The reaction
`CH_(3)COOH (l) + C_(2) H_(5) OH (l) hArr CH_(3) COOC_(2)H_(5) (l) + H_(2) O (l)`
was carried out at `27^(@)C` by taking one mole of each of the reactants. The reaction reached equilibrium when `2//3` rd of the reactants were consumed. Calculate the free energy change for the reaction `(R= 8.314 " JK "^(-1) " mol "^(-1)).`

Explanation

To calculate the free energy change (ΔG) for the given reaction at equilibrium, we will follow these steps:

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

### Step 2: Initial Moles of Reactants
Initially, we have:
- 1 mole of acetic acid (CH₃COOH)
- 1 mole of ethanol (C₂H₅OH)

### Step 3: Determine Moles at Equilibrium
According to the problem, \( \frac{2}{3} \) of the reactants are consumed at equilibrium. Therefore:
- Moles of acetic acid consumed = \( \frac{2}{3} \) moles
- Moles of ethanol consumed = \( \frac{2}{3} \) moles

At equilibrium, the moles of each substance will be:
- Moles of acetic acid left = \( 1 - \frac{2}{3} = \frac{1}{3} \) moles
- Moles of ethanol left = \( 1 - \frac{2}{3} = \frac{1}{3} \) moles
- Moles of ester (CH₃COOC₂H₅) formed = \( \frac{2}{3} \) moles
- Moles of water (H₂O) formed = \( \frac{2}{3} \) moles

### Step 4: Write the Expression for Equilibrium Constant (K)
The equilibrium constant \( K \) for the reaction is given by:
\[ K = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]} \]

Substituting the equilibrium concentrations:
\[ K = \frac{\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)}{\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)} = \frac{\frac{4}{9}}{\frac{1}{9}} = 4 \]

### Step 5: Calculate ΔG using the Equation
The formula for the change in free energy is:
\[ \Delta G = -2.303 \cdot R \cdot T \cdot \log K \]

Where:
- \( R = 8.314 \, \text{JK}^{-1}\text{mol}^{-1} \)
- \( T = 27^\circ C = 300 \, \text{K} \)
- \( K = 4 \)

### Step 6: Calculate log K
\[ \log K = \log 4 \approx 0.602 \]

### Step 7: Substitute Values into ΔG Equation
Now substituting the values:
\[ \Delta G = -2.303 \cdot 8.314 \cdot 300 \cdot 0.602 \]

Calculating this:
\[ \Delta G \approx -2.303 \cdot 8.314 \cdot 300 \cdot 0.602 \approx - 3,458.31 \, \text{J/mol} \approx -3.46 \, \text{kJ/mol} \]

### Final Answer
\[ \Delta G \approx -3,458.31 \, \text{J/mol} \text{ or } -3.46 \, \text{kJ/mol} \]