Half life of first order reaction is 20 minutes. What is the time taken to reduce the initial concentration…
Half life of first order reaction is 20 minutes. What is the time taken to reduce the initial concentration of the reactant to $\frac{1}{10}$ th ?
- $6.6 \mathrm{~min}$
- $66.56 \mathrm{~min}$
- $150 \mathrm{~min}$
- $79 \cdot 68 \mathrm{~min}$
Solution
For first order reaction,
$k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{20}=0.0346 \mathrm{~min}$
Here, if $[A]_{0}=1,[A]_{1}-\frac{1}{10}, t_{1}/10=?$
$\begin{array}{l}
k=\frac{2.303}{t} \log \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{1}} \\
\therefore t_{1 / 10}=\frac{2.303}{0.0346} \log 10=66.56 \mathrm{~min}
\end{array}$
Asked in: MHT CET 2020 (12 Oct Shift 2)
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