If 1 mole of acetic acid and 1 mole of ethyl alcohol are mixed together and the reaction proceeds to the equilibrium, the concentration of acetic acid and water are found to be 1/3 and 2/3 mole respectively. Calculate the equilibrium constant.

Explanation

To solve the problem, we need to analyze the reaction between acetic acid (CH₃COOH) and ethyl alcohol (C₂H₅OH) to form an ester (ethyl acetate, CH₃COOC₂H₅) and water (H₂O). The reaction can be represented as follows:

\[ \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \]

### Step 1: Write down the initial concentrations
Initially, we have:
- 1 mole of acetic acid (CH₃COOH)
- 1 mole of ethyl alcohol (C₂H₅OH)
- 0 moles of ethyl acetate (CH₃COOC₂H₅)
- 0 moles of water (H₂O)

### Step 2: Set up the change in concentrations
Let \( x \) be the amount of acetic acid and ethyl alcohol that reacts to form the products at equilibrium. The changes in concentrations will be:
- Acetic acid: \( 1 - x \)
- Ethyl alcohol: \( 1 - x \)
- Ethyl acetate: \( x \)
- Water: \( x \)

### Step 3: Use the given equilibrium concentrations
From the problem, we know:
- Concentration of acetic acid at equilibrium = \( \frac{1}{3} \) moles
- Concentration of water at equilibrium = \( \frac{2}{3} \) moles

### Step 4: Set up equations based on equilibrium concentrations
From the equilibrium concentration of acetic acid:
\[ 1 - x = \frac{1}{3} \]
This implies:
\[ x = 1 - \frac{1}{3} = \frac{2}{3} \]

From the equilibrium concentration of water:
Since water is formed in the same amount as \( x \):
\[ x = \frac{2}{3} \]

### Step 5: Calculate the equilibrium concentrations of all species
Now we can find the equilibrium concentrations:
- Acetic acid: \( 1 - x = 1 - \frac{2}{3} = \frac{1}{3} \)
- Ethyl alcohol: \( 1 - x = 1 - \frac{2}{3} = \frac{1}{3} \)
- Ethyl acetate: \( x = \frac{2}{3} \)
- Water: \( x = \frac{2}{3} \)

### Step 6: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]} \]

### Step 7: Substitute the equilibrium concentrations into the expression
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)}{\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)} \]

### Step 8: Calculate \( K_c \)
Calculating the above expression:
\[ K_c = \frac{\frac{4}{9}}{\frac{1}{9}} = 4 \]

### Final Answer
The equilibrium constant \( K_c \) is 4.