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Arrhenius Equation

Consider the ...

Question

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 (in $J m o l^{- 1}$ ). If the pre-exponential factor of the forward reaction is $4$ times that of the reverse reaction, the absolute value of $Δ G^{\circ}$ (in J mol $^{- 1})$ for the reaction at $300 K$ is _______.

[Given ln (2) = 0.7, RT = 2500 J mol $^{- 1}$ at 300 K and G is the Gibbs energy].

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Solution

$A (g) + B (g) ⇌ A B (g)$

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

$\frac{A_{f}}{A_{b}} = 4$

$Δ G^{o} = - R T l n K_{e q}$

$K_{f} = A_{f} e^{- (E_{a})_{f} / R T}$

$K_{b} = A_{b} e^{- (E_{a})_{b} / R T}$

$K_{e q} = \frac{K_{f}}{K_{b}} = \frac{A_{f}}{A_{b}} \times e^{- \frac{(E_{a})_{f}}{R T}} \times e^{+ \frac{(E_{a})_{b}}{R T}} = 4 \times e^{\frac{(E_{a})_{b} - (E_{a})_{f}}{R T}}$

$K_{e q} = 4 \times e^{2}$

$Δ G^{o} = - R T \times l n (4 \times e^{2})$

$Δ G^{o} = - R T (l n 4 + 2 l n e)$

$Δ G^{o} = - R T (2 \times 0.7 + 2)$

$Δ G^{o} = - R T (1.40 + 2)$

$Δ G^{o} = - R T (3.40)$

$Δ G^{o} = - 2500 \times 3.40$

$Δ G^{o} = - 8500$ J.

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