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How to determine the surface reaction rate and equilibrium constant for a reaction between a solid site species and gas species
Consider the following reaction mechanism 1:
S1: $\ce{C-H (s) + H (g) <=> C^\star (s) + H2 (g)}$
S2: $\ce{C^\star (s) + H (g) => C-H (s)}$
S3: $\ce{C^\star (s) + C2H2 (g) => C-H (s) + H(g) + 2C(g)}$ Note that 2 C(g) in reality is not a gas but let's assumed it is for now.
S4: $\ce{C^\star (s) + C2H6 (g) <=> C-H (s) + C2H5(g)}$
The pre-exponential factor A in [cm3/(mol.s)][cm3/(mol.s)], n and activation energy E in cal/mol are respectively for each reaction:
S1: A = 4.2E+13,    n=0,     E=13,000
S2: A = 2.0E+13,    n=0,     E=0
S3: A = 8.0E+07, n=1.56,    E=3,800
S4: A = 2.1E+11,    n=0,     E=4,400
$\ce{C-H (s)}$ are $\ce{C^\star (s)}$ are the surface concentration on a soot (nucleated carbon) particle. $\ce{C-H(s)}$ is hydrogen occupied carbon and $\ce{C^\star}$ is unoccupied carbon.
The volumetric reactions rates can be determine using the following equation since the gas species are dependent on the availability of site species  [2] (I don't have any issues up to this stage):
r = kiCi$\frac{f_sX_s}{N_A}$,            i=1 to 4       ...... eq. 1
Where,
r = reaction rate, mol/m3.s
C = molar concentration of gaseous species in mol/m3
fs = ratio of number density of active surface sites on catalyst to that of soot.
Xs = number density of surface sites on soot particles 1/m2.
NA = Avogadro's number 1/mol.
Now this is where I am having problems:
According to [2] it was provided that the number density of Cs-H can be approximated to be 2.3 x 1015 cm-2 based on experimental data. Estimated by considering one site in a
rectangular area with one side being half-width of the benzene ring.
If this number was divided by Avogadro's number we get the site concentration which is in agreement with the site concentration found in other sources (for example 1) equivalent to 3.82 x 10-8 kmol/m2.
The software that I am using deals with the site species using the following equations:
R*j,r = ApnrYjRj,r       ...... eq. 2
Rj,r = Rkin,r(pn - $\frac{R_{j,r}}{D_{0,r}}$)N       ...... eq. 3
Where,
R*j,r = rate of particle species depletion kg/s.
Ap = particle surface area m2.
Yj = mass fraction of surface species $j$ in the particle.
nr = effectiveness factor (dimensionless).
Rj,r = rate of particle surface species reaction per unit area kg/m2.s.
Pn = bulk partial pressure of the gas phase species (Pa).
D0,r = diffusion rate coefficient for reaction  $r$.
Rkin,r = kinetic rate of reaction $r$ (units vary).
N = apparent order of reaction $r$.
The software does not know the number density or site concentrations of any of the site species and deals with them as mass species. I can only enter mass fractions with values 0 to 1 depending on the fraction of each species present in the particle. However, I know that the particle at t = 0 s is 100 nm in diameter. Therefore, based on the initial diameter and the site concentration of 3.82 x 10-8 kmol/m2 for Cs-H, I can work out the mass fractions for each species at t = 0 s assuming Cs* is zero initially (explanation why Cs* is zero is at the end of this post and should not affect anything here). For this reason I defined 3 mass fractions (these represent Yj in eq. 2):

Cs-H (the mass species that is occupied by hydrogen) = 0.012 at t = 0 s.

Cs* (the mass species that is unoccupied by hydrogen) = 0 at t= 0 s.

Cc (the mass species representing the particle core and does not participate in the reaction) = 0.988 at t = 0 s.

So now for the main questions,
Q1 I am not sure how to use A (pre-exponential factor) with the units mentioned in paper 1 to get Rj,r with units of kg/m2.s for each of the reactions S1 to S4.
Update for Q1: I managed to find some sources mentioning that surface reactions like the above are considered unimolecular which is quite confusing if you have a pre-exponential factor with units of cm3 /mol.s instead of 1/s.
Additionally, I managed to find a paper with the same mechanism for Appel [4] which mentions the following: "The rate coefficient of H combination with surface radical, reaction S1, was lowered from 1 x 1014 cm3/mol.s to
2 x 1013 cm3/mol.s". Does this indicate that the volume here is the volume of the reactor and not the particle itself?
Q2 Also, are there any issues in using the standard equilibrium constant equation when dealing with reverse of reactions S1 and S4?
Surface reaction mechanism S1 to S4 shown in the attachment (S5 to S8 are only third body collision reactions and are not part of the surface reaction mechanism).
Here is why Cs* is assumed zero at t=0s:
Simple answer to is very difficult to measure the number density or site concentration experimentally and it is usually calculated via a steady state assumption using the following [2]:
XCs = $\frac{k_{1}[H] + k_{-4}[C2H5]}{k_{-1}[H2] + k_{2}[H] + k_{3}[C2H2] + k_{4}[C2H6]}$ XCs-H
I will not be using this equation as there no way to incorporate it in the software. I need to allow the reaction rates for each reaction to dictate the mass fractions.
References
1 C. Kou, X. Yuan, Y. Luo, and X. Yuan, “Performance investigation of the solar thermal decomposition of methane reactor considering discrete and deposited carbon particles,” vol. 324, pp. 124401–124401, Sep. 2022, doi: https://doi.org/10.1016/j.fuel.2022.124401.
[2] M. Younessi-Sinaki and F. Hamdullahpur, “Numerical investigation on the number of active surface sites of carbon catalysts in the decomposition of methane,” vol. 60, no. 6, pp. 2228–2234, Feb. 2014, doi: https://doi.org/10.1002/aic.14395.
[3] ANSYS Fluent Theory Guide, https://www.afs.enea.it/project/neptunius/docs/fluent/html/th/node136.htm
[4] J. Appel, H. Bockhorn, and M. Frenklach, “Kinetic modeling of soot formation with detailed chemistry and physics: laminar premixed flames of C2 hydrocarbons,” Combustion and Flame, vol. 121, no. 1–2, pp. 122–136, Apr. 2000, doi: https://doi.org/10.1016/s0010-2180(99)00135-2.

Hello Rayana. If you have the full mechanism, I may be able to help you.

Hi Metal Storm, thanks for the reply. I have edited the post to include that info. Any tips would be greatly appreciated.

If doesn't give me much confidence about the paper you're using, when they in use arrows for mesomerism instead of for an equilibrium. Could you also please cite the full paper. Thank you!

From which article the given table $2$ is extracted ? What is the meaning of the index prime (') attached to $\ce{H}$ or to $\ce{C}$ in the given equations ?

Thanks for the reply @Martin-マーチン. I have added the reference to the post.

Hi @Maurice, thanks for the reply. I have added the reference in the post. Subscript s means (Solid) and refers to the carbon on the edge of particle. The asterisk means unoccupied carbon on the edge. The index prime (') does not have any special meaning and could have been avoided.

@MetalStorm I have edited with additional details. Looking forward to discuss.

@Rayana Thanks Rayana, I will try to answer it this weekend. Quick question, are you solving the state variables as a function of space and time? E.g., every gas has a partial pressure in the form of $P_j = P_j(x,t)$? Or you assume that the partial pressure has a unique value in the whole reactor, i.e., $P_j = P_j(t)$, like CSTR's.

@MetalStorm thanks Metal Storm. If there is any variation in the partial pressure that will be calculated in the background with respect to time.  I would be able to see it through the user interface. However, I'm not sure how easy or hard it is to extract these values with respect to each species to use it in an coded equation.

Hello Rayana. Some questions, I was reading the papers. (1) Why would you want to include the reverse reactions for $1-3$, since both papers list them as irreversible. At the start of your post, you list them as reversible. (2)  The Eq. (3) in your post seems to have a typo, $R_{j,r}$ appears at both sides. (3) The difussion coeffient refers to a species, not a reaction. Why is it dividing there? Where did you get that expression from? (4) Can you modify the code? Why can't you include that steady-state approximation equation?

(1) According to ref. [1] and [2], S1 and S4 are reversible reactions.
(2) Yes, you are right it might be a typo. If you go to ref. [3] and scroll down a little, you'll find that equations 7.3-7 to 7.3-9 are the relevant ones. But it really doesn't matter because these equations can be overwritten by any equation that I code as long as the units is kgmol/m2.s.

I was thinking of overwriting equations 2 and 3 with a simple decomposition equation mentioned here (https://en.wikipedia.org/wiki/Reactions_on_surfaces) where k2 is rate constant for the surface reaction (1/s) and k1 the rate constant for surface adsorption (m3/mol.s). In my case I think I am given k1 but not k2. Not sure how to get k2.

(3) Not sure about the diffusion and it probably doesn't matter because these equations will be overwritten by a rate equation which I can code in. I just provided them thinking they might help or provide any clues.

(4) I can't use that particular steady state approximation given in the paper because it calculates the site concentration of Cs* at steady state rather than a rate. I need to input a rate with units kgmol/m2.s and that rate can calculate the site concentration at steady state.

@Rayana I am still thinking about it, I am going to consult some colleagues in university. Are you in a hurry with your code? Sorry of not being able to answer this weekend, because I have some doubts. I am sure I will make it this week. Are you taking into account the effectiveness factor? Have you seen the process of reaction-diffusion in a catalyst pellet in spherical geometry? You come from chemistry or engineering?

That's okay thanks for your help. No I don't need to consider effectiveness factor. I can overwrite the rate equation with anything else no matter how simple as long as the rate is in kgmol/m2.s. I have not seen the reaction I'm trying to simulate it as close as possible by using kinetic parameters and comparing results with literature. I come from an engineering background and trying to simulate the autocatalytic effect of methane decomposition. Before we reach the mechanism presented in this post I already modelled a 37 step homogeneous mechanism which leads to it show in table 1 of ref [1].


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Pressure before equilibrium = Pressure during equilibrium?
For example, let's say we have the equation $$\ce{PCl_3(g) + Cl_2(g) -> PCl_5(g)},$$ and the temperature is held constant.
Would the pressure in the container when equilibrium is reached be greater than, less than, or equal to the pressure when the $\ce{PCl3}$ and the $\ce{Cl2}$ are initially mixed?
I have a few ideas, but they all come up with different conclusions.

The pressure increases. This is because the volume is decreasing because of less molecules when equilibrium is established and by Boyle's law, the pressure increases. However, container has a constant volume, so I don't think this is the correct answer.
The pressure decreases. This is because less molecules present implies less molecules hitting off the walls of the container.
The pressure remains the same. This is because even though less molecules implies less pressure, they will hit the walls of the container with more force. I'm speculating that more force will cancel out the fewer molecules hitting the walls of the container.

Boyle's law is not applicable here. Boyle's law imply a constant amount of substances, and no chemical reaction.

Try $\Delta (pV) = \Delta (nRT)$

Consider $\pu{1 mol}$ of $\ce{PCl3}$ and $\pu{1 mol}$ of $\ce{Cl2}$ at the beginning for the above reaction. Let at equilibrium $x \text{ mol} $
of each reactants  get converted to products. Then $\ce{PCl5}$ is $x \text{ mol}$ and $\ce{PCl3}$ is $1-x\text{ mol}$ and $\ce{Cl2}$ is $1-x\text{ mol}$.
$$\begin{array}{ccccc}
 \ce{PCl3} & + & \ce{Cl2} & \ce{->} & \ce{PCl5} \\
1 &  & 1 &  & 0 \\
1-x &  & 1-x &  & x \\
\end{array}$$
This means at equilibrium, total number of moles $= 1-x+1-x+x=2-x\text{ mol}$. Initially 2 moles were there and now 2-x. Practically pressure must decrease as all the reactants and products are gaseous.

No, initially it is only PCl3 and Cl2 in the container, no PCl5 is present. Assuming there is 1 mole of Pcl3 and cl2, at equilibrium there would be 2-x moles present. ((1-x) + (1-x) + x). Note that 2-x is less than 2.

I have edited the answer

Consider \ce{} for chemical formulas and reactions.  $\ce{PCl3}$ for $\ce{PCl3}$,  $\ce{PCl3 + Cl2 -> PCl5}$ for $\ce{PCl3 + Cl2 -> PCl5}$.

@Poutnik, thanks for the suggestion

I have also created the array structure for the tabelated content. // SE Markdown supports tables too:   SE meta -  tables

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sort: votes


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Deriving the equation of motion for adiabatic approximation
In the textbook, Theories of Molecular Reaction Dyanmics by Flemming Y. Hansen and Niels E. Henriksen, there is a derivation of the equation of motion for the adiabatic approximation (pgs. 5-7). The author starts by showing that the wavefunction can be written as a product state by separating the slow and fast components (i.e. nuclear and electronic) as shown below,
$ \Psi(r,R, t) = \chi(R,t)\psi(r;R)$
where $\chi(R,t)$ is the nuclear component and $\psi(r;R) $ is the electronic component which depends parametrically on the nuclear position (R).
We take the following time-dependent Schrodinger equation, written as
$ i\hbar \frac{\partial \Psi(r,R,t)}{\partial t} = (\hat{T_{Nuc}} + \hat{H_{e}}) \Psi(r,R,t)$
and plug in the product state to get the following equation
$ i\hbar \frac{\partial \chi(R,t)}{\partial t} = [\hat{T_{Nuc}} + E_{i}(R) + \left< \psi \right|\hat{T_{Nuc}}\left| \psi \right>_0] \chi(R,t)$.
It is said that this term is equal to zero, leaving us with the equation of motion of the adiabatic approximation
$ i\hbar \frac{\partial \chi(R,t)}{\partial t} = [\hat{T_{Nuc}} + E_{i}(R) ] \chi(R,t)$.
Questions

How do we get the $\left< \psi \right|\hat{T_{Nuc}}\left| \psi \right>_0 $ term during the derivation? I am confused on how that arises.

What does this term mean? Is it just the coupling between the nuclear and electronic states?

The term is the diagonal element of the nuclear kinetic energy operator with respect to the adiabatic electronic state basis. The term is obtained when the equation is projected on the adiabatic electronic state $\langle \psi| $ and integrated over electronic coordinates,
$$
i\hbar \partial_t  \chi(R,t)\psi(r;R) =(\hat T_N+ \hat H_e) \chi(R,t)\psi(r;R) \stackrel{\langle \psi|_r}{\Rightarrow} \\
\int dr\psi^*(r;R)i\hbar \partial_t  \chi(R,t)\psi(r;R)=\int dr\psi^*(r;R)(\hat T_N+ \hat H_e) \chi(R,t)\psi(r;R)\\
\langle \psi|\psi\rangle_ri\hbar \partial_t \chi(R,t)  = \big (\langle\psi|\hat T_N|\psi\rangle_r + E(R) \big)\chi(R,t)\\
i\hbar \partial_t\chi(R,t)=\big(\langle\psi|\hat T_N|\psi\rangle_r + \hat T_N + E(R) \big)\chi(R,t)
$$
The expansion of $\hat T_N$ applied to the product of $\chi\psi$, according to the product rule, also leads to a mixed term with first order derivatives $ \sum_{Nuclei}\frac{\hbar ^2}{2M}\langle\psi| \hat P_N|\psi\rangle_r \cdot (\hat P_N \chi(R,t)) $, which I have already omitted since this first order term is exactly zero when using a purely real $\psi$. This is also mentioned in the textbook.
But it should be noted that this mixed first order term only vanishes for the diagonal elements of $\langle \psi |T_N|\psi\rangle$, i.e. when $\psi$ is the same adiabatic electronic state. In the general case, this term leads to nonadiabatic effects, where different adiabatic electronic states become coupled by the motion of the nuclei, inducing nonadiabatic transitions.
The nuclear kinetic energy term $\langle \psi| \hat T_N|\psi\rangle_r $ does not vanish when using a real $\psi$, but is typically small in comparison to $E(R)$, in particular around nuclear equilibrium coordinates. But in principle it is an additional energy term that could/should be included in the Hamiltonian for the nuclear wavefunction, and it is due to the coupling of electronic and nuclear motion.
The term can be calculated if $|\psi\rangle$ is known as function of $R$, but is typically very small due to the inverse dependency on nuclear masses and the rather slow change of $\psi$ with $R$. Sometimes it is included at zero-order I.e. the zero-order Taylor expansion term at some $R_0$ is added as correction to the potential energy $E(R)$, for that reason it is also known as diagonal Born-Oppenheimer correction term. But in most cases it is assumed to be negligible.

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Sequential first order reaction graph in chemical kinetics
In graph given below is a sequential first order reaction
$$\ce{A ->[$k_1$] B ->[$k_2$] C}$$
For $k_2 \gg k_1$ the graph of concentration of $\ce{A}$, $\ce{B}$ and $\ce{C}$ is as follows.

Won't the graph of the concentrations of $\ce{B}$ and $\ce{C}$ be the same as if $\ce{B}$ is formed then only $\ce{C}$ will be formed simultaneously? Then why is $\ce{C}$'s graph different?

The answer to your doubt is directly on your graph.  First some B has to be created. Initially, the rate of C formation is zero.

My doubt is if B is only not formed then will C form? Then why the graph of concentration is different?

Neither of questions is clear. Try to reformulate. // C is formed only from B. If no B is present, no C is being formed.

The concentration profiles of B and C are not the same because B is formed only from A and then B only forms C so the concentration of B rises and falls but C only rises as it does not react to form anything else.

@porphyrin thanku sir

Somewhat related

How much calculus do you know?

@RodrigodeAzevedo sir i know quite a bit

Are the two reactions elementary?  Can one assume mass action kinetics?

See also  Pre-equilibrium approximation for k1 >> k2 and Steady State Approximation for k1 << k2.

https://chemistry.stackexchange.com/questions/40785/meaning-of-steady-state-kinetics/40796#40796

This question is about the rate-determining step of multi-step reactions. Bear in mind that the overall rate of reaction is equal to the slowest step of the reaction, in which case, from $\ce{A}$ to $\ce{B}$ ($k_2 \gg k_1$). Since it is a first-order reaction, the overall reaction rate = $k_1 [\ce{A}]$. The graph of concentration vs time for a first-order reaction should be like $y = \frac1x$.
On the other hand, $\ce{B}$ is the reaction intermediate. $[\ce{B}]$ initially increases when the reactant $\ce{A}$ is gradually consumed. But soon, $[\ce{B}]$ reduces to form $\ce{C}$.
The graph of $[\ce{C}]$ must be symmetrical about the horizontal line $y = \frac12[\ce{A}]$ because as stated earlier, the overall reaction rate $= k_1 [\ce{A}]$.

This is a pseudo first-order reaction. The rate of product formation is:
$$
\mathrm{R} = \dfrac{\mathrm{d}}{\mathrm{d}t}\ce{[C]}= k_2\ce{[B]} \tag{1}\\
$$
Applying steady state approximation:
$$
\dfrac{\mathrm{d}}{\mathrm{d}t}\ce{[B]}= k_1\ce{[A]} - k_2\ce{[B]} \approx 0 \tag{2}
$$
$$
\implies k_1\ce{[A]} \approx k_2\ce{[B]}\\
\implies k_1\ce{[A]} \approx k_2\ce{[B]}
$$
Substituting in Equation (1):
$$
R \approx k_1\ce{[A]}
$$
For the overall reaction $\ce{A -> C}$:
$$
-\dfrac{\mathrm{d}}{\mathrm{d}t}\ce{[A]} = k_1\ce{[A]}\\
\implies \ce{[A]} = \ce{[A]_o}\mathrm{e}^{-k_1t} \tag{3}
$$

Are you assuming mass action kinetics?

@RodrigodeAzevedo it is applicable for low concentrations.

Do you mean (1) is applicable?

@RodrigodeAzevedo (1), (2), and (3).


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Is there any reason why KMnO4 for TLC is stored under basic condition?
my mentor told me that the KMnO4 solution specifically for TLC has added some bases like sodium hydroxide but we don't know why. I try to search on google, and found that the bases are usually by-products in the industrial method when preparing KMnO4 from Wiki.
Which are,
2 K2MnO4 + 2 H2O → 2 KMnO4 + 2 KOH + H2
3 K2MnO4 + 2 CO2 → 2 KMnO4 + 2 K2CO3 + MnO2
But is there any reason why they keep the bases?

KMnO4 is a bit too strong oxidiser to be totally stable in water and high pH lowers redox potentials.

The first reaction is the summary reaction of an electrolytic oxidation of $\ce{K2MnO4}$.

Two reasons come to mind:

It's made that way. Potassium permanganate is made by oxidizing manganese dioxide (available as the mineral pyrolusite) with air in concentrated potassium hydroxide (to form initially potassium manganate (VI), $\ce{K2MnO4}$), and then diluting the solution (which disproportionates the potassium manganate(VI) to $\ce{KMnO4}$ plus part of the original $\ce{MnO2}$). Note that the manganate(VI) formation step does not work with sodium hydroxide, which explains why we usually have the potassium salt to begin with.

We don't want to tempt fate. Potassium permanganate solutions are only meta-stable. But it turns out that their decomposition tends to consume aqueous hydrogen ions. So starving the water of these can be expected to improve shelf life:

$\ce{4MnO4^- + 4H^+ -> 4 MnO2(s) + 2H2O + 3O2}$

may I ask why the Mn(VI) formation cannot be carried out by NaOH?

I have to guess on the explanation. Possibly the sodium compound is somehow stabilized at the Mn(V) stage, where the oxidation stops with sodium hydroxide. The less efficient oxidation of manganese would then cut down on permanganate yield upon disproportionation.


text:
Weakness of S−I and Se−I bonds
After recent question about non-existent pnictide(V) iodides, I stumbled upon a similar issue in oxygen group. For example, $\ce{SI2}$ was reported to be found… but at the temperature as low as $\pu{9 K}$. Well, OK, I heard that sulfur prefers disulfides, but $\ce{S2I2}$ was characterised at $\pu{-90 °C}$. Moreover, situation with selenium iodides is even worse!
One of the reasons for that is weakness of $\ce{S-I}$ and $\ce{Se-I}$ bonds: their respective enthalpies are $\pu{-18 kJ/mol}$ and $\pu{-1.3 kJ/mol}.$ Well, that looks more like values for hydrogen bonds…
Klapötke and Passmore [1, p. 234] provide an explanation for the weakness of these bonds:

The instability of the $\ce{S-I}$ and $\ce{Se-I}$ bonds can be attributed to their very low ionic resonance stabilization energies
as the electronegativity of iodine is about the same as that of sulfur and selenium.

I think this is a partial explanation, at best - ionic component sure makes bonds stronger, but by this logic all non-polar covalent bonds should be weak. I’m not saying I’d expect bonds created by iodine to be strong, but it’s not like we need cryogenic temperatures to get methyl iodide!
Could anyone provide a good explanation as to why these bonds are so weak, maybe describing stereo-electronic effects, or using computational methods?
References

T. Klapötke,  J. Passmore, J. Acc Chem. Res. 1989, 22, 234–240. DOI:10.1021/AR00163A002.

I know it's not what you're looking for, but what's wrong with the good ol' "atom size" explanation? They're all pretty large atoms with minute differences in electronegativity. Or are you looking for a fully non-empirical explanation?

Well, phosphorus has similar size and electronegativity, but PI3 is a common reagent.

It isn't just sulfur iodide. Sulfur in low oxidation states does not form any $\ce{SX2}$ molecules that are stable under ambient or near-ambient conditions. Such molecules tend to dimerize, with loss of some halogen atoms in the case of chlorine or a heavier halogens, forming disulfur compounds. (In the case of $\ce{S2I2}$, Wikipedia reports stability up to a somewhat higher temperature than suggested in the question,
$\pu{-30 °C}).$
What happens is sulfur in the disulfur compounds can act as a pi donor, thereby strengthening the sulfur–sulfur bond while only slightly weakening the sulfur–halogen bond. The specific case of disulfur dichloride is discussed here.
In addition to the halides, the structure of disulfur monoxide, which appears to exist in the Jovian moon Io, allows a similar interaction between the sulfur–sulfur bond and oxygen.

These are all good points and remarks, but not getting into strength (or rather weakness) of these specific bonds. S-I and Se-I appear to be very weak in general, not only weakened in disulphides.


text:
Can acids contain OH-?
Given that a solution contains $\ce{OH-}$ ions can we sufficiently determine the substance is a base? From my understanding, an acid may still contain $\ce{OH-}$ ions. However, the concentration of $\ce{H+}$ ions must exceed this. Is this correct?

The title is misleading as the term acid has 2 meanings, a substance and  a solution. You obvious mean the latter.

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

An acid can have some $\ce{OH-}$ ions, as well as $\ce{H+}$. It is the relative ratio that determines if the substance is acid or alkaline. Think of it as mismatching socks, where you must pair 10,000,000 red with 10,000,001 blue socks. There's one blue left over, so blue wins. (Hmm... in that analogy, think red and blue litmus, and there's one more $\ce{OH-}$ ion.)
Also, a zwitterion has both acidic and basic groups, and one can predominate over another, depending on the the pH.
Amino acids and sulfamic acid are two examples. Some amino "acids", such as arginine, are more basic, while aspartic acid is more acidic. The bumbling biochemist offers a more complete explanation of the behavior of zwitterions.


text:
How to pronounce Aʹα-helix?
The LOV protein structure contains a Jα- and an Aʹα-helix. Would “A prime alpha helix” be the correct pronunciation for Aʹα-helix?
I'm speaking soon in front of the scientists in the field that I really admire and English isn’t my first language. I don’t want to embarrass myself.

Andrea this is not really a chemistry question... Do not worry to much about the pronunciation, if they are as good scientist as you suggest, they will know English is not your mother tongue and would understand you presentation if it is properly prepared.

@PAEP I'm sorry to decline your edit, but there is virtually no reason to use MathJax for the letter α, especially in title. Unicode would do better since alpha should be upright, not slanted (there is no upgreek alternative for MathJax).

I would think the question is pretty much OK for Chemistry.SE, specifically having How to pronounce "potassium manganate(VII)"?, How do you pronounce GAMESS or How to pronounce G°? Personally, having very limited biochem background, I find OP's pronunciation perfectly fine.

Given short videos like Exploring different optogenetic systems: Photocaging authored by EMBL, are there colleagues of yours already in the field presenting results similar enough, perhaps in seminars / remote webinars (for the later I infer from small molecule chemistry and what the e.g. the American Chemical Society offers to its members)?

As andselisk declared in his comment, OP's pronunciation of "A-prime alpha helix" is perfectly fine. I just wanted make a few points here, which would help OP's speech if he/she choose to use my reference here. According to following reference (Ref.1):

N-terminal helix is A'$\alpha$ (pronounce as N-terminal A-prime alpha helix).
Further, C-terminal helix is J$\alpha$ (pronounce as C-terminal J alpha helix).

In between there are 4 more helices (C$\alpha$, D$\alpha$, E$\alpha$, and F$\alpha$).
References:

Vladimir Arinkin, Joachim Granzin, Katrin Röllen, Ulrich Krauss, Karl-Erich Jaeger, Dieter Willbold, and Renu Batra-Safferling, "Structure of a LOV protein in apo-state and implications for construction of LOV-based optical tools," Scientific Reports 2017, 7, Article number: 42971 (10 pages) (DOI: https://doi.org/10.1038/srep42971).

Please don't use long obsolete even by the LaTeX standards two-letter macros such as \bf. I'm not sure MathJax is appropriate here at all. The letter alpha should be upright, not italicized, and it's something you cannot do without hackery in MathJax. Bold text in CM font normally implies a vector, and if you want to emphasize the letter, I'd do it via secondary means such as phonetic transcription in the spelled-out part. Finally, "Hope this'd help." is just noise.

@andselisk Hello andselisk. Should quantities that have subscript/superscript belonging to sum indexes be upright? For example, for the Christoffel's symbol of the 2nd kind, it is $\Gamma_{\mathrm{ij}}^\mathrm{k}$ or $\Gamma_{ij}^k$? For a sum, should we write $\sum_i a_i$ or $\sum_\mathrm{i} a_\mathrm{i}$?

@MetalStorm In both these cases the indices are italicized. However, for example in the symbol for ionic radius $R_\mathrm{i}$ the index is upright because it's a descriptive one. See Which symbols are written in roman (upright) font and which are italicized? and references therein.

@andseliks Hello andselisk, sorry to interrupt you with nothing that has to do with the question. Do you know how to use a package like \pu{} in LaTeX? I use the TexStudio editor. I do chemical engineering, so we write much more numbers than chemistry, and now that I know the \pu{} package it would be nice to have something similar. This page unfortunately shows a package too old that won't be compiled. Thank you very much in advance.

@MetalStorm \pu is the macro for physical units bundled with mhchem for MathJax and is not available in original mhchem, the LaTeX package. If you are compiling a LaTeX document, consider loading siunitx package to help with typesetting quantities correctly.


text:
Can excited electronic states also decay due to collisions?
I was wondering if electronic states could decay due to collisions and not by fluorescence?
Just like vibronic relaxations can happen, when vibrationally excited molecules can lose those extra modes of vibration by collision with solvent particles, so the vibrational energy gets converted to translational energy of the solvent molecules. But is something like this also possible for electronic transitions or is this limited because of the short half-life of those electronically excited states?

Half-life is not necessarily short here.

A He-Ne laser uses collisional relaxation to get Ne back to the ground state.

Absolutely! Look at the fluorescence of quinine in tonic water here: https://chemistry.stackexchange.com/a/130758/79678. Adding ordinary table salt quenches the fluorescence: the chloride ions are the culprits. Collisional quenching of fluorescence is common and well known.

BTW, the reverse can occur, where collisions trigger dropping from a metastable state, emitting radiation. Thermoluminescence is useful for dating pottery and igneous rocks: https://en.wikipedia.org/wiki/Thermoluminescence

Yes, there are 3 main mechanisms, energy transfer (Forster  & Dexter type) which needs only close approach of molecules, electron transfer, usually v close range and spin orbit coupling by heavy atoms/ions (heavy atom effect), as mentioned above by Ed V, also close range. Collisions by solvent do not generally cause deactivation unless by one of these mechanisms, but does take heat away after reaction.


text:
How do I synthesize salicylic acid?
I'm looking to synthesize salicylic acid in my school laboratory for an investigatory project. The Kolbe's reaction looks promising:

Any suggestions on where to get the $\ce{CO2}$ from?
Will the reaction of vinegar + $\ce{NaHCO3}$ be okay? Does this reaction need a suitable temperature and/or catalysts? I tried to search for similar experiments but the others usually use reagents and catalysts inaccessible to me.

Do you have access to solid CO2 (dry ice)?

According to Wikipedia, the reaction is performed at CO2  pressure 100 atm. It may not be a suitable scenario for safety reasons.

Reimer-Tiemann followed by oxidation may be more achievable in a school lab  https://en.wikipedia.org/wiki/Reimer%E2%80%93Tiemann_reaction  see procedure  http://orgsyn.org/demo.aspx?prep=CV3P0463

@Waylander No, I don't.

@AdityaKumarPanda  then look at the R-T procedure which is done with hydroxide and chloroform as IMO you will struggle to get a high enough concentration of CO2 in your reaction mixture.

@Waylander would it be possible to do a riemer tiemann reaction with carbon tetrachloride instead of chloroform? (to skip the oxidation step)

@GauravSaiMaddipati  no - the mechanism requires deprotonation of chloroform by hydroxide as the first step, carbon tetrachloride will not do this. Also Carbon Tetrachloride is a serious carcinogen

@Waylander What is the 'riemer-tiemann carboxylation' being referred to here by the OP?https://chemistry.stackexchange.com/questions/86765/reimer-tiemann-like-reaction-of-catechol-with-diiodomethane

@GauravSaiMaddipati  see my answer below

I would suggest a different starting material: aspirin. The hydrolysis is possible under mild conditions, and the reagents are less dangerous.

Riemer and Tiemann originally reported that using carbon tetrachloride in place of chloroform gives salicylic acid as major product plus p-hydroxybenzoic acid. It is mentioned in this review by Wynberg here and this Indian Journal of Chemistry paper here which has experimental conditions. This seems to have received little attention (which is why I was initially unaware of it), possibly because yields appear low. Other researchers report that the para product predominates  (see the Indian Journal of Chem paper). The catalytic effect of copper metal in giving the para isomer was reported here

Dissolve phenol in a solution of sodium hydroxide (NaOH) and water. Pass carbon dioxide (CO2) gas through the solution while keeping it cool in an ice bath. This forms sodium salicylate. Acidify the solution with dilute sulfuric acid (H2SO4) to precipitate salicylic acid. Filter and wash the resulting salicylic acid crystals. Purify the product by recrystallization if necessary.

This will work PROVIDED you have a good source of CO2 - either compressed gas or dry ice. The OP does not.

Does this work well at normal pressures?


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Can Sodium form Anions?
I am at the beginning of self-studying chemistry and one question I have is quite simple yet I can't seem to get an answer that people agree on, or I am not searching for it properly.
The question is whether Sodium can form an Anion.
I know it can form Na+ but some texts I have read say it CANNOT form Na- yet others say it can.
I am unsure if that is just new information included in some texts and not in others or if there are extenuating circumstances.
Thank you for the help in advance.
P.S.: Since I am here, are there any good textbooks for self-studying chemistry?

It exists: https://en.wikipedia.org/wiki/Alkalide#Examples But you should not take its existence as a sign that it is stable, or somehow easy to make. There is good reason why introductory textbooks teach $\ce{Na+}$ and not $\ce{Na-}$. As for textbooks, check this out: https://chemistry.stackexchange.com/questions/37303/resources-for-learning-chemistry/37306#37306

https://chemistry.stackexchange.com/a/34111/9961


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Why heat capacity at constant volume relates to change in volume
I have been dealing with the establishment of a relationship between $w$, $q$, $\Delta U$ and $\Delta H$. I think the entire derivation is not necessary here so I will elaborate until the concerned part.
The process here is adiabatic. So
$$\delta q=0$$
From the first law of thermodynamics,
$$dU=\delta q + \delta w$$
which in this case simply becomes
$$dU=\delta w \tag{1}$$
For only the work of expansion
$$\delta w=-PdV \tag{2}$$
Putting this value in (1), we get
$$dU=-PdV \tag{3}$$
As is known that for an ideal gas,
$$C_v= \frac{dU}{dT}$$
or
$$dU=C_vdT$$
Putting this value in (3), we get
$$C_vdT=-PdV$$
I am wondering why in this last equation are we relating $C_v$ and $dV$ while $dV$ is 0 as the given specific heat is at constant volume. How to justify that in the last equation, if process is at constant volume, that $C_v$ is being related to $dV$ anyways?
My best guess is that we could have better used $C_p$ instead of $C_v$ as that would have led us to $\frac{dH}{dT}$ which would account for $\Delta U$ as well as $-PdV$. Or maybe it has something to do with the process being quasi-static.

Your confusion comes because you think that you can only write $\Delta U = C_V \Delta T$ only for isochoric processes.
The specific heat at constant volume is defined as
$$ C_V := \left(\dfrac{\partial U}{\partial T}\right)_V \tag{1} $$
Eq.(1), as it is written, denotes $C_V$ as a state function.
The changes in internal energy are always given by
$$ \int_{U_1}^{U_2} \mathrm{d}U = \int_{T_1}^{T_2} C_V(T) \mathrm{d}T \tag{2} $$
if

The fluid is an ideal gas, or
The process is really isochoric, independently if the fluid is an ideal gas or not.

To make sure that the formula works even though the process is not isochoric, take a look at the following image, where you are interested in the $\color{purple}{\text{path}}$ in purple:

Since for an ideal gas $U$ is only a function of $T$, if we plot $U$ vs $V$ we get straight lines. Lets calculate then, by virtue that $U$ is a state function, the change between those two arbitrary points by the $\color{red}{\text{path}}$ in red
\begin{align}
  U(T_2,V_2) - U(T_1,V_1) &=
  [U(T_2,V_2) - U(T_1,V_2)] + \underbrace{[U(T_1,V_2) - U(T_1,V_1)]}_
  {=0 \; (\text{constant $T$}))} \\
  U(T_2,V_2) - U(T_1,V_1) &=
  \underbrace{[U(T_2,V_2) - U(T_1,V_2)]}_{\text{real isochoric process}} \\
  U(T_2,V_2) - U(T_1,V_1) &= \int_{T_1}^{T_2} C_V(T) \mathrm{d}T \tag{3}
\end{align}
In the last derivation we "assumed" that we could only replace $C_V \; \mathrm{d}T$ if the process was isochoric. In Eq. (2) we didn't assume anything, and the result was the same.

I am wondering why in this last equation are we relating $C_V$ and $\mathrm{d}V$ while $\mathrm{d}V$ is $0$ as the given specific heat is at constant volume. How to justify that in the last equation, if process is at constant volume, that $C_V$ is being related to $\mathrm{d}V$  anyways?

If you add that the process is indeed isochoric, in addition to that it was also adiabatic, the first  law will tell you $\mathrm{d}U = \delta q + \delta w = 0 + 0 = 0$. For a simple fluid that has two degrees of freedom, asking for two restrictions leads to this behaviour: nothing has happened to the fluid. You will arrive to the same conclusion if you demand an isentropic and isenthalpic process, or a constant pressure and constant volume process, etc. Ask for two restrictions and you will get always $U_2 = U_1$.

U not Cv is the state function. This simply means that if Cv is known DeltaU can be calculated regardless.

@jimchmst This post on Physics stack exchange asserts that $C_v$ and $C_p$ both are intensive state function. I am confused.

OK CP and Cv are constants, well almost, and have a distinct value at a given T They are not a quantity such as U, H, S, G etc. that are composite of various forms of energy. nomenclature is hard. I suggest staying away from physics and studying chemistry it is more basic. The value of a state function is tht no matter how you get there when you arrive at a defined stat it has a given value. In that respect everything about a substance is a state functon but some things like mass,  MW, Cv, specific volume etc. are not composite  and do not need the title.

For a real gas, U=U(T,V) and $$dU=\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV=nC_vdT+\left(\frac{\partial U}{\partial V}\right)_TdV$$Sp, for a real gas you get:$$nC_vdT+\left(\frac{\partial U}{\partial V}\right)_TdV=-PdV$$or $$nC_vdT=-\left(P+\left(\frac{\partial U}{\partial V}\right)_T\right)dV$$
For an ideal gas where U=U(T) , this reduces to w$$nC_vdT=-PdV$$where $C_v=C_v(T)$

I think what you described is a slight variation of the derivation relevant to the Joule's law but unlike the original derivation of that law, here $P_{ext} \neq 0$.

I don't see the connection at all.


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Non-Arrhenius temperature dependence of bimolecular reaction rates at very high temperatures
Once I have read that in some cases bimolecular reactions can exhibit a maximum as a function of temperature due to the short lifetime of the activated complex at very high temperatures.
At low temperatures the rate is increasing because the reaction is limited by diffusion or by the encounter rate. At high temperatures the rate is decreasing because the lifetime of the bimolecular complex is very short, so that there is no time to overcome the barrier for bond breaking or making. This means the reactants have lots of encounters but they do not have time to react each time they encounter, so that the effective rate is small.
Could any one please indicate any piece of reference or useful terminology for this ? I have read this a while ago from a couple of papers and I can only remember the phenomenology, without any precise information.

The cases you describe are
(a) In reactions with a low activation barrier with respect to thermal energy ($k_BT$) the rate constant will be limited by diffusion together of the two species since they react on first collision. The rate constant is thus limited by the viscosity of the solvent.
(b) If there is a large activation barrier vs $k_BT$ the species still collide just as often as in the diffusion limited case, but most collisions are ineffective because they do not have enough energy to surmount the barrier and so the rate constant is small.
(c) It is a fundamental assumption in transition state theory that reactants and products are maintained in Maxwell-Boltzmann equilibrium. This means that relaxation of products and reactants must be faster than reaction rate, but if the temperature is extremely high this may not be achievable and the reaction rate could drop, vs. that at a lower temp, since there is not time for the product to be stabilised before it recrosses the barrier to form reactants. (This could probably also happen in a low pressure gas where stabilising collisions are rare or effectively absent)


text:
If gaseous oxygen and dissolved oxygen (in water) have the same partial pressure, will they diffuse into solid materials at comparable rate?
To be specific, I am wondering about the dynamics of molecular oxygen diffusion into gas-permeable polymer materials, which is the sensing process of the optical oxygen probe (polymer-encapsulated oxygen-sensitive phosphor).
Although the absolute concentration of dissolved oxygen in water can be 40 times lower than the oxygen concentration in air, the optical response of the probe seemed to be consistent when placed in air and under water (completely equilibrated with air). My guess is that, under equilibrium, the chemical potentials of the gaseous oxygen and the dissolved oxygen are the same. So that the rates of diffusion into the polymer are similar (air-polymer interface vs. water-polymer interface).
I've tried my best to search for textbooks and literature, but cannot find any material that explicitly explains this well. I appreciate it very much if someone can verify my assertion and offer me some references. Sorry for any misconception!
Sincerely,

If the air and water are in equilibrium, then che chemical potential of the oxygen in the air is equal to the chemical potential of the oxygen in the water.  If either is placed in contact with your polymer, oxygen present in the polymer at the interface between the polymer and water or polymer and air will come to equilibrium with the air or water extremely rapidly, and its chemical potential in the polymer at the interface will match that in the air or water.  It then will more gradually diffuse into the rest of the polymer.


text:
Substitution of aryl bromide in synthesis of a chelator
Hello I am currently working on a synthesis of a chelator for the synthesis of a photosensitizer for photodynamic therapy. My instructor wants me to do following synthesis procedure:

My problem is, the conditions for the last step are not yet known. How can ipso-substitution of the bromide be carried out? One version we tried was a copper-catalyzed process (CuI (1 eq), DMEDA (0.2 eq, cat.), NaN3 (2 eq.)) which only worked in 10% yield (fragment could only be seen in ESI).
How can the last reaction be sucessfully done?

Protect your phenols (silyl ethers should be OK). Do a metal-halogen exchange with nBuLi or iPrMgCl to make the aryllithium or Grignard and react this with tosyl azide. Work up with sodium pyrophosphate as described in this JACS paper
Also see this Angewandte paper for a recent example which has useful references

I will try the protection + JACS method. Thanks!

Update: I couldn't do the method. I ordered tosyl azide from Sigma and wow they sure take their time, it's been more than two weeks! I tried the method with this method: https://www.thieme-connect.com/products/ejournals/abstract/10.1055/s-2005-872248 , however, I couldn't get it to work somehow. Either my substrate is too fragile regarding elevated temperatures, or my batch is too big. I tried the bromide and the iodide. Any thoughts (while I am still waiting for tosyl azide)?

Were the phenol groups protected during the reaction?  If they were not they are acidic enough to convert your sodium azide to hydrazoic acid.

Thanks for the reply, no they were not protected, but just in case I used an excess of NaN3. I thought this shouldn’t be a problem. Also the deprotonation should be unfavorable due to the low acidity of the phenol, and the weak basicity of the azide ion, or am I wrong?

pKa of phenols is around 10, and if the azide gets protonated the hydrazoic acid is volatile and will be lost from the reaction meixture. The phenols can also complex your copper catalyst. I think it worth trying the reaction again with the phenols protected.

That is true, the hydrazoic acid has a boiling point of 37°C. I also tried the reaction, stirring at room temperature, but didn't work it up yet; maybe it worked there. I will try to work that up today. I also have a batch of protected phenols sitting at my bench


text:
Why the explosives get formed?
In many reactions mechanisms, we study things like 'it forms explosive mixture and should be discarded off immediately'.
Now my question is 'If those compounds are that much unstable that it will explode, why they form at the first place?' For example, if an explosive is getting formed, atoms might have assembled together and its energy will be increasing. But if that much energy is stored that it will explode, why it gets stored then?

Why should explosive substances not get formed? Until the tank car explosion on June 1, 1958, it was not known that nitromethane was explosive. Very simple chemical substance, but nothing suggested it might be explosive. See https://en.wikipedia.org/wiki/Nitromethane.

Why do buildings get formed (built)? A small force is enough to send them crumbling down. Yet still they are all around. How so?

Here are some useful posts: https://chemistry.stackexchange.com/questions/115307/why-do-most-explosives-have-symmetrical-chemical-structures ... https://chemistry.stackexchange.com/questions/7587/what-makes-substances-explosive ... https://chemistry.stackexchange.com/questions/4267/which-chemical-properties-make-a-substance-explosive .. https://chemistry.stackexchange.com/questions/174713/why-is-prismane-explosive-but-cubane-is-not

Think of chemical changes as a heavy ball rolling down a hill. In some cases, there's a divot near the top of the hill, and the ball gets trapped (a metastable state). However, just beyond that divot is a steep drop-off.
As long as the ball stays in the depression (i.e., there's insufficient energy to activate it), all well and good. For example, the substance 2-methyl-1,3,5-trinitrobenzene is easily produced by the (step-wise) nitration of toluene. This product is quite stable under most conditions; in fact, it took 28 years from Wilbrand's synthesis in 1868 until Häußermann found it explosive!
Once the activation energy is reached, and the ball rolls downhill, it shakes the ground (i.e., causing vibration, and releasing heat), freeing enough energy to activate others that are trapped -- a chain reaction.
The only fly in the ointment is quantum mechanics. Depending how deep the activation well, occasionally, the ball can tunnel out of the divot. Bang! with no obvious activation. For example, nitrogen triiodide has a remarkably low activation energy, causing rapid unscheduled disassembly of the sample and related equipment while drying.


text:
Why does oxidation of gold with oxygen work in cyanide leaching?
Normally gold can't be oxidizes by oxygen, but in the presence of $\ce{CN}^-$ it works.
In general, I have understood that this is due to the fact that complexation $\ce{[Au(CN)_2]^-}$ decreases the concentration of gold and thus also the single potential (Nernst's equation).
But I don't quite understand how the complex formation can occur beforehand? For the complex formation, the gold must be oxidized beforehand, so I'm not really getting anywhere with this explanation... Without oxidation no complexation, without complexation no reduction of concentration, without reduction of concentration no reduction of oxidation potential, without reduction no oxidation of gold by oxygen... Somehow this is a circular argument.
Or does the dissolving process start with a small complex formation (because in chemistry it is always the case that a small part of something is formed, although the energy for this is actually not sufficient) which then slowly favors the oxidation and this then favors the further complex formation so that it becomes more and more? So a slow-starting self-reinforcing process?
To summarize: Oxygen alone cannot oxidize gold. Why then does it work in the formation of the cyanide complex? What happens before? The oxidation or the complex formation, whereby during the complex formation Au+ must already be present, but how is that possible?
My intuition tells me it is about the relation between redox process and thermal + kinetic stability.

It's not like you can ever have bare Au+ in solution - it's always coordinated already during oxidation. Oxygen and cyanide are both coordinated to the surface.

@Mithoron yes, but Oxygen alone can't oxidize Au. So how would it work that it forms a complex "during oxidation" if the oxidation potential is too high beforehand?

I could give you some explanation how in equilibrium there's, like, always a bit of product, but who says it's gonna oxidize it alone? Did you heard about concerted mechanisms?

@Mithoron never heard about "concerted mechanisms"  I am in my 2. chemistry B.Sc. semester.

Then it's about time too look it up ;)

Here is a good discussion, although unanswered: https://chemistry.stackexchange.com/questions/164487/gold-cyanidation-process

Water cannot spontaneously get over the walls of a vessel. But it can do it via a tube to another place below the liquid level. That is commonly used for many liquids in vessels with an upper access.

The mechanism of gold dissolution in oxygenated cyanide solution is not based on direct oxidation by oxygen.
It is based on the big difference of the respective standard reduction potentials of the following first two reactions, leading to the summary third rection:
\begin{align}
\ce{Au(s) + 2 CN-(aq) &-> [Au(CN)2]-(aq) + e-} &E^{\circ}=\pu{−0.6 V}\tag{1}\label{1}\\
\ce{4 e- + 2 H2O(l) + O2(aq) &-> 4 OH-(aq)}  &E^{\circ}=\pu{+0.401 V}\tag{2}\label{2}\\
\ce{4 Au(s) + 8 CN-(aq)  + 2 H2O(l) + O2(aq) &-> 4 [Au(CN)2]-(aq) + 4 OH-(aq)} \tag{3}\label{3}
\end{align}
It means that

The reaction $\ce{Au(s) + 2 CN-(aq) <=> [Au(CN)2]-(aq) + e-}$ is at the standard conditions at equilibrium if the electrode potential (wrt SHE) is $\pu{−0.6 V}$
The reaction $\ce{4 e- + 2 H2O(l) + O2(aq) <=> 4 OH-(aq)}$   is at the standard conditions at equilibrium if the electrode potential is $\pu{+0.401 V}$.

It is obvious the electrode cannot have simultaneous both potentials, therefore some net reactions must be happening.
It is a kind of a single place galvanic cell, where the golden metal is at the same time the cathode and the anode, spending simultaneously both cathodic and anodic materials and using electricity for itself.
Gold  in alkaline cyanide solution is a strong reducer, considering the redox potential value. The Au+ ions  release from the metal is made much easier by reaction with cyanide. This is pushing the metal potential (wrt SHE) toward $\pu{−0.6 V}$ , at which the rate of the backward reaction (gold deposition) at standard conditions would be the same.
OTOH, oxygen as oxidant is being reduced to water by such a low potential, pushing the gold potential toward $\pu{+0.4 V}$, at which the rate of the backward reaction (oxygen formation) at standard conditions would be the same.
The gold metal potential will reach such a potential where is established the steady state of the same rate of both listed reactions $\eqref{1}$ and $\eqref{2}$ wrt the the zero net rate of electron release/capture.
For the same reaction with silver, it even reduces water to hydrogen to keep the potential high enough to dissolve silver, so oxygen is not needed.


text:
Equipment recommendation for atomic absorption spectroscopy (AAS) testing for mercury, lead, and cadmium
I work in a public health laboratory in a small mining town and we have the chance to buy equipment to test for lead and cadmium in wastewater, and mercury in not just wastewater, but also in blood, hair and urine.
It is known that the only method to determine mercury in atomic absorption spectroscopy is the cold vapor method, and that there are specific instruments like this one, that can analyze mercury in all those types of samples (biological or water).
Is there any spectrometer that can be used to measure all three of these metals or do we need to buy one for mercury and another for the other metals?

AAS can measure all three elements, and so the answer is yes. I suggest also the following article that can be of help: https://doi.org/10.1016/j.sab.2006.09.001. It's from 2006, but it seems a good start for your subject.

I think its paywalled there's the references: Christopher D. Palmer, Miles E. Lewis, Ciaran M. Geraghty, Fernando Barbosa, Patrick J. Parsons, Determination of lead, cadmium and mercury in blood for assessment of environmental exposure: A comparison between inductively coupled plasma–mass spectrometry and atomic absorption spectrometry,Spectrochimica Acta Part B: Atomic Spectroscopy,Volume 61, Issue 8,2006,Pages 980-990,ISSN 0584-8547,https://doi.org/10.1016/j.sab.2006.09.001.
(https://www.sciencedirect.com/science/article/pii/S0584854706002527)

This is a tall order. Your first step should be to list present and future samples, analytes with expected amounts and necessary detection limits. There are many possible methods and it might not be possible to get a one instrument.  Some considerations:
Mercury dedicated cold vapor; Solid samples: Energy dispersive Xray [EDAX]; Metals:atomic absorption, AA, inductively coupled plasma emission [ICP], in solution ion chromatography, finally wet chemistry: Hach Company has developed wet chemistry methods for almost everything in water. Versatility is important; who knows what you might find when you look hard.
Compile the list and call the various companies to see who can do it. Be sure to have actual samples run with results that exceed expectations.


text:
Raw green onions are spicy, but heated green onions are sweet. Why is this?
Raw green onions are spicy, but heated green onions are sweet. Why is this? More specifically; what change in the raw green onion's organic chemicals makes it no longer spicy when the green onions are slightly burnt, and is there some chemical change in the burnt green onions that also make it sweeter, or is it just the removal of the spicy compound?

Onions caramelize, becoming sweet as various complex saccharides are broken down to more "tasty" sugars. For instance, in onions, green (scallions), yellow, red or other, chains of fructose are broken into fructose monomer and dissolved in the onion's water (~90% by mass), making the sweet taste accessible to the taste buds. Caramelizing also produces many complex changes in other chemicals, too.
In addition, onion and its relatives contain strong-tasting, strong-smelling, and even lachrymatory compounds, such as syn-propanethial S-oxide. Since these are volatile and not too stable, caramelizing onions for half and hour or more destroys or evaporates them, leaving the pure melt-in-the-mouth caramelized onions.
Sorry, my mouth is watering and I must leave to get a napkin...

I have authentically been drawn to emotional tears at the smell of frying and caramelizing onions. Well written and useful answer, danke!

(+1) And garlic is an allicin wonderland, as a colleague used to say!

This is a wonderful answer, and it is a delight to know that mere LLMs (such as ChatGPT) cannot threaten the intellectual capacity of human scientists and lifelong learners. Thank you so much!


text:
Is vibronically resolved spontaneous emission temperature dependent, following a distribution defined by the canonical density operator?
Does the lifetime of a molecular excited electronic state, for example the $S_1$ state, suffice for thermal equilibration before spontaneous emission occurs?
For example the fluorescence of coumarine, squaraine or cyanine dyes. Does the vibronically resolved emission spectrum of the $S_1$ state follow the distribution expected for emission from  a canonical vibrational density operator of the emissive electronic state if the spectrum is measured at temperatures around 300 Kelvin, lets say in solution.
It is well known that in most cases electronic states relax quickly after excitation using for example a laser, before emission occurs from the $S_1$ state, following Kasha's rule. But it doesn't address the population of the vibrational state of the $S_1$ from which emission occurs. I assume that the vibrational ground of the $S_1$ state dominates in most cases, but what happens at higher temperatures? Is a change of the emission spectrum observable or is the emission spectrum generally insensitive to temperature, which would indicate that the fluorescence lifetime is not sufficient for thermal equilibration to occur.
My main concern is how the "initial conditions" for the simulation of vibronic emission spectra should be chosen. If thermal equilibrium can be assumed, then the density operator for a canonical ensemble is appropriate. I would like to know if this is justified by the rate for thermal equilibration vs fluorescence lifetime.

In solution generally yes, measurements show that vibrational relaxation takes less than 5-10 ps and excited state lifetimes are usually longer than this so a thermal population exists after this time.
In isolated molecules in the gas phase it depends on which vibrational level is excited. If $v=0,1,2$ then and transfer to other levels can be very slow, tens of nanoseconds, but as  $v$ increases this transfer rate constant  to other levels in the same or other states rapidly increases. Fermi Golden rule type of behaviour. There is a huge literature on this from the 1970's, non-radiative transitions. See Heller, Freed, Gelbart J. Chem. Phys. 56, 2309–2328 (1972)
Edit. If $v$ is small in the excited state and there are no collisions then emission comes from this level in the excited state. This is clearly not a Boltzmann distribution, but adding inert gas causes transitions between vibrational levels until Boltzmann dist'n is reached. This can take nanoseconds. See  Proc. Roy. Soc. London. Series A, Mathematical and Physical Sciences, Vol. 340, No. 1623, 1974.


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Lifetime components in phosphorescence decay
I'm studying the phosphorescence decay of diacetyl in acetonitrile (it has a high quantum yield of intersystem crossing). The fluorimeter than I'm employing has microsecond resolution and, since the decay of the phosphorescence isn't too long, I have to measure the IRF (width of 1-2 $\mu$s) in order to perform a reconvolution fit. However, I get a biexponential fitting with a long time (like 20 $\mu$s) and a very short time (around 30 ns). Does that make sense? I thought that the short lifetime could be due to fluorescence, but I get it even in the decay of the righmost peak while using a long pass filter. Moreover, I don't know if it makes sense to have a lifetime with more resolution that the limit of the aparatus.
Does this make sense or has any meaning? Maybe is an artifact or I'm making some mistake with the measurements?
Edit: I'm using the Spectrofluorometer FS5 from Edimburgh Instruments, and employing its software (Fluoracle) to make to fittings. The light source is a flash lamp. By reconvolution/convolution I mean a fitting that takes into account the IRF (instrument response factor), as the lifetime of the decay approaches the duration of the IRF.
The fittings look like this:

Figure 1: monoexponential fitting

Figure 2: biexponential fitting

What is the fluorimeter’s light source? Is it a flash lamp? If so, maybe the short component of the output is an electrical artifact due to the flash lamp.

Please add more details about measuring instrument response function, what is meant by reconvolution? Add a figure that explains what you are doing.

If response function is 1-2 microseconds, you can't measure & quantify 30 ns decay with any reasonable amount of accuracy. So while one might exist, I don't think your data shows that. It must be instrumental as EdV points out. It's a fitting artifact of some kind, but with so little information available (as AChem and porphyrin point out) it's hard to give much of an answer. However, if it was real I would say it's due to nonradiative recombination which is density dependent - excitations kill each other off quickly until the remaining density is so low that they can decay at their own rate.

@uhoh, I added information to the post. I also think its an artifact, but I don't know how to get rid of it. I thought that and artifact of the lamp would be take into account with the IRF.

That fast exponential drop decreases by a factor of 100 in about 1.12 µs. Divide that by 4.61 (i.e. ln(100)) and that decay has a mean life of 0.24 µs. https://i.stack.imgur.com/6lAg5.png If I were you I would (try to?) get the data out somehow as a text file and analyze it with a separate program - I really don't like the fitting results it's giving you. You can consider sending the raw data to Edinburgh Instruments directly as well to see what they think.

Phosphorescent materials, such as “glow in the dark” paints and tape, are readily available. Perhaps make up a test specimen using such a material and see how it behaves in your fluorimeter. If the short lived component is always present, then the evidence for an artifact is stronger.

@uhoh, I just tried fitting in Origin with the Fit Convolution app. I followed the tutorial but I didn't get a good fitting, neither mono or biexponential.

Phosphorescence should be at a longer wavelength than any fluorescence so carefully using a filter or grating should remove any fluorescence. Also measuring at different wavelengths will change percent of any fluorescence still getting through a filter. Is there a tiny bit of scattered light being mixed in with the phorphorescence? (Some types of glass filters do fluoresce but usually you need to excite then directly).
How are you doing the convolution,  Marquardt type non-linear least squares iterative convolution is normal. What sort of chi-squared are you getting for the fit, what type of weighting is used? How does a 1 exponential vs 2 compare? What do the residuals look like and so on. Are you fitting to a background also? However, 30 ns is really too short to be confident that it actually exists if the irf is 1-2 microsecs.

I added some information to the post. I'm just doing the convolution with the software (Fluoracle) of the aparatus, so I have no clue about the type of convolution or the parameters. Would it be better to use Origin or another tool? Moreover, I'm not fitting to a background, I just followed the manual of the aparatus (it said I just needed the IRF and the decay).

Well, now we can see the data there is definitely a lot of a short lifetime there. Have you tried a cell filled with only solvent to check scattering or no cell/empty to check any electrical noise? If all is ok then there is a real short fluorescence / phosphorescence component.

Related to the comment immediately above, the OP could spike a little nitromethane into the solution, to quench both the fluorescence and phosphoresce and see what results. (I assume the solution is oxygen free, more or less.) Also, since a flash lamp is used, how good is the instrument at blocking direct excitation light or “after glow” from the Xe (?) flash lamp?

@porphyrin, I monitored a decay with only solvent (at the same wavelenghts and bandwidths than the sample) and I get some signal. Perferming a convolution fit with this signal and the IRF gives me a lifetime of around 50 ns. However, this signal is very weak (5 min of measurement to get a very weak signal vs 1 min to get the strong signal of the sample), so I don't know if that could be the lifetime that I'm seeing.

So it seems that you do have some scatter coming through which it would be best to totally eliminate if you can,  but further to Ed V's comment you could also purge with a heavy atom gas such as Xe or add say iodide ions to cause heavy atom quenching. The short decays will be unaffected if is scatter. All I can now suggest is that you look at, say, Ru trisbpy which has a long excited state lifetime, or some other well measured species to check how you instrument is working.


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Datasets of Light Absorbance Spectra for Atmospheric Gases
I am currently doing some thermal/aerospace engineering research that involves finding light absorption in different points in Earth's orbit.
What is the best way to find absorbance vs. wavelength spectra data (IR,Vis,UV) for common gases such as O, O2, N2, He, Ar, etc.?
Ironically, I've found this spectra for other molecules, but am struggling with the most common gases on the planet. Help is much appreciated.

The homonuclear diatomics have no (or at least v. v. small) ir absorption (due to selection rules) so you will need to look for Raman spectra to get wavelengths, but they do have uv absorption. There should be lots of sites giving this (Search for images of spectra) otherwise look at the NIST site.   Atomic energy levels can be found easily also.

You will need to use some chemical intuition and the right keywords. He and Ar (and all other noble gases) are monoatomic, so it is good omen that one cannot find their vibrational spectra. Similarly, these gases are colorless, so there will be no absorbance in the visible region. There are no double or triple bonds in noble gases, one will not expect an ordinary UV spectrum. However, you will need to search deep UV or vacuum UV absorption spectrum of these gases which must have been studied to the hilt.
Coming to diatomic gases (oxygen and nitrogen), since there is no dipole moment change during vibration of a homonuclear diatomic gas, we do not expect an infrared spectrum of these gases. These molecules do vibrate, should one look for a Raman spectrum then?
(Liquid) oxygen is somewhat pale blue, so it means that gaseous oxygen will absorb light in the deep red region. We can see those beautiful absorption bands in high resolution solar spectrum. Search high resolution solar spectrum. Diatomic oxygen and nitrogen both absorb deep UV at the cost of being broken or ionized (recall ozone formation).

gaseous O2 has no visible spectrum at least at reasonable low pressure. There is purported weak dimerization in the liquid and solid that absorbs. The blue in water is dfferent that is a stretch overtone so is present in liquid and gas.

Actually, gaseous oxygen at 1 bar pressure does absorb near IR and red, to some extent. These give rise to the so-called “telluric” bands that astronomers and astro-spectroscopists take into account when they acquire spectra. See also https://chemistry.stackexchange.com/q/15035/79678.

And the Fraunhofer A “line” is due to atmospheric oxygen absorption around 760 nm: https://astronomy.stackexchange.com/a/18034/45954. So the answer is fine.

For UV-Vis, one source is The MPI-Mainz UV/VIS Spectral Atlas
of Gaseous Molecules of Atmospheric Interest mentioned in an earlier answer.  Given the (chemical) simplicity of the compounds of interest mentioned, NIST Chemistry WebBook can complement this one; this compilation includes IR data e.g. about gaseous carbon dioxide

(image credit to NIST, screen photo from here)
in formats you can process further (e.g., JCAMP-DX).


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betamechaptoethanol smell
I am doing my masters in science and today one of our neighbors from the other lab came to heat his samples on a heating pad. the samples were in small tubes which contained  Betamechaptoethanol, when he finished heating them, the tube cab suddenly opened because of the temperature. Our lab smelled extremely bad even though I was wearing a mask and lab coat. when I got home I could still smell the betamechaptoethanol. I am so worried about it because my clothes even smell like it. Is the smell toxic? I put my clothes in the basement and still can smell it.

Beta mercaptoethanol?

Almost all thiols are smelly even in ppb levels. For example, $\ce{H2S}$ and garlic. But my question is why some one with MS level education let some one to heat closed container containing somewhat volatile compound? Nonetheless, heating a closed container is a major safety issue regardless of what it contains.

Some sulfur organic compounds have rather a strong odor. For your delectation, may I suggest reading Derek Lowe on the not dissimilar compound, thioacetone. For further documentation, see Grayson Van Beuren.
Perhaps your laboratory neighbor so enjoyed working with beta mercaptoethanol (AKA 2-Mercaptoethanol) he wished to share the pleasure. However, "Due to its diminished vapor pressure, its odor, while unpleasant, is less objectionable than related thiols." On the other hand, he did warm it, all the better to increase vapor pressure and to help you appreaciate it.

They also say Se is much worse than S and Te much worse than Se. It is reportedly enough to hold a piece of Te in a bare hand for a while and it would exclude you from social life for months. Some chemists reportedly commited suicide because if that.

They say "Smell it, and you'll vomit farther than you can see".

You will die a lot sooner if safety rules are not in place and enforced in your lab.


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Why isn't residual standard deviation used in FQY analysis?
I've been measuring relative fluorescence quantum yield (hereafter FQY) for a photosensitive pigment. Most analytical articles and standard operating procedures (SOPs) I've read suggest taking the average of multiple FQY measurements (at least 3) before calculating the standard deviation, which seems good; however, wouldn't comparing residual standard deviations (RSDs) of the linear regression of fluorescence intensity versus the absorbance of the dye standard (e.g. rhodamine B) and dye sample (e.g. hypericin) provide thorougher validation of the calibration fits and thus FQY? In other words, since the relative FQY is calculated from the slopes of these plots, can I glean more information by comparing the RSD of the plots and SD of the FQY?

The experts at NIST are worth knowing about: https://nvlpubs.nist.gov/nistpubs/jres/113/1/v113.n01.a03.pdf.

Even better would calculation  of SD estimation of the slope and intercept and the confidence band of the linear regression.  It then leads to ability to determine the y confidence interval from the given x value and vice versa.

You do need to account for the absorbance of each sample otherwise the yield will be hugely inaccurate, then take the average etc. Normally you would try to make each solution of the same concentration but you could do what you suggest provided you are sure that no aggregation etc. of the molecule is present.
To get relative yield (other than just a number proportional to a voltage from your detector) you also need to measure the fluorescence of a known standard molecule, and at the same wavelength because detection sensitivity will vary with wavelength.

Concern for aggregation is a good point, and you give good general advice for FQY. However, the majority of the post doesn't really answer my question. Namely, is RSD a good statistic to include in a paper for validating results? Is calculating the standard deviation of FQY from RSD and error propagation is sound? You seem very intelligent.

Why not simply quote std deviation of slope of relative yield vs concentration, is this what you mean by RSD ? If you look at residuals to the fit then they should be randomly distributed, but you probably don't have enough data points to check this fully, but a visual check should be ok. At this point I think you need to think about what you actually want to achieve, you may not need very much statistical analysis, your result will only be as good as the standard's yield is, I'm not sure rhodamine b is that good, but you should check on this re photostability, aggregates etc.


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What is the meaning of "maximum" weight produced after series ot reactions?
$4.14$ grams of a piece of lead was made to react with $\ce{HNO3}$ so as to produce lead nitrate. Lead nitrate further reacts with a series of reagents to form $\ce{(NH4)2PbCl6}$. Then find the maximum possible weight in grams of $\ce{(NH4)2PbCl6}$ produced.
What is the meaning of maximum here? My method is to apply Principle of Atom Conservation on lead, but I directly get an answer without using the "maximum" condition.

You cannot produce more than maximum  $m_\text{prod, max}  = m_\text{reac} \cdot  \frac{M_\text{prod}}{M_\text{reac}}$  , unless the product is not pure.   But you can always produce less, if not all the lead formed the product or if there were loses.

The question can be answered assuming all atoms of lead ($\pu{4.14 g}$) end up in $\ce{(NH4)2PbCl6}$.  So you need the molar masses of the two materials, because the number of moles of lead is the maximum number of moles of the product :
$$\frac{m(\ce{Pb})} {M(\ce{Pb})} = %
%
  \frac{m(\ce{(NH4)2PbCl6})} {M(\ce{(NH4)2PbCl6})}$$
$$\frac{\pu{4.14 g}} {\pu{207.20 g mol^{-1}} } = %
%
  \frac{m(\ce{(NH4)2PbCl6})} {\pu{455.99 g mol^{-1}}}$$

 Hence at maximum, you can isolate $m(\ce{(NH4)2PbCl6}) = \pu{9.11 g}$.

Good explanation. Just to point out that you have taken mass as $4.41 \text{g}$ instead of $4.14 \text{g}$.

@Aleph Thank you pointing out.  The error in the equation (and subsequent result) was corrected.

The question makes sense, but it is just needs explaining the use of "maximum" in this context.
Without overthinking the question, it asks what is theoretically the maximum yield by mass assuming there is no loss of reactants or product.  Inevitably:

Some of the raw material may be left (since there is an equilibrium, though it may be very far towards the side of the product).
Some of the reactants and product may be rinsed away in the solvent or subsequent washings.
Some of the product may be left on filter paper.
For volatiles, such as some osmium compounds, some may be lost by evaporation.

The question "tries to make this easier for you," by positing 100% yield.
One could quibble with the question (though it likely won't get more points on a quiz).

The wording, "maximum possible weight," rather than mass, means the product should be measured at the North or South Pole. Weight of a fixed mass varies with position on Earth, due to the Earth's spin, oblateness and altitude of the geographic location. There is a quite noticeable difference, ~0.7%, between weight of a mass at the equator and pole, which must be taken into account with spring and strain-gauge scales. At this writing, a 100 gram gold bar is priced at US\$6,167.57. If it were sold by weight at the North Pole, it would have lost ~US$43 in value at the equator... but it ain't, it's sold by mass. Of course, a balance scale would read the same anywhere on Earth.
On a very theoretical level, the (averaged) atomic weight of lead varies by origin. "As time passes, the ratio of lead-206 and lead-207 to lead-204 increases," therefore "the standard atomic weight of lead is given to only one decimal place." So to calculate the greatest maximum (pardon, please, the pleonasm) yield of product, one could select lead from a location with the most 204Pb, the lightest stable isotope, giving the greatest number of moles of lead so that more of the other reactants are incorporated in the product.

Raise these issues with your instructor at your own risk! Written by one who has been thrown out of class (in a kindly fashion) for arguing with a pedagogue.


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Substructure abbreviations with the SMILES format
I am trying to use the SMILES format for encoding molecules from handwritten logbooks. The molecules often contain abbreviations, such as "Ph" for a Phenyl group. Is there a way to have these abbreviated structures in SMILES?

So far, I didn't encounter an equivalent of Ph in the world of SMILES.
Though SMILES (and SMARTS) represent chemical structures (and reactions) in a reduced, simplified form, I would recommend to stick to clearly sketched structural formulae to keep the content of the (handwritten) logbook easy accessible for your colleagues;* especially for those joining the group/enterprise after your left the organization (the reading of the information stored).  For one, the strings can become lengthy and complex (see e.g. the documentation compiled by Daylight, part theory and part tutorial) and you may be faster to (manually) doodle the structure, than composing the string.
For two, if you turn to programs like openbabel to work with them and other (chemistry relevant) file formats, you will meet multiple SMILES dialects out there.  Staying with said program openbabel for example, it can report SMILES in their canonical, or universal, or in their InChI-fied form (ref. documentation).  Especially with the structural more complex molecules, the output as a string can differ; even more likely if SMILES are assigned to a structure by different programs (e.g., GUI based chemdraw, or programming libraries like RDKit. Already benzene can be expressed (at least) by either c1ccccc1 with implicit aromaticity, or c:1:c:c:c:c:c1, or with Kekulé SMILES C1=CC=CC=C1; each of these forms has its pros and cons (e.g. explicit electron count/localization).
* Don't forget outside chemistry and personal trained in STEM only, a lab's log equally may become relevant e.g. for a patent attorney, too.


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Is BET equation applicable to only monolayer adsorption range?
One of my senior colleagues is stating that although BET adsorption isotherm equation hypothesises about multilayer adsorption, the P/P0 values (usually from 0.05 to 0.3) accountable for BET surface area measurement only correspond to monolayer adsorption - as multilayer adsorption doesn't occur at that P/P0 range.
I am confused about that as I used to know that in BET the total adsorption volume of a gas (i.e., V) at different P/P0 values are taken (it may be monolayer or multilayer adsorption) and we only get the monolayer adsorption volume (Vm) after calculation of the BET equation.

In general, BET is applicable to low pressure range, not necessarily the monolayer adsorption range.
According to BET theory, it is not necessary to have a fully formed first layer for the second to start forming; this is a good assumption and it is indeed what occurs in real cases. Although most of the adsorption will occur on the solid surface before monolayer formation, there is a statistical probability of molecules adsorbing on top of the incomplete monolayer. This is especially true when the C value of the BET equation is less than 2 (where it starts to look like a type III isotherm).
However, the BET derivation does have assumptions that occur at low pressures: the most critical one being that there is no lateral interaction between adsorbed molecules. If you plot the isotherms in the BET-linearized fashion (that is, $\frac{p/p^0}{n(1-p/p^0)}$ as a function of $p/p^0$) you will see that the values deviate from linearity at pressures above $0.1-0.3$.
The best way of selecting the correct range for fitting the BET plot, is the classic paper by Rouquerol:
Rouquerol, J.; Llewellyn, P. L.; Rouquerol, F. Is the BET Equation Applicable to Microporous Adsorbents? In Characterization of porous solids VII; Studies in surface science and catalysis; Elsevier, 2007; pp 49–56.


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Gibbs free energy versus reaction progress plots
I am trying to make sense of these two graphs. They appear to have similar axes, yet display different curves.
Please let me know if this understanding is correct:
In the top image (x-axis = "Reaction coordinate"), this shows the free energy as we convert X equivalents of A to Y equivalent of B (with X and Y relating to the stoichiometric coefficients). So this is looking at an individual reaction essentially.
And for the bottom image (x-axis = "Reaction progress"), this shows the free energy versus the reaction quotient (Q) essentially.
A large part of my confusion is why the bottom image does not indicate activation energy yet the top one does. Does this have to do with Gibbs free energy being a state function. It is true that over the course of the reaction, reactants require some activation energy, but it is not represented in the bottom image since Gibbs free energy is a state function? Perhaps the top image shows the actual "path" taken, while the bottom image just indicates the overall change in Gibbs free energy as Q changes.

https://chemistry.stackexchange.com/questions/173557/what-is-the-basis-for-the-relationship-of-gibbs-energy-and-extent-of-reaction-sh

https://chemistry.stackexchange.com/questions/456/what-does-reaction-coordinate-mean-what-does-it-mean-when-a-reaction-has-two

@ait2001 the diagrams you are asking about have no relation between them. The first refer to the reaction coordinate (chemical kinetics) while the second refers to the reaction progress (chemical thermodynamics). You should be able to differentiate those concepts to understand the difference.

The lower plot is OK, it shows that the Gibbs energy is a minimum at equilibrium, i.e. the slope of free energy vs extent of reaction $\xi$ is zero so $\displaystyle \Delta G_r=\left(\frac{\partial G}{\partial \xi}\right)_{T,p}=0$, then $\Delta G^\text{o}=-RT\ln(K_e)$
The upper plot is far less well defined and really only reminds us that there is a barrier between reactants and products, and the use of Gibbs free energy here really only shows that entropy is considered as well as enthalpy, but using $E_a$ is then confusing as it is an energy not free energy. The equilibrium between A and B is at the minimum of the curve in the lower plot. Recall, that there is an activation energy from B to A but is not shown in the top picture. The reaction coordinate is not defined in any real sense either, it is just used to indicate that bonds are broken and others formed.


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Mobility of coordinate complex ions
Conductivity of ions depend on mobility of ions, but I doubt if a Coordinate complex formed can affect the mobility of that ion.
Let's say we have AgNO3 dissolved in Ammonia water. How can we compare mobility and Ion  Transport number of Ag+ in Water and in NH3-water, so that {Ag(NH3)2}+ complex is formed.
I think Mobility is decreased due to complex formation.

Consider the implicit coordination of $\ce{H2O}$ in both $\ce{Ag+}$ and $\ce{[Ag(NH3)2]+}$ as  $\ce{[Ag(H2O)n]+}$ and $\ce{[Ag(NH3)2(H2O){n-2}]+}$.


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How to find the ligand field splitting of d orbitals in a square planar complex?
I am working with a Iron square planar complex with $\ce{NH2-}$ and $\ce{OH-}$ ligands having cis and trans conformation. My question is how to theoretically/computationally calculate the ligand filed splitting of the orbitals of Fe specifically?

Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial

Are you sure of the existence of this iron complex ? Is it Iron(II) or Iron(III) ?

@Maurice the compound probably doesn’t exist. We are making a toy model for computational reasons only. So it’s a wholly neutral system

Why do you play with non-existent compounds ? Why don't you try to study a cyclic compound made of 6 nitrogen atoms, for example ?

I'm not an expert in this field yet, but I think Using Orca you can calculate the orbital energy numerically. I have used CASSCF as method of calculation.
The input file I've prepared is as following
! def2-TZVPP SlowConv PAtom
! Allpop
! printMOs
! PAl2

%casscf
    nel     6        #Numeber of electrons
    norb    5        #Number of Orbitals in active space
    mult    1,3,5    #Multiplicities (I've got this from Term symbols)
    nroots  11,11,5  #Number of micro states for each Multiplicity
end

*xyz -2 5
    Fe       0.0000000000      0.0000000000      0.0000000000
    O        1.8600000000      0.0000000000      0.0000000000
    O       -1.8600000000      0.0000000000      0.0000000000
    N        0.0000000000      1.9400000000      0.0000000000
    N        0.0000000000     -1.9400000000      0.0000000000
    H       -0.8407460951      2.2854839793      0.5145942634
    H        0.8414809543      2.2855771457      0.5133281192
    H        2.1097954426      0.0911972135     -0.9540592676
    H       -2.1097621967      0.0917132162     -0.9538732789
    H        0.8411680767     -2.2854165305      0.5139464472
    H       -0.8411632857     -2.2854156190      0.5139547890
*

the xyz section has the coordination for atoms in your complex
and the CASSCF block contain the information about the active space
The output file has a section for orbital energy and occupation number of each orbital
For the bond length I've just searched on google for bond length of those atoms with Fe.
By analyzing the combination coefficients of orbitals, it is possible to obtain the energies required for these orbitals and subsequently calculate the splitting phenomenon.
By simply modifying the coordination in the input file, you can explore various configurations and calculate their corresponding properties. To achieve more accurate and reliable results, you may consider enhancing the basis set, which can lead to improved energy estimations

Did you optimise the structure at all? The whole calculation seems a bit aimless and possibly overkill. I'd expect dft to give better starting guesses. But basically you're right though, given enough time and effort, this is the way to evaluate this...


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Free-radical addition of bromine to alkene
If light and radical initiators are present, radical addition of $\ce{HBr}$ to alkenes takes place and outcompetes electrophilic addition of $\ce{HBr}$ to alkenes. Why isn't a similar thing same happening if we use $\ce{Br2}$ instead? Why does the addition of $\ce{Br2}$ to alkenes only take place through electrophilic addition? What is wrong with the following radical mechanism?
$$
\begin{align}
&\text{Initiation:} &\quad \ce{Br2 &->[$h\nu$] 2 Br^.} \tag{R1}\\
&\text{Propagation:} &\quad \ce{CH2=CH2 + Br^. &-> ^.CH2-CH2Br} \tag{R2}\\
& &\quad \ce{^.CH2-CH2Br + Br2 &-> CH2Br-CH2Br + Br^.} \tag{R3}\\
&\text{Termination:} &\quad &\ldots
\end{align}
$$

The addition of bromine on a double bond occurs without light. So the molecule $\ce{Br2}$ is not broken into Bromine atoms. But it reacts with the alkene $\ce{R1R2C=CR3R4}$, producing a triangular cation $\ce{C2R1R2R3R4Br^+}$. And this cation reacts later with the remaining Bromide ion to produce the saturated compound $\ce{C2R1R2R3R4Br2}$.

Well, it normally occurs like that, but what if the environment of reaction was more suitable for radical mechanism?

What sort of environment ?


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Is the line between physisorption and chemisorption species specific?
I was studying about heterogenous catalysis, and my Professor said something along the line, "When the adsorption energy is comparable to (or lower than) heat of evaporation, then it is physisorption. When it is larger than that, it tends to be chemisorption"
By this logic, that meant the energy for chemisorption is different for each chemical species. Is this statement true? Or am I understanding it wrong?
If anyone have a reference on this, please let me know. I've been searching for quite a while.

Why it should be the same?

Just keep in mind that human language and human generated labels do not affect the natural and fundamental phenomena in anyway. If we say electron has a positive charge and proton has a negative charge, will it change the electron or proton in anyway? Chemisorption and physisorption are just labels and there is no sharp boundary line to mathematical define boundary of chemisorption and physisorption. At best, accept these labels qualitatively.

Take it as a conventional rule of thumb.
If desorption enthalpy is not bigger than evaporation enthalpy, we can assume the cohesion forces of adsorbed molecules between each other and with the surface are of the same character as cohesion forces between molecules of the respective bulk liquid. We can then take it as physisorption.
If desorption enthalpy is bigger than evaporation enthalpy, we can assume the cohesion forces of adsorbed molecules with the surface are, at least partially, of other character and there is some exothermic chemical bonding. We can take it then as chemisorption.
But by nature of interactions, they are rather both physical and chemical, so it is rather arbitrary clasification.

If the clasification was defined by some arbitrary border value between physisorption and chemisorption, if would be very formal and both ways of interactions would be present in both groups in much bigger extent than in the above way.


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Predict the products of organic reaction sequence

This is a reaction sequence involving alcohols, phenols, halides, alkenes ,alkanes etc , and the task is to identify all the compounds A to E
The few things I was able to identify :

The compound C gives immediate reaction with the Lucas Reagent (ZnCl₂+HCl)therfore it must be a tertiary alcohol or it must produce a highly stable carbocation.
Reaction of alc.KOH after addition of HBr on C suggests that D must be at least an alkyl
bromide
In the conversion from D to B , step 1 produces an alkene , which then is probably
undergoing a reaction by H⁺ to form isopropyl grp , which then gives the Friedel craft alkylation reaction to form Cumene as suggested by the lewis acid catalyst AlCl₃ and then in next step by looking at the reagents I can predict the reaction to be the Cumene peroxide method of preparing phenols

Therefore B must be phenol or a phenol derivative and E must be a Ketone, Acetone to be precise
Now , LiAlH₄ reduces acetone to isopropyl alcohol
I think in the conversion from D to B using KMnO₄ and NaOH/CaO , first oxidation occurs and then decarboxylation

The problem is the molecular weight of isopropyl alcohol will not come out to be 32 , and I am not able to guess the structures accurately as there might be possibilities of some extra non participating groups on B ,C and D


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Why is there a non-uniformity in even the reasons that explain exceptions in the trends in chemistry?
I have been studying the periodic table and several properties of atoms like the Ionization Energy, Atomic Radii, Electron gain enthalpy, Electronegativity etc.
Now, each property has somewhat of a "continuous trend" for left to right(periods) and top to bottom(groups) but there are "a lot" of exceptions as well. There are a few reasons that explain those exceptions but I have been wondering why those reasons do not apply to other elements of the periodic table as well.
For eg- Neon, Beryllium and Nitrogen have the least Electron Affinity in their period with the reason being stated that they have half and fully filled orbitals but on the other hand in the period just below them one would expect to see the same patter of exceptions but now Magnesium, Argon have the least Electron Affinity in their period and Phosphorus which should've been with  Magnesium and Argon is in the middle of the trend of increasing Electron Affinity. This shows that even the pattern in the reasons for exceptions break and there are more and more such examples.
So, finally I want to ask that why is there a disuniformity in even the reasons that explain exceptions in trends in chemistry? Why do the reasons just apply to some elements conveninetly and not to others? Is there any way to explain every single exception there is?(in atleast the trends of the terms I mentioned above like Electron Affinity and Electronegativity).

I am not able to understand why my question is downvoted, I don't think my question is that bad and its just out of curiosity that why exceptions behave the way they do. If we know that there is an exception then there must be a way to explain it somehow.

Your question may have been downvoted because nobody knows how to answer it. It is probably easier to downvote than to admit a failure. I may be wrong. It was just a hypothesis.

The question "Why is not there uniformity of reasons for exceptions?" is of the same kind as "Why are there exceptions?" in the first place. Uniformity requirement assumes there is no place for the 2nd level of exceptions, or that the reasoning cannot  apply just for a single case. Atoms are too complicated  systems to be described by a single level uniform exception reasoning. With so many bound objects, they would be complicated even without quantum effects.

Could someone add a 5th CV before there's half a dozen of "what is this chemistry all about answers" here? Question is blatantly too broad and vague. I suggest to go with things like that to forums, not a Q&A site that is about specific things.

Exceptions exist for periodic trends and the explanations for those exceptions vary (if such explanations exist at all) because individual atoms are highly intricate and complex systems that are often not amenable to simple models, and complex models can't always be connected to straightforward verbal explanations.
A single electron around a nucleus is sufficiently simple that the solution to the Schrödinger equation for hydrogen-like atoms and the development of the associated quantum theory was possible. One of the greatest accomplishments in 20th century physics was this ability to explain with great mathematical precision a huge number details about an object in the real world, one hydrogen-like atom. It wasn't until the early 21st century that very good mathematical models were developed to account for the additional level of complexity resulting from the presence of the second electron in helium. None of those mathematical models of helium can or should be taught to students in high school at the level where verbal explanations are given for periodic trends.
When mathematical theory is sufficiently well developed, we may be able to mimic every physical property of individual multi-electron atoms, but I don't believe verbal explanations will ever be available in the future for every detail about every periodic trend because the mathematical theory will be too complex for such explanations.

Ah, so the reason there is not a unified explanation for these exceptions(or may never be) is that there are more than one factors involved in more and more complex systems which makes it harder to give a simple explanation neglecting all the complex mathematics and factors actually present in those exceptions.
Thanks for taking your time out and answering my question!

Indeed, the amazing thing is that hydrogen-like orbitals are useful at all beyond maybe helium.

It is certainly a good thing to be curious, and if you are, people will want you to show that you are motivated and able to look for solutions and bring new ideas. Your question contains several properties, you should have limited the question or your research to one particular property. Some of them are, while others are really pathological. The properties that arise from orbital energies, such as ionisation energies and electron affinities, arise from a complicated ground state of a many-body electronic system. Until now, there is no specific method that can accurately and efficiently predict these properties for all systems. An intuitive understanding is limited to rules and a few exceptions. If you try to work out a rule for these exceptions, you will arrive at another rule with another set of exceptions, and so on. The most accurate description will be an infinite number of rules, each rule described by a specific well-behaved function that maps the configuration of your element to a specific result. The methods for solving this kind of problem are worked out in this way.
As this reasoning is well known, and your question consists of many different properties, this may be one reason why you have probably received more downvotes than answers.
As this argument is well known, and your question has many different properties, this may be the main reason why you have probably received more downvotes than answers, rather than an inability to answer.

Thanks for informing me where I went wrong with asking my question and I do admit that I should've sticked with one property rather than many different properties but I thought that these properties all have some sort of connection in the sense they all have a part in the ability to chemically bond for atoms and therefore may have a "unified explanation for exceptions"(if it exists).

Chemistry is complicated; simple trends should not be expected to be universal
The expectation that trends in bulk chemical properties should be simple or clean assumes that only one or a very small number of factors cause the trends. But, if that were true, chemistry would be a very dull subject (which is the same as saying it would be physics and a particularly simple for of physics).
In reality, most observed chemical properties, even of elements never mind molecules) are the result of multiple interacting factors. The wavefunctions of electrons in atoms, for example, are only mathematically soluble for the simplest possible atom (a single hydrogen atoms with just one electron). In anything more complex, there are multiple interactions between electrons and then other interactions between multiple nuclei and multiple electrons. Given that, universally simple trends should not be expected.
Without a complete model of all the factors involved in electron-electron and electron-nucleon interactions it is impossible from first principles to know which particular interaction dominates in each case. We can observe some rules and trends, but we can't easily know when other factors will become important enough to break the trend.
This is one reason why chemistry is a rich enough subject to be interesting.