1. Kinetics Reaction Rates

2. Kinetics The study of reaction rates. Thermodynamics will tell us whether or not the reaction will occur Spontaneous reactions (ΔG = neg) are reactions that will happen - but we can’t tell how fast. Diamond will spontaneously turn to graphite – eventually. Kinetics is about how fast

3. Review- Collision Model of Chemical Reactions Particles must collide with enough energy to break bonds EFFECTIVE COLLISIONS

4. Factors that affect the rate of reaction Increase number of effective collisions  increase rate of reaction Temperature Concentration Surface area Pressure Agitation Catalyst

5. Reaction Rate Rate = Conc. of A at t 2 -Conc. of A at t 1 t 2 - t 1 Rate =  [A] Dt Change in concentration per unit time Consider the reaction N 2 (g) + 3H 2 (g)  2NH 3 (g)

6. As the reaction progresses the concentration H 2 goes down ConcentrationConcentration Time [H 2 ] N 2 + 3H 2 → 2NH 3 Δ[H 2 ]/time = neg

7. As the reaction progresses the concentration N 2 goes down 1/3 as fast Δ[H 2 ]/time = neg ConcentrationConcentration Time [H 2 ] [N 2 ] N 2 + 3H 2 → 2NH 3 Δ[N 2 ]/time = neg Δ[H 2 ]/time = neg

8. As the reaction progresses the concentration NH 3 goes up 2/3 times ConcentrationConcentration Time [H 2 ] [N 2 ] [NH 3 ] N 2 + 3H 2 → 2NH 3 Δ[NH 3 ]/time = pos

9. Reaction Rates are always positive Over time –the concentration of products increases –the concentration of reactants decreases So Δ[P]/time is positive But Δ[R]/time is negative, so the rate would be expressed as - Δ[R]/time

10. If rate is measured in terms of concentration of Hydrogen -Δ[H 2 ] - mol H 2 time L time and N 2 + 3H 2  2NH 3 Then using the stoichiometry of the reaction -Δmol H 2 1 mol N 2 L time 3 mol H 2 Then -Δmol H 2 2 mol NH 3 L time 3 mol H 2

11. Calculating Rates Average rates are taken over a period of time interval Instantaneous rates are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time

12. Average Rate – Find the slope of the line between the two points Note: slope = ConcentrationConcentration Time D[H 2 ] DtDtDtDt Note: slope = Δy / Δx Δconcentration / Δtime THE RATE

13. Instantaneous Rate – Find the slope of the line tangent to that point ConcentrationConcentration Time  [H 2 ] D t D t d[H 2 ] dt dt

14. INSTANTANEOUS RATES 2NO 2  2NO + O 2

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

18. Rate Laws Reactions are reversible. As products accumulate they can begin to turn back into reactants. Early on the rate will depend on only the amount of reactants present. We want to measure the reactants as soon as they are mixed. In addition, as the reaction progresses the concentration of reactants decrease This is called the initial rate method.

19. Rate laws show the effect of concentration on the rate of the reaction The concentration of the products do not appear in the rate law because this is an initial rate. The order (exponent) –must be determined experimentally, –cannot be obtained from the equation Rate Laws

20. You will find that the rate will only depend on the concentration of the reactants. (Initially) Rate = k[NO 2 ] n This is called a rate law expression. k is called the rate constant. n is the order of the reactant -usually a positive integer. 2 NO 2 2 NO + O 2

21. NO 2  NO + ½ O 2 The data for the reaction at 300ºC Plot the data to show concentration as a function of time. Time (s) [NO 2 ] (mol/L) 0.00 0.01000 50.0 0.00787 100.0 0.00649 200.0 0.00481 300.0 0.00380

22. NO 2  NO + ½ O 2 To determine the rate law for the reaction, we must find the exponent in the equation. Rate = k [NO 2 ] x Experiment [NO 2 ] MRate, M s -1 10.0600.0360 20.0300.0090 30.0200.0040

23. NO 2  NO + ½ O 2 To determine the rate law for the reaction, we must find the exponent in the equation. Rate = k [NO 2 ] x Exp [NO 2 ] MRate, M s -1 10.0600.0360 20.0300.0090 30.0200.0040 2x [NO 2 ]4x rate

24. NO 2  NO + ½ O 2 The exponent for the concentration of NO 2 must be 2 to show the effect of changing the concentration of NO 2 on the rate. Rate = k [NO 2 ] 2 Exp [NO 2 ] MRate, M s -1 10.0600.0360 20.0300.0090 30.0200.0040 3x [NO 2 ]9x rate

25. NO 2  NO + ½ O 2 Rate = k [NO 2 ] x Pick two experiments to plug into the equation. Experiment [NO 2 ] MRate, M s -1 10.0600.0360 20.0300.0090 30.0200.0040

26. NO 2  NO + ½ O 2 Use the rate and one of the experiments to calculate the rate constant. Experiment [NO 2 ] MRate, M s -1 10.0600.0360 20.0300.0090 30.0200.0040

27. 2ClO 2 + 2OH -  ClO 3 - + ClO 2 - + H 2 O Determine the rate law for the reaction. Calculate the rate constant. What is the overall reaction order? Experiment[ClO 2 - ] M[OH - ] MRate, M s -1 10.0600.0300.0248 20.0200.0300.00276 30.0200.0900.00828

28. Types of Rate Laws Differential Rate Law - describes how rate depends on concentration. Method to determine- change initial concentration and measure effect on rate Integrated Rate Law - describes how concentration depends on time. Method to determine- measure the concentration of reactants as function of time For each type of differential rate law there is an integrated rate law and vice versa.

29. Determining Rate Laws The first step is to determine the form of the rate law (especially its order). Must be determined from experimental data. For this reaction 2 N 2 O 5 (aq) 4NO 2 (aq) + O 2 (g) The reverse reaction won’t play a role because the gas leaves

30. [N 2 O 5 ] (mol/L) Time (s) 1.000 0.88200 0.78400 0.69600 0.61800 0.541000 0.481200 0.431400 0.381600 0.341800 0.302000 Now graph the data

31. To find rate we have to find the slope at two points We will use the tangent method. [N 2 O 5 ] (mol/L) Time (s)

32. At.80 M the rate is (.88 -.68) = 0.20 =- 5.0x 10 -4 (200 - 600) -400

33. At.40 M the rate is (.52 -.32) = 0.20 =- 2.5 x 10 -4 (1000-1800) -800

34. At 0.8 M rate is 5.0 x 10 -4 At 0.4 M rate is 2.5 x 10 -4 Since the rate is twice as fast when the concentration is twice as big the rate law must be.. First power Rate = -D[N 2 O 5 ] = k[N 2 O 5 ] 1 = k[N 2 O 5 ] Dt We say this reaction is first order in N 2 O 5 The only way to determine order is to run the experiment.

35. The method of Initial Rates This method requires that a reaction be run several times. The initial concentrations of the reactants are varied. The reaction rate is measured just after the reactants are mixed. Eliminates the effect of the reverse reaction.

36. An example For the reaction BrO 3 - + 5 Br - + 6H +  3Br 2 + 3 H 2 O The general form of the Rate Law is Rate = k[BrO 3 - ] n [Br - ] m [H + ] p We use experimental data to determine the values of n,m,and p

37. Initial concentrations (M) Rate (M/s) BrO 3 - Br - H+H+H+H+ 0.100.100.108.0 x 10 -4 0.100.100.108.0 x 10 -4 0.200.100.101.6 x 10 -3 0.200.100.101.6 x 10 -3 0.200.200.103.2 x 10 -3 0.200.200.103.2 x 10 -3 0.100.100.203.2 x 10 -3 0.100.100.203.2 x 10 -3 Calculate the rate law and k

38. Integrated Rate Law Expresses the reaction concentration as a function of time. Form of the equation depends on the order of the rate law (differential). Changes Rate =  [A] n  t We will only work with n = 0, 1, and 2

39. First Order For the reaction 2N 2 O 5  4NO 2 + O 2 We found the Rate = k[N 2 O 5 ] 1 If concentration doubles rate doubles. If we integrate this equation with respect to time we get the Integrated Rate Law [N 2 O 5 ] 0 = initial concentration of N 2 O 5 [N 2 O 5 ] t = concentration of N 2 O 5 after some time (t) l n = natural log = - kt l n ( ) [N 2 O 5 ] 0 [N 2 O 5 ] t

40. First Order rate = k[N 2 O 5 ] For the reaction 2N 2 O 5  4NO 2 + O 2 Rearrange this equation to y = mx + b form –Since- log of quotient equals difference of logs SO. l n[N 2 O 5 ] t = - k t + l n[N 2 O 5 ] 0 - l n[N 2 O 5 ] t = l n ( ) [N 2 O 5 ] 0 [N 2 O 5 ] t = kt l n ( ) [N 2 O 5 ] 0 [N 2 O 5 ] t l n[N 2 O 5 ] t y mx b + =

41. First Order Linear form integrated rate law l n[N 2 O 5 ] t = - k t + l n[N 2 O 5 ] 0 = kt l n ( ) [N 2 O 5 ] 0 [N 2 O 5 ] t ymxmxb + = l n[N 2 O 5 ] t t m = slope = -k

42. Graphing natural log of concentration of reactant as a function of time If you get a straight line, you may determine it is first order Integrated First Order First Order = - kt l n ( ) [R] 0 [R] t

43. Half Life The time required to reach half the original concentration. [R] t = [R] 0 /2 when t = t 1/2 If the reaction is first order = - kt l n ( ) [R] 0 [R] t Since [R] t = [R] 0 /2 [R] t = [R] 0 equals 2 l n (2) = kt 1/2

44. Half Life t 1/2 = 0.693 / k The time to reach half the original concentration does not depend on the starting concentration. An easy way to find k

45. The highly radioactive plutonium in nuclear waste undergoes first-order decay with a half-life of approximately 24,000 years. How many years must pass before the level of radioactivity due to the plutonium falls to 1/128th (about 1%) of its original potency?

46. Second Order Rate = -  [R]/  t = k[R] 2 integrated rate law 1/[R] t = kt + 1/[A] 0 y= 1/[A] m = k x= tb = 1/[A] 0 A straight line if 1/[A] vs t is graphed Knowing k and [A] 0 you can calculate [A] at any time t

47. Second Order Rate = k [R] 2 integrated rate law 1/[R] t = k t + 1/[A] 0 y= m x + b t 1/[R] Slope = k

48. Second Order Half Life [A] = [A] 0 /2 at t = t 1/2

49. Zero Order Rate Law Rate = k[A] 0 = k Rate does not change with concentration. Integrated [A] = -kt + [A] 0 When [A] = [A] 0 /2 t = t 1/2 t 1/2 = [A] 0 /2k

50. Most often when reaction happens on a surface because the surface area stays constant. Also applies to enzyme chemistry. Zero Order Rate Law

51. Time ConcentrationConcentration

52. ConcentrationConcentration  A]/  t tt k =  A]

53. Zero orderFirst orderSecond order Rate law Graph [A] vs time Integrated rate law

54. Zero orderFirst orderSecond order Linear graph with time Integrated rate law Rearrange to y=mx+b

55. Summary of Rate Laws

56. Reaction Mechanisms The series of steps that actually occur in a chemical reaction. Kinetics can tell us something about the mechanism A balanced equation does not tell us how the reactants become products.

57. 2NO 2 + F 2 2NO 2 F Rate = k[NO 2 ][F 2 ] The overall equation is a summary. In order for this reaction to occur as written, three molecules must collide simultaneously An unlikely event. So a reaction mechanism is proposed showing a series of likely steps Reaction Mechanisms

58. 2NO 2 + F 2 2NO 2 F Rate = k[NO 2 ][F 2 ] The proposed mechanism is NO 2 + F 2  NO 2 F + F (slow) F + NO 2  NO 2 F(fast) F is called an intermediate It is formed then consumed in the reaction Reaction Mechanisms

59. Each of the two reactions is called an elementary step. The rate for a reaction can be written from its molecularity. The reaction rate reflects the stoichimetry. Molecularity is the number of pieces that must come together. Elementary steps add up to the balanced equation Reaction Mechanisms

60. Unimolecular step involves one molecule - Rate is first order. Bimolecular step - requires two molecules - Rate is second order Termolecular step- requires three molecules - Rate is third order Termolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule.

61. A  products Rate = k[A] A+A  productsRate= k[A] 2 2A  productsRate= k[A] 2 A+B  productsRate= k[A][B] A+A+B  products Rate= k[A] 2 [B] 2A+B  products Rate= k[A] 2 [B] A+B+C  products Rate= k[A][B][C] Molecularity and Rate Laws For an elementary step – the rate law reflects the stoichiometry

62. Reaction Mechanisms Proposed series of elementary steps. –Describing each collision The rate of the reaction is determined by the rate limiting step. The overall experimentally determined rate law must be consistent with the rate law that reflects the rate limiting step.

63. The proposed mechanism is NO 2 + F 2  NO 2 F + F (slow) rate = k [NO 2 ][F 2 ] F + NO 2  NO 2 F (fast) rate = k [F][NO 2 ] Since the first step is the slow step, it is the rate limiting step and thus the rate of the overall reaction depends upon the rate of this step. Note: The rate of this step is also the rate of the overall reaction. 2NO 2 + F 2  2NO 2 F Rate = k[NO 2 ][F 2 ]

64. The steps must add up to the overall reaction. 2 NO 2 Cl → 2 NO 2 + Cl 2 Proposed steps NO 2 Cl → NO 2 + Cl (slow) NO 2 Cl + Cl → NO 2 + Cl 2 (fast) Write the overall rate law that would be consistent with this proposed reaction mechanism.

65. How to get rid of intermediates They can’t appear in the rate law. Slow step determines the rate and the rate law Use the reactions that form them If the reactions are fast and irreversible the concentration of the intermediate is based on stoichiometry. If it is formed by a reversible reaction set the rates equal to each other.

66. Formed in reversible reactions 2 NO + O 2 2 NO 2 Mechanism 2 NO N 2 O 2 (fast) N 2 O 2 + O 2 2 NO 2 (slow) rate = k 2 [N 2 O 2 ][O 2 ] k 1 [NO] 2 = k 1 [N 2 O 2 ] (equilibrium) Rate = k 2 (k 1 / k -1 )[NO] 2 [O 2 ] Rate =k [NO] 2 [O 2 ]

67. Formed in fast reactions 2 IBr I 2 + Br 2 Mechanism IBrI + Br (fast) IBr + Br I + Br 2 (slow) I + II 2 (fast) Rate = k[IBr][Br] but [Br]= [IBr] because the first step is fast Rate = k[IBr][IBr] = k[IBr] 2

68. Collision theory Molecules must collide to react. Concentration affects rates because collisions are more likely. Must collide hard enough. Temperature and rate are related. Only a small number of collisions produce reactions.

69. Potential EnergyPotential Energy Reaction Coordinate Reactants Products

70. Potential EnergyPotential Energy Reaction Coordinate Reactants Products Activation Energy E a

71. Potential EnergyPotential Energy Reaction Coordinate Reactants Products Activated complex

72. Potential EnergyPotential Energy Reaction Coordinate Reactants Products EE }

73. Potential EnergyPotential Energy Reaction Coordinate 2BrNO 2NO + Br Br---NO 2 Transition State

74. Terms Activation energy - the minimum energy needed to make a reaction happen. Activated Complex or Transition State - The arrangement of atoms at the top of the energy barrier.

75. Arrhenius Said that reaction rate should increase with temperature. At high temperature more molecules have the energy required to get over the barrier. The number of collisions with the necessary energy increases exponentially.

76. Arrhenius Number of collisions with the required energy = ze -E a /RT z = total collisions e is Euler’s number (inverse of ln) E a = activation energy R = ideal gas constant (in J/K mol) T is temperature in Kelvin

77. Problem with this Observed rate is too small Due to molecular orientation- they have to be facing the right way written into equation as p the steric factor.

78. O N Br O N O N O N O N O N O N O N No Reaction O N Br O N

79. Arrhenius Equation k = zpe -E a /RT = Ae -E a /RT A is called the frequency factor = zp k is the rate constant ln k = -(E a /R)(1/T) + ln A Another line !!!! ln k vs 1/T is a straight line With slope E a /R so we can find E a And intercept ln A

80. Arrhenius Equation- Another line !!!! l n k = -(E a /R)(1/T) + ln A y = m x + b l n k 1/T slope = -(E a /R)

81. Activation Energy and Rates The final saga

82. Mechanisms and rates There is an activation energy for each elementary step. Activation energy determines k. k = Ae - (E a /RT) k determines rate Slowest step (rate determining) must have the highest activation energy.

83. This reaction takes place in three steps

84. EaEa First step is fast Low activation energy

85. Second step is slow High activation energy EaEa

86. EaEa Third step is fast Low activation energy

87. Second step is rate determining

88. Intermediates are present

89. Activated Complexes or Transition States

90. Catalysts Speed up a reaction without being used up in the reaction. Enzymes are biological catalysts. Homogenous Catalysts are in the same phase as the reactants. Heterogeneous Catalysts are in a different phase as the reactants.

91. How Catalysts Work Catalysts allow reactions to proceed by a different mechanism - a new pathway. New pathway has a lower activation energy. More molecules will have this activation energy. Does not change  E Show up as a reactant in one step and a product in a later step

92. Pt surface HHHH HHHH Hydrogen bonds to surface of metal. Break H-H bonds Heterogenous Catalysts

93. Pt surface HHHH Heterogenous Catalysts C HH C HH

94. Pt surface HHHH Heterogenous Catalysts C HH C HH The double bond breaks and bonds to the catalyst.

95. Pt surface HHHH Heterogenous Catalysts C HH C HH The hydrogen atoms bond with the carbon

96. Pt surface H Heterogenous Catalysts C HH C HH HHH

97. Homogenous Catalysts Chlorofluorocarbons (CFCs) catalyze the decomposition of ozone. Enzymes regulating the body processes. (Protein catalysts)

98. Catalysts and rate Catalysts will speed up a reaction but only to a certain point. Past a certain point adding more reactants won’t change the rate. Zero Order

99. Catalysts and rate. Concentration of reactants RateRate Rate increases until the active sites of catalyst are filled. Then rate is independent of concentration