Activation energy \((E_a)\) and rate constants (\(k_1\) and \(k_2\)) of a chemical reaction at two different temperatures (\(T_1\) and \(T_2\)) are related by:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → ln (k2/k1) = (Ea/R) [1/T1 − 1/T2]
The Arrhenius equation in logarithmic form is
\( \displaystyle \ln k = \ln A - \frac{E_a}{RT} \)
Writing it at the two temperatures and subtracting the first from the second:
\( \displaystyle \ln k_2 - \ln k_1 = \frac{E_a}{RT_1} - \frac{E_a}{RT_2} \quad\Rightarrow\quad \ln\frac{k_2}{k_1} = \frac{E_a}{R}\left[\frac{1}{T_1} - \frac{1}{T_2}\right] \)
With common logarithms the same relation reads \( \log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left[\frac{1}{T_1} - \frac{1}{T_2}\right] \).
Why the other options are incorrect
Option (1) — Incorrect. The factor 2.303 belongs with log (base 10), not with ln; with ln it should not appear.
Option (2) — Incorrect. The minus sign is wrong; for T2 > T1 the bracket is positive and k2 must be greater than k1, but this form would make k2 smaller.
Option (4) — Incorrect. It has both errors, the factor 2.303 with ln and the wrong sign.
NCERT (2026-27 edition):
Temperature Dependence of the Rate of a Reaction – “… Subtracting equation (3.20) from (3.21), we obtain
\( \ln k_2 - \ln k_1 = \displaystyle\frac{E_a}{RT_1} - \displaystyle\frac{E_a}{RT_2} \)
\( \ln \displaystyle\frac{k_2}{k_1} = \displaystyle\frac{E_a}{R}\left[\displaystyle\frac{1}{T_1} - \displaystyle\frac{1}{T_2}\right] \)”