For the reaction, `N_(2)O_(4)(g)hArr 2NO_(2)(g)` the degree of dissociation at equilibrium is 0.l4 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 solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction given is:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

### Step 2: Define the degree of dissociation
Let the degree of dissociation (α) be 0.14. This means that 14% of the initial moles of \( N_2O_4 \) dissociate into \( NO_2 \).

### Step 3: Set up the initial and equilibrium moles
Assume we start with 1 mole of \( N_2O_4 \) and 0 moles of \( NO_2 \):
- Initial moles:
- \( N_2O_4 = 1 \)
- \( NO_2 = 0 \)

At equilibrium, the moles will be:
- Moles of \( N_2O_4 \) at equilibrium = \( 1 - \alpha = 1 - 0.14 = 0.86 \)
- Moles of \( NO_2 \) at equilibrium = \( 2\alpha = 2 \times 0.14 = 0.28 \)

### Step 4: Calculate total moles at equilibrium
Total moles at equilibrium:
\[ \text{Total moles} = \text{moles of } N_2O_4 + \text{moles of } NO_2 = 0.86 + 0.28 = 1.14 \]

### Step 5: Calculate partial pressures
Given that the total pressure \( P = 1 \) atm, we can calculate the partial pressures:
- Partial pressure of \( NO_2 \):
\[ P_{NO_2} = \left(\frac{\text{moles of } NO_2}{\text{total moles}}\right) \times P = \left(\frac{0.28}{1.14}\right) \times 1 = 0.2456 \text{ atm} \]

- Partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = \left(\frac{\text{moles of } N_2O_4}{\text{total moles}}\right) \times P = \left(\frac{0.86}{1.14}\right) \times 1 = 0.7544 \text{ atm} \]

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

### Step 7: Substitute the values into the \( K_p \) expression
Substituting the values we found:
\[ K_p = \frac{(0.2456)^2}{0.7544} \]

### Step 8: Calculate \( K_p \)
Calculating \( K_p \):
\[ K_p = \frac{0.0604}{0.7544} \approx 0.0799 \]

### Step 9: Round off the answer
Rounding off \( K_p \) gives:
\[ K_p \approx 0.08 \]

### Final Answer
The value of \( K_p \) is approximately **0.08**.

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