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During the propagation step 2, the alkyl radical always react with the halogen molecule according to my textbook. However, isn't the alkyl radical react with the halogen radical formed in the initiation step much better since it lowers the potential energy?

If there is a chance that the 2 radicals combine, won't the chain reaction terminate rapidly as less and less halogen radical is formed?

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    $\begingroup$ That's indeed possible, and that's indeed one of the ways how the chain terminates. But it happens very rarely, because the radicals are very few and far between. $\endgroup$
    – Ivan Neretin
    Apr 10, 2020 at 15:18

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You must realize that the number of atoms created in the first step is smaller than one percent of the whole number of Halogen molecules. So the first reaction produces a tiny amount of HCl and a tiny amount of an alkyl radical. The probability that such a radical touches the second halogen radical is extremely small. It will collide much more frequently another halogen molecule, and produce a new halogenated compound plus a new halogen radical, etc.

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