1. 13-1 CHEM 102, Spring 2012, LA TECH CTH 328 9:30-10:45 am Instructor: Dr. Upali Siriwardane e-mail: upali@latech.edu Office: CTH 311 Phone 257-4941 Office Hours: M,W 8:00-9:00 & 11:00-12:00 am; Tu,Th,F 8:00 - 10:00 am.. Exams: 9:30-10:45 am, CTH 328. March 21, 2012 (Test 1): Chapter 13 April 18, 2012 (Test 2): Chapter 14 &15 May 14, 2012 (Test 3): Chapter 16 &18 Optional Comprehensive Final Exam: May 17, 2012 : Chapters 13, 14, 15, 16, 17, and 18 Chemistry 102(01) Spring 2012

2. 13-2 CHEM 102, Spring 2012, LA TECH Chapter 13. Chemical Kinetics 13.1 Reaction Rate 13.2 Effect of Concentration on Reaction Rate 13.3 Rate Law and Order of Reaction 13.4 A Nanoscale View: Elementary Reactions 13.5Temperature and Reaction Rate: The Arrhenius Equation Equation 13.6Rate Laws for Elementary Reactions 13.7Reaction Mechanisms 13.8Catalysts and Reaction Rate 13.9Enzymes: Biological Catalysts 13 ‑ 10Catalysis in Industry

3. 13-3 CHEM 102, Spring 2012, LA TECH How do you measure rates? How do you measure rates? Rates are related to the time it required to decay reactants or form products. The rate reaction = change in concentration of reactants/products per unit time Average rate rate of reaction = –  [reactant]/  t Instantaneous rate rate of reaction = – d[reactant]/dt

4. 13-4 CHEM 102, Spring 2012, LA TECH Rate of Appearance & Disappearance 2 N 2 O 5 (g) -----> 4 NO 2 (g) + O 2 (g) Disappearance is based on reactants rate = -(  [N 2 O 5 ]/  t Appearance is based on products rate =  [NO 2 ]/  t rate =  [O 2 ]/  t Converting rates of Appearance. rate = (  [NO 2 ]/  t = - 4/2  [N 2 O 5 ]/  t  [O 2 ]/  t = - 1/2  [N 2 O 5 ]/  t

5. 13-5 CHEM 102, Spring 2012, LA TECH Measuring Rate a A --> b B Based on reactants rate = -(1/a)  [A]/  t Based on products rate = +(1/b)  [B]/  t  [A]= [A] f - [A] I Change in A  t= t f - t i Change in t

6. 13-6 CHEM 102, Spring 2012, LA TECH Reaction of cis-platin with Water

7. 13-7 CHEM 102, Spring 2012, LA TECH Disappearance of Color

8. 13-8 CHEM 102, Spring 2012, LA TECH Gas buret Constant temperature bath An example reaction where gas is produced

9. 13-9 CHEM 102, Spring 2012, LA TECH Time (s)Volume STP O 2, mL 0 0 3001.15 6002.18 9003.11 12003.95 18005.36 24006.50 30007.42 42008.75 54009.62 6600 10.17 7800 10.53 Here are the results for our experiment. Time vs. volume of gas

10. 13-10 CHEM 102, Spring 2012, LA TECH 2 N 2 O 5 (g) -----> 4 NO 2 (g) + O 2 ( g)

11. 13-11 CHEM 102, Spring 2012, LA TECH Graph of 2 N 2 O 5 (g) ---> 4 NO 2 (g) + O 2 (g)

12. 13-12 CHEM 102, Spring 2012, LA TECH Graph

13. 13-13 CHEM 102, Spring 2012, LA TECH a) Temperature b) Concentration c) Catalysts d) Particle size of solid reactants Factors that affect rates of chemical reactions

14. 13-14 CHEM 102, Spring 2012, LA TECH Effect of Particle Size on Rate

15. 13-15 CHEM 102, Spring 2012, LA TECH Chemical Kinetics Definitions and Concepts a) rate law a) rate law b) rate constant b) rate constant c) order c) order d) differential rate law d) differential rate law c) integral rate law c) integral rate law

16. 13-16 CHEM 102, Spring 2012, LA TECH Every chemical reaction has a Rate Law The rate law is an expression that relates the rate of a chemical reaction to a constant the rate of a chemical reaction to a constant (rate constant-k) and concentration of reactants raised to a power. The power of a concentration is called the order with respect to a particular reactant. Rate Law

17. 13-17 CHEM 102, Spring 2012, LA TECH Rate Law E.g. A + B -----> C rate  [A] l [B] m rate  [A] l [B] m rate = k [A] l [B] m ; k = rate constant rate = k [A] l [B] m ; k = rate constant [A] = concentration of A [A] = concentration of A [B] = concentration of B [B] = concentration of B l = order with respect to A l = order with respect to A m = order with respect to B m = order with respect to B l & m have nothing to do with stoichiometric coefficients l & m have nothing to do with stoichiometric coefficients

18. 13-18 CHEM 102, Spring 2012, LA TECH Rate Constant E.g. A + B -----> C E.g. A + B -----> C rate  [A] l [B] m rate  [A] l [B] m rate = k [A] l [B] m ; rate = k [A] l [B] m ; k = rate constant k = rate constant proportionality constant of the rate law proportionality constant of the rate law Larger the k faster the reaction Larger the k faster the reaction It is related inversely to t ½ It is related inversely to t ½

19. 13-19 CHEM 102, Spring 2012, LA TECH Decomposition Reaction

20. 13-20 CHEM 102, Spring 2012, LA TECH Rate Law E.g. E.g. 2 N 2 O 5 (g) -----> 4 NO 2 (g) + O 2 (g) 2 N 2 O 5 (g) -----> 4 NO 2 (g) + O 2 (g) rate  [N 2 O 5 ] 1 rate  [N 2 O 5 ] 1 rate = k [N 2 O 5 ] 1 ;k = rate constant rate = k [N 2 O 5 ] 1 ;k = rate constant [N 2 O 5 ] = concentration of N 2 O 5 [N 2 O 5 ] = concentration of N 2 O 5 1 = order with respect to N 2 O 5 1 = order with respect to N 2 O 5 Rate and the order are obtained by experiments Rate and the order are obtained by experiments

21. 13-21 CHEM 102, Spring 2012, LA TECH Order The power of the concentrations is the order with The power of the concentrations is the order with respect to the reactant. E.g. A + B -----> C E.g. A + B -----> C If rate law: rate = k [A] 1 [B] 2 If rate law: rate = k [A] 1 [B] 2 The order of the reaction with respect to A is one (1). The order of the reaction with respect to B is two (2). Overall order of a chemical reaction is equal to the sum of all orders (3).

22. 13-22 CHEM 102, Spring 2012, LA TECH Method of initial rates The order for each reactant is found by: Changing the initial concentration of that reactant. Holding all other initial concentrations and conditions constant. Measuring the initial rates of reaction The change in rate is used to determine the order for that specific reactant. The process is repeated for each reactant. Finding rate laws

23. 13-23 CHEM 102, Spring 2012, LA TECH Initial rate

24. 13-24 CHEM 102, Spring 2012, LA TECH How do you find order? A + B -----> C A + B -----> C rate = k [A] l [B] m ; rate = k [A] l [B] m ; Hold concentration of other reactants constant If [A] doubled, rate doubled 1st order, [2A] 1 = 2 1 x [A] 1, 2 1 = 2 1st order, [2A] 1 = 2 1 x [A] 1, 2 1 = 2 b) If [A] doubled, rate quadrupled 2nd order, [2A] 2 = 2 2 x [A] 2, 2 2 = 4 2nd order, [2A] 2 = 2 2 x [A] 2, 2 2 = 4 c) If [A] doubled, rate increased 8 times c) If [A] doubled, rate increased 8 times 3rd order, [2A] 3 = 2 3 x [A] 3, 2 3 = 8 3rd order, [2A] 3 = 2 3 x [A] 3, 2 3 = 8

25. 13-25 CHEM 102, Spring 2012, LA TECH Rate data

26. 13-26 CHEM 102, Spring 2012, LA TECH Determining order

27. 13-27 CHEM 102, Spring 2012, LA TECH Determining K, Rate Constant

28. 13-28 CHEM 102, Spring 2012, LA TECH Overall order

29. 13-29 CHEM 102, Spring 2012, LA TECH Units of the Rate Constant (k) 1 first order: k ─── s -1 first order: k = ─── = s -1 s L second order k ─── second order k = ─── mol s mol s L 2 L 2 third order k ─── third order k = ─── mol 2 s mol 2 s

30. 13-30 CHEM 102, Spring 2012, LA TECH First order reactions

31. 13-31 CHEM 102, Spring 2012, LA TECH Rate Law Differential Rate Law Integral Rate rate  k [A] 0  [A]/  t =k ; ([A] 0 =1)[A] f -[A] i = -kt rate  k [A] 1  [A]/  t = k [A] ln [A] o /[A] t = kt rate  k [A] 2  [A]/  t = k [A] 2 1/ [A] f = kt + 1/[A] i Differential and Integral Rate Law

32. 13-32 CHEM 102, Spring 2012, LA TECH Integrated Rate Laws

33. 13-33 CHEM 102, Spring 2012, LA TECH Graphical method Order R ate Law Integrated Rate LawGraph X vs. time Slope 0rate = k [A] t = -kt + [A] 0 [A] t -k-k 1 rate = k[A] ln[A] t = -kt + ln[A] 0 ln[A] t -k-k 2 rate=k[A] 2 = kt +k 1 [A]t 1 [A]0 1 [A]t

34. 13-34 CHEM 102, Spring 2012, LA TECH Graphical Ways to get Order

35. 13-35 CHEM 102, Spring 2012, LA TECH First-order, Second-order, and Zeroth-order Plots

36. 13-36 CHEM 102, Spring 2012, LA TECH Finding rate laws 0 order plot 1st order plot 2nd order plot As you can see from these plots of the N2O5 data, only a first order plot results in a straight line. As you can see from these plots of the N2O5 data, only a first order plot results in a straight line. Time (s) [N2O5] 1/[N2O5] ln[N2O5]

37. 13-37 CHEM 102, Spring 2012, LA TECH This plot of ln[cis-platin] vs. time produces a straight line, suggesting that the reaction is first-order. Comparing graphs

38. 13-38 CHEM 102, Spring 2012, LA TECH First Order Reactions A ----> B

39. 13-39 CHEM 102, Spring 2012, LA TECH t 1/2 equation t 1/2 equation  0.693 =k t 1/2 0.693 0.693 t 1/2 = ---- t 1/2 = ---- k

40. 13-40 CHEM 102, Spring 2012, LA TECH The half-life and the rate constant are related. t 1/2 = t 1/2 = Half-life can be used to calculate the first order rate constant. For our N 2 O 5 example, the reaction took 1900 seconds to react half way so: k = = = 3.65 x 10 -4 s -1 0.693k 0.693 t 1/2 0.693 1900 s Half-life

41. 13-41 CHEM 102, Spring 2012, LA TECH A Nanoscale View: Elementary Reactions Most reactions occur through a series of simple steps or elementary reactions. Elementary reactions could be unimolecular - rearrangement of a molecule bimolecular - reaction involving the collision of two molecules or particles termolecular - reaction involving the collision of three molecules or particles

42. 13-42 CHEM 102, Spring 2012, LA TECH 2NO 2 (g) + F 2 (g) 2NO 2 F (g) If the reaction took place in a single step the rate law would be: rate = k [NO 2 ] 2 [F 2 ] Observed: rate = k 1 [NO 2 ] [ F 2 ] If the observed rate law is not the same as if the reaction took place in a single step that more than one step must be involved Elementary Reactions and Mechanism

43. 13-43 CHEM 102, Spring 2012, LA TECH Elementary Reactions A possible reaction mechanism might be: Step oneNO 2 + F 2 NO 2 F + F (slow) Step twoNO 2 + F NO 2 F (fast) Overall 2NO 2 + F 2 2NO 2 F slowest step in a multi-step mechanism the step which determines the overall rate of the reaction rate = k 1 [NO 2 ] [ F 2 ] rate = k 1 [NO 2 ] [ F 2 ] Rate Determining Step

44. 13-44 CHEM 102, Spring 2012, LA TECH This type of plot shows the energy changes during a reaction. This type of plot shows the energy changes during a reaction. Reaction profile of rate determining step HH activation energy Potential Energy Reaction coordinate

45. 13-45 CHEM 102, Spring 2012, LA TECH What Potential Energy Curves Show Exothermic Reactions Endothermic Reactions Activation Energy (E a ) of reactant or the minimum energy required to start a reaction Effect of catalysts Effect of temperature

46. 13-46 CHEM 102, Spring 2012, LA TECH Exothermic reaction Endothermic reaction Examples of reaction profiles

47. 13-47 CHEM 102, Spring 2012, LA TECH High activation energy (kinetic) Low heat of reaction (thermodynamic) Low activation energy (kinetic) High heat of reaction (thermodynamic) Examples of reaction profiles

48. 13-48 CHEM 102, Spring 2012, LA TECH Unimolecular Reaction cis-2-butene trans-2-butrne

49. 13-49 CHEM 102, Spring 2012, LA TECH Bimolecular Reaction I - + CH 3 Br ICH 3 + Br -

50. 13-50 CHEM 102, Spring 2012, LA TECH Orientation Probability: Some Unsuccessful Collisions I - + CH 3 Br ICH 3 + Br -

51. 13-51 CHEM 102, Spring 2012, LA TECH Arrhenius Equation: Dependence of Rate Constant (k) on T Rate constant (k) k = A e -E a /RT k = A e -E a /RT A = frequency factor: A = p x z A = frequency factor: A = p x z E a = Activation energy R = gas constant T = Kelvin temperature p = collision factor z = Orientation factor

52. 13-52 CHEM 102, Spring 2012, LA TECH Energy Distribution Curves: Activation Energy

53. 13-53 CHEM 102, Spring 2012, LA TECH An alternate form of the Arrhenius equation: k = A e - E a /RT ln k = + ln A If ln k is plotted against 1/T, a straight line of slope - Ea/RT is obtained. Activation energy - E a The energy that molecules must have in order to react. ( ) 1T1T Ea R - Arrhenius Equation: ln form

54. 13-54 CHEM 102, Spring 2012, LA TECH Calculation of E a k = A e- E a /RT ln k = ln A - E a /RT log k = log A - E a / 2.303 RT using two set of values log k 1 = log A - E a / 2.303 RT 1 log k 2 = log A - E a / 2.303 RT 2 log k 1 - log k 2 = - E a / 2.303 RT 2 + E a / 2.303 RT 1 log k 1 / k 2 = E a / 2.303 R[ 1/T 1 - 1/T 2 ]

55. 13-55 CHEM 102, Spring 2012, LA TECH Reaction rates are temperature dependent. Here are rate constants for N 2 O 5 decomposition at various temperatures. T, o C k x 10 4, s -1 20 0.235 25 0.469 30 0.933 35 1.82 40 3.62 45 6.29 k x 10 4 (s -1 ) Temperature ( o C) Rate vs Temperature plot

56. 13-56 CHEM 102, Spring 2012, LA TECH ln k T -1 Calculation of E a from N 2 O 5 data

57. 13-57 CHEM 102, Spring 2012, LA TECH Collision Model Three conditions must be met at the nano-scale level if a reaction is to occur: the molecules must collide; they must be positioned so that the reacting groups are together in a transition state between reactants and products; and the collision must have enough energy to form the transition state and convert it into products.

58. 13-58 CHEM 102, Spring 2012, LA TECH Effect of Concentration on Frequency of Bimolecular Collisions

59. 13-59 CHEM 102, Spring 2012, LA TECH Transition State: Activated Complex or Reaction Intermediates an unstable arrangement of atoms that has the highest energy reached during the rearrangement of the reactant atoms to give products of a reaction

60. 13-60 CHEM 102, Spring 2012, LA TECH Catalyst A substance which speeds up the rate of a reaction while not being consumed Homogeneous Catalysis - a catalyst which is in the same phase as the reactants Heterogeneous Catalysis- a catalyst which is in the different phase as the reactants catalytic converter solid catalyst working on gaseous materials

61. 13-61 CHEM 102, Spring 2012, LA TECH Catalysts Lowers E a

62. 13-62 CHEM 102, Spring 2012, LA TECH Catalyzed & Uncatalyzed Reactions

63. 13-63 CHEM 102, Spring 2012, LA TECH Conversion of NO to N 2 + O 2

64. 13-64 CHEM 102, Spring 2012, LA TECH Catalytic Converter catalyst catalyst H 2 O (g) + HCs  CO (g) + H 2(g) (unbalanced) catalyst catalyst 2 H 2(g) + 2 NO (g)  N 2(g) + 2 H 2 O (g) catalyst catalyst HCs + O 2(g)  CO 2(g) + H 2 O (g) (unbalanced) HCs + O 2(g)  CO 2(g) + H 2 O (g) (unbalanced) catalyst catalyst CO (g) + O 2(g)  CO 2(g) (unbalanced) CO (g) + O 2(g)  CO 2(g) (unbalanced) catalyst = Pt-NiO HCs = unburned hydrocarbons

65. 13-65 CHEM 102, Spring 2012, LA TECH Enzymes: Biological catalysts Biological catalysts Typically are very large proteins. Permit reactions to ‘go’ at conditions that the body can tolerate. Can process millions of molecules every second. Are very specific - react with one or only a few types substrates of molecules (substrates).

66. 13-66 CHEM 102, Spring 2012, LA TECH The active site Enzymes are typically HUGE proteins, yet only a small part is actually involved in the reaction. The active site has two basic components. catalytic site binding site Model of trios-phosphate-isomerase Model of trios-phosphate-isomerase

67. 13-67 CHEM 102, Spring 2012, LA TECH Relationship of Enzyme to Substrate

68. 13-68 CHEM 102, Spring 2012, LA TECH Enzyme Catalyzed Reaction

69. 13-69 CHEM 102, Spring 2012, LA TECH Maximum Velocity for an Enzyme Catalyzed Reaction

70. 13-70 CHEM 102, Spring 2012, LA TECH Enzyme Activity Destroyed by Heat

71. 13-71 CHEM 102, Spring 2012, LA TECH Reaction Mechanism A set of elementary reactions which represent the overall reaction

72. 13-72 CHEM 102, Spring 2012, LA TECH Mechanism Oxidation of Iodide Ion by Hydrogen Peroxide

73. 13-73 CHEM 102, Spring 2012, LA TECH Rate Law of Oxidation of Iodide Ion by Hydrogen Peroxide Step 1. HOOH + I -  HOI + OH - slow step - rate determining step, suggests that the reaction is first order with regard to hydrogen peroxide and iodide ion rate = k[HOOH][I - ]

74. 13-74 CHEM 102, Spring 2012, LA TECH Mechanisms with a Fast Initial Step 2 NO (g) + Br 2(g)  2NOBr (g) rate experimental = k[NO] 2 [Br 2 ]

75. 13-75 CHEM 102, Spring 2012, LA TECH Mechanism of NO + Br 2 Rate = k[NOBr 2 ][NO]

76. 13-76 CHEM 102, Spring 2012, LA TECH Rate Constants for NO + Br 2 Step +1(forward), rate constant k 1 Step -1(backward), rate constant k -1 Step 2, rate constant k 2 rate Step+1 = rate Step-1 + rate Step2 k 1 [NO][Br 2 ] = k -1 [NOBr 2 ] - k 2 [NOBr 2 ]

77. 13-77 CHEM 102, Spring 2012, LA TECH Relationships of Rate Constants k 1 [NO][Br 2 ] ~ k -1 [NOBr 2 ] thus [NOBr 2 ] = (k 1 /k -1 )[NO][Br 2 ] substituting into rate = k 2 [NOBr 2 ][NO] rate = k 2 ((k 1 /k -1 )[NO][Br 2 ])[NO] rate = (k 2 k 1 /k -1 )[NO] 2 [Br 2 ]

78. 13-78 CHEM 102, Spring 2012, LA TECH Chain Mechanisms chain initiating step - the step of a mechanism which starts the chain chain propagating step(s) the step or steps which keeps the chain going chain terminating step(s) the step or steps which break the chain

79. 13-79 CHEM 102, Spring 2012, LA TECH Chain Mechanisms combustion of gasoline in an internal combustion engine chain initiating step - additives which generate free radicals, particles with unpaired electrons chain propagating step(s) - steps which generate new free radicals chain terminating step(s) - steps which do not generate new free radicals