Free Radical Reaction

[caters]

1. The problem statement, all variables and given/known data
Lets say you have 1 liter of 2 mol/L methane and the same amount of chlorine. Lets also say that both are liquids since those are most likely to react. Now the only way they can both be liquids is if the temperature is as cold as an antarctic winter so this is not aqueous. Gases more often bump the wrong way and solids don't react unless it is oxidization or dissolving.

Now the initiation step is forming the first molecule of HCl and Methyl.

Now the methyl and chlorine atom really want to react and for chloromethane

Now here are the questions.

How much chloromethane, dichloromethane, trichloromethane, and tetrachloromethane will there be?

How much of the more complicated alkanes like ethane and propane will there be?

How many molecules made up of more complicated alkanes and chlorine will there be?

Will at some point the chlorine go back to its normal state?



2. Relevant equations
CH₄ + Cl₂ = HCl + CH₃Cl(this continues up to tetrachloromethane)
2 CH₃Cl = Cl₂ + C₂H₆(this can continue for much longer than the previous one can)

3. The attempt at a solution
2 M CH₄ + 2 M Cl₂ = 2 M HCL + 2 M CH₃ + 2 M Cl
2 M Cl + 2 M CH₃ = 2 M CH₃Cl
2 M CH₃Cl * 2 CH₃Cl = 1 M C₂H₆ + 1 M Cl₂

2 M HCl = 2 M H₂ + 2 M Cl₂

This obviously can't happen because than we have more chlorine than we started out with. Why? well that 1 M Cl₂ from ethane + 2 M Cl₂ from HCl is = 3 M Cl₂ and we started with 2 M Cl₂. Just like the number of each element the molarity has to be balanced. This is where I am stuck is figuring out the molarity of each compound at each step of the process not the compounds themselves.

Edited April 9, 2014 by caters

[aelek]

*Cl + CH4 -> HCl + *CH3

 

*CH3 + Cl2 -> CH3Cl + *Cl

 

CH3Cl + *Cl -> HCl + *CH2Cl

 

*CH2Cl + Cl2 -> CH2Cl2 + *Cl

 

The reactions should go on like this to form the other two methyl chlorides, until all of the Cl2 is used up. Also, methyl radicals should react to form ethan:

 

*CH3 + *CH3 -> C2H6

 

If I got it right, 4 Cl2 molecules will give you one molecule of each methyl chloride.

Edited May 19, 2014 by aelek

[caters]

yeah but I am having trouble with this because I want to figure out the molarity of these compounds from the molarity of both methane and chlorine being 2M but I have trouble when I get to 2 chloromethanes reacting to form chlorine and ethane and 2 HCLs reacting to form chlorine and hydrogen because I get 3M Cl2 when I started with 2M Cl2 and that is not right.

 

What happens is first HCL forms, then the second chlorine atom bonds to the carbon and forms chloromethane. 2 of these chloromethane react to form ethane and it continues making more complex alkanes by a factor of 2.