Why is carbon tetrachloride more stable than methane?

Question

When a mixture of methane and chlorine is exposed to ultraviolet light – typically sunlight – a substitution reaction occurs and the organic product is chloromethane. However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by chlorine atoms. As in the end carbon tetrachloride is formed, it means that $\ce{CCl4}$ is more stable than $\ce{CH4}$.

My doubt is that we know the $\ce{C-H}$ bond is stronger than the $\ce{C-Cl}$ bond, then how can we say that $\ce{CCl4}$ is more stable than $\ce{CH4}$? So why does the chlorine radical replace the hydrogen radical from methane? As this happens it means that $\ce{CCl4}$ is more stable than $\ce{CH4}$, but why?

This are the reactions I am talking about:

\begin{align} \ce{CH4 + Cl2 &-> CH3Cl + HCl}\\ \ce{CH3Cl + Cl2 &-> CH2Cl2 + HCl }\\ \ce{CH2Cl2 + Cl2 &-> CHCl3 + HCl }\\ \ce{CHCl3 + Cl2 &-> CCl4 + HCl} \end{align}

Here methane is taken in excess, why is methane taken in excess?

Answer 1

It is impossible to compare the stability of methane and carbon tetrachloride directly as the do not have a common frame of reference. At the same time you cannot claim that the dichlorine molecule is less stable than the hydrogen chloride molecule.

The only true observation that you can make is that the reaction of methane and chlorine to chloromethane and hydrogen chloride is (under certain conditions) favourable. $$\ce{CH4 + Cl2 -> H3CCl + HCl}$$ Let's take some average bond strengths from Chemistry Libre Texts.^[1] \begin{array}{lr} \text{Bond} & \text{Energy}/\pu{kJ mol-1}\\\hline \ce{H-Cl} & 427\\ \ce{C-H} & 413\\ \ce{C-Cl} & 339\\ \ce{Cl-Cl} & 239\\ \end{array}

You break one $\ce{C-H}$ and one $\ce{Cl-Cl}$ bond and you form one $\ce{C-Cl}$ and one $\ce{H-Cl}$ bond:

$$\pu{+413 kJ//mol} + \pu{+239 kJ//mol} + \pu{-339 kJ//mol} + \pu{-427 kJ//mol} = \pu{-114 kJ//mol}$$

This reaction releases energy (within this crude approximation).
Within this framework a mixture of the reactants would be (thermodynamically) less stable than a mixture of the products.
This would (in this averaging approximation) be true for every following chlorinations.

Another part of this is the mechanism, which is a radical mechanism. Without going into much detail and here are some of the steps that will be involved: \begin{align} \ce{Cl2 &->[h\nu] 2Cl}\\ \ce{CH4 + Cl &-> CH3 + HCl}\\ \ce{CH3 + Cl2 &-> H3CCl + Cl}\\ &\vdots\\ \ce{H3CCl + Cl &-> H2CCl + HCl}\\ \ce{H2CCl + Cl2 &-> H2CCl2 + Cl}\\ &\vdots\\ \end{align}

It becomes a matter of probabilities and concentrations and you would typically expect a mixture of all species. Using excess methane will probably give you the highest amount of chloromethane. The following reactions may also happen to some extent, providing that path: \begin{align} \ce{H2CCl2 + CH3 &-> H3CCl + H2CCl}\\ \ce{H2CCl + CH4 &-> H3CCl + CH3} \end{align}

There are probably studies about this reactions with very accurate numbers, but for the time being and for the purpose of this answer this should be enough of a demonstration.

1. Bond Energies. (2020, August 22). Retrieved September 15, 2021, from https://chem.libretexts.org/@go/page/1981