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

Chemistry · Chemical Kinetics

JEE Main 2025 — 4 April, Evening Shift — Question 7

Half life of zero order reaction $A \rightarrow$ product is 1 hour, when initial concentration of reactant is $2.0 \mathrm{~mol} \mathrm{~L}^{-1}$ . The time required to decrease concentration of $A$ from 0.50 to $0.25 \mathrm{~mol} \mathrm{~L}^{-1}$ is:

Option A:
4 hour
Option B:
0.5 hour
Option C:
60 min
Option D:
15 min
Correct

Answer: D

Step-by-step solution

For zero order reaction:

$C_{t}=C_{0}-k t$ and $t_{1 / 2}=\frac{C_{0}}{2 k}$

So, $k=\frac{2}{2 \times 1}=1 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1}$

$\mathrm{C}_{\mathrm{t}}=\mathrm{C}_{0}-\mathrm{kt}$

$0.25=0.5-1 \mathrm{t}$

$\mathrm{t}=0.25 \mathrm{~h}=15 \mathrm{~min}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 4 April, Evening Shift
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Integrated Rate Laws