Question

Write an equilibrium expression for each chemical equation involving one or more solid or liquid reactants or products. \begin{equation} \begin{array}{l}{\text { a. } \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q)} \\ {\text { b. } 2 \mathrm{KClO}_{3}(s) \rightleftharpoons 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)}\end{array} \end{equation}\begin{equation}\begin{array}{l}{\text { c. } \mathrm{HF}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q)} \\ {\text { d. } \mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q)}\end{array}\end{equation}

Short answer

a) \(K = \frac{[\mathrm{HCO}_3^-][\mathrm{OH}^-]}{[\mathrm{CO}_3^{2-}]}\)\, b) \(K = [\mathrm{O}_2]^3\)\, c) \(K = \frac{[\mathrm{H}_3\mathrm{O}^+][\mathrm{F}^-]}{[\mathrm{HF}]}\)\, d) \(K = \frac{[\mathrm{NH}_4^+][\mathrm{OH}^-]}{[\mathrm{NH}_3]}\)

Step-by-step solution

- Understanding Equilibrium Expressions

An equilibrium expression for a chemical equation involves the concentrations of the aqueous and gaseous species present in the reaction. Pure solids and liquids do not appear in the equilibrium expression.

- Writing the Equilibrium Expression for Reaction a

For the reaction \(\mathrm{CO}_{3}^{2-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{HCO}_{3}^{-}(aq) + \mathrm{OH}^{-}(aq)\), write the equilibrium constant expression (K) excluding the pure liquid water: \[K = \frac{[\mathrm{HCO}_3^-][\mathrm{OH}^-]}{[\mathrm{CO}_3^{2-}]}\]

- Writing the Equilibrium Expression for Reaction b

For the reaction \(2 \mathrm{KClO}_{3}(s) \rightleftharpoons 2 \mathrm{KCl}(s) + 3 \mathrm{O}_{2}(g)\), solids are not included in the equilibrium expression, only the oxygen gas is included: \[K = [\mathrm{O}_2]^3\]

- Writing the Equilibrium Expression for Reaction c

For the reaction \(\mathrm{HF}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_3\mathrm{O}^{+}(aq) + \mathrm{F}^{-}(aq)\), again the pure liquid water is excluded: \[K = \frac{[\mathrm{H}_3\mathrm{O}^+][\mathrm{F}^-]}{[\mathrm{HF}]}\]

- Writing the Equilibrium Expression for Reaction d

For the reaction \(\mathrm{NH}_{3}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(aq) + \mathrm{OH}^{-}(aq)\), we exclude water to write the equilibrium expression: \[K = \frac{[\mathrm{NH}_4^+][\mathrm{OH}^-]}{[\mathrm{NH}_3]}\]

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, leading to no net change in the amount of reactants and products. This dynamic state doesn't mean the reactants and products aren't moving; rather, their concentrations remain constant over time. To visualize this, imagine a busy street where the number of cars entering is equal to the number leaving, keeping the traffic constant even though the cars are in motion.

An important aspect of equilibrium is that it can only be achieved in a closed system, where no substances are added or removed. External factors such as temperature, pressure, and concentration changes can shift the equilibrium, which is described by Le Châtelier's principle.

Aqueous Solutions in Equilibrium

When dealing with aqueous solutions in equilibrium, it's essential to consider that not all components of the solution will appear in the equilibrium expression. Pure liquids, such as water (\text{H\(_2\)O}(l)), and solids are left out because their concentrations do not change during the reaction. Their presence is constant and thus does not affect the equilibrium position.

For instance, salt in water (\text{NaCl}(aq)) will dissociate into ions (\text{Na\(^+\)} and \text{Cl\(^-\)}), which take part in equilibrium. However, the water itself, being a solvent in large excess, does not appear in the equilibrium expression. This concept helps simplify the process of writing equilibrium expressions for reactions involving aqueous solutions by focusing on only the species whose concentrations change.

Writing Equilibrium Constant Expressions

To write an equilibrium constant expression, identify the reactants and products that are in the aqueous or gaseous state—these will appear in your expression. The equilibrium constant (K) quantifies the ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients in the balanced chemical equation.

For example, for a generic reaction where \text{aA} + \text{bB} \rightleftharpoons \text{cC} + \text{dD}, the equilibrium expression is:\[K = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}\]
Each concentration is expressed in moles per liter (Molarity). Solids and liquids are omitted as their concentration remains unchanged. By mastering the construction of these expressions, you can predict how changes in conditions will shift the equilibrium, helping you better understand reaction dynamics in chemical systems.