1. 02/24/10 Lecture

2. Announcements Corrections to Monday’s Powerpoint Lecture Slides Posted (Slides 3, 9 and 15) I have posted last semester’s Exam 1 plus the key Exam 1 – Topics –Ch. 1, 3, and 4 (all topics covered in reading/lectures) –Ch. 6 (Sections 1 through 4 expected)

3. Today’s Lecture Chapter 6 Material –Thermodynamics Definitions Determination of signs of ΔS, ΔH, ΔG Relationship between ΔG, Q and K Le Chatelier’s Principle –Solubility Product Problems

4. Thermodynamics Some Definitions: H = Enthalpy S = Entropy G = Gibbs Free Energy T = Absolute temperature

5. Thermodynamics 1.ΔH is related to heat of reaction -if a reaction produces heat, ΔH < 0 and reaction is “exothermic” -a reaction that requires heat has ΔH > 0 and is endothermic 2.ΔS is related to disorder of system -If the final system is “more random” than initial system, ΔS > 0

6. Thermodynamics Entropy Examples: (Is ΔS > or < 0?) H 2 O(l) ↔ H 2 O(g) H 2 O(s) ↔ H 2 O(l) NaCl(s) ↔ Na + + Cl - 2H 2 (g) + O 2 (g) ↔ 2H 2 O(g) N 2 (g) + O 2 (g) ↔ 2NO(g) ΔS > 0 ΔS < 0 ΔS > 0

7. 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) are favored at all T processes that have Δ H > 0 and Δ S > 0 are favored at high T

8. Example question The reaction N 2 (g) + O 2 (g) ↔ 2NO(g) has a positive  H. Under what conditions is this process spontaneous? - all temperatures - low temperatures - high temperatures - never

9. Thermodynamics ΔG and Equilibrium ΔG = ΔG° + RTlnQQ = Reaction Quotient (for A ↔ B, Q = [B]/[A]) At equilibrium, ΔG = 0 and Q = K ΔG° = -RTlnK

10. Thermodynamics Example Question: The ΔG° for the reaction Ca 2+ + 2OH - => Ca(OH) 2 (s) is -52 kJ/mol Determine K at T = 20°C for Ca(OH) 2 (s) => Ca 2+ + 2OH -

11. Section 6 – 2: Le Châtelier’s Principle The position of a chemical equilibrium always shifts in a direction that tends to relieve the effect of an applied stress Types of stresses: –Addition/removal of reactant/product –Change in volume (gases) or dilution (aqueous solutions) –Changes in temperatures Can take intuitive or mathematical approaches to solving problems

12. Le Châtelier’s Principle Intuitive Method –Addition to one side results in switch to other side –Example: Mathematical Method AgCl(s) ↔ Ag + + Cl - Additional Ag + results in more AgCl When Q>K, ΔG>0 (toward reactants) When Q<K, ΔG<0 (toward products) Example: Q = [Ag + ][Cl - ] As Ag + increases, Q>K

13. 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

14. Le Châtelier’s Principle Stress Number 1 Example: CaCO 3 (s) + 2HC 2 H 3 O 2 (aq) ↔ Ca(C 2 H 3 O 2 ) 2 (aq) + H 2 O(l) + CO 2 (g) 1. Add HC 2 H 3 O 2 (aq) 2. Remove CO 2 (g) 3. Add Ca(C 2 H 3 O 2 ) 2 (aq) 4. Add CaCO 3 (s) No effect because (s)

15. Le Châtelier’s Principle Stess Number Two: Dilution Side with more moles is favored at lower concentrations Example: HNO 2 (aq) ↔ H + + NO 2 - If solution is diluted, reaction goes to products If diluted to 2X the volume: So Q<K, products favored

16. Le Châtelier’s Principle Stess Number Two: Dilution – Molecular Scale View H+H+ NO 2 - Concentrated Solution H+H+ NO 2 - H+H+ H+H+ H+H+ Diluted Solution – dissociation allows ions to fill more space H+H+ NO 2 - H+H+ H+H+ H+H+ H+H+

17. 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 + ↔ H 2 O(l) + heat Increase in T

18. Solubility Product Problems Importance: - gravimetric analysis - chemical separations (e.g. selective removal of Mg 2+ or Ca 2+ to determine single ion in water hardness titration)

19. 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) ↔ Mg 2+ + 2OH - Initial0 0 Change +x +2x Equilibriumx 2x

20. Solubility Product Problems - Solubility of Mg(OH) 2 in water Equilibrium Equation: K sp = [Mg 2+ ][OH - ] 2 K sp = 7.1 x 10 -12 = x(2x) 2 = 4x 3 (see Appendix F for K sp ) x = (7.1 x 10 -12 /4) 1/3 = 1.2 x 10 -4 M Solubility = moles Mg(OH) 2 dissolved = x Solubility = 1.2 x 10 -4 M Conc. [OH - ] = 2x = 2.4 x 10 -4 M

21. Solubility Product Problems - Solubility of Mg(OH) 2 in Common Ion If we dissolve Mg(OH) 2 in a common ion (OH - or Mg 2+ ), 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? K sp = 7.1 x 10 -12

22. 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 MgCl 2.

23. Solubility Product Problems Precipitation Problems What occurs if we mix 50 mL of 0.020 M BaCl 2 with 50 mL of 3.0 x 10 -4 M (NH 4 ) 2 SO 4 ? Does any solid form from the mixing of ions? What are the concentrations of ions remaining?

24. Precipitations Used for Separations Example: If we wanted to know the concentrations of Ca 2+ and Mg 2+ in a water sample. EDTA titration gives [Ca 2+ ] + [Mg 2+ ]. However, if we could selectively remove Ca 2+ or Mg 2+ (e.g. through titration) and retitrate, we could determine the concentrations of each ion. Determine if it is possible to remove 99% of Mg 2+ through precipitation as Mg(OH) 2 without precipitating out any Ca(OH) 2 if a tap water solution initially has 1.0 x 10 -3 M Mg 2+ and 1.0 x 10 -3 M Ca 2+.