The rate of a reaction quadruples when temperature changes from 27^oC…

The correct answer is Option (1) → \(38.04\, \ kJ/mol\)

Since the rate quadruples, the rate constant also becomes four times: k₂/k₁ = 4. Using the Arrhenius equation for two temperatures:

    \( \log \displaystyle\frac{k_2}{k_1} = \displaystyle\frac{E_a}{2.303R}\left(\displaystyle\frac{T_2 - T_1}{T_1T_2}\right) \)

T₁ = 27 + 273 = 300 K, T₂ = 57 + 273 = 330 K, so T₂ − T₁ = 30 K.
0.602 = \( \displaystyle\frac{E_a}{2.303 \times 8.314} \times \displaystyle\frac{30}{300 \times 330} \)
E_a = 0.602 × 2.303 × 8.314 × 3300
E_a = 38039 J mol⁻¹ ≈ 38.04 kJ mol⁻¹

Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
Exercises – “3.30 The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.”