Question

Consider the benzene (1) + ethanol (2) system which exhibits an azeotrope at 760 mmHg and 68.24°C containing 44.8 mole % ethanol. At 68.24°C, the vapor pressures of pure benzene and ethanol are 519.7 mmHg and 503.5 mmHg respectively. Find the activity coefficients of benzene and ethanol at the azeotrope and Calculate the two parameters (A) for the Redlich / Kister expansion.

Answer 1

In this system Low Pressure i.e 760 mm of Hg(1 atm), so it is okay to assume ideal gas and modified Raoult’s law can be applicable.

Also, azeotrope : x1=y1 and x2=y2.

y₁P = γ₁x₁P_{1 sat} => P = γ₁P_{1 sat} => γ₁ = P/P_{1 sat} = 760/519.7 = 1.462 (Ans)

y₂P = γ₂x₂P_{2 sat} => P = γ₂P_{2 sat} => γ₂ = P/P_{2 sat} = 760/503.5 = 1.51(Ans)

Parameter for Redlich-Kister expansion A = G^E/(x₁x₂RT)

=>On solving in terms of γ₁ => Ax₂² =ln γ₁

                                                 =>A= ln(1.462)/(0.448)² = 1.8924 (Ans)