JEE Challenger
More from Chemical Kinetics
ChemMedAdv 2022 • Paper 1

Match Rate Expressions of Decomposition with Concentration Profiles

Match the rate expressions in LIST-I for the decomposition of X\text{X} with the corresponding profiles provided in LIST-II. Xs\text{X}_s and kk are constants having appropriate units.

Question Diagram 1

Options

A

\text{I} \rightarrow \text{P}; \text{II} \rightarrow \text{Q}; \text{III} \rightarrow \text{S}; \text{IV} \rightarrow \text{T}

B

\text{I} \rightarrow \text{R}; \text{II} \rightarrow \text{S}; \text{III} \rightarrow \text{S}; \text{IV} \rightarrow \text{T}

Correct
C

\text{I} \rightarrow \text{P}; \text{II} \rightarrow \text{Q}; \text{III} \rightarrow \text{Q}; \text{IV} \rightarrow \text{R}

D

\text{I} \rightarrow \text{R}; \text{II} \rightarrow \text{S}; \text{III} \rightarrow \text{Q}; \text{IV} \rightarrow \text{R}

Topics & Concepts

Chemical KineticsRate of ReactionOrder of Reaction

Step-by-Step Solution

To determine the correct matching between the rate expressions in LIST-I and the concentration profiles in LIST-II, let us analyze each expression step-by-step:

1. Item (I):

rate=k[X]Xs+[X]under all possible initial concentrations of X\text{rate} = \frac{k[\text{X}]}{\text{X}_s + [\text{X}]} \quad \text{under all possible initial concentrations of } \text{X}

  • Low concentration regime ([X]≪Xs[\text{X}] \ll \text{X}_s): rate≈kXs[X]\text{rate} \approx \frac{k}{\text{X}_s}[\text{X}] The rate increases linearly with initial concentration [X]0[\text{X}]_0.

  • High concentration regime ([X]≫Xs[\text{X}] \gg \text{X}_s): rate≈k[X][X]=k\text{rate} \approx \frac{k[\text{X}]}{[\text{X}]} = k The rate becomes constant (independent of initial concentration [X]0[\text{X}]_0) and reaches saturation.

  • Graphical Profile: A plot of Rate vs initial concentration [X]0[\text{X}]_0 starts linearly from the origin and curves to reach a horizontal asymptote (saturation plateau). This matches profile (R).

  ⟹  I→R\implies \text{I} \rightarrow \text{R}

2. Item (II):

rate=k[X]Xs+[X]where [X]≪Xs\text{rate} = \frac{k[\text{X}]}{\text{X}_s + [\text{X}]} \quad \text{where } [\text{X}] \ll \text{X}_s

  • Since [X]≪Xs[\text{X}] \ll \text{X}_s, the rate law simplifies to: rate=−d[X]dt=kXs[X]=k′[X](where k′=kXs)\text{rate} = -\frac{d[\text{X}]}{dt} = \frac{k}{\text{X}_s} [\text{X}] = k' [\text{X}] \quad \left(\text{where } k' = \frac{k}{\text{X}_s}\right)
  • This represents a first-order reaction.
  • For first-order kinetics:
    • Integrated Rate Law: ln⁡[X]=ln⁡[X]0−k′t\ln[\text{X}] = \ln[\text{X}]_0 - k't, which gives a linear plot of ln⁡[X]\ln[\text{X}] vs time (Profile T).
    • Half-life: t1/2=ln⁡2k′t_{1/2} = \frac{\ln 2}{k'}, which is independent of initial concentration [X]0[\text{X}]_0 (Profile Q).

3. Item (III):

rate=k[X]Xs+[X]where [X]≫Xs\text{rate} = \frac{k[\text{X}]}{\text{X}_s + [\text{X}]} \quad \text{where } [\text{X}] \gg \text{X}_s

  • Since [X]≫Xs[\text{X}] \gg \text{X}_s, the rate law simplifies to: rate=−d[X]dt=k[X][X]=k\text{rate} = -\frac{d[\text{X}]}{dt} = \frac{k[\text{X}]}{[\text{X}]} = k
  • This represents a zero-order reaction.
  • For zero-order kinetics:
    • Integrated Rate Law: [X]=[X]0−kt[\text{X}] = [\text{X}]_0 - kt, which means a plot of [X][\text{X}] vs time is a straight line with a negative slope (Profile S).
    • Half-life: t1/2=[X]02k∝[X]0t_{1/2} = \frac{[\text{X}]_0}{2k} \propto [\text{X}]_0, which is a straight line passing through the origin (Profile P).

  ⟹  III→S\implies \text{III} \rightarrow \text{S}

4. Item (IV):

rate=k[X]2Xs+[X]where [X]≫Xs\text{rate} = \frac{k[\text{X}]^2}{\text{X}_s + [\text{X}]} \quad \text{where } [\text{X}] \gg \text{X}_s

  • Since [X]≫Xs[\text{X}] \gg \text{X}_s, the denominator Xs+[X]≈[X]\text{X}_s + [\text{X}] \approx [\text{X}], so: rate=−d[X]dt=k[X]2[X]=k[X]\text{rate} = -\frac{d[\text{X}]}{dt} = \frac{k[\text{X}]^2}{[\text{X}]} = k[\text{X}]
  • This represents a first-order reaction.
  • For first-order kinetics:
    • Integrated Rate Law: ln⁡[X]=ln⁡[X]0−kt\ln[\text{X}] = \ln[\text{X}]_0 - kt, which corresponds to a straight line of ln⁡[X]\ln[\text{X}] vs time with a negative slope (Profile T).

  ⟹  IV→T\implies \text{IV} \rightarrow \text{T}

Conclusion:

Matching the results:

  • I→R\text{I} \rightarrow \text{R}
  • III→S\text{III} \rightarrow \text{S}
  • IV→T\text{IV} \rightarrow \text{T}

Comparing with the given choices, Option (B) corresponds to these correct matches.

Correct Answer: (B)

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