For the reaction A rightarrow products. The concentration of A at 10 minutes is \ \ \ \ \ × 10 -3 mol\ L -1 (nearest integer). The reaction was started with 2.5 mol L -1 of A. From the plot: slope = 76.92 (appropriate units).

Chemistry · Chemical Kinetics

JEE Main 2025 — 2 April, Morning Shift — Question 19

For the reaction $A \rightarrow$ products. The concentration of $A$ at 10 minutes is $\_\_\_\_\_$ $\times 10^{-3}$ $\mathrm{mol\ L^{-1}}$ (nearest integer). The reaction was started with $2.5 \mathrm{~mol} \mathrm{~L}^{-1}$ of A.

From the plot: slope $= 76.92$ (appropriate units).

Answer: 2435

Numerical answer — enter this value.

Step-by-step solution

For zero-order: $t_{1/2} = \dfrac{[A]_0}{2k}$

\dfrac{t_{1/2}}{[A]_0} = \dfrac{1}{2k} = 76.92

k = \dfrac{1}{2 \times 76.92} = 0.00650\ \mathrm{mol\ L^{-1}\ min^{-1}}

[A] = [A]_0 - kt

[A] = 2.5 - (0.00650)(10) = 2.435\ \mathrm{mol\ L^{-1}}

2.435 = 2435 \times 10^{-3}

Thus, the answer is $2435$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Integrated Rate Laws