One mole of ethanol is treated with one mole of ethanoic acid at `25^(@)C`. Half of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be

Options

A. `1`
B. `2`
C. `3`
D. `4`

Explanation

To solve the problem, we need to determine the equilibrium constant (K) for the reaction between ethanol and ethanoic acid, given that half of the acid changes into ester at equilibrium.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction between ethanol (C₂H₅OH) and ethanoic acid (CH₃COOH) to form an ester (ethyl acetate, CH₃COOC₂H₅) and water (H₂O) can be written as:
\[ C_2H_5OH + CH_3COOH \rightleftharpoons CH_3COOC_2H_5 + H_2O \]

2. **Set Up Initial Concentrations:**
We start with 1 mole of ethanol and 1 mole of ethanoic acid. Initially, the concentrations are:
- [C₂H₅OH] = 1 M
- [CH₃COOH] = 1 M
- [CH₃COOC₂H₅] = 0 M
- [H₂O] = 0 M

3. **Determine Change at Equilibrium:**
According to the problem, half of the acid changes into ester at equilibrium. This means:
- Change in [CH₃COOH] = -0.5 M (since half of 1 mole reacts)
- Change in [C₂H₅OH] = -0.5 M (stoichiometrically equal to the acid)
- Change in [CH₃COOC₂H₅] = +0.5 M (produced)
- Change in [H₂O] = +0.5 M (produced)

4. **Calculate Equilibrium Concentrations:**
At equilibrium, the concentrations will be:
- [C₂H₅OH] = 1 - 0.5 = 0.5 M
- [CH₃COOH] = 1 - 0.5 = 0.5 M
- [CH₃COOC₂H₅] = 0 + 0.5 = 0.5 M
- [H₂O] = 0 + 0.5 = 0.5 M

5. **Write the Expression for the Equilibrium Constant (K):**
The equilibrium constant expression for this reaction is:
\[ K = \frac{[CH_3COOC_2H_5][H_2O]}{[C_2H_5OH][CH_3COOH]} \]

6. **Substitute the Equilibrium Concentrations into the K Expression:**
Plugging in the equilibrium concentrations:
\[ K = \frac{(0.5)(0.5)}{(0.5)(0.5)} = \frac{0.25}{0.25} = 1 \]

### Final Answer:
The equilibrium constant \( K \) for the reaction is \( 1 \).

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