The equilibrium constant `(K_(p))` for the decomposition of gaseous `H_(2)O`
`H_(2)O(g)hArr H_(2)(g)+(1)/(2)O_(2)(g)`
is related to the degree of dissociation `alpha` at a total pressure P by

Options

A. `K_(p)=(alpha^(3)P^(1//2))/((1+alpha)(2+alpha)^(1//2))`
B. `K_(p)=(alpha^(3)P^(3//2))/((1-alpha)(2+alpha)^(1//2))`
C. `K_(p)=(alpha^(3//2)P^(2))/((1-alpha)(2+alpha)^(1//2))`
D. `K_(p)=(alpha^(3//2)P^(1//2))/((1-alpha)(2+alpha)^(1//2))`

Explanation

To find the relationship between the equilibrium constant \( K_p \) for the decomposition of gaseous \( H_2O \) and the degree of dissociation \( \alpha \) at a total pressure \( P \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The decomposition of water can be represented as:
\[ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \]

### Step 2: Define the initial conditions
Assume we start with 1 mole of \( H_2O \). The degree of dissociation is represented by \( \alpha \), which indicates the fraction of \( H_2O \) that has dissociated.

### Step 3: Determine the moles at equilibrium
At equilibrium:
- Moles of \( H_2O \) remaining = \( 1 - \alpha \)
- Moles of \( H_2 \) produced = \( \alpha \)
- Moles of \( O_2 \) produced = \( \frac{\alpha}{2} \)

### Step 4: Calculate the total moles at equilibrium
The total number of moles at equilibrium is:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2} \]

### Step 5: Calculate the partial pressures
Using the total pressure \( P \), we can find the partial pressures of each component:
- Partial pressure of \( H_2 \):
\[ P_{H_2} = \left( \frac{\alpha}{1 + \frac{\alpha}{2}} \right) P \]
- Partial pressure of \( O_2 \):
\[ P_{O_2} = \left( \frac{\frac{\alpha}{2}}{1 + \frac{\alpha}{2}} \right) P \]
- Partial pressure of \( H_2O \):
\[ P_{H_2O} = \left( \frac{1 - \alpha}{1 + \frac{\alpha}{2}} \right) P \]

### Step 6: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) is given by the expression:
\[ K_p = \frac{(P_{H_2})^1 \cdot (P_{O_2})^{\frac{1}{2}}}{(P_{H_2O})^1} \]

### Step 7: Substitute the partial pressures into the \( K_p \) expression
Substituting the values we calculated:
\[ K_p = \frac{\left( \frac{\alpha}{1 + \frac{\alpha}{2}} P \right) \cdot \left( \frac{\frac{\alpha}{2}}{1 + \frac{\alpha}{2}} P \right)^{\frac{1}{2}}}{\left( \frac{1 - \alpha}{1 + \frac{\alpha}{2}} P \right)} \]

### Step 8: Simplify the expression
After substituting and simplifying, we find:
\[ K_p = \frac{\alpha^{\frac{3}{2}} \cdot \frac{1}{\sqrt{2}}}{(1 - \alpha)(1 + \frac{\alpha}{2})} \]

### Final Relationship
Thus, we can express \( K_p \) in terms of \( \alpha \) and \( P \).