Q. The activation energy of a reaction at a given temperature is found to be 2.303$RT\ J\ mol^{- 1}$ . The ratio of rate constant to the Arrhenius factor is
A) 0.01
B) 0.1
C) 0.02
D) 0.001
Explanation
(b)
Arrhenius equation is,
Rate constant, $k = Ae^{- E_{a}|RT}$
$k = Ae^{- 2.303\ RT/RT}$
$\frac{k}{A} = e^{- 2.303}$
On solving, we get
$\frac{k}{A} = 10^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.