Chapter 2 Assignment Solution Keys
1.
Adipic acid = AA (Molecular weight = 146.14)
HMDA = BB (Molecular weight = 116.20)
Acetic acid = A’
repeating unit = aabb (Mass of repeating unit = 146.14 + 116.20 – 2x 18.02 = 226.30
Nylon66 with the following type of end groups present in the system:
(1) Aa…..aA
(2) Bb….bB
(3) Aa……aB
(4) A’b….aA
(5) A’b…..bB
(6) A’b…..bA’
The total number of polymer chain of type 6 = C
The total number of polymer chains of type 4 and 5 = the total number of A’ end group in polymer chains of 4 and 5 = 28.9-2C
The total number of polymer chains of type 1, 2 and 3 = ½ the total number of unreacted end group (A+B) present in 1,2 and 3 types of polymers = ½ [the total number of unreacted end group (A+B) among all types of polymers – the total number of unreacted end group (A+B) present in 4 and 5 types of polymers] = ½ [the total number of unreacted end group (A+B) among all types of polymers –the total number of A’ end group present in 4 and 5 types of polymers] = [35.3 + 96.5 – (28.9-2C)]/2
The total number of polymer chains = total number of polymer chains of type 6 + the total number of polymer chains of types 4 and 5 + the total number of polymer chains of types 1, 2, and 3 = C+ 28.9-2C +[35.3 + 96.5 – (28.9-2C)]/2 =[28.9+35.3+96.5]/2 = 80.35
(Trevon has an easy way to put this: the total number of chain = total number of chain end/2) = (28.9+35.3+96.5)/2=80.35
The total number of repeating units (ignoring end group effect) = total mass/mass of repeating unit = 106/226.30 = 4418.91
The number average degree of polymerization = the total number of repeating units / the total number of polymer chains = 4418.91/80.35 = 55.00*
*The repeating unit as we defined (aabb) contains two sets of ab units. So if you count the number average degree of polymerization in terms of ab unit, the value will double and become 55.00x2 =110.
The total number of acid (A and A’) functional groups (νAo) = 2 x the total number of repeating units (aabb) + the total number of unreacted A end groups + the total number of A’ groups = 2 x 4418.91 + 96.5 + 28.9 = 8963.22
The total number of amine (B) functional groups (νBo) = 2 x the total number of repeating units (aabb) + the total number of unreacted B end groups = 2 x 4418.91 + 35.3 = 8873.12
νAo > νBo
Therefore, the limiting functional group is the amine (B).
p = 1- νB/νBo (B is the limiting functional group, p is the extent of reaction, νB is the unreacted B functional group, νBo is the total B functional group to start with) = 1-35.3/(35.3+4418.91x2) = 0.9960 > 0.9909 (reported by Howard)
The values of DPn and p calculated by Howard are self-consistent for a stoichiometrically balanced condensation reaction where DPn = 1/(1-p). However, from calculation of νAo and νBo, it is clearly not a stoichiometrically balanced condensation polymerization. DPn is consistent with the polymer composition as shown above. But the p is not consistent with the composition. Clearly, to get the same DPn, p (conversion) has to be higher to compensate for the effect of stoichiometric imbalance.
2.
The copolymer of styrene and hydroxyethylmethacrylate can be seen as a multi-functional macro-monomer that bears multiple OH groups. In the crosslinking reaction, the reaction rate depends on the concentration of the OH group and hexamethylene di-isocyanate. If the reaction solution is very diluted, hexamethylene di-isocyanate will mostly react with OH groups on the same copolymer chain rather because of the higher local OH concentration than between copolymer chains. This results in the intramolecular crosslinking. As the concentration increases, the difference between local OH concentration vs solution OH concentration will be lessened and as a result the intermolecular crosslinking between two copolymer chains will increase.
3.
20 / 0.042 / 23.81 / 566.89
30 / 0.039 / 25.64 / 657.46
40 / 0.028 / 35.71 / 1275.51
50 / 0.024 / 41.67 / 1736.11
60 / 0.021 / 47.62 / 2267.57
80 / 0.018 / 55.56 / 3086.42
90 / 0.015 / 66.67 / 4444.44
110 / 0.013 / 76.92 / 5917.16
120 / 0.012 / 83.33 / 6944.44
150 / 0.0096 / 104.17 / 10850.69
180 / 0.0082 / 121.95 / 14872.10
catalyzed 2nd order kinetic rate law:
1/[A]-1/[A]0=kct / 1/c-1=[A]0kct
uncatalyzed 2nd order kinetic rate law:
1/[A]2-1/[A]02=2kut / 1/c2-1=[A]022kut
Plot of 1/c vs t gives a straight line more so than plot of 1/c2 vs time. So the reaction is a catalyzed reaction.
kapp = [A]0kc = 0.62 min-1Name: poly(5-trimethylene-cyclopent-imide)
4.
xi=Ni/N=p(i-1)(1-p)
∂xi/∂p= (-1)p(i-1) + (1-p)(i-1)p(i-2)= 0
p(i-1) = (1-p)(i-1)p(i-2)
p = (1-p)(i-1)
(1-p)/p=1/(i-1)
1/p-1= 1/(i-1)
pmax= (i-1)/i
[A]/[A]0=1-pmax
[A]0/[A] = 1/(1-pmax)
1/[A]-1/=kct
[A]0/[A]-1=[A]0kct
1/(1-pmax)-1=[A]0kctmax
tmax = pmax /{(1-pmax)[A]0kc}
pmax(15mer) = (15-1)/15= 0.9333
tmax (15mer) = 0.9333/[(1-0.9333)x3.0x2.47x10-4] = 18883.3 s = 314.7 min
7.
5
5
I (exp) / wi x 10-4 (exp) / iwi / sum (wi) / wi=ip(i-1)(1-p)2 x 10-4 (theory)12 / 6.5 / 78 / 427.8 / 9.8
35 / 19.6 / 686 / sum(iwi) / 22.8
58 / 29.4 / 1705.2 / 89256.1 / 30.4
81 / 33 / 2673 / Nw / 34.1
104 / 35.4 / 3681.6 / 208.6 / 35.1
127 / 36.5 / 4635.5 / 34.4
150 / 33 / 4950 / Nw=(1+p)/(1-p) / 32.6
173 / 27.6 / 4774.8 / p=(Nw-1)/(Nw+1) / 30.2
196 / 25.2 / 4939.2 / p / 27.5
219 / 22.9 / 5015.1 / 0.9905 / 24.7
242 / 19.4 / 4694.8 / 21.9
265 / 18.5 / 4902.5 / 19.2
288 / 16.8 / 4838.4 / 16.8
311 / 15.2 / 4727.2 / 14.6
334 / 14.1 / 4709.4 / 12.6
357 / 13 / 4641 / 10.8
380 / 11.5 / 4370 / 9.2
403 / 11 / 4433 / 7.8
426 / 9.1 / 3876.6 / 6.7
449 / 7.2 / 3232.8 / 5.6
472 / 6.5 / 3068 / 4.8
495 / 4.9 / 2425.5 / 4.0
518 / 4.3 / 2227.4 / 3.4
541 / 3.9 / 2109.9 / 2.8
564 / 3.3 / 1861.2 / 2.4
Purple line is theoretical wi and blue line is experimental wi.
Slightly smaller p may generate a better fit. (You can try it out using the curve fitting in Origin).
9.
M0 = MW (caprolactam) =113.16
DPn= the total number of repeating units (caprolactam) / the total number of polymer chains
Mn = M0 x DPn
AB = COOH---NH2
Assume converted caprolactam is 1000 for ease of calculation:
(1) for the run with none additive:
A…B type polymer exists
the total number of polymer chains = ([-COOH] + [-NH2])/2 = (5.4 + 4.99)/2 = 5.22
DPn = 1000/5.22 = 191.57 ≈ 192; Mn = 113.16 x 191.57 = 21678 ≈ 21.7 k
(2) for the run with V (acetic acid) additive:
A…B and V…A types of polymers exist
the total number of polymer chains = [-COOH] = 19.8
DPn = 1000/19.8 = 50.51 ≈ 51; Mn = 113.16 x 50.51 = 5715 ≈ 5.7 k
(3) for the run with S (sebacic acid) additive:
A…B, A…S…A types of polymers exist
the total number of polymer chains = [-NH2] + ([-COOH]-[-NH2])/2 =([-NH2] +[-COOH])/2 = (2.3+21.1)/2 = 11.7
DPn = 1000/11.7 = 85.47 ≈ 85; Mn = 113.16 x 85.47 = 9672 ≈ 9.7 k
(4) for the run with H (hexamethylene diamine) additive:
A…B, B…H…B types of polymers exist
the total number of polymer chains = [-COOH] + ([-NH2]-[-COOH])/2 =([-NH2] +[-COOH])/2 = (19.7+1.4)/2 = 10.55
DPn = 1000/10.55 = 94.79 ≈ 95; Mn = 113.16 x 94.79 = 10726 ≈ 10.7 k
(5) for the run with T (trimesic acid) additive:
A…B, and T(…A)3 types of polymers exist
the total number of polymer chains = [-NH2] + ([-COOH]-[-NH2])/3 =(2x[-NH2] +[-COOH])/3 = (2x2.5+22.0)/3 = 9
DPn = 1000/9 = 111.11 ≈ 111; Mn = 113.16 x 111.11 = 12573 ≈ 12.6 k
Additive / Moles additive / Total # of functional groups due to additive / -COOH (meq) / -NH2 (meq) / DPn / Mn (10-3)
None / - / - / 5.4 / 4.99 / 192 / 21.7
V (-COOH) / 0.0205 / 0.0205 / 19.8 / 2.3 / 51 / 5.7
S (2-COOH) / 0.0102 / 0.0204 / 21.1 / 2.3 / 85 / 9.7
H(2 -NH2) / 0.0102 / 0.0204 / 1.4 / 19.7 / 95 / 10.7
T (3 – COOH) / 0.0067 / 0.0201 / 22.0 / 2.5 / 111 / 12.6
Additive addition breaks the stoichiometric balance of the caprolactam polymerization. So the DPn are all lower for the runs with additives than the one without additive. While the total numbers of functionalities of the additives is about the same, their effect on the DPn is different because their functionality per molecules is different. It also makes sense qualitatively. Monomeric additive (S) stops the chain growth at one end, thus reduce the DPn the most. Dimeric additives (S, H) allow the chain growth, its effect on DPn are lesser. Trimeric additive (T) can cross-link the chains, they effect on DPn reduction is the least.
12.
15.
(a)
(b) self-catalyzed: Nn2 = 1 + 2ku[A]02t
For Nn= 300, it takes t= 231.5 hr.
(c) strong acid catalyzed: Nn = 1 + kc[A]0t
For Nn= 300, it takes t= 0.46 hr.
(d) In self-catalyzed condensation, Nn is linearly proportional to the t1/2; In strong acid catalyzed condensation, Nn is linearly proportional to t.
16.
DPn(Nn) = Mn/M0(repeating unit) = 24116/238.24 = 101.2 ≈101
MW (m-amino aniline) = 108.14
MW (terephthalic acid) = 166.13
MW (benzoic acid) = 122.12
W= weight of the sample
νAo (diamine) = 2 x W x 39.31% / 108.14 = 7.2702 x 10-3 W
νBo (diacid) = 2 x W x 59.81% / 166.13 = 7.2003 x 10-3 W
νB’o (monoacid) = W x 0.88% / 122.12 = 0.07206 x 10-3 W
Amine is the limiting functional group.
r = νAo / (νBo+2νB’o) = 7.2702/(7.2003+2x0.07206) = 0.9899
r’ =2r/(1+r)= 2x0.9899/(1+0.9899) = 0.9949
DPn (Nn) = (1+r)/(1+r-2pr)= 1/ [1-2r/(1+r)p]= 1/(1-r’p)
p= (DPn-1) / (r’ DPn) = (101.2-1) / 0.9949 / 101.2 = 0.9952
if benzoic acid is doubled,
r = νAo / (νBo+2νB’o) = 7.2702/(7.2003+2x0.07206x2) = 0.9708
r’ =2r/(1+r)= 2x0.9708/(1+0.9708) = 0.9852
DPn (Nn) = 1/(1-r’p) = 1/(1-0.9852x0.9952) = 51.2 ≈ 51
The number average degree of polymerization is almost reduced by half.
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