Question

Would decreasing the volume of the container for each of the following reactions cause the equilibrium to shift in the direction of the products, the reactants, or not change?

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3O₂(g) ↔ 2O₃(g)
2SO₂(g) + 2H₂O(g) ↔ 2H₂S(g) + 3O₂(g)
2NOBr(g) ↔ 2NO(g) + Br₂(g)

Shift in the direction of the products ____
Shift in the direction of the reactants ____
Equilibrium does not change ____

Answer 1

Decreasing the volume of the container will shift the equilibrium towards the side with fewer gas molecules. For 3O₂(g) ↔ 2O₃(g), it shifts right; for 2SO₂(g) + 2H₂O(g) ↔ 2H₂S(g) + 3O₂(g), it shifts left, and for 2NOBr(g) ↔ 2NO(g) + Br₂(g), there is no shift.

Step-by-step explanation:

The question focuses on how changes in the volume of a container affect the position of equilibrium in various reversible reactions. According to Le Chatelier's principle, an increase in pressure by decreasing the volume of a gaseous system will cause the equilibrium to shift towards the side with fewer moles of gas.

For the reaction 3O₂(g) ↔ 2O₃(g), there are three moles of reactants and two moles of products. When the volume decreases, the equilibrium will shift towards products due to a decrease in moles of gas, following the principle.

In the reaction 2SO₂(g) + 2H₂O(g) ↔ 2H₂S(g) + 3O₂(g), there are four moles of reactants and five moles of products. Decreasing the volume will shift the equilibrium towards the reactants (to the left), as it favors the side with fewer gas molecules.

Finally, for the reaction 2NOBr(g) ↔ 2NO(g) + Br₂(g), there is no change in mole numbers on both sides of the equation, so the equilibrium will not shift with a change in volume.