Q14 / 120 · VITEEE 2017 · Chemistry
The half life period of a first order chemical reaction is $6.93$ minutes. The time required for the completion of $99 \%$ of the chemical reaction will be $(\log 2=0.301)$
- A$23.03$ minutes
- Correct answerB$46.06$ minutes
- C$460.6$ minutes
- D$230.03$ minutes
Solution
For first order reaction
$$
\begin{array}{l}
k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{6.93} \\
k=\frac{2.303}{t} \log \frac{100}{100-99} \\
\frac{0.693}{6.93}=\frac{2.303}{t} \log \frac{100}{1} \\
\frac{0.693}{6.93}=\frac{2.303 \times 2}{t} \\
t=46.06 \mathrm{~min}
\end{array}
$$
Hence, the correct answer is option: 'B'.