Consider the following reversible reaction, $A (g) + B (g) \to AB (g)$
The activation energy of the backward reaction exceeds that of the forward reaction by 2RT (in Jmol ^-1).If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value of ${ΔG}^{0} (in {jmol}^{- 1})$ for the reaction at 300 K is___
(Given $In (2) = 0.7, RT = 2500 J {mol}^{- 1}$ at 300 K and G is the Gibbs energy)

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answer is 8500.

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Detailed Solution

$\begin{array}{l} A (g) + B (g) ⇌ AB (g) \\ E_{ab} = E_{af} + 2 RT and A_{f} = 4 A_{b} \end{array}$
Now, Rate constant of forward reaction
$k_{f} = A_{f} e^{- {Ea}_{f} / RT}$
Rate constant of reverse reaction
$k_{b} = A_{b} e^{- E_{ab} / RT}$
$Equilibrium k_{eq} = \frac{k_{f}}{k_{b}} = \frac{A_{f}}{A_{b}} e^{- (E_{af} - E_{ab}) / RT}$
$k_{eq} = 4 e^{2 RT / RT} = 4 e^{2}$
$Now, {ΔG}^{0} = - RTln k_{eq} = - 2500 \ln (4 e^{2})$
$\begin{array}{l} = - 2500 (\ln 4 + \ln e^{2}) = - 2500 (1 .4 + 2) \\ = - 2500 \times 3 .4 = - 8500 J / mol \end{array}$