When alcohol `(C_(2)H_(5)OH(l)) "and acetic acid" (CH_(3)COOH(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_(2)H_(5)OH(l)+CH_(3)COOH(l)hArrCH_(3)COOC_(2)H_(5)(l)+H_(2)O(l)`is:

Options

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

Explanation

To solve the problem, we need to determine the equilibrium constant \( K_c \) for the reaction between ethanol (C₂H₅OH) and acetic acid (CH₃COOH) to form an ester and water. Here’s a step-by-step breakdown of the solution:

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

### Step 2: Define Initial Concentrations
Assume we start with \( X \) moles of each reactant (C₂H₅OH and CH₃COOH). Therefore, the initial concentrations are:
- \([C_2H_5OH] = X\)
- \([CH_3COOH] = X\)
- \([CH_3COOC_2H_5] = 0\)
- \([H_2O] = 0\)

### Step 3: Determine Changes at Equilibrium
Given that 33% of each reactant is converted into products, we can calculate the changes:
- Change in C₂H₅OH and CH₃COOH: \( -0.33X \)
- Change in CH₃COOC₂H₅ and H₂O: \( +0.33X \)

### Step 4: Write Equilibrium Concentrations
At equilibrium, the concentrations will be:
- \([C_2H_5OH] = X - 0.33X = 0.67X\)
- \([CH_3COOH] = X - 0.33X = 0.67X\)
- \([CH_3COOC_2H_5] = 0 + 0.33X = 0.33X\)
- \([H_2O] = 0 + 0.33X = 0.33X\)

### Step 5: Write the Expression for \( K_c \)
The equilibrium constant \( K_c \) is given by the formula:
\[ K_c = \frac{[CH_3COOC_2H_5][H_2O]}{[C_2H_5OH][CH_3COOH]} \]

Substituting the equilibrium concentrations:
\[ K_c = \frac{(0.33X)(0.33X)}{(0.67X)(0.67X)} \]

### Step 6: Simplify the Expression
This simplifies to:
\[ K_c = \frac{0.33^2 X^2}{0.67^2 X^2} \]
The \( X^2 \) cancels out:
\[ K_c = \frac{0.33^2}{0.67^2} \]

### Step 7: Calculate the Numerical Value
Calculating \( K_c \):
\[ K_c = \frac{0.1089}{0.4489} \approx 0.2439 \approx \frac{1}{4} \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction is:
\[ K_c \approx \frac{1}{4} \]