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: Write the balanced equation
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 concentrations
Assume we start with 1 mole of ethyl alcohol (\(C_2H_5OH\)) and 1 mole of acetic acid (\(CH_3COOH\)). Therefore, the initial concentrations are:
- \([C_2H_5OH] = 1 \, \text{mol}\)
- \([CH_3COOH] = 1 \, \text{mol}\)
- \([CH_3COOC_2H_5] = 0 \, \text{mol}\)
- \([H_2O] = 0 \, \text{mol}\)

### Step 3: Define the change in concentrations at equilibrium
Let \(x\) be the amount of water formed at equilibrium. Then, at equilibrium, we have:
- \([C_2H_5OH] = 1 - x\)
- \([CH_3COOH] = 1 - x\)
- \([CH_3COOC_2H_5] = x\)
- \([H_2O] = x\)

### Step 4: Calculate the total number of moles at equilibrium
The total number of moles at equilibrium is:
\[ \text{Total moles} = (1 - x) + (1 - x) + x + x = 2 \]

### Step 5: Use the mole fraction of water
We are given that the mole fraction of water (\(H_2O\)) at equilibrium is 0.333. The mole fraction is defined as:
\[ \text{Mole fraction of } H_2O = \frac{\text{moles of } H_2O}{\text{total moles}} = \frac{x}{2} \]
Setting this equal to 0.333, we have:
\[ \frac{x}{2} = 0.333 \]
Multiplying both sides by 2 gives:
\[ x = 0.666 \]

### Step 6: Calculate the equilibrium concentrations
Now we can find the equilibrium concentrations:
- \([C_2H_5OH] = 1 - 0.666 = 0.334\)
- \([CH_3COOH] = 1 - 0.666 = 0.334\)
- \([CH_3COOC_2H_5] = 0.666\)
- \([H_2O] = 0.666\)

### Step 7: Write the expression for the equilibrium constant (K)
The equilibrium constant \(K\) for the reaction is given by:
\[ K = \frac{[CH_3COOC_2H_5][H_2O]}{[C_2H_5OH][CH_3COOH]} \]
Substituting the equilibrium concentrations into the expression:
\[ K = \frac{(0.666)(0.666)}{(0.334)(0.334)} \]

### Step 8: Calculate the value of K
Calculating the above expression:
\[ K = \frac{0.666^2}{0.334^2} = \frac{0.443556}{0.111556} \approx 3.97 \]
Rounding to the nearest integer gives \(K \approx 4\).

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

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