At T K,the following equation is obtained fo | vedclass

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(A) According to the Arrhenius equation: $k = A e^{-E_a / RT}$.Taking natural logarithm on both sides: $\ln k = \ln A - \frac{E_a}{RT}$.Rearranging th

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$2.303 x R$

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(A) According to the Arrhenius equation: $k = A e^{-E_a / RT}$.Taking natural logarithm on both sides: $\ln k = \ln A - \frac{E_a}{RT}$.Rearranging the terms: $\ln \left(\frac{k}{A}\right) = -\frac{E_a}{RT}$.Converting natural logarithm to base $10$: $2.303 \log \left(\frac{k}{A}\right) = -\frac{E_a}{RT}$.Dividing by $2.303$: $\log \left(\frac{k}{A}\right) = -\frac{E_a}{2.303 RT}$.Given equation: $\log \left(\frac{k}{A}\right) = -\frac{x}{T}$.Comparing the two equations,we get: $\frac{E_a}{2.303 R} = x$.Therefore,the activation energy is: $E_a = 2.303 x R$.

At $T \ K$,the following equation is obtained for a first order reaction: $\log \frac{k}{A} = -\frac{x}{T}$. The activation energy for this reaction is equal to $(R = \text{gas constant})$

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