The hydrolyiss of ethyl acetate in the pressure of sofium hydroxide is a first order reaction. T or F

Answer

To determine whether the hydrolysis of ethyl acetate in the presence of sodium hydroxide is a first-order reaction, let's analyze the reaction step by step.

### Step 1: Understand the Reaction
The hydrolysis of ethyl acetate (CH3COOCH2CH3) in the presence of sodium hydroxide (NaOH) can be represented as follows:

\[ \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{C}_2\text{H}_5\text{OH} \]

### Step 2: Identify the Reactants
In this reaction, the reactants are ethyl acetate and sodium hydroxide. The products formed are sodium acetate (CH3COONa) and ethanol (C2H5OH).

### Step 3: Determine the Rate Law
The rate of a reaction can be expressed in terms of the concentrations of the reactants. For this reaction, the rate law can be written as:

\[ \text{Rate} = k[\text{CH}_3\text{COOCH}_2\text{CH}_3]^m[\text{NaOH}]^n \]

Where \( m \) and \( n \) are the orders of the reaction with respect to ethyl acetate and sodium hydroxide, respectively.

### Step 4: Analyze the Reaction Order
In the case of the hydrolysis of esters like ethyl acetate in the presence of a strong base such as sodium hydroxide, the reaction is typically second-order overall. This is because both reactants (ethyl acetate and sodium hydroxide) are involved in the rate-determining step of the reaction.

Thus, if we assume \( m = 1 \) and \( n = 1 \), the overall order of the reaction would be:

\[ \text{Overall order} = m + n = 1 + 1 = 2 \]

### Step 5: Conclusion
Since the overall order of the reaction is 2, the statement that the hydrolysis of ethyl acetate in the presence of sodium hydroxide is a first-order reaction is **false**.

### Final Answer
The statement is **False**.

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