1. Chemical Kinetics  2009, Prentice-Hall, Inc. Chapter 14 Chemical Kinetics John D. Bookstaver St. Charles Community College Cottleville, MO Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

2. Chemical Kinetics  2009, Prentice-Hall, Inc. Kinetics In kinetics we study the rate at which a chemical process occurs. Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs).

3. Chemical Kinetics  2009, Prentice-Hall, Inc. Factors That Affect Reaction Rates Physical State of the Reactants –In order to react, molecules must come in contact with each other. –The more homogeneous the mixture of reactants, the faster the molecules can react.

4. Chemical Kinetics  2009, Prentice-Hall, Inc. Factors That Affect Reaction Rates Concentration of Reactants –As the concentration of reactants increases, so does the likelihood that reactant molecules will collide.

5. Chemical Kinetics  2009, Prentice-Hall, Inc. Factors That Affect Reaction Rates Temperature –At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy.

6. Chemical Kinetics  2009, Prentice-Hall, Inc. Factors That Affect Reaction Rates Presence of a Catalyst –Catalysts speed up reactions by changing the mechanism of the reaction. –Catalysts are not consumed during the course of the reaction.

7. Chemical Kinetics  2009, Prentice-Hall, Inc. Reaction Rates Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.

8. Chemical Kinetics  2009, Prentice-Hall, Inc. 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)

9. Chemical Kinetics  2009, Prentice-Hall, Inc. Reaction Rates The average rate of the reaction over each interval is the change in concentration divided by the change in time: C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) Average rate =  [C 4 H 9 Cl]  t

10. Chemical Kinetics  2009, Prentice-Hall, Inc. Reaction Rates Note that the average rate decreases as the reaction proceeds. This is because 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)

11. Chemical Kinetics Suppose that at one point in the reaction, the concentration of A is 0.4658 M and 125s later the concentration has fallen to 0.4282M. During this time, what is the average rate of reaction?  2009, Prentice-Hall, Inc.

12. Chemical Kinetics  2009, Prentice-Hall, Inc. Reaction Rates A plot of [C 4 H 9 Cl] vs. time for this reaction yields a curve like this. 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)

13. Chemical Kinetics  2009, Prentice-Hall, Inc. Reaction Rates All reactions slow down over time. Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

14. Chemical Kinetics  2009, Prentice-Hall, Inc. Reaction Rates and Stoichiometry In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Thus, 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

15. Chemical Kinetics  2009, Prentice-Hall, Inc. Reaction Rates and Stoichiometry What if the ratio is not 1:1? 2 HI (g)  H 2 (g) + I 2 (g) In such a case, Rate = − 1212  [HI]  t =  [I 2 ]  t

16. Chemical Kinetics  2009, Prentice-Hall, Inc. Reaction Rates and Stoichiometry To generalize, then, for the reaction aA + bBcC + dD Rate = − 1a1a  [A]  t = − 1b1b  [B]  t = 1c1c  [C]  t 1d1d  [D]  t =

17. Chemical Kinetics  2009, Prentice-Hall, Inc. Concentration and Rate One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration.

18. Chemical Kinetics Given the reaction, 2A + B  2C + D At some point in the reaction, [C] = 0.2885M and 153s later, the concentration of C is 0.3546M. What is the average rate of reaction during this time period? What is the average rate of formation of D during this time?  2009, Prentice-Hall, Inc.

19. Chemical Kinetics Given the reaction, 2A + B  3C + D If –Δ[A]/Δt is 2.10 E-5 M/s, What is the rate of formation of C?  2009, Prentice-Hall, Inc.

20. Chemical Kinetics Given A + 2B  C + 3D If the rate of disappearance of B is -6.2E-4 M/s, what is the rate of reaction, the rate of disappearance of A and the rate of formation of D?  2009, Prentice-Hall, Inc.

21. Chemical Kinetics If the initial concentration of hydrogen peroxide is 0.2546M and the initial rate of reaction is 9.32 E- 4M/s. What will be the concentration of hydrogen peroxide remaining from its decomposition at t=35s. (reaction given for 1 mole of hydrogen peroxide decomposition)  2009, Prentice-Hall, Inc.

22. Chemical Kinetics Given the same reaction, the initial concentration of hydrogen peroxide is 0.1108M and 12s later the concentration is 0.1060M. What is the initial rate of this reaction in both M/s and M/min.  2009, Prentice-Hall, Inc.

23. Chemical Kinetics  2009, Prentice-Hall, Inc. Concentration and Rate If we compare Experiments 1 and 2, we see that when [NH 4 + ] doubles, the initial rate doubles. NH 4 + (aq) + NO 2 − (aq) N 2 (g) + 2 H 2 O (l)

24. Chemical Kinetics  2009, Prentice-Hall, Inc. Concentration and Rate Likewise, when we compare Experiments 5 and 6, we see that when [NO 2 − ] doubles, the initial rate doubles. NH 4 + (aq) + NO 2 − (aq) N 2 (g) + 2 H 2 O (l)

25. Chemical Kinetics  2009, Prentice-Hall, Inc. Concentration and Rate Likewise, when we compare Experiments 5 and 6, we see that when [NO 2 − ] doubles, the initial rate doubles. NH 4 + (aq) + NO 2 − (aq) N 2 (g) + 2 H 2 O (l)

26. Chemical Kinetics  2009, Prentice-Hall, Inc. Concentration and Rate This means Rate  [NH 4 + ] Rate  [NO 2 − ] Rate  [NH + ] [NO 2 − ] which, when written as an equation, becomes Rate = k [NH 4 + ] [NO 2 − ] This equation is called the rate law, and k is the rate constant. Therefore,

27. Chemical Kinetics  2009, Prentice-Hall, Inc. 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. Since the rate law is Rate = k [NH 4 + ] [NO 2 − ] the reaction is First-order in [NH 4 + ] and First-order in [NO 2 − ].

28. Chemical Kinetics  2009, Prentice-Hall, Inc. Rate Laws Rate = k [NH 4 + ] [NO 2 − ] The overall reaction order can be found by adding the exponents on the reactants in the rate law. This reaction is second-order overall.

29. Chemical Kinetics aA + bB  cC + dD The rate law: Rate = k [A] m [B] n k = rate constant m = experimentally determined #, the value shows the order of A n = experimentally determined #, the value shows the order of B m + n = overall order of the reaction  2009, Prentice-Hall, Inc.

30. Chemical Kinetics Rate = k[A] o [B] 1 m = 0; rxn is 0 th order in A n = 1; rxn is 1 st order in B m + n = 0 + 1; overall rxn is 1 st order Since anything [ ] 0 = 1  simplify Rate = k[B] 1 = k[B]  2009, Prentice-Hall, Inc.

31. Chemical Kinetics H 2 SeO 3 + 6I - + 4H +  Se + 2I 3 - + 3H 2 O Given the rate = k[H 2 SeO 3 ] [I 3 - ] 3 [H + ] 2  2009, Prentice-Hall, Inc.

32. Chemical Kinetics Units of k Rxn orderUnits of k 0 th order 1 st order 2 nd order 3 rd order n th order  2009, Prentice-Hall, Inc.

33. Chemical Kinetics A + B  C Given rate = k[A] 0 [B] What does this mean?  2009, Prentice-Hall, Inc.

34. Chemical Kinetics A + B  C Given rate = k[A] 0 [B] 2 What does this mean? –Double A –Double B –Half B –Triple B  2009, Prentice-Hall, Inc.

35. Chemical Kinetics  2009, Prentice-Hall, Inc.

36. Chemical Kinetics A  B We find that when [A] has fallen to half its initial value, the reaction proceeds at the same rate. What order is the reaction?  2009, Prentice-Hall, Inc.

37. Chemical Kinetics H 2 O 2  H 2 O + ½ O 2 For the decomposition of 1 mole of hydrogen peroxide, use k = 3.66 E-3 1/s to determine the instantaneous rate of the 1 st order decomposition of 2.05M H 2 O 2 ?  2009, Prentice-Hall, Inc.

38. Chemical Kinetics Experiment[NO][Cl 2 ] Initial Rate (M/s) 10.01250.02552.27E-5 20.01250.05104.55E-5 30.02500.02559.08E-5 Determine the reaction order through experimental data 2NO + Cl 2  2NOCl By comparison write the rate laws (differential rate law)

39. Chemical Kinetics Experiment[NO][Cl 2 ] Initial Rate (M/s) 10.01250.02552.27E-5 20.01250.05104.55E-5 30.02500.02559.08E-5 Determine the reaction order through experimental data 2NO + Cl 2  2NOCl By comparison write the rate laws (differential rate law) Suppose there is a hypothetical experiment 4, which has [NO]=0.0500M and [Cl 2 ]=0.0255M what is the rate in this experiment?

40. Chemical Kinetics Determine the reaction order through experimental data S 2 O 8 -2 + 3I -  2SO 4 -2 + I 3 - By comparison write the rate laws. What is the order for each reactant and overall? Solve for the rate constant. What would be the initial rate of reaction if [S 2 O 8 -2 ] = 0.083M and [I - ] =0.115M Experiment[S 2 O 8 -2 ][l - ] Initial Rate (M/s) 10.0380.0601.4E-5 20.0760.0602.8E-5 30.0760.1205.6E-5  2009, Prentice-Hall, Inc.

41. Chemical Kinetics  2009, Prentice-Hall, Inc.

42. Chemical Kinetics Concentration and Reaction Rate  2009, Prentice-Hall, Inc.

43. Chemical Kinetics The rate expression will always have the form Initial rate = k [A] m [B] n [C] p k = rate constant [A] = conc. of reactant A [B] = conc. of reactant B [C] = conc. of catalyst (rare on AP/IB) m = order of rxn for reactant A n = order of rxn for reactant B p = order of rxn for catalyst  2009, Prentice-Hall, Inc.

44. Chemical Kinetics Initial rate = k [A] m [B] n [C] p Exponents can be zero, whole numbers, or fractions (rare) Must be determined by experimentation The rate constant, k –Temperature dependent –Determined experimentally  2009, Prentice-Hall, Inc.

45. Chemical Kinetics  2009, Prentice-Hall, Inc. 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.

46. Chemical Kinetics Zeroth order half-life  2009, Prentice-Hall, Inc.

47. Chemical Kinetics Zeroth order half-life [A] t = -kt +[A] o (not in calculus – trust me)  2009, Prentice-Hall, Inc.

48. Chemical Kinetics Zeroth order half-life [A] t = -kt +[A] o This looks like the equation for what??? - line Graph [A] vs. t will produce a straight line, with slope = -k and y-intercept of [A] 0  2009, Prentice-Hall, Inc.

49. Chemical Kinetics  2009, Prentice-Hall, Inc. First-Order Processes So how do we straighten the line???

50. Chemical Kinetics First order half-life  2009, Prentice-Hall, Inc.

51. Chemical Kinetics  2009, Prentice-Hall, Inc. First-Order Processes When ln P is plotted as a function of time, a straight line results. Therefore, –The process is first-order. –k is the negative of the slope: 5.1  10 -5 s −1.

52. Chemical Kinetics  2009, Prentice-Hall, Inc. First-Order Processes ln [A] t = -kt + ln [A] 0

53. Chemical Kinetics In the first order decomposition of hydrogen peroxide, k = 3.66E-3 1/s and [H 2 O 2 ] 0 = 0.882M –Determine the time at which [H 2 O 2 ] = 0.600M –Determine the [H 2 O 2 ] after 225s  2009, Prentice-Hall, Inc.

54. Chemical Kinetics The decomposition of NH 2 NO 2 has a rate law of rate = k[NH 2 NO 2 ]. k = 5.62E-3 1/min at 15 o C and [NH 2 NO 2 ] o = 0.105 –At what time is [NH 2 NO 2 ] = 0.0250M –What is [NH 2 NO 2 ] after 6.00hrs –What is the rate of reaction after 35 mins, if the [NH 2 NO 2 ] 0 =0.0750M  2009, Prentice-Hall, Inc.

55. Chemical Kinetics  2009, Prentice-Hall, Inc. 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.

56. Chemical Kinetics  2009, Prentice-Hall, Inc. 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, then, the half-life does not depend on [A] 0.

57. Chemical Kinetics What is the rate constant of the first order decomposition of N 2 O 5 if the t 1/2 = 120s  2009, Prentice-Hall, Inc.

58. Chemical Kinetics How long will it take for an initial [N 2 O 5 ] to reach 1/16 of its value? Find the mass of dinitrogen pentoxide remaining after a 4.80g sample has decomposed for 10mins. Determine the time for 90% of a N 2 O 5 sample to undergo decomposition.  2009, Prentice-Hall, Inc.

59. Chemical Kinetics The half-life of the first order decomposition of SO 2 Cl 2 at 320 o C is 8.75 hr. What is the value of the rate constant? What is the pressure of sulfuryl chloride 3.00 hr after the start of the reaction, if the initial pressure of sulfuryl chloride is 722mmHg?  2009, Prentice-Hall, Inc.

60. Chemical Kinetics  2009, Prentice-Hall, Inc. Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A, we get (the math is a little more complicated) 1 [A] t = kt + 1 [A] 0 also in the form y = mx + b

61. Chemical Kinetics  2009, Prentice-Hall, Inc. Second-Order Processes So if a process is second-order in A, a plot of vs. t will yield a straight line, and the slope of that line is k. 1 [A] t = kt + 1 [A] 0 1 [A]

62. Chemical Kinetics  2009, Prentice-Hall, Inc. Second-Order Processes The decomposition of NO 2 at 300°C is described by the equation NO 2 (g) NO (g) + 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 1212

63. Chemical Kinetics  2009, Prentice-Hall, Inc. Second-Order Processes Plotting ln [NO 2 ] vs. t yields the graph at the right. 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].

64. Chemical Kinetics  2009, Prentice-Hall, Inc. Second-Order Processes Graphing ln 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]. 1 [NO 2 ]

65. Chemical Kinetics  2009, Prentice-Hall, Inc. 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

66. Chemical Kinetics Find the rate constant, and the half-life of the decomposition of 1.00M NO 2  2009, Prentice-Hall, Inc.

67. Chemical Kinetics (2 nd order) A  B, takes 55s for the [A] to fall to 0.40 from [A] o = 0.80M What is the k for this reaction? How long would it take for [A] to fall to 0.20M? to 0.10M?  2009, Prentice-Hall, Inc.

68. Chemical Kinetics In a process where Cl atoms combine to form Cl 2­, the reaction follows a second order rate law. The first half-life is 12ms. How long will it take for the [Cl] to drop to 1/8 of its initial value?  2009, Prentice-Hall, Inc.

69. Chemical Kinetics Integrated Rate Laws Use graphical methods to determine the order of a given reactant. The value of the rate constant k is equal to the absolute value of the slope of the best fit line which was decided by performing 3 linear regressions and analyzing the regression correlation coefficient r.  2009, Prentice-Hall, Inc.

70. Chemical Kinetics Integrated Rate Law: [ ] / time Create 3 graphs –Time always goes on the x-axis –Graph 1: [A] on y-axis –Graph 2: ln[A] on y-axis –Graph 3: 1/[A] on y-axis [A] -1 –Alphabetic order: Concentration = 0 order Natural log = 1 st order Reciprocal = 2 nd order  2009, Prentice-Hall, Inc.

71. Chemical Kinetics Integrated Rate Law: [ ] / time –Alphabetic order: Concentration = 0 order Natural log = 1 st order Reciprocal = 2 nd order –Choose the graph that forms a straight line. (r value closest to ±1) –The equation formed by line allows you to predict either [ ] at a time or a time for a specific [ ].  2009, Prentice-Hall, Inc.

72. Chemical Kinetics Integrated Rate Law: [ ] / time y = mx + b Zero order: [A] = -kt + [A] o First order: ln[A] = -kt + ln[A o ] Second order1/[A] = kt + 1/[A o ] *** |slope| = k *** Rate expression =Rate law Rate k[A] order determined from graph  2009, Prentice-Hall, Inc.

73. Chemical Kinetics  2009, Prentice-Hall, Inc.

74. Chemical Kinetics [A]Time(min) 0.800 0.608 0.3524 0.2040  2009, Prentice-Hall, Inc. Determine the order of this reaction using the given experimental data.

75. Chemical Kinetics  2009, Prentice-Hall, Inc. Temperature and Rate Generally, as temperature increases, so does the reaction rate. This is because k is temperature dependent.

76. Chemical Kinetics  2009, Prentice-Hall, Inc. The Collision Model In a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they collide with each other.

77. Chemical Kinetics  2009, Prentice-Hall, Inc. The Collision Model Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.

78. Chemical Kinetics  2009, Prentice-Hall, Inc. Activation Energy In other words, there is a minimum amount of energy required for reaction: the activation energy, E a. 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.

79. Chemical Kinetics http://phet.colorado.edu/en/simulation/re actions-and-rateshttp://phet.colorado.edu/en/simulation/re actions-and-rates  2009, Prentice-Hall, Inc.

80. Chemical Kinetics  2009, Prentice-Hall, Inc. Reaction Coordinate Diagrams It is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile.

81. Chemical Kinetics  2009, Prentice-Hall, Inc. Reaction Coordinate Diagrams The diagram shows the energy of the reactants and products (and, therefore,  E). The high point on the diagram is the transition state. The species present at the transition state is called the activated complex. The energy gap between the reactants and the activated complex is the activation energy barrier.

82. Chemical Kinetics  2009, Prentice-Hall, Inc. 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.

83. Chemical Kinetics  2009, Prentice-Hall, Inc. Maxwell–Boltzmann Distributions As the temperature increases, the curve flattens and broadens. Thus at higher temperatures, a larger population of molecules has higher energy.

84. Chemical Kinetics  2009, Prentice-Hall, Inc. Maxwell–Boltzmann Distributions If the dotted line represents the activation energy, then as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. As a result, the reaction rate increases.

85. Chemical Kinetics  2009, Prentice-Hall, Inc. 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

86. Chemical Kinetics  2009, Prentice-Hall, Inc. Arrhenius Equation Svante Arrhenius developed a mathematical relationship between k and E a : k = A e where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction. - E a RT

87. Chemical Kinetics  2009, Prentice-Hall, Inc. Arrhenius Equation Taking the natural logarithm of both sides, the equation becomes ln k = - ( ) + ln A 1T1T y = m x + b Therefore, if k is determined experimentally at several temperatures, E a can be calculated from the slope of a plot of ln k vs.. EaREaR 1T1T

88. Chemical Kinetics Derivation of Arrhenius  2009, Prentice-Hall, Inc.

89. Chemical Kinetics Rate constants for the first order decomposition of acetone dicarboxylic acid are k = 4.75E-4 1/s at 293K and k = 1.63E-3 1/s at 303K. What is the activation energy?  2009, Prentice-Hall, Inc.

90. Chemical Kinetics The decomposition of di-tert-butyl peroxide is a first order reaction with a half-life of 320 minutes at 135 o C and 100 minutes at 145 o C. Calculate the activation energy of this reaction.  2009, Prentice-Hall, Inc.

91. Chemical Kinetics Decomposition of ethylene oxide at 625K is first order with k=0.012min -1 and an E a of 218 kJ/mol. Find the rate constant at 525K. Find the temperature at which the rate constant k=0.0100min -1  2009, Prentice-Hall, Inc.

92. Chemical Kinetics A reaction has a first order rate constant of 4.82E-5 1/s at 25 o C and 1.41E-2 1/s at 70.0 o C. Calculate the rate constant at 90.0 o C.  2009, Prentice-Hall, Inc.

93. Chemical Kinetics  2009, Prentice-Hall, Inc. Reaction Mechanisms The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism.

94. Chemical Kinetics  2009, Prentice-Hall, Inc. Reaction Mechanisms Reactions may occur all at once or through several discrete steps. Each of these processes is known as an elementary reaction or elementary process.

95. Chemical Kinetics  2009, Prentice-Hall, Inc. Reaction Mechanisms The molecularity of a process tells how many molecules are involved in the process.

96. Chemical Kinetics  2009, Prentice-Hall, Inc. 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.

97. Chemical Kinetics  2009, Prentice-Hall, Inc. 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)

98. Chemical Kinetics  2009, Prentice-Hall, Inc. 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.

99. Chemical Kinetics  2009, Prentice-Hall, Inc. 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)

100. Chemical Kinetics  2009, Prentice-Hall, Inc. 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)

101. Chemical Kinetics  2009, Prentice-Hall, Inc. Fast Initial Step The rate of the overall reaction depends upon the rate of the slow step. The rate law for that step would be Rate = k 2 [NOBr 2 ] [NO] But how can we find [NOBr 2 ]?

102. Chemical Kinetics  2009, Prentice-Hall, Inc. Fast Initial Step 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. Therefore, Rate f = Rate r

103. Chemical Kinetics  2009, Prentice-Hall, Inc. 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 ]

104. Chemical Kinetics  2009, Prentice-Hall, Inc. 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 ]

105. Chemical Kinetics  2009, Prentice-Hall, Inc. Catalysts Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. Catalysts change the mechanism by which the process occurs.

106. Chemical Kinetics  2009, Prentice-Hall, Inc. Catalysts One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.

107. Chemical Kinetics  2009, Prentice-Hall, Inc. 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.