text:
Normalization of a free particle in 1 dimension
I am having difficulty normalizing the wavefunction for a free particle in 1 dimension.
I have sent an attachment of my working out.
My attempt at the normalisation
Free electron:
$$-\frac{\hbar^2}{2m}\frac{\mathrm d^2 \psi(x)}{\mathrm dx^2} = E\,\psi(x)$$
\begin{align}
\psi(x) &= A\sin(kx) + B\cos(kx) \\
\psi(x) &= \psi^*(x)
\end{align}
$$\therefore \int^\infty_{-\infty} \left[ A\sin(kx) + B \cos(ks)\right]^2 = 1$$
$$\int^\infty_{-\infty}\left[A^2\sin^2(kx) + AB\sin(kx)\cos(kx) + AB\cos(kx)\sin(kx) +B^2\cos^2(kx)\right]\mathrm dx  = 1$$
Since, $\sin^2 x + \cos^2 x = 1$,
$$\implies \int^\infty_{-\infty}\left(A^2 + B^2 + 2AB\sin(kx)\cos(kx\right)\mathrm dx) = 1$$
Since, $\sin(2x) = \sin x\cdot\cos x$,
$$\int^\infty_{-\infty}\left(A^2 + B^2 + 2AB\sin(2kx)\right)\mathrm dx = 1$$
$$\int^\infty_{-\infty}\left(A^2 + B^2\right) + AB\sin(2kx)dx = 1$$
\begin{align}
A^2 + B^2 + AB\int^\infty_{-\infty}&\sin(2kx)dx \\
&=A^2+ B^2 +AB\left[-\frac{1}{2}cos(2kx)\right]^\infty_{-\infty} \neq 1
\end{align}
Could someone please let me know of my mistake?

@ToddMinehardt The question explicitly says "for a free particle in 1 dimension" and "free electron" so there is no box, and the integral really is over all space. But I do agree the bounded cases is much easier, in fact the integral that is being attempted here is not well defined - see https://physics.stackexchange.com/questions/165373/normalizing-the-solution-to-free-particle-schr%C3%B6dinger-equation

To answer the question, over and above the errors noted in the other answer the integral is just not well defined - it is infinite. There are two ways to see this

The "common sense" method. Consider
$$ \int^\infty_{-\infty} \left[ A\sin(kx) + B \cos(kx)\right]^2dx$$
The quantity under the integral is positive semi-definite - it is always greater than or equal to zero. It describes oscillations all of which are positive. Each oscillation bounds a finite positive area between it and the x axis. As the bounds of the integral are all space there are an infinite number of oscillations. An infinite number of positive finite areas evaluates to an infinite area. Thus your integral is not well defined.

Maths - assuming A and B are real (but removing this restriction makes no difference to the final conclusion)

$$\left[ A\sin(kx) + B \cos(kx)\right]^2 = \left[A^2\sin^2(kx) + 2AB\sin(kx)\cos(kx) +B^2\cos^2(kx)\right]$$
The way to evaluate the integral is to use the identities
$$ 2sin(\theta)cos(\theta) = sin(2\theta) $$
$$ sin^2(\theta) = {{(1-cos(2\theta))}\over2}$$
$$ cos^2(\theta) = {{(1+cos(2\theta))}\over2}$$
Thus we can write the integral as
$$
\int^\infty_{-\infty} \left[A^2{{(1-cos(2kx))}\over2}+B^2{{(1+cos(2kx))}\over2}+ABsin(2kx)\right]dx
$$
The first term contains
$$
\int^\infty_{-\infty} \left[ A^2\right]dx
$$
This is infinite. Similarly the second term for $B^2$. Thus the integral is not well defined [yes this is slightly hand-waving but avoids a discussion of $cos(\infty)$ and similar]. So what you are doing wrong is trying to evaluate an integral that doesn't have an answer.
What you are running into is the fact that not all wavefunctions are normalisable. See Normalizing the solution to free particle Schrödinger equation in the physics stack exchange forum for more details

Your attempt has multiple issues, most of which are very basic.
The first issue that I noticed was how you expanded and simplified your integral.

\begin{align}
\int^\infty_{-\infty}[A^2\sin^2(kx) &+ AB\sin(kx)\cos(kx) \\
&+ AB\cos(kx)\sin(kx) +B^2\cos^2(kx)]\mathrm dx  = 1  \tag{1}
\end{align}
Since, $\sin^2 x + \cos^2 x = 1$,
$$\implies \int^\infty_{-\infty}\left(A^2 + B^2 + 2AB\sin(kx)\cos(kx\right)\mathrm dx) = 1$$

Error in equation($1$): $A^2\sin^2 x + B^2\cos^2 x \neq A^2 + B^2$
If you were to simplify equation (1), you can not use the above statement, for a very simple reason. That is because even if $\sin^2 x + \cos^2 x = 1$:
$$A^2\sin^2 x + B^2\cos^2 x \neq A^2 + B^2$$
Instead it should be:
$$\int^\infty_{-\infty}\left[A^2\sin^2(kx) + B^2\cos^2(kx) + 2AB\sin(kx)\cos(kx)\right]\mathrm dx= 1$$
The second issue is in the actual solving of the integral itself:

$$\int^\infty_{-\infty}\left(A^2 + B^2 + AB\sin(2kx)\right)\mathrm dx = 1$$
$$\implies A^2 + B^2 + AB\int^\infty_{-\infty}\sin(2kx)dx  = 1$$

Two issues here: $A^2 + B^2$ are not being multiplied and so cannot be just taken outside the integral, this means that the solution is wrong, also a $2$ went missing. Notice where?
$$\int^\infty_{-\infty}\left(A^2 + B^2 + \color{red}{(2?)}AB\sin(2kx)\right)\mathrm dx = 1$$
These are just the two issues in the solution. Hope this is enough.


text:
What are the gaseous outputs of electrolysis of Water along with NaOH or KOH?
Theoretically electrolysis of water yields Hydrogen and Oxygen in gaseous form.
But to enhance conductivity we have to add an electrolyte like $\ce{NaOH}$ or $\ce{KOH}$.
Does this produce any other gases along with Oxygen? Does $\ce{Na+}$ or $\ce{OH^-}$ ions also come out as some form of gas? Or practically speaking can people breathe the oxygen which comes out of electrolysis?
What are the gaseous outputs of electrolysis of Water along with $\ce{NaOH}$ or $\ce{KOH}$?

Per your third question: it's done that way in submarines and on the ISS. Also, you might want to break-up this question into more concise ones. There is a chance this might be close for lacking focus.

@ToddMinehardt all I want to know if can we breathe electrolysis generated oxygen safely. I'm concerned about electrolyte generating toxic gases. You are welcome to edit. Thanks a lot

@ToddMinehardt what electrolyte do they use in Submarine

@NilayGhosh Thank you so much. Regarding the oxygen shortage in India, can people use electrolysis on distilled water to create breathable Oxygen??

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sort: votes


text:
Volume of Oxygen Produced By Electrolysis of Water
I am trying to estimate the volume of oxygen that will be produced by electrolysis of water at normal temperature and pressure with a current of $\pu{50 A}$. The following is my calculation:
The reaction is
$$\ce{2H2O -> O2 + 4H+ + 4e-}$$
Therefore, to produce $\pu{1 mol}$ of $\ce{O2}$ ($\pu{24 L}$), we need to move $\pu{4 mol}$ of electrons through the circuit and this is $4F = 4 \times 9.65 \times 10^4 = \pu{3.86 \times 10^5 C}$ of charge. With a current of $\pu{50 A}$, we move $50 \times 60 = \pu{3000 C}$ of charge per minute. Therefore, we produce $\frac{24}{3.86 \times 10^5} \times 3000 = \pu{0.186 L}$ of $\ce{O2}$ per minute.
Is this calculation correct? If it is, is there any way to improve the yield of oxygen?
Disclaimer: I am not a chemist and it has been years since I studied physical chemistry. The reason for this computation is to decide if electrolysis is a quick and feasible method to produce oxygen to help mild COVID patients at home (since there is a severe shortage of oxygen and concentrators where I live).

Your calculation is correct : $0.186 $L per minute. The electrolysis is not a quick method for producing large amounts of oxygen gas. On the other hand, the electrolysis of water would produce Hydrogen at the other electrode. How to separate these two gases ?  Furthermore, it is extremely difficult to conduct any electrolytic operation with such a high currant ($50 A$). You would need huge electrodes, because whatever the nature  of the electrode used, the current is highly dependent on the surface of the electrode. The upper limit is of the order of $\pu{0.1 - 1 A/dm^2}$.

Thanks for checking my calculation @Maurice. The plan was to use large quantities of wrapped up stainless steel mesh to make the electrodes and have tubes enclosing the electrodes to separate the gases at the source. And since a large amount of heat would be generated, we were planning to setup a cooling system enclosing the electrolysis chamber. But considering the small amount of oxygen produced, we will switch to using a zeolite based method.

Thank you very much for your idea and help.
I think that your calculations are correct.
But I think you'd better to try other simpler choices to make oxygen.
One way is air separation. If you separate nitrogen from air, you gain oxygen.
This can be done by "Zeolite oxygen generators". You can read these site about them.
1- https://www.zeochem.com/our-applications/medical-oxygen-generation
2- https://www.jalonzeolite.com/product-item/molecular-sieve-jlox-500/
There are many important factors that you have to consider about electrolysis of water that I know some of them:
First water has low conductivity. So, you have to add a strong inert electrolyte to make charge transfer between electrodes. This electrolyte could be any soluble salt, an acid or any substance that solves in water and then dissociates to ions. But, the electrolyte must not interacts to the main reaction and should be inert and surely it should be absolutely safe and not harmful for patient. Sodium or potassium sulfate or nitrate can be used.

Next, you have to choose an inert electrode to conduct electricity to the solution and then the gas be produced on it. Platin and gold can be use. But, they are expensive. Another choice can be carbon electrode.
And also take care about Hydrogen that produced on the negative electrode (cathode), that is flammable and can be exploded by spark.
And, so many other considerations to make oxygen efficiently and, next purify it to be suitable for hospital usage.
I wish you health.

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text:
Is the cyclic hexenone drawn correctly in this depiction?
I'm learning organic chemistry from Clayden's book (2e). I've noticed they draw the cyclic hexenone below like this:

However, since the carbon on the carbonyl group is sp2 hybridized I am wondering if this is correct. Shouldn't this just be planar and look like this:

Ie, more of an envelope conformation? Or am I not grasping something correctly, like is it maybe not convenient to draw it like an envelope? The below structure was made in ChemSketch and using the 3D optimization function.

ChemSketch is not a thing for making 3D structures, at least realistic ones.

So, yeah, Clayden's right.

@Mithoron Actually, I can see an envelope now if I Iook at it from a different angle. But because it is not mentioned before it confused me, since you were looking at cyclohexenes from the "front" in all depictions of a regular cyclohexene before in the book like shown here: https://i.imgur.com/vMNnAIK.png . So I think it is a bit misleading in the depictions above, which give the impression that the carbonyl carbon is at an angle up (right conformation in the figure) vs down (left conformation in the figure).with the next carbon.

Simply, three carbons in the ring are $\mathrm{sp^2}$ while other three are $\mathrm{sp^3}$. Thus, it can't be planner without high strain ($109^\circ$ going on $120^\circ$).

Anyway in the red draws the carbonyl C doesn't really depart from planarity.


text:
Alcohol (ethanol) distillation: mystery congener
I regularly home-distil small amounts of potable alcohol (ethanol), mainly from fermented sugar(s) solutions (the 'wash').
My still is a box standard, $4.5\mathrm{L}$ electrically heated (safety!), air-cooled pot still, which produces exactly $1$ single stage distillation ('one Theoretical Plate') Starting from a wash of about $17\text{ %}$ ABV, I can distil that to (initial ABV) about $65\text{ %}$ ABV.
About $1.0\mathrm{L}$ is distilled off from $1$ gallon that way and the distillate then diluted with water to about $37\text{ %}$ ABV.
Distillers (home, pro, moonshiners) use a system to get the most and the best out of their distilled washes and it goes as follows:

The "Heads": the first part ($5 - 10\text{ %}$) of the distillate is often discarded because it contains non-ethanol low boiling congeners like methanol, ethanal, acetone and low boiling esters (it's a commonly held but erroneous belief that methanol and ethanol can be separated by means of a single distillation - see the $\text{MeOH/EtOH}$ VLE). These congeners give the distillate a 'synthetic' and undesired taste profile.
The "Heart": this is the good tasting part of the distillate, that is kept as an alcoholic beverage.
The 'Tails' (or 'faints'): the last part of the distillate contains very little alcohol, as well as higher boiling compounds like higher alcohols and $\text{C8 - C12}$-based esters. The high boiling congeners are also referred to as "fusel oils".

Apologies for the lengthy preamble.

Due to the pandemic and restrictions on the hospitality sector I obtained several gallon of out-of-date $4.5\text{ %}$ ABV apple cider of a very well known brand, from a friendly publican.
I first consumed a couple of liters as such and it tasted fine and caused no adverse effects. The product was indistinguishable from 'in date' virgin cider. But my purpose was always to distill the ethanol of out it.
So, using my standard procedure I distilled several gallon of the out of date cider, obtaining about $2\mathrm{L}$ of distillate at about $35\text{ %}$ ABV. Heads were discarded and the distillate was run over activated carbon (a practice highly recommended by home distillers and moonshiners, even though I'm not sure it serves a real purpose)
And here's the rub: the 'heart' smelled and tasted awful. The taste is hard to describe but keywords would be 'spicy', 'hot', 'sharp', 'solvent-like'.
I then took the step of redistilling the $2\mathrm{L}$ of product, discarding the first $300\mathrm{ml}$ and keeping a 'heart' of about $1.1\mathrm{L}$. Alas it changed very little, if anything: the redistilled alcohol was still basically undrinkable!
The contrast between the redistilled alcohol and cheap commercial gin is quite stark.
I measured the paper $\text{pH}$ of both and they were both essentially neutral.
That redistilling didn't remove the offending compound(s?) suggests that  its BP is close to that of ethanol or that it forms an azeotrope with it.
The smell and taste of the mystery congener(s?) doesn't really comply with what is usually suspected to be in the 'heads':

$\text{MeOH}$ smells and tastes very much like alcohol (and in any case should only be present in very small amounts)
acetone and ethanal probably are too low boiling to remain partly in the 'heart'. Ethanal has a pleasant odour anyway.
all the low boiling esters have pleasant, fruity flavours.

My questions are:

What could be the problem congener{s)?
How to further investigate to find out what might be the problem congener{s)? I don't have access to modern analytical techniques like IR, GC, GC-MS etc.

I do have access to $\text{TLC}$ and a $1\mathrm{L}$ lab still with an $800\mathrm{mm}$ Vigreux column.

Have you ever considered that your liquid may contain ethyl acetate or acrolein ? Their boiling point are 77°C and 53°C, which is not far from ethanol boiling point.

+1 for ethyl acetate. It's a known component of fermented beverages and is described as imparting a fruity flavor when present at very low concentrations, but more of a solvent-like flavor when the concentration increases (as it would during your distillation). And the boiling point is so close to ethanol that it would be very hard to separate them.

acrolein would impart a truly disgusting taste I think. Not sure where it would originate from: oxidised glycerol maybe? EtAc would fit better but why the high concentrations? Thank you both!

People turn apple cider all the way to apple cider vinegar, so during the process, there will be a time when the amount and ratio of ethanol and acetic acid are optimally balanced for making ethyl acetate - but it doesn't seem to happen. However, given enough time (and oxygen and bacteria), maybe different things happen. Interesting that the expired cider didn't have the nasty taste, but the concentrate did. I wonder if you actually made the ethyl acetate by your heating and distillation procedure.

Ethyl acetate presence gives nail polish remover/ solvent flavor/aroma. This is not related to volatile acidity. It's known to be produced by Pichia (and probably other microbes) in the presence of oxygen.
Source: https://www.homebrewtalk.com/threads/need-a-little-help-with-cider-flaws.666728/

The CRC Handbook shows a ternary azeotrope of water (9.0%), ethanol (8.4%) and ethyl acetate (82.6%) that boils at 70.23 °C vs the water-ethanol azeotrope that boils at 78.17 °C (96% ethanol). The boiling points are sufficiently close that a one-plate column might not do a significant separation, but the heads still should taste worse.
Suggestion: if the apple cider is acidic from a little acetic acid, neutralize to pH 7 or 8 before distillation ($\ce{NaHCO3}$). This might reduce the possibility of ethyl acetate formation. Another experiment would be to take fresh apple cider and see if you can get good tasting alcohol from it.
TLC of volatile solvents is difficult, but you might be able to detect different spreading patterns for the heads, the nasty heart, and some other heart (which tastes OK).

I don't think the ethyl acetate was produced during the distillation. Microbes produce it during fermentation by reaction of ethanol with the activated acetate in the form of acetyl-CoA. As long as the concentration is low, the flavor is described as fruity and pear-like, and for whatever reason, the harsh solvent flavor does not become apparent until it reaches higher concentrations, as would happen during a distillation.

@Andrew: Yeah, that's what I would have thought. I would not suggest those conditions to synthesize it. Do you think the microbes could have been activated by the heat to increase whatever EtOAc was already there?

The microbes were probably filtered out of the original cider, and even if not, they'd die very quickly upon heating to the temps used in distillation.

Thank you for the comprehensive answer. Looks like EtOH is the one to look at. It's a shame I have such trouble identifying it...

The azeotrope info is very useful too.

Like 'Jamie Gaidis' I was convinced at least some of the suspected $\ce{EtAc}$ is formed during the distillation step but presently I no longer believe that.
Instead I believe that compound is formed during fermentation and then concentrated into the distillate as the ternary azeotrope that 'Jamie Gaidis' suggests.
My evidence:
Firstly, this page of Home Distiller is awash with 'chemistry salad', half-truths, lies and misconceptions re. chemistry but still allows a chemist to read between the lines and conclude that:

$\ce{EtAc}$ is formed during fermentation and can be removed from the fermented 'wash' (or wine or beer) by means of bicar ($\ce{NaHCO3}$), washing soda ($\ce{Na2CO3}$) or lye ($\ce{NaOH}$), in an irreversible alkaline de-esterification (saponification).

Secondly, this empirical study shows convincingly that low levels of $\ce{EtAc}$ can be saponified by low concentration $\ce{NaOH}$, at near-RT temperatures and in  short reaction times. The conditions described in that paper are probably not a million miles removed to what happens in my factor-space.
Empirical:
The $\mathrm{pH}$ of the cider used below was about $3 - 4$ (still carbonated)

Evaluating $\ce{NaHCO3}$:

To $2\mathrm{L}$ of virgin apple cider, about $8\mathrm{g}$ of $\ce{NaHCO3}$ (baking powder) was added and stirred in with a high speed stick mixer, which minutes before had been used to decarbonate the cider.
Without further ado, the cider was then distilled, with the first distillate (the 'heads') coming over after $30 - 40\text{ minutes}$.
The result was quite astonishing: even the 'heads' were much less 'aromatic', 'solventy' than those obtained without the bicar treatment and were in fact drinkable.

Evaluating $\ce{NaOH}$:

To $3\mathrm{L}$ of virgin apple cider, about $12\mathrm{g}$ of $\ce{NaOH}$ (drainage unblocker, small pellets) was added and stirred in with a high speed stick mixer, which minutes before had been used to decarbonate the cider.
Again, this concoction was distilled immediately after mixing.
Here too a distinct improvement was noted but subjectively I'd say the 'heads' of run $\text{#1}$ were marginally better but the 'heart' of run $\text{#2}$ was very good.
There was one marked difference: the liquid left in the pot with the $\ce{NaOH}$ run was much darker than its $\ce{NaHCO3}$ counterpart.

I now need to test the use of these alkalis on my own $15 - 17\text{ %}$ washes.


text:
Change in Enthalpy when Pressure is constant
Okay so let me start from the beginning.
I know that  $ΔH= ΔU + p ΔV$ ( at constant pressure).
Expanding  $ΔU$,
$$ΔH= q + pΔV + pΔV ~~~~~\text{(q at constant pressure)} \tag{1}$$
Next my chemistry book continued to say that in expansion work done by gas is negative, so the PV work terms cancel out, i.e,
$$ΔH= q - p ΔV + p ΔV$$
Therefore, heat at constant pressure is equal to the change in enthalpy.
Doubt:
Now my question is, in (1) what is the difference between both the PV terms ?
Can't the equation  turn out to be:  $$ΔH= q - p ΔV - p ΔV= q - 2p ΔV $$?
I am a high schooler and I know that I am missing something important, so can someone help me out?
EDIT:

We may write equation (6.1) as
$∆ U = q -p∆V$
at constant pressure, where
$q$
is heat absorbed by the system and $–p∆V$
represent expansion work done by the system.

This is what is written in my book. That is, work done in expansion is negative.
Reference - class 11 chemistry ncert book

You expanding for delta U is wrong. For constant p,  W = - p. Delta V.

@Poutnik I have taken care of that in the very next line.

Expansion work done by gas is not negative. But in U and H context, positive work is the one done to the system, not by the system. See also 2 historically opposite notations, Delta U = Q + W versus Q - W.

According to the convention in chemistry, isn't the work done BY the gas ON the surroundings (BY the system) negative ? i.e the gas is expanding.....but anyway how does this answer my question ? basically all that I am asking is the physical intuitive difference between the pv work for which enthalpy is defined and the pv work of internal energy. I will check with the convention.

For more on the pV sign convention see https://chemistry.stackexchange.com/questions/66088/why-do-chemistry-and-physics-have-different-sign-convention-in-thermodynamics or https://chemistry.stackexchange.com/questions/139100/what-is-the-correct-expression-of-pressure-volume-work?noredirect=1&lq=1  For more on enthalpy definition as relates to heat at constant p see eg https://chemistry.stackexchange.com/questions/39988/does-enthalpy-equal-heat-when-pv-work-is-done?rq=1

See also https://en.wikipedia.org/wiki/Work_(thermodynamics)#Sign_convention

I tried to interpret what you wrote in Eq 1 but you should check this. It's unclear why you write both pV terms with positive signs, which is why you have received many comments regarding the sign convention. Also worth learning how to format math expressions using mathjax, see eg https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference

Your book uses the "chemistry sign convention", that is,  $w=-p_\textrm{ex} \Delta V$, where $p_\textrm{ex}$ is the external pressure.
However the enthalpy is not defined in terms of the external pressure but rather the pressure $p$ in the system: $H=U+pV$. From the definition of enthalpy it follows that $\Delta H= \Delta U+ \Delta (pV)$, or, at constant system pressure, that $ \Delta H = \Delta U+ p \Delta V = q + \left(-p_\textrm{ex} + p \right) \Delta V$. Only when the condition of mechanical balance ($p_\textrm{ex} = p$) is satisfied at the beginning and end of a process can we say that $ \Delta H = q$.
For more on the $pV$ sign convention see other posts. For more on enthalpy definition as relates to heat at constant pressure see eg this post.

On the sign convention:
If you choose to write $U=q+w$ ("chemistry convention") then for constant applied (external)  pressure $w=-p_\textrm{ex} \Delta V$. Then the sign of the work is regarded as negative when the system expands, doing work on the surroundings ($\Delta V>0$, $w<0$) or positive when the surroundings does work on the system ($\Delta V<0$, $w>0$).
If you choose to write $U=q-w$ ("physics convention") then for constant applied (external)  pressure $w=p_\textrm{ex} \Delta V$. Then the sign of the work is regarded as positive when the system expands, doing work on the surroundings ($\Delta V>0$, $w>0$) or negative when the surroundings does work on the system ($\Delta V<0$, $w<0$).

Note that these days physics becomes accepting the chemistry convention as well, so it becomes rather scientific (+) versus engineering (-) convention.
All that may create a confusion mess, so it is adviced always making oneself sure which one is used.

@Buck Thorn♦ That was the clarification i was looking for thanks a lot ;)


text:
Mechanism for removal of trimethylsilyl directly bonded to carbons
I recently came across a scheme for the synthesis of panacene (Sabot et al., 2008OrgLett4629):

In this synthesis, the step from 14 to 9 in removes  a trimethylsilyl ($\ce{TMS}$) group directly bonded to an aryl ring with $\ce{TBAF}$ (tetra-n-butylammonium fluoride) and $\ce{CsF}$ with $\ce{DMF}$ as the solvent.
I understand that silicon has a strong affinity for Fluoride and am familiar with the mechanism for deprotection of silyl ethers. For silyl ethers the $\ce{RO^-}$ acts as a leaving group from the 5-coordinate silicon. Here, however, the aryl anion seems like a bad leaving group. So, what is the mechanism for this?

By TMS I assume you mean trimethyl silyl ? You should at least mention what the acronyms mean, because not everyone might be familiar with every aconym. I don't know what TBAF means for example. Also, the paper is behind paywall, so please consider posting a diagram of the synthesis scheme here, otherwise people might not know what you are asking about.

For future reference: Your question is easier to access (and thus, more likely to receive helping answers / comments) if you may illustrate the reaction by a scheme.  The Org. Lett. paper contains multiple instances where $\ce{C-SiMe3}$ bonds are broken (e.g., equally from a terminal acetylene by moderation of $\ce{AgNO3}$, NBS in acetone), thus it would be a time saver for the reader of your question to see all at once.

@ShoubhikRMaiti It is a guess that this added scheme is about the reaction the OP refers to.

@Buttonwood Thanks, I will take that into account in my future questions. The image is correct.


text:
titrating a weak acid with strong base in the present of buffer
I would like to get a neutral salt solution for a weak acid whose pKa is 3.42. This weak acid is in the form of solution (now is 20% w/w), thus I cannot directly dissolve the powder into the buffer etc. And I would like the resulting solution to be as concentrated as possible.
I am thinking to titrate the acid with NaOH (1M) or other bases (that only generates the salt of the acid and water). After calculation, I found that 2.0998 ml of NaOH will give me around pH 7.4 (which is what I want as it is to be used in a biological system). The problem is that: it is very difficult to accurately add 0.0998 ml of NaOH. And I found that 0.0001 ml of NaOH deviation will lead to huge changes in pH, however, I want to make sure the final pH is between 7.2 - 7.4.
Another consideration is I would want to avoid using pH meter because I would like to maintain the solution as sterile as possible (which I will prepare everything in a biological hood).
In short, I would like to know how to prepare a neutral solution for that acid precisely, e.g. through calculations of the exact volume of base needed or use diluted NaOH (though I still want the resulting salt solution to be as concentrated as possible). I would like to know whether it would help if I add some buffer to the system (e.g. PBS that is often used in biological experiments)?

I can't say I get what you really want there. If you added enough buffer it would give you any pH you'd want, but it would hardly be a solution of acid, but rather its salt.

Is it possible to use a pH meter?

If you make $\pu{1 M}$ solution of pure conjugate base of your acid, its $\mathrm{pH}$ would be roughly 8.7.

Phosphate Buffer Saline (PBS) is a neutral buffer in physiological $\mathrm{pH}$.  Thus, I'd expect if you use it to dissolve the pure conjugate base of your acid, the solution $\mathrm{pH}$ might be between 7.2 - 7.4 as you expected. You may need to use a  $\mathrm{pH}$-meter to check the  $\mathrm{pH}$. You can adjust the  $\mathrm{pH}$ using either dilute $\ce{NaOH}$ or $\ce{HCl}$ solutions.

Thank you for all of your replies, I have edited the question and hope that has clarified the problem. In short, I would like to avoid using pH meter and I cannot dissolve any "powder".@Mathew Mahindaratne

I may be missing something, but a rough reality check says that if you are thinking of adding 2.0998 mL of 1 N NaOH solution, you have ~0.0021 moles of base, which for a good buffer, should have about ~0.0021 moles of acid. If your acid solution is 20% by weight, and guessing 200 for a molecular weight, your total acid volume is 2.0998 mL also. It seems to me that your accuracy problem with the base is in addition to the accuracy problem with measuring the acid.
Unless the acid is very expensive (isotope?), scale up! In your sterile hood, make 2 batches of the buffer solution, starting with 50 mL of the 20% acid in each vessel (of about 200 mL capacity). Add 51 mL of 1 N NaOH solution to one and 49 mL of 1N NaOH solution to the other. Measure the pH of both by using an eyedropper to take out a few mL for a regular pH meter, or just one drop to place on a micro pH meter, such as

https://news.thomasnet.com/fullstory/pocket-ph-meters-utilize-non-glass-silicon-chip-sensor-608720
One of the 100 mL solutions may be exactly what you want. If not, it would be nice it their two pHs straddle your desired pH. Using an eyedropper, transfer some of one solution into the other, stir, and check the pH by withdrawing a drop to test on the micro pH meter. The pH electrode will never be immersed into the buffer. When you get one of the solutions to the pH you need, you can still save the residue for further adjustments.
Some micro pH meters claim to be able to measure as little as 5 microliters, so if the acid is very expensive, you could scale down.
Calculations are very fine for exams and theorizing, but when it comes down to needing an exact pH near 7, you may have to take other variables into consideration, including the preparation, i.e., bringing the calculations into reality. If you calculate, you are guessing you will get it right. If you measure, you'll know if you got it right. (Assuming you calibrate the pH meter, and it works as expected, etc., etc.)

Thanks for your thorough answer! And yes, the acid is quite expensive... It seems using a pH meter is necessary for my problem, though I always find calibrate the pH meter is super complicated. But I will definitely try!
Another follow-up question will be: if I ultimately use the neutralized solution as a concentrated stock solution, and the final concentration is let's say 100-fold diluted in PBS/ culture medium. Do you think the buffer function in these solvents can tolerate the extreme pH (e.g. pH 10 if I over titrate, or pH 3 if I added a little bit less NaOH)?


text:
How can I determine the amounts of potassium bicarbonate and potassium sulfate in a solution?
I have an aquaeous solution of potassium bicarbonate and potassium sulfate. I want to find out the proportions of both compounds in the solution. What would be an efficient procedure?
Note that I do not have an analytical balance or other fancy equipment. I can do simple procedures like titrations.
One idea I had is that I could titrate it with an acid of known molarity. That would react with the potassium bicarbonate and when the pH became neutral as shown by an indicator, that would tell me how much KHC03 I had.
The sulfate is a more tricky problem.

Hint: how could you easily remove the bicarbonate ions? Measure what comes out. Then get the dry weight of what remains and you can calculate proportions.

@DrMoishePippik I don't have an analytical balance.

There must be some sulfates that are insoluble in water, hmm...


text:
Why isn't entropy dimensionless?
The usual units are J/K, however
the only reason temperature has different units than energy is because we didn't used to know temperature was just the average kinetic energy of atoms/molecules.  Hence, using a more reasonable system, the units would be J/J which just cancel.
I get that we're stuck with having different units for temperature now, just as we're stuck with the inconvenient convention that electrons are negative and protons positive.  But since we already have a constant in the definition of entropy anyway, why haven't we defined it in a way as to convert the temperature units to energy and so make entropy dimensionless?
Or, better yet, why not, for the purposes of thermodynamics problems, just use subsume the Boltzmann constant into the units for temperature in the first place?  That would make certainly make for a less messy system of units and make the units of both temperature and entropy better match the definitions of the quantities.  I get that we can't just switch the temperature unit system for everyone, but I'm rather surprised scientists working in thermodynamics (thermodynamicists?) haven't adopted that unit convention within the field.

https://chemistry.stackexchange.com/questions/62794/why-can-entropy-be-measured-in-joules-per-kelvin

https://chemistry.stackexchange.com/questions/133521/why-do-the-first-and-second-laws-of-thermodynamics-not-contradict-each-other

While OP's is correct with the idea. Thing is she asks why it wasn't changed and that's hardly a chemistry question.

There are even "natural units", but are so incredibly impractical that are scarcely  heard about.

Anyway it seems to me that T and E are different quantities even when linked by the Boltzmann constant. At least number of particles or mass must link the two, isn't it?

It is not true that temperature is just the average kinetic energy of atoms/molecules. E.g. gas molecules with different numbers of atoms may have different amounts of translational, rotational, and vibrational kinetic energy while having the same temperature (the existence of solids shows that we cannot restrict us to translational kinectic energy).


text:
What is the derivation for the expression of depression in freezing point?
In my textbook, it is written that the expression for depression in freezing point is
$$\Delta T_\mathrm f=K_\mathrm f\cdot m$$
where $\Delta T_\mathrm f$ is the freezing point depression (defined as positive value),  $K_\mathrm f$ is a constant of the solvent, and $m$ is the molality of the solute.
But how did this expression come about? There is no derivation given in my textbook. Can someone please rigorously prove this expression?

That's likely because you are using a basic introductory textbook.  Getting into the derivation is usually considered beyond the scope of an introductory chemistry/biochemistry class.  Most physical chemistry textbooks will show it.  If you do the derivation, you'll end up with $\Delta T_f \approx \frac{R T_f^2}{\Delta H_f} m$ And since $\frac{R T_f^2}{\Delta H_f}$ is a constant, it's replaced with $K_f$, giving $\Delta T_f \approx m K_f$

If you want to see what a derivation looks like, you can find one here: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Physical_Chemistry_(Fleming)/07%3A_Mixtures_and_Solutions/7.06%3A_Colligative_Properties  But this derivation won't make sense to you unless you understand the starting expression for chemical potential as a function of mole fraction, which is something you will learn if and when you take physical chemistry.

See these answers: https://chemistry.stackexchange.com/questions/61783/verification-of-variants-of-the-ebullioscopic-formula/61818#61818 ... https://chemistry.stackexchange.com/questions/61750/is-the-ebullioscopic-constant-temperature-dependent/74481#74481

The relationship can be determined experimentally. Take any substance, dissolve it in a liquid with some known concentration and record the new freezing point. The new freezing point is linearly correlated with the concentration of the solute regardless of the identity of the solute. See Raoult's 1882 study determining Kf for benzene: https://gallica.bnf.fr/ark:/12148/bpt6k30518.image.f187.langEN

To derive it, you have to know how an equilibrium depends on concentration and on temperature.
At the freezing point, pure liquid and the solid are at equilibrium. If you lower to concentration of the liquid by adding solute, you disturb the equilibrium. If you lower the temperature by the right amount, you are back at equilibrium.
The relationship you get is:
$$x_{A}=\frac{\Delta H_{m}}{RT^{\text{2}}}\ \Delta T$$
This is from a General Chemistry text that also does not derive it. With some algebra, you can express $K_f$ in terms of the enthalpy of fusion comparing this and the equation given by the OP.

Can someone please rigorously prove this expression?

Without knowing what established results of thermodynamics could be used in the derivation, one would have to develop multiple chapters of a thermodynamics textbook. Calling it a proof would probably be overstating it.

The most rigorous treatment comes from Edward Guggenheim's book "Thermodynamics: An Advanced Treatment for Chemists and Physicists" (ISBN 978-0444869517).
Let's assume an equilibrium between a liquid solution and the pure solid solvent 1. Let's also assume pressure is constant. At equilibrium between liquid mixture at its freezing point and the pure solid at temperatur $T$ we assume the chemical potential $\mu$ is equal:
$\mu^{\rm mixture}_1(T) = \mu^{\rm solid}_1(T)$
Let's say $T^{\rm liquid}$ is the equilibrium temperature of the pure liquid (the freezing point of the pure liquid), so we obtain:
$\mu^{\rm liquid}_1(T^{\rm liquid}) = \mu^{\rm solid}_1(T^{\rm liquid})$
If we divide equation $(2)$ by $(1)$, we get:
$\displaystyle \frac{\mu^{\rm liquid}_1(T^{\rm liquid})}{\mu^{\rm mixture}_1(T)} = \frac{\mu^{\rm solid}_1(T^{\rm liquid})}{\mu^{\rm solid}_1(T)}$
This can be rewritten to:
$\displaystyle \frac{\mu^{\rm liquid}_1(T^{\rm liquid})}{\mu^{\rm mixture}_1(T)} = \left( \large \frac{\frac{\mu^{\rm liquid}_1(T)}{\mu^{\rm liquid}_1(T^{\rm liquid})}}{\frac{\mu^{\rm solid}_1(T)}{\mu^{\rm solid}_1(T^{\rm liquid})}} \right)$
Taking the logarithms:
$\displaystyle \ln\left(\frac{\mu^{\rm liquid}_1(T^{\rm liquid})}{\mu^{\rm mixture}_1(T)}\right) = \ln\left(\frac{\mu^{\rm liquid}_1(T)}{\mu^{\rm liquid}_1(T^{\rm liquid})}\right) - \ln\left(\frac{\mu^{\rm solid}_1(T)}{\mu^{\rm solid}_1(T^{\rm liquid})}\right)$
We now apply the following equation (which we can obtain from the Guggenheim square):
$\displaystyle \frac{\partial \mu_1}{\partial T} = \frac{- H_1}{RT^2}$
We get
$\displaystyle \ln\left(\frac{\mu^{\rm liquid}_1(T)}{\mu^{\rm liquid}_1(T^{\rm liquid})}\right) = - \int_{T^{\rm liquid}}^{T} \frac{H_1^{\rm solid}}{RT^2} dT$
and
$\displaystyle \ln\left(\frac{\mu^{\rm solid}_1(T)}{\mu^{\rm solid}_1(T^{\rm liquid})}\right) = - \int_{T^{\rm liquid}}^{T} \frac{H_1^{\rm liquid}}{RT^2} dT$
Combining those equations results in:
$\displaystyle \ln\left(\frac{\mu^{\rm liquid}_1(T^{\rm liquid})}{\mu^{\rm mixture}_1(T)}\right) = - \int_{T^{\rm liquid}}^{T} = - \int_{T^{\rm liquid}}^{T} \frac{{\Delta}_f H_1^{\rm \ominus}}{RT^2} dT$
In this equation, ${\Delta}_f H_1^{\rm \ominus}$ is the proper enthalpy of fusion of the pure solvent.
We now apply the definition of the chemical potential $\mu$:
$\mu_1 = \mu_1^{\ominus} + RT \phi {\sum}_s r_s$
Here, $\phi$ denotes the osmotic coefficient of the solution at its freezing point. We can rewrite the equation to:
$\displaystyle \phi {\sum}_s r_s = \left< {\Delta}_f H_1^{\rm \ominus} \right> \left( \frac{\frac{1}{RT}}{\frac{1}{RT^{\rm liquid}}}\right)$
In this equation, $\left< {\Delta}_f H_1^{\rm \ominus} \right>$ is the average value of ${\Delta}_f H_1^{\rm \ominus}$ over the reciprocal temperature interval $1/T^{\rm liquid}$ to $1/T$. Because ${\Delta}_f H_1^{\rm \ominus}$ is always positive, it follows $T<T^{\rm liquid}$ (The freezing point of the solution is always below that of the pure solvent if the solid phase is pure solvent).
Approximation: For dilute solutions, $\Delta T = T^{\rm liquid}-T << T^{\rm liquid}$, so we get
$\displaystyle \phi {\sum}_s r_s = \frac{{\Delta}_f H_1^{\ominus}(T^{\rm liquid}-T)}{RT^2}$
We rearrange this equation to:
$T^{\rm liquid}-T = \displaystyle \phi {\sum}_s r_s \frac{RT^2}{{\Delta}_f H_1^{\ominus}}$
For calculations, we prefer the molalities $m_s$ over solute-solvent mole ratios $r_s$:
$m_s = r_s/r^{\ominus}$
We obtain:
$T^{\rm liquid}-T = \displaystyle \phi {\sum}_s m_s \left(\frac{r^{\ominus}RT^2}{{\Delta}_f H_1^{\ominus}}\right)$
The factor $\left(\frac{r^{\ominus}RT^2}{{\Delta}_f H_1^{\ominus}}\right)$ is the cryoscopic constant of the solvent.


text:
Why does the non-volatile solute go into the liquid part of the the solution when a solution containing non-volative solute is freezed?
My teacher gave the following diagram in the class notes. He said that when a solution containing a non-volative solute (for e.g.  a solution of $\ce{NaCl}$ in water) is freezed, then there will be some depression in freezing point and some part of the solution will freeze and some part won't freeze due to the presence of the non-volatile solute. My teacher said that the non-volatile solute which is trapped in the solid solvent will gradually move into the solution which has not freezed and make it more and more concentrated until finally we get a pure solid solvent on one side and a highly concentrated solution containing the non-volatile solute on the other side. But why does the non-volatile solute go into the part of the solution which has not freezed? I mean it is trapped in the solid solvent, right? So, why does it go into the liquid part? In the diagram $\ce{NaCl}$ is moving from frozen water to liquid water and eventually all of the $\ce{NaCl}$ from the frozen water will go into the liquid part. But why does this happen? What is the reason behind this? Please explain. I am so confused. Please help (In the diagram the arrows show the movement of the $\ce{NaCl}$).

The solute stays in solution, the lattice of the solid does not accommodate it.

It is hard to follow quite what is going on. If the NaCl/water/ice is frozen it is at a lower temperature than the liquid water, so some, if not all this frozen mush will melt and the whole become a NaCl solution.

there is a phase diagram in this answer. https://chemistry.stackexchange.com/questions/116302/where-does-the-energy-come-from-to-lower-the-temperature-of-a-brine-solution/116316#116316 What happens depends where you are on this plot.

I invite you to use the first comment by Karsten Theis for an easy to grasp sense of the story.


text:
Confusion regarding reaction pathway followed in reaction with lucas reagent
Recently I came across a problem involving Lucas reagent. It asks what is the major product when 2-methylbutanol is treated by HCl in presence of anhydrous ZnCl2? The answer says major product is 2-chloro-2-methylbutane. The reason it gives is 1,2 hydride shift. But my question is whether Sn1 reaction will even occur? I knew primary alcohols always follow Sn2 mechanism.Hence no rearrangement would take place.Can someone clarify what's happening?
Thank you

Lucas reagent always gives SN1. It's designed in a way to give SN1, that's the only reason it can differentiate between primary, secondary and tertiary alcohols.

@Nisarg Bhavsar But my book says and I have verified it on internet that primary alcohols do follow Sn2 mechanism

That is overdone by the highly acidic conditions made by adding HCl. Even ZnCl2 acts as a Lewis acid and thus increases the chances of SN1 reaction. Furthermore the Lewis acid nature of ZnCl2 also decreases the nucleophilicity of chloride ions thus even curbing the SN2 pathway. Lucas reagent is designed to follow SN1 pathway and it efficiency does so.


text:
Phase of reaction between glucose and potassium cyanide to form amygdalin
I'm trying to write a detective-fiction novel, and I was wondering if potassium cyanide reacts with sugar only if the glucose for the amygdalin to form is dissolved in liquid or does it react even if the glucose is still sugar, like sugar cubes?
For example if potassium cyanide was mixed in a sugar canister. Will there be amygdalin or will there be amygdalin only if the potassium and sugar from the canister are mixed in a glass of water?
I can't find the answer in google, and I'm not that good with chemistry.

The glycoside amygdalin is a product of enzymatic biosynthesis. Its hydrolysis produces 2 molecules of glucose, benzaldehyde and hydrogen cyanide. Mixing potassium cyanide and sugar ( = sucrose = combined glucose + fructose) does not produce amygdalin. Neither you would get it, mixing glucose, benzaldehyde and hydrogen cyanide.


text:
Aren't isothiocyanate and thiocyanate ligands monodentate?
I will keep it brief; if I generalize what an assignment is telling me, then a metal complex with 2 (iso)thiocyanate ligands would be named "bis(isothiocyanate)...metal(oxidation state) or bis(thiocyanate)...metal(oxidation state). From what I can tell (1) (2), (iso)thiocyanate ligands are exclusively monodentate... so why do they get the bis-tris-tetrakis-etc. prefix? I thought that those were reserved for polydentate ligands (3) or ligands already containing one or more di-tri-tetra-etc. prefixes. Thanks!

Thiocyanate ligand is technically ambidentate ligand.

The answer to your title question is: Yes, they are monodentate.
But I think you have a misconception about when to use which type of counting: It does not depend on whether the ligand is polydentate or not.
The IUPAC guide states that you use the bis-tris-tetrakis-etc. prefix when you want to avoid confusion:

The  prefixes are ‘di’, ‘tri’, ‘tetra’,etc., for  use  withnames for  simple  entities,  or ‘bis( )’, ‘tris( )’, ‘tetrakis( )’,etc., for names for most entities  which themselves contain multiplicative  prefixes or locants.  Care must  also  be  taken  in  situations when  use  of  a simple multiplicative prefix may be misinterpreted, e.g., tris(iodide) must  be  used  for  $\ce{3I−}$ rather  than  triiodide  (which  is  used  for  $\ce{I3^−}$), and  bis(phosphate)  rather  than  diphosphate  (which  is  used  for $\ce{P2O7^{4−}}$).
Source: https://iupac.org/cms/wp-content/uploads/2018/05/Inorganic-Brief-Guide-V1-3.pdf

So while in the case of thiocyanates and isothiocyanates I don't see that there is any confusion avoided by using bis( ), it is certainly not wrong.


text:
Disproportionation reaction of dimeric sulphur monochloride
I read that group-16 monohalides exist in dimeric forms ex. $\ce{S2Cl2,Se2Cl2,Se2Br2}$ etc.
It also states that these dimeric halides undergo disproportionation reactions and gives an example as follows: $\ce{2Se2Cl2 -> SeCl4 + 3Se}$
Now I thought what would be the disproportion reaction if we consider the case of $\ce{S2Cl2}$ Since $\ce{S(VI)}$ is more stable than $\ce{S(IV)}$ would the disproportion reaction in case of $\ce{S2Cl2}$ yield $\ce{SCl4}$ or $\ce{SCl6}$?
EDIT:  As pointed out in the comments. $\ce{SCl6}$ doesnt exist hence the same question is now considering the case of decomposition of $\ce{S2F2}$ with the essence of question still remaining the same

(1) Here is a different pathway for diselenium dichloride disproportionation which yields $\ce{SeX2}$ and $\ce{Se_{n}X2}$ (2) In this patent, it is  mentioned "at higher temperatures, sulfur monochloride tends to disproportionate to form sulfur and sulfur dichloride" (3) SCl6 doesn't exist.

Didn't consider that $\ce{SCl6}$ doesn't exist but what if we talk about $\ce{S2F2}$ ?

With S2F2 you'll have a similar reaction.

@Ashish Wikipedia says that it decompose to sulfur tetrafluoride ($\ce{S^{(IV)}F4}$) and elemental sulfur when heated. Also, a 1974 paper (I don't have access) suggested that it disproportionate to sulfur difluoride ($\ce{S^{(II)}F2}$) and sulfur.

But isn't VI more stable O.S for sulphur ? And considering the fact that we say that "SF6 is exceptionally stable" shouldn't that form ? Or is it that SF6 is just relatively more stable than other hexafluorides (halides) and not more than SF4 ? Or does it not depend on stability at all ?

O.S. is a made-up concept made by humans. It does not account for stability of a particular compound. SF6 is stable not because it has sulfur in its +6 O.S. but it is due to steric reasons. Sulfur is surrounded by six fluorine atoms so it has no space for sulfur to get attacked, hence you do not see it getting hydrolyzed. This can also explain inexistence of SCl6 because you can't fit six chlorine atoms to sulfur. See Ivan's answer to get an idea.

Compiling my comments into an answer, sulfur monohalide disproportionate to sulfur dihalide and/or sulfur tetrahalide and elemental sulfur.
Wikipedia has the reaction:
$$\ce{2S2F2 ->[180 ^\circ C] SF4 + 3S}$$
Also, there is a reaction scheme from a 1974 entry1 which indicates formation of octasulfur and sulfur difluoride as sideproduct.
$$\ce{12S2F2 -> 3S8 + 8SF4}$$
$$\ce{8S2F2 -> S8 + 8SF2}$$
It also further noted that this disproportionation reaction is useful in the synthesis sulfur tetrafluoride and the decomposition of two moles of $\ce{S2F2}$ into one mole of $\ce{SF4}$ and sulfur yields $\pu{\Delta H^\circ = -24.6 kcal/mole}$.
Furthermore, sulfur difluoride may disproportionate to sulfur tetrafluoride and sulfur or some other oligomer/polymer of sulfur or back to sulfur monofluoride or isomerize to various sulfur compounds like $\ce{FSSF}$,  $\ce{SSF2}$ or $\ce{SF3SF}$. These compounds are extremely unstable and prone to hydrolysis:
$$\ce{16SF2 -> 8SF4 + 8S8}$$
$$\ce{3SF2 -> SF4 + S2F2 isomers}$$
The stability of $\ce{SF2}$ can be considered thermodynamically. The decomposition of two moles of $\ce{SF2}$  to yield $\ce{SF4}$ and $\ce{S8}$(gas) is $\pu{-75.7 kcal/mole}$. Another reason for this decomposition is polymerization which forms new sulfur-fluorine bonds. Also, it is noted that $\ce{SF4}$ is thermodynamically unstable and disproportionate to sulfur hexafluoride and sulfur:
$$\ce{24SF4 -> 16SF6 + S8~~~~\Delta H^\circ = -62.9 kcal/mole}$$
So, sulfur compounds in these intermediate oxidation state are found to be thermodynamically unstable and it is also influenced by its kinetic stability.
Sulfur monochloride disproportionation occurs in a similar way as noted in this patent:

Elevated temperature as used herein refers to temperatures between
about 75 and about 150C; and preferably between about 85 and about
138C. Higher temperatures can be used but the sulfur monochloride will
tend to disproportionate to form sulfur and sulfur dichloride. lf
higher amounts of sulfur in the sulfur monochloride product are not
objectionable, higher temperatures can be utilized

Notes and References

Advances in Inorganic Chemistry and Radiochemistry, Academic Press, 1974
There is a different pathway for disproportionation of selenium monochloride/bromide where it is seen to yield $\ce{SeX2}$ and $\ce{Se_{n}X2 (n =3,4)}$ unlike the general pathway you mentioned which yields $\ce{SeX4}$ and elemental selenium. See this paper for more details.


text:
Osmotic Pressure of Aqueous Solution
The osmotic pressure of an aqueous solution at $25$ Celsius is $92.0$ $kPa$. Calculate the boiling point
of the solution. (Assume that the density of the solution is $1.0$ $g/mL$, ignore the mass of the
solute)
Using the formula $\pi = cRT = \rho g h$, we can find the concentration value and height; however, how one should proceed to evaluate the boiling point using the ebullioscopic constant and morality of the solution? Are there any suggested ideas for solving it, considering all the givens are outlined in the statement?

Are there any ideas except for the formula $\pi = cRT$

Is there any information about the molar mass or molecular formula of the compound?

No, unfortunately. That was the part where I was stuck basically. No additional information is given.

@M.L In fact, according to the statement, it has been mentioned that the mass of the solute should be ignored. And regarding the molar mass, I guess we can find it using the $\pi = MRT$ formula. What do you think?

I edited the problem statement; would be appreciated if anyone could suggest further ideas!

Some solutions are very amoral.  ( morality of the solution )

One of the first ideas coming to mind is searching. As it is very simple topic described on internet or in textbooks many times on many places.

@Poutnik I had already searched before posting this question here. So, be sure that such problems are not anywhere!

Not sure what the problem is--you can calculate the solute concentration from the osmotic pressure. Then you can calculate the boiling point elevation from the concentration. You don't need the molar mass, since osmotic pressure and boiling point elevation both (to first order) depend only on conentration.

@Snowflake Relations osmotic pressure->concentration and concentration+density+E.c->boiling point shift ARE nearly anywhere. The given density 1 g/mL and omitting solute mass is great help. You can consider in this special case numerical equality of molarity and molality.


text:
NMR spectrum chemical shift confusion
Recently started on NMR spectrum, and I'm slightly confused about chemical shifts and their connection to the resonance frequency and magnetic field.
From what I know, the "new" way of doing NMR means keeping the radio-frequency constant and increasing the external magnetic field in small increments.
Now, I believe a greater electron density around the protons (more shielded) will protect the positive nucleus from the external magnetic field and thus, the positive nucleus experiences a lesser net magnetic field. Hence, there is a lower resonance frequency which should result in a smaller chemical shift (the magnetic field will determine the energy differences between the spins).
From what I understand, because of experiencing a lesser net magnetic field, we increase the external magnetic field so that we can get a reading (this apparently gives rise to the idea of high field on the more shielded side). However by increasing the external magnetic field, won't we increase the energy differences between the spins and cause the resonance frequency to be higher which conflicts with the idea that we have a lower resonance frequency?

It looks like you are describing continuous wave (CW) and not pulsed NMR. The latter is generally considered the "new" way of doing NMR.

It looks like you are describing continuous wave (CW) and not pulsed NMR. The latter is generally considered a more modern way of doing NMR as it was developed after the CW technique. The pulsed technique was also accompanied by the development of FT processing which resulted in a Nobel price being awarded to Richard Ernst.
In CW NMR you usually "sweep" the field - increase the external field $B_0$ in small increments, as you explain - while keeping constant the frequency generated by a "read" RF coil. Coincidence between the read frequency and the Larmor frequency $\omega = \gamma \left(1-\sigma \right)B_0$ - the resonance condition - results in a signal registered with a pickup coil. Here $\sigma$ is the chemical shielding which reduces the effective magnitude of the applied field $B_0$  at the nucleus.

by increasing the external magnetic field, won't we increase the energy differences between the spins and cause the resonance frequency to be higher which conflicts with the idea that we have a lower resonance frequency?

This is exactly the point of increasing the field. The Larmor frequency is related to the splitting in the energy levels, and you need a greater field to bring about a splitting in a more shielded nucleus, so that the resonance condition is satisfied.
The resonance frequency in such a CW experiment is always $\omega = -\gamma B_1$, but as the field at the shielded nucleus is $\left(1-\sigma \right)B_0$ and the shielding opposes the applied field, it is necessary to apply a larger field ("go upfield") to excite a more shielded nucleus.

Normally the external magnetic field is held constant and an r.f. pulse is applied to generate the nmr signal. The electrons in the molecule allow the external field and the magnetic dipole due to nuclear spin to interact with one another. This interaction causes a local magnetic field that can add to or oppose the external field and so a particular nucleus has a resonant frequency that is slightly different to one in a different environment, for example methyl protons vs aromatic ones. This difference in frequency with respect to a standard compound (such as tetramethysilane TMS) is the chemical shift. The shift is often measured in parts per million change in frequency and a positive ppm corresponds to de-shielding and in this case the nucleus experiences a larger magnetic field than the reference TMS which is taken to be 0 ppm. In a de-shielded nucleus the local magnetic field due to local motion of the electrons adds to the external field, and opposes this when shielded causing a negative chemical shift.
(The chemical shift $\displaystyle \delta_{ppm}=\frac{v-v_{ref}}{v_{ref}}\cdot 10^6$. The $v$ are frequencies, e.g. $v_{ref}=60\cdot 10^6$ Hz. The reason for using a ratio is so that the chemical shift from nmr machines of different frequency can be compared, 100 vs 300 MHz for example).
The r.f. pulse contains many frequencies so can be used measure lots of different nmr transition frequencies.)


text:
Determine vmax and enzyme concentration (Michaelis–Menten)
I am given that the enzyme concentration is $\pu{15 nM}$ and the following data:
$$
\begin{array}{rl}
\hline
S/\pu{mM} & V/\pu{mM s^{-1}} \\
\hline
1 & 0.202 \\
2 & 0.368 \\
5 & 0.729 \\
10 & 1.08 \\
20 & 1.43 \\
50 & 1.77 \\
100 & 1.92 \\
\hline
\end{array}
$$
And calculated that the Michaelis–Menten constant was $K_\mathrm{M} = 9.41$ (unsure about the units).
I then want to calculate $v_\mathrm{max}$ and $[\ce{E}]_0$ for the same reaction given that the reaction rate is $\pu{0.101 mM s-1}$ at a substrate concentration of $\pu{24 mM}.$
I know that once I have $v_\mathrm{max}$ I could just use $v_\mathrm{max} = k_\mathrm{b}[\ce{E}]_0$ to calculate the enzyme concentration, but I am unsure as to how I can calculate $v_\mathrm{max}$ from one set of data. Could someone give me a hint of how to begin this problem?

In your text book you should have that $K_m=(k_{-1}+k_2)/k_1$ so this gives you the units. You could try a double reciprocal (Lineweaver-Burk) plot $1/[S],;vs;1/V$ assuming $V$ is the initial rate.

Michaelis-Menten kinetics is given by the equation:
$$V = V_\mathrm{max}\frac{S}{K_M+S} \tag1$$
Where $V_\mathrm{max} = k_\mathrm{cat}\cdot [\ce{E_T}]$ and $K_M = \dfrac{k_\mathrm{on} +  k_\mathrm{cat}}{k_\mathrm{off}}$ (using the conventional description of Briggs and Haldane's derivation of the Michaelis-Menten equation; Ref.1):
$$\ce{E + S <=>[$k_\mathrm{on}$][$k_\mathrm{off}$] [ES] ->[$k_\mathrm{cat}$] E + P} \tag2$$
The derivation of the equation has taken a few assumption. Accordingly, as a rule of thumb, the enzyme kinetics follow Michaelis-Menten equation if $K_M \gt 5 \times [\ce{E_{T}}]$. Since calculated $K_M$ is $\pu{0.939 mM}$ (see Vinícius Godim's answer elsewhere) and $[\ce{E_{T}}]$ of the first kinetics is $\pu{15.0 nM}$, this requirement is satisfied.
However, OP's confusion is how to calculate the $[\ce{E_{T}}]$ of the second kinetics using the same enzyme and substrate in different set.  OP didn't aware about the following fact of the enzyme kinetics: If you have used same enzyme and substrate under similar conditions (temperature, etc.) the second time, the constants $K_M, k_\mathrm{on}, k_\mathrm{off}$, and $k_\mathrm{cat}$ you found in the first time are going to be the same.
Thus, for second kinetics, $K_M = \pu{0.939 mM}$ and $k_\mathrm{cat} = \dfrac{V_\mathrm{max-1}}{[\ce{E_{T1}}]} = \dfrac{\pu{2.10 mM s-1}}{\pu{15.0 \times 10^{-6} mM}} = \pu{1.40 \times 10^{5} s-1}$.
Therefore, since it has been calculated that $V_\mathrm{max-2} = \pu{0.141 mM s-1}$ (see Vinícius Godim's answer elsewhere), $[\ce{E_{T2}}]$ can be calculated as follows:
$$V_\mathrm{max-2} = k_\mathrm{cat}[\ce{E_{T2}}] \ \Rightarrow \ [\ce{E_{T2}}] = \frac{V_\mathrm{max-2}}{k_\mathrm{cat}} = \frac{\pu{0.141 mM s-1}}{\pu{1.40 \times 10^{5} s-1}} = \pu{1.01 \times 10^{-6} mM}\\ = \pu{1.01 nM}$$

Late edit:
I think it is beneficial to show how to find $K_M$ and $V_\mathrm{max}$ from the given set of kinetic data. The method is called Lineweaver–Burk plot, which is a plot of $\dfrac{1}{[\ce{S}]}$ versus $\dfrac{1}{V}$:

Since the equation of the plot is:
$$\frac{1}{[V]} = \frac{K_M}{V_\mathrm{max}}\cdot \frac{1}{[\ce{S}]} + \frac{1}{V_\mathrm{max}}$$
which is a straight-line equation ($y = mx + c = 4.4754x + 0.4767$). Thus, positive intercept is equal to $\dfrac{1}{V_\mathrm{max}}$, which is $0.4767$, and slope is equal to $\dfrac{K_M}{V_\mathrm{max}}$, which is $4.4754$. When resolve you would find, $K_M = \pu{9.39 mM}$ and $V_\mathrm{max} = \pu{2.10 mM s-1}$.

Reference:

George Edward Briggs and John Burdon Sanderson Haldane, "A Note on the Kinetics of Enzyme Action," Biochem. J. 1925, 19(2), 338-339 (DOI: 10.1042/bj0190338).

Michaelis-Menten kinetics is given by
$$V = V_{\max} \frac{S}{K_M + S}$$
Taking the inverse of both sides to linearise the equation,
$$\frac{1}{V} = \frac{1}{V_\max} +\frac{K_M}{V_{\max} S}$$
Taking $y = 1/V$ and $x = 1/S$ in $y = A + Bx$ leads by linear regression to $A = 0.4767$, $B = 4.4754$ (not including units). Therefore,
$$V_\max = \frac{1}{A} = \pu{2.10 mMs^{-1}}$$
$$K_M = B V_\max = \pu{9.39 mM}$$
For the new situation with different $V$ and $S$ and necessarily different $[E_0]$ , $V_{\max}$ would change because $V_{\max} = k_b [E_0]$ but $K_M$ remains constant, thus
$$V_{\max} = V \frac{K_M + S}{S} = \pu{0.101mM s^{-1}} \frac{(\pu{9.39 mM} + \pu{24 mM})}{\pu{24 mM}} = \pu{0.141 mM s^{-1}}$$
You can then calculate $k_b$ by $k_b = V_{\max}/E_0$.

Thank you, I didn't realize that K$_M$ would remain constant which really made it clearer! But I want to calculate [E]$_0$ as well, how can I do that if I don't have k$_b$?

You can calculate $k_b$ for the first set of data by $k_b = V_{max}/[E_0]$ since $[E_0]$ is given (initial enzyme concentration). For the second case, calculate $[E_0]$ by $V_{\max}/k_b$ since $k_b$ is constant.


text:
Debye–Waller factors for tin(IV) oxide
How I could find the bulk Debye–Waller factor for tetragonal rutile type $\ce{SnO2}$ at room temperature?
I have been doing some online digging with no luck. If anyone has a reference I could use that would be awesome.

I'm sure that typical values are known for Sn and O atoms in lots of structures. Looking online is v. v.  limited, you need to properly search the literature :)


text:
pH sensor showing incorrect measurement after using on acidic solution
I recently acquired an electrode pH sensor for a project of mine. After calibrating the sensor I used it on tap water and then lemon water to test their respective pH. After finding the pH of the lemon water, I tried the sensor on distilled water to confirm my calibration once...but the value was coming out to be greater than 7. I tried dipping the sensor in it's provided storage solution (3N KCl) but still there was only a minimal reduction in the error. Can anyone with the knowledge of pH sensors explain the problem I am facing and any way of overcoming this problem so that I can test multiple solutions together without affecting the accuracy.

Why do you expect distilled water to have pH 7 ? Think twice..... BTW, I assume you performed at least 2 point calibration bracketing the measured pH range.

At room temperature shouldn't distilled water have a pH of 7 or at least close to 7. In my case the pH value was coming out to be 7.87. That's why I thought that something was wrong. Secondly, I calibrated the pH sensor by following a tutorial provided with it. This was done by shorting it and then adjusting the offset potentiometer until the output value read was 2.5V.

The point is, distilled water is never just distilled water. And even if it was, it would not be, when you put in the pH probe. True calibration must be done against standard pH solutions, otherwise you do not know if it provides the correct output.

Another point is that distilled water can get a little acidic due to CO2 absorption. Hard to get high pH unless vials aren't clean. I don't have an easy explanation, too. Follow the instructions carefully and rinse the bulb immediately after use.

The glass electrode used to measure the pH of a solution is aa long tube terminated by a small glass bulb which is the most important (and the only sensitive) part of the electrode. When dipped into an aqueous solution, this bulb gets a potential which is transformed into a pH value by the pHmeter. But in order to work properly, the glass bulb must be covered by a thin layer of a sensitive gel that should not be touched, rubbed or removed. This thin outer layer adsorbs some ions from the solution, provided the solution is not too concentrated. This absorption is reversible. But, if the glass electrode is dipped into concentrated solutions of acids or of bases, the layer is saturated and may be even destroyed, and will not be able to work properly in the future.
If this accident has happened, the electrode is not able to work properly. The gel must be renewed and synthesized again. The only way to do it is to dip the electrode in a 3 molar KCl solution, and wait a couple of days, depending on the concentration of the acid or base. After this operation, the gel layer will be repaired. It must be checked by dipping the electrode in a known pH buffer.

Thanks a lot now I understood the problem. But just one small thing...In the answer you have mentioned that the gel layer may get destroyed with concentrated solutions of acids or bases. But in my case I was using just a solution of lemon water that was mildly acidic only.

@Aditi Mittal. You are right. Your trouble is not due to an effect of high acidic concentration. I must admit that I don't see any simple explanation.


text:
Number of possible microscopic states and their probabilities
I'm having trouble with the following problem:

Boltzmann defined entropy as a measure of the number of possible microscopic states of a system, $W$,
as:
${S = k_BlnW}$
Here, $k_B$ is the Boltzmann constant. Hereafter, it is assumed that neither energy nor particles are
exchanged between the system of interest and the surrounding environment, and that the system is in
equilibrium at a constant temperature. In addition, the gas atoms and molecules (henceforth referred
to as ‘molecules’) are assumed to behave as an ideal gas.

Figure 1. Example of a microscopic state (4,6) in which ten molecules are distributed in two chambers separated by a wall.
Removing the boundary wall that separates the two chambers in Figure 1 allows the molecules to move
throughout the entire container. Assume that it is possible to measure the arrangement of the molecules
instantly after the removal of the wall. If this measurement is performed a sufficiently high number of
times, the probability that a state with an arrangement (n,m) will be found can be calculated.
1) Calculate the arrangement ($n^*$, $N$ − $n^*$) with the highest probability of occurrence for $N = 10$ and $N = 100$, as well as the corresponding probability of occurrence for each state.
2) Calculate the probability of observing a state in which $n^*$ is in the range $n^* – 0.02 N ≤ n^* ≤ n^* + 0.02 N$ for
$N = 10$ and $N = 100$. Here both $N$ and $n^*$ are integers.

I was able to calculate the first part and got the following results using a spreadsheet.
For $N = 10$:

$W(5,5) = 252$ and $P(5,5) = 0.226$
For $N = 100$:

$\ce{W(50,50) = 1.0*10^{29}}$ and $P(50,50) = 0.0788$
On the other hand, I don't understand what the mentioned range in the second part of the problem means and I'm therefore unable to solve it:

$n^*$ is in the range $n^* – 0.02 N ≤ n^* ≤ n^* + 0.02 N$

Some notes I had easily at hand may help. You have already calculated the probability of obtaining the chance with a number $k$ of type of ball (or molecule) out of a total on $n$ is
$$\displaystyle p=\frac{n!}{k!(n-k)!}\frac{1}{2^n} \tag{25c}$$
This distribution is a maximum when $k=n/2$. This can be seen with a straightforward argument. The factorial terms are symmetric, $k!(n-k)!=(n-k)!k!$ and always positive.  When $k=0$ or $k=n$ the probability is very small tending to zero; i.e $1/2^n$ is small so there must be a maximum somewhere in the range $0\to n$. The symmetric nature ensures that this will be at $k = n/2$. This can also be determined by differentiation. It is not possible to differentiate a factorial, as $n$ is discrete, but replacing $n!$ with the Sterling approximation $(n!=n^ne^n)$ and setting the derivative in $n$ to zero allows the maximum to be found.
The second feature of this probability distribution is that it becomes extremely narrow as $n$ increases. This is shown in the figure which shows the normalised probability $p/p_{max}$ vs $k/n$ which is a fraction between $0$ to $1$. At large $n$ the distribution becomes so narrow, or peaked at $n/2$, that almost all the probability is described by that at $n/2$, i.e. this value is so great compared to others that $p_{max}$ is in effect greater than all others combined.

The normalised probability function  $\displaystyle p=\frac{n!}{k!(n-k)!}\frac{1}{2^n}$ vs $k/n$ for different $n$. The the data only exists for integer values as shown as circles for $n=20$, at larger $n$ a line is drawn as the points become too congested. The curve for $n=10000$ is in red. At large $n$ the distribution becomes very narrow and the maximum value at $n/2$ is greater than all others combined and the distribution can be approximated by its maximum value.
You next need to choose $k/n$ corresponding to 2% either side of $k/n=0.5$ and calculate $p/p_{max}$.

This problem is the 5th Problem taken from the list of the 30 Preparatory Problems published by the Japanese organizers of the 53th International Chemistry Olympiad in July 2021. It is used for training the candidates of this Olympiad. As these 30 problems may be used in different countries for selecting their national candidates, I don't think it is advisable to give an answer or a solution to this problem 5.

This answer was part of a document I wrote some years ago and has nothing to do with any competition.  Besides I firmly believe that on this site we should not censor ourselves because someone might or might not be in a competition.

@Maurice that seems a problem for the organisers rather than SE. One can argue that every question here is written by someone sitting in a class or at home but during assignments. Writing such a neat question won't be feasible during a competition, anyway.


text:
Can't aluminium chloride do electrophilic substitution as an electrophile?
In halogenation reactions of benzene, $\ce{AlCl3}$ is used as a catalyst in order to form electrophile $\ce{E^+}$.
So I was wondering that $\ce{AlCl3}$ is a Lewis acid (that fact being also used in mechanism) and so an electrophile. Why it can't just do electrophilic substitution reaction on the benzene itself? Isn't it strong enough or any other problem there doing that?

The bonds in $\ce{AlCl3}$ are not ionic. Solid $\ce{AlCl3}$ is made of three covalent bonds $\ce{Al-Cl}$ around the aluminum atom. This does not respect the octet rule. This requirement may be obtained by making a dimer $\ce{Al2Cl6}$. Anyway, all $\ce{AlCl3}$ wants to do is getting and attaching a chloride $\ce{Cl-}$ ion by a covalent bond. The result would be a $\ce{AlCl4^-}$ ion, where $\ce{Al}$ is surrounded by 4 doublets, and so it respects the octet rule. Why and how should $\ce{AlCl3}$  attack the benzene ring ?

@Maurice OP's point is entirely valid. AlCl3 should be in principle able to bind to rings, either via sigma or pi-complex. Probably it does to some extent, but  1) dimerisation you mentioned can make it harder 2) AlCl3 is a hard acid and arenes are soft 3) the binding doesn't make products, perhaps because resulting Al-C bond is too fragile? I'm not sure about details, but it's rather interesting.

The reaction wouldn't happen because you forget that $\ce{Cl2}$ is also present in the solution (because $\ce{AlCl3}$ is a strong lewis acid). An acid-base reaction is way faster than an aromatic electrophilic substitution.
Once the acid base has happened I doubt that $\ce{AlCl4-}$ would still behave as an electrophile.
Footnote: $\ce{AlCl3}$ is a strong lewis acid, this is why we don't conduct friedel-craft reaction in water which is why we always see anhydrous $\ce{AlCl3}$ mentioned in this reaction. Most commonly preferred media is $\ce{CS2}$ .

+1 , Thanks a lot  ; but I didn't get this "So even in case of absence of Cl2 this won't work" , Can you kindly explain this case briefly ?

Would also like to mention that " Anhydrous" AlCl3  is generally mentioned with the reaction.

@YashAgrawal yes you are right, I had overseen that we can't use water in friedel-craft reaction. I have made an edit in response to that

The OP asks why doesn't AlCl$_3$ react with benzene - without Cl$_2$!
So, what does aluminum chloride do in benzene? It is slightly soluble in benzene, forming an oily compound (Ref 1, 2).  So if you mix AlCl$_3$ with benzene, the "reaction" just goes to a complex and awaits further reactant. Perhaps the Cl atom on an alkyl chloride or on a Cl$_2$ molecule is more electron-donating than the benzene ring and favors formation of an alkyl cation in some form.
C$_6$H$_6$ + AlCl$_3$ + RCl (or Cl-Cl) --> C$_6$H$_6$ + RAlCl$_4$ (or Cl$^+$ + AlCl$_4$$^-$)
A quick way to see that the reaction of benzene with AlCl$_3$ has nothing going for it, is to imagine that AlCl$_3$ (or Al$_2$Cl$_6$) does react with C$_6$H$_6$. I could postulate two possibilities:

C$_6$H$_6$ + AlCl$_3$ --> C$_6$H$_5$AlCl$_2$ + HCl or

C$_6$H$_6$ + AlCl$_3$ --> C$_6$H$_5$AlCl$_2$H$^+$ + Cl$^-$

When you look at the first equation, you see immediately that HCl is going to attack the phenyl-aluminum bond to regenerate the original reactants.
When you look at the second equation, you see that the complex cation is so acidic that it will either rearrange to form the original reactants, or will protonate some other phenyl group. The possibility that the chloride anion could lie above the phenyl-aluminum bond and preferentially attack the carbon to yield chlorobenzene is just preposterous.
Ref 1. https://pubs.rsc.org/-/content/articlelanding/1951/tf/tf9514701287/unauth#!divAbstract
Ref 2. https://en.wikipedia.org/wiki/Aluminium_chloride

Thanks! , I think I got why AlCl3 wouldn't react with benzene alone as two possibilities you mentioned are pretty thermodynamically unfavourable.


text:
High symmetry points and x-coordinates
Is it possible to work out the x-coordinates related to high symmetry points?
The software I'm using doesn't provide me with that, so I was wondering if there is a way to manually figure it out, as I would like to add high symmetry point labels to an electronic band structure plot.
I found a similar question: https://lists.quantum-espresso.org/pipermail/users/2016-June/035548.html but I don't quite understand whats happening in the equation, if someone could please explain it that would be appreciated.
Thanks

It would help is you gave an example, a molecule or crystal ? Generally I would say 'yes' it is possible.


text:
Relationship between Enthalpy and Gibbs energy
I am a little stuck understanding the difference between Gibbs free energy and enthalpy. Both are measured in Joules and hence some sort of energy.
In a chemical reaction where C --> A + B
If A and B are more stable, this means energy will be released. This ΔPE is enthalpy(ΔH), correct? Enthalpy can be calculated from bond strengths after all.
In that case what does a reaction coordinate with Gibbs free energy tell us? ΔG is the change in free energy, but what is "free energy"?. Clearly that has to be different than the potential energy of the bonds since that is enthalpy (ΔH).
I appreciate any help!

Please take a look at some existing questions and answers, for instance: https://chemistry.stackexchange.com/questions/112815/change-in-gibbs-free-energy-vs-change-in-enthalpy?rq=1 Similar questions have been asked here many times before. If you still find these don't help post a new question or request that this question be reopened.

Also, in case you haven't please take the tour: https://chemistry.stackexchange.com/tour

See also: https://chemistry.stackexchange.com/questions/10683/what-is-the-difference-between-enthalpy-of-formation-and-gibbs-free-energy-of-fo?rq=1

See also: https://chemistry.stackexchange.com/questions/151090/entropy-changes-in-electrolytic-galvanic-cell


text:
Cahn-Ingold Prelog rules concerning tie breaking on second level

comparing the cahn-ingold prelog priority of the above groups, I should first compare the atoms attached to the first carbon.
So we get:
(A): {O,C,H}
(B): {O,C,H}
(C): {O,C,H}
(D): {O,C,H}
therefore we have to consider the second level, and this is where I get confused:
Do I compare it this way:
(A):{C,H,H,H}
(B):{O,H,H,H}
(C):{F,C,H,H}
(D):{C,H,H,H}
I did this by rearranging all the "second level atoms" in descending order of atomic numbers.
since A and D are still in a tie, I rearrange all the "third level atoms" in descending order of atomic numbers:
(A):{H,H,H}
(D):{F,H,H}
so the priority is C>B>D>A
Is that correct?

In your examples you have made the decision because you have{CHO}. Three different groups.. If you had{CCO) for example, you would have to go to a second level to break the C C tie. That is if I understand your question correctly. Take a look here: ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/…

look, all i can say is that $ D > C > A $

why ?

Lets compare A and C
$ OCH$3 and $ H $ are common to both. one has CH3 and the other has CH2F. of course, C > A.

compare A and D in the same way and find that D > A

compare D and C now.
there is clear proof that D > C because it has F in two places while C has H in those two exact positions.
thus it is seen that D > C > A

now see B and C
in B : after O we have CH3 and after C we have { H , H , F }
in C : after O we have H and after C we have { H , H , O }

looking at second layer after oxygens : B > C
looking at second layer after carbons : C > B.
thus even i am confused when it comes to B
but definitely D > C > A

You have avoid B on your final grading. What about B?

i said i dont know about B. this is all i can say and it is pretty fair that i at least provided a solution till here


text:
What is the difference between ΔH = ΔU +Δ(PV) and ΔU = q + w?
Here, ΔH = Enthalpy change, ΔU = Change in internal energy, P = pressure and V = volume.
I know that ΔH(Enthalpy) is heat given/taken to/from system at constant pressure. But I have been practicing questions recently on this topic and came to this confusion.
Let me first put out the question I solved:
"One mole of an ideal gas (Cv = 3/2*R) is heated at constant pressure reversibly at 1 atmosphere from 25°C to 100°C. Calculate ΔU and ΔH, Take R =2 cal/K.mol ".
I proceeded to solve the question,
Since
ΔH = nCpΔT = 1 * 5/2 * R * 75 = 375 cal.(Correct answer) and
ΔU = nCvΔT = 1 * 3/2 * R * 75 = 225 cal.(Correct answer)
Now I found out the work done in the process = -PΔV = -nR(T2 - T1) = -150 cal.
Calculation for work done is as follows:
W = -Pext(V2-V1) = -P[(nRT2/P)-(nRT1/P)] = -nR(T2-T1) = -12(373-298) = -150 cal.
But if I try to find out the ΔU using the equation  ΔH = ΔU +Δ(PV),
ΔU = 375 - (-150) = 525 cal.(Wrong)
But if I use first law of thermodynamics, ΔU = q + w,
ΔU = 375 + (-150) = 225 cal, which is the correct answer.
So, my question is why should I use first law of thermodynamics here even though the process is taking place at constant pressure??

PDelta V=+150, not -150

You have made a mistake in sign of work as $P\Delta V=150$. Also, first law of thermodynamics is valid everywhere, not just at constant pressure.

@LightYagami, Can you please refer the post again, I have added the calculation for work done, to my knowledge I have learnt the formula for work done as -PextΔV. Do correct me if I am wrong.

@Srini First you have written $-P\Delta V=-150$ which is fine. But when you did the calculation for $\Delta U$, you are subtracting -150, where 150 had to be subtracted.

@LightYagami, Oh got it thank you so much!!

You can find useful  Notation basics and How can I format math/chemistry expressions on Chemistry SE.
See also upright vs italic and Math SE MathJax tutorial.

The first law of thermodynamics refers to all thermodynamic processes performed on closed systems.  It tells us that for all possible process paths between two thermodynamic equilibrium states, although the heat and work done on the system may vary with respect to these process paths, their sum is always the same, and equal to a thermodynamic state function called the internal energy.  The relation $\Delta H=\Delta U+\Delta (PV)$ defines a secondary thermodynamic function known as enthalpy, which is a physical property of the system, also independent of process path.

Your trouble comes from the two opposed concepts used in the world to define the work  $\pu{p\Delta V}$.
For theoretical scientists, the work is positive when work is done on the system, when the gas is compressed by an external force. And in such a compression, the volume decreases, so $\pu{\Delta V}$ is negative. So in order for the work to be positive, the work must be : $\pu{w = - p\Delta V}$. So that the total change of internal energy is $\pu{\Delta U = q + w}$.
For practical scientists, and specially for engineers, the work is considered positive when the gaz expands, because the gaseous system is working like a machine that must produce work when heat is given to it. In this case, the work is : $\pu{w = + p\Delta V}$. As a consequence, the internal energy is what remains in the system when heat has been given to it and some work has been produced. So here : $\pu{\Delta U = q - w}$.
Finally both concepts leads to the same final result : $\pu{\Delta U = q - p\Delta V}$

Thank you so much, got a clearer vision about this .


text:
Intramolecular Cannizzaro reaction product
I came across the following question :

My reasoning was as follows :

It can't be option (c) because there is no chance of losing a carbon during Cannizzaro's reaction
It can't be option (b) because basically option (c) gives (b) finally
It can't be option (a) or option (d) because we can't have all carb acid groups or all alcohol groups. we need a combination of both.

So, are the options wrong?

Your reasoning looks correct

I too think there is some mistake in the options, none of them seems correct. What is the given answer?

given answer includes 2 cooh groups and 2 ch2oh groups alternately. but that isnt in the options. the key says (b).

Your reasoning is correct and the problem is with the publisher as they made  a mistake while compiling these previous year questions.
I found out the original 2003 question paper(pdf). The question you mentioned had the following options:

It is clear form the options that the answer is C(which is the product you expected) which can also be confirmed by the key attached at the end of the pdf.

Yes, the reasons you have given for eliminating options seem correct...I think the answer should be the two Ph groups both with COOH and CH2OH (since it has to have carboxylic acid and alcohol groups in the product) so the structure will be similar to option (C) but with CH2OH in place of the OH groups.

I have a little confusion as to how (c) and (b) are the same thing though, maybe check your calculations once again, I solved twice but messed up somewhere and got varying answers

option (c) will convert spontaneously to (c). look at the stability of the products. there is lots of conjugation and it is cyclic and completed. moreover, alcohol and carboxylic acids undergo intramolecular condensation to esters very fast


text:
Why are the two bonds in sulphur dioxide identical?
Triatomic molecules are either linear or bent. If we analyse the $\ce{SO2}$ molecule,  it turns out that it is bent.
I know that $\pi$ bonds do not alter shape, but merely the bond lengths.
Now since one of the bonds between S and O is a p$\pi$-p$\pi$ bond and the other bond is a d$\pi$-p$\pi$ bond, the two $\pi$ bonds are not same, so the bond length should not be the same, but my book says the bond lengths are the same which is quite puzzling. I thought maybe its due to resonance but there's no way resonance takes place here.  I read this question,but it don't answer many questions as:
Why d-orbitals of sulphur cannot be used  ?
And  in answer by @Kenny Lau ,
what's specify reason for 'Sulfur sharing two electrons among itself and the other two oxygen atoms.'
So, that makes me think then, what could  be the reason for same bond length in $\ce{SO2}$?
It is not clear as to why d-orbitals of sulphur cannot be used and in the answer to this question @Kenny Lau does not specify reason for 'Sulfur sharing two electrons among itself and the other two oxygen atoms.'

Sulfur does not use it's d-orbitals for bonding in most cases.

Please see: https://chemistry.stackexchange.com/questions/51168/hybridization-of-sulfur-in-sulfur-dioxide

Sorry, but it does not answer my question. It is not clear as to why d-orbitals of sulphur cannot be used and in the answer to this question @Kenny Lau does not specify reason for 'Sulfur sharing  two electrons among itself and the other two oxygen atoms.'

If you're using a model with resonance, why is there no resonance here?

@Zhe : okay right I am convinced by resonance after reading Maurice's answer, so by resonance that answers my previous question, but now in process of finding hybridisation I hit two more questions.

There are two ways to look at the $\ce{SO2}$ molecule. The valence bond way is this:

$\ce{SO2}$ is a bent molecule with C2v symmetry point
group. A valence bond theory approach considering just s and p
orbitals describes the bonding in terms of resonance between two
resonance structures:

The sulfur–oxygen bond has a bond order of 1.5 and does not invoke d
orbital participation. In terms of electron-counting formalism, the
sulfur atom has an oxidation state of +4 and a formal charge of +1.
(Wikipedia)

However, there is a molecular orbital approach which seems to give a deeper appreciation of the bonding. The picture is this:

(Picture source with detail explanation)
This is my simple explanation:

The sulfur is hybridized sp$^2$-p. The bond angle is 119$^o$ - you
can hardly get closer to 120$^o$!

We will assume that the oxygens are also hybridized sp$^2$-p.

Then the sigma bonding is designated in the diagram by the numbers 1 and 2 for the sigma bonds between sulfur and oxygen. Lone pairs are designated by numbers 3 (for S), and 4, 5, 6 and 7 (for O).
The pi system has three molecular orbitals, one from each atom and four electrons: one from each oxygen, two from sulfur. The lobes in these orbitals are drawn at the lower right side of the picture. Orbital 8 has the lowest energy, with all lobes bonding. The non-bonding pi orbital (#8) has reversed signs for the lobes of oxygens and therefore a node at sulfur. The antibonding pi orbital (#10) has two nodes and is unoccupied.
So the final count is two filled sigma bonds and one filled pi bond; five lone pairs and one filled non-bonding pi orbital. The formal bond order is 1.5, same as in the valence bond explanation with resonance.
The only thing I might change in the diagram is the relative energies of the sulfur orbitals and the oxygen orbitals: I would think they might be closed to the same energy. The diagram might be drawn as it is in order to make it clearer and give more room for all the numbers.
We just don't need d orbitals to explain anything, but their involvement cannot be entirely eliminated.

@Buck Thorn: Thank you for the "edit":) I saw the POST box (again) and thought I had not clicked on it or my mouse was sick.


text:
What is the reason for decomposition of layered cathode materials?
Zhang [1] and related publications claim oxygen can evolve from layered metal oxide cathode materials when charging the Li-ion battery.
Is it the high positive potential given to the cathode while charging pulling the electrons from the chemical bonds of the cathode materials and thus cause the decomposition to evolve oxygen?
If not, what is the cause of the evolution of oxygen?
Reference

Zhang, S. S. Insight into the Gassing Problem of Li-Ion Battery. Front. Energy Res. 2014, 2. DOI: 10.3389/fenrg.2014.00059.

As the particular portion of your linked document says, it is like so:
$$\ce{Li_xMO2 − 4δ e− -> Li_{x−4δ}MO_{2−2δ} + 4δ Li+ +δ O2}$$

A cathode is the electrode where a reduction happens, whatever its working mode (cell or electrolysis). So when recharging a Li-ion battery, the $\ce{Li+}$ ion are going to the cathode to be reduced: reduction happens on the Li-graphite electrode (cathode). And no oxygen can be formed on this electrode. Oxygen may be produced by electrolysis at the other electrode made of cobalt and manganese oxides. These oxides are oxidized to a higher number of oxidation. But of course, these oxides are not very stable, they may lose part of their oxygen as $\ce{O2}$.
To summarize, the lithium electrode is always the negative pole, whatever its working mode. The lithium atom works as an anode when the cell is working (producing $\ce{Li+}$ ions and electrons), and as a cathode when the cell is being recharged. The other electrode is always the positive pole, whatever its working mode. Its metallic oxides are reduced when the cell is producing current (cathode). When being recharged, this electrode is an anode, as its oxides are oxidized to a higher oxidation number.

@Maurice You are generally right. But  to avoid confusion of terms in context of rechargable cells, assignment of the cathode/anode is conventionally decided to be taken always as if in the galvanic mode, not in the electrolytic mode. Therefore, the assignment is clear all the time.

@Poutnik. Are you sure of this interpretation ? I have never heard it before ! Can you justify it ?

@Maurice Is not it justified by the description ? // It is like if you talk the left and the right bank of the river. One approach is relative, which is on you left or right hand. The other approach is absolute, in geographical sense, along the stream. // Notice yourself whenever there is talked Li-Ion cell, when an anode is talked, it is always the graphite intercalate, and the cathode is always the metal oxide. as description of the cell  structure takes precedence over momentary functionality. // I would agree it is rather practical engineering approach than approach of phys. chem.

@Maurice It would make little sense, talking like "The electrode which is a cathode at charging or an anode at discharging" // Like "The river bank which is the left bank looking downstream or the right bank , looking upstream". It is just the left bank.


text:
How to find the reaction rate in an enzyme reaction?
First time posting questions here, so let me know if I need to edit anything.
I need to decide $dC/dt$ and $dP/dt$ when the reaction is like this, where I only know we must have 2 substrates $S$ to react with the enzyme $E$ at the same time:
$$2S + E \overset{k_a}{\longrightarrow} C \overset{k_c}{\longrightarrow} 2P + E, \quad C \overset{k_b}{\longrightarrow} 2S + E$$
Is $dC/dt = 2k_a SE - (k_b + k_c)C$? Is $dP/dt = 2k_cC$? Thanks in advance.
Edit: alright, so $dC/dt$ should be $k_a S^2 E - (k_b + k_c)C$, right? Now suppose I take $E_0 = E + C$ to be the total enzyme concentration, and transform $dC/dt$ as $$dC/dt = k_a E_0 S^2 - (k_a S^2 + k_b + k_c) C.$$ By quasi-equilibrium approximation, let $$dC/dt = 0 \Rightarrow C = (k_a E_0 S^2)/(k_a S^2 + k_b + k_c).$$ Then in this equilibrium, $$dP/dt = 2k_bC = (2k_a k_b E_0 S^2)/ (k_a S^2 + k_b + k_c).$$ So if we want to find out $V_m$ and $K_m$ s.t. $$\lim_{S \to \infty} dP/dt = V_m, \quad dP(S = K_m)/dt = V_m/2,$$ then we would need $V_m = 2 k_b E_0, K_m = \sqrt{(k_b + k_c)/(k_a)}$. Any mistakes so far?

This is very close to the Michaelis-Menten scheme. The rate equations are ok, you now need  to assume that C is  at steady state dC/dt=0 and solve the equation for C. Look up the MM scheme for how to do this if you are not sure.

Sure thanks. May I also confirm that $V_m$ and $K_m$ in this case would be $2k_c E_0$ and $(k_b + k_c)/(2k_a)$ respectively? $E_0$ is the total enzyme concentration which is assumed to be constant.

IMO one of the best derivations of the Michaelis-Menten equation, together with a lucid explanation of why the steady-state hypothesis is superior to the 'equilibrium' assumption of Michaelis and Menten, is given by Haldane (1930) in his book 'Enzymes'.  The first chapter of his book is available here, made available through DC's Improbable Science.  (See also here).

I did not notice  before (sorry) but you should have $dC/dt=k_aS^2E-(k_b+k_c)C=0$ at steady state.

@tomd I don't have any more info about any $k$ or any substances in this problem. How is it now?

dC/dt is surely equal to kaEo[S][S]−(kb+kc)C, and this is where one of the difficulties of treating what essentially is a two-substrate mechanism as a single substrate mechanism arise? [Modified comment: thanks to @porphyrin]

It looks good to me. (It is probably worth pointing out that you are assuming (I think) the formation of a ternary complex between enz and two S (in the nomenclature of Cleland, an ordered Bi-Bi mechanism where both substrates are the same? This might not be the case: the enz could bind one S, do something to it producing a modified enz, take on board another S where the modified enz does something different to the second S to bind - in the nomen. of Cleland a 'ping-pong' mechanism (but I know of no examples where both substrates are the same!), and a different rate law will apply).

One further thought (and you probably know all about this): If the steady-state hypothesis is assumed, the initial rate (in the forward direction, say) for the ordered Bi-Bi mechanism cannot be defined in terms of just Km and Vmax values. One must define one other kinetic constant (this is not true if the equilibr. assumption is made). Dalziel calls this phiAB (or phi12), and it has dimensions of a reciprocal third-order rate constant. Other authors (including Cleland) call it something like KiA KmB. This might be a difficulty?

@tomd just confirmed the reaction forms a ternary complex instead of 1S-E complex first

Problem solved. Should I close this question or...?


text:
What is the typical resistance of a piece of 1 cm^2 of carbon paper?
I have been using carbon paper (Carbon sheet 99.8% from STERM CHEMICALS) to load catalysts for fuel cell study. Just now, I realized that for a 1 cm^2 piece the resistance is about 3 ohms!
Is that a typical value? I can't find any info on Google or the seller's webpage.
Thanks!

If there is no published resistance for that carbon sheet, then the best choice is to do exactly what you have done, i.e., measure the resistance per square. However, that specification is in given in the Strem catalog! It might be difficult o find if you do not spell the company's name correctly.
Two notes:

Resistance of a uniform, homogeneous sheet is measured per square, not per cm2. Whether the sheet is 1 mm on a side, or 100 km on a side, the resistance from one edge to the other would be the same; in your case, 3 Ω. [Consider that making the separation wider between edges is exactly compensated by the increased length of the conducting sheet.]
"Carbon paper" referred, in English, to paper with a mix of lampblack carbon and wax, though now it also applies carbon microfiber paper. For clarity, I suggest using the name Strem uses in it's catalog.