text:
Trying to understand the chemistry in an iron carbon battery
I am trying to understand the chemistry that occurs in an iron carbon battery during charging.
The negative electrode is iron, the positive electrode is carbon. The electrolyte is iron(II) chloride. I understand that ferrous chloride dissociates in water to $\ce{Fe^2+}$ and $\ce{Cl-}$ ions. During charge the $\ce{Fe^2+}$ ions are attracted to to the negative iron electrode and the $\ce{Cl-}$ ions are attracted to the positive carbon electrode. The $\ce{Fe^2+}$ ions combine with electrons from the power supply to form metallic iron that plates on to the iron electrode.
My problem is understanding what happens at the positive carbon electrode. I'm told that iron(II) chloride loses electrons to the positive electrode and becomes iron(III) chloride. Fine, but how does this happen if all the $\ce{Fe^2+}$ ions are at the negative electrode? How does this happen if the iron(II) chloride has dissociated in to ions, there is no iron(II) chloride anymore?
If the power supply drives all the $\ce{Fe^2+}$ ions to the negative electrode and $\ce{Cl-}$ ions to the positive electrode how can any $\ce{Fe^2+}$ ions even be at the positive electrode to somehow recombine with $\ce{Cl-}$ ions back into iron(II) chloride that then loses electrons to the positive electrode and becomes iron(III) chloride?

https://www.sciencedirect.com/science/article/pii/S2468067219300318

Ions in solution

[OP] I understand that ferrous chloride dissociates in water to $\ce{Fe^2+}$ and $\ce{Cl-}$ ions.

Yes. The same goes for the iron(III) species.

[OP] I'm told that iron(II) chloride loses electrons to the positive electrode and becomes iron(III) chloride.

No, iron(II) loses electrons and becomes iron(III). The chloride is less relevant, and could be exchanged for a different ion.

[OP]  How does this happen if the iron(II) chloride has dissociated in to ions, there is no iron(II) chloride anymore?

The reactant is $\ce{Fe^2+(aq)}$, not $\ce{FeCl2(s)}$.
Ion movement

[OP] During charge the $\ce{Fe^2+}$ ions are attracted to to the negative iron electrode and the $\ce{Cl-}$ ions are attracted to the positive carbon electrode.

Yes, but a small fraction, only. Otherwise the repulsion between like ions would be too great.

[OP] My problem is understanding what happens at the positive carbon electrode. I'm told that iron(II) chloride loses electrons to the positive electrode and becomes iron(III) chloride.

Just $\ce{Fe^3+(aq)}$. There might be a double layer, but no solid iron(III) chloride.

[OP] Fine, but how does this happen if all the $\ce{Fe^2+}$ ions are at the negative electrode?

There are plenty of cations in the bulk of the solution, and plenty near either electrode.

I apologize for the delay in responding. I would like to thank you very much for the answer to my question. It has been very helpful to me. I now have a much better understanding of the chemistry. My main stumbling block was the assumption that the ions would segregate to each electrode far more than actually happens. Many thanks. I will now be attempting to understand how to determine the specific energy output by the iron carbon battery by measuring the mass of the iron active material and in some way taking into account the role played by the electrolyte as an active material. Thanks again.

The fully charged cell has an iron anode in a solution of ferric chloride. When it discharges, iron dissolves to produce ferrous ion while ferric ion (the oxidizer) is reduced to ferrous ion. Chloride ion is essentially a spectator ion which happens to be especially good at assisting corrosion of iron. So as the cell discharges, the solution takes 2 ferric ions to 3 ferrous ions by dissolution of one iron atom. The iron electrode is the anode and is the source of electrons to the external circuit. When all the ferric ions have been reduced to ferrous, the cell is discharged and there is no voltage. This is the condition stated in the second sentence: “The electrolyte is iron(II) chloride.”
Charging reverses this: the iron electrode is made negative by an outside source, that is, it becomes the cathode during recharge. Ferrous ions are attracted to this iron cathode and plate out on the iron, tending to restore the original condition of the electrode.
At the same time, electrons are being withdrawn from the solution thru the carbon electrode. This is where it gets a little tricky: the standard potential to produce Cl2 from Cl- is 1.36 V; but it only takes 0.77 V to produce Fe+++ from Fe++. So you can convert essentially all the ferrous ions to ferric ions before liberating any Cl2 gas. It seems a bit counterintuitive to get electrons from a positive ion rather than from a chloride ion, but those chloride ions are very reluctant to give up an electron, and this is what makes the battery so interesting. Other anions have been used also, like sulfate. The reactions are not all so simple, but the battery is cheap, and economy is attractive.

I would like to thank you very much for the answer to my question. I have a much better appreciation for the cell reactions. Drawing my attention to the standard potentials of chloride to chlorine versus fe3+ to fe2+ were the key for me to understand why no chlorine gas was forming. I should have realized this myself. My next task is to understand how to determine the cell's specific energy output. I would usually measure the mass of the active materials but here it seems the mass of the electrolyte, as an active material itself, must be considered. Any suggestions? Thank you, again.

There would be at least 3 ways to determine the energy output of this cell. 1) Fe + 2 FeCl3 (aq) --> 3 FeCl2 (aq). Look up heats of formation (CRC Handbook) for both sides of the equation.  2) Electrochemically (theoretical). Standard potential, corrected for concentration, times current (1 Faraday = 96500 amp sec).  3) My favorite (experimental): start with a discharged cell (FeCl2 solution), charge it and record the actual voltage vs current vs time. Then discharge it, recording the same. You can get the efficiency this way, and how many cycles of utility. Compare with theoretical.

Yes, I see the energy stored and output from the cell reactions could be calculated from 1) and 2). I have spent a lot of time on 3). I have built and cycled (charge/discharge) several iron carbon cells. I have experimentally determined cell out power versus time (mW/h) and efficiency (Ein vs Eout). My problem is how to determine the watt/hours per kg of the cell. For simpler batteries I would measure the mass of active materials. In this case the electrolyte acts as active material. Should I simply take the mass of the electrolyte add the mass of iron and calculate the watt/hours per kg?

Is there a way, other than this comment bar, to discuss this cell more fully?

My "expertise" has run out, I'm afraid. I'm running now on wisdom acquired by age. For your question about Whr/kg, your final cell would consist of electrodes, electrolyte, any containment and separators. All those should be included for a realistic Whr/kg.

Your expertise has provided me with a much better understanding of the iron carbon chemistry. Thanks again.

I have tried to understand the chemical reactions occurring in this Iron-Carbon cell. It seems to me that this cell will have difficulty to work at all. Here is my interpretation.
The initial solution contains $\ce{Fe^{2+}}$ and $\ce{Cl^-}$ ions. During charging, the $\ce{Fe^{2+}}$ ions are moving to the negative electrode made of Iron, where they are discharged and transformed into metallic Fe. Simultaneously, the $\ce{Cl^-}$ ions are moving to the positive electrode made of Carbon. When touching this electrode, they are discharged, converted into Chlorine molecules $\ce{Cl_2}$, but this molecule is a powerful oxidant which reacts easily with $\ce{Fe^{2+}}$ ions and transform them into $\ce{Fe^{3+}}$ according to :$$\ce{2 Cl^- -> Cl_2 + 2e^-}$$ $$\ce{Cl_2 + 2 Fe^{2+} -> 2 Fe^{3+} + 2 Cl^-}$$ Another explanation may be stated at this Carbon electrode, which avoids the presence of $\ce{Cl_2}$. If  the existence of $\ce{Cl_2}$ is denied, the two previous equations may be simply replaced by the oxidation of $\ce{Fe^{2+}}$ into $\ce{Fe^{3+}}$ $$\ce{Fe^{2+} -> Fe^{3+} + e- }$$ Whatever the correct phenomena happening at this Carbon electrode, the result is the same : the ions $\ce{Fe^{2+}}$ are gradually consumed and disappear on both electrodes, and new  $\ce{Fe^{3+}}$ ions will appear at the Carbon electrode, and then in the whole solution. In the long range, the ions $\ce{Fe^{2+}}$ will be slowly replaced by $\ce{Fe^{3+}}$ in the solution. Unfortunately $\ce{Fe^{3+}}$ cannot exist in solution if some metallic Iron is present. $\ce{Fe^{3+}}$ ions react spontaneously with $\ce{Fe}$ according to : $$\ce{2 Fe^{3+} + Fe ->  3 Fe^{2+}}$$
As a consequence, the "charging process" will have no chemical effect, as the composition of the solution remains more or less constant during the "charging" process. The only way for this cell to work is preventing $\ce{Fe^{3+}}$ ions to be dissolved at the Carbon electrode. Maybe this ion is simply hydrolyzed and transformed into a precipitate like $\ce{Fe(OH)_3}$ by reaction with water, as for example :  $$\ce{Fe^{3+} + 3 H_2O -> Fe(OH)_3  +  3 H^+}$$ But this reaction will only happen at relative high pH values, which is improbable as this reaction produced $\ce{H+}$ ions. So my reasoning must be flawed. Where ?

I apologize for the delay in responding. I would also like to thank you so very much for the answer to my question. My main problem was thinking that the ions involved would be isolated to the immediate areas around the electrodes. Having built several of these cells, I have observed the formation over time of a light brown precipitate in the electrolyte. This may indeed be iron hydroxide via reaction with water. With a charged battery, the Fe3+ reaction with the iron electrode that forms Fe2+ may form a barrier to further interaction with remaining Fe3+ ions? Many thanks.


text:
Reaction between Copper nitrate and ascorbic acid
I am confused on the products of reaction between copper nitrate and ascorbic acid. When I searched this online, one of the reactions I got was: $$\ce{2Cu(NO3)2 + C6H8O6 <=> 2CuO + 4H2O + CO2}$$
I am really confused as to how they determined that to be the product. My thinking was that the ascorbic acid breaks into $$\ce{HC6H7O6 <=> H+ + C6H6O6-}$$ and so the hydrogen ion, whose activity is higher than copper, replaces the copper to form nitric acid. I was thinking that this could be a double displacement reaction and so the copper ascorbate would become a precipitate. What am I thinking wrong here?

You have somehow managed nitrogen to annihilate with carbon and oxygen. You should really take your time for enumeration of chemical equations. BTW, such reaction would need high temperature and nitrate would decompose before it would be effective to fully oxidize organic matter. Definitely not happening in solution.

@Poutnik copper ascorbate?

Yes, would copper ascorbate + nitric acid product be possible?

Where is the nitrogen on the R.H.S.?

Nitrogen from the $\ce{NO3-}$ combining with the hydrogen ion of ascorbic acid to form nitric acid?

It is always good to cite the link when you mention that you saw a redox reaction online. It is wrong anyway.
Aqueous Cu(II) is not an strong enough oxidizing agent that it will convert organic matter into carbon dioxide. In other words, it is a weak oxidizing agent. Ascorbic acid is oxidized to dehydroascorbic acid and Cu(II) should be reduced to Cu(I). If you heat the solution strongly, Cu(I) may convert into Cu$_2$O.
Ascorbic acid is indeed an acid, but there is nothing which prevents its conjugate base to behave like a reducing agent. Many organic acids can behave like that.

I am really confused as to how they determined that to be the product.

By conducting the experiment. There is no other simple way to predict what will happen in a complex redox reaction.

Would the way I am thinking about it make sense i.e. you get $\ce{Cu(NO3)2 + C6H8O6 <=> HNO3 + CuC6H7O6}$ or would the copper precipitate out? Is there a way to theoretically predict the product of this reaction with some confidence without an experiment?

"Is there a way to theoretically predict the product of this reaction with some confidence without an experiment? –" I am afraid not. Chemistry has not progressed enough to predict the products theoretically from scratch. May in another 50 or 100 years this will be possible for complex reactions as yours. Today, you need to do an experiment to find out what the products are.

@M.Farooq Come now... There are many methods within thermodynamics, reaction kinetics and quantum chemistry which not only tell us which products occur with which probability but also enables us to model the properties of these products. We have been able to predict fairly well the outcomes of reactions for 300 years now - So while we do not have a common method, or methods for all kinds of reactions, I find your statement unnecessarily defeatist... Suffice to say it is a complicated problem, this predicting business, and there is no single perfect solution found yet.

@StianYttervik, Of course the human mind has been predicting the outcomes of new reaction  by interpolation and extrapolation of known chemical reactions, my point was can an ab initio calculation predict the outcome of this reaction in the OP yet. I don't think so.

would a reaction I proposed be a possibility?

Possibly, if you do a combined PhD in mathematics, theoretical physics and theoretical chemistry.

While written  'answer' albeit it does not explain the complete mechanism, but we do agree on the action of ascorbate part,  and you did sort of miss the fundamental underlying REDOX. And, one might mention, it does not address either the cited creation of CuO (which is not Cu2O), or the curious fate of the MIA nitrate ion. There is also a lack of references...

The equation written in the OP is not correct to begin with. Why should one argue about a wrong equation? I am not here to spoon-feed students with links and references to already well established facts and reactions. They should find the reaction by reading books and reading the literature (if they are advanced students). There is no excuse to anyone's laziness and relying on web only.  Printed books still contain a treasure of information.

It is unfortunate you think this way and believe people do not do their due diligence before turning to help and guidance

@user510, Just look at the statistics of how many questions are downvoted and closed in this forum because people haven't shown enough efforts. This comment is not aimed at any student like you. The gentleman, who is apparently not a student, above was demanding in the comments that I add references to common factual points. Similarly, the OP did not correct the original equation, which is completely wrong. The fact is that people have stopped reading books and expect that one provide links to each and everything.

In my assessment, an advanced reaction involving a cyclic REDOX reaction occurring with Copper ions, which I have previously discussed in this work 'Generation of Hydroxyl Radicals from Dissolved Transition Metals in Surrogate Lung Fluid Solutions' by Edgar Vidrio relating to disease (cancer) likely induced from progressive damage to  DNA (in the new equation this is now an attack on the nitrate ion,...) arising from the transition metal copper interacting with oxygen (in the new reaction this is replaced by HNO3,...) leading to the powerful hydroxyl radical.
That reaction becomes cyclic (at least, to a limited degree depending on the respective concentration of Ascorbic acid) per Vidrio's to quote:
“Similar reactions can occur with Cu, Cr and Ni. Furthermore, biological chelators and reductants can greatly enhance the production of ROS (Burkitt et al., 1991; Engelmann et al., 2003; Wenk et al., 2001). For example, in the presence of ascorbate (Asc), a biological reductant, the oxidized form of the transition metal produced by the Fenton reaction can be reactivated (R2 and R3), thus allowing additional ROS to be produced.
$\ce{Fe(III) + Asc (n) → Fe(II) + Asc (n') (R2)}$
$\ce{Cu(II) + Asc (n) → Cu(I) + Asc (n') (R3)}$ “
which cycles the cupric back to cuprous to feed a REDOX, which, for our new reaction involves aqueous cupric nitrate and Ascorbic acid, that is detailed in this reference, "Fenton chemistry in biology and medicine" by Josef Prousek, to quote reaction (15) on page 2330, a general depiction of Fenton-type reactions, to quote:
"For Fe(II) and Cu(I), this situation can be generally depicted as follows [20,39],
$\ce{ Fe(II)/Cu(I) + HOX → Fe(III)/Cu(II) + ·OH + X- }$ (15)
where X = Cl, ONO, and SCN. "
In the current context, the hydrolysis of cupric nitrate introduces HNO3, which interacts with cuprous (created from the Ascorbic acid presence) reforming cupric and, importantly, always hydroxyl radicals. The latter radical can attack nitrogen oxides compounds which can further progressively be broken down. Some expected reactions:
$\ce{ .OH + NO3- -> OH-  + ·NO2 }$
$\ce{ .NO2 + .NO2 -> N2O4 }$
$\ce{ N2O4 + 2 H2O <=> HNO2 + HNO3}$
$\ce{ 2 Cu(I) + HONO -> 2 Cu(II) + ·OH + N2O2^{2-} }$
etc., so the nitrate, nitrite,..., could be progressively decomposed (even to NH3), which may account for its notable absence in the cited text.
Lastly, the REDOX consumes H+ and select radical reactions could add OH-, all of which could eventually lead to either a basic copper salt, or Cu(OH)2. The latter on warming, forms CuO, as was claimed.

source: Stack Exchange

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

sort: votes


text:
How much time liquid mercury takes to evaporate?
I asked before about a thermometer that I broke yesterday. To summarize, I broke a mercury thermometer in my bedroom, which is a place where is difficult to have sure that I get rid of all mercury, but I cleaned the large amount without touching it.
However, I have more specific questions now. I discovered that I, by my lack of attention, have spread mercury along the house. I found a very tiny drop (really difficult to see) on the floor near to my desktop, which is in the living room, not in my bedroom. I also found a few same little drops in the bedroom, but it was really tiny. So, since I can assume that I will not be able to get rid of all mercury now, I have the following question:

How much time mercury takes to evaporate?

I think that wait the mercury to evaporate it is the only way that I can get rid of them now. Also, I have to say that I live in a country which doesn't have a service to check the amount of mercury in the air of the house neither get rid of the mercury.
Edit:
I found what maybe is more mercury tiny drops under a table in living room, but this time I tried to estimate the size of the possible drops.

Three really tiny mercury beads (zoom in if you could not clearly see the white points)

I know is it difficult to see, but I placed a piece of adhesive tape on the floor, which has the same size as my thumb's nail (about 1.4cm).
I don't know if is a good way to measure something, but I cut one of the three beads in a imagem editor and placed copies of it from on side until another. It takes about 30-40 copies to made the path. So, 1.4cm/30 results in each bead being at least about 0.47mm.
So, with this information, how many of that beads are needed to make the air dangerous?

The risk and recommendations are a bit OT here, but the rate of evaporation that is to be expected is interesting. The vapour pressure of Hg at 300K is around 1 Pa.

I deleted the third one to fit better to the rules. Sorry for that.

You really should find a good way to decontaminate. Letting it disappear by evaporation is akin to saying "let me breath dangerous levels of mercury vapour for a decade". We don't "let it evaporate" even in a well ventilated laboratory: we clean it up as fast as possible.

@matt_black I'm very worried now... As I said, I found some tiny drops even in a room which is not where I broke the thermometer, so I don't think I can clean all that mercury now. I cleaned the major part of mercury and never did or never will vacuum anything here. How can I decontaminate the house?

@karl even a 0.1g drop is enough to contaminate the air in a large house (say 500 cubic meters) above the minimal safe level for the best part of a year (crudely). So evaporating slowly is not a big gain.

@Mycroft You probably need to talk to a professional clean up organisation and double check the EPA advice here.

@matt_black You were right, about a year. Dispersed in smaller drops less than a year, but higher levels.

If I had no other option, professional help out of the question, I would buy 20m of flexible tube, put the vacuum cleaner outside and down the wind, and start vacuuming. But don´t cite me on this. ;-) Otherwise, it won´t kill you in a few weeks if you ventilate the flat twice per day, and in four weeks, Corona should be mostly over in most countries. If anyone is pregnant in your house, do something quick.

Nobody is pregnant, but I didn't get the idea of the vacuum cleaner. Is not supposed to DON'T use a vacuum cleaner?

With a long hose, and the vaccuum cleaner outdoors! OR, much better: Get you high-risk flatmates out on a hike in the countryside, and let the decontamination people do their work. The risk of a contact infection via dry surfaces is effectively zero. That "santizier" frenzy is an idiotic nonsense.

@matt_black I already read this link, and my country doesn't have a specialized service to deal with mercury... By the way, even it has a service like this, I live with two persons which are in a group of risk about COVID-19 and we are not allowing anybody to come inside since two months ago...

@Karl Later, I will add on the post some information about what "decontamination people" think about mercury in my country.

@matt_black
I found an article which says small drops of mercury will evaporate until 2 weeks (https://www.ncbi.nlm.nih.gov/pubmed/23067112?log$=activity) and another one which says that a house with cross-contamination by a thermometer didn't show 3 weeks after the spill (https://www.tandfonline.com/doi/full/10.1080/15563650701664731?src=recsys [at "Small spill investigations" section]). Could you please check if am I correct?

@Mycroft Well, it makes sense. If you do the math with drops of half a millimeter size, you land somewhere in that region. I like that report in the second paper you link. It fits to what the Swiss say (link in my answer), and generally says collect carefully, ventilate properly for a few weeks, and don´t panic.

@matt_black Matt, since you are the one with most pessimistic calculations, could you see the edit on my topic, please?

Sulphur powder (flowers of sulphur) is normally used to clear up mercury spills. You could try to get some (Amazon sell it) , just liberally  spread it over the spill area,  rub it into cracks etc.

How did this end for you @Mycroft? I’m in a similar predicament!

At https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1367823/ Baughman (2006) writes Illinois Department of Public Health "staff receive many calls about broken thermometers. After a broken thermometer incident, IDPH, using a mercury vapor analyzer, has never found airborne Hg0 concentrations > 1 μg/m3."
ATSDR gives 1 μg/m3 as the target cleanup level for residential environments, and 0.2 μg/m3 as the chronic level of exposure for which adverse effects would not be expected.

I found this statement in a paper about $\ce{Hg}$ contamination from dental work:

The evaporation rate of elemental mercury at room temperature ($\pu{20 ^\circ C}$) is approximately $\pu{50 \mu g\:cm-2h-1}$ (range of $\pu{40-60 \mu g\:cm-2h-1}$).

They cited the following work as a source for that number:

Gary N. Bigham, Wanyu R. Chan, Manuel Dekermenjian, Ali Reza, "Indoor Concentrations of $\ce{Hg}$ Vapor Following Various Spill Scenarios," Environmental Forensics 2008, 9(2-3), 187-196 (https://doi.org/10.1080/15275920802121975)

German regulations (as an example) say that 0.035 µg/l of metallic mercury vapour in air pose no risk under lifelong exposition, and 0.35µg/l can be harmful under long (months?) exposition for some people. The math is full of imponderables, esp. the actual amount that was spilled and size of drops.
Swiss officials https://www.toxinfo.ch/quecksilberhaltige-fieberthermometer say that a single broken fever thermometer can lead to intoxications, but the risk can usually be averted by collecting as much as possible and ensuring proper, regular ventilation.
The following publication suggests that a properly cleaned up spill from a fever thermometer should become undetectable within several weeks, but the airborne Hg concentration can be significant in that time, and regular ventilation is highly recommendable.

E. Martin Caravati et al. Elemental mercury exposure: An evidence-based consensus guideline for out-of-hospital management, Clinical Toxicology 2008, 46:1, 1-21, https://doi.org/10.1080/15563650701664731

But one thing is quite clear: A single, large blob that you miss (under furniture, wooden floorboard etc.) can stay for a very long time, and lead to potentially harmful concentrations in the air. There are a lot of reports about little mercury puddles that were found under floorboards during laboratory renovations, even decades after the last mercury instrument had been thrown out.

I have sure the tiny drops that I have mentioned weren't bigger than 1mm.

Well, are they gone by now?

Actually, no. I just found a couple of tiny drops near to my desktop (which is not the area that I broke the thermometer). They were about 1mm each.

@Karl your calculations are way, way out. I reckon that drop would take years to evaporate and this corresponds to my experience of mercury in a lab. It is likely that lots of small particles will persist and lead to dangerous levels of mercury vapour in a domestic environment (if you ever vacuum the carpet you will make this much worse). Even in labs with good ventilation we take major precautions to clean up mercury spills by, for example, sprinkling everything with flowers of sulfur).

@matt_black Ups. I knew that massive drops (hidden under ancient wooden laboratory floors) basically never vanish. Found the error, I missed a factor 4 for the surface, but still I come to ~400h for a 0.1g drop. Let´s say it touches the floor with half it´s surface, so 800h, for a drop of 2.4mm diameter (which I assumed doesn´t shrink until it´s gone, but that doesn´t change much). Is the value I cited unreasonable?

@karl i think your calculations are out by a factor of at least 1,000 (did you confuse micro-grams with milli-grams?). I got something like 50,000hr for that drop to evaporate.

@matt_black More like a factor hundred, but yea, shit. A ball of 0.1g Hg has a surface of 0.18 cm^2. That leads to an initial loss rate of 9µg/h, 11000h, or a year. German regulations require immediate action for levels of more than 0.35µg/L in the air, 0.1g that evaporate within a year would get you in that region, approximately. So I would say no reason to panic, but only just not. The question is how much Hg was lost, actually.

@Karl Apparently a typical thermometer has several grams.

@matt_black Well, a 40cm long lab thermometer with a range of 200K, yes, a common fever thermometer should contain much less.

I have read the most high amount of mercury that a fever thermometer has is 3g, and the most are in 0.7g and 1.5g.

As the comments in the other post say, there is no need to worry. You'll probably get more mercury and in a more dangerous form by eating fish and seafood and even more if you should smoke. To stay on the safe side, open the windows in your house from time to time, as you probably are doing anyway.

Totally disagree. Small mercury droplets dispersed widely indoors will lead to worrying levels of mercury vapour in the air and this will persist for years and will be worse if you ever vacuum as this disperses the particles into smaller droplets and puts them in a warm environment. Gas utilities (whose domestic meters used to use mercury in valves which occasionally spilled) have been prosecuted for this.

Please see the official guidelines by NHS (UK) https://www.nhs.uk/common-health-questions/accidents-first-aid-and-treatments/can-a-broken-thermometer-or-light-bulb-cause-mercury-poisoning/

@imalipusram Those guidelines cover both thermometers and mercury-containing CFL bulbs or fluorescent tubes. But bulbs don't contain enough mercury to cause long term contamination. Thermometers do. And their advice on broken thermometers isn't not to worry: it is to contact experts who know how to decontaminate things like carpet.

I am a chemist and have worked in several labs in which the balls of elemental mercury in the corners on the floors, and in the corners on the benchtops had been there for 50 years or more, perhaps since those buildings were erected. These were the reports of longstanding faculty who indicated the balls had been there when they arrived. If a ball was in our way, we simply rolled it away. The balls never got smaller and no one ever got sick. The vapor pressure of mercury appears to be very low, either that, or something happens on the surfaces that are exposed to air to prevent evaporation--again, those balls never evaporated or, according to my eyes, got smaller.
In the 1800's, and before, people ate mercury as a curative for constipation and other problems. From what I've read, it came out the other end in the same form that it went in the mouth. Digestively speaking, for humans the dangerous forms of mercury are its salt forms; these precipitate our proteins. It seems when fish and shellfish ingest mercury it gets converted to a form that is dangerous to humans--which is probably an ionic/salt form. I can understand the general public's fear of elemental mercury because of the way the story is being told, but I do not understand the scientific community promoting this story.

I'm sorry, but this is an irrelevant anecdote that doesn't address the big problem. Just because the visible drops of mercury don't visibly shrink does not imply that dangerous levels of mercury vapour are not present. Plenty of lab workers have been harmed by vapour inhalation over long periods because mercury spills were not cleared. The quality of ventilation also matters so perhaps you worked in a drafty lab where vapour never built up, but you can't rely on that. Mercury contamination is a big danger.

Yea, labs have fume hoods with flow rates of hundreds of cubic meters per hour.

authors: user1205901 - Слава Україні

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


text:
Why is energy released when lattice formation takes place?
Whenever an ionic bond formation takes place it does so by forming a lattice and a subsequent amount of energy is released. What are the cause of release of energy and where does this energy come from and what is its form?
I am really confused. Any help is appreciated.

See Born–Landé equation and the Madelung constant


text:
Reaction of vinylacetylene (but-1-en-3-yne) with HCl
Vinylacetylene contains both a double bond and a triple bond. We know that alkenes are more reactive to electrophillic addition reactions, so I thought that the $\ce{H+}$ from $\ce{HCl}$ would get abstracted via the double bond.
However, the final answer says the product is 2-chloroprene, which is possible only if the triple bond abstracts it.
My teacher said that it was because of the carbocation stability and the abstraction of $\ce{H+}$ by triple bond forms a stabler carbocation and leads to conjugation and hence is favoured.
But then Wikipedia says that a 1,4 addition takes place and an allene rearrangement takes place.This means that the triple bond is the one that reacts first , but in a 1,4 fashion to form an allene which then rearranges. I was unable to find a mechanism for this allene rearrangement, why is this 1,4 addition favoured, and why it takes place. Also, what is the correct explanation: the first one or the second one?

The allene rearrangement is catalysed by CuCl

@Waylander. Yes but  how does the rearrangement happen? Also is the explanation given by my teacher correct? He does not talk about this rearrangement at all.

Beyond a shadow of a doubt, the second explanation, involving an allene rearrangement is the correct one. As you said, alkenes are more reactive towards electrophilic addition and hence the addition of HCl happens first on the double bond to give 3-chloro-but-1-yne. What follows next is an intramolecular rearrangement, through a cyclic five-membered transition state, which is catalysed by CuCl with NH4Cl. It encompasses the abstraction of a hydrogen from the C4, the migration of the chloride from C3 to C2, following the formation of a double bond at C3 and C4. The product is referred to as chloroprene, which gives neoprene on polymerisation, a useful polymer. However the other explanation offered by your facilitator is used by many as a trick to remember the final product, (including, me!). Hope that helps!

Please consider changing the image to a better readable one.

I can't find a source. I was taught this by my O-chem teacher, who always looks through 4-5 books, before a lesson, which is why I have reason to believe it is correct. However feel free to scan through pdfs of Clayden, Paula Bruce and Nick Greeves.

I wonder why triple bond was not attacked by proton but the double bond. can you please clarify my doubt?

@Swastik Double bonds are more reactive to electrophiles compared to triple bonds as the pi electrons are more tightly held (due to sp charecter) in triple bonds than double bonds Also an unstable vinyl cation is formed.


text:
Production of hypoclorous acid by salt water electrolysis: requesting tips
I wish to use commonly available perfume ultrasonic nebulizers/diffusers to "fog" my working environment with hypoclorous acid, to obtain full coverage (ish) disinfection, albeit slowly.
I also wish to avoid falling victim to the price gauging of everything disinfectant-related during this particular period.
I did my research, and I noticed the most accessible method of HOCl synthesis is by eletrolyzing an ideally saturated solution of NaCl.
I am no chemist, but my work is health related, and I am a DIY enthusiast.
These are the info and tips I got from my research:

First and foremost, avoid stainless steel electrodes (or rather, anode) as the release of hexavalent chromium with fxck you up
You might use two different cups where the same saturated salt solution is poured, and obtain electrical continuity by using an "electrode bridge" between the two, so there is a better separation (albeit non complete) of chlorine and NaOH production
by this reaction, other than HOCl, you will also obtain unaltered gaseous chlorine (which gives the characteristic chlorine smell when standing close to the vat), and chloridric acid (HCl); to reduce the amount of chrloridric acid produced in the reaction you can add "some" acid into the solution, usually vinegar is suggested, but I suppose lemon juice (to give a better smell) and citric acid (to give no smell at all) could work as well
graphite anodes are the cheapest among the "cleanest" anodes (including here gold and platinum), which will release the least "pollutants" in the solution, even if they do release CO, CO2, and flakes of graphite; I experimented that you can use reclaimed old batteries as graphite electrodes, but since they are tiny, they will fleck off really fast
I noticed that if you use common iron electrodes, the solution will be heavily polluted with rust particles (FeO) and coloured with a strong orange, but if left to rest for the night, there will be a very good sedimentation process, and the morning after you will be left with a clear liquid in the vat, and an orange deposit on the bottom, which makes for way faster separation
Using a bench power supply, I can get it to the max 5A output quite easily by increasing the output voltage (so inputting at least 50W worth of power into the electrolysis reaction), since I use large iron bars that offer larger contact surface with the solution, and after a short while I notice that the unsoluted NaCl resting on the bottom of the vat will start dissolving
I tested dropping a small quantity of hydrogen peroxide in the obtained solution, and no foaming occurs, which I would expect to happen with NaOCl

Now come the questions:

Will having the FeO polluted solution rest overnight to sediment, alter the content, as in reduce its HOCl concentration? Is it light-unstable, should I cover it or place it in a dark container?
Is there an easy method to determine the HOCl "concentration"? Using quotation marks here, as I do not expect to have precise values, just a layman idea of "how much HOCl really is in there"
Should I be worried about the presence of HCl in the solution when I place it in my ultrasonic fogger? Also, if I do not separate the vats of anode and catode, hence obtaining a solution containing both HOCl and NaOH, should I be worried about fogging? I am talking about a populated rooms, not a vacated one.
Will pumping 50W+ in the vat create the conditions for a dangerous outcome, or is the heating up of the solution the only consequence?
If I keep the current flowing in the solution even after all the dissolved salt has been electrolyzed, will I mess up anything? I notice that the bubbling at the electrodes goes on for a while even after all the unsoluted salt on the bottom of the vat dissolves, and I cannot really hang out in front of the vat for the whole time to check on it, so I'd rather leave it on and fuggedaboudit until some kind of alarm I set sounds.
Are the any inaccuracies in the premises I made above, before my questions?

Nothing else comes to mind for now, thank you for your attention!

I understand that being a newcomer I might have failed observing some kind of "rule", yet I tried to make this question the most complete and useful possible by adding all information I had (researched)... so I would at least expect a comment explaining why you downvoted it.

I would fear to breathe a strong oxydizing fog more than I fear coronavirus. Neither will do any good to your lungs. Stick to washing your hands and using a mask. Inhaling hypoclorous acid would probably kill you or get you severely ill faster than any virus.

every source I found for HOCl safety states that it's breathable, and I think everyone breathes it when under a swimming pool dome, also the compound is naturally produced by animals as a defense mechanism. I am not worried about inhaling a HOCl fog itself, more about inhaling something else that might be in the solution due to improper synthesis methods :)

Cool idea employing the ultrasonic nebulizers/diffusers.
However, as I have elsewhere cited this work, 'Generation of Hydroxyl Radicals from Dissolved Transition Metals in Surrogate Lung Fluid Solutions' by Edgar Vidrio relating to lung cancer likely induced from dust particles (containing transition metals) interact with O2 leading to reactive oxygen species and powerful/problematic hydroxyl radicals, be aware that what you are working with (HOCl versus H2O2) is as (or more) problematical from a health perspective, so good ventilation with a gas mask is strongly recommended if you elect to proceed.
My thoughts, you do not have to worry about the HOCl concentration (which can be dilute when sourced from electrolysis) if you managed to imbed say a chemical/electrochemical hydroxyl radical presence (which can be tested by its relative stain removing ability).
A suggested path would be to prepare a HOCl/NaOCl mix from the action of CO2 or NaHCO3 on aqueous NaOCl (also available as Chlorine Bleach). This could be converted into a mist/fog, forming a de facto so-called 'Bleach battery' droplet. This is a variation of the 'Bleach Battery' based on HOCl in place of NaOCl, fine carbon soot for the noble cathodic zone (normally copper metal), and for the anode, employ metal particles for the anodic zone (like Al, Zn, Mg, Fe,...).
The battery cell creates electrons and some solvated electrons. In the presence of air (more precisely, oxygen interacting with the mist), I would expect some superoxide radical formation:
$\ce{ O2 + e- (aq) <=> .O2- (aq) }$
which could further react with the hypochlorous acid forming hydroxyl radicals:
$\ce{ .O2- (aq) + HOCl (aq) <=> O2 (g) + .OH (aq) + Cl- (aq)}$ Reference here.
Note: The above reaction could already be occurring in your NaCl/H2O electrolysis cell, if accompanied by O2 creation (or existing dissolved oxygen in the water that could be removed by boiling). While this would be disinfecting in the electrolysis cell itself, there would likely be little HOCl left to extract.
Also, in presence of HCO3-, the longer existing and more selective carbon radical anion may be formed:
$\ce{ HCO3- (aq) + .OH (aq) <=> H2O + .CO3- (aq) }$
In addition, the interaction of a transition metal particle with HOCl will also directly form the hydroxyl radical at pH > 5 (hence the use of an alkaline HOCl/NaOCl mix which also lowers the risk of chlorine gas formation). Then, per this reference, "Fenton chemistry in biology and medicine" by Josef Prousek, to quote reaction (15) on page 2330, a general depiction of Fenton-type reactions, to quote:
"For Fe(II) and Cu(I), this situation can be generally depicted as follows [20,39],
$\ce{ Fe(II)/Cu(I) + HOX → Fe(III)/Cu(II) + ·OH + X- }$ (15)
where X = Cl, ONO, and SCN. "
My last comment, this disinfecting process results in a deposit of NaCl/Carbon, and you should avoid all aluminum alloys which will be disfigured by HOCl, and, of course, don't forget the safety precautions. Do not introduce any organics as there is the possibility of chloro-organic creation.

You completely lost me after the second paragraph; I do not have one hundredth of your know-how, so what I took from your answer (which was still very useful) is: "use safety precautions while you electrolyze salt water to obtain HOCl; do not worry about final concentration, you will get a solution powerful enough". After that, my comprehension was blurred.
I just tried testing the stain removal ability with a black marker on plexiglas; my "HOCl solution" removed marker better after some seconds soaking and then frictioning, than what plain salted water did. I'd upvote you if I could!

Ephestione's procedure is not clear.

He wants to produce hypochlorous acid $\ce{HOCl}$ by electrolysis of $\ce{NaCl}$. Unfortunately $\ce{HOCl}$ cannot be obtained directly by $\ce{NaCl}$ electrolysis. The  $\ce{NaCl}$ electrolysis produces $\ce{NaOH + H_2}$ at the cathode, and $\ce{Cl_2}$ at the anode (+$\ce{O_2}$ as an impurity). So $\ce{HOCl}$ is not produced here.  Of course a small proportion of $\ce{Cl_2}$ reacts with water to produce $\ce{HClO}$ (and HCl) according to : $$\ce{Cl_2 + H_2O <=> HOCl + H^+ + Cl^-}$$ But a high $\ce{HOCl}$ concentration cannot be obtained this way, because the equilibrium is strongly displaced to the left hand side. If more $\ce{HOCl}$ is to be obtained, some $\ce{NaOH}$ must be added, so as to destroy $\ce{H^+}$, but it will also destroy $\ce{HOCl}$.
He wants to separate the anode and the anode compartments, using what he called an "electrode bridge". Never heard about such a thing ! Well. With usual wet junctions made with agar + $\ce{KNO_3}$ solution, the currant going through the cell is rather low. The yield will be weak for reasonable amounts of time, like one or two days, because of the high resistance of the junction, which would limit the currant drastically.  Anyway it does not make sense to separate the two compartments. Using a unique solution with two electrodes, the Chlorine gas produced at the anode would soon react with the NaOH produced at the cathode, yielding a significant amount of $\ce{ClO^-}$ ions by the equation $$\ce{Cl_2 + 2 OH^- ->  ClO^- + Cl^- + H_2O}$$ With this unique solution, a high yield of $\ce{ClO^-}$ ions is produced, which may be acidified to produce significant amounts of $\ce{HClO}$.

Final remarks :
1) if  Ephestione wants to obtain a significant amount of $\ce{HClO}$ in solution, why not use common bleach, which a solution of $\ce{NaClO + NaCl}$, and add some acidic solution ? It would be quicker, safer, without any problems of Iron Oxides.
2) $\ce{HCl}$ is called "hydrochloric acid" and not "chloridric acid".
 "Chloridric" has no meaning. This reminds me of the word "acide chlorhydrique" which is the French translation of "hydrochloric acid".
3) Exposing hypochlorites and $\ce{HClO}$ to air for more than one day, would destroy  them because of the action of atmospheric $\ce{CO_2}$ that reacts according to :
$$\ce{OCl^- + Cl^- + H_2O + 2 CO_2  ->  Cl_2 + 2 HCO_3^-}$$
3) Rust is not $\ce{FeO}$ but $\ce{Fe_2O_3}$ or a a hydrated derivate of $\ce{Fe_2O_3}$

Thanks!

I didn't find it documented anywhere! So I use commercial bleach, add NaCl, and some vinegar, and I obtain HOCl without elecrolysis?

I realized it after the editing time window was already closed; I'm italian, and we say "acido cloridrico", much like french, that's why

So I need to place the HOCl solution in a sealed container far from sunlight to store it (after having placed it in another sealed container for sedimentation if I elecrolyzed it)

Thank you for correcting, I guessed my best and googled the rest, last time I properly studied chemistry was 25 yrs ago

@Ephestione. No ! Don't add NaCl to Bleach. Use it as such ! Bleach is the result of electrolysis of NaCl solutions with big electrodes which are nearly touching one another. So there is plenty of NaCl in the bleach. If you want it, you may add some acidic solution like vinegar.

Per a source, to quote: "AQUAOX™ generated HOCL is pH neutral, super-oxidized water generated by electrolysis of a dilute salt solution passing through an electrolytic cell " at https://www.aquaox.com/technology/ . So, it is possible to make dilute HOCl from the electrolysis of salt water.

The main reseller of HOCl ready solution here in Italy produces it with electrolysis (and sells it for 140€ per 20L...) According to their website, their HOCl solution contains 0.05% of hypoclorous acid in 99.95% of electrolyzed water. Also, electrolysis is the "most affordable" method for a DIYer to produce HOCl that I found by researching online, I never stumbled upon "add acid to bleach" method and that's why I am confused, unless you meant to electrolyze bleach+acid as well. My aim is to obtain a solution safe to nebulize in occupied rooms.


text:
Finding new pH after adding potassium hydroxide to citrate buffer

Calculate the $\mathrm{pH}$ change of citrate buffer containing $\pu{10.26 mmol}$ of citric acid and $\pu{6.98 mmol}$ citrate caused by adding $\pu{4.72 mmol}$ of potassium hydroxide. $\mathrm{p}K_\mathrm{a} = 3.06.$

I'm trying to calculate it like initial $\mathrm{pH}$
$$\mathrm{pH_0} = 3.06 - \lg\frac{\pu{10.26 mmol}}{\pu{6.98 mmol}} = 2.89,$$
then if we add $\pu{4.72 mmol}$ $\ce{KOH},$ it's $1:3$ to citric acid. Final $\mathrm{pH}$ will be
$$\mathrm{pH_1} = 3.06 - \lg\frac{\pu{10.26 mmol} - \pu{1.57 mmol}}{\pu{6.98 mmol} + \pu{1.57 mmol}} = 3.05,$$
but this answer is incorrect. Where is the mistake?

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Never omit units in your calculations. Note that the accepted decimal separator in English used for scientific publishing is a point, not a comma; also, "acid" is denoted with a lowercase letter "a", not "k" (e.g. used in Russian language).

What's the answer given?

the answer is: 0.492

can you share your calculations?

$\mathrm{p}K_\mathrm{a} = 3.06.$

Citric acid is a triprotic acid:
$$\ce{H3C6H5O7 <=>[K_1] H2C6H5O7- <=>[K_2] HC6H5O7^2- <=>[K_3] C6H5O7^3-}$$
The pKa given is for the second step. Citric acid is the fully protonated species, and citrate the fully deprotonated one. When you add the two, they neutralize to give the intermediate species.

I'm trying to calculate it like initial $\mathrm{pH}$

The easiest way is to calculate the amount of protons available (3 x 10.26 mmol = 30.78 mmol) and the total amount of organic species (17.24 mmol). There would be no citrate (17.24 mmol of protons to protonate all with one proton, 13.54 mmol protons left), and no citric acid. 17.24 mmol - 13.54 mmol = 3.70 mmol would be monoprotonated and 13.54 mmol would be diprotonated. You can plug those numbers in (assuming that your volume is not crazy high or low) to get the initial pH.

then if we add $\pu{4.72 mmol}$ $\ce{KOH},$ it's $1:3$ to citric acid.

No, because citrate is not a major species before or after adding the KOH. Instead, you would deprotonate 4.72 mmol of the diprotonated species to get 4.72 mmol more of the monoprotonated species.
$$\mathrm{pH_1} = 3.06 - \lg\frac{\pu{13.54 mmol} - \pu{4.72 mmol}}{\pu{3.70 mmol} + \pu{4.72 mmol}} = 3.04$$

but this answer is incorrect. Where is the mistake?

I think both answers are wrong, for the reasons given above.


text:
How can fluorite structure exist?
From the information from Google the radius of $\ce{Ca^2+}$ cation is $100 \,\pu{pm}$ and of $\ce{F-}$ anion is $132\, \pu{pm}$, so essentially $\ce{F-}$ ion should form the Bravais lattice (FCC/CCP) as it is larger in size and $\ce{Ca^2+}$ should go to the tetrahedral voids but in reality it is just opposite.
How can smaller sized calcium cation form the Bravais lattice?
(The data has been taken form first results form Google search)
Please do correct me if I am conceptually wrong somewhere.

Smaller ion does not always go to interstices, as smaller ion sometimes tend to increase stability of lattice, also in case of CaF2, it's not uncommon to get complex isometric forms. Crystal twinning is rather much more common...

The "lattice" atoms do not have to be the largest.  They only have to be large enough to fit the "interstitial" atoms around them, and when the coordination number is less than 12 (in this case it's 8) this can be done with the "lattice" atoms slightly smaller.


text:
The source for the N-N bond dissociation energy of 240 kJ/mol in the table
Authoring an exam question, I made the mistake of taking a bond dissociation energy from a table I found on the internet. It comes in different styles, but is always labeled Table 7.1. It gives a bond dissociation energy of $\pu{240 kJ/mol}$ for the - admittedly uncommon - nitrogen:nitrogen single bond. A student was puzzled that the enthalpy of formation for hydrazine, estimated based on this value, was so far off from the experimental value; the student then found a value for the bond dissociation energy of $\pu{159 kJ/mol}$ in a more rigorous search.
The General Chemistry textbooks available to me from home have values close to $\pu{159 kJ/mol}$ (OpenStax Chemistry: $\pu{160 kJ/mol}$; Kotz 6th ed (2006): $\pu{163 kJ/mol}$; Radel (1994): $\pu{170 kJ/mol}$). I am not worried about the small differences - it depends which set of molecules containing $\ce{N-N}$ bonds was used, and a substructure search of commercially available compounds shows that there is quite a variety, with many having partial double-bond character (the following sample are 3 from the 100 or so just from Sigma Aldrich):

My question is: Where did Table 7.1 originate, and where did the value for $\ce{N-N}$ originate?
Here is a selection of links to variations of table 7.1, with the first example as picture:

https://wps.prenhall.com/wps/media/objects/4678/4790506/ch07_02.htm
https://www.chegg.com/homework-help/questions-and-answers/use-bond-dissociation-energies-table-calculate-approximate-h-kilojoules-reaction-acethylen-q27780023
https://www.chegg.com/homework-help/questions-and-answers/average-bond-dissociation-energies-d-kj-mol-h-h-4362-h-c-410-h-f-570a-h-ci-432a-h-br-3662--q36015212
https://cdn.clutchprep.com/problem_images/65691-36fd337a4e3e837c.png

The table is from McMurry's Chemistry textbook. I cannot tell the exact edition; for instance, your table appears as Table 7.2 in 7th edition from 2015 (ISBN 978-0-321-94317-0), also listing $D(\ce{N-N}) = \pu{240 kJ/mol}.$

@andselisk Strangely, the 8th edition has a value of 140 kJ/mol. (overshooting in the other direction to compensate, or fixing a single-digit typo?).

Strange indeed. For the record, Table 7.1 in 6th edition from 2012 also has $D(\ce{N-N}) = \pu{240 kJ/mol}.$

Your question about the source of Table 7.1 is correctly answered by the andselisk's comment. The value of $\pu{240 kJ\:mol-1}$ for $\ce{N-N}$ bond is probably initiated from Encyclopedia of Inorganic Chemistry, which has listed bond energies of wide variety of bonds and their respective bond lengths. It has given two values for $D_\circ$ of $\ce{N-N}$ bond: ~$\pu{167 kJ\:mol-1}$ for $\ce{N2H4}$ in general and $\pu{247\pm 13 kJ\:mol-1}$ for $\ce{H2N-NH2}$ in particular.
The reason is given in the very first paragraph of the article. In their words:

There are two different ways to define the bond energy even for the simplest diatomic molecule (see the diagram below). $\Delta D_{\circ'}$ is the dissociation energy measured from the very bottom of the potential energy well. However, the molecule possesses zero-point energy and thus, the experimentally measured dissociation energy $\Delta D_{\circ}$ is somewhat less:

In poly atomic compound there are other considerations. For example, the mean bond energy of $\ce{N-H}$ is the average of three bond energies associated with sequential fission of the three $\ce{N-H}$ bonds in ammonia ($\ce{NH3}$). These three values are different from each other and also different from their average, the mean bond energy of $\ce{N-H}$. Using this mean bond energy of $\ce{N-H}$ in $\ce{NH3}$ and the heat of atomization of $\ce{H2N-NH2}$, a value of $\pu{159 kJ\:mol-1}$ has been found for the $\ce{N-N}$ single bond energy. Then again, using the mean bond energy of $\ce{N-F}$ in $\ce{NF3}$ and the heat of atomization of $\ce{F2N-NF2}$, a value of $\pu{172 kJ\:mol-1}$ has been found for the $\ce{N-N}$ single bond energy. These two values are in somewhat good agreement. However, the dissociation of the two molecules are quite different:
$$\ce{H2N-NH2 -> 2 H2N^.} \qquad \pu{247 kJ\:mol-1}$$
$$\ce{F2N-NF2 -> 2 F2N^.} \qquad \pu{88 kJ\:mol-1}$$
Therefore, it is warned that care should be exercised in using any values of the bond energies.
Source:
R. Bruce King (Editor in chief), "Bond Energies," Encyclopedia of Inorganic Chemistry; 2nd Edition, John Wiley & Sons, Ltd.: New York, NY, 2005 (ISBN: 978-0-470-86078-6). Bond Energies is available online: https://doi.org/10.1002/0470862106.id098

So once the N-N bond in hydrazine is broken, the N-H bonds are much weaker than they would be in ammonia? I like talking about bond dissociation energies in a General Chemistry setting, but always say it is a crude estimate. Even so, I think it is helpful in e.g. understanding how elemental oxygen is a reagent in many exothermic reactions while elemental nitrogen is not (at least not with the reaction partners that are sitting out on the shelf - I guess if they reacted with elemental nitrogen, they would not be sitting there anymore).


text:
Calculation of Degree of Dissociation from Ostwald's Dilution Law
A "weak" electrolyte, $\ce{A+B-}$, ionizes in solution as:
$$\ce{AB <=> A+ + B-}\tag{1}$$
$$  K_d =\dfrac{\ce{[A+][B-]}}{\ce{[AB]}}=\dfrac{(\alpha c_{0})(\alpha c_{0})}{(1-\alpha )c_{0}}=\dfrac{\alpha ^{2}}{1-\alpha }\cdot c_{0}\tag{2}$$
For a weak electrolyte the Degree of Dissociation, $\alpha$, is less than 1 and can be calculated using two different equations.

Equation (2) can be rearranged to a quadratic equation which could be solved exactly with the standard quadratic formula.

$$ c_0 \cdot \alpha^2  + K_d\cdot \alpha -K_d = 0\tag{3}$$
$$\alpha = \dfrac{-K_d + \sqrt{K_d^2 - 4\cdot c_0 \cdot (-K_d) }}{2\cdot c_0} = \dfrac{-K_d + \sqrt{K_d^2 + 4\cdot c_0 \cdot K_d }}{2\cdot c_0}\tag{4}$$

If we assume that there is little dissociation and that $\ce{[AB] \approx c_0}$ then $1- \alpha \approx 1$ and we obtain an approximate answer that is easier to calcualte:

$$  \alpha' \approx \sqrt{\dfrac{K_d}{c_{0}}}\tag{5}$$
If equation (5) does not give the true value of the Degree of Dissociation. How do you know when it is good enough?
To make it clearer, say the approximate value from formula (5) is 0.15. That is not the true value of the degree of dissociation, but would the approximate solution be good enough?

I just edited the question to pinpoint what I think Limbo Productions was trying to ask. // User Limbo Productions if I missed interpreted your question please either edit the current version, or revert back to your original question.

@MaxW The flaw with the question is that it doesn't state what is known beforehand, or what would be regarded as a tolerable error. You provide a nice explanation of how the approximate $\alpha$ provides an upper bound on the true $\alpha$, and that is certainly useful. But otherwise this question seems pointless. If you know the values of $K_d$ and $c_0$ and of the tolerated error, then it is trivially found if the approximation is good enough. Maybe MFarooq is onto something regarding the point of the question, but we'll probably never know (and honestly I don't understand his explanation).

@BuckThorn - I took the OP's question to mean "in general". So for whatever problem comes along The tolerable error and $K_d/c_0$ would be for a particular known. Then for that particular problem, how is the OP to decide if the approximation is good enough? // My take away from MFarooq's point was the total effort to (1) make the approximate calculation and (2) to do a check is almost as great as the effort to solve the quadratic which gives the exact solution. So why bother with the approximation? Just solve the quadratic and be done with it.

@MaxW If that was MFarooq's point I completely agree.

The OP asked a simple question to which I gave a complex answer. My daughter called such answers "Dad answers".
The OP essentially asked "If the dilute solution approximation for $\alpha$ is used when calculating, how do you know how good the approximation is?"
In his answer user porphyrin focused on the pertinent idea.
Dilute Solution Approximation
Now if we assume that $(1-\alpha) \approx 1$ meaning that very little of the compound dissociated, then we get the standard approximation, $\alpha'$:
$$  \alpha^{'} \approx \sqrt{\dfrac{K_d}{c_{0}}}\tag{1}$$
Exact Solution
The equation below is an exact solution even though it doesn't allow $\alpha$ to be calculated directly. You would have to guess at $\alpha$ values and iterate until you found a solution.
$$\dfrac{\alpha}{\sqrt{1 - \alpha}} = \sqrt{\dfrac{K_d}{c_{0}}}\tag{2}$$
Estimated % Error
Now let's calculate an estimated % error as:
$$\pu{Est \% Error} = 100 \times \dfrac{\dfrac{\alpha'}{\sqrt{1 - \alpha'}} - \alpha'}{\alpha'}\tag{3}$$
Quadratic Solution
The exact value of $\alpha$ can be calculated directly using a quadratic equation. The solution is given below.
$$\alpha = - \dfrac{K_d}{2\cdot c_0} + \sqrt{\left(\dfrac{K_d}{2\cdot c_0}\right)^2 + \dfrac{K_d}{c_0}}\tag{4}$$
Data Table
\begin{array}{|c|c|c|c|c|} \hline
(\alpha^{'})^2=K_d/c_0 & \alpha (quadratic) & approx & true\ \% \ error & est.\ \% \ error \\ \hline
1.00000000 & 0.61803399 & 1.00000000 & 61.80339887 &  \\ \hline
0.81000000 & 0.58192705 & 0.90000000 & 54.65856100 & 216.22776602 \\ \hline
0.64000000 & 0.54162637 & 0.80000000 & 47.70329614 & 123.60679775 \\ \hline
0.49000000 & 0.49663670 & 0.70000000 & 40.94810050 & 82.57418584 \\ \hline
0.36000000 & 0.44641839 & 0.60000000 & 34.40306509 & 58.11388301 \\ \hline
0.25000000 & 0.39038820 & 0.50000000 & 28.07764064 & 41.42135624 \\ \hline
0.16000000 & 0.32792156 & 0.40000000 & 21.98039027 & 29.09944487 \\ \hline
0.09000000 & 0.25835623 & 0.30000000 & 16.11874208 & 19.52286093 \\ \hline
0.04000000 & 0.18099751 & 0.20000000 & 10.49875621 & 11.80339887 \\ \hline
0.01000000 & 0.09512492 & 0.10000000 & 5.12492197 & 5.40925534 \\ \hline
0.00810000 & 0.08604108 & 0.09000000 & 4.60119879 & 4.82848367 \\ \hline
0.00640000 & 0.07686397 & 0.08000000 & 4.07996803 & 4.25720703 \\ \hline
0.00490000 & 0.06759286 & 0.07000000 & 3.56123125 & 3.69516947 \\ \hline
0.00360000 & 0.05822699 & 0.06000000 & 3.04498988 & 3.14212463 \\ \hline
0.00250000 & 0.04876562 & 0.05000000 & 2.53124512 & 2.59783521 \\ \hline
0.00160000 & 0.03920800 & 0.04000000 & 2.01999800 & 2.06207262 \\ \hline
0.00090000 & 0.02955337 & 0.03000000 & 1.51124937 & 1.53461651 \\ \hline
0.00040000 & 0.01980100 & 0.02000000 & 1.00499988 & 1.01525446 \\ \hline
0.00010000 & 0.00995012 & 0.01000000 & 0.50124999 & 0.50378153 \\ \hline
0.000001 & 0.0009995 & 0.00100000 & 0.05001250 & 0.05003753 \\ \hline
0.00000001 & 9.9995E-05 & 0.00010000 & 0.00500012 & 0.00500038 \\ \hline
1E-10 & 9.99995E-06 & 0.00001000 & 0.00050000 & 0.00050000 \\ \hline
\end{array}
So the Est % Error from equation (3) works quite well.

When the Est % Error is 5% or less, $\alpha' \le 0.1$, the Est % Error is quite good.
So for the approximate equation when $\alpha^{'} = 0.1$ you get 1 significant figure, 0.01 gets two significant figures and so on. The improvement doesn't continue forever of course - this is chemistry not math. To get three significant figures or more you'd have to be a fastidious experimenter.
For the approximate equation when $\alpha' = 0.7$ the estimated error is 82.6% which is much too large, but an 82.6% error still tells you that the  approximation $\alpha^{'}$ is no good.
To make user M. Farooq's point in another way, by the time you calculate the approximation from equation (1) and the Est % Error from equation (3), you've pretty much done enough work to calculate the exact answer using equation (4). So why bother with the approximation?

EDIT A comment by user Buck Thorn "I never thought about the usefulness of the estimate to bracket the true value" made me think.
It turns out that we can bracket the true value $\alpha$ using a simple calculation of the approximate value $\alpha'$:
$$\alpha'\cdot(1-\alpha') \lt \alpha < \alpha'\tag{5}$$
\begin{array}{|c|c|c|c|} \hline
(\alpha^{'})^2=K_d/c_0 & \alpha (quadratic) & \alpha'\cdot(1-\alpha') & \alpha' \\ \hline
1.0000000000 & 0.6180339887 & 0.0000000000 & 1.0000000000 \\ \hline
0.8100000000 & 0.5819270490 & 0.0900000000 & 0.9000000000 \\ \hline
0.6400000000 & 0.5416263691 & 0.1600000000 & 0.8000000000 \\ \hline
0.4900000000 & 0.4966367035 & 0.2100000000 & 0.7000000000 \\ \hline
0.3600000000 & 0.4464183905 & 0.2400000000 & 0.6000000000 \\ \hline
0.2500000000 & 0.3903882032 & 0.2500000000 & 0.5000000000 \\ \hline
0.1600000000 & 0.3279215611 & 0.2400000000 & 0.4000000000 \\ \hline
0.0900000000 & 0.2583562262 & 0.2100000000 & 0.3000000000 \\ \hline
0.0400000000 & 0.1809975124 & 0.1600000000 & 0.2000000000 \\ \hline
0.0100000000 & 0.0951249220 & 0.0900000000 & 0.1000000000 \\ \hline
0.0081000000 & 0.0860410789 & 0.0819000000 & 0.0900000000 \\ \hline
0.0064000000 & 0.0768639744 & 0.0736000000 & 0.0800000000 \\ \hline
0.0049000000 & 0.0675928619 & 0.0651000000 & 0.0700000000 \\ \hline
0.0036000000 & 0.0582269939 & 0.0564000000 & 0.0600000000 \\ \hline
0.0025000000 & 0.0487656226 & 0.0475000000 & 0.0500000000 \\ \hline
0.0016000000 & 0.0392079992 & 0.0384000000 & 0.0400000000 \\ \hline
0.0009000000 & 0.0295533748 & 0.0291000000 & 0.0300000000 \\ \hline
0.0004000000 & 0.0198010000 & 0.0196000000 & 0.0200000000 \\ \hline
1.000000E-04 & 0.0099501250 & 0.0099000000 & 0.0100000000 \\ \hline
1.000000E-06 & 0.0009995001 & 0.0009990000 & 0.0010000000 \\ \hline
1.000000E-08 & 0.0000999950 & 0.0000999900 & 0.0001000000 \\ \hline
1.000000E-10 & 9.999950E-06 & 9.999900E-06 & 0.0000100000 \\ \hline
\end{array}

I was hoping for a more rigid explanation rooted in mathematics but I guess it has to come down to comparing values and noticing when they start to get reasonably close to each other. Less satisfying but I will take it. Thank you for your help I appreciate it.

@l1mbo - I'm befuddled as to what else you might want. It is easy to calculate the exact error. Let $\alpha$ be the solution from the quadratic solution, and $\alpha'$ be the solution from the approximation. Then: $$\pu{% Error} = 100\times (\alpha' -\alpha)/\alpha)$$ But obviously if you're going to calculate $\alpha$, which is exact, why bother calculating $\alpha'$?

[I know this is 1.5 yrs late, I just happened to be going through my old posts on stack exchange so I stumbled on this, so I'll leave this here for anyone referring to this in the future] what I wanted was, in hindsight, extremely simple. The exact expression for $\alpha = - \dfrac{K_d}{2\cdot c_0} + \sqrt{\left(\dfrac{K_d}{2\cdot c_0}\right)^2 + \dfrac{K_d}{c_0}}\tag{4}$ can be rewritten as function only of $\alpha'$ by replacing $\dfrac{K_d}{c_0}=\alpha'^2$.

Then, we can replace all $\alpha$ in the % error $= 100\times (\alpha' -\alpha)/\alpha)$ and thus % error can be written entirely as a function of $\alpha'$. Finally when we graph the error function we can see at what $\alpha'$ the error becomes greater than whatever value we see fit (5% error here) https://imgur.com/a/LsX7YlS This was 'more rigid' explanation I sought. (No idea why I struggled to figure out this rather elementary answer then)

As an alternative and simple estimation, assume $\alpha \lt\lt 1$ and expand $\displaystyle \frac{K_d}{c_0}=\frac{\alpha^2}{1-\alpha}$ to give $\displaystyle \frac{K_d}{c_0}=\alpha^2(1+\alpha+\alpha^2+\alpha^3+\cdots)$ and compare the result to your approximation by adding correction terms.

Hmmm... You could a search and iteratively solve for $\alpha$ with that equation. However in order to solve for $\alpha$ directly you need some function of $K_d/c_0$. It would seem that using the first three terms would give you some idea of the error.

dah... Don't need the series. Let $\alpha = \sqrt{K_d/c_0}$ and compare $\alpha$ and $\alpha /\sqrt{1-\alpha}$ to get some idea of what the error is. It won't be exact, but in the ballpark.

yep:  that was what I thought.

@porphyrin, The student is talking about the  "our formula derived with approximations will not give us the true value of the degree of dissociation" ($sic$), he is wondering about the validity of the formula itself, not its quadratic approximations. It is the definition of alpha that is the Achilles heel of this whole experiment in terms of conductance measurements. I know how this experiment is taught in South Asia so I can feel what he is trying to ask although quite vaguely. We used make the students solve the whole quadratic equation, which is not a big deal.

The question was about how you could tell if the dilute approximation was any good. If $\alpha = \sqrt{K_d/c_0}$ then comparing $\alpha$ and $\alpha/\sqrt{1-\alpha}$ gives a good indication. // Solving the quadratic will always give the exact answer of course.

@M.Farooq Just to make everything clear, I thought I should specify what exactly I was told. We derive the simpler formula for the approximate value of alpha which should be used when alpha<0.1. But when solving questions alpha is an unknown so we don't know whether it is less than 0.1. To check this we put the values of the constants in the simpler approxformula. If alpha is <0.1 from this then we continue with the same. Otherwise we go back and solve the full quadratic. As to why I am emphasizing on solving problems instead of experiments is because I am studying for an engineering entrance.

@l1mbo, I can understand your background for test preparation. If your sole question is related to solving conductivity problems using Ostwald's dilution law, you should solve the full quadratic equation. Do you think you will have time to do all the Taylor series approximations in MCQs? If your question is about solving the dissociation constant problems in general, then it is a separate question. Post a new question.

Further elaboration to the answer. I recall teaching this experiment more than a decade ago (slightly rusty now). I think the approximation was is not mathematical (A) or (B), as one has to solve the entire quadratic equation by all means.
$$ \alpha = \sqrt{\dfrac{K_d\cdot (1-\alpha )}{c}}\tag{A}$$
and
$$ \alpha^{'} \approx \sqrt{\dfrac{K_d}{c}}\tag{B}$$
As one can see in equation (A) and (B), K$_d$ has a very strong dependence on concentration of acetic acid taken for measurement. Certainly, during the experiment we cannot take an infinitely dilute weak acid solution, so we do the experiment with sufficiently dilute acid but of finite concentration. The fundamental or key issue is the way
$\alpha$ is defined. It is given by
$$ \alpha = \frac{\Lambda}{\Lambda_0}\tag{C}$$
where the numerator indicates the conductivity at concentration $c$ and the denominator shows the conductivity at zero concentration.
The accuracy of the dissociation constant is dependent on this measurement and the validity of the equation (C). This expression is not valid for strong electrolytes (hence strong bases and acids are not used for dissociation constant experiments)
So apparently, the student is confused why only weak electrolytes can be measured by conductivity: The reason is the validity of equation (C).
He also assumes that only conductivity experiment is used to determine the dissociation constant of an acid and base. Since we are determining Kd from a conductivity measurement, how do we know beforehand that the acid/base is weak. This circular argument is not true at all. The weakness of an acid or a base in water is easily determined by a simple pH measurement. For example, pH of 0.01 M acetic acid is higher than pH of 0.01 M HCl, which indicates that acetic acid is not fully dissociated. Conductivity is not the only way to determine the dissociation constants, it is one of the possible ways.

Granted other ways may exist, but that doesn't mean we do not determine dissociation solely on the basis discussed above.

Equation (B) is the typical shortcut to an answer assuming that there is is little disassociation. If $\alpha^{'} \le 0.1$ the answer will be within 5%, if  If $\alpha^{'} \le 0.010$ the answer will be within 1%. So you don't always have to solve the quadratic, though the quadratic will always give the exact answer.

" only weak acids or bases can be measured by conductivity measurements." Isn't the point that you don't have to worry (as much) about activity coefficients in weakly dissociating electrolytes? In any case, I find the last paragraph confusing.

I was referring to the the determination of dissociation constants by conductivity measurements. Only relatively weak acids or bases can be employed for such experiments. I have tried to clarify the the last paragraph.

I rearranged the answer and moved the equations around. However I kept the original numbering so that the comments make sense.

Consider a binary electrolyte AB which dissociates reversibly into A+ and B− ions with a nominal concentration of $c_0$. Ostwald noted that the law of mass action can be applied to such systems as dissociating electrolytes. The equilibrium state is represented by the equation:
$$\ce{AB <=> A+ + B^-}\tag{1}$$
If $\alpha$ is the fraction of dissociated electrolyte, then $\alpha \cdot c_0$ is the concentration of each ionic species. $(1-\alpha)$ must, therefore be the fraction of undissociated electrolyte, and $(1- \alpha )\cdot c_0$ the concentration of same. The dissociation constant may therefore be given as
$$  K_d =\dfrac{\ce{[A+][B-]}}{\ce{[AB]}}=\dfrac{(\alpha c_{0})(\alpha c_{0})}{(1-\alpha )c_{0}}=\dfrac{\alpha ^{2}}{1-\alpha }\cdot c_{0}\tag{2}$$
So far I've copied the Wikipedia article on the Law of dilution almost exactly. Now I'll depart that derivation.
From equation (2) let's start with:
  $$  K_d = \dfrac{\alpha ^{2}}{1-\alpha }\cdot c_{0}\tag{3}$$
Dilute Solution Approximation
now if we assume that $(1-\alpha) \approx 1$ meaning that very little of the compound dissociated, then we get the standard approximation, $\alpha'$:
$$  \alpha^{'} \approx \sqrt{\dfrac{K_d}{c_{0}}}\tag{6}$$
Exact Solution - Odd Form
Now let's play around with Equation (3) to form an exact solution in an odd sort of form.
$$  \alpha^2 = \dfrac{K_d\cdot (1-\alpha )}{c_{0}}\tag{4}$$
and solving for the square root:
$$  \alpha = \sqrt{\dfrac{K_d}{c_{0}}\cdot (1-\alpha )}\tag{5}$$
Now Equation (5) is in an odd form, but it is exact. Knowing $K_d/c_0$ one could guess at values of $\alpha$ until a value was found that would satisfy the equation.
Remember that we're looking for values of $K_c/c_0 << 1$. Comparing equations (6) and (5) it is obvious to the casual observer that equation (6)  must always give a value that is larger than equation (5). Hence if $ 1 \gg \alpha^{'}$ then $ 1 \gg \alpha^{'} \gt \alpha$
Exact Solution - Exact Quadratic Solution
Equation(5) is in the wrong form to directly solve for $\alpha$ of course. So let's go back to equation (3) and rearrange to a reduced quadratic equation which could be solved with the standard quadratic formula.
$$ \alpha^2 +\dfrac{K_d}{c_0}\cdot \alpha -\dfrac{K_d}{c_0}= 0\tag{7}$$
$$\alpha = - \dfrac{K_d}{2\cdot c_0} + \sqrt{\left(\dfrac{K_d}{2\cdot c_0}\right)^2 + \dfrac{K_d}{c_0}}\tag{8}$$
Data Table
Now for various $K_d/c_0$ values let's calculate the approximate solution, the exact solution, and the % Error.
\begin{array}{|c|c|c|c|} \hline
\dfrac{K_d}{c_0} & Approximation & Quadratic & \% Error \\
& Equation (6) & Equation (8) &  \\ \hline
1.00000000 & 1.00000000 & 0.618034 & 61.803399 \\ \hline
0.81000000 & 0.90000000 & 0.581927 & 54.658561 \\ \hline
0.64000000 & 0.80000000 & 0.541626 & 47.703296 \\ \hline
0.49000000 & 0.70000000 & 0.496637 & 40.948101 \\ \hline
0.36000000 & 0.60000000 & 0.446418 & 34.403065 \\ \hline
0.25000000 & 0.50000000 & 0.390388 & 28.077641 \\ \hline
0.16000000 & 0.40000000 & 0.327922 & 21.980390 \\ \hline
0.09000000 & 0.30000000 & 0.258356 & 16.118742 \\ \hline
0.04000000 & 0.20000000 & 0.180998 & 10.498756 \\ \hline
0.02647076 & 0.16269836 & 0.150000 & 8.465257 \\ \hline
0.02250000 & 0.15000000 & 0.139171 & 7.780856 \\ \hline
0.01000000 & 0.10000000 & 0.095125 & 5.124922 \\ \hline
0.00010000 & 0.01000000 & 0.009950125 & 0.501249992 \\ \hline
1.000000E-06 & 0.00100000 & 9.995001E-04 & 5.001250E-02 \\ \hline
1.000000E-08 &  0.00010000 & 9.999500E-05 & 5.000125E-03 \\ \hline
1.000000E-10 & 0.00001000 & 9.999950E-06 &  5.000013E-04 \\ \hline
1.000000E-12 & 0.00000100 & 9.999995E-07 & 5.000001E-05\\ \hline
\end{array}
(1) So for this equation when the approximate amount of dissociation is 0.1 you get 1 significant figure, 0.01 gets two significant figures and so on. The improvement doesn't continue forever of course this is chemistry not math. To get three significant figures you'd have to be a fastidious experimenter.
(2) Notice that the exact quadratic solution is always lower than the approximate solution.
(3) The OP asked how the exact solution of 0.15 dissociation would compare with the approximate solution. From the table the corresponding value is about 0.163, an error of about +8.5%.
Exact Enough Solution - Using Linear Series Solution
In case the hand-waving argument above didn't persuade you that the approximate solution is always bigger that the exact solution, let's explore a solution using a series where an increasing number of terms could be used to get a solution as exact as desired.
Rearranging equation (8), an exact equation, by pulling out a $\sqrt{\dfrac{K_d}{c_0}}$ term:
\begin{align*}
\alpha
&= \sqrt{\dfrac{K_d}{c_0}} \left( \sqrt{\dfrac{K_d}{4\cdot c_0} + 1} - \sqrt{\dfrac{K_d}{4\cdot c_0}} \right)\tag{9} \\
&= \sqrt{\dfrac{K_d}{c_0}} \left( \sqrt{\dfrac{K_d}{4\cdot c_0} + 1} - \sqrt{\dfrac{K_d}{4\cdot c_0}} \right)\cdot \left(\dfrac{\sqrt{\dfrac{K_d}{4\cdot c_0} + 1} + \sqrt{\dfrac{K_d}{4\cdot c_0}}}{\sqrt{\dfrac{K_d}{4\cdot c_0} + 1} + \sqrt{\dfrac{K_d}{4\cdot c_0}}} \right)\tag{10} \\
&=\dfrac{\sqrt{\dfrac{K_d}{c_0}}}{\sqrt{\dfrac{K_d}{4\cdot c_0} + 1} + \sqrt{\dfrac{K_d}{4\cdot c_0}}} \tag{11}
\end{align*}
Now, using the Taylor series for the denominator yields an approximation, which is quite good.  (Many thanks to user Claude Leibovici on the
Mathematics site for the heavy lifting here...)
$${\sqrt{\dfrac{K_c/c_0}{4} + 1} +\sqrt{\dfrac{K_c/c_0}{4}} }=1+\frac{\sqrt{K_c/c_0}}{2}+\frac{K_c/c_0}{8}+O\left((K_c/c_0)^{2}\right)\tag{12}$$ which makes
$$\alpha^{"} \approx \frac {\sqrt{K_c/c_0}}{1+\frac{\sqrt{K_c/c_0}}{2}+\frac{K_c/c_0}{8}}\tag{13}$$
Obviously the more terms the better the approximation, but three terms is good enough for anything that we chemists would do. Again remember that we're solving for dilute solutions where $K_c/c_0 \ll 1$. Also notice that all the terms are positive so the more terms used, the smaller the value of the dissociation fraction.
\begin{array}{|c|c|c|c|c|} \hline
K_c/c_0 & Quadratic & \sqrt{K_c/c_0} & \frac {\sqrt{K_c/c_0}}{1+\frac{\sqrt{K_c/c_0}}{2}} & \frac {\sqrt{K_c/c_0}}{1+\frac{\sqrt{K_c/c_0}}{2} + \frac{K_c/c_0}{8}} \\ \hline
1 & 0.61803399 & 1.00000000 & 0.66666667 & 0.61538462 \\ \hline
0.81 & 0.58192705 & 0.90000000 & 0.62068966 & 0.58017728 \\ \hline
0.64 & 0.54162637 & 0.80000000 & 0.57142857 & 0.54054054 \\ \hline
0.49 & 0.49663670 & 0.70000000 & 0.51851852 & 0.49601417 \\ \hline
0.36 & 0.44641839 & 0.60000000 & 0.46153846 & 0.44609665 \\ \hline
0.25 & 0.39038820 & 0.50000000 & 0.40000000 & 0.39024390 \\ \hline
0.16 & 0.32792156 & 0.40000000 & 0.33333333 & 0.32786885 \\ \hline
0.09 & 0.25835623 & 0.30000000 & 0.26086957 & 0.25834230 \\ \hline
0.04 & 0.18099751 & 0.20000000 & 0.18181818 & 0.18099548 \\ \hline
0.01 & 0.09512492 & 0.10000000 & 0.09523810 & 0.09512485 \\ \hline
0.0001 & 0.00995012 & 0.01000000 & 0.00995025 & 0.00995012 \\ \hline
0.000001 & 0.00099950 & 0.00100000 & 0.00099950 & 0.00099950 \\ \hline
0.00000001 & 0.00010000 & 0.00010000 & 0.00010000 & 0.00010000 \\ \hline
1E-10 & 0.00001000 & 0.00001000 & 0.00001000 & 0.00001000 \\ \hline
1E-12 & 0.00000100 & 0.00000100 & 0.00000100 & 0.00000100 \\ \hline
\end{array}

I am confused, what is the conceptual difference between equation (2) (3) and (8)? It seems like a circular argument.

@M.Farooq - It is the difference between equations 5 and 6 that is important. Note alpha and alpha-prime are calculated and that alpha-prime > alpha always.

I still don't understand how the central question of why we think it's okay to use alpha- prime which we derive through an approximation instead of alpha to judge whether the approximation was correct in the first place

@LimboProductions, read both answers carefully especially the last paragraph of my answer. You have a misconception regarding this circular argument.

@LimboProductions - I agree that I need to make the argument clearer. The gist is that the approximation assumes that little dissociation has occurred so the fraction dissociated is much less than 1. So if the approximation says that 1% has dissociated, the exact solution will show that slightly less has. So it is reasonable to use the approximate value to test the hypothesis.

@MaxW, Then if the approximation shows that dissociation is, say, 13%, but as the real value will be lesser, let's say it is 8-9%. Here, in reality we can use the approximation but we would think otherwise. (I am assuming we have no way to calculate how much lower would the actual value be, or even an ballpark estimate, so correct me if I am wrong in assuming this)

@LimboProductions - I'm trying to work out a series expansion now for the exact expression. Also the exact value is easily obtained by solving a quadratic equation.

Ok, I've done a major revision to the answer, Hopefully the revised version is easier to understand.

+1 I found this a useful answer, I never thought about the usefulness of the estimate to bracket the true value. But I do wonder what is the usefulness of the final exercise. Eq 11 is certainly not an improvement over Eq 9 in terms of simplicity.

@BuckThorn - The point of the last section was to go from equation (8) which is the quadratic solution to equation (13) - never intended to stop at equation (9). Even with just two terms in the denominator equation (13)  works very well for $K_d/c_0 < 0.5$. Using all three terms in equation (13) gives you near perfect values up to $K_d/c_0 = 1$.

I think you changed the equation labels. My question remains: the final series expansion does not seem an advantage  over the exact solution (it seems even more complicated to compute). And the advantage of the simplified formula is that you can quickly evaluate it in your calculator or even your head (back when people knew how to estimate square roots in their heads). The advantage is more or less null in our day of broadly available computing power.

@BuckThorn - I did number the last two equations, otherwise the equations have kept their numbering. In one of the revisions I did move the equations around, but I kept the numbering. // I added a second answer which is much more the the OP's question. My final point there was if you're going to use the approximation, then check the error bound, why not just use the quadratic to begin with since it gives the exact answer for a pittance of extra work.


text:
Stability order of p-quinone, o-quinone and diphenoquinone
Question

Determine stability order:

Answer

 1 > 2 > 3

Thoughts
I found this question quite tricky and confusing. According to my concepts and logic

3 should be non-aromatic, while 1 and 2 are antiaromatic.
Also there is a lone pair repulsion in 2.

After racking my brain I up came with the answer of 3 > 1 > 2 (arguably most convincing for me), which is wrong according to the actually given answer.
So, while my answer for 1 > 2 is correct, I still have no explanation for 3.
Kindly help me on what I am missing or if I am wrong somewhere.

I guess answer given is wrong, because II (ortho-quinone) is highly unstable, at least less stable than I (para-quinone)...

@Zenix the ans given says 1>2>3. Yeah and agreed that 2 is less stable than 1 but what about the 3rd one?.I am confused about 1and 3 and their relative order along with an explanation.

"Benzoquinone is not anti-aromatic, it is not a simple cyclic array of 4 pi electrons. In fact, anti-aromatic compounds distort to some other geometry so they are no longer anti-aromatic, hence more stable." - @ron

@Zenix ohk so I wasn't aware of that...but will u add something about 3???

I currently don't know anything more. I will try to answer it after revising quinones. Not sure, if will be able to..


text:
Limiting the minimum percentage of oxygen in atmosphere maintained by an oxygen absorber / scavenger?
I'm aware that oxygen in the atmosphere of a sealed package can be readily removed by using oxygen absorber/scavenger such as granulated iron and sodium chloride but my understanding is that these bring the O2 content all the way down from 21% only limited by the permeability of the packaging to oxygen and the ratio of scavenger to package volume to something approaching 0%.
My question is if there is a way to formulate an oxygen scavenger that would be limited by some other processes to higher floor of say 5% to 10%... Without having to measure and calculate appropriate ratio of scavenger to package on a per use basis ?

So are you looking for a fully passive solution, not including active flows of gases?

Just to add, this is meant for semi permeable packaging such as paper or plastic where the ongoing reactions of the scavenger compensate for leakages.

Some of us might suggest that all packaging is semi permeable at some level. "Hermetic" seals are allowed to have a real, measurable leak rate...

Yes fully passive is my preference - failing that I will have to include electronics to activate some reaction via heat or otherwise periodically to maintain the atmosphere.

Was watching some YouTube on high vacuum set-ups for playing with plasmas just yesterday so I'm very cognizant of how permeable most packages would be :)

You would have to find something that bonds non covalently to oxygen like hemoglobin. Equillibrium concentration will depend on the temperature. You have to find such substances and their equillibrium constants for reaction X + O2 = Y, and plug them into the equillibrium constant formula to find molar concentration of oxygen at equillibrium

Oxygen absorbers have the purpose to prevent food spoilage by oxidation (of fatty acids, flavors, etc.), so the desire is to remove oxygen as much as possible and also to bind small amounts which may diffuse into the package upon storage.
In order to achieve a certain defined oxygen containing atmosphere, consider flooding the package with an artificial gas mixture, e.g. 5% oxygen, 95% nitrogen. Should be easy to obtain from scuba dive shops who are filling bottles with homemade nitrox blends if they are willing to give you a custom mixture (which in this case is not suitable for diving).
Or just mix air and nitrogen or carbon dioxide (from a soda machine) by adjusting the flow.

Thanks for your response but what I had in mind is more something that you could place in the package just like a traditional scavenger packet and it would actively regulate the internal atmosphere by one or more chemical processes

You'd need a reaction where free oxygen is in an equilibrium with bound oxygen and the status of the equilibrium is almost completely on the side of the oxygen. I think to find such a reaction is difficult to impossible.
Of course, small bags to add would be a simple and convenient solution on the small scale. For production, the gas mixture might be cheaper on the long run, as there are packaging machines that can exactly do that for you, e.g. modified vacuum bag sealers which aren't really expensive.


text:
What is the relationship between entropy and heat capacity?
The molar heat capacity of hydrogen gas and deuterium gas are nearly the same, $\pu{28.8 J K-1 mol-1}$ and $\pu{29.2 J K-1 mol-1}$, respectively, but the absolute entropy of deuterium ($\pu{145.0 J K-1 mol-1}$) is significantly larger than that of hydrogen ($\pu{130.7 J K-1 mol-1}$).
I have seen the equation that associates the heat capacity to entropy, and I understand the equipartition theorem and different modes of motion contribute to the heat capacities. Since the heat capacities are fairly the same, what other factors can affect the entropy in these species?

The entropy depends on the mass. The translational part is given by the Sakur-Tetrode eqn., the rotational part depends on the moment of inertia, and the vibrational part on the vibrational frequency. In each case the partition function $z$ has to be calculated then the entropy via $\displaystyle S=R\ln(z)+RT\left(\frac{\partial z}{\partial T} \right)_V -R\ln(N)+R$ (for a perfect gas). The physical reason is that for the heavier molecule there are more energy levels up to a given energy that can be occupied and so more ways of populating these levels and so more entropy.
The heat capacity is the slope of internal energy $U$ with temperature (at constant volume) $\displaystyle C_V=\left({\partial U}{\partial T}\right)_V$ and the internal energy is related to the partition function as $\displaystyle \frac{U}{T}=RT\left(\frac{\partial z}{\partial T}\right)_V$ and so also to the entropy. The classical thermodynamic relationship is $\displaystyle S=S_0+\int_0^T \frac{C_p}{T}dT$ where $C_p$ is a function of temperature.

I believe that the Sakur-Tetrode is applicable only for monatomic gases, if the problem would be associated with the species like Helium and Neon, then the concept of the mass would be the reasons for the deviation in entropy, as you correctly state the mass is playing a vital role in that equation.

Yes it was derived for an ideal monoatomic gas, but this is only because all the energy is assumed to be kinetic, i.e. $p^2/(2m)$. We can use this just to calculate the translational kinetic energy  part to the entropy then add the other terms.

Based on the difference in mass of the two isotopes you expect a difference of $\pu{14.4 J mol-1 K-1}$ $\left(\frac{5}{2} R\ln(2)\right)$ in the entropy (computing the rotational and translational contributions - $R\ln(2)$ and $\frac{3}{2} R\ln(2)$ respectively- for a rigid rotor ideal gas using basic statistical mechanics). That's in line with what you state in the problem.
On the other hand, the heat capacity is computed by taking the temperature derivative of the logarithm of the partition functions, eliminating any mass dependence. When you do the computation this leads to the conclusion that the ideal heat capacities of the two gases should be identical - at least at this level of theory. So the similarity in heat capacities is at least suggestive that practice does not deviate much from this theory. More interesting is then the fact that they do differ.
As for the question in the title, the answer is simply that the derivative of the entropy w.r.t. temperature is related to the heat capacity as follows: $$\left(\frac{\partial S}{\partial T}\right)_V=\frac{C_V}{T}$$

The entropy at absolute zero is zero. If the heat capacities are only 3% different from each other over the entire temperature range, and we integrate from zero kelvin to room temperature, why do we get entropy values that are about 10% different from each other? @Lackofgoodpeople

@KarstenTheis The expressions I used are applicable at high T. More importantly, as you drain the systems of energy by lowering T molecules lose their translational and rotational degrees of freedom (becoming vibrational instead) during condensation and crystallization. You are left with zero point energy.

@KarstenTheis The more appealing explanation for why mass has this effect on the entropy is what porphyrin wrote, namely that the isotopes result in different energy level gaps and so different state occupation statistics.

The relative entropies of two substances at a temperature $T'$ are not determined by their relative heat capacities at $T'$.  Rather,
$$\text {Absolute entropy at } T' \equiv S(T') =\Delta S_{0 \rightarrow T'}$$
$$= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T} = \int_{0}^{T'}\frac{C_p(T)}{T} dT.$$
I.e., for each substance, you need to consider how $C_p(T)$ varies with $T$ from $T=0$ to $T=T'$.  Just because the heat capacities are about the same at $T'$ doesn't mean they won't be different between $T=0$ and $T=T'$.
Intutively, the way to think about this is that the entropy is determined by how the energy levels are populated;  $\frac{C_p(T)}{T}$ at any given temperature tells us the ways in which the existing populations are able to change as the temperature is increased; and the integral of $\frac{C_p(T)}{T}$ gives us the entropy by summing up these changes.
In addition, hydrogen and deuterium have different freezing and boiling points.  Thus they will not always be of the same phase.  This will cause further differences between their heat capacities.
To get an idea of the complexities involved in determining the relative heat capacities of these substances at lower temperatures, take a look at these two NIST publications, which give the equations of state for hydrogen and deuterium, respectively:
https://tsapps.nist.gov/publication/get_pdf.cfm?pub_id=832374
and
https://www.nist.gov/system/files/documents/2018/03/12/2fundamental_equation_of_state_for_deuterium.pdf


text:
(Mis)understanding on MOT
In VBT, one can understand the example of hydrogen like this - Two hydrogen atoms come closer to each other, and their orbitals overlap, stabilize, and ultimately result in a bond. This is not too difficult to understand, and can easily be visualized and seen in an intuitive manner.
MOT however, is different. Instead of the two orbitals overlapping each other to stabilize the molecule, they both mix together to form a bonding and anti bonding orbitals. Alright, the electrons from both the Hydrogen atoms go into the lowest energy orbital, which is the bonding orbital.
My main doubt however, comes from the electrons of both atoms being SHARED in a covalent bond. When two orbitals overlap in the same phase, they add up to each other (BMO). And when they overlap in opposite phase, they cancel each other (AMBO). So, two hydrogen atoms overlap in the same phase, and they make a bond and so stabilize each other and form a bonding molecular orbital. Wouldn't this mean that both the electrons are shared between the two atoms and hence both their orbitals are filled? Is this a false understanding of how bonds are formed?
Does this have to do with the mathematical equations themselves? They're as of now, too advanced for my course.

Once you use the AOs to form the MOs, the AOs no longer exist. It is like, if you use two oranges to make two cups of orange juice, the oranges no longer exist (if they do, please tell me the secret so that I can set up an infinite orange juice factory). So it is not particularly useful to think about "both their orbitals are filled" – both their orbitals are no longer things. Once you form MOs, it is just "the bonding MO is filled". And note that the bonding MO is lower in energy than the original AOs, so the process of forming these MOs stabilises the system, i.e. it is a bond.

And if we're taking a look at the individual atom's (in the molecule) electronic configurations, they all will have filled orbitals, right? Like bringing back the orange juice and stuffing them up into the peel of the individual oranges?

Well, no, not really; it would take four electrons to fill up both the 1s orbitals, and you only have two to play with. It's just that it's more stable for those two electrons to be in a bonding MO, rather than in their own AOs. So the orange analogy kind of ends there, unless we stretch it even further, but it becomes kind of silly. (For example: you could have two half-juicy oranges, and together you could squeeze out one full cup of juice and one empty cup of juice. But you can't stick the juice back into both oranges to make them both full. Oranges = AOs, cups = MOs, juice = electrons)


text:
Is electrical work equal to the change in enthalpy or internal energy?

A $\pu{0.5 A}$ current from a $\pu{20 V}$ supply is passed through a water sample. $\pu{0.9 g}$ of water is vapourised at $p = \pu{1 atm}$ and $T = \pu{373 K}$. Calculate $\Delta U$ and $\Delta H$ for the vapourisation of $\pu{1 mol}$ of water.

The flow of current is doing work on the system:
$$w = Vq = VIt$$
where $w$ is work, $V$ is voltage, and $q$ is charge, $I$ is current, and $t$ is time. I thought that the work done, $w$, will be equal to the change in internal energy:
$$\Delta U = w$$
From this, I could calculate $\Delta H$ using the relation
$$\Delta H = \Delta U + \Delta n_g RT$$
where $\Delta n_g$ denotes the number of moles of gaseous products minus number of moles of gaseous reactants. Here we are assuming that water vapour behaves as an ideal gas and that the volume of the liquid is negligible compared to the volume of gas formed.
My issue is that, according to the solution, the work done is equal to $\Delta H$, not $\Delta U$.
It got me thinking about how enthalpy is equal to heat supplied at constant pressure, which could provide the energy for vapourisation, much like the electrical work in this case.
Why is $w = \Delta H$ and not $w = \Delta U$?

Work done by electricity = internal energy change + work done against external pressure = Delta H

Isn't change in Internal energy is equal to work done on the system?

Not if the system has done work itself. Or, you should express the work done on system as delta W = U.I.t - p.delta V

Doesn't the problem statement include the amount of time that the current was running?  This seems to be missing from the problem statement you provided.

In this problem, there are two types of work that are done, and no heat Q is transferred.  The two types of work are VIt and $P\Delta V=(\Delta n)R T$.  So the change in internal energy is $$\Delta U=VIt-P\Delta V$$So the change in enthalpy is:  $$\Delta H=\Delta U+\Delta (PV)=VIt$$


text:
Why is a gas less soluble in an aqueous solution of electrolytes than pure water?
Please help me out! This was given as a fact in my textbook but I don't understand the logic behind it.

Gist of the matter...
When an ionic salt dissolves in water the ions and water molecules don't just bounce around independently. The ions have a charge and water molecules are polar. So the water molecules form a cluster, which is a number of concentric layers of  water molecules, around the ions. A gas molecule dissolved in the water can't get into a cluster. Hence there is effectively less water in which the gas can dissolve.


text:
Molecular orbital diagram of CO and charge localisation
MO diagram of CO
My question concerns the interpretation of the Molecular Orbital of CO. I think I find it clear how you build it but I have some concerns about how you rationalize it.
Particularly, from what I read the reason we envision carbon as being partially negatively charged in CO is that the HOMO of the molecule lies closer to the carbon orbitals. However, from the MO diagram, it can be seen that the p-oxygen orbitals are closer to the HOMO than both of the orbitals of carbon.
My question is: what we use as a benchmark to judge that the HOMO is closer in to energy carbon: are we looking at average energy or something completely different and I completely misinterpreted the argument?

I would be inclined to rationalise this using resonance rather than MO diagrams. The lone pair on the O can be donated to make a triple bond with a formal negative charge on the C and a formal positive charge on the O. See https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Chem1_(Lower)/09%3A_Chemical_Bonding_and_Molecular_Structure/9.04%3A_Polar_Covalence for a brief discussion of this

CO is a terribly complex molecule. There is a pen and paper approach you can go for qualitatively determining a trend. I'm confident that is not possible without the proper mathematical/computational tools. I have demonstrated the first approach here: How to rationalise with MO theory that CO is a two-electron donor through carbon? I hope that this question will answer your query: How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory?