Given below is an expression for the rate constant of a first-order reaction occurring at a certain temperature $T$ (K): $\ln k = 14.34 - \frac{1.25\times10^{4}}{T}$ 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 $\ln k=14.34-\frac{1.25\times10^{4}}{T}$ gives $\frac{E_a}{R}=1.25\times10^{4}$. $E_a=1.25\times10^{4}\times R=1.25\times10^{4}\times1.987=24837.5\ \text{cal mol}^{-1}=24.84\ \text{kcal mol}^{-1}$.