Question

Write the equilibrium law corresponding to \(K_{c}\) for each of the following heterogeneous reactions. (a) \(2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\) (b) \(2 \mathrm{NaHSO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{SO}_{2}(g)\) (c) \(2 \mathrm{C}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+\mathrm{CO}_{2}(g)\) (d) \(\mathrm{CaCO}_{3}(s)+2 \mathrm{HF}(g) \rightleftharpoons \mathrm{C}_{2}(g)+\mathrm{CaF}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (e) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{CuSO}_{4}(s)+5 \mathrm{H}_{2} \mathrm{O}(g)\)

Short answer

\(K_{c}^{(a)} = \frac{[CO]^2}{[O_2]}\), \(K_{c}^{(b)} = [H_2O][SO_2]\), \(K_{c}^{(c)} = \frac{[CH_4][CO_2]}{[H_2O]^2}\), \(K_{c}^{(d)} = \frac{[C_2][H_2O]}{[HF]^2}\), \(K_{c}^{(e)} = [H_2O]^5\)

Step-by-step solution

Title - Understanding the Equilibrium Law

The equilibrium law for a chemical reaction, often expressed as the equilibrium constant (\(K_c\)), describes the ratio of the concentrations of products to the concentrations of reactants, each raised to the power of their stoichiometric coefficients, at equilibrium. For heterogeneous reactions (reactions that involve more than one phase), the concentration of pure solids and pure liquids are not included in the expression because their concentrations do not change.

Title - Writing the Equilibrium Expression for Reaction (a)

For the reaction \(2 C(s) + O_2(g) \rightleftharpoons 2 CO(g)\), write the equilibrium expression considering only the gaseous species. Solid carbon is omitted from the expression. \[K_c = \frac{[CO]^2}{[O_2]}\]

Title - Writing the Equilibrium Expression for Reaction (b)

For the reaction \(2 NaHSO_3(s) \rightleftharpoons Na_2SO_3(s) + H_2O(g) + SO_2(g)\), write the equilibrium expression considering only the gaseous species. Solid sodium hydrogen sulfite and solid sodium sulfite are omitted. \[K_c = \frac{[H_2O][SO_2]}{1}\] Note that the denominator is '1' because there are no reactants in the gaseous phase.

Title - Writing the Equilibrium Expression for Reaction (c)

For the reaction \(2 C(s) + 2 H_2O(g) \rightleftharpoons CH_4(g) + CO_2(g)\), write the equilibrium expression considering only the gaseous species. Solid carbon is omitted. \[K_c = \frac{[CH_4][CO_2]}{[H_2O]^2}\]

Title - Writing the Equilibrium Expression for Reaction (d)

For the reaction \(CaCO_3(s) + 2 HF(g) \rightleftharpoons C_2(g) + CaF_2(s) + H_2O(g)\), write the equilibrium expression including the gaseous reactants and products, while omitting the solids. \[K_c = \frac{[C_2][H_2O]}{[HF]^2}\]

Title - Writing the Equilibrium Expression for Reaction (e)

For the reaction \(CuSO_4 \cdot 5 H_2O(s) \rightleftharpoons CuSO_4(s) + 5 H_2O(g)\), write the equilibrium expression considering only the gaseous species. The hydrated and anhydrous copper sulfate solids are not included. \[K_c = \frac{[H_2O]^5}{1}\] Again, the denominator is '1' because there are no reactants in the gaseous phase.

Key concepts

Equilibrium Constant

The equilibrium constant, represented as (K_c), is fundamental to understanding chemical equilibrium. It quantifies the balance between products and reactants in a reaction that has reached equilibrium, indicating whether the forward or reverse reaction is favored. In essence, when we talk about equilibrium constant, we're referring to the stability of the reaction environment at a given temperature.

For the correct formulation of the equilibrium constant, one must consider only the concentrations of the entities that change during the reaction - typically gases and aqueous solutions. Solids and liquids have constant concentrations and are therefore omitted from the (K_c) expression. This can sometimes confuse students, but remember: the equilibrium state considers the dynamic but stable situation where the rate of the forward and reverse reactions are equal, and thus, the concentrations of reactants and products remain constant over time.

Heterogeneous Reactions

Heterogeneous reactions involve reactants and products that are in different phases, such as solids, liquids, and gases. A classic example includes reactions between a gas and a solid, or between a solid and a liquid. A key point to remember is that, in writing expressions for equilibrium constants, the concentration of reactants and products in the solid and liquid phase are not included because their concentration is essentially unchanged; they are present in a large excess compared to gases or solutions.

When dealing with heterogeneous reactions, it's crucial only to account for the species whose concentrations can fluctuate - typically the gases or species in solution. Solid carbon in a reaction, like in the provided examples (a) and (c), does not alter its concentration during the reaction, thus it does not appear in the equilibrium expression. This simplification helps in focusing on the reactive species that actually participate in establishing the equilibrium.

Stoichiometry

The term stoichiometry refers to the quantitative relationship between the amounts of reactants and products in a chemical reaction. It's like the recipe for the reaction, dictating the proportions of each ingredient required. In the context of chemical equilibrium, stoichiometry dictates the exponents in the equilibrium expression: each concentration is raised to the power of its respective coefficient in the balanced chemical equation.

It's this aspect of stoichiometry that allows us to correctly formulate the equilibrium expressions for a reaction. For example, in reaction (a) from our initial problem, the (2) coefficient in front of (CO) means that the concentration of (CO) will be squared in the equilibrium expression. Getting this right is important for accurately describing the system and making correct predictions about the behavior of the reaction at equilibrium.

Equilibrium Expressions

Equilibrium expressions are the mathematical representations of the equilibrium state of a reaction. They are written based on the concentrations of the products divided by the concentrations of the reactants, each raised to the power of its stoichiometric coefficient from the balanced equation. A solid understanding of how to write these expressions is crucial, as they provide the groundwork for calculations related to reaction yields and the extent of chemical processes.

In our provided examples, note how we only include gaseous and aqueous species, omitting solids and liquids. This is because the equilibrium expression is based upon activities or effective concentrations, and for pure solids and liquids, these values are constant and thus do not affect the equilibrium condition. Always recall that the equilibrium constant is a reflection of a dynamic balance, and the equilibrium expression provides a snapshot of this balance quantitatively.