1. Outline:2/23/07 è è Today: Finish Chapter 16 è Chem Dept Seminar – today@4pm è CAPA 10 – deadline moved to Mon. Ù Chemical Equilibrium: LeChâtelier’s principle  G and K eq relationship

2. Types of Equilibrium Constants: Lots of different names…. K eq, K H, K sp, K a, K b, K f, K c, K p … All the same idea: Extent of rxn… K eq is a number that describes the ratio of products/reactants…

3. LeChâtelier’s Principle: A system reacts to change in the direction that minimizes the change. Change in concentrations Change in pressures Change in temperatures Demo

4. LeChâtelier’s Principle Demo: Which way will the reaction shift when water is added? Co(H 2 O) 6 2+ (aq) + 4Cl   CoCl 4 2  (aq) + 6H 2 O Toward reactants!!

5. LeChâtelier’s Principle Demo: Which way will the reaction shift when it is heated? Toward products!!  H > 0 (endothermic) Co(H 2 O) 6 2+ (aq) + 4Cl   CoCl 4 2  (aq) + 6H 2 O Co(H 2 O) 6 2+ (aq) + heat  CoCl 4 2  (aq)

6. Practice problem:  K eq = [Pb 2+ ][Cl  ] 2 More PbCl 2 is added: More H 2 O is added: Solid NaCl is added: Solid KNO 3 is added:  PbCl 2(s)  Pb 2+ (aq) + 2 Cl  (aq)  More dissolves    More precip. Nothing

7. One last link to thermodynamics:  G =  G o + RTlnQ At equilibrium:  G = 0 ; Q =K eq 0 =  G o + RTlnK eq or  G o = RTlnK eq or K eq = e  Gº/RT There is a relationship between  G o and K eq ! (see CAPA-11)

9. Example: Use the thermodynamic tables to find K eq for the following rxn: N 2 + 3 H 2  2 NH 3 at 298 K  G o (kJ/mol) 0 0  16.45  G o rxn =  32.9 kJ/mol K eq = e  G o /RT = e 32,900/(8.315*298) = 585000 Watch units!

10. Example (non std. temp): Use the thermodynamic tables to find K eq for the following rxn: N 2 + 3 H 2  2 NH 3 at 773 K  H o (kJ/mol) 0 0  46.1  S o (J/molK) 192 131 192  o rxn =  92.2 kJ/mol  S o rxn =  198.8 J/mol K  G T rxn =  o rxn  T  S o rxn = 61.5 kJ K eq = e  G T /RT = e  61,500/(8.315*773) = 7.0  10  5

11. Example (non std. temp): Use the thermodynamic tables to find K eq for the following rxn: N 2 + 3 H 2  2 NH 3  H o (kJ/mol) 0 0  46.1  S o (J/molK) 192 131 192  o rxn =  92.2 kJ/mol  S o rxn =  198.8 J/mol K Exothermic reaction + heat K eq @ 773K= 7.0  10 -5 K eq @ 298K= 585000

12. Example: K eq to  G o You can also find  G o given K eq N 2 + 3 H 2  2 NH 3 at 773 K K eq = 7.0  10  5  G o =  RTlnK eq  G o =  (8.314 J/k mol)(773 K) ln(7.0  10  5 ) =61.5 kJ/mol

13. Let’s practice some more: CAPA-11: problem #3 CAPA-11: problem #3 Some reaction… calculate K eq from  G… at 220 K! K eq = e  Gº/RT  G T =  Hº  T  Sº K eq = e  Hº/RT+  Sº/R

14. Let’s practice some more: CAPA-11: problem #7 CAPA-11: problem #7 2A  B + C If 1.00 atm of A initially, and 0.24 atm of C at equilibrium…whats K eq ? 0.24 atm C means 0.24 atm B at equil. K eq = (0.24)(0.24)/ (1.00 .48) 2 0.24 atm C means 0.48 atm A decays

15. Let’s practice some more: CAPA-11: problems #9 & #10 CAPA-11: problems #9 & #10 A + B  C K c = 710 calculate [C]/[B] at equilibrium if: 0.01 M A is added to 0.20 M B What must you do first? K is big: take reactants over to products

16. Let’s practice some more: CAPA-11: problem #9 CAPA-11: problem #9 A + B  C K c = 710 0.01 0.20 0.00 Initial 0.00 0.19 0.01 Initial  0.01  0.01 +0.01 + x + x  x Change x (0.19+x) (0.01  x) Equilibrium K=710 = (0.01  x) / x (0.19+x) x is small… x = 7.41e-5

17. Let’s practice some more: CAPA-11: problem #10 CAPA-11: problem #10 A + B  C K c = 710 7.41e-5 0.19 0.01 Initial Does reaction quotient change? Dilute everything by 10.0… K=710 = (0.01) / (7.41e-5)(0.19) = (0.001) / (7.41e-6)(0.019) ? Q = 7100 Yes!

18. Let’s practice some more: CAPA-11: problem #9 CAPA-11: problem #9 A + B  C K c = 710 7.41e-6 0.019 0.001 Initial Towards more reactants… Q > Keq…which way does it go? + x + x  x Change (7.41e-6+x) (0.019+x) (0.001  x) Equil Q = 7100

19. Let’s test out those keypads…

20. What is the equilibrium expression for: BaS (s) + 2 O 2(g)  BaSO 4(s)12345 1. [O 2 ] 2 2. [BaSO 4 ]/[BaS][O 2 ] 2 3. [BaS][O 2 ] 2 /[BaSO 4 ] 4. 1/[O 2 ] 2 5. None of the above

21. The K p expression equals 400.0 for: CO (g) + H 2 O (g)  CO 2 (g) + H 2(g) If 4 atm each of CO & H 2 O are put into a container what’s the final pressure of CO ? 1. 0.04 atm 2. 0.2 atm 3. 5 atm 4. 80 atm 5. None of the above

22. Have a great week-end!