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

Options

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

Explanation

To determine 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: Understand the Reaction
The reaction involves the decomposition of silver oxide (\( Ag_2O \)) into silver (\( Ag \)) and oxygen gas (\( O_2 \)). It is important to note that \( Ag_2O \) and \( Ag \) are solids, while \( O_2 \) is a gas.

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

\[ K_p = \frac{(P_C)^{c} (P_D)^{d}}{(P_A)^{a} (P_B)^{b}} \]

where \( P \) represents the partial pressures of the gases, and \( a, b, c, d \) are the stoichiometric coefficients.

### Step 3: Identify Gaseous Species
In our reaction, the only gaseous species is \( O_2 \). The solids \( Ag_2O \) and \( Ag \) do not contribute to the \( K_p \) expression. Therefore, their partial pressures are considered to be unity (1).

### Step 4: Write the \( K_p \) Expression for the Given Reaction
For the reaction \( 2Ag_2O(s) \rightleftharpoons 4Ag(s) + O_2(g) \), the \( K_p \) expression simplifies to:

\[ K_p = \frac{(P_{O_2})^{1}}{(1)^{0} \cdot (1)^{0}} = P_{O_2} \]

### Step 5: Conclusion
From the expression derived, we can conclude that the partial pressure of \( O_2 \) is equal to \( K_p \):

\[ P_{O_2} = K_p \]

### Final Answer
The partial pressure of \( O_2 \) in this reaction is \( K_p \).

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