The time required for completion of 93.75 % o | vedclass

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(C) For a first order reaction,the integrated rate equation is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.Given that $93.75 \%$ of the reaction is

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$x / 4$

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(C) For a first order reaction,the integrated rate equation is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.Given that $93.75 \%$ of the reaction is complete,the remaining concentration is $[A]_t = [A]_0 - 0.9375[A]_0 = 0.0625[A]_0$.Substituting this into the rate equation for $t = x$:$k = \frac{2.303}{x} \log \frac{[A]_0}{0.0625[A]_0} = \frac{2.303}{x} \log(16) = \frac{2.303}{x} \log(2^4) = \frac{2.303 	imes 4 	imes \log(2)}{x} = \frac{4 	imes 0.693}{x}$.We know that the half-life $t_{1/2}$ is given by $k = \frac{0.693}{t_{1/2}}$.Equating the two expressions for $k$:$\frac{0.693}{t_{1/2}} = \frac{4 	imes 0.693}{x}$.Therefore,$t_{1/2} = \frac{x}{4}$ minutes.

The time required for completion of $93.75 \%$ of a first order reaction is $x$ minutes. The half life of it (in minutes) is

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