A reaction has a half-life of 1 , text{min}. | vedclass

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(D) For a first-order reaction,the time required for $n \%$ completion is given by $t = \frac{1}{k} \ln \frac{[A]_0}{[A]_t}$.Given $t_{1/2} = 1 \, 	e

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$10$

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(D) For a first-order reaction,the time required for $n \%$ completion is given by $t = \frac{1}{k} \ln \frac{[A]_0}{[A]_t}$.Given $t_{1/2} = 1 \, 	ext{min}$,we know $k = \frac{\ln 2}{t_{1/2}} = \frac{0.69}{1} = 0.69 \, 	ext{min}^{-1}$.For $99.9 \, \%$ completion,$[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$.$t_{99.9} = \frac{1}{k} \ln \frac{[A]_0}{0.001[A]_0} = \frac{1}{k} \ln 1000 = \frac{1}{k} \ln 10^3 = \frac{3 \ln 10}{k}$.Substituting the values: $t_{99.9} = \frac{3 	imes 2.3}{0.69} = \frac{6.9}{0.69} = 10 \, 	ext{min}$.

$A$ reaction has a half-life of $1 \, \text{min}$. The time required for $99.9 \, \%$ completion of the reaction is ......... $\text{min}$. (Round off to the nearest integer)
[Use: $\ln 2 = 0.69, \ln 10 = 2.3$]

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$A$ reaction has a half-life of $1 \, \text{min}$. The time required for $99.9 \, \%$ completion of the reaction is ......... $\text{min}$. (Round off to the nearest integer)[Use: $\ln 2 = 0.69, \ln 10 = 2.3$]