Why is the reaction for sodium hydrogen carbonate:

$$\ce{2 NaHCO3(s) -> Na2CO3(s) + H2O(g) + CO2(g)}$$

but not:

$$\ce{4 NaHCO3(s) -> 4 Na + 2 H2O(g) + 4 CO(g) + 3 O2(g)}$$


$$\ce{2 NaHCO3(s) -> 2 NaOH(s) + 2 CO2(g)}$$

And how could you determine this without actually performing the experiment? I was thinking it may have to do with the fact that carbonate is a polyatomic ion and therefore stays that way, but I'm not sure.

  • $\begingroup$ Chem+Math Expression formatting reference: MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles. $\endgroup$
    – Poutnik
    May 14 at 4:41
  • $\begingroup$ Could you elaborate your reasoning? $\endgroup$
    – ananta
    May 14 at 5:37
  • $\begingroup$ Related: chemistry.stackexchange.com/questions/141689/… $\endgroup$
    – Poutnik
    May 14 at 6:12
  • $\begingroup$ This practical discusses exactly this with an additional reaction $\ce{2NaHCO3 (s) -> Na2O (s) + 2CO2 (g) + H2O (g)}$ $\endgroup$
    – ananta
    May 14 at 8:09
  • 1
    $\begingroup$ The stability of carbonate anion (it being a polyatomic anion) is not the factor at play here, since, if I may be a little vague, in the thermal decomposition of sodium bicarbonate, carbonate anion does change to carbon dioxide moiety. $\endgroup$
    – ananta
    May 14 at 13:37

2 Answers 2


Releasing $\ce{CO2}$ from $\ce{Na2CO3}$ and especially $\ce{O2}$ from $\ce{Na2O}$ or $\ce{CO2}$ requires much higher temperature than temperature of $\ce{NaHCO3}$ decomposition.

There are three major ways to know that:

  • You learn the chemistry and behaviour patterns of elements and substances, so you know it without the need to memorize it.
  • You calculate the energetic outcome (reaction enthalpy) and thermodynamic preference (reaction Gibbs energy) to predict if such reaction can be expected.
  • With no such knowledge, you have to memorize it or search for it.

Generally, there is no shortcut to learn chemistry.


Summary: all these reactions are possible under the right conditions (for example, temperature, pressure, phase, and catalyst); however, for thermal decomposition at atmospheric pressure, the first reaction to become spontaneous (at about $\pu {400 k}$, called the thermal decomposition temperature of $\ce{NaHCO3(s)}$), is the decomposition of $\ce{NaHCO3(s)}$ into $\ce{Na2CO3(s)}$, $\ce{H2O(g)}$, and $\ce{CO2(g)}$.

There is no (accurate and entirely) theoretical way to predict a reaction (as of now). However, we can check if a reaction is possible based on thermodynamic-, to what extent it occurs based on equilibirum-, and weather it happens in real-time based on kinetic-analysis.

Note: This are all semi-empirical analysis methods since the values of thermodynamic potentials, equilibrium constants and kinetic rate coefficients are often obtained experimentally.

I will do the thermodynamic analysis here.

Thermodynamic Analysis

First, let us convert all your reaction-components into standard states in the simplest stoichiometries.


At $\pu{298.15 k}$, the reactions are.

$$ \ce{2 NaHCO3(s) -> Na2CO3(s) + H2O(l) + CO2(g)} \tag{1} $$

$$ \ce{4 NaHCO3(s) -> 4 Na + 2 H2O(l) + 4 CO(g) + 3 O2(g)} \tag{2} $$

$$ \ce{NaHCO3(s) -> NaOH(s) + CO2(g)} \tag{3} $$

At $\pu{400 k}$, the reactions are:

$$ \ce{2 NaHCO3(s) -> Na2CO3(s) + H2O(g) + CO2(g)} \tag{1} $$

$$ \ce{4 NaHCO3(s) -> 4 Na + 2 H2O(g) + 4 CO(g) + 3 O2(g)} \tag{2} $$

$$ \ce{NaHCO3(s) -> NaOH(s) + CO2(g)} \tag{3}$$

This is because of the phase transition $\ce{H2O(l) \xrightarrow{T>\pu{373.15 K}} H2O(g)}$


The following table shows the standard enthalpies and free energies of formation for each component at $\pu {298.15 K}$ and $\pu {400 K}$.$^1$ All values are in $\pu{kJ mol^{-1}}$

Chemical $\Delta_\text{f} H^\circ$($\pu {298.15 K}$) $\Delta_\text{f} H^\circ$($\pu {400 K}$) $\Delta_\text{f} G^\circ$($\pu {298.15 K}$) $\Delta_\text{f} G^\circ$ ($\pu {400 K}$)
$\ce{O2(g)}$ $\pu{0}$ $\pu{0}$ $\pu{0}$ $\pu{0}$
$\ce{H2O(g)}$ $\pu{-242.847}$ $\pu{-223.951}$
$\ce{H2O(l)}$ $\pu{-285.830}$ $\pu{-237.141}$
$\ce{Na(s)}$ $\pu{0}$ $\pu{0}$ $\pu{0}$ $\pu{0}$
$\ce{NaHCO3(s)}$ $\pu{-950.810}$ $\pu{-953.885}$ $\pu{-852.851}$ $\pu{-819.095}$
$\ce{NaOH(s)}$ $\pu{-425.931}$ $\pu{-428.299}$ $\pu{-379.737}$ $-363.788$
$\ce{Na2CO3(s)}$ $\pu{-1030.768}$ $\pu{-1135.717}$ $\pu{-1048.005}$ $\pu{-1019.363}$
$\ce{CO(g)}$ $\pu{-110.541}$ $\pu{-110.129}$ $\pu{-137.180}$ $\pu{-146.354}$
$\ce{CO2(g)}$ $\pu{-393.505}$ $\pu{-393.580}$ $\pu{-394.364}$ $\pu{-394.646}$


Now, let us calculate heats and free energies of reactions (1), (2), and (3) using $\Delta_r H^\circ = \sum_\text{products} \Delta_f H^\circ - \sum_\text{reactants} \Delta_f H^\circ$ and $\Delta_r G^\circ = \sum_\text{products} \Delta_f G^\circ - \sum_\text{reactants} \Delta_f G^\circ$ (considering the stoichiometry). All values are in $\pu{kJ mol^{-1}}$

Reaction $\Delta_r H^\circ$ ($\pu{298.15 k}$) $\Delta_r H^\circ$ ($\pu{400 k}$) $\Delta_r G^\circ$ ($\pu{298.15 k}$) $\Delta_r G^\circ$ ($\pu{400 k}$)
1 $\pu{119.517}$ $\pu{135.626}$ $\pu{26.192}$ $\approx \pu{0}$
2 $\pu{1657.56}$ $\pu{1755.526}$ $\pu{1359.666}$ $\pu{1249.894}$
3 $\pu{131.374}$ $\pu{132.006}$ $\pu{78.750}$ $\pu{60.661}$

Results and Discussion

This tells us that all the reactions are endothermic and non-spontaneous except for reaction (1) at $\pu{400K}$ for which $\Delta_r G = 0.230 \approx 0$. This is (nearly) called the thermal decomposition temperature of $\ce{NaHCO3}$ to convert to $\ce{Na2CO3}$.

This also tells us is that the reactions are endothermic and non-spontaneous at ambient temperature. This why $\ce{NaHCO3}$ is required to be heated for the decomposition. According to reference$^2$ the thermal decomposition temperature of $\ce{NaHCO3}$ is $\pu{373.70 k}$, but our value is closer to $\pu{400 k}$. You can obtain even better (negative values of $\Delta_r G$ for reaction (1) at higher temperatures.


  1. Thermodynamic Data of Pure Substances. Ihsan Barin, Gregor Platzki. Third edition. VCH Verlagsgesellschaft mbH, Weinheim (Federal Republic of Germany), VCH Publishers, Inc., New York, NY (USA)
  2. Thermal Decomposition of Sodium Hydrogen Carbonate and Textural Features of Its Calcines. Miloslav Hartman, Karel Svoboda, Michael Pohořelý, and Michal Šyc. 10.1021/ie400896c

P.S. Could someone check the data and calculations once more (because the decomposition temperature didn't match with that of reference [2]). I just want to be sure. What is causing the $\Delta_r G$ to be slightly positive even above the thermal decomposition temperature$^2$ of $\ce{NaHCO3}$?


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