Q. The activation energy of a reaction is 9 kcal/mol. The increase in the rate constant when its temperature is raised from 295 to 300 K is approximately
A) 10%
B) 50%
C) 100%
D) 28%
Explanation
(d)
$\log\frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303R}\ \left( \frac{T_{2} - T_{1}}{T_{1}T_{2}} \right)$
$\log\frac{k_{2}}{k_{1}} = \frac{9000}{2.303 \times 2}\left( \frac{5}{295 \times 300} \right)$
$\log\frac{k_{2}}{k_{1}} = 0.1103$
$\frac{k_{2}}{k_{1}} = 1.288,\ k_{2} = 1.288\ k$
$ie,$ increase by 28.8%
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.