Question

Consider the equilibrium reaction:

2 SO₂(g) + O2(g) ⇌ 2 SO₃(g)

You place 1.00 move each of SO₂, and O₂, in a flask and find that at equilibrium the concentration of SO₂, is 0.075 mol/L. What are the equilib concentrations of O₂, and SO₃?

Answer 1

The equilibrium concentrations of O₂ and SO₃ in the reaction 2SO₂(g) + O₂(g) ⇒ 2SO₃(g) are 0.5375 mol/L for O₂ and 0.925 mol/L for SO₃, based on the stoichiometry and the given equilibrium concentration of SO₂.

Step-by-step explanation:

To calculate the equilibrium concentrations of O₂ and SO₃ in the equilibrium reaction 2SO₂(g) + O₂(g) ⇒ 2SO₃(g), you must first establish a ratio based on the stoichiometry of the balanced equation. Using an ICE table (Initial, Change, Equilibrium), if you start with 1.00 mole of SO₂ and O₂ initially, the change in SO₂ concentration to reach 0.075 mol/L at equilibrium means (1.00 - 0.075) mol/L of SO₂ reacted. Since the reaction describes a 2:1 mole ratio between SO₂ and O₂, this implies that (1.00 - 0.075)/2 = 0.4625 mol/L of O₂ reacted.

Thus, the equilibrium concentration of O₂ would be 1.00 - 0.4625 = 0.5375 mol/L. Since 2 moles of SO₃ are produced for every 2 moles of SO₂ that reacts, the equilibrium concentration for SO₃ is the same as the amount of SO₂ that reacted, which is 1.00 - 0.075 = 0.925 mol/L. It's important to note that this assumes the volume of the flask does not change.