For a first order reaction A rightarrow P,t_{ | vedclass

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(C) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 10 	ext{ days}$,so $k = \frac{0.693}{10

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(C) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 10 	ext{ days}$,so $k = \frac{0.693}{10} = 0.0693 	ext{ day}^{-1}$.For $1/4^{th}$ conversion,the amount of reactant remaining $[A]_t$ is $1 - 1/4 = 3/4$ of the initial concentration $[A]_0$.Using the first order integrated rate equation: $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.$t = \frac{2.303}{0.0693} \log \frac{1}{3/4} = \frac{2.303}{0.0693} \log(1.333)$.$t = 33.23 	imes 0.1249 \approx 4.15 	ext{ days}$.Thus,the time required is approximately $4.1 	ext{ days}$.

For a first order reaction $A \rightarrow P$,$t_{1/2}$ (half-life) is $10 \text{ days}$. The time required for $1/4^{th}$ conversion of $A$ (in days) is:

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