课后习题答案 第三-九章 - the book of polymer chemistry and its solution mannual PDF

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3 Ri=2fkd[ ]I =Rp !; therefore, f=Rp! (2kd)[ ]ISince [I] is given in g L- 1 , these concentrations must be divided by 164 g mole- 1 ; likewise, kd =1 × 10 - 5 sec - 1 .The average value of f = 0 ± 0.3 For poly(methyl methacrylate):For polystyrene: (C 8 H 8 )n(C 4 H 6 N)mr= cpm polymer cpm initiator=...

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Hiemenz and Lodge, Polymer Chemistry, 2nd Edition , Solutions Manual

3.1

R i = 2fk d[ I] =

Rp !

; therefore, f =

Rp !

(2k d )[I]

-1

-1

Since [I] is given in g L , these concentrations must be divided by 164 g mole ; likewise, kd = -5

-1

1.08 × 10 sec . -1

[I](mole L ): 3.39 × 10 f

-4

0.516

calc:

-3

3

-3

-3

1.52 × 10

1.52 × 10 6.10 × 10

9.15 × 10

1.52 × 10

0.478

0.524

0.553

0.521

0.515

-2

The average value of f = 0.518 ± 0.024 3.2

For poly(methyl methacrylate):

r=

n

cpm poly ! 104 cpm I

m!

5nr 4

4,440

2.13

1.18

3,120

3.12

1.22

2,980

3.01

1.12

1,470

6.27

1.15

1,240

7.93

1.23

913

10.7

1.22

894

10.8

1.21

Average:

1.19

For polystyrene: (C8H8)n(C4H6N)m r=

4m 4m cpm polymer ! = cpm initiator 4m + 8m 8n 4

n

r × 10

3683

2.64

1.95

1130

8.96

2.02

1100

9.27

2.04

1000

9.99

2.00

971

11.80

2.28

Average:

2.05

m = 8nr/4

1

Hiemenz and Lodge, Polymer Chemistry, 2nd Edition , Solutions Manual

3.3

[M ] = !R p

2k t 2k t with = 6.48 " 104 kp kp

also '1 ! k p $'1 Rp ! kp $ with # & = 2.33 [M ] = 1 2 # 1 2& #" (2k t )1 2 &% R i #"(2k t ) &%

3.4

-1

4

1/2

Run

[M] = 6.48 x 10 τRp(mole L )

5

16.1

16.3

6

13.5

13.1

8

13.5

13.4

12

17.8

18.1

13

14.3

14.5

[M] = 2.33 Rp/Ri

-1

= (mole L )

! fk $1 2 d kp = 12 #k & " t % [M][ I] Rp

Rp 12

[M ][ I]

(

)

! 102 L1 2mole"1 2 min"1 :

8.38; 8.25; 8.64; 8.37; 8.33; 8.54; 8.13; 7.77; 8.41; 8.28 -3

Average:

8.31

-1

Since f = 1.0 and kd = 5.8 × 10 min ,

8.31 ! 10"2 = = 1.09 L1 2mole"1 2 min"1 12 "3 1 2 (5.8 ! 10 ) kt kp

3.5 τ is roughly constant (1.8 – 2.3) over the first 30% conversion to polymer. At higher conversions, the polymer radicals are larger (on average) than at low conversions. This leads to a slight increase in the viscosity, and given that the termination of active radical chains is diffusion limited, the rate constant of termination (kt) drops accordingly. This decrease in termination rate accounts for the increased lifetime observed for radicals at higher conversions. This is a rather mild example of the auto-acceleration, known as the Trommsdorff effect, discussed in Section 3.4B.

3.6

2 i"2

N n = ! ixi = ! i(i "1)(1 " p) p

=

(1 " p)2 2 i"1 i"1 ! i p " !ip p

(

)

2

Hiemenz and Lodge, Polymer Chemistry, 2nd Edition , Solutions Manual

2 (1 ! p)2 $ " " i' (1 ! p) $ " " " 1 ' " 1 i p & p ) & p #p ! # ) = ! "p p % "p "p p % "p "p 1! p "p 1! p ( (

=

1 % (1 ! p)2 + 2p(1 ! p) ! (1 ! p)2 2 (1 ! p)2 " (1 ! p)2 + 2p(1 ! p) '' = $$ ! = . 2 4 2 p # 1! p (1 ! p) & p(1 ! p) (1 ! p) We can find wi as in ini ix wi = i = = i ! ini N n ! n i N n Therefore N w = ! iw i = =

1" p 2 1" p 2 ! i (i "1)p i"2 (1 " p) 2 = ! i (i "1)p i"1 (1 " p)2 2 2p

1! p " 2 1! p " p # i(i !1)p i!2 (1 ! p) 2 #i (i !1)p i (1 ! p) 2 = 2p "p 2p "p

The term in the sum is Nn from before, so Nw =

1! p 2p(1! p) + p 2 2(1! p) + p 2 ! p 1! p " 2 2 = = = p 1! p 1! p p 2p "p 1! p (1 ! p)2

not quite right yet. 3.7

3.8

Assuming p = 1: N !=2 w "3 Nn

(1 ! p) =

2!" Nn

p

0.20

0.00159

0.99841

0.34

0.00180

0.99820

0.46

0.00228

0.99772

0.48

0.00250

0.99750

For disproportionation: xi = (1 ! p )p i! 1 (eqn 3.7.16)

For combination: x i = (i ! 1)(1 ! p )2 p i! 2 (eqn 3.7.26) Derivation of number average degree of polymerization expression:

3

Hiemenz and Lodge, Polymer Chemistry, 2nd Edition , Solutions Manual

"

"

"

N n = $ ixi = ! $ i (1# p ) p i#1 + (1# ! )$ i (i # 1)(1# p ) 2 p i# 2 i =1

i =1

i =1

"

"

i= 1

i= 1

N n = ! (1# p ) $ip i#1 + (1# ! )(1# p ) 2 $i (i # 1) p i# 2 = ! (1# p )

Nn =

1 2 + (1# ! )(1# p ) 2 2 (1 # p ) (1 # p ) 3

! 2(1 #! ) 2 #! + = 1# p 1# p 1# p

Note that: !

) i( i "1) p

i" 2

=

i= 1

d # ! i"1 $ d # 1 2 $ ip & = = ) % % 2 & 3 dp ' i= 1 ( dp ' (1 " p) ( (1 " p)

Now derive the equation for the PDI. Given: wi =

ixi Nn

Therefore: !

!

1 Nw = Nn Nn

" iw

i

i =1 !

=

"w

i

i= 1

1 Nn

i 2 xi

"N i=1 !

n

ixi

"N i= 1

n

!

"i x 2

=

1 Nn

i

i =1 !

" ix

i

i= 1

Substituting: " $ " 2 # % i# 1 (1 ) (1 ) ! ! # + # i p p i 2 (i # 1)(1 # p )2 p i 2 ' * * & Nw 1# p i= 1 & i=1 ' = " Nn 2 # ! & ! " 2 i# 2 ' i# 1 ! i (1 # p ) p + (1 # )* i (i # 1)(1 # p ) p &( * ') i =1 i =1

" " $ % 2 i #1 2 p i p p i 2( i #1) pi # 2 ' (1 ) (1 )(1 ) ! ! # + # # * * Nw 1# p & i=1 i=1 & ' = " " Nn 2 # ! & ! (1 ) i #1 i #2 ' 2 # p * ip + (1# ! )(1# p ) * i (i # 1) p &( ') i=1 i= 1

$ 1+ p 1 3(1 + p ) , % 2 + & ! (1 # p) (1 p) 3 + (1 # !)(1 # p) - (1 p) 3 + (1 p) 4 . ' # # Nw 1# p & / # 0' = 1 2 ' Nn 2 # ! & 2 ! (1# p ) + (1# ! )(1# p) 2 3 & ' (1# p ) (1# p ) ( )

Note that: !

) i2 (i " 1) p i"2 = i= 1

1 3(1+ p ) d # ! 2 i"1 $ d # 1+ p $ i p &= = + ) % % 3& 3 4 dp ' i= 1 ( dp ' (1 " p ) ( (1 " p ) (1 " p )

4

Hiemenz and Lodge, Polymer Chemistry, 2nd Edition , Solutions Manual

Now simplify terms and obtain the final expression: " 1+ p 3(1+ p) % # $ 1& p + (1 & ! ) ) + '! 2 2 2*( N w 1 & p ' (1 & p ) + (1 & p) (1 & p) , ( = 1 2 ( Nn 2 & ! ' ! + (1& ! ) ' ( 1& p 1& p . 4 + 2p # " 1+ p + (1 & ! ) ! ' Nw 1& p 1& p 1 & p ( ! (1 + p ) + (1 & ! )(4 + 2 p ) ' (= = Nn 2 & ! ' (2 & ! )(2 & ! ) ! + 2(1& ! ) ( '(. N w 4 & 3! & ! p + 2 p = (2 & ! )2 Nn

3.9

With α = 0.65 and p = 0.99754, the values of the first and second terms and their sums (=

wi/wtot) are listed below for various values of i:

i

4

4

first term x 10 second term x 10 wi/wtotx10

200

4.82

0.64

5.46

400

5.89

1.56

7.45

500

5.75

1.81

7.66

600

5.40

2.15

7.55

700

4.92

2.28

7.20

800

4.40

2.33

6.73

1000

3.36

2.23

5.59

1200

2.46

1.96

4.42

1400

1.76

1.63

3.39

1600

1.23

1.30

2.53

1800

0.84

1.01

1.85

2000

0.57

0.76

1.33

2500

0.21

0.35

0.56

3000

0.07

0.15

0.22

4

5

Hiemenz and Lodge, Polymer Chemistry, 2nd Edition , Solutions Manual

0.0008 0.0007 0.0006 0.0005

w i

0.0004 0.0003 0.0002 0.0001 0 0

500

1000

1500

i

2000

2500

3000

3500

The term with the α weighting factor gives the contribution of termination by disproportionation; the term weighted by (1– α), the combination contribution. It follows, therefore, that the weight fraction i-mer produced by combination is given by 3 i–2

0.5i(i–1)(1–p) p

.

However, see the next problem for a different view! 3.10 The error in this expression arises because while it is ok to use the weighting factors α and (1 – α) for adding the mole fractions, it is not ok for weight fractions. Recalling that weight fractions can be obtained from mole fractions simply as wi = ixi/Nn, the correct expression should be

wi =

!i(1 " p)pi"1 + (1" !)i(i "1)(1 " p) 2 p i"2 !i(1 " p)2 p i"1 (1 " !)i(i "1)(1 " p) 3 p i" 2 + = 2"! 2"! (2 " !) /(1 " p) 0.0008 0.0007 0.0006 0.0005

w i

0.0004 0.0003 0.0002 0.0001 0 0

500

1000

1500

i

2000

2500

3000

3500

The dashed line in the plot shows that this expression is noticeably different, and would require finding new values of p and α. 6

Hiemenz and Lodge, Polymer Chemistry, 2nd Edition , Solutions Manual

3.11

1 [SX] 1 = + Csx ! tr ! [M ]

(a)

1 1 [ SX] ; + 0.46 ! 10"4 = 3709 6580 [M ]

[SM] = 2.56 [M ]

(b)

1 1 + Csx (0.492) ; = 510 6670

CSX = 36.8 × 10

-4

(d)

1 1 ! tr = 147 = + (114.4 x 10"4 )(0.583) ; ! tr 6670 1 1 ! = 920 = + 125.2 "10#4 (0.772) ; 93 ! The values of ! are fairly constant in all solvents except chloroform.

3.12

Plot ! tr versus [CCl4]/[styrene] at each temperature. Slope gives value of CMX and

(c)

(

)

"1

"1

intercept gives ! . Each set of data gives good straight-line plots from which the following parameters are obtained: At 60 °C CMX = slope:

9.1 × 10

intercept:

12 × 10

υ:

8.3 × 10

-3

At 100 °C -3

20.8 × 10

-5

20 × 10

-5

3

5.0 × 10

3

Since CMX is the ratio of two rate constants, it can be analyzed by the Arrhenius method to give an apparent activation energy (Eapp*) which, in turn, equals Etr* – Ep*. E app * " 1 C 1 % 1n MX,1 = ! $ ! ' R # T1 T2 & C MX,2

1n

21 ! 10"3 "3

9 !10

E app* =

="

E app * # 1 1 & " ( % 8.314 $ 373 333'

(0.847(8.314) "4

(3.22 ! 10 )

= 22,000 J mole"1 = 22 kJ mole"1

υ decreases with increasing temperature as expected for thermally initiated polymerizations.

7

Hiemenz and Lodge, Polymer Chemistry, 2nd Edition , Solutions Manual

0.003

0.0025

1/!

tr

0.002

0.0015

0.001

0.0005

y = 0.00012256 + 0.0089773x R= 0.99957 y = 0.00020501 + 0.021696x R= 0.99896

0 0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

[CCl ]/[styrene] 4

3.13 For the free radical polymerization of isobutylene, the transfer to monomer is too favorable: +

+

+

3.14 O O

O

2

H

+

O

O

H C

H

(a)

.

R

+

.

R

(b) H R

H

H

H

.

.

+

R

H

R

H R

+ (c)

8

Hiemenz and Lodge, Polymer Chemistry, 2nd Edition , Solutions Manual

H H

R

H

H

H

.

R

H

+

+

(d) 3.15

3.16 Note: polymerization is run at 60 ºC, which is not explicit in the problem statement (a) R 1.6" 10!10 Ms ! 1 !9 ! 1 fk d = i = = 2.0 " 10 s 2[I ] 2(0.04 M ) # fk [I R p = k p[ M ] % d ' kt kp kt

1/ 2

=

1/ 2

]$ & (

Rp [ M ]( fkd [ I ])

1/ 2

=

Rp #R$ [M ]% i & ' 2(

1/ 2 =

6.4 "10 !7Ms ! 1 !10

!1

# 1.6 "10 Ms $ 2.0 M % & 2 ' (

1/ 2

= 0.036 ( Ms) !1/ 2

9

Hiemenz and Lodge, Polymer Chemistry, 2nd Edition , Solutions Manual

(b)

!=

Rp Ri

=

6.4 # 10"7 Ms" 1 = 4000 1.6 # 10"10 Ms" 1

(c) 1 1 1 = + CM = + 0.85 # 10 "4 = 3.35 # 10 " 4 4000 ! tr !

! = N n = 2.99# 10 3 3 5 M n = N n M 0 = 2.99 # 10 (104 g / mol ) = 3.10 # 10 g / mol

(d) 1 1 [T ] = + CT [M ] ! tr ! # $ # 1 1 $ [M ] % 1 1 & 2.0 M & [T ] = % " & =% " = 0.521 M ' 77 g / L CCl 4 "3 ) ! tr ! * C T % 40000 g / mol 4000 & 9.0( 10 % 104 g / mol & ) *

(e)

Eapp * = E p * +

E d * Et * ! 2 2

Values presented in text: Ep * = 26,000 J/mol (at 60 ºC) Et * = 8,000 J/mol (at 60 ºC) Ed * = 146,900 J/mol (at 100 ºC) Assume the activation energies at 100 ºC are equivalent to those presented in the text. Thus, Eapp * = 95,500 J/mol Also presented in the text: kp = 1.65 x 102 (Ms)-1 (60 ºC) kt = 6.0 x 107 (Ms)-1 (60 ºC) kd = 8.8 x 10-7 s-1 (100 ºC) Use these values to solve for the prefactors Ap, At, and Ad 2

!1

k p = 1.65" 10 (Ms ) = Ap e

!26000 8.314(333)

Ap = 1.98" 106 (Ms )!1 7

!1

!8000 8.314(333)

kt = 6.0" 10 (Ms ) = At e At = 1.08 "109 ( Ms) !1

!146900

k d = 8.80 " 10!7 s! 1 = Ad e8.314(373) Ad = 3.29 " 1014 s !1 Find kd at 60 ºC to determine f:

10

Hiemenz and Lodge, Polymer Chemistry, 2nd Edition , Solutions Manual

14

! 146900 !1 8.314(333)

k d = 3.29 "10 s e

= 2.98 " 10!9 s! 1

9 1 fk d 2.0 " 10 ! s ! = = 0.671 9 1 kd 2.98" 10! s ! Calculate Rp at 100 ºC: 1/ 2 E app * " Ad # + ln( f [I ]1/ 2[M ]) ln R p = ln Ap $ % ! RT & At '

f=

" 3.29 ( 1014 # ln R p = ln[1.98 (10 $ 9 % & 1.08 (10 '

1/ 2

6

]!

95500 + ln[(0.671)(0.04)1/ 2 (2)] 8.314(373)

R p = 2.5( 10! 6 Ms! 1

(f) 1/ 2

d[ M ] k p " Ri # ! = 1/ 2 $ % [ M ] dt kt & 2 ' 1/ 2 d[ M ] k p " Ri # ! = $ % dt [ M ] kt 1/ 2 & 2 ' [M ]

t kp " R # d[ M ] )[M ] ! [ M] = )0 kt1/ 2 $& 2i %' 0

1/ 2

dt

1/ 2

ln

kp " R # [M ] = 1/ 2 $ i % [M ]0 kt & 2 '

t !8000

kt = 1.08 ( 10 9( Ms) ! 1 e8.314(373) = 8.19 ( 10 7( Ms) ! 1 ! 26000

k p = 1.98( 106 (Ms )! 1 e 8.314(373) = 4.52 ( 102 (Ms )! 1 t = 18000 s

Calculate Ri @ 100 ºC: R i = 2 fk d [I ] = 2(8.80" 10!7 s ! 1 )(0.671)(0.04 M ) = 4.72" 10! 8Ms ! 1

ln

[ M ]0 4.52 "102 ( Ms) !1 (4.72 "10 !8 Ms !1 )1/ 2 = (18000 s ) [M ] (8.19 "107 ( Ms) !1 )1/ 2 2

[M ]0 2.0 M = 1.15 # [M ] = = 1.74 M [M ] 1.15

p=

[ M ] 0 ! [ M ] 2.0 !1.74 = = 0.13 [ M ]0 2.0

11

Hiemenz and Lodge, Polymer Chemistry, 2nd Edition , Solutions Manual

4.1

In example 4.1, at some indicated time, the polymer was terminated and analyzed for molecular weight. For ideal living anionic polymerization, the conversion p should have a relationship with time as

p= 1− exp( −kp[I]ot)

(1)

And p is derived indirectly from Nn assuming the following relation is accurate: p=Nn[I]0/[M]0

(2)

In the real system, due to some non-ideality, equation (2) might be not correct. If the relation between conversion and time was the object, the conversion should be calculated directly based on the actual yield of polymers.

4.2

There are four requirements for living polymerization, which lead to Poisson distribution:

1.

A living polymerization proceeds in the absence of transfer and termination reactions.

2.

All active chain ends must be equally likely to react with a monomer throughout the polymerization (equal reactivity and good mixing of reagents at all times).

3.

The rate of initiation needs to be much more rapid than the rate of propagation.

4.

Propagation must be essentially irreversible Nn

PDI(Exp.)

PDI(Poisson)

PDI(p)/PDI(e)

36.3

1.06

1.026

0.968

198

1.02

1.005

0.985

324

1.02

1.003

0.983

413

1.01

1.002

0.992

479

1.008

1.002

0.994

528

1.006

1.002

0.996

593

1.005

1.002

0.997

• If condition 1 is not satisfied, for example, a living chain transfers to another polymer chain, there will be a relatively big increase in PDI. •

If condition 2 is not satisfied, PDI will slightly increase for further polymerization,

• If condition 3 is not satisfied, PDI difference will be smaller as the reaction goes on, because the slow initiation effect will be reduced as the polymer becomes bigger. • If condition 4 is not satisfied, which means that adding a monomer is reversible, the PDI will increase as the polymerization goes on, and finally approaches 2.

Based on the results in example 4.1, condition 3 is most probably not satisfied. 4.3 The key factor that comes into play in non-polar solvents is ion-pairing or clustering of the living ends. Ionic species tend to be sparingly soluble in hydrocarbons, as the dielectric constant of the medium is too low. The rate of polymerization is therefore first order in monomer concentration 1

Hiemenz and Lodge, Polymer Chemistry, 2nd Edition , Solutions Manual

and has a (1/n) fractional dependence on initiator concentration, n is the aggregation number. For the living polymerization of styrene in cyclohe...