The hydrolysis of ethyl acetate `CH_(3)COOC_(2)H_(5) + H_(2)O overset(H^(+))to CH_(3)COOH + C_(2)H_(5)OH` is a reaction of

Options

A. Pseudo first order
B. Second order
C. Third order
D. Zero order

Explanation

To determine the order of the hydrolysis of ethyl acetate, we can follow these steps:

### Step 1: Write the Reaction
The hydrolysis of ethyl acetate can be represented as:
\[ \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \overset{H^+}{\rightarrow} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \]

### Step 2: Identify the Reactants
In this reaction, the reactants are:
1. Ethyl acetate (\( \text{CH}_3\text{COOC}_2\text{H}_5 \))
2. Water (\( \text{H}_2\text{O} \))

### Step 3: Consider the Concentration of Water
In hydrolysis reactions, water is often used in excess. This means that its concentration remains relatively constant throughout the reaction. Therefore, we can treat the concentration of water as a constant.

### Step 4: Write the Rate Law Expression
The general rate law expression for a reaction involving two reactants is:
\[ \text{Rate} = k[\text{A}][\text{B}] \]
Where \( \text{A} \) and \( \text{B} \) are the reactants. However, since water is in excess, we can simplify the expression:
\[ \text{Rate} = k'[\text{CH}_3\text{COOC}_2\text{H}_5] \]
Where \( k' = k[\text{H}_2\text{O}] \) (since \([\text{H}_2\text{O}]\) is constant).

### Step 5: Determine the Order of the Reaction
From the rate law expression, we see that the rate depends only on the concentration of ethyl acetate. This indicates that the reaction behaves as a first-order reaction with respect to ethyl acetate.

### Step 6: Conclusion
Since the reaction is first-order with respect to ethyl acetate and the concentration of water does not affect the rate, we classify this reaction as a **pseudo first-order reaction**. Therefore, the order of hydrolysis of ethyl acetate is **pseudo first order**.

### Final Answer
The hydrolysis of ethyl acetate is a pseudo first-order reaction.

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