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Molecular dynamics (Mac OSX user): what software can I use to plot RDFs for NPT simulations? VMD only works for NVT (constant volume)
I would be very grateful if someone could recommend a way to plot RDFs for NPT ensemble molecular dynamics simulation results.
I use CP2K to run NVT simulations and usually use VMD to process results, but I don't see a way in VMD to deal with NPT results, where the volume is different at each time step.  I'm hoping that someone can recommend a piece of software that can deal with my CP2K output (i.e., a position file in terms of atomic xyz coordinates and a file that contains volume output) on a Mac or Linux system.
I have spent several hours online looking for a solution, but haven't even come across discussions of this problem.  This seems odd so maybe I'm using the wrong search terms.  If anyone can point me to online discussions of this issue, that would be great.
I'm also not having much luck tagging this question!  If anyone knows of any molecular dynamics-related tags I could use to reach people or perhaps of a more appropriate StackExchange site, I would welcome any suggestions.  Thank you.

What happens if you attempt to use the existing processing tool to analyze the NPT data (pretending it's NVT)?

@Buck Thorn, thanks.  Let me know if my explanation isn't clear and I'll rewrite it.  VMD is set up to deal with NVT-type simulations where the volume is constant.  There's a GUI where a sinlge volume value can be entered.  So there's no way that I can see of importing my cell volume file and using a different volume value at each time step.  however, VMD has a sort of coding language and can be used via a terminal.  So I am wondering if there's a way to change the volume that way.  I have so far averaged my volumes for each simulation and have used those, but wonder if my RDFS are meaningful.

@Buck Thorn, there don't seem to be any other tags that would be appropriate.  "Molecular dynamics" and "ensemble" and related terms don't seem to be established as tags.

That's an important clarification. I'm guessing you mean this gui: http://www.ks.uiuc.edu/Research/vmd/plugins/gofrgui/. The gui invokes the tcl command "measure gofr" with additional modifiers as described here: http://www.ks.uiuc.edu/Research/vmd/current/ug/node138.html While somebody else might have the script you are looking for I can recommend familiarizing yourself with the tcl scripting language in vmd as not everything is "canned" and you may want to process using your homebrew macros.

I don't know how gofr works but you can try inspecting it. It probably sets the max r cutoff based on the box size (or volume I guess). In NPT the volume will vary but you may get away with using the smallest volume sampled in the ensemble from which rmax is computed, but that is just a guess.

I'll keep the tag because it's the only one I can see that relates to the problem.  people who are familiar with quantum chemistry may have run into the problem.

I see that CP2K is a QM sim package, but the question is about VMD. But ok, I see that the userbase extends beyond atomistic sims.

Thanks for those links. I'll go through them.  I have used the tcl scripting language before and am slowly getting up to speed, but it would be great if there were a straightforward way to do this without having to spend the time it would take to reach a level that would allow me to write my own code.  If anyone out there knows of an easier way, that would be great.

What are your thoughts on using an average volume (computed from the volumes at each time step) in the existing GUI VMD method for calculating RDFs?

I'm not sure. My intuition tells me using a lower bound is safer but I'd look for the script gofr first. Depends also how much the volume is varying. For an incompressible substance it might not make a difference.

Thanks for your opinion on that, Buck Thorn.  I will look for a script, but so far have had no luck!  I am just about ready to submit a paper where I've used an average volume for my RDFs, but suddenly started feeling uncomfortable.  I'm not a mathematician and appreciate any opinions anyone has.  Appreciate your time.

The volume is probably only used to select a cutoff. If the cutoff is greater than the box size than presumably it has to apply pbcs. I also haven't found the script, I think it's actually a compiled subroutine.

I could very well be talking nonsense because this sort of thing goes beyond my expertise, but my concern is that the volume is used to determine how far apart atoms in the box are from each other.  So therefore, if using an average volume is somehow a problem (if it is squishing atoms too close together or making them be too far apart from each other), I might be calculating bond lengths that are incorrect.  That will mean that I won't correctly identify stable complexes in my system.  Anyway, I will keep searching and considering your comments.  Thanks.

My guess is it assumes the same units across ensemble snapshots and doesn't rescale, but again it's a guess. If I'm wrong then you do have something to worry about.

I recommend testing on a NVT sim result using two different volumes and other gofr parameters to see if they make a difference.

Thanks.  I'll have to think about what you said.

Thank you, Buck, for all your suggestions.  You're right that a good way to check the influence of the volume is to try varying it.  I did that by using the maximum and minimum volumes of the fluctuations in one simulation.  There was some variation, so the volume does matter!
I was lucky enough to hear about "travis," an amazingly simple-to-compile code (even for Macs) available online for free.  I thought I'd share that information here because other people are bound to run into this same problem.  This code, travis, takes my trajectory file and my volume file and then asks a number of questions interactively.  It outputs a nice CSV file that can be plotted.
Thanks again for your help, Buck.  Really helped me think this through.  I'd like to ask the moderators or whoever is in charge of tagging to please add some molecular dynamics-related tags if possible!  Thank you.

It's cool that you found this software, that looks very useful for dissecting your simulations. I would be careful interpreting the effect of altering the volume setting in the gofr gui. I found the associated c routines which it presumably calls, you'll find them in the VMD source code (Measure.C).

regarding "molecular dynamics" tags, I took the freedom to add one to your question yesterday.

Reading the help page for "measure", at http://www.ks.uiuc.edu/Research/vmd/current/ug/node138.html, it says: "The size of the unitcell has to be stored in the trajectory file or has to be set manually for all frames with the molinfo command. The command uses by default only the current active frame for both selections." Not sure how this impacts your analysis. I take back a previous comment regarding pbcs etc as "usepbc=0" looks to be the default. I have to say again that you may get more mileage from posting a question at the vmd users forum.

Thanks a lot, Buck!  That's good advice.  I wasn't aware that  there are VMD C code routines.  They'll be easier for me to understand than the VMD tcl script.

Thank you also for adding the MD tags!  I think your comment above means that in future, I need to set something in CP2K before running it that includes the effects of a varying unit call in my trajectory output.  I think I know how to do this now.  Unfortunately, I have many simulations that I can't redo, so I'm over the moon to know about travis.  I was very worried I would not be able to use my existing simulations.  Anyway, thank you so much for your help.  Really calmed me down and put me on the right track.


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HPLC Columns For Compounds Not Soluble In Non- Or Low Polar Solvents
Can someone recommend HPLC columns for compounds which are not soluble in no- or low polar solvents (hexane, isopropanol, toluene, and dichloromethane)?

I think what you are looking for is reverse phase hplc which employs a polar mobile phase.

Try searching Google Scholar and write HPLC separation and the names of your compounds. You will find plenty of publications and of course column names. If your compounds are new, you should test the solubility in MeOH, ACN, and 50:50 MeOH/H2O and 50:50 ACN/H2O. Do they dissolve in such solvents systems? If they do, you would use C-18 RPLC columns. There are literally hundreds of columns with similar C-18 chemistry.
If your compound is highly polar (carboxylates, amines, phosphates etc.), even bare silica column can do the separation. This technique is called HILIC. I have written several articles on HILIC. Try Google Scholar for reviews on HILIC.

Very detailed articles. I just found what I need there. Many thanks! Great forum, Love~

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Calcium carbonate and hydrochloric acid - data analysis
I conducted an experiment in which $\pu{10mL}$ of dilute hydrochloric acid was added to a beaker, and this acid was heated up to room temperature ($\pu{22 ^\circ C}$) using a Bunsen burner. Then, $\pu{0.25 g}$ of calcium carbonate was then added to the beaker, and the time taken for it to dissolve (stop bubbling) was recorded. This was repeated at 3 other temperatures: $\pu{50^\circ C}$, $\pu{75^\circ C}$, and $\pu{100^\circ C}$.
Here is the graph with the results:

So, my question is: why is there a 2-second difference between $\pu{50^\circ C}$ and $\pu{75^\circ C}$, and then a 6-second difference between $\pu{75^\circ C}$ and $\pu{100^\circ C}$? This is despite the temperature difference ($\pu{25^\circ C}$) staying the same - so why does the reaction rate change? I think this might have something to do with enzyme activity, but I am not sure.
And why is there such a drastic difference between $\pu{22^\circ C}$ and $\pu{50^\circ C}$, a whole 48 seconds? Yes, the temperature difference is $\pu{28^\circ C}$, not $\pu{25^\circ C}$, but that reduction is much more significant than at all the other temperatures. I think it might be something to do with the reactant concentration changing as the experiment progresses - again, I am not sure.
Any responses to this are greatly appreciated.

Enzyme? Why are you mentioning enzyme activity

I am not sure either! I was searching the internet about this and that was all I could find.

This whole experiment makes IMO little sense. This is heterogeneous reaction - one might think it's simple, but it's virtually impossible to analyse without data you'd not think about. Even if you had almost identical monocrystals, kinetics depend on things like CO2 solubility, convection, nucleation sites for CO2...

@BlakedeRaat  Three replications per temperature are better than one, of course, but still are very very few, statistically speaking.  Equally, if you would have more data per temperature, an indication of estimated error / standard deviation would help to infer if changes in reaction time are significant, or not.  «Normal» spreadsheet programs sadly often do not provide this (keyword box-whisker plots, e.g. https://en.wikipedia.org/wiki/Box_plot,  http://lithoguru.com/scientist/statistics/StatReview_Lecture2&3.xlsx) and if interested, you should venture out programs dedictated to this task.

I have plotted your data as time(t) vs temperature(T) in kelvin and then fitted the plot with this equation:
$$ t = \frac{C}{e^{-F/T}}$$
where C and F are just parameters used for the fitting.
The Arrhenius equation gives the reaction rate relationship with temperature:
$$ k = \frac{d|P|}{dt} = Ae^{-Ea/(Kb*T)} $$
If you then integrate you can find the relationship between t(time) and temperature and this gives you an equation similar to the first one that I used to fit your data.
Actually, your data fits reasonably well with the model:

But what is the name of the first equation, the one with t =? I want to research a bit more about this. In particular, what is the name of the variable: −/? What is the C variable or constant called?

Also, how did you create this graph?

The equation with t doesn't have a name, I just derived it  from the differential  arrehnius equation but I inglobated more constants within the same letters C and F for the fitting.

I used origin and defined the first function (the one with t). The program varies the parameters C and F until it finds the best values that interpolate your experimental data. As you see this equation well interpolate your data suggest that the model is correct.

What is the program you used to model this?

The fact that 50 and 75 degrees have a similar time is within the experimental error.  The values more close to the predicted by theory should be read in the red fitted line. As you see temperature and time is not a linear relationship

Origin is the program

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Ventilating combustion byproducts in water heaters
I'm designing a system that will heat water by burning natural gas which is commonly available. I would like to have a slow flame because water will need to reach $\mathrm{70^\circ C}$ and stay there. Having in mind that even weak winds can extinguish the flame, as well as the system, will be put together with other objects and systems in a tight space I decided it would be a good idea to build a chamber around the flame nozzle that's directly attached to the water container. I will build a small opening in the chamber to which I will attach a tube that will feed atmospheric air into the chamber through some kind of a filter (probably a piece of cloth) to keep it clean. I read that the byproducts of burning natural gas are $\ce{CO2}$ and water vapour which raises 2 questions:

Do I have to worry about humidity increasing in the chamber and eventually water forming on the inside, in extreme conditions water in the system can be as low as $\mathrm{-20^\circ C}$ (I call it water but it will be actually antifreeze)

Do I have to worry about the chamber filling up with $\ce{CO2}$ and eventually putting the flame out which will then be unable to be lit again. Is there anything I can do such that the byproduct will exit the chamber naturally without me having to do any work for it?

Edit:
So far from the comments, it was made clear that it would be better to use a condensing boiler design efficiency-wise, however, due to construction complications that I'd like to avoid I'll stick with only the hot half of the design in question where the water will pass through tubing inside the chamber where it will be heated, instead of applying heat to the container directly. There will be an exhaust that will utilize natural convection to expel the exhaust gasses in addition to the inlet.

For better efficiency, you might want to consider a design like this: https://de.wikipedia.org/wiki/Brennwertkessel#/media/Datei:Brennwert.svg

@Loong thank you very much for the reference. I'm definitely going to be using this design which is probably well known but for a guy like me with 0 experience is a gem. I'm curious how exhaust gasses leave the system on their own but it's probably well explained on the internet.

Well turns out I will need a fan as well as it adds some complexity to the build so I'm probably going to sacrifice some efficiency and stick with the initial plan.

If you are burning gas you must have an exhaust to allow the combustion products to exit the space. Otherwise the carbon dioxide will extinguish the flame quickly. Hence the design of every burner ever made that works.

@matt_black but would it be possible to arrange it in a way so that the exhaust wants to leave on its own without adding fans. I'm assuming if the inlet is on the bottom and the exhaust is on the top of the chamber then it could work the way I want it to?

@php_nub_qq Yes. Exhaust gases are hot and, therefore, natural circulation can be exploited to make them leave the space because to gases will rise. You only need a fan if you need stronger forced circulation (which you might do if you want to achieve a more powerful flame than natural circulation allows.) That's why domestic gas boilers have fans.

@matt_black Gas boilers need a fan if their exhaust temperature is very low, i.e. if they are built to be very efficient. See the link in loongs comment.

Rather than design something that offers danger of overheating, carbon monoxide poisoning, gas and over-pressure explosion, etc., why not use an off-the-shelf "instant" tankless water heater, available from sources such as Home Depot, Amazon or Lowes?
Find one certified for your area. Making your own is a tankless task!

Its not much different than heating water on a gas stove. Obviously I left out some details such as there will be temperature monitoring, which are not related to the question, however I'm really interested if you could dig deeper on the carbon monoxide and explosion aspects of your answer. Anyway to answer your question you can simply look at the prices of those and consider the $20 I will have to pay for my 5 liter gas bottle + change for tubing and a spare motor for valve control (the bottle comes with a valve which requires very little effort). For the ignition I will use a $5 spark plug.

Additionally I can not afford to waste electric power for heating because my battery is not powerful enough and they are still quite expensive for my pocket, hence the whole gas thing going on.


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Thermodynamic vs kinetic basicity
I'm writing something related to basicity and came across few papers that distinguish thermodynamic and kinetic basicity.
"In particular, the study of hydrogen bonding (H-bonding) within proton sponges has provided rigorous knowledge into the inter-related roles of symmetry, resonance, and aromaticity in H-bonding, as well as a better understanding of kinetic versus thermodynamic basicity and the emergence of a more reﬁned view of the entropic and enthalpic factors contributing to H-bond strength." ( DOI:10.1021/acs.joc.5b01743 )
"There is a tendency, however, that sterically demanding proton sponges with high thermodynamic basicity with strongly protected protonation sites often have a low kinetic basicity." ( DOI:10.1021/ja052647v )
"Thus, this compound class has been the subject of considerable scientific interest challenging chemists from differentfieldstofindsuitablemethodsfortheirsynthesisand handling, evaluate their basicity strength, and understand the principles underlying the thermodynamic and kinetic foundation of their basicity."( DOI:10.1002/anie.201612446)
"The thermodynamic basicity of all newproton pincers was measured by nmr titration experiments with corresponding neutral organic super bases of known pK>25 in CD3CN, while their kinetic basicity was measured by dynamic nmr proton exchange measurements." (Activating Unreactive Substrates: The Role of Secondary Interactions, Edited by Carsten Bolm and F. Ekkehardt Hahn)
So, my questions are:
1) What is the exact distinction between these two terms?
2) Is thermodynamic basicity the one mentioned in common basicity defintions? (arrhenius, lewis, bronsted-lowry)
3) Why are they measured differently?
4) Are they denoted diferrently, do they have different units and is there some general mathematical formula for each that would help me get deeper understanding and distinction between these two terms?
5) Also, how are these two mesaures of basicity related to gas phase proton affinity and intrinsic basicity and pKb?

I have worked on a computational project where I had to study the sites of deprotonation in this molecule:

When deciding between mechanisms I had to consider both thermodynamic and kinetic. For example, I had this case: deprotonations of two different hydrogens(bond with N Blu atoms in the picture)were similarly thermodynamically favored (similar free Gibbs energy of reaction) but one H was towards the inside of the ring and the other pointed towards the outside, therefore, because kinetically favored, the mechanism proposed was the one with the external H removed.
Your questions:

What is the exact distinction between these two terms?

A Thermodynamically favored base means that the Gibbs energy of the rection where an H+ binds to the protonation site(eg.lone pair of the N) is very negative.
However, if this protonation site is difficult to access, then this reaction is very slow or even impossible and a less thermodynamically favored protonation reaction may occur instead(if the molecule has more than one protonation sites or other bases are present).

Is thermodynamic basicity the one mentioned in common basicity defintions? (arrhenius, lewis, bronsted-lowry)

yes, and the kinetic effect is evident only in limited numbers of cases.

Why are they measured differently?

To make it easy and avoid competitive reaction considerations, suppose you have only one type of protonation site. If you want to know how easy is for the proton to access that site, you have to determine the speed of the reaction(dynamic experiment). If you want to know only about thermodynamic(energies involved) then you let the reaction go to equilibrium and just measure the pKa. If the deprotonation site is very inaccessible then this thermodynamic experiment may take very long because you have to wait it reaches the equilibrium. In this case, if it takes more than your experiment time scales than you will measure an apparent pKa that is lower than expected because your reaction has not reached the equilibrium yet. This apparent lower pKa is due to the kinetic effect.

Are they denoted differently, do they have different units and is there some general mathematical formula for each that would help me get deeper understanding and distinction between these two terms?

I don't know if there are other ways to express this other than kinetic and thermodynamic acid/base.
Thermodynamic: you wait for the reaction to reach the equilibrium and using the $\Delta G$ of reaction determine the pKa.
$ΔG = -RT\ln K_\mathrm{a}$
 $pKa = -log(K_\mathrm{a})$
 In case of a difficult to access protonation site you may not wait for enough and measure a lower pKa but this is an apparent pKa because the definition of it is purely thermodynamic.

Also, how are these two mesaures of basicity related to gas phase proton affinity and intrinsic basicity and pKb?

Not expert of these techniques, I worked only theoretically but as usual, the solvent affects the value of the pka. The most stable conformation in the gas phase may not be the most stable in solution anymore and there's the solvation energy contribution to consider when you determine the Gibbs energy of the protonation reaction. Therefore, the measurements in solution may not be comparable with the gas phase.
Finally, I can give you some example of very evident kinetic effect:
 In the molecule below the lone pair of the nitrogen is difficult to access for an H3O+ molecule and the pKa measured(not the thermodynamic pKa) will be different depending on how much you wait during the titration experiment. I don't know the timescales for this molecule.

or you can have a base trapped in a fullerene molecule:

This is the most extreme case with an infinite equilibrium time. The pKa of the molecule trapped is still the same but you will experimentally measure the pKa of your solvent and conclude it is not basic.

So, if i have a compound with high thermodynamic basicity and low kinetic basicity, that means protonation is energetically favorable, but occurs at a very low rate? What exactly kinetic basicity is then?? Measure of rate of reaction? And thermodynamic basicity is not about rate but about energy?

yes, that's correct. Thermodynamic basicity is bout energy and kinetic basicity about the rate of protonation reaction. And yes, low kinetic basicity, means protonation is energetically favorable but occurs at a very low rate.


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Confusion about the electrolyte in an electrochemical cell
So the cathode is magnesium and the anode copper. The question asks why silver nitrate would not be a good electrolyte. I am confused. Also is the anode always inert in these cells?

Both magnesium and copper are spontaneously oxidized by silver ions, so this just deposits silver and produces heat. Also, if this was to be a galvanic cell, magnesium would be the anode and copper would be the cathode. See the Daniell cell for a similar situation, with Zn in place of Mg.

As Ed said, that both copper and magnesium will reduce silver ions to elemental silver (you have to justify why in your answer. Hint: Look at the electrochemical series and compare the position of Mg, Cu, and Ag half cells. Another fundamental aspect is that one cannot apply Nernst equation if the metal dips in anything else besides its ions.
For instance, think little dip deeper, can we calculate the potential of copper electrode dipping NaCl solution (wrt to some reference cell?). In this case sodium ions will not react with copper or magnesium.


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Will Chlorodifluoromethane (R22 refrigerant) react with calcium metal?
I am attempting to recycle or find a use for chlorodifluoromethane, however I do not want to risk venting it into the atmosphere or exposing myself to hydrofluoric acid. Ideally, I would like to produce chlorine/fluorine salts for safe storage/disposal.
Edit: Upon further research, calcium fluoride appears to be a non-toxic salt of fluorine. MSDS warns that R22 will react with alkali metals as well as calcium, magnesium, etc. What conditions are required in the following reaction?
$$\ce{HCFCl2 + Ca -> CaF2 + ...}$$
Where all the fluorine is consumed by calcium. I am not concerned with other chloride salts since fluorine toxicity is my main concern.

Have you checked the Safety Data Sheet for it?

Yes, apparently it will react with Al, Zn, Mg, Ca, etc. and decompose at temperatures above 250C. Unfortunately, the details behind these reactions is unclear. I assume that F will attack an electron donor before Cl, but the potential to produce HF makes me not want to try this reaction.

I don't know where you are located, but I would consider giving it to an HVAC contractor. They usually don't charge as the refrigerant is rather valuable.

Unfortunately, I have the refrigerant in a 'non-refillable' vessel. In TX, they will only accept it if it is in a reclamation tank and if its of a certain purity, neither of which I can provide


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Reaction between chlorine gas and oxygen at 600 degrees Centigrade
Chlorine gas will not normally react with oxygen gas directly at STP. However, would the two react at standard pressure and 600 degrees Centigrade?

No.$\mathstrut$

Why always 600 degrees Celsius? There are several questions here that involve that specific temperature.

AP chem, apparently it's a common temp in our teacher's lectures and on the class practice sheets.


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Was it CO2 or Muriatic Acid?
We had a malfunction with the chlorination system at our pool.  There was a reaction between the Calcium Hypochlorite tablets and either $\ce{CO2}$, Carbonic Acid, or Muriatic Acid.  There was a definite malfunction that led undetermined amounts of $\ce{CO2}$ (or a mixture of that gas dissolved into water, carbonic acid?), into the tank where hypochlorite tablets are dissolved. The tablets became very warm, a yellow/greenish gas was observed, and there was an extreme chlorine type smell that was easily observable from at a hundred feet away.  We are talking $\pu{150 lbs}$ of calcium hypochlorite tablets.  Is it possible for the $\ce{CO2}$ to react this violently with the hypochlorite? Or, is it more likely someone poured some muriatic acid in the water tank or on the tablets?

Just for more clarification... this was at a public pool.  The two operators were out of town.  The tablets (not powder) are in a hopper.  The hopper sits on top of a tank which sprays a small amount of water up at the tablets to dissolve them.  There is supposed to be 50 lbs of tablets in the hopper but there was more like 150 (lots of fuel).  The "yellow/green" vapor that escaped from the lid when opened  was witnessed by several people (not me) The FD was eventually called as the fumes continued to increase. Next morning the tablets were warm and the plastic tray they sit in was deformed.

It is almost certain that the gas you saw was chlorine (Cl2). It is a VERY TOXIC gas, so stay away!!
Hypochlorite releases chorine gas when pH is too low. That is why bleach must not be mixed with acid. Never!
In this case it is probably Muratic (hydrochloric) acid that has been added to the hypochlorite in excess. The acid is used to adjust the pH in the pool water, because hypochlorite in itself will raise pH to an unpleasant level.
It is unfortunately a fairly common accident seen in chlorination systems.

Note that chlorine can be created just by adding water.
$$\ce{Cl- + ClO-  + H2O <=> Cl2 ^  + 2 OH-}$$

Enough $\ce{CO2}$ dissolved in wet tablets of calcium hypochlorite ( mixed calcium hypochlorite, chloride and hydroxide)  would speed up the above, as it would release hypochlorous acid $\ce{HClO}$, which is weaker than carbonic acid.
$\ce{HClO}$, then oxidizes chlorides to free chlorine.
The summary reaction is like this:
$$\ce{  CO2 + Ca^2+  +  Cl- +  ClO- ->  Cl2 ^ + CaCO3 v}$$

The reaction with hydrochloric ( obsoletely muriatic) acid $\ce{HCl}$, would be the most dramatic, if it was present :
$$\ce{  2 H+  +  Cl- +  ClO- ->  Cl2 ^ + H2O}$$

Calcium hypochlorite is supposed to be added to water. If a little water is added to the hypochlorite, it can start to oxidize and produce stronger chloric acid. The SDS says to avoid water and moisture. To test this, try adding just a few drops of water to a pile of powder or a pill and see if a reaction begins.
Unless you have a vandal, it is not likely that someone would be pouring acid into your hypochlorite. There might be a possibility of moisture getting in thru a small splash of rain or condensation from temperature changes and high humidity; tight closure of containers is advised.
Here's a link to a video where other additives are added. https://www.youtube.com/watch?v=RZOl48Ppanw


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Relation between First Law of Thermodynamics and Ideal Gas Law
Thermodynamics has always been a tough thing for me. There are lots of assumptions in this subject (those assumptions, I know, are necessary, I know the science of thermodynamics is a very practical science).
First Law of Thermodynamics states mathematically:
$$\Delta U=Q+W$$
(with proper sign conventions must be used). This is just a law of conservation of energy and a very straightforward equation, but when we come to chemical thermodynamics this equation changes its form and becomes:
$$\Delta U=Q+p\,\Delta V$$
My intuition says as soon as pressure and volume comes in any equation it becomes specifically for gases. So, my first question is:
Why thermodynamical equations are just for gases?
Let's imagine an isothermal expansion of a gas (that simple piston and gas experiment) under a constant pressure, now work $W$ is
$$W=p\,\Delta V$$
but if use ideal gas Law equation i.e.
$$pV=nRT$$
$$p\,\Delta V = \Delta nRT + nR\,\Delta T\tag1$$
since the expansion is isothermal therefore $\Delta T = 0$ and I can think that during expansion no atom or molecule has been annihilated therefore $\Delta n = 0$, so after all we get
$$p\,\Delta V = 0$$
$$W=0$$
I want to know my mistakes in above consideration.
There is a question in my book:

A swimmer coming out of a pool is covered with a film of water weighing $18\ mathrm g$. How much heat must be supplied to evaporate this water at $298\ \mathrm K$? Calculate the internal energy change of vaporization at $100\ \mathrm{^\circ C}$. $\Delta_\mathrm{vap}H^\circ = 40.66\ \mathrm{kJ\ mol^{-1}}$ for water at $373\ \mathrm K$

My book give its solution like this
$$\ce{H2O(l) -> H2O(g)}$$
Amount of substance of $18\ \mathrm g$ of $\ce{H2O(l)}$ is just $1\ \mathrm{mol}$. Since, $\Delta U=Q-p\,\Delta V$, therefore,
$$\Delta U=\Delta H-p\,\Delta V $$.
$$\Delta U=\Delta H-\Delta nRT$$
$$\Delta U=40.66 \times 10^3\ \mathrm{J\ mol^{-1}}-1\ \mathrm{mol}\times8.314\ \mathrm{J\ K^{-1}mol^{-1}}\times373\ \mathrm K$$
$$ \Delta U=37.56\ \mathrm{kJ\ mol^{-1}}$$
I have a lot of problems with this solution which goes directly to the foundations of science of thermodynamics. (I must say it's because of these books that science becomes a rotten subject, these books destroy the real essence of science).
How is $\Delta n=1\ \mathrm{mol}$?
Why temperature is taken as $373\ \mathrm K$ and not $298\ \mathrm K$, since the process starts at $298\ \mathrm K$ we should use it?
At $373\ \mathrm K$ the process becomes an isothermal one (latent heat) so $\Delta U$ ought to be zero, if we think the process of vaporization starts from $373\ \mathrm K$.
Any help will be much appreciated. Thank you.

Why thermodynamical equations are just for gases?

They are not. The equation $\Delta U = q + P\Delta V$  applies to any phase (gas, liquid, solid...) when only pV work is done. In the particular form of the equation you present, the pressure is in addition constant during the work.
Gases are (1) an easy way to introduce thermodynamics concepts because they allow for a simplified analysis and may exhibit dramatic behavior, and (2) they provide a link to understanding the behavior of other phases. They also happen to be inherently important for practical and historical reasons during the development in the science.

I want to know my mistakes in above consideration.

There are no mistakes. Consider the ideal gas law $$V=\frac{nRT}{p}$$ If you assume that $p$, $n$ and $T$ are constant, then the dependent variable $V$ will also be constant.
What's probably causing the confusion is that the water is undergoing a phase change. We assume that the liquid does no work, that the change in volume is only due to the formation of vapor. In practice $\pu{\Delta n=+ 1 mol}$ for the gas, $\pu{\Delta n= - 1 mol}$ for the liquid, and $\pu{\pu{\Delta n= 0}}$ for all of the water. We ignore the work done when reducing the amount of liquid because it is small compared to that done when the gas is formed (the change in volume of gas is much greater).

How is Δn=1 ?

See the answer to the previous question. You are converting $\pu{18 g}$ of liquid water into vapor. Since the molecular weight of water is $\pu{18 g/mol}$, you are converting $\pu{1 mol}$ of water.

Why temperature is taken as 373 K and not 298 K, since the process starts at 298 K we should use it?

I agree, if the work is actually performed at a lower temperature than $\pu{373 K}$, so this is an estimate. It is assumed that water is "boiling off" the skin at $\pu{373 K}$ and $\pu{1 atm}$ vapor pressure (the boiling point of water at $\pu{1 atm}$ of pressure is $\pu{373 K}$). Maybe not an accurate portrayal of what is going on, but it gets you to practice the theory.

I have lot of problems with this solution which goes directly to the foundations of science of thermodynamics. (I must say it's because of these books that science becomes a rotten subject, these books destroy the real essence of science).

The problems more generally are that (1) we have an intuition about the way the world works, based on everyday observations, and this intuition sometimes misinforms us; some of the effort of education is to develop a more accurate intuition; and (2) when teaching science, practice problems are sometimes abstract and don't reflect real life situations except approximately. For instance, when you dry yourself after swimming, you are far from a thermodynamic equilibrium, with air currents, sun heating your skin, and a low water vapor pressure all playing a potential role. Modeling all this is beyond the scope of an introductory course. Probably the most important approximations here (assuming a closed system free of mass flow) are that the enthalpy of vaporization is constant over a broad temperature range and/or (as you rightly pointed out) that the water evaporates at $373 K$.

Your answer has cleared all my doubts except the one. Your answerI want to know my mistakesis a very nice one but I'm having trouble in understanding it. I humbly request you to please elaborate it and tell me why $P\Delta V $ will not become equal to zero during an isothermal expansion.

@adeshmishra Because you assume $\Delta n=0$ which is not true. The amount of gas increases during evaporation.

@BuckThorn That’s what I want to know: How the amount of gas increases during expansion? I think number of moles doesn’t increase only volume increases.

@adeshmishra For the gas, $\Delta \pu{n=+ 1 mol}$. For the liquid, $\Delta \pu{n=- 1 mol}$. For all of the water, $\Delta \pu{n= 0 mol}$.


text:
Division, multiplication, addition and subtraction of significant figures
$1.$After adding two or more numbers, the number of digits after the decimal in the result=Least no. of digits after the decimal in the numbers being added.
$2.$After multilplying two numbers, no. of significant figures in a result= no. of significant figures in a number with least no. of significant figures, out of the numbers operated.
Why these two rules are made? How they help us? Why only these, not something else?

This is a basic question, and not really about chemistry. However, a quick internet search shows that most treatments of significant figures are in introductory chemistry textbooks, and mostly they don't explain the ideas behind the rules. Thus, I think it is a valid question.

Everything you want on the topic is right here: https://en.wikipedia.org/wiki/Propagation_of_uncertainty . Lots of links to explore. Bottom line: the simplest significant figures rules given in undergrad textbooks, and others of that ilk, do an OK job for most purposes, but they are not definitive.

But all the things mentioned there in wikipedia are really complex. I was unable to get anything it has.

There is also a much simpler wikipedia article: search for “Significant figures” in wikipedia. Bottom line: significant figures rules are just useful rules of thumb. They are attempts to use a single number to convey both a “best estimate” and its uncertainty. This is not really possible, but the simple significant figures rules do an OK job of approximating what is actually correct.

Why these two rules are made?

These rules are approximations of more sophisticated error propagation treatment. For addition and subtraction, for example, if the first number is a ± 0.1 and the second is b ± 0.001, and the numbers are independent of each other and normally distributed, the absolute error of the sum is $$\sqrt{0.1^2 + 0.001^2}.$$ That comes out to just about 0.1. On the other hand, if both have a uncertainty of ±0.1, the uncertainty of the sum is actually a bit higher, approximately ±0.14.
If you want to get a feel for error propagation, you can always try out the highest and lowest estimate for your numbers (e.g. for 14.2 ± 0.1 that would be 14.3 and 14.1) and plug them into your calculation. If you are calculating with two numbers, you get four combinations, all with different answers. The spread in those answers gives you an idea of how the errors propagate. This is especially useful if you need to take a logarithm or raise something to a power because most textbooks leave out the rules for those.

How they help us?

They save use some time compared to the more rigorous error propagation treatment. This comes at the expense of sometimes writing too many digits, or losing some information by writing too few.

Why only these, not something else?

There are lots of other options. My favorite for introductory labs is to say use four significant figures in intermediate calculations, and three significant figures for results. Typically, there is a source of error larger than 1% somewhere in the experiment, so there is no loss of data. On the other hand, writing down three significant figures is not too onerous, and does not use up too much space.

How you got the formula for absolute error of the sum?

Let me consider 14.2+2.3=(14.1+2.1)=16.2 or 14.2+2.3=(14.3+2.4)=16.6. Here, absolute error is 0.2, not equal to 0.14, as mentioned by you. Please spot the mistake here.

What you are doing is the sum of the errors rather than the quadratic sum of the errors. The latter is based on the fact that the variance (which is the square of the standard deviation) of a sum or difference is the sum of the variances. So, first square 0.1, then add it to itself (for the second equal error), then take the square root of the sum. This is the quadratic sum and it is 0.1 times the square root of 2. If you just sum 0.1 and 0.1, you get the ordinary sum, which is 0.2. Both Karsten and @Poutnik (his second equation) correctly show the quadratic sum being used.

Your calculation is a worst-case scenario. In many case errors go in opposite direction and cancel out partially. The derivation of the square root formula involves a double integral over the two Gaussian distributions of actual values, so it would warrant a separate question.

If you want to follow all the math, then you need to learn about expectation values and then this heavily upvoted link is excellent: https://stats.stackexchange.com/a/31181/247352 . But it is way easier to just go with the simplified rules of thumb! One last thing: the variance of a sum is the sum of variances only if the two random variables are uncorrelated, which we are tacitly assuming here.

As mentioned by you that 0.2 uncertainty after the sum is a worst case scenario. You mean that the sum could have atmost 0.2 error, no more than this. Then why we take it to be 0.14, since it could be more or less than this?? Please note that i am unknown to the topics like variance, deviation, etc. BUT then too, i think that the sum should contain max. of 0.2 uncertainty.Then why it makes sense for us to take it equal to 0.14 as in actual it could be more or less than this.

Significant figures were made to aid us in rounding to an appropriate number. We can only base our rounding on the least significant out of given values; or else we aren't guaranteed an accurate result.
For example, in Olympic swimming, there are often lots of ties. Modern technology could definitely track down to thousandths of a second or below, but since the pool's tolerance (and therefore its length for a given lane) fluctuates, there is statistically no way to know that the "winner" didn't have a shorter length to swim. As such, we can only make our time as significant as we know the length of the pool.

Here is an article about why swimming contests currently use 0.01 seconds as the shortest time interval: https://deadspin.com/this-is-why-there-are-so-many-ties-in-swimming-1785234795

If quantities are added or subtracted:
Their absolute maximal errors are additive.
$$e_{A+B}=e_{A}+e_{B}$$
The squares of their absolute  mean errors ( like standard deviation or confidence intervals ) are additive.
$$(\sigma_{A+B})^2=(\sigma_{A})^2+(\sigma_{B})^2$$
If quantities are multiplied or divided:
Their relative maximal errors are additive.
$$e_{\mathrm{rel},A+B}=e_{\mathrm{rel},A}+e_{\mathrm{rel},B}$$
The squares of their  relative mean errors are additive.
$$(\sigma_{ \mathrm{rel}, A+B})^2=(\sigma_{ \mathrm{rel}, A})^2+(\sigma_{  \mathrm{rel},B})^2$$

As consequence, if simplified :
For the addition/subtraction case, absolute errors are significant.
Therefore the last valid digit  position of the most errorneous  quantity matters.
$$(1.0\pm 0.1) + (2.00\pm 0.01) = (3.0\pm 0.1)$$
For the multiplication/division case, relative errors are significant.
Therefore the number of valid  positions of the most errorneous  quantity matters.
$$(1.0\pm 0.1) * (200\pm 1) = (200\pm 20)$$

Let me consider 14.2+2.3=(14.1+2.1)=16.2 or 14.2+2.3=(14.3+2.4)=16.6. Here, absolute error is 0.2, which comes out to be 0.14 according to your formula. Why not 0.2?? ( I had no idea about standard deviation or something like that. But got some idea from karsten Theis' answer. It says that in addition of above mentioned numbers, there is an error propagation of 0.14, but not 0.2, which should be, as per me.)

As was said, positions and valid digit numbers are simplifications. Absolute maximum error or absolute mean error ? Your example has errors in its errors. Get them right. Btw, the formulas are not mine, they are well known formulas of maximum and mean error propagation.

if there is (14.2 +- 0.2 ) + ( 5.6+-0.1)=(19.7+-0.3), it is for maximum absolute error 0.2+0.1=0.3.


text:
Is Octet Expansion appropriate when regular Octet bonding is possible (such as using dative bonds)?
I was taught to use Occam's razor whenever in doubt over an octet expansion, being told it never happened when, for example, dative bonds could resolve the issue with a regular octet. This seems to fit what users here say about the recent paradigm shift on "hypervalent" molecules that "it is now known, that for almost all molecules that were described this way the contribution of d orbitals is negligible (<1%).", but could anyone explain exactly why an element wouldn't expand it's octet unless it absolutely has to? What advantage is there to not involving 'd' orbitals over involving them in bonding? Or is it just Occam's razor and I'm misunderstanding this maxim.

No. There is no octet expansion, the concept has been disproved. The d-orbitals simply have a too high energy to be involved in bonding.

Then what theory has superseded it so I can be more informed, how are moleules like sulfur hexafluoride explained?

In short: Molecular Orbital Theory (but a modern interpretation of Valence Bond theory comes to the same conclusions). Related: Can an atom have more than 8 valence electrons? If not, why is 8 the limit? Unfortunately I have to add that the accepted answer (by Terry) is also beyond wrong. Also: How are the hybrid orbitals of sulfur hexafluoride shaped? What is the highest possible expanded octet?


text:
Chloride Process TiO2 recovery
In the chloride process (the process used to create titanium oxide from titanium ore), how is TiO2 collected? What prevents the TiO2 from building up on the walls of the reactor chamber?
It's my assumption that the solid titania produced in a TiCl4+O2 reaction (assuming that the TiCl4 and O2 are all gases at 200 C) would plate out on the walls of the reactor chamber (since it is the one solid in the reaction). When manufacturing titania, how do manufacturers prevent the titania from plating out on the walls? Also, how would they collect it?

Can yoe be a bit more specific about where you see the problem, because at the moment this looks like a homework question.

@Karl Sure thing, it's my assumption that the solid titania produced in a TiCl4+O2 reaction (assuming that the TiCl4 and O2 are all gases at 200 C) would plate out on the walls of the reactor chamber (since it is the one solid in the reaction). When manufacturing titania, how do manufacturers prevent the titania from plating out on the walls? Also, how would they collect it?

Its a fine powder with very high melting point. I dont see it sticking to the wall, and would collect it with a cyclone, like those filterless vacuum cleaners, the brand name of which I will not mention. ;)


text:
Entropy changes during transfer of heat between two bodies
Consider two metal blocks, one at 500K and another at 300K, they are bought to contact with each other until they reach a common temperature.
Find the entropy change of each object and the total change considering there is no change in the volume of the metal blocks in the process and no the other interactions excluding the heat transfer between the two blocks.Also is this process irreversible, why?
also
$C_v= 10 J/K$ for both the metal blocks
My attempt:
This process is irreversible, as once it reaches the equilibrium temperature it cannot go back to the initial state without external intervention that is to say heat wont flow from cold body to hot body and also when two bodies are at the same temperature heat doesn't flow from one to the other to increase the temperature of one and to decrease the other.
now to compute the $\Delta S$ we can perform the same process very very slowly such that the two systems are always at equilibrium.
thus we can write $dS=\frac{\delta q_{rev}}{T}=\frac{dq}{T}=\frac{C_vdT}{T}$
thus for system 1: $\Delta S_1=C_v ln\frac{T_f}{T_i}$
and thus $\Delta S_{TOTAL}=C_v ln\frac{T_{f}}{T_{i1}}+C_v ln\frac{T_{f}}{T_{i2}}$
My questions :
I have a feeling my reasoning is not rigorous enough and my definitions are going in circles .
and my analytic solution is wrong as the i'm basically using $\delta q$ of the irreversible process and using the same information in the assumed reversible process.

This answer is correct.  After determining the final temperature Tf, you have gone back to the initial state of each block and, to get its change in entropy, you have subjected it to an alternate reversible process in which, instead of contacting it with the other block, you have contacted it with an infinite continuous sequence of constant temperature reservoirs running from its initial temperature to its final temperature.  At least, that is what your equation implies.  This is definitely the correct way to determine the entropy change.

The difference between the irreversible process and the reversible process is in the surrounding. For the former, there is no surrounding, and the entropy of the universe increases. For the latter, the surrounding has a net decrease in entropy (you are heating up the hot reservoirs while cooling down the cold reservoirs), offsetting the net increase of the two blocks such that the entropy of the universe stays constant.

@KarstenTheis. With all respect, my take on this is entirely different.  In my judgment, there is no need to consider the surroundings at all.  My focus, instead, is on the two blocks.  I put each of them separately through an alternative reversible process between the same two end states as the irreversible process.  The difference between the reversible and irreversible processes is that, in the irreversible process, all the heat transfer takes place at the intermediate interface temperature of 400 K while in the reversible process, this is not the case.

I agree with you, there is no need to consider the surrounding to get the correct answer. I was trying to address the OPs struggle to see why it is fine to run a reversible process to get an answer for an irreversible process. It is kind of surprising because for the reversible process, there is no overall increase in entropy, yet the result for the blocks (increase in entropy) is correct. Does that make sense?

@KarstenTheis. Yes, it does make sense.  In the pair of  reversible paths, net entropy is transferred from the surroundings (reservoir sequences) to the blocks.  This makes up for the entropy that was generated within each of the blocks in the irreversible process.

Here is a primer containing a simple cookbook recipe for determining the change in entropy for any irreversible process:  https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/  The article also includes worked examples.  Enjoy!!!

It is worthwhile to mention that entropy is a state function here


text:
Why does the weaker C–H bond have a higher wavenumber than the C=O bond?
My understanding is that a stronger bond has a higher wavenumber in IR spectrum. But why does the C–H vibration have a higher wavenumber than the C=O vibration? The latter is a double bond, so I think it should be a stronger bond than the C–H single bond.

A property of the harmonic oscillator is that the oscillation frequency, $\omega$, is dependent not only on $k$ (the spring constant) but also on the mass $m$ of the object:
$$\omega = \sqrt{\frac{k}{m}}$$
We can crudely model a chemical bond as a two-body harmonic oscillator, which largely obeys the same rule, except that the mass must be replaced with the reduced mass $\mu$, defined by:
$$\frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2}$$
where $m_1$ and $m_2$ are the masses of the two bodies. So, for a C–H bond, we have
$$\frac{1}{\mu_\ce{CH}} = \frac{1}{\pu{12 u}} + \frac{1}{\pu{1 u}} \implies \mu_\ce{CH} = \pu{0.923 u}$$
and for a C=O bond, we have
$$\frac{1}{\mu_\ce{CO}} = \frac{1}{\pu{12 u}} + \frac{1}{\pu{16 u}} \implies \mu_\ce{CO} = \pu{6.86 u}$$
From J. Chem. Phys. 1946, 14 (5), 305–320 the force constant $k$ for the C=O bond in formaldehyde is approximately $2.46$ times that for the C–H bond in methane (the units are in the old cgs system, so I am lazy to quote the actual values).
So, we can come up with a very rough theoretical estimate. The wavenumber used in IR spectroscopy, denoted by $\bar{\nu}$, is directly proportional to the (angular) frequency $\omega$.
$$\begin{align}
\frac{\bar{\nu}_\ce{CH}}{\bar{\nu}_\ce{CO}} &= \frac{\omega_\ce{CH} / 2\pi c}{\omega_\ce{CO} / 2\pi c } \\
&= \frac{\omega_\ce{CH}}{\omega_\ce{CO}} \\
&= \sqrt{\frac{k_\ce{CH}}{k_\ce{CO}} \cdot \frac{\mu_\ce{CO}}{\mu_\ce{CH}}} \\
&= \sqrt{\frac{1}{2.46} \cdot \frac{\pu{6.86 u}}{\pu{0.923 u}}} \\
&= 1.74
\end{align}$$
This is not too far from the experimental value of $\pu{2900 cm-1}/\pu{1730 cm-1} = 1.68$ (using values in the middle of the typical ranges for alkanes and aldehydes).

According to Skoog, Analytical Chemistry:
Using classical mechanics, assuming a diatomic molecule, the frequency of vibration $\nu$ may be described by
$$\nu = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}} $$
where $k$ represents the force constant of this bond, and $\mu$ the reduced mass of the particles bond together, defined as
$$\mu = \frac{m_1 \cdot m_2} {m_1 + m_2}$$
An equivalent quantum chemical / mechanial description includes $c$, the speed of light, yielding a result expressing the radiation in wavenumbers:
$$\bar{\nu} = \frac{1}{2\pi \cdot c} \sqrt{\frac{k}{\mu}} $$
So (1), with the same reduced mass $\mu$, the observed frequency increases with increasing force constant $k$.  And (2) $\nu$ equally increases while lowering the reduced mass $\mu$.
Skoog further mentions as typical force constant of single bonds the range of $3 \times 10^2 \ldots 8 \times 10^2\,\pu{N m}^{-1} $ with an average of $5 \times 10^2\,\pu{N m}^{-1}$; while stating $1 \times 10^3\,\pu{N m}^{-1}$ and $1.5 \times 10^3\,\pu{N m}^{-1}$ for typical double, and triple bonds, respecitively.

Consequently, for a carbonyl $\ce{C=O}$ double bond, assuming as the mass $m_1$ of a carbon atom of
$$m_1 = \frac{12 \times 10^{-3}\,\pu{kg / mol} } {6.0 \times 10^{23}\,\pu{atoms / mol}} \times 1\,\pu{atom} = 2.0 \times 10^{-26}\,\pu{kg}$$
and for oxygen a mass $m_2 = 2.7 \times 10^{-26}\,\pu{kg / atom}$,
yields a reduced mass $\mu = 1.1 \times 10^{-26}\,\pu{kg}$.  Using the above mentioned equation determines as vibration frequency (in wavenumbers) a value of
$$\bar{\nu} = 5.3 \times 10^{-12}\,\pu{s \cdot cm}^{-1} \sqrt{\frac{1 \times 10^3\,\pu{N \cdot m^{-1}}}{1.1 \times 10^{-26}\,\pu{kg}}} = 1.6 \times 10^3\pu{cm}^{-1}$$.


text:
Reduction of ester / eneone in presence of epoxide
I want to reduce an ester and an eneone in the presence of a terminal epoxide.
I read on here @jerepierre did an ester in the presence of an epoxide with DIBAL.
I know the DIBAL works for my ester, but am not sure if it will also open the epoxide; reports are conflicting.
Any advice on reagents or conditions to consider or references to read are greatly appreciated!

You have set yourself a tricky selectivity problem. From what I can find in the literature, Dibal-H reduces some epoxides under some conditions, often dependent on what neighbouring groups are present for the Al to bind to. I can only suggest that you try the reaction on a small scale at low temperature with a strictly controlled amount of Dibal-H. If this fails then I can only suggest you reconfigure your synthesis to avoid such an issue.

Thanks @Waylander!
Subsequently I have found epoxide ring openings with DIBAL tend to be from 0 C to reflux. and a reference (2014, Comprehensive Organic Synthesis, 2nd Edition. Chapter 8.10 Reduction of Carboxylic Acids and their Derivatives to Alcohols,. Ethers, and Amines. GE Arnott) had an example of this; Davies, K. A., et al. (2011). "Oxygen-containing analogues of juvenile hormone III." Tetrahedron Letters 52(18): 2302-2305

Example is from this paper; Davies, K. A., et al. (2011). "Oxygen-containing analogues of juvenile hormone III." Tetrahedron Letters 52(18): 2302-2305.
Ester in diethyl ether at -78°C, d/w DIBAL, stir 1 hr. -40°C for 2.5 hr. Rochelle's salt workup.


text:
Do different oxidation states of the same element have the same emission spectrum?
I looked at the flames of copper (I) chloride and of copper (II) chloride through a spectroscope and they looked the same. The flame colour is the same too. But since they have different oxidation states, and therefore different electron configurations, should their emission spectra be slightly different?
Would this logic also apply, for example, to elemental sodium in a discharge tube, vs sodium chloride burning in a flame? Would their emission spectra differ?

I just read that samples of copper (I) chloride usually are contaminated with copper (II) chloride, which explains seeing the same colour flame and spectrum.
However, I still do not know the answer to my question and would very much like to.

Once in the flame, you have atoms and ions in excited states. The oxidation states of species before they were introduced into the flame are history.

So the lines seen are a combination of the photons emitted by unionized atoms and whatever mixture of ions is present?
Does this apply to the discharge tube case too?

Yes. Sample introduction involves several steps, but what you get in the flame or plasma or discharge tube is excited atoms, excited ions, perhaps some excited high temperature species, and unexcited versions of these. It is generally very complicated and lots of factors are in play.

As for sodium, the famous yellow sodium "D" lines, at 589.0 and 589.6 nm, are the same whether you burn Na in air, have it in a sodium vapor street lamp, have it in a discharge tube or simply heat a piece of soda glass tubing in a flame. It is just excited Na atoms that produce those emission lines.

Thank you, I understand what you are saying, but I am still somewhat confused. In the case of sodium, would an excited Na ion and an Na atom have the same exact energy levels or not? Removing an electron from an atom is a fairly "violent" act, and it would affect the remaining electron cloud, thus changing the energy of the available energy levels. Would this not then result in different transitions? I am thinking of Na and Na+ as being quite fundamentally different, which perhaps is the source of my erroneous thinking.

A sodium atom and a sodium ion have different spectra. The sodium atom, unexcited, cannot emit light, but it can absorb light, as in atomic absorption spectroscopy. An electronically excited sodium atom can emit light. A sodium ion is simply another species that can be excited, etc.

Interesting question. Keep in mind that the elemental emission spectrum in a flame or plasma and even a discharge does not remember its history in solution or a solid phase.
The punchline is that the emission spectrum is dependent on the elements gas phase chemistry in the flame/plasma/discharge. Prof. Ed has explained you the example of sodium. Let us say we have the following
(a) a block of element sodium
(b) a block of sodium chloride crystal
(c) solution of sodium chloride in water
If you introduce (a), (b) and (c) in the flame, the flame will be colored yellow in each case, which means that the emission is coming from a common emitter. That emitter is a elemental sodium atom excited by high temperature in the gas phase. Thus atomic emission spectrum is a fingerprint of the element.
You may ask that you introduced Na(+) in the flame in the case of b and c. Flames can easily reduce an ion to the elemental state.
How to see the sodium ion spectrum: As we just said, the emission spectrum is independent of this original state. You can only cause ionization by increasing the temperature. This is the way to see the spectrum of an ion. If we were using a high temperature flame/electrical discharge, we would start seeing Na(+) spectrum along with elemental sodium spectrum.
Coming to your particular example: You introduced copper (I) and copper (II) into a flame and they all colored it beautiful blue-green. The reason is that if the flame temperature is low, compounds cannot fully dissociate into atoms (not enough energy to break the bonds). In such cases, very simple diatomic or triatomic molecules are formed in the flame which emit their characteristic colors. In the case of copper, CuCl is formed in the flame. CuOH may be formed as well. Whether you introduced Cu(I) or Cu(II), as a chloride, it does not remember its solution phase or solid phase history.
If you were indeed using a high temperature flame, you will never ever such a blue green coloration, because this time, the emission is from Cu atoms (in UV). Hope that clarifies your confusion.

Thank you, that is quite clarifying. I am only confused by one thing: Flames can easily reduce an ion to the elemental state. Na(+) is far more stable than Na... why would it reduce itself to the unfavourable unionized form? You and Ed said the lines in the visible emission spectrum of sodium are from excited atoms... I just don't get why not also ions if the substance being introduced to the flame is actually entirely Na(+).

Like @M. Farooq said, the chemistry in the flame is not the chemistry of what was introduced. If an aerosol of saltwater is nebulized into the flame, the droplets lose water and become tiny salt particles. Those melt and vaporize. Then you have a complicated mixture, highly temperature dependent, of atoms, ion and electrons: a plasma. The Saha equation deals with the relative abundances of the various species. So sodium atoms are the prevalent species and atomic emission from it is what you see. It might help to think of a related system: magnesium acetate. If this is introduced to the flame

then you will have magnesium atoms and singly ionized magnesium atoms, even though magnesium has 2+ oxidation number in the original magnesium acetate solution. You can easily search for the energy level diagrams for magnesium (“Mg I”) and singly ionized magnesium (“Mg II”) and see that they differ a lot. But there is very little doubly ionized magnesium ion in the flame. And the acetate ions are busted up, as expected.

Thank you Ed, I understand. I think I am just surprised to find out that in the very hot environment of the flame, "stable" ions would spontaneously become "unstable" atoms, due to my more pedestrian understanding of where and how atoms and ions exist (i.e. Mg II is more favourable than Mg, at all times).

@Rafael, ED has already explained. You have to think of the chemistry of what happens after the solvent dries up in the flame. Yes, Na+ is more thermodynamically stable at room temperature in an isolated surrounding, but flame chemistry (= hot gas phase reaction) generates neutral Na atoms, and under those reaction conditions, Na+ is not preferred. The spectra of ionized atoms (by using very temperatures) were known in the 1920s. Amazing, isn't it?

@M.Farooq, I understood Ed's explanation. I was simply saying that Na+ turning into Na in a high-energy environment goes against conventional wisdom. Amazing, isn't it? -- yes it is! That is what I was trying to say.


text:
accurate measurement from inaccurate tools
It is a very general question. Is there a good way to systematically increase the accuracy and precision of a measuring tool using only mathematical means?  For example, average 10 measurements can create a better one. Or using two independent tools.  I don't know if measuring theory or statistical quality control or other subject can help.

such a situation would be very context specific, but have a look at kalman filters as an example where a measurement could serve as adding information (i.e. an 'update') to already known information, rather than as the sole information by itself.

@Tasos Papastylianou Yes, the Kalman filter could be helpful, especially if the real value is not really constant, but is possibly drifting, like temperature, atmospheric pressure, speed, etc.

Generally, improvement of measurement process has much better positive impact on results, than sophisticated mathematical processing of low quality data. But, sometimes we are forced to data what we have.
There are 3 quantities:
resolution = the smallest difference of directly measured value, distinguishable by the tool,
precision = Level of uncertainty of repeated measurement of the same value, due random fluctuances. It is affected by the resolution as well, if comparable with precision.
accuracy = Level of agreement of the estimated and real value, due the tool/method bias.
There can be a very precise but inaccurate tool/method and vice versa,
even if one usually comes with the other.

Precision can be improved by measurement repetition. The arithmetic mean of $n^2$ measurements has $n$ times better precision than a single measurement.
The estimation of the standard deviation of normal Gauss distribution of measurements is
$$s=\sqrt{\frac{\Sigma ( x - \overset{-}x )^2}{n-1}}$$
The estimation of the standard deviation of normal Gauss distribution of real value estimation by arithmetic mean is:
$$s_{\overset{-}x}=\sqrt{\frac{\Sigma ( x - \overset{-}x )^2}{n(n-1)}}$$
For a method consisting of multiple measurement steps, each with  their own error, the error propagation rule has to be followed.
If 2 quantities, subjected to random errors, are added or subtracted, the squares of their standard deviation estimations are additive
$$(s_{A\pm B})^2=(s_{A})^2+(s_{B})^2$$
If quantities are multiplied or divided, their relative standard deviation estimations are propagated:
$$\left(\frac{s_{A\cdot B}}{A\cdot B}\right)^2 = \left(\frac{s_{A}}{A}\right)^2 + \left(\frac{s_{B}}{B}\right)^2$$
$$\left(\frac{s_{A/B}}{A/ B}\right)^2 = \left(\frac{s_{A}}{A}\right)^2 + \left(\frac{s_{B}}{B}\right)^2$$
If we e.g. determine the difference of the standard solution volumes $A - B$,
where $A=20.00 \pm 0.03$, $B=10.00 \pm 0.04$,
then $A-B=10.00 \pm 0.05$,
as $\sqrt{0.03^2 + 0.04^2}=0.05$.

Accuracy is harder to improve by mathematical means. It can be improved by measurement bias analysis, by calibration of tools or methods and by combination of more tools / methods / labs.
There is an additional decrease of the measurement method precision due statistical error of calibration via rules above.
$$(s_\mathrm{rel,method})^2 =  (s_\mathrm{rel, meas})^2+(s_\mathrm{rel, cal})^2$$
More tools/methods/labs lead to the  "mean of the means" and partially decreases the possible bias,
similarly as averaging of measurements improve precision.
But it is more costly.

Nice answer, +1. "Accuracy is harder to improve": I'm not so sure about this, though. IMHO noise/random uncertainty is easy to tackle in theory only. In practice, interesting/important sources of noise often turn out to be quite costly or even impossible to tackle as well.

Well, better precision is generally easier to achieve than better accuracy by mathematical means in the context of the question.

Your non-trivial question has a yes and a no. It is not possible to increase the accuracy of a dataset just by mathematical means. If you knew the systematic error in your data then you can do the arithmetic to correct the error. There is no a priori way to ascertain if your measurement tool lacks accuracy until and unless we have a reference for comparison. The simplest example is that of an electronic balance. If we weigh something, there is no way to confirm if the displayed value is correct or not. One can only check that by weighing a reference weight, a known mass. SI got rid of these physical references lately. One might argue that mass can now be measured with Planck's constant, yes, we still need to calibrate/determine the local value of "g" in that area.
Now coming to the second part: Can we improve the precision? This is where mathematics can help. The simplest way is that you make a lot of measurements and assume that there is no systematic error. Lack of precision means you have a lot of noise in the measurement. A very advanced technique is Hilbert's transform to reduce noise in the measurement process. You can also apply a smoothing function on the data to reduce noise.

"SI got rid of these physical references lately": The primary reference is no longer a physical artifact, but there is a method (mise en practique) to calibrate secondary standards: https://www.nist.gov/programs-projects/nist-mise-en-pratique-new-kilogram-definition. Here is a paper about the method: https://ieeexplore.ieee.org/abstract/document/6898427

To add to @Poutnik's and @M.Farooq's answers:

Is there a good way to systematically increase the accuracy and precision of a measuring tool using only mathematical means?

No, because what you can and should do in terms of maths/statistics depends very much on your application and data generation/measurement processes.

Using only mathematical means is not going to work because you'll need to know where your error comes from in order to efficiently tackle it.
That is, besides the difference in treatment between systematic and random errors (see Poutnik's answer), there usually isn't only one error but several sources of error/uncertainty. Besides the mathematical/statistical tools to tackle the errors, knowledge about the application background and your measurement instruments is needed to identify possible/likely sources of error.
Once you have them, you can either measure their contribution experimentally - this is where maths/statistics can first be used and find out which are the few most important ones. (If you know the situation sufficiently well to judge this without the need for detailed experimentation, that's fine, too).

Repeated measurements vs. pooling results from different methods: this will also depend on what your main source of error is. If it is measurement noise then repeated measurements help (usually according to the $n^2$ rule Poutnik cited for the sources of error which your repetitions cover). You'll need to do your repetitions at the correct "level" to address the actual source of the noise though:
Assume you have e.g. field sampling error of standard deviation 3 and instrument noise of standard deviation 1, and they just add up. Total error (error propagation for additive, independent sources of noise) is $\sqrt{3^2 + 1^2} \approx 3.16$ (standard deviation).

You decide to 100 runs of the instrument, reducing instrument error standard deviation by a factor of $\frac{1}{\sqrt{100}} = \frac{1}{10}$. Thus, total error now is $\sqrt{3^2 + 1^2\cdot\frac{1}{10}} \approx 3.02$, so an improvement of almost 5 %. I.e., about 95 % of total error is field sampling error.
If instead you add a 2nd field sample, total error is $\sqrt{3^2\cdot\frac{1}{2} + 1^2} \approx 2.34$, i.e. total error is improved by a quarter.
This level of improvement is already impossible to achieve with improvements in the instrument part.

In conclusion, pooling results from different methods can be beneficial if your sample (matrix) is complicated so you expect different bias (e.g. masking, ...) for the different methods. Usually, one would not just average those results but again use chemical and application knowledge to interpret and weight these results wrt. their reliability and about the sample composition.

Possibly helpful literature fields (unfortunately, to my knowledge this information is quite spread throughout various fields):

A Measurement theory/metrology textbook will give your basic concepts.
Quality control literature has e.g. experimental designs to measure the size of various contributors of random uncertainty.
In statistics, so called mixed models can deal with situations where systematic and random influencing factors occur.
Chemometrics literature also has relevant contributions, particularly with respect to multivariate data.
Literature about your instrumental method that discusses the various noise sources for this type of instrumentation.
Application literature may tell you about particular application-related sources of error.
Sampling has its own set of literature

Experimental design also comes into the fray.  If you have an apparatus that is not quantitatively accurate, you can apply it to your experimental ssmple and to a control for ccx which you have a reference value of what you are trying to measure.  You then calibrate your results to the known reference.  I've used this approach at work,  to compare acid pickling times of differently processed samples of a steel product we're developing.

... and in doing so, please make sure the noise on your calibration results is really negligible compared to the noise on sample measurements. Otherwise, you turn a random uncertainty (noise, variance) into a new bias!


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How to find the temperature relationship between the isotherms in a compressibility factor (Z) vs pressure graph?
The following graph denotes the variation of the compressibility factor (Z) with pressure at different temperatures for a real gas. Simply each of the curves represents an isotherm.

Now, suppose we are not given the temperatures and we are asked to find the temperature relationship between the isotherms, how to prove the topmost curve corresponds to the higher temperature and the lowermost curve corresponds to the lowest temperature? Is there any mathematical proof for this or is this just an experimental result?

If might be possible to use statistical thermodynamics to show it.

@ChetMiller, Thanks for your comment. I don't know that topic. I thought of proving it like what we used to do for Boyle's law isotherms. I tried like that, but it was unsuccessful.

Boyle's law assumes z = 1

@ChetMiller, I was trying to say, for the isotherms of Boyle's law, we use to prove the higher curves represent greater temperatures by some manipulations. In the same way, I need to know how to prove the higher curves in the above graph represent higher temperatures. I know Boyle's law is for ideal gases and the above graph is for real gases. Sorry, I didn't give it clearly the previous time. Hope you understood what I am trying to say.

I'm not sure what you suggest is always true. At higher pressure the curves cross and the higher temperature curve becomes lower than at a lower temperature. You can see this beginning to happen in your plot. If you make a general 'corresponding states' plot, (using reduced values for a van -der-waals gas makes the plot  common to may gasses) you can see the curves cross. Thus a general 'proof' is not possible.

@porphyrin, Thank you for your comment. I understood your point. But in my book it is given "if we are given such isotherms, we must conclude the higher curves represent higher temperatures." So, based on your comment I think the statement is not applicable to all pressures. If you find time,kindly see the following "accepted" answer does prove something related to my issue but neglecting the intersection and crossing of curves. I thought I was done with that. But now realised the importance of your statement.

Probably, I think they forgot to mention the phrase "at lower pressures". Then the statement would be correct.

Yes, but lower pressure means different values for different gasses. The answer https://chemistry.stackexchange.com/questions/68984/deriving-an-alternative-expression-for-the-van-der-waals-equation-using-given-pa?rq=1 gives the reduced values for the van-der waals equation. From these you can make your own general plot.

The compression factor $Z$ is defined as the ratio of the real molar volume with respect to the theoretical ideal
$$Z = \frac{V_m}{V_m^{ideal}}$$
where if $Z$ = 1, the gas behaves ideally.
If we want to get some sort of understanding for how this function behaves when you change the Pressures and Temperatures of your system, we need to substitute into this equation our variables of interest. To take the simplest approach, lets just use the ideal gas law as a substitute for $V_m$ (from a quantitative sense this doesn't make sense to do but qualitatively we can still derive some relationships therefrom)
$$Z = \frac{RTP^{-1}}{V_m^{ideal}}$$
What this shows us is that $Z \propto T$. Thus, an increase in $T$ demonstrates an increase in $Z$, which is the qualitative understanding we're looking for herein.
To note, using a more realistic model for $V_m$ gets complicated; for example if you wanted to use the Van der Waals equation
$$P = \frac{RT}{V_m-b} - \frac{a}{V_m^2}$$
instead of the ideal gas equation, the $V_m$ term doesn't reduce too nicely because of that $\frac{a}{V_m^2}$ term. Gets even worse for the Virial equations.

The seminal Van der Waals equation of state
P=( RT/V-b)- a/V^2
where v = 1/ρ is molar volume. It can be rearranged by expanding 1/(v - b) into a Taylor’s series:
Z= 1+ (b- a/RT)1/V + (b^2/v^2) +( b^3/v^3)+........
The second virial coefficient has roughly the correct behavior, as it decreases monotonically when temperature is lowered. The third and higher virial coefficients are independent of temperature, and are definitely not correct, especially at low temperatures but can be considered here. Now as temp T increase so do Z.


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Copper Tetrafluoroborate neutralisation/precipitation to reduce landfill
I have 25L drums of Copper Tetrafluoroborate waste solution.
Can it be neutralised, precipitated in some way, to minimise the amount of waste that goes to landfill?
also, landfills prefer solid waste instead of liquid, so is there any cost effective way to reduce Copper Tetrafluoroborate to a solid?

This would seem to be a question of your local/state/national hazardous waste disposal laws, rules, and regulations (in addition to any institutional added requirements).

Depending on the quantity / concentration of the tetrafluoroborate, the time and equipment at your disposition, you might be interested in the following publication, where authors claim electrocoagulation and -decomposition with sacrificial electrodes of aluminum.  At least the abstract is accessible freely: https://www.sciencedirect.com/science/article/abs/pii/S0043135419301629


text:
How to formulate the 1 dimensional time-dependent Schrodinger’s equation from a given function?

Assuming $f(t) =e^{−iωt}$, formulate the 1 dimensional time-dependent Schrodinger’s equation.

I started solving the problem as
\begin{align}
  E &= hf, \tag1\label{eq:1}\\
  E &= ħw, \tag2\label{eq:2}\\
  \frac{\partial f(t)}{\partial t}
    &=-iωf(t). \tag3\label{eq:3}
\end{align}
Then I solved \eqref{eq:2} and \eqref{eq:3} to get
$$\frac{-iE}{ħ}f(t)=\frac{\partial f(t)}{\partial t}. \tag4 $$
Now I am not sure how to continue from here to get the time-dependent Schrodinger’s equation.

The derivation you're showing is the time derivation. The time dependent wavefunction is $\Psi (x, t) = f(t)\psi(x) = e^{\frac{-iEt}{\hbar}}\psi(x)$ and setting t=0 gives you $\Psi (x, t) = \psi(x)$ which is where the time-independent equation comes from. IDK how you get that going backwards thought without just knowing that.

Unless what you're actually looking for is this piece: $f(t) = e^{\frac{-iEt}{\hbar}}$ and that just comes from knowing that the only function whos derivative yields the original function multiplied by some constant is the exponential function.


text:
What is the hybridization of chromium in chromate and dichromate ions?
I was recently reading about the transition metals and some important compounds of them. I came across a rather interesting diagram in my book... a tetrahedral structure for the chromate ion. The only tetrahedral structure I know arises from an sp3 hybridization of the central atom, but as far as chromium is concerned, I think that a d3s would be more practical(?).. But then again, which specific orbitals of 3d would hybridize to give a tetrahedral structure?(this is beyond me, i seriously do not understand the shapes of the d orbitals)

Do not use hybridization with transition metals.  Correction:  Best not to use it for anything at all.

but then again, how do you figure out shapes of stuff without that? like... normal covalent compounds, then we have the coordination spheres, how do you predict shape of such stuff WITHOUT taking hybridization into account?

By using a text that minimizes this hybridization stuff and m.j maximizes molecular orbital theory.

im kind of new to this molecular orbital theory stuff, will surely check it out

but then @OscarLanzi is it so that valence bond theory and VSEPR cannot explain everything?

VSEPR is a very good model, but you have to ignore the hybridisation postrationalisation, which came much later. It doesn't get everything right, but it allows prediction on very simple reasoning. It's good for a first guess approach. Valence bond theory is much often taught in a completely abridged (beyond the useful) manner. It's a full quantum chemical model. The exact treatment of a molecule with it results in the same exact wave function as with molecular orbital theory. Beyond that, there is currently no theory to explain everything, or as it is also sometimes called "World Formula".

@Martin-マーチンwould you please suggest some further reading related to the different models for bonding?

One other point to point out: While it may be tempting to assume a hybridisation and then deduce a geometry, it should actually be the opposite: observe a geometry and then deduce a hybridisation.

The d orbitals are commonly grouped into two groups because that is how they transform under both octahedral (including a g subscript) and tetrahedral (no g subscript) symmetry:

$\mathrm{t_{2(g)}}: \mathrm d_{xy}, \mathrm d_{xz}, \mathrm d_{yz}$
$\mathrm{e_{(g)}}: \mathrm d_{z^2}$ and $\mathrm d_{x^2-y^2}$

In tetrahedral environments, the p orbitals also transform as $\mathrm t_2$ while in octahedral ones they transform as $\mathrm{t_{1u}}$. By analogy to main group tetrahedral molecules, it would be easy to assume that just the three $\mathrm t_2$ d orbitals take part in the formation of tetrahedral complexes (throw in one s orbital for completion of the four).
In reality, all orbitals that transform as $\mathrm t_2$ will be represented in the final complex in some way or another, so that whatever bonding and antibonding orbitals of $\mathrm t_2$ symmetry you end up with will always include some contribution from the original p orbitals. Furthermore, outside of hydride complexes the ligands will also have available orbitals for π-type bonds which will complicate the picture further. The final MO for chromate will look something like the following:

Figure 1: tetrahedral $\ce{[ML4]}$ complex with ligand π contributions.
The metal orbitals on the left are d, s, p from bottom to top. The ligand orbitals on the right are the oxygens’ p orbitals (all 12 of them).
Don’t worry, these schemes will come to haunt you at one point when studying chemistry.

Thank you... exactly what i was looking for..


text:
Can mechanical agitation increase the reaction rate?
The specific example I saw recently was cleansing of concrete (portland cement) film from brick using a diluted muriatic acid. It appeared that the more scrubbing was used, the quicker the cleaning process would go. That seems to match up with everyday experience.
However what is not clear is how that mechanical agitation was specifically improving the reaction rate.
More generally, are there a 'class' of reactions or reagents where mechanical agitation will alter the reaction rate? And is that affecting the actual reaction or more just the process efficiency?

If the reaction is heterogeneous (multiple phases involved), mechanical agitation helps. If the reaction is homogeneous (all reactants and products in the same phase), mechanical agitation usually does not help. An exception is when the process is auto-catalytic (i.e. starts in one corner and then propagates).
In the example mentioned, agitation will help to transport concrete away from the brick surface and to bring fresh acid to the surface. Scrubbing also helps if there is no chemical reaction and the removal is mechanical in nature.

In the concrete example I'm pretty sure that's an actual reaction - it produces a gas aside from the fact that scrubbing alone would not be able to clean portland cement from brick :)  Thanks for answering generally not only to my example.

Butt it is the same. You need fresh acid to get in contact to the wall . The reaction as its rate as a virtual system. The point is that mechanical agitation ensures the conditions to be renewed homogeneously.

At the molecular scale, a reaction will not become faster by stirring or agitation.  At larger, macroscopic scale you may however increase the overall rate because agitation mixes the reactants with each other and hence increases the propability over time that they are exposed to each other.  Especially in the case of heterogenous reactions like the one described by you, you equally have to consider transport of matter across matter by diffusion.  See Diffusion laws, for a starter.  Even less quick are transport rates passing phase boundaries like crystallite grains to liquid.
This is why brushing the surfaces both brought new reactants into contact, and removed products formed which otherwise could have hampered the reaction to progress.


text:
What is the mechanism of oxidation of phenol to benzoquinone?
I looked up the mechanism in a lot of organic chemistry books, including Clayden, Klein, Solomons, McMurry but I couldn't find the mechanism of oxidation of a phenol. I also looked it up in pharmacognosy books, but with no luck. All I could find in these books is the following scheme, with no mechanism:

It seems like other oxidizing agents can be used, such as hydrogen peroxide and silver oxide.
Unlike primary and secondary alcohols, phenol doesn't have an hydrogen atom attached to the carbon directly bound to oxygen. My guess is that the double bond of phenol will react with the oxidizing agent, but not sure of how this will work.

http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0103-50532008000800007

does the same thing happen with HNO3 and H2SO4 ??? https://chemistry.stackexchange.com/q/181338/144409

The reaction is used industrially to make hydroquinone (source: wikipedia), but often using hydrogen peroxide as the oxidation agent in the presence of a catalyst.
A study on catalytic hydroxylation of benzene comes to the following conclusion:

experimental evidences have allowed first discarding Fenton-like mechanisms and later proposing the plausible competition between an electrophilic aromatic substitution pathway with the alternative rebound (hydrogen abstraction) route.

The Fenton-like mechanism refers to addition of a free oxygen radical species to benzene, which they reject based on experiments with radical scavengers. Instead, they hypothesize formation of a radical oxygen:catalyst complex. This either reacts with benzene in an electrophilic substitution, or abstracts a hydrogen from benzene, which then rebounds and reacts with the oxygen still attached to the catalyst.
Although the study I am citing is for hydroxylation of benzene instead of phenol, the paper also talks about substituted benzenes.

[OP:] My guess is that the double bond of phenol will react with the oxidizing agent, but not sure of how this will work.

You can't pin down a single double-bond in an aromatic system. What you can say is that the existing hydroxyl group will steer the reaction to make an ortho- or para- product. Once hydroquinone is formed, it oxidizes quickly to benzoquinone under the oxidizing conditions.

[OP:] What is the mechanism of oxidation of phenol to benzoquinone?

For the reactions conditions given by the OP, I did not manage to find a mechanism. However, the mechanisms I discussed probably are sufficiently general that they apply in this special case as well. That is just a guess, though.


text:
When to permute two times to find optical configuration
I'm learning about S and R configuration, and I know that in case where the 4th substituant is not behind we should permute it with the one which is behind. then we find the new configuration and use the opposit, if we find it R then the original configuration is S.
My question concerns the case when should I make 2 permutations ? Is that even possible ? because this will give back the original configuration but I dont know what are the cases which I use only one permutation and what are the cases where I apply more than one permutation, this seems very confusing.

Why, that's simple. You should use two permutations whenever this pleases you. Nobody cares how you arrived at the correct orientation (with the 4th substituent facing back).

If you switch any two substituents, that will change the stereochemistry. If you do a second switch (any two), you are back to the original stereochemistry. You can try it out with substituent 4 in the back, and seeing what switching around 1, 2, and 3 do to the clockwise or counterclockwise orientation (i.e. whether it is R or S).
Two switches (for example switching 2 and 4 and then switching 3 and 2, see animated GIF) amounts to a rotation around the bond of the "unswitched substituent (here substituent 1). This does not change the stereochemistry.
If you are spatially gifted, you can just do the rotation in your head to get substituent 4 in the back. If not, you can do one switch (realizing you just interchanged R and S) or two switches (R stays R, and S stays S). It is personal preference (oops, I think I just said the question should be closed because the answer is opinion-based).
If you want to try it with 3D models yourself, check out http://proteopedia.org/wiki/index.php/R/S_nomenclature.


text:
How do I solve this problem on mass balance from blast furnace iron making?
Question

An ironmaking blast furnace produces hot metal of the composition: $\pu{93.6\%}$ $\ce{Fe},$ $\pu{2.1\%}$ $\ce{Si},$ $\pu{3.6\%}$ $\ce{C},$ $\pu{0.7\%}$ $\ce{Mn}.$ The composition of the iron ore is $\pu{78\%}$ $\ce{Fe2O3},$ $\pu{9\%}$ $\ce{SiO2},$ $\pu{5\%}$ $\ce{Al2O3},$ $\pu{1\%}$ $\ce{MnO},$ $\pu{7\%}$ moisture. Coke rate is $\pu{850 kg/thm}.$ Composition of the coke is $\pu{80\%}$ $\ce{C},$ $\pu{10\%}$ $\ce{SiO2},$ $\pu{5\%}$ $\ce{CaO}.$ Enough limestone is charged to make $\pu{45\%}$ $\ce{CaO}$ in slag and composition of limestone is $\pu{100\%}$ $\ce{CaCO3}.$
Calculate (i) Weight of ore used per tonne of hot metal production; (ii) weight of slag produced per tonne of H.M.; (iii) weight of limestone used to produce given slag composition.

Doubt
In such problems all I need to balance is iron, silica, $\ce{MnO}$ etc. However, in case of $\ce{MnO}$ the solution in my book reads $\ce{MnO}$ from ore equals $\ce{Mn}$ in hot metal plus $\ce{MnO}$ in slag.
My question is for balancing $\ce{MnO}$: why do I need to drag in $\ce{Mn}$ content? The $\ce{Mn}$ in iron ore should be equal to $\ce{Mn}$ in slag. In fact, I feel that any content apart from $\ce{MnO}$ should not be dragged into picture. The rest of the contents like alumina, iron etc. quite comply to this argument. I would require a vivid justification to this.

I would think for (i) you only need to consider iron, and for (ii) you only need to consider calcium. Does all the iron end up in the hot metal? Does the calcium from coke and from limestone all end up in the slag?

For (ii), I would need help determining the coke: ore mass ratio from the coke rate, though.

@KarstenTheis Thankyou for your help.It is the Manganese that I am having doubt with.While they write MnO input=MnO output,MnO output equals "Mn in hot metal+MnO in slag."That is what the solution reads.Please explain why is it not MnO in hot metal.

"The Mn in iron ore should be equal to Mn in slag." vs
"the solution in my book reads MnO from ore equals Mn in hot metal plus MnO in slag."
It seems that the calculation desired involves breaking down the MnO in ore to Mn + O, and putting some of the Mn into the metal; the rest of the Mn goes into slag, as MnO. The hot metal would contain only metallic Mn, not MnO. It shouldn't be a huge adjustment in the answers.

Please tell me something..while balacing MnO why am I not taking the $O_2$ that needs to be liberated into consideration.If MnO in iron ore give Mn in Hot Metal+MnO in slag.The oxygen should go somewhere.But my solution neglects that completely.

Iron ore has Fe2O3. The O goes off with C as CO/CO2. The MnO in iron ore has to lose its O just like the Fe2O3 does. Consider Al2O3 and SiO2 to keep their oxygen; likewise H2O just evaporates. The coke has CaO and SiO2: they keep their oxygen. The overall process takes oxygen-containing metallic compounds and coke and limestone turns them into hot metal and slag and CO/CO2 + H2O. Consider the O in MnO (some of it) to go off as CO/CO2, just like the O in Fe2O3.


text:
Material that absorbs only in NIR and emits in visible
This Thorlabs VRC4 card here uses a material that absorbs only NIR and emits in visible. The spec sheets don't list what chemical this is. Do you know materials that show similar emission properties even with weak laser excitation?

The pdf says ".. of a UV light beam and its focal ..", but the specs make it clear its actually IR. ;) One should say this must be an IR laser, and the visible light is just a faint glow.

Not possible except for non linear phenomena. See comment above.

@Karl Yep, they have a typo

@Alchimista Im not sure it must be an actual nonlinear effect. See my comment to the answer below.

@user1155386 Dont you want to ask about possible processes for this behaviour?

related: https://chemistry.stackexchange.com/questions/77410/is-it-possible-for-a-substance-to-absorb-a-longer-wavelength-of-em-wave-and-emit?rq=1

@Karl TPA is a third order process.

@Alchimista TPA?

@Karl two photons absorption but I wanted to say second harmonic generation

@Alchimista Ja, but under a reasonably strong laser source, you might have a well populated $n=2$ state. From which you could easily get into a $n=3$ state, which decays to $n=1$ via visible fluoresence. You just need a substance where the lowest electronically excited state is within the IR energy range. Im a bit confused about the spectrum of this sensor sheet, which shows three IR absorption peaks.

@Karl The link doesn't work. But I see what you mean, it seems you have in mind something like azulene that in spite of being deeply blue and absorbing along the vis, has some transitions that from nir lead to vis emissions but again is a subtle mechanism based on states lifetime and I would say classifies as TPA. Also your mechanism should be non linear, at least if one does not start the analysis from the already excited state. But I think it is hardly what the op had in mind. In my opinion op could be satisfied with the answer by matt_black. And for details the link you have provided.

@Alchimista matt_black's answer does answer my question but partly. Green laser pointers employ strong IR sources that are then fed to SHG crystals. This card from Thorlabs seems to work even with weak to moderate laser excitation. So I was looking for chemicals that showed the same behavior but even with weak excitation. I edited the question to reflect this.

Check out https://www.rp-photonics.com/laser_viewing_cards.html, the IR light triggers phosphorescence of the card after it has been "charged" with visible light.

@Alchimista I updated the link, check under "graphs" there (VRC4 card).

@Paul Interesting link! Although the description specifically states: "As it is not necessary to charge the active region of the card before use, either CW or pulsed incident light will generate emission, even when the card is used in a darkened room." Maybe this does not employ some charging process.

It seems that the card either use a tpa mechanism or relies on the excitation of already excited states, as suggested by @Karl. I am not sure now if the latter classifies as non linear process, as Karl pointed out. First order parameters should be involved. However also a sequential abs mechanism implies that several photons are absorbed to leads to one photon emission.

Also posting to Physics SE might lead to an answer readily than here.

Such materials not only exist but are common
Most people will have seen an actual demonstration of materials that exploit the phenomenon of a substance that absorbs NIR light and emits visible light. Most Green laser-pointers are based on their use.
These usually rely on a semiconductor laser outputting near-infra-red (NIR) light at 1063nm wavelength which then passes through a crystal of KDP (mono potassium phosphate) or KTP (potassium titanyl phosphate) which emits green light at 532nm (which is the second harmonic of the original laser frequency). The mechanism involves non-linear optical processes (which usually require a fairly high intensity of light so won't happen as easily as phosphorescence, for example).

Like you said, the efficiency with SHG materials isn't that great, requiring high power lasers. I was surprised to see this card working with nominal power lasers as well, and I wasn't sure what kind of material this used.

Have you looked at the light curve they show? The card works at three different IR wavelenghts, but emits only at one VIS wavelenght. Also they explicitly say it is based on fluorencence. Could this be a double photon (or sequential) absortion, followed by vibrational relaxation and fluorencence?

@Karl Aren't double photon absorption cross sections quite small? Know of any material that has a large 2 photon absorption cross section?

@user1155386 Yes, they are,, I guess. Thats why I added the sequential absortion. The upper state of a first fluorescent transition is well populated under laser light, so you could easily get into a higher excited state.