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


which has differential


At equilibrium,


It seems like the equilibrium conditions are



(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?

  • $\begingroup$ If they were not constant, this wouldn't be equilibrium. See also Chemical equilibrium $\endgroup$
    – Roger V.
    Commented Apr 14, 2023 at 11:27

2 Answers 2


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.

  • $\begingroup$ Thanks, 1. So during thermodynamic equilibrium, the potential drops $P^1_n-P^2_n$ are very small and remain constant throughout time? 2. Even though the small amount of extensive quantities $\delta K_n$ are transferred between subsystems, $K_n$ remains constant throughout the whole system? $\endgroup$
    – Ray Siplao
    Commented Apr 14, 2023 at 2:23
  • $\begingroup$ It is NOT assumed that $P^1_n-P^2_n$ is small, instead it is assumed that the variation of $K_n$ by $\delta K_n$ causes in the function $P_n=P_n(K_1, K_2., K_3, ...)$ to be $P_n=P_n(K_1+\delta K_1, K_2+\delta K_2., K_3+\delta K_3, ...)$ is insignificant so that the sum representing the process work lost sum is insignificant. Contrast this with the usual formulation of the Gibbs-Duhem equation according to which $\sum_n K_ndP_n =0$ expresses a 1st order differential relationship between two neighboring equilibrium points. Gibbs-Duhem is a static differential not involving a process. $\endgroup$
    – hyportnex
    Commented Apr 14, 2023 at 2:34

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 $N_1d\mu_1=N_2d\mu_2=0$.

  • $\begingroup$ Thank you! I think $d\mu = - S dT + V dP$ should be $d\mu = - s dT + v dP$ where $s=S/N$ and $v=V/N$. $\endgroup$
    – Ray Siplao
    Commented Apr 14, 2023 at 21:49
  • $\begingroup$ According to this equation $d\mu = - s dT + v dP$, it seems like once the mixture is in mechanical and thermal equilibrium, it is already in chemical equilibrium, right? $\endgroup$
    – Ray Siplao
    Commented Apr 14, 2023 at 22:31
  • 1
    $\begingroup$ @RaySiplao (i) The use of capitals or lower letters for total or molar properties is not a universal convention. After all, wirh $N=1$ we obtain $s=S$ etc. (ii) Regarding the second comment about equilibrium, the way the thought experiment works is this: we partition $N$ between under constant $T$ and $P$ and assume each part to be in internal equilibrium but not necessarily in equilibrium with each other. The latter condition corresponds to the minimization of $G$. This is the only way that equilibrium thermodynamics can be used to analyze a non-equilibrium state. $\endgroup$
    – Themis
    Commented Apr 14, 2023 at 22:51
  • $\begingroup$ @RaySiplao On a second thought, I decided to edit may answer to say $d\mu=-sdT+vdP$, since this situation involves both molar ($s$, $v$) and total ($G$, $N$) properties. $\endgroup$
    – Themis
    Commented Apr 14, 2023 at 22:55
  • $\begingroup$ I am thinking that under conditions of constant $T,P$, chemical potential is a function of $T,P$, ie. $\mu=\mu(T,P)$. The equilibrium condition $\mu_1=\mu_2$ is satisfied iff both parts are in thermal and mechanical equilibrium with temperature and pressure $T_0,P_0$ so that $\mu_1(T_0,P_0)=\mu_2(T_0,P_0)$. $\endgroup$
    – Ray Siplao
    Commented Apr 14, 2023 at 23:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.