1. Outline:2/2/07 n n Chem Dept Seminar - today @ 4pm n n Exam 1 – two weeks from today… n Outline Quiz #3 Chapter 15 - Kinetics (cont’d): - determining reaction order - integrated rate law calcs

2. Quiz #3 n Please put away all books and papers. n If you don’t have a calculator, just set up the equations correctly…

3. Quiz #3 n Please turn your quiz over and pass it to your right.

4. Quiz #3 n Reaction was endothermic… =  H products -  H reactants Reaction increased entropy… Reaction increased entropy… =  S products -  S reactants  G  =  H  - T  S   G  =  H  - T  S  spontaneous when  G = 0 spontaneous when  G = 0  =  G  + RT ln Q

5. Quiz #3 n Reaction was endothermic… =  H products -  H reactants Reaction increased entropy… Reaction increased entropy… =  S products -  S reactants  G  =  H  - T  S   G  =  H  - T  S  spontaneous when  G = 0 spontaneous when  G = 0  =  G  + RT ln Q

6. When do you use  G =  G  + RT ln Q ? Calculate  G reaction for the reaction @ 25°C: C 2 H 5 OH (l) + 3O 2(g)  2CO 2(g) + 3H 2 O (l) if p=0.25 bar for each gaseous substance.  G rxn =  G o rxn  RT ln Q  G o rxn  1325 kJ/mol (from appendix D) R  kJ/mol K T  K

7. Problem 14.40 (c): Calculate  G reaction for the reaction @ 25°C: C 2 H 5 OH (l) + 3O 2(g)  2CO 2(g) + 3H 2 O (l) if p=0.25 bar for each gaseous substance. What is  Q? Q = (0.25) 2 /(0.25) 3 Q  Q   G rxn = -1325  2.48 ln 4 = -1322

8. How many problems of this type have you already worked out on your own.... 1. I have done dozens or more… 2. I have done many, but I still miss some… 3. I have done a couple from the book & CAPA 4. I have done the CAPA ones… 5. Never seen this type of problem before.. 12345

9. Practice Problems: Chapter 14 n 14.11, 14.15, 14.17, 14.19, 14.23, 14.25, 14.27, 14.31, 14.35, 14.37, 14.38, 14.41, 14.43, 14.49, 14.51, 14.53, 14.55, 14.57, 14.61, 14.65, 14.67, 14.71, 14.75, 14.77, 14.79, 14.81, 14.91, 14.101, 14.103

10. Back to Chapter 15…

11. The rate equation cannot be predicted, it can only be measured empirically. n n Calculate k from initial rates n n Use the integrated form of the rate eqn. to solve for concentration (Section 15.4) There are two forms to know: First order: ln[A] = ln[A] o  k t Second order: 1/[A] = 1/[A] o  k t 15-3 15-5

12. k Can use data to find k and reaction order How do you find the reaction order?

13. Plot data

14. Plot both…. ln[A] = ln[A] o  k t 1/[A] = 1/[A] o  k t Only one will be truly linear…. Rate = k [NO 2 ] 2 slope

15. Example: 1st or 2nd order decay?

16. Try both plots…. Rate = k [cis-AB]

17. Rate = k p NO 2 p Cl2 0.000040 atm/s = k (0.1 atm) 2 (0.1 atm) k = 0.040 1/atm 2 s Know the “order” from the isolation experiments: e.g. double p NO and see what happens to rate… Don’t forget to practice lots of problems from the back of the chapter in the textbook Worksheet #4

18. Summary: n n Can build a rate law from observed data: Rate = k [A] m [B] n n n m, n depend only on the chemical reaction under consideration…. n n Can use integrated rate laws to predict rates, concentrations at various times, etc. 15-3 15-5 There are two forms to know: First order: ln[A] = ln[A] o  k t Second order: 1/[A] = 1/[A] o  k t (initial rates)

19. The integrated forms of the rate laws are important: Can predict [concentration] as a function of time! Example:  The decomposition reaction of NO 2 is second- order in [NO 2 ], with a rate constant of 1.51 M  s . If the initial concentration is 0.041M, when will the concentration = 0.010 M?

20. n “Rate Law” of a reaction: Rate = k [NO 2 ] 2 The integrated solution:  1/[NO 2 ]  1/[NO 2 ] o = k t  1/0.010  1/0.041 = 1.51 M -1 s -1 t t = ? 50 sec