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Recovery of Urea from aqueous solution
When adding Urea to water, the reaction is highly endothermic. I was playing with the idea of using that for a Desktop cooling system, but for that, I'd of course have to recover that Urea from the water somehow. Obviously, the easiest way would be to simply boil away the water, but given that this is supposed to feed into a cooling system involving sensible electronics, I unfortunately don't have that option. Since I'm myself better at engineering that at chemistry, I decided to ask here: Which other seperation methods are there that leave both the water and the urea "intact"? Ideally fast ones, but that is not really a requirement, it'd just make things easier for me. Can I use reverse osmosis? Electrochemical methods? I'd really appreciate some help. Thanks!

Reverse osmosis requires a rather high pressure. And it is a very slow process.  Why do you refuse any boiling process ?

@Maurice because sensitive technology and high temperatures are typically a very bad match. I'm sure you could theorethically make it work with proper thermal insulation, but in the end my PC is pretty importent to me, and it's simply not something I'd like to risk if I can avoid it.

Buying a cooling system on air fans only, or with an intermediate water cooling system; they have been around for quite some time.  These could be an option ready-to-go.

I think this cooling method is a bad idea, I second Buttonwood.

An interesting fact: Some datacenter-grade servers have even direct oil cooling, designed to be immersed in an oil bath ( from the local interview with a datacenter person ).

The problem with a process as you have suggested is, plainly, energetics. Specifically, any physical or chemical process that involves a change of state follows the equation
$$\Delta G = \Delta H - T\Delta S\tag{1}$$
where $\Delta G$ is the Gibbs free energy, $\Delta H$ is enthalpy, $T$ is temperature in kelvin and $\Delta S$ is entropy. A reaction or process will only occur if $\Delta G$ ends up being negative (exergonic). In a simplified model, the enthalpy part describes the heat that is transferred either into or out of the system, i.e. the temperature change you observe. If $\Delta H$ is positive – i.e. the reaction/process is endothermic or it absorbs heat from its surroundings – that must be offset by $T\Delta S$ in an appropriate manner to give an overall exergonic process.
The dissolution of urea in water is exergonic but endothermic, meaning $\Delta H>0$ and thus $\Delta S >0$. If you now want to reverse that process to recover your urea, this will necessarily mean that your reverse-process has $\Delta S<0$, meaning you need to find a sufficiently exothermic process to offset $-T\Delta S$ being positive.
Unfortunately for you, $G, H$ and $S$ are state functions, meaning their values depend only on the current state of the system, not the way the system took to get there. So any process of getting back from dissolved urea to solid urea and water will need to overcome the same barrier making the whole thing rather complex if not impossible at any small scale.
Furthermore, you are up against the Second Law of Thermodynamics, meaning that for any spontaneous process $\Delta S > 0$ – I hope you already see how this is negatively impacting your idea.
Of course, nothing in this answer states that it is impossible to separate a solution of urea in water. Obviously, distillation (evaporation and subsequent capture) will recover the pure starting materials. Reverse osmosis may help, I’m no expert on that. (Electrochemistry likely won’t help as that would involve chemical reactions which would modify urea to something else and you would have to reverse that &endash; difficult. Furthermore, all simple electrochemical processes I can think of in five minutes will lead to something that you won’t be able to turn back into urea in any simple way.) However, the first part of this answer shows that whichever process you use it is going to be really energy-intensive.
One potential solution to all this would be a one-way system where you constantly purchase water and urea and dispose of the dissolved urea. This isn’t the cheapest way of doing things but I would imagine that it would be more cost-effective than urea recovery.
On the other hand, cheap, non-energy-intensive cooling methods exist, have been engineered and used in the wild for decades such as water cooling or air cooling with fans.

@MatNX, as you state, "sensitive technology and high temperatures are typically a very bad match." However, there is no magic way to "make" cold, only to move heat from one location to another. Presumably, since you state you have an engineering background, you're familiar with the laws of thermodynamics. Any heat you remove from an IC will be displaced somewhere else, so you must provide a heat-sink, e.g., large volumes of air or lesser volumes of water.
While you could find some way to re-solidify the urea, that releases heat in the process -- more than was removed (2nd law: you can't break even).
There are many ways to move heat, though:

Passive heat pipes, as used in aerospace, as well as in desktop PC's.

Peltier coolers, useful for cooling a small sensor, such as an IR module (though these are inefficient and create yet more heat outside the enclosure).

Phase-change refrigerators, for the hard-core over-clocker.

Magnetocaloric refrigeration, for the quantum computer working at liquid He temperature.

Thanks for the indepth answer! I probably  should have clarified that I don't want to magically remove the heat, but in normal watercooling systems, theheated fluid is cooled using regular fans before it is cycled back. I want to replace that part specifically, having a container with urea that the water flows through. The basic idea is simple: Imitate an instant cold pack, having the endothermic reaction absorb heat, and then remove it in aome other form, be it pressure or electricity. Think fuel cell, H2 produces electricity instead of exploding.

First, water has a higher heat capacity then a system of water + urea, so you'll lose cooling capacity after the water is saturated. Second, this would be a one-shot use of the urea -- again, once the water is saturated with urea, it can hold no more and needs to be dumped. Third, you want to regenerate the plain water and urea, so it's not a one-shot deal? That requires energy, with concomitant waste heat. Thermodynamics is a harsh science; TANSTAAFL, as Prof. Heinlein states.


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How to determine the amount of CO2 sodium hydroxide absorbs?
I first created a solution of $\ce{NaOH}$ and water, 5g of $\ce{NaOH}$, and 100ml of water, with a pH of around 14. I then added $\ce{CO2}$ (mixing baking soda, citric acid, and water) to the solution. The pH then dropped to 11 forming sodium carbonate.
I would like to know how much $\ce{CO2}$ was absorbed by the solution. Can I do this with the data I already have?

I know 1 mole of co2 reacts with 2 moles of NaOH so do I just do that?

You need to do a titration with phenolphthalein and methyl orange.

What is co2 and ph ? Why  Sodium with the capital S ?

I that just a ph indicator? I already found the ph using ph strips

No, you cannot determine precisely with the data you collected (for example, a pH value 'around 14'). Some of the generated carbon dioxide gas also may have escaped or not reacted with the high pH solution.
Remember, pH is a log scale and lose of precision can be significant. However, you may argue, within the measurement error of the pH scale, you are approximately correct. Also, you are assuming correct calibration of the pH device. Mention the nature/degree of possible errors in your write-up.
Also, correct procedure matters, so if you did not add pure carbon dioxide gas directly, meaning you added the citric acid plus Baking Soda into the aqueous $\ce{NaOH}$, some neutralization by reactions with the sodium hydroxide by the citric acid and Baking Soda. The latter will skew the presumed amount of $\ce{CO2}$ actually released.
If appropriate, some added work. For example, test by adding dissolved calcium chloride, in excess, to the solution. Proceed to collect, dry and weigh the recovered calcium carbonate. Moles of the latter equate to moles of consumed carbon dioxide, albeit, expect some $\ce{CaCO3}$ collection loss.

Thank you for your answer, but for clarification, I did add the co2 directly, sorry I didn't say correctly. I'm ok with just getting an average and not an accurate amount of absorbed co2. Since I added 5g of NaOH(along with 100ml of water) and sodium carbonate requires 2 NaOH and 1 co2 molecule, is it safe to assume 2.5g of cpo2 was absorbed?

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Definition of an internal reference
Warning: extreme pedantry.
I am performing Gutmann-Beckett experiments, basically utilizing an NMR-tube with my sample and a 31P-lewis base (e.g. triethylphosphine oxide) inside another one with a standard/reference (e.g. trimethylphosphate). This is classified as an internal reference, but I am unsure about that. Wouldn't an internal reference be in direct contact with the sampled solution?
Since we need a constant 31P signal to reference, I understand that the reference has to be isolated, and wether this is "physically" by glassware or by an extensive shielding residue is irrelevant (when the signal is constant); but somehow this is not what I would call an internal reference.

Is there an error in your wording?  I'm used the other way around (the sealed capillary of e.g., trifluoro acetic acid is plunged into the greater NMR tube containing the sample solution). In the example of TFA, the separation prevents reaction with the analyte. In other cases ($\ce{CDCl3}$ plus the added [an option!] TMS), standard and analyte share the same solution in one sample in common (then per se an inner standard). Really external would be instruct the spectrometer software «with operational frequency x, use  frequency y (outside the spectrometer) to compare with».

We basically use the original method Beckett used (Beckett, M.A. 1996. Polymer 37, 4629–4631) where the sample-tube is inside the reference containing tube. It makes sense the way you explain it, but here we use two different solvents for the two tubes. I am inclined to call that an external reference, since in analytical chem., iirc, a separate HPLC run with the standard in the same matrix as the sample is an external standard, e.g. Maybe calling it internal is due to the constraints of NMR-methods?

Though your sample and standard are recorded simultaneously, the two experience slightly different magnetic field (separate containers shielding each other). But (more importantly:) molecules of one don't interfere with those of the other as they would for a homogeneous solution of TMS in $\ce{CDCl3}$ in one tube. Thus my take for a separate analysis, and hence an external reference. (cf. Orange Book, internal standard/section 9.2.3.3).

The other Colour Books of IUPAC are mentioned briefly in section «nomenclature» in the chemistry.se resource page here.

Thank you very much, that clarified my issue!

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Do exothermic reactions also absorb heat?
As I have read that every reaction(endothermic or exothermic)requires activation energy.
So is it correct to say that even exothermic reactions absorb heat?
I am asking this because it is always taught that exothermic reactions release heat .

Every exothermic reaction releases net heat - by definition.
Every reaction absorbs energy, e.g. thermal energy from heat, for its activation, but the amount of released energy is for exothermic reactions bigger than absorbed energy.

Note that in the gas phase also barrierless reactions do occur (typically involving ions and neutrals). Although these reactions still have to overcome the centrifugal barrier, the $s$-wave collision is truly barrierless.

@Paul  I guess any short enough and simple enough answer has its exceptions.

You are right of course, it was just a small addition to an excellent answer ;)

@Poutnik. Are you sure that all reactions absorb energy for its activation ? Do $\ce{AgCl}$ or $\ce{BaSO4}$ precipitations have activation energies ? Do neutralization reactions (like $\ce{H^+ + OH- -> H2O}$) have activation energies ?

@Maurice See my last note. The answer was intended for the introduction level. For your specific cases, I imagine some energy is needed to break ion hydration layers of ordinary ions, or for charge hopping for H+ and OH-.

@Poutnik. OK. You may be right.

Exothermic/endothermic (or exergonic/endergonic, if you equally account for changes in entropy) only compares initial state prior, and final state after the reaction.
So the height to overcome that this reaction happens (the activation energy) is irrelevant.  It equally is irrelevant if this energy was invested by heat (e.g., Bunsen burner/fire, as in roasting metals), light (laser, UV; as in photocylizations), addition/removal of electrons (electrochemistry, redox-reactions), etc.

[OP] As I have read that every reaction(endothermic or exothermic)requires activation energy.

Yes, and this activation energy is for a single set of reactants reacting to form product.

[OP] So is it correct to say that even exothermic reactions absorb heat?

No. Heat is a bulk transfer of thermal energy from one object to the next.
If you search for activation energy on the internet, you might find strange examples like lighting a candle with a match. This does not really explain activation energy. Activation energy is necessary for every single reaction, not just to "start the reaction". In the example of a candle, the combustion reaction still needs an activation energy after the candle is lit. The temperature of the system is high enough, though, that it is available.
What happens in general is that the reaction mixture has sufficient energy already (no need to provide it from the outside). You just have to wait long enough that molecules with particularly high kinetic energy collide. That means that typically the vast majority of collisions does not lead to a reaction.
In general, the picture of a chain reaction that just needs to be started is incorrect as well. Two reactants colliding don't "know" whether products have been formed elsewhere. The only thing they "know" is whether there is sufficient energy available to reconfigure the nuclei and electrons from reactants to products, i.e. cross the activation barrier.

[OP] I am asking this because it is always taught that exothermic reactions release heat.

That is also not quite correct. Exothermic reactions release heat or do work (or some combination). If you make sure that they don't do work (i.e. no electrochemical reaction, no mechanical work), then it is correct to predict a release of heat (or warming of the reaction mixture, which is a heat transfer to the solvent if there is a solvent). If you have a gas phase reaction in an insulated vessel, there is no transfer of heat - the thermal energy of the reaction mixture will simply increase.
So if you want to know whether a reaction is exothermic or not and measure the reaction enthalpy, you have to make sure there is no (non-PV) work, that the reaction mixtures returns to the same temperature, and that you "catch" all the heat the reaction gives off.
Examples of descriptions of activation energy that don't hit the mark

This energy is called activation energy. For example, activation energy is needed to start a car engine. Turning the key causes a spark that activates the burning of gasoline in the engine. The combustion of gas won't occur without the spark of energy to begin the reaction.

The reaction requires a certain temperature of the reaction mixture, and it requires reactants to be in the gas phase. The mechanism of gasoline combustion is complex and probably involves radical formation and radical chain reactions. The chain reaction steps, while probably faster the the radical formation steps, also have an activation barrier. That means that even when the engine is running, the molecules that react still need to get their activation energy, and there might be collisions that do not result in a reaction (even one as explosive as the combustion of gasoline in the cylinder of a engine).

Reactions require an input of energy to initiate the reaction; this is called the activation energy (EA). Activation energy is the amount of energy required to reach the transition state. The source of the activation energy needed to push reactions forward is typically heat energy from the surroundings.

"Heat energy" is not a technical term. It is either thermal energy or heat transfer. The source of activation energy is the thermal energy of reactants coming together. The activation energy is not needed to push reactions forward (the direction of the reaction is not governed but the height of the activation barrier).

In chemistry and physics, activation energy is the minimum amount of energy needed to start a chemical reaction. Reactants often get activation energy from heat, but sometimes energy comes from light or energy released by other chemical reactions. For spontaneous reactions, the ambient temperature supplies enough energy to achieve the activation energy.

Again, there is a misconception about the start of the reaction. Every single molecule that reacts needs activation energy, not only the first couple.

The activation energy of a chemical reaction is kind of like that “hump” you have to get over to get yourself out of bed. Even energy-releasing (exergonic) reactions require some amount of energy input to get going, before they can proceed with their energy-releasing steps. This initial energy input, which is later paid back as the reaction proceeds, is called the activation energy and is abbreviated $E_A$.

This is almost not incorrect. When the reaction has passed the transition state, it is almost done. So this is different than getting our of bed compared to the rest of the day. Also, getting out of bed usually is downhill, sitting up in bed is what requires the center of mass to be lifted.

Interestingly, IUPAC has disconnected their definitions of endothermic and exothermic from the direction of heat flow:   https://chemistry.stackexchange.com/questions/122102/why-are-iupacs-definitions-of-exo-and-endothermic-disconnected-from-the-direct

@Karsten Theis  "What happens in general is that the reaction mixture has sufficient energy already (no need to provide it from the outside). You just have to wait long enough that molecules with particularly high kinetic energy collide. That means that typically the vast majority of collisions does not lead to a reaction" this paragraph of your answer says that exothermic reactions already have enough energy but it takes time to get activated . So I concluded that exothermic reactions do not absorb energy . Am I correct?


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What is the order of basicity of amines?
My textbook claims that -

Greater is the stability of the substituted ammonium cation, stronger should be the corresponding amine as a base. Thus, the order of basicity of aliphatic amines should be: primary > secondary > tertiary.

Now I have a problem with the first line.

Greater is the stability of the substituted ammonium cation, stronger should be the corresponding amine as a base.

The substituted ammonium cation is a conjugate acid (because its lost a proton/$H^+$). So if the conj. acid is more stable then the corresponding amine should be more acidic and hence less basic. So is the textbook wrong or am I missing something here?
Citation: Page 399, Chapter 13, NCERT Chemistry Part 2

"Greater is the X, stronger is the Y" is grammatically incorrect. You should write "The greater the stability of the ammonium cation, the stronger the corresponding amine is as a base"; or "The more stable an ammonium cation is, the more basic the corresponding amine is"; or even just "A more stable ammonium cation leads to a more basic amine".

Thats exactly what's written in the textbook. I literally copy-pasted the text. Besides, your comment does not answer my Q.

Comments aren't supposed to answer questions. I didn't attempt to answer. Your book is at least factually correct, though. Surely you are familiar with conjugate bases of weaker acids being more stable. The same is true here, except that base and acid are swapped.

@orthocresol I thought this is a place where ppl ask questions and get their doubts clarified. I know its not a homework solution website but asking doubts and seeking for a solution should not be looked down upon. And the irony about your comment is that you said that my textbook has wrong grammar, but you essentially just copy-pasted those lines and changed a couple of words (which I dont think made any difference).

@everyone
All I want is clarification. What makes an primary amine more basic than secondary and tertiary amine. What does that first line mean? I am most definitely sure that I am misinterpreting something here. And I've asked for help regarding that in the Q as well. I dont see a reason why someone needs the entire textbook to tell what the meaning of one line is. But whatever, I still gave it. Now please. Answer my question. I rlly need to clarify this doubt. Thanks very much to whoever who acknowledged and accepted my request.

(1) Nobody here is looking down on you. I made it a point in my first comment to give alternative suggestions, instead of just criticising the wording. If you didn't like me saying "you should write", then replace it with "your book should have written", which I admit would be more accurate. (2) Yes, I only changed a couple of words; but a couple of words can and will make a noticeable difference to someone who is fluent in English. Sometimes a single punctuation mark makes a difference. (3) My second comment in fact touched on the actual chemistry at hand; but you ignored it.

@orthocresol With reference to (3). Sorry about that. I didnt read it. Can you just elaborate a bit on that. Like Im just having a lot of confusion with conjugate acids/bases and how they imply about acidic/basic strength of a base/acid. Thank you.

@HarshDarji I am sorry if you are taking my comment in other ways but (1)you are coming across as someone desperate for an answer. Answers take time and you have to be patient. You can't expect an answer after 1 hour of posting. Sometimes answers takes days or even weeks to answer depending on the level of question. (2) It's good to correct grammatical error in the questions as it is part of editing. If there is an grammatical error in the textbook quote, just mention and just get done with it. Don't need to extend the conversation (3) While commenting, keep in mind the tone you're using.

I would urge the moderators (and OP) to clean up the comment section as it has nothing to do with the question.

The sum of the $\mathrm{p}K_\mathrm{b}$ value of a base and the $\mathrm{p}K_\mathrm{a}$ value of its conjugate acid is a constant (at a constant temperature), and is equal to $\mathrm{p}K_\mathrm{w}$ (which is $14$ at room temperature). The more stable the conjugate acid, the higher the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid $\implies$ lower $\mathrm{p}K_\mathrm{b}$ of the amine $\implies$ more basic amine.
(For those who are reading this without reading the textbook, the quoted lines are part of an explanation of the order of basicity of amines; obviously the order of basicity of aliphatic amines does not follow the order primary > secondary > tertiary, as the quoted lines seem to imply out of context)

But doesnt high stability of conjugate acid imply high acidic nature of the corresponding compound? And hence low basic nature?

If by "corresponding compound" you are talking about the base, then no; high stability of the conjugate acid does not imply high acidic nature of the base.

So weaker conjugate acid (ie. more stable substituted amm. cation) => its a strong base. Which means its not very stable. Right?

You should be more specific as to what exactly you are referring to by "its"... do you want to talk in stackexchange chat? It might be easier to clear up your confusion there instead of leaving multiple comments..

"Its" = strong base


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Usage of AC to measure the conductivity of ionic solutions
there was this statement in my chemistry textbook....

However, for measuring the resistance
of an ionic solution we face two problems. Firstly, passing direct current
(DC) changes the composition of the solution. Secondly, a solution cannot
be connected to the bridge like a metallic wire or another solid conductor.
The first difficulty is resolved by using an alternating current (AC) source
of power<

I don't understand how using AC over Dc would help, I mean electric current would pass ( although less than it would have otherwise) but still it could change the composition of a solution by electrolyzing it. So why AC?

Aside of electrolyte changes, electrode processes caused by DC significantly affect the potential difference applied to the solution itself and therefor the passing current.

The voltage applied is very small and the current passed is hardly enough to do more than rearrange the ions at the surface of the electrodes. The electrode double layers act more like capacitors, and the conductivity (inverse of resistance) is controlled by the movement of ions thru the electrolyte. Whatever electrode process happens during one half cycle (almost nothing) is completely reversed during the next half cycle. If you beef up the voltage (and current) and slow down the frequency, you could get some change in composition (like gas evolution).

Since AC can be capacitively coupled to the solution, metallic electrodes don't  need to contact the solution -- they can be outside an insulating  container, e.g., glass or PTFE, avoiding introducing metal ions.
At moderate AC frequencies, ions won't migrate an appreciable distance in one direction before returning to that position on the opposite half of the cycle. Steam vaporizers running at 50 or 60 Hz, using electrodes immersed in salt water, do not produce hydrogen and chlorine, just hot water and steam, for example.

Thank you, didn't think about capacitative tendencies in the solution

@PriyanshuChoubey If this answer helped you, make sure to accept the answer by clicking the checkmark.


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Caffeine molecule representation and orientation
Almost all the representations of the caffeine molecule that I've come across have the two rings drawn next to each other on the horizontal axis.
Any particular reason for this?
Would it be sacrilege to draw the caffeine molecule (or for that matter, any molecule), oriented in a way that makes it more aesthetically pleasing or maybe even funny? For example, turn the caffeine molecule 90° to make it look like a penguin?
Person with very basic chemistry background here. Wondering if there are IUPAC rules or conventions about structure orientation.

Wondering if there are IUPAC rules or conventions about structure orientation.

There are recommendations, which aren't really rules per se (in fact even the IUPAC nomenclature books are all technically recommendations). For "how to draw structures", the appropriate recommendations are:
Brecher, J. Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008). Pure Appl. Chem. 2008, 80 (2), 277–410. DOI: 10.1351/pac200880020277.
which is free to read online. The relevant section is GR-3.4.2, which states (emphasis mine)

For ring systems containing at least two fused rings, that system should be presented in its preferred orientation. For most practical purposes, it is sufficient to orient the ring system so that:

The fused ring system is oriented more or less horizontally. A symmetry axis (or plane) should be vertical or horizontal.
If the system contains multiple “rows” of rings, any “extra” rings should be positioned so that they are toward the right and the top of the ring system.
Larger rings should be toward the left and smaller rings toward the right.
Heteroatoms should be toward the right and, with slight preference, toward the bottom.

So, generally, according to these specifications, caffeine would usually be drawn with both rings lined up "in a row". Of course, nothing is stopping you from drawing it differently if you wanted to do so. However, it would almost certainly look out of place in formal scientific writing.
[NB: Some people like to think that IUPAC are being a pain in the back when they make such recommendations. The truth is that by far most of these recommendations already exist as "common sense" or "conventions" to chemists, and they already obey most of them subconsciously. Furthermore, these recommendations aren't enforced by anybody, anywhere, so deviations are always possible (and do occur regularly).]

+1 Follow IUPAC, unless you have a good reason not to. And I like rules that contain the words "more or less". :D

+1 It's also worth mentioning that Brecher has been heavily involved in ChemDraw, so the recommendations also come from real-world experience in trying to make "nice looking" diagrams.


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Solubility of multiple solutes and limits of solubility during freezing
I'm interested in modeling the freezing of solutions of water and I'm not really sure what happens in water after the solubility of a given material is reached. I know that as ice crystals form they form into pure crystals of water so during freezing solute concentration of the liquid phase increases. I also know that the solubility of sugar is around 80g/100g and I know that the solubility of salt is about 36/100g (room temperature assumed, I realize solubility might change with tempature but I'm going to ignore that).
My understanding of these terms is such that if I add 81g of sugar to 100g of water, I should see about 1g of sugar participate out or never dissolve. Likewise I imagine that as I froze a solution of sugar and water, after enough of the water had frozen the sugar would start to precipitate out once the ratio of sugar to water had reached 80 to 100 (assuming temp didn't affect solubility). Please correct my understanding if this part is wrong.
What I don't think I know is what would happen if I started with a solution of sugar AND salt. What would the relative concentration of the liquid phase be (or equivalently, what would the ratio of the precipitates be?). How does adding in a miscible fluid like alcohol change this?
If it helps to give a more specific answer I'm primarily only interested in the following solutes:
ethanol, mono and disaccharides (e.g. sugars), glucose polymers, and salts such as sodium and potassium chloride.

If they are ideal solutions (which many are not), it is "easy". The less ideal, the harder it gets. Modeling ternary phase diagrams is hard.

Is there somewhere that I can get existing data/phase diagrams then? Like a book containing a ton of empirical data or something?

BTW, this separation is rate-dependent. Fresh sea ice contains a lot of salt, for example, and the Czochralski crystallization technique is used to purify semiconductors using a s l o w l y traveling region of melting and solidification.

Does this answer your question? Saturation of multiple solutes


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Exceptions to the molar mass trend in London dispersion forces in molecules: do they exist?
"In general, larger molecules have higher boiling points than smaller molecules of the same kind, indicating that dispersion forces increase with mass, number of electrons, number of atoms or some combination thereof. "(https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/physprop.htm#:~:text=In%20general%2C%20larger%20molecules%20have,atoms%20or%20some%20combination%20thereof.)
My understanding of the statement above is that larger molecules (here I want to talk more specifically about larger molecules which are also molecules with larger molar masses than other molecules it's compared with) are likely to initiate more intermolecular interactions between other exact same molecules, creating stronger intermolecular interactions.
Are there any trends of boiling points of molecules not following the tendency in London dispersion forces that molecules with larger molar masses generally have higher boiling points than molecules with smaller molar masses? Please give examples if there are any. Thanks in advance.

"likely to initiate more intermolecular interactions" -- No, there aren't more interactions; the interactions are just stronger. As for exceptions, water is probably the obvious one.

Recommend you just study more chemistry. The exceptions make up the interesting part.


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Name of particles in a suspension
What is the proper name given to the particles in a suspension? Suspensae perhaps? When these particles are brought out of suspension, and precipitate down to the bottom of the mixture, what is that process called, and do they take on a different proper name when no longer in suspension?

I don't know if there is a general term for suspended particles, but solid and liquid particles suspended in the atmosphere are known as particulates:  https://en.wikipedia.org/wiki/Particulates .
The process of particulates coming out of a suspension is known as settling. That distinguishes suspensions from a colloids: Those in a suspension settle out with time (due to gravity), while those in a colloid do not.

Do generic precipitate or floc (https://en.wikipedia.org/wiki/Flocculation) apply?

IUPAC's Gold Book states for suspension only «A liquid in which solid particles are dispersed.»

+1 for "suspensae", quite ingenuitive. There is also obsolete "suspensoid", but it refers to the system, not just particles.

@andselisk Even more so that there is the adjective (suspensus, suspensa, suspensum) and verb (suspendere) in Latin, though maybe no suspensa as noun in the a-declension.  (Изучение латинского языка долго было фундамент всех изучений.)

“Solid particles” is arguably the better term on its own, especially when size distribution is provided alongside with it.
Alternatively, there is “dispersed phase” and its synonym “discontinuous phase” [1, p. 41]:

discontinuous phase In a dispersion or emulsion, the phase which is dispersed as particles or droplets in the continuous phase.

Note, however, that “dispersed phase” is a sloppy term as its applicability relies on properties match between bulk and dispersed phases and it doesn't necessarily refer to a suspension and may equally be used for particles or droplets [2, pp. 605–606]:

The name dispersed phase for the particles should be used only if they have essentially the properties of a bulk phase of the same composition.
[…]
In a suspension solid particles are dispersed in a liquid; a colloidal suspension is one in which the size of the particles lies in the colloidal range.

As for the case when “particles are brought out of suspension”, “precipitation” is not the proper term as it is applicable for solutions, not sols or suspensions [3, p. 260]:

precipitation
Formation of a solid material (a precipitate) from a liquid solution in which the material is present in amounts greater than its solubility in the liquid.

“Agglomeration” and “aggregation” are more correct, yet, strictly speaking, non-interchangeable terms [1, p. 5]:

agglomerate n A collection of primary particles or aggregates joined at their edges or corners, in such a way that the specific surface area is not markedly different from the sum of the areas of the constituent particles. —v the act of agglomeration.
aggregate n A collection of primary particles joined at their faces, with a specific surface area significantly less than the sum of the areas of the constituent particles. Aggregates are distinguished from agglomerates by the increased difficulty in separation. —v the act of aggregation.

Keep in mind that a colloidal particle itself could be an aggregate [2, p. 609], and there are several congregating processes for denoting particles that are “no longer in suspension”, such as flocculation/coagulation (for colloidal suspensions), sedimentation and sintering [2, p. 609–610]:

When a sol is colloidally unstable (i.e. the rate of aggregation is not negligible) the formation of aggregates is called coagulation or flocculation.
These terms are often used interchangeably, but some authors prefer to introduce
a distinction between coagulation, implying the formation of compact aggregates, leading to the macroscopic separation of a coagulum; and flocculation, implying the formation of a loose or open network which may or may not separate macroscopically.
In many contexts the loose structure formed in this way is called a floc.
While this distinction has certain advantages, in view of the more general (but not universal) acceptance of the equivalence of the words coagulation and ﬂocculation, any author who wishes to make a distinction between them should state so clearly in his publication.
[…]
Sedimentation is the settling of suspended particles under the action of gravity or a centrifugal field. If the concentration of particles is high and interparticle forces are strong enough, the process of sedimentation may be better described as compaction of the particle structure with pressing out of the liquid. This particular kind of settling is also called subsidence.
Sediment is the highly concentrated suspension which may be formed by the sedimentation of a dilute suspension.
Coalescence is the disappearance of the boundary between two particles (usually droplets or bubbles) in contact, or between one of these and a bulk phase followed by changes of shape leading to a reduction of the total surface area.
[…]
Coalescence of solid particles is called sintering.

References

Becher, P. F. Dictionary of Colloid and Surface Science, 2019 reprint by by CRC Press, Taylor&Francis Group.; M. Dekker: New York, 1990. ISBN 978-0-8247-8326-6.
Everett, D. H. Manual of Symbols and Terminology for Physicochemical Quantities and Units, Appendix II: Definitions, Terminology and Symbols in Colloid and Surface Chemistry. Pure and Applied Chemistry 1972, 31 (4), 577–638. DOI: 10.1351/pac197231040577.
Gamsjäger, H.; Lorimer, J. W.; Scharlin, P.; Shaw, D. G. Glossary of Terms Related to Solubility (IUPAC Recommendations 2008). Pure and Applied Chemistry 2008, 80 (2), 233–276. DOI: 10.1351/pac200880020233.

A colloidal suspension or a even a suspension consists of two phases: A dispersed phase and a dispersing medium. See this Table as an example:

Table 3.3. Combinations of two phases in colloidal systems.

Dispersed phase
Dispersion medium
Example

Solid
Solid
Alloys

Solid
Liquid
Starch in water

Solid
Gas
Dust in air

Liquid
Solid
Butter and cheese

Liquid
Liquid
Oil in water

Liquid
Gas
Clouds

Gas
Solid
Pumice stone

Gas
Liquid
Froth on beer

Ref: Biophysics, Principles and Techniques by M Subramanian, 2019
If the suspension settles, it is a floc or an agglomerate.
Please note that term dispersed phase is not sloppy usage. It is still widely used in high quality research journals. IUPAC recommends that
https://goldbook.iupac.org/terms/view/C01174

The name dispersed phase for the particles should be used only if they
have essentially the properties of a bulk phase of the same
composition.

Remember that definitions of chemical terms are human explanations, which are neither perfect nor rigid (to the extent of being "brittle").

I probably should've pointed out that "dispersed phase" is a sloppy term for what OP is asking about, not sloppy per se. Sorry if this wasn't evident from the context. I'm sure these high quality research journals are able to use IUPAC definitions correctly and precisely. Although I'm not ready to agree with "widely used" as a pro argument — this "billion flies can't be wrong" level of persuasion always puts me off.


text:
When should the Steady State Approximation be used?
In class, I learned that we should use the steady state approximation when there are multiple slow steps or if the rate determining step is not known (shown on this Chem Libre link).
However, further research (another Chem Libre link) shows that the steady state approximation is only valid when the second step is much faster than the first step, causing the intermediate to be used up immediately.
But isn't this just the same as assuming that the first step is the rate determining step? Why use the steady state approximation at all if it is only valid in cases where we could just use the more simple procedure arising from assuming the first step is rate determining?

Why not to mathematically experiment with  sequential reactions and different ratios of their kinetic constant? You can then make your own conclusions. Ideal scenario is choosing of radioactive decay series.

@Poutnik assuming you do not calculate on an 12 kg sphere of 239-Pu... A less steady state is hard to find =P

@StianYttervik I have not said the steady state is found in all cases. :-P BTW, 12 kg 239Pu is outside of spontaneous decays series :-)

You can have a fast first step that is an equilibrium reaction and the a slow second step. This is described in the second cited textbook portion further down under enzyme catalysis.


text:
Can a reaction produce more products than reactants?

Q:What volume of propane is burnt for every $100 \mathrm{~cm}^{3}$ of oxygen in the reaction?
$$
\mathrm{C}_{3} \mathrm{H}_{8}+5 \mathrm{O}_{2} \rightarrow 3 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O}
$$
(Gas volumes measured under the same conditions).

My solution:
For every 5 volumes of oxygen, 1 volume of propane is burnt. Therefore,
Volume of propane bumt for every $100 \mathrm{~cm}^{3}$ of oxygen $=\frac{1}{5} \times 100=20 \mathrm{~cm}^{3}$.
If we take the total composition, the volumes of reactant are in 1:5 ratio which produced products in ratio 3:4. In other words, the volumes of reactants are $\pu{20 cm^3}$ and $\pu{100 cm^3}$ which produced $\ce{60 cm^3}$ and $\ce{80cm^3}$ products respectively.
Sum of volume of reactants = $\pu{120cm^3}$
Sum of volume of products = $\pu{140cm^3}$
How are products produced more than reactants ?

Ever heard of decomposition reactions? One dollar decomposes into four quarters. How is it possible, when $4>1$? Well, just like that.

@IvanNeretin K. I understand that part. Like we can have two parts of CH4 = CO2 + H2. But Volume means like the amount of space a substance takes. Volume of 140 is taking more space than volume of 120.

Those are gases. A gas mostly consists of empty space anyway.

@IvanNeretin K. I understand that. Thank you. Also , about how it gets broken into parts. But still , how can the products be more than reactants. In any reaction. Or even less sometimes. There must be sth from where they either gained sth extra or lost. It’s a clear observation to any person doing these type of Q. How do I understand this part.

@IvanNeretin Even though gas takes v little space but it can’t violate law of conservation of energy.

Sure. The products can't have more energy than the reactants. But they can have more volume. There is no conservation law for that. Volume is not energy. They don't even start from the same letter.

It seems like you're wondering how the products can have a greater volume than the reactants without violating conservation of mass. So: In a chemical rxn, the number of atoms is conserved. That preserves conservation of mass. But, as you can see from your balanced chemical equation, the number of molecules isn't conserved (diff # on LHS vs. RHS). And the volume of a gas depends on the number of particles, which means the number of molecules, not the number of atoms (unless the gas is monatomic).  So if there are more gas molecules in the products than the reactants, the volume will increase.

@IvanNeretin Ok , So you mean mass of reactants = mass of products. Volume can be different. I agree. Just like how a tall person & healthy can take different volume or space in air but can have same mass. Right ?

@theorist K. Atoms of LHS = RHS but molecules are not.

@S.M.T Correct.

@theorist Would u like to write the answer so I can accept the Q to close it or shall I do it.

@IvanNeretin Would you like to write the answer sir or shall I write it.

Write it yourself, this is acceptable and even encouraged.

Sure, I can write an answer.

@theorist K sir. Even I can do it. Np in that. Just that it is your help , knowledge. I have written 1 answer regarding mass. You can write about the atoms point which u pointed our or as u feel. Shall I call you sir or ma’am ?

@IvanNeretin Now I'm considering maltreating a dollar banknote with a hammer until it finally decomposes into four quarters ^^'

It seems like you're wondering how the products can have a greater volume than the reactants without violating conservation of mass.
So: In a chemical reaction, the number of atoms is conserved. That preserves conservation of mass. But, as you can see from your balanced chemical equation, the number of molecules isn't conserved: There are more molecules on the RHS than the LHS. And the volume of a gas depends on the number of particles, which means the number of molecules, not the number of atoms (unless the gas is monatomic).
This can be seen directly from the ideal gas law:
$$V = \frac{n R T}{p}$$ which means that, at constant $T$ and $p$, $V$ is proportional to $n$, where $n$ is the number of particles.
So if there are more gas molecules in the products than the reactants, the volume will increase.

I had a Q. In most of the Q’s I solved regarding mole concept , H2O is always taken as water(liquid) & not steam H2O. In certain Q , reactions are not given. Is there a condition when I have to assume H2O to me steam or liquid ?

*me = be in last comment.

I was assuming all your products and reactants were gaseous based on the volume calculations you gave.  Obviously if the water were liquid, the volume of the products would be less than that of the reactants.   To know for sure the phases of the reactants and products, you need the person that supplies the equation to either specify all the phases, or specify the temperature at which the reaction occurs.

Have you ever heard of explosions? Explosions are chemical reactions of solid or liquid substances which produce a huge amount of gas in less than one second. For example, nitroglycerin is a liquid with the formula $\ce{C3H5N3O9}$, which makes dynamite. One mole of nitroglycerin weighs $227$ g and has a volume $142$ mL. The explosion corresponds to an equation like $$\ce{2 C3H5N3O9 -> 5 CO +  CO2 + 5 H2O + 6 NO}$$ There are doubts about the relative amounts of  $\ce{CO}$ and $\ce{CO2}$ as a function of the ratio $\ce{NO}$ and $\ce{NO2}$. Anyway this is not the point here. The main point here is the fact that $2$ moles nitroglycerine ($0.284$ L) produces $6 + 3 + 3 + 3 = 15$ moles gas, which occupies, at $0$°C and $1$ atm, a volume of $\pu{15 · 22.4 L = 336 L}$.
$0.285$ litres of a liquid produces a gas volume more than $1000$ times bigger, during the explosion.
$\pu{1 cm^3}$  liquid produces $\pu{1183 cm^3}$ gas. But the mass is conserved.


text:
Aldol Cyclization Cascade Reaction between Two Ketones in Basic Alcoholic Solution
My professor gave a hint that the reaction scheme is a multiple carbonyl reaction which involves cyclization. He emphasizes that the reactivity of carbonyl compound in terms of C-H acidity and carbonyl reactivity will not be obeyed if cyclization is involved.

My initial guess is the formation of a vinylogous carbonyl compound for B by the elimination of one methoxy group. Subsequently, it react with the β-dicarbonyl compound, C. My another guess is that C reacts as methylene component either at the α-carbon or at the methyl group and attacks B by vinylogous substitution with the elimination of second methoxy group.
Next, intramolecular aldol reaction and cyclization take place. However, A (C12H16O4) and C (C10H10O2) react and give a compound with 20 carbon atoms by the elimination of two methoxy groups.
The carbonyl and methylene components are played around but I am very inclined that B is a vinylogous carbonyl compound and sp2-carbon will not undergo further deprotonation. However, I could not get a compound with 19 carbon atoms. My desperate guess is the formation of ester group with the oxygen atom acting as the O-anion instead of C-anion from the enolate during cyclization. Basic hydrolysis of ester will then give COOH group, yet methyl ester does not exist. Another guess is retro-aldol reaction which seems not possible...
Any suggestion for the reaction scheme is really appreciated. Thank you.
(No reaction condition is specified other than basic alcoholic solution)

Which is the most acidic site in C? 1,3 dicarbonyl species usually react through the methylene group between the two carbonyls. Draw this product and then consider that you have in the product another site alpha to a carbonyl group that can do a 2nd aldol condensation.

I am inclined to think that there is a typo in the molecular formula of E

After confirming with my professor, he noted that there is a mistake in the molecular formula. E should be C20H18O4. To whom it may concern, sorry for taking your time in considering the problem.


text:
Saturation of air within isolated system with water and ethanol
I am planning an experiment in an isolated system that contains two droplets.
One droplet is water and the other one is a 40% ethanol-water mixture.
The isolated compartment is filled with air at 20°C.
I understand that for ensuring that the water droplet volume remains constant during the course of the experiment, the air in the compartment needs to be fully saturated according to the vapor pressure of water at 20°C (100% relative humidity).
Under these conditions, my understanding is that the ethanol from the 40% ethanol-water droplet is not yet in equilibrium with respect to the evaporation/condensation rate in the fully water-saturated air, thus decreasing the droplet volume. Is that correct?
If it is correct, that would mean that I would also have to saturate the air with ethanol before the experiment. I can find the vapor pressure of ethanol for 20°C. But how would I calculate the resulting concentration of ethanol in the air in kg/m^3 or ppm?
Also, how would the total air pressure within the compartment be influenced by the saturation?

Why would you need that? It is not like you can buy yourself some air with such-and-such concentration of ethanol, really.

There would evaporation of ethanol + condensation of water at 40% ethanol drop AND evaporation of water and condensation of ethanol at the pure water drop. In the and, there would be the same concentration of ethanol in both drops of unsure volumes, in equilibrium with water/ethanol vapor partial pressure ratio.

You can use formal equality of ethanol vapor concentration and pure vapor density: $$[\ce{C2H5OH}]=\rho_{\ce{C2H5OH}}=\frac {p_{\ce{C2H5OH}}M_{\ce{C2H5OH}}} {RT},$$ assuming ideal vapor behavior.

@IvanNeretin: To ensure that the droplet volume remains constant over the course of the experiment. The idea of course is not to buy the air, but to ensure that the equilibirum according to the vapor pressure is reached before the start of the experiment

@Poutnik: That sounds logical. I did not think of this. Thank you!

Note that for an isolated system, the final temperature is unknown, due ethanol+water evaporation and condensation.

I'll put it differently: where are you going to get the air with known concentration of ethanol?

I don't think there is any combination of ethanol partial pressure and water partial pressure such that the gas phase would be in equilibrium with both droplets.


text:
Why does increasing electrolyte concentration increase the rate of electrolysis?
I'm a Chemistry teacher and have students who are performing an experiment to determine the effect of copper ion concentration on the rate of solid copper formation at the cathode in an electrolytic cell.
I've had students do this experiment in the past and they invariably find that an increase in concentration results in an increase in the rate of copper formation. At first glance, I assumed the theory underpinning this experiment could be explained quite simply using collision theory...
Put simply, an increase in copper ion concentration means that there is an increased number of copper ions at the surface of the cathode. This, in turn, results in copper ions being reduced to solid copper at an increased rate due to a larger number of successful collisions between copper ions and electrons per unit time.
After doing some reading online, however, I can't seem to find a clear-cut description of why increasing electrolyte concentration increases the rate of electrolysis. Is the relationship not as straightforward as I think it is? If not, how could I explain it without overwhelming my students (remembering that they're school students, not PhD students).

There are electrode reactions controlled by electron transfer(slow ones) or by diffusion(fast ones).
Depending on choice of forced electrode potentials, electrolyzer geometry and ion concentration, many reactions can be arranged to be electron-transfer limited or diffusion limited.
If the cathode potential is decreased below its equilibrium potential, the cathode current density starts to exponentially grow (the start of the polarographic step).
With the further decreasing of the cathode potential, the current density does not grow infinitely, but starts to be limited by cation electro-migration, with cations starting to be depleted at the electrode surface.(the end of the polarographic step).
Finally, further decrease of potential leads to the electrode current plateau unless there is yet other electrochemical system able to be electro-reduced.
Electro-deposition of copper is charge-tranfer controlled at low current densities and diffusion controlled at high current densities.
There is direct relation between ion concentration and the current density limit, caused by ion diffusion. In the limiting state, all ions arriving at the cathode due electro-migration, are immediately consumed by the reaction.
There is electrode adjacent electrolyte transition layer where cation concentration of bulk electrolyte starts to drop toward zero when approaching the electrode.

At equilibrium, the cathode reduction and oxidation currents cancel each other to the zero net current.
Note that in context of electrode processes, more important than absolute currents are current densities.
When the electrode potential lower than the equilibrium one is externally applied, the cathode increases reduction current and decreases oxidation current. This decreases the inner cathode potential until the potential difference across the wire causes the current equal to the net redox electrode current.
On the electrolyte side, the net redox current causes cation (or generally reduced object ) depletion, until the extra electrostatic gradient caused by ion displacement forms electromigration current equal to the cathode net redox current.
If the forced potential is too low and the redox net current too high, even almost total depletion is not able to cause high enough electro-migration current. As consequence, the initial net redox current is not sustainable and decreases to match the maximum migration current the electrolyte is able to provide. This is concentration dependent. The wire current adapts to the sustainable net redox current by the inner electrode potential shift toward the external forced potential.

Thanks for your answer. Out of interest, if a reaction is controlled by electron transfer, what does that mean for the effect of concentration on the rate?

Assuming anode and cathode are Cu, the number of Cu atoms oxidized equates the number of Cu ions reduced.  Because as such the solution remains neutral.  But the more charge carriers in the solution, the higher the rate until either rate of oxidation or reduction limit the overall rate (depends on the electrodes' surface, the applied electrical potential, etc.).

Effect of concentration on electron transfer controlled reaction is small. It is like 2 reactions in a serie: diffusion + electron transfer. When one is much slower than the other, it does not matter much how fast is the fast one.

Right. So I'm still a little bit confused here. Can the increase in deposition rate with increasing concentration (for copper electrolysis) be attributed to there being more copper ions at the electrode (i.e., higher frequency of collisions)? Is an increased rate of diffusion to the cathode (from the bulk solution) also responsible? I'm assuming here that limiting current density hasn't been reached.

Surface concentration of copper ions is diffusion controlled. Electrode reaction cannot go faster than is the ion arrival rate. you have also to distinguish the momentary rate and a steady rate. The former can be much  faster than the latter.

Thanks again. I'm going to simplify slightly here, but I'd be grateful if you could tell me whether this interpretation of what you've said is correct... Because the rate of product formation is much faster than the rate of diffusion to the electrode surface, increasing the concentration increases the concentration gradient. This means that the ions diffuse to the electrode surface at a faster rate, which means the rate of deposition increases. Thus, it's not the concentration of ions at the electrode surface that changes with concentration, but the rate at which they move there. Sound right?

Rather then diffusion due concentration gradient, it is migration due electrostatic gradient. Continuous removal of cations causes charge disbalance.

Right. And increasing the concentration of cations in solution effectively increases the charge in the bulk solution, which causes a greater electrostatic gradient?

Let us continue this discussion in chat.

The rate of electrolysis is the amount of copper $n\ce{(Cu})$ deposited at the cathode per second (in mol/s). This amount $n\ce{(Cu})$ is proportional to the time $t$ and to the current $I$ (in Amperes) according to the Faraday's law : $n = It/zF$. Now the current $I$ depends on the voltage $U$ according to Ohm's law : $I = U/R $ where $R$ is the resistance of the solution. This resistance $R$ is such that its inverse $1/R$, called conductance $G$, is equal to the conductivity $\sigma$ times the cell constant $K$ , so that $1/R = \sigma K$, where $K$ is an experimental geometrical constant depending on the distance between the electrodes. Then the conductivity $\sigma$ is equal to the concentration $c\times \Lambda $, where $\Lambda$ is a (nearly) constant for all $\ce{CuSO4}$ solution : $\Lambda$ ($\ce{CuSO4})$ =  $55+80 = 135$$ ~ \pu{S cm^2 mol^{-1}}$. Combining all these formula leads to $$n_{Cu}(t) = \frac{It}{zF} = \frac{U}{R}\frac{t}{zF} = U\times \sigma K~ \times \frac{t}{zF} = U\times c\times \Lambda \times K~ \times \frac{t}{zF} = \frac{135 U K t}{zF}\times c$$ or : $$I = 135 ~UKc$$ As a consequence, if the voltage $U$ is maintained constant in a given cell, the current $I$ is proportional to the concentration $c$ of the solution.

It seems to me that you assume I is proportional to U and $\sigma$ what is obviously false. At least, and just as the first approximation,  there must be subtracted voltage extrapolated for zero current.

@Poutnik. What do you have against Ohm's law: U = RI ?

Applying it on electrolytic cell is kind of stretch. Better approximation is I = ( U - U0)/R, and still just a course approximation. If a cell is to be approximated by an electronic system, it would be quite a complex system, not reducible to a single resistor. U=RI is applicable for conductivity measurement by AC voltage of small amplitude.

@Poutnik What is the value of $\pu{U_o}$ in an electrolysis of a $\ce{CuSO4}$ solution between two copper electrodes ?

Make the graph I=f(U) and U_0 will be the intersection of the graph tangent and axis X. Even if it was initially zero for shared electrolyte, it would soon becomes nonzero due migration gradients. Now, take the case of charging of the Li-on cell.

Copper ion concentration is a variable, of course. However, the rate of copper deposition is DIRECTLY related to the current being passed. Well, except for the possible complexities addressed below.
If increasing the copper ion concentration increases the rate of copper deposition, it is likely because the power supply is not regulated. The resistance of the solution drops, the voltage does not, more current flows, and the result is that more copper deposits.
In any experiment, all the variables ought to be held constant except the one (or two) that you wish to vary. In this experiment, copper ion concentration is one variable, temperature another, applied voltage, applied current, stirring, electrode area, and on and on. Many of these variables cause effects which are better explained by reference to more fundamental variables. The one variable most fundamental in explaining the rate of copper deposition is the rate of electron flow, which is measured by the current.
There is a situation where I believe this anomalous result could occur even if you hold the current constant. If the anode were an inert material, like a carbon rod or platinum, if you have some, then, over some time period, in the dilute copper sulfate solution (assuming sulfate), the initial current would be carried to the cathode solely by copper ions, because there are essentially no other cations (pH near neutral). However, after enough current has passed and deposited copper, the copper ion concentration drops and is replaced by H$^+$ ions. When these carry the current, they evolve H2 gas, and the rate of copper deposition decreases. Aha! But you claim that you see no (or almost no) gas bubbles at the cathode. OK, perhaps you have set the current so low that the H2 dissolves before bubbles can form and rise. H2 could then evolve from the surface slowly and escape without a trace.
There's another effect that can reduce the possibility of H2 evolution: suppose you have a magnetic stirrer in the vessel (not absolutely necessary, but it helps). The O2 evolved at the  anode might react with H2 in the solution, or at the electrodes (especially platinum, if you used that). This causes a reduction of evidence at the electrodes that electrolysis is occurring - an example of Faradaic (or coulombic) inefficiency. This would be totally unexpected (unless you need it to explain the results).
Now if you do the exact same experiment with a more concentrated copper sulfate solution, the generation of H$^+$ ions will be slower, so the deposition of copper will not be reduced so much during the same time of observation. Strictly speaking, it is not the higher concentration that has caused the change, but rather the extra amount of copper ion that supplies the extra copper metal. The exact same effect could be caused by incereasing the amount of dilute solution so that the copper ion is not depleted as rapidly as in the smaller  amount of solution.

Let's imagine that all variables are held constant except for the copper ion concentation and the current increases (which often happens). Isn't the increase in current DUE to the reaction occurring at a faster rate, and not the other way around?

"In an electrolytic cell a current is passed through the cell by an external voltage, causing an otherwise nonspontaneous chemical reaction to proceed. In a galvanic cell the progress of a spontaneous chemical reaction causes an electric current to flow." from Wikipedia. Vandelay specified an electrolytic cell. An applied voltage is  forcing the reaction. A decrease in solution resistance is what allows (or forces) more current if the voltage is held constant. The increase in current IS the reaction being forced (or allowed) to occur at a faster rate.

Thanks for your response. So the current is essentially just a manifestation/measure of the reaction rate? (That gels with my understanding, by the way, based on Faraday's law, but I've read some annoyingly contradictory things online.)

@Vandelay: I think you've got it. It can get complicated because you have cations going one way, anions going the other, electrons going in and coming out - but not individually crossing the electrolyte as in a wire. Not to mention the complexities of the electrode processes and the change in concentrations and double layers and diffusion, etc, etc. These can all be addressed, but the simplest measurements are voltage, current, pH, possible gas evolution, Cu deposit weight, and concentration. Measuring these should enable a good explanation of your results.


text:
Synthesis of sodium ferricyanide
I need to synthesize sodium ferricyanide, $\ce{Na3Fe(CN)6}$, for some benchtop experiments where I don't want any potassium ions in solution. The way I am doing that right now is oxidizing $\ce{Na4Fe(CN)6}$ using sodium permanganate, $\ce{NaMnO4}$, and filtering out the $\ce{MnO2}$ that forms. However, upon reviewing the SDS for sodium ferrocyanide, I found that strong oxidizing agents are an incompatible material (I am not sure why). Is there a better way to go about doing this synthesis?

How do you define a "better" way? What aspect of your present method is making you unsatisfied?

Usually this operation is made by bubbling gaseous chlorine $\ce{Cl2}$ into a sodium ferrocyanide solution.  You will obtain sodium ferricyanide plus sodium chloride, which usually does not disturb further reactions.

@orthocresol I edited the original post. The SDS mentions strong oxidizing agents are incompatible with sodium ferrocyanide.

@OscarLanzi you can buy $\ce{NaMnO4}$ commercially.

As @Maurice said, it can be made by bubbling gaseous chlorine into a sodium ferrocyanide solution. You will obtain sodium ferricyanide plus sodium chloride, which will not disturb further reactions.
$$\ce{2Na4[Fe(CN)6] + Cl2 -> 2Na3[Fe(CN)6] + 2NaCl}$$
There is also a patent which mentioned electrolytic oxidation of sodium ferrocyanide:
$$\ce{2Na4[Fe(CN)6] + 2H2O -> 2Na3[Fe(CN)6] + 2NaOH + H2}$$
You can find  a detailed description of the setup in the patent.

Your method can work but you have to find a way to separate the insoluble MnO2.

This seems like a good way to do it and I see that it's listed as a common reaction done with potassium ferrocyanide to form potassium ferricyanide. However $\ce{Cl_2}$ is a strong oxidizing agent, and those are supposed to be incompatible with $\ce{Na_3[Fe(CN)_6}]$ according to the SDS.

Is there a way to do this and know the exact concentration of NaCl formed? Because I suppose I won't know how much of $\ce{Cl_2}$ I passed reacted with the system?


text:
How to safely transfer P2O5 waste from a round bottom flask to waste bottle?
The $\ce{P2O5}$ is used as a neat desiccant in the round bottom flask as part of a vacuum system and is mainly solid with some liquification after use. Waiting for it all to liquify isn't practical, adding water directly seems dangerous, and trying to scoop it out is very difficult and dangerous for possible contact/exposure reasons.

Yes, indeed, dangerous. You should read the safety sheets before running a reaction, they usually contain the information for a proper disposal. And please tell us what RBF is.

The safety data sheet is at best vague and at worst useless, especially in specific cases like removal from a round bottom flask (RBF) where the substance is now in 2 phases and is mechanically difficult to manipulate out of the flask.

What type of chemistry is performed prior to the vacuum system? This may guide what to expect in this trap (no pun intended). Isn't there a colleague of yours you may contact who knows what is allowed in your local jurisdiction and instructed you in advance? As side note, https://organicchemistrydata.org compiled some resources to ease retrieving MSDS data here e.g., this, and this.

It depends on the amount of substance. If there is only a couple of grams, water can be slowly added, drop by drop. Stop for a while  if the substance becomes too hot. Two hours should be enough to get rid of this stuff.

Fot an amateur: stay away and give the task someone professional. For a chemist this is a standard case. Of course a couple of grams (like in the previous comments) can be safely disposed of with a heavy current of tapwater. For a few hundred grams leave the RBF within a humid area for a couple of days, best for weeks then drain with again a heavy stream of water. With more then a few hundred grams you should have a professional who knows the best way to dispose of. You should respect chemicals and no way do something you are not experienced with. Be careful.


text:
Incompatibility of Helmholtz Double Layer and Redox electrochemistry
In an electric double layer capacitor(EDLC) (or any situation with a Helmholtz double layer) what prevents the ions from being reduced or oxidized like the ions in an electrochemical cell?
If it is because the potential is not high enough, what energy barrier is preventing the reaction?
In the negative ion case, is it that even though the ion has a net negative charge and as a whole is attracted the the positive electrode the electron feels a stronger attraction the the nucleus than the the electrode?
Is this what determines the maximum voltage of a supercapacitor?
How does all of this relate to the inner and outer layers?
Sorry for so many questions in one post but I think they are all related and I hope that they provide context for my confusion.

Ok. I have searched and read pretty deeply. EDLC and Redox electrochemistry are covered for sure but I did not find any discussion of why they are not conflicting ways of describing what happens at the liquid solid interface of a conductor and an ionic solution. I could certainly have missed it. Maybe I didn't search the correct terms. If you find such a discussion please let me know I would be happy to be wrong and have missed an explanation.  Clearly I'm no expert hence the need to ask the question.

In hopes of helping others that have the same question, I think a have a partial answer. From https://www.sciencedirect.com/science/article/pii/B9780124095472111497
"Traditionally, electrochemical double-layer theory has been concerned with the so-called ideally polarizable interfaces, at which by definition no electrochemical reaction takes place and hence no direct current passes through the interface."
I still have not found a definition of "ideally polarizable interfaces" that helps me but I think the point is that any voltage applied if below the redox potential.
I hope this helps. I will keep working on it.

More definitions an ideally polarizable electrode has no Faradic current. "Faradaic current is the current generated by the reduction or oxidation of some chemical substance at an electrode. https://en.m.wikipedia.org/wiki/Faradaic_current


text:
What is the order of decreasing stability among following molecules?

I am almost certain that (IV) is most stable and (III) is least stable. How to compare between (I) and (II)? Is it something related to angle strain?
I am also confused if you can compare between compounds having different amounts of carbon molecules. (I) and (II) have 7 carbon molecules so they would have more bond energy. Is it possible that (I) and (II) could be more stable than (IV) because of this reason?

You are right, it is a flawed question. For a direct comparison, you need molecules with the same molecular formula like I and II.

@KarstenTheis Thanks! How to compare between I and II?

https://chemistry.stackexchange.com/a/31808/72973

Is the answer option c?


text:
Chemical equilibrium textbook question
One textbook question that I came across is:

How, the answer

Doesn't make sense to me because when you add 3M of B, that's the initial position and not the equilibrium position. When you add 3 moles of B, the equilibrium shifts to the right by Le Chatelier's Principle and so the concentrations of of A decreases and C increases, which makes sense with the last statement "A possible new equilibrium condition... because k=2". Is the answer incorrect, or am I missing something here?

How much A reacted, how much B, and how much C was formed?

The solution is humbug. It does not state any volume change and it is clearly not an equilibrium state.

Also proposing this solution means someone should not hire unqualified people for this kind of job (teaching), simply calculate the new euqilibrium condition given K = 2 from the given new condition to reach the desired values.

Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of at least essential parts.
See also formatting math/chemistry expressions/equations.

As Poutnik suggests, please transcribe text within images into typed text.

The volume is assumed constant (1 L). It is not specified whether the reaction occurs in the gas phase, that is largely irrelevant, and neither pressure nor temperature information is provided, that is also not relevant to the problem. The equilibrium constant is a concentration equilibrium constant.
The question does not require you to seek an algebraic  solution for the computation of the equilibrium concentrations under the new conditions, after addition of more B. The objective is that you should use your understanding of chemical equilibrium and in particular stoichiometry, as well as some intuition (or rather an understanding of the mathematical relationship between concentration and amount of substance under constant volume) to assess whether a proposed new equilibrium in which $\pu{[B]= 3 M}$ can be achieved from the new starting conditions.
When volume is constant the concentration $C_i$ and amount of substance $n_i$ are strictly linearly proportional: $$C_i = \frac{n_i}V = \text{constant} \times n_i$$ In the present problem this proportionality is particularly simple because $\pu{V= 1 L}$ and it is possible to think in terms of concentration or amount of substance equally well if you choose units of mol and molar (remembering that the quantities employ different units).
In order to arrive at a concentration of  $\pu{[B]= 3 M}$ from  the starting concentration of $\pu{[B]= 4 M}$ you would have to reduce the concentration by  $\Delta \pu{[B]= -1 M}$ which would simultaneously cause changes $\pu{\Delta [A]= -1 M}$ and $\pu{\Delta [C]= +1 M}$ resulting in final concentrations $\pu{[A]= 1 M}$ and $\pu{ [C]= 5 M}$. As this final concentration of C is inconsistent with an equilibrium point, we can logically conclude that $\pu{[B]= 3 M}$ cannot be achieved.
Unfortunately it is a somewhat poorly formulated question and largely unhelpful answer key.

Statement 1.

Consider the reaction $\ce{A(g) + B(g) <-> C(g)}$ at equilibrium in a 1-L container with [A] = 2 M, [B] = 1 M, and [C] = 4 M.

This statement is neither true nor false, I guess we will assume it as a given.  It's worth noting that the reaction changes the number of moles of gas in the container.  This is a hint that changes in pressure will change the equilibrium of the reaction.  It's also worth noting that we are given no information on whether $\ce{A}$, $\ce{B}$, $\ce{C}$, or mixtures thereof, are ideal gases.  Since we aren't told to assume ideality, we won't.  All we know is that at this particular equilibrium point, there is a total concentration of 7 M of gas.
Statement 2.

To this 1-L container you add 3 moles of B.

This again is neither true or false.  Or at least, if it's false, the rest of the question is meaningless so let's assume it's true.
Statement 3.

A possible new equilibrium is [A] = 1, [B] = 6, and [C] = 3....

This is possible, but would require massive deviations from ideality of the three gases or their mixture.  With 10 M of gas total in the container, the pressure will increase relative to the initial equilibrium with only 7 M total.  For ideal gases, this should result in an increase in the concentration of C, since it lies on the side of the equation with the smaller number of total moles of gas (2 equivalents on the LHS and one on the RHS).  The proposed new equilibrium has less C than the initial equilibrium.  Thus it is not feasible, at least for ideal gases.  But since we don't know if the gases are ideal, this is premature.  We are left concluding that this equilibrium may be possible, but only with very, very, strong and unusual departures from ideal gas behavior.
Also, we aren't told anything about the enthalpy of the reaction or how the container is insulated.  Is it held at constant temperature?  Or is it impermeable to heat flow, i.e. the reaction occurs adiabatically?  If the reaction releases or absorbs heat, when we perturb the equilibrium, the temperature will change.
Statement 4.

... because in both cases $K$ = 2.

There is no reason to care particularly about the value of $K$ for this reaction, since the equilibrium depends on pressure and temperature, not just the concentrations of the reactants and product.  Adding more moles of B will change the pressure and the temperature.  $K$ is not a meaningful quantity for this reaction.
Summary
The answer you were given makes very little sense.


text:
Why does current flow in an electrolytic cell?
Many explanations of current flow in electrolytic cells will state that electrons flow from the external power source into the cell (via the cathode). But surely this current flow would stop if there were no reduction at the cathode (which removes electrons).
Doesn't this therefore mean that current is explained by a simultaneous combination of 'push' forces and 'pull' forces? You can't have current being generated in an electrolytic cell without a reaction occurring, right?
On a related note, it's also often said that the rate of electrolysis depends on current. But, based on my understanding, current is simply a measure of the rate of electrolysis: you can't have current without first having a redox reaction occurring in the electrolytic cell. Am I right in thinking this?

You can find many useful info targeted to education  by googling with parameter site:chem.libretexts.org or by browsing the site.

Let's take as an example a $1$ M $\ce{CuCl2}$ solution in water. If two platinum or charcoal electrodes are dipped into this solution, they will not react with this solution. If a small voltage (< $\pu{1.02 V}$} is applied on the electrodes, nothing happens. If now a higher continuous voltage (>$1.02$ V) is applied, electrolysis proceeds. At the cathode (negative pole), the cation $\ce{Cu^{2+}}$ is reduced into metallic copper $\ce{Cu}$. At the anode (positive pole), the ions $\ce{Cl-}$ are oxidized into some gaseous chlorine $\ce{Cl2}$  according to the equations $$\ce{Cu^{2+} + 2 e^- -> Cu}\tag {1} $$ $$\ce{2 Cl^- -> Cl2 + 2 e-}\tag{2}$$
Let's consider the situation a couple of minutes after the beginning of the electrolysis. The cathode is at least partly covered by metallic copper. The anode is at least partly covered by adsorbed bubbles of gaseous chlorine. If the outer voltage is maintained above $1.02$ Volt, the electrolysis goes on.
But if the outer voltage is < $1.02$ Volt, or even zero Volt (situation where the electrodes are just connected by a simple wire), the previous reactions will be reversed : the cell will produce current and work as a galvanic cell. The current will be reversed and go from $\ce{Cl2}$ to $\ce{Cu}$. This cell $\ce{Cu/Cl2}$ is a good example of the push-pull phenomena discussed by Michael.
Let's summarize : when two electrodes, one made of copper $\ce{Cu}$ and one containing adsorbed chlorine $\ce{Cl2}$, are related by an external wire, and are also dipped into a $\ce{CuCl2}$ solution (or even into pure water), they have a tendency to react according to the following scheme $$\ce{Cu -> Cu^{2+} + 2 e^-}\tag{3}$$ $$\ce{Cl2 + 2 e^- -> 2 Cl^-}\tag{4}$$ The electrodes both "feel like" reacting this way. They have a natural tendency to spontaneously push electrons from $\ce{Cu}$ to $\ce{Cl2}$ through the external wire. The measured voltage between the electrodes is then $\ce{1.02 V}$ (or a little less, due to Ohm's law).
If we replace the external wire by another cell $\ce{Cu/Cl2}$, both would like to send their current through the other one. But, as they are identical, nothing happens. No current is emitted.
If the external wire is removed and a "stronger" cell is connected between the two electrodes for example a cell $\ce{Zn/Cl2}$, our "weaker" $\ce{Cu/Cl2}$ cell cannot work any more according to ($3$) and ($4$). It is obliged to react in the "pull" mode. Reluctantly, the cell must accept working according to ($1$) and ($2$} which is contrary to its natural desire. As ever, the strongest imposes its will to the weaker.


text:
Titration of a polyprotic acid with a strong base
My book says:

in the titration of a polyprotic acid we have more than one equivalence point which represents the neutralization of each step of the acid ionized.

The reaction might be:
$$\ce{H2CO3 + NaOH -> NaHCO3 + H2O}$$
$$\ce{NaHCO3 + NaOH -> Na2CO3 + H2O}$$
Now my questions are:

Why does the reaction of second step in the ionization of acid
does not start until its first step has been consumed totally?
Even if it does (may be), why we do not consider it?

I think of a possible answer, which I have applied to and got fairly well results. I think It might be that the reaction of multiple steps proceed simultaneously but we can, in a good manner, break up all these steps into independent ones such that they start only after the previous one has completely taken, and do calculations with it, and then make it overlap such that it remains the same as it was.
I call this the "Overlapping of Independent steps to work simultaneously".
Is it a correct method to think of?

My guess is that it is more energetically favorable for the first reaction to occur than the second.

Avoid over-formatting your text. It is barely readable. I have edited your question 2. "My book says:" What book?

Each acid ionization step is an equilibrium reaction, involving pH. This topic is described in many ways on many places. Search.

See e.g. https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Monoprotic_Versus_Polyprotic_Acids_And_Bases/Polyprotic_Acids_And_Bases

@KeSHAW. In my opinion, the question (1) has no meaning, because the word "not" should be removed from this question (1). I would rewrite here your question (1) :  "Why does the reaction of second step in the ionization of acid does start until its first step has been consumed totally?" Do you see what I mean ?


text:
Why is there an increase in the pH at the start of a titration between a weak acid and a strong base?
Why is there an increase in the pH at the start of a titration between a weak acid and a strong base even though there is acid already present to neutralize it? My intuition says that there should not be even a slight increase in pH until all of the weak acid is dissociated and neutralized by the base

The StackExchange network  expects that  you  have thoroughly searched and thought  about the topic and have provided explicit compact summary of partial answers/ideas/thoughts you have got until now.   That will  prevent responders to tell you what you already know or what can  be easily found. Effort not shown may be considered as effort not done and such a question may get closed.
How do I ask a good question?.

What  does the equation pH = pKa + log([A-]/[HA]) say to you?

@Poutnik . more acid will dissociate as pH increases.

@NikhilVerma It has also the other side - increasing acid dissociation ratio by acid neutralization increases pH,

Because the weak acid, in this case, is not all that weak (the curve looks like one that would be made with acetic acid). There is a small supply of dissociated, aqueous hydrogen ions with which the base reacts first, before establishing its buffer equilibrium with the weak-acid molecule and its anion. The initial increase in pH then corresponds to exhausting this hydrogen-ion supply.

The curve shown in the image is a classical example of a weak acid titration you can find a lot of them in literature. Why do you say that the reason if that the acid is not so weak? Can you find a curve of a weak acid and a strong base titration where the pH at the beginning doesn't rise?

Whenever the addition of small amounts of strong base changes the ratio of weak acid to weak conjugate bases a lot, the slope will be high. This is the case at the beginning (when the concentration of weak base is low) and near the equivalence point (when the concentration of weak acid is low). When the ratio of weak acid to weak base is 1:1 (at the half equivalence point), buffering is optimal and the slope is minimal. All of this ignores the volume changes when adding the strong base.
In effect, the strong base reacts mostly with hydronium ions in the beginning (steep slope), mostly with the weak acid when nearing the half-equivalence point (shallow slope), and with nothing past the equivalence point (steep slope until you reach extremely basic pH).

Let's start from the basic formula:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A^-}]}{[\ce{HA}]} = \mathrm{p}K_\mathrm{a} + \log\frac{\mathrm{[\ce{A}]_0} + n}{[\ce{HA}]_0 - n}\tag{1}$$
The derivative of $\log x$ is $1/x$. So when introducing $n$ mol $\ce{NaOH}$, we get a curve with a slope equal to:
$$\frac{\mathrm {dpH}}{\mathrm dn} = \frac{[\ce{HA}]}{[\ce{A^-}]} = \frac{[\ce{H^+]}}{K_\mathrm{a}}\tag{2}$$
At the beginning of the titration, when $n$ is small, the acidic concentration $[\ce{H+}]$ is high. So the slope of the curve is steep. When $n$ grows bigger, $[\ce{H+}]$ becomes smaller. As a consequence, the slope of the curve decreases.

The derivative is not quite complete, you are missing the inner term (and the n in numerator and denominator sums or differences. According to Wolfram Alpha, d/dx(log((a + x)/(b - x))) = (a + b)/((a + x) (b - x)). This assumes natural log.

@Karsten Theis. Your derivation is correct. But it is not necessary. The question was not to get a numerical value for the slope. It was only to understand qualitatively why the initial slope pH vs n(NaOH) decreases at the beginning of the titration. So my derivation is valid. And if the whole calculation is done with the initial values (following your method), the final value of $\frac{d\mathrm pH}{dn}$ is the same as mine.

The $\mathrm{pH}$ is a dynamic equilibrium linked with the presence of both $\ce{OH-}$ and $\ce{H3O+}$ ions.
$$
\mathrm{pH} + \mathrm{pOH} = 14
$$
When you are removing $\ce{H3O+}$ the equilibrium shifts.
So you will have an increase in the $\mathrm{pH}$. Hence, you can't avoid perturbing this equilibrium when adding one of the two species, of course, the highest deviation from the equilibrium will be after that you consume all the $\ce{H3O+}$ that's why we use a $\mathrm{p}$ function. In fact, that's why we use a $\log$ function, the variation is not so high seen from the point of view of concentration on a linear scale.

The question can be quite reduced to Why z changes when x/y changes (and vice versa) in z = a.(x/y)?

@Poutnik probably the question is not well posed, there is an increase of the pH in every titration between an acid (week or strong) and a base. So the answer can not be because is not a weak acid as it seems from the other answers.

Sure. The main difference is concave initial curvature for weak acids versus convex one for strong acids.


text:
Absolute configuration of sulfur in phenyl styryl sulfoxide

Provide absolute configuration to sulfur atom.

I know that oxygen atom will get the highest priority by Cahn–Ingold–Prelog rule, but what about the next priority? I believe it should be phenyl, but all my classmates say it is the alkene phenyl.
Please provide an explanation for the sulfur configuration within the the R/S system.

Were you taught some kind of rules for resolving this kind of ambiguity?

Also, did you look for other resources on the Internet, which explain how to resolve ties when the first atom directly bonded to the stereocentre is the same?

Also, what makes you believe it is phenyl? Did you follow the CIP rules, or did you make a guess, or something else?

Yes I followed the rules. Since there are 2 carbons attached to Sulphur it's a tie for priority

Now, Phenyl carbon has only carbons attached to it whereas the alkene carbon has one Hydrogen atom attached to it
Thus, phenyl carbon should get priority

Phenyl gets the second highest priority. To break the tie, check the level of substitution at the carbon bonded to sulfur. Duplicate the doubly bonded carbon. Hence in phenyl it is bonded to three carbon atoms (C,C,C) whereas in alkenyl phenyl it is bonded two carbons and one hydrogen (C,C,H). Phenyl carbon is highly substituted and hence gets highest priority.
Lowest priority is lone pair. Counterclockwise, lowest priority is behind the plane hence 'S' configuration.

Since it would be a substantial edit of your answer, this only is a suggestion to consider the addition of links to highly instructive examples in answers provided by user user55119 in the following sequence: general handling of cyclic structures, complexity level +1, and complexity level +2.


text:
AlCl3 and water
While reading Clayden's (et al) Organic Chemistry I stumbled upon a reaction on page 120 that I just fail to wrap my brain around. It describes a reaction of AlCl3 with water and states that, quote, "Eventually alumina (Al2O3) is formed."
Here's the said reaction

Now...
I was trying to understand what the intermediates are between the AlCl3H2O and Al2O3 and the best I can do (and bear with me, I might be wrong) is an idea that the negative [Al]- can transfer an electron to Cl which then deprotonates the [H2O]+ group leaving HCl as a by-product. Repeat the cycle three times and I get Al(OH)3 with an empty π orbital as in the starting aluminium chloride. I guess maybe it can be hydrolysed further to give a [Al(OH)4]- ion but then what?
To add to my confusion a quick scout on google tells me that AlCl3 can decompose into Al2O3 but it requires a lot of heat, well above the boiling point of water, which is nowhere mentioned in the textbook. Besides if I'm correct that HCl is a by-product here, then from what I understand it would quickly react with Al2O3 in presence of heat to form AlCl3 again, eventually reaching some sort of equilibrium.
All in all I would appreciate some help in understanding what's going on here.

You don't learn Inorganic chemistry from a book on Organic chemistry. Sure, organic chemists have their own use for AlCl3, and know a thing or two about its properties, but their perspective is distorted. That being said, in reality AlCl3 won't hydrolyze all the way to Al(OH)3, unless you apply a good deal of heating.

For the purpose learning organic chemistry, the point to retain should be: water (or, any molecule with an $\ce{-OH}$ group) reacts with the Lewis acid $\ce{AlCl3}$ to quench it.  In other words, count this former $\ce{AlCl3}$ out, it can no longer catalyze your Friedel-Crafts reaction. And not only $\ce{-OH}$ groups with an «acidic proton» undergo this reaction, only that the typical work-up uses water.


text:
Why can one set the Gibbs free energy change to 0 when calculating the standard free energy change?
I am trying to understand the following equation:
$$\Delta G = \Delta G^{\circ} + RT \ln\left(\frac{[C][D]}{[A][B]}\right)$$
for a reversible reaction with reactants A and B and products C and D.
The way I understand it is that the Gibbs free energy change $\Delta G$ is built up by two components. One is the standard Gibbs free energy change $\Delta G^{\circ}$ which is something we can measure in the lab so to say and use as a reference point. The second term accounts for the given conditions such as actual temperature and reactant and product concentrations which deviate from those used to calculate/measure $\Delta G^{\circ}$.
My confusion stems from the following. In the textbook I am reading (Biochemistry by Berg et al) they say that:
"A simple way to determine $\Delta G^{\circ}$ is to measure the concentrations of reactants and products when the reaction has reached equilibrium. At equilibrium, there is no net change in reactants and products; in essence, the reaction has stopped and $\Delta G$ = 0." They then set $\Delta G = 0 $ and solve for $\Delta G^{\circ}$.
So they appear to be doing the backwards process of what I described above. Also why is this allowed at all? Isn't setting $\Delta G = 0 $ akin to enforcing a statement about the spontaneity of the reaction? The way I understood $\Delta G$ was that if it is positive the reaction is not spontaneous and if $\Delta G$ is negative the reaction is spontaneous. By extension if $\Delta G=0$ the reaction is at equilibrium, so I guess this means that $K=1$? Those should be things intrinsic to a given reaction, so how can we just set $\Delta G$=0? Also does $\Delta G$ change throughout a reaction at all? I mean, it clearly does, because it is a process that evolves in time but I thought that when we say $\Delta G$ we always take initial and the final free energy, so in that sense it is a constant and not an evolving quantity.
As you have probably gathered from the question I have no chemistry background so any help would be greatly appreciated.

Isn't setting $\Delta G = 0$ akin to enforcing a statement about the spontaneity of the reaction?

Yes, it is. But they stipulate that the reaction is at equilibrium, which tells you that neither the forward nor backward reaction is spontaneous, which is precisely what allows you to assert that $\Delta G = 0$.* In other words, when a system is at equilibrium, we have $\Delta G = 0$.

By extension if $\Delta G = 0$ the reaction is at equilibrium, so I guess this means that $K = 1$?

No, that's not right, because we have $\Delta G^\circ = -RT \ln K$ (note the standard sign). $\Delta G$ has no direct relation to $K$.
One way of putting it is that $\Delta G^\circ$ and $K$ are properties of the reaction. That is, once you write down a chemical equation, these two are fixed, immutable constants.† However, $\Delta G$ (and $Q$) depend on the exact composition of the reacting system, i.e. they depend on the concentrations of reactants and products at any given point in time.

Also does $\Delta G$ change throughout a reaction at all?

Yes, it does. As the reaction proceeds towards equilibrium, it goes from whatever value it is initially at, to zero.

I thought that when we say $\Delta G$ we always take initial and the final free energy

... Kind of. You need to be careful here because the difference of a free energy should have units of free energy, traditionally kJ. However, $\Delta G$ is usually given in units of molar free energy, i.e. kJ/mol.
The reason is because what you're calling $\Delta G$ is actually not the difference between the free energy of products and free energy of reactants. It's actually a lazy shorthand for $\Delta_\mathrm{r}G$, the reaction Gibbs free energy. Actually, $\Delta_\mathrm{r}G$ is the difference between the chemical potentials of reactants and products in the reaction, multiplied by their stoichiometric coefficients.
I wrote about this difference before, see What is the difference between ΔG and ΔrG?, although I must apologise as it's a bit convoluted. (In fairness to myself, the topic in general is pretty convoluted.)

* A quick and dirty definition of spontaneity is that if you set something up and the reaction naturally goes in the net forward direction, the forward reaction is spontaneous. A formal definition would involve $\Delta G$, but this is not suitable in the present case, as it would be a circular argument.
† To be precise, they're constant for a given set of reaction conditions. $K$, and $\Delta G^\circ$, will change with temperature or pressure, for example. But that's not under consideration here.

Thank you for the quick response. I will also check the other question you linked out. That clears it up I guess, I just found it a bit strange for some reason that we calculate the standard free energy change from an assumption on the free energy change.
If it is not too much to ask, could you give an example of how this equation is used in the lab so to say? It obviously can be used in many ways but what is the prototypical research application from your experience?

@terraregina Er, that's kind of hard to discuss! It really depends on what kind of chemistry you're doing. Often, determining $\Delta_\mathrm{r}G^\circ$ (note the standard symbol again) is only a means to an end, in the sense that you use it to in turn equilibrium concentrations / binding constants / etc which are more interesting. It also depends on what kind of values you are expected to output - if you're studying drug binding then you want binding constants, but if you're just benchmarking some computational method, you could just report $\Delta G$'s and be done with it.

@terraregina You can imagine how to do this yourself.  Suppose you have a reaction corresponding to the expression you wrote above ($\ce{A + B->C + D}$) and measure that, at equilibrium, [A] = 2 M, [B]= 7 M, [C] = 3 M, and [D] = 5M.  Assuming $T=298 \pu{K}$, what would be $\Delta G^{\circ}$?  I'd highly enourage you to try this calculation. I think after you do it, things will become much clearer to you.

That is, once you write down a chemical equation, these two are fixed, immutable constants. $K$ does change with temperature though!

@AVS That's fair, thanks.