The rate constants of the reaction at 100 K and 200 K are 0.01 s^-1 a…

The correct answer is Option (4) → 1152.65 J

The Arrhenius equation written for two temperatures links the rate constants with the activation energy:

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

Here \( k_1 = 0.01\,\mathrm{s^{-1}} \) at \( T_1 = 100\,\mathrm{K} \), \( k_2 = 0.02\,\mathrm{s^{-1}} \) at \( T_2 = 200\,\mathrm{K} \), and \( R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}} \).

    \( \displaystyle \log 2 = \frac{E_a}{2.303 \times 8.314}\left(\frac{200 - 100}{100 \times 200}\right) \)

    \( \displaystyle 0.3010 = \frac{E_a}{19.147} \times \frac{1}{200} \)

    \( \displaystyle E_a = 0.3010 \times 19.147 \times 200 \approx 1152.65\ \mathrm{J\,mol^{-1}} \)

So the activation energy is 1152.65 J (per mole).

Shortcut: The rate constant rises with temperature, so the activation energy must be positive; this rules out Options (1) and (3) at once.

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.”