The rate constant for a first order reaction becomes 2 times when the temperature is raised from 330K to 380K. Then energy of activation for the reaction will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 14.45 kJ/mol
Use the Arrhenius equation for two temperatures:
\( \displaystyle \log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left[\frac{T_2 - T_1}{T_1T_2}\right] \)
Given k2/k1 = 2, T1 = 330 K, T2 = 380 K, R = 8.314 J K−1 mol−1.
\( \displaystyle 0.3010 = \frac{E_a}{2.303 \times 8.314}\left[\frac{50}{330 \times 380}\right] \)
\( \displaystyle E_a = \frac{0.3010 \times 2.303 \times 8.314 \times 125400}{50} \)
\( E_a = 5.763 \times 2508 = 14454\ \text{J mol}^{-1} \approx 14.45\ \text{kJ mol}^{-1} \)
Note: NTA picked this question from an NCERT (2026-27) intext question and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Intext Questions – “3.8 The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate Ea.”