$$\ce{CaCO3(s) <--> CaO(s) + CO2(g)}$$ $$\Delta H^\circ = \pu{178 kJ/mol_{rxn}}$$

The reaction system represented above is at equilibrium. Which of the following will decrease the amount of $\ce{CaO(s)}$ in the system?

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Lets examine all the choices and their effects on equilibrium position:

Increasing the volume of the reaction corresponds to a decrease in pressure. The equilibrium position would shift to produce more gas molecules. In this case, it will shift to the right.

Lowering the temperature effectively decreases reactants since the reaction is endothermic. The equilibrium position would shift to the left.

Removing $\ce{CO2(g)}$, a product, would shift the equilibrium position to the right.

Removing $\ce{CaCO3(s)}$ has no effect on equilibrium position since it is a solid.