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

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Solution

$A (g) + B (g) ⇌ A B (g)$
$(E_{a})_{b} - (E_{a})_{f} = 2 R T$
$\frac{A_{f}}{A_{b}} = 4$
$Δ G^{\circ} = - R T l n K_{e q}$
$K_{f} = A_{f} e^{\frac{- (E_{a})_{f}}{R T}}$
$K_{b} = A_{b} e^{\frac{- (E_{a})_{b}}{R T}}$
$K_{e q} = \frac{K_{f}}{K_{b}} = \frac{A_{f}}{A_{b}} \times e^{\frac{(E_{a})_{f}}{R T}} \times e^{+ \frac{(E_{a})_{b}}{R T}} = 4 \times e^{\frac{(E_{a})_{b} - (E_{a})_{f}}{R T}}$
$K_{e q} = 4 \times e^{2}$
$Δ G^{\circ} = - R T \times l n (4 \times e^{2})$
$Δ G^{0} = - R T (l n 4 + 2 l n e)$
$Δ G^{0} = - R T (2 \times 0.7 + 2)$
$Δ G^{0} = - R T (1.40 + 2)$
$Δ G^{0} = - R T (3.40)$
$Δ G^{0} = - 2500 \times 3.40$
$Δ G^{0} = - 8500 J$

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