Question

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 =\(\frac{k[X]}{Xs+[X]}\) under all possible initial concentration of X	P
II	rate =\(\frac{k[X]}{Xs+[X]}\) where initial concentration of X are much less than Xs​	Q
III	rate =\(\frac{k[X]}{Xs+[X]}\) where initial concentration of X are much higher than Xs​	R
IV	rate =\(\frac{k[X]^2}{Xs+[X]}\),where initial concentration of X is much higher than Xs​	S
		T

 

• A. I → P; II → Q; III → S; IV → T
• B. I → R; II → S; III → S; IV → T
• C. I → P; II → Q; III → Q; IV → R
• D. I → R; II → S; III → Q; IV → R

Answer 1

The Correct Option is A

1. \( \text{rate} = k[\text{X}] \) (Expression I):

This is a first-order rate equation under all possible concentrations. The graph for this case shows initial concentration of X vs time.
The corresponding profile is P.

2. \( \text{rate} = k[\text{X}]^2 \) (Expression II):

- This represents a second-order rate equation when the initial concentration of X is much smaller than \( X_{\infty} \). The graph shows **small concentration of X**. - The corresponding profile is **Q**.

3. \( \text{rate} = k[\text{X}] \) (Expression III):

This is a first-order rate equation where the initial concentration of X is much higher than \( X_{\infty} \). The graph shows **initial concentration of X vs time** and typically exhibits exponential decay.
The corresponding profile is S.

4. \( \text{rate} = k[\text{X}]^n \) (Expression IV):

This represents a general rate equation where the initial concentration of X is much higher than \( X_{\infty} \). The graph shows concentration of X vs time with a quadratic decrease for higher-order reactions.
The corresponding profile is T.

Final Answer:

The correct matches are:

• I → P: \( \text{rate} = k[\text{X}] \), initial concentration of X vs time graph
• II → Q: \( \text{rate} = k[\text{X}]^2 \), small concentration of X
• III → S: \( \text{rate} = k[\text{X}] \), initial concentration of X vs time showing exponential decay
• IV → T: \( \text{rate} = k[\text{X}]^n \), concentration of X vs time for high concentrations showing quadratic decrease

Correct Option:

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