For the reaction ( A arrow ) 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.

Correct answer: 0 aligned & t_ 1/2 [A]_0 Order = zero \& t_ 1/2 = A_0 2 K Slope = 1 2 K =76.92 \& K=(1)/(2 × 76.92) \& [A]_ 10 =-Kt+A_0=-(1)/(2 × 76.92) × 10+2.5=2.435 \& =2435 × 10^ -3 mol / L aligned )

JEE Main 20252 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Morning Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2025 · 2 April · Morning Shift · Chemistry · Chemical Kinetics

For the reaction ( A arrow ) products. The concentration of A at 10...

For the reaction (

\mathrm{A} \rightarrow

) products.
For the reaction ( A arrow ) products. The concentration of A at 10 minutes is _______ ( × 10^ -3 ~mol ~L^ -1 )

The concentration of A at 10 minutes is _______ (

\times

10^{-3}

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

) (nearest integer).
The reaction was started with (2.5

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

) of A.

Correct answer

Step-by-step solution

\begin{aligned}& \mathrm{t}_{1/2} \propto[\mathrm{A}]_0 \Rightarrow \text{ Order }=\text{ zero } \& \mathrm{t}_{1/2}=\frac{\mathrm{A}_0}{2 \mathrm{ K}} \Rightarrow \text{ Slope }=\frac{1}{2 \mathrm{ K}}=76.92 \& \mathrm{ K}=\frac{1}{2 \times 76.92} \& {[\mathrm{A}]_{10}=-\mathrm{Kt}+\mathrm{A}_0=-\frac{1}{2 \times 76.92} \times 10+2.5=2.435} \& =2435 \times 10^{-3} \mathrm{ mol} / \mathrm{L}\end{aligned}

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