Animated Solution for Chemistry - Chemical Kinetics: Match the rate expressions in LIST-I for the decomposition of X with the corresponding profiles provided in LIST-II. ${\text{X}}_s$ and $k$ constants having appropriate units.

This is a beautiful problem that tests your ability to connect mathematical rate laws with their graphical representations. It's not just about memorizing formulas; it's about understanding how approximations change the fundamental nature of a reaction's kinetics.

Before we dive into the complex rate expressions, let's decode the graphs provided in LIST-II. Each graph is a classic signature of a specific reaction order.

Graph (P) shows the half-life ($t_{1/2}$) directly proportional to the initial concentration ($[\text{X}{]}_0$). This is the hallmark of a Zero Order reaction, where $t_{1/2}=\frac{[\text{X}{]}_0}{2k}$.

Graph (Q) shows a constant half-life, completely independent of the initial concentration. This is the defining characteristic of a First Order reaction, where $t_{1/2}=\frac{\ln 2}{k}$.

Graph (R) is fascinating. It shows the rate initially increasing linearly with concentration (like a first-order reaction) but eventually flattening out to a constant value (like a zero-order reaction). This represents a Mixed Order profile, commonly seen in enzyme kinetics (Michaelis-Menten) or surface catalysis (Langmuir adsorption).

Graph (S) plots concentration $[\text{X}]$ versus time as a straight line with a negative slope. This linear decrease is exactly what we expect for a Zero Order reaction.

Graph (T) plots $\ln [\text{X}]$ versus time as a straight line with a negative slope. This logarithmic decay is the classic signature of a First Order reaction.

Now, let's tackle the rate expressions in LIST-I by applying the given approximations.

Expression (I): This expression is valid under all possible initial concentrations. It is the exact mathematical form of the mixed-order curve we discussed. Therefore, it perfectly matches Graph (R). Furthermore, at very high concentrations ($[\text{X}]\to \infty$), the rate approaches a constant $k$, exhibiting zero-order behavior, which links it to Graph (P).

Expression (II): Here, the condition is $[\text{X}]\ll {\text{X}}_s$. Because $[\text{X}]$ is so small, we can approximate the denominator ${\text{X}}_s+[\text{X}]\approx {\text{X}}_s$. The rate simplifies to: This is a pure First Order reaction! Therefore, it matches the first-order graphs, (Q) and (T).

Expression (III): This time, the condition is $[\text{X}]\gg {\text{X}}_s$. The constant ${\text{X}}_s$ becomes negligible in the denominator, so ${\text{X}}_s+[\text{X}]\approx [\text{X}]$. The $[\text{X}]$ in the numerator and denominator cancel out: The rate is constant, meaning it's a Zero Order reaction. This corresponds to graphs (P) and (S).

Expression (IV): With the condition $[\text{X}]\gg {\text{X}}_s$, the denominator again simplifies to $[\text{X}]$. One power of $[\text{X}]$ cancels out: This simplifies beautifully to a First Order reaction! So, it matches graphs (Q) and (T).

By systematically applying these approximations, we have successfully mapped every complex rate law to its fundamental graphical behavior. The final mapping is: - (I) $\to$ (P), (R) - (II) $\to$ (Q), (T) - (III) $\to$ (P), (S) - (IV) $\to$ (Q), (T)