Solute Partitioning and Excess Chemical Potentials

As we will see in succeeding sections, the chemical potential for the species $i$ can be separated into two parts:

$$\mu_i = \mu_i^{id}({\bf r}) + \mu_i^{ex}({\bf r}).$$ (1.50)

Each of the terms on the rhs of Eq. 1.50 can vary throughout the system, but the sum must be a constant at equilibrium. The first (ideal) term is the chemical potential for a particle which does not interact with any of the other particles in the system. We will find that it is given by

$$\mu_i^{id}({\bf r}) = k_BT \ln \rho_i({\bf r}) \Lambda_i^3.$$ (1.51)

where $\rho_i({\bf r})$ is the number density of component $i$ at the location ${\bf r}$. The second (excess) term of Eq. 1.50 is due to interactions of the chosen component with the rest of the system. Since $\mu_i$ is a constant, we can obtain a simple formula for the partition coefficient for a component between different parts of a physical system (for example between two coexisting phases):

$$K_i = \frac{\rho_i(II)}{\rho_i(I)} = e^{-\Delta \mu_i^{ex}/k_BT}$$ (1.52)

where

$$\Delta \mu_i^{ex} = \mu_i^{ex}(II) - \mu_i^{ex}(I)$$ (1.53)

If one of the phases (say I) is an ideal gas in equilibrium with a condensed phase, then $\mu_i^{ex}(I) = 0$, and $\mu_i^{ex}(II)$ yields the solubility of $i$ in the phase $(II)$.