} = N_{c \, initial} + N_{a \, initial} + N_{b \, initial} = N_A + N_B = 2 N_A$$

plus as you wrote (but with "final!)

$$N_{a \, final} = N_{b \, final}$$