Tutorial on Monday and
Wednesday
• No. 2 (Chapter 22: The rates of chemical
reactions) by Bryan Lucier
• March 7 & 9 (from 2:30 to 3:50pm, Essex
Hall room 182)
Kinetic isotope effect
• Kinetic isotope effect: the decrease in the rate of a chemical
reaction upon replacement of one atom in a reactant by a heavier
isotope.
• Primary kinetic isotope effect: the kinetic isotope effect observed
when the rate-determining step requires the scission of a bond
involving the isotope:
~
k (C  D)
 CH 1 / 2 

hc
v
(C  H ) 
with
e

) 
1  (
k (C  H )
2kT
 CD


• Secondary kinetic isotope effect: the variation in reaction rate
even though the bond involving the isotope is not broken to form
product:
k (C  D)
 e 
k (C  H )
~
with
~
 CH 1 / 2 
hc {v  (C  H )  v (C  H )} 

) 
1  (
2kT

CD


Kinetic isotope effect
(primary)
Kinetics isotope effect
(secondary)
22.8 Unimolecular reactions
•
The Lindemann-Hinshelwood mechanism
A reactant molecule A becomes energetically excited by collision with
another A molecule:
A + A → A* + A
d [ A*]
 k1 [ A]2
dt
The energized molecule may lose its excess energy by collision with
another molecule:
A + A* → A + A
d [ A*]
  k1' [ A][ A*]
dt
The excited molecule might shake itself apart to form products P
A* → P
d [ A*]
  k 2 [ A*]
dt
The net rate of the formation of A* is
d [ A*]
 k1 [ A] 2  k1' [ A][ A*]  k 2 [ A*]
dt
• If the reaction step, A + A → A* + A, is slow enough to be
the rate determining step, one can apply the steady-state
approximation to A*, so [A*] can be calculated as
d [ A*]
 k1[ A]2  k1' [ A][ A*]  k 2 [ A*]  0
dt
Then
[ A*] 
k1[ A]2
k1' [ A]  k2
The rate law for the formation of P could be reformulated as
d [ P]
k1k 2 [ A]2
 k2 [ A*]  '
dt
k1[ A]  k 2
Further simplification could be obtained if the deactivation of A* is
much faster than A*  P, i.e., k1' [ A ][ A*]  k2 [ A*] then k1' [ A]  k2
d [ P ] k1 k 2
 ' [ A]
dt
k1
'
in case k1[ A]  k2
d[ P ]
 k1 [ A]2
dt
• The equation
reorganized into
d [ P]
k1k 2 [ A]2
 k2 [ A*]  '
dt
k1[ A]  k 2
can be
d [ P]
k1k 2 [ A]
( '
)[ A]
dt
k1[ A]  k 2
• Using the effective rate constant k to represent
k
k1k 2 [ A]
k1' [ A]  k 2
• Then one has
'
1
k1

k
k1k 2
1

k1[ A]
The Rice-Ramsperger-Kassel
(RRK) model
• Reactions will occur only when enough of required energy
has migrated into a particular location in the molecule.

E 

P  1 
E 

*
s 1
s 1

E 
 kb
kb ( E )  1 
E 

*
• s is the number of modes of motion over which the energy
may be dissipated, kb corresponds to k2
The activation energy of combined reactions
• Consider that each of the rate constants of the following reactions
A + A → A* + A
(E1)
A + A* → A + A
(E1’)
A* → P
(E2)
has an Arrhenius-like temperature dependence, one gets
k1 k 2 A1e  E 1/ RT A2 e  E2 / RT

'
'
k1
A1' e  E1 / RT

A1 A2
A1'
e
( E1  E2  E1' ) / RT
Thus the composite rate constant also has an Arrhenius-like form
with activation energy,
E = E1 + E2 – E1’
Whether or not the composite rate constant will increase with
temperature depends on the value of E,
if E > 0, k will increase with the increase of temperature
Combined activation energy
• Theoretical problem 22.20
The reaction mechanism
A2 ↔ A + A (fast)
A + B → P (slow)
Involves an intermediate A. Deduce the
rate law for the reaction.
• Solution:
Chain reactions
• Chain reactions: a reaction intermediate produced in one
step generates an intermediate in a subsequent step, then that
intermediate generates another intermediate, and so on.
• Chain carriers: the intermediates in a chain reaction. It could be
radicals (species with unpaired electrons), ions, etc.
• Initiation step:
• Propagation steps:
• Termination steps:
23.1 The rate laws of chain
reactions
• Consider the thermal decomposition of acetaldehyde
CH3CHO(g)
→
CH4(g) + CO(g)
v = k[CH3CHO]3/2
it indeed goes through the following steps:
1. Initiation:
CH3CHO → . CH3 + .CHO
ki
v = ki [CH3CHO]
2. Propagation: CH3CHO + . CH3 → CH4 + CH3CO.
Propagation: CH3CO. → .CH3 + CO
3. Termination:
.CH3 + .CH3 → CH CH
3
3
kp
k’p
kt
• The net rates of change of the intermediates are:
d [ .CH 3 ]
 ki [CH 3CHO]  k p [ .CH 3 ][CH 3CHO]  k ,p [CH 3CO. ]  2kt [ .CH 3 ]2
dt
d [CH 3CO. ]
 k p [ .CH 3 ][CH 3CHO]  k ,p [CH 3CO. ]
dt
• Applying the steady state approximation:
0  k i [CH 3CHO]  k p [ .CH 3 ][CH 3CHO]  k ,p [CH 3CO. ]  2k t [ .CH 3 ]2
0  k p [ .CH 3 ][CH 3CHO]  k ,p [CH 3CO. ]
• Sum of the above two equations equals:
ki [CH 3CHO ]  2kt [.CH 3 ]2  0
• thus the steady state concentration of [.CH3] is:
 k
[.CH 3 ]   i
 2k t
•
1/ 2



[CH 3CHO ]1 / 2
The rate of formation of CH4 can now be expressed as
 k
d [CH 4 ]
 k p [ .CH 3 ][CH 3CHO]  k p  i
dt
 2k t



1/ 2
[CH 3CHO]3 / 2
the above result is in agreement with the three-halves order
observed experimentally.
• Example: The hydrogen-bromine reaction has a complicated rate
law rather than the second order reaction as anticipated.
H2(g) + Br2(g) → 2HBr(g)
Yield
k[ H 2 ][ Br2 ]3 / 2
v
[ Br2 ]  k '[ HBr ]
The following mechanism has been proposed to account for the above
rate law.
1. Initiation:
Br2 + M → Br. + Br. + M
ki
2. Propagation:
Br. + H2 → HBr + H.
kp1
H. + Br2 → HBr + Br.
kp2
3. Retardation:
H. + HBr → H2 + Br.
kr
4. Termination:
Br. + Br. + M → Br2 + M*
kt
derive the rate law based on the above mechanism.
• The net rates of formation of the two intermediates are
d[ H . ]
 k p1 [ Br . ][ H 2]  k p 2 [ H . ][ Br2 ]  k r [ H . ][ HBr]
dt
d [ Br . ]
 2k i [ Br2 ][ M ]  k p1[ Br . ][ H 2 ]  k p 2 [ H . ][ Br2 ]  k r [ H . ][ HBr]  2k t [ Br . ]2 [ M ]
dt
• The steady-state concentrations of the above two intermediates can
be obtained by solving the following two equations:
k p1[ Br . ][ H 2]  k p 2 [ H . ][ Br2 ]  k r [ H . ][ HBr]  0
2k i [ Br2 ][ M ]  k p1 [ Br . ][ H 2 ]  k p 2 [ H . ][ Br2 ]  k r [ H . ][ HBr]  2k t [ Br . ]2 [ M ]  0
k
[ Br . ]   i
 kt
[H ] 
.
•
1/ 2



[ Br2 ]1 / 2
k p1 (ki / kt )1 / 2 [ H 2 ][ Br2 ]1 / 2
k p 2 [ Br2 ]  kr [ HBr ]
substitute the above results to the rate law of [HBr]
d [ HBr ]
 k p1 [ Br . ][ H 2 ]  k p 2 [ H . ][ Br2 ]  k r [ H . ][ HBr ]
dt
1/ 2
3/ 2
d [ HBr ] 2k p1 (k i / k t ) [ H 2 ][ Br2 ]

dt
[ Br2 ]  (k r / k p 2 )[ HBr ]
continued
• The above results has the same form as the empirical rate law, and
the two empirical rate constants can be identified as
k
k  2k p  i
 kt
1/ 2



k, 
kr
k p2
• Effects of HBr, H2, and Br2 on the reaction rate based on the
equation
1/ 2
3/ 2
d [ HBr ] 2k p1 (k i / k t ) [ H 2 ][ Br2 ]

dt
[ Br2 ]  (k r / k p 2 )[ HBr ]
• Self-test 23.1 Deduce the rate law for the
production of HBr when the initiation step
is the photolysis, or light-induced
decomposition, of Br2 into two bromine
atoms, Br.. Let the photolysis rate be v =
Iabs, where Iabs is the intensity of absorbed
radiation.
• Hint: the initiation rate of Br. ?
Exercises 23.1b: On the basis of the
following proposed mechanism, account
for the experimental fact that the rate law
for the decomposition
2N2O5(g) → 4NO2(g) + O2(g)
is v = k[N2O5].
(1) N2O5 ↔ NO2 + NO3
k 1, k 1’
(2) NO2 + NO3 → NO2 + O2 + NO k2
(3) NO + N2O5 → NO2 + NO2 + NO2 k3