Question

2. (60 pts.) The vapor pressures of pure species A and pure species B in the solid and in the liquid phase are given by the following integrated Clausius-Clapeyron equations: Pure solid A: 2800K In(p) = 9 - Pure liquid A: In(p) = 7 - 2000K Pure solid B: In(p) = 5-2 2600K T Pure liquid B: In(p) = 3 - - 1000K where p is the vapor pressure in torr and T is the temperature in K. All latent heats associated with phase changes of the pure species are constants. For both species, experiment indicates that the p-T solid-liquid phase boundary (i.e., the fusion line) is a nearly-vertical line for a wide range of pressures. Species A and B are completely miscible both in the liquid and in the solid phase. The resulting liquid and solid solutions of A and B are ideal solutions at all compositions. A liquid solution of A and B of unknown composition is cooled at constant atmospheric pressure. Experiment indicates that freezing is completed at T = 500 K. Calculate the unknown composition of the liquid solution.

Answer 1

Clausius-Clapeyron equation: ln(p) = ln(A) - (Ea/R)*(1/T)

Compare all the given equations with the above equation.

For the pure solid A: Ea/R = 2800 K, i.e. Ea = 2800 K * 8.314 J/mol.K = 23279.2 J/mol or 23.279 kJ/mol

For the pure liquid A: Ea/R = 2000 K, i.e. Ea = 2000 K * 8.314 J/mol.K = 16628 J/mol or 16.628 kJ/mol

For the pure solid B: Ea/R = 2600 K, i.e. Ea = 2600 K * 8.314 J/mol.K = 21616.4 J/mol or 21.616 kJ/mol

For the pure liquid B: Ea/R = 1000 K, i.e. Ea = 1000 K * 8.314 J/mol.K = 8314 J/mol or 8.314 kJ/mol

Here, more activation energies are required for solids than for liquids.

And the activation energy for solid A is more than that for solid B, which indicates that the solid A is more packed than solid B.

And liquid A requires more activation energy than liquid B.