Question

Given the enthalpy and entropy change of a natural phase transition, derive an expression for the normal boiling point of any substance? Hint: consider that the change in Gibbs' free energy is zero for two phases in equilibrium.

Answer 1

The hint tells you exactly what to do. Consider the relationship between the Gibbs' free energy, enthalpy, and entropy:

`DeltaG = DeltaH - TDeltaS`

At equilibrium (more specifically, phase equilibrium at constant temperature and pressure!), `DeltaG = 0`, so at liquid-vapor phase equilibrium:

`DeltaG_(vap) = 0 = DeltaH_(vap) - T_bDeltaS_(vap)`

`=> color(blue)(T_b = (DeltaH_(vap))/(DeltaS_(vap)))`,

where `T_b` is the boiling point.

That is, you are somewhere on the green curved dotted line on a phase diagram:

Once you obtain the temperature in `"K"`, convert to `""^@ "C"` by subtracting `273.15` from the `"K"` temperature.