Partial pressure of `O_(2)` in the reaction
`2Ag_(2)O(s) hArr 4Ag(s)+O_(2)(g)` is

Options

A. `K_(rho)`
B. `sqrt(Krho)`
C. `3sqrt(K_(rho)`
D. `2K_(rho)`

Answer

To find the partial pressure of \( O_2 \) in the reaction

\[ 2Ag_2O(s) \rightleftharpoons 4Ag(s) + O_2(g), \]

we can follow these steps:

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

The equilibrium constant \( K_p \) for the reaction can be expressed in terms of the partial pressures of the gaseous products and reactants. For the given reaction, the expression for \( K_p \) is:

\[ K_p = \frac{(P_{O_2})^1}{(P_{Ag})^4 \cdot (P_{Ag_2O})^2} \]

### Step 2: Identify the states of the substances

In the reaction, \( Ag_2O \) and \( Ag \) are solids, while \( O_2 \) is a gas. The important point to note is that the concentrations (or partial pressures) of solids do not appear in the equilibrium expression because they remain constant throughout the reaction.

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

Since the solids do not affect the equilibrium expression, we can simplify the \( K_p \) expression to:

\[ K_p = P_{O_2} \]

This means that the partial pressure of \( O_2 \) is equal to the equilibrium constant \( K_p \).

### Step 4: Conclusion

Thus, the partial pressure of \( O_2 \) in the reaction is:

\[ P_{O_2} = K_p \]

### Final Answer

The correct option for the partial pressure of \( O_2 \) is \( K_p \).

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