t_{1/4} can be taken as the time taken for th | vedclass

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(B) For a first order reaction,the rate equation is $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.Given that the concentration drops to $3/4$ of its

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$0.29$

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(B) For a first order reaction,the rate equation is $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.Given that the concentration drops to $3/4$ of its initial value,$[A]_t = \frac{3}{4} [A]_0$.Therefore,$t_{1/4} = \frac{2.303}{K} \log \frac{[A]_0}{\frac{3}{4} [A]_0} = \frac{2.303}{K} \log \frac{4}{3}$.$t_{1/4} = \frac{2.303}{K} (\log 4 - \log 3) = \frac{2.303}{K} (0.602 - 0.477) = \frac{2.303}{K} 	imes 0.125$.$t_{1/4} \approx \frac{0.2878}{K} \approx \frac{0.29}{K}$.

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

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