For the equilibrium at `298 K, N_(2)O_(4)(g) hArr 2NO_(2)(g), G_(N_(2)O_(4))^(ɵ)=100 kJ mol^(-1)` and `G_(NO_(2))^(ɵ)=50 kJ mol^(-1)`. If `5` mol of `N_(2)O_(4)` and `2` moles of `NO_(2)` are taken initially in one litre container than which statement are correct.

Options

A. reaction proceeds in forward direction
B. `K_(c)=1`
C. `DeltaG=-0.55 KJ, DeltaG^(ɵ)=0`
D. At equilibrium `[N_(2)O_(4)]=4.84 M` and `[NO_(2)]=0.212 M`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the Reaction and Given Data
The equilibrium reaction is:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

Given:
- Standard Gibbs free energy of formation for \( N_2O_4 \): \( G^\circ_{N_2O_4} = 100 \, \text{kJ/mol} \)
- Standard Gibbs free energy of formation for \( NO_2 \): \( G^\circ_{NO_2} = 50 \, \text{kJ/mol} \)
- Initial moles of \( N_2O_4 = 5 \, \text{mol} \)
- Initial moles of \( NO_2 = 2 \, \text{mol} \)

### Step 2: Calculate Standard Gibbs Free Energy Change (\( \Delta G^\circ \))
The standard Gibbs free energy change for the reaction can be calculated using:
\[ \Delta G^\circ = \sum G^\circ_{\text{products}} - \sum G^\circ_{\text{reactants}} \]
For our reaction:
\[ \Delta G^\circ = 2 \cdot G^\circ_{NO_2} - G^\circ_{N_2O_4} = 2 \cdot 50 \, \text{kJ/mol} - 100 \, \text{kJ/mol} = 100 - 100 = 0 \, \text{kJ/mol} \]

### Step 3: Calculate the Reaction Quotient (\( Q \))
The reaction quotient \( Q \) is given by:
\[ Q = \frac{[NO_2]^2}{[N_2O_4]} \]
Since we have 5 moles of \( N_2O_4 \) and 2 moles of \( NO_2 \) in a 1 L container:
\[ Q = \frac{(2)^2}{5} = \frac{4}{5} = 0.8 \]

### Step 4: Calculate Gibbs Free Energy Change (\( \Delta G \))
Using the equation:
\[ \Delta G = \Delta G^\circ + RT \ln Q \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} = 0.008314 \, \text{kJ/(mol K)} \)
- \( T = 298 \, \text{K} \)

Calculating \( RT \):
\[ RT = 0.008314 \cdot 298 = 2.478 \, \text{kJ/mol} \]

Now substituting into the equation:
\[ \Delta G = 0 + 2.478 \ln(0.8) \]
Calculating \( \ln(0.8) \):
\[ \ln(0.8) \approx -0.2231 \]
Thus:
\[ \Delta G = 2.478 \cdot (-0.2231) \approx -0.552 \, \text{kJ/mol} \]

### Step 5: Determine Spontaneity and Equilibrium Constant (\( K_c \))
Since \( \Delta G < 0 \), the reaction is spontaneous in the forward direction.

For equilibrium, we know:
\[ \Delta G^\circ = -RT \ln K_c \]
Since \( \Delta G^\circ = 0 \):
\[ 0 = -RT \ln K_c \implies K_c = 1 \]

### Step 6: Calculate Equilibrium Concentrations
Let \( x \) be the amount of \( N_2O_4 \) that dissociates at equilibrium:
- Moles of \( N_2O_4 \) at equilibrium: \( 5 - x \)
- Moles of \( NO_2 \) at equilibrium: \( 2 + 2x \)

Setting up the equilibrium expression:
\[ K_c = \frac{(2 + 2x)^2}{(5 - x)} = 1 \]

Assuming \( x \) is small compared to 5, we can simplify:
\[ \frac{(2 + 2x)^2}{5} \approx 1 \]
Solving gives:
\[ (2 + 2x)^2 = 5 \implies 2 + 2x = \sqrt{5} \implies 2x = \sqrt{5} - 2 \implies x \approx 0.618 \]

Calculating concentrations:
- \( [N_2O_4] = 5 - x \approx 4.382 \, \text{mol/L} \)
- \( [NO_2] = 2 + 2x \approx 3.236 \, \text{mol/L} \)

### Conclusion
The correct statements are:
1. \( \Delta G < 0 \) indicates the reaction is spontaneous.
2. \( K_c = 1 \) at equilibrium.
3. The calculated equilibrium concentrations are approximately \( 4.382 \, \text{mol/L} \) for \( N_2O_4 \) and \( 3.236 \, \text{mol/L} \) for \( NO_2 \).