1. Chapter 13 Chemical Kinetics Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Copyright  2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e Why We Study Chemical Kinetics If the speed of a chemical reaction is too slow, the product will not be economical. If too fast, there is the potential for run-away reaction, explosion….

2. Defining Rate Rate = change per given time period speed is a rate = distance traveled, miles, per unit time, 1 hour (miles/hour) Chemical reaction rate = reactant’s concentration decrease per unit time product’s concentration increase per unit time For reactants, a negative sign is placed in front of the definition, since reactants are disappearing 2

3. at t = 0 [A] = 8 [B] = 8 [C] = 0 at t = 0 [X] = 8 [Y] = 8 [Z] = 0 at t = 16 [A] = 4 [B] = 4 [C] = 4 at t = 16 [X] = 7 [Y] = 7 [Z] = 1 3

4. at t = 16 [A] = 4 [B] = 4 [C] = 4 at t = 16 [X] = 7 [Y] = 7 [Z] = 1 at t = 32 [A] = 2 [B] = 2 [C] = 6 at t = 32 [X] = 6 [Y] = 6 [Z] = 2 4

5. at t = 32 [A] = 2 [B] = 2 [C] = 6 at t = 32 [X] = 6 [Y] = 6 [Z] = 2 at t = 48 [A] = 0 [B] = 0 [C] = 8 at t = 48 [X] = 5 [Y] = 5 [Z] = 3 5

6. Hypothetical Reaction Red  Blue In this reaction, one mole of Red turns into one mole of Blue The number of moles will always total 100 The rate of the reaction can be measured as the speed of loss of Red moles over time, or the speed of gain of Blue moles over time 6

7. Hypothetical Reaction Red  Blue 7

8. The Rate of Appearance of Blue vs. Time Red  Blue 8

9. Reaction Rate Changes Over Time Over time, the reaction rate generally slows down Due to reactants decreasing in concentration The reaction stops when the reactants run out OR when the system has reached equilibrium For many reactions, the coefficients of the balanced equation are not all the same For these reactions, the change in the number of molecules of one substance is a multiple of the change in the number of molecules of another 9

10. Reaction Rate and Stoichiometry For example: H 2 (g) + I 2 (g)  2 HI (g) For every 1 mole of H 2 used, 2 moles of HI are made The rate of change will depend on the coefficients, and will therefore be different To make rates be consistent, the concentration change of each substance is multiplied by 1 over their coefficient 10

11. 11 The Instantaneous Rate = the change in concentration at a given time = slope at one point of a curve Most accurate value found by taking the slope of a line tangent to the curve at that particular point first derivative of the function Average Rate = the change in concentration over a set time period Gives a linear approximation of a curve Less accurate than instantaneous rate

12. Hypothetical Reaction Red  Blue 12 Rate=  Blue  sec = 29-16 mole 10-5 sec = 2.6 mol/sec Rate=  Blue  sec = 29-0 mole 10-0 sec = 2.9 mol/sec

13. H2H2 I2I2 HI Stoichiometry tells us that for every 1 mole/L of H 2 used, 2 moles/L of HI are made. At 60 s, we used 0.699 moles of H 2. Therefore the amount of HI made is 2(0.699 moles) = 1.398 moles. The average rate is the change in the concentration in a given time period In the first 10 s, the  [H 2 ] is −0.181 M, so the rate is 13 Assuming a 1 L container, at 10 s, we used (1-0.819)mol =0.181 moles of H 2. Therefore the amount of HI made is 2(0.181 moles) = 0.362 moles.

14. average rate in a given time period =  slope of the line connecting the [H 2 ] points; and ½ +slope of the line for [HI] the average rate for the first 10 s is 0.0181 M/s the average rate for the first 40 s is 0.0150 M/s the average rate for the first 80 s is 0.0108 M/s 14

15. H 2 (g) + I 2 (g)  2 HI (g) Using [H 2 ], the instantaneous rate at 50 s is Using [HI], the instantaneous rate at 50 s is 15

16. Example 13.1: For the reaction given, the [I  ] changes from 1.000 M to 0.868 M in the first 10 s. Calculate the average rate in the first 10 s and the  [H + ]. H 2 O 2 (aq) + 3 I  (aq) + 2 H + (aq)  I 3  (aq) + 2 H 2 O (l) 16 solve the rate equation for the rate (in terms of the change in concentration of the given quantity) solve the rate equation (in terms of the change in the concentration for the quantity to find) for the unknown value

17. 2 mol NOBr:1 mol Br 2, 1 mol = 109.91g, 1 mL=0.001 L Solve: Conceptual Plan: Relationships: 240.0 g NOBr, 200.0 mL, 5.0 min.  [Br 2 ]/  t, M/s Given: Find: If 2.4 x 10 2 g of NOBr (MM 109.91 g) decomposes in a 2.0x10 2 mL flask in 5.0 minutes, find the average rate of Br 2 production in M/s 2 NOBr(g)  2 NO(g) + Br 2 (l) 17  M  mole Br 2 Rate  min  s } 5.45 M Br 2, 3.0 x 10 2 s  [Br 2 ]/  t, M/s 240.0 g NOBr, 0.2000 L, 5.0 min.  [Br 2 ]/  t, M/s

18. Measuring Reaction Rate Requires the experimenter to measure the concentration of at least one component in the reaction mixture at many points in time There are two ways of approaching this problem 1.for fast reactions that are complete in less than 1 hour, it is best to use continuous monitoring of the concentration 2.for slow reactions occurring over time, sampling of the mixture at various times can be used when sampling is used, often the reaction in the sample is stopped by a quenching technique 18

19. 19 Continuous Monitoring Polarimetry – measures the change in the degree of rotation of plane-polarized light caused by one of the components over time Spectrophotometry – measures the amount a particular wavelength of light absorbed by one component over time the component absorbs its complementary color Total pressure – the total pressure of a gas mixture is stoichiometrically related to partial pressures of the gases in the reaction

20. 20 Sampling the Reaction Mixture at Specific Times At specific times during the reaction, a sample from the reaction mixture (an aliquot) is drawn off and quantitatively analyzed By titration for one of the components By gravimetric analysis ( a set of methods in analytical chemistry for the quantitative determination) Gas chromatography can measure the concentrations of various components in a mixture for samples that have volatile components separates mixture by adherence to a surface

21. Factors Affecting Reaction Rate the kind of reactant molecules small molecules tend to react faster than large molecules Some chemicals are more reactive than others  e.g. potassium metal is more reactive than sodium ions react faster than molecules  no bonds need to be broken and the physical state they are in gases tend to react faster than liquids, which react faster than solids powdered solids are more reactive than “blocks”  more surface area for contact with other reactants 21

22. Factors Affecting Reaction Rate: Temperature Increasing temperature increases reaction rate chemist’s rule of thumb—for each 10°C rise in temperature, the speed of the reaction doubles A mathematical relationship between the absolute temperature and reaction speed speed was discovered by Svante Arrhenius (reference slide 79) 22 Using Catalysts substances that affect the speed of a reaction without being consumed Most catalysts are used to speed up a reaction; these are called positive catalysts  those used to slow reactions are negative catalysts. Homogeneous catalysts are present in the rxn phase Heterogeneous are present in a different phase

23. Factors Affecting Reaction Rate: Reactant Concentration The higher the concentration of reactant molecules, the faster the reaction Increases the frequency of reactant molecule collisions Concentration of solutions depends on the Molarity (solute-to-solution ratio) higher gas pressure = higher concentration gas concentration depends on gas partial pressure 23

24. The Rate Law of a Reaction Reaction rate = rate of change of a reactant or product with respect to time. Rate laws take into consideration all reactants to give a total reaction rate the rate law MUST be determined EXPERIMENTALLY The rate of a reaction is directly proportional to the concentration of each reactant raised to a power For the reaction aA + bB  products the rate law would be: k is the proportionality factor called the rate constant n and m are the order for each reactant (usually an integer, but not always) 24

25. Reaction Order The order is not the molar coefficient in front of the reactant (though it can be that value) The sum of the orders (the exponents) on each reactant in the rate law is called the order of the reaction The rate law for the reaction 2NO(g)+O 2 (g)  2NO 2 (g) is Rate = k[NO] 2 [O 2 ] The reaction is second order with respect to [NO], first order with respect to [O 2 ], and 2+1= third order overall 25

26. Sample Rate Laws Normally the rate depends only on reactants, but the bottom reaction is a special case where the product also affects the rate. Hg 2+ is a negative catalyst; increasing its concentration slows the reaction. 26

27. Rate = k[NO][O 3 ] Solve: Conceptual Plan: Relationships: [NO] = 1.00 x 10 −6 M, [O 3 ] = 3.00 x 10 −6 M, Rate = 6.60 x 10 −6 M/s k, M −1  s −1 Given: Find: Example: The rate equation for the reaction of NO with ozone is Rate = k[NO][O 3 ]. If the rate = 6.60 x 10 −5 M/sec when [NO] = 1.00 x 10 −6 M and [O 3 ] = 3.00 x 10 −6 M, calculate the rate constant Rate, [NO], [O 3 ]k 27

28. Rate = k[acetaldehyde] 2 Solve: Conceptual Plan: Relationships: [acetaldehyde] = 1.75 x 10 −3 M, k= 6.73 x 10 −6 M −1 s −1 Rate, M/s Given: Find: What is the reaction rate for the decomposition of acetaldehyde when the rate constant is 6.73x10 −6 M −1 s −1 and the [acetaldehyde] =1.75x10 −3 M? The rate law is Rate= k[acetaldehyde] 2. k, [acetaldehyde]Rate 28

29. Finding the Rate Law: Initial Rate Method 29 3 Cases for the reaction A → Products, where rate = k[A] n doubling the initial concentration of A doubles the initial reaction rate doubling the initial concentration of A does not change the initial reaction rate doubling the initial concentration of A quadruples the initial reaction rate The rate law, depends on the concentration of the reactants Changing the initial concentration of a reactant will therefore affect the initial rate of the reaction

30. Rate = k[A] n If a reaction is zero Order (n=0), the rate of the reaction is constant (rate = k[A] 0 = k ● 1 = k {a constant} ) changing [A] has no effect on the reaction rate If a reaction is 1 st Order (n=1), the rate is directly proportional to the [reactant] doubling [A] will double the rate of the reaction If a reaction is 2 nd Order (n=2), the rate is directly proportional to the square of the [reactant] doubling [A] will quadruple the rate of the reaction 30

31. Determining the Rate Law When there Are Multiple Reactants Each reactant’s order is determined separately Changing initial [reactant] on reactant at a time, shows the effect of each reactant’s concentration on the rate 31

32. Example 13.2: Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below Comparing Expt #1 and Expt #2, the [NO 2 ] changes but the [CO] does not 32

33. Example 13.2: Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below 33

34. Example 13.2: Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below 34

35. Example 13.2: Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below 35

36. Example 13.2: Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below n = 2, m = 0 36

37. Example 13.2: Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below Substitute the concentrations and rate for any experiment into the rate law and solve for k Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1.0.10 0.0021 2.0.200.100.0082 3.0.20 0.0083 4.0.400.100.033 37

38. Practice – Determine the rate law and rate constant for the reaction NH 4 + + NO 2 −  N 2 + 2 H 2 O given the data below 38

39. Practice – Determine the rate law and rate constant for the reaction NH 4 + + NO 2 −  N 2 + 2 H 2 O given the data below Rate = k[NH 4 + ] n [NO 2  ] m 39

40. Finding the Rate Law: Graphical Methods The rate law must be determined experimentally Graphing [reactant] vs. time is used to determine effect of concentration on reaction rate Involves using calculus to determine the area under a curve (beyond the scope of this course) We will use graphs to determine rate laws for several simple cases 40

41. Reactant Concentration vs. Time A  Products 41 insert figure 13.6

42. Integrated Rate Laws For the reaction A  Products, the rate law depends on the [A] Applying calculus to integrate the rate law gives an equation showing the relationship between the [A] and the reaction time called the Integrated Rate Law 42

43. Zero Order Reactions Rate = k[A] 0 = k Anything to the zero power = 1 Rate = k = a constant Integration gives [A] = −kt + [A] initial  recall y=mx+b, the equation of a straight line Graphing [A] vs. time is a straight line slope = −k and y intercept = [A] initial [A] = −kt + [A] initial (y = mx + b) k has the units Molarity/sec Rate has same units 43

44. Zero Order Reactions [A] initial [A] time slope = −k 44

45. First Order Reactions Rate = k[A] 1 = k[A] Integration gives: ln[A] = −kt + ln[A] initial Graph ln[A] vs. time gives straight line with slope = −k and And the y intercept = ln[A] initial used to determine the rate constant When Rate = M/sec, then k = s –1 t ½ = 0.693/k The half-life of a first order reaction is constant 45

46. ln[A] initial ln[A] time slope = −k 46 First Order Reactions

47. Half-Life The half-life, t 1/2, of a reaction is the length of time it takes for the concentration of the reactant to fall to ½ its initial value The half-life of the reaction depends on the order of the reaction t ½ = [A initial ]/2k 47

48. The Half-Life of a First-Order Reaction Is Constant 48

49. Rate Data for C 4 H 9 Cl + H 2 O  C 4 H 9 OH + HCl 49

50. C 4 H 9 Cl + H 2 O  C 4 H 9 OH + 2 HCl 50

51. C 4 H 9 Cl + H 2 O  C 4 H 9 OH + 2 HCl 51

52. C 4 H 9 Cl + H 2 O  C 4 H 9 OH + 2 HCl slope = −2.01 x 10 −3 k = 2.01 x 10 −3 s -1 52

53. Second Order Reactions Rate = k[A] 2 1/[A] = kt + 1/[A] initial Graph 1/[A] vs. time gives straight line with slope = k and y–intercept = 1/[A] initial used to determine the rate constant When Rate = M/sec, then k = M −1  s −1 t ½ = 1/(k[A 0 ]) 53

54. 1/[A] initial 1/[A] time slope = k 54 Second Order Reactions

55. Rate Data For 2 NO 2  2 NO + O 2 55

56. Rate Data For 2 NO 2  2 NO + O 2 56

57. Rate Data Graphs for 2 NO 2  2 NO + O 2 57

58. Rate Data Graphs for 2 NO 2  2 NO + O 2 58

59. Rate Data Graphs for 2 NO 2  2 NO + O 2 59

60. 60 -stop

61. Example 13.4: The reaction SO 2 Cl 2(g)  SO 2(g) + Cl 2(g) is first order with a rate constant of 2.90 x 10 −4 s −1 at a given set of conditions. Find the [SO 2 Cl 2 ] at 865 s when [SO 2 Cl 2 ] initial = 0.0225 M 61 the new concentration is less than the original, as expected [SO 2 Cl 2 ] init = 0.0225 M, t = 865, k = 2.90 x 10 -4 s −1 [SO 2 Cl 2 ] Check: Solution: Conceptual Plan: Relationships: Given: Find: [SO 2 Cl 2 ][SO 2 Cl 2 ] init, t, k

62. Practice – The reaction Q  2 R is second order in Q. If the initial [Q] = 0.010 M and after 5.0 x 10 2 seconds the [Q] = 0.0010 M, find the rate constant 62 [Q] init = 0.010 M, t = 5.0 x 10 2 s, [Q] t = 0.0010 M k Solution: Conceptual Plan: Relationships: Given: Find: k[Q] init, t, [Q] t

63. Graphical Determination of the Rate Law for A  Product Plots of [A] vs. time, ln[A] vs. time, and 1/[A] vs. time allow determination of whether a reaction is zero, first, or second order Whichever plot gives a straight line determines the order with respect to [A] if linear is [A] vs. time, Rate = k[A] 0 if linear is ln[A] vs. time, Rate = k[A] 1 if linear is 1/[A] vs. time, Rate = k[A] 2 63

64. Complete the Table & Determine the Rate Equation for the Reaction A  2 Product 64 What will the rate be when the [A] = 0.010 M?

65. Complete the Table & Determine the Rate Equation for the Reaction A  2 Product 65

66. 66

67. 67

68. 68

69. Because the graph 1/[A] vs. time is linear, the reaction is second order, 69 Rate  k[A] 2 k  slope of the line  0.10 M –1  s –1 Complete the Table & Determine the Rate Equation for the Reaction A  2 Product

70. Relationship Between Order and Half-Life For a zero order reaction, the lower the initial [reactant], the shorter the half-life t 1/2 = [A] init /2k For a 1 st order reaction, the half-life is independent of the concentration t 1/2 = ln(2)/k For a 2 nd order reaction, the half-life is inversely proportional to the initial concentration – increasing the initial concentration shortens the half-life t 1/2 = 1/(k[A] init ) 70

71. Example 13.6: Molecular iodine dissociates at 625 K with a first- order rate constant of 0.271 s −1. What is the half-life of this reaction? 71 the new concentration is less than the original, as expected k = 0.271 s −1 t 1/2 Check: Solution: Conceptual Plan: Relationships: Given: Find: t 1/2 k

72. Practice – The reaction Q  2 R is second order in Q. If the initial [Q] = 0.010 M and the rate constant is 1.8 M −1s −1 find the length of time for [Q] = ½[Q] init 72 [Q] init = 0.010 M, k = 1.8 M −1  s −1 t 1/2, s Solution: Conceptual Plan: Relationships: Given: Find: t 1/2 [Q] init, k

73. Changing the temperature changes the rate constant of the rate law Svante Arrhenius investigated this relationship and showed that: R = gas constant, 8.314 J/(molK)T = temperature in kelvin A = frequency factor, which depends on a) how often molecules collide when all concentrations = 1 mol/L and b) whether molecules are properly oriented when they collide. E a is the activation energy, the extra energy needed to start the molecules reacting The Effect of Temperature on Rate 73

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75. Activation Energy and the Activated Complex There is an energy barrier to almost all reactions The activation energy is the amount of energy needed to convert reactants into the activated complex (the transition state) The activated complex is a chemical species with partially broken and partially formed bonds, which is very high in energy 75

76. Isomerization of Methyl Isonitrile methyl isonitrile rearranges to acetonitrile for the reaction to occur, the H 3 C─N bond must break, and a new H 3 C─C bond form 76

77. As the reaction begins, the C─N bond weakens enough for the C  N group to start to rotate Energy Profile for the Isomerization of Methyl Isonitrile 77 the frequency is the number of molecules that begin to form the activated complex in a given period of time the activation energy is the difference in energy between the reactants and the activated complex the activated complex is a chemical species with partial bonds

78. The Arrhenius Equation: The Exponential Factor The exponential factor in the Arrhenius equation is a number between 0 and 1 It represents the fraction of reactant molecules with sufficient energy to make it over the activation energy barrier the higher the activation energy barrier, the fewer molecules that have sufficient energy to overcome it That extra energy comes from converting the kinetic energy of motion to potential energy in the molecule when the molecules collide increasing the temperature increases the average kinetic energy  therefore, increasing the temperature will increase the number of molecules with sufficient energy to overcome the energy barrier  therefore increasing the temperature will increase the reaction rate 78

79. 79

80. Arrhenius Plots The Arrhenius Equation can be algebraically solved to give the following form: this equation is in the form y = mx + b where y = ln(k) and x = (1/T) a graph of ln(k) vs. (1/T) is a straight line (−R)(slope of the line) = E a, (in Joules) e y intercept = A (units are the same as k) 80

81. Example 13.7: Determine the activation energy and frequency factor for the reaction O 3(g)  O 2(g) + O (g) given the following data: 81

82. Example 13.7: Determine the activation energy and frequency factor for the reaction O 3(g)  O 2(g) + O (g) given the following data: 82

83. Example 13.7: Determine the activation energy and frequency factor for the reaction O 3(g)  O 2(g) + O (g) given the following data: 83 slope, m = 1.12 x 10 4 K y–intercept, b = 26.8

84. Arrhenius Equation: Two-Point Form If you only have two (T,k) data points, the following forms of the Arrhenius Equation can be used: 84

85. Example 13.8: The reaction NO 2(g) + CO (g)  CO 2(g) + NO (g) has a rate constant of 2.57 M −1 ∙s −1 at 701 K and 567 M −1 ∙s −1 at 895 K. Find the activation energy in kJ/mol 85 most activation energies are tens to hundreds of kJ/mol – so the answer is reasonable T 1 = 701 K, k 1 = 2.57 M −1 ∙s −1, T 2 = 895 K, k 2 = 567 M −1 ∙s −1 E a, kJ/mol Check: Solution: Conceptual Plan: Relationships: Given: Find: EaEa T 1, k 1, T 2, k 2

86. It is often said that the rate of a reaction doubles for every 10 °C rise in temperature. Calculate the activation energy for such a reaction. 86 Let T 1 = 300 K, T 2 = 310 K And k 2 = 2k 1

87. Practice – Find the activation energy in kJ/mol for increasing the temperature 10 º C doubling the rate 87 most activation energies are tens to hundreds of kJ/mol – so the answer is reasonable T 1 = 300 K, T 1 = 310 K, k 2 = 2 k 1 E a, kJ/mol Check: Solution: Conceptual Plan: Relationships: Given: Find: EaEa T 1, k 1, T 2, k 2

88. Collision Theory of Kinetics For most reactions to take place, the reacting molecules must collide with each other on average about 10 9 collisions per second Once molecules collide they may or may not react, depending on two factors: 1.whether the collision has enough energy to "break the bonds holding reactant molecules together"; and 2.whether the reacting molecules collide in the proper orientation for new bonds to form 88

89. Effective Collisions Kinetic Energy Factor For a collision to lead to overcoming the energy barrier, the reacting molecules must have sufficient kinetic energy so that when they collide the activated complex can form 89

90. Effective Collisions Orientation Effect 90

91. Effective Collisions Collisions in which these two conditions are met (and therefore lead to reaction) are called effective collisions The higher the frequency of effective collisions, the faster the reaction rate When two molecules have an effective collision, a temporary, high energy (unstable) chemical species is formed – the transition state, (aka, the activated complex) 91

92. Collision Theory and the Frequency Factor of the Arrhenius Equation The Arrhenius Equation includes a term, A, called the Frequency Factor The Frequency Factor can be broken into two terms that relate to the two factors that determine whether a collision will be effective 92

93. Collision Frequency The collision frequency, z, is the number of collisions that happen per second The more collisions per second there are, the more collisions can be effective and lead to product formation 93

94. Orientation Factor The orientation factor, p, is a statistical term relating the frequency factor, A, to the collision frequency, z For most reactions, p < 1 Generally, the more complex the reactant molecules, the smaller the value of p For reactions involving atoms colliding, p ≈ 1 because of the spherical nature of the atoms Some reactions actually can have a p > 1 They generally involve electron transfer 94

95. Orientation Factor The proper orientation results when the atoms are aligned in such a way that the old bonds can break and the new bonds can form The more complex the reactant molecules, the less frequently they will collide with the proper orientation reactions where symmetry results in multiple orientations leading to reaction have p slightly less than 1 For most reactions, the orientation factor is less than 1 95-stop

96. Molecular Interpretation of Factors Affecting Rate – Reactant Nature Reactions in solution generally occur faster than in pure substances mixing gives greater particle contact allowing more effective collisions per second some solutions break bonds that need to be broken 96

97. Molecular Interpretation of Factors Affecting Rate – Reactant Nature Similar materials undergo similar reactions at different rates because the reactants have a higher initial potential energy, therefore a lower Ea CH 4 +Cl 2  CH 3 Cl+HCl CD 4 +Cl 2  CD 3 Cl+DCl C─H bonds in CH 4 are weaker & less stable than C─D bonds in CD 4 So the reaction of CH 4 +Cl 2  CH 3 Cl+HCl starts at a higher PE than CD 4 +Cl 2  CD 3 Cl+DCl, therefore, its Ea is lower and it reacts ~12x faster than CD 4 97

98. Molecular Interpretation of Factors Affecting Rate – Reactant Nature For the reaction CH 4 + X 2  CH 3 X + HX (where X 2 = F 2 and Cl 2 ) This reaction occurs ~100x faster with F 2 than with Cl 2 because the bond strength of the F-F bond < the Cl-Cl bond the activation energy for the reaction with F 2 is 5 kJ/mol but for Cl 2 is 17 kJ/mol 98

99. Molecular Interpretation of Factors Affecting Rate – Temperature Increasing the temperature raises the average kinetic energy of the reactants increases the number of molecules with sufficient kinetic energy to overcome the activation energy 99 Concentration Reaction rate generally increases as [reactant] or partial pressure increases except for zero order reactions More molecules means more molecules with sufficient kinetic energy for effective collision distribution the same, just bigger curve

100. Reaction Mechanisms Most reactions occur in a series of stepwise reactions involving 1 or 2 molecules Reactions involving 3 molecules are rare Probability for > 3 molecules colliding with proper orientation and sufficient energy is negligible A series of reactions that occur to produce the overall observed reaction is called a reaction mechanism Knowing the rate law of the reaction helps us understand the sequence of the reactions in the mechanism 100

101. An Example of a Reaction Mechanism Overall reaction: H 2(g) + 2 ICl (g)  2 HCl (g) + I 2(g) Mechanism: 1.H 2(g) + ICl (g)  HCl (g) + HI (g) 2.HI (g) + ICl (g)  HCl (g) + I 2(g) Each reaction in this mechanism is an elementary step, they cannot be broken down into simpler steps the molecules actually interact directly in this manner without any other steps 101

102. Elements of a Mechanism: Intermediates HI is a product in Step 1, but a reactant in Step 2 HI is made but then consumed HI does not show up in the overall reaction Materials that are products in an early mechanism step, but then a reactant in a later step, are called intermediates 102 1)H 2 (g) + ICl(g)  HCl(g) + HI(g) 2)HI(g) + ICl(g)  HCl(g) + I 2 (g) H 2 (g) + 2 ICl(g)  2 HCl(g) + I 2 (g)

103. Molecularity The number of reactant particles in an elementary step is called its molecularity Unimolecular, bimolecular, termolecular (rare) 103

104. Rate Laws for Elementary Steps Each step in the mechanism is like its own little reaction – with its own activation energy and rate law The rate law for an overall reaction must be determined experimentally But the rate law of an elementary step can be deduced from the equation of the step H 2 (g) + 2 ICl(g)  2 HCl(g) + I 2 (g) 1) H 2 (g) + ICl(g)  HCl(g) + HI(g) Rate = k 1 [H 2 ][ICl] 2)HI(g) + ICl(g)  HCl(g) + I 2 (g) Rate = k 2 [HI][ICl] 104

105. Rate Determining Step In most mechanisms, one step is slower than the other steps the slowest step has the largest activation energy the step determines the rate of the overall reaction Products cannot form any faster than this step allows  called the rate determining step  The rate law of the rate determining step determines the rate law of the overall reaction 105

106. Another Reaction Mechanism The 1 st step is slower than the 2 nd step its activation energy is larger rate determining step The rate law of the 1 st step is the rate law of the overall reaction 106 1. NO 2(g) + NO 2(g)  NO 3(g) + NO (g) Rate = k 1 [NO 2 ] 2 Slow 2.NO 3(g) + CO (g)  NO 2(g) + CO 2(g) Rate = k 2 [NO 3 ][CO] Fast NO 2(g) + CO (g)  NO (g) + CO 2(g) Rate obs = k[NO 2 ] 2

107. Validating a Mechanism To validate (not prove) a mechanism, two conditions must be met: 1.Elementary steps must sum to the overall reaction 2.The rate law predicted by the mechanism must be consistent with the experimentally observed rate law 107

108. Mechanisms with a Fast Initial Step An early step occurs rapidly, but a subsequent slow step acts like a dam that prevents reaction completion. The early step will reach an equilibrium state  the forward and reverse reaction rates of the early step will be equal  the products in this step are intermediates  the intermediates will be present in significant amts. 108

109. An Example 1.2 NO (g)  N 2 O 2(g) Fast (equilibrium established) 2.H 2(g) + N 2 O 2(g)  H 2 O (g) + N 2 O (g) Slow Rate = k 2 [H 2 ][N 2 O 2 ] 3.H 2(g) + N 2 O (g)  H 2 O (g) + N 2(g) Fast k1k1 k−1k−1 2 H 2(g) + 2 NO (g)  2 H 2 O (g) + N 2(g) Rate obs = k [H 2 ][NO] 2 109 Rate = k 2 [H 2 ][N 2 O 2 ] Rate = k 2 [H 2 ] k 1 /k -1 [NO] 2 Rate = k 2 k 1 /k -1 [H 2 ][NO] 2

110. Example 13.9: Show that the proposed mechanism for the reaction 2 O 3(g)  3 O 2(g) matches the observed rate law Rate = k[O 3 ] 2 [O 2 ] −1 1.O 3(g)  O 2(g) + O (g) Fast 2.O 3(g) + O (g)  2 O 2(g) Slow Rate = k 2 [O 3 ][O] k1k1 k−1k−1 110

111. Mechanism Determine the overall reaction, the rate determining step, the rate law, and identify all intermediates of the following mechanism 1.A + B 2  AB + BSlow 2.A + B  ABFast 111

112. Practice – Mechanism Determine the overall reaction, the rate determining step, the rate law, and identify all intermediates of the following mechanism 1.A + B 2  AB + BSlow 2.A + B  ABFast Overall reaction is 2 A + B 2  2 AB B is an intermediate The first step is the rate determining step Rate = k[A][B 2 ] 112

113. Catalysts Catalysts are substances that affect the rate of a reaction without being consumed They provide an alternative mechanism for the reaction having a lower Ea Catalysts are consumed in an early mechanism step, then made in a later step mechanism without catalyst O 3(g) + O (g)  2 O 2(g) V. Slow mechanism with catalyst Cl (g) + O 3(g)  O 2(g) + ClO (g) Fast ClO (g) + O (g)  O 2(g) + Cl (g) Slow 113

114. Molecular Interpretation of Factors Affecting Rate – Catalysts heterogeneous catalysts hold one reactant molecule in proper orientation for reaction to occur when the collision takes place and sometimes also help to start breaking bonds homogeneous catalysts react with one of the reactant molecules to form a more stable activated complex with a lower activation energy 114

115. Energy Profile of Cl Catalyzed Reaction of ozone 115

116. Mechanism Overall reaction isHQ 2 R 2  Q 2 R + HR R − is a catalyst Q 2 R 2 − is an intermediate The second step is the rate determining step Rate = k[Q 2 R 2 − ] = k[HQ 2 R 2 ][R − ] Determine the overall reaction, the rate determining step, the rate law, and identify all catalysts and intermediates of the following mechanism 1.HQ 2 R 2 + R −  Q 2 R 2 − + HRFast 2.Q 2 R 2 −  Q 2 R + R − Slow 116

117. Catalysts Homogeneous catalysts are in the same phase as the reactant particles Cl (g) in the destruction of O 3(g) Heterogeneous catalysts are in a different phase than the reactant particles solid catalytic converter in a car’s exhaust system 117

118. Catalytic Hydrogenation H 2 C=CH 2 + H 2 → CH 3 CH 3 118

119. Enzymes Biological molecules are large and complex most biological reactions require a catalyst to proceed at a reasonable rate Protein molecules that catalyze biological reactions are called enzymes Enzymes work by adsorbing the substrate reactant onto an active site that orients the substrate for reaction 119 1)Enzyme + Substrate  Enzyme─Substrate Fast 2)Enzyme─Substrate → Enzyme + ProductSlow

120. Enzyme–Substrate Binding: the Lock and Key Mechanism 120

121. Enzymatic Hydrolysis of Sucrose 121