HW5-Copolymers- Canvas-solutions

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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