Problem 55 - Chemistry The Science in Context

Question

Mixing aqueous solutions of sodium bicarbonate and calcium chloride results in this reaction: \(2 \mathrm{NaHCO}_{3}(a q)+\mathrm{CaCl}_{2}(a q) \rightleftharpoons\) $$ 2 \mathrm{NaCl}(a q)+\mathrm{CO}_{2}(g)+\mathrm{CaCO}_{3}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell) $$ Write the \(K_{\mathrm{c}}\) expression for the reaction.

Step-by-Step Solution

Answer

\(2 \mathrm{NaHCO}_{3} (\mathrm{aq}) + \mathrm{CaCl}_{2} (\mathrm{aq}) \rightleftharpoons 2 \mathrm{NaCl} (\mathrm{aq}) + \mathrm{CO}_{2} (\mathrm{g}) + \mathrm{CaCO}_{3}(\mathrm{s}) + \mathrm{H}_{2} \mathrm{O}(\ell)\) Answer: \(K_c = \frac{[\mathrm{NaCl}]^2 \cdot [\mathrm{CO}_{2}]}{[\mathrm{NaHCO}_{3}]^2 \cdot [\mathrm{CaCl}_{2}]}\)

Step 1: Identify the products and reactants in the balanced chemical equation

For the given balanced chemical reaction: \(2 \mathrm{NaHCO}_{3} (\mathrm{aq}) + \mathrm{CaCl}_{2} (\mathrm{aq}) \rightleftharpoons 2 \mathrm{NaCl} (\mathrm{aq}) + \mathrm{CO}_{2} (\mathrm{g}) + \mathrm{CaCO}_{3}(\mathrm{s}) + \mathrm{H}_{2} \mathrm{O}(\ell)\) The reactants are \(\mathrm{NaHCO}_{3}\) and \(\mathrm{CaCl}_{2}\). The products are \(\mathrm{NaCl}\), \(\mathrm{CO}_{2}\), \(\mathrm{CaCO}_{3}\), and \(\mathrm{H}_{2}\mathrm{O}\).

Step 2: Write the expression for the equilibrium constant, Kc

The equilibrium constant, \(K_c\), is the ratio of the concentrations of products raised to the power of their stoichiometric coefficients divided by the concentrations of reactants raised to the power of their stoichiometric coefficients. However, we should not consider concentrations of solids or pure liquids in the \(K_c\) expression. For the given balanced reaction, the Kc expression is: \(K_c = \frac{[\mathrm{NaCl}]^2 \cdot [\mathrm{CO}_{2}]}{[\mathrm{NaHCO}_{3}]^2 \cdot [\mathrm{CaCl}_{2}]}\) Note that we did not include the concentrations of \(\mathrm{CaCO}_3\) (solid) and \(\mathrm{H}_2\mathrm{O}\) (liquid) in the \(K_c\) expression because their concentrations remain constant throughout the reaction.

Key Concepts

Chemical Equilibrium

The concept of chemical equilibrium is pivotal in understanding why some reactions do not proceed to completion and instead reach a state where the reactants and products remain constant over time. In the context of the given reaction between sodium bicarbonate and calcium chloride,

chemical equilibrium signifies that the forward reaction (the formation of products) and the reverse reaction (the reformation of reactants) occur at the same rate. This dynamic situation results in no noticeable change in the concentration of either the reactants or the products, even though both reactions are still proceeding.

To visualize this, imagine a busy intersection where an equal number of cars enter and leave, maintaining a consistent traffic flow—neither building up nor emptying. Similarly, at equilibrium, the system's composition stabilizes, and the reaction seems to have 'stopped', even though it still actively proceeds at a molecular level.

Stoichiometry

Stoichiometry plays a fundamental role when dealing with chemical reactions, as it involves the quantitative relationships between reactants and products. It's the math behind chemistry, allowing us to predict the amounts of substances consumed and produced in a given reaction.

In the reaction at hand, stoichiometry dictates the numerical coefficients in the balanced chemical equation, implying that 2 moles of sodium bicarbonate react with 1 mole of calcium chloride to produce 2 moles of sodium chloride, 1 mole of carbon dioxide, 1 mole of calcium carbonate, and 1 mole of water. These proportions are essential when writing the equilibrium constant expression because they determine the exponents in the equation. For example, since 2 moles of NaCl are produced, the concentration of NaCl is squared in the equilibrium constant \(K_c\) expression.

Solubility Rules

Solubility rules are a set of guidelines that predict whether a compound will dissolve in water, forming an aqueous solution, or if it will remain insoluble. These rules are based on the compound's anion and cation, and they help us understand how substances will behave when mixed.

In our exercise, calcium carbonate \(CaCO_3\) is formed as a precipitate, indicating it is not soluble in the aqueous solution according to solubility rules. This fact is utilized in the step-by-step solution to justify excluding \(CaCO_3\) from the equilibrium constant expression. Knowing the solubility of compounds helps us understand not only which components to include in equilibrium calculations but also what to expect in terms of a reaction's physical outcome—a key aspect of predicting and observing chemical behavior.

Reaction Quotient

The reaction quotient \(Q\) serves as a predictive tool for understanding the direction in which a reaction must shift to achieve equilibrium. It is calculated using the same formula as the equilibrium constant \(K_c\), but unlike \(K_c\), \(Q\) can be determined at any point during the reaction, not just at equilibrium.

The values of \(Q\) and \(K_c\) can then be compared to predict whether the reaction will proceed forward \(QK_c\), or is already at equilibrium \(Q=K_c\). For instance, if \(Q\) is determined in the reaction before reaching equilibrium and found to be less than the known \(K_c\), it indicates that more product must be formed for equilibrium to be established. Understanding \(Q\) aids in grasping the dynamic nature of chemical processes and provides insight into the behavior of reactions under various conditions.