The equilibrium constant for the reaction,
`2SO_(2)(g)+O_(2)(g)hArr 2SO_(3)(g)`
at 1000 K is `"3.5 atm"^(-1)`. What would be the partial pressure of oxygen gas, if the equilibrium is found to have equal moles of `SO_(2)` and `SO_(3)`? (Assume initial moles of `SO_(2)` equal to that of initial moles of `O_(2)`)

Options

A. 0.285 at,
B. 3.5 atm
C. 0.35 atm
D. 1.87 atm

Answer

To solve the problem, we need to determine the partial pressure of oxygen gas (O₂) at equilibrium, given that the equilibrium constant (Kp) for the reaction is 3.5 atm⁻¹ and that at equilibrium, the moles of SO₂ and SO₃ are equal.

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
\[ 2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g) \]

2. **Define the initial conditions:**
Let the initial moles of SO₂ and O₂ be \( A \). Therefore, the initial pressures can be expressed as:
- \( P_{\text{SO}_2}^0 = A \)
- \( P_{\text{O}_2}^0 = A \)
- \( P_{\text{SO}_3}^0 = 0 \)

3. **Set up the changes at equilibrium:**
Let \( x \) be the change in pressure due to the reaction proceeding to the right. Since 2 moles of SO₂ react with 1 mole of O₂ to produce 2 moles of SO₃, we can express the equilibrium pressures as:
- \( P_{\text{SO}_2} = A - 2x \)
- \( P_{\text{O}_2} = A - x \)
- \( P_{\text{SO}_3} = 2x \)

4. **Use the condition that at equilibrium, the moles of SO₂ and SO₃ are equal:**
Given that the moles of SO₂ and SO₃ are equal at equilibrium, we can set:
\[ A - 2x = 2x \]
Solving for \( x \):
\[ A = 4x \quad \Rightarrow \quad x = \frac{A}{4} \]

5. **Substitute \( x \) back into the expressions for pressures:**
- \( P_{\text{SO}_2} = A - 2\left(\frac{A}{4}\right) = A - \frac{A}{2} = \frac{A}{2} \)
- \( P_{\text{O}_2} = A - \frac{A}{4} = \frac{3A}{4} \)
- \( P_{\text{SO}_3} = 2\left(\frac{A}{4}\right) = \frac{A}{2} \)

6. **Write the expression for the equilibrium constant \( K_p \):**
The expression for \( K_p \) is given by:
\[ K_p = \frac{(P_{\text{SO}_3})^2}{(P_{\text{SO}_2})^2 \cdot (P_{\text{O}_2})} \]
Substituting the equilibrium pressures:
\[ K_p = \frac{\left(\frac{A}{2}\right)^2}{\left(\frac{A}{2}\right)^2 \cdot \left(\frac{3A}{4}\right)} = \frac{\frac{A^2}{4}}{\frac{A^2}{4} \cdot \frac{3A}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

7. **Set \( K_p \) equal to the given value:**
We know \( K_p = 3.5 \) atm⁻¹, thus:
\[ \frac{4}{3} = 3.5 \]
This leads us to find \( A \) in terms of \( K_p \):
\[ P_{\text{O}_2} = \frac{3A}{4} = \frac{3 \cdot 1}{4 \cdot 3.5} = \frac{3}{14} \approx 0.214 \text{ atm} \]

8. **Final calculation for \( P_{\text{O}_2} \):**
To find the partial pressure of oxygen gas:
\[ P_{\text{O}_2} = \frac{1}{3.5} \approx 0.285 \text{ atm} \]

### Conclusion:
The partial pressure of oxygen gas at equilibrium is approximately **0.285 atm**.