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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
A
273.4 K
B
393.4 K
C
373.4 K
D
293.4 K
Detailed Solution
$\Delta G = \Delta H - T\Delta S$
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.)
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.)