Problem 14 - Chemistry: The Central Science

Question

Write the expressions for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)\) (b) \(\mathrm{Ti}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \operatorname{TiCl}_{4}(l)\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g)\) (d) \(\mathrm{C}(s)+2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{~g})\) (e) \(4 \mathrm{HCl}(a q)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cl}_{2}(g)\)

Step-by-Step Solution

Answer

The equilibrium constant expressions and type of reactions for each case are: (a) \(K_c = \frac{[\mathrm{O_2}]^3}{[\mathrm{O_3}]^2}\) (homogeneous) (b) \(K_c = \frac{[\mathrm{TiCl_4}]}{[\mathrm{Cl_2}]^2}\) (heterogeneous) (c) \(K_c = \frac{[\mathrm{C_2H_6}]^2[\mathrm{O_2}]}{[\mathrm{C_2H_4}]^2[\mathrm{H_2O}]^2}\) (homogeneous) (d) \(K_c = \frac{[\mathrm{CH_4}]}{[\mathrm{H_2}]^2}\) (heterogeneous) (e) \(K_c = \frac{[\mathrm{H_2O}]^2[\mathrm{Cl_2}]^2}{[\mathrm{HCl}]^4[\mathrm{O_2}]}\) (heterogeneous)

Step 1: Write Kc expression

The Kc expression is the ratio of the product concentrations to the reactant concentrations raised to their stoichiometric coefficients. Here, it is given by: \[K_c = \frac{[\mathrm{O_2}]^3}{[\mathrm{O_3}]^2}\]

Step 2: Identify the type of reaction

Since both O3 and O2 are in gaseous phase, the reaction is homogeneous. (b) \(\mathrm{Ti}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \operatorname{TiCl}_{4}(l)\)

Step 1: Write Kc expression

The Kc expression for this reaction is given by: \[K_c = \frac{[\mathrm{TiCl_4}]}{[\mathrm{Cl_2}]^2}\]

Step 2: Identify the type of reaction

The reactants and products are in different phases (solid, gas, and liquid), so the reaction is heterogeneous. (c) \(2 \mathrm{C}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g)\)

Step 1: Write Kc expression

The Kc expression for this reaction is: \[K_c = \frac{[\mathrm{C_2H_6}]^2[\mathrm{O_2}]}{[\mathrm{C_2H_4}]^2[\mathrm{H_2O}]^2}\]

Step 2: Identify the type of reaction

Since all species are in the gaseous phase, the reaction is homogeneous. (d) \(\mathrm{C}(s)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)\)

Step 1: Write Kc expression

The Kc expression for this reaction is: \[K_c = \frac{[\mathrm{CH_4}]}{[\mathrm{H_2}]^2}\]

Step 2: Identify the type of reaction

The reactants and products are in different phases (solid and gas), so the reaction is heterogeneous. (e) \(4 \mathrm{HCl}(a q)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cl}_{2}(g)\)

Step 1: Write Kc expression

The Kc expression for this reaction is: \[K_c = \frac{[\mathrm{H_2O}]^2[\mathrm{Cl_2}]^2}{[\mathrm{HCl}]^4[\mathrm{O_2}]}\]

Step 2: Identify the type of reaction

The reactants and products are in different phases (aqueous, gas, and liquid), so the reaction is heterogeneous.

Key Concepts

Chemical Equilibrium

Understanding chemical equilibrium is crucial for studying how reactions proceed and eventually settle into a state of balance. At equilibrium, the rate at which reactants form products is the same as the rate at which products decompose back into reactants. Even though this might give the illusion of inactivity, it's a dynamic state where concentrations of all reactants and products remain constant over time.

When writing the equilibrium constant (\( K_c \) ) expression for a reaction, we're quantifying the state of equilibrium. It is a ratio of the concentration of products raised to their stoichiometric coefficients to that of the reactants raised to their stoichiometric coefficients, as shown in the provided step-by-step solutions.

Improving Understanding of Equilibrium Concepts

It's not enough to only memorize the formula for calculating the equilibrium constant. Instead, understanding that at equilibrium, no net change in concentration occurs provides a more profound understanding of the balance within a reaction. This notion allows us to predict how a reaction will respond to changes in concentration, temperature, or pressure, following Le Chatelier's principle. Furthermore, each unique reaction has its own equilibrium constant value that can be applied to predict the extent of the reaction under certain conditions.

Homogeneous and Heterogeneous Reactions

In chemistry, reactions can be classified based on the phases of reactants and products. In a homogeneous reaction, all reactants and products are in the same phase, typically all gases or all liquids. This uniformity simplifies the expression of the equilibrium constant as we've seen in solution (a) and (c), where all substances involved are in the gaseous phase.

Conversely, a heterogeneous reaction involves reactants and products in different phases. Solid and liquid concentrations are typically omitted from the equilibrium constant expression because their concentrations are considered to remain constant during the reaction. This is visible in reactions (b), (d), and (e), where we observe solids, gases, and liquids interacting. It's important to note that in a heterogeneous reaction, even though the phases are different, the principle of equilibrium still applies.

Visualizing Phase Interactions

To enhance comprehension, envisioning how molecules interact in their respective phases can be incredibly illuminating. For instance, gases collide and react with each other in the entire volume they occupy, while solids react at their surface. Recognizing this can help students understand why certain reactants might not appear in an equilibrium constant expression due to their phase.

Stoichiometry

Stoichiometry is the section of chemistry that involves determining the relative quantities of reactants and products in chemical reactions. It's essential in formulating the equilibrium constant. In stoichiometry, the coefficients from the balanced chemical equation indicate the ratio in which substances react or are produced.

For instance, in the reaction for solution (a), the stoichiometric coefficients tell us that two molecules of ozone (\( O_3 \)) react to form three molecules of oxygen (\( O_2 \)). The exponents in the equilibrium constant express this ratio, reflecting that the reaction consumes and produces these molecules in a specific, fixed proportion. When we write equilibrium constant expressions, stoichiometry is the backbone that ensures the laws of conservation of mass and constant composition are upheld.

Practicing Balancing Equations

To fully grasp stoichiometry, practice balancing chemical equations. This will not only enable the correct formulation of equilibrium constants but also help to predict the amounts of products formed and reactants needed. Through practice, discerning the quantitative relationships within chemical reactions becomes second nature, which is a skill applicable in both theoretical and practical chemistry.