Why does sodium hydrogen carbonate decompose into sodium carbonate, water vapour, and carbon dioxide?

Why is the reaction for sodium hydrogen carbonate:


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


but not:


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


or:


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


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.




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.


 3

**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 $\ce{CO2(g)}$.




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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.




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Thermodynamic Analysis
----------------------


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


### Reactions


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 $\ce{H2O(l) \xrightarrow{T>\pu{373.15 K}} H2O(g)}$


### Data


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


### Calculations


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.


### References


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](https://doi.org/10.1021/ie400896c)




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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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