Q39 / 160 · TS EAMCET 2014 · Chemistry
In a first order reaction, the concentration of the reactant decrease from $0.6 \mathrm{M}$ to $0.3 \mathrm{M}$ in $15 \mathrm{~min}$. The time taken for the concentration to change from $0.1 \mathrm{M}$ to $0.025 \mathrm{M}$ in minutes is
- A$1.2$
- B$12$
- Correct answerC$30$
- D$3$
Solution
Since, concentration of the reactant decreases from $0.6 \mathrm{M}$ to $0.3 \mathrm{M}$ (i.e., halved) in 15 minutes, therefore half time for this reaction be $15 \mathrm{~min}$.
$$
\text { i.e. }, t_{1 / 2}=15 \mathrm{~min}
$$
Now for a first order reaction
$$
k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{15} \mathrm{~min}^{-1}
$$
Again, $\quad k=\frac{2.303}{t} \log \frac{[A]_0}{[A]}$
$$
\begin{aligned}
\therefore \quad \frac{0.693}{15} & =\frac{2.303}{t} \log \frac{0.1}{0.025} \\
t & =\frac{2.303 \times 15}{0.693} \times \log 4 \\
& =\frac{2.303 \times 15}{0.693} \times 0.6020=30 \mathrm{~min}
\end{aligned}
$$
Alternate method
$\because$ Reaction is of first order with $t_{1 / 2}=15 \mathrm{~min}$
$$
\begin{aligned}
\therefore 0.1 \mathrm{M} \stackrel{15 \mathrm{~min}}{\longrightarrow} \frac{0.1}{2} & =0.05 \mathrm{M} \stackrel{15 \mathrm{~min}}{\longrightarrow} \frac{0.05}{2} \\
& =0.025 \mathrm{M}
\end{aligned}
$$
$\therefore$ Total time $=15+15=30 \mathrm{~min}$
Hence, the correct answer is option: 'C'.