Question

5 moles of $SO_{2}$ and 5 moles of $O_{2}$ are allowed to react. At equilibrium, it was found that 60% of $SO_{2}$ is used up. If the partial pressure of the equilibrium mixture is one atmosphere, the partial pressure of $O_{2}$ is

• A. 0.82 atm
• B. 0.52 atm
• C. 0.21 atm
• D. 0.41 atm

Hint:

Partial pressure = Mole fraction $\times$ Total pressure.

Answer 1

The Correct Option is D

Step 1: Moles at Equilibrium
$SO_{2}$ used = 60% of 5 = 3 moles. Reaction: $2SO_{2} + O_{2} \rightleftharpoons 2SO_{3}$. Remaining $SO_{2} = 5 - 3 = 2$. $O_{2}$ used = $3/2 = 1.5$. Remaining $O_{2} = 5 - 1.5 = 3.5$. $SO_{3}$ formed = 3.
Step 2: Total Moles
Total moles at equilibrium = $2 + 3.5 + 3 = 8.5$.
Step 3: Partial Pressure
$p_{O_{2}} = (\text{moles of } O_{2} / \text{total moles}) \times P_{\text{total}}$. $p_{O_{2}} = (3.5 / 8.5) \times 1 = 0.41$ atm.
Final Answer: (d)