Search any question & find its solution
Question:
Answered & Verified by Expert
The rate constant of a reaction at temperature $200 \mathrm{~K}$ is 10 times less than the rate constant at $400 \mathrm{~K}$. What is the activation energy of the reaction?
Options:
Solution:
2139 Upvotes
Verified Answer
The correct answer is:
$230.3 \mathrm{R}$,
At $\mathrm{T}_{1}=200 \mathrm{~K}$, if $k_{1}=k$
then at $\mathrm{T}_{2}=400 \mathrm{~K}, k_{2}=10 k$
$$
\begin{array}{l}
\log \frac{10 k}{k}=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{400-200}{400 \times 200}\right) \\
\mathrm{E}_{\mathrm{a}}=921.2 \mathrm{R}
\end{array}
$$
then at $\mathrm{T}_{2}=400 \mathrm{~K}, k_{2}=10 k$
$$
\begin{array}{l}
\log \frac{10 k}{k}=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{400-200}{400 \times 200}\right) \\
\mathrm{E}_{\mathrm{a}}=921.2 \mathrm{R}
\end{array}
$$
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.