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. (A) `K=(alpha_(2)P^(1//2))/((1+alpha)(2-alpha)^(1//2))`
B. (B)`K=(alpha^(3//2)P^(1//2))/((1+alpha)(2+alpha)^(1//2))`
C. (C) `K=(alpha^(3)P^(1//2))/(sqrt2)`
D. (D) `K=(alpha^(3)P^(3//2))/((1-alpha)(2+alpha)^(1//2))`

Explanation

To solve the problem of finding 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 equation
The decomposition of water vapor can be represented as:
\[ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \]

### Step 2: Set up the initial moles
Assuming we start with 1 mole of \( H_2O \):
- Initial moles of \( H_2O = 1 \)
- Initial moles of \( H_2 = 0 \)
- Initial moles of \( O_2 = 0 \)

### Step 3: Define the change in moles at equilibrium
Let \( \alpha \) be the degree of dissociation. At equilibrium:
- Moles of \( H_2O = 1 - \alpha \)
- Moles of \( H_2 = \alpha \)
- Moles of \( O_2 = \frac{\alpha}{2} \)

### Step 4: Calculate the total moles at equilibrium
Total moles at equilibrium can be calculated as:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2} \]

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

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

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

### Step 8: Simplify the expression
After simplifying the above expression, we can find:
\[ K_p = \frac{\alpha^{3/2} P^{3/2}}{(1 - \alpha)(1 + \frac{\alpha}{2})} \]

### Final Result
Thus, the relationship between \( K_p \) and \( \alpha \) at total pressure \( P \) is:
\[ K_p = \frac{\alpha^{3/2} P^{3/2}}{(1 - \alpha)(1 + \frac{\alpha}{2})} \]

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