In the esterification
`C_(2)H_(5)OH(l)+CH_(3)COOH(l)" an "hArr CH_(3)COOC_(2)H_(5)(l)+H_(2)O(l)`
equimolar mixture of alcohol and acid taken initially yields under equilibrium, the water with mole fraction `=0.333`. The equilibrium constant. Is

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction
The reaction given is:
\[ C_2H_5OH(l) + CH_3COOH(l) \rightleftharpoons CH_3COOC_2H_5(l) + H_2O(l) \]

### Step 2: Set Up Initial Conditions
Assume we start with 1 mole of each reactant (ethyl alcohol and acetic acid). Therefore, at the start:
- Moles of \( C_2H_5OH = 1 \)
- Moles of \( CH_3COOH = 1 \)
- Moles of \( CH_3COOC_2H_5 = 0 \)
- Moles of \( H_2O = 0 \)

### Step 3: Define Change at Equilibrium
Let \( x \) be the amount of moles that react at equilibrium. Thus, at equilibrium:
- Moles of \( C_2H_5OH = 1 - x \)
- Moles of \( CH_3COOH = 1 - x \)
- Moles of \( CH_3COOC_2H_5 = x \)
- Moles of \( H_2O = x \)

### Step 4: Total Moles at Equilibrium
The total number of moles at equilibrium will be:
\[ \text{Total moles} = (1 - x) + (1 - x) + x + x = 2 \]

### Step 5: Mole Fraction of Water
Given that the mole fraction of water at equilibrium is 0.333, we can express this as:
\[ \text{Mole fraction of } H_2O = \frac{x}{\text{Total moles}} = \frac{x}{2} \]
Setting this equal to 0.333:
\[ \frac{x}{2} = 0.333 \]
Thus,
\[ x = 0.666 \]

### Step 6: Calculate Moles of Reactants at Equilibrium
Now substituting \( x \) back into the moles of reactants:
- Moles of \( C_2H_5OH = 1 - 0.666 = 0.334 \)
- Moles of \( CH_3COOH = 1 - 0.666 = 0.334 \)
- Moles of \( CH_3COOC_2H_5 = 0.666 \)
- Moles of \( H_2O = 0.666 \)

### Step 7: Write the Expression for the Equilibrium Constant
The equilibrium constant \( K \) for the reaction is given by:
\[ K = \frac{[\text{Products}]}{[\text{Reactants}]} \]
Substituting the equilibrium concentrations:
\[ K = \frac{[CH_3COOC_2H_5][H_2O]}{[C_2H_5OH][CH_3COOH]} \]
Substituting the moles:
\[ K = \frac{(0.666)(0.666)}{(0.334)(0.334)} \]

### Step 8: Calculate the Value of \( K \)
Calculating the above expression:
\[ K = \frac{0.666^2}{0.334^2} = \frac{0.443556}{0.111556} \approx 3.97 \]
Thus, rounding gives us:
\[ K \approx 4 \]

### Final Answer
The equilibrium constant \( K \) is approximately 4.

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