melting and boiling ...really equilibrium?
I think what you are asking is this:  Equilibria for chemical reactions typically* (see note at end) require specific ratios of products to reactants (as expressed by the equilibrium constant).  By contrast, equilibria for phase transitions don't require specific ratios of products to reactants.   [For instance, at the phase transition between ice and water, ice and water can be in equilibrium with any relative amounts of ice and water.] Why is this? 

Consider a gas-phase chemical reaction, say

$$\ce{A(g) + B(g) &lt;=&gt; C(g)}.$$

At equilibrium, the chemical potentials of all species are the same.  Associated with this is an equilibrium constant, 

$$K_p(T) = \frac{\displaystyle\frac{p_\ce{C}}{p^\circ}}{\displaystyle\frac{p_\ce{A}}{p^\circ}\cdot\frac{p_\ce{B}}{p^\circ}},$$

which specifies a strict relationship between the partial pressures of the reactants and products.

Alternately, consider a solid-liquid phase transition, 

$$\ce{A(s) -&gt; A(l)}.$$

At the melting temperature, the chemical potentials of the two phases are the same.  However, by contrast with the example gas-phase reaction, this is true independent of the relative amounts of $\ce{A(s)}$ and $\ce{A(l)}.$

The question then arises: why the difference between the two?  The answer is that the particular mathematical form of the equilibrium constant—and the attendant constraint it imposes on the relative amounts of reactants and products—arises from the entropy of mixing, and there is no entropy of mixing in a phase transition!  At equilibrium, in the example gas-phase reaction, the entropy of mixing just balances the relative free energies of the reactants and products.  That's why we have an equilibrium.  If it weren't for the entropy of mixing, reactions would always go completely to the side whose pure components had the lower free energy (which is exactly what happens in a phase transition).   

By contrast, there is no entropy of mixing term in a phase change, because the difference in phases keeps the two components separated.  And without an entropy of mixing term, the relative amounts of the two phases become irrelevant to their relative chemical potential, and thus irrelevant to the equilibrium state (by which I mean any relative amount of the two phases is allowed at the phase transition temperature).

This also explains why, when we are (say) somewhat above the melting temperature, we don't have a small amount of solid in equilibrium with a large amount of liquid (as would be the case if this behaved like a chemical reaction).  The reason for this is there's no entropy of mixing to favor this mixed state over the liquid alone, i.e., over the pure substance with the lower chemical potential (which, above the melting temperature, is the liquid state).  [Or more accurately, while there is an entropy of mixing, because the two phases aren't mixed at a molecular level—phase is a bulk property (i.e., a collective property of many molecules), not a molecular one—the entropy of mixing term (which is essentially an exchange entropy) is negligible in this case.]

N.B.: Just for clarity, let me emphasize that, as a substance is melting or boiling, it is not at equilibrium.  It's not at equilibrium during a phase transition until there is no macroscopic change in the system, i.e., until it settles into a fixed ratio of one phase to another (and that it can be at equilibrium at any ratio). 

*Incidentally, this is also why, when we have a mixed-phase chemical reaction where one phase can't mix with the others, that phase isn't included in the equilibrium expression.  For instance, for

$$\ce{CaCO3(s) &lt;=&gt; CaO(s) + CO2(g)},$$ we find that

$$K_p(T) = \frac{p_\ce{CO2}}{p^\circ}$$