JEE Main 2025 Apr 2 Shift 1, Chemistry Q25: Reactions of Various Orders

JEE Main2025Apr 2, Shift 1Chemistry

For the reaction A $\to$ products.

The concentration of A at 10 minutes is ________ $\times 1 0^{- 3}$ mol $L^{- 1}$ (nearest integer).

The reaction was started with 2.5 mol $L^{- 1}$ of A.

Your answer

Solution

Order determination: Since $t_{1/2}$ is directly proportional to $[A]_{0}$ , the reaction is zero order. For a zero-order reaction:

$t_{1/2} = \frac{[ A ] _{0}}{2 k}$

The slope of the $t_{1/2}$ vs $[A]_{0}$ plot is $\frac{1}{2 k}$ :

$\frac{1}{2 k} = 76.92 \Rightarrow k = \frac{1}{2 \times 76.92} = 6.5 \times 1 0^{- 3} mol L^{- 1} min^{- 1}$

Integrated rate law (zero order):

$[A]_{t} = [A]_{0} - k t$

At $t = 10$ min with $[A]_{0} = 2.5$ mol/L:

$[A]_{10} = 2.5 - \frac{10}{2 \times 76.92} = 2.5 - 0.065 = 2.435 mol L^{- 1}$

$[A]_{10} = 2435 \times 1 0^{- 3} mol L^{- 1}$

$∴$ Required answer = 2435.

Concept behind this question

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