The half-life period of a first-order chemica | vedclass

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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 $t_{1/2} = 6.93 \ min$,so $k = \frac{0.693}{6.93} =

Vedclass · Accepted Answer

$46.06$

Vedclass · Answer

(A) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 6.93 \ min$,so $k = \frac{0.693}{6.93} = 0.1 \ min^{-1}$.The integrated rate equation is $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.For $99\%$ completion,$[A]_0 = 100$ and $[A]_t = 100 - 99 = 1$.Substituting the values: $t = \frac{2.303}{0.1} \log \frac{100}{1}$.$t = 23.03 	imes \log(10^2) = 23.03 	imes 2 = 46.06 \ min$.

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

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