Derivation of Duhem–Margules equation

Duhem–Margules Equation

Duhem–Margules equation is a thermodynamic statement of the relationship between the two components of a single liquid where the vapour mixture is regarded as an ideal gas.

Let consider a binary liquid mixture of two component in equilibrium with their vapor at constant temperature and pressure. Then from Gibbs–Duhem equation is
n_Adμ_A + n_Bdμ_B = 0 -----(equation-1)
Where n_A and n_B are number of moles of the component A and B while μ_A and μ_B is their chemical potential.

dividing equation-1 by n_A + n_B we get-
n_A/(n_A + n_A) dμ_A + n_B/(n_A + n_B) dμ_B = 0
or, X_Adμ_A + X_Adμ_B = 0 -----(equation-2)

Now the chemical potential of any component in mixture is depend upon temperature, pressure and composition of mixture. Hence if temperature and pressure taking constant then chemical potential-
dμ_A = (dμ_A/dX_A)_T,P . dX_A -----(equation-3)
and
dμ_B = (dμ_B/dX_B)_T,P . dX_B -----(equation-4)

Putting these values in (equation-2), then-
X_A(dμ_A/dX_A)_T,P . dX_A + X_B(dμ_B/dX_B)_T,P . dX_B = 0 -----(equation-5)

we know that sum of mole fraction of all component in the mixture is unity i.e., X_A + X_B = 1
Hence, dX_A + dX_B =0
So, equation -5 becomes
X_A(dμ_A/dX_A)_T,P = X_B(dμ_B/dX_B)_T,P -----(equation-6)

We know that the chemical potential of any component in mixture is
μ = μ_o + RTlnP
where P is partial pressure of component. By differentiating this equation with respect to the mole fraction of a component we get-
dμ/dX = RT(dlnP/dX)

So for two component A and B-
dμ_A/dX_A = RT(dlnP_A/dX_A)
dμ_B/dX_B = RT(dlnP_B/dX_B)

Now substituting these values in equation-6 we get-

X_A(dlnP_A/dX_A) = X_B(dlnP_B/dX_B) -----(equation-7)

or, (dlnP_A/dlnX_A)_T,P = (dlnP_B/dlnX_B)_T,P -----(equation-8)

Equation-7 and 8 are two forms of Duhem–Margules equation. and these equations are used in the derivation of Konowaloff's rule which is useful in the binary liquid mixtures by distillation.

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