Consider a reversible reaction, $A (g) + B (g) ⇌ A B (g)$ . The activation energy of the backward reaction exceeds that of forward reaction by 2RT $(i n J . m o l e^{- 1})$ . If the pre exponential factor of the forward reaction is 4 times that of the reverse reaction the absolute value of $Δ G^{0}$ for the reaction at 300K is X + 0.5 KJ.mol^-1
(Given ln2= 0.7, RT = 2500J $m o l e^{- 1}$ at 300K and G = Gibbs energy)

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Detailed Solution

It is reversible reaction $A (g) + B (g) A B (g)$

$E a_{b} = E a_{f} + 2 R T and A_{f} = 4 A_{b}$

Rate constant for forward reaction $k_{f} = A_{f} e \frac{- E a_{f}}{R T}$

Rate constant for back words reaction $k_{b} = A_{b} e \frac{- E a_{b}}{R T}$

Equilibrium constant

$K e q = \frac{k_{f}}{k_{b}} = \frac{A_{f}}{A_{b}} e^{- (E a_{f} - E a_{b}) / R T}$

$k e q = 4 e^{2 R T / R T} = 4 e^{2}$

$Δ G^{0} = - R T \ln K e q = - 2500 (\ln 4 + \ln e^{2})$

$= - 2500 (1.4 + 2) = - 2500 \times 2.8$

$Δ^{0} ≅ - 8 K J m o l e$

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