1. Lecture 51/31/05 Wednesday afternoon lab section Do pre-lab exercise 2 Hand in pre-lab 1 on Wed. Use pre-lab 1 values to do the calculations at the end of the lab 1 You only need pre-lab values and the equation that is given in the lab book mellisa.farrow@umassmed.edu or will be in office, room 210, on Tuesday afternoon from 4-5

2. Volume or pressure change If you increase the volume, do the reaction shift to the right or to the left? PCl 5 (g) ↔ PCl 3 (g) + Cl 2 (g) CaO (s) + CO 2 (g) ↔ CaCO 3 (s) N 2 (g) + O 2 ↔ 2NO (g) H 2 O (l) + CO 2 ↔ H 2 CO 3 (aq)

3. Change in Temperature Actually works by changing K (numerically) But conceptually easier to consider heat as reactant or product Direction of change depends on whether exothermic vs. endothermic Exothermic  heat as product Raising temperature  adds to product  shift to left Endothermic  heat as reactant Raising temperature  adds to reactant  shift to right

5. Effect of a catalyst Increases the rate at which reaction gets to equilibrium Doesn’t change the equilibrium concentrations How does it affect K? Many industrial processes use heterogeneous catalysts

6. Example For the following reaction: ΔH˚ = 2816 KJ/mol 6 CO 2 (g) + 6 H 2 O (l) ↔ C 6 H 12 O 6 (s) + 6 O 2 (g) How is the equilibrium yield of C 6 H 12 O 6 affected by: Increasing P CO2 ? Increasing temperature? Removing CO 2 ? Decreasing the total pressure? Removing part of the C 6 H 12 O 6 ? Adding a catalyst?

7. C (s) + CO 2 (g) ↔ 2 CO (g) P total = 1 atm Endothermic or exothermic? Calculate K at 850 ˚C T (˚C)CO 2 (mol %)CO (mol %) 8506.2393.77 9501.3298.68 10500.3799.63 12000.0699.94

8. Precipitation reactions Reactions that result in the formation of an insoluble product Pb(NO 3 ) 2 + NaI  2 Na + + 2 NO 3 - + PbI 2 (s) Solubility The amount of a substance that can be dissolved in a given quantity of solvent Any substance with a solubility < 0.01 M is considered insoluble PbI 2 (s) ↔ Pb +2 (aq) + 2I - (aq)

9. Solubility Rules p. 151 Soluble CompoundsExceptions NO 3 - None C 2 H 3 O 2 - (acetate)None Cl -, Br -, I - Compounds with Ag +, Hg 2 2+, Pb 2+ SO 4 2- Compounds with Sr 2+, Ba 2+, Hg 2 2+, Pb 2+ Group 1A and NH 4 + Insoluble CompoundsExceptions S 2- Group 1A and NH 4 +, Ca 2+, Sr 2+, Ba 2+ CO 3 2-,PO 4 3-, C 2 O 4 2-, CrO 4 2- Group 1A and NH 4 + OH - Group 1A and NH 4 +, Ca 2+, Sr 2+, Ba 2+

11. Ionic Equations Molecular Equation: Pb(NO 3 ) 2 (aq) + 2 NaI (aq) ↔ 2 NaNO 3 (aq) + PbI 2 (s) Complete Ionic Equation: Pb 2+ (aq) + 2NO 3 - (aq) + 2 Na + (aq) + 2I - (aq) ↔ 2NO 3 - (aq) + 2 Na + (aq) + PbI 2 (s) Net Ionic Equation: Pb 2+ (aq) + 2 I - (aq) ↔ PbI 2 (s)