Q. A reactant $(A)$ forms two products : $A\overset{k_{2}}{\rightarrow}B,\ Activation\ energy\ E_{a_{1}}$ $A\overset{k_{2}}{\rightarrow}C,Activation\ energy\ E_{a_{2}}$ If $E_{a_{2}} = 2\ E_{a_{1}}$, than $k_{1}$ and $k_{2}$ are related as
A) $k_{1} = 2k_{2}{e_{E}}_{a_{2}}/RT$
B) $k_{1} = k_{2}{e_{E}}_{a_{1}}/RT$
C) $k_{2} = k_{1}{e_{E}}_{a_{2}}/RT$
D) $k_{1} = Ak_{2}{e_{E}}_{a_{1}}/RT$
Explanation
(b)
$A\overset{\ \ \ k_{1}\ \ \ \ \ \ }{\rightarrow}B,$
$A\overset{{\ \ \ \ k}_{2}\ \ \ \ }{\rightarrow}C,$
By Arrhenius equation,
$R_{1} = A^{'}e^{- E_{a}1/RT}$and $k_{2} = A^{'}e^{- E_{a}2/RT}$
($A^{'}\$is Arrhenius constant) (Since, $E_{a2} = 2E_{a1}$)
$\therefore k_{2} = A^{'}e^{- 2E_{a'}|RT}$
$\frac{k_{1}}{k_{2}} = \frac{A^{'}e^{- E_{a}1|RT}}{A^{'}e^{- 2E_{a}|RT}} = e^{E_{a}1|RT}$
$\therefore k_{1} = k_{2}e^{E_{a}1/RT}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.