JEE Advanced · Chemistry · 18. Chemical Kinetics
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 _______
[ \(k\) and \(k^{\prime}\) represent corresponding rate constants]
- A 4.47
- B 4.17
- C 25.4
- D 20.5
Answer & Solution
Correct Answer
(B) 4.17
Step-by-step Solution
Detailed explanation
\( \mathrm{A}+\mathrm{R} \rightarrow \text { Product } \)
\( \mathrm{t}=0 \mathrm{~A}_0 10 \mathrm{~A}_0 \)
\( \mathrm{t}=\mathrm{t} 0.6 \mathrm{~A}_0 9.6 \mathrm{~A}_0 \)
\( \text {Rate }=\mathrm{k}[\mathrm{A}][\mathrm{R}] \)
\( \text {Rate }_1=\mathrm{k}\left(0.6 \mathrm{~A}_0\right) \times 9.6 \mathrm{~A}_0 \)
\( \mathrm{~A} \quad+\mathrm{R} \rightarrow \text { Product } \)
\( \mathrm{t}=0 \mathrm{~A}_0 10 \mathrm{~A}_0(\text { excess }) \)
\( \mathrm{t}=\mathrm{t} 0.6 \mathrm{~A}_0 10 \mathrm{~A}_0 \)
\( \text {Rate }=\mathrm{k}^{\prime}[\mathrm{A}], \mathrm{k}^{\prime}=\mathrm{k}[\mathrm{R}] \)
\( \text {Rate }_2=\left(\mathrm{k} \times 10 \mathrm{~A}_0\right) \times\left(0.6 \mathrm{~A}_0\right) \)
\( 100 \times \frac{\Delta \text { Rate }}{\text { Rate }_1}=\frac{(0.6 \times 10-0.6 \times 9.6}{0.6 \times 9.6} \times 100\) \(=4.1666\)
\( \mathrm{t}=0 \mathrm{~A}_0 10 \mathrm{~A}_0 \)
\( \mathrm{t}=\mathrm{t} 0.6 \mathrm{~A}_0 9.6 \mathrm{~A}_0 \)
\( \text {Rate }=\mathrm{k}[\mathrm{A}][\mathrm{R}] \)
\( \text {Rate }_1=\mathrm{k}\left(0.6 \mathrm{~A}_0\right) \times 9.6 \mathrm{~A}_0 \)
\( \mathrm{~A} \quad+\mathrm{R} \rightarrow \text { Product } \)
\( \mathrm{t}=0 \mathrm{~A}_0 10 \mathrm{~A}_0(\text { excess }) \)
\( \mathrm{t}=\mathrm{t} 0.6 \mathrm{~A}_0 10 \mathrm{~A}_0 \)
\( \text {Rate }=\mathrm{k}^{\prime}[\mathrm{A}], \mathrm{k}^{\prime}=\mathrm{k}[\mathrm{R}] \)
\( \text {Rate }_2=\left(\mathrm{k} \times 10 \mathrm{~A}_0\right) \times\left(0.6 \mathrm{~A}_0\right) \)
\( 100 \times \frac{\Delta \text { Rate }}{\text { Rate }_1}=\frac{(0.6 \times 10-0.6 \times 9.6}{0.6 \times 9.6} \times 100\) \(=4.1666\)
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