n mole each of `H_(2)O(g),H_(2)(g) "and" O_(2)(g)` are mixed at a suitable high temperature to attain the equilibrium `2H_(2)O(g)hArr2H_(2)(g)+O_(2)(g)`. If `y` and mole of `H_(2)O(g)` are the dissociated and the total pressure maintained is `P`, calculate the `K_(P)`.

Answer

To solve the problem step by step, we will follow the process of establishing the equilibrium expression and calculating the equilibrium constant \( K_p \).

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the dissociation of water vapor is:
\[ 2 \text{H}_2\text{O}(g) \rightleftharpoons 2 \text{H}_2(g) + \text{O}_2(g) \]

### Step 2: Define the initial moles and changes
Let \( n \) be the initial number of moles of each gas:
- Initial moles of \( \text{H}_2\text{O} = n \)
- Initial moles of \( \text{H}_2 = n \)
- Initial moles of \( \text{O}_2 = n \)

Let \( y \) be the moles of \( \text{H}_2\text{O} \) that dissociate at equilibrium.

### Step 3: Calculate the equilibrium moles
At equilibrium, the moles of each species will be:
- Moles of \( \text{H}_2\text{O} = n - y \)
- Moles of \( \text{H}_2 = n + y \)
- Moles of \( \text{O}_2 = n + 0.5y \)

### Step 4: Calculate the total moles at equilibrium
The total number of moles at equilibrium is:
\[ \text{Total moles} = (n - y) + (n + y) + (n + 0.5y) = 3n + 0.5y \]

### Step 5: Calculate the mole fractions
The mole fractions for each gas at equilibrium are:
- Mole fraction of \( \text{H}_2\text{O} \):
\[ X_{\text{H}_2\text{O}} = \frac{n - y}{3n + 0.5y} \]
- Mole fraction of \( \text{H}_2 \):
\[ X_{\text{H}_2} = \frac{n + y}{3n + 0.5y} \]
- Mole fraction of \( \text{O}_2 \):
\[ X_{\text{O}_2} = \frac{n + 0.5y}{3n + 0.5y} \]

### Step 6: Write the expression for \( K_p \)
The expression for the equilibrium constant \( K_p \) in terms of partial pressures is given by:
\[ K_p = \frac{(P_{\text{H}_2})^2 \cdot P_{\text{O}_2}}{(P_{\text{H}_2\text{O}})^2} \]
Using the mole fractions and the total pressure \( P \):
\[ P_{\text{H}_2} = X_{\text{H}_2} \cdot P, \quad P_{\text{O}_2} = X_{\text{O}_2} \cdot P, \quad P_{\text{H}_2\text{O}} = X_{\text{H}_2\text{O}} \cdot P \]

### Step 7: Substitute the mole fractions into the \( K_p \) expression
Substituting the mole fractions into the \( K_p \) expression:
\[ K_p = \frac{\left(\frac{n + y}{3n + 0.5y} \cdot P\right)^2 \cdot \left(\frac{n + 0.5y}{3n + 0.5y} \cdot P\right)}{\left(\frac{n - y}{3n + 0.5y} \cdot P\right)^2} \]

### Step 8: Simplify the expression
This simplifies to:
\[ K_p = \frac{(n + y)^2 \cdot (n + 0.5y)}{(n - y)^2} \cdot \frac{P^2}{(3n + 0.5y)^2} \]

### Final Expression for \( K_p \)
Thus, the final expression for \( K_p \) is:
\[ K_p = \frac{(n + y)^2 \cdot (n + 0.5y)}{(n - y)^2 \cdot (3n + 0.5y)^2} \cdot P^2 \]