1. 1 Chemical Kinetics Rates at which chemical reactions occur. Main goal is to understand the steps by which a reaction takes place: Reaction Mechanism

2. As we will see, there are several factors which affect the rate of a chemical reaction including: 1.the nature and concentrations of the reactants. 2.the temperature of the reaction system. 3.the presence of a catalyst and 4.the surface area of the reactants or catalyst. 2

3. 3 Reaction Rate Change in concentration of a reactant or product per change in time. Rate = ∆[reactants] Rate = ∆[products] ∆t ∆t ( )Sign is used if the [ ] is decreasing Units usually molar/second concentration ∆[ ] = [final] – [initial]

4. 4 2NO 2 (g)  2NO(g) + O 2 (g) in a flask at 300C, measure change in concentration as it decomposes. Calculate the average rate it changes in the first 50 sec. (-)means decreasing

5. 5 Instantaneous rates Value at a particular time: slope of the line tangent to the curve at that point. ss

6. 6 Rates and Stoichiometry Relative rates (need to look at the coefficients) 2NO 2 (g)  2NO(g) + O 2 (g) Must include reference:  disappearance of reactant  appearance of product

7. 7 2NO 2 (g)  2NO(g) + O 2 (g) Rate of consumption = Rate of production =2(rate of production of NO 2 of NOof O 2 ) Rate of O 2 production is half that of NO ∆[NO] ∆t = 4PH 3  P 4 + 6H 2

8. 8 Example: 2A + 3B  C + 4D What is the rate of B, C and D in reference to one mole of A? - Δ[A] -Δ[B] Δ[C] Δ[D] Δt Δt Δt Δt = = = 3/23/2 1/21/2 2

9. 9 Rate Laws 2NO 2 (g)   2NO(g) + O 2 (g)  Reactions are reversible  Choose conditions where the reverse has negligible contributions.  Study at a point soon after they are mixed before product builds up.  Reaction rate will depend only on concentration of the reactants.

10. Two key points 1.The concentration of the products do not appear in the rate law because this is an initial rate. 2.The order (exponent) must be determined experimentally, can’t be obtained from the equation Rate Laws

11. So, what is a Rate Law? Algebraic expression of the relationship between concentration and the rate of a reaction at a particular temperature. Constant of proportionality in the expression is given the symbol k and is referred to as the specific rate constant for the reaction 11

12. 12 Rate Laws ( ignore reverse) Rate = k[NO 2 ] n k = rate constant (constant of proportionality) n = rate order (must be determined experimentally not from a balanced equation) can be an integer including 0. [product] does not appear in the rate law

13. 13 Types of Rate Laws Differential Rate Law: expresses how rate depends on concentration. (rate law) Integrated Rate Law: expresses how concentration depends on time.

14. 14 Determining the Form of the Rate Law *how a reaction occurs* 1.Need to determine n (order) 2N 2 O 5 (aq)  4NO 2 (aq) + O 2 (g) N 2 O 5 = 0.90M 5.4 x 10 -4 mol/L·s = 0.45M 2.7 x 10 -4 mol/L·s ½ the concentration, ½ the rate

15. 15 Slope of the tangent to the curve

16. 16 This means the rate of the reaction depends on concentration of N 2 O 5 to the first power. The differential rate law is: ***first order reaction doubling the concentration doubles the reaction rate. 5

17. Reaction Rate and Concentration Click in this box to enter notes. Copyright © Houghton Mifflin Company. All rights reserved. Go to Slide Show View (press F5) to play the video or animation. (To exit, press Esc.) This media requires PowerPoint® 2000 (or newer) and the Macromedia Flash Player (7 or higher). [To delete this message, click inside the box, click the border of the box, and then press delete.]

18. 18 Method of Initial Rates (experimentally determining the form of the rate law) Initial Rate: the “instantaneous rate” just after the reaction begins before the [initial] of reactants have changed significantly. Just after t=0 1.Run the reaction several times 2.Vary initial concentrations 3.Measure reaction rate just after reaction is mixed

19. 19 NH 4 + (aq) + NO 2 - (aq) →N 2 (g) + 2H 2 O(l) Double the NO 2 ; double the initial rate

20. 20 Rate 2 = 2.7 x 10 -7 = (.0100) m = (2.0) m Rate 1 1.35 x 10 -7 (.0050) m =2.00 = (2.0) m Value of m=1 First order reaction in terms of NO 2 -. In terms of NO 2 -

21. 21 Rate 3 = 5.4 x 10-7 = (.200) n = (2.0) n Rate 2 2.7 x 10-7 (.100) n =2.00 = (2.0) n Value of n = 1 and the value of m=1 First order in both NH 4 and NO 2 Rate law is : Rate = k[NH 4 + ][NO 2 - ] In terms of NH 4 +

22. 22 Overall Reaction Order Sum of the order of each component in the rate law. rate = k[NH 4 + ][NO 2 - ] The overall reaction order is 1 + 1=2.

23. 23 Calculate k 1.35x10-7=k(0.100mol/L) (0.0050mol/L) k=2.7 x 10 -4 L/mol.s

24. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 24 Sample 12.1 The reaction between bromate ions and bromide ions in acidic aqueous solution is given by the equation BrO3-(aq) + 5Br-(aq) + 6H+(aq) →3Br2(l)+ 3H2O(l)

25. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 25 Determine the order for all three reactants, the overall reaction Order, and the value of the rate constant. Rate = k[BrO 3 -] n [Br-] m [H+] p Determine the values of n, m, and p by comparing the rates from the various experiments.

26. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 26 n= 1 m= 1 p= 2 The rate is 1 st order in BrO 3 - and br -, and 2 nd order in H +. The overall reaction order is n + m + p = 4 Rate law is written as; Rate = k[BrO 3 - ][Br - ][H + ] 2 K= 8.0 L 3 /mol 3 s

27. 27 Integrated Rate Laws So far looked at rate as a funx of the [reactant]. Also useful to express the [reactant] as a funx of time, give the (differential) rate law for the reaction.

28. 28 2N 2 O 5 (soln) → 4NO 2 (soln) + O 2 (g) Since the rate depends on the [N 2 O 5 ] to the 1 st power, 1 st order reaction This means if the [N 2 O 5 ] doubled, the rate would double. Put into different form using calculus

29. 29 First-Order Rate Law Integrated first-order rate law is ln[A] =  kt + ln[A] o For aA  Products in a 1 st -order reaction, **Concentration as a function of time**

30. 30 Integrated first-order rate law is ln[A] =  kt + ln[A] o 1.The equation shows how [A] depends on time. 2.Is in the form y=mx + b, where a plot of y vs. x is a straight line with slope m and intercept b. y=ln[A]x=tm=-kb=ln[A] 0

31. 31 For the reaction: aA→products The reaction is first order in A if a plot of ln[A] versus t is a straight line. If it is not a straight line it is not 1 st order. 3.This integrated rate law for a first-order reaction also can be expressed in terms of ratio of [A] and [A] 0 as follows;

32. 32 2N 2 O 5 (soln) → 4NO 2 (soln) + O 2 (g) Using data, verify that the rate law is first order in [N 2 O 5 ], And calculate the value of the rate constant. [N 2 O 5 ] (mol/L)Time (s).10000 0.70750 0.0500100 0.0250200 0.0125300 0.00625400

33. 33 Figure 12.4: A plot of ln[N 2 O 5 ] versus time. This verifies that the reaction is 1 st order.

34. 34 Since the reaction is 1 st order, the slope of the line equals -k, where ln[N 2 O 5 ] = -kt + ln[N 2 O 5 ] 0 y = mx + b

35. 35 Using the data given, calculate [N 2 O 5 ] at 150s after the start of the reaction. [N 2 O 5 ] (mol/L) Time (s).10000 0.70750 0.0500100 0.0250200 0.0125300 0.00625400 k =6.39x10 -3 s - ln[N 2 O 5 ] = -kt + ln[N 2 O 5 ] 0

36. 36 Half-Life of a First-Order Reaction t 1/2 = half-life of the reaction k = rate constant For a first-order reaction, the half-life does not depend on concentration. Time required to reach [1/2]

37. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 37 Figure 12.5: A plot of [N 2 O 5 ] versus time for the decomposition reaction of N 2 O 5. t 1/2 =100s

38. 38 Formula for t 1/2 is derived from integrated rate law: aA→ product By definition when t=t 1/2, If the reaction is 1 st order in [A]

39. 39 Half-life first order reaction A certain first-order reaction has a half-life of 20.0min. a. Calculate the rate constant for this reaction. b. how much time is required for this reaction to be 75% complete.

40. 40 1. 2.Using the integrated rate law in the form If the reaction is 75% complete, 75% of reactant has been consumed, leaving 25% in the original form. This means that

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42. 42 Another way to solve.. t 1/2 is at 50% completion If the [1mol/L] 0, after one half-life the [0.50mol/L]. One more half-life produces [0.25mol/L]. 25% of reactant is left. Two half-lives = 2(20.0min)=40.0min

43. 43 Second-Order Rate Law For aA  products in a second-order reaction, Integrated rate law is *Doubling the concentration quadruples the reaction rate.

44. 44 1.A plot of 1/[A] versus t will produce a straight line w/slope = k 2.Shows how [A] depends on time and can be used to calculate [A] at any time t, provided k and [A] 0 are known.

45. 45 Half-Life of a Second-Order Reaction t 1/2 = half-life of the reaction k = rate constant A o = initial concentration of A The half-life is dependent upon the initial concentration.

46. 46 Half life of second order reaction At t=t 1/2, Then the equation becomes

47. 47 Butadiene reacts to form its dimer according to the equation: 2C 4 H 6 (g)→C 8 H 12 (g) (two identical molecules combine) The following data were collected for this reaction at a given temp : [C 4 H 6 ](mol/L)Time (+,- s) 0.010000 0.006251000 0.004761800 0.003702800 0.003133600 0.002704400 0.002415200 0.002086200 a.Is the reaction 1 st or 2 nd order? b.What is the value of the rate constant for the reaction? c.What is the half-life for the reaction under the conditions of this experiment?

48. 48 Figure 12.6: (a) A plot of ln[C 4 H 6 ] versus t. (b) A plot of 1n[C 4 H 6 ] versus t.

49. 49 a. Is it first or second order? Write the rate law. 2nd

50. 50 b. What is the rate constant? 2 nd order: plot 1/[C 4 H 6 ] vs. t and get a straight line with slope of k

51. 51 c. Half-life? t1 /2 =1/(6.14x10 -2 L/mol.s)(1.000x10 -2 mol/L) = 1.63x10 3 s

52. 52 Difference between ½-life of a 1 st order and 2 nd order reaction 2 nd order Depends on both k and [A] 0 Each ½ life is double of the preceding one 1 st order Depends only on k Constant time for each ½-life

53. 53 Zero order reaction Rate =k[A] 0 =k(1)=k The rate is constant (does not change with concentration) Integrated rate law : [A]=-kt+[A] 0 Plot of [A] versus t gives a straight line of slope -k

54. 54 Half-life [A]=[A] 0 /2 when t=t 1/2, so

55. Figure 12.7: A plot of [A] versus t for a zero- order reaction.

56. 56 Zero order Metal surface or an enzyme is required for the reaction to occur.

57. 57 The decomposition reaction 2N 2 O(g) → 2N 2 (g) + O 2 (g) takes place on a platinum surface. Concentration has no effect on the rate

58. More Complicated Reactions BrO 3 - + 5 Br - + 6H +  3Br 2 + 3 H 2 OBrO 3 - + 5 Br - + 6H +  3Br 2 + 3 H 2 O For this reaction we found the rate law to beFor this reaction we found the rate law to be Rate = k[BrO 3 - ][Br - ][H + ] 2Rate = k[BrO 3 - ][Br - ][H + ] 2 To investigate this reaction rate we need to control the conditionsTo investigate this reaction rate we need to control the conditions

59. Rate = k[BrO 3 - ][Br - ][H + ] 2 We set up the experiment so that two of the reactants are in large excess. [BrO 3 - ] 0 = 1.0 x 10 -3 M[BrO 3 - ] 0 = 1.0 x 10 -3 M [Br - ] 0 = 1.0 M[Br - ] 0 = 1.0 M [H + ] 0 = 1.0 M[H + ] 0 = 1.0 M As the reaction proceeds [BrO 3 - ] changes noticeablyAs the reaction proceeds [BrO 3 - ] changes noticeably [Br - ] and [H + ] don’t[Br - ] and [H + ] don’t

60. Assume that through the reaction:Assume that through the reaction: [Br - ] = [Br - ] 0 [H + ] = [H + ] 0 Therefore the rate law can be written: Therefore the rate law can be written: 60 Rate = k[BrO 3 - ][Br - ][H + ] 2

61. Rate = k[BrO 3 - ][Br - ] 0 [H + ] 0 2Rate = k[BrO 3 - ][Br - ] 0 [H + ] 0 2 k’ = k[Br - ] 0 [H + ] 0 2 k’ = k[Br - ] 0 [H + ] 0 2 Rate = k’[BrO 3 - ] This is called a pseudo first order rate law. k = k’ [Br - ] 0 [H + ] 0 2 Rate = k[BrO 3 - ][Br - ][H + ] 2

62. 62 A Summary 1.Simplification: Conditions are set such that only forward reaction is important. 2.Two types: differential rate law integrated rate law 3.Which type? Depends on the type of data collected - differential and integrated forms can be interconverted.

63. 63 A Summary (continued) 4.Most common: method of initial rates. 5.Concentration v. time: used to determine integrated rate law, often graphically. 6.For several reactants: choose conditions under which only one reactant varies significantly (pseudo first-order conditions).

64. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 64

65. 65 Reaction Mechanism 4 The series of steps by which a chemical reaction occurs. (bond making and bond breaking) 4 Must be determined by experiment! 4 Must agree with overall stoichiometry AND the experimentally determined rate law. A chemical equation does not tell us how reactants become products - it is a summary of the overall process.

66. 66 Molecularity: how the molecules collide Elementary steps: the steps in the reaction mechanism that add up to the balance equation The rate for a reaction can be written from its molecularity.

67. 2NO 2 + F 2  2NO 2 F Rate = k[NO 2 ][F 2 ] The proposed mechanism is: NO 2 + F 2  NO 2 F + F (slow) F + NO 2  NO 2 F(fast) 2NO 2 + F 2  2NO 2 F F is called an intermediate it is formed then consumed in the reaction Reaction Mechanisms Elementary steps

68. Formed in reversible reactions 2 NO + O 2  2 NO 2 Mechanism 2 NO  N 2 O 2 (fast and reversible) N 2 O 2 + O 2  2 NO 2 (slow) rate = k[N 2 O 2 ][O 2 ] Substitute [NO] 2 for [N 2 O 2 ] Determining! No intermediates What is the rate determining step in this video clip

69. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 69 Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why? Arrhenius: An activation energy must be overcome. Mechanism Sim

70. Factors that Affect Reaction Rate 1.Concentrations of reactants More reactants mean more collisions if enough energy is present 2.Temperature Collision Theory: When two chemicals react, their molecules have to collide with each other with sufficient energy for the reaction to take place. Kinetic Theory: Increasing temperature means the molecules move faster.

71. Collision Theory Particles have to collide to react. Have to hit hard enough Things that increase this increase rate High temp – faster reaction High concentration – faster reaction Small particles = greater surface area means faster reaction Rates and collisions

72. However… Has been found that the rate of a reaction is much smaller than the calculated collision frequency in a collection of gas particles. Therefore only a small fraction of the collisions produces a reaction. WHY? Copyright©2000 by Houghton Mifflin Company. All rights reserved. 72

73. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 73 Arrhenius Equation 4 Collisions must have enough energy to produce the reaction (must equal or exceed the activation energy (threshold energy)). 4 Orientation of reactants must allow formation of new bonds.

74. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 74 Figure 12.11: (a) The change in potential energy as a function of reaction progress for the reaction 2BrNO  2NO + Br 2. The activation energy Ea represents the energy needed to disrupt the BrNO molecules so that they can form products. The quantity ∆E represents the net change in energy in going from reactant to products. (b) A molecular representation of the reaction.

75. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 75 Figure 12.12: Plot showing the number of collisions with a particular energy at T 1 and T 2, where T 2 > T 1.

76. Temperature Dependence of the Rate Constant **10  increase will double the rate** k = A exp( -E a /RT ) E a is the activation energy (J/mol) R is the gas constant (8.314 J/Kmol) T is the absolute temperature A is the frequency factor Ln k = - -E a R 1 T + lnA (Arrhenius equation) 13.4

77. ln(k) = -E a /RT + ln(A)

78. Catalysts Speed up a reaction without being used up in the reaction. Enzymes are biological catalysts. Homogenous Catalysts are in the same phase as the reactants. Heterogeneous Catalysts are in a different phase as the reactants.

79. Ways to speed up a reaction: 1.Catalysts Speed up reactions by lowering activation energy 2.Surface area of a solid reactant Bread and Butter theory: more area for reactants to be in contact 3.Pressure of gaseous reactants or products Increased number of collisions 79

80. 80 Catalysts allow reactions to proceed by a different mechanism - a new pathway with a lower activation energy. More molecules will have this activation energy. Does not affect ∆E Show up as a reactant in one step and a product in a later step

81. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 81 Figure 12.16: Effect of a catalyst on the number of reaction-producing collisions.

82. 82 Heterogeneous Catalysis (gas on a solid surface) 1. Adsorption and activation of the reactants. 2. Migration of the adsorbed reactants on the surface. 3. Reaction of the adsorbed substances. 4. Escape, or desorption, of the products. Steps:

83. Pt surface HHHH HHHH Hydrogen bonds to surface of metal. Break H-H bonds Heterogenous Catalysts Hydrogenation of unsaturated fats

84. Pt surface HHHH Heterogenous Catalysts C HH C HH

85. Pt surface HHHH Heterogenous Catalysts C HH C HH The double bond breaks and bonds to the catalyst.

86. Pt surface HHHH Heterogenous Catalysts C HH C HH The hydrogen atoms bond with the carbon

87. Pt surface H Heterogenous Catalysts C HH C HH HHH catalyst sim

88. Homogenous Catalysts (same phase) Chlorofluorocarbons (CFCs) catalyze the decomposition of ozone. Enzymes regulating the body processes. (Protein catalysts)

89. Ostwald Process Hot Pt wire over NH 3 solution Pt-Rh catalysts used in Ostwald process 4NH 3 (g) + 5O 2 (g)  4NO (g) + 6H 2 O (g) Pt catalyst 2NO (g) + O 2 (g)  2NO 2 (g) 2NO 2 (g) + H 2 O (l)  HNO 2 (aq) + HNO 3 (aq) 13.6

90. Catalytic Converters 13.6 CO + Unburned Hydrocarbons + O 2 CO 2 + H 2 O catalytic converter 2NO + 2NO 2 2N 2 + 3O 2 catalytic converter

91. Enzyme Catalysis 13.6

92. Catalysts and rate Catalysts will speed up a reaction but only to a certain point. Past a certain point adding more reactants won’t change the rate. Zero Order

93. Catalysts and rate. Concentration of reactants RateRate Rate increases until the active sites of catalyst are filled. Then rate is independent of concentration

94. This reaction takes place in three steps

95. EaEa First step is fast Low activation energy

96. Second step is slow High activation energy EaEa

97. EaEa Third step is fast Low activation energy

98. Second step is rate determining

99. Intermediates are present

100. Activated Complexes or Transition States

101. Catalysts Speed up a reaction without being used up in the reaction. Enzymes are biological catalysts. Homogenous Catalysts are in the same phase as the reactants. Heterogeneous Catalysts are in a different phase as the reactants.

102. How Catalysts Work Catalysts allow reactions to proceed by a different mechanism - a new pathway. New pathway has a lower activation energy. More molecules will have this activation energy. Does not change  E Show up as a reactant in one step and a product in a later step

103. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 103 Figure 12.15: Energy plots for a catalyzed and an uncatalyzed pathway for a given reaction.

104. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 104 Chapter 12(b) Chemical Kinetics (cont’d)

105. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 105

106. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 106 Figure 12.9: A molecular representation of the elementary steps in the reaction of NO 2 and CO.

107. Figure 12.10: A plot showing the exponential dependence of the rate constant on absolute temperature.

108. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 108 Figure 12.13: Several possible orientations for a collision between two BrNO molecules. Orientations (a) and (b) can lead to a reaction, but orientation (c) cannot.

109. Figure 12.14: Plot of ln(k) versus 1/T for the reaction 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g). The value of the activation energy for this reaction can be obtained from the slope of the line, which equals -E a /R.

110. Figure 12.17: Heterogeneo us catalysis of the hydrogenatio n of ethylene.

111. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 111 Figure 12.18: The exhaust gases from an automobile engine are passed through a catalytic converter to minimize environmental damage.

112. Figure 12.19: The removal of the end amino acid from a protein by reaction with a molecule of water. The products are an amino acid and a new, smaller protein.

113. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 113 Figure 12.20: (a) The structure of the enzyme carboxypeptidase-A, which contains 307 amino acids. The zinc ion is shown above as a black sphere in the center. (b) Carboxy-peptidase-A with a substrate (pink) in place.

114. Figure 12.21: Protein-substrate interaction. The substrate is shown in black and red, with the red representing the terminal amino acid. Blue indicates side chains from the enzyme that help bind the substrate.