1. 1 Chapter 14 Chemical Kinetics: Principles of Reactivity

2. 2 Kinetics Studies the Reaction rates - How fast the reaction occurs (the change in reactant and product concentration as a function of time) Reaction Mechanisms – The detailed pathway taken by atoms and molecules as the reaction proceeds (how the reaction occurs). Our goal is to understand chemical reactions at the molecular level.

3. 3 The Rate Laws: Factors Reaction Rate  Concentration Concentration –As concentration increases, the reaction rate increases. Physical State of Reactants – Reactants must mix and collide in order to react.. Temperature – As temperature increases, the reaction rate increases. Reaction Rate µ Temperature. Catalyst – A substance that increases the rate of a reaction without being consumed in the process.

4. 4 Reaction Rates Write an equation for a)The rate of change of distance with time b)Rate of change of speed with time c)The rate of change of concentration with time.

5. 5 Rate of Reaction Rate of Reaction: positive quantity that expresses how concentration of a reactant or product changes with time. A  B Note: [ A ] mean conc. in mol/L (M) Units? = M/s

6. 6 RATES OF REACTIONS For General reactions A → B, the graph will be: Focus on: average rate instantaneous rate initial rate

7. 7 Reaction Rates In this reaction, the concentration of butyl chloride, C 4 H 9 Cl, was measured at various times. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

8. 8 Reaction Rates: Average Rate The average rate of the reaction over each interval is the change in concentration divided by the change in time: Average rate =  [C 4 H 9 Cl]  t C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) We monitor the appearance of the product

9. 9 Reaction Rates: Average Rate The average rate decreases as the reaction proceeds. Reason: as the reaction goes forward, there are fewer collisions between reactant molecules. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

10. 10 Calculating Average Rate C 2 H 4 (g) + O 3 (g)  C 2 H 4 O (g) + O 2 (g)

11. 11 Reaction Rates: Instantaneous speed A plot of concentration vs. time. The slope of a line tangent to the curve at any point is the instantaneous rate at that time. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

12. 12 Reaction Rates: Initial Rate All reactions slow down over time. Best indicator of the rate of a reaction is the instantaneous rate near the beginning. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

13. 13 Reaction Rates and Stoichiometry In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. The rate of disappearance of C 4 H 9 Cl is the same as the rate of appearance of C 4 H 9 OH. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) Rate = -  [C 4 H 9 Cl]  t =  [C 4 H 9 OH]  t

14. 14 Rates of Reactions and Stoichiometry For every 1 mole of H 2 or I 2 that react, 2 moles of HI is formed.For every 1 mole of H 2 or I 2 that react, 2 moles of HI is formed. So that rate of formation of HI is two times faster than the depletion of H 2 or I 2. Write the Rate Laws: H 2 (g) + I 2 (g)  2 HI (g) or

15. 15 13.1 Reaction Rates and Stoichiometry General Rule for Every reaction: aA + bBcC + dD Rate = − 1a1a  [A]  t = − 1b1b  [B]  t = 1c1c  [C]  t 1d1d  [D]  t =

16. 16 Reaction Rates Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.

17. 17 The Rate Laws Determining Rate Laws Experimentally Rate laws are determined by measuring initial reaction rates at varying initial reactant concentrations. Done by following: –color change (spectroscopy); –precipitation formation (turbidity, absorbance); –gas formation (pressure monitoring); –conductometry; etc.

18. 18 Concentration and Rate The rate of a reaction changes with changes in concentration. Rates of reactions increase as concentrations increase since there are more collisions occurring between reactants.

19. 19 Concentration and Rate Comparing Experiments 1 and 2, when [NH 4 + ] doubles, the initial rate doubles. NH 4 + (aq) + NO 2 − (aq) N 2 (g) + 2 H 2 O (l)

20. 20 Concentration and Rate Likewise, comparing Experiments 5 and 6, when [NO 2 − ] doubles, the initial rate doubles. NH 4 + (aq) + NO 2 − (aq) N 2 (g) + 2 H 2 O (l)

21. 21 Concentration and Rate This means Rate  [NH 4 + ] Rate  [NO 2 − ] Rate  [NH + ] [NO 2 − ] or Rate = k [NH 4 + ] [NO 2 − ] This equation is called the rate law, and k is the rate constant. Units of k?

22. 22 Rate Laws A rate law shows the relationship between the reaction rate and the concentrations of reactants. The exponents tell the order of the reaction with respect to each reactant. This reaction is First-order in [NH 4 + ] First-order in [NO 2 − ]

23. 23 Concentration and Rate The rate law: examines the dependence of reaction rate on the concentration of reactants and is given as the rate expression. Determined EXPERIMENTALLY. The mathematical expression for the Rate Law: Rate = k [A] m [B] n –[A] & [B] represent the reactants. –The exponents m and n are called “reaction orders”. –The proportionality constant k is called the rate constant. –The overall reaction order is the sum of the reaction orders. –The overall “order of reaction” is therefore… –m + n + ….

24. 24 Units of k Rate expression Units of k Change in [A] Change in [B] Total change in rate Order of reaction

25. 25 Rate Law Determination Determined only by doing an experiment Isolation method: change the concentration of one reactant at a time, while keeping the concentration of all other reactants constant. There is no relationship between the order of the reaction and stoichiometry.

26. 26 Relation Between Reactant Concentration and Time Integration of rate laws allows one to calculate the concentration of any reactant or product at any point in time during the reaction progress.

27. 27 Integrated Rate Laws: 1 st Order A  B Using calculus to integrate the rate law for a first-order process gives us ln [A] t [A] 0 = −kt Where [A] 0 is the initial concentration of A. [A] t is the concentration of A at some time, t, during the course of the reaction.

28. 28 Integrated Rate Laws: 1 st Order Manipulating this equation produces… ln [A] t [A] 0 = −kt ln [A] t − ln [A] 0 = − kt ln [A] t = − kt + ln [A] 0 …which is in the form y = mx + b

29. 29 Integrated Rate Laws: 1 st Order Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be -k. ln [A] t = -kt + ln [A] 0 2 H 2 O 2 (aq)  2 H 2 O (l) + O 2 (g) Rate = k[H 2 O 2 ]

30. 30 First Order (Gases as Reactants) Consider the process in which methyl isonitrile is converted to acetonitrile. CH 3 NCCH 3 CN

31. 31 First Order (Gases as Reactants) This data was collected for this reaction at 198.9°C. CH 3 NCCH 3 CN

32. 32 First Order (Gases as Reactants) When ln P is plotted as a function of time, a straight line results. Therefore, –The process is first-order. –k is the negative slope: - 5.1  10 -5 s −1.

33. 33 First Order (Gases as Reactants) For gaseous substances: A(g) → product PV = nRTand Substitute in integrated equation:

34. 34 Second Order

35. 35 13.3 Integrated Rate Laws: Second Order Reaction * A  B Rate Law: Integrated Rate Law: k depends on the initial conc. of reactant [A] 0

36. 36 Second-Order Processes Integrating the rate law for a process that is second-order in reactant A, we get 1 [A] t = −kt + 1 [A] 0 also in the form y = mx + b

37. 37 Second-Order Processes If a process is second-order in A, a plot of 1/[A] vs. t will yield a straight line, and the slope of that line is -k. 1 [A] t = −kt + 1 [A] 0

38. 38 Second-Order Processes The decomposition of NO 2 at 300°C is described by the equation NO 2 (g) NO (g) + 1/2 O 2 (g) and yields data comparable to this: Time (s)[NO 2 ], M 0.00.01000 50.00.00787 100.00.00649 200.00.00481 300.00.00380

39. 39 Second-Order Processes Graphing ln [NO 2 ] vs. t yields: Time (s)[NO 2 ], Mln [NO 2 ] 0.00.01000−4.610 50.00.00787−4.845 100.00.00649−5.038 200.00.00481−5.337 300.00.00380−5.573 The plot is not a straight line, so the process is not first-order in [A].

40. 40 Second-Order Processes Graphing ln 1/[NO 2 ] vs. t, however, gives this plot. Time (s)[NO 2 ], M1/[NO 2 ] 0.00.01000100 50.00.00787127 100.00.00649154 200.00.00481208 300.00.00380263 Because this is a straight line, the process is second- order in [A].

41. 41 Zero Order Reaction Rate Law Integrated Rate Law Most often when reaction happens on a surface because the surface area stays constant. Also applies to enzyme chemistry.

42. 42 Zero Order Reaction Zero Order Reaction: Rate = k[A] 0 = k 2 NH 3 (g)  N 2 (g) + 3 H 2 (g) Rate = k [NH 3 ] 0 = k [A]t = concentration of [A] after some time, t k= reaction rate constant in units of M/s t= time in seconds [A] 0 = initial concentration of A y = mx +_b m = - k

43. 43 Half-Life Half-life is defined as the time required for one-half of a reactant to react. Because [A] at t 1/2 is one-half of the original [A], [A] t = 0.5 [A] 0.

44. 44 First Order Half-Life

45. 45 Half-Life For a first-order process, this becomes 0.5 [A] 0 [A] 0 ln = −kt 1/2 ln 0.5 = − kt 1/2 −0.693 = − kt 1/2 = t 1/2 0.693 k NOTE: For a first-order process, the half-life does not depend on [A] 0.

46. 46 Second-Order Half Life

47. 47 Second Order: Half-Life For a second-order process, 1 0.5 [A] 0 = kt 1/2 + 1 [A] 0 2 [A] 0 = kt 1/2 + 1 [A] 0 2 − 1 [A] 0 = kt 1/2 1 [A] 0 == t 1/2 1 k[A] 0

48. 48 Integrated Rate Laws: Summary OrderRate LawEquationt 1/2 1R = k[A] 2R= k[A] 2 1/[A] = 1/[A] 0 +kt 1/k[A] 0 0R = k[A] 0 [A] = -kt + [A] 0 [A] 0 /2k

49. 49 Temperature and Rate Generally, as temperature increases, so does the reaction rate. This is because k is temperature dependent. In a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they collide with each other.

50. 50 Collision Model The collision model assumes that in order for molecules to react they must collide. The greater the number of collisions the faster the rate. The more molecules present, the greater the probability of collisions and the faster the rate. The higher the temperature, the more energy available to the molecules and the faster the rate. Complication: not all collisions lead to products. In fact, only a small fraction of collisions lead to product.

51. 51 The Collision Model Molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. For the reaction: Cl + NOCl  NO + Cl 2 There are two possible ways that Cl atoms and NOCl molecules can collide; one is effective and one is not.

52. 52 Activation Energy Arrhenius: molecules must possess a minimum amount of energy to react. Why? –In order to form products, bonds must be broken in the reactants. Bond breakage requires energy. –Molecules moving too slowly, with too little kinetic energy, don’t react when they collide. Activation energy, E a : is the minimum energy required to initiate a chemical reaction. − E a will vary with the reaction. Next we will look at an example of E a.

53. 53 Activation Energy Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.

54. 54 13.4 Activation Energy Consider the rearrangement of methyl isonitrile: –In H 3 C-N  C, the C-N  C bond bends until the C-N bond breaks and the N  C portion is perpendicular to the H 3 C portion. This structure is called the activated complex or transition state. –The energy required for the above twist and break is the activation energy, E a. –Once the C-N bond is broken, the N  C portion can continue to rotate forming a C-C  N bond.

55. 55 13.4 Reaction Coordinate Diagram

56. 56 Reaction Coordinate Diagrams The diagram: shows the energy of the reactants and products (and, therefore,  E). Transition state: The high point on the diagram is the Activated complex: t he species present at the transition state is called the. The energy gap between the reactants and the activated complex is the activation energy barrier.

57. 57 Reaction Coordinate Diagram Exothermic reaction ΔH reactants ΔH products

58. 58 Maxwell–Boltzmann Distributions Temperature is defined as a measure of the average kinetic energy of the molecules in a sample. At any temperature there is a wide distribution of kinetic energies.

59. 59 Maxwell–Boltzmann Distributions As the temperature increases, the curve flattens and broadens. Thus at higher temperatures, a larger population of molecules has higher energy.

60. 60 Maxwell–Boltzmann Distributions If the dotted line represents the activation energy, as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. As a result, the reaction rate increases.

61. 61 Maxwell–Boltzmann Distributions This fraction of molecules can be found through the expression where R is the gas constant and T is the Kelvin temperature. f = e −E a /RT

62. 62 Arrhenius Equation Svante Arrhenius developed a mathematical relationship between k and E a : k = A e −E a /RT where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction.

63. 63 Microscopic View of Reaction Rates Arrhenius Equation Ea can be determined by finding k for a reaction experimentally at several temperatures.

64. 64 Arrhenius Equation The Arrhenius equation relates rate constant, molecular collisions, activation energy and temperature

65. 65 Reaction Mechanism

66. 66 Reaction Mechanisms Reaction mechanism: the sequence of events (path of the reaction) that describes the actual process by which reactants become products is called the. Mechanisms provide a very detailed picture of which bonds are broken and formed during the course of a reaction.

67. 67 Multistep Mechanisms In a multistep process, one of the steps will be slower than all others. The overall reaction cannot occur faster than this slowest, rate-determining step.

68. 68 Reaction Mechanisms Case #1: The first step is a slow step, the following steps are fast steps Case #2: The first step is a fast step and the following step is a slow step.

69. 69 Slow Initial Step The rate law for this reaction is found experimentally to be Rate = k [NO 2 ] 2 CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. This suggests the reaction occurs in two steps. NO 2 (g) + CO (g)  NO (g) + CO 2 (g)

70. 70 Slow Initial Step A proposed mechanism for this reaction is Step 1: NO 2 + NO 2  NO 3 + NO (slow) Step 2: NO 3 + CO  NO 2 + CO 2 (fast) The NO 3 intermediate is consumed in the second step. As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.

71. 71 Fast Initial Step The rate law for this reaction is found to be Rate = k [NO] 2 [Br 2 ] Because termolecular processes are rare, this rate law suggests a two-step mechanism. 2 NO (g) + Br 2 (g)  2 NOBr (g)

72. 72 Fast Initial Step A proposed mechanism is Step 2: NOBr 2 + NO  2 NOBr (slow) Step 1 includes the forward and reverse reactions. Step 1: NO + Br 2 NOBr 2 (fast)

73. 73 Fast Initial Step The rate of the overall reaction depends upon the rate of the slow step. Step 2: NOBr2 + NO  2 NOBr (slow) The rate law for that step would be Rate = k 2 [NOBr 2 ] [NO] But how can we find [NOBr 2 ]?

74. 74 Fast Initial Step Step 2: NOBr 2 + NO  2 NOBr (slow) NOBr 2 can react two ways: –With NO to form NOBr –By decomposition to reform NO and Br 2 The reactants and products of the first step are in equilibrium with each other (fast step). Therefore, Rate f = Rate r Step 1: NO + Br 2 NOBr 2 (fast)

75. 75 Fast Initial Step Because Rate f = Rate r, k 1 [NO] [Br 2 ] = k −1 [NOBr 2 ] Solving for [NOBr 2 ] gives us k1k−1k1k−1 [NO] [Br 2 ] = [NOBr 2 ]

76. 76 Fast Initial Step Substituting this expression for [NOBr 2 ] in the rate law for the rate-determining step gives k2k1k−1k2k1k−1 Rate =[NO] [Br 2 ] [NO] = k [NO] 2 [Br 2 ]

77. 77 Road Map for Reaction Mechanisms

78. 78 Catalysis

79. 79 Catalysis A catalyst is a substance that changes the rate of a chemical reaction without itself undergoing a permanent chemical change in the process. There are two types of catalyst: Heterogeneous--one that is present in a different phase as the reacting molecules. Homogeneous-- one that is present in the same phase as the reacting molecules.

80. 80 The gases exhausted from an automobile engine pass through a catalytic converter where air pollutants such as unburned hydrocarbons (CxHy), CO, and NO are converted to CO2, H2O, N2, and O2. The photo shows a cutaway view of a catalytic converter. The beads are impregnated with the heterogeneous catalyst.

81. 81 Catalysis: Homogenous Catalyst Example: Hydrogen peroxide decomposes very slowly in the absence of a catalyst: 2H 2 O 2(aq)  2H 2 O (l) + O 2(g) In the presence of bromide ions, the decomposition occurs rapidly in an acidic environment: 2Br – (aq) + H 2 O 2(aq) + 2H + (aq)  Br 2(aq) + 2H 2 O (l) Br 2(aq) + H 2 O 2(aq)  2Br – (aq) + 2H + (aq) + O 2(g) Br – is a homogeneous catalyst because it is regenerated at the end of the reaction. The net reaction is still…2H 2 O 2(aq)  2H 2 O (l) + O 2(g)

82. 82 Catalysts and Reaction Rates

83. 83 Catalysts and Reaction Rates Catalysts operate by lowering the overall activation energy, E a, for a reaction…(It lowers the “hill”.) However, catalysts can operate by increasing the number of effective collisions. A catalyst usually provides a completely different mechanism for the reaction. Intermediates are being generated (see decomposition of H 2 O 2, Br 2 is the intermediate. When a catalyst adds an intermediate, the activation energies for both steps must be lower than the activation energy for the uncatalyzed reaction.

84. 84 Heterogeneous Catalysts Solid catalyst is a heterogeneous catalyst (catalytic converter in cars) Many industrial catalysts are heterogeneous. Mechanism of operation: The first step is adsorption (the binding of reactant molecules to the catalyst surface). Adsorption occurs due to the high reactivity of atoms or ions on the surface of the solid. Molecules are adsorbed onto active sites on the catalyst surface. The number of active sites on a given amount of catalyst depends on several factors such as: - The nature of the catalyst. - How the catalyst was prepared. - How the catalyst was treated prior to use.

85. 85 Heterogeneous Catalysis One way a catalyst can speed up a reaction is by holding the reactants together (adsorption has to occur) and helping bonds to break. Less energy is required to break the bonds. Example: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)

86. 86 Enzyme Catalysts Enzymes are biological catalysts. There may be as many as 30,000 enzymes in the human body. (Ex: Lactase) Most enzymes are protein molecules with large molecular masses (10,000 to 10 6 amu). Enzymes have very specific shapes. Most enzymes catalyze very specific reactions. The substances that undergo reaction at the active site on enzymes are called substrates. A substrate locks into an enzyme and a fast reaction occurs. The products then move away from the enzyme.

87. 87 Enzymes Enzymes are catalysts in biological systems. The substrate fits into the active site of the enzyme much like a key fits into a lock.

88. 88