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Standard XII

Chemistry

Arrhenius Equation

Consider the ...

Question

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

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Solution

Arrhenius equation states that
$k = A e^{\frac{- E_{a}}{R T}}$
Hence, for forward reaction: $k_{f} = A e^{\frac{- E_{a_{f}}}{R T}}$
backward reaction: $k_{b} = A e^{\frac{- E_{a_{b}}}{R T}}$
$K = \frac{k_{f}}{k_{b}} = \frac{A_{f}}{A_{b}} e^{\frac{- E_{a_{f}} + E_{a_{b}}}{R T}}$
$K = 4 e^{2 R T} R T = 4 e^{2}$
We know that
$Δ G = - R T l n K$
$Δ G = - 2500 \times l n (4 e^{2})$
$Δ G = - 2500 \times (2 + 2 l n 2) = - 8500 J / m o l$

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Arrhenius equation states that k=Ae−EaRT Hence, for forward reaction: kf=Ae−EafRT backward reaction: kb=Ae−EabRT K=kfkb=AfAbe−Eaf+EabRT K=4e2RTRT=4e2 We know that ΔG=−RTlnK ΔG=−2500×ln(4e2) ΔG=−2500×(2+2ln2)=−8500J/mol