For the gasesous reaction, `2NO_(2)hArrN_(2)O_(4)`, calculate `DeltaG^(@)` and `K_(p)` for the reaction at `25^(@)C`. Given `G_(fN_(2)O_(4))^(@)` and `G_(fNO_(2))^(@)` are `97.82` and `51.30kJ` respectively. Also calculate `DeltaG^(0)` and `K_(p)` for reverse reaction.

Options

A. `6.88 atm`
B. `0.332 atm`
C. `3.12 atm`
D. `0.145 atm`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the reaction and identify the given data
The reaction given is:
\[ 2 \text{NO}_2 \rightleftharpoons \text{N}_2\text{O}_4 \]

Given data:
- \( G_f^\circ (\text{N}_2\text{O}_4) = 97.82 \, \text{kJ/mol} \)
- \( G_f^\circ (\text{NO}_2) = 51.30 \, \text{kJ/mol} \)

### Step 2: Calculate \( \Delta G^\circ \) for the reaction
Using the formula:
\[ \Delta G^\circ = G_f^\circ (\text{products}) - G_f^\circ (\text{reactants}) \]
For the reaction:
\[ \Delta G^\circ = G_f^\circ (\text{N}_2\text{O}_4) - 2 \times G_f^\circ (\text{NO}_2) \]
Substituting the values:
\[ \Delta G^\circ = 97.82 \, \text{kJ/mol} - 2 \times 51.30 \, \text{kJ/mol} \]
\[ \Delta G^\circ = 97.82 \, \text{kJ/mol} - 102.60 \, \text{kJ/mol} \]
\[ \Delta G^\circ = -4.78 \, \text{kJ/mol} \]

### Step 3: Calculate \( K_p \) for the forward reaction
Using the relationship between \( \Delta G^\circ \) and \( K_p \):
\[ \Delta G^\circ = -RT \ln K_p \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} = 0.008314 \, \text{kJ/(mol K)} \)
- \( T = 25^\circ C = 298 \, \text{K} \)

Rearranging the equation to find \( K_p \):
\[ K_p = e^{-\Delta G^\circ / (RT)} \]
Substituting the values:
\[ K_p = e^{-(-4.78 \, \text{kJ/mol}) / (0.008314 \, \text{kJ/(mol K)} \times 298 \, \text{K})} \]
Calculating:
\[ K_p = e^{4.78 / (0.008314 \times 298)} \]
\[ K_p = e^{4.78 / 2.478} \]
\[ K_p = e^{1.93} \approx 6.88 \, \text{atm} \]

### Step 4: Calculate \( \Delta G^\circ \) for the reverse reaction
For the reverse reaction:
\[ \Delta G^\circ_{\text{reverse}} = -\Delta G^\circ_{\text{forward}} = +4.78 \, \text{kJ/mol} \]

### Step 5: Calculate \( K_p \) for the reverse reaction
Using the relationship:
\[ K_p' = \frac{1}{K_p} \]
Substituting the value of \( K_p \):
\[ K_p' = \frac{1}{6.88} \approx 0.145 \, \text{atm} \]

### Final Results
- \( \Delta G^\circ \) for the forward reaction: \( -4.78 \, \text{kJ/mol} \)
- \( K_p \) for the forward reaction: \( 6.88 \, \text{atm} \)
- \( \Delta G^\circ \) for the reverse reaction: \( +4.78 \, \text{kJ/mol} \)
- \( K_p \) for the reverse reaction: \( 0.145 \, \text{atm} \)