Question

It is desired to prepare polyester with the number average molecular weight, Mn 5000 g/mol by reacting of 1 mol butane-1,4 diol (HO-CH-CH2-CH2-CH2-OH) with 1 mol of adipic acid (HOOC-CH2-CH2-CH2-CH2-COOH) (a) Write chemical equation illustrating synthesis of the polyester. (b) Calculate the value of the monomer conversion at which the reaction should be stopped to obtain this polymer, assuming perfect stoichiometric balance. What would be polydispersity index for the polymer obtained? (c) Assuming that 0.5 mol % of the diol is lost to polymerization by a side reaction (dehydratation to a non-reactive substance), what would be the value of Mn if the reaction was carried out to the same extent as in part (b) (d) For the polymerization process value of polymerization rate constant is 1.5102 L(mol s) at 100 degree C. How long would it take to prepare the polyester with the number average molecular weight, Mn5,000 g/mol at 100 degree C? There is no diol loss for this part of the problem Hints: All equations for step-growth polymerization include concentrations of functional groups reacting and not concentrations of monomers. The polymerization is a second order reaction.

Answer 1

Molar mass of Butane 1.4 diol = 90 g/mol Molar mass of Adifric acid = 146 g/mol Total = 236 g By the condensation polymerization polyester is formed. No. of H_2 O molecule in eliminated = No. of each monomer times 2 - 1 If x is the no. of monomer then: x times 236 - [(n - 1) times 18] = 500 [Mx = 5000 theta/mol] x = 24.91 TildeEqual 25 Poly disparity index = Mw/Mn = 5018/5000 = 1.0036 0.5 mol % 25 mole diol = 12.5 mole Therefore Mn = 12.5 times 236 - 23 times 18 = 2536 g/mole Therefore Mn TildeEqual 2500 g/mol \r\n This polymerization is 2^nd order Therefore Time required t = x/k middot a (a - x) = 5000/1.5 * 10^-2 times 5900 (5900 - 5000) x = 5000 g/mol a = 236 times 25 == 5900 g/mol k = 1.5 times 10^-2 L/mol s = 0.063 sec