Rate constant of a first order reaction follows following equation \(\log k = 8.0-\frac{5000}{T}\). The \(E_a\) for the reaction is: (Given \(R=8.3 JK^{-1}mol^{-1}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(95.57\, kJ\, mol^{-1}\)
The Arrhenius equation, k = A e−Ea/RT, in base-10 logarithmic form is:
\( \displaystyle \log k = \log A - \frac{E_a}{2.303\,R\,T} \)
Comparing it term by term with the given equation, log k = 8.0 − 5000/T:
\( \displaystyle \frac{E_a}{2.303\,R} = 5000\,\mathrm{K} \)
\( E_a = 5000 \times 2.303 \times 8.3 = 95\,574\,\mathrm{J\,mol^{-1}} = 95.57\,\mathrm{kJ\,mol^{-1}} \)
(The constant term gives log A = 8.0, so A = 108 s−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 – “3.27 The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 – 1.25 × 104 K/T. Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?”