The activation energy of a first-order reacti | vedclass

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(B) Let the rate constant without a catalyst be $k_1$ and with a catalyst be $k_2$.Activation energy without catalyst $(E_1) = 30\,kJ\,mol^{-1}$.Activ

Vedclass · Accepted Answer

$11$

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(B) Let the rate constant without a catalyst be $k_1$ and with a catalyst be $k_2$.Activation energy without catalyst $(E_1) = 30\,kJ\,mol^{-1}$.Activation energy with catalyst $(E_2) = 24\,kJ\,mol^{-1}$.Temperature $(T) = 25 + 273 = 298\,K$.According to the Arrhenius equation:$\ln(\frac{k_2}{k_1}) = \frac{E_1 - E_2}{RT}$$\log(\frac{k_2}{k_1}) = \frac{E_1 - E_2}{2.303RT}$$\log(\frac{k_2}{k_1}) = \frac{(30 - 24) 	imes 10^3\,J\,mol^{-1}}{2.303 	imes 8.314\,J\,K^{-1}\,mol^{-1} 	imes 298\,K}$$\log(\frac{k_2}{k_1}) = \frac{6000}{5705.84} \approx 1.0515$$\frac{k_2}{k_1} = 	ext{antilog}(1.0515) \approx 11.26$Thus,the rate of reaction increases by approximately $11$ times.

The activation energy of a first-order reaction at $25\,^{\circ}C$ is $30\,kJ/mol$. In the presence of a catalyst,the activation energy of the same reaction at $25\,^{\circ}C$ becomes $24\,kJ/mol$. The rate of the reaction in the presence of the catalyst will be how many times the original rate?

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