The rate constants of a reaction at 200 K and 300 K are 0.02 s−1 and…

The correct answer is Option (2) → 3.4 kJ/mol

The Arrhenius equation written for two temperatures removes the frequency factor A and links the two rate constants to the activation energy:

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

Here T₁ = 200 K, k₁ = 0.02 s⁻¹; T₂ = 300 K, k₂ = 0.04 s⁻¹; R = 8.314 J K⁻¹ mol⁻¹.

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

    \( \displaystyle \frac{T_2-T_1}{T_1T_2} = \frac{300-200}{200\times300} = \frac{1}{600}\ \mathrm{K^{-1}} \)

    \( \displaystyle E_a = 2.303\,R\times\log 2\times\frac{T_1T_2}{T_2-T_1} \)

    \( \displaystyle E_a = 2.303\times8.314\times0.3010\times600\ \mathrm{J\,mol^{-1}} \)

    \( \displaystyle E_a = 3458\ \mathrm{J\,mol^{-1}} \approx 3.4\ \mathrm{kJ\,mol^{-1}} \)

Shortcut: 2.303 × 0.3010 ≈ 0.693 (= ln 2), so E_a ≈ 0.693 × 8.314 × 600 ≈ 3457 J mol⁻¹.

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 – “… Example 3.9 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.
Solution
\( \displaystyle \log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_1T_2}\right] \)
\( \displaystyle \log\frac{0.07}{0.02} = \left(\frac{E_a}{2.303\times8.314\ \mathrm{J\,K^{-1}mol^{-1}}}\right)\left[\frac{700-500}{700\times500}\right] \)
0.544 = E_a × 5.714 × 10⁻⁴/19.15
E_a = 0.544 × 19.15/5.714 × 10⁻⁴ = 18230.8 J …”