Chemical Kinetics
CHAPTER 14
Chemistry: The Molecular Nature of Matter, 6th edition
By Jesperson, Brady, & Hyslop
CHAPTER 14 Chemical Kinetics
Learning Objectives:
 Factors Affecting Reaction Rate:
o Concentration
o State
o Surface Area
o Temperature
o Catalyst
 Collision Theory of Reactions and Effective Collisions
 Determining Reaction Order and Rate Law from Data
 Integrated Rate Laws
 Rate Law  Concentration vs Rate
 Integrated Rate Law  Concentration vs Time
 Units of Rate Constant and Overall Reaction Order
 Half Life vs Rate Constant (1st Order)
 Arrhenius Equation
 Mechanisms and Rate Laws
 Catalysts
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Molecular Nature of Matter, 6E
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CHAPTER 14 Chemical Kinetics
Lecture Road Map:
① Factors that affect reaction rates
② Measuring rates of reactions
③ Rate Laws
④ Collision Theory
⑤ Transition State Theory & Activation Energies
⑥ Mechanisms
⑦ Catalysts
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Molecular Nature of Matter, 6E
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CHAPTER 14 Chemical Kinetics
Mechanisms of
Reactions
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Molecular Nature of Matter, 6E
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Mechanisms
Overall vs Individual Steps
Sometimes rate law has simple form
– N2O5  NO2 + NO3
Rate = -
d [N2O5 ]
dt
= k 1 [N2O5 ]
– NO2 + NO3  N2O5
d [NO2 ]
Rate = = k 2 [NO2 ][NO3 ]
dt
But others are complex
– H2 + Br2  2 HBr
Rate = -
d [H2 ]
dt
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
=
k [H2 ][Br2 ]1/2
1+
k ¢[HBr]
[Br ] 5
Mechanisms
Overall vs Individual Steps
Some reactions occur in a single
step, as written
Others involve a sequence of steps
o Reaction Mechanism
o Entire sequence of steps
o Elementary Process
o Each individual step in mechanism
o Single step that occurs as written
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Molecular Nature of Matter, 6E
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Mechanisms
Overall vs Individual Steps
o Exponents in rate law for elementary process are equal to
coefficients of reactants in balanced chemical equation for that
elementary process
o Rate laws for elementary processes are directly related to
stoichiometry
o Number of molecules that participate in elementary process
defines molecularity of step
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Molecular Nature of Matter, 6E
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Mechanisms
Unimolecular Process
o Only one molecule as reactant
o H3C—NC  H3C—CN
o Rate = k[CH3NC]
o 1st order overall
o As number of molecules increases, number that rearrange
in given time interval increases proportionally
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Molecular Nature of Matter, 6E
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Mechanisms
Bimolecular Process
o Elementary step with two reactants
o NO(g) + O3(g)  NO2(g) + O2(g)
o Rate = k[NO][O3]
o 2nd order overall
o From collision theory:
o If [A] doubles, number of collisions between A and B will
double
o If [B] doubles, number of collisions between A and B will
double
o Thus, process is 1st order in A, 1st order in B, and 2nd order
overall
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Molecular Nature of Matter, 6E
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Mechanisms
Termolecular Process
o Elementary reaction with three molecules
o Extremely rare
o Why?
o Very low probability that three molecules
will collide simultaneously
o 3rd order overall
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Molecular Nature of Matter, 6E
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Mechanisms
Elementary Processes
Molecularity Elementary Step
Rate Law
Unimolecular
A  products
Rate = k[A]
Bimolecular
Bimolecular
A + A  products
A + B  products
Rate = k[A]2
Rate = k[A][B]
Significance of elementary steps:
o If we know that reaction is elementary step
o Then we know its rate law
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Molecular Nature of Matter, 6E
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Mechanisms
Multi-step Mechanisms
o Contains two or more steps to yield net reaction
o Elementary processes in multi-step mechanism must always
add up to give chemical equation of overall process
o Any mechanism we propose must be consistent with
experimentally observed rate law
o Intermediate = species which are formed in one step and
used up in subsequent steps
o Species which are neither reactant nor product in overall
reaction
o Mechanisms may involve one or more intermediates
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Molecular Nature of Matter, 6E
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Mechanisms
Example
The net reaction is:
NO2(g) + CO(g)  NO(g) + CO2(g)
The proposed mechanism is:
NO2(g) + NO2(g)  NO3(g) + NO(g)
NO3(g) + CO(g)  NO2(g) + CO2(g)
1
2NO2(g) + NO3(g) + CO(g)  NO2(g) + NO3(g) + NO(g) + CO2(g)
or
NO2(g) + CO(g)  NO(g) + CO2(g)
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Molecular Nature of Matter, 6E
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Mechanisms
Rate Determining Step
o If process follows sequence of steps, slow step determines
rate = rate determining step.
o Think of an assembly line
o Fast earlier steps may cause intermediates to pile up
o Fast later steps may have to wait for slower initial steps
o Rate-determining step governs rate law for overall
reaction
o Can only measure rate up to rate determining step
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Molecular Nature of Matter, 6E
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Mechanisms
Example: Rate Determining Step
(CH3)3CCl(aq) + OH–(aq)  (CH3)3COH(aq) + Cl–(aq)
chlorotrimethylmethane
trimethylmethanol
o Observed rate = k[(CH3)3CCl]
o If reaction was elementary
o Rate would depend on both reactants
o Frequency of collisions depends on both concentrations
o Mechanism is more complex than single step
o What is mechanism?
o Evidence that it is a two step process
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Molecular Nature of Matter, 6E
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Mechanisms
Rate Determining Step as Initial Step
Step 1: (CH3)3CCl(aq)  (CH3)3C+(aq) + Cl–(aq) (slow)
Step 2: (CH3)3C+(aq) + OH–(aq)  (CH3)3COH(aq) (fast)
o Two steps each at different rates
o Each step in multiple step mechanism is elementary process,
so
o Has its own rate constant and its own rate law
o Hence only for each step can we write rate law directly
o Observed rate law says that step 1 is very slow compared to
step 2
o In this case step 1 is rate determining
o Overall rate = k1[(CH3)3CCl]
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Molecular Nature of Matter, 6E
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Mechanisms
Mechanisms with Fast Initial Step
1st step involves fast, reversible reaction
Ex. Decomposition of ozone (No catalysts)
Net reaction: 2O3(g)  3O2(g)
Observed Rate 
k [O 3 ]2
[O2 ]
Proposed mechanism:
O3(g)  O2(g) + O(g)
(fast)
O(g) + O3(g)  2O2(g) (slow)
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Molecular Nature of Matter, 6E
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Mechanisms
Is the Mechanism Rate Law Consistent?
o Rate of formation of O2 = Rate of reaction 2
= k2[O][O3]
o But O is intermediate
o Need rate law in terms of reactants and products
o and possibly catalysts
o Rate (forward) = kf[O3]
o Rate (reverse) = kr[O2][O]
o When step 1 comes to equilibrium
o Rate (forward) = Rate (reverse)
o kf[O3] = kr[O2][O]
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Molecular Nature of Matter, 6E
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Is the Mechanism Rate Law Consistent?
Mechanisms
o Solving this for intermediate O gives:
[O] =
k f [O3 ]
k r [O2 ]
o Substitution into rate law for step 2 gives:
Observed Rate =
o Rate of reaction 2 = k2[O][O3] =
o where
k obs =
k 2k f
k [O3 ]2
[O2 ]
k 2k f [O3 ]2
k r [O2 ]
kr
o This is observed rate law
o Yes, mechanism consistent
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Molecular Nature of Matter, 6E
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Group
Problem
The reaction mechanism that has been proposed for the
decomposition of H2O2 is
1. H2O2 + I– → H2O + IO– (slow)
2. H2O2 + IO– → H2O + O2 + I– (fast)
What is the expected rate law?
First step is slow so the rate determining step defines
the rate law
rate=k [H2O2][I–]
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Molecular Nature of Matter, 6E
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Group
Problem
The reaction: A + 3B → D + F was studied and the
following mechanism was finally determined:
1. A + B  C
(fast)
2. C + B → D + E
(slow)
3. E + B → F
(very fast)
What is the expected rate law?
Rate Step 2=k2[C][B]
Rate = kobs[A][B]2
Rate forward = kf[A][B]
Rate reverse = kr[C]
kf[A][B] = kr[C]
[C]= kf[A][B]/kr
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Molecular Nature of Matter, 6E
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CHAPTER 14 Chemical Kinetics
Catalysts
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Molecular Nature of Matter, 6E
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Catalyst
Definition
o Substance that changes rate of chemical reaction
without itself being used up
o Speeds up reaction, but not consumed by reaction
o Appears in mechanism, but not in overall reaction
o Does not undergo permanent chemical change
o Regenerated at end of reaction mechanism
o May appear in rate law
o May be heterogeneous or homogeneous
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Molecular Nature of Matter, 6E
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Catalyst
Activation Energy
o By providing alternate
mechanism
o One with lower Ea
o Because Ea lower, more
reactants and collisions
have minimum KE, so
reaction proceeds faster
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Molecular Nature of Matter, 6E
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Catalyst
Activation Energy
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Molecular Nature of Matter, 6E
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Catalyst
Homogeneous Catalyst
• Same phase as reactants
Consider : S(g) + O2(g) + H2O(g)  H2SO4(g)
S(g) + O2(g)  SO2(g)
NO2(g) + SO2(g)  NO(g) + SO3(g) Catalytic pathway
SO3(g) + H2O(g)  H2SO4(g)
NO(g) + ½O2(g)  NO2(g)
Regeneration of catalyst

Net: S(g) + O2(g) + H2O(g)  H2SO4(g)
• What is Catalyst?
– Reactant (used up) in early step
– Product (regenerated) in later step
• Which are Intermediates?
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Molecular Nature of Matter, 6E
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Catalyst
o
o
o
o
Heterogeneous Catalyst
Exists in separate phase from reactants
Usually a solid
Many industrial catalysts are heterogeneous
Reaction takes place on solid catalyst
Ex. 3H2(g) + N2(g)  2NH3(g)
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Molecular Nature of Matter, 6E
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Catalyst
H2 and N2
approach
Fe catalyst
H2 and N2
bind to Fe
& bonds
break
Heterogeneous Catalyst
N—H
bonds
forming
N—H
bonds
forming
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
NH3
formation
complete
NH3
dissociates
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Enzymes: Superoxide Dismutase
Catalyst
H
N
O
OH2
OH
NHis
OAsp FeIII
O2 -
NHis
OH
Asp 156 O
O2-
Fe
N
His 73
NH
H+
NH
OH2
NHis
OAsp FeIII
His 160
His 26 N
O2
NHis
N
NHis
OAsp FeII
NHis
NHis
NHis
NHis
OH2
HOOH
H+
2H+
OAsp FeII
NHis
NHis
O2-
O2NHis
* O2- hydrogen bonds to residues in secondary
coordination sphere, positioning it near Fe(II),
Miller, Anne-Frances. “Fe Superoxide Dismutase” Handbook of Metalloproteins. John Wiley & Sons, Ltd, Chinchester, 2001
Rodrigues, J. V; Abreu, I. A.; Cabelli, D; Teixeira, M. Biochemistry 2006, 45, 9266-9278.
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