Question

The enthalpy of vaporisation of a liquid is $30kJ/mol$ and the entropy of vaporisation is $75Jmo{l}^{-1}{K}^{-1}$. What is the boiling point of the liquid at $1atm$?

Answer 1

Given,
$\Delta H$ = $30kJ/mol=30,000J/mol$
$\Delta S=75Jmo{l}^{-1}{K}^{-1}$
At the boiling point, the rate of the movement of molecules into the gas phase from the liquid equals the rate of movement of molecules into the liquid phase from the gas phase. So the two phases remain in equilibrium at the boiling temperature.
liquid $\rightleftharpoons$ vapour
$\therefore \Delta G=0$, for the boiling process
We know,
$\Delta G=\Delta H-T\Delta S$
$\therefore T=\frac{\Delta H}{\Delta S}=\frac{30,000Jmo{l}^{-1}}{75J{K}^{-1}mo{l}^{-1}}=400K$