text:
General Method For Calculation of Enthalpy Change for a chemical reaction
1. Enthalpy of reaction defined as the enthalpy change observed in a constituent of a thermodynamic system when one mole of substance reacts completely.
2. Enthalpy of formation defined as the enthalpy change observed in a constituent of a thermodynamic system when one mole of a compound is formed from its elementary antecedents.
3. Enthalpy of combustion defined as the enthalpy change observed in a constituent of a thermodynamic system when one mole of a substance burns completely with oxygen.
4. Enthalpy of hydrogenation defined as the enthalpy change observed in a constituent of a thermodynamic system when one mole of an unsaturated compound reacts completely with an excess of hydrogen to form a saturated compound.
5. Enthalpy of atomization defined as the enthalpy change required to atomize one mole of compound completely.
6. Enthalpy of neutralization defined as the enthalpy change observed in a constituent of a thermodynamic system when one mole of water is formed when an acid and a base react.
7. Standard Enthalpy of a solution defined as the enthalpy change observed in a constituent of a thermodynamic system when one mole of a solute is dissolved completely in an excess of solvent so that the solution is at infinite dilution.
8. Enthalpy of fusion defined as the enthalpy change required to completely change the state of one mole of substance between solid and liquid states.
9. Enthalpy of vaporization defined as the enthalpy change required to completely change the state of one mole of substance between liquid and gaseous states.
10. Enthalpy of sublimation defined as the enthalpy change required to completely change the state of one mole of substance between solid and gaseous states.
11. Lattice enthalpy defined as the energy required to separate one mole of an ionic compound into separated gaseous ions to an infinite distance apart (meaning no force of attraction).
12. Enthalpy of mixing defined as the enthalpy change upon mixing of two (non-reacting) chemical substances.
These definitons were taken from the wikipedia article for enthalpy.
Each enthalpy seems to come with its own formula.
Is there a general approach for solving enthalpy change-related questions for reactions instead of having to memorize a formula for each and every case?
If there is an obvious point that I'm missing kindly point it out.

Look up: born-haber cycle. you normally do not have to calculate all of these formulas.

You can use the algebraic form for the definition of enthalpy and derive the formula for each case yourself. Besides, about two thirds of the entropies can be found using tabulated data and a set of linear equations.

Now I understood, thanks for pointing it out!

@TanyaGupta Keep in mind you can answer your own question.


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Equilibrium constant of a reaction whose order of forward and reverse reaction is not same
Suppose, a reaction is like,
$$\ce{aA + bB <=> cC + dD}$$
and that A, B, C, D all are gas.
Now it is known that:
$$K_c = \dfrac{[A]^a[B]^b}{[C]^c[D]^d}$$
Now if the forward reaction is second order and the reverse reaction is in first-order somehow then, how will this change.
I read some articles online but it was not actually clear that if all the reactants or products are in the same state like gas or aqueous then how $K_c$ or $K_p$ (for gas) will change depending on the rate of reaction or if they don't depend on that totally.
It will be helpful if one example is provided.

I'm not sure I understand but would not $K_c=k_f/k_r=[C]/([A][B])$ be an example, where the concentrations are the equilibrium values and $k_f, k_r$ are rate forward and reverse constants. (Normally for gas phase species we would use $K-p$ and use partial pressures instead of concentration.)

I'll point out that the equilbrium expression is inverted. For $K_c$ it should be reactants over products.

I'll try and answer your question, but a "full" answer would take a book.
Given the reaction:
$$\ce{aA + bB <=> cC + dD}$$
Then assuming an elementary reaction in the gaseous state the concentration equilibrium constant always has products over reactants and will be:
$$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$
The above equation relies on two very specific assumptions:

The reaction is an elementary reaction which often isn't true and the actual coefficients must be determined experimentally since the reaction occurs in steps.
That concentrations can be used instead of activities.

Formally using activities explains part of your question immediately. For instance if A, B and C are gases and D is a liquid or solid, then the activity of D would be unity by definition.
Now breaking down the equilibrium expression into two rate equations we have:

$r_\mathrm{f}$ - Forward reaction rate
$r_\mathrm{r}$ - Reverse reaction rate
$\ce{a^*, b^*, c^*}$ and $\ce{d^*}$ = experimentally determined coefficients which may or may not be equal to the stoichiometric coefficients
$k_\mathrm{f}$ and $k_\mathrm{r}$ are constants for the forward and reverse reactions respectively.

Now using concentrations instead of activities:
\begin{align}
r_\mathrm{f} &= k_\mathrm{f} \ce{[A]^{a^*}[B]^{b^*}}\\
r_\mathrm{r} &= k_\mathrm{r} \ce{[C]^{c^*}[D]^{d^*}}
\end{align}
and at equilibrium by definition:
\begin{align}
r_\mathrm{f} &= r_\mathrm{r}\\
k_\mathrm{f} \ce{[A]^{a^*}[B]^{b^*}} &= k_\mathrm{r} \ce{[C]^{c^*}[D]^{d^*}}
\end{align}
so:
$$K_c = \frac{k_\mathrm{f}}{k_\mathrm{r}}= \frac{\ce{[C]^{c^*}[D]^{d^*}}}{\ce{[A]^{a^*}[B]^{b^*}}}$$
To determine the coefficients the experimentalist can manipulate the experiment to simply the kinetic expression. For example let $\ce{[A] \gg [B]}$ then $\ce{[A]^{a^*}}$ is essentially a constant.
Does this answer your question?

Should you not have $K_c=k_f/k_r=..$ in the last eqn instead of $K_c=r_r/r_f..$ ?

@porphyrin - You are correct. Thank you for pointing out the error. I've fixed my mistake.

My question was what if the reverse reaction is the first-order reaction (suppose it depends on C). Then $K_c$ = $\frac{[C]^c}{[A]^a[B]^b}$ because as you wrote $r_f$=$r_r$ or not?

@user104796 - yes at equilibrium $\mathrm{r_f} = \mathrm{r_r}$.  (1) for the reverse reaction to be first order it would have to depend on $\ce{[C]}$, not $\ce{[C]^c}$. (2) to eliminate  $\ce{[D]^d}$ then make a reaction mixture of A, B and D where  $\ce{[D] \gg [A] + [B] }$ so that  $\ce{[D]^d}$ is constant and the reverse reaction would be pseudo-first order.

I have got it. It helps me a lot. Thank you.

source: Stack Exchange

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Estimating BDE Values
I'm having trouble understanding how to "estimate" the value of the BDE for the compound shown below (I believe it is a secondary benzylic C-Br value) shown below (the 58 kcal/mol). We are supposed to be able to estimate the value from the table values shown below. My teacher said to use the C-H values and C-Br values from the table below to estimate it, but I don't understand how they are able to indicate the value. Any helpful drawings would be nice since I'm new to organic chemistry and have trouble just understanding the names! Thank you.

Welcome to ChemSE.  We would like to see how you arrived at your values. We like to help but loath doing homework.

@user55119 Thanks! Sorry if I wasn't clear, but the red is what I am trying to understand - the values are not mine; it is what the teacher said. Specifically, I am confused about how they got the 58 kcal/mol estimation.

Adding a reference for the source of that table might help. Also, ask your teacher if possible, it does look off.

@BuckThorn All the information I have is that it comes from Pearson Education : / . My teacher wants us to use this table to calculate our BDE values. She said the estimate can have a range of 55-60

What you have shown in red is the second propagation step in the benzylic free radical bromination of ethylbenzene. The bond dissociation energy  (BDE) of bromine is +46 kcal/mol while the overall heat of reaction of the step is -12 kcal/mol. Thus, the formation of the C-Br bond must be -58 kcal/mol. The BDE for the C-Br bond in (1-bromoethyl)benzene is +58 kcal/mol.
However, the BDE for bromine is not listed in your tables. So how can one estimate the BDE of the C-Br bond in (1-bromoethyl)benzene using the data in the tables? A simple ratio and proportion will suffice. In equation A compare the primary C-H bond of ethane to that of toluene. This ratio is compared to the ratio of the BDE of the secondary C-H bond of propane and its counterpart in ethylbenzene, which is the unknown "x". An estimated value of
82 kcal/mol is obtained. In equation B, this ratio compares secondary C-H bonds with secondary C-Br bonds. The estimate for the BDE of the C-Br bond in (1-bromoethyl)benzene is the value of "y" which is equal to 59 kcal/mol. Pretty close, wouldn't you say?

thank you so much for the response! this technique definitely works. Just a clarifying question, how did you know how to make the fractions? For instance, why did you choose to put CH3CH2H on top of PhCh2H?

Doesn't matter as long as the two ratios are in the same order. 2:4::1:x. So, x=2. But for 4:2::x:1, the answer is still 2.

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Mechanism of ring opening of amine combined with formylation
What is the mechanism for this reaction? The tertiary amine is converted into an amide and the ring gains a double bond. I can't figure out how the C-O double bond would be formed.

Oh, that's the easy part, but can you figure out what happens earlier? That's more interesting part.

Find the nucleophile. Find the electrophile. Combine.

Hint. The amide is a formyl amide. Where does the H attached to the carbonyl come from?

Can chloroform react with sodium hydroxide?

This reaction is very similar to Reimer-Tiemann reaction which also consists of the reagent $CHCl_{3}$ + $NaOH$. The mechanism of that reaction is believed to have dichlorocarbene intermediate too. A good point to remember is that the dichlorocarbene intermediate is often used to form cyclopropane or carbonyl groups.

This is the mechanism for the reimer tiemann reaction. We can see the similarity in the this reaction and the one you have mentioned previously.

Mm, that's better. Although the arrows still suggest an intramolecular proton transfer which I don't like...

@orthocresol I am not sure if I should discuss this here, but I will be honest, I am a 12th grade student and hence couldn't understand the terms you used earlier. I don't understand what correction there should be here. Can you please elaborate in simple terms, or can we discuss it in chat?

To deprotonate a X–H bond, the base (let's say B) needs to come from the opposite side of the hydrogen, i.e. X–H···B. That's essentially like an SN2 reaction, and the reason is the same: you need to get to the σ* orbital. This makes an intramolecular deprotonation essentially impossible, because that lone pair that is supposedly doing the deprotonation can't reach the σ* orbital. More likely that the negative charge picks up a proton from somewhere else, and the proton on the ring is lost some other way.

Apart from that nitpick, your mechanism is fine, though; your edit fixed the other things I pointed out. But I'd still ask you to type out your text, instead of attaching a picture of written text. It's easier for everybody else to read, and can be searched for.

I will grant that that the σ* orbital seems very difficult  align. I just thought that the aromaticity would drive it to be essentially SN1 type acid base reaction. I guess the arrows seem to direct people into thinking this is SN2( in which case should I just get rid of them and add equilibrium arrows?) I will update this answer to text format once my major exams get over (because I don't mainly know how to do TeX and draw custom images.) That being said thanks for your inputs.

Yes it’s the arrows that are the issue. Removing them is fine.


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What type of spectroscopy can display several vibrational energy levels around the electronic transition?

My question is illustrated as the figure. As far as I know, due to the Franck-Condon principle, the absorption spectrum is related to the ground state to the excited state with the same geometry. Here I marked as $|X \nu_0\rangle \rightarrow |A \nu_m \rangle$, where $\nu_0$ is the ground vibrational state, $\nu_m$ is the vibrational state in the first electronic excited state with the same geometry as the ground state.
The emission spectrum is related to the excited state to the ground state with the same geometry. Here it is labelled as $|A \nu_0 \rangle \rightarrow |X \nu_n \rangle$, where $\nu_n$ is the vibrational state with the ground electronic state geometry.
The difference between absorption and emission spectrum is known as Stokes shift.
My question is, is there any type of spectroscopy has the resolution for many vibrational states around the electronic states? E.g.,  $|X \nu_0\rangle \leftarrow \rightarrow |A \nu_0\rangle$ ,  $|X \nu_0\rangle \leftarrow  \rightarrow |A \nu_1\rangle$ , $|X \nu_1 \rangle \leftarrow  \rightarrow |A \nu_0\rangle$  etc. If yes, what is the name of it?

I think it depends what state of matter are you investigating. In gas phase, the vibrational structure of electronic transitions is visible in UV-VIS spectroscopy
for example gas phase benzene. https://www.researchgate.net/figure/Absorption-spectra-of-benzene-with-different-spectral-resolution_fig3_231116208

Thank you so much for your link. My background is simulation and not familiar with experimental techniques. Is my understanding correct, that at room temperature, the molecules are at a distribution of vibration states, hence, in a lab, we can see all these vibrational structures, namely, peaks?

At room temperature one can assume a molecule is at the electronic ground state and its vibrational energies are distributed due to thermal energy. If you are in gas phase, you can see rotational structure too. I have never done any theoretical simulations of molecular spectrum. But I like experimental spectroscopy. People mostly study gas phase spectrum.

As well as UV visible spectroscopy (using a spectrophotometer for example) , fluorescence also exhibits vibrational levels. For the best resolution vapour phase spectra are best; search for 'benzene vapour absorption' spectrum and look for 'images'.

You can also include Raman as for all begins with a quasi-resonant or not excitation.


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How can change in entropy be the same for all processes if the entropy production $\sigma$ is present for irreversible processes?
From the definition of entropy change,
$$S_2-S_1=\left ( \int_{1}^{2} \frac{\delta Q}{T}\right )_{int.rev}$$
From the closed system entropy balance, we have
$$S_2-S_1=\left ( \int_{1}^{2} \frac{\delta Q}{T}\right )_{b}+\sigma $$
where $\sigma$ is the entropy produced within the system, vanishing to zero in the absence of irreversibilities. I don't quite understand how the entropy change between two states is the same for all processes. Is it the case that the entropy transfer in the case of internal irreversibilities present is lower than that of the entropy transfer in an internally reversible process between these same two states, and the entropy production makes up for this difference?

Yes, that is exactly correct.  In both cases, the heat transfer takes place at the boundary temperature.  What you have stated is the essence of the Clausius inequality.

Seems the post and commentaries could be restructured and gain a lot from a more elaborate answer. Otherwise this will just "hang" in perpetuity as an unanswered  "yes/no" question.

@BuckThorn How should I reconstruct it? I would love someone to expand on it!

Well, I meant that it would be nice to have an open-shut post with a clean answer.@ChetMiller evidently knows his stuff but aimed for conciseness and answered your "yes/no" question with a "yes, you're right" comment.. I think if he could expand on the answer it would be nice, but ultimately the fate of the post is up to him, you and the wisdom of the crowds.

The way that you describe it is exactly correct.  This is the essence of the Clausius Inequality.

When a system changes from state 1 to state 2, the change in entropy of the system is the same for all processes, because entropy is a state function.
The difference between a reversible and an irreversible process is that, in the latter case, there is net entropy production for the universe, i.e., for the system+surroundings.
Since the entropy change for the system is independent of the process, the difference in entropy production between rev. and irrev. processes is seen in the entropy change of the surroundings.
More broadly, since path differences (which in turn means differences in heat flow and/or work flow) have no effect on the final state of the system, they instead always manifest themselves as differences in the final state of the surroundings.

Makes perfect sense, thank you. I also have another question. Does entropy vary within different areas of a system? Is there like an entropy gradient of some sort?

@CalebWilliamsUIC You'll want to post that as a separate question.

Alright! I will make another question at some point today.


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Which formula for kinetic energy of 1 mol of a gas is universally applicable?
I have read 2 formulas for kinetic energy of 1 mol of gas - 3/2 RT and 1/2fRT . But if I equate them then f comes out to be 3 . But f is different of mono, di, polyatomic gases . Where am I wrong in my deduction that 3/2RT = 1/2 fRT ? Do these 2 formulas produce same result for a particular gas?

"Acceptable" is not a thing that a formula may or may not be. You might want to go with "applicable", but then again, it is not comparable in terms like "more" or "less". $3/2$ is applicable when $f=3$, as you correctly deduced; otherwise it isn't.

Ek=3/2RT is universally applicable as the translational part of kinetic energy of an ideal gas. I.e. molecular rotation and vibration is not included,

Simply put, it is used when you don't yet know about any gases other than monoatomic.

Thanks guys .. but I want to know if  can I equate these 2 equations

Yes when they are equal, and no when they aren't.

Thank you .. In what cases they are not equal?

So the formula 1/2 fRT is universally applicable but 3/2RT is only applicable for monoatomic gases ... Am I right ?

@DeboditRay Ask yourself rather, how you get $f$ for a given gas.


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Electrolysis of Nacl with aluminium Electrodes
I conducted an experiment to find out the difference in products/ the final solution whilst using different electrodes. On using copper I obtained a teal blue soluble salt I believe to be $\ce{CuCl2}$ yet upon using aluminum I obtained a silvery sludge of a solution. The aluminum anode was definitely oxidized by chlorine ions in the solution. Also there was notable effervescence at the cathode of what can only be hydrogen gas.
Could someone please tell me what aluminum compound remains in the Liquid and shed some light on the anode and cathode reactions. I'm an amateur so I had some difficulty understanding the outcome.

You are getting Al(OH)3, and "solution" is a wrong word to describe it.

Yes sorry. I forgot to mention it is insoluble

Aluminum pieces are always covered by a thin, waterproof and nearly monomolecular layer of alumina, or aluminum oxide $\ce{Al2O3}$. This layer prevents the aluminum metal to get in touch with water. Because, pure and non-protected aluminum would immediately react with water according to : $$\ce{2 Al  +  6H2O  -> Al(OH)3 (s) +  3 H2 (g)}$$ This reaction can only occur when the alumina layer has been removed. And electrolysis is a way of destroying this layer. There are also other ways, like adding $\ce{CuCl2}$ or $\ce{HgCl2}$ in the solution. But rubbing or scratching aluminum is not such a way, because removing alumina mechanically will allow oxygen from the air to immediately reoxidize the metal and regenerate the alumina layer.
So it is difficult to understand what is happening in the electrolysis of an aqueous solution with aluminum electrodes, because there are two sorts of chemical reactions happening simultaneously : the chemical reaction of aluminum on the solution without electricity, and the electrolysis of the solution.


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How do I prepare citric acid solution (from anhydrous powder) of accurate concentration, given that the solution volume expands when water is added?
I attempted to prepare saturated citric acid solution from anhydrous powder. According to online data, the solubility limit was 59.2 w/w at 20 °C, i.e. 59.2 g citric acid in 100 mL water. To ensure full saturation (as in the case of preparing brine solution), I ended up weighing 67 g of the powder and tried adding 100 mL of water to it.
However, when I had added about 67 mL of water, the 100 mL Schott bottle was already full, and all of the powder dissolved (estimated 5.22 M solution). Since there is an apparent volume expansion, how then should this solution (or a citric acid solution of any other desired concentration) be prepared?

Add some water till the solid stuff dissolves, then gradually add more and more until you hit 100ml.

Concentrations are always given in [amount] per volume of solution, NOT per volume of solvent.

The "expanding solution" problem is irrelevant if you use standard techniques
First, you need to realise that a standard solution is a known amount of solid in a known volume of solution not a known amount of solid and a known amount of the solvent. You don't need to know how much solvent you use. So you need to target getting a result that occupies, say, 100mL not a solution made from 100mL of water.
There is an easy way to achieve this. Use a volumetric flask with an accurate volume marking. Add the solid to the flask. Fill the flask slowly (while shaking enough to ensure the solid dissolves before the flask is full). Keep adding small increments of the solvent until the total volume reaches the graduation. You now have an exact volume of a solution containing an exact amount of the solid. At no point do you need to worry about any volume expansion or contraction.

That is directly applicable with the goal to reach the given mass or molar concentration m/V or n/V. But not so directly if data are provided as solubility, mass fraction or molality. You need to know the target solution density. If not tabelized, you need to prepare the solution first and measure it.

Was the OP trying to prepare a "standard solution"? The expanded question asked for a solution of "any other desired concentration", which seems more like a mass to mass ratio, not mass to volume.

Darn it! I just realized why there are two paths recommended for the calculations! Some of us chemists are thinking in moles per liter, while some of us (well, me) are (is) thinking in terms of %. In commercial practice, where we deal with hundreds or thousands of gallons, % is the way to go, even if you do the preparation on a lab scale. Academics might prefer volumetric flasks. I always had to standardize the flasks by weight - they never seemed accurate enough, or convenient enough, because I always had to add it to more stuff or more stuff to it.

@JamesGaidis In bulk commercial preparation, you usually have a well-tested recipe specifying the amounts to mix in convenient units and whoever developed the recipe will have checked that the recipe works. In a lab you usually want accuracy (in g/L or something like that). Hence volumetric flasks and weighed solids.

One thing is usual solubility in g/100 mL, other thing is solubility as percentual mass fraction.
(Improperly used) 59.2 w/w obviously means 59.2 weight percent, as this solubility data – noted properly – is listed in wikipedia: Citric acid.
You need 59.2 g of the acid and 40.8 g of water ( about 40.8 mL or approx. 41 mL ), in total 100 g. The volume will be less than 100 mL, due the higher than water density.
Recalculated to solubility it is approximately $\pu{145 g/100 mL}$
In the case an acid monohydrate is used instead, the stoichiometric mass of crystal water has to be obviously subtracted from the acid mass and added to the water mass.
That can be recalculated for any target total mass and particular volume can be taken from this stock solution.

comments to comments:
@matt_black Sure, there are multiple ways to prepare saturated solutions.

One is the preparation of arbitrary mass of w/w% solution according to provided data.
Other is recalculation of w/w% to g/100 mL and dissolving in this nominal volume.
Other is reaching equilibrium of stirred solution with an excess of acid.
Other is to obtain density of saturated solution, calculating the needed acid mass as percentage of the total volume mass. And then not fully filling up into given volume. After near full dissolution, mixing/filling up in few steps until the nominal volume.
Other is to calculate the acid mass equivalent to the acid volume with the filling volume as a complement. But this is not recommended, being tricky, not applicable to well soluble and quickly dissolving compounds. As it implies filling is done before significant dissolution occurs, because there is volume change during dissolution.

Actually, nonconforming expressions such as "w/w" should not be used at all.

@Loong I do not object, it was used by the OP. Unless you mean in MathJax. I was in doubt, if I should use it. It could be plain text as well.

I am aware that "w/w" was used in the question; no need to change your answer. My comment was more a reflex reaction to your explanation "59.2 w/w means 59.2 weight percent" because actually such things don't mean anything since they must not be used.

You don't need to know the amount of water at all if you use standard techniques and volumetric flasks.

This is an example of a Schott bottle:

Using the volume markings would not be very accurate, so you would weigh the water.
Citric acid solution is indeed less dense than you would calculate from the volumes and densities of water plus citric acid. 30% solution has a density of 1.13, but the arithmetically calculated density is 1.1995 from citric acid density = 1.665 and water = 1 (CRC Handbook, 62 ed). There were no data for more concentrated solutions.
As Poutnik points out, 59.2% w/w means 59.2% of the final solution is citric acid - yes, it's that soluble! But a better way to make the solution and fill the bottle to the 100 mL mark (if that is important) is to prepare the solution in another flask, by weight. Since you want 100 mL, which will weigh 113 grams, I suggest you go for 125 grams, to have a little extra. 125 x 0.592 = 74 grams citric acid; add 51 grams of warm water (to insure more rapid and complete solution) and swirl till clear. (Swirl the flask, not yourself.)
When the solution is clear, fill the Schott bottle. What to do with the excess solution? Do not fret. Citric acid solution is available commercially as a toilet bowl cleaner. 12 grams of saturated citric acid will not do much cleaning for a really dirty toilet bowl, but at least it won't damage the ecology any more than we normally do.


text:
Effects of dispersed nitrous oxide on the body
I'm not sure if this question would be more appropriate in Biology SE, but I wanted to know if the euphoric effects of $\ce{N2O}$ are limited to when it is concentrated in the gaseous phase, or whether it can also affect individuals in cases where it may be mixed with other compounds as a sol (in essence, as the propellant in the whipped cream of ice-cream).
Maybe that might explain how ice cream is associated with happiness (in most cases and in a majority of people), the same way $\ce{N2O}$ is associated with euphoria. (The focus here being $\ce{N2O}$, not ice-cream-related dopamine reception in the brain.)
Researched a but but unable to find anything relevant.

Why are you looking for an additional theory as to why people like ice cream? Dopamine release due to sugar or fats triggering the pleasure system is more than enough to explain why people enjoy whipped cream and ice cream. Occam's Razor answers your question with a: "No, that will not explain anything, it is well enough explained already".

@Stian Yttervik I'm not looking for a theory, just for a possible answer since I find it interesting and I learn better by connecting new things to things I already know. I'm more interested in learning about N2O, not about why people like ice-cream. I've edited my post to make that more clear, thanks.

What matters is N2O concentration in blood as consequence of N2O exposure. Amount of N2O in the consumed cream is small, and eventual exposure to it distributed across long time. So my guess is the effect is negligible.


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NaCl electrolysis with copper electrodes
Considering the aqueous $\ce{NaCl}$ is already concentrated. If we were to use copper electrodes instead of inert graphite electrodes:
Would copper react with the evolved chlorine gas to give $\ce{CuCl2}$ in solution, whilst still giving hydrogen gas at the cathode? Since in my experimentation, I obtained a teal blue green colored salt that was soluble in water. I do not have appropriate lab equipment to test for the identity of the salt, so I am hoping someone here could shed some light on the reactions taking place and the identity of the obtained substance. Thanks.

Electrolysis of sodium chloride solution produces results at each electrode. At the cathode, water is reduced to $\ce{OH-}$ and makes the solution more alkaline. Copper hydroxide is a blue, bulky precipitate, feebly acidic, and soluble in concentrated alkali hydroxide solutions. So if you got copper ions into solution at the anode, you could form some cupric hydroxide which is reported to form blue polymeric species in solution [1].
Crystals from concentration of this solution will likely be $\ce{NaCl}$ contaminated with some $\ce{CuCl2}$. Examination with a microscope could show whether $\ce{CuCl2}$ has formed, as the blue-green rhombic dihydrate (because, after all, it is in water), or, if you actually did enough electrolysis to do more than color the water, whether you precipitated the yellow-green hexagonal basic salt $\ce{CuCl2.Cu(OH)2}$ or the hydroxide or some mixture [2].
Although cuprous chloride is highly insoluble in water, it does form complexes in concentrated chloride solutions: the main species in $1\rm{M}$ $\ce{KCl}$ is
$\ce{CuCl2^-}$ [1]. So you really have a complex situation with anode/cathode, $\ce{OH-}$/$\ce{Cl2}$, $\ce{Cu^0}$/$\ce{Cu+}$/$\ce{Cu^{+2}}$, $\ce{CuCl2}\cdot2\ce{H2O}$/$\ce{Cu(OH)2}$, and polymeric complexes with $\ce{OH-}$ and $\ce{Cl-}$, and various combinations of products depending on how long you do the electrolysis and, of course, whether you emit $\ce{O2}$ or $\ce{Cl2}$.
Of course, that's what makes chemistry so much fun. There are many variables, some of which you can control.
[1]: Advanced Inorganic Chemistry, Cotton and Wilkinson, 1966
[2]: CRC Handbook, 62$^{\rm nd}$ ed.


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Recovering from a bottle of ferrous chloride
My bottle of ferrous chloride powder/crystals was years in shelf and it turned dark brown. If I break up the clumps I can see a little bit of yellow layers. I suspect it's a combination of iron oxide and ferric chloride.
I want to convert it back to green ferrous chloride and I'm looking for cheap ways to do that. My first idea is to dissolve it and stir it with nails, but I find it hard to dissolve in water. I'm not sure if adding HCl helps but it's too expensive for converting the whole 500g.
Better options or ideas?

As you suspect, your oxidized stuff is a mixture of $\ce{FeCl3 + Fe2O3}$, more or less hydrated. Anyway, it will not get dissolved in water, or even in dilute $\ce{HCl}$ solutions. The only possibility is using rather concentrated $\ce{HCl}$ solution. Maybe not the most concentrated solution ($36$%). Try with a mixture $1+1$ and $1 g$ oxidized mixture. And if you succeed, add some nails, or some steel wool. If the solution gets discolored, the transformation into $\ce{FeCl2}$  is over. Take care to prevent contact with air.

Nails and steel wool contain other acid-soluble metals.  I'd rather buy it fresh.


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Basis of mole concept?
We all have read that 1 mole of any substance contains $6.022 \times10^{23}$ atoms included in the substance. And:
$$Mole=\frac{weight}{molar\,\,mass}$$
Where molar mass is gram atomic mass such that when the atomic mass of the element (expressed in unified mass) is expressed in grams, then that amount contains $6.022 \times10^{23}$ atoms of that element. For example for oxygen, ($unified\,\,mass=16u$) when we take $16$ grams of it, it has $6.022 \times10^{23}$ atoms of oxygen.
My doubt is that how come all these things come together? My teacher just told to memorise. But I can't understand how come we got the Avogadro's constant, and how it's related to gram atomic mass.

Have you done a quick research of this part of history of chemistry? E.g. starting with history of Mole, atomic mass unit(Dalton) and Avogadro number, following eventual references ?

Note that you shall not mix quantities and units as in your wrong equation "Mole = weight/molar mass". Furthermore, descriptive terms or names of quantities shall not be arranged in the form of an equation.

Also note that the concept of "gram atomic mass" is no longer used.

A mole is simply a number. There is no magic in equations involving it. It was originally defined as the number of atoms in 16g of oxygen (later revised to 12g of carbon and recently defined as an exact number). The original definitions were designed to make it easy to compare the mass and atomic counts with the techniques available.

If I understand your question, you would like to know how the Avogadro number is measured. There are no simple and immediate answer.
One of the possibility is using X-rays to measure with precision the distance $d$ between two atoms in a cubic crystal like $\ce{Si}$, where the atoms $\ce{Si}$ are regularly alined along the three axes Ox, Oy, Oz.
As the density of the silicium is known, the volume $V$ of a cube containing $1$ mole ($\pu{28.09 g}$ $\ce{Si}$) may be calculated. Its edge has a length $a = V^{1/3}$ and contains $x$ atoms. As $d$ is known, $x$ may be calculated $x = a/d$. This cube contains ${x^3}$ atoms ; ${x^3}$ is the Avogadro number.


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Arrange in the order of their reactivity of SN2 reaction

Here in case of (1) it is more favorable towards SN1 reaction as it will create benzyl carbocation which is very stable.
(2) Here it is more favorable towards SN1 as it is first of all allyl carbocation and in resonance with the benzene and the double bond.
(3) It is allyl carbocation, so it is less favorable to SN1 than (1) and more favorable to SN2.
(4) It is the most favorable towards SN2 reaction.
Hence the order of reactivity towards SN2 reaction should be
(4) > (3) > (1) > (2)
But the answer is
(2) > (1) > (3) > (4)
Why?

Are you sure the question is about $SN_2$ and not $SN_1$? Because your classification is correct with $SN_2$

Yeah i am sure i can send the question picture if you want

The order you've given is absolutely correct with respect to SN2, I believe, the question needs to be authenticated once again, as the order for SN1 would be the exact reverse of this, which is as given in the answer to your question.


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Total entropy change for a spontaneous reaction

For the oxidation of iron,
$$\ce{4 Fe(s) + 3 H2O(g) -> 2 Fe2O3(s) + 3 H2(g)}$$
the entropy change is $ \pu{–549.4 J K^{-1} mol^{-1}}$ at $\pu{298 K}.$ In spite of the negative entropy change of this reaction, why is the reaction spontaneous? $(\Delta_\mathrm{r}H^\circ$ for this reaction is $\pu{–1648 kJ mol^{–1}})$

In this question the solution given is that we consider the total entropy change for the reaction, i.e. the entropy change for a system plus the entropy change of surroundings, and that comes out to be positive.
But shouldn't the entropy change for surroundings (calculated by $q_\mathrm{rev}/T)$ be the negative of the entropy change for the system (heat absorbed by the surroundings is same as the heat emitted by system)? Thus, shouldn't the total entropy change be zero?

your logic is on the right track, but the heat emitted is the $\Delta H$ term, not the $\Delta S$.

This number is the entropy change of the system, not including the surroundings.  Please specify for us your understanding of the initial state and final state for this change.

Does this answer your question? Is ΔS of a system related to temperature and change in enthalpy?

The initial state of your system consists of 4 moles of $\ce{Fe(s)}$ and 3 moles of water vapor at $\pu{298 K}$ and $\ce{1 bar}$ in separate containers. The final state of your system consists of 2 moles of $\ce{Fe2O3(s)}$ and 3 moles of $\ce{H2(g)}$ at $\pu{298 K}$ and $\pu{1 bar}$ in separate containers.
The $\Delta H^\circ$ and $\Delta S^\circ$ correspond to a process for taking this system reversibly from the initial state to the final state. They don't include the changes in enthalpy and entropy of the surroundings in executing this process.
Do you have an idea of what such a process would be?

Yes I understand your point but what I am asking is that if the heat lost by system is the same magnitude as the heat gained by surroundings then shouldn't the entropy change be same for both of them, just different in the sign?

@ADITYAMITTAL11E Heat gain of surrounding is not only of the same magnitude, but is the same as the heat loss of the system. But entropy change is not limited to heat exchange between the system and surrounding. Imagine an isolated system with zero heat transfer, where is placed a brick with thermal gradient. The entropy of system would grow, even if the heat exchange is zero. Similarly, other cases of entropy change are other spontaneous processes like chemical reactions.

@Poutnik If heat exchange is not the sole contributor to change in entropy then why is change in entropy calculated by Qrev/T

@ADITYAMITTAL11E The change in entropy is calculated as dS=dQrev/T  IF AND ONLY IF the only change leading to a change of entropy is reversible heat exchange.  If there is ongoing any irreversible spontaneous process in the system, it is not valid, respectively, it may be valid as the heat transfer contribution to the system entropy change.

@Poutnik I think the OP is asking about what would happen if the change is carried out reversibly.  I think he is wondering why $\Delta S$. is not equal to $\Delta H/T$.  OP, is that a correct interpretation of what you are asking?

@ChetMiller I am not asking about this interpretation, I understood partially from Poutnik's comment about the entropy change for the system. But just one last thing I wanted to ask from his comment was that if the process in itself is spontaneous i.e. irreversible then how come we are calculating entropy change for the surroundings using dQrev/T because shouldn't that become invalid as well like in the case of the system itself.

I don't understand your question.  What are you saying the change in entropy of the surroundings is equal to?

I am saying that if the process in itself is not reversible then how are we calculating the entropy change for surroundings with $\frac{-\Delta_rH^0}{T}$ cause that would only be applicable for a reversible heat exchange as Poutnik mentioned in his comment.

@ADITYAMITTAL11E  One thing is reversibility of processes within the system and other thing is reversibility of exchange of Q a/o W between the system and surrounding.  If the latter is reversible, than dS_surr=-dQ/T, regardless of if processes within the system are reversible or not. Imagine reversible heating of 2 otherwise same systems. system 1 has 2 L of O2 + 2 L of N2 properly mixed, system 2 has them separated and just connected.

Oh ok now I got it, thanks a lot for your time :)

Under constant pressure and temperature, we have
$$-\frac{\Delta G}{T} = \Delta S_{tot} = \Delta S_{sys} + \Delta S_{surr}$$
hence, a negative $\Delta G$ tells us that the total entropy gets increased (= spontaneous process). A common way for textbooks to justify the definition of $G$ is by specifying $\Delta S_{surr} = -\frac{\Delta H_{sys}}{T}$, which is the entropy change by reversible heat transfer between the system and the surroundings. The part that is probably irritating you is that $\Delta S_{surr}$ is not neccesarily equal to $-\Delta S_{sys}$. We can show why this is the case by rationalizing the entropy changes.
Entropy $S$ correlates with the number of accessible microstates (= positions in the phase space, which consists of all position and momenta coordinates for all particles). Let's consider a simple dimerization:
$$\ce{2 A <=> A_2}$$
We change from two independent particles to a single particle. Hence, such a reaction decreases our accessible portion of the position subspace and entropy will decrease. This change is our $\Delta_r S$. Now besides changes in entropy, we can also have changes in the total potential energy. Let's say the bond formed by the dimerization releases a certain amount of energy $\Delta_r H$. What happens to this energy? Well, we convert potential energy to kinetic energy (= heating up the system). Such a process will increase entropy, since we gain access to higher momenta: $\Delta_{pot\to kin} S$. So the total change in entropy for the system would be $\Delta_r S + \Delta_{pot\to kin} S$. We have one problem though. We initially specified a constant temperature experiment. We can transfer the heat $\Delta_r H$ to the surroundings of course. The entropy change for the system will then just be $\Delta_r S$, while we have $\Delta_{pot\to kin} S=-\frac{\Delta_r H}{T}$ for the surroundings.
Disclaimer: Our considerations for entropy are very simplified here. The actually change in entropy for a dimerization is more complex in reality, since it also includes internal degrees of freedom like vibrations.

$\Delta_r \mathrm{H}$ and $\Delta_r \mathrm{S}$ are independent of each other
Processes can be exothermic or endothermic, and they can show a decrease or increase in entropy. You can't predict $\Delta_r \mathrm{H}$ from $\Delta_r \mathrm{S}$ or vice versa. You can combine the two to calculated the change in Gibbs energy, and this tells you about the direction the process will go in the absence of non-PV work.
$\Delta_r \mathrm{H}$ and $\Delta \mathrm{S}_\mathrm{surr}$ are linked under certain circumstances
In the absence of non-PV work, all energy exchange between reaction and surrounding will be through PV work and through heat exchange. With these restrictions and when there are no differences in the temperature before and after the reaction, $\Delta_r \mathrm{H}$ will be equal to the heat transfer to the surroundings. If the surroundings are a giant reservoir, its temperature will not change much (for the ideal case, not at all), and the heat exchange will almost be reversible (for the ideal case, reversible). In that case,
$$\Delta \mathrm{S}_\mathrm{surr} = - \frac{\Delta_r \mathrm{H}}{T}$$
$\Delta_r \mathrm{S}$ and $\Delta \mathrm{S}_\mathrm{surr}$ are independent of each other
Because $\Delta_r \mathrm{S}$ and $\Delta_r \mathrm{H}$ are independent of each other, $\Delta_r \mathrm{S}$ and $\Delta \mathrm{S}_\mathrm{surr}$ are, too. The one thing that connects them is that processes for which the sum of the entropy changes is smaller than zero don't happen because that would go against the second law of thermodynamics.

Thus, shouldn't the total entropy change be zero?

No, it shouldn't. If it is, you have reached equilibrium and there would not be any changes in any state functions at all. The reaction in question, however, is described as spontaneous. When a spontaneous reaction proceeds, the total entropy will increase.


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How to use dispersion correction with M06-2x in Gaussian 09 or 16?
I would like to perform an optimization calculation using the functional M06-2x using the BJ dispersion correction in Gaussian 0.9. However, I can't find the parameters for this functional as I found for example for B3LyP. Has anyone worked with this?

I thought M06-2X is already parameterized to reproduce the dispersion energy without adding empirical dispersion?

I also thought so but I didn't know if it was specifically for Becke Jonhson's correction

No you misunderstand, sorry if I didn't write it clearly. Becke-Johnson correction is a part of empirical dispersion correction scheme devised by Stefan Grimme and others (read https://manual.q-chem.com/5.2/Ch5.S7.SS2.html). It is added separately to the DFT calculation, i.e. it is evaluated after the SCF step. Whereas M06-2X returns the dispersion energy due to its extensive parameterization i.e. the parameters in the exchange-correlation functions of M06-2X are fitted so that they will give right answers, at least in principle.

It it possible to apply a zero-damped D3 dispersion correction to M06-2X results in Gaussian with the keyword emp=gd3. The parameters are: s6 = 1.0, sr,6 = 1.619, s8 = 0.00.
There are no empirical parameters available for a Becke-Johnson damped D3 correction with M06-2X, either in Gaussian or elsewhere (AFAIK). The Grimme group keeps a list here: https://www.chemie.uni-bonn.de/pctc/mulliken-center/software/dft-d3/functionalsbj
The M06-2X functional includes medium-range correlation contributions by a flexible functional form and extensive parameterization. So it's not "dispersion-corrected" (i.e. correct asymptotic behavior) in the same way as the a posteriori corrections such as D3.


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How to identify the aromatic compound with the sum formula C7H6Cl2?

An aromatic compound (A), $\ce{C7H6Cl2}$, gives $\ce{AgCl}$ on boiling with alcoholic $\ce{AgNO3}$ solution and yields $\ce{C7H7OCl}$ on treatment with $\ce{NaOH}$. (A)  on oxidation gives a mono-chlorobenzoic acid which affords one mono-nitro derivative. Find the Compound (A).

Here from $\ce{C7H6Cl2}$ we can say it is a benzene derivative. Now as only one chlorine atom reacts with $\ce{AgNO3}$, one chlorine atom is directly connected to benzene and one is connected to methyl group which is connected to benzene. But how to find where the $\ce{-CH2Cl}$ is situated, i.e. ortho, meta or para?

Well ! Which chlorobenzoic acid can react with nitric acid and afford only one nitro-derivate ? Think a while ! It is not difficult !

Well ! para-chlorobenzoic acid is the only possible solution. The nitro group will be attached next to the Cl atom, and that is the only reasonable possibility.

Then what is the problem with ortho product

No problem ! This is the consequence of the electrophilic attack on the benzene ring already substituted. You should know these rules about ortho-, para- or meta- substitutions. I will not develop them here.

Hints to find the structure of (A):
(1) The given compound (A), is aromatic, hence, it is highly likely to have only one benzene ring (deduced from it's index of hydrogen deficiency/degree of unsaturation)
(2) Only those halides which can give a stable carbocation at the carbon to which the halogen atom is attached, can give precipitate of AgCl with AgNO3 solution. Therefore, one of the chlorine atoms is benzylic, as you mentioned.
(3) Now, oxidation of (A) yields a monochloro-benzoic acid, which on nitration, yields only one mono-nitro derivative. Out of the possible three options: ortho, meta, and para. now, chlorine atom on a benzene ring is ortho-para directing, while the COOH group is meta directing. Therefore, if the initial compound (A) would have -CH2Cl situated para to the Cl, it's obvious that, on oxidation, the COOH group would be formed from the CH2Cl group. Now, there is only two symmetric positions, at which both COOH and Cl will direct NO2+, hence, as both are symmetrical, they can be considered to be the same molecule. For that very reason, it only makes sense for (A) to be para substituted.

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Reaction of 1,3-dimethyl-2,5-dintrobenzene with NaOMe/MeOH

I have looked through articles and closest I feel is in this question there will be a nucleophilic aromatic substitution. Now, a lot of questions are available where there is a halogen and OMe group substitutes the halogen.
My take on this is maybe due to SIR effect (steric inhibition of resonance) The NO2 group present in between the two methyl groups would be substituted by OMe? Will this be the approach?

You are correct that the nitro group can be displaced by methoxide, this has been reported examples here.
My instinct is that the less hindered nitro group is the one that will be displaced.


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Show the mechanism of the Pinacol-Pinacolone Rearrangement reaction

In the reactant, due to $\ce{H+}$ ion, a positive charge will be created by removing water in place of the rightmost $-\ce{OH}$.
But then how will it give such a product. I cant see any way to make the cyclic pentane first.
There would be a Pinacol-Pinacolone Rearrangement but how ?


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Why is isopropyl alcohol (isopropanol) used in rubbing alcohol instead of 'regular' propanol?
Is isopropanol (or 2-propanol) a better solvent than ordinary propanol (or n-propanol, or 1-propanol)? This is strongly implied from what I have read, but I can't seem to find a place that distinctly says so. Several places have said that mixing well with a wide range of substances makes it a better germ-killer.
Also, is it slightly less volatile than 1-propanol?  If so, does this make it more useful?

Imagine water as a ship, holding an alcohol molecule by OH as an anchor, which tries by its lipophilic end to attach to an eventually lipophilic object. On which alkohol would you put your bet ? Additionally – but this is more valid for longer chain alkohols – linear chains of n-alkohols bind their molecules together, forming colloidial micelles. Try e.g. replacing isoamylalcohol ( $\ce{C5}$ ) for extraction from water by n-amylalkohol - there is big visual difference. Isoamylalcohol would form rather clear, quickly separating mixture. n-amylalcohol would form opaque, stable emulsion.

Also isopropyl alcohol became a kind of reference for reasons negligible at small desk scale. One is cost, as for it can be sold at its chemical and market price with no needing of making it disgusting, as it would be for ethanol. Another one is flammability. It makes a big difference if instead of one bottle you think of shipping and entire chemical plant. And so on. If one has to clean microscopy slide, or disinfect hands, there are sure differences, but of no practical impact.

I suspect one element is also that propanols are made industrially by hydration of propylene, and isopropanol is from propylene is simpler (ie less expensive) to make because the markovnikov addition favors that position.

Let's compare the two compounds.
Solubility: Both compounds are miscible with water, so there is no difference. With the straight chain, 1-propanol may be slightly more lipophilic, and thus better able to act as a surfactant. However, 1-propanol is a really poor surfactant.
Volatility: Isopropanol has a lower boiling point (82.5 degrees C) than 1-propanol (97 degrees C), so isopropanol should evaporate more quickly. This is an important feature when cleaning.
Flammability: Both compounds have similar flammability.
Safety: Both compounds are hazardous to humans, but isopropanol is less so. Below are the three GHS hazard statements for each compound:
1-propanol

Highly flammable liquid and vapor
Causes serious eye damage
May cause drowsiness or dizziness

Isopropanol

Highly flammable liquid and vapor
Causes serious eye irritation
May cause respiratory irritation

While both are highly flammable, 1-propanol poses more serious risks for eye exposure (damage is worse than irritation). 1-Propanol also can also cause drowsiness or dizziness, while isopropanol does not.
Cost: All of these are well and good, but the likely reason that isopropanol was chosen was cost and availability. Isopropanol is easily obtained by direct hydration of 1-propene using water and acid, while 1-propanol requires hydroformylation of ethene via the oxo process. Isopropanol can also be produced by the reduction of acetone, which is a byproduct of several other chemical syntheses (for example, phenol). I'm not going to provide cost data because price is dependent on supplier, purchaser, and location. However, for equivalent grade, isopropanol should be cheaper than 1-propanol.


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Aluminum electrodes for NaCl
Upon using aluminium Electrodes for a concentrated solution of NaCl, i conducted electrolysis on the solution for nearly 4-5 hours. There was definitely a gas possibly hydrogen being evolved at the cathode. The anode was also dissolving at a fast rate. After the electrolysis i evaporated the remainder of the water and was left with a silvery sludge.
Could someone clear me up on the identity of the sludge in question and the anode reaction?

You already asked that and I already told you that it is Al(OH)3.

you were not satisfied with the answer given https://chemistry.stackexchange.com/questions/146322/electrolysis-of-nacl-with-aluminium-electrodes?

Donald Trump or my cat just would not take "no" as an answer. Well, apparently some people just would not take Al(OH)3 as an answer.

At the anode, two reactions may happen simultaneously : One due to the electrolysis, and another one due to the reaction of Al on water. The reaction due to the electrolysis is probably $$\ce{Al(s)  -> Al^{3+}(aq) + 3 e-}$$ followed by the hydrolysis reactions : $$\ce{Al^{3+} + H2O -> [Al(OH)]^{2+}(aq) + H^+(aq)}$$ $$\ce{[Al(OH)]^{2+} + H2O -> [Al(OH)2]^+ (aq) + H+ (aq)}$$ $$\ce{[Al(OH)2]^+ + H2O -> Al(OH)3 (s) + H+(aq)}$$ On the other side, the simple reaction of $\ce{Al}$ with water without any electricity  is $$\ce{2Al(s) + 6 H2O -> 2 Al(OH)3 (s) + 3 H2(g)}$$ followed by the reaction happening at the anode $$\ce{H2  ->  2 H+ + 2 e-}$$ As a consequence, the final result in both cases is the anodic production of a slurry of gelatinous $\ce{Al(OH)3}$(s), and $\ce{H+}$(aq) in solution.


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Destroying micellar particles using electrolytes
I learnt that micellar particles formed by ionic detergents forms due to the negative charges of H2O molecules are facing the solution, hence micellar particles repel each other.

I also learnt that micellar particles are destroyed upon the addition of an electrolyte, but I would like to know how.
My attempt: when aqueous cation and anion forms (from electrolyte), it means a lot of H2O molecules are 'consumed' to surround those ions. Therefore, fewer H2O molecules will surround the micellar particles, hence they will aggregate together (hydrophobic interaction?) and form suspensions.
I know my attempt has flaws e.g. when fewer H2O molecules surround the micellar particles, those Na+ are still present on the surface hence still repel each other, so I may be wrong.
I would like to know the real rationale behind it, thank you.

"Colloid" encompasses a great number of possible things. It would help if you narrowed your question to something more specific, say "micellar particles formed by ionic detergents" which your figure suggests is in question.

@BuckThorn Modified. Thanks for the suggestion


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Entropy of free expansion and isothermal process
Let me tell what I understand of the concepts on which the question is based first.
Heat is a flow of energy that takes place due to a temperature gradient.
When doing reversible isothermal expansion of an ideal gas, even though there is no net change in the system's temperature before and after the process, each time we take a differential pressure off the system, there is a differential work done by the gas which creates a differential temperature difference in the system and surrounding and so the differential heat flow, which is the added to get the total heat exchanged in the process.
Then there is the case of free expansion. In this case, even though the gas expands, as it is done against $0$ pressure there is no work done in any step, so there is no, not even differential, flow of heat in any of the steps involved in the process. Also, as the final and initial temperatures are the same in this process, we can safely say that there is a possible way for reaching the final state
achieved in free expansion, from the same initial states, isothermally.
Question: Let's take two process into account for going from state $(P_1, V_1, T)$ to state $(P_2,V_2,T)$. One via an isotherm and other via free expansion. Entropy for the first process (isothermal) is going to be $\Delta S=\int \frac{dq}{T}$ . As T is constant, the integral becomes $\Delta S=\frac{Q}{T}$. As for the second process (Free Expansion), $dQ$ is $0$ for every step and $T$ is constant for every step. this yields $\Delta S=0$ . But entropy is a state function, so the value for both processes $1$ and $2$ should be the same. So what is going on here? Is there something wrong with my understanding of free expansion?

Free expansion is an irreversible. process, and, for an irreversible process, the change in entropy is not $\int{\frac{dq}{T}}$, where T is the temperature at the system boundary through which the heat flows.  The change in entropy for an irreversible process path must also include entropy generation within the system (which is not present for a reversible path).

Could you please provide a reason for 'the change in entropy for an irreversible process must also include entropy generation within the system'. I am guessing it has something to do with how we cannot predict the internal processing of an irreversible path or how there is a sudden change in the arrangements of particles inside the system which leads to disorder and how entropy can be related to disorder. But could you provide a formal explanation please. Thank you.

In the irreversible process, there are transport processes occurring within the system itself that generate entropy.  These processes include heat conduction with finite temperature gradients, viscous dissipation involving finite fluid velocity gradients, and molecular diffusion, involving finite concentration gradients.  All these transport processes generate entropy..  In addition,, chemical reactions at finite rates result in entropy generation.


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Copolymer purity, what does AN / MA / MS mean?
I would like to buy some polyacrylonitrile (PAN) powder from GoodFellow and I have the choice between these two products:

AN316010, quality : Copolymer (99.5% AN / 0.5% MA)
AN316020, quality : Copolymer (93.9% AN / 5.8% MA/ 0.3 % MS)

What I do not understand is the meaning of these three acronyms. What is the abbreviation hidden behind AN, MA and MS?
This may be an obvious question but I was unable to find any answer so far. I guess this may be somewhat linked to a certain degree of purity of the product, but since polymers can also be sold with different Monomer / Polymer ratios, I wanted to be sure...

My best guess for now is that AN may be AcryloNitrile, but I have no idea as for MA and MS... MA may be Methacrylic Acid, but could also be Methylmethacrylate Acrylonitrile or Maleic Anhydride (see here). Any help to answer this question is more than welcome.
Although in this particular case I may simply contact the supplier to have my answer I also wonder:
Is there a general way to solve these chemical identification issues or is it up to each supplier to name these types of compounds as they see fit?

In general, I would contact the supplier for any questions about a product's composition, since that's the only way to be absolutely sure that you get the right answer. In this case, AN is definitely acrylonitrile since the product is called polyacrylonitrile. I'm fairly sure MA is methyl acrylate rather than methacrylic acid or methylmethacrylate, since methyl acrylate is commonly used as a co-monomer in polyacrylonitriles. MS might be methylstyrene, but I'm really not sure.

Ok, thanks for the advice. Indeed, my question was in the meaning "Is there a general way to know such kind of thing or do each supplier have their own formulation?" I will edit my question to be more explicit and also contact the supplier...

I contacted GoodFellow directly and they gave me the following answer:

93.9% AN: 93.9% AcryloNitrile
5.8% MA: 5.8% Methyl Acrylate
0.3% MS: 0.3% Methyl Sulfonate

I guess different manufacturers may name their products as they see fit. In particular, for polymers that can share the same abbreviations, only the context and a good background knowledge of their synthesis pathway may allow one to decide...
There are, however, IUPAC guidelines for polymer abbreviations available here[1]. In addition several other websites offer their own abbreviation list e.g. here and there.

[1]. J. He et al., Abbreviations of polymer names and guidelines for abbreviating polymer names (IUPAC Recommendations 2014), 2014. DOI: 10.1515/pac-2012-1203


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Why a catalyst is not consumed in a reaction when all steps are reversible?
It is said that a catalyst speeds up the rate of a reaction but is not consumed (assuming no side reactions take place). Suppose we have the following reaction:
$$\ce{A + B <=> C} $$
catalysed by $\ce{X}$ in the following steps:
$$\ce{A + X <=> A-X} \tag{1}$$
$$\ce{A-X + B <=> C + X} \tag{2}$$
Each step is reversible so every step has an equilibrium constant $K$. So an amount of $\ce{X}$ should be present as the intermediate $\ce{A-X}$ even after the catalysed reaction has reached equilibrium. That means that the amount of catalyst that we have added to catalyse the reaction is not the same after the reaction has reached equilibrium. So why we say that a catalyst is not consumed through the reaction?

In the way you have written it X is not a catalyst but forms an intermediate species A-X that then reacts with B to make C. If X were a catalyst X would be released intact as C is formed.

@porphyrin Thanks for pointing it out, I have edited it.

X has 2 active forms: X and A-X, so it is not consumed.

What if you remove C?

@Poutnik it might be. But counting this way no reaction consume something. A-X could be written as Y.

Not if there is no X <-> Y. Said by other way, some reaction steps regenerate the original form of a catalyst.

There are reactions where the "catalyst" is regenerated and therefore does not need to be present in large stoichiometric quantities. A small amount of the catalytic intermediate compound will be present even when the reaction is near completion. But the main point is that the catalyst doesn't need to be present in anything like the same quantities as the main reactants. Each catalyst molecule is reused many times so we say it is not "consumed" even when we never recover it all.

By saying Catalysts are not consumed by reactions. is meant there is no stoichimetric ratio to reactants, consuming catalysts and forming from them catalytically inactive compound.
By other words, some of reaction steps regenerates the original form of a catalyst, consumed by a prior step, so the net consumption is negligible.
The transition between active states of a catalyst is not considered a catalyst consumption, as it does not affect its catalytic activity.
Catalysis means the catalyst is reversibly switching between its 2 – chemically different – active states, producing an alternative and faster reaction path. Each form catalyzes one of the mutually opposite direactions of the net reaction. It is like a kind of a ferry to move compounds between reactant and product shores, bypassing a faraway bridge. The ferry is not consumed if ends at the opposite shore.

Note that things get little complicated, if some of products or intermediates catalyse a part or whole net reaction. Not being particularly aware of the intermediate case, but for the product, the famous case is oxidation of oxalate by permanganate in acidic environment:
$$\ce{5(COO)2^2-(aq) + 2 MnO4-(aq) + 16 H3O+(aq) -> \\ ->10 CO2(g) + 2 Mn^2+(aq) + 24 H2O}$$
$\ce{Mn^2+(aq)}$ as the product acts as a catalyst, forming catalytic system $\ce{Mn^3+(aq)/Mn^2+(aq)}$ with a similar redox potential as $\ce{MnO4-(aq)/Mn^2+(aq)}$.
$$\ce{4 Mn^2+(aq) + MnO4-(aq) + 8 H3O+(aq) <=> 5 Mn^3+(aq) + 12 H2O}$$
$$\ce{(COO)2^2-(aq) + 2 Mn^3+(aq)  -> 2 CO2(g) + 2 Mn^2+(aq) }$$
The effective result is an autocatalytic reaction. There is sometimes applied a small amount of Mn^{+II} salt to speed up the reaction, or it is waited until initial addition of $\ce{KMnO4}$ creates enough $\ce{Mn^2+}$.

But in your example $\ce{MnO4-}$ is ultimately consumed. Is that a good example of catalysis? Is the point that $\ce{Mn^{2+}}$ is the catalyst and $\ce{MnO4-}$ in excess (reagent)?

@Poutnik So could we say that for a catalyst initial concentration equals $\displaystyle\sum_{i}^{} c_i$ where $c_i$ corresponds to every chemical species in the catalysed reaction that catalyst is part of?

@buckthorn Sure, I have thought it is obvious and was explicitly described. Mn2+ is catalyst. MnO4- is not.

@anton It can be said like that. I do not really understand why the catalyst role brings such troubles.


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How do we explain NMR shielding in filled valence shells?
It is universally recognized that NMR measures the shielding of nuclei from magnetic fields by electrons.  The electrons move, producing an eddy current that produces an opposing magnetic field.
It is also fairly widely said that a filled valence shell contains electrons making every possible motion within a certain range of energy.  For example the p orbitals are added up using +1 and -1 values for angular momentum, and the electron spins are paired.  MO theory seems, in most cases, to have the same paradox.
Now I know you can look into it and find more detailed math for NMR tensors, but at the intuitive level: what is going on here?  (Are the filled orbitals being hybridized, slightly, with ordinarily unfilled complex-valued higher-energy orbitals to permit the electrons to take on some degree of net overall angular momentum in relation to the axis of the magnetic field, or is it something else?)


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How much should I scan an aromatic dihedral?
Suppose I have a napthalene-like molecule (attached), and I would like to scan (relaxed) its aromatic dihedral (highlighted atoms). I used CGENFF to obtain its dihedral parameters as follows,
HGA4 CG2DC2 CG311 CG331 0.2000 3 0.00
As shown, the dihedral force constant is 0.2, periodicity is 3, and angle is 0.
However, the penalty factor is high >50. Hence, in order to obtain new parameters and following the typical charmm protocol using MP2/6-31G*, should I perform a full scan +/- 180 degrees? +/- 180 seems unphysical (as compared to a rotatable bond) for a dihedral that belongs to a ring. If so, how much should I lower this value to?

That is a molecule with far more saturation than naphthalene, but it will still be pretty rigid, so exactly how far do you think it could deform without breaking any bonds?

@matt_black not very far at all, I suppose. Any ideas?


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Planes of symmetry of a molecule
Are we limited to use only xy, yz and zx planes to check for planes of symmetry of a molecule? Because I see a plane of symmetry in  the following molecule through the two shown H atoms or OH groups. I am aware of the 3D structure shown by this projection. And since this molecule is Chiral, and has enantiomeric mirror image, this existence of a plane of symmetry appears contradictory.

This question would be more suited to [chemistry.se]?

Mother Nature uses whatever axes she wants, and is not limited to our frame of reference, so, yes.

Isn't this the meso isomer of this chemical?  I'll admit that I'm rusty on what 3D structure is understood by this diagram, but if the main carbon chain is "zig-zagging" towards & away from the viewer, then the molecule is superposable on itself via a point reflection about the center of mass (and therefore via a plane reflection as well.)

@MichaelSeifert No, this is not a meso, and no, it has neither point nor plane symmetry.

I think both answers so far fail to address properly the utter otherworldliness of the premise in question. You have a plane of symmetry (approximate, of course); are you only allowed to walk along the xy, yz and zx planes of the coordinate system which is official in your jurisdiction? I guess no. Well, neither are molecules.

@Neretin: The question was based on the wrong assumption that a symmetry plane is present in the molecule. Clarified this, symmetry operations can move an object in a 2D or a 3D space; in the last case fall all the n-screw axes with n=3,4,and 6. Coming back to the inspected molecule the only symmetry element that I see in this configuration is the inversion. If you would cut out the CH3 groups you would have a 2-fold screw axis, but a symetry element is present if it involve all the molecule. Finally, an 'approximate' plane of symmetry is a misleading concept.

Group theory, tensors describing the physical properties are based on the exact fullfilling of the symmetry properties.Otherwise it would be meaningful to define a tetragonal structure as 'approximately' cubic and so on, but we all know that the story is different (and we are lucky with this; on the contrary it would be a great caos)

I don't see planes of symmetry through H atoms or OH groups. Actually, these planes are not present. It is easy to check this by looking at the position of the CH3 group after reflection in one of these planes: you find CH3 at a completely different position (none of the original positions is mapped and hence you can distinguish the starting configuration from the final one).

@gryphus's answer is correct for the statement that there is no mirror plane in the structure of the question.  Independent that this is a late comment to the answer, the reader of such a Fischer projection should not hesitate to construct a model of the underlying structure (here obviously with quite some non favourable steric interactions) to get the molecule a 3D object:

It then becomes easier to recognize that regardless how one rotates the substitutents around the two sterogenic centers marked (S,S), there will be no conformation including a mirror plane.

To answer the general part of the question: no, we are not limited to the Cartesian planes and axes. The obvious examples are the ammonia and the methane milecules: the former has several reflection planes, positioned at $120^\circ$, whereas the latter has multiple reflection planes and also the rotation axes (one can easily google both).
Regarding the specific molecule mentioned in the question: not only does it not have the indicated symmetry plane, as discussed by @gryphus, but it is generally a bad practice to treat whole groups (OH and CH3 in this case) as single points, as well as to consider a 3D molecule as planar.


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Confusion regarding the mechanism of acidic/basic buffer solutions
While learning about acidic and basic buffer solutions we were told that they can resist pH change if we add small amounts of acid or base by neutralizing the $\ce{H+}$/$\ce{OH-}$ ions from the added acid or base. But I am confused about the fact that if we add a small amount of base to an acidic buffer then after neutralization of the excess $\ce{OH-}$ ions from the base by the $\ce{H+}$ ions from the buffer, shouldn't the concentration of $\ce{H+}$ ions in the solution decrease, thereby making the solution acidic and reducing the $\mathrm{pH}$. (vice-versa for the addition of acid to a basic buffer)

Buffers resisting pH change does not mean pH is constant. It means pH changes a lot less than without them being present. See the Henderson-Hasselbalch ( not -bach) equation on Libretexts or on  Wikipedia and also buffer solution. It was question of 1-2 minutes of search and copy/paste. You may want to pay more attention to searching before asking.

While learning about acidic and basic buffer solutions we were told that they can resist pH change if we add small amounts of acid or base by neutralizing the $\ce{H+/OH−}$ ions from the added acid or base.

Yes, the pH changes less when you add small amounts of acid or base to a buffer, and more if you add it to pure water or an unbuffered aqueous solution.

But I am confused about the fact that if we add a small amount of base to an acidic buffer then after neutralization of the excess $\ce{OH−}$ ions from the base by the H+ ions from the buffer, shouldn't the concentration of $\ce{H+}$ ions in the solution decrease,

Yes, that is what happens. As a consequence, the weak acid/base reaction of the buffer  substances is no longer at equilibrium, and there will be a net reaction to replenish some of the $\ce{H+}$ that was lost.

thereby making the solution acidic and reducing the pH. (vice-versa for the addition of acid to a basic buffer)

This is probably a typo. When you add a small amount of base, the pH would go up, buffered or not. In the presence of buffer, it goes up a little, and in the absence of buffer, a lot.

(vice-versa for the addition of acid to a basic buffer)

Buffers stabilize the pH near the $\mathrm{p}K_\mathrm{a}$ of the conjugate weak acid in "both directions", so even if the pH of a solution is slightly acidic, it is able to buffer against addition of acid or base.

This is linked to the fact that in a weak acid-base pair, the respective concentrations are linked to each other through their ratio.
We have
$$K_\mathrm{a} = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{AH}]},$$
therefore
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}.$$
In water, the reaction is written
$$\ce{AH + H2O -> A- + H3O+}$$
If we add a little strong acid, then we shift the equilibrium a little in the direction of formation of the acid but as the concentrations are related to each other by the $K_\mathrm{a},$ the consequence on the $\mathrm{pH}$ is attenuated.
For a strong acid, the addition of another strong acid contributes totally to the increase in $\mathrm{pH}$ because there is no balance that counterbalances the addition of the acid (as is the case for a weak acid).


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Magnetic properties of potassium manganate(VII)
According to me, $\ce{MnO4^2-}$ is paramagnetic and $\ce{MnO4^-}$ is diamagnetic.
On the basis of this, the corresponding potassium salts inherit these magnetic properties, i.e. $\ce{K2MnO4}$ is paramagnetic and $\ce{KMnO4}$ is diamagnetic.
However, according to my class teacher $\ce{KMnO4}$ is paramagnetic.
Is potassium manganate(VII) dia- or paramagnetic?

Unanswered duplicate: https://chemistry.stackexchange.com/questions/140123/why-is-potassium-permanganate-paramagnetic?r=SearchResults

@Nilay_Ghosh Both are different questions

See this paper, it is a complex story "Magnetic Studies on Potassium Permanganate"  https://iopscience.iop.org/article/10.1088/0370-1328/79/2/318

In this article It's say that Potassium permanganate is Dimagnetic https://www.chemedx.org/video/paramagnetism-oxidation-states-manganese-potassium-permanganate#:~:text=Manganese%20-%20Potassium%20Permanganate-,Paramagnetism%3A%20Oxidation%20States%20of%20Manganese%20-%20Potassium%20Permanganate,unpaired%20electrons%20per%20Mn%20atom.

In the relatively authentic paper I showed you, it says the same thing. KMnO4 crystal is diamagnetic but there are finer details of feeble paramagnetism. Your teacher is not right.

@saketkumar It is diamagnetic, not dimagnetic.

For eventual writing and formatting of chemical formulas or equations, see how to use MathJax with mhchem extension . Note the preferred plain text text for titles.

Paramagnetism in KMnO4 is normally attributed to the Van Vleck Mechanism - see https://iopscience.iop.org/article/10.1088/0370-1328/79/2/318 (as mentioned above) and https://en.wikipedia.org/wiki/Van_Vleck_paramagnetism

Actually it is diamagnetic but due to a phenomenon called charge transfer spectrum (CTS), oxygen transfers one electron to manganese and $\ce{KMnO4}$ as a whole becomes paramagnetic. This is also the reason why it shows a characteristic colour while diamagnetic complexes are colourless due to absence of d-d transitions.

What would be the percentage of the excited state from charge transfer, in let's say a solution of KMnO4 ? Because a significant number of the molecules have to be in excited state before the bulk material gains paramagnetic properties.


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How tetraamminecopper(II) is formed?
I'm trying to get a sense of how $\ce{Cu^2+}$ and $\ce{NH3}$ are bonded together because it doesn't make any!
I understand that $\ce{NH3}$ has a lone pair of electrons that can be donated (right?), but since there are four molecules of $\ce{NH3}$, then there are 4 pairs, and $\ce{Cu^2+}$ only needs two!
the reaction in my textbook was written like this:
$$\ce{CuSO4.4NH3 <=> [Cu(NH3)4]^2+ + SO4^2-}$$
I would appreciate any clarification.

You are aware that element symbols start with capital letters, aren't you ? It is rather  $$\ce{[Cu(H2O)4]^2+ + 4 NH3 <=> [Cu(NH3)4]^2+ + 4 H2O}$$You can find useful  How can I format math/chemistry expressions on Chemistry Stack Exchange

I assume you know what a coordination complex is, right?

There are many ways to describe chemical bonds. Copper II ion binds to some anions with ionic bonds in solids, sometimes in covalent bonds (Ref 1).
In aqueous solution, copper II ions are surrounded (hydrated) with six water molecules, with the oxygen ends pointing in, toward the positive copper ion. Water is acting as a Lewis base. Ammonia is a stronger Lewis base - after all, it is basic in water. So an ammonia molecule will replace a water molecule in a hexa-aquated copper ion. Eventually, you get $Cu(NH_3)_4)^{2+}$.$(H_2O)_2$. Oh, I forgot, sometimes we leave out the water molecules, since we just assume that they are there - everybody knows that, since the teacher has explained that (unless the teacher hasn't explained that).
In the equation of the question, only four of the actual six water molecules are shown, because they are important for describing the interaction with $NH_3$, and the other two are just a complication, because I know you would ask, why don't they get replaced too, to give $Cu(NH_3)_6^{2+}$? After all, nickel, cobalt and chromium III do! Well, zinc doesn't; it forms a tetra-ammonia complex. And iron, manganese and aluminum don't even dissolve to form soluble ammonia complexes in alkaline solution. Reference 3 suggests that it is an equilibrium situation, but there are deeper explanations which you can explore later.
Ref 1. https://aip.scitation.org/doi/10.1063/1.1677079
Ref 2. https://www.chemguide.co.uk/inorganic/transition/copper.html
Ref 3. https://www.chemguide.co.uk/inorganic/complexions/aquanh3.html#top

Answer too long, too complicated, a lot of unnecessary information. If you want to be precise, you have to describe it with orbitals and coordination spheres. Citing papers seems "intelligent" but doesn't say anything at all.

I think this attempts to answer the question but is not very well written.

Yes, you are right. This is not a covalent bond it is a coordination bond. The copper has a electron pair deficit (Lewis Base) and the Nitrogen (Lewis Acid) donates the elctrons pair which makes the molecule more stable in the end.

The answer is a a bit short, and unfortunately defines Lewis acids and bases incorrectly. Lewis bases have electron pairs; Lewis acids accept them. I give you credit for trying.

I made it correct, no reason to downvote.

@deuti I think you misunderstood the correction from James' comment. Copper accepts electrons, so it would be a Lewis acid; Nitrogen donates electrons, so it would be a Lewis base.

@deuti: I didn't downvote, but didn't feel an upvote was appropriate, so I gave you verbal credit for trying. So keep on - everybody makes misteaks.


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Why is quinone mono oxime more stable than 4-nitrosophenol?
When we do nitrosation of phenols, we get para-nitrosophenol as the major product, which tautomerises to quinone mono oxime. This article states that it largely exists in the quinonoid form.
My question is, why? I know that nitroso form is less stable than the oxime form, and enol is less stable than keto, but on the other hand we are losing the aromaticity also. Why is being non aromatic getting more favourable than being aromatic in this case?
I read the post shared by @Safdar in the comments, but this answer there says that both forms are aromatic. Is this true?

Possible dupe: https://chemistry.stackexchange.com/questions/125617/stability-of-4-pyridone-vs-4-pyridinol ? I guess the same principle applies here..

@Safdar I agree that it is similar, but in that post both rings are aromatic, but I doubt that it quinonoid form is aromatic in this case, as you can see in the link I have given.

Two factors come into play: (1) the inherent difference in resonance energy is not as great as you might think, and (2) the greater tendency for a dissociable proton to bind more tightly to an oxime function rather than a phenol function appears to counterbalance the reduced $\pi$-electron bonding difference.
Less than meets the eye
Suppose you were to use the Hückel model to estimate the pi-electron binding energy for several eight-carbon conjugated systems.  By that method I find the following results:

For the conjugated straight chain $\ce{CH2=(CH=)_6CH2}$ the total $\pi$-electron binding energy works out to $9.52\beta$, where in this theory $2\beta$ represents one isolated $\pi$ bond.

For styrene, with a $\pi$-electron pattern matching the nitroso-phenol, the result is $10.53\beta$.  A difference of about one $\beta$ unit is typical when an aromatic ring is compared with the opened ring.

For para-xylylene whose quinonoid $\pi$-electron structure matches the oxime tautomer described here, we get $9.92\beta$.

Of course the aromatic structure is most stable, but not all non-aromatic structures are created equal.  The Hückel-model calculations indicate that even though the quinonoid structure is typically called non-aromatic, in fact it captures a significant part (about $40$% based on the numbers) of the extra $\pi$-electron stabilization we usually associate with an aromatic ring.  The energetic difference between the two competing $\pi$-electron configurations is thereby less than the "aromatic" and "non-aromatic" labels would seem to imply.
Swimming upstream
Now let's see where the dissociable proton is more stable.  To do that we look up the $pK_a$ values typically associated with each prospective acidic functiin:

Phenol gives $pK_a=9.95$.

Acetoxime, $\ce{(CH3)_2C=NOH}$, gives $pK_a=12.2$.

The oxime function is clearly holding the proton more tightly; placing the acidic proton there instead of on the phenol function saves $2.2$ powers of ten in the equilibrium constant.
We cannot directly compare this with the Hückel-model calculations from the previous section, because that model is only semi quantitative and a more complex calculation would be needed to capture the effects of the heteroatoms in the conjugated structure.  But the results do suggest that the combination of the quinonoid structure capturing part of what is usually aromatic $\pi$-electron stabilization with the preference of a dissociable proton to bind to the oxime function leads the oxime tautomer to win out over the nitroso-phenol.


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Schrodinger's Equation and Wave Function
So I understand that there exists the shrodinger's equation, which on solving,gives the wave function of an electron. The wave function as I understand, gives all possible information about an electron(any extra information on wavefunction will be appreciated).
Could someone actually solve a shrodinger's equation for an electron, and vaguely describe the process involved? Also after the wavefunction is obtained, could you explain to me what exactly does the wavefunction depict, and actually graph it?
I am having a lot of trouble grasping the concept. Any help will be greatly appreciated

In a nutshell the Schrödinger equation gives the probability for finding an electron in 3D space around a nucleus. The nasty part is that it only works for a single electron. With two electrons you have a problem analogous to the three body problem of classical physics.

@Max It is incorrect. The SE gives the wave function, which is more information than just the probability density. Also, IT "WORKS" for 2, 3 etc. electrons, too, only you cannot solve it analytically, which is a completely different story.

I am not sure that this site is the best resource for this question. The solution of the Schrodinger equation for a single electron in different potentials is generally a fairly standard part of any quantum mechanics or quantum chemistry course, however, it is rather long and needs university mathematics to get some idea of it. From start, it can be a half semester if you are not familiar with quantum mechanics. What we can briefly, but generally tell is something you most probably already have seen (eg. orbitals for H atom), or if you try to read it and have specific questions on details.

I would start by looking into the particle in a box, a common starting point in QM courses. See eg https://en.wikipedia.org/wiki/Particle_in_a_box

There is nothing special or unusual about this type of differential eqn. Standard maths techniques are used to solve it after we have defined the problem at hand, e.g. particle in a box or H atom etc.  We interpret the square of the result of solving it $\psi^*\psi$ as the probability of finding the particle at a certain place.

Question starts about orbitals as |phi|^2 but the thread turns to the significance of the WF itself as E/position as well as probability amplitude, thus giving a probability distribution function. Note that some of the physicists there reserve the term orbital exclusively to the way function, while for me it seems a mere linguistic choice. As a chemist, when I look at the shape of a molecule, I think of orbitals as occupying space and where they are, not of the underlying wavefunction. https://physics.stackexchange.com/questions/607552/what-do-atomic-orbitals-represent-in-quantum-mechanics

@porphyrin It is all nice, but people have no idea about first year university chemistry or physics are generally the people who cannot solve partial differential equations or have no broader concepts on what is Hamilonian mechanics.

I suspect the problem is not mathematical. It is in the meaning of the wave function. Well. I will now show you how I explain it qualitatively in my high school classes. I state that the electron is like a violin string, but a string with three dimensions that is vibrating in the fourth dimension. As I know this last words are not understandable, I start with one dimension and pass slowly to two and three dimensions.
Let's start from a horizontal string whose end is fixed on a wall, and the other end is in your hand. If you put some energy to the string, it will vibrate in a stationary vibration. The string looks like half a sine curve, with the maximum in the middle of the curve. If you want to describe the curve with words and without drawings, you can say that the curve is like a half-sinusoid whose amplitude has the same sign all over its length, part time positive, part time negative.
Now if you put more energy in the string, it will vibrate in another mode, with a node in the middle. When one half of the string is vibrating upwards, the other half is vibrating downwards. With still more energy you can have it vibrate with two nodes. To describe the shape of the string as before without drawings, you can say that the string is a sinusoid with one or several nodes. So that you need one number $n$, the number of nodes, to describe its movement : $n$ is the first quantum number. $n = 1$ for the first vibration, because the only point not moving is the end of the string in the wall. $n = 2$ for the case with one node in the middle of the string.
Now let's consider the string in two dimensions. This is not a string any more, but a two-dimensional plate like a drum. If you hit it softly in the middle, it will vibrate with all its points moving simultaneously upwards then downwards. The only points of no vibrations are the periphery of the drum. In two dimensions, points of no vibration are replaced by lines of no vibration. Here there is only one such line : the number of nodal lines is $n = 1$.
If you put more energy by hitting the center of the plate, you may produce a new vibration mode if the center of the plate is moving upwards (or downwards) when the outer part is moving in the opposite direction. In this mode no vibration occurs on the small circle between these. This circle is a nodal line. There are two concentric circles on this mode. So  $n = 2$. We may describe its amplitude by drawing curves of equal amplitude on the plate, like level curves on a map.
But it is going to be more difficult, because there is another type of vibration on the plate. If you hit the plate not in the center, but elsewhere, you may produce a plate vibration where one half of the plate moves upwards and one half downwards. Here the nodal line is a diameter. And there are two nodal lines : the diameter and the periphery. Here $n = 2$. As a consequence, two vibration modes exist with the same quantum number. We must introduce a second quantum number, which is called $l$, and which is the number of straight nodal lines. So this type of vibration is given by $n = 2$ and $l = 1$. Hopefully you understand that other vibration modes do exist with the drum, with arbitrary values of $n$ and $l$ < $n$
Let's summarize. An object with one dimension vibrates in another dimension, and one quantum number is enough to describe its vibration. An object with two dimensions is vibrating in the third dimension, and two quantum numbers are required to describe its vibration. The number of dimension is equal to the number of quantum numbers.
Now let's go to the electron, which has $3$ dimensions, and which may be considered as if vibrating in a fourth dimension, called $\Psi$. As this dimension cannot be represented in our world, all we can do is to give different numerical values to three quantum numbers for all vibrating modes, and represent surfaces which correspond to equal numerical values of the local "vibration" of the electron. These surfaces of equal value of $\Psi$ are spheres for $l = 0$, the famous dumb-bells with two lobes for $l = 1$, because of $1$ nodal plane. Here is how it goes.
Here is how I explain the wave function to my high school students. They usually have some difficulty understanding. But all have told me that this explanation had helped them sudying quantum mechanics later on in college.

While this can have a value explaining why there is a need for numbers describing the vibrations of string and surface, it says absolutely nothing on the meaning of the Schrödinger function.

@Alchimista. It is right. It does not mention Schrödinger equation. it just explains the nature of the wave function.


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Which electrons are in HOMO in a peptide bond?
We've recently studied the absorption of UV light by polypeptides, and found that the peptide bonds itself absorhs UV light at around 210nm. I'm confused about what feature causes this to happen.

What question are you asking? Your title is asking about the HOMO and the body is asking about absorption. Yes they are related, but you need to flesh out the link.

The title is even not really clear as well having profound implications. You should reformulate the question entirely.