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

Options

A. `K_(p)`
B. `sqrt(K_(p))`
C. `root(3)(K_(p))`
D. `(K_(p))^(2)`

Explanation

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

\[ 2 \text{Ag}_2\text{O}(s) \rightleftharpoons 4 \text{Ag}(s) + O_2(g) \]

we will follow these steps:

### Step 1: Understand the Reaction
The reaction involves the decomposition of solid silver oxide (\( \text{Ag}_2\text{O} \)) into solid silver (\( \text{Ag} \)) and gaseous oxygen (\( O_2 \)).

### Step 2: Identify the Phases
In this reaction:
- \( \text{Ag}_2\text{O} \) is a solid (s)
- \( \text{Ag} \) is also a solid (s)
- \( O_2 \) is a gas (g)

### Step 3: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) for a reaction is defined in terms of the partial pressures of the gaseous products and reactants. For this reaction, since only \( O_2 \) is a gas, we can write:

\[ K_p = \frac{P_{O_2}}{1} \]

where \( P_{O_2} \) is the partial pressure of \( O_2 \) and the solids do not appear in the expression because their activities are considered to be 1.

### Step 4: Relate \( K_p \) to \( P_{O_2} \)
From the expression derived in Step 3, we can see that:

\[ K_p = P_{O_2} \]

Thus, the partial pressure of \( O_2 \) is equal to \( K_p \).

### Conclusion
The partial pressure of \( O_2 \) in the given reaction is:

\[ P_{O_2} = K_p \]

### Answer
The correct option is \( K_p \).

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