Chemical Kinetics
Chapter 12
H2O2 decomposition in
an insect
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H2O2 decomposition
catalyzed by MnO2
1
Objectives
• Understand rates of reaction and the
conditions affecting rates.
• Derive a rate of reaction, rate constant,
and reaction order from experimental
data
• Use integrated rate laws
• Discuss collision theory and the role of
activation energy in a reaction
• Discuss reaction mechanisms and their
effect on rate law
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Chemical Kinetics
• We can use thermodynamics to tell if a
reaction is product- or reactant-favored.
• 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.
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Reaction Mechanisms
• A reaction mechanism is a
sequence of events at the
molecular level that control the
speed and outcome of a
reaction.
• The reaction mechanism is our
goal!
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Reaction Rates
Section 12.1
• 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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Factors Affecting Reaction Rates
6
• Physical State of the Reactants
–
–
Gas, liquid or solid – how molecules are able to interact with
each other
Solids react faster when surface area is greater so fine
powders react faster than big chunks
• Concentration of Reactants
–
As the concentration of reactants increases, so does the
likelihood that reactant molecules will collide.
• Temperature
–
At higher temperatures, reactant molecules have more kinetic
energy, move faster, and collide more often and with greater
energy.
• Catalysts
–
–
Speed up reaction by changing mechanism.
Catalysts don’t get used up themselves
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Concentrations & Rates
Mg(s) + 2 HCl(aq)  MgCl2(aq) + H2(g)
0.3 M HCl
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6 M HCl
Factors Affecting Rates
• Physical state of reactants
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Factors Affecting Rates
Catalysts: catalyzed decomp of H2O2
2 H2O2  2 H2O + O2
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Factors Affecting Rates
• Temperature
Bleach at 54 ˚C
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Bleach at 22 ˚C
10
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.
Time
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Determining a Reaction Rate
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Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
Average Rate, M/s
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The average rate of
the reaction over
each interval is the
change in
concentration
divided by the
change in time:
13
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
• Note that the
average rate
decreases as the
reaction proceeds.
• This is because as
the reaction goes
forward, there are
fewer collisions
between reactant
molecules.
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Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
• A plot of concentration
vs. time for this reaction
yields a curve like this.
• The slope of a line
tangent to the curve at
any point is the
instantaneous rate at that
time.
Δ[𝑨] 𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒎𝒐𝒍𝒂𝒓𝒊𝒕𝒚
=
Δ𝒕
𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒕𝒊𝒎𝒆
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Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
• The reaction slows
down with time
because the
concentration of the
reactants decreases.
[A] Change in molarity

t
Change in time
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Reaction Rates and Stoichiometry
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
• In this reaction, the
ratio of C4H9Cl to
C4H9OH is 1:1.
• Thus, the rate of
disappearance of
C4H9Cl is the same as
the rate of appearance
of C4H9OH.
Rxn
-[C4H9Cl]
=
Rate =
t
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[C4H9OH]
t
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Reaction Rates
18
2NO2(g)  2NO(g) + O2(g)
[C4H9Cl] M
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In this reaction, the
concentration of
nitrogen dioxide,
NO2, was measured
at various times, t.
Reaction Rates and Stoichiometry
• What if the ratio is not 1:1?
2NO2(g)  2NO(g) + O2(g)
• 2NO can be made from 2NO2
consumed, but only 1 O2 is produced.
∆ 𝑵𝑶𝟐
−
∆𝒕
=
∆ 𝑵𝑶
∆𝒕
=𝟐
∆ 𝑶𝟐
∆𝒕
Read as: the rate of consumption of
NO2 is the same as the rate of
production of NO. This is because
their coefficients are the same.
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20
Reaction Rates
But since the
coefficient for oxygen
is ½ of the other two,
it’s rate is of
production rate is half
as fast. Or 2 x rate O2
= rate of NO
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Reaction Rates
Reaction Rate and Stoichiometry summary:
• For the reaction
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
we know 𝐑𝐚𝐭𝐞 =
Δ[𝑪𝟒𝑯𝟗𝑪𝒍]
−
Δ𝒕
=
Δ[𝑪𝟒𝑯𝟗𝑶𝑯]
Δ𝒕
• In general for
aA + bB  cC + dD
Rxn 𝐑𝐚𝐭𝐞 =
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𝟏Δ 𝑨
−
𝒂 Δ𝒕
=
𝟏Δ 𝑩
−
𝒃 Δ𝒕
=
𝟏Δ 𝑪
𝒄 Δ𝒕
=
𝟏 Δ[𝑫]
𝒅 Δ𝒕
Reaction Rates Practice
1. What is the relative rate of disappearance of
the reactants and relative rate of appearance
of the products for each reaction?
A. 2O3(g)  3O2(g)
B. 2HOF(g)  2HF(g) + O2(g)
2. In the synthesis of ammonia, if
Δ[H2]/Δt =-4.5 x 10-4 mol/L•min, what is rate
with respect to NH3? wrt N2?
N2(g) + 3H2(g)  2NH3(g)
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Concentrations & Rates
2NO2(g)  2NO(g) + O2(g)
Rate of reaction is proportional to [NO2]
We express this as a RATE LAW
Rate of reaction = k [NO2]n
where k = rate constant
k is independent of conc. but increases with T
n is the order of the reactant
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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:
• sum of m, n and p are the reaction order
• can be 0, 1, 2 or fractions like 3/2
• must be determined by experiment! They
are not simply related to stoichiometry!
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Ch. 12.3 - Interpreting
Rate Laws
Determining Rate by inspection (quick method):
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 _2_
• If m = 2, rxn. is 2nd order in A.
Rate = k [A]2
Doubling [A] increases rate by ___4_____
• If m = 0, rxn. is zero order.
Rate = k [A]0
If [A] doubles, rate _Stays the same_
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The method of Initial Rates
• This method requires that a reaction
be run several times.
• The initial concentrations of the
reactants are varied.
• The reaction rate is measured just
after the reactants are mixed.
• Eliminates the effect of the reverse
reaction.
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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
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Disappear of CH3CHO
(mol/L•sec)
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Deriving Rate Laws
Determination of Rate by inspection:
Here the rate goes up by __4___ when initial conc.
doubles. Therefore, we say this reaction is
________2nd_________ order.
So, the Rate of rxn = k [CH3CHO]2
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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Deriving Rate Laws
Determination of Rate by rigorous math:
Rate = k[A]n
𝑹𝒂𝒕𝒆 𝒐𝒇 𝑹𝒆𝒂𝒄𝒕𝒊𝒐𝒏 𝟐 𝟎. 𝟎𝟖𝟏
𝒌 𝟎. 𝟐𝟎 𝒏
=
=
𝑹𝒂𝒕𝒆 𝒐𝒇 𝑹𝒆𝒂𝒄𝒕𝒊𝒐𝒏 𝟏 𝟎. 𝟎𝟐𝟎
𝒌 𝟎. 𝟏𝟎 𝒏
Note: For easier math, put the faster rate in the
numerator.
𝒏
[𝟎. 𝟐𝟎]
𝟎. 𝟐𝟎
𝟒=
=
𝒏
𝟎. 𝟏𝟎
𝟎. 𝟏𝟎
4=2n
n=2
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𝒏
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Determination of Rate Law Practice 30
For the reaction
+
BrO3 + 5Br + 6H  3Br2 + 3H2O
• The general form of the Rate Law is
Rate = k[BrO3-]n[Br-]m[H+]p
• We use experimental data to determine the
values of n, m, and p
• We will choose trials that vary only one of
the concentrations to determine the order
based upon that reactant.
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Determination of Rate Law Practice
Initial concentrations (M)
BrO30.10
0.20
0.20
0.10
Br0.10
0.10
0.20
0.10
H+
0.10
0.10
0.10
0.20
Rate (M/s)
0.8 x 10-3
1.6 x 10-3
3.2 x 10-3
3.2 x 10-3
1) Determine the rate law wrt each reactant
2) Determine the rate constant, k (with units)
3) Determine the overall order of reaction
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Ch. 12.4 – Integrated Rate
Law (Conc vs time)
What is concentration of reactant as function of
time?
Consider FIRST ORDER REACTIONS
The rate law is
𝚫𝑨
𝐑𝐚𝐭𝐞 = −
= 𝒌[𝑨]
Δ𝒕𝒊𝒎𝒆
And the units of k are sec-1.
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33
Integrated Rate Law
• Expresses the reaction
concentration as a function of time.
• Form of the equation depends on
the order of the rate law
(differential).
Rate = [A]n
t
• We will only work with n=0, 1, and 2
• Changes
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Concentration/Time
Relations
Integrating - (∆ [A] / ∆ time) = k [A], we get
ln is natural
logarithm
[A]
ln
= - kt
[A]0
[A]0 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
35
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)  4 NO2(g) + O2(g)
Time (min)
0
1.0
2.0
5.0
[N2O5] (M)
1.00
0.705
0.497
0.173
Rate = k [N2O5]
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ln [N2O5]
0
-0.35
-0.70
-1.75
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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.
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Plot of ln [N2O5] vs.
time is a straight
line!
38
Using the Integrated Rate Law
Plot of ln [N2O5] vs. time is a
straight line!
Eqn. for straight line:
y = mx + b
ln[N2O5 ] = -kt +
conc at
time t
rate const
=-slope
ln[N2O5]0
conc at
time=0
All 1st order reactions have straight line plot
for ln [A] vs. time.
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40
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.
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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.
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Half-Life
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?
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Half-Life
47
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
NOTE: For a first-order
- 0.693 = - k · t1/2
process, the half-life
t1/2 = 0.693 / k does not depend on [A]0.
So, for sugar,
t1/2 = 0.693 / k = 2100 sec = 35
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min
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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49
Second Order
•
•
•
•
•
•
•
Rate = -Δ[A]/Δt = k[A]2
integrated rate law
1/[A] = kt + 1/[A]0
y= 1/[A]
m=k
x= t
b = 1/[A]0
A straight line if 1/[A] vs t is plotted
Knowing k and [A]0 you can calculate [A] at
any time t
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Determining rxn order
The decomposition of NO2 at 300°C is described
by the equation
NO2 (g)
NO (g) + 1/2 O2 (g)
and yields these data:
Time (s)
0.0
50.0
100.0
200.0
300.0
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[NO2], M
0.01000
0.00787
0.00649
0.00481
0.00380
50
Determining rxn order
Graphing ln [NO2] vs. t
yields:
• The plot is not a straight
line, so the process is
not first-order in [A].
Time (s)
0.0
50.0
100.0
200.0
300.0
[NO2], M
0.01000
0.00787
0.00649
0.00481
0.00380
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ln [NO2]
-4.610
-4.845
-5.038
-5.337
-5.573
Does not fit:
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Second-Order Processes
A graph of 1/[NO2] vs. t
gives this plot.
Time (s)
0.0
50.0
100.0
200.0
300.0
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[NO2], M
0.01000
0.00787
0.00649
0.00481
0.00380
1/[NO2]
100
127
154
208
263
• This is a straight
line. Therefore,
the process is
second-order in
[NO2].
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53
Zero Order Rate Law
• Rate = k[A]0 = k
• Rate does not change with
concentration.
• Integrated [A] = -kt + [A]0
• When [A] = [A]0/2 then t = t1/2
• t1/2 = [A]0 /2k
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Zero Order Rate Law
• Most often when reaction happens
on a surface because the surface
area stays constant.
• Also applies to enzyme chemistry.
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C
o
n
c
e
n
t
r
a
t
i
o
n
k = [A]/t
[A]
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t
Time
Rate Laws Summary
Rate Law
Integrated
Rate Law
Plot that
produces a
straight line
Relationship of
rate constant
to slope of
straight line
Half-Life
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Zero Order
First Order
Second Order
Rate = k
[A] = -kt +
[A]0
Rate = k[A]
ln[A] = -kt +
ln[A]0
Rate = k[A]2
[A] versus t
ln[A] versus t
1
versus t
[ A]
Slope = -k
Slope = -k
Slope = k
[ A]0
t1/ 2 
2k
0.693
t1/ 2 
k
1
1
 kt 
[ A]
[ A]0
1
t1/ 2 
k [ A]0
Ch. 12.5 - Reaction Mechanisms
Mechanism: how reactants are converted to
products at the molecular level.
RATE LAW  MECHANISM
experiment 
theory
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Reaction Mechanisms
•
The molecularity of a process tells how many
molecules are involved in the process.
•
The rate law for an elementary step is written
directly from that step.
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Reaction Mechanisms
• If a reaction is an elementary step,
then we know its rate law.
• All of the molecules involved in an
elementary step are defined in the
rate law.
• The full molecularity has to be
included.
• The sum of the elementary steps in a
mechanism must give the overall
balanced equation for the reaction.
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Molecularity Example
• Given the reactions and their rate
laws below, which is the only one
that can be an elementary step?
A) 2A  P
Rate = k[A]
B) A + B  P
Rate = k[A][B]
C) A + 2B  P
Rate = k[A]2
D) A + B + C  P Rate = k[A][C]
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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.
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Mechanisms
Most rxns. involve a sequence of elementary steps.
2 I- + H 2 O 2 + 2 H +  I 2 + 2 H 2 O
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!
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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.
More on intermediates coming up.
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Rate Laws and
Mechanisms
64
NO2 + CO reaction:
Rate = k[NO2]2
Two possible
mechanisms
Two steps: step 1
Single step
Two steps: step 2
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Slow Initial Step
NO2 (g) + CO (g)  NO (g) + CO2 (g)
• The rate law for this reaction is found
experimentally to be
Rate = k [NO2]2
• CO is necessary for this reaction to occur, but
the rate of the reaction does not depend on its
concentration.
• This suggests the reaction occurs in two steps.
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Slow Initial Step
66
• A proposed mechanism for this reaction is
Step 1: NO2 + NO2  NO3 + NO (slow)
Step 2: NO3 + CO  NO2 + CO2 (fast)
• The NO3 intermediate is consumed in the second step.
• As CO is not involved in the slow, rate-determining step, it
does not appear in the rate law.
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Fast Initial Step
• The rate law for this reaction is found
(experimentally) to be
• Because termolecular (aka trimolecular)
processes are rare, this rate law suggests a twostep mechanism.
67
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Fast Initial Step
• A proposed mechanism is
Step 1 is an equilibriumit includes the forward and reverse reactions.
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Fast Initial Step
• The rate of the overall reaction depends
upon the rate of the slow step.
• The rate law for that step would be
• But how can we find [NOBr2]?
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Intermediates
• [NOBr2] is an intermediate, an unstable
molecule of low, unknown
concentration
• Intermediates can’t appear in the rate
law.
• Remember: only reactants (and
products, rarely) can appear in a rate
law.
• Slow step determines the rate and the
rate law.
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Intermediates
• Use the reactions that form them
• If the reactions are fast and
irreversible the concentration of
the intermediate is based on
stoichiometry.
• If it is formed by a reversible
reaction set the rates equal to each
other.
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Fast Initial Step
• NOBr2 can react two ways:
– With NO to form NOBr
– By decomposition to reform NO and Br2
• The reactants and products of the
first step are in equilibrium with
each other.
• Therefore,
Ratef = Rater
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Fast Initial Step
• Because Ratef = Rater ,
k1 [NO] [Br2] = k−1 [NOBr2]
Solving for [NOBr2] gives us
k1/k-1 [NO] [Br2] = [NOBr2]
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Fast Initial Step
Substituting this expression for [NOBr2] in the
rate law for the rate-determining step gives
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Ch. 12.6 – A Model for Kinetics
Molecules need a minimum amount of energy to react.
Visualized as an energy barrier - activation energy, Ea.
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
77
Trans
Activation energy barrier
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MECHANISMS
78
Energy involved in conversion of trans to cis butene
energy
Activated
Complex
+262 kJ
cis
-266 kJ
4 kJ/mol
trans
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Collision Theory
Reactions require
(a) activation energy and
(b) correct orientation.
O3(g) + NO(g)  O2(g) + NO2(g)
1. Activation energy
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2. Activation energy
and geometry
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Mechanisms
O3 + NO reaction occurs in a single ELEMENTARY step.
Most others involve a sequence of elementary steps.
Adding elementary steps gives NET reaction.
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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  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
Iodine clock reaction.
H2O2 + 2 I- + 2 H+ 2 H2O + I2
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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  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.
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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
 straight line.
slope = -Ea/R
Arrhenius Equation Example
• Calculate the Ea from the data
below.
• T(⁰C) k(s-1)
• 189.7 2.52 X 10-5
• 198.9 5.25 X 10-5
• 230.3 6.30 X 10-4
• 251.2 3.16 X 10-3
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Arrhenius Equation Example
• 1st step  change T(⁰C) to T(K), find
1/T and change k to ln k.
T(K)
1/T
ln(k)
462.9
0.002160
-10.589
472.1
0.002118
-9.855
503.5
0.001986
-7.370
524.4
0.001907
-5.757
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89
Arrhenius Plot
-4
0.00185
-5
0.0019
0.00195
0.002
0.00205
0.0021
0.00215
ln k
-6
-7
y = -19080x + 30.582
R² = 0.9995
-8
-9
-10
-11
1/T
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0.0022
Arrhenius Equation Example
90
2nd Step  find the slope (If not using graphing
calculator or excel)
Slope = (-5.757 – (-9.855))/(0.001907 – 0.002118)
= 4.098/(-0.000211)
= -19422 K
3rd Step  Multiply the slope by -8.314 J/mol K
to get Ea
= -19422 s-1 * -8.314 J/mol K = 161472 J = 161 kJ
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More on the Arrhenius Equation
Determining the Activation Energy
• If we do not have a lot of data, then we recognize
Ea
ln k1  
 ln A and
RT1
Ea
ln k2  
 ln A
RT2
 Ea
  Ea

ln k1  ln k2   
 ln A    
 ln A 
 RT1
  RT2

k1 Ea  1 1 
ln

  
k2
R  T2 T1 
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Example Problem
• The Ea of a certain reaction is
65.7 kJ/mol. How many times
faster is the rxn at 50⁰C than
at 0⁰C?
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Ch. 12.7 - CATALYSIS
93
Catalysts speed up reactions by altering the
mechanism to lower the activation energy barrier.
Dr. James Cusumano, Catalytica Inc.
PLAY MOVIE
What is a catalyst?
PLAY MOVIE
Catalysts and the environment
PLAY MOVIE
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Catalysts and society
CATALYSIS
In auto exhaust systems — Pt, NiO
2 CO + O2  2 CO2
2 NO  N2 + O2
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CATALYSIS
2.
Polymers:
H2C=CH2  polyethylene
3.
Acetic acid:
CH3OH + CO  CH3CO2H
4.
Enzymes — biological catalysts
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CATALYSIS
Catalysis and activation energy
MnO2 catalyzes
decomposition of H2O2
2 H2O2  2 H2O + O2
Uncatalyzed reaction
Catalyzed reaction
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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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AP Exam Practice
• 2010 AP Exam #3b-d
• 2009B AP Exam #2
• 2009 AP exam #3d-e
• 2008 AP Exam #3d-f
• 2008B AP Exam #2a-g
• 2005B AP Exam #3
• 2005 AP Exam #3
• 2003 AP Exam #3a-c
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