text:
Effect of carbon dioxide impurities on the Haber–Bosch process
The Haber–Bosch process is a highly exothermic reaction that is used to produce ammonia from nitrogen and hydrogen. The nitrogen is typically obtained from the air using a membrane separation process, which results in a product that is 95-99.5% pure. However, there are still small amounts of impurities present, such as oxygen and $\ce{CO2}$.
Oxygen can be easily reduced by incoming hydrogen, so it is not a major concern. However, $\ce{CO2}$ is an inert gas that can inhibit the reaction between nitrogen and hydrogen. The effect of $\ce{CO2}$ on the Haber–Bosch process is not well-understood, but it is thought that even small amounts of $\ce{CO2}$ can have a significant impact on the yield of ammonia.
I am interested in learning more about the effect of $\ce{CO2}$ impurities on the Haber–Bosch process. In particular, I would like to know:

What is the mechanism by which $\ce{CO2}$ inhibits the reaction between nitrogen and hydrogen?
How does the concentration of $\ce{CO2}$ affect the yield of ammonia?
Are there any methods to remove $\ce{CO2}$  from the nitrogen feedstream before the Haber–Bosch process?

I believe that this question is of interest to the chemistry community, and I hope that you will consider answering it.
Thank you for your time.

Your question is interesting but it will become even better if you add your own search results.

Refer to Dalton’s law on partial pressures and how this will effect the pressure of the produced gas in a le chat type interaction


text:
Why does Clausius inequality exist if entropy is a state function?
In Atkin's Chemical Principles: The Quest for Insight, the Clausius inequality is said to be
$$\Delta S \geq \frac qT$$
where equality applies to a reversible process.
Why would the change in entropy be different for a reversible process compared to an irreversible process if entropy is a state function, and the initial and final properties of the system remain the same?

Review the search results site:stackexchange.com OR site:libretexts.org Clausius inequality entropy state function

q would be different

Consider $\frac qT  \leq  \Delta S$

The variation of the entropy for a reversible process is indeed $\Delta S_{rev} = \frac{\Delta Q_{rev}}{T}$, which is a trivial thermodynamic fact. During the process, even if the initial and final elements are conserved, the number of microstates of the system can change. If the initial state has $\Omega_1$ microstates and the final state has $\Omega_2$ microstates, the variation in entropy will be $\Delta S_{irr} = k_b \log \frac{\Delta_2}{\Delta_1}$. The latter is crucial, if there is no microstate created $\Omega_2 = \Omega_1$, the process is reversible so that $\Delta S = \Delta S_{rev}$. However, when microstates are created, $\Delta S > \Delta S_{rev} = \frac{\Delta Q_{rev}}{T}$, the process is irreversible, because $\Delta S_{irr} > 0$ cannot be destroyed because it contributes to the free energy.
At $T=0\text{ K}$, the increase in entropy has no noticeable effect $\Delta S_{irr}$, there is no need to create new microstates, some processes  are reversible but at a finite temperature $T>0$ the increase in  entropy reduces the free energy more or less considerably with $-T\Delta S$, a disordered system is more favorable and the processes are less reversible.
An example will be identical polarized dipoles (either electric or magnetic) at $T=0\text{ K}$, some will flip to form antiparallel configuration at $T>0 \text{ K}$, creating more information and reducing the free energy.

You're right and the textbook is also right because you're considering two different things. You're considering a process that's a complete thermodynamic cycle where the state of the system (which is the substance itself in chemistry texts) is the same before as after the process. On the other hand, the textbook is considering only the part of a thermodynamic cycle when the substance's entropy is changing due to the flow of heat into or out of the substance.

I don't think OP was referring to a cycle in which the end state is the same as beginning state, but rather comparing reversible and irreversible processes that convert the system from the same starting state to the same ending state, which comes down to a difference in w vs q for the processes.

The entropy of a given state is constant. However, returning to a given states usually requires that the entropy of the universe increases. Imagine the last slice of bread on your plate; it has an entropy.  Replace it tomorrow with an identical slice; it has the same entropy. What happened to the entropy of the universe? In everyday life this is reflected in the efficiency of energy use in an operation. Greater efficiency less entropy gain.

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Ways and ideas to study ym rough surfaces with scanning tunneling microscopy
We want to study porous carbon (with nanoparticles, e.g., FeO, Ni, Au, La, Ir, Ag ...) via STM. But we run into the problem that the surface is very rough. As seen in Figure 1.

Thus there is an extremely high chance that we break the tip while analyzing the sample. Is there a way to study "real" surfaces with STM? Or is it just a method that is confined to model systems?
There, of course, are some papers about STM and rough surfaces. However, they are mostly ancient and look at nm rough surfaces, as shown below.

I am thank full for any ideas to analyze rough surfaces and methods to make carbon smoother.
1: Study of structural and electrochemical characteristics of Co-based hypo–hyper d-electrocatalysts for hydrogen evolution - Scientific Figure on ResearchGate. Available from: https://www.researchgate.net/figure/SEM-photographies-of-catalysts-support-TiO-2-250-C-on-Vulcan-XC-72-10000-b_fig2_222022834 [accessed 9 Jul, 2023]
2: Effects of Indium-Tin-Oxide Surface Treatment on Organic Light-Emitting Diodes - Scientific Figure on ResearchGate. Available from: https://www.researchgate.net/figure/AFM-images-of-ITO-surface-subjected-to-different-polishing-conditions-a-as-received-R_fig2_273595496 [accessed 9 Jul, 2023]

Would you please clarify the terms "ym rough surfaces" and "nm rough surfaces"? From the context it also looks like "tip" and "tipp" should be switched. It wouldn't hurt to explain what this mysterious "something" on your carbon really is. Please do not refer to RG, post graphics directly and cite the source including DOI. Have you tried AFM to determine the roughness of your surface?

We would have the option to use AFM. But so far, we have used SEM and TEM, which should be sufficient to see that this is a little bit troublesome. The paper mentioned is not from us but it shows a system that is related at least on the surface to our system in question.

Start with a very smooth substrate. Crystal silicon for example.

Do you mean carbon nanoparticles with metal on silicon? Of course, that could be an option. But that is a big downgrade of the experiment.

@PM Olympus claims the contactless 3D laser confocal scanning technology with resolution in depth to some 10s of nm  (a specifications table) - a bit like focus stacking I guess. This would not scratch on your sample, nor destroy the tip of an AFM.

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Why decomposition of gas on metal surface at high pressure considered 0 order
While reading my textbook it says that when ammonia decompose on a platinum surface at high pressure, the reaction is 0 order because at high pressure the metal surface get saturated with ammonia molecules and adding more ammonia will not change the rate of reaction. However, high pressure do not necessarily mean large number of molecules per unit volume instead temperature might be the reason for high pressure and molecules per unit volume remain same. If we do thought experiment with small number of molecules at high temperature colliding with metal surface then we can see there is still some space for additional molecules could take if added and the reaction will no longer be a 0 order despite the pressure is high. Is my textbook wrong?

your textbook is essentially correct. I guess your textbook relies on the fact that you are  working in conditions when you saturate the surface of the catalyst.  If the density of molecule is not enough to saturare the surface, of course you will not observe zero order kinetics. Somebody more familiar with the process could confirm this, but I guess your textbook is decribing the conditions that the catalyst is normally used and it is given you an example of a zero order process.

@PAEP but my book don't just simply said that at high pressure it would be 0 order reaction and reason they give is that at high pressure molecule concentration around platinum is high and it would saturated it's surface. But high temperature could also be the reason behind high pressure keeping the concentration of molecules same so in this condition how could metal surface get saturated

You seem confused. The point is that this reaction occurs on a
surface not in a volume. it is easy to fill all the surface sites where the reaction can occur if the pressure is high enough. Hence there is no relationship with the bulk concentration. You can't force more molecules into the available reaction sites when they are already saturated.

@matt_black Saturation occur because molecules cover whole surface of metal at high pressure but suppose a metal can accomodate 10 molecules and we have only 5 molecules so still there is some space left on metal if we think all of them collide with metal at same time. So now if we increase pressure by increasing temperature then no matter how much pressure we apply the metal surface cannot be saturated and the statement that decomposition of gas at metal surface at high pressure is 0 order reaction is not necessarily always valid.

@HarjotDhillon The original claim doesn't say the reaction will always be zero order. Your logic is correct. But the point of mentioning the conditions was to point out that the surface would be saturated. Under much lower pressure the reaction would not be zero order. The point is that, under conditions where the surface is saturated, the reaction is zero order.

Chemistry frequently uses simple models from boundary conditions. So look at it from the opposite point of view. "High pressure" means that the catalyst is saturated, and "low pressure" means that it is not. "High pressure" is easy to model since it is zero order. "Low pressure" on the other hand could be almost any order.

It is also generally assumed in catalytic reactions that the amount of catalyst is much less than the reactant

Well textbook mostly tries to explain a concept in a broader and simple way rather than diving deep into different conditions in which reactions could occur (without giving much details of the conditions in which reaction is occurring). I don't know if our textbooks are same , but in the topic chemical kinetics this is mentioned as just an example of zero order reaction.
While studying such type of reactions in which information is incomplete, try to take that the other properties of system (which are not mentioned) at STP. This could help you to get the meaning of 'high pressure' given in the statement and in what context the example is trying to explain you zero order reactions.
I hope my answer helps


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do catechol and quinol give iodoform test just like resorcinol?
do catechol and quinol give iodoform test just like resorcinol?
as ..in case of resorcinol we do have a active methyl after tautomerization which is highly acidic could some kind of mechanism be applicable for para hydroxy phenol and ortho hydroxy phenol

No they do not as they cannot undergo the retro-aldol to give an open chain compound as resorcinol does - see scheme below

Image sourced from  here  - (and yes the open chain product responsible for the positive iodoform test is missing a double bond)


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Software to draw one-dimensional PES including vibrational energy levels
I am trying to represent the difference between the an anharmonic and harmonic treatment of vibrations. The easiest way to do this is to present an example of a harmonic and anharmonic potential energy surface (PES).
I would like to present a figure similar to the one attached below (from Herzberg),$^1$ where we have the morse oscillator function presented with the vibrational energy levels indicated.

I have some spectroscopic data from an undergraduate experiment I did about $\ce{I2}$, and have the data points (or the equation) to draw the PES, and the energies of the vibrations in order to plot their values.
Is there any software which is able to generate something like this for me, or are there any tricks people know where this can be done in other graphing software (preferably Excel or Igor Pro)? The difference between the spacing of each successive overtone in an experimental vs. harmonic spectrum is part of the explanation, so it's important to illustrate that in the figure, if at all possible.
EDIT I have since come across a Python script which will do the job (see answer below); however, I'm sure that other ways to generate these diagrams that people know would be appreciated (particularly for people who do not know much about Python, such as myself...)!

G. Herzberg, Spectra of Diatomic Molecules, Krieger Publishing Company, Malabar, Florida, Reprint edn., 1989.

In general, this is most easy using software like R or python, but in Excel you can do it in the following way: Assume column A contains $r$ values and column B contains your potential values $V(r)$. Using =IF($B1>En0, NA(), En0) for the values in column C where En0 is your first energy. Use the other columns for the additional eigenvalues and plot with Scatter.

@Paul what do you mean by my 'first energy'? Do you mean the first vibration (i.e. $\upsilon = 1$)?

@isolated_matrix, could be any vibration (note that vibrations with $v=0$ also exist). The cell formula only provides a plottable result if the potential energy is lower than the eigenenergy so that the horizontal level energies are bounded by the potential.

@Paul so, if I understand this right, I would actually not only need a column C, but also columns D, E, F, etc., so that each vibration has its own column?

Perhaps the question is suitable for mattermodelling.se (example).

@isolatedmatrix indeed, every level requires its own column.

@Buttonwood that did occur to me;I saw that exact post on this exchange, and a similar comment on that, but that comment also said it would be appropriate on either exchange (assume we call these different 'topics' exchanges?). I also thought it would be something relatively common in chemistry, which is why I'm so surprised there are no programs which do it automatically!

@Buttonwood I don't think it fits there. The question is about how to draw a Morse potential from experimental values. This should have all the information you need: https://chem.libretexts.org/Courses/Duke_University/CHEM310L_-_Physical_Chemistry_I_Lab_Manual/05%3A_Molecular_Spectroscopy_of_Iodine/5.03%3A_Potential_Energy_Curves

https://pubs.acs.org/doi/10.1021/ed051p282 I'm sure you'll find a copy of that somewhere. Oh, and you can use any program that can draw a function. I recommend gnuplot for science stuff.

I wrote a program that generated the figure here: https://mattermodeling.stackexchange.com/a/1005/5 and you can see some more examples here: https://arxiv.org/pdf/1509.07041.pdf I would have to find it to post it here.

@user1271772 that is perfect, and exactly what I am looking for! If you can find it, and post it as an answer, I’d definitely accept it

@Buttonwood I have all of that information, I know about the Morse potential, etc.

@Martin-マーチン I actually came across that article (I was looking for articles on anharmonicity because I don’t understand perturbation theory as well as I would like), but I’ve got all the data I need ($\omega_e$ and $\omega_e x_e$, etc.). I will look up gnuplot (although I was hoping to avoid a command-line based program, if I could avoid it)

@isolatedmatrix by when do you need this?

@user1271772 Don't worry about it – I found a python script which will do the job

I have made this basic python 3 script which runs in a Jupyter notebook (or in Visual Studio Code). With different molecules you will need to adjust the plot scale. The  first 10 levels are plotted then every third one. You can adjust all this.
The code plots the potential, energy levels and wavefunctions.
# import all python add-ons etc that will be needed later on
%matplotlib inline.  # for jupyter notebook only
import numpy as np
import matplotlib.pyplot as plt
from scipy.special import eval_genlaguerre   #associate Laguerre polynom
init_printing()
plt.rcParams.update({'font.size': 14})   # set font size for plots

# HCl  and Iodine
# Assoc Laguerre polys & Morse potential.  n is quantum number
# wavefunct not normalised

fig = plt.figure(figsize=(5,5))

hbar =   1.05449e-34  # Js/2pi
c    =   2.9979e8     # m/s
nm   =   1.0e-9       # nanometres
amu  =   1.66043e-27  # atomic mass unit

mol  = 1

if mol == 0:

    m1   = 127            # iodine
    m2   = 127
    we   = 128.0          # frequency cm^{-1}
    xewe = 0.834          # anharmionicity no units
    reB  = 0.3016         # equilib bond length nm
    xm1  = 2              # used to limit plot length
    xm0  =0.8
else:
    m1   = 1              # HCl
    m2   = 35
    we   = 2989.7         # frequency cm^{-1}
    xewe = 52.05          # anharmionicity no units
    reB  = 0.123746       # equilib bond length nm
    xm1  = 3.5
    xm0  = 0.5

mass = m1*m2/(m1 + m2) # reduceD mass

fac3 = hbar/(amu*2*np.pi*100*c*nm**2)        #  factor to get constants sorted

DeqB = we**2/(4.0*xewe)                      # Morse dissn energy cm^{-1}

k    = 4.0*DeqB/we
beta = we*np.sqrt(mass/(2.0*fac3*DeqB))      # 1/nm

d = k/2.0
zeta = lambda x: np.exp(-beta*x)
psi  = lambda n,x : np.exp(-zeta(x)*k/2.0)*\
                np.exp(-beta*x*(k-2.0*n-1.0)/2.0)*eval_genlaguerre(n,(k-2*n-1),k*zeta(x))

En = lambda n:  we*(n+0.5)-xewe*(n+0.5)**2     # energy levels  cm^{-1}
nm = 0
while En(nm + 1) > En(nm):
    nm = nm + 1
print('max Q number',nm)

V = lambda x: DeqB*(1-np.exp(-beta*x) )**2     # Morse potential  cm^{-1}

xm = lambda E: -np.log(1-np.sqrt(E/DeqB) )/beta+reB  # find limits of potential
xp = lambda E: -np.log(1+np.sqrt(E/DeqB) )/beta+reB

x = np.linspace(xm0*reB, xm1*reB, 500)           # nm

for n in range(11):                              # plot first 10 levels
    E0 = En(n)
    n0 = max(abs(psi(n,x-reB)) )*3/we            # normalise on plot
    plt.plot([xm(E0),xp(E0)],[E0,E0],color='red',linewidth=0.5)
    plt.plot(x,psi(n,x-reB)/n0+E0,linewidth=1,color='grey')
    plt.text(x[-1],E0,str(n),fontsize=10)

for n in range(11, nm, 3):                       # plot some higher q number wavefunctions gaps of 3
    E0 = En(n)
    n0 = max(abs(psi(n,x-reB)) )*3/we            # normalise on plot
    plt.plot([xm(E0),xp(E0)],[E0,E0],color='red',linewidth=0.5)
    plt.plot(x,psi(n,x-reB)/n0+E0,linewidth=1,color='grey')
    plt.text(x[-1],E0,str(n),fontsize=10)

plt.plot(x,V(x-reB),color='blue',linewidth=1)
plt.axhline(DeqB,linewidth=1,color='red')
plt.ylim([0,1.05*DeqB])
plt.xlim([x[0],x[-1]])
plt.xlabel(r'$r\; /\; nm$')
plt.ylabel(r'$Energy\; /\;cm^{-1}$')

plt.tight_layout()
plt.show()

An example plot is shown below

This is great! I'm totally adapting this for my class to show the effects of the anharmonic oscillator.

@GeoffHutchison the scripts I linked below do something similar (including plotting selected wave functions), so they might also be worth checking out?

Really like the look of this! Although I'm a little confused where the HCl comes into it? Or is that just another example of how the script can be used?

Just an example; one with a large force  constant  and one with a smaller one.

I have come across two Python scripts which will do the job, written by Christian Hill. The links to them are:
Visualizing vibronic transitions in a diatomic molecule
The Morse oscillator
For example, here is a diagram for the $\ce{O2}$ PES generated by one of the codes linked above (I edited it slightly to remove the wave functions which you can overlay on the vibrational levels):

They are not exactly what I was looking for (the curve could extend a little further towards the lower $r$ values to show how the energy keeps increasing beyond the dissociation energy), but it does plot the vibrational transitions the way I want.


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Is chemistry finally seeing the end of the Haber-Bosch nitrogen fixation process?
Do recent developments in alternatives to the Haber process for nitrogen fixation have good prospect of replacing it?
Some recent headlines have suggested that the century-old Haber process for turning nitrogen into ammonia might be seeing some competition. For example this report claims:

Scientists at Australia's Monash University claim to have made a critical breakthrough in green ammonia production that could displace the extremely dirty Haber-Bosch process, with the potential to eliminate nearly two percent of global greenhouse emissions.

is this a reasonable prospect given the chemistry involved? Given what we know about the chemistry and economics of Haber or its alternatives, is replacing the process something chemists can reasonably hope for?

I would not say the process is more dirty than production of car Li-ion batteries, powered by fossil fuel power plants.

@Poutnik How dirty it is in comparison to other processes isn't really the issue. I think chemists can agree it is hugely energy and capital intensive and that any alternative that reduced either of those inputs would be relatively better.

As someone who worked on nitrogen fixation via electrochemical methods: It is unlikely to replace Haber-Bosch in the near and even further future. The Haber-Bosch process is very well optimized, scales well and the infrastructure is established. Currently electrochemical production methods have higher CAPEX and OPEX for most applications and even the better established field of CO2RR is nowhere close to large scale commercial applications. It has however potential in decentralized production, most ammonia comes from just a few production plants. So in the middle of the outback maybe?

@Noah I'd like to hear more detail not least because headline writers have tended to make big claims on opex and capex implications of the electrochemical methods (presumably without any analysis).

This article is clickbait. HB is just equilibrium chemistry, the thermodynamics will never change. Since those guys came up with this, there were a century of chemists trying to replace this with something more efficient. Is nice that people are still trying, but I don't see it happening in my lifetime.

As with many proposed green processes, this method relies on electricity to replace direct use of fossil fuels as an energy source. The electricity must itself come from some primary energy source, so we need green (or at least greener) electricity to work as advertised.
In addition, the Haber-Bosch process may also be made green(er), as hydrogen has also been the object of green production. If green hydrogen production on a large scale were to become successful, then it could also be used as the fuel to sustain the reaction temperature as well as the reducing agent in the H-B process.
A secondary issue is that lithium has limited availability. Magnesium is also a good nitride former, and is much more abundant; so the electrolytic method would be more competitive if it were carried out with magnesium as the discharge anode. A magnesium/magnesium nitride-based method is proposed by Hu et al..
Reference

Yang Hu, George Z. Chen, Lin Zhuang, Zhiyong Wang, Xianbo Jin (2021). "Indirect electrosynthesis of ammonia from nitrogen and water by a magnesium chloride cycle at atmospheric pressure".
Cell Reports Physical Science,
Volume 2, Issue 5,100425,
ISSN 2666-3864,
https://doi.org/10.1016/j.xcrp.2021.100425.


text:
How do the orbitals in atoms synchronize their phases during the formation of a bond?
Suppose I have two hydrogen atoms, each with one electron in their 1s orbital. Then we can represent the wave functions of these two as say $|\Psi\rangle_1$ and $|\Psi\rangle_2$ for each of the two atoms respectively. Since we have two identical atoms, $|\Psi\rangle_2$ is just $|\Psi\rangle_1$ translated to a different point in space with some phase difference, so $|\Psi\rangle_2 = e^{i\theta}|\Psi\rangle_1'$ where the $'$ denotes that we've shifted the wavefunction in space.
We know that if the two atoms have wave functions in phase with each other ($\theta = 0$) then when they come together the orbitals will combine with each other and a $\sigma$-bond will be formed. Equally if the wave functions are completely out of phase with each other then the anti-bonding orbital of the combined system would have 2 electrons while the bonding orbital would have 0 and since the anti-bonding orbital has higher energy this transformation is not energetically favourable so no reaction occurs.
I'm curious about what happens when the phase is not equal to either $0$ or $\pi$, which is a case that happens with probability $1$. Clearly the atoms do end up reacting in real life so the phases must get synchronized somehow before the reaction takes place. I'd like to ask how exactly this happens.

Does this answer your question? Significance of phase of atomic orbitals

Not really, the answer on that question (or the linked one from that answer) doesn't deal with the formation step.

If you're really interested in how H2 forms, than you're missing the point. Molecule is initially in excited state and needs to lose surplus energy quickly.

I'm not interested in how H2 forms, say I have molecule 1 in phase 0 and molecule 2 in phase $\pi/2$. This means that the wavefunction of molecule 2 has an $i$ in front of it so when you add the wavefunctions together you'd get no interference at all (the first one acts in the real dimension, the second one acts in the imaginary dimension). I'm interested in how exactly the phases get synchronized so the reaction can take place.

A specific pair of molecules or atoms does not need any synchronization because it doesn't matter if the pair doesn't react. In a mole of the atoms there will be 6*10^23  of them, each with different phases and, given how frequently they will collide there will be an even bigger number of collisions. Given the phases will all be different, some of the collisions will lead to reactions without any need to synchronise any individual pair. This is why even complex molecules where orientation and many vibrational models might need to have specific values still have some reaction rate.

@matt_black The probability that any two atoms colliding have the exact same phase (as a real number) is exactly equal to 0. That means no matter how many collisions happen there will always be some (non-zero) phase imbalance. Now this imbalance may well be tiny, but it will be non-zero.

@HadiKhan When your math contradicts what we observe, my first assumption is that the math is wrong. Exactly zero to what precision? Is that precision realisable in the real world?

Exactly zero to whatever precision you care to name: I mean the probability is equal to 0 as a number. Note that this is not the same as being impossible, zero probability events happen all the time, but any specific event you care to name (and can describe in finite length) that has zero probability will never happen. Just like how the probability that a ruler is exactly equal to 15.0000000... cm or any other length is equal to 0, but it certainly has some length. See e.g. here for a more thorough explanation: https://math.stackexchange.com/questions/3666392/probability-zero-vs-impossible

And this then means that the probability that any two colliding atoms are exactly in phase is 0, because that requires the phase of the second atom to exactly equal the phase of the first atom, which is a real number, and $\mathbb{P}(\theta = r) = 0$ for any real number $r$ if $\theta$ is uniformly distributed..

In the simplest case when two atoms approach one another their wavefunctions are changed due to mutual interaction, (nuclear-nuclear, electron-electron and nuclear-electron) so phase of isolated and non-interacting wavefunctions no longer matters in the way you suggest. To solve this problem the coordinate system is changed to, say, an elliptical one with each atom at the foci. The integrals of these interactions are then solved as the internuclear distance changes.

The expression $|\Psi\rangle_2 = e^{i\theta}|\Psi\rangle_1'$, for $\theta \neq k \pi$ ($k \in \mathbf{Z}$) is likely to reach to a complex function. This situation is unlikely under an attracting potential because the position tends to be more precise (requires a real wave function) at the expense of the momentum (which requires a complex wave function), even if the exact wavefunction is unknown.
On the other hand if we consider a translation operator $\hat{T}$ acting on a $s$ orbital of one hydrogen atom of a site $i$ transferring the orbital to another site $j$ in a chain of $n > 0$ hydrogen atoms having $a$ as interatomic distance in a $x$ axis $\mathbf{R} = l a \hat{e_x}$ with $l=1,2,...,n$. The case $n = 2$ is the hydrogen molecule $H_2$. Generally, $\hat{T}$ and the Hamiltonian $\hat{H}$ do not commute :  $[\hat{H},\hat{T}] \neq 0$ this means it is "unlikely" that they share the same eigenvectors because of the repulsion $V_{ee}=+\sum_{ij}\frac{1}{|r_i-r_j|}$.
We can consider the case $V_{ee}$ can be ignored so that $[\hat{H},\hat{T}] = 0$, the operator $\hat{T}$ gives $\hat{T}|\Psi\rangle_1 = e^{i\mathbf{k}\mathbf{R}} |\Psi\rangle_1$, if $k=\frac{2\pi}{a}$, therefore $\hat{T}|\Psi\rangle_1 = e^{i 2\pi l} |\Psi\rangle_1 = |\Psi\rangle_2 $. Here the phase shift is due to the translation operator and can only be a positive integer (1). A similar result compared to the expression above, but there is no anti-bonding.
This result is due to the fact that $\hat{T}$ is a generator of bonding orbitals, transferring the wave function from one atom to another. The state described above is similar to a singlet state, where $V_{ee}$ is ignored because there is an exchange interaction (attractive) $V_{x}$ which lowers significantly the impact of $V_{ee}$, for $n=2$, $\Psi_B \propto (|\Psi\rangle_1 + |\Psi\rangle_2)$. Now for the antibonding is due to the fact that $V_{x}$ is only a pairwise interaction, it means the exchange do not act on more than two states, this anti-bonding state has the impact of $V_{ee}$ and $[\hat{H},\hat{T}] \neq 0$, the state is not an eigenvectors of $\hat{T}$, the only possible choice is $\Psi_A \propto (|\Psi\rangle_1 - |\Psi\rangle_2)$.
I did not answer the question directly by I hope it helps.


text:
Calculating Rotational Temperature from an FTIR Spectrum
I am trying to calculate the rotational temperature of a sample of gas phase HCN from an FTIR spectrum. It's a 10 Torr sample in a 10 cm gas cell, held at room temperature. In theory, I should be able to make a semi-log plot of the rotational peak intensities vs. the energy of the transition and pull the rotational temperature from the slope according to the Boltzmann distribution using the following general equation:
$$ ln(\frac{A_{J}}{g_{J}}) = -\frac{\varepsilon_{J}}{k_{B}T} $$
Where $A_{J}$ is the peak intensity for a given transition, $\varepsilon_{J}$ is the energy of the transition, and $g_{J}$ is its degeneracy.
However, whenever I try to make this calculation, regardless of the the units I use for energy or whether I'm taking into consideration more complex factors, like Honl-London, I get a nonsensical rotational temperature, usually on the order of 5-20 Kelvin. Does anyone know where I might be going wrong?
Additionally, I'm including pastebin links with the portion of raw FTIR data I am working with, and the peak assignments, with heights, energies, J states, degeneracies, and FWHM
FTIR Data
Peak Assignments

This is from a paper on HCN in planet-forming disks - and I know nothing about astro*, so this might be far off base - but the rotational temps for $\ce{HCN}$ range from 11 to 21 K (+/- some). I believe you need the high-temperature limit even at room temperature but could be wrong. TLDR: I don't think you're that far off.

The relative population for a (rigid rotor)diatomic is given by the weighted Boltzmann distribution $\displaystyle (2J+1)\exp(-Bhc(J(J+1)/(k_BT))$ where $B$ is in cm$^{-1}$ and $c$ the speed of light in cm/s and $k_B$ the Boltzmann const. This should give an approx temperature but for an accurate value you will need transition probabilities (Honh-London) and also the fact that it is the population difference between levels that matters but for rotational levels in v=0 to v=1 this may be a small effect.

@porphyrin Using your equation gives a much better result of 317 K, although it's still off by ~20 K. I have two different literature methods for calculating Honl-London factors but that results in rotational temperatures greater than 400 K, which makes me think I'm using them incorrectly. How would I incorporate them in?

The fit to the equation I showed needs two parameters, (temperature and amplitude) to get a good fit by non-linear least squares. However,  you probably  need to account for centrifugal distortion and possibly vib-rot coupling and so use a different energy function. You will find the equations in most phys. chem. and spectroscopy textbooks. Try this before any more fancy stuff.


text:
Is gallium slippery at room temperature in an inert atmosphere?
Ice is slippery when it isn't too far below it's melting point because the surface molecules are less loosely bound than the bulk and form a thin liquid layer.
Pressure melting is a much smaller effect since it takes about 10 MPa of pressure per degree below freezing, which is more than ice skates can apply.
Presumably this "edge effect" isn't unique to water. Gallium at 20C (melting point about 30C) should also have such a surface layer. But most metals oxidize if exposed to air and/or water, forming an oxide layer which (in this case) melts at much higher temperatures. But if gallium at 20C is placed in argon and the oxide layer is removed, would the underlying metal feel slippery to a gloved hand?

Gallium has a strong tendency to supercool, so... it would likely be fully liquid.

@Mithron: If melted in the first place yes, but if it is warmed from a cold temperature or given a seed crystal it won't be liquid below 29.7C.

My instinct is that this surface disorder effect is more important for water than for gallium, since the water molecule and the way it packs is more complex than the gallium atoms. What this ultimately would entail, though, I don't know. Surface kinetics/thermodynamics sure can get messy...


text:
Open-circuit potential of an electrochemical cell
I'm studying electrochemistry without a physics background--never taken it for extenuating reasons--for a research project using Bard's Electrochemical Methods. I'm having trouble understanding the concept of open-circuit potential (OCP): the electric potential measured in a cell when disconnected from a circuit. First off, why can you measure OCP at all? It's a result of thermodynamics--i.e. the sum of the standard potentials of the half reactions at the given electrodes. But  wouldn't the lack of a current preclude any measurement of potential? I mean, a voltmeter, with its very high resistance, still allows some current flow.  Since you can measure OCP, does that mean the half-reactions are taking place in small quantities?
Second, Wikipedia describes OCP as "the voltage that must be applied to a solar cell or a battery to stop the current." If a given cell has an OCP of 0.5V, that means I apply 0.5V to stop the current once a circuit is completed? But if there exists 0.5V without a current, applying an external 0.5V won't then create a total potential of 1V? In my head I want to apply -0.5V to negate the current flow.
Thank you in advance. The inclusion of mathematical formulae in your answer would be much appreciated!

The answer to your first question is yes. The higher the input impedance of the voltmeter or electrometer, the better for minimizing having the half reactions take place. For the rest, see my answer here: https://chemistry.stackexchange.com/a/153076/79678. You apply the external potential to oppose the cell potential, not augment it.

@EdV Ahh ok. So if you have two electrodes just sitting there, nothing's going on. But measuring the OCP requires completing the circuit. Without a complete circuit, the difference in potential still exists, but no current = no rxn. It's as if I have a zipline at the top of a hill. The potential to zipline down exists, but without actually connecting the thing, I can't ride it.

Well said! With the typical voltmeter mode in a digital multimeter, the input resistance is 10 M ohm. Using this to measure an approximation of the open circuit potential is fine for most purposes since, e.g., 0.5 V divided by 10 M ohms is only 50 nA. But for a standard reference electrode, e.g., Ag/AgCl, no more than a couple of pA can be drawn without polarizing the electrode. So then you use an electrometer voltmeter, commonly used with pH electrodes. They typically have 1 T ohm input resistance.

Basically you've have the right ideas.
The open-circuit potential of an electrochemical cell is the theoretical voltage that one would measure as the current flow approaches zero. Think of plotting small current flows against measured voltage and then extrapolating to zero current flow. Pragmatically though common voltmeters will only measure a few significant figures. So a microamp current flow is essentially zero current flow. This isn't math where pi can be calculated to millions of digits.
For the solar cell you're right. If the cell output is +0.5 volts then you'd have to apply -0.5 volts to stop the current flow if the voltage is applied across the cell itself. Two +0.5 sources in parallel would only give +0.5 volts output however. You'd have to arrange the solar cell and an additional +0.5 volts in series to get +1.0 volts output.

Ok so OCP is kind of like the limit as i -> 0. In reality you can never reach 0 b/c you need at least a tiny bit of current to measure the voltage.

GMoss - Yes but you can't measure voltage to some huge number of significant figures. So to accurately measure about 0.5 volts of a battery to 3 or 4 significant figures a microamp of current would be small enough to be essentially zero current.


text:
Effects of Hydrogen Peroxide Concentration in the Briggs-Rauscher Reaction and Workarounds for Dilution
I am an IB chemistry student attempting the Briggs-Rauscher reaction, which requires a diluted solution of hydrogen peroxide. The experiment guidelines suggest diluting 400 mL of 30% hydrogen peroxide to 1 L, resulting in a 12% concentration. However, the highest hydrogen peroxide concentration available in my lab is 6%. I would like to understand the effects of hydrogen peroxide concentration on the Briggs-Rauscher reaction and seek possible workarounds or adjustments to achieve the desired reaction conditions.
Question:
How does the concentration of hydrogen peroxide affect the Briggs-Rauscher reaction? Specifically, what are the implications of using a 6% hydrogen peroxide solution compared to the recommended 12% concentration?
Are there any known workarounds or adjustments that can be made to accommodate the lower concentration of hydrogen peroxide (6%) available in my lab? If so, what are they, and what modifications should I make to the experimental procedure?
Note: I have already searched the Chemistry Stack Exchange and other resources but have not found specific information regarding the effects of hydrogen peroxide concentration in the Briggs-Rauscher reaction or suitable workarounds for dilution.


text:
Calculating salinity from sodium and chloride in water
let me start by saying that I am definitely no chemist, so this question might (or might not) seem trivial to all you professionals out there.
I have recently found a brand of carbonated water here in Norway that I really like. It is quite salty, compared to your regular stuff, and I want to know the exact salinity (in parts per thousand) of this water. The minerals contained are as follows (in milligrams per liter):

sodium 455
magnesium 31
calcium 28
potassium 16
chloride 667
bicarbonate 300
sulphate 11
silicon 6

As the chloride forms bonds not only to the sodium, I would assume it isn't as easy as just adding the chloride and the sodium, and I do not know how much NaCl this would equate to. Can you help me?

You can't really balance cations and anions with the information given to get precise compound concentrations. However some "back of the envelope" calculations can be done.
To compare the various ions you need to reduce the concentrations to a common unit. A convenient one would be millimoles per liter. To get millimoles per liter you need to divide the concentration of milligrams per liter by the molar mass of each of the different kinds of ions. (A  mole is just a count of atoms like a dozen. However the mass of an atom is vanishingly small so you need a lot of them to get a convenient mass. So one mole actually has about $\mathrm 6.022*10^{23}$ atoms.)
sodium 455/22.99 = 19.8 millimoles/liter
chloride 667/35.4 = 18.8 millimoles per liter
So the sodium and chloride are roughly the same about 19 millimoles per liter. This is well below what is considered safe levels for drinking water in the USA.

Edit
Since we're doing back of the envelope calculations, the milligrams of sodium and chloride correspond to roughly the same number of atoms in the water. So the mass of sodium and chloride can be added to get:
$\mathrm{0.455 + 0.667 \approx 1.1}$ grams NaCl per liter.
Since a liter of water has a mass of about 1 kilogram, that's 1.1 ppt of NaCl.

Hmm, but is not salinity defined a the mass of dissolved salts per the unit mass (approx volume) of water? Like 500 mg/kg or 500 mg/L? In such a case, he could just summarize masses of provided ions to get the approx. number.

Thank you. As I understand it, being a non-chemist, I still don't know how I go from mmol/L of NaCl to mg/L, would that simply be reversing the calculation? Thus, I end up with approx. 500 mg/L, or a salinity of about .5 ppt?

Scrap that, recalculations considering the molar weight of water gives me a salinity of 0.34 ppt of NaCl. Sounds reasonable?

@Poutnik - I didn't know what the OP really wanted. It seems now that he wanted ppt of NaCl which I added to the answer.

@MaxW Thank you! This would actually mean this water would be considered brackish, as it is >.5 ppt. Interesting.


text:
Perfect amount of HCl to precipitate 1.000g of silver alloy
I am trying to figure out the perfect amount of HCl to drop silver out of a solution for a specific alloy.
This alloy is 79.5% - 80.25 % Ag.
We digest this (size of the sample is usually 1.000g or 0.999g) alloy in 10 mL of a 30% HNO3, once fully dissolved 60 mL's of DIH2O is added then 2.5 mL of HCl is added to precipitate AgCl. (Side note: the amount of AgCl that is ideal for this procedure is around 1.06 g.)
We have been noticing that we have to refilter the sample because AgCl precipitates out of filtrate the next day.
Any suggestions on how I could figure out the perfect amount of HCl added to precipitate out all of the AgCl without adding excess HCl?
I would love to not have to refilter the filtrate anymore.
We use concentrated HCl for this procedure.
(Another side note: full composition of this alloy is 80% Ag 15% In 5% Cd)

(1) I wonder more basically why a second filtration would be needed. Perhaps particle size increases? (2) To avoid two filtrations could you let the ppt's sit for 12-24 hours? Filter after sitting and see if you get more ppt on the third day.

Well we filter the samples for a second time because of AgCl precipitating out after the first filtration. I believe we add too much HCl so when the sample is dropping AgCl out some of the AgCl turns into the Ion AgCl2 which is soluble in excess HCl. We use concentrated HCl so it would make sense. We use 25 mL crucibles that are fitted with a 21 mm quantitative microfiber filter.

I understand that the excess $\ce{HCl}$ added would likely form $\ce{AgCl_2^-}$ but I don't understand why that anion would decompose to $\ce{AgCl}$ after a day. The only thing that I can think of is Ostwald ripening, but I'd wonder how much more $\ce{AgCl}$ that could create after a day.

Honestly thats the only thing I can think of as well its the only thing that makes sense.  The weight of the first filter AgCl is usually around 1.0554g and it is ideally supposed to be around 1.06 for a good Ag percentage. We usually get around .003 - .008 g of AgCl after the second filtration. And to answer part of your first question we usually let the sample sit on our hot plate for around 24 hours and bring it down and let it cool for around 6-7 hours then filter the sample.

dah... You first dissolve with $\ce{HNO3}$, then dilute with water and then add the $\ce{HCl}$. So the final acid mixture would seem to be a dilute solution of aqua regia. Wouldn't the $\ce{HCl}$ reacti with the $\ce{NO3^-}$, as it does with "proper" aqua regia, but more slowly since the solution is dilute?

It is just a dilute solution of HNO3- and HCl (miniscule amounts not able to make A.R.) with all sorts of metals in it. It's really interesting though since the silver is present the HCl doesn't react with the NO3- but instead goes for the silver precipitating AgCl. Aqua Regia is 3 parts HCl to 1 part HNO3-.

One gram silver alloy contains $0.800$ gram silver, which is $\frac{0.800}{107.87} = 0.785 $ mmol $\ce{Ag}$. To be transformed into $\ce{AgCl}$, the same amount of $\ce{Cl}$ atoms, and the same amount of $\ce{HCl}$ is needed. The problem now is to find the volume of $\ce{HCl}$ solution containing $0.785$ mmole $\ce{HCl}$. Unfortunately the concentration of used $\ce{HCl}$ solution is not given. So the problem cannot be solved.
Edit : As Matt grives $\ce{HCl}$ concentration as $12.1$ M, the problem can be solved. $0.785$ millimole $\ce{HCl}$ is contained in a volume : V = $\frac{0.785·10^{-3} \mathrm{mol}}{12.1 \mathrm {mol/L}} = 0.0648$ mL. It is a rather small volume, about two or three drops.

Sorry I completely forgot. It is concentrated HCl. 12.1 M HCl


text:
What is causing rust-like stains on my stainless steel appliances?
Two weeks ago we discovered that our new Kohler faucet was leaking under the sink in our lanudry room adjacent to our kitchen. We shut off the water and have not used the sink.   It apparently was also leaking into the household cleaning chemical tray including Comet Cleanser for some time.  The entire compartment under the sink was significanlty rusted, including hinges and metal components of the new sink.
Concurrently, we noticed a rust-like coating on our stainless steel microwave above the sink.  In the past two weeks, the coating has spread to our Stainless refrigerator (3 feet away through a doorway), the stove (5' away) and other stainless surfaces, some up to 10' away in a separate room.
There is no noticeable odor or other indications.  The coating can be removed with spray stainless steel cleaner.   Other surfaces do not seem to be impacted.
Any suggestions on what could be causing this chemical reaction?

Please post a picture.

What is a Kohler faucet ? Why is it under the sink ? What is a chemical tray ? What is Comet Cleanser ? Why do you speak of a new sink ? Why new ? Why do you put a microwave above the sink ? How are the stainless pieces related ? Are they related ?

If you're familiar with the Comet smell, that may be the culprit.
Comet contains a relatively volatile, odorous bleaching compound, trichloroisocyanuric acid. This compound is a form of chlorine bleach. As such, when it reacts with stains or perchance with metals it can reach via evaporation and travel through the air, it releases chloride ions -- which then break down the passive layer that's supposed to keep stainless steel from rusting.
To minimize this condition, thoroughly rinse this cleanser off and drain it after cleaning.

I don't think that trichloroisocyanuric acid would readily sublime, and chloride ions don't go floating through the air. So I'd think the decomposition of trichloroisocyanuric acid with water (or humid air) would yield chlorine gas  rather than chloride ions.

The fact that it is odorous implies some vapor pressure, and I do not claim the chloride ions flost through the air. The latter would form at the reaction site.


text:
Significance of Character tables in d orbital splitting
When we have a molecule say (Co(NH2)6) hypothetically any octahedral molecule it has the point group of Oh and the character table shows the quadratic function such as ($z^2,2z^2-x^2-y^2$) ($xy, xz, xy$). These let you know about the breaking of degeneracy of orbitals. But for some groups like $C_2v$ there are functions like $x^2,y^2$ which don't really correlate to any orbital. What is the significance of such functions?
Moreover to aforementioned $C_2v$ group doesn't have $x^2-y^2$ function so we don't know how the degeneracy breaks. Is there any method to find it using group theory?

These are operators that show how the sym species behave under the action of these function. For example x, y, z (linear translations) show how species react to optical (dipole) transitions and squared terms, xy, x^2 etc. for Raman transitions. You should consult a text book to get examples.

@porphyrin I was able to get some info of Raman Transitions and all but the thing that I'm looking for the is the relation of these functions to the corresponding d orbitals, such as I've mentioned $d_{z^2}$ and all. I wanted to know the correlation between d orbitals and the functions of $x^2$ and $y^2$

Some of these function will correlate to the orbital symmetry, if there is a d orbital in C2v, Td, Oh, etc. it would transform as e.g. z^2, xy, etc., but the functions x^2 etc. do not have to correlate to orbitals but can do to other things that transform as a squared function  such as Raman scattering.

For the $\mathrm{d}_{x^2 - y^2}$ orbital, note that both $x^2$ and $y^2$ transform as $\mathrm{A_1}$ under $C_\mathrm{2v}$ symmetry: http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=402&option=4
Any linear combination of $x^2$ and $y^2$, including $x^2 - y^2$, thus also transforms as $\mathrm{A_1}$.


text:
Why can we use stoichiometric coefficients in the equilibrium-constant expression?
I just studied the chapter on chemical kinetics on Coursera, wherein I was repeatedly admonished not to use the stoichiometric coefficients in the rate law formula. I was told that this formula has to be experimentally determined instead. The only exceptions are elementary reactions where one can indeed use the stoichiometric coefficients in the rate law formula.
The equilibrium constant is $$\displaystyle K_c=\frac{k_f}{k_r}$$ However, for some reason, one can use the stoichiometric coefficients in the equilibrium constant expression regardless of the reaction mechanism. Why is this?

For the reaction $\ce{A=B}$ , write down $dA/dt$ then set this to zero at equilibrium and solve for $K_e=[B]/[A]$

Typically you write that the rate of change of species $j$ is $\mathrm{d}C_j/\mathrm{d}t = \sum_k \nu_{j,k} r_k$, where $\nu_{j,k}$ is the stoichiometric coefficient of species $j$ in reaction $k$. This formalism is independent of how you write $r_k$, it can be thermodynamically reversible, a catalyst-type rate law, etc. Only if it is thermodynamically reversible, then you recuperate that $K_\mathrm{eq,k} = \prod_j a_j^{\nu_{j,k}}$

@porphyrin If it's a first-order reaction, $-dA/dt=k[A]$. So if, at equilibrium, $dA/dt=0$, would it mean that $[A]=0$? I'm getting confused.

To have equilibrium there must be a reverse reaction so $d[A]/dt=-k_f[A]+k_r[B]$.

So $-k_f[A]+k_r[B]=0\implies k_f[A]=k_r[B]\implies k_f/k_r=[B]/[A]=K_c$ But my question remains: How did we assume that the forward and reverse reactions are elementary so that the powers in the rate law (1 in this case) are the coefficients of the balanced chemical equation?

When molecules react you have to assume a scheme then test it against experiment. If it is simple, such as I gave, then you get that result, if not then you have to work it through.


text:
Adiabatic process (reversible and irreversible) and implications for delta U
I saw this post by Chet Miller regarding the work done during an adiabatic reversible and irreversible expansion processes. He had answered to the OP that dU = CvdT = work done during the process. I had a followup question:
Since the dT would be different for each of these processes, the dU would be different as well (assuming constant Cv). Hence, if there is a reversible path and irreversible path between 2 states, the dU for each of these transformations would be different and this would contradict dU being a path independent state function. What am i missing here?
The confusion kind of ties in what one of the MIT lecturers emphasized that dU = (ONLY) reversible work for an adiabatic process.
Thanks

No need to worry so much about them. Actually, $dU$ and $dT$ are infinitely small and arbitrary. We are doing so in order to integrate to get the actual change in $U$.

There is no contradiction here. Take another example to understand it. $P$ and $V$ are both state functions but $-PdV=w$ is a path function even though work can be done both reversibly and irreversibly.

There is no adiabatic reversible path between the same two end states as an adiabatic irreversible process .

@ChetMiller Hello Chet Miller. Since you answered my comment in one of my posts, I was wondering if you were still interested in an answer on your question. Just to thanks your comment in my question.

@Metal Storm Did you disagree with the answer I gave in that thread?

@ChetMiller No, it is perfectly fine. Since you wanted to treat the non-ideality, I was interested on solving the system using an equation of state, and generalize it.

Chet, thanks for replying. Could you elaborate on your point  (and the reason) a bit more, if you don't mind?  I would like to save your explanation, much like your original post. Thanks

Proscionexium, thanks for replying. The MIT prof. mentioned dU=dW (only for reversible) and said that it is a confusing but important point, He didn't elaborate much.

The moderators have closed our discussion of the focus problem.  Please go to PhysicsForums.com where I have started a thread on the focus problem.  Please register and then click on Classical Physics.  Then click on Thermodynamics and see the thread I started.

In an adiabatic irreversible process, entropy is generated within the system even though there is no exchange of heat with the surroundings.  In a reversible process, the only way for the entropy of the system to increase in going between the same initial and final state is to exchange heat with the surroundings.  This is the only way entropy can change in a reversible process.  So there is no adiabatic reversible path between the same two end states as those for an adiabatic irreversible process.

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

Let consider reversible and irreversible adiabatic compression of an ideal gas.
If they share the same initial state, they will not reach the same final state unless one of them is combined with some non-adiabatic process.
The irreversible process applies bigger external pressure than system pressure.
\begin{align}
p_\text{irrev,ext} &\gt p_\text{rev}\\
\delta W_\text{irrev} = -p_\text{irrev,ext} \mathrm{d}V &\gt -p_\text{rev} \mathrm{d}V = \delta W_\text{rev}\\
\end{align}
Therefore the volume rate of providing work is higher.
\begin{align}
\left(\frac{\partial W_\text{irrev}}{\partial V}\right)_Q &\gt \left(\frac{\partial W_\text{rev}}{\partial V}\right)_Q\\
\left(\frac{\partial U_\text{irrev}}{\partial V}\right)_Q &\gt \left(\frac{\partial U_\text{rev}}{\partial V}\right)_Q\\
\left(\frac{\partial T_\text{irrev}}{\partial V}\right)_Q &\gt \left(\frac{\partial T_\text{rev}}{\partial V}\right)_Q
\end{align}
The irreversible compression would reach the final temperature and internal energy of the reversible process at bigger volume and lower pressure.
Or, it would reach the final volume of the reversible process at bigger temperature, pressure and internal energy.


text:
Qualitative explanation for equal heat capacities in ideal gases
Hello I was wondering the following thing. As you all know from statistical thermodynamics, if only translational states are available for a given ideal gas (such as He, Ne, Ar, Xe), the isochoric heat capacity sums up to 1.5 R.
The derivation of this value is clear for me, however, I have a little bit of a doubt why all ideal monoatomic gases should have the same isochoric heat capacity, that is, the amount of energy to heat up a sample by one kelvin. Wouldn't it take a bit more energy to heat up heavier elements, or is this compensated by the fact that temperature is proportional to kinetic energy, so the heavier elements have the same temperature even though they have lesser velocity according to the Maxwell-Boltzmann distribution?

So your picture would be the following. Imagine we throw $\pu{100 kJ}$ of heat to "$N_\mathrm{A}$" atoms of $\ce{He}$, and separately to $\ce{Xe}$, both at the same initial temperature. Their final temperature will be the same. But it is not intuitive how this can be true, becuase $\ce{Xe}$ is much bigger, so you'd think that it needs more energy to get to a higher temperature. Right?

Indeed, it would seem that way

Isn't that because we need mass and velocity for kinetic energy?

Heavier particles do move slower than lighter particles at the same temperature.

However, this does not affect the net kinetic energy because of equipartition of energy, which states that (translational) kinetic energy of ideal gases is fixed at a given temperature.
Thermodynamics
$$
\left(\dfrac{\partial U}{\partial T}\right)_V = C_V
$$
According to equipartition of energy, $\dfrac{1}{2}\mathrm{R}T$ is the average molar energy available to each degree of freedom. Since monoatomic ideal gases have only three (translational) degrees of freedom, the energy is given by $\dfrac{3}{2}n\mathrm{R}T$. Substituting in Eq. (1):
$$
\begin{align}
C_V &= \left(\dfrac{\partial \dfrac{3}{2}n\mathrm{R}T}{\partial T}\right)_V
&= \dfrac{3}{2}n\mathrm{R} \tag{1}
\end{align}
$$
Note: The mass of the gas plays no role in the heat capacity of a monoatomic ideal gas.
Kinetic Theory
According to the kinetic theory, the kinetic energy of (independent and identical) particles is given by:
$$
\mathrm{KE} = \sum_{i=1}^{N}\dfrac{1}{2}mv_{i}^2 \tag{2}
$$
Equating Eqs. (1) and (2):
$$
\begin{align}
\dfrac{3}{2}n\mathrm{R}T &= \sum_{i=1}^{N}\dfrac{1}{2}mv_{i}^2\\
\implies \dfrac{3}{2}\dfrac{m}{M}\mathrm{R}T &= \sum_{i=1}^{N}\dfrac{1}{2}mv_{i}^2\\
\implies \dfrac{3}{2}\mathrm{R}T &= \sum_{i=1}^{N}\dfrac{1}{2}Mv_{i}^2\\
\implies \sum_{i=1}^{N}v_{i}^2 &= \dfrac{3\mathrm{R}T}{M}
\end{align}
$$
Note: The velocities are weighted with molecular masses, meaning that heavier particles do move slower on average.
This is a graph of the type $y = \dfrac{c}{x}$, as visualized earlier.

Yeah the derivation is clear for me, however I was asking how one can explain this qualitatively

@Mäßige i hope the latest edit addresses your comment.


text:
Inorganic chemistry, Atomic Mass
How can we determine the atomic mass of an unknown pure sample of an element without using modern techniques like spectroscopy etc... basically I'm interested to know that which physical methods or techniques of wet chemistry used by scientists like Berzilius to find atomic masses of elements without modern day techniques !!
Often methods like Dulong-petite method needs specific heat of solids but again we have some elements like carbon and beryllium which don't have constant heat capacity...so how techniques like crystallization and filtration were used to determine atomic masses?

Really this is a long comment...
In gist, mass and volume were early techniques to determine relative mass between particular elements. But your question is very, very broad. It would take a book to really sort your question out.

Not all the elements were discovered at the same time, and chemical
techniques improved over the ages.
The periodic table was worked out over decades of work.
Realize too that not only did early scientists have to figure out
that a particular element existed, but it was a massive amount of
work to figure out how to get that particular element relatively pure.
You'll also have to throw in scientists figuring out that there
are isotopes.

So the periodic table and atomic masses weren't figured out in some straight line path. It was more like square dancing - up to the middle and back.

ok I understand the process is long but can you comment about discovery of atomic masses of few elements like carbon and chlorine?

For carbon, see section "History and etymology" at https://en.wikipedia.org/wiki/Carbon For chlorine see https://en.wikipedia.org/wiki/Chlorine


text:
Cell Diagram or Representation of an Electrochemical Cell
What is meant by anode half cell and cathode half cell in Cell diagram , I know that at Anode oxidation takes place and at cathode reduction takes place But the main concept which is troubling me is what is meant by half cell of both anode and cathode

The anode is producing electrons. It does not work as an ordinary chemical reaction, since it produces electrons, which are not weighable stuff. The chemical equation taking place at the anode contains electrons, which never occurs in usual chemical reactions. For this reason, this equation is called a half-equation.
The cathode is consuming electrons coming from the anode. It does not work like an ordinary chemical reaction, since it consumes electrons, which are not weighable stuff like a metal or an acid. The chemical equation taking place at the cathode contains electrons, which never occurs in usual chemical reactions. So this equation is called a half-equation.
A half-equation can never exists alone. It must be accompanied by another half-equation. The same number of electrons are created on the anode side, which are consumed in the cathode side.


text:
Table of molecular orbital extents — what does it even mean?
In molecular orbital theory, the term molecular orbital extents was mentioned, although I can't really picture it. Does molecular orbital extent refer to the spatial distribution of electrons in a molecule?
When I calculate molecular orbital extents with $\ce{H2O}$, the following table is provided.
  Orbital extents (a.u.):
        MO          <x^2>          <y^2>          <z^2>          <r^2>
   0 A1  0   0.0177335990   0.0177345751   0.0314235271   0.0668917013
   1 A1  1   0.5165227329   0.7453494022   0.5948204224   1.8566925575
   2 B2  0   0.4227875842   1.8458686046   0.6860187318   2.9546749206
   3 A1  2   0.4563255952   0.6542639495   1.4243758427   2.5349653874
   4 B1  0   1.3086927738   0.4362309246   0.4500334749   2.1949571734
   5 A1  3   3.2682125957   5.9000542358   4.8761186497  14.0443854813
   6 B2  1   2.8796959432  10.2335899558   4.1395581510  17.2528440501
   7 B2  2   1.2848895351   4.1475258156   2.3869042266   7.8193195772
   8 A1  4   2.2181684269   3.7911494454   2.5513702157   8.5606880879
   9 B1  1   2.8880726336   0.9626908779   0.9764934282   4.8272569397
  10 A1  5   1.0167358798   1.0037029230   2.9703269612   4.9907657641
  11 B2  3   1.1288038413   3.4979906480   1.5412110158   6.1680055051
  12 A1  6   1.5715312779   1.8762037282   1.5440486064   4.9917836126
  13 A1  7   1.0293258031   3.4022681941   1.8912304400   6.3228244372
  14 B2  4   0.8946420761   3.9039643575   1.9319712799   6.7305777135
  15 B1  2   0.8650504801   0.2883501600   0.3021527104   1.4555533504
  16 A1  8   0.2888876394   0.3407682209   0.8571143170   1.4867701773
  17 B2  5   0.2835432585   0.9003917411   0.3971506399   1.5810856395
  18 A1  9   0.0413406458   0.0433229266   0.0554579364   0.1401215088

However, I am not sure what it means. If someone is familiar with this topic, please teach me.

I don't know which program you are using, but it might be worth seeing if you can visualise the orbitals somehow (in, Avogadro, ChimeraX, Jmol, or any other program you have access to)

I see, I should visualize it!

For something like water, though, you can just google 'water molecular orbitals' and look at the images... it took me less than 10 seconds to find this: https://chem.libretexts.org/@api/deki/files/77758/water-mo.png?revision=1

I recommend using Avogadro. It's open source, and reasonably user friendly.

What is the origin of the table?

The background to all this might be localisation methods for MOs, in particular the Foster-Boys method - see https://en.wikipedia.org/wiki/Localized_molecular_orbitals

It looks like $<x^2>$ relates to the variance of the electron distribution, i.e. a measure how "big" the molecular orbital is in each direction. The final column seems to be the sum of the variances in x, y and z.
The first MO is almost exactly equal to the inner atomic orbital of the oxygen atom, and has the smallest extent. (Strangely, it is not spherical with equal variances in the three directions.)
The header says that values are in units of a.u., yet you would expect the square of a length if the values were the variance. On the other hand, if the extent were defined as the square root of the variance, I would not expect the values to add up to the last column.
I'm not sure orbital extent is a widely used technical term. I found one paper that talks about orbital variance, though:

The orbital variance (or spread) of orbital | p〉is a measure of the spatial extent of that orbital and thus of the locality of the orbital.

Source: https://pure.au.dk/ws/files/38005281/JChemPhys_134_194104.pdf

Isn't the last column the sum of the other three just because, $<r^2> = \int \psi^\hat{r}^2\psi ; \mathrm{d}\tau = \int \psi^(\hat{x}^2 + \hat{y}^2 + \hat{z}^2)\psi ; \mathrm{d}\tau = <x^2> + <y^2> + <z^2>$?

@MetalStorm Yes, that makes sense. But then (a.u.) does not make sense unless it is meant to say "appropriate atomic units", which is a0 squared in this case, I guess.

Yes, that was my same question...


text:
Why my mineral water from different sources do not mixed well?
Background:
In 8 hours range, I have bought cold mineral water (brand A), then refill the plastic bottle with cold mineral water (brand B) from my office. As far i know, brand B has high pH while brand A is almost neutral. After that, I refill again with another mineral water (brand C). After few minutes, I'm surprised the inner-wall of my plastic bottle is filled with micro bubbles. These bubbles seems have sticky property because they are hard to clean even on hard tap. Even they do not burst easily when in the surface almost like bubble soap.
Other property:

Has normal taste, slightly base
Has no smell
Has no color

Question:
Is it possible for mineral water from different sources to not mixed well then generates (yes, they generate!) hard-to-burst bubble? Or something has contaminated my relatively new plastic bottle?
Supporting Image:
Legends:

Fogs
Floating bubbles. Haven't burst in 5 minutes
Bubbles
Bottle condition after hard-tap

Generally, it is not a good idea to mix different kinds of mineral water.

@Poutnik Why that? :D

"... brand B has high pH ... "  Extremely unlikely if you say that "brand A is almost neutral"

@Karl For similar reasons as it is generally not good idea to mix 2 solutions unless you really know what you are doing. :-)

@Poutnik Good general precaution for chemisty, but a bit out of proportion for drinking water.

@Karl Mineral water is not (ordinary) drinking water. It would stay drinkable after mixing, but some combinations of ions may lead to undesired effects of reaction, sedimentation, gas evolving, disbalancing composition for desired health effects.

@Poutnik Sorry, but that's ridiculous nonsense! Please give me ANY example of two supermarket mineral waters (where I can look up the analysis online) that would give the effects you describe.

@Karl You say by other words it is ridiculous nonsense that behaviour of dissolved molecular entities could be possibly affected by pH shift or redox potential shift, or being combined with other molecular entities, leading to precipitation or forming soluble but stable complexes.

Let us continue this discussion in chat.

The bubbles, floating or not floating, are made of $\ce{CO2}$ which was initially dissolved in one of the mineral waters (physically dissolved or combined as $\ce{HCO3^-}$ ion). By mixing it with a slightly more acidic water, this $\ce{CO2}$ is liberated and forms tiny bubbles.

Thank you! Glad to know

Is the analytical result spectral or putative? Could the bubbles be insoluble, therefore not much affected by tapping? That is, unaffected by water as a solvent. CO2 is assumed because Brand C is assumed to be acidic and the (assumed?) high pH of Brand B is assumed to be from bicarbonates - but it could be due to Mg(OH)2. Temperature is reported vaguely. Could the temperature of Brand C be somewhat colder, therefore likely to warm up and produce micro bubbles of air because it had dissolved more air to begin with? Air bubbles would have the same visual spectra as CO2 bubbles.

James Gaidis asks whether the bubbles are insoluble. This question has no meaning. All bubbles contains an insoluble gas. If the gas were soluble, the bubbles would get dissolved and thus disappear. Furthermore, $\ce{Mg(OH)2}$ is a solid. It is nearly insoluble. But it is not a gas.


text:
Biocompatible, high refractive index medium for 780 nm?
I am trying to find a medium for optics experiments on living cells. I have been using yeast cells, saccharomyces cerevisiae and saccharomyces boulardii, and need a medium with refractive index (RI) greater than $1.4$ at a wavelength of $780 \,\pu{nm}$ (and temperature $25^\circ \pu{C}$) and in which the cells are still able to live and reproduce. I have been trying to mix their YPD medium with substances such as polyethylene glycol (PEG) and glicerin at different mass fractions, but to no avail: either the cells died or the RI was still not high enough. The “tuning” substance would need to have a RI larger than $1.4$ because it would need to be mixed with YPD (which has basically the same RI as water), and it also would have to be hydrophilic to some degree (preferably shouldn’t be too viscous, but that can be overlooked). Does anyone know about other alternatives? I have heard of iodixanol through microscopy papers, could this work?

First the refractive index of PEG varies by molecular weight and with time, as it gels. You might try using low-MW PEG, and letting it stand mixed with the YPD culture media. It is one of the most innocuous chemicals.
Your thoughts on trying Iodixanol is certainly reasonable, and supported by the literature, but you'd have to test it with your yeast. Actually, that would be a good subject for a paper to help others using high-resolution optical (well, near-IR, possibly barely visible to dark-adapted eye) microscopy, and if you do publish it, make that an answer to replace this, and add citation to it!
You could also try some other liquids to increase the R.I., but likely a high concentration would kill yeast by osmosis, if not outright poisoning. Here are some hydrophilic fluids  that might be relatively nontoxic. BTW, the metabolic products of ethylene glycol make it poisonous to mammals, but I don't know about yeasts.
See table for R.I. of glucose/water mixtures.
R.I.  Chemical
1,463 Ethylene Glycol
1.433 Propylene Glycol
1.421 Propylene Carbonate
1.473 Glycerol
1.400 Glucose 40%, 60% water


text:
Class vs Functional Group: Imprecise Language?
Alcohols, carboxylic acids, esters, alkanes etc are classes of organic compounds, not functional groups, no? (alkanes contain no functionl groups!)
In other words, a class is a type of compound, while functional groups specifically refer to a group of atoms (or more generally a bonding sequence).
However, it appears that both are mixed up very frequently. So are alcohols both the class and the functional group? Or is this an example of bad language?
Examples:

Alcohol is a functional group that you are probably familiar with.

From:
https://kpu.pressbooks.pub/organicchemistry/chapter/2-3-functional-groups/
Instead, the table in the above link says "class" instead.

The sentences below sounds correct however (perhaps if a possessive is added: alcohol's, ether's, thiol's):

The alcohol functional group involves an oxygen atom that is bonded to one hydrogen atom

The ether functional group consists of an oxygen atom that forms single bonds with two carbon atoms.

The thiol functional group contains a sulfur atom bonded to a hydrogen atom.

https://chem.libretexts.org/Courses/University_of_Kentucky/UK%3A_CHE_103_-Chemistry_for_Allied_Health(Soult)/Chapters/Chapter_4%3A_Structure_and_Function/4.4%3A_Functional_Groups

I do see certain exceptions like ether referring to both the class, and the functional group (formally, alkoxy group). Perhaps this happens often for more complicated compounds; then it slowly encroached upon simpler ones....

Phenyl vs phenol functional group?
What is a functional group?
Inter-classification of organic chemistry functional groups (carboxyl vs alcohol & ether vs ester)
Glossary of class names of organic compounds and reactivity intermediates based on structure (IUPAC Recommendations 1995)
(http://dx.doi.org/10.1351/pac199567081307)
10.1 - What are the 'functional groups' and 'classes' of organic compounds?
https://youtu.be/tL_KSmwIm2I
An article provides nice intuition on the 'functional group approach'
https://www.chem.fsu.edu/chemlab/chm1046course/functional.html

I think the term class is sometimes also used to refer to the compounds containing a specific functional group because LibreTexts has used yet another word family that adds to the confusion.

@Proscionexium I see that family is an even broader classification according to the elements present in the functional group

In some cases, there is a clear correspondence between class and functional group. Alcohols are a class of compounds that all contain a hydroxyl functional group. The carbonyl functional group is an example where it gets more complicated. If a single carbonyl functional group is located at the end of a chain, we get an aldehyde, otherwise a ketone. Also, if the carbonyl functional group is combined with additional heteroatoms (i.e. something other than carbon or hydrogen), we use new names for these functional groups such as ester or amide or thioester.
There are also classes of compounds that contain multiple functional groups such as carbohydrates or dicarboxylic acids. In the end, the nomenclature has to support the amazing complexity and diversity of organic compounds that exist or are dreamed up.

However, it appears that both are mixed up very frequently.

Yes, sometimes it is easier to say alcohol-group instead of hydroxyl. In any field, there will be jargon and day-to-day usage that diverges a bit from the official terms. Also, as organic chemistry is developing, there might be functional groups or compound classes that are novel, and language describing them might change over time.

I think I have two points which can near what answer your question.

The term functional group mainly refers to a group of atoms which imparts certain physical and chemical properties when attached to an organic molecule.
So based on these alcohols, acids, esters etc. are called functional groups.

Classes as said by @Proscionexium, is used to refer to a group of compounds having similar chemical properties (because of functional groups as stated before) but may have different physical properties (which could be due to different molecular mass etc). For example, consider a series C(n)H(2n+1)-COOH, all the compounds can be termed as a class of mono carboxylic acid as they show similar chemical properties.

I hope my answer could clear up your confusion .

+1, but I would strike that functional groups impart physical properties. Of course they do in the end, but the main point is chemical properties. Ethanol and decanol hardly share any physical properties.


text:
How to effectively replace DMSO with LC–MS-compatible solvent for the samples from a well plate?
I want to do LC–MS analysis of the samples from a 384-well plate where each compound (at around 1–10 μM) is dissolved in 10–15 μL of DMSO. I am worried about the DMSO dwarfing the analyte’s signal, as well as the challenges from ion suppression caused by the DMSO.
So, I’d like to evaporate DMSO and replace it with something like ethanol/water (1:1) before passing it on to my autosampler. Does anyone have any experience doing this? The DMSO removal protocols I found are suited for removing a much larger quantity from a single sample.

Have you tried to dilute-and-shoot? Could you share additional information about the analyte and MS? Do you know why DMSO was chosen in the first place and whether it can be replaced with something else in the future shall there be more samples? I presume it might a biological extract with DMSO used as cryoprotectant. Also, do you have access to a high-performance freeze-dryer? Freeze-drying DMSO is a pain, but for small quantities like this it might be worth considering (check out volatility of analyte).

These samples were generated from some colleagues' internal HTS libraries that they had in the back of their freezer, so are all small drug-like compounds -- some solo compounds, some in mixtures.  This is in the planning phase so we haven't actually tried the dilute-and-shoot yet. The compounds of interest are going to be around 100nM.

Why not just rely on chromatography to separate the analytes from the DMSO?  Sure, there might be a huge DMSO peak in your chromatogram, but DMSO isn't retained all that well on C18 chromatography resins.  If most of your analytes are more hydrophobic than DMSO they should be separable from the DMSO during chromatography.  If so, then DMSO won't lead to ion suppression.

If you have 10 mM solutions, after 100x dilutions you still have 100 uM.  If you inject 1 uL of that dilution, it's still 100 pmol on-column.  Should be easily detectable by any decent LC-MS setup.

The standard way to remove water and replace with volatile solvents for microscopy is with a dehydration series, e.g., replacing water with ethanol, and then ethanol with acetone. To more thoroughly remove solvents, critical point drying is used, which should greatly reduce the DMSO in the sample, leaving $\ce{CO2}$. However, if you're analyzing volatile components, then this would badly change the sample!
If there is a microscopy lab at your institution, you might try to use some of their apparatus for dehydration.
However, rather than put valued samples though the treatment, first make a control sample, with known chemicals plus DMSO solvent, and evaluate via LCMS before and after the drying treatment to see how effective drying is and if it is even needed.

I'm afraid using dehydration series could result in even greater substance and, consequently, signal loss that that from DMSO the suppressor. I'm not sure how to use critical point drying for the well plate and not to face a carry-over or cross-contamination problem. Are there any models suited to use with well plates, especially with the larger ones that OP is using (384 wells)?


text:
Resonance energy of organic compounds
I was solving a problem in organic chemistry, where I had to find the resonance energy of a compound with some information which was given in the question itself. Basically, what I had to do was, to sum up all the energies from the information given and then subtract it with the experimental value of energy given. My question was that if the law of conservation of energy was to be followed while solving the problem then where was the energy which I obtained as a difference of the experimental and theoretical data used? And if it is just a calculation of experimental error, then why was it called as resonance energy, was it related to its name in anyway.

What I understand from your question is that you were given different bond energies of the various bonds constituting a compound which was showing some resonance and also you were given the experimental value of the energy of that compound.
Now as a compound is formed when different atoms form bonds together so, the total energy of that compound should be equal to the sum of all the bond energies of bonds between different atoms of that compound but this is not the case (most of the time in the real world).
Every compound tries to go to the lowest energy as possible and one way to do so is by doing resonance.

Resonance is the phenomenon of delocalisation of bonding electrons of a bond between many atoms of a compound rather than just two atoms.

By doing resonance, the electrons in a compound are more evenly distributed and delocalised resulting in more stability of the compound and thus a lower energy. So, the actual value of the energy of a compound is lower than just the sum of bond energies of all its bonds and the decrement in the value is equal to the energy it released in the form of resonance energy. Yes! It is released in the surroundings.
I hope it helped.

As a guide if you are familiar with the quantum model for a particle in a box you will know that the levels becomes lower in energy as the box gets larger. The same is true for a particle on a ring as the ring's radius increases. So, if you have three isolated double bonds, vs. the same bonds with delocalised energy (double - single - double -single  etc., and you may call this resonance if you wish) then you can see that the latter has lower energy.


text:
Why sulfuric acid is used in esterification of alcohol and carboxylic acid anhydride?
I have a question regarding why sulfuric acid is used in esterification of alcohol and carboxylic acid anhydride.
I know that in case of esterification of alcohol and carboxylic acid, sulfuric acid remove water formed as product and it keeps reaction to proceed in forward direction.
$$\ce{CH3OH + CH3COOH <=>[H2SO4] CH3OCOCH3 + H2O}$$
But in case of esterification of alcohol and carboxylic acid anhydride there is no water formed so what is the real reason of using sulfuric acid in this case.
$$\ce{CH3OH + (CH3CO)O2 <=>[H2SO4] CH3OCOCH3 + CH3COOH}$$

Concentrated sulfuric acid is anhydrous. All the other common concentrated acids contain water.

@MaxW  But what is the purpose of sulfuric acid in second reaction if there is no water being released that it could remove.

What is unclear about the word "catalyst"?

@Mithoron How sulfuric acid promote reaction even if there is no water to be removed? How it catalyse reaction? What's the the process?

@MaxW Not completely anhydrous. @ OP Your "water removal" is a misnomer, at best. What acids do? Protonate. Then again, with anhydride, this reaction should go pretty well even without it being additionally protonated.

To start, we should consider the physical state of reactants. We mostly take aqueous solutions of methanol and anhydride for reactions. So one of the reasons to use sulfuric acid can be to remove water as soon as product is formed as in the presence of water, the reaction is highly reversible.
Second reason could be that the reactions proceed with maximum efficiency (rate) in an acidic medium (it acts as catalyst also), so in order to maintain acidic buffer sulfuric acid is used.
I hope my answer helps.

The second reason sounds good but the first reason that sulfuric acid would remove water is bit strange as in aqueous medium there is a lot of water and sulfuric acid cannot just evaporate entire glass of water and leave behind only products and also acid anhydride would just react with water so this reaction cannot be done in aqueous medium.


text:
Hydration of alkenes: is phosphoric acid a gas at these temperatures?

Ethene is mixed with steam and passed over a catalyst consisting of solid silicon dioxide coated with phosphoric(V) acid. The temperature used is 300°C and the pressure is about 60 to 70 atmospheres.

Source: https://www.chemguide.co.uk/physical/catalysis/hydrate.html
But isn't phosphoric acid vapourized at this temperature? I know that phosphoric acid is a gas at 158°C, assuming atmospheric pressure. Perhaps the high temperature increases the boiling point, so that phosphoric acid remains an liquid at this temperature?

Sloppy writing of that "chemguide" (is that a reputable source?). What they mean is silicon dioxide that is acidic on the surface because it was treated with phosphoric acid.

Phosphoric acid definitely is not a gas at $\pu{158 ^{\circ}C}$. It undergoes boiling decomposition at $\pu{212^{\circ}C}$, releasing water and condensating to pyrophosphoric acid
$\ce{2 H3PO4 -> H2O + H4P2O7}$ and eventually higher polyphosphoric acids.
$[\ce{\bond{-}O-P(=O)(OH)\bond{-}}]_n$
Such condensation progressively occurs even at room temperature if concentration of phosphoric acid is higher than 85% (about equimolar acid:water ratio) what is the reason the commercial product has this concentration.
Being initially coated by phosphoric acid does not mean phosphoric acid is present  at operational conditions. It would be already polyphosphoric acid.


text:
Acetone resistant elastomer
I am in search of an elastomer that can be dissolved in solution and applied and cured at room temperature, but that has good chemical stability to acetone once cured. Does any such elastomer exist?

From this chart at Mykin, EPDM, Butyl and Kalrez (FFKM, a Viton derivative) are marked compatible with acetone.
As for applying in a solvent:

EPDM dissolves in toluene and trichloroethylene, but it is not stated if the polymer is usable after evaporation.
Butyl rubber can be dissolved in hexane and other solvents, and is sold as liquid butyl rubber.
Viton-based adhesives are available.

However, you'd need to research if any of these are suitable in your specific application, and if any are cost-effective.


text:
Can diols be oxidised to hydroxyacids?
Conversion of ethanol to ethanoic acid
$
\ce{CH3CH2OH_{(aq)}
  \xrightarrow[KMnO^{-}_{4}]{NaOH}
  CH3CHO_{(aq)} + H2O_{(l)}
}\\
$
$\ce{CH3CHO_{(aq)} + H+_{(aq)}
  \xrightarrow[KMnO^{-}_{4}]{HCl}
  CH3COOH_{(aq)}
}$
Are the steps above correct
Can I apply the above steps when the starting reagent is a diol?
I can't find any information about this on google.
Diol to hydroxy-acid
$
\ce{
  ethane-1,2-diol_{(aq)}
  \rightarrow
  \alpha-hydroxyethanoic acid_{(aq)}
}
$
Equation is not balanced, just shows the reaction I want to do.

This will give oxalic acid.

Can you please guide me on how I can produce a hydroxyacid? Is the first reaction valid for producing a carboxylic acid? Also could you write an answer so I can give credit? Compared to other stackexchanges, chem is very quick to downvote without explanations, perhaps you could explain the downvotes?

You want an explanation? If this was a homework, then it would be lame. For actual synthesis, it doesn't make sense at all. I think Waylander explained why.

yep it is for synthesis, I understand him now

can you see my last question to waylander and help me as to how I can avoid downvotes as I need to keep asking question (graduated chemistry 10 yrs ago)

Firstly no-one runs this reaction with ethane-1,2-diol (ethylene glycol) because glycolic acid is widely available. However if you want to run the oxidation of a primary alcohol to carboxylic acid in the presence of an alcohol elsewhere in your molecule you need to develop a protection strategy so that the alcohol you want to keep as an alcohol is temporarily turned into something else which will not react and can then be easily returned to alcohol. This is an important part of synthetic organic chemistry. There are whole books written about this e.g. Greene's Protection Groups in Organic Synthesis  here.
A suitable group for alcohol protection under basic permanganate oxidation conditions would be the tetrahydropyran ether, more details here
If you have 2 primary alcohols then you have a challenge in finding conditions so that only one reacts to form the protection group.

thanks but I'm more interested in using the diol to make glycolic acid, where I live I can't find glycolic acid, whereas ethan-1,2-diol is readily available and cheap. Using your source can I apply the same steps for the alcohol to carboxylic acid for a diol to hydroxyacid?

Also I don't understand why you would want the the temporary molecule (aldehyde?) to turn back in to the alcohol.

It should be possible to make the mono-THP ether of ethylene glycol from a 1.05:1 mixture of ethylene glycol and dihydropyran where the bulk of the material is mono-protected and the unprotected material is easily removed after the oxidation.

You want to be able to turn the temporary molecule back to the alcohol because you need the alcohol group in the final molecule or for a further transformation

ah I see so this cuts out the step of second oxidation? I am still confused alcohol + [O] -> aldehyde + water then aldehyde + [O] -> carboxylic acid that's how I understand the process so can you explain further so I understand your reasoning of turning the aldehyde back into the alcohol

sorry aldehyde + H^{+} + [O] -> carboxylic acid I think you're using a more advanced procedure, that's why I don't understand

basic permanganate takes the primary alcohol direct to carboxylic acid. If you want the aldehyde use a different oxidation agent

ok I understand the protection now, it's to protect 1 of the hydroxyl groups so with a diol there is a difference thank you

Please accept the answer if you are happy with it.

of course, I just have 1 questions before I delve in to your sources, when protecting the hydroxyl group, how can I be sure that both will not be protected?

I graduated chemistry 10 yrs ago, online sources had nothing on diol conversion, I have some textbooks but because I've been out of chemistry so long (I code now) I still need guidance. Any advice on why this question got so many downvotes and how I can avoid it in the future, as I'd like to keep asking questions. Protection never would've crossed my mind and I may have missed it despite reading my textbooks. Any reasons you might have downvoted it? I'd like to keep in touch with you as you didn't just scoff the question off like most people.

Let us continue this discussion in chat.


text:
What is the shape of tetraamminenickel(II) complex?
I have basic knowledge of crystal field theory (CFT) and how to predict geometries in complexes with a coordination number of 4.
For $\ce{[Ni(NH3)4]^2+}$, we find that the metal center, $\ce{Ni^2+}$, has a $\ce{d^8}$ system.
I believe that $\ce{NH3}$ should behave as a weak field ligand here so a high spin complex should be formed with tetrahedral geometry.
I am making this assumption because, to my knowledge, $\ce{NH3}$ behaves like a weak field ligand in hexaammine complexes of $\ce{Cr^2+}$, $\ce{Mn^2+}$, $\ce{Fe^2+}$ and $\ce{Co^2+}$ and high spin complexes are formed.
I am not sure about this, since $\ce{NH3}$ lies in the middle of the spectrochemical series (its a middle field ligand?). Is my analysis correct?

What shape do you think it would be if it were low spin? What shape is the platinum analogue? Why is it different?

I think square planar for both? Because $\ce{Pt}$ belongs to the $\ce{5d}$ series so pairing of electrons in lower orbitals is energetically favorable in most cases. But how is that related to the complex in question?

In both complexes the metals are d8. In both complexes they have 4 ammonia ligands. But the structure is different. So maybe the fact that the ligand is ammonia is not the main deciding factor between the two structures? Maybe it is the metal?

You might also want to look at https://chemistry.stackexchange.com/questions/40880/why-is-pdcl42-square-planar-whereas-nicl42-is-tetrahedral


text:
Heat increase after pressure drop in polymer filter
I would like to calculate the heat increase in polymer filter. A polymer filter (mesh size p.ex.20 µm) decreases the mass pressure of the polymer flow.
How it works: The polymer is melted in an extruder, this builds up some mass pressure. This pressure is then enhanced by a melt pump to e.g. 150 bar. The polymer is then filtered (seee image below). Thereby the melt pressure drops to e.g. 80 bar. I imagine that there should be a temperature increase in the filter due to the pressure drop.
How can I calculate the temperature increase?
Side info:
When other materials expand (as gas in a refrigerator), the temperature decreases. This is because the intermolecular distance increases. Thereby the atoms potential energy increases. I am just mentioning this to stress that polymer melt is almost incompressible, so this effect is not valid here.
Reference: https://www.physicsforums.com/threads/why-do-gases-cool-on-expansion.470311/


text:
Reduction of La(NO3)3 to La(NO2)3
I'm writing a master's thesis on some lanthanide compounds, in particular those with $\ce{Ce^{+3}}$ and $\ce{La^{+3}}$, although my focus has now shifted to lanthanum, but I ran into a problem when I could not precipitate my anion, which is a coordination compound, with lanthanum so to solve this problem I modified my reaction setup to start with $\ce{La(NO2)3}$ instead of $\ce{NaNO2}$ so that when I want to precipitate the coordination compound I don't have those pesky sodium ions around.
The thing is, I only have a $\ce{La(NO3)3}$ in store, so I was wondering if there is a way to reduce the nitrate ion to nitrite without changing the lanthanide oxidation number.

Be careful to distinguish lanthanum $\ce{La}$ and lanthanides $\ce{Ln}$ as the element group from $\ce{Ce}$ to $\ce{Lu}$.  The former has redox potential $\ce{La/La^3+}$ about $\pu{-2,37 V}$, similar to $\ce{Mg}$. So no worries for $\ce{La^3+}$ to be reduced. It may matter for lanthanides forming $\ce{Ln^2+}$ ions. Check the respective redox potentials for $\ce{Ln^3+/Ln^2+}$ at link, the same for $\ce{NO3-/NO2-}$ and choose the reduction agens with suitable reduction kinetics for nitrates.

You have shifted the question from lanthanides to lanthanum, but not thoroughly. You have to choose: either lanthanum(La), either lanthanides(Ln). For La, reduction of lanthanum ions will not occur.

I've done so because I was asked to be more specific about the lanthanide. If possible a general response that serves for most if not all lanthanides would please me, if not, I'm satisfied with only La and Ce, since this are the focus of my thesis.

Zinc powder can reduce nitrate to nitrite, see Morita and Nakamura[1]. Lanthanum is much higher than zinc in the EMF series, so a side reaction of reducing the lanthanum ions will not occur. However, the reduction process may introduce zinc ions to the solution. Based on Pourbaix-diagram data, maintaining a near-neutral pH between 6 and 8 is suggested for precipitating the zinc as hydroxide while maintaining good solubility of the lanthanum ions.
Reference

Morita, Emi & Nakamura, Eiko. (2008). "Reduction of Nitrate to Nitrite by Using Zinc Powder for Determination of Total Nitrogen in Sea Water". Bunseki Kagaku. 57. 777-781. 10.2116/bunsekikagaku.57.777.


text:
Scrubbing methanol and ethanol from a gaseous mixture
I have a gaseous mixture (air) at a temperature of approximately 30 degrees Celsius in which the following gases are present:

Oxygen 0 - 25%
Carbon Dioxide 0 - 20%
Methanol 0 - 100 ppm?
Ethanol 0 - 100 ppm?
Ethylene 0 - 20 ppm

I want to measure the concentration of each of the gases individually. That is not a problem with Oxygen and Carbon Dioxide but becomes a problem with methanol, ethanol and ethylene because of the cross sensitivity most sensors have for these gases.
Methanol and Ethanol are in liquid state at the temperature of 30 degrees Celsius. In the air the Ethanol and Methanol is present as vapor.
Would it be possible to design a 'scrubber' in which we bubble the air mixture through a container of for example liquid methanol to get rid of the methanol vapor and sequentially through a container of liquid ethanol to get rid of the ethanol vapor.
I can imagine that ethanol might also dissolve in the methanol liquid? And another issue would be the volatility of the ethanol and methanol liquid since we want to prevent new vapor getting into the air stream after the scrubbing.
Do you think this would be a feasible direction to investigate further? Another way to go would be to scrub it through water maybe? If we would be able to separate those two gasses we would then be able to determine all the concentrations individually by measuring the original air mixture and the scrubbed air mixture with two different sensors.

A scrubber made by passing the gas through the liquids will probably increase their concentration in the gas.

Thanks for your reply. I was indeed wondering what the effect would be and thought it could go either way depending on the ratio between volatility and mass transfer. I am a bit rusty on this subject so thanks again for helping me out.

Try a trap filled with a high boiling point liquid, such as glycerol or propylene glycol, which would absorb the organic gases... but would require periodic replacement or heating to drive off vapors.
If you are not concerned by the slight increase in $\ce{CO2}$ that this would create, and if there is sufficient oxygen, a hot mesh of copper, or better, of a platinum metal, would catalytically oxidize the three contaminants.
Also, a cold trap, e.g. with Peltier cooling, could greatly decrease ethanol and methanol, though it would not decrease low-boiling ethylene.

Thank you for your answer. I will look into the cold trap method. To be able to measure the gasses individually it is important that we only extract methanol and ethanol. By measuring both the original gas mixture and the mixture without ethanol and methanol we can then determine the ratio of the different elements in de original mixture. We have found one PID sensor that is responsive to ethylene and ethanol but not to methanol and we can use an electrochemical sensor that is responsvie to all three gasses.