text:
Thermodynamics of chemical reactions "at constant pressure"

Chemical reactions occur at constant temperature and pressure.

Consider a gaseous, equilibrium reaction: $\ce{2NO2(g) <=> N2O4(g)}$. Most questions/textbooks formulate such questions by stating: The reaction happens at ($T$) temperature and ($P$) pressure.
This gives the impression that the surrounding Temperature and pressure are constant.

The first confusion that then arises is: For thermodynamic quantities such as enthalpy that involve pressure, which pressure should we use, internal or external?

After reading some chem.SE answers, it seems to me that $P_{sys}=P_{surr}$ is a basic assumption. However this doesn't seem to go well with the example I mentioned: Since the pressure of the system will be given by $(n_1)RT/V + (n_2)RT/v$, where $n_1$ and $n_2$ are the moles of $\ce{NO2}$ and $\ce{N2O4}$. Clearly, as the reaction progresses, ($n_1+n_2$) changes, and thus,the pressure of the system changes. If the surrounding pressure is kept constant, this beaks the "assumption": $P_{sys}=P_{surr}$.
Am I missing something?

Constant temperature and pressure with variable summary molar amounts means variable volume.

i dont understand the term "summary"

Or rather total molar amount $n_\mathrm{tot} = n_{\ce{NO2(g)}} + n_{\ce{N2O4(g)}}$. By other words, $V = n_\mathrm{tot}RT/p$, implying ideal gas behaviour.

i still dont understand what exactly your statement means. What does "summary mean"?

But you do understand what "sum" means, don't you ? BTW, in the first comment summary was used as adjective, possibly incorrectly.

please write an answer @Poutnik. I really cant decipher what the comments mean

There is nothing to decipher. See the 2nd comment.

@satan29  Poutnik means "variable total molar amounts".   I agree that "summary" is confusing, since (as Poutnik himself suspects), he wasn't using the word correctly.  I'm guessing English isn't Poutnik's first language (Russian, perhaps, based on the name?).  [His English is certainly far better than my French, which is my best second language!]

What I do not get is even if I corrected myself using total instead of summary, and used explicit formula, he remained confused, like if it was his decision. I am Czech. Poutník mean a wanderer or a pilgrim, in outdoor or religious sense. Similarity to the Russian word Sputnik ( cotravelling mate of Earth ) is not accidental, the old Slavic word root "puť" means "a journey". OTOH, there is no relation to the English verb "to pout".

You have correctly stated the pressure of the system is $$p = \frac {(n_1 + n_2)RT}{V}$$
But at the end of your question you incorrectly imply the volume is constant and the pressure changes. That is not true. Constant pressure means the external constant pressure ( like the atmospheric one ) keeps the system pressure constant. Imagine there is a massless, frictionless piston, ensuring $p_\mathrm{ext} = p_\mathrm{sys}$. It is similar as a thermostatic water bath keeps the system temperature constant.
So less confusing is to rewrite the ideal gas state equation :
$$V = \frac {(n_1 + n_2)RT}{p}$$
There are 4 variables: $p, V, T, n_\mathrm{tot}=n_1 + n_2$.
But there are just 3 degress of freedom = 3 independent variables.
Either $V, n_\mathrm{tot}, T$ are given, and $p=f(V, n_\mathrm{tot},T)$,
either $p, V, T$  are given, and $n_\mathrm{tot}=f(p,V,T)$,
either $p, n_\mathrm{tot}, T$  are given, and $V=f(p,n_\mathrm{tot},T)$,
either $p, n_\mathrm{tot}, V$  are given, and $T=f(p,n_\mathrm{tot},V)$.
We have already given 2 variables: $p, T$, with $n_\mathrm{tot},V$ remaining free.
But there is the equilibrium relation
$$K_p = \frac {p_{\ce{N2O4}}} {({p_{\ce{NO2}} )}^2}=\frac {p \cdot ( 1 - x_{\ce{NO2}}) } {({p \cdot x_{\ce{NO2}} )}^2}$$
$$K_p \cdot p \cdot {({x_{\ce{NO2}} )}^2} + x_{\ce{NO2}} - 1 = 0$$
Solving the quadratic equation we would finally get $n_\mathrm{tot}$, so all 3 degrees of freedom are saturated with given values for $p, T, n_\mathrm{tot}$.
So in our case, the following applies:
$p, n_\mathrm{tot}, T$  are given, and $V=f(p,n_\mathrm{tot},T)=\dfrac {n_\mathrm{tot}RT}{p}$

your first paragraph explains it perfectly. I am sorry that i didn't realize this earlier, through your comments. +1.


text:
why does lithium fluoride have a higher lattice energy than lithium iodide?
Based on my knowledge, lattice energy is proportional to the multiplication of the charge of the ions, divided by the sum of the radius of ions, as follows:
$$\text{Lattice energy} \propto \dfrac{q_+ \cdot q_-}{r_+ + r_-}$$
Since iodide has a larger radius than fluoride, by this equation lithium iodide would have less exothermic lattice energy (lower lattice energy).
However, I also learned that an ionic bond with a higher covalent character is stronger (i.e. more exothermic lattice energy). Since the electron cloud of iodide ion is larger and is more polarisable, lithium iodide should have a greater covalent character than lithium fluoride and thus its lattice energy should be more exothermic (higher lattice energy)

The iod atom or ion is too large to make good orbital overlap with basically any partner. Covalent iodides are invariably rather unstable. Same for bromides, or e.g. sulfides, compared to oxides.

The real issue here is that the formula quoted is for the lattice energy as calculated within the ionic model, which by definition has no covalent component whatsoever. A good answer will talk about the difference between a  simple ionic model and a more complex one that includes "covalent" effects, and the need to know enough chemistry to understand which model is best to apply in a given situation. I'm sure the down voters of this good question will find time to discuss fully this once they get around to justifying their down votes. I hope to find time to give such an answer.

An ionic bond with higher covalent character is stronger - This doesn't make sense. LiI has less ionic character than LiF due to more polarizability of I-. But more the bond length, less the bond strength.

source: Stack Exchange

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

sort: votes


text:
How to bond polymer to polymer (TPU on PMMA)
TL;DR I wanted to bond TPU films to PMMA slab. I  have tried thermal bonding but it is was not practical (I do not know why?). Can anyone please give me a method to bound them?
"Thermal bonding" means we melted the surface and tried sticking them together. We want to bond TPU (thermoplastic polyurethane) to PMMA, for microfluidic projects, with in submilimitere features on PMMA slab and a very thin film of TPU (50 micron). We want to decrease the time and temperature if it is thermal bonding because the temperature used in references is more than 105°C which maybe deform micro patterns on PMMA slab. We also appreciate other ways you may propose.

"Thermal bonding" means you melted the surface and tried sticking them together?

Yes @Karl and I want to the ways to improve it

That´s tbe. Most polymers do not mix. Judging from your comment below, your are trying to do a scientific project. What did your literature research yield? You are certainly not the first to try this!

@Karl, we found researchers typically use it in the pressures much lower than our desired amount. We want to work in higher pressures. In most researches, people use thermal bonding with some chemical treatment but it is very tedious, time consuming and against rapid production. Please let me know if there's need any more information that I can provide for you.

related: https://chemistry.stackexchange.com/q/32060/102629

Sorry to say but you question is a bit vague because without knowing the project, it's pretty hard to help. Anyway, here is some general information.
Some of the sophisticated methods of joining these together include solvent bonding, vibration welding, friction welding and induction welding or pressure sensitive adhesives [ (Meth)acrylic copolymers are one of the most widely used polymer classes for the production of pressure-sensitive adhesives]. Acrylic requires an acrylic solvent adhesive. Cyanoacrylate (super glue) may work, depending on the project. UHU glue is very useful with all kinds of plastics, so you might wanna try that.  I myself have tried superglue and it works with both materials. But I think for your project, non adhesive methods woll be an optimal choice.
All the best.

Thanks for your help
We want to bond TPU (thermoplasticl polyurethane) to PMMA, for microfluidic projects, with in submilimitere features on PMMA slab and a very thin film of TPU (50micron). We want to decrease the time and temperature if it is thermal bonding because the temperature used in references is more than 105 which maybe deform micro patterns on PMMA slab, we also appreciate other ways if you propose.

Hello, as I have already given you a brief account of other options, I won't repeat them. But here is a paper that might help you. https://drive.google.com/file/d/1333pQJA7eiVa-sE__7ZYGK2vBy4w4bnP/view?usp=drivesdk

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


text:
why does the partial pressure not change?
Suppose I have the equilibrium in a closed container at constant temperature:
$$\ce{2CaSO4(s) -> 2CaO(s) + 2SO2(g) + O2(g)}$$
If the volume of the container is halved, the reaction should move towards that side having lesser number of gaseous moles and hence the partial pressures of $\ce{O2}$ and $\ce{SO2}$ must change but it is given that they won't change.
How do you solve this question?

$$ \text{You can use \ce{} to automatically format chemical compounds.}$$ Since the temperature is not said to have been changed, can we assume a change in pressure. is there anyother information given?

no other information has been given.Thanks for helping out to format chemical compounds

What is the exact question? Could you specify the source? Currently you have a statement and your attempt. no question to solve :)

Perhaps they mean that the ratio of SO2 and O2 won´t change.

the source-FIITJEE AITS

actual question-First this reaction was given and they stated that enthalpy change for this reaction is positive and it occurs at 1600K in a container and they asked to pick the correct options.In one of the options it was given-'If volume of the container is halved,partial pressure of O2 will remain the same" and it was given correct.In another option it was given-"The partial pressure of so2 will change if volume of container is doubled" and it was given as incorrect option

Of course the pressure won't change; it is determined by the equilibrium alone.

The given explanations are correct, but there is a slightly different way of looking at it which makes it easier to understand. While
2CaSO4(s)⟶2CaO(s) + 2SO2(g) + O2(g)
is quite a correct way to describe the equilibrium, it is worthwhile to remember that since we have an equilibrium, we probably also have some CaO(s) on the left:
2CaSO4(s) + nCaO(s)⟶2CaO(s) + 2SO2(g) + O2(g).
Then, mentally halving the volume of the container doubles the pressure (mentally), which then causes a back reaction with the CaO on the left side, leaving the partial pressures unchanged because they are dependent on the nature of the equilibrium constant.
Putting the CaO on the left side is like completing the mental picture of a seesaw (a metaphor for equilibrium), which otherwise has a weight on one side and not on the other.

The partial pressures initially increase, when the volume is decreased.
But they return to the original values, when the equilibrium is re-established, implying temperature is kept constant.


text:
Poisoned catalysts
CaCO3 in rosenmund reduction, lead in Lindlar's catalyst, or steel in Raney nickel all act as poisins but the carbon in Pd/C is just for support. How do the others affect the catalytic activity but not carbon?

Because carbon maximizes the surface area of palladium which is incorporated with it and thus helps in increasing the activity of the reaction. Other ones are just poisons which reduce catalyst activity.

Have you checked why lead poisons the Lindlar cat? There is a specific, individual chemical reaction behind all catalyst poisons.

Read this Review.


text:
Can I use potassium acetate instead of potassium carbonate?

In the synthesis of N-ethoxycarbonyl L-proline methyl ester from L-proline and ethyl chloroformate, I use $\ce{K2CO3}$ to neutralize $\ce{HCl}$. Can I use $\ce{CH3CO2K}$ instead of $\ce{K2CO3}$? I think I cannot use it because it reacts with $\ce{HCl}$ forming acetic acid which can hydrolyse the ester. Is it correct?

The important thing the pot carb is doing is stopping the proline nitrogen from getting protonated as the reaction progresses by the HCl that is generated by the reaction. Pot. acetate will not do this.

Think of this way: $\mathrm{p}K_\mathrm{b}$ of proline $\ce{NH}$ is 3.60, thus it is already protonated (zwitterionic). $\mathrm{p}K_\mathrm{b}$ of $\ce{AcO-}$ is 9.25 (weakly basic) and that of $\ce{CO3^2-}$ is 3.67 (much better base than $\ce{AcO-}$ and comparable with proline). Thus, you must use $\ce{K2CO3}$ as the base here.

Stay with the carbonate.
In the course of the reaction $\ce{HCl}$ is formed. The aim is to remove this from the reaction mixture both as soon as well as complete as reasonably possible.  Compare

The addition of sodium acetate will increase the complexity of your reaction mixture.  $\ce{HCl}$ is soluble in methanol (reference), and upon dissociation ($\ce{HCl <=> H+ + Cl-}$), the $\ce{H+}$ and the acetate $\ce{OAc-}$ join each other to generate the buffer system of $\ce{HOAc <<=> H+ + OAc-}$.  Think about the $\ce{HOAc}$ as a reservoir, it may release $\ce{H+}$ at any moment you probably do not intend; a prominent example may be an aqueous workup.
The quantity of $\ce{HOAc}$ generated may be small -- in comparison of the methanol used as solvent of reaction.  Still it may complicate an aqueous workup and following extraction, because this organic substance solutes so well in water, but fairly enough in organic solvents such as hexanes, or chloroform, too.  So your extraction of the product may be affected.  A distillation (bp 118 C (pure acetic acid) instead of 65 C (pure methanol) at ambient pressure) ahead of the workup may remove this from your reaction mixture.

with the alternative

addition of $\ce{K2CO3}$.  Darn efficient removal of $\ce{H+}$ for once and forever, because the quench along  $\ce{K2CO3 + 2H+ -> 2K+ + H2O + CO2}\uparrow$ is practically irreversible.  In addition, it allows you to monitor the advancement of the reaction since $\ce{K2CO3}$ is not well soluble in methanol.


text:
Can I use calcium chloride instead of sodium sulfate as drying agent?
In the synthesis of N-ethoxycarbonyl L-proline methyl ester from L-proline and ethyl chloroformate,
I dry the organic phase over sodium sulfate. Can I use calcium chloride instead of sodium sulfate?
I searched for information and I found that calcium chloride can contain basic impurities such as $\ce{Ca(OH)2}$ and $\ce{CaCl(OH)}$. Does someone know if it is actually the case?
If it would be correct I think that it cannot be used because the impurities can hydrolyse the ester.

Why you want to change the drying reagent?

Similar to @MathewMahindaratne's comment: if you know the drying agent suits your needs well enough, keep it.  In particular here, sodium sulfate is a neuter one (as is calcium sulfate, too) is a safe option because while drying the organic layer, the agent will neither protonate, nor deprotonate your product. Thus, if the protocol states $\ce{Na2SO4}$, stick to it.
This is in contrast to calcium chloride which not only will contain traces of calcium hydroxyde when taken from the jar, but upon contact with additional water (a.k.a. while drying your organic solution) will create new $\ce{Ca(OH)2}$ and thus is considered a basic (and potentially deprotonating) drying agent.  Note, $\ce{MgSO4}$ is considered as an acidic drying agent.  Using the wrong salt solutions or drying agents during the workup actually may cleave off (e.g., protecting) groups and thus complicate the isolation and purification of your product.

Yes, theoretically speaking, you CAN use it (not that you should), but only with suitable organic solvent and it also depends on the grade of Calcium chloride you have. I'll try to keep my answer as short and precise as possible.
Two things to consider, as I said before, are the grade of Calcium chloride and the organic solvent used.
• Calcium chloride can be contaminated with Calcium hydroxide etc. but samples from Sigma Aldrich are extremely pure, 99.9%. If your reagent is not that pure or you just want to be sure, you can do two things:

Treat Calcium chloride with dilute Hydrochloric acid, it will convert any Calcium hydroxide to Calcium chloride, then crystallize it and grind the crystals and place in a dessicator at 30°C for 12hrs (time will vary with the amount of calcium chloride you are using). You can also use more sophisticated techniques like drying the Calcium chloride in presence of dry Chlorine gas.
You can try these two more sophisticated methods of Calcium chloride purification: crystallization from aqueous solutions and joint precipitation of impurities from aqueous solutions onto inorganic collectors (hydrated Aluminium or Zirconium oxides).

• Now about organic solvents. It's insoluble in liquid Ammonia, Ethyl acetate and DMSO. It's sparingly soluble in Acetone. It's soluble in Alcohols and Water. If you are using one of the first four solvents, you are good with Calcium chloride otherwise, it's gonna cause some problem. Also, as others have pointed out, Calcium chloride can form Calcium hydroxide if your reagents includes Water.
I hope I was able to help or at least guide you in the right direction. All the best.


text:
How to get or understand a sample in chemistry without knowing its chemical name or compound or formula
How to get or understand a sample in chemistry without knowing its chemical name or compound or formula? I was thinking about whether there is a way to understand what the chemical compound sample that one might have. He/She doesn't know what its formula or its chemical properties or anything .He/She just got that one and was thinking what was that chemical and was interested to know about its structure. Would it be possible?
I am totally new at Stack Exchange so I doesn't know is this the proper way to ask.
And I am 15 year old so I doesn't know if these are taken in my upcoming classes but when I googled it or searched in many websites,I was not able find answers,So I am asking here.I have seen that we can guess the compound using many experiments but so also if the compound has been not found yet in human history would we be able to understand it?
Edit: This question had been closed stating it need to be more focused. I just wanted to clarify that what I wanted to know was whether there is a specific procedure to identify the chemical structure of the compound from just a sample alone.

It is always possible. If you look at periodic table, all of those elements are found by someone. They were not known before. Same goes to all the existing chemicals (millions of them). Even just now, someone may find a new natural product, which is not known before.

First you analyse the compound, then you know its properties and composition, and can derive the name and formula. And yea, you need to study a lot of chemistry before you can do that, or understand how it works. And, yes, if chemists can get their hands on a substance, they can analyse it well enough to understand its properties, and to be able to recognise it again later.

I wonder how you know compound without knowing compound ;-> This title is really bad.

@Mithoron What the OP is trying to say is how to know a compound when only a sample is given...

@DeepakMS Yes, I found it out by reading the stuff that's not in title or bold - quite not like it should be. BTW there was enough of q. on this extremely broad topic already.

Learning the unknown sample is called analysis and is a subject of analytical chemistry.
But your question is probably not about this.
You mean "can i guess something without knowing chemistry at all?"
And answer is yes!
Look at the food salt.
Do you know it without any knowledge of chemistry? Yes, you do!
But how much?
You know what it looks like,
you know its taste,
you know that it's soluble in water...
and probably that's all.
Now suppose you have a sample of something not found yet in human history.
You can easily know it similar to knowing food salt. Just check:
How it looks like?
What it's taste?
Drop it in water - is it soluble?
Now you "know" this substance just like you "know" food salt, without any chemistry!
But I guess you already suspect that there is something wrong with this knowledge.
It's very incomplete.
Chemist can learn more about food salt or your magic unknown sample.
Most of that serious knowledge require understanding of sample structure
at microscopic levels, and using theories that connect the structure to properties.
That was all about "knowing the substance".
Another question is knowing what a substance is, which is rather philosophy than chemistry.
Does a substance exist? What is existance? Does a substance have hidden fundamental purpose? Was it "created" or "appeared by itself"?
These questions are probably more important than chemistry, and you may be the one who one day will guess some answers here.

Just a suggestion tasting isn't always the best idea and it might give some wrong ideas to a novice

I upvoted because it kindly answer the genuine question posed by OP. I would have skipped the philosophical end.

Now that our minds is used to working in a certain direction, we would skip this for ourself.
But I would like someone to tell me about this when I was 15.


text:
How does ammonium nitrate explode on its own?
I thought ammonium nitrate was an oxidizer that needed to be mixed with fuel to form a high explosive (e.g., ANFO). But apparently there have been accidental explosions involving just the "fertilizer". Are these explosions also detonations? What is the chemical formula for the process?
$$\ce{NH4NO3 -> ???}$$
Part of my motive for asking is the news today (Aug 4, 2020) of an explosion in Beirut. Initial reports say it was caused by "2750 tons of stored ammonium nitrate".

Things like that did happen before. Look up Oppau, also Texas City.

Fuel is added because it has surplus oxygen. Still $\ce{NH4^+}$ reduces $\ce{NO3^-}$  good enough, by itself.

see https://en.wikipedia.org/wiki/Ammonium_nitrate_disasters for the two mechanisms

as far as i know it was fertiliser, without fuel added. I don't think even those who permitted huge quantities of NH4NO3 to be stored in port for years would permit storage of a serious explosive which is ANFO

@FrancisL. The videos show an enormous explosion. You're saying ANFO would be worse? I guess that's a separate question if I want to quantify that.

0.74/0.42 = 1.76 times as powerful https://en.m.wikipedia.org/wiki/TNT_equivalent

I'm unable to comprehend how it can detonate absent a fuel mixture. The chemistry I'm seeing suggesting it can does not seem legit. It is not a low explosive, it is an oxidiser. Add fuel and it is a high explosive. Absent fuel just an oxidiser. There has to have been fuel involved, some kind of fuel for the explosion to happen. I will have to see straight Ammonium Nitrate explode before I will believe this. Prove it, because I do not believe .

From amm. Nitrate disasters : "Large stockpiles of the material can also be a major fire risk due to their supporting oxidation, a situation which can easily escalate to detonation." and so in this case, it was the fire that spread onto amm. nitrate which caused it to detonate.

@MuhammadAbbass As many of the answers state, ammonium nitrate is both an oxidiser and an explosive. You have no need to witness an explosion - things like this have occurred before: https://en.wikipedia.org/wiki/Texas_City_disaster

@IvanNeretin Side note:  The one in Toulouse (southern part of France) in September 2001 was forcefully enough that the plant's name, «AZF» was banned as letter combination on the car plates. (https://en.wikipedia.org/wiki/AZF_(factory), https://en.wikipedia.org/wiki/Vehicle_registration_plates_of_France).

@MuhammadAbbass I already mentioned that while nitrate anion is oxidising, ammonium cation is reducing.

@MuhammadAbbass, when ammonium nitrate detonates, it doesn't undergo combustion (which would require a fuel), it undergoes decomposition (which doesn't).

@Buttonwood: and almost 100 years after the Oppau explosion, the BASF plant still has the Trichterstraße [crater road] (Bombentrichter is German for bomb crater, Trichter alone would usually be funnel).

Reportedly there were fireworks stored in the same warehouse as the AN in Beirut: https://www.youtube.com/watch?v=LNDhIGR-83w

It is known that ammonium nitrate decompose exothermically when heated to form nitrous oxide and water. This paper1 notes that the irreversible decomposition of ammonium nitrate occurs at the temperature range of $\pu{230-260 ^\circ C}$.
$$\ce{NH4NO3 ->[t >230 ^\circ C] N2O + 2H2O}$$
They also further noted that beyond $\pu{280 ^\circ C}$, $\ce{NH4NO3}$ is  capable of rapid, self-accelerating decomposition (to the point of detonation).
But at the detonation temperature, $\mathrm{t_d}$ (the temperature at which compounds detonate), ammonium nitrate fully decomposes to nitrogen, oxygen and water releasing a tremendous amount of energy.
$$\ce{2NH4NO3 ->[t_d] 2N2 + O2 + 4H2O}$$
In the context of Beirut explosion, the question that raised was "when did ammonium nitrate reached detonation temperature, and why did it suddenly explode?". According to a news report from cnet.com:

When heated to above 170 degrees Fahrenheit, ammonium nitrate begins
to undergo decomposition. But with rapid heating or detonation, a
chemical reaction can occur that converts ammonium nitrate to nitrogen
and oxygen gas and water vapor. The products of the reaction are
harmless -- they're found in our atmosphere -- but the process
releases huge amounts of energy.[...]
Additionally, in the explosion, not all of the ammonium nitrate is
used up and exploded. Some of it decomposes slowly creating toxic
gases like nitrogen oxides. It's these gases that are responsible for
the red-brown plume of smoke seen in the aftermath of the Beirut
explosion, Rae said.

So, my theory is that ammonium nitrate started heating (from the fire) releasing all sorts of nitrogen oxides (the red fumes). This fire further accelerated the reaction, further heating the remaining ammonium nitrate to the point of detonation and that's when ammonium nitrate exploded instantaneously releasing tremendous amount of energy which send shockwaves around the site along with a white mushroom shaped cloud (from @DDuck's comment) which could probably be nitrogen and/or water vapours where the humid air (water vapor laden air) condensed due to the explosion(@StianYttervik) with release of nitrogen. It is a sad and quite devastating incident.
References

On the Thermal Decomposition of Ammonium Nitrate. Steady-state Reaction
Temperatures and Reaction Rate By George Feick and R. M. Hainer, 1954 (PDF)
Reaction Rates of Ammonium Nitrate in Detonation
Melvin A. Cook, Earle B. Mayfield, and William S. Partridge
The Journal of Physical Chemistry 1955 59 (8), 675-680
DOI: 10.1021/j150530a002 (PDF)
https://en.wikipedia.org/wiki/2020_Beirut_explosions

The white is likely condensation of humid air due to the pressure wave at the shock front though.

I'm accepting this answer but I'd accept more than one if the site allowed it. Mathew's answer below has some more reaction formulas with enthalpy changes and other info.

Condensation or Wilson cloud. https://en.wikipedia.org/wiki/Condensation_cloud

@Nilay Do you consider your theory of the ammonium nitrate exploding on its from heat in Beirut more likely than it being combined with a fuel? In all other cases I know of a contaminant acted as the fuel source: the containers and/or debris from the fire (including smoke and soot) and/or anything that molten AN made contact with as the structures around it burned away. In Beirut, there was a significant and long fire prior to the explosion, so another theory involving contamination could also be possible, I think, since there was plenty of fuel sources, and the suggestion of gradual heating.

@JasonC There are many factors, not just heat. The whole situation is "heat + fuel + oxidation + decomposition + instability + contamination + smoke + recklessness = explosion". I just emphasized on the heat part and other answers explored the other factors, like oxidation + fuel part is explained in Alchimista's answer, the thermodynamics is explained in Mathew's answer. You explained the contamination part, Jan explained the instability of AN. All the answers combined answers OP's question.

Ammonium nitrate ($\ce{NH4NO3}$) is widely used in the fertilizer industry and is one of the most concentrated forms of nitrogen fertilizer (35% of $\ce{N}$). At the same time, it has also been widely used as an explosive material for detonation in mines. Because of its explosiveness, $\ce{NH4NO3}$ is associated with various hazards including fire and explosion, which have occurred repeatedly in the past (more than 70 incidents during 20th century, more than half of them occurred in the US soil). Regardless, $\ce{NH4NO3}$ is not considered a flammable or combustible material at ambient temperature and pressure (Ref.1). However, it is a strong oxidizing agent that can detonate under certain conditions such as temperature, fire, confinement, and presence of impurities (e.g., $\ce{KCl}$), which can be act as a promoter to detonate (Ref.2).
To use as an explosive or a blasting reagent, $\ce{NH4NO3}$ is mixed with fuel oil, which is called ammonium nitrate fuel-oil (ANFO; Ref.1). According to Ref.2, during the explosion, following exothermic reaction would take place (hydrocarbon is represented by $\ce{CH2}$):
$$\ce{3NH4NO3 + CH2 -> 3N2 + 7 H2O + CO2} \quad \Delta H = \pu{-4017 kJ/kg} \tag1$$
Interestingly, this can be compared with TNT, the heat of combustion of which is $\Delta H = \pu{-4196 kJ/kg}$. Without fuel oil, can be detonated under certain conditions.  It is believed that the vaporization of molten $\ce{NH4NO3}$ leads to the formation of ammonia and nitric acid, which could initiate the decomposition of $\ce{NH4NO3}$ through following reaction:
$$\ce{NH4NO3 <=> HNO3 + NH3} \quad \Delta H = \pu{176 kJ/mol} \tag2$$
At higher temperatures (i.e., between $\pu{170 ^\circ C}$ and $\pu{280 ^\circ C}$) exothermic irreversible reactions (equations $(3)-(5)$) occur:
$$\ce{NH4NO3 -> N2O + 2H2O } \quad \Delta H = \pu{-59 kJ/mol} \tag3$$
$$\ce{NH4NO3 -> 1/2N2 + NO + 2H2O } \quad \Delta H = \pu{-2597 kJ/mol} \tag4$$
$$\ce{NH4NO3 -> 3/4N2 + 1/2NO2 + 2H2O } \quad \Delta H = \pu{-944 kJ/mol} \tag5$$
If the material is suddenly heated up, there will be explosive decompositions as shown in equations $(6)$ and $(7)$):
$$\ce{2NH4NO3 -> 2N2 + O2 + 4H2O } \quad \Delta H = \pu{-1057 kJ/mol} \tag6$$
$$\ce{8NH4NO3 -> 5N2 + 4NO + 2NO2 + 16H2O } \quad \Delta H = \pu{-600 kJ/mol} \tag7$$
Keep in mind that all of these reactions except for $(2)$ are exothermic. Also, most products are gases. I attached PDF file if Ref.2 if a reader is interested in how explosions happen during right conditions (otherwise, it is a broad field to explain). For instance, the reaction $(3)$ can be made more exothermic ($\pu{789 kJ/mol}$) with more gaseous products, if some oxidisable fuel is added such as $\ce{C}$ (Ref.3):
$$\ce{2NH4NO3 (s) + C (s)  -> 2N2 (g) + CO2 (g) + 4H2O (g)} \tag8$$
It is evident from the past incidents involving $\ce{NH4NO3}$ that the presence of impurities and environmental conditions have a huge effect on the detonation of $\ce{NH4NO3}$ during storage. For example, one of the deadliest industrial incidents in US history occurred on April 16, 1947, in Texas City, Texas where an $\ce{NH4NO3}$ explosion involving $\pu{2300 tons}$ of $\ce{NH4NO3}$ caused 581 fatalities and thousands of injuries. The fire was caused by the initial explosion of $\ce{NH4NO3}$ on a ship, which resulted in subsequent chain reactions of fires and explosions in other ships and facilities nearby. The exploded $\ce{NH4NO3}$ was coated with (carbon based) wax to prevent caking (See equation $(8)$ above). After this accident, the new technologies and safe practices introduced in the 1950s eliminated the use of wax coatings (Ref.2).
References:

Guy Marlair, Marie-Astrid Kordek, "Safety and security issues relating to low capacity storage of AN-based fertilizers," Journal of Hazardous Materials 2005, 123(1–3), 13-28 (https://doi.org/10.1016/j.jhazmat.2005.03.028).
Zhe Han, "Thermal Stability Studies of Ammonium Nitrate," Ph.D. Dissertation, Texas A&M University, TX, 2016 (PDF).
Alex M. Djerdjev, Pramith Priyananda, Jeff Gore, James K. Beattie, Chiara Neto, Brian S. Hawkett, "The mechanism of the spontaneous detonation of ammonium nitrate in reactive grounds," Journal of Environmental Chemical Engineering 2018, 6(1), 281-288 (https://doi.org/10.1016/j.jece.2017.12.003).

First,
ammonium nitrate is a kind of mixture between an oxidizer - the nitrate part - and a reducer - the ammonium one. This is at the core of your question.
The direct decomposition correctly mentioned in the answers is nevertheless a process in which something gets oxidised and something gets reduced.
In ammonium nitrate basically you have all you need - the "fuel“ and the “oxygen“ analogues of what is involved in a standard, explosive or not, combustion.
Still the other answers are valid and more detailed from a chemical mechanicist viewpoint. One point out the presence of NO2 clearly seen by its red brownish colour before the second powerful explosion.
But the straight answer to your question is that the oxidizer and the reducing stuffs are already within the salt.

Side note: ammonium nitrate can decompose by mechanical shock, so there were enough conditions to trigger the second powerful blast.

The main point of this answer is to use the 2013 West Fertilizer Company explosion (USCSB animation for context) as an example of scenarios that could lead to AN detonation, and also to show that the situation can become very complex and unpredictable.

Anything in the vicinity can become fuel, especially if a fire is already in progress. This includes the containers, impurities, soot and debris from the fire, etc. Plus, ammonium nitrate's melting point is ~337 F, meaning it can become molten, possibly escaping its container, and readily mixing with fuel sources.
The US CSB West Fertilizer Explosion final report, section 4.3, outlines three possible scenarios under which the 2013 explosion in West, Texas could have occurred. Section 4.2 outlines general contributing factors.
These aren't the only ways it can explode, but they are a few examples of the types of conditions that could lead to explosion.
You should definitely read the report; my brief summary below leaves a lot of relevant analysis out.
So from section 4.2, contributing factors (FGAN = fertilizer grade ammonium nitrate):
Contamination

In fire situations, the behavior of FGAN is unpredictable, in part because of the number of endothermic
and exothermic decomposition reactions that take place with increasing temperature. FGAN
decomposition reactions beyond the first step have yet to be uniquely defined, and subsequent
decomposition reactions of FGAN can only be assumed. When contaminants are added to AN, the
decomposition reactions become increasingly more complex. Possible sources of contamination in an
FGAN storage area can include ignitable liquids, finely divided metals or organic materials, chloride
salts, carbons, acids, fibers, and sulfides. These contaminants can increase the explosive sensitivity of
FGAN.
The molten FGAN at the WFC likely came in contact with contaminants that were stored in the fertilizer
warehouse or were produced during the fire that preceded the explosion. Seed materials, zinc, and other
organic products, including the wood-constructed bins, were present near the FGAN storage area or could
have come in contact with molten FGAN. During the fire, soot from the smoke and also collapsing wood
and roofing material might have mixed with the FGAN pile.

Poor Ventilation

The limited ventilation increased the quantity of soot in the smoke and the potential
contamination of the FGAN pile. ...
At some point around 5 to 6 minutes before the detonation, the character of the fire changed, according to
eyewitness accounts and photographic evidence (Figure 40). This change was most likely caused by
increased ventilation through an opening low in the building, possibly when the fire burned through the
seed room doors or the roof. The fire also might have been enhanced by oxidizing gases from the heated
FGAN pile...
The additional ventilation caused a marked decrease in dark smoke and probably was accompanied by a
major increase in heat radiation inside the fertilizer building because of increased oxygen availability to
the burning wood and other fuels. With the dark smoke inside of the structure reduced, radiant heat
would reach the surface of the FGAN in the bin, and the increased airflow through the building would
greatly increase the radiant heat flux by raising the temperature of the burning wood. The surface of the
FGAN, covered with soot or molten asphalt, would absorb the heat flux and cause a very rapid heating of
the surface of the FGAN pile. The very hot and contaminated surface of the pile was then sensitive to
detonation.

And from section 4.3, a few detonation scenarios:

Scenario 1: Detonation from the top of the FGAN pile.
Scenario 2: Detonation in heated FGAN along exterior wall exposed to fire.
Scenario 3: Detonation in elevator pit that spread to main FGAN bin

Scenario 1: Detonation from top of pile

Based on the location of the pile and the properties of the bin along with the circumstances of other fire induced incidents, one possible scenario is that a period of contamination with soot and other organics
(possibly including molten asphalt and plastic dripping from the burning composite shingle roof and PVC
drop pipe from the elevator mechanism) was followed by about 5 to 6 minutes of intense radiant heating
from the flames above and adjacent to the main FGAN bin. During this time, a layer of very hot,
contaminated, and sensitive liquid FGAN could have built up on the pile. The foaming FGAN likely
produced oxidizing gases, and those mixed with flammable smoke to produce a detonable gas cloud over
the FGAN pile in the main bin and possibly in an adjoining bin linked to the main bin through a series of
holes cut in the partition between the bins. The cloud consisted of powerful oxidizers that would be
expected when FGAN undergoes thermal decomposition—such as NO2, O2, and HNO3 as wells as fuelrich smoke and pyrolysis products off-gassing from the molten FGAN. The gas cloud then might have
ignited from above, undergoing a gas-phase deflagration-to-detonation transition (DDT) in the
confinement of the bin.

Scenario 2: Detonation along fire line
This scenario involved heating of the FGAN through the walls and is noted as being highly unlikely, so to keep this short I'm not going to quote it here. See section 4.3.2 for details.
Scenario 3: Detonation in elevator pit

Another possible detonation scenario focuses on the elevator pit near the FGAN bin. A fiberglass lid
covered the pit, and the floor sloped away from the pit to prevent runoff from entering it, but the fire
might have melted the cover, and FGAN remnants could have been in the pit. ...
If the
detonation began in the pit, then the most feasible mechanism would be a collapse of the west wall of the
bin, spilling FGAN into a mixture of burning rubber from the melted elevator belt and residual FGAN in
the bottom of the pit. The mass of the falling FGAN, combined with the strong confinement of the
concrete pit walls, might have provided the conditions for a solid phase DDT beginning in the bottom of
pit and spreading into the main pile.

TLDR
So yeah, the TLDR here is that conditions in a fire can be extremely complicated and unpredictable, giving rise to a lot of opportunities for contamination and detonation.
In particular:

AN can melt and the liquid can do unpredictable things.
The containers can be destroyed by fire allowing AN to escape into unpredictable places.
Even if the AN was contaminate free under normal conditions, anything in the area can become a fuel source including the containers themselves, fire debris, smoke, soot, collapsed building parts, etc.

In Beirut we saw that there was already a fire burning for a significant amount of time before the explosion, as well as a smaller explosion that occurred < 30 seconds before the main one. There were also flashes and bangs and a lot of other stuff going on there (reportedly there were fireworks stored in the same warehouse). It was at a seaport, too, meaning there were probably a lot of nearby things to act as fuel sources.
It is very, very conceivable that the AN became both heated enough and contaminated enough during this time to detonate.
Here is a list of other AN accidents that you could research on your own to find out about other scenarios that can lead to detonations. Most notable:

BASF, Oppau, Germany, 1921
Texas City, Texas, USA, 1947
AZF, Toulouse, France, 2001
Ryongchŏn, North Korea, 2004
Tianjin, China, 2015

Also, you may find some of the theories regarding the 1988 PEPCON accident in Nevada, USA interesting as well. That was not ammonium nitrate (it was ammonium perchlorate, another oxidizer), but the possible scenarios are similar and it also illustrates the complexity of those kinds of situations.

Reportedly there were fireworks stored in the same warehouse as the AN in Beirut: https://www.youtube.com/watch?v=LNDhIGR-83w

When one thinks of fire and dissects the requirements of a fire into a fuel and an oxidiser, this is – in chemical terms – only a different way of determining an oxidising agent and a reducing agent in a redox reaction.
Ammonium nitrate – like other nitrates such as potassium nitrate which is also known as salpetre – is an oxidising agent through the nitrate anion in which the nitrogen atom is in the $\mathrm{+V}$ oxidation state. This happens to be the highest oxidation state nitrogen is found in in stable compounds; high oxidation states of electronegative atoms tend to be unstable and react as oxidising agents (see also perchlorate $\ce{ClO4-}$ with chlorine in the $\mathrm{+VII}$ oxidation state, for which a similar argument can be made). Thus, the nitrogen in nitrate will, as part of a redox reaction reduce its oxidation state by taking up electrons.
Examining ammonium nitrate reveals that there is another nitrogen atom in the ammonium cation which is in the $\mathrm{-III}$ oxidation state. As this is nitrogen’s lowest known oxidation state, it cannot act as an oxidising agent but it can act as a reducing agent – see for example the Ostwald process in which ammonia (also in the $\mathrm{-III}$ oxidation state) is burned in an oxygen atmosphere ultimately leading to nitric acid.
Thus, we actually have both an oxidising and a reducing agent in the salt: the former is the nitrate anion and the latter is the ammonium cation. Under normal conditions, these will not react (hence why ammonium nitrate is shelf-stable, can be purchased from chemical suppliers and the MSDS does not include explosive as a warning sign). However, they are capable of reacting and they have thermodynamically favourable reaction pathways as Matthew’s answer nicely outlines.
Interestingly, ammonium nitrite ($\ce{NH4NO2}$) is far more labile and very hard to prepare as the two ions tend to react to form nitrogen gas and water even in solution at room temperature.
$$\ce{NH4+ + NO2- -> N2 + 2 H2O}$$
As for the question whether the disaster of Beirut was an deflagration or a detonation: which term is applicable depends solely on the rate of decomposition and the resulting velocity of the front. If decomposition happens at supersonic rates, this leads to a shockwave characteristic of a detonation. If decomposition is subsonic, you are observing a deflagration.

edit: my original assesment was wrong
It does decompose ENDOTHERMICALLY, so it's not self propagating, and is not a low explosive:
NH4NO3 - > 2H2O + N2O
However a passing shockwave can induce this decomposition in a large quantity of ammonium nitrate at once, which is a detonation. The starting shockwave could have come from another explosive or from a fuel fire that was intensified by the oxidizing properties of ammonium nitrate. Not sure about exactly how the latter happens, but this is not the first time it did
https://en.m.wikipedia.org/wiki/Ammonium_nitrate_disasters

So ammonium nitrate by itself is considered a low explosive, but has potential to detonate (ie, be a high explosive) under the right conditions? It's hard to imagine something worse than the explosion I saw in videos, but I wonder what the difference would have been if it were combined with fuel oil? More powerful, or simply easier to detonate, or both?

please see my edited answer, also yes, an ANFO explosion would be even more powerful and easier to detonate

I guess one way of looking at things can be that ammonium nitrate decomposing into nitrous oxide and water in an endothermic fashion. However, as per this source, in the presence of a heated catalyst, $\ce{N2O}$ will decompose exothermically into nitrogen and oxygen, at a temperature of approximately 1,070 °F (577 °C). So I guess the exothermicity of this second decomposition, and obviously the oxidizing properties of the products can create the huge detonations observed.

I object. N2O can be obtained by slow controlled decomposition. An explosion goes straight to N2 and O2.

Oh, I see...So the explosion derives its exothermicity directly from the initial breakdown of ammonium nitrate to N2 and O2 ? @IvanNeretin

Yes. It can be greatly enhanced by addition of any reducing agent to react with O2. Short of that, NH4NO3 is a relatively weak explosive, but it comes in great quantities, hence the effect.


text:
6463-T6 aluminum oxidation and role of water and ethanol vapor
Our issue...in a nutshell:    We use 6463-T6 aluminum as a sampling "cell" for analysis of water and ethanol vapor mixtures in both wet and dry gas variations. As I understand it, the aluminum oxide covering our cells surface is being or has been previously damaged and causing micro-abrasions of its surface. This allows water vapor to "attach" to the cells surface which negatively affects our testing process. Incomplete removal of the water or ethanol vapor within the cell is a failure. We have looked at methods of cleaning the cells to essentially create a newer cleaner oxidized surface. Acid baths, boiling in vinegar, etc.. now we are considering how the metal surfaces pH value can affect the quality or this aluminum oxidation. We are kind of stuck and any assistance on testing protocols or processes to achieve our goal would be appreciated.

Ethanol slowly reacts with Al to form an ethoxide. Though the cell could be anodized, https://www.wikihow.com/Anodize-Aluminum , to passivate it, the ethoxide reaction might still interfere with critcal measurements.

Dr Moishe
thanks for the info. will keep it in mind. But it doesnt appear to address my main issue.
this comment system will not allow me to fully comment and explain...not enough characters allowed. I will have to break it into mulitple comments

thanks for the info. will keep it in mind. But it doesnt appear to address my main issue.
A properly cleaned cell assy will have no issues with ethanol, water vapor, nitrogen gas or ethanol/nitrogen gas mixes. My testing process verifies that the system measures these vapors/gases and also purges out back to or close to the baseline value established at test start. Essentially test no 1 goes like this:

system purge to stable baseline value
stable baseline measurements taken
filter measurements at baseline taken
water vapor introduced and stable value measured
system purge to stable value
baseline measurements taken
filter measurements taken
same process repeats for nitrogen gas.
calculations occur checking various aspects/ratios etc... of test conducted. looking at relationships between water vapor and baseline, and nitrogen vapor and baseline.

test no 2 uses a similar process (purging, zeroing, ambient testing, etc...) to introduce a high value ethanol/nitrogen gas mix and then monitor it for a value drop after the gas introduction ceases. Issues are high value drops of the gas mixture that occur within the first 10 and 20 seconds after introduction ceases.

The system is monitored and tested with a known verified reference cell assembly to ensure no mechanical, electrical or other issues are injected into the test process.

Our problem is essentially an unknown. We could assume that the aluminum cells oxidation layer has been damaged/changed some way that allows for the vapors to be "trapped" within the cell (pitting??) and thus causing incorrect readings. I have found white residue and white splotches on the cells interior. A cleaning process using boiling vinegar and rinsing seemed to work, but its not a repeatable trusted process. So we are looking at other methods of cleaning.

We dont have the resources to do a full on metallurgical analysis and full failure analysis....we know the cells are "dirty?" in some way and that cleaning them fixes the issue. We just need to find a good repeatable cleaning method. And we think we should stick with an acidic cleaning approach as acids attack/prefer the oxide layer over the aluminum whereas alkaline cleaning attacks/prefers the aluminum vice the oxide layer.

It is better to edit your question, rather than  add a comment.


text:
Kinetics of inversion of cane sugar
Question 1
The formula for concentration of cane sugar as a function of time is given by
$$k = \left(\frac{2.303}{t}\right) \log\left[\frac{\alpha(0) - \alpha(\infty)}{\alpha(t) - \alpha(\infty)}\right]$$
where $\alpha(t)$ is the optical rotation of the solution at any time t. This was the formula given in my textbook, but I feel it is wrong.
The optical rotation in the formula is the rotation of the solution, which includes rotation due to external factors which is taken care of by the $\alpha(\infty)$ term but it also includes rotation due to glucose and fructose , whereas the formula would only be valid if $\alpha(t)$ denoted only the rotation of cane sugar + external factors.
The reasoning given in the book is that the rotation due to glucose and fructose included in the external factors, but that is just wrong. The fundamental assumption is that rotation due to external factors remains constant and hence gets cancelled out, but if you include glucose and fructose in external factors, then their concentration and hence rotation is changing with time. Also they do not form a racemic mixture, so their rotations do not get cancelled out.
Question 2
Another question given in my textbook was to find the optical rotation if the reaction was 50% complete. Given rotations of cane sugar = 34, glucose = 10 , fructose = -20
The solution was $$\left(\frac{17+5-10}{3/2}\right)$$
Notice the 3/2 which represent the total moles in the denominator. If this is correct then the formula becomes even more incorrect as optical rotation also depends on total number of moles.
What is the correct formula? I am also not able to understand the calculation for total optical rotation as given in second question. Why did they divide by 3/2 and how does the optical rotation depend on the number of moles?

Your formula is wrong. It is $\alpha (t) - \alpha (\infty)$ and not $\alpha (0) - \alpha (t)$ in the denominator

Yes edited that

Did you try using $\frac{n}{n-x}$ to prove this formula on your own? The equation for $k$ of a first order reaction

Which text book is this from?

@Safdar this is from P bahadur - Concepts of physical chemistry. I did try to prove this formula but I couldn't figure out how to remove the rotation of glucose and fructose from my expression.

It seems as though your issue with the formula arises due to the fact that there is a feeling of not knowing what is considered to be factors that are internal or if they are external factors. To prove the formula, the method would be as follows.
The formula for the given reaction would be:
$$\ce{C12H22O11 + H2O ->  C6H12O6 (glucose) +C6H12O6 (fructose)}$$
Drawing a modified version of a RICE table with optical rotation, and the case at t=$\infty$, we get:
\begin{array}{|l|c|} \hline
\text{R} &\ce{C12H22O11   &C6H12O6(glucose)  &C6H12O6 (fructose)} &\text{Optical Rotation} \\ \hline
\text{I} & n & 0& 0 & \alpha_{s}n+\mathrm E \\
\text{C} & -x &  x&  x & - \\
\text{E} & n-x &  x & x & \alpha_{s}(n-x)+ \alpha_{g}(x) + \alpha_{f}(x)+ \mathrm E \\
\infty & 0 & n & n & (\alpha_{g}+\alpha_{f})(n) +\mathrm E \\ \hline
\end{array}
Now, for a general first order reaction, the formula would be:
$$k=\dfrac{2.303}{t}\log \frac{n}{n-x}$$
Therefore, we need to find the value of $\dfrac{n}{n-x}$. Using the given data, we get:
$$\frac{\alpha(\mathrm I)-\alpha(\infty)}{\alpha(\mathrm t)-\alpha(\infty)}=\frac{n}{n-x}$$
Therefore the formula given is true, even considering an external factor. You consider glucose and fructose to be internal.
$$k=\dfrac{2.303}{t}\log \frac{\alpha(\mathrm I)-\alpha(\infty)}{\alpha(\mathrm t)-\alpha(\infty)}$$

I think $\frac{n}{n-x}$ = $\frac{\alpha_s(n)}{\alpha_s(t)}$

In your expression rotation of glucose and fructose is also appearing whereas it should only be of cane sugar. Also how to calculate total rotation at $E$ (Question 2)

Why should it not appear in the expression? In a solution the net optical rotation is said to be the sum of the all components internal and external. $\alpha$ is observed optical rotation.

Because in first order reaction rate is directly proportional to concentration of solute, which is assumed to be directly proportional to optical rotation of solute

@ManitAgarwal-Elpsycongroo What we are able to observe in a reaction is the final optical rotation of the entire solution. And that is what we use to check the rate. We cannot observe only the rotation due to one component of a mixture in said mixture.

isn't that wrong?

Also, in the reaction, the solvent is water and so sucrose,glucose and fructose all are solutes.

Let us continue this discussion in chat.


text:
Applying Bohr's model for Helium atom
Consider the following atomic model for Helium. Helium contains 2 electrons revolving in circular orbit around the nucleus,in such a way that there is minimum possible repulsion between them.Calculate the first ionisation energy of this Helium.Will there be any difference in the ionisation energies calculated and experimentally found?

Bohr's model is applicable only to monoelectron species. So it may work in He+ ion.

Now that's an interesting question. Why don't you go and compare these two numbers?

@IvanNeretin I have calculated the value and I got 29.62 eV but the experimentally found value is 24.6 eV. Could you explain me why that difference occurs?

Yes. It means the model is not applicable.

To be strict it doesn't mean the model is not applicable - it has just been applied after all. It means it is not very accurate.

Can you show some work on how you calculated the value? It's not very helpful to ask what is wrong when we don't know what you did.

Nothing is wrong. Even applying a more accurate model hardly will bring you to the experimental values. I personally known active researchers still calculating hydrogenoid systems :)

What this exercise suggests boils down to placing the two electrons on opposite sides of the nucleus. In this way, the repulsion is minimal because the distance between both electrons is maximal. We can safely assume there will be a difference between the energy calculated with this simplified extension to Bohr's model and the experimental value, otherwise Bohr or his peers would have adopted it. The inability to satisfactorily calculate the ionisation energy of Helium with any such model was a major driver for scientists in the early 20th century to adopt a new quantum mechanical description of the atom.
Bohr's theory tells us that the ionisation energy for one-electron atoms varies with the square of the nuclear charge, i.e. $E=Z^2$ in Rydberg units (1Ry = 13.6eV). So the ionisation energies of $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$ are respectively 1, 4, 9, ...
The two-electron model described here can be simplified to a one-electron problem because the particles are assumed to be neatly aligned at all times. To this end, we define an effective nuclear charge $Z_\rm{eff}$ which sums the charge of the protons in the nucleus (+2 for Helium) and that of the other electron (-1), weighed by the square of the distance : since the opposite electron is twice as far as the nucleus, its repulsive force is four times smaller than the attractive force of one proton in the nucleus. So we get $Z_\rm{eff} = 2-\frac{1}{2^2} = \frac{7}{4}$. The ionisation energy then becomes $E=Z_\rm{eff}^2 = \frac{49}{16}$ = 3.06 Ry (or 41.6 eV).
The experimental value is 1.81 (24.6 eV), which is significantly lower.
Apparently the electron-electron repulsion in the Helium atom is (much) stronger than this simple model suggests.

I think we should go by the Slater's rules?

The exercise at hand seems to be seeking a calculation for a thought experiment. Slater's empirical rules do not apply to that thought experiment.


text:
Which one of the two is the stronger base: methoxide ion or hydroxide ion?
If we consider the experimentally determined pka values, it looks like hydroxide ion should be the right answer but in many books it has been mentioned that all alcohols are less acidic than water implying that all alkoxide ions are stronger bases than hydroxide ion. This has created a lot of confusion.

Could you mention where you saw the $\mathrm pK_\mathrm{a}$ values and what the values were?

Yes sir. I got the values from Solomon and Fryhle organic chemistry book. pka of methanol is given as 15.5 and that of water as 15.7

https://chemistry.stackexchange.com/questions/35068/is-methoxide-a-better-nucleophile-than-hydroxide#:~:text=Alkoxides%20tend%20to%20be%20stronger,to%20more%20easily%20donate%20electrons.&text=Methoxide%20is%20a%20relatively%20unhindered%20base%2C%20as%20is%20hydroxide.

If methanol is more acidic than water does that not mean that it's conjugate base (methoxide) is less basic than the conjugate base of water (hydroxide)?

@Safdar Alright I do not know how much I am correct. I shall get back to this after some more reading. Apparently, NMR spectroscopy has proved that CH3- is electron widrawing in methanol. I am not sure. https://chemistry.stackexchange.com/questions/42535/is-methanol-more-acidic-than-water Otherwise the reason of basicity is same. CH3O- has I+ effect of the CH3- group causing it to be more basic than OH-.

We have $\ce{MeO-}$ and $\ce{HO-}$
$\ce{-CH3}$ has a +I effect. Therefore it destabilizes the methoxide ion. $\ce{-H}$ has a comparatively lesser +I
This leads to the conclusion that $\ce{OH-}$ is more acidic than $\ce{CH3O-}$.
This seems like a valid argument, but the order varies in gaseous/aqueous phases, because we consider polarizabilty and solvation effect, rather than inductive effect there

Question in the title.. :). That has to be fixed in the given answer though.


text:
What exactly is lattice energy?
I was going through my chemistry textbook (Chemistry, 10th Ed. by Raymond Chang) when I encountered this explanation of lattice energy.

9.3 Lattice Energy of Ionic Compounds
We can predict which elements are likely to form ionic compounds based on ionization energy and electron affinity, but how do we evaluate the stability of on ionic compound? Ionization energy and electron affinity are defined for processes occurring in the gas phase, but at 1 atm and 25 °C all ionic compounds are solids. The solid state is a very different environment because each cation in a solid is surrounded by a specific number of anions, and vice versa. Thus, the overall stability of a solid ionic compound depends on the interactions of all these ions and not merely on the interaction of a single cation with a single anion. A quantitative measure of the stability of any ionic solid is its lattice energy, defined as the energy required to completely separate one mole of a solid ionic compound into gaseous ions (see Section 6.7).

In this text, it seems like the writer is implying that for an ion pair in the gaseous state, ionisation energy (IE) and electron affinity (EA) can be used to measure it's stability. But even an ion pair in the gaseous state is held by an electrostatic force that gives it more stability, so more energy will be needed to separate the ion pair and that energy  would be more than IE and EA. What does the writer mean, and what is the correct definition of lattice energy?

"Gas phase" is supposed to mean "zero contact, infinite distance between ions" in this context.

In other words, that sentence about the gas phase and 1 atm/25°C is utter bullshit. Can you please give the book title and author?

It begins by explaining that the relative EA and IE of two atoms can be used to estimate whether they form an ionic compound. The EA and IE are often combined to compute an electronegativity parameter, which is compared between atoms. In this case the question is about the nature of the bond between the atoms, not about the stability of the resulting solid.

@Karl so are you implying that the text is incoherent?

I said the text is just plain wrong. IE and EA are properties of single atoms, not gas molecules (like $\ce{Cl2, F2}$). As such, they are also obivously independent of temperature or pressure. That text is just ridiculous.

The last sentence is just as wrong. The lattice energy is the one needed to separate the ions to infinite distance. That´s not the same as vaporising them. The difference won´t be large, but Mr. Chang is still lying to students for no good reason. Perhaps he thought they were too stupid to grasp the idea of a theoretical value, or was trying to be practical, whatever. Schoolbook authors often do that, and its also disgusting there, but a university professor?

In summary: IE and EA are just partial terms in the formation enthalpy of solid ionic compounds, that includes also LE and atomization enthalpy of elements. Their values alone can give a hint, but do not provide the final word.

Notice that the text says nothing explicit about ion pairs.
IE + EA relate to free gaseous unbound ions. Like:
$$\ce{M(g) -> M+(g) + e-(g)}\tag{IE}$$
resp.
$$\ce{X(g) + e-(g) -> X-(g)}\tag{EA}$$
The lattice energy (LE) is the energy released by forming solid ionic compounds from free ions:
$$\ce{ M+(g) + X-(g) -> MX(s)}\tag{LE}$$
Gaseous ion pairs have already released a part of the lattice energy.
Note that for ion pairs, the pairing energy may be smaller than the sum of IE+EA, especially for big ions, so forming pairs may not be favoured.

A picture is worth a thousand words.

In the picture above, start with a mole of solid sodium and half a mole of Cl$_2$ gas, in the ground state. Then gasify the sodium with 107.3 kJ of heat, then ionize it with 496 kJ of energy, then dissociate Cl$_2$ to Cl radicals with 121.7 kJ, then allow the electron from the sodium to combine with the chlorine radical. You get heat released: 384.8 kJ.
The picture shows only one NaCl pair, so understanding how this energy is "released", and what it does to the system becomes a little mysterious, but if you realize that 6 x 10$^{23}$ ion pairs are all doing this at the same time, it's a bit easier.  This is called the Born-Haber cycle.
One ion-pair doesn't have a lattice energy, but the mole of NaCl does, because one ion-pair doesn't have an extended lattice. When you add everything up, the lattice energy (the blue arrow on the right, pointing downward) comes out to -787.8 kJ/mol. http://derekcarrsavvy-chemist.blogspot.com/2016/12/lattice-energy-1-calculating-lattice.html
The calculation suggests that NaCl is more stable, energetically, than a mixture of solid sodium and chlorine gas, which is in agreement with observation. Also less reactive.


text:
Electrophilic substitution on Borazole
As Borazole ($\ce{B3N3H6}$) is aromatic, it can show electrophilic aromatic substitution. Now, my question is when borazole undergoes electrophilic aromatic substitution reaction, on which atom does the electrophile attack? Is it boron or nitrogen? Also, how to compare it's rate with the same type of reaction when carried on benzene? Should I consider stability of arenium ion (Wheland intermediate) formed when comparing the rates?

Borazole isn't really that aromatic if compared to benzene. I'd expect electrophiles to add to nitrogen.

Chiavarino, et al. [1] report  that where electrophilic substitution occurs with carbocations (borazole more often undergoes addition), it does so on nitrogen.  Nucleophiles such as methanol prefer boron.
We can explain that result in terms of both molecular orbitals and the Wheland intermediate.  In the molecular orbital explanation, recall the familiar delocalized pi orbitals in the benzene ring.  With the polar bonds of borazole the occupied bonding orbitals whose electrons will add to an electrophile are shifted towards the nitrogen atoms, so the electrophile will tend to bond there rather than on the boron (the latter has instead mostly the empty antibonding pi orbitals and is thus favored in nucleophilic substitution).
In terms of the intermediate, adding the electrophile to the ring leaves the remaining pi electrons in a deficient condition -- only four electrons cover the remaining five conjugated atoms, with most of the electron deficiency falling on atoms ortho and para to the attack site.  Attacking on nitrogen puts the less electronegative boron atoms at these deficient sites leading to a more stable EAS intermediate.
Reference

Barbara Chiavarino, Maria Elisa Crestoni, Simonetta Fornarini, "Electrophilic Substitution of Gaseous Borazine," J. Am. Chem. Soc. 1999, 121(11), 2619-2620 (https://doi.org/10.1021/ja983799b).


text:
Why is the overall change in entropy of Photosynthesis positive?
Net Reaction:
$$\ce{6CO2(g) +6H2O(l) -> 6O2(g) + C6H12O6(aq)}$$
The net reaction shows that 12 moles become 7 moles; 6 of gas,and 1 complex molecule.
This implies a decrease in entropy over the net reaction.
However, from reading further sources, it seems that the change in entropy is positive, but they don't justify it. Why is the overall change in entropy positive, over the series of reactions that make up this process? Could this be linked to the role of UV Light in this reaction?

The state of glucose should be solid or aqueous.

Just looking at the net reaction, you would think the entropy decreases. Do you have any evidence that is increases? If you study photosynthesis vs glycolysis (the reverse reaction), you will learn that glycolysis is the direction toward equilibrium (, and that it is exothermic in that direction). It would be great if you could write more about your question to make sure it does not get closed.

State of H2O should be liq, and are you including the photon absorbance?

What do you sketch lower entropy and stores energy. It is something apparently deeply linked to life. This is a field which poses a big of puzzles and misunderstanding. Note that this is true if we look at compartments. But still quite peculiar of living things and their products. Overall photosynthesis increase the universe entropy as for hundred photons impinging not all are absorbed and not all are entering the energy balance of the reaction you have wrote. Overall efficiency is few percent, or may be 0.3%, I've forgot the precise value.

Light is a reactant not a catalyst here

To answer this question, we need to be clear whether we are talking about the change in entropy of the system, the surroundings and the Universe.
With regards to the entropy of the Universe, the answer is direct: photosynthesis is a process we observe to happen on a macroscopic scale. Therefore, by the second law of thermodymamics, photosynthesis must cause the entropy of the Universe to increase, period.
With regards to the entropy of just the chemical reagents, if we consider those as our system, then indeed the entropy of formation of glucose is negative, equal to approximately $\mathrm{-40\ cal\ mol^{-1}\ K^{-1}}$.
The way to reconcile these facts is to note that there is a significant increase in entropy which comes from the destruction of high-energy visible photons from the Sun, whose energy is scattered into chemical bonds and multiple low-energy far infrared photons.
In a simplified analysis, if three photons of $\mathrm{700\ nm}$ light are absorbed to generate one molecule of glucose and excess energy is radiated away with black body spectrum at $\mathrm{300\ K}$, then photosynthesis is allowed by the second law of thermodynamics, so long as the photosynthetic efficiency is less than 88%. The observed efficiency of photosynthesis using red light is on the order of 59%, so this requirement is cleared with room to spare.
Depending on the level of detail you want, it gets complicated to work through all the checking and balancing of entropy changes at every individual microscopic step, but the overall results must hold as stated above.
References:
Negative entropy and photosynthesis. Wesley Brittin and George Gamow, Proceedings of the National Academy of Sciences, 1961, 47 (5), 724-727. DOI: 10.1073/pnas.47.5.724
Entropy balance in photosynthesis. Wolfgang Yourgrau and Alwyn Van Der Merwe, Proceedings of the National Academy of Sciences, 1968, 59 (3), 734-737. DOI: 10.1073/pnas.59.3.734
Entropy production and the Second Law in photosynthesis. Robert S. Knox and William W. Parson, Biochimica et Biophysica Acta (BBA) - Bioenergetics, 2007, 1767 (10), 1189-1193. DOI: 10.1016/j.bbabio.2007.07.004


text:
XPS/EDS vs FTIR/Raman
So I have been studying the four methods the past couple of weeks. To my understanding, XPS is a very surface-sensitive technique. It can give information about the composition of the surface but also of the bulk material via depth profiling. I think it is mainly used for qualitative analysis, i.e. find out the elements of the sample.
The same goes for EDS/EDX (in SEM), it is used to identify samples. Strictly speaking, EDS is not a surface sensitive technique I think, because the spectral resolution is not as good as XPS. XPS looks at the first few layers of atoms, while EDS looks more at the bulk of the sample.
Now my question is: can't we do the same with FTIR or Raman? Except of course that with FTIR or Raman it would be difficult to study metals, whereas with EDS/XPS it is easy.
What are the advantages of FTIR or Raman w.r.t EDS/XPS or the other way around?
I know that there are handheld devices for FTIR and Raman, so they would be easier to use to identify unknown samples and require less sample preparation than XPS or EDS.
To me it seems that all four can be used interchangeably, except for maybe XPS, because it is so surface-sensitive. I know it depends on the applications, but in general, can either of these be used to identify unknown materials?
Summary of the question by the OP from their comments:  I mean to say that either of the methods can be used to identify unknown samples. But of course, it depends on the sample and what you want to know from the sample to know which method works best. While I agree that the methods are different and answer different questions regarding the sample, I don't agree that they are "completely" different. To some extent they do answer the question of what the unknown sample is. Right?

FTIR and Raman microscopy exists, you can buy such instruments off the shelve. Other than that, I think you should keep on studying, because IR/Raman and XPS have basically nothing in common. Completely different techniques that answer completely different questions about your sample.

I think my question isn't very clear. I mean to say that either of the methods can be used to identify unknown samples. But of course, it depends on the sample and what you want to know from the sample to know which method works best. While I agree that the methods are different and answer different questions regarding the sample, I don't agree that they are "completely" different. To some extent they do answer the question of what the unknown sample is. Right?

@Paperreader Keep in mind that, energetically speaking, comparing the three methods with each other, XPS is a high energy / short wavelength method.  Energy is used to excite (core / inner) electrons of atoms to a higher, previously unoccupied level and their radiant relaxation is observed.  By comparison, IR and Raman are soft energy/long wavelength; recording the change of dipole moment of bond vibrations (IR), or the change of polarizability of bonds (Raman).

@Paperreader If you have studied those methods, then you should know what results they give. Do you?

Buttonwood, thanks for the comment, I'll keep that in mind.
Karl, I do know the different results that they give. But I am talking about identification of unknown elements. All four can do it, but I guess it just depends on the application which one is best suited.

@Paperreader, Raman and FTIR are not element identification techniques. You might find isotopic patterns but only happens in very high resolution IR.

@Paperreader Identifying pure elements, well. That´s not something chemists usually do, unless the label on a bottle has come off. ;) And all metals obviously look very similar in IR.

Ah sorry, wrong formulation. FTIR/Raman identify unknown molecules. Thanks for the clarification!

I think your main question is related to the difference? Have you heard a famous poem, The Blind Men and the Elephant? It is a story, I'd rather copy from Wikipedia

It is a story of a group of blind men who have never come across an elephant before and who learn and conceptualize what the elephant is like by touching it. Each blind man feels a different part of the elephant's body, but only one part, such as the side or the tusk. They then describe the elephant based on their limited experience and their descriptions of the elephant are different from each other. The moral of the parable is that humans have a tendency to claim absolute truth based on their limited, subjective experience as they ignore other people's limited, subjective experiences which may be equally true.

So replace these men with these analytical techniques and the sample with the elephant. XPS /EDS can see what Raman/FTIR cannot and vice versa.
XPS/EDS are elemental techniques and they will tell you what elements are present. Sometimes XPS spectrum, for an expert, can also reveal what is the oxidation state of that element, if you do high resolution scans. Each sample may take hours and hours.
Raman and FTIR can used to probe surfaces (they are fancy surface vibrational spectroscopy techniques) and it will tell you the molecular information, and even the angle at which the molecule is standing on the surface.
From the Raman spectra of an unknown it will be very hard to guess what elements are present.
In short, these are complementary techniques. Someone who is trying to study surfaces would use all of them - if needed.

Thanks for your answer. My question was a bit too broad I think, but this helps me answer it.

Indeed, the one you like is the one you have in your lab. You will use it even if it is suboptimal for what you are looking for.


text:
Huckel's rule and aromaticity
I am learning to determine whether a monocyclic conjugated alkene is aromatic or anti-aromatic or not.
question 1:
cyclobutadiene: according to Huckel's rule, 4 * 1 = 4, it should be antiaromatic, like what is said here. However, Wikipedia says the 2 pi bonds have a 'more double bond' character, making it a rectangular shape (electrons not delocalized). Therefore, it is more like a conjugated diene, hence non-aromatic. What is the truth?
question 2:
planar / non-planar: for [18]annulene, it is aromatic (4 * 4 + 2 = 18). From Wiki,  ''it is large enough to comfortably accommodate six hydrogen atoms in its interior, allowing it to adopt a planar shape, thus satisfying Hückel's rule.'' So in this molecule, there is an extremely small steric effect. So I am wondering, is size of the inner space also a factor in determining whether a cyclic molecule is aromatic or not? If the inner lumen of the molecule is larger, it can follow Huckel's rule? Like Cyclodecapentaene, as the 2 hydrogens cannot be in planar (not comfortably accommodate) (figure 3 in Wikipedia). It is non-aromatic rather than aromatic (do not follow Huckel's rule).
question 3:
cyclooctatetraene is non-aromatic rather than antiaromatic (4 * 2 = 8) since it avoids by changing into a 'tub' shape (avoiding being the unstable planar conformation). I am wondering if other supposedly antiaromatic molecules can change their conformation to be non-aromatic (become more stable) as cyclooctatetraene does? Are there any criteria for changing the shape, or cyclooctatetraene is the only example? If yes, why? If no, what monocyclic conjugated alkene molecule is actually antiaromatic then?
Thank you very much.

In all three questions, you are probably missing the following:
for coaxial pi orbitals to form, C-C bonds must be at certain angle, ideally 60 degrees, as in benzene.
Cyclobutadiene cannot do that. Long cycles in some conformations can, but not all of them can also become flat so that these pi orbitals can effectively overlap.

The orbitals in cyclobutadiene still overlap in the rectangular shape. It would be even more antiaromatic if you somehow forced it to be quadratic. To become nonaromatic, it would have to adopt a twisted shape, but double bonds (i.e. sp2 atoms) want to be planar, so that is also not possible.

@Karl I agree that the conformation of cyclobutadiene is impossible to be changed. However, for larger cyclic molecules which supposedly to be antiaromatic (e.g. cyclooctatetraene),  in general, can they all twist into a non-planar shape to avoid being antiaromatic (more unstable)? Thanks

@sa7 OK so for larger cycles, apart from using Huckel's rule, I need to check the C-C bond angles to see whether coaxial pi orbitals can form, hence to check whether it can be planar (flat), and eventually conclude whether it is aromatic or not? Thanks

I wonder why you apparently don't get that asking about there different compounds is two questions too many...

Not to say they are already covered https://chemistry.stackexchange.com/questions/83605/h-nmr-spectroscopy-of-18annulene https://chemistry.stackexchange.com/questions/18557/is-cyclobutadiene-antiaromatic

234ff, yes, Huckel rule has sense only for planar cycles (so that p-like orbitals may overlap) and also C-C-C angle must be close to 120 degrees (or else p-orbitals would "bend" from ideal direction, which is parallel to each other).

@sa7 That's not correct. While Huckel says they should be flat, it's not exactly necessary. And this text about angle makes no sense. It can be even 60 and that's not a problem.

@Mithoron OK thanks it is better for me to repost in separate questions, I have voted to close this question.

@Mithoron no, it should be flat and 120 degrees, for single reason: these are optiomal conditions for close and coaxial p-orbitals on C atoms, to let them overlap and form a pi-orbitals, which is underlying reason of Huckel rule.

It's not better to repost in separate questions - it would duplicate this and other questions.

@sa7 Of course 120 is optimal for unsaturated compounds, but it has nothing to do with p orbital overlap and everything with ring strain - cyclopropenyl cation is aromatic just right.

Steric effects play an important role in determining the planarity, aromaticity of a molecule. Cyclobutadiene is sterically unstable(90 degrees angle between bonds), hence having high energy, and thus  anti-aromatic.
A lot of compounds change their conformations for reducing their overall energy. Check for the conformations of glucose, for instance. A compound will always prefer its low energy conformation, which occurs when Cyclooctatetraene is non-planar depicted in the right hand perspective drawing

(reference)
This compound looks like the big brother of benzene, which may incorrectly induce the notion of it being aromatic, or, by Huckel rule, anti-aromatic. But the inner angle of a regular octane shape is 135 degress, while a much stable angle nearby it is 120 degrees(an sp2 config.), which only occurs in the tub config. of Cyclooctatetraen. 120 degrees is not achieved, but ~126 degrees is achieved.
Frozen (at 129 K) cyclooctatetraene's structure was characterized by single crystal diffraction analysis by Claus and Rüger. You find the corresponding interactive model here.

Cyclobutadiene distorts to a rectangular geometry, so it's not exactly a perfectly antiaromatic square molecule. Also, I feel like "high energy" isn't a sufficient condition for being antiaromatic, much like how "low energy" doesn't quite mean aromatic.

I think the usual understanding of the words aromatic and antiaromatic, when used in conjunction with Huckel's rule, is something along the lines of the Dewar resonance energies, cf. first part of my answer https://chemistry.stackexchange.com/a/82690/16683. Colloquially one would say that benzene is aromatic because its π system is more stable than hexatriene.

It is rectangular, which means it is sterically hindered(90 degrees angle), while a double bonded molecule should have angle nearby 120 degrees. It is very sterically hindered. Thus it will have a high energy: the sense that it will be 'unstable'. Stability is a criteria or a result of aromaticity. Benzene is less reactive in many substitution reactions.

@losnihciL A double bond primarily needs to be planar. If you twist the two sides of a double bond out of plane, it becomes antibinding. There is no sterical hindrance in cyclobutadiene, the problem is electronic, not steric.

@Karl Steric hindrance is there. The hindrance is not caused by atoms/molecules, but by bonds, which are too close to each other to be repelling in nature, and thus reactive. That is why cyclobutadiene exists as a dimer in isolated state.

@losnihciL thanks. So for other antiaromatic molecules classified from Huckel's rule, they tend to undergo conformation changes to lower their energies (from being antiaromatic) to achieve a much stable bond angle (120 degrees)  like Cyclooctatetraene?

Huckel's rule is not applicable to many compounds. Yes, conformation changes are seen in a lot of compounds, the most stable conformer is usually found in a higher quantity than other unstable conformers. 120 degrees is for sp2 hybridized compounds, while around 109 degrees for sp3, and approx 180 degrees for sp compounds.

That is utter nonsense. Steric hindrance is, by definition, the electrostatic repulsion between atoms that are pushed close by the backbone of the molecule they are part of.

According to webster dictionary, steric means relating to, or involving the spatial arrangement of atoms in space. In cyclobutadience the spatial arrangement of C atoms is such that the angle between C=C-C bond is 90 degress, much unstable...

I recommend you learn chemistry from appropriate sources, not merriam webster. And yes, an sp2 hybridised carbon atom with a 90° bond angle is very unhappy, things being the way they are. This is however  NOT called  "steric hindrance". Total misnomer.


text:
Yield of possible products from acid catalysed hydration of an alkene with a phenyl group

Here, (d) may be the major product due to high temperature.
My questions are:

Will a kinetically favoured (minor) product (c) form?

What shall be the order of yield of products?

d>c>b>a........

@user600016 why c over b? Also, do you not agree with the answer given below in that case? If so, please take the time to formulate a proper answer or comment on the given answer on why it would be incorrect.

Why do you think c) is the kinetic product? I mean what is your reasoning behind that. Also, acid catalysed hydration reactions are reversible, so the kinetic product would not form in a high yield.

@ShoubhikRMaiti (1) Usually migratory aptitude of phenide > methanide. Hence, (c) will be kinetically favoured. If you have specific data for the order of migratory aptitude for acid catalysed dehydration of alcohols; please share. (2) Yes, because the reaction is thermodynamically favoured, the yield of kinetically favoured will be very low (or it won't from, I don't know).

A basic principle in OChem says that stability of intermediate is the key factor in deciding the major product, in general. Another factor is stability of final product.

Will a kinetically favoured (minor) product (c) form?

It will form, as far as it's intermediate is known to exist (and it exists). However, that will be present in very minor amount, as the intermediate of it will not be as stable as that of (b) and (d). Additionally, the stable intermediate of (c) has 7 α-H, while that of (a) has 2 α-H, so product (c) will be major than product (a).

Here, (d) may be the major product due to high temperature.

The intermediates of (b) and (d) are equally stable. So, here we use the second factor. By looking at the structures of product (b) and product (d), we can say that (d) is major product due to extended conjugation and hyperconjugation with 9 α-H. Really?
The answer is YES. It is known that acid-catalyzed hydration reaction is highly exothermic$1$, which is why it is favored at low temperatures! (In accordance with the "Le Chatelier's Principle", a system will consume heat when temperature of surrounding is raised).
Now, I think, it's pretty clear what the order of yield of products should be :)

References
$1$: https://en.wikipedia.org/wiki/Hydration_reaction#Alkenes

In one of your statement, you are comparing the stability of final products i.e. an alcohol and an alkene. Can we really compare two different class of compounds based on conjugation and hyperconjugation?

I don't think that they can't be compared just bcoz they contain two different class. IMO, anything which contributes to stability can be used to decide major product. In addition to this, the side chain in (b) isn't participating in delocalization with ring, unlike many other benzene deriv., so the comparison becomes logical. I would suggest you to take more time. Another example, where stability of conjugate base of two different classes are compared: acidity of phenol and benzoic acid.


text:
Choise of solvent for extraction
In the synthesis of (S)-Diphenyl(pyrrolidine-2yl)methanol, I need to extract the product using an organic solvent and water. Which organic solvent is the most suitable in this case?

Ethyl acetate: I think that it is miscible with water and it can form an emulsion, for this reason I would not choose it.
Hexane: it is an apolar solvent so maybe the product is not very soluble in it.
Ethanol: I think I cannot use it because it is miscible with water.
Diethylether: it is an apolar solvent so maybe the product is not very soluble in it.
Dichloromethane: I think that this is the best choice because it is not miscible with water and the product should be soluble in it.

Is my reasoning correct?

Is it diphenyl- or biphenyl-?

It's very likely to be soluble in ether. In my experience, ether is preferred when it works. It's easy to remove, and not a huge ecological problem. DCM is OK, but chlorinated solvents aren't the best for the environment.

@MathewMahindaratne I confirm that it's diphenyl-

@Zhe Could you please detail how can you tell it is likely to be soluble in ether?

Ether is more polar than you think. I have no rationale that would hold water, but I've done this long enough that I have a feeling.

The perspective about ethyl acetate should be revisited.  EtOAc is slightly soluble in water; not infinite (as e.g. ethanol), nor this little as hexane(s).  It is greener than halogenated solvents such as dichloromethane, and the boiling point higher than the for dichloromethane or diethyl ether*) is an advantage during summer.  Thus, EtOAc is an often used polar solvent of extraction.
Equally recall the partition coefficient -- multiple extractions with a small volume of organic are more efficient than one with a large volume as long as after shaking the organic and the aqueous layer separate well from each other. Typically, for the isolation of your raw-product later to be purified by crystallization or chromatography, three extractions just «good enough» may suffice for you.  Except you intend a Craig-Extraction to separate similar compounds from each other (as here), a precursor of the countercurrent chromatography.
The same logic of using multiple small volumes instead of one large is applied when washing the organic phase, e.g. by brine, and back extraction.  Indeed, the presence of salts in the aqueous phase may break an emulsion during an extraction; prior to application, it should be tested on small (e.g., test tube) scale ahead though.
*) Compared to diethyl ether, consider MTBE ether as alternative: it is much less prone to the formation of peroxydes, has a higher boiling point, and is cheaper (per L) than the former, too.

What do you think about using hexane?

@Violet I have no hands-on experience with this compound.  The pyrrolidine and the methanol moiety add polarity to the molecule which is still relatively small thus speculate your product dissolves better («more g product per L solvent») in general purpose solvent EtOAc, than in hexane(s). Speculating you used MeOH as solvent which anyway has to be distilled off at the rotary evaporator, prior to an extraction, check at test tube scale how well a tip of the spatula of the dry «cake» of raw product dissipates in a) 1 mL water + 1 mL hexanes, compared to b) 1 mL water and 1 mL EtOAc.

@Violet Equally I would prefer EtOAc over hexanes because the identification and integration of its signals in 1H NMR (and comparison [e.g., estimate of purity] with your product) is much easier, than with hexane(s).  Check e.g., https://sdbs.db.aist.go.jp/sdbs/cgi-bin/cre_index.cgi for their spectra.  In addition, labs purchase hexane by the 1 L bottle or buy as a blend of isomeric hexanes, the later may be even more difficult to sort out in the 1H-NMR and recycle within the lab, than «just» EtOAc.  Possibly your group already has some experience and guide you along MTBE, EtOAc, Et2O, too

(S)-(-)-$\alpha,\alpha$-Diphenyl-2-pyrrolidinemethanol is a chiral ligand, which is commercially available (e.g., for $\pu{53.00 USD}$/$\pu{5 g}$ from Spectrum; Manufacturer: TCI). That is much cheaper price compared to other vendors and the trouble preparing it (assuming this is not the project you are assigned for). To be honest, I didn't have experience with this compound ever, but read a lot of articles about its stereoselectivity. Here what I found online search:

(S)-(-)-$\alpha,\alpha$-Diphenyl-2-pyrrolidinemethanol is a solid with melting point of around $\pu{77-80 ^\circ C}$ (you can find vendors and corresponding melting points at ChemSpider).

(S)-(-)-$\alpha,\alpha$-Diphenyl-2-pyrrolidinemethanol is soluble in chloroform ($\ce{CHCl3}$). Therefore, I expect it to be soluble in dichloromethane ($\ce{CH2Cl2}$). Therefore, your choice is justified. However, keep in mind that some compounds make emulsion with $\ce{CH2Cl2/H2O}$ systems more than they do in $\ce{EtOAc/H2O}$ systems (based on my 30+-year experience in organic labs).

Based on your scale of synthesis, I'd say you may able to purify this compound by dissolving it in $\pu{2 M}$ $\ce{HCl}$ solution (making its $\ce{HCl}$ salt), then neutralized it to precipitate it as free base again, and recrystallize the solid using methanol or ethanol.
If you are afraid of possible racimization (and synthesis is on a small scale), you can extract the final reaction mixture with $\ce{CH2Cl2/H2O}$.


text:
Dipole moments of cis and gem alkenes
The procedure for comparing dipole moments for cis- and trans-alkenes is quite easy, but how to compare dipole moments of

1,1 dichloroethene (1); and
cis-dichloroethene (2)?

My teacher told me that 1 would have more dipole moment as both chlorines are attached to a single carbon leading to more polarisation of bond, but I was not entirely convinced as the chlorine atoms in 1 would make a $60^\circ$ angle with the horizontal (probably a little larger due to steric repulsions between the two chlorine atoms) whereas the chlorine atoms in 2 would make a only a $30^\circ$ angle with the vertical (where the dipole moment is directed) so 2 should have a greater dipole moment.
Which one is the correct answer? Also is this different for different compounds (e.g., 1,1 diflouroethene or  1,1 diflourobutene etc.)?

Yes, your reasoning is correct, based on bond angles. Checked the data at CCCBDB!

Nice website...


text:
Ozonolysis of alkene

(from Wikipedia)
I am having trouble in the last step where the alkene ozonide is reduced to ketones / aldehydes specifically in the presence of water. My teacher is telling me that water will be abstracting an oxygen atom to form $\ce{H2O2}$.
But why? Both oxygen atoms already have a complete octet, and further are repelled by their electron clouds. Why does water abstract an oxygen atom at all?

I was always under the impression that you needed a mild reducing agent, like dimethyl sulfide. I could see the same mechanism working with water, but am rather skeptical, because water is a pretty bad reducing agent, and on top of that if you put H2O2 in there you will get carboxylic acids (that's the "oxidative workup")...

Agree with orthocresol, a workup with just water makes no sense. It´s probably going to happen if you leave the trioxolane on the shelf with water for a week, and you get a wild mixture of ketones/aldehydes and acids. Not synthetically useful.

Not as far as I'm aware. You need to introduce a reducing agent, usually a sulfide, sometimes a phosphine. I think your teacher is in error.

When the ozonide is reduced to Ketone/Aldehyde, We're concerned with what is called the "reductive workup" . We can summarize it like this:

If we look a little further into the mechanism (of a different alkene):

Step (e) is what we're concerned with. Water is not having an impact on this reaction in any manner. The main compound here that's assisting this reaction is $\ce{(CH3)_2S}$  which is a mild reducing agent.

Source of the pictures:
Reaction summary
Reaction mechanism

I agree. If you work up an ozonolysis reaction mixture with just water, you get the ozonide. DO NOT DO THIS - ozonides are explosive.

I'll keep that in mind when I enter the lab (I won't be doing so in the near future :P )


text:
The minima of minimum boiling azeotropes
This is a very basic doubt, so I'll be really brief about it. The discussion I raise is particularly about minimum boiling azeotropes but it's understood that whatever I ask is about the general corresponding points.
In a minimum boiling binary azeotrope phase diagram, particularly one with mole fraction on the x-axis and vapour pressure on the y axis, I understand that the composition of the azeotrope is found by taking the point of intersection of the graph representing vapour pressure as a function of the liquid composition, and the graph representing vapour pressure as a function of the vapour composition. My doubt if it is absolutely necessary within theoretical scope that that particular composition yields the minimum possible boiling point, (and hence maximum pressure) possible from either side of the phase diagram. (By which I mean vapour composition and liquid composition graphs.)

If that's brief, I don't wanna see your verbose ;)

What is wrong with $\mathrm{d}p(x)/\mathrm{d}x = 0$  for $x = x_{\mathrm{azeo}}$ ?

@Poutnik I am asking if that's what the theory says.