Why are free radicals so reactive?
Radicals are so reactive because they require so much energy to form. When we speak about radical reactivity, "more reactive" generally means a more exothermic hydrogen atom abstraction step. The more exothermic this step, the more the transition state resembles the reactants (Hammond's postulate), with a correspondingly lower character of the carbon-centered radical product. This makes the reaction less sensitive to the stability of the carbon-centered radical.

This being said, some radicals do have an activation energy, and some don't. The $E_a$ is essentially zero for all hydrogen atom abstractions by fluorine atom, and the free energy barrier arises solely from the pre-exponential factor of the Arrhenius equation ($k=Ae^{-E_{a}/(RT)}$), which is near 13 for the halogenations listed below.



The activation energies for abstraction by chlorine atoms are also exceedingly small, but there is a slight preference for more highly substituted $\ce{C-H}$ bonds. Lastly, the activation energies for abstraction by bromine atoms are substantial, clearly showing a preference for abstraction of 3°, 2°, and 1° $\ce{C-H}$ bonds in that order. It is because of these differences in activation energies that we observe the varying regioselectivities of the different halogenation reactions$^{[1]}$.




$[1]$ Anslyn, E. V.; Dougherty, D. A. Modern Physical Organic Chemistry; University Science: Sausalito, CA, 2006.