Question

The equilibrium vapour pressures of ethanol and chloroform at 45oC are 230.33 mbar and 578.0 mbar respectively. The Henry’s law constants for ethanol at 45oC is 1030.3 mbar. A non‐ideal solution of the two liquids has a mole fraction of ethanol in the liquid phase of 0.115.  The equilibrium vapour pressure of chloroform above the solution is 533.7 mbar and the equilibrium vapor pressure of the ethanol is 73.95 bar.

(a) Find the mole fraction of ethanol in the vapour phase in equilibrium with the solution at 45oC.

(b) Calculate the Raoult’s law activity and activity co‐efficient for each liquid.

(c) Calculate the Henry’s law activity and activity co‐efficient for the ethanol in the solution. Does ethanol more closely obey Raoult’s law or Henry’s law? Briefly explain.

(d) What is the equilibrium vapor pressure of ethanol in the Henry’s law standard state?

(e) Determine the total vapor pressure if the solution were ideal.

(f) Calculate the Gibbs energy of mixing of a solution containing 10.00 moles in total of both liquids and a solution mole fraction of ethanol of 0.115. Find the excess Gibbs energy.

Answer 1

(a) According to Henry's law:

The molefraction of ethanol (X_ethanol) = P_ethanol/K_H = 230.33 mbar/1030.3 mbar = 0.224

(b) According to Raoult's law: P_A = P_A^o*X_A = 73.95 bar * 0.115 = 8.50425 bar

P_B = P_B^o*X_B = 0.5337 bar * 0.0.885 = 0.4723245 bar

Where A = ethanol and B = chloroform

(c) The partial pressure of ethanol (P_ethanol) = X_ethanol * K_H = 0.115 * 1.0303 bar = 0.1184845 bar = 118.48 mbar

Therefore, ethanol more closely obey Henry's law because the vapor pressure of ethanol is close to the equilibrium vapor pressure of ethanol (230.33 mbar)

(d) Equilibrium vapor pressure of ethanol in the Henry's law standard state = 0.1184845 bar ~118.48 mbar