One mole of ethyl alcohol was treated with one mole of acetic acid at `25^@C`. Two-third of the alcohol change into ester at equilibrium. The equilibrium constant for the reaction will be

Options

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

Explanation

To find the equilibrium constant for the reaction between ethyl alcohol (C2H5OH) and acetic acid (CH3COOH) to form an ester (ethyl acetate, CH3COOC2H5) and water (H2O), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction can be represented as:
\[ \text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \]

### Step 2: Set up the initial concentrations
Initially, we have:
- Ethyl alcohol (C2H5OH): 1 mole
- Acetic acid (CH3COOH): 1 mole
- Ester (CH3COOC2H5): 0 moles
- Water (H2O): 0 moles

### Step 3: Determine the change in concentrations
According to the problem, two-thirds of the alcohol reacts to form the ester. Therefore:
- Change in ethyl alcohol: -2/3 moles
- Change in acetic acid: -2/3 moles (since the ratio is 1:1)
- Change in ester: +2/3 moles
- Change in water: +2/3 moles

### Step 4: Calculate the equilibrium concentrations
At equilibrium, the concentrations will be:
- Ethyl alcohol: \( 1 - \frac{2}{3} = \frac{1}{3} \) moles
- Acetic acid: \( 1 - \frac{2}{3} = \frac{1}{3} \) moles
- Ester: \( 0 + \frac{2}{3} = \frac{2}{3} \) moles
- Water: \( 0 + \frac{2}{3} = \frac{2}{3} \) moles

### Step 5: Write the expression for the equilibrium constant (Kc)
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{C}_2\text{H}_5\text{OH}][\text{CH}_3\text{COOH}]} \]

### Step 6: Substitute the equilibrium concentrations into the expression
Substituting the equilibrium concentrations into the expression:
\[ K_c = \frac{\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)}{\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)} \]

### Step 7: Simplify the expression
Calculating the above expression:
\[ K_c = \frac{\frac{4}{9}}{\frac{1}{9}} = 4 \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction is **4**.

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