What is the free energy change for the conversion of 1 mole of water into steam at $373.2 \mathrm{~K}$. The…

What is the free energy change for the conversion of 1 mole of water into steam at $373.2 \mathrm{~K}$. The heat of vaporization $\left(\Delta \mathrm{H}_{\mathrm{v}}ight)$ of water of $373.2 \mathrm{~K}$ is $9.1 \mathrm{kcal} \mathrm{mol}^{-1}$. The entropy change is $25.5 \mathrm{cal} / \mathrm{mol} \mathrm{deg}$.
  1. $-401.6 \mathrm{cal} / \mathrm{mol}$
  2. $-416.6 \mathrm{cal} / \mathrm{mol}$
  3. $516.5 \mathrm{cal} / \mathrm{mol}$
  4. $-516.5 \mathrm{cal} / \mathrm{mol}$

Solution

$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$
$\therefore \Delta G=9100-373.2 \times 25.5=-416.6 \mathrm{cal}$
$\mathrm{mol}^{-1}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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