Gibbs Free Energy

Question

The ∆G° of vaporization for benzene at 298 K and 1.00 atm is 4.883 kJ/mol. Calculate the pressure, in atm, of benzene vapor in equilibrium with benzene liquid at 298 K.

Answer 1

This question is a bit strange, but I'm going to give it a shot.

We know that ∆Gº = -RT ln K

We can write an equilibrium expression for the vaporization of benzene:

benzene(l) <==> benzene(g)

We can then write:

K = Pressure of the gas / Pressure of the liquid = P(g) / P(l) and set the P(l) = 1 atm and solve for P(g)

K = P(g) / 1 = P(g)

Substitute this into the ∆Gº equation:

4883 J / mol = - (8.314)(298) ln K

4883 = -2478 ln P(g)

ln P(g) = -1.97

P(g) = 0.14 atm

This is my best attempt at solving this. Maybe some else will weigh in.