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 Gibbs free energy change and the equilibrium constant (K). The formula we will use is:

\[ \Delta G° = -RT \ln K \]

Where:
- \( R \) is the universal gas constant, which is approximately \( 8.314 \, \text{J/mol·K} \)
- \( T \) is the temperature in Kelvin
- \( K \) is the equilibrium constant

### Step-by-step Solution:

1. **Identify the values:**
- Given \( K = 4 \)
- Temperature \( T = 298 \, \text{K} \)
- Gas constant \( R = 8.314 \, \text{J/mol·K} \)

2. **Calculate \( \ln K \):**
- First, we need to calculate the natural logarithm of \( K \):
\[ \ln K = \ln 4 \]
- The value of \( \ln 4 \) can be calculated as:
\[ \ln 4 \approx 1.386 \]

3. **Substitute the values into the formula:**
- Now substitute \( R \), \( T \), and \( \ln K \) into the equation:
\[ \Delta G° = - (8.314 \, \text{J/mol·K}) \times (298 \, \text{K}) \times (1.386) \]

4. **Perform the multiplication:**
- Calculate the product:
\[ \Delta G° = - (8.314 \times 298 \times 1.386) \]
- First, calculate \( 8.314 \times 298 \):
\[ 8.314 \times 298 \approx 2477.572 \, \text{J/mol} \]
- Now multiply by \( 1.386 \):
\[ 2477.572 \times 1.386 \approx 3434.644 \, \text{J/mol} \]

5. **Calculate ΔG°:**
- Since we have a negative sign in front:
\[ \Delta G° \approx -3434.644 \, \text{J/mol} \]

6. **Convert to kilojoules:**
- To convert joules to kilojoules, divide by 1000:
\[ \Delta G° \approx -3.434644 \, \text{kJ/mol} \]

7. **Final answer:**
- The standard free energy change (ΔG°) is approximately:
\[ \Delta G° \approx -3.434644 \, \text{kJ/mol} \]

### Summary:
The standard free energy change for the reaction is approximately \(-3.434644 \, \text{kJ/mol}\).