Q. According to Arrhenius equation, the rate constant $(k)$ is related to temperature $(T)$ as
A) $In\frac{k_{2}}{k_{1}} = \frac{E_{a}}{R}\left\lbrack \frac{1}{T_{1}} - \frac{1}{T_{2}} \right\rbrack$
B) $In\frac{k_{2}}{k_{1}} = - \frac{E_{a}}{R}\left\lbrack \frac{1}{T_{1}} - \frac{1}{T_{2}} \right\rbrack$
C) $In\frac{k_{2}}{k_{1}} = \frac{E_{a}}{R}\left\lbrack \frac{1}{T_{1}} + \frac{1}{T_{2}} \right\rbrack$
D) $In\frac{k_{2}}{k_{1}} = \frac{E_{a}}{R}\left\lbrack \frac{1}{T_{1}} + \frac{1}{T_{2}} \right\rbrack$
Explanation
(a)
Arrhenius equation is written as
$k = Ae^{- E_{a}/RT}$
Taking logarithm, above equation may be written as
In $k = In\ A - \frac{E_{a}}{R} \times \frac{1}{T}$
$\therefore In\ k_{1} = In\ A - \frac{E_{a}}{R} \times \frac{1}{T_{1}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
$In\ k_{2} = In\ A - \frac{E_{a}}{R} \times \frac{1}{T_{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Subtracting the Eq. (i) from Eq. (ii)
$In\ k_{2} - In\ k_{1} = \frac{E_{a}}{R}\left\lbrack \frac{1}{T_{1}} - \frac{1}{T_{2}} \right\rbrack$
$In\frac{k_{2}}{k_{1}} = \frac{E_{a}}{R}\left\lbrack \frac{1}{T_{1}} - \frac{1}{T_{2}} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.