Q. The rate constant $\left( k^{'} \right)$ of one of the reaction is found to be double that of the rate constant $\left( k^{'}' \right)$ of another reaction. Then the relationship between the corresponding activation energies of the two reaction $\left( E_{a}^{'}\ and\ \ E_{a}^{'}'\ \right)$ can be represented as
A) $E_{a}^{'} > E_{a}^{'}'$
B) $E_{a}^{'} = 4E^{'}'$
C) $E_{1}^{'} = E_{a}^{'}'$
D) $E_{a}^{'} < E_{a}^{'}'$
Explanation
(d)
$\because$ Rate constant $\left( k^{'} \right) >$ rate constant $(k^{'\ '})\$
Greater the rate constant lesser will be the activation energy
$\therefore E_{a}^{'} < E_{a}'\ '$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.