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(A) For a first order reaction,the integrated rate equation is $t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$.Given that $99 \%$ of the reaction is comp

Vedclass · Accepted Answer

$t = \frac{4.606}{k}$

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(A) For a first order reaction,the integrated rate equation is $t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$.Given that $99 \%$ of the reaction is complete,the remaining concentration $[R]$ is $100 \% - 99 \% = 1 \%$ of the initial concentration $[R]_0$.So,$[R] = 0.01 [R]_0 = \frac{[R]_0}{100}$.Substituting these values into the equation:$t = \frac{2.303}{k} \log \frac{[R]_0}{[R]_0 / 100} = \frac{2.303}{k} \log(100)$.Since $\log(100) = 2$,we get $t = \frac{2.303 	imes 2}{k} = \frac{4.606}{k}$.

If the rate constant for a first order reaction is $k$,the time $(t)$ required for the completion of $99 \%$ of the reaction is given by

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