When ethyl alcohol `(C_2H_5OH(l))` and acetic acid `(CH_3COOH(l))` are mixed together in equimolar ratio at `27^@C` , 33% of each is converted into ester.Then the `K_C` for the equilibrium, `C_2H_5OH(l))+CH_3CHOOH(l) hArrCH_3COOC_2H_5(l)+H_2O(l)`

Options

A. 4
B. `1/4`
C. 9
D. `1/9`

Answer

To find the equilibrium constant \( K_C \) for the reaction

\[ C_2H_5OH(l) + CH_3COOH(l) \rightleftharpoons CH_3COOC_2H_5(l) + H_2O(l) \]

when ethyl alcohol and acetic acid are mixed in equimolar ratios at \( 27^\circ C \) and 33% of each is converted into ester, we can follow these steps:

### Step 1: Define Initial Concentrations
Assume we start with \( 1 \, \text{mol} \) of ethyl alcohol \( (C_2H_5OH) \) and \( 1 \, \text{mol} \) of acetic acid \( (CH_3COOH) \).

### Step 2: Calculate the Amount Converted
Given that 33% of each reactant is converted into the products, we can calculate the amount converted:
- Amount of \( C_2H_5OH \) converted = \( 0.33 \, \text{mol} \)
- Amount of \( CH_3COOH \) converted = \( 0.33 \, \text{mol} \)

### Step 3: Calculate the Equilibrium Concentrations
Now we can find the equilibrium concentrations:
- Remaining \( C_2H_5OH = 1 - 0.33 = 0.67 \, \text{mol} \)
- Remaining \( CH_3COOH = 1 - 0.33 = 0.67 \, \text{mol} \)
- Amount of ester \( (CH_3COOC_2H_5) \) formed = \( 0.33 \, \text{mol} \)
- Amount of water \( (H_2O) \) formed = \( 0.33 \, \text{mol} \)

### Step 4: Write the Expression for \( K_C \)
The equilibrium constant \( K_C \) is given by the expression:

\[ K_C = \frac{[CH_3COOC_2H_5][H_2O]}{[C_2H_5OH][CH_3COOH]} \]

Substituting the equilibrium concentrations into the expression:

\[ K_C = \frac{(0.33)(0.33)}{(0.67)(0.67)} \]

### Step 5: Simplify the Expression
Calculating the values:

\[ K_C = \frac{0.1089}{0.4489} \approx \frac{1}{4} \]

### Final Result
Thus, the equilibrium constant \( K_C \) is approximately \( \frac{1}{4} \).