Theoretical Models for Chemical Kinetics
•Collision Theory - highly energetic molecular
collisions produce chemical reactions.
•Transition State Theory - examines changes in
molecular structure during a reaction. It postulates an
“activated complex” in transitory equilibrium with
reactants. TS theory provides a powerful way to
think about chemical reaction kinetics.
•Thus far we have derived rate laws and rate constants on the basis of empirical
observations of macroscopic properties. These facts do not rely upon, or require,
any knowledge or theory of molecular structure.
•To understand chemical kinetics, we need to consider what is happening at the
molecular level. The temperature dependence of chemical reactions provides a
useful clue.
Arrhenius Equation
• In 1889 Svante Arrhenius observed that the rate constant for
chemical reactions fit the following equation.
k = Ae
-Ea /RT
•R =8.314 Jmol-1K-1
•T =temp in Kelvin
•A is the pre-exponential factor and carries the units of k
•Ea is the activation energy in kJ/mole
•Given k vs T we can obtain A and Ea OR given A & Ea we can
compute k at any temperature.
Diffusion and Activation
The Pre-Exponential factor A is related to collision frequency and
orientation.
Molecules have to get together in order to react. In solution or in the gas phase the
diffusion controlled rate constant is typically about 1011 M-1s-1 rate = k[A][B]
For two gases (P = 1 atm) rateD = 1011[1/22.4)2] = 106 M/s
For two solutes at 1 M
rateD = 1011 [1]2 M/s
millimolar rate = 1011 [.001]2 = 105 M/s
Eact is related to the energetics of collisions.
Collisions must have enough energy to weaken or break bonds.
average kinetic energy = 1/2mv2 = 3/2 kBT (T = Kelvin)
Typically only a very small fraction of collisions are energetic enough for reaction
(1 in 1010 for gases at 1 atm would give rate = 10-4 M/s ).
Boltzmann population
•Kinetic Molecular Theory (6.7) tells us that the
average kinetic energy of molecules increases with
temperature AND is distributed as shown.
1/2mv2 = 3/2 kT : where v2 is avg squared
speed
-The minimum energy required for a
reaction to occur is indicated by the
arrow in the figure. The fraction of
molecules possessing this energy will be
greater at T2 than at T1
- Collisions continually re-establish the
distribution at constant T.
- at 298 K RT = NkT= 2.5kJ/mol
while Ea is typically 50-100 kJ/mol
* Don’t confuse Boltzmann’s constant
k = 1.38 x 10-23 J/K per molecule with
the rate constant. R is just a molar
version of Boltzmann’s. k = R/N
• The rate of reaction is also limited by the orientation
of molecules at the time of collision
Example: N2O(g) + NO(g) → N2(g) + NO2(g)
k = Zpe-Ea/RT where Z = collision freq. & p = fraction of
collisions with suitable orientation.
Transition State Theory
A+B
(Eyring 1935)
AB*
• The activated complex (AB*) is a transitory molecule. The reaction rate is
proportional to the concentration of activated complexes.
• TS theory lets us think about reaction rates much like equilibria. The higher
the energy (less stable) of the activated complex, the slower the reaction.
• The TS lies at a maximum in energy and does not have a finite lifetime. A
reactive intermediate lies at a local minimum and has a finite lifetime.
Example: N2O(g) + NO(g) → N2(g) + NO2(g)
Transition State Theory
•The red line is the
reaction progress. It
is not time and it is
not fraction reacted.
It traces the route
taken by a molecule
to get from reactants
to products. One
molecule makes the
trip in ~ 10-13 sec, the
duration of a sticky
collision.
•The pass (transition state) is the maximum altitude along the
minimum altitude path from Calgary to Vancouver. Trains do
not go over mountain peaks and neither do chemical reactions.
Transition State Theory
This N3O2 activated
complex lies at the
maximum energy
along this path. The
molecule gains
energy as we
break one bond
and releases it as
we form the new
bond.
Thermodynamics deals only with reactants and products.
Kinetics concerns only reactants and the transition state.
Applying TS Theory
Potential Energy Surfaces
- A topographical map displays contours of equal altitude allowing a 2
dimensional picture of altitude changes.
- Reaction progress can be represented on a multidimensional potential
energy surface. Advanced theoretical methods can compute this surface
for simple reactions such as : H2 + Br HBr + H
Multiple pathways, intermediates and TS’s can be identified on this
surface.
.
Reaction
Profile diagrams are a useful way of displaying and explaining the
energetics of chemical reactions. Try the following
• Show the effect of catalase on the reaction : H2O2 H2O + ½ O2
• Sketch the profile for the uphill reaction : H2O H2 + ½ O2
•
Does the reverse rxn take a different path or give a different TS than the
forward rxn?
Potential Energy Surface for
contours are energy in kcal/mole
H2 + Br HBr + H
1 = reactants 2 = TS 3 = products
PE vs reaction path for H2 + Br HBr + H
Ea = +18 kcal/mole
∆Ho = +15 kcal/mole
Applying the Arrhenius Equation
k = Ae
-Ea /RT
1. Graphical : A plot of lnk vs 1/T is linear:
ln(k) = ln A - Ea/RT
2. k is given at two temps :
k 2 Ea  1 1 
ln
=
ln (k1) = lnA - Ea/RT1
 - 
k1
R  T1 T2 
ln (k2) = lnA -Ea/RT2
note : ln(x/y) = ln x - ln y
3. Ea and k1 given; find k2 at T2.
•HINT: Use 1000/T for
kJ/mole answer. Make sure
the higher T has the larger k.
Example 1. N2O5 → N2O4 + ½ O2
Plot of ln k versus 1/T : slope of -Ea/R and
intercept = ln(A)
(see Lab # 1)
Ea = - slope X R in J/mol
Ea = 12000 X 8.314/1000
= 106 kJ/mol
Units K X J/mol/K = J/mol
J/mol / 1000 J/kJ= kJ/mol
remember T decreases as 1/T
increases and rates almost
always increase with T.
*For log(k) : Ea = -slope x 2.303 R
Example 2. Cricket chirping roughly doubles for every 10 °C
increase in temperature. What Ea does this correspond to?
k 2 Ea  1 1 
ln
=
 - 
k1
R  T1 T2 
ln 2 = Ea/R [ 1000/300 -1000/310)]
Ea = .693 X 8.314 / (3.333-3.225)
Ea = 53.4 KJ/mol.
HINT#1: use 1000/T to convert to kJ/mole and
avoid math blunders.
HINT#2: You need to pick a reasonable
temperature to complete the problem.
HINT #3: A ratio can often remove a variablein this case A is not required.
Reaction Mechanisms
• The detailed step-by-step pathway by which a
reaction occurs is called the reaction mechanism
• a plausible reaction mechanism must be consistent
with the
1) stoichiometry of the overall reaction
2) experimentally determined rate law
• The steps in a mechanism are called elementary
reactions.
Elementary Reactions
1) The number of reactant molecules involved in an elementary
rxn is called the molecularity. Examples:
unimolecular
H2 → 2H
rate = k [H2]
bimolecular
H + H → H2
rate = k[H]2
CH3I + OH- → CH3OH + I-
rate = k[CH3I][OH-]
2) Note that for elementary reactions the molecularity is the same
as the kinetic order in the rate law.
3) Mechanisms typically consist of several elementary steps and
the kinetic order is often not related to the overall reaction
stoichiometry.
4) intermediates produced in an elementary reaction do not appear
in the net chemical reaction or the rate law. Intermediates are
produced by one elementary reaction and consumed by another.
d[I]/dt = 0 is the steady state approximation.
5) The rate of the overall reaction is largely determined by the
slowest step- the rate-determining step = RDS.
TIP: When deriving the rate law for a given mechanism:
1. First find the RDS and write the rate expression for it.
2. Then replace any species which is not a primary reactant by
using equilibrium relations or by using the steady state approx.
for intermediates. The rate law should only include the “stuff”
you measured out- not something produced after mixing.
3. Ignore entirely, fast reactions occurring after the RDS.
4. The kinetic orders tell you how many of each reactant are
involved before and including the RDS.
5. Reversible steps will require that you use #2.
Example 1: H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)
rate of reaction = kexpt [H2][ICl]
Consider the following mechanism:
(1) Slow:
k1
H2 + ICl → HI + HCl
bimolecular
k2
(2) Fast:
HI + ICl → I2 + HCl
Overall: H2 + 2 ICl → I2 + 2 HCl
From rds Rate = k1 [H2][ICl]
bimolecular
consistent with stoich.
consistent with experiment.
* fast steps following rds are irrelevant to the rate law !
Example # 1
H2 + 2 ICl → I2 + 2 HCl
• distinguish reaction intermediates and transition states (activated complexes)
Example 2.
2 NO(g) + O2(g) → 2 NO2(g)
rate =- d[O2]/dt = -1/2 d[NO]/dt = k[NO]2[O2]
A one-step termolecular process is highly unlikely.
Instead consider a rapid preequilibrium giving a small amt of N2O2
Fast:
Slow:
2 NO
k-1
k1
N2O2
k2
N2O2 + O2 → 2 NO2
Overall: 2 NO + O2 → 2 NO2
Rate = k2[N2O2][O2] = k1k2/k-1 [NO]2[O2]
At equilibrium: rate
forward = rate reverse
k1[NO]2 = k-1 [N2O2]
[N2O2] = k1/k-1 [NO]2
Example #3.
Cl2 + CHCl3 HCl + CCl4
k1 = 4800 s-1
k-1
Cl2
2 Cl
k1
k2
Cl + CHCl3 HCl + CCl3
k3
CCl3 + Cl CCl4
Cl2 + CHCl3 HCl + CCl4
P-15-95
k-1 = 3600 M-1s-1
k2 = 0.013 M-1s-1
k3 = 270 M-1s-1
Cl atoms recombine much more
rapidly than they attack chloroform.
Rapid pre-eq. with k2 the rds.
rate = k2 [CHCl3][Cl]
[Cl]2 = [Cl2]k1/k-1 from pre -eq
rate = k2 (k 1/k-1)1/2[Cl2]1/2 [CHCl3]
Steady-State Approximation
The rapid pre-equilibrium is a special case of a more general
approach to intermediates.
Very reactive intermediates never build up to significant
concentrations because they are consumed as rapidly as they are
produced. Consider the O atom in the mechanism below. O is
formed in step 1 and removed in step 2 .
k-1
O3 O2 + O
d[O]/dt = k1 [O3] - k-1 [O][O2] - k2[O][O3] = 0
k2
O + O3 2 O2
2 O3 3 O2
solve for [O]: [O] = k1[O3]/(k-1[O2] + k2[O3])
and plug into : rate = k2 [O3][O] = k1k2[O3]2 / (k-1[O2] + k2[O3] )
at low [O2] rate = k1[O3] rate depends only on ozone fission
at high [O2] rate = k1k2/k-1[O3]2/[O2] •a result we get by assuming
a rapid pre-equilibrium.
Steady-State Approximation
Applying the steady state approx to example # 2 we obtain the more
general rate law which reduces to the simpler form if k-1 >> k2 [O2].
Fast:
k-1
2 NO k
1
Slow:
N2O2
k2
N2O2 + O2 → 2 NO2
Overall: 2 NO + O2 → 2 NO2
d[N2O2]/dt = 0 = k1[NO]2 -k-1[N2O2] - k2[N2O2][O2]
N2O2] = k1[NO]2/ (k-1 + k2[O2] )
Rate = k2[N2O2][O2] = k1k2[NO]2 [O2]/(k-1 + k2[O2])
Catalysis
• A catalyst speeds up a reaction by providing an alternate
reaction pathway with a lower activation energy.
• A catalyst has NO EFFECT on the thermodynamics of the
overall reaction.
A catalyst participates in a chemical reaction, but is neither generated nor consumed.
Stratospheric NO from the SST may threaten the ozone layer.
O3 + hV O2 + O photochemical rxn. in stratosphere
O3 + O 2 O2
bimolecular thermal process
NO catalyzes the thermal process
NO + O3 NO2 + O2
NO2 + O NO + O2
net
O3 + O 2 O2
does not include the catalyst.
* molecules with an odd number of electrons are called radicals. Radicals are usually
very reactive. NO & NO2 are rare examples of stable radicals.
Homogeneous Catalysis - single phase (eg. gas or aqueous)
examples : Mn2+ catalyzes H2O2 decomposition to O2 and H2O
carbonic anhydrase catalyzes H2CO3 CO2 + H2O
NO catalyzes ozone destruction
metal ions catalyze many oxidation processes
many organic reactions are subject to acid or base catalysis
Homogeneous Catalysis
1. Acid-Catalyzed Decomposition of Formic
Acid
HCOOH(aq) → H2O(l) + CO(g)
• in uncatalyzed reaction, H atom must move from one part
of the HCOOH molecule to another before the C-O bond
can break - high activation energy for this atom transfer
• in the catalyzed reaction, H+ from solution can add
directly to this position - lower activation energy
Acid Catalysis
HCOOH → CO + H2O
Rate = k [HCOOH]
K
HCOOH + H+
k2
HCOOH2+
fast
HCOOH2+ → HCO+ + H2O slow
kcat >> k
HCO+ → CO + H+
rate = Kk2 [HCOOH][H+]
fast
Enzymatic Catalysis
P450 catalyzes the oxidation
of camphor by O2. The
polypeptide chain is
shown as ribbons for
easier viewing. The exact
location of every atom
has been established by
X-ray diffraction.
Life relies on the exquisite
control of reaction rates
made possible by the
thousands of enzymes
that make up the human
genome.
“What in the world isn’t
chemistry?”
Enzyme Kinetics
• the Michaelis Menton mechanism is common to
many enzyme-catalyzed reactions:
k
E+S
ES
k
1
-1
k2
ES
E+P
E - enzyme, S - substrate, P - product
• the steady-state approximation applied to ES yields
(after some manipulation) the following rate law:
k2 [E 0 ][S]
rate of reaction =
K + [S]
[E0] = [E] + [ES] = Etotal and K = (k-1+ k2)/k1
k2 [E 0 ][S]
rate of reaction =
K + [S]
• at low [substrate], K >> [S]
the rate is first order in [S] (rate
= k2/K [S] [Eo]
• at high [substrate], [S] >> K
the rate is zero order in [S].
(rate = k2[E0])
the rate no longer depends on
[S] . We have a unimolecular
rxn of ES.
Heterogeneous Catalysis
Solid surfaces- especially metals or metal oxides are important
heterogeneous catalysts.
- Pt metal catalyzes olefin hydrogenation:
H2 + H2C=CH2 C2H6
- a Pt-Rh catalyst is used in the Ostwald process for nitric acid
production:
Pt-Rh
H2O
O2
NH3 + O2 NO NO2
HNO3
in this case the catalyst speeds up oxidation to NO, otherwise the
more favorable oxidation to N2 would predominate. The ability to
control rates either by conditions or catalysis is often crucial to
obtaining an optimal result.
Surface Characteristics
Solids have a definite interior structure but at the surface atoms have incomplete
valencies. These sites are where molecules are absorbed.
Many solids acquire a coating of oxide, hydroxide, or water and are not effective
catalysts. Many substances can “poison” a surface.
Since reactions occur only at the surface of solids, the surface area is an important
characteristic.
The smaller the particle size, the greater the surface area available for catalysis.
Porous solids (sponges) have large surface areas.
Heterogeneous Catalysis proceeds typically via 4 Basic steps.
1) adsorption of reactants. If bonds are broken it is called chemisorption.
2) diffusion of reactants along the surface is accomplished by hopping to adjacent
surface sites.
3) reactions occur on the surface to form adsorbed products
4) desorption of products releases the products and frees up a surface site.
2 CO(g) + 2 NO(g) → 2 CO2(g) + N2(g)
A car’s catalytic converter makes use
of RXNS on a Rh surface
a) adsorption of CO and NO
b) diffusion and dissociation of NO
c) combination of CO and O to form
CO2, N atoms to form N2, along with
desorption of products
Explosions- kinetics out of control
Thermal Explosions - involve reactions which are highly exothermic under
conditions where the heat is not dissipated. Thus the rate increases rapidly along
with the temperature.
- In 1935 a barge loaded with FGAN exploded in Bay City Texas destroying the
city, killing 511. Huge parts of the barge were found over a mile inland. FGAN
was also the primary component of the Oklahoma City bomb.
{FGAN = fertilizer grade ammonium nitrate}
NH4NO3 N2 + 2 H2O + ½ O2
Chain Branching - reactions which produce more reactive intermediates
than they consume produce uncontrollable increases in rate.
Nuclear Fission : 235U + n X + Y + 3 n + Energy
Hydrogen Explosions H2 2 H
initiation
H + O2 OH + O branching
branching
O + H2 OH + H
OH + H2 H2O + H propagation