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What is primary valency

This is from an adapted version of Concise Inorganic Chemistry by J.D Lee.The part on primary valency is confusing me.It mentions that primary valency is the oxidation of the central metal (a number) ,on the other hand I have read that primary valencies are non directional (this description suggests primary valencies are a physical thing).In compounds like $\ce{[Co(H2O)4Cl2]Cl}$ the primary valency is 3 but 2 of the chloride are directional.Can a ligand be both primary and secondary valency simultaneously?

Yes negative ligands satisfy both the coordination number and oxidation state of central metal atom simultaneously.

Does this answer your question https://chemistry.stackexchange.com/questions/7602/werners-theory-and-complex-salts ?

@Sir Arthur7 No

I am sorry to see that they are still teaching these concepts of the 19th century as facts. The concept of primary valence and secondary valence is a translation of Werner's terminology (Nobel) as Hauptvalenz and Nebenvalenz. Haupt means something "main" and neben means something "extra, supplementary". What is valence in old school terminology? It is the combining power of an element.
Suppose you have $\ce{Co^{3+}}$, how many counter anions do you expect? $\ce{Cl^-}$? $\ce{CoCl3}$. This is the primary valence of 3. Standard electrostatics, that is, the number of positive charges must equal negative charges. Charged particles cannot stay alone under normal daily conditions.
However, Werner found that compounds of definite composition can also be found such as $\ce{[Co(NH3)6]Cl3}$. But cobalt could only take three chlorides with it? How come there is additional ammonia, that is a part of the molecular "Komplex".   He invoked the idea of Nebenvalenz, i.e. supplementary valence to explain the additional ammonia molecules built into this molecule. Indeed this was a complex idea for him and he named it as such Komplexverbindung (complex compound).

I have read that primary valencies are non directional (this
  description suggests primary valencies are a physical thing).

Delete this from your mind. This does not mean anything today.

Ok ,so primary valence is the oxidation state in today's terminology ?

Yes, oxidation state=primary valence=Hauptvalenz of Werner.

@M.Farooq: I think the quote in the question is incorrect. This is based on my understanding of a different book (NCERT). Shouldn't the primary valence of the coordination compound be the charge on the complex part (within square brackets) instead of the oxidation state of the central atom?


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Proving a thermodynamic relation between $(\partial H/\partial T)_p$ and $(\partial U/\partial T)_V$
(This question is taken from Problem 1.1(b) of the book Chemical Thermodynamics: Principles and Applications.)$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$

Prove $$\pd HTp = \pd UTV + \left[ V - \pd HpT \right] \pd pTV. \tag{1}$$

First, I expressed $H$ as  $U + PV$, and took the partial derivative against $T$, keeping pressure constant:
$$\pd HTp = \pd UTp + p\pd VTp \tag{2}$$
Expressing $U$ as a function of $(T, V)$:
$$\pd UTp = \pd UTV + \pd UVT \pd VTp \tag{3}$$
Using the well-known identity that $$\pd pVT \pd TpV \pd VTp = -1, \tag{4}$$
$$\pd UVT \pd VTp = -\pd UpT \pd pTV = \left[ V - \pd UpT - V\pd ppT \right]\pd pTV \tag{5}$$
This, of course, means that
$$\pd HTp = \pd UTV + \left[ V - \pd HpT \right]\pd pTV + p\pd VTp \tag{6}$$
Based on my working, there is an additional $$p\pd VTp$$ term. Is my working incorrect? I can’t find my mistake.$\endgroup$

You might want to make a new command for a partial derivative to save yourself some pain. I can give you an example in a while...

put at the start of your post: $\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$ and at the end of your post $\endgroup$... and between those lines you can now use $\pd{V}{T}{p}$ (or even \pd VTp although I don't recommend making the latter a habit, unless you understand why it works without braces).

I also recommend display style mathematics, i.e. $$ ... $$ or if really necessary $\displaystyle ...$.

Could you elaborate on what display style is? I’m not too familiar with it.

I (very cautiously) think the question might be incorrect. My working out of the RHS (albeit by a different approach) is yielding an inconsistency too. Can someone check and confirm?

I believe you make an error when writing the expression for $\left(\frac{\partial V}{\partial T}\right)_p$

@orthocresol Why does it work without braces?  And why did you use {\!#3} instead of just {#3} in defining the command?  I tried it both with and without the added \! in my TeX editor, and it gave the same results both ways (for both with and without braces).

@theorist Truthfully, I don't fully understand it myself... but it has something to do with how TeX scans for arguments. Usually it just reads in one thing at a time (either a letter or a control sequence like \alpha) and treats that as an argument. Grouping several things together with curly braces allows you to pass more than one character at a time. That also means that if you only want a single-character argument, it's unnecessary to enclose it in braces (although having braces is a lot more familiar for most people) [...]

[...] the \! is just cosmetic. You probably know it is a "negative space". I find that without it, the subscripted value is too far away from the brackets; but that's just personal preference, really. If you're interested in the braces, there's always Knuth's good old TeXBook which I think explains this behaviour very carefully.

@orthocresol I really like this -- avoiding braces is very nice for efficiency.   I might post a question on TeX SE asking if there is any chance it could cause problems (assuming you hold to the one character restriction).  If I do I'll link it here.

@orthocresol:  See  https://tex.stackexchange.com/questions/525656/is-it-a-problem-to-implement-a-user-defined-command-without-using-braces

@theorist, I should have but didn’t warn you that it would 99% be a duplicate of something. TeX.SE has been going strong for many years and has serious experts over there, including multiple people involved in LaTeX kernel development. Consequently, nearly everything a rookie like me might think of asking has been asked. :)

@orthocresol  That's OK.  I did search TeX SE extensively (using various search terms) before posting and was unable to find a similar question, so posting was in fact useful because it both gave me an answer an a link to a similar question.

$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$
I would start from
$$ dH = \pd HTp dT + \pd HpT dp$$
This gives rise to
$$ \pd HTV = \pd HTp  + \pd HpT \pd pTV$$
in which you can recognize some components of the solution. Since
$$dH = dU +PdV + VdP $$
so that
$$ \pd HTV = \pd UTV  + V \pd pTV$$
it follows that
$$ \pd UTV  + V \pd pTV = \pd HTp  + \pd HpT \pd pTV$$
which is readily rearranged to obtain the desired equation.
$\endgroup$

Thanks for your answer! Could I inquire where I went wrong in my incorrect proof? I’m not too sure why my first expression is wrong.

@ANZGCFlyingFalcon It looks like you flubbed the substitution of Eq 4 into Eq 3 (see the middle expression in Eq 5).

If I were to take the RHS of (5) and multiply it by (4), wouldn’t I get the LHS? (Obviously the sign would become +ve but you get what I mean.)

To get $(\frac{\partial U}{\partial p})_T$ I recommend starting from the differential form of U=H-PV and taking the derivative wrt p at constant T. I'm not sure how you arrive at your result, it is hard for me to follow.

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Why do diplumbenes (Pb=Pb double bonded compounds) break apart in hydrocarbon solution?
In group 14, the double bond becomes increasingly weaker down the group. Our lecturer told us that many distannenes ($\ce{>Sn=Sn<}$) and almost all diplumbenes ($\ce{>Pb=Pb<}$) dissociate into stannylene ($\ce{R2Sn:}$) and plymbylenes ($\ce{R2Pb:}$) when dissolved in hydrocarbon solution.
The weak nature of $\ce{Sn=Sn}$ and $\ce{Pb=Pb}$ bond has been explained by the $\mathrm{6s-6p}$ energy gap, which makes the $\mathrm{p-p}$ pi-bond unfavourable; instead the $\ce{sp^2}$  orbitals of each plumbylene moiety overlaps with the empty p orbitals of the other. Mixing between $\sigma^*$ and $\pi$ has also been used to explain the weak double bond for distannenes and diplumbenes.
If the double bond is really weak, then why don't these compounds exist as plumbylene monomers in solid state as well? How does dissolving in hydrocarbon make the plumbylene monomer more stable than the dimer diplumbene?
Hydrocarbons can only provide weak van der Waals interactions, so how can they make a large difference to the equilibrium between dimer and monomer?

It's just driving back dimerisation equilibrium with diluting. The lower concentration the more monomer percentage, and in pure form conc. is max.

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Why are stable ionic charges so strongly related to the group number?
Why do lighter elements have ionic charges so strongly related to the group number? For example, why does $\ce{Al}$ only show +3 ionic charge? The basic concept of stability in an ionic compound is that lattice energy is greater than the sum of ionisation energy and electron gain enthalpy. Why can't $\ce{Al}$ show +1 or +2 states? I understand that the +3 state is in a 'sweet spot' since going from +3 to +4 requires a large amount of energy whereas going from +2 to +3 doesn't require much. Besides, a higher charge results in more lattice energy. Hence, it makes sense that the element would want to have the highest charge possible. This explains the abundance of ions in those states but not the absence of others. Why is it that the energy calculations always work out for lighter elements so that they only have the highest charge possible? Is it just chance or am I missing some deeper principle here?

Al does show variety of oxidation states.

So does magnesium.

I'm asking about ionic charge, not oxidation states.

Then magnesium can have one charge per atom.  See the refeeence in my other comment above.


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Pressure turning out to be zero?
I was studying for my freshman chemistry course, and ran into a mind-boggling contradiction.
We know, at equilibrium, entropy is at a maximum:
$$\left(\frac{\partial S}{\partial V}\right)_U=0$$
From the first law
$$dU=TdS-pdV$$
which leads to
$$\left(\frac{\partial U}{\partial S}\right)_V=T$$
From the cyclic rule
$$\left(\frac{\partial U}{\partial V}\right)_S\cdot\left(\frac{\partial V}{\partial S}\right)_U\cdot\left(\frac{\partial S}{\partial U}\right)_V=-1$$
$$\left(\frac{\partial U}{\partial V}\right)_S=-\left(\frac{\partial U}{\partial S}\right)_V\cdot\left(\frac{\partial S}{\partial V}\right)_U$$
Using the result ($\left(\frac{\partial U}{\partial S}\right)_V=T$) it gives
$$\left(\frac{\partial U}{\partial V}\right)_S=-T\cdot\left(\frac{\partial S}{\partial V}\right)_U$$
But we know that $\left(\frac{\partial S}{\partial V}\right)_U=0$, so
$$\left(\frac{\partial U}{\partial V}\right)_S=0 \tag{1}\label{eq1}$$
So far so good.
Writing the total differential of $U$
$$dU=\left(\frac{\partial U}{\partial S}\right)_VdS+\left(\frac{\partial U}{\partial V}\right)_SdV$$
and comparing it to
$$dU=TdS-PdV$$
We get:
$$\left(\frac{\partial U}{\partial V}\right)_S=-P$$
Now I'm in real trouble, because equation $\eqref{eq1}$ says this quantity is zero, and even in this general series of thermodynamic equations, I have reached the conclusion that
$$P=0\,\,\,\,\,\,\text{(?!)}$$
I would appreciate if someone could help identify what the fundamental problem is? I've checked and rechecked my calculations and assumptions, but I'm sure I'm missing something.

I think the cyclic rule is not valid in this case. $\frac{\partial V}{\partial S}=\frac{1}{0}$ so I suspect some of the manipulations there are faulty.

@Tyberius I see your point. But I see another problem here (you reminded me by disallowing $\frac{1}{0}$): the derivative $\left(\frac{\partial U}{\partial S}\right)_V$, which has to be $T$ at all costs, would also take a similar form, leading to supposedly infinte T. :(

The problem begins with your first statement:
$$\left(\frac{\partial S}{\partial V}\right)_U=0$$
Because in fact:
$$\left(\frac{\partial S}{\partial V}\right)_U=\frac{p}{T},$$
which is clearly non-zero (except in limit of zero pressure or infinite temperature).
More generally, being at equilibrium does not equate to a partial derivative being zero!  Rather, being at equilibrium equates to a partial derviative being defined (i.e., they are only defined at equilibrium).  Thus, whether or not a partial derivative equals zero has nothing to do with equilibrium, but rather with the specific nature of the system. And, most typically, partial derivatives are non-zero; it's only for certain specialized cases that a partial would be zero.
Consider, for instance, the total differential for $U=U(T, V,n)$:
$$\mathrm{d}U =  \left(\frac{\partial U}{\partial T}\right)_{\!V,\, n_i}\,\mathrm{d}T+\left(\frac{\partial U}{\partial V}\right)_{\!T,\, n_i}\,\mathrm{d}V +\sum_i \left(\frac{\partial U}{\partial n_i}\right)_{\!T,\, V,\, n_{j\ne i}}\,\mathrm{d}n_i,$$
where the last term allows for chemical reactions and/or matter flow into or out of the system.
Note that state functions ($U, H, A, G, S...$) are only defined at equilibrium, and thus these partial derivatives are likewise only defined at equilibrium.  What the above partial derivatives tell us, then, is how the state function (in this case, $U$) changes as we infinitesimally change the system from one equilibrium state to a new equilibrium state by varying $T$, $V$, or one of the $n_i$.
Thus, far from being zero at equilibrium, these partials typically need to be non-zero at equilibrium, unless the state function is independent of the change in the corresponding independent variable, which only applies in special cases.
One such special case is the internal energy of an ideal gas, which depends only on $T$ and $n$. Hence, for a single ideal gas, or a mixture of ideal gases:
$$\left(\frac{\partial U}{\partial V}\right)_{\!T,\, n_i}=0$$

If I understood you correctly, derivatives of $S_{sys}$ don't have to be zero at equilibrium(although $S_{sys}$ shouldn't be changing)? So there's no condition applicable to the $S_{sys}$ function to signify equilibrium? I will be very grateful if you could help clean up this mess for me.

Ok, I think I had a revelation. dS is zero, but dV could also be zero, so the p.d. might be a non-zero limit. So I think the only query I have now is "How do I apply the $dS_{univ, equil}=0$ to a given system?" $S_{univ}$ has always sounded like an abstract concept to me. Also say I'm not allowed to use conditions on Gibbs energy yet.

@WilliamR.Ebenezer  I rewrote the secnd part of my answer in a way that I hope makes things clearer. As to the connection between equilibrium and entropy maximization, that requires a completely separate answer. I think you will find several questions and answers on this subject if you search this site. And, of course, as an experienced user of this site, you know that if you are unsatisfied with those answers, you can write a new question that specifically addresses your confusion.

Thanks for your added clarifications. I was not intending to ask for help in comments that could only be met by entire answers, but I agree that it is a separate topic altogether. I'll dig for a bit and ask a new question if required. Thanks again for your time.

$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$
If you write
$$dS=\pd SVT dV + \pd STV dT$$
then
$$\pd SVU=\pd SVT  + \pd STV \pd TVU \tag{1}$$
Next use
$$dT = \pd TUV dU + \pd TVU dV$$
to obtain
$$\pd TUS = \pd TUV  + \pd TVU \pd VUS$$
or
$$\pd TVU = \pd UVS \left[ \pd TUS - \pd TUV \right] \tag{2}$$
Substituting Eq. (2) in Eq. (1) and solving for $\pd SVU$
$$\pd SVU = \pd SVT + \pd STV \pd UVS \left[ \pd TUS - \pd TUV \right] \tag{3}$$
Therefore if $$\pd SVU = 0$$ then
$$\pd UVS = -p = \frac{-\pd SVT \pd TSV}{ \left[ \pd TUS - \pd TUV \right]}$$
While this is not extremely illuminating (and I haven't hit on the reason for the discrepancy with the OPs approach) it does not result in an internal contradiction.
$\endgroup$

Thanks to @orthocresol for the nice partial derivative notation shortcut. Saved a lot of time!


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Bonding in molten silicon dioxide
Solid silicon dioxide has a giant lattice structure with single covalent bonds, while in the gas phase it has a molecular carbon dioxide-like structure. But what about in the liquid phase?
Would it be formed of molecules as well, or are there still only single covalent bonds that form and reform? Or is it more like metallic bonding with delocalized electrons? Or something else entirely?

You might look at https://link.springer.com/content/pdf/10.1007/s11663-014-0092-1.pdf and see if that helps.

It´s surely not metallic. ;)

:/ Just about like in glass, only less and less interlinking bonds with rising temperature. Close to boiling point there would a lot of small molecules, I guess.

@Mithoron - a light perusal of the thermodynamic modeling literature suggests that the pure SiO2 liquid remains almost exclusively O-Si-O entities, unlike other oxides that will have a variety of components (one could imagine O-Si, O or O2, etc. being conceivable, but they are not really seen in the liquid). Now, add another oxide and things start getting interesting quickly.

@JonCuster I'm not sure what you mean. It's clearly a variety of structures comprising varying number of SiO4 tetrahedra.  What would be " O-Si, O or O2, etc." though? If there was Na+, the units would be just smaller and more or less anionic. With, Pb, it would indeed be more interesting.

@Mithoron - well, it gets pretty esoteric quickly unless one loves phase diagram modeling. In a variety of pure binary phase diagrams, as well as complex oxide systems, one has to take into account a variety of possible short-range ordering that may be energetically favorable. For example, in Na${2}$O-Al${2}$O${3}$, the model of the liquid is NaO${0.5}$, AlO${1.5}$, and an "associate" NaAlO${2}$. At a given temperature and composition, the relative ratios of the three entities can be calculated. Other oxides may have multiple species in the melt as indicated by XRD measurements.


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How far does the delocalisation (resonance) effects for substituents extend for a conjugated system?
Let's say we have a conjugated system like this one:

Now, the delocalisation of the benzene electrons with the nitrate system means that the ortho, and para positions on the immediate benzene ring would be more electron deficient than the meta position. This effect will be seen in 1H NMR and 13C NMR signals for the protons or carbons on the ortho/para/meta position. Also, the electrophilic substitution on those positions will be affected.
Now as I understand, the resonance effects of the $\ce{-NO2}$ group is due to the overlap of N 2p orbitals with the benzene ring p orbitals, which skews the sizes of molecular orbitals, and makes certain carbons more electron-deficient than others.
However, are resonance effects independent of distances over which they act? I am not sure if distance would have any role on the skewing of the molecular orbitals.
Does the distant benzene ring, and the double bond in between also show the resonance effects from the nitro group on the first ring? If so, is it as strong, or is it weaker? Can this be seen in NMR spectra? or the reactivity of the distant benzene ring?

The effect on the reactivity usually becomes clear if you take an electrophile and try to let it attack in different position.

Whatever rule, the fact that a system has a high degree of effective conjugation (delocalisation) does results in the substituent exerting an effect on the whole. Of course one practically sets limits, but having the substituent, and particularly one involved in the pi system by mesomery results in the double bond being affected. It has an enhanced single bond character. How easy is it seen by hydrogenation I don't know, nor I know on the change of the nmr. They might be little. But the UVVis spectrum will be there to show that the whole system has changed.

@Karl, Do you mean using QM calculations, or experimentally?

@Alchimista, Is there a ballpark figure for where we practically set that limit for the mesomeric effect? (number of bonds etc.)

Take the phenyl ring at the other end. Its reactivity might be very similar with or without the substituent at the other side. Note that practice is complicated and textbooks provide neat example. Here performing a reaction would already give a moderately complex mixtures of products. Likely their analysis requires a preliminary separation. At the end end of the process you might find substantially the same amount of each compounds. In textbook case you would be asked about the activating/orienting effect of a generic ethylene bridge, for instance. The effect you correctly thought of might be

... more or less weaker. But there is one and it can be dramatic on the properties of the congiugsted skeleton, therefore I mentioned the UV Vis abs spectrum. Think or look at some common dyes and indicators to get even a more evident picture. Push pull system are extremes of the ideas above.

@Karl meant in principle. That is why I felt to comment.

Neither. I mean analogous to the basic textbook pen and paper experiments that shows you why some substituents direct into meta and others into ortho and para.

"is there a ballpark figure for where we practically set that limit for the mesomeric effect? (number of bonds etc.)". There might be one in and for specific treatments and properties. But I am not aware of. As far as extended linear conjugated systems are concerned, a theoretical treatment goes as far as infinite chain (!). In practical case torsion reduces the extent of delocalistion, so the effect does not reach the other side of the molecule if the latter is long and flexible enough.

..... What I wanted to tell you that there are strong effect in a molecule such as that in your sketch. Not necessarily evident on the reactivity at the phenyl ring, yet more evident or even striking in El. absorption or in the CC str freq. of the methylene bridge.


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What is the symmetry of dxy orbital?
Considering the sign of orbital and assuming the $z$-axis as a principal axis, for me, it looks like that it has two perpendicular $C_2$ axes that penetrate the lobes, so I think it is $C_\mathrm{2v}$. But the book (Inorganic chemistry by Miessler and Tarr) says it is $D_\mathrm{2h}$.

Note that $C_\mathrm{2v}$ is a subgroup of $D_\mathrm{2h}$. This means that you're just overlooking some of the symmetry elements. In addition to the $C_2$ axes penetrating the lobes "end to end", you also have a $C_2$ axis perpendicular to the orbital plane through its center.
Your $C_\mathrm{2v}$ guess is also inconsistent because that point group only has one $C_2$ axis, but you said yourself that you already found two.

Thank you for your comment. I misunderstood the criteria for C and D groups. Now it makes sense!

Imagine that the $x$ axis and $y$ axis lobes have a  different color (phase). Locate the principal axis, that of highest rotational symmetry, if there is more than one just choose one. A $C_2$ or 180 degree rotation will make the orbital indistinguishable from its starting position if this points out of the plane of the orbital (i.e. I choose this to be along $z$ if orbital is in $xy$-plane). Any 'D' point group has a 2-fold, $C_2$ or 180 degree rotation axis, perpendicular to the principal axis so the $\mathrm{d}_{xy}$ belongs to the  $D_\mathrm{2h}$  point group. You can check other symmetry elements by looking at the point group table. A $C_\mathrm{2v}$ molecule, i.e. one with symmetry of water molecule, has no 2-fold axis perpendicular to the principal axis. If you do not distinguish $x$ and $y$ with 'color' the orbital will have $D_\mathrm{4h}$ symmetry.

OMG I just misunderstood that if there is a perpendicular C2 axis, then it goes to C group. Thank you for your kind explanation & correction.


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Why is the voltage produced by a voltaic cell affected by temperature as described by the Nernst Equation?
I understand why changing the concentrations of the electrolytes in a voltaic cell affects the voltage. Given a zinc-copper cell for example, if I were to increase the concentration of the zinc sulphate electrolyte, the equilibrium between zinc ions and zinc metal would shift towards the side of the zinc metal. This would cause the reduction potential for zinc ions to increase, decreasing the overall cell potential. Please correct me if I’m wrong as this is how I understand it!
However, I don’t quite understand why temperature affects the voltage the way it does. According to the Nernst equation, if the reaction quotient is greater than one (meaning that the zinc sulphate concentration is greater than the copper sulphate concentration in this example), increasing temperature decreases voltage. Conversely, if the reaction quotient is less than one, increasing temperature increases voltage.
I am familiar with Le Chatlier's principle, but if I were to apply it to this scenario, given that the redox reaction taking place is exothermic, it would suggest that increasing temperature would always decrease the voltage produced since it would cause a shift towards the reactant side. This, of course, does not correspond with the relationship predicted by the Nernst equation.
I have looked all over the internet but I could not find any answer to this seemingly straightforward question. Any help is greatly appreciated!

Did you notice the temperature $T$ term in the Nernst equation? Try to derive the equation and what is the origin of $T$ there. Once you have done the mathematical derivation of the equation, one can think about temperature qualitatively.

Your mistake is assuming that the equilibrium constant K and the voltage are linearly related, so that a shift in K towards reactants will reduce voltage. The potential is a function of TlnK, so if T increases more than lnK decreases, the potential will actually increase.

Both terms of the difference in the Nernst equation depend on the temperature.
$$E = E^\circ - \frac{R T}{z F} \ln Q$$
The second term obviously depends on temperature because it contains the temperature (and the other quantities are either constants or not temperature-dependent). The first term, however, is temperature-dependent as well. If you consider the relationship to the standard Gibbs energy of reaction, this becomes clear:
$$Δ_rG^\circ=−z F E^\circ $$
Rather than figure out the temperature-dependence by looking at the two terms, it is easier to forget the Nernst equation and directly relate the cell potential at non-standard conditions to the Gibbs energy of reaction under the same conditions:
$$Δ_rG=−z F E $$
At constant pressure, the partial derivative of the the Gibbs energy of reaction $Δ_rG$ with respect to the temperature is $-Δ_rS$. (Perhaps surprisingly, the temperature-dependence of the equilibrium constant depends on the sign of $Δ_rH$ instead, i.e. whether the reaction is endothermic or exothermic, see Van't Hoff equation).

You are correct in understanding that changing the concentration of electrolytes in a voltaic cell affects the voltage through the Nernst equation. As for the effect of temperature, it is important to remember that the Nernst equation takes into account the effect of temperature on the equilibrium constant (K) of the reaction, not just the effect of temperature on the reaction itself.
While an exothermic reaction would suggest that increasing temperature would shift the equilibrium towards the reactant side, it is not always the case. In fact, the direction of the shift depends on the overall enthalpy change (ΔH) of the reaction and the entropy change (ΔS). If the reaction is exothermic and the entropy change is positive, increasing temperature would actually shift the equilibrium towards the product side, resulting in an increase in voltage.
In the case of a zinc-copper cell, the Nernst equation predicts that increasing temperature would decrease the voltage if the zinc sulphate concentration is greater than the copper sulphate concentration (Q>1). This is because the reaction would be more product-favored at higher temperatures, and the excess of zinc ions would cause a decrease in the cell potential. On the other hand, if the copper sulphate concentration is greater than the zinc sulphate concentration (Q<1), increasing temperature would increase the voltage as the reaction would be more reactant-favored at higher temperatures.
Overall, the effect of temperature on the voltage of a voltaic cell depends on the specific reaction involved, and cannot be predicted solely based on Le Chatelier's principle.

Consider that even when exist a "driving force" to move ions to electrodes, it is necesary that particles stay not static, they need some capacity to move to change direction and take way to electrodes, and what gives this capacity to move, Energy, and in any particle system this energy distribution follows Maxwell-Bolzmann and this means that temperature counts.. e^-(1/kT)


text:
Why does strength of bond matter if melting only affects the the intermolecular of the molecules
So as I've been told, when a substance melts, the actual bonds of the substance aren't broken, only the IMFs (inter-molecular forces). So why is it that metal groups decrease in melting points going down when they should be increasing because more shells exist thus ions in the lattice will be larger so more IMFs therefore it takes more energy to overcome. Instead, I'm being told that since the metallic bonds get weaker due to less attraction between electron sea and the positive ion, melting point decreases.
Could someone please help me figure out what I'm missing here?

Depending whether you have a molecular, metallic or ionic compound, the independently moving particles in the liquid state are molecules, atoms and ions, respectively.
So for a molecular solid, you have to break the intermolecular forces to turn it into a liquid (or make them non-persistent so that the interaction partners can change over time).
For a metallic solid, you have to break the metal bonds intermittently so that the atoms can move in the liquid.
For an ionic solid, you have to break the ionic interactions intermittently so that the ions can move in the liquid.
Finally, for a network covalent solid, you would have to break covalent bonds to liquify it. For this reason, there is no liquid state of diamonds.

So why is it that metal groups decrease in melting points going down [...]

According to Jim Clark:

The strength of a metallic bond depends on three things:

The number of electrons that become delocalized from the metal
The charge of the cation (metal).
The size of the cation.

"no liquid state of diamonds"? https://chemistry.stackexchange.com/questions/6834/is-it-possible-to-melt-diamonds-into-a-liquid

@Mithoron Diamond describes a specific structure of a carbon allotrope; in contrast, NaCl describes a compound that has a distinct structure in the solid, liquid and gas state. So you can definitely obtain a liquid phase of carbon, but why call it diamond?

Yeah, that's true. Even more, diamond turns into other phases of carbon before melting occurs (at least at reasonable pressure).

why is it that metal groups decrease in melting points going down

This isn't quite true for all metals. I think the only metals that behave like like this are the alkali metals. Group 6 metals increase in melting point. Calcium has a higher melting point than Magnesium. Etc.

more shells exist thus ions in the lattice will be larger so more IMFs
  therefore it takes more energy to overcome

I assume that shells here mean the electron shells. An atom having more electron shells usually causes it to have higher polarizability. And an atom that has higher polarizability has stronger dispersion forces. As a result, it will have higher melting point in comparison.  Noble gases illustrate this. These compounds are inert and only interact with each other through weak intermolecular forces. Thus, radon has the highest melting point and neon has the lowest melting point. So you are right in this regard.
Metal atoms, however, share electrons with each other through something known as the metallic bond. Which is not an intermolecular force. A metallic bond is more like a covalent bond. There are a variety of factors that influence the strength of a metallic bond, as noted in Karten's answer.

the metallic bonds get weaker due to less attraction between electron sea and the positive ion

This isn't true for several metals either. Galium has a lower melting point than Indium.
And many transition metals do not obey this trend either.

It's imho abusing terminology to call interatomic or electron-nuclear attraction forces "molecular", but the attraction of an electron (shared or not) for neighboring nuclei is certainly what holds a material together. A metallic bond by any other name holds things together just as well.


text:
Observing bands which I shouldn't be!
I performed the iodine spectrum experiment to observe vibronic transitions. I heated the iodine crystals to produce vapors (at a little above room temperature) and observed absorption spectrum with a diffraction grating and a spectrometer. But I am getting some quirky results as indicated following.

In Ref 1 and Ref 2, it is stated that near room temperatures, $I_2$
is in the ground electronic state and in the lowest vibrational state too ($X,
v''=0$ state), so transitions are observed only from thsat state.
Ref 2, Ref 3, Ref 4 and Ref 5 state that from $X, v''=0$ state, Franck-Condon principle tells that transitions can be observed only to $B, v'$  states with $v'$ ranging around $20-50$, and they also cite the standard values of energies corresponding to particular $v'$ values so as to numerate the observed bands accordingly.

Now, at near room temperatures, I after numerating my bands, observed that I was seeing dark bands for $v'=0$ onwards! (I after assigning the standard band, numerated the lower energy bands backwards.) So, one of the above points have to be false!
Questions: Please tell what can possibly go wrong? If I'm getting some contributions from $X, v''>0$ states, how to identify them (since Ref 4 clearly states that the bands with transitions from $x, v''>0$ states "should not be included in this numeration" in for calculating the parameters from Birge-Sponer plot)?

There are lots of things that can potentially go wrong in an experiment so it would help if you posted a spectrum. The ground state vibrational frequency is $\approx 214$ cm$^{-1}$ and $k_BT \approx 210$ cm$^{-1}$ at room temperature so you expect to see absorption from $v \gt 0$. It helps in the assignment if you look for the wavenumber where dissociation occurs.

@porphyrin Thanks! I hadn’t thought about that. But can you please help then how to assign $v’$s to my observed wave numbers? Or point me to a good resource?

It would be much preferable, if you would provide the references in human readable form and link through the DOI to them.


text:
Chemical Formation Of A Virus
Many articles and books classify virus as a non living thing which only replicate itself inside a living host, yet some describe virus as a living organism. Virus was first discovered on the year 1892 by Dmitry Ivanovsky and Martinus Beijerinck. Since the year it was discovered, virus has different source of it's formation.
I am curious about the formation of a virus. Is there a chemical reaction that will make a virus to became a 'virus'? What are the chemicals responsible for the formation of virus? Are the temperature and geography affect the formation of a virus? On the other hand, what chemical properties classify virus as a non living thing?
Any help is highly appreciated. Thank You.

We cannot give you a microbiology 101 lecture here. Try to read up e.g. on wikipedia how viruses work, then come back if you have specific questions. But I think biology.SE would be a better place to ask. And no, there is no single chemical reaction that makes a virus. A virus is simply a piece of freefloating DNA, and every virus type was obviously first produced by some malfunctioning cell.

@Karl. Thank You. I posted the question here because I am curious whether a virus is produced via chemical reaction.

First, viruses are formed in other cells by the chemical processes of those cells. Viruses "hijack" the cell's means for reproducing viral RNA or DNA and for the protein coat of the virus.
Second, define "alive"... there is some argument on whether a virus is living or not, so that depends on your definition.
Third, what does geography have to do with viral reproduction?
Try to focus your question so you can get a useful answer.

This should have been a comment

@DrMoishe Pippik. I am just curious about the geography, whether it affects the host of the virus, which will also the affect the formation of the virus itself.

@HYDR0GEN You are aware there are very many different viruses and virus types? That every species in flora and fauna has their own selection of viral maladies?

@Karl. Yes, I am aware of that.


text:
How to assign initial and final states of iodine transitions in its spectrum?
In an experiment, I gently heated iodine crystals to create vapors and then observed it absorption spectrum with a spectrometer. The temperatures throughout were moderate (maximum $40^\circ\text{C}$). I observed a series of dark bands from red region to blue-green region with spacing between consecutive bands decreasing from red end towards to blue end.
I calculated the wavelengths corresponding to these bands beginning from the first resolvable band in the red region and ending at the $30^{th}$ reading (without missing a single dark band in between). The data is as follows:
$$\begin{array}{c|c}
 \text{Colour Region} & \lambda/\text{nm} \\ \hline
\text{Red} & 624.884 \\
& 624.884 \\
& 621.831 \\
& 617.552 \\
\text{Orange} & 612.581 \\
& 609.518 \\
& 606.070 \\
& 601.798 \\
& 597.629 \\
& 594.171 \\
& 591.095 \\
\text{Yellow} & 586.862 \\
& 584.243 \\
& 581.853 \\
& 577.225 \\
\text{Yellow-Green} & 575.218 \\
& 571.819 \\
& 567.954 \\
& 565.246 \\
\text{Green} & 562.924 \\
& 560.291 \\
& 557.579 \\
& 554.399 \\
& 551.684 \\
& 550.053 \\
& 548.422 \\
& 547.490 \\
& 544.693 \\
& 541.271 \\
& 540.027 \\
& 538.548 \\
\text{Blue-Green}
\end{array}$$
Now, in order to obtain information about the parameters like bond dissociation energy, convergence limit, etc., I need to know the quantum numbers which correspond to each observed transition wavelength which are the initial and final electronic states and the vibrational states.
Question: How to find these? I know that generally, sources characterize some
standard states, like this says that $541.2\text{ nm}$ corresponds to $v''=0\rightarrow v'=27$. But what am I supposed to do with the larger wavelengths? Cuz the sources say that Franck-Condon prohibits $v''=0$ to low $v'$ transitions. But nevertheless, I do am observing larger wavelengths. Are these from $v''>0$?
How can I even be sure that these are $X\rightarrow B$ transitions? Cuz of operating temperatures?
Please provide some proper approach!

The difficulties that I am facing in this assignment of states are also elucidated in this post of mine.

If you transform the wavelength into energy, you will find that the peaks are equidistant.


text:
General justification for steady state approximation
I want to understand the assumptions that have to be fulfilled in order for the steady state approximation to work. I did some research myself and was able to deduce assumptions of the form $k_1\gg k_2$ for some systems, by solving the system exactly and taking the limiting case in which the exact system reduces to the approximative system. (Later I found this answer doing the same thing Meaning of steady state (kinetics)) This approach will not work for most systems though, as no closed exact solution exists for most non-linear differential equation systems.
I want to develop a clear chemical intuition in which case it is justified to use the approximation. Most literature seems to just tell the reader that it's fine to use the approximation in the reaction thats discussed. This is unsatisfying. From a mathematical standpoint this approximation is all but trivial.

This steady-state approximation is always wrong. It is only justified by common sense. In chemistry, you will never be and stay exactly at a point where the rate of the reaction producing a compound is equal to the rate of destruction of this compound. But you may imagine being "not too far" from this ideal position. That is the trick used for understanding the kinetics of some chain reactions.

Obviously it's never exact but I want to develop an understanding to what kind of processes it's an acceptable approximation. Surely there must be papers comparing numerical solutions to steady state approximations and surely there must be 1 or 2 handy rules for estimating weather it's OK to use steady state approximation based on the underlying chemistry.

Whether it’s okay is going to depend also on what you are using the result for, so one can’t make general rules. A good way to get the feel for the inaccuracy is to plot product vs time for SS approx and actual. To simulate actual, you don’t need an explicit solution, just iterate the differential equations at short time intervals

Related: https://chemistry.stackexchange.com/q/110794

In addition to k1>>k2, there is also the condition k2t>>1, because it only holds after a brief induction period (initially, the rate of change of the intermediate is greater than 0, but by the time k2t>>1, it has become 0). You can see that from this image:https://chem.libretexts.org/@api/deki/files/50967/Reaction2a.bmp?revision=1&size=bestfit&width=356&height=318
Note that initially, d[B]/dt>0, but it quickly becomes 0. This condition often isn't mentioned because the induction period is typically very short.

I would add that in terms of chemical intuition, it's about how unstable/reactive the intermediate formed is. The approximation is valid if the intermediate is an unstable species which therefore reacts as soon as it's formed (examples might include a free radical), whereas if the intermediate is relatively stable (like a benzylic cation stabilised by resonance) then it will be less accurate.

Let me know when (or if) you want the bounty awarded and to whom. I assume I have to do the actual awarding (first time for me), but you decide to whom the bounty goes (if anyone). We still have two days before the deadline, so just tell me what you want to do by then or earlier.

@EdV right now I would give it to Jeffrey Wang, for the valid answer, that there is no general justification. His answer give a nice and relevant example of a case analysis that leads to the approximation in a special case.

@Andrew I think this approach should also be made into an answer, so it is easier for others to find. Obviously this is also no general justification though.

Thanks! I will award Jeffrey Wang the bounty in 10 minutes!

Thanks for drawing attention on the question! @EdV

No very satisfying justifications are available unless you do a case-by-case analysis. But PSSA does not come from common sense. In general, pseudostationary state approximation (PSSA) is justified by the theory of differential equations.
In fact, the pseudostationary state concentrations represent the zeroth approximation
to a set of particular solutions of the correct chemical kinetic equations, which are called principal solutions. After the initial period, the concentrations of the intermediates approach their principal solution rather than their pseudostationary state values.
Therefore, the accuracy or usefulness of the pseudostationary state assumption depends upon the magnitude of the differences between the principal solution and the pseudostationary state concentrations.
OK, that sounds useless. But after doing some maths (like finding the principle solutions), some (obviously very ambiguous) rules are indeed justified mathematically.

Induction period should be short. (As you can see PSSA solutions cannot satisfy the initial conditions)
Destruction of the intermediates should be fast.

For example, in $\ce{A ->[$k_1$] B ->[$k_2$] C}$ if $k_2\gg k_1$ then the conditions above are met while if $k_1\gg k_2$, they aren't.

By the way, PSSA originates from Michaelis-Menton theory for enzyme kinetics. PSSA in it is well justified while the conditions above are not necessarily met. Consider
$$
\ce{E + S <=>[$k_1,k_{-1}$] ES ->[$k_2$] P}.
$$
Let $[\ce{S}]=s,[\ce{E}]=e,[\ce{ES}]=c,[\ce{P}]=p,e+c:=e_0,s+p+c:=s_0$. Furthermore, let $\tau=k_1e_0t,u(\tau)=s/s_0,v(\tau)=c/e_0$. The kinetic equations can be written as
$$
\begin{cases}
\frac{\mathrm du}{\mathrm d\tau}=-u+(u+K-\lambda)v\\
\color{red}{\varepsilon}\frac{\mathrm dv}{\mathrm d\tau}=u-(u+K)v
\end{cases}
$$
where  $\lambda=k_2/\left(k_1s_0\right),K=\left(k_{-1}+k_2\right)/\left(k_1s_0\right),\color{red}{\varepsilon=e_0/s_0\ll 1}$. Therefore, PSSA can be applied (i.e. setting $u-(u+K)v=0$, justified by perturbation methods) without loss of much accuracy. This example indicates that more often than not, a case-by-case analysis is required.

Finally, I think to use PSSA safely, one should do some experiments to check (numerical simulations, real experiments, etc.).

Thank you for you input. This is close to what I personally would have answered. There is no general justification and care should be taken when applying PSSA. Eventhough many introductionary courses just take it as given, that the approximation is fine, there are cases where it is probably not.

In the scheme $ A \overset {k_1} \longrightarrow B \overset {k_2}\longrightarrow C$ the exact solution for B is $B=\displaystyle \frac{k_1A_0}{k_2-k_1}\left( e^{-k_1t}-e^{-k_2t} \right) $. To find out how quickly the steady state is reached and by how much it deviates from the true value, the steady state is compared to the exact value by taking a ratio of concentrations.
Species B follows $\displaystyle \frac{dB}{dt}+ k_2B=k_1A_0e^{-k_1t}$  and letting $dB/dt=0$ forms our steady state value $B_{ss}$.   The ratio to measure the extent of deviation will be defined as $\displaystyle R=\frac{B}{B_{ss}} = \frac{k_2}{k_2-k_1}\left( 1-e^{-(k_2-k_1)t} \right)$ and as a measure at steady state we let $R=1$.
Initially when $t=0,\; R_0=1$ and at long times $\displaystyle R_{\infty} =  \frac{k_2}{k_2-k_1}$, so the difference between $R_0$ and $R{\infty}$ indicates the extend or deviation of steady state from the true value.
If, for example, $k_2 = 5k_1$ then $R_\infty-R_0=1/4$ or a 25% deviation. When $dB/dt=0$, species  B has reached its maximum value which takes time $\displaystyle t_m =\frac{1}{k_2-k_1}\ln(k_2/k_1)$ and in this example this takes $\ln(5)/(4k_1)$ seconds, i.e. this time is needed before steady state can be invoked.
This analysis indicates that for steady state to apply $k_2 \gg k_1$, i.e. the steady state applies under conditions that B disappears rapidly once it is formed.  For other schemes the same general principles apply but the analysis will be far more complicated.


text:
What is the color of a FeS4O6 solution
What is the colour of FeS4O6 solution?!

Why do you give us three problems with answers ? What for ?

No the 23rd question FeS4O6 is mentioned I have tried searching online but it's colour or any good info about it is not present hence the question that I have asked..

The picture is just for source purpose

You should rather ask simply about the color of the FeS4O6 solution made by reaction of Fe2+ and thiosulfate ions, without speaking of CuSO4, KI, etc. I just notice that the formule of potassium iodide is wrong. It is not K.I., but it is KI.

Thanks for the answer and I shall keep that mind.

It would be preferred to transcribe any images, so that the text can then be searched. Please also add a custom for the source.

This is a relatively advanced analytical chemistry question and it requires that you remember the color of all common transition metal ions and their very common precipitates. You have to remember the colors of halogens in solution as well.
Instead of worrying about the color of iron thiosulfate (which is bluish green), think about the first step. KI reacts with one the given choices, which renders the solution brown AND there is no precipitate.  Iodine in aqueous solution gives a brown color.
From the choices, see which ion can oxidize iodide (colorless) to free iodine (brown solution)? Only copper (II). Now understand that sodium thiosulfate is a powerful reducing agent. Which ion when reduced forms a colorless solution AND which ion which forms a precipitate with iodide?
Now you have to double check mercury and see if this a possible choice or not.
Tornado reaction

The correct answer is, indeed, in my opinion choice (B) as is noted.
The action of KI on CuSO4 creates, in situ, the very unstable CuI2 (think of the corresponding salt with iron, Ferric iodide, whose very existence is doubted).
Per the web from Wikipedia:

The CuI2 immediately decomposes to iodine and insoluble copper(I) iodide, releasing I2.

This explains the brown coloration. The formed salt Cuprous iodide per Wikipedia:

Pure copper(I) iodide is white..

With respect to the question on ferrous tetrathionate, from an old chemistry journal, it is unstable on creation and readily decomposes with the release of sulfur dioxide and elemental sulfur.

In aqueous solution, the substance $FeS_4O_6$ is dissociated into a mixture of the ion $Fe^{2+}$ and of $S_4O_6^{2-}$. The ion $Fe^{2+}$ is pale green, but so pale that its solutions are considered as colorless. The ion $S_4O_6^{2-}$ is colorless.

This analogy is not correct and should never be used. Lead (II) is colorless, and iodide ion is also colorless. What is the color of their precipitate? Yellow, right. So I do not know why it is yellow ppt (in terms of spectroscopic reasons), but we should not predict colors like this.


text:
Is there a phase diagram available for hydrogen fluoride?
Has a phase diagram for hydrogen fluoride been published in an article on the internet or in a book ?
Or is there a table for HF for different pressures up to 100 atm. and temperatures up to 500$⁰$ C ?
Or, even better, is there a calculator for HF, like this one for steam ?

I checked Reaxys but couldn't really find anything on it. If I find some time I could check the Gmelin in our library tomorrow, they might have it.

@Justanotherchemist That would be fine, i'm mostly interested in the difference between H2O and HF from 100$⁰$ C to 500$⁰$ C. I think that could be just two parallel lines on the diagram.

Might be offtopic here but of course there is an interesting difference as HF is the only common acid that will form H-bonding bridges to itself rather than to water. Still the H-bonding donating and accepting effects are different for H2O and HF which leads to different properties. But I'll check tomorrow.

Why on Earth are you interested in HF to 100 atm. and at 500°C ? HF is such a dangerous compound.. Who would like to heat up to 500°C a sample of gaz like HF compressed to 100 atm ?

@Maurice  I can imagine you wondered ! I would like to investigate if HF could be a usefull lifting gas in the atmosphere of Venus with temperatures up to 462$⁰$ C and pressures up to 92 atm.

Sounds like a terrible idea anyway. Normal air is enough of a lifting gas out there.

@IvanNeretin Yes, but for this idea i need a gas that can be liquified at about "normal" temperature and pressure so a constant height could be maintained.

I´d guess that no imaginable container material would survive this.

@Karl Maybe Teflon or Basalt fiber ?

surely not teflon. at 800 K, its as soft as ... something very soft, not to mention that it is decomposing with out without air.

So I couldn't find anything in the Gmelin either but there was an interesting page on the polymerisation of HF with a diagram. That might also something you'd have to consider at different temperatures and pressures. Sorry for the quality as I could only take a photo.
https://imgur.com/0TkchID

@Justanotherchemist Nevertheless thanks for your effort ! I suppose the polymerisation will not be a problem for temperatures above 100$⁰$ C.

I seriously doubt that HF is the best thing for this your Venus balloon, though that doesn't mean there's no reason for a phase diagram of HF going to high pressures (maybe it could occur naturally on planets or could appear in energetic explosions or collisions), and measuring it might not be as dangerous as some would think, since these kinds of experiments are usually done with tiny quantities anyway.

@H.H.  I now agree with your seriously doubting. H$_2$O is much safer and has about the same lifting capability. But with H$_2$O the balloon would float at about 45 km, I thought maybe with just HF it could float  higher up in the atmosphere. But of course you can combine H$_2$O with He or H$_2$.

Are you trying to design balloons that would naturally stay at a height with human-survivable temperature and pressure conditions to support a human colony or mission or robot probe, and figuring it will float at the height where the lifting gas begins to condense as liquid?

@H.H.  Yes indeed, at 45 km heigtht it would be the condensing temperature of 110⁰ C and about 2 atm. so not well suited for a colony but a well designed robot probe could survive.

Linde Gas contains a vapor pressure curve.  Note that this does not extend to Venusian surface temperatures which are above the critical point.
Hydrogen fluoride may not lift as well as you think.  Its gas phase may associate through hydrogen bonding so molecules could be heavier than just HF.  See the note near the end of the safety information here and note that the association could persist to higher temperature if you apply higher HF pressure than one atmosphere.

Thank you, this was exactly what i was looking for ! Made an edit but couldn't identify the word "peesust"!


text:
Reaction of octahydroazecine with iodine crystals
I really don’t understand why the following reaction would take place first of all.

So this is a sub part of a question wherein we have to compare the basicity of the compounds formed through some reactions. So, I got the rest 3 of them but I have no clue about this one.
My concerns are..

Why would $\ce{I2}$ react in absence of any oxidising agent or catalyst?
Should the presence of $\ce{N}$ influence the reaction.
Is this reaction feasible?

Ans -

You don't need an oxidizing agent, as the iodine I2 is the oxidant you are looking for. And the Nitrogen atom is necessary, because it has an available doublet.

@Maurice Thanks for the response but it does answer my question. First of all in halogens iodine does not react with alkenes directly ( I have been taught this , in simple words it's difficult to force it to react )...it can react in presence of a solvent or such like THF or some others .. ( I found this while doing some  research) ..but since none of them is here why would it react ...and moreover what is the mechanism of the reaction?

If you think iodine does not react with alkenes you have been misled - it does. The solvent should have been specified but the presence of a solvent in this reaction is taken as read.

What would be the mechanism ..? π complex formation like in the case of Cl2 .. and should the solvent  be non polar ..@waylander

Formation of an iodonium ion [3] intermediate. This reaction will occur in most solvents due to the presence of a very good internal nucleophile.

Okay..I ll try solving this . Thank u

Chlorination and bromination of alkenes are very general reactions, and mechanistic study of these reactions provides additional insight into electrophilic addition reactions of alkenes. Although much less detail is known about fluorination and iodination of alkenes, it is believed that iodination follows the similar mechanistic steps of much studied bromination (Ref.1) to give corresponding 1,2-diiodoalkane. In certain compounds, if neucleophilic center is available within vicinity of cyclic halonium center, the last step of nucleophilic attack of halide ion would be over written by much faster intramolecular cyclization as depicted by following mechanism:

This mechanism is well supported by the examples provided in Ref.1 and Ref.2.
References:

Yoshinao Tamaru, Masato Mizutani, Yutaka Furukawa, Shinichi Kawamura, Zenichi Yoshida, Kazunori Yanagi, Masao Minobe, “1,3-Asymmetric induction: highly stereoselective synthesis of 2,4-trans-disubstituted γ-butyrolactones and γ-butyrothiolactones,” J. Am. Chem. Soc. 1984, 106(4), 1079-1085 (https://doi.org/10.1021/ja00316a044).
Yvan Guindon, François Soucy, Christiane Yoakim, William W. Ogilvie, Louis Plamondon, “Diastereoselective Synthesis of 2,3,5-Trisubstituted Tetrahydrofurans via Cyclofunctionalization Reactions. Evidence of Stereoelectronic Effects,” J. Org. Chem. 2001, 66(26), 8992-8996 (https://doi.org/10.1021/jo010873r).

Much of what I anticipated... Thanks for the clarification.

So only the rate of reaction in affected if the reaction occurs with I2 ...


text:
Estimation of ground state energy from wave function

Given the wave functions of a particle in a 1D box below determine (for each pair) if the left or right function gives a better estimation of the ground state energy if the variational method is used. No calculaltions are needed.

I am having some problem understand how to solve this question. We are not allowed to use any graphing calculators and the tip says "try to plot some points and compare it with the exact wave function", which I don't find helpful.
All help is truly appreciated!

Probably you are supposed to calculate the numerical value of all those wave functions for a couple of x values, like x = 0.2, then x = 0.5, then x = 0.8. This being done, you should intercompare these results (f(0.5):f(0,2 for example) and then compare with the ground state solution (sinusoid). The numerical values relative to one another maybe similar to what is obtained with the sinus function.

The variational method uses a linear combination of functions as an estimate of the true wavefunction. It works by adjusting coefficients in the linear combination to minimize the energy of the trial wavefunction. This is the best estimate of the true wavefunction using the restricted basis set. The variational principle states that adding more (new) basis functions to the linear combination always improves the estimate.
In your problem (or according to your instructions), you don't really optimize anything (except for the last comparison), because the normalization condition fixes the value of the $c_i$ coefficients, but you can estimate which of the two functions appears most like the true wavefunction, as this should return a minimum energy.
The ground state solution has the form $$\psi \propto sin(\pi x)$$
when x is in units of the box length, with bounds at $x=0$ and $x=1$ (and null wavefunction outside these bounds). This function has nodes at $x=0$ and $x=1$ and is concave with a single maximum at $x=\frac12$. By evaluating the functions at the bounds, in the midpoint and at one or more other points you should get an idea about the shape of each function and thereby which are better. It would of course be easier to plot them. Consider to the extent possible symmetry and whether bounds are satisfied (which they must for the solution to be at all physically reasonable - remember that the energy blows up if the wavefunction samples points out of bounds.
The last comparison is a "trick question". There you have to take into account the variational principle which is mentioned in the first paragraph of this answer.

What insight allows you to determine whether $\psi_3$ or $\psi_4$ is better? The other seem obvious.

@user5713492 the true wavefunction has even symmetry

Yeah, but...
https://www.wolframalpha.com/input/?i=int_0%5E1+x%5E2%281-x%5E2%29+d%5E2%2Fdx%5E2%28x%5E2%281-x%5E2%29%29dx
https://www.wolframalpha.com/input/?i=int_0%5E1+%28x%5E2%281-x%5E2%29%29%5E2dx
https://www.wolframalpha.com/input/?i=int_0%5E1+x%5E2%281-x%29%5E2+d%5E2%2Fdx%5E2%28x%5E2%281-x%29%5E2%29dx
https://www.wolframalpha.com/input/?i=int_0%5E1+%28x%5E2%281-x%29%5E2%29%5E2dx ... OK, I guess you're right but I think you can compose examples where a wave function with wrong symmetry is better.

@user5713492 Sorry but I cannot decrypt your comment. In words, what's your point?

Just using Wolfram Alpha to do the integrals quickly.

@user5713492 Ok, you evaluated expectation values for the wavefunctions (energies). I'm pretty sure $\psi_4$ is the better function based on symmetry, and if I evaluated your Wolfram results correctly ($E = <\psi^* H \psi>= -N\int \psi^* (\partial \psi/ \partial x^2) dx$) then that is indeed the correct answer. Thanks for the nice Wolfram example btw.

I agree though that symmetry alone does not seem a sufficient condition, but in this case it is a good guide.

Where asymmetric trial function wins:
https://www.wolframalpha.com/input/?i=int_0%5E1+%28%28x%2B1%29x%281-x%29%282-x%29%2B%281%2F2%29x%281%2F2-x%29%281-x%29%29d%5E2%2Fdx%5E2%28%28x%2B1%29x%281-x%29%282-x%29%2B%281%2F2%29x%281%2F2-x%29%281-x%29%29+dx
https://www.wolframalpha.com/input/?i=int_0%5E1+%28%28x%2B1%29x%281-x%29%282-x%29%2B%281%2F2%29x%281%2F2-x%29%281-x%29%29%5E2+dx
https://www.wolframalpha.com/input/?i=int_0%5E1+%28x%281-x%29%29d%5E2%2Fdx%5E2%28x%281-x%29%29+dx
https://www.wolframalpha.com/input/?i=int_0%5E1+%28x%281-x%29%29%5E2+dx

@user5713492 nice example :-) Of course, knowing the true wavefunction is symmetric would usually point the search toward symmetric solutions, but your point is made.

@BuckThorn: I like your answer very much, and appreciate you putting so much time into writing it with care. I find it unusual that it was closed as "off-topic" because I think it's perfectly on topic, so it's a pity that your answer cannot be chosen as "accepted". Would you mind helping support our proposal to start a stack exchange just for materials modeling and quantum chemistry? https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=NDY4N2I5MzBmODViNGQwYWM2MTBmMzVkMzc0OWU0ZjQ5NDU2YTVhYzQyMmEyY2IwYjdjYzlmZDMwMWQyZmJhNGBP4m1cf3-vYfnwUjsCFf90E8N7syRuU35ETDRpKo-U0

Hi Buck Thorn! It seems that your Area51 account isn't linked properly to your Stack Exchange Network account, which might cause you not to get an invitation to participate in the private beta stage, can you try to see if this answer:
https://area51.meta.stackexchange.com/questions/29706/please-make-sure-you-link-your-other-stack-exchange-accounts helps to fix the problem? On the commitment list you should show up as having 10k reputation but it's listing you as only 51 (which is the default for people that have confirmed their email address).

This says something about signing into Area51 first and then committing, rather than the other way around: https://area51.meta.stackexchange.com/a/29985/190792.

That's probably why it doesn't know which Stack Exchange account to link! Perhaps we can wait until you have the normal browser available again, and you can try with your usual email address. Thanks!

Have you tried it with your original email address now?

@BuckThorn: People have found it to work in Chrome's in-cognito mode. I also find it very frustrating that SE hasn't fixed this, because we have ~900 members in the Materials Modeling Stack Exchange facebook group, but because it's so hard to sign in properly, we have only 76 that are actually signed into Area51.... so this makes it extremely hard for us to reach the minimum requirements to get launched! If we don't meet the minimum requirements within about 2 more months, they'll permanently delete our proposal.

Hi @BuckThorn, have you had any luck? A user this morning was having problems but it got fixed (his Area51 account was completely deleted and re-created from scratch, and this time it connected properly to the rest of the SE network).

Hi @BuckThorn, in this answer: https://area51.meta.stackexchange.com/questions/29933/persistent-bugs-hindering-many-of-our-users-from-participating/29985#29985 a user got it to work by clicking "log out" on Area51, then she was told that her Area51 account would be permanently deleted. Then she gave instructions and screenshots for what she did next, which worked. Some people got it to work when they used Chrome's in-cognito mode, or when they deleted their browser's cookies. Is it helpful? Please try!

It looks like it's working now !!!

I think the only step missing now is to click the confirm link in the email they sent you.


text:
membrane for a reverse osmosis desalination experiment
We are second year cram schoolers , we are currently working on a project tackling water desalination by reverse osmosis and are required to do an experience. During the process, we encountered some difficulties regarding applying pressure and the fabrication of membranes . Given our unfavorable circumstances (low budget , lack of material ), we were wondering if someone could give us some insights on how to proceed. Although , we have tried to diy one with accessible materials and this is what we could come up with so far .
First ,we took an unglazed flowerpot and we filled its hole with cement, we filled it with a 0.1 mol / L potassium ferrocyanide solution. Simultaneously, we soak it in a 0.1 mol / L copper sulphate solution. After waiting several hours, the two products are going to spread in the pores, one from the inside, the other from the outside. In the middle of the pores, a chemical reaction had taken place , forming a precipitate of copper ferrocyanide which closes the large pores and transforms the pottery into a semi-permeable membrane. Finally, we rinsed the pottery well with water, and left it to soak for several days so as to remove the initial reagents.
However, we didn't meet our expectations and we don't know what is it that we're doing wrong . How thick should the cement layer be ? what are the rinses so as to remove the reagents ? and how could we simulate the pressure in a compartment ? and is there any other ways to make a membrane to work with ?

What do you mean by "we didn't meet our expectations" ? To get a reverse osmosis you need a high pressure over the salt solution. Usually it is done by closing the flowerpot by a stopper which must be  watertight and leakproof. Easy to say ! The stopper must be drilled and equipped by a long tube many meters high filled with salt water. If the salt solution is 0.5 M, the tube must be 10 meters high, and the pure water goes through drop by drop


text:
Why fibres are strong only in one direction?
Why fibres are strong only in the direction they are oriented?Why they break if we pull at right angle to their orientation?

Here each blue line represents combination of polymeric chains.

Could you clarify if you refer to an individual molecule, or a composite of them? I suspect that the answer might change.

Each blue line represents combination of polymeric chain.

Fibres are made of long linear molecules, in which the atoms are bound by covalences C-C or C-O. Usually it is a long C-C-C-C-C-C-O-C-C-C, with one O from time to time. Such bonds are hard to break. But paralleled fibres are hold together by van der Wals forces, which are rather weak. They are not difficult to  get separated. That is why a fiber can be easily split in two in the sense of the length, and only in this sense.


text:
Are there any theorems about packing of atoms in molecules subject to the Lennard-Jones potential?
Related to the problem of minimization of the Lennard-Jones potential in a molecule with $n$ atoms, where $n$ is, say, between $5$ and $10$, questions arise such as (but not limited to) the following:

Is it possible to determine a lower bound on the radius of the molecule?
Is it possible to determine an upper bound on the radius of the molecule?
Is it possible to improve the upper and lower bounds using symmetry considerations?

I consider molecules to consist of point-like atoms, so the radius in question is the smallest radius of a sphere containing all points.
If there are theorems regarding similar questions, that would facilitate determining a reasonable search domain for the optimal configuration.

IMHO, "cluster" is the correct word for it, rather than "molecule".

It's not possible to answer this without knowing the structure of the molecule, since both the connectivty and the rigidity will affect the radius of gyration.

All I can offer is that if such things exist, a good place to look would be the work of David Wales at Cambridge. He has also done extensive global optimizations (i.e. searches for all minima) of LJ clusters of various sizes. If such a thing existed though, it would be very useful to clarify if the proposed global minima for the bounds you describe.


text:
DIY chemistry - Calcium from bone
I've gotten really into this subject lately from people who make their own cleaning agents and I was surprised how simplistic (although time consuming) everything was.
I am interested in making slaked lime. I've been putting wood ashes in my garden bed lately and it's been doing wonders but if I had a steady supply of home made lime that would do wonders.
I cook a lot on the weekends and I usually have a lot of bones left over. Some are cooked and some are uncooked. Could I use any to make quicklime with? I was trying to figure out a way to soften bones without leeching the calcium from them. I know I could use something more practical like egg shells but I have an abundance of bones lol
Thank you for any and all help.

Bones contains calcium hydroxyphosphate. When heated it will not produce any quicklime

https://en.m.wikipedia.org/wiki/Bone_ash - "Bone ash is a white material produced by the calcination of bones. Typical bone ash consists of about 55.82% calcium oxide, 42.39% phosphorus pentoxide, and 1.79% water."
https://www.sciencedirect.com/science/article/pii/S2214509520300036 - "The calcium oxide present in bone ash reacts with water to form calcium hydroxide: {CaO (s) + H2O (l) → Ca (OH)2 (s)}."  molecule of carbon dioxide (CO2), leaving quicklime."


text:
Do ions in a salt bridge come from the half cells in electrochemistry?
I understand the basics of electrochemsitry and galvanic cell, where there are half cells where either an oxidation or reduction reaction occur, and electrons moving between the wires provide the current.
What confuses me is the salt bridge. In this diagram here, nitrate ions, which are from the solution of a half cell, go into the salt bridge.

However, other diagrams (and answers like those to Do the salt bridge ions have to match the ions in the half-cells solutions?) do not state that the ions have to be the same (you don't need nitrate ions in the salt bridge if you have a silver nitrate solution for example).

So do ions like nitrate from the solution in a half cell enter the salt bridge?

Of course they do enter. How else would the charge be transferred?

But in the second picture the salt bridge has KCl. So would sulfate ions be moved into the salt bridge?

Of course they would. Again, how else would the charge be transferred?

@IvanNeretin I disagree. If the concentration in the salt bridge is higher than that of the half cells, you could have anions from the salt bridge diffuse into the oxidation half cell (making up the missing negative charges lost by electrons leaving through the wire) and cations from the salt bridge diffuse into the reduction half cell (neutralizing the negative charge electrons from the wire bring in).

You must think electrically, without thinking first to the chemistry. In the anodic solution first, the reaction produces positive ions $\ce{Mg^2+}$ or $\ce{Zn^2+}$. It does not really matter here. The problem is that this cannot stay like this. Either the positive ions have to move where negative ions are available. Or new negative ions must arrive in the solution. Or both.
The simplest way for solving this difficulty is to mix the two solutions. Because a similar difficulty appears on the cathodic region : positive ions ($\ce{Ag+}$ or $\ce{Cu^2+}$) are here discharged, leaving an excess of positive ions around the cathode. By mixing the solutions the excess negative ions produced around the cathode move to the anode where they are needed. But mixing the solutions is the least thing to do, because now $\ce{Ag+}$ or $\ce{Cu^2+}$ ions can touch the $\ce{Zn}$ or $\ce{Mg}$ anode. And the chemical reaction will go on in the contact between positive ion and metal. The same electrons of the half-reaction are produced but they will not go through the long external circuit form the anode to the cathode. They prefer the shortest circuit direct from metal to ion in solution, and metallic $\ce{Ag}$ will  soon be formed on the $\ce{Mg}$ anode (or $\ce{Cu}$ on the $\ce{Zn}$ plate). No current will be emitted in the outer circuit. It is the end of the cell. So it is absolutely necessary to prevent the $\ce{Cu^2+}$ or $\ce{Ag+}$ ions to touch the anode, so the two compartments must be separated.
That is why we introduce the salt bridge. It must contain ions that do not react with any ion contained in the two compartments. Their nature is not important. Usually it is $\ce{K+}$ and $\ce{NO3-}$.
So let's go back to what's happening electrically in solution when the cell is working. There is an excess of positive ions around the anode, and an excess of positive ions around the cathode. To compensate this situation, the negative ions of the bridge are attracted by the positive ions in excess at the anode, and the positive ions of the bridge are attracted by the cathode. Both quit the bridge and get mixed in the solutions. This allow the cell to work.
After some time, the bridge is loosing all its stuff, so some positive ions of the anode ($\ce{Zn^2+}$or $\ce{Mg^2+}$) may enter the bridge and replace the initial positive ions $\ce{K+}$that have moved to the cathode. And symmetrically some negative ions ($\ce{NO3-}$ or $\ce{SO4^2-}$) from the anode will enter the bridge to replace the initial ions $\ce{NO3-}$.

Sorry if this is a basic question, but don't ions like potassium react with things like sulfate? Or is that not a reaction?


text:
What is the "mechanism" of the decomposition of hydrogen peroxide?
In my classwork on the decomposition reaction of hydrogen peroxide by catalase, I wrote the following equation:
$$\ce{2H2O2 -> 2H2O + O2}$$
However, my teacher said that I should go further in depth and write down the (quote unquote) "mechanism" of this reaction. I'm not sure what he is talking about, can anyone help me?
EDIT: I am an IB student studying chemistry in Standard Level so the answer to the above problem should remain within the scope of the IB syllabus. Here's more information from my teacher:
“Although we show reactions in one neat equation, the majority of reactions occur in a series of steps. (…) There’s three step to that (the one above) reaction and catalase will act on one of those steps to reduce the activation energy and therefore make the reaction faster”.

According to the Wikipedia entry for Catalase the mechanism is not currently known however some details are, there is an interaction with Fe(III) bound to the enzyme producing a Fe(IV) species with a bound oxygen. More details here: https://en.wikipedia.org/wiki/Catalase

Probably the teacher would like to know whether the two O atoms of the future of O2 are coming from the same H2O2 molecule or from two different molecules. H2O2 may loose one O atom produce H2O, and the free O join another one coming from the next same situation in the next H2O2 molecule. Another mechanism would be to suppose that one H2O2 looses 2 H to produce O2, and then the two H are reacting with the the next H2O2 to produce two H2O. You may imagine a third mechanism.

See:https://chemistry.stackexchange.com/questions/116006/why-and-how-does-hydrogen-peroxide-decompose-in-the-presence-of-light

@Maurice that may seem like a possible solution. My teacher also mentioned that the reaction occurs in three steps. See edit above.

Related: Mechanism for Hydrogen Peroxide decomposition?

Below is one way to write it that shows what the enzyme does and avoids the uncertainty of the details:
$$\ce{E + H2O2 -> E-O + H2O}\tag{1}$$
$$\ce{E-O + H2O2 -> E-O2 + H2O}\tag{2}$$
$$\ce{E-O2  -> E + O2}\tag{3}$$
In words, the enzyme binds to the first molecule of hydrogen peroxide, release one molecule of water and holds on to the other oxygen atom (via the iron bound to heme bound to the enzyme). Then, the second molecule of hydrogen peroxide binds to the enzyme, which again release water and holds on to the second oxygen atom. Finally, the enzyme releases the two oxygen atoms in the form of the dioxygen molecule.
Alternatively, you could write the following three steps, giving even less information about what is happening in the enzyme:
$$\ce{E + 2H2O2 -> [E\cdot H2O2\cdot H2O2]}\tag{A}$$
$$\ce{[E\cdot H2O2\cdot H2O2]-> [E\cdot H2O\cdot H2O\cdot O2]}\tag{B}$$
$$\ce{[E\cdot H2O\cdot H2O\cdot O2] -> E + 2H2O + 2O2}\tag{C}$$
In words, the enzyme first binds to the substrates, then catalyzes the reaction, and finally releases the products. While this in general might be a good way of describing the action of an enzyme, it is not quite correct for catalase, which releases one product before the second substrate binds. However, this might be the answer you are looking for at the IB level. I would ask the teacher for further clarification.


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Nano coatings reaction with Al bottles
I have a question regarding nano coatings based on the formula "modified $\ce{SiO2}$".
I have been told that $\ce{Al}$ bottles react with substances with a $\mathrm{pH}$ value below 3 or above 9. When the substances are not within these limits, it will form $\ce{Al^3+}$ which will cause the $\ce{Al}$ bottle to corrode, if I understand it correctly.
My question is therefore, when does this corrosion begin, how much time does it take, and can it ruin my nano coating effects?
My supplier of the $\ce{Al}$ bottles also has a bottle that they call "Chemical pesticide insecticide agrochemical pharmaceutical Aluminum bottle", does this bottle create any difference regarding the possible reaction with $\ce{Al}$? Perhaps it corrodes at a lower or higher $\mathrm{pH}$ value?
Also when my supplier of the coatings indicates the products has a chemical stability between 1-13, what does that mean?

A cited a pH value below 3 or above 9 is in line with the usually citation for the stability of the Al2O3 protective layer. Per a source on the effect of pH, to quote:

It is well known that aluminum resistance is related to the thin and compact layer of naturally formed oxide on aluminium surface, but this oxide layer is stable only in pH range 4-8. Lower or higher pH values caused prominent destroying of protective layer and thus the significant metal dissolution [15].

So, outside the pH range, your nano SiO2 coating formulation may be subject to attack. The presence of active Aluminum may serve to introduce some new chemistry.
Most likely, in my opinion, the unprotective Al can react even with water to form hydrogen and aluminum hydroxide. The latter reacts with so-called reactive silica per Science Direct:

Silica forms complex precipitates with iron, aluminum, and magnesium hydroxides.

which disrupts the SiO2.
Another but unlikely path starts with some atomic hydrogen ingressed into the Aluminum metal (see, for example, this Science Direct discussion and also a 2008 thesis "Alkaline dissolution of aluminum: surface chemistry and subsurface interfacial phenomena", by Saikat Adhikari discussing surface atomic hydrogen formation on Aluminum). Suggested reaction mechanics could then be if such a reaction is favorable (unlikely):
$\ce{.H ⇌ H+ + e-}$
$\ce{4 .H + SiO2 (nano) -(?)-> Si + 2 H2O}$
potentially directly destroying the nano SiO2. Here relatedly to the reaction above is a hydrometallurgy source discussing a commercial leaching application with $\ce{.H}$ on Al metal acting on $\ce{PbS}$ at the bottom of Page 218. However, the heat of formation of PbS is 10 times less than SiO2, hence this reaction path is not likely.
Proposed Solution: Test Copper vessels for suitability (see this reference).

Hi AJKOER,
Thank you for your detailed answer.
So regarding the pH values within the limits i should be good to go. Regarding the pH values below the limits it would form silica, which could possibly disrupt the SiO2, that would cause the effect of my nano coatings to wear off?
My products has a shelf life on around one year, will the reaction take effect within this year?
Thank you very much for your answer.


text:
Is the evaporation (liquid to gas conversion) of ethanol dependent on how full you fill it up in a sealed bottle?
It seems intuitive to think that in a bottle (which is air-tight), if you fill up ethanol (or any other volatile liquid) to very full extent, thereby leaving very little headspace, the amount of ethanol evaporated would be reduced. In contrast, if there is more headspace, then more room is allowed for ethanol vapor and hence evaporation would increase. This can also be explained in terms of the smaller increase in partial pressure of ethanol vapor per molecule evaporated with a larger headspace, thereby allowing more time before the forward (evaporation) and backward (condensation) reaction reaches equilibrium.
I would like to know if I'm on track and if not, what did I miss? I do not have the right materials to test it myself and obviously this would not be sophisticated enough to be properly studied in a scientific research (hence I cannot find a specific reference).

It's sealed, i.e. after sufficient time, in equilibrium. The rates are the same.

Yes, you’ve got the right idea.


text:
How to derive the steady-state solution for simple two-step reaction with differential equations?
Consider this simple two step reaction, a variant of a Michaelis-Menten type of problem, where $\ce{A}$ and $\ce{B}$ reversibly bind to make $\ce{AB}$, and $\ce{AB}$ and $\ce{C}$ reversibly bind to make $\ce{ABC}$:
\begin{align}
\ce{A + B  &<=> AB}\\
\ce{AB + C &<=> ABC}
\end{align}
Assume all the rate constants are equal to $k$ for simplicity, and that $[\ce{A}]_0$, $[\ce{B}]_0$, $[\ce{C}]_0$ are initial values for $\ce{A}$, $\ce{B}$, $\ce{C}$, and $[\ce{AB}]_0 = [\ce{ABC}]_0 = 0$.
What is the proper way to derive the steady-state concentration of $\ce{ABC}$ as a function of initial values? here is an attempt to get $\mathrm{d}[\ce{AB}]/\mathrm{d}t$:
$$\frac{\mathrm{d}[\ce{AB}]}{\mathrm{d}t} = ([\ce{A}][\ce{B}]k + [\ce{ABC}]k) - ([\ce{AB}]k + [\ce{AB}][\ce{C}]k)$$
The first parenthesis of the sum are the two ways to make $\ce{AB}$ and second parenthesis are the two ways to lose $\ce{AB}$.
We can set $\mathrm{d}[\ce{AB}]/\mathrm{d}t = 0$ and try to solve for $[\ce{ABC}]$, but I am not sure how this helps and where initial values come in.

Does ABC form a product? If not, there isn’t a steady state. The system approaches equilibrium and then concentrations remain constant after equilibrium is reached

@Andrew: it doesn't form a product. why isn't there a steady state? i would like to analyze the relationship between initial values of A and B and the maximal amount of ABC they yield, i.e. when d[ABC]/dt = 0.

“Steady state” usually refers to a system in flux, where reactant and product concentrations are changing, but an intermediate concentration is constant. It sounds like you are talking about an equilibrium where no concentrations are changing? Or are you interested in the time evolution of the approach to equilibrium?

To elaborate - the reason for my confusion is that you asked about the "steady-state concentration of ABC". Here, ABC is the final product, and products don't have constant steady-state concentrations; only intermediates do. The products will only have a constant concentration when the system reaches equilibrium. Equilibrium concentrations can be determined using the equilibrium constants with no time factor involved.

If you are only interested in the final concentration of ABC, then remove the term "steady-state" completely. Your question is "what is the equilibrium concentration of ABC?".

Based on the discussion in comments, the OP is interested in the equilibrium concentration of the species ABC in this reaction scheme:
$$\ce{A + B + C <=>[k_1][k_2] AB + C <=>[k_3][k_4] ABC}$$
Equilibrium is reached when the rates of the forward and backward reactions of each step are equal. Thus, we have have two equations describing the equilibrium:
$$k_1\ce{[A][B]}=k_2\ce{[AB]}$$
$$k_3\ce{[AB][C]}=k_4\ce{[ABC]}$$
We also know from mass conservation that
$$\ce{[A]_o}=\ce{[A] + [AB] + [ABC]}$$
$$\ce{[B]_o}=\ce{[B] + [AB] + [ABC]}$$
$$\ce{[C]_o}=\ce{[C] + [ABC]}$$
These five equations can be manipulated to find an expression for $\ce{[ABC]}$ in terms of $\ce{[A]_o}$, $\ce{[B]_o}$, and $\ce{[C]_o}$.

can you say more about why steady state is wrong terminology? in equilibrium we will have d[ABC]/dt = 0 so mathematically why can't we solve for [ABC] as function of initial values that way?

Generally, we use "steady-state" to describe a system in which reactant and product concentrations are still changing, but an intermediate concentration remains essentially constant. Here, you are interested in a product. Its concentration is only constant at the end, when all of the concentrations are constant. Mathematically, you will note that I did use the fact that d[ABC]/dt=0 in setting up the two kinetic equations. It just isn't a scenario we would describe as a steady-state.

You must have made a mistake in your algebra. I started with C=Co-ABC. Then AB=(k4ABC)/(k3C). Substitute the expression for C into that, and you've got AB in terms of ABC. Then A and B are both done the same way. Eg A=Ao-AB-ABC and substitute the expression you got above for AB and you have A in terms of only ABC. Finally, subsitute into the first kinetic equation k1A.B=k2AB and isolate ABC. It's a big mess, ultimately quartic in ABC, but once you put in values for the initial concentrations and rate constants, easily solvable.


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How can I eliminate the harmful gas produced from the heating/manipulation of galvanized steel?
When welding on steel with a galvanized coating, hexavalent chromium is yielded from the chemical reaction. This gas is something that I would like to eliminate before I begin welding, possibly with some liquid or mixture. Any ideas?

Galvanized steel is steel with a zinc coating, such as by dipping the steel into molten zinc. Where would chromium come from? Stainless steels have chromium content, but they would not be galvanized. Could you please elaborate in your question?

Examing the possible incidental formation of hexavalent chromium and its corresponding chemistry is an important topic, in my opinion.

Galvanizing by definition is zinc coating. The zinc boils off ( as 1700 F)  is approached and immediately oxidizes to ZnO , a white powder. ZnO has the unusual characteristic of causing zinc chills = brass founders ague = oxide shakes,  IF inhaled while less than one hour old . After that is is pretty harmless and used in various lotions and creams and rubber products.  So avoid the fumes when welding. However , one recovers from "chills" in about 24 hours so not toxic ( I can verify no lasting affects as I had "chills" many decades ago). There is another potential problem ; zinc can cause liquid metal embrittlement of steel . This occurs where the zinc is not exposed to air as if two surfaces are placed together and the edges sealed by welding. I don't know where you think chromium is coming from other than steel made with scrap ( the majority) typically contains about 0.04 % tramp chrome ( and Ni ,and Cu , and Mo). So essentially all steel  welders in the world are exposed to this tramp chrome.

My speculation on the possible mechanics of redox reactions that could move Cr into higher oxidation states upon welding with galvanized-steel. Understanding the chemistry may be instrumental in providing guidance.
To begin, welding creates, in addition to heat, CO2 and water vapor (from the flame), which could act on the Zn coating creating hydrogen (and Zn(OH)2/ZnO). Some of the hydrogen may be ingressed on the metal surface forming atomic hydrogen (preferably cited as the hydrogen radical, $\ce{.H}$). Possible surface chemistry interaction of the latter directly with oxygen forming the hydroperoxyl radical :
$\ce{.H + O2 -> .HO2}$
Also, if there is a galvanic cell resulting in any solvated electrons, interaction with oxygen can produce the superoxide radical anion:
$\ce{O2 + e-  -> .O2-}$
which exists in atmospheric conditions as the acidic hydroperoxyl (or called the perhydroxyl radical $\ce{.HO2}$). The latter could react with a transition metal (including Fe and Cr) as follows:
$\ce{M(n) + .HO2 -> M(n + 1) + HO2-}$
$\ce{H+ + HO2- -> H2O2}$
where any formed hydrogen peroxide could feed a classic Fenton reaction also moving $\ce{M(n) to M( n + 1)}$. Per this 2011 source, "Advanced Oxidation Process Based on the Cr(III)/Cr(VI) Redox Cycle", by Alok D. Bokare and Wonyong Choi. To quote from the abstract:
"Oxidative degradation of aqueous organic pollutants, using 4-chlorophenol (4-CP) as a main model substrate, was achieved with the concurrent H2O2-mediated transformation of Cr(III) to Cr(VI). The Fenton-like oxidation of 4-CP is initiated by the reaction between the aquo-complex of Cr(III) and H2O2, which generates HO• along with the stepwise oxidation of Cr(III) to Cr(VI). The Cr(III)/H2O2 system is inactive in acidic condition, but exhibits maximum oxidative capacity at neutral and near-alkaline pH. Since we previously reported that Cr(VI) can also activate H2O2 to efficiently generate HO•, the dual role of H2O2 as an oxidant of Cr(III) and a reductant of Cr(VI) can be utilized to establish a redox cycle of Cr(III)–Cr(VI)–Cr(III). As a result, HO• can be generated using both Cr(III)/H2O2 and Cr(VI)/H2O2 reactions, either concurrently or sequentially. The formation of HO• was confirmed by monitoring the production of p-hydroxybenzoic acid from [benzoic acid + HO•] as a probe reaction and by quenching the degradation of 4-CP in the presence of methanol as a HO• scavenger. The oxidation rate of 4-CP in the Cr(III)/H2O2 solution was highly influenced by pH, which is ascribed to the hydrolysis of CrIII(H2O)n into CrIII(H2O)n-m(OH)m and the subsequent condensation to oligomers. The present study proposes that the Cr(III)/H2O2 combined with Cr(VI)/H2O2 process is a viable advanced oxidation process that operates over a wide pH range using the reusable redox cycle of Cr(III) and Cr(VI)."
Possible Implied Solution: As the Cr(III)/H2O2 system is inactive in acidic condition, try employing a sulfur-rich fuel gas mix, forming SO2 on combustion, to maintain an acidic pH.


text:
Effect of pressure on viscosity of liquids and gases
My textbook says the following on viscosity and pressure:

increase in pressure decreases the viscosity of water but for other liquids it increases. Viscosity of gases does not change much

No further explanation is provided.
I have two doubts regarding this:

How does the viscosity of water decrease? Does it have something to do with the anomalous expansion of water?
How does the viscosity of gases not change much on increasing the pressure? The gas molecules should have an increase in kinetic energy, contributing to more collisions and hence more viscosity

These two issues are quite different, shouldn't be in one post, please have this in mind in the future.

Ok will do in the future. I can't post a new question (M. Farooq has already written an excellent answer) so for now I'll tweak the title a bit.

One of the best websites for all advanced water properties is a website maintained by Dr. Martin Chaplin http://www1.lsbu.ac.uk/water/physical_anomalies.html#Pvisc
Keep in mind that viscosity is a really complicated subject and water is among the unique solvents. In general viscosity of liquids increases with pressure. At pressure extremes, let us say, 100,000 psi, many liquids become solids. Water also becomes a solid at really really high pressures. So when you talk about viscosity, you should talk about what pressure and what temperature.
From the website:

Water's pressure-viscosity behavior [534, 2890] can be explained by
  the increased pressure (up to about 100-200 MPa) causing deformation,
  so reducing the strength of the hydrogen-bonded network, which is also
  partially responsible for the viscosity. This reduction in
  cohesiveness more than compensates for the reduced void volume. It is
  thus a direct consequence of the balance between hydrogen-bonding
  effects and the van der Waals dispersion forces [558] in water; with
  hydrogen-bonding prevailing at lower temperatures and pressures. At
  higher pressures (and densities), the balance between hydrogen-bonding
  effects and the van der Waals dispersion forces is tipped in favor of
  the dispersion forces and the remaining hydrogen bonds are stronger
  due to the close proximity of the contributing oxygen atoms [655].
  Viscosity, then, increases with pressure.

Second question on the viscosity of gases
The textbooks seems to be talking about a limited range of P. Viscosity of gases is dependent on very low or very high pressures. However, for ideal gases and normal pressures, it is only dependent on the temperature. Yes, it is quite counterintuitive. Also viscosity of gases increases with temperature. This is another surprise. https://en.wikipedia.org/wiki/Viscosity#Pure_gases
You can try posting this second query in Physics forum. I don't know the derivation of that Wikipedia expression.

Thanks for the answer. Could you also answer the second question? (Regarding the viscosity of gases)

I added the second part. You will have look up the mathematical derivation of the viscosity relationship.


text:
Density of water increases from 273 to 277 K due to increase in entropy?
In a set of questions that I was solving, the answer to one of the questions asking why the density of liquid water increases from 0 to 4 degrees C had the above answer: due to an increase in entropy.
This website does explain this a bit with two thermodynamic equations, but I am unable to make sense of how entropy comes into the picture here. Please explain how the increase in density of water is due to an increase in entropy.

IMHO, increase of entropy is not the cause of density increase, but both are consequences of the common cause. The less dense water structure near freezing point are less random and have lower entropy. If more organized structure were more compact as usually, increasing entropy would be toward lower density.

https://chemistry.stackexchange.com/questions/81756/why-is-density-of-water-maximum-at-4-degree-celsius/105390#105390 and the papers referenced in the answer (mine) suggest that entropy has a role to play (but isn't the only factor).

It's very hard to see what entropy has to do with this. The entropy of liquid water increases monotonically with temperature. So, attributing a density maximum to entropy alone does not make sense. Also, this question is unanswerable as there is no agreed upon cause of the density maximum in liquid water.

At $0°$C, the water is not yet made of independent $H_2O$ molecules. It is still a mixture  of hexagones containing 6 molecules $H_2O$. And these hexagones are separated by independent $H_2O$ molecules. There is still a certain order in the water at 0°C. These hexagones are destroyed between $0°$C and $4$°C. These hexagones are more cumbersome than 6 molecules $H_2O$, because there is a hole in the center of the hexagones. That is why the density of water at $0$°C is lower than at $4°$C.

This does neither answer the question, nor is it entirely correct. At least to the extend from what I understood from the page the OP linked.

@martin you may be right, but do you have an alternative we can upvote?

@Oscar Unfortunately I can't dig further into the ressource the OP linked, even though it appears very thorough, and probably a very good read. At least the images provided there look stunning. I believe water, as a system in itself, is incredibly complex, and I would love to have someone with more experience on the matter write an answer. Or maybe someone who has the time and the passion to read and understand and excerpt the linked website.


text:
Kinetic Energy Evaluation Integral Evaluation Program
I'm reading Ostlund's Modern Quantum Chemistry. In Appendix A, the kinetic energy integral is evaluated using the Gaussian Basis functions to be

$$
\left(A\left| -\frac{1}{2}\nabla^2 \right| B\right) =
\alpha\beta/(\alpha + \beta)[3 - 2\alpha\beta/(\alpha + \beta) |\mathbf{R}_A - \mathbf{R}_B|^2][\pi/(\alpha + \beta)]^{3/2} \\
\times \exp [-\alpha\beta/(\alpha + \beta)|\mathbf{R}_A - \mathbf{R}_B|^2]
\tag{A.11}\label{kin-en.int}
$$

So, in the integral evaluation the Gaussian functions themselves are not used, but in the computer program  he is evaluating the integral using
T11=T11+T(A1(I),A1(J),0.0D0)*D1(I)*D1(J)

The function T() is calculating the equation above \eqref{kin-en.int}.
I can't understand why he is multiplying by D1 and D2 which are the Gaussian functions themselves.
$$g_\mathrm{1s}(\alpha) = (2\alpha/\pi)^{3/4}\mathrm{e}^{-\alpha\mathbf{r}^2}$$

Just a comment, if you are looking to write your own integral code, I highly recommend taking a look at the following link https://joshuagoings.com/2017/04/28/integrals/

The equation for the Kinetic Energy matrix element that you quote is for two unnormalised 1s Gaussians. The d factors contain the normalisation factors and the contraction coefficients - look more carefully at the code, you have got it slightly wrong what the d's represent.
Talking of the code please, Please don't use that as a model for your own Fortran - that style is getting on for 50 years out of date!

Thank you Ian for your answer, I'm using another java version, I just wanted to quote the exact line from the book.

He is using T11,T22 and T12. I understand that T11 is the kinetic energy of the first atom, T22 is for kinetic energy of the second atom, what is T12 used for?. Thank you

Best to ask a separate question but (without looking at the book) think matrices is my guess