For the reaction `N_(2)O_(4)(g)hArr2NO_(2)(g)`, the degree of dissociation at equilibrium is `0.2` at `1` atm 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 \( N_2O_4(g) \rightleftharpoons 2 NO_2(g) \) given that the degree of dissociation \( \alpha \) is 0.2 at a total pressure of 1 atm, we can follow these steps:

### Step 1: Define Initial Conditions
Assume we start with 1 mole of \( N_2O_4 \) initially.

### Step 2: Determine Moles at Equilibrium
At equilibrium, if \( \alpha \) moles of \( N_2O_4 \) dissociate, then:
- Moles of \( N_2O_4 \) remaining = \( 1 - \alpha \)
- Moles of \( NO_2 \) formed = \( 2\alpha \)

Substituting \( \alpha = 0.2 \):
- Moles of \( N_2O_4 \) remaining = \( 1 - 0.2 = 0.8 \)
- Moles of \( NO_2 \) formed = \( 2 \times 0.2 = 0.4 \)

### Step 3: Calculate 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: Calculate Mole Fractions
- Mole fraction of \( N_2O_4 \):
\[ \chi_{N_2O_4} = \frac{0.8}{1.2} \]
- Mole fraction of \( NO_2 \):
\[ \chi_{NO_2} = \frac{0.4}{1.2} \]

### Step 5: Calculate Partial Pressures
Using the total pressure \( P = 1 \) atm:
- Partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = \chi_{N_2O_4} \times P = \frac{0.8}{1.2} \times 1 = \frac{0.8}{1.2} \]
- Partial pressure of \( NO_2 \):
\[ P_{NO_2} = \chi_{NO_2} \times P = \frac{0.4}{1.2} \times 1 = \frac{0.4}{1.2} \]

### Step 6: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]

Substituting the values:
\[ K_p = \frac{\left(\frac{0.4}{1.2}\right)^2}{\frac{0.8}{1.2}} = \frac{\frac{0.16}{1.44}}{\frac{0.8}{1.2}} = \frac{0.16 \times 1.2}{0.8 \times 1.44} \]

### Step 7: Simplify the Expression
Calculating the numerator and denominator:
\[ K_p = \frac{0.192}{1.152} = \frac{192}{1152} = \frac{1}{6} \]

### Conclusion
Thus, the equilibrium constant \( K_p \) for the reaction is:
\[ \boxed{\frac{1}{6}} \]