The reaction rate of a self-catalyzed polyesterification reaction is given as -(d[COOH])/dt=k[COOH]²[OH] If [M] is the initial concentration of hydroxyl and carboxyl monomers, then the degree of polymerization, is given by

Q73 · CSIR-NET Chemistry, June 2019

Paper: CSIR-NET June 2019 · Subject: Physical Chemistry · Chapter: Polymer Chemistry · Topic: Polymerisation Kinetics · Marks: 2 · Difficulty: Medium

The reaction rate of a self-catalyzed polyesterification reaction is given as

\[ -\frac{\mathrm{d}[\mathrm{COOH}]}{\mathrm{dt}}=\mathrm{k}[\mathrm{COOH}]^{2}[\mathrm{OH}] \]

If [M] is the initial concentration of hydroxyl and carboxyl monomers, then the degree of polymerization, <N> is given by

(a)$\langle\mathrm{N}\rangle=2[\mathrm{M}]_{0}^{2} \mathrm{kt}$

(b)$\langle\mathrm{N}\rangle^{2}=2[\mathrm{M}]_{0}^{2} \mathrm{kt}$

(c)$\langle\mathrm{N}\rangle^{2}=2[\mathrm{M}]_{0}^{2} \mathrm{kt}+1$

(d)$\langle\mathrm{N}\rangle^{2}=2[\mathrm{M}]_{0} \mathrm{kt}+1$

Answer

Answer: C ✓ checked by 4AB · confidence high

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation

−d[M]/dt = k[M]³ integrates to 1/[M]² − 1/[M]₀² = 2kt. With ⟨N⟩ = [M]₀/[M], this gives ⟨N⟩² = 2[M]₀²kt + 1 (c).

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