Problem 33 - General Chemistry: Principles and Modern Applications

Question

For one of the following reactions, \(K_{c} K_{p}=K .\) Identify that reaction. For the other two reactions, what is the relationship between \(K_{c}, \bar{K}_{\mathrm{p}},\) and \(K ?\) Explain. (a) \(2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})\) (b) \(\mathrm{HI}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{I}_{2}(\mathrm{g})\) (c) \(\mathrm{NH}_{4} \mathrm{HCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(1)\)

Step-by-Step Solution

Answer

For reaction (b), \(K_{c}=K_{p}=K\). For reactions (a) and (c), the relation is expressed by the equation: \(K_{p} = K_{c} \times (RT)^{Δn}\).

Step 1: Identify Δn for Reactions

Determine the change in the number of moles of gas (Δn) in each reaction. For (a), Δn = (2 moles of SO3) - (2 moles of SO2 + 1 mole of O2) = -1. For (b), Δn = (0.5 mole of H2 + 0.5 mole of I2) - 1 mole of HI = 0. For (c), Δn = (1 mole of NH3 + 1 mole of CO2) - 0 moles of NH4HCO3 = 2.

Step 2: Evaluate Relationship Between \(K_c\), \(K_p\), and \(K\) For Each Reaction

Next, the relationship between \(K_c\), \(K_p\), and \(K\) must be evaluated for each reaction. For reaction (a) with Δn = -1, the values of \(K_c\) and \(K_p\) will differ and are related by the ideal gas law, as the reaction involves a change in the number of gas molecules. For reaction (b) with Δn = 0, \(K_c = K_p = K\), as the reaction does not involve a change in the number of gas molecules. For reaction (c) with Δn = 2, the values of \(K_c\) and \(K_p\) will differ and are related by the ideal gas law.

Step 3: Determine The Reaction For Which \(K_{c}=K_{p}=K\)

The only reaction for which \(K_{c}=K_{p}=K\) is the one with Δn=0, which is reaction (b). For reactions (a) and (c), \(K_{c}\) and \(K_{p}\) do not equate because they involve a change in the number of moles of gas. For these reactions, the relationship between \(K_{c}\), \(K_{p}\), and \(K\) is expressed by the equation: \(K_{p} = K_{c} \times (RT)^{Δn}\).

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted as either \(K_c\) or \(K_p\), is a crucial concept in studying chemical reactions. It allows us to understand the balance point of a chemical system at equilibrium, where the rates of the forward and reverse reactions are equal.

- **\(K_c\)** refers to the equilibrium constant calculated using the concentrations of reactants and products, typically in a reaction happening in solution.- **\(K_p\)** is used when dealing with gaseous reactions and is calculated with partial pressures of the gases involved.- These constants provide insight into the relative amounts of products and reactants at equilibrium. A larger \(K\) value indicates a greater proportion of products relative to reactants.
Understanding equilibrium constants helps predict the direction of the reaction, revealing whether it tends to favor the products or reactants at a given set of conditions.

Kc and Kp Relationship

In the study of gaseous reactions, the relationship between \(K_c\) and \(K_p\) provides deeper insight into how temperature and molecular changes affect reaction equilibria. This relationship is governed by the formula: \[ K_p = K_c \times (RT)^{\Delta n} \] where:- **\(R\)** is the ideal gas constant.- **\(T\)** is the temperature in Kelvin.- **\(\Delta n\)** is the change in the number of moles of gas during the reaction.
This equation reveals that if \(\Delta n\) (the change in moles of gas) is zero, like in reaction (b), \(K_c\) equals \(K_p\), because the \((RT)^{\Delta n}\) term becomes one. For reactions with nonzero \(\Delta n\) like (a) and (c), \(K_p\) will differ from \(K_c\), adjusted by \((RT)^{\Delta n}\).

Thus, mastering \(K_c\) and \(K_p\) relationships is crucial in predicting how equilibrium constants are impacted by changing conditions, especially in industrial processes.

Gaseous Reactions

Gaseous reactions are unique since they can extend and compress, influencing equilibrium and reaction rates. These reactions occur completely under gas laws, often involving pressure as a key factor.

- All reactions where gases are reactants or products require careful consideration of partial pressures, as these take the place of concentrations used in non-gaseous equilibria.- Equilibrium is studied under closed conditions where no external particles influence the system.

The exercise presents reactions like \(2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g)\), showcasing how the number of gaseous moles before and after the reaction affects application of the equilibrium constant expressions. Recognizing these patterns simplifies predictive understanding of reaction behavior under varying conditions.

Stoichiometry

Stoichiometry is the foundational concept allowing us to understand and balance chemical equations, ensuring those equations reflect how substances react in the real world. It involves quantifying exact proportions of reactants and products.

- This concept is crucial for writing balanced chemical equations, which are necessary for calculating equilibrium constants like \(K_c\) and \(K_p\).- Reactions provide stoichiometric coefficients such as those in \(2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g)\), essential in determining \(\Delta n\), the change in moles through a reaction.
Understanding stoichiometry ensures precise identification of how quantities of materials shift during a reaction, reflecting accurately in calculations like that of equilibrium constants. It's a fundamental building block for further chemical analysis.