For a chemical reaction with a rise in temperature by 10°C, the rate constant becomes nearly
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2 times
Raising the temperature increases the number of molecules with energy equal to or greater than the activation energy. For most reactions, a rise of 10° nearly doubles the rate constant (temperature coefficient ≈ 2).
The Arrhenius equation shows why. For a typical activation energy of about 50 kJ mol−1, between 298 K and 308 K:
\( \displaystyle \log\frac{k_{308}}{k_{298}} = \frac{E_a}{2.303R}\left(\frac{T_2 - T_1}{T_1 T_2}\right) = \frac{50000}{2.303 \times 8.314} \times \frac{10}{298 \times 308} \approx 0.29 \)
\( \displaystyle \frac{k_{308}}{k_{298}} \approx 10^{0.29} \approx 2 \)
So the rate constant becomes nearly 2 times.
NCERT (2026-27 edition):
Temperature Dependence of the Rate of a Reaction
