Search any question & find its solution
Question:
Answered & Verified by Expert
A plot of \( \frac{1}{T} \mathrm{Vs} . \mathrm{k} \) for a reaction gives the slope \( -1 \times 10^{4} \mathrm{~K} . \) The energy of activation for the
reaction is
(Given \( R=8.314 J K^{-1} \mathrm{~mol}^{-1} \) )
Options:
reaction is
(Given \( R=8.314 J K^{-1} \mathrm{~mol}^{-1} \) )
Solution:
2050 Upvotes
Verified Answer
The correct answer is:
\( 83.14 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
Arrhenius equation,
\( k=A e^{-\frac{E_{a}}{R T}} \rightarrow(1) \)
Taking natural log of Eq. (1), we get
\( \ln k=\ln A-\frac{E_{a}}{R} \times \frac{1}{T} \)
By plotting In \( \mathrm{k} \) against \( 1 / \mathrm{T} \), a straight line with slope \( =-\frac{E}{R} \) and intercept \( \ln \mathrm{A} \).
Now,
slope \( =-\frac{E_{a}}{R} \) \( -1 \times 10^{4}=-\frac{E_{a}}{8.314} \) \( E_{a}=8.314 \times 1 \times 10^{4} \) \( E_{a}=83.140 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
\( k=A e^{-\frac{E_{a}}{R T}} \rightarrow(1) \)
Taking natural log of Eq. (1), we get
\( \ln k=\ln A-\frac{E_{a}}{R} \times \frac{1}{T} \)
By plotting In \( \mathrm{k} \) against \( 1 / \mathrm{T} \), a straight line with slope \( =-\frac{E}{R} \) and intercept \( \ln \mathrm{A} \).
Now,
slope \( =-\frac{E_{a}}{R} \) \( -1 \times 10^{4}=-\frac{E_{a}}{8.314} \) \( E_{a}=8.314 \times 1 \times 10^{4} \) \( E_{a}=83.140 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
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.