Q. For reaction $A \rightarrow B,$ the rate constant $k_{1} = A_{1}e^{- Ea_{1}/RT}$ and for the reaction $X \rightarrow Y,$ the rate constant $k_{2} = A_{2}e^{- Ea_{2}/RT}.$ If $A_{1} = 10^{8},\ A_{2} = 10^{10}\$and$\ Ea_{1} = 600\ cal/mol,\ Ea_{2} = 1800\ cal/mol,\$then the temperature at which $k_{1} = k_{2}$ is $(R = 2\ cal/k\ mol)\$
A) $\frac{200}{3.506}\ K$
B) $\frac{300}{2.303}\ K$
C) $\frac{400}{6.506}\ K$
D) $\frac{200}{5.204}\ K$
Explanation
(b)
$A_{1}\ .e^{- E_{a1}/RT} = A_{2}\ \bullet e^{- E_{a2}/RT}$
$\frac{A_{2}}{A_{1}} = e^{\left( E_{a_{2}} - E_{a_{1}} \right)/RT}$
$10^{2} = \exp\left( \frac{1200}{RT} \right)$
$2\ln 10 = \frac{1200}{2T}$
$T = \left( \frac{600}{4.606} \right)\ K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.