What causes hydrogen abstraction in the radical chain mechanism of alkane halogenation? In the first propagation step of the radical chain mechanism of alkane halogenation, why does the halogen actually abstract hydrogen from the alkane? My organic chemistry textbook so far has went into details on MOs, hybridized orbitals, and general thermodynamics of reactions but has been lacking in the by and large why of reactions. Is the $\ce{C-H}$ bond broken by sufficient collisions from halogen free atoms and the hydrogen goes to the halogen as a result of $\ce{F}$/$\ce{Cl}$/$\ce{Br}$ being more electronegative than carbon, or what should cause this to happen?
why does the halogen actually abstract hydrogen from the alkane?


Look at the two tables of bond energies on this webpage.  The second or bottom table shows the $\ce{C-H}$ bond energies for various carbon containing compounds.  In the top table we see that the H-Cl bond strength is 103 kcal/m.

If we start with a chlorine radical, then if we break a C-H bond and make a Cl-H bond, the energetics of the reaction will be

$$\ce{\Delta H=bonds~ broken~-~bonds~made}$$
$$\ce{\Delta H=C-H ~bond~ broken~-~H-Cl ~bond~made}$$
$$\ce{\Delta H=C-H ~bond~ broken~-~103~ Kcal/m}$$

Look at the top row in the bottom table again.  As long as our chlorine radical breaks any C-H bond that has a dissociation energy less than 103 kcal/m, our reaction will be exothermic, and the reaction will continue.

Edit


  As I understand, the first propagation step in the free-radical
  monochlorination of methane is slightly endothermic but is driven to
  completion by the exothermic nature of the second propagation step


The first propagation step involves breaking a $\ce{C-H}$ bond (103 kcal/m) and making an $\ce{H-Cl}$ bond (103 kcal/m), so it is roughly thermoneutral according to the above link.

$$\ce{CH3-H + Cl{.} -&gt; CH3{.} + H-Cl}$$

$$\ce{CH3{.} + Cl-Cl -&gt; CH3-Cl + Cl{.}}$$

The second propagation step involves breaking a $\ce{Cl-Cl}$ bond (58 kcal/m) and making a $\ce{CH3-Cl}$ bond (81 kcal/m), so it is quite exothermic.

These steps are consistent with my earlier argument.  You are correct that using free energies would be better (more accurate) than enthalpies, but the Tables I could find were based on enthalpies - I suspect they provide us with a reasonably accurate "back of the envelope" calculation and view of what is going on.

As to the actual mechanism of the reaction, once we've created a chlorine radical, we've created a reactive species.  It is reactive because it doesn't have an octet of electrons.  By undergoing the above-reactions it can regain its octet and become a stable entity.  

Here is a diagram of a reaction coordinate (note we are using free energy as the Y-axis in this diagram).  Generally the more exothermic (exergonic for free energy) a reaction, the lower the activation energy required to reach the transition state and pass over from reactants to products.  So our reactive chlorine radical bounces around probing different parts of the reaction coordinate (e.g. different $\ce{C-H}$ bonds, different angles and approaches to these bonds) until it encounters a bond from the right direction, right angle, etc, such that the collision has enough energy to pass over the top of our reaction coordinate and become a product.