`K_(p) " for the following reaction will be equal to " 3 Fe (s) + 4 H_(2)O (g) hArrFe_(3)O_(4) (s) + 4 H_(2) (g)`

Options

A. `(p_(H_(2)))^(4)(p_(Fe_(3)O_(4)))`
B. `p_(H_(2))/(p_(H_(2)O))`
C. `(p_(H_(2)))^(4)/((p_(H_(2)O))^(4))`
D. ` ((p_(H_(2))) (p_(Fe_(3)O_(4))))/(p_(F_(e)))`

Answer

To determine the equilibrium constant \( K_p \) for the given reaction:

\[ 3 \text{Fe (s)} + 4 \text{H}_2\text{O (g)} \rightleftharpoons \text{Fe}_3\text{O}_4 \text{(s)} + 4 \text{H}_2 \text{(g)} \]

we will follow these steps:

### Step 1: Identify the reaction components
The reaction consists of:
- Reactants: 3 moles of solid iron (Fe) and 4 moles of gaseous water (H₂O).
- Products: 1 mole of solid magnetite (Fe₃O₄) and 4 moles of gaseous hydrogen (H₂).

### Step 2: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) is expressed in terms of the partial pressures of the gaseous components of the reaction. The general formula for \( K_p \) is:

\[ K_p = \frac{(P_{\text{products}})^{\text{coefficients}}}{(P_{\text{reactants}})^{\text{coefficients}}} \]

### Step 3: Apply the formula to the given reaction
For our reaction, the gaseous products are H₂ and the gaseous reactants are H₂O. The solids (Fe and Fe₃O₄) do not contribute to the \( K_p \) expression because their activities are considered to be 1.

Thus, the expression for \( K_p \) becomes:

\[ K_p = \frac{(P_{\text{H}_2})^4}{(P_{\text{H}_2\text{O}})^4} \]

### Step 4: Simplify the expression
Since the coefficients for both H₂ and H₂O are 4, we can simplify the expression to:

\[ K_p = \frac{P_{\text{H}_2}^4}{P_{\text{H}_2\text{O}}^4} \]

### Conclusion
The final expression for \( K_p \) for the reaction is:

\[ K_p = \frac{P_{\text{H}_2}^4}{P_{\text{H}_2\text{O}}^4} \]