The equilibrium constant for the decomposition of water `H_2O(g) hArr H_2(g)+1/2O_2(g)` is given by : (`alpha`=degree of dissociation of `H_2O` (g) p=Total equilibrium pressure)

Options

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

Answer

To solve the problem regarding the equilibrium constant for the decomposition of water \( H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \), we will follow these steps:

### Step 1: Write the Reaction and Define Variables
The reaction is:
\[ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \]

Let:
- \( y \) = initial pressure of \( H_2O \)
- \( \alpha \) = degree of dissociation of \( H_2O \)
- \( p \) = total equilibrium pressure

### Step 2: Determine the Changes in Pressure
At equilibrium:
- Pressure of \( H_2O \) = \( y(1 - \alpha) \)
- Pressure of \( H_2 \) = \( y\alpha \)
- Pressure of \( O_2 \) = \( \frac{1}{2} y\alpha \)

### Step 3: Calculate Total Pressure at Equilibrium
The total pressure at equilibrium \( p \) can be expressed as:
\[ p = \text{Pressure of } H_2O + \text{Pressure of } H_2 + \text{Pressure of } O_2 \]
Substituting the pressures:
\[ p = y(1 - \alpha) + y\alpha + \frac{1}{2} y\alpha \]
Simplifying this gives:
\[ p = y(1 - \alpha + \alpha + \frac{1}{2}\alpha) = y(1 + \frac{1}{2}\alpha) \]

### Step 4: Solve for Initial Pressure \( y \)
From the equation derived above:
\[ p = y\left(1 + \frac{1}{2}\alpha\right) \]
Rearranging gives:
\[ y = \frac{2p}{2 + \alpha} \]

### Step 5: Write the Expression for the Equilibrium Constant \( K \)
The equilibrium constant \( K \) is given by:
\[ K = \frac{(P_{H_2})^{1} \cdot (P_{O_2})^{\frac{1}{2}}}{(P_{H_2O})^{1}} \]
Substituting the pressures at equilibrium:
\[ K = \frac{(y\alpha)^{1} \cdot \left(\frac{1}{2}y\alpha\right)^{\frac{1}{2}}}{y(1 - \alpha)} \]
This simplifies to:
\[ K = \frac{(y\alpha) \cdot \left(\frac{1}{2}y\alpha\right)^{\frac{1}{2}}}{y(1 - \alpha)} \]
\[ K = \frac{y\alpha \cdot \frac{1}{\sqrt{2}}(y\alpha)^{\frac{1}{2}}}{y(1 - \alpha)} \]
\[ K = \frac{y^{\frac{3}{2}} \alpha^{\frac{3}{2}}}{\sqrt{2}(1 - \alpha)} \]

### Step 6: Substitute \( y \) into the Expression for \( K \)
Substituting \( y = \frac{2p}{2 + \alpha} \):
\[ K = \frac{\left(\frac{2p}{2 + \alpha}\right)^{\frac{3}{2}} \alpha^{\frac{3}{2}}}{\sqrt{2}(1 - \alpha)} \]
This leads to:
\[ K = \frac{(2p)^{\frac{3}{2}} \alpha^{\frac{3}{2}}}{\sqrt{2}(1 - \alpha)(2 + \alpha)^{\frac{3}{2}}} \]

### Conclusion
The final expression for the equilibrium constant \( K \) is:
\[ K = \frac{(2p)^{\frac{3}{2}} \alpha^{\frac{3}{2}}}{\sqrt{2}(1 - \alpha)(2 + \alpha)^{\frac{3}{2}}} \]