1. Agenda P. 354- Rates of Reactions P. 365- Factors Affecting Rates
HW: pg 360 #1-10,
pg 364 #1-12, pg 371 #11-20
pg 377 #25-30

2. RATES OF REACTION

3. Example Identify EA, DH and transition state.
What are the values of EA and DH?
Endothermic or exothermic?
What are EA and DH for the reverse rxn?

4. Rates of Reaction
The rate of a chemical reaction is the speed at which the reaction occurs (i.e. speed at which the reactants are used or products are produced).
Rate can be measured by the:
1) change in mass of reactants or products.
2) change in pH
3) change in conductivity (ion production)
4) change in colour (intensity of colour)
5) change in temperature
6) production of a gas

5. Calculating Average Rate of Reaction
Average rate of reaction is the change in concentration in a given time period.
Mathematically,
rate= ∆[C] ∆t
rate= [𝐶] 𝑓𝑖𝑛𝑎𝑙 − [𝐶] 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑡 𝑓𝑖𝑛𝑎𝑙 − 𝑡 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
[C] is concentration in mol/L and t is time (usually seconds)
The unit for rate is mol/L s
Rate is expressed in terms of the amount of a product produced or the amount a reactant is consumed.
Reactants are used and products produced based on their molar ratios.

6. Sample Problem: Nitrogen dioxide gas decomposes to nitrogen monoxide and oxygen gases. You began the reaction with 0.10 mol of NO2 in a 10.0 L container. After 150 s, [NO] is mol/L. a) Write a balanced chemical equation. b) Determine the average rate of reaction with respect to NO. c) Express the average rate of reaction with respect to the other 2 gases.

7. 2 NO2(g)  2 NO(g) + O2(g)
rate= ∆[C] ∆t
rate NO = ( mol L −0 mol L ) (150 s−0 s)
rate NO =3.0 x10-5 mol/L s
c) rate O 2 = 1 2 rate NO
= 1.5x10-5 mol/L s
rate N O 2 =− rate NO
= -3.0x10-5 mol/L s

8. Factors Affecting Rates
The factors that will affect the rate of a reaction are:
1) Temperature
2) Surface Area
3) Concentration
4) Catalyst
5) Chemical Nature of Reactants
6) State of the Reactants

9. A Model for Chemical Kinetics
A + B  C
3 things affect the reaction rate:
1. Rate of collisions between A and B
2. Fraction of collisions having correct orientation
3. Fraction of collisions having sufficient energy to cause reaction

10. Collision Theory
The KINETIC MOLECULAR THEORY may be used to explain how the various factors will affect the rate of a chemical reaction.
As the particles in the reactants move around, they collide with each other.
Most collisions do not result in anything but a few will cause the bonds in the existing molecules to break apart and new bonds will form to make new molecules.
Collisions that result in the formation of products are called EFFECTIVE COLLISIONS.

11. To be effective a collision must occur at the correct orientation and with a minimum amount of energy.
The idea of effective collisions is called the COLLISION MODEL and states that the rate of a reaction is affected by the number of effective collisions between reactant molecules.
According to the Collision Model, the rate of a reaction may be increased by increasing the number of total collisions or by increasing the number of effective collisions by decreasing the activation energy.

12. Collision Theory In other words:
The RATE OF REACTION depends on the frequency of collisions and the fraction of effective collisions.

13. The Factors Affecting Rate of Reaction and the Collision Theory
Concentration (Pressure of Gases)
The greater the number of particles the greater the number of total collisions. Since the total number of collisions has increased the number of effective will also increase therefore the rate of the reaction will increase.
NOTE: As a reaction proceeds, it tends to slow down.

14. Surface Area 2. Surface Area (Size of Solid Particles)
Reagents must come in contact to react.
If the size of a solid particle is decreased, there will
be more surface area available for collisions to
occur.

17. 3. Temperature of Reactants
The temperature of a substance is a measure of the average kinetic energy of the particles. The higher the temperature, the faster the particles are moving which will increase the chance for collisions.
Collisions will be more frequent and collisions will have higher energy.
Increased frequency and energy increase the rate of reaction.

18. Recall

20. Temperature
As the temperature is increased from T1 to T2 the KE of the particles increases and more particles achieve the activation energy F REACTION

21. 4. State or Phase of the Reactants
Reactions where all reactants are in the same state (homogeneous reactions) will occur at a faster rate than reactions where reactants are in different states (heterogeneous). This is because the reactants will have a greater opportunity of colliding. The state of the reactants will also affect the rate. The relative rates are:
Gases  fastest Liquids/ Solutions  fast Solids  slow
Note: Stirring increases the rate of reaction

22. State or Phase of the Reactants
FASTEST Reaction
Gas - Gas reactions
Aqueous reactions
Solid in a liquid
Solid - Solid
SLOWEST
Powdered solids can be explosive

23. Nature of the Reactants
5. Nature of Reactants
The type of reactants determine the activation energy
needed. The higher the activation energy, the slower
the rate of reaction. Endothermic reactions are much
slower than exothermic reactions because they tend
to have higher activation energies.
Rate depends on reactivity of reactants.
Ex. Au <<< less reactive than Na.
Some reactions are complex multi-step processes.

24. 6. Presence of a Catalyst
A catalyst works by providing an alternative pathway for the reaction which has a lower activation energy. Lowering the activation energy will only increase the number of effective collisions. The number of total collisions will not be affected.
Stabilizes the activated complex which reduces the activation energy of the reaction.
With a lower activation energy, more particles have the required amount of energy to react.
End of Day 1

25. Catalysts bend or stretch bonds to make them easier to break / react
reduce EA (make transition state easier)
bring two reactants close together
provide a microenvironment for reactions

26. 3. Catalysts
Catalysts lower the EA of a reaction.

27. Catalysts

28. Ex. Pt Catalyst of C2H4 + H2 ® C2H6

29. Catalysts
inhibitors - bind with the reactant or the catalyst to prevent the reaction from occurring and reducing reaction rate

30. 1. Collision Theory Summary
Factors affecting effective collisions:
molecule orientation
molecule energy
Collision Theory Animation
Collision Theory Applet

31. AGENDA P. 366 - Energy Diagrams and Activation Energy
P Maxwell-Boltzmann
HW: P. 371 # 11-20
P.374 #19-24
P.379 # 1-12

32. Catalyst

37. Potential Energy Diagrams
a) Activation Energy (Ea): the minimum amount of potential energy required for a reaction to occur b) Ea for the reverse reaction c) ΔH Activated Complex: reaction intermediate (collision state)

38. Potential Energy Diagrams and Reaction Rates
The lower the activation energy, the greater the rate of reaction.
In general, exothermic reactions tend to have greater rates than endothermic reactions.
Many exothermic reactions are self sustaining.

39. Catalysts increase rate by providing an alternative pathway with a lower activation energy.

40. Maxwell-Boltzmann Distributions
Kinetic Energy

41. Maxwell-Boltzmann Distributions and Reaction Rates
To increase the rate of reaction, the fraction of particles with the activation energy or greater must increase.
Increase Temperature

42. With a Catalyst

43. Activated Complex
Is the short-lived, unstable structure formed during a successful collision between reactant particles.
Old bonds of the reactants are in the process of breaking, and new products are forming
Ea is the minimum energy required for the activation complex to form and for a successful reaction to occur.

44. Fast and slow reactions
The smaller the activation energy, the faster the reaction will occur regardless if exothermic or endothermic.
If there is a large activation energy needed, that means that more energy (and therefore, time) is being used up for the successful collisions to take place.

45. 1. For each of the diagrams above, explain the reaction in complete sentence statements, in terms of the changes in potential energy and kinetic energy over the duration of the reaction. Include the relative reaction rate for each reaction: slow, medium, or fast.
2. The decomposition of potassium chlorate is an exothermic reaction, but which only takes place when the potassium chlorate is strongly heated. Draw the enthalpy –RP diagram for this reaction.

47. Sample Problem
The following reaction has an activation energy of 120kJ and a ΔH of 113kJ.
2NO2  2NO + O2
Draw and label a potential energy (activation energy) diagram for this forward reaction.
Calculate the activation energy for the reverse reaction (if the reaction went backwards)

48. Another Problem
The following hypothetical reaction has an activation energy of 70kJ and a ΔH of -130kJ
A + B  C + D
Draw and label a potential energy diagram for the reaction
Calculate the activation energy for the reverse reaction.

49. Practice:
The following hypothetical reaction has an Ea of 120kJ and a ΔH of 80kJ
2A + B  2C + D
Draw and label a potential energy diagram for this reaction.
What type of reaction is this?
Calculate the activation energy for the reverse reaction.
Calculate the ΔH for the reverse reaction.

50. This illustration shows an activation energy diagram for a endothermic reaction. The total reaction has three steps. It is a multi step reaction . Most reactions occur through a series of steps called a mechanism. This illustration shows a three step mechanism. The step with the highest activation energy controls the rate of the reaction. It looks like step three is the slowest step with the highest activation energy.

51. Ep RP

52. Homework
Review
Do pg 392 #1-69
P.398 #1-25

53. Agenda P.383Rate Law Expressions
P.380- Difference between 0. 1st, 2nd order reactions
HW: P. 387 # 1-11, Worksheets

54. The potential energy diagram below shows the uncatalyzed reaction (solid line) and the corresponding catalyzed reaction (dotted line).
What is the activation energy for the forward uncatalyzed reaction?
Potential Energy (kJ/mol)
Reaction Progress 20 38 50 59 75
Uncatalyzed
Catalyzed

55. Rate Laws and Order of Reaction
There is a mathematical relationship between the rate of reaction and the factors affecting the reaction.
This relationship must be determined empirically (must analyze experimental data).
The RATE LAW for a reaction describes the relationship between rate (r) and the product of the initial concentrations of the reactants raised to some exponential values.

56. given: a X + b Y  products
r α[X]m [Y]n
The exponents m and n describe the relationship between rate and initial concentration and must be determined experimentally.
The value of m and n may be a whole number, zero or a fraction and do not have to equal the coefficients from the balanced chemical equation (i.e. a and b).

57. To determine the rate of a reaction mathematically, a constant k, called the rate constant must be introduced into the relationship.
A RATE LAW EQUATION for the reaction is then written as:
r = k [X]m [Y]n
The value for the rate constant must also be determined experimentally.

58. Orders of Reaction
The exponents in the rate law equation are called the orders of reaction.
The sum of all the individual orders (exponents) is referred to as the overall order of reaction.
Given: 2X + 2Y + 3Z  products
r = k[X]1[Y]2[Z]0
The order of reaction for X is 1, Y is 2 and Z is zero and the overall order is 3.

59. What does the order of reaction mean?
For 1st order, if concentration is doubled, rate is doubled (21)
For 2nd order, if concentration is doubled, rate is quadrupled (22).
For 3rd order, if concentration is doubled, rate is increased by a factor of 8 (23).

60. For zero order, if concentration is doubled, there is no change in the rate (20)
If a reaction is zero order for one of the reactants, that reactant is left out of the equation.

61. Handout - Table 1 Concentration Factor Change
Reaction Rate Factor Change
Rate Law Exponent on the Concentration 2x
no change 3x 4x
21 = 2 1
31 = 3
41 = 4

62. Handout - Table 1 Concentration Factor Change
Reaction Rate Factor Change
Rate Law Exponent on the Concentration 2x
22 = 4 2 3x
32 = 9 4x
42 = 16
23 = 8 3
33 = 27
43 = 64

63. Determining A Rate Law Equation
Rate law equations are determined by analyzing data on changes in concentration and its affect on rate.
The first step is to determine the orders of each reactant and then substitute the data for one trial into the equation to find the value for k.

64. Example #3
Identify the order of the reaction and units of k: a) rate = k[N] b) rate = k[D]1/2[E]2

65. REACTION RATE Identify the order of the reaction and units of k:
REACTION ORDER:
Identify the order of the reaction and units of k:
a) rate = k[N]
b) rate = k[D]1/2[E]2
1st order, s-1
2.5 order, L1.5 mol-1.5 s-1

66. rate = k[H2SeO3]x[I-]y[H+]z
REACTION RATE
RATE LAW EQUATION:
Write out the rate law equation for:
H2SeO3 + 6 I- + 4 H+  Se + 2 I H2O
rate = k[H2SeO3]x[I-]y[H+]z
EXAMPLE 1

67. rate = k[H2SeO3]x[I-]y[H+]z
REACTION RATE
RATE LAW EQUATION:
EXAMPLE 1
rate = k[H2SeO3]x[I-]y[H+]z
At 0°C, k = 5.0 x 105
x = 1, y = 3, z = 2
Rewrite the rate law.
What is the unit for rate?
What are the units for k in this case?
L5 • mol-5 • s-1

68. Units for k
k = L overall order – 1 /mol overall order -1 s

69. rate = 5.0x105L5mol-5s-1[H2SeO3]1[I-]3[H+]2
REACTION RATE
RATE LAW EQUATION:
EXAMPLE 1
rate = 5.0x105L5mol-5s-1[H2SeO3]1[I-]3[H+]2
Determine the rate of reaction at 0°C given:
[H2SeO3] = 2.0x10-2 M
[I-] = 2.0x10-3 M
[H+] = 1.0x10-3 M

70. REACTION RATE RATE LAW EQUATION: EXAMPLE 1
rate = 5.0x105L5mol-5s-1[2.0x10-2M]1[2.0x10-3M]3[1.0x10-3M]2
= 5.0x105L5mol-5s-1[2.0x10-2mol L-1]x[8.0x10-9mol3 L-3]
x[1.0x10-6mol2 L-2]
= 8.0x10-11mol L-1 S-1
= 8.0x10-11 mol/L∙s
Therefore the rate of the reaction at 0°C is 8.0x10-11 mol/L∙s

71. rate = k[HI]2 = 2.5x10-4 mol L-1s-1
REACTION RATE
RATE LAW EQUATION:
EXAMPLE 2
The rate law for the decomposition of HI is:
rate = k[HI]2 = 2.5x10-4 mol L-1s-1
When [HI] is M, what is the value of the rate constant?

72. rate = k[HI]2 = 2.5x10-4 mol L-1s-1
REACTION RATE
RATE LAW EQUATION:
EXAMPLE 2
rate = k[HI]2 = 2.5x10-4 mol L-1s-1
rate = k[HI]2
(2.5x10-4 mol L-1s-1) = k
( mol L-1)2
8.0x 10-2 mol-1 L s-1 = k
8.0x 10-2 L mol-1 s-1 = k
Therefore the value of the rate constant is 8.0x 10-2 L mol-1 s-1

73. Rate Problem Sample Problem
The data below was collected for the reaction;
A(aq) B(aq) C(aq)  E(s) + F(aq)
Determine the rate law equation, including the value of k.
[A]
[B]
[C]
Initial Rate
0.10
4.0
0.20
16.0
0.40
32.0
0.30
12.0

74. 4 mol/Ls = k[0.1 mol/L]1/2[0.1mol/L]2[0.1mol/L]1
r = k[A]x[B]y[C]z
r = k[A]1/2[B]2[C]1
4 mol/Ls = k[0.1 mol/L]1/2[0.1mol/L]2[0.1mol/L]1
4 mol/Ls = k[0.1mol/L]3.5
4 mol/Ls = k x x10-4 mol3.5/L3.5
k = L2.5/mol2.5s
r = L2.5/mol2.5s [A]1/2[B]2[C]1

75. Initial Concentration Initial Rate of Formation of Products (mol/L•s)
Example
Initial Concentration
(mol / L)
Initial Rate of Formation of Products (mol/L•s)
[A]
[B]
0.10
0.20
0.40
0.30
0.60
2.40
5.40
Order for [A]:
Order for [B]:
Order of overall rxn:
k =

76. Initial Concentration Initial Rate of Formation of Products
Example
Initial Concentration
(mol/L)
Initial Rate of Formation of Products
(mol/L•s)
[SO2Cl2]
0.100
2.2 x 10-6
0.200
4.4 x 10-6
0.300
6.6 x 10-6
Order for [A]:
Order for [B]:
Order of overall rxn:
k =

77. Homework
Worksheet
pg 382 #1-4

78. Agenda P. 383 -387 - Reaction Mechanisms
HW: P. 386 # 1-3; P. 387 # 1-9

79. Reaction Mechanisms
A reaction mechanism is step or series of steps that make up a reaction.
The steps in a reaction mechanism are referred to as elementary steps.
Molecularity refers to the number of reactant molecules (ions, atoms or molecules) involved in an elementary step or making up an activated complex.
Unimolecular: steps involving 1 molecule
Bimolecular: steps involving 2 molecules
Termolecular: steps involving 3+ molecules
Simple reaction will have one step where complex reactions will consist of a number of steps.

80. Lower molecular steps tend to be faster than ones with more molecules.
The rate determining step is the slowest step in the reaction mechanism and will have the largest activation energy.
Often it is difficult to determine the actual mechanism of an reaction. In this case it must be determined through experimentally determined relationships.

81. Consider the production of NO2 from NO and O2.
2 NO + O2  2 NO2
This reaction consists of elementary steps.
1) 2 NO  N2O2 (fast, molecularity 2)
2) N2O2 + O2  2 NO2 (slow, molecularity 2
N2O2 is a reaction intermediate which is short lived and difficult to isolate.

82. Using Reaction Mechanisms to Determine the Rate Law Equation
The rate law expression can be determined from the reaction mechanism.
The rate determining step is the elementary step that determines the rate law expression.
The rate of a reaction is proportional to the concentrations of the reactants in the rate determining step raised to their molar coefficients.
If any reaction intermediates appear in this expression, you must use the other steps to eliminate these species.
Only reversible equations, denoted by ⇌ can be used to replace reaction intermediates.

83. Example 1:
The reaction, A(g) + 2 B(g)  C(g) + D(g) occurs by the mechanism below. Use that mechanism to determine the rate law expression. Step 1 A + B ⇌ AB (fast) Step 2 AB + B → C + D (slow)

84. Solution: Step 2 is the rate determining step therefore overall rate is proportional to the concentrations of the reactants in that step r= k[AB][B] AB does not appear in the overall equation therefore use step 1 to replace AB AB=[A][B] Sub this into the rate equation above r=k([A][B])[B] r=k[A][B]2

85. Example 2:
The reaction between NO and H2 is believed to occur in the following three-step process. Write the overall equation for this reaction and the rate law expression.
Step NO(g) ⇌ N2O2(g) (fast)
Step 2 N2O2(g) + H2(g) → N2O(g) + H2O(g) (slow)
Step 3 N2O(g) + H2(g) → N2(g) + H2O(g) (fast)

86. Solution: Overall Equation: 2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g) Step 2 is the rate determining step therefore overall rate is proportional to the concentrations of the reactants in that step r= k[N2O2][H2] N2O2 does not appear in the overall equation therefore use step 1 to replace N2O2 because this is the reversible step. N2O2=[NO]2 Sub this into the rate equation above r=k([NO]2)[H2] r=k[NO]2[H2]

87. Homework
pg 387 #3-5, 8, 9
worksheet

88. Review
CLINICS: P. 392 # 1-47, ;
P. 398 # 1- 25; P. 403 # 1-53, 74; P. 410 # 1- 25