text:
Validly of the equation, du= nCv dt, for an adiabatic process
In an adiabatic process, there is no heat exchange, hence, $Q=0$.
The equation $\mathrm dU=nC_V\,\mathrm dT$ is valid for every type of process given the gas is ideal, where $C_V$ is heat at constant volume required to raise the temperature of a system by 1°, q.
We know that $Q=0$ for an adiabatic process,
So $\mathrm dU$ = Work(adiabatic)
And we define, $W$ (for an adiabatic process)$=nC_V\,\mathrm dT$
But how can $C_V$ have a numerical value for an adiabatic process, there is no heat exchange, no matter how much heat we provide, temperature of system would not increase, shouldn’t $C_V=0$?
But then the equation won’t be valid.
Is the equation , $\mathrm dU=nC_V\,\mathrm dT$ even valid for an adiabatic process?
I don’t get it, where am I wrong? Please help me out, it’s my first time on here, apologies for any error.

Where do you think the gas takes the energy to do the work? There is no gas elastic energy as if you release a compressed rubber. It is taken for kinetic energy of molecules = inner thermal energy, what means T lowering so W = n.C_v.Delta T. (both sides negative)

In thermodynamics, $C_V$ is defined in terms of U, not in terms of q.

Exact meaning of constant volume heat capacity; Why is Cv used in the adiabatic expansion of Carnot Cycle to calculate internal energy?

Consider a closed system for which $\mathrm{d}n=0$.
Then, the definition of $C_v$ is given by $\left(\dfrac{\partial U}{\partial T}\right)_V$.

$C_v$ is an intrinsic property of any material and it doesn't depend
on the process involved.

You could interpret it as following:
If heat was provided at constant volume, it would raise the temperature of the system, and $C_v$ is a measure of this phenomenon.
In an adiabatic process, the work is still equal to $\int_{T_1}^{T_2}C_v\mathrm{d}T$. The temperature is changing in adiabatic processes because of the work and not the heat, which can both change the temperature of the system.


text:
How do I reduce this ketone to two double bonds on the carbo chain?
What reactants should I use to get from 1,5-diphenylpentan-3-one to ((1E,3E)-penta-1,3-diene-1,5-diyl)dibenzene (picture for reference)? Either in 1 step or in multiple steps. I have tried $\ce{NaBH4}$ in acid, then $\ce{POCl3}$ in pyridine, but that only gives one double bond. Is there a way to get both double bonds ?

This can be done (on paper) by alpha-selenation of the starting ketone followed by selenoxide elimination to form the alkene (Wikipedia).
The product from this can be reduced by Luche reduction ($\ce{NaBH4/CeCl3}$) to the allylic alcohol.
Finally this can undergo acid catalysed dehydration (PTSA/toluene Dean-Stark reflux) to give the diene.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
How do I make this terminal alkyne into a methyl alkene on the third carbon?
How do I get from but-3-yn-1-ylbenzene to (3-methylbut-3-en-1-yl)benzene? I have tried $\ce{NaNH2}$ in $\ce{CH3Br}$ and then $\ce{H2}$ with a Lindlar catalyst, but that adds the methyl group at the other end of the double bond. How do I make the methyl group add to the third carbon instead?

Reacting it with LiCuMe2 followed by aqueous workup might work. The more obvious route is a mercury catalysed hydration and Wittig.

@EpichinoM2  is correct. What the question setter is expecting to see is oxymercuration to the methyl ketone followed by Wittig. Not a reaction that any H&S officer will let you do in a lab though.

@waylander: Health & Safety might object to oxymercuration of an alkene (stoichiometric) but certainly not with an alkyne (catalytic). They see "mercury" and panic.

@user55119  My use of Al/Hg amalgam as a reducing agent upset them greatly

@Waylander: I once tried to have H&S dispose of a diffusion pump containing mercury and wrapped in asbestos. Too complex a problem. I put the Hg in a separate container and that was a solvable situation.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
By what mechanism does amylase catalyze hydrolysis?
I've recently been learning bits and pieces of starch metabolism, and am wondering if anyone could explain exactly what is going on when alpha or beta amylase "cleaves" the 1-4 glycosidic bonds?
I understand that amylase, a water molecule and a starch chain are inputs of the reaction, and somehow a 1-4 bond gets cleaved producing 2 smaller glucose polymers; one of the new terminal residues gains a hydrogen and the other gains a hydroxide. In alpha amylase it cleaves any 1-4 bond along the glucose chain, while beta amylase cleaves the second bond in from the terminus yielding a maltose.
I also understand that, as catalysts, they speed up a reaction that already will occur. What I would like to know is what properties of these proteins allow them to do this?

An article on amylase on Proteopedia.org states:

The mechanism involved includes catalyzing substrate hydrolysis by a double replacement mechanism, forming a covalent glycosyl-enzyme intermediate and hydrolyzed through oxocarbenium ion-like transition states. One of the carboxylic acids in the active site acts as the catalytic nucleophile during the formation of the intermediate. A second carboxylic acid operates as the acid/base catalyst, supporting the stabilization of the transition states during the hydrolysis.

So the enzyme uses a carboxylate nucleophile to form a covalent intermediate, which then gets hydrolyzed by water.


text:
Calculating concentration quotient and determining direction of reaction for equilibrium involving gases
I'm currently working on a chemistry problem involving chemical equilibrium, and I'm having some trouble figuring out how to calculate the concentration quotient $Q$ and determine the direction in which the reaction will proceed.
At a certain temperature, when nitrogen dioxide $(\ce{NO2})$ is in equilibrium with nitrogen monoxide $(\ce{NO})$ and oxygen $(\ce{O2}),$ the equilibrium constant is $\pu{0.50 M}.$ A container with a volume of $\pu{0.50 dm3}$ initially contains $\pu{0.020 mol}$ of $\ce{NO2},$ $\pu{0.10 mol}$ of $\ce{NO},$ and $\pu{0.010 mol}$ of $\ce{O2}.$
Now, I need to calculate the concentration quotient $Q$ to determine whether the system is at equilibrium or not. If $Q$ is greater than the equilibrium constant, the reaction will proceed to the left, and if $Q$ is less than the equilibrium constant, the reaction will proceed to the right.
However, I'm not sure how to calculate $Q$ for this particular system. Could someone please walk me through the steps involved? Also, if you could let me know which direction the reaction is likely to proceed based on the value of $Q,$ that would be great.

You have merely just verbosely described the task in the hand. But you have not described what you have tried to do. Like learning what is the an equilibrium constant,  a reaction quotient, what is their relations and what the 0.50 M is hinting to you.


text:
Why did it take so long to figure out what made Roman concrete so durable?
I'm no chemist, but it feels like there should have been more than enough motivation to discover the recipe for strong concrete and it should have been discovered ages ago. Why were the details only discovered in 2023?

It does seem like that knowledge should have been regained faster, in hindsight, which is always 20/20. However, the fact that the concrete was self-healing, filling cracks over hundreds of years,  was only found recently: https://www.scientificamerican.com/article/ancient-roman-concrete-has-self-healing-capabilities/ Now we can take concrete steps to make long-lasting structures.

Talking about motivation doesn't sound on topic - there wasn't all that much of it apparently. Complexity of analysis of purposes and effects of specific ingredients does. I think some editing is needed here.

related https://chemistry.stackexchange.com/questions/127204/how-does-gypsum-concrete-work https://chemistry.stackexchange.com/questions/31776/can-compounds-found-in-concrete-leach-into-soil-against-gravity https://chemistry.stackexchange.com/questions/23505/how-can-concrete-set-underwater

It is always good to add enough context to a question.

Tl;DR: Researchers who were investigating on this topic didn't consider a key component in the concrete mix. They were also puzzled by the mixing technique. These two reason caused the delay in pinpointing the actual reason for the durability of the Roman buildings.
Long Answer:
Researchers indeed have spent decades trying to figure out the secret of the ultradurable construction material in the Ancient Roman buildings that endured especially harsh conditions like weathering, seismic activity and constant volcanic activity. For many years, they have assumed that the key to the ancient concrete’s durability was based on only one ingredient: pozzolanic material such as volcanic ash from the area of Pozzuoli, on the Bay of Naples. This specific kind of ash was shipped all across the Roman empire to be used in construction, and was described as a key ingredient for concrete in accounts by architects and historians at the time. But they forgot to consider one more key material: lime clasts.
An international team of researchers led by the Massachusetts Institute of Technology (MIT) found that not only are the materials slightly different from what we may have thought, but the techniques used to mix them were also different. The smoking guns were small, white chunks of lime called lime-clasts can be found in in the otherwise well-mixed concrete. The presence of these chunks had previously been attributed to poor mixing or materials, but that did not make sense to them. Admir Masic, one of the material scientists quote:

The idea that the presence of these lime clasts was simply attributed to low quality control always bothered me.
If the Romans put so much effort into making an outstanding construction material, following all of the detailed recipes that had been optimized over the course of many centuries, why would they put so little effort into ensuring the production of a well-mixed final product? There has to be more to this story.

This new observation suggested that these tiny lime clasts gave the concrete a self-healing capability. When cracks form in the concrete, they preferentially travel to the lime clasts, which have a higher surface area than other particles in the matrix. When water gets into the crack, it reacts with the lime to form a solution rich in calcium that dries and hardens as calcium carbonate, gluing the crack back together and preventing it from spreading further.
You can find the final observations in their published paper:

Linda M. Seymour et al. Hot mixing: Mechanistic insights into the durability of ancient Roman concrete Sci. Adv.9, (2023). DOI:10.1126/sciadv.add1602

Other references:

https://www.history.com/news/the-secrets-of-ancient-roman-concrete
https://www.sciencealert.com/we-finally-know-how-ancient-roman-concrete-was-so-durable
https://news.mit.edu/2023/roman-concrete-durability-lime-casts-0106


text:
Why is the phrase "abstraction" used in radical reactions?
Often, I see the phrase "abstraction" used to describe certain radical reactions. Why is this specific term "abstraction" used in the case of radicals? This may be more of a chemical history / linguistics question.
To give more context, the term substitution is used in some reactions, and it describes how the molecules change, an atom substituted for another. Is abstraction being used in this same sense to describe the process of the reaction, or just used to differentiate these reactions since they involve radicals?

It is rather a question  for dictionary search :  4/ the act of taking away or separating; withdrawal: The sensation of cold is due to the abstraction of heat from our bodies.

In $\ce{CH4 + Cl^. ->CH3^. + HCl}$, there is the abstraction of  $\ce{H}$ atom from $\ce{CH4}$.

There is no fancy history behind abstraction, it is simply the Latin derivative. One can look up the etymology in any decent dictionary and see that

classical Latin abstract-, past participial stem of abstrahere to drag
away, to appropriate, take away, to set free, to separate, to deduct,
subtract, to exclude, to turn aside, divert, in post-classical Latin
also to summarize (from the subscribed version of the Oxford English Dictionary).

When we say the radical abstracts a proton from a certain compound, it simply means that the radical took away (dragged away) a proton from that neutral or even charged compound. Although, I do not like this anthropomorphism at atomic or molecular scale. This "abstraction" can be classically seen in gas phase reactions of alkanes with chlorine in the presence of light.


text:
Pentose phosphate pathway mechanism: glucose-6-phosphate converting to 6-phosphoglucono-delta-lactone
I am trying to write a mechanism of the first step in the pentose phosphate pathway where G6P is converted to PPP with cofactor NADP+ and His residue.
I wrote this first mechanism where I did not open the ring.

I did not receive a full score for that (7.5 out of 8). So I did my second attempt where I tried to open the ring. However, I am not sure how to proceed from there to reform the ring, or if the ring opening step is correct or not. As the comment mentioned this mechanism is not balanced. So I made a new edit.

Edit:
Here is the new attempt:

Your ring-opening reaction is not balanced.

Lactones are cyclic esters. When you open the ring, you get a carboxylic acid (this is a hydrolysis, in contrast to opening a sugar ring). With 7.5 of 8 points for your first attempt, you might be almost correct and lost 0.5 points for some detail you got wrong. Maybe there is another active site residue that you are not showing?

@Karsten My professor left a note "OK" on the page. So I am thinking the mechanism is not rigorous enough. And the only thing I can think of is I should open the ring first. And only His residue was provided in the problem. When you say "hydrolysis", so there should be a water molecule leaving when I open the ring?

@ananta Ah, I see it! I have written a new mechanism. Could you check it please?

The reaction seems balanced now. Also, I recommend removing the previous (incorrect) mechanism entirely so new readers don't get confused. Also note the names "D-glucopyranose 6-phosphate(2-)" and "6-O-phosphonato-D-glucono-1,5-lactone(2-)" from M-CSA

Pentose phosphate pathway (PPP) AKA pentose phosphate shunt is the name of the reaction mechanism (pathway) and not the product.

Maybe they just wanted you to deprotonate the histidine as the last step to regenerate the enzyme to its initial state and get the stoichiometry right (NADPH + $\ce{H+}$).

Reaction Mechanism
BioLibreTexts shows the following mechanism for the reaction
$$\ce{D-glucopyranose 6-phosphate(2-) \\-> 6-O-phosphonato-D-glucono-1,5-lactone(2-)}
$$

Note that this mechanism was sourced from M-CSA$^1$.
Points to note

I doubt if the ring (completely) opens during the reaction.
$\ce{NADPH}$ is produced from $\ce{NADP+}$ during the reaction.
$\ce{His240-methylimidazole}$ moiety abstracts a proton from the substrate, eliminating a hydride that is added to $\ce{NADP+}$) and returns to original (deprotonated form) upon abstraction of the proton by a water molecule.

Regarding point 2, Wikipedia states:

NADPH is produced from NADP+. The major source of NADPH in animals and
other non-photosynthetic organisms is the pentose phosphate pathway,
by glucose-6-phosphate dehydrogenase (G6PDH) in the first step.

Regarding point 3, the $\ce{His240-methylimidazole}$ moiety acts catalytically, undergoing serial protonation and deprotonation.

References

Mechanism and Catalytic Site Atlas (M-CSA): a database of enzyme reaction mechanisms and active sites, Antonio J. M. Ribeiro, Gemma L. Holliday, Nicholas Furnham, Jonathan D. Tyzack, Katherine Ferris1 and Janet M. Thornton Nucleic Acids Research, 2018, Vol. 46, Database issue, 10.1093/nar/gkx1012

His 240 is one of the amino acid residues of the protein. 240 is the 240th building block in the protein. Here, they are showing the side chain only, attached to the label “his 240”. This is common (if not entirely logical) nomenclature in biochemistry.


text:
Is there a quantitative definition of inductive effect?
Usually, Inductive effects are compared using the electronegativities and distances. However, this only works in compounds with similar structures. For example, we can predict the following orders of acidity:
$$
\ce{2-fluoropropanoic acid} \underset{K_\mathrm{a}}{\gt} \ce{3-fluoropropanoic acid}\\
\ce{3-fluoropropanoic acid} \underset{K_\mathrm{a}}{\gt} \ce{3-chloropropanoic acid}\\
$$
However, besides these (very valid) qualitative reasonings, I want to know if there is a quantitative way to measure inductive effect. I am aware that the concept of group electronegativities also exists, and that is not what I am referring to.

https://chemistry.stackexchange.com/questions/63273; https://chemistry.stackexchange.com/questions/12309

Maybe the Hammett equation? You can distinguish resonance effects from other effects by comparing substitutions on aromatic vs saturated systems.

@Karsten the Hammett equation is quite general and does quantitatively describe combined inductive, mesomeric, and other effects upon substitution of benzene/benzoic-acid derivatives but does not quantitatively describe inductive effect alone.

If you use a parent molecule that is not aromatic, you can't have a mesomeric effect. So studying the distinct effect of substituents with different parent molecules does differentiate between mesomeric and inductive effects, e.g. this paper

Not an organic chemist myself, can this effects not be accounted for using quantum chemical methods and NBO analysis of the wavefunction?

Since Inductive Effect is a permanent effect involving generation of dipoles within molecules, there could be a way to measure it quantitatively. According to Wikipedia, relative inductive effects can be measured, it is unclear on how exactly could it be done.
However, it does mention that inductive effects can be calculated through something called the Hammett equation that makes relationship between reaction rates and equilibrium constants with respect to substituent (the one due to which inductive effect generates).

The Hammett equation is quite general and does provide a quantitative description of (combined inductive, mesomeric, and other) substituent effects, but it only works for benzene/benzoic-acid derivatives.


text:
Effect of phosphoric acid on chrome plating
Wondering whether phosphoric acid a suitable agent to clean chrome plated bathroom figments - a metal base shower head

Vinegar could be enough, as it is mostly lime scale.

Consider the solubility of the resulting salt, if you use phosphoric acid on calcium lime, you get a variation on calcium phosphate (orthophosphates, di- and monohydrogen phosphates, di- and polyphosphates. Check out the solubility in water (though the solubility in $\ce{H3PO4}$ might be higher).
Consider Poutnik's suggestion to use vinegar, and the solubility of $\ce{Ca(CH3COO)2}$.
Phosphoric acid is great as a rust remover (or converter, if the phosphate film adheres), or removing other metal corrosion, though.

Thanks for the replies.
I have used vinegar in the past.
Having a proprietary phosphoric acid based rust remover I wondered whether this would be suitable for cleaning but obviously not wishing to damage the - what I assume to be - chrome on brass. And on a weaker phosphoric acid solution - I did buy some cheap carbonated sugar drink with the intention of possibly using this as a cleaning bath.
I have no usable chemistry knowledge or means of conducting tests.
Since the phosphoric acid may come with risk I may stick with vinegar.


text:
Order of Second Ionisation Potential of C, N, O and F
Here's what I believe:
The electronic configurations of the elements after losing 1 electron are:
N+ -> 1s2 2s2 2p2
O+ -> 1s2 2s2 2p3
F+ -> 1s2 2s2 2p4
So, on losing 1 e- more, the value of the second ionisation potential would be in the order:
F<N<O (as F would be obtaining a stable half filled p-orbital which would contribute to greater stability)
But, most books I've referred suggest that the order of ionisation potentials is N<F<O.
Could you explain how N is more stable despite F having half-filled p-orbitals?

A fluorine nucleus definitely doesnt have the same impact with a nitrogen nucleus.

Wikipedia's ionization energy table somewhat splits the difference. Ignoring a difference of less than 1% between O and F as likely within experimental uncertainty, that table gives
N(2856) < O(3388) = F(3374) < Ne(3952).
where all numbers are in kJ/mol. The inclusion of neon in this sequence indicates that the second ionization of fluorine is indeed lowered by the favorable half-filled shell configuration. But this effect is only enough to balance one additional nuclear charge in the same period comparing fluorine with oxygen. With fluorine versus nitrogen, the larger difference in nuclear charge (two instead of one) wins out.


text:
Is it critical to match vacuum pump oil to specific pump?
We have rotary vane vacuum pumps of many makes, models, and ages, but superficially they seem remarkably similar. Each has a recommended pump oil (that is, if we can find the documentation); often each different, branded oils. For general purpose use, is it really necessary to keep stocks of the correct oil for each pump? Is there a single pump oil that will work reasonably well for all?

If one looks at, e.g., Krytox oils, one will find a wide range of viscosities. You need to match that, at the least. Compatibility with any seals is also needed. A chat with an application engineer, or a talk with a local supplier (should you have one) sounds like a good idea.

In principle yes but if you are just wanting milli-torr pressures any general rotary pump oil should be ok, just purge the pump  first get get rid of whatever has dissolved in the oil by previous users. If its to back a diffusion pump then you really will want the new clean oil.

Yes, it is critical to match vacuum pump oil to the specific pump. Different vacuum pumps have varying operational requirements and design considerations. Pump manufacturers provide recommendations for compatible oils to ensure optimal performance, longevity, and sealing properties. Using the recommended oil type and viscosity helps maintain efficient lubrication, proper sealing, and prevents potential damage to the pump's internal components. Always refer to the manufacturer's guidelines to ensure the correct oil is used for your specific vacuum pump model.


text:
Is dimethyl amine a sigma-donor or acceptor or pi-donor or acceptor?
I would argue that the $\ce{-NMe2}$ is a $\pi$-donor because the nitrogen atom has an unbound pair of electrons. At the same time it is a sigma-donor as the two methyl groups are good sigma donors. Is that correct?
As an example: N,N-dimethylaniline

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

You are right about -$\ce{NMe2}$, or -$\ce{NR2}$ in general, being a $\mathrm{\pi}$-donor ligands. References$^{1,2}$ do support this.
-$\ce{NMe2}$ is also a $\mathrm{\sigma}$-donor ligand, but not because of the methyl groups. Quoting ChemLibreTexts$^1$:

(For) $\mathrm{\pi}$-donors, in addition to donating electron density to a
metal via a $\mathrm{\sigma}$-bond, electrons may be provided to the
metal via a $\mathrm{\pi}$-symmetry interaction. $\mathrm{\pi}$-donor
ligands include -$\ce{X}$ (halide), -$\ce{NR2}$ amide, -$\ce{S^{2-}}$
sulfide (S2–), -$\ce{O^{2-}}$ oxide, and -$\ce{OR}$.

References

ChemLibreTexts
Lecture Notes by Matt Law (and Alan Heyduk), Department of Chemistry, University of California, Irvine


text:
How would you turn a thermite reaction into an electrochemical cell?
A previous question asked about oxidation and reduction agents in the following reaction (not sure about the appropriate physical state of the oxides, solid or liquid, but shown here as liquid):
$$\ce{Fe2O3(l) + 2 Al(l) -> 2 Fe(l) + Al2O3(l)}$$
The false claim in a video was that because oxygen gets transferred, oxygen is the oxidation agent. We usually determine what gets reduced and what gets oxidized by looking at oxidation states. However, these are a fairly arbitrary construct.
Sometimes, it helps to show the reaction split into two components. Here is a naive way:
$$\ce{Fe2O3(l) -> 2 Fe(l) + 3/2 O2}$$
$$\ce{2 Al(l) + 3/2 O2 -> Al2O3(l)}$$
So aluminum gets oxidized, and iron oxide gets reduced. However, there is no elemental oxygen in the real reaction, so this is not a clean argument. In this scheme, you can see that we start with oxygen in an oxide, go to elemental oxygen, and then back to an oxide, so the oxygen is doing (net) nothing, which is confusing as well but helps to debunk the false claim in the video.
More useful would be to split this redox reaction into half reactions. Because you can "see" the electrons in the half reactions, it makes it clear that this is indeed a redox proces. To make it realistic, these would exclude water and elemental oxygen, so this would happen at high temperatures with molten salts and (depending on the temperature) either solid or liquid metal electrodes:
$$\ce{Fe2O3(l) + 6e- -> 2 Fe(s) + 3 O^2-(ms)}$$
$$\ce{2 Al(s) + 3 O^2-(ms) -> Al2O3(l) + 6 e-}$$
The phase designation (ms) is supposed to stand for a molten salt.
My question is how you would realize these two half reactions in an electrochemical cell. How would you build the electrodes (would they be solid or liquid), what would you use as salt bridge or semi-permeable membrane to connect the two ion-containing solutions so that ions can be transferred, and what would you use as solvent (some molten salt)?
I found analogous iron and aluminum chloride molten salt batteries e.g. in this paper (https://doi.org/10.1007/BF01030182), but nothing yet on an electrochemical cell based on the oxides.

As the thermite reaction is not elementary, there can be elementary oxygen. There could be similar E. diagrams for Al as Ellingham diagrams for carbon

@Poutnik I did not know about Ellingham diagrams. In this paper they have elementary reactions releasing dioxygen but find them unlikely because of a positive Gibbs energy of reaction even at high temperatures.

The chemical potential of oxygen changes between the oxides, so it is driving the reaction (with aluminum as accomplice). Those first reactions could be used as cells. The oxygen dianion looks brutally unstable. [edit] "However, there is no elemental oxygen in the real reaction, so this is not a clean argument." That's the point with half reactions, Hess's law and much of thermodynamics: additivity of individual steps.

@Karsten Hello Karsten. Your question is about which chemical materials to use as a cathode, anode, electrolyte, etc.; for this system?

@MetalStorm yes, and the temperature that makes it feasible


text:
Are there compounds distance functions to define a compounds space metrics
Is there exist something like a distance function for compounds, computing a number for any two given molecules? Such a function would measure dissimilarity at the atomic structure and bonds levels, allowing the definition of a metrics on the compounds space.

Why, of course there is. You might want to read about QSAR.

Also called molecular descriptors.

See "Why is Tanimoto index an appropriate choice for fingerprint-based similarity calculations?"

@IvanNeretin the main reason would be to determine potential products in chemical reactions.

To determine products of chemical reactions, you need to code chemistry (all of it).

There are numerous (too many to count) distance fucntions to define a compound's space metrics.
One of the most commonly used, and one of the first such, index is called the Wiener Index, represented by $W$$^1$. It is a topological index; according to Wikipedia:

In the fields of chemical graph theory, molecular topology, and
mathematical chemistry, a topological index, also known as a
connectivity index, is a type of a molecular descriptor that is
calculated based on the molecular graph of a chemical compound.
Topological indices are numerical parameters of a graph which
characterize its topology and are usually graph invariant. Topological
indices are used for example in the development of quantitative
structure-activity relationships (QSARs) in which the biological
activity or other properties of molecules are correlated with their
chemical structure.

Weiner index $W$ is$^1$

the sum over all bonds of the product of the number of vertices on
each side of the bond (Wiener, 1947a; Wiener, 1947b; Wiener, 1947c)
and called at the beginning path number. Usually, the Wiener index is
defined and calculated as the sum of all topological distances in the
$\ce{H}$-depleted molecular graph (Hosoya, 1971).

$\ce{H}$-depleted molecular graph means hydrogen atoms are not considered. It is obtained from the $\ce{H}$-depleted distance matrix $D$:
$$
W = \mathcal{W}(D)= \dfrac{1}{2}\sum_{i=1}^{A}\sum_{j=1}^{A}d_{ij}\text{,}
$$
where $\mathcal{W}$ is the Wiener operator and $d_{ij}$ is the topological distance. Essentially, it is the sum of all distances between all pairs of atoms, except hydrogen atoms. It has been correlated with properties of compounds such as heat of vaporization and boiling points, especially, for hydrocarbons.
Reference
Handbook of Molecular Descriptors by Prof. Dr. Roberto Todeschini, Dr. Viviana Consonni, WILEY-VCH

Additionally, is there any sort of edit distance between compounds (i.e.: what has to be changed in one compound to obtain another)?


text:
Can alkyl groups show -I effect?
Inductive effect is the partial shifting of the sigma-bonding electrons towards more EN (take as electronegativity or electronegative in this text) atom so why we will say that alkyl groups will always show +I effect.
For the case of boron (EN 2.05) and carbon (EN 2.55), will the alkyl group show +I or -I effect?

You are right, if we attach a low electronegativity element (such as boron) to carbon, the alkyl group does show -I effect, withdrawing electrons from the less electronegative element.
However, we usually say that less electronegative element ($\ce{B}$, in this case) is showing +I effect because the alkyl moiety is treated as the primary and the other groups are treated as secondary or substituents. Thus, these effects are calculated with respect to the alkyl moiety.
From Literature
Regarding inductive effect Poon and Mayer (2002)$^\text{1}$ said:

The degree of p-donation decreases across the first row from NH2 to F while the inductive effect due to the difference in electronegativity of the atoms increases in the same order.

Although no direct statement has been made regarding inductive effect of methyl group, the same arguement maybe extended to say that methyl group shows a (weak) -I effect.
Regarding the statement about hyperconjugation, the authors said:

The pseudo p-orbital on the methyl group in $\ce{H2BCH3}$ is involved in a weaker hyperconjugation interaction with the vacant p-orbital on B.

Ab-initio
Since, inductive effect has not been directly mentioned and charge analysis is absent from the referenced study, I performed some calculations on my own at B3LYP level theory on the molecule methylborane $\ce{H2BCH3}$. The calculations provide some insights into the properties of the molecule.

Remark
Corresponding Molecular Orbital

Sigma bond between boron and carbon is polarized towards carbon showing -I effect of methyl group.

Methyl group is involved in hyperconjugation with boron showing +H effect of methyl group.

Element
Mulliken Charge

$\ce{B}$
-0.35

$\ce{C}$
0.00

$\ce{H}$s
positive

Thus, +H effect of methyl group is compensating for its -I effect fulfilling the electron-deficiency on boron.
Reference

Poon, C. and Mayer, P. M (2002). Electron-spin conservation and methyl-substitution effects on bonds in closed- and open-shell systems — A G3 ab initio study of small boron-containing molecules and radicals. Canadian Journal of Chemistry, 80. 10.1139/v01-185

Then BH3 should be more stable than BR3 where R is an alkyl group. But in Hydroboration of an alkene, B is breaking bond from H and forms with R, that should mean that B which is electron deficient is getting relatively more electron density. And if we consider that alkyl groups show +I effect then this fits the logic but we have discussed about their EN values suggesting -I effect of Alkyl groups here.

We also need to consider hyperconjugation for $\ce{BR3}$, for example, $\ce{B(CH3)3}$ has 9 hyperconjugative structures involving the empty $\mathrm{p}$ orbital of $\ce{B}$.

If you could attach some sources then I can mark your answer as accepted.

@D13G do the latest edits address your comment?

I will try to understand it within few days and then I will give you the feedback. And Thanks for your update.


text:
Cannot identify C[Mg+]
I am not able to identify the molecule C[Mg+]. I came across it a number of times in the USPTO-50k dataset, in reactions such as the one shown below:
CON(C)C(=O)c1cn(-c2ccc(Cl)cc2)c(-c2ccc(Cl)cc2Cl)n1.C[Mg+]>>CC(=O)c1cn(-c2ccc(Cl)cc2)c(-c2ccc(Cl)cc2Cl)n1

I searched in PubChem and MolView and couldn't identify the compound. The closest thing I found was magnesium carbide, with SMILES [CH3-].[CH3-].[Mg+2]. Is it the same compound?
Any ideas what C[Mg+] is or where I can look next? Also RDKit accidentally generated a similar molecule once, C[MgH]. Does this one exist too?

Mg+ does not exist. c[Mg+] does not exist ! And I do not understand your language, starting with  CON(C)C(=0)c1cn... etc. What do you want to do with this compound ?

@Maurice This is SMILES.

One explanation for C[Mg+] is the one of of methylmagnesium as in the Grignard reagent of methylmagnesium chloride (C[Mg+].[Cl-]), or the softer methymagnesium bromide (C[Mg+].[Br-]).  Obviously, metal organic compounds with bonds which sometimes differ from «pure covalent» as e.g. in pyridine (c1ccncc1), or «clearly ionic» (as in the example of sodium acetate, [O-]C(=O)C.[Na+]) can represent a limitation of this reduced representation.
Maurice' comment shows that SMILES are better known for the representation of chemical compounds, than for chemical reactions (indicated by >>).  So I copy-pasted the string into ChemDraw's sample page as illustration:

Edit: For comparison, the reading by Marvin (after an additional click to disable the Thiele rings)

Pardon my basic questions, I am new to chemistry. When you say "methylmagnesium as in the Grignard reagent of methylmagnesium chloride" you mean the compound in this case is methylmagnesium chloride? Can methylmagnesium exist on its own?
In the smiles representation you gave, chloride appeared to be disconnected from methylmagnesium. Is that what you mean by "metal organic compounds [...] can represent a limitation of this reduced representation." ? As in SMILES are not good for representing this type of compound?

In methylmagesium written as C[Mg+], the plus sign assigns there is a positive charge. In inorganic chemistry, $\ce{Mg^{2+}}$ is the typical form; here, (formally) there already is one compensation. But you still need an other negative counterion (an anion) for charge compensation because methylmagnesium alone will not exist, and can not be stored in a bottle for the shelf. A chloride, or bromide are chemically simple, stable (commercially available) examples to resolve this, and are reagents known for this type of reaction.

The bonding in organometallic compounds can be all within a spectrum of non-polar covalent bonds, polar covalent bonds (both partners participate, though one attracts the electrons stronger, than the other); or «ionic bonds» (similar to «now all binding electrons  are handed over to one partner only», ions stick together because of charge attraction between cation and anion). To some degree, it depends on the human/the algorithm if the bond is perceived polar covalent, or ionic. E.g., ChemDraw converts "methylmagnesium chloride" still the covalent way, and hence assigns Cl[Mg]C as SMILES.

Hence, consistency in the application of rules becomes important, even more so if multiple algorithms/programs harvest the data (see Daniel Lowe's thesis at Cambridge UK Extraction of chemical structures and reactions from the literature (https://doi.org/10.17863/CAM.16293) as an entry), or multiple databases are joined. This is (in part) the less appealing, less often advertised part of data curation before extraction of knowledge from the data really can start. SMILES can be good, InChI (standard, and the non-standard options) can be good as long as creators and users agree on rules.

See e.g. Big Data from Pharmaceutical Patents: A Computational Analysis of Medicinal Chemists’ Bread and Butter in 2016JMedChem4385, or Annotated Chemical Patent Corpus: A Gold Standard for Text Mining in 2014PlosOne (the later open access) as applications potentially relevant for your work ahead, and http://chemdataextractor.org/ .


text:
What is the basis for the relationship of Gibbs energy and extent of reaction shown in this graph?
How this graph of Gibbs free energy between reaction coordinate is made because because
∆G = ∆G° + RT ln Q
and since ∆G° and T is constant hence the graph should be like that of ln(x)?

Source: https://wisc.pb.unizin.org/chem109fall2020ver03/chapter/day-31/

There's no reaction coordinate there, just composition of mixture.

This ChemLibreText page may be helpful.

If you search "∆G = ∆G° + RT ln Q", you will find a lot of sources that do not distinguish Gibbs energy change and Gibbs energy of reaction properly, e.g. here

I don't think this is homework-like because the textbooks and other readily available sources are so unclear on this topic. Thus, it is not easy to figure this out through independent research. Granted, the y-axis says energy and not change in energy, so if the OP had noticed, that would have been a starting point for some more prior thoughts to add to the question.

The Gibbs energy for a system at a fixed pressure and temperature, is only a function of the concentration of the species. We write it in a differential form
$$ dG = \sum_j \mu_j dn_j \tag{1} $$
Our task is to plot this $G$ versus a variable that can keep track of the reaction, this is the extent of reaction $\xi$. By stoichiometry, we know in general that $n_\pu{j} = n_\pu{j}^0 + \nu_j \xi$, or $dn_\pu{j} = \nu_\pu{j} d\xi$, so we get
\begin{align}
  dG &= \sum_\pu{j} \mu_\pu{j} \nu_\pu{j} d\xi \\
  dG &= \sum_\pu{j} \nu_\pu{j} [\mu_\pu{j}^0 + RT\ln(\hat{f}_\pu{j})] d\xi \\
  dG &= \left[\sum_\pu{j} \nu_\pu{j} \mu_\pu{j}^0 + \sum_\pu{j} \nu_\pu{j}
  RT\ln(\hat{f}_\pu{j})\right] d\xi \tag{2} \\
\end{align}
where $\hat{f}_j$ is the fugacity coefficient of species $j$. We identify the first sum as the standard Gibbs energy of reaction and continue with Eq. (2)
\begin{align}
  dG &= \left[\Delta_\pu{r} G^\circ + RT \ln\prod_\pu{j}
  (\hat{f}_j)^{\nu_j}\right] d\xi \tag{3}
\end{align}
Without making some assumptions on the behavior of the species, this is as far as we can get. But the idea is simple, we need to integrate Eq. (3). However, this may be hard! The fugacity coefficient is a complex function of the composition, and this is why we never integrate this function to get $G$ and find its minimum. Rather, we go back, and just solve $\sum_j \nu_\pu{j} \mu_\pu{j} = 0$.
We will make an exception and do it the hard way, so it is a rather tedious trip. I will do an example so you get the idea.
Example: Let's consider the formation of hydrogen iodide in the gas phase
$$ \ce{\frac{1}{2}H2(g) + \frac{1}{2}I2(g) -> HI(g)} \tag{4} $$
The system is formed initially by a number of moles of $n_\ce{H2}^0$ and $n_\ce{I2}^0$, which I will consider equal, so $ n_\ce{H2}^0 = n_\ce{I2}^0 = n^0$. By stoichiometry we link the moles of the species to the extent of the reaction
\begin{align}
  n_\ce{H2} &= n^0 - \frac{1}{2}\xi \\
  n_\ce{I2} &= n^0 - \frac{1}{2}\xi \tag{5-7} \\
  n_\ce{HI} &= \xi
\end{align}
the total number of moles is constant in this case, $n = \sum_\pu{j} n_\pu{j} = 2n^0$. So, combining this with Eqs. (5-7) we can calculate the molar fractions of the species
\begin{align}
  y_\ce{H2} &= \frac{n^0 - \xi/2}{2n^0} \\
  y_\ce{I2} &= \frac{n^0 - \xi/2}{2n^0} \tag{8-10} \\
  y_\ce{HI} &= \frac{\xi}{2n^0}
\end{align}
If we assume that the mixture behaves as an ideal gas, then the fugacity coefficient is $\hat{f}_\pu{j} = y_jP/P^\circ$

$P$ is the pressure of the system, that we maintained constant when writing Eq. (1).
$P^\circ$ is the standard pressure, more precisely, is the standard pressure of the species $j$ when the standard Gibbs energy of formation was measured.

We combine Eqs. (8-10) with Eq. (3)
\begin{align}
  dG &= \left[\Delta_\pu{r} G^\circ + RT \ln \prod_\pu{j} \left(\frac{y_jP}{P^\circ}\right)^{\nu_j}\right] d\xi \\
  dG &= \left\{ \Delta_\pu{r} G^\circ + RT \ln\left[
  \dfrac{\dfrac{y_\ce{HI}P}{P^\circ}}{
   \left(\dfrac{y_\ce{H2}P}{P^\circ}\right)^{1/2}
   \left(\dfrac{y_\ce{I2}P}{P^\circ}\right)^{1/2}} \right]\right\} d\xi \\
  dG &= \left\{ \Delta_\pu{r} G^\circ + RT \ln\left(
  \dfrac{y_\ce{HI}}{y_\ce{H2}^{1/2}y_\ce{I2}^{1/2}} \right)\right\} d\xi \\
  dG &= \left\{ \Delta_\pu{r} G^\circ + RT \ln\left[
  \dfrac{\dfrac{\xi}{2n^0}}{\left(\dfrac{n^0 - \xi/2}{2n^0}\right)^{1/2}
  \left(\dfrac{n^0 - \xi/2}{2n^0}\right)^{1/2}} \right]\right\} d\xi \\
  dG &= \left\{ \Delta_\pu{r} G^\circ + RT \ln\left[
  \dfrac{\dfrac{\xi}{2n^0}}{\dfrac{n^0 - \xi/2}{2n^0}} \right]\right\} d\xi \\
  dG &= \left\{ \Delta_\pu{r} G^\circ + RT \ln\left(
  \frac{\xi}{n^0 - \xi/2}\right)\right\} d\xi \\
  \int_{G(\xi = 0)}^{G(\xi)} \; dG &= \int_{0}^{\xi}
  \Delta_\pu{r} G^\circ \; d\xi + RT \int_{0}^{\xi} \ln\left(
  \frac{\xi}{n^0 - \xi/2}\right) \; d\xi \\
  G(\xi) - G(\xi = 0) &= \Delta_\pu{r} G^\circ \xi +
   RT \bigg[2n^0 \ln(2n^0 - \xi) + \xi \ln\left(\frac{\xi}{n^0 - \xi/2}\right)\bigg]_{0}^{\xi}
\end{align}
$$ \boxed{
   G(\xi) = G(\xi = 0) + \Delta_\pu{r} G^\circ \xi +
   RT \bigg[2n^0 \ln\left(\frac{2n^0 - \xi}{2n^0}\right) + \xi
   \ln\left(\frac{\xi}{n^0 - \xi/2}\right)\bigg]} \tag{11}
$$
We did all the steps, the only thing that I did is not extend in finding the primitive of the second integral. Secondly, I also skipped that the second logarithm is not continuous on $\xi=0$, but can be seen easily that applying L'Hôpital's rule the limit of that function exists (similar like proving that $\lim_{x\rightarrow0} x \ln(x) = 0 $). Note that Eq. (11) predicts:

If $\xi=0$ then $G = G(\xi = 0)$.
If $\xi=2n^0$, the maximum extent of reaction by Eqs. (8-10), the final term diverges to $\infty$, because both logarithm terms tend to $\infty$.

The following data will be needed to obtain the standard Gibbs energy of reaction:
\begin{array}{c|c}
  \text{Substance} & \Delta_\pu{f} G^\circ \pu{(kJ/mol)} \\ \hline
  \ce{H2} & 0 \\
  \ce{I2} & 0 \\
  \ce{HI} & 1690 \\
\end{array}
\begin{equation}
  \Delta_\pu{r} G^\circ = \sum_\pu{j} \nu_\pu{j} \Delta_\pu{f} G^\circ \rightarrow
  \Delta_\pu{r} G^\circ = 1690 \; \pu{\frac{J}{mol}}
\end{equation}
And finally, we have the desired plot of Eq. (11), which I just put to exemplify $n^0 = 1 \; \pu{mol}$:

Last comment The idea still remains in the integration of Eq. (3), but a primitive may not exist. In this case, just numerically integrate the function many times. So, at a point $\xi_1$ you will get $G(\xi_1) - G(\xi = 0 )$, and so on; and repeat until the maximum extent of reaction possible (minus some tiny value, because $G \rightarrow \infty$ when the reactants are depleted).
References
The thermodynamic data was obtained here:

Atkins, P. and De Paula, J., "Physical Chemistry", 9th ed, W.H. Freeman and Company New York (2016).

Even though the formula is different, the shape of the curve you show and that I show in my answer are the same. Knowing the expression for G(Xi) and deriving the expression of the Gibbs energy of reaction is the easier direction, mainly because differentiation is a bit easier than integration.

@Karsten Hello Karsten. If the formulas are different, then one must be incorrect, since the equilibrium can only be one. I tried to make an effort in showing where $G$ comes from and what $G$ is the one that has to be plotted. In your case, it seems to me that you made an analogy with the Gibbs energy of mixing of ideal gases, and the equations are similar. But the process of mixing is not the one we are interested, it is the chemical reaction.  Do you agree with me?  E.g., if the reaction was $\ce{A + 2B -> 5C}$, how would you proceed?

@Karsten When I read the replies, to me it seems that ananta knew where to head with the math, but could not resolve it. You had the intuition of the formula, but it just appeared from nowhere without proof, and made an analogy with the mixing. I just imagined the scenario of a cooperative response, where we can merge our thoughts, and give the ultimate answer to resolve this obscure diagram and please the OP. What do you think? haha

I updated my answer (top of answer) to explain why mixing together different reaction mixture is not only an analogy to a chemical reaction, but a rigorous derivation of the Gibbs energy of reaction (because the Gibbs energy is a state function).

I opened a chat room to discuss.

Your derivation is v. nice, but it is made simpler if one recognises that $\Delta_r G$ is a derivative  where $\displaystyle  \Delta_r G= \left( \frac{ \partial G}{ \partial \xi} \right)_{T,P}$

@porphyrin Hello porphyrin. And where that whould have helped me?

Well it would not have helped you but $\Delta_r G=\Delta G^\text{o}+RT\ln(Q)$ is v. well known but students often do not realise that $\Delta_r G$ is a differential, an unfortunate notational thing in thermodynamics.

What an interesting question! I was perplexed when I first read this. And then I remembered what my professor told me in college, that is, "when in doubt, read." So, I read and found an acceptable answer to your problem.
What is $G$?
We will begin with the definition of Gibbs energy $G$ for a closed homogeneous system:
$$
\mathrm{d}G = -S\mathrm{d}T + V\mathrm{d}P
$$
At constant temperature:
$$
\mathrm{d}G_{n} = V\mathrm{d}P
$$
Where the subscript indicates the value/s which is/are kept constant. Since our system is gaseous:
$$
\begin{align}
&(\mathrm{d}G)_{T,n} = \dfrac{nRT}{P}\mathrm{d}P\\
\implies &\int_{G_1}^{G_2} = \int_{P_1}^{P_2} \dfrac{nRT}{P}\mathrm{d}P \\
\implies & G_2 - G_1 = nRT\ln{\dfrac{P_2}{P_1}}\\
\implies & G_2 = G_1 + nRT\ln{\dfrac{P_2}{P_1}}
\end{align}
$$
Say $P_1 = 1 \mathrm{bar}$, then $G_1=G^{\ominus}$ and, for simplicity, $G_2 = G$ and $P_2 = P$, then we obtain the following.
$$
G = G^{\ominus} + nRT\ln{P} \tag{1}
$$
What is $\Delta_r G$?
By definition, Gibbs free energy of reaction is related to the extent of reaction $\xi$ by the following formula.
$$
\Delta_r G = \left( \dfrac{\partial G}{\partial \xi}\right)_{T,P}
$$
Two Component System
For simplicity, I am going to assume a simple gas-phase reaction with $1\ \mathrm{mol}$ of $\ce{A}$ and $\ce{B}$ combined.
$$
\ce{A(g) -> B(g)}
$$
For such a system we define the Gibbs energy as follows.
$$
\mathrm{d}G = -S\mathrm{d}T + V\mathrm{d}P + \mu_\ce{A}\mathrm{d}n_\ce{A} + \mu_\ce{B}\mathrm{d}n_\ce{B}
$$
At constant temperature and pressure, we obtain the following.
$$
(\mathrm{d}G)_{T,P} = \mu_\ce{A}\mathrm{d}n_\ce{A} + \mu_\ce{B}\mathrm{d}n_\ce{B} \tag{2}
$$
Now, here, you will have to do some work yourself, and read about $\xi$, but for our simple reaction $\ce{A -> B}$, we obtain the following.
$$
\begin{align}
\Delta_r G &= \left( \dfrac{\partial G}{\partial \xi}\right)_{T,P}\\
&= \dfrac{\partial }{\partial \xi}(\mu_\ce{A}\mathrm{d}n_\ce{A} + \mu_\ce{B}\mathrm{d}n_\ce{B})\\
&= \mu_\ce{B} - \mu_\ce{A}
\end{align}
$$
We know that $\mu_\ce{A} = \left( \dfrac{\partial G}{\partial n_\ce{A}}\right)_{T,P}$ and $\mu_\ce{B} = \left( \dfrac{\partial G}{\partial n_\ce{B}}\right)_{T,P}$. Using Eq. (1) we obtain the following.
$$
\Delta_r G^* = RT\ln{P_\ce{B}} - RT\ln{P_\ce{A}} = RT\ln{\dfrac{P_\ce{B}}{P_\ce{A}}} = RT\ln{Q} \tag{3}
$$
The superscript $^*$ will make sense soon. If we plot this, we obtain the following graph.

The Problem
We have combined an equation (Eq. 1) of single-component system with that (Eq. 2) of a two-component system, which is only valid if the two components are separated into two one-component systems. This would look something like the following.

The Solution
Consider a two-component system to begin with.
$$
\mathrm{d}G = -S\mathrm{d}T + V\mathrm{d}P + \mu_\ce{A}\mathrm{d}n_\ce{A} + \mu_\ce{B}\mathrm{d}n_\ce{B}
$$
Define $\Delta_r G$ accordignly.
$$
\begin{align}
\left( \dfrac{\partial G}{\partial \xi}\right)_{T,P} &= \dfrac{\partial }{\partial \xi}(\mu_\ce{A}\mathrm{d}n_\ce{A} + \mu_\ce{B}\mathrm{d}n_\ce{B})\\
&= \mu_\ce{B} - \mu_\ce{A} \tag{4}
\end{align}
$$
The differential of the above equation is as follows.
$$
\mathrm{d}\left( \dfrac{\partial G}{\partial \xi}\right)_{T,P}  = \mathrm{d}\mu_\ce{B} - \mathrm{d}\mu_\ce{A}
$$
Consider the Gibbs-Duhem Equation at constant $T$ and $P$.
$$
\mathrm{d}\mu_\ce{B} = -\dfrac{n_\ce{A}}{n_\ce{B}}\mathrm{d}\mu_\ce{A}
$$
By substituting into Eq. (4), since $n_\ce{A} + n_\ce{B} = 1$, we obtain the following relation.
$$
\mathrm{d}\left( \dfrac{\partial G}{\partial \xi}\right)_{T,P}  = -\dfrac{\mathrm{d}\mu_\ce{A}}{n_\ce{B}}
$$
Also, since we have taken only one mole of $\ce{A}$ and $\ce{B}$ combined, $\xi = n_\ce{B}$.
$$
\mathrm{d}\left( \dfrac{\partial G}{\partial \xi}\right)_{T,P}  = -\dfrac{\mathrm{d}\mu_\ce{A}}{\xi} = \dfrac{\mathrm{d}\mu_\ce{B}}{1-\xi}
$$
Here onwards, it is semi-empirical analysis that is going to help us. If someone could carry on the derivation theoretically, I would greatly appreciate it.

UPDATE: In the comments to this and other answers, the question came up why the Gibbs energy of mixing is related to chemical reactions. In a chemical reaction, reactant turns into product (or vice versa). The amounts of species change, as do their concentrations. So if I want to compare the Gibbs energy at two points in the reaction, I have to consider both the change in amounts (no matter how the reaction is set up) and, unless each species has their own phase, the Gibbs energy of mixing.
For finding the difference in Gibbs energy, it does not matter whether the reaction actually happens, or if I just add species in the right amounts to reflect the state before and after (because the Gibbs energy is a state function). To be relevant for a reaction, I do have to make sure my two states have the same set of atoms and the same charge (just like a balanced equation). Using this method, I could even experimentally measure how far a reaction mixture is from equilibrium for a hypothetical reaction that is so slow it never happens (until you find a suitable catalyst).
END OF UPDATE.

[OP] How this graph of Gibbs free energy between reaction coordinate is made because because ∆G = ∆G° + RT ln Q and since ∆G° and T is constant hence the graph should be like that of ln(x)?

You are confusing $\Delta G$ with $\Delta_r G$. The first is a generic change in Gibbs energy between two states, whereas the second one is the derivative of G with respect to the extent of reaction $\xi$. One hint are the dimensions. $\Delta G$ has dimensions of energy, the extent of reaction $\xi$ has dimensions of amount of substance, and $\Delta_r G$ (as the derivative of the first with respect to the second), dimensions of energy per amount of substance.
The curve shows $G$, the Gibbs energy, with an arbitrary zero point. In the simplest case of a reaction $\ce{A(g) -> B(g)}$, the shape of the curve is
$$\frac{G_\mathrm{relative}}{R T} =  \xi \cdot \ln \frac{\xi}{n} + (1 - \xi) \cdot \ln \frac{(1-\xi)}{n} + \xi \cdot \frac{\Delta G}{R T}$$
The logarithm terms represent the entropy of mixing, and $\Delta G$ the difference in Gibbs energy of pure reactant and product.
Here is an example:

Source: https://www.desmos.com/calculator/kmrjrjat7x
If you follow the link, you can play with the parameters, having the equilibrium (where the slope is zero) closer to pure reactants or products.
To get $\Delta_r G$, you have to take the slope of this curve. You can explore this here.
Often in textbooks, this curve is sketched rather than plotted, and the slope for $\xi = 0$ and $\xi = 1$ is too small. Below is an attempt to overlay a curve of the correct shape on the textbook figure:

As you can see, the equilibrium position is a bit off, and the curvature of the overlayed calculated curve is larger than that of the sketched curve (and, as stated above, the slope at the two extremes is different). As a caveat, the shape does change for reactions with more than one reactant and product, and then also depends on whether one of the reactant (or products) is in excess or if you have stoichiometric amounts.

Here is a reference for the Gibbs energy of mixing.

The reference does not mention the Gibbs energy of reaction, only of mixing.

@ananta My formula had a dimensional error. I edited it to fix it.

@ananta Let's discuss this in  chat.


text:
NMR and dienes to isolate certain isotopes?
I am looking to achieve isotope separation using transition states.
In the rxn of dienes with halogens, based on the temperature, the dienes can create a thermodynamic product or a kinetic product. The kinetic product is reversible, while the thermodynamic product is not. This is because thermodynamic products have a higher energy need for transition states.
In the context of transition states, I see some possibility of having two different products, each having one type of isotope, by selectively exciting one isotope while leaving the other isotopes untouched.
I look to achieve this by the following protocol:

Dissolve the isotopes in some liquid.
Put the liquid-isotope solution in NMR machine.
Make NMR machine "tip over" only one variety of isotope repeatedly to gain thermal energy.
Maintain temperature of liquid so that other isotopes not tipped over can only maintain the kinetic product.
Make NMR machine energize target isotope enough so that it can produce thermodynamic product.
Put in the diene to react with isotopes.
Seperate the now permanent thermodynamic product, with some pre-existing method to separate isomers.
Yield the target isotope.

Is this method viable? I am proposing this method with the assumption that thermodynamic and kinetic products can be produced based on the individual energies of halogens, which I am not entirely sure of.
In addition, would this mechanism be viable for reactants other than halogens (using a different TS and rxn, of course)? I am interested in enriching heavier metals, such as uranium (where conveniently, 235 reacts with NMR while 238 does not).
Thank you.

There are at least two glaring problems:

Look at the energy scales on which NMR operates (for example, take the resonance frequency $\nu$ and calculate the associated energy $h\nu$)... and compare it to, for example, the thermal energy $k_\mathrm{B}T$. You'll find that the energy absorbed by nuclear spins in a typical NMR experiment is incredibly negligible. We're talking several orders of magnitude too small.

Even if you can somehow put in enough energy via NMR excitation and convert it into thermal (kinetic) energy, this energy will be dissipated amongst every molecule in your reaction sample. You can't have some molecules in your solution be at a higher temperature than the rest, not in a meaningful manner, at least.

Look up the difference of the isotopic masses between hydrogen and deuterium (a factor of about 2),* and the two isotopes of uranium (difference here $\approx1.5\%$), e.g. in NIST's table of atomic weights.  From this, and the examples in Wikipedia's article about the kinetic isotopic effect where the effect correlates with difference of the isotopic masses concerned, you may infer for reactions evolving uranium at conditions typically seen in the lab, the effect likely will be too low to be of practical value.
Similar to orthocresol's argument, NMR spectrometers excite samples with energy in the range of radio frequency.  The energy per photon of this radiation is too small to excite a molecule beyond the typical thermodynamic, or photochemical activation threshold of a reaction.  However, NMR kinetic studies can vary the temperatures of samples to accelerate / slow reactions (keyword Arrhenius equation), or to render some reaction pathways practically inaccessible (barrier of activation energy). Practically, because in perspective of statistical thermodynamics, not all molecules in a sample carry the same energy.  Instead, there is some tailed (Boltzman) distribution of energy among the molecules.
* Use of deuterium in chemicals and pharmaceuticals has its place, though.  See e.g., the youtube video  How "Heavy Hydrogen" makes Drugs work: Deuterium in Pharmaceuticals, Organic Chemistry & Synthesis

@YoungJunLee If you have an afternoon: Reaction Monitoring by NMR getting the most from the data Webinar as a tutorial (to replicate, you would need both their basic software + a separate plug-in), or Time- and site-resolved kinetic NMR for real-time monitoring of off-equilibrium reactions by 2D spectrotemporal correlations in 2022NatureCommun833 as an open access publication to showcase an application. (Work with 2D NMR spectra requires an other plug-in.)


text:
Does Peracetic Acid react with nickel to create Nickel (II) Acetate?
I am planning to create a solution of nickel (II) acetate to use for electroplating. From my research, I found that peracetic acid (which I am using to make copper (II) acetate by placing copper wire into the acid) also dissolves nickel. I am wondering if this will work with nickel, or if I will need to change the procedure.

Nickel acetate is available from chemical suppliers such as Sigma-Aldrich, so consider that option.

Nickel metal is much more resistant to corrosion than copper metal; a solution of acetic acid and hydrogen peroxide that dissolves copper will not necessarily dissolve nickel metal. If you have access to rather concentrated acetic acid (as in >60%) paired with hydrogen peroxide, I say try it, but if only have 5-15% stuff (which works for copper) you will most likely have to find a new way to dissolve the nickel. The author of this page says hydrochloric acid (unspecified concentration, but I assume it is relatively high) with hydrogen peroxide works, and other sources seem to say that sulfuric acid and nitric acid can also be used. All the sources I found recommended adding hydrogen peroxide when creating the desired nickel solution.
So, to answer your question, yes, you can probably dissolve nickel metal with peracetic acid, but it will have to be of a high concentration. Peracetic acid is dangerous at high concentrations, however, so I cannot recommend this as I do not have direct lab experience with it. With that said, I cannot even recommend any of the alternatives, as these all involve acids that are toxic and would have to be procured in high concentrations (HCl, HNO3, H2SO4, and the like). In any case, if you have prepared your peracetic acid solution with only copper metal in mind, it is likely that it will not be sufficiently strong to dissolve nickel metal as well.
This is beyond your question, but if I were you, I would purchase nickel salts (nickel chloride is my recommendation) online and use these for your electroplating. If you go down this route, keep in mind that inhaling nickel salts is a real health concern and a mask/ventilation apparatuses should be used when handling such powders. Hope this helps.

August J Sasaki observes that copper metal is more easily soluble in peracetic acid than nickel metal despite nickel being thermodynamically higher in the EMF series. Here we look into how that could have been predicted.
We don't need much thermodynamic power. A common mineral acid such as hydrochloric acid may require a metal with good reducing power to react, but peacetime acid with 8ts peroxide function does not need all that. Kinetic factors dominate, so we look at a significant kinetic factor.
Nickel atoms are more tightly bound. Nickel is both harder and higher-melting (1455°C versus 1085°C, from the respective Wikipedia articles on these elements). Higher hardness and melting point implies more activation energy to dissolve the metal, even where the thermodynamics is favorable.
Thus the properties of nickel as a metal imply that it would be more difficult to dissolve kinetically than copper despite the latter being more noble. We should expect to have to modify the d8ssolution procedure accordingly, as described in August's answer


text:
How to get continuous spectral data into intensities of channels centered around wavelengths?
Currently, I am looking into using a spectral sensor from SparkFun in order to obtain spectroscopic data to feed into a trained neural network, however this sensor gives intensities based on inputs from channels centered around different wavelengths. The product can be seen here: https://www.sparkfun.com/products/14351
In order to generate the data to train our algorithm, we plan to take spectroscopic data of random samples of dirt and add traces of our target chemical's spectrum (i.e., add together and renormalize). What is a good way to get our continuous spectrum data into the format given by the sensor?
I was considering the using the Fourier transform on the continuous data and using the values located at the frequencies at which the channels are centered to "convert" the data to the proper format, but I'm not sure if this is physically correct. I know that $f=\frac{c}{\lambda}$, so maybe there needs to be a constant factor thrown in somewhere.
I'm not very well versed in chemistry, so I would appreciate some advice about this situation or more resources which could help enlighten me on this topic. Thank you in advance!

It isn't clear what you intend to use as input data. What is the source of the "continuous" data?

Based on the information in the link, the near-IR sensors can only give a response to channels: 610nm, 680nm, 730nm, 760nm, 810nm and 860nm, each with 20 nm FWHM. The sad news is that one cannot extract more information from less, not even by any fancy transform. The information is lost permanently because of a discontinuity of wavelengths. So, this data cannot generate a continuous spectrum of a sample all the wavelengths from 610 to 860 nm.
Perhaps all one can do is to look at the intensity patterns from the six channels and use that for training. Each channel output is a single number corresponding to the intensity to each nominal $\lambda$ with 20 nm bandwidth. So this sensor will generate six numbers for each analyzed sample. That is all. There is no way to interpolate in between the wavelengths, at least in NIR spectroscopy.

I was going to comment to the OP that there is no such thing as a continuous spectrum in practice. When they write "I was considering the using the Fourier transform on the continuous data and using the values located at the frequencies at which the channels are centered to "convert" the data to the proper format" what do they even mean by proper format?

@BuckThorn, My understanding is that the OP is thinking to collect the data at those six wavelengths ranging from 610 to 860 nm. He is wondering is there was a way to fill in the remaining wavelength info from this data so as to make a continuous spectrum. The answer is no.

(+1) And the original collected continuous spectra are apparently reflectance spectra, not absorbance or transmittance spectra. So each such spectrum would be processed to six average (over their respective 20 nm FWHM intervals) values for use as training data.

My interpretation of the goal here is to take a "continuous" spectrum (or high-resolution) obtained with a good spectrometer and to convert that spectrum into a format comparable to that obtained with the SparkFun sensor.
The procedure of turning continuous data or rather, discretely sampled continuous data, into a new discrete spectrum with amplitudes at a new set of specific pre-selected frequencies is called re-sampling or binning, depending on the sampling rate of the input data.
If the input spectrum is oversampled, such that all of the desired frequencies are available at the same or comparable resolution, then it becomes a question of averaging over an appropriate interval (window) around each desired frequency. How to do this is not necessarily trivial (because the instrument has a filter of a particular shape over the window of each bin) but binning, meaning breaking up the original spectrum into bins and summing the contents of each bin, is one way to do this. Applying a polynomial interpolation function over each bin as a means of averaging (smoothing) is another. The results should be similar, the output for each bin (frequency) being a weighted average.
Here there is a registry issue in that neither resolution nor sampling rate match between the input and output. The easiest solution is to interpolate the input spectrum, applying appropriate smoothing (such as with a Savitsky-Golay interpolator) before resampling. Most numerical packages make this relatively easy to do. This is not a problem if the original "continuous" spectrum has high-resolution.
You also need to normalize the spectra somehow to compare them. If the input spectra vary in origin or how they were acquired or in the nature of the samples this might be an issue. IR spectral intensity, as I understand it, is proportional to concentration but peak intensities also depend on vibrational force constants which can vary between samples. I actually don't know what is the best way to do this part (re-normalization). It might involve some trial and error and visiting the literature.

I may not be looking hard enough, but I can’t find any procedures for spectral resampling online, only Iibraries. Am I on the right track in saying that I need to multiply by a Gaussian centered at the desired wavelength with the appropriate FWHM and then integrate over all wavelengths?

If you read my comment under your question and the first paragraph of my answer hopefully it is clear that it is not entirely clear what you want to do but I think I may have gotten it backwards. It sounds like you want to convert a low resolution "spectrum" (a set amplitudes at a small set of frequencies) and modify that so that you can add it to a higher resolution (complete) spectrum. Yes, one way to do this is to generate Gaussian (or whatever is an appropriate lineshape for IR spectroscopy) and add it to the high-res existing spectrum. I suspect however that ....

the limited data from the sparkfun sensor is leaving out a lot of information, so your "baked" spectrum will not provide a good representation of the actual high res spectrum that would be generated by the mixture.

Originally yeah, without the info I have now, I wanted to go from low to high res, but now it’s pretty clear that it’s more feasible to go from high to low res and train on the low res data

I guess what I’m asking for is clarification on what “averaging” over the appropriate interval, where the Gaussian approach I suggested is more like a weighted sum

Just to be clear then: you want to take a high-res spectrum and generate a set of values at selected frequencies that reflects the data obtained with the sparkfun sensor?

BTW what you wanted to do originally is actually not a (completely) terrible idea for some types of spectroscopy, specifically nmr. For instance, organic chemists typically report where a new compounds generates signal in the form of a short table. The same is possible with IR. But your data is not in this form. For any compound, the sensor is always sampling at a (too) small number of the same frequencies.

A final comment. It looks like Achem's answer addressed your original question. Therefore you may want to (1) edit your post to make that clear (2) accept his answer if indeed it addressed the question (3) post a new question with the new problem as you now see it. Finally, this might be a question for https://dsp.stackexchange.com/ There is something called the instrument filter function which seems relevant to what you are wondering.

There is also https://electronics.stackexchange.com/ . That might actually be the best SE site for your question, since they may be able to tell you what the filter function for the sparkfun device is expected to (or does) look like.

Yes you got it exactly right; that's my new objective. Thank you for your help, I will try my luck with the resources you've referenced!


text:
Avogadro law and reactions in a gaseous phase
When $\pu{100 mL}$ of hydrogen combines with $\pu{50 mL}$ of oxygen, why is $\pu{100 mL}$ of water produced and not $\pu{150 mL}$? As $100+50=150$
I understand that $\ce{2 H2 + O2 -> 2 H2O}$, when we balance the equation. But it is Avogadro's Law that confuses me here.

Neither books, neither magazine articles try   to put near all the info to their titles.

@Poutnik Why, some of them do. "The Life and Strange Surprizing Adventures of Robinson Crusoe, Of York, Mariner: Who lived Eight and Twenty Years, all alone in an un-inhabited Island on the Coast of America, near the Mouth of the Great River of Oroonoque; Having been cast on Shore by Shipwreck, wherein all the Men perished but himself. With An Account how he was at last as strangely deliver'd by Pyrates."

What exactly confuses you?

@IvanNeretin Hehe, You are almost right, but I miss there the recipe for the homemade cheese or pottery for making corn wheat. :-) // Seriously and generally, one should not assume implicit presence of generalisms in statements.  :-)

The Avogadro law says by other words that
At the same pressure and temperature, the number of gas molecules in the given volume doesn't depend on type of molecules.
As there are created 2 molecules from 3, the final volume (at the same conditions as initial ones) shrinks to 2/3 of the initial one (assuming the ideal gas behaviour of water vapor with pressure low enough and temperature high enough to prevent vapor condensation).

Do you (by your background in meteorology) have a reference about the molar volume of water (as gas at STP) at hand to complement the visualization of molar volumes of the elements (https://www.webelements.com/periodicity/molar_volume/) with the one of hydrogen ($\pu{11.42 cm^3/mol}$, reference), and oxygen ($\pu{17.36 cm^3/mol}$, reference)? I speculate this could extend nicely the answer with a paragraph: «However, as real gases at STP (esp. water), the reaction of H and O would instead yield xy mL of water only.

@Buttonwood Well, water would probably end as liquid water, but you know how these oversimplified school tasks are like.  Real water at STP has molar volume of 1 mol  of liquid water at STP, some 18.0x mL/mol. But if air with saturated vapor partial pressure at STP is considered, It can be very well approximated as ideal gas.

@Buttonwood I cannot remember we needed vapor molar volume.

$100$ mL gas A + $50$ mL gas B gives $150$ mL total gas if there has not been any chemical reaction between the molecules of the gases, whatever the nature of the gases A and B. If there has been a chemical reaction between them, the volume is not conserved. Only the mass is conserved.


text:
Molecule similarity measurement
I'm not a chemistry specialist, I'm more from informatics, and I have a question concerning molecule similarity measurement.
Let's taking these two molecules:
mol1: c1(cccnc1)C(O)=O
mol2: O=C(O)c1ccncc1
If I compute the Tanimoto similarity of these two molecules, I find 0.42, now, if I use ChemMine tool, it shows:
AP Tanimoto: 0.636364
MCS Tanimoto: 0.6364
My first question is: what's the difference between Tanimoto, AP Tanimoto and MCS Tanimoto ? Which one is the 'true' similarity measure ?
Secondly, let's take a third molecule, mol3: NC(=O)c1cccnc1, if I compute similarities between mol1 and mol3, I have:
Tanimoto: 0.633
AP Tanimoto: 0.636364
MCS Tanimoto: 0.8000
So I find that mol3 is more similar to mol1 than mol2, is it correct ?
But if we look these three molecules representations (the picture shows mol2, mol1, mol3):

We can see that mol1 and mol2 differs just by the position of one N atom, and mol1 and mol3 have one difference atom (OH != H2N), so my last question is: is it 'chemically' true to say that one atom difference is more similar than on atom at a difference place ?
The reason of this last question is because, intuitively, my non-chemical mind find the mol2 more similar than mol3 to mol1, as there is just one atom on the other side of a bond, whereas in mol3 there is an atom which is different.

Tanimoto scores depend on the underlying fingerprints, e.g. https://doi.org/10.1186/s13321-015-0069-3


text:
Stability of beryllium isotopes
I have been reading about isotopes and their abundance on Wikipedia. It states that lithium has 2 stable isotopes, beryllium has 1 stable isotope (monoisotopic and mononuclidic) and boron has 2 stable isotopes. Although the total number of radioisotopes is much larger than the stable ones in each of these three, that is not the point of the question.
The actual question is if lithium and boron have 2 stable isotopes each, why does beryllium, which comes between lithium and boron, has only one stable isotope? What factor is making this happen here?

Beryllium would have had 2 stable isotopes and beryllium-8 would have been the ideal candidate if two helium-4 nuclei had not been even better energetical choice. Note that there exists no stable nucleus with 5 or 8 nucleons. It creates a little trouble for massive stars how to fuse helium-4 in heavier elements.

Note that there is multiple monoisotopic elements, but near all are odd-proton count cases, that have max 2 stable isotopes (that is the rule), and one option is claimed by the lower energy isotope of its greedy even-proton count neighbors. Tc and Pm have double bad luck - both their stable candidates are stolen.

Related: Why is tin-112 stable, but indium-112 radioactive?

If you actually label the stable isotopes of the elements involved ($\ce{^6Li,^7Li,^9Be,^{10}B,^{11}B}$) it becomes evident that $\ce{^8Be}$ is the missing stable isotope. What happens, as alluded to in the comments, is that pairs of alpha particles do not bond very well, much like pairs of noble gas atoms in ordinary chemistry except perhaps worse.
We often hear about "magic numbers" involving "filled nuclear shells" from physicists seeking stable superheavy elements, but these concepts extend down to lighter atoms and $\ce{^4He}$, an alpha particle, is actually the first of the highly stabilized, "doubly magic" nuclei. The stabilization of $\ce{^4He}$ by this property turns out especially effective in preventing pairing by nuclear forces unless you bring in either a three-way interaction (forming another relatively strongly stabilized $\ce{^{12}C}$ nucleus), or other particles such as an additional neutron.
So the two alpha-particle components of $\ce{^8Be}$ don't hold together; you need at least an additional neutron to hold the nucleus together against fission (a neutron works better than a proton because it avoids additional electrostatic repulsion). The stable isotopes of beryllium and boron (and heavier elements) have the additional particles they need to remain bound. Meanwhile for lithium the double-alpha splitting doesn't arise because you can't split off two alpha particles with only three protons.


text:
Is HC(=NH)OH a stable compound?
The molecule $\ce{HC(=NH)OH}$ has been shown to derive from hydrogen cyanide and water (compound 7). Is it a stable compound?

source: Das, T.; Ghule, S.; Vanka, K. Insights Into the Origin of Life: Did It Begin from $\ce{HCN}$ and $\ce{H_2O}$? ACS Cent. Sci. 2019, 5, 1532–1540. https://doi.org/10.1021/acscentsci.9b00520 (open access).

It is not that difficult to add an illustration to the question to provide some context of the question (especially if the source in question is an open access publication).

@GeoffHutchison The initially used SMILES string C(=N)O is understood by obabel 3.1.1, ChemDrawJS (which assigns to this structure N=CO), or molcalc.org.  However the form now used by @Mihoron of HC(=NH)OH is not understood by said obabel (contrasting to the other programs).

You are considering the following molecule.

As stated in the referenced study:

Values have been calculated at the B3LYP-D3/TZVP+COSMO(ε =
80.0)//RI-CC2/TZVP+COSMO(ε = 80.0) and the B3LYP-D3/TZVP+ COSMO(ε = 80.0)//RIMP2/TZVP+COSMO(ε = 80.0) (values shown in parentheses) levels of theory in kcal/mol.

Regarding the accuracy of calculations, Becke, A.D.$^1$ (2014) says, "with dispersion interactions in place, the future looks bright," albeit with warnings "we cannot, quite yet forget the basics," "DFT heaven is probably unattainable," and "DFAs, local or nonlocal, will never be exact," and ends with the note "The next fifty (forty-three) years will be as interesting as the first."
However, the study (referred in the question) suggests that many compounds in the suggested mechanism have been observed experimentally:

The use of the nanoreactor produces as an output many different
pathways to new species from the starting reactants. In most of the
pathways, formaldehyde 3, urea 26, formaldimine 2, and
glycolonitrile 4 were seen to be formed as intermediates. This
suggests that these species were the key intermediates en route to the
formation of the target molecules, as has also been noted by
experimentalists.

Quick (and even less accurate) calculations using MolCalc shows $\Delta_f H = \mathrm{-122.76 kJ mol^{-1}}$, a negative value, suggesting it is a (somewhat) stable molecule.
$$
\ce{C(s) + \dfrac{3}{2}H2(g) + \dfrac{1}{2}N_2(g) \xrightarrow{\Delta H \approx -122.76 \mathrm{kJ\ mol^{-1}}} HOCH=NH}
$$
However, I couldn't find any results for "synthesis of formimidic acid" and thus it could be a (somewhat unstable) intermediate. As has been pointed out in other answers, the tautomer (formamide) is more stable*.

*Approximate heats of formation calculated using MolCalc.
References

Axel D. Becke; Perspective: Fifty years of density-functional theory in chemical physics. J. Chem. Phys. 14 May 2014; 140 (18): 18A301. 10.1063/1.4869598

I wouldn't call anything with B3LYP accurate. DFT isn't accurate in general. But with proper diligence and calibration it provides quite some insight. That being said, local minima on very approximate levels tend to also be local minima on higher levels. So the basic gist still applies.

@Martin-マーチン I was aware that DFT calculations do show 10-20% error, but this is often attributed to ill-consideration of effects such as solvation, intermolecular interactions, conformational aspects, etc. We must remember that some potentials often perform good in specialized circumstances (eg, a potential may be good for haloalkanes but not for ketones); however, computational chemistry is moving towards a "generalized" method. Statistical/classical thermodynamics and quantum mechanics are the guiding principles of computational chemistry and we do rely on those with confidence, don't we?

In my opinion, B3LYP is (among) the worst offender(s). At least they include some dispersion and some solvent effects. To my knowledge, there still isn't a generally applicable solvent model. To sum it up: if it is right, it's basically happy coincidence of error cancellation.

I have partially read the paper. It uses a terrible level of theory and old software, too. It doesn't correctly cite all the methods it is actually using. Maybe they produce a full list in the supporting information. I wouldn't be very surprised if you obtain something with a better suited method of calculation.

I was a bit imprecise. DFT is of course exact. Density functionals, or density functional approximations (DFA) are not exact. There is a wonderful article by Becke about this. The error in the calculations is problematic and it is within their design. A method that isn't accurate will never become accurate. DFAs perform well, or well enough, but that has nothing to do with accuracy. And without calibration, you won't even realise whether a DFA produces garbage. Accurate methods are expensive and I don't see computational chemistry migrating towards those anytime soon.

@Martin-マーチン Becke (assuming you are referring to this article) does say "with dispersion interactions in place, the future looks bright" albeit with warnings "we cannot, quite yet forget the basics," "DFT heaven is probably unattainable," and "DFAs, local or nonlocal, will never be exact," and ends with the note "The next fifty (forty-three) years will be as interesting as the first."

Another point of order: I have down voted this answer. This is due to the fact, that the following statement is incorrect: These are (very) accurate calculations and shouldn't be underestimated. I will gladly reverse this, when a more appropriate wording is used.

@Martin-マーチン I hope the current edits do include relevant comments regarding the accuracy of calculations. Sorry for verbatim copying my previous comment, but I thought that it fulfills the purpose quite well.

Let us continue this discussion in chat.

@ananta out of curiosity, the heat of formation that I obtain with MolCalc for C(=N)O is $\mathrm{-125.71 kJ mol^{-1}}$.

@nightcod3r I calculated it again, and I am still getting $\pu{-122.76 kJ mol^{-1}$. I wonder how you calculated it: What was your starting structure? Which browser did you use? Are there any custom settings in MolCalc.

@ananta, on firefox 113.0.1 (this should not affect, hopefully) just entered the C(=N)O SMILES and calculated the properties, which shows "Heat of Formation -125.71 kJ mol-1". I've been looking for settings as well, but could not find any.

@nightcod3r I entered SMILES and obtained the same value of $\pu{-125.71}$;; apparently, different results are obtained on MolCalc when entering SMILES and drawing the structure (which I did).

@ananta it should not be relevant for the scope of the question (at the end of the day, it's just an approximation). I've learned that MolCalc uses GAMESS to do the math server-side, and would like to reproduce it in our own system, is $\Delta_f H$ related to any of these magnitudes in this table?

@nightcod3r I once tried installing GAMESS and it was a nightmare, but it is surely worth another try. The table "summarizes the program's present capabilities for obtaining wavefunctions, applying correlation treatments, and computing derivatives." $\Delta_f H$ is related to all these (whichever combination is used for the calculation). Perhaps, in MolCalc, different combination of settings is used when calculating using SMILES input vs structure input.

@ananta seems to be related to the G3MP2 method (see log for methane, consistent with the $\mathrm{-74.6 kJ mol^{-1}}$ in Wikipedia), but again MolCalc is given a rather different result: $\mathrm{-54.45 kJ mol^{-1}}$.

@nightcod3r let us continue this discussion in chat.

Compound 7 is the imidic acid of formic acid (hence formimidic acid).  However as indicated by $\Delta{}G$, its tautomer formamide 8 is thermodynamically favoured.  Some activation ($\Delta{}G^\#$) has to be invested, though.


text:
Successive deprotonation - how far can it go?
It has been written that among the equilibria of the dissolution  of $\ce{SO_2}$ in water, the dissociation of sulphur dioxide into $\ce{HSO_3^-(aq) + H_3O^+(aq)}$ is a complete dissociation. See the answer of S R Maiti to this question:
Can physical absorption happen despite the formation of products?
However, in principle, couldn't $\ce{HSO_3-(aq)}$ deprotonate once more? According to this site, it does.
If it doesn't, why would it? In general, how far does deprotonation go?

It depends on the pH value of the solution / the (remaining) presence of a base sufficiently strong if you have a significant concentration of $\ce{H2SO3}$ (note the note), $\ce{HSO^-_3}$, and $\ce{SO^{2-}_3}$ because they are connected by chemical equilibria:
$$\ce{H2SO3 + H2O <=> H3O+  + HSO^-_3}$$
$$\ce{HSO^-_3 + H2O <=> H3O+  + SO^{2-}_3}$$
Note: Similar to $\ce{CO2}$ in water / $\ce{H2CO3}$, there however is an other (more) important equilibrium to consider, about $\ce{SO2}$ merely dissolved as a gas in water
$$ \ce{SO2 + H2O <<=> H2SO3} \quad K \ll 10^{-9} $$
Hence the description $\ce{H2SO3 <=> H+ + HSO^-_3 <=> 2 H+  + SO^{2-}_3}$ with
$$K_1 = \frac{[\ce{H+}] [\ce{HSO^-_3}]} {[\ce{H2SO3}]} = 1.54 \times 10^{-2}$$
and
$$K_2 = \frac{[\ce{H+}] [\ce{SO^{2-}_3}]} {[\ce{HSO^-_3}]} = 1.02 \times 10^{-7}$$
for aqueous solutions at $\pu{18 ^\circ{}C}$ (values by Holleman) has to be read with $[\ce{H2SO3}]$ as the total of the concentration of the anhydride ($\ce{SO2}$) plus the one of undissociated acid $\ce{H2SO3}$ you can't isolate from water.  (For comparison, values by Housecroft for $\pu{298 K}$: $\mathrm{p}K_a(1) = 1.82$, $\mathrm{p}K_a(2) = 6.92$)  Hence, sulfurous acid is a "special aqueous acid" (to cite Jolly, $\ce{H2O + SO2 <=> H+ + HSO^-_3}$, $K_1 = 1.2 \times 10^{-3}$).

Though a showcase about $\ce{H3PO4}$, Wikipedia's diagram about polyprotic acids describes this with the following plot:

(source: Acid dissociation constant)
With the thermodynamic values in hand, it is possible to compute these dissociations / plot the corresponding curves by help of a computer.  Curtipot is an example freely available since 1992 to work with MS Excel.
edit: substitute sulfuric acid ($\ce{H2SO4}$, initial form of the answer) by sulfurous acid ($\ce{H2SO3}$).
references:
Band 1 Grundlagen Und Hauptgruppenelemente; Holleman, A. F., Ed.; De Gruyter, 2016. https://doi.org/10.1515/9783110495850, page 649 (German)
Housecroft, C. E.; Sharpe, A. G. Inorganic Chemistry, 2nd ed.; Pearson Prentice Hall: Upper Saddle River, N.J, 2005.  Here, table 15.8 / p. 458, accessed on archive.org
Jolly, W. L. Customized modern inorganic chemistry, 2nd ed.; MacGraw Hill: New York, 1998. pp. 226 accessed on archive.org (requires their free library card)

Note that, in contrary to carbonic acid, detection of presence of sulfurous acid (Raman IR) has not been confirmed in water solutions, but just hydrated SO2.  $\ce{SO2(aq) + H2O <=> HSO3-(aq) + H+(aq)}$

@Poutnik interesting contrast with carbon dioxide, which does seem to form carbonic acid as a minor species. Molecular sulfurous acid does exist in the gas phase, however.

@OscarLanzi It is similar for carbonic acid too. Many unstable things are willing to exist in gas or vacuum, where  they are bored, nothing to react with.

@Poutnik p 296 of Miessler (by 2014) introduces sulfites as conjugate base to hydrogensulfites when SO2 is dissolved in water (with a login on archive.org). Housecroft (by 2005) on p. 457 describes sulfourous acid with 2 pKa the next page's table (on archive.org).  German Holleman (https://doi.org/10.1515/9783110495850 by 2016) on page 649 with cross-reference to CO2 and H2CO3(aq).  So there is an edit.

Citric acid represents a surprising example of medium-dependent deprotonation. Ordinarily this acid, $\ce{H3C6H5O7}$, has three dissociable hydrogen atoms from its carboxyl groups. But a fourth dissociation is possible from an alcoholic hydroxyl group under strongly basic conditions. Silva et al. 1 have measured the dissociation constant of the alcoholic hydroxyl group in citric acid, obtaining $\pu{pK}_{a4}=14.4$. Thus in $0.1$ molar sodium hydroxide solution several percent of citrate ions would undergo the extra dissociation to form $\ce{C6H4O7^{4-}}$. This is appreciably more acidic than is typically found in alcoholic hydroxyl groups, and a similar enhancement is observed in the alcoholic hydroxyl protons of malic and lactic acids. The easier displacement of this hydrogen facilitates chelation of the anions on these acids with metal ions in biochemical systems.
The authors identify a hydrogen-bonding interaction between the deprotonated alcoholic oxygen and the methylene hydrogens attached to the carboxylate groups. Negative conjugation between the methylene $\ce{C - H}$ bonds and the carboxylate $\pi^*$ orbitals polarizes the former and makes the methylene hydrogens sufficiently protic to form such hydrogen bonds. The interaction may be represented in the case of citrate by a cobtributing valence-bond structure:

Reference

Silva, A.M.N., Kong, X. & Hider, R.C. (2009). "Determination of the pKa value of the hydroxyl group in the α-hydroxycarboxylates citrate, malate and lactate by 13C NMR: implications for metal coordination in biological systems". Biometals 22, 771–778. https://doi.org/10.1007/s10534-009-9224-5


text:
How to obtain the value of the gas constant R = 8.20574 × 10⁻² L atm K⁻¹ mol⁻¹
While reading a chemistry book, I saw a table of values of the gas constant for different units.
For example,
$$R=8.20574\times 10^{-2}\ \text{L}\cdot\text{atm}\cdot \text{K}^{-1}\cdot \text{mol}^{-1}\tag{1}$$
Now, $R$ comes from the ideal gas law
$$R=\frac{PV}{nT}$$
My question is how arrive at the number in (1).
Here are the calculations I came up with.
Suppose $n$ is in the unit mole and $T$ is in the unit $K$.
Let's check the unit of $PV$.
$$\mathrm{Pa\cdot m^3}$$
$$\mathrm{Pa\cdot 10^3 L}$$
$$\mathrm{\frac{101325\cdot Pa}{101325}\cdot 10^3 L}$$
$$\mathrm{atm \cdot \frac{10^3}{101325}L}$$
$$\mathrm{atm\cdot 0.009869232667\cdot L}$$
$$\mathrm{9.86\cdot 10^{-3} atm\cdot L}$$
According to these calculations, $R$ would have the unit
$$\mathrm{9.86\cdot 10^{-3}\ L\cdot atm\cdot K^{-1}\cdot mol^{-1}}$$
Now, clearly, given that (1) is present in both the book and the internet, I surmise there must be an error in the calculations above. But what is it?

Chemistry SE  (in contrary to some other SE sites) strongly recommends plain text question titles for index/search reasons and due possible displaying issues in question title lists.

Try $R=\pu{8.20574E-2 L atm K-1 mol-1}$ displaying $R=\pu{8.20574E-2 L atm K-1 mol-1}$.  //  \pu{} and \ce{} are involved in the optional MathJax module mhchem, loaded by default on CH SE. It can be loaded explicitly via \require{mhchem}, but IMHO only \ce   (chemical equations/expresions) works on other sites. \pu may have some conflict and never worked for me e.g. on Physics SE. //  Formatting of chem expressions

@Poutnik I used unicode superscripts in my edit of the title. This sometimes works (not all superscript characters are available).

@Karsten Unicode should work, but it may cause hard time to some browsers during editing or copypasting. Like replacing some more exotic characters like various flavours of white spaces or hyphens/dashes by some universal ones. AFAIK Chrome is famous by that. IMHO, the safe bet is rather using plain ASCII if possible.

It seems like the content of your book is more than 40 years old if it still uses "atm". You might want to consider finding a better book.

@Loong The book mentions the unit atm when talking about units used in reporting pressure. The book is Chemical Principles, by Atkins, Jones, Laverman, and is used by the MIT OCW course that I am following.

My question is how arrive at the number in (1).

You will never arrive to the value of $R$ by paper. The value of $R$ is a universal physical constant. It appears for all gases, independent of their chemical identity, when the pressure is low enough. It necessitates experimental data to be verified. It would be pretty cool if you could do that, like if we could arrive to the value of Planck's constant $h$ by a mere inspection of his black-body radiation formula.
Nevertheless, lets try to obtain this value for a substance. The approach is simple:

We put a fixed amount of moles of water in a device.
We mantain the pressure $p$ and temperature $T$.
We measure the resulting volume. I will use here the molar volume, this is, this measured volume divided by the amount of moles of $v = V/n$.

Thus, the ideal gas law states that if we graph the pressure $p$ vs the molar volume $v$ we get
\begin{align}
  p = RT\left(\frac{1}{v}\right) \tag{1} \\
\end{align}
Eq. (1) predicts that a plot of $1/v$ vs $p$ should give an increasing linear function with a slope of $RT$.
Lets use isothermal data, along three different temperatures, and do linear regressions. We take care of three points:

I will be cautious and pick on purpose high temperatures, so the ideal gas law will be approached better.
I will pick the lowest pressures available for the same reason.
I will put everything in SI units to get  the value in $\pu{[J]/[mol]·[K]}$.

The data extracted is:
\begin{array}{c|c c c}
 p(\pu{Pa}) & V (\pu{m^3/mol}) \; (573.15 \; K) & V (\pu{m^3/mol}) \;
 (673.15 \; K) & V (\pu{m^3/mol}) \; (773.15 \; K) \\ \hline
  1000     & \pu{5.59554} & \pu{6.42793} & \pu{7.25446} \\
  10000    & \pu{0.55955} & \pu{0.64259} & \pu{0.72582} \\
  20000    & \pu{0.27959} & \pu{0.32121} & \pu{0.36282} \\
  30000    & \pu{0.18646} & \pu{0.21420} & \pu{0.24194} \\
  40000    & \pu{0.13984} & \pu{0.16065} & \pu{0.18141} \\
\end{array}
The results are as follows:

The results are encouraging, it seems that we have obtained straight lines. The fitted line has a slope $m$ and y-intercept $b$ of:
\begin{array}{c|c c c}
 \text{Temperature} & 573.15 \; \pu{K} & 773.15 \; \pu{K} & 873.15 \; \pu{K} \\ \hline
  m \; [\pu{J/mol}] & 5593.4 & 6457.9 & 7256.8 \\
  b \; [\pu{Pa}]    & 0.33051 & -1.0556 & 0.81970 \\
\end{array}
Now, by Eq. (1), the constant should be
$$ \boxed{R = \frac{m}{T}} \tag{2} $$
We tabulate the results of Eq. (2) and the porcentual error with respect to the value informed by CODATA ($\pu{8.314462618 \; J/mol K}$)
\begin{array}{c| c c c}
 \text{Temperature}  & 573.15 \; \pu{K} & 773.15 \; \pu{K} & 873.15 \; \pu{K} \\ \hline
  R \; [\pu{J/mol K}] & 8.3093 & 8.3114 & 8.3111 \\
  \% E_\pu{r} & 0.062 \% & 0.036 \% & 0.040 \% \\
\end{array}
If what I did you think is fun, you can find $PVT$ data over the internet of other substances, and try this approach. You can then corroborate that $R$ is still the same. There are many other ways to get $R$, some are more simpler and elegant (like physicists would do). This is the one that came through my head when reading.
References

R Value of NIST.
The steam tables come from any thermodynamic book at the back of it. I used "Introduction to Chemical Engineering Thermodynamics", J.M. Smith, H.C. Van Ness, M.M. Abbott, McGraw-Hill, 7th edition, 2004.

Your analysis does not consider that one of $n$, $p$, $V$, $T$ variables is dependent on the other three. By other words, with 4 variables and 1 equation, we get 3 degrees of freedom and 1 of 4 variables dependent. R is the constant involved in this dependency.
$R=\pu{8.20574E-2 L atm K-1 mol-1}$ is numerically equal to the volume (L) of $\pu{1 mol}$ of an ideal gas at $T=\pu{1 K}$ and pressure $p=\pu{1 atm}$.
It could be recalculated from $R=\pu{8.31446 J K-1 mol-1}$, numerically equal to the volume in $\pu{m3}$ for $n=\pu{1 mol}$, $T = \pu{1 K}$, $p = \pu{1 Pa}$.

Here is my new understanding based on comments and answers to the original question.
In the original question, the value in (2) is derived by starting with 1 Pa multiplied by 1 cubic meter, then we divide by 1 mol multiplied by 1 K. But then this starting point has the gas constant at a value of 1, which is experimentally wrong.
If we start with the correct value of
$$8.3145 \mathrm{\frac{J}{mol}}$$
$$=8.3145\mathrm{\frac{kg\cdot m^2}{K\cdot mol\cdot s^2}}$$
$$=8.3145\mathrm{\ Pa\cdot m^3\cdot\frac{1}{K\cdot mol} }$$
and proceed from here, then we can use the original calculations for converting $\mathrm{Pa\cdot m^3}$ and we reach
$$8.3145\cdot 9.86\cdot 10^{-3} \mathrm{\frac{L\cdot atm}{K \cdot mol}}$$
$$=8.20577\cdot 10^{-2}\mathrm{\frac{L\cdot atm}{K \cdot mol}}$$

Its that R is written in those units and have to be applied in the units as written. Use whichever value of R has the most common units to help avoid stoichiometry. But you have to use the units as written. Also you misread the table. Should be first two divided by second two. Not all 4 multiplied. So R has to be written as a constant in the units of 0.0821(L * atm / K * mol) “0.0821 Latm/Kmol”
You cant use 0.0821. But also whatever equation that you tried won’t work because its nonsensical. Imagine it as it it’s supposed to be read. “0.0821 L*atm / mol * K = (x)L * (y)atm / (n)mol * (t)K
Where all variables are written in ().
You’re missing every single variable.
Ok i read your latest response.
If everything is 1. Then it reads as
.0821Latm / Kn = 1L1atm / 1K1mol
Cross cancel whatever units can be cross canceled.
You end up with
0.0821 L•atm / K•mol = 1L•1atm / K•mol
Or just R=PV/nT
You’re not applying the units to the constant which is why you’re confused.