1.0 mole of ethyl alcohol and 1.0 mole of acetic acid are mixed. At equilibrium, 0.666 mole of ester is formed. The value of equilibrium constant is

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between ethyl alcohol (ethanol) and acetic acid to form an ester (ethyl acetate) and water can be represented as follows:

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

### Step 2: Set up the initial moles and changes
Initially, we have:
- Ethyl alcohol: 1.0 mole
- Acetic acid: 1.0 mole
- Ester: 0 moles
- Water: 0 moles

At equilibrium, we are given that 0.666 moles of ester are formed.

Let \( x \) be the amount of ester formed. Thus, at equilibrium:
- Moles of ester = \( x = 0.666 \) moles
- Moles of ethyl alcohol remaining = \( 1.0 - x = 1.0 - 0.666 = 0.334 \) moles
- Moles of acetic acid remaining = \( 1.0 - x = 1.0 - 0.666 = 0.334 \) moles
- Moles of water = \( x = 0.666 \) moles

### Step 3: Write the expression for the equilibrium constant
The equilibrium constant \( K_c \) for the reaction is given by the formula:

\[ K_c = \frac{[\text{C}_2\text{H}_5\text{COOCH}_3][\text{H}_2\text{O}]}{[\text{C}_2\text{H}_5\text{OH}][\text{CH}_3\text{COOH}]} \]

### Step 4: Substitute the equilibrium concentrations
Since we are dealing with moles and assuming the volume of the reaction mixture is constant, we can directly substitute the moles into the expression:

\[ K_c = \frac{[0.666][0.666]}{[0.334][0.334]} \]

### Step 5: Calculate the equilibrium constant
Now, we can calculate \( K_c \):

\[ K_c = \frac{0.666 \times 0.666}{0.334 \times 0.334} \]

Calculating the numerator and denominator:

- Numerator: \( 0.666^2 = 0.443556 \)
- Denominator: \( 0.334^2 = 0.111556 \)

Now, divide the two:

\[ K_c = \frac{0.443556}{0.111556} \approx 3.973 \]

### Conclusion
The value of the equilibrium constant \( K_c \) is approximately 4.0.