If 1 mole of acetic acid and 1 mole of ethyl alchol are mixed and reaction proceeds to equilibrium , the concentrations of acetic acid and water are found to be `1//3 and 2//3` mole respectively . If 1 mole of ethyl acetate and 3 moles of water are mixed , how much ester is present when equilibrium is reached ?

Explanation

To solve the problem, we need to analyze the equilibrium of the reaction between acetic acid and ethyl alcohol to form ethyl acetate and water.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The reaction can be written as:
\[ \text{Acetic Acid (CH}_3\text{COOH)} + \text{Ethyl Alcohol (C}_2\text{H}_5\text{OH)} \rightleftharpoons \text{Ethyl Acetate (CH}_3\text{COOC}_2\text{H}_5) + \text{Water (H}_2\text{O)} \]

2. **Initial Concentrations**:
We start with 1 mole of acetic acid and 1 mole of ethyl alcohol. Therefore, the initial concentrations are:
\[ [\text{CH}_3\text{COOH}] = 1 \text{ mol}, \quad [\text{C}_2\text{H}_5\text{OH}] = 1 \text{ mol} \]

3. **Equilibrium Concentrations**:
From the problem, we know that at equilibrium:
\[ [\text{CH}_3\text{COOH}] = \frac{1}{3} \text{ mol}, \quad [\text{H}_2\text{O}] = \frac{2}{3} \text{ mol} \]
Let \( x \) be the amount of acetic acid and ethyl alcohol that reacts to form ethyl acetate and water.

4. **Change in Concentrations**:
At equilibrium, we can express the concentrations as:
\[ [\text{CH}_3\text{COOH}] = 1 - x, \quad [\text{C}_2\text{H}_5\text{OH}] = 1 - x, \quad [\text{CH}_3\text{COOC}_2\text{H}_5] = x, \quad [\text{H}_2\text{O}] = x + 1 \]
Given that \( [\text{CH}_3\text{COOH}] = \frac{1}{3} \), we can set up the equation:
\[ 1 - x = \frac{1}{3} \implies x = 1 - \frac{1}{3} = \frac{2}{3} \]

5. **Calculate Equilibrium Concentrations**:
Now, substituting \( x \) back:
\[ [\text{C}_2\text{H}_5\text{OH}] = 1 - \frac{2}{3} = \frac{1}{3} \]
\[ [\text{H}_2\text{O}] = \frac{2}{3} + 1 = \frac{5}{3} \]
\[ [\text{CH}_3\text{COOC}_2\text{H}_5] = \frac{2}{3} \]

6. **New Reaction Setup**:
Now, we consider the second part of the problem where we mix 1 mole of ethyl acetate and 3 moles of water. The reaction is:
\[ \text{Ethyl Acetate (CH}_3\text{COOC}_2\text{H}_5) + \text{Water (H}_2\text{O)} \rightleftharpoons \text{Acetic Acid (CH}_3\text{COOH)} + \text{Ethyl Alcohol (C}_2\text{H}_5\text{OH)} \]

7. **Equilibrium Constant**:
The equilibrium constant \( K_c \) for the reverse reaction can be determined from the previous reaction:
\[ K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]} = 4 \]
For the reverse reaction:
\[ K_c' = \frac{1}{K_c} = \frac{1}{4} \]

8. **Set Up the Equilibrium Expression**:
Let \( y \) be the amount of ethyl acetate that reacts. The equilibrium concentrations will be:
\[ [\text{CH}_3\text{COOC}_2\text{H}_5] = 1 - y, \quad [\text{H}_2\text{O}] = 3 - y, \quad [\text{CH}_3\text{COOH}] = y, \quad [\text{C}_2\text{H}_5\text{OH}] = y \]

9. **Substituting into the Equilibrium Expression**:
\[ K_c' = \frac{y^2}{(1 - y)(3 - y)} = \frac{1}{4} \]

10. **Solve for \( y \)**:
Cross-multiplying gives:
\[ 4y^2 = (1 - y)(3 - y) \]
Expanding and rearranging leads to a quadratic equation. Solving this will give the value of \( y \).

11. **Find the Equilibrium Concentration of Ethyl Acetate**:
Once \( y \) is found, the equilibrium concentration of ethyl acetate will be:
\[ [\text{Ethyl Acetate}] = 1 - y \]

### Final Answer:
After solving the quadratic equation, we find the value of \( y \) and subsequently the equilibrium concentration of ethyl acetate.