(B) For a first-order reaction, the rate constant $k$ is given by: $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{20} \text{ min}^{-1}$ The time required f

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(B) For a first-order reaction, the rate constant $k$ is given by: $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{20} \text{ min}^{-1}$ The time required for a first-order reaction is: $t = \frac{2.303}{k} \log\left(\frac{[A]_0}{[A]_t}\right)$ Given $[A]_t = \frac{[A]_0}{10}$, so $\frac{[A]_0}{[A]_t} = 10$. $t = \frac{2.303 \times 20}{0.693} \log(10) = \frac{46.06}{0.693} \times 1 \approx 66.46 \text{ min}$. The closest option provided is $66.56 \text{ min}$.

Class 12 Chemistry Chemical Kinetics First Order reaction

Easy

The half-life of a first-order reaction is $20 \text{ min}$. What is the time taken to reduce the initial concentration of the reactant to $\frac{1}{10}$th of its original value (in $\text{ min}$)?

MHT CET 2020 All MHT CET

A
$6.6$
B
$66.56$
C
$150$
D
$79.68$

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Chemical Kinetics

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The half-life of a first-order reaction is $20 \text{ min}$. What is the time taken to reduce the initial concentration of the reactant to $\frac{1}{10}$th of its original value (in $\text{ min}$)?

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