Chemistry Reference
In-Depth Information
Thus:
Ce
4
þ
½
d
½
M
=
d
t ¼ k
1
k
2
k
4
K½
D
½
M
=ðk
2
þk
3
Þðk
5
½
M
þk
6
½
DM
þk
7
Ce
4
þ
Þ
½
M
þk
9
½
(6.16)
If the termination step is presented mainly by Eq. (
6.5
), the termination would be
predominantly by chain transfer of the polymeric radical to monomer as:
Ce
4
þ
k
5
½
M
k
6
½
DM
þk
7
½
M
þk
9
½
(6.17)
Equation (
6.16
) becomes:
d
½
M
=
d
t ¼ k
1
k
2
k
4
K½
D
½
Ce
4
þ
½
M
=ðk
2
þk
3
Þðk
5
Þ
(6.18)
In the presence of excess substrate, the disappearance rate of [Ce
4+
]
T
shows
pseudo-first order with respect to the total Ce
4+
concentration.
Ce
4
þ
T
=
d
t ¼ k
0
½
Ce
4
þ
T
d
½
(6.19)
where:
k
0
¼ k
1
K½
D
=ð
1
þK½
D
Þ
(6.20)
As the experiment gave
k
0
t
1, by rearranging and integrating equations (
6.18
)
and (
6.19
), and assuming that [D] is constant, we obtain:
T
D
Ce
0
4
þ
½
M
M
½ ¼
k1k2k4
=
k5 k2
ð
þ
k3
Þ
ð
1
þ
KD
½
Þ
½
t
(6.21)
or:
Ce
0
4
þ
T
½
ψ ¼
100
k
1
k
2
k
4
=k
5
ðk
2
þk
3
Þð
þK½
Þ½M
0
½
t
1
D
D
(6.22)
where [M
0
], [Ce
0
4+
]
T
, and
are the initial concentration of MMA, total ceric ion
and graft polymerization (%), respectively. If a plot of
ψ
ψ
versus
t
gives a straight
line, the termination must be caused by Eq. (
6.5
).
In the case of bimolecular termination, as in Eq. (
6.6
), assuming that:
Ce
4
þ
k
6
½
DM
k
5
½
M
þk
7
½
M
þk
9
½
(6.23)
Equation (
6.16
) becomes:
1
=
2
d
t ¼ðk
1
k
2
k
4
2
K½
Ce
4
þ
=ðk
2
½
=
½
þk
3
Þðk
6
ÞÞ
½
d
M
D
M
(6.24)
