Match the rate expressions in LIST-I for the decomposition of X with the corresponding profiles provided in LIST-II. X s and k are constants having appropriate units. List I List II --- --- I. rate = frac k[X] X s + [X] under all possible initial concentrations of X . (P) II. rate = frac k[X] X s + [X] where initial concentrations of X are much less than X s . (Q) III. rate =frac k [ X ] X s +[ X ] where initial concentrations of X are much higher than X s (R) IV. rate =frac k [ X ] 2 X s +[ X ] where initial concentration of X is much higher than X s (S) (T)

I rightarrow P ; II rightarrow Q ; III rightarrow S ; IV rightarrow T

I rightarrow R ; II rightarrow S ; III rightarrow S ; IV rightarrow T

I rightarrow P ; II rightarrow Q ; III rightarrow Q ; IV rightarrow R

I rightarrow R ; II rightarrow S ; III rightarrow Q ; IV rightarrow R

Chemistry · Chemical Kinetics

JEE Advanced 2022 — Paper 1 — Question 45

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

List I	List II
I. $\text{rate} = \frac{k[X]}{X_{s} + [X]}$ under all possible initial concentrations of $X$ .	(P)
II. $\text{rate} = \frac{k[X]}{X_{s} + [X]}$ where initial concentrations of $X$ are much less than $X_{s}$ .	(Q)
III. $\text { rate }=\frac{\mathrm{k}[\mathrm{X}]}{\mathrm{X}_{\mathrm{s}}+[\mathrm{X}]}$ where initial concentrations of $X$ are much higher than $\mathrm{X}_{\mathrm{s}}$	(R)
IV. $\text { rate }=\frac{\mathrm{k}[\mathrm{X}]^{2}}{\mathrm{X}_{\mathrm{s}}+[\mathrm{X}]}$ where initial concentration of $X$ is much higher than $X_{s}$	(S)
	(T)

Option A:
I $\rightarrow \mathrm{P} ; \mathrm{II} \rightarrow \mathrm{Q} ; \mathrm{III} \rightarrow \mathrm{S} ; \mathrm{IV} \rightarrow \mathrm{T}$
Correct
Option B:
I $\rightarrow \mathrm{R}$ ; II $\rightarrow \mathrm{S}$ ; III $\rightarrow \mathrm{S}$ ; IV $\rightarrow \mathrm{T}$
Option C:
I $\rightarrow \mathrm{P} ; \mathrm{II} \rightarrow \mathrm{Q} ; \mathrm{III} \rightarrow \mathrm{Q} ; \mathrm{IV} \rightarrow \mathrm{R}$
Option D:
I $\rightarrow \mathrm{R}$ ; II $\rightarrow \mathrm{S}$ ; III $\rightarrow \mathrm{Q}$ ; IV $\rightarrow \mathrm{R}$

Answer: A

Step-by-step solution

$\mathrm{I} \rightarrow$ Rate $=\frac{\mathrm{k}[\mathrm{X}]}{\mathrm{X}_{\mathrm{s}}+[\mathrm{X}]}$

If [ X ] is low, then it follows first order kinetic ie. $\mathrm{t}_{1 / 2}$ is constant (OA part in graph)[0pt] If [X] is high,

then it follows zero order kinetics. So half life $\left(\mathrm{t}_{1 / 2}\right)$ varies linearly with [X] as shown in graph.

$I \rightarrow P$ $\mathrm{II} \rightarrow$ Rate $=\frac{\mathrm{k}[\mathrm{X}]}{\mathrm{X}_{\mathrm{s}}+[\mathrm{X}]}$

If $[X]$ is less than $X_{s}$ , then reaction follows $I^{\text {st }}$ order kinetics as shown by the graph $Q$ and $T$

$\mathrm{II} \rightarrow \mathrm{Q}, \mathrm{T}$ III $\rightarrow$

Rate $=\frac{\mathrm{k}[\mathrm{X}]}{\mathrm{X}_{\mathrm{s}}+[\mathrm{X}]}$

If $[\mathrm{X}] \gg \mathrm{X}_{\mathrm{s}}$ ,

then reaction follows zero order kinetics, as show by the graph (S) III $\rightarrow$ S IV $\rightarrow$

Rate $=\frac{\mathrm{k}[\mathrm{X}]^{2}}{\mathrm{X}_{\mathrm{s}}+[\mathrm{X}]}$

If $[\mathrm{X}] \gg\left[\mathrm{X}_{\mathrm{s}}\right]$ , then reaction follows first order kinetics,

as show by the graph $(\mathrm{Q}),(\mathrm{T})$ $\mathrm{IV} \rightarrow \mathrm{Q}, \mathrm{T}$

Answer key and solution verified before publishing.

Practise Chemical Kinetics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Advanced 2022
Paper: Paper 1
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Rate Laws and Rate Constant