Atkins & de Paula:
Elements of Physical Chemistry: 5e
Chapter 11:
Chemical Kinetics:
Accounting for the Rate Laws
End of chapter 11 assignments
Discussion questions:
• 2, 5
Exercises:
• 5, 13, 14, 18, 25
Use Excel if data needs to be graphed
Time for me to “nag” you!
• Did you:
– Read the chapter?
– Work through the example problems?
– Connect to the publisher’s website &
access the “Living Graphs”?
– Examine the “Checklist of Key Ideas”?
– Work assigned end-of-chapter exercises?
• Review terms & concepts that you
should remember from previous
chapters or courses
Reaction schemes
• All chemical rxns proceed toward a
state of equilibrium
• The reverse rxn becomes increasingly
important as the rxn proceeds
• Many rxns proceed through a series of
intermediates
• These intermediates may be important
Rate Laws
• Rate laws are always determined
experimentally.
• Reaction order is always defined in
terms of reactant (not product)
concentrations.
14.2
Rate Laws
•
The order of a reactant is not related to the
stoichiometric coefficient of the reactant in the
balanced chemical equation.
F2 (g) + 2ClO2 (g)
2FClO2 (g)
1
rate = k [F2][ClO2]
•
The stoichiometric coefficient of the reactant in
the balanced chemical equation is the
exponent in the K expression.
aA + bB
cC + dD
Kc =
[C]c[D]d
[A]a[B]b
14.2
Digression: Rate law and Keq
• Keq is concerned with the initial and
final conditions of the rxn (overall)
• Rate law, order, “kinetics” are
concerned with everything in between
• Let me show you what IUPAC says
about the rate law and Keq
Rate law and Keq
The equilibrium law or the equilibrium
constant are in the first instance derived
from the thermodynamics of the
reaction, the value of the equilibrium
constant Kc depending entirely upon the
characteristics of the reactants and
these of the reaction products.
--continued
Rate law and Keq
Rate of reaction and reaction
mechanism are kinetic aspects of
the reaction. They have nothing to
do with the expression for Kc or with
the value of Kc. It would be simplistic
and imprudent to assume all sorts of
unrealistic reaction equations to
obtain an expression for Kc.
Rate law and Keq
• Quoted from IUPAC’s web site
• http://www.iupac.org/
• Specifically:
http://www.iupac.org/didac/Didac%20Eng/
Didac02/Content
/E06%20-%20E07%20-%20E08.htm
Order & molecularity
The overall reaction order is the sum of the
powers of the concentrations of all of the
substances appearing in the experimental rate
law for the reaction; hence, it is the sum of the
individual orders (exponents) associated with a
given reactant (or product). Reaction order is
an experimentally determined, not theoretical,
quantity, although theory may attempt to
predict it.
Quoted from another book by Atkins…
Order & molecularity
Molecularity is the number of reactant molecules participating in an elementary reaction. This concept has
meaning only for an elementary reaction, but reaction
order applies to any reaction. In general, reaction order
bears no necessary relation to the stoichiometry of the
reaction, with the exception of elementary reactions,
where the order of the reaction corresponds to the
number of molecules participating in the reaction; that is,
to its molecularity. Thus for an elementary reaction,
overall order and molecularity are the same and are
determined by the stoichiometry.
Quoted from another book by Atkins…
Profile of an exothermic rxn
• Fig 11.1 (258)
• Ea is greater for the
reverse rxn than for
the forward rxn
• So the Rf increases
less sharply (than the
Rr) as T increases
• As the T is raised, the
products are favored
The approach to equilibrium
• Derivation 11.1 (259)
• You should work through this
derivation
The approach to equilibrium
• Fig 11.2 (259)
• The approach to
equilibrium of a
rxn that is 1st
order in both
directions
• In this example,
k = 2k’, so at
equilibrium the
ratio of b:a is 2:1
AB
Relaxation methods
• Some externally applied influence
shifts the rxn equilibrium abruptly
– Temperature jump
– Pressure jump
• The adjustment back to equilibrium
is called “relaxation”
Relaxation methods
Fig 11.3 (259)
Relaxation to a
new equilibrium
following a sudden change of
temperature
from T1 to T2
Relaxation methods
• When a sudden temperature jump
shifts the equilibrium of a 1st order rxn:
• x=
x0e–t/
1
and  = ka + kb
• x is the shift in equilibrium at the new t
• x0 is the shift in equilibrium immediately
after the temperature change
•  is the relaxation time
• t is the time (s)
Consecutive reactions
Common examples are radioactive
decay chains and biochemical rxn
sequences (e.g., gylcoysis, Krebs
cycle, etc)
The Uranium Decay Series
Glycolysis
Consecutive reactions
• Fig 11.4 (263)
• Concentrations in
consecutive rxns,
AIP
– A = reactant
– I = intermediate
– P = product
• At any time the
sum [A] + [I] + [P]
is a constant
Elementary reactions
• Many reactions occur as a series of
steps called “elementary reactions”
• Molecularity is the number of
molecules coming together to react
– Unimolecular
– Bimolecular
– What’s next?
Elementary reactions
Fig 11.5 (263)
A unimolecular
elementary rxn
Elementary reactions
Fig 11.6 (264)
A bimolecular
elementary rxn
The formulation of rate laws
• Assuming that a chemical rxn is a
series of elementary rxns…
• One of these elementary rxns is the
rate determining step
• An acceptable rate law for an overall
rxn is expressed only in terms of the
concentrations of the species in the
overall rxn, not in terms of an
intermediate
(264)
The formulation of rate laws
• Consider this rxn:
• 2 NO(g) + O2(g) d 2 NO2(g)
• Experimentally: rate = k[NO]2[O2]
(which is 3rd order overall)
• How can we explain this? Is it a
termolecular rxn (2 NOs and 1 O2)?
• Let’s see…
(264)
Proposed mechanism: 2 NO(g) + O2(g) d 2 NO2
Step 1. NO + NO d N2O2
Rate of formation of N2O2 = ka[NO]2
Step 2. N2O2 d NO + NO
Rate of decomposition of N2O2 = k’a[N2O2]
Step 3. N2O2 + O2 d NO2 + NO2
Rate of consumption of N2O2 = 2kb[N2O2][O2]
Thus the net rate of formation of N2O2 is =
ka[NO]2 – k’a[N2O2] – 2kb[N2O2][O2]
Does text have an error in this last eqn?
Proposed mechanism: 2 NO(g) + O2(g) d 2 NO2
Net rate of formation of N2O2 =
ka[NO]2 – k’a[N2O2] – 2kb[N2O2][O2]
• Solving this eqn involves solving a very
difficult differential eqn and gives a
complex expression
• This drives us to consider…
“The Steady-State Approximation”
(265)
The steady-state approximation
• Our assumption: the concentrations of
all intermediates remain constant and
small throughout the rxn (except at the
very beginning and at the very end)
• So, the net rate of formation of N2O2 = 0
• So, ka[NO]2 – k’a[N2O2] – 2kb[N2O2][O2] = 0
• Solving for [N2O2]
• Solving for [NO2] gives 11.10, which
accounts for overall 3rd order rxn
(265)
The rate-determining step
• If step 3 is much faster than steps 1
and 2, (this is the case if the [O2] is
high) then k’a can be ignored
• Now the rate law simplifies to 11.12
• [NO2] =
2kakb[NO]2[O2]
kb [NO]2[O2]
= 2ka[NO]2
• Now the rxn is 2nd order and [O2] does
not appear in the rate law
The rate-determining step
The rate determining step (RDS)
must have two characteristics:
1. It must be the slowest step, and
2. It must be a crucial part of the path
from reactants to products
The rate-determining step
• Fig 11.7 (266)
• Diagrammatic
illustration of
the RDS
• Heavy lines =
fast steps
• Thin lines =
slow steps
The rate-determining step
• Fig 11.7 (266)
• Heavy lines =
fast steps
• Thin lines =
slow steps
• In (b), the thin
line is not a
bottleneck
because there
is an alternate
route
Kinetic control
Consider these competing rxns:
• A + B d P1
Rate of formation of P1 = k1[A][B]
• A + B d P2
Rate of formation of P2 = k2[A][B]
[P2]
[P1]
=
k2
k1
 Represents kinetic control
Kinetic control
If a rxn is allowed to reach equilibrium,
then thermodynamics (rather than
kinetic considerations) determines the
proportions of products, and the ratio
of concentrations is controlled by
standard Gibbs energies of reactants
and products
(267)
Reaction profile
Fig 11.7 (267)
Multistep
mechanism in
which the first
step is the RDS
(highest Ea)
Unimolecular reactions
• Many gas-phase rxns proceed by two gas
molecules colliding but are 1st order
• These collisions are bimolecular
events and the rxn should be 2nd order
• Explain!
• The Lindemann mechanism posits that
the RDS is unimolecular
(267f)
Activation control & diffusion control
Now we consider
reactions in solution
Activation control & diffusion control
• kd is the rate constant (of diffusion)
• Diffusion-controlled limit—the rate of
the rxn is controlled by the rate at
which reactants diffuse (kd)
• Activation-controlled limit—the rate of
the rxn depends on the rate at which
energy accumulates in the “encounter
pair” (ka)
Activation control & diffusion control
• Now read next-to-last paragraph, p.269
• “A lesson to learn from this analysis is
the concept of the rate-determining step
is rather subtle. Thus, in the…”
Activation control & diffusion control
• Vis-a-vis the rate of diffusion in a liquid,
the rate constant (kd) is related to the
coefficient of viscosity, 
• How are kd and  related?
8RT
kd =
3
•  is the Greek lowercase letter eta
Diffusion
This section relates directly to the
lab experiment you’ve worked on
this semester
(270)
Diffusion
• Terms: diffusion, random walk, flux (J)
• Rate of diffusion  Concentration
gradient
Number of particles passing through a window
• J=
Area of window  Time interval
• J = – D  Concentration gradient (11.18b)
• D is the diffusion coefficient (area/time)
• This is Fick’s first law of diffusion
Diffusion
• Fig 11.9 (271)
• The flux of
solute particles
is proportional
to the concentration gradient
• Gradient is
negative; flux
is positive
Diffusion
• Fig 11.18 (281) Further Information
Diffusion
• Diffusion is often aided by convection
• The diffusion eqn below lets us predict the
rate at which the concentration of a solute
changes in a “non-uniform” solution
• Fick’s second law:
Rate of change of concentration in a region =
D  curvature of the concentration in the region (11.19)
• So, what’s a “wrinkle”?
271
Diffusion in a non-uniform solution
• Fig 11.10 (272)
• Nature abhors
a wrinkle
• The phrase
“non-uniform
solution” is an
oxymoron
Table 11.1 Diffusion coefficients at 25°C, D/(10-9 m2 s-1)
Ar in tetrachloromethane
3.63
C12H22O11 (sucrose) in water
0.522
CH3OH in water
1.58
H2O in water
2.26
NH2CH2COOH in water
0.673
O2 in tetrachloromethane
3.82
(270)
Diffusion in water
• Fig 11.11 (273)
• Temperature
dependence of the
viscosity of water
• Water becomes
more fluid as
temperature
increases
• How does this affect
diffusion of solute
particles?
Catalysis
• Homogeneous catalysis, catalyst and
the rxn mixture are in the same phases
(e.g., gas, solution, etc.)
• Heterogeneous catalysis, catalyst and
the rxn mixture are in different phases
• (e.g., hydrogenation on a Pt surface)
273
Catalysis
• Fig 11.12 (273)
• Catalysts
function by
providing an
alternate
pathway with a
lower Ea
• Does a catalyst
speed up both
forward and
reverse rxns?
Enzyme-catalyzed reactions
• Fig 11.13
• The MichaelisMenten model
• When [S] < KM,
the rate is  [S]
• When [S] > KM,
the rate is independent of [S]
274
Enzyme-catalyzed reactions
• Fig 11.14
• Lineweaver-Burk
plot of an enzymecatalyzed rxn
• KM is a special
case of Keq
Analyzing a Lineweaver-Burk plot
• Fig 11.15 (276)
• Part of Ex 11.1
Enzymes
• Enzymes are biological catalysts
• Virtually all enzymes are proteins
(polyamino acids)
• A few are RNAs
• Since you are not biology majors…
Enzymes
• Fig 11.16
• In competitive
inhibition, the
substrate and
inhibitor
compete for
the active site.
Enzymes
Fig 11.17
In one type of noncompetitive inhibition,
the inhibitor attaches
to the enzyme, but
not at the active site;
this changes the 3D
shape of the enzyme,
rendering it inactive
The structure of chain reactions
• An intermediate in one step generates
a reactive intermediate in a later step,
which generates another reactive
intermediate in a still later step, which
generates….
• These are called “chain carriers”
The structure of chain reactions
•
•
•
•
•
•
Initiation step
Propagation step
Branching step
Retardation step
Inhibition step
Termination step
The rate laws of chain reactions
• Often free radical rxns
• Often have complicated rate laws
Key
Concepts
Key
Concepts
Key
Concepts
The End
…of this chapter…”
Explosions
• Part of Box 11.2 (278)
Let’s derive the equilibrium constant from first principles
aA + bB
cC + dD
ratef  [A]a[B]b
rater  [C]c[D]d
ratef = kf [A]a[B]b
rater = kr [C]c[D]d
Law of Mass Action
Now define equilibrium: at equilibrium, ratef = rater
If ratef = rater then kf [A]a[B]b = kr [C]c[D]d
Collect constants on the left and variables on the right:
[C]c[D]d
kf
=
kr
[A]a[B]b
OR
Kc =
[C]c[D]d
[A]a[B]b
15.1