1. 1 © 2006 Brooks/Cole - Thomson Chemistry and Chemical Reactivity 6th Edition John C. Kotz Paul M. Treichel Gabriela C. Weaver CHAPTER 15 Principles of Reactivity: Chemical Kinetics © 2006 Brooks/Cole Thomson Lectures written by John Kotz

2. 2 © 2006 Brooks/Cole - Thomson Chemical Kinetics Chapter 15 H 2 O 2 decomposition in an insect H 2 O 2 decomposition catalyzed by MnO 2

3. 3 © 2006 Brooks/Cole - Thomson We can use thermodynamics to tell if a reaction is product- or reactant- favored.We can use thermodynamics to tell if a reaction is product- or reactant- favored. But this gives us no info on HOW FAST a reaction goes from reactants to products.But this gives us no info on HOW FAST a reaction goes from reactants to products. KINETICS — the study of REACTION RATES and their relation to the way the reaction proceeds, i.e., the Reaction MECHANISM.KINETICS — the study of REACTION RATES and their relation to the way the reaction proceeds, i.e., the Reaction MECHANISM. Chemical Kinetics

4. 4 © 2006 Brooks/Cole - Thomson Reaction Mechanisms The sequence of events at the molecular level that control the speed and outcome of a reaction. Br from biomass burning destroys stratospheric ozone. (See R.J. Cicerone, Science, volume 263, page 1243, 1994.) Step 1:Br + O 3 ---> BrO + O 2 Step 2:Cl + O 3 ---> ClO + O 2 Step 3:BrO + ClO + light ---> Br + Cl + O 2 NET: 2 O 3 ---> 3 O 2

5. 5 © 2006 Brooks/Cole - Thomson Reaction rate = change in concentration of a reactant or product with time.Reaction rate = change in concentration of a reactant or product with time. Three “types” of ratesThree “types” of rates –initial rate –average rate –instantaneous rate Reaction Rates Section 15.1

6. 6 © 2006 Brooks/Cole - Thomson Determining a Reaction Rate Blue dye is oxidized with bleach. Its concentration decreases with time. The rate — the change in dye concentration with time — can be determined from the plot. Blue dye is oxidized with bleach. Its concentration decreases with time. The rate — the change in dye concentration with time — can be determined from the plot. Screen 15.2 Dye Conc Time

7. 7 © 2006 Brooks/Cole - Thomson Determining a Reaction Rate Active Figure 15.2

8. 8 © 2006 Brooks/Cole - Thomson ConcentrationsConcentrations and physical state of reactants and products (Screens 15.3-15.4)and physical state of reactants and products (Screens 15.3-15.4) Temperature (Screen 15.11)Temperature (Screen 15.11) Catalysts (Screen 15.14)Catalysts (Screen 15.14) Factors Affecting Rates Section 15.2

9. 9 © 2006 Brooks/Cole - Thomson Concentrations & Rates Section 15.2 0.3 M HCl6 M HCl Mg(s) + 2 HCl(aq) ---> MgCl 2 (aq) + H 2 (g)

10. 10 © 2006 Brooks/Cole - Thomson Physical state of reactantsPhysical state of reactants Factors Affecting Rates

11. 11 © 2006 Brooks/Cole - Thomson Catalysts: catalyzed decomp of H 2 O 2 2 H 2 O 2 --> 2 H 2 O + O 2 2 H 2 O 2 --> 2 H 2 O + O 2 Factors Affecting Rates

12. 12 © 2006 Brooks/Cole - Thomson TemperatureTemperature Factors Affecting Rates Bleach at 54 ˚CBleach at 22 ˚C

13. 13 © 2006 Brooks/Cole - Thomson Iodine Clock Reaction

14. 14 © 2006 Brooks/Cole - Thomson Concentrations and Rates To postulate a reaction mechanism, we study reaction rate andreaction rate and its concentration dependenceits concentration dependence

15. 15 © 2006 Brooks/Cole - Thomson Concentrations and Rates Take reaction where Cl - in cisplatin [Pt(NH 3 ) 2 Cl 3 ] is replaced by H 2 O

16. 16 © 2006 Brooks/Cole - Thomson Concentrations & Rates Rate of reaction is proportional to [Pt(NH 3 ) 2 Cl 2 ] We express this as a RATE LAW Rate of reaction = k [Pt(NH 3 ) 2 Cl 2 ] where k = rate constant k is independent of conc. but increases with T

17. 17 © 2006 Brooks/Cole - Thomson Concentrations, Rates, & Rate Laws In general, for a A + b B --> x X with a catalyst C Rate = k [A] m [B] n [C] p The exponents m, n, and p are the reaction orderare the reaction order can be 0, 1, 2 or fractionscan be 0, 1, 2 or fractions must be determined by experiment!must be determined by experiment!

18. 18 © 2006 Brooks/Cole - Thomson Interpreting Rate Laws Rate = k [A] m [B] n [C] p If m = 1, rxn. is 1st order in AIf m = 1, rxn. is 1st order in A Rate = k [A] 1 If [A] doubles, then rate goes up by factor of 2 If [A] doubles, then rate goes up by factor of 2 If m = 2, rxn. is 2nd order in A.If m = 2, rxn. is 2nd order in A. Rate = k [A] 2 Rate = k [A] 2 Doubling [A] increases rate by 4 If m = 0, rxn. is zero order.If m = 0, rxn. is zero order. Rate = k [A] 0 Rate = k [A] 0 If [A] doubles, rate is not changed

19. 19 © 2006 Brooks/Cole - Thomson Deriving Rate Laws Expt. [CH 3 CHO]Disappear of CH 3 CHO (mol/L)(mol/Lsec) 10.100.020 20.200.081 30.300.182 40.400.318 Derive rate law and k for CH 3 CHO(g) --> CH 4 (g) + CO(g) from experimental data for rate of disappearance of CH 3 CHO

20. 20 © 2006 Brooks/Cole - Thomson Deriving Rate Laws Rate of rxn = k [CH 3 CHO] 2 Here the rate goes up by ______ when initial conc. doubles. Therefore, we say this reaction is _________________ order. Now determine the value of k. Use expt. #3 data— 0.182 mol/Ls = k (0.30 mol/L) 2 k = 2.0 (L / mols) k = 2.0 (L / mols) Using k you can calc. rate at other values of [CH 3 CHO] at same T.

21. 21 © 2006 Brooks/Cole - Thomson Concentration/Time Relations What is concentration of reactant as function of time? Consider FIRST ORDER REACTIONS The rate law is Rate = - (∆ [A] / ∆ time) = k [A]

22. 22 © 2006 Brooks/Cole - Thomson Concentration/Time Relations Integrating - (∆ [A] / ∆ time) = k [A], we get [A] / [A] 0 =fraction remaining after time t has elapsed. Called the integrated first-order rate law.

23. 23 © 2006 Brooks/Cole - Thomson Concentration/Time Relations Sucrose decomposes to simpler sugars Rate of disappearance of sucrose = k [sucrose] Glucose If k = 0.21 hr -1 and [sucrose] = 0.010 M How long to drop 90% (to 0.0010 M)?

24. 24 © 2006 Brooks/Cole - Thomson Concentration/Time Relations Rate of disappear of sucrose = k [sucrose], k = 0.21 hr -1. If initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M? Use the first order integrated rate law ln (0.100) = - 2.3 = - (0.21 hr -1 ) time time = 11 hours

25. 25 © 2006 Brooks/Cole - Thomson Using the Integrated Rate Law The integrated rate law suggests a way to tell the order based on experiment. 2 N 2 O 5 (g) ---> 4 NO 2 (g) + O 2 (g) Time (min)[N 2 O 5 ] 0 (M) ln [N 2 O 5 ] 0 01.000 1.00.705-0.35 2.00.497-0.70 5.00.173-1.75 Rate = k [N 2 O 5 ]

26. 26 © 2006 Brooks/Cole - Thomson Using the Integrated Rate Law 2 N 2 O 5 (g) ---> 4 NO 2 (g) + O 2 (g) Rate = k [N 2 O 5 ] Data of conc. vs. time plot do not fit straight line. Plot of ln [N 2 O 5 ] vs. time is a straight line!

27. 27 © 2006 Brooks/Cole - Thomson Using the Integrated Rate Law All 1st order reactions have straight line plot for ln [A] vs. time. (2nd order gives straight line for plot of 1/[A] vs. time) Plot of ln [N 2 O 5 ] vs. time is a straight line! Eqn. for straight line: y = mx + b y = mx + b Plot of ln [N 2 O 5 ] vs. time is a straight line! Eqn. for straight line: y = mx + b y = mx + b

28. 28 © 2006 Brooks/Cole - Thomson Properties of Reactions page 719

29. 29 © 2006 Brooks/Cole - Thomson Half-Life Section 15.4 & Screen 15.8 HALF-LIFE is the time it takes for 1/2 a sample is disappear. For 1st order reactions, the concept of HALF-LIFE is especially useful. Active Figure 15.9

30. 30 © 2006 Brooks/Cole - Thomson Half-Life Section 15.4 & Screen 15.8 Sugar is fermented in a 1st order process (using an enzyme as a catalyst). Sugar is fermented in a 1st order process (using an enzyme as a catalyst). sugar + enzyme --> products sugar + enzyme --> products Rate of disappear of sugar = k[sugar] k = 3.3 x 10 -4 sec -1 What is the half-life of this reaction?

31. 31 © 2006 Brooks/Cole - Thomson Half-Life Section 15.4 & Screen 15.8 Solution [A] / [A] 0 = fraction remaining when t = t 1/2 then fraction remaining = _________ Therefore, ln (1/2) = - k t 1/2 - 0.693 = - k t 1/2 t 1/2 = 0.693 / k t 1/2 = 0.693 / k So, for sugar, t 1/2 = 0.693 / k = 2100 sec = 35 min Rate = k[sugar] and k = 3.3 x 10 -4 sec -1. What is the half- life of this reaction?

32. 32 © 2006 Brooks/Cole - Thomson Half-Life Section 15.4 & Screen 15.8 Solution 2 hr and 20 min = 4 half-lives Half-life Time ElapsedMass Left 1st35 min2.50 g 2nd701.25 g 3rd1050.625 g 4th1400.313 g Rate = k[sugar] and k = 3.3 x 10 -4 sec -1. Half-life is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min (140 min)?

33. 33 © 2006 Brooks/Cole - Thomson Half-Life Section 15.4 & Screen 15.8 Radioactive decay is a first order process. Tritium ---> electron + helium Tritium ---> electron + helium 3 H 0 -1 e 3 He 3 H 0 -1 e 3 He t 1/2 = 12.3 years If you have 1.50 mg of tritium, how much is left after 49.2 years?

34. 34 © 2006 Brooks/Cole - Thomson Half-Life Section 15.4 & Screen 15.8 Solution ln [A] / [A] 0 = -kt [A] = ?[A] 0 = 1.50 mgt = 49.2 y Need k, so we calc k from: k = 0.693 / t 1/2 Obtain k = 0.0564 y -1 Now ln [A] / [A] 0 = -kt = - (0.0564 y -1 ) (49.2 y) = - 2.77 = - 2.77 Take antilog: [A] / [A] 0 = e -2.77 = 0.0627 0.0627 = fraction remaining Start with 1.50 mg of tritium, how much is left after 49.2 years? t 1/2 = 12.3 years

35. 35 © 2006 Brooks/Cole - Thomson Half-Life Section 15.4 & Screen 15.8 Solution [A] / [A] 0 = 0.0627 0.0627 is the fraction remaining! Because [A] 0 = 1.50 mg, [A] = 0.094 mg But notice that 49.2 y = 4.00 half-lives 1.50 mg ---> 0.750 mg after 1 half-life 1.50 mg ---> 0.750 mg after 1 half-life ---> 0.375 mg after 2 ---> 0.375 mg after 2 ---> 0.188 mg after 3 ---> 0.188 mg after 3 ---> 0.094 mg after 4 ---> 0.094 mg after 4 Start with 1.50 mg of tritium, how much is left after 49.2 years? t 1/2 = 12.3 years

36. 36 © 2006 Brooks/Cole - Thomson MECHANISMS A Microscopic View of Reactions Sections 15.5 and 15.6 Mechanism: how reactants are converted to products at the molecular level. RATE LAW ----> MECHANISM experiment ---->theory

37. 37 © 2006 Brooks/Cole - Thomson Activation Energy Molecules need a minimum amount of energy to react. Visualized as an energy barrier - activation energy, E a. Reaction coordinate diagram

38. 38 © 2006 Brooks/Cole - Thomson MECHANISMS & Activation Energy Conversion of cis to trans-2-butene requires twisting around the C=C bond. Rate = k [trans-2-butene]

39. 39 © 2006 Brooks/Cole - Thomson MECHANISMS Activation energy barrier CisTrans Transition state

40. 40 © 2006 Brooks/Cole - Thomson Energy involved in conversion of trans to cis butene trans cis energy Activated Complex +262 kJ -266 kJ MECHANISMS See Figure 15.14 4 kJ/mol

41. 41 © 2006 Brooks/Cole - Thomson Mechanisms Reaction passes thru a TRANSITION STATE where there is an activated complex that has sufficient energy to become a product.Reaction passes thru a TRANSITION STATE where there is an activated complex that has sufficient energy to become a product. ACTIVATION ENERGY, E a = energy req’d to form activated complex. Here E a = 262 kJ/mol

42. 42 © 2006 Brooks/Cole - Thomson Also note that trans-butene is MORE STABLE than cis-butene by about 4 kJ/mol. Therefore, cis ---> trans is EXOTHERMIC This is the connection between thermo- dynamics and kinetics. MECHANISMS

43. 43 © 2006 Brooks/Cole - Thomson Effect of Temperature Reactions generally occur slower at lower T.Reactions generally occur slower at lower T. Iodine clock reaction, Screen 15.11, and book page 705. H 2 O 2 + 2 I - + 2 H + --> 2 H 2 O + I 2 Room temperature In ice at 0 o C

44. 44 © 2006 Brooks/Cole - Thomson Activation Energy and Temperature Reactions are faster at higher T because a larger fraction of reactant molecules have enough energy to convert to product molecules. In general, differences in activation energy cause reactions to vary from fast to slow.

45. 45 © 2006 Brooks/Cole - Thomson Mechanisms 1.Why is trans-butene cis-butene reaction observed to be 1st order? As [trans] doubles, number of molecules with enough E also doubles. As [trans] doubles, number of molecules with enough E also doubles. 2.Why is the trans cis reaction faster at higher temperature? Fraction of molecules with sufficient activation energy increases with T. Fraction of molecules with sufficient activation energy increases with T.

46. 46 © 2006 Brooks/Cole - Thomson More About Activation Energy Arrhenius equation — Rate constant Temp (K) 8.31 x 10 -3 kJ/Kmol Activation energy Frequency factor Frequency factor related to frequency of collisions with correct geometry. Plot ln k vs. 1/T ---> straight line. slope = -E a /R

47. 47 © 2006 Brooks/Cole - Thomson More on Mechanisms Reaction of trans-butene --> cis-butene is UNIMOLECULAR - only one reactant is involved. BIMOLECULAR — two different molecules must collide --> products A bimolecular reaction Exo- or endothermic?

48. 48 © 2006 Brooks/Cole - Thomson Collision Theory Reactions require (a) activation energy and (b) correct geometry. O 3 (g) + NO(g) ---> O 2 (g) + NO 2 (g) O 3 (g) + NO(g) ---> O 2 (g) + NO 2 (g) 2. Activation energy and geometry 1. Activation energy

49. 49 © 2006 Brooks/Cole - Thomson Mechanisms O 3 + NO reaction occurs in a single ELEMENTARY step. Most others involve a sequence of elementary steps. Adding elementary steps gives NET reaction.

50. 50 © 2006 Brooks/Cole - Thomson Mechanisms Most rxns. involve a sequence of elementary steps. 2 I - + H 2 O 2 + 2 H + ---> I 2 + 2 H 2 O 2 I - + H 2 O 2 + 2 H + ---> I 2 + 2 H 2 O Rate = k [I - ] [H 2 O 2 ] Rate = k [I - ] [H 2 O 2 ]NOTE 1.Rate law comes from experiment 2.Order and stoichiometric coefficients not necessarily the same! 3.Rate law reflects all chemistry down to and including the slowest step in multistep reaction.

51. 51 © 2006 Brooks/Cole - Thomson Mechanisms Proposed Mechanism Step 1 — slowHOOH + I - --> HOI + OH - Step 2 — fastHOI + I - --> I 2 + OH - Step 3 — fast2 OH - + 2 H + --> 2 H 2 O Rate of the reaction controlled by slow step — RATE DETERMINING STEP, rds. RATE DETERMINING STEP, rds. Rate can be no faster than rds! Most rxns. involve a sequence of elementary steps. 2 I - + H 2 O 2 + 2 H + ---> I 2 + 2 H 2 O 2 I - + H 2 O 2 + 2 H + ---> I 2 + 2 H 2 O Rate = k [I - ] [H 2 O 2 ] Rate = k [I - ] [H 2 O 2 ]

52. 52 © 2006 Brooks/Cole - Thomson Mechanisms Elementary Step 1 is bimolecular and involves I - and HOOH. Therefore, this predicts the rate law should be Rate  [I - ] [H 2 O 2 ] — as observed!! The species HOI and OH - are reaction intermediates. 2 I - + H 2 O 2 + 2 H + ---> I 2 + 2 H 2 O Rate = k [I - ] [H 2 O 2 ] Rate = k [I - ] [H 2 O 2 ] Step 1 — slowHOOH + I - --> HOI + OH - Step 2 — fastHOI + I - --> I 2 + OH - Step 3 — fast2 OH - + 2 H + --> 2 H 2 O

53. 53 © 2006 Brooks/Cole - Thomson Ozone Decomposition Mechanism Proposed mechanism Step 1: fast, equilibrium O 3 (g)  O 2 (g) + O (g) Step 2: slowO 3 (g) + O (g) ---> 2 O 2 (g) 2 O 3 (g) ---> 3 O 2 (g)

54. 54 © 2006 Brooks/Cole - Thomson CATALYSIS Catalysis and activation energy Uncatalyzed reaction Catalyzed reaction MnO 2 catalyzes decomposition of H 2 O 2 2 H 2 O 2 ---> 2 H 2 O + O 2

55. 55 © 2006 Brooks/Cole - Thomson Iodine-Catalyzed Isomerization of cis-2-Butene