Ina `0.25` litre tube dissociation of 4 moles of NO takes place. If its degree of dissociation is `10%`. The value of `K_(p)` for reaction `2NO hArrN_(2) + O_(2)` is:

Options

A. `(1)/((18)^(2))`
B. `(1)/((8)^(2))`
C. `(1)/(16)`
D. `(1)/(32)`

Explanation

To solve the problem, we need to find the value of \( K_p \) for the reaction:

\[ 2 \text{NO} \rightleftharpoons \text{N}_2 + \text{O}_2 \]

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Initial moles of NO = 4 moles
- Volume of the tube = 0.25 L
- Degree of dissociation (\( \alpha \)) = 10% = 0.1

2. **Calculate Moles at Equilibrium:**
- The change in moles of NO due to dissociation:
\[ \text{Change in moles of NO} = 4 \times \alpha = 4 \times 0.1 = 0.4 \text{ moles} \]
- Moles of NO at equilibrium:
\[ \text{Moles of NO at equilibrium} = 4 - 0.4 = 3.6 \text{ moles} \]
- Moles of N2 produced:
\[ \text{Moles of N2} = \frac{0.4}{2} = 0.2 \text{ moles} \]
- Moles of O2 produced:
\[ \text{Moles of O2} = 0.2 \text{ moles} \]

3. **Calculate Equilibrium Concentrations:**
- Concentration of NO:
\[ [\text{NO}] = \frac{3.6 \text{ moles}}{0.25 \text{ L}} = 14.4 \text{ M} \]
- Concentration of N2:
\[ [\text{N2}] = \frac{0.2 \text{ moles}}{0.25 \text{ L}} = 0.8 \text{ M} \]
- Concentration of O2:
\[ [\text{O2}] = \frac{0.2 \text{ moles}}{0.25 \text{ L}} = 0.8 \text{ M} \]

4. **Calculate \( K_c \):**
- The expression for \( K_c \) is:
\[ K_c = \frac{[\text{N2}][\text{O2}]}{[\text{NO}]^2} \]
- Substituting the equilibrium concentrations:
\[ K_c = \frac{(0.8)(0.8)}{(14.4)^2} \]
- Calculate \( K_c \):
\[ K_c = \frac{0.64}{207.36} = \frac{1}{324.6} \approx 0.00308 \]

5. **Convert \( K_c \) to \( K_p \):**
- The relationship between \( K_p \) and \( K_c \) is given by:
\[ K_p = K_c(RT)^{\Delta n} \]
- Where \( \Delta n \) is the change in moles of gas:
\[ \Delta n = (1 + 1) - 2 = 0 \]
- Since \( \Delta n = 0 \):
\[ K_p = K_c \]
- Therefore, \( K_p \approx 0.00308 \).

### Final Answer:
The value of \( K_p \) for the reaction is approximately \( 0.00308 \).