What is `DeltaG^(ɵ)` for the following reaction? `1/2 N_(2)(g)+3/2 H_(2)(g) hArr NH_(3)(g), K_(p)=4.42xx10^(4)` at `25^(@)C`

Options

A. `-26.5 kJ mol^(-1)`
B. `11.5 kJ mol^(-1)`
C. `-2.2 kJ mol^(-1)`
D. `-0.97 kJ mol^(-1)`

Explanation

To find the standard Gibbs free energy change (ΔG°) for the reaction given the equilibrium constant (Kp), we can use the following relationship:

\[ \Delta G^{\circ} = -RT \ln K_p \]

### Step-by-Step Solution:

1. **Identify Constants**:
- The universal gas constant \( R = 8.31 \, \text{J/(mol·K)} \).
- The temperature \( T = 25^\circ C = 298 \, \text{K} \).
- The equilibrium constant \( K_p = 4.42 \times 10^4 \).

2. **Convert Kp to Natural Logarithm**:
- We need to calculate \( \ln K_p \).
- First, calculate \( K_p \):
\[ K_p = 4.42 \times 10^4 \]
- Now, find \( \ln K_p \):
\[ \ln K_p = \ln(4.42 \times 10^4) \]

3. **Use the Change of Base Formula**:
- To convert from natural logarithm to base 10, we can use:
\[ \ln K = 2.303 \log_{10} K \]
- Therefore:
\[ \ln(4.42 \times 10^4) = 2.303 \log_{10}(4.42 \times 10^4) \]

4. **Calculate Logarithm**:
- Calculate \( \log_{10}(4.42 \times 10^4) \):
\[ \log_{10}(4.42 \times 10^4) = \log_{10}(4.42) + \log_{10}(10^4) = 0.646 + 4 = 4.646 \]
- Now, substitute back to find \( \ln K_p \):
\[ \ln K_p = 2.303 \times 4.646 \approx 10.688 \]

5. **Calculate ΔG°**:
- Now substitute \( R \), \( T \), and \( \ln K_p \) into the ΔG° equation:
\[ \Delta G^{\circ} = - (8.31 \, \text{J/(mol·K)}) \times (298 \, \text{K}) \times (10.688) \]
- Calculate:
\[ \Delta G^{\circ} = - (8.31 \times 298 \times 10.688) \approx -26,500 \, \text{J/mol} = -26.5 \, \text{kJ/mol} \]

6. **Final Answer**:
- Therefore, the standard Gibbs free energy change for the reaction is:
\[ \Delta G^{\circ} \approx -26.5 \, \text{kJ/mol} \]