Chemical Potential and Equilibrium

In the example of two connected boxes containing species at different concentrations, in the case where the material can be exchanged between the boxes, then we can use classical thermodynamics to show that at equilibrium, the chemical potential in both boxes must be equal.

Let's label the amount of substance in box 1 as \( n_1 \), and in box 2 as \( n_2 \), with total moles fixed: \( n_1 + n_2 = \text{constant} \). The total Gibbs energy of the system is the sum of the Gibbs energies of each box: $$ G = G_1 + G_2 $$ This directly implies that an infinitesimal change in the Gibbs energy can be expressed as the sum of the differentials of the two sub-systems. $$ \mathrm{d}G = \mathrm{d}G_1 + \mathrm{d}G_2 $$

The differential changes in Gibbs energy for each box can be written in terms of the number of moles of the diffusing species:

$$ \mathrm{d}G = \left. \frac{\partial G_1}{\partial n_1} \right|_{T,P} \mathrm{d}n_1 + \left. \frac{\partial G_2}{\partial n_2} \right|_{T,P} \mathrm{d}n_2 $$

Since the total number of particles is conserved, a change in one box corresponds to the opposite change in the other:

$$ \mathrm{d}n_2 = -\mathrm{d}n_1 \quad \Rightarrow \quad \mathrm{d}n = \mathrm{d}n_1 = -\mathrm{d}n_2 $$

Defining the chemical potential as:

$$ \mu_i = \left. \frac{\partial G_i}{\partial n_i} \right|_{T,P} $$

we obtain:

$$ \mathrm{d}G = (\mu_1 - \mu_2)\, \mathrm{d}n $$

Now, from the second law of thermodynamics, we know that at constant temperature and pressure, a spontaneous process must reduce the Gibbs energy: \( \mathrm{d}G \leq 0 \), with \( \mathrm{d}G = 0 \) only at equilibrium.

This demonstrates that equal chemical potential is the condition for equilibrium, just as equal temperature defines thermal equilibrium.