1
Chapter 15 — Kinetics
Chemical Kinetics: The Rates of
Chemical Reactions
Thermodynamics – does a reaction take place?
Chapter 15
Kinetics – how fast does a reaction proceed?
Chemical Kinetics:
The Rates of
Chemical Reactions
Chemical Kinetics will now provide information about
the arrow!
Reactants
Products
This gives us information on HOW a reaction occurs!
Jeffrey Mack
California State University,
Sacramento
Chemical Kinetics
Rate of Reactions
Kinetics is the study of how fast (Rates)
chemical reactions occur.
Important factors that affect the rates of
chemical reactions:
• reactant concentration or surface area in
solids
• temperature
• action of catalysts
Our goal: Use kinetics to understand chemical
reactions at the particle or molecular level.
Determining a Reaction Rate
Blue dye is oxidized with bleach.
Its concentration decreases with time.
The rate — the change in dye conc.
with time — can be determined from
the plot.
• Reactants go away with time.
• Products appear with time.
• The rate of a reaction can be measured by
either. In this example by the loss of color
with time.
Reaction Rate & Stoichiometry
Rate =
D [reactant ]
change in concentration D [product ]
=
=change in time
D time
D time
In general for the reaction: aA + bB
Rate of reaction  
1  A
a t

1  B
b t

 cC + dD
1   C
1  D

d t
c t
Dye Conc
reactants go away with time therefore the negative sign…
Time
A
B
Reaction rate is the change in the concentration of
a reactant or a product with time (M/s).
2
Chapter 15 — Kinetics
Determining a Reaction Rate
Determining a Reaction Rate
The rate of appearance or disappearance is
measured in units of concentration vs. time.
moles
Rate =
L
= Ms1 or Mmin1 etc...
time
There are three ―types‖ of rates
1.initial rate
2.average rate
3.instantaneous rate
Reaction Rates
Reaction Rates
The concentration of a reactant
decreases with time.
Reactant concentration (M)
Reactant concentration (M)
Initial rate
Time
Time
Average Rate =
Time
Y [concentration]

X
time
Reaction Rates
Reactant concentration (M)
Reactant concentration (M)
Reaction Rates
Instantaneous rate (tangent line)
Time
3
Chapter 15 — Kinetics
Reaction Rates
Problem:
Consider the reaction:
2NO (g)  O2 (g)  2NO2 (g)
Reactant concentration (M)
Initial rate
Instantaneous rate (tangent line)
During the beginning stages of the
reaction, the initial rate is very close to
the instantaneous rate of reaction.
Over a period of 50.0 to 100.0 s, the concentration of
NO(g) drops from 0.0250M to 0.0100M.
a) What is the average rate of disappearance of NO(g)
during this time?
Time
Practice example
Problem:
Consider the reaction:
2NO (g)  O2 (g)  2NO2 (g)
Over a period of 50.0 to 100.0 s, the concentration of
NO(g) drops from 0.0250M to 0.0100M.
a) What is the average rate of disappearance of NO(g)
during this time?
RATE= -
Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (g)
1 [NO]
2 t
(0.0100M  0.0250M)
100.0 s  50.0 s

1 mol of reaction
= 1.50104 Ms1
2 moles NO (g)
recall…
Ms1 
rate = -
[CH4]
[CO2] 1 [H2O]
[O2]
=
=-1
=
t
t
2 t
2 t
M
s
Reaction Conditions & Rate
• There are several important factors that will
directly affect the rate of a reaction:
• Temperature
• The physical state of the reactants
• Addition of a catalyst
All of the above can have a dramatic impact
on the rate of a chemical process.
Reaction Conditions & Rate:
Temperature
Bleach at 54 ˚C
Bleach at 22 ˚C
4
Chapter 15 — Kinetics
Reaction Conditions & Rate:
Physical State of Reactants
Reaction Conditions & Rate:
Catalysts
Catalyzed decomposition of H2O2
2 H2O2  2 H2O + O2
Effect of Concentration on Reaction
Rate: Concentration
Mg(s) + 2 HCl(aq)  MgCl2(aq) + H2(g)
0.3 M HCl
6 M HCl
The Rate Law Expression
The rate of reaction must be a function of concentration:
• As concentration increases, so do the number of
collisions…
• As the number of collisions increase, so does the
probability of a reaction.
• This in turn increases the rate of conversion of
reactants to products.
• The relationship between reaction rate and
concentration is given by the reaction Rate Law
Expression.
• The reaction rate law expression relates the rate of a
reaction to the concentrations of the reactants.
Reaction Order
For the general reaction:
aA + bB  cC + dD
Rate  k [A]x [B] y
Reaction Orders
A reaction order can be zero, or positive
integer and fractional number.
Order
Name
Rate Law
0
zeroth
rate = k[A]0 = k
1
first
rate = k[A]
x and y are the reactant orders determined
2
second
rate = k[A]2
from experiment.
0.5
one-half
rate = k[A]1/2
x and y are NOT the stoichiometric
1.5
three-half
rate = k[A]3/2
coefficients.
0.667
two-thirds
rate = k[A]2/3
Each concentration is expressed with an order (exponent).
The rate constant converts the concentration expression into
the correct units of rate (Ms 1).
5
Chapter 15 — Kinetics
Reaction Order
Reaction Constant
1
Rate (Ms-1 ) = k[A][B] 2 [C]2
To find the units of the rate constant, divide the rate units by
the Molarity raised to the power of the overall reaction order.
Rate (Ms1 )  k [A]x [B] y
Overall order:
or
1 + ½ + 2 = 3.5 = 7/2
Ms1  k  Mx My
Ms1  k  M xy
sevenhalves order
note: when the order of a reaction is 1 (first
order) no exponent is written.
Reaction Constant
• The rate constant, k, is a proportionality
constant that relates rate and concentration.
• It is found through experiment that the rate
constant is a function temperature.
• Rate constants must therefore be reported at
the temperature with which they are
measured.
• The rate constant also contains information
about the energetics and collision efficiency
of the reaction.
EXAMPLE: The reaction,
k=
Ms1
 xy
 M
1 x  y  1
s
M
if (x+y) = 1
k has units of s-1
if (x+y) = 2
k has units of M-1s-1
EXAMPLE: The reaction,
2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g)
is experimentally found to be first order in H2 and third order in NO
a) Write the rate law.
EXAMPLE: The reaction,
2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g)
2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g)
is experimentally found to be first order in H2 and third order in NO
is experimentally found to be first order in H2 and third order in NO
a) Write the rate law.
a) Write the rate law.
Rate(Ms-1) = k [H2] [NO]
3
b) What is the overall order of the reaction?
Rate(Ms-1) = k [H2] [NO]
3
b) What is the overall order of the reaction?
Overall order = 1 + 3 = 4
c) What are the units of the rate constant?
“4th order”
6
Chapter 15 — Kinetics
Rate Law Expression
EXAMPLE: The reaction,
2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g)
Rate = k [A]x [B]y
is experimentally found to be first order in H2 and third order in NO
a) Write the rate law.
Rate(Ms-1) = k [H2] [NO]
If the rate doubles when [A] doubles and [B] stays
constant, the order for [A] is?
3
b) What is the overall order of the reaction?
“4th order”
Overall order = 1 + 3 = 4
c) What are the units of the rate constant?
Rate =
M
= k ´ M ´ M3 = k ´ M4
s
k=
M
= M-3 s-1
s ´ M4
Rate Law Expression
Rate Law Expression
Rate = k [A]x [B]y
If the rate doubles when [A] doubles and [B] stays
constant, the order for [A] is?
one… 1
Rate = k [A]x [B]y
If the rate doubles when [A] doubles and [B] stays
constant, the order for [A] is?
one… 1
Rate(2) = 2 ´ Rate(1)
x
Rate(2) 2 ´ Rate(1)
[2A]x [B]y é 2A ù
=
=2=
= ê ú = 2x
x
y
Rate(1)
Rate(1)
[A] [B]
ëAû
log(2) = log(2)x = x ´ log(2)
x=
Determining a Rate Equation
Determining Reaction Order: The
Method of Initial Rates
The reaction of nitric oxide with hydrogen at 1280 °C
is as follows:
Initial rate
Reactant concentration (M)
log(2)
=1
log(2)
Instantaneous rate (tangent line)
During the beginning stages of the
reaction, the initial rate is very close
to the instantaneous rate of reaction.
2NO (g) + 2H2 (g)  N2 (g) + 2H2O (g)
From the following experimental data, determine the
rate law and rate constant.
Trial
[NO]o (M)
[H2]o (M)
Initial Rate
(Mmin-1)
1
0.0100
0.0100
0.00600
2
0.0200
0.0300
0.144
3
0.0100
0.0200
0.0120
7
Chapter 15 — Kinetics
Determining Reaction Order: The
Method of Initial Rates
The reaction of nitric oxide with hydrogen at 1280 °C
is as follows:
Determining Reaction Order: The
Method of Initial Rates
The reaction of nitric oxide with hydrogen at 1280 °C
is as follows:
2NO (g) + 2H2 (g)  N2 (g) + 2H2O (g)
Notice that in Trial 1 and 3, the initial concentration of
NO is held constant while H2 is changed.
2NO (g) + 2H2 (g)  N2 (g) + 2H2O (g)
This means that any changes to the rate must be due to the
changes in H2 which is related to the concentration of H 2 & its
order!
Trial
[NO]o (M)
[H2]o (M)
Initial Rate
(Mmin-1)
Trial
[NO]o (M)
[H2]o (M)
Initial Rate
(Mmin-1)
1
0.0100
0.0100
0.00600
1
0.0100
0.0100
0.00600
2
0.0200
0.0300
0.144
2
0.0200
0.0300
0.144
3
0.0100
0.0200
0.0120
3
0.0100
0.0200
0.0120
Determining Reaction Order: The
Method of Initial Rates
Determining Reaction Order: The
Method of Initial Rates
2NO(g) + 2H2(g)  N2(g) + 2H2O(g)
y
y
0.0120M min-1 [0.0200]
é 0.0200 ù
=
y = ê
-1
ú
[0.0100]
0.00600M min
ë 0.0100 û
The rate law for the reaction is given by:
Rate(M/min) = k [NO]x [H2]y
Taking the ratio of the rates of Trials 3 and 1 one finds:
2.00 = éë2.00ùû
Take the log of both sides of the equation:
x
y
k [NO](3)
[H2 ](3)
Rate (Trial 3)
=
x
(1)
log
y
2 (1)
Determining Reaction Order: The
Method of Initial Rates
Similarly for x:
Rate(M/min) = k [NO]x[H2]y
Rate(2)

Rate(1)
x
y
k [NO](2) [H2 ](1)
x
(1)
y
2 (2)
k [NO] [H ]
k [0.0200]x [0.0300]
0.144

0.00600
k [0.0100]x [0.0100]
 2
24 
 2
x
xlog  2   log  8 
x
= [2.00]
y
y
Plugging in the values from the data:
k [0.0100]x [0.0200]y
[0.0200]y
0.0120 M / min


[0.0100]y
0.00600 M / min k [0.0100]x [0.0100]y
2.00
log(2.00) = log [2.00]
k [NO] [H ]
Rate (Trial 1)
y
log(2.00) = y ´ log(2.00)
y=
log(2.00)
log(2.00)
y=1
Rate(M/min) = k [NO]x[H2]
Determining Reaction Order: The
Method of Initial Rates
The Rate Law expression is:
2NO(g) + 2H2(g)  N2(g) + 2H2O(g)
Rate = k[NO]3 [H2 ]
The order for NO is 3
 3
The order for H2 is 1
8
x=3
The over all order is 3 + 1 =4
8
Chapter 15 — Kinetics
Determining Reaction Order: The
Method of Initial Rates
Concentration–Time Relationships:
Integrated Rate Laws
The Rate constant
• It is important know how long a reaction must
proceed to reach a predetermined
concentration of some reactant or product.
• We need a mathematical equation that
relates time and concentration:
• This equation would yield concentration of
reactants or products at a given time.
• It will all yield the time required for a given
amount of reactant to react.
Rate(M/min) = k
[NO]3[H
2]
To find the rate constant, choose one set of data and solve:
0.0120 Mmin1  k  0.0100 M  0.0200 M
3
k
0.0120 Mmin1
0.0120 Mmin1

 0.0100 M  0.0200 M  0.0100   0.0200  M4
3
3
k  6.00  105 M3 min1
Integrated Rate Laws
For a zero order process where ―A‖ goes onto
products, the rate law can be written:
A  products
Rate(Ms1 )  
[A]
 k[A]0 = k
t
k has units of Ms1
For a zero order process, the rate is the rate
constant!
Integrated Rate Laws
A  products
[A]
Rate(Ms )  
 k[A]0 = k
t
Zero order kinetics
1
This is the ―average rate‖
If one considers the infinitesimal changes in
concentration and time the rate law equation
becomes:
d[A]
=k
Rate(Ms-1 ) = dt
This is the ―instantaneous rate‖
Integrated Rate Laws
Zero order kinetics
Integrated Rate Laws
Zero order kinetics
what’s this look like?
[A]o  [A]t = kt
[A]t = kt + [A]o
d[A]
=k
dt
[ A]o
y = mx + b
t
d[A] = -k ò dt
0
where [A] = [A]o at time t = 0 and [A] = [A] at time t = t
[A]t  [A]o = k(t  0) = kt
[A]t  [A]o = kt
[A]t (mols/L)
ò
[ A ]t
rearranging…
 the y-intercept is [A]o
a plot of [A]t vs t looks like…
k has units of M×(time)1
slope = k
t (time)
Conclusion: If a plot of reactant
concentration vs. time yields
a straight line, then the reactant
order is ZERO!
9
Chapter 15 — Kinetics
Integrated Rate Laws
For a first order process,
the rate law can be written:
[A]
Rate(Ms )  
 k[A]
t
A  products
Integrated Rate Laws
First order kinetics
[A]
[A]ot
1
This is the ―average rate‖
Taking the exponent to each side of the equation:
If one considers the infinitesimal changes in
concentration and time the rate law equation becomes:
Rate(Ms1 )  
d[A]
 k[A]
dt
This is the ―instantaneous rate‖
[A]t
 ekt
[A]o
ln  [A]t  [A]o e kt 
Taking the natural log of both sides…
[A]t  [A]o ekt
Integrated Rate Laws
First order kinetics
ln[A]t =  kt + ln[A]o
so a plot of ln[A]t vs t looks like…
y = mx + b
ln[A]t  ln  [A]o e kt   ln[A]o  ln  e kt  = ln[A]o  kt
k has units of (time)1
 the y-intercept is ln[A]o
ln[A]t
rearranging…
or
Conclusion: The concentration of a reactant governed by
first order kinetics falls off from an initial concentration
exponentially with time.
Integrated Rate Laws
First order kinetics
 [A]t 
ln 
  kt
 [A]o 
d[A]
 k 0t dt
[A]
slope = k
Conclusion: If a plot of natural
log of reactant concentration vs.
time yields a straight line, then
the reactant order is FIRST!
ln[A]t =  kt + ln[A]o
t (time)
Problem:
Problem:
The decomposition of N2O5(g) following 1st order kinetics.
If 2.56 mg of N2O5 is initially present in a container and
2.50 mg remains after 4.26 min, what is the rate constant
in s1?
The decomposition of N2O5(g) following 1st order kinetics.
If 2.56 mg of N2O5 is initially present in a container and
2.50 mg remains after 4.26 min, what is the rate constant
in s1?
Begin with the integrated rate law for a 1st order process:
[A]t  [A]o ekt
[N2O5 ]t
[N2O5 ]o
[A]t
 ekt
[A]o
= e-kt
Wait… what is the volume of the container???
Do we need to convert to moles?
10
Chapter 15 — Kinetics
Problem:
Problem:
The decomposition of N2O5(g) following 1st order kinetics.
If 2.56 mg of N2O5 is initially present in a container and
2.50 mg remains after 4.26 min, what is the rate constant
in s1?
The decomposition of N2O5(g) following 1st order kinetics.
If 2.56 mg of N2O5 is initially present in a container and
2.50 mg remains after 4.26 min, what is the rate constant
in s1?
[N2O5 ]t
= e-kt
[N2O5 ]o
Check it out!
You don’t need the
volume of the container!
1g
1 mol

103 mg 108.0 g
2.50 mg
xL


1g
1 mol
2.56
mg
2.56 mg  3

10 mg 108.0 g
xL
2.50 mg 
N2O5 t
N2O5 o
N2O5 t
N2O5 o
k = 9.3 105 s1
Integrated Rate Laws
Second order kinetics
Second order kinetics
1
A  Products
 A t
k has units of M1s1
Rate(Ms )  
d A 
dt
 k A
Integrating as before we find:
1
1
 kt 
 A t
 A 0
Summary of Integrated Rate Laws
 kt 
1
 A 0
so a plot of 1/[A]t vs t looks like…
y = mx + b
2
 the y-intercept is 1/[A]o
1/[A]t
1
2.50 m
2.56 m
taking the natural log and substituting time in
seconds:
 2.50 mg 
ln 
  k   256 s 
 2.56 mg 
Integrated Rate Laws
Rate = k[A]2
 ekt 
k has units of M1s1
slope = k
t (time)
Conclusion: If a plot of one over
reactant concentration vs. time
yields a straight line, then the
reactant order is second!
Half-life & First-Order Reactions
Half-life of a
reaction is the
time taken for the
concentration of a
reactant to drop
to one-half of the
original value.
11
Chapter 15 — Kinetics
Reaction Half-Life: 1st Order Kinetics
[A]0
at time = t½
2
For a first order process the half life (t½ ) is found
mathematically from:
[A]t 
(1) ln  A t  kt  ln  A 0
Start with the integrated
rate law expression for a 1st
order process
(2) ln  A t  ln  A 0  kt
Bring the concentration
terms to one side.
  A t 
(3) ln 
  kt
 [ A]0 
Express the concentration
terms as a fraction using the
rules of ln.
Reaction Half-Life: 1st Order Kinetics
 2  A 0 
ln 
  kt
  A 0 
ln  2   kt 1
2
t1 
2
Reaction Half-Life: 1st Order Kinetics
  A 0 
(4) ln 
  kt
 [A] 

 A 0
(5) ln 
 [A]0

2
exchange [A] with [A]0 to reverse
the sign of the ln term and cancel
the negative sign in front of k

  kt Substitute the value of [A] at the
1

2 half-life

t1 =
2
ln2 0.693
=
k
k
Problem:
A certain reaction proceeds through first order kinetics.
The half-life of the reaction is 180. s.
What percent of the initial concentration remains after 900.s?
1
2
The half-life is
independent of the
initial concentration!
ln2 0.693

k
k
So, knowing the rate constant for a first order
process, one can find the half-life!
Problem:
Problem:
A certain reaction proceeds through first order kinetics.
The half-life of the reaction is 180. s.
What percent of the initial concentration remains after 900.s?
A certain reaction proceeds through first order kinetics.
The half-life of the reaction is 180. s.
What percent of the initial concentration remains after 900.s?
Using the integrated rate law, substituting in the value of k and
900.s we find:
Using the integrated rate law, substituting in the value of k and
900.s we find:
t1 
2
ln2
k
k
ln2
ln2

t1
180.s1
2
k = 0.00385 s-1
12
Chapter 15 — Kinetics
Problem:
Problem:
A certain reaction proceeds through first order kinetics.
The half-life of the reaction is 180. s.
What percent of the initial concentration remains after 900.s?
A certain reaction proceeds through first order kinetics.
The half-life of the reaction is 180. s.
What percent of the initial concentration remains after 900.s?
Using the integrated rate law, substituting in the value of k and
900.s we find:
Using the integrated rate law, substituting in the value of k and
900.s we find:
t1 
2
ln2
k
k
[A]t
 e kt
[A]0
ln2
ln2

t1
180.s1
k = 0.00385 s-1
t1 
2
2
[A]t
 e0.00385 s
[A]0
1
 900.s
= 0.0312
ln2
k
k
ln2
ln2

t1
180.s1
k = 0.00385 s-1
2
[A]t
 e kt
[A]0
[A]t
 e0.00385 s
[A]0
1
 900.s
= 0.0312
Since the ratio of [A]t to [A]0 represents the fraction of [A] that
remains, the % is given by:
100  0.0312 = 3.12%
Reaction Half-Life: 1st Order Kinetics
Elements that decay via radioactive processes
do so according to 1st order kinetics:
Element:
238U
Half-life:
 234Th + 
4.5  109 years
14C
 14N + 
5730 years
131I
 131Xe + 
8.05 days
Reaction Half-Life: 1st Order Kinetics
Recall that that the rate constant for a 1st order
process is given by:
k
[A]t
 ekt
[A]0
ln(2)
t1
Tritium decays to helium by beta () decay:
3
1
H
[A]  [A]  ekt
t
0

0.693
 49.2 years
12.3 years
= 0.094 mg

e 
0
-1
3
2
He
The half-life of this process is 12.3 years
Starting with 1.50 mg of 3H, what quantity remains
after 49.2 years.
Solution:
Begin with the integrated rate law expression for 1st
order kinetics.
[A]t
 ekt
[A]0
Reaction Half-Life: 1st Order Kinetics
Notice that 49.2 years is 4 half-lives…
49.2
4
12.3
After 1 half life:
2
[3H]t = 1.50 mg  e
Reaction Half-Life: 1st Order Kinetics
1.50 mg
2
= 0.75 mg remains
0.75 mg
= 0.38 mg remains
2
0.38 mg
After 3 half life's:
= 0.19 mg remains
2
After 2 half life's:
After 4 half life's:
0.19 mg
= 0.094 mg remains
2
13
Chapter 15 — Kinetics
A Microscopic View of Reaction
Rates
Arrhenius: Molecules must posses a minimum
amount of energy to react. Why?
(1) In order to form products, bonds must be
broken in the reactants.
(2) Bond breakage requires energy.
(3) Molecules moving too slowly, with too little
kinetic energy, don’t react when they collide.
The Activation energy, Ea, is the minimum
energy required to initiate a chemical reaction.
Ea is specific to a particular reaction.
A Microscopic View of Reaction
Rates
When one writes a reaction all that is seen are
the reactants and products. This details the
overall reaction stoichiometry.
How a reaction proceeds is given by the
reaction mechanism.
Activation Energy
Activation Energy
The progress of a chemical reaction as the reactants
transform to products can be described graphically by a
Reaction Coordinate.
The reaction of NO2 and CO (to give NO and CO2) has an activation
energy barrier of 132 kJ/mol-rxn. The reverse reaction (NO + CO2 
NO2 + CO) requires 358 kJ/mol-rxn. The net energy change for the
reaction of NO2 and CO is 226 kJ/mol-rxn.
Activation Energy
The temperature for a
system of particles is
described by a
distribution of energies.
Potential Energy
Transition State
reactants
Eact
In order for the reaction
to proceed, the reactants
must posses enough
energy to surmount a
reaction barrier.
HRXN
products
Reaction Progress
Activation Energy
Molecules need a minimum amount of energy to react.
Visualized as an energy barrier - activation energy, Ea.
At higher temps, more
particles have enough
energy to go over the
barrier.
E > Ea
E < Ea
Since the probability of a
molecule reacting
increases, the rate
increases.
Reaction coordinate
diagram
14
Chapter 15 — Kinetics
Activation Energy
Activation Energy
Orientation factors into the equation
Orientation factors into the equation
The orientation of a molecule during collision can have a
profound effect on whether or not a reaction occurs.
In some cases, the reactants must have proper
orientation for the collision to yield products.
The reaction
occurs only when
the orientation of
the molecules is
just right…
This reduces the
number of collisions
that are reactive!
When the green atom collides with the green atom on the
molecule, a reactive or effective collision occurs.
The Arrhenius Equation
When the green atom collides with the red atom on the
molecule, this leads to a non-reactive or ineffective
collision occurs.
The Arrhenius Equation
-Ea
Arhenius discovered that most reaction-rate data
obeyed an equation based on three factors:
k = Ae RT
(1) The number of collisions per unit time.
k is the rate constant
T is the temperature in K
(2) The fraction of collisions that occur with the
correct orientation.
Ea is the activation energy
R is the ideal-gas constant
(8.314 J/Kmol)
(3) The fraction of the colliding molecules that
have an energy greater than or equal to Ea.
From these observations Arrhenius developed the
aptly named Arrhenius equation.
The Arrhenius Equation
Temperature Dependence of the Rate Constant:
Increasing the temperature of a reaction generally
speeds up the process (increases the rate)
because the rate constant increases according to
the Arrhenius Equation.
A is known the frequency or pre–exponential factor
In addition to carrying the units of the rate constant, ―A”
relates to the frequency of collisions and the orientation of a
favorable collision probability
Both A and Ea are specific to a given reaction.
Effect of Temperature
Reactions generally
occur slower at lower T.
In ice at 0 oC
Room temperature
Rate (Ms-1) = k[A]x[B]y
-Ea
k = Ae RT
As T increases, the value of the exponential part of
the equation becomes less negative thus increasing
the value of k.
Iodine clock reaction.
H2O2 + 2 I- + 2 H+  2 H2O + I2
15
Chapter 15 — Kinetics
The Arrhenius Equation
The Arrhenius Equation
Determining the Activation Energy
Determining the Activation Energy
Ea may be determined experimentally.
First take natural log of both sides of the Arrhenius equation:
One can determine the activation energy of a reaction by
measuring the rate constant at two temperatures:
Writing the Arrhenius equation for each temperature:
-Ea
ln
lnk  
k = Ae RT
Ea
 ln A
RT
lnk1 = -
ln k
y = mx + b
A plot of ln k vs
of 
1
will have a slope
T
Ea
and a y-intercept of ln A.
R
Determining the Activation Energy
æk ö æ E
ö æ E
ö
ln ç 2 ÷ = ç - a + ln A ÷ - ç - a + ln A ÷
ø è RT1
ø
è k1 ø è RT2
Knowing the rate constants at
two temps yields the activation
energy.
æk ö E æ 1 1 ö
or
ln ç 2 ÷ = a ´ ç - ÷
è k1 ø R è T1 T2 ø
Knowing the Ea and the rate
constant at one temp allows one
to find k(T2)
Ea
+ ln A
RT2
æk ö æ E
ö æ E
ö
lnk 2 - lnk1 = ln ç 2 ÷ = ç - a + ln A ÷ - ç - a + ln A ÷
ø è RT1
ø
è k1 ø è RT2
Problem:
The activation energy of a first order reaction is 50.2 kJ/mol at
25 °C. At what temperature will the rate constant double?
(1)
k 2 = 2k1
(2)
æk ö
æ 2k ö
E æ1 1ö
ln ç 2 ÷ = ln ç 1 ÷ = ln(2) = a ç - ÷
R è T1 T2 ø
è k1 ø
è k1 ø
(3)
50.2 kJ/mol 103J
Ea
´
=
= 6.04 ´ 103 K
J
1kJ
R
8.314
mol × K
Catalysis
Problem:
The activation energy of a first order reaction is 50.2 kJ/mol at
25 °C. At what temperature will the rate constant double?
(4)
lnk 2 = -
Subtracting k1 from k2 we find that:
1
T
The Arrhenius Equation
Ea
+ ln A
RT1
æ 1
1ö
- ÷
ln(2) = 0.693 = 6.04 ´ 103 K ´ ç
è 298K T2 ø
Catalysts speed up reactions by altering the
mechanism to lower the activation energy barrier.
Dr. James Cusumano, Catalytica Inc.
algebra!
(5)
1
= 3.24 ´ 10-3 K -1
T2
T2 = 308 K
A 10 °C change of
temperature doubles
the rate!!
What is a catalyst?
Catalysts and the environment
Catalysts and society
16
Chapter 15 — Kinetics
Catalysis
Catalysis
In auto exhaust systems — Pt, NiO
2 CO + O2  2 CO2
2 NO  N2 + O2
2. Polymers:
H2C=CH2  polyethylene
3. Acetic acid: CH3OH + CO  CH3CO2H
4. Enzymes — biological catalysts
Catalysis
Catalysis and activation energy
Iodine-Catalyzed Isomerization of
cis-2-Butene
MnO2 catalyzes
decomposition of H2O2
2 H2O2  2 H2O + O2
Uncatalyzed reaction
Catalyzed reaction
Iodine-Catalyzed Isomerization of
cis-2-Butene
Reaction Mechanisms
The overall stoichiometry of a chemical reaction is
most often the sum of several steps:
(1)
(2)
(3)
2AB
A2B2 + C2
A2B + C2
 A2B2
 A2B + C2B
 A2 + C2B
2AB + A2B2 + A2B + 2C2  A2B2 + A2B + A2 + 2C2B
Net:
2AB + 2C2  A2 + 2C2B
The sequence of steps (1-3) describes a possible
“reaction mechanism”.
17
Chapter 15 — Kinetics
Reaction Mechanisms
Reaction Mechanisms
The overall stoichiometry of a chemical reaction is
most often the sum of several steps:
(1)
(2)
(3)
2AB
A2B2 + C2
A2B + C2
 A2B2
 A2B + C2B
 A2 + C2B
2AB + A2B2 + A2B + 2C2  A2B2 + A2B + A2 + 2C2B
Net:
2AB + 2C2  A2 + 2C2B
The species that cancel out (not part of the overall
reaction) are called “reaction intermediates”.
Reaction Mechanisms
slow (1)
fast (2)
fast (3)
2AB
A2B2 + C2
A2B + C2
 A2B2
 A2B + C2B
 A2 + C2B
2AB + 2C2  A2 + 2C2B
The rate of the overall reaction can never be faster
than the “slowest step” in the mechanism.
If reaction (1) is the slowest of the three steps in the
mechanism…
Then it is known as the “rate determining step”
Reaction Mechanisms
Consider the following reaction:
2NO2(g) + F2(g)  2FNO2(g)
If the reaction proceeded by the overall reaction, the
rate law for the reaction would be 3rd order overall.
The actual rate law is found to be:
Rate = k[NO2][F2]
Indicating that the slowest step in the mechanism is
a bimolecular reaction between NO2 and F2.
(1)
(2)
(3)
2AB
A2B2 + C2
A2B + C2
 A2B2
 A2B + C2B
 A2 + C2B
2AB + 2C2  A2 + 2C2B
Each step in the mechanism is called an “elementary step”.
The number of reactants in an elementary step is called the
―molecularity‖.
In this example each step (1-3) is a bimolecular process.
(2 reactants)
A2  2A
is a unimolecular process (1 reactant)
2A + B  is a termolecular process (3 reactants)
Reaction Mechanisms
slow (1)
fast (2)
fast (3)
2AB
A2B2 + C2
A2B + C2
 A2B2
 A2B + C2B
 A2 + C2B
2AB + 2C2  A2 + 2C2B
The slowest step controls the rate of the reaction.
It determines the rate law!
Rate (Ms-1) = k[AB]2
The rate law is not based on the overall reaction:
Rate (Ms-1) = k[AB]2[C2]2
Reaction Mechanisms
2NO2(g) + F2(g)  2FNO2(g)
Rate = k[NO2][F2]
To explain the observed kinetics, a possible
mechanism is proposed:
18
Chapter 15 — Kinetics
Reaction Mechanisms
2NO2(g) + F2(g)  2FNO2(g)
Rate = k[NO2][F2]
Since the 1st step in the reaction is the slow step, it determines
the kinetics and rate law for the reaction:
Reaction Mechanisms
Validating a Reaction Mechanism:
A mechanism is a proposal of how the reaction proceeds
at the molecular level.
1. The individual elementary steps must sum to yield the
overall reaction with correct stoichiometry.
2. The predicted reaction rate law must be in agreement with
the experimentally determined rate law.
3. Note that there may be more than one mechanism that is in
agreement with the reaction stoichiometry and kinetics.