The time required for a first-order reaction | vedclass

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(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $90\%$ completion,$[A]_t = [A]_0 -

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$3.3$ times of $t_{1/2}$

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(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $90\%$ completion,$[A]_t = [A]_0 - 0.90[A]_0 = 0.10[A]_0$.Substituting the values: $k = \frac{2.303}{t_{90\%}} \log \frac{[A]_0}{0.10[A]_0} = \frac{2.303}{t_{90\%}} \log(10) = \frac{2.303}{t_{90\%}}$.Thus,$t_{90\%} = \frac{2.303}{k}$.We know that the half-life $t_{1/2} = \frac{0.693}{k}$,so $k = \frac{0.693}{t_{1/2}}$.Substituting $k$ in the expression for $t_{90\%}$: $t_{90\%} = \frac{2.303}{0.693} 	imes t_{1/2} \approx 3.32 	imes t_{1/2}$.Therefore,the time required is approximately $3.3$ times of $t_{1/2}$.

The time required for a first-order reaction to be $90\%$ complete is approximately .........

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