When ethanol and acetic acid are mixed together in equimolar proportions, equilibrium is attained when `2//3rd` of acid and alcohol are consumed. The equilibrium constant for the reaction is

Options

A. 0.4
B. 4
C. 40
D. 0.04

Answer

To find the equilibrium constant for the reaction between ethanol (C2H5OH) and acetic acid (CH3COOH), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between ethanol and acetic acid can be represented as:
\[ \text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \]

2. **Initial Moles**:
Since ethanol and acetic acid are mixed in equimolar proportions, let's assume we start with 1 mole of each:
- Moles of C2H5OH = 1
- Moles of CH3COOH = 1

3. **Consumption of Reactants**:
According to the problem, \( \frac{2}{3} \) of both the acid and alcohol are consumed at equilibrium. Therefore, the moles consumed are:
- Moles of C2H5OH consumed = \( \frac{2}{3} \)
- Moles of CH3COOH consumed = \( \frac{2}{3} \)

4. **Moles at Equilibrium**:
The moles remaining at equilibrium will be:
- Moles of C2H5OH left = \( 1 - \frac{2}{3} = \frac{1}{3} \)
- Moles of CH3COOH left = \( 1 - \frac{2}{3} = \frac{1}{3} \)

5. **Moles of Products Formed**:
Since \( \frac{2}{3} \) of the reactants are consumed, the moles of products formed (CH3COOC2H5 and H2O) will be:
- Moles of CH3COOC2H5 = \( \frac{2}{3} \)
- Moles of H2O = \( \frac{2}{3} \)

6. **Equilibrium Constant Expression**:
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[\text{Products}]}{[\text{Reactants}]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]} \]

7. **Substituting Values**:
At equilibrium, the concentrations can be substituted as:
\[ K_c = \frac{\left(\frac{2}{3}\right) \times \left(\frac{2}{3}\right)}{\left(\frac{1}{3}\right) \times \left(\frac{1}{3}\right)} \]
Simplifying this gives:
\[ K_c = \frac{\frac{4}{9}}{\frac{1}{9}} = 4 \]

8. **Final Result**:
Thus, the equilibrium constant \( K_c \) for the reaction is:
\[ K_c = 4 \]