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Would microelectrodes reach a steady state current if convection is negligible and only pure diffusion occurs?
Assuming a two dimensional Cottrell type problem with a micro-electrode and a high enough applied potential so that concentration of the consumed redox species would be zero near the surface. Naturally the concentration profile would have a semi-spherical symmetry and boundary layer.
In the case of linear diffusion, the boundary layer grows infinitely if we consider a semi-infinite domain.
Is this still the case for semi-spherical diffusion with a micro-electrode in the middle of the cell? or does it reach a steady state?

Yes. It does reach a steady state.

We consider the electrochemical academic reaction
$$\ce{O + ne^- <=> R} \tag{R1} $$
to keep things simple, let $C$ be the concentration of species $\ce{O}$.
The concentration of species $\ce{O}$ is found by solving Fick's second law in cylindrical coordinates
\begin{equation}
  \frac{\partial C}{\partial t} = D\nabla^2 C =
  \frac{\partial^2 C}{\partial r^2} + \frac{1}{r}\frac{\partial C}{\partial r} +
  \frac{\partial^2 C}{\partial z^2} = 0 \tag{1}
\end{equation}
where $D$ is the diffusion coefficient of species $\ce{O}$, and $\partial C/\partial t = 0$ because we are interested in the steady-state solution. The semi-infinite problem has the following boundary conditions:

Far away in the radial direction from the UME the concentration is equal to the bulk concentration (denoted with $C^*$), which gives a Dirichlet boundary condition
$$ \lim_{r \to \infty} C(r, z) = C^* \tag{2}$$
Far away in the axial direction from the UME the concentration is equal to the bulk concentration, which gives a Dirichlet boundary condition
$$ \lim_{z \to \infty} C(r, z) = C^* \tag{3}$$
Zero flux of matter in the axial direction, because no reaction takes place outside the UME, which gives a Neumann boundary condition
$$ \frac{\partial C}{\partial z}(r, 0) = 0 \quad \forall r>r_0  \tag{4} $$
where $r_0$ is the radius of the UME.
There is a uniform concentration over the face of the electrode (denoted by $C_0$), which gives a Dirichlet boundary condition
$$ C(r, 0) = C_0 \quad r \leq r_0 \tag{5}$$

Newman solved this problem by a suitable variable transformation. We will define a concentration change and two new variables $\xi$ and $\eta$ which satisfy
\begin{align}
  \Delta C &= C^* - C(r, z) \tag{6} \\
  r &= r_0[(1 + \xi^2)(1 - \eta^2)]^{1/2} \tag{7} \\
  z &= r_0 \xi \eta \tag{8}
\end{align}
Don't worry about the geometrical meaning of these coordinates, because at the end we will come back at the cylindrical coordinates. The partial differential equation along with the boundary conditions take the following form
\begin{align}
  \frac{\partial}{\partial \xi}\left[(1 + \xi^2)
  \frac{\partial \Delta C(\xi,\eta)}{\partial \xi}\right] +
  \frac{\partial}{\partial \eta}\left[(1 - \eta^2)
  \frac{\partial \Delta C(\xi,\eta)}{\partial \eta}\right] &= 0 \tag{9} \\
  \lim_{\xi \to \infty} \Delta C(\xi, \eta) &= 0 \tag{10} \\
  \frac{\partial \Delta C}{\partial \eta}(\xi,0) &= 0 \tag{11} \\
  \Delta C(0, \eta) &= C^* - C_0 \tag{12} \\
\end{align}
The good news is that this PDE problem has a solution by separation of variables. Omitting the details we arrive at the concentration profile
\begin{equation}
  \Delta C(\xi,\eta) = \left[1 - \frac{2}{\pi}\tan^{-1}(\xi)
  \right](C^* - C_0) \tag{13}
\end{equation}
Now we need to find the steady-state current density. This can be evaluated by Faraday's law
\begin{align}
  j &=  nFD\frac{\partial C(r,z)}{\partial z}\bigg\vert_{(r,0)} \\
  j &= -nFD\frac{\partial \Delta C(r,z)}{\partial z}\bigg\vert_{(r,0)} \\
  j &= -nFD\left[\frac{\partial \Delta C(\xi,\eta)}{\partial \xi}
  \frac{\partial \xi}{\partial z}\right]_{(0,\eta)} \tag{14} \\
\end{align}
Eq. (14) needs the calculation of two partial derivatives. The first one comes from Eq. (13) while the third one from Eq. (8), and both are pretty direct
\begin{align}
  \frac{\partial \Delta C}{\partial \xi} &= -
  \left(\frac{2}{\pi}\right)\frac{C^* - C_0}{1 + \xi^2} \tag{15} \\
  \frac{\partial \xi}{\partial z} &= \frac{1}{r_0\eta} \tag{16}
\end{align}
additionally, we put $\eta$ as a function of $r$ and $\xi$. This is again direct with Eq. (7) by just isolating $\eta$
\begin{align}
  \eta &= \sqrt{\frac{1 + \xi^2 - (r/r_0)^2}{1 + \xi^2}} \tag{invert} \\
  \frac{1}{\eta} &= \sqrt{\frac{1 + \xi^2}{1 + \xi^2 - (r/r_0)^2}}
  \tag{multiply by 1/$r_0$} \\
  \frac{1}{r_0\eta} &= \frac{1}{r_0}\sqrt{\frac{1 + \xi^2}{1 + \xi^2 - (r/r_0)^2}}
  \\
  \frac{1}{r_0\eta} &= \sqrt{\frac{1 + \xi^2}{r_0^2[1 + \xi^2 - (r/r_0)^2]}} \\
  \frac{1}{r_0\eta} &= \sqrt{\frac{1 + \xi^2}{r_0^2(1 + \xi^2) - r^2}} \\
  \frac{1}{r_0\eta} &= \sqrt{\frac{1 + \xi^2}{r_0^2 + r_0^2\xi^2 - r^2}}
  \tag{17} \\
\end{align}
Combining Eqs. (14)-(17)
\begin{align}
  j &= -nFD\left[-\left(\frac{2}{\pi}\right)\frac{C^* - C_0}{1 + \xi^2}
  \sqrt{\frac{1 + \xi^2}{r_0^2 + r_0^2\xi^2 - r^2}}\right]_{(0,\eta)}  \\
  j &= -nFD\left[-\left(\frac{2}{\pi}\right)(C^* - C_0)
  \sqrt{\frac{1}{r_0^2 - r^2}}\right] \rightarrow
  j = \frac{2nFD(C^* - C_0)}{\pi\sqrt{r_0^2 - r^2}} \tag{18}
\end{align}
We need one last step, because we are searching for the current. Since the current density depends on the radial coordinate, and the element of surface is $2\pi r\mathrm{d}r$, integrating yields
\begin{align}
  i &= \int j \, \mathrm{d}S \\
  &= \int_0^{r_0} \left(\frac{2nFD(C^* - C_0)}{\pi\sqrt{r_0^2 - r^2}}\right)
  \, (2\pi r\mathrm{d}r) \\
  &= 4nFD(C^* - C_0) \int_0^{r_0} \frac{r\mathrm{d}r}{\sqrt{r_0^2 - r^2}} \\
  &= 4nFD(C^* - C_0) [-\sqrt{r_0^2 - r^2}]_0^{r_0} \\
  &= 4nFD(C^* - C_0) (-0 + \sqrt{r_0^2}) \rightarrow
  \boxed{i = 4nFDr_0C^*} \tag{19}
\end{align}
Eq. (19) is the value of the steady-state current under thec conditions you are serarching. The applied potential is so high, that the concentration $C_0 \approx 0$. Thus, by only knowing the concentration of $\ce{O}$ far away from the electrode, its diffusion coefficient, and the radius of the UME you can calculate its steady-state current.

References
A discussion about this problem can be found in Chapter 5 of the following book:

A. J. Bard, L. Faulkner, and H. S. White, Electrochemical Methods: Fundamentals and Applications, John Wiley & Sons, New York, 2022.

Newman's paper is free to download here:

https://iopscience.iop.org/article/10.1149/1.2424003


text:
Can you clearly explain what osmotic pressure is?
There are two main questions I'm struggling to get my head around.

What exerts osmotic pressure? Is it the molecules of the solvent (water) or solute particles (sugar) ?
Where is it exerted? On the semipermeable membrane?

If we imagine the U-shaped tube scenario, does the osmotic pressure become zero when the the height of water in both tubes stop changing?

Review the guide How to ask and Asking FAQs to prevent clarification requests, objections,  downvoting or closure.

See this partial answer: https://chemistry.stackexchange.com/a/52080/81

and this answer https://chemistry.stackexchange.com/questions/167189/osmotic-pressure-definition-clarification/167286#167286

Osmotic pressure has two definitions. First it is a physical phenomena, namely the movement of water through a semipermeable membrane by osmosis to dilute a concentrated solution. It is also a numerical value, and it is the highest level attained by water having crossed the membrane by osmosis.

By definition osmotic pressure is the pressure that we need to exert on solution so that no osmosis occurs.
So osmotic pressure is applied by solution on piston.
Osmotic pressure is defined when both liquids have same height so it would be meaningless to define osmotic pressure when both liquids have different heights.

But, if both arem contain liquids of the same height and no external pressure, there is no osmotic pressure. Osmotic pressure is measured by looking at the different heiights.

@matt_black if I want to measure osmotic pressure of 5% solution and some solvent flows in solution arm then my solution would no longer be 5%. Measuring height and taking into account  the dilution is very tedious. So better way is to keep solution and solvent arm at same height and measuring how much pressure I need to keep it so.

@matt_black This is the more general answer. It accommodates the fact that solution composition changes with the dilution occurring due to the pressure difference if flow is allowed, and can therefore consider nonideal conditions.

@HarjotDhillon Yes, that might be more convenient, but your last sentence claims "Osmotic pressure is defined when both liquids have same height" which is not correct. You are talking about a convenient way to measure the pressure not its definition.

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Is 2,5-dimethylidenecyclopent-3-en-1-ide aromatic?
Ah yes, the age-old question.
I understand from previous posts on this website that aromaticity can't be defined with using four rules. But I was curious about this particular anion:

It certainly follows all the rules (6 π electrons, cyclic, planar, fully conjugated), but is it really aromatic?

Fully conjugated? This is cross-conjugated - no more aromatic than quinone.

Pendant pi bonding means you can't rely on the 4n+2 rule. Some species with pendant pi bonds are aromatic, but it is much harder to predict than just applying the 4n+2 rule which applies only to things that are exactly conjugated cycles.

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text:
Work from a spontaneous endothermic reaction?
Some reactions are spontaneous ($\Delta_\mathrm{r}G < 0$) although they are endothermic ($\Delta_\mathrm{r}H > 0$), e.g. the plastic packs that once cracked they get cold.
I saw that it is often this spontaneous endothermic reaction taking place here
$$\ce{NH4NO3(s) –>[H2O] NH4+(aq) + NO3-(aq)} \tag{1} $$
more exactly, because of  $\Delta_\mathrm{r}G^\circ = \Delta_\mathrm{r}H^\circ - T\Delta_\mathrm{r}S^\circ$ this reaction will only be spontaneous at high temperatures.
So my question is: given the above exergonic reaction starting in standard conditions and a high enough temperature, could I extract work from this reaction while it reaches equilibrium?
In other words, in $\Delta_\mathrm{r}G^\circ = \Delta_\mathrm{r}H^\circ - T\Delta_\mathrm{r}S^\circ$ only the term $T\Delta_\mathrm{r}S^\circ$ contributes to the Gibbs free energy. Can we get work with this part only, or is it simply an indicator of the direction of the reaction? If the answer is yes, then how? For example, could I somehow get an electrode potential in a suitable electrochemical cell? And if not, is $\Delta_\mathrm{r}G^\circ$ simply an indicator of the direction of the reaction, but no “real” energy?
Does it all has anything to do with the fact that if we create a cold spot in a hot place, there is going to be kinetic energy transferred to the system due to warm air filling the hole?  Or see in the Carnot heat engine, one can increase the efficiency by reducing the temperature of the cold bar…

The reaction of NaHCO3 in an acidic solution of HCl is endothermic and it produces a gas that can produce work.

Cool! I guess that answers my question. Thank you very much!

Even without volume work, the work can be done converting part of heat flow from neighbourhood to the system to work, e.g. by thermocouples or Stirling engine.

Thanks for your answers! This was very helpful.

@Buck Thorn, yes I did, its just that on my screen (chrome on an Imac)  $\circ$ is so small it is almost a dot, I cannot tell if its roman or italic or o or 0. I actually don't mind what it is just that it be bigger so that it looks sensible!

@porphyrin If it's any comfort  o or 0 are rendered almost identically on my system (0 is a tad bigger but lacks anisotropy which is somewhat bizarre). There are certainly many possible solutions but I find copy pasting into another editor as necessary to check suffices.


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Grade 12 chemistry question about Schrodinger's standing wave?
this is my first time ever using a website like this but I have something I'm curious about. My chemistry teacher showed this today (img below) and discussed how quantum of energy that hits the electron has to be an integer otherwise decay occurs. What is stopping this from happening more often? Why aren't things decaying a lot and what are the factors to this?

Good question, at the start of quantum mechanics Bohr had the same question, 'why have'n't all electrons in atoms crashed into the nucleus',  and to fit the experimental data he had to postulate that no decay can occur. This is a huge difference between quantum and classical behaviour. Quantum mechanics is based on postulates such as this- but amazingly it always works- no error has yet been found. 'This is the way' to quote the Mandalorian!

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.

prophyrin, do you know why no decay occurs howverr and why only a integer is consumed? could it be related to the compton effect?

No not related to Compton effect. It is a postulate, we accept it (and Quantum Theory) because by doing so we find that we can explain all experimental observations made so far.

I see that makes a lot more sense thank you, it seems you and maurices answers are very different though. He answered how an electron falling into the nucleus would not decay the atom. You state it's impossible, can you give me some clarification on whether you're right, he's right or you're both somehow right?

You should probably try to find an into to Quantum theory something that explains a bit of history and principles without much math. Try 'Introducing Quantum Theory' by J. McEvoy & Oscar Zarate, publ Totem books in the US/ Icon Books in the UK.

Do you know examples that use math? I find that even if I can't understand everything, if i get the general idea, it would help me better understand. I will definitely look into the book, thank you for all your help :)

There are lots with maths, most any undergraduate physics or physical chemistry textbook.

If an electron "falls" on the nucleus, it cannot produce anything. The electron has not enough mass to produce a neutron, as the mass of the neutron is bigger than the sum of the masses of electron + proton. It is even not stupid to state that in a hydrogen atom, the electron has already fallen on the nucleus.

Maurice, wouldn't the electron shielding from an electron right beside the nucleus have negative effects?


text:
In an unsaturated solution, is the product of ions still equal to Ksp?
Suppose I have a solution of a sparingly soluble compound, and consider its solubility in g/L. If I have less g/L of solution than that amount, it means the solution is unsaturated, but what happens in this case? Will all of the compound dissolve such that there is no solid left? And what happens to Ksp?
Thank you.

No. In this case the product of concentrations will be less than $K_{sp}$.
$K_{sp}$ denotes the product of the concentration of the solute that has dissolved when the solution is saturated.

A better way to say it is that equilibrium constants apply when a system is at equilibrium.

No.
In an unsaturated solution the product of concentration of ions gives ionic product(I).
Ksp is the product of maximum concentration of ions.
So, if I < Ksp : then there will be no solid left (unsaturated solution)
if I > Ksp : then the salt will start precipitating (super saturated solution)


text:
Why are Mo(NR2)3 complexes isolobal with the nitrogen atom in N2
In a Molybdenum complex of the form Mo(NR2)3, the Mo is in the oxidation state of +3, leaving it with 3 d electrons. When combined with an additional 2*3 electrons from the ligands, this leaves it with 9 electrons surrounding the metal. This is 9 less than the full 18, and so it could be said to have 9 frontier orbitals?
Why then, does is this complex considered isolobal with the nitrogen atom in N2, which has a full 8 valence electrons and so 0 frontier orbitals?

The $\ce{NR2}$ groups also act as pi donors with their formally "lone" electron pairs coupling to the molybdenum $4d$ and $5p$ orbitals (compare with the mesomeric effect of such groups in electrophilic aromatic substitution). Therefore the complex has fifteen electrons accessing the molybdenum valence orbitals. Under the threefold symmetry of the complex the three remaining vacant states split into an $a_1$ state and an $e$ pair, matching the symmetries of the bonding orbitals (apart from a center of inversion) in $\ce{N2}$.


text:
Alternate definition of joule Thomson coefficient
We know that Joule-Thomson coefficient is defined as
$$\mu_{jt}=\left(\frac{\partial T}{\partial P}\right)_H$$
Using the cyclic rule for partial derivatives it can be written as
$$\mu_{jt}=-\frac{\left(\frac{\partial H}{\partial P}\right)_T}{\left(\frac{\partial H}{\partial T}\right)_P}$$
Which is equivalent to
$$\mu_{jt}=-\frac{\left(\frac{\partial H}{\partial P}\right)_T}{C_p}$$
My question is can we write simmilarly
$$\mu_{jt}=-\frac{\left(\frac{\partial U}{\partial P}\right)_T}{C_v}$$


text:
Doubt regarding definition of enthalpy
We define enthalpy change as heat released at constant pressure but it is also defined for process where pressure varies. How do I understand this?

Please provide examples. The way to understand it is that H is always defined for a thermodynamic system, and $\Delta H$ for a change between two systems. It doesn't matter if they have the same p and T. But if they do and only reversible pV work is done (no other type of work) then the heat released is  equal to the enthalpy change.

It is not the definition but equality, that is true just conditionally, for reversible isobaric changes with zero non-volume work.
If the process is not isobaric or reversible or non-volume work is done then, generally, $Q \ne \Delta H$ for given two states.
The equality is true if and only if $W = - p \Delta V$.
By other words, how could be the change of any state variable defined by a path dependent variable?

@ManRow Here we have another example...

@MetalStorm Mentioning users in comments works (unless changed recently) only to users who previously explicitly participated in Q, A or comments. (I tried it once but I was informed about it by a mod.)

Oh, thanks. I have contacted the user. Didn't know about that, I thought it worked...

@MetalStorm An example which also supports $\Delta H = Q$ for all isobaric processes (between equilibrium states) regardless of reversibility, since we can always say isobaric $W = -P_{ext}\Delta V$ where $P_{ext}$ is constant and well-defined, which happens to be the case in all isobaric situations regardless of reversibility (barring non-PV work and other such cases).

@Manrow Be aware that from TD and state variables POV, isobaric process need not have constant pressure, just the initial and final pressure must be the same. Therefore, Q and W can vary with different paths of irreversible processes, but the sum Q + W is constant. For irreversible compressions p_ext > p_int, the opposite for irreversible expansion.

Enthalpy(H) is measure of heat contend in the system. It is a state function i.e. it depends on the initial and final state of the system.
We cannot measure the absolute value of enthalpy, we can only measure the change in enthalpy.

H is measure of energy of the system, modified by the pV term. It is not limited to thermal energy. Heat is not thermal energy, but its exchange. That is why we have $Q$, but not $\Delta Q$.


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Stability of n-Dodecyl-beta-D-maltoside in buffer solutions for assay experiments
I have a buffer system which contains multiple salts and in addition DTT and n-Dodecyl-beta-D-maltoside.

I understand that DTT is hydrolysing quite fast and this component must be used always fresh to keep it's reductive capacity. I am wondering however, what your experiences are with the stability of n-Dodecyl-beta-D-maltoside buffer solutions? The solid substance must be stored at -20 degrees. It is quite tedious to always weigh in all the buffer components because of the instability of some components. I had therefore the idea to maybe mix 1 l of all stable components and than take for example 50 ml and add the DTT as a solid substance (or liquid stock?) and maybe also the DDM surfactant if it is not long term stable. This would save time and resources, because the anorganic buffer salts are usually stable and the premix could also be sterilized in an autoclave. I know, that the concentration would be slightly shifted, because adding DTT and DDM (10 mM and 0,04 wt% in final solution for example to 50 ml of buffer pre mix would also slightly alter the volume of the solution, but I assume the error would be under 1%. Do you think this is feasible? What are your experiences with stability?
In terms of the stability I have also a different question: To bring one inhibitor in solution in buffer me tried to adjust the pH of the buffer a bit. When it got too basic, the solution led to white flaky precipitate. I would account this to the maltoside, which is not stable in solution in pH 9?

Are you wondering if you can make buffer stock  without DTT  and DDM and store this until needed? If so what else is in the stock? Seems this is more important. Why discuss DDM/DTT at all? BTW searched DDM and found "Stock solutions are stable for up to 3 months at 4°C." on the Sigma-Aldrich website

Other components are Tris HCl, MgCl2 and ZnCl2 but these are stable and sterilizable anyway. I am thinking about DTT and DDM to save money (DDM is relatively expansive in comparison to the other ingredients). So the only thing to add freshly would be DTT. Would you use it as solid or from a stock solution to save time?

DDM is nonionic so don't see why solubility would be very sensitive to pH. Sigma-Aldrich says: "This product is soluble in water (50 mg/ml), yielding a clear to very slightly hazy, colorless solution. Sonication may be required for complete solubilization." Cayman chemical says don't keep in solution for more than a day before use.

I would need to try this again with miliq, the DDM and just a bit of base. I observed this multiple times already, but not sure if this relates to the surfactant. I guess a valid approach would be to prepare the buffer premix concentrated 1.01x and than add 1/100 part in volume of DTT of 1M stock in miliq that was previously frozen from alliqots to achieve 10mM. Than add the DDM. Volume error would be limited to DDM than solely ans because of the concentration of 0.04wt% relatively small. Agree?

By batch is from avanti lipids, and neither on the website or the container they say anything about stability of wet solution...

Your mixing protocol seems correct. DDM might slowly hydrolyze, but I'd be surprised if it's much outside of high T or extremes of pH. I would worry more about the DTT stability since it is air sensitive. I also wonder if the metal solubility might not be sensitive to pH and Tris (the latter should help, high pH will precipitate magnesium hydroxide).

Ahh, didn't think of Mg precipitation :) Yes, this seems plausible and the appearance looks familiar.


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Can glycerol slow down the kinetics of an inhibitor going to a binding pocket of a target enzyme?
I am currently doing enzymatic inhibition assays and in my current setup I observe that a commercial inhibitor is showing inhibition as expected. Still, the inhibition is not as strong as the literature claims. Ic50 is ~20x higher.
I am wondering if the glycerol, ca. 8 percent from my protein stock may slow down the kinetics that determine the binding of the inhibitor to the protein?
Is this a known problem? If so, I could overcome this by longer pre incubation time? In the future I want do a complete buffer exchange from storage conditions to assay conditions anyway to eliminate glycerin and all other components that are not supposed to be in the final buffer by a spin column, to eliminate this possible source of deviation. Still, at this stage my interest was, if the assay will work in my lab at all as a proof of concept (which it does).

Is it 8% glycerol by volume or mass? Can you provide more detail about the volumes mixed, and perhaps about what's in the sample.

It is by volume. The sample is quite complex and contains buffer salts, a surfactant (DDM), DTT, two proteins, two substrates and one peptide. The two proteins are injected into the matrix without substrate and they contain 52 Percent glycerol each, their volume about 1 and 9 microliters working in 60 microliters in total on 384 well plate.

I am wondering if the glycerol, ca. 8 percent from my protein stock may slow down the kinetics that determine the binding of the inhibitor to the protein?

When you change the solvent, a lot of things change, including the kinetics of a reaction and the equilibrium constants of ligand binding. What the exact reason is for the discrepancy of your observations and the literature value is impossible to say without a full set of experiments. However, it is much easier to remove the glycerol and try to reproduce the conditions used in the literature.

Is this a known problem?

The viscosity of 8% glycerol is very different from that of water. You would expect everything to slow down (both inhibitor and substrate binding/dissociation). I don't think there is a general rule of thumb that inhibitors are less potent in the presence of glycerol. There is no obvious theoretical reason why pre-incubation would change the situation.

I did the same experiment without and with additional 10 min preincubation time. Before I almost did not see any inhibition at all but the increased time had a great effect. So, I would assume this is a plausible working hypothesis at the moment.

The result might be about macroscopic mixing. If your inhibitor is sitting in one corner of the vial and the enzyme and substrate in another, the inhibition will be minimal. Also, even with pre-incubation, you don't reproduce the literature values, so it is not the entire explanation. With the scarcity of data, there are multiple plausible working hypotheses. I'm assuming you don't want to run your assays in high glycerol in any case, so removing it is annoying but probably necessary.

I would think so, and I can visually see the smearing of the glycerol when the protein is injected to the wells. I mix every well with the pipette tip until the smearing disappears, but obviously I can't tell by visual inspection if all density and viscosity gradients are homogenized. Preincubation takes place on a plate shaker. I know when I mix the protein after concentration of the äkta fractions with glycerol it takes some minutes after the mixture is homogenized. Also, more importantly without inverting of the falcon tube there are two layers that keep separated. So, definitely a factor.

Reverse engineering the composition of your sample: here 8% is ambiguous, but assume it means the volume ratio of the original substances is 8:92.
Glycerol has density ~1.261 g/cm3 ($\pu{20 ^\circ C}$), and water ~1.00 g/cm3, so the implied glycerol content by mass is 9.9%, and the dynamic viscosity of the mixture can be computed with an online calculator as 1.29 cP ($\pu{20 ^\circ C}$). The viscosity of water is 1.00 cP ($\pu{20 ^\circ C}).$
A ~30% increase in viscosity will not have an extravagant effect on the diffusion coefficient. Assuming a Stokes model ($D = k_\mathrm{B}T/6\pi\eta a$) $D$ will be reduced by a factor of ~23%. The diffusion coefficient is related to the mean square displacement as $\left <r^2 \right > \propto \tau \times D$ so that the time required to diffuse some rms distance is $ \propto   \eta$ and will increase by ~30%. This is not going to be significant, or at least would not account for an order of magnitude or more difference in kinetics.
If the supplier ships the proteins in glycerol it is safe to assume that this is not going to have a negative effect on stability. If the explanation for the difference in the assay is due to the solvent and not, say, to faulty proteins, it might be due to specific interactions.
References
The online viscosity calculator references the following:

Cheng (2008) Ind. Eng. Chem. Res. 47 3285-3288.
Volk and Kähler (2018) Experiments in Fluids 59 75.

I expressed the proteins by myself. Glycerol is usually used to protect the protein from damaging during freezing and to stabilize specific conformations, so freezing without glycerol is often the problem, not having it.

If the proteins are supposed to be folded and you have enough you might check stability in various ways. If you already did this, next thought is what else are you doing to check the proteins are ok (functional)?

@raptorlane What are you measuring to determine IC50?

I do a flouresence activity assay. The product will react with a probe that will become fluorescently active.

I did nonlinear analysis now and the value is not that terrible far away from the literature, but the data points show some spreading. Actually, I can very well think of glycerol might be the problem. Not that it completely "destroys" diffusion, but the mixing will properly differ quite significant from well to well I would assume.

@raptorlane What is the spread in your assay? (is this a kinetic assay, that is, are you measuring a rate?) You can predict fairly exactly what effect you might have on viscosity due to variation in glycerol concentration. For instance, if one well has an effective glycerol concentration of 16%/volume, then viscosity = 1.7 cP resulting in a 31% decrease in diffusion "speed" (relative to no glycerol).

I am measuring the development of fluorescence over time. Here described: https://chemistry.stackexchange.com/questions/179997/calculate-ic50-from-fluorescence-kinetic To be more precise I compare the initial slope of the uninhibited protein that equals 100% activity to the slope at a specific inhibitor concentration to determine the inhibition. Variation means in this context for example that a higher inhibitor concentration is creating less inhibition or strong outliers in triplets. Or when the inhibition is increasing as an overall trend, than decreasing and than increasing again.

Ok, so you have within well (over time) and between well scatter. You might eventually apply some statistics to this (ANOVA), maybe if you think there are some real differences.

The scattering per well is ok. But triplets deviation is too high. But the general trend works, with increasing inhibitor concentration, the activity drops as expected. I account this as mentioned to mixing and pipetting error at the moment. I will redo the experiments after glycerin removal.


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Reaction of cyclohexene with HCO3H
Reaction of cyclohexene with a peroxy acid like mCPBA gives an epoxy acid according to Prilezhaev Reaction.
I think it should work with any per acid according to the Mechanism here. But HCO3H gives anti diols, as given in my textbook.
Is hydrolysis of epoxide going on?

$\ce{H2CO3}$ is not a peracid.

@Maurice  I think the OP is refering to performic acid as epoxidation reagent

@OscarLanzi  performic acid is moderately oil-soluble according to this paper  https://pubs.acs.org/doi/pdf/10.1021/acs.iecr.1c02212

According to this patent

Performic acid is unsatisfactory as an epoxidizing agent for cyclohexene because the reaction leads preferentially to hydroxy-formoxy cyclohexane with small amounts and thus very low yields of the desired cyclohexene oxide.

So the initial nucleophilic species is formate anion which will readily hydrolyse in work-up

If making cyclohexene oxide is desired, there are better methods than trying performic acid with its potential side reaction(s). The Wikipedia article suggests:

Industrially, using a heterogeneously catalyzed reaction, such as combining cyclohexene with oxygen on silver.

In the laboratory, using magnesium monoperoxyphthalate, $\ce{Mg^{2+}}\ce{[o-C6H4(CO2^-)(C(O)O2^-)]}$, as the active oxygen source.


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How are oxidation states of bioinorganic molybdenum/tungsten complexes determined?
Here a screenshot from our lecture on bioinorganic molybdenum complexes, dealing with the catalytic cycle of xanthine oxidase:

It is said that the only biologically relevant oxidation states of Mo are Mo(IV) and Mo(VI). That said, for the life of me, I cannot figure out why molecule 1 (or the other ones for that matter) in the catalytic cycle should have the oxidation state +VI.. I mean there are two oxo ligands and three thiol residues so, if the molecule is neutral it should be +VII, or am I missing something here?

Who says it's neutral? They clearly don't write charges at all in this picture.

There are charges

Doesn't write overall charges of the species, you can only tell using ox. states and the ones on sulfate.

Okay I see so the species would be charged aswell


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What are the resonating structures of toluene?
I initially made the resonating structures by first converting toluene into a toluene carbocation. But the resonating structures of a neutral molecule should be neutral. So what are the correct resonating structures of toluene?

The resonances in toluene should be roughly the same as in benzene, because it is hard to believe that no bond or a double bond may occur by resonance between the phenyl cycle and the methyl group, whatever the situation in the rest of the molecule.

The motion of electrons inside the benzene ring is a component of the resonance structures in toluene. Toluene contains a methyl group (-CH3) attached to the benzene ring. The benzene ring's pi electrons are uniformly distributed in the major resonance structure, resulting in an arrangement of alternating single and double bonds. As a result of the electron density delocalization, a more stable configuration is produced.
The double bonds are displayed around the benzene ring in various positions in the minor resonance structures. These structures however, add less to the overall picture of toluene and are therefore less significant. The electrons in the pi bonds are distributed throughout the whole benzene ring rather than being fixed in one place, which results in a more stable and even distribution of electron density.
Toluene is a combination of all these resonance structures. The molecule's nature can be described as a resonance hybrid, as the electronic structure is a combination of these forms.


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How accurate are potential energy curves taught in high schools? What does a "real" curve look like?
I am in high school taking AP Chemistry and I am wondering how accurate the curriculum is from a quantum-mechanical perspective--specifically, how accurate are potential energy curves?

The way chemistry is taught in high schools, we use Coulombic forces to explain phenomena such as ionization energies and intermolecular forces. Potential energy curves such as the one above seem to be assuming that atoms in a chemical bond are attracted by a purely inverse-square-law force and experience some sort of angular momentum. This would be completely plausible at a macroscopic scale such as a binary star system, but I'm confused as to how accurate this approximation would be at such a small scale. I've searched for a more rigorous alternative to potential energy curves but as far as I have seen it's just a superficial analogy designed to develop an elementary understanding of bonds.

Why is the Coulombic interaction not plausible at small scales?

Search for the Morse potential, it has more or less the shape you show, approximates the harmonic potential near to the bottom of the well. It is widely used to define potential energies and is a solution to the Schroedinger equation. Actual potentials are more complex  but have the same sort of shape.

buck thorn, while coulombic interaction is plausible at small scales, quantum mechanics is far favored at the level of, say, diatomic molecules, no?

If you want to have a go at your own calculation see the code here https://chemistry.stackexchange.com/questions/174913/software-to-draw-one-dimensional-pes-including-vibrational-energy-levels/174985#174985

@RaphaelEsquivel One still uses the Coulombic potential for each single particle in (nonrelativistic) quantum mechanics.

That is far beyond what I learned in high school!
Those look like pretty accurate potential energy curves for diatomic molecules (as far as their shape is concerned—I cannot speak for the numbers). If you look at almost any book or review article discussing spectroscopy of diatomic molecules, those are the sorts of potential energy surfaces (PESs) you will see (see, for example, this question).
The cool thing is that these PESs can be calculated theoretically or determined experimentally (the classic experiment for undergraduate chemistry students is to analyse the UV-vis spectrum of $\ce{I2}$ and plot the potential energy surface based on the vibrational transitions observed for $\ce{I2}$ in several electronic states (here is an example of the experiment). In fact, just to prove it, here is my experimental PES for when I did the $\ce{I2}$ lab:

The PES is not simply an inverse square rule, it is actually a result of something people cannot actually model accurately yet—anharmonicity. For a diatomic, the closest we can get to the PES is a variation on a Morse potential (as said by Buck Thorn), which looks like:
$$V(r) = D_e(1-e^{-\beta(r-r_e)})^2$$
where
$$\beta = \omega_e\sqrt{\frac{2\pi c \mu}{D_eh}}$$
where $r - r_e$ is the deviation from the equilibrium bond length, $D_e$ is the dissociation energy, $c$ is the speed of light, and $\mu = \frac{m_am_b}{m_a + m_b}$ is the reduced mass of the molecule. For polyatomic molecules, the equations become infinitely more complex.
In summary, you are not being misled by your teacher, they are just covering the initial concepts, which is likely all you need to know to pass your exams. To alleviate your concerns, I can say that the experimental PES above was derived from entirely quantum mechanical principles (observation of discrete vibrational and electronic transitions). And, while I am fuzzy on the maths, I suspect the angular momentum you are referring to in your question is called the vibrational angular momentum. If you want to look that up you can, but, again, it is probably not important for you to know about.

Angular momentum tends to be associated with rotational, not vibrational motions. There are rovibrational energy levels, however: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Rotational_Spectroscopy/Rovibrational_Spectroscopy But I agree that this is a more advanced topic than might be covered in HS.

@BuckThorn I know, I only really know about it in the context of rotational (or electronic) spectroscopy, but I am certain I've heard of 'vibrational angular momentum' somewhere as well. It looks like it is a thing, though, although I'm not good enough at the maths to know whether it is referring to the effect of rotation on the vibration, or the vibrations themselves: https://onlinelibrary.wiley.com/doi/10.1002/jcc.26762

I believe that VAM is a way of dealing with the QM equivalent of ro-vibrational coupling, or coriolis forces. The term VAM was foreign to me.

@BuckThorn that would make sense!

One approach to bridge classical and QM descriptions (consistent with the correspondence principle that these descriptions should merge as you change the energy) is to transform the classical potentials into operator form by substituting the classical variables describing position and momentum into equivalent QM operators. Beyond conceptual differences in interpreting the nature of the solutions provided by QM (quantization, delocalization, etc) and adding some details that account for the effect of relativity (spin), this approach leaves the classical form of the potentials mostly unaltered. In essence the position coordinates get little hats. The treatment of electromagnetic interactions, for instance by using a Coulombic potential, are very similar in the quantum realm and in the classical realm.
There are many aspects of "reality" brushed aside during introductory treatments, and sometimes they do involve the shape of the potential. For instance, atoms bonded in a diatomic molecule may be assumed to vibrate harmonically (like an ideal spring or pendulum, except one obeying quantum mechanics). That harmonic approximation however assumes that the potential between the atoms is quadratic (parabolic), whereas the reality is closer to another, since a quadratic potential does not allow atoms to dissociate (they are stuck in the parabola for all eternity). A better approximation is for instance a Morse potential.

authors: Vladimir F Героям слава


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Hydrogen sulfide in acetone?
In the lecture slides for Analytical Chemistry I am going through right now, for the qualitative analysis of metal cations, in the hydrogen sulfide separation group, there is something that wonders me.
It says to add hydrogen sulfide in acetone (to an acidic solution). I tried to find papers on the solubility of $\ce{H2S}$ in acetone, but found nothing appropriate. I also found no description of the procedure that would use acetone except this one. I then wondered if maybe what they mean is using acetone to produce thioacetone for safer storage, or maybe something like thioacetamide to produce the sulfide in situ? Or maybe they really meant acetate as acid buffer?
The $\ce{H2S}$ acetone is mentioned in two places in the cation separation. The lecture mostly it focuses on modern techniques like mass spectrometry, so I am wondering if they had a typo there. I might write the professor, but I already wrote them a lot of things so I am a bit hesitant.

Produce thioacetone for safer storage? I guess you dunno much about thioacetone.

I truly don't. Thanks for pointing that out! I just checked what I could have meant, and it is thioacetamide.

The solubility of both ammonia and hydrogen sulfide in various polar solvents is reported by Short et al1. The table below, from this reference,  gives values as volumes of gas per unit volume of solvent (L) and mole fraction at one atmosphere (x).

See also this answer, in which the solubility of hydrogen sulfide in water is reported greater than that of methane because of the polarizability of the large sulfur atom.
Reference

Irena Short, Ashok Sahgal, and Walter Hayduk (1983). "Solubility of ammonia and hydrogen sulfide in several polar solvents". J. Chem. Eng. Data 28, 1, 63–66.
https://doi.org/10.1021/je00031a019

Thank you a lot! Both for the reference and your other answer which gives a very helpful explanation!


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Synthesis Sec - Butyl Bromide from alcohol
The question is to make Sec-butyl Bromide from alcohol. From what i have read, 2° and 3° alkyl bromide and alkyl iodide cannot be formed with Conc H2SO4 as it will make it into alkene.
I have tried
P4 + 6Br2 = 4PBr3
But-2-ol + PBr3 = 3 2-bromo butane + H3PO3
Is there any other way?

A good way is to form a sulfonate of the alcohol (tosylate, mesylate, triflate etc) forming a good leaving group then react that with a source of bromide anion.
The standard way of forming a tosylate is with p-toluenesulfonyl chloride in the presence of a base, pyridine catalyses the reaction so is often used as the base. There are many ways to react the tosylate with a source of bromide. I prefer using phase-transfer catalysis in dichloromethane/aqeous potassium bromide with tetrabutylammonium bromide catalysis, this works for mesylates too.


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1,4 (conjugate) addition in unsaturated aldehydes and ketones
When carbonyl compounds react with grignards, they usually give a 1,2 addition (direct addition) product.
However, under certain conditions, they give conjugate addition (1,4 addition) product as the major product.
I tried to understand the reason behind this, but I couldn't find it anywhere. What are the conditions for such a reaction to take place, and what is the mechanism behind it?

Have a look here: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Conjugate_Addition_Reactions

@waylander Thanks a lot! The explanation was extremely lucid. Can you post this link as an answer so that I can accept it?

Borrowed from here

1,2 vs. 1,4 addition

Whether 1,2 or 1,4-addition occurs depends on multiple variables but mostly it is determined by the nature of the nucleophile. During the addition of a nucleophile there is a competition between 1,2 and 1,4 addition products. If the nucleophile is a strong base, such as Grignard reagents, both the 1,2 and 1,4 reactions are irreversible and therefore are under kinetic control. Since 1,2-additions to the carbonyl group are fast, we would expect to find a predominance of 1,2-products from these reactions.
If the nucleophile is a weak base, such as alcohols or amines, then the 1,2 addition is usually reversible. This means the competition between 1,2 and 1,4 addition is under thermodynamic control. In this case 1,4-addition dominates because the stable carbonyl group is retained.


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Multiplicity and signals of cis−[(Ph3P)2PtF2] in two NMR spectra
I am interested in the number of the signals and their multiplicities in the  $\ce{^{31}P\{^1H\}}$ and $\ce{^{19}F\{^1H}\}$ NMR spectra of $\ce{[\textit{cis-}(Ph3P)2PtF2]}$. It is given that $\ce{Pt}$ has $I=1/2$.
I believe that in both spectra we have a triplet signal. Furthermore, I wonder if in the $\ce{^{31}P\{^1H\}}$ spectrum the signal can be split by $\ce{F}$ and, similarly, if in the $\ce{^{19}F\{^1H}\}$ $\ce{NMR}$ spectrum the signal can be split by $\ce{P}$.


text:
Would ferrous ion give the flame test?
My question to ask is "Does $\ce{Fe^2+}$ ion give flame test?"
Background: Last month, I gave the JEE exam and a question was as follow:

Number of metal ions characterized by flame test among the following is ___:
$\ce{Sr^2+,Ba^2+,Ca^2+,Cu^2+,Zn^2+,Co^2+,Fe^2+}$

The answer provided by them was 4 which I'm sure is incorrect considering they only considered the first 4 ions given in the list but the answer should be 5 or 6
A videographic/reputed source would be appreciated.

Google is your friend - https://edu.rsc.org/download?ac=14876

Zinc, cobalt and iron salts do not give a color to the flame, whatever the method used, and whatever the chosen salt. Try to do it yourself ! It is easy. Anybody can do it. You may see a yellow color  in the flame. It is due to some sodium impurities, because the test for sodium is extremely sensitive. It is positive even with trace amounts of sodium compounds. Looking at the flame with a spectroscope may help you be convinced that the yellow color is the famous Na D line (and even a double line, sometimes difficult to show that it is double)

They do give a color in flame test: https://en.wikipedia.org/wiki/Flame_test. Can you kindly explain from where does the sodium impurity comes from?

@Nilay Ghosh. I have read the wikipedia reference. They say that all elements produce a flame test. I would be pleased to see how they have seen for example zinc producing a flame. Under which conditions ? And what is its spectrum ? With a usual Bunsen flame, zinc does not produce a colored flame. Try if you doubt. Furthermore this wikipedia text has mistakes. For example, they state that the violet potassium flame is invisible through cobalt glass. It is wrong. But it is true for the yellow sodium line. Sodium is not potassium.

@Nilay Ghosh. All substances have one or two ppm of some sodium impurity.


text:
Really though, why can't a equilibrium manifest in a open system?
What my teacher said today: "Equilibrium is not attainable in an open system because then there the concentrations of the species in the system are undergoing continuous change, thus the reverse and forward rates can never be equal"
Thought Experiment that contradicts my teacher:
However, consider the tank with an equal-sized hole at the bottom and top at which water is being added in at the same rate. The macroscopic amount of water in the tank is constant and the system is open. But it is an equilibirum!
So is my teacher wrong or what?

That is not an equilibrium, but a steady state. There would be no flow at equilibrium.

https://en.wikipedia.org/wiki/Steady_state Once the water is gone it is gone. That's irreversible, therefore not equilibrium.

But, IMHO, if an open system was a part of a bigger, closed system and if this closed system was in equilibrium, I suppose we could say the open system as its part is in equilibrium too. But it would be rather very specific case.

A (lossless) confined current in a superconductor might be considered an equilibrium system.

Semantics

@BuckThorn Would not it be rather a frozen non-equilibrium state? And it would be a closed system.

@Poutnik I refer to the electrons flowing in the conductor. The example is not perfect as there are surely radiative losses (if not resistive, for a perfect conductor). A cross-section would be an open system.

@Poutnik Actually, ideally there might not even be radiative losses: https://physics.stackexchange.com/questions/387498/a-question-related-to-energy-losses-in-superconductors. Scratch that statement about non-thermal electrons.

Listen to and discuss this with your teacher.


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Why do stable carbocations form majority of the products?
In my A-level chemistry course, one of the topics is on alkenes in which we have to learn that the major product of $\ce{H-\color{green}{X}}$ reactions with unsymmetrical alkenes is the one formed via the most stable intermediary carbocation (one with most alkyl groups attached). I want to know why it is the most stable carbocation that causes for the majority of the products and not the least stable since surely the carbocation that has the positive charge least equally distributed would cause for the most reactions as it would be the most polar.

The protonation of alkenes is a reversible process so the most energetically stable cation gets formed most. this reversibility is much faster than the capture of the cation by X-

similar: https://chemistry.stackexchange.com/questions/30893/why-are-tertiary-carbocations-the-most-reactive-if-theyre-already-stable?rq=1


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Why this problem happens on LC-MS chromatogram?
I used to analyse my compounds on LC-MSdevice with 95% water/5% acn +0,1 formic acid. My column is hypersilgold (C18). When I did my acquisition, I obtain this chromatogram sometimes, and it doesn't depend of my sample, because when I inject again the same sample, it's OK.
Generally, when this problem happens, it disappears after 5-10 injection in a row.

Can you give us any information about the origin or the nature of "your compounds". Is it a biological extract ?

It's for non-targeted screening, m/z beetween 50 and 700 m/z. Indeed, it was biological extract.

Give information about your compound. Do you use a gradient? When it disappears after multiple injections, the problem can be related to impurities that are included in the mobile phase and that accumulate during ideal state.  During gradient run, these impurities may be mobilized.Try, if the background will gradually disappear with continues blanks without any sample injections.

I use gradient. Actually, this background disappears after few injections no matter what the matrix is. So, the result is the same if i use blank or sample, after few injections the background will gradually disappear.


text:
What is the pKa of choline chloride?
I am looking for the pKa of choline chloride, which should be higher than 7 (around 10?), but I cannot find any reference value.

According to the European Chemical Agency you can conclude, ignoring most of the lengthy and somewhat strange analysis and focusing on the choline group, that pKa =10.6 at 25 °C (predicted "using the OASIS Consensus").

Here a $p\mathrm{K_a}$ of 11.2 is reported, making the cation a slightly weaker acid than hydrogen cyanide, but pretty strong for an alcohol.

We have an ionic compound. We would expect the cation to have tge same pKa regardless of the counterion.

Yet tge pKa of the anion in your example remains the same. The choline cation will behave similarly.


text:
Confusion regarding the relation between Gibbs free energy and reaction quotient
According to the Wikipedia page on the Nernst's equation, the change in Gibbs free energy is
$$ \Delta G = \Delta G^\circ + kT\ln Q \tag{1} $$
But, according to many other sources and also to my book it is given as
$$ \Delta G = \Delta G^\circ + RT\ln Q \tag{2} $$
I am confused whether to use $k$ or $R$, and also Eq. (2) doesn't make sense dimensionally ($RT$ has units of $\pu{J mol-1}$, but Gibbs free energy just has units of $\pu{J}$).

You know that $\pu{R = N_ok}$, where $\pu{N_o }$ is Avogadro number. So both formula are correct.

That is the confusion . Why don't we write R/N in the 2nd equation? Why only R?

There is no common nomenclature to distinguish molar quantities (J/mol) and extensive quantities (J). Sometime, bold typeface is used, or $\Delta_r G$. Usually, you have to infer from the context. The wikipedia article has both versions, without switching the symbol, which is confusing and not a good idea.

It's just slightly "sloppy" notation where the same symbol is used to represent different quantities.
You can understand the 1st equation as giving you the change in Gibbs free energy per molecule (unitless), hence the unit of $\Delta G$ and $\Delta G^\circ$ is J.
However, the 2nd equation is about the molar change in Gibbs free energy, ie. the change per mole of substance, hence the unit of $\Delta G$ and $\Delta G^\circ$ is J/mol. You could write the 2nd equation more "precisely" by including a subscript m, which is used for representing molar quantities:
$$\Delta G_{\rm m} = \Delta G^\circ_{\rm m} + RT \ln Q.$$

The section in the wikipedia article starts with "Quantities here are given per molecule, not per mole, and so Boltzmann constant k and the electron charge e are used instead of the gas constant R and Faraday's constant F.", so it is sloppy reading rather than writing.

@Karsten, I don't think I completely understand your comment :) You commented it yourself that there is no common nomenclature – and that is the point I am trying to make with my "sloppy" notation explanation. I havent focused on this particular Wiki article, but on the usage in general ...


text:
Which position do retro aldol reactions occur in (comparing glucose and fructose)?
Regarding glycolysis, my text states:

What is the biochemical rationale for the isomerization of glucose 6-phosphate to fructose 6-phosphate and its subsequent phosphorylation to form fructose 1,6-bisphosphate? Had the aldol cleavage taken place in the aldose glucose, a two-carbon and a four-carbon fragment would have resulted. Two different metabolic pathways, one to process the two-carbon fragment and one for the four-carbon fragment, would have been required to extract energy.

I am trying to understand this and I found this diagram. It seems that glucose undergoes a [4+2] retro aldol reactions but fructose undergoes [3+3]. Why is this? What determines which position a retro aldol decomposition occurs in? Is there any particular reason glucose could not undergo [3+3] decomposition?


text:
Is acetophenone a saturated compound?
In many of the sites in online, acetophenone is mentioned as an unsaturated compound because of the presence of C=C bond in the phenyl group. But, many of my teachers say that it is a saturated compound because it does not decolourise Bromine water and pink colour of Bayer's reagent. Which is true and why?

You need better teachers! Acetophenone, as you correctly say, is unsaturated due to the benzene ring

@Waylander But, why Acetophenone does not decolourise neither Bromine water nor the pink colour of Bayer's reagent?

Because of the extra stability of the conjugated double bonds compared to an isolated double bond. Acetophenone does react with bromine under basic conditions in a bromoform reaction

@Waylander Thanks a lot

This case explains a problem with using "saturated" versus "unsaturated". The boundary in real chemistry is not as distinct as the words would seem to imply because the behavior of the compound may depend on the conditions to which the compound is exposed.
In acetophenone both the carbonyl group and the phenyl group can undergo addition, and thus  show unsaturation, under appropriate conditions. Carbonyl groups do not readily add hydrogen as $\ce{H2}$, but they can be hydrogenated through reaction with a hydride-source nucleophile such as $\ce{NaBH4}$ or "alkanated" with alkyl nucleophiles such as a Grignard reagent. As for the phenyl group, aromatic rings are resistant to addition, but for example benzene can be hydrogenated at elevated temperatures on platinum catalysts. For instance, Lin and Vannice[1] study this catalytic reaction between 44 and 83°C.
Reference

S.D. Lin, M.A. Vannice, "Hydrogenation of Aromatic Hydrocarbons over Supported Pt Catalysts .I. Benzene Hydrogenation", Journal of Catalysis,
Volume 143, Issue 2, 1993, Pages 539-553, ISSN 0021-9517, https://doi.org/10.1006/jcat.1993.1297.


text:
Is there any uv-vis sensor that installs on a pipe like other measurement tools like flowmeter or temperature sensor?
I am looking for an industrial UV-vis sensor that is installed on a pipe like other measurement instruments like temperature and flow transmitters.
I researched this subject and I found one model but not too many.
It looks like it's not common, and it's quite expensive.
I would like to hear from you if you have any experience.

Can I find a cheaper alternative sensor to analyze what's inside a pipe?


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Charge of a complex ion
Consider $\ce{[Cr(H2O)6]^3+}$, the charge on the Chromium before the water molecules react is $3+$, clearly each water ligand contributes one lone pair, meaning the formal charge should be $3-6=-3$ (each bonding pair is identical to having an extra singular electron in the shell) , but the charge is actually $+3$. What am I doing wrong?

Following your way of looking at it if each water gives one electron to the chromium what is the resulting charge on each water? And so what does that make the total charge on the complex?

Charge is +1 which gives net charge of +3, so it works. What is the standard way of working out charge on a complex ion?

Well, you just pointed out that formal charge isn't terribly useful. In complexes ox. state or "valency" is written for central atoms. Overall charge is fine, but actual partial charges are non-trivial to find.

You are doing nothing wrong, but you are confusing the formal charge with the oxidation state. The formal charge is the charge that an atom would have if all the bonding electrons were equally shared between the atoms. The oxidation state is the hypothetical charge that an atom would have if all the bonds were ionic, meaning that the more electronegative atom takes all the bonding electrons.
In the [Cr(H2O)6]3+ complex, the formal charge of Cr is -3, as you correctly calculated. This means that Cr has three more electrons than its valence shell can accommodate. However, the oxidation state of Cr is +3, because we assume that all the bonds are ionic and that the oxygen atoms of the water molecules take all the bonding electrons. This means that Cr has lost three electrons from its valence shell.
The formal charge and the oxidation state are not the same thing, and they do not have to be equal. The formal charge is a way of keeping track of the electrons in a covalent bond, while the oxidation state is a way of assigning charges based on electronegativity differences. The formal charge is useful for predicting the reactivity and stability of a molecule, while the oxidation state is useful for balancing redox reactions and describing the electron transfer processes.
I hope this explanation was clear and helpful. If you want to learn more about the formal charge and the oxidation state, you can check out the web links that I have provided. Thank you for your attention.
Reference:
(1) Chemistry of Chromium - Chemistry LibreTexts. https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_%28Inorganic_Chemistry%29/Descriptive_Chemistry/Elements_Organized_by_Block/3_d-Block_Elements/Group_06%3A_Transition_Metals/Chemistry_of_Chromium/Chemistry_of_Chromium.
(2) Introduction to Crystal Field Theory - Chemistry LibreTexts. https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_%28Inorganic_Chemistry%29/Crystal_Field_Theory/Introduction_to_Crystal_Field_Theory.
(3) What is the oxidation state of chromium in Cr(H2O)6 (3+)? - Socratic. https://socratic.org/questions/what-is-the-oxidation-state-of-chromium-in-cr-h2o-6-3.


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Why does the nitrito ligand have a higher Δo than water in the spectrochemical series?
I collected some data to determine the value of $\Delta_\text{o}$ for a some transition metal complexes (attached) and I'm very confused as to why $\Delta_\text{o}$ for the $\ce{ONO-}$ ligand is higher than $\ce{H2O}$. I would think that the $\ce{ONO-}$ is a better $\pi$ donor than $\ce{H2O}$ so $\text{t}_{2\text{g}}$ would be raised higher in energy and therefore $\Delta_\text{o}$ would be lower.
Similarly, I am confused as to why $\ce{ONO-}$ is not a $\pi$ acceptor and does not have a similar value for $\Delta_\text{o}$ as $\ce{NO2-}$ or why $\ce{NH3}$ is higher than $\ce{ONO-}$?
It is most likely that my thinking is wrong but I would really appreciate it if someone could explain this to me!

I personally wouldn't think that a 1kJ mol-1 difference has any significance within the accuracy expected from the model being applied


text:
Testing whether an unknown powder is wheat gluten (at home)?
Last week, I ordered wheat gluten to make seitan. I received some powder and attempted a recipe, which mixes 100g of water with 100g of gluten to form a sticky dough. However in my case, I only got a thick fluid. Even after adding 50g more gluten, I was unable to form pieces from the result. Almost all recipes for seitan use a one-to-one mixture (or even more water), so I have the assumption, that the powder is actually not gluten. Of course I did contact the company and they sent a replacement. However, the powder seems to be exactly the same.
Now I am wondering whether its possible to properly check whether the powder is gluten or not. There are some test kits online, but they are very costly, probably because they are intended to find even small traces of gluten for allergen tests. Maybe there is a cheap option or even something I could attempt at home with easy-to-get chemicals?
I am aware of the biuret test to check for proteins, but the company does sell lots of supplements and powdered "super foods", so I guess a protein test would be positive for half their stock.
Any other ideas?

Try a different supplier, presumably. I can't imagine it's very expensive or difficult to obtain? Probably simpler than checking gluten content. You could however check whether it is starch: https://en.wikipedia.org/wiki/Iodine%E2%80%93starch_test

Sure, but I am still interested in finding out whether this stuff is gluten at all, or something entirely different. The latter would also be quite interesting regarding food safety.

Also compare similar question at https://www.reddit.com/r/chemistry/comments/1aqj46f/testing_whether_an_unknown_powder_is_wheat_gluten/

@Alduno As for the seitan as such, https://cooking.stackexchange.com/ could be an audience suitable, because there is a tag seitan (as are gluten-free and vital-wheat-gluten, too).

You may want to add details about what kind of budget you have. What you have done (a dissolution test) is about the simplest test you can run, and fairly effective for detecting some forms of adulteration, for instance in honey.

The biuret works fine for an isolated (pure) protein in solution; but the test is not specific to a particular protein. Grains/flour of wheat are a complex mix of chemicals which would require some digest and separation before the analysis per se can be started, e.g. a antigen/antibody ELISA test (example). Perhaps one of the regional interests groups AOECS compiles can hint to a lab they trust and collaborate. I doubt the task can be done «at home».

This might sound silly, but depending on what country you are in, the product might contain composition information (protein, carbs, simple sugars, salt etc). I assume that since you haven't posted this, that such information is not available, but perhaps you don't trust the product and therefore did not include it. If it is available, how about posting it?

Mix with rice flour and see if you get a reasonably sticky dough? A chemical analysis would start with electrophoresis I guess. DOSY would work, too. Nothing you can do at home.

I think you should approach any engineering or chemistry related institutions nearby your location, if they have facility of "mass spectrometry" then they can compare the sample against pure gluten to tell how much of gluten is there. This is the most effective way I know.
There's a another way which is to use Mark–Houwink equation, to determine the molecular weight of polymer (ie your sample) and compare it against molecular weight of gluten.You can easily get procedure 'to determine the molecular weight of a high molecular weight polymer by viscosity measurement'. It requires only a Oswald Viscometer(easily available) and a stop watch. I had used this method to determine molecular weight of polystyrene and results were satisfactory. The only problem is that I am not sure if you can get the data required(value of a constant a for solution of gluten in some solvent) to plug in MHS equation to get molecular weight. I you can then this is the most practical way to solve the problem.
If you can't do both of these, then comparing results of preliminary tests like checking solubility in different solvents or flame test (is flame sooty or not) for given sample against pure gluten may work, but it's not an fool proof way and there's a possibility that you get incorrect result.

Not bad, but gluten is a wild mixture of (mostly) proteins, varying between sources.


text:
Determine equilibrium stability transition of a reaction (function of temp and concentrations)
I am trying to calculate and plot the stability of $\ce{Fe3O4 - FeS2}$ based on the chemical equation:
$$ \ce{Fe3O4(s) + 6SO2(g) + 16CO(g) -> 3FeS2 (s) + 16CO2(g)} \tag{1} $$
I want to plot the transition from $\ce{Fe3O4}$ to $\ce{FeS2}$, in a plot of the concentration/fugacity of $\ce{SO2(g)}$ as a function of temperature.
I have available the following thermochemical data $a_i$ for each individual molecule's formation reaction,
$$ \ln(K) = \frac{a_1}{T} + a_2 \ln T + a_3 + a_4 T + a_5T^2 \tag{2} $$
where the $a$'s are the fit coefficients I have available for each of the species above, and $\ln(K)$ is
$$ \ln\overline{K}_i(T) = -\frac{\Delta_\mathrm{r}G_i(T)}{RT} \tag{3} $$
How can I produce an image like this? (Complete paper, not relevant for calculation)
Edit: I'll be online and working on this all day, so if you have comments/further questions please comment them and I'll respond as soon as possible!
for reference, the fit coefficients for each species
species_data = {
    "Fe3O4": {"a1": 4.0276612308074633e+05, "a2": -9.4661523582307066e+00, "a3": -6.9502410801855703e+01, "a4": 1.5085130307144656e-02, "a5": -1.9444510158593611e-06},
    "FeS2": {"a1": 1.3624235215287484e+05, "a2":-5.2169286395159462e+00 , "a3": -2.4724164596373562e+01, "a4":7.2462091584085467e-03, "a5": -1.2248091281633105e-06},
    "SO2": {"a1": 1.3624235215287484e+05, "a2":  -5.2169286395159462e+00, "a3": -2.4724164596373562e+01, "a4":  7.2462091584085467e-03, "a5": -1.2248091281633105e-06 },
    "CO": {"a1": 1.2899777785630804e+05, "a2": -1.7549835812545211e+00 , "a3": -3.1625806804795502e+00  , "a4": 4.1336204683783961e-04 , "a5":  -2.3579962985989574e-08},
    "CO2": {"a1": 1.9284060511825030e+05, "a2": -2.7725941201250066e+00 , "a3": -1.4374508628203330e+01, "a4": 9.5527761423722102e-04, "a5":  -5.3273135070236439e-08},
}

I may be able to help, but something is suspicious. Firstly, could you send the original research paper from which your image comes from? Secondly, the concept of equilibrium constant $K$ refers to a chemical reaction, not to a particular species. There is no such thing as an 'equilibrium constant for oxygen', but we could have a $K$ for that reaction you posted. Thus, what does those $K$'s refer to? To the formation reaction of each species?

@MetalStorm thanks for responding! You are right, the constant K refers to the formation reaction of each species (I'll edit the post for clarity)

First we apply Gibbs' phase rule so we know the number of intensive variables we need to fix so that the equilibrium state of the system is established. We have

Five different species, i.e. $N = 5$.
Three phases, one gas phase and two solid phases of magnetite and iron(ii) disulfide. Thus, $\pi = 3$.
One chemical reaction, i.e. $r = 1 $.

So we have
\begin{align}
  F &= 2 - \pi + N - r \\
    &= 2 - 3 + 5 - 1 \rightarrow F = 3 \tag{1}
\end{align}
and we need to specify $3$ intensive variables.

Since the activity of a pure solid phase is one, the law of mass action yields
\begin{align}
  K &= \frac{a_\ce{FeS2}^{3}\hat{f}_\ce{CO2}^{16}}
            {a_\ce{Fe3O4}\hat{f}_\ce{SO2}^{6}\hat{f}_\ce{CO}^{16}} \\
  &= \frac{\hat{f}_\ce{CO2}^{16}}{\hat{f}_\ce{SO2}^{6}\hat{f}_\ce{CO}^{16}} \tag{2}
  \\
\end{align}
The fugacity of a species in the gas phase is a function of the pressure, temperature, and composition. I will assume that the gas phase behaves ideally. If this is not valid, just say in the comments and we can take into account the non-ideality. For the ideal case, the fugacity is equal to the partial pressure
\begin{align}
  K &=
  \dfrac{\left(\dfrac{y_\ce{CO2}p}{p^\circ}\right)^{16}}
        {\left(\dfrac{y_\ce{SO2}p}{p^\circ}\right)^{6}\
         \left(\dfrac{y_\ce{CO}p}{p^\circ}\right)^{16}} \\
  &= \left(\frac{p^\circ}{p}\right)^6
           \frac{y_\ce{CO2}^{16}}{y_\ce{SO2}^6y_\ce{CO}^{16}} \tag{3}
\end{align}
where $p^\circ = \pu{1 bar}$ is the standard pressure. Finally, we combine Eq. (3) with the relation $ y_\ce{CO2} = 1 - y_\ce{CO} - y_\ce{SO2}$
\begin{equation}
  \boxed{K(T)
   = \left(\frac{p^\circ}{p}\right)^6
   \frac{(1 - y_\ce{CO} - y_\ce{SO2})^{16}}{y_\ce{SO2}^6y_\ce{CO}^{16}}} \tag{4}
\end{equation}
Eq. (4) is a non-linear equation with $4$ unknowns which are $p$, $T$, $y_\ce{CO}$ and $y_\ce{SO2}$. Once we set three of them, we can solve for the remaining one, which agrees with Gibbs' phase rule. The simplest algorithm in my opinion would be:

Set a temperature $T$. This allows the calculation of $K$ because you have the Gibbs' free energy of formation for each chemical species.
Set the pressure $p$ of interest.
Set one molar fraction.
Solve the equation. You will have now the x value which is $T$, and the y value can be calculated with $\hat{f}_\ce{SO2} = y_\ce{SO2}(p/p^\circ)$.

thanks for your response!! I agree/understand up to Eq. (4). I am then confused by the algorithm you've laid out. Not sure what you mean by set a temperature, if i'm trying to vary T in the end. So unsure how to calculate K. I know I can get Delta G for the reaction, by using the parametrisation for theGibbs' free energy of formation for each chemical species, and subtracting the products'-reactants'.  Unsure how to proceed from there.

Set a temperature = Give a value to the temperature, e.g. $T = \pu{350 K}$. Once you solve the equation, then you do the same for $T = \pu{360 K}$, and so forth. If you solve the equation for $10$ temperatures, you will have $10$ points for the curve.

ok cool that makes sense. Similar Q about the y_CO, since that is an unknown and I'd be solving for y_SO2?

@learning_physics Yes, that's right.

The problem is that if you mix $\ce{Fe3O4, SO2, CO}$, you will never obtain $\ce{FeS2}$, whatever the temperature and the pressure. The activation energy is too high.

@Maurice Hello Maurice. I answered directly the question, but I didn't bother to check its characteristics. Why do you say the activation energy is too high? By just looking at the reaction?

@Metal Storm. First a reaction between three products is rare. Second I know that $\ce{Fe3O4}$ is used to produce metallic iron in the steel industry by heating it tumor than 1000°C. I also know that $\ce{FeS2}$ can not be made in lab as it need high temperature and high pressure to be made in geological times, out of any oxygen around. And here, there is plenty of oxygen atoms, contained in oxides of course. But it must be extremely difficult (nearly impossible) to synthesize $\ce{FeS2}$ out of such a mixture, even at high temperature. This is why the activation energy must be high.

@Maurice I completely agree with all your observations. In fact, the curves that the OP shows in the paper support your claim (even though they refer to other reactions). The fugacity of oxygen is extremely low (and $\ce{O2}$ is a product in those reactions), which means that in equilibrium the reaction barely took place. However, the OP asked how to calculate the curve, so I didn't bother on the feasibility. I will try to edit the post and add your remarks, when  I do some calculations. Thanks.


text:
Favorability of a reaction with positive entropy change and negative enthalpy change
Suppose a reaction had a positve entropy change and a negative enthalpy change.
According to Le Chatelier's principle, the reaction should be less favorable at higher temperatures, as heat is released. The same is applicable when looking at a Van't Hoff plot:

However, the entropy change is also positive, and as
$\Delta G=\Delta H-T\Delta S$
one would expect the gibbs energy to decrease at higher temperatures, therefore increasing the reaction's equilibrium constant:
$K=e^{\frac{-G^{0}}{RT}}$
So what is going on here, why do those two statements contradict each other?

Using your first approach, van't Hoff :

$$
\begin{eqnarray}
&& \ln \frac{k_2}{k_1} = -\frac{∆H^\circ}{R} (\frac{1}{T_2} - \frac{1}{T_1})
\end{eqnarray}
$$
For an exothermic reaction $\Delta H <0$
Now , Let  $T_2>T_1$   therefore   $K_2<K_1$ i.e. equilibrium constant decreases.

Using your second approach,

$$
\begin{eqnarray}
\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ
\end{eqnarray}
$$
$$
\begin{eqnarray}
\Delta G = \Delta H - T \Delta S
\end{eqnarray}
$$
$$
\begin{eqnarray}
\Delta G = \Delta G^\circ + RT \ln(Q)
\end{eqnarray}
$$
At equilibrium $ \Delta G = 0 $ and $Q= K_{eq}$  and  hence
$$ \begin{eqnarray}
&& K=e^{\frac{-\Delta G^\circ}{RT}} =e^{\frac{-(\Delta H^\circ - T \Delta S^\circ)}{RT}}
\end{eqnarray} $$
For Temperature $T_1$ and Equilibrium Constant $K_1$
$$
\begin{eqnarray}
K_1=e^{\frac{-(\Delta H^\circ - T_1 \Delta S^\circ)}{RT_1}}
\end{eqnarray}
$$
Similarly, For Temperature $T_1$ and Equilibrium Constant $K_1$
$$
\begin{eqnarray}
K_2=e^{\frac{-(\Delta H^\circ - T_2 \Delta S^\circ)}{RT_2}}
\end{eqnarray}
$$
Dividing the above two equations, you get
$$
\begin{eqnarray}
&& \frac{k_2}{k_1} = e^{-\frac{∆H^\circ}{R} (\frac{1}{T_2} - \frac{1}{T_1})}
\end{eqnarray}
$$
The Van't Hoff Equation !!! And it would give the same result as before.
I didn't conclude Dependence of $K_{eq}$ and T from $\Delta G^\circ$ because if a reaction is exothermic, it only tells that $\Delta H^\circ <0$ and doesn't tell anything about $\Delta G^\circ$ ..

This answer does not answer the question. The van't Hoff equation describes the change of an equilibrium constant with temperature and is not in question. The "reaction" questioned with different signs for enthalpy and entropy will not reach equilibrium as long as those conditions persist.

A reaction with a negative enthalpy change must lose heat with a decrease in entropy for equilibrium there is an equal reverse reaction that absorbs heat with an increase in entropy.  A reaction that both releases heat [exothermic] and increases entropy  cannot reach equilibrium. It must be either a composite reaction or a steady state with continuous energy input .

@jimchmst I think heat has correctly pointed it out jim in the comment to your answer ,,,I'll give an example ,,,Take Delta H∘ to be -7kJ ,,,, Delta S∘ to be + 1.314 J ,,, T = 1000 K ,,R = 8.314 J mol⁻¹K⁻¹ ,, and Keq to be e (2.71828) ,,,,As in this example Delta G is zero ,,,,Delta H∘ is negative and Delta S∘ is positive,,, and the reaction can reach equilibrium

The question that must be answered is whether those conditions can exist in reality. the primary consideration is that at equilibrium DeltaG EQUALS ZERO  =deltaH -Tdelta S delta H  equals TDelta S  your example Delta H= -7  ;  TDelta S = Some plus value THEY DO NOT HAVE the SAME SIGN THEY are NOT EQUAL. Also the standard values do not apply to equilibtium They are the energy changes from a defined standard state to equilibrium.

Equilibrium means reaction rates are the same in each direction that is only possible if the energy and entropy changes are opposing at equilibrium.

Equilibrium must be approachable from either direction. The other direction is not from the standard states.

Yeah you're correct ...but now I'm really tired of this question jim ...after my exams ...I'll sit one day and rectify my error as soon as I get time ...but not now ...thanks

van't Hoff equation is the formal form of Le Chatelier's principle, as in they are the same. A main feature of both these concepts is that they are local properties of temparature and are applicable only in small temparature ranges when enthalpy is independent of temperature.
From van't Hoff equation:
$$
\begin{eqnarray}
&& \frac{\textrm{d}}{\textrm{d}T}\ln{K}=\frac{1}{K}\frac{\textrm{d}K}{\textrm{d}T}=\frac{\Delta H^{\circ}}{RT^{2}} \\\\
&& \frac{\textrm{d}K}{\textrm{d}T}=\left(\frac{K}{RT^{2}}\right)\Delta H^{\circ}
\end{eqnarray}
$$
As you can see, the sign of the derivative depends on enthaply change. The caveat here is that, enthalpy itself is temparature dependent and ultimately the sign on the derivative depends on the temparature. Hence, Le Chatelier's principle is local.
Temparature dependence of enthalpy is given by Kirchhoff's Law: $\Delta H=C_{\textrm{p}}\Delta T$

In this answer I will try to resolve your apparent contradiction first by showing arithmetically why only 1 outcome is true and in the next part try to convince your intuition why this is the case.
Arithmetic proof
As alluded to already in other responses, there is no contradiction between the Van't Hoff equation and the equation relating $K$ and $\Delta G$. Given:
\begin{align}
K &= \exp\left({-\Delta G^° \over RT} \right). \label{5-7}\tag 1\\
\end{align}
Assuming that $\Delta H^\circ$ and $\Delta S^\circ$ are invariant with temperature, we can differentiate the expression with respect to T to obtain:
\begin{align}
  \frac{\mathrm{d}K}{\mathrm{d}T} &=
    \frac{\Delta H^\circ\exp\left(
      \frac{\Delta S^°}{R}-\frac{\Delta H^°}{RT}
    \right)}{RT^2}
  \label{four}\tag 2\\
\end{align}
Since $RT^2 >0$ and $\exp(\frac{\Delta S^°}{R}-\frac{\Delta H^°}{RT}) >0$, the sign of $\frac{\mathrm{d}K}{\mathrm{d}T}$ is only dependent on $\Delta H^\circ$. For exothermic reactions, increasing T will lead to decreasing $K$. Similarly, for endothermic reactions, increasing T will lead to increasing $K$. In other words, Le Chatelier's principle and the Van't Hoff equation are correct.
Intuition
This result might contradict our intuition. For a reaction with $\Delta H^\circ < 0$ and $\Delta S^\circ > 0$, we would expect, given the maths, that $K$ decreases with increasing temperature. However, as you mentioned, the negative $\Delta S^\circ$ term would mean $\Delta G^\circ$ decreases with increasing temperature. This gives rise to an apparent contradiction: at increasing temperatures, $G^\circ$ of products decreases relative to reactants, meaning (at standard state) products are more stable than reactants, yet the position of equilibrium is shifting toward the reactants.
The key to this conundrum lies in the $RT$ term and more specifically what temperature actually means. Notice that in equation (1), changing $T$ does not only change $\Delta G^\circ$ but also $RT$.
Let us consider just enthalpy first
For the sake of discussion let us consider the following: a closed box filled with inert gas with some molecules of A(g) which can isomerise to B(g):
$$\ce{A(g) <=> B(g)}$$
This system has constant total energy $E_{total}$. We consider each molecule of A or B as a smaller subsystem with energy $e_A$ and $e_B$ respectively. The energy of the rest of the system with respect to that one molecule is therefore $E_{total} - e$.
Consider that the isomerisation reaction is exothermic such that when $\ce{A(g) -> B(g)}$, heat is released such that $e$ decreases and $E_{total} - e$ increases. Consider also, for now, that the reaction involves no change in entropy ($\Delta S^\circ = 0$).
Hopefully it is intuitive that at equilibrium at some temperature $T_1$, more molecules of B than A exist, therefore $K > 1$. You can think of it this way: the B isomer contains stronger intramolecular bonds than A is therefore more stable than A. This is also why the reaction is exothermic.
Here is another more useful way to appreciate why $K > 1$. We consider the system as a microcanonical ensemble, such that every microstate occurs with equal probability. Let us rewrite the equation for $K$. Since $\Delta S^\circ = 0$:
\begin{align}
K &= \exp\left({-\Delta H^° \over RT} \right)
\end{align}
\begin{align}
K &= \frac{\exp\left({-H^°_B \over RT} \right)}
{\exp\left({-H^°_A \over RT} \right)}
\end{align}
\begin{align}
K \propto \frac{\exp\left({-H^°_B \over k_BT} \right)}
{\exp\left({-H^°_A \over k_BT} \right)} \tag{3}
\end{align}
Now, by considering the equation for Boltzmann's distribution,
\begin{align}
K \propto \frac{\text{Probability(molecule exists as B isomer)}}
{\text{Probability(molecule exists as A isomer)}}
\end{align}
\begin{align}
K \propto \frac{\Omega(E_{total} - e_B)}
{\Omega(E_{total} - e_A)}
\end{align}
Where $\Omega(E)$ is the number of microstates accessible to the macrostate with energy $E$.
From the re-expression we can appreciate that the equilibrium constant depends on the probability that the molecule exists as either A or B, which depends on the number of microstates available. Here is the key: because $e_A > e_B$, then $(E_{total} - e_A) < (E_{total} - e_B)$, such that $\Omega(E_{total} - e_A) < \Omega(E_{total} - e_B)$. Therefore, by virtue of the fact that the reaction is exothermic, $K>1$.
Now, there are 2 ways to appreciate that increasing the temperature of the system causes $K$ to decrease.
The simplest way is to appreciate that from equation (3), increasing T will cause both the numerator and denominator to approach 1, meaning $K$ decreases from $K > 0$ to $K=1$. But how does this work intuitively?
Let us consider the thermodynamic definition of temperature. For our system,
$$T = \frac{dE}{dS}$$
It is easier to consider the definition of temperature as:
$$\frac{1}{T} = \frac{dS}{dE}$$
Therefore, higher temperatures mean that the same increase in energy results in a smaller increase in entropy (in other words, a smaller increase in the number of possible microstates). In other words, a small decrease in energy results in lower entropic cost. Applying this to our system, we see that at higher and higher temperatures, $E_{total}$ increases, such that $(E_{total} - e_B) \approx (E_{total} - e_A) \approx E_{total}$. In other words, as temperature increases, the relative entropic cost of the molecule existing as A or B becomes negligible, to the extent that
$$\text{Probability(molecule exists as B isomer)} \approx \text{Probability(molecule exists as A isomer)}$$
such that $K=1$.
The above is generalizable to endothermic reactions. In the case where $\Delta H^\circ > 0$, at equilibrium at some temperature T more A than B molecules will be present such that $K<1$. As temperature increases, relative entropic cost of converting between A and B decreases such that $K=1$
Now let us consider entropy as well
Without considering entropy, the apparent contradiction does not exist. Therefore, let us consider that the reaction
$$\ce{A(g) -> B(g)}$$
Involves a positive entropy change. For the sake of simplicity, consider that this entropy change does not affect the entropy of the rest of the system. For example, the A isomer forms a closed, rigid ring while the B isomer forms a long linear molecule. (It is not always the case that $\Delta S^\circ$ of a reaction does not affect entropy of the system. Consider kosmotropic ions that reorganise solvent molecules into an ordered clathrate cage.)
How does this affect what have concluded so far for enthalpy? The answer: It doesn't change anything. As we increase temperature, the relative energies of A and B are still defined by the $\Delta H^\circ$. Therefore, increasing temperature means the relative entropic cost of converting between A and B diminishes, such that $K$ decreases.
The difference here is that $\Delta S^\circ$ is not affected by temperature in the same way. The positive entropy change that accompanies $\ce{A(g) -> B(g)}$, and therefore the increased number of possible microstates accessible to B isomers relative to A isomers remains the same at higher temperatures. Therefore, as $T\rightarrow\infty$, $K$ will no longer approach $K=1$, but will instead approach some value above 1, proportional to the magnitude of entropy increase.
Indeed, this is the reason why:
$$\lim_{T \to \infty} K
= lim_{T \to \infty} \exp(\frac{T\Delta S^\circ - \Delta H^\circ}{RT})
= \exp(\frac{\Delta S^\circ}{R})$$
The limit of $K$ as $T \rightarrow \infty$ is only dependent on $\Delta S^\circ$ and not on $\Delta H^\circ$.
Conclusion
In other words, we should, in a way, consider enthalpy and entropy as separately contributing to the equilibrium constant. Instead of saying: exothermicity means K decreases with increasing temperature and endothermicity means K increases with increasing temperature, we should see it as: at finite temperatures, exothermicity pushes the equilibrium constant toward $K>1$ and endothermicity pushes the equilibrium constant toward $K<1$, but with increasing temperature the influence of $\Delta H^\circ$ on $K$ diminishes. On the other hand, the influence of $\Delta S^\circ$ on the equilibrium constant does not wane with increasing temperature.
The contradiction is resolved by considering that the equilibrium constant is not solely dependent on the relative free energies of the reactants and products, but also on the entropy of the rest of the system, which is dependent on the temperature.
Of course, the above is highly simplified compared to real-life chemical reactions. Many assumptions were taken, but the goal was to help you appreciate, intuitively, why there is in fact no contradiction.

The equilibrium constant and standard free energy of a reaction describe the free energy change from the Standard states of Reactants AND Products to equilibrium. Explicit in this definition is that the reaction can reach equilibrium.
At equilibrium delta G = Zero  = delta H -Tdelta S. Therefore: delta H = Tdelta S. A system to reach equilibrium must have the same sign for delta H and delta S. When the signs are different the reaction must be composite or possibly a steady state but not equilibrium.  Therefore, the concept of equilibrium constant does not apply.
A system with a negative deltaH and a positive entropy change might be described as a system with a positive feedback.

This is incorrect. For most chemical reactions we have to consider $\Delta_\mathrm{r} G = \Delta_\mathrm{r} H^\circ - T\Delta_\mathrm{r} S^\circ+ RT\ln Q$. The $RT\ln Q$ term denotes the entropy of mixing. Whatever the combination of $\Delta H$ and $\Delta S$ signs, equilibrium can always be achieved because the $ \Delta_\mathrm{r} G^\circ$ term can always be balanced by the $RT\ln Q$ term. In other words, the instantaneous change in free energy converting from products to reactants is balanced by the need to maximise entropy of mixing.

You would exclude the $RT\ln Q$ term, as in your answer, only in reactions such as solid to liquid phase changes. In such a reaction, then it is indeed the case that at equilibrium $\Delta_\mathrm{r} H^\circ = T\Delta_\mathrm{r} S^\circ$. Solid to liquid phase changes are always endothermic and positive entropy change.

If your answer were true, you would run into all sorts of problems. Consider, for example, the dissociation of any weak acid which has opposite signs for $\Delta H^\circ$ and $\Delta S^\circ$. (see some examples here: https://en.wikipedia.org/wiki/Acid_dissociation_constant#Thermodynamics). If it was true that such reactions had no equilibrium, it would be impossible to set up buffer solutions with these acids since adding equal amounts of acid and conjugate base would, according to your "positive feedback" cause one species to completely convert into the other.

Thanks for reading! Now think things thru. First Standard free energies are an artifact so we can make tables of energies. At equilibrium delta G =zero, not deltaG[0] =zero. For a reaction to proceed to equilibrium heat must be added or removed; at constant T and P Delta H =Q delta S = Q/T Both have the same sign. Standard free energies are   from equilibrium to a defined standard state . True equilibria are from actual conditions. and can be approached from either direction to equilibrium. In one direction the reaction is driven by removal of energy, the other by increase of entropy.

For an exothermic reaction the obligatory increase in entropy is the entropy of the universe; for the endothermic reaction the increase in entropy is in the system. In either case there is a net increase in entropy. An interesting aspect is that equilibrium must be approachable from either direction. This means that the path to equilibrium is spontaneous from both directions. This is only possible if H and S have the same sign.

I think it all comes down to how much the change in temperature affects the numerator and the denominator of the exponent in the equation for equilibrium constant. Afterall temperature is "hidden" in ΔG and directly present in RT expression. Increase of temperature causes decrease of ΔG and because for given example it is negative (negative ΔH and positive ΔS) so when decreasing it gets more negative and the minus sign makes the numerator larger, but on the other hand increase of the temperature maked the denominator greather as well.
Now it would make sense to consider how strong is the influence of temperature change on the ΔG value. In the reasoning below I will neglect influence of the temperature on ΔH and ΔS.
We have the equation:
ΔG = ΔH − TΔS
I would say that this influence of temperature is determined by value of ΔS.
Just out of curiosity I took a look in thermodynamic tables and I noticed that for examples given there ΔH values (in J/mol) are three to four orders of magnitude greater than ΔS values (in J/(mol K)).
Let's think of an extreme example:
Let's say we have a reaction that has ΔH = −100 (kJ/mol) and ΔS = 100 (J/(mol K)). Then for two temperatures T1 = 300 K and T2 = 600 K we have:
ΔG1 = −130000 (J/mol) and ΔG2 = −160000 (J/mol)
Looks like for doubling the denominator increase of numerator would be less than two times. Equlibrium constants values that we would got would be:
K1 = 4.32 * 10^(22) and K2 = 8.51 * 10^(13)
So equilibrium constant got smaller with increase of the temperature.

G=-RTlnKeq =H-TS; H is negative and T positive; increasing T with no change in H and S makes G more negative and Keq larger. The combination of negative deltaH and positive delta S cannot reach equilibrium.

When I do test calculations Keq gets smaller with increase of T for positive S and negative H (when totally neglecting their temperature dependence). It works like that for small and big temperature changes and different H and S combinations. Looks to me that it can even be seen from the plot on the top of this question: when T is increased then (1/T) gets smaller and with (1/T) getting smaller lnKeq decreases. So in other words with increase of T we have decrease of lnKeq. If lnKeq decrease this means that Keq decrease.

G = H-TS;  G = -RTlnKeq if S were positive increasing T makes G more negative causing an increase in Keq. This situation would cause a positive feedback loop with eventual catastrophe.

I would like to avoid going in circles. You can take a calculator and see for yourself that this is true: when using -RTlnKeq = ΔH∘ - TΔS∘ for negative ΔH∘ and positive ΔS∘ increase of T leads to decrease of Keq. Looking at your answer for this question I've noticed something that needs clarification. I mean that you, essentially, wrote that at equilibirium 0 = ΔH∘ - TΔS∘. And that is not true (except maybe for some specific examples). In general: ΔG = ΔG∘ + RTlnQ At equilibrium ΔG = 0 and Q = Keq and we get ΔG∘ = -RTlnKeq If you think some other theory applies - please give me the reference.

At equilibrium DELTA G = ZERO not DELTAG[0]  This means that to reach equilibrium deltaH and delta S must have the same sign. If deltaH is negative and delta S is positive the reaction will be increasing in only one direction  with increasing T, a positive feedback. A metaphor is putting more fuel on the fire a definite positive entropy event. A negative deltaH combined with a positive delta S makes delta G more negative with increasing T therefore making Keq larger. Thanks for reading my answer but please read what I said.

In your answer you wrote: "At equilibrium Delta G = Zero; Delta H - TDelta S = Zero; so Delta H =TDelta S." If by "Delta H" you mean ΔH and not ΔH∘; by "Delta S" you mean ΔS and not ΔS∘ -then it looks ok for me, but from the whole body of your answer and the discussion we have it seems to me that you meant standard values. At quilibrium when ΔG = 0 then ΔH = TΔS but these are not standard values (not ΔH∘ and not ΔS∘). For standard values we have ΔG∘ = ΔH∘ - TΔS∘ and ΔG∘ = -RTlnKeq. Equilibrium constant is calculated from equations taking into account standard values of thermodynamic functions.

Equilibrium constants define the energy change from the standard states to equilibrium. My answer does mention Keqs or standard states. My answer states that to reach equilibrium deltaH and delta S must be the same sign. If they are not the reaction is either composite or the situation is steady state .  Stand values are the free energy change from and to the standard states and equilibrium at the standard T and P. If a reaction cannot reach equilibrium it cannot have a Keq.

At equilibrium Delta G = Zero; Delta H - TDelta S = Zero; so Delta H =TDelta S. This means if a reaction can reach equilibrium the change in enthalpy and the change in entropy have the same sign and are in opposition to each other. This means that the reaction in one direction is energy driven and in the other entropy driven. My interpretation when the energy and entropy signs are opposite is that the reactions are probably inadequately described and are composite or steady state rather than capable of attaining equilibrium.
The process of a reaction with a negative DeltaH and a Positive Delta S means that the reaction CANNOT reach equilibrium and the concept of equilibrium constant is meaningless.


text:
Chlorine monoxide formation
In World War 1, the chlorine gas was handled by some soldiers by damping their hankerchiefs into sodium carbonate solution. The article chlorine monoxide suggests that the reaction release chlorine monoxide at 20-30 degree Celcius. I wonder how is the gas released at this temperature since the concentration of sodium carbonate was not given.
From this paper, it says:

Under the optimal conditions of reaction, sodium carbonate, containing approximately 10% of water, is allowed to fall slowly down a reaction tower and contacted with an upward flow of chlorine diluted to about 25% in a gas such as air, nitrogen, or carbon dioxide. The humidity in the reaction zone must be maintained at around 10%, either by saturating the incoming gas with water or by injecting steam into the reactor at regular intervals. The temperature of the reaction zone is preferably maintained at about room temperature (20-30 ᵒC), but may be raised to as high as 200 ᵒC.

So, I figure that under small amount of water, chlorine gas could form HClO via
Cl2 + H2O -> HCl + HClO
My question is:

What is the mechanism of Cl2O formation?
How does the concentration of sodium carbonate and temperature would affect the formation of Cl2O compound?

$\ce{Cl2O}$ is simply formed by dehydration of $\ce{HClO}$ according to : $$\ce{2 HClO -> Cl2O + H2O}$$ $\ce{Cl2O}$ can be considered as an anhydride of the acid $\ce{HClO}$.
$\ce{Na2CO3}$ is used to push the reaction of chlorine on water to the right-hand-side, as $\ce{Na2CO3}$ destroys $\ce{HCl}$ as soon as produced, which prevents back-reaction to occur. The equations are : $$\ce{Cl2 + H2O <=> HCl + HClO}$$
$$\ce{Na2CO3 + 2 HCl -> 2 NaCl + H2O + CO2}$$

A simple and good answer. But, assuming the excess of carbonate, I would rather write $\ce{Na2CO3 + HCl ->  NaHCO3 + NaCl}$.

@Poutnik.OK. Thank you. I know and I have hesitated to show "your" reaction, with the formation of $\ce{NaHCO3}$

Hehe, I have hesitated to edit your answer, I let it often on the question author if he/wants to use it.

So, regardless of the concentration of Na2CO3, chlorine monoxide would always be produced, wouldn't it?

@Seiji. I am not a specialist of this chemistry. Apparently the concentration of $\ce{Na2CO3}$ is not crucial. Its total amount is more important. It must be sufficient to destroy all $\ce{HCl}$ produced. But for example, I do not know if and how $\ce{Cl2O}$ reacts with $\ce{Na2CO3}$. Sorry !

@Seiji: Perhaps the gist is that (effectively) the salt NaClO stays in solution (or is bound to the hygroscopic Na2CO3)?