Question

consider the following reaction, which is at equilibrium. What effect will reducing the volume of the reaction mixture have? CuS (s) + O2 (g) ⇋ Cu (s) + SO2 (g) Consider the following reaction, which is at equilibrium. What effect will reducing the volume of the reaction mixture have? CuS (s) + O2 (g) ⇋ Cu (s) + SO2 (g) The reaction will shift to the left in the direction of reactants. The reaction will shift to the right in the direction of products. The equilibrium constant will decrease. No effect will be observed. The equilibrium constant will increase.

Answer 1

Answer:

No effect will be observed.

Explanation:

Let´s consider the following reaction.

CuS(s) + O₂(g) ⇋ Cu(s) + SO₂(g)

What effect will reducing the volume of the reaction mixture have?

We need to take into account Le Chatelier's Principle: if a system at equilibrium suffers a perturbation, it will react to counteract that perturbation. According to Boyle's Law, if the volume is reduced then the pressure increases. According to Le Chatelier's Principle the system will try to reduce pressure by shifting the equilibrium towards where there are fewer moles of gases. In this case, there is 1 mol of gases on the left side and 1 mol of gases on the right side. That´s why no effect will be observed.

Answer 2

There will be no effect to be observed after reducing the volume of the reaction mixture.

What is volume?

The volume of any substance is the three-dimensional space taken by the substance.

When the volume of the system decreases, the equilibrium shifts in favor of the direction that produces lesser moles of gas.

When the volume of the system grows, the equilibrium shifts in favor of the direction that produces more moles of gas.

The reaction is an equilibrium, there will be no effect on the volume.

Thus, the correct option is d, No effect will be observed.  

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