A first order reaction has the rate constant of 90 s-1. How much time…

The correct answer is Option (4) → 3.3 x 10^-2 s

For a first order reaction, the time needed for the reactant concentration to fall from [R]₀ to [R] is given by the integrated rate equation:

    \( \displaystyle t=\frac{2.303}{k}\log\frac{[R]_0}{[R]} \)

This time depends only on the fraction of reactant left, not on the actual starting concentration.

Here the concentration falls to 1/20 of its initial value, so [R] = [R]₀/20 and the ratio [R]₀/[R] = 20.

    \( \displaystyle \log 20=\log 2+\log 10=0.3010+1=1.3010 \)

    \( \displaystyle t=\frac{2.303}{90}\times1.3010 \)

    \( \displaystyle t=\frac{2.996}{90}=0.0333\ \mathrm{s} \)

    \( \displaystyle t\approx3.3\times10^{-2}\ \mathrm{s} \)

Shortcut: Since 2.303 × 1.3010 ≈ 3, the time is t ≈ 3/90 = 1/30 s ≈ 3.3 × 10⁻² s.

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 – “… 3.16 The rate constant for a first order reaction is 60 s⁻¹. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?”