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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Advanced 2025 · Paper 2 · Chemistry · Chemical Kinetics

Consider a reaction A+R arrow Product. The rate of this reaction is...

Consider a reaction A+R→A+R \rightarrow Product. The rate of this reaction is measured to be k[A][R]k[A][R]. At the start of the reaction, the concentration of R,[R]0R,[R]_0, is 10 -times the concentration of A,[A]0A,[A]_0. The reaction can be considered to be a pseudo first order reaction with assumption that k[R]=k′k[R]=k^{\prime} is constant. Due to this assumption, the relative error (in \%) in the rate when this reaction is 40%40 \% complete, is _______ [ kk and k′k^{\prime} represent corresponding rate constants]

Correct answer

0

Step-by-step solution

A+R→ Product t=0A010A0t=t0.6A09.6A0 Rate =k[A][R] Rate 1=k(0.6A0)×9.6A0A+R→ Product t=0A010A0( excess )t=t0.6A010A0 Rate =k′[A],k′=k[R] Rate 2=(k×10A0)×(0.6A0)100×Δ Rate  Rate 1=(0.6×10−0.6×9.60.6×9.6×100=4.1666\begin{aligned} & \qquad \quad \mathrm{A}+\mathrm{R} \rightarrow \text{ 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 \\ & \text{ Rate }=\mathrm{k}[\mathrm{A}][\mathrm{R}] \\ & \text{ Rate }_1=\mathrm{k}\left(0.6 \mathrm{ A}_0\right) \times 9.6 \mathrm{ A}_0 \\ & \qquad \quad \mathrm{ A} \quad+\quad \mathrm{R} \rightarrow \text{ Product } \\ & \mathrm{t}=0 \quad \mathrm{ A}_0 \qquad 10 \mathrm{ A}_0(\text{ excess }) \\ & \mathrm{t}=\mathrm{t} \quad 0.6 \mathrm{ A}_0 \quad 10 \mathrm{ A}_0 \\ & \text{ Rate }=\mathrm{k}^{\prime}[\mathrm{A}], \mathrm{k}^{\prime}=\mathrm{k}[\mathrm{R}] \\ & \text{ 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\end{aligned}
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