JEE Advanced 2022 — Paper 1 — CHEM-III solution

JEE Advanced2022Paper 1CHEM-III

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)

Which one of the following options is correct?

A
I $\to$ P; II $\to$ Q; III $\to$ S; IV $\to$ T
B
I $\to$ R; II $\to$ S; III $\to$ S; IV $\to$ T
C
I $\to$ P; II $\to$ Q; III $\to$ Q; IV $\to$ R
D
I $\to$ R; II $\to$ S; III $\to$ Q; IV $\to$ R

Solution

This is Michaelis-Menten-type kinetics. Two limits matter:

Limit A: $[X] ≫ X_{s}$ . The denominator $X_{s} + [X] \approx [X]$ , so rate $\approx k$ . Zero-order kinetics: $[X]$ decreases linearly with time, and the half-life is proportional to $[X]_{0}$ (a straight line through the origin).

Limit B: $[X] ≪ X_{s}$ . The denominator $\approx X_{s}$ , so rate $\approx \frac{k}{X _{s}} [X] = k^{'} [X]$ . First-order kinetics: $ln [X]$ vs $t$ is a straight line with negative slope, and half-life is independent of $[X]_{0}$ .

(I) No limit imposed. As $[X]$ varies from very large to small, rate transitions from zero-order to first-order behaviour. The rate vs $[X]$ plot shows a saturating curve, and the half-life vs $[X]_{0}$ plot in the high-concentration zero-order regime is a straight line through the origin. The single best match capturing the full behaviour is P (half-life linear in $[X]_{0}$ , the zero-order saturating regime).

(II) $[X] ≪ X_{s}$ : pseudo-first-order. Half-life is constant. II $\to$ Q.

(III) $[X] ≫ X_{s}$ : zero order. $[X]$ vs $t$ is a straight line with negative slope. III $\to$ S.

(IV) $rate = \frac{k [ X ] ^{2}}{X _{s} + [ X ]}$ with $[X] ≫ X_{s}$ : rate $\approx k [X]$ , i.e., first-order. $ln [X]$ vs $t$ is linear with negative slope. IV $\to$ T.

The correct matching is option (A).

← Back to full paper

Practice more CHEM-III

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →