1. Outline:2/7/07 n n Extra Seminar – today @ 4pm n n Exam 1 – one week from Friday… n Outline Chapter 15 - Kinetics (cont’d): - integrated rate law calcs = Half-life calcs - Arrhenius Rate equation

2. 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

3. Half-life example ln(A t /A o ) =  kt ln(7.14/12)=  k (6 days) k = 0.0865 day  1 ln(1/2) =  kt 1/2  0.693 = .0865 (t 1/2 ) t 1/2 = 8.0 day Radiation Biology ( 131 I): Time (days) 0 3 6 9 12 Mass 131 I (  g) 12 9.25 7.14 5.50 4.24 What is the half-life of 131 I?

4. Half-life example ln(A t /A o ) =  kt ln(1/2)=  k (22.3 yr) k = 0.0311 yr  1 ln(A 100 /10000) =  0.0311 yr  1 (100 yr) A 100 /10000 = exp(  3.11) A 100 = 446 atoms Given t 1/2 of 22.3 years for 210 Pb, and 10,000 atoms deposited on top of the lake sediment in 1905, how much 210 Pb left now?

5. Summary of Chapter 15 so far: n n Rates depend upon concentrations of the species involved. n n There is a relationship between number of species involved and rate of reaction (e.g. unimolecular, bimolecular…) 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….

6. Summary of Chapter 15 so far: n n Can use integrated rate laws to predict rates of reaction, concentrations at various times, etc. e.g. half-life n n Can use empirical data to determine rate law.

7. Let’s try some keypad questions…

8. Determine the rate law for the reaction: 2ClO 2(aq) + 2 OH  (aq)  ClO 3  (aq) + ClO 2  (aq) + H 2 O (l) Experiment [ClO 2 ] [OH  ] Initial Rate of rxn 1 0.12M 0.12M 2.07 x 10  4 M/s 2 0.12M 0.24M 8.28 x 10  4 M/s 3 0.24M 0.24M 1.66 x 10  3 M/s 1. Rate = k [ClO 2 ] 2 [OH  ] 2 2. Rate = k [ClO 2 ][OH  ] 2 3. Rate = k [ClO 2 ] 2 [OH  ] 4. Rate = k [ClO 2 ][OH  ]

9. Determine the value of the rate constant for: 2ClO 2(aq) + 2 OH  (aq)  ClO 3  (aq) + ClO 2  (aq) + H 2 O (l) Experiment [ClO 2 ] [OH  ] Initial Rate of rxn 1 0.12M 0.12M 2.07 x 10  4 M/s 2 0.12M 0.24M 8.28 x 10  4 M/s 3 0.24M 0.24M 1.66 x 10  3 M/s 1. k = 0.0144 M  1 s  1 2. k = 0.0144 M  2 s  1 3. k = 0.120 M  1 s  1 4. k = 0.120 M  2 s  1 5. None of the above

10. DEMO n k also depends on temperature and Arrhenius Activation energy….

11.   k = A e  E a /RT k = Rate law constant A = constant (frequency factor) E a = Arrhenius Activation Energy R = 8.315 J/mol K T = Temperature n k also depends on temperature and Arrhenius Activation energy…. 15-6

12. What does Ea mean in terms of rate?   k = A e  E a /RT e.g. big E a  e  (big) = small k = small rate! e.g. small E a  e  (small) = big k = big rate!

13. Problems to try: Chapter 15 n 15.13, 15.17, 15.19, 15.21, 15.23, 15.25, 15.27, 15.29, 15.31, 15.33, 15.39, 15.45, 15.47, 15.51, 15.57, 15.59, 15.61, 15.63, 15.65, 15.67, 15.69, 15.73, 15.77, 15.81, 15.85, 15.87, 15.89, 15.91, 15.95, 15.103

14. 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

15. Problems to try: Chapter 13 n 13.1, 13.3, 13.5, 13.7, 13.9, 13.11, 13.15, 13.19, 13.21, 13.35, 13.43, 13.45, 13.47, 13.49 Problems to try: Chapter 12 n 12.3, 12.5, 12.7, 12.9, 12.11, 12.17, 12.19, 12.29, 12.31, 12.33, 12.35, 12.39, 12.41, 12.43, 12.49, 12.51, 12.53, 12.55, 12.63