For which of the following reactions the molecularity and order of the reaction are respectively two and two

Options

A. Ester hydrolysis in acid medium
B. Inversion of cane sugar in acid aqueous solution
C. Hydrolysis of ethyl acetate in caustic soda aqueous solution.
D. Decomposition hydrogen peroxide in acid solution

Explanation

To determine which of the given reactions has both molecularity and order equal to 2, we need to analyze each reaction based on the definitions of molecularity and order.

### Step-by-Step Solution:

1. **Understanding Molecularity and Order**:
- **Molecularity**: This is the number of reactant molecules involved in the rate-determining step of a reaction. It is always a whole number and can be 0, 1, 2, etc.
- **Order**: This is the sum of the powers of the concentration terms in the rate law expression. It can be a whole number, fraction, or zero.

2. **Analyzing Each Reaction**:
- **Option 1: Ester Hydrolysis in Acidic Medium**:
- Reaction: CH3COOCH3 + H2O → CH3COOH + CH3OH
- Rate Law: Rate = k[CH3COOCH3][H2O]
- Here, water is in excess, making the reaction pseudo-first-order.
- **Molecularity**: 2 (two reactants: CH3COOCH3 and H2O)
- **Order**: 1 (since the rate law effectively depends on one reactant due to excess water)

- **Option 2: Inversion of Cane Sugar**:
- Reaction: C12H22O11 + H2O → C6H12O6 + C6H12O6
- Rate Law: Rate = k[C12H22O11]
- Water is in excess, making it pseudo-first-order.
- **Molecularity**: 2 (two reactants: C12H22O11 and H2O)
- **Order**: 1 (since the rate law effectively depends on one reactant due to excess water)

- **Option 3: Hydrolysis of Ethyl Acetate in Caustic Soda**:
- Reaction: CH3COOC2H5 + NaOH → CH3COONa + C2H5OH
- Rate Law: Rate = k[CH3COOC2H5][NaOH]
- Both reactants are involved in the rate law.
- **Molecularity**: 2 (two reactants: CH3COOC2H5 and NaOH)
- **Order**: 2 (since both reactants contribute to the rate law, each raised to the power of 1)

- **Option 4: Decomposition of Hydrogen Peroxide in Acidic Solution**:
- Reaction: 2H2O2 → 2H2O + O2
- Rate Law: Rate = k[H2O2]
- **Molecularity**: 2 (since two molecules of H2O2 are involved in the overall reaction)
- **Order**: 1 (since the rate law depends only on H2O2)

3. **Conclusion**:
- The only reaction where both molecularity and order are equal to 2 is the **hydrolysis of ethyl acetate in caustic soda** (Option 3).

### Final Answer:
The reaction for which the molecularity and order are respectively two and two is the **hydrolysis of ethyl acetate in caustic soda**.