Chemical kinetics is the study of the speed of chemical reactions and factors that affect the reaction rate. It provides information about reaction feasibility, timescales, and reaction mechanisms. The rate of a reaction can be examined by measuring changes in reactant or product concentrations over time. Reaction rates are determined experimentally and may follow zero-order, first-order, pseudo-first order, or second-order rate laws depending on the rate-determining step. Factors like temperature, concentration, physical state, and catalysts influence reaction rates.

1. Chemical Kinetics Study of speed with which a chemical reaction occurs and the factors affecting that speed Provides information about the feasibility of a chemical reaction Provides information about the time it takes for a chemical reaction to occur Provides information about the series of elementary steps which lead to the formation of product

2. Rate Data for A + B -> C

3. A + B -> C C A B

4. The Rate of a Chemical Reaction The speed of a reaction can be examined by the decrease in reactants or the increase in products. a A + b B -> c C + d D Where m and n are determined experimentally, and not necessarily Equal to the stiochiometry of the reaction

5. Reaction A -> 2 B A A B B B B A B B B B A A A B B B B A A A B B B B A B B B B 22 22 = 6.022 x 10 molecules B 6.022 x 10 molecules = A in a 1.00 L container in a 1.00 Liter container 1 mol/L 2 mol/L

6. Average Rate Rate of A disappearing is Let’s suppose that after 20 seconds ½ half of A disappears. Then And Rate of B appearing is Then

7. Average Rate Law for the General Equationa A + b B -> c C + d D For Example: N2O5 (g) -> 2 NO2 (g) + ½ O2 (g)

8. Determination of the Rate Equation Determined Experimentally Can be obtained by examining the initial rate after about 1% or 2% of the limiting reagent has been consumed.

9. Consider the Reaction:CH3CH2CH2CH2Cl (aq) + H2O (l) -> CH3CH2CH2CH2OH (aq) + HCl (aq)

10. Average Rates,

11. Average Rates,

12. Average Rates,

13. Average Rates,

14. Instantaneous Rate or initial rate at t=0 s Instantaneous Rate at t = 500 s

16. Order of Reaction Zero order –independent of the concentration of the reactants, e.g, depends on light First order - depends on a step in the mechanism that is unimolecular Pseudo first order reaction – one of the reactants in the rate determining step is the solvent Second order – depends on a step in the mechanism that is bimolecular Rarely third order – depends on the step in the mechanism that is termolecular

17. Data from the hydrolysis of n-butyl chloride

18. IF Zero Order

19. Therefore, the reaction is not zero order

20. If Second Order

21. Therefore, the reaction is not second order

22. IF First Order Reaction

23. First Order Plot

24. First Order Plot

25. Slope

26. Slope

27. Rate of the Reaction

28. For the ReactionN2O5 (g) -> 2 NO2 (g) + ½ O2 (g) The rate can be used to explain the mechanism

29. (1) Slow Step . (2) Fast Step .

30. Sum of the two steps: N2O5 -> 2 NO2 + ½ O2 or 2 N2O5 -> 4 NO2 + O2

31. ApplicationMechanism of a Chemical Reaction (a) Suggest a possible mechanism for NO2 (g) + CO (g) -> NO (g) + CO2 (g) Given that (b) Suggest a possible mechanism for 2 NO2 (g) + F2 (g) -> 2 NO2F (g) Given that

32. Factors Affecting the Rate of a Chemical Reaction The Physical State of Matter The Concentration of the Reactants Temperature Catalyst

33. For A Reaction to Occur Molecules Must Collide Molecules must have the Appropriate Orientation Molecules must have sufficient energy to overcome the energy barrier to the reaction- Bonds must break and bonds must form

34. A Second Order Reaction

35. Rate Constant “k” Must be determined experimentally Its value allows one to find the reaction rate for a new set of concentrations

36. The following data were collected for the rate of the reaction Between A and B, A + B -> C , at 25oC. Determine the rate law for the reaction and calculate k.

37. From Experiments 1 and 2 Solution A: Divide equation (1) into equation (2)

38. Solution B: Subtract equation (2) from equation (1)

39. From Experiments 4 and 5 Solution A: Divide equation (1) into equation (2)

40. Solution B: Subtract equation (2) from equation (1)

41. Rate Constant k

42. Rate Constant kFrom Experiment 3

43. Rate Constant kFrom Experiment 1

44. Your Understanding of this ProcessConsider the Data for the Following Reaction: Determine the Rate Law Expression and the value of k consistent With these data.

45. From Experiments 1 and 2 Solution : Divide equation (1) into equation (2)

46. From Experiments 2 and 3 Solution : Divide equation (1) into equation (2)

47. Rate Expression

48. AssignmentDetermine the Rate Law for the following reaction from the given data: 2 NO (g) + O2 (g) -> 2 NO2 (g)

49. Relationship Between Concentration and Time First Order Reaction

50. A plot of Gives a Straight line The Following Reaction is a First Order Reaction: Plot the linear graph for concentration versus time and obtain the rate constant for the reaction.

51. Data for the Transformation of cylcpropane to propene

52. slope = 2 x 10-5 s-1

53. Data for the Transformation of cylcpropane to propene

54. -slope = -2 x 10-5s-1 Slope = 2 x 10-5 s-1

55. Relationship Between Concentration and Time Second Order Reaction

56. A plot of Gives a Straight line The Following Reaction is a Second Order Reaction: 2 HI (g) -> H2 (g) + I2 (g) Plot the linear graph for concentration versus time and obtain the rate constant for the reaction.

57. Data for the Transformation of hydrogen iodide gas to hydrogen and iodine

58. slope = 30. L mol-1 min-1

59. Graphical Method for Determining the Order of a Reaction First Order Reaction: y = ln (ao – x); x = t; slope = -k; and the intercept is lnao or y = ln (ao /(ao – x)); x = t; slope = k; and the intercept =0 Second Order Reaction: y = 1/ (ao – x); x = t; slope = k; and the intercept = 1/ ao Zero Order Reaction: y = x; x = t; slope = k and the intercept = 0 or y = ao – x ; x = t; slope = -k and the intercept = ao

60. Zero Order Reaction

61. Application of the Graphical Method for Determining the Order of a Reaction N2O5 (g) -> 2 NO2 (g) + ½ O2 (g) Tabulate the data so that each order may be tested

62. Data tabulation to determine which order will give a linear graph

63. Test for zero order reaction Not linear; therefore, the reaction is not zero order

64. Test for first order reaction Linear; therefore, the reaction is first order

65. Test for second order reaction Non-linear; therefore, the reaction is not second order

66. Half Life for a First Order Reaction

67. Half Life for a Second Order Reaction

68. Half Life for a Zero Order Reaction

69. Application of Half Life The rate constant for transforming cyclopropane into propene is 0.054 h-1 Calculate the half-life of cyclopropane. Calculate the fraction of cyclopropane remaining after 18.0 hours. Calculate the fraction of cyclopropane remaining after 51.5 hours.

70. Half-Life Fraction of cyclopropane Remaining after 18.0 hours Fraction of cyclopropane Remaining after 51.5 hours

71. Effect of Temperature on the Reaction Rate Arrhenius Equation

72. Use the Following Data to Determine the Eactfor

74. Effect of a Catalyst on the Rate of a Reaction Lowers the energy barrier to the reaction via lowering the energy of activation Homogeneous catalyst- in the same phase as the reacting molecules Herterogeneous catalyst – in a different phase from the reacting molecules

75. Example of a Homogeneous Catalyst

77. Example of Heterogeneous Catalyst

78. An interesting problem: The reaction between propionaldehyde and hydrocyanic acid have been observed by Svirbely and Roth and reported in the Journal of the American Chemical Society. Use this data to ascertain the order of the reaction and the value of the rate constant for this reaction.

80. Check to determine first order in HCN

81. Check to determine first order in propionaldehyde

82. So close; therefore, let’s take another approach. Let [HCN] = ao and [propionaldehyde] = bo Then,

85. Let’s construct the data in a different format

86. Slope = 0.678; therefore, k = 0.678 M-1min-1

87. Mechanism:

88. Steps 1 and 2 are fast equilibrium steps; and step 3 is the rate determining step

91. Revisit the kinetics for 2 NO (g) + O2 (g) -> 2 NO2 (g)