Transcript Lecture 19 - University of Windsor

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: theintermediates 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
v = ki[CH3CHO]
2. Propagation: CH3CHO + . CH3 → CH4 + CH3CO.
Propagation: CH3CO. → .CH3 + CO
3. Termination:
.CH3 + .CH3 → CH CH
3
3
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 3 CO . ]
 k p [ .CH 3 ][CH 3 CHO ]  k ,p [CH 3 CO . ]
dt
kp
k’p
• 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
 2kt



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 ]  kr [ H . ][ HBr ]
dt
1/ 2
3/ 2
d[ HBr ] 2k p1 (k i / kt ) [ 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 / kt ) [ 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
23.2
Explosions
•
Thermal explosion: a very rapid reaction arising from a rapid increase of
reaction rate with increasing temperature.
•
Chain-branching explosion: occurs when the number of chain centres
grows exponentially.
•
An example of both types of explosion is the following reaction
2H2(g)
+
O2(g)
→ 2H2O(g)
1. Initiation:
H2
→ H. + H.
+ .OH → H. +
2. Propagation
H2
H2O
kp
3. Branching:
O2 + .H → O + .OH
O + H2 → .OH + H.
kb1
Kb2
4. Termination
H. + Wall → ½ H2
H. + O2 + M → HO2. +
kt1
kt2
M*
The explosion limits of the H2 + O2
reaction
• Analyzing the reaction of hydrogen and oxygen (see preceding
slide), show that an explosion occurs when the rate of chain
branching exceeds that of chain termination.
Method: 1. Set up the corresponding rate laws for the reaction
intermediate and then apply the steady-state approximation.
2. Identify the rapid increase in the concentration of H.
atoms.
d[ H . ]
 v init  k p [ .OH ][ H 2 ]  kb1 [ H . ][O2 ]  k[O][ H 2 ]  k t 1 [ H . ]  k t 2 [ H . ][O2 ][ M ]
dt
d [ .OH ]
  k p [ .OH ][ H 2 ]  kb1 [ H . ][O2 ]  kb 2 [O][ H 2 ]
dt
d[O]
 kb1[ H . ][O2 ]  kb 2 [O][ H 2 ]
dt
Applying the steady-state approximation to .OH and O gives
 k p [ .OH ][ H 2 ]  kb1[ H . ][O2 ]  kb 2 [O][ H 2 ]  0
k b1 [ H . ][O2 ]  k b 2 [O ][ H 2 ]  0
kb1[ H . ][O2 ]
[O] 
kb 2 [ H 2 ]
2kb1[ H . ][ O2 ]
[ OH ] 
k p[ H 2 ]
.
Therefore,
d[ H . ]
 v init  (2k b1 [O2 ]  k t 1  k t 2 [O2 ][ M ])[ H . ]
dt
we write kbranch = 2kb1[O2] and kterm = kt1 + kt2[O2][M], then
d[ H . ]
 v init  (k branch  k term )[ H . ]
dt
At low O2 concentrations, termination dominates branching, so kterm >
vinit
kbranch. Then [ H . ] 
(1  e ( kterm  kbranch ) t ) this solution corresponds
kterm  kbranch
to steady combustion of hydrogen.
At high O2 concentrations, branching dominates termination, kbranch >
vinit
kterm. Then [ H . ] 
( e ( kbranch  kterm ) t  1)
 kterm  kbranch
This is an explosive increase in the concentration of radicals!!!
• Self-test 23.2 Calculate the variation in radical
composition when rates of branching and
termination are equal.
• Solution:
d[ H . ]
 v init  (k branch  k term )[ H . ]
dt
kbranch = 2kb1[O2] and kterm = kt1 + kt2[O2][M],
d[H . ]
 vinit
dt
The integrated solution is [H.] = vinit t
Polymerization kinetics
• Stepwise polymerization: any two monomers
present in the reaction mixture can link together
at any time. The growth of the polymer is not
confined to chains that are already formed.
• Chain polymerization: an activated monomer
attacks another monomer, links to it, then that
unit attacks another monomer, and so on.
23.3 Stepwise polymerization
• Commonly proceeds through a condensation reaction, in which
a small molecule is eliminated in each step.
• The formation of nylon-66
H2N(CH2)6NH2 + HOOC(CH2)4COOH →
H2N(CH2)6NHOC(CH2)4COOH
• HO-M-COOH + HO-M-COOH → HO-M-COO-M-COOH
• Because the condensation reaction can occur between
molecules containing any number of monomer units, chains of
many different lengths can grow in the reaction mixture.
Stepwise polymerization
• The rate law can be expressed as
d [ A]
  k[ A]2
dt
• Assuming that the rate constant k is independent
of the chain length, then k remains constant
throughout the reaction.
[ A]0
[ A] 
1  kt [ A]0
p
[ A]0  [ A]
kt[ A]0

[ A]0
1  kt[ A]0
• The degree of polymerization: The average
number of monomers per polymer molecule, <n>
n
[ A]0
1

[ A] 1  p