For a first-order reaction with rate constant k, what fraction of the initial concentration of the reactant is left after a time equal to 1/k?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1/e
In a first-order reaction the concentration of the reactant falls exponentially with time:
\( \displaystyle [\mathrm{R}] = [\mathrm{R}]_0\,e^{-kt} \)
\( \displaystyle \text{At } t = \frac{1}{k}: \quad \frac{[\mathrm{R}]}{[\mathrm{R}]_0} = e^{-k \times \frac{1}{k}} = e^{-1} = \frac{1}{e} \)
This is about 0.37, so about 37% of the reactant is left.
NCERT (2026-27 edition):
3.3.2 First Order Reactions – “Taking antilog of both sides”
\( \displaystyle [\mathrm{R}] = [\mathrm{R}]_0\,e^{-kt} \qquad (3.14) \)