JEE Advanced 2025 Paper 2, Chemistry Section 3 Q4: Reactions of Various Orders

JEE Advanced2025Paper 2Chemistry Section 3

Consider a reaction $A + R \to$ Product. The rate of this reaction is measured to be $k [A] [R]$ . At the start of the reaction, the concentration of $R$ , $[R]_{0}$ , is 10-times the concentration of $A$ , $[A]_{0}$ . The reaction can be considered to be a pseudo first order reaction with assumption that $k [R] = k^{'}$ is constant. Due to this assumption, the relative error (in %) in the rate when this reaction is 40% complete, is ______.

[ $k$ and $k^{'}$ represent corresponding rate constants]

Your answer

Solution

Start with $[A]_{0}$ and $[R]_{0} = 10 [A]_{0}$ . At 40% conversion of $A$ , $[A] = 0.6 [A]_{0}$ and (the true) $[R] = 10 [A]_{0} - 0.4 [A]_{0} = 9.6 [A]_{0}$ .

True rate (full second-order law): $Rate_{true} = k [A] [R] = k (0.6 [A]_{0}) (9.6 [A]_{0})$ .

Pseudo first-order rate uses $k^{'} = k [R]_{0} = k (10 [A]_{0})$ held constant, so $Rate_{pseudo} = k^{'} [A] = k (10 [A]_{0}) (0.6 [A]_{0})$ .

Relative error:

$\frac{Rate _{pseudo} - Rate _{true}}{Rate _{true}} = \frac{0.6 \times 10 - 0.6 \times 9.6}{0.6 \times 9.6} = \frac{10 - 9.6}{9.6} = \frac{0.4}{9.6}$

Expressed as a percentage: $\frac{0.4}{9.6} \times 100 = 4.1666... \approx 4.17%$ .

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JEE Advanced 2025 Paper 2, Chemistry Section 3 Q4: Reactions of Various Orders

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