Question

If $x$ is the temperature (in${}^{o}C$) at which liquid $H_{2}O$ will be in equilibrium with its vapours ($\Delta Hand\Delta S$ for vaporization are $40.7kJ{mol}^{-1}$ and $0.11kJ{mol}^{-1}{K}^{-1}$ respectively) , then $x$ is:

Answer 1

$\Delta G=\Delta H-T\Delta S$
At equilibrium, $\Delta G=0$
The equation reduces to,
$\Delta H=T\Delta S$
$T=\frac{\Delta H}{\Delta S}=\frac{40.7}{0.11}=370K={97}^{o}C$
Hence, $x=97$ is the answer.