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(A) For a first-order reaction,the integrated rate equation is $K = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t}$.Given that the concentration becomes $3/

Vedclass · Accepted Answer

$0.29$

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(A) For a first-order reaction,the integrated rate equation is $K = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t}$.Given that the concentration becomes $3/4$ of its initial value,$[R]_t = \frac{3}{4} [R]_0$.Substituting the values: $t_{1/4} = \frac{2.303}{K} \log \frac{[R]_0}{\frac{3}{4} [R]_0} = \frac{2.303}{K} \log \frac{4}{3}$.$t_{1/4} = \frac{2.303}{K} (\log 4 - \log 3) = \frac{2.303}{K} (0.6020 - 0.4771) = \frac{2.303}{K} 	imes 0.1249 \approx \frac{0.2877}{K} \approx \frac{0.29}{K}$.

For a first-order reaction,the time taken for the concentration of the reactant to reach $3/4$ of its initial value is $t_{1/4}$. If the rate constant for the reaction is $K$,then $t_{1/4}$ can be expressed as: (in $/K$)

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