For vaporization of water at 1 atmospheric pr | vedclass

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Question

(C) According to the Gibbs equation:$\Delta G = \Delta H - T \Delta S$When $\Delta G = 0$,the system is at equilibrium,so:$\Delta H = T \Delta S$Given

Vedclass · Accepted Answer

$373$

Vedclass · Answer

(C) According to the Gibbs equation:$\Delta G = \Delta H - T \Delta S$When $\Delta G = 0$,the system is at equilibrium,so:$\Delta H = T \Delta S$Given:$\Delta H = 40.63 \ kJ \ mol^{-1} = 40.63 	imes 10^{3} \ J \ mol^{-1}$$\Delta S = 108.8 \ J \ K^{-1} \ mol^{-1}$Substituting the values:$T = \frac{\Delta H}{\Delta S}$$T = \frac{40.63 	imes 10^{3} \ J \ mol^{-1}}{108.8 \ J \ K^{-1} \ mol^{-1}}$$T \approx 373.43 \ K$Thus,the temperature is approximately $373.4 \ K$.

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 Gibb's energy change $(\Delta G)$ for this transformation will be zero,is ............ $K$. (in $.4$)

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