Reaction Mechanisms

1
Reaction Mechanisms A simple or elementary
reaction is one that takes place in a single step
or collision. The molecularity of an elementary
reaction is the number of simultaneously
colliding or reacting species, i.e., the number
of reactants in the elementary reaction. What
is the molecularity of the elementary
reaction? NO (g) O2 (g) -----gt NO2 (g) O
(g) Elementary reactions are usually classified
as unimolecular, bimolecular, termolecular,
etc. Do you think that termolecular reactions are
common?
2

• Many, if not most, reactions are composite or net
  reactions that occur via a reaction mechanism,
  i.e., a sequence of elementary reactions which
  describe the molecular pathway by which the net
  reaction occurs. For example the formation of
  phosgene from the reaction of carbon monoxide and
  chlorine
• CO (g) Cl2 (g) --------gt COCl2 (g)
• is a composite reaction thought to occur via the
  mechanism
• Cl2 (g) ltgt 2 Cl (g)
• Cl (g) CO (g) ltgt COCl (g)
• COCl (g) Cl2 (g) ------gt COCl2 (g) Cl (g)
• The fundamental problem in chemical kinetics is
  to deduce the detailed molecular path by which a
  reaction occurs, i.e., its mechanism, from
  experimental rate data. The general approach is
• to postulate a mechanism based on reasonable
  chemical assumptions,
• to derive a rate law for the reaction based on
  the postulated mechanism,
• and finally to test the derived rate law
  against the experimental data.

3

• The following considerations need to be kept in
  mind when deriving a rate law from a proposed
  mechanism
• The development of the rate law usually begins
  with the rate determining reaction or step in the
  reaction mechanism, which is the slowest step in
  the set of reactions that comprise the mechanism.
  You can, if you want, relate the rate written in
  terms of some species in the rate determining
  step to other species in subsequent faster steps.
  Fast steps that follow the rate determining step
  are, however, often ignored.
• The rate law should ultimately be expressed
  only in terms of chemical species which appear in
  the net or composite reaction. Intermediates,
  which 1st appear as products in early mechanistic
  steps and are consumed as reactants in later
  mechanistic steps, should not appear in the final
  version of the derived rate law.
• One approach is to assume that all fast steps
  that preceed the rate determining step come to
  equilibrium and then to use the equilibrium
  constants for these steps to eliminate
  intermediates in the development of the rate law.
• A second approach is to write steady state
  expressions for intermediates and then use these
  expressions to eliminate these intermediates from
  the final rate law. In steady state the amount
  of a species is not changing with time and time
  derivatives involving this species can be set
  equal to zero.
• In any elementary reaction the order of the
  reaction with respect a particular chemical
  species is equal to the stoichiometric
  coefficient for that species in that elementary
  reaction.

4
Consider the mechanism for the formation of
phosgene from the reaction of carbon monoxide and
chlorine introduced earlier k1
Cl2 (g) ltgt 2 Cl (g)
k-1 k2 Cl (g) CO
(g) ltgt COCl (g) k-2 k3
COCl (g) Cl2 (g) ------gt COCl2 (g) Cl
(g) What role does COCl (g) play in this
mechanism? What is the role of Cl (g) in the last
two steps? Lets assume that the 3rd step is the
slow step and write the initial form of the rate
law in terms of the formation of phosgene d
COCl2 / dt k3 COCl Cl2 Since the 3rd
step is slow, the 1st two steps have time to
reach equilibrium. When the 2nd step is at
equilibrium we can write forward rate of step 2
reverse rate of step 2 k2 Cl
CO k-2 COCl This expression can be used
to eliminate the intermediate COCl from the
developing rate law d COCl2 / dt
k3 (k2 / k-2) CO Cl Cl2 k3 K2 CO
Cl Cl2 What is K2 in this expression?
5
Now use the equilibrium condition on step 1 to
eliminate Cl (g) from the rate law, since it does
not appear in the net reaction This rate
law is found to agree with the experimentally
determined rate law Z. Phys. Chem., 10399
(1922) d COCl2 / dt knet CO Cl23/2
Does agreement of the experimental data with
the rate law derived from the proposed mechanism
prove that the proposed mechanism is correct? If
you think so, derive the rate law for the
formation of phosgene using the mechanism Cl2
(g) ltgt 2 Cl (g) Cl (g) Cl2 (g) ltgt
Cl3 (g) CO (g) Cl3 (g) ------gt COCl2 (g)
Cl (g)
6
If the reaction of hydrogen and iodine to form
hydrogen iodide H2 (g) I2 (g) -----gt 2 HI
(g) proceeds by the mechanism k1
I2 (g) ltgt 2 I (g) (fast)
k-1 k2 I (g)
H2 (g) ltgt H2I (g) (fast) k-2
k3 H2I (g) I2 (g) ------gt 2 I (g)
I (g) (slow) what is the rate law for the
formation of hydrogen iodide? ½ d HI / dt ?
7
Benson and Axworthy J. Chem. Phys., 261718
(1957) proposed the following mechanism for the
thermal (as opposed to photochemical)
decomposition of ozone k1 O3 (g) M
(g) ltgt O2 (g) O (g) M (g)
k-1 k2 O (g) O3 (g) ------gt
2 O2 (g) What is the net reaction associated
with this mechanism? M, sometimes called a
chaperon, provides the energy needed for the
initial decomposition of ozone in step 1 and can
be any molecule in the gas phase (including those
involved in the reaction). An initial rate law
for the decomposition of ozone can be written
as d O3 / dt - k1 O3 M k-1 O2
O M - k2 O O3 Could you have produced
this expression? By assuming that the
intermediate O (g) is in steady state 0 d
O / dt k1 O3 M - k-1 O2 O M
- k2 O O3 we can obtain an expression for
O O k1 O3 M / ( k-1 O2 M k2
O3 ) that can be used to eliminate O from the
rate law to give d O3 / dt - 2 k1 k2
O32 M / ( k-1 O2 M k2 O3 ) Could
you derive this equation?
8
What form does this rate law take a high
pressures and how does the rate of decomposition
of ozone depend on the oxygen concentration under
this condition? What form does this rate law
take at low oxygen or high ozone pressures and
how does the rate of decomposition of ozone
depend on the oxygen concentration under this
condition?
9
For the mechanism k1 H HNO2
ltgt H2NO2 k-1 k2
H2NO2 Br- -------gt ONBr H2O
k3 ONBr fNH2 -------gt fN2
H2O Br- where f stands for a phenyl
group a. Write the overall reaction. b. Derive a
rate law based on this mechanism for the
formation of fN2, assuming that both H2NO2 and
ONBr are in steady state. c. What additional
assumption must be made in order to obtain
agreement with the experimental rate law d
fN2 / dt k H HNO2 Br- and what
does this imply about step 1 in the
mechanism? d. How is k in the experimental rate
law related to the rate constants in the
mechanism? e. What role does Br- play in the
mechanism? Sketch how the reaction rate would
be expected to vary with increasing Br-.
10

• Take Aways
• The fundamental problem in chemical kinetics is
  to deduce the detailed molecular path by which a
  reaction occurs, i.e., its mechanism, from
  experimental rate data. The general approach is
• to postulate a mechanism based on reasonable
  chemical assumptions,
• to derive a rate law for the reaction based on
  the postulated mechanism,
• and finally to test the derived rate law
  against the experimental data.
• A reaction mechanism is a sequence of elementary
  reactions which describe the molecular pathway by
  which the net reaction occurs.
• A simple or elementary reaction is one that takes
  place in a single step or collision.
• The molecularity of an elementary reaction is the
  number of simultaneously colliding or reacting
  species, i.e., the number of reactants in the
  elementary reaction.
• Elementary reactions are usually classified as
  unimolecular, bimolecular, termolecular, etc.

11

• The following considerations need to be kept in
  mind when deriving a rate law from a proposed
  mechanism
• The development of the rate law usually begins
  with the rate determining reaction or step in the
  reaction mechanism, which is the slowest step in
  the set of reactions that comprise the mechanism.
  You can, if you want, relate the rate written in
  terms of some species in the rate determining
  step to other species in subsequent faster steps.
  Fast steps that follow the rate determining step
  are, however, often ignored.
• The rate law should ultimately be expressed
  only in terms of chemical species which appear in
  the net or composite reaction. Intermediates,
  which 1st appear as products in early mechanistic
  steps and are consumed as reactants in later
  mechanistic steps, should not appear in the final
  version of the derived rate law.
• One approach is to assume that all fast steps
  that preced the rate determining step come to
  equilibrium and then to use the equilibrium
  constants for these steps to eliminate
  intermediates in the development of the rate law.
• A second approach is to write steady state
  expressions for intermediates and then use these
  expressions to eliminate these intermediates from
  the final rate law. In steady state the amount
  of a species is not changing with time and time
  derivatives involving this species can be set
  equal to zero.
• In any elementary reaction the order of the
  reaction with respect a particular chemical
  species is equal to the stoichiometric
  coefficient for that species in that elementary
  reaction.

12
The steady state approach to deriving rate laws
from reaction mechanisms is a more general
approach and often contains the results of the
rate determining approach as a limiting case. M
O (g) O (g) ------gt O2 (g) M M in the
above reaction is a chaperon, any chemical
species, including a reactant or product, that
either removes excess energy and stabilizes the
product (as in the above example) or provides the
energy needed for reaction to occur.