Q. If $k_{1} =$ rate contant at temperature $T_{1}$ and $k_{2} =$rate constant at temperature $T_{2}$ for a first order reaction, then which of the following relations is correct? $(E_{a}\ :activation\ energy)$
A) $\log\begin{array}{r}
k_{1} \\
- \\
k_{2}
\end{array} = \frac{2.303\ E_{a}}{R}\left( \frac{T_{2} - T_{1}}{T_{1}T_{2}} \right)$
B) $\log\begin{array}{r}
k_{2} \\
- \\
k_{1}
\end{array} = \frac{\ E_{a}}{2.303R}\left( \frac{T_{2} - T_{1}}{T_{1}T_{2}} \right)$
C) $\log\begin{array}{r}
k_{2} \\
- \\
k_{1}
\end{array} = \frac{\ E_{a}}{2.303R}\left( \frac{T_{1}T_{2}}{T_{2} + T_{1}} \right)$
D) $\log\begin{array}{r}
k_{1} \\
- \\
k_{2}
\end{array} = \frac{E_{a}}{2.303\ R}\left( \frac{T_{1}T_{2}}{T_{2} - T_{1}} \right)$
Explanation
(b)
For a first order reaction, Arrhenius equation is given as
$k = A.e^{- E_{a}/RT}$
where, $E_{a} =$energy of activation
$\therefore\ \frac{k_{2}}{k_{1}} = \frac{e^{- E_{a}/RT_{1}}}{e^{- E_{a}/RT_{2}}}$
$= e^{E_{a}/RT_{2} - E_{a}/RT_{1}}$
$= e^{\frac{E_{a}}{R}}\left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right)$
$= e^{\frac{E_{a}}{R}}\left( \frac{T_{2} - T_{1}}{T_{1}T_{2}} \right)$
Taking log on both the sides, we get
$\log{\frac{k_{2}}{k_{1}} = \log{e^{\frac{Ea}{R}}\left( \frac{T_{2} - T_{1}}{T_{1}T_{2}} \right)}}$
$= \frac{E_{a}}{2.303R}\left( \frac{T_{2} - T_{1}}{T_{1}T_{2}} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.