Ammoina dissociates into `N_(2)` and `H_(2)` such that degree of dissociation `alpha` is very less than 1 and equilibrium pressure is `P_(0)` then the value of `alpha` is [if `K_(p)` for `2NH_(3)(g)hArrN_(2)(g) + 3H_(2)(g) + 3H_(2)(g)` is `27xx10^(-8)P_(0)^(2)]` :

Options

A. `10^(-4)`
B. `4xx10^(-4)`
C. `0.02`
D. can't be calculated

Explanation

To solve the problem, we need to analyze the dissociation of ammonia (NH₃) into nitrogen (N₂) and hydrogen (H₂) and calculate the degree of dissociation, α, given the equilibrium pressure \( P_0 \) and the equilibrium constant \( K_p \).

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The dissociation of ammonia can be represented as:
\[ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g) \]

2. **Define Initial Moles:**
Assume we start with 1 mole of ammonia (NH₃).

3. **Calculate Moles at Equilibrium:**
If the degree of dissociation is α, then:
- Moles of NH₃ remaining = \( 1 - \alpha \)
- Moles of N₂ formed = \( \frac{\alpha}{2} \) (since 2 moles of NH₃ produce 1 mole of N₂)
- Moles of H₂ formed = \( \frac{3\alpha}{2} \) (since 2 moles of NH₃ produce 3 moles of H₂)

4. **Total Moles at Equilibrium:**
The total moles at equilibrium will be:
\[ \text{Total moles} = (1 - \alpha) + \frac{\alpha}{2} + \frac{3\alpha}{2} = 1 + \alpha \]

5. **Calculate Partial Pressures:**
The partial pressures at equilibrium can be expressed as:
- \( P_{NH_3} = \frac{(1 - \alpha)}{(1 + \alpha)} P_0 \)
- \( P_{N_2} = \frac{\frac{\alpha}{2}}{(1 + \alpha)} P_0 = \frac{\alpha P_0}{2(1 + \alpha)} \)
- \( P_{H_2} = \frac{\frac{3\alpha}{2}}{(1 + \alpha)} P_0 = \frac{3\alpha P_0}{2(1 + \alpha)} \)

6. **Write the Expression for \( K_p \):**
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_{N_2} \cdot (P_{H_2})^3}{(P_{NH_3})^2} \]
Substituting the partial pressures:
\[ K_p = \frac{\left(\frac{\alpha P_0}{2(1 + \alpha)}\right) \cdot \left(\frac{3\alpha P_0}{2(1 + \alpha)}\right)^3}{\left(\frac{(1 - \alpha) P_0}{(1 + \alpha)}\right)^2} \]

7. **Simplify the Expression:**
Simplifying the expression gives:
\[ K_p = \frac{\frac{\alpha P_0}{2(1 + \alpha)} \cdot \frac{27\alpha^3 P_0^3}{8(1 + \alpha)^3}}{\frac{(1 - \alpha)^2 P_0^2}{(1 + \alpha)^2}} \]
This simplifies to:
\[ K_p = \frac{27\alpha^4 P_0^2}{16(1 + \alpha)^2(1 - \alpha)^2} \]

8. **Approximate for Small α:**
Since α is very small (α << 1), we can approximate \( 1 + \alpha \approx 1 \) and \( 1 - \alpha \approx 1 \):
\[ K_p \approx \frac{27\alpha^4 P_0^2}{16} \]

9. **Set Equal to Given \( K_p \):**
We know \( K_p = 27 \times 10^{-8} P_0^2 \). Setting the two expressions for \( K_p \) equal:
\[ \frac{27\alpha^4 P_0^2}{16} = 27 \times 10^{-8} P_0^2 \]

10. **Solve for α:**
Dividing both sides by \( P_0^2 \) and simplifying:
\[ \frac{27\alpha^4}{16} = 27 \times 10^{-8} \]
\[ \alpha^4 = \frac{16 \times 10^{-8}}{27} \]
\[ \alpha^4 = \frac{16}{27} \times 10^{-8} \]
Taking the fourth root:
\[ \alpha = \sqrt[4]{\frac{16}{27} \times 10^{-8}} = 2 \times 10^{-2} = 0.02 \]

### Final Answer:
\[ \alpha = 0.02 \]