consider the reaction represented by the following equation: 2SO$_2$ + O$_2$  ⇌ 2SO$_3$  A decrease in the pressure of the reaction would

Analyze the question: The reaction is 2SO₂ + O₂ ⇌ 2SO₃. We are asked what happens when the pressure is decreased.

Concept: Le Chatelier's Principle When a system at equilibrium is disturbed, it shifts in the direction that opposes the change.

• Increasing pressure favours the side with fewer moles of gas.
• Decreasing pressure favours the side with more moles of gas.

Solution:

Count the moles of gas on each side:

• Left (reactants): 2 mol SO₂ + 1 mol O₂ = 3 mol
• Right (products): 2 mol SO₃ = 2 mol

When pressure is decreased, the system tries to increase the pressure again. It does this by moving to the side with more gas molecules, which is the left side (3 mol).

So the equilibrium position shifts to the left, and SO₃ decomposes to form more SO₂ and O₂.

Checking the options:

• A. "Decrease in the concentration of SO₂": wrong, because the shift to the left actually increases SO₂.
• B. "No effect": wrong, because the numbers of moles of gas differ on the two sides.
• C. "Shift the equilibrium position to the left": correct.
• D. "Shift to the right": wrong, because that is what an increase in pressure would do.

Exam tip: Count only the gas moles on each side. Then remember: high pressure → fewer moles; low pressure → more moles. This is why the Contact Process for making sulphuric acid uses relatively high pressure to favour SO₃ formation.