For vaporization of water at $1$ atmospheric pressure, the values of $\Delta H$ and $\Delta S$ are $40.63 ext

Question

For vaporization of water at $1$ atmospheric pressure, the values of $\Delta H$ and $\Delta S$ are $40.63	ext{ kJ mol}^{-1}$ and $108.8	ext{ J K}^{-1}	ext{ mol}^{-1}$, respectively. The temperature when Gibbs energy change ($\Delta G$) for this transformation will be zero, is:

Accepted Answer

According to the Gibbs-Helmholtz equation, $\Delta G = \Delta H - T\Delta S$.
When the Gibbs energy change is zero ($\Delta G = 0$), the system is at equilibrium. At this point:
$\Delta H = T\Delta S$
$\Rightarrow T = \frac{\Delta H}{\Delta S}$
Given:
$\Delta H = 40.63	ext{ kJ mol}^{-1} = 40630	ext{ J mol}^{-1}$ (Note: Convert kJ to J to match the unit of $\Delta S$)
$\Delta S = 108.8	ext{ J K}^{-1}	ext{ mol}^{-1}$
Substituting the values:
$T = \frac{40630	ext{ J mol}^{-1}}{108.8	ext{ J K}^{-1}	ext{ mol}^{-1}} = 373.4375	ext{ K} \approx 373.4	ext{ K}$.

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