The equilibrium constant for the reaction
`2SO_(2(g)) + O_(2(g)) rarr 2SO_(3(g))`
is 5. If the equilibrium mixture contains equal moles of `SO_(3)` and `SO_(2)` , the equilibrium partial pressure of `O_(2)` gas is

Options

A. 0.2 atm
B. 2 atm
C. 0.02 atm
D. 0.04 atm

Answer

To solve the problem, we need to find the equilibrium partial pressure of \( O_2 \) for the reaction:

\[ 2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} \]

Given that the equilibrium constant \( K_p \) for this reaction is 5 and that the equilibrium mixture contains equal moles of \( SO_3 \) and \( SO_2 \), we can denote the partial pressures of \( SO_2 \) and \( SO_3 \) as \( P_{SO_2} \) and \( P_{SO_3} \) respectively.

### Step 1: Set up the equilibrium expression

The equilibrium constant \( K_p \) for the reaction can be expressed as:

\[ K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 \cdot (P_{O_2})} \]

### Step 2: Define the partial pressures

Since it is given that the equilibrium mixture contains equal moles of \( SO_2 \) and \( SO_3 \), we can denote:

\[ P_{SO_2} = P_{SO_3} = x \]

### Step 3: Substitute into the equilibrium expression

Substituting \( P_{SO_2} \) and \( P_{SO_3} \) into the equilibrium expression gives:

\[ K_p = \frac{(x)^2}{(x)^2 \cdot (P_{O_2})} \]

### Step 4: Simplify the expression

This simplifies to:

\[ K_p = \frac{x^2}{x^2 \cdot P_{O_2}} = \frac{1}{P_{O_2}} \]

### Step 5: Set the equation equal to the given \( K_p \)

We know from the problem that \( K_p = 5 \):

\[ \frac{1}{P_{O_2}} = 5 \]

### Step 6: Solve for \( P_{O_2} \)

Rearranging the equation gives:

\[ P_{O_2} = \frac{1}{5} = 0.2 \, \text{atm} \]

### Conclusion

The equilibrium partial pressure of \( O_2 \) is \( 0.2 \, \text{atm} \).

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