Question

Consider the benzene (1) + ethanol (2) system which exhibits an azeotrope at 760 mmHg and...

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.

Homework Answers

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 = γ1x1P1 sat => P = γ1P1 sat => γ1 = P/P1 sat = 760/519.7 = 1.462 (Ans)

y2P = γ2x2P2 sat => P = γ2P2 sat => γ2 = P/P2 sat = 760/503.5 = 1.51(Ans)

Parameter for Redlich-Kister expansion A = GE/(x1x2RT)

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

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

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