
Polyethylene is used in many items, including water pipes, bottles, electrical insulation, toys, and mailer envelopes. It is a
The R •� species (called a radical) reacts with an ethylene molecule (M) to generate another radical:
The reaction of
This step can be repeated with hundreds of monomer units. The propagation terminates when two radicals combine:
The initiator frequently used in the
This is a first-order reaction. The half-life of benzoyl peroxide at

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
The relation between rate constant k and half-life is
Answer to Problem 135AP
Solution:
a)
b)
c)
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:
The reaction of
The propagation terminates when two radicals combine as follows:
a) The rate constant
The relation between the rate constant and half-life is given as follows:
The value of
is given as
The rate constant for the reaction is
b) The activation energy
The half-life of benzoyl peroxide is
The rate constant for the reaction at
There are two values of rate constants
Here,
Substitute the values
The activation energy for the reaction is
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:
The rate law for the propagation step is as follows:
The rate law for the propagation step is as follows:
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
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
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- The grading is not on correctness, so if you can just get to the correct answers without perfectionism that would be great. They care about the steps and reasoning and that you did something. I asked for an extension, but was denied the extension.arrow_forwardShow your work and do something that is reasonable. It does not have to be 100% correct. Just show something that looks good or pretty good as acceptable answers. Something that looks reasonable or correct would be sufficient. If you can get many of them correct that would be great!arrow_forwardShow your work and do something that is reasonable. It does not have to be 100% correct. Just show something that looks good or pretty good as acceptable answers. Something that looks reasonable or correct would be sufficient. If you can get many of them correct that would be great!arrow_forward
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- = 1 = 2 3 4 5 6 ✓ 7 8 ✓ 9 =10 Devise a synthesis to prepare the product from the given starting material. Complete the following reaction scheme. Part 1 of 3 -Br Draw the structure for compound A. Check Step 1 Step 2 A Click and drag to start drawing a structure. × ↓m + OH Save For Later S 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privaarrow_forwardPredict the products of this organic reduction: 田 Check AP + + H2 Lindlar catalyst Click an drawing 2025 McGraw Hill LLC. All Rigarrow_forward70 Suppose the molecule below is in acidic aqueous solution. Is keto-enol tautomerization possible? • If a keto-enol tautomerization is possible, draw the mechanism for it. Be sure any extra reagents you add to the left-hand sid available in this solution. • If a keto-enol tautomerization is not possible, check the box under the drawing area. : ☐ Add/Remove step Click and drag to st drawing a structure Check Save For Late. 2025 McGraw Hill LLC. All Rights Reserved. Terms of Usearrow_forward
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