Duhem–Margules equation

The Duhem–Margules equation, named for Pierre Duhem and Max Margules, 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:

[math]\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} }[/math]

where P_A and P_B are the partial vapour pressures of the two constituents and x_A and x_B are the mole fractions of the liquid.

Derivation

Duhem - Margulus equation give the relation between change of mole fraction with partial pressure of a component in a liquid mixture.

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

[math]\displaystyle{ n_A \mathrm{d} \mu_A + n_B \mathrm{d} \mu_B = 0 }[/math]

(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, then

[math]\displaystyle{ \frac{n_A}{n_A+n_B} \mathrm{d}\mu_A + \frac{n_B}{n_A+n_B}\mathrm{d}\mu_B = 0 }[/math]

Or

[math]\displaystyle{ x_A \mathrm{d} \mu_A + x_B \mathrm{d} \mu_B = 0 }[/math]

(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

[math]\displaystyle{ \mathrm{d} \mu_A = \left( \frac{\mathrm{d}\mu_A}{\mathrm{d}x_A} \right)_{T,P}\mathrm{d}x_A }[/math]

(3)

[math]\displaystyle{ \mathrm{d} \mu_B = \left( \frac{\mathrm{d}\mu_B}{\mathrm{d}x_B} \right)_{T,P}\mathrm{d}x_B }[/math]

(4)

Putting these values in equation (2), then

[math]\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 }[/math]

(5)

Because the sum of mole fraction of all component in the mixture is unity i.e.,

[math]\displaystyle{ x_1 + x_2 = 1 }[/math]

Hence

[math]\displaystyle{ \mathrm{d}x_1 + \mathrm{d}x_2 = 0 }[/math]

so equation (5) can be re-written:

[math]\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} }[/math]

(6)

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

[math]\displaystyle{ \mu = \mu_0 + RT \ln P }[/math]

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

[math]\displaystyle{ \frac{\mathrm{d}\mu}{\mathrm{d}x} = RT \frac{\mathrm{d} \ln P}{\mathrm{d}x} }[/math]

So we have for components A and B

[math]\displaystyle{ \frac{\mathrm{d}\mu_A}{\mathrm{d}x_A} = RT \frac{\mathrm{d} \ln P_A}{\mathrm{d}x_A} }[/math]

(7)

[math]\displaystyle{ \frac{\mathrm{d}\mu_B}{\mathrm{d}x_B} = RT \frac{\mathrm{d} \ln P_B}{\mathrm{d}x_B} }[/math]

(8)

Substituting these value in equation (6), then

[math]\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} }[/math]

or

[math]\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} }[/math]

this is the final equation of Duhem–Margules equation.

Sources

• 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.