At the temperature $T(\mathrm{~K})$ for the reaction $X_2 \mathrm{O}_4(l) ightarrow 2 X \mathrm{O}_2(g)…

At the temperature $T(\mathrm{~K})$ for the reaction $X_2 \mathrm{O}_4(l) ightarrow 2 X \mathrm{O}_2(g) \Delta U=x \mathrm{~kJ} \mathrm{~mol}^{-1}$, $\Delta S=y \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. Gibbs energy change for the reaction is (Assume $X_2 \mathrm{O}_4, X_2$ are ideal gases) $\Delta U=x \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta S=y \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
  1. $1000 x+2 R(T-y) \mathrm{J} \mathrm{mol}^{-1}$
  2. $1000 x+T(2 R-y) \mathrm{J} \mathrm{mol}^{-1}$
  3. $x+T(2 R-y) \mathrm{J} \mathrm{mol}^{-1}$
  4. $x+2 R(T-y) \mathrm{J} \mathrm{mol}^{-1}$

Solution


But, $\quad H=U+p V$ or $H=U+n R T$
From equation (i) and (ii) we get $$ \Delta G=\Delta U+\Delta n R T-T \Delta S $$ But, for given reaction. $\Delta n=2, \Delta U=1000 \times \mathrm{J} / \mathrm{mol}$ and $$ \Delta S=y \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} $$ $\begin{aligned} \therefore \quad \Delta G & =1000 x+2 R T-y T \\ & \Delta G=1000 x+T(2 R-y) \mathrm{J} \mathrm{mol}^{-1}\end{aligned}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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