In a container equilibrium `N_(2)O_(4) (g)hArr2NO_(2) (g)` is attained at `25^(@)C`. The total equilibrium pressure in container is 380 torr. If equilibrium constant of above equilibrium is `0.667` atm, then degree of dissociation of `N_(2)O_(4)` at this temperature will be:

Options

A. `(1)/(3)`
B. `(1)/(2)`
C. `(2)/(3)`
D. `(1)/(4)`

Explanation

To find the degree of dissociation of \( N_2O_4 \) at 25°C given the equilibrium reaction \( N_2O_4 (g) \rightleftharpoons 2NO_2 (g) \), total equilibrium pressure of 380 torr, and equilibrium constant \( K_p = 0.667 \) atm, we can follow these steps:

### Step 1: Convert Total Pressure to Atmospheres
The total pressure given is 380 torr. We need to convert this to atmospheres since the equilibrium constant is given in atm.

\[ P = \frac{380 \text{ torr}}{760 \text{ torr/atm}} = 0.5 \text{ atm} \]

### Step 2: Set Up the Initial and Equilibrium Conditions
Assume we start with 1 mole of \( N_2O_4 \) and let \( \alpha \) be the degree of dissociation. At equilibrium:
- Moles of \( N_2O_4 \) = \( 1 - \alpha \)
- Moles of \( NO_2 \) = \( 2\alpha \)

Total moles at equilibrium:
\[ \text{Total moles} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

### Step 3: Calculate Partial Pressures
The partial pressures of the gases can be expressed using the mole fractions:
- Partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = \left( \frac{1 - \alpha}{1 + \alpha} \right) P \]
- Partial pressure of \( NO_2 \):
\[ P_{NO_2} = \left( \frac{2\alpha}{1 + \alpha} \right) P \]

### Step 4: Substitute into the Equilibrium Expression
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]

Substituting the expressions for partial pressures:
\[ K_p = \frac{\left( \frac{2\alpha}{1 + \alpha} P \right)^2}{\left( \frac{1 - \alpha}{1 + \alpha} P \right)} \]

This simplifies to:
\[ K_p = \frac{4\alpha^2 P^2}{(1 - \alpha)(1 + \alpha) P} \]
\[ K_p = \frac{4\alpha^2 P}{1 - \alpha} \]

### Step 5: Substitute Known Values
Substituting \( K_p = 0.667 \) atm and \( P = 0.5 \) atm:
\[ 0.667 = \frac{4\alpha^2 (0.5)}{1 - \alpha} \]
\[ 0.667 = \frac{2\alpha^2}{1 - \alpha} \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ 0.667(1 - \alpha) = 2\alpha^2 \]
\[ 0.667 - 0.667\alpha = 2\alpha^2 \]
\[ 2\alpha^2 + 0.667\alpha - 0.667 = 0 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = 0.667, c = -0.667 \):
\[ \alpha = \frac{-0.667 \pm \sqrt{(0.667)^2 - 4 \cdot 2 \cdot (-0.667)}}{2 \cdot 2} \]
\[ \alpha = \frac{-0.667 \pm \sqrt{0.444889 + 5.336}}{4} \]
\[ \alpha = \frac{-0.667 \pm \sqrt{5.780889}}{4} \]
\[ \alpha = \frac{-0.667 \pm 2.404}{4} \]

Calculating the two possible values:
1. \( \alpha = \frac{1.737}{4} \approx 0.43425 \)
2. \( \alpha = \frac{-3.071}{4} \) (not valid since degree of dissociation cannot be negative)

### Final Result
Thus, the degree of dissociation \( \alpha \) is approximately:
\[ \alpha \approx 0.434 \]