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 find the equilibrium partial pressure of \( O_2 \) in the reaction:

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

with an equilibrium constant \( K_p = 5 \) and equal moles of \( SO_2 \) and \( SO_3 \), we can follow these steps:

### Step 1: Define the variables
Let the initial pressure of \( SO_2 \) be \( P \) and the initial pressure of \( O_2 \) be \( P \). Since we are told that the equilibrium mixture contains equal moles of \( SO_2 \) and \( SO_3 \), we can denote the change in pressure of \( SO_2 \) and \( O_2 \) at equilibrium.

### Step 2: Set up the changes in pressure
Assume that at equilibrium, the pressure of \( SO_2 \) decreases by \( 2x \) (since 2 moles of \( SO_2 \) are consumed for every mole of \( O_2 \) and \( SO_3 \) produced), and the pressure of \( O_2 \) decreases by \( x \). The pressure of \( SO_3 \) increases by \( 2x \).

Thus, we can express the pressures at equilibrium as:
- \( SO_2: P - 2x \)
- \( O_2: P - x \)
- \( SO_3: 2x \)

### Step 3: Use the equilibrium condition
Since we know \( SO_2 \) and \( SO_3 \) are equal at equilibrium, we can set:
\[ P - 2x = 2x \]

### Step 4: Solve for \( x \)
From the equation \( P - 2x = 2x \), we can rearrange it to find \( x \):
\[ P = 4x \quad \Rightarrow \quad x = \frac{P}{4} \]

### Step 5: Substitute \( x \) back into the equilibrium pressures
Now we can substitute \( x \) back into the expressions for the pressures at equilibrium:
- \( SO_2: P - 2\left(\frac{P}{4}\right) = P - \frac{P}{2} = \frac{P}{2} \)
- \( O_2: P - \left(\frac{P}{4}\right) = P - \frac{P}{4} = \frac{3P}{4} \)
- \( SO_3: 2\left(\frac{P}{4}\right) = \frac{P}{2} \)

### Step 6: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 \cdot (P_{O_2})} \]

Substituting the equilibrium pressures:
\[ 5 = \frac{\left(\frac{P}{2}\right)^2}{\left(\frac{P}{2}\right)^2 \cdot \left(\frac{3P}{4}\right)} \]

### Step 7: Simplify the equation
This simplifies to:
\[ 5 = \frac{\frac{P^2}{4}}{\frac{P^2}{4} \cdot \frac{3P}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

### Step 8: Solve for \( P \)
Cross-multiplying gives:
\[ 5 \cdot \frac{3}{4} = 4 \quad \Rightarrow \quad P = \frac{4}{15} \]

### Step 9: Find the partial pressure of \( O_2 \)
Now, substituting back to find the partial pressure of \( O_2 \):
\[ P_{O_2} = \frac{3P}{4} = \frac{3 \cdot \frac{4}{15}}{4} = \frac{3}{15} = 0.2 \]

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

\[ \boxed{0.2} \]