1. Dr. Namphol Sinkaset Chem 201: General Chemistry II
Ch. 15: Chemical Kinetics
Dr. Namphol Sinkaset
Chem 201: General Chemistry II

2. I. Chapter Outline Introduction The Rate of a Chemical Reaction
Reaction Rate Laws
Integrated Rate Laws
Temperature and Rate
Reaction Mechanisms
Catalysis

3. I. Introduction
Molecules/atoms are constantly moving; in some collisions, electrons in one are attracted to nuclei in another.
Some reactions are quick (explosions) while others are slow (rusting of iron).
If we understand what contributes to the rate, we can control the reaction.

4. I. Introduction Temperature influences how fast reactions occur.
Biochemical reactions are subject to the same rules as other chemical reactions.

5. I. Introduction
We will begin by examining the collision model of chemical reactions.
Three factors that influence how fast reactions occur:
Reactant concentrations
Temperature
Structure and orientation of colliding particles

6. II. Reaction Rates
Rates are generally change of something divided by change in time.
Reaction rates are no different.
The rate of a reaction can be written with respect to any compound in that reaction.
However, there can only be one numerical value for a rate of reaction.

7. II. Average Rates of Reaction
H2(g) + I2(g) 2HI(g)

8. II. General Reaction Rates
aA + bB  cC + dD

9. II. Some Rate Data
If we plot average rate data as a function of time, we see that the reaction rate constantly changes.
Thus, rate depends on concentration of reactants!

10. II. Measuring Rates
Need a chemical “handle” to see how fast a reaction takes place.
One way is to measure how quickly the color of a solution changes.*

11. III. Rate Laws
If the rate depends on concentration of reactants, then we should be able to write an equation.
A rate law describes the mathematical relationship between the concentration of reactants and how fast the reaction occurs.

12. III. A Simple Rate Law
Consider a decomposition reaction where A  products
If the reverse reaction is negligible, then the rate law is: Rate = k[A]n.
k is called the rate constant
n is called the reaction order

13. III. Reaction Orders
The reaction order, n, determines how the rate depends on the concentration of the reactant. For the previous reaction, if…
n = 0, zero order, rate is independent of [A]
n = 1, first order, rate is directly proportional to [A]
n = 2, second order, rate is proportional to the square of the [A]

14. III. Reaction Orders and Rate
The rate law for the decomposition can then be either:
Rate = k[A]0 = k
Rate = k[A]1
Rate = k[A]2
Each will have a different type of curve when graphed.

15. III. Determining Orders
Reaction orders can only be determined by experiment!!
Reaction orders are not related to the stoichiometry of a reaction!
If reaction orders match a reaction’s stoichiometry, it is just a coincidence.
Therefore, orders cannot be determined without experimental data!

16. III. Sure-fire Method [A] (M) Initial Rate (M/s) 0.10 0.015 0.20 0.060
For the reaction, A  Products, we have the following data:
[A] (M)
Initial Rate (M/s)
0.10
0.015
0.20
0.060
0.40
0.240

17. III. More Complex Reactions
What if we have a more complicated reaction like: aA + bB  cC + dD?
Writing the general rate law is easy. Simply include all reactants, each with its own order.
Rate = k[A]m[B]n
If there are more reactants, there are more terms in the rate law.

18. III. Example Reaction 2H2(g) + 2NO(g)  N2(g) + 2H2O(g)
After looking at experimental data, the rate law was found to be Rate = k[H2][NO]2.
We say the reaction is 1st order in H2, 2nd order in NO, and 3rd order overall.
Note that Rate always has units of M/s, so the units on k will depend on the rate law.
What are the units of k for the rate law above?

19. III. Steps for Finding Rate Law
Pick two solutions where one reactant stays same, but another changes.
Write rate law for both w/ as much information as you have.
Ratio the two and solve for an order.
Repeat for another pair of solutions.
Use any reaction to get value of k.

20. III. Sample Problem
Determine the complete rate law for the reaction CHCl3(g) + Cl2(g)  CCl4(g) + HCl(g) using the data below.
[CHCl3] (M)
[Cl2] (M)
Initial Rate (M/s)
0.010
0.0035
0.020
0.0069
0.0098
0.040
0.027

21. III. Sample Problem Sometimes, rate laws can be found by inspection.
Determine the rate law for the reaction 2NO(g) + 2H2(g)  N2(g) + H2O(g) using the data below.
[NO] (M)
[H2] (M)
Initial Rate (M/s)
0.10
0.20

22. IV. Concentration and Time
Study and elucidation of rate laws allow the prediction of when a reaction will end.
An integrated rate law for a chemical reaction is a relationship between the concentrations of reactants and time.
Integrated rate laws depend on the order of the reaction; thus, we examine each separately.
We will only consider reactions with one reactant.

23. IV. 1st Order Integrated Rate Law

24. IV. 1st Order Integrated Rate Law
Notice this equation is in y = mx + b form.
A plot of ln[A] vs. t for a 1st order reaction yields a straight line with m = -k and b = ln[A]0.

25. IV. 2nd Order Integrated Rate Law

26. IV. 2nd Order Integrated Rate Law
Again, this equation is in y = mx + b form.
A plot of 1/[A] vs. t yields a straight line with slope equal to k and y-intercept equal to 1/[A]0.

27. IV. Zero Order Integrated Rate Law

28. IV. Zero Order Integrated Rate Law
Yet again in y = mx + b form!
Plot of [A] vs. t results in a straight line with slope equal to -k and b = [A]0.

29. IV. Reaction Half Lives
The half-life, t1/2, of a reaction is the time required for the concentration of a reactant to decrease to half its initial value.
Half life equations depend on the order of the reaction.

30. IV. 1st Order Reaction Half Life

31. IV. 1st Order Reaction Half Life
Notice that the half life doesn’t depend on reactant concentration!
Unique for 1st order.
The half life for a 1st order reaction is…*

32. IV. 1st Order Half Lives

33. IV. 2nd Order Reaction Half Life

34. IV. 2nd Order Reaction Half Life
For 2nd order, the half life depends on initial concentration.
As concentration decreases, half life…*

35. IV. Zero Order Reaction Half Life

36. IV. Zero Order Reaction Half Life
We see that for zero order reactions, the half life depends on concentration as well.
As concentration decreases, half life…*

37. IV. Rate Law Summary

38. V. Temperature and Rate
In general, rates of reaction are highly sensitive to temperature – a 10 °C increase in T increases rate 2x to 3x.
If Rate = k[A]n, where does the temperature factor in?
It’s in the constant k!
Generally, increasing temperature increases k.

39. V. The Arrhenius Equation
Note that R is the gas constant, and T is temperature in kelvin.

40. V. Parameters of Arrhenius Eqn.
We can describe the physical meanings of the aspects of the Arrhenius equation by considering a specific reaction.

41. V. Activation Energy
To get to product state, reactant must go through high-energy activated complex, or transition state.
Even though reaction is exo overall, it must go through an endo step.
Higher Ea means…*

42. V. Frequency Factor
The frequency factor (A) represents the number of approaches to the activation barrier per unit time.
For this reaction, it represents how often the NC part of the molecule vibrates.
Note that not all approaches result in reaction due to not having enough energy.
A frequency factor of 109/s means that there are 109 vibrations per second of the NC group.

43. V. Exponential Factor
The exponential factor is a number between 0 and 1 that represents the fraction of molecules that successfully react upon approach.
An exponential factor of 10-7 means that 1 out of every 107 molecules has enough energy to cross the energy barrier.

44. V. Exponential Factor & Temp
Since exponential factor = e-Ea/RT, temperature has a huge influence.
As T  0, the factor goes to 0, and as T  ∞, the factor goes to 1.
Thus, higher temperatures mean more successful approaches because the molecules have more energy to overcome the activation barrier.

45. V. Finding A and Ea

46. V. Arrhenius Plots
If we have kinetic data at various temperatures, we can plot ln k vs. 1/T.
We should get a straight line with m = -Ea/R and b = ln A.

47. V. Two-Point Form

48. V. Sample Problem
The decomposition of HI has rate constants of k = /M·s at 508 °C and k = /M·s at 540 °C. What is the activation energy of this reaction in kJ/mole?

49. V. The Collision Model
In the collision model of a reaction (2 reactants), two factors influence whether a reaction will occur.
Energy of the collision
Orientation of the collision

50. V. Collision Theory & Arrhenius Equation
p is the orientation factor, a number between 0 and 1 (1 means any orientation will work for the reaction).
z is the collision frequency, i.e. how many collisions per second.

51. V. Effective/Ineffective Collisions

52. VI. How Reactions Occur
Most chemical reactions occur through several small steps, not one big step.
A chemical equation typically shows the overall reaction, not the intermediate steps.
e.g. H2(g) + 2ICl(g)  2HCl(g) + I2(g) only shows what’s at the beginning and what you end up with.

53. VI. Reaction Mechanisms
A reaction mechanism is a series of individual chemical steps through which an overall chemical reaction occurs.
A proposed mechanism for the reaction H2(g) + 2ICl(g)  2HCl(g) + I2(g) is:
Step 1 H2(g) + ICl(g)  HI(g) + HCl(g)
Step 2 HI(g) + ICl(g)  HCl(g) + I2(g)

54. VI. Elementary Steps
The reactions in a mechanism are called elementary steps; what’s implied in these steps is exactly what happens.
Proposed reaction mechanisms must add up to the overall reaction!
Does the previous mechanism add up?
Species that are formed in one step and then consumed in another are known as intermediates. What is/are the intermediate(s) in the previous mechanism?

55. VI. Elementary Step Rate Laws
Elementary steps are characterized by their molecularity, i.e. the # of reactant particles involved in the step.
Rate laws for elementary steps can be written directly from their stoichiometry!
e.g. If A + B  C + D is an elementary step, then the rate law for this step is: Rate = k[A][B].

56. VI. Energy Diagram, 2-Step Mechanism

57. VI. The Rate-Determining Step
The slow step in the mechanism will determine the overall rate of reaction.
This step is known as the rate-determining step.
It’s the bottleneck of the reaction.

58. VI. Valid Mechanisms Valid mechanisms satisfy 2 criteria:
Elementary steps add up to overall reaction.
Rate law predicted by mechanism must be consistent with experimental rate law.
Note that a valid mechanism is not a proven mechanism.

59. VI. Example Consider the reaction: NO2(g) + CO(g)  NO(g) + CO2(g).
Experimentally, Rate = k[NO2]2. This implies it’s not a single-step reaction. Why?
Is the mechanism below valid?
NO2(g) + NO2(g)  NO3(g) + NO(g) Slow
NO3(g) + CO(g)  NO2(g) + CO2(g) Fast

60. VI. Rate Laws w/ Intermediates
Rate laws must always be written from the rate-determining step.
However, rate laws cannot contain intermediates.
Rate laws from other steps can be used to substitute for intermediates.
We look at fast first steps.

61. VI. Fast 1st Steps
When the 1st step is fast, its products will build up and reverse reaction starts.
Eventually, an equilibrium is set up.
Thus, for A + B  C + D (Fast), we can write A + B  C + D.
Rate = k[A][B] and Rate = k-1[C][D].
At equilibrium, k[A][B] = k-1[C][D].
This can be used to rewrite rate laws.

62. VI. Sample Problem
What is the overall reaction and rate law for the mechanism below? Identify the intermediates as well.
Cl2(g)  2Cl(g) Fast
Cl(g) + CHCl3(g)  HCl(g) + CCl3(g) Slow
CCl3(g) + Cl(g)  CCl4(g) Fast

63. VII. Catalysts
We know we can change reaction rates by changing the temperature or changing reactant concentrations.
However, there are limits to these tactics. What are these limits?*
If available, can use catalysts, substances that increase reaction rate, but aren’t used up in the reaction.

64. VII. Catalytic Destruction of O3
Catalyzed:
Cl(g) + O3(g)  ClO(g) + O2(g)
ClO(g) + O(g)  Cl(g) + O2(g)
Uncatalyzed:
O3(g) + O(g)  2O2(g)
O3(g) + O(g)  2O2(g)
Atomic chlorine from photodissociated CFC’s is the catalyst.

65. VII. How Do Catalysts Work?
Catalysts provide a lower-energy mechanism for the reaction.

66. VII. Types of Catalysts
There are homogeneous and heterogeneous catalysts.

67. VII. Hydrogenation of Ethene

68. VII. Biological Catalysts
A biological catalyst is called and enzyme.
An enzymes has an active site into which a specific substrate fits – like a lock and key.

69. VII. Sucrose  Glucose + Fructose