For the equilibrium, `N_(2) O_(4) hArr 2NO_(2) , (G^(@)N_(2)O_(4))_(298)` = 100 kJ/mole and `(G^(@) NO_(2))_(298)` = 50 kj/mole . When 5 mole / lit of each is taken, the value of `Delta ` G for the reaction.

Options

A. 3.99 kJ `"mole"^(-1)`
B. 39.9 kJ `"mole"^(-1)`
C. 3.99 j `"mole"^(-1)`
D. 39.9 J `"mole"^(-1)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the reaction and identify the standard free energy changes.
The equilibrium reaction is:
\[ N_2O_4 \rightleftharpoons 2NO_2 \]

Given:
- Standard free energy of formation for \( N_2O_4 \) at 298 K, \( \Delta G^\circ (N_2O_4) = 100 \, \text{kJ/mol} \)
- Standard free energy of formation for \( NO_2 \) at 298 K, \( \Delta G^\circ (NO_2) = 50 \, \text{kJ/mol} \)

### Step 2: Calculate the standard change in free energy (\( \Delta G^\circ \)) for the reaction.
Using the formula:
\[ \Delta G^\circ = \sum \Delta G^\circ_{\text{products}} - \sum \Delta G^\circ_{\text{reactants}} \]
For our reaction:
\[ \Delta G^\circ = 2 \times \Delta G^\circ (NO_2) - \Delta G^\circ (N_2O_4) \]
Substituting the values:
\[ \Delta G^\circ = 2 \times 50 \, \text{kJ/mol} - 100 \, \text{kJ/mol} = 100 \, \text{kJ/mol} - 100 \, \text{kJ/mol} = 0 \, \text{kJ/mol} \]

### Step 3: Calculate the reaction quotient (\( Q \)).
The reaction quotient \( Q \) is given by:
\[ Q = \frac{[NO_2]^2}{[N_2O_4]} \]
Given that the concentrations are 5 moles per liter for both \( N_2O_4 \) and \( NO_2 \):
\[ Q = \frac{(5)^2}{5} = \frac{25}{5} = 5 \]

### Step 4: Use the Gibbs free energy equation to find \( \Delta G \).
The equation relating \( \Delta G \) to \( \Delta G^\circ \) and \( Q \) is:
\[ \Delta G = \Delta G^\circ + RT \ln Q \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} = 0.008314 \, \text{kJ/(mol K)} \)
- \( T = 298 \, \text{K} \)

Substituting the values:
\[ \Delta G = 0 + (0.008314 \, \text{kJ/(mol K)} \times 298 \, \text{K}) \ln(5) \]
Calculating \( \ln(5) \):
\[ \ln(5) \approx 1.609 \]
Now substituting this back:
\[ \Delta G = 0.008314 \times 298 \times 1.609 \approx 3.994 \, \text{kJ/mol} \]

### Step 5: Final answer.
Thus, the value of \( \Delta G \) for the reaction is approximately:
\[ \Delta G \approx 3.994 \, \text{kJ/mol} \]