# 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.

Jul 13, 2017

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

$\Delta G = \Delta H - T \Delta S$

At equilibrium (more specifically, phase equilibrium at constant temperature and pressure!), $\Delta G = 0$, so at liquid-vapor phase equilibrium:

$\Delta {G}_{v a p} = 0 = \Delta {H}_{v a p} - {T}_{b} \Delta {S}_{v a p}$

$\implies \textcolor{b l u e}{{T}_{b} = \frac{\Delta {H}_{v a p}}{\Delta {S}_{v a p}}}$,

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 $\text{K}$, convert to $\text{^@ "C}$ by subtracting $273.15$ from the $\text{K}$ temperature.