The standard free energy of formation of NO(g) is 86.6 kJ/ mol at 298 K what is the standard free energy of formation of ` NO_(2) g` at 298 k? `K _(p)= 1.6 xx 10^(12)`

Options

A. `86600 + R(298)` ln `(1.6 xx 10^(12))`
B. `86600 - ("ln " (1.6 xx 10^(12)))/(R(298))`
C. `0.5 [2 xx 86,600 - R(298) " ln "(1.6 xx 10^(12))]`
D. `R(298) " ln " (1.6 xx 10^(12)) - 86600`

Answer

To find the standard free energy of formation of NO2 (g) at 298 K given the standard free energy of formation of NO (g) and the equilibrium constant Kp, we can follow these steps:

### Step-by-Step Solution:

1. **Write down the given data**:
- Standard free energy of formation of NO:
\[ \Delta G^\circ_f \text{ (NO)} = 86.6 \, \text{kJ/mol} = 86,600 \, \text{J/mol} \]
- Equilibrium constant \( K_p \):
\[ K_p = 1.6 \times 10^{12} \]

2. **Write the balanced chemical equation**:
The formation of NO2 from NO and O2 can be represented as:
\[ 2 \text{NO(g)} + \text{O}_2(g) \rightarrow 2 \text{NO}_2(g) \]

3. **Use the relationship between free energy and equilibrium constant**:
The standard free energy change (\( \Delta G^\circ \)) for the reaction can be related to the equilibrium constant \( K_p \) using the equation:
\[ \Delta G^\circ = -RT \ln K_p \]
where \( R \) is the universal gas constant (8.314 J/(mol·K)) and \( T \) is the temperature in Kelvin (298 K).

4. **Calculate \( \Delta G^\circ \)**:
Substitute the values into the equation:
\[ \Delta G^\circ = - (8.314 \, \text{J/(mol·K)})(298 \, \text{K}) \ln(1.6 \times 10^{12}) \]

5. **Calculate \( \ln(1.6 \times 10^{12}) \)**:
Using a calculator, we find:
\[ \ln(1.6 \times 10^{12}) \approx 27.631 \]

6. **Substitute back to find \( \Delta G^\circ \)**:
\[ \Delta G^\circ = - (8.314)(298)(27.631) \]
\[ \Delta G^\circ \approx - 8.314 \times 298 \times 27.631 \approx - 68899.5 \, \text{J/mol} \approx -68.9 \, \text{kJ/mol} \]

7. **Set up the equation for the formation of NO2**:
The standard free energy of formation of NO2 can be expressed as:
\[ 2 \Delta G^\circ_f \text{ (NO}_2) = 2 \Delta G^\circ_f \text{ (NO)} + \Delta G^\circ \]
Rearranging gives:
\[ \Delta G^\circ_f \text{ (NO}_2) = \frac{1}{2} \left( 2 \Delta G^\circ_f \text{ (NO)} + \Delta G^\circ \right) \]

8. **Substituting known values**:
\[ \Delta G^\circ_f \text{ (NO}_2) = \frac{1}{2} \left( 2 \times 86600 \, \text{J/mol} - 68899.5 \, \text{J/mol} \right) \]
\[ = \frac{1}{2} \left( 173200 - 68899.5 \right) \]
\[ = \frac{1}{2} \left( 104300.5 \right) \approx 52150.25 \, \text{J/mol} \approx 52.15 \, \text{kJ/mol} \]

### Final Result:
The standard free energy of formation of NO2 (g) at 298 K is approximately:
\[ \Delta G^\circ_f \text{ (NO}_2) \approx 52.15 \, \text{kJ/mol} \]