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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 $\ce{NaHCO3(s)}$!
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  • $\begingroup$ Why the downvote? I merely happened to skip the possibility of carbonate formation : / Never said a "4th possibility" could not exist! : ) $\endgroup$ – ManRow Sep 17 at 23:16
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    $\begingroup$ Downvotes without any reason have no uses, it only serves the ego of the person doing it. $\endgroup$ – M. Farooq Sep 18 at 0:14
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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:

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

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  • $\begingroup$ Yep, and the first reaction would also happen in a solid, thereby generating at least some amount of $\ce{H2O}$. $\endgroup$ – Curt F. Sep 17 at 17:25
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    $\begingroup$ @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. $\endgroup$ – Poutnik Sep 18 at 1:07

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