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What scientific and technical challenges prohibit the industrial use of the Sabatier reaction?
It seems that the Sabatier reaction solves the biggest problem of climate change — it does create clean gas and hydrogen fuel from the captured carbon dioxide and water (from melting ice).
$$\ce{CO2 + 4H2 -> CH4 + 2H2O}$$
Currently, there is temporal overproduction of clean energy and problems with its storage. E.g., some countries (in Eastern Europe) have stopped support for new solar panel parts because the current electricity systems cannot cope with the overproduction of solar electricity, especially during the spikes. Why is there no industrial deployment of this reaction?
I can imagine several challenges that can be too difficult to overcome:

$\ce{CO2}$ is plentiful but there may be no process to capture it from the air and such sucking can have a bad impact on local regions (all the $\ce{CO2}$ is sucked from some region and there is no sufficient amount left for plants and crops)

The reaction requires $3\,\pu{MPa}$ pressure and it may be impossible to achieve such high pressure.

The reaction requires some catalysts. While $\ce{Ni}$ is mentioned, it itself is costly and may require replacement with costly rare earth catalysts.

I have no idea about all of this. But the idea seems to miraculous not to think about it.
There is already a more general question that was asked 10 years ago. My question is more specific and the situation has changed. Especially - my question exclude the energy as challenge. While the cost of energy can be high (e.g., in NordPool 2022), generally there are solar panel solutions that are not being adapted because of the energy storage and flow unreliability problem. So, the cost of energy stems mostly from the problem of storage and not from the global availability of electricity. As I said, some countries have withdrawn support for solar panels and explicitly states that no new solar panels can be connected to the grid. So, the energy challenge should be considered as solved in case of my question.
Note added: while the energy storage is an important point and application of methane production, my question is concerned about methane production proper. We can and we should abandon the fossil fuels but - as I understand - methane is used as important input in other chemical processes and manufacturing (non-energy use, especially in Germany) and that is why it may be required to produce it because of its own merit. It is though, interesting, that the wiki page of non-energy of methane https://en.wikipedia.org/wiki/Methane#Chemical_feedstock does not give any valuable information. It gives many hints how methane is used for the production of H2 which is used further for the fertilizers, but such use of methane can be replaced by other input and that is why methane can be avoided in that use as well. But in any case - I guess that there may be chemical processes which require methane without alternatives and that is why methane production can be beneficial.

You got the reaction wrong.  It is: $$\ce{CO2 + 4H2 -> CH4 + 2H2O}$$ while hydrogen is often produced by $$\ce{CH4 + H2O -> CO + 3H2}$$

Thanks, indeed. I corrected the reaction - the correction required moving one O from the left side to the right. Sadly, now it looks a bit less appealing. But anyway - H2 can be produced from the electrolysis and it is really the process that requires the electricity only.

It would consume a lot of energy. It would make sense only with excess of energy from other than fossil fuel sources.

My question explicitly said that there is overproduction of electricity from solar panels in our country, so much, that the energy companies have prohibited any connections from the solar installations to the grid. Electricity is not the problem now, the storage of the electricity is the problem and the mentioned reaction gives one solution to it.

Then it could be better to stay at hydrogen....

Somewhat related

See https://www.hklaw.com/en/insights/publications/2021/03/hydrogen-production-from-excess-renewables for a discussion of using hydrogen to store excess energy from renewables. No need to involve the CO2

For your now narrowed question specific to methane, the answer is simple - extracting natural gas has historically been much less expensive than making CH4 by any other means, so there was no incentive to get it any other way.  Current events in Europe might change that calculation, depending on where the price of nat gas ends up long term.

The primary barrier to all related reactions is their economics
There are a number of ways to produce useful synthetic hydrocarbons from various carbon-containing sources. The Sabatier reaction might look like an attractive way to use up carbon dioxide and generate one useful hydrocarbon feedstock, but it suffers from the same economic barriers as those other reactions.
The biggest problem with all of them is the amount of energy that is required to drive the reaction forward. Despite this some of the alternatives have been used in bulk to create hydrocarbons in restricted circumstances. Usually because war or international sanctions inhibiting access to oil or gas prevents access to the (far cheaper) natural sources of the critical hydrocarbon products. So WW2 germany and apartheid-era south Africa both used some of these reactions to make fuel, despite their expense.
The Sabatier reaction faces even more economic problems that those other processes. It needs a concentrated source of carbon dioxide: that costs energy and capital. The core reaction also uses a great deal of both energy and capital equipment. And the product, methane, is–even now–relatively cheap and available from the oil industry. So, even if the energy were very cheap, it is hard to see the economics stacking up to make it the best way of producing methane.
Plus, we already have a known way to convert carbon dioxide into useful feedstocks: growing plants. They don't even need us to concentrate the carbon dioxide. Even the energy needed to convert biomass into methane would be less than the cost of the Sabatier reaction (were methane really the desired product).


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What does heating Pd do in this organic reaction?

I thought the answer to this question is 7 but the answer is 8. What happens in last step?

After finding the intermediate products in the above process, the reactant in the final step comes out to be the following compound,

Now, the catalyst used in the final step was $\ce{Pd}$. According to this abstract,

Supported gold, palladium and gold–palladium catalysts have been used to oxidatively dehydrogenate cyclohexane and cyclohexenes to their aromatic counterpart. The supported metal nanoparticles decreased the activation temperature of the dehydrogenation reaction. We found that the order of reactivity was Pd ≥ Au–Pd > Au supported on TiO2.

So, by this abstract it is clear that $\ce{Pd}$ catalyst is used for dehydrogenation of cyclohexenes. Accordingly the final step is given below,

Usage of $\ce{Pd(II)}$ catalyst:
According to this abstract,

A palladium(II) catalyst system has been identified for aerobic dehydrogenation of substituted cyclohexenes to the corresponding arene derivatives.

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Acetylation of Secondary amines
I am trying to protect a second degree amino group by doing acetylation and although I did find workup mechanism in DCM, water and NaOH using acetyl chloride, its a custy reaction scheme because Acetyl chloride with water just doesn't feel right to me.
So what I do is that I make a hydrochloride salt of the secondary amine and dissolve it in water and DCM layer and under vigorous stirring, I add some Acetyl Chloride and NaOH simultaneously. But Acetyl Chloride and water is a not too compatible mixture so I am looking for an alternative pathway.
If anyone has ever performed such a reaction in a non-aqueous solution, I'd appreciate the help you can provide with solvents and conditions for the same. In the meanwhile, I am trying to do it with some basic tertiary amine based solvents like NMP, disubstituted piperazine etc. I appreciate any inputs on this

A standard procedure from my lab uses acetyl chloride and pyridine in DCM.
Dissolve your substrate amine and 1.1 eq of pyridine in DCM. Stir under nitrogen and cool in an ice bath. Add dropwise a solution of 1.05 eq acetyl chloride in DCM. Monitor by TLC and leave it until no starting amine is detected (Ninhydrin visualisation is useful here). If the reaction is slow adding a few mgs of DMAP should speed it up.
Wash the organic phase with dilute acid then dilute base, a third wash with aq CuSO4 will remove the last traces of pyridine. Dry, filter and evaporate.

Hi sorry its late my system got fried. I got some questions. 1 Can I use Triethylamine instead of pyridine because both are tertiary amines and Triethylamine is lot easier to source for me than Pyridine and someday I'll have to scale this and pyridine is frowned on in the industry and 2 My amine is in its hydrochloride salt form. Should I free base it first or just add 2.2 eq of Triethylamine in order to freebase the secondary amine in situ? I really want to avoid an ether extraction again due to scale up issues. Thanks!

Pyridine works better as it forms pyridinium acetate as an active intermediate, otherwise you can use 2.2eq TEA. You may need to run it dilute as TEA.HCl has limited solubility in DCM. A few mgs of DMAP can replace the catalytic effect of pyridine.

So elevated temperature and longer reaction time with dilute DCM (I'll use like 100ml for 10 g) should help achieve results with TEA and I'll add 1-2% DMAP as catalyst. Am I going in the right direction?

I don't think you need elevated temperature initially unless tlc shows the reaction is not proceeding. I would use  more DCM, maybe 250ml

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Why should I use thioacetamide instead of H2S when trying to separate cations from the H2S-group of cations?
I thought I shouldn't use $\ce{H2S}$ because of the smell/toxicity but was told there is a different reason. Does anyone know?

You may find these papers helpful: here and here. However, thioacetamide is also a carcinogen and is known to induce acute or chronic liver disease.

$\ce{H2S}$ is a gas which is an ideal choice for carrying out qualitative analysis of cations in the beginning chemistry laboratory. It makes fine precipitates, easy to filtrate quantitatively. It has been used for decades in the previous century. But it has a disadvantage : its odor. Its odor of rotten eggs is painful and even unbearable. And the gas is also toxic. Its use is only possible if an important ventilation system is available in the lab, and if working in a hood. This is feasible in chemistry labs. But the problem is not solved for good. Because the gas is usually released on the roof of the building, and despite the high dilution factor, its odor may be perceptible by the neighbors in the next buildings. They are usually angry about the chemistry lab buildings of their neighborhood.
Thioacetamide gets slowly decomposed into $\ce{H2S}$ in the solution to be tested. Usually it is added to the solution to be tested in the minimum amount just necessary for the test. The odor is weak and bearable, even with awkward students.
As a difference, $\ce{H2S}$ is produced nearly continuously in a unique Kipp's apparatus installed in a hood where all students have to go for their analysis. Much $\ce{H2S}$ is lost during these experiments, and rejected in the outer atmosphere. This is why there is a tendency to replace $\ce{H2S}$ by thioacetamide in schools.

"H2S is a gas which is an ideal choice (...)" - not at all, being gaseous is itself a big complication.

@Mithoron. OK. Maybe it is not ideal. I should have said that it has the advantage of not adding any additional substance in the solution to be analyzed.


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What chemical reactions takes place in purifying acetylene?
Acetylene produced by water on calcium carbide method contain several impurity compound, that are harmful for welding or other purposes of acetylene and needs to be absorbed from the gas mixture before filling the gas into cylinders.
Different stages is considered in a acetylene plant in order to decrease impurities.
Mainly, these impurities are Phosphine, Hydrogen sulfide, Ammonia, Arsine, water and traces amount of some other compounds. Water is absorbed by several dryer stages and ammonia by a water scrubber. The other gases are absorbed in purifier stage. Purifier absorbing material can be wet or dry. One formulation sample of dry absorbing material contains following ingredients.
Diatomaceous Earth –  35.8 % (mass percentage)
(Contains approx. 35% crystalline silica (Cristobalite))
Ferrous Chloride, 35% solution in water - 62.7 %
(Contains small amounts of ferric chloride, copper, acid)
Copper Sulfate - 1.1 %
Manganese Dioxide - 0.3 %
Mercury Dichloride - 0.023 %
Can you please direct me with the chemical reactions that take place in absorbing gases and the role of each chemical ingredient?
As mercury chloride is an environmentally hazard, what will happen if we eliminate it from the formulations?
Can a formulation be found, without mercury chloride?
In acetylene code of practices it is mentioned that, Purifying material can be regenerated several times before disposal. Regeneration is done by passing air through purifier for several hours. What reactions are taking place in regeneration process?
Thank you very much.

The main impurities of $\ce{C2H2}$ are : 1) $\ce{H2}$ destroyed by $\ce{MnO2}$, 2) $\ce{NH3}$, destroyed by copper sulfate, mercury chloride and  and ferrous chloride, 3) $\ce{PH3}$, destroyed by copper sulfate into copper phosphide, 4) $\ce{H2S}$ by copper and mercury, and 5) heavier hydrocarbons adsorbed physically on diatomaceous earth. Iron(II) is here specially for removing tiniest traces of oxygen from the air. The equations are : $$\ce{2 MnO2  +  H2  ->  Mn2O3 + H2O}$$ $$\ce{Cu^{2+} + 4 NH3 ->  [Cu(NH3)4]^2+}$$ $$\ce{HgCl2 + 2 NH3 ->  HgNH2Cl + NH4^+ + Cl-}$$ $$\ce{Fe^{2+} + 2 NH3 + 2 H2O -> Fe(OH)2 + 2 NH4^{+}}$$ $$\ce{3 Cu^{2+}  +  2 PH3  ->  Cu3P2 + 6 H+}$$ $$\ce{Cu^{2+} + H2S  ->  CuS  +  2 H+ }$$ $$\ce{HgCl2 + H2S -> HgS + 2 H+  +  2 Cl^-}$$ $$\ce{4 Fe^{2+} + O2 + 2 H2O -> 4 Fe(OH)^{2+}}$$
Mercuric chloride is used to remove all sorts of thioderivatives in mixtures, like thiols (sometimes called mercaptans, meaning prompt to fix mercury) and dithianes. This reaction exploits the high affinity of $\ce{Hg(II)}$ for sulfur ligands.
Edit : Mercury can be removed from the given mixture. $\ce{H2S}$ will be removed by copper, but not thiols and not to the same level.
Edit : These reactions are not reversible

Thanks for your attention. Can these reaction be  reversed by passing air from purifier?

Just technical note: Dotted or numbered markdown  indentation of the answer points is easy and good looking.

Yes, Hg(II) has a high affinity for sulfur ligands. However in one of EIGA publications (Environmental_Impacts_of_Acetylene_Plants) It said that purifier OFTEN contains mercury chloride. That's why I thought maybe there an be a greener formulation. Here is the exact note.

Solid purification media that use ferric chloride (FeCl3) and ferric oxide (Fe2O3) as active agents often
contain catalysts such as mercuric chloride (HgCl2) or cupric chloride (CuCl2). These materials shall be
disposed of in accordance with relevant waste regulations.

@Reza. Mercury is the most sensitive reagent for removing sulfur compounds. If you accept that your maximum amount of $\ce{H2S}$ is higher than with mercury, you can replace it by copper ions.

Thanks so much. The concentration of hydrogen sulfide has to be lower than 500 ppm. We are going to examine the capacity of a mixture without mercury chloride to see can it decrease the concentration of hydrogen sulfide to this amount or not. As this type of absorbing mixture(without mercury chloride) is not found in market, we have to prepare it by our own. Do you have any recommendation about the grind size and purity of required chemicals or any other things that has to be considered in making such a purifying mixture? thanks.


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Biochemically speaking, why aren't the cell membranes of eukaryotic, multicellular organisms made of more heat-resistant stuff?
TL;DR: there are lipids out there that are more heat-resistant than those in current cell membranes, yet current cell membranes don't have them. Why?
I've cobbled together various miscellaneous sources stating glycerophospholipids - glycerol-based phospholipids - are the majority of lipids in cell membranes.
According to this source, which itself cites lecture notes from one of Roland Faller's lectures at UC Davis, and is fairly well backed up by this table and this study (cmd-F the phrase "It has been shown that DLPC-DAPC") - the following phosphatidylcholines - phospholipids which are a subset of glycerophospholipids, have choline as their head, and are common components of cell membranes - undergo a phase change (henceforth described as "melting") from gels to liquids at the following temperatures:

These results are consistent with Wikipedia's table on lipid bilayer phase behavior, and would imply that:

Dimyristoyl phosphatidylcholine/DMPC has a tail length of 14 alkanes - carbon atoms chained to other carbon atoms via single bonds, with hydrogen atoms taking up their other electrons.
Dipalmitoyl phosphatidylcholine/DPPC has a tail length of 16 alkanes.
Distearoyl phosphatidylcholine/DSPC has a tail length of 18 alkanes.
Diarachidoyl phosphatidylcholine/DAPC has a tail length of 20 alkanes.

According to said Wikipedia page, the temperature at which lipid bilayers melt is determined by the strength of the Van der Waals interactions between their tails, which increase in strength as the tails increase in length (and significantly decrease in strength if there are any carbon=carbon double bonds in their alkane chains). This, too, is consistent with the results in the table: the more alkanes, the more temperature resistance there is.
I've found many sources (more than I can cite) stating cell membranes melt/dissolve at temperatures between 40°C and 55°C. This would suggest that many cell membranes:

Are made of the middle two phospholipids: either dipalmitoyl phosphatidylcholine/DPPC or distearoyl phosphatidylcholine/DSPC.

Are made of phospholipids that are either 16 or 18 alkanes in length.

Melt/dissolve between 40°C and 55°C because that's when dipalmitoyl phosphatidylcholine/DPPC and distearoyl phosphatidylcholine/DSPC melt/dissolve.

However, diarachidoyl phosphatidylcholine/DAPC, also on that table, melts at 64.1°C and starts melting at 62.1°C, since it has more alkanes, a longer tail, and, therefore, stronger Van der Waals interactions. Moreover, very-long-chain fatty acids can have up to 35 alkanes/carbons, which suggests they'd require exceptional temperatures to melt. According to Parchem, for instance, geddic acid - 34 alkanes in length - melts at 93-95°C, about twice as much as existing cell membrane lipids. Although geddic acid isn't a phospholipid (it has no phosphorous, for instance), it seems that the melting temperatures of lipids are based off their tail lengths, not off whether or not they have a phosphorous-based head.
As such, some fatty acids out there are clearly much more heat-resistant than those in cell membranes, such as:

Diarachidoyl phosphatidylcholine/DAPC, which has the same hydrophilic head and hydrophobic tail structure that more conventional phospholipids do - it's just longer. It melts at 64.1°C.

Geddic acid. Although lacking the hydrophilic head structure of glycerophospholipids, it could probably be grafted to such a head; it appears to me, at least, that the hydrogen-oxygen bond on its tip could have its hydrogen removed and subsequently have its oxygen atom bonded to a phosphorus atom to attach it to a choline head, making an ester out of it. It melts at 93-95°C.

These substances are much more heat-resistant than the lipids actually found in cell membranes. Why aren't cell membranes made out of these substances rather than the ones they're actually made out of? You'd think that heat-resistant cell membranes would be an evolutionary advantage that would be better at propagating themselves than less heat-resistant cell membranes.
Is it because, as the Wikipedia page on very-long-chain fatty acids points out:

Unlike most fatty acids, VLCFAs are too long to be metabolized in the
mitochondria, in the endoplasmic reticulum (ER) in plants and must be
metabolized in peroxisomes.

That would suggest that they're simply too large to be produced, and that DPPC and DSPC are compromises: heat-resistant enough to be effective, but small enough to actually be built without using peroxisomes to do so. Is this the case?
This might be more of a question for Stack Exchange Biology, but I figured this might be a chemistry-related thing rather than a biology-related thing.

Warm-blooded animals have the luxury of being able to optimize function without having to "pay much attention" to temperature. If you want to learn about how to make high-temperature membranes, look at bacteria or archaea that survive very high temperature.

Here is a nice article showing amazing membrane lipid adaptations: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5487899/

@KarstenTheis Certainly - if I recall correctly, those organisms use ether bonds rather than ester ones, which radically increases the durability of their cells at the cost of something to do with flexibility and transmembrane proteins. I'm pretty sure, however (correct me if I'm wrong) that ether bonds don't work with multicellular organisms, which is why I'm referring to ester-bonded lipids here. I suppose I'll change the title, though.

@KEY_ABRADE, cross-posting the exact same question is generally not advised on StackExchange.

@Domen Certainly, but I legitimately don't know whether this falls under chemistry or biology. It might be an evolutionary reason or based off of the size of mitochondria (biology) or it might be due to how longer alkane tails interact with cellular chemistry or how cell membranes of longer-tailed lipids are less permeable (chemistry). It probably falls under one more than the other, but the boundary between the two is so thin in this case that I can't see which side of it this question is supposed to be on.

Fluidity is important to the function of a membrane. Membrane that work at high temp don’t work well at lower temps

@Andrew Why? What does the length of a lipid's tail have to do with how functional a membrane made out of said lipids is?

If it was designed by humans, I would advice to ask the designer, why this way. As it was not...

For essentially the same reason that it affects the heat stability. A more rigid membrane is not a good thing. Read up on membrane fluidity and membrane function

This question is a cross-post from Biology Stack-Exchange.

It is a basic feature of the process of evolution that redundant features not relevant to survival tend to disappear over time (as effort spent on unnecessary functions is wasteful: this is why primates with ready access to fruit can no longer make their own vitamin C). Why would any living thing not residing on a hot deep sea vent or a volcano need to be very heat resistant? Only the small number of creatures in such environments use membranes that are more heat resistant.

The evolution of organisms tends to drive out features that are not necessary for survival
The right question to ask here is: why would they be heat resistant?
The key features of most organisms have evolved to support their survival in the environment they live in. The majority or normal organisms live in relatively temperate regimens (often 10° to 30°) where heat resistance is irrelevant to their survival. If you were a fireman you might wish for heat-resistant skin but most people are not firemen and there are very few advantages to having that feature (and being people they can equip themselves with fireproof clothing).
This generalises across most living things: the features they have are matched to the environment they live in. Anything else is redundant and costs effort they do not need to expend. Let's imagine some species of insect that had two variants: one heat resistant and one not. The heat resistant variant would need to consume more food to create the different membranes than its less resistant variant. That is energy that can't be used to grow faster or breed faster. In a normal environment it would be rapidly outcompeted by its less hardy cousin.
But there are heat resistant creatures. Hot deep sea vents often support thriving communities of beasts that can resist hot and toxic environments (past the normal boiling point of water!) They do have all the adaptations needed to resist heat including heat resistant membranes. The reverse is true in extremely cold environments where many fish, for example, have blood and cells that are, in effect, full of anti-freeze to prevent ice crystals damaging their cell membranes.
But neither set of extremophile adaptations exist beyond those environments. They can't compete with the "normal" creatures that lack the (expensive) adaptations.
So heat and cold-resistant adaptation are possible. But, in most environments not useful. So most creatures or plants don't have them. The only places we see them are in environment where they are useful or even essential for survival. They would be a complete waste of effort anywhere else.
Most humans do not walk around wearing bullet proof clothing all the time, despite the survival advantage in a bullet-rich environment. That would only be useful in a war zone or, perhaps, an american high school.

The disadvantage of heat-resistant membranes is not just the "cost". They actually perform poorly at "normal" temperatures because they are too rigid. Some organisms that live in a wide range of temperatures (and don't have thermoregulation) adjust their membrane composition to maintain fluidity at whatever the current temp is.

@Andrew This is probably true. But it doesn't make much difference to the central argument. Rather it strengthens the central drive to eliminate costly features that don't confer any benefit by adding to the cost.


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How would one calculate the Heat Capacity Ratio for a multiple compound Gas?
So I'm attempting to calculate the heat capacity ratio (known also as adiabatic index) of a gas with several concentrations of different compounds such as N2, H2O, CO2, CH4 etc. And while I have an idea (I show below) I haven't actually found any confirmation that this is how the ratios work. Perhaps there is some special way the gases interact changing the heat capacity ratio of the entire gas that someone knows of?
What I'm assuming I should do is take the ratios of each and multiply them by the percentage of the weights within the gas. For example if my gas was 20% CO2 and 80% N2, and CO2 had a ratio of K=1.2, and N2 had a ratio of K=1.33 (just as placeholder values), then my calculation would be
Total ratio of my gas : (0.2)(1.2) + (0.8)(1.33) = 1.304
I'm operating at about temperature ranges of 1000°K - 3000°K, and pressures from 455 Megapascal and under. However I'm still basically assuming ideal gas (I really don't know how you would do this if not anyway, if anyone has an idea I am open to suggestions)
Separate portion of my question, does anyone know what the heat capacity ratios of the compounds I named above would actually be? Maybe someone here has a detailed chart of heat capacity ratios for different compounds as a function of temperature? The wiki has a table but it's not enough for my purposes.
I have them down as the following using the algorithm from the degrees of freedom wiki (including vibrational modes) and plugging them into
heat capacity ratio = 1 + [2/(degrees of freedom)]

N2 : k = 1+ 2/[3 + 2 + (3x2 - 5)] = 1.33 (linear algorithm)
H2O : k = 1+ 2/[3 + 3 + (3x3 - 6)] = 1.22 (non-linear algorithm)
CO : k = 1+ 2/[3 + 2 + (3x2 - 5)] = 1.33 (linear algorithm)
CH4 : k = 1+ 2/[3 + 3 + (5x3 - 6)] = 1.13 (non-linear algorithm)
(k is heat capacity ratio)
EDIT : I just found out that vibrational degrees of freedom contribute twice according to the equipartition theorem? And thus twice for the specific heat ratio? Meaning I would have multiply the vibrational brackets by 2? So my new ratios would be (?) :
N2 : k = 1+ 2/[3 + 2 + 2(3x2 - 5)] = 1.285 (linear algorithm)
H2O : k = 1+ 2/[3 + 3 + 2(3x3 - 6)] = 1.167 (non-linear algorithm)
CO : k = 1+ 2/[3 + 2 + 2(3x2 - 5)] = 1.285 (linear algorithm)
CH4 : k = 1+ 2/[3 + 3 + 2(5x3 - 6)] = 1.083 (non-linear algorithm)
Would love if anyone can confirm any of this or answer my question

Yes, vibrational modes contribute twice (kin + pot. E contribution). You will need also to count with nonlinear dependance of vibrational mode contribution on T, due its significant quantization.

The heat capacity is a measure of translational, vibrational and rotational energy in molecule. Heat capacity is $U/T$. The translational energy is always $U=3RT/2$ ($R$ is gas constant, $T$ temperature). A diatomic molecule has rotational energy $RT$ and vibrational energy $RT$. This means that the constant volume heat capacity for a diatomic molecule is $C_V=7RT/2$ provided the temperature is high enough.
For a polyatomic the rotational energy is $3RT/2$ and vibrational energy $xRT$ where $x$ is the number of vibrations, $3N-6$ for a molecule of $N$ atoms and $3N-5$ if it is linear. This means that the heat capacity is $C_V=(x+3)R$. The constant pressure heat capacity is given by $C_P-C_V=R$ and the ratio $\gamma=C_P/C_V$ is often quoted as adiabatic index, $1.66\cdot$ for an atom, $1.29$ for a diatomic and $\displaystyle \frac{x+4}{x+3}$ for a polyatomic.
At $1000$K the temperature is probably so high that many vibrational levels are populated and these expressions will be accurate ( take their limiting values) if not  you will need to consult a phys. chem. text book and work out the partition functions $Z$ for the molecules and this means knowing the frequencies vibrational motion. These are tabulated and the calculation is not hard just a bit tedious. You can then calculate how the heat capacity varies with temperature. If you cannot find the equations I can post them. (The translational and rotational energies are so small that their limiting values can be assumed without error at $1000$ K and also usually at room temperature.)
Edit:
In general the partition function for energies $E_i$ is
$$\displaystyle Z=\sum_i g_ie^{-E_i/k_BT}$$
The partition functions you need are for vibrations. Each vibration's energy is $E_i= hc\omega_i(n_i+1/2)$ assuming harmonic motion and frequency $\omega$ in wavenumbers. ($h$ is the Planck constant and $c$ the speed of light, $k_B$ Boltzmann constant, $g_i$ is the degeneracy, $1$ for most vibrations). For a molecule with many vibrations with frequencies $\omega_1$ having quanta $n_1$ etc.
$$\displaystyle Z=\sum_{n_1}\sum_{n_2}\cdots e^{-(u_1n_1+u_2n_2\cdots)}$$
where $u_i=hc\omega_i/k_BT=1.4387\omega_i/T$.
This can be simplified to
$$\displaystyle Z=\prod_i\sum_{n_i} e^{-u_i n_i}$$
and as all thermodynamic functions depend on $\ln(Z)$ this becomes
$$\ln(Z)=\sum_i\ln\left( \sum_{n_i}e^{-u_i n_i}\right)$$
Luckily the exponential summation has a closed form as
$$\displaystyle \sum_{n_i}e^{-u_i n_i}=\frac{1}{1-e^{-u_i} }$$
The final thing you need is the connection for an ideal gas between the partition function and the heat capacity and this is
$$\displaystyle C_V=RT\left(2\frac{d \ln(Z)}{dT} +T\frac{d^2 \ln(Z)}{dt^2}\right)$$
You can probably get accurate enough values for the derivatives numerically but an algebraic solution looks straightforwards but messy.
I get for one vibration
$$\displaystyle C_V=\frac{R s^{2} e^{\frac{s}{T}}}{T^{2} \left(e^{\frac{2 s}{T}} - 2 e^{\frac{s}{T}} + 1\right)}$$
with $u=s/T$ but you should check this.

I will be attempting to calculate the partition functions as you said, I will definitely comment if I have trouble finding the correct expressions. What about my first portion of the question, was I correct in weighting the adiabatic index for each component of the gas to give the overall adiabatic index of the gas?

I have been attempting to calculate the partition function, however there seems to be many equations for different types of partition functions. Discrete classical vs discrete quantum, distinguishable vs non-distinguishable etc. I am unsure what equations to be using for my purposes. I would greatly appreciate you posting the equations you were referring to in pursuit of the heat capacity.

I have edited my answer. yes add each part so treat the gasses as not interacting.

I forgot I meant add and make average.

Refering to "The heat capacity is a measure of translational, vibrational and rotational energy in molecule." does it go without saying that it follows from Avogadro's law that all gases no matter what their specific mass is have identical heat capacity if you do not consider the above? Would that make up some other question, not even related?


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Why is radial probability density exactly zero at the center?
I have read that the radial probability density is exactly zero at $r = 0$ because the radial probability density is
$$P(r) = \psi^2\,\mathrm dv,\tag{1}$$
where $\mathrm dv$ is the volume of spherical shell of infinitesimal width, i.e. the volume between sphere of $r + \mathrm dr$ and $\mathrm dr$ radius, and
$$\mathrm dv = 4\pi r^2\,\mathrm dr,\label{q-eqn-02}\tag{2}$$
and thus when $r = 0,$ $\mathrm dv = 0$ and the radial probability density is exactly zero. But \eqref{q-eqn-02} is an approximation as the the actual volume of spherical shell of an infinitesimally small width is equal to
\begin{align}
\mathrm dv &= \frac 4 3\pi (r + \mathrm dr)^3 - \frac 4 3\pi r^3 \\
  &= \frac 4 3\pi (r^3 + 3r^2\,\mathrm dr + 3r\,\mathrm dr^2 + \mathrm dr^3 − r^3) \\
  &= \frac 4 3\pi (3r^2\,\mathrm dr + 3r\,\mathrm dr^2 + \mathrm dr^3).\tag{3}
\end{align}
Thus when $r = 0,$ $\mathrm dv = 4/3\pi\,\mathrm dr^3$ instead of $\mathrm dv = 0,$ and $\mathrm dr^3$ approaches zero, but it is not exactly zero. The radial probability density at $r = 0$ should be approximately equal to zero instead of exactly zero as stated in the books. What am I missing?

Comments are not for extended discussion; this conversation has been moved to chat.

Poutnik's comment is right on the money.  I'll try to put this in intuitive terms.  In calculus we routinely add up an infinite number of zeroes to get a finite sum, so it may seem like you can work with squares or cubes of dr and still get something other than zero.  Of course in a sense you can ... if you're doing multiple integration and are focused only on the rate a rate increases.  But here we are simply integrating over the first dr.  We allow that one zero into our math, but if you multiply by dr again, the effect really is zero.
Another way to think of it: we're adding up the entire volume of the sphere, by looking at the sum for an infinite number of radius values.  While the volume for the area at any given radius, plainly evaluated, is zero, they represent shells of an infinite number of points.  Each one covers an area in the shape of the surface of a sphere.  But at r=0, there is no area - there's just that one single point.  For any other r, you would have to cut up the spherical shell two different ways into an infinite number of tiny bits to get pieces that small.  And if you did that, you'd reduce the rate by which the volume of the sphere goes up ... to zero.
Okay, one more way, because this does confuse students and it can be hard to explain it sensibly.  Suppose I want to start adding up the volume, just the volume, of a sphere or spherically symmetrical distribution.  Now suppose I give a kid a crayon and ask him to color in a picture of an s orbital, and he diligently scrawls around with the crayon, while a sophisticated AI turns that into an inferred 3D wiggly line.  Taking the time at which any point was drawn as a parameter t, the computer even extrapolates after the tyke grew tired, and before he began, so that there is an (x,y,z) point specified for every real-number time t from negative infinity to infinity.  Now whatever that curve is, whether it passes through (0,0,0) or not, whether it repeats itself or not, the calculation of the volume starts the same way - by unfolding it into a simple line along one dimension t.  And it ends the same way, by noting any geometric line has volume zero.  Now here you're interested in a probability density, but the point is: the density is never infinite, and so it can multiply by the volume of the line to get only one thing: zero.  No matter what the scribble, even though it was done for an infinite time, the particle is still certainly somewhere else when you look!  That's what having three real-number coordinates to play with will do for you.

$\mathrm{d}P=|\Psi|^2\mathrm{d}V$
$r=0 \implies x=y=z=0 \implies \mathrm{d}V=0,$
giving just a point with zero volume.
Therefore $\mathrm{d}P=0$ as  well, as $|\Psi|^2$ is finite.
We can also express it as
$\mathrm{d}P=|\Psi|^24\pi r^2\mathrm{d}r$
which is for r=0 exactly zero, however dr is or is not zero.


text:
How to depict small molecules such as D-glucose based on isomeric SMILES
The isomeric SMILES for D-glucose is C([C@@H]1[C@H]([C@@H]([C@H](C(O1)O)O)O)O)O.
The Haworth projection looks like this:

I can parse the SMILES and draw the diagram, but the question is how to translate the CW/ACW of isomeric SMILES into up/down in the Haworth projection.

In the example given, the SMILES @@/@/@@/@ matches up/down/up/down, reading anti-clockwise around the ring, but I am far from confident that will always be the case.
[The SMILES string does not treat the last carbon as a stereo centre.]

Isn't this question basically the same as "How to draw a Haworth projection"? Because I believe the clockwise/anticlockwise notation uses the same convention as the Fischer projection

@Szgoger: Perhaps it is, but I haven't been able to confirm this is so. See edit.

Wikipedia link states:

Configuration at tetrahedral carbon is specified by @ or @@. Consider the four bonds in the order in which they appear, left to right, in the SMILES form. Looking toward the central carbon from the perspective of the first bond, the other three are either clockwise or counter-clockwise. These cases are indicated with @@ and @, respectively (because the @ symbol itself is a counter-clockwise spiral).

So I suppose your question is equivalent to asking if a SMILES string always maps to a Haworth projection in such a way that switching between @ and @@ strictly corresponds to alternating between opposite 'sides' of the pseudo-3D representation, and vice versa.
I am not sure, either.
What I can show you however (hence my answer) is that if you use python rdkit.Chem to translate your SMILES to a molecule, you get this:
m = Chem.MolFromSmiles('C([C@@H]1[C@H]([C@@H]([C@H](C(O1)O)O)O)O)O')

But if you then translate m back to SMILES (which makes it 'canonical' by default):
s = Chem.MolToSmiles(m)

'OC[C@H]1OC(O)[C@H](O)[C@@H](O)[C@@H]1O'

Now there are two consecutive @@ notations, although no consecutive carbons in the ring have OH substituents on the same side of the 'plane' (the hemiacetal OH in your projection should be flat, not down, as its configuration is indeed not specified in the SMILES).

Thank you for your answer. My sense is that the SMILES rule based on the order in which elements are added does not reliably translate to any particular geometric configuration. I think your example tends to confirm that, but doesn't really solve my problem.


text:
Understanding uni- and termolecular reactions in combustion kinetics
I'm trying to understand and incorporate 9-species 19-reaction H2/O2 combustion mechanism into my numerical solver. The reaction mechanism in question is O'Conaire. It seems that the paper is in some standard chemical-kinetics-paper format and thus some basic things are omitted. I have no background in chemistry and after few hours of googling there are still several unclear things that I hope you could help me with.
It is probably illegal to reproduce significant part of the aforementioned paper, so I'll have to only refer some thing here.
So far I assume the following things:

The partial reaction rate for density of $i$-th gas component in $k$-th reaction can be calculated via Arrhenius and acting masses laws:
$$
\frac{d}{dt}\left(\frac{\rho_i}{\mu_i}\right)_k = \omega_i = A T^n e^{\left(\frac{E_a}{RT}\right)}\left(\frac{\rho_1}{\mu_1}\right)^a\left(\frac{\rho_2}{\mu_2}\right)^b
$$
where $\mu_i$ is $i$-th component molar mass, $A$ is pre-exponent constant,  $T$ is absolute temperature, $n$ is temperature exponent and $E_a$ is activation energy; $\rho_1, \rho_2$ are densitites of relevant reactants. Division of densities by molar masses result in molar concentrations:
$$
\left[\rho_i\right] = \frac{kg}{m^3}, \quad \left[\mu_i\right] = \frac{kg}{mol}, \quad \left[\frac{\rho_i}{\mu_i}\right] = \frac{mol}{m^3}.
$$
For example, in the "$O+H_2 = H + OH$" reaction (number 2 in the paper), the rate for $H_2$ would be
$$
\frac{d}{dt}\left(\frac{\rho_{H_2}}{\mu_{H_2}}\right) = -5.08 \times 10^4 T^{2.67} e^{\left(\frac{6.292}{RT}\right)}\left(\frac{\rho_O}{\mu_O}\right)\left(\frac{\rho_{H_2}}{\mu_{H_2}}\right).
$$
with proper units. The minus sign is due to that $H_2$ is consumed during this reaction.

All of the reactions given in the paper are considered elementary and thus exponents for molar concentrations all equal unity. Except maybe reactions like "$O + O + M = O_2 + M$" (see the questions 2, 4, 5 below).

For a particular gas component, final rate is the sum of all reaction rates that the component participate in, with "-" sign if it is consumed and with "+" sign if it is produced.

So, the questions are the following:

Are my assumptions correct?
What if there is there are two of the same molecule/atom in the left side of the reaction, what would the molar concetration exponent be? Is it just $1+1 = 2$?
What is the physical sense of negative $E_a$? Is it that participating radicals are so chemically active that the reaction would be very fast even at very low temperatures?
For reactions involving third bodies (denoted "M"), so called "efficiency factors" are provided. What are these?
How to properly take into account third body concentrations?
In an older paper, reactions with "M" are called pressure dependendent. Does that mean that all other reactions are considered independent of pressure?
What are the Troe parameters (given for reacxtion 9 and 15)
What do $k_O$ and $k_{\infty}$ in the older paper mean?
What would be the good textbook or article to read to understand these things better?

Access to the reference is expensive for those, like me, who are not really interested in this paper. And why not replace $\pu{\frac{\rho}{\mu}}$ by simply the concentration $c$ ?

Note that questions should be understandable standalone, without following any links. By other words, even if the links do not work. Links should be needed only for further, more detailed reading, for full context or as the source reference.


text:
How do I quantify the carbonate system and its pH speciation?
I did Environmental Water Chemistry as part of my undergraduate course where we quantified pH-speciation for the full carbonate system and I got it right according to my tutor. My speciation profile is as seen below:

I've used this graph in other applications and have used it as a template with other speciation diagrams (i.e. chlorine and ammonia speciation etc.) but have recently found out that what I did for the carbonate system is actually wrong!
I find this topic extremely fascinating and was reading about it in Masters et al. "Introduction to Environmental Engineering and Science" (p. 69) where I found their carbonate speciation plot as seen below:

Note that the carbonic acid trends ("$H_2CO_3$" in mine, "$\alpha_0$" in Masters et al) do not have the same curves, this is especially noticeable at the high pH values where the trend declines to zero in Masters et al but not mine. The carbonate species ("$CO_3^{2-}$" in mine, "$\alpha_2$" in Masters et al) also begin at different points. What's going on?!
Note, as I understand it, the system at 25◦C is defined by the following equations:
• Water-based Ions ($H^+\,and\,OH^-$): Formed from the dissociation of water with changing
system pH.
$H_2O_{l} \rightleftharpoons H_{(aq)}^+ + OH_{(aq)}^-$
• Carbonic Acid ($H_2CO_3$): Diprotic acid (which can lose two protons or hydrogen ions)
formed from aqueous $CO_2$ dissolution.
$CO_{2(aq)} + H_2O_{(l)} \rightleftharpoons H_2CO_{3(aq)} \rightleftharpoons H_{(aq)}^+ + HCO_{3(aq)}^-$
• Bicarbonate Ion ($HCO_3^-$): As above reaction, the bicarbonate ion is a protic acid formed
from the dissolution of $H_2CO_3$.
$H_2CO_{3(aq)} \rightleftharpoons H_{(aq)}^+ + HCO_{3(aq)}^-$
• Carbonate Ion ($CO_3^{2-}$): Formed from the dissolution of HCO−3 (assumed no dissolution from CaCO3(s) if the aqueous solution is in contact with calcaerous waters (i.e. water-bodies containing limestone
based corals).
$HCO_{3(aq)}^- \rightleftharpoons H_{(aq)}^+ + CO_{3(aq)}^2-$
The equilibria for all the above reactions can be determined as per the below set of equations
respectively:
$[H_{(aq)}^+] [OH_{(aq)}^-] = K_w = 1 \times 10^{-14} mol^2/L^2$
$\frac{[H_{(aq)}^+] [HCO_{3(aq)}^-]}{[CO_{2(aq)}]} = K_1 = 4.47 \times 10^{-7} mol/L$
$\frac{[H_{(aq)}^+] [CO_{3(aq)}^{2-}]}{[HCO_{3(aq)}^-]} = K_2 = 4.68 \times 10^{-11} mol/L$
Using these relationships I derived the trends in my spreadsheet which Ive tried going back to do the logic for but I just can't seem to get the same answer as the book and I don't know why...
The trends I put in my spreadsheet were based on the following equation rearrangements (although I know their wrong from cross-referencing with the book but annoyingly don't know how I originally derived them as it was some years ago):
pH vs $H_2CO_3$ Speciation = $LOG(\frac{[H^+][CO_3^2]}{[H^+]+[K_1]})$
pH vs $CO_3^{2-}$ Speciation = $LOG(\frac{[K_2][HCO_3^-]}{[H^+]})$
pH vs $H_2CO_3$ Speciation = $LOG(\frac{[K_1][H_2CO_3^2]}{[H^+]})$ for pH 0 - 8.2 and $LOG(\frac{[H^+][CO_3^2]}{[H^+]+[K_2]})$ for pH 8.2 - 14
I don't understand what's going on and have spent DAYS trying to figure out why and where I've gone wrong... please help!
I've also attached a picture from the previous page of the book for context!

Your answer looks great but I can't see how you did it! What expressions did you use for each species concentration? I tried reformatting my spreadsheet with the alpha equations from the book and it was still wrong for me. Could you please provide an extended answer with all the details?

Thank you for the source. Unfortunately, I haven't found any good derivations of the alpha expressions and I've looked at multiple books and online. I'll keep looking for a complete answer.

Why? They weren't bad responses and at least somehow helps answer the question? There are no other helpful answers at the moment anyway.

Hi @EdV, do you have any idea of where I could possibly find those derivations you mentioned? I haven't had any help on here so far unfortunately..

Cheers Ed V, I've downloaded the links and screenshotted your code! I use Engineering Equation Solver (EES) so I'll try and get something going in there, if I do and it works I'll post the answer here unless someone else gets around to it sooner :)

This is a one-man show. All the questions, 5 comments and the answer are from the same person. The only text written by somebody else is just a remark starting by : "Great ! The answer is helpful for future folks".

I'm not sure if your comment Maurice is a criticism? Either way, yes, I thought @EdV deleting his comments might end up having things in this scenario where it makes the whole thing seem strange and hard to follow. It's why I originally asked why he was doing it in the first place.

I am puzzled : future folks will not understand anything about what remains from this long one-man-show.

Either way, I have provided all the proof and calculations in the question and answer which I researched for the past month or so. The answer is now public and people can easily follow it where before they could not. So @Maurice, I don't think your statement "future folks will not understand anything about what remains from this long one-man-show" is false. All good :).

It's important to know that this is called a Bjerrum plot, if only because you could have saved a lot of time getting to the right answer - in fact, the Wikipedia link details the carbonate system you're looking at.

Yes. That's certainly important to know, i wish I found that link earlier, it would've saved me a lot of time and effort.

The difference between the two plots seems to be coming from the differing concentrations of H2CO3. It was pointed out to me by a colleague that my 1M $CO_3^{2-}$ basis is fairly high and not the right "starting point". The actual concentration is a result of equilibrium between atmospheric $CO_{2(g)}$ and $CO_{2(aq)}$, governed by the Henry’s law which I had not included in the calculations and seems to be the better "starting point" which makes a lot more of the system make sense.
A more appropriate starting point is by assuming a partial pressure of $CO_{2(g)}$ ($p_{CO_2}$) of 0.0004 atm and noting the Henry's constant ($K_{H,CO_2,25^\circ  C}$) for $CO_{2(g)}$ being 29.41 atm/(mol/L).
From this and the equilibrium equations mentioned in the original question, I get these following system of equations for pH 0-14:
$[H^+] = 10^{-pH}$
$[OH^-] = \frac{K_w}{[H^+]}$
$[CO_{2(aq)}] = \frac{p_{CO_2}}{K_{H,CO_2,25^\circ C}} = 1.36E-05$ (Constant at all pH)
$[HCO_{3(aq)}^-] = \frac{(K_1\times [CO_{2(aq)}])}{[H^+]}$
$[CO_{3(aq)}^{2-}] = \frac{(K_2\times [HCO_{3(aq)}^-])}{[H^+]}$
Where the molar balance of $CO_{2(aq)}$, termed "TOTCO2" can be defined as:
$TOTCO2 = [CO_{2(aq)}] + [HCO_{3(aq)}^-] + [CO_{3(aq)}^{2-}]$
And the calculated a charge balance ($CB$) on the system where $z$ is the species' ionic charge:
$CB = \sum{z\times[positive\,ions]}=\sum{z\times [negative\,ions]}$
The CB is thus;
$ [H^+] = [OH^-] + [HCO_{3(aq)}^-] + 2[CO_{3(aq)}^{2-}]$
The relative speciations (a [mol%]) are thus:
$ a_{CO_2} = \frac{[CO_{2(aq)}]}{TOTCO2}$
$ a_{HCO_{3}^-} = \frac{[HCO_{3(aq)}^-]}{TOTCO2}$
$ a_{CO_{3}^{2-}} = \frac{[CO_{3(aq)}^{2-}]}{TOTCO2}$
Which gives the following graph:

Taking the logs of all speciation equations gives the graph as seen in the book previously which now, unlike in the original question, aligns completely as seen below:

Happy days.


text:
How do you calculate the heat capacity ratio for a multi-compound gas?
Let's say I have a heat capacity ratio of 1.4 N2 and a heat capacity ratio of 1.2 for O2 (just random placeholder values no significance), and I wanted to calculate the heat capacity ratio of their mixture, similar to air let's say 80% N2, 20% O2 (again placeholder values).
How would I go about calculating the heat capacity ratio of air in total? I have seen people say that I should find the molar fractions of N2 and O2, and then multiply the fractions by the heat capacity ratio.
BUT, I have also seen a solution manual from Thermodynamics: An Engineering Approach, instead use the mass fractions of N2 and O2, and then multiply the fractions by the specific heat capacity at a constant pressure. Then once you find the specific heat capacity at a constant pressure of the mixture, you then use Cv=Cp - R, to find the specific heat capacity at a constant volume, and then do Cp/Cv to get the heat capacity ratio of the mixture.
Can anyone confirm the correct method? And unfortunately I have checked, you do not get the same answer.

Placeholders are the best expressed by  variables instead of random values.  // For mass fractions and specific heat, Cp=Cv+R is not applicable .Check dimensions of the formula.

I would use 78% N2, 21% O2 and 1% Ar. It's more complicated, but besides more closely approximating real air it introduces a component that actually has a different ratio when pure.

Could you please clarify how this is different from your previous question? I can see the text is different, but this seems to be the same underlying question, to the point where you've just used the same title, verbatim. In other words: why shouldn't this extra text just been edited into your previous question?

First off, in reality both nitrogen and oxygen have the same ratio, $7/5=1.4$. A more realistic example would be to include the argon component in the air, as argon has a ratio of $5/3\approx1.67$.
When you calculate the heat capacity, which fractions you use depends on the basis for the heat capacity. If you are calculating on a mass basis, you use mass fractions; if you are calculating on a molar basis, you use mole fractions.
The ideal gas relation $C_p-C_v=R$ applies on a molar basis, so for calculating the heat capacity ratio using a molar basis is preferred. Therefore use mole fractions in your mixing rule.
We can develop a formula for the overall heat capacity ratio given the molecule fractions and individual ratio of each component. For a pure gaseous compound we have
$C_p-C_v=R$
$C_p/C_v=\gamma$
and from these we easily obtain
$C_v=\dfrac{R}{\gamma-1}$
$C_p=\dfrac{R\gamma}{\gamma-1}$
Therefore for a multicomponent gas with mole fractions $x_i$ we render
$\bar{\gamma}=\dfrac{\Sigma\dfrac{x_i\gamma_i}{\gamma_i-1}}{\Sigma\dfrac{x_i}{\gamma_i-1}}.$
We now subtract $1$ to get
$\bar{\gamma}-1=\dfrac{\Sigma x_i}{\Sigma\dfrac{x_i}{\gamma_i-1}}$
and thus a weighted harmonic mean formulation:
$\dfrac{1}{\bar{\gamma}-1}={\Sigma\dfrac{x_i}{\gamma_i-1}}.$

I think I should have clarified, the Cp - Cv = R relation they use is for Cp being mass specific heat capacity, [kJ/(kg x K)], and so is Cv. And R isn't the universal gas constant but the specific gas constant so Rs, which is the universal R divided by the molecular weight (kg/kmol)of the gas. So Rs = R/molecular weight. Which pops out units of [kJ/(kg)(K) ] which matches the Cp and Cv. I think you can do it your way as well, just wanted to clarify the books method. I will actually just link it here  : https://www.sfu.ca/~mbahrami/ENSC%20461/Suggested%20Problems/Chapter%2013/13_60.pdf

@onetwo what are you talking about?

I mean when you mention "The ideal gas relation Cp−Cv=R applies on a molar basis" however they (the link) use it on a mass basis. In essence I just want to verify that the way they reach their heat capacity ratio is reasonable. That is the main point of my entire question. Sorry for the confusion

The partial molar enthalpy and partial molar internal energy of a component in an ideal gas mixture is identical to the corresponding properties of the pure component at the same temperature and pressure equal to its partial pressure in the mixture.  From this, it follows that $$C_{p,mixture}=\sum{y_i}C_{p,i}$$$$C_{v,mixture}=\sum{y_i}C_{v,i}$$where the C's are the molar heat capacities of the pure components, and y's are the mole fractions in the gas.


text:
How to safely alloy magnesium castings into aluminum?
I want to dissolve previously sand cast magnesium to molten aluminum to create an aluminum magnesium alloy. Of course I want to do this as safely, and inexpensively, as possible. I have an electric (resistive) furnace, so no flames.
Of course the normal response is to use an inert atmosphere but this will likely make the process uneconomical. I have 2kg of chunky magnesium casting. (old 70s VW beetle transmission housing) In the end I would like to have an Al-Mg alloy with about 7% being magnesium.
My initial thought was to get the aluminum up to temperature, preheat the magnesium chunk to about 500°C then submerge the magnesium into the aluminum and have it dissolve.
If the magnesium were hot enough to ignite under normal conditions, I wondered if the aluminum would prevent runaway ignition because it would be absorbing the heat from the reaction. Remember there will be 14x as much aluminum (by weight) compared to the magnesium being alloyed.
Of course I would do a limited volume test in order to try out the process. Maybe 300ml total volume. And the reaction would take place in an open top insulated furnace.
Does this sound feasible?

Heating magnesium in air seems not a good idea, and aluminum which also reacts very exothermically with oxygen is not going to be a good heat sink. Can you use a nitrogen atmosphere? It may still react but won't initiate combustion.

Magnesium burns in nitrogen too.

But can the nitrogen initiate combustion?

Temperature initiates combustion.

Bulk magnesium does not ignite easily. Molten magnesium in a crucible covered with a flux is not much worse than molten aluminium. Manipulations of the surface might cause local burning, but this can be minimised by blowing argon over the surface. This is way less worrying than you might expect.

If the magnesium is held below the surface of the molten aluminum, would there still be a significant reaction? As far as a covering flux, that is doable, but if it came to also using argon, that would be cost prohibited. Just not worth all the effort. Might as well just buy new alloy material.

I found a paper that might be informative to me question.  ---   A review on ignition mechanisms and characteristics of magnesium alloys --- https://www.sciencedirect.com/science/article/pii/S2213956720300335

Magnesium melts at a lower temperature than Al, so just drop the Mg alloy into the molten aluminum and keep it covered with the aluminum. Note that as you solidify you will not end up with a consistent alloy composition.

My thought was that if it were submerged that it would not have oxygen to react with. I did find an example of this on http://forums.thehomefoundry.org/index.php?threads/alloying-aluminum-for-casting-and-machinability.164/.  =>[My last mag fire was an Audi valve cover.
About the same time I threw it in my big furnace it registered that it was lite, it ignited before i could get it back out, The furnace had a pool of about 50 lbs of aluminum so I submerged it in the pool, by keeping it submerged it contained most of the fire but that batch of aluminum was about 4% mag when i had it scanned.]

"Note that as you solidify you will not end up with a consistent alloy composition." If the solution was mixed well (stirred gently below the surface oxide layer" why could it not be consistent through the mix? That seems to be a normal procedure for alloying copper into aluminum. Is there some reason it will not work with magnesium?


text:
Does a solvent with the following properties exist?
I am looking for a solvent with these properties:

It's aprotic.
At 1 atm, it has a liquid range of at least 0 to 50 degrees Celsius.
Water can dissolve in it, with a solubility of at least 10g/100g.
It has a high flash point, preferably 100 degrees Celsius or higher. (Or better, be non-combustible.)

Does a solvent with these properties exist? If it does not exist, what's the reason?

There's probably several options. Triethyl phosphate seems to fit the bill, from a cursory search. Propylene carbonate is even better.

@nicolau when I click on the WP pages for the compounds above, their links to the safety data sheets do not work. Therefore turn the comment to an answer with a summary of the required miscobility and combustibility properties.

N-Methyl-2-Pyrrolidone  https://www.m-chemical.co.jp/en/products/departments/mcc/c4/product/1201005_7922.html

Turns out propylene carbonate actually isn't sufficiently miscible with water at 25 °C, so it doesn't work.

There are many examples which clear the requirements. Here is a non-exhaustive list. I do not claim these data are absolutely accurate - you should cross-check everything if it matters to you.
A couple of notes:

Data for solubility of water in solvents are significantly sparser than for solvents in water. I have collected the latter values, but most of these are miscible with water in any proportion, and if not, they should still clear the requirement.
Few organic solvents are non-flammable, and the ones that are typically have poor miscibility with water. To get high flash points, you'll need high-boiling solvents.
I am not advertising Sigma-Aldrich/Merck, they simply have most of the examples and the necessary data.

N,N′-Dimethylpropylene urea (DMPU)

Melting point at 1 atm: -20 °C
Boiling point at 1 atm: 246 °C
Flash point: 121 °C
Solubility in water at 25 °C: Fully miscible in any proportion

Trimethyl phosphate

Melting point at 1 atm: -46 °C
Boiling point at 1 atm: 197 °C
Flash point: 150 °C
Solubility in water at 25 °C: 100 g solvent per 100 g water

Triethyl phosphate

Melting point at 1 atm: -56 °C
Boiling point at 1 atm: 215 °C
Flash point: 130 °C
Solubility in water at 25 °C: 50 g solvent per 100 g water

Dihydrolevoglucosenone (Cyrene™)

Melting point at 1 atm: -18 °C
Boiling point at 1 atm: 227 °C
Flash point: 108 °C
Solubility in water at 25 °C: Fully miscible in any proportion

Triethylene glycol dimethyl ether (triglyme)

Melting point at 1 atm: -45 °C
Boiling point at 1 atm: 216 °C
Flash point: 113 °C
Solubility in water at 25 °C: "Very soluble" (likely fully miscible in any proportion)

Some near misses of interest:
Propylene carbonate

Melting point at 1 atm: -55 °C
Boiling point at 1 atm: 240 °C
Flash point: 132 °C
Solubility in water at 25 °C: 7.5 g water in 100 g solvent
Note: Fully miscible with water in any proportion above 61 °C

Hexamethylphosphoramide (HMPA)

Melting point at 1 atm: 7 °C
Boiling point at 1 atm: 232 °C
Flash point: 144 °C
Solubility in water at 25 °C: Fully miscible in any proportion
NOTE: Carcinogenic

gamma-Butyrolactone

Melting point at 1 atm: -45 °C
Boiling point at 1 atm: 204 °C
Flash point: 98 °C
Solubility in water at 25 °C: Fully miscible in any proportion

N-Methyl-2-pyrrolidinone (NMP)

Melting point at 1 atm: -24 °C
Boiling point at 1 atm: 202 °C
Flash point: 91 °C
Solubility in water at 25 °C: Fully miscible in any proportion


text:
From an electrochemical perspective, why do Li-ion batteries charge with constant-current and constant-voltage phases?
I've recently been looking at battery charging and management circuits, and battery manufacturers often specify a constant-current, constant-voltage and sometimes even a startup phase to battery charging. I imagine there must be some electrochemical or materials science explanation for why this charging pattern is most desirable. Presumably, this method best preserves battery life/ suppresses unwanted phase transitions/ etc., but calls for a mechanistic explanation.

The constant current phase controls and limits the heat released by charging. High temperature causes faster cell ageing by acceleration unwanted sidereactions. These are mainly reactions of trace metallic lithium with carbonate solvent, happening especially when almost charged.
The constant current phase monitors charging voltage and is stopped when it reaches the voltage threshold, determined by the manufacturer. This voltages should never be crossed. this happens typically at 70-80% of the current cell capacity. Sooner of aged cells due higher internal resistance.
After reaching voltage threshold, the charging pattern is switched to the constant voltage with the above threshold value. This phase ensures avoidance of side reactions like metallic lithium deposition, or irreversible oxidation on metal oxide electrode, deteriorating the cell and releases gases.
The cell charge is being topped until the charging current drops below chosen current threshold.

Feedback response:
CA-only charging brings constant controlled stress on being charged cells, but would end prematurely for Li-Ion cells.
CV-only charging would destroy or heavily damage Li-Ion cells, unless already at low enough current charging curve.
Those charging parameters are qualitatively general for all rechargeble cells. The optimal values for charging current ( in numerical multiples of capacity ) and max charging voltage ( or alternatively charging stop detections) differ across cell chemistries.
These data are better to get on cell manufacturer pages.
Even within Li-Ion/Pol cell family, there is multiple cell electrolyte + electrode chemistries and cell designs, focused on capacity, power or longevity, with various maximum current/capacity and voltage values. Some chemistries like NiCd/MH do not have strict voltage threshold, so charge stop detection is managed by different electronic ways, what is out of scope.

Thanks for the explanation! Any technical references for the optimality/ advantages of the CC-CV charging method you describe?

Feedback in answer extension.


text:
Chemical potential vs derivative of free energy
Let $x$ be the concentrations of species in a chemical system, and let $G(x)$ be the Gibbs free energy. My understanding is that the chemical potentials should be given by
$$
\mu_k = \frac{\partial G}{\partial x_k}.
$$
Now Kondepudi and Prigogine seem to say that for an ideal solution,
$$
G(x) = \sum_k x_k\mu^0_k + RT\sum_k x_k\log x_k
$$
(equation 8.4.1 of the second edition, with a slight change in notation, see below). But if I differentiate that with respect to $x_k$, I get
$$
\mu_k = \mu^0_k + RT(\log x_k + 1)
$$
and not the usual expression for the chemical potential in an ideal solution, which is
$$
\mu_k = \mu^0_k + RT\log x_k.
$$
On the other hand, if I would take the free energy to be
$$
G(x) = \sum_k x_k\mu^0_k + RT\sum_k x_k(\log x_k - 1)
$$
then I would get the expected formula for the chemical potential.
So what's going on here? Is Kondepudi and Prigogine's formula for the free energy incorrect or am I making a conceptual mistake somewhere?
(One slight thing I'm unsure of: Kondepudi and Prigogine actually use $\mu^*$ instead of $\mu^0$ in this formula. But I think the idea is that $\mu^*=\mu^0$ for an ideal solution. Certainly I've seen this formula with $\mu^0$ instead of $\mu^*$ before.)

Your definition is incorrect.  Chemical potential is the partial derivative of G with respect to the number of moles of a species, holding T, P, and number of moles of all other species constant.  You have replaced the number of moles of the various species with mole fraction.  in terms of mole fraction, you third equation is correct.

@ChetMiller that sounds like it might be the beginnings of an answer. However, I haven't knowingly used mole fractions anywhere - my $x_k$'s are supposed to be numbers of moles. Can you point to where I've implicitly used mole fractions instead?

Well the equations should be $$\mu_k=\frac{\partial G}{\partial N_k}$$ and $$\mu_k=\mu_k^0+RT\ln{\omega_k}$$where $N_k$ s the number of moles of species k and $\omega_k$ is the mole fraction of species k

@ChetMiller so when you said "in terms of mole fraction, you third equation is correct," did you mean to say "in terms of mole number"? Then it would make sense to me.

No, I meant what I said.

@ChetMiller your second comment contradicts your first one, so I assume you meant what you said in one of them but not the other. It would be helpful to know which one.

I don't see any contradiction if, in your 3rd equation, xk is mole fraction.


text:
Why is the excluded volume in van der Waals 4 times the actual volume of the gas molecule?
I am having trouble understanding why the excluded volume of a particle is four times its actual volume.
According to Wikipedia,
"The excluded volume b is not just equal to the volume occupied by the solid, finite-sized particles, but actually four times the total molecular volume for one mole of a Van der waals' gas. To see this, we must realize that a particle is surrounded by a sphere of radius 2r (two times the original radius) that is forbidden for the centers of the other particles. If the distance between two particle centers were to be smaller than 2r, it would mean that the two particles penetrate each other, which, by definition, hard spheres are unable to do."
I do understand that the centre of one particle will not be able to enter the sphere of radius 2r surrounding the other particle. But shouldn't atleast half the volume of the particle be able to enter the sphere?

This is from the perspective of a single particle, bearing in mind that on average, the other particles are not "entering each other's spheres" because it is a gas. Should the gas condense, the packing will be tighter, i.e. each particle will use up less space than a sphere with radius 2r (compare densest packing of spheres).

Imagine the centre molecule to be stationary. The collision partner approaches from where you are viewing and goes straight into the image. The collision will occur if the partner is within a disc of radius $2R$ as shown in the image, that is within the hemisphere of radius $2R$ (i.e. hemisphere facing you) which is $4$ times the volume of the spherical molecule with radius $r$.

2 balls and nothing else, taking 1/4 of volume in the double size free sphere, is a model of a minimal free space for free molecule moving.
Anything less and balls get blocked or locked by each other. For distances L > 2r, molecules pass each other. For L < 2r, they block each other.
For the pressure-aware gas state equation, it effectively means like if the own molecular volume were like 4 times more than it really is.
Note that this has much more free space than the tight ball arrangement. It is obvious that much more free space is needed for balls freely passing around neighbor balls.

Aren't the balls actually blocking each other? Both of them were travelling on their own paths when they collided or blocked each other right?

Yes, they are, typically 10E10 times per second for air. The point is, not the neighbour ones on regular bases. The mean free flight path in air is 70 nm, several hundred times the molecule size and several dozen times the mean molecule distance.// Take the tightest equal ball arrangement and calculate how much you have to expand it if balls are to be just able to pass their neighbours. It could be a nice homework.

So like if we take those two balls and label the point where they touch as P, then we assume that both of the balls could be spinning in any direction around P and thus they form a sphere of radius 2r. Is this the case?

What spinning ? Imaging 2 tightly passing balls.


text:
Sampling Groundwater for CFCs
Possibly stupid question from a non chemist (field hydrologist). I am trying to sample groundwater for CFCs for the purpose of age dating water. I am planning on sampling with a peristaltic pump (such that the water doesn't come into contact with the atmosphere) using Viton tubing. However, the short length (~ 6") of tubing that fits into the pump itself is not Viton (it's silicon), as I couldn't find any Viton with a large enough wall thickness to work. My question is this: could the water running through that short length for that short an amount of time potentially contaminant the sample? Thanks and my apologies if this is not the right place for this question.

You did not mention detection technique. If you are detecting CFCs, don't use any sampling device that itself contains fluorinated compounds like Viton. Alternatively, you can do a control test, by soaking Viton tubing for more than 2-3 days in water and get it analyzed.


text:
How to measure pottery glaze leaching?
I am looking for a way to know if glazes from pottery leach chemical compound in water.
Those are usually metal ions: cobalt, manganese, chrome, etc.
The procedure would be to place hot water in the vessel with possibly an acid to help dissolve the compound faster.
Is there a tool that can give me a reliable reading of the elements and amount dissolved in the solution?
The tool need to be very sensitive (in the ppm range)
What are the tools they use to do water analysis, especially of toxic metals?
Edit: I am interested to know the instrument needed, maybe also the protocols if someone has knowledge about it, not if it’s hard or expensive or for ’certified individuals’.
At the moment most artisanal pottery is never tested, and I think that most ceramic importation are not tested either and although most people have moved away from the worst offenders lead, barium, uranium there is still many recipes floating around with high level of chrome, cobalt, manganese, etc.
Furthermore, each recipe is different in the composition and firing so if someone use 2% of cobalt carbonate and get a good result, another glaze with the same amount might get a very different result.
Some potters just look the other way or don’t know. Some put a lemon slice on the pottery or use the dishwasher and look if there is visible difference on the surface of the glaze. I never heard of anyone actually doing a real test.
Thanks

For private person and home available means, it is extremely demanding. While for companies with trace metal water analysis - like for water treatment companies - it is a piece of cake, e.g. by AAS. But it would not be cheap.

If you want to determine if a glaze results in a vessel safe for human consumption, you will need professional support (for more than just the analysis).


text:
Aproximation of H2 combusion limits
I'm looking for a relatively simple approximation for combustibility limits of H2/O2/N2 gas mixtures over temperature/pressure. (It doesn't need to be conservative.)
I've found a fair bit of data for H2 mixtures at room temperature and above (rocketry in particular tends to care a fair bit about this), but very little for lower temperatures.
Squinting at figure 3 of MODELING THE RISK OF FIRE/EXPLOSION DUE TO OXIDIZER/FUEL LEAKS IN
THE ARES I INTERSTAGE, for instance, gives the following approximations for H2/O2/N2 mixtures (presumably at standard pressure, though the source doesn't say):
(v/v)% O2 must be >= 7.25 - 0.00648*T_K, and
(v/v)% H2 must be >= 5.06 - 0.00667*T_K
...but this is from data in the range of 70F-900F, and as such is probably terrible at lower temperatures. Ditto, this doesn't take pressure into account at all (though from what I've seen the effects of pressure on H2 combustibility are weird. A minimum in LFL at ~0.3 bar, an intermediate maximum at ~20 bar, and then asymptotic from there.)


text:
Is cyclobut‐3‐ene‐1,2‐diide aromatic?
I think cyclobut‐3‐ene‐1,2‐diide is non-aromatic since there are two sp3 carbons (carrying negative charge), but my source says it is aromatic. Why?

https://www.masterorganicchemistry.com/2017/03/03/is-this-molecule-aromatic-some-practice-problems/

https://chemistry.stackexchange.com/a/18566/9961

simplified answer-the carbons are not sp3 because lone pairs delocalized by resonance are not counted in the hybridization of the atom

Just because a carbon atom is negatively charged and has a lone pair, it need not be $sp^3$ hybridized. The lone pair could be on a $\pi$-bonding $p$ orbital with the atom hybridized $sp^2$. If you allow for this possibility, you find that the given ion can satisfy the $4n+2$ rule for aromaticity, which the text apparently is claiming.
Whether the ion is actually a stable aromatic species is more complicated. Formation of a highly strained ring, the high charge density, and a relatively low amount of resonance stabilization for four-member aromatic rings all work against forming a strongly stabilized ring (the cyclopropenyl cation is also highly strained, but it has less charge density and greater resonance stabilization). Known cyclobutadiene-dianion compounds are stabilized by using substituents that allow some dispersion of the charge and closely associating the counterions above or below the ring. See for instance References 1 and [2].

Structure of a lithium cyclobutadienide complex, from Ref. [2]. Unlabeled atoms in the ring substituents and solvent molecules are carbon; hydrogen atoms are not shown.
References

Akira Sekiguchi, Tsukasa Matsuo & Hidetoshi Watanabe (2001). "Cyclobutadiene Dianion Dilithium: a New Aromatic Ring System**. Phosphorus, Sulfur, and Silicon and the Related Elements 168:1, 51-58, https://doi.org/10.1080/10426500108546530.

Akira Sekiguchi, Tsukasa Matsuo, and Hidetoshi Watanabe (2000). "Synthesis and Characterization of a Cyclobutadiene Dianion Dilithium Salt:  Evidence for Aromaticity". J. Am. Chem. Soc. 122, 23, 5652–5653.
https://doi.org/10.1021/ja0004175


text:
Is it possible to have different isomers of compounds with SP2 hybridized carbons based on the orientation of the Pi bond?
I was building molecular models with my son who is in high school to teach him about molecular geometries and he asked me a question I had never considered before.  We built Acetone, using plastic models and I mentioned that in reality there were not two equal bonds between C and O, but a Sigma bond running straight between them and a pi bond that actually existed both "above and below" the Sigma.  He asked me if the Pi bond had to be in a plane perpendicular to the atoms in the molecule as it appears to be with the limitations of plastic ball and stick models. Does the Pi bond actually lie oriented in the plane perpendicular to the atoms, or is it in the plane of the atoms, or are BOTH "conformations" possible, and if so does that constitute two different isomers of Acetone with slightly different properties?

I think the differentiation between sigma and pi-molecular orbitals are done in terms of symmetry. In the real molecule, there is no separate orientation, there is just electron density between the two atoms (which we call bond). So, there can't be any "conformer" based on orientation of pi-bonds.

Isomers are not possible in this example. At the high school/introductory college level the $\pi$-bond is taught to be orthogonal to the plane of the carbons and oxygen. The nonbonding pair of electrons on oxygen lie in the plane of the molecule.

@user55119 That is what I had remembered or assumed, having taken organic chemistry over 30 years ago, though your answer is not consistent with S R Maiti's response above.


text:
Titration of a weak acid with a strong base
I recently learned that when titrating a strong base like NaOH into a weak acid, like Acetic Acid, a buffer region/zone develops and can be illustrated as follows:
CH₃COOH + OH⁻→ CH₃COO⁻ + H₂O
I've also been told that the hydroxide anions from the titrant will almost solely consume hydronium produced from partially dissociated acetic acid (reaction below)
CH₃COOH + H₂O → CH₃COO⁻ + H₃O⁺
But if a buffer system does in fact develop, that would mean that at least a portion of the acetic acid is consumed by the hydroxide anions.
My question is can someone explain sequentially what happens? Does the hydroxide consume all of the weak acid first to overwhelm the buffer and then consume hydronium?

See right here for the battle between acetic acid and sodium hydroxide: https://chemistry.stackexchange.com/a/133874/79678.

Reaction of strong/weak acids with strong/weak bases is an evergreen topic and you can find plenty of related info on this site ( See related ) and generally on internet.


text:
Spontaneity criterion for phase transformations and chemical reactions
Why is the Gibbs free energy (G) considered a spontaneity criterion for phase transformations and chemical reactions? Why are other thermodynamic parameters such as enthalpy (H), entropy (S), and Helmholtz free energy (A) not accepted as spontaneity criteria?

That is just convention. It would be confusing to have multiple, sometimes conflicting criteria for a single term.

All S, G and A can be used for spontaneity evaluation, depending on system conditions (isolated, closed isobaric, closed isochoric)

2nd Law of thermodynamics - The Gibbs free energy change of the system is equal to the opposite of the temp times the change in entropy of the universe. Since the universe will change in the way that increases its entropy (second law), that term is positive, as is the temperature (in K), so the Gibbs free energy change must be negative in order to get the opposite.

Spontaneity is simply a system not at equilibrium (this is regardless of direction) with a possible mechanism to reach equilibrium.
Before getting upset note that reactions involving only standard conditions do not reach equilibrium in either direction, so one direction will be spontaneous and the other not. This is incorrect both directions are from the standard states to equilibrium [JG].
For a multistep reaction each elementary step should be considered in turn. At equilibrium forward and reverse reactions are at the same rate and the chemical activities satisfy the equilibrium constant.  The equilibrium constant value determines the extent of reaction in each direction.
The energetics of a reaction are determined by the energy change and the entropy change, not by only one of the two. The functions that combine these properly are the Gibbs free energy
$$G = H - TS, \tag{1}$$
especially useful at constant pressure; and the Helmholtz free energy
$$A = E - TS, \tag{2}$$
especially useful at constant volume.
There are cases where either $H$ or $S$ is much larger and seems to dominate the process. But if one thinks it through, that is true when the conditions are far from equilibrium. At equilibrium
$$\Delta G = 0 = \Delta H - T\,\Delta S\quad\implies\quad \Delta H = T\,\Delta S. \tag{3}$$
This means at equilibrium an infinitesimal change in either will perturb the equilibrium in the appropriate direction with an appropriate change in the other to satisfy the equilibrium condition. If energy is added, $\Delta H > 0$; bonds break increasing particles, $\Delta S > 0$; $\Delta G = 0.$

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. ALL CAPS shouldn't be used for emphasis on a regular basis (use italics for normal emphasis and bold text for strong emphasis). Square brackets and braces have different meaning and are not universally interchangeable in prose or math. Structuring the text with paragraphs improves readability, too.

Why was this answer downvoted? If for formatting give me a break! If for erroneous content please enlighten me.

Most users have a lot to process in real life, and liberating them from putting extra unnecessary work to decipher posts written using own standards is the least an author can do. On SE sites we strive towards highest possible quality and straightforward Q&A experience. If you don't want to improve your communication after you were explicitly guided several times, then you should no longer wonder why the content is downvoted. Sorry, but at the end of working day interpreting a non-formatted wall of text isn't something many are inclined to do, so I won't even tell if this is correct or not.

I will correct one error in my post above. Reactions written in the standard states do attain equilibrium so the forward reaction has the delta G for that direction and the reverse the negative delta G. In both cases the energy is the from the standard states to equilibrium.

Thank you in advance. Please feel free to ask if something is unclear, I think the links should give a decent start. Markdown and MathJax are simple, but could be cumbersome. Yet they are ubiquitous markup tools, so I think learning them is worth investing some time. Sorry if I sounded harsh, I could be too straightforward at times.

Andselisk. unless my posts look different from what others see they are not unreadable. I know I am remiss in formatting, but I fail to see that others are held to your supposedly high standards. I try to give accurate helpful answers and appreciate being brought to task when wrong or incomplete. My computer skills are archaic.

Fair enough, I did some barebone formatting and tried to logically split the text for better readability. Internet users have smaller attention span than those who primarily read books, plus they often use small mobile screens which cram text a lot, so paragraphs are a bit shorter. I'm afraid I don't know what "[JG]" stands for or what meaning it conveys. Anyway, please feel free to peek at the source code and maybe you could gradually implement some of the markup in your posts. Thank you, the answer is indeed a nice one now that I spent time parsing it.

Enthalpy $H$ is not a valid criterion for spontaneous reactions. Spontaneous reactions may be exothermic or endothermic. The famous reaction of $\ce{NaHCO3 + HCl}$ is endothermic, but it is spontaneous.
It is the same for entropy. Some reactions produce entropy. Dimerization reactions like
$$\ce{A + A -> A2}$$
lead to a decrease of entropy.
So the only valid criterion for spontaneity is the Gibbs free energy. In all cases, a spontaneous reaction occurs if $\Delta G < 0.$


text:
Calculating pourbaix diagram of Fe-H2O system at different temperatures
I have been trying to calculate the pourbaix diagram for Fe-H2O at different temperature for my high school research but I am facing a lot of problems. I have several question regarding the calculation of equations for the diagram.
1) Horizontal line
Correct me if I am wrong.
Finding the horizontal line for the diagram is not a problem. I have the standard electrode potential of the ion, lets consider the equation
Fe3+ (aq) + e-  = Fe2+ (aq)
(I should always consider the eqaution in reduction form right?)
and thus using Nernst equation, I get;

Now as I will be considering the concentration to be 10^-6, thus log 1 = 0. Thus,
The electrode potential will be equal to standard electrode potential of 0.771 for Fe2+.
Thus I get my horizontal line, Y = 0.771
2. The vertical line.
I am having issues here as thermodynamic properties like free energy, entropy, enthalpy are sensitive to temperature change.
I) Where do I find the standard thermodynamic properties of aq ions at 10^-6 M? The CDC handbook has the values for 1M.
II) Gibbs free energy for the formation of Fe2O3 is mentioned as -742.2 kJ/mol. But some papers have mentioned it to be -8242 J/mol for the following equation
2Fe3+ + 3H2O → Fe2O3 + 6H+
So the gibbs free energy will vary for the compound depending upon the reactants, right? If so then how do I find the standard gibbs free energy at 25 degree C for;
Fe(OH)2 + 2 H+  = Fe2+  + 2 H2O
Fe3O4 + 2e-  + 8 H+ = 3 Fe2+   + 4 H2O
Fe2O3 + 2e-  + 6H+  = 2Fe2+  + 3H2O
Similarly the values for enthalpy and entropy will be different for different equations. Once I am able to find the standard enthalpy and entropy for these reactions at 25 degree C, I can use the following formulas to get enthalpy and entropy at desired temperature.

where heat capacity, C can be derived from the following formula (right?)

I search on google and there are different variations to this formulas, why? Also where can I find the values of a,b,c,d,e constands for iron. Do they change for every compound of iron or are same for all species of iron?
After I successfully, find all of these values I will be able to find the gibbs free energy at a given temperature by

Next, I can use

to find the equilibrium constant. All of this just to find the equilibrium constant for the reaction. Is there any other way to find K for the reaction at different temperatures?
After that I can finally calculate the pH or the X value for pourbaix diagram for the equation

using;

3. Slope
If I can figure out these 2 equations then the slope will be easier to do as it is just the combination of the above 2 equations. I will be using the following equation for finding the slope.

please let me know if there is an easier way to approach this problem. I would be really grateful if you could answer my questions as I do not have a supervisor to help me for a long time now and the deadline is coming close.

Is $\pu{10^{-6}}$ the concentration of both ions ? Why do you choose such small concentrations as $\pu{10^{-6}}$ M ? $\ce{Fe^{2+}}$ ions are so sensitive to oxygen that at the tiniest traces of oxygen, some $\ce{Fe^{2+}}$ ions  will be oxidized. And the ratio will change.

Thanks for replying, Yes for both the ions. I choose that because some research papers said that generally the species have a very small concentration of 10e-6. Should I consider 1M instead?

No ! $1$ M is too concentrated. Maximum solubility of iron(II) sulfate is $0.56$ M. I would choose $0.1$ M.

Can you help me with other problems


text:
Retrosynthesis of (2E)-1-(2,4-dimethylphenyl)-3-(thiophen-2-yl)prop-2-en-1-one
I am struggling to find a way to disconnect this molecule and I am unsure of the order:

First, I was thinking of disconnecting the C=C of the α,β-unsaturated carbonyl. Then for the thiophene group disconnecting the formyl substituent, then disconnecting the thiophene either side of the sulfur atom to give a 1,4-diketone.
Then for the benzene substituent: disconnecting the carbonyl first (via Friedel–Crafts acylation), as the methyl groups will be ortho/para-directing.
Would this be the correct order?

Synthesis of compound T11 in https://doi.org/10.1002/ardp.201100005 should be a good hint.

There are several approaches to this molecule, but I think the key disconnection is the E-double bond. This looks like it can be formed by a Horner-Emmons-Wadsworth reaction between the widely commercially thiophene-2-carboxaldehyde and a ketophosphonate derived from 2,4-dimethylacetophenone. HEW reactions give predominantly E-alkenes which is the reason for the choice.
The ketophosphonate can be prepared in good yield from the commercially available 2,4-dimethylbenzoyl chloride by reaction with diethyl phosphonoacetic acid/MgCl2/Et3N, method described here.
It is possible to prepare both starting materials, but it is an important principle of synthetic chemistry that your time as a researcher is more valuable than the cost of advanced materials. If a starting material is available, buy it.


text:
Why is active sec-pentyl group called so?
My teacher as well as Wikipedia — Pentyl group use the following names for the “pentyl” groups:

Group name
Structure with attachment point

1
active sec-pentyl

2
sec-pentyl

3
active pentyl

If a group is attached to the second carbon of pentyl, then it is called “active sec-pentyl” (1). If it is attached to the third carbon, then it is called “sec-pentyl” (2). In case 3 it is called “active pentyl”.
What is the difference between them, and why are they called so?

Where are these terms from?

@orthocresol Really, you never heard these!! Am I reading wrong things, because when I search sec pentyl chloride on google, i get results. https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(McMurry)/03%3A_Organic_Compounds-_Alkanes_and_Their_Stereochemistry/3.03%3A_Alkyl_Groups here it says we have sec butyl but sec pentyl doesn't exist, but i didn't got reason. And i have read these names as common names. I am literally confused.

Oh, sec-pentyl is fine, even if uncommon (the usefulness of the sec- prefix probably stops at butyl, because for pentyl onwards there are plenty of possible isomeric candidates which can lay claim to the 'sec' description). But 'active', that's a new one.

@orthocresol Wikipedia considers sec-pentyl as 1-methylbutyl and active pentyl as 2-methylbutyl. Are they referring to "active" as in optically active?

@NilayGhosh I see the answer has taken a similar approach at least...

Active! the word optically is omitted like oxide is omitted from carbon. sec-butanol from fermentation fusel oils is usually optically active whence the term active. #2 is definitely ambiguous. It is best that these names be used with care.

I, as orthocresol♦ questioned in the comments section, don't know where these terms came from. Thus, any attempt to rationalize these terms would be a speculation. For clarity, it could be argued that each of the structures of these alkyl radicals are drawn in such a way that each of them is attached to a functional group such as hydroxyl or halide or etc. For sake of an argument, let's assume the functional group is hydroxyl, thus the compound from the first radical would be 2-pentanol (or pent-2-ol). Thus, let's try to give a meaningful rationalization to the terminology in the question:

Why it is called 'active sec pentyl' (case 1): If you carefully look at the alkyl radical in case 1, you would realize it is a secondary radical so that it is a sec-pentyl group. Thus, the compound is optically active alcohol (in our case), pent-2-ol, which can exists in both (R)- and (S)-forms. Thus, I assume 'active' part comes to indicate optically active compounds.
Why it is called 'sec pentyl' (case 2): If you carefully look at the alkyl radical in case 2, you would realize it is also a secondary radical so that it is also a sec-pentyl group. Thus, the relevant compound in our case is pent-3-ol, which can exists in only one form, because it is not optically active. Thus, I assume 'active' part is dropped from 'active sec pentyl' to indicate the radical forms only the non-optically active compounds, yet it is a 'sec pentyl'.

The compound related to 'case 3' is 3-methylbutan-1-ol, which is  optically active so that it has been given the terminology, 'active pentyl'.


text:
In a 20wt% ammonia solution, what are the species in solution?
I have bought a 20wt% ammonia solution. It is described as having a specific gravity of 0.92.
I worked out its Molarity earlier today, and I think it was approximately 10M
My question can I assume that the solute remains effectively as ammonia molecules because the dissociation constant is so small?
In my ammoniated solution I have:
NH3  + H2O  and I want to know what concentration of NH4$^+$ and OH$^-$ I have.
So my equation is NH$_3$ + H$_2$O    <<<>>>   NH$_4^+$ + OH$^-$
I have a dissociation constant of ~ 10$^{-5}$.
So if I assume that from 1 mole of ammonia,  x moles of ions form I get
10$^{-5}$ = $\frac{x^{2}}{1-x}$, so x = 3.16 x 10$^{-3}$  i.e.there are 0.003 moles of ammonium ions in solution.
Could someone please check?  If this is the case, I am going to ignore their presence
Thank you

Look at the distribution diagrams of ammonia.

So you are saying ammonia forms reactive ion species?  Would you be able to look over my amended question?  If I have the dissociation constant right, is my estimate of ions in solution OK?

Thanks for your question. Your equation and calculations are correct.
First, I have to say that the variable in equilibrium equations is concentration. By multiplying concentration into volume of your solution, you can compute the number of moles of the ion or extra. So in your example of ammonia solution, by considering the concentration of ammonia be 1M and equilibrium constant be equals to 1.8 * 10^-5,  the concentration of ammonium ion would be calculated 0.00423M and the concentration of remaining ammonia after dissociation would be (1-0.00423= 0.99577M). We can express this amount of dissociation by dissociation degree. I think this is what you are looking for. It is the ratio of dissociated solute to the entire solute that had been dissolved. In the 1M Ammonia solution, the dissociation degree is 0.42% (just a note, in ammonia solution, ammonia doesn't dissociate. Actually, water will dissociate and acts like an acid by giving a proton to Ammonia.)
If I understand correctly, your claim was this that, Whenever the dissociation constant of an acid or base be law, the dissociation degree will be law too(most of the solute remain undissociated). That seems to be correct, but it's not!
Actually, the dissociation degree not only depends on equilibrium constant, but also to solute concentration. The equation bellow demonstrate the relation between dissociation degree, initial solute concentration and equilibrium constant that is called the law of dilution (https://en.wikipedia.org/wiki/Law_of_dilution).

Where the square brackets denote concentration, and c0 is the total concentration of electrolyte and the degree of dissociation of a weak electrolyte is α.
Let's see how α percent change by concentration of ammonia solution. I have solved that equation for several concentrations as you can observe below.
11M (~20% weight percent)  :  0.128%
10M         :  0.135%
9M          :  0.142%
8M          :  0.15%
7M          :  0.161%
6M          :  0.17%
1M          :  0.42%
0.1M        :  1.34%
0.01M       :  12.6%
As you can observe the degree of dissociation is increased by dilution of ammonia solution. So, for your propose you have to consider concentration too.
One last thing is, these calculations are only assumptions of what concentration is going to be measured. Divisions may observe when another salt for example is in solution. For more precise computing of the degree of dissociation and extra, you need to consider other possible equilibrium too.
Take care about reactions with ammonia, that is capable to form explosive and or toxic compounds in some circumstances.
A site about degree of dissociation:
https://www.nagwa.com/en/explainers/950137956236/

Thank you Reza.  That was very kind of you to post such a detailed reply.  I will read it carefully but I am very busy at the moment.  I think it will be useful to me.

Your welcome, I'm glad that I could be of help.