text:
Does 1,4-dimethylcyclohexane have a chiral center?
My textbook says that 1,4-dimethylcyclohexane (cis or trans) does not have a chiral centre, since two groups on the tertiary carbon (that are the part of the ring) are identical, and thus the compound cannot be called a meso compound. But it also says that it has two stereogenic centres (C-1 and C-4), since interchange of groups around them creates a geometrical isomer.
The IUPAC definition of chiral center is:

An atom holding a set of ligands in a spatial arrangement which is not superposable on its mirror image. A chirality centre is thus a generalized extension of the concept of the asymmetric carbon atom to central atoms of any element, for example N+abcd, Pabc as well as Cabcd.

This definition of chiral center is also equivalent to (for coordination number 4) saying that if exchange of any 2 ligands on the carbon atom creates a stereoisomer, then that carbon is chiral center. So according to this, 1,4-dimethylcyclohexane has two chiral center and can be called a meso compound.
But my textbook says the opposite. So, my question is, which one is correct?
(My basics regarding chirality and stereochemistry is clear but the confusion in this question arose as I think that my textbook considered that since the two groups on the tertiary carbon seem to be identical, so it is not a chiral center, but I think it is wrong.)

There are only 2 stereoisomers here, RR and SS are identical, and RS and SR are identical.

@NisargBhavsar if you are able to define R and S, then the carbon must be chiral, and that's what my question is about.

No time for an answer, but see pseudo-asymmetric centre. Google, or searching on SE, will throw up many descriptions.

pseudo-asymmetric centres are created when 2 of the 4 components of a carbon are same structurally but have different chirality correct me if I am wrong, @orthocresol but I don't think "pseudo-chirality" applies in this particular case.

@napstablook That definition is correct and applies precisely here. To go into why would be way beyond a comment, but in short, you need to construct the full digraph which involves "breaking up" the ring: there is a similar example in https://chemistry.stackexchange.com/a/35511/16683.

I would still argue that that definition is improbable here keeping in mind that both the carbons would be pseudo-asymmetric, I think the condition clearly wants to apply chirality on other carbons in 'expanded structures'  rather than depend on "pseudo-chirality" of another carbon to call it pseudo-asymmetric. But it is entirely possible that I am misinterpreting the definition.

@NisargBhavsar: The descriptors should be r/s, and not R/S, because both centers are stereogenic and achirotopic.

As it is said in the definition of IUPAC, for a carbon atom to be chiral it should be attached to 4 different groups:

In case of 1,4-dimethylcyclohexane, you will see that the two groups on the right and on the left are identical. So in this case it's not Cabcd, it's Cabbc. If you alter one group (let's say the one on the right), these groups will no longer be identical, so you will have 4 non-identical groups attached to a carbon atom, and this atom will become a chiral centre.

Meso-compounds are compounds that have chiral centre but they also have a plane of symmetry so that they are still superimposable with their mirror image. Following is the example of a meso-compound from McMurry's book (Ref.1):

So a meso-compound must have a chiral centre by definition, but if a compound has a chiral centre, that doesn't necessarily mean that it's a meso-compound. And the answer to your question is no, 1,4-dimethylcyclohexane doesn't have any chiral centre.

References:

John E. McMurray, In Organic Chemistry, Ninth Edition; Cengage Learning: Boston, MA, 2016, pp. 133 (ISBN-13: 978-1-305-08048-5).

Keep in mind that you should provide all details of the references. Look at my editing as an example, which is in ACS format.


text:
Gibbs free energy of mixing in a control volume
I'm trying to model the amount of energy lost due to mixing of a $\mathrm{HCl}$ solution using the Gibbs free energy of mixing. However, I really like a second opinion on the model and the results, because think the energy is high when compared to the theoretical available energy density. The model results in $\sim7\textrm{ kWh/m}^3$ change in Gibbs energy from mixing compared to theoretical available energy of $11.1\textrm{ kWh/m}^3$  for $1\textrm{ M HCl}$ in a acid base flow battery$^1$. Is this change in Gibbs energy then all transformed in heat if we assume no energy is used for work?
Model

A concentration $c$ ($\mathrm{M}$) is flowing out of the control volume which models a tank with flow rate $F$ ($\mathrm{L/s}$). A constant amount of moles per seccond is taken out of the outcoming concentration, called the mole rate $J$ ($\textrm{mol/s}$). The returning concentration $c_r$ ($\mathrm{M}$) flows back in the control volume with the same flow rate $F$ ($\mathrm{L/s}$) .
Mass balance for the control volume can be written as
$$
V\frac{dc}{dt} = Fc_r - Fc
$$
With returning concentration $c_r$
$$
c_r = \max\left(\frac{J}{F} + c,\:0\right)
$$
Note that the returning concentration $c_r$ is capped at $0$ to make sure negative concentrations don't occur. Concentration after mixing is written as
$$
c_m= c+\frac{dc}{dt}\Delta t
$$
Homogenous mixing of the concentration in the control volume is assumed after time 2. Therefore we can say that
$$
\Delta G_{mix} = G_2 - G_1 = G_{m} - (G_{c}+G_{d})
$$
Where $G_m$, $G_c$, and $G_d$ are the Gibbs energy after mix, of the concentrated part, and the diluted part respectively ($\textrm{J}$). Here Gibbs energy is:
$$
G = \sum_i \mu_in_i
$$
Where $\mu_i$ is the chemical potential of component $i$ in the solution ($\textrm{J/mol}$), and $n_i$ is the number of moles of component $i$ in the solution. The chemical potential for ideal solutions without pressure change and charge transport can be written as
$$
\mu_i = \mu_i^0 + RT\ln x_i
$$
Where $\mu_i^0$ is the standard chemical potential ($\textrm{J/mol}$), $R$ the universal gas constant $[8.314\textrm{ J/mol.K}],$ $T$ temperature ($\textrm{R}$), $x_i$ the mole fraction$^2$.
Replacing $n$ by $cV$, and substituting $\mu_i$ in the $\Delta G_{mix}$ results in
$$
\Delta G_{mix} = \sum_{i}\left( \left[ c_mVRT\ln x_{m} \right]_{mix} - \left[ c(V-F\Delta t)RT\ln{x_c}  \right]_{con} - \left[ c_r(F\Delta t)RT\ln{x_r} \right]_{dil} \right)
$$
Note that the standard chemical potential $\mu_i^0$ is elliminated.
Replacing the summation for $2$, because we have two ions with the same concentration in the system, and writing $\Delta G_{mix}$ in differential form leads to the final equation of
$$
\frac{dG_{mix}}{dt} = 2FRT(c\ln{x_c} - c_r \ln{x_r})
$$
To calculate the mole fractions for $HCl$ a linear interpolation is used
$$
x_{HCl}(c) = 0.0169 c + 0.0004
$$
Thus, making the mole fraction per ion
$$
x_{i}(c) = \frac{0.0169 c + 0.0004}{2}
$$
Parameters

$t_{f}=20\cdot(3600\textrm{ s})$, final time (run time of $20\textrm{ h}$)
$V = 1000\textrm{ L}$
$T = 293 \textrm{ K}$, room temperature.
$F=5V / t_{f} \approx 0.14\textrm{ L/s}$, flushing the entire volume every 2 hours.
$J = -2c_0V/t_f\approx0.028\textrm{ mol/s}$, if all the way linear then from $1\textrm{ M}$ to $0\textrm{ M}$ in $10\textrm{ h}$.

Inital conditions

$c_0 = 1\textrm{ M}$

Results
Concentration over time:

Energy change over time:

References

W.J. van Egmond, et al. Performance of an environmentally benign acid base flow
battery at high energy density; International Journal of Energy Research Volume 42, Issue 4, 2018 (PDF)
Kitty Nijmeijer, Sybrand Metz. Chapter 5 Salinity Gradient Energy; Sustainability Science and Engineering Volume 2, 2010 (link)

the equation you used for chemical potential is for an ideal gas, not an aqueous solution.  In addition, this system is really, in essence a closed system.

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text:
Why vapor pressure is unaffected by change in atmospheric pressure
I am a high school student and I am very confused in a topic about "vapor pressure of liquid". Why doesn't a change in atmospheric pressure affect vapor pressure? I think intuitively that vapor pressure should decrease on increasing the atmospheric pressure. Say I have an open bowl of water at temperature T and 1atm pressure and the vapor pressure at temperature T is P1. Now I increase the atmospheric pressure over the bowl keeping the temperature same (we can do it by pushing more air by some external means towards liquid surface). Because a higher atmospheric pressure is pushing on the liquid, doesn't that mean less evaporation and hence less vapor pressure for same temperature?
But on the other hand, I also think that vapor pressure is defined when there is an equilibrium between the liquid and vapor phase of water and equilibrium constants are only affected by temperature. Therefore, if vapor pressure changes with a change in external pressure, than it would mean the equilibrium constant would have to change?
Please prove me wrong (if I am) with some intuitive explanation at a high school level.

The vapor pressure does not depend on the pressure. It only depends on the temperature.

@Maurice The vapor pressure is proportional to the gas pressure. What is pressure independent is the saturated vapor pressure, and event that is not exactly true.

Beside reading the answer, you have to see it as an attitude of the liquid...

Please check also many of additional posts linked from the duplicate. This question has been asked many times before in different guises.

The equilibrium is independent on the pressure above the liquid because the mechanism of creating it has nothing to do with the other gases above the liquid
It may seem intuitive that a higher gas pressure above a liquid would "push" liquid vapour back into the liquid. But not if you understand how gases work or how a liquid/vapour equilibrium is established.
First, the molecules of gases (unless very far from ideality) don't interact much. So there is no "push". Most of the space in a gas if free space and, while molecules bounce off each other, that's about it for their interactions. Also worth noting is that those interactions redistribute the kinetic energy among all the gas molecules in a specific statistical distribution.
Second, the equilibrium between a liquid and its vapour is reached when the number of liquid molecules having enough energy to escape the liquid, match the number of gas molecules having insufficient  energy to stay in the gas (or, think of the number of molecules striking the liquid surface being equal to the number escaping from the liquid surface because they have enough energy to escape). The only role any gas molecules have here is to distribute the kinetic energy evenly in the gas according to the statistical distribution for a given temperature. That distribution is independent of the content of the gas (all gases are–approximately–the same).
Since the equilibrium between vapour and gas only depends on the distribution of kinetic energy in the liquid and in the vapour, there is no role for the other gases present or for pressure. The same equilibrium between the two states will be achieved in a vacuum or in a high-pressure vessel.
To be fair this is an approximation. This point of view ignores the relationship between boiling and pressure (but that doesn't depend on the other gases).

Could you enlighten me why you want to keep a question that was asked like a dozen times and it doesn't make it any better? We don't need so many dupes, by any means.

@Mithoron I didn’t check for dupes. If it is, close it and I’ll agree.

Say I have a bowl of water let in a box having air at 1 atm. So sooner an equilibrium will get establish between its vapors and liquid then we call it saturation vapor pressure, now I have a 2nd identical bowl kept I a box having very less air (or vaccum) we know at room temperature water will start boiling so more vapors will form so how can we say that vapor pressure will remain the same as in previous case?

@VirenderBhardwaj boiling occurs when the pressure is lower than the equilibrium vapour pressure. What that equilibrium pressure is doesn't change.

I assume you consider the saturated vapor pressure at 100% relative humidity. (As at 50% rel.humidity,  vapor pressure = 0.5 saturated vapor pressure).
Be aware that ability to contain certain amount of vapor, having respective vapor pressure, is property of space, not of a gas.
In first approximation, presence of air or other gas has no effect of evaporation equilibrium.
In the second approximation, the saturated vapor pressure very slightly increases with the total pressure.

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text:
Association constant calculation in 3:1 complexes
I have a question about calculating the association constant related to 3:1 host-guest complexes. I am using the UV-vis titration method, I know the stoichiometry from the molar fraction method which I mentioned earlier is 3 to 1. Below are the formulas that I used to calculate the concentration of complex and the association constant ($\ce{H}$=Host, $\ce{G}$=Guest, $\ce{C}$=Complex):
$$\ce{aH + bG<=>C}$$
$$K=\frac{[\ce{C}]}{[\ce{H}]^a[\ce{G}]^b}$$
$$[\ce{H}]_0=[\ce{H}]+a[\ce{C}]$$
$$[\ce{G}]_0=[\ce{G}]+b[\ce{C}]$$
I had some issues with understanding how to use these equations.

What exactly is the definition of $\ce{[G0]}$ (the initial or total concentration of guests) practically?
Is it the concentration of prepared metal ion solution? Or do I have to calculate it from the first mole of ionic solution added into the host solution? Or is it the moles of guest divided by the volume of the host solution?
Can I eliminate this parameter from the formula by assuming that at the endpoint all of the guests are consumed?

By eliminating $\ce{[G0]}$ from the formula, the concentration of complex became negative, which doesn't make sense as the stoichiometry of the host is 3 so it makes a huge effect on the value of the association constant and makes it too high value.

[OP] Can I eliminate this parameter from the formula by assuming that at the endpoint all of the guests are consumed?

No. At equilibrium, all species are present. If one of the reactants is depleted, the reaction went to completion, distinct from reaching equilibrium.
Sometimes, you can neglect a species in a sum or a difference, but only in that sense. For the calculation of the equilibrium constant, you need to know all the equilibrium concentrations, and saying that one of them is very little does not help.

What exactly is the definition of [G0] (the initial or total concentration of guests) practically?

It is both the initial concentration of guest (assuming there is no complex at the time) and the total concentration of guest. If you are mixing two components of known concentration, you have to account for the dilution that goes along with mixing them.

Is it the concentration of prepared metal ion solution?

No, because when you add the other component, you are diluting the metal ion solution.

Or do I have to calculate it from the first mole of ionic solution added into the host solution?

I don't understand that idea.

Or is it the moles of guest divided by the volume of the host solution?

No, it is not. That ratio would tell you whether host or guest is in stoichiometric excess.

Can I eliminate this parameter from the formula by assuming that at the endpoint all of the guests are consumed?

No, not from the equilibrium constant expression. If the equilibrium lies far on the side of the complex, you could estimate the concentration of complex at equilibrium from the initial concentration of the limiting reactant. You could also estimate the concentration of reactant in excess remaining at equilibrium. However, you need to know the concentration of the limiting reactant at equilibrium to figure out the equilibrium constant.

By eliminating [G0] from the formula, the concentration of complex became negative, which doesn't make sense as the stoichiometry of the host is 3 so it makes a huge effect on the value of the association constant and makes it too high value.

That's good critical thinking. If you get a nonsense result in a calculation, you either made a mistake in setting up the calculation, or made a numerical error. In this case, it is a mistake in setting up the calculation.

I am using the UV-vis titration method

If you are monitoring the concentration of the complex, you have to try to find conditions where your titration curve is not a straight line (i.e. the complex forms stoichiometrically and the reaction looks like it goes to completion). You can accomplish that by lowering the concentrations, or by changing the temperature in the direction that makes the complex less stable. On the other hand, if you are measuring the "free" host (i.e. the host gives a spectroscopic signal while the guest and the complex do not), you would use the guest in excess (so you can easily calculate its equilibrium concentration) and would try to detect how much (very little) of the host is not complexed at equilibrium.


text:
How to estimate the temperature needed to overcome an activation energy barrier?
Trying to elucidate a reaction pathway, I calculated the energies for different transition states possible for that reaction (computational chem. DFT). I found activation free energies ranging from 29 to 39 kcal/mol, and wanted to understand if all those transition states are possible in the reaction conditions that were used in previous papers. That is, given that this reaction is known to occur at around 355 K, is it possible to overcome those free energy barriers of 29 to 39 kcal/mol?
I know I can calculate the rate constant for a given temperature using the Eyring equation but I'm not sure how to relate the rate constant to the reaction "feasibility".

Related: How to relate a reaction barrier to the time the reaction needs to proceed?

You could convert the rate constant($k$) to half-life($t_{1/2}$) which would give you an idea of the time scale required for the reaction to finish at a certain temperature.
The equation to obtain half-life from $k$ is different depending on the order of the reaction. For an unimolecular reaction($\ce{A->P}$), it is:
$$t_{1/2}=\frac{\text{ln 2}}{k}$$
For a bimolecular reaction ($\ce{2A->P}$),
$$t_{1/2}=\frac{1}{k[\text{A}]_0}$$
where $[\text{A}]_0$ is the concentration of A at the beginning of the reaction. See this page for more.
The half-life($t_{1/2}$) is defined as the time required for the initial concentration of the reactant to be reduced to half of its initial value.
The following table is from Shriver and Atkins' Inorganic Chemistry, and it shows the representative time scales for various ligand exchange reactions in metal complexes. (You can get an idea of the feasibility of the reaction from the timescales, and that is applicable to all types of reactions)

Anything with a half-life of less than a minute is fast. Half-lives in the range of minutes to hours are moderately fast reactions. Hours to days is slow. Reactions with half-lives more than weeks is intractable (without catalyst/heating).
Reference:

Shriver and Atkins' Inorganic Chemistry, P. Atkins, T. Overton, J. Rourke, M. Weller, F. Armstrong, M. Hagerman, 5th ed., 2010, W.H. Freeman and Company, p509

From the Eyring equation, we can simply calculate the $k$ value for it.
\begin{align}
k &= \frac{k_\mathrm{b} T}{h}\exp\left(\frac{-\Delta G^\ddagger}{RT}\right)\\
k_\mathrm{b} &= \pu{1.38E-9 J K^-1}\\
T &= \pu{355 K}\\
h &= \pu{6.626E-34 J s}\\
R &= \pu{8.3145 J K^-1 mol^-1}
\end{align}
Here, I am assuming the average of $\pu{29 kcal/mol}$ and $\pu{39 kcal/mol}$ to be $\Delta G^\ddagger = \pu{34 kcal mol^-1} = \pu{142256 J mol^-1}$
If we plug in the values into the equation, we get a $k$ value equal to about $\pu{8.64 x 10^-9 s^-1}$ which is quite small.
If we take for example a Gibbs free energy of activation of $\pu{23 kcal mol^-1}$, that would equate to a half-life of 2 hours at room temperature based on $t_{1/2} = \frac{\ln2}{k}$ for first-ordered reactions. If we use the same formula for the $k$ we calculated at $\pu{355 K}$ for $\Delta G^\ddagger = \pu{34 kcal mol^-1}$, we get a half-life of $\pu{8.02 x 10^7 s}$ which is about $2.54$ years. Therefore, the reaction is likely not going to proceed quickly at $\Delta G^\ddagger = \pu{34 kcal mol^-1}$ and can be considered unfeasible as the reaction goes extremely slowly.
If we instead take the lower end with $\Delta G^\ddagger \pu{= 29 kcal mol^-1}$, the half-life is less than a day at $\pu{355 K}$ which is reasonable. However, anything after $\Delta G^\ddagger = \pu{30 kcal mol^-1}$ are essentially unfeasible at $\pu{355 K}$. Even with $\Delta G^\ddagger = \pu{30 kcal mol^-1}$ at $\pu{355 K}$, the half-life is over 3 days and it will only increase exponentially as $\Delta G^\ddagger$ increases.
Note that $\Delta G^\ddagger$ is temperature dependant and the calculations above are done assuming that $\Delta G^\ddagger$ is constant at the given temperatures. Therefore, the theoretical values I calculated may not perfectly match the experimental results of the reaction pathway you are studying. However, it should hopefully give a general idea of the rate and the feasibility of reactions for the range of $\Delta G^\ddagger$ values you derived.

G_a is temperature dependent; it'll rise with rising temperature. Other reactions possibly take over in the meantime. Anything > 100 kJ/mol and STP needs a bit more persuasion. As for DFT, without calibration, up to 80 kJ/mol gives you a guesstimate but nothing more.

@M.L. Note that you are actually dealing with the change in the Gibbs free energy ($\Delta G$) not the absolute value of Gibbs free energy ($G$) here.

@Martin-マーチン Yes, I forgot to take that into account. I'm not an expert in this area but what do you think I could say to fix my answer? Should I say that we assume delta G is independent of temperature?

I've linked a question (above) that covers some more ground, have a look at that maybe it inspires you to improve/include some points. I think making sure that a reader will see the approximate nature of what you have written is important. The numbers at the end are just too tempting if taken at face value. Other than that, I think your answer is still fine; especially the part with the "hundred year" half-life.


text:
The ambiguity regarding aufbau principle
According to a Blog https://ericscerri.blogspot.com/2012/06/trouble-with-using-aufbau-to-find.html
written by Dr.Eric Scerri,

On moving from the $\ce{Sc^3+}$ ion to that of $\ce{Sc^2+}$ it is plain to see that the additional electron enters a 3d orbital and not a 4s orbital as the sloppy scheme dictates. Similarly on moving from this ion to the $\ce{Sc^1+}$ ion the additional electron enters a 4s orbital as it does in finally arriving at neutral scandium atom or Sc.

According to it, the last electron enters the 4s orbital. Why is Scandium a d-block element, then?

Please do use MathJax where it is appropriate (e.g. to typeset chemical formulae using $\ce{...}$) and don't use it where it is inappropriate (e.g. to place emphasis on text by simply wrapping it in $...$). Please consider reading FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange?.

Is the configuration of Sc [Ar] 4s2 4p1?

@Poutnik,it is indeed,[Ar]4s2 3d1,and as per the blog,the last electron enters 4s,so why it isn't a s block element?

Why do you think element blocks are named according to where the last formally added electron enters?

@Poutnik,isn't it the case?if not please explain me correct information

Seems to be a duplicate (in effect) of Blocks on the Periodic Table.

If the last electron place principle had been applied, the second f block of actinoids would have been destroyed as there is 6d-5f energy switching in the first half of actinoids. The first actinoids behaving more like d-block transition metals. E.g. There is no lanthanoid with oxidation number +VI like uranium, which may remind rather Cr-Mo-W.

@Poutnik,so what is the correct reason of the naming of blocks in periodic table?

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

@Poutnik,I am extremely sorry for that,If i had not put enough efforts on understanding the topic,or the concept,but I have searched a lot that why are elements categorized in block as S,P,D,F,but Every where,there is reasoned to follow aufbau rule and subshell which is filled last in the electronic configuration determines its block,,but why is this so,aufbau principle correctly describes overall configuration,but what was reason behind accepting the aufbau for dividing into blocks,what does the blocks of element indicate,their valence electron?if so then it is incorrect in case of scandium?

@Poutnik,thanks a lot,after a Bit surfing and thinking over your comment again, I got the answer,Those are the DIFFERENTIATING electrons which decide the block of the element in periodic table

I got the answer after a short surfing,The answer to this question is pretty simple,the blocks are made ,by taking the basis of classification as the Differentiating electrons,wherever the differentiating electron exists in a subshell,that is the block of that Element.
Differentiating electrons simply mean that the electron which differentiates the element's Electronic configuration from the preceeding element's Electronic Configuration,
for example
Though the last electron enters in 4s subshell of Scandium atom, it is a 3d electron which makes the difference between the atom and the previous one.Since Ca is $[Ar] 4s^2$ and Sc is $[Ar]3d^1 4s^2$


text:
How exactly does a buffer made up of a weak acid and its conjugate base work upon addition of hydroxide ions?
Consider a buffer solution containing a weak acid and its conjugate base ($c_\mathrm{total} = \pu{0.1 mol L-1}$):
$$\ce{HA <=> H^+ + A^-}$$
When $\ce{H+}$ ions are added (from a strong acid), the position of the equilibrium shifts to the right and $\ce{A^-}$ combines with most of the $\ce{H+}$, preventing a significant decrease in $\mathrm{pH}$.
Along the same line, my text book says upon the addition of a strong base, the $\ce{OH-}$ react with $\ce{H+}$ in the reaction mixture and reduce their concentration, causing the $\ce{HA}$ to ionize and replace most of them, which prevents a significant decrease in $\mathrm{pH}$.
However, If you add $\ce{OH-}$ ions from a strong base, say $c = \pu{0.01 mol L-1}$ to a buffer with $\mathrm{pH} = 5$, there obviously aren't enough $\ce{H+}$ ions in the reaction mixture to react with the base. How will the buffer function in this case?
I can think of two ways:

The $\ce{OH-}$ ions will keep depleting $\ce{H+}$ from the reaction mixture, driving the equilibrium to the right until all $\ce{OH-}$ ions are neutralized, and then the system will try to regain equilibrium.
The $\ce{OH-}$ ions react with $\ce{HA}$ instead, and then the partial increase in pH occurs from the shifting of the equilibrium to the left.

Is one of these correct? Or would the buffer just not work at all if $\ce{OH-}$ added is more than $\pu{E-5 mol L-1}$?

The principle is simple. All equilibrium constants are to be honoured simultaneously, if the solution is in equilibrium. It includes all respective acidity/basicity constants and water auto-dissociation constant. // You confuse equilibrium and reaction quotient. A reaction is at equilibrium when reaction quotient is equal to the equilibrium constant.

You may find the answers in here useful.

Many confrontation-like questions either A either B lead to compromise-like answers Both A AND B. OH- reacts with both H+ AND HA, simultaneously honouring all equilibrii expressed by Ka, Kb and Kw.

@Poutnik I'm not asking if its either A or B; all I did was mention 2 possible explanations I could think of to show my previous work/researh

Which one of the 2 (if any) is correct? implies you think both cannot be correct at the same time. If not then I am sorry for misinterpretation, but it could be formulated better. :-)

@KarstenTheis Yes, it does actually. Should I delete my question?

@OVERWOOTCH No, it will remain as duplicate to make the other question easier to find. Also, it will preserve the answer and the comments.

During neutralization of the buffer by a strong base solution, there are ongoing three reversible reactions, maintaining three equilibria:
$$
\begin{align}
\ce{H2O + A- &<=> OH- + HA} &\quad K_\mathrm{b} &= \frac{\ce{[OH-][HA]}}{\ce{[A-]}} \tag{1}\\
\ce{H2O  &<=> OH- + H+} &\quad K_\mathrm{w} &= \ce{[H+][OH-]} \tag{2}\\
\ce{HA &<=> H+ + A-} &\quad K_\mathrm{a} &= \frac{\ce{[H+][A-]}}{\ce{[HA]}} \tag{3}
\end{align}
$$
$$K_\mathrm{w} = K_\mathrm{a} \cdot K_\mathrm{b}\tag{4}$$
Therefore, added $\ce{OH-}$ reacts with both $\ce{HA}$ and $\ce{H+}$, reaching again equilibria according to $K_\mathrm{b}$ and $K_\mathrm{w}$.
Additionally, $\ce{HA}$ autodissociation equilibrium makes balance according to $K_\mathrm{a}.$

If there are 3 reversible reactions that have to be accounted for, does this mean there's no easy way to calculate the change in pH of the solution upon the addition of a strong base? I'm only familiar with using a single ICE table

There is relatively easy way, and even very easy one with few simplifying assumptions. See https://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation


text:
Limitations on the 18-electron rule
The 18-electron rule and the corresponding methods for counting the total valence electrons of transition metal complexes are among the most useful basic tools in modern inorganic chemistry, particularly in its application to organometallic species.
To explain the limitations of this rule, the coordination compounds of transition metals are divided into three broad classes, the characteristics of which are described in Rasmussen, S. C.  ChemTexts 2015, 1 (1), No. 10, DOI: 10.1007/s40828-015-0010-4.
For example, the coordination compound $\ce{[Zn(en)3]^2+}$ does not satisfy this rule, since it contains 22 electrons. On the other hand, the compounds $\ce{[PtF6]^2-}$ and $\ce{[Fe(CO)4]^2-}$ do obey this rule, since both species have 18 electrons. Based on the previous article, I could not explain why the first species does not obey the rule and the other two do. I suppose it is linked to the nature of the ligand and the central metal atom, but I am not sure.

Zinc often isn't considered a transition metal

In the notes with which I study it does appear as such. @Mithoron

This earlier edit should not have been accepted, for a couple of reasons: (1) no need to use MathJax for plain numbers in the title. MathJax in the title should be reserved for very special situations, where there is absolutely no alternative. (2) Chemical formulae should be written using $\ce{...}$, not $...$. Please consider reading: FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange?.

The $8$-electron and the $18$-electron rules are rules. They are often useful. But they have plenty of exceptions. Look at $\ce{ NO}$, $\ce{NO2, ClO2, B2H6, MnO2, CuSO4, K3[Fe(CN)_6]}$, etc.

I understand that there are exceptions to the rule, but I do not know how to explain its origin based on the article I have included in the question. That is the reason for my query. @Maurice


text:
Determining Boiling Point and Melting Point by Graph
This is the plot of $\Delta G^{^\circ}$ vs $T$ for $\ce{Pb-> PbO}$

I understand that the points where there is a sudden change in the slope of the graph, there is a phase change but exactly what are the reactions going at all the four points shown and also how can we justify whether it is a point of melting or boiling and what further can we conclude from this graph about entropy? Specifically, I am more concerned about the main reaction and what are the individual reactions at every point: $$\ce{Pb(s) + \frac{1}{2}O_2(g) -> PbO(s)}$$

Have you looked up the melting and boiling points ?

What is the context and origin of the graph shown?  Cite the source to add detail to your question, please.  For the points shown, there may be phase transitions solid -> solid, solid -> liquid ($\approx \pu{328 ^\circ{}C}$), liquid -> gaseous ($\approx \pu{1744 ^\circ{}C}$), the later two one may look up, e.g., in Wikipedia's properties box about the element. Yet that is not all yet, as there may be indeed onsets of oxidation reactions.  Note, lead and oxygen may e.g., yield $\ce{PbO}$, or $\ce{PbO2}$, or mixed oxides (like minium, formally Pb II and Pb IV are present simultaneously).

What do you report in ordinate ? Is it a Gibbs energy of formation ? Of which product ? Why do the points seem equidistant on the graph ?

@Buttonwood, Yes, but specifically only about PbO

@Maurice, gibbs energy of formation of PbO

@Community Independent of Maurice' good note about the points' distance along T (about $\approx \pu{200 K}$ above $\pu{1535 ^\circ{}C}$ are anticipated as considerably shorter than the $\approx \pu{650 K}$ above $\pu{888 ^\circ{}C}$), there is too few information here.  Is it accidental coincidence, or intent (if so, why?) that $\ce{PbO}$'s melting point ($\pu{888 ^\circ{}C}$) and boiling point ($\pu{1477 ^\circ{}C}$) apparently equally are in this diagram of (still) unquoted origin? How (if so) did you compute it?

PbO exists under two crystallographic systems : tetragonal and rhombic. The point at $327$°C is probably the transition between these two forms. $888$°C is the melting point, and $1749$°C the boiling point. The problem is how to explain the point at $1535$°C in the liquid state, which is accompanied by an increase in stability  ...

At least two points is clear: $\pu{327 ^\circ C}$ is the melting point of $\ce{Pb}$ and $\pu{888 ^\circ C}$ is the melting point of $\ce{PbO}$.

@MathewMahindaratne, won't the 888 C point be boiling point of Pb? Maybe we can also lookout for the equation of gibbs free energy and entropy

Good point in a sense of which I didn't think about that. It clears the last break: Boiling point of $\ce{Pb}$ is $\pu{1749 ^\circ C}$. According to Wikipedia, melting point of $\ce{PbO}$ is $\pu{888^\circ C}$ and its boiling point is $\pu{1477 ^\circ C}$. That is the only odd number in your graph.


text:
Claisen condensation of methyl benzoate and ethyl propionate
I am planning to preform the Claisen condensation between a benzoate ester (methyl benzoate is the only one I can get) and a propionate ester (ethyl propionate is my only option). The idea is to prepare this ketoester:

Will this work with ethyl instead of methyl propionate?

If ethyl propionate is your only option, then your product is ethyl ketoester. In the scheme, you have used methyl propionate and got methyl ketoseter. Yes, it work with ethyl esters as well, but there is a possibility to get transesterification.

What solvent is this in? Run it in MeOH and you will get the methyl ketoester

To be honest, it doesn't matter what the ester is. Well, my stoichiometry will be a bit off, but I am just hydrolyzing the ester anyway.

And I will likely be using MeOH.

The ester is likely to decarboxylate after hydrolysis. Is that your intention?

It is my intention; I plan to make propiophenone


text:
Can we have an equilibrium constant for a reaction at constant volume?
Suppose we have a sealed container (fixed volume) and we introduce two gases $\ce{A}$ and $\ce{B}$. The two reactants
$$\ce{A + B <=> C}$$
form another gas $\ce{C}$ with equilibrium constant $K$. We know that a closed system at constant $(T,P)$ a system reaches equilibrium where Gibbs free energy $G$ takes its minimum value. Can a reaction reach equilibrium in a closed container (fixed volume)?
Although the system can be coupled to a heat bath (providing the constant temperature condition), by fixing the volume, the necessary condition of constant pressure for $G$ minimization is lost. Can the equilibrium state now be predicted?
I mean if the piston was movable (e.g. coupled to an external pressure of $1$ atm) the system would reach equilibrium and we could find the equilibrium state (e.g. the gas composition) by using the equilibrium constant $K$.
I am asking this question because in many sites I have seen the following Le Chatelier's Principle:

When there is a decrease in volume, the equilibrium will shift to favor the direction that produces fewer moles of gas.

I can't understand why we are free to make such a statement if the pressure is not constant. I am not asking about Le Chatelier's Principle. I just gave that example to justify the motivation behind the question.
Another example is when we want to predict the vapor pressure of a liquid (again sealed container). We put some liquid in the closed container and after a while equilibrium is established where vapor pressure is related to an equilibrium constant $K_
\text{p}$.
In summary I want to understand what let us use equilibrium constants under non-constant pressure conditions.

Can a reaction reach equilibrium in a closed container (fixed volume)? Sometimes even much faster, like small amount of explosive in a rigid cavity.

The equilibrium constant $K_P$ is just that, constant (at constant T). If the pressure changes then the extent of reaction alters to compensate for this. see https://chemistry.stackexchange.com/questions/153545/how-does-the-value-of-the-equilibrium-constant-give-information-about-the-extent/153551#153551

@porphyrin But the system is not at constant pressure.

I checked the link. My question is about if it is proper for a reaction or physical change (like liquid turning into vapor) to use Helmholtz vs Gibbs free energy for the derivation of equilibrium constants.

Exactly, not at constant pressure, thus the degree of dissociation changes just as in the example I linked to.

So@porphyrin the equilibrium constant is only temperature and not pressure dependent as we know because of the scaling of the partial gas pressures. But why does one use the Helmholtz energy for explosives e.g like outlined in the answer to this question, is it even correct?

Yes.  The universal condition for equilibrium is that, given the current constraints on the system, no further spontaneous change can take place.  I.e., that the entropy of the universe (system + surroundings) is maximized.
At constant temperature and pressure, a maximization of the entropy of the universe corresponds to a minimization of the Gibbs free energy (G) of the system, and we have:
$$\Delta G^{\circ} = - R T \ln K_p$$
By contrast, at constant temperature and volume, a maximization of the entropy of the universe corresponds to a minimization of the Helmholtz free energy (A) (sometimes also symbolized using "F") of the system.  Here, we have:
$$\Delta A^{\circ} = - R T \ln K_V$$
Indeed, these are why we use the Gibbs and Helmholtz free energies—we can't typically measure the change in the entropy of the universe to determine the point at which it's maximized, so we instead use the minimzation of the Gibbs or Helmholtz free energies of the system (which can be measured) as surrogates for it.
[You're probably going to ask me if that standard state for A includes standard pressure.  I'm going to need to think about that.]

So in the example of the reaction $\ce{A + B <=> C}$ Le Chatelier Principle still applies but the equilibrium constant (and how the equilibrium will shift) will depend on $\Delta A^{\circ} = - R T \ln K_V$? Does the same apply for the vapor liquid equilibrium?

Yes to both.  For the 2nd, imagine you have a sealed container that (by volume) is half liquid water and half gaseous water.  Now suppose you inject some gaeous water into it.  By Le Chatelier's principle, that will shift the rxn to towards the liquid side:  Gaseous water will be converted to liquid water until the pressure of the gaseous water returns to the equilibrium vapor pressure.  Interestingly, in this case, adding gaseous water will result in a net reduction in the total amount of gaseous water in the container, because the volume available for the gaseous water will be reduced!


text:
Would the hydrogen-bonded polymer hexaamminecobalt(III) hexakis(trifluoromethyl)cobaltate(III) be soluble in water?
It is clear that both the hexaamminecobalt(III) ion and the hexakis(trifluoromethyl)cobaltate(III) ion are both low spin (the former is already low spin, and the trifluoromethyl ligand is clearly softer than the ammine ligand) and therefore valence isoelectronic to each other, since the Jahn-Teller effect cannot occur. Since their topological connectivities are the same, they must both be of the "same shape" and therefore be able to form hydrogen bonded networks using the ammine protons and trifluoromethyl fluorides.
Question: Would the solvation properties (dipole moments, hydrogen bonding abilities, and the like) of water sufficient to solubilize the hydrogen-bonded polymer $\ce{[Co(NH3)6][Co(CF3)6]}$? I'm asking this, since the very reason that "$\ce{HF}$ is a weak acid in dilute aqueous solution" is that the $\ce{H3O+}$ and $\ce{F-}$ hydrogen bond is too strong for the $\ce{H3O^+}$ to dissociate and "display its acidity".

Do we know that the proposed salt exists?

The cation already exists. As for the anion, hydrofluorination of the known cobalticyanide anion should do the trick.

Ammonium and hydroxide ions exist, but except perhaps under extreme high-pressure conditions like the interior of Neptune there is no ammonium hydroxide. I ask about the salt as a whole, not its parts.

Check their shapes. They perfectly fit in a halite structure supported by H-F hydrogen bonds.


text:
What happens in a unimolecular reaction?
Molecularity is defined as the number of molecules required to collide simultaneously to form a product in an elementary reaction.
In a unimolecular elementary reaction, do two molecules of the same species collide or does a single molecule dissociate?

?!? If two molecules collide, you have a bimolecular reaction.

Just for future reference, As homework is considered literal homework, exams, self-study questions, puzzles, worked examples etc.

@Karl Is reaction $\ce{A + A ->[collision \Delta E ] A^{*} + A -> B + C + A}$ unimolecular ?

Isn’t that the unimolecular, bimolecular etc. defined by the no. Of distinct species (eg. NaCl AND KOH, and not NaCl and NaCl) ?

Ionic compounds are not the best example.  $\ce{2 N -> N2}$ is bimolecular. Generally, bimolecular elementary reaction is A + B -> products different to  A, B, with A possible be the same as B.

Just found the right link https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/03%3A_Rate_Laws/3.02%3A_Reaction_Mechanisms/3.2.01%3A_Elementary_Reactions

@Poutnik elementary reaction

@Karl  It is conventionally still considered as elementary reaction $\ce{A -> B + C}$. Collisions providing activation energy are not considered as reactions. Instead of the 2nd A, it could be a helium atom.

@Poutnik If A* is an identifiable intermediate (as your first eq. above explicitly says!), then A-->B+C cannot be an elementary reaction. You're barking up the wrong tree ;)

@Poutnik The elementary reaction in that case is $\ce{A^* -> B +C}$ !

@Karl  Lakedaimonians would say  IF....   A* is not identifiable intermediate, it is just A with higher then ground energy. Compounds have dozens of different energy states. Many reactions use A as implicit A*, A -> B + C is always A* -> B + C as A does not have enough energy. // My point since the beginning was that unimolecular reaction can be triggered by molecular collision as one of possible form to reach activation energy. I have thought it was clear.

@Poutnik My point was you shouldn't call it A* and put it in an equation without square brackets if it's just thermal. And I don't know if thermal decomposition in the gas phase is actually 100% unimolecular. Probably not.

@Karl I have not talked about any particular reaction. Is $\ce{A ->[thermal collision with A] B + C}$ better ?

Well, a certain fraction of "A" particles in the thermodynamical ensemble is always above the energy limit in your thought experiment. To discuss it, you need to decide if thermal collisions happen at a far higher rate than the inverse lifetime of your [A*] state, or not. In the former case, you are clearly unimolecular, and A* is not worth mentioning. In the former, your gas is not even in thermodynamic equillibrium (or the reaction is superfast, if you're in a more dense state, i.e. you won't have time to investigate the kinetics. ;))

Let us continue this discussion in chat.

Often other inert molecules collide with the reactant molecules, allowing the reaction to take place, but since those inert molecules are not counted in the rate law, the reaction is considered "unimolecular".

What is called a unimolecular reaction in textbooks is normally (a) an isomerisation such as cis $\ce{<=>}$ trans isomerisation or dissociation $A\to$ products or in the gas phase (b) the reaction $\ce{A + M<=> A^* + M, A^\to }$products usually described by the Lindemann scheme or more advanced by RRKM theory. Here $A^$ is a collisionally activated molecule.

Truly collisionless unimolecular reaction are rare, as even highly unstable molecules must be triggered somehow to gain the reaction activation energy.
This is frequently obtained as an energy gain by thermal collision with other molecule, or by a photon absorption. Collisions are usually much more frequent, but photons are usually much more energetic than thermal collisions.


text:
How do the terms in a rate law relate to effectiveness and number of collisions of reactants?
In chemical kinetics, we use rate law to find the instantaneous rate of reaction.
Let us consider a simple reaction
$$\ce{aA + bB -> products}$$
where the rate law is
$$\mathrm{rate} = k[\ce{A}]^a[\ce{B}]^b$$
I think that the rate constant tells us about the effectiveness of collisions for the reaction at a particular temperature; and the concentration terms raised to the powers of stoichiometric coefficients represents the total number of collisions.
Is my reasoning correct? If not, what is the correct explanation?

Sounds about right. Then again, reactions are rarely simple, and the actual rate law might differ from $k[A]^a[b]^b$, but that's another story.

@IvanNeretin I am not sure how concentration terms raised to the powers of stoichiometric coefficients represents the total number of collisions, could you please prove some mathematical approach to figure this out.
Also, I know about the complex reaction and how it differs but I want to know the basic idea behind the rate law thus assuming it is a simple reaction.
Thanks for your clarification.

Rate laws such as this only show how many reactant species are needed to produce the product according to mass balance. They tell us nothing about how the reaction actually happens. The actual mechanism for any reaction has to be determined by experiment. You can imagine that if, say, $a=b=2$  the chance of four solute species coming together at the same small instance and in the same place is so absolutely tiny, that it just does not happen. What tends to happen is that two species form an intermediate and this reacts again and so on. (There are whole books on this topic)

@RodrigodeAzevedo Because it works but I am unable to understand how?

I have mentioned it is a simple reaction meaning elementary.

As others have pointed out, the power terms are not necessarily equal to the stoichiometric coefficients. In a single-step (elementary) reaction - yes, the concentration terms represent the number of collisions that can occur. Think of it as probability - the probability of 2 species of A coming together at a spot is proportional to $A^2$.

@TRC but the probability is always less than 1 but concentration can be anything.

I'm saying it's proportional to the concentration. The proportionality constant gets included in the rate constant.

Yes I know that but I need a mathematical proof for the same.

Try examining the Arrhenius Equation and see if that gives you more information about the rate constant.
In general, the rate of reaction (in terms of the reactants) is a function of how many successful collisions occur per unit time. There are also unsuccessful collisions. You can also write the rate of reaction in terms of the products. Since the products are not always 1:1 with the reactants, this wouldn't necessarily equate to the number of successful collisions.
In gas phase reactions, inert species can influence the rate of reaction. Since they aren't colliding with the reactants to form the products, the concentrations can't always be equal to the number of collisions.

Thanks but need a more clear explanation

Your reasoning is not correct. The concentration terms raised to the powers of stoichiometric coefficients do not represent the total number of collisions. An inert species does not participate in the reaction but can be included in the rate law. Therefore, the concentrations raised to the stochiometric coefficient can't possibly represent the number of collisions. I think you are on the right track with the rate constant. You should read about and examine the Arrhenius Equation, I think this may give you more insight about the rate constant.

Refer to this article from chem.libretexts.org for more information about collision theory.

I got the concept with the rate constant but can't understand what concentration term really means.


text:
Effect of metal identity on CO bond length in carbonyl complexes
In organometallic carbonyl complexes, a back-donation effect occurs: the σ molecular orbital of CO yields electron density to an orbital of the appropriate metal atom, and in turn, a d orbital of appropriate symmetry yields electron density to the antibonding π* orbitals of CO.
On the basis of this model, the properties of carbonyls (e.g., the vibrational energy of the C–O bond or its bond length) can be explained. For example, in the following table, I understand that all of the bond lengths are longer than in free CO, because of this backdonation which decreases the C–O bond order. However, how can the differences between the three metal complexes be explained?

Species
C–O bond length (pm)

$\ce{CO}$
112.8

$\ce{[Ti(CO)6]^2+}$
116.0

$\ce{[Cr(CO)6]}$
114.0

$\ce{[Ni(CO)4]}$
113.0

There we go, that's a bit cleaner, and more to the point.

Related: How to rationalise with MO theory that CO is a two-electron donor through carbon?

Let's look at the molecular orbital diagram calculated with JSmol.

As you know, the principal effect is back-donation from the metal to the π-antibonding orbitals of the carbon monoxide. But not all 3d transition metals are created equal. Earlier ones like titanium, with lower electronegativity and lower ionization energy, are better at this back-donation than later ones.


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If something is not hydrophilic, is it necessarily hydrophobic?
Is it possible to create hydrophobic water?

Two species are phobic within some medium under some conditions when they don't here together (cohere if they're the same species; adhere otherwise).
To be clear: phobic species don't really repel so much as they're not very attracted to each other.
Species can become self-phobic. Of course, it's important to remember that members of the species aren't repulsing others of their own kind so much as just aren't very attracted to them. And then since they're not really attracted, they do the natural thing and just tend to disperse, filling the volume containing them. In other words, they vaporize.
Water can be vaporized, e.g. through heating.

In the question I linked above the third answer says that since the water molecules do not attract each other in gaseous state they are hydrophobic. Even if the water molecules are not hydrophilic are they thus necessarily hydrophobic?

Otoh. water condenses from the gas phase, so it must be hydrophilic. ;) You see these categories make no sense for gases.

To address the title itself, being hydrophilic/hydrophobic is rather being somewhere on a continuous gray scale than inside of either black either white box.

Two water drops 1 mm apart do not attract each other, if I omit gravity and electrostatic charges. Are they hydrophobic ? No, they are not. The relevant intermolecular forces are contact ones, decreasing very fast with high power order of distance.
Water molecules in a gaseous phase have still strong attraction during their collisions, what reflects in vapor being far from an ideal gas. The estimated boiling point of water, assuming absence of hydrophilic hydrogen bonds, is about $\pu{-120 ^{\circ}C}$. This gives some idea what makes pure water vapor to condensate at $\pu{+100 ^{\circ}C}$ ( at $\pu{101 325 Pa}$ ).
Hydrophilicity means having comparable strength of intermolecular bonds as water has, to itself as well as to water molecules. Mutual mixing often releases energy ( ethanol + water ) and is supported by entropy increase too.
Hydrophobicity means having weak intermolecular bonds, to itself as well as to water molecules. Breaking mutual strong bonds between water molecules to create weak bonds with the hydrophobic substance is not energetically favored.


text:
Towards a hard Lewis base like water which is the stronger Lewis acid, silicon tetrachloride or tin (IV) chloride?
On Pauling scale, both $\ce{Si}$ and $\ce{Sn}$ have electronegativity value of $\mathrm{1.9}$. The difference is in size of d-orbitals which accept lone pair of electrons. $\ce{Si}$ has vacant 3d-orbitals whereas $\ce{Sn}$ has vacant 5d-orbitals.
What will be the Lewis acid strength order?
My Attempt:
In case of $\ce{SiCl4}$ there can be back-bonding between vacant 3d orbitals of $\ce{Si}$ and filled 3d orbitals of $\ce{Cl}$ (though I did not know, how efficient is the back-bonding between 3d-3d orbitals). This introduces partial double bond character in $\ce{Si-Cl}$ bonds, which makes it a weaker acceptor of lone pair of electrons.
In case of $\ce{SnCl4}$ there can be no back-bonding between vacant 5d orbitals of $\ce{Sn}$ and filled 3d orbitals of $\ce{Cl}$. This makes it a stronger acceptor of lone pair of electrons.
${^1}$Hence, the Lewis acid strength order: $\ce{SnCl4 > SiCl4}$
(Though it is not explicit in the mentioned reference.)
It is to be noted that, in case of group- 13 elements, the strength of Lewis acidity order is: $\ce{BCl3 > AlCl3 > GaCl3}$
This is due to difference in electronegativity and size of central atoms.

Reference:

Page 328-329, 4th Edition (Paperback), Shriver & Atkins- Inorganic Chemistry (Oxford)

Why this question is receiving request to close? Whoever wants to close, at least they can comment the reason!

Not one of the close voters, but it would probably help if you explained what you think the result should be and the reasoning/research you used to come to that conclusion.

@Tyberius You can check now.


text:
Hydrogen concentration measurement without gas chromatography
Please suggest ways to measure $\ce{H2}$ concentration in a sample of gas containing $\ce{HCl},$ $\ce{CH4},$ $\ce{N2},$ $\ce{CO2}$ and trace amounts of chlorosilanes without GC.
The problem is that I don't have access to another carrier gas like argon and helium. I need a sensor that can measure $\ce{H2}$ in a corrosive gas mix. Expected concentration $\phi(\ce{H2})\approx 10\,\%.$

What instruments do you have access to?

Do you have access to mass spectroscopy?  Some of them may be portable (Wikipedia entry).

Ordinary mass spectrometry is really bad for hydrogen until and unless the user has tons of money to spend ;-)

https://www.sciencedirect.com/science/article/pii/S1044030502004312

@ToddMinehardt: we have a GC with a TCD detector using H2 as carrier gas. We can technically move to a different carrier gas, but that is not preferred in this case, as this is a dedicated GC for trace methane measurement in a H2 stream. The gas sample in question has ~ 10% H2 content, hence GC seems to be an overkill. Was wondering if there are other portable cost effective techniques for such bulk H2 quantification.

There are a lot of classical analytical methods but they are tedious and not so user friendly. Unfortunately, no simple alternative to GC. The problem is that you have chlorosilanes. They must be removed from the analysis mixture because they will react with your GC columns and change their column chemistry-forever!

Is there a way to leverage the small size of the hydrogen? I remember that conventional balloons lose hydrogen, but keep air (Graham's law).

How about letting the hydrogen react in a fuel cell with a PEM? E.g. https://www.researchgate.net/publication/245147741_Separation_of_hydrogen_from_a_hydrogenmethane_mixture_using_a_PEM_fuel_cell

There are commercial, calibrated, hydrogen detectors, such as those from GasSensing or from Mcmaster-Carr, but you'd have to check with the supplier or manufacturer to determine if the other gases give a false reading, or, particularly silanes, poison the sensing element.
You can also make your own device using a sensing element, such as SGAS701 from Renesas, or, for US$5, the Hydrogen Gas Sensor - MQ-8 from SparkFun.
Be aware of some limitations of these sensors:

Those catalytically measuring the heat generated from hydrogen combustion require oxygen in the stream.
Some gases, such as silanes, may poison the sensor or build up a coating on it preventing gases from contacting it.
Other gases, such as methane, $\ce{CH4}$, may give false readings.

You likely will need to contact the vendor's support for specifics; the online information, sch as that for the ATI electrochemical sensor, does not cover the specifics of your situation.
BTW, you might mention the desired accuracy and repeatability for these measurements.

The problem is not detection, I am looking to quantify hydrogen vol fraction.

@user112894, as mentioned, what is desired accuracy and repeatability for these measurements? The sensors mentioned are quantitative, but without knowing your specific needs, you will not get a useful answer.

"The temperature of liquid nitrogen (−195.8 °C) can readily be reduced to its freezing point −210 °C (−346 °F; 63 K) by placing it in a vacuum chamber pumped by a vacuum pump." Wikipedia
A gas sampling tube:

or perhaps even this:

with a pressure meter

attached could be used to take a sample, then  the total pressure could be read after cooling to condense all the gases except hydrogen.
I realize this is crude, but it would be fairly cheap, except for the liquid nitrogen and pump.


text:
Advantages of p-toluenesulfonic acid over sulfuric acid in the formation of acetals
I am currently working on an undergraduate experiment to form a cyclic acetal.
We were asked to use p-toluenesulfonic acid as a catalyst rather than sulfuric acid.
This seems to be common practice when forming an acetal, but the reasons given are not satisfactory to me.
Being a solid, it is more easy to handle, I can understand that. But another reason given is that it is less dehydrating than sulfuric acid. In the formation of acetals, water is being formed as a product, thus dehydrating effects of the acid would be productive. In this case, it would not make a difference since I am using a water separator, but still I am wondering why dehydrating effects of the acid are not desired.
Is there any other reason I am missing? If so, I would also be happy about literature references.

pTsOH is better soluble in organic solvents than sulfuric acid, and not as strong dehydrating than sulfuric acid. Do you recall sulfuric acid's action on sugar example?  Thus it is better to have some pTsOH well dissolved in the reaction mixture remove the water by Dean-Stark distillation. See if your school has access to Paquette's enyclopedia (there is an online-version, too [campus license?]) and continue from there.


text:
Making bioplastic at high school level chemistry using PLA/Starch blends
I'm currently working on a high school chemistry essay about biodegradable plastics. I know that there is an easy way to synthesize Thermoplastic Starch (TPS), a starch-based bioplastic using glycerol and other materials at home (https://www.wikihow.com/Make-Bioplastic). But, I was wondering if there would also be a way to make Polylactic acid (PLA)-based bioplastic using the same process. I know that PLA bioplastic synthesis usually goes through extrusion or melt blending process, but since conducting these processes are not possible at my school due to equipment limitation, I just wonder if doing the same process described for starch-based could produce PLA-bioplastics. I got PLA pellets in hand, so would crushing them into powders and repeating the same process, or maybe mixing it with starch, be a valid procedure?

Personal Opinion: if it were up to me, I would focus on mixing different ratios of the ingredients listed in your source. I would then text some physical properties of each plastic and try to explain differences using "chemistry knowledge". However, this might be a bit too much for just an essay. So if no experiment is needed, search for a real-life experiment that does this and go into detail about their results.

I think you are in correct track in preparing Polylactide (PLA)-starch bioplastic, in a short. The question is actually an opinion-based but interesting one. So, even though there is high possibility that I may get down-voted, I'd like to give few of my opinions.
I think you may have to tryout few steps. I agree with the crushing PLA pellets into powder would be advantages because it increases the surface area. I think it is also important to prepare Thermoplastic Starch (TPS) as one of your reagents, rather than having PLA, starch, and glycerol together and heat. I'm telling this because it has been done in literature (peer-review) before with polystyrene (Ref.1):

Conventional plastics has a large impact in increasing the environment’s pollution. That’s why the interest has turned towards novel partially and completely biodegradable polymers. In this work, blends of polystyrene and thermoplastic starch with glycerol and Buriti (Mauritia flexuosa L.) oil as plasticizers were prepared. Samples were analyzed using TG/DTG and DSC techniques. The TG results indicated that the blends with Buriti oil are thermally more stable than those with glycerol. The DSC analysis that Buriti oil provides a higher degree of plasticization of PS, compared to the blends plasticized using glycerol under the studied conditions.

According to the proceedure in Ref.1:
Preparation of thermoplastic starch: TPS was obtained by mixing starch powder, water and glycerol or Buriti oil in $50:15:35 \ (mass/vol/vol)$ ratio according to Ramsay et al. (Ref.2) but after some modifications. The constituents were mixed for $\pu{30 min}$ to obtain a paste which was transformed to TPS by heating at $\pu{95 ^\circ C}$ in water bath with continuous stirring for $\pu{30 min}$.
Preparation of the blends: PS and TPS were mixed in different ratios $(0.9:0.1, 0.7:0.3, 0.5:0.5$ and $0.3:0.7,$ $(w/w\%)$ and then stirred $(\pu{3 h}, \ \pu{50 ^\circ C})$ with hot ethyl acetate. The blends were prepared by conventional-casting technique using Teflon$^\mathrm{TM}$ dishes as casting surfaces for at least $\pu{24 h}$ at room temperature and vacuum (between $6.6 \times 10^2$ – $\pu{13.2 \times 10^2 Pa}$). The obtained material, with $200$ to $\pu{300 \mu m}$ thickness was stored far from light to avoid the decomposition of the oil.
The Thermogravimetric analysis (TG) and DTG analysis were depicted in following image:

The authors mentioned that to convert starch to thermoplastic starch (TPS)
and allow its proper blending with PS, it is necessary to use a plasticizer agent like water, glycerin, sorbitol, etc., at high temperature $(\pu{90–180 ^\circ C})$ and shear (Ref.3).
As you can see from the image we have different blends behaving differently according to the ratio. Thus, it is safe to say that PLA may act similar to PS. However, you may need to try few ratios and blend in different temperatures before you conclude the results in your review. Good Luck!
References:

Daniela Schlemmer, E. R. de Oliveira, and M. J. Araújo Sales, "Polystyrene/thermoplastic starch blends with different plasticizers: Preparation and thermal characterization," Journal of Thermal Analysis and Calorimetry 2007, 87(3), 635–638 (DOI: https://doi.org/10.1007/s10973-006-7776-y).
Bruce A. Ramsay, Vincent Langlade, Pierre J. Carreau, and Julianna A. Ramsay, "Biodegradability and mechanical properties of poly-(beta-hydroxybutyrate-co-beta-hydroxyvalerate)-starch blends," Appl. Environ. Microbiol. 1993, 59(4), 1242-1246 (DOI: https://doi.org/10.1128/aem.59.4.1242-1246.1993).
P. Aggarwal, D. Dollimore, and K. Heon, "Comparative thermal analysis study of two biopolymers, starch and cellulose," Journal of Thermal Analysis 1997, 50, 7-17 (DOI: https://doi.org/10.1007/bf01979545).


text:
Linear algebra and Hess's law?
I am a high school student taking chemistry who was tasked to solve a large set of Hess's Law problems. Naturally I tried to find a way to get my computer to carry out this repetitive piece of work for me.
I tried to model the problem mathematically and figured out that Hess's Law problems basically had me finding the correct linear combinations.
Suppose we have a problem like this
\begin{align}
\ce{4 NH3 + 3 O2 &->6 H2O + 2N2} &\qquad &\Delta_\mathrm{r}H_1 = \pu{-1516 kJ/mol} \tag{1} \\
\ce{2 H2 + O2 &-> 2 H2O} &\qquad &\Delta_\mathrm{r}H_2 = \pu{-572 kJ/mol} \tag{2} \\
\end{align}
And our goal is to try to find the enthalpy of formation of ammonia. How would I express this as a matrix and solve it to find the linear combination. I assume we have to use the row reduce echelon method here. Just not sure how.

Searching for  linear algebra tutorial gives a lot of interesting links.

A generally useful and efficient method is the Gauss elimination method, getting the triangular  matrix. Many software platforms, including Excel worksheet  functions, provide matrix and vector operations to directly solve the set of n linear equations of n variables. But for n=2, the manual substitution approach is easy.

It gets set up as the part of the Gauss' algorithm itself. You nulify the numbers below the diagonale by substrating (multiplied ) other rows.

Formulate the equation of the Hess law for each reaction and then transform it to the algebraic equation set.

Both. Coefficients for the same reaction on the same row of A. Coefficients for the same variable ( unknown Delta H ) to the same column of A. Absolute terms of f(x)=0 to the separate column b.

Maybe this video will help you. https://www.youtube.com/watch?v=BvcX1n35U5A The resulting system can be easily solved by inspection or with the help of matlab, maple. You could also make a small program to do everything.

To convert this into a generic linear algebra problem you'd rewrite it in the form $Ax=b$ where $A$ is a matrix of stoichiometric coefficients of size $m \times n$; $x$ is a vector of length n of unknown fitting parameters (the heats you want to determine) and $b$ a known vector of length n.
For the example problem the set of equations can be written in matrix form as
$$\left[\begin{array}{rrrrr} 0 & -4 & -3 & 6 & 2\\ -2 & 0 & -1 & 2 & 0 \end{array}\right]\left[ \begin{array}{l} \Delta_\mathrm{f} H(\ce{H_2(g)})\\\Delta_\mathrm{f} H(\ce{NH_3(g)})\\\Delta_\mathrm{f} H(\ce{O_2(g)})\\\Delta_\mathrm{f} H(\ce{H_2O(l)})\\\Delta_\mathrm{f} H(\ce{N_2(g)}) \\  \end{array}\right] = \left[ \begin{array}{c } \Delta _\mathrm{r}H_1\\\Delta_\mathrm{r}H_2  \end{array}\right]$$
Since many of the terms in the vector of heats on the left-hand-side are known and equal to zero this can be rewritten as
$$\left[\begin{array}{rrlll} -4 & 6 \\ 0 & 2 \end{array}\right]\left[ \begin{array}{c } \Delta_\mathrm{f} H(\ce{NH_3(g)})\\\Delta_\mathrm{f} H(\ce{H_2O(l)})   \end{array}\right] = \left[ \begin{array}{c } \Delta_\mathrm{r} H_1\\\Delta_\mathrm{r} H_2  \end{array}\right]$$
(alternatively the original equation can be written as a $5 \times 5$ array by including additional trivial equations defining the heats of formation of the diatomic gases).
Solution of the above system should be easy, for instance in MATLAB:
A = [-4 6; 0 2];
b = [-1516 -572]';
x = A\b

The problem can of course be solved by inspection. The second equation can be solved for $\Delta_\mathrm{f} H(\ce{H_2O(l)})$, which can be inserted into the first to solve for $\Delta_\mathrm{f} H(\ce{NH3(g)})$.

 >>> import numpy as np; A = np.array([[-4, 6], [0,2]]); b = np.array([-1516, -572]); np.linalg.solve(A,b)```


text:
What is the correct molecular orbital diagram for the d orbitals in platinum for the tetraammineplatinum(II) complex?
I am trying to construct the diagram of $\mathrm{d}$ orbitals in the  tetraammineplatinum(II) complex. According to the angular overlap model, $\mathrm{d}$ orbitals are going to look like those in the picture. However, planar square complexes are drawn completely differently according to other approaches and this makes me doubt the validity of my diagram.
To arrive to this diagram, I used the following table:[citation needed]

As $\ce{NH3}$ is a σ only ligand, I am not taking into account the π interactions, for which there is a separate table. In the square planar shape, the ligands in the 1 and 6 positions are absent, so they do not contribute to the rise of the d orbital energies, only the 2, 3, 4, and 5 positions do.  According to this, the $\mathrm{d}_{z^2}$ orbital should be raised in energy by $(1/4)+(1/4)+(1/4)+(1/4)=1 \mathrm{e}_\sigma$, and $\mathrm{d}_{x^2-y^2}$ should be raised by $(3/4)+(3/4)+(3/4)+(3/4)= 3 \mathrm{e}_\sigma$. The rest of the orbitals should stay the same energy they are as their coefficients are zero.

Is this a correct diagram for a planar square coordination complex, with sigma only ligands, such as $\ce{[Pt(NH3)4]2+}$?

What is the angular overlap model? If you are considering crystal field theory then your diagram doesn't look right. Take a look at the diagram in this page: https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Crystal_Field_Theory

It's a way of predicting the energies of the d orbitals in a complex. I read about it in an Inorganic Chemistry textbook by Gary. L. Miessler.

Hmm I have never heard of that method, but I feel that both methods should give the same result. Perhaps you could share all of the steps that you did in the calculation?

@ShoubhikRMaiti I could show but I do not see any point in that if you are not familiar with the theory of Angular overlap model already, no offense. I do acknowledge that this diagram not the same that is given by CFT. However, I was thinking that maybe this incompatibility is due to CFT being a simplistic approach compared to angular overlap model.

@Holyshmarckel Even though Shoubhik isn't familiar with the concept, you should add the details. In a couple of days this question this question will get many views and there is a high likelihood that someone who is familiar with the concept will find it.

I also hadn't come across AOM previously, possibly because it fits in a somewhat small niche between the mostly qualitative crystal/ligand field theory and the more quantitative semiempirical electronic structure methods like Huckel theory.  Based on question 10.15 of this solution sheet, you have the right idea. While this question was for a square pyramidal splitting, it seems clear from your table that in going to square planar, the xy/xz/yz orbitals would be unaffected and z^2 would shift down by 1 sigma as you suggest.

@Tyberius What I drew is definitely what AOM predicts to be the d orbital energies of metal & sigma only ligand complex. Your solutions sheet supports that and my math makes sense. But is it a correct diagram empirically or according to LFT?

@Holyshmarckel yes, if you used LFT, you would get the same qualitative result. This is because $t_{2g}$ orbitals are nonbonding (since there are only $\sigma$ interactions), so their energies won't change when going from octahedral $\ce{[Pt(NH3)6]^2+}$ to square planar $\ce{[Pt(NH3)4]^2+}$. But as you remove the ligands on $z$-axis, there will be less overlapping with $d_{z^2}$ so it will decrease in energy because $e_{g}^*$ is antibonding in this case.

@Tyberius Ligand field theory is only an application of molecular orbital theory, so from a purely theoretical point of view it should be complete. The question then becomes how do you approach it and how do you (approximately) solve it. In general, you could be exact, but phenomenological you don't need to be. Crystal field theory on the other hand uses orbitals without describing bonding. From what I have gathered so far, the angular overlap method is (again) rooted in LFT (=MOT), so it must produce the same result (at least approximately or qualitatively).

I have inserted a placeholder for a proper citation. I have to be brutally honest here: There are two ways this question may be read, and in both ways, there need to be more details/ more context about the AOM. Apparently this technique is niche or at least not common, so an overview and links must be included. One way of reading the question would then be: Does AOM correctly reproduce the MO diagram for the complex? The other one is: What is the MO diagram for the complex, I have used AOM to solve it. As of now, I personally would consider this question off-topic. cc @SRMaiti

This looks like a fun and interesting read: Suta, M.; Cimpoesu, F.; Urland, W. The angular overlap model of ligand field theory for f elements: An intuitive approach building bridges between theory and experiment. Coord. Chem. Rev. 2021, 441, 213981. DOI: 10.1016/j.ccr.2021.213981.

@Tyberius maybe you could write the answer for this question? I know you gave it in comments but as this question is bountied I think someone didn't see it here

@Martin-マーチン I interpreted this question as "does AOM correctly reproduce the MO diagram for the complex" which then ties into "what is the MO diagram of the complex?" as you mention. I personally don't feel this question is off-topic imho, despite AOM being a somewhat niche.

@SRMaiti that is the whole point of the comment: we shouldn't have to fall back to interpretation of the question, it should be clear from the start. To me it isn't, and that is why I'd consider it rather off-topic. And that is only tangential to AOM being niche. That's just contributing to the confusion. It would be a great question, if there was an actual source for those tables and a brief explanation. But anyway, my opinion doesn't count here.

Yes, you have correctly produced the diagram according to the angular overlap model, and the result is qualitatively in agreement with what you would get from other models. The reason it looks different from some canonical diagrams that you may find in textbooks for square planar complexes where dxy is raised in energy is because of the lack of $\pi$ interactions in this complex. It is the $\pi$ interactions that affect the dxy in the more general case.
Also, a general comment on the AOM since it seems not to be widely known. For what it's worth, it is described in the lecture notes that Richard Holm distributed for his undergraduate Inorganic Chemistry class at Harvard University in the 90's. He describes it thus:

The Angular Overlap Model (AOM) is a simple first approximation to the full MO model and embodies all of the characteristics of metal-ligand interactions important to understanding of the principles of structure, magnetism, and absorption spectra (color). Its primary value is the rapid establishment of d-orbital degeneracies (or lack thereof) and relative energies. It provides a guide to the metal-ligand orbital interactions as a basis of geometric and electronic structure and reactivity of MLn complexes. The model utilizes two parameters:

$e_\sigma$: $*$ a parameter defining energies of d-orbitals involved in $\sigma$-bonding in a given molecule
$*$ depends on the overlap integral and $d_{z^2}$/L$\sigma$ energy difference; always positive

$e_\pi$: $*$ a parameter defining energies of d-orbitals involved in $\pi$-bonding in a given molecule
$*$ +ve --$\pi$ donor
$*$ -ve --$\pi$ acceptor

This text is taken from the 1996 version of the lecture notes for Chemistry 40: Inorganic Chemistry written by R.H. Holm. The pages are not numbered. Given Holm's stature in the inorganic chemistry community and the general reputation of the Harvard chemistry department, I would assume that the approach was more widely accepted and taught (at least at that time), but I do not have evidence of that. Perhaps someone can add a reference to a commercially published text?
UPDATED REFERENCE: The AOM is also described in detail (with many references) in
Cotton, F.A., and Wilkinson, G.  Advanced Inorganic Chemistry 4th Edition, John Wiley & Sons, 1980, p.652-655.
Given the popularity of this text, I would argue that AOM was more than a niche model, though quite possibly no longer taught.

+1 But perhaps adding diagrams (MO or AOM) of the sigma-only and sigma+pi bonding case would make the answer more easily understandable?

I'm sorry, but I cannot take any book seriously, if they abbreviate positive and negative in that form.

@Martin-マーチン as noted, it's lecture notes, not a book. If you want formal writing, read the cited section in Cotton and Wilkinson.

The Angular Overlap model is still appreciated, at least at the research level. I don't know how much it is taught. See this very recent Coordination Chemistry Reviews "The angular overlap model of ligand field theory for f elements: An intuitive approach building bridges between theory and experiment" https://doi.org/10.1016/j.ccr.2021.213981


text:
$\Delta \bar{h}_{vap}$ and $\Delta \bar{s}_{vap}$ from vapor pressure vs. temperature data
I'm trying to work on the following two problems:

The answers are given below:
4.2: 57.7 kJ/mol, 122.6 J/(K-mol), 12 kJ/mol
4.3: -58.8 kJ per mole of dimer
For 4.2, I used the equation
$$\frac{\mathrm{d} lnP}{\mathrm{d} T}=\frac{\Delta \bar{h}_{vap} }{\bar{R}T^2}$$
and made a least squares line using linear algebra, taken the natural log of the pressure and tabulating it versus the temperature, and taking the slope of that least squares line to be the slope $\frac{\mathrm{d} lnP}{\mathrm{d} T}$. After doing that, in the equation
$$\frac{\mathrm{d} lnP}{\mathrm{d} T}=\frac{\Delta \bar{h}_{vap} }{\bar{R}T^2}$$,
I took $T$ to be the average temperature over the range of data we were given, and that is 386.425 K.
So for $\Delta \bar{h}_{vap}$, I got 59.132 kJ/mol, which is not far off the mark from what the textbook answers are. The problem is that the problem never gives a $T_{vap}$ or an identity of the liquid. When I tried to find $\Delta \bar{s}_{vap}$ from the equation
$$\Delta \bar{s}_{vap} = \frac{\Delta \bar{h}_{vap} }{T_{vap}}$$,
I got 153.02 J/(K-mol), which is very far off the book's answer of 122.6 J/(K-mol). Finding $\Delta \bar{g}_{vap} $ will just be a consequence of finding the molar enthalpy and entropy of vaporization. I'm confused as to how to solve this problem.
For 4.3, I honestly don't know which equations to use or how to figure out the equilibrium constant.

Note that plain text titles are preferred on Chemistry SE site.// The symbols for enthalpy and entropy are  $H$ and $S$ // You can find useful  upright vs italic

It would have been better to fit ln(p) vs 1/T.  What do you get when you do it that way?  The change in G should be zero at all temperatures.

Using their data for the changes in H and S, the corresponding T would be 470 K, which doesn’t make sense.

@ChetMiller Can you explain why ln(P) vs. 1/T is a better fit? I've been trying to process that for a little while.

@ChetMiller Also, does the problem itself make sense? Or is it made poorly.

I would say it is poorly posed.  It should not be asking for a value of delta G, since the starting equation is based on delta G being zero.

$$\frac{d\ln{p}}{d(1/T)}=-\frac{\Delta H}{R}$$ is the same as your starting equation.

@ChetMiller I ended up getting 62.268 kJ/mol using ln(P) vs. 1/T.

Very nice.  Did you plot a graph using a semi-log and a linear scale to see how well it conformed to a straight line?  What was the average value of 1/T for this range of data?  If you multiply the $\Delta H$ value by the average value of 1/T, what do you get for $\Delta S$ over this range?

Regarding 4.3, the equilibrium constant is the partial pressure of the dimer divided by the square of the partial pressure of the monomer.

@ChetMiller I did plot a graph but I didn't use any of the scales. I looked at it and it seems to follow more of an exponential trend. The average value of 1/T is .0026 $K^{-1}$. Multiplying that by the change in molar enthalpy of vaporization yields and molar entropy of vaporization of 172.3 J/(K-mol).

Here is a plot of p on a logarithmic scale vs 1/T on a linear scale.  It is almost a perfect straight line.  The slope is $-\Delta H/R$


text:
Is there a way to experimentally determine if a copolymer has a random monomer distribution or a blocky monomer distribution?
I prepared a styrene-butadiene copolymer with a polar modifier. The expectation is that the copolymer is "random". Is there a way to confirm this using standard characterization techniques?

I've got this kind of confirmation but my polymer - a conjugated one -  was soluble and perhaps short enough to be characterised via HNMR. Otherwise NMR on polymers usually requires at least a specialist (of polymers, or NMR, or both).

Homework! What different properties will stat-PS-PB and block-PS-PB have?


text:
Why does conductivity of a solution of sodium in liquid ammonia decrease with a decrease in temperature?
What happens to the conductivity when a solution of sodium in liquid ammonia at $\pu{-33^{\circ}C}$ (Boiling point of ammonia) is further cooled?
I expected the solution to behave as any other solution that conducts through ions due to the $\ce{[Na(NH3)_x]+}$ and the solvated electrons $\ce{[e(NH3)_y]-}$, such that conductivity increases on decreasing the temperature due to lesser collisions between ions resulting in higher mobility.
However it seems to be the opposite. I tried going through some papers via Google Scholar however I couldn't make out much from the preview alone

A solution of sodium in liquid ammonia contains $\ce{Na^+}$ cations but no anions. It just contains solvated electrons. This may be the reason why its conductivity decreases by cooling it further down.

Could it be related to the lower solubility of Na at the lower temperature, or a phase transition? https://www.researchgate.net/figure/Phase-diagram-of-Na-NH-3-The-experimental-data-on-the-locus-of-the-phase-separation-are_fig1_237048938

@Maurice the solvated electrons don't behave as anions?

Ion pairs Na+(solv)e-(solv) may be more preferred at lower temperature.  As conditions for forming ion pairs are substance specific.

The ion mobility decreases with temperature, hence the conductivity goes down, period. Watery solutions usually conduct by protons changing their hydrogen bonds on one side into covalent, and vice versa, i.e. it's actually electrons who are moving. Totally differenct concenpt.


text:
How do molecules with opposite rotational photoisomerization bonds behave?
If a molecules bond change upon isomerization with the respective bonds straining the molecules as they orient themselves opposite from each other, would the molecule undergo some kind of decomposition?
To expand, and make an example. If there are two bonds in a molecule that undergo photoisomerization that twist the molecule is opposite directions, effectively straining the molecule.
Edit: I am having a lot of hard time explaining this verbally, so I appreciate your patience and time.

Shouldn't be more logical to think that there is not such a molecule, or better, nothing happens? At least not with the photon you have in mind.

What do you mean by 'opposite rotational photoisomerisation' ?


text:
What is the effect of H2O2 on SPIONs?
I have a system contains super paramagnetic iron oxide nanoparticles (SPIONs) either Fe3+ or Fe2+. I need to introduce hydrogen peroxide to the system but i am not sure about its effect to the SPIONs if it can degrade SPIONs or make any change?

$\ce{H2O2}$ is at ambient conditions a liquid and a capable oxidizer.  Not only would it be likely that it transfers $\ce{Fe^{2+}}$ into $\ce{Fe^{3+}}$, the very same ions likely catalyze the decomposition of $\ce{H2O2}$.  This is why I'm going to vote to close the question.

This answer is barely more than a comment. If you add hydrogen peroxide to nanoparticles of an iron oxide such as hematite or magnetite, the hydrogen peroxide will decompose to H2O and O2. Smut (Fe3O4) on stainless steel from etching with FeCl3 is dislodged when treated with H2O2; oxygen bubbles up and knocks the iron oxide off.
However, SPIONs are not just nanoparticles of paramagnetic oxides, but are stabilized in the dispersion, coated by surfactants and treated with drugs for delivery to specific locations (Ref 1). The individual particles start out quite small:

and are engineered to be stable in the blood system, but whether and to what extent they are sufficiently protected to withstand chemical attack from hydrogen peroxide is a question to be determined by experiment.
I would suggest adding the H2O2 to the SPION concoction and look for bubbling - but even if all the H2O2 decomposes, it need not signify the total decomposition of all the SPION.
Ref 1. Advanced Drug Delivery Reviews 63 (2011) 24–46; M. Mahmoudi et al.


text:
Query related to ionic equilibrium constant
I know that [H+][OH-] = kw =10^-14 at 25 degree.But while calculating the pH of 10^-8 M HCl, this equation is used for total concentration of [H+] and [OH-] in the solution. Why is this equation valid in all kinds of aqueous solutions? I think that this equation should be valid only for the dissociation of water, and the amount of H+ and OH- released by water should be 10^-7 mol/l each in any situation.Can someone clear this confusion?

Well I'll point out that $K_w$ is a pseudo constant. The value does depend on temperature, pressure and ionic strength among other factors.  The gist is that $K_w$ is for the equilibrium when a water molecule splits as so: $$\ce{H2O <=> H+ + OH-}$$ The concentration of $\ce{H2O}$ is left out of the mathematical expression for the equilibrium since the molarity of water is assumed to be a constant for dilute solutions.

Why should be the equilibrium of water auto-dissociation the exception, compared to other equilibrium reactions? It is about coexistence of $\ce{H+}$ ($\ce{H3O+}$) and $\ce{OH-}$ ions. There is no rule their concentration must be equal.
The net rate of water dissociation/recombination in pure water or diluted solutions is
$$\frac{\mathrm{d}[\ce{H+}]}{\mathrm{d}t}=A - B \cdot [\ce{H+}]  \cdot  [\ce{OH-}].$$
So the condition for equilibrium, when both $\ce{[H+]}$ and $\ce{[OH-]}$ are constant, is not the equality of concentrations, but the product of concentrations being equal to $K_\mathrm{w}$.
$$\frac{\mathrm{d}[\ce{H+}]}{\mathrm{d}t}=A - B \cdot K_\mathrm{w}=0$$

If $\ce{[H+][OH-]} > K_\mathrm{w}$, ion recombination is faster than water autoionization and the value of  $\ce{[H+][OH-]}$ decreases ( and vice versa ), therefore  $\ce{[H+][OH-]}$  converges to  $K_\mathrm{w}$.
At different temperature, or at concentrated solutions the particular value $\pu{e-14}$ is not valid anymore, but the principle remains.

But Why then is the equation  [H+][OH-] = kw =10^-14 valid in all kinds of aqueous solutions ( whether acids or bases)?


text:
how to show the coordinate bond in avogadro?
I have this structure figure 1, I want to visualize by Avogadro, but the shape is not as I want like in figure 2 where I used Molden, so how can I make Avogadro draw the coordinate bonds.

What Avogadro version? What format are working with (I hope it's not *.xyz)? Could you post the file content? Have you tried to manually connect atoms? And why [tag:organic-chemistry] tag?

Sorry for Tag organic chemistry
the version is 1.2.0
it is not xyz file, it is different (file from CASPT2 Calculation) as I want to visualize the molecular orbital.

The first representation is ball and stick, because atoms represented as spheres, and bonds in rod-like shape.  The lower illustration is an example of stick, where atoms and bonds share the visualization as a stick. The question is not limited to the visualization of coordination compounds, nor the file type read by the program. If you use Avogadro 1.2.0 (e.g., from Linux Xubuntu 18.04 LTS)* then

go Settings -> Toolbars -> Tools to enable the relevant toolbar which show a button «Display settings»

a click on this button opens a new GUI.  Here deselect the entry «Ball and Stick», and select instead «Stick».  With default parameters (via the small wrench symbol at the right hand side you may adjust them), having both options enabled will yield an image like using «Ball and Stick» only.

*) You may be interested to join the Avogadro discourse to stay informed about recent developments of the program (e.g., announced release 1.9.4 a few days ago, entering debichem; or a glimpse about an improved Avogadro2 anticipated for fall 2021 here).

Thank you for your answer, I still have the same problem but ok, I will complete working with molden or gmolden.

For files types anticipated to be read, Jmol has lists (example) and test files (repository). I don't know CASPT2 you mentioned, nor if there a similar public repository of a test output (atom location + orbital's information).


text:
Can't fit an exponential curve to the decomposition of hydrogen peroxide
I have carried out an experiment on the $\ce{MnO2}$ catalyzed decomposition of hydrogen peroxide.   I would have expected the reduction in concentration to be exponential but when I fit an exponential model to the data the residuals are not random.  My data curve is too curved.  So in other words, the decomposition is faster at the beginning of the experiment and slower at the end of the experiment than I would expect if the process was exponential.   It's highly unlikely that it's measurement error - if it was the residuals would be closer to random.
Am I correct in believing the process should follow a negative exponential?   I did wonder if the discrepancy might be due to catalyst contamination which would explain why the exponential model is less curved than the data.
EDIT:  I have also fitted the data to a reciprocal to model a second order reaction and the reciprocal of concentration squared - just for fun.    The residuals retain their smile under all models so I think that M.Farooq is correct and I didn't control the temperature as well as I should have.  However, the second order fit is not as good as the exponential one on the whole (SE of residuals is 0.025 for the fist order model and 0.1673 for the second order but it's marginal )so I think it is first order.  I've run out of manganese dioxide so I'll have to try it again when I get hold of some.   This Paper indicates that the decomposition of hydrogen peroxide in the presence of a solid catalyst is indeed first order.

par(mfrow=c(1,2))
dat<-read.csv("H:\\code_library\\R\\joe_chem.csv", header=TRUE)
Conc=dat$Conc
model<-lm(log(Conc) ~ reading_time, data = dat)
y<-function(x,m,c){
                    exp(m*x+c)
                  }
plot(dat, main="Conc vs Time",xlab="Time(s)",ylab="Concentration")
legend(90,1,legend="model:  y")
lines(dat$reading_time, y(dat$reading_time, coef(model)[2], coef(model)[1]),col="blue")
plot(dat$reading_time, residuals(model), main="Residuals",xlab="Time(s)", ylab="Residuals")
summary(model)

Call:
lm(formula = log(Conc) ~ reading_time, data = dat)

Residuals:
      Min        1Q    Median        3Q       Max
-0.030231 -0.021338 -0.005329  0.014697  0.079083

Coefficients:
               Estimate Std. Error t value Pr(>|t|)
(Intercept)  -1.871e+00  6.533e-03  -286.4   <2e-16 ***
reading_time -1.784e-03  2.048e-05   -87.1   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.02477 on 54 degrees of freedom
Multiple R-squared:  0.9929,    Adjusted R-squared:  0.9928
F-statistic:  7586 on 1 and 54 DF,  p-value: < 2.2e-16

Please consider checking out FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange?. (Apart from the use of the $\ce{...}$ macro, which is covered there, you shouldn't type a zero 0 when you mean an oxygen O.) And in general, we try not to use MathJax in titles, unless it's absolutely necessary.

Thanks for the pointer - I'll take a look.

If you posses access to the raw data, and because the dots in the plot of $c = f(t)$ don't overlap each other, may you replot the diagram with solid dots (e.g., black).  This could ease accessing the data from the illustration, e.g., by WebPlotDigitizer.  What program did you use to plot and fit the data?  And while working on the figure, may you add the equation describing for your fit blue curve and some quality indicators (e.g., correlation $r^2$, residual errors)?

Could you also mention how you measured concentration on the y-axis?

I can imagine several reasons for the curve not to follow a single exponential decay: The simplest one is that as H2O2 begins to decompose, it generates a lot of heat, which further accelerates the rate of reaction. If you did not control the temperature very strictly...then fitting the curve is of no use.

Clearly the model you used is incorrect as the residuals are not evenly distributed. You should  fit with $ae^{-kt}+b$ and vary $a,k,b$ to fit, in which case you would need to have some estimate of the background $b$.  Only when the residuals have no systematic deviation can you sensibly apply statistical tests such as std dev of fit or $\chi^2$. Never, never,  use $R^2$ it is far, far too insensitive for chemical data but is ok when there is just a general trend.

Hi.  I edited the question to show the R code I used.

What does a graph of Conc. vs time look like if the Conc. scale is logarithmic?

Hi Chet.  Broadly linear but it deviates at the tails.  Kind of that you'd expect from the residuals.

Who says it's exponential? That the reaction is first-order?
If the rate is "too high" early in the reaction when concentration is high, but "too low" later on when concentration is low, that indicates you have a higher concentration dependence than the first-order you assumed when you expected an exponential curve.
If we were to assume second-order then we would have
$\dfrac{dc}{dt}=-kc^2$
Solving by standard methods leads to the result
$\dfrac{1}{c}=\dfrac{1}{c_0}+kt.$
So, if $1/c$ is a linear function of time (and it looks that way when I try to pick points off the graph) you have a second-order reaction rate law, not first-order.
PS -- next time show units on the graph. You will need them to extract the rate constant if, indeed, the reaction is other than first-order.

Again to the point to emphasize for the OP is that the temperature must be held constant in a kinetics experiment. "MerelyUseful" did you control the temperature? If not, it has to be repeated.

I would still try the second -order model. I assume the concentration was low enough so that the water solvent sank the heat and kept temperature reasonably within range, although it would have been nice to show units onthe graph.

This is a typical mistake in student graphs. People forget to put the units because they are so comfortable with their own data, the units are "understood".

To my shame I can't remember the units - I think mol/dm-3.    When I've got a few spare minutes I'll try it using the second order model.

I am thinking that while not perfect, it will be much more accurate. Since multiple H2O2 molecules, or fragments thereof, would have to combine to get a single molecule of O2, a reaction order greater than 1 should come as no surprise.

See my edits.  The model fit for a second order reaction is not as good so I think it is first order but it looks like temperature creep messed about with the rate as the reaction progressed.

You should compare not the residuals but the error is concentration itself for both cases. This is the only fair comparison because $\ln c$ in the first-order model and $1/c$ in the second order model have different units, you can compare things only with the same units. This is the basis for my claim that the second-order model will be improved.

I don’t think it will make any difference.  The shape of the residual is almost invariant under first and second order models so there must be something else causing the deviation.  I agree with M Farooq that it’s  temperature.  No matter what model I assume, I’m still going to get heteroskedastic residuals.  The clincher though is the research paper I found that says it’s first order.

You will see that the second order model flips and is smaller if you actually compute concentration errors instead of residuals in ln c or 1/c. Again you need to put things into the same units before you compare.