text:
Temperature effects on pKa and pKw
I am an industrial electrician apprentice who needs to learn more about pH-measurements and how the temperature effects the measurement itself (I am in charge of all the pH measurements on the powerplant I work at and I have no one to ask about this stuff as no one knows a thing about pH measurements).
So far I have learned how the temperature affects the electrochemical and physical properties of the combination electrode and how this can easily by adjusted for by incorporating an automatic temperature compensation that keeps the NERNST equation updated in real time.
So far so good.
I have, however, also read about the auto-ionisation of water and the constant pKw and how a change in temperature affects the span of the pH scale: With a change in temperature comes a change in hydronium content due to a shift in equilibrium as predicted by Le Chatelier's Principle. However, with a change in hydronium content comes an equal change in hydroxide content and so a change in this constant DOES NOT change the acidity of a solution. In other words, this particular constant decides the SPAN of the pH scale and because of this also the point of neutrality (7 at 25 degree celcius in pure water and less at higher temperatures).
And then there is the so called pKa that says something about the strength of an acid via the acids point of equilibrium in water. This dissociation constant is also temperature dependent although it is my understanding that this constant have nothing to do with the location of the neutrality point. It's simply a constant that changes with temperature and so the solution (unlike with pKw) DOES become more/less acidic as the temperature changes.
Is this correct?
If so, could you please give me an answer to the following questions I have?

Why does a solution become less temperature dependent the more acidic it is? Basic solutions are extremely temperature sensitive, neutral solution still significant sensitive, but highly acidic solutions don't seem give a damn about temperature. Why is that?

It is my understanding that pKw is only important in pure water and highly diluted solutions. But what am I to make about this? If I have a tank of an acidic solution (say pH 1) and the temperature is 50 degrees celcius, does this mean that the point of neutrality hasn't shifted as much away from pH 7 as it would had done in  pure water? Or is it because there is simply so much hydronium present that the pKa constant takes up proportionally more importance than pKw and so the autoionisation becomes a non issue to consider in every practical sense?

Thanks in advance for any help you may provide me with

Not quite sure what you are trying to say? Could you elaborate?
The temperature transmitters I am working with has two different temperature compensation systems. The first one has to do with the NERNST equation and is easily compensated for via an build-in PT1000 temperature sensor. The other type of temperature compensation is the pH-compensation on the measured solution from its temperature dependency. This compensation is not as easy to incorporate as it requires either a pH-temperature coefficient or matrix (or the NEN6411 algorithm).

It is. But what has that to do with anything? As said, the transmitter easily compensates for temperature effects on the measuring chain via real time updates to the NERNST equation.

Ok, I now see what you were trying to say. Yes, I am aware that especially alkaline solutions coupled with higher temperatures means shorter life span of the pH-sensor. This is also why I routinely calibrate my sensors and toss them out whenever the asymmetry potential along with it's time constant (first order system response time) becomes too large.
But what has this to do with my initial question? Thanks

Also, yes I see that Nernst is a name and not an acronym. However, in much of my reading material it is for some reason always written as NERNST. No clue why though.

pKw shifts are actually quite substantial in relation to many of the chemical processes we have at the power plant. With no solution temperature compensation  it could mean millions of dollars due to unoptimized control loops and unoptimized dosing of chemicals.

I see, sorry for misinterpreting you. But why exactly is it that the pKw is more important in a basic solution compared to a strong acidic solution? I still fail to understand this.

The pH electrode has nothing to do with the temperature effects I am trying to understand. The temperature effects on the pH I am talking about is only in relation to the solution itself. So again, could you in a clear way explain to me why pKw is important in basic solutions and insignificant in acidic solutions? Remember that I am no chemist so it is paramount that the explanation doesn't take shortcuts. Also, does the neutrality point still change with temperature even though we are dealing with highly acidic solutions? Thanks

Neutrality point changes with T, but you have either highly acidic, either neutral solution.

The background of the topic is, that for the function
$$y = \frac {a(T)}{x}$$
$y$ does depend on $T$, but $x$ does not.
$\mathrm{pH}$ of basic solutions is temperature dependent via $\mathrm{p}K_\mathrm{w}$ temperature dependance.
Additionally, $\mathrm{pH}$ of weak acid/base solutions is temperature dependent via $\mathrm{p}K_\mathrm{a}$ or $\mathrm{p}K_\mathrm{b}$ temperature dependance.
There is also additional, difficult to quantify factor of temperature dependency of activity coefficients.

The $\mathrm{pH}$ of strong acid solutions ( like hydrochloric or sulphuric acids ) of molar concentration $c \pu{[ mol/L]}$ is approximately(*)
$$\mathrm{pH}=-\log{c}$$
( For sulphuric acid, it is more complicated due 2 acidic hydrogens, where the second one is not strongly acidic ( in sense it has considerable $\mathrm{p}K_\mathrm{a}$ )).
The $\mathrm{pH}$ of strong base solutions ( like sodium hydroxide) is approximately
$$\mathrm{pH}=\mathrm{p}K_w(T) + \log{c}$$
because concentration of hydroxide ions $\ce{[OH-]}$ is the same as the concentration of e.g. sodium hydroxide. Concentration of hydronium ions $\ce{[H3O+]}$ then follows the above $y = \frac {a(T)}{x}$, i.e. $\ce{[H3O+]} = \frac {K_\mathrm{w}(T)}{\ce{[OH-]}}$
The $\mathrm{pH}$ of weak acid solutions ( like acetic acid ) is approximately
$$\mathrm{pH}=0.5 \cdot (\mathrm{p}K_\mathrm{a}(T) - \log(c))$$
The $\mathrm{pH}$ of weak base solutions ( like ammonia or sodium carbonate ) is approximately
$$\mathrm{pH} = \mathrm{p}K_\mathrm{w}(T) - 0.5 \cdot (\mathrm{p}K_\mathrm{b}(T) - \log {c})$$
The neutral $\mathrm{pH}$ is then
$$\mathrm{pH}_\mathrm{neutral}=\frac {\mathrm{p}K_\mathrm{w}}{2}$$
Note that if there was an OH- electrode, situation would be the opposite. Strongly acidic solutions would have temperature dependent pOH and strongly bases would not.

For the strong acid/base $T$ dependency, see the table:

t[deg C]
pKw
Solution
pH
pOH

25
14
0.1 M HCl
1
13

25
14
0.1 M NaOH
13
1

50
13.27
0.1 M HCl
1
12.27

50
13.27
0.1 M NaOH
12.27
1

(*): The equations are not exact, as there plays the role also the ionic strength, activity coefficients and for weak acid/base cases the simplification of $\ce{pH}$ calculation.

Thank you for an easily understandable answer.
I am aware of the fact that pH is a concept concerning the activity of the hydronium ions and not the concentration itself - and that this discrepancy grows in magnitude that more acidic a solution becomes.
One more question:
Is a solution with pH 1 at 50 deg C less acidic than a solution with pH 1 at 25 deg C? I mean, because at 50 dec C a pH 1 is closer to being neutral than a pH 1 at 25 deg C. True?

I would say acidity is the same, but alkality grows. But there is the major effect caused by exponential growth of kinetic rates of chemical reactions with temperature, so it is rather a moot point. By other words, more diluted acid at higher temperature can have the same effect during given time intervals as more concentrated one at lower temperature.

Why does a solution become less temperature dependent the more acidic it is? Basic solutions are extremely temperature sensitive, neutral solution still significant sensitive, but highly acidic solutions don't seem give a damn about temperature. Why is that?

Is this statement based on theory or on your own experience?

It is my understanding that pKw is only important in pure water and highly diluted solutions.

No. It is certainly important in solutions of strong bases, where hydroxide is a major species and the hydronium concentration (or activity) depends on the pKw.

If I have a tank of an acidic solution (say pH 1) and the temperature is 50 degrees celcius, does this mean that the point of neutrality hasn't shifted as much away from pH 7 as it would had done in pure water?

Are you asking whether the pH of an acidic solution is less temperature-dependent than that of pure water? Without knowing which acids and bases are in that solution it would be difficult to say. If you have 100 mM of a strong acid at room temperature and heat it up, the amount of hydronium would not change. The concentration would change because the volume of the solution changes a bit. The activity would change more because the activity coefficient would also change.

Or is it because there is simply so much hydronium present that the pKa constant takes up proportionally more importance than pKw and so the autoionisation becomes a non issue to consider in every practical sense?

For a solution of a weak acid, the pKa makes a big difference. For a solution of a strong acid, neither pKa nor pKw makes a big difference.

[OP in comments] With no solution temperature compensation it could mean millions of dollars due to unoptimized control loops and unoptimized dosing of chemicals.

If I were about to waste millions of dollars, I would not trust the answers in an online forum, I would hire a chemist. They would ask what the role of the pH is in the reactions you want to sustain or avoid, and then figure out how temperature and pH play a role in this (e.g. is the the hydronium or the hydroxide activity relevant for the process, or maybe both).

Well, it works both ways...

t[deg C]
pKw
Solution
pH
pOH

25
14
0.1 M HCl
1
13

25
14
0.1 M NaOH
13
1

50
13.27
0.1 M HCl
1
12.27

50
13.27
0.1 M NaOH
12.27
1

Surely it does. :-)


text:
Why is 11-oxabicyclo(4.4.1)undeca-1,3,5,7,9-pentaene aromatic?

I have been told by my instructors that for a compound to be aromatic, it must be planar. In this given molecule the oxygen seems to be out of the plane of the conjugated system: how can it still be aromatic, then?

Aromaticity is a product of the conjugated bonds in a system, not of the molecule as a whole.

Please use titles that are descriptive, not cryptic. "This compound" is 1,6-oxido[10]annulene (traditional name) or 11-oxabicyclo(4.4.1)undeca-1,3,5,7,9-pentaene (PIN) and not only the oxygen bridge, but also the ring is not planar. Read about aromaticity of 1,6-methano(10)annulene, the ideas are applicable here as well.

Related; https://chemistry.stackexchange.com/questions/86734/applying-huckels-rule-to-non-planar-compounds/86737#86737

I point out that the Wikipedia pages on [10]-annulene and methano annulene contain conflicting info about the ring planarity. Still the less distorted and thus aromatic is the second one.

The aromatic system should be planar. The oxygen is not part of the aromatic system.
Please note, of course, that your drawing does not capture the 3-dimensional geometry of the system. You would still have to analyze the carbon skeleton to see if it satisfied the planarity constraint enough to be aromatic.

source: Stack Exchange

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

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text:
How are these three things equal?
In this answer, I don't understand how this step worked.

Using $\mu_i = \mu^\circ_i + RT\ln \frac{P_i}{\pu{1 bar}}$
\begin{align}
\Delta G &= (c\mu^\circ_\ce{C} + d\mu^\circ_\ce{D} - a\mu^\circ_\ce{A} - b\mu^\circ_\ce{B}) + RT \ln\frac{P_\ce{C}^c P_\ce{D}^d}{P_\ce{A}^a P_\ce{B}^b}\\
\Delta G &= \Delta G^\circ + RT\ln Q
\end{align}

How does $P_i=\dfrac{P_\ce{C}^c P_\ce{D}^d}{P_\ce{A}^a P_\ce{B}^b}$? And how does that equal the quotient $Q$?
If this is wrong, then please provide me the correct way to proceed from there.

Refer to the balanced equation and combine. The $i$ will refer to the exact species of interest (i.e., A, B, C, or D).

You have, for each compound, $\mu_i = \mu^\circ_i + RT\ln(P_i/P^\circ)$.
So, replacing each potential in $\Delta G$ by its expression, assuming $P^\circ = 1$ bar, we get :
$$\Delta G = c(\mu^\circ_C + RT\ln(P_C)) + d(\mu^\circ_D + RT\ln(P_D)) - a((\mu^\circ_A + RT\ln(P_A)) - b(\mu^\circ_B + RT\ln(P_B))$$
Then we develop and group standard potentials and logs:
$$\Delta G = (c\mu^\circ_C + d\mu^\circ_D - a\mu^\circ_A - b\mu^\circ_B) + RT(c\ln(P_C) + d\ln(P_D) - a\ln(P_A)-b\ln(P_B))$$
Then, using calculation rules on logarithms:
$$\Delta G = (c\mu^\circ_C + d\mu^\circ_D - a\mu^\circ_A - b\mu^\circ_B) + RT\left(\ln({P_C}^c)+\ln({P_D}^d)-\ln({P_A}^a)-\ln({P_B}^b)\right)$$
And finally, since $-\ln(x) = \ln(1/x)$ and $\ln(x)+\ln(y)=\ln(xy)$,
$$\boxed{\Delta G = (c\mu^\circ_C + d\mu^\circ_D - a\mu^\circ_A - b\mu^\circ_B) + RT\ln\left(\frac{{P_C}^c{P_D}^d}{{P_A}^a{P_B}^b}\right)}$$
Then, try expressing the quotient of the reaction in terms of partial pressures, assuming standard pressure is 1 bar and you'll find out it gives the inside of the log!

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text:
Acid-Base Titration of ammonium dihydrogenphosphate?
How can I determine the concentration of ammonium dihydrogenphosphate solution?
In this compound, the ammonium cation and the dihydrogen anion are both weak acids.
I have a weak solution of this salt, $C_A\approx 0.05\mathbf{M}$, that I need to quantitatively determine.
My first thought was acid-base titration because it's cheap and easy but the presence of two weak acids complicates things.
Would it be possible to determine the concentration of ammonium and dihydrogen in one equivalence point, as $2\times C_A$?

The pH can be measured vs. the amount of $0.1$ M NaOH added to your solution with a pHmeter. Then two titrations would be visible : first the titration of dihydrogenphosphate ($p\ce{K_a = 7.21}$) and then the titration of ammonium ($p\ce{K_a = 9.22}$). If you don't have a pHmeter, you would probably use a colored indicator. But the end of the first titration will not be determined with precision, as the two $p\ce{K_a}$'s are not different enough. You will only be able to determine the end of the titration of the two acids together. To do this, a rather basic indicator is needed, like Alizarine yellow whose $p\ce{K_a = 10.5}$.

Thank you for your answer which is much along my own thinking. I do have a pH meter and have used it for AB titrations a lot (but rather a long time ago). So that settles it and I'll accept your answer.


text:
Identifying which lone pair is more basic?
In the the compound RCONH2 which lone pair is more basic? Is it the lone pairs on Oxygen atom or the lone pair on Nitrogen atom.
I believe that as O is more electronegative than N, the lone pair on O is of lower energy. Hence, the O's lone pairs are less basic. (But if one draws the resonance structure for the compound, a formal negative charge is found on the oxygen atom and the Nitrogen lone pairs are engaged in making bond)
So which atom should react first with an neutral electrophile?

Your analysis shows some correct lines of reasoning, but the overall reasoning is unclear. The way to solve the problem is to look at the two structures that could be created from reaction at the two respective nucleophilic sites. More importantly, do not confuse basicity in general with nucleophilicity. Those are separate concepts, and you can find examples of non-basic nucleophiles and non-nucleophilic bases.


text:
Why does the pH of a weak acid not increase by 1 when diluted by a factor of 10?
Strong acid pH increases by one unit when diluted by a factor of 10, but why do weak acids not?

HINT - What is the difference between a strong acid and a weak acid?

Why does the pH of water not increase by 1 when diluted by a factor of 10? Oh, and BTW, how come pure water has a pH?

Because 1010/10 <> 11/1

In a weak acid  $\ce{HB}$ solution, with a nominal concentration $c$, a tiny amount of its molecules are dissociated into $\ce{H^+}$ and $\ce{B^-}$. Let's call this concentration $[\ce{H^+}]$ = $\ce{[B^-]}$ = $x <<c$, so that the following approximation can be made : $c - x$= $c$. The dissociation equilibrium constant $K_a$ of this weak acid $\ce{HB}$ can be approximated by$$K_a\ce{= \frac {[H^+][B^-]}{$c - x$} = \frac{$x$^2}{$c-x$} = \frac{$x$^2}{$c$}}$$ $$\ce{[H^+] = x = \sqrt{K_ac}}$$ $$\ce{$p$H = \frac{1}{2} (log$K$_a - log $c$)}$$ Look at the coefficient $1/2$ before the logarithm. If the concentration $c$ is multiplied by $10$, the log c increases by $1$, but the $p$H changes by $1/2$.


text:
What is the role of imidazole in the Appel Reaction with iodine (I2)? Why are other bases not used?
I presume the role is to soak up the HI that is formed. Why not other bases such as triethylamine? Is this because the quaternary ammonium salt could be formed? Would other non-nucleophilic or inorganic bases work (for instance, k2co3 or DBU)?
http://www.commonorganicchemistry.com/Rxn_Pages/Alcohol_to_Halide/Iodide/Appel.htm

related: https://chemistry.stackexchange.com/a/9669/102629


text:
Why are isomers of parathion less active acetylcholinesterase inhibitors than paraoxon?
Parathion itself has been found to be a very weak inhibitor of acetylcholinesterase. It normally requires metabolic activation and the conversion into paraoxon in the body to actually start exhibiting anticholinesterase activity. However, the study cited below found that active isomers of parathion, O,S-Diethyl O-4-nitrophenyl thiophosphate and O,O-diethyl S-4-Nitrophenyl thiophosphate (Compounds II and III in the cited paper, respectively), are less active inhibitors of AChE than paraoxon. I would have thought that the isomers of parathion would be more active than paraoxon due to thioethyl and 4-nitrothiophenyl being better leaving groups than 4-nitrophenyl, and thus the two active isomers of parathion would more readily form an adduct with the enzyme. I would presume that they also wouldn't need metabolic activation due to already having an oxygen atom double-bonded to the phosphorus. However, my prediction does not seem to hold true. So, why exactly are the active isomers of parathion (Reference numbers 7 and 8 in the table in the cited study) less active AChE inhibitors than paraoxon (Reference number 9 in the table in the cited study)?
Reference
Diggle, W. M., & Gage, J. C. (1951). Cholinesterase inhibitionin vitrobyOO-diethylO-p-nitrophenyl thiophosphate (Parathion, E 605). Biochemical Journal, 49(4), 491–494. doi:10.1042/bj0490491

Background
Paraoxon is an irreversible inhibitor. It forms an adduct with the catalytic residue Ser203 of AChE, along with release of p-nitrophenol as leaving group. The resulting covalent phosphoester conjugate is quite stable, but also reversible when using an appropriate oxime before the complex ages. See Franklin et al. 2016 for more details.
To facilitate the discussion below, here are the key molecules:

Regarding the OP's questions
Parathion is known to be metabolized to Paraoxon in the liver, likely though the action of cytochrome P450. (Chambers et al., 1991)
A phosphorothioester bond is indeed expected to be more susceptible to hydrolysis than a phosphoroester bond. Such a P-S labile linkage is ~106-fold more labile than P-O at neutral pH, according to Kuimelis-RG and McLaughlin-LW, 1995. However, Xu-Y and Kool-ET, 1998 noted the importance of unfavorable interactions of a studied enzyme with the sulfur atom, “because of the increased bond lengths or relatively poor hydrogen bond accepting ability of the sulfur relative to oxygen.” Indeed the P-S bond length is ~0.4 Å longer than P-O bond. (Knight et al., 1991)
The enzyme’s active site environment, that is beyond how good the leaving group is may help better understand the inhibition results of the molecules listed above.
Structural environment in AChE
The co-crystal structure of AChE with paraoxon (see Franklin reference above) deposited with RCSB as entry 5HF5.pdb provides the context of how the residual adduct binds to the AChE active site. The Franklin paper provides nice descriptive panels of this structure, but I thought I’d prepare the following image to help give more context of this interaction. In the following panel of the co-crystal structure, the residual adduct covalently linked to Ser203 is colored with cyan carbons:

a) The oxygen of P=O accepts a Hydrogen bond from multiple amides. Given that a sulfur atom does not form as good Hydrogen bonds as an oxygen, then this enzymatic environment will not provide as good stabilization to a sulfur as it does to an oxygen. In addition, as a sulfur atom has a larger radius than an oxygen atom, there may not be enough room for a sulfur atom. These two factors may begin to explain why parathione is not as good a substrate as paraoxon.
b) One of the ethoxy oxygen accepts a hydrogen bond from His447. Having a sulfur atom at that position, such as with Compound II, would bring it too close to the imidazole, leading to steric hindrance. In addition, such a sulfur atom would not form as a productive hydrogen bond as the oxygen. Similarly, the other ethoxy forms close van der Waals interactions with Phe297 and Trp236, there may not be enough room to accommodate a larger sulfur atom. This may begin to explain why Compound II is not as good as paraoxon.
c) Similar to the above, the environment may be too restrictive to accommodate Compound III for the productive formation of an adduct.
In summary, the immediate environment of the active site may be more optimized for paraoxon than the thiol analogs; this may help explain the published SAR of the active isomers of parathion.
Edited for clarity.


text:
Transition metals with closed shells
From wiki page about valence electrons:

An atom with a closed shell of valence electrons (corresponding to an
electron configuration $s^2p^6$ for main group elements or $d^{10}s^2p^6$ for
transition metals) tends to be chemically inert.

My question is, which are the transition metals with $d^{10}s^2p^6$? Looking at the periodic table, I do not see anyone.
After this paragraph, wiki says that valence orbitals for d-block elements (transition metals) are (see table):

ns (n-1)d np

I though one of the characteristics of transition metals is that they do not have electrons in last p orbital.

Your premise Wikipedia's statement about transition metals seems to be flawed.
(1) It isn't possible to have a transition metal that has the electronic structure $\mathrm{(n-1)d^{10} ns^2np^6}$. Such an electronic structure would imply a noble gas.
(2) Gold $\mathrm{[Xe]4f^{14} 5d^{10} 6s^1}$ and platinum $\mathrm{[Xe]4f^{14} 5d^{9} 6s^1}$ are the most noble metals and neither even has a $\mathrm{(n-1)d^{10} ns^2}$ configuration.

Granted these configurations are for the free atom, not the bulk metal...

I'm sorry, but I do not understand this answer. You cite a phrase not in the question nor in the wiki page, where is it from ? Moreover, answer starts by "your premise". About what premise are you talking ?

@pasabaporaqui - I tried to clarify the answer. I'm not sure what you meant by You cite a phrase not in the question nor in the wiki page, where is it from ?

Thanks. At least in my computer, the phrase "Granted these configurations ..." appears as a citation (gray bar at left). But I do not see the source of this cite.

@pasabaporaqui - the phrase "Granted these configurations ..." is from my own knowledge, not a quote. I don't understand your confusion. Atoms in a bulk metal are bonded together, they are not free atoms.

The zinc atom $\ce{Zn}$ has the configuration $\ce{(Ar) 3d^{10} 4s^2}$ or, if you prefer : $\ce{1s^2 2s^2 2p^6  3s^2 3p^6 3d^{10} 4s^2 }$. The same final configuration is repeated with the cadmium atom $\ce{Cd}$, and with the mercury atom $\ce{Hg}$. Is it not what you are looking for ?

Zinc does not seem to be inert ... .

@Oscar Lanzi. I agree. Of course zinc is not inert.  But the question was not about the element to be inert or not. The question was whether a metal does exist with the configuration $d^{10} s^2 p^6$.

Wiki seems (?) to talk about a (n-1)d ns np configuration, that could have relation with closed shell and inert atom. The examples on this answer seems to be about "(n-1)p (n-1)d ns"


text:
Reactivity of Silicon
$$\ce{SiO2 + 2C ->[2200°C] Si + 2CO}$$(Ref)
In this reaction, C is apparently displacing Si from SiO2. Does this mean C is more reactive than Si? But Si is more electropositive than C. What is the explanation for this? Where will Si be placed in the reactivity series?

It depends. What temperature?

Reactivity is not a thing at all. Also, CO2 does not form in your reaction, nor does your reference say so.

Let's compare the boiling points of these substances. Silicon is boiling at $2355°$C. Carbon is boiling at $4200°$C. $\ce{SiO2}$ is boiling at $2590°$C. So at $2200°$C, $\ce{CO}$ is the only gaseous substance. Whatever the change of enthalpies or the electronegativities, its formation is favored.

That is not a reaction that happens. That is a summary of a set of reactions that happen, at (at least 2) different equilibria and different temperatures.
The direct reactions are as follows, at "steady state" solids if not specified:
$$\ce{2SiO_2 + SiC <-> 3SiO(g) + CO(g)}$$
$$\ce{SiO_2 + Si(l) <-> 2SiO(g)}$$
$$\ce{SiO(g) + SiC <-> Si(l) + CO(g)}$$
$$\ce{SiO(g) + 2C <-> SiC + CO(g)}$$
There is no reasonable answer to your question, because that reaction does not happen in a single step, it is not displacing anything. Though, the fact that you have to pour a spectacular number megawatts into the process for it to continue does suggest that $\ce{Si}$ is not "more reactive" though.


text:
Hell-Volhard-Zelinski Reaction
What is the role of red phosphorus in hvz reaction?
Initially I think that red phosphurus is nothing but a catalyst. But it is not a catalyst.

https://chemistry.stackexchange.com/a/90666/102629

TL;DR: When $\ce{PBr3/PCl3}$ is not available as catalyst, red phosphorus is used which is reacted to bromine or chlorine to form the catalyst in situ.
Long answer:
Hell–Volhard–Zelinsky halogenation reaction halogenates carboxylic acids at the α-carbon.  The reaction is initiated by addition of a catalytic amount of $\ce{PBr3}$, after which one molar equivalent of $\ce{Br2}$ is added. $\ce{PBr3}$ replaces the carboxylic -OH with a bromide, resulting in a carboxylic acid bromide. The acyl bromide can then tautomerize to an enol, which will readily react with the $\ce{Br2}$ to brominate a second time at the α position.

What can be done if $\ce{PBr3}$ is not available?

You can use red phosphorus. It is reasonably air stable (whereas white phosphorus combust in air almost spontaneously). Phosphorus reacts with bromine to give phosphorus tribromide which now becomes the catalyst for the reaction.
$$\ce{2P_{(red)} + 3Br2 -> 2PBr3}$$

Why exactly $\ce{PBr3/PCl3}$ is needed and why an extra molar equivalent of $\ce{Br2/Cl2}$ is used?

When left out in the air, $\ce{PBr3}$ will readily combine with any water vapor present to form a new $\ce{P-O}$ bond along with $\ce{HBr}$. The strength of the $\ce{P-O}$ bond [about $\pu{140 kcal/mol}$] makes $\ce{PBr3}$ very useful for swapping out $\ce{C-O}$ bonds for $\ce{C-Br}$ bond and that's how carboxylic acids gets converted to acyl bromides.
However, $\ce{PBr3}$ by itself is not sufficient to perform the HVZ reaction. A molar excess of $\ce{Br2}$ is added to the already present $\ce{PBr3}$, which perform the halogenation of the alpha-carbon. That's why the catalyst is always mentioned as ($\ce{PBr3 + Br2}$).
References

https://en.wikipedia.org/wiki/Hell%E2%80%93Volhard%E2%80%93Zelinsky_halogenation
https://www.organic-chemistry.org/namedreactions/hell-volhard-zelinsky-reaction.shtm

That generates another question: Can we use phosphorous triiodide in Hell-Volhard-Zelinsky reaction?

I see it written at some places like "catalytic amount of red phosphorous" or other words which suggest "catalyst". I know you've mentioned it as a reagent in your answer, but can you please confirm whether using the word catalyst in this case for red phosphorous is wrong or not?

@Ashish Yes, you are correct with the usage of term "catalytic amount". Any substance or mixture for use in chemical analysis and that causes chemical reaction is called reagent. That also includes catalyst. In this context, "catalyst" should be a better term because it is causing catalysis. I have edited appropriately.


text:
Understanding strong acid titrated with weak base graph

I don’t understand why the conductivity doesn’t change after the equivalence point. To my understanding, at the equivalence point there is $\ce{NH3}$, $\ce{NH4+}$ and $\ce{Cl-}$ in the solution. As more $\ce{NH3}$ is added it only partially dissociates, thus conductivity stays the same. But doesn’t the concentration of $\ce{Cl-}$ ions also contribute to the conductivity, and its conductivity decreases are more aqueous $\ce{NH3}$ is added as it becomes more dilute? Where is the issue in my understanding. Thanks for the help.

I would not pay the chart too much attention, as it seems to me rather just illustrative, comparing trends before and after the equivalence point.

Source of image: https://www.tau.ac.il/~chemlaba/Files/conductometry-titrations.pdf

You are missing an important point. Conductometric titration is not just plotting the conductance after adding the titrant. One cannot plot conductance as it is, because you are significantly diluting the solution.  You have to apply a correction to take dilution into account. Once you apply a correction, the curve will remain flat because further addition of ammonia will not enhance conductivity.
See for example,

Conductometric titrations
Throughout a titration the volume of the solution is always increasing, unless the
conductance is corrected for this effect, non linear titration curves result. The
correction can be accomplished by multiplying the observed conductance either
by total volume (V+V') or by the factor (V+ V')/V, where V is the initial volume
of solution and V' is the total volume of the reagent added. The correction
presupposes that the conductivity is a linear function of dilution, this is true only
to a first approximation.

Thank you. We learnt it very briefly at school and this helped me make sense of it.

After equivalence point you add more and more NH3 solution (remember NH4OH does not exist). So you add a negligible amount of ions. As a consequence, the conductance does not change significantly. This is what your diagram shows.

But wouldn’t conductivity decrease because of lowered Cl - concentration?

@Subbota It would and will, but see the other comment. Its decrease is insignificant, compared to replacing highly mobile hydronium ions by relatively slow ammonium ions.


text:
What about sodium triphosphate in drinking water?
Recently, at my place, a sort of filter for hard water was installed. They told me that it contains sodium triphosphate which is dissolved into all the water that enter home, i.e. both cold and hot water to prevent calcium carbonate to accumulate into the plumbing system, dishwasher, etc.... From what I can understand, there is no way this device is considering the flow rate in order to decide how much sodium goes into the water. It just happens.
According to what I googled, sodium triphosphate is used as an additive in many foods and, even though is not considered harmful, high levels may lead to health risks. Also, I have found some links (not in english) stating that such devices should be installed in the hot water plumbings since sodium triphosphate and calcium carbonate interact only at certain temperatures. According to these sources, drinking cold water is not harmful since no chemical reaction took place (but still sodium is present in the water, I believe) and, also, once I use the same cold water for cooking it gets hot and a reaction may take place.
I am quite puzzled and I don't know what to do with this filter. I can move it to hot water only, or I can keep it where it is (both cold and hot) if it's not harmful but I couldn't find any clear information. Is there something that may help?
EDIT 1:
this is the exact product I am using.
EDIT 2:
its chemical formula is Na5 P3 O10
EDIT 3:
this seems to be the device that was installed and this is one of the components that is taking care of the dosing (that is equal to 4-5ppm of sodium triphosphate, apparently)

How is this filter installed ? Is it on a bypass of the main pipe ? How is the flow controlled ? Trisodium phosphate is a bit soluble in water. So, without control, it should be quickly dissolved and eliminated.

it is installed at the very entrance of the water inside the house and, as far as I know, the flow is not controlled in any way.
Also, If it is quickly dissolved...what is the purpose of having it in the first place?

If the sodium triphosphate is in big chunks, it will not be quickly dissolved. But I am afraid it is not sodium triphosphate. By reading your reference, it is sodium polyphosphate. There are several sodium polyphosphates. For example, the sodium tripolyphosphate is poorly soluble and is known to be useful for removing calcium ions from tap water. It is not a dangerous substance. You can maintain it in your filter. It will make a complex ion with calcium, so there will not be any calcareous deposits in your tubing and other equipment. It seems to be a good idea.

@Maurice I added the exact chemical formula!

It may not have immediate negative effects, but it is even more aggressive to enviroment by the phosphate load of waste water than the washing formulas for washing machines. As the latter treat just water used for machine washing. The former treat all the water.

@Maurice Sodium triphosphate:  Sodium triphosphate (STP), also sodium tripolyphosphate (STPP), or tripolyphosphate (TPP) .....  Not to be confused with Trisodium phosphate Na3PO4. Note that sodium diphosphate is not Na2HPO4 either, but Na4P2O7.

@Poutnik the fact that it does not have immediate negative effects is not reassuring. Also, the fact that it is harmful for the environment and is worse than a washing machine soap...well that is even more not reassuring.

@Poutnik, yes that is what I found by googling it. It looks like it should be applied to hot water only, to avoid exposure with drinking water (assuming one only uses cold water for drinking and cooking). Doing so still harms the environment, though.

yeah well, a washing machine is replaceable while health is not. I have zero doubt about my priorities, in this case. Lets see if I can find something else or someone has something to add.

I use TSP as garden fertilizer, plants do well  ,so no problem in limited amounts. I have acidic soil so I want alkaline fertilizer.

@blacksmith37 Trisodium phosphate (TSP) is different than sodium triphosphate. TSP is a great fertilizer, which is why its use as an ingredient in detergents has been regulated in the USA due to the fact that it causes algae blooms in lakes and ponds downstream from where waste treatment effluent is released.

It appears to be safe, provided that you are putting it into one of the delivery systems that it’s made for which I found through the link you provided.
Here is one of the delivery systems:
https://www.pineco.com/eng/products/dosing/polyphosphate-dosers/spillo-polyphosphates-doser-dm
Per the specs, it delivers 3ppm, which means that you’re only introducing 1ml per 333L of water, which is approximately what the average person uses per day.
Contributing to phosphate load shouldn’t be a concern as someone (with “normal” diet &renal function) has a concentration of 10-300ppm of phosphorus in their urine!
This company also sells water filters for regionally specific water issues like high arsenic, and water softeners with different ion exchange resins based on the ions present in the water. They appear to have a solid understanding of drinking water chemistry, but it’s your domestic water, not mine, so caveat emptor

Hm, but 3 ppm can hardly make any difference for hard water. E.g. if we have a hard water 25 dGH ( old unit, old school ), it means 250 mg of CaO equivalent.
And as 1 CaO is equivalent to 1 Na5P3O10, forming complex, it is approx 250*(523+331+10*16)/(40+16)=1640 mg of Na5P3O10, so 1640 ppm. It would be less, as there is no need to cover all calcium.

ok! so yes, I checked which device is the one I have installed and edited the question accordingly. I haven't thought about checking on it.

also, I wanted to add that you considered 333l of water per day but we only use roughly 7/8l for actual drinking and cooking hence the amount of phosphorus should be even lower, right?

@Poutnik yes, agreed this won’t do much as a primary water softener. The website mentions that it can be installed after the water softener to reduce scaling in pipes as water softeners aren’t 100% effective.

@gabt that 333L is average household use for ALL water, including washing & showering to understand how much phosphate was being introduced to the waste water as far as environmental impact is concerned. You’re only consuming a trace amount from the water you drink added by the system as I imagine most of it is reacting with the water before it leaves the tap.

@JackSpencer ok, so...if it already reacted, then what? I guess I am still drinking some residues that are as not-healthy as the original compound.


text:
Activation energy and degrees of freedom
According to the Arrhenius equation the rate of a reaction increases as at higher temperature a greater percentage of molecules have translational kinetic energy greater than or equal to activation energy Wikipedia.
I was recently introduced in molecular potential surfaces and I read that some reactions can proceed when excess kinetic energy is stored into vibrationals degrees of freedom than translational. But according to Wiki it seems that only translational kinetic energy matters. First, I want to understand if the factor:
$$e^{\frac{-E_a}{RT}}$$
takes into account only the percentage that have the necessary kinetic energy in translational degrees of freedom or in general the percentage with at least this energy irrespective if this energy is stored in a combination of various degrees of freedom? Also for two reactants $\ce{A}$ and $\ce{B}$ does it calculate the percentage of "pairs" $\ce{A}$, $\ce{B}$ that have this amount of energy or just the percentage of the population with at least this energy?
Secondly, if some reactions speed up depending in which type of degrees of freedom (vibrational or translational) are excited, does this mean that the Arrhenius equation even for an elementary bimolecular reaction holds only if the reactants have only translational degrees of freedom (like collision of two atoms $\ce{A}$ and $\ce{B}$ to form $\ce{A-B})?$

In my opinion, the Arrhenius activation energy is an experimental value determined by comparing different reaction rates versus temperature. It is useful for determining reaction rates at any desired temperature. But the nature of this activation energy is not specified. It can be vibrational, translational. Who cares ?

You may be interested in the answer here https://chemistry.stackexchange.com/questions/139196/why-does-activation-energy-not-change-with-temperature-change/139222?

This depends upon the theory used to model the reaction; there are two major ones.
Collision Theory
If your molecules may be assumed to be spherical particles with no rotational or vibrational DOF, then collision theory appropriately describes your system. In this case, only translational energy is accounted for. The rate expression may then be derived from the Maxwell-Boltzmann kinetic theory of gases, where the energy of each particle is $\frac{1}{2}mv^{2}$.
I. Filot has a decent explanation of this.
Transition State Theory
Unfortunately, most reaction systems are much more complicated; the reactants experience not only translational, but also rotational and vibrational DOF. This must be accounted for in the rate expression. Again, I. Filot has a good explanation for this.
A general derivation includes the following:
Consider a bimolecular reaction:
$$A+B \rightarrow C$$
We assume there exists a transition state (TST), which defines the highest point along a minimum energy path on the potential energy surface. Moreover, we assume that our reactants are quasi-equilibrated with this TST; the rate of the reaction depends upon the frequency of reactants overcoming this barrier $(k_{f})$. Then:
$$A+B \leftrightarrow C^{\ddagger} \rightarrow C$$
Assuming this reaction occurs in the gas phase under ideal conditions, we can approximate their fugacity using the partial pressure of each reactant. Let us also assume that the reference partial pressures are unity. We derive an equilibrium constant between the TST and the reactants:
$$K^{\ddagger}=\frac{P_{C^{\ddagger}}}{P_{A}P_{B}} = exp(-\Delta G^{\ddagger}/RT) = exp(-\Delta H^{\ddagger}/RT)exp(\Delta S^{\ddagger}/R))$$
Our rate expression becomes:
$$r = K^{\ddagger}k_{f} P_{A}P_{B} = [(k_{f}) exp(\Delta S^{\ddagger}/R))exp(-\Delta H^{\ddagger}/RT)]P_{A}P_{B} = kP_{A}P_{B}$$
Our rate constant is:
$$k = (k_{f}) exp(\Delta S^{\ddagger}/R))exp(-\Delta H^{\ddagger}/RT) = (A)exp(-\Delta H^{\ddagger}/RT) \equiv (A)exp(-E_{a}/RT)$$
Where the partition functions defining the enthalpy and entropy include translational, rotational and vibrational DOF.
Therefore, to answer your questions:
"First, I want to understand if the factor $e^{-E_{a}/RT}$ takes into account only the percentage that have the necessary kinetic energy in translational degrees of freedom or in general the percentage with at least this energy irrespective if this energy is stored in a combination of various degrees of freedom?
No, it takes into account all DOF. In transition state theory, both the activation energy and pre-exponential factor consider the translational, rotational and vibrational DOF of the transition-state irrespective of the DOF providing the energy (This is directly tied to the canonical partition function, which is used in defining the enthalpy and entropy. It only cares about the total microstate energy and not from which DOF it comes from).
"Secondly, if some reactions speed up depending in which type of degrees of freedom (vibrational or translational) are excited, does this mean that the Arrhenius equation even for an elementary bimolecular reaction holds only if the reactants have only translational degrees of freedom?"
Reactions do not speed up depending on which type of DOF are excited. However, certain DOF speed up the reaction much more than other DOF. For example, based on the derivation I have provided, the pre-exponential factor scales exponentially with the transition-state entropy.
$$S_{total} = S_{translations}+S_{rotations}+S_{vibrations}$$
The DOF with the largest contribution to the entropy are translations. In theory, we may increase the rate of the reaction by increasing the vibrational entropy of our transition state. But, we may increase it to a much larger extent by increasing it's translational entropy.

Thank you for your answer. For A and B does it take into account the percentage of pairs that have this energy or the individuals A and B? Also, I still can't get why a reaction speed up more when for example we use IR radiation to excite the reactants vibrationally. Does it speed it up more because for a brief time the vibrational DOF are more excited than Maxwell-Boltzmann distribution (it takes a little time before the system equilibrates again)? So if the reaction is more sensitive to vibrational DOF it will speed up more (than if we just heated the reactants)? I will check out the book.

They are not mutually exclusive quantities when referring to the rate. When you say "pairs", what you are referring to are transition states (i.e AB contacting each other) Per the TST theory derivation, the rate constant depends on microstate populations among AB, A and B. Remember: the Gibb's free energy barrier, is (del_G of AB - del_G of A - del_G of B). We can use IR to populate A and B into higher energy levels, thereby lowering the activation barrier. On the contrary, if IR were to "excite" AB, we would increase the barrier.


text:
Which is more reactive in E1, benzyl or allyl
I'm don't know the english terminology for chemistry so sorry for probably butchering some terms.
My question is, which of these is more reactive in an E1 reaction:

The bottom two are the intermediates.
To determine the most reactive molecule, we look at the most stable intermediate.
When trying to stabilize the positive charge, the benzyl has 3 resonances and the allyl has only 1.
But in the resonance forms of the benzyl, it loses its aromaticity.
Which effect is more important, not losing aromaticity during resonance forms, or having more resonance forms?
Thanks for your help!
Note:
A similar question was asked here: Is benzyl chloride or allyl chloride more reactive?, but my question is only specific to this E1 reaction, while the other question is a general reactivity question (and has no answer...).


text:
How is 4H‐1,4‐oxazocine aromatic?
I’ve found 4H‐1,4‐oxazocine in my chemistry book and I’m perplex as it states as aromatic:

I can count six electrons on p orbitals which could satisfy Hückel’s $n = 1,$ but the oxygen is sp3-hybridized. Even if we’d consider one lone pair as one sp2-bonding, the second lone pair would add 2 electrons to the p system, counting a total of 8 electrons, which doesn’t satisfy Hückel’s rule.

So what about the nitrogen lone pair?

@Waylander so basically I could count n=10 considering O as $sp^2$ with a lone pair on the pi system?

Didn't we do the same in furan? Didn't we do the same in pyrrole? Well, now put the two together.

@IvanNeretin Thanks! I actually had this preconception I’ve read somewhere that n=number of aromatic rings, thus in my reasoning this compound couldn’t have more than 6 e^- in its pi system.

@pseuronimo n is not the number of aromatic rings, nor could it be since the 4n+2 rule is really accurate only for a single ring.  Rather it's an optimal value that depends on ring size.  Typically n=0 for a three-atom ring, 1 for 4-7 atoms, 2 for 8-9 atoms.  Aromaticity is usually insignificant for larger rings than that.

@OscarLanzi I'd say n=0 for 3-4, 1 for 5-8, 2 for 8-11 or 12.

I am more conservative; aromatic dication rings and rings with ten or more atoms are less common and may not always be energetically favored; for example $\ce{C_{11}H_{11}^+}$ is not an aromatic ring.

According to Huckel's rule four criteria for aromaticity
The molecule needs to be (1) planar, (2) cyclic, (3) fully conjugated, and has (4) 4n+2 electrons.
Your molecule does not seem to be planar (b/c of N and O atoms) but assuming that it has a nearly planar conformation it satisfies all other criteria.
You only need to count one of oxygen's lone pair and the only nitrogen lone pair:
$10=4*2+2$
So it could be aromatic (if it is nearly planar)


text:
How reliable is spectrometry in the determination of the oldest material Startdust?
After reading one of the latest findings by scientists, Stardust was considered the oldest material on earth. However having not much of an experience with chemistry, I was intrigued upon how scientists were able to find such finding therefore from my understanding, scientists used spectrometry to determine composition of stars to then determine the age of the material(hopefully I am not wrong). However, reading one of the news article, it said that “ stardust that predated the formation of our solar system by billions of years, scientists said on Monday.” This gabbed my attention because it says that scientists found that it was on Monday but the question that came to mind is, how reliable is spectrometry in the determination of such exact findings? Please let me know what you think?
Thanks a lot.
News article: https://www.aljazeera.com/news/2020/1/14/stardust-oldest-material-on-earth-found-inside-meteorite

It's a matter of radioisotopic dating, with it's own issues. Mass spectroscopy data tell isotopic composition precisely.

You have to read this original paper in order to get the idea of uncertainties involved in measurements. The estimation was based on Neon isotopes. https://www.pnas.org/content/117/4/1884


text:
Would the initial step of this mechanism be exothermic or endothermic? (Protonation)

My understanding is that since an O-H bond is being formed, the protonation step must be exothermic because energy is released upon bond formation. However, if I were to draw an energy/reaction coordinate diagram for this mechanism, the more reactive protonated alcohol would be higher up in energy than the unprotonated alcohol so I'm not sure if the initial protonation step would be exothermic or endothermic.
This problem's been plaguing me for days now.

Hi @Beowulf8991, can you please draw the reaction coordinate diagram for this reaction so we can gain a better understanding of your problem? Thanks and happy learning.


text:
Food-grade antifungal preservatives for alcohol-containing aqueous non-food product that won't UV-yellow when dried?
I need a broad-spectrum antifungal (yeast and mold) agent for an aqueous non-food product (pH $\simeq$ 6) that contains 20% v/v alcohol (ethanol or isopropanol).  A key requirement is that it not UV-yellow if it comes into contact with textiles (e.g., clothing) after drying.
I would prefer it to be a food-grade preservative—or, barring that, something not on the California Prop. 65 list.    What preservative (or combination of preservatives) would work, and at what concentration should I use them?
Sodium benzoate and potassium sorbate can be effective anti-fungal agents when used in combination.  However, these would not work here, since they don't have significant antimicrobial activity at pH $\simeq$ 6.  In addition, potassium sorbate is known to yellow (at least in solution) when exposed to UV; and since sodium benzoate is aromatic, it might UV-yellow as well.
I've contacted both Sigma and Lonza (which has a specialist preservative division).  Sigma said they can't make a recommendation at that level of detail, and Lonza said they would need to send me to their business office to set up a research contract for that.

This is a question about formulation/toxicology. It may be better suited for biology or medical sciences SE. With regard to that, it has to be GRAS and is intended for topical use (will not be ingested)?

It will not be ingested, correct. Biology SE and Medicine SE would know what compounds would be antifungal, but they won't know what won't yellow when exposed to UV, and particuarly how yellowing might be textile-dependent.  And the latter two points are the trickier parts, which is why I asked it here.  Indeed, I suspect if I posted this on either of those sites, they'd recommend it be moved to Chemistry SE.

Is the preservative for preservation of the formula, or of the target textil ? Natural or man-made fibres ?

@Poutnik For the preservation of the formula.  Both natural (silk, linen, wool, cotton) and man-made textiles.

You say it is a non-food product but it must be food for the fungi. What elements are the in the sample, and what metabolite is in it for the fungi to digest and get hold of some Gibbs energy? I didn't think yeast would grow at 20% ethanol or isopropanol.

@KarstenTheis  (1) Many non-edible products can support fungal growth, e.g., cosmetics.  (2) The formulation is proprietary, so I'm not able to disclose the other ingredients.  (3) I've read that at least 50% ethanol is needed to serve as a general antimicrobial, but I'm not sure about the specifics for yeast and mold.

@theorist Brewer's yeast undergoes apoptosis when alcohol levels reach ~15%, which sets a limit on the alcohol content of wine and beer (distilled spirits do not have that limit). See e.g. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3354062/ I'm sure other microorganisms have a higher tolerance, but alcoholic drinks were one way of storing sugar-containing liquids without refrigeration before we had other methods and preservatives.

@KarstenTheis  What about mold?  According to http://www.milkthefunk.com/wiki/Mold "Mold is tolerant of both hops and alcohol levels above what is in beer (most are killed at ~28% ABV [alcohol by volume], but some can survive up to 50-60% ABV), so do not rely on hops or alcohol to prevent mold growth."  They don't give a primary reference, but perhaps I can find one....


text:
N-hydroxysuccinimide as a good leaving group
In the synthesis of carboxylic acid derivatives (esters, amides), I often see N-hydroxysuccinimide used in conjunction with a coupling reagent (EDC, DCC). The reason given for this (from online) is that N-hydroxysuccinimide is a good leaving group.
Can anyone explain why this is so? I am unable to think of any reasons for why N-hydroxysuccinimide has a weak conjugate base compared to a general alkoxide.
Thank you.

Off the top my head, having an electron-withdrawing nitrogen attached to the oxygen is going to help stabilize the negative charge on the conjugate base.

@Zhe but if I'm not wrong, isn't the electron donating effects of nitrogen stronger than it's electron withdrawing effects? For example how the amino group NH2 has electron donating effect rather than withdrawing.


text:
Effect of transition energy gap and frequency of microwave on appearance of CW EPR spectrum
Consider continuous-wave EPR for species with only one unpaired electron in the following argument.
The interaction energy in the presence of external magnetic field of strength B is given by
$$E = Bg\mathrm{m_s}\mu_\mathrm{B} - Bg_{\mathrm{N}}\mathrm{m_I}\mu_\mathrm{N} + a\mathrm{m_sm_I}$$
Taking the specific selection rules for EPR transition into account,
$$\Delta E = Bg\mu_\mathrm{B} + a\mathrm{m_I}$$

Just to check my understanding.

The figure below illustrates splitting of states in the presence of magnetic field, when the only unpaired electron is coupled to a spin-half nucleus$^{[1]}$ :

assuming $a$ and nuclear $g$ to be positive and isotropic.
Using the last equation above,
EPR I: $\Delta E = Bg\mu_\mathrm{B} -\frac{1}{2}a$
EPR II: $\Delta E = Bg\mu_\mathrm{B} +\frac{1}{2}a$
So EPR II transition energy gap is bigger. Then shouldn't the EPR II signal appears at lower field in the EPR spectrum? Logically, if the energy gap is already big, to make the gap coincides with the frequency of incident microwaves, the external magnetic field only need to be increased relatively little. But the opposite is shown in the EPR spectrum inserted. Did I understand the concept wrongly?

This second figure encountered in the same reference book$^{[2]}$ baffles me:

The figure shows that, when the frequency of incident microwaves is increased, the separation between two lines in a multiplet also increases. Doesn't that imply the hyperfine splitting constant will also change? But I read hyperfine interaction is field-independent.$^{[2]}$

Chechik V, Carter E, Murphy D. Electron Paramagnetic Resonance. Oxford: OUP Oxford; 2016.
1 page 12
2 page 36


text:
Why is Specific heat of Ammonia higher than water?
The specific heat of water is lower than that of $\ce{NH3}$, but why?
$\ce{H2O}$ forms more hydrogen bonds if I am right, besides there's more electronegativity for Oxygen.
But still, $\ce{NH3}$ has higher Specific heat. I found this link saying $\ce{NH3}$ has lower molar mass, but its just by one gram so it shouldn't be having much effect right?

https://chemistry.stackexchange.com/questions/26651/why-is-the-specific-heat-of-water-high#:~:text=Liquid%20ammonia%20has%20a%20higher,just%20for%20the%20same%20reason.&text=The%20heated%20water%20will%20contribute,a%20lot%20to%20water%20molecules. May refer to the link for all clarification.

$\ce{H20}\ $  is unknown. You may mean $\ce{H2O}$. Be aware chemists may be quite sensitive for O vs 0 replacement in chemical formulas.

Consider the effect of lower molar mass and of counts of vibration degrees of freedom on the specific heat capacity.

The link in the question states that ammonia has a higher specific heat than water. This is true, but it is important to compare the specific heat values under identical (or very similar) conditions.

The isobaric specific heat of ammonia (gas) is 2.1636 kJ/kgK at 1 atmosphere and temperature = 26.9 C. https://www.engineeringtoolbox.com/ammonia-heat-capacity-specific-temperature-pressure-Cp-Cv-d_2016.html At -33.6 C (boiling point of $NH_3$ at 1 atm), the specific heat of $NH_3$ gas is 2.2946 kJ/kgK, not much different from the gas at 26.9 C; however, for liquid $NH_3$ at 1 atm and -33.6 C, the specific heat is 4.4464 kJ/kgK, a large increase from the gas at that temperature.

Liquid water has one of the highest specific heats among common substances, about 4182 J/(K kg) at 20 °C https://en.wikipedia.org/wiki/Specific_heat_capacity At 100 C, liquid water has a specific heat of 4.2157 kJ/kgK (implied pressure = 1 atm). Again, not much difference because of temperature. https://www.engineeringtoolbox.com/specific-heat-capacity-water-d_660.html Now we find water vapor data in a different place: https://www.engineeringtoolbox.com/water-vapor-d_979.html where gaseous $H_2O$ at 216 F (102.2 C) has a specific heat of 1.890 kJ/kgK, a large drop, even bigger than in the ammonia system.

So although it is often stated that water has one of the highest specific heats among common substances, it seems that either ammonia is not considered common for this issue, or yes, it's true that water has one of the highest specific heats...And that is true of the gas and the liquid states!
But then among all the data, which are truly interesting, there is one other thing that stands out: the heat of vaporization of water is 40.8 kJ/mol, whereas for ammonia, it is only 23.5 kJ/mol. So water can still be considered an extreme molecule.

I believe you are comparing the specific heat of gaseous ammonia to that of liquid water. Presumably the linked page was referring to the difference in specific heat  when both are in the same state. Liquid ammonia has a higher specific heat than liquid H2O, and gaseous ammonia has a higher specific heat than gaseous H2O. The same trend is observed with specific heat of PH3 vs SH2.

@Andrew: Thank you for your comment! You are quite correct, and I will seriously edit my answer because it is currently a non-answer.


text:
How to prevent sedimentation of an EDTA suspension?
I tried to create a suspension of EDTA powder in glycerin with 1:1 ratio. I keep facing problem of EDTA sedimentation. I've tried other ratios also but this was the only one which formed least amount of sediment.
When I used coarse powder, the sediment formed within 24 hours but when finely ground powder was used, the same occurred but after 3 days. I even tried heating the powder & glycerin together on low & medium heat but EDTA sediment still forms but with varied time.
I'd like that the suspension remain consistent irrespective of how much time goes by.
I read on the internet that anti-settling agents (like xanthan gum & microfibrillated cellulose) can be used but I haven't tried that yet.

Is shaking/mixing before usage a problem ? There is probably no way but increasing the viscosity, what may affect the medical usage. Sedimentation may be affected by surfactants, but so does the medical usage as well. Mixing is the safest way.

Shaking & mixing is quite difficult once the powder settles due to viscosity of the glycerin especially once the mixture is placed in the test tube.

Be aware clarification because of incomplete or missing information means extra effort on both sides. Provide all possibly relevant background when asking.

Please, clarify. I've already mentioned all the details in the question...

You have not mentioned there why mixing is not easily applicable. Is applicable reversing orientation during storage ? Any additives may affect the medical application, but you provided no medical context. Somebody may advice the way how to stabilize the suspension, which would work, but would be totally not applicable for the purpose of the suspension. And there is still hanging danger of misguiding with medical consequences, at the edge of being off-topic for this site.

If you must know, I'm researching for a biochemistry paper. As the paper is confidential, I can't divulge much information about it due to institutional policy. However, the part for which I couldn't find any reasonable solution was the EDTA sedimentation therefore I put forth my query here to find an answer. If people here are unable to help, I'll ask elsewhere.

We're still in the experimental stage so whatever is told will be thoroughly tested before it's even considered for trials. Don't worry about it. This conversation hasn't been very fruitful. There are plenty of other chemistry forums out there. I'll ask my question there. Thank you for your replies.

Easy there, fella. Not knowing something is one thing but personal attacks... That's low. And I mean really low...

Thanks for the kind words. And btw I never made any personal comments.

Well, "conversation with you was not very fruitful" and "You were not helpful either. " are at the same level. :-) I.e. None meant personally.

Let us continue this discussion in chat.

I see 4 possibilties to avoid EDTA sedimentation:

Mechanical ways like shaking, mixing or container turning - but you say it is not applicable.

Increasing viscosity by thickening glycerol to  decrease the rate of sedimentation. Gel stages could probably even stop the sedimentation, but may not be applicable. You may want to try

Carboxymethyl cellulose (CMC)
Polyvinyl alcohol(PVA)
Polyvinylpyrrolidone/polyvidone/povidone (PVP)
Polyethylene glycol(PEG)

Other way would have been adjusting the glycerol density to match EDTA density if EDTA had not had too low density $\pu{0.86 g/cm^3}$ ( while glycerol $\pu{1.26 g/cm^3}$). OTOH, if there was a chance to replace glycerol by ethanol 73 w/w %, having the same density as EDTA has ... but I guess  there is not.

It may be worthy to try some emulsifiers, but EDTA particles may be too large and emulsifiers may not be applicable for the suspension purpose.

Soy lecithin is another emulsifier and thickener
Pickering stabilization – uses particles under certain circumstances
Sodium phosphates - not directly an emulsifier, but modifies behavior of other molecules, e.g. caesin
Mono- and diglycerides - a common emulsifier found in many food products (coffee creamers, ice-creams, spreads, breads, cakes)
Sodium stearoyl lactylate
DATEM (diacetyl tartaric acid esters of mono- and diglycerides) – an emulsifier used primarily in baking
Simple cellulose – a particulate emulsifier derived from plant material using only water

Mechanical mixing wasn't an option as after sedimentation EDTA in Glycerol becomes quite hard to mix.

Lowering density was an option but I'm unsure of how that can be achieved? Would appreciate more clarification...

I haven't tried cellulose yet. Will do that & get back to you. But (just a thought) don't you think cellulose won't work as it's water soluble. Will it mix into Glycerol?

@SohailSA aboue density:  You may have noticed the 3rd conditional - a conditional not achivable in past, present or future. The only option would be replacing glycerol by less dense solvent, what is not probably applicable. Celullose is not soluble. You have not commented reversing orientation.

Will using xanthan gum be of any help?

No experience with that.

Can't replace Glycerol as the experiment requires a material that can also act as a lubricant. Can you suggest any other solvent that could serve as a good alternative?

You know, advising to a black box is like an ordering a meal in a restaurant where waiters would not say what they have, but only what they do not have if you ask for it. Ordering by trials and errors would get old pretty fast.

Let us continue this discussion in chat.

My apologies. I was using this forum for the first time therefore didn't understand when you asked me to continue through chat.

It is rather forced by the site policy and administrators/moderators, as comments below Qs/As are not intended for discussion. Note the StackExchange sites generally, and so does Chemistry SE are NOT (discussion) forums, but Questions/Answers sites, aiming to high quality of both. So there is big difference between ordinary forums like Quora or topical forums on 1 side and SE on other side.


text:
Why do the units of rate constants change, and what does that physically mean?
Just to be clear, I do understand that the units of the rate constant k is selected to make the equation dimensionally consistent. That is not what I am asking.
It should be the case that k 'means' something physically, which is implied by its dimensions/units. And I would also assume that what k 'means' should not be dependent on the order of the reaction, since I cannot see any obvious reason why it should. But we know that the overall order of a reaction changes the units of the rate constant, so I must be wrong.
My question is as stated: Why do the dimensions of the rate constant change depending on different reaction orders, and what does that physically mean?
As an extension to this, I would also like to understand how the same thing works for the equilibrium constant, where the stoichiometric coefficients of the species or the number of products/reactants change the units of the constant. Again, I cannot see the reason why these would change what the constant means dimensionally.
Sorry if this an obvious question. I am not a chemist.
Edit:
I wanted to clear up a few things.
First of all, as I understand I was expecting there to be a deeper meaning to the dimensions of the rate constant even though there isn't always an obvious one, since it is an empirical equation which merely simplifies a process rather than describing the fundamental workings of it. But I still want to understand intuitively (if there is indeed an intuitive reason) why the units are different for each overall order. As in the examples Maurice gives in his answer, a reaction of order zero produces dimensions of amount per time whereas a first order reaction produces [ratio] per time and so on for other real number orders. It just seemed strange to me that a constant in the equation, which seems to be doing the same job regardless of order, express something in completely different dimensions depending on order.
After reading your answers, this is what I thought: In a zeroth order reaction where the rate is constant, amount per time seems to be the only reasonable way to describe the rate since ratio per time inevitably produces a curve. Similarly for a first order reaction, amount per time wouldn't work because this value changes constantly. So, the constant is expressed in terms of the units which can describe the reaction given its nature. This was stated/suggested in some replies but thinking of it over an example is what cleared it up (if it is correct) and thank you for your responses. Perhaps this was obvious but a constant with varying units was not something I was familiar with.

The units of the "equilibrium constants" $K_c$ or $K_p$ have no meaning. The One True Equilibrium Constant $K$ (note the lack of subscript) is defined in terms of thermodynamic activities and is dimensionless, see Which equilibrium constant is appropriate to use? or Unit of the equilibrium constant: contradiction of Bridgman's theorem?

think he's talking about rate constants in kinetics and not in static equilibria

@Bertram Thanks, but the second last paragraph certainly talks about equilibria. I don't pretend that my short comment is good enough to be an answer that addresses the complete question, I just wanted to post a couple of extra links that addressed that part.

I post these two links to pdf. Although I didn't went through them carefully they are about chemical education and are almost certainly of quality. I hope no to infringe any copyright or so... https://pubs.acs.org/doi/pdf/10.1021/ed067p88.1 http://www.ijpce.org/index.php/IJPCE/article/download/51/57

The situation is weirder that the OP realizes. Only in elementary reactions will the exponents be integers. For many reactions evaluated experimentally the exponents would be some real number like 1.27.

I seems the number of answers here rises with lack of clarity as usual.

@Mithoron :)) I think this is rooted in the way we, even the chemists, start learning. But I won't recommend teaching chemical potential and activities before of solutions and concentrations.

see the answer here https://chemistry.stackexchange.com/questions/145046/can-rate-constant-have-units-as-1-time

@porphyrin It's not really the same question and although one of the answers sort of comes close, I'm not satisfied by that explanation. But perhaps I am expecting there to be a deeper meaning to the units than there actually is.

@Aerovolo to the units of k yes, you put a mean that isn't there. This is general with all constants except natural ones, by the way. A somehow more substantial issue is about K. Indeed, this two days "discussion" makes me rethink. I am still on the line "use (conceptually at least) mole fractions and everything is fixed" but I am not sure anymore. I must go through the paper I have linked above (the one from ACS).

Maybe it would make more sense if rate constants would have different symbols depending on whether it is a zero, first or second order reaction.
As an analogy, we might be asking "how big" an object is depending on the length of it. Depending whether the object is, say, a square, a cube or a hypercube, the answer would be in terms of its area, volume or 4D-volume, and the units (and dimensions) would be different in each case.
For the equilibrium constant, we avoid the issue by switching from concentrations to either concentrations divided by a standard concentration (this is a simplification) or to activities (the former multiplied by an activity coefficient that accounts for non-ideality).

Why do the dimensions of the rate constant change depending on different reaction orders, and what does that physically mean?

They have to change with the chosen system of units so that the equations are consistent. We could choose instead to have a dimensionless measure of concentration when we set up the kinetic equations. This would describe the reality equally well, but the kinetic constants would always have the same dimensions (1/time).
According to this argument, there is no physical meaning in rate constants having distinct dimensions depending on the order of reaction.

The physical reason why the dimensions of k differ for different reactions is the different mechanism behind the reaction.
A process that only involves the internal vibration or rearrangement of one species differs from a process in which two or more equal or different species must first diffuse and collide*.
It seems intuitive to me that this should be reflected via power relationship between concentrations/activities and reaction rate,
or alternatively
as for the rate of reaction has dimensions of amount of substance over time, the dimensions of k forcedly must change for reactions of different order.
Finally, concerning the equilibrium constant K, consider that at equilibrium the rate of forward and back reactions are equal. This shows that the true equilibrium constant is a dimensionless quantity kf / kb.
Apparent dimensions occur because, operatively, concentration is used, but molar ratio units could be chosen instead.
*note that different mechanisms pose different upper limits to the reaction rate.

Hmm, I don't agree re. eqm constant $K$. Both the forward and backward rate constants $k_\mathrm{f}$ and $k_\mathrm{b}$ have units which include concentration and don't necessary cancel out; the units of $k_\mathrm{f}/k_\mathrm{b}$ are analogous to $K_c$, which is in general not dimensionless.

@orthocresol isn't  the same, you can express kinetic laws in terms of mole fraction as well. Actually what is true is alway the activity, both expressing equilibrium as well as the kinetics. Physically there is no reason to use [M].

Maybe a numerical example may help you compare the different orders. Let's take first an example of a reaction of zeroth order : the combustion in air of a candle containing $n$ $mol$ wax. At every second, the same amount of wax is burned. The rate of the reaction is $$r = -\frac {dn}{dt} = k_o·n^0 = k_o$$ If the candle contains $1.44$ mole wax and burns in $2$ hours ($7200$ s), the rate constant is $$k_o = r = \frac{1.44 mol}{7200 s} = 2·10^{-4} mol/s$$ Here $k_o$ is the number of moles burning per second. This number is constant during the whole chemical reaction. Its unit is $mol/s$. If, instead of studying a candle, we are working in solution, $k_o$ is defined in $mol·L^{-1} s^{-1}$
Let's compare this result with a first order reaction, namely the decay of a radioactive isotope . The rate law is : $$r = - \frac{dn}{dt} = k_1·n^1 = k_1· n$$ This can be written so $$- \frac{dn}{n ·dt} = k_1 $$ You should observe that $k_1$ is not the number of moles reacting per second. It is the proportion or the fraction of the initial number of moles that decays per second. $k_1$ has not the same dimension as $k_o$. The unit of $k_1$ is $s^{-1}$.

This answer comes close to what I was trying to address. Is there an intuitive reasoning to 'why' the constant expresses something different (e.g. amount per time, ratio per time, whatever per time etc.) for different orders, and why should it not be the same? Does it simply have to do with different reaction mechanisms for different orders? I'm aware my wording isn't very clear but thank you for the answer.

@Aerovolo are you comfortable with the fact that rate has always dimensions of amounts of substance over time, because it is what you are looking for? At least this would fix one halve of your question.

@Alchimista Even though the units always have a term relating to amount (or concentration directly) over time, its power changes surely? (e.g. mol^0 and mol^-1 as given) So wouldn't it express something slightly different for different powers (ratio for 0, amount itself for -1, and something else for other real numbers which I'm not entirely sure)? I was wondering what the powers or the fact that they change for different orders signify; if it does at all.

I don't understand fully the example you give, but say, if something requires two things colliding, it can even be a billiard pool, the number of  a red and a white ball collisions change by adding just balls of one colour or both. If the balls have an internal mechanism that causes them to broken in pieces, you can change nothing. The number of balls keeping crashing (as compared to the total of course) will keep its pace. Note that all this is due to the use of concentration as Mol.

@Aerovolo rereading your comment... You must fix Rate not k, while thinking on this. Rate is surely amount of substance over mole, it cannot be anything else. For k, refer to my answer and comments. And also M. Farooq answer that puts my billiard pool in a formal context. A being related to the speed of the balls. The rest mechanism dictated.

The simplest way to approach this is that the rate constant has whatever units that make the rate have the correct units.
Generally, the rate is in units of molar per second, so based on how you're multiplying concentrations (in molar) in the rate law, you should be able to figure out the appropriate units for the rate constant.

I'll add to this response that different rate laws have different differential equations they are associated with, as the physics of the reactions are distinct.  Naturally the units of the rate constant would depend on which differential equation modeled the scenario.

Well, it may not be appropriate to say that the rate constant, k, has no physical meaning. It is the only factor in the rate equation which is temperature dependent, so larger k means faster rate of reaction. Another way of looking at the units of k is in terms of Arrhenius equation.
$$
k(T)=A e^{-E_{\mathrm{a}} / R T}
$$
The factor $A$ is called the frequency factor and it has the same dimensions as the rate constant, because the exponential term is dimensionless. You can see why $A$ is called the frequency factor for any order.
For order $(m + n)$, the rate constant has units of mol$^{1−(m+n)}·L^{(m+n)−1}·s^{−1}$, (see Wikipedia on Reaction rate constant)
so you can associate the units of $k$ as in terms of the units of the frequency factor- roughly the frequency of collisions. It is called frequency because of the second$^{-1}$ factor.


text:
What is the proper use of the term "pi-donor ligand"?
Robert Crabtree's The Organometallic Chemistry of the Transition Metals (7e) describes a pi-donor ligand as being a pi-donor "as a result of the lone pairs that are left after one lone pair has formed the M-L sigma bond". This, of course, makes sense, and the author goes on to explain that this pi-donation is a destabilizing effect.
Later in the chapter, he goes on to call ethylene a pi-donor because it forms the primary M-L pi-bond via the donation of pi-electrons. Of course, I understand that both types of ligands described are technically donating pi-electrons, but I find it very confusing that the term pi-donor is used to describe two entierly different ligand effects -- one that is not the primary bond and destabilizes the M-L bonding interaction and one that is the actually M-L bonding interaction.
My question is wheather or not there are better terms for differentiating the primary bonding and the secondary interactions. For example, I find it confusing to call ethylene a "pi-donor and pi-acceptor ligand" since forms its primary bond through pi-donation and back bonds via pi-acceptance.

Personally I’d use “pi-coordination” rather than “pi-donation” for the ethylene example

@Mithoron: I should elaborate that it is destabilizing towards the M-L bonds, not the lone pairs of the pi-donor. Is this what you meant?

** Destabilizing the M-L bonds due to electron-electron repulsions between the t2g set and the electrons in the lone pairs

Huh, I guess you're right then, with this. Still, I don't exactly get your point with the question. What's wrong with one group being sigma donor / pi donor and the other pi donor / pi acceptor?

@Mithoron: I just thought the similarity between the two concepts was confusing and could get mixed up, but I thought Andrew's comment cleared up my concern (eg. I would rather refer to ethylene as having pi-coordination and being a pi-acceptor than saying it is a pi-donor and a pi-acceptor)


text:
Properties of inconel
I am doing a project on the physics of the spaceX raptor engine. I believe it is made out of a certain type of inconel alloy. I want to understand why this alloy has the properties that it does, but I don't know anything about materials science. I am a senior physics major.
I am looking for a good introductory text/texts that will give me insights into the properties of inconel so I can explain why it doesn't flare at high temperature+pressure in my paper.
Any other advice would be welcome as well. Thanks for the help!

The inconel article in wikipedia might give you a start.

The Wiki  article is a start , I would not rely on it if you must have correct info  ,I see Specialty Metals acquired Inco , Huntington  and Wiggins.  Inco developed most info ( not Wiggins) ,they did have  a connection with Wiggins.

There are several Inconels , the basic one is 600. Because the Inco and Huntington divisions are now gone or acquired ,it would be difficult to get literature . But they produced copious amounts , you may be able to find some ; I have about a 6" stack of brochures I have been too lazy to discard. American Society for Metals Handbooks should be easy to find ( there are several issues). ASM is a common and comprehensive source. What do you mean "Flare" ?

The 600 Inconels do not age harden; The 700 Inconels do age harden. All have good strength and oxidation resistance at high temperatures . Some useful above 2000 F. If you are talking about a rocket ,I guess you do not need blade and vane alloys ( for turbines). 800 and 900 are Incoloys  ; 400 and 500 are Monels
Addenda
Looking at SpaceX info they are using something like Inconel 750 ( formerly X , and X750 ), so it is age hardened for strength due to pressure in the engine. Supposedly they have a modification named SX 500 . Apparently by "flare" they means rapid oxidation because of the oxygen in the engine , but they do not explain.

This is useful thanks. I assumed that flare must have been a technical term, but I guess not. I think it probably means that in the combustion chamber, the oxidizer and metal oxidize and ablate so rapidly that it is essentially combustion?

I think "flare" was created by someone in the SpaceX  public relations department ; no reflection on SpaceX impressive accomplishments. Any turbine hot stage alloy would resist rapid oxidation in the rocket although high oxygen partial pressure can cause various problems.  As NASA found out when they made the poor decision to use 100% O in the capsule and killed 3 astronauts on the ground ( see Gus Grisom , et.al.)

Ya, I'm pretty sure there is a certain section of the engine that combusts a super oxygen rich mixture and uses the hot exhaust to drive a turbine that pressurizes the combustion chamber.


text:
Why "monoxide" but not "diodine"?
When naming compounds, sometimes when there are two vowels in a row the second is elided: this happens for example with "mono-oxide", which becomes "monoxide" instead.
Why is this not always applied, e.g. with "diiodine"? Why aren't the repeated i's removed to make it "diodine"?

As I see it, there's two possible reasons: (1) It's a bit more awkward to say monooxide than diiodine, hence the dropping of the extra "o". You also see that in "monopsony", which combines "mono" with "opsony" (buyer). [My MacOS disctionary defines it as "a market situation in which there is only one buyer."] (2) More commonly-used words get modified, through usage, more often than than less-commonly used words, and monoxide is more common than diiodine (though that wouldn't explain monopsony!).

One more example: EDTA (ethylenediaminetetraacetic acid)

Note that the "o" in "mono" is often elided not only when double "o" results: e.g. monatomic.

We do have cyclooctane, though.

The fact that "mono" is disyllabic probably has something to do with it. We can analyze "monoxide" as "mono" being reduced to "mon". If we had "diodine", it would be hard to analyze that as "di" being reduced to "d"; prefixes don't get reduced to just a consonant. A vowel prefix can be pronounced by itself, but a consonant prefix can only alter the syllable of the main word, rather than being pronounced in its own right.

Discussion in ELU.SE: Why "monatomic" and not "monoatomic"?

Both "monooxide" and "monoxide" are used in the literature, yet "monoxide" is being  used more often (Google Books Ngram Viewer).
Although this is an accepted elision, it is not the preferred one, and must not set a precedent for other cases when multiplicative prefix ends with the same vowel as the root word begins with, such as "diiodine".
According to the current version of Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 [1, p. 31]:

IR-2.7 ELISIONS
In general, in compositional and additive nomenclature no elisions are made when using
multiplicative preﬁxes.
Example:

tetraaqua (not tetraqua)
monooxygen (not monoxygen)
tetraarsenic hexaoxide

However, monoxide, rather than monooxide, is an allowed exception through general use.

Further, from section IR-5.2 Stoichiometric names of elements and binary compounds [1, p. 69]:

The multiplicative preﬁxes precede the names they multiply, and are joined directly to them without spaces or hyphens. The ﬁnal vowels of multiplicative prefixes should not be elided (although ‘monoxide’, rather than ‘monooxide’, is an allowed exception because of general usage).
[…]
Examples:
[…]

$\ce{NO}$ nitrogen oxide, or nitrogen monooxide, or nitrogen monoxide

References

IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005. ISBN 978-0-85404-438-2. IUPAC website

@andselik, monooxide is not an entry in the OED, not even with a dash. I don't think it was accepted as legit by English linguists. And di-iodide occurs with a dash there.

@M.Farooq I wonder how much of that comes down to linguists having heard of monoxides.

@J.G. I think it’s actually a matter of phonetics. English generally does not like doubled vowels outside of certain specific cases (and almost all words that have a doubled vowel that is not either a diphthong or collapsed to a single vowel sound (like in aardvark) are specifically loan words), so ‘monooxide’ would end up either pronounced just like ‘monoxide’ or as if it were two words (or possibly as if the ‘oo’ were an ‘oa’).

@J.G., I did not mention the context, the OED (Oxford English Dictionary) is a monumental work describing collecting all the English words since 13th century. It think in print, it is 20 volumes (it is not the free internet version). You will find all common chemistry terms, ancient and recent. If monooxide is not listed, but monoxide is, then it means that the linguists did not accept monooxide as a valid spelling even though if it were used by some.

@M.Farooq OED is not an authoritative source for chemical nomenclature and terminology, IUPAC is. "Di-iodide" adds inconsistency and is plain illiterate. Zumdahl's textbooks also force "tetroxide" and "monoxide", but it's a personal preference that's not backed up by any guidelines or standards. To the English linguists admiring monoxide: You Have No Power Here.jpeg.

@M.Farooq I know what the OED is, I went to Oxford. In any case it's a decsriptive dictionary, so it's not telling us monooxide is invalid, it's telling us whoever prepared it hasn't read enough chemistry papers to have encountered it. They certainly won't be able to put any of these terms in their IUPAC context.

@J.G., I think we have to stick to IUPAC rather than OED as you and andselisk say.  The problem with anonymous Q&A sites is that it this gives us no clue who were are talking to. It could be a student or very well experienced respected scientist. Good for you that you were trained at such a good place. I still respect the level of work done by the OED board even for scientific terminologies. There are only three languages in the world who have this level of thoroughness with their monumental dictionaries of English, German and Urdu. I am not aware of others, if they exist in > 20 volumes.

Although it seems like a question of English language phonetics, monoxide is not an exception but a general trend.
Mon(a)oxide (vowel "o" dropped),
Dioxide (no "a" vowel in the prefix)
Trioxide (no "a" vowel in the prefix)
Tetr(a)oxide (vowel a dropped)
Pent(a)oxide (vowel a dropped).
The accepted spelling of diiodide is di-iodide in the grand Oxford Dictionary. It is also consistent with the other halides
difluoride,
dichloride,
dibromide, and
di-iodide
Nothing seems unusual about it.
As to the English language, someone wrote why b-u-t is pronounced differently from p-u-t? Some times we have to accept things as they are.

On the contrary, the fact that this dropping of the vowel is not uncommon is all the more reason to consider it a linguistics question.  And as to your "Some times [sic] we have to accept things as they are" advice, I disagree.  Questions in linguistics are just as valid to pursue as questions in chemistry.  What you said is like someone saying, in response to the question "Why is the sky blue?" that "sometimes we have to accept things as they are", to which any chemist or physicist would respond -- "hang on a sec, there is an explanation for that."

Well, that's a bit of a straw man. I wasn't saying this should get a funded study! I was simply saying it's a question worth asking on a linguistics forum. Just because we, as non-linguists, don't have an rigorous explanation, doesn't mean the question should be dismissed as not worth pursing. It's not something you or I have the background to dismiss. It could have a fascinating explanation.  We need to have enough respect for other fields to appreciate that, even though we might not understand the value in pursing a question, an expert in that field might know something we don't.

@M.Farooq No, a beaker does not have a beak. The word is derived from ancient Greek βῖκος, a jug or wine jar. Calling your logic a "straw man" is insulting to straw, IMO.

@alephzero, IMO, your "IMO" does not count! I was talking about the beaker in jest! Sorry that you missed it.

"Monooxide" was previously used to describe compound with single oxygen before being replaced with "Monoxide". Instance of "Monooxide" can be found in papers from 20th century which later became obsolete. Here is an example:

Reference:

Official Gazette of the United States Patent and Trademark Office: Patents, Volume 1055, Issue 2, U.S. Department of Commerce, Patent and Trademark Office, 1985

OED says for mono-: "before a vowel or h usually mon-." There's no such notation for di- (probably because a prefix that short can't afford to be reduced by 50%!). So there isn't a general rule in English or in chemistry that demands that identical vowels coalesce, but there is one that says that "mono-" is liable to drop its final "o" before any vowel, as in monarch, monaural, or monocle. This doesn't always happen (and it appears to be less common in modern scientific contexts), but it's still a regular feature.


text:
Can you separate CO2 into carbon and oxygen using pyrolysis?
I recently read about a pyrolysis process that separates methane into Carbon and Hydrogen. The basic technology/process is you have a vertical column of super-heated liquid metal and you inject methane into the bottom of the column. The methane breaks down into hydrogen and solid carbon. Are there any similar processes that separate CO2 into solid carbon and oxygen using liquid metals and/or salts etc.?

https://chemistry.stackexchange.com/questions/915/how-can-carbon-dioxide-be-converted-into-carbon-and-oxygen

Not energetically feasible at all.

The current modeling of future power system operations suggests there will be a large amount of excess generation which is "spilled" and/or dumped because there is no practical use for the electricity. This is called over-build and spill. It's a new/old strategy in power system operations. There will be many cases where we can dump this power into a thermal reservoir of some sort and then gradually draw off the reservoir to accomplish some task. It could be drying wood or making powdered milk or possibly pyrolysis. I'm asking if there's a known pyrolysis process that accomplishes this.

By this I mean a pyrolysis process that breaks up CO2 into carbon and oxygen using a liquid metal bath and/or a salt and or some other thermal + catalyst driven process.

The answer is No ! Unfortunately ! $\ce{CO2}$ cannot be transformed into $\ce{C + O2}$ using liquid metals or salts. If it would have been feasible, somebody would have already done it somewhere.

Standard enthalpy of formation of carbon dioxide is -393.5 kJ/mol. Standard enthalpy of formation of both oxygen and graphite is 0 J/mol. Standard entropy of carbon dioxide is 213.8 J/mol-K (NIST). Standard entropy of oxygen is 205 J/mol-K (Chem LibreTexts). Standard entropy of graphite is 5.69 J/mol-K (ChemLibreTexts).
Everything below is a lot of fairly inexact approximations/assumptions because I didn't want to use a better model or search literature for empirical adjustments and/or constants.
Assuming constant T,P (which I doubt would be true under reactor conditions) G = H - TS. Approximating $\Delta H, \Delta S$ as constant with T, $\Delta G = (0 - (-393500)) - (T \cdot (5.69 + 205 - 213.8)) = 393500 + T \cdot (3.11)$. Then $\Delta G < 0$ regardless of temperature. Catalysts also do not affect thermodynamics, so this process can never occur spontaneously. Increasing pressure of the reaction chamber would not help appreciably either, as the reaction has 1 mol gas on each side. While assuming $\Delta H, \Delta S$ are constant with T is quite incorrect as conditions become more extreme, the approximation should suffice to show that the reaction does not occur easily. Additionally, even if one were to account for the perturbations in $\Delta H, \Delta S$ with T and P, these conditions would be expensive to maintain in a reactor. (Note that this analysis does not even account for the activation energy/kinetics problem! I did not look up the activation energy of the problem, but it is entirely possible that the extreme conditions would still not allow the process to occur at a reasonable rate.)
As a result, this reaction is (probably) practically only driven forward by direct energy input (aka a coupled reaction/external energy). This often, if not always, involves some energy input, such as light or a coupled chemical reaction, that directly impacts the electron states of the participant compounds. These changes in electron states change the identity/properties and stability of compounds, and, when applied effectively, they can help drive the reaction forward. However, relevant changes to electron states using heat not only often involve very high temperatures, but the effect of heat on electron states is also a highly statistical process. Thus, it may be relatively inefficient. Thus, pyrolysis, while it may be theoretically possible, would be incredibly impractical at achieving separation of carbon dioxide. (Also, to compare to your methane example, methane combusts readily, carbon dioxide does not. Methane has a standard enthalpy of formation of -74.87 kJ/mol.)
Finally, if you still need more reasons to look at a different way to drive this reaction, I have not mentioned: energy/cost to continuously scrape/collect the produced carbon, purify/collect/pressurize an inlet carbon dioxide stream, (if you actually want to run the reaction, you'll probably want) catalytic surfaces to encourage carbon deposition, possible packed/fluidized bed catalysts, heat exchangers, and much, much more.
ELI5: Methane combusts/reacts readily and is an "unstable" compound. Carbon dioxide is very stable and does not react easily. Higher temperatures can make some reactions that do not happen at a lower temperatures occur, but this particular reaction receives no such benefit. Heat and catalysis can also drive forward the speed of a reaction, but the reaction has to be able to occur first. Pyrolysis, as you described it, involves catalysts and heat. As a result, pyrolysis cannot be used to separate carbon dioxide into carbon and oxygen.
tldr: Excepting conditions extreme enough to perturb $\Delta H_{rxn}, \Delta S_{rxn}$, conditions of constant T,P never favor breakdown of carbon dioxide into elemental carbon and diatomic oxygen. Pyrolysis cannot drive this reaction.

Thanks for all of that. How about this? What if you had two stages of reactions. Stage 1: Fe3O2 is heated to 1500C driving off oxygen. The resulting FeO is moved to CO2 chamber where it absorbs oxygen from the CO2. Result is CO and cooled Fe3O2.

Stage 2: Pyrolyze the carbon monoxide. I realize this may sound like a silly waste of fuel but I'm only asking if it's technically possible? This may sound strange but there's a high probability we'll "spill" upwards of 25 to 40% of our total electricity production in the future. This won't really start to happen for 20 years but this is where we are headed. Dumping this production into a thermal reservoir and then pulling off the thermal energy could be a good way to utilize some of this spill.

Aha.... Here is the sort of thing I was looking for. https://www.thechemicalengineer.com/news/turning-co2-into-carbon-black/ You make methane first by combining CO2 and hydrogen. Pyrolyze the methane into carbon and hydrogen. Repeat cycle.

Here is a more technical description. Of course, you have to go outside of America to et the answer - to Germany, in this case. SE is notoriously American-centric. If it was not invented/discovered/developed in America, it does not exist. https://www.sek.kit.edu/downloads/Datenblatt_NECOC_ENG.pdf


text:
Calculating change in Gibbs Free Energy for a specific type of problem
Can anyone describe what is going on here?
$$Si(liq.)=Si_{(1 weight percentFe)}$$
I have come across this statement quite a few times in Metallurgical Thermodynamics.All that they ask for is change in Gibbs free energy for the process.I am completely unable to figure why this particular model is so important for a sepecific question.It would be great if someone could interpret both the analytical and the mathematical analogy to this.

It's not clear what "Metallurgical Thermodynamics" refers to. Please add a complete citation so that the context could be checked. Also, if $$Si(liq.)=Si(1 weight percent Fe)$$ is exactly how your source presents it, I'd say it's just poor formatting or the author doesn't know what they are writing about.

@andselisk This is from Gaskell.I do not know why they are not conting the space during formatting.

You may want to tinker with formatting. For instance, $$\ce{Si\textrm(liq)<=>Si{(1 \textrm{weight percent Fe})}}$$ is written as "\ce{Si\textrm(liq)<=>Si{(1 \textrm{weight percent Fe})}}"

Is it not more convenient to write : $\ce{0.99 Si(liq) + 0.01 Fe -> Si_{0.99} Fe_{0.01}}$ ?

@Maurice Correct but I feel the person wants to know why that is working

It is written in my book,"In situations in which the pure component exists in a physical state which differs from that of the solution of the Henrian standard state may be more convenient than the Raoultian standard state. Such situations include the solution of a solid in liquid solvent.The Henrian Standard Statte is obtained from consideration of Henry's law,which strictly being a limiting law obeyed by the solute B at infinite dilution."