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 t2 -Conc. of A at t1 t2- t1
Rate = D[A] Dt
Change in concentration per unit time
For this reaction
N2 + 3H NH3

3. As the reaction progresses the concentration H2 goes down
Time

4. As the reaction progresses the concentration N2 goes down 1/3 as fast
Time

5. As the reaction progresses the concentration NH3 goes up.
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
Concentration
D[H2] Dt
Time

8. Instantaneous slope method.
Concentration
D[H2]
D 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 N2 + 3H NH3
-D[N2] = -3D[H2] = 2D[NH3] Dt Dt Dt

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. Rate Laws Two key points
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 NO + O2
You will find that the rate will only depend on the concentration of the reactants.
Rate = k[NO2]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 NO2 2 NO + O2 The rate of appearance of O2 can be said to be.
Rate' = D[O2] = k'[NO2] Dt
Because there are 2 NO2 for each O2
Rate = 2 x Rate'
So k[NO2]n = 2 x k'[NO2]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 N2O5 (aq) NO2 (aq) + O2(g) The reverse reaction won’t play a role

16. Now graph the data [N2O5] (mol/L) Time (s) 1.00 0 0.88 200 0.78 400
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 (.98 - .76) = 0.22 =- 5.5x 10 -4 (0-400) -400

19. At. 40 M the rate is (. 52 -. 31) = 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 = -D[N2O5] = k[N2O5]1 = k[N2O5] Dt
We say this reaction is first order in N2O5
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 BrO3- + 5 Br- + 6H+ 3Br2 + 3 H2O
The general form of the Rate Law is Rate = k[BrO3-]n[Br-]m[H+]p
We use experimental data to determine the values of n,m,and p

23. Now we have to see how the rate changes with concentration
Initial concentrations (M)
Rate (M/s)
BrO3-
Br- H+
x 10-4
x 10-3
x 10-3
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 = D[A]n Dt
We will only work with n=0, 1, and 2

25. First Order For the reaction 2N2O5 4NO2 + O2
We found the Rate = k[N2O5]1
If concentration doubles rate doubles.
If we integrate this equation with respect to time we get the Integrated Rate Law
ln[N2O5] = - kt + ln[N2O5]0
ln is the natural log
[N2O5]0 is the initial concentration.

26. First Order General form Rate = D[A] / Dt = 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

29. Half Life The time required to reach half the original concentration.
If the reaction is first order
[A] = [A]0/2 when t = t1/2

30. Half Life The time required to reach half the original concentration.
If the reaction is first order
[A] = [A]0/2 when t = t1/2
ln(2) = kt1/2

31. Half Life
t1/2 = 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 = -D[A] / Dt = 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 = t1/2

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 = t1/2
t1/2 = [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. Concentration
Time

37. Concentration
k =
D[A]/Dt
D[A] Dt
Time

38. More Complicated Reactions
BrO Br- + 6H Br2 + 3 H2O
For this reaction we found the rate law to be
Rate = k[BrO3-][Br-][H+]2
To investigate this reaction rate we need to control the conditions

39. Rate = k[BrO3-][Br-][H+]2
We set up the experiment so that two of the reactants are in large excess.
[BrO3-]0= 1.0 x 10-3 M
[Br-]0 = 1.0 M
[H+]0 = 1.0 M
As the reaction proceeds [BrO3-] changes noticably
[Br-] and [H+] don’t

40. Rate = k[BrO3-][Br-][H+]2
This rate law can be rewritten
Rate = k[BrO3-][Br-]0[H+]02
Rate = k[Br-]0[H+]02[BrO3-]
Rate = k’[BrO3-]
This is called a pseudo first order rate law.
k = k’ [Br-]0[H+]02

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 2NO2 + F2 2NO2F Rate = k[NO2][F2]
The proposed mechanism is
NO2 + F NO2F + F (slow)
F + NO NO2F (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 NO2
Mechanism
2 NO N2O2 (fast)
N2O2 + O NO2 (slow)
rate = k2[N2O2][O2]
k1[NO]2 = k-1[N2O2]
rate = k2 (k1/ k-1)[NO]2[O2]=k[NO]2[O2]

49. Formed in fast reactions
2 IBr I2+ Br2
Mechanism
IBr I + Br (fast)
IBr + Br I + Br2 (slow)
I + I I2 (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. Potential
Energy
Reactants
Products
Reaction Coordinate

52. Activation Energy Ea Reaction Coordinate Potential Energy Reactants
Products
Reaction Coordinate

53. Reaction Coordinate Activated complex Potential Energy Reactants
Products
Reaction Coordinate

54. Potential
Energy }
Reactants DE
Products
Reaction Coordinate

55. Reaction Coordinate Br---NO Potential Br---NO Transition State Energy
2NO + Br 2
Reaction Coordinate

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-Ea/RT
z = total collisions
e is Euler’s number (opposite of ln)
Ea = 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 O O O O O O O O O No Reaction N N N N Br Br Br Br Br N N Br Br N Br

61. k = zpe-Ea/RT = Ae-Ea/RT
Arrhenius Equation
k = zpe-Ea/RT = Ae-Ea/RT
A is called the frequency factor = zp
ln k = -(Ea/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- (Ea/RT)
k determines rate
Slowest step (rate determining) must have the highest activation energy.

64. This reaction takes place in three steps

65. Ea
First step is fast
Low activation energy

66. High activation energyEa
Second step is slow
High activation energy

67. Ea
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 E

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

74. Heterogenous Catalysts
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
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. Rate
Rate increases until the active sites of catalyst are filled.
Then rate is independent of concentration
Concentration of reactants