1. Copyright  2011 Pearson Education, Inc. Chapter 13 Chemical Kinetics Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

2. Copyright  2011 Pearson Education, Inc. Ectotherms Lizards, and other cold-blooded creatures, are ectotherms – animals whose body temperature matches their environment’s temperature When a lizard’s body temperature drops, the chemical reactions that occur in its body slow down as do all chemical reactions when cooled This causes the lizard to become lethargic and to slow down Chemical kinetics is the study of the factors that affect the rates of chemical reactions Such as temperature Tro: Chemistry: A Molecular Approach, 2/e2

3. Copyright  2011 Pearson Education, Inc. Chemical Kinetics The speed of a chemical reaction is called its reaction rate The rate of a reaction is a measure of how fast the reaction makes products or uses reactants The ability to control the speed of a chemical reaction is important 3Tro: Chemistry: A Molecular Approach, 2/e

4. Copyright  2011 Pearson Education, Inc. Defining Rate Rate is how much a quantity changes in a given period of time The speed you drive your car is a rate – the distance your car travels (miles) in a given period of time (1 hour) so the rate of your car has units of mi/hr 4Tro: Chemistry: A Molecular Approach, 2/e

5. Copyright  2011 Pearson Education, Inc. Defining Reaction Rate The rate of a chemical reaction is generally measured in terms of how much the concentration of a reactant decreases in a given period of time or product concentration increases For reactants, a negative sign is placed in front of the definition 5Tro: Chemistry: A Molecular Approach, 2/e

6. Copyright  2011 Pearson Education, Inc. 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 6Tro: Chemistry: A Molecular Approach, 2/e

7. Copyright  2011 Pearson Education, Inc. 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 7Tro: Chemistry: A Molecular Approach, 2/e

8. Copyright  2011 Pearson Education, Inc. 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 8Tro: Chemistry: A Molecular Approach, 2/e

9. Copyright  2011 Pearson Education, Inc. Hypothetical Reaction Red  Blue In this reaction, one molecule of Red turns into one molecule of Blue The number of molecules will always total 100 The rate of the reaction can be measured as the speed of loss of Red molecules over time, or the speed of gain of Blue molecules over time 9Tro: Chemistry: A Molecular Approach, 2/e

10. Copyright  2011 Pearson Education, Inc. Hypothetical Reaction Red  Blue 10Tro: Chemistry: A Molecular Approach, 2/e

11. Copyright  2011 Pearson Education, Inc. Hypothetical Reaction Red  Blue 11Tro: Chemistry: A Molecular Approach, 2/e

12. Copyright  2011 Pearson Education, Inc. Reaction Rate Changes Over Time As time goes on, the rate of a reaction generally slows down because the concentration of the reactants decreases At some time the reaction stops, either because the reactants run out or because the system has reached equilibrium 12Tro: Chemistry: A Molecular Approach, 2/e

13. Copyright  2011 Pearson Education, Inc. Reaction Rate and Stoichiometry In most reactions, the coefficients of the balanced equation are not all the same H 2 (g) + I 2 (g)  2 HI (g) 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 for the above reaction, for every 1 mole of H 2 used, 1 mole of I 2 will also be used and 2 moles of HI made therefore the rate of change will be different To be consistent, the change in the concentration of each substance is multiplied by 1/coefficient 13Tro: Chemistry: A Molecular Approach, 2/e

14. Copyright  2011 Pearson Education, Inc. Average Rate The average rate is the change in measured concentrations in any particular time period linear approximation of a curve The larger the time interval, the more the average rate deviates from the instantaneous rate 14Tro: Chemistry: A Molecular Approach, 2/e

15. Copyright  2011 Pearson Education, Inc. Hypothetical Reaction Red  Blue 15Tro: Chemistry: A Molecular Approach, 2/e

16. Copyright  2011 Pearson Education, Inc. H2H2 I2I2 HI Stoichiometry tells us that for every 1 mole/L of H 2 used, 2 moles/L of HI are made. Assuming a 1 L container, at 10 s, we used 0.181 moles of H 2. Therefore the amount of HI made is 2(0.181 moles) = 0.362 moles. 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 16Tro: Chemistry: A Molecular Approach, 2/e

17. Copyright  2011 Pearson Education, Inc. 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 17Tro: Chemistry: A Molecular Approach, 2/e

18. Copyright  2011 Pearson Education, Inc. Instantaneous Rate The instantaneous rate is the change in concentration at any one particular time slope at one point of a curve Determined by taking the slope of a line tangent to the curve at that particular point first derivative of the function  for you calculus fans 18Tro: Chemistry: A Molecular Approach, 2/e

19. Copyright  2011 Pearson Education, Inc. 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 19Tro: Chemistry: A Molecular Approach, 2/e

20. Copyright  2011 Pearson Education, Inc. 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) 20 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 Tro: Chemistry: A Molecular Approach, 2/e

21. Copyright  2011 Pearson Education, Inc. Practice – If 2.4 x 10 2 g of NOBr (MM 109.91 g) decomposes in a 2.0 x 10 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) 21Tro: Chemistry: A Molecular Approach, 2/e

22. Copyright  2011 Pearson Education, Inc. 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: Practice – If 2.4 x 10 2 g of NOBr decomposes in a 2.0 x 10 2 mL flask in 5.0 minutes, find the average rate of Br 2 production 2 NOBr(g)  2 NO(g) + Br 2 (l) 22  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 Tro: Chemistry: A Molecular Approach, 2/e

23. Copyright  2011 Pearson Education, Inc. Measuring Reaction Rate To measure the reaction rate you need to be able to measure the concentration of at least one component in the mixture at many points in time There are two ways of approaching this problem 1.for reactions that are complete in less than 1 hour, it is best to use continuous monitoring of the concentration 2.for reactions that happen over a very long 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 23Tro: Chemistry: A Molecular Approach, 2/e

24. Copyright  2011 Pearson Education, Inc. 24 Continuous Monitoring Polarimetry – measuring the change in the degree of rotation of plane-polarized light caused by one of the components over time Spectrophotometry – measuring the amount of light of a particular wavelength 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 Tro: Chemistry: A Molecular Approach, 2/e

25. Copyright  2011 Pearson Education, Inc. 25 Sampling the Reaction Mixture at Specific Times At specific times during the reaction, drawing off aliquots, (samples from the reaction mixture), and doing quantitative analysis titration for one of the components gravimetric analysis 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 Tro: Chemistry: A Molecular Approach, 2/e

26. Copyright  2011 Pearson Education, Inc. Methods for Determining Concentrations in a Mixture 26Tro: Chemistry: A Molecular Approach, 2/e

27. Copyright  2011 Pearson Education, Inc. Methods for Determining Concentrations in a Mixture 27Tro: Chemistry: A Molecular Approach, 2/e

28. Copyright  2011 Pearson Education, Inc. Methods for Determining Concentrations in a Mixture 28Tro: Chemistry: A Molecular Approach, 2/e

29. Copyright  2011 Pearson Education, Inc. Factors Affecting Reaction Rate: Nature of the Reactants 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  e.g. potassium metal is more reactive than sodium ions react faster than molecules  no bonds need to be broken 29Tro: Chemistry: A Molecular Approach, 2/e

30. Copyright  2011 Pearson Education, Inc. 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  for many reactions There is a mathematical relationship between the absolute temperature and the speed of a reaction discovered by Svante Arrhenius, which will be examined later 30Tro: Chemistry: A Molecular Approach, 2/e

31. Copyright  2011 Pearson Education, Inc. Factors Affecting Reaction Rate: Catalysts Catalysts are 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 catalysts used to slow a reaction are called negative catalysts. Homogeneous = present in same phase Heterogeneous = present in different phase How catalysts work will be examined later 31Tro: Chemistry: A Molecular Approach, 2/e

32. Copyright  2011 Pearson Education, Inc. Factors Affecting Reaction Rate: Reactant Concentration Generally, the larger the concentration of reactant molecules, the faster the reaction increases the frequency of reactant molecule contact concentration of gases depends on the partial pressure of the gas  higher pressure = higher concentration Concentrations of solutions depend on the solute-to-solution ratio (molarity) 32Tro: Chemistry: A Molecular Approach, 2/e

33. Copyright  2011 Pearson Education, Inc. The Rate Law The rate law of a reaction is the mathematical relationship between the rate of the reaction and the concentrations of the reactants and homogeneous catalysts as well 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 have the form given below n and m are called the orders for each reactant k is called the rate constant 33Tro: Chemistry: A Molecular Approach, 2/e

34. Copyright  2011 Pearson Education, Inc. Reaction Order The exponent on each reactant in the rate law is called the order with respect to that reactant The sum of the exponents on the reactants is called the order of the reaction The rate law for the reaction 2 NO(g) + O 2 (g)  2 NO 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 third order overall 34Tro: Chemistry: A Molecular Approach, 2/e

35. Copyright  2011 Pearson Education, Inc. Sample Rate Laws The bottom reaction is autocatalytic because a product affects the rate. Hg 2+ is a negative catalyst; increasing its concentration slows the reaction. 35Tro: Chemistry: A Molecular Approach, 2/e

36. Copyright  2011 Pearson Education, Inc. 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 is 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 36Tro: Chemistry: A Molecular Approach, 2/e

37. Copyright  2011 Pearson Education, Inc. Practice – The rate law for the decomposition of acetaldehyde is Rate = k[acetaldehyde] 2. What is the rate of the reaction when the [acetaldehyde] = 1.75 x 10 −3 M and the rate constant is 6.73 x 10 −6 M −1s −1 ? 37Tro: Chemistry: A Molecular Approach, 2/e

38. Copyright  2011 Pearson Education, Inc. 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: Practice – What is the rate of the reaction when the [acetaldehyde] = 1.75 x 10 −3 M and the rate constant is 6.73 x 10 −6 M −1s −1 ? k, [acetaldehyde]Rate 38Tro: Chemistry: A Molecular Approach, 2/e

39. Copyright  2011 Pearson Education, Inc. Finding the Rate Law: the Initial Rate Method The rate law must be determined experimentally The rate law shows how the rate of a reaction depends on the concentration of the reactants Changing the initial concentration of a reactant will therefore affect the initial rate of the reaction 39 if for the reaction A → Products then doubling the initial concentration of A doubles the initial reaction rate then doubling the initial concentration of A does not change the initial reaction rate then doubling the initial concentration of A quadruples the initial reaction rate Tro: Chemistry: A Molecular Approach, 2/e

40. Copyright  2011 Pearson Education, Inc. Rate = k[A] n If a reaction is Zero Order, the rate of the reaction is always the same doubling [A] will have no effect on the reaction rate If a reaction is First Order, the rate is directly proportional to the reactant concentration doubling [A] will double the rate of the reaction If a reaction is Second Order, the rate is directly proportional to the square of the reactant concentration doubling [A] will quadruple the rate of the reaction 40Tro: Chemistry: A Molecular Approach, 2/e

41. Copyright  2011 Pearson Education, Inc. Determining the Rate Law When there Are Multiple Reactants Changing each reactant will effect the overall rate of the reaction By changing the initial concentration of one reactant at a time, the effect of each reactant’s concentration on the rate can be determined In examining results, we compare differences in rate for reactions that only differ in the concentration of one reactant 41Tro: Chemistry: A Molecular Approach, 2/e

42. Copyright  2011 Pearson Education, Inc. 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 42Tro: Chemistry: A Molecular Approach, 2/e

43. Copyright  2011 Pearson Education, Inc. 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 43Tro: Chemistry: A Molecular Approach, 2/e

44. Copyright  2011 Pearson Education, Inc. 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 44Tro: Chemistry: A Molecular Approach, 2/e

45. Copyright  2011 Pearson Education, Inc. 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 45Tro: Chemistry: A Molecular Approach, 2/e

46. Copyright  2011 Pearson Education, Inc. 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 46Tro: Chemistry: A Molecular Approach, 2/e

47. Copyright  2011 Pearson Education, Inc. 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 47Tro: Chemistry: A Molecular Approach, 2/e

48. Copyright  2011 Pearson Education, Inc. 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 48Tro: Chemistry: A Molecular Approach, 2/e

49. Copyright  2011 Pearson Education, Inc. 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 49Tro: Chemistry: A Molecular Approach, 2/e

50. Copyright  2011 Pearson Education, Inc. Finding the Rate Law: Graphical Methods The rate law must be determined experimentally A graph of concentration of reactant vs. time can be used to determine the effect of concentration on the rate of a reaction This involves using calculus to determine the area under a curve – which is beyond the scope of this course Later we will examine the results of this analysis so we can use graphs to determine rate laws for several simple cases 50Tro: Chemistry: A Molecular Approach, 2/e

51. Copyright  2011 Pearson Education, Inc. Reactant Concentration vs. Time A  Products 51 insert figure 13.6 Tro: Chemistry: A Molecular Approach, 2/e

52. Copyright  2011 Pearson Education, Inc. Integrated Rate Laws For the reaction A  Products, the rate law depends on the concentration of A Applying calculus to integrate the rate law gives another equation showing the relationship between the concentration of A and the time of the reaction – this is called the Integrated Rate Law 52Tro: Chemistry: A Molecular Approach, 2/e

53. Copyright  2011 Pearson Education, Inc. 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 53Tro: Chemistry: A Molecular Approach, 2/e

54. Copyright  2011 Pearson Education, Inc. Zero Order Reactions Rate = k[A] 0 = k constant rate reactions [A] = −kt + [A] initial Graph of [A] vs. time is straight line with slope = −k and y–intercept = [A] initial t ½ = [A initial ]/2k When Rate = M/sec, k = M/sec [A] initial [A] time slope = −k 54Tro: Chemistry: A Molecular Approach, 2/e

55. Copyright  2011 Pearson Education, Inc. First Order Reactions Rate = k[A] 1 = k[A] ln[A] = −kt + ln[A] initial Graph ln[A] vs. time gives straight line with slope = −k and y–intercept = ln[A] initial used to determine the rate constant t ½ = 0.693/k The half-life of a first order reaction is constant When Rate = M/sec, k = s –1 55Tro: Chemistry: A Molecular Approach, 2/e

56. Copyright  2011 Pearson Education, Inc. ln[A] initial ln[A] time slope = −k 56Tro: Chemistry: A Molecular Approach, 2/e

57. Copyright  2011 Pearson Education, Inc. The Half-Life of a First-Order Reaction Is Constant 57Tro: Chemistry: A Molecular Approach, 2/e

58. Copyright  2011 Pearson Education, Inc. Rate Data for C 4 H 9 Cl + H 2 O  C 4 H 9 OH + HCl 58Tro: Chemistry: A Molecular Approach, 2/e

59. Copyright  2011 Pearson Education, Inc. C 4 H 9 Cl + H 2 O  C 4 H 9 OH + 2 HCl 59Tro: Chemistry: A Molecular Approach, 2/e

60. Copyright  2011 Pearson Education, Inc. C 4 H 9 Cl + H 2 O  C 4 H 9 OH + 2 HCl 60Tro: Chemistry: A Molecular Approach, 2/e

61. Copyright  2011 Pearson Education, Inc. 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 61Tro: Chemistry: A Molecular Approach, 2/e

62. Copyright  2011 Pearson Education, Inc. 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 t ½ = 1/(k[A 0 ]) When Rate = M/sec, k = M −1  s −1 62Tro: Chemistry: A Molecular Approach, 2/e

63. Copyright  2011 Pearson Education, Inc. 1/[A] initial 1/[A] time slope = k 63Tro: Chemistry: A Molecular Approach, 2/e

64. Copyright  2011 Pearson Education, Inc. Rate Data For 2 NO 2  2 NO + O 2 64Tro: Chemistry: A Molecular Approach, 2/e

65. Copyright  2011 Pearson Education, Inc. Rate Data For 2 NO 2  2 NO + O 2 65Tro: Chemistry: A Molecular Approach, 2/e

66. Copyright  2011 Pearson Education, Inc. Rate Data Graphs for 2 NO 2  2 NO + O 2 66Tro: Chemistry: A Molecular Approach, 2/e

67. Copyright  2011 Pearson Education, Inc. Rate Data Graphs for 2 NO 2  2 NO + O 2 67Tro: Chemistry: A Molecular Approach, 2/e

68. Copyright  2011 Pearson Education, Inc. Rate Data Graphs for 2 NO 2  2 NO + O 2 68Tro: Chemistry: A Molecular Approach, 2/e

69. Copyright  2011 Pearson Education, Inc. 69Tro: Chemistry: A Molecular Approach, 2/e

70. Copyright  2011 Pearson Education, Inc. 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 70 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 Tro: Chemistry: A Molecular Approach, 2/e

71. Copyright  2011 Pearson Education, Inc. 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 71Tro: Chemistry: A Molecular Approach, 2/e

72. Copyright  2011 Pearson Education, Inc. 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 72 [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 Tro: Chemistry: A Molecular Approach, 2/e

73. Copyright  2011 Pearson Education, Inc. 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 73Tro: Chemistry: A Molecular Approach, 2/e

74. Copyright  2011 Pearson Education, Inc. Practice – Complete the Table and Determine the Rate Equation for the Reaction A  2 Product 74 What will the rate be when the [A] = 0.010 M? Tro: Chemistry: A Molecular Approach, 2/e

75. Copyright  2011 Pearson Education, Inc. Practice – Complete the Table and Determine the Rate Equation for the Reaction A  2 Product 75Tro: Chemistry: A Molecular Approach, 2/e

76. Copyright  2011 Pearson Education, Inc. 76Tro: Chemistry: A Molecular Approach, 2/e

77. Copyright  2011 Pearson Education, Inc. 77Tro: Chemistry: A Molecular Approach, 2/e

78. Copyright  2011 Pearson Education, Inc. 78Tro: Chemistry: A Molecular Approach, 2/e

79. Copyright  2011 Pearson Education, Inc. Practice – Complete the table and determine the rate equation for the reaction A  2 Product Because the graph 1/[A] vs. time is linear, the reaction is second order, 79 Rate  k[A] 2 k  slope of the line  0.10 M –1  s –1 Tro: Chemistry: A Molecular Approach, 2/e

80. Copyright  2011 Pearson Education, Inc. Relationship Between Order and Half-Life For a zero order reaction, the lower the initial concentration of the reactants, the shorter the half-life t 1/2 = [A] init /2k For a first order reaction, the half-life is independent of the concentration t 1/2 = ln(2)/k For a second 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 ) 80Tro: Chemistry: A Molecular Approach, 2/e

81. Copyright  2011 Pearson Education, Inc. 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? 81 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 Tro: Chemistry: A Molecular Approach, 2/e

82. Copyright  2011 Pearson Education, Inc. 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 82Tro: Chemistry: A Molecular Approach, 2/e

83. Copyright  2011 Pearson Education, Inc. 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 83 [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 Tro: Chemistry: A Molecular Approach, 2/e

84. Copyright  2011 Pearson Education, Inc. Changing the temperature changes the rate constant of the rate law Svante Arrhenius investigated this relationship and showed that: R is the gas constant in energy units, 8.314 J/(molK) where T is the temperature in kelvins A is called the frequency factor, the rate the reactant energy approaches the activation energy E a is the activation energy, the extra energy needed to start the molecules reacting The Effect of Temperature on Rate 84Tro: Chemistry: A Molecular Approach, 2/e

85. Copyright  2011 Pearson Education, Inc. 85Tro: Chemistry: A Molecular Approach, 2/e

86. Copyright  2011 Pearson Education, Inc. 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 aka transition state The activated complex is a chemical species with partially broken and partially formed bonds always very high in energy because of its partial bonds 86Tro: Chemistry: A Molecular Approach, 2/e

87. Copyright  2011 Pearson Education, Inc. 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 87Tro: Chemistry: A Molecular Approach, 2/e

88. Copyright  2011 Pearson Education, Inc. 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 88 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 Tro: Chemistry: A Molecular Approach, 2/e

89. Copyright  2011 Pearson Education, Inc. The Arrhenius Equation: The Exponential Factor The exponential factor in the Arrhenius equation is a number between 0 and 1 Tt represents the fraction of reactant molecules with sufficient energy so they can make it over the energy barrier the higher the energy barrier (larger activation energy), 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 of the molecules 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 89Tro: Chemistry: A Molecular Approach, 2/e

90. Copyright  2011 Pearson Education, Inc. 90Tro: Chemistry: A Molecular Approach, 2/e

91. Copyright  2011 Pearson Education, Inc. 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 (−8.314 J/mol∙K)(slope of the line) = E a, (in Joules) e y–intercept = A (unit is the same as k) 91Tro: Chemistry: A Molecular Approach, 2/e

92. Copyright  2011 Pearson Education, Inc. 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: 92Tro: Chemistry: A Molecular Approach, 2/e

93. Copyright  2011 Pearson Education, Inc. 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: 93Tro: Chemistry: A Molecular Approach, 2/e

94. Copyright  2011 Pearson Education, Inc. 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: 94 slope, m = 1.12 x 10 4 K y–intercept, b = 26.8 Tro: Chemistry: A Molecular Approach, 2/e

95. Copyright  2011 Pearson Education, Inc. Arrhenius Equation: Two-Point Form If you only have two (T,k) data points, the following forms of the Arrhenius Equation can be used: 95Tro: Chemistry: A Molecular Approach, 2/e

96. Copyright  2011 Pearson Education, Inc. 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 96 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 Tro: Chemistry: A Molecular Approach, 2/e

97. Copyright  2011 Pearson Education, Inc. Practice – 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. 97 Hint: makeT 1 = 300 K, T 2 = 310 K andk 2 = 2k 1 Tro: Chemistry: A Molecular Approach, 2/e

98. Copyright  2011 Pearson Education, Inc. Practice – Find the activation energy in kJ/mol for increasing the temperature 10 º C doubling the rate 98 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 Tro: Chemistry: A Molecular Approach, 2/e

99. Copyright  2011 Pearson Education, Inc. Collision Theory of Kinetics For most reactions, for a reaction to take place, the reacting molecules must collide with each other on average about 10 9 collisions per second Once molecules collide they may react together or they may not, 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 99Tro: Chemistry: A Molecular Approach, 2/e

100. Copyright  2011 Pearson Education, Inc. 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 100Tro: Chemistry: A Molecular Approach, 2/e

101. Copyright  2011 Pearson Education, Inc. Effective Collisions Orientation Effect 101Tro: Chemistry: A Molecular Approach, 2/e

102. Copyright  2011 Pearson Education, Inc. 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 activated complex 102Tro: Chemistry: A Molecular Approach, 2/e

103. Copyright  2011 Pearson Education, Inc. 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 103Tro: Chemistry: A Molecular Approach, 2/e

104. Copyright  2011 Pearson Education, Inc. Collision Frequency The collision frequency 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 104Tro: Chemistry: A Molecular Approach, 2/e

105. Copyright  2011 Pearson Education, Inc. Orientation Factor The orientation factor, p, is a statistical term relating the frequency factor to the collision frequency 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 generally involve electron transfer 105Tro: Chemistry: A Molecular Approach, 2/e

106. Copyright  2011 Pearson Education, Inc. 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 106Tro: Chemistry: A Molecular Approach, 2/e

107. Copyright  2011 Pearson Education, Inc. Molecular Interpretation of Factors Affecting Rate – Reactant Nature Reactions generally occur faster in solution than in pure substances mixing gives more particle contact particles separated, allowing more effective collisions per second forming some solutions breaks bonds that need to be broken Some materials undergo similar reactions at different rates either because they have a (1) higher initial potential energy and are therefore closer in energy to the activated complex, or (2) because their reaction has a lower activation energy CH 4 + Cl 2  CH 3 Cl + HCl is about 12x faster than CD 4 + Cl 2  CD 3 Cl + DCl because the C─H bond is weaker and less stable than the C─D bond CH 4 + X 2  CH 3 X + HX occurs about 100x faster with F 2 than with Cl 2 because the activation energy for F 2 is 5 kJ/mol but for Cl 2 is 17 kJ/mol 107Tro: Chemistry: A Molecular Approach, 2/e

108. Copyright  2011 Pearson Education, Inc. Molecular Interpretation of Factors Affecting Rate – Temperature Increasing the temperature raises the average kinetic energy of the reactant molecules There is a minimum amount of kinetic energy needed for the collision to be converted into enough potential energy to form the activated complex Increasing the temperature increases the number of molecules with sufficient kinetic energy to overcome the activation energy 108Tro: Chemistry: A Molecular Approach, 2/e

109. Copyright  2011 Pearson Education, Inc. Molecular Interpretation of Factors Affecting Rate – Concentration Reaction rate generally increases as the concentration or partial pressure of reactant molecules increases except for zero order reactions More molecules leads to more molecules with sufficient kinetic energy for effective collision distribution the same, just bigger curve 109Tro: Chemistry: A Molecular Approach, 2/e

110. Copyright  2011 Pearson Education, Inc. Reaction Mechanisms We generally describe chemical reactions with an equation listing all the reactant molecules and product molecules But the probability of more than 3 molecules colliding at the same instant with the proper orientation and sufficient energy to overcome the energy barrier is negligible Most reactions occur in a series of small reactions involving 1, 2, or at most 3 molecules Describing the series of reactions that occurs to produce the overall observed reaction is called a reaction mechanism Knowing the rate law of the reaction helps us understand the sequence of reactions in the mechanism 110Tro: Chemistry: A Molecular Approach, 2/e

111. Copyright  2011 Pearson Education, Inc. 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) The reactions in this mechanism are elementary steps, meaning that they cannot be broken down into simpler steps and that the molecules actually interact directly in this manner without any other steps 111Tro: Chemistry: A Molecular Approach, 2/e

112. Copyright  2011 Pearson Education, Inc. H 2 (g) + 2 ICl(g)  2 HCl(g) + I 2 (g) 1)H 2 (g) + ICl(g)  HCl(g) + HI(g) 2)HI(g) + ICl(g)  HCl(g) + I 2 (g) Elements of a Mechanism Intermediates Notice that the HI is a product in Step 1, but then a reactant in Step 2 Because 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 112Tro: Chemistry: A Molecular Approach, 2/e

113. Copyright  2011 Pearson Education, Inc. Molecularity The number of reactant particles in an elementary step is called its molecularity A unimolecular step involves one particle A bimolecular step involves two particles though they may be the same kind of particle A termolecular step involves three particles though these are exceedingly rare in elementary steps 113Tro: Chemistry: A Molecular Approach, 2/e

114. Copyright  2011 Pearson Education, Inc. Rate Laws for Elementary Steps Each step in the mechanism is like its own little reaction – with its own activation energy and own 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] 114Tro: Chemistry: A Molecular Approach, 2/e

115. Copyright  2011 Pearson Education, Inc. Rate Laws of Elementary Steps 115Tro: Chemistry: A Molecular Approach, 2/e

116. Copyright  2011 Pearson Education, Inc. Rate Determining Step In most mechanisms, one step occurs slower than the other steps The result is that product production cannot occur any faster than the slowest step – the step determines the rate of the overall reaction We call the slowest step in the mechanism the rate determining step the slowest step has the largest activation energy The rate law of the rate determining step determines the rate law of the overall reaction 116Tro: Chemistry: A Molecular Approach, 2/e

117. Copyright  2011 Pearson Education, Inc. Another Reaction Mechanism The first step in this mechanism is the rate determining step. The first step is slower than the second step because its activation energy is larger. The rate law of the first step is the same as the rate law of the overall reaction. 117 NO 2(g) + CO (g)  NO (g) + CO 2(g) Rate obs = k[NO 2 ] 2 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 Tro: Chemistry: A Molecular Approach, 2/e

118. Copyright  2011 Pearson Education, Inc. Validating a Mechanism To validate (not prove) a mechanism, two conditions must be met: 1.The 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 118Tro: Chemistry: A Molecular Approach, 2/e

119. Copyright  2011 Pearson Education, Inc. Mechanisms with a Fast Initial Step When a mechanism contains a fast initial step, the rate limiting step may contain intermediates When a previous step is rapid and reaches equilibrium, the forward and reverse reaction rates are equal – so the concentrations of reactants and products of the step are related and the product is an intermediate Substituting into the rate law of the RDS will produce a rate law in terms of just reactants 119Tro: Chemistry: A Molecular Approach, 2/e

120. Copyright  2011 Pearson Education, Inc. An Example 1.2 NO (g)  N 2 O 2(g) Fast 2.H 2(g) + N 2 O 2(g)  H 2 O (g) + N 2 O (g) SlowRate = 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 120Tro: Chemistry: A Molecular Approach, 2/e

121. Copyright  2011 Pearson Education, Inc. 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 121Tro: Chemistry: A Molecular Approach, 2/e

122. Copyright  2011 Pearson Education, Inc. 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 122Tro: Chemistry: A Molecular Approach, 2/e

123. Copyright  2011 Pearson Education, Inc. 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 reaction2 A + B 2  2 AB B is an intermediate The first step is the rate determining step Rate = k[A][B 2 ], same as RDS 123Tro: Chemistry: A Molecular Approach, 2/e

124. Copyright  2011 Pearson Education, Inc. Catalysts Catalysts are substances that affect the rate of a reaction without being consumed Catalysts work by providing an alternative mechanism for the reaction with a lower activation energy 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 124Tro: Chemistry: A Molecular Approach, 2/e

125. Copyright  2011 Pearson Education, Inc. Ozone Depletion over the Antarctic 125Tro: Chemistry: A Molecular Approach, 2/e

126. Copyright  2011 Pearson Education, Inc. Molecular Interpretation of Factors Affecting Rate – Catalysts Catalysts generally speed a reaction They give the reactant molecules a different path to follow with a lower activation energy 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 126Tro: Chemistry: A Molecular Approach, 2/e

127. Copyright  2011 Pearson Education, Inc. Energy Profile of a Catalyzed Reaction polar stratospheric clouds contain ice crystals that catalyze reactions that release Cl from atmospheric chemicals 127Tro: Chemistry: A Molecular Approach, 2/e

128. Copyright  2011 Pearson Education, Inc. Practice – Mechanism 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 128Tro: Chemistry: A Molecular Approach, 2/e

129. Copyright  2011 Pearson Education, Inc. Practice – Mechanism reactionHQ 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 129Tro: Chemistry: A Molecular Approach, 2/e

130. Copyright  2011 Pearson Education, Inc. 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 130Tro: Chemistry: A Molecular Approach, 2/e

131. Copyright  2011 Pearson Education, Inc. Types of Catalysts 131Tro: Chemistry: A Molecular Approach, 2/e

132. Copyright  2011 Pearson Education, Inc. Catalytic Hydrogenation H 2 C=CH 2 + H 2 → CH 3 CH 3 132Tro: Chemistry: A Molecular Approach, 2/e

133. Copyright  2011 Pearson Education, Inc. Enzymes Because many of the 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 133 1)Enzyme + Substrate  Enzyme─SubstrateFast 2)Enzyme─Substrate → Enzyme + ProductSlow Tro: Chemistry: A Molecular Approach, 2/e

134. Copyright  2011 Pearson Education, Inc. Enzyme–Substrate Binding: the Lock and Key Mechanism 134Tro: Chemistry: A Molecular Approach, 2/e

135. Copyright  2011 Pearson Education, Inc. Enzymatic Hydrolysis of Sucrose 135Tro: Chemistry: A Molecular Approach, 2/e