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.

The correct matching option is I → P; II → Q; III → S; IV → T.

Let us systematically analyze each rate expression in LIST-I and match it with its corresponding graphical profile in LIST-II based on reaction kinetics:

Analysis of (I):
Given rate expression:
$\text{rate}=\frac{k\left[\text{X}\right]}{{\text{X}}_s+\left[\text{X}\right]}$
under all possible initial concentrations of X.
When initial concentration [X]₀ is small ([X]₀ << X_s), denominator ≈ X_s, so rate = (k/X_s)[X], which is first-order kinetics (half-life is constant).
When initial concentration [X]₀ is very large ([X]₀ >> X_s), rate ≈ k, which is zero-order kinetics (half-life is directly proportional to initial concentration [X]₀).
Therefore, as [X]₀ increases, the overall half-life (t_1/2) increases linearly from a constant value at very low concentration to an increasing function at higher concentration.
This corresponds to profile (P), which shows a plot of half-life (t_1/2) vs Initial concentration of X with a positive slope starting from a non-zero intercept.

Analysis of (II):
Given rate expression:
$\text{rate}=\frac{k\left[\text{X}\right]}{{\text{X}}_s+\left[\text{X}\right]}$
where initial concentrations of X are much less than X_s ([X] << X_s).
In this limit:
$\text{rate}=\frac{k}{{\text{X}}_s}\left[\text{X}\right]=k^{\text{'}}\left[\text{X}\right]$
This is a standard first-order reaction.
For a first-order reaction, the half-life is given by:
$t_{1/2}=\frac{\ln 2}{k^{\text{'}}}$
Thus, half-life is completely independent of the initial concentration of X.
This corresponds to profile (Q), showing a horizontal line for half life (t_1/2) vs Initial concentration of X.

Analysis of (III):
Given rate expression:
$\text{rate}=\frac{k\left[\text{X}\right]}{{\text{X}}_s+\left[\text{X}\right]}$
where initial concentrations of X are much higher than X_s ([X] >> X_s).
In this limit:
$\text{rate}=-\frac{d\left[\text{X}\right]}{dt}=k$
This represents a zero-order reaction.
Integrating the rate equation gives:
$\left[\text{X}\right]={\left[\text{X}\right]}_0-kt$
A plot of [X] versus time yields a straight line with a negative slope (-k).
This corresponds to profile (S).

Analysis of (IV):
Given rate expression:
$\text{rate}=\frac{k{\left[\text{X}\right]}^2}{{\text{X}}_s+\left[\text{X}\right]}$
where initial concentrations of X are much higher than X_s ([X] >> X_s).
Since [X] >> X_s, the term (X_s + [X]) can be approximated as [X]:
$\text{rate}=\frac{k{\left[\text{X}\right]}^2}{\left[\text{X}\right]}=k\left[\text{X}\right]$
This represents a first-order reaction with integrated rate law:
$\ln \left[\text{X}\right]=\ln {\left[\text{X}\right]}_0-kt$
A plot of ln[X] versus time produces a straight line with a negative slope (-k).
This corresponds to profile (T).

Hence, the correct matching sequence is:
I → P
II → Q
III → S
IV → T