Chemistry Reference
In-Depth Information
f
2
r
1
f
1
1
1
P
12
5
(9-50)
f
2
5
r
1
ð½
M
1
=½
M
2
Þ
1
1
r
2
f
2
f
1
1
r
2
P
22
5
(9-51)
r
2
f
2
5
½
M
1
=½
M
2
Þ
1
r
2
f
1
f
1
1
½
M
1
=½
M
2
P
21
5
(9-52)
r
2
f
2
5
½
M
1
=½
M
2
1
r
2
To determine the distribution of sequence lengths of each monomer in the
polymer, an M
1
unit in the copolymer is selected at random. If this unit is part of
a sequence of
n
i
M
1
units, reaction (9-2) would have to have been repeated
(
n
i
2
1) times. Since the probability of one such event is
P
11
, the probability that
it occurs (
n
i
5
n
i
2
1
1) times is
ð
P
11
Þ
:
If the M
1
sequence is exactly
n
i
units long the
(
n
i
2
1) reactions of M
1
with M
i
must be followed by reaction (9-3). This place-
ment has the probability
P
12
5
P
11
. [Reactions (9-2) and (9-3) represent the
only alternatives available to radical M
i
and therefore
P
11
1
1
2
P
12
must equal 1.]
We conclude, then, that the probability that the original M
1
unit was part of a
sequence of
n
i
such units is
P
n
i
2
1
But the probability that the sequence
contains
n
i
M
1
units is also the fraction of all M
1
sequences that contain
n
i
units.
That is to say, it is the number distribution function
N
(M
1
,
n
i
) for M
1
sequence
lengths:
ð
1
P
11
Þ:
2
11
P
n
i
2
1
11
P
12
P
n
i
2
1
11
N
ð
M
1
;
n
i
Þ
5
ð
1
P
11
Þ
5
(9-53)
2
Similarly, the fraction of all M
2
sequences that contain exactly
n
j
units is
P
n
i
2
1
22
P
n
i
2
1
22
N
ð
M
2
;
n
j
Þ
5
ð
1
P
22
Þ
5
P
21
(9-54)
2
The number average length of M
1
sequences
N
1
is
N
N
1
5
N
ð
M
1
;
n
i
Þ
n
i
(9-55)
n
i
5
1
[Thi
s
is completely analogous to the definition of number average molecular
weight M
n
in Eq. (2-6). A number average of a quantity is always the sum of pro-
ducts of values of that quantity times the corresponding fraction of the whole
sample which is characterized by the particular value.] Substituting
Eq. (9-53)
into
Eq. (9-55)
,
N
P
11
N
1
P
12
n
i
P
n
i
2
1
n
i
P
n
i
11
N
1
5
11
P
12
5
(9-56)
n
i
5
1
The
summation in
Eq.
(9-56)
is of
the
same
form as
the
series
:
P
n
5
1
nx
n
½ð
x
Þ=
x
;
x
)if
x
(here
P
11
)
1
2
which equals 1/(1
2
,
1. Thus,
1
1
P
12
N
1
5
(9-57)
P
11
5
1
2
