The Kp value for `2SO_2(g)+O_2(g)⇌2SO_3` ​ (g) is 5.0atm If the equilibrium pressures of `SO_2` and `SO_3` are equal. What is the equilibrium pressure of `O_2` ​ ?

Answer

To solve the problem, we need to determine the equilibrium pressure of \( O_2 \) given the equilibrium pressures of \( SO_2 \) and \( SO_3 \) are equal and the value of \( K_p \) is 5 atm for the reaction:

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

### Step-by-Step Solution:

1. **Write the expression for \( K_p \)**:
The equilibrium constant \( K_p \) for the reaction can be expressed as:
\[ K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 \cdot (P_{O_2})} \]

2. **Set the equilibrium pressures**:
Since it is given that the equilibrium pressures of \( SO_2 \) and \( SO_3 \) are equal, we can denote their pressures as:
\[ P_{SO_2} = P_{SO_3} = x \, \text{atm} \]
Therefore, we can express the pressures in terms of \( x \):
- \( P_{SO_2} = x \)
- \( P_{SO_3} = x \)
- \( P_{O_2} = y \)

3. **Substitute into the \( K_p \) expression**:
Substitute \( P_{SO_2} \) and \( P_{SO_3} \) into the \( K_p \) expression:
\[ K_p = \frac{x^2}{x^2 \cdot y} \]

4. **Simplify the equation**:
The \( x^2 \) terms cancel out:
\[ K_p = \frac{1}{y} \]

5. **Substitute the known value of \( K_p \)**:
We know that \( K_p = 5 \, \text{atm} \):
\[ 5 = \frac{1}{y} \]

6. **Solve for \( y \)**:
Rearranging gives:
\[ y = \frac{1}{5} = 0.2 \, \text{atm} \]

7. **Conclusion**:
The equilibrium pressure of \( O_2 \) is:
\[ P_{O_2} = 0.2 \, \text{atm} \]