Search any question & find its solution
Question:
Answered & Verified by Expert
The rate constant, $\mathrm{k}$ for a first order reaction, $\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}(\mathrm{g}) \rightarrow \mathrm{C}_2 \mathrm{H}_4(\mathrm{~g})+\mathrm{HI}(\mathrm{g})$ is $\mathrm{xs}^{-1}$ at $600 \mathrm{~K}$ and $4 \mathrm{x} \mathrm{s}^{-1}$ at $700 \mathrm{~K}$.
The energy of activation of the reaction (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) is
Options:
The energy of activation of the reaction (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) is
Solution:
2154 Upvotes
Verified Answer
The correct answer is:
$48.16$
$\log \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_1 \mathrm{~T}_2}\right)$
$\begin{aligned} & \log \left(\frac{4 \mathrm{x}}{\mathrm{x}}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{2.303(8.314)}\left(\frac{700-600}{700 \times 600}\right) \\ & 0.602=\frac{\mathrm{E}_{\mathrm{a}}}{19.147}\left(2.38 \times 10^{-4}\right) \\ & \Rightarrow \mathrm{E}_{\mathrm{a}}=48548.3 \mathrm{~J}=48.54 \mathrm{~kJ} \\ & \approx 48.16 \mathrm{~kJ} .\end{aligned}$
$\begin{aligned} & \log \left(\frac{4 \mathrm{x}}{\mathrm{x}}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{2.303(8.314)}\left(\frac{700-600}{700 \times 600}\right) \\ & 0.602=\frac{\mathrm{E}_{\mathrm{a}}}{19.147}\left(2.38 \times 10^{-4}\right) \\ & \Rightarrow \mathrm{E}_{\mathrm{a}}=48548.3 \mathrm{~J}=48.54 \mathrm{~kJ} \\ & \approx 48.16 \mathrm{~kJ} .\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.