For the reaction, `N_(2)O_(4)(g)hArr 2NO_(2)(g)` the degree of dissociation at equilibrium is 0.14 at a pressure of 1 atm. The value of `K_(p)` is

Options

A. 0.381 atm
B. 0.80 atm
C. 0.762 atm
D. 0.195 atm

Answer

To find the value of \( K_p \) for the reaction \( N_2O_4(g) \rightleftharpoons 2NO_2(g) \) given the degree of dissociation \( \alpha = 0.14 \) at a pressure of 1 atm, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Reaction and Initial Conditions:**
- The reaction is \( N_2O_4(g) \rightleftharpoons 2NO_2(g) \).
- Let the initial amount of \( N_2O_4 \) be 1 mole (since we are looking at mole fractions, we can assume 1 mole for simplicity).

2. **Determine the Change in Moles at Equilibrium:**
- The degree of dissociation \( \alpha = 0.14 \) means that 14% of \( N_2O_4 \) has dissociated.
- At equilibrium, the amount of \( N_2O_4 \) remaining will be \( 1 - \alpha = 1 - 0.14 = 0.86 \) moles.
- The amount of \( NO_2 \) produced will be \( 2\alpha = 2 \times 0.14 = 0.28 \) moles.

3. **Calculate Total Moles at Equilibrium:**
- Total moles at equilibrium = moles of \( N_2O_4 \) + moles of \( NO_2 \)
- Total moles = \( 0.86 + 0.28 = 1.14 \) moles.

4. **Calculate Mole Fractions:**
- Mole fraction of \( N_2O_4 \):
\[ \text{Mole fraction of } N_2O_4 = \frac{0.86}{1.14} \approx 0.7544 \]
- Mole fraction of \( NO_2 \):
\[ \text{Mole fraction of } NO_2 = \frac{0.28}{1.14} \approx 0.2456 \]

5. **Calculate Partial Pressures:**
- Since the total pressure is given as 1 atm:
- Partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = \text{Mole fraction of } N_2O_4 \times \text{Total Pressure} = 0.7544 \times 1 \, \text{atm} = 0.7544 \, \text{atm} \]
- Partial pressure of \( NO_2 \):
\[ P_{NO_2} = \text{Mole fraction of } NO_2 \times \text{Total Pressure} = 0.2456 \times 1 \, \text{atm} = 0.2456 \, \text{atm} \]

6. **Calculate \( K_p \):**
- The expression for \( K_p \) for the reaction is:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]
- Substituting the values:
\[ K_p = \frac{(0.2456)^2}{0.7544} \approx \frac{0.0600}{0.7544} \approx 0.0796 \]

### Final Answer:
\[ K_p \approx 0.0796 \]