For the reaction , `N_2O_4(g)hArr2NO_2(g)` the degree of dissociation at equilibrium is 0.4 at a pressure of 1 atm. The value of `K_p` is

Options

A. 0.64 atm
B. 0.60 atm
C. 0.19 atm
D. 0.762 atm

Answer

To find the value of \( K_p \) for the reaction \( N_2O_4(g) \rightleftharpoons 2NO_2(g) \) given that the degree of dissociation \( \alpha \) is 0.4 at a pressure of 1 atm, we can follow these steps:

### Step 1: Define the initial moles and degree of dissociation
Let’s assume we start with 1 mole of \( N_2O_4 \). The degree of dissociation \( \alpha \) is given as 0.4, which means that 40% of \( N_2O_4 \) dissociates.

### Step 2: Calculate the moles at equilibrium
- Initial moles of \( N_2O_4 \) = 1 mole
- Moles of \( N_2O_4 \) dissociated = \( \alpha \times 1 = 0.4 \) moles
- Moles of \( N_2O_4 \) remaining = \( 1 - 0.4 = 0.6 \) moles
- Moles of \( NO_2 \) produced = \( 2 \times \text{(moles of } N_2O_4 \text{ dissociated)} = 2 \times 0.4 = 0.8 \) moles

At equilibrium:
- Moles of \( N_2O_4 \) = 0.6
- Moles of \( NO_2 \) = 0.8

### Step 3: Calculate total moles at equilibrium
Total moles at equilibrium = Moles of \( N_2O_4 \) + Moles of \( NO_2 \) = \( 0.6 + 0.8 = 1.4 \) moles.

### Step 4: Calculate partial pressures
Using the ideal gas law, we can find the partial pressures:
- Total pressure \( P = 1 \) atm.

Partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = \frac{\text{moles of } N_2O_4}{\text{total moles}} \times P = \frac{0.6}{1.4} \times 1 = \frac{0.6}{1.4} \text{ atm} \]

Partial pressure of \( NO_2 \):
\[ P_{NO_2} = \frac{\text{moles of } NO_2}{\text{total moles}} \times P = \frac{0.8}{1.4} \times 1 = \frac{0.8}{1.4} \text{ atm} \]

### Step 5: Calculate \( K_p \)
The expression for \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]

Substituting the partial pressures:
\[ K_p = \frac{\left(\frac{0.8}{1.4}\right)^2}{\frac{0.6}{1.4}} = \frac{(0.64/1.96)}{(0.6/1.4)} = \frac{0.64 \times 1.4}{0.6 \times 1.96} \]

Calculating the above:
\[ K_p = \frac{0.896}{1.176} = \frac{896}{1176} = \frac{224}{294} = \frac{16}{21} \]

### Final Answer
Thus, the value of \( K_p \) is \( \frac{16}{21} \).