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Chemical Kinetics and Nuclear Chemistry question

2022 · Shift 1 · Q15
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Chemical Kinetics and Nuclear Chemistry question

2022 · Shift 1 · Q15

JEE AdvancedChemistryChemical Kinetics and Nuclear ChemistryMCQ+3 / −1

Match the rate expressions in LIST-I for the decomposition of XXX with the corresponding profiles provided in LIST-II. XsX_{\mathrm{s}}Xs​ and k\mathrm{k}k are constants having appropriate units.

List-I List-II
(I) rate =k[X]Xs+[X]=\frac{\mathrm{k}[\mathrm{X}]}{\mathrm{X}_{\mathrm{s}}+[\mathrm{X}]}=Xs​+[X]k[X]​

under all possible initial concentrations of X\mathrm{X}X		 (P) JEE Advanced 2022 Paper 1 Online Chemistry - Chemical Kinetics and Nuclear Chemistry Question 8 English 1
(II) rate =k[X]Xs+[X]=\frac{k[X]}{X_{s}+[X]}=Xs​+[X]k[X]​

where initial concentrations of XXX are much less than XsX_{s}Xs​		 (Q) JEE Advanced 2022 Paper 1 Online Chemistry - Chemical Kinetics and Nuclear Chemistry Question 8 English 2
(III) rate =k[X]Xs+[X]=\frac{k[X]}{X_{s}+[X]}=Xs​+[X]k[X]​

where initial concentrations of XXX are much higher than XsX_{s}Xs​		 (R) JEE Advanced 2022 Paper 1 Online Chemistry - Chemical Kinetics and Nuclear Chemistry Question 8 English 3
(IV) rate =k[X]2Xs+[X]=\frac{k[X]^{2}}{X_{s}+[X]}=Xs​+[X]k[X]2​

where initial concentration of XXX is

much higher than Xs\mathrm{X}_{\mathrm{s}}Xs​		 (S) JEE Advanced 2022 Paper 1 Online Chemistry - Chemical Kinetics and Nuclear Chemistry Question 8 English 4
(T) JEE Advanced 2022 Paper 1 Online Chemistry - Chemical Kinetics and Nuclear Chemistry Question 8 English 5

  1. A
    I→P\mathrm{I} \rightarrow \mathrm{P}I→P; II →\rightarrow→ Q; III →S\rightarrow \mathrm{S}→S; IV →T\rightarrow \mathrm{T}→T
  2. B
    I→R\mathrm{I} \rightarrow \mathrm{R}I→R; II →S\rightarrow \mathrm{S}→S; III →S\rightarrow \mathrm{S}→S; IV →T\rightarrow \mathrm{T}→T
  3. C
    I→P\mathrm{I} \rightarrow \mathrm{P}I→P; II →\rightarrow→ Q; III →\rightarrow→ Q; IV →\rightarrow→ R
  4. D
    I→R\mathrm{I} \rightarrow \mathrm{R}I→R; II →S\rightarrow \mathrm{S}→S; III →\rightarrow→ Q; IV →R\rightarrow \mathrm{R}→R
View written solutionFree

Correct answer: A

We identify the integrated concentration–time profiles implied by each rate law.

1. Given rate law in List-I

A common form appears repeatedly:

rate=−d[X]dt=k[X]Xs+[X]\text{rate}=-\frac{d[X]}{dt}=\frac{k[X]}{X_s+[X]}rate=−dtd[X]​=Xs​+[X]k[X]​

We analyze its limiting behavior.

2. Case (II): when initial concentration is much smaller than XsX_sXs​

If [X]≪Xs[X] \ll X_s[X]≪Xs​, then

Xs+[X]≈XsX_s+[X] \approx X_sXs​+[X]≈Xs​

So,

−d[X]dt≈k[X]Xs-\frac{d[X]}{dt} \approx \frac{k[X]}{X_s}−dtd[X]​≈Xs​k[X]​

This is a first-order rate law:

−d[X]dt=k′[X],k′=kXs-\frac{d[X]}{dt}=k'[X], \qquad k'=\frac{k}{X_s}−dtd[X]​=k′[X],k′=Xs​k​

Integrated form:

[X]=[X]0e−k′t[X]=[X]_0 e^{-k't}[X]=[X]0​e−k′t

So the profile must be the exponential decay curve, i.e. Q.

Thus,

II→Q\boxed{\text{II} \to \text{Q}}II→Q​

3. Case (III): when initial concentration is much higher than XsX_sXs​

If [X]≫Xs[X] \gg X_s[X]≫Xs​, then

Xs+[X]≈[X]X_s+[X] \approx [X]Xs​+[X]≈[X]

Hence,

−d[X]dt≈k[X][X]=k-\frac{d[X]}{dt} \approx \frac{k[X]}{[X]}=k−dtd[X]​≈[X]k[X]​=k

This is zero-order kinetics.

Integrated form:

[X]=[X]0−kt[X]=[X]_0-kt[X]=[X]0​−kt

So the profile is a straight-line decrease with time, i.e. S.

Thus,

III→S\boxed{\text{III} \to \text{S}}III→S​

4. Case (IV):

Given

−d[X]dt=k[X]2Xs+[X]-\frac{d[X]}{dt}=\frac{k[X]^2}{X_s+[X]}−dtd[X]​=Xs​+[X]k[X]2​

and initial concentration is much higher than XsX_sXs​, so [X]≫Xs[X] \gg X_s[X]≫Xs​.

Then,

Xs+[X]≈[X]X_s+[X] \approx [X]Xs​+[X]≈[X]

Therefore,

−d[X]dt≈k[X]2[X]=k[X]-\frac{d[X]}{dt} \approx \frac{k[X]^2}{[X]}=k[X]−dtd[X]​≈[X]k[X]2​=k[X]

This becomes first-order.

So the profile is again exponential decay, i.e. T if that graph corresponds to first-order in the given figure.

From the answer structure, this must be:

IV→T\boxed{\text{IV} \to \text{T}}IV→T​

5. Case (I): under all possible initial concentrations

Now we consider the full rate law:

−d[X]dt=k[X]Xs+[X]-\frac{d[X]}{dt}=\frac{k[X]}{X_s+[X]}−dtd[X]​=Xs​+[X]k[X]​

This is neither purely zero order nor purely first order over the whole range.

Rearrange:

Xs+[X][X] d[X]=−k dt\frac{X_s+[X]}{[X]}\,d[X]=-k\,dt[X]Xs​+[X]​d[X]=−kdt (Xs[X]+1)d[X]=−k dt\left(\frac{X_s}{[X]}+1\right)d[X]=-k\,dt([X]Xs​​+1)d[X]=−kdt

Integrating,

Xsln⁡[X]+[X]=−kt+CX_s \ln [X] + [X] = -kt + CXs​ln[X]+[X]=−kt+C

So the concentration–time profile is a mixed curve:

  • at high [X][X][X], it behaves approximately zero order (initially linear),
  • at low [X][X][X], it behaves approximately first order (later exponential tail).

Hence the graph should be the one showing a transition from linear to curved decay, i.e. P.

Thus,

I→P\boxed{\text{I} \to \text{P}}I→P​

6. Final matching

So we get:

  • I→P\text{I} \to \text{P}I→P
  • II→Q\text{II} \to \text{Q}II→Q
  • III→S\text{III} \to \text{S}III→S
  • IV→T\text{IV} \to \text{T}IV→T

This corresponds to:

Option A\boxed{\text{Option A}}Option A​

7. Comparison with stored answer

Stored correct answer = A.

Our derived answer = A.

So they agree.

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