text:
Why would a metal form an ionic compound with a nonmetal from a lower period?
Take sodium iodide for example.  I understand both elements can obtain a full outer shell via the electron transfer.  But doesn't the electron move from a lower-energy orbital ($3s$ in sodium) to a higher-energy one ($5p$ in iodine)?  So how could that transfer be energetically favorable?  Yet it is exothermic, with enthalpy of formation of $\Delta H_f = -287\ \text{kJ/mol}$.
I don't understand how having a full outer shell inherently lowers the energy of the system.  I guess that means it's not just about the energy of the electron that is transferred -- it must be that the presence of that electron lowers the energies of the other electrons in the valence shell as well.  Maybe this is partially about pairing of up/down spins, but wouldn't that only matter for filling one orbital, not the full shell?  What's the deal?

5p of some atom is higher in energy that 3s of the same atom. With different atoms, not necessarily so.

@IvanNeretin  Oh okay -- so what if I ask the reverse question -- why would cesium chloride formation be exothermic when the electron is going from $6s$ in cesium (many protons, hence lower energy) to $3p$ in chlorine (fewer protons, higher energy)?

Generally, forming 2 isolated ions from 2 isolated atoms is not thermodynamically favoured, as ionization energies are higher than electron affinities.
But, ion formation is thermodynamically favoured, if the above is backed up by releasing of energy due ion pairing, ionic lattice formation or ion solvation.

I don't understand how having a full outer shell inherently lowers the energy of the system.

This depends on which exact systems are being compared. For your example, the ions with two full valence shells, Na+ and I–, need to be compared to the atoms Na and I where the valence shell in Na has one electron and the valence shell in I is one electron away from being full. Energy changes are about the total energy after an event minus the total energy before the event. You can't just think about the position of that one electron like you can in hydrogen-like atoms.
Some quantum chemistry considerations can be taught to general chemistry students with hand-waving arguments like "noble gas electron configurations are low energy configurations" as long as the reason for those arguments is also taught ("this is due to the Schrödinger model of the atom.") More advanced undergraduate students and graduate students might learn that approximate methods like Hartree-Fock for many-electron atoms are needed instead of finding actual solutions to the Schrödinger equation, like what's possible for hydrogen-like atoms with one electron.
General chemistry students should recognize that the process you're describing requires multiple steps other than the electron transfer. The electron transfer is described by the ionization energy of a sodium atom (496 kJ/mol) and the electron affinity of an iodine atom (-295 kJ/mol). Those two things happening together in the absence of other processes actually would not be energetically favorable. But the enthalpy of formation of sodium iodide also involves converting solid iodine to gaseous I2, breaking the bond in I2 to get individual I atoms, converting solid sodium to gaseous sodium to get individual Na atoms, then applying the ionization and electron affinity values to create two ions before finally considering the lattice energy of letting these gaseous ions come together to form solid NaI (–700 kJ/mol). The value you wrote of –287 kJ/mol for the enthalpy of formation of NaI should be the sum of all those values when Hess's Law is applied.

Essentially, metals will form ionic compounds with non-metals of a lower period because the energy released in the formation of the bond is vastly greater than the energy required to promote an electron.
Reactions which cause an increase in entropy (I.e. kinetic instability) or a decrease in enthalpy (I.e. Thermodynamic instability) will be favourable and will naturally occur.
In this case, only considering enthalpy is particularly relevant.
So a decrease in enthalpy will occur when the energy of the products is less than the energy of the reactants, and if this is the case, the reaction will be favourable and occur naturally.
It is important to note that as we’re talking about an ionic bond, e.g. NaI, only one electron from the sodium atom is given to the iodide atom.
So you’re correct, that promoting this electron requires energy (or is an endothermic process in itself) because energy levels are quantised, and so the electron must be promoted to a higher level to be a part of the iodine ion.
However, the energy released by the formation of the ionic bond is vastly greater than the energy required to promote the electron, and so the net change of the reaction will be that the energy released is greater than the energy supplied, and so the reaction is exothermic, leads to a decrease in enthalpy, and so is favourable.
The presence of the additional electron does not decrease the energy of the other valence electrons as they are quantised and do not change their energy level.
The concept of bond formation releasing energy can be conceptually difficult to grasp, so it is best to think of bond formation as the opposite of bond-breaking, which it inherently is. And since it will obviously require energy to be supplied in order to break a bond, by extension, forming a bond releases energy.


text:
Why does 2,3 dichlorobutane have 2, and not 3, stereoisomers
I understand how and that the R,S and R,R stereoisomers are different.
But I don’t see why the R,R and S,S compounds are stereoisomers and not meso compounds with each other.

Online it says that these two compoundsare different stereoisomers, but I don’t see why they are not identical.
I went onto a 3D molecule visualiser website, and still cannot see how they are different.

It does say that the visualiser website is not always accurate, but I can’t see why these molecules are not identical.
Furthermore, what specific type of stereoisomers are these two structures. Would it be correct to say they’re a member of the set of ‘conformers and rotamers’, which itself is a subset of the set of diastereomers, or are they something else.

Probably a good way to see this is with a Newman projection. Imagine looking at the R,R isomer molecule down the middle carbon-carbon bond axis. Then you have one of the chiral carbons (the proximal atom) facing you and the substituents $\ce{Cl, CH3, H}$ will be arranged around it in clockwise order*. The other (distal) chiral carbon along the viewing axis will have the same substituents in counterclockwise order. If you turn the molecule around to switch the locations of the chiral carbons, you still get the clockwise-front and counterclockwise-back outcome; the mirror image, corresponding to the S,S-isomer, is not superposable on the R,R molecule we have described.
With the meso (R,S) arrangement, two mirror images can be superposed: the Newman projection looking one way has $\ce{Cl, CH3, H}$ in clockwise order at both the proximal and distal locations, but turning the molecule around renders both orders counterclockwise.
Thus our molecule has a pair of enantiomers and one meso isomer. This is a total of three counting the pair of mirror images separately.
*The ordering of substituents ranked 1,3,4 when the rank-2 substituent is directed to the back matches the 1,2,3 order when 4 is directed to the back.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
What is the difference between exciting an electron to a higher bound state and removing it completely (ionizing the atom)?
We are reviewing quantum mechanics and orbitals in my class and I'm confused about the energies to move an electron to an excited state vs. remove it from an atom.
Even though the electron on a H atom occupies the 1s orbital in ground state, it can be excited to the 2s orbital, 2p, etc. and each transition has an amount of energy it requires. We also learned that it takes -13.6eV to remove that electron from the H atom entirely.
If all the higher orbital states are still possible for the electron in the H atom (and the last electron in a polyelectronic atom?), at what point is the electron removed and not just excited to higher and higher energy states? If it's removed when it no longer interacts with the nucleus, does that mean there is a finite number of energy states/orbitals an electron can occupy before it's not interacting anymore? I have a feeling this might be really simple and it's just not "clicking" yet.

We do not really understand what you are looking for. The number of energy states an electron can occupy is infinite. But high energy states are more and more close together, just before the electron is not interacting any more with the nucleus. So what is the problem ?

See also https://en.wikipedia.org/wiki/Rydberg_atom

I’m voting to close this question because it is more a physics question than a chemistry question

Let's take the simplest atom (hydrogen) with the simplest model (Bohr model). The energy of the electron with a principal quantum number $n$ is:
$$ E = \frac{\pu{-13.6 eV }}{ n^2}$$
So the lowest (most tightly bound) electron has an energy of -13.6 eV. This is relative to the unbound electron (which defines a potential energy of zero). So if you excite with more than 13.6 eV, the electron will no longer be bound (the excess energy is "used" as kinetic energy). If you excite with less, it will still be bound. Notice that when you go to higher and higher $n$, the expression for the energy approaches zero.
Here is a diagram that illustrates this:

Source: https://chem.libretexts.org/Courses/Lumen_Learning/Book%3A_General_Chemistry__Lecture_and_Lab_(Lumen)/05%3A_Electronic_Structure_and_Periodic_Properties_of_Elements/5.03%3A_The_Bohr_Model

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How can a molecule with a double bond on one side have a plane of symmetry, and thus, be achiral?

Does norbornene have a plane of symmetry? Is norbornene achiral or chiral?

I think norbornene is chiral since there is a double bond on the only one side. Thus, there is no plane of symmetry. The answer in the book is that it is achiral and has a plane of symmetry. How so?

The plane is not where you think it is.

This site   https://molecule-viewer.com/    has many examples that you can test finding mirror planes and axes etc.

I think this is a case of picture being worth a thousand words. Arguably the easiest way to visualize the plane of symmetry for nonbornene (mirror plane m) is to redraw the molecule in 2D:

Alternatively, one can load the 3D structure in Jmol and make sure the plane of symmetry is here for explicit hydrogen atoms too (feel free to check out my other answer as to how to visualize symmetry elements with Jmol):

Imagine a plane perpendicular to the double bond that cuts the double bond in half. That plane also runs through the carbon atom represented as the vertex at the top of the structure as drawn and runs halfway through the single bond on the other side of the molecule. That's the plane of symmetry for the molecule, there is a half coming out of the computer screen and an identical half going into the computer screen, and the presence of that plane means the molecule is achiral. There are two carbon atoms with four different substituents, but those two stereocenters are part of the larger system of symmetry, so there would be a single meso compound.
If there were a double bond on both sides then there would be two planes of symmetry. The second plane would divide the molecule into a left and right half.


text:
How does the partial pressure of oxygen relate to its concentration in water?
I've read that the partial pressure of oxygen in water will be the same as the partial pressure of oxygen in the atmosphere, but that the saturation of oxygen in water is dependent on factors like salinity and temperature. If the partial pressure of oxygen in water is equal to its partial pressure in the atmosphere, how is it possible that the concentration of dissolved oxygen in water can vary?

Partial pressure in chemistry is only strictly defined for a gas in a mixture of gases. If a chemist were asked for the partial pressure of oxygen dissolved in water, I think the answer would be, "Huh? If it's dissolved in water then it's not in the gas phase any more, so it doesn't really have a partial pressure." Chemistry describes the linear relationship between the partial pressure of a gas above a liquid and the dissolved concentration (not partial pressure) of the gas dissolved in the liquid at equilibrium as Henry's Law.
Other fields, especially biology and medicine, have used the convention for a long time that the partial pressure of a gas dissolved in a liquid can be defined by imagining a liquid/gas interface at equilibrium and then observing the partial pressure of the gas in the gas phase. That's a likely explanation for why "the partial pressure of oxygen in water will be the same as the partial pressure of oxygen in the atmosphere." Under equilibrium conditions, this is true by definition according to this unpleasant and potentially confusing definition used in those other fields.
However, the physical reality is that Henry's law constant varies with temperature and salinity, and equilibrium conditions also often don't exist in the most interesting systems. That's why, for the same partial pressure of oxygen in the gas phase, the concentration of dissolved oxygen in water can vary.


text:
How to calculate composition of alcohol/water mixture to achieve given concentration of alcohol vapors in air for calibrating a DIY breathalyzer?
I made a breathalyzer using a MiCS-5524 gas sensor. I would like to calibrate it so it can be close to accurate. e2v technologies — MiCS Application Note 4, Using MiCS Sensors for Alcohol Detection (PDF) shows a method of heating an alcohol/water solution and testing the air that has been bubbled through the mixture.
If I was to heat a liter of alcohol/water solution to 37 °C, what ratio of alcohol to water would I need to make the air have 10 ppm of alcohol? I don't have much experience with chemistry so please keep that in mind when you explain the solution.

Now this is a case where experimenting on humans does not sound inhumane. Why not use your body as a test input?

@IvanNeretin Human bodies are difficult to be adjusted to produce breath with 10 ppm of ethanol. :-)  Unless evaluated side by side by the DIY and commercial tester. Then the particular ethanol content is not important.

I've done that already to see if it detects alcohol on my breath but its still hard to calculate a fairly precise ppm by just drinking some booze and blowing after a half hour. I was hoping to get a more precise read out so it'll work on persons of different heights and weights.


text:
Reaction of lead with acetic acid and hydrogen peroxide solution
My textbook says lead reacts with acetic acid when there is enough oxygen to oxidize lead:

$$\ce{2 Pb + O2 + 4 CH3COOH -> 2 Pb(CH3COO)2 + 2 H2O}$$

So, I added lead into mixed solution of acetic acid and hydrogen peroxide.  Immediately bubbles began to emit from the surface of the lead piece. After several days the lead chunk has been completely dissolved and pale yellow precipitate was formed. This is not what I expect considering the reaction above. Can the bubbling be attributed to the oxidation of lead by water?
I used rice vinegar as a source of acetic acid. It might have been contaminated with ions like $\ce{Cl-}$.

You quote the reaction of lead with oxygen as oxidant, but you used hydrogen peroxide as oxidant. The two conditions give different outcomes.

This reaction most likely needs to be run in neat acetic acid to give the outcome you want. Rice vinegar is mostly water.

As lead can exist in more oxidation states, it may catalyze the parallel reaction of hydrogen peroxide decomposition to water and oxygen.

@Poutnik It seems you are right. I tried this experiment without vinegar this time and bubbles were emitted too. So, it is highly probable that lead acts as a catalyst.

You must check the composition of the bubbles. Lead will react in acid to give H2 and in the acidic solution the lead did react so there are possibly several concurrent reactions: reaction of lead with acid and water to give hydrogen, oxidation of lead by peroxide to give lead ion and water, decomposition of peroxide to give O2, in acid possible oxidation of Pb++ to PbO2 [barely possible but since nothing is controlled here who knows]. You might try just lead and acid while you are at it. Get some distilled white vinegar, almost pure acetic acid in water. Why are you interested in lead acetate

@jimchmst Thank you for your comment. Yeah, I will buy some pure vinegar soon. The reason I make lead acetate is that I am teaching chemistry as a tutor and I wanted to show the crystal of it to my student. I know it is toxic. I will be very careful.

Another source of more concentrated acetic acid is photo stop bath from a photo supply store.


text:
Why are ions in a mass spectrometer accelerated through electric plates?
I was reading an article on the AP Chemistry course in Khan Academy about mass spectrometry, which had this:

... ions are then accelerated through electric plates and subsequently deflected by a magnetic field

My question is, why do the ions need to be accelerated? Why can they not move without acceleration?

Ions need to reach the detector, not just dwell in ionization unit. Plus there are numerous analyzers such as ToF-MS that utilize velocity of accelerated ions.

@andselisk as porphyrin points out thermal velocity already guarantees non-dwelling; it's mass resolution that drives the need for a well-defined and carefully controlled acceleration. There are no mass analyzers that don't use velocity, be it free motion in a vacuum or diffusion in a gas, liquid, gel, or other medium.

People are not generally mind-readers, so it is difficult sometimes to understand where the confusion that leads to a question lies. In this case you ask "why can they not move without acceleration?" Ions in a gas do move without needing to accelerate them with an external potential source. The reason for the plates has to do with separating different ions. If you take a gas consisting of a mixture of different types of molecules, how to know which are there (without using chemical reactions)?

@BuckThorn so if I edit my question to remove the part where I ask why they can't move with acceleration, and just keep the first question, can my question be considered for reopening?

The use of plates (and the associated electric and magnetic fields) to accelerate and deflect charged particles is what characterizes mass spectrometry as a mass determination tool. Your question is a bit like asking, "why do we need gravity to measure the mass of an object with a balance?" There are in fact other ways to determine mass, it just happens that mass spec uses magnetic and electric fields in a clever way to provide us with a powerful tool. There are other ways of measuring average molecular mass of a gas, for instance by measuring effusion  https://en.wikipedia.org/wiki/Effusion

I will leave it to votes for the question to be opened.

The field helps capture almost all of the ions and accelerates them towards the detector. These ions initially have a statistical distribution of velocities (due to Boltzmann/thermal distribution plus inhomogeneities in the source) and this can lead to wide mass peaks. Accelerating the ions (with the same charge $z$) gives them approximately the same kinetic energy, independent of mass, as $KE=eV=(1/2)mv^2$ where $V$ is the acceleration voltage (typically a few thousand volts) and $v$ the velocity on leaving the accelerating plate. The heavier ions move more slowly than lighter ones and thus  become separated in proportion to $m/z$ while moving along the drift tube. The speed is now such that the initial distribution becomes a smaller fraction of the kinetic energy than it was and is therefore less important. When entering the magnetic field they are deflected by differing amounts (i.e. follow a different radius) and so strike the detector in different places. The change in radius $\delta r$, and so splitting of mass peaks, is proportional to $\sqrt{V}$.

Just a caveat: though perhaps not as prevalent in a typical analytical lab, there are also spherical electrostatic and electric quadrupole mass spectrometers that don't use magnetic fields; they disperse as $\Delta E / E$ rather than $\Delta p / p$ for magnetic dispersion.

@Uhoh and of course also time of flight


text:
Why does emulsions hinder the full extraction of a substance?
I've learend that in processing liquid-liquid extractions, emulsions can appear between two immiscible solvents.
For exmaple, to extract caffeine from green tea leaves;
The caffeine in water can be moved to ethyl acetate with liquid-liquid extraction. There were precautions for not making emulsions by gently shaking the two solvents together in the sepearatory funnel. Why is this important? Don't we re-extract from the extracted water again several times to maximize the extraction of a substance(in this case, caffeine)?
If the emulsions were in the water part while extracting, it doesn't matter because we can re-extract the water again with the seperatory funnel.

You may need much more solvent to achive the same extraction degree, when the solvent "competes" not only with water, but also with the emulgated solvent.

Emulsions are a problem because they make separation of the two imiscible liquids difficult. Imagine doing the extraction in a separatory funnel: you can't remove one layer cleanly if you have an emulsion.


text:
Solubility product of a sparingly soluble salt in presence of other dissolved salts
I read here that solubility product of a salt (say $X^+ Y^- $) remains constant even in presence of another salt with a common ion (say $Z^+ Y^-$).
Though I can intuitively understand that as the concentration of $Y^-$ has increased due to the added salt, and hence by le-Chatelier's principle concentration of $X^-$ has to decrease, is it possible to get a strong proof for the fact that the value of Ksp remains constant? (Earlier, I had thought that Ksp would decrease with addition of another salt, but later realized that the concentration of the common ion due to both the contributing salts is considered in the calculation of Ksp)
Also, would the solubility product of a salt remain constant in the presence of another dissolved salt with no ion common with it? (i.e, would the the Ksp value of a salt XY be the same in water and in a solution of another salt AB ?)

Presence of salts with the common ion decreases the salt solubility via the activity product.
Presence of salts without the common ion increases the salt solubility via the activity coefficients.

Solubility product as a thermodynamic constant is defined for ion thermodynamic activities, like:
$$K_\mathrm{sp} = a(\ce{X+})a(\ce{Y-})$$
which are defined as
$$\mu_i = \mu_i^{\circ} + RT\ln(a_i)$$
where $\mu$ is a chemical potential of a substance, being a partial derivative of the system Gibbs function per its molar amount ( at constant temperature pressure and content of other compounds):
$$\mu_i = \left(\frac{\partial G }{\partial n_i}\right)_{p,T,n_j,j \ne i}$$
If a the solid salt XY is in the equilibrium with its hydrated ions:
$$\ce{XY(s) <=> X+(aq) + Y-(aq)}$$
then
$$\mu(\ce{XY(s)}) = \mu(\ce{X+}) + \mu(\ce{Y-})= \\
\mu^{\circ}(\ce{X+}) + \mu^{\circ}(\ce{Y-}) +
RT(\ln{(a(\ce{X+})}) + \ln{(a(\ce{Y-})}))= \\
\mu^{\circ}(\ce{X+}) + \mu^{\circ}(\ce{Y-}) +  RT\ln{(a(\ce{X+})a(\ce{Y-}))}$$
$$\exp{\left(\frac{\mu{(\ce{XY(s)}) - \mu^{\circ}(\ce{X+}) - \mu^{\circ}(\ce{Y-})  }}{ RT}\right)} = K_\mathrm{sp} =a(\ce{X+})a(\ce{Y-})$$
Standard chemical potentials for ions are defined in such a way that ion activities are numerically equal to concentrations for the latter converging to infinitely diluted solution.
The ratio of the ion activity and its molar concentration is called activity coefficient: $a = \gamma \cdot \frac{c}{c^{\circ}}$
For enough diluted solutions, activity coefficients can be approximated by Debye–Hückel equation (for ionic strength below 0.1 M)

$$\ln(\gamma_i) =
 -\frac{z_i^2 q^2 \kappa}{8 \pi \varepsilon_r \varepsilon_0 k_\text{B} T} =
 -\frac{z_i^2 q^3 N^{1/2}_\text{A}}{4 \pi (\varepsilon_r \varepsilon_0 k_\text{B} T)^{3/2}} \sqrt{10^3\frac{I}{2}} =
 -A z_i^2 \sqrt{I}$$

with $A \approx 0.51$, using the ionic strength $I$:

$$I = \begin{matrix}\frac{1}{2}\end{matrix}\sum_{i=1}^{n} c_i z_i^{2}$$

Be more specific.

Solubility will change but the value of solubility product constant (Ksp) of a sparingly soluble salt will remain the same in the presence of other dissolved salts?

Ksp (activity-wise) will remain, if solutions are diluted enough and water activity can be considered constant. Otherwise, Ksp would change as well.

This is another example of everything you learned in HS, then in College, and then in Grad School; is wrong! Unfortunately or fortunately even after one does research there is still more to learn or unlearn. The site you give gives a thorough if simplified view. Equilibrium constants are constant for a specific reaction at a given temperature if the chemical potentials of the reactants are expressed as activities not necessarily the same as concentrations. At low concentrations of everything in solution concentrations approximate activities closely. This is similar to the use of PV=nRT in gas calculations.
PbCl2[crystals] dissolve in pure water to give hydrated Pb++ ions and hydrated Cl- ions. With crystals of appropriate size and proper mixing an equilibrium is established. Careful analysis shows that the measured amount of Pb and Cl in solution are constant regardless of the method of preparation when expressed as concentrations. The relationship is expressed as the equilibrium constant shortened to the solubility product provided the solid is a pure crystal with chemical activity equal to 1..  Ksp = [Pb++][Cl-]^2. The concentration of chloride is squared because the molecularity of chloride at equilibrium is 2.
If an ionic compound not related to lead chloride is present, the solvation of the lead and chloride ions will change. This change is specified as a change in the chemical activity of the ions and tables of the activity coefficients are found in handbooks and books on quantitative analysis [and are usually ignored]. If the concentration of lead or chloride ions is changed by a change in total ionic strength the amount of PbCl2 that dissolves changes usually to greater dissolution. This is because the activity of the ions in solution has lowered, and more are needed to satisfy the Ksp. There are two possible approaches to handle this: 1. generate a new Ksp for every possible solution [quite a job] or 2. Try to measure activity coefficients as a function of ionic strength for each ion. The second works [and as usual is ignored until necessary].
The common ion effect means adding an ion contained in the solid. This can be done in two ways: 1. electrolytically by using an appropriate cell. The best example of this that I can think of is the operation of the lead acid battery used in cars. Measures had to be taken to prevent PbSO4 from precipitating everywhere. and 2. The addition of a salt containing one of the ions. This has two effects: it increases the concentration of one of the ions and slightly increases ionic strength the first effect is greater so adding a common ion will cause the other ion to be reduced in solution to satisfy Ksp. This means that concentrations approximate activities, best at low concentrations. There can be complications with some salts if the ion added can react further with the salt to give complex ions. This happens with ions such as chloride, cyanide, oxalate.  Silver chloride or cyanide are insoluble in water but are appreciably soluble in excess added chloride or cyanide and cyanide solutions will dissolve gold under mildly oxidizing conditions.


text:
What is the logic behind the steric number formula?
My teacher taught us two methods two calculate steric number for determining hybridisation.
The first method was to count the total number of sigma bonds and add the lone pairs of the central atom. I have no problem with this method.
But we were also given a formula:
Steric No. = 0.5( V + SA (+/-) G)
where
V= no. of valence electrons in central atom;
SA= no. of monovalent surrounding atoms;
G= charge (if charge is positive we subtract it and if negative, we add it)
When I calculate the steric number by this formula it always comes out to be equal to the number calculated by the first method. But I am unable to understand why. What is the logic behind this formula?
EDIT: A similar question has been asked:
The logic for hybridisation formula
But it doesn't answer my question as the answer in the link suggested assumes that I already understand why the formula works for monovalent atoms and is explaining why the formula works if there are polyvalent atoms too. In my case, I don't even know why the formula works for monovalent atoms.

Does this answer your question? The logic for hybridisation formula

@Mithoron the answer in the link suggested assumes that I already understand why the formula works for monovalent atoms and is explaining why the formula works if there are polyvalent atoms too.  In my case, I don't even know why the formula works for monovalent atoms.


text:
Find the percentage of dissociation of nitrogen tetroxide given pressure, temperature, enthalpy and entropy

Given
$$
\begin{align}
\Delta_\mathrm f H^\circ(\ce{N2O4}) &= \pu{9.16 kJ mol^-1}
&\quad
\Delta_\mathrm f H^\circ(\ce{NO2}) &= \pu{33.18 kJ mol^-1}
\\
S^\circ(\ce{N2O4}) &= \pu{304.3 J K^{-1} mol^{-1}}
&\quad
S^\circ(\ce{NO2}) &= \pu{204.1 J K^-1 mol^-1}
\end{align}
$$
for the reaction $\ce{N2O4 <=> 2 NO2},$ find the percentage of dissociation of $\ce{N2O4}$ at $\pu{1 bar}$ and $T = \pu{333 K}.$

I formed the equilibrium:
$$
\begin{array}{ccc}
\ce{&N2O4 &<=> &2 NO2,} \tag{1} \\
&n_0\left(1 - \frac{\alpha}{2}\right) & & n_0\alpha
\end{array}
$$
where $n_0 $ are the initial amount of $\ce{N2O4}$ and $\alpha$ the dissociation value:
$$\alpha = \frac{2x}{n_0}. \tag{2}$$
After doing
$$\Delta G^\circ = -RT\ln K_p \quad\implies\quad K_p = \exp{\frac{-\Delta G^\circ}{RT}} = \pu{7.32E21}, \tag{3}$$
and having to find $\alpha,$ I get stuck with two unknown variables $n_0$ and $\alpha$:
$$K_p = p_\mathrm{tot}\frac{x(\ce{NO2})^2}{x(\ce{N2O4})} = \ldots = p_\mathrm{tot}\frac{n_0\alpha^2}{1 - \alpha/2}. \tag{4}$$
How to proceed?

Ideal gas law, $pV=nRT$, to find $n_0$?

But we don't know the volume?

Your  $K_p$ is vast but $K_p \approx 0.14$ so it looks like you forget that some values are in kJ and others in J . It is easier if you use $1-\alpha$ and $2\alpha$ and get partial pressure as $P_{NO2}=2\alpha P_{tot}/(1+\alpha) $ and $P_{N2O4}=(1-\alpha)P_{tot}/(1+\alpha)$ and $K_p=P_{NO2}^2/P_{N2O4}$.

Let:
$A$ represent $\ce{N_2O_4}$
$C$ represent $\ce{NO_2}$
Then the reaction becomes:
$$\ce{A <=> 2C}$$
Since no initial amounts are given, I'd suggest using mole fractions for the ICE table rather than moles. Assuming we start with pure $A$:
$X_{A^o}=1$
$X_{C^o}=0$
Writing equations for the equilibrium mole fractions of $A$ and $C$:
$X_A=X_{A^o}-y=1-y$
$X_C=X_{C^o}+2y=2y$
Then, we calculate $K_x$ by using its relationship with $K_p$:
$K_x=\frac{K_p}{P^{\Delta n}}$
($\Delta n=1$ for this reaction)
Then we can use the equilibrium expression for $K_x$ and substitute the expressions of the equilibrium molar fractions:
$K_x=\frac{X_C^2}{X_A}=\frac{(2y)^2}{1-y}$
We can then solve for $y$, and finally use its value to find $\alpha$:
$\alpha=\frac{y}{X_{A^o}}$
But since $X_{A^o}=1$, $\alpha$ is simply:
$\alpha = y$
Note 1: The answer will depend on whatever units your calculated $K_p$ value has, so make sure it matches the units of total pressure $P$ when calculating $K_x$.
Note 2: The assumption that we initially start with pure $A$ can be found to be correct or incorrect by evaluating: $X_A+X_C≈1$

The total number of moles of gas is $n_0(1+\frac{\alpha}{2})$.  So the mole fraction of $\ce{NO_2}$ is $\frac{\alpha}{(1+\frac{\alpha}{2})}$ and the mole fraction of $\ce{N_2O_4}$ is $\frac{(1-\frac{\alpha}{2})}{(1+\frac{\alpha}{2})}$.  The corresponding partial pressures are equal to these mole fractions times the total pressure, and do not contain $n_0$.


text:
What is the correct method for converting a concentration given in ppm to molarity?
I need to convert a concentration of dissolved boric acid $(\ce{H_3BO_3})$ in water (at high temperature and pressure) from ppm to mol/L.
Concentration of boric acid: $c_{ppm}=1500\ \ce{ppm}$
Molar mass of boric acid: $M_i=61.83\ \ce{g/mol}$
Density of water at $25\ ^{\circ}\ce{C}$ : $\rho_{25}=1.0\ \ce{kg/L}$
Density of water at $300\ ^{\circ}\ce{C}$ : $\rho_{300}=0.7125\ \ce{kg/L}$
Note that the water density is low due to high temperature ($300\ ^{\circ}\ce{C}$) and pressures.
Method 1: At $300\ ^{\circ}\ce{C}$
$$
c_i = \frac{c_{ppm}\rho_{300}}{1000M_i} = \frac{1500\times0.7125}{1000\times61.83}=0.0173 \ [\ce{mol/L}]
$$
Method 1: At $25\ ^{\circ}\ce{C}$
$$
c_i = \frac{c_{ppm}\rho_{25}}{1000M_i} = \frac{1500\times1}{1000\times61.83}=0.0243 \ [\ce{mol/L}]
$$
Method 2:
I have seen the following method that takes the temperature to be 25C, despite modelling the same high temperature/pressure body of water:
https://i.stack.imgur.com/mS2hV.jpg
Which yields a final concentration of $8.33\times10^{-8}\ [\ce{mol/L}]$
I am convinced that method 2 given in the attached image is incorrect (as the molarity calculated is far lower than expected), but am unsure where the fault lies.
Edit: I have improved formatting and readability.

Attached image is not the standard definition of ppm as used by analytical chemists.

Both European and British Pharmacopoeias: "The expression ‘parts per million’ (or ppm) refers to mass in mass, unless otherwise specified." I don't know other trustworthy sources that would explicitly interpret ppm as anything other than mass fraction, so the screenshot is likely made up and should not be trusted, especially considered the formatting and use of notations.

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Thanks for the information on post formatting- I've updated my question as per the site's guidelines.

This question is confusing to me, and the calculation that you show doesn't look right to me.
You state that the question has to do with a solution of boric acid in water. Fair enough, until you report that the temperature of the solution is 300 °C. This would imply a pressure of at least about 85 atmospheres; any less and the solution would boil, assuming that the solution behaves as water would. This is a fair assumption, as the concentration of boric acid (I'll get to that below) is much too low to affect the physical properties of water all that much.
This isn't wrong. I'm just trying to orient myself to the problem.
The calculation that you reference by link to imgur.com is suspect. Notice that it has conversion factors that undo each other (the molar mass of boric acid). In addition, it uses a density conversion that is valid under standard pressure (1 atm), and not at a minimum of about 85 atm, where the density of water at 300 °C is, as you correctly state, 0.7125 kg/L.
Let's start fresh:
The concentration of the boric acid solution is given as 1500 parts per million. Parts per million is defined as mg of solute/liter of solution. Molarity is defined as moles of solute/liter of solution. The denominators match, so the problem is just one of converting the numerator from milligrams to moles. Which I do, thusly:
(1500 mg H3BO3 / L sol'n) * (1 g / 1000 mg) * (1 mole H3BO3 / 61.83 g H3BO3)
= 0.0243 M H3BO3
Instead of calling it a day, let me consider the calculation on imgur.com. It starts with the expression:
1500 ppm H3BO3 = 1500 E -6 (molecules H3BO3 / molecules H2O)
This seems to imply that "parts per million" refers to a molecule fraction of solute in solvent, which is not used in chemistry, at least not as far as I know. But if we accept this understanding of "parts per million", the logic and calculation becomes this:
By the definition of "mole":
1500 molecules H3BO3 / 1 E 6 molecules H2O is equivalent to:
1500 moles H3BO3 /  1 E 6 moles H2O
So:
(1500 moles H3BO3 /  1 E 6 moles H2O) * (1 mole H2O / 18.01 g H2O) * (1000 g / kg) * (0.7125 kg / L)
= 0.0593 M H3BO3
...which matches the calculation by the formula you stated at the top of your post:
c_i = (1500 * 0.7125) / (1000 * 18.01) = 0.0593 M H3BO3
The omission of units of measure is unsettling to me, but we're all friends here, right?
I should point out that the conversion in the imgur.com file: 1 cm^3 = 1000 dm^3 is wrong. The correct relationship is: 1000 cm^3 = 1 dm^3. The difference between the calculated answer in the imgur.com file and the answer above can be traced to this improper conversion of cm^3 to dm^3, and the density of water at a temperature and pressure (1.0 kg/L at ~25 °C and 1 atm) that does not apply to the conditions (0.7125 kg/L at 300 °C and at least ~85 atm).
I hope this helps, or at least, that I haven't added to your confusion.

ppm is defined as 1 in 1,000,000. Nothing more, nothing less. It does not say what the parts are. It may be typically mass, volume, molar amount.  ppm as mg/L implies mass is considered and density= 1g/mL so it means mg/kg.  Air context considers volume ppm, what is nearly the same as molar ppm.
That all is the reason why using ppm and similar is highly discouraged, as there are possible multiple interpretations unless explicitly excluded.

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Poutnik:
Your point about the ambiguity of the phrase "parts per million" is well-taken, especially in light of the notation "ppt". I can never remember the rule; is it "parts per trillion", while "parts per thousand" has to be written out, or the other way around?
andselisk:
Thank you for the links. I am relatively new to the site, and this information will help me to be a better contributor.

Thank you for your answer. The water is the coolant in the primary circuit of a PWR, and is consequently at high temperatures and pressures, as correctly assumed by you.


text:
Can magnets boil water with no increase in temperature?
In this YouTube video an experiment is done with magnets and water. And at 7:14 they claim that the water boils, but with no increase in temperature. Is this possible?

Sure, just lower the ambient pressure a lot. Nothing to do with magnets and I didn't watch the video, btw.

Actually, if water is at 100 °C, applying heat will not increase water temperature, magnets or not, and it will boil till all gone!

At 7:14 they claim that the water boils but the magnets stay cool. The claim is not that the water, the glass, or the aluminum under the glass stay cool. They don't stay cool. They reach the boiling temperature of water. Then the water boils.

Would be good if you included more details so we don't actually have to watch the video- describe the relevant points.

In the included Czech text is said that rotating (and cold ) magnets are inducing electric eddy currents in the aluminium disk inserted in the vessel.
So the disk is heated up by Ohmic loses and warms up the water.


text:
How to apply Le Chateliers' principle for the reaction between nitrogen and hydrogen to produce ammonia?

$\ce{N2 (g)}$ and $\ce{H2(g)}$ are allowed to react in closed vessel in given temperature and pressure for the formation of $\ce{NH3 (g)}$ according to,
$$\ce{N2 (g) + 3H2 (g) <=> 2NH3 (g) + \pu{22.4 kcal}}$$
If $\ce{He(g)}$ is added at equilibrium at constant pressure, then which is/are correct?
a) Concentration of  $\ce{N2(g)}$, $\ce{H2(g)}$ and $\ce{NH3(g)}$ decrease.
b) Moles of $\ce{NH3(g)}$ decreases.
c) The extent of cooling depends upon the amount of $\ce{He(g)}$ added.
d) Concentration of $\ce{N2(g)}$ and $\ce{H2(g)}$ increases and concentration of $\ce{NH3(g)}$ decreases.

I know that addition of inert gas at constant pressure increases the volume and forces the reaction to proceed towards the side with more gaseous moles (backward in this case). So, option b) and c) are obviously right.
Also, the concentration of all gases decreases suddenly after the inert gas addition due to increase in volume. But I'm unsure of the final state after equilibrium and I can't figure out why option a) is given right.
How can I make a prediction about the concentration of all gases at equilibrium?

Can someone help me out with this?

"Also, the concentration of all gases decreases suddenly after the inert gas addition due to increase in volume." I don't understand your question, haven't you answered it already?

"Also, the concentration of all gases decreases suddenly after the inert gas addition due to increase in volume." Here, I meant that concentration of all gases decreases "just" after inert gas addition, but the concentrations of N2 and H2 will also increase by a certain amount, as the reaction shifts in the backward direction to reach a new equilibrium position. So, my question is, what is the net effect of these changes on the "final equilibrium" concentrations?

See Haber Process.

Reacting in a closed vessel.  At constant pressure one is simply diluting the mixture.  But you might try raising the pressure enough for NH3 to liquefy while N2 and H2 remain gaseous.  The NH3 could then be removed, driving the reaction towards the product.

The problem doesn't seem to be asking what happens to concentrations after equilibrium is reestablished. Rather, it's asking what happens immediately after adding the inert gas.


text:
Hydrogen peroxide generation by UV-Light
I would like to know if it is possible to generate hydrogen peroxide in the atmospheric air with UV-Light or if some catalyzer is needed for that.
If affirmative, is there a specific UV wavelength that is the peak of H2O2 generation?
Thank you

What's the source of hydrogen? The air consists of about 0.00005 % of hydrogen, and of approx. 4 % water at best. It would also be helpful to understand what this $\ce{H2O2}$ generation would be useful for.

@andselisk the reaction would be by the moisture in the air and oxygen, the idea is to use hydrogen peroxide as a disinfectant such as this product: https://rgf.com/products/phi-air-purification/halo-led-whole-home-in-duct-air-purifier/

These products are of no use.

Hydrogen peroxide is surprisingly ubiquituous in the atmosphere, all you need is trace or ultratrace analytical methods. Rain water has a measurable amount of it. This is not due to pollution. It is Nature's composition of rainwater. There must be a purpose just like trace nitric acid in rain water serves as a plant fertilizer. Recently, research at Stanford showed that if pure water is misted in air, measurable quantities of $\ce{H2O2}$ are formed in the mist. Even more surprising fact is that no UV light is needed. Misting water in air in complete darkness also generates $\ce{H2O2}$.
In short, no catalyst or light is needed. All you need is a very fine mist of water being sprayed in the air and you will have a small but measurable concentration $\ce{H2O2}$ in the mist. The phenomenon is not well-understood but there are only conjectures so far for this reproducible experiment. The authors had been studying this for several years.
Further reading: Lee, J.K., Walker, K.L., Han, H.S., Kang, J., Prinz, F.B., Waymouth, R.M., Nam, H.G. and Zare, R.N., 2019. Spontaneous generation of hydrogen peroxide from aqueous microdroplets. Proceedings of the National Academy of Sciences, 116(39), pp.19294-19298.

Nothing spontaneous about it; the atomization forms static electricity, think thunderstorm. here is their comment "In what follows, we report the direct, spontaneous generation of H2O2 from aqueous microdroplets in the absence of applied voltage, catalyst, or any other added chemicals. We also speculate about the nature of the mechanism responsible for these observations."  I get upset when a proper energy balance is ignored; someone has to squeeze the rubber bulb.  The chemistry is exciting tho.

I must apologize for using spontaneous in the vernacular sense because the production of the peroxide is spontaneous in the thermodynamic sense, otherwise it would not form. The energy supplied from imposed KE in the atomization overcomes the probably positive entropy change of a system far from equilibrium since there was no peroxide to start.  All that is needed is a possible mechanism and the micro surfaces seem to provide that. Static electricity can be formidable albeit difficult to control and direct.


text:
Stereochemistry R,S configuration
I'm trying to find the R/S configuration of this molecule right here:

At first I tried to find the rotation and then reverse it (since H is pointing outwards), I got an S configuration (drawn poorly here):

Then I tried to use the enantiomers method. I swapped the Br and the H then tried to find the configuration of the new molecule. And, I got an S configuration (of the resulting new molecule, which leads to the original having R configuration), which is not what I expected:

I don't know what I've done wrong here, the two attempts should lead to the same result (unless I've done something wrong). I tried looking up other R/S config examples using the enantiomers method (switching method) and what I've noticed is that the wedge and the dotted lines are right next to each other, like this for example:

While the molecule above has a straight (?) line in between:

So does that mean I can only switch groups when the wedge and the dotted lines are next to each other? Or was the first method wrong? Please kindly explain. Thank you.
Update 1
I've added a 3D visualization of the molecule. The big red ball is bromine, the gray medium-sized balls are carbon atoms, and the little white ones are hydrogens. I've marked the chiral center (green arrow pointing to it) for clarity. I've also rotated the molecule so that the bromine is pointing outwards.

Have you tried building a 3D model to make sure you have not mistakenly swapped the configuration?

@matt_black I've looked at the ball and stick model of the molecule (on KingDraw), then I tried switching the Br and H around. From what I observed, the molecule seems to have R configuration. I'll leave an image of the molecule on the question asap.

Just my two cents: the way your stereocenter is drawn seems to make it unassignable (it is ambiguous). With that perspective, you could have either two wedged bonds or two dashed bonds to form a tetrahedron. Both your assignments are correct but for different (and unattainable) carbon geometries. The angles are a mess.

I agree, poor drawing! Make the carbon in question look like it is tetrahedral. In your first picture, move the dashed bromine pointing northeast to a position pointing northwest, and still dashed ,just above the bold hydrogen. Now you should have no problems in visualization.

@TheRelentlessNucleophile
Thanks for pointing that out! Yeah the stereocenter does look kinda weird now that I've looked at it. It's more like two lines intersecting than a tetrahedron. I was just curious to see how this molecule would have its configuration assigned (the problem was to draw (2S,3E)-2-bromohept-3-ene, and that was the molecule my prof. drew).

@user55119 So I can freely move the bromine that's pointing inwards (dashed lines) without changing the configuration right? Like this one for example.

As long as it keeps pointing inwards, yes.

Follow Ivan's lead. I would tweek tha methyl group like this. You might be interested in this.


text:
Advantages of boiling opium with a quicklime reaction
Background
Before the First Opium War, the Chinese official Lin Zexu destroyed confiscated opium by mixing it with salt, quicklime, and water and then throwing the result into the sea. I (not a chemist) seek to confirm my understanding of what must have happened (before I mislead my students about it). Ideally, I also want to understand the advantages of this method over "normal" boiling.
My understanding
First, to my understanding, the salt plays no chemically important role. Except perhaps raising the boiling temperature of the water, so that the opium is boiled even more hotly.
Lin must have incited the reaction

$\ce{CaO + H2O —> Ca(OH)2}$

in a receptacle also containing the opium balls.
This, as I have seen, is highly exothermic. Meaning, he now had opium in boiling water. He also got $\ce{Ca(OH)2}$, which is solid and denser than water, meaning it sinks!
What is the advantage of boiling the opium this way vs conventional boiling over fire?
Would the violence and quickness of the reaction allow the water to superheat?
After reading about Portlandite, I thought that Lin‘s goal might have been to bind the opium balls inside the solid $\ce{Ca(OH)2}$, so that everything would sink, once thrown into the sea. Am I correct in now assuming that this would not happen and that the $\ce{Ca(OH)2}$, though solid, would just be a white dust, in no way aiding the sinking of the balls?
In case it offers any clues, this is a picture of the process found on Wikipedia:

Why do you care about a substance which is usually considered as a drug ?

C'mon, quicklime is pretty tame, @Maurice...

It is quite possible particular processes did not matter. The purpose could be just a spectacular destruction by mixing with unpleasant additives and then disposing it in the sea. Simple boiling would be far less dramatic.

I am not familiar with the material properties of opium in 19th-century China, but I am going to assume it was like most other unrefined natural products: full of cellulose and sticky fatty acids.
There are a few reasons quicklime was likely used to destroy the opium.

Quicklime increases the pH of the resulting solution (i.e., make it more basic or alkaline), which would also make the actual narcotic opiate molecules in the opium soluble in water, so when they are thrown into the sea the drugs don't just float around waiting to be picked up (at night, in boats, with nets?).
Bases react with fatty acids to make soaps in a process called saponification. This would  make the opium less sticky so they would be broken up more easily by the mechanical action of the boiling water, and it would increase the solubility of the plant material. This is the same way that detergent works in your dishwasher to dissolve and wash away greasy organic matter.
The calcium ions in quicklime, not found in lye (i.e., caustic soda; NaOH or KOH), will make the boiling water harder and cause less foaming, which would probably make a huge mess otherwise. So not only would the addition of quicklime cause the water to heat and boil without a fire, it would keep it from boiling over.

At the end of the day, all this would make it easier and faster to get rid of more opium.


text:
Does single swap rule apply to in-the-plane bonds?
When forming enantiomers, does the single swap rule (swapping 2 of the atoms directly bonded to a chiral C will give its enantiomer) apply, even when we're switching bonds that are in the plane?
For example, swapping a wedge and a dash gives the enantiomer; swapping the dash and an in-the-plane bond gives the enantiomer etc, but what about two in-the-plane bonds? Does this give the enantiomer too?

Yes it does. Think of it this way: the molecule is real, but the plane only exists in our imagination. Can you imagine a different plane, so that the bonds which were in the plane become wedge-and-dash, and vice versa? Of course you can. Well, the molecule is still the same.

If your molecule has more than one element of chirality, then switching  two bonds at only one center will lead to an optically active diastereomer, not an enantiomer.


text:
Hartree Fock calculations on F2 don't match the traditionally taught MO diagram
I ran some Hartree-Fock calculations on F2 to visualize the MOs, and the energy levels don't match the trend commonly taught (shown here: https://chem.libretexts.org/Courses/Heartland_Community_College/HCC%3A_Chem_161/9%3A_Molecular_Geometry_and_Bond_Theory/9.8%3A_M.O._Theory_and_the_Period_2_Diatomic_Molecules).
Instead, I am seeing the pi_2p lower in energy than the sigma_2p. Not sure why they would disagree?
This trend persists with any basis set.

This is probably due to the lack of correlation effects in HF theory; e.g. the σs and σp orbitals are also switched for N2 as well, in that theory.

Does using something like b3lyp change the ordering?

B3LYP does match the MO theory ordering. Either way, it seems silly to assert something in MO theory that HF doesn't capture. Any argument about s-p mixing that people make in drawing molecular orbitals would be included in HF, and MO theory of course doesn't account for things like correlation.


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Question on Nernst equation
What does the graph of cell potential versus temperature look like? Would that look same as the graph between cell potential and temperature difference?
(Please correct me if I am missing some concept!)

Welcome to CH SE site!
Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of at least essential parts.
Suitable formatting can be done according to formatting math/chem expressions/equations.

By temperature difference presumably you mean how potential changes with temperature (the slope) as suggested in the answer?

Note that having just the photo is still better than not having even just the photo.

Let assume the classical representative of cells – the Daniell's cell:
$$\ce{Zn(s)}|\ce{ZnSO4(aq)}||\ce{CuSO4(aq)}|\ce{Cu(s)}$$
\begin{align}
E_\mathrm{cell} &= \Delta E^{\circ} + \frac{RT}{nF} \ln{\frac{a(\ce{Cu^2+})}{a(\ce{Zn^2+})}}\\
&= - \frac{\Delta G^{\circ}_\mathrm{r}}{nF} + \frac{RT}{nF} \ln{\frac{a(\ce{Cu^2+})}{a(\ce{Zn^2+})}}\\
&= \frac{1}{nF}\left( - \Delta H^{\circ}_\mathrm{r} + T \Delta S^{\circ}_\mathrm{r} + RT \ln{\frac{a(\ce{Cu^2+})}{a(\ce{Zn^2+})}}\right)\\
&= \frac{1}{nF}\left( - \Delta H^{\circ}_\mathrm{r} + T\left( \Delta S^{\circ}_\mathrm{r} + R \ln{\frac{a(\ce{Cu^2+})}{a(\ce{Zn^2+})}}\right)\right)
\end{align}
$$\frac{\mathrm{d}E_\mathrm{cell}}{\mathrm{d}T}=\frac{1}{nF}\left( \Delta S^{\circ}_\mathrm{r} + R \ln{\frac{a(\ce{Cu^2+})}{a(\ce{Zn^2+})}}\right)$$
Equations assume the electrode local equilibrium states and for simplicity the same reactant and product heat capacities, otherwise calculations would be complicated by temperature trends of the reaction enthalpy and entropy.


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Is analytical imprecision normally distributed or log-normally distributed?
In textbooks and scientific litterature, analytical imprecision is often assumed to be normally distributed. I.e., theoretically, if the same chemical sample were analysed over and over again ad infinitum, the measurements would follow a normal distribution with a mean value equal to the true analyte concentration and a standard deviation corresponding to the coefficient of variation (CV) of the analysis.
However, according to some papers such as this one and this presentation by the same author...

[...] the distribution of values attributed to the measurand is sometimes approximately lognormal and therefore asymmetric around the measurement value.

In another paper...

[i]t is argued that (a) there is no theoretical reason why such distributions should be log-normal and there is abundant evidence that they are not; (b) quasi-log-normal distributions can be produced as artifacts by data recording practices; and (c) inordinately large numbers of analytical results would be needed to distinguish a log-normal distribution from a normal distribution.

This may be important because if measurements are normally distributed, the true concentration in a sample is best estimated by making several measurements and calculating their mean. For log-normally distributed measurements, however, the true value is more accurately estimated by the geometric mean, whereas the arithmetic mean will yield a positive bias.
When the CV is low, the normal and log-normal distribution will be very similar in shape. When the CV increases, the difference will be more pronounced. In addition to the difference in shape, the normal distribution opens up to the possibility that measurements will sometimes have negative values, whereas the log-normal will always yield positive values.
Are there good reasons to believe that measurements are normally rather than log-normally distributed? If measurements can indeed end up having negative values, how does that happen analytically?
Addition: I am particularly interested in how the above question pertains to UHPLC-MS/MS within the field of analytical pharmacology.

I'm not  at all sure that a general answer is possible. The answer depends on what is being measured and how the measurement is done. Eg precise measurement of a long length will look normal even though it isn't (it can't be negative). Some measurements (like decay counts in radioactivity) are nothing like normal distributions. The experimentalist has to think carefully about the detail to be sure what assumption to apply.

It may depend on the relative distance of confidential interval and limit of quantification. Closiness may effectively make both sides asymmetrical. The test of distribution symmetry may offer the decision factor. (If there are not theoretical reasons for other distributions.)

I have encountered log normal distributions when instrument errors are low and sample variation is high but is bound at one end by detection limits. The variation is only on the high side of the measurements.

As per Lippmann from the book "The Calculus of Observations", "Everybody believes in the exponential law of errors: the experimenters, because they think it can be proved by mathematics; and the mathematicians, because they believe it has been established by observation" Source https://mathworld.wolfram.com/NormalDistribution.html

I think you have already answered your own query "(c) inordinately large numbers of analytical results would be needed to distinguish a log-normal distribution from a normal distribution."

The correct answer in most cases is likely neither normal nor log normal but rather something more complicated resulting from multiple effects including but not limited to nonlinearity of method, limits of detection, absolute bounds, and decreased noise at higher measured values

While this is relevant to chemistry, this question applies more generally to all branches of physical science that employ instrumentation, so you may want to ask this at the stat SE site: https://stats.stackexchange.com/

Sometimes, the same measurement is reported two different way, e.g. transmission and absorption in UVvis spectroscopy. Depending which one you use, the distribution will be different for the displayed value.

The comparison between absorbance transmission and can be difficult because the machine measures light intensity and calculates absorbance. Every FTir I have used has difficulty in accurately measuring 0%T; Almost all will record negative transmissions for moderately thick samples that slowly approach 0% for very thick samples or for complete beam blockage. I think this is because moderately thick samples reemit absorbed radiation that is absorbed as the samples become larger and reabsorb the light.  This is an addition error above the instrument noise.

Ran out of space. This means that for low transmitting frequencies the zero is artificially high giving a falsely high absorbance that unfortunately is given a false maximum absorbance by the software if the transmission as measured is negative.  The point is that instrument performance must be understood in how the instrument errors affect variation in the sample measurements. This sounds trivial but I have encountered several publications quoting these assigned absorbances as actual absorbance readings

Thank you all for your contributions. It would seems that there is no simple answer to my question. Perhaps I can narrow down the problem by adding that I am particularly interested in UHPLC-MS/MS within the field of analytical pharmacology.

There is surely an answer, but perhaps there is too much to chew on in this question. For instance, start with the concept "true concentration". See eg https://pubs.acs.org/doi/pdf/10.1021/ac971793j

More recently: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6999182/

The terminology in this debate is confusing.  Things can be normally distributed or lognormally distributed; but also, error can be multiplicative or additive.  These are different concepts.  If $y_{measured} = (1+\epsilon_m)y_{true} + \epsilon_a$, where $\epsilon_m$ is multiplicative error, and $\epsilon_a$ is additive error, we still haven't said whether these $\epsilon$ values are both normally distributed, both lognormally distributed, or neither.  I suspect that most of these debates are really about whether $\epsilon_m \ll \epsilon_a$ or vice-versa, not about log-normality.

As the long series of comments illustrates, there is no universal answer, it is highly situation specific. However it seems that you got concerned by reading Ramsey et al. papers and presentations. It is true that his observations are correct, but look at his samples: lead in contaminated soils. Why should we expect it to be normally distributed? A contaminated piece of land will indeed show a skew because the contaminated area will have more lead near the contamination site, especially if it is coming from a very heterogeneous waste (landfill). For example, if someone discards a piece of lead in the trash, the local area will have more lead than the surrounding area. The whole argument of log-normal is about heterogeneous samples and long-range heterogeneity. If we were analyzing sodium in the surface seawater, chances of skew are nil and normal distribution is expected.
Now that the focus of the question is on pharmaceutical analysis and UHPLC MS/MS type work, for all practical purposes the normal distribution is used by all. This is because the analysis pertains to precisely controlled compositions and sample preparation processes of active pharmaceutical ingredients (APIs). We should be worried if a certain XYZ company prepares tablets in a skewed way. Tablets prepared in the morning always have a different composition than those prepared in the evening. There will be a red-flag about this skew.
When examining solids by Raman spectroscopy, we might observe skewness in analytical results for tablets. The laser spot size would be a few microns, and the laser would be rastered on the entire tablet. In any case, if the API solids are not homogeneous mixed or are intentionally segregated, then the analytical results will be skewed. For liquid samples, we should not expect skewness.
Some other points:

Are there good reasons to believe that measurements are normally
rather than log-normally distributed? If measurements can indeed end
up having negative values, how does that happen analytically?

Yes for all practical purposes in UHPLC MS/MS. Negative concentration has no physical significance. Such values have to be discarded or the analytical method re-checked! By definition, concentration must be positive. In other words, the limit of detection of any analyte should not be centered at zero. The "blank" readings may be distributed near zero. In short, the three selected papers represent a "skew" in the analysis because you have discarded hundreds of other paper which believe in normal distribution.
As per Lippmann quotes in his book "The Calculus of Observations", "Everybody believes in the exponential law of errors: the experimenters, because they think it can be proved by mathematics; and the mathematicians, because they believe it has been established by observation" Source mathworld.wolfram.com/NormalDistribution.html
BTW, there is an expert in SE who has authored an entire book on the Limit of Detection. I hope he can also add his valuable insight.

There is one overwhelming reason why errors can never be normally distributed which is so obvious that I hesitate to mention it.
Suppose (to fix our ideas) that we analyse a sample and the analysis says that the sample contains 1ppm of lead.
Then normally-distributed error would imply that there is a non-zero probability that the real concentration of lead in the sample is less than 0ppm. Which is impossible.

You have to explore the concept of limit of detection in analytical chemistry, which addresses your exact concern. There is a nice book "Limits of Detection in Chemical Analysis" by Edward Voigtman "https://www.wiley.com/en-us/Limits+of+Detection+in+Chemical+Analysis-p-9781119188971"

Here, I think it's relevant to distinguish between imprecision and uncertainty. While uncertainty is definitely related to imprecision (and also bias), it is not synonymous with it. I completely agree that any distribution describing the uncertainty of the measurement should not contain negative values because the true value must be positive. However, it is conceivable that a measurement can be negative despite the true value being positive, for example if the random error/imprecision is normally distributed.


text:
Slow evaporation versus vacuum evaporation
I have a salt in water and I want to evaporate the water. I'm wondering, what will be differences (if any) in the character of the crystals in a few scenarios:

Slow evaporation - let the solution sit in open air and allow natural evaporation
Use vacuum to evaporate the water quickly
Vacuum + heat to evaporate/boil the water quickly

Thanks in advance.

This is an excellent situation for some careful experimentation to determine which method of evaporation comes closest to giving the crystals that you want. Evaporation rate depends on the vapor pressure of the solvent, a function of temperature, its mole fraction in the solution and surface area. Condensation location depends on physical constraints of the vapor and partial pressure of the vapor.

may be vulnerable to collect dust by prolonged contact to air, if not protected. It may be preferable as slow evaporation leads to big, more pure and well built crystals and vice versa.

may be a good trade off between evaporation speed and agressiveness of the method.  Rotating evaporation vacuum devices are frequent way to go.

may violate salt stability, promoting decomposition or formation of lower hydrates. There may be mechanical salt loses and contamination of equipment by the salt or the liquid. Generally is boiling a bad idea, unless the time is priority.


text:
Why doesn't hydroxide associate with water analogously to proton forming hydronium?
I'm a novice at chemistry. I've heard that an acid needs water to show its acidic properties, that's why nitric acid needs to be associated with water to show its acidic properties:
$$\ce{HNO3 + H2O <=> NO3-  + H3O+}$$
$\ce{H+}$ does not exist freely and is always associated with another molecule like $\ce{H2O}$.
But then you would expect that the hydroxide anion would need to associate with water to show its basic properties. However, reactions never show that. In the case of sodium hydroxide, it just shows the $\ce{Na+}$ cation and the $\ce{OH-}$ anion dissociate in solution. But by that logic, nitric acid in water should be $\ce{NO3- + H+}$, which shouldn't be the case.
Why does $\ce{H+}$ need to associate with $\ce{H2O}$ to become $\ce{H3O+}$, but $\ce{OH-}$ can remain as it is and not as
$$\ce{NaOH + H2O <=> Na+ + H3O2-}?$$

It does, and so what? Like every single ion is hydrated, but it's only shown when there's a reason for that.

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in titles.

See also https://en.wikipedia.org/wiki/Hydronium // Note that strong acids need not water to manifest their acidic properties. They manifest it by reacting with water as a base, because they are stronger acids than H3O+(aq).

$\ce{H}_3\ce{O}_2^-$ is documented here. Note the two parts aren't joined by a covalent bond.

@J.G. Pubchem isn't a good source. There are good reasons to think there's symmetric hydrogen bond in this anion. That's as good as covalent bond, albeit weak, highly polar 3c4e one. https://pubs.aip.org/aip/jcp/article-abstract/78/5/2498/791466/The-structure-of-H3O2?redirectedFrom=fulltext

One could argue that, even though all the following representations are wrong in aqueous solution, $\ce{H+}$ is much more wrong than $\ce{H3O+}$, while $\ce{OH-}$ is only somewhat more wrong than $\ce{H3O2-}$.
Almost uniquely in Chemistry, the idealized chemical species $\ce{H+}$ contains no electrons - it's a bare proton. Because a bare proton has a +1 charge concentrated in a volume around $10^{15}$ times smaller than virtually all other ions, it has a comparatively gigantic charge density, and therefore behaves very differently. Due to its extreme polarizing power, true isolated $\ce{H+}$ is supremely reactive and can force the formation of a covalent bond with virtually anything, even unreactive or non-basic species such as $\ce{CH4}$, $\ce{N2}$, $\ce{H2}$, $\ce{He}$, etc. Therefore, if $\ce{H+}$ is ever generated outside of a dilute gas phase, it will instantaneously combine with something, and whatever the resulting ion is, it will almost always have lost the vast majority of the reactivity inherent to isolated $\ce{H+}$. Therefore, it kind of makes sense to avoid writing $\ce{H+}$ (with no solvation implied) except in very special circumstances.
In the real world, it's most convenient to write $\ce{H+\color{red}{(aq)}}$ and $\ce{OH^-\color{red}{(aq)}}$ in chemical reactions to make stoichiometry simple and to get the point of the chemistry across, recognizing that solvation is a factor and whose details can be brought into the discussion if required.


text:
What is an ideal binding agent for wood ash?
I have some wood ash left over after extracting potash. I want to make biocomposites as per the below link. https://materiom.org/recipe/59
The recipe uses agar agar which is a food grade material and quite expensive. Since this will not be used as food, I do not need to use food grade material. Hence, looking for a substitute.
As per my understanding agar agar and cornstarch give the binding effect differently. One gives while cooling, other while heating. Can I use corn starch instead of agar agar? Any other binding agent is also fine. Any other recipes for wood ash biocomposites are also fine.
Is the calcium propionate really necessary?
On a related note: a lot of wood ash is calcium carbonate. Some of it is calcium oxide. Will the biocomposite work if it is calcium oxide?
Appreciate the assistance

Calcium (and Na and K) propionate is a common, approved food preservative, added to comestibles such as bread and cakes. It is safe to assume the only purpose is to preserve the finished item. 2. If the ash is fresh, the Ca will be mostly as oxide, but slightly dampened and exposed to air, in a few days it should convert to the carbonate.


text:
How can Planck’s equation and the wave equation simultaneously coexist with contradicting each other?
The equations:
E=hf (Planck’s equation, where E is the energy, h is the Planck’s constant and f is the frequency) and
V=fλ (where v is the speed (in a vacuum), f is frequency and λ is the wavelength) Can be combined to form the equation:
E = hv/λ Which is often used in calculations involving the absorption and emission spectrums of elements.
However, as atoms emit photons when electrons fall to lower levels, and all frequencies of light travel at the same speed, this equation is often shown as:
E=hc/λ
The emission spectrum of atoms is often shown like the image below:

From the diagram as energy is on the y axis, and energy and frequency are proportional as per Planck’s equation, which would indicate that the ‘longer arrows’ on the diagram are of higher frequency as they are of greater energy change.
HOWEVER, as the diagram also shows the quantum energy levels gradually getting closer together, would a ‘longer arrow’ not also indicate a greater wavelength as the electrons has dropped/moved up a higher distance, in which case wouldn’t a shorter arrow, with a shorter wavelength lead to a greater frequency.
Essentially, I am confused about whether an arrow of shorter distance means a greater frequency because wavelength and frequency vary inversely, or if an arrow of longer distance means a greater frequency because energy and frequency are directly proportional.

What is on the y axis? Energy. Therefore, the arrow length is proportional to the photon energy and frequency and reciprocal to the photon wavelength.

What matters for the emitted photon is not the absolute energy level of the levels but the energy difference between the levels. That is what the length of the arrows shows.

"HOWEVER, as the diagram also shows the quantum energy levels gradually getting closer together, would a ‘longer arrow’ not also indicate a greater wavelength" <- no, why? You seem to understand the relation of wavelength, freq and energy, but here you seem confused.

Essentially, I am confused about whether an arrow of shorter distance
means a greater frequency because wavelength and frequency vary
inversely, or if an arrow of longer distance means a greater frequency
because energy and frequency are directly proportional.

The diagram you show is called an energy level diagram or sometimes a Grotrian diagram. As Poutnik pointed out, read the axes. The vertical-axis has E(nergy) written on it. The horizontal lines simply show the energy level number. So the arrow length shows the magnitude of energy change in going from one energy level to the other. You can give the arrow a pure geometric meaning, the length of an arrow show the "distance"  between two lines in "energy units" from the starting point to the end point.
Longer arrow: Larger energy change and therefore corresponding higher frequency light emitted or absorbed (ultraviolet side)
Shorter arrow: Smaller energy change, therefore corresponding lower frequency light emitted or absorbed (red side)
This chart from NASA may help you visualize better Link:


text:
Chemical kinetics for cyclic reversible reaction

Find $[\ce{C}]/[\ce{A}]$ for the following system at equilibrium:

I know that at equilibrium
$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-3}}{k_3} = K_3, \tag{1}$$
but my teacher told me there was another way to express it with rate constant and gave the hint that the numerator is the sum of three terms:
$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-3}}{k_3} = \frac{\ldots + \ldots + \ldots}{\ldots}. \tag{2}$$
So, I use steady state approximation (SSA) to solve this problem:
$$
\begin{align}
\frac{\mathrm d[\ce{A}]}{\mathrm dt} &= k_3[\ce{C}] + k_{-1}[\ce{B}] - (k_1 + k_{-3})[\ce{A}] = 0 \tag{3} \\
\frac{\mathrm d[\ce{B}]}{\mathrm dt} &= k_1[\ce{A}] + k_2[\ce{C}] - (k_{-1} + k_{-2})[\ce{B}] = 0 \tag{4} \\
\frac{\mathrm d[\ce{C}]}{\mathrm dt} &= k_{-2}[\ce{B}] + k_{-3}[\ce{A}] - (k_2 + k_3)[\ce{C}] = 0 \tag{5}
\end{align}
$$
$$[\ce{C}] = \frac{k_{-2}[\ce{B}] + k_{-3}[\ce{A}]}{k_2 + k_3} \tag{6}$$
$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-2}[\ce{B}]/[\ce{A}] + k_{-3}}{k_2 + k_3} \tag{7}$$
Since
$$\frac{[\ce{B}]}{[\ce{A}]} = \frac{k_1}{k_{-1}}, \tag{8}$$
$$
\begin{align}
\frac{[\ce{C}]}{[\ce{A}]} &= \frac{k_{-2}k_1/k_{-1} + k_{-3}}{k_2 + k_3} \\
  &= \frac{k_{-2}k_1 + k_{-1}k_{-3}}{k_{-1}k_2 + k_{-1}k_3}. \tag{9}
\end{align}
$$
I don't get the correct answer term just as my teacher said. I get two or four terms in numerator, so my answer is likely wrong. Where am I mistaken?
$
% \documentclass{article}
% \usepackage{chemfig}
% \begin{document}
%
% \schemestart
% A
% \arrow(A--C){<=>[$k_{-3}$][$k_{3}$]}[-60,1.25,,]
% C
% \arrow(@A--B){<=>[${k_{-1}}$][${k_{1}}$]}[-120,1.25,,]
% B
% \arrow(@B--@C){<=>[$k_{2}$][$k_{-2}$]}[,1.25,,]
% \schemestop
%
% \end{document}
$

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in titles.

(1) does not reflect indirect parallel equilibrium A <=> C via B.

with equations (3), (4) and (5), you have three equations and three unknowns. No need for equation (8) (which may not be true).

There is missing molar amount inventory equation, like [A]+[B]+[C] = c_tot, giving 4 equations for 3 variables and 1 degree of freedom for rate constants.// For (8) there is the same objection as for (1).

P.S.: ..what can be intuitively verified as one can multiply all rate constants by the same positive real number and still get the same solution.

By solving the equations without using eqn 1 or 8 I found $$\displaystyle \frac{C}{A}=\frac{(k_1+k_{-3})k_{-2} +k_{-1}k_{-3} }  { (k_2+k_3)k_{-1}  + k_3k_{-2}} $$

@porphyrin Thank you!!! It seems like the correct answer term just as my teacher said!! Could you tell me how  you solve this question ?Thanks for a lot!!

Get B from any equation then substitute into another equation if you want C/A and simplify. You can only get ratios from the three rate equations in this case, you need $c_{tot}=A+B+C$ to get individual concentrations.

@porphyrin Thanks you!!! Now I know how to solve it!!!Thanks for a lot!!

Equations 8 and 1 hold for "direct equilibria". Note that you go from A to C with B as an intermediary. This means that B influences your equilibrium from A to C, so the constants pertaining to B will also appear in the expression of the equilibrium constant. Same reasoning for A to B. C influences that equilibrium just as B does to that between A and C.

Steady state equations (4) an (5) should have k2 and k-2 swapped.

Using the steady state approximation (SSA), we can obtain the following:
$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-2}\frac{[\ce{B}]}{[\ce{A}]} + k_{-3}}{k_2 + k_3} \tag{1}$$
$$\frac{[\ce{B}]}{[\ce{A}]} = \frac{(k_1 + k_{-3})[\ce{A}] - k_3\frac{[\ce{C}]}{[\ce{A}]}}{k_{-1}} \tag{2}$$
$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-2}\left((k_1 + k_{-3}) - k_3\frac{[\ce{C}]}{[\ce{A}]}\right) + k_{-1}k_{-3}}{k_{-1}(k_2 + k_3)}\tag{3}$$
$$k_{-1}(k_2 + k_3)\frac{[\ce{C}]}{[\ce{A}]} = k_{-2}(k_1 + k_3) - k_{-2}k_3\frac{[\ce{C}]}{[\ce{A}]} + k_{-1}k_{-3} \tag{4}$$
$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-2}(k_1 + k_{-3}) + k_{-1}k_{-3}}{k_{-1}(k_2 + k_3) + k_{-2}k_3} \tag{5}$$
If you want to show three terms in the numerator (to match the teacher's hint), you can also write it this way:
$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_{-2}k_1 + k_{-2}k_{-3} + k_{-1}k_{-3}}{k_{-1}k_2 + k_{-1}k_3 + k_{-2}k_3} \tag{6}$$

I don't think it makes sense to use a steady-state approximation. At equilibrium, every reaction is at equilibrium. There is no net turnover from A to B that is compensated by net turnover from B via C to A, or something like that. Using the identity $$\frac{k_1 k_2 k_3}{k_{-1} k_{-2} k_{-3}} = 1$$ you can probably simplify to the expected simple fraction of two rate constants. Given that the problem statement asked for the complicated expression, it is the correct answer, I guess.

[from the comments to the OP's question] Equations 8 and 1 hold for "direct equilibria". Note that you go from A to C with B as an intermediary. This means that B influences your equilibrium from A to C, so the constants pertaining to B will also appear in the expression of the equilibrium constant. Same reasoning for A to B. C influences that equilibrium just as B does to that between A and C.

The equilibrium between A and C is not influenced by the mutual reaction to B. This is the same as adding a catalyst. It changes the kinetics, but not the equilibrium constant.
The OP used a steady-state equation (bracket denotes equilibrium concentration):
$$
\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k_3[\ce{C}] + k_{-1}[\ce{B}] - (k_1 + k_{-3})[\ce{A}] = 0 \tag{3}$$
At equilibrium, all steps are at equilibrium, so there are two independent equations instead of just the combined one:
$$ k_3[\ce{C}]  - k_{-3}[\ce{A}] = 0 \tag{3a}$$
$$ k_{-1}[\ce{B}] - k_1 [\ce{A}] = 0 \tag{3b}$$
Also, the three equilibrium constants are linked, as are the six rate constants:
$$\frac{k_1 k_2 k_3}{k_{-1} k_{-2} k_{-3}} = 1$$
or simpler
$$k_1 k_2 k_3 = k_{-1} k_{-2} k_{-3}\tag{A}$$
So if we take the solution (corrected for the swapped $k_2$ and $k_{-2}$)
$$\frac{[\ce{C}]}{[\ce{A}]} = \frac{k_2 k_1 + k_2 k_{-3} + k_{-1}k_{-3}}{k_{-1}k_{-2} + k_{-1}k_3 + k_2 k_3} \tag{6}$$
we can expand the first term in the numerator with $\frac{k_{-3}}{k_{-3}}$ and use (A) to eliminate the $k_{-1}k_{-2}$ term in the denominator, we get:
$$\frac{[\ce{C}]}{[\ce{A}]} =
\frac{\frac{k_2 k_1}{k_{-3}} k_{-3} + k_2 k_{-3} + k_{-1}k_{-3}}{\frac{k_1 k_2}{k_{-3}} k_3 + k_{-1}k_3 + k_2 k_3} = \frac{k_{-3}}{k_3}$$
This reiterates that at equilibrium, there is no flux "around the triangle". Every single reaction is at equilibrium, and there is no need for a steady-state approximation.

To troubleshoot the answers, I set the kinetic constants to 5, 6, 7, 10, 1, 21, so that $$5 * 6* 7 = 10 * 1 * 21$$. This made it easier for me to quickly check whether expressions are correct or not, and how to factor out $$\frac{k_{-3}}{k_3}$$ in the last bit of the answer.

The 'eq' subscripts seem excessive as square brackets already imply equilibrium concentration. Any other amount-of-substance concentration is denoted with $c$ anyways.

@andselisk I was not aware of that convention. Do you have a source for that? I have seen brackets (always square) for defining the reaction quotient as well. Of course, this is always an approximation, using concentration divided by standard state concentration instead of activities. I have also seen brackets used to define rate laws (where concentrations are typically not a equilibrium.

I'm not saying the usage is logical. For rate laws, the values have units, for the equilibrium constant, they don't, so the usage is a bit of a mess.

It is a bit confusing that you use the eq subscript in equations 3a and 3b, but not in equations 3 or 6, both of which also are true only for equilibrium concentrations. Nonetheless, you've nicely shown that the expressions derived from the two methods are equivalent.

@Andrew Now I explicitly defined the brackets and left out the inconsistent "eq" subscript. I'm still interested whether there is a convention about the meaning of brackets, though.

I don't think brackets generally indicate equilibrium concentrations. My understanding is that brackets just denote concentration and that when you are solving for the equilibrium condition, you can either use the subscript (which is rare) or preface the solution with, "At equilibrium. . . ". For example, you could split equation 3 into the first equality (with d[A]/dt), which is true for all concentrations, and then explicitly state the condition of equilibrium results in the second equality (to 0), and with that condition stated, you don't need the eq subscript.

@Andrew Your right, of course. I guess I was confused.


text:
Why doesn't Mn2+ ion form ammine complexes in aqeous medium?
Cations of 3d- elements particularly, the last four ones such as Cu2+ Zn2+ Co2+/Co3+and Ni2+ form metal-ammine complexes readily in aqeous solution, while the cations Mn2+ Cr3+ and Fe3+ usually dont.
In the case of Fe3+, I learnt that since it is a highly charged cation, behaving as an acid it normally doesn't form complexes with ammonia. But, Mn2+ is not so acidic. Still, it doesn't form the complexes. What could be the reason behind it?

I guess it has more to do with Lewis acidity (see HSAB theory) of the ion rather than pH acidity since complexes are about sharing electrons rather than exchanging protons and electrons. For example, ammonia is Lewis base and a ligand because of the free electron pair, but ammonium (NH4+) does not behave like that at all. Iron is a borderline Lewis acid so it may just be the ammine complex exists but is unfavored. It can be unfavored in water as the complex formation would require large activation energy but supplying the energy as heat would break the compex...


text:
Has dilead hexahydride been synthesised?
Dilead hexahydride is listed on Wikipedia as a compound composed of two lead atoms, and six hydrogen atoms:
  H  H
  |  |
H-Pb-Pb-H
  |  |
  H  H

As the wikipedia page for this compound does not exist, and Google scholar, along with Google, show no results on it, it's suggested that this is an artificial chemical that might have to be created in a similar manner to plumbane (PbH4).
How is dilead hexahydride synthesised?
My secondary question is, what are it's properties/characteristics? (if any info can be obtained?)

Even PbH4 was only made about 20 years ago. I doubt diplumban is gonna be made anytime soon.

https://doi.org/10.1002/chem.200400525 predicts low $\ce{H3Pb-PbH3}$ bond dissociation energy, so let me state the obvious: diplumbane should be quite unstable. Wiberg's Anorganische Chemie, 2017, says $\ce{Pb2H6}$ is an unknown compound.

This blows lead based life forms out of the water! Is there still hope for silicon, germanium, and tin?

@jimchmst Umm, not exactly. Tetraethyllead was added to fuel, so I think there could be plenty of chain derivatives, just not much of binary hydrides, that are somewhat unstable even for silicone.

Plumbane: https://chemistry.stackexchange.com/a/162199/17368

Apparently diplumbane itself is not well enough known to appear  in amy casual Google search. According to andselisk in a comment citing Wiberg's Anorganische Chemie (2017), $\ce{Pb2H6}$ is unknown.
However, some substituted diplumbanes appear to be more stable. hexaphenyldiplumbane, $\ce{Pb2(C6H5)6}$, appears to have known spectra, including the complex pattern below for protons in $\ce{CDCl3}$ solvent1. The complexity of the pattern suggests the phenyl groups are not equivalent in this compound. Since lead (and other Group 14 elements) has larger atoms than carbon, the structure follows the substituted ethane model unlike "hexaphenylethane"; the same is true with other Group 14 elements besides carbon in place of lead.

Reference

John Wiley & Sons, Inc. SpectraBase; SpectraBase Compound ID=1WQFC6VQS6Z SpectraBase Spectrum ID=6esVxgAw9qV

https://spectrabase.com/spectrum/6esVxgAw9qV (accessed Oct 8, 2022).

This is not an answer to the question, while the observation of Pb2Ph6 is interesting, it does not answer the question about Pb2H6.

Apparently, protonated diplumbane $\ce{Pb2H7+}$ have been predicted to be energetically favorable to varying degrees.

@NuclearChemist please see my update. On this basis the claim should be resolved.

@NilayGhosh I saw this. One notable feature: the proton inserts itself into the lead-lead bond si that the protonated species no longer has a direct bond between the lead atoms.

The Pb2H7 ion is interesting, maybe it has a three centre two electron bond. But the species is clearly not something with a Pb-Pb bond. I reason that Pb2H7 could be reasoned to relate to MeOMe while Pb2H6 could be argued to relate to ethane.


text:
API to get GHS info given substance / CAS #
Is there an API that can provide the GHS info for a given substance / CAS #?
If not, which website is the most extensive to get GHS info for a given substance / CAS # in human-readable format?

It looks like you already found this prior question, but just to point out this related post to future readers: https://chemistry.stackexchange.com/q/16583/41556


text:
Most stable monoatomic ion of nitrogen
I often see that the most stable monoatomic ion of nitrogen is N$^{3-}$ (for example on Khan Academy), and I remember being taught something similar, along the lines of atoms wanting complete octets.
However, the $1^\text{st}$, $2^\text{nd}$, and $3^\text{rd}$ electron affinities of nitrogen are, according to [1]:
Nitrogen (kJ/mol)
$-7$,  $+673$, $+1070$
The positive (endothermic) $2^\text{nd}$ electron affinity implies that the anions with negative charge greater than one are inherently unstable, in the sense that if I bring an electron near an N$^-$ in the vacuum of outer space, then they would not want to bind.  And if I had an N$^{2-}$ atom, out in space, then it would spontaneously decompose to N$^-$ and an electron.
My interpretation of "most stable ion" is that it should not involve any stabilising complexation with a (variable) cationic component.
Is N$^{3-}$ the most stable monoatomic ion?  Is there some additional completing octet-related favourability that compensates?  I would have thought that if there were, that that would turn up in the energy measurement. If I am tutoring people in chemistry, is there a better way to describe the drive of nitrogen to complete its octet, or should the way we teach the non-metals be revamped by actually looking at the electron affinities for each? (Oxygen for example, has negative both first and second electron affinities.)

By your definition and the data you provide N- is the most stable monoatomic ion of nitrogen in that it is the only ion which will not spontaneously decompose to the uncharged atom and an electron. However that is all but irrelevant for chemistry - it refers to a completely isolated gas phase nitrogen atom. The higher charged ions are stablised in the solid and liquid phase by the favourable electrostatic potential due to the rest of the lattice, and a the greater gain in energy due to the higher charge offsets that lost due to the electron affinity. I guess you would call this "complexation".

There are no known stable 2- free ions.

Nitride in the liquid phase?

Wikipedia gives a melting point in vacuum for aluminium nitride. On the other hand not totally convinced of the ionic model in that case, but it shouldn't be a total disaster

@IanBush There is big difference between free and lattice-bound ions, with the latter stabilized by the released lattice energy. Even isolated Na+(g)  +  Cl-(g)  are energetically unfavoured, unless at least bound in a ionic pair.

@JonCuster the dodecahydrido-closo-dodecaborate cluster dianion is known to be stable in vacuo.

@KanghunKim Monotopic dianion?

@Poutnik yes, see https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.osti.gov/servlets/purl/1528908&ved=2ahUKEwiolNH9rdX6AhVcp1YBHSM1AlMQFnoECA4QAQ&usg=AOvVaw1YM1EkTRF-ezddAt5Knazb

@KanghunKim  Assuming well delocalized. Interesting.

@Poutnik Yes, that is my point

@KanghunKim - I meant atomic ions. You reference a molecule.

@JonCuster for atomic dianions, they are still unknown AFAIK

@KanghunKim - correct, there are no stable ones. Good old atomic physics…

The correct answer is:
Both nitrogen(-III) and nitrogen(-I) are unstable (with respect to electron ejection) per se, and exist in ionic crystals only due to the lattice energies outweighing the energy for spontaneous electron ejection.
Then,
The former becomes more stable due to the valence orbitals being fully filled, while the latter ends up taking away electrons from the cation(which is, vide supra, endothermic per se, but is compensated for by the growth of lattice energies).
Note: An even bigger difference occurs between oxygen(-I), which is stable per se, and oxygen(-II), which is unstable per se. The latter can be stabilised by lattic energies in crystals, while the former ends up taking electrons away from the cation by the same mechanism as above.

So what is meant when we are taught (I was taught this in multiple schools/universities) that atoms “want” to get their nearest noble gas configuration (ex. O wants to form O$^{2-}$)?  Is this just plain wrong?

@FurrierTransform that holds in MOLECULES or SOLIDS, not IN VACUO.

But surely that depends on the molecule or solid and is a properly of the differences in electronegativity of the pair of atoms (or more for a conjugated system), and not a property of an individual element.  So what I am hearing is that what we were taught is just plain wrong.

"Both nitrogen(-III) and nitrogen(-I) are unstable (with respect to electron ejection) per se" Doesn't the 1st electron affinity of nitrogen indicate that nitrogen(-I) is stable?

The values in the OQ have different signs for the three consecutive EA's of nitrogen, which is not possible as all three are actually endothermic.

That the source of the EAs in the question (https://bcs.whfreeman.com/WebPub/Chemistry/raynercanham6e/Appendices/Rayner-Canham%205e%20Appendix%205%20-%20Electron%20Affi%20nities%20of%20Selected%20Nonmetals.pdf)  might be incorrect wasn't obvious from the question and seems rather important in order to make sense of your answer. It may be worth mentioning in your answer? And link to a better source? Even https://en.wikipedia.org/wiki/Electron_affinity_(data_page)#Elements


text:
How to balance iodometric equations using ion electron method?
I found out the reductant, oxidant, oxidation half reaction, reduction half reaction; balanced ions (adding $\ce{H2O},$ $\ce{H+},$ electrons) and charges:
$$\ce{K\overset{-1}{I} + \overset{+2}{Cu}SO4 -> K2SO4 + \overset{+1}{Cu}_2I2 + \overset{0}{I}_2}$$
$$
\begin{align}
\ce{KI &-> K+ + I + e-} \tag{ox} \\
\ce{CuSO4 + 8 H+ + e- &-> Cu+ + S^6+ + 4 H2O} \tag{red} \\
\hline
\ce{KI + CuSO4 + 8 H+ &-> K+ + I + Cu+ + S^6+ + 4 H2O} \tag{redox}
\end{align}
$$
Products should come across to $\ce{K2SO4 + Cu2I2 + I2}.$ But adding the half reactions I get $\ce{I + K+ + Cu+ + 4H2O + S^6+}.$ Water isn't present in the actual products.
Also, what do I do with the $\ce{H+}$ ions in the left? Where did it go wrong and how to proceed further?

I think you're trying to balance the equation following some recipe too closely, and you're throwing away the actual chemistry in the process, which is making the problem way harder than it should be. Both potassium and sulfate ions are spectators in the reaction, so you might as well just disregard them, and that would simplify things a lot - for one, it wouldn't have caused you to make the mistake of adding $\ce{H2O}$ as a product, or add $\ce{H+}$ to compensate for that. "$\ce{S^{6+}}$" also doesn't exist as an ion.

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Many mistakes. First, it is not $\ce{CuSO4}$ that reacts, because in solution $\ce{CuSO4}$ does not exist. It is transformed into $\ce{Cu^{2+}}$ and $\ce{SO4^{2-}}$. Then, $\ce{Cu^{2+}}$ is reduced into $\ce{Cu^+}$ without any other change. Second, the ion $\ce{SO4^{2-}}$ is not modified at all, although you changed it into a non-existant $\ce{S^{6+}}$ ion. Third, in solution the substance $\ce{KI}$ is transformed into two ions : the ion $\ce{K^+}$ that does not react, and the ion $\ce{I^-}$  which is transformed into $\ce{I2}$ (and not $\ce{I}$) in the reaction. So start again

First issue is the incorrect formula for copper(I) iodide. $\ce{Cu2I2}$ is a common moiety in dicopper(I) iodide complexes (C.N. 4) with various ligands, but the correct formula unit for copper(I) iodide is $\ce{CuI}.$ Unlike mercury(I) iodide with $\ce{Hg2I2}$ formula unit depicting $\ce{Hg2^2+}$ moiety, there are no $\ce{Cu-Cu}$ linkages in copper(I) iodide.
Second, assigning states of aggregation would help to identify spectator ions and write net ionic equation \eqref{a-rxn-r2}. For simplicity, copper(I) iodide assumed to exist as precipitate due to its low solubility, and iodine as soluble compound due to formation of triiodide ion with potassium iodide $(\ce{KI3(aq)})$:
$$
\begin{align}
\ce{\overset{+2}{Cu}SO4(aq) + K\overset{-1}{I}(aq) &-> K2SO4(aq) + \overset{+1}{Cu}I(s) + \overset{0}{I}_2(aq)}\label{a-rxn-r1} \tag{R1} \\
\ce{\overset{+2}{Cu}^2+(aq) + \overset{-1}{I}^-(aq) &-> \overset{+1}{Cu}I(s) + \overset{0}{I}_2(aq)}\label{a-rxn-r2} \tag{R2}
\end{align}
$$
Third problem is improper half-reactions which contain non-existing chemicals in solution such as $\ce{I}$ or $\ce{S^6+}$. It also looks like you are mixing up dissociation and redox reactions, which is unnecessary at this point once you've came up with the net ionic equation. Correct half-reactions and balanced net ionic equations are as follows (states of aggregation which have already served their purpose are omitted for clarity):
$$
\begin{align}
\ce{\overset{+2}{Cu}^2+ + I- + e- &-> \overset{+1}{Cu}I} &|\times 2 \tag{red} \\
\ce{2\overset{-1}{I}^- &-> \overset{0}{I}_2 + 2 e-} & \tag{ox} \\
\hline
\ce{2 Cu^2+ + 4 I- &-> 2 CuI + I2} \tag{redox}
\end{align}
$$
To complete the full balanced reaction you need to add spectator ions — potassium and sulfate — with corresponding coefficients:
$$\ce{2 CuSO4(aq) + 4 KI(aq) -> 2 K2SO4(aq) + 2 CuI(s) + I2(aq)}\label{a-rxn-r3}\tag{R3}$$

Shouldn't $\ce{KI3}$ be a separate entity in the final equation? See Jan's answer to What happens when potassium iodide is added to a copper (II) salt?.

@NilayGhosh $\ce{KI3}$ is implied as $\ce{I2(aq)}$ for simplicity since it's not the major point of the question and it exists among the reactants and the products, too.


text:
How exactly does oil drying work (specifically with the case of linoleum)?
I understand that it’s a form of autoxidation, and so with the presence of oxygen, the oxygen reacts with a double bond of the unsaturated fat to create a hydro peroxide, however I don’t really understand the processes that follow and why they happen.

Readily found on Wikipedia: https://en.wikipedia.org/wiki/Drying_oil