Half life of zero order reaction A arrow product is 1 hour, when initial concentration of reaction is 2.0 ~mol ~L ~L^ -1 . The time required to decrease concentration of A from 0.50 to 0.25 ~mol ~L^ -1 is:

Correct answer: C. 15 min For zero order reaction aligned & Half life =(A_o)/(2 k) \& 60 =(2)/(2 k) \& k=(1)/(60) M / aligned Now aligned & A_ t =A_ o -kt & t= A_ o -A_ t k & =(0.5-0.25)/(1 / 60) & 0.25 × 60 aligned t=15 min

JEE Main 20254 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Evening Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2025 · 4 April · Evening Shift · Chemistry · Chemical Kinetics

Half life of zero order reaction A arrow product is 1 hour, when...

Half life of zero order reaction

\mathrm{A} \rightarrow

product is 1 hour, when initial concentration of reaction is

2.0 \mathrm{~mol} \mathrm{~L} \mathrm{~L}^{-1}

. The time required to decrease concentration of A from 0.50 to

0.25 \mathrm{~mol} \mathrm{~L}^{-1}

is:

Options

A0.5 hour
B4 hour
C15 min
D60 min

Correct answer

C. 15 min

Step-by-step solution

For zero order reaction

\begin{aligned}& \text{ Half life }=\frac{A_o}{2 k} \& 60 \min =\frac{2}{2 k} \& k=\frac{1}{60} \mathrm{M} / \min\end{aligned}

Now

\begin{aligned} & \mathrm{A}_{\mathrm{t}}=\mathrm{A}_{\mathrm{o}}-\mathrm{kt} \\ & \mathrm{t}=\frac{\mathrm{A}_{\mathrm{o}}-\mathrm{A}_{\mathrm{t}}}{\mathrm{k}} \\ & =\frac{0.5-0.25}{1 / 60} \\ & 0.25 \times 60\end{aligned}

\mathrm{t}=15 \mathrm{ min}$

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