Chemistry · Chemical Kinetics

JEE Advanced 2025 — Paper 2 — Question 41

Consider a reaction $A+R \rightarrow$ 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 $k[R]=k^{\prime}$ is constant. Due to this assumption, the relative error (in %) in the rate when this reaction is $40 \%$ complete, is $\qquad$ .[0pt] [ $k$ and $k^{\prime}$ represent corresponding rate constants]

Answer: 4.16

Numerical answer — enter this value.

Step-by-step solution

$\quad \mathrm{A} \quad+\mathrm{R} \rightarrow$ Product

$\mathrm{t}=0 \quad \mathrm{~A}_{0} \quad 10 \mathrm{~A}_{0}$

$\mathrm{t}=\mathrm{t} \quad 0.6 \mathrm{~A}_{0} \quad 9.6 \mathrm{~A}_{0}$

Rate $=\mathrm{k}[\mathrm{A}][\mathrm{R}]$

Rate $_{1}=\mathrm{k}\left(0.6 \mathrm{~A}_{0}\right) \times 9.6 \mathrm{~A}_{0}$

$\mathrm{A}+\mathrm{R} \rightarrow$ Product

$\mathrm{t}=0 \quad \mathrm{~A}_{0} \quad 10 \mathrm{~A}_{0}$ (excess)

$\mathrm{t}=\mathrm{t} \quad 0.6 \mathrm{~A}_{0} \quad 10 \mathrm{~A}_{0}$

Rate $=\mathrm{k}^{\prime}[\mathrm{A}], \mathrm{k}^{\prime}=\mathrm{k}[\mathrm{R}]$

Rate $_{2}=\left(\mathrm{k} \times 10 \mathrm{~A}_{0}\right) \times\left(0.6 \mathrm{~A}_{0}\right)$

$100 \times \frac{\Delta \text { Rate }}{\text { Rate }_{1}}=\frac{(0.6 \times 10-0.6 \times 9.6}{0.6 \times 9.6} \times 100=4.1666$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2025
Paper: Paper 2
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Rate Laws and Rate Constant