The rate constant of a reaction increases from 0.02 s^-1 to 0.04 s^-1…

The correct answer is Option (2) → \(17.289\, kJ\, mol^{-1}\)

The Arrhenius equation relates the rate constants at two temperatures to the activation energy:

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

Data: \( k_1 = 0.02\ \mathrm{s^{-1}} \) at \( T_1 = 500\ \mathrm{K} \); \( k_2 = 0.04\ \mathrm{s^{-1}} \) at \( T_2 = 600\ \mathrm{K} \); \( R = 8.314\ \mathrm{J\,K^{-1}\,mol^{-1}} \).

    \( \displaystyle \log\frac{0.04}{0.02} = \log 2 = 0.3010 \)

    \( \displaystyle \frac{T_2 - T_1}{T_1\,T_2} = \frac{100}{500 \times 600} = \frac{1}{3000}\ \mathrm{K^{-1}} \)

    \( \displaystyle E_a = 2.303 \times 8.314 \times 0.3010 \times 3000 \)

    \( \displaystyle E_a \approx 17290\ \mathrm{J\,mol^{-1}} = 17.29\ \mathrm{kJ\,mol^{-1}} \)

So the activation energy is 17.289 kJ mol⁻¹.

Shortcut: \( 2.303 \times 0.3010 = 0.693 \), so \( E_a = 0.693 \times 8.314 \times 3000 \approx 5.76 \times 3000 \approx 17.3 \) kJ mol⁻¹. Options (1) and (3) are in cal and far too small, so only the kJ values need checking.

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

NCERT (2026-27 edition):
Temperature Dependence of the Rate of a Reaction – “The rate constants of a reaction at 500K and 700K are 0.02s^–1 and 0.07s^–1 respectively. Calculate the values of E_a and A.”