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

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

Vedclass · Accepted Answer

$46.06$

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(B) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 6.93 \, 	ext{min}$,so $k = \frac{0.693}{6.93} = 0.1 \, 	ext{min}^{-1}$.The integrated rate equation for a first-order reaction is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $99 \%$ completion,$[A]_0 = 100$ and $[A]_t = 100 - 99 = 1$.Substituting the values: $0.1 = \frac{2.303}{t} \log \frac{100}{1}$.$0.1 = \frac{2.303 	imes 2}{t}$.$t = \frac{4.606}{0.1} = 46.06 \, 	ext{min}$.

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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