Answer to Question #101470 in Physical Chemistry for Ashish

Question

Give the thermodynamic derivation of the distribution law.

Answer 1

The thermodynamic derivation of the distribution law is based upon he principle that if there are two phases in equilibrium (i.e. two immiscible solvents containing the same solute dissolved in them), the chemical potential* of a substance present in them must be same in both the phases.

From thermodynamics, we know that the chemical potential of a substance is a solution given by

$\mu={\mu}^0+RTlna$

Where ${\mu}^0$ is the standard chemical potential and ‘a’ is the activity of the substance (solute) in the solution.

Thus for the solute in liquid A, we have

$\mu_A={\mu}_A^0+RTlna_A$

Similarly for the solute in liquid B we have

$\mu_B={\mu}_B^0+RTlna_B$

But as already stated, since the liquids A and B are in equilibrium,

$\mu_A=\mu_B$

$\therefore$ ${\mu}_A^0+RTlna_A={\mu}_B^0+RTlna_B$

$ln(a_A/a_B)=(\mu^0_B-\mu^0_A)/RT$

Further at a given temperature, $\mu^0_A$ and $\mu^0_B$ are constant for given substance in the particular solvents. Hence at constant temperature, we have

$ln(a_A/a_B)=constant$

$a_A/a_B=constant$

This is the exact expression of the distribution law. However, if the solutions are dilute, the activates are equal to the concentrations so that the expression is modified to

$C_A/C_B=constant$

Which is the original form of the distribution law.