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John C. Kotz
Paul M. Treichel
John Townsend
http://academic.cengage.com/kotz
Chapter 15
Principles of Reactivity: Chemical Kinetics
John C. Kotz • State University of New York, College at Oneonta
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3
Chemical Kinetics
Chapter 15
PLAY MOVIE
H2O2 decomposition in
an insect
© 2009 Brooks/Cole - Cengage
H2O2 decomposition
catalyzed by MnO2
Chemical Kinetics
• We can use thermodynamics to tell if
a reaction is product- or reactantfavored.
• But this gives us no info on HOW FAST
reaction goes from reactants to products.
• KINETICS
— the study of REACTION
RATES and their relation to the way the
reaction proceeds, i.e., its MECHANISM.
• The reaction mechanism is our goal!
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Reaction Mechanisms
The sequence of events at the molecular
level that control the speed and
outcome of a reaction.
Br from biomass burning destroys
stratospheric ozone.
(See R.J. Cicerone, Science, volume 263, page 1243, 1994.)
Step 1:
Br + O3 f BrO + O2
Step 2:
Cl + O3 f ClO + O2
Step 3:
BrO + ClO + light f Br + Cl + O2
NET:
2 O3 f 3 O2
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Reaction Rates
Section 15.1
• Reaction rate = change in
concentration of a reactant or
product with time.
• Three “types” of rates
–initial rate
–average rate
–instantaneous rate
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Determining a Reaction Rate
Dye Conc
PLAY MOVIE
See Chemistry Now , Chapter 15
Time
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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.
7
Determining a Reaction Rate
See Active Figure 15.2
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9
Factors Affecting Rates
• Concentrations
• and physical state of reactants
and products
• Temperature
• Catalysts
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Concentrations & Rates
Section 15.3
Mg(s) + 2 HCl(aq) f MgCl2(aq) + H2(g)
0.3 M HCl
6 M HCl
PLAY MOVIE
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11
Factors Affecting Rates
• Physical state of reactants
PLAY MOVIE
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Factors Affecting Rates
Catalysts: catalyzed decomp of H2O2
2 H2O2 f 2 H2O + O2
PLAY MOVIE
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Catalysts
1. CO2(g) e CO2 (aq)
See Page 702
2. CO2 (aq) + H2O(liq) e H2CO3(aq)
3. H2CO3(aq) e H+(aq) + HCO3–(aq)
• Adding trace of NaOH uses up H+. Equilibrium shifts to produce
more H2CO3.
• Enzyme in blood (above) speeds up reactions 1 and 2
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Factors Affecting Rates
• Temperature
Bleach at 54 ˚C
Bleach at 22 ˚C
PLAY MOVIE
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Iodine Clock Reaction
1. Iodide is oxidized to iodine
H2O2 + 2 I- + 2 H+ f 2 H2O + I2
2.
I2 reduced to I- with vitamin C
I2 + C6H8O6 f C6H6O6 + 2 H+ + 2 I-
When all vitamin C is depleted, the I2
interacts with starch to give a blue
complex.
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Iodine Clock Reaction
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Concentrations and Rates
To postulate a reaction
mechanism, we study
• reaction rate and
• its concentration
dependence
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Concentrations and Rates
Take reaction where Cl- in cisplatin
[Pt(NH3)2Cl3] is replaced by H2O
Rate of change of conc of Pt compd
Am't of cisplatin reacting (mol/L)
=
elapsed time (t)
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Concentrations & Rates
Rate of change of conc of Pt compd
Am't of cisplatin reacting (mol/L)
=
elapsed time (t)
Rate of reaction is proportional to [Pt(NH3)2Cl2]
We express this as a RATE LAW
Rate of reaction = k [Pt(NH3)2Cl2]
where k = rate constant
k is independent of conc. but increases with T
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Concentrations, Rates, &
Rate Laws
In general, for
a A + b B f x X with a catalyst C
Rate = k [A]m[B]n[C]p
The exponents m, n, and p
• are the reaction order
• can be 0, 1, 2 or fractions
• must be determined by experiment!
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Interpreting Rate Laws
Rate = k [A]m[B]n[C]p
• If m = 1, rxn. is 1st order in A
Rate = k [A]1
If [A] doubles, then rate goes up by factor of __
• If m = 2, rxn. is 2nd order in A.
Rate = k [A]2
Doubling [A] increases rate by ________
• If m = 0, rxn. is zero order.
Rate = k [A]0
If [A] doubles, rate ________
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Deriving Rate Laws
Derive rate law and k for
CH3CHO(g) f CH4(g) + CO(g)
from experimental data for rate of disappearance
of CH3CHO
Expt.
[CH3CHO]
(mol/L)
1
0.10
0.020
2
0.20
0.081
3
0.30
0.182
4
0.40
0.318
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Disappear of CH3CHO
(mol/L•sec)
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Deriving Rate Laws
Rate of rxn = k [CH3CHO]2
Here the rate goes up by ______ when initial conc.
doubles. Therefore, we say this reaction is
_________________ order.
Now determine the value of k. Use expt. #3 data—
0.182 mol/L•s = k (0.30 mol/L)2
k = 2.0 (L / mol•s)
Using k you can calc. rate at other values of [CH3CHO]
at same T.
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Concentration/Time
Relations
What is concentration of reactant as function of
time?
Consider FIRST ORDER REACTIONS
The rate law is
Ę[A]
Rate = = k [A]
Ętime
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Concentration/Time
Relations
Integrating - (∆ [A] / ∆ time) = k [A], we get
ln is natural
logarithm
[A]
ln
= - kt
[A]0
[A] at time = 0
[A] / [A]0 =fraction remaining after time t
has elapsed.
Called the integrated first-order rate law.
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Concentration/Time Relations
26
Sucrose decomposes to simpler sugars
Rate of disappearance of sucrose = k [sucrose]
If k = 0.21 hr-1
and [sucrose] = 0.010 M
How long to drop 90%
(to 0.0010 M)?
Glucose
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Concentration/Time Relations
Rate of disappear of sucrose = k [sucrose], k = 0.21 hr-1. If
initial [sucrose] = 0.010 M, how long to drop 90% or to
0.0010 M?
Use the first order integrated rate law
 0.0010 
-1
ln 
=
(0.21
h
)t

 0.010 
ln (0.100) = - 2.3 = - (0.21 hr-1)(time)
time = 11 hours
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Using the Integrated Rate Law
The integrated rate law suggests a way to tell
the order based on experiment.
2 N2O5(g) f 4 NO2(g) + O2(g)
Time (min)
0
1.0
2.0
5.0
[N2O5]0 (M)
1.00
0.705
0.497
0.173
ln [N2O5]0
0
-0.35
-0.70
-1.75
Rate = k [N2O5]
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Using the Integrated Rate Law
2 N2O5(g) f 4 NO2(g) + O2(g) Rate = k [N2O5]
Data of conc. vs.
time plot do not fit
straight line.
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Plot of ln [N2O5] vs.
time is a straight
line!
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Using the Integrated Rate Law
Plot of ln [N2O5] vs. time is a
straight line!
Eqn. for straight line:
y = mx + b
ln [N 2O5] = - kt + ln [N 2O5]o
conc at
time t
rate const
= slope
conc at
time = 0
All 1st order reactions have straight line plot
for ln [A] vs. time.
(2nd order gives straight line for plot of 1/[A]
vs. time)
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Properties of Reactions
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Half-Life
HALF-LIFE is
the time it
takes for 1/2 a
sample is
disappear.
For 1st order
reactions, the
concept of
HALF-LIFE is
especially
useful.
See Active Figure 15.9
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Half-Life
• Reaction is 1st order
decomposition of
H2O2.
© 2009 Brooks/Cole - Cengage
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Half-Life
• Reaction after 1
half-life.
• 1/2 of the reactant
has been
consumed and 1/2
remains.
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Half-Life
• After 2 half-lives
1/4 of the reactant
remains.
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Half-Life
• A 3 half-lives 1/8
of the reactant
remains.
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Half-Life
• After 4 half-lives
1/16 of the
reactant remains.
© 2009 Brooks/Cole - Cengage
Half-Life
Sugar is fermented in a 1st order process (using an
enzyme as a catalyst).
sugar + enzyme f products
Rate of disappear of sugar = k[sugar]
k = 3.3 x 10-4 sec-1
What is the half-life of this reaction?
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Half-Life
Rate = k[sugar] and k = 3.3 x 10-4 sec-1. What is the halflife of this reaction?
Solution
[A] / [A]0 = fraction remaining
when t = t1/2 then fraction remaining = _________
Therefore, ln (1/2) = - k · t1/2
- 0.693 = - k · t1/2
t1/2 = 0.693 / k
So, for sugar,
t1/2 = 0.693 / k = 2100 sec = 35
© 2009 Brooks/Cole - Cengage
min
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Half-Life
Rate = k[sugar] and k = 3.3 x 10-4 sec-1. Half-life
is 35 min. Start with 5.00 g sugar. How much is
left after 2 hr and 20 min (140 min)?
Solution
2 hr and 20 min = 4 half-lives
Half-life Time Elapsed Mass Left
1st
35 min
2.50 g
2nd
70
1.25 g
3rd
105
0.625 g
4th
140
0.313 g
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Half-Life
Radioactive decay is a first order process.
Tritium f electron + helium
3H
0 e
3He
-1
t1/2 = 12.3 years
If you have 1.50 mg of tritium, how much is left
after 49.2 years?
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Half-Life
Start with 1.50 mg of tritium, how much is left after 49.2
years? t1/2 = 12.3 years
Solution
ln [A] / [A]0 = -kt
[A] = ?
[A]0 = 1.50 mg
t = 49.2 y
Need k, so we calc k from:
k = 0.693 / t1/2
Obtain k = 0.0564 y-1
Now ln [A] / [A]0 = -kt = - (0.0564 y-1)(49.2 y)
= - 2.77
Take antilog: [A] / [A]0 = e-2.77 = 0.0627
0.0627 = fraction remaining
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Half-Life
Start with 1.50 mg of tritium, how much is left after 49.2
years? t1/2 = 12.3 years
Solution
[A] / [A]0 = 0.0627
0.0627 is the fraction remaining!
Because [A]0 = 1.50 mg, [A] = 0.094 mg
But notice that 49.2 y = 4.00 half-lives
1.50 mg
f 0.750 mg after 1 half-life
f 0.375 mg after 2
f 0.188 mg after 3
f 0.094 mg after 4
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Half-Lives of Radioactive Elements
Rate of decay of radioactive isotopes given in
terms of 1/2-life.
238U f 234Th + He
4.5 x 109 y
14C f 14N + beta
5730 y
131I f 131Xe + beta
8.05 d
Element 106 - seaborgium - 263Sg
0.9 s
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MECHANISMS
A Microscopic View of Reactions
Mechanism: how reactants are converted to
products at the molecular level.
RATE LAW f
MECHANISM
experiment f theory
PLAY MOVIE
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Activation Energy
Molecules need a minimum amount of energy to react.
Visualized as an energy barrier - activation energy, Ea.
PLAY MOVIE
Reaction coordinate
diagram
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MECHANISMS
& Activation Energy
Conversion of cis to trans-2-butene requires
twisting around the C=C bond.
Rate = k [trans-2-butene]
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MECHANISMS
Cis
Transition state
48
Trans
Activation energy barrier
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MECHANISMS
49
Energy involved in conversion of trans to cis butene
energy
Activated
Complex
+262 kJ
cis
-266 kJ
4 kJ/mol
trans
© 2009 Brooks/Cole - Cengage
Mechanisms
• Reaction passes thru a
TRANSITION STATE
where there is an
activated complex
that has sufficient
energy to become a
product.
ACTIVATION ENERGY, Ea
= energy req’d to form activated complex.
Here Ea = 262 kJ/mol
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MECHANISMS
Also note that trans-butene is MORE
STABLE than cis-butene by about 4 kJ/mol.
Therefore, cis f trans is EXOTHERMIC
This is the connection between thermodynamics and kinetics.
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Effect of Temperature
• Reactions generally
occur slower at
lower T.
In ice at 0 oC
Room temperature
PLAY MOVIE
Iodine clock reaction. See
Chemistry Now, Ch. 15
H2O2 + 2 I- + 2 H+
f 2 H2O + I2
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
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Activation Energy and Temperature
Reactions are faster at higher T because a larger
fraction of reactant molecules have enough energy to
convert to product molecules.
In general,
differences in
activation
energy cause
reactions to vary
from fast to slow.
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Mechanisms
1.
Why is trans-butene e cis-butene reaction
observed to be 1st order?
As [trans] doubles, number of molecules
with enough E also doubles.
2.
Why is the trans e cis reaction faster at
higher temperature?
Fraction of molecules with sufficient
activation energy increases with T.
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More About Activation Energy
Arrhenius equation —
Rate
constant
k  Ae
Frequency factor
Temp (K)
-E a / RT
Activation 8.31 x 10-3 kJ/K•mol
energy
Frequency factor related to frequency of collisions
with correct geometry.
Ea 1
ln k = - ( )( ) + ln A
R T
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Plot ln k vs. 1/T f
straight line.
slope = -Ea/R
More on Mechanisms
A bimolecular reaction
Reaction of
trans-butene f cis-butene is
UNIMOLECULAR - only one
reactant is involved.
BIMOLECULAR — two different
molecules must collide f
products
PLAY MOVIE
Exo- or endothermic?
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Collision Theory
Reactions require
(a) activation energy and
(b) correct geometry.
O3(g) + NO(g) f O2(g) + NO2(g)
1. Activation energy
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
2. Activation energy
and geometry
PLAY MOVIE
57
Mechanisms
O3 + NO reaction occurs in a single ELEMENTARY step.
Most others involve a sequence of elementary steps.
Adding elementary steps gives NET reaction.
PLAY MOVIE
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Mechanisms
Most rxns. involve a sequence of elementary
steps.
2 I- + H2O2 + 2 H+ f I2 + 2 H2O
Rate = k [I-] [H2O2]
NOTE
1.
Rate law comes from experiment
2.
Order and stoichiometric coefficients not
necessarily the same!
3.
Rate law reflects all chemistry down to
and including the slowest step in multistep
reaction.
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Mechanisms
Most rxns. involve a sequence of elementary steps.
2 I- + H 2 O 2 + 2 H + f I2 + 2 H 2 O
Rate = k [I-] [H2O2]
Proposed Mechanism
Step 1 — slow
HOOH + I-
f HOI + OH-
Step 2 — fast
HOI + I-
f I2 + OH-
Step 3 — fast
2 OH- + 2 H+ f 2 H2O
Rate of the reaction controlled by slow step —
RATE DETERMINING STEP, rds.
Rate can be no faster than rds!
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Mechanisms
2 I- + H2O2 + 2 H+ f I2 + 2 H2O
Rate = k [I-] [H2O2]
Step 1 — slow HOOH + I- f HOI + OH-Step
2 — fast
HOI + I- f I2 + OHStep 3 — fast 2 OH- + 2 H+ f 2 H2O
Elementary Step 1 is bimolecular and involves I- and
HOOH. Therefore, this predicts the rate law should be
Rate  [I-] [H2O2] — as observed!!
The species HOI and OH- are reaction
intermediates.
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Rate Laws and
Mechanisms
62
NO2 + CO reaction:
Rate = k[NO2]2
PLAY MOVIE
Two possible
mechanisms
Two steps: step 1
PLAY MOVIE
Single step
PLAY MOVIE
Two steps: step 2
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Ozone Decomposition
over Antarctica
2 O3 (g) f 3 O2 (g)
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64
Ozone Decomposition
Mechanism
2 O3 (g) f 3 O2 (g)
Proposed mechanism
Step 1: fast, equilibrium
O3 (g) e O2 (g) + O (g)
Step 2: slow
O3 (g) + O (g) f 2 O2 (g)
Rate = k
© 2009 Brooks/Cole - Cengage
[O3 ]2
[O2 ]
CATALYSIS
65
Catalysts speed up reactions by altering the
mechanism to lower the activation energy barrier.
Dr. James Cusumano, Catalytica Inc.
See Chemistry Now, Ch 15
PLAY MOVIE
What is a catalyst?
PLAY MOVIE
Catalysts and the environment
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
Catalysts and society
CATALYSIS
In auto exhaust systems — Pt, NiO
2 CO + O2 f 2 CO2
2 NO f N2 + O2
PLAY MOVIE
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CATALYSIS
2.
Polymers:
H2C=CH2 ---> polyethylene
3.
Acetic acid:
CH3OH + CO f CH3CO2H
4.
Enzymes — biological catalysts
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CATALYSIS
Catalysis and activation energy
MnO2 catalyzes
decomposition of H2O2
2 H2O2 f 2 H2O + O2
PLAY MOVIE
Uncatalyzed reaction
Catalyzed reaction
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68
Iodine-Catalyzed Isomerization of
cis-2-Butene
See Figure 15.15
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Iodine-Catalyzed Isomerization of
cis-2-Butene
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