1. Unit 5: Chemical Kinetics
Chapter 14
Day 1: Slides 2-21
Day 2: Slides 21-35
James F. Kirby
Quinnipiac University
Hamden, CT
Edited By Myriah Day

2. Warm Up
What are some factors you can name that control or effect how fast a reaction takes place?

3. Agenda Lecture Notes: Chemical Kinetics
“Guided” Guided Inquiry: Rate of Reaction

4. Chemical Kinetics = Rate of Reactions
If you’re making something, you might think making it last is a good thing.
But what if you’re making a pesticide with known detrimental impacts on human health? Then you may only want it to stay intact for a few days after it has been applied before it decomposes into products that aren’t harmful.

5. Chemical Kinetics = Rate of Reactions
Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism, a molecular-level view of the path from reactants to products.

6. Factors that Affect Reaction Rates
Physical state of the reactants
Reactant concentrations
Reaction temperature
Presence of a catalyst

7. Physical State of the Reactants
The more readily the reactants collide, the more rapidly they react.
Homogeneous reactions are often faster.
Heterogeneous reactions that involve solids are faster if the surface area is increased; i.e., a fine powder reacts faster than a pellet or tablet.

8. Reactant Concentrations
Increasing reactant concentration generally increases reaction rate.
Since there are more molecules, more collisions occur.

9. Temperature
Reaction rate generally increases with increased temperature.
Kinetic energy of molecules is related to temperature.
At higher temperatures, molecules move more quickly, increasing numbers of collisions and the energy the molecules possess during the collisions.

10. Presence of a Catalyst
Catalysts affect rate without being in the overall balanced equation.
Catalysts affect the kinds of collisions, changing the mechanism (individual reactions that are part of the pathway from reactants to products).
Catalysts are critical in many biological reactions.

11. Write at least one question in the margin.
Chunk Notes
AND
Write at least one question in the margin.
TIME: 2 MINUTES

12. Reaction Rate
Rate is a change in concentration over a time period: Δ[A]/Δt:
Δ means “change in” (final-initial)
[A] means molar concentration of substance A
t represents time.
Types of rate measured:
average rate
instantaneous rate
initial rate

13. Following Reaction Rates
Rate of a reaction is measured using the concentration of a reactant or a product over time.
In this example, [C4H9Cl] is followed.

14. Following Reaction Rates
The average rate is calculated by the – (change in [C4H9Cl]) ÷ (change in time).
The table shows the average rate for a variety of time intervals.

15. Plotting Rate Data
A plot of the data gives more information about rate.
The slope of the curve at one point in time gives the instantaneous rate.
The instantaneous rate at time zero is called the initial rate; this is often the rate of interest to chemists.
This figure shows instantaneous and initial rate of the earlier example.
Note: Reactions typically slow down over time!

16. Relative Rates
As was said, rates are followed using a reactant or a product. Does this give the same rate for each reactant and product?
Rate is dependent on stoichiometry.
If we followed use of C4H9Cl and compared it to production of C4H9OH, the values would be the same. Note that the change would have opposite signs—one goes down in value, the other goes up.

17. Relative Rates and Stoichiometry
What if the equation is not 1:1?
Do Sample Exercise 14.3

18. Relative Rates and Stoichiometry
In general, for the reaction
a A + b B  c C + d D
The rate is given by
Rate = -1 Δ [A] = -1 Δ [B] = 1 Δ [C] = 1 Δ [D]
a Δt b Δt c Δt d Δt

19. Your Turn!
What will the relative rates be for: 2 O3 (g) → 3 O2 (g) Remember, in general, for the reaction a A + b B  c C + d D Rate = -1 Δ [A] = -1 Δ [B] = 1 Δ [C] = 1 Δ [D] a Δt b Δt c Δt d Δt

20. Write at least one question in the margin.
Chunk Notes
AND
Write at least one question in the margin.
TIME: 2 MINUTES

21. –0.750 M/min. –0.225 M/min. –2.25 M/min. 2.25 M/min.
Consider the reaction X2 + 3 Y2 3 XY3. If Δ[X]Δt is –0.750 M/min, then Δ[XY3]/Δt is
–0.750 M/min.
–0.225 M/min.
–2.25 M/min.
2.25 M/min.
Answer: d

22. Switch notebooks with a table partner
Switch notebooks with a table partner. Is there any information that they have that you don’t? If so, add it!

23. Determining Concentration Effect on Rate
How do we determine what effect the concentration of each reactant has on the rate of the reaction?
We keep every concentration constant except for one reactant and see what happens to the rate. Then, we change a different reactant. We do this until we have seen how each reactant has affected the rate.

24. An Example of How Concentration Affects Rate
Experiments 1–3 show how [NH4+] affects rate.
Experiments 4–6 show how [NO2−] affects rate.
Result: The rate law, which shows the relationship between rate and concentration for all reactants:
Rate = k [NH4+] [NO2−]

25. Rate Law and Order of Reaction
The exponents tell the order of reaction with respect to each reactant.
In our example from the last slide:
Rate = k [NH4+] [NO2−]
The order with respect to each reactant is 1. (It is first order with respect to NH4+ and NO2−.)
The overall reaction is second order (1 + 1 = 2; we just add up all of the reactants’ orders to get the reaction’s order). This reaction is said to have an overall reaction order of 2, or is 2nd order overall

26. Rate Law and Rate Constant
Rate = k [NH4+] [NO2−]
The rate law equation shows how rate depends on concentrations.
For the general reaction
a A + b B  c C + d D
The rate law generally has the form of
Rate = k [A]m [B]n
What is k? It is the rate constant. It is a temperature- dependent quantity.
The exponents m and n are typically small whole numbers. If we know these exponents, the “orders”, we can figure out some of the steps of complex reactions.

27. Learning Check: Share with a neighbor
Write the general form for a rate law that has reactant A and B and has this equation:
A + B  C

28. Context
In 2000, over 20 million kilograms of the pesticide 1,3- dichloropropene (1,3-D) were applied to crops in the U.S. [Before it’s use?] Scientists investigated the rate of decomposition of 1,3-D in acidic, basic, and neutral solutions as well as in soil. For each case they generated plots of the amount of intact 1,3-D persisting versus time and found that the reaction could be characterized as pseudo first order. Knowing the order allowed them to determine the half life of intact 1,3-D. In acidic solutions, the half life was 8 days. In basic solutions, it was four days. Experimentally determined data like this is vital to the ability of society to use chemicals more wisely – to improve not only the quality of life for humans, but to preserve the health of our ecosystems.

29. Warm Up
Take a moment and discuss with a partner what you think the most important “chunk” of information was last class.

30. Rate Law and Rate Constant
Once we know the rate law for a reaction and the reaction rate for a set of reactant concentrations, we can calculate k.
Example page 622
Once we have both the rate law and the value of k for a reaction, we can predict the reaction rate for any set of concentrations.

31. Reaction Orders: Exponents in the Rate Law
The rate law generally has the form of
Rate = k [A]m [B]n
The exponents in the rate law equation are called reaction orders
These numbers tell us exactly how the rate will be affected by reactant concentrations.
In the previous example,
Rate = k [NH4+] [NO2−]
The rate doubles when the concentration of NH4+ doubles, triples when it triples, etc. All because the rate law shows that the exponent for [NH4+] is one.

32. first. second. third. d. fourth.
Reaction: A + B  C + D Rate = k[A][B] The overall order of this reaction is
first.
second.
third.
d. fourth.
Answer: b

33. first. second. third. d. fourth.
Reaction: A + B  C + D Rate = k[A][B] The overall order of this reaction is
first.
second.
third.
d. fourth.
Answer: b

34. Reaction Orders: Exponents in the Rate Law
But what if the exponent for either reactant concentration was not one? If the order for any reactant is 2, then doubling the concentration of that reactant will do what to the rate? What if you triple the concentration, what will happen to the rate?
Although the exponents in the rate law equation are sometimes the same as the coefficients in the balanced equation, this is not always the case. For any reaction, the rate law must be determined experimentally.

35. and write at least one question in the margin.
Chunk Notes
and write at least one question in the margin.

36. Warm Up 4/1 Take out notes for me to check
The following data were measured for the reaction of nitric oxide with hydrogen:
2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g)
Determine the rate law for this reaction
Calculate the rate constant
Calculate the rate when [NO] = M and [H2] = M
Experiment
[NO]
[H2]
Initial Rate (M/s 1
0.10
1.23 x 10-3 2
0.20
2.46 x 10-3 3
4.92 x 10-3

37. First Order Reactions
Some rates depend only on one reactant to the first power.
These are first order reactions.
The rate law becomes:
Rate = k [A]

38. First Order Reactions rate = −Δ [A] / Δt So: k [A] = −Δ [A] / Δt
rate = k [A]
rate = −Δ [A] / Δt
So: k [A] = −Δ [A] / Δt
Rearrange to: Δ [A] / [A] = − k Δt
Integrate: ln ([A] / [A]o) = − k t
Rearrange: ln [A] = − k t + ln [A]o
Note: this follows the equation of a line: y = m x + b
So, a plot of ln [A] vs. t is linear.

39. An Example: Conversion of Methyl Isonitrile to Acetonitrile
The equation for the reaction:
CH3NC → CH3CN
It is first order.
Rate = k [CH3NC]

40. Finding the Rate Constant, k
Besides using the rate law, we can find the rate constant from the plot of ln [A] vs. t.
Remember the integrated rate law:
ln [A] = − k t + ln [A]o
The plot will give a line. Its slope will equal −k.

41. Half-life ln [A] = − k t + ln [A]o ln 2 = k t½ or t½ = 0.693/k
Definition: The amount of time it takes for one-half of a reactant to be used up in a chemical reaction.
First Order Reaction:
ln [A] = − k t + ln [A]o
ln ([A]o/2) = − k t½ + ln [A]o
− ln ([A]o/2) + ln [A]o = k t½
ln ([A]o / [A]o/2) = k t½
ln 2 = k t½ or t½ = /k

42. and write at least one question in the margin.
Chunk Notes
and write at least one question in the margin.

43. Share one thing you learned
Table Talk
Share one thing you learned
Share one thing you are not sure about or confused about

44. 10 seconds 69 seconds 100 seconds d. 690 seconds
The value of the rate constant (k) for a first-order reaction is sec–1. What is the half-life of this reaction?
10 seconds
69 seconds
100 seconds
d. 690 seconds
Answer: b

45. 10 seconds 69 seconds 100 seconds d. 690 seconds
The value of the rate constant (k) for a first-order reaction is sec–1. What is the half-life of this reaction?
10 seconds
69 seconds
100 seconds
d. 690 seconds
Answer: b

46. Second Order Reactions
Some rates depend only on a reactant to the second power.
These are second order reactions.
The rate law becomes:
Rate = k [A]2

47. Solving the Second Order Reaction for A → Products
rate = k [A]2
rate = − Δ [A] / Δ t
So, k [A]2 = − Δ [A] / Δ t
Rearranging: Δ [A] / [A]2 = − k Δ t
Using calculus: 1/[A] = 1/[A]o + k t
Notice: The linear relationships for first order and second order reactions differ!

48. An Example of a Second Order Reaction: Decomposition of NO2
A plot following NO2 decomposition shows that it must be second order because it is linear for 1/[NO2], not linear for ln [NO2].
Equation:
NO2 → NO + ½ O2

49. Half-Life and Second Order Reactions
Using the integrated rate law, we can see how half-life is derived:
1/[A] = 1/[A]o + k t
1/([A]o/2) = 1/[A]o + k t½
2/[A]o −1/[A]o = k t½
t½ = 1 / (k [A]o)
So, half-life is a concentration dependent quantity for second order reactions!

50. Zero Order Reactions
Occasionally, rate is independent of the concentration of the reactant:
Rate = k
These are zero order reactions.
These reactions are linear in concentration.

51. Select the incorrect statement.
The exponents in a rate law are taken from the coefficients in the balanced reaction.
For a first-order reaction, the plot of ln[A] versus time is linear.
For a zero-order reaction, the plot of [A] versus time is a parabola.
The slope of 1/[A] versus time is the half-life of a first-order reaction.
Answer: c

52. Select the incorrect statement.
The exponents in a rate law are taken from the coefficients in the balanced reaction.
For a first-order reaction, the plot of ln[A] versus time is linear.
For a zero-order reaction, the plot of [A] versus time is a parabola.
The slope of 1/[A] versus time is the half-life of a first-order reaction.
Answer: c

53. zeroth first second d. third
If tripling the concentration of reactant A multiplies the rate by a factor of nine, the reaction is _______ order in A.
zeroth
first
second
d. third
Answer: c

54. zeroth first second d. third
If tripling the concentration of reactant A multiplies the rate by a factor of nine, the reaction is _______ order in A.
zeroth
first
second
d. third
Answer: c

55. Factors That Affect Reaction Rate
Temperature
Frequency of collisions
Orientation of molecules
Energy needed for the reaction to take place (activation energy)

56. Temperature and Rate
Generally, as temperature increases, rate increases.
The rate constant is temperature dependent: it increases as temperature increases.
Rate constant doubles (approximately) with every 10 ºC rise.

57. Frequency of Collisions
The collision model is based on the kinetic molecular theory.
Molecules must collide to react.
If there are more collisions, more reactions can occur.
So, if there are more molecules, the reaction rate is faster.
Also, if the temperature is higher, molecules move faster, causing more collisions and a higher rate of reaction.

58. The Collision Model
In a chemical reaction, bonds are broken and new bonds are formed.
Molecules can only react if they collide with each other.

59. Orientation of Molecules
Molecules can often collide without forming products.
Aligning molecules properly can lead to chemical reactions.
Bonds must be broken and made and atoms need to be in proper positions.

60. Energy Needed for a Reaction to Take Place (Activation Energy)
The minimum energy needed for a reaction to take place is called activation energy.
An energy barrier must be overcome for a reaction to take place, much like the ball must be hit to overcome the barrier in the figure below.

61. zeroth first second d. third
If tripling the concentration of reactant A multiplies the rate by a factor of nine, the reaction is _______ order in A.
zeroth
first
second
d. third
Answer: c

62. zeroth first second d. third
If tripling the concentration of reactant A multiplies the rate by a factor of nine, the reaction is _______ order in A.
zeroth
first
second
d. third
Answer: c

63. Warm Up
Below are graphs of how the concentration of a reactant, A, changed over time during a reaction. Based on the graphs, what order reaction is this?

64. Today’s Plan Kinetics of Crystal Violet Fading
Read: New Lab Handout
Answers to Pre-Lab Together
Introductory Activity (in lab)
Guided Inquiry Design Discussion/Help

65. Kinetics of Crystal Violet Fading
Concentration
(mol/L)
Volume of
2.5 x10-5 CV (mL)
Volume of distilled water
2.5 x10-5 = 25 uM
10.0 mL
0.0 mL
2.5 x10-6 = 2.5 uM
5.0 x10-6 = 5.0 uM
7.5 x10-6 = uM
10.0 x10-6 = 10.0 uM
12.5 x10-6 = 12.5 uM
5.0 mL
Pre-Lab 1.

66. Kinetics of Crystal Violet Fading
Guided Inquiry Design and Procedure
Part A
CV+ + OH-  CVOH
What percent of OH- ions will remain at the end of a reaction if the initial concentrations of each reactant were equal (1:1 ratio)?

67. WARM UP 4/12 Please take out a computer from the computer cart.
Open the file I sent you and your team over

68. Kinetics of Crystal Violet Fading
Review:
CV+ + OH-  CVOH
purple colorless
Rate = k[CV+]m [OH-]n

69. Kinetics of Crystal Violet Fading
Rate Law:
Rate = k’ [CV+]m
where k’ = k [OH-]n
The constant k’ is a new “pseudo” rate constant. The new pseudo rate law is valid when the concentration of OH- ions is much greater than the concentration of CV+ ions. Under these conditions, the [OH-] term in the original rate law will not change much over the course of the reaction and may be treated as a constant in the rate equation.

70. Kinetics of Crystal Violet Fading

71. Kinetics of Crystal Violet Fading
BIG IDEA:
The more color, the more concentrated, the more absorbance.
Absorbance = Concentration

72. Part B: Procedure Outline
NaOH conc. needs to be changed, not CV+, so that the effect on k’ can be calculated and the order of reaction with respect to OH- determined

73. Part B: Procedure Outline
NaOH conc. needs to be changed, not CV+, so that the effect on k’ can be calculated and the order of reaction with respect to OH- determined
Colorimeter must be zeroed with a blank, equal volumes M NaOH and water

74. Part B: Procedure Outline
NaOH conc. needs to be changed, not CV+, so that the effect on k’ can be calculated and the order of reaction with respect to OH- determined
Colorimeter must be zeroed with a blank, equal volumes M NaOH and water
To make 10 mL of 0.01 M NaOH, measure 5.0 mL of 0.02 M NaOH in a serological pipet and…

75. Part B: Procedure Outline
NaOH conc. needs to be changed, not CV+, so that the effect on k’ can be calculated and the order of reaction with respect to OH- determined
Colorimeter must be zeroed with a blank, equal volumes M NaOH and water
To make 10 mL of 0.01 M NaOH, measure 5.0 mL of 0.02 M NaOH in a serological pipet and…
Clean the pipet with small portions of crystal violet solution. Measure 10 mL of 25 uM crystal violet solution…

76. Part B: Procedure Outline
NaOH conc. needs to be changed, not CV+, so that the effect on k’ can be calculated and the order of reaction with respect to OH- determined
Colorimeter must be zeroed with a blank, equal volumes M NaOH and water
To make 10 mL of 0.01 M NaOH, measure 5.0 mL of 0.02 M NaOH in a serological pipet and…
Clean the pipet with small portions of crystal violet solution. Measure 10 mL of 25 uM crystal violet solution…
After the absorbance vs. time data are recorded, the absorbance values need to be converted into concentration, then separate data columns for ln[CV+] and 1/[CV+] need to be created and analyzed to determine k’ for the new concentration of NaOH, 0.01 M.

77. Transition State (Activated Complex)
Reactants gain energy as the reaction proceeds until the particles reach the maximum energy state.
The organization of the atoms at this highest energy state is called the transition state (or activated complex).
The energy needed to form this state is called the activation energy.

78. Reaction Progress
Plots are made to show the energy possessed by the particles as the reaction proceeds.
At the highest energy state, the transition state is formed.
Reactions can be endothermic or exothermic after this.

79. Distribution of the Energy of Molecules
Gases have an average temperature, but each individual molecule has its own energy.
At higher energies, more molecules possess the energy needed for the reaction to occur.

80. The Relationship Between Activation Energy & Temperature
Arrhenius noted relationship between activation energy and temperature: k = Ae−Ea/RT
Activation energy can be determined graphically by reorganizing the equation: ln k = −Ea/RT + ln A

81. Law vs. Theory
Kinetics gives what happens. We call the description the rate law.
Why do we observe that rate law? We explain with a theory called a mechanism.
A mechanism is a series of stepwise reactions that show how reactants become products.

82. Reaction Mechanisms
Reactions may occur all at once or through several discrete steps.
Each of these processes is known as an elementary reaction or elementary process.

83. Molecularity
The molecularity of an elementary reaction tells how many molecules are involved in that step of the mechanism.

84. Termolecular?
Termolecular steps require three molecules to simultaneously collide with the proper orientation and the proper energy.
These are rare, if they indeed do occur.
These must be slower than unimolecular or bimolecular steps.
Nearly all mechanisms use only unimolecular or bimolecular reactions.

85. What Limits the Rate?
The overall reaction cannot occur faster than the slowest reaction in the mechanism.
We call that the rate-determining step.

86. What is Required of a Plausible Mechanism?
The rate law must be able to be devised from the rate-determining step.
The stoichiometry must be obtained when all steps are added up.
Each step must balance, like any equation.
All intermediates are made and used up.
Any catalyst is used and regenerated.

87. A Mechanism With a Slow Initial Step
Overall equation: NO2 + CO → NO + CO2
Step 1: NO2 + NO2 → NO + NO3 (slow)
Step 2: NO3 + CO → NO2 + CO2 (fast)

88. A Mechanism With a Slow Initial Step
Overall equation: NO2 + CO → NO + CO2
Step 1: NO2 + NO2 → NO + NO3 (slow)
Step 2: NO3 + CO → NO2 + CO2 (fast)
NO2 + NO2 + NO3 + CO → NO + NO3 + NO2 + CO2

89. The sum of the two steps = overall equation
Overall equation: NO2 + CO → NO + CO2
Step 1: NO2 + NO2 → NO + NO3 (slow)
Step 2: NO3 + CO → NO2 + CO2 (fast)
NO2 + NO2 + NO3 + CO → NO + NO3 + NO2 + CO2

90. Notice the intermediate?
Overall equation: NO2 + CO → NO + CO2
Step 1: NO2 + NO2 → NO + NO3 (slow)
Step 2: NO3 + CO → NO2 + CO2 (fast)
NO2 + NO2 + NO3 + CO → NO + NO3 + NO2 + CO2

91. Rate law: Rate = k [NO2]2 Step 1: NO2 + NO2 → NO + NO3 (slow)
Overall equation: NO2 + CO → NO + CO2
Step 1: NO2 + NO2 → NO + NO3 (slow)
Step 2: NO3 + CO → NO2 + CO2 (fast)
NO2 + NO2 + NO3 + CO → NO + NO3 + NO2 + CO2

92. Rate law: Rate = k [NO2]2 Step 1: NO2 + NO2 → NO + NO3 (slow)
Since the slow step determines the rate…
Rate = k [NO2]1 [NO2]1
Rate = k [NO2]2

93. A Mechanism With a Slow Initial Step (completed)
Rate law: Rate = k [NO2]2
Since the first step is the slowest step, it gives the rate law.
If the first step is the rate-determining step, the coefficients on the reactants side are the same as the order in the rate law!
If you add up all of the individual steps (2 of them), you get the stoichiometry.
Each step balances, this is a plausible mechanism.

94. Catalysts
Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction.
Catalysts change the mechanism by which the process occurs.

95. Types of Catalysts Homogeneous catalysts Heterogeneous catalysts
Enzymes

96. Homogeneous Catalysts
The reactants and catalyst are in the same phase.
Many times, reactants and catalyst are dissolved in the same solvent, as seen below.

97. Heterogeneous Catalysts
The catalyst is in a different phase than the reactants.
Often, gases are passed over a solid catalyst.
The adsorption of the reactants is often the rate-determining step.

98. Enzymes Enzymes are biological catalysts.
They have a region where the reactants attach. That region is called the active site. The reactants are referred to as substrates.

99. Lock-and-Key Model
In the enzyme–substrate model, the substrate fits into the active site of an enzyme, much like a key fits into a lock.
They are specific.