2 BaO2(s) ⇄ 2 BaO(s) + O2(g) ΔH° = 162 kJ/mol_rxn A sealed rigid vessel contains BaO2(s) in equilibr — Equilibrium Chemistry Question

STEPS:

1. Identify the goal of the question: The objective is to determine which change will shift the equilibrium to the left to increase the amount of $\text{BaO}_2(s)$ (the reactant) in the vessel. According to Le Chatelier's Principle, a system at equilibrium will shift to counteract any applied stress.
2. Examine the state of matter of the substances:
* $\text{BaO}_2(s)$ and $\text{BaO}(s)$ are pure solids.
* $\text{O}_2(g)$ is a gas.
* Because pure solids do not change their concentration/density and have an activity of 1, they do not appear in the equilibrium constant expression. The equilibrium constant expression for this reaction is simply:
$$K_p = P_{\text{O}_2}$$
3. Analyze the thermodynamic parameter ($\Delta H^\circ$): The enthalpy change of the reaction is $\Delta H^\circ = +162\text{ kJ/mol}_{rxn}$. Because $\Delta H^\circ$ is positive, the forward reaction is endothermic (it absorbs heat). We can conceptually treat "heat" as a reactant on the left side of the equation:
$$\text{Heat} + 2\text{ BaO}_2(s) \rightleftharpoons 2\text{ BaO}(s) + \text{O}_2(g)$$
4. Determine the effect of temperature on the equilibrium:
* According to Le Chatelier's Principle, lowering the temperature is equivalent to "removing heat" from the system.
* To counteract this stress and regenerate the lost heat, the system shifts in the exothermic direction (to the left, toward the reactants).
5. Conclude which option is correct: The shift to the left consumes the products and produces more of the solid reactant $\text{BaO}_2(s)$. Thus, lowering the temperature increases the amount of $\text{BaO}_2(s)$ present, making Option D the correct choice.

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WHY_OTHERS_WRONG:

• A is incorrect: Removing a small amount of $\text{O}_2(g)$ decreases the partial pressure of a product. The system will shift to the right (forward direction) to produce more $\text{O}_2(g)$, which would consume and decrease the amount of $\text{BaO}_2(s)$.
• B is incorrect: $\text{BaO}(s)$ is a pure solid. Adding or removing a pure solid does not change its active concentration or alter the reaction quotient ($Q$). Therefore, removing $\text{BaO}(s)$ causes no shift in the equilibrium.
• C is incorrect: Adding an inert gas like Helium ($\text{He}$) to a rigid vessel increases the total pressure, but it does not change the volume of the container or the partial pressure of the reacting gas ($\text{O}_2$). Because the partial pressure of $\text{O}_2$ remains constant, there is no shift in the equilibrium position.