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

Options

A. `R (298)In (1.6 xx 10^12) - 86600`
B. `866000+ R (298) In (1.6xx10^(12))`
C. `866000+(1.6xx10^(12))/(R(298))`
D. `0.5[2xx86600- R(298) In (1.6xx 10^(12))]`

Answer

To find the standard free energy of formation of \( NO_2(g) \) at 298 K, we can follow these steps:

### Step 1: Write the Reaction
The formation of \( NO_2(g) \) from \( NO(g) \) and \( O_2(g) \) can be represented by the following balanced chemical equation:
\[ 2 NO(g) + O_2(g) \rightarrow 2 NO_2(g) \]

### Step 2: Identify Given Data
From the question, we know:
- The standard free energy of formation of \( NO(g) \) is \( \Delta G_f^0 (NO) = 86.6 \, \text{kJ/mol} \)
- The equilibrium constant \( K_p \) for the reaction is \( 1.6 \times 10^{12} \)

### Step 3: Use the Relationship Between \( \Delta G^0 \) and \( K_p \)
The relationship between the standard free energy change (\( \Delta G^0 \)) and the equilibrium constant (\( K_p \)) is given by the equation:
\[ \Delta G^0 = -RT \ln K_p \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} \) (convert to kJ by dividing by 1000)
- \( T = 298 \, \text{K} \)
- \( K_p = 1.6 \times 10^{12} \)

### Step 4: Calculate \( \Delta G^0 \) for the Reaction
First, we need to calculate \( RT \ln K_p \):
1. Convert \( R \) to kJ:
\[ R = 8.314 \, \text{J/(mol K)} = 0.008314 \, \text{kJ/(mol K)} \]
2. Calculate \( \ln K_p \):
\[ \ln(1.6 \times 10^{12}) \approx 27.63 \]
3. Now, calculate \( RT \ln K_p \):
\[ RT \ln K_p = 0.008314 \times 298 \times 27.63 \approx 69.00 \, \text{kJ} \]

Now, substitute this value into the equation for \( \Delta G^0 \):
\[ \Delta G^0 = -69.00 \, \text{kJ} \]

### Step 5: Relate \( \Delta G^0 \) to the Formation of \( NO_2 \)
Using the equation for the standard free energy change of the reaction:
\[ \Delta G^0 = \sum \Delta G_f^0 \text{(products)} - \sum \Delta G_f^0 \text{(reactants)} \]
For our reaction:
\[ \Delta G^0 = 2 \Delta G_f^0 (NO_2) - [2 \Delta G_f^0 (NO) + 0] \]
Substituting the known values:
\[ -69.00 = 2 \Delta G_f^0 (NO_2) - 2 \times 86.6 \]
\[ -69.00 = 2 \Delta G_f^0 (NO_2) - 173.2 \]
\[ 2 \Delta G_f^0 (NO_2) = -69.00 + 173.2 \]
\[ 2 \Delta G_f^0 (NO_2) = 104.2 \]
\[ \Delta G_f^0 (NO_2) = \frac{104.2}{2} = 52.1 \, \text{kJ/mol} \]

### Final Answer
The standard free energy of formation of \( NO_2(g) \) at 298 K is:
\[ \Delta G_f^0 (NO_2) = 52.1 \, \text{kJ/mol} \]