In: Chemistry
A closed container at 58.1˚C and 285 mm Hg contains a mixture of n-heptane and cyclohexane. Both liquid and vapor phases exist in equilibrium, and the vapor phase contains 32 mol % n-heptane. a) Calculate the mole fractions of the liquid phases. b) Now, the pressure in the container doubles, will the saturation pressure of n-heptane increase, decrease, or stay the same? Justify your answer
(a) : Given temperature, T = 58.1 C = 58.1 + 273 = 331.1 K
We can calculate the vapor pressure of pure components by applying Antoine equation.
For n-heptane: A = 4.02832, B = 1268.636, C = - 56.199
logP0h = A - B / (C+T) = 4.02832 - 1268.636 / ( - 56.199 + 331.1K)
=> P0h = 0.260 atm = 0.260 atm x ( 760 mm Hg / 1 atm) = 198 mm Hg
For cyclohexane: A = 4.13983, B = 1316.554, C = - 35.581
logP0c = A - B / (C+T) = 4.13983 - 1316.554 / ( - 35.581 + 331.1K)
=> P0c = 0.484 atm = 0.484 atm x ( 760 mm Hg / 1 atm) = 368 mm Hg
From Raoult's law in binary solution,
285 mm Hg = Xh x P0h + (1 - Xh) x P0c
=> 285 mm Hg = Xh x (198 mm Hg) + (1 - Xh) x (368 mmHg) ------- (1)
=> 170 x Xh = 83
=> Xh = 0.488
Hence mole fraction of heptane = 0.488 (answer)
mole fraction of cyclohexane = 0.512 (answer)
(b) If we double the pressure inside the container, the saturation pressure of vapor pressure will also increase as from Raoult's law the pressure inside the container is directly proportional to saturation pressure (vapor pressure)