The half life of a first order reaction having rate constant, k = 1.7…

The correct answer is Option (2) → 11.3 hrs

For a first order reaction the half-life does not depend on the initial concentration; it depends only on the rate constant:

    \( \displaystyle t_{1/2} = \frac{0.693}{k} \)

Substituting \( k = 1.7 \times 10^{-5}\ \mathrm{s^{-1}} \):

    \( \displaystyle t_{1/2} = \frac{0.693}{1.7 \times 10^{-5}\ \mathrm{s^{-1}}} = 4.08 \times 10^{4}\ \mathrm{s} \)

The options are in hours, so the seconds are divided by 3600:

    \( \displaystyle t_{1/2} = \frac{4.08 \times 10^{4}\ \mathrm{s}}{3600\ \mathrm{s\,h^{-1}}} = 11.3\ \mathrm{h} \)

So the half-life is 11.3 hours.

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 – “Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s^–1 (ii) 2 min^–1 (iii) 4 years^–1”