text:
Why is enthalpy and not heat released by system used while calculating entropy of surroundings?
While finding the sum of change in entropy of the universe and thus defining Gibbs free energy, why is the change in entropy of surrounding the negative of enthalpy of the system divided by the temperature?
I mean why are we considering the pressure to be constant ( thus considering enthalpy) in all systems while calculating Gibbs free energy ? Shouldn’t a more general form of entropy of surroundings be negative of heat released by system divided by the temperature?
I was thinking that maybe it is because all closed and open systems are constant pressure systems and change in entropy of surroundings is zero for isolated system but I don’t know, I am just guessing.
So, could someone please help me in understanding why enthalpy and not just heat in general is considered for any system while calculating Gibbs free energy?
Any help will be greatly appreciated!

Because $q=\Delta H$ only if only expansion work is done and the pressure of the system is constant. In that case, for an isothermal process, in which T is the same for system and surroundings, the entropy of the surroundings can be equated with the change in enthalpy divided by T. Which is why the applicability of the integrated formula $\Delta G = \Delta H -T \Delta S$ is restricted to constant p and T. You could write $q_p$ in place of $\Delta H$ the equation but only when work is restricted to pV work and when T and p are constant. It would not hold generally since sometimes work other than expansion work is performed. In general $\Delta H = q_p+w_\textrm{non-pV}$ at constant T and p.
To show this more explicitly, for a small (differential) change:
$$dU=dq+dw$$
$$dU=dq+dw_{\textrm{non-pV}}+dw_{\textrm{pV}}$$
$$dH = dU+d(pV) = dq+dw_{\textrm{non-pV}}+dw_{\textrm{pV}} + pdV+Vdp$$
At constant pressure of the system $dw_{\textrm{pV}}=-pdV$, $Vdp=0$ and
$$dH = dq+dw_{\textrm{non-pV}}$$
Further derivation is possible* to show how this relates to $G$, by assuming constant T and considering reversibility, but this should suffice to show that $dH$ and $dq$ (or $\Delta H$ and $q$) are not in general the same.

*For a reversible process $dq=TdS$ and
$$dH =  TdS+dw_{\textrm{non-pV,rev}}$$
Finally the differential expression for $G$ at constant $T$ follows (since then $dG=dH-d(TS)=dH-TdS$):
$$dG=dH - TdS = dw_{\textrm{non-pV,rev}} $$


text:
Only one unique adiabatic path between two states
My Textbook says:

Two states A and B can never lie both on a reversible as will as irreversible adiabatic path. There lies only one unique adiabatic path linkage between two states A and B.

I don't understand what the book means. Why cannot there be both reversible and irreversible adiabatic paths between two states? And if only one path is possible, then how do we know which one?
What I understood is: In adiabatic process: ΔU=W. Now since ΔU is state function, it will mean that Wirreversible=Wreversible, which is not possible. But mathematical proof aside, is there any example/physical explanation to this?

Think longer, and make some calculations. Then come back.

Wrev <> Wirr.  But the question elaboration is better placed to the question. More space and editing preview available. Comments are for a short feedback, not for supplementing additional question elaboration.

For an infinitesimal adiabatic process $dS_\mathrm{surroundings} = 0$ since no heat is exchanged with the surroundings. But $dS_\mathrm{system} = 0$  is only true for a reversible process, since only then  $dS_\mathrm{system} = dq/T$. For an irreversible process $dS_\mathrm{system} \neq dq/T =0$. Therefore
$$dS_{\mathrm{irrev}}\neq dS_{\mathrm{rev}}=0$$
This means that the state accessed with a reversible process through an integral change $\int dS_{\mathrm{rev}}=0$ cannot be accessed directly through an irreversible adiabatic process, for which $\Delta S = \int dS_{\mathrm{irrev}} \neq 0$.
Now since S is a state function, the terminal states of the two processes must therefore differ (QED).
As regards the second part (how do we know which is the reversible), you can use the approach just outlined to conclude that for the reversible process $\Delta S =0$.

so can we say that irreversible adiabatic process is not directly possible between any two given states?

If you have an irreversible adiabatic process between two end states (i.e., thermodynamic equilibrium states), there is no reversible adiabatic process between these same two end states.

and vice-versa?

Yes, of course.

Suppose that in a given transformation, ${\Delta U}$ is defined, and cannot be changed. But the same ${\Delta U}$ is the sum of two terms $Q$ and $W$ that can be changed and modified arbitrarily. If one of these terms, here the heat $Q$, is chosen to be zero, the other term $W$ cannot vary and stays equal to $W_{rev}$.

Sure but why only $Wrev$. why not $Wirreversible$?

It is up to you. It can be reversible work or irreversible work. It does not matter here. The only important thing is that this work is equal to the change of internal energy.

source: Stack Exchange

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

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text:
Calculate quantity of ions eliminated by acids
I have limescale $(\ce{CaCO3})$ and a solution of hydrochloric acid at 30% concentration of hydrogen chloride $(\ce{HCl}).$
This acid-base equation
$$\ce{HCl + H2O -> H3O^+ + Cl^-}$$
with the following data:
$m_\mathrm{solution} = \pu{1180 g},$ $m_\ce{HCl} = \pu{354 g},$ $M_\ce{HCl} = \pu{36.5 g mol-1},$ $n_\ce{HCl} = \pu{9.70 mol}.$
But now I need to calculate how much of $ \ce{CO3^{2-}} $ is eliminated by the $\ce{H3O^+}.$ I've written the molar concentration of $\ce{H3O^+}$ which is $\pu{9.70 mol L-1}$ and I calculated the $\mathrm{pH}$ of the solution:
$$\mathrm{pH} = -\log_{10}9.70 = 0.986$$
But I have no idea how to go from there.

The term chloridric acid is highly unusual, called usually hydrochloric acid. // Search for the CaCO3 vs HCl equivalence on the molar amount level.

You should guess that  $\ce{1 mol CO3^{2-}}$ reacts with  $\ce{2 mol H3O+}$ to produce $\ce{CO2 + H2O}$. So how many moles $\ce{CO3^{2-}}$ are reacting with $\ce{9.7 mol }$ acid ?

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


text:
Identification of salts observed in steam vaporiser

Recently I observed some precipitated salts in steam vaporizer after completion of vaporization.
I am guessing that some of the white particles may be calcium carbonate ($\ce {CaCO3}$) formed from calcium bicarbonate dissolved in water by the reaction
$$\ce {Ca(HCO3)2(aq) \rightarrow CO2(g) + H2O(l) + CaCO3(s)}$$
I am curious to know what may be the other salts and how are they formed.

I don't think we would be able to adequately answer this, too many unknowns. Besides, what makes you think it's salt? From the look of it, it might be rust, particles of tubing/paint/ion exchange resin or who knows what else.

The only way for answering your question is to make a qualitative analysis of this substance. You would need a lab, some glassware and some chemicals: Without our knowing if you have got this material, we cannot do much more.

As complete vaporization was done, there would be salts from both carbonate and permanent hardness and also other salts, mostly sodium ones, aside of minor constituents.

You might get quite close to the composition of the deposit by reading the bottle label or by searching data about the water delivered by your municipality or the company serving you.


text:
Why is nitrogen gas collected over water?
In Duma's method, we need to collect Nitrogen gas over water and subtract the aqueous tension. But why can we not just collect it over the air?

Ever seen any air? Other gases, maybe? Guess what: they mix.

It is possible only if density difference is huge, and even then it is just matter of time.

Perhaps more importantly, you won't be able to see the border between the two, because it won't be there.

Dumas's method for determination of nitrogen (and usually then for estimation of protein content) is fairly old (1831 Ref 1) and has been modified to be quick and automatable. In the original method, the equipment was arranged to collect the gaseous nitrogen after passing thru various traps. The picture (Ref 2) shows a modern arrangement of the original equipment used by Dumas. The inflow marked CO$_2$ represents the gaseous product of the ignition of the unknown specimen, and N$_2$ is collected over aqueous KOH (to remove CO$_2$).

In the picture below, modern, automated, Dumas nitrogen determination takes well-homogenized samples, heats them in a high-temperature furnace where combustion takes place at over 1000 °C in the presence of pure oxygen. This produces mostly H$_2$O, CO$_2$ and N$_2$ as several oxides. This gas mixture passes through a reduction chamber containing copper heated to around 650 °C. This stage converts nitrogen oxides into N$_2$ and collects excess O$_2$. Various traps remove the residual H$_2$O and CO$_2$. Total N$_2$ content is measured by a thermal conductivity detector. The N$_2$ gas is not collected over water - it is essentially measured (not collected) over air, because the flow thru the TCD (thermal conductivity detector) can measure how much N$_2$ is going by.

Now if you don't have one of these (expensive?) automatic measuring devices, you can go back to the older way and just collect the N$_2$ over water (actually over KOH solution).
Ref 1. https://www.velp.com/en-ww/dumas-method-1.aspx
Ref 2. https://www.brainkart.com/article/Dumas-method---Estimation-of-nitrogen_36468/

then why does nitrogen gas come out to be moist?

In the first diagram, the exit gas is collected over KOH solution. The vapor over that solution contains H2O molecules, which exert a vapor pressure. If you collect over plain water, there will be a higher vapor pressure, which you can correct for. If you had efficient water traps in the line, and collect over mercury, you could assume that there is no water in the collected gas. In the second diagram, there are 2 water traps (WT1 and WT2), so the TDC doesn't measure any H2O.


text:
Mechanism of formaldehyde / phenol condensation
The production of bakelite using this method, but with an acid as the catalyst is simple enough; a protonation of the carbonyl oxygen, followed by an electrophilic substitution to the ring. The subsequent cross-linkage formation is the same for both cases, but I can't figure out how the initial reaction
steps would work with a base as the catalyst. Google searches and the texts I have yield nothing.
How would the mechanism proceed with a base as the catalyst here?

Presuming that the reaction is base-catalyzed, "push" electrons toward the carbonyl by forming the anion of a phenol. In the acid-catalyzed reaction, the protonated formaldehyde is a "hot" electrophile that can react with a neutral phenol nucleophile. The base mechanism is similar to an aldol condensation.

@user55119: okay, so there's electromerism on the formaldehyde, yielding an electrophilic carbon, and deprotonation of the phenol yields a phenoxide ion that joins with the carbonyl carbon. But that doesn't result in anything like an elementary unit of Novolac. Where did I go wrong?

Keep going once the addition occurs. Make the phenoxide again, beta-eliminate hydroxide. Now you have the equivalent of a conjugated unsaturated ketone. Do a Michael addition of another phenoxide. Now you have a methylene between two aromatic rings. Keep doing this until all ortho and para positions are substituted or you get bored.

@user55119: I got a deprotonated phenoxymethanol doing that, but that isn't a conjugated unsaturated ketone. Moreover, Novolac has hydroxyls ortho to the methylene bridges, which don't seem to come up here.

@Harry Holmes: Here is a simplified example of how the formation of Bakelite may occur under base-catalyzed conditions to incorporate formaldehyde as -CH2- at the ortho and para positions of phenol. There are a myriad of permutations as to the order of condensations. I have selected one that illustrates the Michael-like addition that seems to have eluded you.
Phenoxide 1 adds to formaldehyde at the para position in what amounts to the first step in an aldol addition. The addition is likely adding to the ortho position of phenoxide as well. Species 3 may condense with phenoxide at this point but I have chosen to incorporate two more equivalents of formaldehyde at the ortho positions of 3 to form 4. Now elimination of hydroxide from any of the three positions of 4 leads to a Michael acceptor, such as 5, which reacts with phenoxide or any substituted phenoxide having an open ortho or para position. Structure 6 simply illustrates the mechanism for incorporating formaldehyde. The o-methylenedienone 5 is the product of an aldol condensation wherein phenoxide is the enolate.


text:
Generate fit of virial expansion to osmotic pressure data and obtain second virial coefficient
So I have some osmotic pressure data that I would like to obtain the second virial coefficient ($B_{22}$) for, but I am getting increasingly more confused on how the theory works. I know you can rewrite the expansion in the following manner:
$\Pi (T, c) = RT(c + B_{22}c^{2} + B_{23}c^{3} + ...)$
where $\Pi$ is the osmotic pressure, $T$ the temperature in K, $c$ the concentration and $R$ the gas constant. I would like to do (presumably) a least-squares fit but my maths is a little rusty. Can you do this in Mathematica or Python and if so, is there any literature on this that I can reference? Apologize if this is a stupid question.

Yes, if you know how to use software such as Mathematica or Python to perform a least-squares fit then you can fit a polynomial in c to the data to extract virial expansion coefficients. Unfortunately that is not a chemistry problem, that is a programming problem.


text:
On the molybdenum(III) oxide-catalysed oxidation of alkanes to carbonyls
The following reaction is specified in my notes;
$$\ce{CH4 ->[Mo2O3, \Delta]HCHO}$$ The intermediate steps involve the dehydration of an alcohol to yield the final product.
Wouldn't this only work if there's acid in the reaction mixture? Or is it that water functions as a proton donor, inducing the protonation of the hydroxyl and causing the dehydration?

Is it not a reaction $\ce{CH4 + O2}$ ?

@Maurice: it was shown with $\ce{[O]}$ above the arrows in the reaction steps.

Sorry, There is no [O] above the arrow. There is only a molybdenum oxide

@Maurice: Yes, not in this simplified equation, but when you write it out, apparently you use nascent oxygen. It's the controlled oxidation of alkanes, to be specific.

The nature of the oxidizing agent should absolutely be given. Because it is the very heart of the problem. How is this "nascent oxygen" produced ? UV ? $\ce{H2O2}$ ? Spark ?

@Maurice: It's not specified, and I can't find the reaction online. I suppose it's not generally known.

Does this answer your question? Is there any evidence, *any evidence at all*, that nascent hydrogen actually exists?

The above depicts a concept, rather than an equation for that there is no balance of atoms on the left and right of the arrow.  Because if $\ce{Mo2O3}$ is said to be used as catalyst rather than a (stoichiometric) reagent, then the oxygen in formaldehyde must be of different origin than $\ce{Mo2O3}$.
On industrial scale, formaldehyde is prepared from methanol by an  heterogeneous redox-reaction.  The oxygen needed may come from oxygen gas, or simply air.  Thus, the last step known as Formox process is
$$ \ce{2 CH3OH + O2 -> 2 CH2O + 2 H2O} $$
(from wikipedia)
$\ce{Mo2O3}$ is deployed ease the advancement of the reaction, i.e., to perform the wanted reaction at lower temperature (heating costs money) than without this help, and to further increase rate or / and yield of said reaction compared performing this conversion at room temperature.

So oxidising alkanes up the oxidation ladder isn't a thing?

@HarryHolmes Compared to other compounds, saturated hydrocarbons / alkanes are not very much reactive.  On occasion, you encounter them under the name «paraffin», derived from the Latin «parum» (little) and «afffinis» (related) to just describe their low reactivity in comparison to functionalized and unsaturated hydrocarbons.  It is easier e.g., to chop long alkane chains into smaller alkenes (e.g., in a steam cracker) and then to oxidize these fragments (e.g., Wacker process with ethylene to yield acetaldehyde, $\ce{MeCHO}$). Formally, the remove of hydrogen from alkanes is an oxidation, too.


text:
What is the average distance of two different types of molecules of different concentrations in solution?
Concerning the simplest approximation for intermolecular distance in solution:
$d=(\frac{10^{24}}{cN_A})^{1/3}=1.18c^{−1/3}$
This applies for one type of molecule in solution. But, how to account for two different molecules A and B of different concentrations?
My case: I have molecule A of 0,4 uM and molecule B of 16x0,4 = 6,4 uM. The average distance A-A is therefore determined as 160 nm, average distance of B-B as 63,5 nm and average distance of any two molecules as 62,3 nm.
Is it possible to approximate the average distance of A-B as a weighed average of A-A and B-B, therefore (160+16*62,3)/17 = 69,2 nm, since any given molecule is 16-times more likely to be of type B than type A?
Thank you for your answers!

What exactly are we talking about, to begin with? Is it the average distance from A to the nearest B, or from B to the nearest A? These two are going to be different.

In this case, what matters is the average distance between A (the molecule of lower concentration) to B. Thanks a lot!


text:
How to find the largest cyclic substructure with RDKit?
I have a bunch of structures and of each I would like to find the largest cyclic substructure.
I'm not a chemist, so I don't know if there is a common terminology for this. I hope, I made it clear what I'm looking for.
Example:

Each structure on the top (1,2,3) as input should return the bottom structure (4).
There is the rdkit.Chem.MCS module which will return the maximum common substructure of a set of molecules.
But that's not what I'm looking for.
Any ideas? And hints where or which module to look at...? Which term to search for?
I'm aware that there is a RDKit mailing list, however, I'm asking here first, because of the possibility to easily attach images. If nobody can help me here, I will try there. Maybe it already goes into "heavier" programming which then would be more a topic for StackOverflow.

Not that this question is off-topic for CSE, but the rules of thumb are heavy_programming → SO; heavy_programming + chemistry → MMSE. If you think MMSE could be a better place for your question, we can migrate it there.

@andselisk well, then the question is: does this require "heavy programming" or is it just a "simple" existing module or function which I haven't found yet?

Yeah it's definitely on-topic but you might have better luck at Matter Modeling eventually. No harm letting it sit here for a couple of days (there is some overlap between the user base anyway) but if you don't get any answers we can migrate it.

@theozh Sorry, I'm afraid I cannot answer this. I used RDkit once for something I already forgot what it was.

@theozh Contributions to RDKit's mailing list -- both questions, answers, and comments -- are retained in searchable archives.  They may be accessed by anyone, regardless if you already are a subscriber, or not.  One of the threads potentially relevant to your question is this one.  Or, is your question to identify the largest «contiguous set of rings» in these structures?

@Buttonwood thank you for the link. I checked it but I guess it is not what I am looking for. And if I understand correctly, there is a limit of 20 atoms(?). I guess your definition largest «contiguous set of rings» sounds good. Is there an easy to use function for this?

@theozh From your comment, my guess you aim for one of the scaffold analyses out there.  If 4 represents the structure you would expect for the others to be displayed, you run for the Murcko scaffold (but not the Murcko skeleton).  The file format of your structures is not evident, but if it is in SMILES strings, RDKit documents one approach here.  An alternative to work from the CLI / a more GUI based approach is DataWarrior, outlined here.

In the RDKit Cookbook is an example how to find the indices of ring systems.
RDKit Cookbook Example
Having the indices, it is simple to find the largest ring.
from rdkit import Chem
from rdkit.Chem import Draw
from rdkit.Chem.Draw import IPythonConsole
IPythonConsole.molSize = (400, 300)
IPythonConsole.drawOptions.addAtomIndices = True

m = Chem.MolFromSmiles('C1=CC2C(C=C1)=c1ccccc1=C2C1C=C(c2ccccc2)c2cccc3c4ccccc4nc1c23')
m

# the cook book example
def GetRingSystems(mol, includeSpiro=False):
    ri = mol.GetRingInfo()
    systems = []
    for ring in ri.AtomRings():
        ringAts = set(ring)
        nSystems = []
        for system in systems:
            nInCommon = len(ringAts.intersection(system))
            if nInCommon and (includeSpiro or nInCommon>1):
                ringAts = ringAts.union(system)
            else:
                nSystems.append(system)
        nSystems.append(ringAts)
        systems = nSystems
    return systems

ringSys = GetRingSystems(m)
ringSys

[{16, 17, 18, 19, 20, 21},
 {13, 14, 15, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35},
 {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}]

Highlighting the largest ring for proof.
highlightAtoms = max(ringSys, key=len)
Draw.MolToImage(m, size=(400,300), highlightAtoms=highlightAtoms)

I have some difficulties to replicate your work.  May you have a look into the chat in «The Periodic Table» here, please?  Maybe it is only a minor detail missing.

@Buttonwood It could be the RDKit version. Here is my configuration if that helps: Windows 10, Python 3.7.10, Jupyter 6.3.0, RDKit 2021.03.1 (from conda-forge)

@rapelpy , Great! That's what I was looking for. I guess I wouldn't have found this without your help. Thank you very much! Works fine for me. I have: Win10, Anaconda, Python 3.6.12, RDKit 2020.09.01

@rapelpy , Ah, l was too excited about finding/marking the largest ring system. But, well, now I still have to find a way to get the structure (i.e. SMILES) from these largest ensemble of rings. I need to dig into the RDKit documentation. I hope there will be another cookbook example for this task.


text:
What is an appropriate way to write a recursive SMARTS that excludes trifluoromethyl groups as a potential R group?
C[CX4&!$(C(F)(F)F)] is the smarts I've tried. It does not exclude the trifluoride in CC(F)(F)F.

+1 Although you might get better answers on Matter Modelling SE

If I see it correctly, you want to exclude C(F)(F)F, but find all other C?

By the way did you post the same question on Matter Modelling SE, from a different user account? : https://mattermodeling.stackexchange.com/questions/4859/excluding-certain-sp3-hybridized-carbons-with-smarts . If not, it is asking the same question about excluding trifluoromethyl so it might be helpful. If you have asked the other question too, and this answer satisfies your question, then feel free to close the other one, because cross-posting is usually discouraged.

I'm not sure if I correctly understand your intent.  My assumptions are

you have a set of molecules, e.g., as a list of SMILES strings (say FC(c1ccccc1)(F)F, CCCC(F)(F)F, CC(Cl)(Cl)Cl, and CCCO) in a file like probe.smi
you want to identify the molecules without the $\ce{-CF3}$ group

If so, I would i) search for the molecules with said group in a first step and then ii) remove this sub-set of positives from the set to know about the molecules without the $\ce{-CF3}$ group.

OpenBabel's positives: obabel -ismi probe.smi -s'C(F)(F)F' -osmi -O positives.smi.  In the present example, this yields only FC(c1ccccc1)(F)F and CCCC(F)(F)F, regardless of the different notation of the trifluormethyl group in the entries concerned.

The molecules lacking the $\ce{-CF3}$ group are just the complementary sub-set to the sub-set of the positives.  To start, one could opt for a Python script like:

#!/usr/bin/python3
"""Report molecules lacking the -CF3 group."""

def reporter():
    """Report .smi of molecules without -CF3 group."""
    # prepare lists about the molecules:
    all_mols = []  # all molecules, no discern
    positives = []  # molecules with -CF3
    negatives = []  # molecules without -CF3

    with open('probe.smi', mode='r') as source:
        for line in source:
            all_mols.append(line.strip())

    with open('positives.smi', mode='r') as source:
        for line in source:
            positives.append(line.strip())

    # truncate the set to yield the sub-set about molecules without -CF3 group:
    set_all_mols = set(all_mols)
    set_positives = set(positives)
    set_negatives = set_all_mols.difference(set_positives)

    # convert the sub-set (back) into a list of entries:
    negatives = list(set_negatives)

    with open('negatives.smi', mode='w') as newfile:
        newfile.write('\n'.join(negatives))

if __name__ == "__main__":
    reporter()

which in file negatives.smi correctly reports CC(Cl)(Cl)Cl and CCCO as molecules without the $\ce{-CF3}$ group.

Of course it would be nice to identify how to negate OpenBabel's search filter (-s'C(F)(F)F') to shorten the route traveled here.


text:
Safety concerns (Cyanide): Dicyandiamide + Polyamine condensation reaction
I am attempting to synthesize a poly-amine polymer by condensation between dicyandiamide (also known as cyanoguanidine and dicyandiamine) and diethyltriamine. The polymer is a kind of dye fixative agent for cellulosic fabrics.
The reaction liberates ammonia. Once no more ammonia is liberated, it can be considered complete.
My lab setup is a three neck flask with stirring and a temperature between 110-160 ºC. One neck has the thermometer, the other the stirrer, and the third a tube that goes to a flask filled with acidic water, to trap the ammonia. A bit of vapor escapes through the stirrer as the seals are not perfect. The trapping flask is also exposed.
Since the reaction involves dicyandiamide and heat, I was fearful that it could also liberate some kind of toxic cyanide-like gas. In the literature there is no mention of any kind of toxic emission, however, I take safety seriously. So I wanted to know if anyone can assure me that there are no toxic cyanide-like emissions. And if they were, any advice is appreciated.
Literature:

https://patents.google.com/patent/US5632781A/en
https://patents.google.com/patent/US4764585A/en
https://patents.google.com/patent/US2649354A/en

Well, accidents do happen. If you have a fume hood and sufficient experience with such reactions, you can handle that danger. If you don't, I recommend you leave the job to someone who has both.


text:
Why do we want to operate at dew point or bubble point when doing flash calculations?
I'm trying to dimension a flash drum for a separation between water and isobutane. I now the total inlet flow (F), the composition of the inlet flow ($z_1$ for isobutane and $z_2$ for water) and the total pressure (P). My supervisor told me that if my inlet flow to the flash was gas I would want my flash drum to operate at dew point temperature. And vice versa I would like my flash drum to operate at bubble point if my inlet flow was liquid. Therefore I have to do a dew or bubble point calculation for the temperature. Why is this true? According to her this is something you always want you flash drum to operate at.
If I had known the outlet molar flow of vapor (V) and liquid (L) I would have used the following iterative model:

We know F, $z_i$, P, V and L
Guess T
Calculate $P_1°$ and $P_2°$ from a model (as an example Antoines equation)
Calculate $x_i=\frac{Fz_i}{V\frac{P_i°}{P}+L}$
Check if $\sum x_i=1$. If not true redo from point 2. If true you have found the flash temperature.

However this isn't possible since I don't know the outlet flows V and L.
So to recap my question is: Why does the flash need to operate at dew or bubble point temperature? Is it because it's most cost  efficient to do so? Do I get the best separation at this temperature (and if so why)? Any other reasons why this temperature is the best to operate at?

Well, the flash is going to take place at equilibrium between the vapor and liquid.  So you are setting the drum pressure (which is less than the feed pressure) and determining the split between liquid and vapor under the condition that the enthalpy change across the inlet is zero.  You adjust the drum pressure to get the split you want.


text:
How does (E)CA glue ("super glue", "crazy glue", etc.) cure?
I use CA glue† very frequently in my shop, but I'm not entirely sure how it works. I've noticed three main modes of curing, I've emphasized the key observations:

When dispensed, it will remain fluid until even a small amount of pressure is applied. With pressure, it will set and cure in 5 - 45 seconds. Often, depending on the materials, it remains completely clear after curing.
If a drop of it is left out undisturbed (also applies to glue that may have been squeezed out of the joint), it will cure after a relatively long time, on the order of 20 - 60 minutes. If it cures in this way, it will develop a white crystalline substance (ranging from on-the-surface-only to the entire drop). This structure is hard and brittle.
It will cure in the bottle after a very long time, on the order of months or years (depending on whether or not the bottle has been opened). When it cures this way, it remains clear. It also becomes hard and brittle, but seems to have a higher strength than the white crystalline form.

So; here's what I know about it and some other factoids I found in research:

The active chemical in most consumer-grade CA glues† is ethyl 2‑cyanoacrylate (CAS 7085‑85‑0).
Generally, it's present in the glue at a concentration of 60% to 100% by weight.
It emits very strong, irritating, noxious fumes as it cures, but is odorless after cure.
I know that curing is some type of polymerization and is a reaction with water.
† It seems "CA" has either become a misused colloquialism, or is used to refer to the entire class of cyanoacrylate glues (e.g. "ECA" appears to be the technically correct abbreviation for ethyl 2‑cyanoacrylate). I actually didn't even know there were different types until right this moment (hence the awkwardly placed †'s, heh).

So my question is, basically: What is going on here? In other words: How, exactly does it cure? Why does pressure seem to be involved? What is the white crystalline substance and why does it only form in certain situations? What are the fumes? What is the reaction here? All the details.
I'm looking for a detailed answer, as much as possible -- for example, I already know from the Wikipedia page that it polymerizes in presence of water. What I'd like to know is some more advanced information about how and why that reaction takes place, energies involved, intermediates, crystal structure, what causes my observations, etc. I'm not a chemist but hit me with all the jargon, I could stand to learn a few new words.
I'm actually not particularly interested in how it reacts with the materials it is applied to or how it achieves adhesion, I'm much more interested in the curing process itself.

There is a simple observation that explains the process. The cyanoacrylate can undergo a chain reaction to form a strong polymer. That reaction is tuned to be triggered by interaction of the glue with water. Most surfaces have adsorbed water in enough quantities to initiate the reaction and, once started, it proceeds until the whole mixture has been polymerised into a strong polymer forming a bond. I'll leave the details to others to fill in.

Look up “anionic olefin polymerization”


text:
Why is the melting point of magnesium nitride less than that of beryllium nitride?
Why is melting point of $\ce{Mg3N2}$ lower than that of $\ce{Be3N2}$ ?
I have two reasonings:

The lattice energy of $\ce{Be3N2}$ is higher than that of $\ce{Mg3N2}$ so accordingly the melting point of the former should be higher
Both of them form ionic nitrides. Since $\ce{Mg3N2}$ has more ionic character it should have higher melting point than $\ce{Be3N2}$

Since both of these are contradicting, I wish to ask which one is correct? And if both are correct in some way, which one should be prioritized over the other?
Can this be determined or is it experimental?

Melting point in a compound with a lattice structure correlates more closely with lattice energy than with ionic character, and lattice energy tends to be greater when you use smaller atoms.  Among alkaline earth nitrides it is not just beryllium versus magnesium that shows this effect; it keeps going.  Melting point data below are from the Wikipedia articles as of April 30, 2021:
$\ce{Be3N2} \pu{= 2200°C}$
$\ce{Mg3N2} \pu{= 1500°C}$
$\ce{Ca3N2} \pu{= 1195°C}$
$\ce{Sr3N2} \pu{= 1200°C}$
The alkaline earth oxides show a similar trend, for the same reason.  Here the lattice energy of beryllium oxide is slightly depressed because each atom has a coordination number of only four whereas magnesia and its heavier congeners have coordination number six; thus the peak melting point is at magnesia:
$\ce{BeO} \pu{= 2507°C}$ -- four-coordinate
$\ce{MgO} \pu{= 2852°C}$ -- six-coordinate, same for the following oxides
$\ce{CaO} \pu{= 2613°C}$
$\ce{SrO} \pu{= 2531°C}$
$\ce{BaO} \pu{= 1923°C}$


text:
Products of the reaction between ethoxyethene and HI
Problem

The products formed in the reaction
$$\ce{CH2=CHOCH2CH3 + HI ->}$$
are
(a) $\ce{CH2=CHI}$ and $\ce{CH3CH2I}$
(b) $\ce{CH2=CHI}$ and $\ce{CH3CH2OH}$
(c) $\ce{CH3CHO}$ and $\ce{CH3CH2I}$
(d) $\ce{CH3CH(I)OCH2CH3}$

Answer

 (c) $\ce{CH3CHO}$ and $\ce{CH3CH2I}$

Question
Does ether cleavage occur or electrophilic addition of $\ce{HI}$ to the double bond take place or both? I'm equally inclined towards both (c) and (d).

d is formed reversibly, then CH3CH(+)OCH2CH3 is cleaved by iodide.

The substrate is an enol ether, which is very sensitive to acid. And HI is a very strong acid, which can cleave usually stable normal ether linkages.

@MathewMahindaratne Why is the cleavage reaction favored over the electrophilic addition reaction? The cleavage reaction will proceed via breaking of a sigma bond whereas E+ addition proceeds via breaking a pi bond, so shouldn't E+ addition give a thermodynamically more stable product?

$$\ce{H2C=CH-O-CH2CH3 ->[H+] H3C-CH=O^+!-CH2CH3 ->[I-]  H3C-CHO + I-CH2CH3}$$ The intermediate is more stable by resonance as shown.

Here is a mechanism for the cleavage. After proton transfer, the carbocation has a resonance contributor. Nucleophilic attack can occur on the ethyl group because there is a good leaving group.

The electrophilic addition can happen, with iodide attacking the carbocation. However, it is reversible. Iodide is a good leaving group, and the carbocation is resonance stabilized. The cleavage reaction is much less reversible.

Thanks Ben, your answer made things absolutely clear


text:
How is chirality center defined for an atom with 5 substituents?
I understand that carbon atoms normally only have up to 4 substituents and the R and S chirality centers are defined for 4 substituents. Is there a similar convention of 'chirality' for an atom with 5 substituents?
The SMILES website (section 3.3.4 General Chiral Specification) shows that general 'chirality' for 5 and 6 substituents exists. But the text there focuses on how to convert them to SMILES strings. Is there any references about the classifications and nomenclatures?
https://www.daylight.com/dayhtml/doc/theory/theory.smiles.html
Sorry I am not a chemistry major and my nomenclature may not be correct.

I don't understand why it's downvoted. They do exist as shown in the link. Conceptually you can easily see such chirality variants could exist

Chirality is not about bonds at all.

@IvanNeretin How should I phrase it? 4 connected objects (what's the chemical term for that)?

Yes, I got that. A carbon with four different substituents is a typical example of chirality, that's for sure. But it is neither sufficient nor necessary for chirality. Ditto for the cases with more bonds. Chirality is not a property of one center at all. It is a property of a molecule as a whole.

I see, thanks @IvanNeretin I think my question is about the classification of chirality center when there are 5 substituents.

Your question implies that chirality is caused by certain centers. It is not.

I think you are not actually asking about chirality, but rather configuration. Notice l-tartaric acid (O[C@H]([C@H](C(O)=O)O)C(O)=O); d-tartaric acid (O[C@@H]([C@@H](C(O)=O)O)C(O)=O); meso-tartaric acid (O[C@@H]([C@H](C(O)=O)O)C(O)=O). The @'s and @@s are not meant to be flags which indicate that the carbon is asymmetric (otherwise there wouldn't be two different flags just to indicate the presence of an asymmetric carbon). They're intended to describe, or fully specify, the configuration of these carbons, i.e. the exact distribution of the substituents in space.

@orthocresol .And. -- to support your point that @ are about the configuration of stereogenic centers -- C[S@](c1ccccc1)=O about (S)-, and C[S@@](c1ccccc1)=O about (R)-methyl phenyl sulfoxide.

Chirality depends not so much on coordination number, but on the symmetry of the substituents.  Meaning you have to pay attention to coordination geometry.  In the case of four-coordination you can have four different substituents and yet no chirality... if you have square planar instead of tetrahedron coordination.
For five-coordination you lack a mirror plane and thus have a chiral center in the following cases:

Square pyramidal geometry: if all four substituents on the base are different, or if just one pair of these basal substituents are identical and the identical pair is on adjacent vertices.  The axial substituent does not have any impact because all possible mirror planes pass through it.

Triangular bipyramid geometry: if all three equatorial substituents are different and the two axial substituents are different from each other.  You may still have chirality if one axial substituent happens to be the same as one equatorial one.

Suggested change: «not so much on coordination but on the symmetry of the substituents» to «not so much on coordination number, but on the symmetry of the substituents».  Because so far I understood the priorities of substitutents according to the CIP rules to eventually assign (R) or (S) as not constrained to four atoms around one in the centre.  If the rules equally account for a lone electron pair (the chiral sulfoxides) as then lowest  priority among the other three atoms, why should this not work for 5?

The group ranking methodology you see for tetrahedral four-coordination needs some patching when we pass th five-coordination, because as we see above, in both cases we have to allow point chirality with identical attached groups.  And with six or more it's going to get worse.

I agree that four (or five, six) atoms around one center in common as number is not a sufficient criterion for a chiral pointgroup.  Its recognition requires the background a flow chart like this summarizes and some programs assist to see (e.g., Jmol when issuing calculate pointgroup;, show pointgroup; (table), and draw pointgroup; (display in the structure) on its CLI, with adjustable parameter thresholds).


text:
Are solvent effects the reason for the methoxide ion to be a stronger nucleophile than the thiomethoxide ion?
Between $\ce{CH3O-}$ and $\ce{NC-}$, $\ce{NC-}$ is a better nucleophile than $\ce{CH3O-}$ in a polar protic solvent, while $\ce{CH3O-}$ is a better one in a polar aprotic solvent, because in a polar aprotic solvent $\ce{CH3O-}$ is free to move and $\ce{O-}$ has more electron density.
Does the same reason apply for $\ce{CH3O-}$ and $\ce{CH3S-}$, is $\ce{CH3O-}$ a stronger nucleophile than the latter one in polar aprotic solvent?

I think you're overestimating influence of solvents.


text:
What happens if you add pure food grade lactic acid to fresh pasteurized milk?
What happens if you add pure food grade lactic acid that is free of bacteria to fresh pasteurized milk? What chemical reactions would take place? What does it taste like straight after mixing and a short time later (not so much later that bacteria could multiply and spoil the milk) Does it smell and taste like spoil milk? Or like yogurt?
I'm asking the question as part of my amateur research into spoiled milk. I am not especially interested in yogurt. I am especially interested in how similar or not fresh milk with lactic acid added is to slightly spoiled milk.
Is fresh milk with lactic acid added anything like spoiled milk in taste or smell?

The sour smell of yogurt, cheese, sweaty socks,.... comes from acetic acid and many other fermentation byproducts. Lactic acid is basically free from scent, tastes simply sour.

What is the down voting for?

I did not vote and do not know why others voted, but maybe a reason is to be found in the old Latin expression: de gustibus non disputandum est. People have no way of knowing how other people perceive particular smells and tastes. Just my two cents.

@EdV One should not argue about matters of taste. One of my favorite maxims. But I asked what chemical reactions would take place, and how similar to sour milk the taste of the mixture would be. The disparaging remarks about sour milk were just for clarity and to make the text more readable. Anyway, thanks for hazarding a guess.

I just checked ebay and you can buy food grade lactic acid in reasonable quantities, i.e., fifty to several hundred grams. Not too expensive, either. So, in the spirit of experimentation, why not do the taste test and let enquiring minds know how it goes? ;-)

I find the question interesting but it seems mostly about the taste of a particular food. You might really want to give seasoned advice SE a try: https://cooking.stackexchange.com/questions/100389/how-acidic-does-a-mixture-have-to-be-for-milk-to-curdle

BTW the taste of lactic acid according to various unofficial sources is negligible eg: https://www.reddit.com/r/chemistry/comments/j15maj/what_does_lactic_acid_smell_like/ So the effect would be primarily on the tarteness and the consistency (it would lead to coagulation)

@BuckThorn Asking on Seasoned Advice seems like a good idea.

Likely, you'd get a yogurt-like mixture, and, eventually, cottage cheese: "Cheesemakers create cottage cheese by heating... milk, then mixing it with an acid -— which could be lactic acid-producing bacteria cultures or an acid like rennet, vinegar, or lemon juice. This acid causes the curds and whey to separate."
There are differences between just adding lactic acid and letting lactic-acid producing bacteria do the work.

The bacteria consume some of the milk sugars, reducing sweetness.
The bacteria have other waste products, which produce accustomed flavors of yogurt.

As for skipping the pre-heating step, I don't think it will greatly change flavor, but it might allow growth of undesirable microbes and it would likely increase the time for the curds to coagulate.
It seems that you'd need to experiment to see if you like the taste of the product.

The vile smell and taste of sour milk is surely unmistakable. I don't AFAIK have access to lactic acid, otherwise of course I would try the experiment.Interesting point about the bacteria consuming sugar and reducing sweetness that way. I hadn't thought of that. On the other hand, adding sugar to sour milk does not make it any less revolting.

It depends on which bacteria colonize the milk first... The "vile smell" is likely from butyric-acid producing bacteria, which may even be desirable in some aged cheeses. If you're lucky, though, the a more pleasant lactobacillus may make yogurt or buttermilk. Don't sour on making your own cheese!

Which bacteria and which aged cheeses? I'm surprised to hear that anyone would want to eat cheese that tastes of sour milk. I'm focused on sour milk right now, but making my own cheese is interesting, especially if I could use sour milk to do it.


text:
Is Chromium(VI) fluoride able to be synthesized properly?
Chromium(VI) fluoride, also known as chromium hexafluoride is a possible chemical compound with a chromium atom attached to 6 fluorine atoms. It has been unsuccessfully synthesized a number of times by heating up chromium to 400 °C in a fluorine atmosphere at 20 megapascals of pressure and freezing it as it was formed but this was still a failure as this instead produces chromium(V) fluoride or chromium pentafluoride. Is the following possible?
The predicted reaction was:
$$\ce{Cr + 3 F2 -> CrF6}$$
Instead, the following reaction was happening:
$$\ce{2 Cr + 5 F2 -> 2 CrF5}$$

I assume that CrF6 is such a strong oxidizer that it easily oxidizes CrF4 to give two CrF5.

So what is the question? Your reference states clearly, "CrF6 has yet to be synthesized." It does not state it is impossible. Proving a negative is a  bit difficult.

Apparently, chromium favors oxide instead of fluoride ligands at such a high oxidation state.

Check these papers: https://pubs.acs.org/doi/10.1021/ic00052a007 ... https://pubs.acs.org/doi/10.1021/ic00026a002 .... https://pubs.rsc.org/en/content/articlelanding/1984/c3/c39840001355/unauth#!divAbstract

Also, some papers from the late 19th century reported that CrF6 is a deep-red vapor which condensed to a blood-red liquid at low temperature. It is made by reacting lead chromate, calcium fluoride and sulfuric acid($\ce{PbCrO4 + 3CaF2 + 4H2SO4 -> PbSO4 + 3CaSO4 + 4H2O + CrF6}$) It strongly fumes in air and decomposes on contact with water/moisture to form chromium trioxide and hydrofluoric acid. (But then again, this is from a very old source, so this info. maybe outdated).

Both steric and electronic factors inhibit the formation of chromium hexafluoride.

Steric factors. Chromium, like most transition metals, commonly forms octahedral complexes. But in the fourth period where chromium lies these octahedral complexes tend to occur with relatively low oxidation states like $+3$ for chromium. The metal center would be smaller in an oxidation state as high as $+6$, and possibly we would need a fifth or higher period metal to get the center large enough to continue to coordinate octahedrally. So we can get hexafluorides of molybdenum and tungsten but as yet not chromium.

Electronic factors. In this answer early transition metals are identified as strong pi acceptor and therefore prefer oxo complexes (with a better pi-donor ligand) over fluoro complexes. This does not prevent the synthesis of a hexafluoride (as we saw above with molybdenum and tungsten) or perhaps even a fluoro-cation (which would avoid the steric issues that a neutral hexafluoride would have with chromium). But it does imply that such compounds are likely to break down at the drop of an oxide-bearing hat. The respective Wikipedia pages for $\ce{MoF6}$ and $\ce{WF6}$ indicate that these hydrolyze in water, the tungsten compound specifically being reported to form oxyfluorides and ultimately the oxide. Presumably $\ce{CrF6}$ would react similarly with water, to form chromic acid and then, since that acid is strong, the anions $\ce{Cr2O7^{2-}}$ and $\ce{CrO4^{2-}}$.


text:
Should rates and rate constants be expressed in terms of pressure or concentration?
If an elementary reaction was conducted in the gaseous phase, say $$\ce{X (g) -> Y(g)},$$ would the formal definition of the rate law be $\mathrm{rate} = kP_\ce{X}$ or $\mathrm{rate} = k[\ce{X}]$?

The ambiguity doesn't lie so much in the definition of the rate, because as long as the constants $a$ and $b$ in $ap_\ce{X}$ and $b[\ce{X}]$ are chosen appropriately, the two expressions are entirely equivalent. So you can define the rate either way, and there is no physical difference, not even a mathematical difference. It is somewhat akin to asking whether you should factorise the number $12$ as $2 \times 6$ or $4 \times 3$. It doesn't matter how you factorise $12$, it's still the same number.
The ambiguity is actually in the definition of the rate coefficient, $k$. The constants $a$ and $b$ above, though directly proportional to each other, are obviously different: and only one of them can be the "true" rate coefficient $k$.
Before launching into the answer proper I should mention that there is no physical difference between using the pressure and the concentration. At the end of the day, it doesn't change the actual rate at which your system is reacting: it's just an arbitrary choice as to what physical constants you want to include in the rate coefficient $k$.
That said, the IUPAC Gold Book seems to come down quite strongly on the side of the concentration, meaning that $k$ is the same as the second constant $b$. They do not make any reference to the phase of the reaction, so it should probably be assumed that they intended it to hold for any general phase. (If anyone knows of a more specialised definition, please point it out!)

order of reaction, $n$ (DOI: 10.1351/goldbook.O04322)
If the macroscopic (observed, empirical or phenomenological) rate of reaction ($nu$) for any reaction can be expressed by an empirical differential rate equation (or rate law) which contains a factor of the form $k[\ce{A}]^\alpha[\ce{B}]^\beta\cdots$ (expressing in full the dependence of the rate of reaction on the concentrations $[\ce{A}]$, $[\ce{B}], \cdots$)
[...]
The proportionality factor $k$ above is called the ($n$th order) 'rate coefficient'. Rate coefficients referring to (or believed to refer to) elementary reactions are called 'rate constants' or, more appropriately 'microscopic' (hypothetical, mechanistic) rate constants.

This approach is quite consistent throughout. For example, the rate of reaction is also expressed in terms of concentrations, at least under constant-volume conditions:

For the general chemical reaction:
$$\ce{aA + bB -> pP + qQ + \cdots}$$
occurring under constant-volume conditions, without an appreciable build-up of reaction intermediates, the rate of reaction $\nu$ is defined as:
$$\nu = -\frac{1}{a}\frac{\mathrm{d}[\ce{A}]}{\mathrm{dt}} = -\frac{1}{b}\frac{\mathrm{d}[\ce{B}]}{\mathrm{dt}} = \frac{1}{p}\frac{\mathrm{d}[\ce{P}]}{\mathrm{dt}} = \frac{1}{q}\frac{\mathrm{d}[\ce{Q}]}{\mathrm{dt}}$$
The quantity $$\dot{\xi} = \frac{\mathrm{d}\xi}{\mathrm{d}t}$$ defined by the equation:
$$\dot{\xi} = -\frac{1}{a}\frac{\mathrm{d}n_\ce{A}}{\mathrm{dt}} = -\frac{1}{b}\frac{\mathrm{d}n_\ce{B}}{\mathrm{dt}} = \frac{1}{p}\frac{\mathrm{d}n_\ce{P}}{\mathrm{dt}} = \frac{1}{q}\frac{\mathrm{d}n_\ce{Q}}{\mathrm{dt}}$$
(where $n_\ce{A}$ designates the amount of substance $\ce{A}$, conventionally expressed in units of mole) may be called the 'rate of conversion' and is appropriate when the use of concentrations is inconvenient, e.g. under conditions of varying volume.

One tangential question I can anticipate is: what's the difference between the expressions $r = -(1/a)(\mathrm{d}[\ce{A}]/\mathrm{d}t)$ and $r = k[\ce{A}]$? The answer is: the former is a definition. The rate is literally defined by how fast the reactants disappear, which is a universal truth, applicable to all reactions. The latter is an empirically observed expression, which only holds true for a small subset of reactions (i.e. first-order ones), and it says that the rate at which the reactant disappears is directly proportional to its concentration.

@Orthocresol   Can we conclude that  $r = -(1/a)(\mathrm{d}[\ce{A}]/\mathrm{d}t) = k[\ce{A}]$

@AdnanAL-Amleh Well, if (and only if) that first-order dependence is indeed observed, then yes you can equate them: and then you can integrate with respect to $t$ to get $[\ce{A}]_t = [\ce{A}]_0 \mathrm{e}^{-akt}$. If $\ce{A}$ is the only reactant, then it's quite typical to set its stoichiometric coefficient $a$ to be just $1$. The resulting equation should be familiar to you if you have studied the integrated rate equations.

@orthocresol Thank you for the answer! More specifically, if the reaction rate is given to the reader as bar/hour, would this imply that the rate would be written in terms of partial pressure? I'm confused because P and n/V are just a factor of RT away.

@ilovechemistry You could always convert between pressure and concentration as you say. But if the rate is given in terms of pressures, then it's by far easier to just stick to pressures throughout. It won't be the IUPAC-sanctioned "rate coefficient", but it's just a proportionality factor, so not exactly a huge deal (and doesn't affect the physical interpretation of the system).


text:
Why negative charge delocalises away from electronegative atoms
In the mechanism for the Wolff-Kishner reduction, after the $\ce{OH-}$ abstracts a proton, the resulting negative charge delocalises away from the nitrogen atom to the carbon, and the subsequent protonation occurs to the carbon.

Why does the negative charge delocalise away from the more electronegative nitrogen to carbon, for ostensibly a longer time period (for the chances of the proton settling with the carbon to be high)?

How does this happen, when the molecule that results from protonation here doesn't have resonance stabilisation, but there is resonance if it was on Nitrogen that the proton attached?

A similar question arises in Aldol condensation; why is the carbanion the molecule that moves the reaction forward and not the enolate? Since the enolate has the negative charge on the oxygen, I would have assumed it would be the prevalent species in the reaction mixture.

Remember that these steps are reversible and not irreversible as shown. The hydrazone anion likely has most of the negative charge on nitrogen. If it gets protonated on nitrogen, the deprotonation starts over again. Only C-protonation leads to the diimide which leads to the W-K product. As to the aldol condensation, how do we know that the oxygen of the enolate doesn't add to another carbonyl group? Such a species does not lead to productive products and is likely reversible. Reaction at carbon leads to a productive pathway --- the aldol condensation.

@user55119: oh, yeah, the W-K question is meaningless, got it. For aldol condensation, I assume you mean the reaction conditions are tweaked so as not to let these products form. Is that so?

I don't think you should consider the negative charge as "delocalising away". It doesn't happen like that, once you deprotonate, the negative charge becomes delocalised over the whole pi-system, so that there is some partial negative charge on C and some partial negative charge on N. Now the proton can be attracted to either of those, if it goes to N, it just returns to the original structure, and if it goes to C you get the final structure that you drew.

@ShoubhikRMaiti: I get the logic for the Wolff-Kishner case; that it would be the same product either way; that's what I meant by my comment. Could you elaborate on the aldol case?

On the aldol case too the negative charge is delocalised between the oxygen and alpha-carbon. Different electrophiles prefer different ends. Traditionally this is explained with HSAB theory, where carbon is the soft nucleophile and attacks softer electrophiles whereas the oxygen is hard nucleophile and attacks hard electrophiles. Examples of hard electrophiles include Si in $\ce{Me3SiCl}$, Me in $\ce{Me3O+}$ etc. In those cases the product forms where the oxygen attacks the electrophile

@ShoubhikRMaiti: the explanation's great, but it's just that apparently there's this.

Yes of course there are problems with each piece of theory that tries to predict a lot of things. You can either deal with each problem case by case, or you can use a rule of thumb which may not work everytime. HSAB is that: it's a rule of thumb, and it's still used in chemistry quite a lot, regardless of what one research paper says.


text:
What does acidity of para-alkyl substituted phenols depend on?
Problem

Arrange the following organic compounds in decreasing order of acidity.

Answer

 S, R, Q, P.

Question
According to my knowledge the answer should be P, Q, R, S. As we proceed from P to S we are increasing the number of methyl groups, and so the number of electron donating groups increases, decreasing the polarity of of the C−O bond, which is the opposite to the answer. This obviously means that whatever I'm thinking is the right reason is the wrong logic, but I'm not sure how.

The major factor is hyperconjugation and not inductive effect.

The reason provided by ML cannot explain the difference in acidities since the +I effect doesn't come into play normally for the para positions.
A more likely reason is hyperconjugation (or no bond resonance) where the number of hydrogens directly attached to the benzyl carbon decreases as it goes from P to S.
This means that the number of destabilizing structures that are produced via this resonance decreases as you go down the list. This leads to greater stability in the deprotonated form compared to each other.
Therefore the correct order becomes S, R, Q, P.

For alkyl groups attached to the phenol ring, there are two main points that your reasoning does not take account of -

Inductive effect is distance dependent. As the distance increases, its effect becomes very small.
Even if the group were near, the general trend is that hyperconjugation has stronger effect than inductive effect.

You can refer the following links for an explanation on hyperconjugation -
https://chem.libretexts.org/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hyperconjugation
https://en.wikipedia.org/wiki/Hyperconjugation
More the number of hydrogen atoms attached to the benzyl carbon, more is the number of hyperconjugative structures possible and more is the withdrawal of positive charge (or equivalently, donation of negative charge) into the benzene ring.
Phenol forms phenoxide ion on losing H+ and the phenoxide tries to delocalize its negative charge into the ring. If hyperconjugation already donates electrons into the ring, the extent of delocalization decreases, making phenoxide less stable and hence phenol less acidic.
As a result, less the number of benzylic hydrogen, more is the stability of phenoxide and more the acidity. That explains the order S > R > Q > P.
NOTE : Hyperconjugation, like resonance (mesomeric effect), operates only at the ortho and para positions. In case the same groups were attached at meta, then hyperconjugation would have no role and your order based on inductive effect would be correct.

Are you implicitly saying hyperconjugation dominates the delocalization of electron density of O? Why should hyperconjugation even take place when a phenoxide ion is required to be formed? Your answer makes sense, but I have a doubt about this now.

I meant to say that due to hyperconjugation the electron density in the benzene ring increases. As a result the extent of delocalization of electron density of the O atom is decreased. As to why hyperconjugation takes place, it can't be controlled (as far as I know) and takes place regardless of any other substituents.


text:
If an atom had 200 protons, how much neutrons would it need to have to be most likely stable and why?
I know there is a specific proton/neutron ratio but i'm unaware if it applies to atoms with 86 protons and beyond.

Such extrapolation is a moot point as nobody would be able to synthetise it. If magically succeeded, It would have t1/2 like 10^-23 s, decaying probably to 3 fractions.

BTW, it would fit better to Physics SE.

Theoretical physics states that no atom can have more than $137$ protons.

Related the-last-elements-atomic-number

Uh ... $10^{57}$?


text:
What does the horizontal axis on cyclic voltammograms mean?
In cyclic voltammograms (like the one in the attached image) what exactly is the horizontal axis' voltage refering to? Is it the measured voltage between the reference electrode and working electrode? If so, and assuming the measurements are made with a standard voltmeter, which electrode is considered the cathode and which is considered the anode (as in, which is attached to the positive lead and which is attached to the negative lead)?
I was also wondering if these voltages would represent standard electrode potentials or would that depend on the conditions of the data collection

The best is to review the context of the chart in the original source, as we can just guess.

Of course is the potential at which the working electrode is kept at that moment of the swap. What else it could/shall be? The working electrode can be the cathode or anode, depending on the process going on. The voltages are those against your reference, it can be a standard or a quasi standard or whatever, electrode calibrated versus a well known redox system. For the actual values plotted, one can choose ref. and state it. Figures have usually captions, and are rarely appended to vacuum but comes with a text. I do not think this deserves an answer other than this comment.

Please look at my potentiostat answer here. The working electrode is clamped at virtual ground, i.e., held to less than a millivolt or so of ground (circuit zero potential reference). A voltmeter would not be attached between the reference and working electrodes! So the abscissa is the potential relevant to the electrochemically active species, i.e, going more positive is the oxidative direction and going more negative is the reductive direction.

You may find many helpful information in the handy primer «A Practical Beginner’s Guide to Cyclic Voltammetry» by Elgrishi et al. (e.g. experimental set up, IUPAC vs. US convention to plot a CV, etc) in 2018JChemEduc197.  It is one of the journal's most read publications and open access / available without institutional subscription.


text:
Electrolysis for rust removal: is "line of sight" a showstopper?
So I am using electrolysis to remove rust from a fuel tank.  The tank is the cathode and a sacrificial anode is inserted where the fuel pump normally goes.  It's not practical to put it anywhere else.
However, the fuel tank has internal baffles, apparently for anti-slosh.  They have continuity with the tank, i.e. they are part of the cathode.
It cannot be denied that the the tank is getting a lot cleaner, and formerly-rusty areas are becoming shiny.  And it's difficult to tell for sure, but subjectively, it seems like it's not doing so well in parts of the tank placed "in shadow" by the baffle, i.e. the baffle interrupts line-of-sight to the anode.  This is more obvious where there are holes in the baffle.

My setup, and what seems to be happening.
So my question is: Is line-of-sight a requirement for electrolysis?  Does electrolysis happen through the electrolyte (acting on all wetted surfaces) or does it only use the most direct path between anode and cathode?
Electrical current takes all paths in proportion to their conductance... so will the path through the electrolyte around the baffle eventually have effect?

Direct visibility is not required but the cell geometry may significantly affect the electrostatic potential field, electromigration, diffusion, local availability of active components and reaction rates.

Your setup looks like the baffle will prevent adequate treatment of the tank wall. Have you considered something like a rust conversion treatment, like Naval Jelly, which you could slosh around (if you buy enough)? It's mostly phosphoric acid with a thickener.


text:
Calculating the ionic strength of a histidine solution
I'm trying to repeat the calculation of the ionic strength of a solution containing histidine as a buffer I recently read in a paper. In their paper, the authors calculate the ionic strength of a 20 mM solution of histidine at pH 5.8 to be 7.7 mM. They also state that the average charge of Histidine at this pH is equal to + 0.61.
For my calculation I used to Henderson-Hasselbalch equation to calculate the distribution of charged species at the pH specified,
pH = np.arange(0.1, 14, 0.001)

def macroscopic(pH, pka1, pka2, pka3):
    return ((-1)**2 + 0 * 10**(pka1 - pH) + (1)**2 * 10**(pka1+pka2 - 2*pH) + (2)**2 * 10**(pka1+pka2+pka3 - 3*pH)) / (1+10**(pka1 - pH) + 10**(pka1+pka2 - 2*pH) + 10**(pka1+pka2+pka3 - 3*pH))
def DDD(pH, pka1, pka2, pka3):
    return 1 / (1+10**(pka1 - pH) + 10**(pka1+pka2 - 2*pH) + 10**(pka1+pka2+pka3 - 3*pH))
def PDD(pH, pka1, pka2, pka3):
    return 10**(pka1 - pH) / (1+10**(pka1 - pH) + 10**(pka1+pka2 - 2*pH) + 10**(pka1+pka2+pka3 - 3*pH))
def PPD(pH, pka1, pka2, pka3):
    return 10**(pka1+pka2 - 2*pH) / (1+10**(pka1 - pH) + 10**(pka1+pka2 - 2*pH) + 10**(pka1+pka2+pka3 - 3*pH))
def PPP(pH, pka1, pka2, pka3):
    return 10**(pka1+pka2+pka3 - 3*pH) / (1+10**(pka1 - pH) + 10**(pka1+pka2 - 2*pH) + 10**(pka1+pka2+pka3 - 3*pH))'''

Where the values of pKa1, 2 and 3, used by the authors, are equal to 9.17, 6.0 and 1.82 respectively.
I then calculate the ionic strenght of the solution according to
def ionicstrength(pH, pka1, pka2, pka3):
    return 0.5*(DDD(pH, 9.17, 6.0, 1.82)*BuffermMConc*(-1)**2 + PDD(pH, 9.17, 6.0, 1.82)*BuffermMConc*(0)**2 + PPD(pH, 9.17, 6.0, 1.82)*BuffermMConc*(1)**2 + PPP(pH, 9.17, 6.0, 1.82)*BuffermMConc*(2)**2)
#def averagecharge(pH, pka1, pka2, pka3)

where the variable BuffermMConc is equal to the mM concentration of the buffer, in this case 20.
When I plot the results I get the following graph,

Where the dashed line represents the ionic strength and the coloured lines the distribution of histidine species. For pH 5.8 I obtain an ionic strength of 6.13 mM, which compares to the ionic strength of 7.7mM determined by the authors. The authors also report the average charge of histidine, which is identical to the value I determined. This would indicate that the distribution of charged species is probably correct in my calculation. I cant  figure out why my ionic strength calculation is different.
Any help would be greatly appreciated.


text:
First or second order reaction
I have that $100$ mM of K$_3$Fe(CN)$_6$ is dissolved in equimolar of the organic ion (which I assume is HCN). In the solution Fe(CN)$_6$$^{3-}$ reacts to form Fe(CN)$_6$$^{4-}$.
The formation of Fe(CN)$_6$$^{4-}$ as a function of time is measured in a Na$_2$SO$_4$ (which does not take part in the reaction) solution of different concentrations.
One of the concentrations of Na$_2$SO$_4$ = $0$ and I want to determine the rate constant from this and:
$$\begin{array}{c|c}
t/\pu{min} & \ce{[Fe(CN)_{6}^4-]}/\pu{mM} \\ \hline
10\ & 20.3\\
20 & 33.7\\
30\ & 43.3\\
40\ & 50.4 \\
50\ & 56\\
60\ & 60.4 \\
\hline
\end{array}$$
I assumed that the reaction would be second order since the reactants are equimolar and so I plotted using the equation:
$$ \frac{1}{[A]}= kt + \frac{1}{[A_0]}$$
When I plotted using this I get that the linear equation is: $−5.78×10^{−4}*x + 0.0463$
and R$^2$ = $0.771$.
So I tried using a first order equation and got that the linear equation is: $0.0204*x + 3.01$
and R$^2$ = $0.878$.
Doesn't this mean that my reaction should actually be first order? I don't see how it could be that and would appreciate it if someone could tell me if it is first or second or what I am doing wrong. Thank you!
EDIT
I did plot the curves and now when I look at them I do think that the first order line is a bit curved while the second order is more straight. It is really hard to see and my professor usually looks at the R-value which is why I thought that would help me determine the order here. But does that mean the reaction is second order?
Here are my graphs:

It is always a good idea to plot data before any curve fitting is done. Perhaps plot your data and see what it may reveal.

You could also try doing a similar analysis to check if your points correspond to a first order reaction. Also, what makes you think that your organic ion is HCN? Thinking about the process would get you closer to understanding what's happening here.

Using $R^2$ is a very, very  poor discriminator of the goodness of a fit, basically never use it. Look at fit and see it is curved vs the data, then plot residuals, (calc data-real data)/real data  for example and see if they are random, look at size of  error in gradient and error in intercept.

@EdV I have added my plots in my question now but I am still a bit unsure of the order..

@Ezze I really just went on a similar problem that I had seen where RuCl$_2$ and FeCl$_2$ were dissolved and then the organic ion was HCl. I didn't really find anything online and thought that it probably should work with HCN. Am I wrong?

I'd be really surprised if either HCl or HCN could be called an organic ion, and I hope you agree with me on this one. The wording of the exercise is a bit weird for sure, but to me it looks like you might want to work assuming pseudo first order conditions.

neither fit is good. You are measuring the appearance of product if it is first order then $\displaystyle [P]=a_0(1-e^{-kt})] $ where $a_0$ is initial amount of reactant and $k$ rate constant.

Use the 2nd order scheme $2A \to B$ with initial amount of $A=A_0$ so $B=A_0-A$ and B is the product you observe.  The rate of change is $\displaystyle \frac{1}{2}\frac{dA}{dt}=-k_2A^2$ and I'm not going to use [ ] for concentrations just $A,\,B,\,A_0$. Also the 2 in the equation is incorporated into the rate constant just for convenience.
Solving the equation gives $\displaystyle A=\frac{A_0}{A_0k_2t+1}$ and so the product follows $\displaystyle B=\frac{A_0^2k_2t}{A_0k_2t+1}$. This is shown as the red line in the figure below. I just guessed the rate constant a few time to fit  and as you can see the fit is good. You can also linearise this equation by isolating $A_0k_2t$.
The first order plot is also shown (grey line) and this is poor. $B=A_0(1-e^{-kt}$)


text:
The effect of dilution on the rate of a system in equilibrium
I'm attempting to answer a question from my textbook, however the answer that the textbook provides doesn't make sense to me.
The question:

Laura, a VCE chemistry student, loves graphs and wanted to represent an equilibrium reaction by using graphs. The reaction she chose was:
$\ce{Fe^{3+}(aq) + SCN-(aq) <=> FeSCN^{2+}(aq)  \ \ \ \ \ \ \Delta H=-\nu e}$
Draw the rate-time graph for the reaction when the system is diluted at constant temperature.

In my mind, this dilution would instantly decrease the concentration of the reactants and products. But as it decreases the concentration of each species present by the same factor this will not disturb equilibrium, and simply result in a slower rate of reaction.
Given my assumptions I drew a graph where the forwards and reverse reaction both dropped vertically then continued on at the same, lower rate of reaction in equilibrium.
But the answer provided by the textbook is as follows

I just came across another question similar to this, and supposedly the system partially opposes the increase in concentration by favouring the reaction that produces the least amount of particles (in the same way as a change in pressure would cause the system to react). But if this is the case, and the concentrations are still balanced after the dilution, wouldn't it change the equilibrium constant? Which I have been told only temperature can change.
I'm assuming then that my understanding is flawed somewhere in regards to the system remaining balanced after the dilution, but I can't seem to see how diluting the system would unbalance the concentrations as they're all being decreased by the same factor.
Is the answer provided by the textbook wrong? Or is there something that I'm missing?

the value of $K_\mathrm{eq}$ doesn't change but $\ce{[Fe^3+]}$, $\ce{[SCN-]}$ and $\ce{[FeSCN^2+]}$ all change. Unless the change is very precise, there would be a change in point of equilibrium..

The rate forwards and back are the same at equilibrium  and differ before this is attained. Let the reaction be $A+B=C$. If each concentration is suddenly halved then the reaction is not necessarily any longer at equilibrium and in this case the rates forwards and back differ until a new equilibrium is reached. The figure shows this. The red curve is the forwards rate $k_1[A][B]$ and the green one $k_{-1}[C]$ the reverse. As you see the new equilibrium takes a short while to re-establish. (You can ignore the scales they depend on the particular rate constants and concentrations used but the behaviour is general.)
The figure you show is not fully supported by the calculation. The red curve (forwards rate)  agrees but the blue one does not.  If you do not halve the product concentration you get the  blue curve in the figure.

@ porphyrin :Do the rate of the back reaction more than the rate of forward reaction before new equilibrium ?Why?

@AdnanAL-Amleh ...because the number of particles is higher for the back reaction (2) compared to the forward reaction (1).

Never mind I've figured it out. I was previously thinking that the ratios of the concentrations were all that mattered. But looking at the equilibrium constant of any equation where there is an imbalance of moles on either side of the equation, that isn't the case.
The concentration ratios will remain constant after the dilution, but that doesn't necessarily mean the system will remain in equilibrium.

"If you do not halve the product concentration you get the blue curve in the figure." Are you referring to the OPs figure?

yes, the OP's figure


text:
Can we use silver nitrate to distinguish between chloride and carbonate salts?
Initially, we are given a solution containing two salts, they can be either carbonate or chloride salts (we do not know their composition initially).
Generally, $\ce{AgNO3}$ is used as a confirmatory test of $\ce{Cl-}$ (a white precipitate is obtained on adding silver nitrate to solution containing the chloride salt).
$$\ce{NaCl + AgNO3 -> NaNO3 + \underset{white}{AgCl} \downarrow}$$
But
$$\ce{Na2CO3 + 2AgNO3 ->
 2NaNO3 + \underset{brownish white}{Ag2CO3} \downarrow}$$
Therefore, can we use $\ce{AgNO3}$ to directly identify chloride in the solution or do we need to remove the possible carbonate ions to confirm the presence of chloride before performing the test as mentioned above?
(A similar question may arise in the identification of bromide and iodide ions but I have ignored it for now as they are rarely asked in exams. Still, I would appreciate it if someone could answer this as well.)
The question has been asked mainly because the silver nitrate test is generally taught to be performed without mentioning the above precautions and it is also a very famous confirmative test.

You should work on your English sentences a little bit anyway, If you are asking how to test for $Cl^-$ reliably then ,You can add dil. $HNO_3$ to move the equilibrium to the carbonic acid side through application of Le-Chatelier Principle so as to have minimal [${CO_3}^{2-}$]in the solution which would not bring about any unreliability in the confirmation of $Cl^-$

Yes, Moreover I think there will be some other ways as well for this but to me this is the simplest.

My general understanding of your question is you are given sodium chloride and sodium carbonate and you are to react them with silver nitrate in order to distinguish them. Right? In that case, it might work. Silver carbonate becomes more and more yellow and brown when excess of silver nitrate is used. Silver chloride remains white. Also, silver carbonate is soluble in nitric acid whereas silver chloride is not.

The test for Chloride ion is silver nitrate in an Acidified solution [with nitric acid]. This removes carbonate (as carbonic acid), cyanide[careful] and sulfide (as hydrogen sulfide). (Use a well-ventilated hood as the products of acidifying both sulfide and cyanide are poisonous.) Sulfate must be absent, if present it can be removed with barium nitrate (as barium sulfate).  The silver chloride formed is solubilized by addition of ammonia; insoluble precipitate remaining may be bromide or iodide. The exact procedure to follow is given in reputed indepth Qualitative Analysis lab books.


text:
In an equilibrium, why are partial pressures not affected by the coefficients of the balanced equation?
It's probably easiest for me to explain in an example. For the following reaction: $$\ce{2NOBr(g) <=> 2NO(g) + Br2(g)}$$
At equilibrium, if the partial pressure of $\ce{NOBr}$ was 6 atm, the partial pressure of $\ce{NO}$ was 4 atm, and the partial pressure of $\ce{Br2}$ was 3 atm. Why is the equilibrium constant: $$Kp = \frac{3\cdot 4^2}{6^2}$$ instead of the individual pressures multiplied by their coefficients like in molar solubility: $$Kp = \frac{3\cdot \left(2\cdot 4\right)^2}{\left(2\cdot 6\right)^2}$$

And your reasoning behind the latter way is .. ?

@Poutnik Like how solving for molar solubility of Ag2CO3 would give Ksp = (2x)^2 * x

But the 2 in () is there not because of coefficient itself, but because there is twice as big equilibrium concentration because of that coefficient, assuming all comes from Ag2CO3 dissolution.  If concentrations are general, with external ion source, coefficients play no role in (), just in the exponent.

One way of thinking about this is writing the original equation
$$\ce{2NOBr(g) <=> 2NO(g) + Br2(g)}$$
as
$$\ce{NOBr(g) + NOBr(g)  <=> NO(g) +NO(g) +  Br2(g)}$$
Now it should be clear you have to multiply, not add concentrations.


text:
pH of an aqueous solution with acetic acid and sodium acetate

An aqueous solution of $\pu{1L}$ contains $\pu{0.2 mol}$ of acetic acid and $\pu{0.2 mol}$ of sodium acetate ($K_\mathrm{a} = 1.8 \times 10^{-5}$). Compute the $\mathrm{pH}$ .

I'm trying to help my little sister in solving this exercise... however, a lot of years passed away since the time I attended the school...
Our (unsuccessful) attempt:
$\ce{CH3COOH <=> H+ + CH3COO-}$
$K_\mathrm{a} = \frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \frac{x\cdot x}{0.2} = 1.8 \cdot 10^{-5} \rightarrow \mathrm{pH} = 2.72$
Then
$\ce{CH3COONa -> Na+ + CH_3COO-}$
$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$
\begin{align}
K_\mathrm{b} = \frac {K_\mathrm{w}}{K_\mathrm{a}} &= 5.56 \cdot 10^{-10} \\
 &= \frac{[\ce{OH-}][\ce{CH3COOH}]}{[\ce{CH3COO-}]}  = \frac{x^2}{0.2} \\
 \implies \mathrm{pOH} &= 4.97 \\
\implies \mathrm{pH} &= 9.02
\end{align}
Hence, $\mathrm{pH}_\text{solution}= (9.02+2.72)/2 = 5.87$
However, the result should be $4.74$.

It is much simpler, if you realize that the ratio of concentrations of acetic acid and acetate anion is 1, so Ka=[H+].

Because the acid and conjugate bases are equimolar - and because the equilibrium constant is small, obviating the need to solve a quadratic if you approached this in another way - the Henderson-Hasselbalch equation is:
$$\pu{pH} = {\rm p}K_\mathrm{a} + {\rm log}_{10}\left({[A^-]\over[HA]}\right)$$
$$\pu{pH} = 4.74 + {\rm log}_{10}\left({0.2\ {\rm M}\over 0.2\ {\rm M}}\right)$$
which reduces to
$$\pu{pH} = 4.74$$


text:
Does the 3D conformation of rings affect chirality?
During the chirality analysis of various molecules containing rings, we assume for convenience that the rings are completely flat/planar and if the molecule has a plane of symmetry, it is achiral. However, does the fact that say, 6-membered rings actually exist in a chair or boat conformation have any effect on our final answer (it has a plane of symmetry if it is completely planar, but does not if it is in its 3D conformation)?
I'm a little confused because the odd, puckered conformation of various rings don't seem like they would have a plane of symmetry. Take for example, the molecule below. In the 2D drawing, we see 1 plane of symmetry, it goes through $\ce {Br}$ and cuts through the ring in a perpendicular fashion. However, in the 3D drawing, the plane we identified earlier doesn't seem like a plane of symmetry anymore.

Of course, in the 2D picture, the molecules would seem to be equivalent no matter if the bond were wedged or dashed, but why do they appear to be different if they are drawn in 3D?

Not sure whether your 3D drawings are accurate, but that doesn't matter anyway. In short, if the structures appear to be different in 3D, then they are different, but that doesn't matter either, because they easily change from one conformation to another.

Any chiral conformation of bromocyclopentane is matched by an equal population of its enantiomer. Thus, there is a racemate.

No matter how you turn or twist these rings to change from one conformer to an other: the configuration of the stereogenic center remains either (R), or (S) and the atom's connectivity does not change.  Some conformations may not be easy accessible (higher energy, thus less likely populated [in parlance of statistical thermodynamics]). But a change of the configuration between (R) and (S) of a stereogenic center however is a change of constitution, only possible by breaking old and establishing new bonds (on/off binary) costing more energy than the wiggles and twists about conformers.

@user55119 How does it work so that the ever fluctuating concentrations have equal concentrations of enantiomers at all times? I may be interpreting your comment incorrectly but wouldn't this mean bromocyclopentane is chiral (even though it's racemic), which isn't true?

You can draw bromocyclopentane with 4 ring carbons in a plane with C5-Br as the "flap of the envelope". With the flap up or down, there are two achiral conformations having a plane of symmetry. Any other conformation, and there many of them, is chiral as a pair of enantiomeric conformations. All of these conformations are rapidly interconverting. It is the chiral conformations that prevail and, as racemates, account for the majority of the  lack of optical rotation. The achiral conformations, having a plane of symmetry, also do not contribute to optical rotation.

...continued. Your 2nd structure is chiral. But there is an equal amount of its isoenergetic mirror image. The same is true of the 3rd structure. Go back to meso-tartaric acid in the eclipsed conformation having a plane of symmetry. Rotate CW about the C2-C3 bond by 60 deg. giving a chiral conformation of a meso cmpd. It is statistically probable that rotation occurs in the CCW direction, forming the mirror image. Now you have a racemate of conformations. This may help: https://chemistry.stackexchange.com/questions/48995/is-2r-3s-butane-2-3-diol-chiral/87415#87415

Wonky ring structures could, in principle, create chirality, but only if you could force the ring to get stuck in a wonky conformation. This might be possible with very bulky groups that could force one side of the ring to bend differently to the other. But how to do that with symmetrical bulky substituents (if they are not symmetrical then the molecule might already be chiral ignoring the ring conformation).

And, the other problem with your example, is that the rings have a plane of symmetry even in chair and boat-like confirmations. The plane is perpendicular to the ring: we don't have to pretend the ring is flat.


text:
Regarding Hard and Soft nucleophiles

In the above reaction, I was told that the reason organocadmium reacts only with aryl chloride segment is because it is very less reactive and reacts only with most reactive sites. This was consistent with my previous beliefs that we always react most electrophilic and most nucleophilic reagents.

In this reaction also there are two electrophilic sites and I was expecting a major product from carbonyl site but opposite was true. The explanation was given that soft nucleophile prefers soft electrophiles while hard nucleophiles prefer hard electrophiles. The explanation seemed credible to me but contradictory to previous one.
Where am I going wrong?
I don't know if I have made my doubt clear enough. I am new to the community and any suggestions on improving the questions are welcome.

HSAB theory is pretty dodgy. The "true" explanations for some of these things are usually rather more complicated, so people came up with this notion of "hard" and "soft" to explain some phenomena. The problem is that it's quite tautological: the HSAB concept only seems to make sense when applied precisely to these phenomena. Take it with a huge pinch of salt whenever you see it, and don't try to stretch it beyond the contexts that it's applied to. See e.g. https://onlinelibrary.wiley.com/doi/full/10.1002/anie.201007100 for a refutation of HSAB arguments.

For what it's worth, there's a good underlying question in here, i.e. why does HSAB theory fail in some cases. It's a shame that the example used, i.e. Me2Cd plus this 1,4-aromatic, is made up. If someone can find a real example, that would make this much better. (I strongly suspect that organocadmiums have been out of favour for... a while now.)


text:
Will dioxygen difluoride set fire to sand?
I've heard that dioxygen difluoride (a.k.a FOOF) would set fire or react with nearly everything including sand ,bricks or even ice. Is this true?

It generally explodes first. But chlorine trifluoride burns all that stuff, which is a reason it was used in V-2 rockets in WW II.

@EdV ClF3 was not used in V2 rockets though it was investigated as a self-igniting flamethrower. the problem was it tended to ignite the user, the apparatus, the manufacturing equipment and everything else, making it a little impractical.

@matt_black Thanks, I will have to check my source, which I think was John D. Clark’s Ignition! book. Always fun to go through again.

V2 propelent was ethanol + liquid oxygen. But ClF3 can light up most of stuff, like sand, glass, concrete or human body.

@Poutnik I was thinking of chlorine trifluoride as the Hell on Earth payload. Just that.

@EdV Ignition was my source too! And always worth reading for the chemist's equivalent of bedtime story books that warn kinds what not to do.

I don't think this is an exact duplicate, though it clearly overlaps a little.

@Bruh Self-vandalism of posts is frowned upon. If you really want to edit/add something, please go ahead but be aware of doing a "constructive editing". If you want this question to reopen, explain how this question is different from the dupe question (in the question body and not in comment section). Otherwise, leave it as it is.

It is hard to be sure as it is too reactive to test
FOOF has been made, though no sane person would want to repeat the most recent synthesis (see the paywalled article reporting it) as it involves passing oxygen a fluorine through a red hot metal tube. An earlier paper that tested its reactivity stored it as a solid at 90K to stop it from, well, exploding. Derek Lowe, in his hilarious blog "Things I Won't Work With" describes some of the (very low temperature) reactions:

And he’s just getting warmed up, if that’s the right phrase to use for something that detonates things at -180C (that’s -300 Fahrenheit, if you only have a kitchen thermometer). The great majority of Streng’s reactions have surely never been run again. The paper goes on to react FOOF with everything else you wouldn’t react it with: ammonia (“vigorous”, this at 100K), water ice (explosion, natch), chlorine (“violent explosion”, so he added it more slowly the second time), red phosphorus (not good), bromine fluoride, chlorine trifluoride (say what?), perchloryl fluoride (!), tetrafluorohydrazine (how on Earth. . .), and on, and on. If the paper weren’t laid out in complete grammatical sentences and published in JACS, you’d swear it was the work of a violent lunatic. I ran out of vulgar expletives after the second page. A. G. Streng, folks, absolutely takes the corrosive exploding cake, and I have to tip my asbestos-lined titanium hat to him.

Many other reactivity experiments were not possible because they are just too impractical with something so dangerously explosive at temperatures any normal person might want to conduct reactions at (anything much above liquid nitrogen is bad for FOOF).
So, I suspect, we don't know whether it would set fire to sand as nobody has ever been able to make enough of it to test that reactivity, or, at least, test it and survive.
FOOF's–far more benign–relative chlorine trifluoride, which is commercially available and has several uses, most certainly does set sand on fire. As rocket scientists know. This account is from John D Clarke's Ignition:

It is hypergolic with every known fuel, and so rapidly hypergolic that no ignition delay has ever been measured. It is also hypergolic with such things as cloth, wood, and test engineers, not to mention asbestos, sand, and water—with which it reacts explosively.

So, given ClF3 is far less reactive than FOOF, it seems likely that FOOF, too, would set fire to sand. But with any luck we will never know for sure as nobody will be so monumentally, suicidally insane to try making enough of it to find out.

(+1) Ah, wonderful answer! This is a nice example of what a high quality answer looks like! And Derek Lowe’s blog is sweet: the iron man PPE and Bunsen’s cacodyl cyanide episode are hilarious in a chemistry way!


text:
Does steam from nuclear reactor does work on the system or to the system?
I was doing research on nuclear reactors and was wondering when the steam from the fission pushes the steam turbine does it do work on the system or to the system?

What are your thoughts?

Do you work on anything or to anything ?


text:
Internal Energy Changes in Adiabatic Reactor
If we consider adiabatic reactor (one which doesn't exchange energy and matter with surroundings) with some reaction in liquid phase, energy balance for such reactor is written as heat of reaction being equal to time derivative of enthalpy at any point in time. Since reaction is in liquid phase during reaction there are negligible changes in volume and pressure of reactor so that change in enthalpy is approximately equal to change in internal energy. Problem here is that how can internal energy (or enthalpy in this case) change in adiabatic reactor since it doesn't exchange energy with surroundings? Internal energy is a sum of all microscopic kinetic energy and potential energy of interaction of all bonds in the system. During chemical reaction interatomic potential energy is converted to microscopic kinetic energy (or vice versa) because of which temperature changes, but change in each type of energy must be equal due to energy conservation. Since this is the case internal energy doesn't change after reaction in adiabatic reactor. If so, why does equation of energy balance say it does change as temperature of system is changed during reaction?

Here is how it is properly supposed to play out for the reaction in your video:
Assuming an ideal solution, the enthalpy of the reactor contents can be represented as:
$$H=n_Ah_a+n_Bh_B$$ where the n's are the numbers of moles of the two species and the h's are the enthalpies of the pure species at the same temperature as the reactor mixture:
$$h_A(T)=h_A(T^0)+ C_{PA}(T-T^0)$$
$$h_B(T)=h_B(T^0)+ C_{PB}(T-T^0)$$where $T_0$ is a reference temperature typically taken as 298 K and $h(T^0)$ is so-called heat of formation of the species (from the pure elements at $T^0$).  Taking the derivative of the reaction mixture enthalpy with respect to time yields:
$$\frac{dH}{dt}=n_A\frac{dh_A}{dt}+n_B\frac{dh_B}{dt}+h_A\frac{dn_A}{dt}+h_B\frac{dn_B}{dt}$$But, $$n_A\frac{dh_A}{dt}+n_B\frac{dh_B}{dt}=(n_AC_{PA}+n_BC_{PB})\frac{dT}{dt}$$and$$\frac{dn_B}{dt}=-\frac{dn_A}{dt}=r$$where r is the rate of reaction.  Therefore, $$\frac{dH}{dt}=(n_AC_{PA}+n_BC_{PB})\frac{dT}{dt}+r(h_B(T)-h_A(T))$$But, from Hess' law, the heat of reaction at temperature T is $$\Delta H_r(T)=h_B(T)-h_A(T)=h_B(T^0)-h_A(T_0)+(C_{PB}-C_{PA})(T-T^0)$$Therefore, for the reaction mixture, $$\frac{dH}{dt}=(n_AC_{PA}+n_BC_{PB})\frac{dT}{dt}+r\Delta H_r(T)$$If, for an adiabatic reactor, we set this equal to zero, we obtain:
$$0=(n_AC_{PA}+n_BC_{PB})\frac{dT}{dt}+r\Delta H_r(T)$$or$$(n_AC_{PA}+n_BC_{PB})\frac{dT}{dt}=r(-\Delta H_r(T))$$This is basically the same equation as in the video

Thank you. I would like if people would derive energy balances for reactors in more detail (like you did here) and not just write end equation since it can create misunderstandings like it did for me. I have only one question what is connection between enthalpy of A or B and enthalpy of formation of A or B? Since enthalpy of formation is heat which is exchanged with surroundings for formation of compound from elements at constant T and P. What does such heat have to do with enthalpy of component A or B at some temperature?

Basically, why is enthalpy of component A or B at reference temperature equal to enthalpy of formation of that component at same temperature?

That’s the definition of heat of formation.

Yes, that is what I don't quite understand. Enthalpy of formation is heat which is exchanged with surroundings at constant temperature and pressure when compound is created from elements in most stable state. Enthalpy of component is sum of U + PV of component at some temperature relative to some reference. Why would this two quantities be equal?

Sorry, I can't teach an entire course in thermo here.  Your background right not is too limited for me to spend the necessary time.  What I hope is that, even without understand that, you still get the gist of what I was trying to show in this derivation.

Okay, I understand. Derivation really helped, thank you.

You omitted the energenics associated with breaking chemical bonds and making new chemical bonds.  If this contributes a negative change to the internal energy (exothermic reaction), then the temperature would have to rise to offset this effect so that the overall change in internal energy would be zero.  If it contributes a positive change to the internal energy (endothermic reaction), then the temperature would have to drop to offset this effect so that the overall change in internal energy would be zero.  Saying this in a different way, when you have a chemical reaction occurring, the internal energy can change even if the temperature is held constant.

Yes I agree, but if internal energy change is zero in our case why does every book in which I was looking say that it changes in adiabatic reactor? Change in internal energy was used to calculate temperature difference. As I said energy balance says that heat of reaction equals time derivative of enthalpy (or internal energy in this case) at every point in time. Ths video from LearnChemE channel from Colorado University is only one example (you can skip to the part where he writes energy balance for adiabatic reactor): https://youtu.be/--cZwa2TXhY

I don't know what books you are using, but, if they say that, they are incorrect.  In your video, the lecturer is incorrect with regard to the sign of the heat of reaction on the right hand side of the equation.  It should be a minus sign (see Fogler's book on reactor design).  If you move that term to the other side of the equation, it is going to start looking a lot more like dH/dt = 0 for the reaction mixture (where they are assuming that the reaction is occurring at constant pressure).

Regarding reaction enthalpy in the video, I think minus sign denotes the fact that exothermic reactions ,which tend to increase temperature of reactor, have negative enthalpy of reaction - nothing too important. But, issue I have is that he says enthalpy changes in adiabatic reactor with temperature. Derivative of enthalpy with time is different from zero. How can this be since we agree that enthalpy (or internal energy in this case) can't change in such reactor?

Who says that he says that?  If he does, he is wrong.  The derivative of the enthalpy with respect to time is zero.  That's basically what his equation amounts to.

Did you see the equation he wrote? He connected time derivative of enthalpy with change of with emperature (its time derivative). dH/dt = [Nacp(a) + Nbcp(b)]*dT/dt = heat of reaction

The left hand side of the equation is not the rate of change of enthalpy.  The left hand side minus the right hand side is the rate of change of enthalpy, which is zero.

Well, dH/dt is rate of change by definition, but only if it is the only thing on the left side of equation. What you said is rate of change of enthalpy minus the right side of equation, not only rate of change.

The change in enthalpy is the sum of two terms, one involving temperature change and the other involving the heat of reaction.  It is NOT, NOT, NOT just the temperature change that determines the change in enthalpy.

Enthalpy of a mixture is a function of both the temperature and the chemical composition of the mixture (which for a mixture undergoing chemical reaction is changing).

For a pure substance, H = H(T,P), but for a mixture of substances, $H = H(T, P, N_A, N_B, N_C,...)$ where the N's are the numbers of moles of the various chemical species in the mixture (which change when a chemical reaction takes place).

In this example, enthalpy of mixing isn't taken into account. It seems we can't reach agreement. Will ask my professor on university this question. Thank you .

In this example, ideal solution behavior is assumed, so enthalpy of mixing is zero.  But, when the chemical composition of the mixture changes, its enthalpy would change as a result of breaking and forming chemical bonds, even if the temperature of the mixture remained constant.


text:
Can more than 1 mol of gas dissolve in 1 mol of H2O?
$\pu{1176 mL}$ of ammonia can dissolve in $\pu{1 mL}$ of water at $\pu{0℃, 1 atm}$, according to wikipedia. Equivalently, about $\pu{0.94 mol}$ of $\ce{NH3}$ dissolve in $\pu{1 mol}$ of $\ce{H2O}$.
I wonder that is there any gas such that more than $\pu{1 mol}$ of the gas can dissolve in $\pu{1 mol}$ of $\ce{H2O}$.


text:
Comparing the boiling points of common organic molecules
The order I was provided with is(comparing molecules of similar molecular mass from each);
alkanes< ethers< esters< aldehydes< ketones<< alcohols<< carboxylic acids<< amides
Where << is supposed to signify a considerable hike.

Here, how do you compare esters and ethers?
I assume the higher position of the carbonyls with respect to those are due to the keto-enol tautomerism they have(with the enol forms  having hydrogen bonding).But how are ketones placed below alcohols?
Why are amides at the very top? Nitrogen being less electronegative than oxygen, I would've thought the hydrogen bonding in those wouldn't be as strong as in the carboxylic acids.

Much can be attributed to dipole moments. Ethers typically have smaller ones. 2. The enol form is typically negligible. Again you can think about dipole moments (polarity). Alcohols can H-bond 3. Amides have large dipole moments (polarity) and can H-bond. See a solvent polarity table eg https://people.chem.umass.edu/xray/solvent.html

The values are okay, but it's the explanation I needed. I'll take your word for it, thanks!

You may also want to refer to a the relative order of the strength of different intermolecular interactions: induced-dipole-induced-dipole<dipole-induced-dipole<dipole-dipole<H-bond

Substituted amides can H-bond less of course. Compare formamide with DMF and you see the effect of methyl-substitution (less available sites for H-bonding) on the bp.