Match the entries of column I with appropriate entries of column II and choose the correct option out of the four options (a), (b), ( c ), (d) given at the end of each question.
`{:(,,"Column I",,,,,,"Column II"),((A),,"Zero order reactions",,,,(p),,t_(1//2)prop(1)/([A]_(0))),((B),,"First order reactions",,,,(q),,t_(100%)=[A]_(0)//k),((C),,"Second order reactions",,,,(r),,"Involves at least two reactants"),((D),,"Pseudounimolecular reactions",,,,(s),,[A]=[A]_(0)e^(-kt)):}`

Options

A. `A-s,B-q,C-p,D-r`
B. `A-p,B-r,C-q,D-s`
C. `A-q,B-s,C-p,D-r`
D. `A-r,B-p,C-s,D-q`

Answer

To solve the problem of matching the entries from Column I with those from Column II, we will analyze each entry step by step.

### Step 1: Analyze Zero Order Reactions (A)
- **Zero Order Reactions**: The half-life (t_(1/2)) for zero-order reactions is given by the formula:
\[ t_{1/2} = \frac{[A]_0}{k} \]
This shows that the half-life is directly proportional to the initial concentration \([A]_0\).
- **Match**: This corresponds to option (p) in Column II, which states:
\[ t_{1/2} \propto \frac{1}{[A]_0} \]
Thus, **A matches with p**.

### Step 2: Analyze First Order Reactions (B)
- **First Order Reactions**: The half-life for first-order reactions is independent of the initial concentration and is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
The relationship can also be expressed in terms of the natural logarithm:
\[ [A] = [A]_0 e^{-kt} \]
- **Match**: This corresponds to option (s) in Column II, which states:
\[ [A] = [A]_0 e^{-kt} \]
Thus, **B matches with s**.

### Step 3: Analyze Second Order Reactions (C)
- **Second Order Reactions**: The half-life for second-order reactions is given by:
\[ t_{1/2} = \frac{1}{k[A]_0} \]
This indicates that the half-life is inversely proportional to the initial concentration.
- **Match**: This corresponds to option (q) in Column II, which states:
\[ t_{100\%} = \frac{[A]_0}{k} \]
Thus, **C matches with q**.

### Step 4: Analyze Pseudounimolecular Reactions (D)
- **Pseudounimolecular Reactions**: These reactions appear to be bimolecular but behave as unimolecular due to the presence of a large excess of one reactant. They involve at least two reactants.
- **Match**: This corresponds to option (r) in Column II, which states:
\[ \text{Involves at least two reactants} \]
Thus, **D matches with r**.

### Summary of Matches
- A → p
- B → s
- C → q
- D → r

### Final Answer
The correct option based on the matches is:
- **(A) p, (B) s, (C) q, (D) r**