Chemistry and Chemical Reactivity
6th Edition
1
John C. Kotz
Paul M. Treichel
Gabriela C. Weaver
CHAPTER 15
Principles of Reactivity: Chemical Kinetics
Lectures written by John Kotz
©2006
2006
Brooks/Cole
Thomson
©
Brooks/Cole
- Thomson
2
Chemical Kinetics
Chapter 15
H2O2 decomposition in
an insect
© 2006 Brooks/Cole - Thomson
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!
© 2006 Brooks/Cole - Thomson
3
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 ---> BrO + O2
Step 2:
Cl + O3 ---> ClO + O2
Step 3:
BrO + ClO + light ---> Br + Cl + O2
NET:
2 O3 ---> 3 O2
© 2006 Brooks/Cole - Thomson
4
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
© 2006 Brooks/Cole - Thomson
5
Determining a Reaction Rate
Dye Conc
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.
Screen 15.2
Time
© 2006 Brooks/Cole - Thomson
6
Determining a Reaction Rate
Active Figure 15.2
© 2006 Brooks/Cole - Thomson
7
Factors Affecting Rates
Section 15.2
• Concentrations
• and physical state of
reactants and products (Screens
15.3-15.4)
• Temperature (Screen 15.11)
• Catalysts (Screen 15.14)
© 2006 Brooks/Cole - Thomson
8
Concentrations & Rates
Section 15.2
0.3 M HCl
6 M HCl
Mg(s) + 2 HCl(aq) ---> MgCl2(aq) + H2(g)
© 2006 Brooks/Cole - Thomson
9
10
Factors Affecting Rates
• Physical state of reactants
© 2006 Brooks/Cole - Thomson
11
Factors Affecting Rates
Catalysts: catalyzed decomp of H2O2
2 H2O2 --> 2 H2O + O2
© 2006 Brooks/Cole - Thomson
12
Catalysts
CO2 in H2O
Add dye
Add NaOH
1. CO2(g) Æ CO2 (aq)
Page 698
2. CO2 (aq) + H2O(liq) Æ H2CO3(aq)
3. H2CO3(aq) Æ H+(aq) + HCO3–(aq)
• Adding trace of NaOH uses up H+. Equilibrium shifts
to produce more H2CO3.
• Enzyme speeds up reactions 1 and 2
© 2006 Brooks/Cole - Thomson
13
Factors Affecting Rates
• Temperature
Bleach at 54 ˚C
© 2006 Brooks/Cole - Thomson
Bleach at 22 ˚C
Iodine Clock Reaction
1. Iodide is oxidized to iodine
H2O2 + 2 I- + 2 H+ -----> 2 H2O + I2
2.
I2 reduced to I- with vitamin C
I2 + C6H8O6 ----> C6H6O6 + 2 H+ + 2 I-
When all vitamin C is depleted, the I2 interacts
with starch to give a blue complex.
Page 705 in CCR
© 2006 Brooks/Cole - Thomson
14
Iodine Clock Reaction
© 2006 Brooks/Cole - Thomson
15
16
Concentrations and Rates
To postulate a reaction
mechanism, we study
• reaction rate and
• its concentration
dependence
© 2006 Brooks/Cole - Thomson
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)
© 2006 Brooks/Cole - Thomson
17
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
© 2006 Brooks/Cole - Thomson
18
Concentrations, Rates, &
Rate Laws
In general, for
a A + b B --> 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!
© 2006 Brooks/Cole - Thomson
19
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 ________
© 2006 Brooks/Cole - Thomson
20
Deriving Rate Laws
Derive rate law and k for
CH3CHO(g) --> 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
© 2006 Brooks/Cole - Thomson
Disappear of CH3CHO
(mol/L•sec)
21
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.
© 2006 Brooks/Cole - Thomson
22
Concentration/Time
Relations
What is concentration of reactant as function of
time?
Consider FIRST ORDER REACTIONS
The rate law is
[A]
Rate  = k [A]
time
© 2006 Brooks/Cole - Thomson
23
Cisplatin
Concentration/Time Relations
Integrating - (∆ [A] / ∆ time) = k [A], we get
[A] / [A]0 =fraction remaining after time t
has elapsed.
Called the integrated first-order rate law.
© 2006 Brooks/Cole - Thomson
24
Concentration/Time Relations
25
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
© 2006 Brooks/Cole - Thomson
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 M
ln
0.010 M
= - (0.21 hr-1) t
ln (0.100) = - 2.3 = - (0.21 hr-1) • time
time = 11 hours
© 2006 Brooks/Cole - Thomson
26
Using the Integrated Rate Law
The integrated rate law suggests a way to tell
the order based on experiment.
2 N2O5(g) ---> 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]
© 2006 Brooks/Cole - Thomson
27
Using the Integrated Rate Law
2 N2O5(g) ---> 4 NO2(g) + O2(g) Rate = k [N2O5]
Data of conc. vs.
time plot do not fit
straight line.
© 2006 Brooks/Cole - Thomson
Plot of ln [N2O5] vs.
time is a straight
line!
28
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)
© 2006 Brooks/Cole - Thomson
29
30
Properties of Reactions
page 719
© 2006 Brooks/Cole - Thomson
31
Half-Life
Section 15.4 & Screen 15.8
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.
Active Figure 15.9
© 2006 Brooks/Cole - Thomson
32
Half-Life
• Reaction is 1st order
decomposition of
H2O2.
© 2006 Brooks/Cole - Thomson
33
Half-Life
• Reaction after 1
half-life.
• 1/2 of the reactant
has been
consumed and 1/2
remains.
© 2006 Brooks/Cole - Thomson
34
Half-Life
• After 2 half-lives
1/4 of the reactant
remains.
© 2006 Brooks/Cole - Thomson
35
Half-Life
• A 3 half-lives 1/8
of the reactant
remains.
© 2006 Brooks/Cole - Thomson
36
Half-Life
• After 4 half-lives
1/16 of the
reactant remains.
© 2006 Brooks/Cole - Thomson
Half-Life
Section 15.4 & Screen 15.8
Sugar is fermented in a 1st order process (using
an enzyme as a catalyst).
sugar + enzyme --> products
Rate of disappear of sugar = k[sugar]
k = 3.3 x 10-4 sec-1
What is the half-life of this reaction?
© 2006 Brooks/Cole - Thomson
37
Half-Life
38
Section 15.4 & Screen 15.8
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
© 2006 Brooks/Cole - Thomson
min
Half-Life
Section 15.4 & Screen 15.8
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
© 2006 Brooks/Cole - Thomson
39
Half-Life
Section 15.4 & Screen 15.8
Radioactive decay is a first order process.
Tritium ---> 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?
© 2006 Brooks/Cole - Thomson
40
Half-Life
Section 15.4 & Screen 15.8
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
© 2006 Brooks/Cole - Thomson
41
Half-Life
Section 15.4 & Screen 15.8
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 ---> 0.750 mg after 1 half-life
---> 0.375 mg after 2
---> 0.188 mg after 3
---> 0.094 mg after 4
© 2006 Brooks/Cole - Thomson
42
Half-Lives of Radioactive Elements
Rate of decay of radioactive isotopes given in
terms of 1/2-life.
238U --> 234Th + He
4.5 x 109 y
14C --> 14N + beta
5730 y
131I
--> 131Xe + beta
8.05 d
Element 106 - seaborgium
263Sg
© 2006 Brooks/Cole - Thomson
0.9 s
43
MECHANISMS
A Microscopic View of Reactions
Sections 15.5 and 15.6
Mechanism: how reactants are converted to
products at the molecular level.
RATE LAW ----> MECHANISM
experiment ----> theory
© 2006 Brooks/Cole - Thomson
44
Activation Energy
Molecules need a minimum amount of energy to react.
Visualized as an energy barrier - activation energy, Ea.
Reaction coordinate
diagram
© 2006 Brooks/Cole - Thomson
45
MECHANISMS
& Activation Energy
Conversion of cis to trans-2-butene requires
twisting around the C=C bond.
Rate = k [trans-2-butene]
© 2006 Brooks/Cole - Thomson
46
MECHANISMS
Cis
Transition state
47
Trans
Activation energy barrier
© 2006 Brooks/Cole - Thomson
MECHANISMS
48
Energy involved in conversion of trans to cis butene
energy
Activated
Complex
+262 kJ
cis
-266 kJ
4 kJ/mol
trans
See Figure 15.14
© 2006 Brooks/Cole - Thomson
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
© 2006 Brooks/Cole - Thomson
49
MECHANISMS
Also note that trans-butene is MORE
STABLE than cis-butene by about 4 kJ/mol.
Therefore, cis ---> trans is EXOTHERMIC
This is the connection between thermodynamics and kinetics.
© 2006 Brooks/Cole - Thomson
50
Effect of Temperature
• Reactions generally
occur slower at
lower T.
In ice at 0 oC
Room temperature
Iodine clock reaction, Screen
15.11, and book page 705.
H2O2 + 2 I- + 2 H+ --> 2 H2O + I2
© 2006 Brooks/Cole - Thomson
51
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.
© 2006 Brooks/Cole - Thomson
52
Mechanisms
1.
Why is trans-butene <--> cis-butene
reaction observed to be 1st order?
As [trans] doubles, number of molecules
with enough E also doubles.
2.
Why is the trans <--> cis reaction faster at
higher temperature?
Fraction of molecules with sufficient
activation energy increases with T.
© 2006 Brooks/Cole - Thomson
53
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
© 2006 Brooks/Cole - Thomson
54
Plot ln k vs. 1/T
--->
straight line.
slope = -Ea/R
More on Mechanisms
A bimolecular reaction
Reaction of
trans-butene --> cis-butene is
UNIMOLECULAR - only one
reactant is involved.
BIMOLECULAR — two different
molecules must collide -->
products
Exo- or endothermic?
© 2006 Brooks/Cole - Thomson
55
Collision Theory
Reactions require
(a) activation energy and
(b) correct geometry.
O3(g) + NO(g) ---> O2(g) + NO2(g)
1. Activation energy
© 2006 Brooks/Cole - Thomson
2. Activation energy
and geometry
56
Mechanisms
O3 + NO reaction occurs in a single ELEMENTARY step.
Most others involve a sequence of elementary steps.
Adding elementary steps gives NET reaction.
© 2006 Brooks/Cole - Thomson
57
Mechanisms
Most rxns. involve a sequence of elementary
steps.
2 I- + H2O2 + 2 H+ ---> 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.
© 2006 Brooks/Cole - Thomson
58
Mechanisms
Most rxns. involve a sequence of elementary steps.
2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O
Rate = k [I-] [H2O2]
Proposed Mechanism
Step 1 — slow
HOOH + I- --> HOI + OH-
Step 2 — fast
HOI + I- --> I2 + OH-
Step 3 — fast
2 OH- + 2 H+ --> 2 H2O
Rate of the reaction controlled by slow step —
RATE DETERMINING STEP, rds.
Rate can be no faster than rds!
© 2006 Brooks/Cole - Thomson
59
Mechanisms
2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O
Rate = k [I-] [H2O2]
Step 1 — slow HOOH + I- --> HOI + OHStep 2 — fast HOI + I- --> I2 + OHStep 3 — fast 2 OH- + 2 H+ --> 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.
© 2006 Brooks/Cole - Thomson
60
Rate Laws and
Mechanisms
61
NO2 + CO reaction:
Rate = k[NO2]2
Two possible
mechanisms
Two steps: step 1
Single step
Two steps: step 2
© 2006 Brooks/Cole - Thomson
Ozone Decomposition
over Antarctica
© 2006 Brooks/Cole - Thomson
2 O3 (g) ---> 3 O2 (g)
62
63
Ozone Decomposition
Mechanism
2 O3 (g) ---> 3 O2 (g)
Proposed mechanism
Step 1: fast, equilibrium
O3 (g) Æ O2 (g) + O (g)
Step 2: slow
O3 (g) + O (g) ---> 2 O2 (g)
© 2006 Brooks/Cole - Thomson
[O3 ]2
Rate = k
[O2 ]
CATALYSIS
64
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 the environment
Catalysts and society
© 2006 Brooks/Cole - Thomson
CATALYSIS
In auto exhaust systems — Pt, NiO
2 CO + O2 ---> 2 CO2
2 NO ---> N2 + O2
© 2006 Brooks/Cole - Thomson
65
CATALYSIS
2.
Polymers:
H2C=CH2 ---> polyethylene
3.
Acetic acid:
CH3OH + CO --> CH3CO2H
4.
Enzymes — biological catalysts
© 2006 Brooks/Cole - Thomson
66
CATALYSIS
Catalysis and activation energy
MnO2 catalyzes
decomposition of H2O2
2 H2O2 ---> 2 H2O + O2
Uncatalyzed reaction
Catalyzed reaction
© 2006 Brooks/Cole - Thomson
67
Iodine-Catalyzed Isomerization of
cis-2-Butene
Figure 15.16
© 2006 Brooks/Cole - Thomson
68
Iodine-Catalyzed Isomerization of
cis-2-Butene
© 2006 Brooks/Cole - Thomson
69