# Thermodynamic Derivation of Distribution Law

## Thermodynamic Derivation of Nernst's Distribution Law

The thermodynamic derivation of the distribution law is based upon the principle that if there are two phases in equilibrium (i.e. two immiscible solvents containing the same solute dissolved in them), the chemical potential(μ) of a substance present in them must be same in both the phases.From thermodynamics, we know that the chemical potential (μ) of a substance is a solution given by-

μ = μ

^{o}+ RT lna

Where μ

^{o}is the standard chemical potential and 'a' is the activity of the solute in the solution.

Thus for the solute in liquid A, we have-

μ

_{A}= μ

^{o}

_{A}+ RT lna

_{A}

Similarly for the solute in liquid B we have-

μ

_{B}= μ

^{o}

_{B}+ RT lna

_{B}

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

μ

_{A}= μ

_{B}

or, μ

^{o}

_{A}+ RT lna

_{A}= μ

^{o}

_{B}+ RT lna

_{B}

or, RT lna

_{A}− RT lna

_{B}= μ

^{o}

_{B}− μ

^{o}

_{A}

or, ln(a

_{A}/a

_{B}) = (μ

^{o}

_{B}− μ

^{o}

_{A})/R -----(equation-1)

At a given temperature, μ

^{o}

_{A}and μ

^{o}

_{B}are constant for given substance in the particular solvents. Hence at constant temperature, equation (1) becomes-

ln(a

_{A}/a

_{B}) = Constant

or, a

_{A}/a

_{B}= Constant -----(equation-2)

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 (equation-2) is modified as-

C

_{A}/C

_{B}= Constant -----(equation-3)

(equation-3) is the original form of the distribution law.

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