Enthalpy change and boiling point?

Question

For CH3OH the enthalpy of vaporization is 38.0 kJ/mol. The entropy of vaporization is 113 J/(K mol).
Calculate the normal boiling point (Kelvin).

Answer 1

The equation you need to use is DG=DH-TDS (where D=delta).

At the boiling point, the free energy change (DG) is 0. So, you can substitute DH and DS, and solve for T. Make sure you express both DH and DS in either J/mol OR kJ/mol.

Answer 2

At the boiling point, delta G = 0, so:

delta G = 0 = delta H - (T x delta S).

So T = delta H/delta S, and the only thing to watch out for is to take care of the units, because entropy is in joules, and enthalpy is in kilojoules.