text:
Definition of ionisation energy
My lecturer has defined ionization energy or ionization potential of a species A as the molar internal energy change, ΔU, for the reaction: $\ce{A (g) -> A+ (g) + e-}$ at $\pu{0 K}$, with reactants and products being in their standard states.
How can reactants and products be in their standard states when they must be gaseous?!
He then goes on to say that we often assume  $\Delta U (\pu{0 K})\approx \Delta H (\pu{298 K})$. I presume he means that the molar internal energy change of the above process at $\pu{0 K}$ is approximately equal to the enthalpy change of the above process at $\pu{298 K}$? Why is this approximation valid?

I am wondering how some teachers can teach simple topics in a dry, complicated and "bookish" way. https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/02%3A_Basic_concepts%3A_molecules/2.4%3A_The_Octet_Rule_and_Isoelectronic_Species/24.D%3A_Periodic_Trends. They explain it here "This energy change is generally considered equivalent to the enthalpy change at 298 K (ΔH298 K). Estimates of the error suggest < 10 kJmol-1 which when compared to typical IE values often in their thousands, is insignificant."

It is common practice to define the standard state of a gas when describing its thermodynamic properties. For instance, the free energy of an ideal gas can be defined relative to the free energy in its standard state (the gas at a partial pressure of 1 bar and at the same temperature), as
$$G = G^\circ + RT\log\left( \frac{p}{p^\circ} \right)$$
where $p^\circ= \pu{1 bar}$.
In practice it may be impossible to achieve the gaseous state described in the OP, not least because absolute zero is not experimentally accessible. However, there are workarounds that might allow one to extrapolate to $\pu{0 K}$ and to standard conditions (a pressure of 1 bar). First the easy part, the pressure: if you have a very dilute gas, then intermolecular interactions become unimportant in describing its behavior, it becomes practically ideal. As such, its behavior at any other pressure can be described by using standard equations applicable to ideal gases. It does not matter that the state described is not real, it can be used as a hypothetical reference point. For instance, if we know the free energy under ideal gas conditions at a particular pressure, we can compute the free energy in the standard state as follows:
$$G^\circ = G - RT\log\left( \frac{p}{p^\circ} \right)$$
For an ideal gas, the internal energy $U$ is independent of pressure, so in the given example there is no need to perform any computation at all, provided we are at the right T. The problem is of course that we are not at the right T. In practice it may be possible to either assume that the experiment was conducted sufficiently close to $T=\pu{0 K}$, so that we can assume the equality holds, or to extrapolate to that temperature by performing experiments over a range of T.
As for the approximation $\Delta U (\pu{0 K})\approx \Delta H (\pu{298 K})$, I'll give a roundabout justification, since I am not an expert on the subject. There are two points that can be made. First, it is probably ok to assume $\Delta U (\pu{0 K})\approx \Delta U (\pu{298 K})$ if it can be assumed that the electronic transition is not sensitive to differences in the distribution of (mainly vibrational) thermal motions at $\pu{0 K}$ and $\pu{298 K}$.
Second, $$\Delta H = \Delta U + \Delta (pV) = \Delta U + \Delta n RT $$ However here we have a conundrum. The reported IE ignores the kinetic energy of the electron (it assumes that the emitted electron has no kinetic energy). But the thermodynamic expression assumes thermal equilibrium, so which is it? Well, at zero Kelvin there is no translational energy, so when considering $\Delta U(\pu{0 K})$ we can safely ignore any thermal contribution from the electron. So we are ignoring the thermal properties of the released electron, and then since $\Delta n =0$ (ignoring the electron) we have that $$\Delta H = \Delta U$$


text:
Contribution of vibrational degrees of freedom in linear triatomic molecule

The ratio $\displaystyle\frac{C_p}{C_v}$ for linear triatomic molecules including vibration contribution is given by ____.

Attempt:
We are aware that $C_v = \frac{fR}{2}$ and $C_p = C_v+ R$
Hence, $\gamma = 1 + \frac 2f$
Now, for $f$ we have:

$2$ rotational degrees of freedom
$3$ translational degrees of freedom
$3N - 5 = 3\times 3-5 = 4$ vibrational degrees of freedom (formula available here)

Thus, $f = 2+3+4 = 9$
And $\gamma = 1+ \frac 29  = 1.22$
But answer given is $\gamma = 1.154$
Someone in chat room told me that each degree of freedom of vibrational actually has two times contribution so we would have $2\times 4  = 8$ vibrational degres of freedom and a total of $f = 13$, which gives $\gamma = 1.154$. However then I see a contradiction here. If vibrational indeed contributed $2$ per degree then degree of freedom of methane would be $3+3+2\times(3\times 5-6) = 24$ not $3+3+ (3\times 5 - 6) = 15$.
Can someone please clear the confusion here?

Each degree of freedom contributes $R/2$; so each vibration has $R/2$ from potential energy and the same from the kinetic energy making $R$ in total. This is based on the equipartition of energy and we add $R/2$ for each 'squared term' in the energy meaning the squared velocity in kinetic energy as $mv^2/2$ and $x^2$ in the harmonic potential, $kx^2/2$.

VibrationAL KE is a reasonable quantity at high temperature, usually ignored at stp

@Chemist, yes this is true, but what is a 'high temperature' depends on the molecule, one with heavy atoms will have some low frequency vibrations, also bending vibrations are usually of low frequency. The heat capacity then depends smoothly on temperature, see https://chemistry.stackexchange.com/questions/110944/are-the-vibrational-modes-of-co2-active-at-room-temperature/110947#110947. (The partition function is needed and then calculate $(\partial U/\partial T)_V$ vs $T$. )

You have to be careful to make a distinction between the coordinates needed to describe a molecule, and the degrees of freedom which the molecule can actually access. For this purpose, one might choose to define the number of degrees of freedom as the minimum number of spatial coordinates needed to fully describe the positions of atoms relative to a fixed reference frame. This will be $3N$ total coordinates which corroborates the other answer you link. We then recognize that motion for non-linear (linear) systems, motion in six (five) of these coordinates correspond to translation and rotation. Therefore, the other $3N-6$ ($3N-5$) coordinates describe the vibrations of the system.
Now, if one is concerned with the total energy of a system, then we should keep track of the Hamiltonian. In a classical context, the Hamiltonian is a function of all coordinates which are capable of holding energy in one form or another, $H(\mathbf{q},\mathbf{p})$. For molecules this means we must keep track of $3N$ spatial coordinates, $\mathbf{q}$, and $3N$ momentum coordinates, $\mathbf{p}$.
Now, if we want to know about the average kinetic energy of atom 1 in the $x$ direction, we perform an ensemble average
$$
\langle T_{x,1}\rangle=\frac{1}{2m}\frac{\int p_{x,1}^2e^{-\beta H(\mathbf{q},\mathbf{p})}d\mathbf{q}d\mathbf{p}}{\int e^{-\beta H(\mathbf{q},\mathbf{p})}d\mathbf{q}d\mathbf{p}}
$$
It is a nice exercise to show that if all the degrees of freedom in the Hamiltonian are quadratic, then this separates into lots of integrals over gaussians and the top and bottom will cancel in all coordinates except for the $x$ direction motion of atom 1. This integral will give you the average energy as $1/2kT$.
But, since the integral over the quadratic momentum and the quadratic position as in the harmonic potential are exactly the same, we have just shown that every type of energy which depends quadratically on a coordinate has an average energy of $1/2kT$.
Therefore, if we have a particle in 1-D subject to a harmonic potential, the Hamiltonian is,
$$
H=\frac{p^2}{2m}+\frac{1}{2}kx^2
$$
which means the average energy is $kT$ since there are two coordinates which the Hamiltonian depends on quadratically.
So, hopefully now it is clear why we double to count of vibrations and arrive at the answer,
$$
\gamma=1+\frac{2}{13}=1.154
$$
I know this was very long-winded and overly detailed, but I hope this gives you a little better picture into where the equipartition theorem comes from and how this convenient counting of coordinates we do is really a coincidence of the fact a harmonic potential and kinetic energy are each quadratic in their respective coordinates.

As an aside, the distinction I have made between coordinates and degrees of freedom is completely arbitrary. Indeed, people often refer to both a harmonic potential and the kinetic energy as quadratic degrees of freedom. I prefer this language, but then one must be careful to clarify between degrees of freedom of position and degrees of freedom of momentum because this is more coordinates than are needed to describe the system completely, so it seems there are more degrees of freedom than there really are.

source: Stack Exchange

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text:
Exception in trend of increasing ionisation energy across a period
One exception to the trend of increasing ionisation energy across a period in e.g. period 2 is going from N to O. There is no loss of exchange energy in I1 of O, but there is 2K lost in I1 of N, hence more energy is required to remove the valence electron of N.
If Zeff increases from N to O, and n stays constant, then
$$E=-R_H \frac{{Z_{eff}}^2}{n^2} $$
must become more negative, and I = -E, which implies that I increases, when in actual fact it decreases.
Is this equation therefore flawed?

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text:
Jacobi coordinates - definition
I became quite confused about Jacobi coordinates. Until now I've been used to them as the coordinate system, where coordinates are vectors pointing to the center of mass of previous "subsystem", as we can see in the picture

Also, they're described in the same way in the Wikipedia.
But today I've met two colleagues, who were talking about Jacobi coordinates in 3-atomic system as about "vector-vector-angle", looking like this

with descriptor set being $\left(R,r,\cos^2(\Theta)\right)$.
Is this coordinate system also called Jacobi coordinates? And if it isn't, what's its real name?


text:
Orbital size of Carbon and Oxygen in Methanal
I was given an introductory question into Carbonyl chemistry today and I wanted to ask a question about Methanal.
The following picture was shown below of Methanal:

and I noted that the orbital sizes differed for both Carbon and Oxygen in the top and bottom diagrams. I thought one of these diagrams must be wrong.
I then knew that as Oxygen has more protons than Carbon, then its atomic radius is smaller, meaning the orbital size must be smaller too, so I thought that the bottom diagram represented the size of the p orbitals better both for anti bonding and bonding.
Is the diagram for $\pi$ bonding drawn correctly in terms of orbital size?

You hardly ever need to worry about the shape of the bonding orbital. The HOMO is probably the one (or linear combination of the ones) that contains the lone pairs. The LUMO below does have the right size since the two factors are in the same direction. Of course the real molecular orbitals span the whole molecule, so this is just a simplified representation of what two specific contributions might look like. For example, see: https://www.quora.com/What-do-the-C-O-molecular-orbitals-of-acetaldehyde-look-like-compared-with-those-of-formaldehyde

The orbital size in the picture shows the coefficient in the linear combination of atomic orbitals, i.e the pi bonding MO has a larger contribution from the oxygen atomic orbital, and the pi anti-bonding MO has a larger contribution from the carbon atomic orbital.
Here is another picture for a different molecule where they use the same convention, but also show the coefficients explicitly (source):

Is the diagram for π bonding drawn correctly in terms of orbital size?

An orbital is not a "thing" so it does not have a size. It is a 3-dimensional function. The picture just shows the symmetry of that function and the node (where it is zero). For a slightly more realistic depiction, you can check out molcalc.

I’ll try to keep this explanation simple- if you want a more comprehensive example, please consult books like Orbital Interactions in Chemistry.
Oxygen is more electronegative, and hence the Oxygen p orbitals experience a greater effective nuclear charge and have a lower orbital potential energy than the carbon p orbitals. As such, this is effectively analogous to the bonding situation in compounds like HF or LiH - the bonding orbital has a greater contribution from the lower-energy atomic orbital (Oxygen) while the antibonding orbital has a greater contribution from the higher-energy atomic orbital (carbon).
In the specific context of your question, in the case of such qualitative MO diagrams the individual AOs are assumed to be of the same size, and the size of the AOs in the MO “cartoon” simply represents the respective contributions. That’s why the MO diagram in your question is drawn as shown.
This is all a bit hand-wavy, so I suggest you read up on “electronegativity perturbation” if you’re interested.


text:
How to programmatically identify a chemical compound in my program?
I have a small set (a hundred of items) of chemical compounds in XYZ formats in my DB.
How to quickly identify that a given chemical is or isn't in this DB?
I don't want to just scan through them, or do something else of an ad-hoc nature.
Is there an algorithm that allows to do this?

Are you using database software, or are you programming in a general purpose programming language?

In general, you can't because a chemical formula is not an unambiguous way to identify a compound.

@KarstenTheis I use a general purpose programming language.

@Zhe I corrected this in the question: I use XYZ coordinates, not formulas.

How do you know one compound matches another? Same coordinates, same coordinates allowing for translation and rotation, same set of atoms with RMSD below threshold?

You might be able to convert your structures to smiles or something similar, i.e. InChI and then match that, or is hash. Obviously there are drawbacks, and you might lose information along the way, but this should be fairly quick, and the are libraries you might be able to use.

Cartesian coordinates capture a snapshot of a molecule in time. Molecules do adapt different conformations. Some equivalent connections are invalid because they cannot be accessed (atropisomers) and others have equivalent connections but the exact nature of the 3D connectivity matters (stereochemistry). This is a non-trivial problem. @Martin-マーチン 's suggestion can get you a good way there, but I don't think there is a general solution.

How to identify multiple molecules in a single xyz file?

Either way, you need to define what it means for a chemical to be 'in' your database. Do you care about the 3D coordinates (given that you have XYZ) or just the atom types and connectivity?
Short answer. use a cheminformatics library that provides structure matching or can encode your molecules (query and targets) as canonical SMILES, InChI, etc.
See Is there a way to use free software to convert SMILES strings to structures?
Long answer. Use a series of filters (or questions):

Are there any molecules in my database with the same number of atoms?
Do any of these have the same atom types?
Finally, are these atoms connected to the same neighbours?

The third one is a little trickier, but is still a quick way to determine isomorphism. If you  need true structure matching, see the short answer! If you need 3D matching, then again it's going to be much quicker to use a library method.

A series of filters, from cheap to more expensive, makes a lot of sense. For 1. and 2., you could write the sum formula in a string and use a dictionary (hash) to retrieve matches (built into many programming languages). For 3., you could use algorithms relating to distance geometry or, once you have a connectivity graph, use graph matching algorithms.


text:
Determination of pKa and molar absorptivity coefficient using spectrophotometry

An indicator BPB (Brophenol Blue) is used and dissolved in solutions of various known $\mathrm{pH}$. Then, the absorbance of each solution at 590 nm was measured and the question asks for molar absorption coefficient of the $\ce{HIn}$ form and $\ce{In-}$ form at the 590 nm wavelength.
Derive the relation of absorptivity coefficient $\varepsilon(\ce{HIn})$ and $\varepsilon(\ce{In-})$ to the total concentration of indicator and also the $K_\mathrm{a}$ for the indicator.
Hint: plot the value of $A$ vs $\mathrm{pH}$.

I don't know how to interpret the resulting graph. I tried to derive a relation, but it is not linear with respect to $\mathrm{pH}$. Here I'm determining the relations of absorbance to $\mathrm{pH}$, trying to interpret the graph of $A$ vs $\mathrm{pH}$, but the relations are not linear.
$$[\ce{HIn}] + [\ce{In-}] = c(\text{BPB})$$
$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{In-}]}{[\ce{HIn}]} \quad\to\quad [\ce{In-}] = K_\mathrm{a}\frac{[\ce{HIn}]}{[\ce{H+}]}$$
$$
\begin{align}
A &= \varepsilon(\ce{HIn})\cdot c \\
  &= b\cdot\left(\varepsilon(\ce{HIn})\cdot [\ce{HIn}] + \varepsilon(\ce{In-})\cdot [\ce{In-}]\right) \\
  &= b\cdot\left(\varepsilon(\ce{HIn})\cdot[\ce{HIn}] + \varepsilon(\ce{In-})\cdot K_\mathrm{a}\frac{[\ce{HIn}]}{[\ce{H+}]}\right) \\
  &= b\cdot[\ce{HIn}]\left(\varepsilon(\ce{HIn}) + \varepsilon(\ce{In-})\cdot\frac{K_\mathrm{a}}{[\ce{H+}]}\right) \\
  &= \frac{b\cdot c(\text{BPB})}{1 + \frac{K_\mathrm{a}}{[\ce{H+}]}}\left(\varepsilon(\ce{HIn}) + \varepsilon(\ce{In-})\cdot\frac{K_\mathrm{a}}{[\ce{H+}]}\right)
\end{align}
$$

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text:
How do I calculate the hydronium ion concentration if I know the strontium hydroxide concentration?
This is my first question so I would appreciate feedback on how to ask better questions. Forgive me if I don't know how to properly format math/chem. The guide is confusing.
I was taking a practice test on Brønstead–Lowry acids and bases and I came across a question that has totally stumped me:

Calculate the $\ce{[H3O+]}$ in a $\pu{0.010 M}$ solution of $\ce{Sr(OH)2}$.

Here are my steps:
Write a balanced equation:
$$\ce{Sr(OH)2 <=> Sr^2+ + 2OH-}$$
Initial concentrations:
$$\ce{[Sr(OH)2]} = \pu{0.010 M}$$
$$\ce{[Sr^2+]} = 0$$
$$\ce{[OH-]} = 0$$
Change in Concentration:
$$\ce{[Sr(OH)2]} = (0.010 - x)~\pu{M}$$
$$\ce{[Sr^2+]} = +x$$
$$\ce{[OH-]} = +2x$$
I am not sure whether $\ce{OH-}$ loses 2 moles to every one of $\ce{Sr(OH)2}$.
Assume $x\ll\pu{0.010 M}$. Now I will set up a $K_\mathrm{a}$ expression:
$$K_\mathrm{a} = \ce{\frac{[H3O][OH]}{[Sr(OH)2]}}$$
$$\frac{K_\mathrm w}{K_\mathrm b} = \frac{x\cdot 2x}{0.010}$$
I don't know what to now.  I feel like I have probably started the problem wrong.

As to learning how to format chem, I suggest you learn by opening up the "edit" option for posts containing them, and try imitating it on your posts.

Strontium hydroxide is a strong base, so you can calculate $[\ce{OH-}]$ as $\pu{0.02 M}$, then use
$$K_\mathrm{w} = \ce{[OH-][H3O+]}$$
$$1\cdot 10^{-14} = 0.02\cdot [\ce{H3O+}] \quad\to\quad [\ce{H3O+}] = \pu{5e-13 M} \quad\to\quad \mathrm{pH} = 12.3$$

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text:
Why do old black and white photos fade?
I understand why photos based on organic dyes would fade (discussed previously). But photos based on metallic silver fade as well such as like the one shown below. As a metal (a fairly stable one at that) I would think silver should be resilient to fading but still old black and white photos do fade.
What causes old black and white photographs to fade?

Source: www.photo-restoration.co.uk

https://en.wikipedia.org/wiki/Photographic_paper#Black-and-white_prints

Consider all the parts.It is not just the silver. The silver is in a medium that is mounted to a backing. All the parts are vulnerable to various ravages like temperature, humidity, UV, chemical pollution, physical abrasion etc ... To understand why any one photo faded, you would need to know what it is made of and precisely what it was subjected to.


text:
How can we tell if an ion is amphiprotic
If I had the following ions:
$$\ce{HCO3^-}$$
$$\ce{H2PO4^-}$$
$$\ce{CH3COO^-}$$
Which ones would be amphiprotic?
Looking at my $ K_A$ chart, it looks like each ion is formed by the disassociation of weak acids, so It looks like all of them are amphiprotic because they could all technically act as a bronstead-lowry base.
The answer is the first two ions, but why?
Thanks for answering :)

Shouldn't the third choice be just $\ce{CH3COO-}$?

yes, I fixed that :)

How does each anion change if you add $\ce{H+}$ and if you add $\ce{OH-}$?

To be amphiprotic means that the chemical species can donate or accept H+ ions.
$$\ce{H2CO3<=>[H+] HCO3^- <=>[OH-] CO3^{2-}}$$
$$\ce{H3PO4 <=>[H+] H2PO4^- <=>[OH-] HPO4^{2-}}$$
$$\ce{CH3COOH <=>[H+] CH3COO^- <=>[OH-] \text{No Change}}$$
Thus since the acetate anion can't donate a proton, it is not amphiprotic.

Good question. If you show these three ions (as written) to a super intelligent being person who never took chem or physics, they will not be able to tell you which ion is amphoteric. In short, there is no simplr short-cut! This requires some knowledge of some formula writing conventions.
For simple inorganic ions, if H is written with a p-block element, that H is ionizable and it is an acid in water
HCl, HBr, H2SO4, H2S, H3PO4, H2CO3
From HCl, we can remove one H+
From H2SO4, we can remove two H+, one at a time
From H3PO4, we can remove three H+, one at a time.
So if you happen to see ions of these acids, SO4(2-), you can immediately guess from the minus two charge that this ion can accept two protons.
If you see, HPO4 (2-), you should be able to guess that HPO4(2-) can accept one or two proton(s), however it can also lose the last one.
For small organic ions, it is slightly tricky, the ionizable H is written in the end with "COO" group
CH3COOH, HCOOH


text:
Obtaining atomic form factor from electron density?
I know that atomic form factors for x-ray scattering are obtained by Fourier transform of the electronic charge density.
I am looking for a citable elementary source (book, review, article) which covers the systematic and explicit derivation of the relationship between electronic charge density and the atomic form factor, to which people unfamiliar with the connection can be referred to for better context.

Something like this: http://web.physics.ucsb.edu/~fratus/phys103/LN/Scattering.pdf

Or 7.3.33 in https://ocw.mit.edu/courses/physics/8-06-quantum-physics-iii-spring-2018/lecture-notes/MIT8_06S18ch7.pdf ?

try IUCR  Texts on Crystallography -2, ed C. Giacovazzo publ. IUCR/OUP, chapter 3. for a brief but clear examination of scattering off electrons and atoms, and R. James, 'The Optical Principles of the Diffraction of X-rays vol II' , publ. Bell 1965, chapter 3 which is an older but still excellent book on all the basics.

The International Union of Crystallography (IUCr) compiled a freely accessible  Online Dictionary.*  To cite a section from the entry about the atomic scattering factor (often anonymously used for atomic form factor):

The scattering from a crystal of an X-ray beam results from the interaction between the electric component of the incident electromagnetic radiation and the electrons in the crystal. Tightly bound electrons scatter coherently (Rayleigh scattering); free electrons scatter incoherently (Compton scattering). The scattering process from atomic electrons in a crystal lattice has both coherent and incoherent components, and is described as Thomson scattering.
The scattering amplitude from a neutral atom depends on the number of electrons ($Z$ = the atomic number) and also on the Bragg angle $\theta{}$ – destructive interference among waves scattered from the individual electrons reduces the intensity at other than zero scattering angle. For $\theta{} = 0$ the scattering amplitude is normally equal to $Z$. However, the scattering factor is modified by anomalous scattering if the incident wavelength is near an absorption edge of the scattering element.
The X-ray scattering factor is evaluated as the Fourier transform of the electron density distribution of an atom or ion, which is calculated from theoretical wavefunctions for free atoms.

Scattering factor of stationary C and Fe atoms plotted as a function of Bragg angle for incident X-ray wavelength of $\pu{0.70930 Å}$. Ticks on the horizontal axis correspond to Bragg angle increments of 10 degrees; ticks on the vertical axis are increments of 5 electrons.

The lemma equally picks three references from the International Tables (chiefly IT) for additional information about the theory and listing the values.  You may find it in print in groups active in e.g., (chemical) crystallography, earth sciences, solid state physics, structure biology; or in research libraries with the corresponding subscription:

C. Colliex, J. M. Cowley, S. L. Dudarev, M. Fink, J. Gjønnes, R. Hilderbrandt, A. Howie, D. F. Lynch, L. M. Peng, G. Ren, A. W. Ross, V. H. Smith Jr, J. C. H. Spence, J. W. Steeds, J. Wang, M. J. Whelan and B. B. Zvyagin. International Tables for Crystallography (2006). Vol. C, ch. 4.3, pp. 259-429 doi:10.1107/97809553602060000593.  (preview of the content)
Intensity of diffracted intensities: P. J. Brown et al., A. G. Fox, E. N. Maslen, M. A. O'Keefe and B. T. M. Willis. International Tables for Crystallography (2006). Vol. C, ch. 6.1, pp. 554-595 doi:10.1107/97809553602060000600 (preview of the content)
Neutron techniques: I. S. Anderson, P. J. Brown, J. M. Carpenter, G. Lander, R. Pynn, J. M. Rowe, O. Schärpf, V. F. Sears and B. T. M. Willis. International Tables for Crystallography (2006). Vol. C, ch. 4.4, pp. 430-487 doi:10.1107/97809553602060000594 (preview of the content)

* This reference already is among the site's resources for learning chemistry.

Here is a possible source: Modern X-ray analysis on single crystals: a practical guide by Peter Luger, De Gruyter, Inc. 2014.
The relevant excerpt describing how to estimate the density of an atom j, and with that, the scattering factor (really scattering function) is below.

So atomic scattering factors are derived by taking the Fourier transform of the computed electron density of a stationary atom (the density has spherical symmetry, so the scattering factor is a function that only depends on $s = \sin \theta / \lambda$).

Thanks for your time. But this is not what I am looking for. The atomic scattering factors are calculated by taking a Fourier transform of the charge density. But this relation itself has an analytical derivation which starts with differential scattering cross section. That is what I am interested in.


text:
Redox Combustion Reactions
Consider a combustion reaction like the complete combustion of ethane. Here is what I have for the complete combustion as well as the oxidation and reduction:
$$\ce{\overset{-3}{C}_2\overset{+1}{H}_6 + \frac{7}{2} \overset{0}{O}_2 -> 2 \overset{+4}{C}\overset{-2}{O}_2 + 3 \overset{+1}{H}_2\overset{-2}{O}}$$
$$
\begin{align}
\ce{2 H2O + C2H6 &-> 2 CO2 + 10 H+ + 10 e-}\tag{ox}\\
\ce{\frac{1}{2} O2 + 10 H+ + 10 e- &-> 5 H2O}\tag{red}
\end{align}
$$

Why is it that when writing the reduction half equation for the reduction of oxygen that we don't consider oxygen being reduced into carbon dioxide? i.e. why isn't the reduction equation something of the form $\ce{O2 -> H2O + CO2}~?$

Also the oxidation reaction suggests that water is required initially for the reaction to proceed, is this true?

Edit
I realise my reduction half equation is not balanced and so I now have this equation:
$$\ce{\overset{-3}{C}_2\overset{+1}{H}_6 + \frac{7}{2} \overset{0}{O}_2 -> 2 \overset{+4}{C}\overset{-2}{O}_2 + 3 \overset{+1}{H}_2\overset{-2}{O}}$$
$$
\begin{align}
\ce{2 H2O + C2H6 &-> 2 CO2 + 10 H+ + 10 e-}\tag{ox}\\
\ce{\frac{7}{2} O2 + 14 H+ + 14 e- &-> 7 H2O}\tag{red}
\end{align}
$$
which is clearly not correct as the reduction and oxidation half equations do not add together to give the combustion reaction and the reduction equation does not have the $\ce{CO2}$ product as well. I guess my question now stands as how might I write (or begin to write) the half equations for a combustion reaction?
Edit 2
I realise I wrote my equations wrong above
$$
\begin{align}
\ce{4 H2O + C2H6 &-> 2 CO2 + 14 H+ + 14 e-}\tag{ox}\\
\ce{\frac{7}{2} O2 + 14 H+ + 14 e- &-> 7 H2O}\tag{red}
\end{align}
$$
But my question still remains. Why do we not consider the reduction of $\ce{O2}$ into $\ce{CO2}$ as well?

Such combustion is a different type of reaction than a typical redox reaction. For combustion reactions there are a multitude of intermediate products.

Don't use this redox approach for balancing organic reactions. It will make miserable. The approach is not only wrong, but it is confusing you for unnecessary reasons. Use simple algebra or do trial guesses for practice. If a teacher is using this way, please tell him/her that this is not a good way.

I'm not sure what you mean by simple algebra or trial guesses. I don't quite understand what I have done wrong. Could you please elaborate

I do not know if this method is good or bad but you could always proceed through balancing the (mostly) reduction reaction ie for oxygen, as in combustion oxygen (surely) does reduce. @sabhoque : Your reduction reaction is a wrong one: 1) It is not balanced, 2) Not all oxygen atoms which  are the reduced products (CO2 and H2O both) appear in your half reaction.

@glucose how did I not see that it was not balanced. I have edited my question and am now unsure of how to write the half equations

Consider the fact that O(0) as in O2 is being reduced to O(-2) as in H2O and CO2 in the reduction half reaction.

I will have to strongly back M. Farooq's comment. It is really not advisable to use redox approach in organic reactions.

Note that near all redox reactions can be expressed as redox half reactions with related standard redox potentials, which are equivalent to the Gibbs energy of the half  reaction, oxidizing the hydrogen. But many of them have only formal significance in electrochemistry, as such reactions due slow kinetics may not be ever observable and inert electrodes like Pt may not measure such a potential.

A fuel cell that uses ethylene as a fuel is not inconceivable.  In that case, the proposed half reactions (if balanced) are reasonable models for the reaction, protons and all.  Direct methanol fuel cells are a thing after all.

I disagree with M. Farooq and William R. Ebenezer.  Organic reactions can be modeled this way, and at least some of the time, it is useful to do so.

Your redox equations look awkward (and wrong, too) because you are trying apply the half-reactions method as if it  were happening in aquatic media to the combustion reaction, which effectively occurs in the gas phase.
Of course, there should be no requirements for water to be present.
As it's been noted in the comments, ionic half-reactions method is not the most elegant way of find coefficients in this case.
If you really want to use it, I'd suggest to write the elements undergoing redox reactions separately from the corresponding compounds (and yes, we do consider oxygen being reduced into carbon dioxide too):
$$
\begin{align}
\ce{2 \overset{-3}{C} &→ 2 \overset{+4}{C} + 14 e-} &|\cdot 2 \tag{ox} \\
\ce{\overset{0}{O}_2 + 4 e- &→ 2 \overset{-2}{O}} &|\cdot 7 \tag{red} \\
\hline
\ce{4 \overset{-3}{C} + 7 \overset{0}{O}_2 &→ 4 \overset{+4}{C} + 14 \overset{-2}{O}}
\end{align}
$$
resulting in a balanced equation:
$$\ce{2 C2H6 + 7 O2 → 4 CO2 + 6 H2O}$$
Note that in general there is an average oxidation number of carbon involved, and it might as well be fractional.

This is a very interesting approach that I have not seen before. It seems to suggest that both reactants atomize and undergo their respective half reactions or are redox half equations generally not an indicator of reaction mechanisms?

@sabhoque No, half-reactions method is a simplistic construct helping to illustrate electron transfer during the redox process (and to balance the reaction along the way) and has little to do with the mechanism. Whereas one would include oxo- and hydroxocomplexes in "typical" red+ox half-reactions (e.g. $\ce{SO4^2-}$ or $\ce{[Cr(OH)6]^3-}$) occurring in aqueous phase as the units that do participate in the process, the same cannot be done in the case of combustion as there are tons of radicals generated, and it's overall easier to track oxidation numbers of the elements instead.


text:
Where does copper form?

 I thought zn and cu both lose 2e- to pt but someone is telling me it's wrong]

If both Zn and Cu were to lose electrons, what would receive these electrons?

the cu2+ from cuso4?

Which one? We have two separate compartments with CuSO4.

the cuso4 for both platinum?wouldnt pt a and pt b both receiving2e- since cu and zn are oxidized when connected with pt?

You might want to check again the oxidation state of Pt, and whether it can be reduced.

@donnny consider also respecting electrostatic neutrality. Separating of few coulombs would create very mighty force.

Zn doesn't lose electrons to Pt: it rarely participates in cell reactions (it's called inert electrode!)

Unless Pt is just passing them to $\ce{Cu2+}$


text:
Mechanism for Friedel-Crafts Acylation in Synthesis of Sertraline
In the first step of the synthesis of Sertraline, aluminium chloride catalyst was used in a Friedel-Craft's reaction to give the following product. My question is how does the exact mechanism work? 1-napthol does not seem look like an electrophile to me, and I do not understand how AlCl3 is able to turn 1-napthol into an electrophile. The best guess I have is that the phenol tautomerise to give a ketone, and AlCl3 forms a complex with it. Yet again, I do not understand how the γ carbon is made electrophilic using AlCl3. Any help would be appreciated.

Even if we assume that the $1$-napthol is getting attacked somehow by $1,2$-dichlorobenzene, instead of substituted $\alpha$-tetralone in the product, shouldn't the product contain $\alpha , \beta$-unsaturated ketone ? The only plausible mechanism generates that by a conjugate addition.

Note that this is not a Friedel-Crafts acylation. In that reaction, the Lewis-acid is a catalyst. 2 equiv. of aluminium chloride here is definitely not catalytic. In other words, you need a superstoichiometric amount of the Lewis acid to get this reaction to go. Almost certainly coordination of the OH is involved as the first step. But the aluminum chloride is likely not regenerated in this reaction.

Also, very importantly, look at the severely reduced unsaturation of the ring. You're going to need a hydride from somewhere.

The alleged original RSC source for this reaction was not accessible online.1. However, Olah et al. have reported2 that α-naphthol 1 in superacid medium forms a dication species 4. The use of excess AlCl3 provides a mechanism for the formation of superacid 3. Benzylic dication 4 undergoes substitution of o-dichlorobenzene 5 which ultimately forms the sertraline precursor, tetralone 8. For reaction conditions, see ref. 3. No reducing agent is required.

1) Kavitha, C. V.; Mantelingu, K.; Sarala, G.; Naveen, S.;
Anandalwar, S. M.; Prasad, J. S.; Rangappa, K. S. J. Chem. Res. 2006,
11, 730.
2) Koltunov, K. Y.; Chernov, A. N.; Prakash, G. K. S.; Olah, G. A. Chem. Pharm. Bull. 2012, 60, 722.
3) Lee, S. H.; Kim, I. S.; Qing, R. L.; Dong, G. R.; Jeong, L. K.; Jung, Y. H. J. Org. Chem., 2011, 76, 10011.


text:
Proving a relation for fugacity of component i in term of independent variables (T,V)
In Molecular Thermodynamics of fluid phase equilibria by Prausnitz et al. [1] the authors recommend to use the following equation which gives the fugacity of component $i$ in terms of independent variables $V$ (volume) and $T$ (temperature):
$$RT\ln\phi_i = \int_v^\infty\left(\left(\frac{\partial p}{\partial n_i}\right)_{T, V, n_i} - \frac{RT}{V}\right)\mathrm{d}V - RT\ln z$$
I know that I should use chemical potential definition to prove this equation, but there are many different ways to calculate it like using Helmholtz free energy or Gibbs free energy.
I would appreciate someone give me an idea or any references to help me proving this equation.
References

Prausnitz, J. M.; Lichtenthaler, R. N.; Azevedo, E. G. de. Molecular Thermodynamics of Fluid-Phase Equilibria, 3rd edition; Prentice Hall: Upper Saddle River, N.J, 1998. ISBN 978-0-13-977745-5.

See my new addendum.

I haven't been able to work out all the details of the mathematics yet, but the derivation of this equation must start out from the following equation:
$$nRT\mathrm{d}\ln{\phi} = (V - V^{ig})\mathrm{d}P,$$
where $\phi$ is the fugacity coefficient of the mixture, $n$ is the total moles of the mixture, $V$ is the volume of the mixture $znRT/P$ and $V^{ig}$ is the volume of the mixture under ideal gas conditions: $nRT/P$.  Once the equation is integrated to get $RTn\ln{\phi}$, we get the fugacity of the $i$th component of the mixture by evaluating the following partial derivative:
$$RT\ln{\phi_i} = RT\frac{\partial (n\ln{\phi})}{\partial n_i}$$
Hope that this helps a little.
ADDENEUM
If we write, $$d(nV)=\left(\frac{\partial (nV)}{\partial n_i}\right)_{p,n_k}dn_i+\left(\frac{\partial (nV)}{\partial p}\right)_{ni,n_k}dp$$
then it follows that:  $$\left(\frac{\partial (nV)}{\partial n_i}\right)_{p,n_k}=-\left(\frac{\partial (nV)}{\partial p}\right)_{ni,n_k,V}\left(\frac{\partial (p)}{\partial n_i}\right)_{n_k,V}$$
The rest is easy.  The subscript $n_i$ in the OP equation is a typo.  It should be $n_k$, representing all other species being held constant.

It very much looks like one of the thermodynamic equations after being subjected to some maxwell relation partial derivation. Or perhaps starting from the definition of fugacity and then just iterating towards T and V as variables. Oh and BTW, fugacity and ideal gas are mutually exclusive. Fugacity is a property only real gases have.

That's not my take on this.  The first equation I wrote represents the basic relationship for the Residual Gibbs free energy of a mixture, and the second relationship follows directly from the definition of a partial molar quantity of a species in solution, in this case $\phi_i$.  Getting the final equation in the form that they have written it must involve writing VdP=d(PV)-PdV, but I still haven't been able to work out the mathematics.

Actually, the partial molar quantity is $\ln{\phi_i}$


text:
What defines a discharged battery?
In the case of an AA Alkaline battery, what actually happens inside the battery once it is fully discharged? What components of the battery change as we use it?

I think perhaps discharged is a highly relative term in reference to the surrounding chemical environment.

what actually happens inside the battery once it is fully discharged?

Alkaline batteries use the exchange of electrons from zinc to manganese dioxide to produce electricity
$$\begin{array}{rcl}\\
\ce{Zn (s) +2 OH- (aq)}& \ce{->}& \ce{ZnO (s) + H2O (l) + 2 e-}\qquad \text{(anode)}\\
\text{(high potential anode)}\quad \ce{2e-}&\ce{->[load][\ ]}&\ce{2e-}\quad\text{(low potential cathode)}\\
\ce{2MnO2 (s)+ 2e- + H2O (l)} &\ce{->}&\ce{Mn2O3 (s) + OH- (aq)}\qquad \quad\text{(cathode)}\\
\end{array}$$
typically a battery voltage would decline according to:
$$\mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF}\ln Q$$
but in alkaline batteries all of the products and reactants are solids, which without going into heavy detail of the thermochemistry means that $Q = 1$ thus $\frac{RT}{nF}\ln Q = 0$ which means cell potential remains fairly constant ($\mathscr{E} = \mathscr{E}^{\circ}$). But this is only in theory as we can see alkaline batteries do lose voltage:

This is partly because the internal resistance of the cell will increase with time meaning that for a given external load the total load will increase with time reducing current and apparent voltage. But even with no load voltage drops, the the zinc oxide will dissolve in alkaline medium meaning the reactions depart from ideal equations:
$$\ce{ZnO (s)  ->[+ H2O (l)] Zn(OH)2 ->[+2OH- (aq)] [Zn(OH)4]^2-}$$
this is not to say that $\mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF}\ln Q$ is an unreliable equation, but it does not account well for the complexity of the system and does not apply well to alkaline batteries. Additionally it cannot explain the vertical slope of the discharge curve. What can explain the vertical slope is that once all of either zinc or manganese dioxide is consumed, there is no more matterial to react and the battery is considered fully discharged.

A discharged battery means that cell reaction has achieved equilibrium. When this system is at equilibrium no (useful) work can be done by the battery. It cannot drive or pump electrons in the circuit. In other words, people say that its Gibbs free energy is zero.

I think you can improve this answer by incorporating the Nernst equation.

Good idea Zhe. I find MathJax difficult. Feel free to edit.

I don't quite feel comfortable editing to say something different, but I will add an addendum as a separate answer.

@M.Farooq You cannot rely on other's edits and avoid proper markup forever. You'll just have to bite a bullet and get to know formatting of scientific documents. This may also help you to get started with $\mathrm{\LaTeX}$ when you start publishing your work. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Latex is good, I have heard. Somehow never needed it for my own publications. "This may also help you to get started with LATEX when you start publishing your work." andselisk, I have significant publishing experience for the last ten years :-)

I don't know how you managed not to use LaTeX for publications in natural sciences, but that's none of my business:) Unfortunately, you cannot ditch MathJax on Chemistry.SE. Really, MathJax isn't the hardest ting in the world, maybe you could just try adding it gradually in your answers while others fix the issues if there'll be any – it's still better than asking others to do the core writing for you:)

It depends on the field. What is your field of specialization? MS Word has a decent professional equation editor. Most of good science was also done when we had typewriters :-). LaTex is not the end of the world. I also review articles for ACS, Elsevier, Springer etc. Most articles are written in Word, and then converted into pdf. Anyway, this is just a distraction for the OP.

Mine mostly gravitates towards inorganic/solid state chemistry; to deal with publishers who are only accepting MS Word files, I'd send a pandoc'ed DOCX version, and if they are unhappy (they usually are, who would've thought), they get a PDF. It's entirely a personal choice, but I find LaTeX much more easier, cross-platform and cross-versions compatible, robust and transparent, and it rarely gets in a way.

I have heard LaTex makes good looking CVs. It is a personal choice. I know a (old) colleague who uses Excel all the time-even very complicated simulations. I told him of MATLAB but he doesn't budge. I believe science is getting all about presentation. I would love the old time where the focus was science not decoration.

@M.Farooq : A note to you also... Could you please read my comment to Zhe's answer.

@glucose, You are right, a discharged battery does not mean that the voltage is zero.

@M.Farooq : which implies that the Gibbs free energy of cell is also not zero... since the cell is in equilibrium with its surrounding... meaning the system of cell and discharger component are in equilibrium...

@glucose, I believe the story is far more complex for real batteries. A battery can be dead (as well) if the internal resistance becomes too large during the discharge process. Zhe and I were talking about the simple Cu/Zn cell. See Zhe's point as well. What do you think?

@M.Farooq : Complex batteries: agree... I do agree with Zhe's  point that 'There is no potential to push electrons through any circuit' (irrespective of surrounding)... And even your point: 'When this system is at equilibrium no (useful) work can be done by the battery. It cannot drive or pump electrons in the circuit' ... But I disagree when it comes to Gibbs free energy or potential of the particular cell reaction(caution: Not cell system ie  body + surrounding) being zero.

What is the value of Gibbs free energy at equilibrium? It is supposed to be zero because a system at equilibrium cannot do any useful work. Let us not discuss real cells which are quite complex, but what is your take with a simple Cu/Zn cell? During discharge a time will come when E will be zero. What would be the corresponding G value?

The discussion's going on great and I would really like to continue but (extremely sorry) actually I am travelling... So would continue the discussion after probably a day or two, or even at times in between...

Better to start a new thread/question once you come back. Nobody reads comments so people will miss your point of view.

To extend @M.Farooq's answer, let's look at this a slightly different way.
From the Nernst equation:
$$\mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF}\ln Q$$
As the reaction progresses, the value of $Q$, the reaction quotient, increases because product concentrations are higher and reactant concentrations are lower. As a result, the cell potential $\mathscr{E}$ decreases.
When $\mathscr{E}$ reaches zero, then battery is fully discharged. There is no potential to push electrons through any circuit.

AFAIK there are only two accepted notations for the standard state: uppercase plimsoll "$^⦵$" and a circle "°". "∅" looks like an "empty set symbol" and doesn't make any sense here in this context.

I reverted to using \varnothing because there is no simple LaTeX symbol and I had various instances where the plimsoll unicode was not supported. I'll use circ instead, though to be fair, the empty set symbol (and that's what I'm using) is just a rotated plimsoll. @andselisk

May I ask you where the "⦵" Unicode symbol isn't supported? I've been using it thorough with MathJax and XeLaTeX for the past couple of years and it always worked for me just fine.

It appears as a small rectangle on this Android tablet.

@badjohn Interesting, your Android must be lacking fonts with the plimsoll symbol. Personally, I've never experienced this issue on any of the Android devices (I used AOSP since version 4 and later LineageOS), but it's good to know some vendors might cut off support for special characters.

@andselisk It's rather old.  A Samsung Galaxy Tab S SMT-700 running Android 6.0.1.

Hmm, my comment didn't post. Yeah, it doesn't show up on my Pixel book, which is one of the newer models. My windows desktop and Linux workstation are both able to show it.

@Zhe I don't think it's a hardware issue, rather a firmware one (poor font selection/lack of Unicode support). I installed "pure" stock Androids (often non-official builds by the community) on all of the devices I used, starting with an 8-year old Samsung Galaxy S2, and I never had any problems with missing characters or tofus. I'm quite surprised to see this sort of problems emerging on one of the latest Google's devices which costs nearly $1000 :(

Well, it didn't cost me anything, and the problem may just be that I need to install a missing font somewhere.

@Zhe : I would like to say that if a battery is discharged it does not necessarily mean that potential of battery is zero, it is infact a relative term wherein the potential of battery is less than or equal(in some cases may be zero also) to the potential of the (chemical) surrounding.

@glucose This is true.


text:
Numbering starts from the side that has more locants or from the side that has a more preferable group?

I came across this question while practicing iupac nomenclature.
Should we number the parent chain from left to right (shown in blue) because there are three cholrine atoms attached to C1 while there are only two alcohol groups attached to C2?
Or should we start the numbering from the right side instead (shown in orange) as alcohol group is preferred over the chlorine group?

Alcohol function has a preference over chloro function. Therefore numbering in orenge is the correct one (name ending with -1,1-diol).

But isn't there a rule that says that numbering starts from the side that has more substituents?

Look here for related answer.

Concerning the numbering of locants for substituents, the most important simplified criteria are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The $\ce{-OH}$ group is the principal characteristic group in this case. It is expressed as suffix ‘ol’.
Therefore, low locants are assigned first to the $\ce{-OH}$ groups according to Rule (c), which yields ethane-1,1-diol.
After that, a low locant is assigned to the chloro groups according to Rule (f); however, the only possible numbering left is 2,2,2-trichloroethane-1,1-diol.


text:
Electrolysis of water: Which equations to use? (IB Chem)
There is a list of standard electrode potentials at 298 K from the p. 23 of IB Data Booklet 2016. Which of the following equations (forward/backward reactions), from the two possible ones involving the discharge of hydrogen gas and the other two with oxygen gas discharge, should I use for the oxidation and reduction of water in electrolytic cells?
$$
\begin{array}{cc}
\hline
\ce{\text{Oxidized species} <=> \text{Reduced species}} & E^⦵(\pu{V}) \\
\hline
  \begin{align}
    \ce{H2O(l) + e- &<=> 0.5 H2(g) + OH-(aq)} \\
    \ce{H+(aq) + e- &<=> 0.5 H2(g)} \\
    \ce{0.5 O2(g) + H2O(l) + 2 e- &<=> 2 OH-(aq)} \\
    \ce{0.5 O2(g) + 2 H+(aq) + 2 e- &<=> H2O(l)}
  \end{align}
&
  \begin{array}{r}
   -0.83 \\
    0.00 \\
   +0.40 \\
   +1.23
  \end{array}
\\
\hline
\end{array}
$$
(Unless the use of any of these equations cannot be generalized — for a concise explanation of why this is so and what to do then I would be equally grateful.)

For the acidic electrolysis, use the reactions where $\ce{H+}$ occurs.
As $\ce{OH-}$ is not available in considerable amount there as a reagent, neither it is created as a product.
Generally, for a reaction choice, apply the principle of availability and stability, allowing for a reagent to exist in (relative) abundance.
$\ce{OH-}$ or anions of weak acids like $\ce{ClO-}$ do not survive in acids. Acids do not survive in hydroxides.
But note that using reactions with half of a molecule is not necessery.
$$\begin{align}
\ce{O2(g) + 4H+(aq) + 4e-  &<=> 2 H2O(l)}\\
\ce{2H+(aq) + 2e- &<=> H2(g)}
\end{align}$$
For the alkaline electrolysis, similarly, use the reactions where $\ce{OH}$- occurs.
$$\begin{align}
\ce{2 H2O(l) + 2e- &<=> H2(g) + 2 OH^-(aq)}\\
\ce{O2(g) + 2 H2O(l) + 4e- &<=> 4 OH^-(aq)}
\end{align}$$

Thank you, @Poutnik. Could you also explain to me how I could apply this knowledge to, for example, the electrolysis of aqueous sodium chloride? A textbook example uses equations from both categories you mentioned.

Apply the same principle of availability and stability.

@Poutnik, I suppose that by "availability" you mean concentration of the salt, but I am not sure how to tackle "stability"... Does "stability" have to do with electrode potential values? I am too verdant to figure this out by myself.

@w_w I mean ability to survive. $\ce{OH-}$ or anions of weak acids like $\ce{ClO-}$ do not survive in acids. Acids do not survive in hydroxides. Generally, choice duch a reaction form, where reactants on both sides may exist in abundance.

I forgot to ask: If we have a neutral solution can we act and choose the equation as if we want to keep it neutral or is this not necessarily the case?

At neutral solutions it is rather arbitrary what reaction we choose, as ion equilibrium reactions cancel the difference.


text:
Questions in Equilibrium
I have two inquiries concerning equilibrium:

Does $K_c = 5.0$ indicate that the equilibrium mixture contains both products and reactants at an approximately equal amount? I know that $K_c \approx 1.0$ is said to be so, however I am unsure of if $5.0$ would be considered to be approximately $1.0$ in this case.
Is a reaction at equilibrium if the forward reaction does not change? I believe it would as an indication of equilibrium is that both the forward and reverse reactions occur at the same rate.

Could you please clarify the line "Is a reaction at equilibrium if the forward reaction does not change"? What do you mean by a reaction not changing?

I suppose the OP wants confirmation the equilibrium means a dynamic equilibrium, with the equal rate of the opposite reactions. Yes, it is . By "reaction does not change" he probably means the concentrations are constant.

I mean to say that the forward reaction maintains a constant rate, which can be said as "the forward reaction does not change".

You won't get quicker or better answers if you spam Chemistry.SE with the same questions. This is not a paid service, there is no staff constantly sitting on-line and answering all the questions, and you cannot demand a fast and accurate response (plus you already got one answer in less than an hour, and in addition to that I already voted to reopen your first question). Please visit Help Center to familiarize yourself with the basic principles of this site.

If you observed the answer that was given and my following comment, you would have realized that this "answer" was not much of an answer but rather a repetition of the question itself. There is no need for me to familiarize myself with this site. I am already one who knows well of the site's guidelines. Also, there is no need to be so quick to aggression. I am not indicating this as a paid service at all. I am unsure of why anyone would interpret my comment in such a way. We are not fighting here. This a platform for learning. Period.

If were familiar with the rules, you would've not re-posted the same question just because you were dissatisfied with the first answer given and want a quicker response. I feel pity for you if you see this as aggression or fight, I'm just informing that what you've done wasn't correct; also, as I see it, the answer was as decent as the question itself.

@James It is very unusual the rate of reaction being constant. Do you have a particular equilibrium in mind ?

@andselisk, as any other human would do you are trying to defend yourself. Remarking "the answer was as decent as the question itself" is not the appropriate response. The question was in fact detailed in every sense of the word. If you were to be an experienced with the StackExchange clear guidlines, insulting is not allowed in the book. I will not go on if you maintain this manner. There is only one clear purpose of my inquiry. Unless you have a meaningful answer to my inquiry which has been thought through, you comments will not help either you or me. It is as simple as that.

@James You overestimate the quality and clarity of your question. Additionally, reposting the question instead of elaborating it can be considered rather rude to those who responded.

I do not overestimate the quality of my question. @Poutnik, my question is very straightforward. Any professor would be shocked at you recent reply remarking "It is very unusual the rate of reaction being constant. Do you have a particular equilibrium in mind?". This is a very primitive question. Let's start from the beginning.

This is not a laboratory result but rather a hypothetical question which can be answered by the understanding of chemistry. Secondly, I have no particular equilibrium in mind. I am quite surprised that someone is unsure of what equilibrium I am referring to. Just to make it clear, I am referring to the equilibrium of reactions in which the rate of the reverse and forward reactions are equal.

There are no other equilibriums but those with the equal rates of the opposite reactions.

@James correct me if I'm wrong but I choose to disagree. A specific example of an equilibrium refers to an equilibrium example with reactants and products for example the equilibrium formed by iron thiocyanate ions in solution. Without a specific example it is difficult to tell if an equilibrium constant 5.0 corresponds to roughly similar amounts of reactants and products

@James You are quite arrogant for somebody who asks questions with very simple answers you should already know.

@Poutnik, this is the very reason I am surprised by your response. If there is only one equilibrium, which I knew of before however I did not know of your extent of knowledge, why would you ask "Do you have a particular equilibrium?". You have frankly defeated yourself. If this is such an easy question, why can you not think properly answer it. Your responses have led to nothing more than confusion upon yourself. Not in a way of offense but rather a piece of advise, you should definitely revise the basic chemistry concepts before responding to any questions.

I have asked about particular equilibrium of context of particular reaction. For chemists, it is clear. No chemist would imply I ask about various kinds of equilibrium. BTW, I graduated in Analytical chemistry 30 years ago.

@James You should revise your question in the first place. Formulation "reaction does not change" is rather sloppy.

A reaction is in equilibrium,
if the reaction quotient $Q=K_{\rm c}$,
For reaction $$\ce{A <=> B}$$ $K_{\rm c}=5$ means concentration ratio 1:5. It does not seem to me to be approximately the same concentrations.
The equilibrium constant is the thermodynamical quantity, determining the position of equilibrium. It's value does not say if the reaction is in equilibrium.
About the same concentrations, it means rather the same product of concentrations of reagents and products.
If there is a reaction $$\ce{A + B <=> C + D}$$ with $K_{\rm c}=1$ and if concentrations are $$c_{\rm A}=1000c_{\rm B}=100c_{\rm C}=10c_{\rm D}$$
we can hardly speak about equal concentrations at equilibrium. as $$K_c=1=\frac{10\cdot 100}{1000\cdot 1}$$
Additionally, if a reaction is not symmetric in counts of reagents of both sides, equilibrium concentration ratios change with concentrations.
If there is a reaction $$\ce{A <=> 2 B}$$
The same concentrations during equilibrium with $K_{\rm c}=1$ occur only for $c_{\rm A}=1$.
If we consider concentrations of $\ce{A}$ and $\ce{B}$, the following concentrations are in equilibrium:

A     B
0.0001 0.01
0.01 0.1
1 1
100 10
10000 100

Yes, you think right, at equilibrium, the rates of the forward and reverse reactions are equal.
If
$$\begin{align}
rate_{\rm forward}&=const\\
rate_{\rm backward}&=f([products])\\
\end{align}$$
("Forward reaction does not change"=has a constant rate, as was explained)
..then as reaction is progressing, the backward reaction rate increases, until it matches the forward rate and reaction is at equilibrium ( regardless of the $K_{\rm c}$ value.)

If you are familiar with the concept of the equilibrium constant, which I expect you are, you would know that if Kc, which is the denotation for the equilibrium constant, is approximately 1.0, the reaction is said to be at equilibrium. The issue arises to what is meant by "approximately". This is in fact what my question is asking.

@James The Kc value and reaction being at equilibrium are 2 independent things. Kc value is matter of reaction thermodynamics, describing the position of the equilibrium state with minimal value of Gibbs energy. Being at equilibrium is matter of reaction kinetics. The Kc value does not say if the reaction already reached equilibrium or not.

@James A reaction can be in equilibrium for any value of Kc, e.g $10^{52}$


text:
How to write electronic configuration of p-orbitals?
For example, the electronic configuration for the element boron can be writen as:
1s2 2s2 2p1
However, it is can be written more accurately by taking the different p-orbitals into account:

The electronic configuration with the different p-orbitals starting from boron ascending in atomic number:

B - 1s2 2s2 2px1
C - 1s2 2s2 2px1 2py1
N - 1s2 2s2 2px1 2py1 2pz2
O - 1s2 2s2 2px2 2py1 2pz2
F - 1s2 2s2 2px2 2py1 2pz2

Each type of 'p-orbital' can hold a maximum of 2 electrons. But the way they are assigned to which orbital is not always systematic e.g. the electrons do no wait till, pX orbital is full to be able to fill the next one (I know this has something to do with energy levels and trends in periodic table).
Can someone tell me how this works and how I can use it to be able to write electronic configurations for p-orbitals. It is much easier just using the summation of  px + py + pz = pX; but my teacher sometimes writes it out fully...

Please explain what exactly you want to know in 'how this works'; do you want to know how the Hund's rule works out or the nomenclature system?

Remember also that even the "more accurate" notation you've used is somewhat artificial.  Consider, e.g., "C: $1s^2 2s^2 2p_x^1 2p_y^1$.    This notation may suggest the two 2p electrons occupy the $2p_x \text{ and } 2p_y$ orbitals, leaving the $2p_z$ orbital empty.  But that's not the case; the electrons can't tell the difference among them –  they're equivalent orbitals.  Instead, the two 2p electrons are evenly distributed across the $2p_x, 2p_y, \text{ and } 2p_z$ orbitals. For this reason carbon, as well as all other isolated atoms, have spherically symmetric electron distributions.

@theorist; I’m not sure what you mean by “electrons can’t tell the difference among them” - doesn’t that go against Hund’s rule?

(cont.) that before filling up one orbital that is halfly occupied it first fills orbitals with similar energy; I don’t believe the assignment of e- is purely random ... what I don’t understand is why the p values don’t add up to the total summative p value for Nitrogen, Oxygen and Fluorine. F has 3e-s in its 2p shell, O has 4e-s in its 2p shell etc.

What we label as $p_x, p_y, \text{ and } p_z$ are arbitrary (e.g., you could switch what you label as $p_y \text{ and } p_z$). In the absence of an external field, they're degenerate.  The two 2p electrons in carbon thus occupy a linear combination of those three orbitals.  And physically, considering $e^-–e^-$ repulsion, it makes sense that the two electrons would want to spread their density among all three orbitals, rather than being confined to two of them. It's been a while since I've played with the details of Hund's rule, so perhaps someone else could address your Hund's rule question.

Electrons don't pair up in orbitals before each is singly occupied by Hund's rule of maximum multiplicity.
Writing out px, py, pz fully or using p^n notation (n=no of electrons in a p orbital) is a matter of context and circumstance rather than the standard convention; you might use the 'full version' when explaining bonding through hybridization but omit it while stating the electronic configuration of an atom generally. Both notations are quite interchangeable.

Hi Kartik, "if it is a matter of preference" then how in exam situations are we expected to discern the element or ion from its e- config using the Cartesian axes of pX, pY & pZ? For example what element has the following e. config: 1s2 2s2 2px1 2py1 2pz2. I would need to know this works and how the e- are assigned to be able to do this..or even write them out myself.

@aesking It's mainly a matter of context (you can't go without coordinate representations here)! Px, py, pz can be preferably used also in writing the electronic configuration of any atom in general, standalone.

@Kartix I see. Sifting through past exam papers we don't get coordinate representations! (Is there no way?). Or can you exemplify it in a specific context so I have a better understanding. Thanks.

@aesking You've to write orbital diagrams in your answers the way the image shows in the question; exams may or may not give them. I am still not clear on what exactly you want to know: notation or the chemistry of orbitals? Well, ChemSE always discourages an exam-oriented approach.

Both? If possible. Notation is linked to the behaviour (or the chemistry of orbitals); as orbital is by definition the mathematical probability of finding an e- in a given volume of the atom with specific energy. As supposedly, electrons behave like a wave. In terms of this, that's the only grasp of understanding I have on this matter. Rather what I want to know is how this is mathematically represented.., to put simply. Knowing the chemistry of orbitals would be useful; however.

@aesking Try reading here: https://chem.libretexts.org/Under_Construction/Essential_Chemistry_(Curriki)/Unit_1%3A_Atomic_and_Molecular_Structure/1.4%3A_Electron_Configuration_and_Orbital_Diagrams; https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Electronic_Configurations/Hund's_Rules.

Let us continue this discussion in chat.

Thanks for the second link. I will read into Hund’s rule more and get back to you.


text:
Can static electricity damage solar cells?
How does static electricity damage common solar cells making them less photo reactive? At watt voltage and/or amps would change the chemistry of the common solar cell?
I want to think that as long as the amps are low with a high voltage the static electricity would not change the silicon?

____________________________________________________________
Related: https://space.stackexchange.com/questions/35661/could-a-solar-panel-be-modified-to-also-collect-a-static-charge-on-mars
From comments: How they should accumulate that static charge?
This similar to the method I would use to generate static electricity but is not limited to this method. Only the body above the line is important for an answer.
They would accumulate a static charge by moving dust and air past the cells like the blades of this helicopter but with more surface area with less speed.

https://space.stackexchange.com/questions/35587/could-electricty-be-generated-from-a-martian-dust-storm
https://defence.pk/pdf/threads/helicopter-static-electricity-phenomenon.75190/

How they should accumulate that static charge?

@Alchimista how does this look?

My comment was about a standard scenario. The copter blade can't be grounded opposite to a non flying solar cells. In general is hard enough to charge something mounted and composed by a semiconductor cabled by metals. Different if you have a special scenario in mind. It seems more suited for physics or engineering

@Alchimista the helicopter is grounded during this. I am only interested if high voltage could chemically change the silicon?

The last is a different question. I don't see how a grounded manufactured consisting of electrical wiring and conductive enough to work as solar cells should accumulate electrostatic charge. If you mean instead if there is a threshold V than certainly yes, but here I mean during solar cell or diode operation.  Move to electrical engineering, I think there is a SE branch.


text:
Grignard reagent and acidic hydrogen
Why does the Grignard reagent not react as a base in case of ketones even though the α-hydrogen atom in case of ketones is acidic? Grignard reagents do react with ethyne to give hydrocarbons, so why not here?
The only thing I can put my finger on is electrophilicity of the carbon.

Grignard reagents certainly can cause enolization. Additionally, those Grignard reagents with beta-hydrogens can effect reduction of ketones to secondary alcohols. See Heathcock's synthesis of copaene for enolization: DOI: 10.1021/ja00969a051

Grignard reagent is a hard base. The HSAB theory states that soft acids react faster and form stronger bonds with soft bases, whereas hard acids react faster and form stronger bonds with hard bases, all other factors being equal (Wikipedia). Thus, Grignard reagent (a hard base) would preferably attack the harder position when it is available and in offer, like in the situation with its reaction with ketone carbonyl carbon even with the presence of $\alpha$-$\ce{H}$s. Keep in mind that the Grignard reagent probably react with $\alpha$-$\ce{H}$s to certain extent, but when the carbonyl group is readily accessible, that reaction proceeds to much faster rate. Yes, Grignard reagents do react with ethyne to give hydrocarbons, but it is a different story if the molecule contains a carbonyl group.

Do you have a reference for carbanions or Grignard reagents being hard, in general?


text:
Which is more stable in gas phase, free radicals or ions?
Answering some kinetics questions I came across this statement:

Free radicals are more stable in the gas phase than ions.

Is this correct or not? Why? Is there a different trend in liquid or solid phase?

I'd assume that the statement refers to the fact that free radicals wouldn't be charged, hence no electrostatic attraction.

The word stable is meaningless without a context. It is misused a lot especially by organic chemists. Stable with respect to what? This should the question in our mind whenever we see this word. Free radicals are extremely reactive but some of them can survive in liquids with a very long lifetime. Our favorite O2 is also a diradical.

It is difficult to comment on such a broad statement because often it depends on the specifics. So instead, I will talk about a single example: Will the HCl molecule dissociate into radicals (atoms, in this case) or ions?
$$\tag{1}\ce{H-Cl -> H+ + Cl-}$$
$$\tag{2}\ce{H-Cl -> H. + Cl.}$$
Under many circumstances, neither (1) nor (2) happens, H-Cl is the major species.
In water, where the dipole of water solvates chloride ions and the hydrogen ion "reacts" with water to form hydronium ions and large structures, HCl ionizes completely (strong acid). In other solvents, it ionizes less but would also act as acid if there is an appropriate base.
$$\ce{HCl(aq) + H2O(l) -> H3O+(aq) + Cl-(aq)}$$
In the gas phase, ions of opposite charge have a strong long-range attraction, so reaction (1) is unlikely. Reaction (2) is endothermic (bond dissociation energy is defined as homolytic cleavage into radicals, and it has a positive value). If you raise the temperature, you favor the products in reaction (2), so at higher temperatures, the concentration of radicals would increase.
To directly compare ions and radicals in the gas phase, you would consider the following reaction:
$$\tag{3}\ce{H. (g) + Cl. (g) <=> H+(g) + Cl-(g)}$$
Taking away an electron from hydrogen corresponds to the ionization energy of hydrogen atoms (1313. kJ/mol, i.e. it takes energy to remove the electron from the electrostatic field of the nucleus). Adding an electron to the chlorine atom corresponds to the electron affinity of chlorine atoms (-359 kJ/mol, i.e. energy is released when an additional electron encounters a neutral chlorine). Overall, the process is endothermic. Different from reactions (1) or (2), however, the number of reactant particles is equal to the number of product particles, so in this case, it is not clear if the change in entropy is positive or negative, but it won't dominate the reaction at modest temperature. So at room temperature, this reaction will not happen, the radicals are favored over the ions.

What is stabler in gas phase, free radicals or ions?

For the example I chose, free radicals are more stable than ions.

Is there a different trend in liquid or solid phase?

For the example I chose, it is the opposite in aqueous solution.


text:
How to obtain spectroscopic constants Omega_e, Alpha_e, B_e, D_e and D_0 from centrifugal constants?
I need to compute a spectroscopic constants of $N_2^+$ states and I would like to do it using LEVEL16 software, due to its high interpolation flexibility.
But it seems, that LEVEL16 can only compute centrifugal distortion constants directly, while I need even the others like $B_e, \omega_e, \alpha_e$ etc.
So, my current LEVEL16 output looks like this (full output is attached in the  very bottom of this question):
Matrix element arguments are powers of the distance  r (in Angstroms)
 Coefficients of expansion for radial matrix element/expectation value argument:
       1.000000D+00 -2.000000D-01  3.000000D-02 -4.000000D-03
 Find  4  potential minima:   Vmin=************************************************
               at mesh points   R =    1.07800    3.58200    3.72200    4.72500
 Find  4  potential maxima:   Vmax=************************************************
               at mesh points   R =    3.54100    3.62900    3.75900   10.00000
 E(v=  0, J=  0)=**********   Bv=  2.0436791  -Dv= -5.7294D-06   Hv= -2.9963D-11
           Lv= -5.2792D-16   Mv=  2.2663D-20   Nv=  4.3228D-25   Ov= -5.1173D-30
 E(v=  0, J=  0)=***********   <M(r)>=  0.8129565017       <KE>=    613.747
           <X** 1>=   1.08679787   <X** 2>=   1.18212033   <X** 3>=   1.28688369

Now, I know, that
$G(v) \approx \omega_e\left(v + \frac{1}{2} \right) + \omega_e x_e\left(v + \frac{1}{2} \right)^2 + \omega_e y_e\left(v + \frac{1}{2}\right)^3 + \omega_e z_e\left(v + \frac{1}{2} \right)^4$
$B_v \approx B_e - \alpha_e\left(v + \frac{1}{2}\right) + \gamma_e\left(v + \frac{1}{2}\right)^2$
$D_v \approx D_e + \beta_e\left(v + \frac{1}{2}\right)$
but I don't see, how can I achieve the active constants here? Let's take $D_e = D_0 + G(v)$ as an example - if I had an energy of a lowest vibrational state, I'd be able to at least determine the dissociation energy, but I don't see $G(0)$ anywhere...
So, is there any parameter in LEVEL16 I've overlooked?  Or did I just forget about some principle?

Complete LEVEL16 output
  Reduced masses below are based on atoms 1 & 2 with charges (+0/2) and (+2/2),
        respectively, with subtraction/addition of 0 and 2 half-electron masses.

Computation of N2+ (B2 SigmaU+) spectroscopic constants
================================================================================
 Generate   ZMU=  7.00126270151(u)   &   BZ= 4.153171609D-01((1/cm-1)(1/Ang**2))
          from atomic masses:  14.00307400443  &   14.00197684461(u)
    Since the molecule is an ion with charge +1
      use Watson's charge-adjusted reduced mass   mu = M1*M2/[M1 + M2 - (+1)*me]
 Integrate from  RMIN=  0.800  to  RMAX=  10.00  with mesh  RH= 0.001000(Angst)

 Potential-1 for  N( 14)- N( 14)
 ================================
 State has  OMEGA= 0   and energy asymptote:   Y(lim)=************(cm-1)
 Perform cubic spline interpolation over the   84 input points
 To make input points Y(i) consistent with  Y(lim),  add  Y(shift)=      0.0000
 Scale input points:  (distance)* 1.000000000D+00  &  (energy)* 8.065600000D+03
             to get required internal units  [Angstroms & cm-1 for potentials]
      r(i)         Y(i)        r(i)         Y(i)        r(i)         Y(i)
   ----------------------   ----------------------   ----------------------
    0.800000*************    2.200000*************    3.650000*************
    0.850000*************    2.250000*************    3.700000*************
    0.900000*************    2.300000*************    3.750000*************
    0.950000*************    2.400000*************    3.800000*************
    1.000000*************    2.450000*************    3.850000*************
    1.050000*************    2.500000*************    3.900000*************
    1.100000*************    2.550000*************    3.950000*************
    1.150000*************    2.600000*************    4.000000*************
    1.200000*************    2.650000*************    4.050000*************
    1.250000*************    2.700000*************    4.100000*************
    1.300000*************    2.750000*************    4.150000*************
    1.350000*************    2.800000*************    4.200000*************
    1.400000*************    2.850000*************    4.250000*************
    1.450000*************    2.900000*************    4.300000*************
    1.500000*************    2.950000*************    4.350000*************
    1.550000*************    3.000000*************    4.400000*************
    1.600000*************    3.050000*************    4.450000*************
    1.650000*************    3.100000*************    4.500000*************
    1.700000*************    3.150000*************    4.550000*************
    1.750000*************    3.200000*************    4.600000*************
    1.800000*************    3.250000*************    4.650000*************
    1.850000*************    3.300000*************    4.700000*************
    1.900000*************    3.350000*************    4.750000*************
    1.950000*************    3.400000*************    4.800000*************
    2.000000*************    3.450000*************    4.850000*************
    2.050000*************    3.500000*************    4.900000*************
    2.100000*************    3.550000*************    4.950000*************
    2.150000*************    3.600000*************    5.000000*************
 ----------------------------------------------------------------------------
 Extrapolate to   X .le. 0.8500  with
   Y=-23872936.002  +1.108309D+08 * exp(-8.496962D+00*X)
 Extrapolate to  X .GE.  4.9500  using
   Y=************ - [ 7.145852D+02/X**1 -3.206362D+03/X**3]
 ----------------------------------------------------------------------------

 Calculate properties of the single potential described above
 Potential-1 uses inner boundary condition of  zero value  at  RMIN

 Eigenvalue convergence criterion is   EPS= 1.0D-06(cm-1)
 Airy function at 3-rd turning point is quasibound outer boundary condition

 Since state-1 has (projected) electronic angular momentum  OMEGA= 0
         eigenvalue calculations use centrifugal potential  [J*(J+1) - 0]/r**2

 For  J=  0, seek the first   1 levels of Potential-1   with   VLIM=***********

 Matrix element arguments are powers of the distance  r (in Angstroms)
 Coefficients of expansion for radial matrix element/expectation value argument:
       1.000000D+00 -2.000000D-01  3.000000D-02 -4.000000D-03
 Find  4  potential minima:   Vmin=************************************************
               at mesh points   R =    1.07800    3.58200    3.72200    4.72500
 Find  4  potential maxima:   Vmax=************************************************
               at mesh points   R =    3.54100    3.62900    3.75900   10.00000
 E(v=  0, J=  0)=**********   Bv=  2.0436791  -Dv= -5.7294D-06   Hv= -2.9963D-11
           Lv= -5.2792D-16   Mv=  2.2663D-20   Nv=  4.3228D-25   Ov= -5.1173D-30
 E(v=  0, J=  0)=***********   <M(r)>=  0.8129565017       <KE>=    613.747
           <X** 1>=   1.08679787   <X** 2>=   1.18212033   <X** 3>=   1.28688369
 -------------------------------------------------------------------------------
 ===============================================================================

is this your complete output? Level prints a table with $v, J, G(v), B_v, D_v$ and etc. on the fort.9 file. Check your input because the "Potential-1 for  N( 14)- N( 14)" table should print your potential energy curve.


text:
Relationship between rate of reaction and pressure
Given the equation
$$\ce{A(g) + B(g) → C(g) + D(g)}$$
with reaction rate given by
$$v = k[\ce{A}]^2$$
How can I prove that reaction rate goes up with pressure? I tried substituting $k$ for $Ae^{-E_A/(RT)}$ which leaves me with
$$v = Ae^{-\frac{E_A}{RT}}[A]^2$$
If I use
$$PV = nRT \implies RT = \frac{1}{n}PV \implies RT = \frac{P}{C}$$
I end up with this:
$$v = Ae^{-E_A\frac{C}{P}}[A]^2$$
This shows there is a relation, but I don't know what to do with the concentration $C$, or whether it was correct to plug in the ideal gas law in the first place.  Any help?

Pressure is directly proportional to concentration via the ideal gas law, so the rate $\sim kp^2$ .

The exponential term in the Arrhenius equation does not imply reaction with other molecules. It applies to ready to react unstable adduct, about to overcome the energy barrier $E_a$ to form the product.
The concentration of this adduct is driven by concentration of reagents and therefore by the pressure.
Therefore the reaction rate equation implicitly involve 2 multiplicative terms:

Concentration of reagent adduct
$$\mathrm{const}\cdot [A]^2 = \mathrm{const}\cdot \left(\frac{p_{\rm A}}{RT}\right)^2$$
Probability the adduct reaches the reaction energy barrier, from the Boltzmann distribution:
$$\exp{\left(-\frac{E_A}{RT}\right)}$$

By other words, the constant $A$ in your equation
$$v = A \exp\left(-\frac{E_A}{RT}\right)$$
implicitly involves concentration, respectively pressure of the reagents.


text:
Why ¹H–¹⁴N coupling is not observed in ¹H NMR
Since the nuclear spin value of $\ce{^{14}N}$ is $I = 1$, we should expect a coupling between $\ce{^1H}$ and $\ce{^{14}N}$. Hence, the proton peak in a primary amine $(\ce{-NH2})$ should split in a triplet ($2 × 1 × 1 + 1 = 3$).
But generally that is not observed. Is this because of lower gyromagnetic ratio of $\ce{^{14}N}$ that leads to very less population difference between ground and excited state, or is there some other reason behind it?

Although not exactly the same question, my answer here addresses this issue, in particular the last paragraph, although one would likely need to read the whole thing.

The highly symmetric ammonium cation does in fact display ¹H-¹⁴N coupling, forming a nice sharp 1:1:1 triplet in the ¹H-NMR spectrum, as shown here. Once the source of much confusion, before an amazing a-ha moment!


text:
How do I decide whether ring will expand in this reaction or not?

Attempt:
After forming the carbocation adjacent to the cyclopropyl ring I had two options,

$\ce{Cl-}$ then and there without any changes
Expand ring, form a new carbocation and add $\ce{Cl-}$ there.

I chose the first option because of exceptional stability of cyclopropyl methyl carbocation. However, answer given is option C which follows mechanism  (2).
Can someone explain how to judge and decide how to proceed after the carbocation formation?

Cyclopentyl ring does not let the cyclopropyl ring to provide stability. Think why? The reason is due the cyclopentyl ring the p-orbital of carbon (containing positive charge) becomes perpendicular to p-oribital on carbon of cyclpropyl ring. So the double bond character cannot be achieved there is a image in my answer see it.

This is a challenging question because you can form a nonclassical carbocation (homoallyl) by protonating farther from the cyclopropyl ring...

Explanation
In case 1. You can see it is a 2° carbocation (or secondary carbonation) with angle strain of ~25° on cyclopropane
In case 2. You can see it is 3° carbocation (or tertiary carbocation)
with angle strain of ~10° on cyclopropane
It is obvious that case 2 will be major intermediate yielding $\ce{1-chlorobicyclic [3,2,0] heptane}$ as main product that is C.
Edit1: After reading your question more thoroughly I found that I missed one thing about 'exceptional' stability of cyclopropyl methyl carbocation.
The stability that cyclopropyl provides to carbocation is due to interaction of p-orbitals which is not possible in your case because p-orbitals of carbon having '+' charge becomes out of plane with respect to p-orbitals of cyclopropyl.
Edit2: See image below for reference

Then in case of cyclopropyl methyl carbocation the p-orbitals of both carbon containing '+' charge and the carbon on ring are parallel
In your case the p-orbitals of carbon containing '+' charge is perpendicular to plane of cyclopentyl ring but the p-orbitals of carbon on cyclopropyl ring is parallel to plane cyclopentyl ring, so stability is not provided by cyclopropyl ring.
To sum up you can say that the cyclopentyl ring does not let cyclopropyl ring to provide stability to carbocation which makes ring expansion a preferred option.
Edit3 (Bonus): I asked to my professor about this, he told me that there are two conformational isomers of cyclopropyl methyl carbocation. One is bisected conformation and other is perpendicular conformation. The bisected one allows the carbocation to be exceptionally stable but the perpendicular one isn't that much stable (like in your case) due to less interaction of p-orbital and the partial double bond character cannot be achieved.


text:
Mechanism of oxidative dearomatisation with hypervalent iodine
The following step was taken from the synthesis of Kinamycin C on SynArchive. It employs the use of a peculiar reagent, that is, bis(acetoxy)iodobenzene (BAIB), also known as phenyliodine(III) diacetate (PIDA). Being a high-valent iodine species, it is used as an oxidising agent in organic chemistry. So clearly, there is some sort of oxidation taken place in this step, along with nucleophilic aromatic substitution involving an $\ce {MeOH}$ molecule. However, I am quite uncertain as to what is the mechanism by which this step takes place.

The conventional mechanism is as follows:

The phenol displaces one acetate group on iodine – this makes the iodine itself act as what is essentially a fancy leaving group. Nucleophilic attack at either the ortho or para position allows the loss of iodobenzene and another molecule of acetic acid. What I can't explain right now is why it chooses to go para rather than ortho; it probably depends on the exact substitution pattern as I have seen examples of both. A good reference for this is Zhdankin, Hypervalent Iodine Chemistry (2013).

Perhaps, the preference for the position of substitution is due to the stronger inductive withdrawal of the methoxy substituent already attached at that position.


text:
Why does electrolysis of aqueous concentrated sodium bromide produce bromine at the anode?
It is often stated that electrolysis of any aqueous concentrated halogen compound will produce the halogen at the anode, despite it having a more postitive standard electrode potential than the oxidation of hydroxide, while when it is dilute oxygen will be produced, but what is the reason for this?
Why does the principle of concentrated solutions during electrolysis only apply for halogens and not any other element. For example, why isn’t sodium produced at the cathode when concentrated sodium chloride is electrolysed?

There is important to know the actual potential depends on actual activities of reagents ( approximately concentrations for dilute solutions).
Additionally, the electrode potential is the thermodynamic quantity. If some net reaction kinetic is involved, other potentials affects the actual potential at the solution border, like the diffusion potential and dielectric layer potential.
We consider reactions
$$\begin{align}
\ce{Br2 + 2e- &<=> 2 Br-}\\
\ce{O2 + 2 H2O + 4e- &<=> 4 OH-}\\
\end{align}$$
The initial bromine activity is very low.
Activity/concentration ratio of ions typically decreases with concentration, as seen in the Debye-Huckel equation,
but for high salt concentrations it often jumps up the roof.
Therefore, activity of bromides is very high and the actual potential gets low.
$$E_{c_{\ce{Br2}}/c_{\ce{Br-}}}=E^{\circ}_{c_{\ce{Br2}}/c_{\ce{Br-}}} + 0.059 \log \left( \frac{a_{\ce{Br2}}}{a_{\ce{Br-}}}\right)$$
OTOH,
The potential for hydroxide oxidation  is pushed up for neutral solutions to the middle between standard redox potentials for acidic and alkaline solutions $$(+0.401 + +1.229)/2=+0.815$$
$$E_{c_{\ce{O2}}/c_{\ce{OH-}}}=E^{\circ}_{c_{\ce{O2}}/c_{\ce{OH-}}} + 0.059\log \left( \frac{a_{\ce{O2}}}{a_{\ce{OH-}}}\right)$$

Very important thing is the oxygen overpotential. Its value is listed in range +0.56 V on Ni to +1.02 V on Au, in contrary to the maximal value for chlorine ( on graphite)  +0.12 V.

For $\ce{NaCl}$ solutions, difference of standard potentials is too high to produce metallic sodium, unless one uses mercury as the cathode, where happens the hydrogen, related to its reaction kinetics. This is used for production of sodium hydroxide, where sodium amalgam reacts with water.


text:
Ionisation of a gas - from gas to plasma
Is it possible to ionise a gas such as Carbon Dioxide? If you continue to heat up a gas eventually the electrons from the gaseous atoms are stripped away and those e- gain energy (endothermic reaction) to become a plasma. A plasma is a sea of gaesous cations and electrons that are not too far apart from each other (unlike a gas).
Having done some research, it is certainly possible:
http://www.dynamicscience.com.au/tester/solutions1/chemistry/isotopes/isotopes4.htm
https://link.aps.org/doi/10.1103/PhysRev.36.472
How do I represent this change from gas to plasma using an equation?
CO2 (g) -> X+ (g) + Xe-
Ionisation of Carbon and Oxide ion:
C (s) -> C+ (g) + e-
O2^2- (g) -> O2- (g) + e-
Related key terms: ionisation, recombination

Plasmas are generated by electrical means (using radiofrequency current) rather than by heating to temperature extremes. Decomposition of CO2  by microwaves plasma was recently published in Nature. They list all the possible reactions and report the species observed. https://www.nature.com/articles/srep18436

@M. Farooq Thanks for the link! are you saying that gases can’t be ionised through photo radiation/ionisation ? Thats why I don’t understand this sweeping statement: “plasmas are generated by electronic means rather than...” Because isn’t the state transition from a gas to plasma also called ionisation and the reverse process plasma to gas, recombination?

Sorry for the confusion, ionization is a very broad term. It can be achieved by anything (radiation, heat, inductive coupling of radiofrequency current etc.). But an ionized gas is not necessarily a plasma however the plasma starts from the ionization process. It has to be sustained by an external energy source. See how it is defined by IUPAC "A gas which is at least partly ionized and contains particles of various types, viz. electrons, atoms, ions and molecules. The plasma as a whole is electrically neutral."

Carbon and oxygen are on the series of nuclear reactions that happen in stars. A big enough ball of carbon dioxide gas would collapse under its own gravity to create a thermonuclear reaction which would create a plasma. Since the temperature would vary in the depth of the gas various species would be present at the various depths.

@ M.Farooq - The paper you provided a link to was published in Scientific Reports, an open access journal from the Nature Research group that also publishes Nature.


text:
Does silver oxide react with hydrogen sulfide?
Electronic connectors are often silver plated. However, silver tarnishes fairly quickly and heavily. There exists a widespread misconception that the tarnishing of silver contacts is harmless, because silver oxide has about the same conductivity as silver itself. The problem however is that silver does not oxidize under normal conditions. The tarnish on the contacts is not silver oxide, but silver sulfide that develops due to the presence of some hydrogen sulfide in the air. Unlike silver oxide, silver sulfide is not a conductor, but a semiconductor with various potential adverse effects for the connection.
I have come across a reference that suggests oxidizing silver contacts before using them in order to prevent the development of the silver sulfide layer. This implies that silver oxide does not react with hydrogen sulfide in the air under normal conditions. Is this claim correct?
EDITS
I have corrected the typo by changing "sulfur dioxide" to "hydrogen sulfide" in the title and body of the question. Thanks for pointing this out in the answer!

Silver sulfide is formed by Ag and hydrogen sulfide not sulfur dioxide. You need reducing conditions where the latter can be reduced to
$$\ce{SO2 +reduction -> H2S}$$
Actually, sulfur dioxide chemisorbs on ultraclean silver surface, however heating can remove it. So this is reversible sorption, as suggested by Lassiter [1].
Just note that Auger electron spectroscopy is done under extremely clean environment. There is no trace of water, oxygen or any other component! Real atmosphere is far more complicated and tons of photochemical reactions occur in the atmosphere. A typical indoor air has plenty of undesirable components. How does $\ce{SO2}$ react with Ag must be another story because we cannot avoid or control other factors.
Coming to the second part of the query: If we have surface layer of silver oxide, will it prevent sulfide formation. In principle, possibly yes, because $\ce{Ag2O}$ is decent oxidizing agent. The moment traces of $\ce{H2S}$ come in contact with the oxide, it will reduce the oxide to elemental silver. This is my personal speculation.
As per Franey et al. [2]:

Polycrystalline silver has been exposed to the atmospheric gases $\ce{H2S}$, $\ce{OCS}$, $\ce{CS2}$ and $\ce{SO2}$ in humidified air under carefully controlled laboratory conditions. $\ce{OCS}$ is shown to be an active corrodant while $\ce{CS2}$ is quite inactive. At room temperature, the rates of sulfidation by $\ce{H2S}$ and $\ce{OCS}$ are comparable, and are more than an order of magnitude greater than those of $\ce{CS2}$ and $\ce{SO2}$. It appears that $\ce{OCS}$ is the principal cause of atmospheric sulfidation of silver except near sources of $\ce{H2S}$ where high concentrations may render the latter gas important. At constant absolute humidity, the sulfidation rate of
  silver by both H2S and OCS decreases from 20 to 40 °C and then increases to 40 to 80 °C.

So may hydrogen sulfide may not be a major culprit!
References

Lassiter, W. S. Interaction of Sulfur Dioxide and Carbon Dioxide with Clean Silver in Ultrahigh Vacuum. J. Phys. Chem. 1972, 76 (9), 1289–1292. https://doi.org/10.1021/j100653a011.
Franey, J. P.; Kammlott, G. W.; Graedel, T. E. The Corrosion of Silver by Atmospheric Sulfurous Gases. Corrosion Science 1985, 25 (2), 133–143. https://doi.org/10.1016/0010-938X(85)90104-0.

Thanks for the correction, yes of course, hydrogen sulfide is exactly what I meant :) Please forget about sulfur dioxide, it was just a silly typo, sorry. I will edit the question momentarily. So, does silver oxide react with hydrogen sulfide? I am asking for a very practical purpose.

Yes it  does, almost instantaneously, especially if traces of water are present. Add a new paragraph in your question with a heading EDITS.

Added, thanks! So the claim is wrong. Oxidizing silver contacts does not protect from silver sulfide developing on top (or in place) of silver oxide. Did I get this right? Thank you!

Interesting, does OCS have a name? Also, this is a bit unclear to me: "The moment traces of $\ce{H2S}$ come in contact with the oxide, it will reduce the oxide to elemental silver.$ - By oxidizing what, hydrogen or sulfur or both? I mean, would this not produce water and silver sulfide? (I've no idea, just asking.) so, even in the best case silver oxide would be only a temporary protection until it's all consumed.

According to Wikipedia carbonyl sulfide is the most abundant carbon sulfur compound in the atmosphere (ppb levels). Typically when hydrogen sulfide is oxidized, the product is elemental sulfur. You will have to read the ref 2 get more details. Silver oxide is an oxidizing agent, hydrogen sulfide is a reducing agent. Again this is speculation. What happens in reality esp. in the presence of light or moisture!

When you mate connectors there is mechanical abrasion. Thus electronic connectors with silver contacts would not be resistant to silver sulfide formation because of this mechanical abrasion which would remove the silver oxide coating. That is why gold plating is often used on higher quality connectors. Gold is resistant to tarnishing of any sort and the gold layer is tick enough to resist mechanical abrasion for a significant number of matings.

Thanks for your answer. I did not want to overcomplicate the question, but a related issue is tarnishing of a bare silver wire introducing undesirable skin effects of semiconductive silver sulfide.