JEE Advanced 2023 Paper 1 · Q12 · Thermodynamics
In a one-litre flask, $6$ moles of $A$ undergoes the reaction $A(g) \rightleftharpoons P(g)$. The progress of product formation at two temperatures (in Kelvin), $T_1$ and $T_2$, is shown in the figure. If $T_1 = 2T_2$ and $(\Delta G_2^\circ - \Delta G_1^\circ) = RT_2 \ln x$, then the value of $x$ is ___.
Reveal answer + step-by-step solution
Correct answer:8
Solution
From the figure: at $T_1$, equilibrium $[P]/[A] = 4/2 = 2$, so $K_1 = 2$. At $T_2$: $[P]/[A] = 2/4 = 1/2$, so $K_2 = 1/2$. $\Delta G^\circ = -RT\ln K$. $\Delta G_1^\circ = -RT_1 \ln 2 = -2RT_2\ln 2$. $\Delta G_2^\circ = -RT_2 \ln(1/2) = RT_2 \ln 2$. Difference: $\Delta G_2^\circ - \Delta G_1^\circ = RT_2 \ln 2 + 2RT_2\ln 2 = 3RT_2\ln 2 = RT_2\ln 8$. So $x = 8$.
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