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Why are IUPAC's definitions of exo- and endothermic disconnected from the direction of heat flow?
I'm wondering if anyone can provide a rationale for IUPAC's definitions of exothermic and endothermic, since they disconnect these terms from the direction of heat flow.
Specifically, here are IUPAC's Gold Book definitions:

exothermic reaction (https://doi.org/10.1351/goldbook.E02269, Last updated: February 24, 2014 (v. 2.3.3)):
  A reaction for which the overall standard enthalpy change $\Delta H^\circ$ is negative.
endothermic reaction (https://doi.org/10.1351/goldbook.E02095, Last updated: February 24, 2014 (v. 2.3.3)):
  A reaction for which the overall standard enthalpy change $\Delta H^\circ$ is positive.

The superscript "$\circ$" indicates standard state, which is $\pu{1 bar}$ for a gas phase reaction (https://doi.org/10.1351/goldbook.S05927 and https://doi.org/10.1351/goldbook.S05921, both last updated: February 24, 2014 (v. 2.3.3))
Suppose we have a gas-phase reaction for which $\Delta H^\circ < 0$.  This is defined by IUPAC as an exothermic reaction.  Correspondingly, if the reaction is carried out isothermally at standard state, $q_p = \Delta H^\circ < 0$, indicating that there will be heat flow out of the reaction mixture. This comports with the common understanding of "exothermic".
But: Suppose we study the same reaction at the same temperature, but a higher pressure.  If $|\Delta H^\circ|$ is sufficiently small, and $\left(\partial ΔH\over\partial p\right)_T$ is sufficiently large and positive then, with a sufficient increase in $p$, we could readily reach a point at which $\Delta H > 0$, and thus where (again, assuming isothermal and isobaric conditions) $q_p > 0$, i.e., a point at which the reaction absorbs, rather than releases, thermal energy.  Yet (according to IUPAC)  we must still call that reaction "exothermic", even under those conditions, because $\Delta H^\circ < 0$.
I.e., according to IUPAC, if a reaction is exothermic at standard state, we must always call it exothermic, regardless of the pressure at which we're studying it, and the direction of heat flow at that pressure.
To my mind, this goes against both the common usage/understanding of the terms exothermic and endothermic within the field of chemistry, and also against the inherent meaning of the words themselves.  For instance, if a group is studying a reaction in Jupiter's atmosphere that has $\Delta H_p > 0$ under those conditions, but $\Delta H^\circ < 0$, they're supposed to call it "exothermic" when they discuss it in a paper.  This is very confusing.
This could be resolved if IUPAC replaced "overall standard enthalpy change $\Delta H^\circ$" with "overall isobaric enthalpy change $\Delta H_p$" in its respective definitions. Given this, can anyone explain why IUPAC chose to tie the definitions of exothermic and endothermic to $\Delta H^\circ$ rather than $\Delta H_p$?
N.B. #1: There's also the broader question of whether, to match standard usage in the field, the definitions of exo/endothermic should be untethered from $\Delta H$ generally and, correspondingly, from isobaric processes. I.e., perhaps the definition needs to be made less restrictive still, accounting for both constant-p and constant-V processes. I don't do any thermochemistry, but don't researchers using bomb calorimeters (constant V rather than constant p) describe reactions that evolve thermal energy under those conditions as exothermic? If so, the criterion used there to determine if a reaction is exo- or endothermic is the sign of $\Delta U_V$ rather than $\Delta H_p$.
Alternately, we could abandon pressure-volume restrictions altogether, and tie the definition directly to heat flow instead of a state function:  'An exothermic reaction is one that, if carried out isothermally, has $q < 0$. This means that: (1) If it is carried out at constant p, $\Delta H < 0$; (2) If it is carried out at constant V, $\Delta U < 0$; (3) if it is carried out adiabatically ($q = 0$), $\Delta T > 0$.'   But I don't think IUPAC would wish to go that far.
N.B. #2: There is an alternate interpretation of IUPAC's definition, which is that endothermic or exothermic are only defined at standard state—i.e., that if a reaction takes place outside of standard state, one cannot use the terms exothermic or endothermic at all.  But this also goes against common practice.

Maybe this definition is supposed to provide a standard example rather than a strict definition, but it certainly is not worded that way. I find it odd.

OMG. I stumbled upon this article: Understanding Endothermic and Exothermic Reactions The first paragraph makes me ask myself "where did she get a PhD from?" In any case and sliding away from the off-topic: The authors make the statement in the paper "Every effort has been made to include what appear to be the most commonly accepted definitions. [...] The reader is warned, however, that alternative definitions of some of the terms, [..], are frequently to be found in the scientific literature." (10.1351/pac199668010149)

I think for the reasoning we would have to go back to the commission's reports, but I was unable to find it (probably not online, the earliest one in their database was from the end of 1996; even old iupac starts 1998). This is certainly an interesting query, but resources on the matter might be difficult to obtain. || On a technical note: the iupac tag was removed, see Do we need the [iupac] tag?

@Martin-マーチン Yeah, I also searched through IUPAC's online resources prior to posting, and wasn't able to find an explanation. The only way to get an answer might be to contact IUPAC directly and get a referral to the appropriate person.  I might do that at some point.  But, since this is time-consuming, I was hoping someone on this site might have been involved with the relevant IUPAC commission and thus already have that institutional knowledge.

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

@Martin You can receive a PhD without a deep understanding of chemistry - it depends which field you get it in. Dr. AMH has a PhD in biomedical sciences and a B.A. in math and physics. This probably means a lack of physical chemistry background. Too bad she is such a prolific writer in a space (the internet) without much competition.

@BuckThorn The chat conversation that was opened by now user7951 has disappeared. Has anyone kept a trace of this conversation? Did it brought more matter to the question?

@TheQuark no, the deleted comments are not relevant to the post.

The "N.B. #2" is a reasonable guess. "Exothermic" and "endothermic" then become just a binary value associated with every reaction seen as a collection of precursor and product species.
Common usage is also often confusing "exothermic" and "exergonic", so it might be actually good idea to start following IUPAC standards.
Same reaction at different conditions is possibly reversing its direction of  heat flow. We may use language differently and say: "Changing the reaction conditions does enough work to overcome exothermic reaction's thermal effects."

This could be resolved if IUPAC replaced "overall standard enthalpy change ΔH∘
" with "overall isobaric enthalpy change ΔHp
" in its respective definitions. Given this, can anyone explain why IUPAC chose to tie the definitions of exothermic and endothermic to ΔH∘
rather than ΔHp
?

Probably just to keep it simple: now we do not need to say anything about pressure, when talking about a reaction.
Note that even isobaric conditions do not guarantee ΔHp<0 to cause heat to flow out of the volume, where the reaction took place. Entropy may change that.
To conclude, these definitions are not really disconnected from the direction of heat flow. They are intended to be understood as intrinsic property of a reaction measured for all species in their standard states. It says "exothermic reaction". "exothermic process" may still be used as something way more general.


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How strong are molecular bonds in layman's terms?
Suppose you had a single sheet of graphite the size of an A4 page. What would it feel like to handle it?
Would your fingers instantly tear through it as you tried to grasp it?
Could you punch your finger through on purpose?
Could you tear it like a sheet of paper?
Could you fold it into a paper aeroplane without breaking it?

https://en.m.wikipedia.org/wiki/Orders_of_magnitude_(force) should put it in scale. I noticed you took the right material to ask the question as for depending on composition you can do the operations you listed on much much ticker samples.

It would feel like nothing. Also, it won't be able to support its own weight and will disintegrate instantly.

When it comes to its own weight is something different. As for macroscopic object. The forces between atoms are absolutely small, but thrir weight is small as well. Otherwise we won't have things as roofs of any material. Size will play a role.  @Ivan Neretin.

You might want to look into AFM techniques and optical tweezers, which deal with such types of forces

Related: https://chemistry.stackexchange.com/questions/31299/ripping-apart-plastic-is-this-a-chemical-or-physical-change

I will assume that by "a single sheet of graphite" you mean graphene.
The link in the comments to the wikipedia page on orders of magnitude of different forces, provided by @Alchimista in the comments, is helpful in getting a perspective on at least a few of your questions. Two useful reference values are the force exerted by a 1-kg weight due to gravity at sea level, ~10 N, and the force required to rupture a typical covalent bond, ~1 nN.
Perhaps the first question would seem to be the easiest to answer, but I found it is not: mechanoreception by sensors on skin is complicated. In the same way you can sense very gentle contact, you may be able to sense the graphite sheet touching your skin. It would probably be an odd feeling. Note also that the sheet would be just barely visible.
The strength of a material is described by more than one property. Perhaps the simplest property to describe is the tensile strength, the stress required to rupture the material by pulling it apart under tension. You can use estimates of the geometry of a sheet of graphite and the C-C bond enthalpy to arrive at the force per unit length of the material require to rupture it:
$$\pu{F_l(N/m)} \approx \frac{\pu{350 kJ/mol}}{(\pu{2.46 Å*1.42 Å})}$$
That's $F_l\approx\pu{16 N/m}$, so you might expect a single layer of graphite that is 1 meter long to be able to hold up a 1-kg weight suspended normal to this dimension, assuming the force is evenly distributed. If we assume the sheet is ~$\pu{2 Å}$ thick then we arrive at an estimate of the tensile strength of ~$\pu{10^11 Pa}$, making it the "strongest material ever tested."
Here I've estimated the number of bonds per unit length normal to the direction of tension as 1 per 2.46 Å based on this figure depicting the dimensions of a sheet of graphite. I also assumed the average force required to break one CC bond is $\Delta_{bond}H/r_{CC}$ and used a low estimate of $\Delta_{bond}H$. You could derive a similar value of $\Delta_{bond}H$ from the heat of sublimation of graphite.
Based on this information I would assume that it would be easy to tear the sheet, but that you would not necessarily tear the sheet by grasping it, provided you did this very carefully. On the other hand, you could certainly punch a hole with your finger. Ignoring subtle issues regarding the geometry at which forces are applied, we can use the previous estimate of the rupture force per unit length and an estimate of the radius of a finger (~1/2 cm) to arrive at the force required to punch the sheet, or $\approx\pu{0.5 N}$, or equivalently the pressure:
$$p=\frac{2F_l}{r_{finger}}$$
or $\approx\pu{7 kPa}$. A fat cat would only exert a pressure of about 300 Pa.
Could you fold it into a paper aeroplane? Apparently graphene is tough enough to be significantly bent out of shape without rupturing (Refs. 1 and 2), so I'd have to guess "yes".

References

Briggs et al., Electromechanical robustness of monolayer graphene with extreme bending. Appl. Phys. Lett. 97, 223102 (2010)
Edges of graphite and refs therein

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Calculating the solubility of aspirin in water
I am trying to investigate the relationship between the solubility of aspirin in water with the relationship to some variables. In my investigation, I seem to be stuck with the calculation of finding the solubility. Since reacting aspirin with aqueous NaOH affects solubility, one possible direction I considered taking is investigating the effects on the solubility by changing the amount of aspirin input. Or maybe keeping the aspirin input constant but change the amount of NaOH in the reaction. Either way, the calculation for the solubility is required, but how does one numerically calculate the solubility upon reacting with NaOH?
One way I found online, is dissolving aspirin into a conical flask. Titrate it with x amount of NaOH and record the final and initial volume of neutralization. Using the data, I would try and find the amount of dissolved aspirin in the flask, and basically, compare the amount with the amount stated for the tablet. But how will this give rise to the actual finding of the solubility? It will allow me to get the dissolved amounts, but is that the quantity of solubility? I heard solubility is grams/liter.

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How to get the SMILES of all compounds on PubChem?
I would like to download all the compounds from ChEMBL and PubChem. For ChEMBL this is easy using their webUI. Nevertheless, for PubChem, it isn't clear how to download all the compounds on the database including their SMILES representations.
I am also not sure how pubchem has 33,000,000 compounds where ChEMBL has "only"  2,000,000.

Have a look at https://github.com/mcs07/PubChemPy. Maybe you can find some application of it. And related for ChemSpider: https://github.com/mcs07/ChemSpiPy

The best way to download bulk data from PubChem is actually FTP, as documented in their documentation.
For example, if you want the unfiltered SMILES of every CID in PubChem, the URL is ftp://ftp.ncbi.nlm.nih.gov/pubchem/Compound/Extras/CID-SMILES.gz
You can also download subsets using the PubChem Structure Download service
And as mentioned above, there are Python and other tools to access both PubChem and ChemSpider through their documented web APIs.

One thing to keep in mind.. PubChem intentionally has duplicates - there are SMILES with no stereochemical information and the same SMILES with correct stereo. So depending on your use, you may want to do some level of standardization, filtering and validation.


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Does salt hydrolysis takes place in buffer solution?
I am a bit confused whether hydrolysis of salt takes place in buffer solutions because the books which I read doesnt consider them.But I think since weak base or weak acid salt is present hydrolysis should take place.

We have a policy which states that you should show your thoughts, effort and attempts to answer your question yourself. It'll make us certain that we aren't doing your homework for you, and that the Q/A is beneficial for broad audience.   As "homework-like questions" are considered literal homeworks, self-study questions, puzzles, worked examples etc.  Please [edit] in your full reasoning or thoughts on this. See Homework.  Otherwise, the question may get closed as "low effort homework-like question class."

Water autohydrolyzes, or autodissociates, or autoionizes, to produce a very small amount of H+ and OH- ions. The product of their concentrations is 10e-14.
Most salts that dissolve in water dissociate into ions. Often, one of these ions will have a strong attraction for H+ or OH- and tends to reduce the concentration of that ion in the solution. But since the autodissociation of water is an equilibrium process, if an ion (like acetate ion) combines with the H+, more water will dissociate to produce H+ and also OH-. When sodium acetate is added to water, it dissociates into Na+ and CH3CO2- ions; the Na+ ion attracts several water molecules around it without a preference for OH-. The CH3CO2- ion exhibits a strong preference for H+ over plain H2O, and so would reduce the H+ concentration in the solution. Water compensates by producing more H+ and OH- to keep their product equal to 10e-14.
Acetate ions continue to absorb H+ (it also has an equilibrium to maintain), and finally the acetate ion may be largely present as acetic acid (HCH3CO2). The extra OH- ions that came from the water hang around increasing the pH.
So, the answer to your question is yes, the salt hydrolyzes; in fact, the hydrolysis of the salt is what allows a buffer to be made. You establish a ratio of concentrations of the salt and its acid to give a desired pH. Instead of adding tiny amounts of acid to affect a tiny equilibrium product like 10e-14, you use the ratio of larger amounts of buffer ingredients, governed by the acid dissociation constant.
If the salt doesn't hydrolyze, like NaCl, which ionizes without changing the pH of water, it isn't useful as a buffer material - you can't use it to affect the concentration of H+ or OH- in water.

Salts do not have any idea they are in a buffer solution.
They interact with $\ce{H2O, H+, OH- }$ according to their respective equilibrium constants and $\mathrm{pH}$ of solution, that determine the concentration ratio of the respective conjugated acid/base pairs.


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How to produce produce gel for ocean?
Chemical Formula    CaCo
Molecular weight    89
Purity  Minimum 50%
Properties / Specification

Physical Appearance White, off white prills.
Calcium Carbonate   50% Min
Magnesium Chloride  2.0% Max
Calcium Hydroxide   0.3% Max
Calcium Sulfate 0.3% Max
Water Insoluble 0.5% Max
pH (5% w/v solution)    8.5-10.5

If the calcium chloride is already anhydrous (as shown by the amount of water, $\ce{H2O}$, by weight percent), i.e. has little water, then there's nothing to do but package it in a manner that when it liquefies from absorbing water from the air, it won't drip on containers below. For an example, with a plastic box or pail below.
If there is substantial moisture alerady in the desiccant, then you'd need to dry it by heat (or find another source of calcium chloride).
Be careful pouring it into bags, as that generates irritating, though not poisonous, dust. Use a respirator mask and goggles if working with large amounts.


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Electrophilic substitution of benzene with conc. HNO₃ and HNO₂
The following question was given in my FIITJEE study material:

$$\ce{Benzene + \text{conc.} HNO_3 + HNO_2 ->}$$ Final product of the reaction is:

,
,
,

I started with the solution by deciding the Lewis acid out of $\ce{HNO_3}$ and $\ce{HNO_2}$. I guess $\ce{HNO_3}$ should be Lewis acid as O.S. of N is higher in it. Hence, I wrote the reaction to generate electrophile:
$$\ce{HNO_3 + HNO_2 -> NO_3^- + H_2O + NO^+}$$
Now, it is a bit clear who's the electrophile. Hence, I thought that first option would be the final product. But, according to the answer key it was wrong.
What I am doing wrong?

The correct answer was:

 Second one (nitrobenzene)

You have conc. HNO3 there, and HNO2 even can't be concentrated. It's basically adding water to mixture doesn't change anything besides kinetics.

Concentrated HNO3 is a nasty oxidising agent and will take the nitroso (NO) group to a nitro (NO2) group. Then again, I'm not even sure how good the reaction of HNO2 with an unactivated benzene ring is. I've personally only seen this reaction mentioned once and it was in the context of phenols. I will happily be corrected, though.

@orthocresol Nitrosation is known with toluene and xylenes in strong acids (J. Chem. Soc., Perkin Trans. 2, 1999, 699–705), so nitrosation of benzene does seem reasonable here.  But yes, the nitrosyl group does not last long in annixidizing acid (the above reference uses nonoxidizing acids).

Never thought I’d see the day where a FIITJEE question would make it to HNQ.

Nitrosyl cation, $\ce{NO^+}$, is a better electrophile than molecular $\ce{HNO3}$ (you don't get $\ce{NO_2^+}$ in this system without sulfuric acid or other strong auxiliary acid) and so you get nitrosobenzene in the substitution reaction.  But then the nitrosyl group with its nonbonding electron pair is an attractive target for oxidation by the nitric acid and thereby is converted to $\ce{NO_2}$.  Thus nitrobenzene.  Then, the nitrobenzene, deactivated by the nitro group, will not react with the relatively mild electrophile $\ce{NO^+}$ anymore, meaning products C and D are never reached (and their nitrosyl groups would not have survived anyway).


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In a galvanic cell,do charges accumulate on electrodes 1st (then redox occur) which produce E in the wire when we connect between electrodes?
We study that when we connect a wire between the terminals of a battery ,an electric field in established between terminals  due to potential difference .Is the potential difference result of accumulated charges on terminals. In other words   when an electrode is put in the solution like Zn in ZnSO4  will Zn electrode will become -vely charged ? If not then how does electric field established into wire at exact moment when we connect wire between terminals (electrodes) which essential for electric current .

You can imagine the cell electrodes as capacitors, that are charged by chemical way.
But the capacitance of these capacitors
$$C=\frac {\mathrm{d}q}{\mathrm{d}E}$$
is very small, so does the accumulated charge $q$.
The charge loses due pushing electrons  through the wires of a closed circuit are balanced by simultaneous recharging by ongoing redox reactions.
If we consider the zinc electrode of the Daniell cell, the reaction :
$$\ce{ Zn -> Zn^2+ + 2 e-}$$
is charging the electrode negatively, while the opposite reaction:
$$\ce{  Zn^2+ + 2 e- -> Zn}$$
is charging the electrode positively.
The electrode gains such a potential, where both reactions are ongoing at the same rate.

"You can imagine the cell electrodes as capacitors, that are charged by chemical way."  So it means  "Before" we connect wire between electrodes , there is charge (may be small) on the each electrode due to their chemical reaction to the solution and when we talk about open circuit "Potential difference" between battery terminal the Potential difference is due to "these charges" . Right ?  (Important point is here that I am talking about scenario when there is no wire connected between terminals .)

Yes, it means that. But the positive/negative electrode does not have necessarily positive/negative charge . Rather more positive/,more negative charge than the other.


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Changing toothpaste to a more stable structure
I am using charcoal toothpaste for an actor to spit bile on stage during a show. The paste is not the correct state for our needs.
I know that you can freeze toothpaste and it becomes "more" solid and stable at room temperature but still not stable enough.
What organic edible compound(s) could I add to the toothpaste to make it more solid?
My previous comment that I want the toothpaste to be more solid is perhaps misleading. When I freeze the toothpaste, as one would expect, it becomes more solid due to the water content. Once removed from the freezer, the toothpaste retains a more solid form than it was in the toothpaste tube, but it does not remain as solid as I want. I was considering adding more xanthan gum powder, a stabilizer found in toothpaste.

Crushed raisins? How solid does it need to be?

Thanks, raisins are not likely a suitable solution. I am not trying to make it more chewy or thick, but rather I need to make the toothpaste physically more stable. I am not sure how to better explain what I am trying to do.

My previous comment that I want the toothpaste to be more solid is perhaps misleading. When I freeze the toothpaste, as one would expect, it becomes more solid due to the water content. Once removed from the freezer, the toothpaste retains a more solid form than it was in the toothpaste tube, but it does not remain as solid as I want. I was considering adding more xanthan gum powder, a stabilizer found in toothpaste.


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Stability of alkenes
The following question was asked on myPAT website:

I started with the solution by counting the hyper-conjugative structures available to each alkene. Based on this concept only, order is:
I > II = III > IV
Then, I noticed that II and III have different carbon count. As, n-hexane (~68 °C) have more thermodynamic stability than n-pentane (~36 °C), therefore the order becomes:
I > II > III > IV
But, option B was incorrect. What I am doing wrong?

The correct option is:

The queston involves heats of formation. Check here: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/thermo/heats-formation.html#Alkenes

@user55119: what should the order be? (acc. to you)

They only want an answer based on hyperconjugation in which case you have a point. As to heats of formation, II and III should be switched. But the atom count makes a difference.

You are erroneously associating the boiling point of the substance with its thernodynamic stability. As mentioned by user55119, the heats of formation are a better indicator for stability.

@TanYongBoon: Okay, I admit that I'm in chaos. I know that lesser the $\ce{\Delta H_f^°}$, greater is the stability. Using this concept and some data from the table (provided by user55119), I can write the stability order based on heat of formation only as: I > II > IV > III. Hence, resultant order comes out as given in option B (as hyperconjugation is more dominant factor). Now what I'm supposed to do with my life?

@rv7 Ok this question probably doesn't need you to figure out the answer using experimental data. You probably just need to first consider the degree of substitution of the alkene C=C bond and then secondly, in the case of the two di-substituted alkenes, the length of the alkyl groups attached to the C=C bond. The first consideration is, of course, legitimate but the second is slightly dubious, in my opinion. A longer alkyl chain does not actually increase the electron-donating ability of the alkyl chain. But in this case, I suppose the question assumed so.

See here for the discussion on this topic: https://chemistry.stackexchange.com/q/47008/44877


text:
Is pH in a charged hydrogel and its supernatant solution constant?
Let us assume we deal with ideal systems without interactions.
The gel phase and the supernatant solution phase are in thermodynamic equilibrium. The supernatant solution shall consist of different ion species $i$.
The gel is in free swelling equilibrium and the total chemical potential of each ion species $i$ is constant throughout the supernatant phase and the gel phase:
$\mu_i^\mathrm{gel}=\mu_i^\mathrm{supernatant}$.
Since the total chemical potential is constant we can just write $\mu_i$ and forget about the phase in which it was determined.
Under the assumption that gel phase is always electroneutral we obtain a Donnanpartitioning of ions:
$c_i^\mathrm{gel} \neq c_i^\mathrm{supernatant}$
compare: https://en.wikipedia.org/wiki/Gibbs%E2%80%93Donnan_effect
The difference in concentration between the two phases is allowed (at same total chemical potential) because the total chemical potential not only has the ideal part but also an electric potential part (needed due to the electroneutrality constraint):
$\mu_i=\mu_i^0+kT\ln(c_i/c_0) +z_i e_0 \Psi$.
Same total chemical potential at different ion concentration therefore gives rise to the Donnan potential (https://en.wikipedia.org/wiki/Donnan_potential):
$\mu_i^\mathrm{gel}=\mu_i^{supernatant} \Leftrightarrow \Psi^\mathrm{gel}-\Psi^\mathrm{supernatant}=\frac{kT\ln(c_i^\mathrm{supernatant}/c_i^\mathrm{gel})}{z_i e_0}$
Now the question is: Is pH constant (case 1) or is it different (case 2) in the gel phase and the supernatant solution phase.

If pH is defined based on the total chemical potential, then it is constant (because chemical potential is constant) and we have the same pH in the gel and in the supernatant phase:
$pH=-\log_{10}(a_H)=-\log_{10}(\exp(\frac{\mu_H-\mu^0}{kT}))$

If we define pH based on the concentrations (or making use of the mean activity coefficient $\hat{\gamma}_i$=1, in which the Donnan Potential cancels), then the pH is different in the gel and in the supernatant phase: $pH^\mathrm{gel}=-\log_{10}(c_H^\mathrm{gel} \hat{\gamma}_H^\mathrm{gel}/c_0) \neq pH^\mathrm{supernatant} =-\log_{10}(c_H^\mathrm{supernatant} \hat{\gamma}_H^\mathrm{supernatant} /c_0),$
where $c_0$ is the standard concentration, i.e. 1mol/l and where the inequality arises from $c_H^\mathrm{gel} \neq c_H^\mathrm{supernatant} $ (due to Donnan partitioning and due to the mean activity coefficients $\hat{\gamma}_H^\mathrm{gel}=1=\hat{\gamma}_H^\mathrm{supernatant}$ for an ideal system.

What is correct? What would we measure when measuring the pH in the gel? Would we measure the same pH as in the supernatant phase (i.e. the first case is correct) or would we measure a pH different from the supernatant phase (i.e. the second case is correct?)

You can view the problem also from a different perspective:
pH is defined by IUPAC (https://goldbook.iupac.org/terms/view/P04524) $pH=−\log_{10}[a_H]$, where $a_H$ is the relative activity https://goldbook.iupac.org/terms/view/A00115): $a_H=\exp(\mu_H-\mu_H^0)$
The question is do I need to use

the total chemical potential in this definition of the activity or
this definition of activity which you also find in literature (e.g. in this article "THE DONNAN EQUILIBRIUM" by Stell and Joslin, https://www.ncbi.nlm.nih.gov/pubmed/19431690):

$\mu_H=\mu_H^0+kT\ln(a_H)+z_H e_0 \Psi \Leftrightarrow a_H=\exp(\mu_H-\mu_H^0-z_H e_0\Psi)$ ?
In case 1) pH would be equal across both phases, in case 2) pH in both phases would be different: we could write $pH^\mathrm{gel}-pH^\mathrm{supernatant}=-\log_{10}(\exp(\mu_H^\mathrm{gel}-\mu_H^\mathrm{supernatant}-z_H e_0 (\Psi^\mathrm{gel}-\Psi^\mathrm{supernatant}))
=-\log_{10}(\exp(-z_H e_0 (\Psi^\mathrm{gel}-\Psi^\mathrm{supernatant}))$

You would be measuring the activity,  which is equal across the sample provided again that it is in equilibrium and that your pH probe is not perturbing that thermodynamic equilibrium.

According to IUPAC $pH=-\log_{10}(a_H)=-\log_{10}(\frac{c_H}{c^{ref}} \gamma_H)$ where $\gamma_H$ is the activity coefficient of H. However, only mean activity coefficients are measurable  and therefore gamma_H is the mean activity coefficient. The Donnan potential (arising already purely from the electroneutrality condition) cannot be part of the mean activity coefficient since it cancels in the computation.
The electrochemical potentials of H are equal in both phases due to electrochemical equilibrium. The Donnan potential gives rise to a difference in chemical potentials between both phases and therefore $a_H$ is different in both phases, i.e. also pH is different in both phases.
Update
See also suspension effect for experimental evidence in a similar situation https://www.degruyter.com/document/doi/10.1351/pac200779010067/html#:~:text=Guidelines%20for%20potentiometric%20measurements%20in%20suspensions%20Part%20A.,2002%22%20%5B%20Pure%20Appl.%20Chem.%2074%2C%202169%20%282002%29%5D Guidelines for potentiometricmeasurements in suspensionsPart A. The suspension effect(IUPAC Technical Report)

Note that activity coefficient takes into account the non-ideality due to electrostatic interactions. For instance, you can look up for the activity coefficient of ions in the Debye-Hückel theory. This indicates that the electric part of the hydronium ion potential is already included in its activity coefficient.
So, since the activity coefficient of a species is the same across co-existing phases, pH (according to pH=−log10(aH) definition) should be the same, too.

One really needs to be careful here. Surely the activity coefficient accounts for electrostatic interactions within a phase. However (to my understanding) the https://en.wikipedia.org/wiki/Donnan_potential is a Galvani-Potential difference which is not accounted for in the definition of pH (which makes use of the mean activity coefficient). Therefore, pH should be different in both phases.


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Magnesium Bicarbonate and Ascorbic Acid Reaction
30 ml. Magnesium Hydroxide USP is added to 2 l. carbonated water to make magnesium bicarbonate to be used as a highly bio-absorbable Mg dietary supplement. I'm told the result is 90 mg. elemental Mg per each 4 oz. of this "Magnesium Water."
If one gram of ascorbic acid is added to a small amount of this Mg solution that has been diluted with filtered tap water, the solution becomes fizzy releasing CO2.  I didn't measure the amount of Mg solution or the amount of water, but it was approximately 2 tablespoons of dilute solution with one gram (measured) vitamin c crystals. I do understand this release of CO2 is due to the acid-base reaction.
How are the ascorbic acid and the magnesium changed in this solution? Is there still any vitamin C in the solution? What form of Mg is the result?
This is not a homework question. It's just for personal curiosity, but I only took two semesters in chemistry in the early 80's so my knowledge of the subject is very limited. Sorry if this is the wrong place for such a question. You can delete it.

It is a common reaction between an acid and a salt of a weaker acid. Have you ever tried mixing of solutions of a baking or washing soda and vinegar/citric acid ? It is the same kind of reaction.Note that neutralized ascorbic acid is much less stable, vulnerable to fast oxidation.

Ascorbic acid reacts with bicarbonate:
$$\ce{H2Asc  + 2 HCO3-  -> Asc^2- + 2 H2O + 2 CO2 ^}$$
Ascorbate as a strong reduction agent reacts with oxygen:
$$\ce{2 Asc^2-  + O2 + 2 H2O -> 2 DHA + 4 OH-}$$
where $\ce{DHA}$ is dehydroascorbic acid.
If the ascorbic acid solution is alkalized by sodium hydroxide, the solution slowly turns yellow as the result of the oxidation progress.

What happens to the Magnesium in the reaction? Does the reaction produce Magnesium Ascorbate?

Nothing happens to magnesium. It is a bystander in the reaction. It's complementary anion has been replaced, ascorbate instead of bicarbonate.

As the general rule, ascorbic acid should avoid basic environment like bicarbonates, if it's stability is the concern.


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How does different pH affect copper electroplating
I'm currently investigating the effect on the concentration of hydrogen ions in the copper sulfate solution on the amount of copper deposited on the cathode within a certain amount of time.
I've been trying to find an answer online but have had no luck in getting a clear cut answer. I was wondering if there is anyone here willing to provide me with some insight or a suggestion for the method to carry out this experiment.

If your solution is too acidic, then hydrogen starts reducing instead of copper. If your solution is not acidic enough, your copper will hydrolyze and fall out of the solution. It is as simple as that.

Is there a more in-depth explanation as to why that happens?

Not really. Why? Chemistry has no depth. If you dig deeper, you'll find yourself in physics.

No cause it's for a lab report and I kinda need some more explanation for it.

Then write both reactions (those of electrolysis and hydrolysis) and call it a day.


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How can batteries reach equilibrium if the reactions are irreversible?
My book says that the batteries we have considered so far (Daniell cells) have a low life span because they reach equilibrium too quickly.
In the next paragraph, it states they are also made by irreversible reactions and we call them "primary batteries."
How can a reaction reach equilibrium without being reversible?

The path to an equilibrium state does not need to be reversible. What do you think would stop it from reaching equilibrium?

So what is an equilibrium state? I thought the only way to reach it is being a reversible reaction ( the book also says that the Daniell cells are in dynamic equilibrium Zn+Cu2+  <------> Zn2+ + Cu). Plus, here( https://www.siyavula.com/read/science/grade-12/chemical-equilibrium/08-chemical-equilibrium-01) says that an equilibrium is a state of a reversible reaction.

Sorry, I have deleted it.

So is my book wrong?

Maybe you can include the passage from the book, or a link to the text as an image (in whatever language) as a temporary solution. It is not possible to pass judgement based on the info you have provided. You have now two questions asking the same thing, no? If so please remove one of them.

Okay, here is the link where it explains the potential difference and the reversible reaction that happens in the Daniell cell: https://drive.google.com/file/d/184nNSsQLRjJ9JRISgoGFMRgYAWhmHtzM/view                                                 this other image shows the passage where it says about the irreversible reactions: https://drive.google.com/file/d/1400cc24HHLbDNmRbdjA0ryPWv4v4u0h1/view?usp=sharing                        obs: I have highlighted the passages with thr red rectangles.

It's good to look at the original: it basically says that the discharging process of the Daniell cell is considered irreversible. The answer Karl gave you in https://chemistry.stackexchange.com/questions/122169/question-about-reversible-reactions-and-batteries is correct. Basically it is not possible to bring the cell to its original state - to recharge it completely - because of side reactions.

Your confusion probably stems from the definition of a reversible path between two states: an infinite sequence of steps along a continuum of equilibrium states, such that at each step the equivalence condition of the second law of thermodynamics, $\mathrm dS_\text{total}=0$, is satisfied.
However, there is no requirement that a path between two states must be reversible. Many textbook problems are based on comparing the entropy changes associated with reversible and irreversible paths. For the reversible paths, $\Delta S_\text{total}=0$, for the (spontaneous) irreversible ones, $\Delta S_\text{total}\gt0$.
Your textbook explains that the discharging process of a Daniell cell is considered irreversible. The usage of the term irreversible in the book is rather similar to the common usage, which means "impossible to bring back to its original state, for practical reasons.". The answer that Karl has given you is correct. Basically it is not possible to bring the cell to its original state. The chemical and physical integrity of the recharged electrodes and electrolyte are altered compared to the original structure. There are entropic costs associated with cycling the cell.


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What dictates cathode vs anode nomenclature use?
what was the justification for the use the cathode and anode terminology in IEF?
There is no redox taking place in IEF, just proteins interacting with an electric field

Who was responsible for this naming system and how can we change it?

Michael Faraday was responsible for the terms anode and cathode more than hundred years ago.
All the confusion regarding the nomenclature will vanish if you do not associate electrostatic signs with these two terms. One should identify the electrode labels with the redox processes rather than the signs.

Cathode: The electrode where reduction occurs
Anode: The electrode where oxidation occurs

These definitions are true whether you have a electrolytic or galvanic cell. This eliminates the need to call the negatively charged electrode as a cathode in an electrolytic cell or the positively charged electrode as the cathode in the galvanic cell. These definitions are self-consistent and hence nobody bothered to changed them.
Isoelectric focusing is "sort" of an an electrolytic "cell" hence its cathode is has a electrostatic negative sign because if we were to remove this gel and replace it with just salt water, we will see hydrogen evolution (= reduction of water) at this electrode.

thanks for the answer, I still dont understand why we use that terminology in IEF, there is no oxidation or reduction taking place, just interaction with an electric field. How can oxidation (loss of negatively charged electrons) result in anything other than a positive charge? the two things seem very closely intertwined

Yes, as I said if you dip the same two electrodes used in IEF in an aqueous salt solution (i.e., remove the gel), you would  see reduction occurring at the electrode with a negative charge. Therefore that electrode is the cathode.

I think the point of confusion for me is that to my understanding electolysis requires an electrical current to drive a reaction that is non-spontaneous (putting electrons on a negative species) are you saying that the cathode and anode used in IEF is made via electrolysis?  I appreciate you taking the time to engage in conversation with me on this

Think of it this way: There are two electrodes X and Y in isoelectric focussing. $Disconnect$ the leads and let us say you dip these electrodes in a salt water. If electrode X shows evolution of hydrogen, you will call electrode X, the cathode. If electrode X shows oxygen, you will label electrode X as the anode.

Okay now I get it. Thank you very much. this is why you called IEF "sort of" an electrolytic cell. the cathode and anode are analogous, but bc of the gel they dont change

The good mnemonic tool to remember is :

Anode = anabasis, electrons would be going upwards from the electrode to the wire = oxidation,  ( Xenophon, Anabasis, 404BC, "The journey upwards(to north))
Cathode = cathabasis(the journey downwards), electrons would be going downwards from the wire to the electrode = reduction

That means a cathode/anode is the electrode where reduction/oxidation occurs, what means the flipped signs for a cell being a source/recipient of energy.
A cathode is:

the negative electrode during electrolysis  ( electrons going down the wire, causing reduction)
the positive electrode during the cell provision of electricity( electrons going down the wire, causing reduction)

And vice versa for an anode.

Anabasis/catabasis has application in other domains as well. E.g. ababatic/cathabatic processes in synoptical meteorology are processes related upward/downward convection of air,like at the front and the back side of the cold front of the 2nd kind.


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Why is E°(cell) not ∆E°cell?
$E^\circ_\mathrm{cell}$ is the EMF of the cell for a reaction with respect to standard hydrogen electrode (SHE). So why don't we represent $E^\circ_\mathrm{cell}$ or $E_\mathrm{cell}$ by $\Delta$ symbol, in the same manner in which we represent $\Delta H^\circ$, $\Delta G^\circ$, $\Delta S^\circ$, etc., with respect to initial absolute zero temperature?

By convention, we write $V$ or $E$ to represent a difference in potential. Adding a $\Delta$ would refer to the difference in potential difference.


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Removing or hiding (white) magnesium carbonate residues on natural rock surfaces
As noted in this question, magnesium carbonate is often used by rock climbers to help mitigate the effect of sweating whilst climbing. While this is not a major issue on artificial walls, the residue left by climbers on natural rock is unsightly and has lead to climbing bans or adverse public perceptions.

The above picture shows a hand hold used by climbers (ca. 20cm wide) that is covered in a white residue from climbers chalk. Other likely components of this residue include salts from sweat, oils/lipids from the climbers skin, and skin particles. The rock feature is protected from the weather by the overhang above which has allowed the residue to build up over many years. Conscientious climbers attempt to brush off as much residue as possible (typically using a soft bristle brush to minimise damage to the rock substrate). Adding water to these residues makes them invisible but they reappear upon drying.
Can anyone suggest method(s) for effectively removing MgCO_3 residues from natural rock surfaces without significantly affecting the underlying rock substrate (typically, oxidised quartzite, granite, sandstone, or limestone)?
Alternatively, can anyone suggest a way to conceal with 'whiteness' of the residue (e.g via some staining process or a specific reaction) to reduce the visual impact?
Any method needs to be effectively non-toxic to prevent significant adverse impacts on the surrounding ecosystems.


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How to achieve Löwdin transformation?
I use Gaussian 09 to obtain the Hamiltonian and overlap matrices, $H_0$ and $S_0,$ where the B3LYP functional and the 6-31G basis set are adopted. Then how to
transform the system to an orthogonal basis using Löwdin transformation?
$$
\begin{align}
H_1 &= S_0^{-\frac{1}{2}}H_0S_0^{-\frac{1}{2}} \\
H &= U^\dagger H_1U
\end{align}
$$

Are you wondering how to perform Löwdin population analysis in Gaussian?

No, I just want to make an orthogonal transformation of the Hamiltonian matrix.

https://chemistry.stackexchange.com/questions/22124/unitary-transformation-of-the-hf-equations/22134#22134


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I want a certain Ion in my solution. How do I choose which chemical to buy?
thanks for taking the time. I am physicist doing experiments in which I need certain Ions (at the moment, Ru(II)) in solution. I have no chemical training excepting what I'm learning right now. What I need is a research strategy on how to find which salts or compounds containing Ru(II) will dissolve in water in such a way, that the Ru ions are available to the biological system that I'm studying. Do you have any resources that you could point my way or any good starting points? Thank you again

There are only so many chemical elements out there, and you couldn't possibly be the first who thought of introducing Ru at biological systems. Look at your predecessors, and do the same.

This may assist  https://www.sciencedirect.com/science/article/pii/S1878535216301198

Thank you you two, I totally agree that I could google procedures for every trace metall I want to try, its more about finding a general strategy for someone new to chemistry. If there isn't one, thats fine too obviously


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Neither Raman nor IR Active vibrational modes
Is there any molecular vibrational mode (or activity) that is neither Raman nor IR active?
In other words. Is there any activity in the molecules that can't be captured by either Raman spectrometer (all wavelenght) or IR absorption spectrometer (all wavelenght)? What are those?

Why, there are many such modes (silent, as they are called).

Please give some examples of what they could be. Thanks.

A molecule's normal modes are determined by its symmetry, the selection rules for transitions by change in dipole (IR) or change in polarisability (raman) which can only be satisfied  by certain normal modes. If you look at point groups tables there are many irreps (symmetry species) with neither ($x,y,z$) for IR, ,or product operators ($xy, x^2$, etc), Raman. The silent vibrational frequency can sometimes be worked out by comparing the measured heat capacity with that produced via the partition function using the measured frequencies, the discrepancy is the unknown vibrational energy.

Yes it is possible. While no normal modes can be both IR and Raman active in molecules with a centre of symmetry, as mentioned in a comment molecules with silent modes exist.
The easiest example is  probably the HCH out-of-plane twisting of ethylene:
Source: http://www.shodor.org/succeed-1.0/compchem/labs/vibrations/ethylene.jpg
Due to the symmetry of the molecule, when the HCH twist out of plane the dipole and polarizability of molecule is unchanged.
Other examples can be found here
https://www.chem.purdue.edu/jmol/vibs/c6h6.html
where the modes of benzene are sketched. The silent modes easily seen are kind of distorted breathing of the ring, that due to the C6 axis of symmetry again let the molecule unchanged both for the electric dipole moment and its polarisability.

Does it mean that if a molecule has a change in dipole moment or has electron cloud polarizability, they are always IR and raman active? Are there change in dipole moment or cloud polarization that can't be detected by any spectrometer?

@Jtl active mode doesn't implies the intensity of the abs. or scatt. to be high. The latter is always small and just lasers made raman possible. The effect is always there, just not detected using a standard light source. More close to your Q: for IR smaller is the el. dipole change weaker is the vibr. band intensity. For Raman the same though the change involves polarisability.In Raman, is even more complicated as coupling of vibrations with the el. transitions can lead to enhancement of bands. Vibrations within a chromophore can be greatly enhanced, so that remaining vibrations " vanishes".

for water/h2o which is both raman active and IR active. Is there a way to influence the molecules without it affecting either the dipole moment or polarization? How specificaly for water/h2o?

"Is there a way to influence the molecules without it affecting either the dipole moment or polarization?".Not sure what you mean. If you mean if there is a normal mode that is silent, no. H2O has 3 normal modes which are both IR and Raman active. Its symmetry isn't high enough.

Using this glass vial on a 532nm raman spectrometer https://imagizer.imageshack.com/img924/4674/00uhyz.jpg  How come at some distance I get this raman curve with some unknown peak at near 500 1/cm, https://imagizer.imageshack.com/img924/9114/MsOBdQ.png  then at some distance, the peak vanishes https://imagizer.imageshack.com/img924/4361/eZs05H.png? What material do you think is the vial? Thanks.

Tentative: That region might sense impurities that are hydrogen bound and vial walls might be involved. Assuming T to be reasonably constant you could see if the band change with time. Moreover I am not sure that the band really vanished if normalized to the main vibrational bands. You are at the detection boundary subtraction of noise and background might be tricky and you have basically two different setups (distance 1 and distance 2).

What do you mean by the "band change with time"? How do you adjust the time? I think by focusing the laser focal point right in the middle of the vial. It can sense just the water and not the wall and this is the second spectra with less disturbance from the wall? note the raman spectra of Boron silicate is exactly that region with the peak, see: https://www.researchgate.net/figure/Raman-spectra-of-borosilicate-glass-at-ambient-condition_fig1_234945622

I mean the time water stayed in the vial. If you are near a vial signal then it just matter of focusing / reflection and I would treat it as cut-off, meaning that region not very informative about water anymore. Note that 1) things are now totally unrelated to the Q 2) people can't guess details


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Calculating the Molecular Orbitals of a Molecular State
I am performing ab initio calculations on a heteronuclear diatomic compound using the MRCI method on the MOLPRO quantum chemistry package. I obtained the molecular states of the compound but I was wondering how to find the molecular orbital configuration of each electronic state in addition to their associated compositions. I tried to include the "print, orbitals" before and after the CI, but I didn't get anything!
Your help and support are much appreciated...

MRCI does not optimize the molecular orbitals. They remain in whatever you used as input orbitals for MRCI (typically from a preceding CASSCF calculation).

I used HF and MCSCF before the CI calculations. Are they generated in the HF or in the MCSCF?

both methods generate orbitals. but what you are interested in are the MCSCF orbitals (assuming your calculation input is set up appropriately). CASSCF is just a special case of MCSCF.


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Does air dissolve in the R-113 (i.e. Freon 113)?
Can anyone tell me whether or not air or N$_2$ dissolves easily in the R-113 (liquid solvent) under the saturation pressure of R-113? If yes, what is the solubility roughly? From this link one can only find its solubility in water.
It will be better if you could provide a link. Thank you in advance!
PS: In Table V of this data provided by @Jon Custer, I found no solubility data for air (or N_$2$) dissolved in R-113.

Well, solubility wouldn't be as good as in perfluorocarbons, but it may be tough to get specific data.

https://pubs.acs.org/doi/pdfplus/10.1021/cr60242a003?src=recsys has data...

@Mithoron a little irrelevant... I knew R113 is a chlorofluorocarbon. Did you mean chlorofluorocarbon $\in$ perfluorocarbon or perfluorocarbon $\in$ chlorofluorocarbon or chlorofluorocarbon $=$ perfluorocarbon?

@JonCuster thank you, sir. I checked Table 5 in that article. But there seems no solubility data of air or N$_2$ in R113 (liquid solvent).

@JonCuster I mean all fluor beats half-chloro/half-fluoro.

While data for air solubility in R-113 is difficult to find, Henry's law constants $k_\text{H}$ are available for $\ce{N2}$ in a series of related refrigerants (Ref. 1) comprised of variously halogenated ethanes. However, these are predicted, not experimental, Henry's law constants: the vapor-liquid equilibrium of binary mixtures containing nitrogen was modeled with an equation-of-state based on a perturbed–Chain (PC) modification of the Statistical Associating Fluid Theory (SAFT). I back-calculated the $k_\text{H}$ from the published temperature coefficients. Assuming a $\ce{N2}$ partial pressure $\pu{p_{\ce{N2}}= 0.78 atm}$ then leads to the following $\ce{N2}$ solubilities (as mole fractions) in the refrigerants in the temperature range $\pu{170-280 K}$:

Evidently the solubilities are generally below $\pu{0.02 mol \%}$, which I believe would qualify as "not high" but not negligible ($\chi=1\times 10^{-4}$ corresponds to $\pu{\approx 1 mM}$ solution concentration).
After more careful examination, I found that Vinš et al. cite a publication by Huang et al. (Ref. 2) which compares simulation and experimental (Ref. 3) Henry's law constants for $\ce{N2}$ in R-113, the value at 298 K being $\pu{k_H= 52.5 MPa}$. This is larger than the values above, which are in the range $\pu{1-15 MPa}$. Since $p_\ce{N2} = k_\text{H} \chi_\ce{N2}$, the larger $k_\text{H}$ for R-113 means nitrogen is even less soluble in this refrigerant.
Note of course the analysis here is restricted to nitrogen, but you can proceed in the same way to analyze other gases for which $k_\text{H}$ are reported.
References

Vinš, V., Hruby, J., Solubility of nitrogen in one-component refrigerants: Prediction by PC-SAFT EoS and a correlation of Henry’s law constants. International Journal of Refrigeration, 34, 2109-2117 (2011).
Huang, Y.L., Miroshnichenko, S., Hasse, H., Vrabec, J., Henry's Law Constant from Molecular Simulation: A Systematic Study of 95 Systems. Int. J. Thermophys. 30, 1791–1810 (2009).
H. Hiraoka, J. H. Hildebrand, J. Phys. Chem. 68, 213 (1964).

Thank you for your reply. In the plot, there appears no curve for R113? What is $k_H$ and $T_r$? Most importantly, did you predict that N$_2$ should be not easily dissolved in saturated R113?

@user55777 I updated my answer to address your question.


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Grignard reagent with sulfur dioxide
I have studied that the Grignard reagent $\ce{RMgX}$ reacts with $\ce{CO2}$ to form $\ce{RCOOH}$ after acid hydrolysis. But does the same apply for sulfur dioxide?
Grignard reagents are good bases and also good nucleophiles.
So the alkyl anion will attack the electron deficient carbon atom in $\ce{CO2}.$
But in the case of $\ce{SO2}$ the following conditions aren't present.
First of all, sulfur is more electronegative than carbon. Secondly, sulfur has a lone pair of electrons which the carbon lacks. So this should further discourage an attack by the Grignard's alkyl anion.
So, what actually will happen when Grignards are treated with $\ce{SO2}?$

I'm pretty sure they will react.

A Grignard reagent is basically (pun intended) a bull in a china shop.  It will go after just about anything having a hint of Bronsted or Lewis acidity.

You get sulfinic acids $\ce{RSO2H}.$
This paper by Dowson et al. [1] states that Grignard plus $\ce{SO2}$ gives sulfinic acids and as a reference quotes an older article by Allen Jr. et al. [2] which states

The reaction between sulfur dioxide and Grignard reagent has been widely used for the preparation of aromatic and aliphatic sulfinic acids.

I would suggest you read [2] and the references it quotes.
References

Dowson, G. R. M.; Dimitriou, I.; Owen, R. E.; Reed, D. G.; Allen, R. W. K.; Styring, P. Kinetic and Economic Analysis of Reactive Capture of Dilute Carbon Dioxide with Grignard Reagents. Faraday Discuss. 2015, 183, 47–65. https://doi.org/10/gf9hvn. (Research Gate — PDF)
Allen, Jr., Paul.; Rehl, W. Richard.; Fuchs, P. E. Sulfone Formation during Sulfination of the Alkyl Grignard Reagent. J. Org. Chem. 1955, 20 (9), 1237–1239. https://doi.org/10/d42z3x.


text:
What makes the odor of flatulence fade away?
If an individual expelled their intestinal gas in an enclosed space, that smell would seemingly fade away. Why is this so? I would think of 3 possible explanations:

The gases dilute in the air. If that space were tightly enclosed and sufficiently small, that smell would persist.
The odorous gases decompose upon contact with air or by exposure to light.
The odor does not fade away but the sense of smell adapts to the constant concentration. If another individual sampled that specimen later, they would experience the original intensity.

Why just flatulence? :)) Likely 1 and 3, but it would be the case for other gases as well. And certainly there are odorous compounds that might decompose as well.

At least for compounds like H2S, the odor goes away as molecules bind to the sensor sites (not an adaption per se, mind you).


text:
How to get rid needle clogging on electrospinning of PLA/PCL/PMMA/Collagen?
For my thesis research, I dissolved PLA and PCL in chloroform: methanol (3:1), PMMA (350.000 MW, 13 wt%) in acetone, and collagen in acetate acid--and then, I mixed them together. The problem is when I tried to electrospinning with this solution, the needle clogging happened every time. It causes uncontinuous fiber synthesis. Fortunately, when I tried to observe some fibers from the solution, the result was 'no beads' appear.
I use needle 21 G with electrospinning parameters 20 kV and flow rate 2 ml/h.
Is there any suggestions for me on how to get rid needle clogging? Thank you.

I'm voting to close this question as off-topic because it's technical issue.


text:
What is the true meaning of Gibbs energy and chemical potential?
After doing some research and reading I found some problems which I will try to state as clearly as possible.
The definition of Gibbs Free Energy says "the greatest amount of mechanical work which can be obtained from a given quantity of a certain substance in a given initial state, without increasing its total volume or allowing heat to pass to or from external bodies, except such as at the close of the processes are left in their initial condition. (I have taken the definition from the wikipedia). The equation of the Gibbs energy $$ G= H -TS$$ is also clear.
But the way I understand it is this: the maximum non-expansion work , like transporting an electron, breaking a chemical bond, moving real life things etc.  which can be obtained from a system . These things made sense during the study of Second Law of thermodynamics but as I moved to Chemical Equilibrium these concepts began to tremble (for me).
In elementary classes it is said that Equilibrium is a state where the composition of reactants and products do not change over time, but in higher chemistry classes it is said that equilibrium is a condition corresponds to $$ \Delta G = 0$$ So, my first question is how do these two concepts mean same thing?  The next thing which causes problem is chemical potentials. If we adhere to the formal meaning of potential i.e. something which is stored and can be used when proper conditions are met, so chemcial potential would mean the potential of substance to react and again this is related to Gibbs energy whose definition I gave above. So, how do chemical potential and Gibbs energy can have any relation?
A question which is off topic over here but I want to mention it, why do we bother so much about standard  things like $$ \mu_\mathrm A = \mu_\mathrm A^\circ + RT \ln(p_\mathrm A)$$ why we wanted to express it in that standard (that little circle) form ?
I want to make myself clear that the conception of Gibbs energy was quite clear to me in the context of thermodynamics, we simply meant it to be the work which can be extracted from a substance, but it all-pervading use has made me to doubt myself, just like in mathematics the number $\mathrm e$ appears in odd places. Even if Gibbs energy (according to the understanding that gave above in bold) is appearing mathematically then also it would be having some  physical meaning because Thermodynamics and Equilibrium are natural sciences and not the mathematics.
Thank you, any help will be much appreciated.

Your equation gives the chemical potential of a species in a gas mixture at a partial pressure of $p_A$ bars in the mixture.  In this equation, $\mu^o_A$ is the molar free energy of the pure species at 1 bar and at the same temperature.

@ChetMiller Your such a small comment has solved my such a big problem. I’m really grateful to you. Can you please tell me one more thing: what is the meaning of chemical potential the way you understand it?

I understand it as the partial derivative of the total Gibbs free energy with respect to the number of moles of the given species holding the number of moles of all other species (and T and P) constant.

@ChetMiller Okay, means chemical potential means that how much the Gibbs energy of a system gonna change if change the number of moles of a particular substance. Am I right?

Yes, you are right. Chemical potential is the partial derivative of the Gibbs energy per molar amount of the compound.

Your first question:
Before the second law was understood it used to be thought that the maximum amount of work that could be extracted from a reaction was $-\Delta H$ but many experiments showed that this was not the case. It is true that, according to the first law, (the law of conservation of energy),  that external work done must be equal to the loss of energy of the system, unless some heat is given to or taken from the surroundings. This was first understood by Gibbs.  In an isothermal reaction working under  reversible conditions, the heat absorbed from the surroundings is $T\Delta S$, if this is positive then the work done will be even greater than the heat of reaction and so heat will be taken from the thermostat.
Second question:
The $\Delta G$ you mention is actually the gradient of the free energy with extent of reaction $(\partial G/\partial \xi)_{T,p}$ at constant temperature and pressure. This differential is sometimes written as $\Delta G'$. The equation in general is $\Delta G'=\Delta G^\text{o}+RT\ln(Q)$ where $Q$ is the ratio of partial pressures, for example, in a gas phase reaction.  At equilibrium a plot of $G$ vs $\xi$ reaches a minimum and the gradient is zero, i.e. $\Delta G'=0$ then $\Delta G^\text{o}=-RT\ln(K_p)$  where $K_p$ is the equilibrium constant and is used instead of $Q$ at equilibrium. This reaction tells us how the position of equilibrium can be determined in terms of standard state free energies of reactants and products at 1 atm pressure.
(The extent of reaction is zero when only reactants are present and is one when one mole of reactants have changed to product)

Thank you, your answer is very nice. Can you please write something on chemical potential too?

@porphyrin When we say that at "equilibrium" it reaches a minimum do we mean to the equilbrium state of both system and surroundings? I understand that $\Delta G'$ must be zero because the entropy (of the isolated system, system + surroundings) should stop increase at equilibrium. Is it that correct or the condition for $\Delta G'=0$ comes from the fact that at chemical equilibrium the chemical potentials are the same?

yes system and surroundings so chemical potential the same.


text:
Electric fields in a galvanic cell
I have read hat there is two types of electric fields exist in galvanic cell.I understand E because of charges accumulated on electrodes but what about other E -field? What does cause this field and what is the direction? Detailed answer with explanation in terms of ions and charge will be helpful.
Full thought and Reasoning : In a galvanic cell when we immerse electrodes in solution,both electrode get charged.This building of charges is the result of chemical reactions on both sides. Now we in this process of building charges on the electrode if we see individual electrode ,lets say on one electrode there is negative charge (-ve) is building due to chemical reaction.This reaction also result in extra metal ions in solution which will be equilibrium with the electrode in joining the it and again separating with it again. Now I have understood till now .I know that there will be electric field between electrode due to these charges built  on electrodes.My question is what about the ions in the solution of negative electrode(-ve ) side  and lack of ions on the positive electrode side ? Will there be any Electric field from Positive to Negative electrode due to these ions (there are in different solution & different beakers )?
Example : Let's say We have a circuit containing battery and a resistor.In that circuit
In above image what is fs  . it is adding to E which means it has to be some kind of Electric field.From where did it come ? That's only my question .Thank you

We have a policy which states that you should show your thoughts, effort and attempts to answer your question yourself. It'll make us certain that we aren't doing your homework for you, and that the Q/A is beneficial for broad audience.   As "homework-like questions" are considered literal homeworks, self-study questions, puzzles, worked examples etc.  Please [edit] in your full reasoning or thoughts on this. See Homework.  Otherwise, the question may get closed as "low effort homework-like question."

I did edit it sir .

Still no thoughts, reasoning and effort to answer your question. You just more or less summarised your previous question and answer to it. Involve basic knowledge of electrostatics.

A hint: The beakers/electrode compartments are connected in sense on ion mobility.

Yes ! by salt bridge .So the ions in these compartment make a Electric field which is directed "due" to excess of Positive ions in one compartment to  lack of Positive ions in other compartment ?

Does displacement of negative and positive charges create electrostatic field ? Does a field cause charges to move ? Does change of charge displacement affect the field ?


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Question About Reversible Reactions and Batteries
So my book says that the batteries we have considered so far( Daniell cells) are low life span, because they reach equilibrium too fast. In the next paragraph, it says they are also made by irreversible reactions and we call them primary battery. I know that the reactants and products of Daniell cell reach equilibrium (Zn+Cu2+ <------> Zn2+ + Cu). How can one reaction reach equilibrium without being reversible?

Can you please identify that book?

I am brazilian, but the book is called Química: Cotidiano e Transformações( Chemistry: Day by Day and Transformations). I can translate some paragraphs if you wish.

A battery has a small shelf life if it discharges internally. Saying that a discharged battery is in chemical equillibrium (or much closer to it than a fresh one) is a true statement, obviously, but nothing else.
If you reverse the polarity of a primary battery, you get some other reaction than the one happening during discharging backwards. That says something about that specific battery type, not about the principal reversibility of the basic schoolbook reaction. Typical case is that the electrolyte is electrolysed (e.g. water to hydrogen and oxygen) instead.
You could sometimes avoid that side reaction by e.g. using a lower voltage for charging, but of course then you can only charge the battery up to a few percent or so.

But a chemical equilibrium means reactants are becoming products and products are becoming reactants. So my book is wrong by defining that the reactions inside a battery are irreversible?

Quite wrong. Primary cells can usually be recharged a bit, only some significant part of the energy goes into their destruction.

Thanks!!! One more thing, what you consider as chemical equilibrium is the dynamic equilibrium of reaction?

@SonimBlenim Can you rephrase that last question? Not sure what you mean.


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Super-absorbent Polymers
I want to cast a cube out of super-absorbent powder (Fake Snow) for an art piece, and make it become shapeless by gradually exposing it to water over 7 days. I have 2 questions:

What is the best type of glue that I can use to stick it together while keeping it exposable to water + keeping them together after expansion?
Would it expand with slowly dripping water over 7 days? (All the videos I watched recommend "pouring water as quickly as possible")

this is two questions.  2. could you elaborate your questions with more detail of what you want?


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qualitatively: Why are thioesters more reactive than regular esters?
In ochem students are taught that shorter bonds (like alkynes) have more energy than longer bonds such as alkanes. However in biochem we are taught that thioesters are very energetic bonds and release alot of energy upon breaking.
Is there a specific mechanism that explains the relationships between bond energy and length?
background: a university student with a chem minor

Related: Atomic orbitals: how is bond strength related to stability, energy, bond length?; Are stronger bonds always shorter?; Bond length and bond strength correlation (possible dupes).

helpful links, a bit over my head tbh, I was hoping for a discussion is more simplified terms

https://chemistry.stackexchange.com/questions/42848/why-is-the-thioester-bond-weaker-than-a-regular-ester-bond

http://guweb2.gonzaga.edu/faculty/cronk/CHEM440pub/thioesters.html
"thioester cannot make as much of a contribution because the required overlap between carbon's 2p orbital and the sulfur 3p orbital is ineffective. The large difference in energy between these two orbitals means they cannot "mix" as well as the 2p orbitals of oxygen and carbon."


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Any way to see "cubic-active" vibrational modes in molecules?
For molecular complexes with a moderate amount of symmetry there are usually vibrational modes that are neither infrared nor Raman active. Sometimes these modes transform as some cubic polynomial in x,y,z. For example the $B_{1u}$ irrep in the $D_{4h}$ group which transforms as $xyz$.
My question is: it possible to measure these kinds of vibrations using some sort of non-linear optics? For example, measuring the absorption of three photons instead of just one like in infrared spectroscopy/reflectivity. In the case of $B_{1u}$ above, this would translate to 3 photons polarized with x,y,z respectively and whose total wavenumber matches a $B_{1u}$ vibration. Or perhaps a more complicated multi-photon version of Raman spectroscopy could do this?

I think you are at the core of some of the nlo phenomena, perhaps THG or stimulated Raman. Can be suitable for Physics SE, too.

The product $xyz$ is an odd function and as such should produce the same selection rules as 1 photon (electric dipole) transitions. Thus CO has a 1 photon transition at 146.7 nm in the vacuum uv that can be more easily reached with a 3-photon transition with a pulsed laser at 440 nm. ( Three photon transitions are also used in bio-imaging where a ir photon is used so as not the excite the sample. Also, as the transition is multiphoton the focus area is smaller than a 1 photon transition and this improves spatial resolution).
As the product $xyz$ can be thought of as $xy$ multiplied by $yz$ then this is like a double Raman transition. Raman involves a change in polarisability. Polarisability is proportional to a volume but the projection for the radiation is 2D hence $xy$ type products in the point group tables. A  Raman type transition may occur between 4 levels. Levels 1 and 4 are  'real' and 2 and 3 'virtual'. The steps are 1 up to 3 , 3 down to 2, and 2 up to 4. This has been demonstrated in doppler free geometry between Na $3^2S_{1/2}\to 3^2P_{1/2}$.

Ah oops, I guess my question is basically moot since all cubic invariants are odd under parity. I assumed they would be too weak to see because of the higher order, but I was wrong. Is there a similar scheme for higher order multipole vibrations that aren't odd under some parity?

Say, for example, quadrupoles?


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Can sulfuric acid attack a hydrocarbon/paraffin? Under what conditions?
Sulfuric acid can attack benzene and other hydrocarbons (when mixed with hydrogen peroxide). Also, chlorine gas and bromine can replace hydrogen(s) in hydrocarbons like methane, ethane, propane,butane etc. in a presence of sunlight and/or mild heating.
Can dil. or conc. $\ce{H2SO4}$ be used to attack hydrocarbons and produce sulfonates? Please cite an example.

These papers seem to be relevant: https://doi.org/10/cskmsr; https://doi.org/10/bcc5dn. Also, conc. sulfuric acid is a potent oxidizer and a dehydrating agent.

TLDR: Sulfuric acid act as catalyst in the isomerization reaction of parafin hydrocarbons and sometimes it produce undesired sideproducts like sulfonate salts, napthenic acid, bisulfite ions etc. from incomplete oxidation. It follows "hydrogen exchange mechanism.

Long answer:
@andselisk cited some papers that shows that sulfuric acid can act as a catalyst for the isomerization reaction of certain paraffin hydrocarbons. It is observed that sulfuric acid act destructively towards these hydrocarbons and isomerizes and some more interesting observation were made i.e which hydrocarbons isomerizes and which don't(emphasis mine):

It is clear from an examination of the experimental results that two
  different reactions took place. Isomerization is the only reaction at
  the milder experimental conditions. Some destructive alkylation
  takes place with 98% sulfuric acid and higher temperatures, as
  indicated by the production of low and high boiling compounds.
  In the experiments with n-octane and neohexane there is no reaction
  even at the boiling point of respective hydrocarbon and 98% sulfuric
  acid concentration. 2,4-Dimethylpentane isomerizes mostly to
  2,3-dimethylpentane, with minor amounts of 2-methylhexane and
  3-methylhexane. The main products of triptane isomerization were
  2-methylhexane and 2,3-dimethylpentane. These results substantiate the
  observations of previous investigators that sulfuric acid is a
  specific catalyst for shifting the methyl group along the hydrocarbon
  chain and that only such paraffin hydrocarbons which contain tertiary
  carbon atoms are susceptible to isomerization with this catalyst.(1)

The mechanism of these reaction were studied previously. Scientist proposed "hydrogen exchange mechanism" from paraffin and sulfuric acid reaction:

Ingold, Raisin, and Wilson were the first to show that hydrogen
  exchange occurs between paraffins and sulphuric acid. It has recently
  been found that such exchange occurs readily only in the cases of
  hydrocarbons that contain hydrogen atoms on tertiary carbon atoms
  (tertiary hydrogens). The mechanism of the exchange process has been
  established by studies of the racemisation of optically active
  paraffins in sulphuric and by a detailed mass-spectrographic
  investigation using mono-deuterated butanes and [13C]isobutane.
The initial step is the formation of a very small concentration of
  tertiary carbonium ions by oxidation $\ce{R3CH + 2H2SO4 -> R3C+ +
 HSO4- + SO2 + 2H2O}$, followed by a chain reaction between the
  carbonium ions and paraffin molecules involving exchange of hydride
  ions between tertiary carbon atoms. An important postulate of the
  mechanism is that the carbonium ion intermediate is able to undergo
  rapid and complete exchange of hydrogen with the sulphuric acid
  solvent, whereas the hydrocarbon molecules are quite unreactive in
  this respect. The chains are thought to be terminated by some
  irreversible side reactions leading to non-volatile products.(2)

In case one is wondering whether some sulfonate salts is formed as a byproduct, then it has been confirmed. The reaction do produce sulfonic acid and other byproducts during the sulfuric acid treatement of wax purification process and these byproduct proved harmful if untreated as they react undesirably(emphasis mine):

Sulfuric acid treatment serves to remove resinous materials, oil,
  unsaturated hydrocarbons, and other unstable substances from crude
  wax, as well as to remove polycyclic aromatic compounds, by treatment
  with 102-103% oleum and subsequent neutralization with caustic.
After sulfuric acid treatment, the wax may contain free sulfuric acid
  or caustic, the presence of which is explained either by incomplete
  neutralization of the acid after treatment or by incomplete washing of
  the caustic used in neutralizing the acid. The presence of either
  acids or alkalis in waxes is extremely undesirable, since such
  materials will cause corrosion of metals under the conditions of
  production, use, and storage. These impurities in commercial paraffin
  waxes are the cause of undesirable processes of oxidation, resin
  formation, and decomposition. Apart from free sulfuric acid,
  derivatives such as sulfonic acids and acid esters may be present in
  paraffin. These are equivalent to the mineral acid in chemical action.
  A negative result from the test of a petroleum product for free acid
  or alkali still does not guarantee that these compounds will not be
  formed in the petroleum product. If the product contains salts of
  sulfonic acids, acid esters, naphthenic acids, or byproduct compounds,
  then the action of high temperature or moisture (hydrolysis) will
  quite probably result in the formation of acidic substances of a
  mineral nature.(3)

References

Reactions of Paraffin Hydrocarbons in the Presence of Sulfuric Acid
V. I. Komarewsky and W. E. Ruther
Journal of the American Chemical Society 1950 72 (12), 5501-5503
DOI: 10.1021/ja01168a034
R. J. GILLESPIE and J. A. LEISTE "THE BEHAVIOUR OF ORGANIC COMUPOUNDS IN
SULPHURIC ACID "(pdf)
DETERMINATION OF ACIDITY OF PARAFFIN WAX by V. S. Glukhov, O. I. Lapitskaya,
E. V. Zosimov, and G. M. Grechko (pdf)
Hydrogen Exchange and Isomerization of Saturated Hydrocarbons with Sulfuric Acid
D. P. Stevenson, C. D. Wagner, O. Beeck, and J. W. Otvos
Journal of the American Chemical Society 1952 74 (13), 3269-3282
DOI: 10.1021/ja01133a017(@andselisk)


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Formation of green precipitation whilst adding potassium fluoride to aqueous solution of copper(II) sulfate

Aqueous copper sulfate solution (blue) gives a green precipitation with aqueous potassium fluoride. Explain these experimental results.

I've read answers like, $\ce{H2O}$ is a weak ligand and fluorine can replace it, but isn't $\ce{H2O}$ a stronger ligand than fluorine according to the spectrochemical series?

Consider the possible products of a double displacement reaction.  Which one is likely to be green?  You may also want to read this WP article.

$\ce{CuSO4}$ exists as $\ce{[Cu(H2O)4]SO4}$ in solution. It is blue in colour due to the presence of $\ce{[Cu(H2O)4]^{2+}}$ ions.
Now if aqueous $\ce{KF}$ is added, the solution turns green due to formation of complexes $\ce{[CuF4]^2-}$ and $\ce{[CuF6]^4-}$(Initially copper(II) fluoride($\ce{CuF2}$) is formed but it formed complexe in present of water). The complete reaction is given by:
$$\ce{[Cu(H2O)4]SO4 + 4KF -> K2[CuF4] + K2SO4 + 4H2O}$$

WP says CuF2 is only slightly soluble in water.  Do we need to resort to complex formation to explain these results, and how is the complex less soluble than the slightly soluble binary fluoride anyway?


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Arrange the following compounds in the increasing order of equilibrium constant for hydration

Arrange in their increasing order of equilibrium constants for hydration:

(a) A < B < C < D < E
(b) A < C < B < E < D
(c) A < C < E < B < D
(d) C < A < B < E < D

How to compare figures E and D? Formaldehyde (figure D) doesn't have steric repulsion. Figure E has steric repulsion along with an electron withdrawing group. Given answer is (b). Does this mean steric hindrance is dominant over inductive effect while comparing? Please explain!

To my knowledge, comparing the steric hindrance of the carbonyl group is not the path you want to follow to answering this question, because the trigonal planar configuration of the carbonyl carbon provides plenty of room in order for the inductive effect to outweigh the steric hindrance. We have to examine the partial positive charge of the carbonyl carbon resulted from the C=O and the other substituents. The C=O bond has a great dipole moment resulting in a partially positive carbonyl carbon. This charge can be stabilized or increased by alkyl substituents (hyperconjugation) or electronegative atoms (inductive effect) respectively. The greater the positive charge on the carbonyl carbon the higher the equilibrium constant for hydration (due to the mechanism). Therefore, we can easily say that (A) < all, because of the two alkyl substituents. C > A, because the chlorine reduces the hyperconjugation effect. B > C (only 1 alkyl substituent). E > B (because of the same reason as the C > A scenario. And lastly D > E, because of the abscense of any alkyl groups (same reason as the B > C scenario). Now, why is B > C and D > E? Why is less alkyl substitution increasing the charge to a greater extent than 1 more alkyl group with an electronegative atom? If the chlorine was attached to the carbonyl carbon itself (attach the chlorine in E to the C=O), it would greatly icrease the charge and in fact this molecule would have the highest Constant of all. However, we have to keep in mind that the inductive effect greatly deminshes the further away you go from the chlorine atom. The carbonyl carbon is 2 σ bonds away from the chlorine (as in C and E), so yes it will feel the induction but the alkyl group will still donate elenctron density via hyperconjugation to a lower extent. In conclusion, the chlorine in C and E is reducing the hyperconjugation effect but it doesn't neutralize it completely. Therefore, B > C and D > E


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How can a methyl cation exist without 8 valence electrons?
So far I have been taught that everything wants to achieve noble gas configuration - 8 electrons in the outer shell.
I came across a methyl cation. Which contains a positive carbon, no lone pairs and 3 bonding pairs of electrons so 6 electrons in total. 6 electrons are obviously less than 8 electrons so how can it exist as it defies apparently the rules I have been taught.

Some say humans can't fly. In fact they can, but not for long. Same thing here.

Methyl cation certainly does exists. What matters about such ions/radicals is how long could they exist stably or alternatively, in what ways and what extent can they be stabilized.
Methyl cation, $\ce{CH_3^+}$ has a sextet of electrons which is not a big problem in its existence. But it is short lived because the extra positive charge is localised in carbon and has no means of either dispersal or diminish it. This makes it short-lived.
But secondary carbocation ($\ce{R_2CH^+}$) gets two alkyl groups to draw electrons from them and be relatively more stable. Tertiary carbocation ($\ce{R_3C^+}$) can draw electrons from all the three alkyl groups and hence is the longest lived and so the most stable.
Please note that noble gas configuration is a widespread rule but not a universal rule. Encyclopedia Britannica asserts that methyl cation is a transient intermediate so it will eventually constitute a compound after reaction.

Some stable molecules exist without $8$ electrons in the outer shell, like $\ce{NO}$ (or $\ce{ClO2}$). Of course they dimerize at rather low temperatures. But at room temperature and usual pressures, these molecules exist and are stable.

The way you put it that everything wants to achieve noble gas configuration is a good rule of thumb for predicting molecules that are stable and long-lived. However, not everything is in equilibrium. A methyl cation certainly exists, and it certainly wants to achieve the noble gas configuration, but it still takes time to do so. So we can conclude that the methyl cation is an unstable, short-lived, highly reactive molecule, which is true. And you are totally right to explain this with the tendency of having a closed electronic shell.

Requiring every atom to simultaneously have a noble gas configuration does not always work.  Cyclopropenyl cation doesn't have that when you draw the Lewis structure(s), the anion does; yet it is the cation which (at least if chilled slightly) is indefinitely stable.

Methyl cations are not unstable by their lonesome but are reactive. Evaluation of stability requires a reaction.


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How long would it take for a tank of same-spin hydrogen atoms to become a tank of H₂?
In the question
Is there an energy cost associated with flipping the spin of an electron?, it is shown that it is very unlikely for two hydrogen atoms to bond if their electrons have the same spin. The reason is the very low rate at which such a system would be able to emit photons. This makes me think that a bottle of $\ce{H1}$ could be prepared if the spin-up hydrogen was separated from the spin-down hydrogen with an apparatus like the one used in the Stern-Gerlach experiment. Is it true that the spin transition can be ignored for "all practical purposes"  (see linked question), or can tanks of spin-aligned $\ce{H1}$ only exist for times uninterestingly short for chemical purposes?

Not long at all.

Assuming the tank survives ...

I doubt this experiment would be a valid "practical purpose". ;) Also spin relaxation typically occurs via energy exchange with the "lattice", not photon emission. Unless your tank had no walls and its content was infinitely dilute, it would present a suitably solid lattice for your electrons.

And wouldnt one stray 21/cm photon self amplify like crazy?

@Karl What if the vessel was lined with a compound that only exposed "spin-saturated" atoms to the gas, atoms whose electrons could not possibly accept any more spin? Is such a material possible?

Hm, nonradiative interactions can change the spin without restrictions, as far as I understand.


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Irreversible Adiabatic Process
While studying about the entropy change of surroundings in irreversible adiabatic process, I came across a sentence.

For infinite reservoir,  all processes are considered reversible

And suddenly it resumes with the equation:
q/T = dS (surrounding)
Can anyone explain the line and how the conclusion was drawn?

Can you provide a reference? What are you using as a convention for the sign of q?

Are you copying the equation verbatim? If using the convention that q>0 when heat is transferred from the system to the surroundings, then the equation makes sense.

q is the heat of the process. From the second law we know I know that q(rev)/T= dS. But to describe it for the surroundings only, the limitation reversible is removed. This is what I thought, but it just confuses me over.

Possible duplicate of Calculating entropy change of surroundings

You'll probably find a partial answer in this post. The reversible/irreversible tag refers to what is happening to the system, not the surroundings. Events in the surroundings are as a rule treated as reversible.
You should also keep in mind that heat is a signed quantity, and that there can be different conventions with regard to what is a positive/negative heat transfer. Usually chemists say q>0 when the system receives energy as heat, and when the surroundings loses energy as heat. Since heat is regarded as the sole source of change in the entropy of the surroundings (assumed to be an inifinitely large reservoir at constant T$^\ast$), we write $$dS_{surroundings} = -\frac{dq}{T}$$ for an infinitesimal heat transfer and $$\Delta S_{surroundings} = -\frac{q}{T}$$ for a finite heat transfer.
$^\ast$ An infinite reservoir can provide or accept energy in the form of heat without temperature change (or any other change in its state).


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Bond length in hydrogen chloride

The interatomic distances in $\ce{H2}$ and $\ce{Cl2}$ are $74$ and $\pu{198 pm},$ respectively. Find the bond length of $\ce{HCl}.$
a. $\pu{136 pm}$
b. $>\pu{136 pm}$
c. $<\pu{136 pm}$

According to me, if $\ce{HCl}$ was a non polar molecule then the bond length would simply be
$$\frac{\pu{74 pm} + \pu{198 pm}}{2} = \pu{136 pm}$$
But as there is polarity the electrostatic force of attraction should reduce the bond length to less than $\pu{136 pm}.$
However, the answer is exactly $\pu{136 pm}.$ I do not understand why.

I suppose they have used the simplified approach, using general covalent atom radius. The true radius is 127.4 pm.


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Why Gibbs energy change of a process can be negative at constant temperature and pressure?
This question has been asked before, but the answers are not clear and satisfactory.
If the pressure and the temperature are constant then G would be always zero. How would then G have a value other than 0?

Why should G be always zero?

The difference in G between two sides of a reaction is not, in general, zero.

G is a function of T and P, but also, in a mixture, the number of moles of each of the species in the mixture.  If these change because of a chemical reaction, G also changes (even at constant T and P).


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How can I know which way to number the carbon atoms?

Please help me, I don't know which is the "first" one and what rules apply here...

There are three rules to follow for assigning locants (numbers) to a compound like this one.

Assign locants to generate the longest possible chain of carbon atoms.

Here is an example of mine:

I can find chains of eight carbon atoms a few different ways. Two of them are below. There is at least one more. Can you find it? How is it similar to one that I highlighted?

The longest chain I can make in your structure is six carbon atoms, and it looks like I can make that chain length multiple ways.

Assign locants to maximize the number of substituents.

The goal is to have the most possible substituents. For all of the ways to make a six-carbon atom chain, I count the same number of possible substituents.
Back to my example, the first way to make a chain of eight carbon atoms has four substituents: three methyl groups and one ethyl group. The second way has three substituents: two methyl groups and one isopropyl group. We choose the first one.
I think that five is the maximum number of substituents that can be reached in your structure, and again, there are multiple ways to achieve that number.

Assign locants to generate the lowest possible combination of locant values for your substituents.

Using my example, we can number the red parent chain two ways:

In the first one, we have a substituent locant set of {2,3,4,5}. In the second one, we have a substituent locant set of {3,5,6,7}. The first set represents the lower combination of locants, so the first set is the way to go.
For your example, since it has a high degree of symmetry, there may be multiple six carbon atom chains with five substituents having the same set of locant values. In that case, choose the one that puts the alphabetically first substituent at the lowest possible value. I suspect that these sets will be equivalent however given the symmetry. If so, then choose any of them as they are all degenerate.

Nice guidance without giving the answer!


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Heat exchange in adiabatic process
I know my teacher told me that

In adiabatic process there is no exchange for heat between system and surrounding.

Then he came to point. From first law of thermodynamics
$$∆U = q + W \tag{1}$$
Since in adiabatic process there is no heat exchange then,
$$∆U = W \tag{2}$$
And then he said as we know
$$∆U = nC_v∆T \tag{3}$$
So,
$$W = nC_v∆T \tag{4}$$
I don't get it. Above equation tells us that the work done when a system is given heat at constant volume equals to $nC_v∆T.$
But how can this be applied in adiabatic process, since there is no exchange of heat?

The above equation does not say that the work done when a system is given heat at constant volume equals $nC_v\Delta T$.  The relationship is not equal to the heat transferred.  It is equal to the change in internal energy.  Did you think that the only way the temperature of a gas can change is by adding or removing heat?

Adiabatic work is done in expense of thermal energy and vice versa. E.g air lifted by 100m adiabatically expands, does the expansion work and cools itself by 0.98 K.

Please note that heat and temperature are completely different concepts. Do not confuse them.

The problem is that we were taught incorrectly in freshman physics.  They told us that $Q=nC\Delta T$, where C is called the heat capacity.  However, they neglected to tell us that this only applies if no work is done (i.e., the volume is essentially constant, as for a solid or liquid).  Otherwise, it gives the wrong answer for the heat Q.
"Heat capacity" is a bit of a misnomer for C.  From what we then learn in thermodynamics, a better term would be the "internal energy capacity," because that is how Cv is more properly defined.  For an ideal gas,  $$nC_v=\frac{\Delta U}{\Delta T}$$ or $$\Delta U=nC_v\Delta T$$This has nothing to do with the heat added unless the volume is constant and no work is done.  In that case, $$Q=\Delta U=nC_v\Delta T$$So the freshman form of the equation is recovered only if no work is done.

Could you please tell me work done in Adiabatic process. Or just write it down on paper and send me it's image

For an ideal gas, the equation for the work in your original post is correct.  However, you can't really use it unless you know the final pressure and volume.  The final conditions depend on the details of how the process is carried out.


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Distillation of oxalyl chloride
May I use water as a condensing medium while distilling out oxalyl chloride?
As oxalyl chloride reacts with water and the products are $\ce{HCl},$ $\ce{CO}$ and $\ce{CO2},$ when there is a leak or breakage of condenser shell/tube separating water and oxalyl chloride, there may be liberation of toxic gases causing the situation more unsafe.

It should be safe enough to use water as your coolant, check that your water hoses a securely attached, conduct the distillation in an uncluttered fume hood if possible.