During the reduction of copper(II) oxide by carbon monoxide, CuO(s) + CO(g) ⇌ Cu(s) + CO2(g), which change will drive the equilibrium toward more Cu(s) production?

• A. Remove CO(g)
• B. Add a catalyst
• C. Increase the volume of the container
• D. Remove CO2(g)

Answer: (D)

Removing the product CO2 from the system shifts the equilibrium to the right, producing more Cu and CO2 to replace what was removed. In this reaction, the solids have activities that don’t change with amount, so the position of equilibrium is governed by the gas-phase species. Lowering the CO2 concentration makes the system favor the formation of more products (Cu) to restore equilibrium, increasing Cu metal production.

Catalysts speed both directions equally, so they don’t change where the equilibrium lies. Increasing volume would only shift the equilibrium if there were a different number of gas moles on the two sides; here there is essentially one mole of gas on each side, so volume change has little effect. Removing CO would also push the reaction to the right, but removing CO2 directly reduces the product concentration and effectively drives more Cu formation.