Problem 16

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) \(\mathrm{Ti}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{TiCl}_{4}(l)\) (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) \(\mathrm{C}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{4}(g)\) (e) \(4 \mathrm{HCl}(a q)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cl}_{2}(g)\) (f) \(2 \mathrm{C}_{8} \mathrm{H}_{\mathrm{t}}(l)+25 \mathrm{O}_{2}(g) \rightleftharpoons 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g)\) (g) \(2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \rightleftharpoons 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(l)\)

Step-by-Step Solution

Verified
Answer
The equilibrium constant expressions for the given reactions are: (a) Homogeneous reaction: \(K_c = \frac{[\mathrm{O}_2]^3}{[\mathrm{O}_3]^2}\) (b) Heterogeneous reaction: \(K_c = [\mathrm{Cl}_2]^2\) (c) Homogeneous reaction: \(K_c = \frac{[\mathrm{C}_2\mathrm{H}_6]^2[\mathrm{O}_2]}{[\mathrm{C}_2\mathrm{H}_4]^2[\mathrm{H}_2\mathrm{O}]^2}\) (d) Heterogeneous reaction: \(K_c = \frac{[\mathrm{CH}_4]}{[\mathrm{H}_2]^2}\) (e) Heterogeneous reaction: \(K_c = \frac{[\mathrm{Cl}_2]^2}{[\mathrm{HCl}]^4[\mathrm{O}_2]}\) (f) Heterogeneous reaction: \(K_c = \frac{[\mathrm{CO}_2]^ {16} [\mathrm{H}_2\mathrm{O}]^{18}}{[\mathrm{O}_2]^{25}}\) (g) Heterogeneous reaction: \(K_c = \frac{[\mathrm{CO}_2]^{16}}{[\mathrm{O}_2]^{25}}\)
1Step 1: Equilibrium Constant Expression
The reaction is homogeneous since both reactants and products are gases. The equilibrium constant expression is: \[K_c = \frac{[\mathrm{O}_2]^3}{[\mathrm{O}_3]^2}\] (b) \(\mathrm{Ti}(s)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{TiCl}_{4}(l)\)
2Step 2: Equilibrium Constant Expression
The reaction is heterogeneous since the reactants and products are in different phases. The equilibrium constant expression excludes solids and liquids, so it is: \[K_c = [\mathrm{Cl}_2]^2\] (c) \(2 \mathrm{C}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g)\)
3Step 3: Equilibrium Constant Expression
The reaction is homogeneous since all reactants and products are gases. The equilibrium constant expression is: \[K_c = \frac{[\mathrm{C}_2\mathrm{H}_6]^2[\mathrm{O}_2]}{[\mathrm{C}_2\mathrm{H}_4]^2[\mathrm{H}_2\mathrm{O}]^2}\] (d) \(\mathrm{C}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{4}(g)\)
4Step 4: Equilibrium Constant Expression
The reaction is heterogeneous since the reactants and products are in different phases. The equilibrium constant expression excludes solids, so it is: \[K_c = \frac{[\mathrm{CH}_4]}{[\mathrm{H}_2]^2}\] (e) \(4 \mathrm{HCl}(a q)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Cl}_{2}(g)\)
5Step 5: Equilibrium Constant Expression
The reaction is heterogeneous since the reactants and products are in different phases. The equilibrium constant expression includes only the gases and aqueous substances: \[K_c = \frac{[\mathrm{Cl}_2]^2}{[\mathrm{HCl}]^4[\mathrm{O}_2]}\] (f) \(2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \rightleftharpoons 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g)\)
6Step 6: Equilibrium Constant Expression
The reaction is heterogeneous since the reactants and products are in different phases. The equilibrium constant expression includes only gases: \[K_c = \frac{[\mathrm{CO}_2]^ {16} [\mathrm{H}_2\mathrm{O}]^{18}}{[\mathrm{O}_2]^{25}}\] (g) \(2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \rightleftharpoons 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(l)\)
7Step 7: Equilibrium Constant Expression
The reaction is heterogeneous since the reactants and products are in different phases. The equilibrium constant expression includes only gases: \[K_c = \frac{[\mathrm{CO}_2]^{16}}{[\mathrm{O}_2]^{25}}\]

Key Concepts

Homogeneous ReactionHeterogeneous ReactionChemical EquilibriumPhases of Matter
Homogeneous Reaction
A homogeneous reaction involves reactants and products that exist in the same phase of matter, usually gases or solutions. An example is the reaction between ozone (O3) and oxygen (O2), both in the gaseous state, which can be represented as 2 O3(g) ⇌ 3 O2(g).

In this case, we can write the equilibrium constant expression, Kc, by using the concentrations of the gases. As both reactants and products are in the same phase, all of them are included in the expression. For the given reaction, the equilibrium constant expression is \[K_c = \frac{[\mathrm{O}_2]^3}{[\mathrm{O}_3]^2}\].

Understanding homogeneous reactions is important because the behavior of the reaction can be more straightforward to predict as all species interact in a single phase. This simplification makes calculations and conceptual understanding of the reaction's dynamics easier.
Heterogeneous Reaction
On the flip side, a heterogeneous reaction involves substances in different phases, such as gases, liquids, or solids. For instance, the formation of titanium tetrachloride from titanium solid and chlorine gas, Ti(s) + 2 Cl2(g) ⇌ TiCl4(l), is a heterogeneous reaction.

When writing the equilibrium constant expression for a heterogeneous reaction, we include only the concentrations of the reactants and products that are in gases or dissolved in solution (aqueous). Solids and liquids are omitted because their concentrations do not change during the reaction. Thus, for this reaction, the expression is \[K_c = [\mathrm{Cl}_2]^2\].

This simplification assumes that the activity of pure solids and liquids is constant and therefore, they do not affect the equilibrium state of the reaction. Understanding this concept is crucial for accurately predicting the progress of a reaction involving different phases.
Chemical Equilibrium
The state of chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the amount of reactants and products. At equilibrium, the concentration of all reactants and products remains constant, though they are not necessarily equal.

For every reversible reaction at equilibrium, we can write an equilibrium constant expression, Kc, which helps in understanding the extent to which a reaction proceeds under given conditions. This expression is a ratio of the concentration of products raised to their stoichiometric coefficients to the concentration of reactants raised to their stoichiometric coefficients. For example, in the synthesis of ethane from ethylene and water vapor, the equilibrium constant expression is \[K_c = \frac{[\mathrm{C}_2\mathrm{H}_6]^2[\mathrm{O}_2]}{[\mathrm{C}_2\mathrm{H}_4]^2[\mathrm{H}_2\mathrm{O}]^2}\].

Grasping this balance is essential for controlling reactions, such as in industrial processes where optimizing yield is vital.
Phases of Matter
Chemical reactions occur in and across different phases of matter: solids, liquids, gases, and plasma. The phase of each reactant and product affects the reaction rate, the form of the equilibrium constant, and the overall thermodynamics.

For example, in the reaction of liquid octane and oxygen gas to form carbon dioxide gas and liquid water, 2 C8H18(l) + 25 O2(g) ⇌ 16 CO2(g) + 18 H2O(l), both liquid and gaseous phases are involved. The equilibrium constant includes only the concentrations of the gases. Phases of matter are vital for predicting the outcome of a reaction, understanding the nature of reactants and products, and during the separation and purification of chemical compounds.