# Question #323a9

Jul 31, 2017

Activation energy

#### Explanation:

The activation energy for the chlorination of methane is very high at room temperature, so you have to either use heat or light to initiate the reaction between $C {l}_{2}$ and $C {H}_{4}$. Overall, the $\Delta$H of the reaction is negative.

$C {H}_{4}$ is more stable than $C {l}_{2}$ as seen with the $\Delta$H values for breaking Cl-Cl and C-H bonds. Since $\Delta$H is smaller for Cl-Cl than C-H, the initiation step will be homolytic cleavage of the Cl-Cl bond. Because $\Delta$H is positive for the Cl-Cl bond, you need a lot of energy for this breakage.

The first propagation step forms a methyl carbocation. This reaction results in a positive $\Delta$H just like in the initiation step. The overall reaction up to this point is endothermic. This propagation reaction is slightly unfavorable. Once this reaction occurs, formation of methyl chloride is very favorable.

It is just getting past the energy barrier that you need to put in a lot of energy for the reaction to occur. Overall, the reaction is exothermic, but if you look solely at the initiation and first propagation steps, you end up with an endothermic reaction.

Here is the potential-energy diagram for the reaction to visually show what I explained.