Chapter 7. Equilbrium

7.1 Introduction to Equilibrium

7.1.1

7.1.2

7.1.3

7.1.4

A) Freezing/melting of benzene

(C₆H₆) C₆H₆(l) ⇌ C₆H₆(s)

B) Dissolution of the weak acid, HCN(aq)

HCN(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CN⁻(aq)

C) The proton transfer between the weak base, HS⁻(aq), and the weak acid, H₂PO₄⁻(aq)

HS⁻(aq) + H₂PO₄⁻(aq) ⇌ H₂S(aq) + HPO₄²⁻(aq)

Detailed explanation here.

PART A. Chemical equilibrium is dynamic because forward and reverse reactions occur continuously at equal rates. This results in constant macroscopic properties (like concentrations) despite ongoing microscopic activity. This balance of opposing processes distinguishes dynamic equilibrium from a static state where no changes occur.

Detailed explanation here.

PART B. At equilibrium, observable properties like concentrations and color remain constant, because the forward and reverse reaction rates are equal, resulting in no net change in the system.

Detailed explanation here.

PART C. The system cannot reach equilibrium if H₂ or I₂ is completely consumed because both are required for the dynamic equilibrium.

Detailed explanation here.

The equilibrium states for acetic acid dissociation and iodine sublimation are dynamic because the forward and reverse processes are continuously occurring at equal rates. This balance maintains constant macroscopic properties (concentrations and amounts) while allowing microscopic activity to persist.

Detailed explanation here.

PART A. The initial concentration of A is 1.0 mol/L.

Detailed explanation here.

PART B. The system reaches equilibrium at approximately 6 s.

Detailed explanation here.

7.2 Direction of Reversible Reactions

7.2.1

PART A. At the start, the forward reaction rate exceeds the reverse reaction rate because only the reactant NO₂ is present in significant amounts.

Detailed explanation here.

PART B. At equilibrium, the rates of the forward and reverse reactions are equal, and the concentrations of NO₂ and N₂O₄ are constant.

Detailed explanation here.

7.3 Reaction Quotient and Equilibrium Constant

7.3.1

7.3.2

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Detailed explanation here.

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Detailed explanation here.

7.4 Calculating the Equilibrium Constant

7.4.1

PART A. The equilibrium constant (Kc) is approximately 51 (rounded to two significant figures).

Detailed explanation here.

PART B. The equilibrium concentration of H₂ will be less than 0.25 M because the reaction strongly favors the formation of HI due to the large Kc.

Detailed explanation here.

7.5 Magnitude of the Equilibrium Constant

7.5.1

The value Kc = 50.0 indicates that the concentration of HI (the product) is much greater than the concentrations of H₂ and I₂ (the reactants) at equilibrium. This means the reaction strongly favors the formation of products, with most of the reactants being converted into HI.

Detailed explanation here.

7.6 Properties of the Equilibrium Constant

7.6.1

7.6.2

7.6.3

PART A. Kc' = ([N₂][H₂]³) / [NH₃]²

Detailed explanation here.

PART B. Kc' = 1.67 × 10⁻⁶

Detailed explanation here.

Kp′ ≈ 35

Detailed explanation here.

Overall Reaction: CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g); Koverall = 15.0

Detailed explanation here.

7.7 Calculating Equilibrium Concentrations

7.7.1

The equilibrium concentrations: [H₂] = 0.05 M; [I₂] = 0.05 M.

Detailed explanation here.

7.8 Representations of Equilibrium

7.8.1

PART A. Diagram 1: Qc = 0.167; Diagram 2: Qc = 6; Diagram 3: Qc = 1.125

Detailed explanation here.

PART B. Diagram 1: Not at equilibrium (Qc < Kc); Diagram 2: Not at equilibrium (Qc > Kc); Diagram 3: At equilibrium (Qc = Kc)

Detailed explanation here.

PART C. Diagram 1: Shift to the right (toward products), Diagram 2: Shift to the left (toward reactants), Diagram 3: N/A

Detailed explanation here.

7.9 Introduction to Le Chatelier's Principle

7.9.1

7.9.2

PART A.

(i) Increasing the temperature: shift toward the products;

(ii) Increasing the pressure by decreasing the volume: shift toward the reactants;

(iii) Removing CO2: shift toward the products, favoring the forward reaction.

(iv) Decreasing the temperature: shift toward the reactants to produce more heat.

(v) Adding neon gas (Ne) at constant volume: no shift occurs

Detailed explanation here.

PART B. Le Châtelier's Principle explains the following shifts: (i) Increasing temperature shifts the equilibrium towards products (endothermic direction) to consume added heat. (ii) Increasing pressure shifts the equilibrium towards reactants (fewer gas molecules). (iii) Removing CO₂ shifts the equilibrium towards products to replenish it. (iv) Decreasing temperature shifts the equilibrium towards reactants (exothermic direction) to generate heat. (v) Adding an inert gas at constant volume causes no shift as partial pressures remain unchanged.

Detailed explanation here.

PART A.

Cl⁻ is added: The solution will turn more blue.

The temperature is increased: The solution will turn more blue.

The solution is diluted with water: The solution will turn more pink.

The temperature is decreased: The solution will turn more pink.

Detailed explanation here.

PART B. Adding NaOH increases pH, reacting with H⁺ and effectively removing water from the equilibrium. To replenish the water, the equilibrium shifts towards the pink Co(H₂O)₆²⁺ reactant.

Detailed explanation here.

7.10 Reaction Quptient and Le Chatelier's Principle

7.10.1

7.10.2

PART A. The number of moles of SO2(g) will remain the same because the total number of gas molecules on both sides of the reaction is equal.

Detailed explanation here.

PART B. The forward reaction is exothermic because increasing the temperature decreases Kc, shifting the equilibrium toward the reactants.

Detailed explanation here.

PART A: the equilibrium constant (Kc) decreases

Detailed explanation here.

PART B: equilibrium constant increases with temperature

Detailed explanation here.

7.11 Introduction to Solubility Equilibria

7.11.1

7.11.2

7.11.3

PART A. BaF₂(s) ⇌ Ba²⁺(aq) + 2F⁻(aq)

Detailed explanation here.

PART B. The molar solubility of BaF2 is 7.21×10−3 mol/L

Detailed explanation here.

PART C. the solubility in grams per liter: 1.26g/L

Detailed explanation here.

PART A. BaCrO₄ will have a higher [CrO₄²⁻] than Ag₂CrO₄.

Detailed explanation here.

PART B.

Ag₂CrO₄ Dissociation: [CrO₄²⁻] = 6.5 × 10⁻⁵ mol/L

BaCrO₄ Dissociation: [CrO₄²⁻] = 1.1 × 10⁻⁵ mol/L

Detailed explanation here.

PART A

[Pb²⁺] = s = 1.68 × 10⁻² mol/L

[Br⁻] = 2s = 2(1.68 × 10⁻²) = 3.36 × 10⁻² mol/L

Detailed explanation here.

PART B. After adding NaBr, [Pb²⁺] will decrease. This is due to the common ion effect, where an increase in Br− reduces the solubility of PbBr2, shifting the equilibrium toward the solid phase.

Detailed explanation here.

7.12 Common-Ion Effect

7.12.1

PART A. The mass of PbI₂(s) in the beaker will increase. According to the common-ion effect, the additional iodide ions increase the concentration of one of the products of the dissolution equilibrium. This shift in equilibrium causes the reaction to shift to the left (toward the reactants) to reduce the disturbance caused by the increased iodide concentration.

Detailed explanation here.

PART B. Adding KI to a saturated PbI₂ solution increases the I⁻ concentration (common ion effect). This shifts the dissolution equilibrium to the left, causing more PbI₂ to precipitate (increasing its mass) and decreasing the concentration of Pb²⁺ in the solution.

Detailed explanation here.

7.13 pH and Solubility

7.13.1

PART A. Adding concentrated NaOH(aq) introduces a large amount of OH⁻ ions, increasing their concentration in the solution. According to Le Châtelier’s principle, the system shifts the equilibrium to the left (toward solid Mg(OH)₂) to reduce the OH⁻ concentration. Consequently, the concentration of Mg²⁺ in the solution decreases, and more Mg(OH)₂ precipitates.

Detailed explanation here.

PART B. Adding HNO₃ introduces H⁺, which reacts with OH⁻, decreasing its concentration. Le Châtelier's principle dictates a shift to the right, dissolving more Mg(OH)₂ and increasing Mg²⁺ concentration to replenish the lost OH⁻.

Detailed explanation here.

7.14 Free Energy of Dissolution

7.14.1

PART A. The dissolution of NaCl exhibits a near-zero or small positive enthalpy change (ΔH°) due to the minimal temperature change upon dissolving, suggesting a slightly endothermic process. Its entropy change (ΔS°) is positive, as dissolving a solid increases disorder by dispersing ions in the water. In contrast, CaF₂ dissolution is clearly endothermic (positive ΔH°) as evidenced by the decrease in solution temperature. Despite its limited solubility, the dissolution of CaF₂ still results in a positive entropy change (ΔS°) because the transition of ions into solution increases disorder.

Detailed explanation here.

PART B. The dissolution of NaCl is spontaneous (negative ΔG°) due to a small positive or near-zero enthalpy change (ΔH°) and a positive entropy change (ΔS°). Conversely, CaF₂ dissolution is non-spontaneous at room temperature (positive ΔG°) because its large positive ΔH° outweighs the positive ΔS°, leading to low solubility. Predicting ΔG° for ionic compounds in water is complex because it depends on the balance between the endothermic lattice breaking (ΔHlattice), the exothermic ion hydration (ΔHhydration), and the entropy change (ΔS). The net enthalpy of solution (ΔH°solution = ΔHlattice + ΔHhydration) and the entropy contribution together determine ΔG°, making accurate prediction difficult.

Detailed explanation here.

PART C. For CaF₂, ΔH° is positive, and ΔS° is positive. As temperature (T) increases, the TΔS° term becomes larger. If the temperature is high enough, the TΔS° term could become larger than the ΔH° term, making ΔG° negative. This would increase the solubility of CaF₂ at higher temperatures.

Detailed explanation here.