Given below is an expression for the rate constant of a first-order reaction occurring at a certain temperature $T$ (K): $\ln k=14.34-\frac{1.25\times 10^4}{T}$ The energy of activation in ${\text{kcal mol}}^{-1}$ for the reaction is: (Given: $k$ in ${\text{s}}^{-1}$, $R=1.987{\text{cal mol}}^{-1}{\text{K}}^{-1}$)

Solution & Explanation

\textbf{Answer:} (A) $24.84$ \textbf{Solution:} From the Arrhenius equation $k=A{e}^{-E_a/RT}$, taking ln gives $\ln k=\ln A-\frac{E_a}{R}\cdot \frac{1}{T}$. Comparing with $\ln k=14.34-\frac{1.25\times 10^4}{T}$ gives $\frac{E_a}{R}=1.25\times 10^4$. $E_a=1.25\times 10^4\times R=1.25\times 10^4\times 1.987=24837.5{\text{cal mol}}^{-1}=24.84{\text{kcal mol}}^{-1}$.