For water,the value of Delta H_{vap} is 40.73 | vedclass

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Question

(A) At equilibrium,the change in Gibbs free energy $\Delta G = 0$. Using the relation $\Delta G = \Delta H - T\Delta S$,we get $\Delta H = T\Delta S$.

Vedclass · Accepted Answer

$100.67 \, ^\circ C$

Vedclass · Answer

(A) At equilibrium,the change in Gibbs free energy $\Delta G = 0$. Using the relation $\Delta G = \Delta H - T\Delta S$,we get $\Delta H = T\Delta S$. Therefore,$T = \frac{\Delta H_{vap}}{\Delta S_{vap}}$. Given $\Delta H_{vap} = 40.73 \, kJ \, mol^{-1} = 40730 \, J \, mol^{-1}$ and $\Delta S_{vap} = 109 \, J \, K^{-1} \, mol^{-1}$. $T = \frac{40730 \, J \, mol^{-1}}{109 \, J \, K^{-1} \, mol^{-1}} \approx 373.67 \, K$. Converting to Celsius: $T(^circ C) = 373.67 - 273 = 100.67 \, ^\circ C$.

For water,the value of $\Delta H_{vap}$ is $40.73 \, kJ \, mol^{-1}$ and the value of $\Delta S_{vap}$ is $109 \, J \, K^{-1} \, mol^{-1}$. At what temperature will water be in equilibrium with its vapor?

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