Determine the temperature at which CH3OH boils according to the reaction below.

Question

Boiling point is the temperature at which this process reaches equilibrium at 1 atm of pressure.)

CH3OH(l) --> CH3OH(g)

H° = 38 kJ/mol

S° = 113 J/(K*mol)

a. 0.34K

b. 3.0K

c. 150K

d. 340K

e. 3000K

I just have no idea what equations or what I need to use for these kinds of problems, any help would be greatly appriciated. Also, I'm not really sure if the H° and S° should have had deltas in front of them, this copy is missing some of the symbols and such. I have the answer, it is D. I just don't know how to get there! Please help!

Answer 1

A system that is at equilibrium will have a free energy of zero. In this case , liquid methanol is at an equilibrium with gaseous methanol and this process has a free energy of zero.

Recall that G=H -TS        (G,H and S should have deltas in front of them)

If G=0, then H=TS

We know that H for this process is 38kJ and S is 113 J/kmol

Plug those values into the equation to get the temperature at which this equilibrium occurs

38 x 10^3 = T(113)     T=336 .28   and using proper sig figs.... T=340