`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 determine the equilibrium constant \( K_p \) for the reaction:

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

we will follow these steps:

### Step 1: Write the expression for \( K_p \)

The equilibrium constant \( K_p \) is defined as the ratio of the partial pressures of the products to the partial pressures of the reactants, each raised to the power of their respective stoichiometric coefficients.

### Step 2: Identify the products and reactants

In the given reaction:
- Products: \( \text{Fe}_3\text{O}_4(s) \) and \( \text{H}_2(g) \)
- Reactants: \( \text{Fe}(s) \) and \( \text{H}_2\text{O}(g) \)

### Step 3: Write the \( K_p \) expression

The general form of the \( K_p \) expression for the reaction is:

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

where:
- \( P_{\text{H}_2} \) is the partial pressure of hydrogen gas,
- \( P_{\text{H}_2\text{O}} \) is the partial pressure of water vapor.

### Step 4: Consider the states of matter

Since \( \text{Fe}(s) \) and \( \text{Fe}_3\text{O}_4(s) \) are solids, their activities are considered to be 1. Therefore, they do not appear in the \( K_p \) expression.

### Step 5: Finalize the \( K_p \) expression

Thus, the \( K_p \) expression simplifies to:

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

This can also be written as:

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

### Conclusion

The correct expression for \( K_p \) for the reaction is:

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