The alkaline hydrolysis of ethyl acetate is represented by the equation
`CH_(3) COOC_(2)H_(5) + NaOH to CH_(3) COONa + C_(2)H_(5) OH`
Experimentally it is found that for this reaction
`(dx)/(dt) = k [CH_(3)COOC_(2)H_(5)][NaOH]`
Then the reaction is

Options

A. Bimolecular and of first order
B. Bimolecular and of second order
C. Pseudo-bimolecular
D. Pseudo- unimolecular

Explanation

To determine the order of the reaction for the alkaline hydrolysis of ethyl acetate, we can analyze the given rate equation and the reactants involved.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction is given as:
\[ \text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{C}_2\text{H}_5\text{OH} \]
Here, ethyl acetate (an ester) reacts with sodium hydroxide (a base).

2. **Write the Rate Law Expression**:
The rate of the reaction is given by:
\[ \frac{dx}{dt} = k [\text{CH}_3\text{COOC}_2\text{H}_5][\text{NaOH}] \]
This indicates that the rate of reaction depends on the concentrations of both ethyl acetate and sodium hydroxide.

3. **Determine the Order of the Reaction**:
In the rate law expression, the concentration of each reactant is raised to the power of its stoichiometric coefficient. Since both reactants are present in the first power:
- The order with respect to ethyl acetate \([\text{CH}_3\text{COOC}_2\text{H}_5]\) is 1.
- The order with respect to sodium hydroxide \([\text{NaOH}]\) is also 1.

Therefore, the overall order of the reaction is:
\[ \text{Order} = 1 + 1 = 2 \]

4. **Conclusion**:
The reaction is a **bimolecular reaction** and is of **second order** overall.

### Final Answer:
The reaction is a **second order reaction**.

---