Kp value for `2SO_(2(g)) + O_(2(g)) hArr 2SO_(3(g)) ` is 5.0 `atm^(-1)`. What is the cquilibrium partial pressure of `O_(2)` if the equilibrium pressures of `SO_(2) and SO_(3)` are equal ?

Options

A. 0.2 atm
B. 0.3 atm
C. 0.4 atm
D. 0.1 atm

Answer

To solve the problem, we need to find the equilibrium partial pressure of \( O_2 \) given that the equilibrium pressures of \( SO_2 \) and \( SO_3 \) are equal. We are also given the equilibrium constant \( K_p \) for the reaction:

\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]

with \( K_p = 5.0 \, \text{atm}^{-1} \).

### Step-by-Step Solution:

1. **Write the expression for \( K_p \)**:
The equilibrium constant \( K_p \) for the reaction is given by the formula:
\[ K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 \cdot (P_{O_2})} \]

2. **Let the equilibrium pressures be defined**:
Since the equilibrium pressures of \( SO_2 \) and \( SO_3 \) are equal, we can denote their pressures as \( P \):
\[ P_{SO_2} = P \quad \text{and} \quad P_{SO_3} = P \]

3. **Substitute the pressures into the \( K_p \) expression**:
Substituting \( P \) into the \( K_p \) expression, we get:
\[ K_p = \frac{(P)^2}{(P)^2 \cdot (P_{O_2})} \]

4. **Simplify the equation**:
The \( (P)^2 \) terms in the numerator and denominator cancel out:
\[ K_p = \frac{1}{P_{O_2}} \]

5. **Rearrange to find \( P_{O_2} \)**:
Rearranging the equation gives:
\[ P_{O_2} = \frac{1}{K_p} \]

6. **Substitute the value of \( K_p \)**:
Now substituting the given value of \( K_p = 5.0 \, \text{atm}^{-1} \):
\[ P_{O_2} = \frac{1}{5.0} = 0.2 \, \text{atm} \]

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