A substance ‘A’ decomposes in solution following first-order kinetics. Flask 1 contains 1 L of 1 M solution of ‘A’ and flask 2 contains 100 mL of 0.6 M solution of ‘A’. After 8.0 h, the concentration of ‘A’ in flask 1 becomes 0.25 M. In what time will the concentration of ‘A’ in flask 2 become 0.3 M?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4.0 h
For a first order reaction the half-life, t1/2 = 0.693/k, does not depend on the initial concentration (or on the volume). Both flasks hold the same substance decomposing under the same conditions, so they have the same k and the same t1/2.
Flask 1: the concentration falls from 1 M to 0.25 M in 8.0 h. After n half-lives:
\( \displaystyle [\mathrm{A}]=\frac{[\mathrm{A}]_0}{2^n} \qquad 0.25=\frac{1}{2^n} \qquad n=2 \)
\( \displaystyle t_{1/2}=\frac{8.0\,\mathrm{h}}{2}=4.0\,\mathrm{h} \)
Flask 2: 0.6 M → 0.3 M is a fall to half, which is one half-life:
\( \displaystyle t=t_{1/2}=4.0\,\mathrm{h} \)
NCERT (2026-27 edition):
3.3.3 Half-Life of a Reaction – “…
\( \displaystyle t_{1/2}=\frac{0.693}{k} \qquad (3.17) \)
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.”