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The rate of a first order reaction doubles when the temperature changes from $300 \mathrm{~K}$ to $310 \mathrm{~K}$. The activation energy of the reaction (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) is
$$
\left(\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \log 2=0.3\right)
$$
Options:
$$
\left(\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \log 2=0.3\right)
$$
Solution:
2970 Upvotes
Verified Answer
The correct answer is:
$53.33$
$\begin{aligned} \log \frac{\mathrm{K}_2}{\mathrm{~K}_1} & =\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right] \\ \log (2) & =\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \times 8.314}\left[\frac{1}{300}-\frac{1}{310}\right]\end{aligned}$
$\begin{aligned} & =\frac{\mathrm{E}_{\mathrm{a}}}{19.15}(0.0001075) \\ \Rightarrow \mathrm{E}_{\mathrm{a}} & =53441.86 \mathrm{~J} \mathrm{~mol}^{-1} \\ & =53.442 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \approx 53.33 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$
$\begin{aligned} & =\frac{\mathrm{E}_{\mathrm{a}}}{19.15}(0.0001075) \\ \Rightarrow \mathrm{E}_{\mathrm{a}} & =53441.86 \mathrm{~J} \mathrm{~mol}^{-1} \\ & =53.442 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \approx 53.33 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$
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