Consider the following reversible reaction,A_ | vedclass

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(A) For the reaction $A_{(g)} + B_{(g)} \rightleftharpoons AB_{(g)}$,the equilibrium constant $K$ is given by $K = \frac{k_f}{k_b} = \frac{A_f e^{-E_a

Vedclass · Accepted Answer

$8500$

Vedclass · Answer

(A) For the reaction $A_{(g)} + B_{(g)} ightleftharpoons AB_{(g)}$,the equilibrium constant $K$ is given by $K = \frac{k_f}{k_b} = \frac{A_f e^{-E_a/RT}}{A_b e^{-E_b/RT}}$.Given $E_b - E_a = 2RT$ and $A_f = 4A_b$,we substitute these into the expression for $K$:$K = \frac{A_f}{A_b} e^{(E_b - E_a)/RT} = 4 e^{2RT/RT} = 4e^2$.The standard Gibbs free energy change is $\Delta G^{\ominus} = -RT \ln K$.Substituting the values: $\Delta G^{\ominus} = -RT \ln(4e^2) = -RT(\ln 4 + \ln e^2) = -RT(2 \ln 2 + 2)$.Given $RT = 2500 \ J \ mol^{-1}$ and $\ln 2 = 0.7$:$\Delta G^{\ominus} = -2500 	imes (2 	imes 0.7 + 2) = -2500 	imes (1.4 + 2) = -2500 	imes 3.4 = -8500 \ J \ mol^{-1}$.The absolute value of $\Delta G^{\ominus}$ is $8500 \ J \ mol^{-1}$.

Similar Questions

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