Q. The rate constant of a reaction increases by 5% when its temperature is raised from $27℃\ to\ 28℃.$ The activation energy of the reaction is
A) $36.6\ kJ/mol$
B) $16.6\ kJ/mol$
C) $46.6\ kJ/mol$
D) $26.6\ kJ/mol$
Explanation
(a)
$\log\frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303\ R} \times \frac{T_{2} - T_{1}}{T_{1}\ \bullet T_{2}}$
$\log\frac{105}{100} = \frac{E_{a}}{2.303 \times 8.314} \times \frac{1}{300 \times 301}$
$E_{a} = 36.65\ kJ$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.