$$\ce{2BaO2(s) <--> 2BaO(s) + O2(g) }$$ $$\Delta H^\circ = \pu{162 kJ/mol_{rxn}}$$

A sealed rigid vessel contains $\ce{BaO2(s)}$ in equilibrium with $\ce{BaO(s)}$ and $\ce{O2(g)}$ as represented by the equation above. Which of the following changes will increase the amount of $\ce{BaO2(s)}$ in the vessel?

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Since the reaction is endothermic, we can treat heat as a reactant. Lowering temperature effectively reduces the concentration or amount of the reactants, leading to a shift in equilibrium position to the left.

Removing $\ce{O2}$ would shift the equilibrium position the right.

Removing the solid $\ce{BaO}$ has no effect on the concentration of that species, and therefore has no effect on the equilibrium position.

Adding $\ce{He}$ gas, which is not involved in the reaction, would not shift the equilibrium position.