The equilibrium constant for the reaction:
`CH_(3)COOH(l) + C_(2)H_(5)OH(l) iff CH_(3)COOC_(2)H_(5)(l)+ H_(2)O(l)` has been found to be equal to 4 at `25^(@)C`. Calcuclate the free energy chnage for the reaction.

Explanation

To calculate the Gibbs free energy change (ΔG°) for the given reaction, we can use the relationship between the equilibrium constant (K) and the standard Gibbs free energy change (ΔG°). The formula is:

\[ \Delta G° = -2.303 \times R \times T \times \log K \]

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Equilibrium constant (K) = 4
- Temperature (T) = 25°C

2. **Convert Temperature to Kelvin:**
- To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
- Therefore:
\[ T = 25 + 273 = 298 \, K \]

3. **Use the Universal Gas Constant (R):**
- The value of the universal gas constant (R) is:
\[ R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \]

4. **Calculate the Logarithm of K:**
- We need to calculate \(\log K\):
\[ \log 4 \approx 0.6020 \]

5. **Substitute Values into the Gibbs Free Energy Formula:**
- Now substitute the values into the formula:
\[ \Delta G° = -2.303 \times 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \times 298 \, K \times 0.6020 \]

6. **Perform the Calculation:**
- First, calculate the product:
\[ -2.303 \times 8.314 \times 298 \times 0.6020 \approx -570.8 \, \text{J/mol} \]
- Convert Joules to kilojoules:
\[ -570.8 \, \text{J/mol} = -0.5708 \, \text{kJ/mol} \]

7. **Final Result:**
- Therefore, the standard Gibbs free energy change for the reaction is:
\[ \Delta G° \approx -3.434 \, \text{kJ/mol} \]

### Summary of the Result:
\[ \Delta G° \approx -3.434 \, \text{kJ/mol} \]