text:
Is the strength of hydrogen bonding greater in hydrogen peroxide or water?
When we compare water and hydrogen peroxide's hydrogen bond strength, which one should have the greater one? I know that peroxide has more hydrogen bonds, but when we compare the strength of each hydrogen bond, which one would be greater?
Online values of the boiling points indicate a greater bond strength in $\ce{H2O2}$. But peroxide has a larger number of hydrogen bonds.
This was a question for comparing bond strength in a test paper.

Who said peroxide has more H bonds?

Really interesting question. // It seems that hydrogen bonds are weaker in H2O2, but that there are more of them. https://www.researchgate.net/publication/230086767_Hydrogen_bonding_in_hydrogen_peroxide_and_water_A_Raman_study_of_the_liquid_state

Online values of the boiling points indicate a greater bond strength in $\ce{H2O2}$.

The boiling point of water is $\pu{100 ^\circ C}$ while that of hydrogen peroxide is about $\pu{150 ^\circ C}$. Here is a list of factors that might contribute to this difference (adapted from source)

Stronger dispersion forces in $\ce{H2O2}$ (larger molar mass)
Higher dipole moment in $\ce{H2O2}$
More hydrogen bond acceptors in $\ce{H2O2}$

As for the number of hydrogen bonds, if we look at the structure of ice, each hydrogen atom of water is involved in a hydrogen bond (each oxygen is involved in two). If we assume that each hydrogen can make one bond at maximum, we would not expect more hydrogen bonds per molecule for  $\ce{H2O2}$. The crystal structure of solid  $\ce{H2O2}$ confirms that.

But peroxide has a larger number of hydrogen bonds.

We can't infer that from the molecular structure alone. As stated above, in the solid state, the number of hydrogen bonds per molecule is the same. However, the density of solid $\ce{H2O2}$ (1.7 g/mL) is much higher than that of ice. This is true of the liquid forms as well (1.45 g/mL vs about 1 g/mL). This might allow for more dispersion forces (better contacts).
In liquid water, there are 3.4 hydrogen bonds per molecule on average (source). The higher density of $\ce{H2O2}$ compared to water suggests that atoms are closer in $\ce{H2O2}$, but you have to consider that the composition of the two substances are different. On a per-atom basis, the available volume is comparable.

Is the strength of hydrogen bonding greater in hydrogen peroxide or water?

The question is a bit vague. If you compare one $\ce{H2O2}$:$\ce{H2O2}$ hydrogen bond to one water:water hydrogen bond, the hydrogen bond length is very similar (at least in the solid phase - source). Because the bond partners are the same atoms and the geometry around the oxygen is not that different in the two molecules, this would suggest that the strength of the bond is similar as well.

If I were to dissolve diatomic oxygen into water, the water would form only a negligible amount of hydrogen bonds with the oxygen molecules.  This is because the oxygen atoms in an oxygen molecule share electrons evenly, and as such are neutral (ignoring the dipoles that may form for a few femptoseconds due to the electrons being temporarily closer to one nucleus than the other).  On the other hand, water molecules form strong hydrogen bonds, as the oxygen in water has a -2 oxidation state, meaning that based on probabilistic sharing or electrons, one would expect 2 more electrons to be closest to the oxygen than it has protons in its nucleus.  The combination of electronegativity relative to hydrogen (which gives the probability of the oxygen having a temporary negative charge at any given moment) and of oxidation state (which gives the expected strength of the temporary negative charge) results in hydrogen bonds that are analogous to (albeit weaker) than the ionic bonds formed by a +1 cation and a -2 anion.
By contrast, if we look at hydrogen peroxide, the two oxygens will share the electrons in their bond equally.  As such, each oxygen has only a -1 oxidation state, and the resulting hydrogen bonds are more analogous to the ionic bonds formed by +1 cations and -1 anions.  As the expected distance between the hydrogen and oxygen in the hydrogen bond is more or less the same, the factor that governs relative strength is the strength of the (temporary) ions involved in the bond, so water should form stronger hydrogen bonds.  Similarly, I would expect stronger hydrogen bonds in ammonia than in hydrazine.
I would hazard a guess that the reason why hydrogen peroxide has more hydrogen bonds than water, as seen from the research quoted by maxW in a comment, is that the expected number of hydrogen bonds made by each hydrogen should increase along with the number of atoms to which the hydrogen bond can form.

Well its an unexpected question and shows the interest of one's in Chemistry.
We can understand or can compare the strength of H. Bonding in water and Hydrogen peroxide by observing their stability in open air.
Take 10 ml of each liquid in an open jar and let them place in open for  about 10 minutes and then see what happens? Measure the volume of both liquids after 10 minutes you will see that there will be more water then Hydrogen per oxide. this is because of the thermal stability of inter-molecular forces like Hydrogen Bonding.
Actually in Hydrogen peroxide polarity is not so prominent as in Water, So we can say that Strength of Hydrogen bond in Water is stronger than in Hydrogen Peroxide

You are absolutely right, it has a higher boiling point because it has more hydrogen bonds.
However the bonds are slightly weaker since the oxygen are closer to each other and repel slightly more.
" The hydrogen bonds are slightly weaker (12–15%) than those in water " See https://www.google.com/amp/s/www.researchgate.net/publication/230086767_Hydrogen_bonding_in_hydrogen_peroxide_and_water_A_Raman_study_of_the_liquid_state/amp#ampshare=https://www.researchgate.net/publication/230086767_Hydrogen_bonding_in_hydrogen_peroxide_and_water_A_Raman_study_of_the_liquid_state
Edit my appologies, maxW was earlier with approximate the same answer but I did not see his comment. Should I delete this awnser so you can get the credits you deserve? (Glad to know we are on the same page :)

You are free to do whatever you feel like. Note, however, this you answer may be considered a link-only (non-) answer. It is better to include the key points of the reference here (as a quote, if need be) and use the link as a source only.

MaxW has posted it as a comment, and while I think the answer could use some more context or explanation and much better formatting, it essentially is an answer. Posting the same information as an answer, which was given already in a comment, is not against the guidelines. (I personally think this should be encouraged.)


text:
How do nucleophiles change Sn1 energy profiles?
Suppose two different nucleophiles react with a single substrate via SN1. Would the energy of activation for the second transition state (nucleophile reacts with carbocation) be less for the better nucleophile? Would the products be of equal energy and produce a 1:1 mixture for both nucleophiles?
I know SN1 is unimolecular so changing the nucleophile would not change the reaction rate. Additionally, since the activation energy for the rate determing step was already surpassed, I believe the products will form in 1:1 mixture where both nucleophiles react equally and both products are of equal energy.

related: https://chemistry.stackexchange.com/q/84040/4945

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Why is the nitrile group not hydrolyzed as well during hydrolysis of imine in Thorpe reaction?

while imine is getting hydrolysed, why does not nitrile group hydrolyse to give carboxylic acid?

Imines are much easier to hydrolyse. Nitrile hydrolysis takes much more forcing conditions (usually heat)

A word on your mechanism: You cannot have $\ce{tBuO-}$ and $\ce{H3O+}$ in the same reaction vessel. And likewise, as soon as you have a few molecules of $\ce{NH4+}$, you will no longer see any $\ce{H3O+}$.

Is the water coming in at the work up stage? It wouldn't normally occur that you have water in with butoxide from the beginning

I couldn't upload the mechanism from the book. Posted it from wikipedia.

It has to do with hybridization. The nitrogen in a nitrile is sp hybridized; imine sp2 and ammonia sp3. The more s-character in hybrid orbitals, the less basic the electron pair (protonated nitrile pKa ~ -10; pKb ~24). Ammonia has the lowest percent s-character (25%, NH4+ pKa 9.25). Imines lie in between although most examples you will find are measured in DMSO and are loaded with biasing phenyl groups.
Compare the carbon series: acetylene (sp, pKa ~25), ethylene (sp2, ~44), methane (sp3, pKa ~50). The closer the electron density is to carbon, the more acidic the C-H bond. If electrons are held closer to carbon, then it is more difficult for homolytic cleavage of a C-H bond. That is, acetylene is willing to give up a proton but not a hydrogen atom. Thus, Bond Dissociation Energy runs in the opposite order as acidity.
The harder it is to protonate, the harder it is to hydrolyze.
There were no conditions provided by user 481311 for the hydrolysis. Imines, such as the acetaldehyde methylimine (CH3CH=NCH3), will hydrolyze in water. As the User illustrates, imines hydrolyze readily in acid medium. These hydrolyses may be conducted at ambient temperature or at mild temperatures. Nitriles, owing to their low basicity, require more vigorous conditions like refluxing aqueous acid. Its all about conditions. I hope this helps.

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Slight variation in NMR integrals of CH protons vs. CH2 or CH3
While acquiring some NMRs I find that there is a consistent pattern in that protons attached to sp2 carbons tend to have slightly smaller integrals than protons attached to sp3 carbons:

(peaks in aromatic region are from a para-disubstituted benzene ring; peak at 4.92 ppm is from a trisubstituted alkene; 4.50 is a NH; the remainder are aliphatic CH2's or CH3's.)
I calibrated the total sum of integrals to 21; one can see from the spectrum that the aromatic and olefinic peaks are slightly smaller than integral values, and the others are slightly larger. Obviously it doesn't hinder the analysis of the spectrum, but it was an interesting trend (the same was observed in multiple spectra).
Am I reading too much into this, or is there a reason behind it? I strongly suspect there is something - perhaps related to the marginally slower relaxation of these protons, since there are no geminal protons (only vicinal protons)?
If it is of relevance - a 60° pulse is used (pulse program zg60). The number of scans ns is 16, and the relaxation delay d1 is 1 second. I'm happy to provide any other necessary acquisition parameters, or to provide the full structure of the compound if it's necessary.

P/S I found something in Findeisen & Berger's 50 and More Essential NMR Experiments, in which they mention that the integral of an aromatic proton in strychnine is smaller than expected as it has the longest relaxation time. Even though the authors used ns = 1, too short a delay between the receiver gain adjust and the scan was used, which led to incomplete relaxation and hence a smaller integral.

Indeed I think all that's going on is that 1 second is too short a relaxation delay; bump it up to 5 s or maybe 10 s for safety and the integration should be essentially perfect, bar random noise. For quantitative NMR standards, it's pretty common to use at least ~5 s. See this Sigma-Aldrich brochure (pdf) for some typical relaxation delays for reference. I once had a very slowly relaxing aryl proton in a zwitterion, but with 25 s relaxation delay, it too integrated perfectly.

Just so I can fix my answer below, can you give the duration of the recorded FID, np x dw?

@Karl not totally sure, but I'd expect it to be around 2 or 3 s (probably 16k complex points, sw ~ 20 ppm = 8 kHz (I think this was acquired at 400 MHz), so dwell time between complex points ~ 125 µs)

That's to be expected: Less protons as spatially close neighbours, less dipolar coupling, slower relaxation.
For a relaxation/recycle delay of $\pu{1 s}$ and $T_1$ of perhaps $\pu{3 s}$, the Ernst angle is already only $44^\circ$, so your measurement is far from optimal anyway. Use a shorter pulse or longer recycle delay.
At one second, your are anyway dangerously close to generating echoes which overlay the signal from further scans, distorting the accumulated spectrum.

d1 is the delay between end of the acquisition and next scan. For the Ernst angle, der relevant delay includes the acquisition, so the number of 44° I give above is wrong.

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How many atoms can be reduced in a single redox reaction?
I am doing high school chemistry and I am busy working with redox reactions.
The explanation in my text book, of how a redox reaction works, left me a bit confused as to how many atoms can be reduced/oxidized in a single reaction equation.
Can only one atom (species) be reduced/oxidized in a single equation or can there be more than one reduced/oxidized in the same chemical equation?
All of the reaction equations I have come across so far have just one atom reduced and one atom oxidized, for example:
$$ \ce {Mg(s) + 2HCl(aq) -> H_2(g) + MgCl_2(aq)} $$
In this reaction only Mg is oxidized and only H is reduced. Is there a reaction in which more than one species of atom it reduced or oxidized?

Redox reactions can be much more complicated then that.

I know they get much more complex. I am not doing very complex ones yet. I just chose a simple one to try illustrate my question, which is a very simple question.

Yes, there can be more and yes/no questions are suitable for this site.

Thanks. That really helps. I think my text book spoke as if only one could be reduced/oxidized at a time because I would not need to do more complex ones yet. Unfortunately I don't have enough reputation to up-vote you :-(

I'd rather point out that the reactions in which multiple elements get oxidized/reduced are a relative minority (in school chemistry, at least). True, you'd encounter pretty complex reactions further on, but their complexity stems from other reasons; despite it, most of the times there would be only one oxidant and one reducing agent.

Of two species, one would usually be easier to reduce/oxidise than the other. As a reducing agent/oxidiser, I would in most cases just leave out the latter, and work on two of the former. Exemptions, as usual, prove the rule.

I'm not sure if I understand your question correctly, but consider the reaction
$$\ce{CH3-SH + 3 O2 -> SO2 + CO2 + 2 H2O}$$
where methanethiol is oxidized with oxygen to sulfur dioxide, carbon dioxide, and water.
Thereby the oxidation states of sulfur and carbon change from $-2$ to $+4$ and the oxidation state of oxygen from $\pm0$ to $-2$. So sulfur and carbon atoms are both oxidized while oxygen is reduced.
I hope this answers your question.

@adventurin: This did answer the question, thank you. The example equation helped. By the way, any suggestions on how to make the question clearer would be appreciated.


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How connecting galvanic batteries in series actually works. "In Series" detection experiment
How I am interested in understanding what is the actual mechanism behind increasing potential in simple galvanic cell connected in series. Almost all explanations I found on internet are more like recursive "explanation from definition". First we define that each cell adds potential difference and then it becomes the actual explanation. I am feeling that this lacks an actual chemical insight of why electrons experience more EMF in such circle. So I am trying my luck here. This duplicate left with no answer.
To make question more concrete consider the following:- suppose you can examine only one cell in galvanic (classical Zn/Cu) battery you do not know where its anode and cathode are connected. However you do know that the circuit is closed. Will it be possible by observation of chemical/electrical/atomical process in it to know that it connected in series or not? In other words will it be possible to determine that this is the only source of EMF? If yes what chemically/electrically changes will it experience vs the condition where it works alone.
And yes I do not have any real background in chemistry or electronics so use simple English :)

No. If you could examine what is going on that cell from inside you won't be able to tell if it is one or one of a series.

Electrons in the wires do not "know" whether there is a particular chemical cell, or more than  one cell... that information is lost. However, one can infer if there are multiple cells in series from the current during discharge, if the cells have different capacities.  For example, voltage from a small carbon-zinc cell in series with one much larger would be ~3 V open circuit, drop fairly quickly to ~1.5 V with increasing current, and then drop more slowly as current increases further.

We don't have to go to discharge actually. The potential accumulated on cathode in disconnected state will be three times more when connected say three cells in batterym than on single one. Is it right?

Not in the case of different cell chemistries: If ~3.2 V is measured, it might be a single Li cell, two fresh Zn-C cells in series or three Ni-Fe cells in series.

Thanks a lot for response. Can you comment on my answer? It is based on your input and comments are limited in size to put it here.

Ok let's reduce the scope to only one classical Zn/Cu battery. (The real reason behind it is that I am not familiar with any other :)) Also we will reduce the characteristics of the circuit to most simple one. Battery (new cells with well known characteristics) and resistor as a load. The answer it my question would be.
Can we by observing only the first cell know if it is acting alone or connected in series....

Yes it is possible by observing the number of electrons in disconnected state on the cathode . It should be times higher than when cell is acting alone.
The reason behind it is that when cell is connected in series they can "borrow" more charge from cells behind before the reaction stops (when circuit is disconnected).

Using ohm's law, for an infinitely large cell, the current will scale with voltage. In real life, for small cells and small resistances, the internal cell resistance cause two cells in series to produce less than two times as much current as for a single cell. You may be confusing current (number of electrons or coulombs moved/second) with potential (electromotive force, voltage).


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Pressure change when volume and temperature both increase
By what factor does the pressure change if:

the temperature rises from 0 to 100 degrees Celsius,
and the volume is doubled?

$$
\pu{0 °C} = \pu{273.2 K}\\
\pu{100 °C} = \pu{373.2 K}
$$
Assuming $pV = nRT$:
Since pressure is proportional to temperature, neglecting the change in volume, the pressure would increase by a factor of $$\frac{373.2}{273.2} = 1.366\ .$$
Now since the volume is doubled as well, can I simply conclude that the pressure increases by a factor of $$\frac{1.366}{2} = 0.683\ ?$$
This is a task from a introduction to gases, I've mainly used the ideal gas law $pV = nRT$, but I wanna make sure my logic isn't flawed from the beginning.

This seems fine.

Cool! Appreciate the assistance


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What is the correct nomenclature for Glucuronyl C5-epimerase?
I see people writing "Glucuronyl C5-epimerase", "Glucuronyl C-5-epimerase", "Glucuronyl-C-5-epimerase" and "Glucuronyl C5 epimerase", which one is correct? Is there a naming convention?

@basseur The question is very clear and concise. It is: ‘should enzyme names like the one above contain hyphens and should there be a hypen between C and 5?’ I think this is answerable and sufficiently on-topic (while it may be even more on topic on [biology.se]).

@Jan: You are right. I guess from my standpoint as a professional chemist I was too narrow-minded to see, that this is not as obvious as it seems to me. My bad.
I actually don't believe, this is a homework question, as it was flagged.

Welcome to chemistry.se! I don't think this is homework, and I am reopening the question. Nevertheless, please see our guidelines on how to ask nomenclature questions.

I have not found this particular enzyme in Enzyme Nomenclature. However, following the Recommendations of the Nomenclature Committee of the International Union of Biochemistry and Molecular Biology, the name should be glucuronyl 5-epimerase. See also this sublist for some analogous names of other epimerases.

In addition to Loong’s answer which includes the formally correct way to write it: if you wish to unambiguously indicate that a carbon atom is epimerised,[1] the only possible choices are those that leave no space between the symbol for carbon and the digit. Atoms are number-named by writing the element symbol directly followed by their number without hyphen or space. So the atom in question is C5, meaning we have a C5-epimerase.

Note:
[1]: Which is superfluous since carbon atoms are practically the only atoms ever epimerised.


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Mixing isopropyl alcohol and propylene glycol
What would happen if I mixed isopropyl alcohol and propylene glycol? I need to winterize the water base of my basketball hoop, and already added some isopropyl alcohol, but I discovered that propylene glycol would be better.
Also, I know isopropyl alcohol is toxic to ingest, but will it contaminate my soil if I drain it?

If you'll mix them, you'll get a mixture of the two, and nothing more (no reactions or something). Both are toxic to ingest, and so will be the mixture. As for the contaminated soil, it will become toxic to ingest as well.

About how much isopropyl alcohol did you add to the water?

I want to point out that two of the close votes (one being mine) went towards ‘too broad’ because the question is asking two unrelated questions.

The two chemicals can be mixed without any reaction between them. However, you should add enough water to avoid fire hazards after the winter. Take a look at this table to find out, what mixture should suffice.
You should never pour solvents onto your soil or down the drain. Contact local authorities to find out, how you can safely dispose of that mixture. You can, of course, reuse the mixture and store it in some plastic cans if you need to empty the basketball hoop.

I accidentally switched the numbers. I revised my answer to give more options.

Thank you for your help. The table is for ethylene glycol, and I would like to use the less toxic propylene glycol. Is the table still valid?

I changed the link in my answer to propylene glycol. Now you should be good to go! :)


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How are N95 masks different from dust masks?
There is a fire raging through my hometown right now. People in the area are advised to be wearing N95 masks to filter out smoke and particulates; and there is misinformation being spread that these masks actually don't work for this purpose. What's the science behind the N95 filter? The Wikipedia page didn't really elaborate.

Get a cartridge filter half mask instead, they are much more comfortable, and provide much better filtration. Obviously these will not protect you from any immediate hazard related to fire, but probably great relief from blown nuisance smoke from a wildfire. https://www.amazon.com/3M-6391-P100-Reusable-Respirator/dp/B001NDN29O/

The short answer to your question "What's the science behind the N95 filter?" is that it is what's called a tortuous path filter, removing particulates from a flow of contaminated air by impact and absorption of the particulate onto the filter material or by trapping particles of sufficient size between the fibers constituting the filter.
The term N95 comes from the NIOSH (National Institute for Occupational Safety and Health, a body of the US government) rating where the 95 means that the material is certified to block 95% of particles 0.3 μm or larger in diameter. The limiting factor toward achieving the stated level of performance during usage is how well the mask seals to the face of the user. Whiskers, dirt and improper fit can all allow a larger portion of particles to circumvent the filter material. The image in the OP shows a mask having a valve to facilitate breathing and maintain a seal by opening during the exhale portion of a breath.

From official 3M Q&A brochure (emphasis mine):

Can N95 respirators filter particles that are really small like smoke, soot and ash?
Yes. Particulate filters employ multiple mechanisms that are effective at filtering a range of particles that include those so small you cannot see them. In fact, as part of their certification process, NIOSH tests particulate respirators against submicron particles in the size ranges of smoke, soot and ash.

I guess if you want to participate in firefighting activities or there is a chance you might need to escape areas with high fire smoke concentration you need an oxygen mask with eye and face protection (or SCBA – self-contained breathing apparatus) instead.

The masks filter particulates, not molecules like carbon dioxide or carbon monoxide. The masks also of course can't furnish oxygen if the fire has deleted it.

Or depleted it! ;)

The principle on which they depend isn't that different. This does not mean, however, that you can wear a dust mask instead of an N95.
From the Review article of "Respiratory Protection" from the New England Journal of Medicine:

The most commonly recommended respirator in the health care setting is the N95: “N” means “not oil-proof,” and “95” means that it is at least 95 percent efficient at filtering particles with a median diameter of greater than 0.3 μm. Higher numbers indicate greater filtering efficiency.

So it is a good mask for particulates. However, according to OSHA, it won't protect you against gases and vapors.

Filtering facepiece respirators filter out particles and do not protect against non-particulate hazards such as gases or vapors.

So it may not be the best mask to wear when you are near the site of the fire.
The best type of respirator is the one which has its own oxygen source. However, these are expensive.


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What are the differences between the Hilbert space and real space representations in chemistry?
I would like to understand the difference between Hilbert space and real space in a molecular code.
My understanding is that the Hilbert space of a system (such as a molecule) can be represented by a set of atomic orbitals (AOs) while the real space represents a system through a grid of points in the specific space of a system. Is this true?
For a more concrete example, consider electronic excitations. Using the entire (orbital) space of a reference system, one can project out (excite) an electron to the virtual space of a reference ground state. However, I'm not sure how this is handled in a "real space" program, or what that even means.

Generally true, but what do you want to make of that? Electron density or position of nuclei are real values, orbitals or wave functions are complex. Nothing more to it.

electronic excitations, using the entire space of a reference system one can project out an electron to the virtual space of a reference ground state. However, conceptually I wanted to distinguish  between these two "spaces". I am not quite sure what people mean by real space codes. Thanks for asking

What do you mean by "molecular code"? I honestly have no idea what you are aiming for. The first part of your comment above is totally incomprehensible. Please update your question.

A Hilbert space is a kind of linear vector space.
In chemistry which encounter it when quantum mechanics, when we can represent wavefunctions by their contributions from different orthonormal single particle states.  It is these single particle states which we build up out our atomic orbitals (AOs).
Example:
Consider a two level systems - minimal basis $\ce{H2}$:
Each atom has a $1s$ AO on its hydrogen: $\phi_1(\textbf{r})$, $\phi_2(\textbf{r})$.  These AOs aren't orthogonal though:
$$
\iiint d^3\textbf{r}\ \phi_1^*(\textbf{r}) \phi_2(\textbf{r}) \ne 0
$$but we can construct orthonormal states:
$$
\psi_\pm(\textbf{r}) = \frac{1}{\sqrt{2}}\left(\phi_1(\textbf{r})\pm\phi_2(\textbf{r})\right)
$$
such that like functions integrate to $1$ and functions of unlike functions vanish:
$$
\iiint d^3\textbf{r}\ \psi_\pm^*(\textbf{r}) \psi_\mp(\textbf{r}) = 0 \\
\iiint d^3\textbf{r}\ \psi_\pm^*(\textbf{r}) \psi_\pm(\textbf{r}) = 1
$$.
These two orthogonal function can be represented as state vectors:
$$
\psi_+(\textbf{r}) = \pmatrix{1\\0},\; \psi_-(\textbf{r}) = \pmatrix{0\\1}
$$
This defines a Hilbert or vector space, where our single particle states $\psi_\pm$ are our vectors and the inner product of two vectors, $\psi_i$ and $\psi_j$, is defined by the integral over all space:
$$
\langle\psi_i\vert\psi_j\rangle = \iiint d^3\textbf{r}\ \psi_i^*(\textbf{r}) \psi_j(\textbf{r})
$$
and is equivalent to the dot product of two state vectors.
In a electronic structure (specifically DFT) code, a "real space code" means using atom orbitals as basis functions rather than plane waves.
The Hilbert space described above is complete - it describes all possible wavefunctions which can be made from the AOs included.  If you included the infinite number of hydrogenic wavefunctions on each atom you would be able to have an exact description of all possible wavefunctions for $\ce{H_2}$ - as the hydrogenic wavefunctions form a complete set, instead you include as many as practical.
An alternative complete set of functions is the infinite set of all plane waves of all frequencies.  In cases where you are performing calculations on a periodic system, such as a solid, plane waves can be a more natural choice as they themselves are also periodic.  In a box of width $L$, the set of waves with wavelengths:
$$\lambda = \frac{2L}{n}$$
for all integral n, form a complete set.  They are more commonly described by their wavevectors, $\textbf{k}$, a vector in what is called reciprocal space, which describes the momentum of the wave.
In DFT, when calculating integrals over density, in a plane wave code you manipulate the density in terms of wavevectors, $\textbf{k}$.  In a real space code, you compose your density from Kohn-Sham orbitals in real space and calculate integrals of the density over quadrature grids in real (Euclidean) space.
Hence "real space" is used as a descriptor, for DFT codes that work with AOs and basis functions defined on atoms in Euclidean space, in contrast to "plane wave" codes which used basis functions defined in position space.

This isn't quite right; the common quantum chemistry packages that use MOs formed from CGTOs are not real space, they're Fock space. Real space programs would evaluate AOs/basis functions directly on a grid.

I find it hard to connect your answer with the math, but it is interesting to read a chemical viewpoint. @pentavalentcarbon , they belong to  Fock spaces, however, they belong to Hilbert spaces too

@user1420303 Yes, that is true, but the point is that they are not necessarily evaluated on a grid. In fact, you could evaluate any type of basis function (GTOs, STOs, plane waves, splines, ...) on a Cartesian grid and it would be real space.

@pentavalentcarbon - It's the quadrature grid bit that I didn't fully appreciate until seeing the comment and googling some more.  I'm more used to wavefunction methods than DFT - hence the slight confusion over the typical usage the term "real space".

Also, don't forget that there is still XC quadrature in Gaussian, Q-Chem, etc., but they are atom-centered spherical grids rather than more evenly-spaced boxes.

@pentavalentcarbon Those are the types of grids I was thinking of.  Aren't atom-centred grids considered "real space"?

By our discussion, yes, though we don't call it a real-space calculation. That terminology is reserved for when all operators ($\hat{T}, \hat{V}, \hat{J}, \hat{K}$) are evaluated on a grid.

Thank you @user1420303.  It is clear now the difference. And now I understand the difference in both wavefunction and DFT worlds. I would say that as long as the space (a molecule) is fully represented either by a basis set or a grid of points I can considered that space as my Hilbert space, in which I can manipulate the electrons (in the molecule) that pertain to that same space.

Also @pentavalentcarbon thanks for help me realize the difference in both worlds wavefunction and DFT.

@Pablor , I really didn't help. In most cases, the system can be described by a vector living in a Hilbert space (state vector), this constitutes a postulate of quantum mechanics. As pentavalentcarbon pointed out, sometimes other spaces becomes necessary or preferable.

A quibble about the Hilbert space. A Hilbert space is a vector space endowed with a scalar product (a bilinear form $X\times X\to \mathbb C$, with some stipulations) and its induced norm. Fock space is a Grassmann or exterior algebra (a vector space endowed with a skew-symmetric bilinear form $X\times X\to X$) that is also a Hilbert space. None of this has anything to do with the  choice of basis for representation (GTO, FE, PW, etc.).


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Chemical Kinetics / Rate Question -- Error in Textbook Question/Answer?
I am reading Kaplan's General Chemistry MCAT Review 2018-2019, and think that one of its chemical kinetic questions is wrong. (Or perhaps I am simply missing something). I have attached an image of the question, which is easier than typing out as it includes the standard table of concentrations at different trials.
The answer reads: "In the first two trials, the concentration of $\ce{XH4}$ is held constant while the concentration of $\ce{O2}$ is multiplied by 4, and the rate of the reaction also increases by a factor of approximately 4..." It ultimately comes up with a rate of $k\ce{[XH4]^2[O2]}$
Am I completely missing something? My first question is, is the book's answer right? If it is not, is the problem solvable (i.e. does the table make any sense at all)? I've tried to come up with rates using the table and encountered the following issues. Please let me know if I am right in approaching the question in this way:

If you try to find the rate with respect to $\ce{XH4}$ using trials 1 and 3, I believe you get an order of 3. However, if you try to find the same rate using trials 1 and 4, you get 4. This isn't normal, right? Maybe an example of a mixed order reaction? Or just a mess up?
How do we interpret the data from trials 3 and 4? The concentration of neither reactant changes and yet the rate doubled. If we ever see something like this in a table, does it suggest that the reaction is 0 order? Or can we not make that leap? And is it even possible to have this data alongside the other data in the table?
It seems to me that, based on Trials 1 and 2, the rate is 0 order with respect to $\ce{[O2]}$. Is that a correct interpretation?

Thank you so much!

From what I can see the table looks very messed up, so I think the question is just wrong. The answer you found that says the O2 concentration is increased by 4, while that doesn't occur anywhere in the table, is a big sign that this question is wrong.

Another sign that the table is wrong, is that [XH4] and [O2] in entries 3 and 4 are the same but still have a different rate. Thus, the same input gives two different outputs: The rate is not a function of the concentration, and the question is unsolvable.

There are so many other things wrong with this table. Obviously, the table suggests that the rate is constant over time. Therefore, it has to be a zeroth order kinetic anyway. Or do they mean the Rate at some point in time, e.g. initial rate. Absurd excercise.

The exercise is absurd. None of the proposed answers can be right, as for any answer A,B,C,D the rate is a function of the concentrations [XH4],[O2] but the table suggests that the rate is constant over time, e.g. 2.0$~$M/min for entry 1. Thus:
$$
R = 2.0~\text{M/min} = \text{constant} \cdot [\mathrm{XH_4}]^0 \cdot [\mathrm{O_2}]^0
$$
and is therefore a zeroth order kinetics.
I assume that the initial rate $R_0$ or a rate at some point $R_{t>0}$ is meant, and the question just heavily confuses things. So from here on, let's assume that $R$ means some rate at some point in time.
But even then, it is not possible to answer the question. This is because entries 3 and 4 have the same domain ([XH4] and [O2]) but a different codomain (different rates R). This means that there is no function mapping from the concentration of the reactants to the rate of the reaction.
However, there obviously is such a functional relation for all answers A,B,C,D.
Therefore, the question is absurd. It is probably meant to be that [O2] is 2.0 in entry 4.

And then it would probably be answer D. But answers are sometimes not so clear when you don't know the question ; )


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Minimum image convention and the atom “seeing” itself
The minimum image convention in molecular dynamics can be stated as:

A particle in the central square must not interact with one of its neighbors in the central square and an image of this neighbor in one of the replicas at the same time but only with the closer one.

What is the basis for this convention? In other words, what is the problem with a central atom seeing another atom from both the image and the central box?

If I look at the simulation box as just part of an infinite bulk material, then I should be able to take the cutoff radius as large as I like, no matter how many images of an atom are seen.

Sure, you are able to take the cutoff radius as large as you like, but still you must take it, and that's how it is done.

MIC is used to prevent extensive calculation for pairwise interactions, and double counting. For VdW-forces you can assume that no significant artefacts are introduced by using a sufficiently large cut-off (~ 1 nm). For electrostatics you might need to take in to account some long-range forces.

Thanks for the comments.
My main question is: what is wrong with an atom both from the image and in the actual box?


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When titrating a weak acid should only undissociated acid molecules be considered?
Let's say we do the titration of a weak acid (acetic acid in this case)  and a strong base (sodium hydroxide) represented by the following equation:
$$\ce{CH3COOH + NaOH -> CH3COONa + H2O}$$
My question is that, since $\ce{CH3COOH}$ is partially dissociated, does $\ce{NaOH}$ react  only with the partially dissociated part or the complete acetic acid solution?

Related: https://chemistry.stackexchange.com/questions/60407/why-will-a-strong-acid-neutralize-as-much-base-as-a-weak-acid/60409#60409

Your formula is nonsense for a titration. Your base is already dissolved, and so stays sodium acetate. OH- reacts with undissociated acetic acid, nothing else. Write up reactions with reactants in the form you actually use them and you will never make such (stupid) mistakes. Insult goes to your teacher, not you, btw.

Acetic acid and acetate are connected by the dissociation equilibrium $(1)$:
$$\ce{CH3COOH + H2O <<=> H3O+ + CH3COO-}\tag{1}$$
So you basically have two acidic particles that can react with hydroxide: hydronium or acetic acid. Hydronium will actually react much faster on a molecular level even though its concentration is significantly lower as it is much more mobile in solution. However, any hydronium consumed is immediately reformed by acetic acid molecules reestablishing the equilibrium.
Therefore, on a very minute level hydroxide reacts with neither acetic acid nor acetate but with hydronium which is produced in the dissociation of acetic acid. In practice, that still means that hydroxide will be added until $n(\ce{CH3COOH})+n(\ce{CH3COO-})=n(\ce{OH-})$. From this point onwards, there will be more hydroxide than acidic particles causing the $\mathrm{pH}$ value to rise significantly, the indicator to change colour and the student to realise that they reached (and passed) the equivalence point.


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Mechanism for decarboxylation
Propose a mechanism for the following transformation.

I gather that this mechanism might be similar to decarboxylation of keto acids:

However, I am unsure of how to begin.

first, addition of water will occur at the ester to form a carboxylic acid which will then be removed after heating.

In acidic media, the ester carbonyl will be protonated and the carbonyl carbon attacked by a water molecule. The intermediate will have a proton abstracted by a water molecule and the ethoxy group will be protonated (it can happen either by intermolecular reaction with an H3O+ or by an intramolecular reaction with a proton migration from the protonated intermediate after the water attack). The carbonyl is restored and an ethanol molecule leaves the structure. A subsequent proton abstraction by water forms the carboxylic acid, which will then suffer decarboxylation following the mechanism you presented above. (I'm sorry about my ChemDraw, I suck at drawing even on a computer).


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Unexpected behavior during preparation of copper hypophosphite
I am using copper hypophosphite for the activation of through holes in printed circuit boards. The circuit board is dipped into activator solution and upon heating, metallic copper is deposited in the holes, rendering them conductive and allowing me to electroplate copper to the desired thickness.
The solution I have prepared works for this purpose, but during its preparation I observe behavior that does not match what others report for the same process. I hope that somebody can suggest what I might be doing wrong.
Here is the recipe:

Mix 140 ml distilled $\ce{H2O}$ and $30\ \pu{g}\ \ce{CuSO4.5H2O}\\ \pu{MW} = 249.677\  \pu{ g/mol}\ \dfrac{30\ \pu{g}}{249.7\ \pu{g/mol}} = 0.12\ \pu{mol}$
Stir until dissolved, then add $22\ \pu{g}\ \ce{Ca(H2PO2)2}\ \text{(Ca hypophosphite)}\\ \pu{MW} = 168.038\ \pu{g/mol}\ \dfrac{22\ \pu{g}}{168.04\ \pu{g/mol}} = 0.13\ \pu{mol}$ The
Calcium hypophosphite is a small excess, so all the $\ce{Cu^2+}$ will become
$\ce{Cu(HPO2)2}$ and all the $\ce{SO4^2-}$ will precipitate as $\ce{CaSO4}$ Ca sulfate (gypsum)
Filter solution (Buchner vacuum filtration)
Rinse with 100 ml distilled water and discard sediment.
Add $40\ \pu{ml}\ 25\% \ \ce{NH4OH}\ (13.4)\ \pu{M}\\ 0.040\ \pu{L} \times 13.4\ \pu{M} =  0.536\ \pu{mol}$

“Solution may thicken and develop white flakes, but as the reaction proceeds the flakes will dissolve and the solution will become a deep blue color.”

And this is where my problem occurs: when I add 40 ml of 25% ammonia, I get a flash of dark blue followed by the development of a green slurry. Adding more ammonia creates a local area of dark blue solution, but mixing causes it to vanish into the green slush. Instead of 40 ml Ammonia, I've had to use 110 ml of Ammonia to dissolve the slurry and convert it from a green slush into a dark blue solution. The solution does work as expected, but I can't understand why my results are (repeatedly) different from those reported by others.
I have four sources for this preparation. One of them is the original Russian patent, one is a Russian electronics enthusiast who shared the technique with the hobbyist community, one is a Russian-speaking American electronics blogger, and one is an engineer at a high-end fabrication company. All of these sources agree on the molar ratios of the chemicals, and three of them have images showing that the addition of 40 ml ammonia will lead to the immediate formation of a dark blue solution. One of them has shared a video of the entire process, including all weighing and measuring, and it's clear that something very different is happening on my bench.
I searched the literature and found suggestions that the green precipitates are likely insoluble basic salts of copper (mostly hydroxides). I found sources indicating that copper hydroxide dihydrate precipitates when stoichiometric amounts of ammonia are added, and dissolves in an excess of ammonia.  However, 40 ml of 25% w/w ammonia should provide more than double the stoichiometric amount needed for the formation of copper hydroxide.
I've verified that all the sources agree on the molar ratios and that the reagents they are using are labeled similarly to mine. I've checked my weights and measurements carefully through multiple trials. I've verified my scale. The ammonium hydroxide is a new bottle that was full upon opening. The calcium hypophosphite is a reagent grade. My copper sulfate pentahydrate is a technical grade intended for home use, but I prepared a saturated solution at two temperatures and verified that the solubility at each temperature matches the published solubility for copper sulfate to the limits of my scale. Apparently there is some likelihood that industrially prepared copper sulfate could be contaminated with small amounts of iron or phosphorus, but I haven’t attempted to analyze for contaminants because I found no suggestion that they could double the requirement for ammonia. Although I rinse my glassware with distilled water, I did not dry it and re-wash, so it’s possible that some small amount of tap water could have consistently contaminated my process.
The temperature in my lab area is 25 – 26 C. I found no mention of temperature in the published process, nor did I find any suggestion in any online source that the solubility of metal hypophosphites in ammonia is strongly temperature dependent.
The Russian blogger says this:

After a lot of experiments, I came to this recipe, in 80 ml of water we dissolve 15 grams of copper sulfate (copper sulfate), add dry calcium hypophosphite with stirring, take it in excess, it's 11 grams, stir for 40 minutes, to form calcium sulphate precipitate, then filter out the precipitate, washed with 20 ml of water is obtained bluish liquid copper hypophosphite and calcium hypophosphite solution was then poured slowly with stirring ammonia water (ammonium hydroxide) to 25%, often it will be 16-25 ml, should get a homogeneous dark solution.

Here are some of the sources for this process:
An instructable on the topic
A video queued up to this point in the process
The improved patent
So what am I doing wrong???

How old is your ammonium hydroxide solution?

"Recipe??". You are not making food. Are you?

As @BenNorris implies, 25% ammonium hydroxide outgasses so much you need to open it under a hood, so titrate a sample to check its strength.

As indicated in my notes, the ammonium hydroxide was a new bottle just received from a reputable supplier, sealed and completely full when opened. I'll do a titration anyway, but the discrepancy wouldn't be explained unless it was down to 8%, which isn't entirely consistent with what I experienced when opening the bottle :)

@NilayGhosh Not food, exactly. But calcium hypophosphite is schedule 1 because it's used to "cook" meth, so I felt "recipe" was appropriate...


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How does the addition of acetone cause a shift from hexaaquacobalt(II) to tetrachloridocobaltate(II)?
For this reaction, at equilibrium,
$\ce{[Co(H_2O)_6]^2+ + 4Cl^- <=> [CoCl_4]^2- + 6H_2O}$
$\ce{[Co(H_2O)_6]^2+}$ and $\ce{[CoCl_4]^2-}$ are red and blue in aqueous solution, respectively.
When acetone is then added, the solution changes from red to blue. What causes this?

I haven't tried this before but it sounds interesting. As the blue chloro-complex is in an equilibrium with the red aqua-complex, I'd suggest the equilibrium is shifted as the water is absorbed by the acetone. As water and acetone are miscible this may take some water from the equilibrium, causing the shift towards the chloro side.
If you use ethanol to dissolve the anhydrous $\ce{CoCl2}$ it will turn blue as well. Yutaka Fukuda mentioned a chloro-ethanol-complex in his book 'Inorganic Chromotropism' for acetone I doubt it will coordinate.
Therefore I'd rather tend to the first one.

Half of this answer is ‘I think’, ‘I assume’ or irrelevant to the question. I was indifferent of it at first but seeing how it gained so many upvotes I have to counter that. Please attempt to find sources and references to back your assumptions.

I think an answer should also include similar cases. Here two possibilities can arise and I know of only this one book mentioned above that considered a coordination of the solvent as well (for this particular equilibrium) so why not give this as an option for someone having a similar question?

Please don’t misunderstand me: I’m all for noting similar or seemingly similar cases once the key case in point has been explained. However, your answer explains the key case with words that suggest you’re just guessing. I could have written a guesswork answer like that but I didn’t because as a scientist I prefer the experimental evidence where I am unsure.

I would rather you consider the problem using concentrations and or activities. I think we need to consider the change in the activity of the water and the chloride in the system.

The great problem is that the ability of a chemical substance to act in some way, is controlled by the activity of the substance and not by the concentration of it. Oftein people consider the concnetration in a first approximation where the activity coefficent of everything is assumed to be one. But for a good prediction or good undertsanding we need the activity coefficents.
for a species X
activity (a) = concentration x activity coefficent (gamma)
We need to understand that as the composition of the solvent changes or the temperture changes the activity coefficents will change.
I think that the answer relates to the relative dielectric constant of the medium, as you replace water with ethanol (or acetone) or increase the temperature then the relative dielectric constant decreases.
By lowering the relative dielectric constant you will make the coordination of the anions to the cations more favourable. I think that this is the reason why a solution of cobalt chloride in ethanol/water is thermochromic. It is pink when cold and blue when hot.
The replacement of the water with a less coordinating solvent will also help to shift the equilibrium in favour of the blue tetrachloro anionic complex.
If you look at the Debye-Huckel equation for the activity coefficient and make sure that you look at what is inside the A term, you will find that when the relative dielectric constant changes that the activity coefficients of ions will change. I think that the change of colour that you see is due to a change in activity coefficient.
The extended Debye-Huckel equation in its full glory is
$$\log (\gamma i) = \left(\frac{-Z_i^2 e^3}{4 \pi \left({\epsilon_0 \epsilon_r k_\mathrm B T}^{1.5}\right)}\right)\frac 1{1 + r_i(8 \pi n e^2/\epsilon_0 \epsilon_r k_\mathrm B T)^{0.5}}(n/2)^{0.5}$$

$Z_i$ is the charge on the ion of interest
$e$ is the electronic charge
$\epsilon_0$ is the vacuum permittivity
$\epsilon_r$ is the relative dielectric constant
$k_\mathrm B$ is Boltzmann’s constant
$T$ is the temperature in kelvin
$n$ is the number of ions in one cubic meter.
$r_i$ is the sum of the radii of the cation and anion which are in contact.

I have done the calculations for the activity coefficient of chloride in mixtures of water and acetone for a 500 μM solution of cobalt(II) chloride at 298 K in mixtures of water and acetone. As the acetone replaces the water the activity coefficient of the chloride goes down, I think that this will be due to the electrostatic interactions of the chloride anions with the cobalt cations (ion pairing):

I repeated the calculation for the activity coefficients of the cobalt ion and the $\ce{CoCl4^2-}$ ion. Bear in mind that the sum of the radii for $\ce{Co^2+}$ and $\ce{CoCl4^2-}$ will be greater than the sum of the radii for $\ce{Co^2+}$ and $\ce{Cl-}$.
Now if we consider the equation
$$K_\mathrm{eq} = a\frac{\ce{CoCl4^2-}}{a(\ce{Co^2+}){a\ce{(Cl-)^4}}}$$
Now if $K_\mathrm{eq}$ is a constant regardless of the solvent system, then as
$$a(\ce{Cl-}) = [\ce{Cl-}] \times \text{activity coefficient of } \ce{Cl-}$$
$$a(\ce{CoCl4^2-}) = [\ce{CoCl4^2-}] \times \text{activity coefficient of } \ce{CoCl4^2-}$$
$$a(\ce{Co^2+}) = [\ce{Co^2+}] \times \text{activity coefficient of } \ce{Co^2+}$$
Then if we have a new equation which is
$$\text{Random greek symbol} = \text{activity coefficient } \ce{CoCl4^2-} / {\text{activity coefficient } \ce{Co^2+} \times (\text{activity coefficient } \ce{Cl-})^4}$$
If we multiply "random greek symbol" with $K_\mathrm{eq}$' we will get a new equilibrium constant for the system.
($K_\mathrm{eq}$' is the $K$ value determined in water where it is an apparent constant - one calculated making the assumption that all activity coefficients are one)
We can get a new equilibrium constant $K$ (solvent composition) which we can plot against the fraction of the solvent mixture which is acetone. Here we have the graph.
Here K' is the true equilibrium constant which would be for a system in water with a zero concentration of cobalt chloride.
We can see as the acetone content increases that the cobalt cations are more likely to react with the chloride anions to form $\ce{CoCl4}$ dianions than they are in the aqueous system.

If you read the papers by people like Bromley[3]The Journal of Chemical Thermodynamics 1972, 4 (5), 669–673. who clearly expresses the ideas of Guggenheim you will find a specific ion interaction term in the activity coefficient equation (Bromley equation). I am sure that the value of the specific ion interaction constant for cobalt and chloride will increase as the water is replaced with acetone.
Sadly I very much doubt if the constants for water/acetone mixtures exist, most of the research effort has gone into aqueous only systems. You could make a lot of the measurements by doing UV/vis spectroscopy on solutions of chloride salts in acetone/water mixtures. But it will be a complex problem, I know that while sodium chloride is a strong electrolyte in water it is a weak one in acetone. I know a classic reaction in organic chemistry for the formation of alkyl iodides is to treat an alkyl bromide or alkyl chloride with sodium iodide in acetone. The larger anion size makes sodium iodide more soluble (and a stronger electroyte) in acetone than sodium chloride/sodium bromide.
If you look at the supporting information for [1], you will see at the end an example of thermochromism. Here mixtures of water, cobalt chloride and the lactic acid / choline chloride eutectic are shown both hot and cold.
This effect has also been reported with nickel chloride in deep eutectic solvents made from choline chloride by some chemists in China. Similar effects were seen for nickel by [2]. They have observed thermochromism which I think is due to a change in the coordination chemistry which in turn is induced by the temperature induced change in relative dielectric constant.

References:

Foreman, M. R. S. J.; Holgersson, S.; Mcphee, C.; Tyumentsev, M. S. Activity coefficients in deep eutectic solvents: implications for the solvent extraction of metals. New J. Chem. 2018, 42 (3), 2006–2012. DOI: 10.1039/c7nj03736h.
Gu, C.; Tu, J. Thermochromic behavior of chloro-nickel(II) in deep eutectic solvents and their application in thermochromic composite films. RSC Adv. 2011, 1 (7), 1220. DOI: 10.1039/C1RA00345C.
Bromley, L. A. Approximate individual ion values of β (or B) in extended Debye-Hückel theory for uni-univalent aqueous solutions at 298.15 K. The Journal of Chemical Thermodynamics 1972, 4 (5), 669–673. DOI: 10.1016/0021-9614(72)90038-9.

I also tried it experimentally now. I prepared solutions of cobalt chloride in the solvents, added enough water to discolor them and then added more solvent until it turned blue again. For acetone this shift does not take much solvent and the color is quite intense. For THF however, which has a lower dielectricity constant it takes a lot of solvent and only gives a pale blue. DMF on the other hand has no sudden jump from pink to blue but a huge gradient where it turns purple inbetween and the color is intense, too. I don't really understand what happens with the THF though.

THF is a better donor than acetone, I think that in some ways THF is a bit like water. It would be nice if you could post some pictures. Depending on your reputation you could edit my answer to add the pictures. Failing that we will find a way of doing it.

I can try that tomorrow when I'm back at the lab. I know that THF can act as a donor here but it was blue as well when I 'dissolved' the cobalt chloride in pure THF, I only added enough of the salt to fully dissolve it in THF so therefore it should turn blue again at some point.

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


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PF3 as a source of fluoride for substitution in alcohols
It is known that $\ce{PX3 (X= Cl, Br, I)}$ and $\ce{PX5 (X= Cl, Br)}$ are used to make halogen derivatives of hydrocarbons from alcohols. My question is why there's no information about using $\ce{PF3}$? At least I couldn't find any, so I assume $\ce{PF3}$ is not used for that.
A wild guess that phosphorus holds fluorine atoms tight, and no "free" fluorides are present for the last steps of the reaction to complete.

related https://chemistry.stackexchange.com/questions/41556/why-cant-pf3-be-used-as-a-lewis-acid-catalyst-in-electrophilic-substitution-of

Yeap, I've seen that, but is this case different?

I think it's quite similar. In halogenation these halides first act as Lewis acids.

Also a good question is whether or not fluoride is a good enough nucleophile.


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Different behavior of bromine radical with alkenes
Let's assume we have propene and it's undergoing two different reactions: a Kharasch reaction and allylic substitution reaction.
In both reaction bromine radicals are formed, and the attacking particle is definitely bromine radical, but in one case it goes to the double bond and addition happens, in the other - it breaks C-H bond, making an allylic radical.
Why in these two similar reaction, where the most "powerful" piece is the Br radical, such different products are formed?
My guess is that even if an allylic radical is formed in the case of using HBr and peroxide, the first molecule it meets is HBr, which makes H add to the radical and frees the Br radical again, so no effective interaction happens in summary. And in the case of allylic bromination the Br radical cleaves the C-H bond, since it is like less stable in the allylic position, and the next molecule this radical meets is either NBS or Br2, depending on the reagent.

To form propylbromide in the above case, a hydrogen source is needed, while to form allylbromide in the bottom case, no hydrogen source is added. This should control, which of the two reactions is observed.

@deusexmachina agree. Although Kharasch reaction sometimes involves using of H2O2 itself. Which is able to produce H radicals and OOH radicals.

This answer will help.


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Is dH(S,P) = dH(T,P) true (always)?
I came across an interesting question with some physical chemistry students today. Based on the following steps, we're uncertain whether the statement in the title is/could be true. Assuming $dN = 0$,
Enthalpy's natural variables are Entropy and Pressure:
$$dH(S,p) = TdS + Vdp \space (1)$$
Enthalpy can be expressed as a total derivative of Temperature and Pressure:
$$ dH(T,p) = \frac{\partial H}{ \partial T}\vert_p \space dT + \frac{\partial H}{\partial p}\vert_T \space dp \space (2)$$
Total Enthalpy is $$H = U + pV = TS - pV + \sum_i \mu_i N_i + pV = TS + \sum_i \mu_i N_i \space (3)$$
Thus, taking partial derivatives in (2),
$$ dH(T,p) = SdT + 0 \space dp \space (4)$$
This means that, if $ dH(S,p) = dH(T,p)$,
$$ SdT = TdS + Vdp \space (5)$$ must be true. Can anyone see something wrong?
This seems to boil down to a math question: are two differentials of the same function (always) equal if expressed by different variables?

I don't agree with you assertion in $(3)$ that $U=TS-pV$.  Where does that come from?

https://en.wikipedia.org/wiki/Internal_energy#Internal_energy_of_multi-component_systems  -- maybe that is a bad assumption since T and P are variable?

What that line $U=TS-pV+\sum_i \mu_i N_i$ actually tells you is that $G=\sum_i \mu_i N_i=U-TS+pV$ where $G$ is the Gibbs free energy.  You've neglected the chemical potential $\mu$ of the different species in your system - which will have a hidden and complex relationship with $T$ and $p$.

True, thanks. Nonetheless, the focus of my question remains. The partial derivatives of H are still the same (no $\mu$ or $N$ dependence)

https://en.wikipedia.org/wiki/Enthalpy#Other_expressions probably provides the relationship you want for the exact differential if $H(T,p)$ with a reference for the derivation - but as a rule $\frac{\partial \mu}{ \partial T}\ne 0$ hence leading your fourth equation to be incorrect.

With regard to chemical potential the answer to https://physics.stackexchange.com/questions/168704/chemical-potential-as-a-function-of-temperature is quite informative also.

I imagine the problem is that $(\partial H/\partial T)|_P = S$, because, as user213305 has mentioned, other thermodynamic variables will depend on $T$ too. On the other hand, the answer to your titular question is yes, because in that case we are just using different variables to describe the same thing.

I see. The dependence of pressure and temperature in $\mu$ should lead to a different result in (4). I don't know what that result is, but if someone can prove it to be $TdS + Vdp$ then I'll be satiated that $dH = dH$ regardless of variables

The reasoning stated is partially correct, but the final relation you arrived to is incorrect. I will try to explain why and write the N dependence explicitly for completeness. The crucial thing is that when one writes an expression such as $$\left(\frac{\partial H}{\partial T} \right)_{P,N}$$
what one really means is "take the partial derivative of $H$ written as a function of $T$, $P$ and $N$ with respect to $T$". When you take the partial derivatives in equation (2) then, you should take them considering $H$ as a function of $T$, $P$ and $N$. You have to consider then the expression
$H = H(T,P,N) = S(T,P,N)T + \mu(T,P) N$. If you differentiate that equation with respect to T and P the result is:
$$\left(\frac{\partial H}{\partial P} \right)_{T,N} = \left(\frac{\partial S}{\partial P} \right)_{T,N} T + \left(\frac{\partial \mu}{\partial P} \right)_{T} N~~~;~~~ \left(\frac{\partial H}{\partial T} \right)_{P,N} = \left(\frac{\partial S}{\partial T} \right)_{P,N} T + S + \left(\frac{\partial \mu}{\partial T} \right)_{P} N$$
If you replace those two relations in your equation (2) the result is:
$$ \mathrm{d}H = \left(\left(\frac{\partial S}{\partial P} \right)_{T,N} T+ \left(\frac{\partial \mu}{\partial P} \right)_{T} N \right) \mathrm{d}P + \;\left(\left(\frac{\partial S}{\partial T} \right)_{P,N} T + S+ \left(\frac{\partial \mu}{\partial T} \right)_{P} N\right)~\mathrm{d}T.$$
You can indeed equate this with your equation (1), which is what you ask in your main question. This is the same thing one ordinarily does when expressing a scalar as a function of different sets of coordinates, for instance $f = f(x,y) = x^2 + y^2$ and $f = f(r,\theta) = r^2$ this means, equating both, that $x^2 + y^2 = r^2$, which is a relationship that must hold if you want both $(x,y)$ and $(r,\theta)$ to refer to $f$ (this last sentence may be a bit tautological, I hope what it means is clear, note that it is certainly not "always" true that $f(x,y)$ and $f(r,\theta)$ are equal, since they are two different functions, albeit expressed with the same letter, for instance, $f(x=1,y=0) = 1$ but $f(r=2,\theta= \pi) = 4$, there must be a specific relation between the coordinates for these to be equal). If you do this you obtain:
$$\left(\left(\frac{\partial S}{\partial P} \right)_{T,N} T+ \left(\frac{\partial \mu}{\partial P} \right)_{T} N \right) \mathrm{d}P + \;\left(\left(\frac{\partial S}{\partial T} \right)_{P,N} T + S+ \left(\frac{\partial \mu}{\partial T} \right)_{P} N\right)~\mathrm{d}T = T~\mathrm{d}S + V~\mathrm{d}P.$$
Note that you would get the same expression even if you considered processes in which $\mu \mathrm{d}N$ wasn't cero, cause both terms would cancel out. If one remembers that $T\mathrm{d}S = T\left(\frac{\partial S}{\partial P} \right)_{T,N}\mathrm{d}P + T\left(\frac{\partial S}{\partial T} \right)_{P,N}\mathrm{d}T$ then this simplifies to:
$$\left(\left(\frac{\partial \mu}{\partial P} \right)_{T} N \right) \mathrm{d}P + \;\left(S+ \left(\frac{\partial \mu}{\partial T} \right)_{P} N\right)~\mathrm{d}T = V~\mathrm{d}P.$$
Dividing through by N:
$$\left(\frac{\partial \mu}{\partial P} \right)_{T} \mathrm{d}P + \;\left(\bar{S}+ \left(\frac{\partial \mu}{\partial T} \right)_{P} \right)~\mathrm{d}T = \bar{V}~\mathrm{d}P.$$
This is true if and only if:
$$\left(\frac{\partial \mu}{\partial P} \right)_{T} = \bar{V}~~;~~\left(\frac{\partial \mu}{\partial T} \right)_{P} = -\bar{S}.$$
These are correct relations and can also be deduced from the Gibbs-Duhem equation: $N\mathrm{d}\mu -V\mathrm{d}P + S\mathrm{d}T = 0$.


text:
Why does pH decrease when diluting a base?

Take a strong base such a sodium hydroxide
Adding water will result in adding a hydroxide ions as well as hydronium ions to the solution
However if the pH decreases this means that the hydronium ion concentration will increase , hence the hydroxide ion concentration decreased
Why is this so ? What about all the hydroxide ions that were added from the water
I understand the mathematics behind it , but it doesn't intuitively make sense

They're all there. But the solution is more dilute and pH only measures concentration...

You are overcomplicating it. The concept of pH and pOH stems from the concentration of $\ce{OH-}$ or $\ce{H+}$ per solvent unit $\ce{H2O}$, like so $\ce{\frac{[OH-]}{[H2O]}}$ or $\ce{\frac{[H+]}{[H2O]}}$. There is also a logarithm involved, but lets keep it simpler than that. You are adding far more of the denominator than the numerator, hence the fraction is becoming smaller and the pH is going towards pH of pure water.


text:
Why does 1,3-dimethylcyclopent-1-ene have geometric isomerism while 1-bromo-2-chlorobenzene doesn't?

The bottom compound has bromine and an R group connected to one of its SP2 double bond carbons, and chlorine and the same R group connected to the other. This fact, it would seem, indicates cis-trans isomerism.
The top compound has a methyl group and an R group connected to one of its SP2 double bond carbons, and a hydrogen atom and a different R' group connected to the other. Conversely, it appears that this precludes cis-trans isomerism.
First, Why does the top compound has geometric isomerism while the bottom compound doesn't given these facts?
Secondly, given that benzene is planar and its substituents and hydrogen atoms are level with the plane of the ring, are the methyl group and the hydrogen atom around the double bond of the top compound also flatly level as in benzene, or are they equatorial like in cyclohexane, slating above or under the plane of the ring?

Benzene ring is planar all right, but cyclopentene isn't.

So, is the cis-trans isomerism a result of the Methyl groups' at positions 1 and 3 possibly being at different sides of the ring's plane? Also, can the Methyl group at position 1 be axial as well as equatorial?

There is no cis/trans isomerism here.

I'm confused. In the assignment it's noted that 1,3-Dimethylcyclopent-1-ene has cis-trans isomerism. From what I could dig up, cycloalkenes can have only a trans configuration. If so, is the methyl group at the double bond perfectly planar or does it slant in such a manner as to form a trans isomer?

IUPAC defines geometric isomerism as an obsolete synonym for cis-trans isomerism. This makes it a subset of diastereoisomerism, meaning that cis and trans isomers exhibit different properties such as melting point or density. The descriptors cis and trans themselves are also obsolete for double bonds and proper nomenclature should instead use (E) and (Z) as stereodescriptors since these are also applicable to tri or tetrasubstituted double bonds.
While linear or branched alkene chains can — as soon as the double bond is substituted asymmetrically — typically exist in both the (E) (or trans) and (Z) (or cis) isomers, this is not true for any cycloalkenes with less than eight carbon atoms. (E)-Cyclooctene is the smallest and most strained (E) configured cyclic alkene. You can use a molecular modelling kit to find out why this is the case.
Therefore, neither of your two compounds can exhibit cis-trans isomerism. Both ring sizes — the six-membered benzene ring and the five-membered cyclopentene ring — are smaller than eight and can therefore not accomodate an (E) configured double bond.

In an additional exercise, we may investigate whether other types of stereoisomerism can apply to the compounds in question — basically this boils down to whether enantiomers can exist. The flat benzene ring is $C_\mathrm s$ symmetric and thus no enantiomers are possible. However, the 3-methyl group on 1,3-dimethylcyclopentene can point both above (R) or below (S) the paper plane.

Neither has geometric isomers.  1,3-Dimethylcyclopent-1-ene is necessarily in its (Z)-configuration, because a five membered ring can't contain a trans double bond.  The smallest stable trans cycloalkene is cyclooctene.

Think about what the (E)-isomer of 1,3-dimethylcyclopent-1-ene would look like, if it existed!

In the assignment it's noted that  1,3-Dimethylcyclopent-1-ene has cis-trans isomerism. I was thinking it could be due to the methyls' at 1 and 3 being on either different sides or on the same side of the ring's plane.

This is not the typical understanding of the term geometric isomerism, which is synonymous (according to the IUPAC) with cis-trans isomerism.  You can't have a trans double bond in a ring that small.

So, the geometric isomerism only concerns with the double bond? Also, will the methyl group at the double bond slant or be planar?

Because the bottom one is a benzene, which is flat and aromatic. All its substituents would be in the same plane.

But, even so, the mentioned double bond of the Benzene resembles an Alkene's double bond, which is two-dimensionally planar as well. Doesn't this qualify it as having geometric isomerism?

Frank it is easy to see that is the same compound. You can virtually put two of these molecules literally on top of each other. Answer by WM6 and Jan are correct. There is an error in your assignment.


text:
Reducing the thickness of a layer of elemental sulfur
I want to make a very thin film of elemental sulfur. While mechanical polishing can bring a film to 10 microns in thickness, but I need to bring my thickness lower (hundreds of nanometers). Although I searched, I was unsuccessful in finding any methods for chemical etching elemental sulfur.
Are there any methods for reducing the thickness of elemental sulfur below 10 microns, preferably to lengths on the order of hundreds of nanometers?

Rather than trying to make a thick layer thinner, how about vacuum deposition of sulfur? In other words, just make a thin layer.

As @MaxW states, vacuum (or vapor) deposition is the way to go. I've produced layers of gold just a few atoms thick, so thin that they are completely transparent or a clear, pale, green, that way.

I don't have access to vapor deposition type coating systems for this process, so I was hoping to find some way to do this in a chemical etch method.


text:
Explanation of bond angles in the aluminium chloride dimer
In order to attain stability $\ce{AlCl3}$ dimerises to $\ce{Al2Cl6}$ whose structure is depicted as:

I want to understand why one of the bond angles is $79^\circ$ and the other $118^\circ$. Is it possible to justify this using Bent's rule?

Related: https://chemistry.stackexchange.com/questions/34842/why-is-bcl3-a-monomer-whereas-alcl3-exists-as-a-dimer/64600

https://chemistry.stackexchange.com/questions/19715/why-does-alcl3-dimerise-to-al2cl6-at-low-temperatures

@NilayGhosh that answer actually fully answers this question.

@StianYttervik it does not. Which part of the answer deals with bond angles?

@abcd I see there were two links, the first one is the one I meant. Here is the important bit for your question: "So after all we have two kinds of bonds in the dimers four terminal X−Cl and four bridging X−μCl bonds. Therefore the most accurate description is with formal charges (also the simplest)." So, if it is correct to describe the dimer with formal charges, then the deviation in bond angles is in perfect compliance with Bent's rule.

Related: https://pubs.acs.org/doi/abs/10.1021/jp9842042

@ApoorvPotnis That paper does make me feel that this question does not at all have a very decent answer...

@StianYttervik I think Martin's calculation does not really address why the other Cl-Al-Cl angle is 118. Sure, bent's rule will explain why the inner Cl-Al-Cl is 79 < tetrahedral, it does not explain why the other angle is larger than tetrahedral (and using bent's rule there will mean those Cl will have more negative formal charges than the bridging Cl)

The bonding in aluminum chloride ought to be considered largely ionic. The electronegativity of $\ce{Al}$ is $1.47$ vs $2.83$ for $\ce{Cl}$ (for comparison: $\ce{C}$ $2.50$; $\ce{Mg}$ $1.23$; $\ce{Na}$ $1.01$). Molecular orbitals and covalent bonding can be added for a fuller picture of electron densities.
Aluminum chloride solid (Melting point $\pu{193^\circ C}$, under pressure) has a cubic close-packed array of chlorine atoms with only $2/3$ of the octahedral holes between every other pair of chlorine planes occupied by the aluminum atoms. At low temperatures in the vapor state, aluminum chloride is a tetrahedrally coordinated dimer, and at very high temperatures it is a trigonal monomer. So aluminum can easily have $3$, $4$ or $6$ coordination. Whatever directionality the bonds have from hybridization of the $\ce{3s}$ and $\ce{3p}$ orbitals is mixed with plain electrostatic (ionic) attraction.
Imagine putting just the 6 chlorine atoms together. There will be two tetrahedral holes for aluminum (see image below). The radius ratio
$$\frac{\ce{Al^3+}}{\ce{Cl-}} = \frac{\pu{50 pm}}{\pu{181 pm}} = 0.276$$
is well within the stability range ($0.225 - 0.414$) for tetrahedral coordination. So you might expect all the $\ce{Cl-Al-Cl}$ bond angles to be $109^\circ$.
But while the void for the aluminum is big enough, the distance between the internal chlorine atoms would be only $\pu{300 pm}$ in this configuration; the chlorine atoms are squeezed (ionic radius $\pu{181 pm}$). In addition, the $\ce{Al-Al}$ distance is only $\pu{250 pm}$ and the $\ce{Al^3+\bond{-}Al^3+}$ repulsion would be extreme. (In the solid, the octahedral holes are farther apart) So both aluminum atoms and the internal chlorine atoms are pushed apart, consistent with imagining more $\ce{p}$ character in the central bonds and therefore less $\ce{p}$ character in the external chlorine atoms, so $\ce{Al}$ goes to $\ce{sp^2}$ (or nearly so). The internal $\ce{Cl-Al-Cl}$ bond angle is even less than $90^\circ$ (less than you would expect for two $\ce{p}$ orbitals), so π bonding needs to be considered.
Aluminum chloride is a strong Lewis acid. Its solubility in organic solvents is due to formation of complexes. It is not soluble in the way that carbon tetrachloride or hexachloroethane are soluble. The salt-like (ionic) character of solid aluminium chloride is still present in $\ce{Al2Cl6}$; the aluminum atoms are surrounded by a cocoon of chlorides. Solid aluminum chloride can be ground into particles and nano particles; $\ce{Al2Cl6}$ is like a pico particle of salt.

(1) Did you mean " At low temperatures in the solid state,"? (in the second paragraph) (2) Also, "So aluminum can easily have 3, 4 or 6 coordination. " do you have an example of a 6 coordination aluminium compound?

In solid aluminum chloride, the aluminum is octahedrally coordinated - in the holes between two close packed chlorine atoms.

Well, JD Lee says that the compound is ionic at low temperatures  or when dissolved in waer.


text:
What are the dangers associated with a copper sulfate CuSO4 + Water H2O solution aprox 5 gallons at room temperature in a room with no ventilation.
Will there be hazardous gases in the room? is the evaporation of this solution harmful in any way to animals or humans?

There should be no problems with evaporation/toxic off-gasing of any kind. Where did you get this solution and how concentrated is it? There should be a Safety Data Sheet if you got it from a reliable source.

The only issue I can see is that as it evaporates it will leave behind a solid residue, which when mechanically disturbed could cause dust that would be unhealthy to inhale. Just make sure you store it in a closed container.

It depends on how it is being used.  For example, dropping hot metal into the solution or electroplating at high current might cause a bit of spray in the air.

Don't drink it.

Of course there can be hazardous gases. If you are boiling the water the hot steam produced could burn you and or suffocate you depending on the rate of production. We live in a world where the SDS sheet for table salt tells you to call a poison hotline if ingested.

The solution will be for a low voltage 0.5v  copper electroplating for apron 3 moths, (open container) it will be in constant motion since we will have a magnetic stirrer,  the container will always be full of the solution or at least 3/4 of the container, any recommendations on the temperature for better results?  No intention to boil, my question is basically will the vapors leaving the container be poisonous or harmful for humans or pets? or the evaporation is simply water, leaving all the copper sulfate CuSO4 in the container as solid residue.

airhuff its Copper Sulfate Pentahydrate 99.9% Crystals + water solution,  lets say high concentration.

To have another person be notified of your comment, you need to place an at sign in front of their username, like @aafco.

When a solution of copper(II) sulphate evaporates, the only gaseous product is water vapour. Like any ionic compound, copper(II) sulphate will be left behind as a solid residue — presumably crystalline — after water evaporation.


text:
Can absolute enthalpy values be used to determine reaction enthalpy?
Our teacher added this question to an exam, and there was a huge disagreement with my classmates and teacher about the correct answer. One side favoured answer 2, the other side favoured answer 4 being the false statement.
I personally selected answer 4, as it did contradict with the other 3: option 1 has $\Delta H < 0$, option 2 has $\Delta H = H_\mathrm{f} -H_\mathrm{i} = \pu{-500kJ}$ and option 3 has $\Delta H = \pu{-500 kJ}$,  so option  4 contradicts the above. However I would like to hear your thoughts.

Which of the following options that refer to the enthalpy of a specific chemical reaction is not correct?

$H_\mathrm{i} > H_\mathrm{f}$
The enthalpy of the system has altered from the initial value of $\pu{1600 kJ}$ ($H_\mathrm{i}$) to the final value of $\pu{1100 kJ}$ ($H_\mathrm{f}$)
$\Delta H = \pu{-500 kJ}$
$\Delta H = \pu{500 kJ}$

You can't measure an absolute value of enthalpy.

Notice also that 1, 3, and 4 all refer to delta values, explicitly or implicitly.

@wolphram That's a legend from dark times before relativity, another outdated theory needing debunking. https://chemistry.stackexchange.com/a/20103/9961

Assuming that 3 are true, the only self-consistent set are 1,2 and 3.
Importantly, statements 3 and 4 contradict one another, so one of them must be the incorrect statement.
Consider a reaction with a change in enthalpy ($\Delta H$) of $\pu{-500 kJ/mol}$, and statement 3 is true:
\begin{align}
  \Delta H &= H_\mathrm{f} - H_\mathrm{i}\\
  \implies H_\mathrm{i} &= H_\mathrm{f} - \Delta H\\
  H_\mathrm{i} &= H_\mathrm{f} + \pu{500 kJ/mol}\\
  \implies H_\mathrm{i} &> H_\mathrm{f}
\end{align}
We've shown that statement 1 is true also.
As wolphram says in the comments, you can't measure an absolute value of enthalpy, but you can measure it relative to an chosen zero enthalpy.  If you chose zero enthalpy such that $H_\mathrm{i} = \pu{1600 kJ/mol}$, then by the first equation above we can say that:
\begin{align}
  H_\mathrm{f} &= \Delta H + H_\mathrm{i}\\
  \implies H_\mathrm{f} &= \pu{-500 kJ/mol} + \pu{1600 kJ/mol}\\
\implies H_\mathrm{f} &= \pu{1100 kJ/mol}
\end{align}
We've shown that statement 2 is true as well.
Statement 4 contradicts statement 3 (as well as the other two), so statement 4 must be incorrect.

We still refer to the referenced values as deltas. All reported values are already referenced this way and those values are indeed reported as $\Delta H$ and not $H$. I agree that your answer and rationale are correct, but I have serious issues with the wording of the question then...

@Zhe The system itself can't have a $\Delta H$ without reference to something, but it can have a $\Delta H_{form}$ (of formation) which, as you state, would be a vastly more physically sensible thing to talk about that $H_f =1100$.  I definitely agree that statement 2 is worded very badly and the general phrasing of the question ought to be serious rethought.


text:
Why can we use standard entropy when determining temperature at which a reaction becomes spontaneous?
In this chemistry problem, $\Delta H$ and $\Delta S$ are given at $298\text{ K}$, and it asks for the temperature at which the reaction becomes spontaneous:

For the reaction $2\text{SO}_3 \rightarrow 2\text{SO}_2 + \text{O}_2$, $\Delta H_r^\circ = +198 \text{ kJ/mol}$ and $\Delta S_r^\circ = 190 \text{ J / mol}\cdot\text{K}$. At what temperature will the forward reaction will become spontaneous?

I understand to solve this problem; the Gibbs free energy equation,
$$\Delta G = \Delta H - T \Delta S,$$
can be used to determine the temperature by assuming $\Delta H$ and $\Delta S$ are constant and solving for $T$ when $\Delta G = 0$. This gives $T = 1042\text{ K}$.
However, I don't understand why is this assumption valid. Isn't it true that $\Delta S_r$ would change at different temperatures? Why is it that the value for $\Delta S$ for this reaction under standard conditions can be used to calculate $\Delta G$ at $1042 \text{ K}$?

yes, $\Delta$G will change with temperature, the values given are likely for 298K but I'm not checking. there is also the matter that the reaction is assumed to occur at 1 atm and no consideration is given to an equilibrium mixture of the two gases.

as an aside, spontaneous is a poor word for thermodynamic reactions given that some "spontaneous" reactions are kinetically hindered from reacting and require a catalyst or concentrated heat for activation.  From this some people get the impression that the reaction will proceed the instant that the system reaches the temperature calculated from $\Delta$G, which is not true (think flash point vs. autoignition temperatures). a more proper term is "thermodynamically favored".

You are correct that entropy and enthalpy should change with temperature. Are you sure enthalpy and entropy are given at 298K? Contrary to what many students (and even teachers!) seem to assume, the standard state for standard molar enthalpy and entropy takes no account of temperature.  Related: https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation

@Tyberius Yes, the values are given at 298K. Could you elaborate on how standard molar enthalpy and entropy take no account of temperature? I thought that entropy would change greatly at different temperatures.

@A.K. I get that $\Delta G$ changes, and yes, the reaction is assumed to be at $1\text{ atm}$. I don't understand why the formula for $\Delta G$ still holds though, even using $\Delta H$ and $\Delta S$ values at 298 K.

@EricZhang I don't mean to suggest that enthalpy and entropy don't depend on temperature. I was trying to say that standard enthalpy/entropy doesn't specify a temperature. You can have standard enthalpies/entropies at different temperatures.

The question implicitly assumes that $\Delta H^\circ$ and $\Delta S^\circ$ are independent of temperature.
The temperature dependences depends on the change in heat capcity ($\Delta C_p$) are given by $\Delta H^\circ(T_2) = \Delta H^\circ(T_1) + \Delta C_p(T_2-T_1)$ and $\Delta S^\circ(T_2) = \Delta S^\circ(T_1) + \Delta C_p \ln(T_2/T_1)$.  (These equations assume $\Delta C$ is constant.)
So if $T_2-T_1$ or $\Delta C_p$ is small then $\Delta H^\circ$ and $\Delta S^\circ$ will be fairly constant.


text:
Do all supercritical fluids dissolve the same substances?
In the case of regular fluids, polar solvents dissolve polar molecules and nonpolar solvents dissolve nonpolar molecules. Is there something analogous for supercritical fluids? In other words, does the shape of the supercritical fluid molecule affect what it dissolves, or do all supercritical fluids dissolve the same set of substances?

Related: Supercritical fluids polarity (no answer yet).


text:
Plastic Softening Compounds
I am trying to identify a compound that could be used to soften a Rubic’s Cube’s plastic to make it turn faster and smoother.
I have read that a number of cubers have used a Boron treatment to accomplish this.  Unfortunately Boron is hazardous and I am looking for a safer alternative that would accomplish the same thing.
Thanks!

What about a VR-headset and any Rubik's Cube game made for Oculus or Vive? No physical object, no friction :)

Don't think that you'd want to "soften" the plastic since that would make it sticky. Rather you'd want to harden and polish the plastic joints.

I used to use molydisulfide (MoS2) dry lubricant - it worked quite well. Disassemble the cube, lightly coat the interior faces, and reassemble.


text:
Are electrons in a helium atom diametrically opposite?
The helium atom is represented with electron opposite on same orbital:

not fully opposite on same orbital and even on two normal orbitals:

But in this case wouldn't the electron crash?
Can you please specify if their real position is known, and if it is not, what is the most plausible solution?
Also, we know from Pauli's principle that they must have different spin:
but isn't in that case the electron with spin up behaving like a positron or a proton? How is that possible?

Electrons are quite totally not what you think they are.

@IvanNeretin, why then are they always represented like that? Even in QM,  they are a probability funcion,but if one electron materializes in one point,is the other supposed to materialize at an opposite point wrt to nucleus, or it can be exiceted anywhere?

They are (not really always) represented like that for the same reason that Earth is represented as a flat disk supported on the back of a huge turtle.

Electrons don't have an orbit. An orbital is not the same as an orbit.

Probably best to try and digest Wikipedia's page on Atomic Orbitals before going further.

That you know of probability functions and still spawning such nonsense is kinda impressive, obviously in a negative way...

@DSVA It's only some of interpretations (especially Copenhagen one) that give this "non-existance".

@Mithoron which one gives no orbitals but rather orbits and how do they get around the problem the Bohr model got?

@DSVA That's not what I meant. I was thinking about: https://chemistry.stackexchange.com/a/13587/9961

Throughout your question it is clear that you are considering electrons as particles. This is not entirely incorrect; for example, the cathode ray tube is best explained by electrons being considered negatively charged particles that can travel through vacuum and be deflected by (electro)magnetic fields. However, particles as small as electrons can also be considered waves as is evident by electron diffraction. The probably most frequently cited connection between the particle nature and the wave nature of electrons and others is the de Broglie wavelength given by equation $(1)$.
$$\lambda = \frac hp\tag{1}$$
This discovery earned de Broglie a Nobel Prize for the theory and Thompson and Davisson for the supporting diffraction experiments.
Quantum mechanics is the underlying theory that attempts to shed light into how electrons and other microscopic particles behave according to their wave nature. The key principle is that each particle is described by a wave function, typically labelled $\Psi$, dependent on space and/or time coordinates. A key part of the theory is the existence of operators which can be applied to the wave function to observe observables — physical parameters such as a particle’s energy or magnetic moment. Operators and observables observe the Schrödinger equation $(2)$ in which the Hamiltonian operator $\hat H$ can be replaced by any other operator to observe a different observable than energy $E$. Naturally, the wave function may be a different one depending on the operator in question.
$$\hat H\Psi = E \Psi\tag{2}$$
Using this equation alongside the derivation of the correct Hamiltonian, the wave function for a singly negatively charged particle rotating and translating freely in a spherical electric field of a given positive charge can be determined. This long and tangled sentence boils down to: we can calculate the wave function of an electron in the vicinity of a nucleus. The nice thing is that the electron’s energy is quantitised and dependent only on the principal quantum number $n$ while other quantum numbers are also required to fully describe the wave function ($l, m_l, s, m_s$). The quantisation of the electron’s energy provides a nice and succinct explanation of the hydrogen spectral lines.
Unfortunately, it is not possible to solve the Hamiltonian for two electrons. The energy of a single electron consists of its kinetic energy (described with the $\hat T$ operator) and its potential energy towards the nucleus ($\hat V_\ce{el-nuc}$). If a second electron is introduced, we now need to consider electron-electron repulsion potential ($\hat V_\ce{el-el}$) which we cannot solve. (In my quantum mechanics lecture, the professor gave the impression that this is a problem proven to be unsolvable but I don’t recall seeing any proof.) Therefore, we cannot calculate the wave functions of the two electrons of a helium atom a priori. However, multiple computer-assisted calculation algorithms exist that allow us to come close.
Whatever the actual solution, the key is that electrons in an atom or molecule should primarily be considered waves, not particles. Their wave function contains the spin orientation quantum number $m_s$ which can be either $+\frac
12$ or $-\frac12$, depending on whether the electron is spin up or spin down. These two wave functions basically operate the same space at the same time. Yet the electrons do not collide because we are dealing with waves, not particles.
You considered the electron with opposite spin to be a positron. That is not correct. A positron has the opposite charge with respect to an electron ($\ce{e+}$). Spin and charge are, however, two different properties. Spin is probably best considered a magnetic moment that would arise if a charged particle were to rotate around itself. However, this macroscopic picture is not fully valid; no macroscopic pictures you draw of microscopic particles are.
Finally, note that all particles are subject to the uncertainty principle. This means that we cannot determine the position of an electron accurately since it would imply maximum uncertainty with respect to impuls. Likewise, we cannot accurately determine an electrons impuls. And another complicating matter is the observer effect: by attempting to observe where an electron is (approximately) we have to influence the system so strongly that it is no longer what we wanted to observe. This is because even very small energies have a significant effect on small scales.

Summary
Your notion of electron particles is not fully supported by the currently accepted (and experimentally validated) theory. Rather, think of them as waves. Waves can interfere but cannot collide. Furthermore, since the wave function does not give a location, we cannot accurately determine the location of an electron.
Spin and charge are two different things.

It's not correct to describe as a wave. The electron exhibits properties that, in aggregate, produce wave-like phenomena.


text:
Calcium concentration in NMDARs e Nitric Oxide Synthase
My question is related to calcium $\ce{Ca^{2+}}$ concentration in NMDA receptors and nitric oxide synthase. I've found a very simple model that describes how the time change of calcium can be related to the NMDA current (http://www.pnas.org/content/98/22/12772.full.pdf, Appendix B), but, however, I would like to find some relationships between the time derivative of $\ce{NO}$ concentration and the time derivative of $\ce{Ca^{2+}}$.
Does exist an expression that could solve this problem?


text:
How does a combustion process produce carbon monoxide?
My question is in the context of combustion heater safety- I want to identify the mechanism by which CO is produced, and hopefully have additional understanding, and clues to the warning signs, to identify unsafe stoves or installations. The primary safety measures will always be CO detectors and good ventilation.
The worst horror stories usually concern charcoal barbecues in confined spaces- solid carbon burning at about 900 °C. Is it a safe educated guess that the main combustion product here is CO, which fails to burn fully to CO2 in depleted oxygen levels? Or is a more complex process at work?
How does the yellow combustion of an oil or gas flame produce CO? Is it a simple matter of poor gas flow in the flue, and depleted oxygen in the feed air?
Thanks for any help

That's right, it is ultimately a matter of incomplete combustion due to the lack of oxygen.

CO2+C->2CO, kind of a comproportionation reaction.

Since your question is in "the context of heater safety", you might get a more specific and relevant answer if you briefly described the nature of the heater of interest and the environment in which it operates.

@airhuff Good point. A bit more context- my primary interest is in outdoor burners, with a slight twist in that I burn wood to charcoal for use as biochar. On the strength of this, I'm being approached for advice from residents of an Off-Grid intentional community on installation and operation of wood-burning stoves to heat cabins- advice which I'm not really experienced to give. I don't want anyone gassed or cremated because of my advice or lack of it, so the best I can do is learn to spot  installations that are really dangerous, and why, before they get lit up in the cold weather.

If combustion occurs in oxygen, (at least) the following reactions are relevant:
$$\ce{C + O2 -> CO2} $$
$$\ce{C + 1/2 O2 -> CO} $$
You can favor the first reaction by having more oxygen in the gas phase, since it required twice $\ce{O2}$ as the first.
However, one of the most important factors is the temperature: high temperature always favors the $\ce{CO}$ product, since reaction wants to proceed fast and react with any available oxygen.
You can clearly see these dependences in the Figure 6 of this article, at low temperatures the CO/CO2 ratio is low, but quickly increases by heating.
More detailed information is easily found in some classical papers, a full PhD thesis on it, or even calculations with modern quantum chemistry methods.


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"Non-stick" action on ceramic-coated pans
I understand how PTFE, or Teflon as it is commonly called, prevents anything from sticking to pans; apparently, anything coming into contact with the PTFE coating has ignorably tiny van der Waals forces with it.
However, there are also ceramic coated pans, with similar properties.
How do ceramic coated pans prevent anything from sticking onto the pan?


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questions for my 13 year old's STEM class. (Users of Potassium)
Would you please answer the following questions for my 13 year old girl's STEM class?

Do you have a story about a time something went wrong using Potassium?
What are some of the precautions you take when using Potassium?
What are some thing Potassium is used for?
Is there a suitable substitute?

Most common use would be making potassium alkoxides by reaction with alcohols.
Have also seen it used as a drying agent for preparing ultra-dry THF. This leaves potentially hazardous residues for the person who has to clean the still. Seen more than one fire because of potassium residues at the bottom of the flask.
There are certainly other drying agents for preparing dry solvents. But if your chemistry requires a potassium compound and not sodium or lithium then that's what you have to do.
It is usually supplies as chunks stored under liq. Paraffin. Handling it under hexanes with appropriate gloves, goggles etc is safe enough. I hope I don't have to tell you to keep it away from water.