2. 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
3. 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]
4. 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 measure-
ment of reaction rates.
5. Change of Concentration in Time
• 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. 0
0
Time
Concentration
AB
[A]0
[A]
[B]
t
6. [A]
[B]
0
0
Time
Concentration
AB
[A]0
t
A B Reaction Rate
• 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]
7. aAbB Rxn Rate
• d[A]/dt = – (a/b)d[B]/dt
is the new stoichiometric
condition.
– Now neither differential
is “the reaction rate.”
– But we can fix this by …
• Rate – (1/a) d[A]/dt
– Because that equals …
• Rate = + (1/b) d[B]/dt
[A]
[B]
0
0
Time
Concentration
aAbB
t
[A]0
8. 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.
9. 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 [A] [A] 2[A]
0 .3000
.5
1 .2714
1.5
2 .2456
–.0286
–.0258
–.0028
–.0014
–.0300
10. 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.
11. 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]0
1–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.
12. 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]0
m) [A]n
13. 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].
14. 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.
15. # 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.
16. The Big Three
• 0th Order: d[A]/dt = – k0 [A]0 = – k0
– or [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
17. Integrated Law Curve Shapes
(same values of k and [A]0)
[A]0
t
0
0
0th order
1st order
2nd order
2t½
t½
½
(½)²
1st order trick:
Curve falls by equal
factors in equal times.
[A] linear with t
confirms 0th order.
Slope = – k
18. Confirming 1st Order
0 t
ln[A]0
ln[A]
0th order
2nd order
A straight line in ln[A] vs. t
1st order!
ln½[A]0
t½ = (ln2)/k
Slope = – k
19. Confirming 2nd Order
0 t
1/[A]
1/[A]0
2nd order!
0th order
1st order
A straight line in 1/[A] vs. t
Slope = k
20. 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.
21. 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.
22. 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!
23. 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!
24. 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.
25. 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.
26. 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?
27. 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.
31. 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
32. 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)
33. Chemical Bobsledding
• The trajectories match
a bobsled run.
• So you can use your
dynamical instincts to
guess the outcome of
collisional encounters!
• E.g., what would a
bobsled coming from
the left do?
H2 + Br
H + HBr*
Lots of HBr
vibration.
(H<0)
34. Forcing Endothermic Reactions
• Since very exothermic
rxns make vibration,
how do we best force
them in reverse?
• Supply vibration in the
endothermic reactants!
H2 + Br
H + HBr*
(H>0)
35. “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.
36. Chemical Reaction Coordinate
• 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,
Ea , and its geometry an
activated complex, ‡.
A+BC AB+C
ABC‡
37. 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.
Ea
H
‡
reactants
products
E
38. Origin of Activation Energy
• In the reaction A+BCAB+C, we have
broken the B:C (Lewis) bond and formed
the A:B one.
– This means that electron spins were A+ BC
and became AB+C.
– But at ‡, they were , implying that
was antibonding even as the bonding slipped
from BC to AB.
39. 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]).
40. 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) ]
41. Ea and Molecular Remainders
– In order to simplify reaction dynamics, we have
reduced reactions to A+BCAB+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?
42. 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!
43. 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.
44. 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’
45. 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.
46. 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.
47. 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.
48. Kinetics of Enzyme Catalysis
– Enzyme+Substrate ESProducts+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
49. 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.
