# Atkins_ Kinetics LH equation

Topics: Polymer, Polymer chemistry, Polymerization Pages: 11 (1148 words) Published: October 10, 2013
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 kb1[ H . ][O2 ]  kb 2 [O ][ H 2 ]  0

[O ] 

kb1[ H . ][O2 ]
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 > v init
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

[ 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...

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