For the reaction `N_(2)O_(4)hArr 2NO_(2(g)),` the degree of dissociation of `N_(2)O_(4)` is 0.2 at 1 atm. Then the `K_(p)` of `2NO_(2)hArr N_(2)O_(4)` is

Answer

To find the equilibrium constant \( K_p \) for the reaction \( 2NO_2 \rightleftharpoons N_2O_4 \) given the degree of dissociation of \( N_2O_4 \) is 0.2 at 1 atm, we can follow these steps:

### Step 1: Understand the Reaction
The reaction is:
\[ N_2O_4 \rightleftharpoons 2NO_2 \]

### Step 2: Define Initial Conditions
Assume we start with 1 mole of \( N_2O_4 \) at time \( t = 0 \):
- Initial moles of \( N_2O_4 = 1 \)
- Initial moles of \( NO_2 = 0 \)

### Step 3: Calculate Moles at Equilibrium
Given the degree of dissociation (\( \alpha \)) of \( N_2O_4 \) is 0.2, this means:
- Moles of \( N_2O_4 \) that dissociate = \( \alpha \times \text{initial moles} = 0.2 \times 1 = 0.2 \)
- Moles of \( N_2O_4 \) remaining = \( 1 - 0.2 = 0.8 \)
- Moles of \( NO_2 \) produced = \( 2 \times 0.2 = 0.4 \)

Thus, at equilibrium:
- Moles of \( N_2O_4 = 0.8 \)
- Moles of \( NO_2 = 0.4 \)

### Step 4: Calculate Total Moles at Equilibrium
Total moles at equilibrium:
\[ \text{Total moles} = 0.8 + 0.4 = 1.2 \]

### Step 5: Calculate Mole Fractions
- Mole fraction of \( N_2O_4 \):
\[ \text{Mole fraction of } N_2O_4 = \frac{0.8}{1.2} = \frac{2}{3} \]

- Mole fraction of \( NO_2 \):
\[ \text{Mole fraction of } NO_2 = \frac{0.4}{1.2} = \frac{1}{3} \]

### Step 6: Calculate Partial Pressures
Given the total pressure \( P = 1 \) atm:
- Partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = \text{Mole fraction of } N_2O_4 \times P = \frac{2}{3} \times 1 = \frac{2}{3} \text{ atm} \]

- Partial pressure of \( NO_2 \):
\[ P_{NO_2} = \text{Mole fraction of } NO_2 \times P = \frac{1}{3} \times 1 = \frac{1}{3} \text{ atm} \]

### Step 7: Write the Expression for \( K_p \)
The expression for \( K_p \) for the reaction \( 2NO_2 \rightleftharpoons N_2O_4 \) is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]

### Step 8: Substitute the Values
Substituting the partial pressures:
\[ K_p = \frac{\left(\frac{1}{3}\right)^2}{\frac{2}{3}} = \frac{\frac{1}{9}}{\frac{2}{3}} = \frac{1}{9} \times \frac{3}{2} = \frac{1}{6} \]

### Step 9: Find \( K_p \) for the Reverse Reaction
Since we need \( K_p \) for the reaction \( 2NO_2 \rightleftharpoons N_2O_4 \), we take the reciprocal:
\[ K_p = \frac{1}{\frac{1}{6}} = 6 \]

### Final Answer
Thus, the value of \( K_p \) for the reaction \( 2NO_2 \rightleftharpoons N_2O_4 \) is:
\[ \boxed{6} \]