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I've come across a series of reactions, where $$\ce{2NaHCO3 <=> Na2CO3 + H2O + CO2}$$ at temperatures above 50 degrees Celsius, and $$\ce{Na2CO3 + H2O <=> NaHCO3 + NaOH}$$ I Have 3 questions about these equations:

  1. Will the above reactions proceed in a boiling Baking Soda solution set to reflux until virtually all of the baking soda is converted to lye and CO2?
  2. If so, would the reaction follow a half-life pattern, over a certain period of time?
  3. How would I calculate the amount of time required to convert a saturated baking soda solution into a solution with a certain concentration of lye?

I appreciate any advice provided.

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    $\begingroup$ The decomposition of heated NaHCO3 is rapid. It is the fundamental reaction for preparing Na2CO3 by the Solvay process, but it is not carried out in solution. It is done with dry powder. It is not necessary to make any rate calculation. $\endgroup$
    – Maurice
    Mar 29 '20 at 19:31
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Actually, if one adds your two reactions:

$\ce{2 NaHCO3 <=> Na2CO3 + H2O + CO2}$

$\ce{Na2CO3 + H2O <=> NaHCO3 + NaOH}$

Upon cancelling, to derive the reaction:

$\ce{ NaHCO3 <=> NaOH + CO2}$

which implies that heated Sodium bicarbonate is a chemically active source of Lye, if one allows the CO2 gas to escape (which answers (a) of your question).

To answer (b) and (c), add Aluminum to the heated aqueous mix and it will dissolve, and you could collect the generated hydrogen gas over water and record volume increase over time. Reaction:

$\ce{2 Al + 6 NaOH -> 2 Al(OH)3 + 3 H2 (g)}$

Now, per this Aluminum shiny finish destroyed in dishwasher, you have an understanding as to why this occurs with Aluminum ware in the presence of NaHCO3, which is a popular spotless-rinse agent employed in dishwasher's washing powder.

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    $\begingroup$ Did you read Maurice's comment? It's quite correct this time, and your post isn't. $\endgroup$
    – Mithoron
    Mar 29 '20 at 23:16
  • $\begingroup$ Note, I stated upon 'canceling' meaning removing compounds appearing on both sides of the reaction. This implies that the liberated H2O (as water vapor from heating the dry powder, for example) remains in a closed system where it could possibly interact with Na2CO3. This is not likely to occur upon decomposing NaHCO3 by heating in open air, which is the context of Maurice comment. $\endgroup$
    – AJKOER
    Mar 30 '20 at 15:11
  • $\begingroup$ Thanks Mithoron, for your point which is certainly worthy of clarification. $\endgroup$
    – AJKOER
    Mar 30 '20 at 15:20
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    $\begingroup$ The point is while bicarbonate easily decomposes when pure is not a proof it will in solution. $\endgroup$
    – Mithoron
    Mar 30 '20 at 17:54
  • $\begingroup$ Your aluminum pan being disfigured in a dishwasher in the presence of NaHCO3 should be all the proof you need. By the way, I have performed the reaction of attacking preheated Al foil (to remove any protective annealing, oils,..) in aqueous NaHCO3 in a vessel with microwave heating which turns out to be vigorous, give it a try, $\endgroup$
    – AJKOER
    Mar 31 '20 at 0:15
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2NaHCO3 --> Na2CO3+H2O+CO2 at temperatures above 50 degrees Celsius: TRUE.

However, Na2CO3+H2O --> NaHCO3+NaOH is a hydrolysis reaction: it only goes to a small extent, maybe 2-5%, and on evaporating the water, it goes back to Na2CO3. Na2CO3 melts at 851C without decomposition.

So, if you take a solution of NaHCO3 (pH ~8.5) and boil it, you will generate a solution of Na2CO3 (pH ~12) by loss of CO2 and if you keep evaporating the water, you will get Na2CO3 as a crystalline powder (and when it cools, it will begin absorbing water from the air to form one or more of its many hydrates).

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  • $\begingroup$ I never planned to evaporate the water: "...a boiling Baking Soda solution set to reflux" $\endgroup$ Mar 30 '20 at 14:35

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