For the reaction `N_(2)O_(4)(g) hArr 2NO_(2)(g)` the degree of dissociation at equilibrium is `0.2` at `1` atmospheric pressure. The equilibrium constant `K_(p)` will be

Options

A. `1//2`
B. `1//4`
C. `1//6`
D. `1//8`

Explanation

To find the equilibrium constant \( K_p \) for the reaction

\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

given that the degree of dissociation \( \alpha \) is \( 0.2 \) at \( 1 \) atmospheric pressure, we can follow these steps:

### Step 1: Initial Moles
Assume we start with 1 mole of \( N_2O_4 \) and no \( NO_2 \).

- Initial moles of \( N_2O_4 \) = 1
- Initial moles of \( NO_2 \) = 0

### Step 2: Moles at Equilibrium
The degree of dissociation \( \alpha \) is given as \( 0.2 \). This means that \( 20\% \) of \( N_2O_4 \) has dissociated.

- Moles of \( N_2O_4 \) that dissociate = \( 1 \times 0.2 = 0.2 \)
- Moles of \( N_2O_4 \) remaining = \( 1 - 0.2 = 0.8 \)

Since 1 mole of \( N_2O_4 \) produces 2 moles of \( NO_2 \):

- 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 3: Total Moles at Equilibrium
Total moles at equilibrium = moles of \( N_2O_4 \) + moles of \( NO_2 \)

\[ \text{Total moles} = 0.8 + 0.4 = 1.2 \]

### Step 4: Mole Fractions
Now, we can calculate the mole fractions of each component:

- 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 5: Partial Pressures
The total pressure \( P \) is given as \( 1 \) atm. We can find the partial pressures:

- Partial pressure of \( N_2O_4 \):

\[ P_{N_2O_4} = \text{Mole fraction of } N_2O_4 \times P = \frac{2}{3} \times 1 \text{ atm} = \frac{2}{3} \text{ atm} \]

- Partial pressure of \( NO_2 \):

\[ P_{NO_2} = \text{Mole fraction of } NO_2 \times P = \frac{1}{3} \times 1 \text{ atm} = \frac{1}{3} \text{ atm} \]

### Step 6: Calculate \( K_p \)
The equilibrium constant \( K_p \) is given by the expression:

\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]

Substituting the values we calculated:

\[ K_p = \frac{\left(\frac{1}{3}\right)^2}{\frac{2}{3}} = \frac{\frac{1}{9}}{\frac{2}{3}} = \frac{1}{9} \times \frac{3}{2} = \frac{1}{6} \]

### Final Answer
Thus, the equilibrium constant \( K_p \) is

\[ \boxed{\frac{1}{6}} \]