•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.

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.

Arrhenius Equation

• In 1889 Svante Arrhenius observed that the rate constant for

chemical reactions fit the following equation.

a-E /RTk = Ae

•A is the pre-exponential factor and carries the units of k

•Ea is the activation energy in kJ/mole

•R =8.314 Jmol-1K-1

•T =temp in Kelvin

•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

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

•Kinetic Molecular Theory (6.7) tells us that the

average kinetic energy of molecules increases with

temperature AND is distributed as shown. * 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 (Eyring 1935)

A + B 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 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.

•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.

Transition State TheoryThis 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 H2 + Br � HBr + H contours are energy in kcal/mole 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

a-E /RTk = Ae

1. Graphical : A plot of lnk vs 1/T is linear:

ln(k) = ln A - Ea/RT

2. k is given at two temps :

ln (k1) = lnA - Ea/RT1

ln (k2) = lnA -Ea/RT2

note : ln(x/y) = ln x - ln y

3. Ea and k1 given; find k2 at T2.

a2

1 1 2

Ek 1 1ln = -

k R T T

•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?

a2

1 1 2

Ek 1 1ln = -

k R T T

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 variable-

in 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 theslowest 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: H2 + ICl → HI + HCl bimolecular

(2) Fast: HI + ICl → I2 + HCl bimolecular

Overall: H2 + 2 ICl → I2 + 2 HCl consistent with stoich.

From rds Rate = k1 [H2][ICl] consistent with experiment.

* fast steps following rds are irrelevant to the rate law !

k1

k2

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: 2 NO � N2O2

Slow: 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

k2

k1

k-1

Example #3. Cl2 + CHCl3 � HCl + CCl4

Cl2 � 2 Clk1

k-1

Cl + CHCl3 � HCl + CCl3

CCl3 + Cl � CCl4

Cl2 + CHCl3 � HCl + CCl4

k2

k3

k1 = 4800 s-1

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]P-15-95

Steady-State ApproximationThe 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 .

O3 � O2 + O d[O]/dt = k1 [O3] - k-1 [O][O2] - k2[O][O3] = 0

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]

at high [O2] rate = k1k2/k-1[O3]2/[O2] •a result we get by assuming

a rapid pre-equilibrium.

rate depends only on ozone fission

k2

k-1

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: 2 NO � N2O2

Slow: 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])

k2

k1

k-1

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

AcidHCOOH(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 + H+ HCOOH2+ fast

HCOOH2+ → HCO+ + H2O slow

HCO+ → CO + H+ fast

rate = Kk2 [HCOOH][H+]

HCOOH → CO + H2O

Rate = k [HCOOH]

kcat >> k

K

k2

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

E - enzyme, S - substrate, P - product

• the steady-state approximation applied to ES yields

(after some manipulation) the following rate law:

[E0] = [E] + [ES] = Etotal and K = (k-1+ k2)/k1

• the Michaelis Menton mechanism is common to

many enzyme-catalyzed reactions:

E + S ES

ES E + P

0[E ][S]

rate of reaction =+[S]

2k

K

k1

k-1

k2

• 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.

0[E ][S]

rate of reaction =+[S]

2k

K

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:

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.

O2H2OPt-Rh

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

O + H2 � OH + H branching

OH + H2 � H2O + H propagation