Equilibrium Constant and Surface Area
In a reaction A(s) giving B(g) + C(g) Increasing the surface area of A increases the rate of the forward reaction, but the backward reaction is between two gases and is unaffected.


This is an assumption that you're making and it is thermodynamically and geometrically questionable.  If increasing the surface area of $\ce{A}$ increases the rate at which it sublimes/gasifies to form $\ce{B + C}$, it means that the reactive molecules of $\ce{A}$ are only those at the surface, in which case it is misleading to denote the reactant as $\ce{A(s)}$.  It is really $\ce{A(surface)}$ which is reacting.  Similarly, if the rate of the reverse reaction is not really dependent on the surface area and surface geometry of $\ce{A(s)}$, then it implies that very quickly molecules of $\ce{B}$ and $\ce{C}$ would condense to form a nanocrystallite of $\ce{A}$ that is one molecule big.  These nanocrystallites of $\ce{A}$ would have, collectively, nearly infinite surface area.  So the backward reaction creates surface area under the assumptions you have made.  You have to account for that created area.

Solid-phase reactions are complicated by phenomena such as crystal nucleation and other mechanisms of crystal growth.  Treating solids as uniformly reactive materials with unit activity is inappropriate for understanding such phenomena.  Another way to say it is that the assumption that solids have unit activity and don't appear in expressions for $K$ or $\Delta G$ is an assumption, one that is inapplicable when dealing with reactions whose rates depend on surface area.