In a first-order reaction, the concentration of the reactant decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 30 minutes
For a first order reaction the half-life depends only on the rate constant:
\( \displaystyle t_{1/2}=\frac{0.693}{k} \)
So the half-life is the same whatever the starting concentration.
From 0.8 M to 0.4 M the concentration halves in 15 minutes, so:
\( \displaystyle t_{1/2}=15\,\mathrm{min} \)
From 0.1 M to 0.025 M the concentration falls to one quarter, which takes two half-lives:
\( \displaystyle 0.1\,\mathrm{M}\xrightarrow{t_{1/2}}0.05\,\mathrm{M}\xrightarrow{t_{1/2}}0.025\,\mathrm{M} \)
\( \displaystyle t=2\times15=30\,\mathrm{min} \)
The time taken is 30 minutes.
NCERT (2026-27 edition):
3.3.3 Half-Life of a Reaction – “The half-life of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration. It is represented as t1/2. … It can be seen that for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species. The half-life of a first order equation is readily calculated from the rate constant and vice versa.”