text:
Can free energy of a spontaneous reaction be positive?
Common Misunderstandings
Some comments and answer have suggested the following:

The data are Inaccurate.
The reaction cannot be carried out under standard conditions.
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.

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.

Context
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 $\pu{500 K}$.

Reaction
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):
$$
\Delta_\text{r} G^{\circ}(T) = \sum_\text{products}\Delta_\text{f} G^{\circ}(T) - \sum_\text{reactants}\Delta_\text{f} G^{\circ}(T)
$$
or:
$$
\Delta_\text{r} G^{\circ}(T) = \sum_i \nu_i \Delta_\text{f} G_i^{\circ}(T)
$$
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 $T$.

Data
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.

Results

$\ce{\Delta_\text{r} G^{\circ}}$ ($\pu{298.15 K}$)
$\ce{\Delta_\text{r} G^{\circ}}$ ($\pu{400 K}$)
$\ce{\Delta_\text{r} G^{\circ}}$ ($\pu{500 K}$)

$\pu{+26.192 kJ mol^{-1}}$
$\pu{+0.230 kJ mol^{-1}}\text{ !?}$
$\pu{-33.544 kJ mol^{-1}}$

Discussion

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?

Question
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.

References

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

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

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.

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

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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:

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.
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:

The ratio of the fugacity and the fugacity at standard state of a pure solid species is $1$ by definition.
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:

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).
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:

[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:

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:

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.
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.
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 $.

References

The heat capacities of gases has been taken from  H. M. Spencer, Ind. Eng. Chem, vol 40, pp. 2152-2154.
The heat capacity of $\ce{NaHCO3}$ has been taken from K. K. Kelley, U.S. BUR. Mines Bull. 584, 1960.
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 $T$.
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}

Nice. Hehe, is there any SE storage limit per 1 Q/A ? :-)

@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

It was not meant as a criticism of your post, rather a joke. It is really a good answer, voted up.

@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.

If you make another edit, you may want to set differential operators in (14)(15) upright. / Upright vs italic

In the expression $\dfrac{C_\mathrm{p,j}}{R} = A_\mathrm{j} + B_\mathrm{j}T+\dfrac{C_\mathrm{j}}{T^2}$, you missed that water boils in the temperature range you have considered, thus the integral is not valid.

@ananta I answer you in the chat.

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.

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.

@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.

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.

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.

(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!

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.

@ananta I’m not impressed.  It  clearly depends simply on defining a reaction as spontaneous if K>1

@ananta I disagree with you on your so-called correction; I used the standard change at 400 K in the calculation.

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 $K>1$.

Sorry, you are right, the calculations do show $K^\circ<1$ and $(\Delta G){T,P}^\circ>0$, that is the problem we are facing. It should be $K>1$ and $(\Delta G){T,P}^\circ<0$. That is the mystery we are trying to resolve. Sorry, you are right, the calculations do show $K^\circ<1$ and $(\Delta G){T,P}^\circ>0$, that is the problem we are facing. It should be $K>1$ and $(\Delta G){T,P}^\circ<0$. That is the mystery we are trying to resolve. All the experiments indicate $K^\circ \pu(400 K)>1$, the theoretical calculation by Metal Storm also indicates the same.

I stand by what I said.

Let us continue this discussion in chat.

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.

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.

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.

It would be interesting to make a chart of T_decomp = f(p_H2O,p_CO2)

Exactly my thoughts once you pointed out the discrepancy in partial pressures. I am looking for that now.

However, I do have (thermodynamic) reasons to believe that the decomposition temperature won't change significantly between $\pu{1 bar}$ and $\pu{2 bar}$.

But between 1 bar and 0.0004 bar for aerial CO2 and whatever for e.g. 50% humidity at 25 Deg C these initial conditions are far from standard ones. It will start decompose sooner than SRGE says.

Let us continue this discussion in chat.


text:
How do I plot ternary phase diagrams using Flory-Huggins solution theory?
I have three components, $A,B,C$, with 3 exchange parameters: $\chi _{AB}, \chi _{BC}, \chi _{AC}$. I want to create a FH ternary diagram to see how such a mixture behaves and how phase separation takes place i.e find points on the ternary plot where I will see $(A+B,C), (A+C,B), (B+C,A), (A+B+C), (A,B,C)$, where $I+J$ represents a mixed phase of $I,J$ and a separation by a comma $","$ indicates the third species is in a different phase.
The ternary spinodal condition is:
$$D=\frac{\partial ^2 f}{\partial \phi _B ^2}\Bigg|_{N,T,\phi_A}\frac{\partial ^2 f}{\partial \phi _A ^2}\Bigg|_{N,T,\phi _B} - \left( \frac{\partial ^2 f}{\partial \phi _B \partial \phi _A} \Bigg| _{N,T}\right)^2 = 0$$
My question is, how do I find the regions on the ternary plot indicated above? How do I find the region where two species are mixed while the third one is not?
I am not quite sure what it means when A+B is in a mixed state and when C is not. Does this mean that $\mu _A = \mu _B \neq \mu _C$?
This is basically the figure I want to make from this paper:

I would appreciate any advice you have for me.

Do you have discrete data points?  If so, did you query the net for plotters of ternary data?  Online like https://www.ternaryplot.com/, https://ternaryplot.netlify.app/, on-site for Windows like past4, or OS independent with Python/plotly?  Wolfram seems to be there, too example.

@Buttonwood, thanks for your reply. my question is more along the lines of how does one obtain these points from the FH free energy.

If it were me, I would run the calculations using fixed C concentrations and vary across A-B concentrations, searching for common tangents to get phase boundaries in the pseudo binary system. Vary C in small steps to get whatever density of points you need…

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text:
How can we find a 3d sdf file if there is no in PubChem?
I'm looking for a 3d sdf file of BBr3.
Though I cheked PubChem, there is no 3d file, but there is a 2d file.
In that case, how can I get a 3d file, are there any other repositry on 3d file?
or below is acceptable?
    mol = Chem.SDMolSupplier("2d-sdf file", removeHs=False)
    # compatible with "removeHs=False"
    mol = Chem.AddHs(mol)

    params = ETKDGv3()
    params.randomseed = 1
    EmbedMolecule(mol, params) # expand mol to 3d

best regards!

You may want to include what you are doing. This doesn't appear to be code that actually runs. Also you might want to include the file you are looking at. The geometry of the module is actually quite simple, there are probably 3d coordinates in a different database.

https://cccbdb.nist.gov/geom1x.asp

BBr3 is symmetrical planar so you can edit the 2D data and add a z column  and  make each value  zero. (Levy & Brockway JACS 1937, p2085,v59). The B-Br bond length is 0.187 nm, so in fact this is all that you need.

Thanks for the comemnt. I automatically download e.g. BBr3 via pubchempy https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/CID/25134/record/SDF?record_type=2d&response_type=display and I'm use above code for structure optimixation before calculating wafe-function. Is this proper way? Sorry for my beginner question.

Looks like ordinary 3d coordinates to me. I really don't understand what problem you are facing. You should [edit] your question to make it more clear. Also, this appears to be a better fit at [mattermodeling.se].

With a SMILES string (BrB(Br)Br), a program like openbabel (its most recent documentation here) can
obabel -:"BrB(Br)Br" --gen3d -osdf -xv2000

generate a conformer (described here)* for you. As such, the approach is not this much different to the one PubChem uses – their 3D conformer preview equally is an affordable guess about an isolated molecule without any influence by a solvent molecule, or other neighbors (e.g., packing as in crystals).
If you don't have a SMILES string in hand, nor an other structure file openbabel can process to yield a SMILES string for you: ChemDraw's sample page allows to draw small molecules which can be exported in this representation (structure -> get smiles) by copy-paste.
Kim, S.; Bolton, E. E.; Bryant, S. H. PubChem3D: Conformer Ensemble Accuracy. J. Cheminform. 2013, 5, 1. https://doi.org/10.1186/1758-2946-5-1 (open access).
* OpenBabel's default is MMFF94 you can adjust to other force fields (entry).  They differ in coverage of atoms / patterns encoded in parameters.  With more flexible molecules, the program's search for multiple conformers is a helpful tool, too.

Dear martin, porphyrin, buttonwood. Thank you for your kind comemnt. I got saficient infomation, probaly. Originally, I just would like to know whether a 2d-sdf file is available for psi4 structure optimization after it is epanded to a 3d model via EmbedMolecule

In this particular case (boron) Open Babel will use UFF since there are no MMFF94 parameters for boron.

@GeoffHutchison Because Open Babel automatically looks up for a force field suitable and the specialty of boron here, I now think the explicit call obabel -:"BrB(Br)Br" --gen3d -osdf -xv2000 --minimize --ff MMFF94 (which, by lack of parameters, fails) vs obabel -:"BrB(Br)Br" --gen3d -osdf -xv2000 --minimize --ff UFF (no error report) is the better approach (for that the limitations are prominently reported). On the other hand I assumed --gen3d and --minimize as two separate actions, where --minimize works with the results obtained by --gen3d. Do you recommend the second approach?

I personally would prefer it, to clear up the question first before answering. Now it'll sit here with a negative score, unappealing and closed, with an answer that in my personal opinion does not really address the actual problem. The pubchem file is ok, there is no need for a different strategy.

@Martin-マーチン If, as moderator, you think the neither one is for the benefit of chem.se I can agree to a clean-up which removes question and answer altogether.

I'd like to emphasize again that I was sharing my personal preference, i.e. in the capacity of an ordinary user. I was somewhat invested in the question and I'm slightly disappointed that this led nowhere. This is of course no fault or yours. In my capacity as a moderator I see no reason to step in at all.

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text:
Does an ionic bond have a dipole?
Is an ionic compound like NaCl considered a dipole? It has a positive side (Na+) and a negative side (Cl-). Or is it true that an ionic bond does not have a dipole because a dipole is, by definition, a polar molecule?
Some people said it has a dipole, because the word dipole describes a separation of charge, so we could call an ionic compound a dipole.
While others said it doesn't have a dipole, because ionic compounds are held tight in a giant lattice, so the electrostatic forces of attraction are in all directions between all the ions. Thus, ionic lattices are not polar.
Does it maybe depend on the state of the ionic compound, whether it's solid, liquid or gas?

It must have a dipole because the di means two. So Na+ and Cl- must make up a dipole.

So all ionic compounds are dipoles whether they're a solid, liquid or gas? And all ionic lattices are polar?

Your question only makes sense for gas phase - there are discrete NaCl molecules in there, even if ionic component of bonding is large.

So, only in gas phase ionic compounds are considered dipoles (they're polar) because they exist as discrete molecules, but when they're in liquid or solid phase, they're not dipoles (they're nonpolar) because they don't exist as discrete molecules?

In molten or solid salt... you could calculate dipoles for each neighbouring Na and Cl, but they all compensate each other - if you consider all neighbours you get zero. Also word "polar" is only good for solvents. Calling molecules like that is a dangerous mental short-cut, that leads to misconceptions.

"A dipole" strongly implies a single thing having two poles. A molecule, like for example HCl, is a single thing. It consists of two atoms sharing a privileged spatial relationship, that stay close to one another (and much closer than to any other atom) for as long as the molecule exists. Because the shared electrons holding them together are unevenly distributed, the molecule has a "plus" and a "minus" side and is called a dipole.
It is a common mistake to see an ionic compound like NaCl and imagine it also contains some sort of fundamental entity, a distinct "molecule" consisting of a single Na and Cl atom and analogous to the real HCl molecule but held together by an ionic rather than covalent bond. The only place you'd see something like this would be in the gas phase, where ion pairs occur in rapid equilibrium with dimers (Na2Cl2) and larger clusters. For the time they exist under conditions you'll probably never encounter, these ion pair could be considered molecules and they are obviously highly polar, so yes, they would be dipoles.
Under ambient conditions though, ionic compounds are usually solids, in which each positive ion is surrounded by many more negative ions than the one (in the case of NaCl) it's "supposed" to bond with according to its empirical formula. Moreover, these opposite charges are, by defintion, symmetrically distributed around the ion (if they weren't, the lattice wouldn't be in equilibrium and would shift until it was). Each individual ion in the bulk of an ionic solid is in a perfectly apolar environment, so it doesn't make sense to talk about dipoles here.
The same logic holds for molten or dissolved salts, just that the ions move about randomly in stead of being stuck in a lattice; they're still doing that as individual ions, not ion pairs, so again there is no single thing to be called "a dipole".

-1 because there's no free ions in gas phase  https://chemistry.stackexchange.com/questions/2511/will-gaseous-ionic-compounds-be-free-moving-ions https://chemistry.stackexchange.com/questions/14174/what-is-sodium-chloride-like-in-gas-state

Not all crystals containing ions have charges that balance out: ferroelectric crystals exist where ions can be polarised the same way magnetic domains can be polarised in ferromagnetic solids, giving the whole crystal a net dipole moment.

@Mithoron You are right and I have edited my answer to reflect this.

Context is important for any coherent answer
The question is more subtle than it looks because the context matters for the answer.
Does an ionic bond have dipole? Yes, by definition: some separation of charge must create a dipole. But does a solid substance, NaCl, have any net dipole: no, as all the constituent dipoles cancel out when averaged over the whole crystal. And, in a typical ionic crystal, it isn't even clear that the idea of a single ionic "bond" makes sense as the bonds are not discrete entities as they would be in a molecular solid.
So, if the context of the question is whether the interaction between single ions of chlorine and sodium creates a dipole, the answer is "yes". But if the question is whether the bulk solid has a dipole, the answer is no. If the question is whether discrete gas-phase units with dipoles exist in NaCl vapour, the answer is maybe (depending on conditions).

As a supplementary answer, the common definition of a dipole is a separation of charges, this separation means different the presence of opposite charge particules or quasiparticles. To be clear a proton (positive charge) and an electron (negative charge) should normally form a dipole (two poles) but the quantum ground state of this system is not a dipole, the elwcrronic distribution is symmetric and there is no dipole moment. Therefore, a dipole should definitely have a dipole moment.
In the case of the polar molecules as free NaCl, there is a dipole moment, so it is a dipole. However the crystal form of NaCl do not have a dipole moment and is not a dipole : the dipole moments of pairs cancel each other. There is no need of a pair of two elements to form a dipole (HCl) even larger molecules (donor-sigma-acceptor) form a dipole.
A set of polarized dipoles is almost always an unstable configuration, they always interact to form a gas, solid but especially liquid etc. where the total dipole moment is zero. The same configuration appears for magnetic dipoles by forming magnetic domains with a zero total magnetization.

A simple explanation can be that a dipole occurs for a molecule.
Keeping this in mind we can say that until and unless the ionic compound maintains a molecular identity, it will have a dipole. This is possible only if the ionic compound exists in a gaseous state; since in solid state it forms a lattice structure where individual molecules are indistinguishable and in molten or aqueous form, the ions dissociate into individual ions.

From the Merriam-Webster dictionary definition of dipole:
ˈdī-ˌpōl

1 a
: a pair of equal and opposite electric charges or magnetic poles of opposite sign separated especially by a small distance

b: a body or system (such as a molecule) having such charges or poles

2: a radio antenna consisting of two horizontal rods in line with each other with their ends slightly separated

Thus, there is ionic bond, like in the mentioned molecular $\ce{NaCl}$, as from definition 1a from that dictionary.
As to the question of whether an ionic bond has a dipole or not, from the dictionary definition b for dipole, the body or system needs to have such charges or poles.  The reason that this can be a little tricky to understand is that a system of bonds together may not contribute any dipole moment, when the individual plus dipoles and minus dipoles of the bonds in the system cancel out, given a net zero dipole moment for the overall molecule.
So it is important then to reference if a, one is describing a specific bond of equal and opposite charges or if b, one is considering a system containing multiple charges, perhaps equal and opposite or perhaps not.

Well the thing is dipole only considered in the case of covalent bond because in covalent bond, the bond is formed by mutual sharing of electrons between two atoms and in which the more electronegative atom attracts the shared pair of electrons known as electronegativity of that element that creates a dipole moment or polarity in a bond but in case of ionic bond there is no such sharing of electrons, instead there is a complete transfer of electrons for an example in NaCl the electron is transferred from sodium to chlorine which cause the formation of cation and anion now there is no sharing but there is a complete transfer and due to these formation of cation and anion they got charges and according to the law opposite charges attract each other and eventually electrostatic bond formed now Fajan's rule comes in action as he suggested that if a cation has a really powerful nucleus that means it has a high polarising power which attract the electron cloud of anion toward itself and cause a little polarity which induce a little covalent character in ionic compounds for an example LiCl is more covalent than NaCl due to the more powerful nucleus of lithium which causes high polarization power to the electron cloud of chlorine. (Feel free to correct me!!!)


text:
What are some efficient approaches for balancing long chemical equations?
I am having some difficulties in balancing chemical equations involving many elements. It takes a lot of time (for me) to balance the coefficients using the hit and trial method. Especially like this one:
$$\ce{K2Cr2O7 + HCl -> KCl + CrCl3 + H2O + Cl2}$$
My problem is: Where should I start? The heavier or the lighter elements? Which is more efficient?
For the above problem, I started with K since it appears first in the equation. I don't think this is a great way for approaching these questions.

You understand it's a redox reaction? Start with the elements that get oxidised and reduced.

Let's start by the atoms of the most complicated molecule : $\ce{K2Cr2O7}$.
Start with $\ce{K}$. There is $2$ $\ce{K}$ on the left hand side. They must appear in the products : $2$ $\ce{KCl}$. We'll take care of the $\ce{Cl}$ atoms later on. Let's finish the starting molecule : $\ce{K2Cr2O7}$
There is also $2$ $\ce{Cr}$ atoms in the reagents. So $2$ $\ce{Cr}$ atom must appear in  the products : $2$ $ \ce{CrCl3}$.
And now there are $7$ $\ce{O}$ atoms in $\ce{K2Cr2O7}$. As a consequence, $7$ $\ce{H2O}$ molecules must appear in the products, with $14$ $\ce{H}$ atoms, and of course $14$ $\ce{HCl}$ molecules in the reagents.
Now the first molecule of the reagents is OK. Let's consider what atoms have been introduced in the products which are NOT YET equilibrated, namely the chlorine atoms.
In total, $8$ $\ce{Cl}$ atoms are included in $2 (\ce{KCl + CrCl3}$). These $8$ chlorine atoms must appear in the reagents : It makes $8$ $\ce{HCl}$ molecules in the reagents. We have already seen that $14$ $\ce{HCl}$ has been produced, and here only $8$ chlorine atoms are needed for producing $2$ $\ce{KCl}$ and $2$ $\ce{CrCl3}$. The $14 - 8 = 6$ chlorine atoms in excess are used to produce the $\ce{Cl2}$ molecules. And to produce exactly  $3$ $\ce{Cl2}$ molecules (with $6$ $\ce{Cl}$). So the final equation is :$$\ce{K2Cr2O7 + 14 HCl -> 2 KCl + 2 CrCl3 + 7 H2O + 3 Cl2}$$ This result can also be obtained with the half-equation method, as suggested in Mithoron's answer.


text:
Why are moles (quantity of a gas) and temperature inversely related?
I'm learning about the combined gas law, and while doing calculations, I found that a doubling in the quantity of moles causes the temperature to be cut in half (all other things being equal).
I don't understand the mechanics of the change. Why does twice as much gas in the same amount of space cause the temperature to drop?
I would assume that there would be no change, since the total heat mass of the gas is unchanged. The particles in the gas are moved closer together, true, and more influenced by intermolecular bonds; but that seems to cut both ways, as a more crowded cluster of particles might produce more collisions? All in all, I understand the formula, but I don't get how the particles are interacting.
Also, is it true to say, mechanistically, that an increase in volume produces an increase in temperature? I don't understand how that would work, but it seems like a parallel to the example above (particles with less space to move have less energy; you can either shrink the container or crowd it, and reduce the temperature accordingly).
I understand how Charles' Law works in the other direction: increasing temperature gives the particles more energy, which causes them to collide harder and more frequently with their container, which will then expand.
But does merely making a container larger cause the gas within to rise in temperature? How would more space increase the total heat mass contained within the system? Where is the extra energy coming from?
A decrease in V causes an increase in P, and an increase in P causes an increase in T, but a decrease in V does not increase T?
I would appreciate your help very much, this is driving me a little crazy.
ps. Down voting questions is rather ungenerous. It's a question, not an assertion. I am confessing that I don't understand. If I know too little to ask the question well, forgive me. Your help is appreciated.

Review the guides Asking and How to ask to prevent not satisfying answers, misunderstanding, objections, question downvoting or closure

For your awareness:
Quantities: length, mass, amount // unit names:  metre, kilogram, mole // unit symbols: m, kg, mol

Okay, where have I gone wrong with that above? The quantity of a gas = the mass or amount of it. Moles are the most common unit used to denote a quantity of gas. Is it wrong to say heat mass? I thought I read that in a definition of enthalpy. I could just as well say total kinetic energy, but the two seem synonymous.

You should include in questions a short conclusion with eventual quotes/links from prior info searching, so others do not write things you already know, or you should know if you searched about it. If nothing was found, write at least what you have searched for. // Aside of general keyword search, search also with the search modifier site:stackexchange.com OR site:libretexts.org OR site:wikipedia.org OR inurl:hyperphysics keywords to find similar topics on the respective sites.

This actually belongs on Physics.SE rather than on Chemistry.SE. And the true answer is that you are using the logical flow of the formula wrongly.

Ahh, okay, naturallyInconsistent, that sounds like an answer! Thank you! Could you explain a little more, even if it's a physics problem?

It's not a forum, but Q&A site - we primarily collect questions and answers that are valuable.

That's what I said? And you didn't answer my question, just implied it was stupid. That's not a very generous attitude for a teacher to have.

Before why, ask if.  As $n = \frac{pV}{RT}$, doubling gas amount is doubling $\frac{pV}{T}$.  There are 2 degrees of freedom, you can independently control 2 of 3 state variables and the 3rd is determined by the other 2.

Okay, that's an answer, and I appreciate it, but it's describing the formula, not the particles. Can you describe what the particles are doing such that moving them closer together decreases the temperature? Either by crowding them in a smaller container or doubling the quantity in the same container.

There is no sense of answering questions that start with false premises.  Amount of a  gas and temperature are not inversely  related.

Well, that's what I expected. But then I did a question for my course and a doubling in the quantity of gas halved the temperature. Not so?

We had a problem in which pressure and volume were unchanged, the number of moles doubled (from 1 to 2), and the initial temperature was 298K. We were asked to calculate final temperature. The result is 149K. When the formula is rearranged, n2 is the denominator, while n1 is the numerator along with T1; so that explains why the temperature will be halved. If all other factors remain constant, and moles are doubled and temperature halves, that seems like an inverse correlation. Where have I gone wrong?

You can ask this question for several intensive and extensive quantities where the relationship between is simply intuitive : why the temperature and the volume are inversely related, why the velocity and the time are inversely related, why a frequency and a mass are inversely related.

You have misunderstood the equation
$$pV = nRT$$
If we double the amount, then to keep in the fixed volume $V$ the initial pressure $p$, you have to set the half of the initial absolute temperature.
$$pV = (2n)R(\frac{T}{2})$$
Because for constant volume and temperature, it would have the double pressure:
$$(2p)V = (2n)RT$$
But that does not mean the doubling the amount decreases temperature.

What is the R? They aren't using that in my course. And ahh, I think I've understood.

And yeah, I was drawing the wrong conclusions, which is what had me so puzzled. That relationship between n and T did not make sense to me (because it was wrong). So, what you're saying is, if you want to keep pressure constant, while doubling quantity, and the container can't change size, then something's got to give, and that's temperature. Alright, thanks.


text:
How can I accelerate iron corrosion in a way that simulates natural processes?
I am attempting to conduct a research project comparing the effect of various current levels on electrolytic rust removal. Somewhat ironically however, I need a way to induce corrosion in a predictable and fast manner so as to have a standard baseline from which to conduct my experiment. Ideally it would somewhat resemble natural corrosion so that the data is applicable outside of a laboratory setting. Thank you in advance to anyone who answers.

Not my area, but have you checked any of the corrosion-relevant ASTM (American Society for Testing and Materials) publications? Corrosion of steels is a big topic, so there may well be standard protocols for preparing test specimens, testing corrosion inhibitors, etc. Best of success!

A general way to speed corrosion is to use a mix of salt, hydrogen peroxide and vinegar (5% acetic acid), https://ctsciencecenter.org/blog/science-at-play-instant-rust/ As @EdV states, this might be used in some ASTM corrosion resistance tests.


text:
Why can't ethoxide ion act as nucleophile?
Consider the reaction of chlorocyclohexane with ethoxide in a suitable inert solvent. The major product according to my textbook is cyclohexene, which implies E2 reaction pathway. Why not SN2?
$\ce{EtO^-}$ is an unstable anion, so why can it act as nucleophile rather than acting as a base?

Ethoxide is a bulky species. So it would undergo elimination rather than substitution. Perhaps you might get a better insight to this by this question.

@Proscionexium Then what about $\ce{NH_2^-}$? It also prefer elimination to above compound, this time it isn't bulkier.

Are you talking about $\ce{R-NH_2^-}$ or just only $\ce{NH_2^-}$?

@Proscionexium just $\ce{NH2-}$

I think maybe because abstracting $\ce{H^+}$ by $\ce{NH_2^-}$ would also release $\ce{Cl^-}$ (forming cyclohexene) just as $\ce{Cl^-}$ could be released by substitution (but remaining only cyclohexane).

@Proscionexium But $\ce{ \Delta H}$ whould be higher for elimination (Removing two $\sigma$ bonds). Also $\ce{EtO-}$ should be a good nucleophile due to its unstablity.

SN2 does occur. What do you think the minor product is?

@Proscionexium It's not bulky - compare with t-butanolate that actually is.

@Waylander I mean why not major product  by SN2.

@Leibniz-Z  There is enough hindrance of SN2 by the cyclohexane ring to favour proton abstraction by ethoxide

@Waylander then according to you $\ce{NH2-},\ce{OH-}$ would guve major product according to SN2.

Not in the case of amide anion. It is a very much stronger base and likely to produce more elimination

@Waylander but why? Since $\ce{\Delta H}$ is higher for elimination. And what about hydroxide ion?

Ethoxide acts as a base only in presence of $\beta$-hydrogens. For secondary and primary alkyl halides, usually, $\text{S}_\text{N}\text{2}$ is favored over $\text{E}_\text{2}$ mechanism.
$\text{E}_\text{2}$ mechanism does not occur in the absence of $\beta$-hydrogens:

$\text{S}_\text{N}\text{2}$ reaction is slow for secondary alkyl halides, further stearic hinderance is present due to the cyclic nature of chlorocyclohexane.

Reference

Solomons, T. W. G., Fryhle, C. B. (2011). Organic Chemistry, 10th ed. John Wiley & Sons.


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Derivation of isobaric heat capacity
As I was thinking about isobaric processes, one thing came to mind: In a reversible isobaric process, the value of Cp (molar isobaric heat capacity) is bigger than Cv (molar isochoric heat capacity) by exactly one R (ideal cas constant). For example, for argon, Cp is 2.5R while for Cv it is 1.5R.
This is due to the derivation, where under the pV integral, pV can be substituted by RT for one mole of the gas. This is only possible though, if the gas is in equilibrium with the outside pressure the whole time:
https://en.m.wikipedia.org/wiki/Isobaric_process
So my question is, would the value of the molar isobaric heat capacity change, if we applied an irreversible isobaric change?
That is, the gas is heated much more quickly that it can expand, so that the pressure in the gas is higher than the outside pressure on the piston, so that the gas and outside pressure are not in equilibrium the whole time.

As you realise if equilibrium is not maintained a different value of $C$ will be measured compared to that measured while equilibrium is maintained i.e. under reversible conditions. The average heat capacity is $C_{av}\equiv q/\Delta T$ and as $q$ is different, reversible vs non-reversible conditions so is $C$.  In the past rates of fast reaction were measured using this effect via the velocity of sound.

@porphyrin Hmm, but as $p_\text{ext}$ is constant, $W=\int{p_\text{ext}\cdot \text{d}V}= p_\text{ext}\cdot \int{\text{d}V}$ is constant for the the same $T_2$ and $T_1$ regardless of reversibility, so does $\Delta(p
V)$. So does $\Delta U(\Delta T)$ and $\Delta H$ and $C_V$. Am I wrong? // UNLESS, the OP means a differential $C_V$ where heat is passed, T is locally raised but  work is not being done proportionally yet.
Hm, then what T would this differential $C_V$ refer to?// But in integral form from initial to final state, I think the former holds.

@Poutnick I didn't read the question carefully enough, was thinking of a reaction where due to rapid heating the reaction equilibrium does not have time to establish itself. Otherwise, I agree with your analysis.

If the gas is initially in equilibrium with the surroundings and, in the final state, it is also in equilibrium with the surroundings, then $$p_{gas,init}=p_{gas, final}=P_{ext}$$then$$W=p_{ext}(V_{final}-V_{init})=p_{gas,final}V_{final}-p_{gas,init}V_{init}=\Delta (PV)_{gas}$$

Okay that makes sense, now consider following case: An ideal gas is originally in equilibrium with surroundings, then the outside pressure is abruptly lowered and kept constant at that smaller value; this is commonly known as an "irreversible isobaric process" although the initial and end gas pressures are not the same; why is it called that then?

Well, for sure, it's a little ambiguous, but it is what people with experience normally call this.  But, for sure, in that case, Q is not equal to $\Delta H=C_p\Delta T$ (even if the final state is at thermodynamic equilibrium).  Run the analysis and see what you get.

Yeah in this case, as the pressure is not constant Cp is not applicable.


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Which reaction to consider for overall rate
So I have a multistep reaction
\begin{align}
\ce{4 NH3 + 5 O2 &-> 4 NO + 6 H2O}\\
\ce{2 NO + O2 &-> 2 NO2}\\
\ce{3 NO2 + H2O &-> 2 HNO3 + NO}
\end{align}
If I know the initial decreasing rate of $\ce{O2}$ and want to find the initial overall reaction rate, can I just look at the coefficient in the rate-determining step or do I need to consider $\ce{O2}$ in the other step too?
Also if I want average rate over a period of time, is it still the same?

As neither of these reactions is elementary, you need to determine the reaction rate order or more generally reaction rate equation for each of them first. Then you may try to determine the rate determining step. I am afraid this cannot be solved without much experimental data. // Additionally, I assume there reaction are not ongoing simultaneously in the same place, but sequentially at different conditions.

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.

The rate determining step approximation is not going to help here. You have in this short and incomplete sequence already two reactions giving a common product.


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How to prepare solution of liquid reagent of the given molar concentration?
I have a liquid reagent with molecular mass $M = \pu{294.08 g mol^-1}.$ I'd like to prepare $V = \pu{400 ml}$ of solution of that reagent with a concentration $c = \pu{100 mM}.$
Does it mean I have to add
$$m = cVM = (\pu{0.1 mol l^-1})(\pu{0.4 l})(\pu{294.08 g mol^-1}) = \pu{11.76 g}$$
of reagent to $\pu{400.00 ml}$ of water, or should I add $\pu{11.76 g}$ of reagent to $\pu{388.24 ml}$ of water so that the total volume is $\pu{400 ml}?$

The amount concentration (molar  concentration, molarity) is defined in IUPAC Gold book as:
$$c_\mathrm{solute}=\frac{n_\mathrm{solute}}{V_\mathrm{solution}}$$
Practical procedure is:

Add solvent to the solute for the total volume to be little less than the target volume.
Dissolve the solute

There is frequent volume contraction, $V_\text{solution} \lt
V_\text{solute} + V_\text{solvent}$.

Be sure no liquid drops are on the flask walls or the plug.
Add the rest of solvent to reach the target volume and make the final mixing.

For precise preparation, this is being done in calibrated conical glass flasks with the narrow neck, marked for the target water level.
Your latter assumed procedure incorrectly assumes the solute density $\pu{1 g/mL}$ and that there is no volume change during dissolution.
Obviously, the procedure may not be that strict if accuracy requirements are low.

@ Pono. Your two propositions are both wrong. You must introduce $11.76$ g solute in a volumetric flask having a volume of $400$ mL (which is not easy to find by the way). Then you add $300$ or $350$ mL water in the flask, and stir until the product is totally dissolved. Then you add enough water to get a final volume equal to exactly $400$ mL. This volume of water is not important to know, but the total amount of added water is probably different from $400$ mL and from $388.24$ mL. Remember the volumes are not additive in chemistry. $a$ mL solute (powder) plus $b$ mL solvant never gives a solution volume equal to $a + b$.


text:
Difference in salinity and/or sugar content: Does it have a name?
Let's say that I put a tea infuser full of high-salt and/or high-sugar content, into a tea pot full of fresh water. I know that the salt and/or sugar will move from the infuser to the tea pot until their salt/sugar levels equalize.
Does the force, the pressure, which moves the water, have a name?
I thought about osmotic pressure, but I think that's just for salt? The system is also trending towards homeostasis, but that's a state, not a force. So, I'm not sure what to call this.

No, it's for sugar etc. too.

It is not osmosis, which applies when water can pass through a barrier, but dissolved compounds cannot.
It is diffusion of dissolved compounds driven by concentration difference and convection driven by liquid density difference. Usually the latter effect does the most of the transport taking a lot of the compound along a long distance.
If the concentration of the dissolved compound changes in some direction, the rates of migration of dissolved compounds between two places is not equal. The net effect causes equalization of the concentration. For more see Fick's laws of diffusion.
Dissolving of salt or sugar forms solution that is denser than water, convecting down, being replaced by water or more diluted solution.
A lot of useful info, presented in form suitable for education, can be found on sites Libretexts.org and Hyperphysics.


text:
How do the ions in electrolytes lead to lower dielectric constants compared to pure water?
I am interested in the salts used in the gel electrophoresis buffers. As I understand it, one of their roles is to change the conductivity of the medium, because otherwise the electric field would be too weak inside the medium to move the negatively charged sample.
How does this work from a mechanistic perspective? I found one lead here, but I don't see how the diagram explains the effect of the rearrangements on the overall electric field. (It seems to imply that the effects on the saline-associated electric field compared to pure water would be net zero to me, since the fields from the charge separation cancel out with the ones from the water.)
Using simplifications like point charges and intuition, is it possible to explain why the dielectric constant becomes lower in saline solutions compared to pure water?

An electric dipole with a dipole moment $\vec{p}$ in an electric field $\vec{E}$ has the energy : $\Delta E = - \vec{p}.\vec{E}$. This energy is minimal if the dipole moment and the field are aligned. If they are aligned by applying a weak field, the compound has a good susceptibility. However, some ionic compounds cannot align their dipoles when a field is applied especially in the case of solids like $\text{NaCl}$ where the dipole moments are antiparallel but the atoms cannot move leading to a low susceptibility and a low dielectric constant.
In the case of $\text{H}_2\text{O}_{(aq)}$, the effective dipole moment of each water molecule can be aligned in the direction of the field because water molecules can move as the hydrogen bonds are weaker than covalent bonds.
When $\text{NaCl}$ is in water, all the ions are spherically surrounded by water molecules : $\text{Na}^+$ ions form hydrogen bonds with $O$ and $\text{Cl}^-$ ions with $H$. If a field is applied along one axis in this spherically symmetric system and the water molecules is polarized as in a pure water, the hydrogens with a charge $\delta^+$ trying to move in the direction of the field will create an electrostatic repulsion with $\text{Na}^+$ ions which are also following the same direction, this process is identical for $\text{Cl}^-$ and $O$. This means that for aligning a water molecule when a field is applied with the presence of ions there is an electrostatic energy cost to pay with the energy $-\vec{p}.\vec{E}$. This means that the field $\text{E}$ should be stronger for a more negative  $\Delta E$ to achieve the same process than in pure water. Therefore a salted water has dropped the water susceptibility more significantly if more ions are present.
A simple approach when we deal with a mixture is to calculate approximately some properties by applying the rules of mixture at the equilibrium :
$$\epsilon_r(\text{H}_2\text{O/NaCl}) \approx \frac{\epsilon_r(\text{H}_2\text{O}+\epsilon_r(\text{NaCl})}{2} \approx 42$$
Aproximately the same result that in your reference. I am not good at drawing but I hope the picture of dipoles in a field and this repulsion is not complex.

@M06-2x Hello M06-2x. In the first line you speak about the change of energy $\Delta E$. Given that the equation comes from electrodynamics, I understand that change of energy as the energy that needs to be given/taken to rotate the dipole along their axis by an angle $\theta$. Is that the same energy you are talking about?

(...) I was surprised when you wrote that the relative permittivity of a mixture can simply be put as the arithmetic mean of the compounds that form the mixture. Do you think that the proportions will modify th equation? Do you have any book that makes this simplification for a problem? Any publication? Thanks in advance.

@MetalStorm  $\Delta E$ will reciprocally be the energy to be given if you want to rotate the dipole in the presence of the field. A field is  needed for rotating an electric dipole the angle depend on the intensity of the field and the susceptibility. The given expression of the permittivity using this rule is generally inaccurate for most of the cases and is more likely to be realistic if the proportion of ions is maximal comparable to water. There are several documentations about the rules of mixtures especially for composite materials or entropic systems.


text:
Question about measurement of half-cell potential
I am reading about electrochemical cells from the chapter: Electrochemical cells, Batteries and Fuel cells in the book 'Physical Chemistry' by Thomas Engel and Philip Reid, 3rd edition.
I came across the following paragraph and I want to know if my understanding of this paragraph is correct or not?
The paragraph is about $\ce{Zn}$ elctrode partially immersed in an aqueous solution of $\ce{ZnSO4}$

Paragraph: Can we measure $\Phi$( half-cell potential) directly? Let us assume we can carry out the measurement using two chemically inert Pt wires as probes. One Pt wire is placed on Zn electrode and the second Pt wire is placed in the $\ce{ZnSO4}$ solution. However, the measured voltage is the difference in electrical potential between a Pt wire connected to a zinc electrode in  $\ce{ZnSO4}$ solution and a Pt electrode in a $\ce{ZnSO4}$ solution, which is not what we want.

My question is whether the author wants to say that the second platinum wire will act as an electrode in zinc sulphate solution and hence the potential difference measured will be between zinc electrode and platinum wire which won't be the potential difference between zinc electrode and zinc sulphate solution?
Any help is appreciated.
Thanks!

See also the topic Absolute electrode potential

@Poutnik. Thanks for pointing about absolute electrode potential. To be sure, absolute electrode potential refers to the potential difference between a point inside the metal and a point outside the electrolyte solution?

@Poutnik, did you read the comment that I wrote below the answer?

You should use $\ce{ZnSO4}$ for $\ce{ZnSO4}$ and other formulas, as element symbol should not be in italics.

Yes, the author wanted to say what you have assumed.

Absolute measurements of half-potentials have no sense. They do not exist. They can be measured with respect to something, like for example the hydrogen electrode. It is similar to measuring the absolute altitude. The absolute altitude does not exist. The altitude is measured with respect to the sea level. The altitude cannot have an absolute value.

So half cell potentials do not exist independently? But in the book it is written that as zinc atoms leave the electrode leaving behind electrons, the zinc electrode acquires a negative charge and a corresponding positive charge builds up in the surrounding solution.  This charging leads to a difference in electric potential $\phi$ between the electrode and the solution, which we call the half-cell potential.

@NatashaJ Direct measurement of the potential difference between the metal and the electrolyte is like an attempt to separate 2 quarks of a pion.  If you succeeded, you would not have 2 quarks, but 2 pairs of quarks, providing during the separation energy for 2 other quarks.  this potential difference has to be determined indirectly, or estimated theoretically.

@Poutnik But we can determine the potential difference between zinc electrode and zinc sulphate solution indirectly if not directly?

@NatashaJ Possibly yes, but I am not aware of the way. Things get very complicated by extreme reactivity of free electrons in water. We cannot measure their work function as there are none. All are chemically bound.

Okay, thank you for being patient with me


text:
How could I desorb the CO2 molecules contained in the pores of activated carbon?
I have CO2 trapped in my activated carbon. What would be the most environmentally sustainable way to release it? I have heard of PSA (Pressure Swing Adsorption), which works when the adsorbent is subjected to a high-pressure environment, allowing it to adsorb CO2. Then, the pressure is reduced, causing the CO2 to desorb from the adsorbent and collect as a concentrated stream. But can that even work on activated carbon? Or is my best bet a temperature swing system (TSS), which requires lots of heat?

The best and simplest way to desorb adsorbed $\ce{CO2}$ molecules is to heat the charcoal to the highest temperature where it does not burn.

In most environments, heating the carbon is most efficient -- low density and moderate electrical resistivity makes it easy and efficient to heat the carbon directly. However, outside Earth's atmosphere, one could just leave it in vacuo, warmed by sunlight, and wait for it to outgas.


text:
Bonding and then breaking radioactive molecules
I was looking at Radon and noticed that it has a half-life of 3.8 days, yet still makes compounds with some other elements.
This got me wondering, are there any examples of chemical processes or industrial processes, where a radioactive material is bonded into a molecule, with the knowledge that its going to decay relatively "soon" (minutes <-> days)?  Ex: I know Radon will decay in 3.8 days, so I bond it with Fluorine, send it somewhere, and then let it decay into its final decay state.
Other reasons might be:

Favorable energy characteristics (ex: exothermic to bond, and then exothermic products after it decays and breaks).
Favorable breakdown results (ex: the decays itself causes a reaction, or the resulting molecule after decay is no longer stable and spontaneously breaks, or the decay results in a desired molecule with the final bonded decay products)
Favorable physical characteristics (ex: good way to transport some material somewhere and then let it decay and naturally separate)

Have not found any examples in searching yet, although may just not know the words to search

Search for "hot atom chemistry" and I think you'll find some relevant results.

While Radon-222 has a half life of 3.8 days, that does not mean that any given Radon-222 atom will decay in 3.8 days. It means that only roughly half will decay in that time and you don't know which.

And do you count nuclear medicine, e.g. for PET scans?

@Anger Density Thank you for the reference on "hot atom chemistry."  Found a couple summary papers and that looks like what I'm considering.  Most stuff is really old though, so I guess it must not be a very active field.  Jon Custer, the PET scan idea is similar, its just using/detecting the decay particle, rather the resulting molecule after decay.

That's rather broad. Some molecules were synthesized via decay. Some isotopes are useful because they make other ones, etc.

In terms of my question, I was mostly interested in the case of isotope + bond -> molecule.  Then isotope decays while part of molecule, resulting in interesting or valuable effect.

So something like "Fludeoxyglucose F18 (FDG) is a positron-emitting radiotracer used with positron emission tomography (PET)" - although that is often made in-situ in the molecule by the O18(p,n)F18 nuclear reaction. The glucose goes to regions of high metabolic activity in the body such as cancer tumors, and the positron emission results in electron-positron annihilation and detection of the gammas to localize the tumor.

That's a fair argument on Fludeoxyglucose F18.  It does specifically go to regions of the body that are relevant, only because it has been bonded into the molecular form, and as a free isotope, it would not have nearly the utility.  And the resulting decay does produce an interesting or valuable effect.


text:
Delta Bond Or Pi Bond, Which Is More Strong?
It is pretty evident that sigma bonds are more stronger than pi bonds. (due to the extent of overlapping)
But when there are 4 lobes of two d-orbitals involved in overlapping as that is the case in formation of 'delta bond', is it more stronger than pi bond ?
[it would of great help, if you can list down the sources backing your answers too]

Usually stronger ....  $\ce{N-N}\ \pu{160 kJ/mol}$ // $\ce{N=N}\ \pu{418 kJ/mol}$ // $\ce{N#N}\ \pu{941 kJ/mol}$

can you kindly elaborate it a bit ?

Can you kindly elaborate a bit what I could kindly elaborate a bit? // 1 sigma bond with 2 pi bonds is almost 6 times stronger than just 1 sigma bond.

mate, the initial question speaks about the comparision between pi and delta bond. You did not compare the so anywhere in the solution.

Comments are not supposed to answer the question, otherwise they would be posted as answers.

Red cars and blue cars, which are faster?

In the case of transition elements, these bonds are involved in the transition from an isolated atom to a transition metal solid. A simple approach for understanding this transition by keeping an intuitive description is to use atomic orbitals, precisely a linear combination of atomic orbitals (LCAO) basis set  of the Tight-binding approximation. A useful picture summarizing this approximation is this Hamiltonian : $\hat{H} = - t \sum_{i,j,\sigma} c_{j \sigma}^\dagger c_{i \sigma} + h.c$. which means that an electron can hop from an atomic site $i$ to another atomic site $j$ with a transfer integral  $t$  as the strength of this motion leadind a bandstructure.
If you do not understand this theory, you just need to keep in mind that the parameter $t$ can be related to the strength of the bond, therefore this approximation can give a quantitative strength of bonds relative to the overlap between orbitals. In the basic application of the theory, the Hamiltonian is made of 10 Slater-Koster (SK)  hopping parameters between $s, p$ and $d$ orbitals : $ss\sigma$, $sp\sigma$, $ss\sigma$,$pp\sigma$, $pp\pi$, $pd\sigma$, $pd\pi$, $dd\sigma$, $dd\pi$, $dd\delta$. The strength of the SK parameters can give a quantitative approximation of the strength of the bond. I suppose that you are mainly interested in $dd\sigma$, $dd\pi$, $dd\delta$. These parameters can be obtained by a fit of a higher level of theory or experimental bandstructure if available. For a modeller, the values can be obtained easily, a reliable source is this book where some pages appears online. Another more complicated source is the NRL tight-bnding.

The sign of the SK parameters is not interesting for your question, only the magnitude matters : we will consider SK parameters as positive. For example, for Chromium (Cr) :

$dd\sigma| \rightarrow 0.84 $ eV (81.46 kJ.mol$^{-1}$)
$dd\pi    \rightarrow 0.50 $ eV (48.22 kJ.mol$^{-1}$)
$dd\delta \rightarrow 0.04 $ eV ( 3.51 kJ.mol$^{-1}$)

Obviously, the $\sigma-$bond is the strongest and the $\delta-$bond is the weakest bond formed but the overlap between two d-orbitals, about 10 times weaker than the $\pi-$bond. This result seems surprising maybe counterintuitive, but it is intuitive according to the picture above. The strength of a bond can be enhanced if the electron lies between the centers of the potential of the nucleus. An overlap has a coulomb cost reduces by the exchange interaction and more significantly by the potential of the nucleus if the overlap is not far from the nucleus : this is the case of a $\sigma-$bond. As shown in the figure above, in the case of $d$-orbital, the overlap $dd\pi$ is greater and has a $\sigma$ bond character but is simply farther from the nucleus :  $dd\pi < dd\sigma$, however $dd\pi \gg dd\delta$ because of a planar overlap $dd\delta$.
The overlap $dd\delta$ is very similar to $pp\pi$, the only case where a $\delta-$bond can have more strength than a $\pi-$bond is when the $\pi-$bond between two $p$ orbitals.
Image source

Interesting answer. I only have a technical note: please cite images the same way you would cite quoted text. Some images may require additional details if they are covered by specific licenses. Please keep that in mind.

You might also want to check out the mhchem package for MathJax. Specifically the \pu{...} macro for units. More help is on [meta].

@Martin-マーチン got it! thanks.


text:
Electrolysis of gasoline (or other uniform liquid hydrocarbon)
Electrolysis, when I usually search about it, is almost exclusively interested in splitting $\ce{H2O}$. However, looking at the Wikipedia article on electrolysis, it seems like it works for a bunch of other molecules. Could one perform electrolysis on gasoline, or some other hydrocarbon? In other words, can one dehydrogenate hydrocarbons using electricity?
Maybe gasoline is too non-uniform. It seems like this would generate a lot of hydrogen.

Interesting question, but how are you going to perform the electrolysis of a non-conductive liquid? Theoretically, you can add supporting electrolytes, but this needs to be specified in the question.

Organic electrochemistry is a thing and is a subject of considerable interest these days. A substantial part of my PhD was in synthetic electrochemistry back in the 80s.

What is the motivation for this question? The production of hydrogen?

@Rodrigo de Azevedo, partially the production of hydrogen, since H2O electrolysis is mainly used for that, and it seems like they're always looking for improvements.  Figured something like methane, or other high H ratio hydrocarbons, would produce better (if it worked).  Partially, also just curiosity as to whether electrolysis would work with something like gasoline or a hydrocarbon.  Per andselisk's note above, apparently would need some way to gain enough conductivity.  (maybe just a close enough charge gap?)

@Waylander, have not read about organic electrochemistry previously, although that's pretty neat from a quick summary like this.  Looks like selective catalysis of chemistry based on electrical input to encourage / discourage certain reactions.  Seems like there could be lots of potential with different AC/DC, voltages, currents, waveforms, frequencies, electrode shapes, ect...

@G.Putnam Please append such motivation to the question itself. Comments are ephemeral


text:
What is the physical significance of the frequency factor in the Arrhenius equation?
As the title says, I am wondering what the physical significance of the frequency/pre-exponential factor in the Arrhenius equation. My idea of it is that:
If the units of $E_a$ are energy per molecule, then the Arrhenius equation can be written
$$k_r=Ae^\frac{-E_a}{k_BT}$$
It is my understanding that, if quantum mechanical effects are not significant, the average kinetic energy of a particle in a system will be $k_BT$; this leads me to interpret the exponential factor as the fraction of particles that possess at least $E_a$. I have confirmed this by showing that, in a system in which the available energy levels are separated by a factor $\epsilon$, so that energy levels are $n\epsilon$, with $n=1,2,…$, the fraction of molecules in states with $E$ at least $n_min\epsilon$ is
$$e^\frac{-i_{min}\epsilon}{k_BT}$$
Which is of the same form as
$$e^\frac{-E_a}{k_BT}$$
So: if all of the particles in a system possess sufficient energy to react, then $k_r=A$. If not all of the particles in a system possess sufficient energy to react, then the rate constant is $A$ scaled by the fraction of the particles that do have sufficient energy.
Still, the constant itself seems ambiguous to me, and I would like some guidance - and perhaps to be pointed in the direction of some reading. Please correct any erroneous chemistry in this post; I am not doing a chemistry degree yet, I am in my first year of A levels.

Hello PhysPhil. You could start reading about the (1) transition state theory and (2) collision theory (or Lewis theory). Both give physical insight into the parameter $A$ of the Arrhenius equation. I think it would be a good place to start, at least it was introduced to me in physical chemistry classes.

A as "frequency factor" is quite intuitively the frequency of collisions with the required molecule mutual orientation, regardless of the collision energy. Note that it is a function of T as well.

Very intuitive. So when $A$ is multiplied by the fraction of particles with sufficient energy to react it returns the frequency of successful collisions. I'm not sure how to close this question, since no one has actually 'answered' it, but this answer and @MetalStorm 's suggestions have given me what I need.


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Why is electrochemical potential defined as sum of electrical potential and apparent electrical potential energy?
Engel and Reid's Physical Chemistry [1, p. 308] defines the electrochemical potential in terms of the charge in units of the electron charge $(z),$ Faraday constant $(F),$ and electrical potential $(\phi)$ as follows:

The electrochemical potential is a generalization of the chemical potential to include the effect of an electrical potential on a charged particle.
It is the sum of the normal chemical potential $\mu$ and a term that results from the nonzero value of the electrical potential:
$$\tilde{\mu} = \mu + zF\phi \tag{11.3}$$

Is the electrical potential $\phi$ due to electric field generated from charge separation between electrode and solution, or is it an externally applied electric field?

Isn't $zF\phi$ electrical potential energy? How can we simply add it to chemical potential?

Reference

Engel, T.; Reid, P. Thermodynamics, Statistical Thermodynamics, and Kinetics: Physical Chemistry, 4th ed.; Pearson: New York, 2019. ISBN 978-0-13-480458-3.

As long as the dimensional analysis is correct, the math or equation will never know the labels of those terms (whatever the humans call the first term or the second term).

I did check the dimensions of both the terms and they do come out to be equal. But we are adding electrical potential energy to chemical potential. Isn't it like adding oranges and apples?

The chemical potential of a substance $i$ in a system is generally defined in relation to the Gibbs energy differential:
$$\mathrm{d}G = \left( \frac{\partial G}{\partial n_i} \right)_{T,p}\mathrm{d}n_i =  \mu_i\ \mathrm{d}n_i \tag{1}$$
The related Gibbs energy differential for a charge is:
$$\mathrm{d}G = \left( \frac{\partial G}{\partial q} \right)_{T,p}\mathrm{d}q = \phi \mathrm{d}q = \phi (zF \mathrm{d}n_i) =   (\phi zF) \mathrm{d}n_i = \mu_{i, E}\cdot  \mathrm{d}n_i \tag{2}$$
Then, it can be summed up:
$$\mathrm{d}G = \tilde{\mu_i}\cdot  \mathrm{d}n_i = \left(\mu_i  + \mu_{i, E}\right)\cdot  \mathrm{d}n_i =\left(\mu_i  + \phi zF\right)\cdot  \mathrm{d}n_i \tag{3}$$
By other words, the electrochemical potential of a charged molecular entity is its chemical potential dependent on local electrostatic potential:
$$\tilde{\mu_i}=\tilde{\mu_i}(\phi)\tag{4}$$
This allows us to modify chemical potentials of molecular entities at the electrode surface by externally forced potential. This is used during electrolysis, when reactions that would not be otherwise spontaneous become spontaneous.

Commenting the feedback:
I have incorrectly evaluated your level from your question. :-)  I think I have not written nothing extraordinary.
The contribution of electrostatic potential can be written as dG/dn the same way as is for the chemical potential. It formally contributes to the summary electrochemical potential.

Regarding the 1st question, it is potential of any origin. It may be forced externally, it may be caused local charge disbalance, it may be caused by the electrode surface dual layer, it may be the bulk solution volume potential, there may be Ohmic potential gradient....
But the potential at electrode/solution boundary is the most important, as that is the place where redox reactions occur.

I see... Thank you very much. Can you comment on my first question (1.)? I just want to have a clear understanding about this topic. Thanks again!!

thank you again!


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Ionization Energy equal to single electron energy
The ionization energy represents the amount of energy needed to detach an electron from an atom or molecule. Is it accurate to state that this electron possesses an energy precisely equivalent to the ionization energy?

And after ionization the energy of an electron is zero ?  An electron does not possess an ionization energy (I) but lies on a state with - I in energy. After ionization it keeps its intrinsic energy.

The ionization energy is the energy deficit the bound electron has with respect to the energy of a free electron in rest.
So, if the ionization energy of a $\ce{H}$ atom is $\pu{13.6 eV}$ and if we take as the reference the energy of a free electron as $\ce{0 eV}$, then the energy of an electron in H atom is $\pu{−13.6 eV}$.
It is consisting of (both mean values) $\pu{−27.2 eV}$ of potential energy and $\ce{+13.6 eV}$ of kinetic energy.
To answer the question, providing just the ionization energy will release the electron from the bound with (approximately) zero kinetic energy.

No, you need to also state the reference state to which you compare the energy content.
What you can state is that you used the ionization energy to detach the electron from atom (where it was previously bounded).
Imagine an electron being detached from an atom in a gravitational field. This electron might have much more energy in the gravitational field than the ionization energy you just used.

But if the energy of the electron consists only of kinetic energy + potential energy through coulomb interaction with the other electrons and nuclei, the statement is correct?

Gravitational potential energy would be negligible enough. And besides, a significant wave-particle duality of an electron will complicate it further.


text:
Do peroxy acids such as peroxynitric acid have anhydrides?
Anhydrides reversibly react with water to form a respective acid or base. For example, $\ce{CO2}$ is the anhydride of $\ce{H2CO3}$:
$$\ce{CO2 + H2O <=> H2CO3}.\tag{R1}$$
Acidic anhydride can be obtained by removing water from acid. Nitroperoxynitrate appears to be the anhydride of peroxynitric acid:
$$\ce{N2O7 + H2O <=> 2 HNO4},\tag{R2}$$
but I couldn't confirm it with a reputable source. Also, can we produce the anhydride of any acid or base by removing water?

Perhaps N2O6 exists - NO3 radical is known, so should dimerise at least a bit.

With peroxy acids, you could rather talk about "andihydroperoxides" or whatever. It's a matter of removing whole substituent, not cleaving bonds however you like.

There's for  example trifluoroperacetic acid, which forms trifluoroperacetic acid anhydride. "[H]ydrogen peroxide – urea can be used to give the peracid. This method involves no water, so it gives a completely anhydrous peracid."
That might make a starting point for a Derek Lowe article similar to that he wrote on peroxides. I'd get a bang out of reading it! And all that lovely fluorine, too...

Acyl peroxides, such as acetyl peroxide, can potentially serve as mixed anhydrides of the peroxy acid and the normal acid:
$\ce{R–C(O)–O–O–C(O)–R +H2O -> R–C(O)–O–OH + HO–C(O)–R}$
(e.g. acyl peroxide: $\ce{R}=\ce{CH3}$)
However, they are rarely reacted in this way because the normal acid is typically not desired. To prevent this, hydrogen peroxide solution is used. In that case the reaction is usually rendered as
$\ce{R–C(O)–O–O–C(O)–R +H2O2 -> 2 R–C(O)–O–OH}$
But the same net reaction would be obtained if the anhydride reacts first with water and then the normal acid is oxidized by the hydrogen peroxide. So from an experimental point of view, whether the acyl peroxide reacts with water or not is a matter of bookkeeping.


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EM Induction from Diffusing Ionic Wave Front
I am studying the Belousov-Zhabotinsky Reaction-Diffusion system in ~1mm wide 1-dimensional channels. In my research we are activating the reduced phase with a silver wire and studying the propagation of the oxidized wavefront through space. I am using the standard malonic acid, sodium bromate, sulfuric acid, and ferroin recipe. If you are interested in the mechanism here it is.
The light blue wavefront is an "activator" species which is negatively charged. My question is whether a diffusing ionic wavefront is known to induce an electromagnetic field, and how one would calculate its order given the molarity or charge of the wavefront. Magnetic fields are already known to affect BZ waves, but my question is whether BZ waves can themselves produce magnetic fields.
Please let me know how I could improve this question.

Certainly, it would produce a detectable EM effect.
Likely, an electrostatic effect could be detected with insulated probes immersed in the mix and a high-impedance (100 MΩ or higher) voltmeter. However, such a setup would be slow to respond, perhaps on the order of seconds, due to low capacitively-coupled current, high impedance, and inevitable circuit capacitance. Don't expect to detect fast transients.
However, detecting external electromagnetic effects might be quite difficult because of the very low amplitude and frequency. You might need to use a superconducting quantum interference device (SQUID). "Most fundamental measurements in biomagnetism, even of extremely small signals, have been made using RF SQUIDS." While that would be way cool, it might be a bit out of budget in most labs. Perhaps there's a physiology lab nearby that might give time on their apparatus?