1. KINETICS The speed of a reaction

2. Kinetics The study of reaction rates. Spontaneous reactions are reactions that will happen - but we can’t tell how fast. Diamond will spontaneously turn to graphite – eventually. Reaction mechanism- the steps by which a reaction takes place.

3. Reaction Rate Rate = Conc. of A at t 2 -Conc. of A at t 1 t 2 - t 1 Rate =  [A] Dt Change in concentration per unit time For this reaction N 2 + 3H 2 2NH 3

4. As the reaction progresses the concentration of H 2 goes down ConcentrationConcentration Time [H 2 ]

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

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

7. Calculating Rates Average rates are taken over long intervals 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.

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

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

10. 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. In our example: N 2 + 3H 2 2NH 3 -  [N 2 ] = -3  [H 2 ] = 2  [NH 3 ]  t  t  t

11. Rate Laws Reactions are reversible. As products accumulate they can begin to turn back into reactants. 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. This is called the Initial rate method and time = 0.

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

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

14. The rate of appearance of O 2 can be said to be. Rate' = D[O 2 ] = k'[NO 2 ] 2 Dt Because there are 2 NO 2 for each O 2 Rate = 2 x Rate' So k[NO 2 ] n = 2 x k'[NO 2 ] n So k = 2 x k' 2 NO 2 2 NO + O 2

15. Types of Rate Laws Differential Rate law - describes how rate depends on concentration. 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. Rate laws can help us better understand reaction mechanisms.

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

17. [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

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

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

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

21. 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 ] Dt We say this reaction is first order in N 2 O 5 The only way to determine order is to run the experiment. So the Rate Law is Rate = k[N 2 O 5 ]

22. The Method of Initial Rates This method requires that a reaction be run several times. The initial concentrations of the reactants are varied. The reaction rate is measured just after the reactants are mixed. Eliminates the effect of the reverse reaction. In our problems, this isn’t a concern.

23. An Example of a Rate Law 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 We use experimental data to determine the values of n, m, and p

24. 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

25. Remember our Formula ! - Δ Concentration order = Δ Rate In words, the change in concentration raised to the order of the exponent = the change in the rate. - Fortunately, most of the time, these can be solved on inspection. Eg. If the concentration is tripled and the rate is tripled, then 3 x = 3 so x = 1. HOWEVER…

26. Sometimes more drastic measures are needed. Namely the law of logarithms. Δ Concentration order = Δ Rate becomes Order x (Log (conc.)) = Log (rate) It is simple and works every time. 2 3 = 8 3 log 2 = log 8 3 x.301 =.903

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

28. First Order For the reaction 2N 2 O 5 4NO 2 + O 2 We found the Rate = k[ N 2 O 5 ] 1 If concentration doubles rate doubles. 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 is the natural log [N 2 O 5 ] 0 is the initial concentration.

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

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

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

32. Half Life The time required to reach half the original concentration. If the reaction is first order [A] = [A] 0 /2 when t = t 1/2

33. Half Life 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

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

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

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

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

38. Most often when reaction happens on a surface because the surface area stays constant. Also applies to enzyme chemistry. The material is said to be at the zeroeth order. Zero Order Rate Law

39. Time ConcentrationConcentration

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

41. 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 be:For 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 conditions.To investigate this reaction rate we need to control the conditions.

42. 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 [Br - ] 0 = 1.0 M [H + ] 0 = 1.0 M As the reaction proceeds [BrO 3 - ] changes noticably [Br - ] and [H + ] don’t

43. This rate law can be rewritten Rate = k[BrO 3 - ][Br - ] 0 [H + ] 0 2Rate = 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

44. Summary of Rate Laws Know your integrated rate laws. Know how to find the order for the reaction Know how to find k after you determine the orders. As always, read the question carefully to make sure you are using all the given information.

45. Reaction Mechanisms The series of steps that actually occur in a chemical reaction. Kinetics can tell us something about the mechanism. A balanced equation does not tell us how the reactants become products. We will always want to find the rate determining step, which will usually be the slowest step in the mechanism.

46. 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) F is called an intermediate It is formed then consumed in the reaction, so it isn’t present in the given equation. Reaction Mechanisms

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

48. Unimolecular step requires one molecule so the rate is first order. Bimolecular step - requires two molecules so the rate is second order. Termolecular step - requires three molecules so the 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. Molecularity

49. A Products Rate = k[A] A + A ProductsRate= k[A] 2 2A ProductsRate= k[A] 2 A + B ProductsRate= k[A][B] A + A + B Products Rate= k[A] 2 [B] 2A + B Products Rate= k[A] 2 [B] A + B + C Products Rate= k[A][B][C]

50. How to get rid of intermediates Use the reactions that form them 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. A picture is worth 1,000 words here, so let’s check out the next slide.

51. Formed in reversible reactions 2 NO + O 2 2 NO 2 Mechanism 2 NO N 2 O 2 (fast) Step 1 N 2 O 2 + O 2 2 NO 2 (slow) Step 2 rate = k 2 [N 2 O 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 ] NOTICE: Step 1 is in equilibrium so the rate of formation of the N 2 O 2 is equal to the rate of formation of the 2 NO.

52. Formed in fast reactions 2 IBr I 2 + Br 2 Mechanism IBr I + Br (fast) IBr + Br I + Br 2 (slow) I + II 2 (fast) What is the rate determining step ? Rate = k[IBr][Br] but [Br]= [IBr] Rate = k[IBr][IBr] = k[IBr] 2

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

54. Potential EnergyPotential Energy Reaction Coordinate Reactants Products

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

56. Potential EnergyPotential Energy Reaction Coordinate Reactants Products Activated complex

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

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

59. Terms 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.

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

61. Arrhenius Number of collisions with the required energy = ze -E a /RT z = total collisions e is Euler’s number (opposite of ln) E a = activation energy R = ideal gas constant T is temperature in Kelvin

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

63. O N Br O N O N O N O N O N O N O N O N O N No Reaction

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

65. Activation Energy and Rates The final saga

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

67. + This reaction takes place in three steps

68. + EaEa First step is fast Low activation energy

69. Second step is slow High activation energy + EaEa

70. + EaEa Third step is fast Low activation energy

71. Second step is rate determining

72. Intermediates are present

73. Activated Complexes or Transition States

74. 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.

75. 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. Do not change  E

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

77. Pt surface HHHH Heterogenous Catalysts C HH C HH

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

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

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

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

82. 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

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