Which of the following is pseudo first order reaction?
I. `CH_(3)COOC_(2)H_(5)+H_(2)O overset(H^(+)) rarr CH_(3)COOH +C_(2)H_(5)OH`
II. `C_(12)H_(22)O_(11)+H_(2)O overset(H^(+)) rarr C_(6) H_(12)O_(6) +C_(6)H_(12)O_(6)`

Options

A. only I
B. only II
C. both I & II
D. none of these

Explanation

To determine which of the given reactions is a pseudo first-order reaction, we need to analyze both reactions based on the concept of pseudo first-order kinetics.

### Step-by-Step Solution:

1. **Understanding Pseudo First-Order Reactions**:
- A pseudo first-order reaction occurs when one reactant is in large excess compared to the other reactants. In such cases, the concentration of the excess reactant remains relatively constant during the reaction, allowing the reaction to be treated as a first-order reaction with respect to the limiting reactant.

2. **Analyzing Reaction I**:
- The first reaction is:
\[ CH_3COOC_2H_5 + H_2O \overset{H^+}{\rightarrow} CH_3COOH + C_2H_5OH \]
- Here, ethyl acetate (CH₃COOC₂H₅) reacts with water (H₂O) in the presence of H⁺ ions.
- In hydrolysis reactions, water is often present in large excess. Therefore, we can consider the concentration of water to be constant throughout the reaction.
- The rate law can be expressed as:
\[ \text{Rate} = k [CH_3COOC_2H_5][H_2O] \]
- Since [H₂O] is constant, we can rewrite the rate law as:
\[ \text{Rate} = k' [CH_3COOC_2H_5] \]
- Here, \( k' = k[H_2O] \), indicating that this reaction behaves as a pseudo first-order reaction.

3. **Analyzing Reaction II**:
- The second reaction is:
\[ C_{12}H_{22}O_{11} + H_2O \overset{H^+}{\rightarrow} C_6H_{12}O_6 + C_6H_{12}O_6 \]
- This reaction represents the hydrolysis of sucrose (C₁₂H₂₂O₁₁) into glucose (C₆H₁₂O₆).
- Similar to the first reaction, water is also present in excess in this reaction.
- The rate law can be expressed as:
\[ \text{Rate} = k [C_{12}H_{22}O_{11}][H_2O] \]
- Again, since [H₂O] is constant, we can rewrite it as:
\[ \text{Rate} = k' [C_{12}H_{22}O_{11}] \]
- Thus, this reaction also behaves as a pseudo first-order reaction.

4. **Conclusion**:
- Both reactions I and II are pseudo first-order reactions because they involve hydrolysis with water in excess, which allows us to treat the concentration of water as constant.

### Final Answer:
Both reactions I and II are pseudo first-order reactions.