The equilibrium constant `K_p` for the reaction `2SO_(2)+O_(2) Leftrightarrow 2SO_(3)` is 2.5 atm`""^(-1).` What would be the partial pressure of `O_2` at equilibrium. If the equilibrium pressures of `SO_(2) and SO_(3)` are equal

Options

A. 304 mm
B. 30.4 mm
C. 0.04 mm
D. 760 mm

Answer

To solve the problem, we need to find the partial pressure of \( O_2 \) at equilibrium given that the equilibrium constant \( K_p \) for the reaction

\[ 2SO_2 + O_2 \leftrightarrow 2SO_3 \]

is \( 2.5 \, \text{atm}^{-1} \) and that the equilibrium pressures of \( SO_2 \) and \( SO_3 \) are equal.

### Step-by-Step Solution:

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

2. **Set the partial pressures**:
Since it is given that the equilibrium pressures of \( SO_2 \) and \( SO_3 \) are equal, we can denote:
\[ P_{SO_2} = P_{SO_3} = x \]
Thus, we can substitute \( P_{SO_2} \) and \( P_{SO_3} \) into the \( K_p \) expression:
\[ K_p = \frac{x^2}{x^2 \cdot P_{O_2}} \]

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

4. **Substitute the given value of \( K_p \)**:
We know \( K_p = 2.5 \, \text{atm}^{-1} \):
\[ 2.5 = \frac{1}{P_{O_2}} \]

5. **Solve for \( P_{O_2} \)**:
Rearranging the equation gives:
\[ P_{O_2} = \frac{1}{2.5} \]
Calculating this gives:
\[ P_{O_2} = 0.4 \, \text{atm} \]

### Final Answer:
The partial pressure of \( O_2 \) at equilibrium is \( 0.4 \, \text{atm} \).

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