Relationship between rate equation and equilibrium constant
At equilibrium it does not matter if the reaction has multiple steps; your equilibrium constant only has to involve the initial reactants and final products.  

For instance, in an S$_\mathrm{N}$1 reaction, obviously the RDS is the departure of the leaving group, but you don't have to (and usually are not in a position to) know the concentration of the intermediate carbocation in order to calculate the equilibrium of the overall reaction.  You can simply write an  equilibrium equation for the initial reactants and final products: $$K=\frac{[\ce{LG}][\ce{product}]}{[\ce{nucleophile}][\ce{substrate}]}.$$

Of course finding $K$ may be not be easy, even if you have access to good thermodynamic data for the reactants and products, because $\Delta G_0$ (and therefore $K$) depends on solvation, temperature and other conditions. But given identical conditions, $K$  is constant over a wide range of concentrations, so you might be able to find it in the literature or measure it with a single experiment.

All this assumes that the RDS is fast enough that you're able to reach  equilibrium before the solvent evaporates, or your reactants and products decompose via other pathways, or you die of boredom!

Note that you can also write an equilibrium equation for the RDS alone:
$$
K'=\frac{[\ce{LG}][\ce{carbocation}]}{[\ce{substrate}]}
$$
But since you seldom know the concentration of the carbocation, this equation isn't as useful in practice (it is of course extremely useful for mechanistic studies).

Edit: This page has a more complete and very nice answer!