text:
Heating NaOH solutions
I am trying to determine the best heat at which to use an NaOH solution to clean a textile fiber. This is being done, carefully, in Asian villages.
The raw material is ribbons of bark peeled from a plant stalk. The fiber adheres to the inside of the bark. We need to loosen the bark, at least. Having it degrade and fully separate would be better as long as that doesn't damage the fiber too much.
At present, we prepare a relatively weak caustic solution and bring it to a boil with bark ribbon in it. It sits and cools overnight and then the material is removed and washed. So, the temperature ranges from ambient (22-34C) to 100C and back to ambient.
I will figure out the % of NaOH we have in the water, but I'm hoping to reduce the quantity we use as it's one of our biggest costs.
I think boiling the solution is causing it to be less powerful in treating the bark/fiber, but I'm not at all sure. I am, obviously, not a scientist!
Thanks.

We use 30kg of NaOH flakes per 1000 liters of water. I'm not sure how to calculate this as a solution.

I don't think boiling the solution is causing it to be less powerful. If anything, it is probably the opposite.

Your initial solution consists of 30 kg NaOH in 1000 kg of water; it will treat a certain amount of bark ribbon. I suspect that the final pH of the solution is much less than the original pH; i.e., the solution gets used up by reacting with the plant material. You mention that one of the biggest costs is for NaOH.
Optimizing the procedure could involve the reaction temperature, as you suggest. Generally, heating makes reactions go faster and more completely. Occasional stirring might be helpful. One way to check on the reaction is to measure the pH with (commercial) paper strips designed for high pH measurements; as the pH drops below some value, the useful part of the reaction may stop.
One way to reduce cost could be to wash the bark with plain water before treating it with NaOH solution. The purpose of the plain water wash would be to remove water soluble sugars, starches and acids that otherwise consume the NaOH, leaving it to do the harder job of disadhering the fibers.
Another way could be to substitute sodium carbonate (soda ash: https://en.wikipedia.org/wiki/Sodium_carbonate) for the sodium hydroxide. Prices vary for many reasons, and it is not clear whether you need the high pH of NaOH if you can boil it and allow the solution to cool slowly. You could consider insulating the kettle to keep it warmer longer. Sodium carbonate is available in many forms, from anhydrous to monohydrate to decahydrate, and technical grades in various purities; you will need a little chemistry to calculate the actual cost of the operation with each of these variations. Sodium carbonate's lower pH would be more gentle to the desired fibers.
I checked previous prices in the US: 1 mole of NaOH was about 3 to 4 times as expensive as a half mole of Na2CO3.H2O (the monohydrate). If the pH of sodium carbonate (~11.5) is adequate for the operation, you may not need the (initial) pH ~14 of your 3% solution (0.75 M). And you could use more sodium carbonate if that helps.
Another possibility for reducing costs is liquid (50%) NaOH. This is usually cheaper than pellets or flakes, and may be more convenient (faster dissolution) but does present an extra hazard of splashes and spills. Eye protection is a requirement, because a splash in the eye means blindness. (Of course, a splash of 3% NaOH is just about as dangerous.)

THANK YOU! James Gaidis. It's so difficult to get help with this. I'm trying to learn the science to make decisions and your answer really helps.
I'm going to read this a couple of times and then get back to you with more questions, I'm sure.
We tried soda ash a long time ago. I don't remember the results, but we were trying it in complete ignorance. Historically, villagers used ash from their fires, so a lower pH base is possible.

Is it possible to speak with you directly? I could really use some help with this as I have so many more questions.

@EllieBelly: I've asked how to do that. I'll get back to you when I find out.

@EllieBelly: Put your email address in your profile. After we establish communications, you can modify your profile and delete your email address.


text:
Driving force in Corey-House reaction
In my textbook of organic chemistry by francis carey 11th edition, the Corey-house synthesis is elaborated on in the topic of organocopper reagents. It is mentioned that as the R-X alkyl halide is added to the [R-Cu-R]- group, the electron count of copper increases from 14 to 16 and the oxidation state increases from +1 to +3. I interpreted this as a "driving force" to make sense of why this reaction occurs and why it was mentioned. the driving force being that as long as the reaction makes the electron count of copper approach 18, it is favoured.
however this driving force does not hold for the reductive elimination part of the Corey-House synthesis where the electron count of copper is reduced from 16 to 14. it was mentioned that this decreases the oxidation state from +3 to +1.
It seems very strange to me that in the oxidative addition, the reaction is favoured due to increase of electron count but the reductive elimination is favoured due to a more preferrable oxidation state of copper. it does not feel fair as it can be seen the other way around, where oxidative addition is not favoured due to change in oxidation state and reductive elimination is not favoured due to change in electron count.
So my questions would be what exactly drives the reaction forward? It does not feel right to use separate driving forces for this scenario.
And instead of oxidative addition and reductive elimination, is it possible that the ligands reversibly bind to the copper? and the R-R' desired product is formed when the reverse reaction occurs,where they are unbound from the copper and bind to each other

This mechanistic fluff is irrelevant as far as "why reaction happens" is concerned - delta G of reaction is.

@Mithoron why is that so?

Because that's how thermodynamics works? Thermodynamic functions of substrates and products (and irreversible steps if there are any) - that's important.

In general metal halide salts have high lattice enthalpies, which make them very stable compared to metal alkyl or alkyl halide reactants. This isn't always true, but most of the times, the driving force is the high lattice enthalpy of the salt i.e. thermodynamics.

@Mithoron sorry i should rephrase. Is it not possible for both the mechanism and ▵G to work together to come up with a good explanation?

source: Stack Exchange

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

sort: votes


text:
Where do MSDS data sheets get their information? (How to write one)
For example looking at Sigma-Aldrich MSDS sheets, where do they cite their studies that found hazard in the chemicals? Or in other words, how does one "write" an MSDS sheet? I need to write one for a solution that I designed but I don't know how I should approach writing it other than just combining MSDS data on the components of the solution individually. Thanks

Industrial Hygiene groups produce various handbooks and other collections of data. So if you solution is just a variety of known chemicals you should be able to get some info. For new chemicals, you would be surprised how little info an SDS actually has to contain...

@JonCuster Yes it's a solution containing well studied chemicals, where could I access these "industrial hygiene group handbooks"? Thank you

Well, your local IH professional is the best bet. One such group is the American Conference of Governmental Industrial Hygienists who put out things like https://www.techstreet.com/standards/2021-threshold-limit-values-tlvs-and-biological-exposure-indices-beis?product_id=2198547 - a good library may have copies, or can get them through inter-library loan if you don't have an IH professional around.

@JonCuster I tried this but I couldn't find some of the chemicals I needed. For example Thymol. Most MSDS cites OSHA 2012 standard saying its hazardous but I can't even find Thymol in the OSHA database or any of these hygiene handbooks

Google a Thymol SDS. I did and got "This chemical is considered hazardous by the 2012 OSHA Hazard Communication Standard (29 CFR 1910.1200)", along with categorization of Acute oral toxicity, and others...

Yes but how can I find the "This chemical is considered hazardous by the 2012 OSHA Hazard Communication Standard (29 CFR 1910.1200)" reference? As in, where did they get that information in the first place? Got a link?

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


text:
Conditions for a substance to show deliquescent or efflorescent behaviour
I have learnt that deliquescent substances are the ones which absorb moisture from the surroundings and convert them to solution and efflorescent substances are the ones which release moisture into the surroundings.
But I am confused about the mathematical relations used to determine whether the substance will be efflorescent or deliquescent ?
My attempt at trying to understand it :

I have read that for deliquescent substances the vapour pressure of the solution formed should be less than the partial pressure of water vapour in air and it should be the other way round for efflorescent substances.

But I am unsure about this as my teacher compared such the behaviour of such substances to the real life example that if we want to come out of a door, the pressure exerted by us should be more than the pressure exerted from outside and this makes me think that for deliquescent and efflorescent substances I should rather be comparing vapour pressure of the solution or the substance with the total atmospheric pressure?

I am really confused about this. I tried searching for this elsewhere but couldn’t be sure about this.
Any help would be greatly appreciated!

Total pressure plays role only if we consider boiling conditions, otherwise partial vapor pressure is considered. Thus applies as well to pure water as to this case.

First forget about atmospheric pressure. The vapor pressure of a substance is typically not significantly affected by the applied pressure, at least under atmospheric conditions. Now consider how you might run an experiment. You take a weighted sample of the dry substance and place it in a sealed container (filled say with an inert gas) together with a drop of water (not in contact). You control the temperature and wait. Some time later you take the sample and again measure its weight. If the weight increased, it is deliquescent, if the weight decreased, it is efflorescent.


text:
Can nitric oxide (NO) reduce xenon fluorides
I came across the following question:

$\ce{XeF4}$ on reaction with gaseous nitric oxide forms xenon gas and (P). Find the total number of lone pairs present in one molecule of P.

As far as xenon chemistry is concerned, I just know that xenon fluorides undergo hydrolysis giving different products like Xe, $\ce{XeOF4, XeO2F2, XeO3}$.
I have no idea that xenon fluorides even react with nitric oxide. I did a lot of digging on Google but could not find a single source/article discussing about the reaction of xenon fluorides with nitric oxide or the other way round.
I would be glad if someone can tell about the products of the given reaction with a link to the source. The answer the question is given as '6 lone pairs'.

I have no idea that Xenon Fluorides even react with nitric oxide.

Xenon fluorides do in fact react with nitric oxide. Following is the abstract of a 1964 paper(See ref.):

The rate of reaction of $\ce{XeF4}$ and $\ce{XeF2}$ with $\ce{NO}$ and with $\ce{NO2}$ has been studied between $\pu{300° - 350°K}$ at pressures between $\pu{0.1 - 30 mm}$. The second‐order rate constant at $\pu{300°K}$ and the activation energies are, respectively: $\ce{XeF4 + NO}$, ($\pu{3 liters/mole sec, 7 kcal/mole}$); $\ce{XeF2 + NO}$, ($\pu{0.3 liters/mole sec, 10 kcal/mole}$). The rate of $\ce{NO2}$ with $\ce{XeF4}$ and $\ce{XeF2}$ was too slow to observe, $\pu{k < 0.01 liter/mole sec}$. The fluorides of xenon react with the oxides of nitrogen much more slowly than $\ce{F2}$ reacts with $\ce{NO}$ or $\ce{NO2}$. By means of a Polanyi—Semenov plot, the reactions of $\ce{XeF4, XeF2}$, and $\ce{F2}$ with oxides of nitrogen were combined to give an estimate of the successive dissociation energies of $\ce{XeF4}$, which are (kcal/mole): $\ce{XeF4→XeF3 + F (48)}$; $\ce{XeF3→XeF2 + F (15)}$; $\ce{XeF2→XeF + F (54)}$; $\ce{XeF→Xe + F (11)}$. From these kinetic and thermochemical data a strong case can be made for the stepwise reduction of $\ce{XeF4}$.

Ref.:Reaction Rates of Xenon Fluorides with Oxides of Nitrogen by Harold S. Johnston and Robert Woolfolk, J. Chem. Phys. 41, 269 (1964); DOI: 10.1063/1.1725632

@ Nilay Ghosh. So basically the Fluorine atoms react to form $F_2$, which has 6 lone pairs. Thanks a lot.

@AmmaarahFatima Yes, fluorine is a byproduct although most of them further reacts with remaining nitric oxide to form nitrosyl fluoride


text:
Radius ratio of octahedral interstitial site in BCC lithium
Problem

Determine the minimum radius of an atom that will just fit into (a) the tetrahedral interstitial site in FCC nickel and (b) the octahedral interstitial site in BCC lithium.

Solution

(a) For the tetrahedral site in FCC nickel $(a_0 = \pu{3.5167 Å}):$
$$r_\ce{Ni} = \frac{\sqrt{2}(\pu{3.5167 Å})}{4} = \pu{1.243 Å}$$
$r/r_\ce{Ni} = 0.225$ for a tetrahedral site. Therefore:
$$r = (\pu{1.243 Å})(0.225) = \pu{0.2797 Å}$$
(b) For the octahedral site in BCC lithium $(a_0 = \pu{3.5089 Å}):$
$$r_\ce{Li} = \frac{\sqrt{3}(\pu{3.5089 Å})}{4} = \pu{1.519 Å}$$
$r/r_\ce{Ni} = 0.414$ for an octahedral site. Therefore:
$$r = (\pu{1.519 Å})(0.414) = \pu{0.629 Å}$$

In the BCC crystal lattice the radius ratio for the octahedral site, as I learnt it is $~0.155,$ which is derived taking into account the distance between two body centered atoms.
However, in the above answer the radius ratio used is $~0.414,$ which as I've learnt is the radius ratio for the octahedral interstitial site in a FCC crystal lattice.
Is the answer given in the text a mistake, or have I understood something wrong?

Please typeset your question: others who might search for this material will be impeded by the image.

It’s a mistake, https://m.youtube.com/watch?v=rRGJ8zoGYEI&feature=youtu.be

Shouldn’t that be ‘maximum’ radius that will just fit? The minimum radius is 0.

Actualluy it was a mistake. The impurities usually fall into the larger interstitial sites of crystal structures. According to this problem solving chapter, the interstitial sites in FCC and BCC crystal structures are described as follows:

For both FCC and BCC crystal structures, there are two different types of interstitial sites. Of these two interstitial sites in each crystal structures, the one site is always larger than the other and is normally occupied by impurity atoms.

Karsten Theis has given excellent video to show which one of the two interstatial sites is the larger than other, with calculations to show that (in here, the radius of the host atom is $R$ while that of the guest or thw impurity atom is $r$). Thus, the octahedral interstitial site with $\frac{r}{R}$ ratio of $0.414$ in FCC is the larger one between two sites, while that in the BCC is the tetrahedral interstitial site with $\frac{r}{R}$ ratio of $0.291$.
Note that the $\frac{r}{R}$ ratio of the tetrahedral interstitial site of FCC is the value of $0.225$. The $\frac{r}{R}$ ratio of the octahedral interstitial site of BCC is equal to the value of $0.154$.
According to the answer given in the solution manual, it has been using both interstitial sites of only FCC crustal structure for the calculations.
The different view of two larger sites in bothe crustal units are shown in following image:

A few typos: 1st line Actually, and in bold cyrstal.


text:
How is it decided that a particular chemical or compound is useful for a very specific purpose?
In textbooks, I often find extremely specific uses of chemicals. Is chemistry research mostly focused towards finding specific use cases?
For example, I know that boric acid kills snails. How do chemists target research for very specific uses of chemicals and compounds? Are these results purposely tested or discovered?
In the case of finding the best compound to kill snails, are different compounds tested or is the snail studied in great detail and compounds then crafted to target a snail-specific set of criteria?

The current body of knowledge of chemistry is a collection of at least a thousand- year old body of experiments done by a large number of people from all parts of the world. Borax (salt of boric acid) was known to alchemists. Modern understanding of chemistry is relatively young (less than 100 years old). Unfortunately it is impossible to trace and associate each and every known application of chemistry to a certain person who studied it.

When you see applications of certain chemicals, there must an (unsung) chemist behind it. Most of the discoveries are not accidental though. Snails or slugs can become pests in gardens. Companies like Bayer or other pesticides companies do invest money in research to find out effective and relatively safe solutions to pests.

The way to particular application is from wide range beginning at old good accidental empirical knowledge to quantum chemistry computation of new drug candidates as the first round preselection before synthesis such compounds.

Beautiful question! I thought this myself, and I think I can partly answer at least for the case of drugs we make as mediciene. So in the body, the enzyme and most important type of biomolecules fit it into very specific places either of the body or into other molecules. Now, once we know this shape , we just need to create a drug which fits there. Search up Key and Lock Mechanism, Also check David Klein's organic chemistry, he has about 10 examples in his book about how  this could be achieved.


text:
What is "super triple phosphate" and is it soluble or not?
I have been working on some detergent formulations and some specifications call for "super triple phosphate". The problem is that the formulations generally require all the inputs to be water soluble and in my experiments with "super triple phosphate" purchased naively, it is not a soluble substance.
So, in other words, I have bought product advertised as "super triple phosphate" which is generally available as a fertilizer, and tried to use it and it does not work because it does not dissolve in water.
So, what is going on here? Some of the possibilities:

The written formulations I am working from are simply incompetent BS
that do not describe real working detergents
There are different kinds of "super triple phosphate", some of which
(fertilizer) are not water soluble and others of which are. In this
case, I need to know what the soluble form is called so I can find
it and order it.
The "super triple phosphate" needs to be processed in some way to
make it soluble

Could it be triple superphosphate $\ce{Ca(H2PO4)2}$ (TSP)? Looks like the manufacturer decided to distort the name.

@andselisk Ah hah, that seems to be the explanation. Some of the formulations I am working from come from a non-English speaker, so apparently they just screwed up the order of words.

FMI, you can find the history and usage of the term "super" in this question and you'll find why it might create confusion. The term had been used since antiquity so there is no proper rationale but has become a staple term in agricultural fertilizers.

Detergent formulations would avoid introducing calcium ions in washing machines or handwashing formulations, so calcium salts are a no-no. Calcium phosphate is a good plant fertilizer, because it slowly release phosphorous in the soil. This is most likely what you purchased-a slow release fertilizer.
On the other hand, sodium triphosphate was a popular additive in laundry detergents. Unfortunately, it is also called trisodium phosphate (TSP). This is water soluble. Regardless, phosphates are discouraged now because they promote algae growth in waters. Whoever wrote triple super phosphate in your formulations was confused by the nomenclature.

Not sure what super triple acid is;  but super phosphoric acid was used in agriculture ( I do not readily find a listing on the net today). It was 107 % phosphate as I remember. Fully water soluble as custom liquid fertilizers were mixed  using it ( by Amoco and no doubt others ). It was stored hot ( 200 F ) to make it more fluid at the individual fertilizer bulk plants. At ambient temperature it was the consistency of heavy syrup. I happen to have waded in about 8" of it to inspect rubber linings in storage tanks. It was shipped in insulated , 316 SS lined, rail cars. There were two types , a "furnace" grade which contained traces of all sorts of stuff like Fl ; I forget the other grade but it was more expensive so not used as much in fertilizers. Today I use TSP as fertilizer and for cleaning and had not noticed it to be very difficult to dissolve in water.


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Numerically solving chemical equilibrium equations
Here are chemical equilibrium equations:
$\begin{cases}
\mathbf{N}^\text{T}\mathbf{X}+\mathbf{C}=\mathbf{Y}\\
\mathbf{N}\ln\mathbf{Y}=\ln\mathbf{K}
\end{cases}$
Here $\mathbf{C}=\begin{pmatrix}c_1\\c_2\\\vdots\\c_n\end{pmatrix}$ are initial concentrations (mol/L) of n types of substances;
$\mathbf{N}=\begin{pmatrix}
\nu_{11}&\nu_{12}&\cdots&\nu_{1n}\\
\nu_{21}&\nu_{22}&\cdots&\nu_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
\nu_{m1}&\nu_{m2}&\cdots&\nu_{mn}\\
\end{pmatrix}$ are coefficients of m chemical reactions (negative for reagents, positive for products, and zero if the substance is not related to the reaction);
$\mathbf{X}=\begin{pmatrix}x_1\\x_2\\\vdots\\x_m\end{pmatrix}$ are amounts of reactions (concentration changes for each reaction);
$\mathbf{Y}=\begin{pmatrix}y_1\\y_2\\\vdots\\y_n\end{pmatrix}$ are final (equilibrium state) concentrations;
$\mathbf{K}=\begin{pmatrix}k_1\\k_2\\\vdots\\k_m\end{pmatrix}$ are equilibrium constants for each reaction.
I need to solve X (m variables) and Y (n variables) by given C, N and K, either analytic or numeric solutions.
For Example
0.1 mol/L ammonia solution, we have 2 equilibria:
(A) $\ce{H2O <=> H+ + OH-}$ (concentration of $\ce{H2O}$ is not considered)
(B) $\ce{NH3 + H2O <=> NH4+ + OH-}$ (concentration of $\ce{H2O}$ is not considered)
Equilibrium constants:

$[\ce{H+}][\ce{OH-}] = 10^{-14}$
$[\ce{NH4+}][\ce{OH-}]/[\ce{NH3}] = 1.77 \times 10^{-5}$

So we have 2 reactions (A and B) and 4 substances ($\ce{H+}$, $\ce{OH-}$, $\ce{NH3}$, and $\ce{NH4+}$), and known values are:
$\mathbf{C}=\begin{pmatrix}c_\ce{H+}\\c_\ce{OH-}\\c_\ce{NH3}\\c_\ce{NH4+}\end{pmatrix}=\begin{pmatrix}0\\0\\0.1\\0\end{pmatrix}$
$\mathbf{N}=\begin{pmatrix}\nu_{\text{A},\ce{H+}}&\nu_{\text{A},\ce{OH-}}&0&0\\0&\nu_{\text{B},\ce{OH-}}&\nu_{\text{B},\ce{NH3}}&\nu_{\text{B},\ce{NH4}}\end{pmatrix}=\begin{pmatrix}1&1&0&0\\0&1&-1&1\end{pmatrix}$
$\mathbf{K}=\begin{pmatrix}k_A\\k_B\end{pmatrix}=\begin{pmatrix}10^{-14}\\1.77\times10^{-5}\end{pmatrix}$
The solutions are:
$\mathbf{X}=\begin{pmatrix}x_A\\x_B\end{pmatrix}=\begin{pmatrix}7.57\times10^{-12}\\1.32\times10^{-3}\end{pmatrix}$
$\mathbf{Y}=\begin{pmatrix}y_\ce{H+}\\y_\ce{OH-}\\y_\ce{NH3}\\y_\ce{NH4+}\end{pmatrix}=\begin{pmatrix}7.57\times10^{-12}\\1.32\times10^{-3}\\0.0987\\1.32\times10^{-3}\end{pmatrix}$
Here I used the approximation of $y_\ce{OH-}=y_\ce{NH4+}$, so just solved a quadratic equation.
But how can I solve a more complicated system like:
\begin{align}
\ce{H2O &<=> H+ + OH-} \\
\ce{NH3 + H2O &<=> NH4+ + OH-} \\
\ce{CO2 + H2O &<=> H+ + HCO3-} \\
\ce{HCO3- &<=> H+ + CO3^2-}
\end{align}
which includes 4 reactions and 7 substances?

I'm trying to get numeric solution via SciPy but failed. Here is my example: https://stackoverflow.com/questions/68302523

Please, consider taking a look at FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange?. Clearly, you already know a good part of it, so specifically I suggest section 3 which describes the mhchem package.

You may find it useful to make your variables be $\ln [\ce{A}]$ and $\ln [\ce{B}]$ rather than $[\ce{A}]$ and $[\ce{B}]$, which should be more robust numerically.

Smith and Missen cover this in their textbook. First of all, get rid of your set of equations. You can generate the minimum necessary set using only the species, dont get trapped into thinking like an experimentalist and thinking you need reaction paths.

I will answer this later. Sympy is friend too, or autodiff.

It is going to take abit, why couldn't you ask for a system of two equations, the principles are all the same... doing the tedious bits on this system is killing me...

@B.Kelly I haven't found the online book Chemical reaction equilibrium analysis: Theory and algorithms by Smith and Missen yet. Can you paste related pictures or screenshots in that book here?

@auntyellow I don't know how to private message, but if you pass me a private message, I will give you my email, and can send you stuff... I am working this out but it is taking forever, and I need to sleep haha

@a-cyclohexane-molecule there are lnY and Y in both equations. Actually in the second equation it looks more likely to calculate if we use $\Pi$ notation rather than logarithm, because they are just polynomial equations.

Based on your linked StackOverflow question, it seems like you're running into issues with converging your solution because you're working with small numbers that vary over orders of magnitude. In your code, the variables you used were concentrations. I'm suggesting that you use log-concentrations because that might improve the stability of the solver.

@B.Kelly I committed this question into my GitHub project here https://github.com/auntyellow/math/blob/main/equilibrium.md , and you can find my email there. Or would you please upload some pages from Smith and Missen's book there?

@a-cyclohexane-molecule Use $\ln$ works better than original. But also need some luck to try initial values. e.g. I changed equation $\mathbf{N}\ln(\mathbf{N}^\text{T}\mathbf{X}+\mathbf{C})=\ln\mathbf{K}$ to $\mathbf{N}\ln(\mathbf{N}^\text{T}e^\mathbf{X}+\mathbf{C})=\ln\mathbf{K}$, then with inital value X=(-1,-1) and X=(-10,-10) fsolve can't converge, while X=(-3,-3) can get correct result.

First, just so you know, you don't need to use a set of reactions that make sense... See Smith & Missen Chemical Reaction Equilibrium Analysis: Theory and Algorithms. It shows how, given a set of species, you can make an infinite number of set of chemical reactions that all satisfy elemental abundance etc...
Part 1: Systems of equations - Enter the Jacobian
Each one of these expressions with the evil equilibrium constant, is a function, a nonlinear function generally, in the case of reactions. For a function, we can perform a taylor expansion to linearize it, and make it workable. We then can use linear algebra to solve the set of linear equations, here is an example of two variables... your problem has 4, but it is all general, just add the extra terms similar to the first two variables, and you will have 4 equations rather than 2

This is the familiar $Ax=b$ where $A$ is the Jacobian, and is a matrix of derivatives, x is what you want... in reactions these are usually your extents of reactions, one per reaction
For example, above, $f(\xi_1, \xi_2, \xi_3, \xi_4) = K - \frac{[H^+][OH^-]}{[H_2O]} = 0$, but put concentrations in terms of $\xi's$ using initial amounts, changes (extents of reactions) possibly via an ice table. The other three equations will be more important, or, at least  2 of them will.
Anyways, you need the 4 equations, one for each reaction, and the Jacobian matrix (matrix of derivatives.). I prefer using Sympy to calculate the derivatives analytically for me, and populate the matrix. Back in the day though, this was done by hand (yikes). Then we have the matrix A which is just numbers, and the vector b, again, just a vector of numbers, and just need to solve Ax=b. Solving Ax=b is trivial using software, you just need to provide a guess for what each extent of reaction is in order to evaluate the b vector (functions) and A matrix (derivatives of the 4 functions), and off it goes (a bad guess will give a bad answer, unfortunately). Solving this once will give you a new vector of extents of reactions... you repeat this, recalculate the b vector (functions) with the new extents of reactions, and the new Jacobian, solve for the new extents of reactions, and keep going until the extents of reactions stop changing.
You can also solve this using blackbox solvers in programs like Julia, Python, Matlab etc. You need to be able to provide them with the functions though, in terms of initial concentrations and extents of reactions.
To solve for x, you just need to solve $x = A^{-1}b$. normally you don't take the inverse, but this is so small, you can just tell the software to take the inverse of A and multiply it by b. There are many ways to solve it though, depending on the software. feel free to comment for more info you are after a Julia, python or matlab method.
I will add some symbolic python code in awhile, I am at my end though... this is not my first version :)
Also, as has been pointed out, there are certainly the possibility of numerical issues. However, this is why the method for creating systems of equations of Smith & Missen is so important, if concentrations get small, you can create a new set of reactions automatically to deal with this. Smith has done this, I am not sure if he ever published that code, I think he at least published the method.
Also, a common method is to solve these using LaGrange multipliers as well. There are many ways to solve systems of equations.
Finally, deriving everything from the chemical potential and avoiding equilibrium constants (instead using reference chemical potentials of the species) really adds a lot of robustness. Proffs teaching first year chem love using equilibrium constants because they don't need to solve anything other than toy problems. The pros don't mess around with kiddie equilibrium constants. Chemical potential or bust.

Your Jacobian approach looks like Newton's method. However, I failed to get the result just as a-cyclohexane-molecule mentioned. See my another post https://stackoverflow.com/questions/68302523

SymPy can't give solution like $(a\ln x_1 + b\ln x_2)=c$ because the result is not analytic. If I replaces with power, e.g. $x_1^a\cdot x_2^b=c$ , if the system is too complicated (Such as $\ce{H3PO4}$ + $\ce{H2CO3}$ + $\ce{NH3}$ + ...), we may encounter the degree larger than 4, which is also unable to be solved.

I would suggest sending Professor Smith an email. I dont think he has anymore copies, but he may have a pdf version he could give you

Numerical i stabilities are a real issue, but there are methods for dealing with them. The textbook covers all of this and more.

And yes i was suggesting a Newton method, for simplicity, but there are better methods .

@auntyellow You could both ask about a copy of the textbook, and pose the problem to him. You will likely get a thorough answer back.

Inspired by Cesareo's answer in MathSE, I tried SciPy's basinhopping and some tips:

I use "Nelder-Mead" as minimization method, although very slow, because the function may be non-differentiable. I haven't tried "Powell"; "CG" sometimes doesn't work and BFGSs don't work at all.
Set large and gradient "bounds" to avoid negative logarithm, and this also help basinhopping .
Set very large iterate count (for both basinhopping and minimize) if system is very complicated, such as McIlvaine buffer.

Here is my code with SciPy:
import numpy as np
from scipy.optimize import basinhopping

'''
# H2O <-> H+ + OH-
# NH3 + H2O <-> NH4+ + OH-
# [H+, OH-, NH3, NH4+]
# 0.1M NH3
C = np.array([0, 0, 0.1, 0])
N = np.array([
    [1, 1, 0, 0],
    [0, 1, -1, 1]
])
K = np.array([1e-14, 1.77e-5])
X0 = np.array([0, 0])
'''

# H2O <-> H+ + OH-
# CO2 + H2O <-> H+ + HCO3-
# HCO3- <-> H+ + CO32-
# [H+, OH-, CO2, HCO3-, CO32-]
# 0.001M CO2 + 0.01M HCO3-
C = np.array([0, 0, 0.001, 0.01, 0])
N = np.array([
    [1, 1, 0, 0, 0],
    [1, 0, -1, 1, 0],
    [1, 0, 0, -1, 1],
])
K = np.array([1e-14, 4.3e-7, 4.8e-11])
X0 = np.array([0, 0, 0])

def concentr(X):
    return np.matmul(N.transpose(), X) + C

def chem_eq(X):
    C1 = concentr(X)
    bounds = 0
    for c1 in C1:
        if c1 <= 0:
            bounds += (1 - c1)*1e+14
    if bounds > 0:
        return bounds
    return np.linalg.norm(np.matmul(N, np.log(C1)) - np.log(K))

def main():
    print('Inequilibrium:', chem_eq(X0))
    print('Concentrations:', concentr(X0))
    # `niter` should be so large for H3PO4
    result = basinhopping(chem_eq, X0, niter = 1000, \
            minimizer_kwargs = {'method': 'Nelder-Mead', 'options': {'xatol': 1e-14, 'maxiter': 1000}})
    print(result)
    X = result.x
    print('Inequilibrium:', chem_eq(X))
    print('Concentrations:', concentr(X))

if __name__ == '__main__':
    main()

Result for 0.1M $\ce{NH3}$:

$[\ce{H+}] = 7.56662656\times10^{-12}$ (pH = 11.12)
$[\ce{OH-}] = 1.32159291\times10^{-3}$
$[\ce{NH3}] = 9.86784071\times10^{-2}$
$[\ce{NH4+}] = 1.32159290\times10^{-3}$

Result for 0.1M $\ce{NH4HCO3}$:

$[\ce{H+}] = 4.35771835\times10^{-8}$ (pH = 7.36)
$[\ce{OH-}] = 2.29477887\times10^{-7}$
$[\ce{CO2}] = 1.01117642\times10^{-3}$
$[\ce{HCO3-}] = 9.97783306\times10^{-3}$
$[\ce{CO32-}] = 1.09905218\times10^{-5}$

btw there is an algorithm called HALTAFALL for equilibrium calculations, which may be much more effective than basinhopping. I found a Java implementation in this project


text:
Double bond or carbanion for nucleophillic attack
I came across this question:

According to me the Hydride ion should take hydrogen from left (unsubstituted carbon) because carbanion will be more stable there and this carbanion will then act as a nucleophile and attack $\ce{CH3I}$ giving product (b).
However according to the answer, hydride ion takes hydrogen from the right (Substituted carbon) and then enol form is made and then the double bond attacks $\ce{CH3I}$. Why so?

Kindly let me know the reason for the downvote so that I know where to work on?

As the medium is not basic the major enol will be formed on the more substituted side.

@NisargBhavsar Thankyou! What would have happened if the medium was acidic?

It is not necessarily true that "hydride ion takes hydrogen from the right (Substituted carbon) ". All you know is that it is the enolate that undergoes methylation. Under the conditions of the reaction, enolate formation may also occur at the less substituted carbon followed enolate exchange with methylcyclohexanone under equilibrating conditions.

The image should be properly oriented (i.e. not rotated 90°) and its source must be cited (we prefer ACS style). Also, chemical formulas must be properly formatted as such. Please visit this page, this page and this one on how to format your posts better with MathJax and Markdown.

First, there is a little mistake in your reasoning. You said that the carbanion is more stable on the unsubstituted side but actually, forming an enolate on the more substituted side will give a more stable substituted alkene. Remember, the negative charge is not concentrated on carbon. It is concentrated on the oxygen atom as the system is conjugated. That is why electrostatic interactions tend to occur on the oxygen while orbital (HOMO-LUMO) interactions occur on the carbon.
Now, an asymmetric ketone can form two regioisomeric enolates, kinetic and thermodynamic:

The kinetic enolate is the one that forms the quickest. Often, it is promoted by LDA, which is a strong but hindered base, at low temperatures, typically $\pu{-78^\circ C}$. Because it is more hindered, it will attack the less hindered and more accessible alpha hydrogen and form the less substituted enolate.
On the other hand, thermodynamic enolates tend to form at higher temperatures with unhindered bases like $\ce{NaH}$.

Why does $\ce{H-}$ not attack $\ce{C=O}$ and form an alcohol? One could argue that acid-base is faster but then why do reagents like $\ce{LiAlH4}$ form alcohols by nucleophilic addition but not $\ce{NaH}$?

NaH is not known as a nucleophile in organic chemistry, rather more as a base. The reason for this is likely due to the poor orbital overlap in terms of the HOMO-LUMO interactions.


text:
Why does the Lewis structure for NH2 have 4 dots and not 3?
Nitrogen has 5 valence electrons.
So I would expect that when bonding with two hydrogens, two electrons would go into those bonds, and three would remain in place on the nitrogen.
Yet, all the diagrams I see for it look like this

Why are there 4 dots and not 3?

As the compound is negatively charged, there is an extra electron present in the compound. Since the extra electron is held on nitrogen, there is 4 electrons on nitrogen instead of just 3. The extra electron will not be held on either hydrogens as it would be too unfavourable.

Why would it be negatively charged? And how do you know this in advance?

It is because in the compound you have drawn in the question, the compound is negatively charged.

I got that from the site linked

You have to start with a least some information. You either know it's an anion from whatever consideration you had before, say it's part of a compound like sodium amide and you have to draw the structure. Or you'll be given structure with the four dots and you'll have to figure out that it requires a negative charge to balance things. If none is given it would be like having ClO2 as Chlorine dioxide versus ClO2(-), the anion of the chlorous acid.


text:
Can I synthesize potassium hydroxide from readily available materials?
Where I live, KOH is not readily available and is cost prohibitive to ship. A number of other materials are available, however. I have read similar questions and their answers here, but most are too general or too specific for me to understand.
I could mix NaOH and KCl, both of which are available, to yield NaCl and KOH. But I'm not sure how I would separate them, as I can't think of any readily available substance that could be used to precipitate out the Na.
Or, I could mix Ca(OH)2 and K2CO3, as that would produce CaCO3(s) and KOH(aq). However... I do not have K2CO3 available. I could make the K2CO3 myself, by baking NaHCO3 into Na2CO3 and mixing it with KCl to produce NaCl and K2CO3, but then I am back at having to precipitate out Na.
I could also make the K2CO3 by combining C and KNO3, but... again... I do not have KNO3 available. I could make KNO3 from KCl and NH4NO3, but that's not available here either.
I feel stuck. KCl, NaOH, and Ca(OH)2 are all available and relatively cheap. NaHCO3 and NaCl are also obviously available and cheap. What else am I missing here? What else is generally available and cheap that I could use to synthesize KOH?

It would be very difficult to separate Na+ from K+ once they are in aqueous solution, because salts of both are highly soluble. I don't see any easy way to do what you wish to do. Why do you need KOH specifically? Can't NaOH do the same job?

I am planning on making liquid soap, which requires KOH. NaOH makes solid soap. I just discovered that potassium bitartrate (cream of tartar) decomposes into potassium carbonate, though, so this may be promising. Currently researching this option.

If you cannot get K2CO3 by easier way, you can get it by extraction of wooden ash, as this was the ancient way to get potash.

@Poutnik I have read that wood ash leachate contains both potassium carbonate and potassium hydroxide. Is this a mistake? Or is there a way to limit it to only one or the other? Or to know the ratio in solution, so one could add an appropriate amount of calcium hydroxide to convert all the K2CO3 to KOH?

I would doubt there is substantial amount of hydroxide, putting aside the equilibrium due carbonate hydrolysis. The needed amount of Ca(OH)2 is to be tested to avoid presence of dissolved calcium .

Related potassium-carbonate-vs-potassium-hydroxide-in-wood-ash

"is cost prohibitive to ship" - Are you in an embargoed country or something? I find it hard to believe that KOH is cheaper to make than to ship just about anywhere in the world.  It's one of the most heavily used industrial and commercial compounds in the world.  Have you really exhaustively searched for a supplier and come up empty?  You ask "What else is generally available and cheap that I could use to synthesize KOH?" - well, KOH is generally available and cheap, so you must be somewhere where generally available things are VERY hard to find.  So where is that?

Wood ashes is made of $5 - 10$% $\ce{Na2CO3}$, $5 - 10$% $\ce{K2CO3}$, $2 - 3$% $\ce{K2SO4}$, $40 - 70$% $\ce{CaCO3}$, $5 - 10$% $\ce{MgCO3}$, $2 - 5$% $\ce{Ca3(PO4)2}$, $2 - 10$% $\ce{SiO2}$$. So it is not easy to obtain KOH out of this mixture..

Anyway, you may discover a process to produce $\ce{KOH}$ in solution. But you will never be able to obtain it as a pure solid, because KOH cannot be separated from water by evaporation. At high enough temperature, $\ce{KOH}$ and water are distillating together. The only way of evaporating $\ce{KOH}$ solution is to do it in a vacuum. This is the way used in the industry to produce pellets of $\ce{KOH}$

@Maurice interesting, I found this:   The major elements of wood ash include calcium (7–33%), potassium (3–4%), magnesium (1–2%), manganese (0.3–1.3%), phosphorus (0.3–1.4%), and sodium (0.2–0.5%). Density of wood ash decreases with increasing carbon content.  https://link.springer.com/content/pdf/10.1007/978-3-540-74294-4_9.pdf

Can't you use bicarbonate to precipitate Na ? Admittedly the yield may be poor when it comes to separating, but it is easier to do.

Electrolysis of potassium carbonate solution is used to produce KOH and CO2. It is for homemade degreaser made in single chamber with platinum-plated electrodes. A similar apparatus is used with NaCl and acetic acid to make hypochlorous acid. Possibly you could use cheaper graphite electrodes? Wikipedia says: Potassium carbonate can be made by burning cream of tartar (potassium bitartate), which produces "pearl ash".

In which country is it necessary to use $\ce{KOH}$ (and not the more familiar $\ce{NaOH}$) without having the possibility of buying it somewhere ? What for ?

Potassium hydroxide lye from hardwood ash. Simple procedure.

If the only potassium salt you have is $\ce{KCl}$, then perhaps the only feasible way to get rid of the chlorine is electrolysis.
Using a cation-selective membrane, electrolysis in water will produce $\ce{KOH}$ (and hydrogen, $\ce{H2}$) at the anode, and chlorine, $\ce{Cl2}$, at the cathode (along with potassium hypochlorite, $\ce(KClO)$).
Though this will produce the $\ce{KOH}$, it also leaves you with hydrogen, chlorine, potassium hypochlorite and other byproducts which must be disposed safely. Further, the process is energy intensive. This could be done as a classroom demonstration, but commercially, you'd need to find good uses for the gases and bleach/disinfectant for any chance at economic feasibility.
If you can obtain potassium carbonate ($\ce{K2CO3}$) or "bicarbonate" (acid or hydrogen carbonate, $\ce{KHCO3}$), it can, with effort, be converted to $\ce{KOH}$ by heating. That said, decomposition temperature is ~1200 °C, which can be achieved in a ceramic kiln - barely. Again, this calls for a great deal of energy, so you'd pay far more for the cost of heating and the carbonate than for $\ce{KOH}$ alone. Also, at 1200 °C, the $\ce{KOH}$ evaporates, so it would have to be condensed on a cooler surface. It might also shorten the life of the kiln. On second thought, this sounds like a project that would be described in Derek Lowe's blog, Things I Won’t Work With.
@Poutnik suggests a displacement reaction with $\ce{Ca(OH)2}$, which would be far easier, but leaves the user to purify his product from remaining calcium -- calcium soaps tend to be insoluble.
In any case, $\ce{KOH}$ is very caustic: "CAUSES SERIOUS EYE DAMAGE. CAUSES SEVERE SKIN BURNS..." Use caution!

What other potassium salts are generally available??

@Jaredbrandt K2CO3 is often available as well.// Just for curiosity - it is  largely used in Dutch processing of cocoa ( cocoa gets from light brown to dark brown by the process ) to produce chocolate from inferior cocoa beans, enhancing cocoa flavour and removing acidity.

Potassium bicarbonate is used in agriculture as a fungicide, and as a soil conditioner to control acidity. It is also used in dry powder fire extinguishers, and as an alternative to sodium bicarbonate (baking soda) in cooking by people on a low-sodium diet. Depending how pure you need it, you should be able to find a source for it somewhere.

I am curious, how would you easily convert K2CO3 to KOH by heating at home , assuming you cannot afford even KOH shipping. The medieval way M2CO3 + Ca(OH)2(from bricklayers) -> CaCO3 + 2 MOH seems to be much easier.

@Poutnik, "a great deal of energy" was stated. Perhaps a pottery kiln? Decomposition temp ~1200 °C, which can be reached in a kiln. Amending answer for that.Admittedly, not easy to do.

Controlling of bubbling melted carbonate ( m.p < 900 Deg C) seems more difficult than checking/precipitating of residual calcium by extra carbonate. Glowing foam could easily cause fire.


text:
Does the nucleophilic substitution in an acid chloride always proceed through a tetrahedral intermediate?
I am a beginner in organic chemistry and would like to ask for the mechanism for the nucleophilic substitution of carbonyl groups from acid chloride - especially regarding the tetrahedral intermediate? From the internet, what I got is:

From Wikipedia, tetrahedral intermediate exists.
From Clayden Organic chemistry (2nd edition), the mechanism of this reaction consists of the formation of the tetrahedral intermediate.

But I also found Schotten-Baumann synthesis of amides:

This website: it seems for the first step, nucleophilic attack of the nitrogen of amine directly results in the loss of the leaving group i.e. no tetrahedral intermediate.
But it seems to me that only for this mechanism, that H+ can be produced (I can't draw a mechanism in which both tetrahedral intermediate and H+ are formed), and this HCl is transited to an aqueous phase and neutralized by NaOH, rendering this biphasic condition useful that 2nd eq. of amine need not be used.

What is the true mechanism for this reaction and why?

It always proceeds through the tetrahedral intermediate for carbonyls. The page you linked just shows this in one step as a shorthand. Do you see the arrow circling around the carbonyl oxygen? That means that a tetrahedral intermediate is formed then collapsed. The two steps are not shown explicitly but they are there.

Oh sorry I overlooked the circle on the around the oxygen, thanks for reminding

If you got the answer to your question, feel free to self-answer!


text:
Reaction mechanism of formation of salicylic acid with Phenol and carbon tetrachloride in basic medium?
With classic Reimer-Tiemann reaction a variation is also given in my lecture notes which use carbon tetrachloride instead of chloroform. But to me, it seems like it would be just a substitution reaction rather than Reimer-Tiemann which proceeds through carbene. Given slow hydrolysis of $\ce{CCl4}$ how such a substitution reaction could be useful, even if we use $\ce{OH-}$ ? Or is there something else going on here?

You're right, as far as I know the $\ce{CCl4}$ reaction first occurs by the phenol (or phenoxide) acting as a nucleophile and substituting one of the chlorine atoms. The remaining three chlorine atoms are hydrolyzed and salicylic acid is formed

Will not the hydroxide be better-attacking species than phenoxide due to steric factors, if so we should expect a phosgene type intermediate.

That never occurred to me... I'm sorry, I'm not knowledgeable enough to provide a satisfying explanation.

Are you asking about abnormal RT reaction? https://chemistry.stackexchange.com/questions/152344/bromochlorofluoromethane-as-a-reagent-in-riemer-tiemann-reaction/152354

@NilayGhosh the A V Gaonkar paper just point to the original German paper. I doubt that mechanisms in CCl4 and CHCL3 are different. How will CCl4 form the carbene ?

No time to answer but I am providing you some links where the mechanism of this reaction is given: https://www.youtube.com/watch?v=zbXqpZHX7wM https://www.youtube.com/watch?v=n22TGgejl-8 https://www.quora.com/What-is-the-mechanism-of-Reimer-tiemann-reaction-of-phenol-and-carbon-tetrachloride-in-potassium-hydroxide (ofc the mechanism is different (no carbene is formed) and it is rather a variation if not an abnormality. So you can call whatever you want).

@NilayGhosh I don't think this is a correct mechanism or even the most probable, as I pointed out to TRC. The attack of OH- seems more probable than phenoxide.

@DeepakArya Want to make it an answer? :)

@NilayGhosh I haven't found any solid literature


text:
Redox reaction between Potassium thiocyanate and iodine
How would you balance the following redox reaction in acidic medium?
$$\ce{KSCN + H2O + I2 -> KHSO4 + HI + ICN}$$
In this reaction, it seems the sulfur is being oxidized but iodine is undergoing both (disproportion?). The tricky part about the problem is the carbon. The carbon is going from a +4 to a +2 but still is not considered a redox atom. Why is this?
Oxidation half reaction:
$\ce{KSCN -> KHSO4}$ (sulfur being oxidized from -2 to +6)
$\ce{I2 -> ICN}$ (iodine has +1 charge)
Reduction half reaction:
$\ce{I2 -> HI}$  (iodine has -1 charge)
$\ce{KSCN -> ICN}$ (here the carbon is going from +4 to +2)

Did you check this explanation? https://www.periodni.com/half-reaction_method.php?eq=KSCN%2BH2O%2BI2%3DKHSO4%2BHI%2BICN

Can you give a source for the reaction itself? I mean, a source that shows that the reaction actually occurs when $\ce{KSCN}$ and $\ce{I2}$ are mixed? I'm not doubting you, just that this reaction is important for an exam of mine and a reliable source showing the same would be helpful :) Thank you.

I did find an ACS journal (its an old journal though) that talks more about the kinetics of this reaction. If you're an ACS member you should be able to view the entire journal. Hope this helps! Here's the link: :https://pubs.rsc.org/en/Content/ArticleLanding/TF/1935/TF9353100868#!divAbstract

I would solve this problem by algebra, without thinking about oxidation numbers. It is quicker. Let the coefficients be $a, b, c, d, e, f$ in front of the reactants and products. It yields $$\ce{a KSCN + b H2O + c I2 -> d KHSO4 + e HI + f ICN}$$ The number of each atom must be the same on both sides of the equation arrow. Counting atoms by order of appearance, it is easy to see that, for potassium and sulfur : $a = d$. Counting $\ce{C}$ and $\ce{N}$ atoms yields also $a = f$.  Now let's admit that $a = 1$, then $d = f = 1$.
Let's consider the remaining $\ce{H, O, I}$ atoms.
Counting $\ce{O}$ atoms yields : $b = 4d = 4$.
Counting $\ce{H}$ yields : $d + e = 2b = 8$. So $b = 4$ and : $e = 8 - 1 = 7$
Counting $\ce{I}$ atoms yields : $2 c = e + f = 7 + 1 = 8$. So : $c = 4$
So the final equation is  $$\ce{ KSCN + 4 H2O + 4 I2 ->  KHSO4 + 7 HI +  ICN}$$

Use molecular species in your half-reactions.
In your case, identify thiocyanate as the reducing agent and, since the cyano function combines with iodine, include enough iodine in that half-reaction to form the iodine cyanide/cyanogen iodide product. This gives
$\ce{SCN^- + (1/2)I2 -> S^{(+6)} + ICN +7 e^-}$
where the +6 oxidation state on sulfur in the bisulfate is rendered equivalent to six positive charges; thus seven electrons on the right.
If we include the hydrogen and oxygen atoms so that the bisulfate ion is a complete species, we would have
$\ce{SCN^- + (1/2)I2 -> HSO4^- + ICN +7 e^-}$ (hydrogen and oxygen unbalanced)
where we still have seven electrons based on the oxidation state differential noted previously. Adding hydrogen ions on the right to balance charges and water on the left to balance hydrogen or oxygen atoms then gives this form for the thiocyanate oxidation:
$\ce{SCN^- + (1/2)I2 +4H2O -> HSO4^- + ICN + 7 H^+ + 7 e^-}$
Combine either thiocyanate oxidation half-reaction with iodine reduction:
$\ce{(1/2)I2 +e^- -> I^-}$
Balancing electrons in the usual way then reveals that the redox reactant ratio is $\ce{(1)SCN^-:4I2}$.  So leave the thiocyanate coefficient at $1$ and put a coefficient of $4$ before $\ce{I2}$, then use atom balances to handle the remaining elements. (If you used the second balanced equation given for thiocyanate oxidation, you have to add just the potassium ions.)


text:
Formation of enamine for unsymmetrical cyclohexanone derivatives
Question:

Answer given: D
My answer: C
My reasoning - I outlined the following steps in the reaction:

Nucleophilic attack of pyrrolidine on the carbonyl, leading to the carbon atom now having one $\ce{-OH}$ group and attached to the nitrogen of pyrrolidine.

Heat is mentioned, so I assumed it is for dehydration. Dehydration, as far as I know, should result in the most stable double bond i.e. the one with maximum alkyl substituents. So I formed an enamine with double bond on the right side.

Nucleophilic attack of enamine's $\ce{C=C}$ on the leftmost carbon atom of $\ce{H2C=CH-CN}$, which is Michael addition.

$\ce{H3O^+}$ to provide proton to the carbanion.

These steps lead to the formation of C. For D to be formed, step 2 must be altered, and the less substituted (and less stable) double bond must be formed. Why does that happen?
I also can see that the nucleophile $\ce{C=C}$ is sterically hindered in step 3. That will lower the yield of step 3. However, that steric hindrance or comfort in step 3 should not affect the formation of $\ce{C=C}$ in previous step 2.
I am also open to the fact that the answer key given might be wrong. But in case the key is right and I am wrong, I would like to know why my step 2 is wrong.

I think the answer key is probably wrong (judging from the other options it seems like you really had to decide between the 2 enols really. $\alpha-H$ calculation makes it clear that we should prefer C. What is this cis-trans it is mentioning in opt. B and D? I don't see an stereogenic center..

@TRC I can see 2 pathways for this reaction, one through additon reaction(in first step) other  through removal of acidic H and my prof had taught, that in such condtions carry out acid base first as this reaction is faster. I also doubt whether acc to your mech dehydration would not happen, I might be wrong about this though.

@Rishi That's not really cis-trans though. I should have added db bond with cis/trans though

@napstablook Thank you for your input. The cis-trans is probably referring to isomerism with respect to the cyclohexane ring. My main textbook uses that term to refer to groups around ring on same side or opposite sides (for example, as in 1,2-dimethylcyclohexane). Not sure if that term has become obsolete now or not.

@Rishi I wasn't able to understand the other pathway you suggested. Where will the acidic H be removed from?

@TRC 1. Removal of acidic H by pyrrolidene from the less substituted alpha carbon then this Nu- reacts as you mentioned in step 3 and follows

@napstablook I just asked for the test paper's solutions from my teacher. I checked and it has the same mechanism as I suggested, only with the step 2 double bond formed on the other side. They have not mentioned any reason for that.

@TRC you can put their mech in the post if it has other noticeable details else the post is fine as it is.

@Rishi I knew only strong bases (like NaOH or LDA) were sufficiently basic to remove the hydrogen from alpha-carbon. In the pathway you mention wouldn't the equilibrium lie largely to the reactant side?

@Rishi No, nothing. It has the same mechanism, just with the structures included.

1 last point, this is a really bad question, the h3o+ at the end will hydrolyse the cyanide. Usually if you have to question the author's intention more than think about the question it is not a good qn. :(

@napstablook True, it should have worked out if they had mentioned mild workup.

@napstablook Absolutely agreed! It's not from a book though, it's from a coaching institute's mock test for a national level exam. I've found a lot of their other questions to be meaningless as well.

@napstablook  the H+ will hydrolyse the enamine very much more easily than the nitrile. Mild aq acid at rt hydrolyses enamines, you need forcing conditions (conc acid, high temp) to hydrolyse nitriles. It is entirely possible to be selective about which group is hydrolysed.

@Waylander Yes, the enamine does get hydrolyzed when $\ce{H3O^+}$ is added - that's why the options again depict cyclohexanone. As for the concentration and temperature, that's another fault of this exam - they give us only half the experimental information and expect us to know what products will be formed.

@Waylander I agree, I still think it should be mentioned though.

According to this site here citing Carey & Sundberg and Morrison & Boyd
"In case of 2-methyl cyclohexanone, this planarity would cause steric clashes between the methyl group and the pyrrolidine hydrogen atoms. Due to this, formation of the less stable 6-position carbanion is preferred and hence forces reactions to proceed at this position."

In turn, "site here" cites Carey & Sundberg and Morrison & Boyd. Referencing primary sources (i.e. these textbooks) is preferred.

The answer is D as @Waylander has noted. Here is the original Stork paper: https://pubs.acs.org/doi/10.1021/ja00885a021

@Waylander I think that while this is a problem in normal situations, I would rather take heat as an assumption that the reaction is under thermodynamic control and only final product matters..


text:
Why is the concentration of gases denoted in form of partial pressure?
In most of the chemical laws of chemistry involving concentration of products or reactants like equilibrium constant, reaction quotient, Henry's law, Nernst equation etc, I often see a similar note stating concentration of gas to be taken as partial pressures of that gas or something similar to that. So, Why is the concentration of gases denoted in form of partial pressure ?
Does it have any relation with Dalton's law?
I don't know why this equation has been closed and it is not the duplicate of this question Please review again.

My guess would be that pressure is much easier to measure for a gas. Of course once you have the pressure, temperature and ideal gas equation would give you the concentration too. Also, concentration term, if replaced by partial pressure, will give you $\ce{K_p}$, not $\ce{K_c}$. Both terms are different.

Considering gas pressure is much more common than gas concentration. There is the relation for ideal gases $p=cRT$. The point is, many gas behaviour quantitatively depends rather on partial pressure than on concentration.

In addition to what TRC states, it can be mentioned that, in thermodynamics, concentration is not a fundamental parameter. Pressure is fundamental : it exists in the definition of the work, which is the very beginning of thermodynamics. $K_\mathrm{p}$ can be directly obtained , and has the correct dimension. On the contrary,$K_ \mathrm{c}$ is indirectly obtained, and the concentration have to be divided by a reference value of concentration

Your question is based on a false premise that concentrations and partial pressures are interchangeable or identical entities, which is not true. Fundamentally different physical quantities cannot "take form" of one another. The question I marked as a dupe is not a literal duplicate, but it explicitly reveals this flaw and the accepted answer by Martin can be used to answer this question of yours as well.

Are you familiar with the derivation of the equation for the equilibrium constant of an ideal gas phase reaction in terms of the standard change in Gibbs free energy?


text:
Reduction of nitrobenzen using LiAlH4
Can you please explain  why isn't nitrobenzene reduced to aniline but turns out as azobenzene on reaction with LiAlH4? Please explain the mechanism.

Duplicate: https://chemistry.stackexchange.com/questions/88616/reduction-of-nitrobenzene-with-lialh4-and-nabh4?r=SearchResults

It is entirely possible you form aniline, but it gets consumed. During the reduction nitrosobenzene would also be formed, and Delmagro et al. [1] report that in basic solution the aniline can couple with the nitrosobenzene to give azobenzene. This coupling depends on the nitrogen in aniline acting as a nucleophile, so under acidic conditions as in a tin/hydrochloric acid reduction it would be suppressed (and thus aniline preserved/protected) by protonating the aniline. See also this answer.
Reference
1.
Jacir Dalmagro, Rosendo A. Yunes, Edesio Luiz Simionatto. "Mechanism of reaction of azobenzene formation from aniline and nitrosobenzene in basic conditions. General base catalysis by hydroxyide ion". Journal of Physical Organic Chemistry 7 (8), 399-402. https://doi.org/10.1002/poc.610070803


text:
Preparation of isotopic CO2 standards from Sn-encapsulated CaCO3
I wish to prepare isotopic CO2 gas standards for highly 13C enriched CO2. The CO2 concentrations will be ~1000 ppm, with the balance either N2 or zero air.
My plan is as follows:

Encapsulate labeled CaCO3 in normal 5x9mm tin capsules (varying quantities of Ca12CO3 and Ca13CO3.)
Place these tin capsules in a borosilicate tube in-line between the balance gas tank and the storage vessel, with quartz wool packed on each side.

3.Thermally decompose the CaCO3 with a bunsen burner, and then very slowly increase the pressure on the balance gas regulator to sweep the CO2 into the storage vessel (max pressure 7 bar, allowing plenty of time for the tube to cool).
The reason why I want to use tin capsules is that I don't want the chance of CaCO3 powder getting stuck on any surfaces while putting it in the tube, as it's crucial that all of the weighed CaCO3 be decomposed in order to arrive at the right isotopic composition. The quartz wool is to prevent any reacted CaO from being swept into the storage vessel.
I have two questions:
Is the tin going to mess in any way with the release of CO2 from the CaCO3, either using zero air or N2 as the balance gas?
Can you think of anything that could go wrong with this setup?

I guess zero air is CO2 free air,or eventually synthetic air.

Do you know that you may purchase analytical standards like "normal" $\ce{CO2}$ gas, $\ce{C^{18}O2}$, $\ce{^{13}C^{18}O2}$, $\ce{^{13}C^{17}O2}$, $\ce{^{13}CO2}$, $\ce{^{12}CO2}$, $\ce{^{13}C^{16}O2}$,$\ce{C^{17}O2}$ from sources like SigmaAldrich? It may be much less hassle for you to purchase such a gas of guaranteed quality in a small pressurized cylinder from a source with the savoir faire than to set up such an experiment and to train yourself.

If I understand correctly, you want to heat some CaCO3 to a temperature where it is decomposed. At this temperature, the glas will be nearly liquid. Don't you think that some gas may leak through a hole in the molten glass ?

Poutnik - Yes, zero air is zero CO2, zero hydrocarbon air.
Buttonwood - Yes, I do know that. I got a set of isotope calibration gases quoted out for me by Airgas (3 in air, 3 in N2), enough for approx a year of my work. It was >$12k. Maurice - Thermal decomposition of CaCO3 occurs at ~800C, borosilicate melts at ~1600C, so no, I'm not too worried. Thermal decomposition of CaCO3 in a test tube is a classic high school chemistry experiment.


text:
How to separate Cu and Ni from alloy containing different metals such as Al, Cu, Ni, (Zn, Pb, Sn < 5%)?
I have an alloy containing different metals such as Al, Cu, Ni, (Zn, Pb, Sn, Ag, Si in small amount < 5%). How can I separate Cu and Ni from this alloy?
I tried to dissolve alloy in sulfuric acid, but it didn't work.
Then I tried to dissolve it in mixture of nitric and sulfuric acids and I managed to dissolve 400 g out of 1000 g. I noticed a thin black layer on alloy piece, which I suppose can be nickel oxide, which can be obstruction in dissolving alloy.
How can I prevent formation of this oxide layer?

Apparently a Ni Al Bronze ( will also contain Fe and Mn, plus tramp elements). Likely one would need to go back to electroplating out the copper as is done in copper production. It is a relatively corrosion resistant alloy so will not dissolve easily.

See: https://chemistry.stackexchange.com/questions/34371/separating-nickel-coin-into-copper-and-nickel-components

First, the alloy $(0.5$ to $\pu{2 g})$ should be treated by $\pu{10 mL}$ nitric acid $32\,\%.$ All metals will get dissolved, except tin and silicon, which will be transformed into insoluble dioxide $\ce{SnO2}$ or $\ce{SiO2}$. Dilute in $\pu{100 mL}$ hot water. $\ce{SiO2 + SnO2}$ will make a gelatinous precipitate, that can be eliminated by filtration. Add $\pu{0.2 g}~\ce{Na2SO4}$ and $\pu{0.2 g}~\ce{NaCl}$ in order to remove lead and silver: lead sulfate $\ce{PbSO4}$ and silver chloride $\ce{AgCl}$ will precipitate, and are eliminated by filtration.
The remaining solution contains aluminum, copper and nickel. If you want to separate copper, you may add $\pu{2 g}$ potassium iodide $\ce{KI},$ that will make a precipitate of copper(I) iodide $\ce{CuI}$ according to the reaction
$$\ce{2 Cu^2+ + 5 I^- -> 2 CuI(s) +  I3^-}\tag{1}$$
So copper may be separated as a white precipitate ($\ce{CuI}$). This reaction is specific. In the final solution, nickel can be separated with dimethylglyoxime. Add $\pu{10 mL}$ of a dimethylglyoxime $(\ce{C4H8N2O2})$ solution $(1\,\%$ in ethanol). Add slowly some ammonia $\pu{2 M}$ up to the $\mathrm{pH}$ is just greater than $7$. An intense red precipitate of nickel dimethylglyoxime $\ce{Ni[C4H5N2O2]}2$ appears and can be filtrated:
$$\ce{2 C4H8N2O2 + Ni^2+ -> Ni(C4H7N2O2)2 + 2 H^+}\tag{2}$$
This reaction is specific, and the other metals do not interfere.

Nice - this was pure analytical chemistry which is not taught today. Which book do you consult Maurice?

@M. Farooq. You are right. Qualitative analysis is not taught any more today. But it was what we did every day in the lab when studying in the university to obtain our diploma in chemistry in the years 1970. We studied in the lab the properties of all elements, including arsenic, mercury, cyanides. There was no accidents. These experiments were taken from books like F. P. Treadwell, Analytical Chemistry, 1924. Difficult to find in present bookshelves.

"This reaction is specific, and the other metals do not interfere". While the reagent is specific to nickel, it can also detect other metal ions which might be an interference like cobalt or palladium. Moreover,  there was a study which discussed removal of heavy metals like lead, iron and zinc from waste water using dmg.

@Nilay Ghosh. Cobalt does also react with dimethylglyoxime, but only at high levels of concentration. It produces a blue color, but the rection is not very sensitive. With nickel, the reaction is visible even at nanomole levels. . Furthermore, cobalt does not produce a precipitate, just a blue color.

Fresenius' analytical scheme is also one possibility, although it is not as direct. More information can be found in Harvey's Analytical Chemistry 2.1 Libretext.  I bring this up to highlight that even though this is a classical technique, it is still mentioned in some analytical textbooks still in active use.

What is the full name of the book which mentions Fresenius' analytical scheme? Thanks.

@M.Farooq Analytical Chemistry 2.1 (yep, that's what Harvey named it).  In addition to the Libretext link in the answer, you can find it here.  FWIW, I use this text for my non-majors quant chem class and it works reasonably well.

Thanks. Oh I thought you had the original translated book of Fresenius.


text:
Decarboxylation of primary amino acids
Since primary amino acids like glycine, alanine can't be decarboxylated under oxidative conditions, what would be the best catalyst for this reaction?
I was thinking about copying decarboxylases with $\ce{Cu^2+}$ catalyst, but not sure how that would work.

Might be a starting point: https://chemistry.stackexchange.com/questions/89054

I've noticed this question before, but this procedure is for alpha amino acids, not sure if that would work for alanine as well.

But Cu+2 is an oxidative condition! Did you mean under some oxidative conditions?

Seconding user55119's comment, see this paper: https://pubs.acs.org/doi/full/10.1021/jf502751n

@steved. Alanine and glycine both are alpha amino acids. Primary refers to the set coded by the genetic code, alpha refers to the position of the amino group with respect to the carboxylic acid group.

Depending on the amino acid of interest and the scale and cost required, you might be able to use a PLP-dependent decarboxylase enzyme. They are typically soluble stable enzymes that can be overproduced easily in recombinant systems.

@KarstenTheis alanine can exist in both forms, I'm specifically looking into methods of β-alanine decarboxylation

Since primary amino acids like glycine, alanine can't be decarboxylated under oxidative conditions, what would be the best catalyst for this reaction?

Decarboxylation of $\alpha$-amino acids is known enzymatic procedure in biological systems. Non-enzymatic decarboxylation of $\alpha$-amino acids is also a long-known reaction, which leads to amines with a range of applications (Ref.1 and 2). Initially, this type of decarboxylation was tried by heating them above $\pu{200 ^\circ C}$, often dissolved or dispersed in an inert solvent. As these reactions are always exothermic, heating is only required to reach the rather high energies of activation (Ref.1).
In enzymatic decarboxylation in which Nature lowers these activation energies by enzymatic reactions has been frequently discussed. For instance, the formation of a Schiff's base between an $\alpha$-amino acid and pyridoxal phosphate as the proved prosthetic group of most of the decarboxylases is postulated by many authors. The easiness of the expected decarboxylation in this way may be understood with the help of the electronic theory of valence (Ref.1). As a consequence, the most commonly used method to decarboxylate $\alpha$-amino acids employs thermolysis of the amino acid in the presence of catalytic amount of an aldehyde (usually, an aromatic aldehydes can act as catalysts). The earliest such attempt has been done by Curtius and Lederer in 1886, when they have obtained decarboxylation of glycine by heating a mixture of glycine and benzaldehyde up to $\pu{130 ^\circ C}$. But the end-product of this reaction was not methylamine as expected, but benzylamine, which could not be understood by the authors at the time (Ref.3):

The driving force of this reaction is the resonance stabilization of the intermediate carbanion:

During a intensive work in this subject, Dose found that in the o- or p-position to the aldehyde group there must be a group with free electrons (such as dimethylamino-group) significant enough to reduce fully the tendency of transamination in the Schiff's base (reverse the electron flow; Ref.1). For instance, when Dose has used p-dimethylaminobenzaldehyde instead of benzaldehyde as the catalyst in valine decarboxylation, isobutylamine was the single product isolated after a method similar to that suggested by Curtius and Lederer in Ref.3 (with benzaldehyde it was $2:1$ of benzyl amine to isobutylamine while it is $1:1$ with anisaldehyde). There is also use of non-aromatic aldehyde as a catalyst as well (Ref.4). In this sequence, the authors have obtained corresponding amines (some of them are optically active) in good to excellent yields. For example, 4-hydroxyproline has given $(3R)$-(-)-3-hydroxypyrrolidine in 93% yield (Ref.4). Note that the elevated temperature is an important factor in this decarboxylation.
Other non-enzymatic methods include irradiation with UV light, heating in diphenylmethane solvent, or thermolysis in a high boiling solvent in the presence of a peroxide catalyst (Ref.2). Since some unnatural $\alpha$-amino acids do not undergo decarboxylation under the above conditions described (e.g., compounds 1,2, and 3), a new general non-thermal procedure is introduced in 2003 (Ref.2):

The method can be described as a ‘one-pot’ combination of two known reactions, oxidative decarboxylation of $\alpha$-amino acids to nitriles induced by N-bromosuccinimide, and the reduction of resulting nitriles to amines effected by sodium borohydride-nickel chloride (Ref.2):

Using this two step one-pot procedure, compounds 1, 2, and 3 were successfully converted to their corresponding polyamines.
References:

Klaus Dose, “Catalytic Decarboxylation of $\alpha$-amino acids,” Nature 1957, 179, 734-735 (DOI: https://doi.org/10.1038/179734b0).
Gilles Laval and Bernard T. Golding, “One-pot Sequence for the Decarboxylation of $\alpha$-amino acids,” Synlett 2003, (4), 542-546 (DOI: 10.1055/s-2003-37512).
Th. Curtius and G. Lederer, “Notiz über Benzylamin,” Berichte der deutschen chemischen Gesellschaft 1886, 19(2), 2462-2463 (DOI:  https://doi.org/10.1002/cber.188601902184).
Hashimoto Mitsunori, Eda Yutaka, Osanai Yasutomo, Iwai Toshiaki, and Aoki Seiichi, “A novel decarboxylation of $\alpha$-amino acids. A facile method of decarboxylation by the use of 2-cyclohexen-1-one as a catalyst,” Chemistry Letters 1986, 15(6), 893-896 (DOI:  https://doi.org/10.1246/cl.1986.893).

This is very useful information. Especially the one using cyclohexenone as a catalyst. But these methods are for α-alanine and unfortunately, and the α-amino group seems vital for the reaction to proceed. Can't find any articles about decarboxylation of β-amino acids.

Regarding decarboxylation by heating above 200°C, could refluxing β-amino acids in naphtalene/nitrobenzene be an acceptable method?

@steve d.: Heating above $\pu{200 ^\circ C}$ is an earlier method for decarboxcylation regardless of it is $\alpha$-amino acid or not. I'm sur it is okay for $\beta$-amino acids as well.

@Mathew Mahindaratne; Are you aware that the condensation of phenylalanine and benzaldehyde in the presence of N-phenylmaleimide affords a [3+2]-cycloaddition product with the loss of CO2/ This result requires a slight tweaking of your decarboxylation mechanism.  Accordingly, the nitrogen should be protonated by -CO2H with the loss of CO2 to afford the dipolar species. Journal of the Chemical Society, Perkin Transactions 1:  Organic and Bio-Organic Chemistry (1972-1999) Issue10, Pages2285-96, 1987 . [From Chem. Absts.]  Fine answer nonetheless.

@user55119: Nice comment and info. I really appreciate the comment. I'll look at the paper on monday and add to the answer.

@Mathew: I looked at the original paper. While they intimated that the process could occur, their examples retained the carboxyl group in the cycloaddition when the rxn. was run in acetic acid. They claim HOAc represses the decarboxylation. Perhaps CA was being overly generous. https://doi.org/10.1039/P19870002285

One paper1 suggested use of pyridoxal as catalyst:

When α-amino acid are heated with pyridoxal in dilute aqueous solution in the absence of metal ions, two closely but independent reaction takes place:
$$
\begin{align}
\ce{RR'CNH2COOH ->[Pyridoxal] RR'CHNH2 + CO2}   & \tag{R1}\\
\ce{RR'CNH2COOH ->[Pyridoxal] RR'C=O + CO2 + Pyridoxamine} & \tag{R2}\\
\end{align}
$$

(R1) is analogous to decarboxylation of amino acid by pyridoxal phosphate enzymes. (R2) is a decarboxylation-dependent transamination which doesn't have any enzymatic analogy. The reactions get partially inhibited if metal ions are used as catalyst along with pyridoxal (activity of pyridoxal on the reaction is reduced).
One more paper2 suggested use of glyoxal/ninhydrin as catalyst. However, oxygen absorption was noted in the reactions.

References

Pyridoxal-Catalyzed Decarboxylation of Amino Acids, G. D. Kalyankar and Esmond E. Snell, Biochemistry 1962 1 (4), 594-600, DOI: 10.1021/bi00910a008
Masao Fujimaki, Nguyen Van Chuyen, Tadao Kurata, Studies on the Decarboxylation of Amino Acids with Glyoxal, Agricultural and Biological Chemistry, Volume 35, Issue 13, 1 January 1971, Pages 2043–2049, DOI: 10.1080/00021369.1971.10860189 (full paper here)

Using pyridoxal might be viable option, but not sure if it also works for β-amino acids like β-alanine. According to the paper, decarboxylation catalyzed by glyoxal/ninhydrin on β-alanine or β-amino-n-butyric acid resulted also in deamination. So alanine produced acetaldehyde.

I swear I read a paper where phenylalanine was decarboxylated in household bleach by heating for 30 minutes at 160 cel?
Wouldn't that work here.

Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center.

«I swear I read a paper» is better substituted by a reference from primary literature (like a journal article) or a textbook including a literature reference.


text:
In an open system, can water vapor escape into the atmosphere before reaching the boiling point?
In a closed system, water vapor would have nowhere to go and atmospheric pressure would not be a factor. It would reach dynamic equilibrium with only water vapor.
But in an open system, I think that the water vapor would be forced back down into the liquid state by the atmospheric pressure until it is heated to the boiling point, at which the water vapor would equal or exceed the atmospheric pressure allowing water vapor particles to escape.
Is my understanding correct?
It just doesn’t seem to be how it works since water vapor can evaporate and escape even before it reaches the boiling point. Or instead, would only fast enough particles be able to overcome and escape the atmospheric pressure and the boiling point would only be when practically all of the water particles have reached that speed
I need help understanding in-depth the process of evaporation and vaporization on a particulate level. Starting from very basic chemistry might help.

Have you ever washed clothes and let them dry in open air? Have you even observed water evaporating at room temperature? Have you ever observed sea, lakes or rivers boiling or being nearly as hot as boiling water ?( aside of geothermal activities) // Liquid evaporation occurs at any temperature, just its rate changes with temperature. // https://en.wikipedia.org/wiki/Evaporation // Remember that a particular molecule does not have temperature.

In contrary to some other Q/A or forum sites, answers on CH SE site are figuratively paid by the user's own effort. When you ask, you are supposed to search and thoroughly think about the topic and to provide explicit compact summary of partial answers or at least ideas or thoughts you have got until then. Effort not shown may be considered as effort not done and such a question may be closed.

Your second explanation is correct. I copy it : *only fast enough particles are able to overcome and escape the atmospheric pressure *

@Maurice Rather: Only fast enough particles are able to overcome the liquid cohesive forces.

@Poutnik. OK for your modification. Of course. I wanted to be positive about OP's question.

Liquid evaporation occurs at any temperature, just its rate changes with temperature. You must have seen washed clothes getting dry in open air, water evaporating at room temperature. OTOH, I guess you have not observed boiling sea, lakes or rivers to obtain clouds - putting aside  geothermal activities.
There are few things for you to understand:
Molecules do not have temperature. For each temperature, there is a wide distribution of molecular energies and speeds. Similarly, the same molecule energy/speed can belong to a wide range of substance temperatures.
Molecules do not deal with pressure. Well, for usual evaporating context. If a molecule would leave liquid or not does not depend on pressure(*), but on liquid intermolecular forces, the molecule energy and velocity. When a molecule leaves liquid, it moves mostly freely with occasional collisions. ( 10 billions per second as the typical value is still occasional for them )
Only fast enough molecules are able to overcome the liquid cohesive forces to escape to a gaseous phase. At any temperature, there are always some molecules with enough energy to leave. But for low temperatures, it would be very tiny amount.
OTOH, even at the boiling point, only minor fraction of molecules has enough energy to escape liquid, as there is not enough energy in liquid for all of them.(**) This should not be surprising, as there is needed about 5.5 times more energy to boil water out, compared to energy for heating the same amount of water from $\pu{0^{\circ} C}$ to $\pu{100^{\circ} C}.$

(*) In fact, increasing of total pressure very sligthly increases the liquid saturated vapor pressure, but the reason for it is out of the question scope.
(**) There are assumed conditions near to thermal equilibrium of molecular energies at given temperature.

@buckthorn There is a large set of statements that are generally true, with lists of extreme conditions, when they are not, but which are not useful to be mentioned at given level. :-) I have meant it for conditions not far from thermal equilibrium of molecular energy distribution.

Agreed. Perhaps I should have taken an example that is more on topic, but thought I'd make a point that boiling is a non-equilibrium process.

Surely boiling is not an equilibrium process, but energy distribution is still close to the equilibrium one. (In fact,all observed processes are just more or less close to equilibrium.)

Never mind, I see your point. If the constraint is that the temperature be at the boiling point.


text:
Can exact bond angles in cycloalkanes be calculated?
I know that bond angles in cycloalkanes are usually different from $\ce{109.5}^{\circ}$ as predicted by the VSEPR theory because of their shape, and a higher deviation will cause a higher ring strain. Can we calculate exact ring angles using ring strain, or other data?

Yes, molecular orbital methods, which rely on quantum mechanics will produce accurate geometry. These are definitely not trivial calculations and so most non-experts  will use commercial software. A standard method is called the Hartree-Fock self consistent field (SCF) approximation.

There may be experimental data for the specific cycloalkane in question, for example by a single-crystal diffraction analysis.  Beside entries in the primary literature, there are databases with the relevant structure data including the atomic coordinates; the later ones may be used to calculate the angles in question.
If experimental data are not (yet) available, one may predict the geometry of the compound in question, and use these atom positions guessed to compute the angles in question.  The accuracy of these predictions depends on the know-how implemented in these programs.  For a quick prediction of smallish motifs seen in organic chemistry, you may draw such molecules, e.g. in MolCalc, run an optimization to yield one reasonable spatial arrangement of the atoms, and eventually measure distances and angles:

The geometry predicted may be (but need not be) the most stable one among all of the conformations of the molecule submitted.  It requires insight in theory and practice with the relevant programs to gauge the results of these predictions, e.g., by comparison with other structures already described.

MolCalc serves as an interface to the programs in the background (GAMESS, JSmol) performing the computation at RHF/STO-3G (orbitals and orbital energies), or PM3 level (all the rest, including optimization of the geometry).
Jensen, J. H.; Kromann, J. C. The Molecule Calculator: A Web Application for Fast Quantum Mechanics-Based Estimation of Molecular Properties. J. Chem. Educ. 2013, 90, 1093–1095; doi 10.1021/ed400164n.

Thanks for the recommendation. Your answer really helps and I'll accept that.


text:
Humidity diffusion in Silicone Elastomer and effects on refraction / transmissivity
There is an issue I'm facing that is very tricky.
A solar cell that is encapsulated in Sylgard 184 Silicone Elastomer with a front borofloat front glass. This cell is used to measure a light intensity.
The cell is enclosed in a chamber that can vary the temperature. The room ambient temperature is 25°C and humidity is 50%.
The cell has a temperature sensor applied to it that measures the cell's temperature.
The cell is shunted with a resistor (which has low temperature coefficient) in order to measure it's Isc value, and the whole is compensated in temperature, and is expected to behave linearly over temperature.
It does behave linearly when we raise the temperature, but when cooling, it goes off the chart and I can't really explain why.
One possibility is that as the cell cools, humidity diffuses into the silicone, which somehow changes its characteristics (refraction/transmission?) when the system cools down.
Perhaps, humidity seeps between the elastomer and the glass (a bonding agent is used) and when cooled, the dew point is reached causing the interface glass->elastomer to become glass->water->elastomer, changing the refractive index of the system.. ?
This graph shows the temperature VS the measured light intensity.  0-45 being the temperature in °C, irradiation being in W/m2

A clear change in behavior is noticed at around 25°C, which is the ambient temperature at which the cell rests when not performing tests.

First, what are the coordinates of the graph? Temperature in °C and cell current in amperes? Light intensity in W/m2? In lux?? Edit the question to show the units. without that, the graph is meaningless.
However, my first assumption is that the bonding of sealant to cell is not effective. More likely than water seeping into the seal, which would happen slowly, would be a break in the silicone-photovoltaic device bond, which could occur rapidly, allowing air to seep in and forming extra interfaces, with concomitant Fresnel reflection at each boundary (air-silicone and air-solar cell).
Compare a test unit before temperature cycling with one after cycling. Is there a slight "silvery" reflective air layer in the cycled unit, or noticeable bubbling of the sealant? Try sliding the sealant atop the cell with a fingertip -- is it as tightly adhered?

The coordinate are given above the graph "temperature VS the measured" 0-45 being the temperature in °C, irradiation being in W/m2

Added, there is nothing visible, and it is repeatable. The cell is encapsulated on an aluminum body with a front glass so I cannot "peel" the sealant.


text:
How can I determine the spin of a molecule?
I was reading statistical mechanics today and came across a line

For $\ce{O2}$ and $\ce{N2},$ we get half-spin, and that is why we get $σ = 2$ in their rotational partition equation.

I didn't understand it quite well! What does this line mean, actually? And how can I determine the spin of $\ce{O2}$ or $\ce{N2}?$
Secondly, how did they state that $σ = 2?$
Reference:
Haque. Physical Chemistry 4, 8th ed.; Shams Publication, 2021; Vol. 1.

Generally, it may be good to quote a broader context and marking the confusing part. // Your own reasoning – based on searching, reading and thinking – is supposed to be present to avoid the question closure for lack of own explicit  effort.

Did you read the box only? Certainly the book introduce that before.

@Alchimista, what box?

Quoting more than one sentence would be helpful. But I'd guess that you're probably looking for something called nuclear spin statistics.


text:
Why is the enthalpy of formation of 2,2-dimethylpentane larger than that of 2,2,3-trimethylbutane?
Firstly, it is commonly known and taught that branched alkanes are more stable than linear ones, and so have larger enthalpies of formation. But why is that the case? Are there some steric interactions at play?
Secondly, even if we just accept that relationship to be true, why are there instances where this relationship does not hold? Specifically, when I was exploring the data on NIST WebBook and I noticed that $\Delta_\mathrm fH^⦵$ is larger for 2,2-dimethylpentane than 2,2,3-trimethylbutane even though the latter is more branched than the former.

Are you sure it is not a value sign issue ?  As ΔfH⦵ should be larger for 2,2-dimethyl Pentane than 2,2,3-trimethyl Butane if the latter is more TD stable because of branching. ( Unless entropy takes decisive role, having lower formation deltaG even with higher formation deltaH ).

The NIST values are (kcal/mol): 2,2-dimethylpentane, -49.29 +/- 0.32; 2,2,3-trimethylbutane, -48.96 +/- 0.27. Considering the error in measurement, the latter may have a more negative heat of formation than the former. @Poutnik

Sorry for the confusion. As user55119 said in his comment, I meant the absolute value of ΔfH⦵ is larger for 2,2,3-trimethyl Butane compared to 2,2-dimethyl Pentane. @Poutnik


text:
Possibility of further chlorination of hexachlorocyclohexane
The reaction of benzene with excess of chlorine in the presence of sunlight produces 1,2,3,4,5,6-hexachlorocyclohexane as the major product. This is also given in the preparation section of the Wikipedia page on lindane.
Also, multiple chlorination of methane is also well-known, specially when chlorine is in excess. For example, it is mentioned at the bottom of this page from MasterOrganicChemistry.
My question: After hexachlorocyclohexane (lindane) is formed, I thought that it is an ordinary haloalkane, comparable to chloromethane, so if chlorine is excess, it should also substitute the remaining six hydrogen atoms from lindane - forming 1,1,2,2,3,3,4,4,5,5,6,6-dodecachlorocyclohexane. Since the reaction is used industrially, what I propose obviously doesn't occur. Why is that so?

+1 Perhaps they don't use excess chlorine, or control the reaction somehow?

@ShoubhikRMaiti I thought so too, maybe they use 3:1 ratio of $\ce{Cl2}$ and benzene for the reaction, but I couldn't find it in the two or three very basic textbooks (or on the net) I have currently. If you can find that they indeed use a stoichiometric amount in any textbook or online source you know of, please feel free to post that as an answer :) Thank you.

Somebody has AGAIN downvoted this question of mine without leaving any reason/comment as to why. Please DO NOT downvote if you cannot give the reason - specially as the question had 6 upvotes, meaning it was certainly not an outright nonsensical question.

@TRC it looks like that this problem has started for quite some time. Even other users are complaining about their posts getting downvoted without any reason. If this problem persists, feel free to contact support team.

Given enough exposure time, the reaction will lead to formation of various products, even the fully chlorinated compound: dodecachlorocyclohexane. Increasing the concentration of chlorine also help in the formation of this product. Doing some literature survey confirms this:

[...] During most of the course of this reaction (photochemical chlorination of benzene), both substitution and addition takes place simultaneously, the final reaction product after long periods approaching the composition of dodecachlorocyclohexane ($\ce{C6Cl12}$)
Photochemical Studies. XVII. The Chlorination of Chlorobenzene; a Comparison with Benzene, Edwin J. Hart and W. Albert Noyes, Journal of the American Chemical Society 1934 56 (6), 1305-1310, DOI: 10.1021/ja01321a016

The photochlorination of aromatic hydrocarbons may result in either addition or substitution or both. The photochlorination of benzene in either liquid phase or gas phase yields mainly hexachlorocyclohexane, but when high chlorine concentration and long exposure time are employed, the product is mainly dodecachlorocyclohexane.
Technique of Organic Chemistry: Catalytic, photochemical, and electrolytic reactions, Arnold Weissberger, Interscience Publishers, 1956


text:
Will Sodium Polyacrylate (SAP) absorb the gas from liquid ammonia?
I'm considering using sodium polyacrylate grains to absorb spills of a relatively strong Ammonium Hydroxide solution. Will the SAP (super absorbent polymer) only absorb the water, and in effect make the remaining solution a higher concentration of NH4OH? Or would some of the dissolved gas also be held by the SAP? I am a mechanical engineer by trade, so I can research any terms you use, but my chemistry background is limited.

Since NH3 is held very tightly by H2O (700 volumes of gaseous ammonia can dissolve in one volume of liquid water), at a guess, most of the NH3 would stay with the water. That said, does NH3 chemically adversely affect SAP? If you can't find that in the literature, perhaps a small-volume experiment would help. You might also include a mild acid, such as citric acid, mixed with SAP, to make a non-volatile salt.

The available literature is conflicting. On the one hand, a website claims that its SAP is unsatisfactory for absorbing ammonia, acetone and isopropanol (Ref 1).
On the other hand, there is a patent that claims an SAP polymer specifically for absorbing ammonia (Ref 2).
The monomers that are used to make the super absorbent polymer may be sodium or ammonium acrylate, or acrylic acid itself. It has been reported that multivalent cations inhibit the swelling of the SAP by tying separate chains together, and it could be that even monovalent cations in large quantities could inhibit swelling.
It does seem that an SAP polymerized from an acid monomer could react with NH3 or NH4OH to give the ammonium salt, which should swell the SAP just as if it were polymerized from the ammonium salt itself.
Since there are so many grades of SAPs, several are surely acid-containing, and would be useful for containing ammonia spills. Other grades, based on polymers of salts of acidic monomers would be unsuitable.
The preferred course of action, then, would be to contact manufacturers or distributors of SAPs and state your requirements.
Ref 1. https://recycphp.com/en/absorption-domestic-liquids-sodium-polyacrylate/
Ref 2. https://patents.google.com/patent/WO2017209339A1/en


text:
Why aren't electrons included in the equilibrium constant expression?
This might be elementary, but why aren't electrons included in the equilibrium constant for a half-reaction?
For example, for
$$\ce{O2 (g) + 4H+ (aq) + 4e- -> 2H2O(l)}$$
we have
$$Q = \frac{(a_\ce{H2O})^2}{(a_\ce{O2})^2(a_\ce{H+})^4}$$
without the electron.
This means that the activity for electron is 1? Why is that so? Is it because of any definition just like solid and pure liquid?
Atkins$^1$ says that because the electrons are 'stateless'. I am not sure what that mean.

[1] Peter Atkins, Julio de Paula, James Keeler. Physical Chemistry (11th Edition). OUP. 2018. Page 218.

You've clearly used some MathJax before; but I think you can use it more efficiently. For one, fractions with nested subscripts/superscripts should really be written out in display style $$...$$, otherwise it's extremely difficult to read. Secondly, you can use $\ce{H2O}$ $\ce{H2O}$ to typeset chemical formulae more easily, rather than using \text{} on every element symbol. Consider reading FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange? for a refresher, or introduction, if you've not seen these before.

Atkins at times generates useless points and unnecessarily complicates matter by using overly fancy language (esoteric). Anyway, he is getting pretty old and most likely his new co-authors added this. Yes, you can assume unity for electron activity.

@M.Farooq I can't get why that textbook had much success. It often rises Qs without giving solid As. In addition, it uses so many difficult to read characters and, especially, breaks the flow in too many boxes, highlights, activities and the like.

Hmm I am not completely sure, but my guess is that the electron is generally provided by an electrode. In that case, the electrode surface, volume etc. are constant throughout the reaction, so its ability to transfer electrons is also constant, so whatever the factor is, we can neglect it by setting it to 1.

@Alchimista, Atkin's was never my favorite for the reasons you state above.