For the reaction `CH_3COOH(l)+C_2H_5(l)hArrCH_3COOC_2H_5(l)+H_2O(l)` the value of equilibrium constant (K) is 4 at 298 K. The standard free energy change `(DeltaG^@)` is equal to

Options

A. `3.473"kJ mol"^(-1)`
B. `-34730"kJ mol"^(-1)`
C. `34730"kJ mol"^(-1)`
D. `-3.434"kJ mol"^(-1)`

Answer

To find the standard free energy change (ΔG°) for the reaction given, we can use the relationship between the equilibrium constant (K) and the standard free energy change, which is expressed by the equation:

\[ \Delta G° = -RT \ln K \]

### Step-by-Step Solution:

1. **Identify the values needed**:
- The equilibrium constant \( K = 4 \)
- The universal gas constant \( R = 8.314 \, \text{J/mol·K} \)
- The temperature \( T = 298 \, \text{K} \)

2. **Substitute the values into the equation**:
\[ \Delta G° = - (8.314 \, \text{J/mol·K}) \times (298 \, \text{K}) \times \ln(4) \]

3. **Calculate \( \ln(4) \)**:
- Using a calculator, we find:
\[ \ln(4) \approx 1.386 \]

4. **Plug \( \ln(4) \) back into the equation**:
\[ \Delta G° = - (8.314) \times (298) \times (1.386) \]

5. **Perform the multiplication**:
- First, calculate \( 8.314 \times 298 \):
\[ 8.314 \times 298 \approx 2477.572 \, \text{J/mol} \]

- Now multiply by \( \ln(4) \):
\[ 2477.572 \times 1.386 \approx 3434.644 \, \text{J/mol} \]

6. **Apply the negative sign**:
\[ \Delta G° \approx -3434.644 \, \text{J/mol} \]

7. **Convert to kilojoules per mole**:
\[ \Delta G° \approx -3.434644 \, \text{kJ/mol} \approx -3.43 \, \text{kJ/mol} \]

### Final Answer:
\[ \Delta G° \approx -3.43 \, \text{kJ/mol} \]