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degree of polymerization

label Chemistry
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. After 130 minutes at 240 degrees Celsius 96% of the COOH and NH2 groups had reacted. Calculate the value of CoK where Co is the concentration of monomer t=0 and k is the rate constant.

Oct 17th, 2017

  1. This is a second order reaction since Monomer 1 + Monomer 2 = Polymer

    The rate low for a second order reaction is: Rate = - ∆ Co/2∆t=k [Polymer]2

    1/[Polymer]= 1/[Monomer] + kt

    [Monomer]/ [Polymer]=1+[Monomer]Kt

    [Monomer]/ [Polymer]t=1/t+[Monomer]k

    [Monomer]/ [Polymer]t - 1/t = [Monomer]k (This is the value that you need to find)

    If 96% was polymerized, then you can let value [Monomer]=1M, and value [Polymer]= 0.96M.

    Thus, (1/(0.96x130min) - 1/130 min = [Monomer]k

    1/124.8 - 0.0077 = [Monomer]k

    0.008 – 0.0077 = [Monomer]k = 0.00031 or 3.1 x 10 -4 sec-

    Good Luck!


Dec 17th, 2014

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