1. Chemical Kinetics The Study of Reaction Rates

2. Chemical Kinetics Kinetics involves the study of several factors that affect the rates of chemical reactions. The final goal is to use all of the data to develop a step-by-step reaction mechanism. The mechanism is a possible path by which reactants become products.

3. Factors Affecting Reaction Rates Scientists typically examine how each of the following factors affect the rate of a particular reaction. Concentration of Reactants Concentration of Reactants Temperature Temperature Solvent (if applicable) Solvent (if applicable) Catalysts (if applicable) Catalysts (if applicable)

4. Reaction Rates The rate of reaction is typically expressed in the rate of disappearance of reactants, or the rate of formation of products. Since the stoichiometry of the reaction is known, the concentration of only one component of the reaction needs to be measured.

5. Reaction Rates Note that the concentration of NO increases by the same amount that [NO 2 ] decreases.

6. Reaction Rates [O 2 ] = ½[NO] So only one component of the reaction need be measured.

7. Reaction Rates For the reaction: 2NO 2 (g)  2NO(g) + O 2 (g) rate of loss of NO 2 = rate of formation of NO = 2 (rate of formation of O 2 ) = 2 (rate of formation of O 2 )

8. Reaction Rates For the reaction: 2NO 2 (g)  2NO(g) + O 2 (g) rate of loss of NO 2 = rate of formation of NO = 2 (rate of formation of O 2 ) = 2 (rate of formation of O 2 ) -Δ[NO 2 ] = Δ[NO] = 2(Δ[O 2 ]) Δt Δt Δt Δt Δt Δt Note the negative sign for the rate of loss of reactant.

9. Reaction Rates -Δ[NO 2 ] = Δ[NO] = 2(Δ[O 2 ]) Δt Δt Δt Δt Δt Δt This relationship can also be seen in a graphical presentation of concentration versus time.

10. 2NO 2 (g)  2NO(g) + O 2 (g) Note that the rate of reaction varies as the reaction proceeds.

11. Reaction Rates: Reaction Rates: 2NO 2 (g)  2NO(g) +O 2 (g) The rate of reaction of NO 2 is most rapid at the beginning of the reaction. It slows considerably as the reaction proceeds.

12. Reaction Rates Reaction rates vary with time, and also depend upon the stoichiometry of the reaction. As a result, we must be very specific in what we mean by a reaction rate. Initial rates are often used. This is the rate of reaction just after the reaction begins. The tangent to the curve during the initial moments of the reaction provides the rate.

13. Reaction Rates This graph for the decomposition of N 2 O 5 to form NO 2 and O 2 shows an initial rate of 5.4 x 10 -4 mol/L-s

14. Reaction Rates The convention for dealing with the stoichiometry of the reaction is that for a general reaction: aA + bB  cC + dD Rate = - 1 Δ[A] = - 1 Δ[B] = 1 (Δ[C]) = 1 (Δ[D]) a Δt b Δt c Δt d Δt a Δt b Δt c Δt d Δt

15. Rate Laws One of the goals of kinetics is to determine the rate law for a reaction. The rate law is the mathematical relationship that shows how the reaction rate depends upon the concentration of reactants. Rate = k[A] x [B] y k is the rate constant, and is highly temperature dependent

16. Rate Laws Rate = k[A] x [B] y x and y are called the order of the reaction with respect to reactant A and B respectively. They will usually have the value of 0, 1 or 2, though other values are possible. Rate laws must be determined experimentally.

17. Rate Laws The rate law of a reaction, along with information about temperature effects and solvent effects can be used to develop a possible reaction mechanism. The reaction mechanism is the step-by-step process by which reactants become products.

18. Determination of Rate Laws All rate laws are experimentally determined. There are two basic methods used: 1. The Method of Initial Rates 2. Graphical Techniques using the Integrated Rate Law Integrated Rate Law

19. The Method of Initial Rates [NO 2 - ] doubles O

20. The Method of Initial Rates [NO 2 - ] doubles Rate doubles O

21. The Method of Initial Rates [NO 2 - ] doubles Rate doubles Rate α [NO 2 - ] 1 O

22. The Method of Initial Rates [NH 4 + ] doublesRate doubles Rate α [NH 4 + ] 1 O

23. The Method of Initial Rates rate α [NH 4 + ][NO 2 - ] or rate = k [NH 4 + ][NO 2 - ] The reaction is first-order in ammonium ion, first-order in nitrite ion, and second-order overall.

24. The Method of Initial Rates

25. [BrO 3 - ] doubles rate doubles rate α [BrO 3 - ] 1

26. The Method of Initial Rates [H + ] doubles rate quadruples rate α [H + ] 2

27. The Method of Initial Rates [BrO 3 - ] doubles[Br - ] doublesrate quadruples rate α [Br - ] 1

28. Method of Initial Rates rate α [BrO 3 - ][Br - ][H + ] 2 The rate law is usually written with the rate constant, k, included: rate =k[BrO 3 - ][Br - ][H + ] 2 The reaction is first order in bromate, first order in bromide, and second order in hydronium ion.

29. Method of Initial Rates rate =k[BrO 3 - ][Br - ][H + ] 2 The data for any reaction trial can be used to calculate the value of k, the rate constant. The reaction rate has the units mol/liter-time, so the units of k depend upon the exponents in the rate law. Usually the value of k for several trials is averaged.

30. The Integrated Rate Law In general, rate laws will be first or second order. In either case, the rate law can be integrated to provide a linear equation of the form: y=mx + b. This permits graphical presentation of the data and determination of the value of the rate constant.

31. The Integrated Rate Law Once concentration versus time data have been collected, the scientist constructs one or more graphs to determine both the order of the reaction and the value of the rate constant.

32. The Integrated Rate Law We need not measure concentration. Any property that is proportional to concentration (intensity of color, pressure or volume of gases, pH, etc.) may be graphed.

33. The Integrated Rate Law- First Order Reactions rate = - d[A] = k[A] 1 dt dt Rearranging, we obtain: -d[A] = kdt [A] [A] When this expression is integrated from time =0 to time t, we obtain: ln[A] = -kt + ln[A] o

34. The Integrated Rate Law- First Order Reactions ln[A] = -kt + ln[A] o This is the integrated rate law for first-order reactions. It has the linear form y=mx+b. If the reaction is first-order, a graph of ln[A] versus time will be linear with a slope equal to –k.

35. The Integrated Rate Law- First Order Reactions The linearity indicates that the reaction is first-order with respect to N 2 O 5. The rate constant (- slope) has the units of (time) -1.

36. The Integrated Rate Law- First Order Reactions ln[A] = -kt + ln[A] o If a graph isn’t linear, a second-order plot must be prepared.

37. The Integrated Rate Law- First Order Reactions The curvature of the graph of ln[A] vs time indicates that this reaction is not first order.

38. The Integrated Rate Law – 2 nd The Integrated Rate Law – 2 nd Order Reactions rate = - d[A] = k[A] 2 dt dt Rearranging, we obtain: - d[A] = k dt [A] 2 [A] 2 When this expression is integrated from time =0 to time t, we obtain: 1 = kt + 1_ 1 = kt + 1_ [A] [A] o [A] [A] o

39. The Integrated Rate Law – 2 nd The Integrated Rate Law – 2 nd Order Reactions 1 = kt + 1_ 1 = kt + 1_ [A] [A] o [A] [A] o This equation has a linear (y=mx +b) form. If a reaction is second-order, a plot of 1/[A] versus time is linear, with the slope equal to the rate constant. The rate constant has the units (M) -1 (time) -1.

40. The Integrated Rate Law – 2 nd The Integrated Rate Law – 2 nd Order Reactions The linearity of the graph of [A] -1 indicates that the reaction is second-order with respect to NO 2.

41. The Integrated Rate Law- 2 nd Order Reactions The linearity of the graph of 1/[A] versus time indicates that the reaction is second-order with respect to C 4 H 6.

42. Zero-Order Rate Laws Most reactions involving a single reactant exhibit first or second-order kinetics. Rarely, a reaction will have zero-order kinetics. In this case, Rate = k[A] 0 = k(1) = k

43. Zero-Order Rate Laws Rate = k[A] 0 = k(1) = k The integrated rate law is: [A] = -kt + [A] o and a graph of [A] versus time is linear.

44. Zero-Order Rate Laws Unlike the rate laws for other reactions, for a zero- order reaction, a graph of [A] versus time is linear. The slope = -k.

45. Graphical Techniques

46. Half-life and 1 st Order Reactions The half-life is the time it takes for the concentration of a reactant to halve. It is represented by the symbol t ½. The half-life for a 1 st -order reaction doesn’t depend upon concentration. First-order reactions have a constant half-life. That is, it takes the same length of time for the concentration to halve throughout the reaction.

47. Half-life and 1 st Order Reactions

48. First order reactions have a constant half-life throughout the course of the reaction.

49. First-Order Half-Life

50. Half-life and 1 st Order Reactions The first order rate law can be rearranged: ln[A] = -kt + ln[A] o Since [A] = ½[A] o at the half-life, ln ([A] o /.5[A] o ) = kt ½ ln(2.00) = kt ½ 0.693 = kt ½ ln [A] o [A] = kt

51. Half-life and 1 st Order Reactions 0.693 = kt ½ or t ½ =0.693 k The half-life is constant throughout a first order reaction. All radioactive decay exhibits first-order kinetics.

52. Half-life Problem A mummy’s shroud has a 14 C activity of 8.9 dis/min/gC. Living things have a 14 C activity of 15.2 dis/min/gC. The half-life of 14 C is 5, 730 years. Estimate the age of the shroud. A mummy’s shroud has a 14 C activity of 8.9 dis/min/gC. Living things have a 14 C activity of 15.2 dis/min/gC. The half-life of 14 C is 5, 730 years. Estimate the age of the shroud.

53. Half-life and 2 nd Order Reactions For a 2 nd order reaction, each successive half- life is twice the previous one.

54. Half-life and 2 nd Order Reactions The half-life of second-order reactions depends upon the concentration of reactant. As the concentration decreases, the half-life increases.

55. Half-life and 2 nd Order Reactions The integrated rate law for a 2 nd order reaction is: 1 = kt + 1_ [A] [A] o [A] [A] o At the half life, [A]= ½ [A] o, and t = t ½. 1 = k t ½ + 1_ 1 = k t ½ + 1_.5[A] o [A] o.5[A] o [A] o

56. Half-life and 2 nd Order Reactions 1 = k t ½ + 1_ 1 = k t ½ + 1_.5[A] o [A] o.5[A] o [A] o 1 = k t ½ 1 = k t ½ [A] o [A] o t ½ = 1/k [A] o The half-life changes with concentration of reactant for a second-order reaction.

57. Rate Laws for Multiple Reactants Many reactions involve more than one reactant. The reaction order for each reactant can be determined graphically by monitoring the concentration of a specific reactant while ensuring that all other reactants are in very large (at least ten-fold) excess. In this way, the rate will depend only upon the reactant studied. The concentrations of other reactants will not change much during the reaction.

58. Reaction Mechanisms The goal of kinetics is to determine the series of steps by which reactants become products. This series of steps is called a reaction mechanism. The reaction mechanism is a series of elementary steps that represent single chemical events such as a collision, decomposition, etc.

59. Reaction Mechanisms Because each elementary step represents a single event, we can write rate laws for each of the steps. For example, if an elementary step involves the collision of molecule A with molecule B, than the rate will depend upon both the concentration of A and the concentration of B. Rate = k[A][B]

60. Reaction Mechanisms Rate laws can only be written for elementary steps. They cannot be written for chemical reactions unless experiments are performed to determine the relationship between rate and concentration.

61. Elementary Steps and their Rate Laws

62. Reaction Mechanisms Scientists study a particular reaction and experimentally determine the rate law. Using other experimental information, they propose a reaction mechanism. The reaction mechanism is a series of elementary steps that represent single chemical events such as a collision, decomposition, etc.

63. Reaction Mechanisms The proposed reaction mechanism must meet two criteria: 1. The proposed mechanism for the reaction must be consistent with the observed rate law. 2. The sum of the elementary steps must give the overall balanced equation for the chemical reaction.

64. Reaction Mechanisms Consider the reaction between NO 2 (g) and CO(g). The balanced equation is: NO 2 (g) + CO(g)  NO(g) + CO 2 (g) The experimentally determined rate law for the reaction is: Rate = k[NO 2 ] 2

65. Reaction Mechanisms The proposed mechanism for the reaction is: NO 2 + NO 2 (g)  NO 3 (g) + NO(g) NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) NO 3 (g) + CO(g)  NO 2 + CO 2 (g) NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) k1k1 k2k2

66. Reaction Mechanisms The sum of the steps must add up to the overall balanced equation: NO 2 + NO 2 (g)  NO 3 (g) + NO(g) NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) NO 3 (g) + CO(g)  NO 2 + CO 2 (g) NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) k1k1 k2k2

67. Reaction Mechanisms The sum of the steps must add up to the overall balanced equation: NO 2 + NO 2 (g)  NO 3 (g) + NO(g) NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) NO 3 (g) + CO(g)  NO 2 + CO 2 (g) NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) NO 2 (g) + CO(g)  NO(g) + CO 2 (g) k1k1 k2k2

68. Reaction Intermediates This is a species that is neither a reactant nor a product. It is produced during the course of the reaction and consumed in a later step. NO 2 + NO 2 (g)  NO 3 (g) + NO(g) NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) NO 3 (g) + CO(g)  NO 2 + CO 2 (g) NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) NO 2 (g) + CO(g)  NO(g) + CO 2 (g) k1k1 k2k2

69. Reaction Mechanisms NO 2 + NO 2 (g)  NO 3 (g) + NO(g) NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) NO 3 (g) + CO(g)  NO 2 + CO 2 (g) NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g)

70. Reaction Mechanisms NO 3 is a reaction intermediate. It is formed and consumed during the course of the reaction.

71. Reaction Mechanisms NO 2 (g) + CO(g)  NO(g) + CO 2 (g) The proposed two-step mechanism provided the overall balanced equation for the reaction. For the mechanism to be valid, the experimental rate law must match the rate law derived from the proposed reaction mechanism.

72. Reaction Mechanisms NO 2 (g) + CO(g)  NO(g) + CO 2 (g) The experimentally determined rate law for the reaction is: Rate = k[NO 2 ] 2

73. Reaction Mechanisms The proposed mechanism for the reaction is: NO 2 + NO 2 (g)  NO 3 (g) + NO(g) NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) NO 3 (g) + CO(g)  NO 2 + CO 2 (g) NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) k 1 and k 2 are the rate constants for the two steps of the mechanism. Many reaction mechanisms contain a rate determining step k 1 and k 2 are the rate constants for the two steps of the mechanism. Many reaction mechanisms contain a rate determining step. k1k1 k2k2

74. Reaction Mechanisms If a multi-step process has one step which is much slower than all of the other steps, the slow step will determine the overall rate. The slow step is called the rate determining step. The overall reaction can be no faster than its slowest step.

75. Reaction Mechanisms The proposed mechanism for the reaction is: NO 2 + NO 2 (g)  NO 3 (g) + NO(g) (slow) NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) (slow) NO 3 (g) + CO(g)  NO 2 + CO 2 (g) (fast) NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) (fast) For this mechanism, the first step is slow compared to the second, and is the rate determining step For this mechanism, the first step is slow compared to the second, and is the rate determining step. k1k1 k2k2

76. Reaction Mechanisms The proposed mechanism for the reaction is: NO 2 + NO 2 (g)  NO 3 (g) + NO(g) (slow) NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) (slow) NO 3 (g) + CO(g)  NO 2 + CO 2 (g) (fast) NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) (fast) The rate of the reaction will be the rate of the first step. Rate = k 1 [NO 2 ] 2 This is the derived rate law. k1k1 k2k2

77. Reaction Mechanisms The derived rate law is: Rate = k 1 [NO 2 ] 2 The observed rate law is: Rate = k[NO 2 ] 2 Since the rate laws match, the proposed mechanism may be the correct mechanism.

78. Reaction Mechanisms Further experimentation and research into related reaction mechanisms would be needed to help confirm the proposed reaction mechanism. In general, reaction mechanisms cannot be proven absolutely.

79. Energy and Chemical Reactions Reaction rates depend upon concentrations of reactants because molecules need to collide with each other in order to form products. Not all molecular collisions lead to product formation. The collisions must have sufficient energy to weaken existing bonds and also the proper orientation to produce product(s).

80. Collision Theory

81. Energy & Reaction Rates Collisions with the proper orientation also need sufficient energy to form products. For all reactions, exothermic or endothermic, a minimum amount of energy is required for the reactants to form products. This minimum amount of energy is called the activation energy.

82. Energy & Reaction Rates The activation energy, E a, is used to help weaken the bonds of reactants and help form the bonds in products.

83. Energy & Reaction Rates

84. Since the kinetic energy of the collisions influences the reaction rate, there is a relationship between temperature and the value of the rate constant.

85. Energy & Reaction Rates The experimentally based Arrhenius equation relates temperature and activation energy to the rate constant. k = Ae -Ea/RT where k is the rate constant; E a is the activation energy (J/mol); R = 8.314 J/K-mol; T is temperature in Kelvins; and A is the frequency factor.

86. Energy & Reaction Rates k = Ae -Ea/RT A, the frequency factor, (or pre-exponential factor) takes into account the frequency of collisions and the fraction with proper orientation to form products.

87. Energy & Reaction Rates k = Ae -Ea/RT The Arrhenius equation is best used in logarithmic form: ln(k) = - + ln(A) This equations has the linear form y = mx +b. A graph of ln k versus 1/T has a slope of -Ea/R. E a 1 R T

88. Determination of E a Chemists usually determine the value of the rate constant for a reaction at several temperatures. A graph of ln(k) versus 1/T will be linear, with the slope equal to – E a /R.

89. Temperature and Reaction Rate Chemical reactions always go faster when the temperature is increased. However, increasing the temperature of a reaction may be costly, or cause unwanted side reactions to occur.

90. Catalysts A catalyst is a substance which provides a lower energy pathway for the reaction. Catalysts alter the reaction mechanism of the reaction. Catalysts allow the reaction to proceed more rapidly at lower temperatures. Catalysts allow the reaction to proceed more rapidly at lower temperatures.

91. Catalysis The reaction will go faster with a catalyst, or proceed at a comparable rate at lower temperatures.

92. Catalysis The reaction goes faster because a greater percentage of molecules have sufficient energy to form products when the activation energy is lower.

93. Catalysts Catalysts are present at the beginning of the reaction. Although they are involved in the reaction, they are regenerated. The net result is that the catalyst speeds up the reaction without being consumed. Biological catalysts are called enzymes. Enzymes allow many complex chemical reactions to occur at body temperature (37 o C).

94. Catalysis

95. Catalysis and the Ozone Layer Freons, or chloroflurocarbons, have been linked to destruction of the ozone layer of the earth. Freons, when exposed to high energy light in the upper atmosphere, form highly reactive chorine atoms. These chlorine atoms convert ozone to oxygen and are regenerated in the process.

96. Catalysis and the Ozone Layer CFCl 3 + hν  CFCl 2 + Cl Cl + O 3  ClO + O 2 ClO + O  Cl + O 2 _____________________ O 3 + O  2 O 2 Ozone is depleted, and the chlorine radical is regenerated.

97. Heterogeneous Catalysis Heterogeneous catalysis usually involves gaseous reactants that react on the surface of a solid catalyst. An example is the reaction of ethylene with hydrogen. H 2 C=CH 2 (g) + H 2 (g)  H 3 CCH 3 (g)

98. Heterogeneous Catalysis H 2 C=CH 2 (g) + H 2 (g)  H 3 CCH 3 (g) The reaction proceeds slowly due to the strength of the bond in hydrogen. The gaseous reactants are collected on the surface of catalysts such as platinum, palladium or nickel, where the reaction occurs.

99. Heterogeneous Catalysis The metal catalyst interacts with the hydrogen molecules and weakens the H-H bond.

100. Heterogeneous Catalysis