Zumdahl’s Chapter 12
Chemical Kinetics
Chapter Contents
• Introduction
• Rates of Reactions
• Differential Reaction
Rate Laws
• Experimental
Determinations
– Initial Rates
– Saturation Methods
• Integrated Rate Laws
–
–
–
–
0th Order
1st Order & ½ Life
2nd Order
Multiple Reactants
• Reaction Mechanisms
• Models for Kinetics
• Catalysis
It’s déjà vu all over again …
• Kinetics of processes have appeared before:
– Kinetic Theory has been invoked several times.
• In the origin of pressure …
• As van der Waal’s pressure correction, (P+a[n/V]2)
– [n / V]2 is a concentration dependence on collision rates
– As a justification for Raoult’s Law …
– In the development of the Mass Action Law …
• kf[A][B] = kr[C][D]  K = kf/kr = [C][D]/[A][B]
A two-pronged approach
• The speed with which chemical reactions
proceed is governed by two things:
– The rate at which reactants come into one
another’s proximity (“collide”) and
– The probability that any given collision will
prove effective in turning reactants to products.
• We look first at the macroscopic measurement of reaction rates.
• Reactants vanish in
time, so [reactant] is a
falling function of t.
• Likewise [product] is
a rising function of t.
• The shape of these
functions tells us
about concentration
dependence.
Concentration 
Change of Concentration in Time
[A]0
AB
[B]
[A]
0
0
t Time 
Concentration 
A  B Reaction Rate
[A]0
AB
[B]
[A]
0
0
t Time 
• Stoichiometry requires
d[A]/dt = – d[B]/dt
• But d[A]/dt can itself
be a function of time.
– It falls rapidly initially.
– Then it approaches its
equilibrium value, as
[A] on the graph,
asymptotically.
– K = [B] / [A]
aAbB
aAbB Rxn Rate
• Rate  – (1/a) d[A]/dt
Concentration 
• d[A]/dt = – (a/b)d[B]/dt
is the new stoichiometric
condition.
– Because that equals …
0
– Now neither differential
is “the reaction rate.”
– But we can fix this by …
• Rate = + (1/b) d[B]/dt
[B]
[A]0
[A]
0
t Time 
aA + bB  cC + dD
– If z is the stoichiometric coefficient of the general
compound Z, and z takes on positive signs for
products and negative signs for reactants:
• Rate = (1/z) d[Z]/dt is “rate of reaction,” M/s
• d[Z]/dt is easy if Z=f(t) is known, but it isn’t.
• All we can measure is [Z]/t and (use the
Fundamental Theorem of Calculus to) approximate
d[Z]/dt as [Z]/t as t 0.
Estimating Experimental Rates
• For reasons soon
apparent, we will
often want the t=0
value of d[A]/dt.
• That requires an
extrapolation of
A/t to t=0 where
it is varying rapidly!
t
0
[A]
–.0286 –.0014
.2714
1.5
2
2[A]
.3000 –.0300
.5
1
[A]
–.0028
–.0258
.2456
Why d[A]/dt at t = 0?
• Ask the question the other way around:
– At t > 0 are there additional complications?
– Sure! At the very least, the reverse reaction of
products to produce reactants changes the rate
of loss of A. An added headache.
• Also [A] is changing most rapidly at t = 0,
minimizing the “small difference of large
numbers” error.
Simplified Rate Laws
– Not “laws” like “Laws of Thermodynamics”
but rather rate “rules” for simple reactions.
• Two versions of the Rate Laws:
– Differential like d[A]/dt = – k [A]n
– Integral like [A]1–n = [A]01–n + (n – 1) kt
• But they must be consistent for the same reaction.
– As these happen to be … iff n  2 of course.
• Rate exponents are often not stoichiometric.
Simplified INITIAL Rate Laws
• Since products are absent at t=0, such laws
include only rate dependence on reactants.
• Simple reactions often give power rate laws.
• E.g., Rate = – (1/a) d[A]/dt = k [A]n [B]m
• The n and m are often integers.
• A’s dependence is studied in excess [B],
since [B]0 will be fixed! So (k[B]0m) [A]n
Reaction Rate Orders
• Rate = k [A]n [B]m
– The n and m are called the “order of the
reaction” with regard to A and B, respectively.
– The reaction is said to have an overall order, O,
that is the sum of the species’ orders, e.g., n+m.
– The significance of overall order is simply that
increasing all [species] by a factor f increases
the reaction rate by a factor f O.
• We find a species’ order by changing only [species].
Determining Reaction Order
• If we use only initial
rates, all [species]
remain at [species]0.
• Then by fixing all
[species] except one,
we find its order by
knowing at least two
initial rates where its
concentrations differ.
[A]
[B]
k [A]n[B]m
0.1
0.1
0.5 M/s
0.2
0.1
2.0 M/s
0.2
0.2
4.0 M/s
• This data is consistent with
n = 2 and m = 1, and we find
k = 500 M–2 s–1 as a bonus.
# expts. must match # unknowns
• In k [A]n [B]m, we had k, n, & m unknown.
• So we needed at least 3 experiments.
• More if we want self-consistency checks!
• This is just like linear equations, in fact:
– ln(k[A]n[B]m) = ln(k) + n ln([A]) + m ln([B])
– So we’ll need at least 3 ln(Rate) experiments in
order to find n, m, and ln(k) unambiguously.
The Big Three
• 0th Order:
– or
d[A]/dt = – k0 [A]0 = – k0
[A] = [A]0 – k0 t
• 1st Order:
d[A]/dt = – k1 [A]1
– or
[A] –1 d[A] = d ln[A] = – k1 dt
– hence ln[A] = ln[A]0 – k1 t
• 2nd Order:
d[A]/dt = – k2 [A]2
– or
[A] –2 dt = – d [A] –1 = – k2 dt
– hence [A] –1 = [A]0–1 + k2 t
Integrated Law Curve Shapes
(same values of k and [A]0)
[A]0
1st order trick:
Curve falls by equal
factors in equal times.
½
2nd order
0th order
(½)²
[A] linear with t
confirms 0th order.
0
0
t½
Slope = – k
1st order
2t½
t
Confirming
st
1
Order
ln[A]0
A straight line in ln[A] vs. t
ln½[A]0
2nd order
ln[A]
Slope = – k
0th order
1st order!
0
t½ = (ln2)/k
t
Confirming
1/[A]
nd
2
Order
1st order
0th order
2nd order!
Slope = k
A straight line in 1/[A] vs. t
1/[A]0
0
t
Caveat
– The 0th Law plot showed [A]0 which
presumes there is no reverse reaction. (The
reaction is quantitative.)
• Indeed all these plots ignore all reactants,
products, and intermediates except A.
– In reality, these shapes can be trusted only
under conditions of initial rate and where A is
overwhelming the limiting reactant.
Multiple Reactants
• What about A + B  P? Rate = k2[A][B]
• where P is any combination of products.
– What’s an integrated law for d[P]/dt=k2[A][B]?
– By stoichiometry, d[A]/dt = d[B]/dt = – d[P]/dt
– Via those substitutions, we can produce …
• kt = { 1 / ([B]0 – [A]0) } 
ln{ [A]0([B]0 – [P]) / [B]0([A]0 – [P]) }
• where “[Z]0 – [P]” is merely [Z] at time t.
What Lies Beneath?
• Reaction orders are most often not equal to
the stoichiometric coefficients because our
reactions proceed in a series (called the reaction
mechanism) of elementary steps!
– If we stumble upon a reaction whose molecules
collide and react exactly as we’ve written it in
one go, the orders are the molecularity, and the
rate can be written from the stoichiometry!
Elementary Steps
– Real reactions most often proceed through
reactive intermediates, species produced in
disappearing when equilibrium is reached. early
steps and consumed in later ones,
• These steps add up to the overall reaction
which never shows the intermediates.
– The rate expressions of elementary steps are
always of the form: k[A]n[B]m… n, m, integer!
Guessing Reaction Mechanisms
– More often than not, we know only what’s in
the overall reaction; the intermediates and thus
the mechanism are a mystery.
• So we postulate a mechanism and confirm
that’s its overall rate matches our reaction’s.
– But many mechanisms meet that criterion!
– We can hunt for evidence of our postulate’s
intermediates in the reacting mixture.
Importance of the Mechanism
• It gives us control!
(insert maniacal laughter here)
– If we know precisely how a reaction proceeds,
we can take steps to enhance or inhibit it!
• To inhibit it, we might add a “scavenger” molecule
that consumes an intermediate efficiently.
• To enhance it, we include extra [intermediate] in the
mixture, assuming it’s a stable species.
• But intermediates are often highly reactive and even
radicals like the •OH in smog chemistry.
Mechanistic Example
• 2 NO + O2  2 NO2 has rate k [NO]2 [O2]
– Might it be elementary? It’s consistent!
– But the T dependence of k suggests otherwise.
– How about a 2-step mechanism (steps a & b)…
• 2 NO  N2O2 with Ka = [N2O2] / [NO]2
• N2O2 + O2  2 NO2 with kb [N2O2] [O2]
– It adds up all right, but what’s the overall rate?
Rate from Mechanism
• N2O2 + O2  2 NO2 has a rate expression
kb [N2O2] [O2], but what’s [N2O2] ?
• If the equilibrium in step a is really fast, it
will be maintained throughout the reaction.
• [N2O2] = Ka [NO]2 can be exploited.
• So step b is (kb Ka) [NO]2 [O2] as hoped.
– And the T dependence turns out OK.
Chain Reactions
• H2 + ½O2  H2O goes by chain reaction:
–
–
–
–
–
–
H2 + O2  HO2• + H•
H2 + HO2•  HO• + H2O
H2 + HO•  H• + H2O
H• + O2  HO• + •O•
•O• + H2  HO• + H•
H• + HO• + M  H2O + M*
initiates
propagates
propagates
branches!
branches!
terminates
Chemical Reaction Potentials
• A + BC  AB + C
• At large RAB, V = VBC
V
RBC
Chemical Reaction Potentials
• A + BC  AB + C
• At large RAB, V = VBC
• At large RBC, V = VAB
Chemical Reaction Potentials
•
•
•
•
A + BC  AB + C
At large RAB, V = VBC
At large RBC, V = VAB
At molecular distances
V is a hypersurface
potential for the ABC
complex.
AB +C
A+BC
D• + H2  DH + H•
• Chemical reaction
potentials have slopes
–dV/dR that are forces
guiding the nuclei.
• Time evolution of
nuclear positions trace
trajectories across the
hypersurface.
• if Isaac Newton’s right
from C.A. Parr and D.G. Truhlar, J. Am. Chem. Soc., 75, 1884 (1971)
Chemical Bobsledding
• The trajectories match
a bobsled run.
H2 + Br
• So you can use your
dynamical instincts to
guess the outcome of
(H<0)
collisional encounters!
• E.g., what would a
bobsled coming from
Lots of HBr
the left do?
vibration.
H + HBr*
Forcing Endothermic Reactions
H2 + Br
• Since very exothermic
rxns make vibration,
how do we best force
them in reverse?
• Supply vibration in the
endothermic reactants!
(H>0)
H + HBr*
“Supplying” Vibration
• Vibration is a form of molecular energy.
• Heating a molecule increases its energy.
• But the Boltzmann distribution of energy
ensures that if a reactive vibrational level is
abundant, so too are dissociative levels!
• The surgical way to supply vibration is with
laser beams tuned to colliding molecules.
Chemical Reaction Coordinate
ABC‡
A+BC
• The geometries and
potential energies that
most efficiently lead to
products are called the
reaction coordinate.
• The highest potential
along this best path is
the activation energy,
AB+C
Ea , and its geometry an
activated complex, ‡.
Activation Energy Diagram
• While the previous
graphic shows the
origin of the reaction
coordinate in multiple
dimensions, it’s most
often given as E vs. .
• Reactants must have at
least Ea in order to
surmount this barrier.
‡
E
Ea
reactants
H
products

Origin of Activation Energy
• In the reaction A+BCAB+C, we have
broken the B:C (Lewis) bond and formed
the A:B one.
– This means that electron spins were A+ BC
and became AB+C.
– But at ‡, they were , implying that  
was antibonding even as the bonding slipped
from BC to AB.
Collision Model of Kinetics
• Rate = k [A] [B] depends upon how often A
meets B and how energetic is their collision.
• Svante “Aqueous Ion” Arrhenius predicted a
form for the rate constant k = A e–Ea / RT
– The Boltzmann term, e–Ea / RT, gives fraction of
collisions whose energy exceeds Ea.
– Arrhenius factor, A, measures frequency of
collision (when multiplied by [A] [B]).
Measuring Ea as a Slope
• Once reaction orders have been determined,
measured rates vs. T give measured k.
– Take natural log of the Arrhenius Equation:
• ln (k) = ln(A) – (Ea / R) ( 1/T )
– Déjà vu: –ln(k) varies with 1/T like K
– Subtracting ln(k1) from ln(k2) cancels lnA and
• ln(k2/k1) = (Ea/R) [ (1/T1) – (1/T2) ]
Ea and Molecular Remainders
– In order to simplify reaction dynamics, we have
reduced reactions to A+BCAB+C.
• What’s the effect of substituents attached to
these atoms? It must have some!
– In other words, the activated complex may be
(stuff)nA…B…C‡(other stuff)m where stuff may
have an effect on Ea.
– If so, can we take advantage of this?
Tinkering with Reaction Sites
• If changing stuff influences electron density
at the heart of A…B…C‡, preferably
weakening B:C while strengthening A:B,
we will lower Ea by lowering H! (cheat)
• But can we have a similar effect while
keeping stuff (and  the molecules and
their thermodynamics) exactly as they are?
– Yes!
Catalysis
• Instead of tweaking stuff on the molecules,
we can tweak just the complex, ‡, having A
meet BC in a molecular environment that
changes ‡’s e– distribution to advantage.
• When AB (and C) leave that catalytic
environment unchanged on their departure,
that is the essence of catalysis.
– Catalyst accelerates rxn w/o being consumed.
A Catalyst’s Dramatic Influence
‡
• Without the catalyst,
the reaction proceeds
slowly over ‡.
• In the presence of a
catalyst at ‡, the rxn
proceeds faster over
the now lowered Ea’.
– G and hence K are
the same either way!
‡
Ea’
H
Ea
Heterogeneous Catalysis
• Added advantages come to a solid catalyst
adsorbing liquid or gaseous reactants.
– Adsorption takes place on the catalyst’s surface
which is 2-d vs. reactants’ natural 3-d phase.
– Migrating on a 2-d (or, given irregularities, 1-d)
surface vastly improves chance of encounters!
– Surface can predissociate reaction site bonds.
– Reactant lone pairs fit in empty metal d shell.
Homogeneous Catalysis
• If instead the catalyst has the same phase as
the reactants, the dimensionality advantage
may be lost … unless
• Catalyst captures reactants in an active site
(like biological enzymes), and releases only
products.
– Sites can be phenomenally reactant-specific!
(Lock-and-key model.) Except for poor Rubisco.
Catalysts as Intermediates
• Homogeneous catalysts can also be
intermediates in reactions as long as they
are reproduced as efficiently as consumed.
• Atomic chlorine’s catalytic destruction of
ozone in the stratosphere:
Cl + O3  ClO + O2
ClO + O  Cl + O2
• Kills “odd oxygen” while maintaining catalytic Cl.
Kinetics of Enzyme Catalysis
– Enzyme+Substrate  ESProducts+Enzyme
• d[ES]/dt = ka[E][S] – ka’[ES] – kb[ES]  0
• [ES]steady state = [E][S] ka / (kb+ka’)
• But [E] = [E]0 – [ES] leads (collecting [ES] terms) to:
•  [ES]steady state = ka[E]0[S] / (kb+ka’+ka[S])
• d[P]/dt = kb[ES]ss = kb[E]0[S] / (KM+[S])
– KM = Michaelis-Menten constant = (kb+ka’)/ka
Catalysis of the Mundane
• Esoteric isn’t a prerequisite for a catalyst.
• Many reactions are catalyzed merely by
acid or base!
– This should come as no surprise because H+(aq)
or rather H3O+ bears a potent electrical field
that can influence neighboring electrons.
– And electron pushing is what Chemistry is all
about. 