The rate constant for the 1st order decomposition of H_2O_2 is given…

The correct answer is Option (2) → \(2.3\, \ kJ\)

The Arrhenius equation k = A e^−E_a/RT, written with common logarithms, is

    log k = log A − E_a/(2.303 RT)

Comparing it with the given equation log k = 15 − (1.2 × 10²)/T, the term that multiplies 1/T is E_a/2.303R:

    E_a/(2.303 R) = 1.2 × 10² K
    E_a = 2.303 × 8.314 J K⁻¹ mol⁻¹ × 120 K
    = 2297.7 J mol⁻¹ ≈ 2.3 kJ mol⁻¹

(log A = 15 gives the frequency factor, A = 10¹⁵ s⁻¹.)

Shortcut: 2.303 R ≈ 19.15 J K⁻¹ mol⁻¹, and 19.15 × 120 ≈ 2300 J.

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 – “3.27 The rate constant for the first order decomposition of H₂O₂ is given by the following equation:

    \( \log k = 14.34 - 1.25 \times 10^4\,\mathrm{K}/T \)

Calculate E_a for this reaction and at what temperature will its half-period be 256 minutes?”