You can do it. Use this as a template.

The methyl radical and chlorine radical are present in very low concentrations (relative to CH4 and Cl2) and the rate should be low for most of the reaction since rate is proportional to concentration. If these radicals do meet each other, they will react of course, it’s just that it’s quite unlikely.
The reaction of the alkyl radical with Cl2 is more favored simply because the concentration of cl2 is much, much higher than that of the Cl radical.

Examine the radical formed through homolytic cleavage of the C2 C-H bond, compared to the radical formed through homolytic cleavage of the C3 C-H bond. What do you notice about its neighbor?
Although this isn’t the best example, you might find this interesting: https://en.wikipedia.org/wiki/Captodative_effect

The unpaired electron of Cl interacts with the s orbital of the hydrogen and there is a transition state where there are partial bonds between Cl and H and carbon. The H-Cl bond is comparable in strength (102 kcal/mol) to the H-C bond in methane (105 kcal/mol) so the process is reasonably favorable. The resulting carbon-based radical then reacts with leftover Cl2. C-H bond dissociation energies in methane are quite high relative to other alkanes, so in other alkanes it will be even more favorable.

Same as written here, just use propane instead of methane. The complication is that you will have two different monochlorination products. 1-chloropropane and 2-chloropropane.

This is for free-radical halogenation of alkanes, although the reaction of CFCs with ozone also proceeds through initiation, propagation and termination steps.

The way to look at it is to compare bond enthalpies and calculate the difference in energy going from starting material to product.

The bond strength of H-Cl is about 106-107 kcal/mol whereas the bond strength of H-CH3 is about 104 kcal/mol. Therefore the “gain” in energy is about 3kcal/mol from the chlorination of methane.

(The C-Cl bond does not factor into this reaction, except as a possible termination step).

That at least gives you an idea of whether or not the process will be energetically favorable overall.

We don’t cover free-radical fluorination because the H-F bond is even stronger than C-H, and furthermore the reaction is so exothermic that combustion tends to take place.
On the other hand free-radical iodination generally fails because H-I bonds are so weak, relative to C-H .

So, in summary: bond strengths are a good guide.

Thank you!