Free Radical reaction of methane and Chlorine

[caters]

Homework Statement

Lets say you have 1 liter of 2 mol/L methane and the same amount of chlorine. Let's 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 and the hydrogen go back to the carbon so that you have just methane, ethane, propane etc?

Homework 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)

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.

 

[Bystander]

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?

This is an "infinite dilution" equilibrium problem, the solution being a function of temperature, pressure, and composition, f(T,P,X_CH4,X_Cl2).

Will at some point the chlorine go back to its normal state and the hydrogen go back to the carbon so that you have just methane, ethane, propane etc?

No.

 

[caters]

Why would it not do that? I mean Cl-Cl bonds are very strong and there has to be a point where HCl stops forming. Plus carbons form bonds with each other in the process which gives ample time for Cl2 to leave the system and hydrogen to bond back to the carbon or leave as a gas.

 

[Bystander]

What's the equilibrium constant for formation of HCl?

 

[caters]

I don't know but surely some hydrogen and CL2 leaves the system in the process so the likelihood of HCl formation decreases as time increases both for that reason I just stated and carbons forming bonds with each other and assuming no carbon leaves the system that means lower chance of significant reaction with the chlorine(by significant,I mean like a step that moves the process forward by 1 or more carbons bonding with other carbons) and a much higher chance of carbons shielding other carbons.

 

[Bystander]

surely some hydrogen and CL2 leaves the system in the process

What process? It's a closed system.

 

[caters]

How do you know this is a closed system? I didn't say it was a closed system.

 

[Bystander]

Lets also say that both are liquids

That's a closed system.

 

[caters]

But that ignores evaporation. Evaporation can happen even when it is super cold. So I don't see how liquid+ liquid= closed system when it is possible(especially for hydrogen) that evaporation occurs. Evaporation is just liquid to gas transition and is not nearly as temperature dependent as boiling vs freezing.