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.

Question

Larry Myers

Chemistry

11 August, 18:33

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.

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Answers (1)

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Ryland Graham · Answer 1

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.