At `27^@C` and 1 atmosphere pressure `N_2O_4` is 20% dissociated into `NO_2` find `K_P`

Options

A. 0.2
B. 0.166
C. 0.15
D. 0.1

Explanation

To find the equilibrium constant \( K_p \) for the dissociation of \( N_2O_4 \) into \( NO_2 \) at 27°C and 1 atmosphere pressure, we can follow these steps:

### Step 1: Write the balanced chemical equation
The dissociation of \( N_2O_4 \) can be represented as:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

### Step 2: Define the degree of dissociation
Given that \( N_2O_4 \) is 20% dissociated, we can denote the degree of dissociation as \( \alpha = 0.20 \).

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

At equilibrium, the moles will change as follows:
- Moles of \( N_2O_4 \) at equilibrium = \( 1 - \alpha = 1 - 0.20 = 0.80 \)
- Moles of \( NO_2 \) at equilibrium = \( 2\alpha = 2 \times 0.20 = 0.40 \)

### Step 4: Calculate total moles at equilibrium
Total moles at equilibrium:
\[ \text{Total moles} = \text{moles of } N_2O_4 + \text{moles of } NO_2 = 0.80 + 0.40 = 1.20 \]

### Step 5: Calculate partial pressures
Using the total pressure \( P = 1 \) atm, we can calculate the partial pressures of \( N_2O_4 \) and \( NO_2 \).

**Partial pressure of \( NO_2 \)**:
\[ P_{NO_2} = \frac{\text{moles of } NO_2}{\text{total moles}} \times P = \frac{0.40}{1.20} \times 1 = 0.333 \, \text{atm} \]

**Partial pressure of \( N_2O_4 \)**:
\[ P_{N_2O_4} = \frac{\text{moles of } N_2O_4}{\text{total moles}} \times P = \frac{0.80}{1.20} \times 1 = 0.667 \, \text{atm} \]

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

### Step 7: Substitute the values into the expression
Substituting the partial pressures:
\[ K_p = \frac{(0.333)^2}{0.667} \]

### Step 8: Calculate \( K_p \)
Calculating the above expression:
\[ K_p = \frac{0.111}{0.667} \approx 0.166 \]

### Final Answer
Thus, the value of \( K_p \) is approximately \( 0.166 \, \text{atm} \).

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