The equation for the reaction is:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Explain why the percentage yield of sulfur trioxide in this reaction is greater if the pressure is higher.

This question relates to Le Chatelier's principle and the effect of pressure on gaseous equilibria. 1. **Count the Moles of Gas:** We look at the balanced symbol equation: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). - On the reactant (left) side, we have 2 moles of SO₂ and 1 mole of O₂, making a total of 2 + 1 = 3 moles of gas. - On the product (right) side, we have 2 moles of SO₃ gas. 2. **Apply Le Chatelier's Principle:** The principle states that if the pressure of a system at equilibrium is increased, the equilibrium will shift to the side with the fewer number of moles of gas, as this will help to reduce the overall pressure. 3. **Conclusion:** In this reaction, the product side has fewer moles of gas (2) than the reactant side (3). Therefore, increasing the pressure will shift the equilibrium position to the right, favouring the forward reaction and increasing the percentage yield of the product, sulfur trioxide (SO₃).