## Exploration on Some Problems about Condensation Polymerization in Polymer Chemistry Teaching

Meng Zhifen,

 基金资助: 河南省高等学校合格基层教学组织建设项目

 Fund supported: 河南省高等学校合格基层教学组织建设项目

Abstract

There are some problems in the condensation polymerization teaching, such as the ratio of function groups, the Polyester linear equilibrium polycondensation kinetic equation, etc. The students were often puzzled with these problems. Based on many years of polymer chemistry teaching experience, these problems were discussed in this paper to provide some reference for peers teaching and students learning.

Keywords： Condensation polymerization ; The ratio of function groups ; The polyester linear equilibrium polycondensation kinetic equation

Meng Zhifen. Exploration on Some Problems about Condensation Polymerization in Polymer Chemistry Teaching. University Chemistry[J], 2021, 36(2): 2001002-0 doi:10.3866/PKU.DXHX202001002

## 1 关于r值的含义

$\overline {{X_n}} = \frac{{1 + r}}{{1 + r - 2rp}}$

r为基团数比，则：$r = \frac{{{N_{\rm{a}}}}}{{{N_{\rm{b}}}}} = \frac{{1 \times 2}}{{1.02 \times 2}} = 0.98$

r为单体物质的量之比，则：$r = \frac{{{N_{\rm{a}}}}}{{{N_{\rm{b}}}}} = \frac{1}{{1.02}} = 0.98$

$r = \frac{{{N_{\rm{a}}}}}{{{N_{\rm{b}}} + 2N'_{\rm{b}}}}$

r为基团数比，计算得：r = 0.985；$\overline {{X_n}}$= 79.4。

r为单体物质的量之比，计算得：r = 0.971；$\overline {{X_n}}$= 50.5。

## 3 关于聚酯化线形平衡缩聚动力学方程的推导与应用

$\overline {{X_n}} \approx \sqrt {\frac{K}{{{n_{\rm{w}}}}}}$

t时，水未排除ccc0cc0c

t时，水部分排除ccc0cNw

$R = - \frac{{{\rm{d}}c}}{{{\rm{d}}t}} = {k_1}{c^2} - {k_{ - 1}}{({c_0} - c)^2}$

$R = - \frac{{{\rm{d}}c}}{{{\rm{d}}t}} = {k_1}{c^2} - {k_{ - 1}}({c_0} - c){N_w}$

$\because p = \frac{{{N_0} - N}}{{{N_0}}} = 1 - \frac{N}{{{N_0}}} = 1 - \frac{c}{{{c_0}}} ； \therefore c = {c_0}\left( {1 - p} \right)$

$- \frac{{{\rm{d}}c}}{{{\rm{d}}t}} = {k_1}c_0^2{(1 - p)^2} - \frac{{{k_1}}}{K}c_0^2{p^2}$

$- \frac{{{\rm{d}}c}}{{{\rm{d}}t}} = {k_1}c_0^2\left[ {{{(1 - p)}^2} - \frac{{{p^2}}}{K}} \right]$

$- \frac{{{\rm{d}}c}}{{{\rm{d}}t}} = {k_1}c_0^2{\left( {1 - p} \right)^2} - \frac{{{k_1}}}{K}{c_0}p{N_{\rm{w}}}$

$- \frac{{{\rm{d}}c}}{{{\rm{d}}t}} = {k_1}c_0^2\left[ {{{\left( {1 - p} \right)}^2} - \frac{{p{N_{\rm{w}}}}}{{K{c_0}}}} \right]$

${k_1}c_0^2\left[ {{{(1 - p)}^2} - \frac{{{p^2}}}{K}} \right] = 0$

${(1 - p)^2} - \frac{{{p^2}}}{K} = 0$

$p = \frac{{\sqrt K }}{{\sqrt K + 1}}$

$\overline {{X_n}} = \sqrt K + 1$

${k_1}c_0^2\left[ {{{\left( {1 - p} \right)}^2} - \frac{{p{N_{\rm{w}}}}}{{K{c_0}}}} \right] = 0$

${\left( {1 - p} \right)^2} - \frac{{p{N_{\rm{w}}}}}{{K{c_0}}} = 0$

$\overline {{X_n}} = \frac{1}{{1 - p}}$

$\overline {{X_n}} = \sqrt {\frac{{K{c_0}}}{{p{N_{\rm{w}}}}}}$

nw = Nw/c0，则式(18)可写为：

$\overline {{X_n}} = \sqrt {\frac{K}{{p{n_{\rm{w}}}}}} \approx \sqrt {\frac{K}{{{n_{\rm{w}}}}}}$

$\overline {{X_n}} = \sqrt {\frac{K}{{P{P_{\rm{w}}}}}}$

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