1. Chemical Kinetics CHAPTER 14 Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson, Brady, & Hyslop

2. CHAPTER 14 Chemical Kinetics Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 2 Learning Objectives:  Factors Affecting Reaction Rate: o Concentration o State o Surface Area o Temperature o Catalyst  Collision Theory of Reactions and Effective Collisions  Determining Reaction Order and Rate Law from Data  Integrated Rate Laws  Rate Law  Concentration vs Rate  Integrated Rate Law  Concentration vs Time  Units of Rate Constant and Overall Reaction Order  Half Life vs Rate Constant (1 st Order)  Arrhenius Equation  Mechanisms and Rate Laws  Catalysts

3. CHAPTER 14 Chemical Kinetics Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 3 Lecture Road Map: ① Factors that affect reaction rates ② Measuring rates of reactions ③ Rate Laws ④ Collision Theory ⑤ Transition State Theory & Activation Energies ⑥ Mechanisms ⑦ Catalysts

4. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 4 Factors that Affect Reaction Rates CHAPTER 14 Chemical Kinetics

5. Kinetics: Study of factors that govern o How rapidly reactions occur and o How reactants change into products Rate of Reaction: o Speed with which reaction occurs o How quickly reactants disappear and products form Kinetics The Speed at Which Reactions Occur Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 5

6. Kinetics The Speed at Which Reactions Occur Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 6 ABAB Reaction rate is measured by the amount of product produced or reactants consumed per unit time. o [B] concentration of products will increase over time o [A] concentration of reactants will decrease over time

7. Kinetics Factors Affecting Reaction Rates Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 7 1. Chemical nature of reactants o What elements, compounds, salts are involved? o What bonds must be formed, broken? o What are fundamental differences in chemical reactivity?

8. Kinetics Factors Affecting Reaction Rates Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 8 2.Ability of reactants to come in contact (Reactants must meet in order to react) The gas or solution phase facilitates this o Reactants mix and collide with each other easily o Homogeneous reaction o All reactants in same phase o Occurs rapidly o Heterogeneous reaction o Reactants in different phases o Reactants meet only at interface between phases o Surface area determines reaction rate o Increase area, increase rate; decrease area, decrease rate

9. Kinetics Factors Affecting Reaction Rates Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 9 3. Concentrations of reactants o Rates of both homogeneous and heterogeneous reactions affected by [X ] o Collision rate between A and B increase if we increase [A] or increase [B ]. o Often (but not always) reaction rate increases as [X ] increases

10. Kinetics Factors Affecting Reaction Rates Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 10 4. Temperature o Rates are often very sensitive to temperature o Raising temperature usually makes reaction faster for two reasons: o Faster molecules collide more often and collisions have more energy o Most reactions, even exothermic reactions, require energy to occur o Rule of thumb: Rate doubles if temperature increases by 10 °C (10 K)

11. Kinetics Factors Affecting Reaction Rates Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 11 5. Presence of Catalysts o Substances that increase rates of chemical reactions without being used up o Rate-accelerating agents o Speed up rate dramatically o Rate enhancements of 10 6 not uncommon o Chemicals that participate in mechanism but are regenerated at the end

12. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 12 Measuring Reaction Rates CHAPTER 14 Chemical Kinetics

13. Rates Measuring Rate of Reaction Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 13 o Rate = ratio with time unit in denominator o Rate of Chemical Reaction o Change in concentration per unit time. o Always with respect to a given reactant or product o [reactants] decrease with time o [products] increase with time

14. Rates Measuring Rate of Reaction Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 14 o Concentration in M units o Time in s units o Units on rate: o [product] increases by 0.50 mol/L per second  rate = 0.50 M/s o [reactant] decreases by 0.20 mol/L per second  rate = 0.20 M/s

15. Rates Rate of Reaction Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 15 Always positive whether something is increasing or decreasing in [X ] o Reactants o Reactant consumed o So  [X ] is negative o Need minus sign to make rate positive o Products o Produced as reaction goes along o So  [X ] is positive o Thus rate already positive

16. Rates Measuring Rate of Reaction Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 16 Coefficients indicate the relative rates at which reactants are consumed and products are formed o Related by coefficients in balanced chemical equation o Know rate with respect to one product or reactant o Can use equation to determine rates with respect to all other products and reactants.  A +  B   C +  D

17. Rates Rate of Reaction: Example Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 17 O 2 reacts 5 times as fast as C 3 H 8 CO 2 forms 3 times faster than C 3 H 8 consumed H 2 O forms 4/5 as fast as O 2 consumed C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (g)

18. Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 18 Clorox bleach is sodium hypochlorite. It should never be mixed with acids, (like vinegar) because it forms chlorine gas: NaClO + 2 HCl → Cl 2 + H 2 O + NaCl If Chlorine gas (Cl 2 ) is formed at a rate of 5.0 x 10 -4 mol/Ls what rate is HCl consumed? HCl: Cl 2 2:1 Therefore HCl will disappear twice as fast as Cl 2 is formed. Rate HCl consumed = 10. x 10 -4 mol/Ls

19. Rates Change of Reaction Rate with Time Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 19 Generally reaction rate changes during reaction, it isn’t constant o Often initially fast when lots of reactant present o Slower and slower as reactants are depleted Why? o Rate depends on the concentration of the reactants o Reactants being used up, so the concentration of the reactants are decreasing and therefore the rate decreases Measured in 3 ways: o instantaneous rate, average rate, initial rate

20. Rates Instantaneous & Initial Reaction Rate Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 20 Instantaneous rate o Slope of tangent to curve at some specific time Initial rate o Determined at time = 0

21. Rates Average Rate of Reaction Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 21 Average Rate: Slope of line connecting starting and ending coordinates for specified time frame

22. Rates Example Reporting Different Types of Rates 22 Rate at any time t = negative slope (or tangent line) of curve at that point Concentration vs. Time Curve for 0.005M phenolphthalein reacting with 0.61 M NaOH at room temperature http://chemed.chem.purdue.edu

23. Rates Example Reporting Different Types of Rates Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 23 [P] (mol/L)Time (s) 0.0050 0.004510.5 0.00422.3 0.003535.7 0.00351.1 0.002569.3 0.00291.6 0.0015120.4 Initial rate = Average rate between first two data points

24. Rates Example Reporting Different Types of Rates Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 24 Instantaneous Rate at 120.4 s (90,0.0028) (160,0.0018)

25. Rates Example Reporting Different Types of Rates Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 25 Average Rate between 0 and 120.4 s

26. Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 26 A reaction was of NO 2 decomposition was studied. The concentration of NO 2 was found to be 0.0258 M at 5 minutes and at 10 minutes the concentration was 0.0097 M. What is the average rate of the reaction between 5 min and 10 min? A. 310 M/min B. 3.2 × 10 –3 M/min C. 2.7 × 10 –3 M/min D. 7.1 × 10 –3 M/min

27. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 27 Rate Laws CHAPTER 14 Chemical Kinetics

28. Rate Laws Rates Based on All Reactants Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 28  A +  B   C +  D o Rate Law or Rate expression o k is the rate constant o Dependent on Temperature & Solvent o m and n = exponents found experimentally o No necessary connection between stoichiometric coefficients ( ,  ) and rate exponents (m, n) o Usually small integers o Sometimes simple fractions (½, ¾) or zero = k[A] m [B] n

29. Rate Laws Rates Based on All Reactants Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 29 Below is the rate law for the reaction 2A +B → 3C rate= 0.045 M –1 s –1 [A][B] If the concentration of A is 0.2 M and that of B is 0.3 M, and the reaction is 1 st order (m & n = 1) what will be the reaction rate? rate=0.0027 M/s  0.003 M/s rate=0.045 M –1 s –1 [0.2][0.3]

30. Rate Laws Order of Reactions Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 30 Rate = k[A] m [B] n Exponents specify the order of reaction with respect to each reactant Order of Reaction o m = 1 [A] 1 1 st order in [A] o m = 2 [A] 2 2 nd order in [A] o m = 3 [A] 3 3 rd order in [A] o m = 0 [A] 0 0 th order in [A] [A] 0 = 1  means A doesn't affect rate Overall order of reaction = sum of orders (m and n) of each reactant in rate law

31. Rate Laws Order of Reactions: Example Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 31 5Br – + BrO 3 – + 6H +  3Br 2 + 3H 2 O x = 1y = 1z = 2 o 1 st order in [BrO 3 – ] o 1 st order in [Br – ] o 2 nd order in [H + ] o Overall order = 1 + 1 + 2 = 4

32. Rate Laws Order of Reaction & Units for k Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 32

33. Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 33 The following rate law has been observed: Rate = k[H 2 SeO][I – ] 3 [H + ] 2. The rate with respect to I – and the overall reaction rate is: A. 6, 2 B. 2, 3 C. 1, 6 D. 3, 6

34. Rate Laws Calculating k Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 34 If we know rate and concentrations, can use rate law to calculate k From Text Example of decomposition of HI at 508 °C Rate= 2.5 × 10 –4 M/s [HI] = 0.0558 M

35. Rate Laws Determining Exponents in Rate Law Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 35 Experimental Determination of Exponents o Method of initial rates o If reaction is sufficiently slow o or have very fast technique o Can measure [A] vs. time at very beginning of reaction o Before it slows very much, then o Set up series of experiments, where initial concentrations vary

36. Rate Laws Determining Rate Law Exponents: Example Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 36 3A + 2B  products Rate = k[A] m [B] n Expt. #[A] 0, M[B] 0, MInitial Rate, M/s 10.10 1.2  10 –4 20.200.10 4.8  10 –4 30.20 4.8  10 –4  Convenient to set up experiments so  The concentration of one species is doubled or tripled  And the concentration of all other species are held constant  Tells us effect of [varied species] on initial rate

37. Rate Laws Determining Rate Law Exponents Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 37 If reaction is 1 st order in [X], –Doubling [X] 1  2 1 –Doubles the rate If reaction is 2 nd order in [X], –Doubling [X] 2  2 2 –Quadruples the rate If reaction is 0 th order in [X], –Doubling [X] 0  2 0 –Rate doesn't change If reaction is n th order in [X] –Doubling [X] n  2 n times the initial rate

38. Rate Laws Determining Rate Law Exponents: Example Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 38 Expt. #[A] 0, M[B] 0, MInitial Rate, M/s 10.10 1.2  10  4 20.200.10 4.8  10  4 30.20 4.8  10  4 Comparing Expt. 1 and 2 o Doubling [A] o Quadruples rate o Reaction 2 nd order in A = [A] 2 2 m = 4 or m = 2

39. Rate Laws Determining Rate Law Exponents: Example Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 39 Expt. #[A] 0, M[B] 0, MInitial Rate, M/s 10.10 1.2  10  4 20.200.10 4.8  10  4 30.20 4.8  10  4 Comparing Expt. 2 and 3 o Doubling [B] o Rate does not change o Reaction 0 th order in B = [B] 0 = 1 2 n = 1 or n = 0

40. Rate Laws Determining Rate Law Exponents: Example Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 40 o Conclusion: rate = k[A] 2 o Can use data from any experiment to determine k o Let’s choose first experiment Expt. #[A] 0, M[B] 0, MInitial Rate, M/s 10.10 1.2  10 –4 20.200.10 4.8  10 –4 30.20 4.8  10 –4

41. Rate Laws Determining Rate Law Exponents: Ex 2 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 41 2 SO 2 + O 2  2 SO 3 Rate = k[SO 2 ] m [O 2 ] n Expt # [SO 2 ] M[O 2 ] M Initial Rate of SO 3 Formation, M s –1 10.250.30 2.5  10  3 20.500.30 1.0  10  2 30.750.60 4.5  10  2 40.500.90 3.0  10  2

42. Rate Laws Determining Rate Law Exponents: Ex 2 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 42 4 = 2 m or m = 2

43. Rate Laws Determining Rate Law Exponents: Ex 2 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 43 3 = 3 n or n = 1

44. Rate Laws Determining Rate Law Exponents: Ex 2 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 44 Rate = k[SO 2 ] 2 [O 2 ] 1 1 st order in [O 2 ] 2 nd order in [SO 2 ] 3 rd order overall Can use any experiment to find k

45. Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 45 Using the following experimental data, determine the order with respect to NO and O 2. A.2, 0 B.3,1 C.2, 1 D. 1, 1 Expt # [NO] M [O 2 ] M Initial Rate M s –1 10.120.25 1.5  10 –3 20.240.25 6.0  10 –3 30.240.50 1.2  10 –2

46. Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 46

47. Example : Method of Initial Rates BrO 3 – + 5Br – + 6H +  3Br 2 + 3H 2 O Expt # [BrO 3 – ] mol/L [Br – ] mol/L [H + ] mol/L Initial Rate mol/(L s) 10.10 8.0  10 –4 20.200.10 1.6  10 –3 30.20 0.10 3.2  10 –3 40.10 0.20 3.2  10 –3 47

48. Ex.: Method of Initial Rates 48 Compare 2 and 3 Compare 1 and 2

49. Ex.: Method of Initial Rates First order in [BrO 3 – ] and [Br – ] Second order in [H + ] Overall order = m + n + p = 1 + 1 + 2 = 4 Rate Law is:Rate = k[BrO 3 – ][Br – ][H + ] 2 49 Compare 1 and 4

50. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 50 Integrated Rate Laws CHAPTER 14 Chemical Kinetics

51. Integrated Rate Laws Concentration & Time Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 51 Rate law tells us how speed of reaction varies with concentrations. Sometimes want to know o Concentrations of reactants and products at given time during reaction o How long for the concentration of reactants to drop below some minimum optimal value  Need dependence of rate on time

52. Integrated Rate Laws First Order Integrated Rate Law Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 52 Corresponding to reactions –A  products Integrating we get Rearranging gives Equation of line y = mx + b

53. Integrated Rate Laws First Order Integrated Rate Law Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 53 Yields straight line o Indicative of first order kinetics o Slope = –k o Intercept = ln [A] 0 o If we don't know already Slope = –k

54. Integrated Rate Laws 2 nd Order Integrated Rate Law Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 54 Corresponding to special second order reaction –2B  products Integrating we get Rearranging gives Equation of line y = mx + b

55. Integrated Rate Laws 2 nd Order Integrated Rate Law Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 55 Yields straight line o Indicative of 2 nd order kinetics o Slope = +k o Intercept = 1/[B] 0 Slope = +k

56. Integrated Rate Laws Graphically determining Order Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 56 Make two plots: 1. ln [A] vs. time 2. 1/[A] vs. time o If ln [A] is linear and 1/[A] is curved, then reaction is 1 st order in [A] o If 1/[A] plot is linear and ln [A] is curved, then reaction is 2 nd order in [A] o If both plots give horizontal lines, then 0 th order in [A]

57. Integrated Rate Laws Graphically determining Order Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 57

58. Integrated Rate Laws Graphically determining Order Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 58 Add examples of graphs

59. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 59 Time, min[SO 2 Cl 2 ], Mln[SO 2 Cl 2 ]1/[SO 2 Cl 2 ] (L/mol) 00.1000-2.302610.000 1000.0876-2.435011.416 2000.0768-2.566613.021 3000.0673-2.698614.859 4000.0590-2.830216.949 5000.0517-2.962319.342 6000.0453-3.094422.075 7000.0397-3.226425.189 8000.0348-3.358128.736 9000.0305-3.490032.787 10000.0267-3.623137.453 11000.0234-3.755042.735 Example: SO 2 Cl 2  SO 2 + Cl 2 Integrated Rate Laws

60. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 60 Reaction is 1 st order in SO 2 Cl 2 Example: SO 2 Cl 2  SO 2 + Cl 2 Integrated Rate Laws

61. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 61 Example: HI(g)  H 2 (g) + I 2 (g) Integrated Rate Laws Time (s) [HI] (mol/L) ln[HI] 1/[HI] (L/mol) 00.1000-2.302610.000 500.0716-2.636713.9665 1000.0558-2.886017.9211 1500.0457-3.085721.8818 2000.0387-3.251925.840 2500.0336-3.393229.7619 3000.0296-3.520033.7838 3500.0265-3.630637.7358

62. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 62 Example: HI(g)  H 2 (g) + I 2 (g) Integrated Rate Laws Reaction is second order in HI.

63. Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 63 A plot for a zeroth order reaction is shown. What is the proper label for the y-axis in the plot ? A. Concentration B. ln of Concentration C. 1/Concentration D. 1/ ln Concentration

64. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 64 Half Life (t 1/2 ) for first order reactions Integrated Rate Laws Half-life = t ½ We often use the half life to describe how fast a reaction takes place First Order Reactions o Set o Substituting into o Gives o Canceling gives ln 2 = kt ½ o Rearranging gives

65. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 65 Half Life (t 1/2 ) for First Order Reactions Integrated Rate Laws Observe: 1. t ½ is independent of [A] o o For given reaction (and T) o Takes same time for concentration to fall from o 2 M to 1 M as from o 5.0  10 –3 M to 2.5  10 –3 M 2. k 1 has units (time) –1, so t ½ has units (time) o t ½ called half-life o Time for ½ of sample to decay

66. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 66 Half Life (t 1/2 ) Integrated Rate Laws Does this mean that all of sample is gone in two half-lives (2 × t ½ )? No! o In 1 st t ½, it goes to ½[A] o o In 2 nd t ½, it goes to ½(½[A] o ) = ¼[A] o o In 3 rd t ½, it goes to ½(¼[A] o ) = ⅛[A] o o In n th t ½, it goes to [A] o /2 n

67. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 67 Half Life (t 1/2 ) Integrated Rate Laws

68. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 68 Half Life (t 1/2 ): First Order Example Integrated Rate Laws 131 I is used as a metabolic tracer in hospitals. It has a half-life, t ½ = 8.07 days. How long before the activity falls to 1% of the initial value?

69. Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 69 The radioactive decay of a new atom occurs so that after 21 days, the original amount is reduced to 33%. What is the rate constant for the reaction in s –1 ? k = 6.11 × 10 –7 s –1 k = 0.0528 day –1

70. Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 70 The half-life of I-132 is 2.295 h. What percentage remains after 24 hours? 0.302 h –1 = k A = 0.0711% A o

71. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 71 Half Life (t 1/2 ): Carbon-14 Dating Integrated Rate Laws

72. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 72 Half Life (t 1/2 ): Second Order Reactions Integrated Rate Laws How long before [A] = ½[A] o ? o t ½, depends on [A] o o t ½, not useful quantity for a second order reaction

73. Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 73 The rate constant for the second order reaction 2A → B is 5.3 × 10 –5 M –1 s –1. What is the original amount present if, after 2 hours, there is 0.35 M available? A 0 =0.40 M

74. Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 74 Add better rate law problem

75. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 75 Collision Theory CHAPTER 14 Chemical Kinetics

76. Collision Theory Reaction Rates Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 76 Collision Theory As the concentration of reactants increase o The number of collisions increases o Reaction rate increases As temperature increases o Molecular speed increases o Higher proportion of collisions with enough force (energy) o There are more collisions per second o Reaction rate increases

77. Collision Theory Reaction Rates Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 77 Rate of reaction proportional to number of effective collisions/sec among reactant molecules Effective collision o One that gives rise to product e.g. At room temperature and pressure o H 2 and I 2 molecules undergoing 10 10 collisions/sec o Yet reaction takes a long time o Not all collisions lead to reaction Only very small percentage of all collisions lead to net change

78. Collision Theory Molecular Orientation Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 78

79. Collision Theory Temperature Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 79 As T increases o More molecules have E a o So more molecules undergo reaction

80. Collision Theory Activation Energy (E a ) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 80 Molecules must possess certain amount of kinetic energy (KE) in order to react Activation Energy, E a = Minimum KE needed for reaction to occur o Get energy from collision with other molecules o If molecules move too slowly, too little KE, they just bounce off each other o Without this minimum amount, reaction will not occur even when correctly oriented

81. Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 81 Summerize with a pretty picture

82. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 82 Transition State Theory CHAPTER 14 Chemical Kinetics

83. Transition State Molecular Basis of Transition State Theory Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 83 KE decreasing as PE increases KE PE KE Is the combined KE of both molecules enough to overcome Activation Energy

84. Transition State Molecular Basis of Transition State Theory Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 84 Reaction Coordinate (progress of reaction) Potential Energy Activation energy (E a ) = hill or barrier between reactants and products Heat of reaction (  H) = difference in PE between products and reactants  H reaction = H products – H reactants Products

85. Transition State Exothermic Reactions Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 85 Reaction Coordinate (progress of reaction) Potential Energy Exothermic reaction Products lower PE than reactants Exothermic Reaction  H = – Products

86. Transition State Exothermic Reactions Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 86 o  H reaction < 0 (negative) o Decrease in PE of system o Appears as increase in KE o So the temperature of the system increases o Reaction gives off heat o Can’t say anything about E a from size of  H o E a could be high and reaction slow even if  H rxn large and negative o E a could be low and reaction rapid

87. Transition State Endothermic Reactions Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 87 Endothermic Reaction  H = +  H reaction = H products – H reactants

88. Transition State Endothermic Reactions Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 88 o  H reaction > 0 (positive) o Increase in PE o Appears as decrease in KE o So temperature of the system decreases o Have to add E to get reaction to go o E a   H rxn as E a includes  H rxn o If  H rxn large and positive o E a must be high o Reaction very slow

89. Transition State Activated Complex Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 89 o Arrangement of atoms at top of activation barrier o Brief moment during successful collision when o bond to be broken is partially broken and o bond to be formed is partially formed Example Transition State

90. Example Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 90 Develop example that is not NO2Cl + Cl

91. Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 91 Draw the transition state complex, or the activated complex for the following reaction: CH 3 CH 2 O - +H 3 O +  CH 3 CH 2 OH +H 2 O

92. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 92 Activation Energies CHAPTER 14 Chemical Kinetics

93. EaEa EaEa Arrhenius Equation Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 93 The rate constant is dependent on Temperature, which allows us to calculate Activation Energy, E a Arrhenius Equation: Equation expressing temperature-dependence of k o A = Frequency factor has same units as k o R = gas constant in energy units = 8.314 J mol –1 K –1 o E a = Activation Energy—has units of J/mol o T = Temperature in K

94. EaEa EaEa Calculating Activation Energy Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 94 Method 1. Graphically Take natural logarithm of both sides Rearranging Equation for a line y = b + mx Arrhenius Plot Plot ln k (y axis) vs. 1/T (x axis)  yield a straight line Slope = -E a /R Intercept = A

95. EaEa EaEa Arrhenius Equation: Graphing Example Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 95 Given the following data, predict k at 75 ˚C using the graphical approach k (M/s)T, ˚CT, K 0.00088625298 0.00089450348 0.000908100398 0.000918150448 ?75348 ln (k) = –36.025/T – 6.908 ln (k) = –36.025/(348) – 6.908 = – 7.011

96. EaEa EaEa Arrhenius Equation: Graphing Example Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 96

97. EaEa EaEa Arrhenius Equation Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 97 Sometimes a graph is not needed o Only have two k s at two Ts Here use van't Hoff Equation derived from Arrhenius equation:

98. EaEa EaEa Arrhenius Equation: Ex Vant Hoff Equation Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 98 CH 4 + 2 S 2  CS 2 + 2 H 2 S k (L/mol s)T (˚C)T (K) 1.1 = k 1 550823 = T 1 6.4 = k 2 625898 = T 2

99. Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 99 Given that k at 25 ˚C is 4.61 × 10 –1 M/s and that at 50 ˚C it is 4.64 × 10 –1 M/s, what is the activation energy for the reaction? E a = 208 J/mol

100. Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 100 A reaction has an activation energy of 40 kJ/mol. What happens to the rate if you increase the temperature from 70˚C to 80 ˚C? A. Rate increases approximately 1.5 times B. Rate increases approximately 5000 times C. Rate does not increase D. Rate increases approximately 3 times

101. Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 101 A reaction has an activation energy of 40 kJ/mol. What happens to the rate if you increase the temperature from 70˚C to 80 ˚C?

102. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 102 Mechanisms of Reactions CHAPTER 14 Chemical Kinetics

103. Mechanisms Overall vs Individual Steps Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 103 Sometimes rate law has simple form – N 2 O 5  NO 2 + NO 3 – NO 2 + NO 3  N 2 O 5 But others are complex – H 2 + Br 2  2 HBr

104. Mechanisms Overall vs Individual Steps Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 104 Some reactions occur in a single step, as written Others involve a sequence of steps o Reaction Mechanism o Entire sequence of steps o Elementary Process o Each individual step in mechanism o Single step that occurs as written

105. Mechanisms Overall vs Individual Steps Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 105 o Exponents in rate law for elementary process are equal to coefficients of reactants in balanced chemical equation for that elementary process o Rate laws for elementary processes are directly related to stoichiometry o Number of molecules that participate in elementary process defines molecularity of step

106. Mechanisms Unimolecular Process Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 106 o Only one molecule as reactant o H 3 C—N  C  H 3 C—C  N o Rate = k[CH 3 NC] o 1 st order overall o As number of molecules increases, number that rearrange in given time interval increases proportionally

107. Mechanisms Bimolecular Process Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 107 o Elementary step with two reactants o NO(g) + O 3 (g)  NO 2 (g) + O 2 (g) o Rate = k[NO][O 3 ] o 2 nd order overall o From collision theory: o If [A] doubles, number of collisions between A and B will double o If [B] doubles, number of collisions between A and B will double o Thus, process is 1 st order in A, 1 st order in B, and 2 nd order overall

108. Mechanisms Termolecular Process Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 108 o Elementary reaction with three molecules o Extremely rare o Why? o Very low probability that three molecules will collide simultaneously o 3 rd order overall

109. Mechanisms Elementary Processes Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 109 MolecularityElementary StepRate Law Unimolecular A  products Rate = k[A] Bimolecular A + A  products Rate = k[A] 2 Bimolecular A + B  products Rate = k[A][B] Significance of elementary steps: o If we know that reaction is elementary step o Then we know its rate law

110. Mechanisms Multi-step Mechanisms Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 110 o Contains two or more steps to yield net reaction o Elementary processes in multi-step mechanism must always add up to give chemical equation of overall process o Any mechanism we propose must be consistent with experimentally observed rate law o Intermediate = species which are formed in one step and used up in subsequent steps o Species which are neither reactant nor product in overall reaction o Mechanisms may involve one or more intermediates

111. Mechanisms Example Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 111 The net reaction is: NO 2 (g) + CO(g)  NO(g) + CO 2 (g) The proposed mechanism is: NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) 2NO 2 (g) + NO 3 (g) + CO(g)  NO 2 (g) + NO 3 (g) + NO(g) + CO 2 (g) or NO 2 (g) + CO(g)  NO(g) + CO 2 (g) 1

112. Mechanisms Rate Determining Step Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 112 o If process follows sequence of steps, slow step determines rate = rate determining step. o Think of an assembly line o Fast earlier steps may cause intermediates to pile up o Fast later steps may have to wait for slower initial steps o Rate-determining step governs rate law for overall reaction o Can only measure rate up to rate determining step

113. Mechanisms Example: Rate Determining Step Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 113 (CH 3 ) 3 CCl(aq) + OH – (aq)  (CH 3 ) 3 COH(aq) + Cl – (aq) chlorotrimethylmethane trimethylmethanol o Observed rate = k[(CH 3 ) 3 CCl] o If reaction was elementary o Rate would depend on both reactants o Frequency of collisions depends on both concentrations o  Mechanism is more complex than single step o What is mechanism? o Evidence that it is a two step process

114. Mechanisms Rate Determining Step as Initial Step Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 114 Step 1: (CH 3 ) 3 CCl(aq)  (CH 3 ) 3 C + (aq) + Cl – (aq) (slow) Step 2: (CH 3 ) 3 C + (aq) + OH – (aq)  (CH 3 ) 3 COH(aq) (fast) o Two steps each at different rates o Each step in multiple step mechanism is elementary process, so o Has its own rate constant and its own rate law o Hence only for each step can we write rate law directly o Observed rate law says that step 1 is very slow compared to step 2 o In this case step 1 is rate determining o Overall rate = k 1 [(CH 3 ) 3 CCl]

115. Mechanisms Mechanisms with Fast Initial Step Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 115 1 st step involves fast, reversible reaction Ex. Decomposition of ozone (No catalysts) Net reaction: 2O 3 (g)  3O 2 (g) Proposed mechanism: O 3 (g)  O 2 (g) + O(g) (fast) O(g) + O 3 (g)  2O 2 (g)(slow)

116. Mechanisms Is the Mechanism Rate Law Consistent? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 116 o Rate of formation of O 2 = Rate of reaction 2 = k 2 [O][O 3 ] o But O is intermediate o Need rate law in terms of reactants and products o and possibly catalysts o Rate (forward) = k f [O 3 ] o Rate (reverse) = k r [O 2 ][O] o When step 1 comes to equilibrium o Rate (forward) = Rate (reverse) o k f [O 3 ] = k r [O 2 ][O]

117. Mechanisms Is the Mechanism Rate Law Consistent? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 117 o Solving this for intermediate O gives: o Substitution into rate law for step 2 gives: o Rate of reaction 2 = k 2 [O][O 3 ] = o where o This is observed rate law o Yes, mechanism consistent

118. Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 118 The reaction mechanism that has been proposed for the decomposition of H 2 O 2 is 1. H 2 O 2 + I – → H 2 O + IO – (slow) 2. H 2 O 2 + IO – → H 2 O + O 2 + I – (fast) What is the expected rate law? First step is slow so the rate determining step defines the rate law rate=k [H 2 O 2 ][I – ]

119. Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 119 The reaction: A + 3B → D + F was studied and the following mechanism was finally determined: 1. A + B  C (fast) 2. C + B → D + E (slow) 3. E + B → F (very fast) What is the expected rate law? Rate Step 2=k 2 [C][B] Rate forward = k f [A][B] Rate reverse = k r [C] k f [A][B] = k r [C] [C]= k f [A][B]/k r Rate = k obs [A][B] 2

120. Terminal Reaction for Superoxide Radical 120 Miller, Anne-Frances. “Fe Superoxide Dismutase” Handbook of Metalloproteins. John Wiley & Sons, Ltd, Chinchester, 2001 Rodrigues, J. V; Abreu, I. A.; Cabelli, D; Teixeira, M. Biochemistry 2006, 45, 9266-9278. Catalyst

121. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 121 Catalysts CHAPTER 14 Chemical Kinetics

122. Catalyst Definition Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 122 o Substance that changes rate of chemical reaction without itself being used up o Speeds up reaction, but not consumed by reaction o Appears in mechanism, but not in overall reaction o Does not undergo permanent chemical change o Regenerated at end of reaction mechanism o May appear in rate law o May be heterogeneous or homogeneous

123. Catalyst Activation Energy Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 123 o By providing alternate mechanism o One with lower E a o Because E a lower, more reactants and collisions have minimum KE, so reaction proceeds faster

124. Catalyst Activation Energy Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 124

125. Catalyst Homogeneous Catalyst Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 125 Same phase as reactants Consider : S(g) + O 2 (g) + H 2 O(g)  H 2 SO 4 (g) S(g) + O 2 (g)  SO 2 (g) NO 2 (g) + SO 2 (g)  NO(g) + SO 3 (g) Catalytic pathway SO 3 (g) + H 2 O(g)  H 2 SO 4 (g) NO(g) + ½O 2 (g)  NO 2 (g) Regeneration of catalyst  Net: S(g) + O 2 (g) + H 2 O(g)  H 2 SO 4 (g) What is Catalyst? –Reactant (used up) in early step –Product (regenerated) in later step Which are Intermediates?

126. Catalyst Heterogeneous Catalyst Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 126 o Exists in separate phase from reactants o Usually a solid o Many industrial catalysts are heterogeneous o Reaction takes place on solid catalyst Ex. 3H 2 (g) + N 2 (g)  2NH 3 (g)

127. Catalyst Heterogeneous Catalyst Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 127 H 2 and N 2 approach Fe catalyst H 2 and N 2 bind to Fe & bonds break N—H bonds forming NH 3 formation complete NH 3 dissociates

128. Catalyst Methane Oxidation to Methanol CH 4  CH 3 OH o C-H has a high bond strength (~ 410 kJ/mol) o Thermodynamically favorable to oxidize CH 4 to CO 2 o Selective catalysts can stop the oxidation at methanol GOAL: Develop a selective and robust catalyst that produces high yields of methanol at low temperatures and pressures o Syngas synthesis of methanol (CO + H 2 ): o High Pressure with ZnO/Cr2O3 catalyst o Cu-Zeolite catalyst o Shilov cycle with a platinum catalyst

129. Catalyst Methane Monooxygenase CH 4  CH 3 OH o Oxidoreductase Enzyme found in methanotrophic bacteria that help cycle carbon in anarobic sediments. o Bioinorganic chemistry tries to replicate the highly specialized chemistry found in nature using smaller molecules that can be synthesized in a lab.

130. Splitting Water Catalyst Requirments: Very Endothermic o Need a minimum of 1.23 V to split water o Kinetically infrared light could do this, but the reaction is very slow o The potential really needs to be at least 3.0 V to utilize the full spectrum of light

131. Catalyst Photosystem II PQ + H2O --> PQH2 + O2 (g) The overall reaction of Photosystem II is the oxidation of water and the reduction of plastoquinone.

132. Splitting Water Catalyst Increase efficiency and decrease activation energy of electrolysis of water with a catalyst that will work at room temperature: Co3+ HPO4-