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Chem. 31 – 2/22 Lecture Announcements I • Upcoming Assignments – – – – Exam 1 – Monday, March 6th (Monday after next) AP2 – Due March 1st (next Wednesday) Next quiz will be from Homework Set 2 (after Exam 1) Lab reports (glassware resubmission – due 2/27; Cl lab report – due 3/8) • Chemistry Seminar – This Friday – Gallo Winery (mentioned because they also have an internship program mid-summer to late fall) Announcements II • Today’s Lecture – Chapter 6 (Chemical Equilibria and Thermodynamics) – Thermodynamics – Le Châtelier’s Principle – Solubility Equilibria Thermodynamics ΔG = Change in Gibbs free energy This tells us if a process is spontaneous (expected to happen) or non-spontaneous ΔG < 0 process is spontaneous (favored) ΔG = ΔH - TΔS (T is absolute temperature) processes that are exothermic (Δ H < 0) and increase disorder (Δ S > 0) are favored at all T processes that have Δ H > 0 and Δ S > 0 are favored at high T Example question The reaction N2(g) + O2 (g) ↔ 2NO(g) has a positive DH. Under what conditions is this process spontaneous? - all temperatures - low temperatures - high temperatures - never Thermodynamics • Example Question: The ΔG° for the reaction Ca2+ + 2OH- => Ca(OH)2(s) is -52 kJ/mol Determine K at T = 20.°C for Ca(OH)2(s) => Ca2+ + 2OH- Le Châtelier’s Principle Intuitive Method Mathematical Method – Addition to one side results in switch to other side DG DG RT ln Q – Example: Q DG RT ln K reaction shifts to reactants (more AgCl(s)) AgCl(s) ↔ Ag+ + ClAddition of Ag+ When Q>K, ΔG>0 (toward reactants) When Q<K, ΔG<0 (toward products) Example: Q = [Ag+][Cl-] As Ag+ increases, Q>K Le Châtelier’s Principle Stress Number 1 Reactant/Products: Addition of reactant: shifts toward product Removal of reactant: shifts toward reactant Addition of product: shifts toward reactant Removal of product: shifts toward product Le Châtelier’s Principle Stress Number 1 Example: CaCO3(s) + 2HC2H3O2(aq) ↔ Ca(C2H3O2)2(aq) + H2O(l) + CO2(g) 1. Add HC2H3O2(aq) 2. Remove CO2(g) 3. Add Ca(C2H3O2)2(aq) 4. Add CaCO3(s) No effect because (s) Le Châtelier’s Principle Stess Number Two: Dilution Side with more moles is favored at lower concentrations Example: HNO2(aq) ↔ H+ + NO2If solution is diluted, reaction goes to products If diluted to 2X the volume: K H NO 2 HNO2 1 1 H NO2 2 Q 2 1 HNO2 2 1 Q K 2 So Q<K, products favored Le Châtelier’s Principle Stess Number Two: Dilution – Molecular Scale View Concentrated Solution H+ NO2- H+ NO2- Diluted Solution – dissociation allows ions to fill more space H+ NO2- H+ NO2- H+ H+ H+ NO2H+ NO2NO2- H+ NO2- H+ NO2NO2- Le Châtelier’s Principle Stress Number 3: Temperature If ΔH>0, as T increases, products favored If ΔH<0, as T increases, reactants favored Easiest to remember by considering heat a reactant or product Example: OH- + H+ ↔ H2O(l) + heat Increase in T Some Le Chatelier’s Principle Examples • Looking at the reaction below, that is initially at equilibrium, AgCl(s) ↔ Ag+(aq) + Cl-(aq) (ΔH°>0) determine the direction (toward products or reactants) each of the following changes will result in a) b) c) d) increasing the temperature addition of water (dilution) addition of AgCl(s) addition of NaCl Ch. 6 – Solubility Problems •Why Solubility is Important • Use in gravimetric analysis (predict if precipitation is complete enough) • Use in precipitation titrations (in next chapter) • Use in separations (e.g. separation of Mg2+ from Ca2+ in tap water for separate analysis) • Understand phase in which analytes will exist •Problem Overview • Dissolution of sparingly soluble salts in water • Dissolution of sparingly soluble salts in common ion • Precipitation problems (and selective precipitation problems) Solubility Product Problems - Solubility in Water Example: solubility of Mg(OH)2 in water Solubility defined as mol Mg(OH)2 dissolved/L sol’n or g Mg(OH)2 dissolved/L sol’n or other units Use ICE approach: Mg(OH)2(s) ↔ Mg2+ + 2OHInitial 0 0 Change +x +2x Equilibrium x 2x Note: x = [Mg2+] = solubility Solubility Product Problems - Solubility of Mg(OH)2 in water Equilibrium Equation: Ksp = [Mg2+][OH-]2 Ksp = 7.1 x 10-12 = x(2x)2 = 4x3 (see Appendix F for Ksp) x = (7.1 x 10-12/4)1/3 = 1.2 x 10-4 M Solubility = 1.2 x 10-4 M = [Mg2+] Conc. [OH-] = 2x = 2.4 x 10-4 M Solubility Product Problems - Solubility of Mg(OH)2 in Common Ion If we dissolve Mg(OH)2 in a common ion (OH- or Mg2+), from Le Châtelier’s principle, we know the solubility will be reduced Example 1) What is the solubility of Mg(OH)2 in a pH = 11.0 buffer? No ICE table needed because, from pH, we know [OH-]eq and buffer means dissolution of Mg(OH)2 doesn’t affect pH. Solubility Product Problems - Solubility of Mg(OH)2 at pH 11 – cont. [H+] = 10-pH = 10-11 M and [OH-] = Kw/[H+] = 10-3 M Ksp = [Mg2+][OH-]2 Moles Mg(OH)2 dissolved = moles Mg2+ [Mg2+] = Ksp/[OH-]2 = 7.1 x 10-12/(10-3)2 [Mg2+] = 7 x 10-6 M Solubility Product Problems - Solubility of Mg(OH)2 in Common Ion Example 2) Solubility of Mg(OH)2 in 5.0 x 10-3 M MgCl2.