Polymerization Kinetics

1
Polymerization Kinetics Polymers are large
macromolecules composed of many generally similar
smaller molecular units known as monomers. For
example condensation polymers are formed when
monomers and smaller polymers react or condense
with each other to form a larger molecule or
polymer with the elimination of a usually much
smaller molecule. An example of an early step in
the formation of a condensation polymer would be
the reaction between glycolic acid monomers
to form an ester and eliminate a molecule of
water. Notice that the product also has one
terminal acid functionality and one terminal
alcoholic functionality and can react just as the
monomer did
to form an even slightly larger polymer (actually
these small molecular weight polymers are called
oligimers). This polymerization process is
repeated in the reacting mixture to eventually
yield a distribution of polymers of different
molecular weights.
2
In the analysis of the kinetics of condensation
polymerization the rate constants in each of the
polymerization steps are assumed to be the same.
Since in each condensation step a carboxylic acid
functionality is lost, the rate law can be
expressed in terms of the decrease in the
concentration of these acid functional groups,
-COOH d -COOH / dt - k -COOH -OH acid
catalyst Could you justify writing the rate law
as we have? Why is the rate proportional to the
concentration of acid catalyst? In the
particular condensation that we are considering
the concentration of acid functionalities is
always equal to the concentration of alcholic
functionalities -COOH -OH and
furthermore the only acid catalyst present is the
carboxylic acid groups on the reacting molecules.
Thus we can write the rate law as d -COOH /
dt - k -COOH3 Separating variables and
integrating this differential rate law from time
zero gives 1 / -COOH2 1 / -COOHo2 2
k t
3
The usual practice is to write these expressions
in terms of the fraction of functional groups
reacted f ( -COOHo - -COOH ) /
-COOHo Solving this expression for -COOH and
using the result to eliminate -COOH from the
integrated rate law yields 1 / (1 - f)2 1
2 k -COOHo2 t
4
A generalized reaction in the polyamidation of
the monomer aminoheptanoic acid H2N(CH2)6COOH in
m-cresol is H--NH(CH2)6CO-n-OH
H2N(CH2)6COOH -----gt H--NH(CH2)6CO-n1-OH
This reaction was followed at 187 oC by titrating
the remaining -COOH functional groups in a 5.00
mL aliquot with 0.100 M KOH time VKOH time
VKOH (min) (mL) (min)
(mL) 0 30.00 220 12.98 20 25.40 240 12.53 40
22.43 260 12.12 60 20.30 280 11.75 80 18.68 3
00 11.41 100 17.40 320 11.09 120 16.35 340 10.
81 140 15.47 360 10.54 160 14.71 380 10.29 18
0 14.06 400 10.06 200 13.49 Calculate the
fraction of -COOH groups reacted and plot some
function of this fraction versus time to obtain a
linear plot and from this plot obtain a value for
the rate constant in M-2 min-1.
5
In a free radical or addition polymerization an
initiator radical attacks the double bond in a
monomer, yielding a monomeric radical that can in
turn attack the double bond on another monomer,
etc. resulting in a growing polymer radical. This
process begins with bond cleavage in an initiator
(sometimes called a sensitizer or catalyst) to
form radicals
The initial radical, R1, adds to the double bond
of the monomer (in this case styrene) to begin
the building the polymer
6
The free radical polymerization mechanism can be
summarized as follows ki I
-----gt 2 R1 kp
R1 M -----gt R2 kp
R2 M -----gt R3 kp
R i M -----gt
R i 1 kt R m
R n -----gt Pm n Can you label these steps
as chain initiating, chain propogating, or chain
terminating? Note that the rate constant, kp, for
all the propogation steps is assumed to be the
same. Assuming that each polymer radical, Rn, is
in steady state, we can write 0 d R1 / dt
2 ki I - kp R1 M - kt R1 S
Ri The sum in the last term reflects the
observation that any polymer radical can
terminate a given polymer (in this case R1). The
general steady state expression for Rn can be
written 0 d Rn / dt kp Rn - 1 M
- kp Rn M - kt Rn S Ri
7
Summing over all of these steady state equations
gives 0 2 ki I - kt ( S Rn )2 Could
you justify this equation? The rate of
polymerization is defined in this case as the
rate at which monomers are consumed rate of
polymerization - d M / dt kp
M R1 kp M R2 kp
M S Rn Using the result of summing over
the steady state equations allows us to write
ultimately for the rate of polymerization rate
of polymerization - d M / dt kp M (2
ki I / kt )1/2
8
The initial rates of polymerization of methyl
methacrylate (MMA)
initiated by the decomposition of the initiator
azobisisobutyronitrile (ABIN) in benzene at 77 oC
as a function of monomer and initiator
concentration are reported in the following table
J. Am. Chem. Soc., 742027 (1952) MMA
ABIN rate MMA
ABIN rate (M) (M)
(M/min) (M) (M) (M/min) (x
104) (x 103) (x 104) (x 103) 9.04 2.35
11.61 4.75 1.92 5.62 8.63 2.06 10.20 4.22 2.3
0 5.20 7.19 2.55 9.92 4.17 5.81 7.81
6.13 2.28 7.75 3.26 2.45 4.29
4.96 3.13 7.13 2.07 2.11 2.49 Given that the
rate constant for the decomposition of ABIN
is ki (1015.2 sec-1) e - (30,800 cal/mole /
R T) determine a value for the rate constant
ratio kp / kt1/2 ?