Q3 / 90 · JEE Main 2012 (26 May Online) · Chemistry
The activation energy for a reaction which doubles the rate when the temperature is raised from $298 \mathrm{~K}$ to $308 \mathrm{~K}$ is
- A$59.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- B$39.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- Correct answerC$52.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- D$29.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
Activation energy can be calculated from the equation.
$$
\begin{aligned}
& \frac{\log K_2}{\log K_1}=\frac{-E_a}{2.303 R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) \\
& \text { Given } \frac{\log K_2}{\log K_1}=2 \quad T_2=308 ; \quad T_1=298 \\
& \therefore \quad \log 2=\frac{-E_a}{2.303 \times 8.314}\left(\frac{1}{308}-\frac{1}{298}\right) \\
& E_a=52.9 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.