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feed ratio of adipic acid and hexamethylene diamine

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feed ratio of adipic acid and hexamethylene diamine

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1. Calculate the number-average degree of polymerization of an equimolar mixture of

adipic acid and hexamethylene diamine for extent of reaction 0.500, 0.800, 0.900,

0.950, 0.970, 0.990 and 0.995.

Solution:




1
Xn 
(1  p)

Where, p is the extent of reaction and X n is number-average degree of polymerization.



Extent of reaction (p) 0.5 0.8 0.9 0.95 0.97 0.99 0.995

Number-average degree of polymerization ( X n ) 2 5 10 20 33 100 200

, 2. Calculate the feed ratio of adipic acid and hexamethylene diamine that should be

employed to obtain polyamide of approximately 15,000 molecular weight at 99.5%

conversion. What is the identity of the end groups of this product? Do the same

calculation for a 19,000-molecular weight.


Solution:




Note that




Molecular weight (Mw) = 146.14g/mol Molecular weight (Mw) = 116.21g/mol

N A (1  1/ r ) / 2 1 r
Xn  
[ N A (1  p)  N B (1  rp)] / 2 1  r  2rp

A repeat unit of Nylon 6,6 can be thought of as being composed of a monomer adipic
acid having reacted with hexamethylene diamine with the release of 2 water molecules.

Thus, molecular weight of a repeat unit

M repeat  146.14 g / mol  116.21g / mol – 2 18g / mol


= 226.35 g/mol

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