JEE Advanced 2023 · Paper 1 · Chemistry Q12

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ChemistryInteger+4Question 46 of 51

In a one-litre flask, 6 moles of A undergoes the reaction $\ce{A(g) <=> P(g)}$ . 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 $\left(\Delta G_2^\ominus - \Delta G_1^\ominus\right) = RT_2\ln x$ , then the value of x is _____.

[ $\Delta G_1^\ominus$ and $\Delta G_2^\ominus$ are standard Gibb's free energy change for the reaction at temperatures $T_1$ and $T_2$ , respectively.]

[P] in mol L^-1 (y-axis) against time in hours (x-axis). Two curves rise from the origin and level off: the T1 curve plateaus at [P] = 4 mol L^-1 and the T2 curve plateaus at [P] = 2 mol L^-1 (dashed guides at 4 and 2).

Your answer

Answer and worked solution

Official answer: 8

Equilibrium constants. A starts at 6 mol L $^{-1}$ .

At $T_1$ : $[\ce{P}]=4$ , $[\ce{A}]=2$ , so $K_1=2$ .
At $T_2$ : $[\ce{P}]=2$ , $[\ce{A}]=4$ , so $K_2=\frac12$ .

Gibbs energies. Using $\Delta G^\ominus=-RT\ln K$ with $T_1=2T_2$ :
$\Delta G_2^\ominus-\Delta G_1^\ominus=-RT_2\ln\tfrac12+2RT_2\ln2=3RT_2\ln2=RT_2\ln8.$

So $x=8$ .

Topics:Equilibrium and Gibbs energy Law of mass action and Kc, Kp

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JEE Advanced 2023 · Paper 1, Chemistry question 12