text:
Role of amalgamated zinc in Clemmensen reduction
What is the role of amalgamated zinc in the Clemmensen reduction mechanism?

Formation of Radical? https://chemistry.stackexchange.com/a/74861/69977

Zinc just grabbed out of a bottle will have a thin oxide passivation layer.  Amalgamating the zinc removes that layer.

And yes, like MaxW said, it is amalgamated to allow it to react to the maximum extent.

A topic for a good another question is, how mercury deals with oxide layers.

As the illustration below would show, you can use either amalgamated zinc  (in a Clemmensen reduction) or hydrazine (in a Wolf-Kishner reduction). This reaction as drawn hides some of the secondary products which should be zinc peroxide and mercury oxide*

Metallic zinc is “activated” by mercury by “dissolving” it into individual atoms of zinc that are distributed evenly in the alloy. You're basically creating a zinc solution.
The reaction of zinc with HCl releases hydrogen gas.
However, when this reaction takes place with zinc amalgam, the individual hydrogen molecules that are formed by the reaction of zinc with HCl remain in a reactive state (sometimes called “nascent hydrogen”) and reacts with the ketone to initiate the reduction reaction.
So the mercury in zinc amalgam serves to “trap” the active hydrogen as it is formed, and allows it to attack the carbonyl compound rather than be released as a gas.
*Disclaimer: I have never actually performed this reaction and I could be dead wrong.

This does not really answer the question of "what is the role of the zinc/merury amalgam". Yes, it is the reductant but why use it in preference to zinc?

I did not understand the question like that, as the requester didn't ask for an explanation of the differences, however I've updated my answer.

I have found an interesting note in the Czech version of the Wikipedia article, that while the amalgam is preferred for alkylarylketones, for dialkylketones is preferred metallic zinc.


text:
Chemical Formulae and the Periodic Table of Elements
Reading about heavy water, I have found that the chemical formula for it can be written 2H2O or D2O.

Deuterium is an isotope of hydrogen that is found in large quantities in water, more than one atom per ten thousand hydrogen atoms has a deuterium nucleus. The isotope is denoted “2H” or “D”, and is normally known as “heavy hydrogen”. Deuterium is used in a number of conventional nuclear reactors in the form of heavy water (D2O), and it will probably also be used as fuel in fusion reactors in the future (Source: ScienceDaily).

Encyclopædia Britannica says the formula can be 2H2O.  Is this correct?
2H (or 2H) = D is interesting to me as D is not in the periodic table of elements.
I thought that formulae should be formed using the symbols within the periodic table.  What are the rules surrounding these chemical formulae?

Yes the Encyclopædia Britannica is correct. The 2 exponent before the H means that the atomic mass number is 2. Since H has one proton, that means there is one neutron also. // Deuterium is often denoted with a D, and Tritium with a T. The bare $\ce{^4He}$ nucleus is also known as an $\alpha$ particle. No other element has multiple symbols for isotopes.

@MaxW - So is it more correct to use superscript 2 or is it down to style requirements in articles?

If i writing about using H isotopes for some purpose then using D and T is fine. If I were just discussing various isotopes (eg $\ce{^{23}Na}$) in general  then I'd use the $\ce{^2H}$ and $\ce{^3H}$ forms for consistency.

According to this entry in the IUPAC Gold book
https://goldbook.iupac.org/html/D/D01648.html
the symbol for Deuterium is H together with the isotopes notation.
However deuterium and tritium are routinely indicated by D and T respectively. Their use is dictated by the framework. Besides nuclear reactions, when only straightforward chemical properties* are considered, it is convenient to use a single letter rather than H and super/sub-scripts.
In organic nomenclature, while writing names for molecules, the presence of deuterium is signalled by the symbol (d).
*note that isotopic effect might however affect chemical properties at fine levels. The toxicity of heavy water is a notable striking example.

source: Stack Exchange

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How do we know if a newly developed chemical compound has any taste or is tasteless?
I was reading this about polystyrene, which says: "Polystyrene is an odorless, tasteless, rigid thermoplastic", and got hung up on "tasteless".
I searched Chemistry.SE for "tasteless" and found this and this, but they don't exactly answer my question in the title, because polystyrene is not considered a poison and is not a natural compound. So for such a highly-used, man-made, industrial-scale-manufactured compound, how do we know its taste (or lack of it)? Was the data acquired in an official way, through intended testing, or only from "accidents" (which I cannot imagine, so just for the sake of argument)?

Not an answer but a fun bit - in 1935 Harold Urey published a short bit in Science on 'Concerning the Taste of Heavy Water' (there was no difference between distilled water and pure heavy water, if you are interested).

I'm trying to remember where I saw this, but I'm pretty sure I ran into a paper from the early 1900s where part of the experimental said something along the lines of "[compound] has characteristic taste of [some class of compounds]". Good thing that class of compounds was not strychnoids...

Historically, many compounds were tasted for sure. For what is happening nowadays, I suspect that first the level of toxicity is determined. Obviously the taste of most of the immense number of new compounds prepared each day will remain unknown :)

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text:
Which has greater N-O bond length: NO or NOCl?
I came across this question. According to me, $\ce{NO}$ should have a shorter bond length as it has a triple bond, while $\ce{NOCl}$ has a double bond.
But the answer is given that $\ce{NO}$ has greater bond length than $\ce{NOCl}$. Why?
Is the answer wrong

Given the data on Wikipedia NO 115 pm, NOCl 114 pm, there is hardly a difference (I did not cross-validate these values, but I'd say they are essentially the same). Could you please point to the source of this question.

The data I found is 115.08+/-0.005 for NO (Herzberg, Diatomic) and 114+/- 2 for NOCl (Cotton & Wilkinson, Inorg Chem) and from this data it is uncertain which NO bond length is greater.

I think that $\ce{NO}$ bond order is 2.5 and $\ce{NOCl}$ exist as $\ce{NO+}$ and $\ce{Cl-}$ and $\ce{NO+}$ has bond order 3 that is why $\ce{N-O}$ bond length in $\ce{NOCl}$ is shorter.

Also relevant:  the nitrogen-chlorine bond lengths:
$\ce{NOCl}=196\text{ pm}$ Wikipedia
$\ce{NCl3}=176\text{ pm}$ Wikipedia
The oxygen in nitrosyl chloride is donating its lone pair to the antibonding orbital in the nitrogen-chlorine linkage, weakening that bond but adding more pi bonding between nitrogen and oxygen to become like a triple bond.
If we were supposed to compare bond lengths using "simple" principles of bonding, this is a poor choice.  Nitrosyl chloride presents more than meets the eye.

Interesting take, but I believe it distracts from the general stupidity of the original question, i.e. determining the difference of things that are essentially the same.

And the "extra" bonding in nitrosyl chloride explains why they are basically the same.  Again, the question does not work for understanding simple bonding principles due to the complexity of nitrosyl chloride.


text:
Why is the reason for color different in these two cases?
I have read in a book (NCERT India) about flame color for salts of alkali metal and color of transition metal complexes.
In the first case, it says that during excitation of electrons to higher energy it absorbs energy and then when it falls back it is emitted again giving color to the flame.
In the second case, it says that during d-d transition it absorbs a wavelength of photon and complementary color is shown. So, why can't there be emission in the second case like the first case.

Because the s and p orbitals always have more effective nuclear charge than d orbitald for a fixed value of principal quantum number... Moreover, the d orbitals are prone to splitting in the presence of applied field,so these d-d transitions are now across a bandgap of energy,which leads to light of certain wavelengths being emitted in the visible range.. Due to more penetration towards the nucleus, p-p transitions are merely releasing exchange energy from degenrate transitions(although p orbitals can split under some special conditions)

There could be emission from the transition metal complexes, but usually there are non-radiative pathways that compete with emissions and remove the energy before it has much of a chance to emit. However, in some crystals, e.g.  Ti sapphire(Ti/Al2O3), Ruby (with Cr ions) and Nd YAG (Yttrium Aluminium Garnet) among others,  there is sufficient emission from some levels to make them into lasers.

These at two different scenarios and two different phenomena. Flame color is due to atomic emission and complexes are colored because of light absorption in the solid state and in solution.
In a high temperature flame, an atom is considered to be a free and independent entity. With heat, the electrons in the atoms are excited to higher energy levels. However, selection rules determine where an electron can land e.g. s electron goes to p, p electron can go to d, however s electron cannot go to d.
In case of transition metals, Hans Bethe (Nobel Prize) developed a theory called called crystal field splitting. In a free gaseous atom, all five d orbitals orbitals have the same energy (called degenerate orbitals), so electronic transitions do not make any sense. It is also forbidden. However in the presence of ligands, the d- orbitals are no longer of the same energy. Electronic transitions are possible within d orbitals. Search crystal field theory if you are interested...

I know about Crystal field splitting but I am not able to understand why can't a d-d transition be reversed and the same wavelength that was absorbed be re-emitted.

There are two ways an excited species lose energy (i) re-emission and (ii) non-radiative loss of energy. Re-emission is somewhat common in excited gas phase atoms and even molecules, on the other hand non-radiative process are quite common in solution. Very few molecules are fluorescent. Transition metals ion also lose energy by non-radiative processes.


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Mixture of water, methanol and dichloromethane
If I have a mixture of 25 vol% methanol in water and I mix it with an equal volume of dichloromethane (which is miscible with methanol but not water) can I roughly say that both phases (i.e. water and dichloromethane) will have the same methanol concentration upon reaching equilibrium? Or will it vary significantly based on the activity of each mixture?

I doubt it. I would expect the MeOH to go into the water phase

When you have two phases they is always some of each component in both phases.  // (solubility of dichloromethane in pure water is 17.5 g/L @ 25 °C) // I'd expect the dichloromethane to contain both water and methanol, and the methanol/water phase will contain a small amount of dichloromethane. In certain proportions I'd expect one phase.

What is needed is the ternary phase diagram for the methanol/water/dichloromethane system. // I looked and couldn't find one. But there must have been studies on this.

This is a partial phase diagram. That seems to indicate that systems with roughly 50% methanol are miscible.
Source: http://shodhganga.inflibnet.ac.in/bitstream/10603/9995/11/11_chapter%206.pdf


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Standard litres to kg conversion for Hydrogen
When looking at hydrogen fuel tanks, they often give the capacity in terms of "standard litres". However, I can't find a definition of what this is.
I assumed this may mean a litre of hydrogen gas at room temperature? If so, is this at standard air pressure or is it compressed to a certain pressure? If so, what is this pressure?
I also thought perhaps it was a conversion to a litre of liquid hydrogen?
Or is it neither of these?
Is there a conversion between standard litres and kilograms for Hydrogen?

Since 1982, STP is defined as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of exactly $10^5$ Pa (100 kPa, 1 bar). At STP a mole of gas occupies 22.7 liters. A mole of hydrogen has a mass of 2.016 grams. Thus $8.88\times10^{-2}$ grams/liter.

Are you talking about hydrogen fuel tanks for cars? or something else?

@MathewMahindaratne, it isn't specifically tanks for cars, although I guess this could be an application. It is merely that if I look at Hydrogen storage tanks on the internet, many companies define the capacity in terms of "standard litres" and I was just curious as to what the definition of one "standard litre" is.

There has been a lot of promising buzz around hydrogen. Many people believe that hydrogen can be used as an energy carrier to replace other fuels like gasoline in the near future. One of the largest challenges to use hydrogen in a wide scale is its storage after hydrogen is produced. Some storage technologies have currently been developed. Yet, according to Florida Solar Energy Center, following three methods have been continuously used:

As a compressed gas (in high-pressure tanks).
As a liquid in dewars or tanks (stored at $\pu{-253 ^{\circ}C}$).
As a solid by either absorbing or reacting with metals or chemical compounds or storing in an alternative chemical form.

The current status of various storage technologies in terms of weight, volume and costs is given below (From: Florida Solar Energy Center):
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
\textrm{Storage Technology} & \textrm{Weight} & \textrm{Volume} & \textrm{Average Cost} \\
\textrm{} & \textrm{(kWh/kg)} & \textrm{(kWh/L)} & \textrm{(USD/kWh)} \\ \hline
 \textrm{Chemical Hydrides} & 1.6 & 1.4 & 8 \\ \hline
 \textrm{Complex Metal Hydrides} & 0.8 & 0.6 & 16 \\ \hline
 \textrm{Liquid Hydrogen} & 2.0 & 1.6 & 6 \\ \hline
\textrm{Compressed Gas (10,000-psi)} & 1.9 & 1.3 & 16 \\ \hline
\end{array}
$$


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What does it mean to calculate hardness in terms of ppm of calcium carbonate?
PPM means per part million or milligram per litre of solution.Calculation of ppm for a given compound say $x \pu{gms}$ of $\ce{Ca(HCO_3)}$ in 1 liter of water would be done like
$$ppm= \frac{1000\cdot x}{1\pu{1L}}$$
But I do not understand by what's meant by in terms of $\ce{CaCO3}$ is it equivalents or weight? This was asked in an exam and the formula is not covered in my textbook.

It is meant as molar equivalent. $\ce{CaCO3}$ has the molar mass about  $\pu{100 g / mol}$, $\ce{Ca(HCO3)2}$ has the molar mass cca $\pu{ 162 g / mol}$. Therefore $\pu{100 g}$ of the former is equivalent to $\pu{ 162 g }$ of the latter. The similar for the concentration expressed in ppm. E.g. if water contains $\pu{1.62 ppm}$ of $\ce{Ca(HCO3)2}$, it  is equivalent to $\pu{1 ppm}$ of $\ce{CaCO3}$

Expressing the hardness of water by ppm CaCO3 is nothing but a conventional convenience in water industry. There is a background behind it.
Basically, you titrate hard water with ethylenediamminetetraacetate (EDTA) and the moles of EDTA consumed are assumed to be consumed by Ca2+ ions only. As you would guess, this is not true because hard water contains many other ions which can form a complex with EDTA especially magnesium is always present. Anions have nothing to do with the calculation of ppm CaCO3. The calculation is rather trivial
moles of EDTA used = moles of Ca2+ (as calcium forms 1:1 complex with EDTA) = moles of CaCO3 and even = moles of Ca(HCO3)2.
One can easily convert moles of CaCO3 into parts per million from the known volume of hard water taken from titration.


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Relationship between the ideal gas constant and entropy
So it dawned on me the other day that the ideal gas constant $$R = \pu{8.31 J mol-1 K-1}$$ has the same units as molar entropy. Is there some deeper meaning behind this or is it just a coincidence that this occurs?

Well, R is in the equation for the entropy change of an ideal gas at constant temperature.  Is that what you were asking?

What is that, equation as it may answer my question. But, i was asking why does R have the units for entropy - Is it representing entropy?

Both entropy and the ideal gas constant are in some sense related to the Boltzmann constant.

Could you elaborate on that please. How are they related?

$\Delta S=R\ln{(V_f/V_i)}=-R\ln{(P_f/P_i)}$

Is V and P referring to pressure and volume for initial and final

Unfortunately, a discussion of why $k_{B}$ is related to entropy may be beyond the scope of this discussion if you don't already know that. You might just need to learn some statistical mechanics.

Yes, it's referring to pressure and volume for an isothermal change of an ideal gas.

The gas constant is equal to Avogadro's constant times Boltzmann's constant, the latter serving as a proportionality constant between the average thermal (kinetic) energy of the particles in an ideal gas and the temperature:
$$\left(\frac{\partial \bar U}{\partial T}\right)_p=\frac{3}{2}k_\mathrm{B}$$
The entropy can be regarded as a proportionality constant between the change in free energy $G$ with change in temperature of a system at constant pressure, since
$$\left(\frac{\partial G}{\partial T}\right)_p=-S$$
Note also that the average entropy of a particle can be written as
$$S=k_\mathrm{B} \log\Omega$$
So both $R$ and $S$ can be regarded as proportionality constants between energy and temperature.
The Wikipedia page on the equipartition theorem may provide some enlightenment on the origin of these proportionalities.

The fundamental equation is Boltzmann's for the entropy $S =k_B\ln(\Omega)$ where $k_B$, Boltzmann's constant, has units J/molecule/K.  When we use $R$ instead of $k_B$ it is trivially because molar units are used to define $S$.
In the equation $\Omega$ is the number of arrangements or configurations (or 'complexions' to use an old word) of distinguishable 'particles' among all the available energy levels. In that sense entropy is a measure of the uniformity of population in these levels.
(If there are $N$ distinguishable particles  then $\displaystyle \Omega =\frac{N!}{n_1!n_2!n_3!\cdots}$ where there are $n_i$ particles in level $i$. The log is evaluated by assuming all the numbers $N,\, n_i$ etc. are large numbers so that Stirling's approximation for $\ln(N!) $ etc. is valid.)

In the ideal gas, $3R/2$ is the thermodynamic entropy of one Mol.
The factor $\frac{3}{2}$ comes from $ST=N\bar{E}=N\frac{3k}{2}T=\frac{3R}{2}T$.
The energy of 1 Mol of ideal gas is either given by $3pV/2$ or $3RT/2$ or $ST$.
(In the ideal gas) the entropy does not depend on temperature.
That is why one writes the $T$ separate in $TS$.
For non-ideal gases $S$ changes with temperature: $E=\iint dSdT$.
For the ideal gass $C_V = S$ (always per Mol here).
$ST = \int dQ_{\text{rev}} = C_V \int dT = S \int dT$.
Because of constant volume, all the energy goes into $T$.
$C_V$ and $S$ stay constant.
On the $pV$ side this amounts to an increase in pressure.


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Definition of Lattice Energy
From previous studies, I recalled in my lecture theatre today that lattice energy was the energy released when an ionic compounds is formed from individual atoms in the gaseous state. I thought this idea would follow through throughout my studies at university.
However, my lecturer this afternoon defined lattice energy as the following:

• For an ionic solid the lattice energy is defined as the energy required to
  completely separate a mole of solid ionic compound into its gaseous ions

I thought this was my definition but in the reverse order.
Which one of us is correct, the lecturer or myself?

There is not necessarily a set convention: https://en.wikipedia.org/wiki/Lattice_energy#Lattice_energy_vs_lattice_enthalpy
However, the sign must match the convention used. For example, the values in the table you provided are all positive which implies you are adding energy to break the lattice.
The definition from you prior experience would require negative values.


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Solvent for caffeine extraction
Can anybody please help me find a way to separate caffeine from coffee that uses a solvent other than DCM, ethyl acetate, or diethyl ether?
This an experiment for science class, but I don't have access to any of the above-mentioned solvents.
Can you suggest any alternatives that are easily obtainable?

Supercritical $CO_2$?

It is kind of silly for us to guess 50 solvents to try and find one that you can use. I suggest you use Google "Extraction of Caffeine" and find a suitable solvent yourself. // I know that you are not going to have the equipment to use supercritical carbon dioxide.

Can you use Acetone?

@Waylander thanks for the reply but i don't think it would work all for the fact that acetone is miscable in water this is the same reason i can't use ethanol aswel if it weren't miscable it would work. Thanks for taking the time to reply!                                                 I apreciate it.

It would be better to re-word your question by stating the names of solvents which are available to you.

Here are three NileRed videos, you may find some helpful solvent information in them, or at least enjoy them with a nice cup of coffee or tea ;-) 1) How to extract Caffeine from Coffee 2) How to extract Caffeine from Tea (Classic DCM Method) 3) How to extract Caffeine from Caffeine Pills Also enjoyable are Periodic Videos How much caffeine in coffee? and Tea Chemistry - Periodic Table of Videos

Your question isn't bad. The wording has been improved, but people who leave "drive-by downvotes" probably won't return to notice and change them, so I"m voting +1 since your question as currently written seems reasonable enough to at least be non-negative!

Following is the list of solvents previously checked for the solubility (From reference 1 below):
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
\textrm{Solvent} & \textrm{Temperature, } \pu{^{\circ}C}  & \textrm{Solubility, } \pu{\%} (w/w) \\ \hline
 \textrm{Water} & 25 & 2.2  \\ \hline
 \textrm{water} & 100  & 66.7  \\ \hline
 \textrm{Ethanol} & 25 & 1.2 \\ \hline
\textrm{ Ethanol} & 60 & 4.5 \\ \hline
\textrm{Ether} & 25 & 0.3 \\ \hline
\textrm{Ethyl Acetete} & 25 & 2.5 \\ \hline
\textrm{Chloroform} & 25 & 18.0 \\ \hline
\textrm{Acetone} & 25 & 2.0 \\ \hline
\textrm{Benzene} & 25 & 1.0 \\ \hline
\end{array}
$$
My suggestion is to use ethanol in $\pu{60 ^{\circ}C}$. However, I'm not sure how ethanol would behave with the presence of tannins. For example, in a separate literature report, $\pu{1.0 mg}$ of tannin was isolated from a $\pu{g}$ of grape skins using 100% ethanol while the amount was increased to about $\pu{5.5 mg}$ when solvent was switched to 100% acetone (Ref.2).
References:

Q. V. Vuong, P. D. Roach, “Caffeine in Green Tea: Its Removal and Isolation,” Separation & Purification Reviews 2014, 43(2), 155-174 (DOI: 10.1080/15422119.2013.771127).
M. O. Downey, R. L. Hanlin, “Comparison of Ethanol and Acetone Mixtures for Extraction of Condensed Tannin from Grape Skin,” South African Journal for Enology and Viticulture 2010, 31(2), 154–159 (DOI: https://doi.org/10.21548/31-2-1412).


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Help with finding the right catalyst and reaction conditions for hydroamination reaction
A friend of mine, final year PhD, and I, final year Bachelor, are focusing on hydroamination reactions. In particular those with secondary amines and substituted alkenes. The reaction in particular is given by

Intermolecular hydroamination of unactivated alkenes with substituted secondary amines seems to be quite difficult. However, additions of cyclic and acyclic secondary alkyl amines to a wide range of alkenes with anti-Markovnikov regioselectivity have been reported. Some of those reactions are redox-neutral and occur readily at room temperature under visible light irradiation.
Our question is: what would be an applicable catalyst for this particular reaction (secondary amine + substituted alkene) and what reaction conditions would be necessary?
Thanks in advance!
EDIT:
The reaction may also concerns the primary amine,

However, we would like to keep the substituted alkene substrate as it is.

I think converting to tertiary halide and then reacting it with secondary amine, while sort of primitive, may be more effective method in this case.

Have you tried looking in the literature?

@Zhe Yes have, the problem we face however is the second methyl group that forms (resulting from the alkene substrate). It causes problems with hydrogen abstraction in certain mechanisms.

@Zhe by this I mean the two methyl groups next to each nitrogen

Why not methylate after?

@Zhe how does one change from C-H to C-CH3?

Your edit addresses this, but I'm talking about methylating the nitrogen after the hydroamination.

You may follow the procedure in this reference. I included the abstract for your convenience:

Abstract: The hydroamination of styrene with aniline catalyzed by phosphine-ligated palladium triflates exhibits a substantial $\ce{^{13}C}$ isotope effect at the benzylic carbon. This supports rate-determining nucleophilic attack of amine on a $\eta^3$-phenethyl palladium complex. Deuterium exchange observations and predicted isotope effects based on DFT calculations support this mechanism. Selectivity in these reactions is determined by the facility of palladium displacement after reversible hydropalladation of the alkene.

The Graphical Abstract:

Reference:
L. K. Vo, D. A. Singleton, “Isotope Effects and the Nature of Stereo- and Regioselectivity in Hydroaminations of Vinylarenes Catalyzed by Palladium(II)−Diphosphine Complexes,” Org. Lett. 2004, 6(14), 2469–2472 (DOI: 10.1021/ol049137a).


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Chemical reaction
Consider the following
Ca(OH)2 + 2NH4Cl → CaCl2 + 2NH3 + 2H2O
(Source: https://chemiday.com/en/reaction/3-1-0-440)
It appears to be double displacement reaction as CaCl2 is formed, but it’s not as NH4OH is not formed instead NH3 is formed
Why NH3 is formed instead of NH4OH ?
Is there intermediate step which transform NH4OH into NH3( I don’t think so but still)

It is postulated that NH4OH as a molecule does not exist in solution. A better way to write might be NH3(aq). As to the evolution of gaseous ammonia, this reaction would only proceed if you heat up (i) a moist paste of NH4Cl and Ca(OH)2 or (ii) strongly heat a solution, otherwise ammonia would remain dissolved in the solution.

$\ce{NH4OH <=>NH3 + H2O}$, that is, they are in equilibrium. Since the solution is weakly ionized, (Kb is ~1.8 * 10-5 at 293 K), there is little $\ce{NH4OH}$ present at a given instant, so ammonia water is only weakly alkaline.
One might say the answer $\ce{NH3 + H2O}$ is 99.998% correct, but $\ce{NH4OH}$ is still 0.002% correct, not entirely wrong.
See also this discussion on what species are present.

As the reaction takes place in water medium ,(More water on RHS) Why can’t we apply le’chatliiers principle and say equilibrium shifts to left hand side. Similar statement is made In the linked answer

I think the debatable and interesting part is the question: Does NH4OH as a molecule exist in solution? It was the same story as the hypothetical H2CO3 but people finally detected it. Not sure about the status of ammonium hydroxide.


text:
How to get the second electron affinity of tellurium?
I need the value of the second electron affinity of Te, i.e. the energy change associated with the attachment of an electron to a $\ce{Te-}$ anion
$$\ce{Te- + e- -> Te^{2-}}$$
In the books (e.g. J. Huheey, E. Keiter, R. Keiter; Inorganic Chemistry: Principles of Structure and Reactivity, 4th Ed.) I find second electron affinities only for certain elements from the periods 2,3,4 of the Periodic Table but not for tellurium.
I found a paper[1] that shows how to calculate the second electron affinities, for different elements, and gives an example how to do it for the oxygen.
The calculations use Coulomb and exchange integrals $\mathrm{J_{xy}}$ and $\mathrm{K_{xy}}$ for two electrons in different 2p orbitals, and with the same spin.
Do I understand correctly, for tellurium, I need the values of these integrals for two electrons in different 5p orbitals, and with the same spin?
Is there any software package where I can calculate these integrals or is there a paper or a book where these values are calculated? Is there another way to calculate the second electron affinity of Te or a paper where it is calculated or measured?
Reference:

Pearson, R. G. Negative electron affinities of nonmetallic elements. Inorg. Chem. 1991, 30 (14), 2856–2858 DOI: 10.1021/ic00014a008.

Calculations of negative ions are less reliable than those of positive or neutral particles, and those of doubly negative ions are worse still. Second electron affinity is a tricky  thing. I guess many elements don't have it at all.

You can certainly always calculate a second electron affinity in as given methodology, if you can measure this, and whether this is reliable is a completely different story.


text:
What is the crystal structure of bismuth oxyhydroxyphosphate (BOHP)?
The recent Chemical & Engineering News article Photocatalyst shreds drinking water contaminant PFOA report the following:

Cates and his colleagues made an accidental breakthrough while testing the photocatalyst bismuth phosphate. After finding that it could break down PFOA as effectively as a previously reported photocatalyst, gallium oxide, they tried to improve its performance by varying the pH of the reaction used to make the catalyst particles. The researchers had hoped this would change the particles’ size and shape, but the process inadvertently produced bismuth oxyhydroxyphosphate (BOHP). This new catalyst has the highest photocatalytic activity for PFOA degradation ever reported.
In lab tests under UV light, BOHP degraded all the PFOA in a 130 μM solution in about an hour, roughly 15 times faster than did bismuth phosphate or gallium oxide. BOHP performed almost as well over five repetitions of the experiment, suggesting it can be reused. To simulate PFOA-contaminated groundwater, the researchers then tested a more dilute solution of 1.2 μM PFOA accompanied by 500 μg dissolved organic carbon/L. In this case, the photocatalyst also degraded all the PFOA, though it took two hours. Strathmann calls it a “promising development” that advances the application of photocatalytic treatment of PFOA.

It also includes this SEM image of microcrystals of the "inadvertently produced bismuth oxyhydroxyphosphate (BOHP)."

The accidentally synthesized BOHP (shown) can break down the toxic industrial contaminant PFOA faster than any other photocatalyst. Credit: Environ. Sci. Technol. Lett.

A quick check with google doesn't send me to the materials Wikipedia page, and I'm wondering if that is because this is an unusual substance or the crystal structure hasn't been solved yet.
Question: What is the crystal structure of bismuth oxyhydroxyphosphate (BOHP)?

Another quick google leads to this paper : https://pubs.acs.org/doi/abs/10.1021/acs.estlett.8b00395?src=recsys&journalCode=estlcu that mentions 'Petitjeanite' which is https://www.mindat.org/min-3329.html

@gilleain That's great, thank you! I won't be able to read the paper one until I get to the library, but the page for Petitjeanite, triclinic, α = 88.28°, β = 115.27°, γ = 110.7° is a great start! Hopefully someone can find a diagram of what the unit cell looks like as in these questions 1, 2, 3

@gilleain for comparison, I've asked about the mineral here: What does the unit cell of petitjeanite look like?.

What is the crystal structure of bismuth oxyhydroxyphosphate (BOHP)?

Not exactly sure what an answer looks like so here is everything.
First off looking at the diffraction pattern from the paper, bismuth oxyhydroxyphosphate matches that of petitjeanite so that is definately the prototype structure.

Using the information about diffraction, from the ICDD Databse (pdf 01-082-8058)we find that petitjeanite is a triclinic system of the $\mathrm P\bar 1$ space group (#2). with lattice parameters of:
$\require{mediawiki-texvc}$
$$a = 9.798\AA; ~ b =  7.250\AA; ~ c= 6.866 \AA; \qquad \alpha = 88.280^\circ \beta = 115.270^\circ ~ \gamma = 110.700^\circ$$
and atomic positions of:
$$\begin{array}{c|lll}
\text{Atom}&x&y&z\\
\hline
\ce{Bi}&~~~0.52327&~~~0.41421&~~~0.7523\\
\ce{Bi}&~~~0.29838&~~~0.74044&~~~0.55549\\
\ce{Bi}&~~~0.07735&~~~0.1609&~~~0.28045\\
\ce{P}&~~~0.6952&~~~0.0172&~~~0.9215\\
\ce{P}&~~~0.0961&~~~0.372&~~~0.7862\\
\ce{O}&~~~0.3777&~~~0.5382&~~~0.4759\\
\ce{O}&~~~0.3066&~~~0.1549&~~~0.5579\\
\ce{O}&~~~0.859&~~~0.0275&~~~0.8943\\
\ce{O}&~~~0.5569&-0.2084&~~~0.8602\\
\ce{O}&~~~0.621&~~~0.1596&~~~0.7526\\
\ce{O}&~~~0.7684&~~~0.1052&~~~1.1811\\
\ce{O}&~~~0.2584&~~~0.5101&~~~0.7497\\
\ce{O}&-0.0468&~~~0.4637&~~~0.6902\\
\ce{O}&~~~0.0209&~~~0.1498&~~~0.6362\\
\ce{O}&~~~0.1695&~~~0.3727&~~~1.0506\\
\end{array}
$$
and last but not least this renders to an image of:

Purple: Bi
Orange: P
Red: O

Oh this is excellent! You've answered from the points of view of several disciplines, how lovely! The last but not least image of the unit cell is what I was hoping for but the list of atom coordinates is helpful too if I want to stay indoors this weekend and make more space-filling models. Visualization is hard for me, I can't figure out the four disembodied (apparently unbonded) orange oxygens(?) not connect to anything, nor why I can't see an outside-the-box oxygen corresponding to 0.7684 0.1052 1.1811 with z>1, but those are challenges for me ;-) Thank you!

@uhoh those are phosphorous. I have ammended a legend. If my crystal maker skills were better there would be more images.


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How can I dissolve photography film?
I am looking to completely dissolve negative film strips. Acetone was my first try and did not work. If anyone knows/ or has any ideas i would be so greatful.

Consider what it is made of (plastic) and see what appropriate solvents are at your disposal or just set fire to it.

AFAIK, films were in last few decades made of acetylcelulose, with the active layer of gelatine. Useful link is https://www.researchgate.net/post/Solvent_for_Cellulose_Acetate

I am very surprised acetone did not work.


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explanation of why interphase occurs in polymers
Im new to chemistry network of stackexchange so please be kind to me!
I heard that interphase is created in clay polymer for 2 reasons.
one of the reasons is the insoluble part of the polymer and the second reason is for second TG.
can some one please explain how these 2 factors cause interphase? or can you recommend articles?
thanks in advance

Do you mean interface? What is "TG"?

Wlcome to chemistry.SE! To me interphase is a biological process. Are you asking about something like a suspension?


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What is the purpose of trimming the bottom corners of TLC plates?
During my chemistry lab course, I was told by my instructor to make 2 small cuts on the bottom corners of the TLC (Thin-Layer Chromatography) plate. Does this make the solvent rise uniformly on the plate? Is this a standard practice in labs? I could not find any information online. In this question the OP has made the cuts as well, but this does not seem to make the solvent rise uniformly.

My guess is so you will remember which is the bottom of the plate. I cannot see how it would improve flow in any significant way.

Never heard of this. Certainly not std practice in any lab I've worked in.

Re: comments, It’s quite a personal thing. Plenty of people I know do it, plenty don’t.

@orthicresol I know plenty of people that don't operate correctly. This is totally another story. One can get decent results, nevertheless. I know lot of  people using ematocrito capillaries to deposite on the plate. It does not make it handbook practice, nor it should recommended. Of course preparing ad hoc capillaries is tedious and requires a higher than usual flame temperature, so many go with far too big capillaries. Etc.

@Alchimista What are ematocrito capillaries? Googling seems of no use. I use glass capillaries with a smaller diameter than melting point capillaries.

Something like melting point capillaries.  That was my meaning. A lot of groups use that (surprisingly! !!). Working approximately does not prevent results. .....

Yes, your question contains the answer. Think about capillarity.
Without the little cuts the eluant would be "sucked" and flow from both the bottom and the left ( and right) edges of the thin plate, resulting in a tilted flow that

tends to move the elute faster along the vertical edges while clumping the spots to the center;
decrease the resolution at the given conditions by deforming and enlarging the spots.

Think of it as a weird bad executed two dimensionally-developed plate, if it helps.
Note that beside resolution, the cuts should be there to guarantee a levelled horizontal and straight flow front, as requisite to obtain reliable and useful ratio front values.
(Good practice: draw the real appearance of the developed plate or even post it to the lab journal, so that a future user get a feeling of the separation whatever it happened. The latter not recommended if a product spots out to be irritant/allergenic).

Left: correct - Right: not horizontal flow frontline due to capillarity at vertical edges.
As OP mentions an example with cutted corners but not levelled flow, obviously cutting the corners but plunging the plate in too much eluant is useless. The idea is to have eluant sucked up just by the bottom edge.

Well I've seen the light now! Never heard of this trick. Surprised it would matter. Is it considered standard practice?

It should be standard practice! Still, a not perfect plate might be of some help, of course. Sometimes you do the same over and over for instance.  But a Rf value shouldn't take into accoubt how far from center a compound/mixture was dropped.

@Night Writer ↑

Would help if TLC plates were sold like this. Cutting your own plates is a good way to have a reportable H&S incident.

Scissors marketing perhaps :) @Waylander. More seriously, normally TLC prepared plates come in a rather large size, or at least these are cheaper,  and then one cuts them in the meaningful size.

Here, when we use 4$\times$8 cm plates there is no need to cut the bottom. But when we cut the cheaper 20$\times$20 cm plates, the elution is much better with the cuts.

@alchimista I still have the scars on my fingers from cutting my own plates.

@Waylander. I have read that before chemists recognised themselves by scars. Apparently drilling corks was dangerous :) Source: Il sistema periodico by P. Levi. It should be translated in many languages. If not done, read it.I love it.

@Raoul Kessels. Genau, is mostly about geometry.

Well, I have always had the impression that when you cut the plate, the aluminium foil bends and silica gets detached from the foil. When put in the eluent, it goes up much faster in this gap then in the center. And since both sides are different, you get an horrible front

I known what you mean. It really goes to practical details.  Let us say that edge effects might be avoid. I would never use a ten cm foil for tre four deposited spots in routines such as reaction following or eluant selection.  Itt is quite some money for even a small but active group.  My answer points to the best practice. Same as I have commented with the calibre of capillaries. But yes the cheapest foils just suffices for individual assessment, either you cut them or use them as they are. Likely they are not very homogeneous in thickness/packing.

Can you cite some book or source which suggests trimming the plate? Also, I am using small plates with dimensions around 4 by 2 cm.

4x2 might be even too short. Now, this really goes to practical details. Again read myt comments, everything is possible/acceptable. Whenever you gett a straight & levelled frontline is fine. They trick helps, especially with a little plate, getting that. No reference. You should see it yourself. Place a "empty" plate in the chamber 30'' or so, pencil mark the front. You should see something like in my sketch.


text:
What do electrons usually do when lead acid battery is idle?
Lead acid batteries operation are governed by Redox reactions for charging and discharging. But, what do electrons usually do when lead acid battery is idle? And why?
Assumptions:

Say more specifically when the lead acid battery is 100% charged and just before we begin to have an intentional discharge on it.
There is no intention to look for a Quantum Mechanics answer
Self-discharge is yet to happen, too

Update: I am not referring to the instant of time between both half reactions. I am only referring to the instant of time just before the whole redox reaction.
Efforts to answer myself but failed:
Half reaction of the redox describes what happens during discharge at the anode where the circuit is about to make most of the electrons necessary for the upcoming load but it does not show the instant of the state of electrons just before discharge:
$$\ce{Pb + SO4^2- -> PbSO4}$$
So, what may the electrons be doing at this time?
Do they just stay in their actual elements till a wire is connected so they can move out?
Or do they just hang out nearby the plates?

Note that at molecular/atomic level, there is no such a thing  like being idle.
If a particular system ( like the particular lead acid half-cell ) is in equilibrium, so macroscopically "nothing happens", there is ongoing dynamic equilibrium.
The rate of the oxidation reaction, providing electrons to the electrode wire, increases exponentially with raising of the electrode pontential.
The rate of the opposite, reduction reaction, absorbing electrons from the electrode wire, decreases exponentially with raising of the electrode potential.
The electrode maintains such a potential, where these rates of the opposite reactions match.
It is the same principle as for gas pressure equilibrium, temperature equilibrium, radiation equilibrium, concentration equilibrium. the both respective opposite processes are ongoing by the same same.
See also wikipedia article about dynamic equilibrium.
The NET redox reaction occurs, if the electrode has other potential than the potential of the redox equilibrium. That happens as a very temporary state just after the electrode is put into the solution, or if an external potential is forced on the electrode, e.g. by the wire. But even if the net redox reaction has zero rate, the opossite half reactions of the equilibrium are still ongoing.
Majority of the motion is due ions leaving and returning to the solid matrice of $\ce{PB, PbO2 , PbSO4}$ to keep the charge (dis)balance to maintain the respective electrode potential.
Both forward and reverse reactions
$$\begin{align}
\ce{Pb + SO4^2- &-> PbSO4 v + 2 e^-\\
PbSO4 v + 2 e^- &-> Pb + SO4^2-}
\end{align}$$
are ongoing at the same speed, effectivelly canceling each other.
The same for the other electrode:
$$\begin{align}
\ce{PbO2 v + SO4^2- + 4 H+ + 2 e-  &-> PbSO4 v + 2 H2O \\
PbSO4 v + 2 H2O &-> PbO2 v + SO4^2- + 4 H+ + 2 e- }
\end{align}$$

I was also referring to the instant just before the other half reaction at the cathode happens. I am not talking about the instant between the two half reactions since they happen almost simultaneously. I will update the question to indicate this clearly. But, I see your answer and will comment on in a separate comment anyways.

But they do happen simultaneously all the time. Just with different rate at different circumstances.

In regards your answer, you said the ions are leaving & returning the solids. This is essentially the reaction but what about just before it? if you meant they are just bouncing back and forth without doing meaningful work, is this a random motion to keep equilibrium? Or do the ions align themselves nearby the electrodes first? I am still not getting what is the "macroscopic state" of situation just before the whole redox reaction.

I thought the redox reaction will not happen till there is a wire from the anode to the cathode. What do you mean by ions leaving and entering the electrode at the same time? I just do not get this. You mean on the surface of the lead (say at the anode for example), ions of $\ce{SO4}$ enters the electrode? then what? if they leave it? for what? I just do not get it...

If you think I need to read about something before understanding your answer, please let me know because I am not familiar to equilibrium notions and a like...

@Gold_Sky See https://en.wikipedia.org/wiki/Dynamic_equilibrium

I moved majority of my comment content to the answer.

Let us continue this discussion in chat.


text:
Oxidation of alcohols using acidified potassium dichromate
What are the products obtained on oxidising primary, secondary and tertiary alcohols using potassium dichromate? I am reading conflicting accounts. My book says that it oxidises all three to carboxylic acids but another book and a site i visited say that primary alcohol is oxidised to carboxylic acid, secondary alcohol to ketone and there is no reaction with tertiary alcohol. Could anyone tell me which is the correct version?

I believe you've got the words oxidized and reduced mixed up, you might want to correct that.

Tertiary alcohol cannot be oxidised (short of forcing conditions which blow the molecule apart) - this is why tBuOH is sometimes used as solvent

@Waylander my book explicitly states that acidified KMnO4 and acidified K2Cr2O7 can oxidise tertiary alcohols (I have added a photo regarding this). Is it impossible in both cases or only for potassium dichromate?

As I said in my comment, this represents forcing conditions that cleave C-C bonds. With control of temperature and reagent quantities, it is routinely possible to oxidise primary and secondary alcohols in the presence of tertiary alcohols. I believe this is what the online source is referring to.

In a sense both are correct. With control of temperature and reagent quantities primary and secondary alcohols can be oxidised in the presence of tertiary alcohols. Indeed acetone and t-BuOH are both used as solvents for such reagents see Jones oxidation.
However it is also true that with concentrated reagents and elevated temperature less activated C-C and C-H bonds can be broken, further reading here. Under strongly acidic conditions tertiary alcohols undergo dehydration and the resulting alkene is oxidised by permanganate across the double bond via the diol to carbonyl compounds more here

I have also studied the same thing that when tertiary alcohols are oxidized, no reaction takes place in the presence of sulfuric acid and Potassium Dichromate(VI).
As no colour change of Potassium dichromate(VI) is seen. Which should become green if the tertiary Alcohol was oxidized.


text:
Charge, multiplicity, level of theory and coordination coordinates of gold compleaxes
Good day,
I need to learn a trick in answering the following questions:

How to come up with cartesian coordinates of a complex say, [Au(CH2)2(S)PPh2]2, to be ran on Gaussian?
The charge and its multiplicity?

Kindly please help.

There are various software to draw cartesian coordinates. Gaussian has its own GaussView, for example. I believe MolView or Avogadro are also popular. You can draw the molecule in any of these, then use the export option to create your coordinate file (export to Gaussian input file or xyz file). Take advantage of the fragment options for things like phenyl ligands. Also, depending on your goals, it's sometimes helpful to align your molecular axis with a principle axis (e.g. z-axis) for easier reading later, if you're using the calculation for more than just basic output parameters.
Furthermore, you can try to find the molecule in an x-ray database and adjust the geometry yourself to approximately match, which can pay dividends later by drastically simplifying the Gaussian calculation, particularly if you are using a larger basis set. You at least want to make sure you're drawing the correct stereoisomer, because the calculation may not correct this on its own.
The charge and multiplicity are manually defined in the input file for Gaussian. I believe you type in the values in the line just above the input coordinates if I recall correctly. You should check the official manual for sample inputs. There's much more syntax to be observed for an input file.

Thank you @Blaise . I have utilized the Avagadro software and worked!!


text:
Why is this reaction not feasible?
The reaction is:$$NH_3+CH_3OCH_3 \rightarrow CH_3NH_2+CH_3OH$$
The mechanism here would be $S_N2$ and since $NH_3$ is a stronger base so it should be a better nucleophile. Why then is this reaction not feasible?
Any help would be appreciated.

Terrible leaving group.

@orthocresol Is it because methanol is a weak acid? Also does that mean strong bases are always good leaving groups?

Energetically also you can easily decide the reaction is highly non-spontaneous. If you consider the reaction mentioned above, the product should be $\ce{CH3NH3^+}$ and $\ce{CH3O^-}$, as both will undergo acid-base reaction. Now you consider the energetics, you are breaking one $\ce{C-O}$ bond, and forming one $\ce{C-N}$ bond. $\Delta H_{C-N} = 305 $kJ/mol, $\Delta H_{C-O} = 358 $ kJ/mol .So, $\Delta H$ for the overall reaction is positive, Entropy change is almost zero, as same no. of species are there on both sides. So, $\Delta G > 0$ and hence it's not spontaneous at all at any temperature.

You will be challenged to find a strong base that is a good leaving group.

I don't think I've ever seen cyanide as a leaving group in an Sn2 reaction.


text:
What is meant by the term composition in chemistry
This may seem trivial at first glance, but It is used a lot in introduction to chemistry chapters and so I need to understand what they mean by composition.
Yes, I have seen the google definitions, I am here because I dint understand them.
And lastly, is chemical composition the same as composition

Composition tells you what components something is made of.
Chemical composition tells you what type of atoms a compound is made of, and how many atoms of each type are in the compound. It is the information contained in a sum formula, for example of the molecule $\ce{H2O}$ (two atom of hydrogen, one atom of oxygen).
In biochemistry, the composition of a protein refers the which amino acids it contains, and how many of each. Similarly, you can discuss the composition of other polymers.
Sometimes, composition does not give you the number of components but only the ratio. For example, I could say a sample of DNA is composed of 56% A and T nucleotides, and 44% C and G nucleotides. Or I could say magnesium chloride crystals contain twice as many chlorine atoms than magnesium atoms, so we write $\ce{MgCl2}$. That makes sense because crystalline magnesium chloride is not made of molecules.


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Equilibrium Constants For Phenol Tautomers
I encountered the following equilibrium constant values for the keto-enol interconversions of phenol in Organic Chemistry (Solomons, Fryhle, Snyder, 3rd Edition):

I wonder what could explain the differences (though I'm not sure whether they're significant) between the equilibrium constants of the two keto-enol interconversions wherein in one case the double bond of carbon with the oxygen and the hydrogens are oriented in ortho position and in para position in the other.
Edit: Can this difference concern the medium the reaction is catalyzed in or directive influence of groups?
Do explanations for equilibrium constant differences exist for other tautomer interconversions or are such queries quite speculative?

Consider what is happening during the tautomerism. More so, consider the movement of the proton during the tautomerism and its position relative to the oxygen.

@LigninPauling By your hint, the high opposite charge separation in the 10^11 Keq tautomer and the relative positions of O and Hs on the unsaturated carbons are factors for equilibrium towards the left-hand tautomer.

I believe that's because in the first step, phenol regains its aromaticity which makes the $K_{eq}$ substantially ($1000$ times) larger than that in the second step where it loses its aromaticity.

Isn't phenol already aromatic? And phenol is non-aromatic in both its keto forms; I didn't get where phenol 'regains' its aromaticity. Do you want to say that in the interconversion with the higher equilibrium constant (cyclohexa-2,4-dienone) the reaction proceeds such that phenol regains its aromaticity?

@Kartik What I'm saying is that phenol is aromatic in its enol form but when you perform the keto conversions we have to break a double bond which makes it lose aromaticity but when you perform the reverse, we re-establish its aromatic properties making it more stable and thus the $K_{eq}$ value is larger. This is also why we hardly ever write phenol in its keto form.

Ohhkay. But how does that explain the 10^3 equilibrium constant difference between cyclohexa-1,3-dienone and cyclohexa-2,4-dienone interconversions with phenol's enol? (The two keto isomers are formed in different interconversions.) The double bond is broken in both cases and the phenol is re-formed in similar mechanisms (I guess it's something to do with directive influence or the like).


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Inductive effect in carbanions
In $\ce{CH3CH2-}$  what should be direction of inductive effect?
If going by order mentioned in my book, it is to be from carbon containing negative charge to CH3, but is not the carbon containing the negative charge more electronegative? So should not the direction be towards it?

The charge on CH2- doesn't say it's electronegative. The direction of inductive effect is from a negatively charged atom to the rest of the uncharged molecule (if there's only one negatively charged atom in a given (canonical) structure) and the effect decreases as we go away from the negative charge. The electronegativity of carbon atoms in organic molecules is more closely related to their hybridization; sp-hybridized carbons are  the most electronegative, followed by sp2- and sp3- hybridized carbons. Thus, your textbook is right.

It seems that you use the term electronegativity incorrectly, electronegativity is the tendency of an atom to draw a shared pair of electrons (or electron density) towards itself. (according to wikipedia)
Carbanion is not electronegative (compared to CH3), it tries to get rid of that extra electron making it relatively electropositive so the electron cloud is being pushed away slightly, towards the methyl group.
So your textbook is right, it’s from $\ce{C-}$ to CH3, i.e electropositive to electronegative.

So it is pushing the negative charge on it or is it pushing sigma electron away from it because of presence of extra electron?

It is pushing away the electron electron cloud/ orbital , towards . Also note that I have used pushing term to make it easy to understand though it’s not precise starement


text:
What is the Gibbs energy of dissolving a solid in pure water?
I am calculating the Gibbs energy of dissolving one mole of solid glucose in pure water such that the final solution has a volume of one liter and a concentration of one mole per liter. I have three different ways of calculating it, and I get two different answers.
The Gibbs energy is a state function, so it should not matter which path I take. At the beginning of the reaction, there is one mole of solid glucose and about one liter of pure water, and at the end of the reaction, there is one mole of glucose dissolved in one liter of solution.
I will use $\Delta_r G$ for the molar Gibbs energy of reaction (dimensions: energy/amount) and $\Delta G$ for the change in Gibbs energy from start to finish (extensive quantity, dimensions: energy).
1. Calculation using Gibbs energies of formation
This corresponds to two processes, turning the reactants into elements and then turning the elements  into products. In the problem, all species are at standard state, so there are no correction terms for concentration.
$$\Delta G_{\text{total}} = \pu{1 mol} \Delta G_f(\text{Glucose(s)}) - \pu{1 mol }  \Delta G_f(\text{(Glucose(aq)})$$
$$= \pu{1 mol }  \Delta_r G^\circ\text{(dissolution)}$$
2. Integrating over Gibbs energy
The reaction starts with no glucose in solution, and then the glucose concentration increases gradually until it reaches the final concentration. During this process, the Gibbs energy of reaction changes because it is concentration-dependent:
$$ \Delta_r G\text{(dissolution)} = \Delta_r G^\circ\text{(dissolution)} + R T \ln(Q)$$
I will use the concentration of glucose divided by 1 mol/L as the integration variable x. Q is equal to x. The amount of glucose dissolved is $c\ V = (x\ \pu{mol/L) }V = x\ \pu{ mol}$. We have to integrate $ \Delta_r G\text{(dissolution)}$ from zero to one:
$$\Delta G_{\text{total}} = \pu{1 mol }\int_0^1 \left[ \Delta_r G^\circ\text{(dissolution)} + R T \ln(x) dx \right] $$
Taking constants and constant factors out of the integral, we get:
$$\Delta G_{\text{total}} = \pu{1 mol } \Delta_r G^\circ\text{(dissolution)} + \pu{1 mol } R T \int_0^1 \ln(x) dx$$
The value of the integral is negative one, so overall we have:
$$\Delta G_{\text{total}} = \pu{1 mol } (\Delta_r G^\circ\text{(dissolution)} - R T) $$
3. Running the reaction at a constant concentration of 1 M
Here, we will use a process that keeps the glucose concentration constant. We place a semi-permeable membrane into the pure water, separating it into two compartments. At the beginning, one compartment (the one in contact with the solid glucose) has a volume of zero. As glucose dissolves, we move the membrane, increasing the volume of the compartment so that the glucose concentration remains at 1 mol/L. We keep doing that over the course of the dissolution reaction until the volume of solution is one liter at the end (and the volume of pure water is zero).
Because all species are at standard state at all time, we can use the standard Gibbs energy of reaction without a term correcting for concentration. This is one component of the total change in Gibbs energy. The other one is work against the osmotic pressure difference between pure water and 1 M glucose:
$$ w = \Pi \times V = \Delta c R T \times V = \pu{1 mol } R T$$
This work represents the difference between dissolving 1 mol glucose into pure water and dissolving 1 mol glucose into a 1 M glucose solution, so we have to add (or subtract?) it to get the Gibbs energy of our original process.
Which calculation is correct, and where are the problems with the other ones?
Calculation 2 and 3 are off by a difference of $\pu{1 mol }R T$ compared to calculation 1.
I have two hunches where the problem might be. First, the concentration of water changes during the reaction (in an ideal solution, it would change by 1 mol/L, I think). I did not include the water in the first calculation, and I wonder if this is related to the discrepancy. Second, I realize that 1 M glucose in water is not an ideal solution, and that I should use activities rather than concentrations. I don't know what happens in an example with much lower concentrations.

For starters the molar mass of glucose is 180g/mol which means your solution is 15 wt% glucose which has a density of ~1.07 g/mL which translates to concentration of about 0.934 M. which is not trivially off

@A.K. - I changed the volume of water so that the final solution has a volume of exactly one liter and a concentration of exactly one molar.

With some hindsight, I should have used something like benzene as solvent and naphthalene as solute to be a bit closer to an ideal solution.

I think the issue with your calculation 2 is that you are missing a term in the expression that you have integrated.
Let's start with the definition that $\Delta_r G = \frac{dG}{d\xi}$. For simplicity, I'm going to replace $\xi$ with $x$ from here on.
For each component of the reaction, the molar free energy $\overline{G}$ is given by $G^\circ + RT\ln a$, where $a$ is the activity, for which we will substitute the concentration in molarity, assuming an ideal solution, etc. Thus, the partial free energy contribution is the product of $\overline{G}$ and the number of moles.
For the aqueous glucose, the concentration is equal to $x$ and for the solid glucose the activity is $1$. The number of moles of the aqueous glucose is also equal to $x$, and the number of moles of solid glucose is $1-x$.
We can now write an expression for the total free energy of the reaction system as a function of $x$:
$$G(x)=xG_{gluc(aq)}^\circ + xRT\ln x + (1-x)G_{gluc(s)}^\circ + (1-x)RT\ln 1.$$
Taking the derivative with respect to $x$, we get
$$\frac{dG}{dx}=G_{gluc(aq)}^\circ + RT\ln x + xRT\left(\frac{1}{x}\right) - G_{gluc(s)}^\circ=\Delta_r G^\circ + RT\ln x + RT.$$
Integration of that expression from $x=0$ to $x=1$ gives the expected result of $\Delta G =\Delta_r G^\circ$ because of the extra $RT$ term that you are missing, which cancels out the extra $-RT$.
For calculation 3, I believe the osmotic work term should be left out, since there is no work done on the surroundings if you treat the entire container as the system. The expanding volume of the glucose solution and contracting volume of the water will move the (frictionless, massless) membrane without any energy input from outside the system.
UPDATE: For calculation 2, one can also deal explicitly with the water to reconcile the disagreement between the known $\Delta_r G$ at standard state and the derived $\frac{dG}{d\xi}$ expression
First, we need to change our reaction to:
$$\ce{Glucose(s) + water(l) -> Glucose (aq) + water (aq)}$$
By "water(aq)" I mean the water in the glucose solution. I'm not sure the correct notation for that.
First, we calculate the standard molar free energy of reaction as $G^\circ(products)-G^\circ(reactants)$:
For the water, I'm going to use a density of 55 mol/L. The exact value doesn't matter. Since we're assuming ideal solutions, one mol of aqueous glucose occupies the same space as 1 mol of water, so to make 1 L of a 1 M glucose solution (ie 1 mole of standard state aqueous glucose), we would add 1 mole of solid glucose to 54 moles of water. The volume of the "aqueous water" in the 1 M glucose solution is therefore 55/54 the volume of pure water. Therefore, the difference in free energy between the aqueous water in a 1 M solution and the same number of moles of pure water at constant temp is given by $-nRTln(V_f/V_i)=-54RTln(55/54)=-0.99RT$.
So our standard free energy change is
$$\Delta_r G^\circ = \Delta_f G^\circ(Glucose(aq)) - \Delta_f G^\circ(Glucose(s)) - 0.99RT,$$
and "calculation 1" multiplies this by 1 mol to get $\Delta G = \Delta_f G^\circ(Glucose(aq)) - \Delta_f G^\circ(Glucose(s)) - 0.99RT$.
We can also see that the free energy of the water as a function of $x$ is $G_{wat}=54G^\circ_{pure}+54RT\ln\left(\frac{54}{54+x}\right)$, where the term in parentheses is the mole fraction of water in the aqueous mixture. We also need to modify the term for the activity of aqueous glucose since the volume is not constant. Instead of being equal to $x$, it is corrected for the volume being less than 1 liter by a factor of $\frac{54+x}{55}$.
Now we do calculation 2 over again. We start with
$$G(x)=x\left(G^\circ_{(glu(aq))}+RT\ln\left(x\cdot\frac{55}{54+x}\right)\right) + (1-x)(G^\circ_{(glu(s))}+RT\ln(1)) + 54G^\circ_{pure}+54RT\ln\left(\frac{54}{54+x}\right)$$
Taking the derivative with respect to $x$,
$$\frac{dG}{dx}=G^\circ_{(glu(aq))} + RT\ln\left(x\cdot\frac{55}{54+x}\right)+ xRT\left(\frac{1}{x}-\frac{1}{54+x}\right) - G^\circ_{(glu(s))} - 54RT\left(\frac{1}{54+x}\right).$$
$$=G^\circ_{(glu(aq))} - G^\circ_{(glu(s))} + RT\left(\ln\left(x\cdot\frac{55}{54+x}\right)+1-\frac{x+54}{x+54}\right)=G^\circ_{(glu(aq))} - G^\circ_{(glu(s))}  + RT\ln\left(x\cdot\frac{55}{54+x}\right),$$
Now, we go back to our equation $\Delta_r G^\circ = \Delta_f G^\circ(Glucose(aq)) - \Delta_f G^\circ(Glucose(s))-0.99RT$, and make a substitution into the above expression to get
$$\Delta_r G = \frac{dG}{dx}=\Delta_r G^\circ + 0.99RT + RT\ln\left(x\cdot\frac{55}{54+x}\right).$$
To check this, we can integrate from $x=0$ to $x=1$, and we get $\Delta G = \Delta_r G^\circ$, so that matches up with calculation 1.
For the $\Delta_r G$ at standard state, things are a little trickier. There is no point in this reaction where the pure water and the aqueous glucose are both present, so if we explicitly include water as a reactant, we can't have all reactants and products at standard state at the same time for any value of $x$. That's the reason for the apparent discrepancy.
In order to determine $\Delta_r G^\circ$ using a function of $x$, we need to have a reaction path for which the standard state occurs at some value of $x$, such as your calculation 3. Another approach that I think will work is to have two beakers of water and add solid first to one and then the other. Halfway through the reaction, you have 1 L of 1 M glucose, 54/55 L of pure water and one mole of solid glucose, so all of the activities are 1. If you redo the above analysis for that reaction, you should get the correct result of $\Delta_r G$ at standard state is equal to $\Delta_r G^\circ$.

Your comments about method 3 are very helpful: G is a state function, so if I find a path where I know how G changes, work and heat transfer does not matter. About method 2: If I understand correctly, you derive the expression $$\Delta_r G = \Delta_r G^\circ + RT\ln x + RT$$ I'm surprised that the RHS is not equal to the LHS for x = 1.

I agree there's something problematic about that result, although $\xi =1$ and $Q=1$ (standard state) do not represent the same state, since solid is present when $Q=1$, but not when $\xi=1$.  To get Q=1, we would need to start with more than 1 mole solid and have $\xi<1$, but I don't think that solves the problem. Dealing with the water explicitly also would only introduce a small correction, so definitely there's more to this. Certainly a really good question.

UPDATE: It looks like -RT is the free energy change of the water, so the +RT term is in the $\Delta_r G$ to correct for that. Explicitly dealing with the water is necessary to reconcile the two. I can add that to the answer or rewrite the answer if you like.

I added a section in the answer to explain where the -RT comes from and why we can't get to $\Delta_r G = \Delta_r G^\circ$ at standard state in this calculation. I think that clears everything up, but I may still be missing something. Thanks for a really thought-provoking question!

Great, thanks for your thoughts sorting this out for me! One more thing: Should $$\pu{\Delta G = \Delta_f G^\circ(Glucose(aq)) - \Delta_f G^\circ(Glucose(s)) - 0.99RT}$$ be $$\pu{\Delta G = \Delta_f G^\circ(Glucose(aq)) - \Delta_f G^\circ(Glucose(s)) - 0.99mol RT}$$ or (not sure which dimensions your Gibbs energy of formation has) $$\pu{\Delta G = 1 mol (\Delta_f G^\circ(Glucose(aq)) - \Delta_f G^\circ(Glucose(s)) - 0.99 RT)}?$$

@KarstenTheis - the latter. The standard Gibbs energies of formation have units of J/mol, as does the 0.99RT term, which is the free energy change of the water per mole of 1 M aqueous glucose solution. So the entire expression is multiplied by 1 mol to get the $\Delta G$ in units of J. I tried to make that clear when I said "multiply this by 1 mol", but including units would have been better. Sorry for the confusion.


text:
Mendeleev periodic table

Mendeleev accurately predicted the properties of gallium(eka aluminium). In his justification of properties he says(according to image given) the melting point of the element would be low. What could have been his insight considering the fact that melting points of  aluminium and indium are 660.3°C and 156.6°C respectively?

What do you think his insight was?

All I could think of is that it's melting point should be intermediate to that of aluminium and indium

Actually it's m.p. is lowest in the group not intermediate, but "low m.p." could be applied maybe even to Al.

I know that it has the lowest melting point in the group ,but in Mendeleev's case he would not have encountered the element right?

Sorry if the question was too basic, but all I was wondering was what Mendeleev was thinking when he mentioned low melting point of "eka-aluminium" ?Also was it relative to all other elements of the periodic table or pertaining only to the particular group?


text:
Ideal gas constant applied to water, lead, and other non-gases
I've asked a number of chemists why the ideal gas constant is in the equation for Gibbs free energy, and all say the same thing: "The name is a relic of the argon gas chamber used to determine energy and temperature at different pressures and temperatures, but it applies to every thermodynamic system." This sounds great and I understand names of constants can be relics, but why can a conversion factor that only appears to pertain to ideal gases be used for other substances? R assumes elastic collisions, where energy is not lost during molecular collisions, but only ideal gases have this property. Wouldn't you use a different conversion factor for something much denser and fluid like water or even more dense like lead to convert from temperature to energy? I've heard that R is used for free electrons as well, which are pretty different from atoms much less gases. Would "R-specific" be best to use everywhere, as I've heard they do in engineering, so maybe every substance has its own temperature-energy conversion factor?

Every substance does indeed have its own temperature-energy conversion factor. It is different for different substances, and generally has nothing to do with R.

YOu should do some research on "kinetic theory of gases" and the "Dulong-Petit Law"

Thank you. Ivan, I see R listed in Gibbs equation, namely the expression RTlnQ. Maybe “this R” is different from “that R” you’re suggesting. Should I take the ideal gas constant and divide by the molecular weight of the substance? I heard that’s “specific R”. Or is there a table of R’s for lots of elements and molecules in various phases? Let’s assume I’m not working with gases at all, like a deep-sea sediment (pure silica fine sand). Could I take R and divide by silica’s MW to get an adjusted energy-temperature conversion factor?

@IvanNeretin I commented below, didn’t hit reply directly. (New to this site.)

No, R from RTlnQ is really the same everywhere, and there are profound reasons for that. I wouldn't call it a temperature-energy conversion factor, though.

@IvanNeretin And profoundly interesting to me. Could you help me find a book or paper I could read about that point about RTlnQ being the same everywhere? I own Gibbs’ long paper from c.1900 but it’s a bit too dense for me.

It is a very good question: Why does R appear everywhere in equations even though it is apparently related to or derived from the ideal gas law? Yes historically R was derived from the limiting gas law in the 1830s.
The beauty of R is that it is related to two physical constants called the Boltzmann constant and Avogadro's constant. Read here
https://en.wikipedia.org/wiki/Gas_constant#Relationship_with_the_Boltzmann_constant
The Boltzmann constant appears everywhere in thermodynamics, semiconductors and statistical mechanics: https://en.wikipedia.org/wiki/Boltzmann_constant

"The name is a relic of the argon gas chamber used to determine energy
  and temperature at different pressures and temperatures, but it
  applies to every thermodynamic system."

I feel this is nothing but fiction. Not sure why people make up stories. William B Jensen traces the history of R in the Journal of Chemical Education.

Thank you! I suppose the next question is does the Boltzmann constant apply to all molecules in all phases? I heard that a mathematical constant is a non-variable value (so any number not a specific number), whereas a physical constant is an immutable value no matter the context. Could a gas like argon at room temp have one Boltzmann constant and a solid like lead at room temp have a different Boltzmann constant? Avagadro’s number sounds like a physical constant whereas the Boltzmann constant sounds like a mathematical constant, subject to the constext, then R responds in kind as their ratio.

The profoundly universal significance of $\rm\bf R$ ultimately stems from Boltzmann's equipartition theorem, which states that (roughly speaking) you are guaranteed to have $\rm kT/2$ of energy per one degree of freedom of whatever nature, or, consequently, $\rm{\bf R}T/2$ per mole of those. It is only in this sense that $\rm\bf R$ may be perceived as a some kind of intermediary between temperature and energy. Real compounds all have different number of degrees of freedom, and hence different temperature-to-energy coefficients (heat capacities, as we call them).
So you see that $\rm\bf R$ is not derived from ideal gases at all, to the point that calling it the "ideal gas constant" might be considered a misnomer. Their relation is of different nature. Like all ideal things, ideal gases are simple; they only have 3 well-defined degrees of freedom per particle, hence the simple formula for their heat capacity. No wonder it is hinged on a universal constant.
Long story short, your bewilderment is misplaced. It is as if we happened to name $\rm CO$ "the Ford gas", because of its presence in the exhaust of Ford cars, and then marveled at its widespread appearance in nature, all the way up to deep space. How could old man Ford put it there? He didn't.
So it goes.

This makes a little more sense. It sounds like the ideal gas law shows gases use R and the Dulong-Petit law shows solids use something closer to 3R. I presume this is because the solid’s molecules have more degrees of freedom (or is it fewer?). I’ve never seen this adjustment in a thermodynamics calculation. Would that mean a liquid like water or a gel have a energy-temperature conversion not of R but between R and 3R? What would the conversion be for an ion in water, maybe around 1.5R because solvated ions move somewhat freely like a gas but bump into water and don’t fly about in a vacuum?

A little?  Thanks for that! Now to the point. You'd better stop using vague terms like "use R" (what does that even mean?) or "energy-temperature conversion" (which is technically OK, but this is not the way people in the field call it). Yes, an ideal gas has heat capacity $c_v={3\over2}R$, which translates to $c_p={5\over2}R$, which grows to ${7\over2}R$ for diatomic gases and then even more. And yes, a solid within a certain idealized model would have $c_v=c_p=3R$, which is pretty much what Dulong and Petit say, and which is more or less true for pretty many elements.


text:
Would electrolysis work with a conductive (but waterproof) divider between anode and cathode?
Note I am not trying to "double fuel efficiency" in a car or something scamey like that. My purpose is to fill balloons with hydrogen and possibly power an HHO torch (also make it safer because pure hydrogen or pure oxygen will not explode).
I found this https://www.thingiverse.com/thing:1200458 HHO generator someone made which seems perfect for my needs except that it outputs HHO together. I cannot find a design for actually useful HHO production that actually separates H2 and O (by useful I mean large volumes of gas, not just a science experiment).
My plan for alterations is to make that design thicker so that it can hold two arrays of plates and split it in halves. I would also add in separate nozzles for each side. I would then print both halves and glue them together with a metal plate (aluminum) in the middle. Would this work? I don't see why it would not however I have not seen a design like this which probably means that I'm wrong rather than everyone else :D

An electrolytic cell divided with an aluminum wall reminds me of an aircraft seen in the "Flight of the Phoenix" (the one they built from the wreckage, I mean). It did fly, but that's not what I would recommend to anyone, unless they are stranded in the desert with no hope of rescue.

A cell with two electrodes separated from each other and an aluminum divider plate in the middle is really two cells: one has an aluminum plate as a cathode, and the other cell has the same aluminum plate as the anode.
In this situation, the solution does not directly connect the external anode and cathode; the aluminum divider separates the solution into two separate cells.
When you pass current (electrons) thru this arrangement, ions travel. Protons (or other cations) travel toward the negative electrode. In the cell attached to the external positive source, protons move away, and discharge at the aluminum. Simultaneously, protons in the adjacent cell move away from the aluminum and discharge at the stainless steel cathode. Hydroxyls (or other anions) do the reverse.
You will need twice the cell voltage to get a high current, and hydrogen will be evolved from the external cathode (-) and the farther side of the aluminum divider plate. Oxygen will be evolved from the other side of this aluminum plate and from the external anode (+).
Now make a small change: let the aluminum divider go from the top of your case to below the water line, say, halfway down. Then ions can travel thru the solution. This is now just one cell. Hydrogen will be evolved at the (one) cathode and oxygen will be evolved at the (one) anode. Your separation of the gas phase by the aluminum divider allows hydrogen to be collected separately from the oxygen. BTW, you might as well use stainless steel for this divider, since it is used for the electrodes.
When collecting the gasses, hydrogen will be evolved at twice the rate of oxygen. If you build up pressure, the cell will respond by pushing liquid from the cathode compartment into the anode compartment to equalize the pressure. You may have to use a check valve or some kind of flow equalizer to keep liquid from spurting  out of one of the gas outlets.

Ahh I see.  I actually planned to make the electrodes aluminum and use baking soda as electrolyte (lye would erode aluminum fast).   I will revise design to have holes drilled in the center sheet near the bottom.    I worry that some bubbles will still get fizz though.   Does the size of the holes make a difference in efficiency?  I assume it would perform as if the plates were spaced farther apart?  Any suggestions for a design that would allow for high volume production of separate gasses?  I did not think about the pressure difference although I did know that hydrogen was created 2x Oxygen.


text:
Sigma bonds, pi bonds and the puzzling symmetry of methane
My A-Level Chemistry textbook assures me that:
a sigma bond is formed when two s orbitals overlap.
a pi bond is formed when two p orbitals overlap.
all single bonds between carbon and any other atom are sigma bonds
(side note) double bonds between carbon atoms consist of a sigma bond plus a pi bond
(side-side note) triple bonds between carbon atoms consist of a sigma bond plus 2 pi bonds.
However, the electronic structure of carbon is [He]2s22p2, implying two s orbitals and two p orbitals. I deduce from this that e.g. methane (CH4) has 2 sigma bonds and 2 pi bonds.
I am pretty sure that my deduction is incorrect, as it contradicts the statement in bold above, but cannot see why.
(In addition, the book states that methane has a completely regular tetrahedral structure with all four C – H bonds exactly the same, at identical angles to each other.)
Edit: Although aspects of this question overlap(!) with Can a s orbital overlap with any p orbital to form a sigma bond?, I am most interested in why what appears to a mixture of s- and p- bonds ends up creating such a symmetrical molecule as CH4.

Sigma bonds can involve any type of orbitals, as long as they have right geometry. Those in methane can be considered  to be made from sp3 from C and s from H.

Now it is just about time that someone utters the word "hybridization".

You'll get your answer here satisfactorily: https://www.chemguide.co.uk/basicorg/bonding/methane.html

@Kartik Many thanks for the link. Lots of really clear explanations there - I spent a couple of fascinated hours reading the site.

I am going to attempt to answer this, based on the link supplied by @Kartik and expanding on @Mitheron’s comment above.
As we know, covalent bonding is the process of “filling up” partially filled outer orbitals by sharing electrons between atoms. Carbon’s structure [He]2s22p2, or, more accurately, [He]2s22px12py1. When bonding occurs, however, two things happen:
1.  One of the 2s electrons gets “promoted” to a p orbital. This is an “excited” state in which the electrons have slightly more energy (though not enough to remove any of them). The configuration therefore becomes [He]2s12px12py12pz1, and voila, we now have 4 unpaired electrons available to bond with other atoms.
2.  The extra energy allows the orbitals to “hybridize”. In effect, the 2s and 2p orbitals morph into 4 identical orbitals of slightly higher average energy. These orbitals are called sp3 orbitals (one part s, three parts p). The electronic structure of the ready-to-bond atom is now [He] (sp3)4.
These four sp3 orbitals of course repel each other equally, and hence end up pointing towards the vertices of an imaginary tetrahedron. Hence the regular structure of CH4 and analogous compounds.
Returning to the question, the sigma bonds formed by carbon are neither from overlapping s or overlapping p orbitals, but from overlapping sp3 orbitals.
The thing that makes a bond a sigma bond is not the type of subshell from which the electrons “originated”. It is the fact that the orbital overlap directly, with the two atomic nuclei and the main concentration of the shared electron cloud in a straight line. Pi bonds, on the other hand, overlap “sideways”, with the main concentrations of the bonding electron cloud off to the sides of the bonded atoms.
Please chime in, people, if I have omitted or misunderstood anything.


text:
Changes over time in sodium hydroxide concentration in a hot water bath used for pretzel manufacture
I make soft pretzels using a hot (190 °F) and dilute (2-4% w/v) sodium hydroxide solution. The bath is kept hot in a open-top stainless steel vessel like a stockpot. I make my pretzel dough and dip the dough in the hot solution for about 20 s prior to baking. As you can imagine, there is a fair amount of evaporation, as well as absorption by the dough of the solution.
Can someone help me theorize what happens to the sodium hydroxide concentration over time? pH is already 14, so pH is not a good measure of concentration. I don’t need to measure precisely, but I would like to understand if concentration is increasing as water evaporates, or if concentration stays the same because the solution is evaporating, or if there is a decrease in concentration by absorption of the pretzel dough.
This will help me understand what I need to do to keep my production going without running out of solution. In other words, do I just add water back to the bath as it evaporates, do I add water and the same concentration of sodium hydroxide I did in the beginning? Or other?

If you measure the density, that might give you a better metric.

Welcome to Chemistry.SE! Questions are the central focus of this site, but we ask that questions demonstrate an attempt to solve/understand the question to maintain quality.  Please take a minute to look over the help center and tour page to better understand our guidelines and question policies.

One posting (https://www.npr.org/sections/thesalt/2014/08/09/338591194/for-a-proper-pretzel-crust-count-on-chemistry-and-memories) suggests lye is a browning agent; so perhaps as long as the pretzels come out brown it is working.

pH may still work for you. Take a sample of the fluid. If pH is 14, then dilute it, take pH, and do the math.

Measuring pH 14 may be tricky, as most of glass of pH electrodes does not like it. Using pH papers may not provide needed resolution.

You need both.

Adding water, as water evaporates, but hydroxide does not.
Adding the hydroxide solution, as it is taken away by pretzels

But I suppose the evaporation is the major effect.
Probably the most convenient way is measuring density, with areometer $\pu{1.0-1.2 g / ml}$
Additionally, hydroxide gradually absorbs carbon dioxide, becoming sodium carbonate. pH paper strips may do good here, if solution looses it's alkality.


text:
Why does the exchange interaction in Hartree-Fock theory lower the total energy?
In Hartree-Fock theory, the expectation value of the total energy can be written as
$$E = \langle\Psi| H |\Psi\rangle = \sum_{a} \langle a| h |a \rangle + \frac{1}{2}\sum_{ab} \big( [aa |bb] - [ab|ba] \big),$$
where the first summation contains the single electron kinetic energy and the interaction energy of the external field. The second summation contains a classical Coulombic term plus an exchange term. The exchange term is introduced by the antisymmetrization of the wavefunction (through a Slater determinant) and takes the form of
$$[ab|ba] = \int \mathrm { d } \mathbf { r } _ { 1 } \mathrm { d } \mathbf { r } _ { 2 } \psi _ { a } ^ { * } \left( \mathbf { r } _ { 1 } \right) \psi _ { b } \left( \mathbf { r } _ { 1 } \right) \frac { 1 } { r _ { 12 } } \psi _ { a } ^ { * } \left( \mathbf { r } _ { 2 } \right) \psi _ { b } \left( \mathbf { r } _ { 2 } \right).$$
This integral is always positive, and so the exchange contribution to the total energy is always negative. Can anyone explain why the exchange contribution to the total energy is negative? In Fermi-Dirac statistics, we can see that the effect of Pauli exclusion of indistinguishable Fermions is an effective outward pressure. Hartree-Fock theory seems to suggest that the exchange interaction lowers the energy of the system, and therefore stabilizes it. Shouldn't the effect of exchange, therefore, be an effective inward pressure?

Would you check if this is answered by https://chemistry.stackexchange.com/a/61185/39153 ? It's quite a bit to read, but the relevant bit is in there: the $[aa|bb]$ term is valid for a Hartree product - it intuitively cannot be valid on its own for a Slater determinant. The latter disallows some combinations of orbitals/orbital forms because it does account for Pauli exclusion. $[aa|bb]$ does not accurately calculate the resulting energy by itself, so $[ab|ba]$ must be included to correctly lower the energy.

I understand how the mathematics of a Slater determinant give rise to the form of the exchange integrals. I am having difficulty reconciling the fact that if we used a Hartree product, which is equivalent to removing exchange antisymmetry/Pauli exclusion, then the total energy is higher. This suggests bonding would be weaker if Fermion exchange antisymmetry did not exist. How could the same principle that is responsible for stronger bonds also cause an effective outward pressure in  Fermi statistics?

@CuriousChemStudent I believe the common explanation of why exchange lowers the energy is that it allows the electron motion to be somewhat more correlated. Same spin electrons will avoid each other due to the exchange interaction, reducing the unfavorable electron-electron repulsion and thus lowering the energy.

Thank you, I get it now! There is an effective outward pressure created by a reduction in effective volume cause by Pauli exchange. The volume that is excluded is entirely the highest energy configurations of phase space. Therefore, the electrons correlated by Pauli exchange do not on average exist in locations of high energy with respect to each other. This causes an effective reduction in energy.

@CuriousChemStudent that's how I've understood it. You could convert your comment to an answer with a bit more detail and I think it would be well received.

@Karsten Theis, since this question did not get any answer after almost a year, and even after 6 days of having a bounty, perhaps you might be interested in committing to a stack exchange more dedicated just to computational and theoretical modeling: https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=YjY4NDU3YTdkODc4MDExNTg3NzEzN2JjODYyNTkzZWRmZjJkMTkzMDJlOGM1OGFhNTFhOWJiMTI4MGYzNGFmOAWUUVpW31KAHeTeh53E9BlpdwxKi3ms8oRyVzIYlFDs0

Can anyone explain why the exchange contribution to the total energy is negative?

I find it misleading that exchange interaction is treated as something that changes total energy of the system. This lowering of energy is actually due to the Hartree-Fock scheme being in principle inexact, and is not really specific to indistinguishable particles.
Let me elaborate. Consider a hypothetical system of distinguishable particles with exactly the same masses, charges and spins (and all their other properties entering the Schrödinger's equation). Let these parameters be equal to those of an electron. Now, if you solve the Schrödinger's equation for this system in the electromagnetic field of an atomic nucleus, you'll get some (spinor-valued) wavefunctions $\Psi_n$, where $n$ enumerates the states. These wavefunctions won't in general be antisymmetric.
But the symmetry of the Hamiltonian, namely its invariance with respect to exchange of a pair of particles (since all their relevant properties are the same), tells us that there's a large degeneracy. If you linearly combine the degenerate states corresponding to the same energy so as to get symmetric $\Psi_k^{\mathrm s}$ and antisymmetric $\Psi_k^{\mathrm a}$ linear combinations, these will still be solutions of the Schrödinger's equation we started with.
If you throw away the symmetric solutions and only leave the antisymmetric ones, these will actually be the solutions for the case where our particles are indistinguishable. You see? We've just arrived at exchange interaction between indistinguishable particles, without any change in energy. What we got is just a possible increase in ground state energy due to losing a bunch of states (and thus renaming one of the formerly excited states to the ground one).
Now, why do we get lowering of states in inexact schemes when we introduce such antisymmetrization? That's directly related (at least for the antisymmetrized ground state) to the variational principle, which tells us that an approximation of a ground state will always have mean energy $E_{\text{approx}}$ (which is the Rayleigh quotient) obeying
$$E_{\text{approx}}\ge E_0,$$
where $E_0$ is the exact ground state energy.
Before introducing the Slater determinant, the trial wavefunction was smooth and generally nonzero at the loci of electron-electron collisions. But the exact (even non-symmetric) eigenfunction should have a local minimum (a cusp, see here for an example) or a (smooth) zero at such loci due to the Coulomb repulsion. After antisymmetrization, many of such loci of collisions will get zeros. This will make the approximation better, and the mean energy lower, closer to the exact eigenenergy.
If instead of Hartree-Fock method we used some other approximation that had begun with exact treatment of electron collisions, sacrificing something other, then such antisymmetrization might not lead to any improvement in estimation of eigenenergies (limit case being very precise numerical treatment yet disregarding Pauli principle, which is basically equivalent to an exact solution for distinguishable particles).

The last paragraph reads to me like it contradicts the rest. The fact that exchange leads to zeros for same spin electron collisions (Fermi holes) should change the total energy.

I like this answer, it definitely gave me a different perspective. This same reasoning can then be applied to correlation energy, guaranteeing exact correlation energy is always negative (or zero). So would Bosons correspond to the symmetric results, meaning that a hypothetical system of Bosons would have lower energy than the same system of Fermions?

@CuriousChemStudent well, if you take the same spinors for bosons as you took for fermions (however unphysical that could be), or if you ignore spin-dependent energy-changing interactions like spin-orbit interaction (or ignore the spin altogether), then yes, the system of bosons will have lower ground state energy.

@Tyberius I've edited the ending. Hopefully it's a bit clearer now.

I have been watching this question with keen interest, but wanted to let someone else go for the bounty -- However it expires tomorrow and no one has answered, so I will give my perspective.

Why does the exchange interaction in Hartree-Fock theory lower the
total energy?

The answer is given in the question:

This integral is always positive, and so the exchange contribution to
the total energy is always negative.

Again, I appreciate that this very simple answer may not be satisfactory for some, which is why I was waiting for someone else to go for the bounty.
A comment by the person who originally asked the question (not the user that offered the bounty) offers an attempt to explain why it might make phenomenological sense for the energy to go down, and I will not repeat that here because it's really them that deserves to get any points that might be awarded for that line of thinking.
However I have a different perspective and it is this:
Feel free to try to make phenomenological models or analogies in your head (such there being an outward "pressure" or "reduction in effective volume") if it helps you in any way (maybe to remember things easier, or for your own peace of mind). But in my experience, such "analogies" using "classical physics" thinking (like pressure and forces) do not in quantum mechanics always work the way you would expect. A good example is quantum entanglement, which has absolutely no classical analog; and another example is quantum spin, which is a bit like classical angular momentum but is not precisely the same.
The best answer really, is that the exchange integral is always positive, and in the equation for the energy it has a negative sign, so it will always reduce the total energy.
You can come up with phenomenological ways to conceptualize this, and perhaps someone else will come up with a valid and convincing argument for why that conceptualization does not work, but the bottom line is that the integral is mathematically always positive, so by definition of the "exchange integral", the "exchange interaction" always lowers the energy here..

Thanks for getting the ball rolling and for your insight!

I'll try to give my interpretation of the "physical" explanation of why exchange would lower the energy.
For the true wavefunction, the motion of all the electrons should be correlated, with the classical view being that the electrons are avoiding each other to minimize repulsion. With Hartree-Fock, we find an approximate wavefunction by solving for 1-electron functions using the average potential of the other electrons rather than having each electron feel the instantaneous potential of all the others. This eliminates most, but not all of that correlation.
Exchange terms arise due to the Pauli Exclusion Principle and the fact that electrons are fermions. This causes same spin electrons to be correlated, as they can't occupy the same space. While these electrons aren't fully correlated as they would be if you did Full CI, the exchange part of the correlation is reproduced exactly by Hartree-Fock. If not for exchange, these electrons could be located close to each other, which would be highly repulsive and increase the energy. So by avoiding these interactions, the energy is reduced and this correlative effect is what leads exchange to always be negative.
There is a little more in this same vein on Physics SE and the sources therein.

I would remove the 'somewhat'. The exchange correlation is recovered exactly by HF as you write they cannot occupy the same space. This is why we mix in HF into hybrid DFA and often term it exact exchange. Good and intuitive answer!

@Martin-マーチン I can see how that phrasing would be confusing, I was trying to say that while they don't have the overall correlation from considering the full instantaneous coulomb interactions, they are still partially correlated due to exchange. I'll change it to make it a little clearer.