Question

how does adding oxygen (o₂) to this reaction change the equilibrium?
2so₂(g) + o₂(g) ⇌ 2so₃(g)
a. the equilibrium shifts right to produce more so₃ molecules.
b. the equilibrium shifts left to produce more o₂ molecules.
c. the equilibrium shifts right because of decreased collisions between so₂ and o₂ molecules.
d. the equilibrium shifts left with an increase in so₂ and o₂ molecules.
e. the equilibrium shifts left because of increased collisions between so₃ and o₂ molecules.

Explanation:

To solve this, we use Le Chatelier's Principle, which states that if a stress is applied to a system at equilibrium, the system will shift to counteract that stress. Here, adding \( O_2 \) (a reactant) is the stress. The system will shift to consume the added \( O_2 \), so it will shift towards the products (right) to make more \( SO_3 \). Let's analyze each option:

• Option A: Adding \( O_2 \) (reactant) increases reactant concentration. The system shifts right to produce more \( SO_3 \) (product) to consume the excess \( O_2 \). This matches Le Chatelier's Principle.
• Option B: Shifting left would produce more reactants, but we added a reactant, so the system should shift to use it, not produce more. Eliminate.
• Option C: Adding \( O_2 \) increases collisions between \( SO_2 \) and \( O_2 \), not decrease. The reason for shifting right is to consume excess \( O_2 \), not collision decrease. Eliminate.
• Option D: Shifting left would increase \( SO_2 \) and \( O_2 \), but we added \( O_2 \), so the system should shift to reduce \( O_2 \) (by making \( SO_3 \)), not increase. Eliminate.
• Option E: \( O_2 \) is a reactant, not a product. Collisions between \( SO_3 \) (product) and \( O_2 \) (reactant) aren't the reason for shift, and shifting left is incorrect. Eliminate.

Answer:

A. The equilibrium shifts right to produce more \( \text{SO}_3 \) molecules.