For vaporisation of water at $1 \mathrm{~atm}$ pressure, the values of $\Delta \mathrm{H}$ and $\Delta S$ are $40.63 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $108.8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, respectively. The temperature when Gibbs energy change $(\Delta G)$ for this transformation will be…

Question

For vaporisation of water at $1 \mathrm{~atm}$ pressure, the values of $\Delta \mathrm{H}$ and $\Delta S$ are $40.63 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $108.8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, respectively. The temperature when Gibbs energy change $(\Delta G)$ for this transformation will be zero, is

Aicharya · Accepted Answer

$\begin{aligned} \Delta \mathrm{G} & =\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S} \ \Delta \mathrm{G} & =0(	ext { given }) \ \Delta \mathrm{H} & =\mathrm{T} \Delta \mathrm{S}, \ \mathrm{T} & =\frac{40.63 	imes 10^3}{108.8}=373.4 \mathrm{~K} \end{aligned}$

For vaporisation of water at $1 \mathrm{~atm}$ pressure, the values of $\Delta \mathrm{H}$ and $\Delta S$…

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