Chemistry Reference
In-Depth Information
Equation (8-29)
was derived for the polymerization in which radicals are gen-
erated by the decomposition of an initiator in a homogeneous reaction mixture.
More generally,
R
1
=
2
M
d
½
M
=
dt
5
ð
k
p
=
k
t
1
=
2
Þ½
M
(8-30)
2
where
R
M
is the rate of formation of monomer-ended radicals.
Equation (8-30)
can be used when initiation is by means summarized in
Section 8.5
in which the
relation between
R
i
and the initiator concentration may differ from that assumed
in these paragraphs or in which the reaction may be started without an initiator
per se
.
Equations (8-29) and (8-30)
show that the instantaneous rate of polymerization
depends directly on the monomer concentration and on the square root of the rate
of initiation.
The ratio
k
p
=
k
1
=
t
will appear frequently in the equations we develop for radi-
cal polymerization. The polymerizability of a monomer in a free radical reaction
is related to
k
p
=
k
1
=
t
rather than to
k
p
alone. From
Eq. (8-30) or (8-29)
, it can be
seen that a given amount of initiator will produce more polymer from a mono-
mer with a higher
k
p
=
k
1
=
t
ratio. Thus, at 60
C, the
k
p
values for acrylonitrile
and styrene are approximately 2000 and 100 liter/mol sec, respectively. The for-
mer monomer does not polymerize 20 times as fast as styrene under the same
conditions at 60
, however, because the respective
k
t
values are 780
10
6
liter/
3
k
1
=
t
for acrylonitrile is 0.07 liter
1
/
2
/
mol
1
/
2
sec
1
/
2
, which is just six times that of styrene. Styrene forms the less reac-
tive radical in this case because the unpaired electron can be delocalized over
the phenyl ring. It adds monomer more slowly as a consequence but it also ter-
minates less readily because its radical is stabler than that of polyacrylonitrile.
10
6
liter/mol sec. Then
k
p
=
mol sec and 70
3
8.3.5
Integrated Rate of Polymerization Expression
If the initiator decomposes in a unimolecular reaction the corresponding rate
expression is first order in initiator [cf. reaction (8-6)]:
d
½
I
=
dt
k
d
½
I
(8-31)
2
5
Integration of
Eq. (8-31)
between [I]
[I]
0
at
t
0 and [I] at
t
gives
5
5
0
e
2
k
d
t
½
I
5
½
I
(8-32)
[For first-order reactions it is often convenient to integrate between [I]
[I]
0
5
at
t
t
1
/
2
, the half-life of the initiator. The half-life is
evidently the time needed for the initial concentration of initiator to decrease
to half its initial value. It is related to the initiator decomposition rate constant
k
d
by
0 and [I]
[I]
0
/
2at
t
5
5
5
t
1
=
2
5
ð
ln2
Þ=
k
d
(8-33)
