Question

Write the equilibrium constant expression for the following equilibria: (a) \(\mathrm{Cu}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \Longrightarrow \mathrm{Cu}(\mathrm{OH})_{2}(s)\) (b) \(2 \mathrm{Au}_{2} \mathrm{O}_{3}(s) \rightleftharpoons 4 \mathrm{Au}(s)+3 \mathrm{O}_{2}(g)\) (c) \(\mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \Longrightarrow \mathrm{HSO}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\)

Short answer

The equilibrium constant expressions for the reactions are: (a) \(K_{c} = \frac{1} {[Cu^{2+}][OH^{-}]^2}\), (b) \(K_{p} = [O_{2}]^3\), and (c) \(K_{c} = {[HSO_{3}^{-}][H_{3}O^{+}]} / {[SO_{2}]}\)

Step-by-step solution

Setting up the equilibrium constant for Reaction (a)

For reaction (a), which is \(Cu^{2+}(aq) + 2 OH^{-}(aq) \rightarrow Cu(OH)_{2}(s)\), the equilibrium constant, \(K_{c}\), is given by the concentrations of the products divided by the concentrations of the reactants. However, as the product is a solid and does not feature in the expression, the equilibrium constant expression only includes the reactants. It is given by \(K_{c} = \frac{1} {[Cu^{2+}][OH^{-}]^2}\).

Setting up the equilibrium constant for Reaction (b)

For reaction (b), which is \(2 Au_{2}O_{3}(s) \rightleftharpoons 4 Au(s) + 3 O_{2}(g)\), the equilibrium constant, \(K_{p}\), is given by the partial pressures of the gases in the products. Since the products side has only \(O_{2}\) as a gas and the reactants are solid, only \(O_{2}\) appears in the equilibrium expression. Therefore, the equilibrium constant expression for this reaction is \(K_{p} = [O_{2}]^3\).

Setting up the equilibrium constant for Reaction (c)

For reaction (c), which is \(SO_{2}(g) + 2 H_{2}O(l) \rightarrow HSO_{3}^{-}(aq) + H_{3}O^{+}(aq)\), the equilibrium constant, \(K_{c}\), is given by a ratio of the products to reactants. However, the reactant \(H_{2}O\) is a liquid and so does not appear in the equilibrium constant expression. Therefore, the expression is given by \(K_{c} = {[HSO_{3}^{-}][H_{3}O^{+}]} / {[SO_{2}]}\).

Key concepts

Understanding Chemical Equilibria

Chemical equilibria refer to the balanced state in a chemical reaction where the rates of the forward and reverse reactions are equal. At this point, the concentrations of the reactants and products remain constant over time. This does not mean the amounts are equal, but that their proportions are steady.

The concept of equilibrium is foundational in chemistry because most chemical reactions do not go to complete conversion of reactants to products. Instead, they settle into a state of dynamic equilibrium where both reactants and products coexist.

An important feature of chemical equilibria is the equilibrium constant (denoted as either \(K_c\) for concentration-based reactions or \(K_p\) for pressure-based reactions), which quantifies the concentrations or pressures of the participating species at equilibrium. \

• For Reactions with Gaseous Products: \(K_p\) is expressed in terms of the partial pressures of gases.
  
• For Reactions in Solution: \(K_c\) relies on the molarity of the substances.
  
• For Solids and Pure Liquids: Their concentrations are constant and thus, omitted from equilibrium expressions.

This understanding can help us predict how changing conditions such as concentration, pressure, or temperature might shift the equilibrium according to Le Chatelier's Principle.

Nature of Aqueous Solutions

Aqueous solutions are crucial in chemical reactions, especially in equilibria, as they involve substances dissolved in water. When a substance is dissolved in water, it often dissociates into ions, influencing the reaction's dynamics.

The term 'aqueous' is denoted by the symbol \((aq)\) in chemical equations and signifies that the substance is in a solution with water. For reactions in aqueous solutions, water often acts as a solvent, but interestingly, its concentration is generally omitted from equilibrium constant expressions.


Features of Aqueous Solutions:

• Water is the solvent, providing a medium for reactants to interact at the molecular level.
  
• The solution's properties, such as conductivity, depend on the presence of ions.
  
• For equilibrium expressions, only the solute species appear, as the solvent concentration is effectively constant.

Understanding how substances behave in aqueous solutions is essential for correctly writing reaction expressions and predicting their behavior in dynamic states of equilibrium.

Writing Reaction Expressions

To express a chemical reaction mathematically, we focus on how reactants convert to products and the conditions at equilibrium. A reaction expression is a mathematical representation that illustrates these changes and allows the calculation of the equilibrium constant.

The general format of an equilibrium reaction expression depends on the types of reactants and products:

• The coefficients from the balanced reaction equation become exponents in the expression.
  
• For the given reactions, solids and pure liquids are excluded, as seen in steps 1 and 2 of the solution.
• Concentration (for \(K_c\)) or pressure (for \(K_p\)) of gaseous and dissolved species are used to formulate the expression.

In reaction (a), for instance, the solid \(Cu(OH)_2\) is omitted from the \(K_c\) expression, and only aqueous ions are considered.

Creating these expressions correctly is vital, as it sets the foundation for calculating how equilibrium might shift under varying conditions.