Does bicarbonation require water?

In an aqueous solution of $$\ce{NaOH}$$ exposed to the atmosphere one can probably expect some degree of $$\ce{NaOH(aq) + CO2(g) -> NaHCO3 (aq)}$$

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,

$$\ce{NaOH(s) + CO2(g) -> NaHCO3 (s)}$$

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

1. Nothing happens—no water (i.e., as a catalyst), no reaction!
2. A passivation layer of $$\ce{NaHCO3(s)}$$ forms on the surface, preventing the rest of the $$\ce{NaOH(s)}$$ inside from reacting^
3. No passivation layer—the $$\ce{NaOH(s)}$$ just keeps reacting away until it's all $$\ce{NaHCO3(s)}$$!
• Why the downvote? I merely happened to skip the possibility of carbonate formation : / Never said a "4th possibility" could not exist! : ) – ManRow Sep 17 at 23:16
• Downvotes without any reason have no uses, it only serves the ego of the person doing it. – M. Farooq Sep 18 at 0:14

Rather

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

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

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

$$\ce{HCO3^{-}\ (aq) + OH- (aq)<=>> CO3^{2-}(aq) + H2O}$$

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

$$\ce{NaHCO3(s) + NaOH(s) -> Na2CO3(s) + H2O}$$

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

$$\ce{CO3^{2-}(aq) + CO2(aq) + H2O <=>> 2 HCO3^- (aq) }$$

• Yep, and the first reaction would also happen in a solid, thereby generating at least some amount of $\ce{H2O}$. – Curt F. Sep 17 at 17:25
• @ManRow Not exactly for that reason, but the thermal decomposition $\ce{2 NaHCO3(s) -> Na2CO3(s) + H2O + CO2(g)}$ occurs very slowly even at room temperature. – Poutnik Sep 18 at 1:07