1. Dr. Namphol Sinkaset Chem 201: General Chemistry II
Ch. 17: Kinetics
Dr. Namphol Sinkaset
Chem 201: General Chemistry II

2. I. Chapter Outline Introduction The Rate of a Chemical Reaction
Factors that Affect Reaction Rates
Reaction Rate Laws
Integrated Rate Laws
Collision Theory
Reaction Mechanisms
Catalysis

3. I. Introduction
Just because something is energetically and entropically favored to happen doesn’t mean that it will happen quickly.
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 is one factor that influences how fast reactions occur.
Biochemical reactions are subject to the same rules as other chemical reactions.

5. 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.

6. II. Rate Data for Decomposition of H2O2
As the concentration of the reactant decreases, logically the rate of the reaction decreases.
These are average rates over the specified time interval.

7. II. Graphical Rate Data
If we plot average rate data as a function of time, we see that the reaction rate constantly changes.
Slopes of tangent lines give us instantaneous rates.

8. II. Relative Rates of Reaction
H2(g) + I2(g) 2HI(g)

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

10. III. Factors that Influence Rate
Chemical nature of reactants
e.g. Ca vs. Na reaction in water
Phase of the reactants
Heterogeneous vs. homogeneous reactions
Concentrations of reactants
Higher concentrations usually faster rates.
Temperature of reactants
Increase of 10 °C doubles reaction rate.
Presence of catalyst
Some substances can increase rates.

11. IV. 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. IV. 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. IV. 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. IV. 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. IV. 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. IV. 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. IV. 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. IV. 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. IV. 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. IV. 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. IV. 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. V. 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. V. 1st Order Integrated Rate Law

24. V. 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. V. 2nd Order Integrated Rate Law

26. V. 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. V. Zero Order Integrated Rate Law

28. V. 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. V. 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. V. 1st Order Reaction Half Life

31. V. 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. V. 1st Order Half Lives

33. V. 2nd Order Reaction Half Life

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

35. V. Zero Order Reaction Half Life

36. V. 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. V. Rate Law Summary

38. VI. 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. VI. Collision Theory Three postulates:
Reaction rate is proportional to rate of collisions.
Orientation of collision must be correct.
Collision must be energetic enough to allow rearrangement of valence e-’s to form new bonds.

40. VI. Effective/Ineffective Collisions

41. VI. Reaction Coordinate Diagram
To get to product state, reactant must go through high-energy transition state.
Higher Ea means…*

42. VI. Energy Distribution
As the temperature increases, more reactant particles will have enough energy to overcome the activation energy.

43. VI. The Arrhenius Equation
This equation relates activation energy to the rate constant of a reaction.
Note that R is the gas constant, and T is temperature in kelvin.

44. VI. Frequency Factor
The frequency factor (A) represents how often collisions with the correct orientation occur.
Remember that not all correctly oriented collisions result in reaction due to insufficient energy.
A frequency factor of 109/s means that there are 109 correctly oriented collisions per second between reactant molecules.

45. VI. Exponential Factor
The exponential factor is a number between 0 and 1 that represents the fraction of molecules that successfully react upon properly colliding (have enough energy).
An exponential factor of 10-7 means that 1 out of every 107 collisions has enough energy to cross the energy barrier.

46. VI. 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 collisions because the molecules have more energy to overcome the activation barrier.

47. VI. Finding A and Ea

48. VI. 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.

49. VI. Two-Point Form

50. VI. 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?

51. VII. 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.

52. VII. 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)

53. VII. Elementary Reactions
The reactions in a mechanism are called elementary reactions; 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?

54. VII. 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].

55. VII. Energy Diagram, 2-Step Mechanism

56. VII. 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.

57. VII. 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.

58. VII. 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

59. VII. 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.

60. VII. 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.

61. VII. 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

62. VIII. 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.

63. VIII. 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.

64. VIII. How Do Catalysts Work?
Catalysts provide a lower-energy mechanism for the reaction.

65. VIII. Types of Catalysts
There are homogeneous and heterogeneous catalysts.

66. VIII. Hydrogenation of Ethylene