HW5-Copolymers- Canvas-solutions
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Dec 6, 2023
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1
Homework
# 5
(Canvas)-
Solutions
Name:
Gower
EMA 3066 Fall ’23
Question 1 :
10 / 10 pts
Suppose you wanted to make diblock copolymer of copoly(acrylonitrile-
block
-
methyl methacrylate), with the blocks having a molecular weight (number average)
of
150,000 g/mol
and 325,000 g/mol, for AN and MMA respectively.
If the
initiator concentration is set at 1.75 mM for both parts of the sequential reaction,
where they were each reacted to full conversion, what monomer concentrations
should you use for AN? The next question asks for the other monomer, MMA
[AN] = ? M
4.9500
Hint: Did you check for proper significant figures?
From Table on pg. 8 of Topic 8- Both PAN and PMMA can be prepared by
anionic polymerization, so we can make block copolymers of controlled block
lengths.
The repeat unit for
PAN
will be -CH
2
-CH(CN)-
MW(repeat) = 3C + 1N + 3H = 3(12.011) + 1(14.01) + 3(1.008) = 53.067 g/mol
Xn = Mn/MW(repeat) = 150,000 g/mol /53.067 g/mol = 2826.6
The repeat unit for
PMMA
is –CH
2
-C(CH
3
)(COOCH
3
)-
MW(repeat) = 5C + 2O + 8H = 5(12.011) + 2(16.00) + 8(1.008) = 100.119 g/mol
Mn = Xn x MW(repeat)
Xn = Mn/MW(repeat) = 325,000 g/mol /100.119 g/mol = 3246.1
With the assumption that they are each reacted to full conversion, so X =1, the
formula Xn = Mo/Io can be used.
Mo(AN) = Xn x Io = 2826.6 x (1.75 x 10
-3
mole/L )= 4.947 mole/L
Mo(MMA) = Xn x Io = 3246.1 x (1.75 x 10
-3
mole/L )= 5.681 mole/L
[AN] = 4.95 M
[MMA] = 5.68 M
80 pts
2
Question 2 :
5 / 5 pts
Continuing the prior problem, what concentration of MMA will be required?
[MMA] = ? M
=
5.68 M
Question 3:
5 / 5 pts
The following computational data is for a hypothetical copolymer of M and P
monomers. You don’t need to do anything with the last two columns that contain
Probabilities (P)
. This is just real world data, where the authors of this paper were
calculating probabilities to predict reactivity ratios or something.
One of the following sets of reactivity ratios was used to calculate this
data.
Which of the sets seems most likely to yield these results?
Note:
r
m
= k
mm
/k
mp
,
r
p
= k
pp
/k
pm
rm = 0.003 and rp = 0.0004
rm = 0.0004 and rp = 5.4
rm = 5.4 and rp = 0.0004
3
Given the stronger incorporation of monomer M, where F
1
(
M
) is more enriched
in M than the feed f
1
(
M
), the r
m
must be much larger than r
p
.
Although there
are two possibilities that show a larger r1 than r2, the one on the top has both
values being very close to zero, so one would predict the chains would have a
strong alternating tendency. That would be seen in the F2 values as being 0.5.
They don’t start at 0.5, so that doesn’t seem the correct answer. However, the
values are not that far off, and thus have a tendency to be between the two
cases 1 and 4.
Question 4 :
7 / 7 pts
Using the same copolymer data from the prior problem, which Case does this
system fit the best?
(i.e., which type of copolymer will this produce)?
Case 1. alternating copolymer
Case 2. large blocks to nearly homopolymer
Case 3. truly random copolymer with composition matching the feed
Case 4. “ideal” copolymer with randomness, and with compositional drift
Case 5. copolymer formed at azeotropic composition
One r is nearly zero, and the other is quite large (although not quite infinity), so it
seems closest to case 4, where r1 = 1/r2. Although not exact, all of the other
cases seem very far removed. Given the copolymer composition is not at 50:50,
is seems not to be alternating. And there clearly is compositional drift, so case 3
and 5 are not likely.
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4
Question 5 :
10 /10 pts
Fill in all blanks with text, not a number:
Given the data below, taken from a paper about the influence of nanoclay on the
copolymerization of styrene (St) and methyl methacrylate (MMA), where the
reactivity ratios for the pure copolymer are
r
St
= 0.290 and
r
MMA
= 0.443:
(A)
Which Case does this system fit the best (spell out the number from one to
five)?
Case:
five
(B)
Does the copolymer prepared with nanoclay (red curve) contain an azeotrope
composition?
Yes or No?
yes
(C) Determine the composition (mole fraction) of styrene in the feed to achieve the
azeotrope composition?
(fill in the blank with just the letter (a - f) indicating which is correct response):
d
5
a)
there is no azeotrope
b)
f
St
= 0
c)
f
St
= 0.20
d)
f
St
= 0.44
e)
f
St
= 0.56
f)
f
St
= 0.66
(D) What is the composition (mole fraction) of styrene in the copolymer at this
condition?
(fill in the blank with just the letter (a - f) indicating which is correct response):
d
a)
there is no copolymer since there is no azeotrope
b)
F
St
= 0
c)
F
St
= 0.20
d)
F
St
= 0.44
e)
F
St
= 0.56
f)
F
St
= 0.66
(E)
What percentage of MMA will be in the copolymer at that same condition?
(fill in the blank with just the letter (a - f) indicating which is correct response):
d
a)
MMA =
0%
b)
MMA =
22%
c)
MMA =
44%
d)
MMA =
56%
e)
MMA =
66%
f)
MMA =
100%
Question 6 :
5 / 5 pts
For the data below, predict which temperature will give the most “ideal”
copolymer, as in randomness of EA and MMA units, for the ethyl-methyl
methacrylate (EA-MMA) system?
6
T = 50ºC
T = 60ºC
T = 65ºC
Case 4 gives ideal copolymers (and case 3). The
ratios are not close to 1, so that knocks out case 3.
The one closest to being reciprocals is the 50C,
where 0.47 x 1.83 = 0.86 (closer than the 0.24 x
2.03 = 0.49)
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7
Question 7
8 / 8 pts
The schematic below shows data from a paper where the authors used a clever
route for making a graft copolymer.
They incorporated “grafted” chains of PDMS
into a mostly methyl methacrylate backbone by copolymerizing with a small
amount of a macro-monomer that contained a short oligomeric chain of PDMS as
the “side unit” (see schematic below).
The macro-monomer is basically just like
the methyl methacrylate (orange), but has a PDMS oligomer (blue, with 30
repeats) rather than a methyl as in the methacrylate side group. In their table of
data below, you can see that they examined a variety of polymerization methods
(with various initiators and temperatures), and xylene was used as a solvent to
control the viscosity.
Although we haven’t covered all these methods of
polymerization, other than FRP (free radical polymerization), given their illustrated
data, you should still be able to determine which of the followings statements is/are
true based on the schematic of the chain distributions.
(There can be more than
one answer):
8
Note
‐
The column for "Distribution" is representing both the dispersity in chain
lengths, as well as homogeneity of macro
‐
monomer incorporation (seen as the
blue PDMS oligomer).
They are showing the blue side groups as being either
homogeneously incorporated (considered perfectly random) versus the blue not
being evenly distributed within the chains.
RAFT and ATRP provide a lower polydispersity index than FRP.
The ATRP method provides both low PDI and improved randomness of
PDMS monomer incorporation.
RAFT seems to follow Case 3, while ATRP seems to follow Case 4.
False: Case 4 will have compositional drift, which seems to be the case for RAFT
(the blue dashes become more enriched at the end of the polymerization), but
not ATRP (where the blue dashes are uniformly distributed along the chain
length), suggesting it follows Case 3.
Click if this description is True: This diagram provides a nice simple
representative of the monomer distribution, but it really is somewhat misleading
because if these reactions proceed by a chain growth mechanism, the whole chains
formed early on will have a different monomer distribution than the entire chains
formed later, as the feed mixture changes with conversion. This is because the
monomers become incorporated into each chain very rapidly during propagation,
while the slow conversion of monomer is from more and more chains being
formed over time.
9
Question 8:
10 / 10 pts
(A) On the plot below, determine the direction of compositional drift for the
following initial values of monomer feed compositions (f
1
):
Fill in the blank with
either upward, downward, none.
a.
f
1
= 0.2
upw ard
b.
f
1
= 0.5
dow nw ard
c.
f
1
= 0.8
dow nw ard
(B)
Does this copolymer have an azeotrope?
Yes
or No:
yes
(C)
If it does have an azeotrope, at what composition does this occur (in mole
fraction)?
Fill in the blank with a letter from the choices (a-f) from the list
below:
c
(a)
f
1
= 0.0
(b)
f
1
= 0.2
(c)
f
1
= 0.4
(d)
f
1
= 0.6
(e)
f
1
= 0.8
(f)
f
1
= 1.0
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10
a) At an initial feed ratio of f
1
= 0.20, the resulting copolymer has a composition
that is significantly lower (F
1
0.08), meaning that is has incorporated less of the
M
1
relative to M
2
compared to what was in the feed solution. Therefore, since M
2
has preferentially reacted, the remaining monomer feed will have more M
1
than it
previously did, so f
1
will increase on the next increment. As this continues, if it is
not corrected for by adding more M
1
, the instantaneous monomer ratio will
continue to drift upward.
Instantaneous Copolymer Composition
A
11
b) At an initial feed ratio of f
1
= 0.5 the resulting copolymer has a composition
that is a lot higher (F
1
0.7), meaning that is has preferentially incorporated M
1
relative to M
2
as compared to the feed ratio. Therefore, the remaining monomer
feed will be reduced in M
1
and enriched in M
2
for this next iteration. As this
continues, if it is not corrected for, the instantaneous feed ratio will drift
downward. Note
‐
0.5 is not at the azeotrope composition in this example.
c) At an initial feed ratio of f
1
= 0.8 the resulting copolymer has a composition
that is higher (F
1
0.95), meaning that is has preferentially incorporated M
1
relative to M
2
as compared to the feed ratio. Therefore, the remaining monomer
feed will be reduced in M
1
and enriched in M
2
for this next iteration. As this
continues, if it is not corrected for, the instantaneous feed ratio will drift
downward.
12
Question 9 :
10 / 10 pts
In considering the copolymerization of two monomers M
1
and M
2
, where an
azeotrope is formed at a ratio of 1 mole of M
1
to 2 moles of M
2
.
If monomer M
1
is
known not to homopolymerize (k
11
= 0):
A) What copolymer composition (F
1
) will result initially when a mixture of 1 mole
of M
1
with 9 moles of M
2
is polymerized?
Hint- use the instantaneous copolymer equation since I have not given a plot. You
will first need to solve for r
2
.
Then you can use the equation to solve for F
1
at any
other composition.
0.1540
We are given the azeotropic composition of f
1
= 1/(1 + 2) = 1/3.
From this, f
2
= (1-
f
1
) = 2/3.
Also, F
1
= f
1
at the azeotrope. We are told M
1
doesn’t homopolymerize,
meaning that k
11
= 0 and therefore r
1
= 0. Given F
1
and f
1
and r
1
, we can use the
copolymer composition equation to find r
2
.
Once we know the r values, we can
then determine copolymer composition (F
1
) at any other f
1
feed composition (i.e.
not just the azeotrope):
r
2
= 0.5
f
1
= 1mol/(1mol + 9mol) =
0.1
f
2
= (1-f
1
) = 0.9
r
1
= 0
r
2
= 0.5
2
2
2
2
1
2
1
1
2
1
2
1
1
1
2
f
r
f
f
f
r
f
f
f
r
F
2
2
2
9
/
4
9
/
4
9
/
2
)
3
/
2
(
)
3
/
2
)(
3
/
1
(
2
0
)
3
/
2
)(
3
/
1
(
0
3
/
1
r
r
9
/
2
)
27
/
4
(
27
/
4
2
r
27
/
2
27
/
4
27
/
6
)
27
/
4
(
2
r
154
.
0
)
9
.
0
)(
5
.
0
(
)
9
.
0
)(
1
.
0
(
2
0
)
9
.
0
)(
1
.
0
(
0
2
1
F
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13
Question 10 :
10 /10 pts
Prior problem continued:
B) If you were make a copolymer starting with an initial mixture of 4 moles of M
1
with 6 moles of M
2
, would copolymer formed at a high conversion contain more or
less
more
of M
1
compared to that formed earlier at a low conversion.
We don’t have the copolymer composition curve, but we can solve it just using the
equation.
f
1
= 4mol/(4mol + 6 mol) = 0.4
f
2
= (1-f
1
) = 0.6
r
1
= 0
r
2
= 0.5
Starting at this initial monomer feed, it is apparent that the initial copolymer has a
bit less M
1
than the monomer feed (copolymer is enriched in M
2
), so M
2
is going to
be consumed faster than M
1
(feed will become enriched in M
1
).
This means that f
1
will increase, and thus F
1
will increase with conversion.
Thus polymer at high
conversion will be
richer
in M
1
than at low conversion (because M
2
becomes more
and more depleted).
This would represent an upward drift on the curve if the
equation were plotted (see below).
C) If instead you were start with an initial mixture of 2 moles of M
1
with 8 moles
of M
2
(the same monomers as part (A)), would copolymer formed at a high
conversion contain more or less
less
of M
1
compared to that formed at low
conversion? On an exam, you would be required to show how you derived your
answer for credit.
Starting at this lower initial monomer feed, we are now below the azeotrope point
where the curve crosses the diagonal.
Now the initial copolymer has a bit more M
1
than the monomer feed, so M
1
is going to be consumed faster than M
2
(feed will
become enriched in M
2
).
This means that f
1
will decrease, and thus F
1
will
decrease with conversion (downward drift for M
1
).
Thus polymer at high
conversion will be richer in M
2
than at low conversion (less M
1
).
f
1
= 2mol/(2mol + 8mol) = 0.2
0.364
14
f
2
= (1-f
1
) = 0.8
r
1
= 0
r
2
= 0.5
D) On an exam, I might ask you this: Sketch the general shape of the polymer
composition curve versus conversion for both of the two monomers if the starting
feed composition was M
1
= 0.33 and M
2
= 0.66.
Answer 1:
more
Answer 2:
less
25
.
0
)
8
.
0
)(
5
.
0
(
)
8
.
0
)(
2
.
0
(
2
0
)
8
.
0
)(
2
.
0
(
0
2
1
F
0
20
40
60
80
100
Degree of conversion
Polymer
composition
0
0.2
0.4
0.6
0.8
1.0
Azeotrope at f
1
= 0.33, so F
1
=0.33. Since the copolymer
incorporates the same
composition of monomer as the
feed, there should be no drift.
So the sketch should show
composition remaining constant
throughout the reaction, with
F1
0.66
0.33
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Naming and drawing ketones
Write the systematic name of each organic molecule:
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to
Explanation
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1.
Component
Container
Stretch Wrap
Pallet
Product
Material
Polypropylene
Polystyrene
Wood
Ca(OCI)2
Weight (kg/pallet)
48
0.5
15-18
725
Based on the above table, answer the following questions.
(a) Discuss the possibility of fire at the calcium hypochlorite Ca(OCI)2 storage warehouse. Include your assumption on how the fire could spread in the storage warehouse.
(b) Recommend solutions to avoid fire occur at the storage warehouse.
(10)
(10)
(Total / Jumlah: 20 )
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