# Why chemical potential changes as a system approaches chemical equilibrium?

Consider a closed system of constant volume $$V$$, constant pressure $$P$$, temperature $$T$$, and Gibbs energy $$G$$ that is in thermal and mechanical equilibrium with surroundings. It is filled with $$N$$ particles of one type.

We divide the system into two components: component 1 which has volume $$V_1$$, pressure $$P$$, temperature $$T$$, particle count $$N_1$$, chemical potential $$\mu_1$$, Gibbs energy $$G_1$$, and component 2 which has volume $$V-V_1=V_2$$, pressure $$P$$, temperature $$T$$, particle count $$N-N_1=N_2$$, chemical potential $$\mu_2$$, Gibbs energy $$G-G_1=G_2$$.

The total change in Gibbs energy for this two-component closed system is

$$dG=dG_1+dG_2=\mu_1 dN_1+\mu_2 dN_2=\mu_1 dN_1-\mu_2 dN_1=(\mu_1-\mu_2)dN_1$$

Suppose the system is initially not in chemical equilibrium $$\mu_1>\mu_2$$, it will decrease the particle count in component 1 $$dN_1<0$$ until $$dG=0$$.

However, I am not sure how $$G_1$$ and $$G_2$$ change as the system approaches chemical equilibrium. In terms of Gibbs energy, chemical potential is defined as

$$\mu=\left(\frac{\partial G}{\partial N}\right)_{T,P}=\frac{G}{N}$$

Since $$G$$ and $$N$$ are extensive quantities, it follows that $$\mu$$ is an intensive quantity. This means the value of $$\mu$$ is independent of particle count. For example, even if the particle count is doubled, $$\mu$$ stays the same.

So I am not sure how it is possible for $$\mu_1$$ to decrease as $$N_1$$ decreases if chemical potential is an intensive quantity.

The chemical potential of component $$1$$ in a mixture with component 2 is $$\tag{1} \mu_1 = \left(\frac{\partial G}{\partial N_1}\right)_{T,P,N_2} \color{red}\neq \frac{G}{N}$$ Notice that equality you wrote applies only in a one-component system. The relationship between the chemical potentials, the Gibbs energy of the mixture and the number of moles is $$\tag{2} N_1 \mu_1 + N_2 \mu_2 = (N_1+N_2) g = G$$ A simple example can be given for the case of an ideal mixture. Then $$\tag{3} \mu_i = g_i + RT \ln \frac{N_i}{N_1+N_2}$$ where $$g_i=G_i/N_i$$ is the molar Gibbs energy of pure $$i$$ at the same temperature and pressure as the mixture. This shows that the chemical potential changes with composition.
• @JimmyYang (i) Yes, your reading of Eq 2 is correct. In fact, the first equation in your post is the differential of my Eq 2. (ii) Eq 3 is a special result for ideal mixtures, namely mixtures whose enthalpy is $H=n_1 h_1 + n_2 h_2$, and their entropy is $S = N_1 s_1+N_2 s_2 - R(N_1\ln\frac{N_1}{N_1+N_2} + N_2 \ln\frac{N_2}{N_1+N_2})$, where $h_i$ and $s_i$ are the properties of the pure components. Ideal gas mixtures and ideal solutions (solutions of chemically similar molecules) satisfy these equations. Set $G = H-T S$ and you will get to Eq (3). Mar 18, 2023 at 19:27
• I see. If each component is filled with a different particle type, then the total Gibbs energy is $G=N_1\mu_1+N_2\mu_2$? I still have doubts about the $N_1+N_2$ factor. Also, the differential for equation 2 is $\mu_1 dN_1+\mu_2 dN_2=GdN_1+N_1dG+GdN_2+N_2dG$ which reduces to $(\mu_1-\mu_2)dN_1=(N_1+N_2)dG=NdG$. I don't see how it results in my equation $dG=(\mu_1-\mu_2)dN_1$. Mar 18, 2023 at 20:08
• (i) You are right (I corrected it), I should have said $(N_1+N_2)g = G$. (ii) According to the Gibbs-Duhem equation (which is easy to prove), $N_1 d\mu_1+N_2 d\mu_2 = 0$. Mar 18, 2023 at 20:22
• Is equation 3 intensive or extensive? I understand that $g_i$ is intensive but not sure about $RT \ln \frac{N_i}{N_1+N_2}$ term. I don't think that equation is intensive since it changes as $N_i$ changes? Mar 18, 2023 at 21:54
Your definition of the chemical potential should be $$\mu_1=\left(\frac{\partial G}{\partial N_1}\right)_{T,P,N_2}$$
• You're right, I forgot to write constant $N_2$. Mar 18, 2023 at 16:07