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(C) Since the concentration of the reactant decreases from $0.6 \ M$ to $0.3 \ M$ (i.e.,halved) in $15 \ min$,the half-life $(t_{1/2})$ of the reactio

Vedclass · Accepted Answer

$30$

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(C) Since the concentration of the reactant decreases from $0.6 \ M$ to $0.3 \ M$ (i.e.,halved) in $15 \ min$,the half-life $(t_{1/2})$ of the reaction is $15 \ min$. For a first-order reaction,the time required for the concentration to change from $[A]_0$ to $[A]$ is given by $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}$. Given $[A]_0 = 0.1 \ M$ and $[A] = 0.025 \ M$,the ratio $\frac{[A]_0}{[A]} = \frac{0.1}{0.025} = 4 = 2^2$. This implies that the concentration reduces to $1/4$ of its initial value,which corresponds to two half-lives $(2 	imes t_{1/2})$. Therefore,$t = 2 	imes 15 \ min = 30 \ min$.

In a first order reaction,the concentration of the reactant decreases from $0.6 \ M$ to $0.3 \ M$ in $15 \ min$. The time taken for the concentration to change from $0.1 \ M$ to $0.025 \ M$ in minutes is

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