For the following reactions:A xrightarrow{700 | vedclass

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(D) The rate constant $k$ is given by the Arrhenius equation: $k = A e^{-\frac{E_a}{RT}}$.For the uncatalysed reaction at $700 \ K$: $k_1 = A e^{-\fra

Vedclass · Accepted Answer

$75$

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(D) The rate constant $k$ is given by the Arrhenius equation: $k = A e^{-\frac{E_a}{RT}}$.For the uncatalysed reaction at $700 \ K$: $k_1 = A e^{-\frac{E_a}{R 	imes 700}}$.For the catalysed reaction at $500 \ K$: $k_2 = A e^{-\frac{E_a - 30}{R 	imes 500}}$.Since the rate remains unchanged,$k_1 = k_2$,which implies:$-\frac{E_a}{700R} = -\frac{E_a - 30}{500R}$.Simplifying the equation:$\frac{E_a}{700} = \frac{E_a - 30}{500}$.$5E_a = 7(E_a - 30)$.$5E_a = 7E_a - 210$.$2E_a = 210 \implies E_a = 105 \ kJ/mol$.The activation energy for the catalysed reaction is $E_a - 30 = 105 - 30 = 75 \ kJ/mol$.

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