Free Radical Reaction of Methane and Chlorine

Question

The problem statement:

Lets say you have $1 L$ of $2 M$ 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 at STP 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 $\ce{HCl}$ and Methyl.

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

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Relevant equations

$\ce{CH4 + Cl2 -> HCl + CH3Cl}$
(this continues up to tetrachloromethane)

$\ce{2 CH3Cl -> Cl2 + C2H6}$
(this can continue for much longer than the previous one can)

The attempt at a solution

$\ce{2 CH4 + 2 Cl2 -> 2 HCL + 2 CH3 + 2 Cl}$

$\ce{2 Cl + 2 CH3 -> 2 CH3Cl}$

$\ce{2 CH3Cl + 2 CH3Cl -> 1 C2H6 + 1 Cl2}$

$\ce{2 HCl -> 2 H2 + 2 Cl2}$

This obviously can't happen because then we have more chlorine than we started out with. Why? Well that $1 M\; \ce{Cl2}$ from ethane plus $2 M\; \ce{Cl2}$ from $\ce{HCl}$ is $ 3 M\; $$\ce{Cl2}$ and we started with $2 M \ce{Cl2}$. Just like the number of each element the molarity has to be balanced. What I find confusing is figuring out the molarity of each compound at each step of the process not the compounds per se.

Answer 1

OK, let's focus on the stoichiometry of your last 4 equations:

$\ce{2 CH4 + 2 Cl2 -> 2 HCL + 2 CH3 + 2 Cl~~~~}$ this looks OK

$\ce{2 Cl + 2 CH3 -> 2 CH3Cl~~~~~~~~~~~}$ this looks OK

$\ce{2 CH3Cl + 2 CH3Cl -> 1 C2H6 + 1 Cl2~~~~}$ this should be $\ce{1 CH3Cl + 1 CH3Cl -> 1 C2H6 + 1 Cl2}$

$\ce{2 HCl -> 2 H2 + 2 Cl2~~~~~~~}$ this should be $\ce{2 HCl -> 1 H2 + 1 Cl2}$

I'm not sure I agree with all of your chemical reactions, but at least now you don't wind up with more chlorine then you started with.