Q16 / 110 · VITEEE 2024 · Chemistry
The activation energy for a reaction is $9.0 \mathrm{~K} \mathrm{cal} \mathrm{mol}^{-1}$. The increase in the rate constant when its temperature is increased from 298 K to 308 K is
- A$10 \%$
- B$100 \%$
- C$50 \%$
- Correct answerD$63 \%$
Solution
According to the Arrhenius equation
$\log K=\log A-\frac{E_a}{2.303 R T}$....(i)
$\frac{\log K_{30 B}}{\log K_{29 B}}=\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_2}\right]\ldots(ii)$
$\begin{aligned} & \frac{\log K_{308}}{\log K_{298}}=\frac{9.0}{2.303 \times 2 \times 10^{-3}}\left[\frac{308-298}{308 \times 298}\right] \\ & or \quad \frac{\log K_{308}}{\log K_{298}}=0.2129\end{aligned}$
$\begin{aligned} K_{308} & =1.632 K_{298} \ldots(iii)\\ \% \text { increase } & =\frac{K_{308}-K_{298}}{K_{298}} \times 100 \\ & \cong 63 \%\end{aligned}$
Hence, the correct answer is option: 'D'.