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 \( N_2O_4(g) \rightleftharpoons 2NO_2(g) \) given the degree of dissociation \( \alpha = 0.2 \) at a total pressure of 1 atm, we can follow these steps:

### Step 1: Define Initial Conditions
Assume we start with 1 mole of \( N_2O_4 \) and no \( NO_2 \) at the beginning (initially at \( t = 0 \)):
- \( N_2O_4 = 1 \) mole
- \( NO_2 = 0 \) moles

### Step 2: Define Change at Equilibrium
At equilibrium, the degree of dissociation \( \alpha = 0.2 \). This means that 20% of \( N_2O_4 \) has dissociated:
- Moles of \( N_2O_4 \) at equilibrium = \( 1 - \alpha = 1 - 0.2 = 0.8 \) moles
- Moles of \( NO_2 \) produced = \( 2 \times \alpha = 2 \times 0.2 = 0.4 \) moles

### Step 3: Calculate Total Moles at Equilibrium
Total moles at equilibrium:
\[ \text{Total moles} = \text{moles of } N_2O_4 + \text{moles of } NO_2 = 0.8 + 0.4 = 1.2 \text{ moles} \]

### Step 4: Calculate Mole Fractions
Now, we can calculate the mole fractions of each species:
- Mole fraction of \( N_2O_4 \):
\[ X_{N_2O_4} = \frac{0.8}{1.2} = \frac{2}{3} \]
- Mole fraction of \( NO_2 \):
\[ X_{NO_2} = \frac{0.4}{1.2} = \frac{1}{3} \]

### Step 5: Calculate Partial Pressures
Using the total pressure \( P_{total} = 1 \) atm, we can find the partial pressures:
- Partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = X_{N_2O_4} \times P_{total} = \frac{2}{3} \times 1 = \frac{2}{3} \text{ atm} \]
- Partial pressure of \( NO_2 \):
\[ P_{NO_2} = X_{NO_2} \times P_{total} = \frac{1}{3} \times 1 = \frac{1}{3} \text{ atm} \]

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

### Step 7: Substitute the Partial Pressures
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 \( K_p \) for the reaction is:
\[ K_p = \frac{1}{6} \]

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