Question

Write equilibrium constant expressions \((K)\) for the following reactions: (a) \(2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \rightleftharpoons\) $$ \begin{array}{l} \text { (b) } 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+3 \mathrm{Cu}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \\ \text { (c) } \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightleftharpoons \mathrm{CaCO}_{3}(s) \end{array} $$

Short answer

Question: Write the equilibrium constant expressions for the following reactions: (a) 2 NO₃⁻(aq) + 8 H⁺(aq) + 3 Cu(s) ⇌ 2 NO(g) + 3 Cu²⁺(aq) + 4 H₂O(l) (b) 2 PbS(s) + 3 O₂(g) ⇌ 2 Pb²⁺(aq) + 2 SO₄²⁻(aq) (c) Ca²⁺(aq) + CO₃²⁻(aq) ⇌ CaCO₃(s) Answer: (a) \(K_{a}=\frac{[\mathrm{NO}]^{2} \cdot [\mathrm{Cu}^{2+}]^{3} \cdot [\mathrm{H}_{2}\mathrm{O}]^{4}}{[\mathrm{NO}_{3}^{-}]^{2} \cdot [\mathrm{H}^{+}]^{8}}\) (b) \(K_{b}=\frac{[\mathrm{Pb}^{2+}]^{2} \cdot [\mathrm{SO}_{4}^{2-}]^{2}}{[\mathrm{O}_{2}]^{3}}\) (c) \(K_{c}=\frac{1}{[\mathrm{Ca}^{2+}] \cdot [\mathrm{CO}_{3}^{2-}]}\)

Step-by-step solution

Write the balanced chemical equation for the reaction

For reaction (a), the balanced chemical equation is given as: $$ 2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \rightleftharpoons 2 \mathrm{NO}(g)+3 \mathrm{Cu}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) $$

Write the equilibrium constant expression

For reaction (a), the equilibrium constant expression \(K\) is defined as: $$ K_{a}=\frac{[\mathrm{NO}]^{2} \cdot [\mathrm{Cu}^{2+}]^{3} \cdot [\mathrm{H}_{2}\mathrm{O}]^{4}}{[\mathrm{NO}_{3}^{-}]^{2} \cdot [\mathrm{H}^{+}]^{8}} $$ Note that since \(\mathrm{Cu}(s)\) is a solid, it is not included in the equilibrium constant expression. 2. Reaction (b) - PbS(s), O₂(g):

Write the balanced chemical equation for the reaction

For reaction (b), the balanced chemical equation is given as: $$ 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Pb}^{2+}(a q) + 2 \mathrm{SO}_{4}^{2-}(a q) $$

Write the equilibrium constant expression

For reaction (b), the equilibrium constant expression \(K\) is defined as: $$ K_{b}=\frac{[\mathrm{Pb}^{2+}]^{2} \cdot [\mathrm{SO}_{4}^{2-}]^{2}}{[\mathrm{O}_{2}]^{3}} $$ Note that since \(\mathrm{PbS}(s)\) is a solid, it is not included in the equilibrium constant expression. 3. Reaction (c) - Ca²⁺(aq), CO₃²⁻(aq), CaCO₃(s):

Write the balanced chemical equation for the reaction

For reaction (c), the balanced chemical equation is given as: $$ \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightleftharpoons \mathrm{CaCO}_{3}(s) $$

Write the equilibrium constant expression

For reaction (c), the equilibrium constant expression \(K\) is defined as: $$ K_{c}=\frac{1}{[\mathrm{Ca}^{2+}] \cdot [\mathrm{CO}_{3}^{2-}]} $$ Note that since \(\mathrm{CaCO}_{3}(s)\) is a solid, it is not included in the equilibrium constant expression.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is crucial for grasping how reactions occur. At equilibrium, the rates of the forward and reverse reactions are equal, and the concentrations of the reactants and products remain constant over time. This does not mean the reaction has stopped, but rather that it is dynamically balanced.

When writing equilibrium constant expressions, it's key to remember that only the concentrations of gases and solutes are included. Pure solids, like copper in the given exercise, and pure liquids, such as water, are omitted because their concentrations do not change during the reaction and are therefore set to one.

This gives us expressions that quantify the balance of a reaction. For example, in reaction (a) in the textbook, the equilibrium constant expression was written as follows:
\[ K_a = \frac{[\mathrm{NO}]^2 \cdot [\mathrm{Cu}^{2+}]^3 \cdot [\mathrm{H}_2\mathrm{O}]^4}{[\mathrm{NO}_3^{-}]^2 \cdot [\mathrm{H}^{+}]^8} \]
Here, the concentration of each ion is raised to the power of its coefficient in the balanced equation, highlighting the relationship between the amounts of reactants and products at equilibrium.

Reaction Quotient

The reaction quotient, 'Q', is used to determine the direction in which a reaction will proceed to achieve equilibrium. It is calculated using the same formula as the equilibrium constant, 'K', but with the initial concentrations of the reactants and products instead of the concentrations at equilibrium.

Comparing 'Q' to 'K' offers insight into the system's status:

• If \( Q < K \), the reaction will proceed forward to produce more products.
• If \( Q > K \), the reaction will proceed in the reverse direction to produce more reactants.
• If \( Q = K \), the reaction is at equilibrium.

For example, if we had initial concentrations for reaction (a), we could calculate 'Q' and predict the direction the reaction will take to reach equilibrium.

Le Chatelier's Principle

Le Chatelier's principle is a guiding concept in chemical equilibrium suggesting that if a dynamic equilibrium is disturbed, the system will adjust to counteract the change and restore a new equilibrium. This principle can be applied in several scenarios:

• Changing the concentration of reactants or products will shift the equilibrium to either the right or left to re-establish balance.
• Varying the temperature can favor either the endothermic or exothermic reaction direction, depending on whether heat is treated as a product or reactant.
• Altering the pressure of a gaseous system will shift the equilibrium toward the side with fewer moles of gas.

Understanding this principle helps predict the effects of stressors on a reaction, like when a certain product or reactant is added or removed, or when the temperature changes, as it tells us the direction in which the equilibrium will adjust.

Solubility Product

The solubility product constant, 'Ksp', is a special type of equilibrium constant that applies to the dissolution of sparingly soluble ionic compounds. Essentially, it represents the extent to which a compound can dissolve in water to form its constituent ions.

Reaction (c) in the exercise is an excellent example, illustrating the dissolution of calcium carbonate:
\[ \mathrm{Ca}^{2+}(aq) + \mathrm{CO}_{3}^{2-}(aq) \rightleftharpoons \mathrm{CaCO}_{3}(s) \]The 'Ksp' for the reaction is given by:
\[ K_{c} = \frac{1}{[\mathrm{Ca}^{2+}] \cdot [\mathrm{CO}_{3}^{2-}]} \]
Here, because the product is a solid, we exclude it from the expression and invert the equation, focusing on the equilibrium between the dissolved ions and the undissolved solid.
A higher 'Ksp' value indicates greater solubility, allowing predictions about how much of the compound can dissolve under given conditions.