1. Chemical Kinetics
Unit 11

2. Overview Factors affecting reactions Collision Model Reaction Rate
Activation energy, activated complex
Exothermic/endothermic reactions
Energy of reactions ∆E
Reaction Rate
Average reaction rate
Instantaneous reaction rate
Kinetics Stoichiometry
Reaction Rate and Concentration
(Differential) Rate laws
Specific rate constant (k)
Writing rate laws
Reaction Order
Rate laws
Integrated Rate Laws
Graphing
Solving (0, 1st, 2nd order)
Half life equations
Rate Constant and Temperature
Arrhenius equation
Catalysis
Homogeneous, heterogeneous, enzymes
Reaction Mechanisms
Identifying intermediates
Rate determining step

3. Chemical Kinetics Thermochemistry = does a reaction take place?
Kinetics = how fast does the reaction happen?
Kinetics: The area of chemistry concerned with the speeds (rates) of reactions

4. Factors Affecting Reaction Rates
Physical state of reactants (surface area)
Concentration of reactants
Temperature at which reaction occurs
Presence of a catalyst
Pressure of gaseous reactants or products

5. Collision Model Molecules must collide to react
Only a small fraction of collisions produces a reaction. Why?

6. Collision Model
Collisions must have enough energy to produce the reaction (must equal or exceed the activation energy).
Reactants must have proper orientation to allow the formation of new bonds.

7. Collision Model

8. Activation Energy
Activation Energy (Ea)– minimum energy required to transform reactants into the activated complex
(The minimum energy required to produce an effective collision)
Example: flame, spark, high temperature, radiation
The lower the Ea, the faster the reaction

9. Activation Energy Activated Complex – transitional structure
All bonds have been broken but no new bonds have formed
Point in reaction of highest energy but lowest stability

10. Reactants  Products + energy
Exothermic Processes
Processes in which energy is released as it proceeds, and surroundings become warmer
Reactants  Products + energy

11. Endothermic Processes
Processes in which energy is absorbed as it proceeds, and surroundings become colder
Reactants + energy  Products

12. Energy of Reactions (∆E)
Same as ∆H
“Heat” and “energy” can be used interchangeably
∆E = ∆Eproducts- ∆Ereactants
- ∆E = exothermic (energy lost)
+∆E = endothermic (energy gained)

13. Reaction Rate A B Because [A] decreases with time, D[A] is negative.
Reaction Rate: the change in concentration of reactants per unit time as a reaction proceeds (M/s)
A B
rate = -
D[A] Dt
D[A] = change in concentration of A over
time period Dt
rate =
D[B] Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is negative.

14. A B
time
rate = -
D[A] Dt
rate =
D[B] Dt
13.1

15. Reaction Rate
Average reaction rate: average rate throughout the course of the reaction
Instantaneous reaction rate: rate of reaction at a specific point in time during the reaction
Think of a car trip where you drove 100 miles in 2 hours
Average speed = 50 mi/hr
Instantaneous speed is speed at specific
point – you don’t go exactly 50 mi/hr
the entire trip

16. Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
slope of
tangent
slope of
tangent
slope of
tangent
average rate = -
D[Br2] Dt
= -
[Br2]final – [Br2]initial
tfinal - tinitial
instantaneous rate = rate for specific instance in time (slope of tangent)

17. Kinetics Stoichiometry
When mole ratios are not 1:1, stoichiometry is used to compare rates of reactions
aA + bB  cC + dD
Rate = = = =
1 ∆[A]
a ∆t
1 ∆[B]
b ∆t
1 ∆[C]
c ∆t
1 ∆[D]
d ∆t

18. Kinetics Stoichiometry
2HI  H2 + I2
If the rate at which H2 appears, ∆[H2]/∆t, is 6.0×10-5 M/s at a particular instant, at what rate is HI disappearing at this same time, - ∆[HI]/∆t?
Rate = =
Therefore, = (6.0×10-5 M/s )
= 1.2×10-4 M/s
1 ∆[HI]
2 ∆t
1 ∆[H2]
1 ∆t 1
1 ∆[HI]
2 ∆t
∆[HI] ∆t

19. Reaction Rate and Concentration
The initial rate of a reaction depends on the concentration of the reactants
Rate changes based on the concentration of reactants
(Differential) Rate Law: an equation that relates reaction rate and concentrations of reactants
Rate a [reactants]
a means “proportional to”

20. Rate Law R = k[A]n[B]m R = reaction rate k = specific rate constant
[A] and [B] = concentrations of reactants
n and m = the order of the reactant (usually 0, 1, 2)
(not related to moles of reactants from the equation)
Overall order of reaction = n + m

21. Rate Laws F2 (g) + 2ClO2 (g) 2FClO2 (g) 1 rate = k [F2][ClO2]
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

22. Specific Rate Constant (k)
Once the reaction orders are known, the value of k must be determined from experimental data
Value of k is for a specific reaction; k has a different value for other reactions, even at the same conditions
Units of k depend on the overall order of the reaction
Value of k does not change for different concentrations of reactants or products
Value of k is for reaction at a specific temperature; if temperature increases, k increases
Value of k changes if a catalyst is present

23. Writing a (differential) Rate Law
Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction:
2 NO(g) + Cl2(g)  2 NOCl(g)
Experiment
[NO]
(mol/L)
[Cl2]
Rate
Mol/L·s 1
0.250
1.43 x 10-6 2
0.500
5.72 x 10-6 3
2.86 x 10-6 4
11.4 x 10-6

24. Writing a Rate Law  R = k[NO]2[Cl2]m
Step 1 – Determine the values for the exponents in the rate law:
R = k[NO]n[Cl2]m
Experiment
[NO]
(mol/L)
[Cl2]
Rate
Mol/L·s 1
0.250
1.43 x 10-6 2
0.500
5.72 x 10-6 3
2.86 x 10-6 4
1.14 x 10-5
In experiment 1 and 2, [Cl2] is constant while [NO] doubles.
The rate quadruples, so the reaction is second order with respect to [NO]
 R = k[NO]2[Cl2]m

25. Writing a Rate Law  R = k[NO]2[Cl2] R = k[NO]2[Cl2]m
Experiment
[NO]
(mol/L)
[Cl2]
Rate
Mol/L·s 1
0.250
1.43 x 10-6 2
0.500
5.72 x 10-6 3
2.86 x 10-6 4
1.14 x 10-5
In experiment 2 and 4, [NO] is constant while [Cl2] doubles.
The rate doubles, so the reaction is first order with respect to [Cl2]
 R = k[NO]2[Cl2]

26. Writing a Rate Law
Part 2 – Determine the value for k, the rate constant, by using any set of experimental data:
R = k[NO]2[Cl2]
Experiment
[NO]
(mol/L)
[Cl2]
Rate
Mol/L·s 1
0.250
1.43 x 10-6

27. Reaction Order Given the rate law R = k[NO]2[Cl2]
We say that the reaction is
2nd order with respect to NO
1st order with respect to Cl2
Overall the order of the reaction is a 3rd order reaction
2 + 1 = 3

28. Example 2 Run # Initial [A] ([A]0) Initial [B] ([B]0)
Initial Rate (v0) 1
1.00 M
1.25 x 10-2 M/s 2
2.00 M
2.5 x 10-2 M/s 3
What is the order with respect to A?
What is the order with respect to B?
What is the overall order of the reaction? 1 1

29. Initial Rate (mol dm-3 s-1)
Example 3
[NO(g)] (mol dm-3)
[Cl2(g)] (mol dm-3)
Initial Rate (mol dm-3 s-1) 
0.250 
1.43 x 10-6
0.500 
2.86 x 10-6
1.14 x 10-5
What is the order with respect to Cl2?
What is the order with respect to NO?
What is the overall order of the reaction? 1 2 3

30. Reaction Order Zero Order 1st Order 2nd Order 3rd Order Rate Law R = k
R = k[A]
R = k[A][B]
R = k[A] 2
R = k[A]3
R = k[A][B][C]
R = k[A] 2[B]

31. Concentration VS Time Integrated Rate Law
Graph data to solve for order of reaction
Zero Order = time vs concentration is linear
1st Order = time vs ln[concentration] is linear
2nd Order = time vs is linear 1
concentration

32. Zero Order Reactions
13.3

33. First Order Reactions
13.3

34. Second Order Reactions

35. Solving an Integrated Rate Law
Time (s)
[H2O2] (mol/L)
1.00
120
0.91
300
0.78
600
0.59
1200
0.37
1800
0.22
2400
0.13
3000
0.082
3600
0.050
Problem: Find the integrated rate law and the value for the rate constant, k

36. Time vs. [H2O2] Time Regression results: y = ax + b a = -2.64 x 10-4
Time (s)
[H2O2]
1.00
120
0.91
300
0.78
600
0.59
1200
0.37
1800
0.22
2400
0.13
3000
0.082
3600
0.050
Time
[H2O2]
Regression results:
y = ax + b
a = x 10-4
b = 0.841
r2 =
r =

37. Time vs. ln[H2O2] Time Regression results: y = ax + b a = -8.35 x 10-4
Time (s)
ln[H2O2]
120
300
600
1200
1800
-1.514
2400
-2.04
3000
-2.501
3600
-2.996
Time
Regression results:
y = ax + b
a = x 10-4
b = -.005
r2 =
r =

38. Time vs. 1/[H2O2] Regression results: y = ax + b a = 0.00460
Time (s)
1/[H2O2]
1.00
120
1.0989
300
1.2821
600
1.6949
1200
2.7027
1800
4.5455
2400
7.6923
3000
12.195
3600
20.000
Time
1/[H2O2]
Regression results:
y = ax + b
a =
b =
r2 =
r =

39. Integrated Rate Law The graph is linear for “time vs ln[H2O2]”
The reaction is 1st order VS VS

40. Concentration VS Time [A] = -kt + [A]0
Rate laws for concentration of reactants versus time are based on the linear equation y = ax + b
Zero Order Reactions (time vs concentration)
[A] = -kt + [A]0
1st Order (time vs ln[concentration])
ln[A] = -kt + ln[A]0
2nd Order (time vs 1/concentration)

41. 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 ?
ln[A] = -kt + ln[A]0
[A]0 = 0.88 M
ln[A] - ln[A]0 = - kt
[A] = 0.14 M
ln[A]0 - ln[A] = kt 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

42. Half-life t½
Time it takes for concentration of reactant to equal ½ of its initial value
A fast reaction will have a short half-life
In a first order reaction, concentration of reactant decreases by ½ each time interval of t½
[A] = ½ [A]0

43. Half-life t½
For zero and second order reactions, half-life changes based on concentration of the reactant
To find equation for half-life, place ½[A] 0 in for [A] in each integrated rate law and put t½ in for t
When [A] = ½[A]0 and t = t½ …
Zero Order: [A] = -kt + [A]0
1st Order ln[A] = -kt + ln[A]0
2nd Order

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

45. Concentration-Time Equation
SUMMARY
Order
Rate Law
Concentration-Time Equation
Half-Life
t½ =
[A]0 2k
rate = k
[A] - [A]0 = - kt t½
Ln 2 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

46. Rate Constant and Temperature
Reaction rate increases with temperature
Molecules move faster
More molecules collide to cause a reaction in a time period
Molecules have greater kinetic energy at higher temperatures
More molecules will have enough energy to reach the activation energy of the reaction at higher temperatures

47. Boltzmann Distribution Curve

48. Rate Constant and Temperature
At higher temperatures (T2), more molecules are able to reach the activation energy

49. Rate Constant and Temperature
Arrhenius Equation: relationship between k and T
Ea is the activation energy (J/mol)
Ln k = -
-Ea R 1 T
+ lnA
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature (K)
A is the frequency factor
Takes into account collision frequency and orientation

50. Graphing Arrhenius
A plot of lnk versus 1 / T produces a straight line with the familiar form y = - mx + b, where
x = 1 / T
y = lnk
m = - E a / R
b = lnA
The activation energy E a can be determined from the slope m of this line: E a = - m · R

51. Catalysis
Catalyst: A substance that speeds up a reaction by lowering activation energy
Homogeneous catalyst: Present in the same phase as the reacting molecules.
Heterogeneous catalyst: Present in a different phase than the reacting molecules.
Enzyme: A large molecule (usually a protein) that catalyzes biological reactions.
Avoids temperature increase in reactions of living organisms 51

52. Ea k ratecatalyzed > rateuncatalyzed
A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. Ea k
uncatalyzed
catalyzed
ratecatalyzed > rateuncatalyzed
13.6

53. Endothermic Reaction w/Catalyst53

54. Exothermic Reaction w/Catalyst54

55. Factors Affecting Reaction Rates
Physical state of reactants (surface area)
Increased surface area means more collisions between reacting molecules
Concentration of reactants
Higher concentration means more molecules (solute) to react meaning more collisions
Temperature at which reaction occurs
Higher kinetic energy
Molecules moving faster so collide more frequently
Presence of a catalyst
Speeds up a reaction without being used up
Lowers activation energy
Pressure of gaseous reactants or products
Increased number of collisions

56. Rate Laws (vocabulary)
Elementary Reaction – rate law is based directly on proportionality
Describes an individual molecular event
Unimolecular – overall rate law is 1st order
Rate = k[A]
Bimolecular – overall rate law is 2nd order
Rate = k[A]2
Rate = k[A][B]
Termolecular – overall rate law is 3rd order
Rate = k[A]3
Rate = k[A] 2[B]
Rate = k[A][B][C]

57. Reaction Mechanisms Reactants  Products
Some reactions occur in one step, others do not
Reaction Mechanism: step-by-step sequence of reactions by which the overall chemical change occurs
A chemical equation does not tell us how reactants become products; it is a summary of the overall process.
The sum of the elementary steps must give the overall balanced equation for the reaction
Reactants  Products
The  sign has represents the reaction mechanism, but gives no indication of the steps in the mechanism

58. N2O2 is detected during the reaction!
Reaction Mechanisms (Example)
For the reaction:
The following steps occur:
2NO (g) + O2 (g) NO2 (g)
Elementary Step 1:
NO + NO N2O2
Overall reaction:
2NO + O NO2 +
Elementary Step 2:
N2O2 + O NO2
N2O2 is detected during the reaction!

59. Reaction Intermediates
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 +

60. Energy Diagrams
For multi-step reactions, each elementary step requires activation energy and has its own transition state
Peaks = transition states
Valleys = intermediates

61. Energy Diagrams (multi-step)

62. Writing 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.

63. The Rate-Determining Step
In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of reaction.
Step 1. NO + Cl2 <----> NOCl2 (fast) Step 2. NOCl2+ NO ----> 2NOCl (slow)
Together they give: 2NO + Cl2 ----> 2NOCl
Step 2 is the rate determining step

64. Identifying the Rate-Determining Step
For the reaction:
2H2(g) + 2NO(g)  N2(g) + 2H2O(g)
The experimental rate law is:
R = k[NO]2[H2]
Which step in the reaction mechanism is the rate-determining (slowest) step?
Step #1 H2(g) + 2NO(g)  N2O(g) + H2O(g)
Step #2 N2O(g) + H2(g)  N2(g) + H2O(g)
Step #1 agrees with the experimental rate law