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1 Chapter 15 — Kinetics
Jeffrey Mack
California State University,
Sacramento
Chapter 15 Chemical Kinetics: The Rates of Chemical Reactions
Chemical Kinetics will now provide information about
the arrow!
This gives us information on HOW a reaction occurs!
Reactants Products
Chemical Kinetics: The Rates of Chemical Reactions
Thermodynamics – does a reaction take place?
Kinetics – how fast does a reaction proceed?
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.
Chemical Kinetics
• 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.
Rate of Reactions
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.
Dye C
on
c
Time
Determining a Reaction Rate
A B
aA + bB cC + dD
Rate of reaction
In general for the reaction:
t
1
c
C
reactants go away with time therefore the negative sign…
A1
a t
B
t
1
b
t
1
d
D
[ ] [ ]product reactant change in concentrationRate
change in time time time
D D= = = -
D D
Reaction Rate & Stoichiometry
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
The rate of appearance or disappearance is
measured in units of concentration vs. time.
Rate =
molesL
time
There are three ―types‖ of rates
1.initial rate
2.average rate
3.instantaneous rate
= Ms1 or Mmin1 etc...
Determining a Reaction Rate
Rea
cta
nt
co
nce
ntr
ation
(M
)
The concentration of a reactant
decreases with time.
Time
Reaction Rates
Rea
cta
nt
co
nce
ntr
ation
(M
)
Time
Initial rate
Reaction Rates
Rea
cta
nt
co
nce
ntr
ation
(M
)
Time
Y [concentration]Average Rate =
X time
Reaction Rates
Rea
cta
nt
co
nce
ntr
ation
(M
)
Time
Instantaneous rate (tangent line)
Reaction Rates
3 Chapter 15 — Kinetics
Rea
cta
nt
co
nce
ntr
ation
(M
)
Time
Instantaneous rate (tangent line)
Initial rate
Reaction Rates
During the beginning stages of the
reaction, the initial rate is very close to
the instantaneous rate of reaction.
Problem: Consider the 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?
2 22NO (g) O (g) 2NO (g)
Problem: Consider the 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?
2 22NO (g) O (g) 2NO (g)
= 1.50104 Ms1
recall…
1 MMs
s
(0.0100M 0.0250M)
100.0 s 50.0 s
1 mol of reaction
2 moles NO (g)
RATE= - [NO]
t 1
2
Practice example
Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
rate = - [CH4]
t = -
[O2]
t 1
2 =
[H2O]
t 1
2 =
[CO2]
t
• 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
Bleach at 54 ˚C Bleach at 22 ˚C
Reaction Conditions & Rate: Temperature
4 Chapter 15 — Kinetics
Reaction Conditions & Rate: Physical State of Reactants
Catalyzed decomposition of H2O2
2 H2O2 2 H2O + O2
Reaction Conditions & Rate: Catalysts
0.3 M HCl 6 M HCl
Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g)
Effect of Concentration on Reaction Rate: Concentration
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.
The Rate Law Expression
aA + bB cC + dD
x and y are the reactant orders determined
from experiment.
x and y are NOT the stoichiometric
coefficients.
x yRate k [A] [B]
Each concentration is expressed with an order (exponent).
The rate constant converts the concentration expression into
the correct units of rate (Ms1).
For the general reaction:
Reaction Order
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]
2 second rate = k[A]2
0.5 one-half rate = k[A]1/2
1.5 three-half rate = k[A]3/2
0.667 two-thirds rate = k[A]2/3
Reaction Orders
5 Chapter 15 — Kinetics
Overall order:
or sevenhalves order
note: when the order of a reaction is 1 (first
order) no exponent is written.
11 22Rate (Ms ) k[A][B] [C]- =
1 + ½ + 2 = 3.5 = 7/2
Reaction Order
To find the units of the rate constant, divide the rate units by
the Molarity raised to the power of the overall reaction order.
1 x yRate (Ms ) k [A] [B]
1Ms x yk M M
x yk M
1Ms
k =
1Ms
x yM
1 x y 1M s
if (x+y) = 1 k has units of s-1
if (x+y) = 2 k has units of M-1s-1
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.
Reaction Constant 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)
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] 3
b) What is the overall order of the reaction?
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.
Rate(Ms-1) = k [H2] [NO] 3
b) What is the overall order of the reaction?
= 4 Overall order = 1 + 3 “4th order”
c) What are the units of the rate constant?
6 Chapter 15 — Kinetics
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.
Rate(Ms-1) = k [H2] [NO] 3
b) What is the overall order of the reaction?
= 4 Overall order = 1 + 3 “4th order”
c) What are the units of the rate constant?
MRate
s=
3k M M= ´ ´ 4k M= ´4
Mk
s M=
´
3 1M s- -=
x yRate = k [A] [B]
If the rate doubles when [A] doubles and [B] stays
constant, the order for [A] is?
Rate Law Expression
x yRate = k [A] [B]
If the rate doubles when [A] doubles and [B] stays
constant, the order for [A] is?
one… 1
Rate Law Expression
x yRate = k [A] [B]
If the rate doubles when [A] doubles and [B] stays
constant, the order for [A] is?
one… 1
(2) (1)
xx y(2) (1) x
x y
(1) (1)
x
Rate 2 Rate
Rate 2 Rate [2A] [B] 2A2 2
Rate Rate [A] [B] A
log(2) log(2) x log(2)
log(2)x 1
log(2)
= ´
´ é ù= = = = =
ê úë û
= = ´
= =
Rate Law Expression
Instantaneous rate (tangent line)
Re
acta
nt co
nce
ntr
atio
n (
M)
During the beginning stages of the
reaction, the initial rate is very close
to the instantaneous rate of reaction.
Initial rate
Determining a Rate Equation
The reaction of nitric oxide with hydrogen at 1280 °C
is as follows:
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
Determining Reaction Order: The Method of Initial Rates
7 Chapter 15 — Kinetics
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
The reaction of nitric oxide with hydrogen at 1280 °C
is as follows:
Notice that in Trial 1 and 3, the initial concentration of
NO is held constant while H2 is changed.
Determining Reaction Order: The Method of Initial Rates
2NO (g) + 2H2 (g) N2 (g) + 2H2O (g)
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
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)
This means that any changes to the rate must be due to the
changes in H2 which is related to the concentration of H2 & its
order!
Rate(M/min) = k [NO]x [H2]y
The rate law for the reaction is given by:
2NO(g) + 2H2(g) N2(g) + 2H2O(g)
Taking the ratio of the rates of Trials 3 and 1 one finds:
0.0120 M/min
0.00600 M/min
x y
x y
k [0.0100] [0.0200]
k [0.0100] [0.0100]
y
y
[0.0200]
[0.0100]
Plugging in the values from the data:
Rate (Trial 3)
Rate (Trial 1) x y
(1) 2 (1)k [NO] [H ]
x y
(3) 2 (3)k [NO] [H ]
=
Determining Reaction Order: The Method of Initial Rates
y0.0200
0.0100
é ù= ê ú
ë û
1
1
0.0120M min
0.00600M min
-
-=
y
y
[0.0200]
[0.0100]
y2.00é ùë û= 2.00
[ ]y
2.00= 2.00 log
Take the log of both sides of the equation:
log(2.00) [ ]y
log 2.00=
log(2.00) y log(2.00)= ´
y = 1 log(2.00)
ylog(2.00)
= Rate(M/min) = k [NO]x[H2]
Determining Reaction Order: The Method of Initial Rates
Rate(2)
Rate(1)
0.144
0.00600
x
x
k [0.0200] [0.0300]
k [0.0100] [0.0100]
24 x
2 3
x
2 8
xlog 2 log 8 x = 3
Similarly for x: Rate(M/min) = k [NO]x[H2]y
k x
(2)[NO] y
2 (1)[H ]
k x
(1)[NO] y
2 (2)[H ]
Determining Reaction Order: The Method of Initial Rates
The Rate Law expression is:
3
2Rate k[NO] [H ] =
The order for NO is 3
The order for H2 is 1
The over all order is 3 + 1 =4
2NO(g) + 2H2(g) N2(g) + 2H2O(g)
Determining Reaction Order: The Method of Initial Rates
8 Chapter 15 — Kinetics
The Rate constant
Rate(M/min) = k [NO]3[H2]
To find the rate constant, choose one set of data and solve:
310.0120 Mmin k 0.0100 M 0.0200 M
1
3
0.0120 Mmink
0.0100 M 0.0200 M
1
3 4
0.0120 Mmin
0.0100 0.0200 M
5 3 1k 6.00 10 M min
Determining Reaction Order: The Method of Initial Rates
• 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.
Concentration–Time Relationships: Integrated Rate Laws
For a zero order process where ―A‖ goes onto
products, the rate law can be written:
A products
1Rate(Ms ) [A]
t
0k[A] = k
k has units of Ms1
For a zero order process, the rate is the rate
constant!
Integrated Rate Laws
This is the ―average rate‖
If one considers the infinitesimal changes in
concentration and time the rate law equation
becomes: 1Rate(Ms )- =
This is the ―instantaneous rate‖
d[A]k
dt- =
Zero order kinetics A products
Integrated Rate Laws
1Rate(Ms ) [A]
t
0k[A] = k
t
o
[A]
[A]d[A]ò
t
0k dt= - ò
d[A]k
dt- =
[A]t [A]o = k(t 0) = kt
and [A] = [A] at time t = t where [A] = [A]o at time t = 0
[A]t [A]o = kt
Zero order kinetics
Integrated Rate Laws
[A]o [A]t = kt
what’s this look like?
y = mx + b
[A]t = kt + [A]o
rearranging…
a plot of [A]t vs t looks like…
[A] t
(mo
ls/L
)
t (time)
the y-intercept is [A]o
Conclusion: If a plot of reactant
concentration vs. time yields
a straight line, then the reactant
order is ZERO!
slope = k
k has units of M×(time)1
Zero order kinetics
Integrated Rate Laws
9 Chapter 15 — Kinetics
For a first order process,
the rate law can be written: A products
1 [A]Rate(Ms ) k[A]
t
This is the ―average rate‖
If one considers the infinitesimal changes in
concentration and time the rate law equation becomes:
1 d[A]Rate(Ms ) k[A]
dt
This is the ―instantaneous rate‖
Integrated Rate Laws
t
o
[A] t0[A]
d[A]k dt
[A] t
o
[A]ln kt
[A]
ktt
o
[A]e
[A]
Taking the exponent to each side of the equation:
or kt
t o[A] [A] e
Conclusion: The concentration of a reactant governed by
first order kinetics falls off from an initial concentration
exponentially with time.
First order kinetics
Integrated Rate Laws
kt
t oln [A] [A] e
tln[A]
Taking the natural log of both sides…
k t
oln [A] e kt
oln[A] ln e
ln[A]t =
rearranging…
= ln[A]o kt
kt + ln[A]o
First order kinetics
Integrated Rate Laws
y = mx + b so a plot of ln[A]t vs t looks like…
ln[A
] t
t (time)
the y-intercept is ln[A]o
Conclusion: If a plot of natural
log of reactant concentration vs.
time yields a straight line, then
the reactant order is FIRST!
slope = k
ln[A]t = kt + ln[A]o
k has units of (time)1
First order kinetics
Integrated Rate Laws
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 s1?
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 s1?
Begin with the integrated rate law for a 1st order process:
ktt
o
[A]e
[A]
kt
t o[A] [A] e
2 5 t
2 5 o
[N O ]
[N O ]kte-=
Wait… what is the volume of the container???
Do we need to convert to moles?
10 Chapter 15 — Kinetics
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 s1?
2 5 t
2 5 o
[N O ]
[N O ]kte-=
Check it out! You don’t need the
volume of the container!
3
2 5 t
2 5 o3
1 g 1 mol2.50 mg
108.0 g10 mgN O 2.50 mg
1 g 1 molN O 2.56 mg2.56 mg
108.0 g10 mg
L
L
x
x
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 s1?
taking the natural log and substituting time in
seconds:
2.50 mg
ln k 256 s2.56 mg
k = 9.3 105 s1
2 5 ktt
2 5 o
N O 2.50 me
N O 2.56 m
Rate = k[A]2
21d A
Rate(Ms ) k Adt
Integrating as before we find:
t 0
1 1kt
A A
A Products
k has units of M1s1
Second order kinetics
Integrated Rate Laws
y = mx + b so a plot of 1/[A]t vs t looks like…
1/[A
] t
t (time)
the y-intercept is 1/[A]o
Conclusion: If a plot of one over
reactant concentration vs. time
yields a straight line, then the
reactant order is second!
slope = k
Second order kinetics
t 0
1 1kt
A A
k has units of M1s1
Integrated Rate Laws
Summary of Integrated Rate Laws
Half-life of a
reaction is the
time taken for the
concentration of a
reactant to drop
to one-half of the
original value.
Half-life & First-Order Reactions
11 Chapter 15 — Kinetics
0t
[A][A]
2
For a first order process the half life (t½ ) is found
mathematically from:
t 0
(1) ln A kt ln A
t 0
(2) ln A ln A kt
t
0
A(3) ln kt
[ ]A
Start with the integrated
rate law expression for a 1st
order process
Bring the concentration
terms to one side.
Express the concentration
terms as a fraction using the
rules of ln.
at time = t½
Reaction Half-Life: 1st Order Kinetics
0
A(4) ln kt
[A]
0
10 2
A(5) ln kt
[A]2
1
2
ln2 0.693t
k k= =
exchange [A] with [A]0 to reverse
the sign of the ln term and cancel
the negative sign in front of k
Substitute the value of [A] at the
half-life
Reaction Half-Life: 1st Order Kinetics
12
ln 2 kt
12
ln2 0.693t
k k
So, knowing the rate constant for a first order
process, one can find the half-life!
The half-life is
independent of the
initial concentration!
Reaction Half-Life: 1st Order Kinetics
12
0
0
2 Aln kt
A
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?
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?
Using the integrated rate law, substituting in the value of k and
900.s we find:
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?
Using the integrated rate law, substituting in the value of k and
900.s we find:
k = 0.00385 s-1 12
ln2t
k 1
12
ln2 ln2k
t 180.s
12 Chapter 15 — Kinetics
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?
Using the integrated rate law, substituting in the value of k and
900.s we find:
k = 0.00385 s-1 12
ln2t
k 1
12
ln2 ln2k
t 180.s
ktt
0
[A]
[A]e
10.00385 900.st
0
[A]e
[A]
s = 0.0312
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?
Using the integrated rate law, substituting in the value of k and
900.s we find:
k = 0.00385 s-1 12
ln2t
k 1
12
ln2 ln2k
t 180.s
ktt
0
[A]
[A]e
10.00385 900.st
0
[A]e
[A]
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%
Elements that decay via radioactive processes
do so according to 1st order kinetics:
Element: Half-life:
238U 234Th +
14C 14N +
131I 131Xe +
4.5 109 years
5730 years
8.05 days
Reaction Half-Life: 1st Order Kinetics
Tritium decays to helium by beta () decay:
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. ktt
0
[A]e
[A]
3 0 3
1 -1 2H Hee
Reaction Half-Life: 1st Order Kinetics
Recall that that the rate constant for a 1st order
process is given by:
12
ln(2)k
t
ktt
0
[A]e
[A]
kt[A] [A] et 0
[3H]t = 1.50 mg
0.693 49.2 years
12.3e years
= 0.094 mg
Reaction Half-Life: 1st Order Kinetics
Notice that 49.2 years is 4 half-lives…
49.24
12.3
After 1 half life: 1.50 mg
2= 0.75 mg remains
After 2 half life's: 0.75 mg
2= 0.38 mg remains
After 3 half life's: 0.38 mg
2= 0.19 mg remains
After 4 half life's: 0.19 mg
2= 0.094 mg remains
Reaction Half-Life: 1st Order Kinetics
13 Chapter 15 — Kinetics
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.
A Microscopic View of Reaction Rates
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 progress of a chemical reaction as the reactants
transform to products can be described graphically by a
Reaction Coordinate.
Po
ten
tia
l E
ne
rgy
Reaction Progress
reactants
products
HRXN
Eact
In order for the reaction
to proceed, the reactants
must posses enough
energy to surmount a
reaction barrier.
Transition State
Activation Energy
The temperature for a
system of particles is
described by a
distribution of energies.
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.
Activation Energy
Molecules need a minimum amount of energy to react.
Visualized as an energy barrier - activation energy, Ea.
Reaction coordinate diagram
Activation Energy
14 Chapter 15 — Kinetics
Orientation factors into the equation
The orientation of a molecule during collision can have a
profound effect on whether or not a reaction occurs.
When the green atom collides with the green atom on the
molecule, a reactive or effective collision occurs.
The reaction
occurs only when
the orientation of
the molecules is
just right…
Activation Energy
Orientation factors into the equation
When the green atom collides with the red atom on the
molecule, this leads to a non-reactive or ineffective
collision occurs.
In some cases, the reactants must have proper
orientation for the collision to yield products.
This reduces the
number of collisions
that are reactive!
Activation Energy
Arhenius discovered that most reaction-rate data
obeyed an equation based on three factors:
(1) The number of collisions per unit time.
(2) The fraction of collisions that occur with the
correct orientation.
(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
aE
RTk Ae-
=
Both A and Ea are specific to a given reaction.
k is the rate constant
Ea is the activation energy R is the ideal-gas constant
(8.314 J/Kmol)
T is the temperature in K
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
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.
Rate (Ms-1) = k[A]x[B]y
aE
RTk Ae-
=
As T increases, the value of the exponential part of
the equation becomes less negative thus increasing
the value of k.
The Arrhenius Equation
Reactions generally occur slower at lower T.
Iodine clock reaction.
H2O2 + 2 I- + 2 H+ 2 H2O + I2
Room temperature
In ice at 0 oC
Effect of Temperature
15 Chapter 15 — Kinetics
Determining the Activation Energy
Ea may be determined experimentally.
First take natural log of both sides of the Arrhenius equation:
aE
RTk Ae-
=ln aE
lnk lnRT
A
ln k
1A plot of ln k vs will have a slope
T
of and a y-intercept of ln A. aE
R
1
T
y = mx + b
The Arrhenius Equation
2 1lnk lnk-
One can determine the activation energy of a reaction by
measuring the rate constant at two temperatures:
Writing the Arrhenius equation for each temperature:
Subtracting k1 from k2 we find that:
2
1
kln
k
æ ö= ç ÷
è ø
a
1
Eln A
RT
æ ö- - +ç ÷
è ø
a
2
Eln A
RT
æ ö= - +ç ÷
è ø
Determining the Activation Energy
1lnk = a
1
E
RT- lnA+
2lnk = a
2
E
RT- lnA+
The Arrhenius Equation
2
1
kln
k
æ ö=ç ÷
è ø
Knowing the rate constants at
two temps yields the activation
energy.
or
Knowing the Ea and the rate
constant at one temp allows one
to find k(T2)
aE
R 1 2
1 1
T T
æ ö´ -ç ÷
è ø
Determining the Activation Energy
2
1
kln
k
æ öç ÷è ø
a
1
Eln A
RT
æ ö- - +ç ÷
è ø
a
2
Eln A
RT
æ ö= - +ç ÷
è ø
The Arrhenius Equation 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?
2 1k 2k=
2
1
kln
k
æ ö=ç ÷
è ø
1
1
2kln
k
æ öç ÷è ø
ln(2)= a
1 2
E 1 1
R T T
æ ö= -ç ÷
è ø
aE
R= 36.04 10 K= ´
(1)
(2)
(3) 50.2 kJ/mol 310 J
1kJ´
J8.314
mol K×
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?
ln(2) 0.693= 36.04 10 K= ´ ´2
1 1
298K T
æ ö-ç ÷
è ø
3 1
2
13.24 10 K
T
- -= ´
(4)
(5)
T2 = 308 K A 10 °C change of
temperature doubles
the rate!!
algebra!
Catalysts speed up reactions by altering the
mechanism to lower the activation energy barrier.
Dr. James Cusumano, Catalytica Inc.
What is a catalyst?
Catalysts and society
Catalysts and the environment
Catalysis
16 Chapter 15 — Kinetics
In auto exhaust systems — Pt, NiO
2 CO + O2 2 CO2
2 NO N2 + O2
Catalysis
2. Polymers: H2C=CH2 polyethylene
3. Acetic acid: CH3OH + CO CH3CO2H
4. Enzymes — biological catalysts
Catalysis
Catalysis and activation energy
Uncatalyzed reaction
Catalyzed reaction
MnO2 catalyzes
decomposition of H2O2
2 H2O2 2 H2O + O2
Catalysis Iodine-Catalyzed Isomerization of
cis-2-Butene
Iodine-Catalyzed Isomerization of cis-2-Butene
The overall stoichiometry of a chemical reaction is
most often the sum of several steps:
(1) 2AB A2B2
(2) A2B2 + C2 A2B + C2B
(3) A2B + C2 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”.
Reaction Mechanisms
17 Chapter 15 — Kinetics
The overall stoichiometry of a chemical reaction is
most often the sum of several steps:
(1) 2AB A2B2
(2) A2B2 + C2 A2B + C2B
(3) A2B + C2 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
(1) 2AB A2B2
(2) A2B2 + C2 A2B + C2B
(3) A2B + C2 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
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”
slow
fast
fast
(1) 2AB A2B2
(2) A2B2 + C2 A2B + C2B
(3) A2B + C2 A2 + C2B
2AB + 2C2 A2 + 2C2B
Reaction Mechanisms
slow
fast
fast
(1) 2AB A2B2
(2) A2B2 + C2 A2B + C2B
(3) A2B + C2 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
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.
Reaction Mechanisms
2NO2(g) + F2(g) 2FNO2(g)
Rate = k[NO2][F2]
To explain the observed kinetics, a possible
mechanism is proposed:
Reaction Mechanisms
18 Chapter 15 — Kinetics
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.
Reaction Mechanisms