Chemistry Reference
In-Depth Information
M
2
M
2
M
½
d
½
polymer
=
dt
k
tc
½
2
k
td
½
k
tr
½
TH
(8-72)
5
1
1
in the presence of chain transfer. If we substitute
Eq. (8-72) into (8-61)
and invert
the resulting expression (for ease in manipulation), we obtain
M
k
p
½
M
1
DP
n
5
k
tc
½
2
k
td
½
k
tr
½
TH
(8-73)
1
1
k
p
½
k
p
½
M
M
M
Then, putting [M
]
R
p
k
p
[M] (from
Eq. 8-13
):
5
1
DP
n
5
k
tc
R
p
k
p
½
2
k
td
R
p
k
p
½
k
tr
½
TH
(8-74)
2
1
2
1
k
p
½
M
M
M
The ratio
k
tr
/k
p
will depend on the particular transfer agent and monomer,
as well as the reaction temperature. The relation given can be generalized by
breaking the last term on the right-hand side of
Eq. (8-74)
into a sum of the
contributions from the various potential chain transfer agents (except polymer).
Then
1
DP
n
5
k
tc
R
p
k
p
½
2
k
td
R
p
k
p
½
k
tr
;
M
k
p
½
M
K
tr
;
I
k
p
½
I
(8-75)
2
1
2
1
1
½
M
½
M
M
M
where
k
tr
,
M
,
k
tr
,
I
,
k
tr
,
S
are the rate constants for the transfer reaction of reaction
(8-67) with monomer, initiator, and solvent (S), respectively, and Ta stands for
any chain transfer agent which is added deliberately for this purpose.
It is customary to define a chain transfer constant
C
for each substance as
the ratio of
k
tr
for that material with a propagating radical to
k
p
for that radical.
Thus,
K
tr
;
M
k
p
K
tr
;
I
k
p
K
tr
;S
k
p
;
K
tr
k
p
C
M
;
C
I
;
C
S
C
(8-76)
In this notation,
Eq. (8-75)
is written
k
tc
R
p
k
p
½
2
k
td
R
p
k
p
½
2
1
C
M
1
C
I
½
1
C
S
½
1
C
½
1
DP
n
5
I
S
Ta
(8-77)
2
1
½
M
½
M
½
M
M
M
Equation (8-77)
is not as formidable in practice as it appears on first glance.
In the bulk polymerization of styrene, for example, [S]
0, [I] is usually
so small as to make transfer to initiator negligible, and the right-hand side of this
equation is left with only three terms.
The following sections review the magnitudes of the various chain transfer
constants, methods for measuring these parameters, and their significance in free-
radical polymerizations.
5
0,
k
td
5
