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. The part of chemistry that is thermodynamics and the part that is kinetics.

3. FACTORS AFFECTING RATE
1. Nature of Reactants
2. Concentration of Reactants
3. Temperature
4. Catalysts
5. Surface Area of Reactants
6. Adding an Inert Gas (NO EFFECT).

4. COLLISION THEORY Molecules must collide to react.
Only two particles collide at a time.
Must have proper orientation.

5. COLLISION THEORY Must collide with enough energy.
Only a small number of collisions produce reactions.
Concentration affects rates because collisions are more likely.
Temperature and rate are related.

6. ENERGY PLOT
Know Ea, transition state, and activated complex. Know exothermic versus endothermic.

7. Potential
Energy
Reactants
Products
Reaction Coordinate

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

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

10. Potential
Energy }
Reactants DE
Products
Reaction Coordinate

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

12. MAXWELL BOLTZMANN DISTRIBUTION

13. MAXWELL BOLTZMANN DISTRIBUTION

14. REACTION RATE t2- t1 Rate = D[A] Dt
Rate = [A] at t2 – [A] at t1
t2- t1
Rate = D[A] Dt
Change in concentration per unit time.
[reactants] decreases with time.
[products] increases with time

15. N H2 → 2NH3
As the reaction progresses the concentration H2 goes down
Concentration
[H2]
Time

16. N H2 → 2NH3
As the reaction progresses the concentration N2 goes down 1/3 as fast.
Concentration
[N2]
[H2]
Time

17. N H2 → 2NH3
As the reaction progresses the concentration NH3 goes up.
Concentration
[N2]
[H2]
[NH3]
Time

18. CALCULATING RATES
Average rates are taken over a range of time. Worry about coefficients.
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.
RATES ARE ALWAYS POSITIVE!

19. AVERAGE SLOPE METHOD
Concentration
D[H2] Dt
Time

20. INSTANTANEOUS SLOPE METHOD
Concentration
D[H2]
D t
Time

21. INSTANTANEOUS RATE

22. 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.
example N2 + 3H → 2NH3
-D[N2] = -3D[H2] = 2D[NH3]
Dt Dt Dt
Negative because [reactant] goes down.

23. 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.

24. 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; they can’t be obtained from the equation.

25. 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.

26. 2 NO NO + O2
Oxygen can appear only half as rapidly as the nitrogen dioxide disappears while NO appears twice as fast as oxygen appears.
Calculate the AVERAGE rate at which [NO2] changes in the first 50.0 seconds given the [NO2] at 0.0s = M and [NO2] at 50s = M .
RATE = −Δ [NO] = −[.0079]−[0.0100]
Δt s
= −[−4.2 × 10−5 M / sec] = 4.2 × 10−5 M /sec

27. 2 NO2 → 2 NO + O2 In terms of NO2: Rate = -Δ[NO2] = k[NO2]n
In terms of O2:
Rate’ = -Δ[O2] = k’[O2]n
So: Rate = 2 x Rate’ OR k[NO2]n = 2k’[NO2]n

28. RELATIVE RATES
We can consider the appearance of products along with the disappearance of reactants.
The reactant’s concentration is declining, the products is increasing.
Respect the algebraic sign AND respect the stoichiometry.
Divide the rate of change in concentration of each reactant by its stoichiometric coefficient in the balanced chemical equation.

29. RELATIVE RATES 2 NO2 → 2 NO + O2 Thus.....
Rate of reaction = - 1Δ[NO2] = 1 Δ[NO] = Δ [ O2]
2 Δtime Δtime Δtime
There are 2NO2 molecules consumed for every O2 molecule produced.

30. ANOTHER EXAMPLE
Using the coefficients from the balanced equation, you should be able to give relative rates. For example: 4 PH3(g) → P4(g) + 6 H2(g)
Initial rate rxn = −¼ 𝛥[𝑃𝐻3] 𝛥𝑡𝑖𝑚𝑒 = + 𝛥[𝑃4] 𝛥𝑡𝑖𝑚𝑒 = 𝛥[𝐻2] 𝛥𝑡𝑖𝑚𝑒

31. PRACTICE ONE
What are the relative rates of change in concentration of the products and reactant in the decomposition of nitrosyl chloride, NOCl? 2 NOCl(g) → 2 NO(g) + Cl2(g)

32. 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.

33. TYPES OF RATE LAWS Reactions are reversible.
When the rate of the forward = the rate of the reverse we have EQUILIBRIUM! To avoid this complication we will discuss reactions soon after mixing--initial reactions rates, and not worry about the buildup of products and how that starts up the reverse reaction.
So reaction rate will only depend on [reactants] and initial [reactants] right after mixing.

34. 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 N2O5(aq) → 4NO2(aq) + O2(g)
The reverse reaction won’t play a role

35. DIFFERENTIAL RATE LAW
Describes how the reaction rate varies with the concentration of various species in a system.
Rates generally depend on reactant concentrations.
To find the exact relation between rate and concentration, we must conduct experiments and collect information.

36. DIFFERENTIAL RATE LAW C aA + bB → xX
Becomes Initial rxn rate = k[A]m[B]n[C]p
NOT COEFFICIENTS
K = temperature dependent

37. DIFFERENTIAL RATE LAW
Exponents can be zero, whole numbers or fractions AND MUST BE DETERMINED BY EXPERIMENTATION!!
THE RATE CONSTANT, k
temperature dependent and must be evaluated by experiment.
Example: rate = k[A]. If k is 0.090/hr when [A] = mol/L
rate = (.0090/hr)(0.018 mol/L) = mol/(L• hr)

38. ORDER OF A REACTION
order with respect to a certain reactant is the exponent on its concentration term in the rate expression.
order of the reaction is the sum of all the exponents on all the concentration terms in the expression.
DETERMINATION OF THE RATE EXPRESSION
aA + bB → xX
initial rate = k[A]om[B]on the little subscript “o” means “original” or at “time zero”.

39. ORDER OF A REACTION
1. Zero order: The change in concentration of reactant has no effect on the rate. These are not very common. General form of rate equation: Rate = k
2. First order: Rate is directly proportional to the reactants concentration; doubling [rxt], doubles rate. These are very common! Nuclear decay reactions usually fit into this category. General form of rate equation: Rate = k [A]1 = k[A]
3. Second order: Rate is quadrupled when [rxt] is doubled and increases by a factor of 9 when [rxt] is tripled etc. These are common, particularly in gas-phase reactions. General form of rate equation: Rate = k [A]2 or Rate = k[A]1[B]1 which has an overall order of two (second order).

40. ORDER OF A REACTION To determine the order of the reaction:
1. Determine skeleton rate law
2. Determine exponents using data from trials.
3. Determine k with its units.
4. Write the rate law.
Adding the orders of each reactant gives the overall order of the reaction.

41. NO(g) + Cl2(g) → NOCl(g) at 295K
EXAMPLE
NO(g) + Cl2(g) → NOCl(g) at 295K
Rate = k[NO]m[Cl2]n
Experiment
[NO]
[Cl2]
Initial Rate (M/s) 1
0.050
1.0 x 10-3 2
0.150
3.0 x 10-3 3
9.0 x 10-3

42. HOW TO SOLVE 1. Skeleton: Rate = k[NO]m[Cl2]n
2. Determine order: Look for two trials where the concentration of a reactant was held constant. Next, focus on the other reactant. Ask yourself how it’s concentration changed for the same two trials. Was it doubled? Was it tripled? Was it halved? Once you have determined the factor by which the concentration of the other reactant was changed, determine how that affected the rate for those same two trials.

43. HOW TO SOLVE
To find the rate order of NO, find two trials where [Cl2] is constant and [NO] changes. From experiment 1 to experiment 3, [NO] triples while [Cl2] is constant. Since the initial rate increases by a factor of 9x, the rate is directly proportional to [NO]2 and the RATE ORDER FOR [NO] IS 2.
Experiment x 10-3 k[0.150]n [0.050]m 9 = 3n
Experiment x 10-3 k[0.050]n [0.050]m n = 2

44. HOW TO SOLVE
To find the rate order of Cl2, find two trials where [NO] is constant and [Cl2] changes. From experiment 1 to experiment 2, [Cl2] triples while [NO] is constant. Since the initial rate triples, the rate is directly proportional to [Cl2]1 and the RATE ORDER FOR Cl2 IS 1.
Experiment x 10-3 k[0.050]n [0.150]m 3 = 3m
Experiment x 10-3 k[0.050]n [0.050]m m = 1

45. HOW TO SOLVE THE RATE LAW EXPRESSION IS RATE = k[NO]2[Cl2]1
OVERAL ORDER: 3
NOTE: The rate orders aren’t always the same as the coefficients.

46. HOW TO SOLVE
3. Determine k and units: Use any experimental data to find the value of k and its units. The units for K are one less that the overall order. Using experiment 1 data, plug the values into the rate law and solve for k
 Rate = k[NO]2[Cl2]1 = 1.0 x 10-3M = k (0.050M)2 (0.050M)1
k = (1.0 x 10-3M/s) / [(0.050M)2 (0.050M)1]
k = 8.0M-2 x s-1 or 8.0L2 / (mol2 x s)
The larger the k, the faster the reaction will be. k only changes with changes in TEMPERATURE. Increasing the temperature will increase k and increase the reaction rate.

47. PRACTICE TWO
In the following reaction, a Co-Cl bond is replaced by a Co-OH2 bond.
[Co(NH3)5Cl]+2 + H2O → [Co(NH3)5H2O]+3 + Cl
Initial rate = k{[Co(NH3)5Cl]+2}m
Using the data below, find the value of m in the rate expression and calculate the value of k.
Exp. Initial Concentration Initial rate
of [Co(NH3)5Cl]+2 (M) mol/(L• min)
× × 10-7
× × 10-7
× × 10-7
× × 10-7

48. PRACTICE THREE
The reaction between bromate ions and bromide ions in acidic aqueous solution is given by the equation: BrO3-(aq) + 5 Br –(aq) + 6 H+(aq) → 3 Br2(l) + 3 H2O(l)
 The table below gives the results of four experiments. Using these data, determine the orders for all three reactants, the overall reaction order, and the value of the rate constant. What is the value of k?
What are the units of k?
Experiment Initial [BrO3-] Initial [Br–] Initial [H+] Measured initial
rate (mol/L•s)
× 10-4
× 10-3
× 10-3
× 10-3

49. PRACTICE THREE
The reaction between bromate ions and bromide ions in acidic aqueous solution is given by the equation: BrO3-(aq) + 5 Br –(aq) + 6 H+(aq) → 3 Br2(l) + 3 H2O(l)
 The table below gives the results of four experiments. Using these data, determine the orders for all three reactants, the overall reaction order, and the value of the rate constant. What is the value of k?
What are the units of k?
Experiment Initial [BrO3-] Initial [Br–] Initial [H+] Measured initial
rate (mol/L•s)
× 10-4
× 10-3
× 10-3
× 10-3

50. TYPES OF RATE LAWS
Differential Rate law – data table contains concentration and rate data
Integrated Rate Law – data table contains concentration and time data. Only one reactant.

51. 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

52. HOW TO SOLVE 1. Set up your axes so that time is always on the x-axis.
2. Plot the concentration of the reactant on the y-axis of the first graph.
3. Plot the natural log of the concentration (ln [A], NOT log[A]) on the y-axis of the second graph.
4. Plot the reciprocal of the concentration on the y-axis of the third graph.
You are in search of linear data.

53. HOW TO SOLVE
If you do the set of graphs in this order with the y-axes being “concentration”, “natural log of concentration” and “reciprocal concentration”, the alphabetical order of the y-axis variables leads to 0, 1, 2 orders respectively for that reactant.
You can now easily solve for either time or concentration once you know the order of the reactant.
Just remember y = mx + b.
Choose the set of variables that gave you the best straight line (r value closest to ±1) and insert them in place of x and y in the generalized equation for a straight line. “A” is reactant A and Ao is the initial concentration of reactant A at time zero [the y-intercept].

54. ZERO ORDER Slope = -k Rate is independent of concentration.
Rate = k[A]0
[A] = -kt + [A0]
Remember that the rate constant is NEVER negative.
This is not a straight line.

55. FIRST ORDER Slope = -k Concentration is a function of time
Rate = k[A]1
ln[A] = -kt + ln[A]0
In the form y = mx + b
This is a straight line.

56. SECOND ORDER Slope = k Concentration is a function of time
Rate = k[A]2
1/[A] = kt + 1/[A]0
In the form y = mx + b
This is not a straight line.

57. PRACTICE FOUR Determine the rate law and calculate the value of k.
The decomposition of N2O5 in the gas phase was studied at constant temperature.
2 N2O5(g) → 4 NO2(g) + O2(g)
The following results were collected:
[N2O5] Time (s)
Determine the rate law and calculate the value of k.
What is the concentration of N2O5(g) at 600s?
At what time is the concentration of N2O5(g) equal to M ?

58. PRACTICE FOUR zero order [A] = −kt + [Ao]
y = mx b
zero order [A] = −kt [Ao]
first order ln[A] = −kt ln[Ao]
second order 1/[A] = kt /[Ao]

59. If the reaction is first order
HALF LIFE – FIRST ORDER
If the reaction is first order
[A] = [A]0/2 when t = t1/2
t½ = ln2/k = 0.693/k

60. HALF LIFE – SECOND ORDER
If the reaction is second order
[A] = [A]0/2 when t = t1/2
t½ = 1/k[A]o

61. PRACTICE FIVE
A certain first-order reaction has a half-life of 20.0 minutes.
Calculate the rate constant for this reaction.
How much time is required for this reaction to be 75% complete?

62. PRACTICE SIX For the reaction of (CH3)3CBr with OH−,
(CH3)3CBr + OH− → (CH3)3COH + Br−
 The following data were obtained in the laboratory.
TIME (s) [(CH3)3CBr]
Plot these data as ln [(CH3)3CBr] versus time. Sketch your graph.
Is the reaction first order or second order? What is the value of the rate constant?

63. PRACTICE SEVEN
Butadiene reacts to form its dimer according to the equation
2 C4H6(g) → C8H12(g)
 The following data were collected for this reaction at a given temperature:
[C4H6] Time (± 1 s)

64. PRACTICE SEVEN
What is the order of this reaction? Explain. Sketch your graph as part of your explanation. Write the rate law expression:
 What is the value of the rate constant for this reaction?
 What is the half-life for the reaction under the conditions of this experiment?

65. HALF LIFE – ZERO ORDER Rate = k[A]0 = k (1) = k
Integrated rate law is [A] = -kt + [A]o
t1/2 = [A]o 2k
Zero-order reactions are most often encountered when a substance such as a metal surface or an enzyme is required for the reaction to occur. The enzyme or catalyst may become saturated and therefore an increase in the [reactant/substrate] has no effect on the rate.

66. SUMMARY OF RATE LAWS
Memorize this!

67. REACTION MECHANISMS
The sequence of bond-making and bond-breaking steps that occurs during the conversion of reactants to products. This is using the rate determining step to relate a reaction mechanism to its rate law expression.
Kinetics can tell us something about the mechanism
A balanced equation does not tell us how the reactants become products.

68. REACTION MECHANISMS Two Prong Test: Must be determined by experiment!
Must agree with overall stoichiometry
Must agree with the experimentally determined rate law
The coefficients from the rate determining step can be used to give rate orders shown in the rate law expression.

69. REACTION MECHANISMS For the reaction NO2(g) + CO(g) → CO2(g) + NO(g)
#1 NO2 + NO2 → NO3 + NO SLOW
#2 NO3 + CO → CO2 + NO2 FAST
NO2 + CO → CO2 + NO
The unstable NO3, an intermediate, cancels out as does 1 NO2

70. REACTION MECHANISMS
The final rate expression must include ONLY those species that appear in the balanced equation for the overall reaction.
Step #1 is the rate determining step, and the rate law equation can be found using its coefficients.:
2NO2 → NO3 + NO
RATE = k[NO2]2 always written from the slow step.

71. 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.

72. REACTION MECHANISMS
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.

73. REACTION MECHANISMS
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]

74. PRACTICE EIGHT
Nitrogen oxide is reduced by hydrogen to give water and nitrogen,
2 H2(g) + 2 NO(g) → N2(g) + 2 H2O(g)
and one possible mechanism to account for this reaction is
2 NO(g) → N2O2(g)
N2O2(g) + H2(g) → N2O(g) + H2O(g)
N2O(g) + H2(g) → N2(g) + H2O(g)
What is the molecularity of each of the three steps? Show that the sum of these elementary steps is the net reaction.

75. REACTION MECHANISMS AND RATE EXPRESSIONS
determined by experiment
the rate of the overall reaction is limited by, and is exactly equal to, the combined rates of all elementary steps up to and including the slowest step in the mechanism
the slowest step is the rate determining step
reaction intermediate - produced in one step but consumed in another.
Catalyst - goes in, comes out unharmed and DOES NOT show up in the final reaction

76. PRACTICE NINE
The balanced equation for the reaction of the gases nitrogen dioxide and fluorine is
2 NO2(g) + F2(g) → 2 NO2F(g)
The experimentally determined rate law is Rate = k [NO2][F2]
A suggested mechanism for the reaction is
NO2 + F2 → NO2F + F Slow
F + NO2 → NO2F Fast
Is this an acceptable mechanism? That is, does it satisfy the two requirements? Justify.

77. 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.

78. HOW CATALYSTS WORK
Catalysts allow reactions to proceed by a different mechanism - a new pathway.
New pathway has a lower activation energy, Ea.
More molecules will have this activation energy.
Do not change E also known as ∆H

79. HETEROGENEOUS CATALYSTS
different phase than reactants, usually involves gaseous reactants adsorbed on the surface of a solid catalyst
adsorption—refers to the collection of one substance on the surface of another
absorption—refers to the penetration of one substance into another; water is absorbed by a sponge

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

81. Heterogenous Catalysts
Pt surface

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

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

84. Heterogenous Catalysts
Pt surface

85. HOMOGENOUS CATALYSTS
exists in the same phase as the reacting molecules.
Chlorofluorocarbons catalyze the decomposition of ozone.
Enzymes regulating the body processes. (Protein catalysts)

86. 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