Duhem–Margules equation

Let us consider a binary liquid mixture of two component in equilibrium with their vapor at constant temperature and pressure. Then from the Gibbs–Duhem equation, we have

${\displaystyle n_{A}\mathrm {d} \mu _{A}+n_{B}\mathrm {d} \mu _{B}=0}$

(1)

Where n_A and n_B are number of moles of the component A and B while μ_A and μ_B are their chemical potentials.

Dividing equation (1) by n_A + n_B, then

${\displaystyle {\frac {n_{A}}{n_{A}+n_{B}}}\mathrm {d} \mu _{A}+{\frac {n_{B}}{n_{A}+n_{B}}}\mathrm {d} \mu _{B}=0}$

Or

${\displaystyle x_{A}\mathrm {d} \mu _{A}+x_{B}\mathrm {d} \mu _{B}=0}$

(2)

Now the chemical potential of any component in mixture is dependent upon temperature, pressure and the composition of the mixture. Hence if temperature and pressure are taken to be constant, the chemical potentials must satisfy

${\displaystyle \mathrm {d} \mu _{A}=\left({\frac {\mathrm {d} \mu _{A}}{\mathrm {d} x_{A}}}\right)_{T,P}\mathrm {d} x_{A}}$

(3)

${\displaystyle \mathrm {d} \mu _{B}=\left({\frac {\mathrm {d} \mu _{B}}{\mathrm {d} x_{B}}}\right)_{T,P}\mathrm {d} x_{B}}$

(4)

Putting these values in equation (2), then

${\displaystyle x_{A}\left({\frac {\mathrm {d} \mu _{A}}{\mathrm {d} x_{A}}}\right)_{T,P}\mathrm {d} x_{A}+x_{B}\left({\frac {\mathrm {d} \mu _{B}}{\mathrm {d} x_{B}}}\right)_{T,P}\mathrm {d} x_{B}=0}$

(5)

Because the sum of mole fractions of all components in the mixture is unity, i.e.,

${\displaystyle x_{1}+x_{2}=1}$

we have

${\displaystyle \mathrm {d} x_{1}+\mathrm {d} x_{2}=0}$

so equation (5) can be re-written:

${\displaystyle x_{A}\left({\frac {\mathrm {d} \mu _{A}}{\mathrm {d} x_{A}}}\right)_{T,P}=x_{B}\left({\frac {\mathrm {d} \mu _{B}}{\mathrm {d} x_{B}}}\right)_{T,P}}$

(6)

Now the chemical potential of any component in mixture is such that

${\displaystyle \mu =\mu _{0}+RT\ln P}$

where P is the partial pressure of that component. By differentiating this equation with respect to the mole fraction of a component:

${\displaystyle {\frac {\mathrm {d} \mu }{\mathrm {d} x}}=RT{\frac {\mathrm {d} \ln P}{\mathrm {d} x}}}$

we have for components A and B

${\displaystyle {\frac {\mathrm {d} \mu _{A}}{\mathrm {d} x_{A}}}=RT{\frac {\mathrm {d} \ln P_{A}}{\mathrm {d} x_{A}}}}$

(7)

${\displaystyle {\frac {\mathrm {d} \mu _{B}}{\mathrm {d} x_{B}}}=RT{\frac {\mathrm {d} \ln P_{B}}{\mathrm {d} x_{B}}}}$

(8)

Substituting these value in equation (6), then

${\displaystyle x_{A}{\frac {\mathrm {d} \ln P_{A}}{\mathrm {d} x_{A}}}=x_{B}{\frac {\mathrm {d} \ln P_{B}}{\mathrm {d} x_{B}}}}$

or

${\displaystyle \left({\frac {\mathrm {d} \ln P_{A}}{\mathrm {d} \ln x_{A}}}\right)_{T,P}=\left({\frac {\mathrm {d} \ln P_{B}}{\mathrm {d} \ln x_{B}}}\right)_{T,P}}$

This final equation is the Duhem–Margules equation.

• Atkins, Peter and Julio de Paula. 2002. Physical Chemistry, 7th ed. New York: W. H. Freeman and Co.
• Carter, Ashley H. 2001. Classical and Statistical Thermodynamics. Upper Saddle River: Prentice Hall.