CaCO3(s) ⇄ CaO(s) + CO2(g) ΔH° = 178 kJ/molrxn The reaction system represented above is at equilibri — Equilibrium Chemistry Question

STEPS:

1. Analyze the thermodynamics of the reaction: The reaction is given as:
\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \quad \Delta H^\circ = 178\text{ kJ/mol}_{\text{rxn}}\
Because the enthalpy change ($\Delta H^\circ$) is positive, the reaction is endothermic (it absorbs heat). We can write heat as a reactant:
$$\text{Heat} + \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$$
2. Determine the direction of the required equilibrium shift: The question asks what change will decrease the amount of $\text{CaO}(s)$ in the system. Since $\text{CaO}(s)$ is a product (on the right side of the equation), we must find a disturbance that shifts the equilibrium to the reverse direction (to the left).
3. Apply Le Chatelier's Principle to temperature changes: According to Le Chatelier's Principle, changing the temperature of a system at equilibrium shifts the reaction to counteract the change:
* Lowering the temperature is equivalent to removing heat (a reactant) from the system.
* The system counteracts this cooling by shifting in the heat-producing (exothermic/reverse) direction to regenerate the lost heat.
* Shifting to the left consumes the products $\text{CaO}(s)$ and $\text{CO}_2(g)$ to produce $\text{CaCO}_3(s)$.
4. Conclude the correct option: Because a shift to the left successfully decreases the amount of $\text{CaO}(s)$ in the container, lowering the temperature (Option B) is the correct choice.

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WHY_OTHERS_WRONG:

• Option A is incorrect: Increasing the volume of the reaction vessel at constant temperature decreases the partial pressure of the gaseous product, $\text{CO}_2$. According to Le Chatelier's Principle, the system will shift toward the side with more moles of gas to restore the pressure. Since the reactant side has 0 moles of gas and the product side has 1 mole of gas ($\text{CO}_2$), the equilibrium will shift to the forward direction (to the right), which *increases* the amount of $\text{CaO}(s)$.
• Option C is incorrect: Removing $\text{CO}_2(g)$ (a product) decreases its concentration. According to Le Chatelier's Principle, the system will shift to the forward direction (to the right) to replace the lost product. This forward shift would *increase* the amount of $\text{CaO}(s)$, not decrease it.
• Option D is incorrect: $\text{CaCO}_3(s)$ is a pure solid. In heterogeneous equilibria, pure solids and liquids do not appear in the equilibrium constant expression ($K_p = P_{\text{CO}_2}$). Changing the amount of a pure solid does not alter the concentrations or partial pressures of the reacting species, meaning removing some solid reactant will not shift the equilibrium and has no effect on the amount of $\text{CaO}(s)$.