The following reaction is performed at 298 K ?
`2NO(g)+O_(2)(g)hArr2NO_(2)(g)`
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)=1.6xx10^(12))`

Options

A. `86600-(ln(1.6xx10^(12)))/(R(298))`
B. `0.5[2xx86,600-R(298ln(1.6xx10^(12))]`
C. `R(298ln(1.6xx10^(12))-86600`
D. `86600+R(298)ln(1.6xx10^(12))`

Answer

To find the standard free energy of formation of NO₂ (g) at 298 K for the reaction:

\[ 2 \text{NO}(g) + \text{O}_2(g) \rightleftharpoons 2 \text{NO}_2(g) \]

we will use the following thermodynamic relationships:

1. The relationship between Gibbs free energy change (\( \Delta G \)) and equilibrium constant (\( K_p \)):
\[ \Delta G^\circ = -RT \ln K_p \]

2. The relationship for the Gibbs free energy change of a reaction:
\[ \Delta G^\circ = \Delta G^\circ_{\text{products}} - \Delta G^\circ_{\text{reactants}} \]

### Step-by-Step Solution:

**Step 1: Identify the given data.**
- Standard free energy of formation of NO(g), \( \Delta G^\circ_{\text{NO}} = 86.6 \, \text{kJ/mol} = 86600 \, \text{J/mol} \)
- \( K_p = 1.6 \times 10^{12} \)
- Temperature, \( T = 298 \, \text{K} \)
- Gas constant, \( R = 8.314 \, \text{J/(mol K)} \)

**Step 2: Write the expression for \( \Delta G^\circ \) for the reaction.**
For the reaction:
\[ \Delta G^\circ = \Delta G^\circ_{\text{NO}_2} - [2 \Delta G^\circ_{\text{NO}} + 0] \]
Since the standard free energy of formation for O₂(g) is zero, we have:
\[ \Delta G^\circ = 2 \Delta G^\circ_{\text{NO}_2} - 2 \Delta G^\circ_{\text{NO}} \]

**Step 3: Substitute \( \Delta G^\circ \) from the first equation.**
From the first equation, we have:
\[ \Delta G^\circ = -RT \ln K_p \]
Substituting this into the equation from Step 2 gives:
\[ -RT \ln K_p = 2 \Delta G^\circ_{\text{NO}_2} - 2 \Delta G^\circ_{\text{NO}} \]

**Step 4: Rearrange to solve for \( \Delta G^\circ_{\text{NO}_2} \).**
\[ 2 \Delta G^\circ_{\text{NO}_2} = -RT \ln K_p + 2 \Delta G^\circ_{\text{NO}} \]
\[ \Delta G^\circ_{\text{NO}_2} = \frac{-RT \ln K_p + 2 \Delta G^\circ_{\text{NO}}}{2} \]

**Step 5: Calculate \( \Delta G^\circ_{\text{NO}_2} \).**
Substituting the values:
\[ \Delta G^\circ_{\text{NO}_2} = \frac{- (8.314 \, \text{J/(mol K)})(298 \, \text{K}) \ln(1.6 \times 10^{12}) + 2(86600 \, \text{J/mol})}{2} \]

**Step 6: Calculate \( -RT \ln K_p \).**
First, calculate \( \ln(1.6 \times 10^{12}) \):
\[ \ln(1.6 \times 10^{12}) \approx 27.63 \]
Now, calculate \( -RT \ln K_p \):
\[ -RT \ln K_p = -(8.314)(298)(27.63) \approx -68606.7 \, \text{J/mol} \]

**Step 7: Substitute back into the equation.**
\[ \Delta G^\circ_{\text{NO}_2} = \frac{-68606.7 + 173200}{2} \]
\[ \Delta G^\circ_{\text{NO}_2} = \frac{104593.3}{2} \approx 52296.65 \, \text{J/mol} \approx 52.3 \, \text{kJ/mol} \]

### Final Answer:
The standard free energy of formation of NO₂(g) at 298 K is approximately \( 52.3 \, \text{kJ/mol} \).