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$t \large\frac{1}{4}$ can be taken as the time taken for concentration of reactant to drop to $\large\frac{3}{4}$ of its initial value. If the rate constant for a first order reaction is $K,$ then $t \large\frac{1}{4}$ can be written as

$(A)\;0.10/K \\ (B)\;0.29/K \\(C)\;0.69/K \\(D)\;0.75/K $
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$K=\large\frac{2.303}{t}$$ \log \large\frac{a}{a-x}$
$\quad= \large\frac{2.303}{t 1/4}$$ \log \large\frac{a}{\Large\frac{3a}{4}}$
$K=\large\frac{2.303}{t1/4}$$ \log \large\frac{4}{3}$
$K= \large\frac{2.303 \times 0.125}{t 1/4}$
$\qquad= \large\frac{0.29}{t1/4}$
Hence b is the correct answer.
answered Apr 3, 2014 by meena.p

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