JEE Advanced 2023 — Paper 1 — CHEM-III solution

JEE Advanced2023Paper 1CHEM-III

In a one-litre flask, $6$ moles of $A$ undergoes the reaction $A (g) ⇌ P (g)$ . The progress of product formation at two temperatures (in Kelvin), $T_{1}$ and $T_{2}$ , is shown in the figure:

If $T_{1} = 2 T_{2}$ and $(Δ G_{2}^{\circ} - Δ G_{1}^{\circ}) = R T_{2} ln x$ , then the value of $x$ is ______.

[ $Δ G_{1}^{\circ}$ and $Δ G_{2}^{\circ}$ are standard Gibb's free energy change for the reaction at temperatures $T_{1}$ and $T_{2}$ , respectively.]

Your answer

Solution

Equilibrium constants. At $T_{1}$ : $[P]_{e q} = 4$ , $[A]_{e q} = 6 - 4 = 2$ , so $K_{1} = 4/2 = 2$ . At $T_{2}$ : $[P]_{e q} = 2$ , $[A]_{e q} = 4$ , so $K_{2} = 2/4 = 1/2$ .

Gibbs energy changes. $Δ G_{1}^{\circ} = - R T_{1} ln K_{1} = - 2 R T_{2} ln 2$ (using $T_{1} = 2 T_{2}$ ). $Δ G_{2}^{\circ} = - R T_{2} ln K_{2} = - R T_{2} ln (1/2) = R T_{2} ln 2$ .

Difference. $Δ G_{2}^{\circ} - Δ G_{1}^{\circ} = R T_{2} ln 2 - (- 2 R T_{2} ln 2) = 3 R T_{2} ln 2 = R T_{2} ln 2^{3} = R T_{2} ln 8.$ Therefore $x = 8$ .

Answer: 8.

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