Why does free silicon react with water?
Is there an explanation for the tendency of silicon compounds to react
  with water?


Reaction with water, oxygen, etc., the answer is the same and has to do with the strength of the $\ce{Si-O}$ bond. Let's compare the following two reactions:

$$\ce{CH4 + \frac{1}{2}O2 -&gt; CH3OH}$$
$$\ce{SiH4 + \frac{1}{2}O2 -&gt; SiH3OH}$$

In both reactions we break an $\ce{X-H}$ bond and an $\ce{O-O}$ bond, while we make an $\ce{X-O}$ bond and an $\ce{O-H}$ bond.  The $\ce{O-O}$ bond breaking and the $\ce{O-H}$ bond making are common to both reactions and therefore cancel out.  Let's compare the only differences, the $\ce{X-H}$ bond breaking and the $\ce{X-O}$ bond making.

The following Table contains the bond energies for these bonds.  

\begin{array}{|c|c|c|c|} \hline
\ce{X} &amp; \text{X-H Bond Strength} &amp; \text{X-O Bond Strength} &amp; \text{Overall Change}  \\ &amp; \text{(kcal/mol)} &amp; \text{(kcal/mol)} &amp; \text{(kcal/mol)}\\ \hline
\ce{C} &amp; 99 &amp; 86 &amp; -13\\ \hline
\ce{Si} &amp; 75 &amp; 110 &amp; 35\\ \hline
\end{array}

Looking at the numbers it becomes clear that the $\ce{Si-H}$ bond is a lot weaker than the $\ce{C-H}$ bond, and the $\ce{Si-O}$ bond is a lot stronger than the $\ce{C-O}$ bond.  Consequently, the reaction involving silicon is more exothermic than the analogous reaction involving carbon.  This greater stability in the $\ce{Si-O}$ product is reflected in the transition state leading to it.  It lowers the energy of the transition state and consequently the reaction involving silicon is much more facile than the same reaction involving carbon.