Common Misunderstandings

Some comments and answer have suggested the following:

  1. The data are Inaccurate.
  2. The reaction cannot be carried out under standard conditions.
  3. Reactions can be spontaneous with a positive free energy of reaction ($\Delta_\text{r}G$) at constant temperature $T$ and pressure $P$.

The following clarifications address these misunderstandings.

1. The Data are Accurate

To support this, I will quote Wiley Online Library and Faraday Transactions from About this Book page of the third edition of Thermodynamic Data of Pure Substances$^2$ by Ihsan Barin and Gregor Platzki (1995):

Wiley Online Library:

This is the revised, extended, up-to-date third edition of the acclaimed reference book.

Faraday Transactions:

This is undoubtedly one of the most extensive sets of data available, covering a remarkable number of compounds... listing values for all of the thermodynamic variables over the entire range of temperatures considered.

2. The Reaction can be Carried out under Standard Conditions

  • To illustrate this, I will show how to perform the reaction in a closed vessel (so no external chemical interactions occur) while maintaining standard (partial) pressures.

thermal decomposition of sodium bicarbonate under standard conditions at 2 bar pressure

  • To prove that the reaction can be carried out under standard conditions, we refer to The Dissociation Pressures Of Alkali Bicarbonates: Part I. Sodium Hydrogen Carbonate.$^1$ by Robert Martin Caven and Henry Julius Salomon Sand (1911), who carried out the decomposition of sodium bicarbonate for $T\text{/}K \in [\approx 373.2, 388.3]$ and found that $\pu{2 bar}$ pressure is achieved at $\approx \pu{386.0 K}$ according to the predicted equation:

$$ \log{P\text{/}\pu{mmHg}} = 11.8185-\dfrac{3340}{T\text{/}\pu{K}} $$

Furthermore, the authors said:

... actual meaaurements have proved that sodium carbonate monohydrate does not exist in the equilibrium mixture under the conditions of our experiments.

supporting that no re-adsorption of evolved $\ce{H2O(g)}$ occurs in the temperature range, in which case:

... the combined pressures of carbon dioxide and water, which are equal, will be less than twice as great as the dissociation pressure of the monohydrate.

supporting that $p_{\ce{CO2}} = p_\ce{H_2O}$. I do not fully understand why "the combined pressures of carbon dioxide and water will be less than twice as great as the dissociation pressure of the monohydrate. Some help here would be greatly appreciated.

  • Some more clarifications are provided at the end of the question.

3. For a Spontaneous (or Irreversible) Reaction at Constant $T$ and $P$, $\Delta_\text{r}G < 0$:

$$ \boxed{(\Delta_\text{r} G_\text{spontaneous})_{T,P} < 0} $$

  • Justification is provided at the end of the question.
  • Unlike this answer, we do not make a distinction between irreversible and a spontaneous processes. Besides, according to it, our reaction is both spontaneous and irreversible.


While answering why does sodium hydrogen carbonate decompose into sodium carbonate, water vapor, and carbon dioxide?, I noticed that $\ce{NaHCO3}$ does not decompose spontaneously at $\pu{298.15 K}$, which is not surprising since, according to Scientific American, the thermal decomposition $T$ of sodium bicarbonate is (at least more than) $\pu{176^{\circ}F}$ or $\pu{353.15 K}$:

Baking soda starts to decompose at temperatures around 176 degrees F. At these temperatures, however, the decomposition will be relatively slow. Fifteen minutes in the oven at 200 degrees F is not enough time to significantly decompose the baking soda.

When you increase the temperature to 400 degrees F the decomposition reaction will happen much faster. Fifteen minutes is enough time to decompose the baking soda into sodium carbonate, water and carbon dioxide. The gaseous products (water and carbon dioxide) will escape into the air, which is why your resulting product should be significantly lighter than what you put into the oven.

Okay! The reaction is slow at $\pu{353.15 K}$ but, as claimed, it is spontaneous. The article Thermal Decomposition of Sodium Hydrogen Carbonate and Textural Features of Its Calcines$^2$ by Miloslav Hartman, Karel Svoboda, Michael Pohořelý, and Michal Šyc (2013) claims that decomposition $T$ of sodium bicarbonate is $\pu{373.70 K}$, which is more than that reported by Scientific American (why?), but both values are less than $\pu{400 K}$, where, surely, the reaction must be spontaneous.

I used Thermochemical Data of Pure Substances$^3$ by Barin I. and Platzki G. (1995), same as the reference that was used by the authors of Reference 1, to calculate the free energy of reaction at $\pu{298.15 K}$, $\pu{400 K}$, and $\pu{500 K}$.


We are considering the reactions:

$T$ Reactants Products
$<\pu{373.15 K}$ $\ce{2 NaHCO3(s)}$ $\ce{Na2CO3(s) + H2O(l) + CO2(g)}$
$>\pu{373.15 K}$ $\ce{2 NaHCO3(s)}$ $\ce{Na2CO3(s) + H2O(g) + CO2(g)}$

at $\pu{298.15 K}$, $\pu{400 K}$, and $\pu{500 K}$.

Calculating Free Energy of Reaction

I used the formula (considering the stoichiometry as well):

$$ \Delta_\text{r} G^{\circ}(T) = \sum_\text{products}\Delta_\text{f} G^{\circ}(T) - \sum_\text{reactants}\Delta_\text{f} G^{\circ}(T) $$


$$ \Delta_\text{r} G^{\circ}(T) = \sum_i \nu_i \Delta_\text{f} G_i^{\circ}(T) $$

where $\nu_i$ is the stoichiometric number of moiety $i$ and $\Delta_\text{f} G_i^{\circ}(T)$ is the corresponding free energy of formation at $T$.


Free energy values are reported in $\pu{kJ mol^{-1}}$:

Chemical $\ce{\Delta_\text{f} G^{\circ}}$ ($\pu{298.15 K}$) $\ce{\Delta_\text{f} G^{\circ}}$ ($\pu{400 K}$) $\ce{\Delta_\text{f} G^{\circ}}$ ($\pu{500 K}$)
$\ce{NaHCO3(s)}$ $\pu{-852.851}$ $\pu{-819.095}$ $\pu{-785.428}$
$\ce{Na2CO3(s)}$ $\pu{-1048.005}$ $\pu{-1019.363}$ $\pu{-990.384}$
$\ce{H2O(g)}$ $\pu{-223.951}$ $\pu{-219.113}$
$\ce{H2O(l)}$ $\pu{-237.141}$
$\ce{CO2(g)}$ $\pu{-394.364}$ $\pu{-394.646}$ $\pu{-394.903}$

Values marked — are not required for the analysis.


$\ce{\Delta_\text{r} G^{\circ}}$ ($\pu{298.15 K}$) $\ce{\Delta_\text{r} G^{\circ}}$ ($\pu{400 K}$) $\ce{\Delta_\text{r} G^{\circ}}$ ($\pu{500 K}$)
$\pu{+26.192 kJ mol^{-1}}$ $\pu{+0.230 kJ mol^{-1}}\text{ !?}$ $\pu{-33.544 kJ mol^{-1}}$


  • The reaction is not spontaneous at temperatures below the thermal decomposition $T$, so $\ce{\Delta_\text{r} G^{\circ}}(\pu{298.15 K}) = \pu{+26.192 kJ mol^{-1}}$ (positive free energy change) makes sense.

  • The reaction is spontaneous at high temperatures, $\ce{\Delta_\text{r} G^{\circ}}$ ($\pu{500 K}) = \pu{-33.544 kJ mol^{-1}}$ (negative free energy change) also makes sense.

  • However, even though I know that the value (at $\pu{400 K}$) is very close to zero, but it is still positive. This is the result that surprised me. It cannot be. Why is this happening?


Why is free energy of reaction at $\pu{400 K}$, where the reaction ought to be spontaneous, positive?

Further Clarifications Regarding Carrying out the Reaction under Standard Conditions

1. Why is the pressure $\pu{2 bar}$?

In order to maintain the partial pressures of both $\ce{H2O(g)}$ and $\ce{CO2(g)}$ at $\pu{1 bar}$ (standard), we keep the total external pressure as $\pu{2 bar}$. At equilibrium:

$$ P_\text{internal} = p_\ce{H2O(g)}+p_\ce{CO2(g)}=\pu{2 bar} = P_\text{external} $$

2. Wouldn't the external pressure of $\pu{2 bar}$ change the $\ce{\Delta_\text{f} G}$ for $\ce{NaHCO3}$ and $\ce{Na2CO3}$?*

The free energy of formation of solids doesn't vary much with pressure. Furthermore, we are only changing the external pressure by $\pu{1 bar}$, which isn't much.

Justification for Definition of a Spontaneous Process or Spontaneity

It is a common misunderstanding that reactions can be spontaneous with a positive free energy of reaction ($\Delta_\text{r}G$) at constant temperature $T$ and pressure $P$.

Note: The links will lead you to general reading resources, which, unfortunately, lack some (important) details. I will look for a better source that puts things together in a more organized manner. For a thorough analysis, you may read Thermodynamics and an Introduction to Thermostatics$^4$ by Herbert Bernard Callen (2015).

Unfortunaly, the IUPAC Gold Book glossary of terms does not contain the terms spontaneous or spontaneity. Therefore, we are going to consider three definitions: that from the second law of thermodynamics, common knowledge in the chemistry, and the dictionary.

1. The Second law of thermodynamics.

According to the second law of thermodynamics the entropy of the universe (at constant internal energy $U$ and volume $V$) for a spontaneous irreversible process is positive:

$$ (\Delta S_\text{universe, spontaneous})_{U,V} > 0 $$

For reversible process, it is equal to zero:

$$ (\Delta S_\text{universe, reversible})_{U,V} = 0 $$

For a nonspontaneous process (or spontaneous in the opposite direction), it is negative:

$$ (\Delta S_\text{universe, nonspontaneous})_{U,V} < 0 $$

2. Commonly understood definition of spontaneous process or spontaneity in chemistry.

The most widely accepted definition of a spontaneous process or spontaneity is in terms of entropy of the universe. A process is spontaneous if and only if the process increases the entropy of the universe at constant $U$ and $V$. In chemistry, because most reaction are carried out at constant $T$ and $P$ and are irreversible, spontaneity of a reaction is, in equivalent terms with the entropy definition, defined in terms of $\Delta_\text{r} G$; fortunately, while making these (very valid) assumptions, we don't have to consider the free energy change of the surroundings. We consider a reaction spontaneous when $\Delta_\text{r} G$ at constant $T$ and $P$ is negative:

$$ \boxed{(\Delta_\text{r} G_\text{spontaneous})_{T,P} < 0} $$

Furthermore, this can be stated in terms of reaction quotient $Q$ and standard equilibrium constant $K^\circ$, the latter of which is, fortunately, defined in the IUPAC Gold Book:

$$ \begin{align} \Delta_\text{r} G &= \Delta_\text{r} G^\circ + n\mathrm{R}T\ln{Q}\\ &= n\mathrm{R}T\ln{\dfrac{Q}{K^\circ}} \end{align} $$

Thus, at fixed $T$ and $P$, the reaction is spontaneous when $Q<K^\circ$.

3. Dictionary definition of spontaneous. According to Merrriam-Webster, one way to define 'spontaneous' is "arising from a momentary impulse." In chemical terms, this refers to our usual $Q<K^\circ$ with a hint to kinetics of the reation, that the rate constant is also highly positive. Since we are not focusing on kinetics, we use the first and, especially, the second definition. However, even the third definition is very much valid for the reaction and conditions we have considered.


  1. Caven, R. M. and Sand, H. J. S. (1911). The dissociation pressures of alkali bicarbonates. Part I. Sodium hydrogen carbonate. J. Chem. Soc., Trans., 99, 1359-1369. 10.1039/CT9119901359

  2. Hartman, M., Svoboda, K., Pohořelý, M., and Šyc, M. (2013). Thermal Decomposition of Sodium Hydrogen Carbonate and Textural Features of Its Calcines. Ind. Eng. Chem. Res., 52, 31, 10.1021/ie400896c

  3. Barin, I. and Platzki, G. (1995). Thermodynamic Data of Pure Substances (3rd edition). Weinheim (Federal Republic of Germany): VCH Verlagsgesellschaft mbH. New York, NY (USA): VCH Publishers, Inc.

  4. Callen, H. B. (2015). Thermodynamics and an Introduction to Thermostatics (2nd edition). New Delhi (India): Wiley India Pvt. Ltd.

  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Chemistry Meta, or in Chemistry Chat. Comments continuing discussion may be removed. $\endgroup$
    – andselisk
    Commented May 18, 2023 at 15:28

4 Answers 4


At the moment I am writing this, the question has $26$ edits and $19$ links to other pages. This is getting out of control, so I am answering to edit number $26$. My approach will be the following: I will study the thermodynamic equilibrium of this particular system, and see what it tells me. Since you data has been regarded as not reliable, I picked other data for the sake of it.

Definitions There was some trouble with the definition of an spontaneous chemical reaction. I will use the definition of Peter Atkins & Julio de Paula, in Physical Chemistry, in chapter $6$ about chemical equilibrium:

  • If $\Delta_\mathrm{r} G > 0$ the reverse reaction is spontaneous.
  • If $\Delta_\mathrm{r} G < 0$ the forward reaction is spontanoeus.
  • If $\Delta_\mathrm{r} G = 0$ the reaction is at equilibrium.

If you accept this as truths, then your question in the title can be answered:

  1. If $\Delta_\mathrm{r} G^0(T) + RT\ln[Q(p,T,\{y_k\})] \color{red}{>} 0 $ for a certain pressure $p$, temperature $T$, and compositions $\{y_k\}$ then the products will spontaneously convert into reactants.
  2. If $\Delta_\mathrm{r} G^0(T) + RT\ln[Q(p,T,\{y_k\})] \color{red}{<} 0 $ for a certain pressure $p$, temperature $T$, and compositions $\{y_k\}$, then the reactants will spontaneously convert into products.

However, you are interested in the decomposition of a unique reactant. So we start the journey here.

Degrees of freedom We have the following chemical reaction $$\ce{2NaHCO3(s) \rightleftharpoons Na2CO3(s) + H2O(g) + CO2(g)} \tag{1} $$ composed of three phases (two solid phases and a gaseous phase), and four substances. The Gibbs phase-rule for reacting systems is \begin{align} F &= 2 - \text{Num. of phases} + \text{Num. of components} - \text{Num of chem rxns.} \\ F &= 2 - 3 + 4 - 1 \\ F &= 2 \tag{2} \end{align} Thus, we need two intensive variables to specify thermodynamically this system.

Stoichiometry We consider a test tube, that we are going to heat, with an initial number of moles of $n_\ce{NaHCO3}^0 = n_0$. The moles in terms of the extent of reaction $\xi$ is $n_\mathrm{j} = n_\mathrm{j}^0 + \nu_\mathrm{j} \xi$ and by Eq. (1) \begin{align} n_\ce{NaHCO3} &= n_0 - 2\xi \\ n_\ce{Na2CO3} &= \xi \\ n_\ce{H2O} &= \xi \\ n_\ce{CO2} &= \xi \\ \tag{3-6} \end{align}

where the total number of moles is $n = \sum_\mathrm{k} n_\mathrm{k} = n_0 + \xi$. Then, the molar fractions $y_\mathrm{j} = n_\mathrm{j}/n$ are \begin{align} y_\ce{NaHCO3} &= \frac{n_0 - 2\xi}{n_0 + \xi} = \frac{1 - 2\psi}{1 + \psi} \\ y_\ce{Na2CO3} &= \frac{\xi}{n_0 + \xi} = \frac{\psi}{1 + \psi} \\ y_\ce{H2O} &= \frac{\xi}{n_0 + \xi} = \frac{\psi}{1 + \psi} \\ y_\ce{CO2} &= \frac{\xi}{n_0 + \xi} = \frac{\psi}{1 + \psi} \\ \tag{7-10} \end{align} where we defined the dimensionless extent of reaction $\psi := \xi/n_0$. Clearly, by Eq. (7), $ 0 \leq \psi \leq 1 $. This is, we start with pure $\ce{NaHCO3}$ and nothing more can happen when it has completely reacted.

Equilibrium The law of mass action in terms of fugacities is \begin{align} K(T) &= \prod_\mathrm{j} \left(\frac{\hat{f}_\mathrm{j}}{f_\mathrm{j}^0}\right)^{\nu_\mathrm{j}} \\ K(T) &= \frac{\hat{f}_\ce{H2O}}{f_\ce{H2O}^0} \frac{\hat{f}_\ce{CO2}}{f_\ce{CO2}^0} \\ K(T) &= \left(\frac{\hat{\phi}_\ce{H2O} y_\ce{H2O} p}{p^0}\right) \left(\frac{\hat{\phi}_\ce{CO2} y_\ce{CO2} p}{p^0}\right) \\ K(T) &= y_\ce{H2O}y_\ce{CO2} \left(\frac{p}{p^0}\right)^2 \tag{11} \\ \end{align} where:

  1. The ratio of the fugacity and the fugacity at standard state of a pure solid species is $1$ by definition.
  2. The fugacity coefficients of the species in the gas solution of $\ce{H2O}$ and $\ce{CO2}$ are unity, thus, the gas phase behaves for simplicity as an ideal gas.

Note that Eq. (11) agrees with the Gibbs phase rule. If we specify $p$ and $T$, we know the gas-phase compositions (combining with the obvious $y_\ce{H2O} + y_\ce{CO2} = 1$). We can specify a molar fraction and $T$, and solve $p$ also. Now we combine Eq. (11) with Eqs. (9-10) so we have a simple equation \begin{align} K(T) &= \left(\frac{\psi}{1 + \psi}\right)^2 \left(\frac{p}{p^0}\right)^2 \\ K(T) &= \left(\frac{\psi}{1 + \psi} \frac{p}{p^0}\right)^2 \\ \sqrt{K(T)} &= \frac{\psi}{1 + \psi} \frac{p}{p^0} \\ \frac{p^0}{p}\sqrt{K(T)} &= \frac{\psi}{1 + \psi} \\ \end{align} $$ \boxed{\psi(p,T) = \dfrac{\dfrac{p^0}{p}\sqrt{K(T)}}{1 - \dfrac{p^0}{p}\sqrt{K(T)}}} \tag{12} $$ Eq. (12) is as far as we can get and the interpretation is very clear, which I summarize in this two items:

  1. This chemical reaction in chemical equilibrium at $p$ at $T$, will have a unique dimensionless extent of reaction $\psi$, that can be calculated by evaluation of Eq. (12).
  2. Clearly, the Eq. (12) has limitations in terms of specified intensive variables. If we have thermodynamic data to evaluate $K(T)$ and we get $\psi = - 0.5$ this is not a real solution.

Evaluation of K(T) We will consider the heat capacity as a function of temperature, that for every substance, it has the following mathematical form $$ \frac{C_\mathrm{p,j}}{R} = A_\mathrm{j} + B_\mathrm{j}T + \frac{C_\mathrm{j}}{T^2} \tag{13} $$ where $j$ denotes the substance.

Unfortunately, this is a long calculation and I will not put it here. If anyone wants to see the whole math I link to this PDF. The issue is that we first have to integrate this $$ \frac{d\Delta_\mathrm{r} H^\circ}{dT} = \Delta_\mathrm{r} C_{p} \tag{14} $$ because the heat capacites are a function of temperature, and then use van't Hoff's equation $$ \frac{d\ln K}{dT} = \frac{\Delta_\mathrm{r} H^\circ}{RT^2} \tag{15} $$ so we have to do two integrations. All the work resumes to \begin{align} K(T) = \exp \bigg\{ &-\frac{\Delta_\mathrm{r} H_0^\circ}{R} \left(\frac{1}{T} - \frac{1}{T_0}\right) + A \left[\ln\left(\frac{T}{T_0}\right) + \left(\frac{T_0}{T} - 1\right)\right] + \notag \\ &\frac{B}{2} \left[(T - T_0) + T_0 \left(\frac{T_0}{T} - 1\right)\right] + \\ & (2C) \left[\left(\frac{1}{T^2} - \frac{1}{T_0^2}\right) - \frac{1}{2T_0}\left(\frac{1}{T} - \frac{1}{T_0}\right)\right]\bigg\} K(T_0) \tag{16} \end{align} where $K(T_0) = \exp[-\Delta_\mathrm{r} G^\circ(T_0)/(RT_0)]$, $A:= \sum_\mathrm{k} \nu_\mathrm{k} A_\mathrm{k}$, and the same for $B$ and $C$.

Results Eq. (16) begs to be displayed in a semilogarithmic plot. We have:

enter image description here

[OP] Okay! The reaction is slow at $\pu{353.15 K}$ but, as claimed, it is spontaneous. The article Thermal Decomposition of Sodium Hydrogen Carbonate and Textural Features of Its Calcines^11 by Miloslav Hartman, Svoboda, K., Pohořelý, M., and Šyc, M. claims that decomposition temperature of sodium bicarbonate is $\pu{373.70 K}$, which is more than that reported by Scientific American (why?), but both values are less than $\pu{400 K}$, where, surely, the reaction must be spontaneous.

At least for our results, it seems that the authors are right. $K$ is around unity on that tempearture, $\pu{373.70 K}$, and below that temperature it if far too low.

Now we will plot the dimensionless extent of reaction $\psi$, i.e., Eq. (12). We do it for two different pressures:

enter image description here

On purpose I overextended the temperature. The values where $\psi>1$ are not physically meaningful, and $\ce{NaHCO3}$ has been totally converted. We can also see that there is some product already at lower temperatures, however, this depends on the ability of: (1) detect the products (like water in the walls of the test tube), (2) how much you approach thermodynamics in the lab (this is, doing this process sufficiently slow). In addition, you would have to try to measure somehow this compositions in the gas phase, not also perceiving them.

Are our results meaningful? No data that I have used here surpasses going to the lab and do a measurement. You can try other data, and get similar results, but you should not get ridiculous results. So, what is decomposition temperature? Well, an operational definition, it will be that temperature where we are very sure that $\ce{NaHCO3}$ will decompose, and the author states that it happens at $\ce{373.15 K}$. Our data shows that, around that temperature, $\pu{80 \%}$ of the test tube has already reacted. It also shows that we will have decomposition at considerably lower temperatures. Without being there and see how was the experiment was made, nothing can be concluded.

A funny thing happened when I sought for the data. For $\ce{NaHCO3}$ the temperature range validity of $C_\mathrm{p}$ is up to $\ce{400 K}$. This makes sense because if we try to measure the heat capacity at higher temperatures, we cannot measure it anymore, because other products will appear and absorb the heat. I imagined that if you use higher pressure, you can change the tendency, and have a lower proportion of products, like in the figure. This is a simple application of Le Châtelier's principle, the higher the pressure, the reaction has more tendency to the side of lower gaseous species (to the left). However, this $C_\mathrm{p}$ is now ascribed not to a standard pressure, but a higher one, so something more needs to be done. I have not found data that solves this issue with a higher temperature range, so it is a good exercise to the brain how to keep the $\ce{NaHCO3}$ intact at higher temperatures. The author managed to pull it off up to $\pu{400 K}$ (the validity of the equation), but could not do for more.

Now we can finally answer your question. Imagine that the pressure is $p$ and some degree of reactants+products gives you a dimensionless extent of reaction $\psi$. You are sitting on temperature $T$ of the graph, then:

  1. If you are above the curve, $\Delta_\mathrm{r} G > 0 $, and the reaction is spontaneous in the reverse way. You will aproach equilibrium trying to hit the curve by forming reactants from products. The other way, formation of more products, will never happen.
  2. If you are below the curve, $\Delta_\mathrm{r} G < 0 $, and the reaction is spontaneous in the forward way. You will approach equilibrium trying to hit the curve by forming products from reactants. The other way, decomposition to more reactant, will never happen.
  3. If you are exactly in the curve, you don't move, the free energy is minimimized, because all those points in the curve are locations of temperature where $\Delta_\mathrm{r} G = 0 $.


  1. The heat capacities of gases has been taken from H. M. Spencer, Ind. Eng. Chem, vol 40, pp. 2152-2154.
  2. The heat capacity of $\ce{NaHCO3}$ has been taken from K. K. Kelley, U.S. BUR. Mines Bull. 584, 1960.
  3. The heat capacity data for $\ce{Na2CO3}$ was taken from the NIST Website here. The equation is too long (like $5$ parameters) so I reduced it to the form that I presented, performing minimum squares over the range of $T$.
  4. The standard enthalpy of reaction and Gibbs free energy were taken from the CRC Handbook of Thermophysical and ThermoChemical Data, D. R. Lide, H. V. Kehiaian, 1994.

\begin{array}{c|ccccc} \pu{Compound} & \Delta_\mathrm{f} H^\circ \; \pu{(kJ/mol)} & \Delta_\mathrm{f} G^\circ \; \pu{(kJ/mol)} & A & B \cdot \pu{10^{-3} (K^-1)} & C \cdot \pu{10^5 (K^2)} \\ \hline \ce{NaHCO3 (s)} & -950.800 & -851.000 & 5.128 & 18.184 & 0 \\ \ce{Na2CO3 (s)} & -1130.680 & -1044.440 & 0.336 & 3.225 & 0.228 \\ \ce{H2O (g)} & -241.818 & -228.572 & 3.470 & 1.450 & 0.121 \\ \ce{CO2 (g)} & -393.509 & -394.359 & 5.547 & 1.045 & -1.157 \\ \end{array}

  • 2
    $\begingroup$ Nice. Hehe, is there any SE storage limit per 1 Q/A ? :-) $\endgroup$
    – Poutnik
    Commented May 21, 2023 at 11:47
  • $\begingroup$ @Poutnik In fact I was going to post something like porphyrin. However, the post became gigantic, and in your answer (in the comments) you said would like to see $T$ as a function of the partial pressures (in this case, in terms of $\psi$). So I just did that. One reason for me to be here is to write in a less obscure way, as a recommendation of my supervisor. So I am practicing. I think everything here is understandable, but it is too long, for the sake of being clear. I need to minimize the function $\endgroup$ Commented May 21, 2023 at 11:55
  • $\begingroup$ It was not meant as a criticism of your post, rather a joke. It is really a good answer, voted up. $\endgroup$
    – Poutnik
    Commented May 21, 2023 at 11:57
  • 1
    $\begingroup$ @Poutnik As you know there are forms $f$ and content $c$, the content may be great, but if it is far too long it doesn't matter. You would define some youtube channels as $f\rightarrow \infty$ and $c\rightarrow 0$. So comments like yours really help me. Thanks. $\endgroup$ Commented May 21, 2023 at 12:02
  • $\begingroup$ If you make another edit, you may want to set differential operators in (14)(15) upright. / Upright vs italic $\endgroup$
    – Poutnik
    Commented May 22, 2023 at 4:38

I've tried to summarise some points made in the comments and in the question's title.

The word 'spontaneous' is perhaps an unfortunate one when chemical reactions are involved. In a process such as the mixing of two non-reactive gasses or diffusion of a dye in a solvent, where any enthalpy changes are insignificant, to talk of spontaneous mixing makes sense. The free energy is negative due to the increase in entropy, and furthermore, in cases such as a dye diffusing in a mobile solvent the process can be followed by eye over a few minutes.

When chemistry is involved 'spontaneous' really only means that the equilibrium constant $K_e \gt 1$, i.e. that more products than reactants are present in the sense that $ [C][D]/[A][B] >1$ if $\ce{A + B\rightleftharpoons C + D}$ (instead of [A] etc., use activities as necessary). Equivalently, the fact that $\Delta G$ is negative in a spontaneous reaction means only that the products are of lower energy than the reactants and as always in thermodynamics, it does not tell us anything about how fast the reaction will be.

Conversely when $K_e\lt 1$ we may call this non-spontaneous but reaction still occurs and some products will still be present, how much clearly depends on the magnitude of $K_e$. Additionally some reactions can be spontaneous at one temperature and non-spontaneous at another, the decomposition of cupric oxide to cuprous oxide and oxygen is an example. At high temperatures only is the reaction spontaneous: this is an example of an endothermic but spontaneous reaction.

The other unfortunate implication of 'spontaneous' is that the reaction can be observed to occur, or will happen on some sort of shortish time scale, and this may be true but is definitely not always so.

Edit: It should also be mentioned that some authors use 'spontaneous' to mean an 'irreversible process' as opposed to a 'reversible' one. An irreversible process is one from which after it has happened we cannot reform the initial state without expenditure of energy from elsewhere.

  • $\begingroup$ What about Q < K versus Q > K conditions ? IF K = 0.5 and if I mix reactants, they would have spontaneous TD tendency to react until Q = K = 0.5. I do agree it makes sense to speak about reaction spontaneity generally wrt K value, but also wrt particular reaction conditions and Q/K. $\endgroup$
    – Poutnik
    Commented May 19, 2023 at 13:57
  • $\begingroup$ @Poutnik I don't quite follow, there is nothing special about $K_e=1/2$. Why worry about $Q$ ? Thermodynamics deals with equilibrium states and we otherwise never know $Q$ unless we are doing some sort of kinetic experiment. Does not $Q \lt \,\gt K_e$ mean that we start at opposite sides of the equilibrium ? Of course a part, but not all, of $\Delta G$ is just due to mixing. $\endgroup$
    – porphyrin
    Commented May 19, 2023 at 15:37
  • $\begingroup$ Nothing special about K=1/2. Just example. Spontaneity has both TD and kinetic aspects. And we do the kinetic experiment, real or thought one. / $\ce{NaHCO3}$ at high enough T in open air can decompose even if $\Delta_r G^\circ \gt 0$ as it never reaches equilibrium until there is just $\ce{Na2CO3}$. Or I am terribly wrong and not seeing it. IT is always possible. $\endgroup$
    – Poutnik
    Commented May 19, 2023 at 17:56

Decomposition temperature can have the standard reaction Gibbs energy still positive, if nonstandard one is already negative.

Thermal decomposition of solids to products partly gaseous (in context of thermal stability) is usually experimentally studied at atmospheric conditions, not with product standard partial pressures.

If a closed vessel is pumped up by pressurised carbon dioxide and water vapor, than sodium bicarbonate decomposition temperature will be obviously higher than for being placed to heated room air.

The reaction cannot be carried out under standard conditions.

It can be carried out under standard conditions, but may not be.

Decomposition temperature at standard conditions may be interesting in context of theoretical studies and tabulated standard thermodynamical values. For practical purposes, standard conditions are not interesting.

  • $\begingroup$ Just because standard conditions, as you say, are not interesting for practical purposes doesn't answer the question, that is, why the standard free energy change is positive when it ought to be negative because the reaction is spontaneous under standard conditions. So, let's forget about the non-standard conditions and try to resolve why a spontaneous reaction, under standard conditions, is showing a positive standard free energy change, shall we? This itself is interesting enough. $\endgroup$
    – ananta
    Commented May 19, 2023 at 5:17
  • $\begingroup$ Who says it is spontaneous at standard conditions while the standard reaction Gibbs energy is positive? If they say it is spontaneous, it is not the same. $\endgroup$
    – Poutnik
    Commented May 19, 2023 at 5:22
  • $\begingroup$ It would be interesting to make a chart of T_decomp = f(p_H2O,p_CO2) $\endgroup$
    – Poutnik
    Commented May 19, 2023 at 5:47
  • $\begingroup$ Exactly my thoughts once you pointed out the discrepancy in partial pressures. I am looking for that now. $\endgroup$
    – ananta
    Commented May 19, 2023 at 5:50
  • 1
    $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Poutnik
    Commented May 19, 2023 at 6:03

The notion that a reaction is spontaneous if $\Delta G^0$ is < 0 is only a rule of thumb, and all it indicates that for a totally gas phase reaction, the equilibrium constant is <1. If there are solid phase also present, such as in the present case, this rule does not necessarily even come close to applying.

The diagram you showed with partial pressures of 1 bar for equilibrium is only correct at the temperature corresponding to $\Delta G^0(T)=1$. At other temperatures, the equilibrium partial pressures are different. For example, at 400 K, where $\Delta G^0=230\ kJ/mole$, the equilibrium constant is $$K=exp{-\frac{\Delta G^9}{RT}}=0.9332$$Therefore the equilibrium partial pressures of CO2 and H2O are $$p=\sqrt{0.9332}=0.966\ bars$$, and the total pressure at equilibrium at 400 K is 1.932 bars. If the pressure is less than this, some of the NaHCO3 will convert to Na2CO3, CO2, and H2O, and, if the total pressure is greater than this, it will fall to this value as some Na2CO3 is converted to NaHCO3.

  • $\begingroup$ (1) The second law of thermodynamics, (2) common knowledge in chemistry, (3) as stated under the section Common Misunderstanding (with justification), (4) answer by Metal Storm and by (5) porphyrin, all agree K>1 is a necessary condition for reaction to be spontaneous (at constant $T$ and $P$). These five ideas, all in complete agreement, ought to convince you that, for reaction (at constant $T$ and $P$, which is the case here) to be spontaneous, the free energy of reaction $(\Delta_\text{r} G)_{T,P}<0$ and $K>1$. It is not just a rule of thumb, it is a fact! $\endgroup$
    – ananta
    Commented May 22, 2023 at 14:52
  • $\begingroup$ Correction: $\Delta_\text{r} G^\circ (\pu {400 K}) = \pu{0.230 kJ mol^{-1}}$. You are also incorrect while calculating the $K^\circ$ value. The calculation $K^\circ = e^{-\dfrac{\Delta G^{\circ}}{RT}}$, as the $^\circ$ indicates, needs to be performed at standard conditions. The data point I have presented ($T=\pu{386 K}$) is empirical and $p_\ce{CO2}+p_\ce{H2O}=\pu{2 bar}$. Above this temperature, pressures will be larger, non-standard, under the conditions the authors did the experiment. $\endgroup$
    – ananta
    Commented May 22, 2023 at 15:29
  • $\begingroup$ @ananta I’m not impressed. It clearly depends simply on defining a reaction as spontaneous if K>1 $\endgroup$ Commented May 22, 2023 at 15:32
  • $\begingroup$ @ananta I disagree with you on your so-called correction; I used the standard change at 400 K in the calculation. $\endgroup$ Commented May 22, 2023 at 15:36
  • $\begingroup$ You don't have to be impressed by it, just accept it. The second law of thermodynamics states that for a spontaneous process $(\Delta S)_{U,V}$>0, this *translates* to $(\Delta G)_{T,P}<0$. Under standard conditions, $(\Delta G)_{T,P}^\circ<0$ and $K>1$. $\endgroup$
    – ananta
    Commented May 22, 2023 at 15:38

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