1. Intro
Kinetics

2. 1) describe that chemical kinetics is the
Learning Target: I can
1) describe that chemical kinetics is the
area of chemistry that deals with the rates of reactions.
2) explain why the study of kinetics is beneficial to people.
3) describe how collision theory, concentration and proper orientation of molecules relates to the rate of a reaction.

3. Do Now (10 min) In groups of 3: Go to a poster
You will have 1 minute to quickly discuss and write down an answer.
You may NOT write the same answer as another group.
Then, each group will assign an artist, and they will have a second minute to draw an image to go with their answer (can be a picture, graph, diagram etc.)

4. Chemical Kinetics
Area of chemistry that deals with rates or speeds at which reactions occur.
T.P.S (Think-Pair-Share) Why is studying this area of chemistry important to society?
Examples of why the study of kinetics is important:
- allows determination of expiration dates on food.
- allows determination of how fast a medicine will work.
- increases efficiency of chemical processes; more product, less waste.

5. Some reactions take only a fraction of a second to occur (explosions)

6. Some Reactions are Very Slow

7. Some reactions take thousands of years (formation of coal).

8. For a rxn to take place.. What must occur?
Molecules must be moving
Molecules must come in contact (collide)

9. Collision Theory
In order for molecules to react, they must be properly oriented.
More collisions provides a better chance of correct orientations.
Higher rate of properly oriented collisions equals faster reaction.

10. Proper Orientation
In this presentation we will look at conditions that affect how fast a reaction will go, this is known as the reaction rate.

11. Notice the Orientation
A chemical reaction will occur when the reactants have enough energy and are in the right alignment to interact to break and form new bonds leading to new products. The more reactions per time the higher the reaction rate.

12. Without correct alignment

13. Not Properly Oriented
No reaction

15. Not Properly Oriented
Again, no reaction

17. Properly Oriented
They come together in the right orientation and energy and presto …

18. They react!
REACTION !

19. Several factors influence the rate of collisions…and therefore, the rate of a reaction.

20. TPS: What is the rxn? What factors could speed up/slow down this rxn?
Increased temperature = increased collision rate = increase rate of reaction.

21. Increase Temperature
Temperature is a measure of the kinetic energy of molecules.
Kinetic Energy = ½ mv2
Temperature related to the SPEED of the molecule
Temperature is a measure of the average kinetic energy of molecules. As you remember from the first half of physical science, kinetic energy is the energy of motion and is expressed as ½ mass time the velocity squared.

22. Concentration
Concentration is the amount of a substance present; the more present the higher the concentration

23. High Concentration of Reactants Much interaction Low Concentration
The more reactants present in a system the greater the chance they will interact leading to reactions and the formation of products.
High Concentration
of Reactants
Much interaction
Low Concentration
of Reactants
Little interaction

24. Oxygen in Atmosphere
Pure Oxygen

25. TPS:
Would you expect medicine in the form of a powder to enter the blood stream slower or faster than the same medicine in tablet form?
Why?
FASTER: More Surface Area!

26. Physical States and Surface Area
A phase is the state the matter is in

27. When solids and gases react, the reaction is limited to the surface of the solid.

28. Grain Elevator Fire
Grinding chemicals with a mortar and pestle increases reaction rate.

29. Gas Gas Gas Gas Different States; Less Interaction
Different States; Less Interaction Possible
The state the reactants are in is also very important in determining how fast a reaction will occur.
Liquid
Liquid
Liquid

30. Gas Same States; More Interaction Possible Gas
If they are in the same phase (gas, liquid or dissolved in water) they will be more likely be able to come in contact and react. Along this line, if the area exposed is increased, (increased surface area) reaction rate can also increase.
Gas

31. Mg (s) + H2O (l)  No Reaction Mg (s) + H2O (g)  MgO + H2
Mg metal reacts with high temperature steam, but not water.
Mg (s) + H2O (l)  No Reaction
Mg (s) + H2O (g)  MgO + H2

32. Catalysts Very Important to Living Organisms!!!
The last of factors that will increase the rate of a reaction on our list is the presents of a catalyst.
Very Important to Living Organisms!!!

33. If you remember from the beginning of this unit a list of symbols were given that represent different conditions for reactions, the last one represent a catalyst.

34. Reactants (Substrates) Catalyst (Enzyme)
The reactants are shaped so they will fit in the enzyme

35. Enzyme reactants Complex
Once inside the enzyme, they are in close proximity and correctly orientated.

36. Enzyme products Complex
They can react at the active site of the enzyme to form the product

37. Products Unchanged Enzyme
They leave the enzyme complex as the new product.
Unchanged Enzyme

38. Reactants Catalyst Enzyme Enzyme reactants Complex Enzyme products
This shows an overview of the enzyme combining with the reactants to form the product
Products
Unchanged Enzyme

39. A catalyst is a substance that is present in a reaction that speeds up the reaction without being changed at the end of the reaction. Enzymes that are found in biological systems are great examples of catalysts. They can be described as lock and keys, where the lock is an enzyme with a specific shape that will hold a key, the reactants, so the reactants can come close together and react to form the products.

40. Enzyme animation: “Lock and Key Model”
Were there any terms in that video that are new to you?

41. Activation Energy-- energy need to get a reaction started
This process reduces the energy of activity, the energy required to get the reactants together so they can react.
Activation Energy-- energy need to get a reaction started

42. Activation Energy
This can be represented by the idea of a child trying to ride a bike over a hill to get the other side.

43. Activation Energy
He needs to put in a given amount of energy to get to the top.

44. Activation Energy
The top is were the energy needs change

45. Activation Energy
After that he can coast down to the bottom.

46. Activation Energy

47. Activation Energy
The presents of the catalyst is like lowering the height of the hill, thus requiring less energy for the reaction to occur.

48. Activation Energy

49. Activation Energy

50. Activation Energy

51. Activation Energy

52. Activation Energy
As a note catalyst can never make a reaction occur that would not normally occur in nature, it can only speed the up the reaction by lowering the activation energy.

53. * SUMMARY of lesson 1: TPS!
*on a molecular level, rxn rates depend on the frequency of collisions between molecules.
*for a collision to lead to a chemical rxn, it must occur with enough energy to break bonds, and with suitable orientation for new bonds to form.
Increase temperature
Increase concentration
Physical State and Surface Area
The presence of a catalyst
As a review, along with correct orientation and energy the reaction rate will depend on, temperature, concentration, phase and the presents of a catalyst.

54. Homework:
Mastering Chem: post-lab: Chapter 14

55. Exit Ticket: (3 min) Answer on a separate sheet of paper:
What are the 4 factors that affect the rate of a chemical reaction?
1.) Most reactions increase in rate if one or more of the reactants have an increase in concentration.
2.) As temperature is increased, the rate of reaction increases.
3.) Rates of reaction can be increased by introducing a catalyst.
Most living organisms depend on enzymes, which are proteins that act as catalysts.
4.) reactions involving solids often increase in rate as surface area is increased.

56. Do Now (10 min) Learning Target: I can
Calculate reaction rates given concentration of reactants and products over time.
Utilize rate graphs to calculate instantaneous rates of reactions.
Do Now: Answer the 6th question on your MC Do Now Sheet

57. A student performs an acid-base titration and plots the experimental results in the graph above. Which of the following statements best explains the experimental findings?

58. Review: Question 6 Options Why is each option partially right?
Choose the most correct answer, and why
A strong acid was titrated with a strong base, as evidenced by the equivalence point at pH = 7
A strong acid was titrated with a strong base, as evidenced by the equivalence point at pH > 7
A weak acid was titrated with a strong base, as evidenced by the equivalence point at pH > 7
A weak acid was titrated with a weak base, as evidenced by the equivalence point at pH approximately 7

59. Now for the details and calculations!
Learning Target: I can
Calculate reaction rates given concentration of reactants and products over time.
Utilize rate graphs to calculate instantaneous rates of reactions.
Now for the details and calculations!

60. The units of a reaction rate are mol L-1 s-1 = M s-1.
The Rate of a Chemical Reaction is the rate of change in the concentration of reactants and/or products per unit time.
Reaction rates are determined by measuring the concentration of one or more chemicals involved at different times during the chemical reaction.
Reaction rates are expressed as the number of moles per liter that react each second during the reaction.
The units of a reaction rate are mol L-1 s-1 = M s-1.

61. Considering Reaction A  B
The reaction rate is a measure of how quickly A is consumed (disappears) or B is produced (appears).
Average rate of reaction can be written
(remember: Δ = final – initial)
This is a measure of the average rate of appearance of B.

62. Average rate can also be written in terms of A:
This is the rate of disappearance of A (negative value makes final rate positive).

63. Notice the decrease in rate as A is used up.
Data for reaction AB
Notice the decrease in rate as A is used up.

64.  Moles A  Moles B

65. Rates in Terms of Concentration
The volume in the reaction vessel remains constant.
Therefore, in analyzing the reaction AB we can actually consider the rate in terms of Molarity.
This gives us the units M/s (molarity per second).
We can consider our moles in the graph to be moles per volume due t the container being of fixed volume.

66. Start with one mole of A at time zero, measure amounts of A and B at given time intervals.

67. Data for Reaction AB
Time A B
0 seconds
1.0 mol
0.0 mol
20 seconds
0.54 mol
0.46 mol
40 seconds
0.30 mol
0.70 mol
Calculate the average rate of appearance of B over the time interval from 0 to 40 seconds.

68. Same Value: Notice negative sign makes value positive.
Lets Try Another!
Time A B
0 seconds
1.0 mol
0.0 mol
20 seconds
0.54 mol
0.46 mol
40 seconds
0.30 mol
0.70 mol
Calculate the average rate of disappearance of A over the time interval from 0s to 40s.
Negative sign needed to make rate = positive #. Rates are always reported as positive values of appearance or disappearance.
Same Value: Notice negative sign makes value positive.

69. You Try One!
Time A B
0 seconds
1.0 mol
0.0 mol
20 seconds
0.54 mol
0.46 mol
40 seconds
0.30 mol
0.70 mol
Calculate the average rate of appearance of B over the time interval from 20s to 40s.
Negative sign needed to make rate = positive #. Rates are always reported as positive values of appearance or disappearance.

70. Consider the reaction between butyl chloride and water:
C4H9Cl(aq) + H2O(l) C4H9OH(aq) +HCl(aq)
Brackets indicate concentration.
Whenever a brackets surround a chemical substance, it indicates the concentration of that substance (usually expressed as molarity)

71. TPS: with your partner determine the avg
TPS: with your partner determine the avg. rate of disappearance of C4H9Cl, over time:
Time (s)
[C4H9Cl] (M)
0.1 50
0.09
100
0.082
150
0.074
200
0.067
250
0.06
300
0.055
350
0.049
400
0.045
450
0.040
500
0.036
550
0.033
600
0.03
650
0.026
700
0.024
750
0.022
800
0.02

72. Show your Work: (Students will volunteer to come to the board and explain their work to the class)

73. Average Reaction Rate reaction rate over a given amount of time.
Example: Change in concentration during the time period of 20s-40s during the reaction (as in previous examples).
Instantaneous Reaction Rate reaction rate at a given moment in time.
Notes: Initial Rate where time = 0 is an instantaneous rate Use kinetic curve to calculate this.

74. Sometimes the unit on the y axis is moles (Liters are assumed).

75. ( ) = Remember the equation for finding the slope (math class).
We use the same equation to determine the instantaneous (for a specific time) rate of a reaction from a kinetic curve. (
) =
Reaction rates have + values; - sign makes rate of disappearance +.

76. Consider AGAIN the reaction between butyl chloride and water:
C4H9Cl(aq) + H2O(l) C4H9OH(aq) +HCl(aq)
Brackets indicate concentration.
Whenever a brackets surround a chemical substance, it indicates the concentration of that substance (usually expressed as molarity)

77. Using the curve created from the data, we can determine the instantaneous rate for any given point on the curve. (slope of tangent line; tangent line touches curve at point of interest).
The instantaneous rate if obtained from the straight-line tangent that touches the curve at the point of interest.
The slope of the tangent give the instantaneous rate of reaction at these times.
Slope is rise over the run.

78. Calculate the rate of disappearance of C4H9Cl at t = 0.
0.06 M – 0.1M
Rate = -
200 s – 0.0 s
The instantaneous rate if obtained from the straight-line tangent that touches the curve at the point of interest.
The slope of the tangent give the instantaneous rate of reaction at these times.
Slope is rise over the run.
= 2 x 10-4 M/s

79. Calculate the rate of disappearance of C4H9Cl at t = 300.
0.02 M – M
Rate = -
600 s – 300 s
The instantaneous rate if obtained from the straight-line tangent that touches the curve at the point of interest.
The slope of the tangent give the instantaneous rate of reaction at these times.
Slope is rise over the run.
= 1 x 10-4 M/s

80. TPS: Calculate the rate of disappearance of C4H9Cl at t = 700s.

81. Time (min) [HCl] (M) [CH3Cl] (M)
Practice Activity: Use the data table below to answer the problem on your index card. If you complete your problem quickly, go to a group that is still working, and offer your assistance.
CH3OH (aq) + HCl (aq)  CH3Cl (aq) + H2O (l)
Time (min)
[HCl] (M)
[CH3Cl] (M)
0.0
1.85
0.491
54.0
1.58
0.990
107.0
1.36
1.28
215.0
1.02
1.60
430.0
0.580
2.03

82. Time (min) [HCl] (M) [CH3Cl] (M)
Practice Activity: Let’s graph!!
CH3OH (aq) + HCl (aq)  CH3Cl (aq) + H2O (l)
Time (min)
[HCl] (M)
[CH3Cl] (M)
0.0
1.85
0.491
54.0
1.58
0.990
107.0
1.36
1.28
215.0
1.02
1.60
430.0
0.580
2.03

83. Now, choose 3 points on your graph to determine the instantaneous rate

84. Answer on a separate sheet of paper
Exit Ticket:
Answer on a separate sheet of paper
How do you determine the instantaneous rate law?

85. Quiz (8 min)
Learning Target: I can…
Explain how stoichiometric relationships in a chemical equation that are not one-to-one, affect the corresponding rates of disappearance /appearance of reactants/products, and over all rate of the rxn; Express the way in which reaction rates depend on reactant concentrations using rate law.
Complete the Quiz silently and independently

86. Recap Quiz
Write the name or the formula for the following polyatomic ions:
a. Sulfate b. CO32-
c. Nitrite d. phosphite
e. BrO3- f. Cl-

87. Recap Quiz Time (sec) [A] [B] 1.00 0.00 10 .50 0.50 20 0.25 0.75
1.00
0.00 10
.50
0.50 20
0.25
0.75
For the disappearance of A over the time interval 0 seconds to 20 seconds   
For the appearance of B over the time interval 10 seconds to 20 seconds

88. Reaction Rates and Stoichiometry
What if the ratio is not 1:1?
H2(g) + I2(g) ⎯⎯→ 2 HI(g)
Only 1/2 HI is made for each H2 used. 88

89. Reaction Rates and Stoichiometry
To generalize, for the reaction
aA + bB
cC + dD
Reactants (decrease)
Products (increase) 89

90. THE RATE LAW USING INITIAL CONCENTRATIONS
Each reaction has its own equation that gives its rate as a function of reactant concentrations.

91. Calculating The Initial Rate Law
The rate law expresses the mathematical relationship of the rate of a reaction to the concentrations of the reactants. The products play no role in the calculations.
aA + bB products
Rate = k [A] x [B] y
Rate = Molarity /sec
k= rate constant and it varies with each reaction, temperature dependent.
[A] [B] = Concentrations of reactants in M (moles/liter)
x and y = Values for the reaction order. They tell us how important each of the reactants is in regard to speed.The higher the number the more value it has to overall speed.
x does not equal a and y does not equal b

92. aA + bB products
Trial
[A] M/S
[B] M/S
Rate M/S 1
0.100
.100
2.00 x10-3 2
0.200
4.00 x 10 -3 3
.200
16.00 x 10 -3

93. Concentration and Rate
Compare Experiments 1 and 2: when [NH4+] doubles, the initial rate doubles. 93

94. Ex: Decomposition of ammonia over tungsten
ZERO ORDER
The rate of change is independent of the concentration of the reactant.
A products
Rate = k [A] 0
Rate = k
Ex: Decomposition of ammonia over tungsten

95. Concentration and Rate
Likewise, compare Experiments 5 and 6: when [NO2-] doubles, the initial rate doubles. 95

96. INTEGRATED RATE LAWS

97. Zero Order Integrated Rate Law
The rate of change is independent of the concentration of the reactant.
[A]t = - kt + [A]0
Y = mx + b
This means that we can graph the concentration as a function of time and it should create a straight line that will give a slope and then you can find the value of -k!

98. First Order Integrated Rate Law
The rate is proportional to the concentration of a single reactant raised to the first power
Using calculus you get the following equation:
ln[A]t = - kt + ln [A]0
Y = mx + b
If a reaction is first-order, a plot of ln[A]t vs t
will give a straight line with a slope of -k.
So, use graphs to determine rxn order.

99. Second Order Integrated Rate Law
The rate is proportional to a either the concentration of a single reactant raised to the second power or two reactants each raised to the first power.
y = mx + b
also in the form 99

100. How are these equations helpful?
You now have a technique for determining the order of a reaction by observing graphs created from experimental data of concentration vs time.
If a simple graph of concentration vs time is a straight line then you know that it is a zero order reaction.
If the natural log of concentration vs time is a straight line then you know that it is a first order reaction.
If the reciprocal concentration vs time is a straight line then you know that it is a second order.

101. The conclusion:
In all cases, once you get a straight line fit , you can then get the rate constant from calculating the slope of the line.
Slope (m) = ΔY ΔX

103. Zero Order Integration
2NH3(g) N2 + 3 H2 (g) [A]t = - kt + [A]0

104. First-Order Integration
When lnCH3NC is plotted as a function of time, a straight line results.
The reaction is first-order, a plot of ln [A]t vs. t will yield a straight line with a slope of -k. .
104

105. Second-Order Integration
So if a reaction is second-order in A, a plot of
vs. t will yield a straight line with a slope of k.
105

106. Determining rxn order
The decomposition of NO2 at 300°C is described by the equation
2NO2 (g)
2NO (g) + O2 (g)
and yields these data:
Time (s)
[NO2], M
0.0
50.0
100.0
200.0
300.0
ln [NO]2
1/ [NO2]
106

107. [NO2] vs Time
The plot is not a straight line, so the process is not zero-order.

108. Determining rxn order Graphing ln [NO2] vs. t yields:
The plot is not a straight line, so the process is not first-order in [A].
Time (s)
[NO2], M
ln [NO2]
0.0
-4.610
50.0
-4.845
100.0
-5.038
200.0
-5.337
300.0
-5.573
Does not fit:
108

109. Second-Order Processes
A graph of 1/[NO2] vs. t gives this plot.
This is a straight line. Therefore, the process is second-order in [NO2].
k= 0.543
Time (s)
[NO2], M
1/[NO2]
0.0
100
50.0
127
100.0
154
200.0
208
300.0
263
109

111. Half-Life
Half-life is defined as the time required for one-half of a reactant to react.
Half-Life is also defined as the rate of decay of a radioactive substance. A radioactive substance is one that will slowly decay into a more stable form as time goes on.
111

112. Determining a Half Life
To determine a half life, t½, the time required for the initial concentration of a reactant to be reduced to one-half its initial value, we need to know:
The order of the reaction or enough information to determine it.
The rate constant, k, for the reaction or enough information to determine it.
In some cases, we need to know the initial concentration, [Ao]
Substitute this information into the equation for the half life of a reaction with this order and solve for t½.

113. Half-Life
For a first-order reaction the formula is as follows and it is also found on the AP Reference sheet in the Kinetics section. A Products
113

114. 1st Order reactions Half-life is constant

115. Half-Life for a Second- Order Reaction
A products
A + B products
115

116. 2nd order reaction half-life is variable.
It increases with decreasing concentration.

117. 117

118. Do Now (4 min) Learning Target: I can…
Determine the reaction order for multistep reactions, and Investigate concentration and temperature as factors that affect rates of reaction through a virtual lab
Do Now: (Answer on your Do Now Sheet)
Using the following reaction profile:
How many intermediates are
formed ?
2) How many transition states?
3) Which step is the fastest?
4) For the reaction, A  D, is
(delta)E positive, negative,
or zero?

119. Finding the Overall Rate Law by Utilizing Reaction Mechanisms

120. Reaction Mechanisms Not all reactions occur in one step.
The step by step sequence of 2 or more simple reactions that combine to form the overall reaction.
A chemical mechanism describes in detail exactly what takes place at each stage of an overall chemical reaction.
120

121. Reaction Mechanism Terms
Complex reaction- overall reaction
elementary steps- series of simpler reactions
that combine to form the complex reaction.
Intermediate-a substance produced in one elementary step and consumed in another.
Catalyst-increases the rate but not consumed by the reaction.It remains unchanged.

122. Multistep Mechanisms
In a multistep process, one of the steps will be slower than all others.
The slowest step is called the rate-determining step.It will be the one used to find the rate law for the overall reaction.
The overall reaction cannot occur faster than this slowest, rate-determining step
122

123. Hypothetical Reactions #1
P + Q B (SLOW)
2) B + Q R (FAST)
What is the overall rate law for this reaction?
What is the complex equation for this reaction?
123

124. Hypothetical Reactions # 2
P + Q B (SLOW)
2) B + Q P + 2R (FAST)
What is the overall rate law for this reaction?
What is the complex equation for this reaction?

125. Practice Reactions 2NO N2O2 (slow) N2O2 + H2 N2O + H2O (fast)
N2O + H H2O +N (fast)
a.) Which is the rate determining step?
b.) Does this reaction have a catalyst?
c.) What is the complex equation for this reaction?
d.) What is the rate law?

126. Answers to practice problem
A) Step one is the rate determining step
B) No catalyst
C) 2NO + 2H N2 + 2H2O
D) rate = k [NO]2

127. Please Remember!!!!
YOU CAN ONLY USE THE COEFFICIENT METHOD TO DETERMINE THE RATE LAW OF ELEMENTARY STEPS.
YOU CANNOT USE THE COEFFICIENTS FROM THE COMPLEX EQUATION AND DO THE SAME THING.

128. Virtual Lab (55 min)
Work independently to complete the virtual lab on the computer
Actively manipulate the concentrations and temperatures in the lab in order to determine the rate law of the reaction
Can sit in pairs to help each other through the complex calculations

129. Answer on a separate sheet of paper
Exit Ticket:
Answer on a separate sheet of paper
How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction:
2O3 (g)  3O2 (g)
b) If the rate at which O2 appears, ∆[O2]/∆t, is 6.0 x 10-5 M/s at a particular instant, at what rate is O3 disappearing at this same time,-(∆[O3]/∆t)?

130. Practice
Grab a laptop
Go to:
Independently work through 3 different problems
Show your work on a separate sheet of paper