Question

Consider the following reaction: 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g). If you decrease the volume of the reaction chamber, you would observe?

1) An increase in the concentration of SO₂(g)
2) An increase in the concentration of O₂(g)
3) An increase in the concentration of SO₃(g)
4) No change in the concentrations of the reactants or products

Answer 1

Decreasing the volume of the reaction chamber for the equilibrium mixture of 2 SO2(g) + O2(g) ⇌ 2 SO3(g) will result in an increase in the concentration of SO3(g) as the equilibrium shifts

Step-by-step explanation:

When the volume of the reaction chamber containing the equilibrium mixture of 2 SO2(g) + O2(g) ⇌ 2 SO3(g) is decreased, the system will respond according to Le Chatelier's Principle. This principle indicates that the equilibrium will shift to minimize the effect of the change. In this case, decreasing volume increases pressure, and the reaction will shift towards the side with fewer moles of gas.

Since the reaction produces 2 moles of SO3 for every 3 moles of reactants consumed (2 SO2 and 1 O2), the equilibrium will shift to the right, towards the products. As a result, there will be an increase in the concentration of SO3(g), while the concentration of SO2(g) and O2(g) will decrease.

A decrease in the volume of the reaction chamber will shift the equilibrium towards the side with fewer moles of gas. In this reaction, the forward reaction produces 2 moles of SO₃(g), while the reverse reaction requires 4 moles of gas. Therefore, decreasing the volume will shift the equilibrium to the right, increasing the concentration of SO₃(g).

So, the correct answer is option 3) An increase in the concentration of SO₃(g).