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dem82 [27]
3 years ago
14

The first step in the free radical mechanism for the preparation of polyethylene is:

Chemistry
1 answer:
kherson [118]3 years ago
3 0

Answer:

The first step in the free radical mechanism for the preparation of polyethylene is:

b. heating an organic peroxide to break the O-O bond

Explanation:

The preparation of ethylene occur by free radical mechanism , which occurs in three steps :

1. Chain Initiation

2. Chain propagation

3. Chain termination.

The first step i.e chain initiation require a radical. The compound are generally peroxide(O-O linkage) that produce radical is :

Benzoyl Peroxide

This compound when heated (with U.V radiation) produce two species of radicals .

(see the attached image)

These RADICALS are highly unstable and very reactive in nature.

They readily react with ethene and initiate the polymerisation.

Thus,

The first step in the free radical mechanism for the preparation of polyethylene is:

b. heating an organic peroxide to break the O-O bond

<u><em>The sequence of Polymerisations are:</em></u>

b. heating an organic peroxide to break the O-O bond

d. propagation of the free radicals

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Metallic pan is provided with a wooden plastic handle give reason​
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3 years ago
A mixture of 75 mole% methane and 25 mole% hydrogen is burned with 25% excess air. Fractional conversions of 90% of the methane
son4ous [18]

Solution :

Consider a mixture of methane and hydrogen.

Take the basis as 100 moles of the mixture.

The mixture contains 75% of methane and 25% of hydrogen by mole and it is burned with 25% in excess air.

Moles of methane = 0.75 x 100

Moles of hydrogen = 0.25 x 100

The chemical reactions involved during the reaction are :

$CH_4+2O_2 \rightarrow CO_2 + 2H_2O$

$CH_4+1.5O_2 \rightarrow CO+2H_2O$

$H_2+0.5O_2 \rightarrow H_2O$

The fractional conversion of methane is 90%

Number of moles of methane burned during the reaction is = 0.9 x 75

                                                                                                   = 67.5

Moles of methane leaving = initial moles of methane - moles of methane burned

                                           = 75 - 67.5

                                           = 7.5 moles

Fractional conversion of hydrogen is 85%

The number of moles of hydrogen burned during the reaction is = 0.85 x 25

                                                                                                   = 21.25

Moles of hydrogen leaving = initial moles of hydrogen - moles of hydrogen burned

                                           = 25 - 21.25

                                           = 3.75 moles

Methane undergoing complete combustion is 95%.

$CO_2$ formed is = 0.95 x 67.5

                       = 64.125 moles

$CO$ formed is = 0.05 x 67.5

                       = 3.375 moles

Oxygen required for the reaction is as follows :

From reaction 1, 1 mole of the methane requires 2 moles of oxygen for the complete combustion.

Hence, oxygen required is = 2 x 75

                                            = 150 moles

From reaction 3, 1 mole of the hydrogen requires 0.5 moles of oxygen for the complete combustion.

Hence, oxygen required is = 0.5 x 25

                                            = 12.5 moles

Therefore, total oxygen is = 150 + 12.5 = 162.5 moles

Air is 25% excess.

SO, total oxygen supply = 162.5 x 1.25 = 203.125 moles

Amount of nitrogen = $203.125 \times \frac{0.79}{0.21} $

                                = 764.136 moles

Total oxygen consumed = oxygen consumed in reaction 1 + oxygen consumed in reaction 2 + oxygen consumed in reaction 3

Oxygen consumed in reaction 1 :

1 mole of methane requires 2 moles of oxygen for complete combustion

 = 2 x 64.125

 = 128.25 moles

1 mole of methane requires 1.5 moles of oxygen for partial combustion

= 1.5 x 3.375

= 5.0625 moles

From reaction 3, 1 mole of hydrogen requires 0.5 moles of oxygen

= 0.5 x 21.25

= 10.625 moles.

Total oxygen consumed = 128.25 + 5.0625 + 10.625

                                        = 143.9375 moles

Total amount of steam = amount of steam in reaction 1 + amount of steam in reaction 2 + amount of steam in reaction 3

Amount of steam in reaction 1 = 2 x 64.125 = 128.25 moles

Amount of steam in reaction 2 = 2 x 3.375 = 6.75 moles

Amount of steam in reaction 3  = 21.25 moles

Total amount of steam = 128.25 + 6.75 + 21.25

                                     = 156.25 moles

The composition of stack gases are as follows :

Number of moles of carbon dioxide = 64.125 moles

Number of moles of carbon dioxide = 3.375 moles

Number of moles of methane = 7.5 moles

Number of moles of steam = 156.25 moles

Number of moles of nitrogen = 764.136 moles

Number of moles of unused oxygen = 59.1875 moles

Number of moles of unused hydrogen = 3.75 moles

Total number of moles of stack  gas

= 64.125+3.375+7.5+156.25+764.136+59.1875+3.75

= 1058.32 moles

Concentration of carbon monoxide in the stack gases is

$=\frac{3.375}{1058.32} \times 10^6$

= 3189 ppm

b).  The amount of carbon monoxide in the stack gas can be decreased by increasing the amount of the excess air. As the amount of the excess air increases, the amount of the unused oxygen and nitrogen in the stack gases will increase and the concentration of CO will decrease in the stack gas.  

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