Kinetics and Thermodynamics in Polymer Chemistry

KP5. Solutions to Selected Problems

 Problem KP1.1.

Two C-C σ bonds and one new π bond are made.  Three old π bonds are lost.

ΔH = bonds broken - bonds made

 ΔH = (3 x 64) - ((2 x 83) + 64) kcal/mol

ΔH = -38 kcal/mol

 

Problem KP1.2.

Tc = ΔH / (ΔS + Rlog[M]) 

Tc = - 7,000 cal mol-1 (-8.6 cal K-1 mol-1  +  1.98 cal K-1 mol-1 log(8.7))

Tc = - 7,000 (-8.6 + 1.98 (0.939)) K

Tc = - 7,000 (-6.74) K

Tc = 1038 K = 765 °C

 

Problem KP1.3.

For each amide bond, a C-N bond and a H-Cl bond are made.  A C-Cl and a N-H bond are lost.

ΔH = bonds broken - bonds made

     = (C-Cl + N-H)  - (C-N + H-Cl)

 ΔH = (81 + 93) - (73 + 102) kcal/mol

ΔH = -1 kcal/mol

 

Problem KP2.1.

75% conversion means 0.75 in terms of fractions.

DP = 1 / (1 - p)

DP = 1 / (1 - 0.75)

DP = 1 / 0.25

DP = 4

 

Problem KP2.2.

slope = 2 [M]0 k = 0.717 s-1

k = 0.717  s-1 / (2 x 17 mol L-1)

k = 0.021 L mol-1s-1

 

Problem KP2.3.

Mn = M0 / (1 - p)

      = 120 g/mol / (1 - 0.99)

      = 120 g/mol / 0.01

      = 12,000 g/mol

 

Mw = M0(1 + p) / (1 - p)

      =  120 g/mol  (1 + 0.99) / ( 1 - 0.99)

      = 120 g/mol  (1.99) / 0.01

      = 23,800 g/mol 

 

D = 1 + p

    = 1 + 0.99

    = 1.99 

 

Problem KP3.1.

Rate = k' [M][I]1/2

slope = k' [I]1/2

0.0024136 s = k' (0.00025)1/2

k' = 0.015 s-1

 

Problem KP3.2.

At steady state:

Rateinit   = Rateterm

 ki [M][I]  =   kt[M+]

Rearranging:

[M+]  = (ki/kt) [M][I]  

 

Problem KP3.3.

Rateprop = kp [M+][M]

Substituting the steady state expression for [M+]:

Rate = (kikp/kt) [M]2[I] 

 

Problem KP3.4.

v = Rateprop/Rateinit

vkp [M+][M] /  ki [M][I] =  (kp / ki)[M+] / [I]

but [M+] is not a known quantity. Alternatively, at steady state, Rateinit = Rateterm

v = Rateprop/Rateterm

vkp [M+][M] /  kt [M+] =  (kp / kt)[M]

 

Problem KP3.5.

 a) v =  [M]0/[I]0 = 4.5 / 1.25 x 10-3 = 3,400

b)  v =  (kp/2f kt kd)([M]/[I]1/2)

     v =  (0.003 / 2(0.5)(0.003)(0.0001))(4.5/(1.25 x 10-3)1/2)  =    (1/0.0001)(4.5/0.035) = 128/0.0001 = 1,280,000

c)  v =  (kp/2f kt kd)([M]/[I]1/2)

     v =  (0.003 / 2(0.5)(0.03)(0.0001))(4.5/(1.25 x 10-3)1/2)  =    (0.01/0.0001)(4.5/0.035) = 128/0.01 = 12,800

d) v =  (kp/2f kt kd)([M]/[I]1/2)

     v =  (0.003 / 2(0.5)(0.003)(0.1))(4.5/(1.25 x 10-3)1/2)  =    (1/0.1)(4.5/0.035) = 128/0.1 = 1,280

 

Problem KP4.1.

The key point is that, when two terms are added together and one is much larger than the other, the sum is approximately the same as the larger of the two terms.  You can ignore the smaller one.

 

Problem KP4.2.

 

 

This site is written and maintained by Chris P. Schaller, Ph.D., College of Saint Benedict / Saint John's University (with contributions from other authors as noted).  It is freely available for educational use.

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Structure & Reactivity in Organic, Biological and Inorganic Chemistry by Chris Schaller is licensed under a Creative Commons Attribution-NonCommercial 3.0 Unported License

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