For the reaction: `N_(2)O_(4)(g) hArr 2NO_(2)(g)`
`(i)" "`In a mixture of `5 mol NO_(2)` and `5` mol `N_(2)O_(4)` and pressure of 20 bar. Calculate the value of `DeltaG` for the reaction. Given `DeltaG_(f)^(@) (NO_(2)) = 50` KJ/mol, `DeltaG_(f)^(@) ((N_(2)O_(4)) =100` KJ/mol and T=298 K. `(ii)` Predict the direction in which the reaction will shift, in order to attain equilibrium
[Given at `T=298 K, 2.303 "RT" = 5.7`KJ/mol.]

Explanation

To solve the problem step by step, we will calculate the Gibbs free energy change (ΔG) for the reaction and then predict the direction of the reaction to reach equilibrium.

### Step 1: Calculate ΔG° (Standard Gibbs Free Energy Change)

The reaction is:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

The standard Gibbs free energy change (ΔG°) can be calculated using the formula:
\[ \Delta G^\circ = \sum \Delta G_f^\circ \text{(products)} - \sum \Delta G_f^\circ \text{(reactants)} \]

Given:
- \( \Delta G_f^\circ (NO_2) = 50 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ (N_2O_4) = 100 \, \text{kJ/mol} \)

Substituting the values:
\[ \Delta G^\circ = [2 \times \Delta G_f^\circ (NO_2)] - [\Delta G_f^\circ (N_2O_4)] \]
\[ \Delta G^\circ = [2 \times 50] - [100] = 100 - 100 = 0 \, \text{kJ/mol} \]

### Step 2: Calculate the Reaction Quotient (Q)

The reaction quotient \( Q \) is defined as:
\[ Q = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]

Given that there are 5 moles of \( NO_2 \) and 5 moles of \( N_2O_4 \) in a total pressure of 20 bar, the partial pressures can be calculated as follows:

Since there are equal moles of \( NO_2 \) and \( N_2O_4 \):
\[ P_{NO_2} = P_{N_2O_4} = \frac{20 \, \text{bar}}{2} = 10 \, \text{bar} \]

Now substituting into the expression for \( Q \):
\[ Q = \frac{(10 \, \text{bar})^2}{10 \, \text{bar}} = \frac{100}{10} = 10 \]

### Step 3: Calculate ΔG using the Gibbs Free Energy Equation

The Gibbs free energy change (ΔG) can be calculated using the equation:
\[ \Delta G = \Delta G^\circ - 2.303 RT \log Q \]

Where:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 298 \, \text{K} \)
- \( 2.303 RT = 5.7 \, \text{kJ/mol} \)

Substituting the values:
\[ \Delta G = 0 - 5.7 \, \text{kJ/mol} \cdot \log(10) \]
\[ \Delta G = -5.7 \, \text{kJ/mol} \cdot 1 = -5.7 \, \text{kJ/mol} \]

### Step 4: Predict the Direction of the Reaction

Since \( \Delta G \) is negative (\( -5.7 \, \text{kJ/mol} \)), the reaction is spontaneous in the forward direction.

### Summary of Results
1. **ΔG** = -5.7 kJ/mol
2. The reaction will shift to the right (towards the formation of \( NO_2 \)) to reach equilibrium.