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 step-by-step, we will analyze the given reaction and the conditions provided.

**Step 1: Write the balanced chemical equation.**
The reaction is:
\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]

**Step 2: Define the initial conditions.**
Let the initial moles of \( SO_2 \) and \( O_2 \) both be \( A \). The initial moles of \( SO_3 \) are \( 0 \) since no product has formed yet.

**Step 3: Define the change in moles at equilibrium.**
Let \( 2\alpha \) be the moles of \( SO_2 \) that react. According to stoichiometry:
- Moles of \( SO_2 \) at equilibrium = \( A - 2\alpha \)
- Moles of \( O_2 \) at equilibrium = \( A - \alpha \)
- Moles of \( SO_3 \) at equilibrium = \( 2\alpha \)

**Step 4: Set up the condition given in the problem.**
We are told that at equilibrium, the moles of \( SO_2 \) and \( SO_3 \) are equal:
\[ A - 2\alpha = 2\alpha \]
This can be rearranged to:
\[ A = 4\alpha \]

**Step 5: Express the equilibrium moles in terms of \( A \).**
From \( A = 4\alpha \):
- \( \alpha = \frac{A}{4} \)
- Moles of \( SO_2 \) at equilibrium = \( A - 2\left(\frac{A}{4}\right) = A - \frac{A}{2} = \frac{A}{2} \)
- Moles of \( O_2 \) at equilibrium = \( A - \frac{A}{4} = \frac{3A}{4} \)
- Moles of \( SO_3 \) at equilibrium = \( 2\left(\frac{A}{4}\right) = \frac{A}{2} \)

**Step 6: Write the expression for the equilibrium constant \( K_p \).**
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 \cdot (P_{O_2})} \]
Where:
- \( P_{SO_2} = P_{SO_3} = P \) (since \( SO_2 \) and \( SO_3 \) have equal moles)
- \( P_{O_2} = P_{O_2} \)

Thus, we can write:
\[ K_p = \frac{P^2}{P^2 \cdot P_{O_2}} = \frac{1}{P_{O_2}} \]

**Step 7: Substitute the value of \( K_p \).**
Given \( K_p = 3.5 \, atm^{-1} \):
\[ 3.5 = \frac{1}{P_{O_2}} \]

**Step 8: Solve for \( P_{O_2} \).**
Rearranging gives:
\[ P_{O_2} = \frac{1}{3.5} \]
Calculating this gives:
\[ P_{O_2} \approx 0.2857 \, atm \]

Thus, the partial pressure of oxygen gas at equilibrium is approximately **0.286 atm**.

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