At a certain temperature `T`, a compound `AB_(4)(g)` dissociates as
`2AB_(4)(g)hArrA_(2)(g)+4B_(2)(g)`
with a degree of dissociation `alpha`, which compared to unity. The expressio of `K_(P)` in terms of `alpha` and total pressure `P` is:

Options

A. `8P^(3)x^(5)`
B. `256P^(3)x^(5)`
C. `4Px^(2)`
D. None of these

Explanation

To solve the problem, we need to derive the expression for the equilibrium constant \( K_p \) in terms of the degree of dissociation \( \alpha \) and the total pressure \( P \) for the reaction:

\[ 2AB_4(g) \rightleftharpoons A_2(g) + 4B_2(g) \]

### Step 1: Write down the initial conditions
Assume we start with 1 mole of \( AB_4 \) and no products:
- Initial moles of \( AB_4 = 1 \)
- Initial moles of \( A_2 = 0 \)
- Initial moles of \( B_2 = 0 \)

### Step 2: Define the degree of dissociation
Let \( \alpha \) be the degree of dissociation of \( AB_4 \). This means that at equilibrium, the amount of \( AB_4 \) that has dissociated is \( 2\alpha \) moles (since 2 moles of \( AB_4 \) dissociate to produce 1 mole of \( A_2 \) and 4 moles of \( B_2 \)).

### Step 3: Calculate the moles at equilibrium
At equilibrium, the moles of each species will be:
- Moles of \( AB_4 = 1 - \alpha \)
- Moles of \( A_2 = \frac{\alpha}{2} \)
- Moles of \( B_2 = 2\alpha \)

### Step 4: Calculate the total moles at equilibrium
The total moles at equilibrium \( n_{total} \) will be:
\[ n_{total} = (1 - \alpha) + \frac{\alpha}{2} + 2\alpha = 1 - \alpha + \frac{\alpha}{2} + 2\alpha = 1 + \frac{3\alpha}{2} \]

### Step 5: Calculate the mole fractions
The mole fractions for each component are:
- Mole fraction of \( AB_4 \):
\[ X_{AB_4} = \frac{1 - \alpha}{1 + \frac{3\alpha}{2}} \]
- Mole fraction of \( A_2 \):
\[ X_{A_2} = \frac{\frac{\alpha}{2}}{1 + \frac{3\alpha}{2}} \]
- Mole fraction of \( B_2 \):
\[ X_{B_2} = \frac{2\alpha}{1 + \frac{3\alpha}{2}} \]

### Step 6: Calculate the partial pressures
Using the total pressure \( P \):
- Partial pressure of \( AB_4 \):
\[ P_{AB_4} = X_{AB_4} \cdot P = \frac{1 - \alpha}{1 + \frac{3\alpha}{2}} P \]
- Partial pressure of \( A_2 \):
\[ P_{A_2} = X_{A_2} \cdot P = \frac{\frac{\alpha}{2}}{1 + \frac{3\alpha}{2}} P \]
- Partial pressure of \( B_2 \):
\[ P_{B_2} = X_{B_2} \cdot P = \frac{2\alpha}{1 + \frac{3\alpha}{2}} P \]

### Step 7: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{A_2}) (P_{B_2})^4}{(P_{AB_4})^2} \]

Substituting the partial pressures:
\[ K_p = \frac{\left(\frac{\frac{\alpha}{2}}{1 + \frac{3\alpha}{2}} P\right) \left(\frac{2\alpha}{1 + \frac{3\alpha}{2}} P\right)^4}{\left(\frac{1 - \alpha}{1 + \frac{3\alpha}{2}} P\right)^2} \]

### Step 8: Simplify the expression
After substituting and simplifying:
\[ K_p = \frac{\left(\frac{\alpha}{2}\right) \left(\frac{2\alpha}{1 + \frac{3\alpha}{2}}\right)^4 P^5}{\left(\frac{1 - \alpha}{1 + \frac{3\alpha}{2}}\right)^2 P^2} \]

### Step 9: Final expression for \( K_p \)
After simplification, we find:
\[ K_p = \frac{8\alpha^5 P^3}{(1 - \alpha)^2} \]

### Conclusion
Thus, the expression for \( K_p \) in terms of \( \alpha \) and \( P \) is:
\[ K_p = 8\alpha^5 P^3 \]