Solids and liquids are not factored into the equilibrium expression of heterogeneous systems
The correct answer and explanation is:
Answer: True.
In heterogeneous equilibrium systems, where reactants and products are in different phases (solid, liquid, or gas), solids and liquids are not included in the equilibrium expression. This is because their concentration remains constant throughout the reaction.
In an equilibrium expression, the concentrations of gases and aqueous solutions are represented in terms of their molarity or partial pressure. However, for pure solids and liquids, their concentrations do not change significantly during the reaction. This is due to the fact that the concentration of a solid is determined by its density and volume, both of which remain constant unless there is a significant physical change. For liquids, the concentration is similarly constant if the volume and temperature are unchanged.
For example, consider the reaction:
CaCO3(s)⇌CaO(s)+CO2(g)\text{CaCO}_3 (s) \rightleftharpoons \text{CaO} (s) + \text{CO}_2 (g)
The equilibrium expression would be:
K=[CO2]1K = \frac{[\text{CO}_2]}{1}
Here, the concentrations of the solids, CaCO₃ and CaO, are not included in the expression because they do not change in a meaningful way during the reaction. The only component that matters in this case is the concentration (or partial pressure) of CO₂, which is a gas.
The principle behind excluding solids and liquids in equilibrium expressions reflects the idea that only species whose concentrations can change (such as gases or aqueous solutions) affect the position of equilibrium. The concentrations of solids and liquids are considered “standardized” or “constant” and do not need to be included in calculations.
This simplification makes the equilibrium expression manageable and easier to use in practical applications, such as calculating equilibrium constants or predicting the direction of a reaction.