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Gibbs energy
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For the reaction: $X_2O_4(l) \rightarrow 2XO_2(g)$, $\Delta U$ = 2.1 kcal, $\Delta S$ = 20 cal $K^{-1}$ at 300 K. Hence $\Delta G$ is:$X_2O_4(l) \rightarrow 2XO_2(g)$; $\Delta n_g = 2 - 0 = 2$
$\Delta H = \Delta U + \Delta n_gRT = 2.1 + 2\times\frac{2}{1000}\times300 = 3.3$ kcal
$\Delta G = \Delta H - T\Delta S = 3.3 - 300\times\frac{20}{1000} = -2.7$ kcal -
For vaporization of water at 1 atmospheric pressure, the values of $\Delta H$ and $\Delta S$ are 40.63 kJ $mol^{-1}$ and 108.8 J $K^{-1}mol^{-1}$, respectively. The temperature when Gibbs energy change ($\Delta G$) for this transformation will be zero is$\Delta G = \Delta H - T\Delta S$
When $\Delta G = 0$: $\Delta H = T\Delta S$
$T = \frac{\Delta H}{\Delta S} = \frac{40.63\times10^3\ J\,mol^{-1}}{108.8\ J\,K^{-1}mol^{-1}}$
T = 373.4 K
(This is the boiling point of water, where liquid and vapour are in equilibrium at 1 atm.)