Reaction Rate. Rate = Conc. of A at t 2 -Conc. of A at t 1. t 2 -t 1. Rate = Δ[A] Δt

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1 Kinetics The study of reaction rates. Spontaneous reactions are reactions that will happen - but we can t tell how fast. Diamond will spontaneously turn to graphite eventually. Reaction mechanism- the steps by which a reaction takes place.

2 Reaction Rate Rate = Conc. of A at t 2 -Conc. of A at t 1 Rate = Δ[A] Δt t 2 -t 1 Change in concentration per unit time For this reaction N 2 + 3H 2 2NH 3

3 As the reaction progresses the concentration H 2 goes down C o n c e n t r a t i o n [H 2 ] Time

4 As the reaction progresses the concentration N 2 goes down 1/3 as fast C o n c e n t r a t i o n [H 2 ] [N 2 ] Time

5 As the reaction progresses the concentration NH 3 goes up. C o n c e n t r a t i o n [H 2 ] [NH 3 ] [N 2 ] Time

6 Calculating Rates Average rates are taken over long intervals 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 Derivative.

7 Average slope method C o n c e n t r a t i o n Δ[H 2 ] Δt Time

8 Instantaneous slope method. C o n c e n t r a t i o n Δ[H 2 ] Δ t Time

9 Defining Rate We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product. In our example N 2 + 3H 2 2NH 3 -Δ[N 2 ] = -3Δ[H 2 ] = 2Δ[NH 3 ] Δt Δt Δt

10 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. This is called the Initial rate method.

11 Two key points Rate Laws The concentration of the products do not appear in the rate law because this is an initial rate. The order must be determined experimentally, can t be obtained from the equation

12 2 NO 2 2NO + O 2 You will find that the rate will only depend on the concentration of the reactants. 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.

13 2 NO 2 2NO + O 2 The rate of appearance of O 2 can be said to be. Rate' = Δ[O 2 ] = k'[no 2 ] Δt Because there are 2 NO 2 for each O 2 Rate = 2 x Rate' So k[no 2 ] n = 2 x k'[no 2 ] n So k = 2 x k'

14 Types of Rate Laws Differential Rate law - describes how rate depends on concentration. Integrated Rate Law - Describes how concentration depends on time. For each type of differential rate law there is an integrated rate law and vice versa. Rate laws can help us better understand reaction mechanisms.

15 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

16 [N 2 O 5 ] (mol/l) Time (s) Now graph the data

17 To find rate we have to find the slope at two points We will use the tangent method

18 At.90 M the rate is ( ) = 0.22 =- 5.5x 10-4 (0-400)

19 At.40 M the rate is ( ) = 0.22 =- 2.7 x 10-4 ( )

20 Since the rate at twice the concentration is twice as fast the rate law must be.. Rate = -Δ[N 2 O 5 ] = k[n 2 O 5 ] 1 = k[n 2 O 5 ] Δt We say this reaction is first order in N 2 O 5 The only way to determine order is to run the experiment.

21 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 bust after the reactants are mixed. Eliminates the effect of the reverse reaction.

22 An example For the reaction BrO Br - + 6H + 3Br 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

23 Initial concentrations (M) Rate (M/s) BrO - 3 Br - H x x x x 10-3 Now we have to see how the rate changes with concentration

24 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 = Δ[A] n Δt We will only work with n=0, 1, and 2

25 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 ln[n 2 O 5 ] = - kt + ln[n 2 O 5 ] 0 ln is the natural log [N 2 O 5 ] 0 is the initial concentration.

26 First Order General form Rate = Δ[A] / Δt = k[a] ln[a] = - kt + ln[a] 0 In the form y = mx + b y = ln[a] m = -k x = t b = ln[a] 0 A graph of ln[a] vs time is a straight line.

27 First Order By getting the straight line you can prove it is first order Often expressed in a ratio

28 First Order By getting the straight line you can prove it is first order Often expressed in a ratio ln [ ] A A [ ] 0 = kt

29 Half Life The time required to reach half the original concentration. If the reaction is first order [A] = [A] 0 /2 when t = t 1/2

30 Half Life The time required to reach half the original concentration. If the reaction is first order [A] = [A] 0 /2 when t = t 1/2 ln ln(2) = kt 1/2 [ ] A [ ] A = kt 12

31 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

32 Second Order Rate = -Δ[A] / Δt = k[a] 2 integrated rate law 1/[A] = kt + 1/[A] 0 y= 1/[A] m = k x= t b = 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

33 Second Order Half Life [A] = [A] 0 /2 at t = t 1/2 1 [A] 0 1 [ A] 0 2 = kt [ A] [ A] = kt 12 1 [A] = kt t = 1 2 k[a] 0

34 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

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

36 C o n c e n t r a t i o n Time

37 C o n c e n t r a t i o n Δ[A] Δt k = Δ[A]/Δt Time

38 More Complicated Reactions BrO Br - + 6H + 3Br H 2 O For this reaction we found the rate law to be Rate = k[bro 3 - ][Br - ][H + ] 2 To investigate this reaction rate we need to control the conditions

39 Rate = k[bro 3 - ][Br - ][H + ] 2 We set up the experiment so that two of the reactants are in large excess. [BrO 3 - ] 0 = 1.0 x 10-3 M [Br - ] 0 = 1.0 M [H + ] 0 = 1.0 M As the reaction proceeds [BrO 3 - ] changes noticably [Br - ] and [H + ] don t

40 Rate = k[bro 3 - ][Br - ][H + ] 2 This rate law can be rewritten Rate = k[bro 3 - ][Br - ] 0 [H + ] 0 2 Rate = k[br - ] 0 [H + ] 0 2 [BrO 3 - ] Rate = k [BrO 3 - ] This is called a pseudo first order rate law. k = k [Br - ] 0 [H + ] 0 2

41 Summary of Rate Laws

42 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.

43 Reaction Mechanisms 2NO 2 + F 2 Rate = k[no 2 ][F 2 ] 2NO 2 F 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

44 Reaction Mechanisms Each of the two reactions is called an elementary step. The rate for a reaction can be written from its molecularity. Molecularity is the number of pieces that must come together.

45 Unimolecular step involves one molecule - Rate is rirst 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.

46 A products Rate = k[a] A+A products Rate= k[a] 2 2A products Rate= k[a] 2 A+B products Rate= 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]

47 How to get rid of intermediates 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.

48 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 ] rate = k 2 (k 1 / k -1 )[NO] 2 [O 2 ]=k[no] 2 [O 2 ]

49 Formed in fast reactions 2 IBr I 2 + Br 2 Mechanism IBr I + Br (fast) IBr + Br I + Br 2 (slow) I + I I 2 (fast) Rate = k[ibr][br] but [Br]= [IBr] Rate = k[ibr][ibr] = k[ibr] 2

50 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.

51 P o t e n t i a l E n e r g y Reactants Reaction Coordinate Products

52 P o t e n t i a l E n e r g y Reactants Reaction Coordinate Activation Energy E a Products

53 P o t e n t i a l E n e r g y Reactants Activated complex Reaction Coordinate Products

54 P o t e n t i a l E n e r g y Reactants Products ΔE Reaction Coordinate

55 P o t e n t i a l E n e r g y 2BrNO Br---NO Br---NO Reaction Coordinate Transition State 2NO + Br 2

56 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.

57 Arrhenius Said the at 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.

58 Arrhenius Number of collisions with the required energy = ze -E a /RT z = total collisions e is Euler s number (opposite of ln) E a = activation energy R = ideal gas constant T is temperature in Kelvin

59 Problems Observed rate is less than the number of collisions that have the minimum energy. Due to Molecular orientation written into equation as p the steric factor.

60 O N O N O N O N Br Br Br Br O N Br Br N O Br Br N O O N Br N O O N Br No Reaction

61 Arrhenius Equation k = zpe -E a /RT = Ae -E a /RT A is called the frequency factor = zp ln k = -(E a /R)(1/T) + ln A Another line!!!! ln k vs t is a straight line

62 Activation Energy and Rates The final saga

63 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.

64 This reaction takes place in three steps

65 E a First step is fast Low activation energy

66 E a Second step is slow High activation energy

67 E a Third step is fast Low activation energy

68 Second step is rate determining

69 Intermediates are present

70 Activated Complexes or Transition States

71 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.

72 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. Do not change ΔE

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

74 Heterogenous Catalysts H H H C C H H H H H Pt surface

75 Heterogenous Catalysts The double bond breaks and bonds to the catalyst. H H H C C H H H H H Pt surface

76 Heterogenous Catalysts The hydrogen atoms bond with the carbon H H H C C H H H H H Pt surface

77 Heterogenous Catalysts H H H C C H H H H H Pt surface

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

79 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

80 Catalysts and rate. R a t e Rate increases until the active sites of catalyst are filled. Then rate is independent of concentration Concentration of reactants

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