Transcript Lecture 19 - University of Windsor

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
23.4 Chain polymerization
• Occurs by addition of monomers to a growing
polymer, often by a radical chain process.
• Rapid growth of an individual polymer chain for
each activated monomer.
• The addition polymerizations of ethene, methyl
methacrylate, and styrene.
• The rate of polymerization is proportional to the
square root of the initiator concentration.
The three basic types of reaction
step in a chain polymerization
I → R. + R. vi = ki[I]
M + R. → .M1
(fast)
(b) Propagation: M + .M1→ .M2
M + .M2→ .M3
░
vp = kp[M][.M]
M + .Mn-1→ .Mn
(c) Termination:
Mutual termination: .Mn + .Mm→ Mn+m
Disproportionation: .Mn + .Mm→ Mn + Mm
Chain transfer:
M + .Mn→ Mn + .M
(a) Initiation:
Influences of termination step on
the polymerization
•
Mutual termination: two growing radical chains
combine. vt = kt ([.M])2
•
Disproportionation: Such as the transfer of a
hydrogen atom from one chain to another,
which corresponds to the oxidation of the
donor and the reduction of acceptor.
vt = kt ([.M])2
•
Chain transfer: vt = ?
• the net rate of change of radical concentration is calculated as
 d [.M ] 


 2 fk i [ I ]  2kt [.M ]2
 dt 

 production
• Using steady-state approximation (the rate of production of radicals
equals the termination rate)
1/ 2
 fk 
[.M ]   i 
 kt 
[ I ]1 / 2
• The rate of polymerization
1/ 2
 fk i 
.
vp = kp[ M][M] = kp[M] k  [I ]1 / 2

t

• The above equation states that the rate of polymerization is
proportional to the square root of the concentration of the initiator.
• Kinetic chain length, v,
• <n> = 2v
num berof m onom erunitsconsum ed
 k[ M ][I ]1/ 2
num berof activatedcentres produced
1
where k  k p ( fki kt ) 1/ 2
2
v
(for mutual termination)
• Example: For a free radical addition polymerization with ki = 5.0x10-5
s-1 , f = 0.5, kt = 2.0 x107 dm3 mol-1 s-1, and kp = 2640 dm3 mol-1 s-1 ,
and with initial concentrations of [M] = 2.0 M and [I] = 8x10-3 M.
Assume the termination is by combination.
(a) The steady-state concentration of free radicals.
(b) The average kinetic chain length.
(c) The production rate of polymer.
Solution: (a)
(b)
1/ 2
 fk 
[.M ]   i 
 kt 
v  k[ M ][ I ]
1 / 2
[ I ]1 / 2
1
where k  k p ( fki kt ) 1/ 2
2
(c) The production rate of polymer corresponds to the rate of
polymerization is vp:
vp = kp[.M][M]
23.5 Features of homogeneous catalysis
• A Catalyst is a substance that
accelerates a reaction but
undergoes no net chemical
change.
• Enzymes are biological
catalysts and are very specific.
• Homogeneous catalyst: a
catalyst in the same phase as
the reaction mixture.
• heterogeneous catalysts: a
catalyst exists in a different
phase from the reaction mixture.
Example: Bromide-catalyzed decomposition of hydrogen peroxide:
2H2O2(aq) →
2H2O(l) + O2(g)
is believed to proceed through the following pre-equilibrium:
H3O+
+ H2O2 ↔
H3O2+ + H2O
[ H 3O2 ]
K
[ H 2O2 ][ H 3O  ]
H3O2+ + Br- → HOBr + H2O
v = k[H3O2+][Br-]
HOBr + H2O2 → H3O+ + O2 + Br-
(fast)
The second step is the rate-determining step. Thus the production rate
of O2 can be expressed by the rate of the second step.
d [O2 ]
 k[ H 3O2 ][ Br  ]
dt
The concentration of [H3O2+] can be solved
[H3O2+] = K[H2O2][H3O+]
Thus
d[O2 ]
 Kk[ H 2O2 ][ H 3O  ][ Br  ]
dt
The rate depends on the concentration of Br- and on the pH of the solution
(i.e. [H3O+]).
• Exercise 23.4b: Consider the acid-catalysed reaction
(1) HA + H+ ↔ HAH+
k1, k1’ , both fast
(2) HAH+ + B → BH+ + AH
k2, slow
Deduce the rate law and show that it can be made
independent of the specific term [H+]
Solution: