Chapter 12
Chemical Kinetics
2
Chapter 12
• Chemical Kinetics: Rates and
Mechanisms of Chemical
Reactions
Chapter Twelve
Chapter 12
Table of Contents
•
•
•
•
•
•
•
12.1
12.2
12.3
12.4
12.5
12.6
12.7
Reaction Rates
Rate Laws: An Introduction
Determining the Form of the Rate Law
The Integrated Rate Law
Reaction Mechanisms
A Model for Chemical Kinetics
Catalysis
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3
ASSIGNMENTS 3-7-13 AP Chem
4
• Stop here Thursday - 3-7-13
• Work on problem sets from handout AND
the separate page...
• HW: ch. 12.1 to 12.4 #1-13 due Monday.
• HW: Read ch. 12 by Mon.
• HW: Read lab report
• TEST next WED. ch. 12
• TURN in last week’s lab report.
• Yesterday’s lab report due next Thursday.
• Kinetics LAB tomorrow - switch rooms
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
5
Chemical Kinetics: A Preview
• Chemical kinetics is the study of:
– the rates of chemical reactions
– factors that affect these rates
– the mechanisms by which reactions occur
• Reaction rates vary greatly – some are very fast
(burning, precipitation) and some are very slow
(rusting, disintegration of a plastic bottle in
sunlight).
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
6
Variables in Reaction Rates
• Concentrations of reactants: Reaction rates
generally increase as the concentrations of
the reactants are increased.
• Temperature: Reaction rates generally
increase rapidly as the temperature is
increased.
• Surface area: For reactions that occur on a
surface rather than in solution, the rate
increases as the surface area is increased.
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
7
The Meaning of Rate
• The rate of a reaction is the change in
concentration of a product per unit of time
(rate of formation of product).
• Rate is also viewed as the negative of the
change in concentration of a reactant per
unit of time (rate of disappearance of
rate of disappearance of reactant
reactant).
or of formation of product
General
raterate
of reaction
=
• The
of reaction
often has the units of
stoichiometric coefficient of that reactant
–1 s–1 or
moles per liter per orunit
time
Lequation
product
in the(mol
balanced
M s–1)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
Section 12.1
Reaction Rates
Reaction Rate
• Change in concentration of a reactant or product
per unit time.
Rate
•
[A] means concentration of A in mol/L; A is the
reactant or product being considered.
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9
If the rate of consumption of H2O2 is 4.6 M/h, then …
… the rate of formation of H2O must also be 4.6 M/h, and …
… the rate of formation of O2 is 2.3 M/h
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Chapter Thirteen
10
2 H2O2  2 H2O + O2
1L
2.960 g O2 (0.09250
mole) produced in 60
s means …
Prentice Hall © 2005
… 0.1850 mol H2O2
reacted in 60 s.
0.1850 mol H2O2/L
Rate =
60 s
= 0.00131 M H2O2 s–1
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
Section 12.1
Reaction Rates
The Decomposition of Nitrogen Dioxide
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Section 12.1
Reaction Rates
The Decomposition of Nitrogen Dioxide
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12
Section 12.1
Reaction Rates
Instantaneous Rate
• Value of the rate at a particular time.
• Can be obtained by computing the slope of a
line tangent to the curve at that point.
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13
14
Average vs. Instantaneous Rate
Instantaneous rate is
the slope of the
tangent to the curve at
a particular time.
We often are interested
in the initial instantaneous rate; for the initial
concentrations of
reactants and products
are known at this time.
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Chapter Thirteen
Section 12.3
Atomic
Rate
Laws:
Masses
An Introduction
Rate Law
• Shows how the rate depends on the
concentrations of reactants.
• For the decomposition of nitrogen dioxide:
•
2NO2(g) → 2NO(g) + O2(g)
•
Rate = k[NO2]n:
 k = rate constant
 n = order of the reactant
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15
Section 12.3
Atomic
Rate
Laws:
Masses
An Introduction
Rate Law
• Rate = k[NO2]n
• The concentrations of the products do not
appear in the rate law because the reaction rate
is being studied under conditions where the
reverse reaction does not contribute to the
overall rate.
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Section 12.3
Atomic
Rate
Laws:
Masses
An Introduction
Rate Law
• Rate = k[NO2]n
• The value of the exponent n must be
determined by experiment; it cannot be written
from the balanced equation.
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17
Section 12.3
Atomic
Rate
Laws:
Masses
An Introduction
Types of Rate Laws
• Differential Rate Law (rate law) – shows how
the rate of a reaction depends on
concentrations.
• Integrated Rate Law – shows how the
concentrations of species in the reaction
depend on time.
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Section 12.3
Atomic
Rate
Laws:
Masses
An Introduction
Rate Laws: A Summary
• Because we typically consider reactions only
under conditions where the reverse reaction is
unimportant, our rate laws will involve only
concentrations of reactants.
• Because the differential and integrated rate
laws for a given reaction are related in a well–
defined way, the experimental determination of
either of the rate laws is sufficient.
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19
Section 12.3
Atomic
Rate
Laws:
Masses
An Introduction
Rate Laws: A Summary
• Experimental convenience usually dictates
which type of rate law is determined
experimentally.
• Knowing the rate law for a reaction is important
mainly because we can usually infer the
individual steps involved in the reaction from
the specific form of the rate law.
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Section 12.3
Atomic
Rate
Laws:
Masses
An Introduction
Videoclip Summary Kinetic #1 - 7 minutes
• Click below to watch videoclip.
• Link for Lesson 12.1-12.2 7 minute video
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21
Section 12.3
The Mole
Determining
the Form of the Rate Law
• Determine experimentally the power to which
each reactant concentration must be raised in
the rate law.
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Section 12.3
The Mole
Determining
the Form of the Rate Law
Method of Initial Rates
• The value of the initial rate is determined for
each experiment at the same value of t as close
to t = 0 as possible.
• Several experiments are carried out using
different initial concentrations of each of the
reactants, and the initial rate is determined for
each run.
• The results are then compared to see how the
initial rate depends on the initial concentrations
of each of the reactants.
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Section 12.3
The Mole
Determining
the Form of the Rate Law
Overall Reaction Order
• The sum of the exponents in the reaction rate
equation.
•
Rate = k[A]n[B]m
•
Overall reaction order = n + m
•
k = rate constant
[A] = concentration of reactant A
[B] = concentration of reactant B
•
•
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25
The Rate Law of a Chemical Reaction
• The rate law for a chemical reaction relates
the rate of reaction to the concentrations of
aAreactants.
+ bB + cC … products
rate = k[A]n[B]m[C]p …
• The exponents (m, n, p…) are determined by experiment.
• Exponents are not derived from the coefficients in the
balanced chemical equation, though in some instances the
exponents and the coefficients may be the same.
• The value of an exponent in a rate law is the order of the
reaction with respect to the reactant in question.
• The proportionality constant, k, is the rate constant.
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Chapter Thirteen
26
The Rate Law
Rate = k[A]1 = k[A]
Reaction is first order in A
Rate = k[A]2
Reaction is second order in A
Rate = k[A]3
Reaction is third order in A
If we triple the
concentration of A in a
second-order reaction,
the rate increases by a
9
factor of ________.
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Chapter Thirteen
27
More About the Rate Constant k
• The rate of a reaction is the change in
concentration with time, whereas the rate
constant is the proportionality constant
relating reaction rate to the concentrations
of reactants.
• The rate constant remains constant
throughout a reaction, regardless of the
initial concentrations of the reactants.
• The rate and the rate constant have the
same numerical values and units only in
Prentice Hall © 2005
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Chapter Thirteen
28
Method of Initial Rates
• The method of initial rates is a method of
establishing the rate law for a reaction—
finding the values of the exponents in the rate
law, and the value of k.
• A series of experiments is performed in which
the initial concentration of one reactant is
varied. Concentrations of the other reactants
are held constant.
• When we double the concentration of a
reactant A, if:
Chapter Thirteen
– there is no effect on the rate, the reaction is zero-
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
29
The concentration of
NO was held the same
in Experiments 1 and
2…
… while the
concentration of Cl2
in Experiment 2 is
twice that of
Experiment 1.
The rate in Experiment 2 is
twice that in Experiment 1,
so the reaction must be first
order in Cl2.
Which two experiments are
used to find the order of the
reaction in NO?
Prentice Hall © 2005
How do we find the value of k
after obtaining the order of the
reaction in NO and in Cl2?
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
Section 12.3
Atomic
Rate
Laws:
Masses
An Introduction
Videoclip Finding Rate Laws from data - 3.5 minutes
• Click below to watch videoclip.
• Link for Lesson 12.3 Rate Law 3.5 minutes video
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Section 12.3
Atomic
Rate
Laws:
Masses
An Introduction
Problems
• Day 1 - Stop here and work practice problems assigned.
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Section 12.4
The Integrated Rate Law
Half-Life of Reactions
Click above when hand appears
to see visual. Be in play mode.
Use my log-in as needed to
access.
dvann@bryantschools.org
“Bryant123”
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Section 12.4
The Integrated Rate Law
First-Order
• Rate = k[A]
• Integrated:
•
ln[A] = –kt + ln[A]o
•
•
•
•
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of A
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34
First-Order Reactions
• In a first-order reaction, the exponent in
the rate law is 1.
• The integrated rate law describes the concentration of a
1 = k[A]
• reactant
Rate =ask[A]
a function of time. For a first-order process:
ln
[A]t
[A]0
= –kt
ln [A]t – ln [A]0 = –kt
ln [A]t = –kt + ln [A]0
Look! It’s an
equation for a
straight line!
y = mx+b
• At times, it is convenient to replace molarities in an
integrated rate law by quantities that are proportional to
concentration.
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General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
Section 12.4
The Integrated Rate Law
Plot of ln[N2O5] vs Time
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36
Half-life of a Reaction
• The half-life (t½) of a reaction is the time
required for one-half of the reactant
originally present½[A]
to be consumed.
0
ln
= –kt½
• At t½, [A]t = ½[A]
,
and
for a first order
[A]
0 0
reaction:
ln (½) = –kt
½
–0.693 = –kt½
t½ = 0.693/k
• Thus, for a first-order reaction, the half-life is a constant; it
depends only on the rate constant, k, and not on the
concentration of reactant.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
Section 12.4
The Integrated Rate Law
First-Order
• Half–Life:
•
•
•
k = rate constant
• Half–life does not depend on the
concentration of reactants.
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Section 12.4
The Integrated Rate Law
Exercise
•
A first order reaction is 35% complete at the
end of 55 minutes. What is the value of k?
•
k = 7.8 x 10–3 min–1
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39
Second-Order Reactions
• A reaction that is second order in a reactant
has a rate law in which the exponent for that
What do we plot
reactant is 2.
vs. time to get a
straight line?
21
1
• Rate = k[A]––––
= kt + ––––
[A]t
[A]0
• The integrated rate law has the form:
• The half-life of a second-order reaction depends on the initial
concentration as well as on the rate constant k:
1
t½ = –––––
k[A]0
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
Section 12.4
The Integrated Rate Law
Second-Order
• Rate = k[A]2
• Integrated:
•
•
•
•
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of A
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40
Section 12.4
The Integrated Rate Law
Plot of ln[C4H6] vs Time and Plot of 1/[C4H6] vs Time
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41
Section 12.4
The Integrated Rate Law
Second-Order
• Half–Life:
•
•
k = rate constant
[A]o = initial concentration of A
• Half–life gets longer as the reaction progresses
and the concentration of reactants decrease.
• Each successive half–life is double the
preceding one.
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Section 12.4
The Integrated Rate Law
Exercise
•
For a reaction aA  Products,
[A]0 = 5.0 M, and the first two half-lives are 25
and 50 minutes, respectively.
a) Write the rate law for this reaction.
–
? rate = k[A]2
– b)
Calculate k.
–
k = 8.0 x 10-3 M–1min–1
c) Calculate [A] at t = 525 minutes.
–
[A] = 0.23 M
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Section 12.4
The Integrated Rate Law
Zero-Order
• Rate = k[A]0 = k
• Integrated:
•
[A] = –kt + [A]o
•
•
•
•
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of A
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45
A Zero-Order Reaction
rate = k[A]0
=k
Rate is independent of
initial concentration
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Chapter Thirteen
Section 12.4
The Integrated Rate Law
Plot of [A] vs Time
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Section 12.4
The Integrated Rate Law
Zero-Order
• Half–Life:
•
•
k = rate constant
[A]o = initial concentration of A
• Half–life gets shorter as the reaction
progresses and the concentration of reactants
decrease.
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Section 12.4
The Integrated Rate Law
Concept Check
•How can you tell the difference among 0th, 1st,
and 2nd order rate laws from their graphs?
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Section 12.4
The Integrated Rate Law
Rate Laws
Click below to watch visual.
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Summary of Kinetic Data
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50
Chapter Thirteen
ASSIGNMENTS 3-7-13 AP Chem
51
• Stop here Thursday - 3-7-13
• Work on problem sets from handout AND
the separate page...
• HW: ch. 12.1 to 12.4 #1-13 due Monday.
• HW: Read ch. 12 by Mon.
• HW: Read lab report
• TEST next WED. ch. 12
• TURN in last week’s lab report.
• Yesterday’s lab report due next Thursday.
• Kinetics LAB tomorrow - switch rooms
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
ASSIGNMENTS 3-11-13 AP Chem
52
• HW ch. 12 handout section 12.1-12.4 due check answers; make corrections with
WORK shown to get credit. HW: ch. 12.1
to 12.4 #1-13 due today.
• Notes 12.4-12.8
• TEST ch. 12 WEDNESDAY.
• CW: Quiz ch. 12 (12 problems - 20 min.)
• If you haven’t read ch. 12, you need to
along with packet handout.
• HW: Lab report due FRIDAY. The other
Prentice
Hall ©report
2005
lab
not formal lab report from EOC Chapter Thirteen
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Section 12.4
The Integrated Rate Law
Exercise
•
A first order reaction is 35% complete at the
end of 55 minutes. What is the value of k?
•
k = 7.8 x 10–3 min–1
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53
Section 12.4
The Integrated Rate Law
Exercise
•
Consider the reaction aA  Products.
[A]0 = 5.0 M and k = 1.0 x 10–2 (assume the units
are appropriate for each case). Calculate [A] after
30.0 seconds have passed, assuming the
reaction is:
a) Zero order
b) First order
c) Second order
4.7 M
3.7 M
2.0 M
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54
Section 12.5
Reaction Mechanisms
Reaction Mechanism
• Most chemical reactions occur by a series
of elementary steps.
• An intermediate is formed in one step and
used up in a subsequent step and thus is
never seen as a product in the overall
balanced reaction.
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56
Reaction Mechanisms
• Analogy: a banana split is made by steps in
sequence: slice banana; three scoops ice
cream; chocolate sauce; strawberries;
pineapple; whipped cream; end with cherry.
• A chemical reaction occurs according to a
reaction mechanism—a series of
collisions or dissociations—that lead from
initial reactants to the final products.
• An elementary reaction represents, at the
level, a single step in the
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Chapter Thirteen
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
57
Molecularity
• The molecularity of an elementary
reaction refers to the number of free
atoms, ions, or molecules that collide or
dissociate in that step.
Termolecular processes
are unusual, for the same
reason that three
basketballs shot at the
same time are unlikely to
collide at the same
instant …
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Chapter Thirteen
Section 12.5
Reaction Mechanisms
A Molecular Representation of the Elementary Steps in the
Reaction of NO2 and CO
• NO2(g) + CO(g) → NO(g) + CO2(g)
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58
Section 12.5
Reaction Mechanisms
Elementary Steps (Molecularity)
• Unimolecular – reaction involving one
molecule; first order.
• Bimolecular – reaction involving the
collision of two species; second order.
• Termolecular – reaction involving the
collision of three species; third order.
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60
The Rate-Determining Step
• The rate-determining step is the crucial step in
establishing the rate of the overall reaction. It is
usually the slowest step. A reaction is only as fast
as its slowest step.
Fast
• Some two-step
mechanisms
have
a
slow
first
step
Mechanism for
followed
by a fast second step, while others have
2 NO + O2  2 NO2
a fast reversible first step followed by a slow Slow
second step.
The rate-determining step (slowest step) determines the rate law and the
molecularity of the overall reaction.
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Chapter Thirteen
Section 12.5
Reaction Mechanisms
Reaction Mechanism Requirements
• The sum of the elementary steps must
give the overall balanced equation for the
reaction.
• The mechanism must agree with the
experimentally determined rate law.
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62
Initial amount
Decomposition of
N2O5 at 67 °C
After one half-life, half
the N2O5 has reacted.
After two half-lives, half of
the remaining N2O5 has
reacted—three-fourths has
been consumed.
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Chapter Thirteen
Section 12.5
Reaction Mechanisms
Decomposition of N2O5
Click below to watch visual.
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63
Section 12.5
Reaction Mechanisms
Decomposition of N2O5
•
2N2O5(g)  4NO2(g) + O2(g)
•
) 3
Step2(1: N2O5
NO2 + NO
(fast)
Step 2: NO2 + NO3 → NO + O2 + NO2
(slow)
Step 3: NO3 + NO → 2NO2
(fast)
•
•
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Section 12.5
Reaction Mechanisms
Concept Check
•The reaction A + 2B  C has the following
proposed mechanism:
–
A+B
D
(fast
equilibrium)
–
D+BC
(slow)
•Write the rate law for this mechanism.
•
rate = k[A][B]2
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66
Theories of Chemical Kinetics:
Collision Theory
• Before atoms, molecules, or ions can react,
they must first collide.
• An effective collision between two
molecules puts enough energy into key
bonds to break them.
• The activation energy (Ea) is the minimum
energy that must be supplied by collisions
for a reaction to occur.
certain fraction of all molecules in a Chapter Thirteen
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Hall A
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67
Importance of Orientation
One hydrogen atom can
approach another from
any direction …
Effective collision; the I
atom can bond to the C
atom to form CH3I
… and reaction will still occur; the
spherical symmetry of the atoms means
that orientation does not matter.
Prentice Hall © 2005
Ineffective collision;
orientation is important
in this reaction.
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
Section 12.6
A Model for Chemical Kinetics
Collision Model
• Molecules must collide to react.
• Main Factors:
 Activation energy, Ea
 Temperature
 Molecular orientations
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69
Distribution of Kinetic Energies
At higher temperature
(red), more molecules
have the necessary
activation energy.
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Chapter Thirteen
Section 12.6
A Model for Chemical Kinetics
Activation Energy, Ea
• Energy that must be overcome to produce
a chemical reaction.
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71
Transition State Theory
• The configuration of the atoms of the
colliding species at the time of the collision
is called the transition state.
• The transitory species having this
configuration is called the activated
complex.
• A reaction profile shows potential energy
plotted as a function of a parameter called
Prentice Hall © 2005
Chapter Thirteen
the progress of the reaction.
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Section 12.6
A Model for Chemical Kinetics
Transition States and Activation Energy
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Section 12.6
A Model for Chemical Kinetics
Change in Potential Energy
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74
A Reaction Profile
CO(g) + NO2(g)
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
CO2(g) + NO(g)
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
75
An Analogy for Reaction Profiles
and Activation Energy
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Chapter Thirteen
Section 12.6
A Model for Chemical Kinetics
For Reactants to Form Products
• Collision must involve enough energy to
produce the reaction (must equal or
exceed the activation energy).
• Relative orientation of the reactants must
allow formation of any new bonds
necessary to produce products.
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Section 12.6
A Model for Chemical Kinetics
The Gas Phase Reaction of NO and Cl2
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77
78
Effect of Temperature on
the Rates of Reactions
• In 1889, Svante Arrhenius proposed the
following expression for the effect of
temperature on the rate constant, k:
• k = Ae–Ea/RT
• The constant A, called the frequency factor,
is an expression of collision frequency and
orientation; it represents the number of
collisions per unit time that are capable of
leading to reaction.
Prentice Hall © 2005
–EGeneral
/RTChemistry 4
th
edition, Hill, Petrucci, McCreary, Perry
Chapter Thirteen
Section 12.6
A Model for Chemical Kinetics
Arrhenius Equation
Note the division sign “/”.
•
•
•
•
A =
Ea =
R =
T =
frequency factor
activation energy
gas constant (8.3145 J/K·mol)
temperature (in K)
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Section 12.6
A Model for Chemical Kinetics
Linear Form of Arrhenius Equation
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Section 12.6
A Model for Chemical Kinetics
Linear Form of Arrhenius Equation
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Section 12.6
A Model for Chemical Kinetics
Exercise
•
Chemists commonly use a rule of thumb that
an increase of 10 K in temperature doubles the
rate of a reaction. What must the activation
energy be for this statement to be true for a
temperature increase from 25°C to 35°C?
•
Ea = 53 kJ
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Section 12.7
Catalysis
Catalyst
• A substance that speeds up a reaction
without being consumed itself.
• Provides a new pathway for the reaction
with a lower activation energy.
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Section 12.7
Catalysis
Catalysis Teaching Videoclip & Example
• Optional
• Click below. Must be logged in cengage to watch.
• http://college.cengage.com/chemistry/discipline/thinkwell
/2944.html
• Good example to show Elephant Toothpaste experiment
demonstration.
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Section 12.7
Catalysis
Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a
Given Reaction
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Section 12.7
Catalysis
Effect of a Catalyst on the Number of Reaction-Producing
Collisions
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Section 12.7
Catalysis
Heterogeneous Catalyst
• Most often involves gaseous reactants
being adsorbed on the surface of a solid
catalyst.
• Adsorption – collection of one substance
on the surface of another substance.
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Section 12.7
Catalysis
Heterogeneous Catalysis
Click picture below to watch animation.
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Section 12.7
Catalysis
Heterogeneous Catalyst
1. Adsorption and activation of the reactants.
2. Migration of the adsorbed reactants on the
surface.
3. Reaction of the adsorbed substances.
4. Escape, or desorption, of the products.
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Section 12.7
Catalysis
Homogeneous Catalyst
• Exists in the same phase as the reacting
molecules.
• Enzymes are nature’s catalysts.
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Section 12.7
Catalysis
Homogeneous Catalysis - Click below for visual.
Requires log-in
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