1. Outline:2/5/07 n n Seminar Report - in front n n Extra Seminar – Wednesday @ 4pm n n Pick up Quiz #3 – from me n n Exam 1 – one week from Friday… n Outline Chapter 15 - Kinetics (cont’d): - integrated rate law calcs

2. Quiz #3 n Average = 7.9

3. 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)

4. Note something weird about k: The definition of rate constant varies from reaction to reaction…. e.g. Rate = k [A] [B] or Rate = k [C] Therefore the units of k vary from reaction to reaction…. e.g. k is in M -1 sec -1 or k is in sec -1 or k is in atm -1 sec -1

5. The integrated forms of the rate laws are important: Can predict [concentration] as a function of time! Can predict rate as a function of time too!

6. e.g. CAPA-7 Question 10 “Popcorn kernels pop unimolecularly….” 29 kernels pop in 10 seconds when 400 total kernels are present. After 100 have popped, how many will pop over the next 10 seconds? = first-order reaction ln(A)  ln(A 0 ) =  kt 29 kernels pop in 10 seconds when 400 total kernels are present. A 0 = 400 (initial number of kernels) t = 10 sec A = 371 (number of kernels at time t)  k = 0.00753 sec -1

7.  1 st ORDER: ln(A)  ln(A 0 ) =  kt and k = 0.00753 sec -1 After 100 have popped, how many will pop over the next 10 seconds? A 0 = 300 (initial number of kernels) t = 10 sec  k = 0.00753 sec  1 A = 278.25   A = 21.75

8. Try Worksheet #5…. Work with somebody nearby and complete “Version 1”…

9. On Worksheet #5….answer (a): 1. Rate = k p SO2Cl2 2. Rate = k p SO2Cl2 2 3. Rate = k p SO2Cl2 3 4. Rate = k 12345

10. Try Worksheet #5….

12. On Worksheet #5….answer (b):12345 1. k = 5.95 /hr 2. k =  5.95 atm/hr 3. k = 0.168 /hr 4. k =  0.168 atm/hr 5. None of the above

13. Worksheet #5…. Rate = k (p SO2Cl2 ) k = 1/5.95 = 0.168 hr  SO 2 Cl 2  SO 2 + Cl 2

14. Practice! CAPA problems (sets 7-8) textbook examples (15.25-15.34)

15. n How do you tell which integrated rate equation to use if you aren’t told?  How do we actually use “kinetics” for anything useful? Like determining mechanisms? A + B  C + D Rate = k [A] n [B] m

16. n Keep one reagent constant (in excess), vary the other reagent…. O 3 + isoprene  prodRate = k [O 3 ] n [iso] m the rate becomes: Rate = k’ [iso] m where k’ = k[ O 3 ] n Exp # 1: Let [ O 3 ]>>[iso]  [ O 3 ] = 5.4  10  4 M [iso] = 2.5  10  M “Isolation experiment”

17. Plot the data…. ln[iso] = ln[iso] o  k t 1/[iso] = 1/[iso] o  k t Only one will be truly linear…. Rate = k’ [iso] 1 m=1, k’ = 4.4 s  1

18. Perform the exp. again and graph data as 1st order to find k” n To find n, you change [O 3 ] but keep [ O 3 ]>>[iso] “Isolation experiment” Exp # 2: Let [ O 3 ]>>[iso]  [ O 3 ] = 2.7  10  4 M [iso] = 2.5  10  6 M [ O 3 ] II = 1/2 [ O 3 ] I k” = 2.2 s  1

19. “Isolation experiment”

20. n That is all there is to it! “Isolation experiment” Rate = k [O 3 ] 1 [iso] 1 k’ = k[O 3 ] so k = k’/[O 3 ] = 8.1  10 +3 M  1 s  1 You’re about to do this…. Chem 114: CV lab!

21. Variation on a theme: Half-life n n Radioactive decay = 1 st order decay   Integrated rate law: ln(A t /A o ) =  kt n half-life is just another way to define the rate constant k. n In the case that A t = 0.5 A o (half is left) t = t 1/2  therefore: ln(0.5) =  kt 1/2