Half life of a first order reaction is 3 min | vedclass

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Question

(C) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 3 \ minute$,so $k = \frac{0.693}{3} = 0.

Vedclass · Accepted Answer

$9.97 \ minute$

Vedclass · Answer

(C) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 3 \ minute$,so $k = \frac{0.693}{3} = 0.231 \ minute^{-1}$.The time $t$ required for a first order reaction is given by $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.If the concentration is reduced by $90 \%$,then $[A]_t = [A]_0 - 0.90[A]_0 = 0.10[A]_0$.Substituting the values: $t = \frac{2.303}{0.231} \log \frac{[A]_0}{0.10[A]_0} = \frac{2.303}{0.231} \log(10)$.Since $\log(10) = 1$,$t = \frac{2.303}{0.231} \approx 9.97 \ minute$.

Half life of a first order reaction is $3 \ minute$. What is the time required to reduce the concentration of reactant by $90 \%$ of its initial concentration?

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