Chemical equilibrium in a multispecie system

In Sec.  we noted that the condition $dS=0$ at equilibrium (or rather the absence of internal constraints that impede particle exchanges between the two systems) implies that the distribution of the particles in the two systems is such that their chemical potentials, defined in , are equal. If there is more than one specie in the system, then the discussion in Sec.  can be repeated separately for each specie, removing the constraint of impermeability to particle transfer one specie at a time. This leads to the definition of a different chemical potential for each specie, and to a generalisation of the fundamental thermodynamic relation to:

$$TdS= dE + PdV - \sum_i \mu_i dN_i,$$ (6.9)

with the sum over $i$ running over the different species in the system, and $\mu_i$ the chemical potential of specie $i$, defined by:

$$\mu_i = -T\left ( \frac{\partial S}{\partial N_i}\right )_{E,V,N_{j \neq i}}.$$ (6.10)

As a result, the fundamental thermodynamic relation for the Gibbs free energy becomes:

$$dG = VdP - SdT + \sum_i \mu_i dN_i.$$ (6.11)

The chemical potentials $\mu_i$ are also obtained as:

$$\mu_i = \left ( \frac{\partial G}{\partial N_i}\right )_{P,T,N_{j \neq i}}$$ (6.12)

and we have:

$$G(P,T,N_1,N_2,\dots) = \sum_i N_i \mu_i.$$ (6.13)

For constant temperature and constant pressure variations, combining Eqs.  and  gives the Gibbs-Duhem relation:

$$\sum_i N_i d \mu_i = 0,$$ (6.14)

which shows that the chemical potentials of the various species are not all independent from each other.

Let us consider now chemical equilibrium between two phases, for example a solid and a liquid. The second law states that if we fix the pressure and the temperature the Gibbs free energy of the system must be minimum and so, using , we obtain a generalisation of :

$$\sum_i \mu_i^s dN_i^s + \mu_i^l dN_i^l = 0,$$ (6.15)

where $\mu_i^s$ and $\mu_i^l$ are the chemical potentials of specie $i$ in the solid and in the liquid, respectively, and $N_i^s$ and $N_i^l$ are the number of particles in the two phases. Since particles of any specie $i$ cannot be created or destroyed but can only move from one phase to another, we must have $dN_i^s = - dN_i^l = dN_i$ and so:

$$\sum_i (\mu_i^s - \mu_i^l) dN_i = 0$$ (6.16)

and since this has to be satisfied for any variation $dN_i$, at equilibrium we must have

$$\mu_i^s = \mu_i^l, \quad \quad \forall i.$$ (6.17)

Suppose that the system is out of equilibrium and, for example, $\mu_i^s > \mu_i^l$ for one particular value of $i$. For any change that drives the system to equilibrium we must have $dG < 0$, implying that $dN_i$ must be negative, i.e. particles move form the solid to the liquid phase, where the chemical potential is lower.