Chapter 14, Problem 135AP

Polyethylene is used in many items, including water pipes, bottles, electrical insulation, toys, and mailer envelopes. It is a polymer, a molecule with a very high molar mass made by joining many ethylene molecules together. (Ethylene is the basic unit, or monomer, for polyethylene.) The initiation step is:

${\text{R}}_{\text{2}}\stackrel{K_1}{\to }2\text{R}\cdot \left(\text{initiation}\right)$

The R •� species (called a radical) reacts with an ethylene molecule (M) to generate another radical:

$\text{R}\cdot \text{+ M }\to {\text{ M}}_{\text{1}}$

The reaction of ${\text{M}}_{\text{1}}\text{•}$ with another monomer leads to the growth or propagation of the polymer chain:

${\text{M}}_{\text{1}}\cdot \text{ + M }\stackrel{k_{\text{p}}}{\to }{\text{ M}}_{\text{2}}\cdot \left(\text{propagation}\right)$

This step can be repeated with hundreds of monomer units. The propagation terminates when two radicals combine:

$\text{M'}\cdot \text{M"}\cdot \stackrel{k_1}{\to }\text{ M' - M" (termination)}$

The initiator frequently used in the polymerization of ethylene is benzoyl peroxide $\left({\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COO}\right)}_{\text{2}}\right]\text{:}$

${{\text{(C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COO)}}_{\text{2}}\to {\text{ 2C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COO}$

This is a first-order reaction. The half-life of benzoyl peroxide at $\text{100°C}$ is 19.8 min. (a) Calculate the rate constant $\left({\text{in min}}^{\text{-1}}\right)$ of the reaction. (b) If the half-life of benzoyl peroxide is 7.30 h, or 438 min, at $\text{70°C,}$ what is the activation energy $\left(\text{in kJ/mol}\right)$ for the decomposition of benzoyl peroxide? (c) Write the rate laws for the elementary steps in the preceding polymerization process, and identify the reactant, product, and intermediates. (d) What condition would favor the growth of long, high-molar-mass polyethylenes?

Interpretation Introduction

Interpretation:

The rate constant, activation energy, and the rate law of reaction are to be determined.

Concept introduction:

At a given temperature, the concentration ratio of concentration of product to concentration of reactant in a chemical reaction is called rate constant. According to the Arrhenius equation, rate constant depends on temperature.

The amount of energy required for the reaction to form products by the formation of an activated complex is called activation energy.

Half-life: The time required for the concentration of a reactant to decrease to one-half its initial value. This period of time is called the half-life of the reaction, written as $t_{1/2}$. Thus, the half-life of a reaction is the time required for the reactant concentration to decrease from ${\text{[A]}}_{\text{0}}{\text{ to [A]}}_{\text{0/2}}$

The relation between rate constant k and half-life is

$k=\frac{0.693}{t_{1/2}}$

Answer to Problem 135AP

Solution:

a) $0.035{\min }^{-1}$

b) $110kJ/mol$

c) $\text{Rate}=k_{\text{i}}\left[{\text{R}}_2\right]$

$\text{Rate}=k_{\text{p}}\left[\text{M}\right]\left[{\text{M}}_1\right]$

$\text{Rate}=k_{\text{t}}\left[{\text{M}}^{\text{'}}\right]\left[{\text{M}}^{"}\right]$

d) A higher rate of propagation step and a lower rate of termination step.

Explanation of Solution

Given information: The initiation step is as follows: ${\text{R}}_2\stackrel{k_{\text{i}}}{\to }2\text{R}\cdot$. The $\text{R}\cdot$ species is a radical that reacts with M as follows: $\text{R}\cdot \text{+M}\stackrel{}{\to }{\text{M}}_1\cdot$

The reaction of ${\text{M}}_1\cdot$ leads to the propagation of the polymer chain as follows: ${\text{M}}_1\cdot \text{+M}\stackrel{k_{\text{p}}}{\to }{\text{M}}_2\cdot$

The propagation terminates when two radicals combine as follows: $\text{M}\text{'}\cdot +\text{M"}\cdot \stackrel{k_{\text{t}}}{\to }\text{M}\text{'}-\text{M"}$. The initiator frequently used in the polymerization of ethylene is benzoyl peroxide and the reaction is as follows:

${\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COO}\right)}_2\stackrel{}{\to }2{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COO}$

a) The rate constant ${\text{(in min}}^{-1}\text{)}$ of the reaction

The relation between the rate constant and half-life is given as follows:

$k=\frac{0.693}{t_{1/2}}$

The value of $t_{1/2}$

is given as $19.8\text{}\text{ min}$. By substituting this value in the above equation, the rate constant is determined as follows:

$\begin{array}{c}k=\frac{0.693}{t_{1/2}}\\ =\frac{0.693}{19.8\min }\\ =0.035{\min }^{-1}\end{array}$

The rate constant for the reaction is $0.035{\min }^{-1}$.

b) The activation energy $\text{(in kJ/mol)}$ for the decomposition of benzoyl peroxide.

The half-life of benzoyl peroxide is $7.30\text{ hr or 438 min at 70}°\text{C}$.

The rate constant for the reaction at $70°C$ is as follows:

$\begin{array}{c}k=\frac{0.693}{t_{1/2}}\\ =\frac{0.693}{438\min }\\ =0.00158{\min }^{-1}\end{array}$

There are two values of rate constants $k_1$ and $k_2$ at temperatures $T_1$ and $T_2$. According to the Arrhenius equation, the activation energy is calculated as follows:

$\ln \frac{k_1}{k_2}=\frac{E_a}{\text{R}}\left(\frac{T_1-T_2}{T_1{T}_2}\right)$

Here, $\frac{k_1}{k_2}$ is the rate constant, $\text{R}$ is the gas constant, $E_a$ is the activation energy, and $T_1\text{ and }{T}_2$ are the absolute temperatures.

Substitute the values $\frac{k_1}{k_2}$, $R$, and $T_1\text{ and }{T}_2$ in the above equation as follows:

$\begin{array}{c}\ln \left(\frac{0.0350{\min }^{-1}}{0.00158}\right)=\frac{E_a}{\left(8.314\text{ J/mol}\text{.K}\right)}\left(\frac{\left(373-343\right)\text{K}}{\left(373\right)\left(343\right)\text{K}}\right)\\ \ln 22.1518=\frac{E_a}{\left(8.314E\text{J/mol}\text{.}\cancel{\text{K}}\right)}\left(\frac{30}{127939\cancel{\text{K}}}\right)\\ E_a=\frac{\left(3.0979\right)\left(8.314\text{ J/mol}\right)\left(127939\right)}{30}\end{array}$

$\begin{array}{c}{E}_a=\frac{3295189.284}{30}\\ =109838.64\text{ J/mol}\\ =1.10\times {10}^5\text{ J/mol}\\ \end{array}$

$E_a=110\text{ kJ/mol}$

The activation energy for the reaction is $110kJ/mol$.

c) The rate laws to be written for elementary steps in the preceding polymerization process and identify the reactant, product, and intermediates.

The rate law for the initiation step is as follows:

$\text{Rate}=k_{\text{i}}\left[{\text{R}}_2\right]$

The rate law for the propagation step is as follows:

$\text{Rate}=k_{\text{p}}=\left[\text{M}\right]\left[{\text{M}}_1\right]$

The rate law for the propagation step is as follows:

$\text{Rate}=k_{\text{t}}\left[{\text{M}}^{\text{'}}\right]\left[{\text{M}}^{"}\right]$

The ethylene monomers are the reactant molecules, and polyethylene is a product in the reaction mechanism. The intermediates are formed in the early elementary step and consumed in the next step. So, the intermediates are radicals of $R$ species, $M^{\text{'}}$ and $M^{"}$.

d) The condition that favor the growth of long, high-molar-mass polyethylenes.

A higher rate of propagation and a lower rate of termination are favoured when the growth of long polymers take place. In the propagation step, the rate law is dependent on the concentration of monomer. If the concentration of ethylene is increased, the propagation rate also increases.

The concentration of radical fragments $M^{\text{'}}$ and $M^{"}$ is low and it can be seen from the rate law of termination, due to which a slower rate of termination takes place. By using a low concentration of initiator $\left(R_2\right)$, it can be accomplished.