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

Short answer

(a) Homogeneous reaction: \(K_c = \frac{[O_2]^3}{[O_3]^2}\) (b) Heterogeneous reaction: \(K_c = \frac{1}{[Cl_2]^2}\) (c) Homogeneous reaction: \(K_c = \frac{[C_2H_6]^2[O_2]}{[C_2H_4]^2[H_2O]^2}\) (d) Heterogeneous reaction: \(K_c = \frac{[CH_4]}{[H_2]^2}\) (e) Heterogeneous reaction: \(K_c = \frac{[Cl_2]^2}{[HCl]^4 [O_2]}\) (f) Heterogeneous reaction: \(K_c = \frac{[CO_2]^{16}[H_2O]^{18}}{[O_2]^{25}}\) (g) Heterogeneous reaction: \(K_c = \frac{[CO_2]^{16}}{[O_2]^{25}}\)

Step-by-step solution

Reaction (a) equilibrium constant and type

For the given reaction: \[ 2 O_3(g) \rightleftharpoons 3 O_2(g) \] All species are in gaseous phase, so it is a homogeneous reaction. The equilibrium constant expression can be written as: \[ K_c = \frac{[O_2]^3}{[O_3]^2} \]

Reaction (b) equilibrium constant and type

For the given reaction: \[ Ti(s) + 2Cl_2(g) \rightleftharpoons TiCl_4(l) \] This reaction involves a solid, a liquid, and a gas phase, so it is a heterogeneous reaction. The equilibrium constant expression only includes the concentration of the gas: \[ K_c = \frac{1}{[Cl_2]^2} \]

Reaction (c) equilibrium constant and type

For the given reaction: \[ 2C_2H_4(g) + 2H_2O(g) \rightleftharpoons 2C_2H_6(g) + O_2(g) \] All species are in gaseous phase, so it is a homogeneous reaction. The equilibrium constant expression can be written as: \[ K_c = \frac{[C_2H_6]^2[O_2]}{[C_2H_4]^2[H_2O]^2} \]

Reaction (d) equilibrium constant and type

For the given reaction: \[ C(s)+2 H_2(g) \rightleftharpoons CH_4(g) \] This reaction involves a solid and a gas phase, so it is a heterogeneous reaction. The equilibrium constant expression only includes the concentrations of the gases: \[ K_c = \frac{[CH_4]}{[H_2]^2} \]

Reaction (e) equilibrium constant and type

For the given reaction: \[ 4 HCl(aq) + O_2(g) \rightleftharpoons 2 H_2O(l) + 2 Cl_2(g) \] This reaction involves aqueous, liquid, and gas phase, so it is a heterogeneous reaction. The equilibrium constant expression only includes the concentration of the gas and aqueous species: \[ K_c = \frac{[Cl_2]^2}{[HCl]^4 [O_2]} \]

Reaction (f) equilibrium constant and type

For the given reaction: \[ 2 C_8H_{18}(l) + 25 O_2(g) \rightleftharpoons 16 CO_2(g) + 18 H_2O(g) \] This reaction involves liquid and gas phases, so it is a heterogeneous reaction. The equilibrium constant expression only includes the concentrations of the gases: \[ K_c = \frac{[CO_2]^{16}[H_2O]^{18}}{[O_2]^{25}} \]

Reaction (g) equilibrium constant and type

For the given reaction: \[ 2 C_8H_{18}(l) + 25 O_2(g) \rightleftharpoons 16 CO_2(g) + 18 H_2O(l) \] This reaction involves liquid and gas phases, so it is a heterogeneous reaction. The equilibrium constant expression only includes the concentrations of the gases: \[ K_c = \frac{[CO_2]^{16}}{[O_2]^{25}} \]

Key concepts

Homogeneous Reactions

Homogeneous reactions are those in which all reacting species exist in the same phase. Commonly, we consider reactions in states like gases and solutions as homogeneous. An instructional example is the reaction involving ozone and oxygen:

\textbf{Example:}
\begin{text}For the decomposition of ozone \(2 O_3(g) \rightleftharpoons 3 O_2(g)\), both reactants and products are in the gaseous state, making it a homogeneous reaction. Its equilibrium constant expression is \(K_c = \frac{[O_2]^3}{[O_3]^2}\), highlighting the importance of only the gaseous reactants and products in defining the reaction's equilibrium dynamics.\textbf{Why is this important?}Understanding homogeneous reactions aids learners in appreciating that the physical state of substances directly influences both a reaction's mechanism and the representation of its equilibrium constant expression.\textbf{Application in Exercises:}In exercises, the equilibrium constant for a homogeneous reaction will include the concentration or pressure of all gaseous or aqueous species within the balanced chemical equation to calculate the equilibrium constant.\textbf{Quick Tip:}When dealing with homework exercises, always check the phase states of reactants and products to determine if a reaction is homogeneous and, consequently, to write the correct \(K_c\) expression.\text

Heterogeneous Reactions

Heterogeneous reactions involve reactants and products that are in different phases, such as solid, liquid, and gas. These differences in phase create additional considerations when writing expressions for the equilibrium constant.

Significant Considerations

In the reaction \(Ti(s) + 2 Cl_2(g) \rightleftharpoons TiCl_4(l)\), we have a solid reactant (titanium), a gaseous reactant (chlorine gas), and a liquid product (titanium tetrachloride). It is crucial to realize that solids and pure liquids do not appear in the equilibrium constant expression because their concentrations do not change during the reaction.

Expression Insights

Therefore, the equilibrium constant for the reaction above is expressed as \(K_c = \frac{1}{[Cl_2]^2}\), excluding the solid Ti and liquid TiCl_4. For students, understanding how to handle phase states is key to correctly setting up and solving chemical equilibrium problems. This knowledge helps prevent common mistakes, such as including the concentration of solids or pure liquids in the equilibrium constant expression.Remember, for heterogeneous reactions, only include the concentrations of gases and aqueous solutions in your equilibrium expressions.

Chemical Equilibrium

Chemical equilibrium is a dynamic state in which the rates of the forward and reverse reactions are equal, leading to no net change in the concentration of reactants and products over time, although both reactions continue to occur. This concept is central to understanding how reactions proceed and how they can be influenced by different conditions.

At equilibrium, the ratio of the concentrations of products to reactants is constant and is represented by the equilibrium constant, \(K\). The value of \(K\) can provide insights into the favorability of a reaction: a large value indicates a product-favored reaction, while a small value indicates a reactant-favored reaction.

Direct application of the equilibrium concept can be seen in the solution exercise where for each reaction, the \(K_c\), or equilibrium constant in terms of concentration, is determined based on the phase state of the reactants and products.Proper application of equilibrium principles not only helps in solving textbook problems but also in predicting the outcome of reactions under different conditions in practical scenarios.

Reaction Phase States

The phase state of reactants and products -- gases (g), liquids (l), solids (s), and aqueous solutions (aq) -- plays a significant role in chemical reactions, including whether the reaction is homogeneous or heterogeneous. When considering equilibrium constant expressions, the phase states dictate which species are included in the equilibrium expression.

Gases and aqueous solutions are included because their concentrations can change and affect the reaction's dynamics. In contrast, solids and liquids generally have constant concentrations due to their large molar volume and, as such, are not included in the equilibrium expression.

A comprehensive grasp of phase states is crucial for students as they engage with complex exercises. For instance, in reaction \((g) \; 2 C_8H_{18}(l) + 25 O_2(g) \rightleftharpoons 16 CO_2(g) + 18 H_2O(l)\), only the gaseous oxygen and carbon dioxide are considered in the equilibrium expression \(K_c = \frac{[CO_2]^{16}}{[O_2]^{25}}\) because the reactant octane and the product water are both in the liquid phase.Understanding these subtleties is imperative for students who aim to accurately study and predict the behavior of reactions under varying conditions.