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?


1 Answer 1


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}$


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