The boiling point of water at $1 \mathrm{~atm}$ is $100^{\circ} \mathrm{C}$. Which among the following is…
The boiling point of water at $1 \mathrm{~atm}$ is $100^{\circ} \mathrm{C}$. Which among the following is true for vaporisation of water at $75^{\circ} \mathrm{C}$ ?
$\Delta G_{\text {vap }}^{\circ}>0$
$\Delta H_{\text {vap }}^{\circ} < 0$
$K_{\text {vap }}=1$
$\Delta S_{\text {vap }}^0 < 0$
Solution
According to standard Gibbs free energy equation
i.e.,
$
\begin{aligned}
& \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} \\
& \Delta G^{\circ}=-T \Delta S^{\circ} \\
& \Delta G^{\circ} \propto-T
\end{aligned}
$
It means $\Delta G^{\circ}$ is inversely proportional to temperature due to negative sign present in Gibbs free energy equation.
Hence, lesser the temperature i.e., $75^{\circ} \mathrm{C}$, greater the value of $\Delta^{\circ} G_{\text {vap }}\left(\Delta G_{\text {vap }}^0>0\right)$.
At $100^{\circ} \mathrm{C}, \Delta G_{\text {vap }}^0$ value is lesser than that at temperature $75^{\circ} \mathrm{C}$.
$\therefore$ Vaporisation of water at $75^{\circ} \mathrm{C}$ has $\Delta G_{\text {vap }}^{\circ}>0$