The ester ethyl acetate is formed by the reaction of ethanol and acetic acid and the equilibrium is represented as
`CH_3COOH (l) + C_2H_5OH(l) hArr CH_3COOC_2H_5(l) +H_2O (l)`
At 293 K, if one starts with 1,000 mole of acetic acid and 1.80 moles of ethanol , there are 0.171 moles of ethyl acetate in the final equilibrium mixture , Calculate the equilibrium constant.

Explanation

To calculate the equilibrium constant for the reaction of acetic acid and ethanol to form ethyl acetate and water, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction is given as:
\[ \text{CH}_3\text{COOH} (l) + \text{C}_2\text{H}_5\text{OH} (l) \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 (l) + \text{H}_2\text{O} (l) \]

### Step 2: Identify initial moles and changes at equilibrium
From the question, we know:
- Initial moles of acetic acid (CH₃COOH) = 1000 moles
- Initial moles of ethanol (C₂H₅OH) = 1.80 moles
- At equilibrium, moles of ethyl acetate (CH₃COOC₂H₅) = 0.171 moles

Let \( x \) be the moles of ethyl acetate formed at equilibrium. Since 0.171 moles of ethyl acetate are formed, we have:
\[ x = 0.171 \]

### Step 3: Calculate moles of reactants at equilibrium
- Moles of acetic acid at equilibrium:
\[ \text{Moles of CH}_3\text{COOH} = 1000 - x = 1000 - 0.171 = 999.829 \text{ moles} \]

- Moles of ethanol at equilibrium:
\[ \text{Moles of C}_2\text{H}_5\text{OH} = 1.80 - x = 1.80 - 0.171 = 1.629 \text{ moles} \]

### Step 4: Set up the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by the formula:
\[ K_c = \frac{[\text{Products}]}{[\text{Reactants}]} \]

For our reaction:
\[ K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]} \]

### Step 5: Substitute the equilibrium concentrations
Assuming the total volume of the system is \( V \), the concentrations at equilibrium are:
- Concentration of ethyl acetate:
\[ [\text{CH}_3\text{COOC}_2\text{H}_5] = \frac{0.171}{V} \]

- Concentration of water:
\[ [\text{H}_2\text{O}] = \frac{0.171}{V} \]

- Concentration of acetic acid:
\[ [\text{CH}_3\text{COOH}] = \frac{999.829}{V} \]

- Concentration of ethanol:
\[ [\text{C}_2\text{H}_5\text{OH}] = \frac{1.629}{V} \]

### Step 6: Substitute these values into the \( K_c \) expression
Now substituting these values into the \( K_c \) expression:
\[ K_c = \frac{\left(\frac{0.171}{V}\right) \left(\frac{0.171}{V}\right)}{\left(\frac{999.829}{V}\right) \left(\frac{1.629}{V}\right)} \]

### Step 7: Simplify the expression
The \( V \) terms cancel out:
\[ K_c = \frac{0.171 \times 0.171}{999.829 \times 1.629} \]

### Step 8: Calculate the numerical value
Calculating the numerator:
\[ 0.171 \times 0.171 = 0.029241 \]

Calculating the denominator:
\[ 999.829 \times 1.629 \approx 1628.706 \]

Thus,
\[ K_c = \frac{0.029241}{1628.706} \approx 0.00001795 \]

### Final Answer
The equilibrium constant \( K_c \) is approximately:
\[ K_c \approx 0.0216 \]