1. Kinetics The Rates and Mechanisms of Chemical Reactions

2. What does the word KINETIC imply to you? Why should we care about KINETICS? What factors affect KINETICS? FASTSLOW

3. We Are Talking About Reaction Rates Speed of any event is measured by a change that occurs per unit of time. The speed of reaction (i.e. reaction rate) is measured as a change in concentration (Molarity; M) of a reactant or product over a certain timescale. ◦Time is the independent variable (x-axis) and concentration is the dependent variable (y-axis) ◦Reaction rate is expressed in M/s

4. The 3 Fundamental Questions of Chemical Reactions 1) What happens? ◦Answer given by balanced chemical equation and stoichiometry 2) To what extent does it happen? ◦Answer deals with chemical equilibrium which we will study in a later unit 3) How fast and by what mechanism? ◦Chemical kinetics

5. Why? Importance Examples Chemical Kinetics is very important for biological (your life), environmental (our lives) and economic (industry) processes. ◦Biological: Large proteins (aka Enzymes) increase the rates of numerous reactions essential to life. ◦Environmental: The maintenance or depletion of the ozone layer depends on the relative rates of reactions that produce or destroy ozone. ◦Economic: The synthesis of ammonia (NH 3 ) from N 2 and H 2 depends on rates of reactions. Fertilizer industries use catalysts to speed up these rates for economic reasons.

6. Schematic: Exothermic Process where energy is released as it proceeds. Heat is given off to surroundings. Reactants  Products + Energy

7. Schematic: Endothermic Process where energy is absorbed as it proceeds. Heat is consumed and surroundings become cooler. Reactants + Energy  Products

8. Factors that affect KINETICS All based on COLLISION THEORY: Collision theory: For a reaction to occur, the atoms or molecules must collide with one another with enough energy (activation energy) and must collide in the right orientation. FACTORS: 1) Concentration of reactants 2) Temperature 3) Presence of a catalyst 4) Surface area 5) Agitation 6) Nature of reactants

9. Factor 1: Concentration All based on COLLISION THEORY: Collision theory: For a reaction to occur, the atoms or molecules must collide with one another with enough energy (activation energy) and must collide in the right orientation. FACTORS: 1) Concentration of reactants If you increase concentration (Molarity), the rate of reaction increases. Why? There are more molecules which increases the number of collisions altogether; however, there are better chances that molecules will collide in the right orientation.

10. Factor 2: Temperature All based on COLLISION THEORY: Collision theory: For a reaction to occur, the atoms or molecules must collide with one another with enough energy (activation energy) and must collide in the right orientation. FACTORS: 2) Temperature Temperature is an averaged kinetic energy of molecules so if you increase temperature, you increase kinetic energy. This means you increase the number of collisions Heat supplies the energy to allow the reaction to proceed (i.e. overcoming the activation energy barrier) Think about: Why do we refrigerate milk?

11. Factor 3: Presence of Catalyst All based on COLLISION THEORY: Collision theory: For a reaction to occur, the atoms or molecules must collide with one another with enough energy (activation energy) and must collide in the right orientation. FACTORS: 3) Presence of a catalyst Catalyst assist a reaction and increase the reaction rate without being consumed in the reaction. Adding a catalyst decreases the activation energy which means more molecules will have enough energy to react. Think about: Catalytic converter, Enzymes

12. Factor 3: Presence of Catalyst

13. Factor 4: Surface Area All based on COLLISION THEORY: Collision theory: For a reaction to occur, the atoms or molecules must collide with one another with enough energy (activation energy) and must collide in the right orientation. FACTORS: 4) Surface Area Increased surface areas of molecules/particles will increase the rate of reaction. This means to break into smaller particle sizes. More places to react give better chances for collisions in the right orientation. How to increase surface area? Grind or crush a mixture of reactants. Ex: A crushed aspirin will enter your blood stream faster than taking it whole.

14. Factor 5: Agitation All based on COLLISION THEORY: Collision theory: For a reaction to occur, the atoms or molecules must collide with one another with enough energy (activation energy) and must collide in the right orientation. FACTORS: 5) Agitation Stirring or shaking a reaction will increase the reaction rate. By stirring or shaking, you are introducing energy into the reaction and thus giving molecules/particles more energy to react (overcome activation energy barrier). Your mechanical energy is converted to kinetic energy.

15. Factor 6: Nature of Reactants All based on COLLISION THEORY: Collision theory: For a reaction to occur, the atoms or molecules must collide with one another with enough energy (activation energy) and must collide in the right orientation. FACTORS: 6) Nature of reactants Reactants whose bonds are weaker have a lower activation energy and thus a higher rate of reaction. All chemical reactions involve bond breaking and bond making. Bond breaking occurs on reactant side. Collisions between reactants that require less kinetic energy are needed to break weaker bonds (i.e. smaller activation energy)

16. Rate Laws and Reaction Order Rate Law: An equation that shows the dependence of the reaction rate on the concentration of each reactant. a A + b B products k is the rate constant. ∆[A] ∆t∆t rate = rate = k[A] m [B] n The values of the exponents (m and n) in the rate law must be determined by experiment; they cannot be deduced from the stoichiometry of the reaction. *[ ] means molarity

17. Rates of Chemical Reactions a A + b Bd D + e E rate = = = 1 4 – ∆[O 2 ] ∆t∆t rate = 1 2 ∆[N 2 O 5 ] ∆t∆t = ∆[NO 2 ] ∆t∆t – 1 b ∆[B] ∆t∆t = – 1 e ∆[E] ∆t∆t = 1 a ∆[A] ∆t∆t 1 d ∆[D] ∆t∆t General rate of reaction: 2 N 2 O 5 (g)4 NO 2 (g) + O 2 (g) The rate of reaction can be measured based on reactants or products. The negative sign in front for reactants indicates they are consumed. Typically rates of reactions are expressed based on reactants.

18. Rates of Chemical Reactions ΔT should be Δt (time, not temperature)

19. Rate Laws and Reaction Order The values of the exponents in the rate law must be determined by experiment; they cannot be deduced from the stoichiometry of the reaction.

20. Determining a Rate Law: The Method of Initial Rates 2 NO(g) + O 2 (g)2 NO 2 (g) [O 2 ] n rate = k[NO] m Compare the initial rates to the changes in initial concentrations.

21. Determining a Rate Law: The Method of Initial Rates [O 2 ] n rate = k[NO] 2 m = 2 The concentration of NO doubles, the concentration of O 2 remains constant, and the rate quadruples. 2 m = 4 2 NO(g) + O 2 (g)2 NO 2 (g)

22. [O 2 ]rate = k[NO] 2 Determining a Rate Law: The Method of Initial Rates Reaction Order with Respect to a Reactant NO:second order O 2 :first order Overall Reaction Order 2 + 1 = 3 (third order) 2 NO(g) + O 2 (g)2 NO 2 (g)

23. Determining a Rate Law: The Method of Initial Rates =k = M s (M 2 ) (M) 1 M 2 s = rate [NO] 2 [O 2 ] [O 2 ]rate = k[NO] 2 Units for this third-order reaction: 2 NO(g) + O 2 (g)2 NO 2 (g) You can pick any experiment or trial and solve for k (should all be the same).

24. Example Problem #1 [HgCl 2 ] [C 2 O 4 2- ]Rate (M/s) 0.1640.153.2 x 10 -5 0.1640.452.9 x 10 -4 0.0820.451.4 x 10 -4 a. Determine the rate law. What is the order of the reaction? b. Determine the rate law constant (specify the units) c. What is the rate when the initial concentrations of both reactants are 0.100 M?

25. Example Problem #1

28. Example Problem #2 [NOCl]Rate M/s 30005980 20002660 1000665 400010640 a. Determine the rate law and order. b. Determine the rate law constant. Specify the units. c. What is the rate when the concentration of NOCl is 8000M?

29. Example Problem #2 [NOCl]Rate M/s 30005980 20002660 1000665 400010640 Determine the rate law and order. Rate = k[NOCl] 2 ; Second Order Determine the rate law constant. Specify the units. For first trial: 6.64x10 -4 (1/Ms) For second trial: 6.65x10 -4 (1/Ms) What is the rate when the concentration of NOCl is 8000M? 42560 M/s

30. Example #3 [A][B]Rate (M/s) 1.501.500.32 1.502.500.32 3.001.500.64 a. Determine the rate law and order. b. Determine the rate law constant. Specify the units.

31. Example #3 [A][B]Rate (M/s) 1.501.500.32 1.502.500.32 3.001.500.64 Determine the rate law and order. Rate = k[A]; first order Determine the rate law constant. Specify the units. k = 0.21 (1/s)

32. Example #4 [CH 3 COCH 3 ][Br 2 ][H + ] Rate (M/s) 0.300.050.05 0.000057 0.300.100.05 0.000057 0.300.050.10 0.000120 0.400.050.20 0.000310 0.400.050.05 0.000076 a. Determine the rate law and order. b. Determine the rate law constant. Specify the units.

33. Example #4 [CH 3 COCH 3 ][Br 2 ][H + ] Rate (M/s) 0.300.050.05 0.000057 0.300.100.05 0.000057 0.300.050.10 0.000120 0.400.050.20 0.000310 0.400.050.05 0.000076 Determine the rate law and order. Rate = k[CH 3 COCH 3 ][H + ]; second order Determine the rate law constant. Specify the units. k = 0.0038 (1/Ms)

34. Example #5 [S 2 O 8 -2 ][I - ]Rate (mol/L/s) 0.0180.0362.6 x 10 -6 0.0270.0363.9 x 10 -6 0.0360.0547.8 x 10 -6 0.0500.0721.4 x 10 -5 a. Determine the rate law and order. b. Determine the rate law constant. Specify the units.

35. Example #5

36. Integrated Rate Laws Graphs, reaction order and concentrations.

37. Graphs: concentration of substance A vs. time Scientists don’t like curves, they like straight lines. SO they have to manipulate to obtain straight line.

38. Determining a Rate Law: The Method of Initial Rates Rate Law Overall Reaction OrderUnits for k Rate = kZeroth orderM/s or M s –1 Rate = k[A]First order1/s or s –1 Rate =k[A][B]Second order1/(M s) or M –1 s –1 Rate = k[A][B] 2 Third order1/(M 2 s) or M –2 s –1 *Units for reaction rates are always M/s

39. Zeroth-Order Reactions Aproduct(s) rate = k[A] 0 = k ∆[A] ∆t∆t – = k For a zeroth-order reaction, the rate is independent of the concentration of the reactant. Calculus can be used to derive an integrated rate law. [A] t concentration of A at time t [A] 0 initial concentration of A y = mx + b [A] t = –kt + [A] 0 t ½ = [A] 0 2k

40. Zeroth-Order Reactions A plot of [A] versus time gives a straight-line fit and the slope will be –k.

41. Half-Life: t 1/2 For zeroth order: [A] t = –kt + [A] 0 [A] t1/2 = [A] 0 2 Half-life: time for A to reduce by 50%

42. First-Order Reactions: The Integrated Rate Law Aproduct(s) rate = k[A] Calculus can be used to derive an integrated rate law. ∆[A] ∆t∆t – = k[A] x y ln= ln(x) – ln(y) Using: [A] t [A] 0 ln = –kt ln[A] t = –kt + ln[A] 0 y = mx + b [A] t concentration of A at time t [A] 0 initial concentration of A

43. First-Order Reactions: The Integrated Rate Law ln[A] t = –kt + ln[A] 0

44. First-Order Reactions: Half-Life Half-Life: The time required for the reactant concentration to drop to one-half of its initial value. Aproduct(s) rate = k[A] [A] t [A] 0 ln = –kt t = t 1/2 = t 1/2 [A] 2 [A] 0 = –kt 1/2 1 2 ln t 1/2 = k 0.693 or

45. Second-Order Reactions Aproduct(s) rate = k[A] 2 Calculus can be used to derive an integrated rate law. ∆[A] ∆t∆t – = k[A] 2 [A] t concentration of A at time t [A] 0 initial concentration of A = kt + [A] 0 1 [A] t 1 y = mx + b

46. Second-Order Reactions 2 NO 2 (g)2 NO(g) + O 2 (g) Time (s)[NO 2 ]ln[NO 2 ]1/[NO 2 ] 08.00 x 10 –3 –4.828125 506.58 x 10 –3 –5.024152 1005.59 x 10 –3 –5.187179 1504.85 x 10 –3 –5.329206 2004.29 x 10 –3 –5.451233 3003.48 x 10 –3 –5.661287 4002.93 x 10 –3 –5.833341 5002.53 x 10 –3 –5.980395

47. Second-Order Reactions 2 NO 2 (g)2 NO(g) + O 2 (g)

48. = kt + [A] 0 1 [A] t 1 Second-Order Reactions Aproduct(s) rate = k[A] 2 t = t 1/2 = t 1/2 [A] 2 [A] 0 Half-life for a second-order reaction: [A] 0 1 = kt 1/2 + [A] 0 2 =t 1/2 k[A] 0 1

49. Second-Order Reactions =t 1/2 k[A] 0 1 For a second-order reaction, the half-life is dependent on the initial concentration. Each successive half-life is twice as long as the preceding one.

51. Formulas Given

52. Example #1 A first order reaction has a half-life of 20 minutes. a. Calculate the rate constant for this reaction. b. How much time is required for this reaction to be 80% complete?

53. Example #1

54. Example #2 N 2 O 5(g)  4NO 2 (g) + O 2 (g) The following results were collected: [N 2 O 5(g) ] Time(s) 0.10000 0.070750 0.0500100 0.0250200 0.0125300 0.00625400 A. Determine the rate constant from the data above. B. Calculate the concentration of N 2 O 5(g) after 175 seconds have passed.

55. Example #2 N 2 O 5(g)  4NO 2 (g) + O 2 (g) The following results were collected: [N 2 O 5(g) ] Time(s) 0.10000 0.070750 0.0500100 0.0250200 0.0125300 0.00625400 A. Determine the rate constant from the data above. First order k = 0.00693 1/s B. Calculate the concentration of N 2 O 5(g) after 175 seconds have passed. 0.0297M

56. Example #3 2C 4 H 6(g)  C 8 H 18(g) The following data were collected: [C 4 H 6 ] Time (s) 0.01000 0 0.006251000 0.004761800 0.003702800 0.003133600 A. What is the order of the reaction? B. What is the rate law constant? C. How long will it take for the [C 4 H 6 ] to be 93% gone?

57. Example #3 2C 4 H 6(g)  C 8 H 18(g) The following data were collected: [C 4 H 6 ] Time (s) 0.01000 0 0.006251000 0.004761800 0.003702800 0.003133600 A. What is the order of the reaction? Second order B. What is the rate law constant? 0.061(1/Ms) C. How long will it take for the [C 4 H 6 ] to be 93% gone? 21,780 s

58. Example #4 A second order reaction has a rate constant of 5.99M -1 s -1. The reaction initially contains 4M [A]. A. How long will it take for the reaction to drop to 0.17M? B. What will the concentration be after an hour?

59. Example #4 A second order reaction has a rate constant of 5.99M -1 s -1. The reaction initially contains 4M [A]. A. How long will it take for the reaction to drop to 0.17M? 0.94s B. What will the concentration be after an hour? 4.63x10 -5 M

60. Example #5 The following data was collected. [A]Time(s) 100 836 49 212 A. Determine the order of the reaction from the data. B. What is the rate law constant? C. What is the half life for the reaction?

61. Example #5 The following data was collected. [A]Time(s) 100 836 49 212 A. Determine the order of the reaction from the data. Zeroth order B. What is the rate law constant? 0.67 M/s (slope) C. What is the half-life for the reaction? 7.46s

62. Transition State: The configuration of atoms at the maximum in the potential energy profile. This is also called the activated complex. Reaction Rates and the Temperature: Collision Theory and the Arrhenius Equation

63. The Arrhenius Equation Activation Energy (E a ): The minimum energy needed for reaction. As the temperature increases, the fraction of collisions with sufficient energy to react increases.

64. The Arrhenius Equation

65. Using the Arrhenius Equation RT EaEa ln(k) = ln(A) – y = mx + b + ln(A) T 1 R –Ea–Ea ln(k) = ln(k) = ln(A) + ln(e –E /RT ) a rearrange the equation Slope intercept formula when you plot ln(k) vs. 1/T : R = 0.008314 kJ/K. mol A: collision frequency factor

66. Using the Arrhenius Equation + ln(A) T 1 R –Ea–Ea ln(k) = t (°C)T (K)k (M –1 s –1 )1/T (1/K)lnk 2835563.52 x 10 –7 0.001 80–14.860 3566293.02 x 10 –5 0.001 59–10.408 3936662.19 x 10 –4 0.001 50–8.426 4277001.16 x 10 –3 0.001 43–6.759 5087813.95 x 10 –2 0.001 28–3.231

67. Using the Arrhenius Equation + ln(A) T 1 R –Ea–Ea ln(k) =

68. Example #1 Determine the activation energy and the collision frequency factor at 700K for the data below. k T (K) 0.011700 0.035730 0.105760 0.343790 0.789810

69. Changes in Temperature will change the rate constant Increasing temperature will increase the rate constant (thus increasing the rate). ln k 1 - ln k 2 = - ( E a / R )( 1 / T 1 - 1 / T 2 )

70. Example #2 The rate constant of a first-order reaction is 0.034 at 298K. The activation energy for this reaction is 50.2 kJ/mol. What is the rate constant at 350K? (if the temperature is given in Celsius, add 273 to change to Kelvin)

71. Example #3 At 45 o C the rate constant of a first order reaction is 0.8. At 135 o C the rate constant is 2.4. What is the activation energy for this reaction?

72. Example #4 The rate of a reaction increased by a factor of 10 when the temperature goes from 25 o C to 50 o C. What is the activation energy of the reaction?

73. Reaction Mechanisms Elementary Reaction (step): A single step in a reaction mechanism. Reaction Mechanism: The sequence of reaction steps that describes the pathway from reactants to products. Possible: Hydrogen reacts with oxygen to make water. H 2 + O 2  H 2 O 2 H 2 O 2  H 2 O + O H 2 + O  H 2 O

74. Same reaction, other approaches O 2  2O H 2 + O  H 2 O H 2  2H H + O 2  OH + O H + OH  H 2 O O + H 2  H 2 O

75. How do you decide which is right? The reaction mechanism must result in the overall balance reaction The slow step must support the rate law

76. Reaction Mechanisms Experimental evidence suggests that the reaction between NO 2 and CO takes place by a two-step mechanism: NO 3 (g) + CO(g)NO 2 (g) + CO 2 (g) NO 2 (g) + NO 2 (g)NO(g) + NO 3 (g) NO 2 (g) + CO(g)NO(g) + CO 2 (g) elementary reaction overall reaction elementary reaction An elementary reaction describes an individual molecular event. The overall reaction describes the reaction stoichiometry and is a summation of the elementary reactions.

77. NO 3 (g) + CO(g)NO 2 (g) + CO 2 (g) NO 2 (g) + NO 2 (g)NO(g) + NO 3 (g) NO 2 (g) + CO(g)NO(g) + CO 2 (g) Reaction Mechanisms Experimental evidence suggests that the reaction between NO 2 and CO takes place by a two-step mechanism: elementary reaction overall reaction elementary reaction A reactive intermediate is formed in one step and consumed in a subsequent step.

78. Reaction Mechanisms Molecularity: A classification of an elementary reaction based on the number of molecules (or atoms) on the reactant side of the chemical equation. termolecular reaction: unimolecular reaction: bimolecular reaction: O(g) + O(g) + M(g)O 2 (g) + M(g) O3*(g)O3*(g)O 2 (g) + O(g) O 3 (g) + O(g)2 O 2 (g)

79. Rate Laws for Elementary Reactions The rate law for an elementary reaction follows directly from its molecularity because an elementary reaction is an individual molecular event. termolecular reaction: unimolecular reaction: bimolecular reaction: O(g) + O(g) + M(g)O 2 (g) + M(g) O3*(g)O3*(g)O 2 (g) + O(g) rate = k[O] 2 [M] rate = k[O 3 ] rate = k[O 3 ][O] O 3 (g) + O(g)2 O 2 (g)

80. Rate Laws for Overall Reactions Rate-Determining Step: The slowest step in a reaction mechanism. It acts as a bottleneck and limits the rate at which reactants can be converted to products. 1)First check to make sure the individual steps will add up to the overall reaction. 2)In elementary steps, the coefficient of the reactants become the power in the rate law. 3)Slow step will always be the basis for comparing experimental to predicted rate law. 4)If slow step is the first, then everything is easy. 5)If not, then the forward rate of step above = reverse rate and some substituting will take place. 6)Intermediates cannot be in rate law.

81. Rate Laws for Overall Reactions NO 3 (g) + CO(g)NO 2 (g) + CO 2 (g) NO 2 (g) + NO 2 (g)NO(g) + NO 3 (g) NO 2 (g) + CO(g)NO(g) + CO 2 (g) fast step overall reaction slow step Based on the slow step: rate = k 1 [NO 2 ] 2 k2k2 k1k1 Initial Slow Step

82. Rate Laws for Overall Reactions N 2 O(g) + H 2 (g)N 2 (g) + H 2 O(g) 2 NO(g) + 2 H 2 (g)N 2 (g) + 2 H 2 O(g) slow step overall reaction fast step, reversible Based on the slow step: rate = k 2 [N 2 O 2 ][H 2 ] k3k3 k -1 Initial Fast Step 2 NO(g)N2O2(g)N2O2(g) k1k1 N 2 O 2 (g) + H 2 (g)N 2 O(g) + H 2 O(g) k2k2 fast step

83. Rate Laws for Overall Reactions rate = k 2 [N 2 O 2 ][H 2 ] intermediate First step: Rate reverse = k –1 [N 2 O 2 ] Rate forward = k 1 [NO] 2 k 1 [NO] 2 = k –1 [N 2 O 2 ] [NO] 2 [N 2 O 2 ] = k –1 k1k1 Slow step: rate = k 2 [N 2 O 2 ][H 2 ] rate = k 2 [NO] 2 [H 2 ] k –1 k1k1

84. H 2 O 2 (aq) + I – (aq)H 2 O(l) + I O – (aq) H 2 O 2 (aq) + I O – (aq)H 2 O(l) + O 2 (g) + I – (aq) 2 H 2 O 2 (aq)2 H 2 O(l) + O 2 (g)Catalyst Since the catalyst is involved in the rate-determining step, it often appears in the rate law. rate = k[H 2 O 2 ][ I – ] overall reaction rate-determining step fast step

85. Example #1 Overall reaction: 2NO 2 + CO  NO 3 + CO 2 Proposed mechanism: NO 2 + NO 2  NO 3 + NO(slow) NO 3 + CO  NO 2 + CO 2 (fast) If the rate law is Rate = k[NO 2 ], is this a possible mechanism?

86. Example #2 For the overall reaction: 2NO + Br 2  2NOBr The proposed mechanism is: NO + NO  N 2 O 2 (fast equilibrium) N 2 O 2 + Br 2  2NOBr(slow) Determine the rate law.

87. Example #3 Overall reaction: 4HBr + O 2  2H 2 O + 2Br 2 Proposed Mechanism: HBr + O 2  HOOBr HOOBr + HBr  2HOBr 2HOBr + 2HBr  2H 2 O + 2Br 2 Rate law: Rate = k[HBr] 2 [O 2 ] Which is the rate determining step?

88. Example #4 From the following mechanism, determine the rate law. Cl 2  2Cl(fast) Cl + CHCl 3  HCl + CCl 3 (slow) Cl + CCl 3  CCl 4 (fast) Overall reaction: Cl 2 + CHCl 3  HCl + CCl 4