Consider the following reversible reaction, $\mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g}) \rightleftharpoons \mathrm{AB}(\mathrm{g})$ The activation energy of the backward reaction exceeds that of the forward reaction by $2RT$ (in $\mathrm{J}\mathrm{mol}^{-1}$). If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value of $\Delta G^{\ominus}$ (in $\mathrm{J}\mathrm{mol}^{-1}$) for the reaction at 300 K is ____. (Given; $\ln(2)=0.7$, $RT=2500\mathrm{~J}\mathrm{mol}^{-1}$ at 300 K and $G$ is the Gibbs energy)

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Chemical Thermodynamics

Consider the following reversible reaction,

A (g) + B (g) ⇌ AB (g)

The activation energy of the backward reaction exceeds that of the forward reaction by

2 R T

(in

J mol^{- 1}

). If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value of

Δ G^{⊖}

(in

J mol^{- 1}

) for the reaction at 300 K is ____.

(Given;

ln (2) = 0.7

R T = 2500 J mol^{- 1}

at 300 K and

G

is the Gibbs energy)

Answer: 8500

Given

(E_{a})_{b} - (E_{a})_{f} = 2 R T

and

A_{f} / A_{b} = 4

. The equilibrium constant

K_{e q} = \frac{k _{f}}{k _{b}} = \frac{A _{f}}{A _{b}} \times e^{- [(E_{a})_{f} - (E_{a})_{b}] / R T} = 4 \times e^{2 R T / R T} = 4 e^{2}

. Then

Δ G^{\circ} = - R T ln K_{e q} = - R T ln (4 e^{2}) = - R T (ln 4 + 2) = - R T (2 ln 2 + 2) = - 2500 (2 \times 0.7 + 2) = - 2500 \times 3.4 = - 8500

J/mol. Absolute value: 8500.

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Numerical answer