1. AP Notes Chapter 15 Principles of Reactivity: Chemical Kinetics

2. The study of how fast chemical reactions occur. The study of how fast chemical reactions occur. How fast species are appearing and disappearing. How fast species are appearing and disappearing. It has nothing to do with the extent to which It has nothing to do with the extent to which a reaction will occur – that is thermodynamics.

3. There is a difference between the probability that molecules will react when they meet each other and once they do react how stable is the product. There is a difference between the probability that molecules will react when they meet each other and once they do react how stable is the product. First, probability of reaction has to do with overcoming an energy barrier or activation energy. First, probability of reaction has to do with overcoming an energy barrier or activation energy. Second, has to do with the differences between reactants and products. Second, has to do with the differences between reactants and products.

4. Thermodynamics: Spontaneity deals with how likely a reaction will take place. Spontaneity deals with how likely a reaction will take place. Spontaneous reactions are reactions that will happen - but we can’t tell how fast. Spontaneous reactions are reactions that will happen - but we can’t tell how fast. Diamond will spontaneously turn to graphite – eventually. Diamond will spontaneously turn to graphite – eventually.

5. Factors Affecting Rates 1.Surface area Powders 2.Temperature Dye 3.Pressure 4.Catalyst /Inhibitor 5. Concentration H 2 0 2 & I 2

6. Reaction Rate The change in concentration of a reactant or product per unit time For a reaction A  B Average Rate = Change in Moles of B Change in time Rate =  [B] = -  [A]  t  t

7. Reaction Rate Rate = Conc. of A at t 2 -Conc. of A at t 1 t 2 - t 1 Rate = Conc. of A at t 2 -Conc. of A at t 1 t 2 - t 1 Rate =  [A] D t Rate =  [A] D t Change in concentration per unit time Change in concentration per unit time Changes in Pressure to changes in concentration Changes in Pressure  to changes in concentration

8. N 2 + 3H 2 2NH 3 As the reaction progresses the concentration H 2 goes down As the reaction progresses the concentration H 2 goes down ConcentrationConcentration Time [H 2 ]

9. N 2 + 3H 2 2NH 3 As the reaction progresses the concentration N 2 goes down 1/3 as fast As the reaction progresses the concentration N 2 goes down 1/3 as fast ConcentrationConcentration Time [H 2 ] [N 2 ]

10. N 2 + 3H 2 2NH 3 As the reaction progresses the concentration NH 3 goes up. As the reaction progresses the concentration NH 3 goes up. ConcentrationConcentration Time [H 2 ] [N 2 ] [NH 3 ]

11. Calculating Rates Average rates are taken over long intervals Average rates are taken over long intervals are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time Instantaneous rates are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time. Derivative and Integrated Rates.

12. Average slope method Average slope method ConcentrationConcentration Time D[H 2 ] DtDtDtDt

13. Instantaneous slope method. Instantaneous slope method. ConcentrationConcentration Time  [H 2 ] D t D t

15. Defining Rate We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product. We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product. In our example In our example N 2 + 3H 2 2NH 3 -  [N 2 ] = -  [H 2 ] =  [NH 3 ]  t 3  t 2  t -  [N 2 ] = -  [H 2 ] =  [NH 3 ]  t 3  t 2  t

16. RATE LAW Mathematical model that contains specific information about what reactants determine the rate and how the rate is controlled

17. Rate Laws Reactions are reversible. Reactions are reversible. As products accumulate they can begin to turn back into reactants. As products accumulate they can begin to turn back into reactants. Early on the rate will depend on only the amount of reactants present. Early on the rate will depend on only the amount of reactants present. We want to measure the reactants as soon as they are mixed. We want to measure the reactants as soon as they are mixed. This is called the Initial rate method. This is called the Initial rate method.

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

19. You will find that the rate will only depend on the concentration of the reactants. You will find that the rate will only depend on the concentration of the reactants. Rate = k[NO 2 ] n Rate = k[NO 2 ] n This is called a rate law expression. This is called a rate law expression. k is called the rate constant. k is called the rate constant. n is the order of the reactant -usually a positive integer. n is the order of the reactant -usually a positive integer. 2 NO 2 2 NO + O 2

20. The rate of appearance of O 2 can be said to be. The rate of appearance of O 2 can be said to be. Rate' = D [O 2 ] = D [NO 2 ] = k'[NO 2 ] D t 2 D t Rate' = D [O 2 ] = D [NO 2 ] = k'[NO 2 ] D t 2 D t Because there are 2 NO 2 for each O 2 Because there are 2 NO 2 for each O 2 Rate = 2 x Rate' Rate = 2 x Rate' So k[NO 2 ] n = 2 x k'[NO 2 ] n So k[NO 2 ] n = 2 x k'[NO 2 ] n So k = 2 x k' So k = 2 x k' 2 NO 2 2 NO + O 2

21. Types of Rate Laws Differential Rate law - describes how rate depends on concentration. Differential Rate law - describes how rate depends on concentration. Integrated Rate Law - Describes how concentration depends on time. Integrated Rate Law - Describes how concentration depends on time. For each type of differential rate law there is an integrated rate law and vice versa. For each type of differential rate law there is an integrated rate law and vice versa. Rate laws can help us better understand reaction mechanisms. Rate laws can help us better understand reaction mechanisms.

22. Determining Rate Laws The first step is to determine the form of the rate law (especially its order). The first step is to determine the form of the rate law (especially its order). Must be determined from experimental data. Must be determined from experimental data. For this reaction 2 N 2 O 5 (aq) 4NO 2 (aq) + O 2 (g) The reverse reaction won’t play a role For this reaction 2 N 2 O 5 (aq) 4NO 2 (aq) + O 2 (g) The reverse reaction won’t play a role

23. [N 2 O 5 ] (mol/L) Time (s) 1.000 0.88200 0.78400 0.69600 0.61800 0.541000 0.481200 0.431400 0.381600 0.341800 0.302000 Now graph the data

24. To find rate we have to find the slope at two points To find rate we have to find the slope at two points We will use the tangent method. We will use the tangent method.

25. At.90 M the rate is (.98 -.76) = 0.22 =- 5.5x 10 -4 (0-400) -400

26. At.40 M the rate is (.52 -.31) = 0.22 =- 2.7 x 10 -4 (1000-1800) -800

27. Since the rate at twice the concentration is twice as fast the rate law must be.. Since the rate at twice the concentration is twice as fast the rate law must be.. Rate = - D [N 2 O 5 ] = k[N 2 O 5 ] 1 = k[N 2 O 5 ] D t Rate = - D [N 2 O 5 ] = k[N 2 O 5 ] 1 = k[N 2 O 5 ] D t We say this reaction is first order in N 2 O 5 We say this reaction is first order in N 2 O 5 The only way to determine order is to run the experiment. The only way to determine order is to run the experiment.

28. The method of Initial Rates This method requires that a reaction be run several times. This method requires that a reaction be run several times. The initial concentrations of the reactants are varied. The initial concentrations of the reactants are varied. The reaction rate is measured bust after the reactants are mixed. The reaction rate is measured bust after the reactants are mixed. Eliminates the effect of the reverse reaction. Eliminates the effect of the reverse reaction.

29. Integrated Rate Law Expresses the reaction concentration as a function of time. Expresses the reaction concentration as a function of time. Form of the equation depends on the order of the rate law (differential). Form of the equation depends on the order of the rate law (differential). Rate =  [A] n  t Rate =  [A] n  t We will only work with n=0, 1, and 2 although negative and fractional exist We will only work with n=0, 1, and 2 although negative and fractional exist

30. General Rate Law aA + bB  cC Rate  -[A] or -[B] r = k[A] m [B] n Where: m = order with respect to A n = order wrt B and n + m = order of reaction Notice we are primarily discussing the reactants

31. For example: aA + bB  cC + dD Zero Order: r = k Zero Order: r = k First Order in A: r = k [A] First Order in A: r = k [A] First Order in B: r = k [B] First Order in B: r = k [B] Second Order in A: r = k [A] 2 Second Order in A: r = k [A] 2 Second Order in B: r = k [B] 2 Second Order in B: r = k [B] 2 First Order in A, first Order in B, for a second order overall: r = k [A][B] First Order in A, first Order in B, for a second order overall: r = k [A][B]

32. Initial Rate Method to establish the exact rate law R 0 = k[A 0 ] m [B 0 ] n but experimental data must provide orders Use ratio method to solve for m and n R 0 = k[A 0 ] m [B 0 ] n

33. Exp Rate [A 0 ] [B 0 ] 1 0.05 0.50 0.20 2 0.05 0.75 0.20 3 0.10 0.50 0.40 R 0 = k[A 0 ] m [B 0 ] n Thus, rate = k[A] 0 [B] 1 or rate = k[B]

34. Where k is the rate constant, with appropriate units. rate = k[B] Using Exp 1:

35. INTEGRATED RATE LAWS Each “order” has a separate “law” based on the integration of the rate expression.

36. You NEED TO KNOW the rate expressions & half-life formulas for 0, 1 st, and 2 nd order reactions. Third order and fractional order reactions exist, but are rare. Nuclear reactions are typically 1 st order, therefore, use the first-order rate law and half-life to solve problems of this type

37. aA  B Zero Order First Order Second Order Rate Law r=k[A] o r=k[A] 1 r=k[A] 2 d[A] = k[A]m y=mx +b y=mx +b [A]=-kt +[A] o ln[A]=-kt+ln[A] o 1 = kt + 1 1 = kt + 1 [A] [A] o Plot for a linear graph [A] vs t ln[A] vs t 1 vs t 1 vs tln[A] slope of linear plot -k-kk half-life t1/2 when t 1/2 = [A] o 2k 2k t 1/2 = 0.693 k t 1/2 = 1 k[A] o k[A] o

38. Most often when reaction happens on a surface because the surface area stays constant. Most often when reaction happens on a surface because the surface area stays constant. Also applies to enzyme chemistry. Also applies to enzyme chemistry. Zero Order Rate Law

39. Time ConcentrationConcentration

40. ConcentrationConcentration  A]/  t tt k =  A]

41. Zero Order Rate Law Rate = k[A] 0 = k Rate = k[A] 0 = k Rate does not change with concentration. Rate does not change with concentration. Integrated [A] = -kt + [A] 0 Integrated [A] = -kt + [A] 0 When [A] = [A] 0 /2 t = t 1/2 When [A] = [A] 0 /2 t = t 1/2 t 1/2 = [A] 0 /2k t 1/2 = [A] 0 /2k

42. For a zero-order reaction, the rate is independent of the concentration. Expressed in terms of calculus, a zero - order reaction would be expressed as:

43. t [B] m = -k

44. Time for the concentration to become one- half of its initial value Half-life 0-Order Reaction

45. First Order For the reaction 2N 2 O 5 4NO 2 + O 2 For the reaction 2N 2 O 5 4NO 2 + O 2 We found the Rate = k[ N 2 O 5 ] 1 We found the Rate = k[ N 2 O 5 ] 1 If concentration doubles rate doubles. If concentration doubles rate doubles. If we integrate this equation with respect to time we get the Integrated Rate Law If we integrate this equation with respect to time we get the Integrated Rate Law ln[N 2 O 5 ] = - kt + ln[N 2 O 5 ] 0 ln[N 2 O 5 ] = - kt + ln[N 2 O 5 ] 0 ln is the natural log ln is the natural log [N 2 O 5 ] 0 is the initial concentration. [N 2 O 5 ] 0 is the initial concentration.

46. General form Rate =  [A] /  t = k[A] General form Rate =  [A] /  t = k[A] ln[A] = - kt + ln[A] 0 ln[A] = - kt + ln[A] 0 In the form y = mx + b In the form y = mx + b y = ln[A]m = -k y = ln[A]m = -k x = tb = ln[A] 0 x = tb = ln[A] 0 A graph of ln[A] vs time is a straight line. A graph of ln[A] vs time is a straight line. First Order

47. By getting the straight line you can prove it is first order By getting the straight line you can prove it is first order Often expressed in a ratio Often expressed in a ratio First Order

48. By getting the straight line you can prove it is first order By getting the straight line you can prove it is first order Often expressed in a ratio Often expressed in a ratio First Order

49. The time required to reach half the original concentration. The time required to reach half the original concentration. If the reaction is first order If the reaction is first order [A] = [A] 0 /2 when t = t 1/2 [A] = [A] 0 /2 when t = t 1/2 First Order Nuclear Decay and Half life

50. The time required to reach half the original concentration. If the reaction is first order [A] = [A] 0 /2 when t = t 1/2 ln(2) = kt 1/2 ln(2) = kt 1/2

51. t 1/2 = 0.693/k t 1/2 = 0.693/k The time to reach half the original concentration does not depend on the starting concentration. The time to reach half the original concentration does not depend on the starting concentration. An easy way to find k An easy way to find k

52. For a 1st-order reaction, the rate is dependent of the concentration. Expressed in terms of calculus, a 1st - order reaction would be expressed as:

54. t ln [A] m = -k

55. Half-life 1 st -order reaction

56. Second Order Rate = -  [A] /  t = k[A] 2 Rate = -  [A] /  t = k[A] 2 integrated rate law integrated rate law 1/[A] = kt + 1/[A] 0 1/[A] = kt + 1/[A] 0 y= 1/[A] m = k y= 1/[A] m = k x= tb = 1/[A] 0 x= tb = 1/[A] 0 A straight line if 1/[A] vs t is graphed A straight line if 1/[A] vs t is graphed Knowing k and [A] 0 you can calculate [A] at any time t Knowing k and [A] 0 you can calculate [A] at any time t

57. Second Order Half Life [A] = [A] 0 /2 at t = t 1/2 [A] = [A] 0 /2 at t = t 1/2

58. For a 2nd-order reaction, the rate is dependent of the concentration. Expressed in terms of calculus, a 2nd - order reaction would be expressed as:

59. t 1/[B] m = k

61. Half-life for 2nd - order

62. General Rate Law aA + bB + cC dD

63. A + B + C  products Exp[A][B][C]Initial Rate 11.25 8.7 22.501.25 17.4 31.253.021.2550.8 41.253.023.75457 53.011.001.15??

64. Exp[A][B][C]Initial Rate 11.25 8.7 22.501.25 17.4 If [B] and [C] are held constant, then any change in rate is caused by [A]. [A] doubles from 1.25 to 2.50 the rate double from 8.7 to 17.4 therefore order for A is 1 st order

65. Exp[A][B][C]Initial Rate 31.253.021.2550.8 41.253.023.75457 If [A] and [B] are held constant, then any change in rate is caused by [C]. [C] doubles from 1.25 to 3.75 the rate quadruples or a squaring of the change from 50.8 to 475 therefore order for C is 2nd order

66. Exp[A][B][C]Initial Rate 11.25 8.7 31.253.021.2550.8 If [A] and [C] are held constant, then any change in rate is caused by [B]. [B] doubles from 1.25 to 3.02 the rate quadruples or a squaring of the change from 8.7 to 50.8 therefore order for B is 2nd order

67. aA + bB + cC  dD A + B + C  products rate = k[A] 1 [B] 2 [C] 2 if we make substitutions with data from experiment 1 to solve for k 8.7 = k[1.25] 1 [1.25] 2 [1.25] 2 k=2.85

68. Exp[A][B][C]Initial Rate 53.011.001.15?? rate = 2.85[3.01] 1 [1.00] 2 [1.15] 2 rate =11.35

69. For the reaction BrO 3 - + 5 Br - + 6H + 3Br 2 + 3 H 2 O For the reaction BrO 3 - + 5 Br - + 6H + 3Br 2 + 3 H 2 O The general form of the Rate Law is Rate = k[BrO 3 - ] n [Br - ] m [H + ] p The general form of the Rate Law is Rate = k[BrO 3 - ] n [Br - ] m [H + ] p We use experimental data to determine the values of n,m,and p We use experimental data to determine the values of n,m,and p Another Complex Reaction

70. BrO 3 - + 5 Br - + 6H + 3Br 2 + 3 H 2 O BrO 3 - + 5 Br - + 6H + 3Br 2 + 3 H 2 O For this reaction we found the rate law to be For this reaction we found the rate law to be Rate = k[BrO 3 - ][Br - ][H + ] 2 Rate = k[BrO 3 - ][Br - ][H + ] 2 To investigate this reaction rate we need to control the conditions To investigate this reaction rate we need to control the conditions

71. Rate = k[BrO 3 - ][Br - ][H + ] 2 We set up the experiment so that two of the reactants are in large excess. 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 noticably As the reaction proceeds [BrO 3 - ] changes noticably [Br - ] and [H + ] don’t [Br - ] and [H + ] don’t

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

73. Initial concentrations (M) Rate (M/s) BrO 3 - Br - H+H+H+H+ 0.100.100.108.0 x 10 -4 0.100.100.108.0 x 10 -4 0.200.100.101.6 x 10 -3 0.200.100.101.6 x 10 -3 0.200.200.103.2 x 10 -3 0.200.200.103.2 x 10 -3 0.100.100.203.2 x 10 -3 0.100.100.203.2 x 10 -3 Now we have to see how the rate changes with concentration

74. Reaction Mechanisms The series of steps that actually occur in a chemical reaction. The series of steps that actually occur in a chemical reaction. Kinetics can tell us something about the mechanism Kinetics can tell us something about the mechanism A balanced equation does not tell us how the reactants become products. A balanced equation does not tell us how the reactants become products.

75. 2NO 2 + F 2 2NO 2 F 2NO 2 + F 2 2NO 2 F Rate = k[NO 2 ][F 2 ] Rate = k[NO 2 ][F 2 ] The proposed mechanism is The proposed mechanism is NO 2 + F 2 NO 2 F + F (slow) NO 2 + F 2 NO 2 F + F (slow) F + NO 2 NO 2 F(fast) F + NO 2 NO 2 F(fast) F is called an intermediate It is formed then consumed in the reaction F is called an intermediate It is formed then consumed in the reaction Reaction Mechanisms

76. Each of the two reactions is called an elementary step. Each of the two reactions is called an elementary step. The rate for a reaction can be written from its molecularity. The rate for a reaction can be written from its molecularity. Molecularity is the number of pieces that must come together. Molecularity is the number of pieces that must come together. Reaction Mechanisms

77. Mechanism must pass two tests to be accepted 1. Sum of mechanism steps produces the net equation. 2. Rate laws of all rate determining steps must be combined to produce experimental rate law.

78. Unimolecular step involves one molecule - Rate is rirst order. Unimolecular step involves one molecule - Rate is rirst order. Bimolecular step - requires two molecules - Rate is second order Bimolecular step - requires two molecules - Rate is second order Termolecular step- requires three molecules - Rate is third order Termolecular step- requires three molecules - Rate is third order Termolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule. Termolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule.

79. A products Rate = k[A] A products Rate = k[A] A+A productsRate= k[A] 2 A+A productsRate= k[A] 2 2A productsRate= k[A] 2 2A productsRate= k[A] 2 A+B productsRate= k[A][B] A+B productsRate= k[A][B] A+A+B Products Rate= k[A] 2 [B] A+A+B Products Rate= k[A] 2 [B] 2A+B Products Rate= k[A] 2 [B] 2A+B Products Rate= k[A] 2 [B] A+B+C Products Rate= k[A][B][C] A+B+C Products Rate= k[A][B][C]

80. How to get rid of intermediates Use the reactions that form them Use the reactions that form them If the reactions are fast and irreversible - the concentration of the intermediate is based on stoichiometry. If the reactions are fast and irreversible - the concentration of the intermediate is based on stoichiometry. If it is formed by a reversible reaction set the rates equal to each other. If it is formed by a reversible reaction set the rates equal to each other.

81. Formed in reversible reactions 2 NO + O 2 2 NO 2 2 NO + O 2 2 NO 2Mechanism 2 NO N 2 O 2 (fast) 2 NO N 2 O 2 (fast) N 2 O 2 + O 2 2 NO 2 (slow) N 2 O 2 + O 2 2 NO 2 (slow) rate = k 2 [N 2 O 2 ][O 2 ] k 1 [NO] 2 = k -1 [N 2 O 2 ] k 1 [NO] 2 = k -1 [N 2 O 2 ] rate = k 2 (k 1 / k -1 )[NO] 2 [O 2 ]=k[NO] 2 [O 2 ] rate = k 2 (k 1 / k -1 )[NO] 2 [O 2 ]=k[NO] 2 [O 2 ]

82. Formed in fast reactions 2 IBr I 2 + Br 2 2 IBr I 2 + Br 2 Mechanism Mechanism IBrI + Br (fast) IBrI + Br (fast) IBr + Br I + Br 2 (slow) IBr + Br I + Br 2 (slow) I + II 2 (fast) I + II 2 (fast) Rate = k[IBr][Br] but [Br]= [IBr] Rate = k[IBr][Br] but [Br]= [IBr] Rate = k[IBr][IBr] = k[IBr] 2 Rate = k[IBr][IBr] = k[IBr] 2

83. Collision theory Molecules must collide to react. Molecules must collide to react. Concentration affects rates because collisions are more likely. Concentration affects rates because collisions are more likely. Must collide hard enough. Must collide hard enough. Temperature and rate are related. Temperature and rate are related. Only a small number of collisions produce reactions. Only a small number of collisions produce reactions.

84. Potential EnergyPotential Energy Reaction Coordinate Reactants Products

85. Potential EnergyPotential Energy Reaction Coordinate Reactants Products Activation Energy E a

86. Potential EnergyPotential Energy Reaction Coordinate Reactants Products Activated complex

87. Potential EnergyPotential Energy Reaction Coordinate Reactants Products EE }

88. Potential EnergyPotential Energy Reaction Coordinate 2BrNO 2NO + Br Br---NO 2 Transition State

89. A + B --> A-B* --> AB A + B A-B* AB HH E a (cat) Reaction Coordinate Energy Misconception about catalysts Catalysts do not take part in reactions

90. Terms Activation energy - the minimum energy needed to make a reaction happen. Activation energy - the minimum energy needed to make a reaction happen. Activated Complex or Transition State - The arrangement of atoms at the top of the energy barrier. Activated Complex or Transition State - The arrangement of atoms at the top of the energy barrier.

91. Arrhenius Said the reaction rate should increase with temperature. Said the reaction rate should increase with temperature. At high temperature more molecules have the energy required to get over the barrier. At high temperature more molecules have the energy required to get over the barrier. The number of collisions with the necessary energy increases exponentially. The number of collisions with the necessary energy increases exponentially.

92. Arrhenius Equation Number of collisions with the required k = Ae ^(-E a /RT) k = rate constant k = rate constant A = total collisions ≈ frequency factor A = total collisions ≈ frequency factor e is Euler’s number (opposite of ln) e is Euler’s number (opposite of ln) E a = activation energy E a = activation energy R = Universal gas constant R = Universal gas constant T = temperature in Kelvin T = temperature in Kelvin Linearized Arrhenius Equation lnk = lnA – (E a /RT)

93. Using the linear equation we can find a number of things Using the linear equation we can find a number of things Plot lnk vs 1/T the slope will give you the activation energy Plot lnk vs 1/T the slope will give you the activation energy If we know Ea and k at one temperature then you can find k at a different temperature If we know Ea and k at one temperature then you can find k at a different temperature ln k2 = E a 1 - 1 k1 R T1 T2 k1 R T1 T2

94. Problems Observed rate is less than the number of collisions that have the minimum energy. Observed rate is less than the number of collisions that have the minimum energy. Due to Molecular orientation Due to Molecular orientation written into equation as p the steric factor. written into equation as p the steric factor.

95. O N Br O N O N O N O N O N O N O N O N O N No Reaction

96. Arrhenius Equation k = zpe -E a /RT = Ae -E a /RT k = zpe -E a /RT = Ae -E a /RT A is called the frequency factor = zp A is called the frequency factor = zp ln k = -(E a /R)(1/T) + ln A ln k = -(E a /R)(1/T) + ln A Another line !!!! Another line !!!! ln k vs t is a straight line ln k vs t is a straight line

97. Mechanisms and rates There is an activation energy for each elementary step. There is an activation energy for each elementary step. Activation energy determines k. Activation energy determines k. k = Ae - (E a /RT) k = Ae - (E a /RT) k determines rate k determines rate Slowest step (rate determining) must have the highest activation energy. Slowest step (rate determining) must have the highest activation energy.

98. This reaction takes place in three steps

99. EaEa First step is fast Low activation energy

100. Second step is slow High activation energy EaEa

101. EaEa Third step is fast Low activation energy

102. Second step is rate determining

103. Intermediates are present

104. Activated Complexes or Transition States

105. Possible Mechanisms a. CO + NO 2  CO 2 + NO b. 2NO 2  N 2 O 4 N 2 O 4 + 2CO  2CO 2 + 2NO (slow)

106. c. 2NO 2  NO 3 + NO (slow) NO 3 + CO  NO 2 + CO 2 (fast)

107. d. 2NO 2  2NO + O 2 (slow) 2 CO + O 2  2CO 2 (fast)

109. Transition State Theory transition state = activated complex A + B ---> A--B* ---> AB

110. A + B --> A-B* --> AB A + B A-B* AB HH EaEa Reaction Coordinate Energy activated complex activation energy Heat of reaction

111. A + B A-B* AB HH EaEa Reaction Coordinate Energy A + B --> A-B* --> AB

112. In General rate(R)  [A] R = k[A] k = rate constant

114. If one examines the relationship between the rate constant, k, and temperature, we find that ln k  1/T

115. 1/T ln k m = -E a /R

116. Arrhenius Equation

117. A = frequency factor R = 8.314 J/mol. K

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

121. How Catalysts Work Catalysts allow reactions to proceed by a different mechanism - a new pathway. Catalysts allow reactions to proceed by a different mechanism - a new pathway. New pathway has a lower activation energy. New pathway has a lower activation energy. More molecules will have this activation energy. More molecules will have this activation energy. Do not change  E Do not change  E

122. Pt surface HHHH HHHH Hydrogen bonds to surface of metal. Hydrogen bonds to surface of metal. Break H-H bonds Break H-H bonds Heterogenous Catalysts

123. Pt surface HHHH Heterogenous Catalysts C HH C HH

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

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

126. Pt surface H Heterogenous Catalysts C HH C HH HHH

127. Homogenous Catalysts Chlorofluorocarbons catalyze the decomposition of ozone. Chlorofluorocarbons catalyze the decomposition of ozone. Enzymes regulating the body processes. (Protein catalysts) Enzymes regulating the body processes. (Protein catalysts)

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

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