1. 1 CHAPTER 11 Rates of Reactions

2. 2 Chemical Kinetics Kinetics is the study of rates of chemical reactions and the mechanisms by which they occur. Reaction rate –increase in concentration of a product per unit time or –decrease in concentration of a reactant per unit time Reaction mechanism –series of steps by which a reaction occurs

3. 3 Chemical Kinetics Thermodynamics tells us if a reaction can occur Kinetics tells us how quickly the reaction occurs –some reactions that are thermodynamically feasible are kinetically so slow as to be imperceptible

4. 4 The Rate of Reaction Consider the hypothetical reaction, A(g) + B(g)  C(g) + D(g) equimolar amounts of reactants, A & B, will be consumed and products, C & D, will be formed as indicated in this graph:

5. 5 [A] = concentration of A in M ( mol/L) Note that rxn does not go to completion

6. 6 Reaction rates –rates at which reactants disappear or products appear The Rate of Reaction

7. 7 Rate of a simple one-step reaction is directly proportional to the concentration of the reacting substance [X] is concentration of X in molarity or moles/L k = specific rate constant

8. 8 The Rate of Reaction For a simple expression like R = k[A] doubling the initial concentration of A doubles the initial rate of reaction halving the initial concentration of A halves the initial rate of reaction

9. 9 The Rate of Reaction Rate Law Expressions must be determined experimentally –cannot be determined from balanced equations –trap for new students of kinetics –most chemical reactions are not one-step reactions Rate law expressions are also called: –rate laws –rate equations –rate expressions

10. 10 The Rate of Reaction Order of a reaction –expressed in terms of either: 1each reactant 2overall reaction For example:

11. 11 The Rate of Reaction

12. 12 The Rate of Reaction

13. 13 The Rate of Reaction Look at the following one step reaction and its experimentally determined rate-law expression

14. 14 The Rate of Reaction Look at the following one step reaction and its experimentally determined rate-law expression because it is a second order rate-law expression –doubling the [A] increases the rate of reaction by a factor of 4 2 2 = 4 – halving the [A] decreases the rate of reaction by a factor of 4 (1/2) 2 = 1/4

15. 15 Factors That Affect Reaction Rates Several factors that can influence the rate of a reaction are:  nature of the reactants  concentrations of reactants  temperature  presence of a catalyst  we will look at each factor individually

16. 16 Nature of Reactants Broad category that includes the different reacting properties of substances. For example: 1Sodium reacts with water explosively at room temperature to liberate hydrogen and form sodium hydroxide.

17. 17 Nature of Reactants 2Calcium reacts with water only slowly at room temperature to liberate hydrogen and form calcium hydroxide.

18. 18 Nature of Reactants 3The reaction of magnesium with water at room temperature is so slow that that the evolution of hydrogen is not perceptible to the human eye.

19. 19 Nature of Reactants However, Mg reacts with steam rapidly to liberate H 2 and form magnesium oxide. Differences due to “nature of the reactants”

20. 20 Concentrations of Reactants Simplified representation of effect of different numbers of molecules in the same volume. –Increase in concentration A(g) + B(g)  Products A B B A B 4 different possible A-B collisions 6 different possible A-B collisions 9 different possible A-B collisions

21. 21 Concentrations of Reactants Example: The following rate data were obtained at 25 o C for the following reaction. What are the rate-law expression and the specific rate-constant for this reaction? 2 A(g) + B(g)  3 C(g)

22. 22 Concentrations of Reactants

23. 23 Concentrations of Reactants

24. 24 Concentrations of Reactants

25. 25 Concentration vs. Time: The Integrated Rate Equation Integrated rate equation relates time and concentration for a chemical or nuclear reaction First Order Reactions 1st order in reactant A & 1st order overall For example: a A  products common for many chemical reactions and all simple radioactive decays

26. 26 Concentration vs. Time: The Integrated Rate Equation where: [A] 0 = mol/L of A at time t=0. [A] = mol/L of A at time t. k = specific rate constant t = time elapsed since beginning of reaction a = stoichiometric coefficient of A in balanced overall equation

27. 27 Concentration vs. Time: The Integrated Rate Equation Solve the first order integrated rate equation for t.

28. 28 Concentration vs. Time: The Integrated Rate Equation Solve the first order integrated rate equation for t. Define the half-life, t 1/2, of a reactant as the time required for half of the reactant to be consumed, or the time at which [A]=1/2[A] 0.

29. 29 Concentration vs. Time: The Integrated Rate Equation At time t= t 1/2, the expression becomes

30. 30 Concentration vs. Time: The Integrated Rate Equation Example: Cyclopropane, an anesthetic, decomposes to propene according to the following equation. The reaction is first order in cyclopropane with k = 9.2 s -1 at 1000 0 C. Calculate the half life of cyclopropane at 1000 0 C.

31. 31 Concentration vs. Time: The Integrated Rate Equation Example: Cyclopropane, an anesthetic, decomposes to propene according to the following equation. The reaction is first order in cyclopropane with k = 9.2 s -1 at 1000 0 C. Calculate the half life of cyclopropane at 1000 0 C.

32. 32 Concentration vs. Time: The Integrated Rate Equation Example: The half-life for the following first order reaction is 688 hours at 1000 0 C. Calculate the specific rate constant, k, at 1000 0 C and the amount of a 3.0 g sample of CS 2 that remains after 48 hours. CS 2 (g)  CS(g) + S(g)

33. 33 Concentration vs. Time: The Integrated Rate Equation Example: The half-life for the following first order reaction is 688 hours at 1000 0 C. Calculate the specific rate constant, k, at 1000 0 C and the amount of a 3.0 g sample of CS 2 that remains after 48 hours. CS 2 (g)  CS(g) + S(g)

34. 34 Concentration vs. Time: The Integrated Rate Equation

35. 35 Concentration vs. Time: The Integrated Rate Equation

36. 36 Concentration vs. Time: The Integrated Rate Equation For reactions that are second order with respect to a particular reactant and second order overall, the rate equation is

37. 37 Concentration vs. Time: The Integrated Rate Equation At t 1/2 [A] = 1/2[A] 0, so

38. 38 Concentration vs. Time: The Integrated Rate Equation Use the common denominator to derive:

39. 39 Concentration vs. Time: The Integrated Rate Equation Solve for t 1/2 : Note that the half-life of a second order reaction depends on the initial concentration of A.

40. 40 Concentration vs. Time: The Integrated Rate Equation Example: Acetaldehyde, CH 3 CHO, undergoes gas phase thermal decomposition to methane and carbon monoxide. The rate-law expression is Rate = k[CH 3 CHO] 2, and k= 2.0 x 10 -2 L/(mol. hr) at 527 0 C. (a) What is the half-life of CH 3 CHO if 0.10 mole is injected into a 1.0 L vessel at 527 0 C?

41. 41 Concentration vs. Time: The Integrated Rate Equation

42. 42 Concentration vs. Time: The Integrated Rate Equation (b) How many moles of CH 3 CHO remain after 200 hours?

43. 43 Concentration vs. Time: The Integrated Rate Equation (b) How many moles of CH 3 CHO remain after 200 hours?

44. 44 Concentration vs. Time: The Integrated Rate Equation (b) How many moles of CH 3 CHO remain after 200 hours?

45. 45 Concentration vs. Time: The Integrated Rate Equation (c) What percent of the CH 3 CHO remains after 200 hours?

46. 46 Concentration vs. Time: The Integrated Rate Equation (c) What percent of the CH 3 CHO remains after 200 hours?

47. 47 Rate equations Table 11.2 is a review of the Rate Equations!

48. 48 Enrichment - Derivation of Integrated Rate Equations For the first order reaction a A  products the rate can be written as:

49. 49 Enrichment - Derivation of Integrated Rate Equations We can use calculus to rearrange the rate equation and get the integrated rate equation

50. 50 Enrichment - Derivation of Integrated Rate Equations The rate equation for a reaction that is second order in reactant A and second order overall. The rate equation is:

51. 51 Enrichment - Derivation of Integrated Rate Equations the second order integrated rate equation

52. 52 Enrichment - Derivation of Integrated Rate Equations For a zero order reaction the rate expression is:

53. 53 Enrichment - Derivation of Integrated Rate Equations Which gives this relationship: the zeroeth order integrated rate equation

54. 54 Enrichment -Rate Equations to Determine Reaction Order Plots of the integrated rate equations can help us determine the order of a reaction. Rearrange the first-order integrated rate equation –laws of logarithms give ln (x/y) = ln x - ln y

55. 55 Enrichment -Rate Equations to Determine Reaction Order The equation for a straight line is: Compare this equation to the rearranged rate-law

56. 56 Enrichment -Rate Equations to Determine Reaction Order Now we can interpret the parts of the equation as follows: –y = ln[A]plot on y-axis –m= -akslope of line –x =tplot on x-axis –b =ln[A] 0 y-intercept

57. 57 Enrichment -Rate Equations to Determine Reaction Order Example: Concentration-versus-time data for the thermal decomposition of ethyl bromide are given in the table below. Use the graphs to determine the rate of the reaction and the value of the rate constant

58. 58 Enrichment -Rate Equations to Determine Reaction Order

59. 59 Enrichment -Rate Equations to Determine Reaction Order Make three different graphs 1[C 2 H 5 Br] vs. t – if linear then reaction is zero order 2ln [C 2 H 5 Br] vs time –if linear then reaction is first order 31/ [C 2 H 5 Br] vs. time –if linear then reaction is second order

60. 60 Enrichment -Rate Equations to Determine Reaction Order Plot of [C 2 H 5 Br] versus time

61. 61 Enrichment -Rate Equations to Determine Reaction Order Plot of ln [C 2 H 5 Br] versus time

62. 62 Enrichment -Rate Equations to Determine Reaction Order Plot of 1/[C 2 H 5 Br] versus time

63. 63 Enrichment -Rate Equations to Determine Reaction Order Note that the only graph which is linear is the plot of ln[C 2 H 5 Br] vs. time. Thus this is a First Order Reaction. Determine the value of the rate constant from the slope of the line on the graph of ln[C 2 H 5 Br] vs. time

64. 64 Enrichment -Rate Equations to Determine Reaction Order Note that the only graph which is linear is the plot of ln[C 2 H 5 Br] vs. time. Thus this is a First Order Reaction. Determine the value of the rate constant from the slope of the line on the graph of ln[C 2 H 5 Br] vs. time

65. 65 Enrichment -Rate Equations to Determine Reaction Order From the equation for a first order reaction we know that: slope = -a k In this reaction a = 1

66. 66 Enrichment -Rate Equations to Determine Reaction Order The integrated rate equation for a reaction that is second order in reactant A and second order overall. This equation rearranges to:

67. 67 Enrichment -Rate Equations to Determine Reaction Order Compare this equation with an equation for a straight line

68. 68 Enrichment -Rate Equations to Determine Reaction Order Now we can interpret the parts of the equation as follows: –y = 1/[A]plot on y-axis –m= a kslope of line –x =tplot on x-axis –b =1/[A] 0 y-intercept

69. 69 Enrichment -Rate Equations to Determine Reaction Order

70. 70 Enrichment -Rate Equations to Determine Reaction Order Note that the only graph which is linear is the plot of 1/[NO 2 ] vs. time. Thus this is a Second Order Reaction. Determine the value of the rate constant from the slope of the line on the graph of 1/[NO 2 ] vs. time

71. 71 Enrichment -Rate Equations to Determine Reaction Order Note that the only graph which is linear is the plot of 1/[NO 2 ] vs. time. Thus this is a Second Order Reaction. You can now determine the value of the rate constant from the slope of the line on the graph Solve in the same way as the previous example

72. 72 Collision Theory of Reaction Rates Three basic events must occur for a reaction to occur the atoms, molecules or ions must: 1collide 2collide with enough energy to break and form bonds 3collide with the proper orientation

73. 73 Collision Theory of Reaction Rates A method to increase the number of collisions and the energy necessary to break and reform bonds is to heat the molecules. Look at the reaction of methane and oxygen: Must start reaction with a match. –Provides the initial energy necessary to break the first few bonds. –Afterwards reaction is self-sustaining.

74. 74 Collision Theory of Reaction Rates Illustrate the proper orientation of molecules that is necessary for reaction. X 2(g) + Y 2(g)  2 XY (g) Some ineffective possible collisions are : XXXX Y YYYY X Y

75. 75 Collision Theory of Reaction Rates An example of an effective collision is: X Y + X Y

76. 76 Transition State Theory Theory postulates that reactants form a high energy intermediate, the transition state, which then falls apart into the products. For a reaction to occur, the reactants must acquire sufficient energy to form the transition state. –Activation energy or E a Mechanical analogy for activation energy

77. 77 Transition State Theory Cross section of mountain Boulder E activation hh h2h2 h1h1 E pot =mgh 2 E pot =mgh 1  E pot = mg  h Height

78. 78 Transition State Theory Potential Energy Reaction Coordinate X 2 + Y 2 2 XY E activation - a kinetic quantity  E  H a thermodynamic quantity

79. 79 Transition State Theory The relationship between the activation energy for forward and reverse reactions is –Forward reaction = E a –Reverse reaction = E a +  E –difference =  E

80. 80 Transition State Theory The distribution of molecules possessing different energies at a given temperature may be represented as

81. 81 Reaction Mechanisms & the Rate-Law Expression Use experimental rate-law to postulate a mechanism. The slowest step in a reaction mechanism is the rate determining step. Elementary step: any process that occurs in a single step. Elementary steps must add to give the balanced chemical equation. Intermediate: a species which appears in an elementary step which is not a reactant or product.

82. 82 Reaction Mechanisms & the Rate-Law Expression Molecularity: the number of molecules present in an elementary step. –Unimolecular: one molecule in the elementary step, –Bimolecular: two molecules in the elementary step, and –Termolecular: three molecules in the elementary step. It is not common to see termolecular processes (statistically improbable).

83. 83 Reaction Mechanisms & the Rate-Law Expression Consider the iodide ion catalyzed decomposition of hydrogen peroxide to water and oxygen.

84. 84 Reaction Mechanisms & the Rate-Law Expression Reaction is known to be first order in H 2 O 2, first order in I -, and second order overall. Mechanism is thought to be:

85. 85 Reaction Mechanisms & the Rate-Law Expression Important notes:  one hydrogen peroxide molecule and one iodide ion are involved in the rate determining step  the iodide ion catalyst is consumed in step 1 and produced in step 2 in equal amounts  hypoiodite ion has been detected in reaction mixture as a short-lived reaction intermediate

86. 86 Reaction Mechanisms & the Rate-Law Expression Ozone, O 3, reacts very rapidly with nitrogen oxide, NO, in a reaction that is first order in each reactant and second order overall.

87. 87 Reaction Mechanisms & the Rate-Law Expression A possible mechanism is:

88. 88 Reaction Mechanisms & the Rate-Law Expression A mechanism that is inconsistent with the rate-law expression is:

89. 89 Reaction Mechanisms & the Rate-Law Expression Experimentally determined reaction orders indicate the number of molecules involved in: the slow step only or the slow step and the equilibrium steps preceding the slow step.

90. 90 Temperature: The Arrhenius Equation Svante Arrhenius developed the relationship among (1) temperature, (2) activation energy, and (3) the specific rate constant.

91. 91 Temperature: The Arrhenius Equation Illustrate the effect of temperature on a reaction, with all other variables in Arrhenius equation remaining constant.  Write the Arrhenius equations for two temperatures (T 2 >T 1 )

92. 92 Temperature: The Arrhenius Equation Illustrate the effect of temperature on a reaction, with all other variables in Arrhenius equation remaining constant.  Write the Arrhenius equations for two temperatures (T 2 >T 1 )

93. 93 Temperature: The Arrhenius Equation Subtract one equation from the other.

94. 94 Temperature: The Arrhenius Equation  Rearrange and solve for ln k 2 /k 1.

95. 95 Temperature: The Arrhenius Equation Consider the rate of a reaction for which E a =50 kJ/mol, at 20 0 C (293 K) and at 30 0 C (303 K).

96. 96 Temperature: The Arrhenius Equation For reactions that have an E a  50 kJ/mol, the rate approximately doubles for a 10 0 C rise in temperature, near room temperature. Consider: 2 ICl(g) + H 2 (g)  I 2 (g) + 2 HCl(g) The rate-law expression is known to be R=k[ICl][H 2 ]

97. 97 Temperature: The Arrhenius Equation For reactions that have an E a  50 kJ/mol, the rate approximately doubles for a 10 0 C rise in temperature, near room temperature. Consider: 2 ICl(g) + H 2 (g)  I 2 (g) + 2 HCl(g) The rate-law expression is known to be R=k[ICl][H 2 ]

98. 98 Catalysts Catalysts change reaction rates by providing an alternative reaction pathway with a different activation energy.

99. 99 Catalysts

100. 100 Catalysts Homogeneous catalysts exist in same phase as the reactants. Catalysts can operate by increasing the number of effective collisions. That is, from the Arrhenius equation: catalysts increase k be increasing A or decreasing E a. A catalyst may add intermediates to the reaction. When a catalyst adds an intermediate, the activation energies for both steps must be lower than the activation energy for the uncatalyzed reaction.

101. 101 Catalysts Heterogeneous catalysts exist in different phases than the reactants. Typical example: solid catalyst, gaseous reactants and products (catalytic converters in cars). Most industrial catalysts are heterogeneous. First step is adsorption (the binding of reactant molecules to the catalyst surface). Adsorbed species (atoms or ions) are very reactive. Molecules are adsorbed onto active sites on the catalyst surface.

102. 102 Catalysts Examples of catalysts include:

103. 103 Catalytic oxidation CO to CO 2 Overall reaction 2 CO (g) + O 2(g)  2CO 2(g) Absorption CO (g)  CO (surface) + O 2(g) O 2(g)  O 2(surface) Activation O 2(surface)  O (surface) Reaction CO (surface) +O (surface)  CO 2(surface) Desorption CO 2(surface)  CO 2(g)

104. 104 Catalysts Enzymes are biological catalysts. 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. Substrates undergo reaction at the active site of an enzyme.

105. 105 Catalysts

106. 106 Catalysts A substrate locks into an enzyme and a fast reaction occurs. The products then move away from the enzyme. Only substrates that fit into the enzyme lock can be involved in the reaction. If a molecule binds tightly to an enzyme so that another substrate cannot displace it, then the active site is blocked and the catalyst is inhibited (enzyme inhibitors).

107. 107 Catalysts Nitrogen gas cannot be used in the soil for plants or animals. Nitrogen compounds, NO 3, NO 2 -, and NO 3 - are used in the soil. The conversion between N 2 and NH 3 is a process with a high activation energy (the N  N triple bond needs to be broken).

108. 108 Catalysts

109. 109 Catalysts An enzyme, nitrogenase, in bacteria which live in root nodules of legumes, clover and alfalfa, catalyses the reduction of nitrogen to ammonia. The fixed nitrogen (NO 3, NO 2 -, and NO 3 - ) is consumed by plants and then eaten by animals. Animal waste and dead plants are attacked by bacteria that break down the fixed nitrogen and produce N 2 gas for the atmosphere.

110. 110 Synthesis Question The Chernobyl nuclear reactor accident occurred in 1986. At the time that the reactor exploded some 2.4 MCi of radioactive 137 Cs was released into the atmosphere. The half-life of 137 Cs is 30.1 years. In what year will the amount of 137 Cs released from Chernobyl finally decrease to 100 Ci? A Ci is a unit of radioactivity called the Curie.

111. 111 Synthesis Question

112. 112 Group Question 99m Tc has a half-life of 6.02 hours and is often used in nuclear medical diagnostic tests. Patients are injected with approximately 10  Ci of 99m Tc that is then directed to specific sites in the patient’s body to detect gallstones, brain tumors and function, and other medical conditions. How long will the patient have a higher than normal radioactivity level after they have been injected with 10  Ci of 99m Tc?