(A) For a first-order reaction,the rate constant $K$ is given by $K = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 6.93 \, \text{min}$,so $K = \frac{0.693}

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(A) For a first-order reaction,the rate constant $K$ is given by $K = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 6.93 \, \text{min}$,so $K = \frac{0.693}{6.93} = 0.1 \, \text{min}^{-1}$.The time $t$ required for $99 \%$ completion is given by $t = \frac{2.303}{K} \log \frac{[A]_0}{[A]_t}$.Here,$[A]_0 = 100$ and $[A]_t = 100 - 99 = 1$.$t = \frac{2.303}{0.1} \log \frac{100}{1} = 23.03 \times \log(10^2) = 23.03 \times 2 = 46.06 \, \text{min}$.

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

The half-life period of a first-order chemical reaction is $6.93 \, \text{min}$. The time required for the completion of $99 \%$ of the chemical reaction will be .......... $\text{min}$ $(\log \, 2 = 0.301)$.

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A
$46.06$
B
$460.6$
C
$230.3$
D
$23.03$

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The half-life period of a first-order chemical reaction is $6.93 \, \text{min}$. The time required for the completion of $99 \%$ of the chemical reaction will be .......... $\text{min}$ $(\log \, 2 = 0.301)$.

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