CHAPTER 12: KINETICS
Chem
1212
Dr. Aimée Tomlinson
Section 12.1
Reaction Rates
Reaction Rate
The rates at which products are formed and reactants are consumed
are connected
They are always represented as concentration change / time
change
For the general case aA + bB  cC + dD the relationships are:
Rate  
1 [ A]
1 [ B] 1 [C ] 1 [ D]



a t
b t
c t
d t
where t = tf - ti, [A] is the concentration of A (moles/L) and
[A] = [A]f - [A]i
we loose reactants as products are formed which is why the rates of
A & B are negative
Example Application of Definition
1. What is the rate relationship between the production of O2 and O3?
2O3(g)  3O2(g)

[O3 ]
1 [O3 ] 1 [O2 ]
3 [O3 ] [O2 ]
2 [O2 ]



or

2 t
3 t
2 t
t
t
3 t
the rate of O2 production is 1.5 times faster than the ate of consumption of O3
the rate of O3 consumption is 2/3 times the production of O2
2. The decomposition of N2O5 proceeds according to the equation:
2N2O5(g)  4NO2(g) + O2(g)
If the rate of decomposition of N2O5 at a particular instant is 4.2 x 10-7
M/s, what is the rate of production of NO2 and O2?
1 [ N 2O5 ] 1 [ NO2 ] [O2 ]


2
t
4 t
t
[ NO2 ] 4 [ N 2O5 ]

 2  4.2 107 M  8.4 107 M
s
s
t
2
t
[O2 ] 1 [ N 2O5 ]

 2.1107 M
s
t
2
t
Rate  
Example: 2 N 2 O5( g )  4 NO2( g )  O2( g )
rate of decomposition of N 2 O5  
rate of formation of O 2 
[ N 2O5 ]   0.0101  0.0120  M

 1.9  10 5 M
s
t
 400  300  s
[O2 ]  0.00049  0.0040  M

 9  106 M
s
t
 400  300  s
rate of formation of NO 2 
[ NO2 ]  0.0197  0.0160  M

 3.7  105 M
s
t
400

300
s


Different Types of Rates
Average Rate
Concentration change over
a time interval (colored
region in plot)
Instantaneous Rate
Slope of the tangent line at a
given time (purple line in plot)
Section 12.2
Rate Law &
Reaction Order
Rate Law
Relates the rate directly to reactant concentrations
For the General case aA + bB  cC + dD, the rate law is:
Rate = k[A]m[B]n
where m and n are determined experimentally
k is called the rate constant
CAUTION!!! Rate law is NOT related to stoichiometry!
2 N 2 O5( g )  4 NO2( g )  O2( g )
CHCl3( g )  Cl2( g )  CCl4( g )  HCl( g )
Rate  k[ N 2 O5 ]
Rate  k [CHCl3 ][Cl2 ]
1
2
Reaction Order
The power to which each reactant is raised
For the General case aA + bB  cC + dD, with
Rate = k[A]m[B]n
“m” is the order of reactant A and “n” is the order of
reactant B
Overall reaction order is the sum of all or m+n in this case
Name the reactant orders and overall reaction order for
Rate  k[CHCl3 ][Cl2 ]
1
2
It is 1st order in CHCl3 and ½ order for Cl2 with 1 ½
order overall
Section 12.3
Experimental
Determination of Rate Law
Illustrative Example
Determine the rate law, the rate constant and the reaction orders for
each reactant and the overall reaction order using the data given below.
Experiment
[A]0
[B]0
Initial Rate M/s
1
2
3
0.100
0.100
0.200
0.100
0.200
0.100
4.0 x 10-5
4.0 x 10-5
16.0 x 10-5
Rate constant k & Overall Order
It is an indicator of the overall rate order of the equation
rate  k  A
n
M
s
?
M
M
 ? M n
s
M
1
?

n
sM
s  M n 1
order
m
k units
zeroth
0
M /s
first
1
1/ s
second
2
1/ M  s
Final Thoughts
What to do when finding ‘m’ isn’t obvious
m
1 1
1
1
    ln    m ln  
4 2
4
2
1
ln  
4

m
2
1
ln  
2
Section 12.4 & 12.5
First-Order Integrated
Rate Law & its Half-Life
First-Order Integrated Rate Law
[ A]
rate  k[ A] 
t
[ A]
  k t
[ A]
[ A ]t
t
d [ A]
  kdt

[ A] o
[ A ]0
ln[ A] |[[ AA]]0t  kdt |t0
ln[ A]t  ln[ A]0  kt
 ln[ A]t   kt  ln[ A]0
• Final equation is the integrated rate law
• Change in reactant concentration over time
may be plotted to get rate law
• We plot ln[A]t versus t:
ln[ A]t  k t  ln[ A]0
y
m x
b
Half-Life for First-Order Reaction
[ A]t
ln
 kt
[ A]0
1
2
[ A]0
ln
 kt 1
2
[ A]0
1
ln  kt 1
2
2
0.693   kt 1
t 1  0.693
2
k
2
Section 12.6
Second-Order Reactions
Second-Order Integrated Rate Law
rate  k[ A]2 
[ A]
t
[ A]
  k t
2
[ A]
[ A ]t
t
d [ A]
  kdt
2

[ A ]0 [ A]
o
1 [ A]t
|[ A]0  kdt |t0
 A
1
1

 kt
 At  A0

1
1
 kt 
 At
 A0
• Final equation is the integrated rate law
• This time we plot 1/[A]t versus t to get linearity
1
 kt  1
[ A]t m x
[ A]
y
b
Section 12.7
Zeroth-Order Reactions
Zeroth-Order Integrated Rate Law
[ A]
rate  k 
 [ A]   k t
t
[ A ]t

[ A ]0
t
d [ A]   kdt  [ A] |[[ AA]]0t  kdt |t0  [ A]t  [ A]0  kt
o
 [ A]t  kt  [ A]0
Summary of Integrated Equations
Section 12.8
Reaction Mechanisms
Definitions
Reaction Mechanisms: how electrons move during a reaction
Intermediate
Species that is produced then consumed
Never partakes or appears in the rate law
Elementary Step
A step that occurs in a reaction
Most reactions have multiple elementary steps
The stoichiometry of these steps CAN be used to get the reaction order
The sum of these steps leads to the overall reaction
Molecularity
Number of reacting particles in an elementary step
Uni- (1 species), Bi- (2 species), Ter- (3 species)
Example Application
Given the mechanism below state the overall reaction, rate law and
molecularity of each step as well as the intermediate.
step1:
NO2  NO2  NO  NO3 elementary
step 2 : NO3  CO  NO2  CO2
NO2  CO  NO  CO2
Rate law for step 1: rate1 = k1[NO2]2
Rate law for step 2: rate2 = k2[NO3][CO]
Both steps have 2 interacting species so bimolecular
Intermediate is NO3 which is produced then consumed
elementary
overall
Section 12.9
Rate Laws & Reaction
Mechanisms
Examples of Molecularity
Molecularity
Elementary step
Rate Law
Unimolecular
A  products
Rate = k[A]
Bimolecular
A + A  products
Rate = k[A]2
Bimolecular
A + B  products
Rate = [A][B]
Termolecular
A + A + A  products
Rate = k[A]3
Termolecular
A + A + B  products
Rate = k[A]2[B]
Termolecular
A + B + C  products
Rate = k[A][B][C]
Example Problem
What is the molecularity & rate law for each of the following?
molecularity
Rate law
H 2  NO  N  H 2 O
bimolecular
rate  k[ H 2 ][ NO ]
N  NO  N 2  O
bimolecular
rate  k[ N ][ NO]
H 2  O  H 2O
bimolecular
rate  k[ H 2 ][O]
Section 12.10
Rate Laws for Overall
Reactions
Rate Determining Step
The slowest step in a mechanism
Just like the slowest person in a relay race – the slowest step in
the mechanism will ruin the speed/time of the reaction
If the slow step is first it is very easy to get the rate law:
k1
step 1: 2 NO2 
 NO  NO3
( slow)
k2
step 2 : NO3 
 NO  O2
( fast )
the rate law would then be: rate  k1 [ NO2 ]2
Example for Fast Step First
What is the rate law for the mechanism below?
k1
step1: NO  O2 
 NO3
( fast )
k2
step 2 : NO  NO3 
 2 NO2 ( slow)
Fast Equilibrium Applied Example
The rate laws for the thermal and photochemical decomposition of
NO2 are different. Which of the following mechanisms are possible
for thermal and photochemical rates given the information below?
Thermal rate = k[NO2]2
Photochemical rate = k[NO2]
slow
a.) NO2 
 NO  O
fast
O  NO2 
NO  O2
fast
b.) NO2  NO2 
N 2 O4
slow
N 2 O4 
 NO  NO3
fast
NO3 
NO  O2
slow
c.) NO2  NO2 
 NO  NO3
fast
NO3 
NO  O2
Section 12.11
Reaction Rates &
Temperatures – Arrhenius
Equation
Arrhenius Equation
Relates rate to energy and temperature
k  Ae
 Ea
RT
Frequency Factor A
Energy of Activation, Ea
Indicative of the number of successful collisions
Energy that must be overcome to form products
Section 12.12
Using the Arrhenius
Equation
Finding the Ea through plotting
Mechanisms & Energy Profiles
slow
NO2  NO2 
 NO  NO3
fast
NO3 

 NO  O2
14_19.jpg
The slow step has the larger Ea since it takes longer to generate more energy
Transition State
Defn: state at which reactant bonds are broken and product
bonds begin to form
Located at the top of the hump for each elementary step in the
reaction profile
Section 12.13 & 12.14
Catalysts
Catalyst
A species that lowers the activation energy of a chemical reaction
and does not undergo any permanent chemical change
it is not present in the overall reaction expression
it is not present in the rate law
it must be consumed and produced in the elementary steps
Two types of catalysts:
Homogeneous: in the same phase as the reactants
Heterogeneous: a different phase from reactants
Example of Catalysis
Uncatalyzed mechanism blue line in the figure
hv
step 1: O3 
 O2  O
step 2 : O  O3 
 2O2
2O3 
 3O2
Ea  17.7 kJ / mol
Cl Catalyzed mechanism red line
14_21.jpg
step 1: Cl  O3 
 O2  ClO
step 2 : ClO  O3 
 Cl  2O2
2O3 
 3O2
Ea  2.2 kJ / mol