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2 years ago

Name the reaction processes that lead to the two types of synthetic polymers.

1 answer:

7 0

The reaction processes that lead to the two types of synthetic polymers are chain reaction polymerization and step reaction Polymerization.

<h3>What is Polymerization Reaction? </h3>

The chemical reaction which involves the addition of monomers to form higher molecular mass molecules is known as polymerization. There are two basically two types of polymerization, one is chain-reaction (or addition) and the other is step-reaction (or condensation) polymerization.

<h3>Chain-reaction (addition) polymerization</h3>

One of the most common and important types of polymer reactions is chain-reaction (addition) polymerization. This type of polymerization involves three step process including two chemical entities. The first which is known simply as a monomer. It is initially exists as simple units. Normally in all cases, the monomers must have at least one carbon-carbon double bond. Ethylene is one of the best known example of a monomer which is used to make a common polymer.

<h3>Step-reaction (condensation) polymerization </h3>

It is second type of polymerization. This polymerization method is used to produce polymers having lower molecular weight as compared to chain reactions and requires higher temperatures to occur.

Unlike chain reaction polymerization, this reactions include two different types of di-functional end groups or monomers which react with one another, forming a chain.

Step wise polymerization produces a small molecular by-product (water, HCl, etc.). The example is the formation of Nylon 66.

Thus, we concluded that the reaction processes that lead to the two types of synthetic polymers are chain reaction polymerization and step reaction Polymerization.

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Explanation:

The chemical equation is as follows.

$\frac{1}{2}Cl_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow ClO(g)$

And, the given enthalpy is as follows.

$\frac{1}{2}Cl_{2}(g) + O_{2}(g) \rightarrow ClO_{2}(g)$ ; $\Delta H$ = 102.5 kJ

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Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.

$\Delta H = \sum \text{bond broken in reactants} - \sum \text{bond broken in products}$

102.5 = $[(\frac{1}{2})x + 498] - [(2)(243)]$

102.5 = $(\frac{1}{2})x + 498 - 486$

102.5 - 12 = $\frac{x}{2}$

x = 181 kJ

Now, total bond enthalpy of per mole of ClO is calculated as follows.

$\Delta H = \sum E(\text{bond broken in reactants}) - \sum (\text{bond broken in products})$

x = $[(\frac{1}{2})181 + (\frac{1}{2})498] - 243$

= 339.5 - 243

= 96.5 kJ

Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.

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