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

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.