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

Options

A. `(P_(H_(2)))^(4)(P_(Fe_(3)O_(4)))`
B. `(P_(H_(2)))/(P_(H_(2O)))`
C. `((P_(H_(2)))^(4))/((P_(H_(2O)))^(4))`
D. `((P_(H_(2)))xxP_(Fe_(3)O_(4)))/(P_(Fe))`

Answer

To find the equilibrium constant \( K_p \) for the reaction:

\[ 3Fe(s) + 4H_2O(g) \rightleftharpoons Fe_3O_4(s) + 4H_2(g) \]

we will follow these steps:

### Step 1: Identify the components of the reaction
In this reaction, we have:
- Reactants: \( 3Fe(s) \) (solid) and \( 4H_2O(g) \) (gas)
- Products: \( Fe_3O_4(s) \) (solid) and \( 4H_2(g) \) (gas)

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

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

### Step 3: Identify the gaseous components
From our reaction, the only gaseous components are \( H_2O(g) \) and \( H_2(g) \). The solids \( Fe(s) \) and \( Fe_3O_4(s) \) do not contribute to the \( K_p \) expression because their activities are considered to be 1.

### Step 4: Write the \( K_p \) expression
For our reaction, the \( K_p \) expression will include only the gaseous components:

\[ K_p = \frac{(P_{H_2})^4}{(P_{H_2O})^4} \]

### Step 5: Conclusion
Thus, the equilibrium constant \( K_p \) for the given reaction is:

\[ K_p = \frac{(P_{H_2})^4}{(P_{H_2O})^4} \]