
Deals with:
› Speeds
› Rates
› Mechanisms

4 Factors Affect Kinetics of a Reaction
Physical States
 Homegenous conditions create a
molecules that will collide more with one
another more readily.
 Heterogeneous conditions limit the area
of contact.
 Medicine!!
1.
Concentration
 Concentration increases the frequency
in which molecules will collide making
the rate increase.
 Steel Wool
2.
Temperature
 At higher temperature the molecules will
collide with a higher energy, increasing
the rate.
 Spoiled Milk
3.
Catalyst
 Affect the kinds of collisions
(mechanism), lowering activation
energy.
 Living Organisms (DNA Polymerase)
4.
The change in the concentration of
reactants and products per unit time.
 Rate = [A] at t2 –[A] at t1
t 2- t1
 Rate = D[A]
Dt

N2 + 3H2  2NH3
Reactants rate will be negative because
they are being used up.
Products rates will be positive because
they are being made.

C
o
n
c
e
n
t
r
a
t
i
o
n
As the reaction progresses the
concentration H2 goes down
[H2]
Time

C
o
n
c
e
n
t
r
a
t
i
o
n
As the reaction progresses the
concentration N2 goes down 1/3 as
fast
[N2]
[H2]
Time

C
o
n
c
e
n
t
r
a
t
i
o
n
As the reaction progresses the
concentration NH3 goes up.
[N2]
[H2]
[NH3]
Time
The rate at a particular instant during the
reaction.
 The slope of the curve at that given time
is determined using a tangent line.


C
o
n
c
e
n
t
r
a
t
i
o
n
Average slope method
D[H2]
Dt
Time

C
o
n
c
e
n
t
r
a
t
i
o
n
Instantaneous slope method.
D[H2]
Dt
Time
 In
our example
N2 + 3H2  2NH3
-D[N2] = -D[H2] = D[NH3]
Dt
3 Dt
2Dt
aA + bB
cC + dD
1 D[A]
1 D[B]
1 D[C]
1 D[D]
= b
= c
= d
Rate =  a
Dt
Dt
Dt
Dt
NH4+(aq) + NO2(aq)N2(g) + 2 H2O(l)
 Comparing Experiments 1 and 2:
Ammonium ion concentration doubles
and so does the rate. Nitrite
concentration stays constant.
General Rate Law:
Rate = k [A]m[B]n
k= rate constant
 Changes with temperature.

m and n = reaction orders
 Small whole numbers.
 Add together to get the overall order of
reaction.
Have to be determined experimentally!!
NOT FROM COEFFICIENTS.
 Most rate laws have reaction orders of
0,1 or 2.
 Reaction order can be a fraction or
negative but this happens very rarely.

If k is 109 or higher, rxn is fast, and 10 or
lower is slow.
 Units for the rate constant depend on
the overall order of the rxn.
 What are the units for a second order
rxn?

We use initial rates because the reaction
slows down as concentration decreases.
 Determine the rate immediately as the
reactants are mixed.
 Again, THE RATE LAW NEEDS TO BE
DETERMINE EXPERIMENTALLY!!!!!!!!

 For
the reaction
BrO3- + 5 Br- + 6H+
3Br2 + 3 H2O
 The
general form of the Rate Law is
Rate = k[BrO3-]n[Br-]m[H+]p
 We
use experimental data to
determine the values of n,m,and p
Initial concentrations (M)
Rate (M/s)
BrO30.10
0.20
0.20
0.10
Br0.10
0.10
0.20
0.10
H+
0.10
0.10
0.10
0.20
8.0 x 10-4
1.6 x 10-3
3.2 x 10-3
3.2 x 10-3
Now we have to see how the rate
changes with concentration
Differential Rate law - describes how
rate depends on concentration.
Integrated Rate Law - Describes how
concentration depends on time.
 For
each type of differential rate law
there is an integrated rate law and
vice versa.
 Rate laws can help us better
understand reaction mechanisms.
For rxn : A products
Differentiated rate law:
Rate =
−Δ[𝐴]
Δ𝑡
= k [A]
Integrated rate law:
ln[A]t – ln[A]o = -kt or ln[A]t/ln[A]o = -kt
[A]t= Concentration at time t
[A]o= Initial Concentration
 We can determine 3 things:
› Concentration of reactants at a given time.
› A time interval for a fraction of reactants to
react.
› A time interval for a reactant concentration
to fall to a certain level.

Normally written as follows:
ln[A]t = -kt + ln[A]o
Now it is written in slope-intercept form and
we know –k is the slope of our line and
initial concentration is our y-intercept.
For rxn: A  products
Differentiated rate law:
Rate =
−∆[𝐴]
=
∆𝑡
k[A]2
Integrated rate law:
1
=
𝐴𝑡
kt +
1
𝐴𝑜
For rxn: A  products
Differentiated rate law::
Rate =
−∆[𝐴]
=
∆𝑡
k
Integrated rate law:
[A]t= -kt + [A]o
The time required for the concentration
of a reaction to reach half its initial value.
[A]t1/2 = ½ [A]o
 Substitute into first-order reaction.
t1/2 = 0.693/k
 Half-life does not depend on initial
concentration and is therefore constant
throughout the reaction.


Substitute into second-order
t1/2=
1
𝑘𝐴
𝑜
This half-life depends on initial
concentration .
 Lower initial concentration, longer halflife.

Most reactions will increase rate as
temperature increases.
 Examples:

› Dough rising room temp vs refrig.
› Plants in warm weather
› Glow sticks

An increase in temperature increases the
rate constant. (approx. doubles for
every 10 degree Celsius)
Reaction rates are affected by
concentration and temperature.
 Molecules must collide to react.
 Concentration increase the number of
collisions
 Temperature increase the force of the
collision.

Atoms must be oriented in the right way
during a collision to form new bonds.
 Talk about steric factors.

O
O
O
O
N
Br
N
Br
N
Br
N
Br
Br
N O
Br
O N
O N Br
Br N O
Br N O
O N Br
No
Reaction
Molecules need to collide with enough
kinetic energy.
 Kinetic energy is converted into
chemical potential energy.
 The amount needed is known as
activation energy (Ea)
 Rate depends on the magnitude of
activation energy.

P
o
t
e
n
t
i
a
l
E
n
e
r
g
y
Activation
Energy Ea
Reactants
Products
Reaction Coordinate
Activated complex
P
o
t
e
n
t
i
a
l
E
n
e
r
g
y
Reactants
Products
Reaction Coordinate
Br---NO
P
o
t
e
n
t
i
a
l
E
n
e
r
g
y
Br---NO
Transition State
2BrNO
2NO + Br
Reaction Coordinate
2
P
o
t
e
n
t
i
a
l
E
n
e
r
g
y
Reactants
}
Products
Reaction Coordinate
DE
 Said
that reaction rate should
increase with temperature.
 At high temperature more molecules
have the energy required to get over
the barrier.
 The number of collisions with the
necessary energy increases
exponentially.

Number of collisions with the required
k=Ae-Ea/RT

k= rate constant
A= frequency factor
 e is Euler’s number (opposite of ln)
 Ea = activation energy
 R = ideal gas constant
 T is temperature in Kelvin

The steps by which a reaction occurs.
 Elementary reactions:

› Reactions that occur in a single step.

Molecularity
› The number of molecules that participate in
an elementary reaction.
› Unimolecular: single molecule
› Bimolecular: two molecules
› Termolecular: three molecules (rare)
A
products
 A+A
products
 2A
products
 A+B
products
 A+A+B
Products
 2A+B
Products
 A+B+C
Products

Rate = k[A]
Rate= k[A]2
Rate= k[A]2
Rate= k[A][B]
Rate= k[A]2[B]
Rate= k[A]2[B]
Rate= k[A][B][C]

The net change represented by a
balanced equation often occurs in
multiple steps.
2 NO + O2  2 NO2
Mechanism
 2 NO
N2O2
 N2O2 + O2  2 NO2

(fast)
(slow)
2
NO + O2
2 NO2
 Mechanism
 2 NO
N2O2
 N2O2 + O2
2 NO2
(fast)
(slow)
The slow step limits the overall reaction
rate.
 Example:

› Toll Plaza
First step is equal to the rate law.
 Does not depend on the intermediate.
 Intermediate is zero order and therefore
absent from the rate law.


Less straight forward

2 NO
N2O2
(fast)
N2O2 + O2
2 NO2
(slow)
rate = k2[N2O2][O2]
k1[NO]2 = k-1[N2O2]
rate = k2 (k1/ k-1)[NO]2[O2]=k[NO]2[O2]




 There
is an activation energy for each
elementary step.
 Activation energy determines k.
 k = Ae- (Ea/RT)
 k determines rate
 Slowest step (rate determining) must
have the highest activation energy.
This reaction takes place in three steps
Ea
First step is fast
Low activation energy
Ea
Second step is slow
High activation energy
Ea
Third step is fast
Low activation energy
Second step is rate determining
 Speed
up a reaction without being
used up in the reaction.
 Enzymes are biological catalysts.
 Homogenous Catalysts are in the
same phase as the reactants.
 Heterogeneous Catalysts are in a
different phase as the reactants.
 Catalysts
allow reactions to
proceed by a different mechanism
- a new pathway.
 New pathway has a lower
activation energy.
 More molecules will have this
activation energy.
 Do not change DE
H H
Hydrogen bonds to
surface of metal.
 Break H-H bonds
H H

H
H
Pt surface
H H
H
H
H
C
C
H
H H
H
H
Pt surface

The double bond breaks and bonds to
the catalyst.
H
H
H
C
H
C
H
H
Pt surface
H H

The hydrogen atoms bond with the
carbon
H
H
H
C
H
C
H
H
Pt surface
H H
 Chlorofluorocarbons
catalyze the
decomposition of ozone.
 Enzymes regulating the body
processes. (Protein catalysts)
 Catalysts
will speed up a reaction but
only to a certain point.
 Past a certain point adding more
reactants won’t change the rate.
R
a
t
e
Rate increases until the active
sites of catalyst are filled.
 Then rate is independent of
concentration

Concentration of reactants