Does bicarbonation require water? In an aqueous solution of $\ce{NaOH}$ exposed to the atmosphere one can probably expect some degree of $$\ce{NaOH(aq) + CO2(g) -&gt; NaHCO3 (aq)}$$

But, what would happen if just a solid block of $\ce{NaOH(s)}$ was placed in some completely anhydrous environment, populated only by excess $\ce{CO2(g)}$ at STP? After all,

$$\ce{NaOH(s) + CO2(g) -&gt; NaHCO3 (s)}$$

is technically a spontaneous process with $\Delta G^\circ = -76.91\ \mathrm{kJ/mol}$, but I can only think of three possibilities:


Nothing happens—no water (i.e., as a catalyst), no reaction!
A passivation layer of $\ce{NaHCO3(s)}$ forms on the surface, preventing the rest of the $\ce{NaOH(s)}$ inside from reacting^
No passivation layer—the $\ce{NaOH(s)}$ just keeps reacting away until it's all $\ce{NaHCO3(s)}$!
Rather 

$$\ce{2 NaOH(aq) + CO2(g) -&gt; Na2CO3 (aq) + H2O}$$

and analogically on the solid $\ce{NaOH}$.

Bicarbonate in aqueous solutions cannot survive the excess of hydroxide, forming carbonate.

$$\ce{HCO3^{-}\ (aq) + OH- (aq)&lt;=&gt;&gt; CO3^{2-}(aq) + H2O}$$

The same is further enforced by dehydration effect of solid hydroxide.

$$\ce{NaHCO3(s) + NaOH(s) -&gt; Na2CO3(s) + H2O}$$

Only in excess of $\ce{CO2}$ Is formed bicarbonate:

$$\ce{CO3^{2-}(aq) + CO2(aq) + H2O &lt;=&gt;&gt; 2 HCO3^- (aq) }$$