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Question: Answered & Verified by Expert
Activation energy $\left(E_a\right)$ and rate constants $\left(k_1\right.$ and $\left.k_2\right)$ of a chemical reaction at two different temperatures $\left(T_1\right.$ and $\left.T_2\right)$ are related by
ChemistryChemical KineticsNEETNEET 2012 (Mains)
Options:
  • A $\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$
  • B $\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$
  • C $\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}+\frac{1}{T_1}\right)$
  • D None of the above
Solution:
1111 Upvotes Verified Answer
The correct answer is: $\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$
According to Arrhenius equation, activation energy $\left(E_a\right)$ and rate constants $\left(k_1\right.$ and $\left.k_2\right)$ of a chemical reaction at two different temperatures ( $T_1$ and $\left.T_2\right)$ are related as
$$
\begin{aligned}
\ln \frac{k_2}{k_1} & =\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \\
& =-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)
\end{aligned}
$$

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