Step 1: Understanding the Concept:
The rate of a chemical reaction is given by the change in concentration per unit time: \( Rate = \frac{d[C]}{dt} \).
The units of rate are always \(mol \cdot L^{-1} \cdot s^{-1}\) (or \(M \cdot s^{-1}\)).
The rate law is expressed as: \( Rate = k[A]^n \), where \(n\) is the order of the reaction.
The units of the rate constant \(k\) must adjust so that the overall product of the right side matches the units of rate.
Key Formula or Approach:
The general units for the rate constant \(k\) of an \(n^{th}\) order reaction are:
\[ [k] = (mol \cdot L^{-1})^{1-n} \cdot s^{-1} \]
Or more simply:
\[ [k] = M^{1-n} \cdot s^{-1} \]
Step 2: Detailed Explanation:
Let's calculate the units for each case:
For Zero Order (\(n=0\)):
\[ [k] = (mol \cdot L^{-1})^{1-0} \cdot s^{-1} = mol \cdot L^{-1} \cdot s^{-1} \]
This matches with (II). For zero-order, the rate is independent of concentration, so \(Rate = k\).
For First Order (\(n=1\)):
\[ [k] = (mol \cdot L^{-1})^{1-1} \cdot s^{-1} = (mol \cdot L^{-1})^0 \cdot s^{-1} = s^{-1} \]
This matches with (III). For first-order, \(k = Rate / [A]\), so the concentration units cancel out.
For Second Order (\(n=2\)):
\[ [k] = (mol \cdot L^{-1})^{1-2} \cdot s^{-1} = (mol \cdot L^{-1})^{-1} \cdot s^{-1} = L \cdot mol^{-1} \cdot s^{-1} \]
This matches with (I). For second-order, \(k = Rate / [A]^2\), leading to inverse concentration units.
Therefore, the correct matching is:
A \(\rightarrow\) II, B \(\rightarrow\) III, C \(\rightarrow\) I.
This corresponds to Option (A).
Step 3: Final Answer:
The correct option is (A) A–II, B–III, C–I.