The rate of a first order reaction is \( 0.06\text{ mol L}^{-1}\text{ s}^{-1} \) after 20 seconds and \( 0.03\text{ mol L}^{-1}\text{ s}^{-1} \) after 30 seconds from the start of the reaction. The half life period of this reaction is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 10\ \text{s} \)
For a first order reaction the rate is proportional to the concentration of the reactant, rate \( = k[\mathrm{A}] \). So the ratio of two rates is the ratio of the concentrations at those times:
\( \displaystyle \frac{r_{20}}{r_{30}} = \frac{k[\mathrm{A}]_{20}}{k[\mathrm{A}]_{30}} = \frac{0.06}{0.03} = 2 \)
The concentration has fallen to half between 20 s and 30 s, that is, in 10 s. The time in which the concentration falls to half is the half-life, and for a first order reaction it does not depend on the starting concentration, so any halving takes the same time.
The same result comes from the integrated rate equation, taking the 20 s value as the start and t = 10 s:
\( \displaystyle k = \frac{2.303}{t}\log\frac{[\mathrm{A}]_{20}}{[\mathrm{A}]_{30}} = \frac{2.303}{10\ \mathrm{s}}\log 2 = \frac{2.303 \times 0.3010}{10\ \mathrm{s}} = 0.0693\ \mathrm{s^{-1}} \)
\( \displaystyle t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0693\ \mathrm{s^{-1}}} = 10\ \mathrm{s} \)
So the half-life is 10 s.