The degree of dissociation of `N_(2) O_(4)` into `NO_(2)` at one atmosphere and `40^(@)`C is 0.310. calculate Kp at `40^(@)`C

Options

A. Kp 0.42 atm
B. Kc =0.003 mole `lit^(-1)`
C. Degree of dissociation at 10 atm pressure at same temperature `alpha` = 0.1025
D. Degree of dissociation `alpha` decreases with increase pressure a given temperature

Answer

To solve the problem of calculating the equilibrium constant \( K_p \) for the dissociation of \( N_2O_4 \) into \( NO_2 \) at \( 40^\circ C \) given a degree of dissociation of \( 0.310 \), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Equation**:
The dissociation of \( N_2O_4 \) can be represented as:
\[ N_2O_4 (g) \rightleftharpoons 2 NO_2 (g) \]

2. **Define Initial Conditions**:
Assume we start with 1 mole of \( N_2O_4 \) and no \( NO_2 \):
- Initial moles of \( N_2O_4 = 1 \)
- Initial moles of \( NO_2 = 0 \)

3. **Calculate Moles at Equilibrium**:
Given the degree of dissociation \( \alpha = 0.310 \):
- Moles of \( N_2O_4 \) that dissociate = \( \alpha \times 1 = 0.310 \)
- Moles of \( N_2O_4 \) remaining = \( 1 - 0.310 = 0.690 \)
- Moles of \( NO_2 \) produced = \( 2 \times 0.310 = 0.620 \)

4. **Determine Total Moles at Equilibrium**:
Total moles at equilibrium:
\[ \text{Total moles} = \text{Moles of } N_2O_4 + \text{Moles of } NO_2 = 0.690 + 0.620 = 1.310 \]

5. **Calculate Partial Pressures**:
The total pressure is given as \( 1 \, \text{atm} \).

- Partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = \left( \frac{0.690}{1.310} \right) \times 1 \, \text{atm} = 0.5267 \, \text{atm} \approx 0.53 \, \text{atm} \]

- Partial pressure of \( NO_2 \):
\[ P_{NO_2} = \left( \frac{0.620}{1.310} \right) \times 1 \, \text{atm} = 0.4735 \, \text{atm} \approx 0.47 \, \text{atm} \]

6. **Calculate \( K_p \)**:
The expression for \( K_p \) is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]
Substituting the values:
\[ K_p = \frac{(0.47)^2}{0.53} = \frac{0.2209}{0.53} \approx 0.416 \approx 0.42 \]

### Final Answer:
Thus, the value of \( K_p \) at \( 40^\circ C \) is approximately \( 0.42 \).

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