Q38 / 200 · Manipal (MET) 2010 · Chemistry
For a first order reaction, the concentration changes from 0.8 to 0.4 in 15 min . The time taken for the concentration to change from 0.1 M to 0.025 M is
- Correct answerA30 min
- B15 min
- C7.5 min
- D60 min
Solution
$\begin{aligned} & T_{50}=15 \mathrm{~min} \\ & \qquad k=\frac{2.303 \log 2}{T_{50}}=\frac{2.303 \log 2}{15}\end{aligned}$
$\begin{aligned}
a & =0.1 \mathrm{M} \\
(a-x) & =0.025 \mathrm{M}
\end{aligned}$
For first order reaction,
$k=\frac{2.303}{t} \log \left(\frac{a}{a-x}\right)$
$\begin{aligned} \frac{2.303 \log 2}{15} & =\frac{2 \times 2.303}{t} \log \frac{0.1}{0.025} \\ & =\frac{2.303}{t} \log 4\end{aligned}$
$\therefore \quad \frac{2.303 \log 2}{15}=\frac{2 \times 2.303 \log 2}{t}$
$\therefore \quad t=30 \mathrm{~min}$
Hence, the correct answer is option: 'A'.