text:
What's the current of an aromatic ring?
Aromatic molecules like benzene have a ring current, created by the movement of electrons around the ring. This is most visible during proton NMR, when the magnetic field of the ring current opposes that of the NMR superconductor.
What's the current though?  How fast are electrons moving around the ring at room temperature?

Well, they don't exactly "have" it.  Ring current can be induced in the ring by magnetic field, like in any loop of conductor. Afaik actual velocities of drift of electrons in electric currents are very small, negligible in comparison with overall speed of electrons.


text:
Reaction Rate for Surface Reactions
Assume reactions S1 to S4 in the image. How can I can determine the reaction rate in mol/m2.s with the given Arrhenius parameters?
Cs' and Cs-H are site species on the surface.
Cs' is just an abbreviation for carbon site and unoccupied by hydrogen.
Cs-H is hydrogen occupied carbon site.
For ease of calculation assume an arbitrary site concentration of site species in mol/m2 and the gas species in mol/m3.

Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of at least essential parts.
// Optionally, here are formatting guides for texts and formulas/equations/expressions.

@Poutnik well it is neither of the above rest assured. This part of a larger project and as a mechanical engineer my chemical background is very limited.

@Poutnik I just asked a question nothing weird about it. Do I really need to explain the background of why I need to ask? Feel free to remove the question if you want.

It is OK for me. Knowledge of background and purpose generally helps to formulate optimal answers, as what is needed sometimes differs from what is asked.

Well I'm modelling the thermal decomposition of methane in a software called ANSYS Fluent. I modelled the homogeneous decomposition using a 37 step process but struggling with the surface reactions. The software doesn't have capability of performing surface reactions efficiently so I need to code those in and to do that I need some basic understanding about the mechanism which is represented in S1 to S4.

S1 - S4 occur due to autocatalysis from the nucleated carbon a product of the reaction itself.

Just technical note: Elaboration of the question and adding details is better to add to the question itself than to post as comments, that have temporal nature. Comments may be moved to site chat rooms or deleted if not relevant anymore. (I also deleted few my obsokete comments).

@Poutnik I didn't get the time to edit my post due to time difference. If I edit it now, would the question open again?

After editing, you have to ask for reopening. If mods or enough of users with close/reopen voting privilege consider it worthy, it will be reopened.

[OP] For ease of calculation assume an arbitrary site concentration of site species in $\pu{mol/m2}$ and the gas species in $\pu{mol/m3}$.

Although this may seem OK, in the theory of heterogeneous-catalyzed chemical reactions, that unit for concentration is not frequently employed. Typically those magnitudes are measured in terms of fractions $\theta_j$. I you have ever seen the BET's derivation or Langmuir's theory, the balances of sites are treated in fractions, not concentrations over a surface.

[OP] How can I can determine the reaction rate in $\pu{mol/m-2s-1}$ with the given Arrhenius parameters?

This is an irritating fact when publishers don't specify in a precise manner the volume or surface in question, which the kinetic rate constants are referred too. This happens too frequently. I will give you a bunch of definitions of rate of reactions:

Rate of reaction based on reacting fluid volume $V$
$$ r_i = \frac{1}{V} \frac{\mathrm{d}N_i}{\mathrm{d}t} \tag{1} $$
Rate of reaction based on the mass of solid $W$, really used in fluid-solid systems, when we have a mass of catalyst $W$
$$ r_i' = \frac{1}{W} \frac{\mathrm{d}N_i}{\mathrm{d}t} \tag{2} $$
Rate of reaction based on the interface of a gas-solid system $S$
$$ r_i'' = \frac{1}{S} \frac{\mathrm{d}N_i}{\mathrm{d}t} \tag{3} $$
Rate of reaction based on the volume of solid $V_s$ in gas-solid reacting systems
$$ r_i''' = \frac{1}{V_s} \frac{\mathrm{d}N_i}{\mathrm{d}t} \tag{4} $$
Rate of reaction based on the volume of the reactor $V_r$, not to be confused to the volume of the fluid phase. This one is rarely used
$$ r_i'''' = \frac{1}{V_r} \frac{\mathrm{d}N_i}{\mathrm{d}t} \tag{5} $$

In my opinion, they gave you the rate in the form of Eq. (4), but you want it in terms of Eq. (3), thus we need two pieces of information:

The surface area of the catalyst per unit mass of catalyst $S_\mathrm{a}$. This can be obtained experimentally by the $\ce{N2}$ test.
The mass density of the catalyst $\rho_\mathrm{c}$, which if you know its nature, I am sure you can find it everywhere.

Once you write every rate of reaction, where the concentrations are divided by the standard concentration, you will end with the rate law in the fashion $ r = k \cdot [...] $, where $k$ has units of $r$. The relationship we need finally is
$$ k''' = (S_\mathrm{a} \rho_\mathrm{c})k'' \tag{6} $$
$$ \pu{\frac{\pu{mol}}{\pu{m3 cat· s}}} =
   \left[{\frac{\pu{m2 cat}}{\pu{kg cat}}}\right]
   \left[{\frac{\pu{kg cat}}{\pu{m3 cat}}}\right]
   \left[\pu{\frac{\pu{mol}}{\pu{m2 cat · s}}}\right] \tag{7}$$

your answer is the most helpful I got from everyone I asked so far. I would have liked to discuss more but the question was closed during the night and I didn't have time to edit it. Is it possible to take our discussion to a previous post which you commented on that includes the same question? I will edit that one instead with the additional info and code. Your help will greatly appreciated.

@Rayana Hello Rayana. Yes, I remember your post, I have been so busy lately... Of course I'd help you. Maybe you can open a chat room?

Thanks for your time. At the moment I'm trying to edit the last post to try include as much information as possible with my findings. I didn't think not including them would be an issue because the data is just too much. I will try my best to consolidate. It seems you were able to capture what I was looking for with limited info which quite impressive. I think you might have the answer once all the info is provided to you. I don't have the 'create a new room' option in the chat page. Not sure why. Will try again once I finish editing the post.

@Rayana I am flattered. Ping me when you are done.

I have edited with old post and added additional details. Looking forward to discuss.

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text:
What does it mean Keq=1?
I’m struggling with a chemistry question related to equilibrium in a reaction. I don’t understand the meaning of $K_\text{eq} = 1$. After some research I found that when $K_\text{eq}$ is 1 it means that the concentration of products and reactants is the same. But this is not true! If my reaction is of the type $\ce{A <=> 2B}$ only if their concentration is $\pu{1 M}$ this is correct, but it is only a special case and not the rule. So what does really mean $K_\text{eq} = 1$? I know in this case neither products nor reactants are favourites, but why? Also the fact that the standard free Gibbs energy is 0 is not clear to me. Maybe it means all the energy it is used for the reaction itself and no energy is left to be converted into work?
If possible I’ll appreciate a numerical example.

Keq=x.y/z=1 for  z=xy, no need for x=y=z=1

The zero standard reaction Gibbs energy at Keq=1 means reactants and products have the same standard Gibbs energy.

@Poutnik so if z = xy products and reactants are not the same quantity. I still don’t understand the physical meaning of Keq = 1.

Is Keq <> 1 OK for you?

@Poutnik the only thing I understood is that if Keq > 1 than products are favorited over the reactans  because the numerator is greater than the denominator and viceversa if Keq < 1.  The definition is clear to me, but I would like to go a little deeper in order to understand in whihch way this fact is related to the quantities of pruducts and reactans.

How is K = x^2/y = 1 related to values x and y? Both x>y and x<y is possible.

@Poutnik I know that, but if x and y are molar concentrations and the meaning of K = 1 is that at the equilibrium x = y, this means that x = y = 1M at the equilibrium, but this is not always true. As yous said x could be greater or smaller than y and K still be equal 1 if x = y^1/2. So K = 1 doesn't mean that x = y at the equilibrium, so what does it mean ?

It means the numerator and the denominator are equal  It also means (for elementary reactions) that the ratio of the kinetic constants of the forward and backward reactions is equal to 1.

@Poutnik thank you for the explanation. My question arose because after a web search I found some sites state that when Keq = 1 the concentration of products is equal to that of reactans. So I can conclude that is not actually true that meaning of Keq.

@Andrea  To familiarize with the syntax of mhchem suitable on chemistry.se for the body of questions, answers, and comments, see here.  Because it is somewhat special not every web browser processes equally well, do not use mhchem it in the title of questions.

In these days I’ve thought a lot about this question, also thanks your comments and answers. I do some conclusion (hoping to say right things).

Standard free Gibbs energy is a measure of the energy available to be converted into work (correct for reversible reactions). 2) standard Gibbs energy is 0 when the equilibrium is reached. 3) K = 1 doesn’t have a special meaning, when it happens, the free Gibbs energy is 0 starting from reactants and finishing to products (no variation) and remains 0 (obviously) at the equilibrium reached.

After some research I found that when Keq is 1 it means that the concentration of products and reactants is the same. [...] If my reaction is of the type $\ce{A <=> 2B}$ only if their concentration is 1M this is correct, but it is only a special case and not the rule.

IUPAC's Green Book discerns two types of equilibria constants $K$.  Either the one which fits $K = - \exp(-\Delta_rG/-RT)$, or the ones which are derived from a basis of pressure, concentration, or molality.  For the later $K$ is a quotient of e.g., corresponding products of concentrations to the power of the corresponding stoichiometry factors.
Let's assume your reaction $\ce{A <=> 2B}$ in diluted solution is balanced, then the constant is defined as
$$K_c = \frac{[\ce{B}]^2} {[\ce{A}]}$$
In this case, at chemical equilibrium, rearrangement of the mathematical equation leads to the general forms of
$$[\ce{B}] = \sqrt{K_c \cdot [\ce{A}]}$$
and
$$[\ce{A}] = \frac{[\ce{B}]^2} {K_c}$$
A detail frequently forgotten by beginners not familiar (yet) with dimensional analysis: the unit of $K$.  If $K$ was derived from $\Delta_rG$, the constant is unit less.  If derived from pressure, concentration, or molality, the sum of the stoichiometric coefficients has to be taken into account to yield a constant in unit of either $\pu{Pa}^{\sum \nu_\text{B}}$ (for pressure), $\pu{(mol m^{-3})}^{\sum \nu_\text{B}}$ (for molar concentration), or $\pu{(mol kg^{-1})}^{\sum \nu_\text{B}}$ (for molality).*
Only for a reaction $\ce{A <=> B}$ the units cancel out each other.  For your above reaction of $\ce{A <=> 2B}$, however
$$K = \frac{(\pu{1 mol/L})^2} {\pu{1 mol/L}} = %
%
\frac{\pu{1^2 mol^2/L^2}} {\pu{1 mol/L}} = %
%
\pu{1 mol/L} $$
* Cohen, E. R. et al. (eds.) Quantities, Units and Symbols in Physical Chemistry, IUPAC Green Book, 3rd edition / 2007; section 2.11 chemical thermodynamics, p. 58.

If the hypothetical reaction were balanced $\ce{3A + 2B <=> 5C + 4D}$
$$K_c = \frac {[\ce{C}]^5 \cdot [\ce{D}]^4} {[\ce{A}]^3 \cdot [\ce{B}]^2}$$
or more general, $\ce{\alpha A + \beta B + \ldots <=> \sigma S + \tau T}$,
$$K_c = %
  %
  \frac {[\ce{S}]^\sigma \cdot [\ce{T}]^\tau \ldots} {%
         [\ce{A}]^\alpha \cdot [\ce{B}]^\beta \ldots}$$

$K_c$ (derived from concentrations in $\pu{mol/L}$), and $K_p$ (derived from partial pressures of gaseous equilibria) are approximations for diluted reactants.  The higher the concentration of molecules, however, the more they can hinder each other during a reaction; at some point, thermodynamic activities are the better descriptions of the reagents.

Note, capital italic $K$ is about the thermodynamic equilibrium constant.  Lower case italic $k$ typically is used to describe the kinetic rates of reactions.  This is convention of nomenclature compiled in IUPAC's Green Book (see resources page).  At high school level, the freely available 4 page summary (link to pdf) likely provides enough information for you.

Note that true thermodynamic equilibrium constants, fitting in $\Delta_r G^{\circ}=-RT\ln{K}$, are unitless, based on unitless quantities like activity or fugacity. Constants expressed for mass or amount concentrations or partial pressures do not fit the above equation and their values are dependent on used quantities and their units.

@Poutnik I agree and it was edited.

The equilibrium constant $ K=1$ has no more significance than that when $K=10$ or $K=0.1$ etc. You will know that the equilibrium shifts as the temperature changes, so consider the gas phase equilibrium
$$\ce{PCl_5(g) = PCl_3(g) + Cl_2(g) }$$
which has $K=0.230$ at $472$ K and at a slightly higher temperature $524$ K, $K=2.69$ and clearly $K=1$ is to be found somewhere in between these two temperatures.  The Gibbs free energies at the two temperatures are $\Delta G^\text{o}=+5.77$ kJ/mol and $-4.31$ kJ/mol. (In 'thermodynamic speak' this means going from non-spontaneous to spontaneous reaction as the temperature increases and although it may seem so it is actually not very revealing).
In terms of the equation expressed as partial pressures (as appropriate for a gas phase reaction)
$$\displaystyle K=\frac{p_{PCl_3}p_{PCl_2} }{p_{PCl_5}}\equiv\frac{p_{PCl_3}^2}{p_{PCl_5}}$$
as $K$ is a constant at a given temperature we only ever know the ratio of the species. In this example $p_{PCl_5}K =p_{Pcl_3}^2$ or  $p_{PCl_5} =p_{Pcl_3}^2$ when $K=1$.
It is necessary to point out at this point that using partial pressure in the gas phase or concentrations in solution is not technically correct, although very practical. We should use activities or equivalently fugacities which are dimensionless. To get around this we must divide each partial pressure or concentration by 1 unit, say 1 atmosphere or 1 mol/dm$^3$.
Using thermodynamics we know that the change in standard reaction free energy is related to the equilibrium constant as $\Delta G^\text{o}=-RT\ln(K)$, which trivially means that $K$ has to be dimensionless (see above), and when $K=1$ then $\Delta G^\text{o}=0$ which using $\Delta G = \Delta H -T\Delta S$ means that only at this temperature does  $\Delta H^\text{o}=T\Delta S^\text{o}$, in other words an exact balance between heat change  and entropy change. However, although it may seem that it is, this has no particular significance, other that it is the tipping point one way or the other in the reaction.

Andrea, you seem to have asked a deep question without a simple answer; I will Try. The problem is in carefully figuring activities.  If there is strict stoichiometry and the coefficients = 1, the activities are equal. In this case there is no change in the number of moles in the reaction. Should a component be augmented activities will change and no longer be equal. A reaction that could fit this description is a simple esterification: ester + water = acid + alcohol.
The simplest example I can think of is an ice-water equilibrium. Pure ice and pure water each have mole fractions of 1.. At equilibrium the Keq = 1/1 = 1. This means that as long as both water and ice are present the temperature is constant [Gibbs phase rule]. Lowering the water activity by adding a solute lowers the melting point [ice melts to remove heat] until the activity of the ice equals that of the water. Keq is still 1 but the activities are lower. [An interesting thought! Should a solute raise the chemical potential of liquid water would the freezing point be increased?] A change in pressure is similar, the activities change, temperature adjusts until the activities are equal.
I am trying to give a rigorous explanation for a more complex system but am having trouble expressing activities in a manner that makes sense, so I am posting this partial answer to put this question back into the queue and give it its due.

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text:
Why does Cr have higher melting point in the 4 period among the transition metal?
There is a lot of answer stating that the low enthalpy of atomisation in Cr is due to the reason of partially filled d orbital, thus having extra stability. This makes the orbital more attracted toward the nucleus and forming a weaker metallic bonding.
But if we consider this as a reason it does not correctly explains the higher melting point among the 4-period transition metal.
Could you explain what could be possibly wrong with either of this explanation?

The answer to a similar question is not sufficient to fully understand it. At least it gives a useful picture where it is noticeable that the melting point depends on the nuclear charge, because most of the elements in the sixth period, where the potential is less shielded by $d$ and $f$ orbitals, have the highest melting point. I will explain the electronic part:
$\ce{Cr}$ forms a solid-state metal, its melting point is related to solid state properties and not directly to the atomic configuration. These properties depend on the structure and the electronic configuration. The solution of the Schrödinger equation in this case does not keep the shape of the atomic orbitals, but delocalized wave function tending to plane waves, some approximations keeps atomic orbitals to describe metals, but they aren't very accurate. Because the solid combines a large number of atoms, the molecular levels have a tiny space giving a continuous band structure. The width of the band structure describes a rate of transfer $t$ of an electron from one site to another; if this rate is high enough, the repulsion can be neglected. $t$ is related to the bond strength, and the bond strength is related to the melting point.
The previous arguments are difficult to put into an analytical form, the most accurate analytical band structure is that of a homogeneous electron gas where the dispersion is quadratic $E \propto k^2$, as the electrons are interacting in a metal, this dispersion relation is not accurate in this case, moreover the $d$ orbitals are localized and do not overlap much (the repulsion is very important as the transfer rate $t$ is smaller). Instead of trying to solve a complex many-body equation to calculate the binding energy, we can use an approximate density of states (DOS). The $d$ orbitals are farther away from the nucleus and have the largest number of electrons in the case of $\ce{Cr}$, so they have the most influence on the bond energy. Since $t$ is small, the bandwidth $W$ is also small and the density of states (DOS) is narrow, a rectangular approximation is often used:

The figure above gives the density of states, where the surface of black part is the number of electrons of the metal filling the bands: $n_d$, the surface of remaining part is the number of empty states at $\bf{T=0}$ K. The energy of a bond is the difference between the sum of the energies of all states from $-W/2$ to the Fermi energy $E_F$ and the atomic energy of the $d$ orbital considered as degenerate.
$$E_{bond} = \int_{-w/2}^{E_F} E \cdot DOS(E) dE - E_{d,at}$$
$E_{bond}$ is similar to the enthalpy of atomisation or a cohesive energy. As $d$ orbitals have $10$ electrons :
$n_e= \int_{-w/2}^{W/2} E \cdot DOS(E) dE =  DOS(E)\cdot W = 10$, $DOS(E) = \frac{10}{W}$ and $n_d=\int_{-w/2}^{E_f} DOS(E) dE = \frac{10}{W} (E_F+W/2)=\frac{10}{W} (E_F+5)$. Therefore $E_{bond}$ gives :
$$E_{bond} = \frac{10}{W} \int_{-w/2}^{E_F} E  dE -  E_{d,at} =\frac{W^2}{20} \left[ (n_d^2-5)^2 -25\right] -  E_{d,at}  $$
The bond energy starts at zero when the filling is zero and decreases to the minimum when the $d$ "band" is half filled and increases to the value when the band is completely filled. The lowest value is a "band" half filled (red dot), it may be surprising but as shown in the figure below ($W/20=E_{at}=1$) that neither chromium, manganese nor vanadium sits at the minimum,  it is a choice due to the charge distribution in a solid compared to isolated atoms which I have tried to explain here.  In fact, the partial charge in a specific orbital of an atom forming a solid is rarely equal to the free atomic charge of this orbital, it is due to the fact that $s$ and $d$ orbitals are involved in the metallic bonds, a partial charge from the $s$ orbital (closer to the nucleus) is transferred to the $d$ band, resulting more charge in the $d$ band and then less bond energy for manganese (who has half-filled $d$ orbitals).  The process occurs with chromium with half-filled atomic orbitals, but the charge in the $d$ band is more than charge of the free atom, the bond energy id higher than the minimal value.  The closer is vanadium, which tends to the minimum and has less electronic repulsion, therefore has the highest melting point, then comes chromium, and so on. This approximation seems crude at first, but in the end it is very efficient.

But if we consider this as a reason it does not correctly explains the higher melting point among the 4-period transition metal.
Could you explain what could be possibly wrong with either of this explanation?

Vanadium has the highest melting point, when I studied this metal in the past, the charge in the $d$ band was not too far from the atomic charge, there are certainly other effects to include and question is about chromium. In the case of $\ce{Cr}$, the considerations above are one approximation to explain the anomaly with manganese, the charge $n_d$ of $\ce{Ce}$ is also similar to the atomic charge of $d$ orbitals, but in the case of $\ce{Mn}$, $n_d > 5 $ and the melting point drops.

Yes, you are right. The melting point of chromium (1907 C) is slightly less than vanadium (1910 C). Sorry for the confusion because the book that I was referring to give this graph. There is still a bit of confusion as some site tells that the melting point of Cr is 1890 C.


text:
Production SnO2 nanoparticles through chemical co-precipitation method: chemical equation of the reaction
In this paper, the production of $\ce{SnO2}$ nanoparticles through the chemical co-precipiation method is described using $\ce{SnCl2·2H2O}$ as a precursor. According to their experimental section,

$\ce{SnO2}$ NPs were prepared by chemical co-precipitation method. 0.27 M $\ce{SnCl2⋅2H2O}$ was added to double destilled (DD) water and ethanol (2.2:1 v/v) in a 1000 mL beaker under continuous magnetic stirring for 60 min in order to obtain a homogeneous solution. 50 mL of $\ce{HCl}$ was added drop wise in order to reduce the precursor and the solution was kept on stirring for another 30 min to allow proper mixing resulting in a clear transparent solution. Later aqueous $\ce{NH3}$ was added dropwise till the pH of the solution became 8.5 and a milky white solution was obtained. The solution was kept on continuous stirring for another 3 h. After this the stirring was stopped and the beaker was kept undisturbed overnight in order to facilitate the particles to settle down completely.
The solution was then washed with DD water and ethanol multiple times so as to remove the residual chlorine content using 125 mm Whatman filter paper (Cat no. 1001 125). The resulting product was covered and left for drying overnight. The dried powder was then divided into three parts and annealed at 400 °C, 500 °C, and 600 °C named as Sn4, Sn5, and Sn6 respectively.

However, the paper focuses mainly in the characterization of the syntetized nanoparticles, and it doesn't describe in further detail the chemical reactions of the process. What would the role of the added $\ce{HCl}$ be? Why should the $\ce{NH3}$ be added, and why should it be added until an 8.5 pH is obtained? What would the chemical reaction of the process be? I would like to know the reason behind these steps in order to fully understand the process, unfortunately the paper doesn't cover them.

Some concepts are wrong in the paper. HCl is not there to reduce anything, it is there to prevent hydrolysis of tin chloride. Addition of ammonia is for controlled hydrolysis of the tin salt. The pH 8.5 must be empirical. They must have done a lot of hit and trials. Did you search other papers?

Tin(II) chloride $\ce{SnCl2}$ is not perfectly soluble in pure water. It is only soluble in aqueous $\ce{HCl}$ solutions. In pure water $\ce{SnCl2}$ is partially hydrolyzed according to the equilibrium $$\ce{SnCl2 + 2 H2O <=>  Sn(OH)2  + 2 HCl}$$ If enough $\ce{HCl}$ is present in solution, $\ce{SnCl2}$ gets dissolved without hydrolysis, and produces $\ce{Sn^{2+}}$ ions. But in the presence of ammonia, $\ce{HCl}$ is destroyed and the $\ce{Sn^{2+}}$ ion is precipitated as $\ce{Sn(OH)2}$ according to the irreversible reaction : $$\ce{Sn^{2+} + 2 NH3 + 2 H2O -> Sn(OH)2(s) + 2 NH4^+}$$ The trouble is that $\ce{Sn(OH)2}$ is easily oxidized by atmospheric oxygen. So in contact with air, $\ce{Sn(OH)2}$ is quickly oxidized into $\ce{SnO2}$ according to : $$\ce{2 Sn(OH)2 + O2 -> 2 SnO2 + 2 H2O}$$ This is why tin(II) chloride is easily transformed and oxidized into insoluble $\ce{SnO2}$.

Very helpful! One last thing, the paper specifies to use tin(II) chloride dihydrate, would it be irrelevant to use non-hydrated tin(II) chloride, as you did in the answer?

@User. No ! All I said about $\ce{SnCl2}$ can be also said about its hydrate $\ce{SnCl2·2H2O}$ or any other hydrate. I used the formula $\ce{SnCl2}$ in my answer because it is simpler than $\ce{SnCl2·2H2O}$


text:
Selective reaction of a terminal alkene from a mixture of terminal and internal alkenes
Isomerisation reaction of a internal alkene results in a mixture of internal and terminal alkene. I want to hydroxylate the terminal alkene alone. Rather than separating it from the mixture, is there any way to directly get the desired yield

So you want to separate or not? (Compare title with the rest...) There are selective catalysts and afair they prefer terminal alkenes, but I'm not sure if for this reaction too.

I mean, you could separate the mixture by vacuum distillation with precise temperature control and long Vigreaux columns.

The reagent you want is disiamyl borane, which is selective for terminal alkenes over internal alkenes as shown by this Organic Syntheses procedure then work up with basic hydrogen peroxide to give primary alcohol.


text:
How to determine uncompensated or solution resistance of a 3 electrode setup for IR correction?
To correct for ohmic losses from the potential drop across the electrolyte due to the distance between the electrodes, electrochemical impedance spectroscopy (EIS) is typically employed to measure this electrolyte resistance (R_s).
Currently I think that the Nyquist plot (where each point has a frequency and a real and imaginary impedance) needs to be fitted to an equivalent Randles circuit, which then allows determination of the R_s. The one in particular should be
What I'm hoping to achieve is something like this, as well as to finding the solution resistance to calculate iR drop:

How could this be done? Here is an example of the data I have:

freq in Hz
Re(Z) in Ohm
-Im(Z) in Ohm

1.0000371E+005
1.2838266E+000
-4.7236100E-001

6.7491828E+004
1.2575362E+000
-2.9951397E-001

4.5557453E+004
1.2489935E+000
-1.9674678E-001

3.0748963E+004
1.2461497E+000
-1.3021792E-001

2.0756010E+004
1.2463084E+000
-8.5121296E-002

1.4010540E+004
1.2458339E+000
-5.3675517E-002

9.4523906E+003
1.2463924E+000
-3.2538280E-002

6.3779312E+003
1.2452079E+000
-2.0627080E-002

4.3063779E+003
1.2464187E+000
-7.8994250E-003

2.9059297E+003
1.2512513E+000
-7.8682322E-004

1.9631409E+003
1.2515490E+000
5.4816552E-003

1.3237295E+003
1.2460785E+000
5.2928450E-003

8.9414160E+002
1.2562760E+000
1.8789139E-002

6.0287787E+002
1.2608212E+000
2.7019683E-002

4.0721527E+002
1.2652799E+000
3.7225980E-002

2.7473624E+002
1.2735685E+000
4.8571289E-002

1.8553433E+002
1.2811457E+000
6.3650906E-002

1.2520035E+002
1.2927077E+000
8.0610186E-002

8.4486298E+001
1.3058180E+000
9.9876016E-002

5.7063412E+001
1.3209573E+000
1.3237002E-001

3.8485222E+001
1.3433586E+000
1.6853021E-001

2.5972410E+001
1.3701465E+000
2.2286178E-001

1.7516815E+001
1.4067172E+000
2.9568654E-001

1.1837119E+001
1.4666675E+000
3.9322603E-001

7.9882402E+000
1.5431337E+000
5.2406406E-001

5.3879313E+000
1.6745517E+000
6.8878186E-001

3.6371040E+000
1.8626976E+000
8.5898042E-001

2.4567609E+000
2.1712506E+000
1.0030348E+000

1.6587046E+000
2.5552261E+000
1.0842789E+000

1.1199110E+000
2.9400694E+000
1.0472580E+000

7.5526887E-001
3.2742524E+000
8.9296573E-001

5.0942224E-001
3.5140378E+000
7.0039696E-001

3.4379947E-001
3.6070650E+000
5.1860118E-001

2.3239036E-001
3.6902442E+000
3.5689598E-001

1.5689009E-001
3.7351005E+000
2.2994351E-001

1.0589487E-001
3.7651119E+000
1.3973090E-001

7.1417451E-002
3.7816873E+000
3.9920006E-002

4.8211023E-002
3.7677062E+000
-8.1531301E-002

3.2556150E-002
3.7300577E+000
-1.7607281E-001

2.1934874E-002
3.6879458E+000
-1.6583309E-001

1.4821894E-002
3.6872365E+000
-1.9536464E-001

1.0009173E-002
3.6411600E+000
-1.3885058E-001

I just looked up “Randles circuit” in wikipedia and the last sentence in the short article was this: “To obtain the Randles circuit parameters, the fitting of the model to the experimental data should be performed using complex nonlinear least-squares procedures available in numerous EIS data fitting computer programs.” So, do you have such a program? If not, then getting such a program would be a good idea even if you program the fit yourself: you would need to check your program’s performance against a program that has been used and debugged by other people. Good luck!

If you cannot fit a randles circuit because a lack of analytical tools, you can always look at the bode plot in the high frequency range and see what the impedance is. This should be the solution resistance. Also reading from the Nyquist plot you can imagine where the semi circle crosses the x-axis. Typically this figure is 'good enough'.

I have reluctantly appended some code written in Python as a Jupyter notebook ( Jupyter notebooks was used via Anaconda and is free to use), reluctantly because the code not v. elegant but should be usable to start with, after a little effort. The code should also run using Visual Studio Code, which is also free to use.
The function to fit is in 'Model'. I used $R_s+R_c/(1+iR_cC\omega)$ and as you can see the fit is poor suggesting that the data fits a more complicated model. You should be able to change the equation used to a more suitable form.
There is computed test data. Towards the bottom of the code there is a True or False switch to change between reading real data and using test data. The real data file should be comma separated in three columns in the order in your post and have no header or other characters etc. You will need to generate your own data file and change the name in this code if needs be. This file should be in the same folder as the code.
# import all python add-ons etc that will be needed later on. Python 3 is used
# the follwing line is only used in Jupyter notebooks
%matplotlib inline

import numpy as np                      # import fast numerical library
import matplotlib.pyplot as plt         # import plotting library
from scipy.optimize import leastsq      # import non-lin least squares routine
plt.rcParams.update({'font.size': 14})  # set font size for plots

def read_data():
    filename ='nyquist-data.txt'  #  The numbers are in 3 columns comma separated, freq, real, imag
                                  #  No other text and no comma at end of lines
    w = []
    rdata = []                    # real floats
    idata = []                    # imag floats
    with open(filename) as f:
        for line in f.readlines():
            f_list = [ float(i) for i in line.split(",") ] # separate out numbers comma separated
            w.append(f_list[0])
            rdata.append(f_list[1])
            idata.append(-f_list[2])
    w = np.array(w)                            # make w into array
    data = np.zeros(len(w),dtype = complex)    # make new array of complex numbers
    data.real = rdata
    data.imag = idata

    a = 5                               # used to limit range of experimental data use [:] for all data
    b = 38
    return w[a:b], data[a:b]            # return data but leave out messy bits either end

#----------------------------
def make_data(Rs,Rc,C):                  #  For testing only. Noise added
    n = 10000                            #  number of data points
    w = np.linspace(0.0,n,n)             # range of frequency, 0 to n, and n points
    rng = np.random.default_rng()
    z = model(w,Rs,Rc,C) + 0.01*(rng.random(n) + 1j*rng.random(n))  # 1j is sqrt(-1)
    return w,z
#-----------------------------

def model(w,Rs,Rc,C):                    # Change equation here as necessary
    Z = Rs + Rc/(1.0 + Rc*C*1j*w )       # 1j is sqrt(-1)
    return Z                             # z is complex array, i.e has real and imag parts
#----------------------------

def resids(params,w,z):
    Rs, Rc, C = params
    diff =  model(w, Rs, Rc, C) - z      # calculates residuals using current set of parameters
    zri = []                             # make new  1D array as real part, imag part, real, imag... etc
    for a, b in zip(diff.real,diff.imag):
        zri.append(a)
        zri.append(b)
    return zri                           # we can only input 1d array to least squares, i.e. not complex array
#----------------------------

def do_calc(params):
    vals = leastsq(resids, params, (w,z) )  # w, z is the data, resids is routine to calc resduals
    Rs = vals[0][0]                         # get fitted parameters from vals
    Rc = vals[0][1]
    C  = vals[0][2]
    sum_resids = sum( abs(model(w, Rs, Rc, C) - z)**2 )
    print('{:s} {:6.3f} {:6.3f} {:6.3f} {:s} {:6.3f}'.format('values Rs, Rc, C are',Rs, Rc, C ,'sum^2 resids', sum_resids) )
    return Rs,Rc,C
#-----------------------------

test_calc = False # True to use test data   # False to read in your data
if test_calc == True :
    Rs, Rc, C = 2.5, 1.5, 0.01   # used to make test data
    w, z = make_data(Rs,Rc,C)    # w is frequency, z complex array real & imag values
else:
    w, z = read_data()

params = [0.5, 2.0, 0.01]        # Rs, Rc, C initial values or fitting

fitRs, fitRc, fitC = do_calc(params)

zre = model(w,fitRs,fitRc,fitC).real  # make data from fitted params to plot on Nyquist plot
zim = model(w,fitRs,fitRc,fitC).imag

plt.plot(zre, -zim, color='red',linewidth=5,alpha=0.3)  # wide red-ish line

plt.scatter( z.real, -z.imag ,s = 5, color='blue'  )    # data points
plt.ylabel('-Z(im)')
plt.xlabel('Z(re)')
plt.tight_layout()
plt.show()

The fit to the data is shown below, the params are  Rs, Rc, C are  $1.311,  2.408,  0.238\; \sum^2\;=  0.232$. The fit is the thick red line, the data points are blue dots. The fit is made to ignore the first few and last few data points but is still a poor fit to the data. The results are not particularly sensitive to the initial guesses.


text:
Are all salts completely dissociated in solution when put in smaller amounts than their molar solubility?
From what I understand, the solubility is given in terms of molar solubility (or $\pu{K_{sp}}$), from which it can be easily calculated). Indeed, the saturation point represents the maximum amount of a slightly soluble substance in an aqueous solution that solubilizes (dissociates).
Therefore, in extremely small amounts even very sparingly soluble salts with extremely small molar solubilities (and $\pu{K_{sp}'}$) can be completely ionized. For instance, $\ce{HgS}$ is one of the least soluble inorganic substances with a molar solubility of about $\ce{10^{-27}}$. Now, if I put up to $\ce{10^{-27}}$ moles of HgS in 1L of water, will it be completely dissolved into $\ce{Hg^{2+}}$ and $\ce{S^{2-}}$?

HgCl2 is reportedly in large extent covalent. Some salts forms ionic pairs, like MgSO4 in sea water. Other may form charged or neutral complexes.

This is an equilibrium process and so to calculate the extent of reaction you can use the usual methods starting with different initial amounts.

A solubility of $\ce{10^{-27}}$ M corresponds to $1$ atom $\ce{Hg}$ and $1$ atom $\ce{S}$ in $10^4$ liters, or in $10$ metric tons. This is practically zero atom per litre. This amount is not detectable.

You cannot put less than $\pu{\frac{1}{N_\mathrm{A}} mol}$ of anything anywhere, as it would be less than 1 object.

Yes. Ksp describes a saturated solution. If from dissolution of a salt stoichiometric quantities of cation & anion will be dissolved ("in solution')

We should not expect salts with multiply charged ions to be completely dissociated in aqueous solution except, perhaps, at very low concentrations. The comments refer to magnesium sulfate, a technical reference for which is Sebastiani et al.[1]. Note that in this case the paired ions are not in direct contact but separated by solvent molecules, so it's an "outer-sphere" association.
Reference
Federico Sebastiani, Ana Vila Verde, Matthias Heyden, Gerhard Schwaab, and  Martina Havenith (2020). "Cooperativity and ion pairing in magnesium sulfate aqueous solutions from the dilute regime to the solubility limit". Phys. Chem. Chem. Phys., 22, 12140-12153.


text:
When creating a separation method in gas chromatography, calibration with compounds mixture or for each separate component?
When creating a separation method in gas chromatography, is there any difference between the calibration with the compounds mixture and the calibration for each separate component? I have 5 components.

Note that there are often mixed standards with 10-20 components.

In gas chromatography, the response factor depends only on the concentration of the component in the solution to be analyzed. It does not depend on the presence of internal standards.


text:
Mechanism of elimination of trans-dibromocyclooctane
I recently did this reaction according to a protocol:

And although the product formation was monitored via NMR and IR techniques, I am unable to comprehend how the product 3 was formed. It should be an elimination reaction, however, the bromines are located in a trans fashion, so the usual reaction path for an E2 elimination (antiperiplanar hydrogen) is not given. How can this elimination happen then?

Syn eliminations do occur as in endo- or exo-2-bromo-[2.2.1]-bicycloheptane.


text:
Paramagnetic liquids at room temperature
So as most inorganic liquids are diamagnetic at room temperature (like water), what are some applications to paramagnetic liquids at room temperature?

One way to generate a paramagnetic liquid is to use an active metal to generate solvated electrons. Burrows et al.[1] describe a solution of lithium in ethylenediamine plus tetrahydrofuran, which can be used in Birch reductions at ambient conditions without handling liquid ammonia.
Reference
JAMES BURROWS, SHOGO KAMO, AND KAZUNORI KOIDE (2021). "Scalable Birch reduction with lithium and ethylenediamine in tetrahydrofuran". Science 374, 741-746. DOI:10.1126/science.abk3099


text:
How do I decide on supra-supra/supra-antara and thermally/photochemically when the educt and product of the sigmatropic rearrangment is given?
Was the given product of this [1,5]-sigmatropic rearrangement reached thermally with orbitals suprafacial-suprafacial or photochemically with orbitals suprafacial-antarafacial?

The same question goes for this Claisen-rearrangement:

I would think both a thermal reaction where orbitals meet suprafacial-suprafacial and a photochemical reaction where orbitals meet suprafacial-antarafacial is possible as neither educts nor products are stereospecific. I am supposed to choose one though, which confuses me. Am I missing something? I hope you can help, as I do not know where else to look anymore.
Thanks in advance!

The first is tautomerisation - why should it have anything to do with sigmatropic?


text:
Heat capacity of (ideal) gases at constant pressure
I've come across this article which says that the constant pressure heat capacity is independent of pressure.
It also gives a mathematical explanation which says that any work of an ideal gas at constant pressure heating is equal to -R for all pressures. How can one rationalize this, because I am having trouble understanding why this needs to be the case.
And also, how can one account for the pressure dependence using a real gas equation?

The work is equal to R for one mole of ideal gas. Remember : one mole ! 2) Real gases do not have an exact equation ...

@Proscionexium If you want to improve the posts, please add proper citations to websites and links. Hiding URLs behind words may look nicer but may only obfuscate the otherwise obvious.

Statistical Thermodynamics
The rationalization of the specific heat capacity at constant volume being only a function of temperature, can be derived from combining quantum mechanics and statistical thermodynamics. I am sure you know the steps:

You solve the time-independent Schrodinger's equation for the $1$-dimensional particle in the box. The result you are interested in, are the the energy eigenvalues $E_n$ that depend on the particular state of the particle, denoted by $n$.
Considering Boltzmann's theory, we calculate the molecular partition function, by summing along all possible states of the quantum system
$$ q = \sum_{n = 1}^\infty \exp(-\beta E_n) \tag{1} $$
before that, you correct the $E_n$, so they are relative to the lowest state, i.e. $n = 1$. This is actually the exact value. However, we approximate the infinite sum to an integral in order to get an analytic expression.
You assume that the translational movements along the $x$, $y$, and $z$ directions are independent. This allows you to write down the molecular partition function as a product of three individual ones, i.e., $q = q_xq_yq_z$.
That's all. Once you have the molecular partition function, all the thermodynamic properties are obtained with calculus. In this case our task is to do
$$ C_\mathrm{V} = 2k_BT\left(\frac{\partial \ln q}{\partial T}\right)_V
   + k_BT^2 \left(\frac{\partial^2 \ln q}{\partial T^2}\right)_V \tag{2} $$
Once you compute Eq. (2), you will get that it is independent of pressure and  temperature. If we take into accounte other modes of movement, this will not hold, and $T$ appears again. Is this a realistic result? Well, we need to test the theory. We search for some substances, and find that in the case of $\ce{Ar}$, the heat capacity is independent of pressure and also on temperature. What is our conclusion? Reality may be complex, but for $\ce{Ar}$, this is all the movement that we need to account for explaining the specific heat capacity at constant volume (if it is a gas). For most gases, this doesn't work, unfortunately.

Classical Thermodynamics

[OP] I've come across this article which says that the constant pressure heat capacity is independent of pressure.

First you need to accept the ideal gas law, that comes from early experimentation by well-known chemists
$$ pV = RT \tag{3} $$
$V$ here is the molar volume.
In addition you need to accept some consequences that derive from Joule's experiments, which he carried out in his basement in Manchester. If you are interested in, you can search them, they were very basic but accurate. One of these consequences is the following:

The internal energy of an ideal gas is only a function of temperature, i.e., $ U = U(T)$.

The main consequence he found is that heat is another form of energy, such as work, and it can change the internal energy of a substance (the $1$st law in fact).
Thereby, by definition of the heat capacity at constant volume
$$ C_\mathrm{V}^\mathrm{ig} := \left[\frac{\partial U(T)}{\partial T}\right]_V =
   \mathrm{function(T)} \tag{4} $$
and that is the proof, no more, no less.
The specific heat capacity at constant pressure also derives from the definition of enthalpy
\begin{align}
  H &:= U(T) + pV \quad \text{(use Eq. 3)} \\
  H &= U(T) + RT \therefore H = H(T) \\
\end{align}
so the enthalpy is just a function of temperature. Again, by the definition of specific heat capacity at constant pressure
$$ C_\mathrm{p}^\mathrm{ig} := \left[\frac{\partial H(T)}{\partial T}\right]_p =
   \mathrm{function(T)} \tag{5} $$
And that's it. Again, no more, no less.

[OP] And also, how can one account for the pressure dependence using a real gas equation?

Unfortunately, EoS's will not tell you the heat capacity of any substance. The values must be obtained by experimentation. The answer should end here. However, something which is not said often, is that it offers information of the residual heat capacity. A residual thermodynamic property is the difference between the real value and the value if the substance behaved as an ideal gas, ie. $M^R(p,T) = M(p,T) - M^\mathrm{ig}(p,T)$. If you are interested in this, you can keep reading.
Similar to statistical thermodynamics, once we have an EoS, the rest is just math. I will exemplify to you with the very basic virial equation in terms of pressure
$$ Z = 1 + \frac{Bp}{RT} \tag{6} $$
and by the fundamental relationship $\mathrm{d}G= V\mathrm{d}p - S\mathrm{d}T$, or in residual form, $\mathrm{d}G^R= V^R\mathrm{d}p - S^R\mathrm{d}T$. At constant temperature we have
\begin{align}
 \require{cancel}
 \mathrm{d}G^R &= V^R \mathrm{d}p \\
 \mathrm{d}G^R &= (V - V^\mathrm{ig}) \mathrm{d}p \\
 \mathrm{d}G^R &= \left(\frac{ZRT}{p} - \frac{RT}{p}\right) \mathrm{d}p \\
 \mathrm{d}G^R &= \frac{RT}{p} (Z - 1) \mathrm{d}p \quad \text{(use Eq. 6)} \\
 \mathrm{d}G^R &= \frac{\cancel{RT}}{\cancel{p}} \frac{B\cancel{p}}{\cancel{RT}} \mathrm{d}p\\
  \mathrm{d}G^R &= B \mathrm{d}p \\
  \int_{G^R(0,T)}^{G^R(p,T)} \mathrm{d}G^R &= B \int_{0}^p \mathrm{d}p \\
  G^R(p,T) - G^R(0,T) &= B(p - 0)  \quad \text{(divide by $RT$)} \\
  \frac{G^R(p,T)}{RT} &= \frac{Bp}{RT} + \frac{G^R(0,T)}{RT} \rightarrow
  \boxed{\frac{G^R(p,T)}{RT} = \frac{Bp}{RT} + C} \tag{7} \\
\end{align}
That the residual Gibbs energy is zero in the limit of zero pressure, is something that we cannot determine, but it can be shown by experimental data that it is not zero. Therefore, we just leave it as letter $C$, that is only a function of temperature (we evaluated the pressure at “$0$” there). There is a relation between $G^R$ and $H^R$, which I will not prove, let me know in the comments if you  want to know where it comes from
\begin{align}
 \require{cancel}
 \frac{H^R}{RT} &= -T\left[\frac{\partial (G^R/RT)}{\partial T}\right]_p
 \quad \text{(use Eq. 7)} \\
 \frac{H^R}{RT} &= -T\left[\frac{\partial (Bp/RT + C)}{\partial T}\right]_p \quad \text{(constant goes away, great!!!)} \\
 \frac{H^R}{RT} &= -T\left[\frac{\partial (Bp/RT)}{\partial T}\right]_p \\
 \frac{H^R}{RT} &= -\frac{pT}{R}
 \left[\frac{\mathrm{d} (B/T)}{\mathrm{d} T}\right] \\
 \frac{H^R}{\cancel{RT}} &= -\frac{pT}{\cancel{R}}
 \left(\frac{1}{\cancel{T}}\frac{dB}{dT} - \frac{B}{T^\cancel{2}}\right) \\
 H^R &= -pT\left(\frac{dB}{dT} - \frac{B}{T}\right) \rightarrow
 \boxed{H^R(p,T) = p\left(B - T\frac{dB}{dT} \right) = p(B-TB')} \tag{8}
\end{align}
We need one more step. By definition of the residual specific heat capacity at constant pressure
\begin{align}
  C_\mathrm{p}^R &= \left(\frac{\partial H^R}{\partial T}\right)_p \quad
  \text{(use Eq. 8)} \\
  C_\mathrm{p}^R &= \left\{\frac{\partial [p(B-TB')]}{\partial T}\right\}_p \\
  C_\mathrm{p}^R &= p\left[\frac{\mathrm{d}(B-TB')}{\mathrm{d}T}\right] \\
  C_\mathrm{p}^R &= p (\cancel{B'} - \cancel{B'} - TB'') \rightarrow
  \boxed{C_\mathrm{p}^R(p,T) = -pT\frac{\mathrm{d}^2B}{\mathrm{d}T^2}} \tag{9}
\end{align}
Eq. (9) ends our journey. Observations:

If the gas is ideal, $B=0$ and the residual heat capacity at constant pressure is zero, so we are happy.
The residual heat capacity at constant pressure is a linear function of the pressure.
Higher the pressure, higher the deviations from the ideal gas heat capacity. This makes sense, however, we need to make sure that $Z = 1 + Bp/RT$ is still a good representation of the non-ideal gas at high pressures (probably not).
When $B$ changes from concave up to concave down, the heat capacity behaves as an ideal gas. If you watch Eq. (7), when $B=0$ the residual Gibbs energy is zero. This is something really funny that happens to thermodynamic functions: the gas is not ideal, because we have an EoS, but the properties sometimes behave as ideal for special circumstances.
We can make measurements of a gas at low pressures, where the gas approximates by the ideal-gas behaviour. For other pressure, we can apply Eq. (9) to calculate the real heat capacity
$$ C_\mathrm{p}(p,T) = C_\mathrm{p}^\mathrm{ig}(T) -
   pT\frac{\mathrm{d}^2B}{\mathrm{d}T^2} \tag{10} $$
the results will be as good as you can estimate numerically the second derivative of $B$ at a certain $T$, a difficult task...

The general temperature- and pressure dependence of the molar enthalpy of a pure real gas is given by $$dH=C_pdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$In the limit of ideal gas behavior (P --> 0), the second term in parenthesis is zero, and both H and $C_p$ are functions only of T:$$H(T,0)=H^{IG}(T)$$and$$C_p(T,0)=C_P^{IG}(T)=\frac{dH^{IG}(T)}{dT}$$
The enthalpy change of a real gas between two arbitrary states $(T_1,P_1)$ and $(T_1,P_2)$ can be obtained by taking advantage of Hess' Law, which recognizes that enthalpy H is a function of state:
$$\Delta H=H(T_2,P_2)-H(T_1,P_1)=\Delta H_A+\Delta H_B+\Delta H_C$$where $$\Delta H_A=H(T_1,0)-H(T_1,P_1)=-\int_0^{P_1}{\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP}$$$$\Delta H_B=H(T_2,0)-H(T_1,0)=\int_{T_1}^{T_2}{C_P^{IG}(T')dT'}$$$$\Delta H_C=H(T_2,P_2)-H(T_2,0)=\int_0^{P_2}{\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP}$$


text:
Are these formulas for the critical value for detection interchangeable?
I am trying to compare equations for calculating the critical value for detection. One is presented by the ISO standards (https://www.sciencedirect.com/science/article/pii/S0969804316306455?via%3Dihub)
where $y^* = k \cdot \sqrt{a\cdot (a+1) \cdot (n_a + n_b) } =k\cdot \sqrt{2n_0} = k\cdot u(0)$

$n_a$, $n_b$ are counts in the right and left side regions of the peak
$k$ is the quantile of the standardized normal distribution
$a=\frac{b}{2l}$ where $b$ is the number of channel defining the region of interest for a spectrum - peak and $l$ is the number of channel at the right and left side of the peak used

It is presented, that $a\cdot (a+1) \cdot (n_a + n_b) =2n_0$, where $n_0$ sands for counts of blank
and other equation used at my workplace (by Obrusník and Kučera, The digital methods of peak area computation and detection limit in gamma-ray spectrometry. Radiochem Radioanal Letters. 1978, 32(3-4), 149-160): $L_D = \frac{k}{2}\sqrt{(GA-NA) \cdot 2 \cdot (N - 2) } $, where

$GA$ gross area
$NA$ net area
$N$ is the number of channels in the section of interest

It seems to me that: $GA -NA = B = n_0$
$a\cdot (a+1) = \frac{b}{2l}\cdot (\frac{b}{2l}+1) = \frac{b}{2l}\cdot(\frac{b+2l}{2l}) = \frac{b}{4l^2}\cdot N$
$\frac{b}{2l} \cdot N \cdot \frac{n_a+n_b}{2l}$, where $\frac{n_a+n_b}{2l}$ is average number of counts for one channel belonging to background.. and therefore $N\cdot \frac{n_a+n_b}{2l} = B =(?) n_0  $
but I didn't even arrive to $k \cdot \sqrt{a\cdot (a+1) \cdot (n_a + n_b) } =k\cdot \sqrt{2n_0}.$
Is there anything I overlooked/didn't understand right?
Thank you


text:
Effect of competitive inhibitor on substrate inhibition
In an enzyme that undergoes substrate inhibition, how would the presence of a competitive inhibitor affect said substrate inhibition? Would the substrate concentration at which substrate inhibition begins be affected? Would the rate of enzyme activity decrease with increasing substrate be affected?

We need to deal with the mechanism of competitive inhibition+substrate inhibition:
\begin{align}
  \ce{E + S  <=> ES}   \quad &K_\text{M} = \frac{\ce{[E]}\ce{[S]}}{\ce{[ES]}} \tag{R1/1} \\
  \ce{ES     -> E + P} \quad &r = k[\ce{ES}] \tag{R2/2} \\
  \ce{E + I  <=> EI}   \quad &K_\text{I} = \frac{\ce{[E]}\ce{[I]}}{\ce{[EI]}} \tag{R3/3} \\
  \ce{ES + S <=> ES2}  \quad &K_\text{S} = \frac{\ce{[ES]}\ce{[S]}}{\ce{[ES2]}} \tag{R4/4}
\end{align}
where:

$\ce{E}$ is the enzyme.
$\ce{ES}$ is the enzyme-substrate complex.
$\ce{I}$ is the inhibitor.
$\ce{EI}$ is the enzyme-inhibitor complex.
$\ce{ES2}$ is the compound generated by the reaction of the enzyme and the enzyme-substrate complex, that is taking away the substrate the enzyme needs.
The $K$'s are the equilibrium constants, e.g., $K_\text{M}$ is the Michaelis's one. In biochemistry, it is standard practice to write them upside down.

Our task is to find an expression for the rate of formation of the product $\ce{P}$, in terms of the substrate concentration $\ce{S}$, the variable we can easily track.
Enzyme balance The initial concentration of enzyme is equal to
\begin{align}
  C_\text{E0} &= \underbrace{[\ce{E}]}_1 +
                 [\ce{ES}] +
                 \underbrace{[\ce{EI}]}_3 +
                 \underbrace{[\ce{ES2}]}_4 \tag{5} \\
\end{align}
The underbraced parts (1), (3), and (4) are obtained via Eqs. (1), (3), and (4)
\begin{align}
  K_\text{M} = \frac{\ce{[E]}\ce{[S]}}{\ce{[ES]}}   &\rightarrow
  \ce{[E]} = \frac{K_\text{M}\ce{[ES]}}{\ce{[S]}} \tag{6} \\
  K_\text{I} = \frac{\ce{[E]}\ce{[I]}}{\ce{[EI]}}   &\rightarrow
  \ce{[EI]} = \left(\frac{\ce{[I]}}{K_\text{I}}\right)\ce{[E]} \tag{7} \\
  K_\text{S} = \frac{\ce{[ES]}\ce{[S]}}{\ce{[ES2]}} &\rightarrow
  \ce{[ES2]} = \frac{\ce{[ES]}\ce{[S]}}{K_\text{S}} \tag{8}
\end{align}
and combining Eqs. (5-8)
\begin{align}
  C_\text{E0} &=
  \frac{K_\text{M}[\ce{ES}]}{[\ce{S}]} + [\ce{ES}] +
  \left(\frac{\ce{[I]}}{K_\text{I}}\right)\color{blue}{\ce{[E]}} +
  \frac{[\ce{ES}][\ce{S}]}{K_\text{S}} \tag{9} \\
\end{align}
In Eq. (9) the enzyme concentration apears again in blue, so we combine Eq. (9) and Eq. (6) again
\begin{align}
  C_\text{E0} &=
  \frac{K_\text{M}[\ce{ES}]}{[\ce{S}]} + [\ce{ES}] +
  \left(\frac{\ce{[I]}}{K_\text{I}}\right)
  \frac{K_\text{M}[\ce{ES}]}{[\ce{S}]} +
  \frac{[\ce{ES}][\ce{S}]}{K_\text{S}} \\
  C_\text{E0} &=
  \color{blue}{\frac{K_\text{M}[\ce{ES}]}{[\ce{S}]}} + [\ce{ES}] +
  \color{blue}{\frac{K_\text{M}\ce{[ES]}}{[\ce{S}]}
  \left(\frac{\ce{[I]}}{K_\text{I}}\right)} +
  \frac{[\ce{ES}][\ce{S}]}{K_\text{S}} \\
  C_\text{E0} &=
  \frac{K_\text{M}[\ce{ES}]}{[\ce{S}]}
  \underbrace{\left(1 + \frac{[\ce{I}]}{K_\text{I}}\right)}_\alpha
  + [\ce{ES}] +
  \frac{[\ce{ES}][\ce{S}]}{K_\text{S}} \tag{10} \\
\end{align}
where we factored out the repeating term in both blue parts of the equation.
The main issue with Eq. (10), is that $\ce{[I]}$ is still there. We could use an inhibitor balance, but unfortunately, we can prove that we can't have a simple final expression. If you want further math just say in the comments. I will name underbraced part with $\alpha$, and name it the inhibitor power. We will explore its consequences further. Eq. (10) now becomes
\begin{align}
  C_\text{E0} &=
  \frac{[\ce{ES}]}{[\ce{S}]}K_\text{M} \alpha +
  [\ce{ES}] +
  \frac{[\ce{ES}][\ce{S}]}{K_\text{S}} \\
  C_\text{E0} &= [\ce{ES}] \left(
  \frac{K_\text{M} \alpha}{[\ce{S}]} + 1 + \frac{[\ce{S}]}{K_\text{S}}\right) \\
  C_\text{E0} &= [\ce{ES}] \left(
  \frac{K_\text{M}\alpha + [\ce{S}]}{[\ce{S}]} + \frac{[\ce{S}]}{K_\text{S}}\right) \\
  C_\text{E0} &= [\ce{ES}] \left(
  \frac{K_\text{S}K_\text{M}\alpha + K_\text{S}[\ce{S}] + [\ce{S}]^2}
  {K_\text{S}[\ce{S}]}\right)\\
  [\ce{ES}]  &=  \left(\frac{K_\text{S}[\ce{S}]}
  {K_\text{S}K_\text{M}\alpha + K_\text{S}[\ce{S}] + [\ce{S}]^2}\right)C_\text{E0}
  \tag{11} \\
\end{align}
Combining Eqs. (2) and (11)
\begin{equation}
  r = \frac{kK_\text{S}[\ce{S}]C_\text{E0}}
  {K_\text{S}K_\text{M}\alpha + K_\text{S}[\ce{S}] + [\ce{S}]^2} \rightarrow
  \boxed{\frac{r}{kC_\text{E0}} = \frac{[\ce{S}]}
  {K_\text{M}\alpha + [\ce{S}] + \dfrac{[\ce{S}]^2}{K_\text{S}}}} \tag{12} \\
\end{equation}
Eq. (12) is exactly what we want, a rate which is only a function of the substrate concentration. I also put it in a dimensionless form, which will be easier to analyze. The plot of Eq. (12) is the following, where I employed for simplicity $K_\text{S} = K_\text{M} = 1 \; \pu{mol/dm3}$:

Observations:

If $\alpha = 1$, the inhibitor power is zero, and $\ce{R3}$ just vanishes. We end with only substrate inhibition.
As $\alpha$ goes up, $\ce{R3}$ is more aggressive. This means that more enzime is consumed and be inactivated. Thus, the reaction goes down for every concentration of substrate.
Any operation with too high concentrations of a substrate is discouraged, since $\ce{R4}$ is favored, and the substrate inhibition is too predominant.
For all cases, we have an optimum value of substrate concentration, where the rate is maximum. This is really obvious in Eq. (12). In the numerator we have a linear function, that will win when $\ce{[S]}$ is low, and in the denominator a quadratic function that will when when $\ce{[S]}$ is high. This is a typical characteristic that substrate inhibition displays. We can prove that this happens in
$$ \frac{\mathrm{d}(r/kC_\text{E0})}{\mathrm{d}[S]} = 0 \rightarrow
\boxed{[\ce{S}]_\text{opt} = \sqrt{K_\text{S}K_\text{M}\alpha}} \tag{13} $$
This is the curve in magenta, thus, the best "ideal" operation will be the one that follows this curve for every instant in time in the biochemical reactor, disregarding the spatial effects.

Thanks for your well-written answer. Please do provide more math.

@user73910 Hello! Thanks. Enumerate the parts that were not so clear to you and I add them at the end of the post.

Could you write out how you got equations 6 and 7?

@user73910 Tell me if it is clearer now.

Yes, it's clear now. Thank you.


text:
How can a volatile solvent make an ideal solution with a non-volatile solid solute?
I ask because to make an ideal solution the intermolecular forces between the molecules of solvent and solute respectively should be similar and after mixing they should also be similar to what they were before mixing. But if we mix solid solute in liquid solvent then individually their respective intermolecular forces will be different from each other, because one component is solid and the other is liquid, hence the intermolecular forces in the solid component should be greater than the intermolecular forces in the liquid component.

Ideal solutions do not exist, like ideal gases do not exist  Both are simplified models of reality.

Also in chemistry you must think of the precision a model predicts for some behavior. Any model good to 4 significant figures would be considered perfect by a chemist. So chemistry isn't like pure math where $\pi$ has been calculated to over a hundred trillion digits.

Once the solid dissolves in the liquid it is present only as molecules, for example solid naphthalene completely dissolving in benzene. The intermolecular forces between the molecules are then of three types solvent-solvent, solute-solute and solvent-solute, how big each is depends on the types of molecules and how important each is to the total energy/entropy of mixing depends on their relative concentrations, usually converted into mole fractions.
You mention an ideal solution. In thermodynamics an ideal gas or ideal solution is one that follows a  defined standard behaviour. The ideal gas is one that obeys the perfect gas law, and this is found to be the case for many gasses but only, for example, at v. low pressure. The ideal solution is one that also follows some standard behaviour but the model of ideality depends on the type of system, divided between non-electrolytes and electrolytes. For non-electrolytes the ideal solution is defined as one whose parameters for the free energy of mixing follow that for an ideal gas, i.e. $\displaystyle \Delta G_{mix}=RT \sum_i  x_i\ln(x_i)$ where $x_i$ is the mole fraction of species $i$ and that $\Delta H_{mix} =0$. The entropy of mixing is therefore $\Delta S_{mix}=-\left( \partial \Delta G_{mix}/\partial T\right)_P=-R \sum_i  x_i\ln(x_i)$.


text:
If gold does not oxidise in air, why doesn't it reform it's lattice?
It is a common fact that without oxidation, a metal can reform it's lattice with a tad bit of pressure.
Gold is often said to be non-oxidising(doesn't oxidise in ambient temperature and pressure) and that only aqua regia can oxidise gold. In that case, why doesn't gold reform it's lattice on applying pressure. I did try it and noticed that gold partially reformed it's lattice if we try to attach the pieces immediately, but even that eventually tore out as I put just a tad bit of force, I'd say something like 20g.wt. So, why does gold not reform it's lattice if it doesn't oxidise, which basically says gold does oxidise?
Edit: Okay, for context, I mean cold welding should be possible with gold if it doesn't oxidise. But that obviously doesn't happen, which means the surface gold does lose electrons. So I mean to say, gold should be considered as oxidisable in air, but slowly instead of saying it doesn't at all, which most resources claim. Am I right, if not, please correct me.

Oxidation, lattice reforming and reattaching 2 metal pieces are 3 different independent things.

But oxidation prevents lattice reforming, which causes us not to be able to reattach 2 metal pieces. For reference, see cold welding on wiki

Or gas adsorption. Such reattaching is best done in vacuum.

Yep, because there is no reaction with air. Which means gold does react, though slowly, with air.

You have not excluded gas adsorption. // Golden foil can have thickness down to lower nanometers, just few atoms thick. If it oxidated, such a foil would stop become a metal soon. BTW, gold oxides are not thermodynamically stable at low temperatures.

So, you’re suggesting that instead of oxidizing, gas adsorption takes place. That actually makes sense.

Wire bonding for semiconductor devices using gold wires and pads is routinely performed.

Gold most definitely does cold-weld in vacuo, or even with moderate pressure, extended clamping time and/or mild heating at 1 atm, That is how gold leads are attached to an IC. If you want to test in the lab, though, whether with gold or any other substance, start with clean surfaces. Clean, lapped gauge-blocks cannot be left wrung together lest they cold-weld.
However, reformation of a crystalline lattice is not necessary for cold welding. For example, glass vacuum tubes can be produced by cold-welding without heat sealing. Glass is noncrystalline.
Richard Feynman stated: "If we try to get
absolutely pure copper, if we clean and polish the surfaces,
outgas the materials in a vacuum, and take every conceivable
precaution... the slider will not fall off — the two
pieces of copper stick together... The reason for this unexpected behavior is that when the atoms in contact are all of the same kind, there is
no way for the atoms to "know" that they are in different pieces
of copper." On a quantum level, can one distinguish an atom in one piece from an adjacent one in what was the other piece?


text:
Equilibrium Solubility of CO2 in Aqueous Solution and its Dependence on H3O+ Concentration
The equilibrium solubility of $\ce{CO2}$ in an aqueous solution is given by three chemical reactions:
$$
\begin{align}
\ce{CO2(g) &<=> CO2(aq)}\label{rxn:R1}\tag{R1}\\
\ce{CO2(aq) + H2O &<=> H2CO3(aq)}\label{rxn:R2}\tag{R2}\\
\ce{H2CO3(aq) + H2O &<=> HCO3-(aq) + H3O+(aq)}\label{rxn:R3}\tag{R3}
\end{align}
$$
Considering a system $E_1$ at equilibrium with constant pressure and variable volume, an increase in the concentration of $\ce{H3O+}$ would shift the equilibrium towards $\ce{CO2(g)}$ and consequently result in a change in volume. Let's denote the new system with the altered equilibrium as $E_2$.
The law of mass action for each reaction is defined as:
$$
K_1 = \frac{\ce{[CO2(aq)]}}{\ce{[CO2(g)]}},\
K_2 = \frac{\ce{[H2CO3(aq)]}}{\ce{[CO2(aq)]}},\
K_3 = \frac{\ce{[HCO3-(aq)][H3O+(aq)]}}{\ce{[H2CO3(aq)]}}
$$
Since $K_1$ is constant at a given temperature, and $\ce{[CO2(g)]}$ remains constant because the pressure is assumed to be constant, $\ce{[CO2(aq)]}$ is also constant. With the same argument, the concentration of $\ce{H2CO3(aq)}$ should be constant. However, the concentration of $\ce{HCO3-(aq)}$ in $E_2$ needs to be lower than in $E_1$ because the concentration of $\ce{H3O+}$ is higher in $E_2$, and $K_3$ is a constant. Therefore, the change in volume can be fully explained by the alteration in the concentration of $\ce{HCO3-(aq)}$.
Does that make sense? It seems quite strange to me that there is no change in concentration for the reactants in $\mathrm{R1}$ and $\mathrm{R2}$

Concentrations of a given molecular entity must be the same in all equations, as it is the same system. // Be aware MathJax is preferred not to be used in CH SE Q Titles.

Add the equations or multiply the $K$'s, then as $\ce{H_3O^+}$ goes up $\ce{HCO_3^-}$ goes down because each $K$ is constant and so is $\ce{CO_2(g)}$.

@porphyrin Hello porphyrin. I was thinking the same but then I got the following. Lets sum them all and "magically" put some protons there so that the $\mathrm{pH}$ goes down. Then, $\ce{HCO3-}$ must go down. But it can only go down with the generation of $\ce{CO2(g)}$ (so the global reaction actually goes to the left), but $\ce{CO2(g)}$ is constant. What am I thinking wrong?

General solution of such systems is solving a set of $N$ nonlinear equations for $N$ variables.
There are concenctrations of $\ce{H+}$, $\ce{OH-}$, $\ce{CO2(aq)}$, $\ce{H2CO3(aq)}$, $\ce{HCO3-(aq)}$, $\ce{CO3^2-(aq)}$ + $p_{\ce{CO2}}$, i.e. $N=7$.

There is 1 equation for total $\ce{CO2}$ inventory
There is 1 equation for charge neutrality.
There are equilibrium equations:

1 equation for $\ce{CO2}$ dissolution
1 equation for $\ce{H2CO3}$ formation
2 equations for $\ce{H2CO3}$ dissociation
1 equation for water auto-ionization

So 7 variables and 7 independent equations.
With constant $\ce{CO2(g)}$ partial pressure, it can be simplified as $[\ce{CO2(aq)}]$  and $[\ce{H2CO3(aq)}]$ would be at equilibrium constant as well.
Exact solving of the equation set often lead to a cubic or more complicated equation for $[\ce{H+}]$. It may be impossible or challenging to resolve analytically.
But can be solved numerically e.g. by the MS Excel solver package, e.g. by minimalization of function:
$$z = \sum_{i=1}^N{\text{(EqLeftSide}_i - \text{EqRightSide}_i)^2}$$
The practical hint: Use for variables and parameters single-letter symbols, common in algebra, for easier manipulation.
Another way is simplification by evaluation of eventual strong inequalities.
E.g.
$$[\ce{HCO3-}] \gg [\ce{CO3^2-}] \implies [\ce{HCO3-}] + [\ce{CO3^2-}] \simeq [\ce{HCO3-}]$$
or
$$|x| \ll 1 \implies \frac{1}{1 \pm x} \simeq 1 \mp x$$
Such simplifications often lead to exact solutions, e.g. a quadratic equation, of an inexact, simplified task.


text:
Is half-life defined for a reaction or for a reactant?
Consider the following chemical reaction.
$$\ce{A + B -> Products}$$
If this is a zero-order reaction with respect to $\ce{A}$ and $\ce{B}$, then is there a half-time for this reaction? Or must we calculate the half-time for species $\ce{A}$ and $\ce{B}$ separately?

The concept of half life is only useful when the reaction is first order with respect to A and does not depend on anything else. In all other cases, though still technically applicable, it does not do any good.

For these reaction order you have
$$ -\frac{d|\ce{A}|}{dt} = -\frac{d|\ce{B}|}{dt} = k|\ce{A}|^0|\ce{B}|^0 = k$$
and the integrated velocity equations have the form
$$ |\ce{A}|_t = |\ce{A}|_0 - kt$$
$$ |\ce{B}|_t = |\ce{B}|_0 - kt$$
Using the definition of $t_{1/2}$ you will have
$$t_{1/2}(\ce{A}) = \frac{|\ce{A}|_0}{2k} $$
$$t_{1/2}(\ce{B}) = \frac{|\ce{B}|_0}{2k} $$
so, as you suggested, unless $|\ce{A}|_0 = |\ce{B}|_0$, the value of $t_{1/2}$ depends on which reactant you are using to estimate its value.

So there is no term like half time of reaction for this case right,?

@OJi There is mathematically infinite number of possible formal half life values, depending on initial concentrations of A and B. Such a parameter is useless but for very particular reaction settings.

No I meant that for overall reaction Is there half life of reaction? Like if initially concentration of A is 2M and B is 4M, then is there some thing like half life of reaction?

@Oji I agree with Poutnik, In the case you mention where both concentrations are different I do not think there is a definition for an overall lifetime of the reaction. If the initial concentrations of the reactants differ, It always would depend on the concentration of one of the reactants. It just will create confusion.

Since one of the reactants will be exhausted first, one could define $$t_{1/2} := \frac{\min\left{[\ce{A}]_0, [\ce{B}]_0\right}}{2k} $$

@PAEP If it was a first order reaction with respect to both A and B, will we use the same method i.e. find half life of A and B seperately using half time formula for A and B of first order reaction?

You can follow the same procedure. However, you would find that you are working with global second order kinetics so the integrated equations are a bit different. Check a textbook.


text:
Change in Gibb's free energy at equilirbrium
I know the basic relation that ΔG=ΔG⁰+RTln(Q) where Q is the reaction quotient, my doubt is that if Q=1 at equilibrium then ΔG=ΔG⁰ but will ΔG=0 or ΔG≠0? As far as I know, generally at equilibrium Q=Keq, ΔG=0.

As $$\Delta_r G^{\circ}=-RT \ln K$$
then
$$\Delta_r G=\Delta_r G^{\circ} + RT \ln Q = RT \ln{\frac{Q}{K}}$$
If there is by the task conditions equilibrium and $Q=1$
then
\begin{align}
K&=1\\
RT \ln K &= 0\\
RT \ln Q &= 0\\
\Delta_r G^{\circ}&=0\\
\Delta_r G&=0
\end{align}

Note:At equilibrium ∆rG is always zero And at Q=1,∆rG is always equal to ∆rG°
Both are different concepts and you are merging them into one.
Let's understand them one by one.
First of all , the equation you have written  is not correct , there must be ∆rG instead  of ∆G. Both are different . ∆G is difference of Gibbs free energy between any two states of system whereas ∆rG is derivative of Gibbs free energy with respect to extend of reaction. Now understand difference between these two terms in more detail.
Let's consider a reaction A -> B and plot the graph of Gibbs free energy vs extent of reaction for this reaction.

a = initial mole of reactant A
Extent of reaction ξ means amount of reactant converted into product like ξ = 0 , no amount of reactant is converted into product hence ξ = 0 represent pure reactant. Similarly ξ = a means all reactant is converted in product , no reactant is left over so ξ = a means pure product.
Although you might be wonder that this graph should be like that-

Yes , this graph of straight line is correct but only when there is no mixing between reactant and product. Mixing involves interaction phenomenon between reactant and product molecules and due to this at Gibbs free energy is lower than expected and correct graph is that curved one.
If you want to further read about mixing - check out these links
Lecture on Gibbs free energy of mixing
Article on Thermodynamics of mixing
Okay so let's get back to our reaction where mixing is involved -

∆G is difference of Gibbs free energy between any two points , if we take X and Y two points then ∆Gxy will represent Gibbs free energy difference between these two points.
∆rG- iIf at any point a infinitesimal amount dξ of reactant A is converted into B and difference in Gibbs free energy due to this small change is dG then ∆rG is dG/dξ.
If still you are confused with ∆G and ∆rG read this answer - https://chemistry.stackexchange.com/a/50585/129242
∆rG°
As we take electrode potential of hydrogen zero as reference for convince because we can't measure it in absolute terms similarly we take ∆rG° (or Q = 1) as reference for a reaction so that we can further calculate variation of ∆rG with the concentration of reactant and product using the formula  ∆rG =  ∆rG° + RTln(Q). ∆rG° simply denotes ∆rG at a specific point when reaction quotient (Q) = 1. ∆rG° has nothing to do with equilibrium.
In this example if we consider reaction elementary then reaction quotient cna be written as Q = [B] / [A] where [•] represent concentration. Also we assume that solution in which reaction is going on has volume 1 litre so concentration = moles.
So reaction quotient will be equal to one Q = 1 when [A] = [B] or ξ = a/2.
Equilibrium
Same like ∆rG° which denotes  a special point in terms of reaction quotient where Q = 1, equilibrium ( at which Gibbs free energy  of system is minimum) is special point (here Y) in terms  of Gibbs free energy  at which ∆rG = 0 , Equilibrium doesn't have any specific value of Q , at equilibrium Q may be equal to one or may not be. Also remember that Q at equilibrium is known as Equilibrium constant Keq.

In this diagram you can clearly see that Q = 1 , where ∆rG = ∆rG° and Q = Keq , where ∆rG = 0 are two different points.
Finally discuss your question.

I know the basic relation that ΔG=ΔG⁰+RTln(Q) where Q is the reaction quotient, my doubt is that if Q=1 at equilibrium then ΔG=ΔG⁰ but will ΔG=0 or ΔG≠0? As far as I know, generally at equilibrium Q=Keq, ΔG=0.

Question is about a special case where Q = 1 at equilibrium .
When Q = 1, ∆rG = ∆rG° and at equilibrium ∆rG = 0 so final conclusion is that in this case ∆rG° will also be equal to 0.
So in our reaction A -> B whose thermodynamics and Equilibrium we have already discussed, if Q = 1 at equilibrium then Gibbs free energy diagram will look like -

Comment if any point is not clear to you.

please consider upvote if you like a answer.


text:
What are the limitations of the Nikolsky-Eisenman (NE) Equation? And what exactly is a more accurate equation to use for modeling selectivity?
I am very curious as to where exactly the Nikolsky-Eisenman (NE) equation fails. I know that the reference I posted here mentions that when the charge of the interfering ion versus the analyte's charge is different, then the equation no longer appears to describe reality accurately. However, I don't understand where the equation might fail on the theorized calibration curves with interferences (assuming we know concentrations of the system as well as all selectivity coefficients).
The reason I am curious is because I built a model of the NE equation in python to attempt to model how each interferent affects the calibration plot of a ion selective electrode (ISE). I used 5 different analytes where two had +1 charges and three had +2 charges. The analyte has a +2 charge. I am curious what equation is more accurate to use to model this type of scenario as I was simply only using the standard NE equation and I couldn't not find the so called permutated or semi-empirical form of the NE equation that claims to be more accurate of reality.
I assuming standard temperatures and conditions for all my simulations and activities = concentrations for simplicity.

https://www.researchgate.net/publication/321532727_Describing_Ion_Exchange_at_Membrane_Electrodes_for_Ions_of_Different_Charge

Please add a proper typed reference citation in place of the image you included.

https://doi.org/10.1002/elan.201700700
does that work?


text:
Can an epoxy resin create a terpene polymer?
I am developing plant-based antifouling coatings that will stop barnacles binding to ship hulls for example. I have started simply by mixing specific terpenes with an epoxy resin (BISPHENOL A DIGLYCIDYL ETHER), then adding the hardener and painting this onto rocks or metal before placing them in the ocean. The results are amazing so far in that I am preventing biofouling compared to an epoxy resin control.
Does the epoxy resin I am using have the ability to alter the molecular structure of terpenes, like create a terpene polymer? The reason I am thinking about this is because of this recent publication (https://pubs.acs.org/doi/10.1021/acs.biomac.2c00721), showing terpene polymerisation that increased anti-bacterial activity.
What do you think? Could BISPHENOL A DIGLYCIDYL ETHER polymerise terpenes?

Well, this paper shows functionalised terpens. You could modify them to fit the reaction of epoxy. For example epoxidise, or add glycidyl group.


text:
Gibbs free energy of transition and activation energy
Let's consider a reaction A (reactant) -> B(product) and activated complex is denoted by C.

This graph ( potential energy vs reaction coordinate ) tells us that reactant need some amount of activation energy (Ea) to convert in product, which has low potential energy which is shown here in terms of enthalpy ∆H. We can assume from this graph that activation represent same kind of potential energy between A (reactant) and C (activated complex ) that Enthalpy ∆H represent between A and B (product).
Now look at another graph of reaction (Gibbs free energy vs extent of reaction)

This graph represents that activation energy is difference between Gibbs free energy of reactant and activated complex or there is also possibility that the activation energy shown here is not arrhenius activation energy Ea but it is Gibbs energy of activation ΔG‡ according to transition state theory.

Q. But to perform a reaction what amount of energy we need to supply to reactants arrhenius activation energy Ea or gibbs free energy of activation ΔG‡ ? I think it's ΔG‡ as defination of Gibbs free energy states - minimum amount of work needed to supply for a non spontaneous reaction (here A -> C ) to be happened but then why arrhenius theory states that - for reactants to transform into products, they must first acquire a minimum amount of energy, called the activation energy Ea ?
And also what these two energies represent physically in terms of bonds , interatomic interactions etc ?
Mathematical equations -
ΔG‡ = ∆H‡ - T∆S‡
ΔG‡ = Ea - RT - T∆S‡   ( ∆H‡ = Ea - RT )

Here are two sources (oxford lecture notes and wikibooks) to look into this topic further than the Libretext source cited in the question.

It is possible that you are mixing several concepts that are related but are not identical. $\mathrm{E_a}$ stems from the Arrhenius equation that describes the dependence of k(T) with temperature. $\mathrm{E_a}$  is an empirical parameter. ...

(cont.) However, for a simple reaction profile that presents a barrier between reactants and products, Fowler and  Guggenheim showed that $\mathrm{E_a}$ roghtly corresponds to difference be the average energy of the molecules that react and the average energy of the reactant molecules (see K. J. Laidler, Chemical Kinetics, Pearson,1987). ...

(cont.) The thermodynamic formulation of the TST allows you to express k(T) as a function of the "thermodynamic" magnitudes that you can estimate through theoretical calculations or experimental data. Be aware that the mathematical expressions that relate $\mathrm{\Delta G^‡}$, $\mathrm{\Delta H^‡}$ and $\mathrm{\Delta S^‡}$ are obtained when you compare the TST expression with the Arrhenius equation. $\mathrm{E_a}$ and $\mathrm{A}$ do not appear as part of the TST theory.

@PAEP To perform a reaction what amount of energy we need to supply to reactants arrhenius activation energy Ea or gibbs free energy of activation ΔG‡ ?

@Karsten the link you have provided (Oxford lecture notes) is not working

@Bharat Prajapat your question is answered in the reference we  have given you. I think that unless you understand what you are asking your question does not make much sense.

By the way you can access Clare Vallance lecture notes through the link provided by @Karsten. I checked it today and yesterday. It is a link to a pdf file, so you need a pdf reader.

@PAEP "However, Ea roghtly corresponds to difference be the average energy of the molecules that react and the average energy of the reactant molecules ". That average energy is what - 1. Internal energy  2. Gibbs free energy or 3. Enthalpy

Guess it is not so simple. You should review your Statistical Thermodynamics and  what is the meaning of the $\mathrm{\Delta U^‡}$, $\mathrm{\Delta H^‡}$,  $\mathrm{\Delta S^‡}$  and $\mathrm{\Delta G^‡}$.

@PAEP I know about each term that's why I am asking that when fowler and guggenheim derived expression for average energy change between reactant and  product molecules,  in which form of energy (  ΔU‡, ΔH‡ and ΔG‡  ) they calculated average energy.

Please note that the activated complex model is outdated and superseded by transition state theory.

To obtain the Fowler and Gugghenheim expression you do not refer to Thermodynamic parameters. It simply did not make sense to partition the energy change in $\mathrm{ΔU^‡}$, $\mathrm{ΔH^‡}$, $\mathrm{ΔS^‡}$ or $\mathrm{ΔG^‡}$. It is well explained in Laidler's books which is a textbook (old but very insightful).

I think you are overstretching the capabilities of the Thermodynamic Formulation of the TST. As @Martin points, it has been superseeded by later more evolved formulations of the theory (btw, some of them described in Laidler´s textbook).

Well yes, sure, there are more evolved models after TST, but I actually wasn't referring to that. One needs to understand TST, especially its shortcomings,before understanding anything further. But you don't need to understand AC to understand TST.


text:
Thermodynamics potential and differences between them
There are 5 most common thermodynamics potential -

But we will discuss about only three of them -

Internal energy (∆U)
Enthalpy(∆H)
Gibbs free energy (∆G)

As Gibbs free energy is also a thermodynamics potential that means it is not related to only energy of bonds and interatomic interactions but also to entropy potential.

1. So now my first question is that what thermodynamics potential actually represent physically not mathmatically?

Now talk about those three thermodynamics potential -
I find it most confusing that how internal energy , enthalpy and Gibbs free energy are different form of energies or potential although looking at the mathmatical expressions we can easily say Gibbs free energy is different from   enthalpy by a additional term -TS or enthalpy is different from internal energy as it has a additional term pV. But how they are different form of energies theoretically?
Let's consider a example of chemical reaction to understand what I am actually asking.

This is most common diagram used in our textbooks which shows  potential energy of system when reactant is converting into product. but the difference of potential energy between reactant and product is ∆U or ∆H or ∆G ? And how  these 3 terms (∆U,∆H,∆G) physically represent 3 different kind of potential energies  between reactant and product ?
Physically means in terms of bond energies, interatomic forces and  interactions , molecular kinetic energy , vibrational energy, rotational energy etc.
Ok finally I address my question in more clarified language - internal energy of a system (here reactants and products) as per it's  defination includes all kinds of kinetic and potential energies of system so when we say there is difference between reactant's and product's internal energy by ∆U , we have included all kind of energy differences between reactant and product. but when we say there is a difference ∆H between enthalpy ( or ∆G between Gibbs free energy ) of reactant and product , does it represent any specific type of potential energy in internal energy difference between reactant and product means does ∆H ( or ∆G) is a specific part of internal energy?

H and F are used under different conditions and automatically account for any pV work.  G was invented because, experimentally the limiting amount of work that could be obtained in reactions was not described by $\Delta H$.  Gibbs first realised that in a reaction heat could be taken from or given to surroundings (via entropy) and so work done is different to that predicted by $\Delta H$. In your picture  potential is not defined, but probably U is meant, but could be energy just in the bonds that break.

Note that the term state function is more common that the thermodynamic  potential, while both are correct.
Considering $U$, $H$, $F$ and $G$ as different forms of energy may be misleading.
The state function internal energy $\ce{U}$ is intuitively well understood as the total system kinetic and potential energy. Formally, the energy E=mc^2 should be included to, but as the interest is usually in $U$ changes, it is not involved, being many orders higher than the rest.
The state function enthalpy $H = U + pV$ (as pV is a state function too) is not other form of energy, but made up state function. At isobaric conditions, enthalpy change is energy added to the system that is used

to increse the system internal energy
to do mechanical volume work on the neighbourhood, due volume changes.

For thermally induced system expansion:
$$\mathrm{d}H = \delta Q - \delta W = C_V \mathrm{d}T + p\mathrm{d}V = C_V \mathrm{d}T + p \left(\frac{\partial V}{\partial T}\right)_p \mathrm{d}T \\
= \left(C_V  + p \left(\frac{\partial V}{\partial T}\right)_p \right)\mathrm{d}T = C_p \mathrm{d}T$$
$\Delta H - \Delta U$ may be formally considered as energy, that system does not store in form its internal energy, but lends/borrows it to/from its neighbourhood via mechanical work.
(Free) Gibbs energy (or "free enthalpy) $G = H -TS = U + pV - TS$ is not other form of energy either, but yet another made up state function. It is the maximal part of enthalpy, that is, at least theoretically, available to do work, allowed by the 2nd law of thermodynamic.
For reversible isothermal and isobaric process, $-\frac{\Delta G_\text{sys}}{T} = \Delta S_\text{sys} + \Delta S_\text{surr}$
Imagine 2 almost identical, isolated systems:

The system A contains 2 thermally connected identical metal blocks at equilibrium, at $\pu{T = 300 K}$ and temperature independent heat capacity.
The system B is identical to the system A, but the blocks have the temperature respectively $\pu{290 K}$ and $\pu{310 K}$.

Both systems have the same internal energy and enthalpy. But the system A has higher entropy $S$ and lower Gibbs energy $G$. The difference is the part of the thermal energy of the blocks that can be used to do some work, e.g. via thermal engines or thermocouples generating electricity.
Note that the part of the thermal energy available to do work has the same form as the part of the thermal energy not available to do work. The difference is in quantity, not quality.

in the question 18 , here  https://books.google.co.in/books/content?id=2eWe_lUQP9IC&pg=PA178&img=1&zoom=3&hl=en&bul=1&sig=ACfU3U0D2XpDwUvrdR5o5d6vf7Xyr3NSFQ&w=1280         author writes that enthalpy is that energy of system which is available  for conversion  in heat but then how can enthalpy be greater than internal enery. as expression of enthalpy has a additional  pv term

When you isobarically heat up the system, part of heat is used by system doing expansion work. the added enthalpy is then bigger than added internal energy. The reverse path converts compression work done on system to the extra heat, so the enthalpy
(absolute) change is again bigger than internal energy change. It can be effectively said that the part of enthalpy that is not internal energy is stored remotely, out of the system, as external energy.

whenever I use word internal energy of system, the thing comes in my mind is total energy of system including  kinetic + potential but when you say enthalpy of system what comes in your mind that which property of system  you are talking about  ?

As above....... //  Or by other, analogy way:   Internal energy = local hard disk.  Enthalpy = local hard disk + cloud storage.  (do not consider their ratio, just qualitatively] Then not all data go to the local disk and not all data come from the local disk.  The data stored as enthalpy are not stored in the local disk by different way. They are stored there by usual way or are not stored there at all, but elsewhere.

By yet other words, the difference between Delta H and Delta U relates to energy interchange between the system and the neighbourhood  behind your back, as W = - p. Delta V..

last comment - I found a question https://chemistry.stackexchange.com/q/175332/129242 although it is not related to what i have asked , but if you answer this question I think that will solve all my doubts regarding Different  types  of energies and their physical significance .

So the change in energy, indeed is a mixture of kinetic and potential energies as you said.
The difference with the free energies in general is that they give you an idea of what the direction of the reaction is going to be (towards the lower $\Delta G$ or $\Delta\mu$ etc.).
This is a very general explanation which gives an idea of what is happening. Of course, many things can happen since the entropy and enthalpy also have to be taken into account.
The difference between all the free energies is dependent on how you define your system (or what it's characteristics are).

in the question 18 , here  https://books.google.co.in/books/content?id=2eWe_lUQP9IC&pg=PA178&img=1&zoom=3&hl=en&bul=1&sig=ACfU3U0D2XpDwUvrdR5o5d6vf7Xyr3NSFQ&w=1280         author writes that enthalpy is that energy of system which is available  for conversion  in heat but then how can enthalpy be greater than internal enery. as expression of enthalpy has a additional  pv term

I wasn't able to access the question, but about what the author wrote, don't forget that energy can be taken from many places like breaking bonds, or from irradiation or maybe, by changing the pressure or volume of the system which comes into account by the PV term.

last comment - I found a question  chemistry.stackexchange.com/q/175332/129242
although it is not related to what i have asked , but if you answer this question I think that will solve all my doubts regarding Different types of energies and their physical significance .


text:
Is it possible to electroplate iron onto nickel in a sodium chloride solution?
Over the years I have salvaged many laptop and power tool batteries and collected a lot of metal strips, but I don't know if they are made of nickel or nickel-plated steel.
I have a piece that I know for sure is pure nickel, and I would like to electrically deposit onto it all the nickel from the salvaged strips using a solution of table salt and thus, effectively, separate the nickel from the iron.
However, I don't know how to make sure that I actually electroplated nickel and not some alloy of nickel and iron or just iron. For this reason, I'm wondering if it is possible to deposit iron onto nickel in a table salt solution. I've tried searching on the Internet, but I've only found information about electroplating nickel onto iron, not the other way around.

If there is both nickel and iron in solution, nickel will deposit with priority. But using high enough voltage= low enough cathode potential, iron may co-deposit.

@Poutnik Do you know how low I should keep the voltage in order to deposit almost pure nickel?


text:
Does stoichiometry matter in extent of reaction calculation?
I was recently introduced to the extent of reaction and learned that it can be calculated using this equation
$$\xi = \frac{n_{A,out}-n_{A,in}}{\nu_A}.$$
If I'm using this formula to calculate the extent of reaction, does it then depend on how I write the stoichometry? For example, reaction $$\ce{2CO + O2 -> 2CO2}$$ could also be written as $$\ce{CO + 1/2 O2 -> CO2}$$ which seems to result in a different value for the extent of reaction according to the formula above.

[OP] If i'm using this formula to calculate the extent of reaction is it then dependent on how i write the stoichometry?

No. The formula is general and, for the $j$th species in a chemical reaction, the amount of $j$ is
$$ n_j = n_{j,0} + \nu_j \xi \tag{1} $$
where $n_{j,0}$ is the initial amount of species $j$.

[OP] Which to me this seems to result in a different value for the extent of reaction according to this formula.

Yes. But the results will be the same. I will illustrate with the case you have given, considering that: (1) there is no carbon dioxide initially, and (2) carbon monoxide is the limiting reactant.
\begin{align}
  \ce{2CO(g) + O2(g) -> 2CO2(g)} \quad & \quad \ce{CO(g) + 1/2O2(g) -> CO2(g)}
  \tag{R1/R2}
\end{align}
By Eq. (1) and making a difference between $\xi_1$ for $\text{R1}$ and $\xi_2$ for $\text{R2}$
\begin{align}
  n_\ce{CO} = n_\ce{CO,0} - 2\xi_1 \quad & \quad
  n_\ce{CO} = n_\ce{CO,0} - \xi_2 \tag{2,3} \\
  n_\ce{O2} = n_\ce{O2,0} -  \xi_1 \quad & \quad
  n_\ce{O2} = n_\ce{O2,0} - \frac{1}{2}\xi_2 \tag{4,5} \\
  n_\ce{CO2} = 2\xi_1 \quad & \quad n_\ce{CO2} = \xi_2  \tag{6,7} \\
\end{align}
Consider as an example where the limiting reactant is depleted, i.e., $n_\ce{CO} = 0$. By Eqs. (2) and (3) we get that the final extents of reaction are
$$ \xi_{1,\text{max}} = \frac{n_\ce{CO,0}}{2} \quad
   \xi_{2,\text{max}} = n_\ce{CO,0} \tag{8,9} $$
But when you go to evaluate the final amounts in Eqs. (4-7) we get the same result
\begin{align}
  n_\ce{O2}  = n_\ce{O2,0} - \frac{n_\ce{CO,0}}{2} \quad & \quad
  n_\ce{O2}  = n_\ce{O2,0} - \frac{n_\ce{CO,0}}{2} \tag{10,11} \\
  n_\ce{CO2} = n_\ce{CO,0} \quad & \quad n_\ce{CO2} = n_\ce{CO,0}  \tag{12,13} \\
\end{align}
Thus, as we can see we are at the same point, however both extents have different bounds
$$ 0 \leq \xi_1 \leq \frac{n_\ce{CO,0}}{2} \quad
   0 \leq \xi_2 \leq n_\ce{CO,0} \tag{14,15} $$
if the reactions are irreversible. In my humble opinion, as a good practice, always have a stoichiometric number equal to $1$ for the limiting reactant. It keeps tracks of mistakes.

Does stoichiometry matter in extent of reaction calculation?

Yes, it does. There are several quantities that depend on the choice of a common multiplicative factor for the chemical reaction equation. The reaction quotient $Q$, the equilibrium constants $K$, and the enthalpy, entropy and Gibbs energy of reaction are other examples.
On the other hand, the measurable quantities do not depend on it. No matter which multiplicative factor you choose, e.g. the equilibrium concentrations or how much of one species is needed to react with a given amount of another species does not change.
If you look at a specific calculation done based on two net equations that differ by a common factor, you will see that the common factor cancels out for measurable quantities, even if the extent of reaction and many other quantities will have different values for the two distinct net equations.


text:
What is the reaction between crystal violet and bleach (sodium hypochlorite) and what is the mechanism of its action against crystal violet?
I'm trying to find the reaction between bleach (sodium hypochlorite) and crystal violet, and need to find the mechanism of its action, as during my lab investigation the bleach did cause the crystal violet to lose its color but at a very slow rate of reaction, this was done by measuring the rate of absorbance using a colorimeter and the independent variable is the temperature.
There have been studies showing its rate of order but not the mechanism at which sodium hypochlorite discolors crystal violet, and there have been some studies claiming that it has no action.

I think you can find this article useful:  Nathanael Kazmierczak and Douglas A. Vander Griend, Improving Student Results in the Crystal Violet Chemical Kinetics Experiment. Journal of Chemical Education 2017 94 (1), 61-66 DOI: 10.1021/acs.jchemed.6b00408


text:
Why does Ascorbic Acid on paper turn brown when heated?
I am finding conflicting information about how ascorbic acid degrades and turns brown when used as "invisible ink" on paper and then heated.  Some say the heat accelerates the natural degradation process when it is exposed to air (oxidation?) and some say that the heat rearranges the molecules.  Can someone explain in some detail what is actually happening?  Does ascorbic acid turn brown when NOT on paper?  What is the chemical process involved?
Thanks.

"Invisible ink" works because of caramelisation - it has barely anything to do with vit. C

@Mithoron So this would be pyrolysis rather than oxidation of the cellulose, and the acid is not what turns brown.

@Karsten Yeah, acid's just a catalyst.

While I did not find sources that directly address your question, what I read about durability of paper seems to say that the acidity of the ascorbic acid would weaken the paper, accelerating the aging process and making it more heat-sensitive.
In the classical "magic ink" experiment, the sender uses lemon juice to write an invisible message, and the recipient heats up the paper to visualize the message. The active ingredient in lemon juice magic ink is citric acid, although it also is a source of vitamin C a.k.a. ascorbic acid. (A lemon has grams of citric acid and only milligrams of ascorbic acid.)
Other acids like acetic acid (from vinegar) or lactic acid (from milk) also work.
The U.S. Library of Congress has a nice article on preserving paper that mentions acid and heat treatment.


text:
IUPAC Name of Carbon Suboxide (C3O2)
I have searched the internet looking for the IUPAC name for carbon suboxide ($\ce{C3O2}$). Its structure is as follows:
$$
\ce{O=C=C=C=O}
$$
I did find one name here (PubChem), but as far as I know, a $\ce{C=O}$ at the terminal position of a carbon chain is not termed as ketone, so I am not sure if 1,2-propadiene-1,3-dione is the right name for the compound.

"Preferred IUPAC name
Propa-1,2-diene-1,3-dione"   Wikipedia

Can't help but wonder why did it get 5 upvotes. -on signifies carbonyl group in general, not necessarily ketone. In this case the functional group is rather enone - the compound is diketene.

I call into question that this compound should be banned according to the recommendations of organic compounds. Is just a more fancy carbon dioxide. Also the iupac tag was dedicated years ago.

And the initial part of the comment should have been: I call into question that this compound should be named according to the recommendations of organic compounds.