1. IIIIIIIVV Ch. 13 – Kinetics

2. A. Reaction Rates n Rate is the amount of change over time. For example: 70 mph n The “reaction rate” is the rate at which reactants disappear and products appear. n The “rate” or speed of chemical reactions depends on external factors.

3. A. Factors Affecting Rxn Rate 1. Surface Area 2. Concentration 3. Temperature 4. Catalysts 5. Nature of Reactants

4. A. Factors Affecting Rxn Rate 1. Surface Area  high SA = fast rxn rate  more SA = more opportunities for collisions n Increase SA by…  using smaller particles  dissolving in water

5. A. Factors Affecting Rxn Rate 2. Concentration  high conc = fast rxn rate  Concentration of gases depends on the partial pressure of the gas. - Higher pressure = higher concentration  more particles = more opportunities for collisions

6. A. Factors Affecting Rxn Rate 3. Temperature  high temp = fast rxn rate  high temp = high KE - fast-moving particles - more likely to reach activation energy

7. A. Factors Affecting Rxn Rate 4. Catalyst  substance that increases rxn rate without being consumed in the rxn  lowers the activation energy » Homogeneous = present in same phase » Heterogeneous = present in different » phase Inhibitor – Slows the rate of the reaction

8. A. Factors Affecting Rxn Rate n 5. Nature of the reactants means what kind of reactant molecules and what physical condition they are in.  Small molecules tend to react faster than large molecules.  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  Certain types of chemicals are more reactive than others. - For example, potassium metal is more reactive than sodium  Ions react faster than molecules. - No bonds need to be broken.

9. B. Activation Energy n Activation Energy (E a )  minimum energy required for a reaction to occur Activation Energy Activated Complex

10. B. Activation Energy n Activation Energy  depends on reactants  low E a = fast rxn rate  high E a = slow rxn rate EaEa

11. Maxwell–Boltzmann Distributions n 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. © 2012 Pearson Education, Inc.

12. B. Collision Theory n Reaction rate depends on the collisions between reacting particles. n Molecules must collide in order to react. n Successful collisions occur if the particles...  collide with each other  have the correct orientation (2 particles)  have enough kinetic energy to break bonds

13. B. Collision Theory n Particle Orientation Required Orientation Successful Collision- Formation of products Unsuccessful Collisions- no products formed

14. F. Reaction Rates n the average rate decreases as the reaction proceeds. n 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) © 2012 Pearson Education, Inc.

15. - Average Rate n The average rate is the change in measured concentrations in any particular time period.  Linear approximation of a curve  Reminder Δ = final reading-initial reading n The larger the time interval, the more the average rate deviates from the instantaneous rate. Average or simple rate =  [R or P]  t

16. - Instantaneous Rate n The instantaneous rate is the change in concentration at any one particular time.  Slope at one point of a curve n The instantaneous rate is determined by taking the slope of a line tangent to the curve at that particular point.  First derivative of the function (for all of you calculus fans)

17. Reaction Rates and Stoichiometry n To generalize, then, for the reaction n (-) is in front of reactants meaning they are disappearing n (+) is in front of products meaning they are appearing aA + bBcC + dD

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

19. F. Reaction Rates n All reactions slow down over time. n Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction. © 2012 Pearson Education, Inc.

20. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Calculate the instantaneous rate of disappearance of C 4 H 9 Cl at t = 0 s (the initial rate). Solution Plan To obtain the instantaneous rate at t = 0s, we must determine the slope of the curve at t = 0. The tangent is drawn on the graph as the hypotenuse of the tan triangle. The slope of this straight line equals the change in the vertical axis divided by the corresponding change in the horizontal axis (that is, change in molarity over change in time). Solve The tangent line falls from [C 4 H 9 Cl] = 0.100 M to 0.060 M in the time change from 0 s to 210 s. Thus, the initial rate is

21. Reaction Rates and Stoichiometry n 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 © 2012 Pearson Education, Inc.

22. F. Reaction Rates n In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. n 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 © 2012 Pearson Education, Inc.

23. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Relating Rates at Which Products Appear and Reactants Disappear Solution Analyze We are given a balanced chemical equation and asked to relate the rate of appearance of the product to the rate of disappearance of the reactant. Solve (a) Using the coefficients in the balanced equation and the relationship given by Equation 14.4, we have: (b) Solving the equation from part (a) for the rate at which O 3 disappears,  [O 3 ]/  t, we have: (a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction 2 O 3 (g)  3 O 2 (g)? (b) If the rate at which O 2 appears,  [O 2 ]/  t, is 6.0  10 –5 M/s at a particular instant, at what rate is O 3 disappearing at this same time,  [O 3 ]/  t? Plan We can use the coefficients in the chemical equation as shown in Equation 14.4 to express the relative rates of reactions.

24. Concentration and Rate One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration. © 2012 Pearson Education, Inc.

25. G. Rate Laws Rate Law: used to determine rate of a reaction for different concentrations. rate = k [ A] x [ B ] y Molar concentrations of the reactants Rate constant Powers of the conc./determined experimentally

26. MARBLE STATUES

27. G. The Rate Law 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) © 2012 Pearson Education, Inc.

28. G. The Rate Law 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) © 2012 Pearson Education, Inc.

29. G. The Rate Law Rate  [NH 4 + ] Rate  [NO 2  ] Rate  [NH 4 + ] [NO 2  ] when written as an equation, becomes Rate = k [NH 4 + ] [NO 2  ] n This equation is called the rate law, and k is the rate constant. © 2012 Pearson Education, Inc.

30. G. The Rate Law n A rate law shows the relationship between the reaction rate and the concentrations of reactants. n The exponents tell the order of the reaction with respect to each reactant. n 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  ] © 2012 Pearson Education, Inc.

31. G. The Rate Law Rate = k[NH 4 + ] [NO 2  ] n The overall reaction order can be found by adding the exponents on the reactants in the rate law. n This reaction is second-order overall. © 2012 Pearson Education, Inc.

32. What are the overall reaction orders for the reactions described in Equations 14.9 and 14.10? H. Examples 14.9 = 1 14.10 = 3/2

33. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Consider a reaction A + B  C for which rate = k[A][B] 2. Each of the following boxes represents a reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures in order of increasing rate of reaction. Solution Box 1 contains 5 red spheres and 5 purple spheres, giving the following rate: Box 1: Rate = k(5)(5) 2 = 125k Box 2 contains 7 red spheres and 3 purple spheres: Box 2: Rate = k(7)(3) 2 = 63k Box 3 contains 3 red spheres and 7 purple spheres: Box 3: Rate = k(3)(7) 2 = 147k The slowest rate is 63k (box 2), and the highest is 147k (box 3). Thus, the rates vary in the order 2 < 1 < 3. Relating a Rate Law to the Effect of Concentration on Rate

34. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Determining a Rate Law from Initial Rate Data The initial rate of a reaction A + B  C was measured for several different starting concentrations of A and B, and the results are as follows: Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when [A] = 0.050 M and [B] = 0.100 M.

35. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward (a) (b) Using the rate law and the data from experiment 1, we have (c) Using the rate law from part (a) and the rate constant from part (b), we have Rate = k[A] 2 [B] 0 = k[A] 2 Rate = k[A] 2 = (4.0  10  3 M  1 s  1 )(0.050 M) 2 = 1.0  10  5 M/s

36. © 2012 Pearson Education, Inc. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward How to Find Exponents of the Reactants Zero Order = conc changes, rates stays the same 1 st Order = conc doubles, rate doubles ect….. 2 nd Order = conc doubles, rate quadruples 2 2 ect… 3rd Order = conc doubles, rate gets 8x faster 2 3 ect.. ½ Order = conc quadruples, rate doubles 4 1/2 ect… CAUTION!!!!! This is how you find the exponents from experimental data, not the overall order!!!!!

37. y = mx + b I. Integrated Rate Laws

38. Using calculus to integrate the rate law for a zero-order process gives us where [A] 0 is the initial concentration of A, and [A] t is the concentration of A at some time, t, during the course of the reaction. © 2012 Pearson Education, Inc.

39. I. Integrated Rate Laws 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, and [A] t is the concentration of A at some time, t, during the course of the reaction. © 2012 Pearson Education, Inc.

40. I. Integrated Rate Laws Manipulating this equation produces ln [A] t =  kt + ln [A] 0 which is in the form y = mx + b © 2012 Pearson Education, Inc.

41. [A] t =  kt + [A] 0 J. Zero-Order Processes If a reaction is zero-order in A, a plot of [A] vs. t will yield a straight line, and the slope of the line will be  k.

42. J. First-Order Processes If a reaction is first-order in A, 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 © 2012 Pearson Education, Inc.

43. J. First-Order Processes Consider the process in which methyl isonitrile is converted to acetonitrile. CH 3 NCCH 3 CN © 2012 Pearson Education, Inc.

44. J. First-Order Processes This data were collected for this reaction at 198.9  C. CH 3 NCCH 3 CN © 2012 Pearson Education, Inc.

45. J. First-Order Processes n When ln P is plotted as a function of time, a straight line results. n Therefore,  The process is first-order.  k is the negative of the slope: 5.1  10  5 s  1. © 2012 Pearson Education, Inc.

46. K. Second-Order Processes Similarly, 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 © 2012 Pearson Education, Inc.

47. J. Second-Order Processes So if a process is second-order in A, a plot of vs. t yields a straight line, and the slope of that line is k. 1 [A] t = kt + 1 [A] 0 1 [A] © 2012 Pearson Education, Inc.

48. K. Second-Order Processes The decomposition of NO 2 at 300C is described by the equation NO 2 (g)NO(g) + O 2 (g) and yields data comparable to this table: Time (s)[NO 2 ], M 0.00.01000 50.00.00787 100.00.00649 200.00.00481 300.00.00380 1212 © 2012 Pearson Education, Inc.

49. K. Second-Order Processes Plotting ln [NO 2 ] vs. t yields a graph that is not a straight line 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 So the process is not first- order in [A]. © 2012 Pearson Education, Inc.

50. K. 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 ] © 2012 Pearson Education, Inc.

51. L. Half-Life n Half-life is defined as the time required for one-half of a reactant to react. n Because [A] at t 1/2 is one-half of the original [A], [A] t = 0.5 [A] 0. © 2012 Pearson Education, Inc.

52. Half-life  Fluorine-21 has a half-life of 5.0 seconds. If you start with 25 g of fluorine-21, how many grams would remain after 60.0 s? GIVEN: t ½ = 5.0 s m i = 25 g m f = ? total time = 60.0 s n = 60.0s ÷ 5.0s =12 WORK : m f = m i (½) n m f = (25 g)(0.5) 12 m f = 0.0061 g

53. L. Half-Life For a first-order process, this becomes = t 1/2 0.693 k Note: For a first-order process, then, the half-life does not depend on [A] 0. © 2012 Pearson Education, Inc.

54. L. Half-Life For a second-order process, = t 1/2 1 k[A] 0 © 2012 Pearson Education, Inc.

55. B. Collision Theory n Reaction rate depends on the collisions between reacting particles. n Molecules must collide in order to react. n Successful collisions occur if the particles...  collide with each other  have the correct orientation (2 particles)  have enough kinetic energy to break bonds

56. B. Collision Theory n Particle Orientation Required Orientation Successful Collision- Formation of products Unsuccessful Collisions- no products formed

57. M. Reaction Mechanism n Reaction Mechanism: The series of steps that show how the reactants are converted to the products

58. M. Reaction Mechanisms n Reactions may occur all at once or through several discrete steps. n Each of these processes is known as an elementary reaction or elementary process. © 2012 Pearson Education, Inc.

59. M. Reaction Mechanisms The molecularity of a process tells how many molecules are involved in the process. © 2012 Pearson Education, Inc.

60. Chemistry, The Central Science, 12th Edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward Solve (a)The first elementary reaction involves a single reactant and is consequently unimolecular. The second reaction, which involves two reactant molecules, is bimolecular. (b) Adding the two elementary reactions gives 2 O 3 (g) + O(g)  3 O 2 (g) + O(g) Because O(g) appears in equal amounts on both sides of the equation, it can be eliminated to give the net equation for the chemical process: 2 O 3 (g)  3 O 2 (g) (c) The intermediate is O(g). It is neither an original reactant nor a final product but is formed in the first step of the mechanism and consumed in the second. It has been proposed that the conversion of ozone into O 2 proceeds by a two-step mechanism: O 3 (g)  O 2 (g) + O(g) O 3 (g) + O(g)  2 O 2 (g) (a) Describe the molecularity of each elementary reaction in this mechanism. (b) Write the equation for the overall reaction. (c) Identify the intermediate(s). Determining Molecularity and Identifying Intermediates

61. M. Multistep Mechanisms n In a multistep process, one of the steps will be slower than all others. n The overall reaction cannot occur faster than this slowest, rate-determining step. © 2012 Pearson Education, Inc.

62. 1. Slow Initial Step n The rate law for this reaction is found experimentally to be Rate = k[NO 2 ] 2 n CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. n This suggests that the reaction occurs in two steps. NO 2 (g) + CO(g)  NO(g) + CO 2 (g) © 2012 Pearson Education, Inc.

63. 1. Slow Initial Step n 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) n The NO 3 intermediate is consumed in the second step. n As CO is not involved in the slow, rate-determining step, it does not appear in the rate law. © 2012 Pearson Education, Inc.

64. 2. Fast Initial Step n The rate law for this reaction is found to be Rate = k[NO] 2 [Br 2 ] n Because termolecular processes are rare, this rate law suggests a two- step mechanism. 2 NO(g) + Br 2 (g)  2 NOBr(g) © 2012 Pearson Education, Inc.

65. 2. Fast Initial Step n 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) © 2012 Pearson Education, Inc. Rate law = k[NO] 2 [Br 2 ]

66. Fast Initial Step n The rate of the overall reaction depends upon the rate of the slow step. n The rate law for that step would be Rate = k 2 [NOBr 2 ] [NO] n But how can we find [NOBr 2 ]? © 2012 Pearson Education, Inc.

67. Fast Initial Step n NOBr 2 can react two ways:  With NO to form NOBr.  By decomposition to reform NO and Br 2. n The reactants and products of the first step are in equilibrium with each other. n Therefore, Rate f = Rate r © 2012 Pearson Education, Inc.

68. Fast Initial Step n Because Rate f = Rate r, k 1 [NO] [Br 2 ] = k  1 [NOBr 2 ] n Solving for [NOBr 2 ], gives us k1k1k1k1 [NO] [Br 2 ] = [NOBr 2 ] © 2012 Pearson Education, Inc.

69. 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 ] © 2012 Pearson Education, Inc.

70. F. Reaction Mechanism 2NO + F 2 2NOF Rxn Mechanism involves 2 elementary steps: Step 1: NO + F 2 NOF 2 Step 2: NOF 2 + NO 2 NOF The slowest elementary step is called the Rate Determining Step. Intermediate product

71. G. Rate Laws Rate Law: used to determine rate of a reaction for different concentrations. rate = k [ A] x [ B ] y Molar concentrations of the reactants Rate constant Powers of the conc./determined experimentally

72. G. Rate Laws NO 2 + O 3 NO 3 + O 2 rate = k [ NO 2 ] [ O 3 ] The rate is directly proportional to the concentrations. If the conc. of one reactant increases by 10, the rate increases by a factor of 10 as well.

73. G. Rate Laws Example Problem n If the rate of a reaction is given by the following rate law: rate = k [NO ] 2 [ O 2 ] And the concentration of each reactant is 1.00 M, what happens when……

74. G. Rate Law Example Problem a). The concentration of Oxygen is doubled? The rate is doubled (x2) b). The concentration of NO is doubled? The rate is x4 because of the exponent.