text:
How does PEDOT:PSS work (mechanism) in Organic Electrochemical Transistors?
I have a question about the fairly fundamental theory behind the application of PEDOT:PSS in an organic electrochemical transistor (OECT) device.
As at least I understand it, the polyethylenedioxythiophene (PEDOT) chain is rich with holes and then is doped with polystyrene sulfonate (PSS) which has sulfonate anions that compensate these holes. Now according to the literature I have read, PEDOT:PSS is conductive (p-type) in its initial state (i.e. no gate voltage is applied). This confuses me a bit because I would have thought compensating these holes by adding PSS would have made the PEDOT less conductive.
Then of course when you apply a gate voltage, cations compensate the PSS anions instead which according to literature makes the PEDOT:PSS less conductive, but I would have thought this would make it more conductive as it liberates the holes in the PEDOT backbone.
What do I seem to be missing here?

You might want to define your acronyms...

Ah sorry, first time poster, wasn't aware of the convention, thanks for the fix!

This is about the use of well known words in semiconductor physics in the relatively new context of organic electronics. This is not always strictly justified, and at least confusing in many occurrences.
I will try not to slip in any deep debate about communities. Just let me say that obviously general physics applies both in classical semiconductors and organic semiconductors*, in particular when the output of a device - rather than its microscopic structure - is considered. It is fine to me to use whatever approach and treat these systems up bottom (semiconductors people do that) or vice versa, but keeping in mind that the constituents are covalent molecules.
Reality is in the middle, and organic semiconductors are likely to be compared to amorphous inorganic semiconductors even in the case of the firsts having rather crystal like structures. Molecules will be always therein, and a kind of hopping processes will be more or less crucial to conduction anyway.
PEDOT:PSS is in fact an ionic compound, a "poly(salt)", let me say. I want to say more than general ionomer, I do NOT want to propose a new name. It will be clear as you read. Be careful of italic.
For those not familiar with the material here is the link to it in Wikipedia: https://en.m.wikipedia.org/wiki/PEDOT:PSS
I would have placed a posive charge already in the structure, sketching a bit more than one EDOT ring.
Let's start with a neutral PEDOT chain itself. It is poly(ethylene-dioxythiophene), a conjugated polymer with quite low energy gap and low oxidation potential.
No drastic oxidative process, and even photoinduced electron transfer process (transfer to whatever available e-acceptor within the sample, purposefully added or not) lead to relatively stable radical-cations delocalised within the pi system along the polymer chain (there is an adapted or adopted name for these too, polarons).
If the anionic species formed are relatively stable too, then long living charges are formed.
In presence of very stable ionic species, either added or formed by the same process, a polimeric salt is obtained. In this polimeric salt charges of one sign are in chain, not in pending functional groups which is a much trivial case for which coincidentally polystyrene sulphonate PSS is an example.
PEDOT:PSS is right  one example of the polimeric salts that I have just introduced. The double dot here means a certain ratio depending on formulation, as the PPS providing the counterions (neutrality) can be in excess. The degree of oxidation can also varies depending on formulation.
You said " As at least I understand it, the PEDOT chain is rich with holes and then is doped with PSS which has sulfonate anions that compensate these holes."
Is the use of "doped", or alternatively "then", that confuses you.
The oxidized PEDOT chain is rich with holes in the sense that is rich of positive radical-cations and they behave as the carriers. However the chain is far to be neutral, is effectively positively charged, as you know.
Actually there are positive charges (positive carriers) by virtue of the presence of the external compensating PSS counterions. The latters are strictly necessary for the so called holes to exist within the PEDOT chains.
This is not the case for inorganic semiconductors, e.g. doped silicon is inherently neutral.
PSS stabilize the charges that in its absence will be either transient or prone to further chemical reactions as for radicals are generally very reactive.
I think this description should help you in getting the subtleties of this field and those of its dictionaries.
I could have simply written that
in conjugated pi system the so called doping is a redox process, transferring electron between the system and a second material
(generally from - to respectively, with notable exceptions of which fullerenes are probably the best known examples).
The carrier in oxidized PEDOT are positive and their motion is the opposite motion of electrons residing on the highest singly occupied MO, that ideally spread all along a perfect chain. In reality along a chain their will be a distribution of segments with different effective conjugated length, each with its corresponding  singly occupied orbital.
Positive charges move i.e. electrons move between these in chain or inter-chain SOMOs, which can be regarded as a band. In such a sense these carriers are called holes.  They move by a deficit of electrons in the normally occupied levels.
As for the convenience of calling  these carriers holes or not is up to you.
The point is that PEDIT:PSS can be regarded as a whole neutral material with mobile positive charges.
Then you said " Then of course when you apply a gate voltage, cations compensate the PSS anions instead which according to literature makes the PEDOT:PSS less conductive, but I would have thought this would make it more conductive as it liberates the holes in the PEDOT backbone. What do I seem to be missing here?".
I do not know what device structure you refer to. The conductivity of PEDOT:PSS is high enough to shorts two metals or doped Si - metal contacts, and to work as antistatic layer.
I believe that the behaviour of your diode will be dictated mostly by a second material with conductivity/mobility compatible to a transistor function.
PEDOT:PSS is a conductor tout court (between metals) or is in general applied*** as a selective contact when its SOMO levels mentioned above match (or mismatch) the relevant conduction levels of other materials.
In a sense the term semiconductor should be reserved to neutral PEDOT (although its bulk conductivity will be rather that of an insulator) while PEDOT:PSS has metal like bulk conductivity but a hole** type conduction mechanism as described above.
*it is clear what one mean saying "organic semiconductors", but I would prefer "extended pi conjugated systems". The same for "doped", I would say "oxidised (or reduced)". And so on. And not, there are not dampling bonds in molecules, even when it happens they are organic semiconductors.
**this should clarify that sometimes an adopted nomenclature is the first solution at hand ;)
***PEDOT:PSS is also reported as smoothing, adhesion promoter, etc all uses not pertinent to this Q.
EDIT: I realised you refer specifically to electrochemical transistors. Well, the above is still answering. Indeed, upon negative bias electrolyte cations infiltrate the material thereby neutralising the PSS anions. For the above discussion this reduces back the PEDOT, diminishing the available amount of carriers and as such the drain current.


text:
Molality of the solution prepared from a hydrated salt

A solution is made up by dissolving $\pu{13.8 g}$ $\ce{Na2CO3 · 10 H2O}$ in $\pu{103.5 g}$ of water. What is the molality of $\ce{Na2CO3}$ in this solution?

I can't solve it. I tried converting to moles with $\pu{13.8 g mol}/\pu{286 g} = \pu{0.04825 mol}$. Then I did $\pu{0.04825 mol}/\pu{0.1035 kg} = 0.466$ but it didn't work.

Because your salt contains waters of hydration, you actually have more than $\pu{103.5 g}$ of water.

Show some respect to the units too, they deserve better than this. I always recommend to start from writing down a complete math equation where you have an unknown on one side (molality) and the other side would contain known variables and constants (masses of salt and water, molar masses). Then plug in the numbers and keep an eye on the units – they should match (if they don't, convert).

Let's denote a hydrated salt as $\ce{M_pA_q*xH2O}$ ($\ce{M}$ – metal, $\ce{A}$ – anion). Then molality $b$ of the aqueous solution of $\ce{M_pA_q}$ is defined as
$$b=\frac{n(\ce{M_pA_q})}{m_\mathrm{tot}(\ce{H2O})}\, ,\label{eq:1}\tag{1}$$
where $n(\ce{M_pA_q})$ – amount of water-free salt; $m_\mathrm{tot}(\ce{H2O})$ – total mass of water.
Note that the amounts of the hydrated and the water-free salt are equal:
$$n(\ce{M_pA_q})=n(\ce{M_pA_q*xH2O})=\frac{m(\ce{M_pA_q*xH2O})}{M(\ce{M_pA_q*xH2O})}\, ,\label{eq:2}\tag{2}$$
where $M$ – molar mass.
Total amount of water is found as the sum of water added to dissolve the salt and the water "stored" within the hydrated salt itself:
$$
\begin{align}
m_\mathrm{tot}(\ce{H2O}) &= m(\ce{H2O})+m_\mathrm{hydr}(\ce{H2O})\\
&= m(\ce{H2O})+n_\mathrm{hydr}(\ce{H2O})\times M(\ce{H2O})\\
&=m(\ce{H2O})+x\times n(\ce{M_pA_q*xH2O})\times M(\ce{H2O})\\
\eqref{eq:2}\quad\Rightarrow\quad&=m(\ce{H2O})+x\times m(\ce{M_pA_q*xH2O})\times \frac{M(\ce{H2O})}{M(\ce{M_pA_q*xH2O})}\label{eq:3}\tag{3}
\end{align}
$$
Now, putting \eqref{eq:2} and \eqref{eq:3} back in \eqref{eq:1}, one can obtain uniform expression for molality of a hydrated salt:
$$b=\frac{m(\ce{M_pA_q*xH2O})}{m(\ce{H2O})\times M(\ce{M_pA_q*xH2O})+x\times m(\ce{M_pA_q*xH2O})\times M(\ce{H2O})}\label{eq:4}\tag{4}$$

 Plugging in the numbers for sodium carbonate decahydrate:
 $$b=\frac{\pu{13.8 g}}{\pu{103.5 g}\times\pu{286.14 g mol-1}+10\times\pu{13.8 g}\times\pu{18.02 g mol-1}}=\pu{4.3E-4mol g-1}$$

A side note regarding the notations and units [1, p. 28]:

The term molal and the symbol $\pu{m}$ should no longer be used because they are obsolete. One should use instead the term molality of solute $\ce{B}$ and the unit $\pu{mol/kg}$ or an appropriate decimal multiple or submultiple of this unit. (A solution having, for example, a molality of $\pu{1 mol/kg}$ was often called a $1$ molal solution, written $\pu{1 m}$ solution.)

Reference

Thompson, A.; Taylor, B. N. Guide for the Use of the International System of Units (SI). NIST Special Publication 811, 2008.

Sodium trioxocarbonate (iv) pentahydrate, a.k.a washing soda, is an efflorescent crystal containing 10 molecules of water of crystallization in its molecular formula.The molar mass of washing soda comprises a:
carbonate portion of 106 g + pentahydrate portion of 180 g = 286 g/mol.
i.e. 1 mole or 286 g of washing soda contains 106 g
                      of sodium trioxocarbonate (iv)
therefore, 13.8 g of washing soda contains
                 (106/286) x 13.8 g = 5.1147 g
                                    = 5.1147/106 mol
                                    = 0.04825 mol
                                  sodium carbonate
Similarly, 13.8 g of washing soda contains
          (180/286) x 13.8 g = 8.6853 g crystal water
On dissolution of washing soda, this 8.6853 g of crystal water adds to the 103.5g of solvent water so that the total mass of water = 8.6853 + 103.5 g
                             = 112.19 g water
This implies 112.19 g of water contains 0.04825 mol
                                   sodium carbonate
But, molality is the amount of substance, in moles, contained in 1000 g or 1 kg of solvent (water in this case).
Thus,
   112.19 g of water contains 0.04825 mol sodium
                                          carbonate
therefore,
  1000 g water contains (1000/112.19) x 0.04825 mol
                                         per 1000 g
                           = 4.300739816 E(-4) mol/kg
                           = 4.3 E(-4) mol/kg
                                     sodium carbonate
        - John Kerry Omoti, Lagos, Nigeria.

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text:
Oxidation of naphthalene with KMnO4

What is the product when naphthalene is oxidized by alkaline (or acidic) solutions of $\ce{KMnO4}$?

Some possible reactions show up in a google search, but they have no references:

Naphthalene to phthalic acid by alkaline $\ce{KMnO4}$:

Another product on oxidation of napthalene by alkaline $\ce{KMnO4}$:

If possible, please include the reaction mechanisms in the answer as well.

Quick search on Reaxys shows that both the products you've already listed have been made, also plus 2-formylbenzoic acid. Some are from 19th-century Chem. Ber. articles, which I'm not inclined to look up.

@orthocresol , Is there a reaction mechanism? If not, can you post this as answer?

will the 2nd reaction happen? As far as i know, the 2nd reaction will only happen to triple bonds.

Here is a reference: https://www.sciencedirect.com/science/article/abs/pii/S0045653510001906

The ketoacid in 2 is oxidized to phthalic acid by high valence manganese. Derivatize ArCOCO2H as ArCOCO2Mn(+7). Move electrons from the sigma C-C bond toward Mn+7.which forms ArCO+ , CO2 and Mn+5. The acylium species is captured by base. Alternatively, base attacks the keto group of the derivatized Mn+7 species, which upon collapse, moves electrons toward Mn+7 forming o-phthalic acid.

jchps.com/issues/Volume%2010_Issue%201/40-0581116.pdf. This link has a mechanism for the first reaction. However, the conditions are not clear. As for the second product, is can be observed in multiple sources (but no mechanism): http://www.orgsyn.org/demo.aspx?prep=cv2p0523 and https://pubs.acs.org/doi/pdf/10.1021/j150083a001. From what I am getting, it is likely the first product is in acidic conditions while the second one is in basic conditions. The mechanism for the second one isn't too clear though.

@ShoubhikRMaiti For #1 in basic media, there are old references like 1907JPhysChem93. For running the reaction in acidic media, you find via google e.g. this, too.

@Shoubhik RMaiti For #2, This may help(I mean maybe.. Just maybe, since it is Russian I think). Since the product in the #2 matches the one on the page 168

I already made a comment about some of what I am about to say but I will provide a partial answer. I say partially because I could not find any mechanism for the second product. However, from literature, what I found was that in acidic conditions, $\ce{KMnO4}$ will oxidize naphthalene into the first product, phthalic acid1. In basic/alkaline conditions, $\ce{KMnO4}$ will oxidize naphthalene into Phthalonic Acid2. The results also agree from other papers. The proposed mechanism for the acidic solution reaction is here (taken from the paper in the first link):

It is essentially the same mechanism for the cleavage of a double bond but it happens in two positions in naphthalene. Note this is the mechanism for $\ce{KMnO4}$ oxidative cleavage:

Also note that the mechanism in the first image was based on a 2017 paper studying the kinetics of the reaction and then formulating a mechanism. So this reaction is still being studied. Along the same lines, I could not find the mechanism for the reaction of $\ce{KMnO4}$ and naphthalene in alkaline conditions. I would assume that its mechanism is still being researched.
References

https://www.jchps.com/issues/Volume%2010_Issue%201/40-0581116.pdf
The Oxidation of Naphthalene to Phthalonic Acid by Alkaline Solutions of Permanganate, R. Arthur Daly, The Journal of Physical Chemistry 1907 11 (2), 93-106, DOI: 10.1021/j150083a001
http://www.orgsyn.org/demo.aspx?prep=cv2p0523
https://www.acros.com/_rainbow/pdf/oxidation_brochure_manga.pdf
http://studymaterial.unipune.ac.in:8080/jspui/bitstream/123456789/8059/1/POLYNUCLEAR%20HYDROCARBONS.pdf

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text:
Deuteration of enone - mechanism?

What happens when the above compound is mixed with D2O (deuterated water)?
As a result of tautomerism, I'd expect the 4H atoms, one alpha to carbonyl carbon, and three at gamma position; to be replaced with deuterium.
However, I'm not clear about the mechanism that'd be followed to achieve the same.
My thoughts:
The first step should involve formation of carbanion by abstraction of hydrogen gamma to carbonyl group. The carbanion is stabilised by extended conjugation, and has three resonance structures.
Now, how does the reaction exactly proceed? Could someone please provide a detailed mechanism, it'd be really helpful. Thanks a lot.

Besides proton transfers, I do not see any significant mechanistic steps here. Your analysis of the resonance seems correct to me.

As TAR86 mentioned, you seem to understand where deuteration occurs in crotonaldehyde. The process is usually conducted under alkaline conditions. All of the steps are reversible and need not occur in the same order as in the diagram. For instance, enolate 2 can deuterate at the α-position prior to γ-deuteration. Hope this helps.


text:
Should I write propanol or propan-1-ol?
Should I write propanol or propan-1-ol?
Personally, I'd go for the former (it being the simpler of the two), but what's the answer according to the IUPAC system of organic nomenclature?

"Propanol" is ambiguous.

Propanol can refer to either propan-1-ol or propan-2-ol. This is why Neretin pointed out propanol as "ambiguous".

Propanol can refer to both propan-1-ol and propan-2-ol.

Earlier we used notations like 1-propanol or 2-propanol. But according to latest IUPAC recommendations, we prefer names like propan-1-ol and propan-2-ol.
For compounds like aldehydes, carboxylic acids and for other compounds involving carbon centered functional groups, we need not mention position because we always consider the carbon of these functional groups as the 1 carbon.
But, for alcohols, as in this case, position should be mentioned for being more specific.

"Carbon centered functional groups" is misleading. Ketones for example definitively need a position specifier, except acetone and butanone. It's just that aldehydes and acids only have one bond free to start an alkyl chain.


text:
What is product obtained when B2O3 is reacted with NaOH?
Will the product be a metaborate ($\ce{BO2-}$) or a borate ($\ce{BO3-}$) or anything else and why?

Might be anything, depending on the conditions.

As @Ivan said, it may form metaborate, borax or a tetrahydroxoborate complex all depending on reaction conditions(temperature, pressure, medium, stoichiometry etc.).

$$\ce{B2O3 + 2NaOH ->[\Delta] 2NaBO2 + H2O}$$

Boron trioxide react with sodium hydroxide to produce sodium metaborate and water. This reaction takes place at a temperature of 400-550°C.

$$\ce{2B2O3 + 2NaOH → Na2B4O7 + H2O}$$

Boron trioxide react with sodium hydroxide to produce sodium tetraborate and water. Sodium hydroxide - diluted solution. Boron oxide is amorphous.

$$\ce{B2O3 + 2NaOH + 3H2O → 2Na[B(OH)4]}$$

Boron trioxide react with sodium hydroxide and water to produce sodium tetrahydroxoborate(III). Boron oxide is amorphous. Sodium hydroxide - concentrated solution. The reaction proceeds at room temperature.

Can't the product be NaBO3?

@Abcd NaBO3 requires borax and hydrogen peroxide. See this: http://www.prepchem.com/synthesis-of-sodium-perborate/


text:
Coffee and argon/nitrogen
I’ve recently been experimenting blanketing coffee with argon and nitrogen. I’m finding that argon has a more positive result. I can’t find any data or information that would confirm this other then taste! Can anyone help me to determine why argon is better to use with roasted coffee? I know argon is less reactive to other things then nitrogen. But is that anything backed up with data including some that is relatable to coffee?

To clarify - you are blanketing roasted coffee beans inside a closed container with argon and nitrogen, separately. Do you know what else (if anything) is in the argon and nitrogen you are using?

Argon 'has a more positive result' in what measurable way?

Argon normally comes in cylinders with a purity ranging from 4.8 to 6.0, which means from 99.998% to 99.9999%. Nitrogen can come in cylinders with similar purities, but it can also come from a nitrogen membrane system which yields purities between 95 and 99%. This last one is then only 2.0. If coffee is very sensitive to oxygen, this 1 to 5% could make a difference.

@JonCuster The OP says that it is the taste. It is not a conventional analytical instrument, but I know that in the perfumery industry there are employed people for their noses, wich are more accurate then GLC in detecting differences.

@RaoulKessels - perhaps, although the question is not clearly worded about that point. A wine or perfume expert would be able to clearly describe what was different between two samples, not just say that the taste or smell had a more positive result.


text:
How can the outermost electrons in silicon be 2s and 2p electrons?
My book says the number of outermost electrons in silicon is $4$. Yes, right. But then it says those are $\ce{2s}$ and $\ce{2p}$ electrons. The electronic configuration of silicon is $\ce{1s^2  2s^2  2p^6  3s^2  3p^2}$.
How are the outermost electrons $\ce{2s}$ and $\ce{2p}$ electrons. Those belong to third shell.
Then it says the maximum number of electrons in the outermost shell is $8$. Correct. But then it says those  are $\ce{2s + 6p}$ electrons. How?

The electron configuration you give is correct, 2 electrons each in 3s and 3p orbitals. Sometimes this is written in shorthand as $\ce{(Ne) 3s^2 3p^2}$ where the (Ne) represents Neon's configuration. I suspect that there is /are typos in your text. (btw, don't hate Chemistry: you can be interested in all sorts of subjects while still having favourites)

Please add citations to your post, it is not helpful to anyone further if the source of this is unclear.

The book means to say that there are 2 electrons in the  $\mathrm{3s}$ subshell and and 2 electrons in the  $\mathrm{3p}$ subshell. You can see this in the electron configuration as

$\mathrm{...3s^2 3p^2}$

and that the maximum configuration has 2 electrons in the  $\mathrm{3s}$ subshell and 6 electrons in the  $\mathrm{3p}$ subshell which would be

$\mathrm{...3s^2 3p^6}$

Concur with this point:
"$\ce{2s}$ and $\ce{2p}$ valence electrons" is not the same thing as "two $\ce{s}$ and two $\ce{p}$ valence electrons".
Your book probably used the latter, but you interpreted it as the former. Though, it is entirely possible that the book wrote it the way you say it had, but unless you quote that particular line and cite the book (as Mod. Martin pointed out), it will be impossible to ascertain that.
So for the record, silicon does have a valence configuration of $$\ce{3s^{2} 3p^{2}}$$
You must be attentive while reading such things ;-)

Now for the second part of your question,

Then it says the maximum number of electrons in the outermost shell is $\ce{8}$

Well, technically, the maximum possible number of electrons in the outermost shell (which is $\ce{n = 3}$ in this case) is $\ce{18}$ and not $\ce{8}$.
That's because the third shell is constituted by $\ce{s, p, d}$ subshells. (Granted, ground-state silicon doesn't use those $\ce{d}$ orbitals, but your post only mentions "outermost shell", so I thought mentioning this should be of help)
The third shell would therefore have one $\ce{s}$ orbital (holds 2 electrons), three $\ce{p}$ orbitals (holds 6 electrons) and five $\ce{d}$ orbitals (holds 10 electrons). So,
$$\mathrm{ 2 + 6 + 10 = 18 }$$


text:
Are the units of rate constants affected if we have more than one substance in the rate law
$$A+B \rightarrow C$$ $$rate = k[A] [B]$$
[A]= x mol of A/L
[B]= y mol of B/L
So why the unit of the rate constant is merely $$L/M$$ not having mol of A or mol of B?

Why the unit of weight is merely a kilogram, and not "kilogram of lead" when you weigh lead, or "kilogram of fluff" when you weigh fluff?

The rate has units of concentration/time  or $\ce{mol dm^{-3}s^{-1}}$. As this is an equation the left had side equals the right hand side both in value and units thus the rate constant $k$ has units  1/(concentration x time) or  $\ce{dm^{3}mol^{-1} s^{-1}}$

The heart of your problem is that you're assuming that there exist different types of the unit mole.
Consider the unit of length one meter. Take two meters of cotton thread, and two meters of jute thread. Place them together and you get a nice square of $1\times1 \text{ <unit of area>}$. Now what should the unit of area be? Should it be $=\text{cotton meter} \times \text{jute meter}$?
Well, it should NOT be $\text{cotton meter} \times \text{jute meter}$ because a meter, being a unit, is same for all substances being measured. A meter of jute is exactly the same length as a meter of cotton. Hence, the unit of area should be $\text{meter}^2$, which it always is.
Similarly, in your solution, you are assuming there exist two different types of the unit mole, one for $x$ and one for $y$. However, there is only one fundamental unit, and it measures the same amount of substance for both $x$ and $y$.
Hope it helps!


text:
Is it a redox reaction if the oxidation state of the element has changed only in one of the products?
I’m not sure on which elements to compare to ensure that this is a redox reaction:
$$\ce{2 H2SO4 (aq) + 2 NaBr (s) -> Br2 (l) + SO2 (g) + Na2SO4 (aq) + 2 H2O}$$
I’m trying to indicate whether this involve oxidation/reduction. I’m looking at the oxidation state of $\ce{S}$.
What do I compare $\ce{H2SO4}$ with: $\ce{SO2}$ or $\ce{Na2SO4}$?
I tried both. Oxidation state of $\ce{S}$ in $\ce{SO2}$ is $+4$ and in $\ce{Na2SO4}$ is $+6$. One of them tells me it’s reduced, and the other tells me that no redox reaction is happening.
Which do I compare with and why? I’m not sure.

I edited the question's title to make it more definitive (feel free to rollback if you think this was wrong). Also, I improved the formatting a bit. If you are interested, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

In a redox reaction you must both an oxidation and a reduction. It is impossible to have only a reduction, or only an oxidation.

One sulphuric acid is used to form sodium sulphate and the other one is used for oxidizing.
So you have a reduction:
$\ce{4H+ + SO4^2- +2e- -> SO2 + 2H2O}$
and an oxidation:
$\ce{2Br- -> Br2 +2e^-}$
The sodium sulphate is formed with:
$\ce{H2SO4 + 2NaBr -> Na2SO4 + 2HBr}$
This last one is not a redox reaction.

@Nilay Ghosh Thank you for editing. I really should learn that if I want to write here from my mobile. Is that the LaTeX mhchem package?

You can find this page of help.


text:
In the mass spectrum of niacin (3-pyridine-carboxylic acid), what ion corresponds to the peak at m/z = 105?
The mass spectrum can be found here.
I am interested to know what ion this is and why the peak at $m/z = 105$ is higher than the peak at $m/z = 106$ (which I believe corresponds to $\ce{C5H4NCO+}$).

From the general interpretation of electron ionization mass spectra, it would seem logical to report the $m/z$ 105 ion as a loss of $\ce{H_{2}O}$, thus $\ce{[C_{6}H_{3}NO]^{\bullet +}}$. As correctly suggested in the question, the peak at $m/z$ 106 is likely arising from a loss of $\ce{OH^{\bullet}}$ through an $\alpha$-cleavage of the $\ce{C-O}$ bond.
So the issue is: why would niacin (or nicotinic acid) undergo this $\ce{H_{2}O}$ loss? It is not observed for other similar molecules such as benzoic acid, which starts by a $\ce{OH^{\bullet}}$ loss. I did find two relevant publications in the literature on this point. Neeter and Nibbering[1] performed $\ce{D}$ labelling experiments, showing that the transferred hydrogen atom originates mostly from the $\alpha$ position (ortho from both the carboxylic acid and the nitrogen atom). Opitz[2] measured the appearance energy (AE) for $m/z$ 105 at 10.94 eV, more than 0.6 eV lower than the AE for $m/z$ 106 at 11.58 eV. He also suggests, based on metastable dissociation data showing loss of CO to $m/z$ 77, two possible structures for the $m/z$ 105 ion. So to answer the question: (a) formation of the $m/z$ 105 ion is thermodynamically favored compared to the formation of the $m/z$ 106 ion.
Based on this (scarce) literature evidence, I would suggest, based on labelling experiments, that some proton transfer occurs between the $\alpha$ cycle hydrogen and the hydroxyl group. This exchange can either move back and forth, or lead to a loss of water through formation of a cyclic structure.

References

Neeter, R.; Nibbering, N. M. M. Mass spectrometry of pyridine compounds. II: Hydrogen exchange in the molecular ions of isonicotinic and nicotinic acid as a function of internal energy. Org. Mass Spectrom. 1971, 5 (6), 735–742. DOI: 10.1002/oms.1210050612.
Opitz, J. Electron-impact ionization of benzoic acid, nicotinic acid and their n-butyl esters: an approach to regioselective proton affinities derived from ionization and appearance energy data. Int. J. Mass Spectrom. 2007, 265 (1), 1–14. DOI: 10.1016/j.ijms.2007.04.014.

Where the 7th carbon atom in $\ce{[C_{7}H_{3}NO]^{\bullet +}}$ appeared since niacin only has 6?

@andselisk : Just edited the formula, sorry for the confusion.

I did quick simulation with MMass (open source, available on Windows, MacOS, Linux) for the isotopic distribution in this region (Tools > Mass Calculator), and it looks like there are two $[\ce{M}-\ce{1e}]^+$ fragments:
$$
\begin{array}{rlr}
\hline
\text{avg.}~m/z & \text{Fragment} & \text{a.i.}\\
\hline
105.09 & \ce{C6H3NO+} & \approx45\%\\
106.10 & \ce{C6H4NO+} & \approx21\%\\
\hline
\end{array}
$$

Addition of the signals from both fragments (blue and green colors) would give nearly ideal peaks distribution reflecting experimental data (red color).
Technical note: MMass can import ASCII data, but NIST provides EI-MS spectra in unsupported JCAMP-DX format. To convert JDX to TXT, I used JDXview (free, Windows only).

I admit that MS is probably not my favorite method of analysis, but it would be nice if the downvoter would explain the down-vote.


text:
terminology for breakdown of liquid dielectric under the application of high voltage
I saw a diagram a while ago of increasing voltage (x-axis) and what seemed to be the chemical decompositions of ethylene glycol as it occurred at various applied voltages. (I am not a chemist- sorry about poor terminology) I want to find that diagram or something like it again, and I am failing.
I hoped to see intermediary products, final products, and intermediary reactions that occur when glycol was used as a dielectric in pulse power applications (we call the product of that breakdown "streamers," which is super general).
Any suggestions on how to word my search for something like that?
thanks for your help.

Off the top of my head, I'd call this "dielectric breakdown"

thanks! someone (whom I work with) suggested "breakdown curve" too.

There are various forms of glow discharge, at low current levels where ohmic heating is not significant, such as plasma glow discharge (usually in a gas, but could also occur in liquid, or perhaps even in  solids) and contact glow discharge electrolysis, http://iopscience.iop.org/article/10.1088/0963-0252/24/6/063001/meta


text:
Can every liquid pure substance be compressed to solid?
Apart form the fact that liquid are often misleadingly considered incompressible,
fact that I recently find out to be true only as an approximation in specific circumstances.
I was curious to know if it's a property of all elements that they can be turned from liquid to solid by applying pressure.
In fact by looking at the different phase diagrams and at the general one, seems an intrinsic property of the elements the capacity of becoming solid under some pressure.
A description of what happens at the atomic level, addressing the reasons for the phenomena would be appreciated too.

Water is no exception: it does become solid under high pressure.

Presuming your mention of water is because of the fact that the solid is less dense then the liquid (at normal atmospheric pressure), it turns out that water really isn't that weird. Elements with the same property include Si, Ge, Bi, and Sb. However, as noted by @IvanNeretin, they all are solids at high pressures, in part at least because they assume more densely packed crystal structures eventually.

From the diagram I see it seems that water can become solid from gas but not from liquid (under high pressure)...Instead it can become liquid from solid..Can you link me to your diagrams?

https://en.wikipedia.org/wiki/Properties_of_water#Melting_point

Consider neutronium...

@DrMoishePippik It may be fluid, but not liquid.


text:
Mechanism for acylation of benzene with succinic anhydride
I've just started with EAS. I wanted to write down a mechanism for acylation of a cyclic anhydride (for some reason the formation of the carbocation is not given in my textbook). So I wish to know whether this mechanism is correct or now. Since this is my first attempt at writing one, I don't know whether it's correct.
The reaction:

What I came up with:

The mechanism shown in the question seems reasonable enough to me. I would note that ring opening before the attack on the $\pi$-system is likely, but not absolutely necessary. Loss of $\ce{AlCl3}$ may not occur until workup. Direct proton transfer seems unlikely, but is a inconsequential detail. Note however, that the electron pushing arrow is drawn opposite to what it needs to be for that transfer.

I didn't quite understand what you meant by 'workup'

Workup = quenching the reaction usually with ice/water

are you sure that the top oxygen's lone pair forms the complex and not the middle ones' ?

Seems more likely to me, but the mechanism would then still work.

Your mechanism is almost correct. In your first intermediate, you have a red arrow in the wrong place. Since this ring is opening up, you want the arrow showing movement of the ring electrons (those between the lower carbonyl and the ring oxygen). Ideally, you would show those electrons becoming the new CO double bond of the aluminum carboxylate.

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


text:
Soxhlet extraction condenser - addition of compound to refluxing setup
I've never used Soxhlet apparatus before and I have troubles understanding how to properly perform the following step from my protocol:

To a gently refluxing suspension of 1 g Lithium aluminium hydride in 100 mL of diethyl ether, there was added 1.2 g of compound as a saturated diethyl ether solution with use of a Soxhlet extraction condenser which is modified allowing the continuous return of condensed solvent (through the thimble). After the addition was complete, the refluxing conditions were maintained."

Soo... How do I do it? I run RBF with condenser on top until reflux stabilizes and then I open it, add Soxhlet with compound and continue refluxing or I set up everything cold (RBF with ether + Soxlet with compound in ether + condenser) and then start refluxing?
Also - I'm not quite sure if I understand the thimble modification - can anyone shed some light on it for me?

Puzzling indeed

Perhaps it is so. ... They place the 1.2 g of compound in the thimble. At beginning.  The gently refluxing do the rest by dropping diethyl ether to the thimble. You get a kind of gentle addition of an almost saturated solution filtered by the thimble dropping down to the reaction flask. They basically innested a syphon at the bottom of the soxhlet .It seems a clever way to slowly add the compound without opening the systems with volatile solvents. This offer the way to keep a fixed volume of solvent all through the reaction.. But this means sacrificing a soxhlet and skill on molding glass. ..

Obviously if compound does not readily dissolve in diethyl ether.

It's such a pity the creator (Sasha Shulgin) of the protocol passed away. He would have loved to chat with you about it I am sure. I just read it from the good book myself and arrived here after puzzling over it as well.

i suspect it is possible that the confusion results from careless ambiguous language describing a standard Soxhlet process. The point is that a saturated ether solution of the compound results from putting the compound in the thimble of a Soxhlet apparatus so the resulting liquid is saturated when it leaves the thimble and is returned to the reaction.

In my opinion the intention is to have a very slow addition of the reagent to the hydride. The compound is dissolved in ether, so a Soxhlet is of no use since it is for solids.
What they are possibly referring to is a liquid-liquid extraction apparatus, specifically the one for extracting with solvents lighter than water:

The saturated solution of the compound is then placed in the thimble and when reflux starts, the solvent is condensed, falls into the tube and returns to the flask carrying over the compound. Normally the tube goes to the very bottom of the thimble. In fact, usually it is a loose tube with a wide opening on the top. How to know that the addition is complete is another story. If the compound is not colored, it will not be easy without opening the system.  In any case, maintaining reflux for a long time will ensure that everything goes into the flask.
I suppose that for some reason it is difficult to perform a slow addition through a funnel in this case. It would be, however, better to use a syringe pump but not everybody has one of those.

Yes but why to call it modify soxhlet?  I really think of something as in my comments. In that case addition is completed when no solid remains in the paper thimble .surely not a clear description anyway....

Modified in the sense that it is not an extraction. It is also said that a "saturated solution" is placed in the thimble, which is not posible in a Soxhlet with a bypassed siphon. Anyway, without some more information we can only guess.

Understand. But I would have written liq liq extractor :) I thought they were forming a saturated solution drop by drop. But in your case, why not to use a dropping fumnel and just a bit more solvent. I just tried to help.

True. As I have said in the answer, I suppose that there might be some problem for using a funnel. This is many times the case when volumes are small. Increasing the volume for having a slower addition might lead to other problems. By using this setup the addition is slow and the volumes remain constant as you have pointed out already.

Thank you for great responses so far!
Indeed - the protocol is quite minimalistic, but it was writen by a proffesional. Which Im not :D The reflux conditions are supposed to last for 48 h, and the compound is yellow so I guess Ill notice the addition. Corect me if Im wrong, but wouldnt it be simpler to omit the thimble completly? Just fill the Soxhlet with saturated solution of compound, set everything and star slow reflux. Refluxed solvent would slowly dilute compound and add it dropwise to LAH. Am I not seeing something?

If you do that, when the Soxhlet is full, the whole volume will be siphoned into the flask adding all the compound at once. Then there is no slow addition, which is what @Alchimista and myself think that is the reason for this non-conventional setup. The Soxhlet is normally used to extract solids. The solid is placed in a bag of filter paper in the thimble. The solvent is refluxed and, after condensing, it fills the thimble. The solid is extracted with the liquid. Then, when the thimble is full, the liquid is discharged into the flask through the siphon.

My last. In my idea the total volume remains 100 ml , that of refluxing hydride slt. I think I will use my setup if I would have a really modified soxhlet, I.e. syphon at botton , or I can even think of other setups let the condensing solv drop to the compound. If I would have an extractor I could use Raul setup. Total vol 100 plus that of tge sat solution.

@Alchimista in the liquid-liquid extractor the volume also remains constant. It flows over from the thimble, so you place the water phase inside and then you must fill it to the top with solvent so that the volume remains constant in the flask. Otherwise, the empty volume would be filled up with distilled solvent from the flask. In this case, the solution with the compound should fill the whole thimble. Since the volume is usually large, it would help to dissolve the product while not increasing the volume of reaction.

Ok ,I  overlooked that  poiny ,now I see.  But I was not taking it as a flaw.  It is just that we are somehow guessing.  Not to say that if compound is rather soluble I would go carefullu with a dropping funnel (it shouldn't be the case else the protocol does not make sense).


text:
Residual catalyst leaching from PETE
I am trying to look for healthy options for 5 gallon water jugs. The problem is all the plastic 5 gallon water jugs on the market are either made of #1 PETE (BPA free) or #7 (which contains BPA). It seems #1 PETE is healthier option but it turns out #1 PETE is known to leach a bit of antimony oxide ($\ce{Sb2O3}$ catalyst, used in its synthesis) into water.
So my question is if I bathe the jug in water, so that the antimony oxide can leach into the water and wash the jug one more time, can all the antimony be removed that way?

@Mithoron Actually, $\ce{Sb2O3}$ is used in PET production as a catalyst, and it's a relevant question IMO.

You should just make a link to source and be careful with capitalisation & punctuation. I didn't know about this Sb2O3 issue, and found it hard to believe. It doesn't seem to be particularly sever, but if it's supposed to be water to drink then metal would be better then plastic.

Let's ignore the safety aspect here, and focus on the 'if i bathe the jug in water so that all the antimony can leach into the water and wash the jug one more time, can all the antimony be removed that way'. In short, no.
There have been a number of studies on antimony in PET bottles. They look at various factors that influence the rate of leaching of Sb from PET bottles. These factors include temperature, UV, colour of bottle, pH, bottle volume and time. There is evidence that leaching is still occurring well beyond 6 months. One study suggests that leaching levels out after about 800 days. Another study calculates that the total levels of Sb in the plastic are very unlikely to be completely leached out.
Factors that have significant influence of amount of Sb leached include:

time stored (major influence)
temperature (higher temps have major influence)
volume of bottle (smaller volume bottles have higher SA/vol ratio and increase leaching)
type/origin of PET (significant differences between the colour of bottles used in one study)

I'm not aware of a study that attempts to find the conditions whereby Sb no longer leaches into the water, but effectively you would need to bathe the the bottle at high temperature for several days at a time, repeatedly, over months. That's just not going to happen.
Some useful references for you include:

doi.org/10.1016/j.watres.2007.07.048
doi.org/10.1016/j.scitotenv.2009.04.025
doi.org/10.1021/es061511+

Thank you, this answer is exactly what i am looking for. It seems like the rate of leaching is not as fast as i thought. Based on your information, if i use this type of plastic to store  water for a weekat a time at 23 degree C, it should not have much of an impact on my health then. But dam i wish there was s safer alternative

Greater concern comes from the storage of water at higher temperatures during transport and warehouse storage before it gets to the consumer. Feel free to accept the answer in order to close the question if has addressed your questions.


text:
How are reaction rate laws derived?
Let's take the reaction $2A + B \rightarrow C$, with the assumption that this is an elementary reaction. If I'm only given this information, how would I derive the following forward rate law from scratch?
$$
\frac{dC}{dt} = k[A]^2 [B]
$$
I tried to find an answer online but most sources jump straight to the above format for a rate law. I also understand that there are many experimental considerations which can affect what is actually observed (e.g. pseudo $0^{th}$rate law behavior) but I just want to know if there's a way to derive these rate laws given just the reaction. Thank you.

They're found experimentally.

So there is absolutely no way to derive them theoretically? It's all empirical?

I just read online that the form for an elementary rate equation can be derived from collision theory. Can anyone confirm that this is a common theoretical approach?

The elementary steps have to be found empirically, but once those are known, the rate laws are obtained from the law of mass action. Unfortunately, I'm not familiar enough with kinetics to say more than that.

It can be theoretical if you can determine the microscopic behavior of the reaction and be able to mathematically model it. Of course, it's pure speculation without experimental data, and you could easily be wrong. If you say I derived this rate law and nature says it's this other one, guess who's wrong.

@a-cyclohexane-molecule: This makes sense. The Law of Mass Action tells us how the rate law should look like, given a stoichiometric reaction. What I'm interested in is if this Law can be supported mathematically from microscopic principles.

@Zhe: Makes sense! I understand. I just wanted to figure out if a mathematical approach was possible, even if experimental approaches are more preferred and reliable.

No, you should think of it as practically impossible.

I think you may need to be more precise in wording your question. You're trying to ask whether or not the law of mass action can be derived, not whether reaction mechanisms can be derived, correct? I think a lot of these comments are talking about the latter.

I mostly agree @a-cyclohexane-molecule's preceeding comment. The question has been misinterpreted; OP does say to assume an elementary reaction. The classical ways of deriving rate laws are collision theory (Trautz–Lewis), and Eyring transition state theory (also Evans–Polanyi). The question is quite broad, though. You (@OP) might include more of your background after following up on these two approaches.

I see! Thank you for your answers. I'm sorry if I worded my question poorly.

@Cain It's alright, grasping the intent of a question is at least a binary system, responsibility laying with both asker and answerer. Be sure to edit your question or ask a new one once you've made the query more specific (like some particular step in a derivation, or clarify what exactly you are looking for: about distributions, partition functions, some assumption, or a specific resource etc.). Good luck!

The initial part of this answer might be helpful.

I am not a professional or anything close, but I think the answer to your question is simply that $C$ is jointly proportional to $A$ and $B$. The following link explains joint-proportion (https://www.mathwords.com/jk/joint_variation.htm). Just in case, the link gets broken in the future:

Here is a source that directly answers the question "just want to know if there's a way to derive these rate laws given just the reaction". The answer is generally no except for single-step mechanism, but the exponents will equal to the stoichiometric coefficients for the rate-determining step. Reference link: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/Rate_Laws/The_Rate_Law , to quote :
"For nearly all forward, irreversible reactions, the rate is proportional to the product of the concentrations of only the reactants, each raised to an exponent. For the general reaction
aA + bB → cC + dD  (11)
the rate is proportional to  [A]m[B]n :
rate = k[A]m[B]n   (12)
This expression is the rate law for the general reaction above, where  k  is the rate constant....
The dependence of the rate of reaction on the reactant concentrations can often be expressed as a direct proportionality, in which the concentrations may be raised to be the zeroth, first, or second power. The exponent is known as the order of the reaction with respect to that substance. In the reaction above, the overall order of reaction is given by the following:
order = m+n  (13)
The order of the chemical equation can only be determined experimentally, i.e.,  m  and  n  cannot be determined from a balanced chemical equation alone (e.g., Equation 11). The overall order of a reaction is the sum of the orders with respect to the sum of the exponents (Equation 13). Furthermore, the order of a reaction is stated with respect to a named substance in the reaction. The exponents in the rate law are not equal to the stoichiometric coefficients unless the reaction actually occurs via a single step mechanism (an elementary step); however, the exponents are equal to the stoichiometric coefficients of the rate-determining step. In general, the rate law can calculate the rate of reaction from known concentrations for reactants and derive an equation that expresses a reactant as a function of time.
The proportionality factor k, called the rate constant, is a constant at a fixed temperature; nonetheless, the rate constant varies with temperature. "


text:
Is dibromine a Lewis acid or base?
Is $\ce{Br2}$ a Lewis acid or base?
As it has both, empty orbitals and lone pairs, so if separately asked without any reaction what would we consider it?

Without any reaction, it is nothing. Within a reaction, it can be either. There is nothing unusual about that.

Ralph Pearson classified all Lewis acids and bases as hard and soft acids and bases. Br2 and Br+ are soft Lewis acids and Br- has properties in between soft bases and hard bases. Hence Br2 is considered to be a soft Lewis acid.
Soft acids have large acceptor atoms of low positive charge, high polarisability and low electronegativity. Hard acids have small acceptor atoms of low polarisability and high electronegativity.

$\ce{Br2 + H2O <=> HBr + BrOH}$ is only one example to show that this general statement cannot be true. It is always dependent on the reaction conditions.

The reaction should be mentioned. But the question was what is Br2 in general considered to be. We say HNO3 is an acid inspite of the well known reactions in which it is a base. Similarly, Br2 in general, is considered to be a soft Lewis acid.

enlighten me in which reactions $\ce{HNO3}$ is a base, please.

In nitration mixture, H2SO4 being a stronger acid abstracts a proton from HNO3 and HNO3 is a base in that reaction. Thus, in the presence of a stronger acid, HNO3 acts as a base.

even though your wording does not make any sense, I get what you mean. In any case, considering dibromine as an acid is questionable at best. Generalisations of this kind just distort chemistry for the learning, and lead to wrong conclusions in the long term. It is clearly both, and neither. You are simply offering a deceivingly simple solution.

I agree and understand but since the question itself was about generalisation, I made use of HSAB theory to find out whether Br2 is an acid or a base from its chemical properties.


text:
Le Chatelier's principle: Are there any exceptions?
The way Le Chatelier's principle is presented in most introductory chemistry books (high-school) is as though it's an indisputable law of the physical world (in the sense that we're never shown an exception, not that its universality is explicitly stated).
This is supposed to differ from the more frequently encountered  half-assed  "laws" like Ohm's Law or Hooke's Law (which are really just generalizations drawn from a few, everyday cases... not real "laws" as such).
Now I'm curious. Is there really no exception to Le Chatelier's principle (for a closed, isolated system)?
Wording this differently: Is there any situation (closed, isolated system) where conditions conducive to Le Chatelier's principle are in place, yet the expected Le Chatelier's "response" (the tendency to oppose change) is not observed upon effecting the said change (in parameters such as concentration, temperature, volume, etc.)?
If there are any exceptions that one comes across in everyday life, I'd prefer to hear those (but since that's extremely unlikely, I suppose just about any exception will do).
Are there any theoretical grounds for exceptions to Le Chatelier's principle?
If Le Chatelier's principle cannot be violated, then why is it so?

What sparked this question, is the following statement from a (school-issued) book of mine:

The dissolution of NaOH in water is exothermic. According to Le Chatelier's principle decreasing the temperature of the solution should favor solubility, however, in reality solubility of NaOH increases with temperature. This is therefore an exception.

However, if it weren't for the sharp users of Chemistry.SE I would've accepted my book's claims (since it corroborates seemingly well with common experience). O:)

I did find this article on the ACS page, but it's paywalled >_<

Most introductory textbooks make overly-broad statements that need to be corrected in later chapters or books - science is an onion where you have to peel back the layers and keep digging to get the the heart of the matter. But, Wikipedia's entry discusses this a bit and at least hints at why there are exceptions.

Related - http://aip.scitation.org/doi/10.1063/1.3261849

If Le Chatelier seems violated, then you're regarding an incomplete  description of the problem. To actually violate Le Chatelier, you would need some seriously perverted thermodynamics i think. I'd be very happy to see an answer with some reasonable proof  for this gut feeling of mine. ;-)

@JonCuster I read that Wikipedia page twice, but couldn't find any reference to any exception of Le Chatelier's principle, not even any hints to it, as you said :/

@GaurangTandon - the key line from Wikipedia to contemplate is "The principle is typically used to describe closed negative-feedback systems, but applies, in general, to thermodynamically closed and isolated systems in nature..." and start thinking about systems that aren't closed and isolated.

" and start thinking about systems that aren't closed and isolated." Well, I was actually interpreting it the other way around. If we can apply Le Chatelier's principle to a system, then it must be closed or isolated. That means if a system is open, we cannot apply Le Chatelier principle to it. That implies that the consideration of an exception to Le Chatelier's based on an open system is invalid, as being closed/isolated is a precondition to be able to apply Le Chatelier's principle in the first place. Though it's only my thoughts, would be glad to hear yours ^_^

An exception is known in the ammonia (Haber) synthesis reaction when extra nitrogen gas is added at constant T, P, and moles of hydrogen & ammonia. See Uline & Corti, J. Chem Educ 2006, v83, p138 for a detailed discussion.  Depending on nitrogen mole fraction the reaction moves to the left not right.

@Jon Well, my post does mention "closed" systems ;-) I've edited it to mention "isolated" as well O:)

@porphyrin Aha! I found that article (just the title really) when I went Googling at the time of posting this question (It's linked at the end of the post). Your link has an abstract (which mine doesn't), and that makes for an interesting read, but since the actual article's paywalled, I can't read up on it more :-(. If you ever have the time to spare, I'd love it if you could post an answer discussing that article's findings (and any other exception if you know it).

@porphyrin Just a note: The original requirements in my post was for an exception in a closed (and isolated) system. I guess adding extra nitrogen to a reaction doesn't count, but seeing as there are no other candidates, I suppose that particular case will do :-)

@GaurangTandon - As is often the case, understanding where the limits are is a very useful thing in science. Or, if we limit the Le Chatelier principle to only applying to systems that the Le Chatelier principle applies to, we don't learn much. Now, how does one perturb a thermodynamically closed and isolated system is a different way of asking the question that illuminates other aspects. In general, yes, we would expect a closed/isolated system to have on global minimum of free energy.

@paracetamol, I  agree that adding  material to a system already at equilibrium is stretching the point a lot. I prefer a more conservative definition of Le Chatelier that is along the lines of  'If the system is in a stable equilibrium then any spontaneous change in its parameters will be counteracted to restore the equilibrium'.  If this were not true then any fluctuation would be amplified and the equilibrium would not  be stable.

@paracetamol As by 2019-Jun-25, the article you indicate as «behind a paywall» is accessible freely.  It is marked by the green padlock of «unpaywall», an add-on to the firefox webbrowser (https://unpaywall.org/ or https://addons.mozilla.org/en-US/firefox/user/12867616/).

There are many exceptions to Le Chatelier's Principle (LCP) so it can certainly be violated. Recently, I have read many articles pertaining to this violation so I believe I should be able to provide an informed answer. Most of the violations of the principle can be seen in gaseous systems. I would not call them "violations", rather, just cases where LCP does not work so well
One of the most classic cases for such an exception, as presented by Cheung (2004), is the addition of nitrogen gas, at constant temperature and pressure, to an equilibrium mixture containing $\ce {N2}$, $\ce{H2}$ and $\ce {NH3}$. The mechanical approach invoking LCP would quickly conclude that there would be an increase in the amount of the gaseous product due to an increase in the amount of one of the reactant gases, favouring the forward reaction. However, this approach fails to consider the instantaneous increase in volume of the system, which would cause the partial pressure of $\ce {H2 (g)}$ to decrease. Now, if we consider the principle again, taking into account both changes, we would realise that the principle would tell us that the equilibrium would shift in both directions. Thus, we see that there is a possibility that LCP cannot give a definite prediction if the number of moles of gaseous products in the balanced equation is not equal to the number of moles of gaseous reactants.
Another example cited by Cheung (2004) is the dimerisation of $\ce {NO2}$. Consider an equilibrium mixture of $\ce {NO2}$ and $\ce {N2O4}$ in a sealed container. When we raise the temperature, we may expect that the position of equilibrium to shift to the left since the backward reaction to form $\ce {NO2}$ is endothermic. However, there is also an increase in the total pressure of the system when the mixture is heated. Again, by LCP, the pressure increase would likely shift the position of equilibrium to the right to the side with less moles of gaseous molecules. This is another case where there are indeterminate predictions given by LCP.
References
Cheung, D. (2004). The Scientific Inadequacies of Le Chatelier's Principle. Hong Kong Science Teachers' Journal, 22(1), 35-43. Retrieved July 27, 2018, from https://www3.fed.cuhk.edu.hk/chemistry/files/LCP.pdf.
Cheung, D. (2009). The Adverse Effects of Le Chatelier's Principle on Teacher Understanding of Chemical Equilibrium. Journal of Chemical Education, 86, 514-518. doi:10.1021/ed086p514
Quilez-Pardo, J., & Solaz-Portoles, J. J. (1995). Students' and teachers' misapplication of le chatelier's principle: Implications for the teaching of chemical equilibrium. Journal of Research in Science Teaching, 32(9), 939-957. doi:10.1002/tea.3660320906

The ammonia reaction is exactly what the paper @paracetamol linked talks about.

Your examples prove the opposite of what you say: Once you look at your system properly, you find that LeChatelier is not violated at all!

@Karl Yes, you are correct. I would not consider them "violations". These cases represent instances where the principle does not work so well (i.e. they indicate limitations of the principle).

I think Le Chatelier's principle is closely related to thermodynamics. Take the Clausius-Clapeyron equation for example. When water freezes into ice the volume increases. Therefore adding pressure helps ice melt into water at lower temperatures, which is how (I was told) skaters can skate on ice with less friction. Le Chatelier's principle is in the same spirit.
When applying Le Chatelier's principle to a chemical equilibrium, one should remember that the "counteract" statement only applies to intensive quantities. For extensive quantities, the shift of equilibrium actually helps with the applied change. For example, let's compare a normal ideal gas with no chemical equilibrium and the $\ce{NO2}$ gas with the equilibrium
$$\ce{2NO2<=>N2O4}.$$
If we add pressure to $\ce{NO2}$, the equilibrium moves forward to counteract the increase of pressure. But in doing so, the volume of the system becomes easier to compress (needing less pressure) than a gas that satisfies the equation $pV=nRT$ with no such reaction.
As for the dissolution of $\ce{NaOH}$ in water, I think there's more than just $$\ce{NaOH->Na+ + OH-}$$ going on. For example, the mixing entropy of $\ce{Na+}$ and $\ce{OH-}$ with the water molecules, the shift of the $\ce{H2O<=>H+ +OH-}$ equilibrium, and effects on the hydrogen bonds in water, etc. I think one (or maybe a computer) can still in principle correctly reproduce the trend of how solubility depends on temperature once all players of the game are identified. The water solution is a much more complicated system than a gas reactor. Thinking of heat as a product and raising temperature as providing more "heat" and applying Le Chatelier's principle this way is an oversimplification of what temperature can do.

Ice skating isn't due to pressure-induced melting. The surface of near-melting ice has a more disordered structure which makes it easy to deform.

@gsurfer04, thanks. I suspected this because skaters move very quickly and I doubt ice would have time to melt. But another experiment that cuts a big piece of ice with an iron wire attached to hanging weights should be genuine.

That's regelation! Cool stuff.


text:
Measuring heat effect of reaction
I did an experiment to determine enthalpy change when Zinc is added to Copper(II) sulphate solution.
I recorded values and plotted a temperature-time graph to obtain values of the temperature change to calculate heat transfer .

The trend line that I obtained is when temperature of the mixture started to drop.
I never knew why I needed to extend that trend line and draw a vertical line at the point when Zinc is added and thus never understood why the difference between the interception point and the point when Zinc is added is $\Delta T$
To summarise, I don’t understand why I need to extend the trend line and the interception point gives the highest ‘T’ value...

In an ideal world you would not have to do that. You'd simply wait until the reaction completes and then make a single measurement of temperature.
Too bad our world isn't ideal. The moment your reaction vessel becomes slightly hotter than the environment, it starts losing heat. If you wait for too long, it will cool all the way down to room temperature, so there will be nothing to measure in the first place. If you make your measurement in the first seconds so as not to let any heat escape, that's too early; the reaction has barely started. If you wait just until completion (which is a few minutes into the process, judging by your graph), that's too late, because by then some amount of heat is already lost.
So what do we do? We account for the loss. We assume that the temperature drop is linear (which of course is just an approximation, but still, it is much better than nothing), make a dozen or so measurements, build that trendline, and extend it to the origin. That's where your temperature would stop if there were no drop.
So it goes.

Thanks alot! One more question from me, I did another exercise where I just recorded the initial and highest temperature it reached to calculate enthalpy change. As compared to the style I went for above (when I recorded the temperature in intervals), why would I get a more accurate $\Delta H$ value for the style I went for above rather than the exercise when I record the initial and highest temperature ?

Is it because of the same explanation ? When we assume heat lost to the surrounding for this but not for the one when I just used highest and initial temperature .

That's right, the highest temperature is reached when some heat is already lost.


text:
Rate of electrophilic bromination in aromatic nucleus
In Clayden,  the increased rate of electrophillic aromatic substitution is explained  by the stabilization of the intermediate through sigma conjugation. In Solomon & Frhyle, the same is explained through inductive effects.
Both  the factors  oppose each other in the following compounds: toluene, ethylbenzene, isopropylbenzene, and t-butylbenzene, respectively.

My question:

What should be the order of electrophillic aromatic substitution for the above molecules ?

Can you cite the relevant chapters, page numbers, and the exact paragraph written in the books, to support your first paragraph? That will also give potential answer posters some additional insight into what the books might have meant. Thanks!

Hyperconjugation is more generally effective than inductive effects, I guess.

@Apoorv see this https://chemistry.stackexchange.com/questions/88752/why-is-the-rate-of-reaction-of-polar-eas-favored-by-inductive-effects-and-the-ra

I  was thinking too simplistically.


text:
Pressure exerted by the mixture of nonreactive gases

$\pu{2.0 L}$ of $\ce{O2 (g)}$ and $\pu{8.0 L}$ of $\ce{N2}$, each at $\pu{0.00^\circ C}$ and $\pu{1.00 atm}$, are mixed together. The nonreactive gaseous mixture is compressed to occupy $\pu{2.0 L}$ at $\pu{298 K}$. What is the pressure exerted by this mixture?

I tried to solve this problem by first using
$$P_\mathrm{tot}=\frac{n_\mathrm{tot}RT}{V}\quad(T=\text{constant}; V=\text{constant})$$
then using
$$V_\mathrm{tot}=\frac{n_\mathrm{tot}RT}{P}\quad(T=\text{constant}; P=\text{constant})$$
Plugging in the numbers for the second equation
$$10=\frac{n_\mathrm{tot}\cdot 8.314\cdot 273}{101325}$$
$$n_\mathrm{tot}=\pu{0.45 mol}$$
Then plugging in the numbers for the first equation
$$P_\mathrm{tot}=\frac{0.45\cdot 8.314\cdot 298}{2}=\pu{6.76 atm}$$
Whereas the answer is $\pu{5.5 atm}$. What am I doing wrong here?

The issue is that you're using two different units of pressure in the same set of calculations. If you plug in $\ce{1 atm}$ instead of $\ce{101325 Pa}$ (since that's the simplest way to go about this) and do everything else the same, you'll end up with ~ $\ce{ 5.46}$ moles.

You did too many unnecessary steps and also didn't use unified units (e.g. liters instead of cubic meters, since you decided to use $R=\pu{8.314 J mol-1 K-1}$). I denoted both total pressures as $P_1$ and $P_2$ for simplicity since we don't need partial pressures here. From the ideal gas law:
$$P_2=\frac{n_2RT_2}{V_2}\tag{1}$$
Since both gases are not reacting at these conditions,
$$n_1=n_2=\frac{P_1V_1}{RT_1}\tag{2}$$
and (1) can be rewritten as following:
$$P_2=P_1\frac{V_1T_2}{V_2T_1}=\pu{1.00 atm}\cdot\frac{\pu{10.0 L}\cdot\pu{298.13 K}}{\pu{2.0 L}\cdot\pu{273.13 K}}\approx\pu{5.46 atm}\tag{3}$$


text:
Let us say that height of HCP is h then what is the distance of tetrahedral voids
In a hexaoal cose-packedstructue et $h$ be the spacing between equivalent basal planes.  Then what are the "heights" of the tetrahedral voids lying between any two adjacent, equivalent basal planes, as distances from either plane?
The book gives the answer $h/8$ and $(7h)/8$. How is this derived?  My (the original asker's) expectation, based on how I understand the hcp structure, was that the heights would be $h/4$ and $(3h)/4$.

@andselisk it might mean the base of a single unit cell

What is the base of a single unit cell then? This is not a crystallographic term by any means and carries no information. I suppose it's an attempt to denote $xy$ plane.

@andselisk the question does not specify any other details... So could you solve it by assuming x-y plane?

Colleagues, please refrain from close-voting. This question deserves an answer which should be broader than the question itself. In particular, it should explain why the given answer of h/8 and 7h/8 is wrong, and in what sense it may be considered right (sort of).

@IvanNeretin sir can the question be actually factual b/c in chemistry first experimental proves are presented then logical reasoning is given? Then either the question's answer is wrong or it may be an 'exception'.

This is a great reason why we should be allowed to proactively vote to prevent closure.  I got my answer in but intend to vote to reopen if the q gets closed.

I down-voted this. Even though there might be people who understand your cryptic wording, it is hard to tell that we are actually talking about a crystal structure, or whatever. As such I think this question is utterly useless except for yourself.In a general sense I think it is unclear. An if you start a question with 'the answer is' - then why are you even asking? If you are asking whether a book is wrong include the source.

I second what @Mart said. Though the question looks less homework close-worthy after the edit, it's still illegible (to me). If someone who actually understands it, it'd be appreciated if they edit this post to make it clearer. Until then, I'm abstaining from voting (to reopen) this.

Actually $h/8$ and $(7h)/8$ are correct ... but not the whole truth.  TL, DR the correct set is $\{h/8,(3h)/8, (5h)/8, (7h)/8\}$.
Here is how to get $h/8$ and $(7h)/8$:
Think of the unit cell as consisting of a "base" having parts of seven atoms, then three middle atoms entirely inside the cell, then the top layer having parts of seven more atoms.  Define $h$ as the height from the middle of the base atoms to the middle of the top atoms.  Then each of the three atoms at height $h/2$ above the base is the common vertex of two tetrahedea.  One has its base at the base of the larger cell, and since the center of the tetrahedron is one fourth of the way from the base to the apex it's height is $(h/2)/4=h/8$.  The second of the two tetrahedea is upside down, with its base at the top of the cell, and the height of its center is $h-(h/8)=(7h)/8$.
Now for the curveball: these cannot be all solutions.  We forgot to include the possibility of voids involving atoms from surrounding cells!  If we include these surrounding atoms we discover that there is a layer of tetrahedral voids at $h/8$ above the mid-level, matching those at $h/8$ above the base, and another layer at $h/8$ below the midlevel matching those just below the top.  These additional voids are hard to "see" because they involve atoms in surrounding cells.  Nevertheless they are there, the complete solution set is $\{h/8,(3h)/8, (5h)/8, (7h)/8\}$, and an equal number of tetrahedral voids (three per cell) lie at each of these heights.

+1 Your answer is more insightful than the question. It would be of great help if you considered editing the question to make it clearer (to the rest of us). Thanks!


text:
calculating ph of a mixture of acids
I came across a question that couldn't find the answer to and i couldn't find the answer anywhere. The question is the following.
You have a soltion created by dissolving HCN and HBrO in water (T = 298.15K) and the pH = 4.816. A titration of 100ml of this solution needs 100ml of 0.2M NaOH to reach its point of equivalence. Wich of the following statements is true:
-the concentration of HBrO equals 1.5 * 10^-1 M in the original solution
-the concentration of HCN equals 5.4 * 10^-2 M in the original solution
-the concentration of BrO- equals 1.5 * 10^-1 M in the titration solution
-the concentration of CN- equals 6.1 * 10^-1 M in the titration solution
I already found the following things:
the concentration of H3O+ in the original solution should be 10^-4.816. I can fill this in in the acidity constants of the reaction to find:
°the concentration of CN- = 6.2*10^-10 / (10^-4.862) * the concentration of HCN
°the concentration of BrO- = 2.8*10^-9 / (10^-4.862) * the concentration of HBrO
and the change of the sum of these two concentrations should equal the concentration of H3O+.
This is the point where i get stuck. I don't really know how I can use the titration to benefit my calculations. It would be a huge help if someone would help me.
I also apologise if I sometimes use wrong/weird terms to describe things. English isn't my mother tongue and I haven't got the slightest clue what the english terms are.

This proved to be a most interesting problem as posed. To solve it we note that we have two things we need to find: the amount of $\ce{HCN}$ and the amount of $\ce{HBrO}$ that went into the original solution. We also have two observations: the pH of the original solution and the amount of $\ce{NaOH}$ that was required to bring the solution to the equivalence point. Given that we have two equations in two unknowns we can get a solution if the two equations are consistent (we can find the point of intersection of crossing straight lines given the slope and intercept of each) but if they are not we cannot (we can't find the point of intersection of two parallel lines). The data here are inconsistent so there is no solution. If the strength of the NaOH solution used for the titration were 0.1 M instead of 0.2 M or if it took 50 mL of 0.2 N instead of the specified 100 to reach equivalence then we can obtain an answer which is that the original solution is 0.117 M in HCN and .081 M in HBrO. To check on this answer we note that the fraction of molecules of a monoprotic acid in a solution at pH that lose their proton is $$f_1(pH, pK_a) = 1 - 1/(1 + 10^{(pH - pK_a)})$$ This comes out of the Henderson Hasselbalch equation. Thus if we have $M_{HCN}$ molar HCN and $M_{HBrO}$ molar HBrO the number of protons released to a liter of solution at pH 4.186 would be $$N_{H^+} = M_{HCN}f_1(4.816,9.21) +  M_{HBrO}f_1(4.816,8.70)$$ with 9.21 and 8.70 being, respectively, the $pK_a$ values for prussic and hypobromous acids. This is the first equation. It is given that $N_{H^+} = 10^{-4.816}$ and if you substitute $M_{HCN} = 0.117$  and $M_{HBrO} = 0.081$ into it you will see that these values satisfy it.
The second equation deals with the titration data. If we have $M_{HCN}$ molar prussic acid at some pH, here pH0 = 4.816,  there are, as we have seen, $M_{HCN}f_1(pH0,9.21)$ moles of dissociated acid molecules in a liter at that pH. At some other pH, call it $pH_i$ the number of dissociated molecules HCN per liter will be $M_{HCN}f_1(pH_i,9.21)$. To get from pH0 to $pH_i$ thus requires that $$dQ_{HCN} = M_{HCN}f_1(pH_i,9.21) - M_{HCN}f_1(pH0,9.21)$$ moles of protons be absorbed by $\ce{OH^-}$ ions. There is a similar requirement for the other acid. The water itself goes from a concentration of hydrogen ions of $10^{-pH0}$ to $10^{-pH_i}$ requiring $10^{-pH0}- 10^{-pH_i}$ additional $\ce{OH^-}$ ions to neutralize them and finally, the $\ce{OH^-}$ ion concentration rises from $10^{(pH0 - 14)}$ to $10^{(pH_i - 14)}$ requiring that $10^{(pH_i - 14)} - 10^{(pH0 - 14)}$ additional. Defining the water's  $\ce{OH^-}$ requirement as
$$ dQ_w =10^{-pH0} - 10^{-pH_i} + 10^{(pH_i - 14)} - 10^{(pH0 - 14)}$$
we can now write the second equation
$$N_{OH^-} = M_{HCN}dQ_{HCN} + M_{HBrO}dQ_{HBrO} + dQ_w $$
At this point we have our second equation but there is a big problem here. We don't know what $pH_i$ is and thus cannot obtain values for $N_{OH^-}$ to compare to the amount of base in the titrant we used. Either the question has been posed with $pH_i$ omitted in error or the poser really wants us to think. We are given the amount of base required to reach the 'equivalence point' and while it is pretty clear what 'equivalence point' means when a single acid (each acid has its equivalent point) but what's the equivalent point of a mix? If the acids are close in strength, as these two are, the composite titration curve exhibits a clear inflection point and I am interpreting equivalence point here to mean inflection point. Fortunately, the inflection point of the titration curve of a mix of acids of similar strength depends can be accurately found from log of the ratio of the strengths as long as -2 < log ratio < 2. Using this we can eliminate the functional dependence of the second equation on $pH_i$. Thus, while the second equation appears to depend on $M_{HCN}$,$M_{HBrO}$ and $pH_i$ in fact the latter is completely determined from the former 2 and so we have a second equation dependent only on $M_{HCN}$ and $M_{HBrO}$ which, along with the first (also a function of these two variables) gives us a system that can be solved as long as we are given reasonable data to plug into the system. 100 mL of 0.2 M NaOH is not reasonable data. 50 mL of 0.2 M is reasonable data. If the titration NaOH was 50 mL we would, as stated earlier, obtain $M_{HCN} = 0.117$  and $M_{HBrO} = 0.081$ from which we can calculate $pH_i = 9.20$  So what would happen if we added 100 mL of 0.2 NaOH? The mixture would come to some pH other than the inflection point pH. If we are told what that is we could still solve the problem and much more easily.
I'm not going to say much about how one solves this pair of equations as, while the first is linear in the unknowns, the second isn't because it is nonlinearly dependent on $pH_i$ which is, in turn, dependent on the unknowns is a non linear fashion.  The 4 derivatives of $N_{H^+}$ and $N_{OH^-}$ are found with respect to the two unknowns are found and arranged in a 2 x 2 matrix. The is inverted and multiplied by the error vector (the difference between what values of $N_{H^+}$ and $N_{OH^-}$ we get using our guess as to what $M_{HCN}$ and $M_{HBrO}} might be) to give a correction for our guess. This is repeated until the errors are very small. It's Newton's method for more than one unknown.
This has been, it seems, more about math than chemistry by I don't know how else one might solve this problem.


text:
What happened to this 10 year old unopened plastic bottle?
A colleague of mine found this unopened Aquarius bottle (non-carbonated drink) that was 10 years past its expiration date.
The bottle was collapsed inward which would suggest that the pressure in the bottle was lower then outside the bottle.
Can someone explain what happened here? Also, what happened to the air inside the bottle?

Not that I doubt you, but as a clarification: Are you sure some idiot didn't squeeze it like that and leave it lying around 10 years ago ;-)

As someone who has never had any of it: Is this drink carbonated?

Nope, that didn't happen.  We opened the bottle and it was still very firmly shut close.  This is not a carbonated drink.

Aha, I see! And by "still very firmly shut close", should I take it that bottle was unopened prior to finding it?

Yes, it was closed all these years

Thanks. I've edited your post, do give it a look to make sure I didn't distort anything :-)

Further clarification: Can you shed any light on where/how the bottle was stored all this time? i.e- was it in a refrigerator, icebox, etc? And could the  bottle have been taken out of storage in those 10 years?

The bottle was stored in a closet for 10 years, it was never taken out

Okay, and the temperatures normally lie in what range? 25-27 degrees Celsius (normal room temp.) or something else?

Regular room temperature. Between 18°c in the winter and 25°c in the summer?

Added ingredients in Dutch/French

The real question is, where did all the liquid go?

After we opened it, the bottle expanded again.  The liquid seemed to be on the same level as when the bottle was bought.  Suggesting no liquid escaped/evaporated/was lost in the process :)

Since the drink is not carbonated, no liquid has been escaping the bottle, and the components don't seem to be able to participate in any kind of chemical reactions involving gaseous products, I assume the gas that's been leaked is air.
I also cannot imagine any kind of chemical reaction involving consumption of this amount of air leading to such a severe bottle deformation. My theory is that air simply escaped over time through the bottleneck. There are 2 parts made of 2 materials with different thermal expansion coefficients:

the bottle itself (polyethylene #1 (PET), or polyethylene terephthalate (PETE));
the screw cap (polypropylene (PP)).

Due to the construction of the joint, both parts basically act as a pressure equalization valve. I suspect that the coefficients of thermal expansion are chosen in such a way that when heated, excess gas would slowly escape in order to minimize the risk of explosion when the storing conditions for the finished product is not met. So, when the bottle is heated, the air is slowly leaking out. Once the temperature drops, the "valve" closes and the process stops. Over the course of 10 years, there surely were many temperature fluctuations making air leave the bottle little-by-little.

If nothing can fix oxygen or perhaps CO2 from some fermentation, it must be something like that.

What do you think of forming CO2 which might selectively permeates out?

I guess if fermentation took place, it would be noticeable, but it looks like OP has no complaints regarding the color or smell. And there is no membrane that can selectively pass $\ce{CO2}$ through; the lid-bottle joint is hardly that selective:)

I agree.  But for selectivity I mean the bottle . I thought for some  reason it could be more permeable to CO2 than nitrogen and oxygen.

Oh. But how? PET is not quite porous, and if it were, then smaller diatomic gases (oxygen and nitrogen) should've escaped first, I think.

I didn't mention this in my OP but the color of this drink is more orange then shown on the picture. So the drink also lost its color in those 10+ years

I was just thinking of permeability because of 10 y is pretty long. Not high science but besides the fact that P lowered I find this rather puzzling. I do not see any trivial chemistry in it. I try to look for pet permeability to CO2 and air. And water too.

Assuming CO2 is formed at expense of O2 or anaerobic processes I found something on this direction http://www.composite-agency.com/archive/Carbon-Dioxide-Diffusion-PET.pdf

@Johan Interesting point. Perhaps that's simply the result of the drink's contents and coloring agents degrading (since it's well past the expiration date). andselisk's argument sounds pretty solid to me. If this answer helped you, consider clicking the "Accept answer" button (below the voting interface) :-)

https://chemistry.stackexchange.com/questions/51695/why-do-sealed-plastic-bottles-of-pop-collapse-as-the-contents-get-old


text:
Why is the rate of reaction of polar EAS favored by inductive effects and the rate of less polar EAS reactions through hyperconjugation?

  1. Reactions involving strong electrophiles (nitration) are
       said to be governed
       by
       Inductive effects thus the order of nitration is
D>C>B>A
  2.Chlorination is a less polar reaction thus chlorination said the be governed by hyper conjugation.
A>B>C>D

My question is,  why?
The factors controlling EAS in both the cases will be  the same for both the substitutes.
Hyperconjugation and inductive effects will both operate without distinction in both the cases.
So why is it that
The rate of reaction of polar EAS favored by inductive effects and the rate of less polar EAS reactions through hyperconjugation.

SOURCE A guidebook to mechanism Peter Sykes

EDIT
I have been asking around for an answer to this question.
No one has been able to answer it conclusively , the primary school of thought   is that

In Highly polar reactions , inductive effects dominate as more electron density is available , hyperconjugation on the other hand is just redistribution of electrons.

First of all is this correct ?   Even if it's correct but how does it explain the opposing rate orders for nitration and bromination?

The only explanation I can think of is that less polar reactions are governed by orbital interactions (early TS) and hyperconjugation increases the coefficient of HOMO at the most reactive sites making it the dominant factor. On the other hand, more polar reactions are governed by stabilization of transition states which have a significant charge development (late TS)  and the inductive effect predominates here.

@Marko Isn't the sigma complex an intermediate and not a TS?

@ApoorvPotnis Yes, the sigma complex is an intermediate.

@Marko Then about which TS are you talking about in your comment? Is it the donation of electrons to the nucleophile or the removal of hydrogen?

@ApoorvPotnis  The donation of electrons to the electrophile.

OP, you should read about Hammond's postulate and early and late transition states (just keep in mind that both chlorination and nitration are exothermic). I dont' think I should write an answer as I am only guessing.

I may be oversimplifying things.

A look at electron density

Let's pick up the extreme cases of the substituted benzenes . $$\ce{A -> ArCH3}$$
$$\ce{D -> Ar(CH3)_3}$$

Case A shows the maximum hyperconjugation. Case D on the other has the maximum +I effect.

For a moment let's imagine that the electrons are coins.
Case D , has more coins, more methyl groups (let's say 3).
Case A, on the other hand has 1 coin , hyperconjugation would simply mean that I am  simply moving that coin in circles. It's simply redistribution of electrons.it doesn't actually increase the density.
So the first point which I'm going to use to answer the question is that -

Inductive effects increases  the electron density, hyperconjugation simply redistributes the electron density.

A look at Coulomb's law
$$\ce{ F \propto Q_1*Q_2}$$

More polar reactants will have a greater charge on them as compared with less polar reactants.
Since inductive effects concentrate  electron density at one area the negative charge concentration will be greater.

HOMO/LUMO
Unoccupied orbitals are high in energy and occupied orbitals are low in energy. For the best overlap the energy gap between the HOMO and LUMO must be small.

Rate Order for more polar reactions
In more polar reactions the electrostatic forces of attractions play the primary role as the polar reactants is highly charged. As established earlier ,inductive effects lead to a electron density thus a greater pull. Homo/Lumo interactions are secondary as the overlapping happens after a collision occurs.

The electrostatic forces are pulling the benzene and reactant. That would by itself increases the number of collisions.
A greater electron density leads to the better stabilization of the +tive charge on the reactant.

inductive effects support more polar reactions

Rate order for less polar reactions
The electrostatic forces of attraction are lower in the case as as the reactant is less charged.
So the control of the reaction is primarily through Homo/Lumo interactions.

Unoccupied orbitals of the reactant are high in energy.( They are Unoccupied).
To bridge the HOMO/LUMO gap the Lumo of benzene carbon must be high in energy. Non bonded electrons or the electrons which are delocalized through hyperconjugation are high in energy as they are held by only one atom.
So the HOMO/Lumo gap reduces in the case of hyperconjugation thus hyperconjugation supports less polar reactions.

I can't say whether it is true for false. But the reasoning and explanation is good.

Nitration (strong electrophile, "hard" Lewis acid) has a low activation energy, so effects on the transition state will have a smaller relative contribution to the rate controlling step.
Chlorination (weaker electrophile, "softer" Lewis acid, more polarizable) has a higher activation energy and slower reaction rate. The transition state is more important in rate determination.
An aromatic compound has a "hard" face and a "soft" face: the sigma bonds (inductive effect) are more stable, less polarizable. The pi system is less stable, lower energy (diagramatically higher energy level!), more polarizable, and more directly connected to the hyperconjugation effect.
Nitration (very polar, "hard" cation) is influenced by inductive effects that make the aromatic ring more negative in the order D>C>B>A. The rate of nitration increases as you go from A to D. Compared to nitration, the transition state in chlorination is more important and thus more stabilized by hyperconjugation. In chlorination, hyperconjugation decreases in the series A>B>C>D, which gives a higher activation energy and slower rate of reaction as you go from A to D.

Why exactly that works? T.S can be stabilized by both inductive and hyper conjugation. Proper explanation will be better.

Inductive and hyperconjugation are part of ground and transition states, but inductive effects are thru the sigma system (lower energy level, more stable) and hyperconjugation is thru p character of the carbon sp3 bonds and pi system of the aromatic compound; this is higher energy level, less stable, closer to the transition state energy. The slower reaction rate of chlorination implies higher transition state energy and therefore relatively more contribution is likely from hyperconjugation effects.

Chlorine is a weak electrophile, while the nitronium ion is a strong electrophile.
Methyl groups are +I groups and they are electron releasing groups.
The NO2 group will attack a negatively charged carbon attack readily.
The +I effect is permanent, and its magnitude decreases with increase in distance, and therefore, the nitronium ion will attack the carbon nearest from the methyl groups.
The Chlorine will not get attracted to to a negatively charged carbon atom as strongly as NO2 will. In hyperconjugation. the charge is delocalized i.e., it is distributed throughout the structure, in various resonance forms. Hence, the effect of hyperconjugation is weak. The Chlorine atom. therefore, will get attracted to to a weakly charged carbon atom, determined by hyperconjugation.
The hyperconjugation won't attract NO2, as the charge is delocalized, and not one specific carbon atom, or there is no carbon atom that is more charged than the others.

I have edited the answer

I've framed the answer using my own limited information, and with the help of my coaching notes. You can refer to Morrison and Boyd, or L.G Wade for deeper understanding and information.

https://crab.rutgers.edu/~alroche/Ch17.pdf

https://crab.rutgers.edu/~alroche/Ch17.pdf  There could be steric hindrance as well. Methyl groups are ortho para directing due to hyperconjugation. In hyperconjugation, the ortho and para carbons have more charge stability.


text:
Why f is multiplied in solubility of salt constant

I understand till we get “f” And also why f is directly proportional to PH. But I don’t understand why f is multiplied in solubility product with solubility of salt of weak acid.
(Sorry for poor english) (couldn’t find link of pdf of this book online)

What is this book?

11th class ncert chemistry “acids bases and salts”

The notations are very poorly picked. Upright "S" stands for both sulfur element and solubility, brackets "[]" are used both for concentrations and to enclose variables, the font makes it impossible to distinguish between { and (, and so on (and this is only from what you photocopied). I've never seen this book before, but I'd rather get something better than this.

Notice that $f$ is equal to the maximum concentration of $\ce{M}$, which is unchanged as part of the acid-base equilibrium. On the other hand, the pH changes the concentration of the counterion in proportion to $f$. Therefore, the second term in the solubility product is scaled accordingly.


text:
Is SbF5 symmetrical?
SbF5 forms a trigonal bipyramidal shape. There are polar bonds present and I assumed that it is non polar because of it's molecular geometry. However, The molecule itself is not entirely symmetrical (unless I'm thinking of symmetry incorrectly). If I'm not incorrect, can an asymmetrical molecule be non polar?

A molecule with two non-coincident axes of rotation cannot be polar.

Why isn't it symmetrical? There are three the equatorial bonds and angle between them is 120° and the axial bonds are there. You have yourself mentioned that it has trigonal bipyramidal shape. Then where is your problem? If you think it is non symmetrical say why. And upload aan image for the same.

@DJKoustav I think OP is confused about level of symmetry. For example, cube versus sphere. But I think the point they are missing is that a molecule does not need very high symmetry to be non-polar. Just enough.

@Zhe exactly.....We just need to vectorially add the dipole moments for each bond to get the net dipole moment. And so a symmetry that yields a net dipole of zero is sufficient enough for a molecule to be considered as symmetrical.

I think I was struggling with picturing it in 3D. When I looked up the shape itself, it just didn't seem entirely symmetrical, but your comments have put it in a different perspective and it clicked. Thanks!

Yes you are thinking of symmetry incorrectly. Just how could it be more symmetrical, pray tell us?

As mentioned in the comments, $\ce{SbF5}$ in the gas phase has a rather high symmetry, the point group looks like $D_\mathrm{3h}$ to me, containing a principal three-fold rotational symmetry axis (along the axial bonds) and three two-fold axis (along each of the equatorial bonds), among other operations. As mentioned in this text, a dipole moment can only coincide with a rotational axis. Since there are four such axis in the molecule, it cannot have a dipole moment.
Note however, that each of the bonds is certainly polar. They just cancel out, as could be proven by vectorial addition of the bond dipole moment.

Also, pentafluoroantimonate $\ce{SbF5^2-}$ would also have another geometry (distorted square pyramid) and quite a significant dipole moment.

As mentioned in Inorganic Chemistry: Principles of Structure and Reactivity
By James E. Huheey, Ellen A. Keiter, Richard L. Keiter, Okhil K. Medhi

The 3D image of SbF5  is as follows:
a side view

a top view

The bond dipole moments in vertical axis cancel out .Similarly the three Sb-F bonds positioned at 120 degrees to each other (top view image ) cancel out each other. Therefore  SbF5 in non polar molecule.

Where the atomic coordinates were taken from?


text:
Why objects usually don't exibit fluorescence/phosphorescence?
We learn in school that the color we perceive is determined by the light that is reflected back at us. Not all wavelengths of light reflect back at our eyes, some wavelengths are absorbed.
I'm confused about this absorption. From my understanding, molecules can absorb light typically rotationally, vibrationally, and electronically. rotationally and vibrationally are associated with the microwave and infrared region. Electronic excitations can be in the visible, but I always had the impression that the excited electrons would collapse down to the ground state and emit light. Thus, I don't think I know any mechanism which visible light can be absorbed without emitting light, albeit, longer wavelength light typically.

Related: https://chemistry.stackexchange.com/q/8099/16683 [As far as I know, they are usually due to electronic transitions. The question of what happens after excitation has been covered several times here, I believe, but the links are tough to find.]

I changed the title to more specific one, feel free to improve further.

You are correct in most of what you write. Absorption in the visible (and uv) parts of the spectrum are of far larger energy that IR and microwave and the lowest electronic transition involves promoting an electron from the highest populated orbital into the lowest unoccupied one. Once the excited state is formed it can emit via fluorescence and phosphorescence, which are both at a slightly lower energy that the light absorbed and this is because higher energy vibrational levels can be populated in the ground state. (Look for picture of a Jablonski diagram).
However, fluorescence is emitted into all directions ($4\pi$ radians due to molecular motion between absorption and emission) and only a small part of this could enter the eyes but also and most importantly there are non-radiative processes that competitively remove the energy from the excited state ultimately populating high vibrational levels in the ground state from which energy is rapidly lost into the surroundings. These non-radiative events mean that for most types of molecules the fluorescence yield is small, which means that only a few % of the excited state's energy ends up as photons. The non radiative processes are called intersystem crossing between singlet and triplet excited state and internal conversion between states of the same spin state. The processes causing these (besides and chemical reaction) are anharmonicity in molecule vibrations and particularly heavy atoms or unpaired electrons as part of the molecule itself or in nearby molecules such as in a solvent. Some molecules may also have symmetry forbidden optical transitions in emission but not in absorption: carotenes are of this type.


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Is it plausible to draw an enthalpy diagram for a nuclear reaction?
For a project I am required to draw a enthalpy diagram for the conversion of thorium-232 to uranium-233 and I am unsure on how to proceed. I understand a certain amount of energy is lost through beta decay, but would beta decay be considered a reaction?

Generally, a chemical reaction only applies to changes in the arrangement of electrons (ie bonds forming/breaking). Beta decay could be described as a nuclear reaction. You can still diagram the transition and how it affects the energy of the particles involved.