text:
Why does balancing the oxidation of dichromate(VI) to chromium(III) require six iron atoms?
The problem:
Balance the following chemical equation using oxidation numbers:
$$ \ce{Fe^2+ + Cr_2O_7^2- -> Fe^3+ + Cr^3+} $$
What I have tried:
$$\ce{\overset{+II}{Fe}^{2+}(aq) + \overset{+VI}{Cr}_2O7^2–(aq) \rightarrow \overset{+III}{Fe}^3+(aq) + \overset{+III}{Cr}^3+(aq)}$$
\begin{align}
\text{Electron transition:}\quad
\ce{\overset{+VI}{Cr}_2 + 6 e- &-> 2 \overset{+III}{Cr^{3+}}} &\quad &|\times 1 \\
\ce{\overset{+II}{Fe} &-> \overset{+III}{Fe} + 1 e-} &\quad &|\times 3
\end{align}
$$\ce{3Fe^{2+}(aq) + Cr2O7^2–(aq) –> 2 Cr^3+(aq) + 3Fe^{3+}(aq)}$$
\begin{align}
\text{Charges:}\quad
&\text{left side} &\quad 3\cdot(+2) + 1\cdot(-2) &= +4 &\quad &|\text{add}~\ce{11 H+} \\
&\text{right side} &\quad 3\cdot(+3) + 2\cdot(+3) &= +15 &\quad &|\text{add}~\ce{11 H+ and 7 O} \\
\hline
&\text{difference} &\quad &= +11
\end{align}
But this doesn't align with the solution for this problem, which is:
$$ \ce{6Fe^2+(aq) + Cr2O7^2-(aq) + 14H+(aq) ->
       6Fe^3+(aq) + 2Cr^3+(aq) + 7H2O} $$
Why am I getting three iron atoms and not six?

Now, why are six iron atoms added?

There is this magical formula 2 x (6-3)=6 x (3-2) // By other words, the sum of  changes  of oxidation numbers must be zero, to honor the charge conservation law.

In the question, you have forgotten there are 2 Cr in Cr2O7^2-. Therefore  1 Cr2O7^2-.  is able to oxidize 6 Fe^2+ and not just 3.

Hmm, alright. Thank you. I thought you were being sarcastic when you said "magic formula". My bad. Thanks =p

I use it often in general form m.n=n.m. As when chemical equations are concerned, many people forget basics of arithmetics of small integers and the commutative law. If they forget it is chemistry, they can see it is simple. so 2 times 3 = 6 times 1.

I would like to go back to your original demand.
On the first line, there was a title : "The problem". The second line is : "Balance the chemical equation using oxidation numbers". The third line is : "What I have tried".
The fourth line is the equation between $\ce{Fe^{2+}}$ and $\ce{Cr2O7^{2-}}$, which is correct, although no stoichiometric coefficients are indicated. Up to here all is right. Nothing to object. The main problem here is to define these coefficients. This is where you made a mistake.
The next two lines start with "Electron transition", follow by the dichromate reduction. Here you correctly find that $6$ electrons are needed when reducing $1$ $\ce{Cr2O7^{2-}}$ ion to $2$ $\ce{Cr^{3+}}$, although the formula of the chromium ions has a $+3$ charge, which was forgotten. Well. This typo has no consequence.
What is more serious is the following line, where you correctly state that the oxidation of $\ce{Fe^{2+}}$ to $\ce{Fe^{3+}}$ produces $1$ electron. $1$ electron par iron ion is correct. But just after this equation you multiply this equation by $3$. Strange decision. Why by $3$ ? There is no reason for multiplying by $3$ the equation $$\ce{Fe^{2+} -> Fe^{3+} + e-}$$ If you multiply this equation by $3$, you produce $3$ electrons. And these $3$ electrons cannot do anything ! To reduce one $\ce{Cr2O7^{2-}}$ ion, $3$ electrons are not useful. Instead of $3$, the ion $\ce{Cr2O7^{2-}}$ requires $6$ electrons. Six ! Not three ! The final equation must bring $6$ electrons ; so it must contain $6$ $\ce{Fe^{2+}}$ ions on the left-hand side.
This is the origin of your mistake. Once you have understood the origin this number $6$, I am sure you will find the rest of the problem, and obtain the final equation.


text:
Anthracycline vs Angucycline
Anthracyclines and angucyclines both have a tetracyclic core. Now, I need to classify urdamycinone B, fridamycin A, fridamycin G, and dehydroxyaquayamycin as anthracycline or angucycline. Both urdamycinone B and dehydroxyaquayamycin contain benz[a]anthraquinone core. While fridamycin A and G both contain anthraquinone core. I have classified urdamycinone B and dehydroxyaquayamycin as anthracyclines. And fridamycin A and G as angucyclines.
Did I classify the compounds correctly? And what are the differences between anthracyclines and angucyclines in terms of their sources and structures?

Have you tried drawing the structures? As far as I can tell, for example, the fridamycins don't have tetracyclic cores.

Yes. Urdamycinone B and dehydroxyaquayamycin have a tetracyclic benz[a]anthraquinone core while fridamycins A and G have a tricyclic anthraquinone core.

source: Stack Exchange

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

sort: votes


text:
Calculate percent composition from XPS?
I have an XPS graph and was wondering how to calculate percent composition. I do have the percent compositions that were determined from the graph, but I don't know how they were found. For reference, I am looking at a N 1s XPS plot, shown below (part C):

I had tried calculating the composition by the areas under the curves for N 1s, but when I did I got an equal area for both the pyridinic (pink area) and graphitic (orange area) Nitrogens, which is not correct as the graphitic Nitrogen has a much smaller composition. Any help would be much appreciated.

The coloured curves are made by least squares fitting of 3  gaussian curves to the data and the area under each calculated. This area then can be made a fraction or percentage of total area.

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text:
Is Liquid Air Safe?
Does Liquid Air mitigate the asphyxiation risks associated with Liquid Nitrogen?
As Oxygen and Nitrogen have marginally different boiling points, do you run the risk of all the nitrogen boiling off first, leaving behind a very flammable, very dangerous pure liquid Oxygen? Assuming you were to store the liquid air in a dewar tank, approximately how long would it take for this to become a concern, if at all?

Effective ventilation reduces the risk. Why would liquid air be better than that?

Since they will boil off at different temperatures, no, it just causes other problems.

Oxygen is flammable only in reducing atmosphere, like H2, CH4 or NH3.

Well, if you are running pure oxygen in non-rated plumbing it will (un)happily make some exciting flames just fine…

@JonCuster, what are these other problems? Really I'm trying to understand whether you could safely have a dewar of liquid air in a confined space without suffocating anyone.

You can have a dewar with LN2 in a confined space without suffocating. Unless you tip it over, or live in there for days. You will get a shortage of oxygen, and too much CO2, in there without the dewar even. Why would you want to do that?

The boil-off rate of N2 cannot be predicted without knowing exactly the conditions of this ill-advised experiment. Quality of the dewar, its surface-to-volume ratio, outside temperature,  ...

@Karl, not really an experiment. I just want to know whether using Liquid Air would be safer than Liquid Nitrogen and what safety consideration I would need to make.

If we agree that asphyxiation is lack of providing cells of the the human organism with oxygen, liquid air is not safe because of its low temperature. Once the isolating buffer of air (Leidenfrost effect) between the liquid air and the gaseous air is off of the alveoli, the tissue will be frozen (and become brittle)* and blood behind these membranes will freeze, too.  Don't forget blood contains much of water, which expands above and below about $\pu{4 ^\circ{}C}$, too.
At atmospheric pressure, the boiling temperature of nitrogen is about $ \pu{-196 ^\circ{}C}$ (ref), the of oxygen about $\pu{-183 ^\circ{}C}$ (ref).  While you argue about liquid air, it actually is a known issue as potential danger when handling liquid nitrogen as a cryogenic.  The very first example which comes to my mind are the Dewar traps installed to protect oil pumps filled with LN instead of dry ice and iPrOH.  The ones filled with LN are cool enough to condense oxygen available in the ambient air.  Because condensed oxygen supports combustion so much better than gaseous oxygen diluted in air (which is about 78% nitrogen, and 21% oxygen), you just don't mindlessly refill these traps with LN again and again.  It isn't safe to await for the blue color of liquid oxygen as an indicator of a risk (and compared to liquid ozone one may trap behind an ozonolysis, its blue is much fainter), nor to use liquid air as a cryogenic.
Running a Linde machine to harvest nitrogen actually requires a number of steps to filter, wash, dry air to remove e.g., dust, pollen, humidity, $\ce{CO2}$, $\ce{N2O}$ before air is liquefied; and uses a distillation column to remove oxygen trapped by the process.  Hence you see these machines more likely in locales with both reliable and affordable supply of electricity (for the compressors) and water (for the cooling) to provide you with purified (colorless) LN from the pressurized truck / local tank for your lab can.
* Some lectures display shatter e.g., a flexible rubber tube just cooled in liquid nitrogen with a hammer.  More seriously though, cryogenic grinding is technique to eventually amend (heat sensitive) samples of polymers and organic materials to an analysis to lower the rates of chemical reactions e.g., if cells are broken up, or because deep frozen samples tend to be less sticky to the grinder.


text:
Do solvated electrons short circuit the electrolysis of liquid ammonia
If attempting the electrolysis of liquid ammonia using NaNH2 as an electrolyte. It is understood that Na+ will form solvated electrons.  I understand that these solvated electrons can dramatically increase the conductivity of the solution.  Will the movement of these electrons aid the production of H2 and N2 in electrolysis.  Or will these electrons travel directly between the electrodes effectively short circuiting the ammonia and hampering the electrolysis.

Na+ can't and won't form solvated electrons, for it has no extra electrons to spare. Metallic Na is another story.

To get solvated electrons in solution, it is necessary to dissolve metallic sodium in liquid ammonia, and to be quick, because the solvated electrons will only exist before the following reaction : $$\ce{2Na + 2NH3 -> 2NaNH2 + H2}$$ or $$\ce{2 e^- + 2 NH3 -> 2 NH2^- + H2}$$

Solvated electrons do not always short-circuit the electrolytic process. In some processes the solvated electrons accumulate in the ammoniacal medium instead, particularly if the anode which supplies the electrons in the first place is made of a soluble metal. As electrons are removed from the metal anode, the metal ions dissolve and intermingle with the electrons that accumulate in the ammonia. In effect we are observing what would ordinarily be an oxidation half-reaction of the anodic metal.
In many cases this metal is magnesium. As this answer describes, magnesium is difficult to dissolve directly into liquid ammonia because it tends to passivate with the formation of a poorly soluble amide, and the magnesium ions are sufficiently Lewis-acidic to solvolyze with the resulting ammonium ions then destroying the solvated electrons. With an electrolytic process we surmount these barriers. Pumping the electrons out of a magnesium anode promotes the otherwise hindered metal dissolution, and the supporting electrolyte stabilizes the magnesium ions against reacting with the solvated electrons (apparently by forming ion pairs). See Combellas et al1.
Reference

C. Combellas, F. Kanoufi, A. Thiébault, "Solutions of solvated electrons in liquid ammonia: Part 1. Chemical properties of magnesium solutions", Journal of Electroanalytical Chemistry ,
2001 499, 1, Pages 144-151.   ISSN 1572-6657, https://doi.com/10.1016/S0022-0728(00)00504-0

I think even inert metal electrodes in an ammonia solution containing an electrolyte (say LiCl) will release free electrons into the ammonia solution. Whether this eventually causes a short circuit is unclear, but the effect is clearly visible as the solution near the electrode releasing electrons goes blue.

@matt_black note I am saying the anode dissolves. The electrons pass into the solution from the cathode which, in an electrolytic generation of solvated electrons, would be inert.


text:
Doubt regarding Gibbs energy and Electrode potential
I have a doubt regarding the relation between Gibbs energy change and E cell, especially when calculating for redox reactions.
This is a question I have seen on a website called "ExamSide questions JEE 2024 Chemistry"

$$\ce{MnO4^-(aq) + H^+(aq) + C2O4^2-(aq)<=>Mn^2+(aq) + H2O(l) + CO2(g)}$$
The standard potentials are:
$E^\circ(MnO_4^-/Mn^{2+}) = +1.51V$
$E^\circ(CO_2/C_2O_4^{2-}) = -0.49V$
Find the $K$ for the given redox reaction.

Now, to calculate the net E cell for this reaction, the answer they have provided is:

$$
E^\circ_{cell} = E^\circ_{OP,anode} + E^\circ_{RP,cathode}
= 0.49 + 1.51 = 2V
$$

Now, if we do the same question using Gibbs energy formula, that is calculating Gibbs energy for each half reaction and adding them up, it goes something like this:

$$
\ce{MnO4^- + 8H^+ + 5e^- -> Mn^2+ + 4H2O}$$
$$\ce{C2O4^2- -> 2CO2 + 2e^-}
$$
Balancing and calculating the net Gibbs energy from each of them gives:
$$\ce{2MnO4^- + 16H^+ + 5C2O4^2- -> 2Mn^2+ + 10CO2 + 8H2O}$$
$$\Delta_\mathrm{r} G^⦵ = \Delta_\mathrm{r} G^⦵(Oxidation) + \Delta_\mathrm{r} G^⦵(Reduction)$$
$$
-n_{net}FE^\circ_{net} = -nFE^\circ_{Oxidation} + -nFE^\circ_{Reduction}
$$
$$
n_{oxidation} = 2, n_{reduction} = 5, and, n_{net} = 10
$$
So,
$$
E^\circ_{net} = \frac{n_{red}E^\circ_{red} + n_{ox}E^\circ_{ox}}{n_{net}}
= \frac{2(1.51)+5(0.49)}{10} = 0.547V$$

which is approx. 4 times lower than given answer.
So, what am I missing or doing wrong?

Both n_ox and n_red are 10. You have to multiple respective half-reactions by 2 resp. 5 to match them.

There is a fundamental misunderstanding as to what 'n' in -nFE represents. It represents the number of electrons and not a multiplier to your reaction stoichiometry. Understand that $E^o$ is an intrinsic value and deltaG is extrinsic the 'n' accounts for the number of electrons.

@zynox I converted your answer into a comment. It looks correct but it was being downvoted, presumably because it requires some elaboration (I don't know the actual reason).

Note that the standard reduction potential is an intensive property, but the standard reaction Gibbs energy is an extensive property scaling proportionally with formal multiplication of reaction coefficients. It still matches the potential, as the electron count changes with the scaled standard reaction Gibbs energy the same way.
To evaluate standard Gibbs reaction energy and the cell potential, both half reactions have to be scaled to the same electron counts, leading to the standard cell potential(voltage) being simple subtraction of standard reduction potentials of both half-reactions and standard reaction Gibbs energy being subtraction of scaled standard  half-reaction Gibbs energies.
In the above particular case for Gibbs energy evaluation:
\begin{align}
\Delta_\text{r} G^{\circ} &= 5 \cdot \Delta G^{\circ}_\text{ox} + 2 \cdot \Delta G^{\circ}_\mathrm{red}=\\
&=5 \cdot (-2 \cdot F\cdot E^{\circ}_\text{ox}) + 2 \cdot(- 5 \cdot F \cdot E^{\circ}_\text{red})=\\
&= -10 \cdot F\cdot E^{\circ}_\text{ox} - 10 \cdot F \cdot E^{\circ}_\text{red}\\
\\
E^{\circ}_\text{cell} &= - \frac{\Delta_\text{r} G^{\circ}}{10F} =  E^{\circ}_\text{ox} +E^{\circ}_\text{red}=\\
&= -(\pu{-0.49 V}) + \pu{1.51 V} =\\
&= \pu{+2 V}
\end{align}


text:
Definition of Gibbs Free Energy and its interpretation

RELATIONSHIP BETWEEN $\Delta G$ & $W_\mathrm{non-PV}$
$dU = dq + dw_\mathrm{py} + dw_\mathrm{non-PV}$
for reversible process at constant T & P
$dU+pdV-TdS = dw_\mathrm{non-PV}$
$dH-TdS = dw_\mathrm{non-PV}$
$(dG_\mathrm{system})_\mathrm{T,P} = dw_\mathrm{non-PV}$
$(dG_\mathrm{system})_\mathrm{T,P} = (dw_\mathrm{non-py})_\mathrm{by system}$
Means free energy change for a process is equal to the maximum possible work that can be derived from the process i.e.,
$\Delta G^\circ= W_\mathrm{max}$ (for a reversible change at constant pressure and temperature)

The derivation shown above considers
$$dU = dq + dw_\mathrm{pv} + dw_\mathrm{non-pv}$$
(1) Is the dq used here different from the general q (as below) used in thermodynamics at definition level - where W stands for mechanical work?
$$ dU = dq - dW $$
(2) Assuming that the term $dw_\mathrm{pv}$ includes $Vdp + pdV$, is $Vdp$ also considered to be mechanical work generally?
(3) What can be an interpretation of $dw_\mathrm{non-pv}$, and, is there a mathematical way to calculate it? Example involving this would be helpful.
Before reading the above text, I was looking at it the following way:
$dG = dH - TdS \\
dS = dq_\mathrm{rev}/T \\
dG = dH - dq ~~\mathrm{[assuming~ the~ process~ to~ be~ reversible]}\\
   =  dU + d(PV) - dU - PdV \mathrm{[assuming~statement~2}~\\ \mathrm{~to~be~wrong~ and~work~to ~only~ mean}~ PdV~\mathrm{ by ~definition}\\
   = Vdp $
I realize that I am completely wrong. Now, I am confused as well. Please help with the same.

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

Denbigh, Principles of Chemical Equilibrium, analyzes all this in a more precise way.  He considers a closed system initially at temperature T and pressure P, and a process for which all heat transfer takes place with a constant temperature reservoir also at T, and with the system in contact with a surroundings at constant external pressure $P_{ext}=P$.  In addition, in the final thermodynamic equilibrium state, the system temperature and pressure are also T and P respectively.
For this situation, the first law of thermodynamics tells us that $$\Delta U=Q-P\Delta V-W_{non\ PV}$$And, for this situation, the second law of thermodynamics tells us that $$\Delta S=\frac{Q}{T}+\sigma$$where $\sigma$ is the (positive definite) entropy generation due to irreversibility.  If we combine these two equations by eliminating Q, we obtain:$$\Delta U=T\Delta S-P\Delta V-W_{non\ PV}-T\sigma$$Rearranging this equation then gives:$$W_{non PV}=-\Delta G-T\sigma$$So, for this situation, for the same initial and final end states, the maximum non-PV work that the system can do is $-\Delta G$ if the process is reversible and less than that if the process is irreversible.

Thank you for your answer. I find it surprising and shocking that most books don't mention the Wnon PV in first law of thermodynamics and σ in second law of thermodynamics as you had mentioned.
@Chet Miller, could you please provide some insights on σ and what Wnon_pv refers to in a real-life scenario, through an example.
In the image we are later referring to this Wnon_pv as the maximum work which can be extracted. Is that right and shouldn't max work actually refer to W_pv+W_non-pv ?

An example of non-PV work is electrical work, where correct enters the system at a higher voltage and leaves at a lower voltage.  This might be associated with an electrolytic chemical reaction,, for example.  For discussion of $\sigma$, see Moran et , Fundamentals of Engineering Thermodynamics.  Bird eo al, in Transport Phenomena, give explicit expressions for the rate of entropy generation per unit volume in terns ov velocity gradients and temperature gradients.

I would like to answer the question of the title : Definition of the Gibbs Free Energy and its interpretation.
The free energy is a concept created by Gibbs in the $19$th century. At that times Gibbs was trying to apply in chemistry the concepts developed in physics and mechanics for spontaneous phenomena. In physics, a spontaneous process happen each time an object has the possibility to fall from a high position to a lower position. And never the contrary. An object cannot spontaneously pass from a lower position to a higher level.
This is related to the concept of potential energy. Each object has a mechanical potential energy $mgh$ which depends on its level $h$ measured above an arbitrary reference level. This reference level which may be the level of the sea (in which case $h$ is called altitude). But, main result for Gibbs is that $mgh$ is a mechanical potential energy. This energy has the property that when an object with a high mechanical potential energy has the possibility to go to a lower level, the transformation is spontaneous. The passage from a low level to a higher level is never spontaneous. Never !
Gibbs was looking to find a potential energy valid for chemistry. For a while he thought that the enthalpy $\ce{H}$ was this potential energy, because in nearly all spontaneous chemical reactions, heat is emitted. So that the molecules pass spontaneously from an initially high $\ce{H}$level to an lower one, and the difference is emitted as heat. This is usually right. But it is not valid for endothermic reactions, which are also spontaneous. So enthalpy is not the energy that Gibbs was looking for. What was the potential energy that Gibbs was looking for, so that in all cases, it decreases in spontaneous processes ?
It is not heat content. Another way of looking at this problem is to say : What is the potential energy that is liberated in all chemical reactions, that is in endo- or exothermic reactions ?
For Gibbs, this strange and new potential energy, called "free energy", is the electrical energy liberated in galvanic cells. Galvanic cells are working by chemical reactions. And the galvanic cell is producing spontaneously electrical energy, whatever the endothermic or exothermic character of the internal chemical reaction. A cell will always produce electricity spontaneously, never the contrary. The chemical reaction may be endo- or exothermic, it must be exoergic. A galvanic cell cannot absorb electrical energy : it would be a non-sense.
So, for Gibbs, all reactants in a cell possess an initially high free enthalpy $G$ ($G$ for Gibbs). And, when spontaneously working, the cell emits an electrical energy Delta $G$ as non-pV work, equal to : Delta $G = - zEF$, where $z$ is the number of electrons liberated in the chemical equation of the cell, $E$ is the tension of the cell, and $F$ is the Faraday (= $96500$ $Cb$).
After this definition, Gibbs decided to behave like the chemists in calorimetry. Using all known galvanic cells, he measured their $E$ values and calculated the corresponding values of free enthalpy of formation of all compounds found in cells. These values are published in tables, like the enthalpies of formation.

Your answer has helped a lot with the conceptual clarity of this topic. However, a few queries
"The chemical reaction may be endo- or exothermic, it must be exoergic. " - I don't understand the difference between exothermic and exoergic. Please help.
In Thermodynamics we completely ignore this electrical energy , so are we saying W_non-pv is 0 is any non-electrical process ?

Extending my doubt towards entropy .Say we have a container and by an adiabatic reversible process you expand it. by definition of entropy ds=q_rev/T. Because the process is adiabatic q_rev=0 and hence ds=0.  My assumption would be that the entropy of the gas increases as there is more randomness with more volume. but doesn't this contradict the fact that q_rev=0 and hence ds=0 ?

based on the answers, comments and discussions, I thought of summarizing what I understood:

For any system, the energy present (requisite for its existence) is called Internal Energy.
I as part of the surrounding can interact with the system by either giving or gaining Heat or by applying/extracting work.
While Heat, Internal Energy and Work help me in altering conditions of a system, say to restore conditions after a particular process is done, they do not give me an Intergrated view of how and if the process is of any use to me (in deriving some productive output)
Why should I do any process? Because it can give me energy which can be used for other applications. This energy is termed as Gibbs Free Energy and also called as usable work (because it can be utilized in any know form, say electrical or magnetic or say to cause movement of something, etc.)
Gibbs Free Energy tells me the usable energy I can extract from a process. As long as it is negative (it releases usable Energy, I can consider my process to be spontaneous, feasible and possible for it to occur)
Now is this Gibbs Free energy same as Internal Energy?
Internal Energy just tells us about the energy content a system can hold. If Iam unable to extract this energy for any external use, there is no point in using Internal Energy to denote whether a reaction would happen or not. (when it happens, I should get some change in the environment in the form of the usable energy I mentioned above). Gibbs Free Energy on the other hand is defined to tell me this, whether a process is spontaneous or not.
But is Gibbs Free Energy same as Work done?
Gibbs Free Energy is known to be Usable work. In a process apart from the Usable energy released, the process may also involve a change in volume. Say, I have a gas in a container and through a process, I increase the size of the piston. The volume is changing and for this new state of gas of high volume, the internal energy is different. The change in Internal Energy is compensated by work done on/by gas to carry this process. Such a work is called W_pv and clearly this is not included in Gibbs Free Energy. (Iam not able to use it for any external use, so there is no point in considering it to denote spontaneity of a process)

UPDATE : Another way of seeing it has come to my mind, please correcct if its wrong.
The first law of thermodynamics can be rearranged to be written as $W_{non-pv} = \Delta U - Q - W_{pv} $
So, any change in one variable, can causes changes in other variables. Different permutations of distribution of the changes leads to different processes. While $W_{pv} $ indicates the work done by changing P and V conditions, it's not necessary that all of it is converted to useful energy or  W$_{non-pv} $, ie. Gibbs Energy.
Thank you BuckThorn, Chet Miller, Maurice for your help through your answers and comments!
I know that topics such as these vary a lot as they are interpreted in different ways. Please let me know if you have a better way of defining this as it might help us all!

It is better to relate Internal energy and Helmholtz energy ( free internal energy), and enthalpy and Gibbs energy ( free enthalpy).  The Gibbs energy is the part of Enthalpy as the hard max limit for work the system may provide. It does not say how it can be done or if it can be done at all at given scenario.


text:
Does Calcium Carbonate dissolve in an aqueous solution of sodium bicarbonate?
A countertop company (Wilsonart) advises the use of sodium bicarbonate to remove limescale (solid  calcium carbonate) from the worktop surface. Acidic solutions are contraindicated because the countertop (which is solid laminate, probably phenolic resin) is sensitive to acid. But, does sodium bicarbonate increase the solubility of calcium carbonate, at room temperature? (It does not seem to work)
If not, are there any other ways of increasing the solubility of calcium carbonate without resorting to acid (or other things that would damage a laminate countertop)?

Sorry carbonate was a typo, fixed now. Meant NaHCO3.

Calcium carbonate cannot be totally insoluble in any solvant.

@Maurice hard water is pretty much a good counter-example. You will get more solid calcium carbonate and not less.

The chemical common sense says that the sodium bicarbonate is not going to make the limescale much more soluble.
On the other hand, the same sodium bicatrbonate is widely used as a mild abrasive (when not fully dissolved in water). This may or may not help in your case (probably not in an aestetically acceptable manner).
On the third hand, whatever your countertop is made of (it is not marble, is it?), it will likely not be vulnerable to the ordinary food-rated acids, e.g. vinegar or lemon juice. This is what you can try with little risk and a great probability of success.

You might try club soda, seltzer; CO2 dissolves CaCO3 converting it into bicarbonate.

@jimchmst this is pretty much traditional (this is how calcium carbonate got into the utility water in the first place). On the other hand, for the club soda to work, one would need a lot of time and a lot of soda - and maybe a high pressure setup of some sort. And still falls in "using acid" category.

Can't upvote yet, but I would if I could @fraxinus


text:
Elements with most commonly occurring isotope being different from the most stable one
Usually, stable isotopes have highest abundances, often much higher than radioactive ones. Are there any elements having most commonly occurring isotope different from the most stable isotope?

What is "most stable"?

Fe-56 is the most stable iron isotope and the most abundant. Ni-62 has the highest binding energy of all  but is not the most abundant Ni isotope. Nuclide formation in stars is complex as is radioactive decay and artificial element manufacture.  There is a lot in Wikipedia.

When isotopes are stable (resistence to spontaneous decay) enough , observationally stable or assumed as stable, their relative abundance is not determined by their stability, but by probability of the path of the nucleogenesis, leading to the considered isotope. If there is formed about 10 times more isotope A than isotope B, that there is about 10 times more of isotope A than B, even if B is more stable.

How about $\ce{^{79}Br}$ and $\ce{^{81}Br}$ $(1:1)$?

Yes.
You're probably thinking about comparing nuclear stability by comparing radioactive decay half-lives, which is a reasonable and intuitive (though somewhat limited) criterion. Already here there are multiple examples which fit the requirement.
For instance, rhenium has two isotopes naturally found on Earth. 62.6% of all rhenium is rhenium-187, which beta-decays to osmium-187 with a measured half-life of 41.2 billion years (about four times the current age of the universe). Meanwhile, the remaining 37.4% of rhenium is "observationally stable" rhenium-185, which is to say it is theoretically expected to undergo spontaneous nuclear decay, but no such decay has yet been measured, suggesting its half-life is likely far higher.
An even more lopsided case is indium. 95.7% of all indium is radioactive indium-115 with a double beta-decays with a half-life of 441 trillion years, compared to 4.3% of observationally-stable indium-113.
Another interesting example is tellurium, which has eight naturally occurring isotopes. Two of these, tellurium-128 and tellurium-130, compose 65.8% of all tellurium, and both have staggeringly huge measured half-lives, greater than a billion times the age of the universe.
There are also a number of other close cases such as rubidium, neodymium, samarium and cadmium, where radioactive isotopes compose >10% of natural abundance, but not enough to beat the predominant observationally-stable isotope.
It's possible to adopt other definitions of stability, such as comparing the nuclear binding energies per nucleon. Under this criterion, then it turns out that for the majority of elements the most stable isotope is not the most naturally abundant. You can conveniently check using this Wolfram applet. If you scroll though the elements, you'll notice that the peaks in the two charts very often don't match perfectly, though they're usually close.
One may ask an interesting question - why do these mismatches exist in the first place? I suspect the answer ultimately comes down to the fact that most naturally-occurring isotopes are generated in violent astrophysical processes which are far from equilibrium, combined with the fact that nuclear processes only allow a limited variety of discrete steps up/down in proton and neutron numbers to be made, and sometimes isotopes get trapped because there is no good combination of steps to change them.

Just as in Chemistry thermodynamics says what can happen, but kinetics controls which of those possibilities actually does.

Tantalum-180 famously has an isomer, $_{73}^{180m}\ce{Ta}$, which isn't the lowest energy (nuclear) isomer having that combination of mass number and atomic number. But it has a long half-life ($\ge 4.5×10^{16}$ years). This isomer is only about $\sim 0.01$% of naturally occurring tantalum, the rest being tantalum-181; but it far exceeds the ground-state tantalum-180 isomer which is known only synthetically.


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Simple explanation of the maximum work theorem
I have been trying to understand the maximum-work theorem for some time. But it seems like an impossible task, and it is a very confusing concept. First, the sign convention of work is different in different resources. Further, phrases like "work is maximum" and "delivery of work" do not specify if it is work done on the system or by the system. Is there a way to explain the maximum work theorem in simple terms. What is the physical basis for the maximum-work theorem and what are the implication.

The simplest way to understand the maximum-work theorem is through an example. Consider compression of a gas. First we are going to compress the gas reversible keeping $P_{\text{ext}}=P_{\text{int}}$, and then compare it with an irreversible compression where $P_{\text{ext}}>P_{\text{int}}$. Points to note before the explanation are:

Work done is given by the area under a $P$-$V$ graph.
Compression work is positive.
During compression, work is done on the gas.

Reversible Compression

The work done is given by area $A$.
Comparison with Irreversible Compression

The work done is given by area $A+B$.

Conclusion: It can be seen that $A+B > A$. Thus, work done on a system is maximum, and, as a corollary, work done on a system is minimum during a reversible process, respectively.

Yes, but if the external pressure is greater than that of the system, what is being pushed against? Shouldn't the work be $-p_{system}\Delta V$?

@BuckThorn the definition of work is $-P_{ext}\Delta V$.

That's the definition, but I'm wondering why that is the definition. After all, when $p<p_{ext}$ only part of the external pressure is overcoming the internal pressure, or? Take the extreme case of pushing into a vacuum. Is any work done?

In Europe, the work is positive when entering a gaseous system, For example in a compression, w > 0, and the volume decreases : Delta V < 0. So w = -p·Delta V. As a consequence, the change of internal enegrgy is DeltaU = Q + w = Q  - p·Delta V. On the contrary, USA scientists think mechanically that adding heat to a gaseous system must produce some work, as the gas volume increases by heating, DeltaV > 0. So : w = + p·Delta V, and the change of internal energy is : DeltaU  = Q- w = Q - p DeltaV. As a whole, Delta U is the same everywhere, but w is positive in USA, and negative in Europe.

@Maurice AFAIK it is not regional. Each of conventions was originated in different activity domains. For long time, physics and engineering used dominantly Q - W convention, while chemistry used mostly Q + W  convention. Later physics gradually shifted to mostly Q + W convention, engineering is possibly more conservative.

@Poutnik. Thank you for the complement of information.


text:
How does more s character shorten bonds and strengthen bonds?
I'm aware that having more s character means that electrons are in the lowest energy state (?) and very close to the positive atomic nucleus. But then what’s the connection between having electrons tightly bound to the nucleus and bond length and strength?

related: https://chemistry.stackexchange.com/questions/152/why-is-the-2s-orbital-lower-in-energy-than-the-2p-orbital-when-the-electrons-in?rq=1


text:
How do we judge whether a bond is a coordinate bond?
We know a coordinate bond refers to a covalent bond in which one atom provides a lone pair and the other provides an empty orbital. When OH- gets a proton and becomes H2O, O provides a lone electron pair and the proton (H+) provides an empty orbital; when Na2O is dissolved by water, the oxygen ion has a similar change; also when H2O gets a proton and becomes H3O+. So, are the bonds in H3O+ all covalent bonds? I think if so, the definition is useless, for many bonds can also be coordinate bonds, e. g. when Mg3N2 is dissolved by acid, the nitrogen ion (N3-) gets 4 protons and form 4 coordinate bonds, just like H3O+.

You are correct. Once  formed, a coordinate bond is no different from any other covalent bond.

And in H3O+ and NH4+ all bonds are equivalent. Coordinate/Dative bonding is solely a tool to help electron counting, it has no separate "reality" from covalent bonding.

In the examples You provided, it is true that the bonds can be seen as both coordinate and just covalent depending only on context. However, there are cases in which taking apart a covalent bond cannot result in one of the species retaining a lone pair intact, instead the molecule rearranges. Even though this kind of situation is rare, the definition is not completely "useless", since it helps also describe a process from which a molecule results along with its structure.


text:
Oxidation of compounds vs elements
I was under the impression that when talking about oxidation/reduction, we are referring to specific elements, and the changes in their oxidation numbers.
However, I saw the following reaction referred to as the oxidation of ammonia:
$$\ce{4 NH3 + 3 O2 -> 2 N2 + 6 H2O}\tag1$$
$\ce{N}$ changes its oxidation state from -3 to 0, whilst $\ce{H}$ stays in +1.  It makes sense that we refer to this as the oxidation of nitrogen. Can we extend this understanding and also say that the compound $\ce{NH3}$ is being oxidized?

Is it correct to talk about compounds being oxidized/reduced in this way?

If so, how do we deal with more ambiguous cases where there is both oxidation and reduction of a compound happening? For example:

$$\ce{2 ClO3- -> 2 Cl- + 3 O2}\tag2$$
In this case $\ce{Cl}$ goes from +5 to -1 and is reduced, whereas oxygen goes from -2 to 0 and is oxidized. Is there a rule for which part of the compound you consider when talking about whether its oxidized or reduced? Or do you say the compound is being both oxidized and reduced?
Or do we consider the net movement of electrons/net change in oxidation state?

Oxidation and reduction are also defined in terms of oxygen gain/loss. In a hypothetical reaction like the one below

$$\ce{CS2 + 2 H2O -> CH4 + SO2}\tag3$$
which element do you consider to be the 'main' part of the compound when talking about oxygen gain/loss? Here you have the carbon gaining hydrogen (reduction) whilst the sulfur is gaining oxygen (oxidation). Do you consider carbon to be the 'main' element and talk about the compound as being reduced due to carbon gaining hydrogen, or do you consider sulfur to be the main element and talk about the compound being oxidized due to it gaining oxygen?

Chem+Math formatting reference:
MathJax Basics / Expressions/formulas/equations / Upright vs Italics / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles.

I took the liberty of numbering your examples so they could be referred to.

Your first impression was correct; it is elements that get oxidized/reduced. Talking of compounds as being oxidized/reduced is an abuse of language. This can be allowed when it creates no ambiguity, and can't otherwise.
To your examples:

$\ce N$ get oxidized, $\ce H$ stays as it was. We talk about the compound as a whole getting oxidized, and that's just fine.
$\ce{Cl}$ gets reduced, $\ce O$ gets oxidized. We simply don't talk about the compound as a whole getting either oxidized or reduced. These words are just not applicable here, they would make about as little sense as words like "vandalized" or "glorified". There is no rule to determine which way it would be, because there is nothing to determine. Neither do we say that a compound is being both oxidized and reduced.
This is no different from the previous example. You may assign the oxidation states all right, and see that the two definitions of oxidation/reduction are in fact equivalent. $\ce C$ gets reduced, $\ce S$ gets oxidized, end of story. There is no main element.

So it goes.

There is a historical change about the word oxidation. Before the middle of the $19$th century, oxidation was a reaction of a substance with oxygen. It was similar to "burning". For example glucose $\ce{C6H12O6}$ was oxidized when "burning" with oxygen $\ce{O2}$ and producing $\ce{CO2 + H2O}$. Today we say that it is not glucose, but the carbon atom which is oxidized in the same reaction. In a similar manner, $\ce{NH3}$ was oxidized when burning with $\ce{O2}$. Today we consider that only the nitrogen atom of $\ce{NH3}$ is oxidized in this reaction.
I agree this definition is somewhat ridiculous today. But you should go back to the times when the only experimental parameter available to the first chemists was the balance.
Same thing for reduction. Before 19th century, a reduction was a loss of oxygen atom and a "reduction" of solid mass. For example, $\ce{CuO}$ was reduced to copper atom when reacting with $\ce{H2}$, $\ce{CO}$ or $\ce{CH4}$, because the mass of the solid stuff ($\ce{CuO}$ becoming $\ce{Cu}$) decreases (= is reduced) between the beginning $\ce{CuO}$ and the final $\ce{Cu}$ of the reaction.
Same thing for potassium chlorate. Before the middle of the $19$th century, potassium chlorate $\ce{KClO3}$ was "reduced" to $\ce{KCl}$ when loosing oxygen $\ce{O2}$, because the mass of the solid stuff was "reduced" or decreased during the transformation.

Oxidation / reduction terminology for compounds is still current.  Not in the sense of combustion, but in the sense of other, less extreme kinds of chemical conversions, such as oxidizing ethanol to acetic acid.

@Jojn Bollinger. Thank you for the remark. You are right. I have not thought of this terminology, alcohols being oxidized to aldehydes and then to acids. Thank you.


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Do d-orbital electrons shield each other or not?

"The shielding of one 4f electron by another is less than that of one d electron by another, and as the nuclear charge increases along the series, there is fairly regular decrease in the size of the entire 4f n orbitals."

"One d electron does not shield another electron from the influence of nuclear charge because d-orbitals differ in direction."

Both of the above statements are from the same chapter and are clearly contradicting.One says that d orbitals do shield each other (and that too better than f orbitals), whereas the other says they don't. Where am I going wrong ?
{Reference- ncert.nic.in/textbook/pdf/lech108.pdf page 220, 11th line from top}

Please quote sources for this to be more useful. You might want to check Slater's rules: https://chem.libretexts.org/Courses/Mount_Royal_University/Chem_1201/Unit_2._Periodic_Properties_of_the_Elements/2.06%3A_Slater%27s_Rules

A reference would be very useful because as presented "another" in the first quote is ambiguous. Another 4f electron, or another electron in general?

reference- https://ncert.nic.in/textbook/pdf/lech108.pdf
page 220, 11th line from top


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Why doesn't HBr show allylic substitution in presence of Peroxide?
HBr reacts with alkenes such as propene in the presence of Peroxide as per the Anti-Markovnikov rule. This is an addition reaction, forming 1-bromopropane as the major product.
This happens due to the formation of Br• (Bromine Free radical) formed due to the weak peroxide linkage.
The free radical attacks the alkene to form a stable 2-degree carbon free radical, hence forming 1-bromopropane.
But at high temperatures, we can observe yet another reaction... allylic substitution.
If we react Br2 with propane at very high temperatures, we get 2-bromopropene (an allylically substituted product)
This is because when Br• takes away H• in the form of HBr, it leaves a resonance-stabilised allylic carbon free-radical.
My question is, why doesn't this happen in the case of HBr in the presence of peroxide? It follows the same mechanism of free radicals, why doesn't it have an allylic substituted product?
(I have a few ideas... I guess allylic substitution happens at high temperatures because it forms less stable product... and is hence endothermic. At room temperature however, endothermic reactions like that would not be favoured... that does make sense to me... just need some more clarification)

I just realised...
(I would find it helpful if someone confirms that I am on the right track)
HBr while doing allylic substitution would form propene itself
Br radical would take away allylic Hydrogen as HBr, leaving behind the allylic carbon free-radical. This would in turn take up Hydrogen from HBr (since H radical is less stable... so that is to be reacted first) forming propene again, which yet another Br radical...
So for any change to be observed, either both radicals would have to be Br (Br2 for instance) or you would simply do an addition. (or I reckon there is a very minor product of allylic substitution during reaction with HBr if somehow Br radical reacts first.)


text:
Can someone help me name this molecule?

I'm having a hard time deciphering it's IUPAC name like I have 4-butanoxy and 3-propanoxy so far, but how do you put it all together with the double-bonded oxygen?
Thank you and really sorry for the trivial question here.

This is an acid derivative. Try to identify the acid first.

The motif in the center is carbonic acid, hence your compound is a carbonate as for instance dimethyl carbonate. (Perhaps the BOC group is known to you as a protection of amines, too.)
To the left of the acid you have a butyl rest, to the right an ethyl.  You sort them alphabetically, and hence name the compound butyl ethyl carbonate, CAS 30714-78-4.

The CAS name list also allows ethyl butyl carbonate, without the alphabetical sorting.

Thank you very much! This was really helpful :DD

@OscarLanzi Perhaps equally allowed (as on the CAS site itself), true. On the other hand, a methyl and an ethyl group on atom 3 of a pentane - this is (preferred?) as 3-ethyl-3-methyl-pentane.  The brief 4-page abstract of the Red Book names MEK as as ethyl methyl ketone, too (page 4, right hand column).

@idontknowhowtocode The Red Book (and other material about chemical nomenclature) is compiled in the resources page.


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Neglected partial pressures dependence of water splitting
I am studying electrochemical water splitting and I have a huge doubt. The standard redox potential for the total reaction is 1.23 V, but this is in standard conditions, where the hydrogen and oxygen partial pessures are 1 bar. Now, if I stick two metallic electrodes in an open pool of water and apply a voltage between them to make water splitting, isn't the redox potential required (even not considering the overpotentials that have to be added to make the kinetics faster and compensate resistive losses) going to be different because of the contribution of the partial pressures of the gases that are not the standard ones, but the ones in the real atmosphere? Or am I getting something wrong? Do they compensate this somehow in electrolyzers? But then why do many experimental groups studying materials for water splitting simply put the electrodes in an open becher? I made a rough calculation using Dalton's principle for their partial pressures and the contribution to the Nernst equation due to the partial pressures of these gases doesn't look negligible compared to the standard redox potential, as the actual partial pressure of hydrogen in the atmosphere is a lot smaller than 1 bar. Thanks in advance for the answer

Ongoing electrolysis very quickly establishes the dissolved gas activities corresponding to the external pressure that is very close to the standard one. So the major deviation from standard potentials are overpotentials (especially oxygen one) and nonstandard ion activities. // Commercial electrolyzers control rather the optimal current, and adjust the electrolytic voltage accordingly, ignoring assumed theoretical values.

In a real electrolysis setup, the electrode lives in its own cloud of bubbles filled with the corresponding gas.
The gas in the bubbles has almost atmospheric pressure because of hydrostatic reasons and is either pure hydrogen or pure oxygen.
One may assume the bubbles being in near-equilibrium with the dissolved gas in the water around the electrode, but pretty much not with the atmosphere. Bubbles are one-way transport.

If you try to really run the electrolysis as a reversible process...
Well, you can abuse the same equation and calculate how much hydrogen partial pressure we could have at 0V between the electrodes and 20kPa oxygen pressure. Because at 0V one does not really need electrodes, this is the lowest possible hydrogen we could have in the atmosphere (we have somewhat more because we do have other sources of hydrogen around)

Thanks! What you said makes a lot of sense in my head. So, can I imagine that in the first instants of the reaction the potential required would be different because I still have no hydrogen and oxygen bubbles, then I change it gradually until the partial pressures are in hydrostatic equilibrium with the external atmosphere? Or did I get it wrong?

The second part of my answer implies that in the first instants the potential required will be virtually zero. It will go up as both hydrogen and oxygen build up around the electrodes and will stabilize as the concentration of both gases rises to the point of forming bubbles and growing these bubbles against the pressure of the surrounding liquid.


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Can phase tranformations occur with change in temperature?
I know that, usually, phase transition occurs with a fixed temperature. Is there a case where this is not followed? Can phase transformations occur with changing/variable temperature?

Second order phase transitions do not have an enthalpy associated with them.

Yes, phase transformations can occur with changing temperature. Consider the example of a liquid-vapor equilibrium in a closed system.
Since the system is closed, heat input definitely increases temperature. Further, more of liquid gets converted into vapor, which is a phase transformation.

The reasoning is that the pressure is changing. For a phase transition to occur at constant temperature, the pressure must remain constant.

Or to occur at constant pressure the temperature remain constant. The Gibbs Phase rule [Freedoms = Components - Phases + 2] explains it nicely one component 2 phases gives one degree of freedom. fixing T or P makes the system invariant.

At a fixed total pressure, phase transformation can take place over a range of temperatures for a multicomponent vapor-liquid equilibrium system.


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How did Sinclair Oil put "liquid nickel" into gasoline, and what were the chemical mechanism(s) by which it was supposed to reduce engine wear?
The April 22, 2022 Veritasium video The Man Who Accidentally Killed The Most People In History is worth a watch, and was the basis for the question Why would tellurium + sodium hydroxide have worked as a good anti-knock gasoline additive (if it wasn't so smelly)?
But today I've come across a different heavy metal additive to gasoline!
I fondly remember the Sinclair Oil dinosaur statues and the free green dinosaur soap we sometimes got with a fill-up. But this circa 1964-1965 (from the reference to the 1964 World's Fair at the bottom) advertisement mentions a new "liquid nickel" additive to gasoline that "reduces engine wear as much as 29%".
Question: How did Sinclair Oil put "liquid nickel" into gasoline, and what were the  chemical mechanism(s) by which it was supposed to reduce engine wear?
I've proactively added the organometallic-compounds tag; if lead was introduced in its tetraethyl form, maybe nickel was made soluble with organics as well (not sure about that tellurium in the linked question though).

From the advertisement:

Liquid Nickel in Sinclair Gasolines is an exclusive Sinclair discovery that:

Reduces engine wear as much as 29%
Combats harmful engine deposits
Restores lost power in old cars
Keeps new cars at peak performance
Sinclair and only Sinclair has Liquid Nickel

Source: US National Park Service's Tourism and Dinosaurs

From a web search I get this quote "deposit modifier used in leaded gasoline was nickel isodecylorthophosphate, which was introduced by. Sinclair Oil" cited to Gibbs, JM "Gasoline Additives - when and why" Vol. 99, Section 4: JOURNAL OF FUELS & LUBRICANTS (1990), pp. 618-638 (21 pages)
Published By: SAE International,

I didn't make that into an answer because I don't have full text access to read the cited reference beyond that short excerpt

Following up, a primary source from Sinclair researchers is Schoen, W. and Pontious, R., "New Gasoline Additive Improves Engine Performance, and Durability," SAE Technical Paper 650941, 1965, https://doi.org/10.4271/650941.

@Andrew I won't be able to read journal articles until Tuesday, but I will definitely check these out when I can. Thanks for your answer post!

In a 1965 paper [1], researchers from Sinclair described the properties of the gasoline additive nickel isodecylorthophosphate. In their words

The outstanding performance features of this nickel additive can be
summarized as follows:
1 . Reduces abnormal deposit-induced ignition phenomena, such as
rumble and wild ping, both in the presence and absence of other
deposit modifiers.
2 . Reduces equilibrium octane requirements.
3 . Modifies combustion chamber deposits without adding to the total
amount of deposit accumulation.
4 . Reduces engine wear.
5 . Prolongs exhaust valve life.
6 . Functions as an effective carburetor deicer.
7 . Functions as a good rust inhibitor.

They don't describe the mechanism by which the deposit modification occurs, though they do attribute the carburetor deicing and rust inhibition effects to the surfactant properties of the additive, and they speculate that the extension of exhaust valve life is to due to the nickel compound "softening" depositions on the valves so that they come off more easily and do not build up.
The introduction to the paper also describes the fortuitous discovery of the effects of alkyl metal salts as resulting from addition of amine alkylphosphate surfactants for deicing and rust inhibition. It was observed that the surfactants reacted with galvanized surfaces in the presence of water and produced a gelatinous precipitate, which was identified as comprised of alkyl zinc compounds that were soluble in gasoline. Concerned that these compounds might have a detrimental effect on engine performance, Sinclair researchers added them to gasoline and found, to their surprise, that there were actually benefits. This prompted a screening of gasoline-soluble metal compounds, from which nickel isodecylorthophosphate emerged as the best candidate for commercialization.
[1]Schoen, W. and Pontious, R., "New Gasoline Additive Improves Engine Performance, and Durability," SAE Technical Paper 650941, 1965, https://doi.org/10.4271/650941.

Another newspaper advertisement tells us the US Patent Number 2,937,992.

The patent tells us that the nickel compound is an oil-soluble nickel dithiophosphate:

A particularly suitable nickel dithiophosphate which can be employed with advantage is that prepared from a di-2-ethylhexyl dithiophosphate obtained by reacting a technical mixture of 2-ethylhexanol with phosphorus pen tasulfide. The nickel dithiophosphate is employed in my improved lubricating compositions in a small or minor amount sufficient to inhibit oxidation and bearing corrosion and in most instances the amount used provides about 0.03 to 0.2% by weight of phosphorus on the basis of mineral oil in which it is incorporated.

The preparation of the nickel di-2-ethylhexyl dithiophosphate base is included:

Basic nickel carbonate (2585 g.) was added portion wise to 13,375 g. of dry di-2-ethylhexyl dithiophosphoric acid at 80 to 132 F. over a period of 3 hours. The solution was dehydrated under vacuum (20-30 mm.) at 185 to 200 F. to give, after filtration, 12,230 g. product, pH 6.8, and acid number (D-974) 20.9.

The patent also presents laboratory engine tests.

This is very interesting, thanks! I'm glad to hear that Sinclair Oil cares about me enough to supplement my heavy metal exposure! :-) The advertisement in the question includes "Liquid Nickel in Sinclair Gasolines..." rather than motor oil, and shows drawings of gasoline pumps with "Nickel" written on them. I would not be surprised if they moved their "liquid nickel" concept from gasoline to oil at some point, but did you happen to come across any information on a gasoline additive?


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Mechanism for elimination in 3-Chloro-1,4-cyclohexadiene
So, I came across this reaction as an example of E2 elimination.

Our teacher told us that this is what we call E2' reaction, as in, the $\ce{OH^-}$ takes the terminal $\ce{H^+}$, followed by resonance and then removal of $\ce{Cl^-}$.
This didn't really make sense to me at first by how he has taught us the mechanism for E2, which is through concerted single step $\beta$-elimination, while this one seems like $1,4$-elimination, and at least three steps by the path he told us.
I tried to make it make sense by the following pathway:

Question: Is this correct? What exactly is happening here? This doesn't seem like resonance to me as that doesn't involve partial charges, I think. Does this happen for even longer chains, like $1,6$ or $1,8$ eliminations?

This is essentially correct. Remember that there is a big energy gain from aromatisation - you are pushing rocks downhill

@Waylander Thanks for the comment. I do understand the compound would gain a lot of stability, relatively, due to aromatic ring. Are there any issues with the mechanism I've proposed?

Mechanism looks OK

@waylander Thanks!

Yes the mechanism is correct. When ever a nucleophile is already present in a compound then its is favourable that intramolecular attack would occur as they are faster than intermolecular attack. In this case the pie electron attacks and removes chlorine and as you have taken a strong base simultaneously protonation will occur in one step. So it is like E2. If you had taken a weak base then the reaction would occur through carbocation path that is
1)SNGP(Neighbouring group participation) then
2)E1
And about longer chains is that longer the chain the intramolecular attack can be slow. Or It is possible that ring formation takes place like 6,5,3 membered ring according to the substrate.

I would assume that the molecule is deprotonated and then loses the chloride ion to give the aromatic product.
Either caroon atom 3 (with the chlorine) or carbon atom 6 (with only a pair of hydrogens attached off the ring) could be reversibly deprotonated; either way a conjugated pentadienide carbanion is formed with the negative charge formally distributed over three atoms. But to go on to the aromatic product there must be a good anionic leaving group from the other one of the two carbons mentioned above. Deprotonating C-3 means a hydride ion has to leave from C-6, which is not favored so the deprotonation ends up getting reversed. Deprotonating C-6 allows C-3 to release its more favorable chloride ion and thus enables formation of the aromatic product.


text:
Why is picric acid not more steam volatile than p-nitrophenol, if intramolecular bonding results in increased steam volatility?
I understand that relative to p-nitrophenol, o-nitrophenol would be more steam volatile (as p-nitrophenol can form intermolecular hydrogen bonds), but my question lies in the fact that its volatile nature is often attributed to its ability to form intramolecular H bonds. How does this contribute to steam volatility?
Also, assuming that it is true that steam volatility is a result of intramolecular H bonding, then picric acid would be more steam volatile than p-nitrophenol, which is not true (as it has a higher boiling point). Doesn't this result directly contradict the assumption?
Just a quick  note, my question doesn't relate to the actual process of steam distillation (was closed b/c it was similar to something related - I'm just curious about why picric acid is not more steam volatile than p-nitrophenol.)

@Mithoron it doesn't actually, my doubt is regarding the effect of H bonding on steam volatility, not the process itself. thanks though! :)

Intramolecular H-bonding decreases availability for intermolecular H-bonding and general polar bond interactions, making the molecule effectively less polar and more volatile.

@Poutnik That makes sense - thanks! I was curious as to how that'd explain the higher boiling point of picric acid when compared to p-nitrophenol? I'd imagine picric acid to undergo intramolecular H bonding and thus, have a lower b.p. and greater volatility.

Picric acid has a much large molecular weight, 229 vs 139 g/mol. That's enough to justify the difference, but that's just one factor. The example of intramolecular H bonding illustrates the potential importance of this effect but it is only another of many factors.

I think it kinda does, you just didn't see it. H-bonding is like last thing that's important, you need to know basics first - why it does anything, which still is not that much.

Ah alright, I guess I should've considered other factors as well, kinda got stuck in the whole H bonding idea. Thanks a lot everyone!


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When light shines on a material, how much of the light interacts with the electrons and the nucleus?
I heard for certain materials, about 99% of the light interacts with the electrons. I heard it also depends on whether it is UV light, IR light, etc. Does the proportion increase for UV but decrease for IR? Is there a formula to derive this?
Similarly, how much of the light will interact with the nucleus? And what happens to the nucleus when light makes contact with it, preferably of lower wavelength energies?

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

Consider that at visual frequencies,

The wavelength of light is, very roughly, three orders of magnitude larger than atoms.
The energy in visible light, very roughly, one eV, is about three orders of magnitude less than needed to disrupt the nucleus.
The mass of nucleons is, again, very roughly, three orders of magnitude greater than that of electrons, yet a protons has the same charge as an electron.

Imagine visible light washing over the whole atom, illuminating all at once. The heavy nucleons barely are affected, but an electron can be raised to a higher energy level. If an electron absorbs a quantum of energy, it is not split evenly among the other particles, but at that instant of observation becomes defined in space.
Ah! But what if the "light" had more energy? Brittanica.com states,

Absorption of gamma rays by nuclei can cause them to eject neutrons or
alpha particles or it can even split a nucleus like a bursting bubble
in what is called photodisintegration.

When light shines on a solid surface, what % of light hits the electrons and what % hits the nucleus?

If we replace "hit" with "scatter from or become absorbed by" and restrict ourselves to visible light, then there is a better way to look at the question.
Light is an oscillating electric field and it's definitely the electrons that are the culprit when something happens to visible light.
We can split electrons in solids up conceptually into two groups; "bound" and "free". This is definitely a simplification but it's a been a very productive one.
Once we do that, we can think of materials as dielectrics, metals, and semiconductors/semimetals. They differ by the density of free electrons.
In dielectrics (think glass, salt crystals, etc.) light can travel a long way before being absorbed. The bound electrons do wiggle back and forth in the oscillating electric field, and that is what makes light move more slowly (think index of refraction) but they usually don't scatter or absorb it. Impurities, defects ("color centers" or F-centers) or other things that perturb the ordered state of bound electrons gradually scatter or absorb light as it passes through (think obsidian (dark  glass) or colorful precious gems).
In metals with their very high density of free electrons, light doesn't stand a chance! Within a very thin layer at the top (a few to a few tens of nanometers, much shorter than a wavelength) the "free" electrons run back and forth in response to the electric field of light, setting up oscillating currents. These exactly cancel the forward going beam so that nothing propagates forward, and at the same time produce an identical beam of light going back out into vacuum. This is what is happening during reflection from metal surfaces.  If the frequency of the light is too high though, there may not be enough electron density to do this, and a single crystal of the metal becomes transparent in the UV.
For more on that see:

Born & Wolf; Alkali metals transparent to UV? Cesium transparent to blue?

The highest frequency that can be reflected for a given free electron density is called the plasma frequency and this concept applies to anything with a significant density of free electrons, including our ionosphere, whose plasma frequency varies between about 15 and 50 MHz depending on what the Sun is doing. Ionosphere; Layers of ionization. Lower frequencies are reflected back to Earth (so we can listen to a short wave radio station thousands of km away) and higher frequencies are transmitted (so we can listen to deep space spacecraft, GPS, and Starlink internet).
In semiconductors things get very interesting and we can play all kinds of games with the band gap. The energy of visible (and near infrared) light can be absorbed in a few microns in silicon because it can kick an electron up from one band to another. But the carrier density is low enough that those can freely be pulled away from each other and collected as charge before scattering off of other carriers. So we have CCD imagers, yay!
The reverse process (charge in, light out) gives us LEDs and semiconductor lasers which revolutionized several fields of science, technology and communications.

semiconductors can also give solar cells where the light is effectively converted to electricity.

@uhoh I only added that because with my reputation level I can only suggedt edits with at least six character changes. I had initially thought that this was the correct way to pluralize acronyms in English, but after your comment I did some searching and learned I was wrong with that. Sorry :(

@GeekyGuy no problem, and thanks for all the useful edits! I really wasn't sure either, I'm a major apostrophe offender; I always write "let's" when I should write "lets". I've been that way my whole life, it's hard-wired.

Matter is energy trapped in a field or fields that give the energy certain properties that we manage to measure or at least acknowledge. Photons seem to be pure energy, however, they have properties that are associated with mass: Kinetic energy associated with length or possibly time of traverse, momentum, an electric and magnetic field, that crazy thing of polarization or entanglement; they are complicated little, or not so little, devils.
Photons encountering matter interact with all the fields that define that matter. If a specific change in an energy level is possible within the time frame of the interaction it might happen. We work with the differences in energy of the fields that define matter not the absolutes; the absolutes are the baseline. A photon does not hit the electrons or the nucleus; it interacts with all the fields defining the matter and effects changes where the energy fits. The photon does not enter the atom looking for an electron to dropkick to a higher level. It perturbs the fields defining the atom; the atom etc. then relaxes into its lowest available quantum state. If none are available, the photon proceeds on its way.

Edit: I've replaced occurrences of the word "hit(s)" with "interact(s) with" as the former term is not to the liking of those who prefer using perfect terminology.

All the radiation interacts with the entirety of the atoms. Think of it as a spatial wave of energy washing over you.
The incoming radiation is absorbed as internal energy, and part of it maybe reflected (or refracted) by the atoms vibrating in sync with the radiation. From there, some of the absorbed radiation gets emitted, and the rest remains as heat. You can read more about what happens to absorbed light from this answer.
If the radiation is high enough, it can result in complete emission of electrons (photoelectric effect or ionizing radiation) or of nucleons (photodisintegration).
The level of excitation or emission of particles is defined by the binding energy of the particle system.
It can be energy gap between electron shells, or excitation energy (for excitation), the threshold energy (for photoelectric emission, in the solid state), the ionization energy (for emission of electrons in the gaseous state), average nuclear binding energy per nucleon (for emission of nucleons), etc.
The structure of the material may also affect the orbital excitation energy in more complex non hydrogen-like molecules due to quantum decoherence.
The answer to your question, "what % of light hits interacts with the electrons and what % hits interacts with the nucleus" is all of it.
But how much is distributed between the electrons and the nucleus is more complicated, and is described by the equations of quantum mechanics. I will quote from the answer I linked earlier:

Heating applies to the whole system (nuclei and electrons). However, due to the large difference of mass between nuclei and electrons (larger than 1800), and taking into account that the correct description of the dynamics at atomic level is provided by quantum mechanics (QM), the way a gain of total energy is distributed among different internal degrees of freedom is quite complex.

Additional Info
The general types of interactions between light and matter involve:
At Lower Energies (At the atomic level):

Absorption (excitation), Reflection, Refraction, Emission (of light)
Photoelectric Emission
Mie Scattering (Coherent and elastic scattering with a forward directionality, for particles sized similar to the wavelength)
Rayleigh Scattering (Coherent and elastic scattering due to polarization, and is more wavelength dependent than Mie Scattering)
Raman Scattering (Incoherent and inelastic scattering due to absorption by molecules)

At Mid Energy Levels (At the single electron level):

Thomson Scattering (Coherent and elastic scattering from a free electron by the electric field; Classical low energy limit of Compton Scattering, caused by accelerating the electron)
Compton Scattering (Incoherent and inelastic scattering from a free or weakly bound electron, caused by absorption and emission by an electron)

At Higher Energies (At the nuclear level):

Pair Production (Emission of a particle-antiparticle pair after absorption by a nucleus)
Photodisintegration (Emission of nucleons, mainly neutrons, but also protons or alpha particles)
Photofission (Nuclear fission)

By mid-energy levels that includes IR and UV? Or is IR for lower-energy levels?

@NealConroy well in this case, lower energy would include UV. Compton Scattering was observed for X-Rays. Thomson scattering is the limit of Compton scattering below ~1.2 EHz. And Pair Production is observed beginning with Gamma Rays. The mass of an electron is 0.511 MeV, so we need twice that to form a pair, so 1.022 MeV (~247 EHz). At this point we start referring to energy in terms of eV and not Hz. X-Rays are 100eV - 100 KeV, and Gamma Rays are upwards. Ionizing Radiation begins at around 10 eV - 33 eV in short-wave UV (UV-C / Far-UV).

Ok, so lower energy levels wavelength almost has no reaction with when it hits the nucleus, only at high energies?

It does not "hit the nucleus".

@jimchmst sure, the orbitals could carry the wave around, but what about high frequency radiation that removes nucleons?

@NealConroy It should also be noted that there is a nuclear shell model, just like the electron shell model, and the nuclear shell can absorb or emit gamma rays, when nucleons transition shells. So that's something to think about when you say radiation "hits" the nucleus. Sure it does not "hit", but it does "interact" at least. I don't see why many people are pointing it out.


text:
Removing DCC and its by-product from Steglich Esterfication
In my small research, I want to synthesize 3 alkyl gallates (ethyl gallate, butyl gallate, and hexyl gallate), and here's my overall reaction using Steglich esterification):

However, from reading some literature, it's mentioned that using DCC in Steglich esterification has drawbacks; it produces a by-product (DCU) and can prove difficult to remove remaining trace amounts.
My questions:

Are there any effective methods to remove the DCC by-product and the remaining trace amounts? If there are, could you tell me in detail about the methods?
Otherwise, could you suggest me a different way/reaction to synthesize alkyl gallate (using easily available chemicals)

Important Note:
I'm aware that I could use DIC or EDC to prevent this problem, but I don't have DIC or EDC, and I'm trying to avoid purchasing them because they're quite expensive in my country.

According to this discussion thread https://www.researchgate.net/post/How_can_I_remove_DCU_dicyclohexyl_urea_by_product_formed_during_estrification_reaction_via_DCC_DMAP_with_carboxylic_acids   dissolve your product mixture in the minimum amount of cold MeCN and filter to remove DCU. Do this a couple of times.

You'd really use THF at 0C? Why not DCM at RT? If you pick right solvent DCU simply precipitates out. Also higher temp. makes reaction faster and lowers risk of oxidation - you'd need to keep O2 off.

Ah, you make a valid point. I chose THF as the solvent and set the temperature to 0°C based on the method outlined in this publication (http://dx.doi.org/10.13005/ojc/340207). I think I've become too reliant on this method.
But, could you elaborate on the risk of oxidation? Like, is gallic acid easy to oxidize in RT or what?

Higher temp. means faster reaction. At RT it still could take a week, but for some reason your procedure is faster. Looks like gallic acid is insoluble in DCM, though.


text:
How to draw endo-bicyclo[2.2.1]hept-5-ene-2,3-dicarboxylic anhydride on ChemDraw?
I'm struggling to draw endo-bicyclo[2.2.1]hept-5-ene-2,3-dicarboxylic anhydride on ChemDraw. I currently have this:

How can I draw it in the chair conformation side on view? I know what it is supposed to look like, just not sure how to draw it on ChemDraw.

First of all, there is no chair conformation in this bicycle framework. It is always have a boat-confirmation with the two edges of the boat to make the bridge head. Following is the step-by-step guidance to draw exo-form of the compound on ChemDraw:

On side panel of the ChemDraw file, you can find various different bicyclo frameworks. Carefully choose the correct bicycloheptane and click on the drawing space. Then follow the instruction to get what you want.


text:
What is the correct way to write sodium tetraphenylborate formula?
In wikipedia, it is written in two different ways.
(C6H5)4BNa and NaB(C6H5)4
Is there not a unique criterion? Or is it chemistry-brach-dependant?

More than two, in that article: SMILES shows bonding. [B-](C1=CC=CC=C1)(C2=CC=CC=C2)(C3=CC=CC=C3)C4=CC=CC=C4.[Na+].
A brief online survey shows [Alfa, Chemical Book, Fisher, Sigma and NIH prefer metal-last, though most also show the SMILES string.
A few outliers, such as Enmore, Grainger and WebQC, still use the metal-first layout.
The answer might be, "The preferred format is the one chosen by the instructor or administrator."


text:
Reaction Mechanism for tert-butyl alcohol and an aldehyde
Consider the reaction between tert-butyl alcohol and acetadelhyde in the presence of $\ce{HBr}$ in aqueous solution:
$$\ce {(CH_3)_3COH(aq) + (CH_3)CHO(aq)} \longrightarrow ?$$
Here is my assumed solution to the reaction mechanism:

Essentially, the $\ce{H+}$ ion removes the alcohol from the tert-butyl alcohol allowing the acetadelhyde to bond with the carbocation. In this process the double bond between the oxygen and a carbon becomes a single leaving a positive region on the respective carbon. This allows the bromide ion to attach forming the molecule in the last step.
Is the product correct? Is the mechanism to get to this product correct?

Mostly, but it's not the product. This is acetalisation, and some bromide is intermediate, at best.

I guess this photo might help? Just that it's HCl and HBr also might work...

This one is the mechanism...

Basically from what I saw @Mithoron, the HCl gas protonates the oxygen of the carbonyl compounds and increases its electrophilicity, thus facilitating the nucleophilic attack of the alcohol. Acetals and ketals are hydrolysed with aqueous mineral acids to yield corresponding aldehydes and ketones respectively. So HBr also would do the same thing...

@HarikrishnanM Not HCl gas, perhaps in solution, but for something like that it would be H2SO4.

In my NCERT reader, its given as gas... here

https://chemistry.stackexchange.com/questions/135162/does-pure-hcl-without-water-is-an-acid

In short, the mechanism posted by the OP is not correct.
Over here, an aldehyde is being combined with a tertiary alcohol, and the reaction is acetalisation.
We'll go through the main crux of the reaction, and the detailed mechanism.

The reaction:
Aldehydes undergo nucleophilic addition reaction in dry HCl gas (HBr also can be considered) with monohydric alcohol to form a hemiacetal. Being unstable, hemiacetals further react and form a gem-dialkoxy compound called acetal.

Mechanism:
HCl gas protonates the oxygen of the carbonyl compounds and increases its electrophilicity, thus facilitating the nucleophilic attack of the alcohol...

followed up by the nucleophilic attack,

and the deprotonation.

Further reaction takes place due to the instability of the acetal (thermodynamic reasons).

This would be sufficient for the OP to draw a conclusion and write the final mechanism.

Sources:

Chemistry, Class 12, NCERT
Acetals


text:
Reason for differences in AChE aging/reactivation kinetics between neutral and charged organophosphorus inhibitors
According to the referenced papers, acetylcholinesterase that has been inhibited by an organophosphate possessing thiocholine as its leaving group is more susceptible to reactivation by oxime compounds compared to AChE inhibited by the neutral analogs of these compounds. One of the authors writes:

Sundwall (1961) reported that P2S was considerably more effective as
an antagonist of the lethal action of 37 SN
+(isopropoxy-2-trimethylammonio-tiomethylphosphino oxide) in mice than it was of those of Sarin and 37 S-N
(isopropoxy-2-dimethylaminothiomethylphosphine oxide).

The author goes on to propose that the reason for this observation is that the phosphonyl group is oriented differently in AChE inhibited by the charged compound than in AChE inhibited by the neutral analog.
My question is: Is there any hard evidence that indicates that this proposed explanation is correct, such as crystallographic data?  Are there any alternative possibilities?
References:
BOSKOVIC, B. (1981). The treatment of soman poisoning and its perspectives. Fundamental and Applied Toxicology, 1(2), 203–213. doi:10.1016/s0272-0590(81)80059-0
BošKović, B., Maksimović, M., & Minić, D. (1968). Ageing and reactivation of acetylcholinesterase inhibited with Soman and its thiocholine-like analogue. Biochemical Pharmacology, 17(8), 1738–1741. doi:10.1016/0006-2952(68)90237-2


text:
Mechanism of oxidation reaction with periodate
What is the mechanism of reaction of periodate with β-hydroxy amine? I tried the to write the mechanism similar to that of reaction with 1,2-diol but I am not able to obtain 2 carbonyl compound and like that in case of diol neither evolution of ammonia.
Here's my work:

Your mechanism is good but you need to remember that the imine product you have drawn will hydrolyse to the carbonyl compound in acidic conditions as you are using periodic acid and there is water present.

Yeah right I didnt think of that thanks


text:
Copper reaction with nitric acid
According to this paper, the green colour from the reaction of copper and nitric acid could be explained:

The color is greenish, probably as a result of a combination of the
pale-blue color of the traces of dissolved salt and the pale yellow
color of the acid. Another reason for the color difference might the
absence of water in the system; hence, no aqua complexes are
formed.

The deep green colour arises when copper contact with nitric acid solution, I think the colour might be due to copper complexes. After consulting my teacher, he said that when nitric acid solution is concentrated, the main component would be [Cu(H2O)4(NO3)2] which is responsible for the colour. I wonder if this is true?

The shift from blue to green could be also caused by dissolved NO2.

My friend once conducted electrolysis of CuSO4 experiment. The cathode turned green also. The device was new and CuSO4 solution was prepared from CuSO4.5H2O. Couldn't explain it

@Shira your green cathode is probably not the same as the copper nitrate in nitric acid. Copper plating at the cathode could be accompanied by reduction of water to hydroxide ion thus precipitating copper hydroxide. Industrial copper platers add acid when plating from a sulfate bath.

Well but copper hydroxide is not green, isn't it? Might have been some faults while experimenting

Blue to green from what I read.


text:
Is methylpropane actually butane?
My notes from my teacher says that 1- chloromethylpropane can be prepared from methylpropane under sunlight with chlorine.
However I am confused as I think there is no such thing as methylpropane because it is actually butane.

https://en.wikipedia.org/wiki/Isobutane#Nomenclature

Ironically if you Wikipedia'd methylbutane, which takes you straight to isobutane, you would have saw the difference, making the 2nd part of your question unnecessary.

Obviously not: it is dimethyl ethane. ;-)

The bottom molecule is butane and the top is methylpropane.
The suffix -ane implies a molecule containing only carbon-carbon and carbon-hydrogen single bonds, which can be seen in both compounds.
The prefix but- implies that the longest continuous carbon chain in the molecule is four carbon atoms long. Note that every carbon atom in butane constitutes such a chain (which is why there is no need to mention substituents - there are none)
However, if you look at methylpropane and try to find the largest continuous chain, it is impossible to include all four carbon atoms. As evident in the numbering, the longest such chain that can be produced is 3 carbon atoms long, (as the fourth carbon atom is always attached to the second one in chain - this is important)  which is indicated with the prefix prop-.
Therefore, we need to add the prefix methyl- to indicate a substituent of one carbon atom bonded to three hydrogen atoms. We usually add a number (such as 2-) before this prefix to indicate which base chain carbon atom the substituent is attached to. In this case, it is not required. To see why, think about how the length of the longest continuous carbon chain would change if the fourth carbon atom was instead attached to the third or first in the original chain, and how this change would affect how we name the molecule.


text:
Temperature dependence of enthalpy in electrochemistry
I've been trying to use Kirchoff's law to determine the difference in the enthalpies of electrochemical reactions at different temperatures for a general electrochemical process. So here's my progress:
$$\Delta_r G = -nFE_\mathrm{cell}$$ And, $$\mathrm{d}G = V \mathrm{d}P - S \mathrm{d}T$$
Considering the reaction to be occurring at moreover constant pressure, the first term vanishes.
Hence $$\Delta_r S = nF \left(\frac{\partial E_{cell}}{\partial T} \right)_P $$
$$\Delta_r H = - nFE_{cell} + nFT \left(\frac{\partial E_{cell}}{\partial T} \right)_P$$
Since $$ \partial (\Delta_r C_{p,m})= \frac{\partial}{\partial T} (\Delta_r H)$$
I obtained;
$$\Delta_r C_{p,m} = nFT \left( \frac{\partial ^2 E_{cell}}{\partial T^2} \right)_{P}$$
But this seems to imply there is no dependence of $\Delta_r C_{p,m}$ on temperature at all since the Nernst equation implies only linear temperature dependence. So, is there any hidden assumption made during this derivation or is it true that for all electrochemical reactions at constant pressure, specific heat, and hence reaction enthalpy, does not depend on temperature at all?  (I'm a new student to electrochemistry so please pardon me if I made any notational errors)

Note that the Mathjax \partial is used for partial derivatives. For exact differentials, there is used upright d,  like $\mathrm{d}G$  for $\mathrm{d}G$. For nonexact differentials of heat or work, there is used Unicode character "dyet" or as a workaround, \delta as  $\delta W$ for $\delta W$.

I will keep that in mind next time onwards, thank you for bringing that to my attention

The Nernst equation implicitly assumes the standard potential (and therefore the general potential too) is function of temperature.
$$\left(\dfrac{\partial^2E^{\circ}_\mathrm{cell}}{\partial T^2}\right)_p \ne 0 \implies \left(\dfrac{\partial^2E_\mathrm{cell}}{\partial T^2}\right)_p \ne 0 \implies \Delta_r C_{p,m} \ne 0$$
The explicit linear dependence of a potential on temperature at the logarithmic term $\dfrac{RT}{nF} \ln{Q}$ describes only potential changes due changing activities of involved compounds. For the given change of activities at constant temperature, the impact on the potential is proportional to absolute temperature.


text:
Number of state variables in the general expression for a state function
In the tenth addition of Physical Chemistry: Thermodynamics, Structure, and Change by Peter Atkins and Julio de Paula, it is given, in section 2E, that the general expression for the differential of internal energy is given by
$$
dU = \left( \dfrac{\partial U}{\partial V}\right)_T dV + \left(\dfrac{\partial U}{\partial T}\right)_V dT
$$
This would imply that $U = U(V, T)$. However, isn't it true that the general expression of a state function depends on, at least, three state variables?

I'm not quite sure what you are getting at by "the general expression of a state function depends on, at least, three state variables". Are you effectively asking why there is no pressure dependence for U in the above? Or dependence n the number of molecules in the system? Or something else? It would be good if you gave an example of what you think is a "general expression" and a reference as to why you think that is the case.

@IanBush it could be either $P$ or $n$, but, afaik, there has to be a third variable.

Please, a reference to support the assertion that the general expression needs 3 variables. I'm not saying you are wrong, its just i cant remember it being specified as such. And In the case of P in the above the equation of state can always in principle be used to eliminate it from the functional form for U.

The third state variable is $n$, the amount of matter, in moles

The definition here is for a pure substance in a closed system.

At constant n Defining or measuring T and P or V defines the other.

You are right, there has to be a 3rd variable, and it is usually mass or particle number. Even for an ideal gas there are 3 independent variables in its thermal equation of state $pV=\tfrac{m}{M}RT$.
When you express the internal energy as function of extensive variables, say volume, entropy and mass as $$U=U(S, V, m) = TS-pV+\mu m \tag{1}$$ then $$dU=TdS-pdV+\mu dm \tag{2}$$ and $$T=T(S,V,m)=\frac{\partial U}{\partial S}, \quad p=(S,V,m)=\frac{\partial U}{\partial p},\quad \mu=\mu(S,V,m)=\frac{\partial U}{\partial m}.$$ Usually, for a closed system, it is assumed that the mass (particle number) is kept constant, $dm=0$, and then you can write $dU=TdS-pdV.$
That ignoring the 3rd variable can lead to nonsensical statements even for a closed is quite evident when you consider the Gibbs-Duhem equation which is a consequence of eq (1) and (2):
$$SdT-Vdp+md\mu =0 \tag{3}$$
If you consider an isothermal process, $dT=0$, you would get $$dT=0 \quad \rightarrow Vdp = md\mu \tag{3a}.$$ But if you ignore the 3rd variable, that is using only $dU=TdS-pdV$ because your system has constant mass, $dm=0$ then it would follow that an isothermal process must also be isobaric, as well, a clearly nonsensical result. The correct relationship, eq (3a), shows that in an isothermal process the mechanical work $Vdp$ is actually done against the system's mass changing its chemical potential, a chemical work. Note that for an ideal gas in which $dT=0$ and $dm=0$, you also have $d(pV)=pdV+Vdp=0$, that is $pdV=-md\mu.$

OTOH, many parameters that affect U are either constant, either are variables dependent on p,T. So their effect can be implicitly involved in dU/dT resp. dU/dp. All depends on what parameters are explicitly said to be constant. dU/dT without n constant implicitly assumes U=f (T, n)=f(T,g(T)).

What follows "The correct relationship" is correct but this statement is not: "because your system has constant mass, dm=0 then it would follow that an isothermal process must also be isobaric." The constraints applied to the definition of U apply to the Gibbs-Duhem equation too.

@BuckThorn the sentence was phrased in the conditional tense with an "if": if one ignores the 3rd variable, not that you should, and writes $U=TS-pV$ with $dU=TdS-pdV$ then one gets $SdT-Vdp=0$ which leads to nonsensical result.

@Poutnik, that is true, and that is why I specifically mentioned the simplest case with which almost all textbooks and courses start, namely, the thermal equation of state $pV=\tfrac{m}{M}RT$ where it is clear that even for an ideal gas there is a 3rd variable and when that is ignored by blindly writing always in the same introductory texts and courses $dU=TdS-pdV$ can lead to nonsensical result.  I am really surprised that anything I wrote above is even controversial.

Yes, I see that now, you are correct.

You may want to check punctuation. I still find it a little hard to follow the logical flow in some of the sentences.

@BuckThorn I suspect that the true but never admitted reason for preferring to write $dU=TdS-pdV$ is because a 2-variable differential form is always integrable and this way it is simpler beyond the obvious of having only 2 and not 3 variables. I have never been a teacher of the subject (thermodynamics), I only recall the pain I felt when learning about it but I think this misleading simplification is a terrible historical legacy. Just look the way novices keep getting confused by $dS=\delta Q/T$, what is $Q$ what is $T$, etc., EM diffraction never appears to be so difficult to beginners.

@hyportnex it's easier to avoid discussion of chemical potential early on, but as you point out this has shortcomings. I'll see if I can think of a way around your argument for why you can't assume U=-pV+TS.

@BuckThorn I think for regular thermoD system, one that is composed of massive particles, the chemical potential is unavoidable. All massive bodies carry entropy. Even electrons in a metal have an electrochemical potential composed of an electric potential (electric charge) and a chemical potential (inertial mass charge); it is truly an eyeopener for an ElecEeng that a voltmeter measures the electrochemical not the electric potential. I know almost nothing about radiative thermoD, maybe there.


text:
Aquatic chemistry, pH and alkalinity
Hope you can help me with an aquatic chemistry question.
I have different types of water of different pH and alkalinity.
pH, alkalinity, HCO3 and CO2 concentrations are known.
I need to calculate the resulting pH from mixing the different water sources together (assuming equal fractions, and complete mixing). I am struggling with how to tackle this, because of the role of alkalinity.
To keep a good overview, let's use two of the sample water sources:

A) pH: 4,9, CO2: 73 umol/L, HCO3: 2 umol/L, alkalinity: 0.13 meq/L
B) pH: 8, CO2: 50 umol/L, HCO3: 1823 umol/L, alkalinity: 1.67 meq/L

My first guess would be to convert pH to [H+] and average those to get the new pH. But then I'm ignoring alkalinity, which is an important factor. Not sure how to incorporate that. Any thoughts?

If you can get your hands on a copy of Water Chemistry by Snoeyink and Jenkins, you'll be able to work this out - hands-down the best book on the topic, IMO. You'll also need to know how you are expressing alkalinity, i.e., as $\ce{HCO_{3}^{-}}$ or $\ce{CaCO3}$. Chapters 4 and 6 are particularly relevant and walk through the mass balance and other equations. If you're interested in learning how a lot of aquatic chemists do it, download PHREEQC and go from there.


text:
Pairing of electrons in Nickel in presence of strong ligand(tetrahedral)
For tetrahedral CFSE, Nickel's last 6 electrons will go in the t2g set of orbitals. Since it is already the higher energy orbital, pairing should NOT happen right? However, all internet sources I found for NiCN4 show that pairing happens in the t2g. Shouldn't electrons not pair and follow hunds rule?

You do not have $t_{2g}$ or for that matter even $t_2$ (there's a "g", meaning centrally symmetric, only if the coordination geometry has that symmetry). You don't have a tetrahedral coordination at all.
When you have a $d^8$ central core with a strong field ligand, both crystal field theory and ligand field theory predict that a strong extra stabilization occurs by going to a square planar configuration, and so it generally does. $\ce{Ni(CN)4^{2-}}$ is actually a classic example of such a square planar coordination.
This square planar coordination no longer has the right symmetry to give a $t_2$ set and an $e$ set; instead the orbitals split in an entirely different way with the net result that one of the orbitals is much more destabilized than all the others. From https://www.jove.com/science-education/11462/crystal-field-theory-tetrahedral-and-square-planar-complexes:

Then the eight electrons go into the four lower-energy orbitals where perforce they are all paired.


text:
Why is reaction rate proportional to the order of reactants?
I know that for some elementary reaction $m\text{A}+n\text{B}\rightarrow\text{C}$, the rate of reaction is given by $-\frac{\text{dA}}{\text{dt}}=k[\text{A}]^m[\text{B}]^n$.
Now if we consider the case $m=1$ and $n=1$ and where A and B are gas molecules, then, from collision theory we know that the rate of reaction can also be written as collision frequency $\cdot$ fraction of molecules with sufficient energy $\cdot$ steric coefficient: $$\sigma\left(\frac{8\text{RT}}{\pi\mu}\right)^{1/2}\text{N}_\text{A}^2[\text{A}][\text{B}]\cdot e^{\frac{-E_{min}}{\text{RT}}}\cdot P$$
Where $\mu=\frac{\text{M}_A\text{M}_{B}}{\text{M}_A+\text{M}_B}$ ($\text{M}_A$ and $\text{M}_B$ are molar masses of A and B, respectively), $N_\text{A}$ is Avogadro's number, and $P$ is the steric coefficient. With this equation, if we let $$k=\sigma\left(\frac{8\text{RT}}{\pi\mu}\right)^{1/2}\text{N}_\text{A}^2\cdot e^{\frac{-E_{min}}{\text{RT}}}\cdot P$$ then we get the general equation for the rate of reaction, rate = $k[A][B]$, and we also understand the relationship between the rate constant, temperature, activation energy, and the pre-exponential factor.
I am wondering if there is any way to explain how the equation can be modified to explain the general form for rates of elementary reactions ($\text{rate}=k[\text{A}]^m[\text{B}]^n$), and why it would be the case that concentrations of A and B are raised to the their stoichiometric coefficients.
Equation from: Atkins, P., & Jones, L. (2009). Collision Theory. In Chemical Principles: The Quest For Insight (5th ed., pp. 591–594). essay, W. H. Freeman and Company.

Put another A instead of B in your equations and you'd see why. If you got so far, what's the point of asking this?

BTW what's the point of using \text in Mathjax, when there's \pu, \mathrm and \ce?

I figured as much, but I just wanted some confirmation. Also would this be sufficient in explaining why the rate of a reaction is proportional (rather than related by some other function) to the order of the reactants? Side note: I just picked up \text first when I was learning Mathjax, not other reason tbh.

Chem+Math MathJax formatting reference:
Basics / Expressions/formulas/equations / Upright vs Italics / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles.

https://chemistry.stackexchange.com/questions/131105/chemical-equilibrium-why-multiplication-rather-than-addition?noredirect=1&lq=1


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What is this gel byproduct of acetylsalicylic acid recrystallization?
I did a very simple acetylsalicylic acid synthesis from salicylic acid and acetic anhydride (with phosphoric acid as a catalyst). After rinsing, filtering and drying the product I dissolved it in ethanol, added some water and recrystallized. After the process, this gel was left in my waste beaker after disposing of the liquid. I thought it might be ethanol based but it wouldn't burn. I'm wondering if anyone knows what it might be? It's worth noting that this gel was only noticed on the flask the recrystallized aspirin was filterd over. Additionally, the flask wasn't steralized so the gel could have formed due to a reaction with some contaminant. Thanks for any information

With phosphoric acid as catalyst used, do you refer to the waxy solid polyphosphoric acid?

This seems to be your phosphoric acid, though if you have rinsed well, it should have left with the mother liquors.
Try dissolving your gel with the aqueous NaOH.


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What could be used to thicken Ferric Chloride solution?
I'm curious if it's possible to thicken Ferric Chloride so I can control it more on a less than horizontal surface ?

Not an answer, but suggestions for you to test the thickeners (and surfactants) added to:

Phosphoric acid-based Naval Jelly® Rust Dissolver (thixotropic polysaccharide, silicon dioxide) and
This HCl-based toilet cleaner formula from Yeser suggests polyoxyethylated oleyl amine
Carbopol® 674 polymer, a lightly crosslinked polyacrylic acid polymer.

Note that $\ce{FeCl3}$ solution, by itself, is not terribly stable, and the added thickener would likely react with it immediately, or catalytically slowly destroy it. Thus, this would be a good experiment to test, and to edit this answer, or at least comment on your results, to help others.

I have the Naval Jelly currently so I'll try that first and acquire the others ! Thank you

The other Naval Jelly ingredients might react badly with the FeCl3. The thing is to get the thickener in it, as with the others.