Do chemical potentials stay constant in chemical equilibrium?

Question

I am uncertain whether chemical potentials stay constant during chemical equilibrium.

Consider a closed container divided into two parts 1 and 2 filled with ideal gas particles. The barrier between parts 1 and 2 allows particle exchange.

The Gibbs energy for parts 1 and 2 are

$$G_1=N_1\mu_1$$ $$G_2=N_2\mu_2$$

The total Gibbs energy for the container is

$$G=G_1+G_2=N_1\mu_1+N_2\mu_2$$

which has differential

$$dG=(\mu_1-\mu_2)dN_1+N_1d\mu_1+N_2d\mu_2$$

At equilibrium,

$$(\mu_1-\mu_2)dN_1+N_1d\mu_1+N_2d\mu_2=0$$

It seems like the equilibrium conditions are

$$\mu_1=\mu_2$$

$$N_1d\mu_1=-N_2d\mu_2$$

(Not sure if it is necessarily true that $N_1d\mu_1=N_2d\mu_2=0$.)

What I can conclude from here is that in chemical equilibrium, 1 and 2 exchange the same number of particles, ie. 1 gains some particles from 2 but gives back the same amount of particles to 2 so that the net change is zero?

Is my reasoning correct? Are there any cases when chemical potentials stay constant during chemical equilibrium?

Answer 1

You are right but your question could be asked for any potential, not just for the chemical potential. In general, denoting the extensive work quantity by $K_n$ and the corresponding work potential $P_n$ for a reversible process you can always write that the total work lost in the process is zero $$\sum_n (P^1_n-P^2_n)\delta K_n=0$$ where $\delta K_n$ is the extensive quantity (1-entropy, 2-electric charge, 3-gravitational mass, 4-molar mass, 5-volume, etc.) is being moved, "dropped", between the corresponding potential (1-temperature, 2-electric potential, 3-gravitational potential, 4-chemical potential, 5-pressure, etc.) of values $P^1_n$ and $P^2_n$. That the total work lost is zero in such complex of processes is a sufficient and necessary condition for the process being reversible.

But because the potentials themselves are functions of all the extensives $K_n$, we must assume that while the extensive quantities $\delta K_n$ are being moved through the potential drop $P^1_n-P^2_n$ the potentials do not change. This can only happen if the quantities being moved are sufficiently small. Of course, in practice, the definition of small is the one that keeps the total work lost zero, or essentially zero. Usually, the potentials are at least piece-wise differentiable functions of the extensives, so a first order change in the latter is a first order change in the former. So as long as the extensives being moved are insignificant relative to the total we can assume that the change in the potentials is small and thus making our process that would be otherwise reversible now to be a little irreversible.

Answer 2

You have performed the minimization of Gibbs incorrectly, although the final conclusion that $\mu_1=\mu_2=0$ is correct.

How to minimize $G$

The minimization of $G$ requires constant $T$ and $P$, so first we need to modify the experiment. We cannot just remove the partition unless (i) both parts are at the same $T$ and $P$ to begin with and (ii) the gas is ideal. However, Gibbs minimization does not require an actual experiment, a thought experiment will do: move molecules between the two parts while maintaining (somehow) conditions of constant $T$ and $P$. We don't need to assume ideality in this case.

Begin with the Gibbs energy of the mixture in the box: $$G = N_1\mu_1 + N_2\mu_2 $$ whose differential is, as you say, $$dG=(\mu_1-\mu_2)dN_1+N_1d\mu_1+N_2d\mu_2$$ since $dN_2=-dN_1$. Now, the differential of $\mu$ is $$ d\mu = - s dT + v dP $$ and since $dT=0$, $dP=0$, we conclude that $d\mu_1=d\mu_2=0$ and $\mu_1=\mu_2$.

A secondary point

(Not sure if it is necessarily true that $N_1d\mu_1=N_2d\mu_2=0$.)

As we see from the above derivation, $d\mu_1=d\mu_2=0$, so in this particular case it is true that $N_1d\mu_1=N_2d\mu_2=0$.