What is the activation energy $(\mathrm{kJ} / \mathrm{mol})$ for a reaction if its rate constant doubles…

What is the activation energy $(\mathrm{kJ} / \mathrm{mol})$ for a reaction if its rate constant doubles when the temperature is raised from $300 \mathrm{~K}$ to $400 \mathrm{~K}$ ? $\left(R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$
  1. $69.8$
  2. $6.92$
  3. $34.4$
  4. $3.44$

Solution

$\begin{aligned} & \text { } \log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \\ & \log \frac{2}{1}=\frac{E_a}{2.303 \times 8.314}\left(\frac{1}{300}-\frac{1}{400}\right) \\ & 0.3010=\frac{E_a}{2.303 \times 8.314}\left(\frac{100}{300 \times 400}\right) \\ & \Rightarrow E_a=0.3010 \times 2.303 \times 8.314 \times 3 \times 400 \\ & =6.916 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$

Asked in: NEET 2016 (Phase 1)

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