Substitution reaction between methane and bromine

[JonnyG]

We can react one mole of $CH_4$ and one mole of $Br_2$ to obtain the following equation:
$CH_4 + Br_2 \rightarrow CH_3 Br + HBr$

A single bromine atom switches places with a single hydrogen atom. Now, if we supply a greater amount of $Br_2$ while keeping the amount of $CH_4$ the same then we can replace more than one of the hydrogen atoms in each of the $CH_4$ molecules with $Br$ atoms. So suppose we react one mole of $CH_4$ with two moles of $Br_2$. Then for each $CH_4$ molecule there are four $Br$ atoms. Thus we may have the equation:
$CH_4 + 2Br_2 \rightarrow CBr_4 + 2H_2$

However, we can also balance the equation in the following manner:
$CH_4 + 2Br_2 \rightarrow CH_2Br_2 + 2HBr$

So what actually happens and why does that happen as oppose to the other one? The first equation seems to me to represent a lower energy state for the products, but the second equation seems to represent a drive toward increasing entropy. As far as I know, a system will drive towards the lowest energy state but also drive toward increasing entropy, unless one of the drives overcomes the other. Now, looking at the electronegativities of hydrogen, carbon and bromine, the difference in electronegativities between bromine and hydrogen is greater than the difference in electronegativities between any other two of the possible atoms and hence the system will tend toward the product $CH_2Br_2 + 2HBr$

I would appreciate if someone could please correct me where I have gone wrong.

 

[Borek]

Reaction mechanism is such that the HBr is one of the products.

Besides, assuming your second reaction is right - at some point you have a mixture of H₂ and Br₂. What do they do when they are mixed?

 

[JonnyG]

$H_2 + Br_2 \rightarrow 2HBr$

Borek, why is the reaction mechanism such that $HBr$ is produced?

 

[Borek]

It is not in the mechanism, it is in thermodynamics. HBr synthesis is exothermic and that's enough in this case.