1. Chapter 14 Chemical Kinetics
Lecture Presentation
Chapter 14 Chemical Kinetics
James F. Kirby
Quinnipiac University
Hamden, CT

2. Chemical Kinetics
In chemical kinetics we study the rate (or speed) at which a chemical process occurs.
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.

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

4. 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.

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

6. 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.

7. 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.

8. Example:
If a heated steel nail were placed in pure O2, would you expect it to burn as readily as the steel wool does?
Yes, the nail and steel wool should have the same composition.
Yes, the O2 is the major contributor to the reaction.
No, the surface area of the nail is not as great as that of the wool.

9. Example:
If a heated steel nail were placed in pure O2, would you expect it to burn as readily as the steel wool does?
Yes, the nail and steel wool should have the same composition.
Yes, the O2 is the major contributor to the reaction.
No, the surface area of the nail is not as great as that of the wool.

10. Reaction Rate
Rate is a change in concentration over a time period: Δ[ ]/Δt.
Δ means “change in.”
[ ] means molar concentration.
t represents time.
Types of rate measured:
average rate
instantaneous rate
initial rate

11. Following Reaction Rates
Rate of a reaction is measured using the concentration of a reactant or a product over time.
(product appears as reactant disappears)
In this example, [C4H9Cl] is followed.

12. 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.
Average rate of appearance of product = ∆ [product]
∆ time
Average rate of disappearance of reactant = - ∆ [reactant]

13. Estimate the number of moles of A in the mixture after 30 s.
~0.6 mol A
~0.4 mol A
< 0.4 mol A
> 0.3 mol A and < 0.4 mol A
Answer: d

14. Estimate the number of moles of A in the mixture after 30 s.
~0.6 mol A
~0.4 mol A
< 0.4 mol A
> 0.3 mol A and < 0.4 mol A
Answer: d

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 disappearance of C4H9Cl and compared it to production of C4H9OH, the values would be the same because there is a 1:1 ratio.
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?
What will the relative rates be for:
2 O3 (g) → 3 O2 (g)

18. In a reaction involving reactants in the gas state, how does increasing the partial pressures of the gases affect the reaction rate?
The effect of increasing the partial pressures of the reactive components of a gaseous mixture depends on which side of the chemical equation has the most gas molecules.
Increasing the partial pressures of the reactive components of a gaseous mixture has no effect on the rate of reaction if each reactant pressure is increased by the same amount.
Increasing the partial pressures of the reactive components of a gaseous mixture increases the rate of reaction.
Increasing the partial pressures of the reactive components of a gaseous mixture decreases the rate of reaction.
Answer: c

19. In a reaction involving reactants in the gas state, how does increasing the partial pressures of the gases affect the reaction rate?
The effect of increasing the partial pressures of the reactive components of a gaseous mixture depends on which side of the chemical equation has the most gas molecules.
Increasing the partial pressures of the reactive components of a gaseous mixture has no effect on the rate of reaction if each reactant pressure is increased by the same amount.
Increasing the partial pressures of the reactive components of a gaseous mixture increases the rate of reaction.
Increasing the partial pressures of the reactive components of a gaseous mixture decreases the rate of reaction.
Answer: c

20. In Figure 14.4, order the following three rates from fastest to slowest: (i) The average rate of the reaction between 0 s and 600 s, (ii) the instantaneous rate at t = 0 s, and (iii) the instantaneous rate at t = 600 s. You should not have to do any calculations.
i > ii > iii
iii > ii > i
ii > i > iii
iii > i > ii
Answer: c
Answer: C

21. Practice Problems: Write rate expressions for the following reactions
I- (aq) + OCl – (aq) → Cl – (aq) + OI – (aq)
3 O2 (g) → 2O3 (g)
4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O(g)

22. Practice Consider the reaction: 4NO2 (g) + O2 (g) → 2N2O5 (g)
If oxygen is reacting at the rate of M/s,
At what rate is N2O5 being formed?
At what rate is NO2 reacting?

23. 14.3 Differential Rate Laws
Dependence of reaction rate on the concentration of reactants is called the rate law, which is unique for each reaction
For a general reaction, a A + b B + c C products
the rate law has the general form
order wrt A, B, and C, determined experimentally
reaction rate = k [A]X [B]Y [C]Z the rate constant

24. The Rate Law
The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.
aA + bB cC + dD
Rate = k [A]x[B]y
reaction is xth order in A
reaction is yth order in B
reaction is (x +y)th order overall
13.2

25. Double [F2] with [ClO2] constant
F2 (g) + 2ClO2 (g) FClO2 (g)
rate = k [F2]x[ClO2]y
Double [F2] with [ClO2] constant
Rate doubles
x = 1
rate = k [F2][ClO2]
Quadruple [ClO2] with [F2] constant
Rate quadruples
y = 1
13.2

26. Rate Laws Rate laws are always determined experimentally.
Reaction order is always defined in terms of reactant (not product) concentrations.
The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.
F2 (g) + 2ClO2 (g) FClO2 (g) 1
rate = k [F2][ClO2]
13.2

27. S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq)
Determine the rate law and calculate the rate constant for the following reaction from the following data:
S2O82- (aq) + 3I- (aq) SO42- (aq) + I3- (aq)
Experiment
[S2O82-]
[I-]
Initial Rate (M/s) 1
0.08
0.034
2.2 x 10-4 2
0.017
1.1 x 10-4 3
0.16
rate = k [S2O82-]x[I-]y
y = 1
x = 1
rate = k [S2O82-][I-]
Double [I-], rate doubles (experiment 1 & 2)
Double [S2O82-], rate doubles (experiment 2 & 3)
k =
rate
[S2O82-][I-] =
2.2 x 10-4 M/s
(0.08 M)(0.034 M)
= 0.08/M•s
13.2

28. For a reaction: BrO3- + 5Br- + 6H+ 3 Br2 + 3H2O
a set of experiments are performed, with different initial conditions.  The experiments are designed to investigate the concentration effect of one of the reactants, while keeping the others constant.   
Experiment
[BrO3-] (M)
[Br-] (M)
[H+] (M)
Initial Rate (M/s) 1
0.10
8.0 x 10 -4 2
0.20
1.6 x 10 -3 3
3.2 x 10 -3 4
Determine the rate law for the reaction
Find the value of the rate constant, k

29. First Order Reactions
First Order reactions depend only on one reactant to the first power.
The rate law becomes:
Rate = k [A]
This is the differential rate law

30. Relating k to [A] in a First Order Reaction
[A]t= concentration at time (t)
[A]0= initial concentration
rate = k [A]
rate = −Δ [A] / Δt
integrated rate law : ln [A]t = − 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.

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

32. 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.

33. The reaction 2A B is first order in A with a rate constant of 2
The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?
[A]0 = 0.88 M
ln[A] = ln[A]0 - kt
[A] = 0.14 M
kt = ln[A]0 – ln[A] ln
[A]0
[A] k = ln
0.88 M
0.14 M
2.8 x 10-2 s-1 =
ln[A]0 – ln[A] k
t =
= 66 s
13.3

34. Half-life
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 2 = k t½ or t½ = 0.693/k

35. First-order reaction A product # of half-lives [A] = [A]0/n 1 2 2 4 38 4 16
13.3

36. 13.3

37. First-Order Reactions
What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? t½
ln2 k =
0.693
5.7 x 10-4 s-1 =
= 1200 s = 20 minutes
How do you know decomposition is first order?
units of k (s-1)
13.3

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

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

40. 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

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

42. 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.

43. Zero Order Reactions Integrated Rate Law [A]t = -kt + [A]0 Half-life

44. Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order
Differential
Rate Law
Integrated Rate Law
Half-Life
t½ =
[A]0 2k
rate = k
[A] = [A]0 - kt t½
ln2 k = 1
rate = k [A]
ln[A] = ln[A]0 - kt 1
[A] =
[A]0
+ kt
t½ = 1
k[A]0 2
rate = k [A]2
13.3

45. Iodine atoms combine to form molecular iodine in the gas phase
I(g) + I(g) → I2 (g)
This reaction follows 2nd order kinetics and has the high rate constant 7.0 x 109 M-1s-1 at 23°C.
If the initial concentration of I was M, calculate the concentration after 2.0 min
Calculate the half-life of the reaction if the initial concentration of I is 0.60 M and if it is 0.42 M
Answers
1.2 x M
2.4 x 10-10s; 3.4 x s

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

47. 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.

48. 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.

49. 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.

50. 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.

51. 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.

52. 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.

53. 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. (determined by ∆E)
Rate depends on Ea
The lower the value of Ea, the faster the reaction

54. 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.

55. 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
k = rate constant
Ea = activation energy
A = frequency factor
R = J/mol-K
T = Absolute Temperature (K)

56. 14.6 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.

57. Reaction Mechanisms Some reactions occur in a single event or step
These processes are known as an elementary reactions
Ex.
NO(g) + O3(g) → NO2(g) + O2(g)

58. Reaction Mechanisms
The molecularity of a reaction is the number of molecules reacting in an elementary step.
Unimolecular reaction – elementary step with 1 molecule
Bimolecular reaction – elementary step with 2 molecules
Termolecular reaction – elementary step with 3 molecules
Ex.
NO(g) + O3(g) → NO2(g) + O2(g) = bimolecular

59. 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.
Nearly all mechanisms use only unimolecular or bimolecular reactions.

60. Rate Laws for Elementary Reactions
For elementary reactions, rate law is determined by molecularity

61. N2O2 is detected during the reaction!
Multistep Mechanisms
Some reactions occur as a series of simple elementary steps or elementary reactions.
The sequence of elementary steps that leads to product formation is the reaction mechanism.
2NO (g) + O2 (g) NO2 (g)
N2O2 is detected during the reaction!
Elementary step:
NO + NO N2O2
Overall reaction:
2NO + O NO2 +
Elementary step:
N2O2 + O NO2
13.5

62. Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation.
An intermediate is always formed in an early elementary step and consumed in a later elementary step.
Elementary step:
NO + NO N2O2
N2O2 + O NO2
Overall reaction:
2NO + O NO2 +
For this profile, is it easier for a molecule of the intermediate to convert to reactants or products?
13.5

63. Rate Laws and Elementary Steps
Unimolecular reaction
A products
rate = k [A]
Bimolecular reaction
A + B products
rate = k [A][B]
Bimolecular reaction
A + A products
rate = k [A]2
Writing plausible reaction mechanisms:
The sum of the elementary steps must give the overall balanced equation for the reaction.
The rate-determining step should predict the same rate law that is determined experimentally.
The rate-determining step is the slowest step in the sequence of steps leading to product formation.
13.5

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

65. What is the equation for the overall reaction?
The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps:
Step 1:
NO2 + NO NO + NO3
Step 2:
NO3 + CO NO2 + CO2
What is the equation for the overall reaction?
NO2+ CO NO + CO2
What is the intermediate?
NO3
What can you say about the relative rates of steps 1 and 2?
rate = k[NO2]2 is the rate law for step 1 so
step 1 must be slower than step 2
13.5

66. 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.

67. A Mechanism With a Slow Initial Step
Overall equation: NO2 + CO → NO + CO2
Rate law: Rate = k [NO2]2
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!
So, the first step of the mechanism begins:
NO2 + NO2 →

68. A Mechanism With a Slow Initial Step (continued)
The easiest way to complete the first step is to make a product:
NO2 + NO2 → NO + NO3
We do not see NO3 in the stoichiometry, so it is an intermediate, which needs to be used in a faster next step.
NO3 + CO → NO2 + CO2

69. A Mechanism With a Slow Initial Step (completed)
Since the first step is the slowest step, it gives 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.

70. A Mechanism With a Fast Initial Step
Equation for the reaction:
2 NO + Br2 ⇌ 2 NOBr
The rate law for this reaction is found to be
Rate = k [NO]2 [Br2]
Because termolecular processes are rare, this rate law suggests a multistep mechanism.

71. A Mechanism With a Fast Initial Step (continued)
The rate law indicates that a quickly established equilibrium is followed by a slow step.
Step 1: NO + Br2 ⇌ NOBr2
Step 2: NOBr2 + NO → 2 NOBr

72. What is the Rate Law?
The rate of the overall reaction depends upon the rate of the slow step.
The rate law for that step would be
Rate = k2[NOBr2] [NO]
But how can we find [NOBr2]?

73. [NOBr2] (An Intermediate)?
Step 1: NO + Br2 ⇌ NOBr2
Step 2: NOBr2 + NO → 2 NOBr
An intermediate should not appear in the overall rate law
NOBr2 can react two ways:
With NO to form NOBr.
By decomposition to reform NO and Br2.
The reactants and products of the first step are in equilibrium with each other.
For an equilibrium (as we will see in the next chapter):
Ratef = Rater

74. The Rate Law (Finally!)
Substituting for the forward and reverse rates:
k1 [NO] [Br2] = k−1 [NOBr2]
Solve for [NOBr2], then substitute into the rate law: Rate = k2[NOBr2] [NO]
Rate = k2 (k1/k−1) [NO] [Br2] [NO]
This gives the observed rate law!
Rate = k [NO]2 [Br2]

75. 14.7 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.

76. Types of Catalysts Homogeneous catalysts Heterogeneous catalysts
Enzymes

77. 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.

79. 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.
Pt(s)
2 SO2(g) → 2 SO3 (g)

80. 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.

81. 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.