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(C) For a first-order reaction,the half-life $(t_{1/2})$ is the time taken for the concentration to reduce to half of its initial value.Given that the

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$30$

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(C) For a first-order reaction,the half-life $(t_{1/2})$ is the time taken for the concentration to reduce to half of its initial value.Given that the concentration decreases from $0.8 \, M$ to $0.4 \, M$ in $15 \, 	ext{minutes}$,the half-life is $t_{1/2} = 15 \, 	ext{minutes}$.To find the time taken for the concentration to change from $0.1 \, M$ to $0.025 \, M$,we note that $0.025 \, M$ is $\frac{1}{4}$ of $0.1 \, M$ (i.e.,$0.1$ $ightarrow 0.05$ $ightarrow 0.025$).This corresponds to two half-lives.Therefore,the total time required is $2 	imes t_{1/2} = 2 	imes 15 \, 	ext{minutes} = 30 \, 	ext{minutes}$.

$S.No.$	Time $/$ $s$	Total pressure $/$ $atm$
$1.$	$0$	$0.5$
$2.$	$100$	$0.512$

For a first-order reaction,the concentration of the reactant decreases from $0.8 \, M$ to $0.4 \, M$ in $15 \, \text{minutes}$. The time required for the concentration to change from $0.1 \, M$ to $0.025 \, M$ is ....... $\text{min}$.

Similar Questions

The following data were obtained during the first order thermal decomposition of $N_{2}O_{5(g)}$ at constant volume:
$2N_{2}O_{5(g)} \rightarrow 2N_{2}O_{4(g)} + O_{2(g)}$

$S.No.$ Time $/$ $s$ Total pressure $/$ $atm$

$1.$ $0$ $0.5$

$2.$ $100$ $0.512$

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