Question

consider the equation below.
caco₃(s)⇌cao(s) + co₂(g)
what is the equilibrium constant expression for the given reaction?
○ $k_{eq}=\frac{cao}{caco₃}$
○ $k_{eq}=\frac{co₂cao}{caco₃}$
○ $k_{eq}=co₂$
○ $k_{eq}=\frac{1}{co₂}$

Explanation:

Step1: Recall equilibrium constant rules

For a reaction at equilibrium, the equilibrium constant expression (\( K_{eq} \)) is written as the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. However, pure solids (s) and pure liquids (l) are not included in the equilibrium constant expression because their concentrations are considered constant.

Step2: Analyze the given reaction

In the reaction \( \text{CaCO}_3(\text{s})
ightleftharpoons \text{CaO}(\text{s}) + \text{CO}_2(\text{g}) \), \( \text{CaCO}_3 \) and \( \text{CaO} \) are pure solids. So, we exclude their concentrations from the equilibrium constant expression. The only gaseous (or species with variable concentration) product is \( \text{CO}_2 \) (with a stoichiometric coefficient of 1) and there are no gaseous reactants (since \( \text{CaCO}_3 \) is solid). Thus, the equilibrium constant expression is \( K_{eq} = [\text{CO}_2] \).

Answer:

\( K_{eq} = [\text{CO}_2] \)