In a one-litre flask,6 moles of A undergoes t | vedclass

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(C) At $T_1 	ext{ K}$: $A_{(g)} ightleftharpoons P_{(g)}$$t=0$: $6 \quad 0$$t=\infty$: $6-4=2 \quad 4$ (from plot)$\Rightarrow K_{c_1} = \frac{4}{2

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$8$

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(C) At $T_1 	ext{ K}$: $A_{(g)} ightleftharpoons P_{(g)}$$t=0$: $6 \quad 0$$t=\infty$: $6-4=2 \quad 4$ (from plot)$\Rightarrow K_{c_1} = \frac{4}{2} = 2$At $T_2 	ext{ K}$: $A_{(g)} ightleftharpoons P_{(g)}$$t=0$: $6 \quad 0$$t=\infty$: $6-2=4 \quad 2$ (from plot)$\Rightarrow K_{c_2} = \frac{2}{4} = \frac{1}{2}$Since $\Delta n_g = 0$,$K_p = K_c$.$\Delta G_2^{\Theta} = -RT_2 \ln K_{p_2} = -RT_2 \ln \frac{1}{2} = RT_2 \ln 2$$\Delta G_1^{\Theta} = -RT_1 \ln K_{p_1} = -R(2T_2) \ln 2 = -2RT_2 \ln 2$$\Delta G_2^{\Theta} - \Delta G_1^{\Theta} = RT_2 \ln 2 - (-2RT_2 \ln 2) = 3RT_2 \ln 2 = RT_2 \ln 2^3 = RT_2 \ln 8$Given $\Delta G_2^{\Theta} - \Delta G_1^{\Theta} = RT_2 \ln x$,therefore $x = 8$.

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