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Why are some chemical reactions reversible when we generally call all chemical changes as irreversible and permanent? (I am new to this subject.)
I just read that physical changes are temporary and reversible while chemical changes are permanent and irreversible. If that is the case, why are some chemical reactions reversible. Does it mean that chemical changes are reversible?

Yeah, that statement is definitely oversimplified.

Remember that reversibility has false intuitive understanding and has several different, context dependent meanings. One outstanding example is the reversibility of a thermodynamic process versus reversibility of a chemical reaction, ongoing within the process.

I am actually pretty old to this subject and still I cannot really answer, because the boundary between physical changes and chemical reactions is indistinct. Essentially anything that involves rearrangement at an atomic or molecular level, even something as seemingly straightforward as melting ice or dissolving sugar into water, has characteristics of a chemical reaction.
Let us compare the melting of ice with something that is obviously a chemical reaction. For the latter, we choose the reaction of elemental carbon at high temperature with carbon dioxide gas, forming carbon monoxide:
$\text{C(s) + CO}_2\text{(g)}\overset {\Delta}{\to} 2\text{CO(g)}$
First off, both processes are advanced by heating the material. True, carbon requires a higher temperature to react at an appreciable rate with carbon dioxide than ice requires to melt, but this amounts only to a difference in degree rather than a difference in kind of process. Moreover, and more to the point, both the "physical change" of melting ice and the "chemical reaction" of carbon with carbon dioxide can be reversed upon cooling; in the latter case it can lead to a real industrial problem of furnace parts being covered with soot in the cooler parts of the furnace.
One might suppose that we could differentiate the above processes hecause the reaction of carbon with carbon dioxide surely involves the breaking and making of bonds. True, the bonds in carbon monoxide are arranged quite differently from those in the carbon and carbon dioxide from which it can be made. But the same holds true with melting ice, which involves the breaking of hydrogen bonds. The molecular orbital description of hydrogen bonds[1] identifies this hydrogen bonding as a form of (delocalized) covalent bonding -- weaker bonds than those rearranged when carbon and carbon dioxide are heated in a furnace, but again there is only a difference in degree and not in the fundamental type of process.
So there is no clear boundary between physical and chemical changes. Physical changes can be reversible, and so can chemical reactions. Chemical reactions may be essentially irreversible, and (as I look at the scar permanently left on my arm after a long-ago surgery) so can physical changes.
Reference

Bo Wang, Wanrun Jiang, Xin Dai, Yang Gao, Zhigang Wang, and Rui-Qin Zhangb (2016). "Molecular orbital analysis of the hydrogen bonded water dimer". Sci Rep. 6: 22099. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4764947/#!po=1.11111

The statement you challenge is a gross oversimplification and generalisation that fails to capture much of the subtlety of chemical reactions.
It seems to be pitched at describing the difference between simple examples such as the melting of water (physical, easily reversible by putting the water in a freezer) and the simple reaction between chlorine and hydrogen to give common salt (chemical, not spontaneously reversible or dinner tables would be far more entertaining).
But the variety of possible chemical reactions is far bigger than that. Many chemical transformations can be reversed easily given the right conditions. Esterification, the reaction of an acid with an alcohol, makes new molecules and water, for example. But to push the reaction to produce the ester, chemists need to remove the water or the reverse reaction just undoes the new bonds. So they have a cool piece of equipment called a [Dean-Stark apparatus][1] that removes water from the reaction to ensure they get the product they want. The whole point of this is a recognition that the underlying reaction is reversible and needs to be pushed in one direction.
The key to reversible reactions is that there is some mechanism that can undo the new bonds created in a chemical change. In many cases there is no such mechanism which makes many chemical changes impossible to reverse, which is why a simplistic view might distinguish physical from chemical change. But this is unfair on the variety of chemical behaviours.
Another example where you might expect reversibility is the conversion of diamond to graphite. Graphite is more stable under normal conditions but the conversion is not seen. This is because the mechanism to convert diamond t graphite would need to break a very large number of chemical bonds in diamond (every carbon connected to other carbons in an infinite tetrahedral grid) to get graphite (every carbon connected to 3 other carbons in an infinite hexagonal array using delocalised multiple bonds). At very high temperatures and pressures (such as carbon dissolved in liquid iron at pressure) many of the bonds in any form of carbon are already broken and the two forms of carbon can interconvert (this is roughly how diamond is formed in nature and laboratories). The reason why reversibility is not normally seen is that, at room temperature, there is no available mechanism to break the bonds.
Sometimes reversibility has tragic real-world consequences. Thalidomide is a chiral molecule one form of which is a mild sedative and is active against some diseases and cancers. The other form is a dangerous teratogen, causing monstrous deformities in the fetus. Unfortunately there is a reaction path in living systems that interconverts the two isomers, hence tragedy for pregnant women.
In short, reversibility is sometimes possible in chemical reactions, if there is an easily accessible mechanism to undo a chemical reaction under the right conditions. It is wrong to say that all chemical reactions are irreversible.
[1]: https://en.wikipedia.org/wiki/Dean%E2%80%93Stark_apparatus#:~:text=The%20Dean%E2%80%93Stark%20apparatus%2C%20Dean,other%20liquid)%20from%20a%20reactor.

All chemical reactions are reversible. But the back reaction may be extremely improbable. For exemple the combustion of hydrocarbons produces $\ce{CO2}$ and $\ce{H2O}$. But if you mix and heat a mixture of these two gases, you will never even obtain one molecule of any hydrocarbon by the back reaction, because the equilibrium constant is rather huge. The reaction seems irreversible.

A chemical reaction is reversible when and if it reaches chemical equilibrium; [A steady state can possibly be a close approximation.]  This is almost restricted to simple, elementary reactions that are controlled by "Microscopic Reversibility"; this means that the forward path and reverse path have the exact same mechanism. The conditions for equilibrium are reasonably simple: 1. The reactants and products must be present and properly mixed. 2. There must be a possible mechanism for the reaction [catalysts are permitted]. 3. There must be sufficient energy present for the reaction to happen at a reasonable rate. At equilibrium the rates of the forward and reverse reactions are the same. If these conditions are present the equilibrium position will be reached from either direction. A corollary to this is that if the equilibrium is gently perturbed by changing a factor in the equation: concentration, T, P, the equilibrium will adjust to new equilibrium conditions or even to a nonequilibrium state if a component is completely removed [Le Chatelier's Principle]. At equilibrium Delta G = 0 and the concentrations [activities] satisfy the equilibrium constant conditions. This means that equilibrium is dependent on both the energy change and entropy change between reactants and products. These ideas are explained rather obtusely in all Physical and some Organic Chemistry Texts. Go to the library, read several and get confused then think hard about it.
Most common reactions seem to be irreversible because the above conditions are not met: Products are not confined, energy is removed, and they are a mix of several if not thousands of elementary reactions. Also some cyclic reactions such as the processes of respiration and photosynthesis are not reversible processes because the mechanisms of the various reactions are entirely different and the energetics are profoundly different. It does seem to be equilibrium though. The difference is that the complicated process is not energy driven but entropy driven by the vast entropy differences [increases] in the energy forms as it passes thru reaction steps. The individual steps in the mechanisms are mostly steady state reactions approaching equilibrium.

Why was this answer downvoted?


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Will the total transition from ground to excited state decrease when two potential energy surface come near to each other?
I am trying to study Tully simple avoided crossing model via mean-field dynamics. I see when two surfaces are near when value of parameter C is 0.0005 (Unavoided/trivial crossings) there is less transition to excited state whereas, when the surfaces are further, i.e. C=0.05 then I see more transition. This seems counter intuitive.Is it the breakdown of Born-oppenheimer interaction and Ehrenfest(meanfield dynamics)?

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Serial reactions or consecutive reactions: rate vs time
Serial reactions, or consecutive reactions, are two or more reactions in which the product of the first reaction becomes the reactant in the next. The simplest case of a serial reaction involves a reagent A turns into B which in turn, again in the reaction environment, turns into P. In the simplest case, all reactions are irreversible reactions of the first order, so we can be write
$$ \mathrm{A} \xrightarrow{k_1} \mathrm{B} \xrightarrow{k_2} \mathrm{P} $$
the rates of these reactions will be
$$
\begin{equation*}
    \begin{cases}
      r_\mathrm{A} = -k_1\ c_\mathrm{A} \\
      r_\mathrm{B} = k_1\ c_\mathrm{A} -\ k_2\ c_\mathrm{B} \\
      r_\mathrm{P} = k_2\ c_\mathrm{B} \\
    \end{cases}
\end{equation*}
$$
Where $k$ are the kinetics constants, and $c$ is the concentration of the various substances. Plotting $r = f (t)$, I get this

The maximum rate of P formation is reached when $r_\mathrm{B}$ is zero. My hypothesis is that the maximum rate of formation of P must be reached when the concentration of B is maximum, therefore at the minimum of the $r_\mathrm{B}(t)$ curve..
Is my guess right, or is the graph right?

I hope this would be helpful.

Remember that a condition for a maximum of a differentiable function is the first derivative is zero. So rB must be zero for B having maximal concentration.

Simple anwser.
Your answer is correct. The concentration of species B reach a maximum when the rate of formation of the intermediate must equals the rate of dissapeance.
Since $r_B$ can be written as
$$ r_B = \frac{d|B|}{dt} =  k_1|A| - k_2|B|$$
this function has an stationary point at $t_{max}$ when $r_B = 0$. $r_B$ is positive for $t < t_{max}$ and negative after reaching the maximum  $t > t_{max}$.
This result is intuitive if you consider that below $t_{max}$ the concentration of B increases ($r_B > 0$) and drecreases above it $t_{max}$.
The rate equations have analytic solutions and are easily integrated numerically. But you must be awere that the analitical solutions of the rate equations differ for $k_1 = k_2$ and $k_1 \neq k_2$.
You can find a detailed discussion of this problem in the textbooks:

J. H. Espenson.  Chemical Kinetics and Reaction Mechanisms, second
edition. Mc Graw-Hill (1995).
J. I. Steinfeld, J. S. Francisco, W. L. Hase. Chemical Kinetics and Dynamics, second ed., Prentice Hall (1999).

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What is the best way to get a small amount of chemical from a 4L bottle?
I have 4L of multiple chemicals but often only need to work with a few mL at a time. What is the best way to go about this procedure without wasting much of my chemicals?
Pipetting out of the large 4L container is not sustainable, pouring seems dangerous, what else?

5ml or 10ml syringe with a long needle

To answer this question, you should provide additional details such as what size of small volume you are talking about, etc. Additionally, are these chemicals sensitive to the environmental conditions?

I would pour some of the bottle into a beaker, with the aid of a funnel, then extract the needed amount form there. True, we then have to dispose of the rest of the beaker contents, but we can still keep most of the bottle intact. Using labeled secondary containers may be a good idea as well.

Thanks everyone! Mathew, I gave a general question and was looking for generalized answers. Serological Pipettes, pouring, and long syringes sound right.

We used 20 years ago for serial few mL pipetting the special commercial pipetting caps mounted on original 2L chemical bottle flasks. E.g. for methanol or n-hexane. It was simple handle drawing up and down, like a pump, with adjusted volume.


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Required induction time for steady-state approximation to hold
In chemical kinetics, since we often have to deal with complex reactions, approximations are introduced in order to make the calculations easier. A widely used simplification is the steady state approximation, which can be applied when an intermediate is present in the reaction. This can be applied when the concentration of the intermediate is considered essentially constant over time. This is true when dealing with very reactive intermediates that transform as soon as they form, so their concentration is practically constant.
Consider a reaction where
$$\ce{A ->[$k_1$] B ->[$k_2$] P}$$
A point where the concentration of the intermediate B is definitely  constant is at the maximum point of the curve $ [\ce{B}] = f (t) $, obtained by setting
$$
\begin{equation}
\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = k_1 [\ce{A}] - k_2 [\ce{B}] = 0
\end{equation}
$$
By applying the approximation of the steady state, the concentration of $\ce{B}$ is given by the expression (after separation of variables and integration)
$$[\ce{B}] = [\ce{A}]_0 \frac{k_1}{k_2} \mathrm{e}^{-k_1t} \tag{1}$$
The exact solution of $(1)$ is
$$[\ce{B}] = \frac{k_1[\ce{A}]_0}{k_2-k_1} \left( \mathrm{e}^{-k_1t} - \mathrm{e}^{-k_2t} \right) \tag{2}$$
$(1)$ and $(2)$ are equivalent when $k_2 \gg k_1$ and $\mathrm{e}^{-k_2t} = 0$. In other words, $\ce{B}$ does not accumulate. If we attempt a physical interpretation of these two conditions:

$k_2 \gg k_1$ guarantees that the intermediate reacts as soon as it is formed, which is intuitive

$\mathrm{e}^{-k_2t} = 0$: this condition is valid when $t \gg k^{-1}_2$, that is, after an induction period equal to $t \approx k^{-1}_2$.

Could anyone help me understand why the condition $t \gg k^{-1}_2$ is required?

Are you confused with the term ``induction period'', or with the mathematical condition?

The statement is contradictory, but $t \approx k_2^{-1}$ is correct up to log factors. From Atkins: The induction period is an interval during which the concentrations of intermediates rise from zero, and during the major part of the reaction, the rates of change of concentrations of all reaction intermediates are negligibly small. By this definition and taking the derivative, we obtain that both the concentration of $[\ce{B}]$ and its derivative attains maxima at $\sim \ln k_2/k_2$ provided that $k_2 \gg k_1$. After this $[\ce{B}]$ and $\mathrm d[\ce{B}]/\mathrm dt$ decrease.

@JeffreyWang I'm not seeing how the units work for the $\ln k_2 / k_2$ term.  What is the implicit ratio or dividend of $k_2$ inside the $\ln$ that makes the units go away?

@CurtF. The exact point is $\ln \frac{k_2}{k_1}/(k_2-k_1)=(\ln k_2-\ln k_1)/(k_2-k_1) \approx \ln k_2/k_2$.

Let's look at your exact solution again.
$$
[\mathrm{B}] = \dfrac{k_1[\mathrm{A}]_0}{k_2-k_1} \left( e^{-k_1\ t} - e^{-k_2\ t} \right) \qquad
$$
Let's think carefully about the criteria for the steady-state approximation, to well, approximate, this exact solution.

$k_2 \ggg k_1$. This is a criterion you mentioned, and for the right reasons.  If $k_2 \ggg k_1$, then $k_2 - k_1 \approx k_2$.  Another way of saying $k_2 \ggg k_1$ is $k_2 \approx M k_1$, where $M$ is a large number like 100 or whatever.

I think you had the right thought about the other term, but a bit too harsh.  We don't need $e^{-k_2t} = 0$, we need $e^{-k_1t} \ggg e^{-k_2t}$.  If you write it that way, then maybe you can see that this will always be true according to the first criterion, unless $t$ is extremely small.

Thus, now we are asking, how large does $t$ have to be for the approximation $e^{-k_1t} - e^{-k_2t} \approx e^{-k_1t}$ to hold?  If $t$ is 0, then we have the "approximation" $e^0 - e^0 \approx e^0$, which reduces to $1 - 1 = 1$, which isn't very good.  But if $k_2 t \approx 1$ or equivalently $t \approx k_2^{-1}$, then the approximation is $e^{-\frac{k_2 t}{M}} - e^{-1} \approx e^{-\frac{k_2 t}{M}}$ or $e^{-\frac{1}{M}} - e^{-1} \approx e^{-\frac{1}{M}}$, which is better.  If $M$ is 100, then it boils down to $0.62 \approx 1$, which is a much better approximation than $0 = 1$.

Hi Curt, can I suggest $e^{-k_2t/M}$ $e^{-k_2t/M}$ over $e^{-\frac{k_2t}{M}}$ $e^{-\frac{k_2t}{M}}$ for inline math: the latter is a bit hard to read, imo. (though to be totally honest I personally like $\exp(-k_2t/M)$ $\exp(-k_2t/M)$ myself since that doesn't lead to any cramped characters)

A minor nitpick in addition to @orthocresol 's comment: ISO 80000-2 suggests to use upright style for operators and constants with a fixed meaning to differentiate them from variables, so please feel free to consider using \mathrm e $\mathrm e$ for Euler number if you find this suggestion reasonable.

Both of those changes sound good to me.  Feel free to edit away!

Approximate solutions are equivalent to solutions are when
$k_2 \ggg k_1$: this condition ensures that the intermediate reacts as soon as it is formed
$e^{-k_2\ t} = 0$: for this to be true, this expression must be checked
$$e^{-k_1\ t} - e^{-k_2\ t} \approxeq e^{-k_1\ t}$$
this expression is verified only after $t$ takes values greater than or equal to the period of introduction. The introduction period is the period necessary for intermediate to reach its maximum concentration. According to the steady state approximation, once the introduction period has elapsed, the concentration of intermediate remains constant and equal to its maximum concentration: therefore, the steady state approximation cannot be applied for values of $t$ less than $t_\mathrm{max}$. The value of $t_\mathrm{max}$ is given by
$$t_{\mathrm{max}} = \dfrac{\ln k_2 - \ln k_1}{k_2 - k_1} \qquad (\bigstar)$$
since we are in the hypothesis $k_2 \ggg k_1$, $(\bigstar)$ boils down to the form
$$t_\mathrm{max} = \dfrac{\ln k_2}{k_2}$$
for the induction period to have elapsed, an equal time interval must have elapsed
$t\approx k^{- 1}_2 $


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Can the word "iso" be used if the quantities are zero in number?
Both $\ce{H+}$ and $\ce{He^2+}$ ions have zero electrons. So can we call them isoelectronic species?
And if two species have a difference in the number of protons and neutrons equal to zero, can we call them isodiaphers?
Finally, I know that only hydrogen has zero neutrons. But if there was any other element with zero neutrons, could we call hydrogen and that element isotones?

The notorious example is the zero isotherm in context of meteorology and Celsius temperature scale. Quantities with zero value difference are considered the same ( isos = the same ). Quantities themselves, or objects with them as an attribute,  form respectively named isosets, isolines or isosurfaces. It is valid even for cases when quantities themselves have zero value.

(cont.) but for particular cases, zero values may lead to trivial, formally defined  but unused isoset.

Yes, for the reasons everyone else said.
“Iso-” establishes an equivalence class. When that’s based on the number of something, 0 is a number, and all the equivalence relations such as the transitive property hold for the set of objects with relevant number = 0.
I cannot take sole credit for being the first to answer this though.


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What will be the final product?

I think water will attack at double bond and epoxy but I don't know how to proceed further, how is the answer B?

Consider what the site of first protonation is.

I think the path below fully shows what happens.


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How does 81-crown-27 complexes itself with alkali metal with its size?
The biggest crown ether experimentally confirmed is 81-crown-27 1
How can this molecule form complexes when the distance between the 27 oxygen atoms which complexes with the cation inside seems extremely large, making coordination seems extremely hard.
Moreover, from my understanding of coordination bonding, it seems that there is an extreme excess of electron sharing from the aforementioned oxygen atoms, to the point that it seems electrostatic repulsion should take over.
The discovery is only observational in nature, so this in not addressed. I might be making wrong assumptions here, so clarify if needed. Thank you.

I guess you're imagining a huge circle with tiny cation inside? Well, I'm imagining rather something like a braid with cation coordinatin' to a little loop

Most crown ethers are very flexible, not rigid, and can reorganise themselves to achieve decent coordination with, in the case of 81-crown-27, probably many ions.

For such enormous macrocyclic polyethers, coordination with metal cations is probably very closely emulated by simply dissolving metal salts in polyethylene glycol solutions (which is in fact alluded to in the abstract of the article).

The article does not directly describe a complex, only the neat ether, so there is no indication of how a complex may be formed. Most likely, an alkali metal cation would be complexed with only some of the oxygen atoms, the rest being in loops along the ligand between the metal-ligand linkages. For example, six oxygen atoms could be connecteded to the metal with three or four others between adjacent metal-ligand linkages.


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Electrolysis of water+salt: Which molecules is the product?
I should show the kids how electrolysis could create $\ce{H2}$.
We used a drinking glass of water mixed with salt. Then we put an 1.5 volt AA battery in the glass and the electrolysis was visible after about 30 sec.
We then added red cabbage juice which after some hours turned from violet to blue/green. So it must obviously had become base.
Is this the reaction?:
$\ce{H2O + NaCl -> H2 + O2 + NaOH + Cl2}$
Or could the nickel in the battery have an impact on the result?
Salt often contains  $\ce{KCl}$ but that might just give $\ce{KOH}$ which also is a base.
I guess the amount of chlorine gas ($\ce{Cl2}$) is so small that it is not a health risk.

Electrolysis reaction
The electrolysis of sodium chloride solution proceeds by the following reaction:$$\ce{2NaCl (aq) + 2H2O (l) -> H2 (g) + Cl2 (g) + 2NaOH (aq)}$$
The absence of a membrane cell
Since you carry out the reaction without a membrane cell, there is no separation of liberating chlorine from forming sodium hydroxide solution. The absence of the membrane cell causes chlorine disproportionation in alkali solution. So, the following reaction occurs under standard conditions (20-25°C):$$\ce{2NaOH (aq) + Cl2 (g) -> NaOCl (aq) + NaCl (aq) + H2O (l)}$$
Heating of the resulting solution leads to further oxidation of $\ce{[OCl]-}$ to $\ce{[ClO3]-}$:$$\ce{3[OCl]- (aq) -> [ClO3]- (aq) + 2Cl- (aq)}$$
Nickel impact
It is needed more information about the battery itself to determine if nickel from it can have an impact on the electrolysis reaction.
References

R.J.D. Tilley (2004). Understanding Solids: The Science of Materials: p. 281.
C.E. Housecroft, A.G. Sharpe. Inorganic Chemistry (4th ed.): "The group 17 elements", p. 615.
N.N. Greenwood, A. Earnshaw. Chemistry of the Elements (2nd ed.): "Industrial Production and Uses of Chlorine", p. 798.

So no O2 will be produced?

@hschou, it depends on a concentration of sodium chloride in the solution. If $\ce{NaCl}$ solution is concentrated then there will be mostly liberation of chlorine. In case of low concentration of $\ce{NaCl}$ in the solution there will be liberation of chlorine and oxygen simultaneously (since the electrolysis of water on the anode). Compare two electrode potentials:$$\ce{2Cl- (aq) -> Cl2 (g) + 2e-, E^\circ= -1.36 V}$$$$\ce{2H2O (aq) -> O2 (g) + 4H+ (aq) + 4e-, E^\circ= -1.23 V}$$References:
W.M. Haynes. CRC Handbook of Chemistry and Physics (95th ed.): Electrochemical Series, 5-80.

See also: https://www.youtube.com/watch?v=1nAe_T5mO2o


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Is there a heat-based alternative to photosynthesis?
I can imagine one main disadvantage of a process using sunlight to metabolize sugars would be that wherever there is little/no sun (nighttime), plants don't grow. Heat is more abundant than sunlight and could theoretically be an alternative (think units buried deep underground near geothermally active areas).
Is there a heat-based (artificial or natural) replacement for photosynthetic processes?
Note: my understanding of chemistry involving molecules with more than a few atoms is severely limited.

You can metabolize sugar without sunlight, our bodies already do (glycolysis). Plants use the sunlight to turn carbon dioxide and water into sugars and oxygen. A very general law (the 2nd law of thermodynamics) implies that the latter reaction is not possible with thermal energy alone. You can use electrical energy, though.

The second law implies nothing. It states that a process requiring an entropy decrease results in a greater entropy increase in the total system. In practicality the plants simply tried a multitude of times until success.

AFAIK, the only possible hypothetical way would be organisms involving thermal gradient.

There are many sources of energy deep underground and underwater other than photosynthesis. Many species living near deep sea vents can use the oxidation of sulfur and other reduced species to drive CO2 fixation. See, for example, this review: Shively, Jessup M., Geertje Van Keulen, and Wim G. Meijer. "Something from almost nothing: carbon dioxide fixation in chemoautotrophs." Annual review of microbiology 52.1 (1998): 191-230.)

@KarstenTheis You're describing the catabolizing of sugar. OP is interested in anabolizing. Both belong to the two-sided coin of metabolizing. Unlike in layspeak, in technical fields metabolism isn't solely breaking molecules down for energy (catabolism), it includes building up from simpler parts (anabolism). So while OP could have used a more specific word, they are not incorrect.

@jimchmst The plants found a way to do this within the constraints of the second law. We can find a different way to do this, but also within the constraints of the second law.

Related: https://chemistry.stackexchange.com/questions/70791/a-non-life-process-making-glucose-from-carbon-dioxide-and-water-using-electricit (thanks @Glorfindel)

First, it's important to distinguish between the synthesis of sugars from carbon dioxide (photosynthesis) and the catabolism of those sugars to provide energy for growth, which is a process similar to that employed by all sugar consuming organisms. Some plants are able to acquire sugar from other sources (typically through parasitism of other plants or fungi) and so are able to grow just fine without photosynthesis.
Similarly, there are quite a few alternative metabolic pathways that organisms can use to provide energy for growth that do not involve metabolism of sugars. Almost any soluble organic molecule can be used by some organism somewhere, and a many organisms are able to oxidize inorganic compounds such as metals or hydrogen or sulfide for energy. (They still need a supply of organic compounds as biosynthetic precursors, but do not derive energy for growth from them.)
I am going to assume, however, that your specific question is whether the synthesis  from carbon dioxide and water of a reduced organic compound which can support life can be achieved with heat (ie infrared radiation) rather than higher energy visible radiation (ie sunlight).
The wavelengths most important for photosynthesis fall between 400 and 700 nm. The infrared radiation emitted even by a quite hot object on earth will have a wavelength of at least a couple of micrometers. It is an unfortunate consequence of quantization of electron energy levels that one cannot substitute a lot of low energy (long wavelength) radiation for a small amount of high energy (short wavelength) radiation in a reaction, even if the amount of energy provided is equivalent. (There are some minor exceptions such as two-photon absorbances, but those are very special cases.) So we can conclude that it is at least not possible to drive existing photosynthetic processes with heat rather than light.
The follow-up question is whether it is possible through other reactions. A logical choice is use of electricity to reduce the carbon dioxide, since electricity can be produced using heat. One of the simplest organic compounds to make from carbon dioxide (and which requires a fairly low voltage) is methanol, which can be produced with voltages around 300 mV. Many bacteria and archaebacteria are able to metabolize methanol. Ethanol is even more easily metabolized and can be produced with similar voltages. Since bacterial metabolic pathways exist for sugar synthesis, one could theoretically engineer a bacterium to direct much of the energy from ethanol or methanol metabolism towards the synthesis of glucose, ultimately achieving the same end as photosynthetic light reactions.
An electric potential can be produced from heat via a standard steam turbine generator, and geothermal heat sources are sufficient for this purpose, so using that electricity to synthesize methanol and feeding that to glucose-synthesizing bacteria would technically achieve your goal of heat-mediated synthesis of sugar from carbon dioxide.

Light and dark reactions are usually defined differently. Both happen in chloroplasts and participate in sugar biosynthesis. Light reaction refers to synthesis of NADPH and dioxygen from NADP+ and water with concomitant phosphorylation of ADP. Dark (or light-independent) reaction refers to the Calvin cycle, i.e. carbon assimilation and sugar biosynthesis, an ATP-dependent reaction. Sugar breakdown in plants does not have a special name (glycolysis, citric acid cycle, electron transfer chain), and works the same way as in non-photosynthetic plants.

See e.g. here

Note: my understanding of chemistry involving molecules with more than a few atoms is severely limited.

An answer to your question requires understanding of concepts in physical chemistry and biochemistry, and careful use of technical terms such as "metabolize" and "heat". Thus, this answer might be a bit frustrating today, but maybe helpful to motivate further study.

I can imagine one main disadvantage of a process using sunlight to metabolize sugars would be that wherever there is little/no sun (nighttime), plants don't grow.

When we say "metabolize a substance", usually we mean starting with that substance and turning it into something else. The process that makes use of sunlight results in the biosynthesis of sugars. The plant (and this is something you would usually not say) is "metabolizing water and carbon dioxide to produce dioxygen and sugar".
Over their lifespan, photosynthetic plants utilize light to grow. However, they make sugars during the day and store them (and distribute them to non-photosynthetic parts of the plant, e.g. the roots). The plant grows just fine in the absence of light as long as there is some light every now and then. So leaves can grow at night, and roots can grow in the absence of sun exposure. In fact, seeds grow to plants using the nutrients (usually mostly starch or fat) stored in the seeds combined with oxygen in air while photosynthesis is not yet possible.
So plants can grow without light, but photosynthetic plants rely on light to capture most of the (Gibbs) energy that they expend.

Heat is more abundant than sunlight and could theoretically be an alternative (think units buried deep underground near geothermally active areas).

"Heat" is a technical term reserved for the transfer of thermal energy. From the context of the question, I would think you mean "thermal energy".
Sunlight, wind, and electricity are more "useful" sources of energy than thermal energy. You can turn sunlight, wind, and electricity into thermal energy, but you can't "extract" thermal energy from a sample to make e.g. electricity. What is possible is to make e.g. electricity using a hot and a cold sample, where the cold sample gets heated up. You see that in coal, gas or oil-fired electrical power stations, that often are next to rivers to have access to cooling water.
How is this related to making sugar and dioxygen from carbon dioxide and water? This direction of the reaction goes away from equilibrium, and you have to do work to force it in this direction. Work is an umbrella term for electrical work, photochemistry, or mechanical work. (The same is true for making heat flow from a cold to a hot sample, like a refrigerator.)

Is there a heat-based (artificial or natural) replacement for photosynthetic processes?

Yes, you could use a cold and a hot reservoir to make some electricity (for example with a turbine), use that to electrolyze water to dioxygen and hydrogen, and let the hydrogen react with the carbon dioxide to make molecules like formaldehyde or methanol or methane. From there, you still have to make sugar. There are probably some organism that are able to do that in the absence of sunlight.
What you can't do is cool down a hot substance, and use the energy to turn carbon dioxide and water into sugar and dioxygen.

[from the comments] There are devices such as photothermovoltaic cells or thermoradiative cells that can input heat and output electricity.

Yes, these hot-of-the-press devices are pretty cool (pun intended).

The radiative heat part would not work if the system were at thermal equilibrium. So there is still heat transferred from a hot to a cold part. When they calculate the efficiency, they get a familiar result.

So while there are some technical advantages of this over a Carnot cycle, the second law still governs this technology.

"you can't "'extract' thermal energy from a sample to make e.g. electricity" There are devices such as photothermovoltaic cells or thermoradiative cells that can input heat and output electricity.

Heat and thermal energy are synonymous. Q in the statement of the first Law is the CHANGE in heat content not the process. The idea that there is more heat than solar energy is quite foolish. No Sun  no Heat check this out every night or in the upcoming solar eclipse. What is needed for a process like photosynthesis is a LOW ENTROPY energy source.

@jimchmst Not according to IUPAC: https://goldbook.iupac.org/terms/view/H02752

Limits on life powered by heat engines
One strict interpretation of your question is whether living organisms could use a heat engine to power their metabolism.  Heat engines convert a temperature difference between two bodies to useful work.  Hypothetical organisms that live this way could use this work to power chemical reactions that aren't prone to "go" without external power, like protein biosynthesis, growth and division, DNA replication, and other fundamental biochemical process.
There's no fundamental physical rule or law that prohibits this, but (a) all known organisms are made of cells, (b) cells are very tiny, and (c) in natural environments, it is very rare to find temperature differences of appreciable magnitude on the length scale of a cell.  Thus its unlikely that there were many environments on early Earth that would favor the evolution of a heat engine.
To get quantitative:

Let's suppose the hot temperature of a heat engine was 70 ° C = 343 K, and the cold temperature 20 ° C = 293 K.

The hot region and the cold region are separated by about 1 μm.

There is a hypothetical cell of around 1 μm in diameter (this is the size of many modern bacteria and archaea) set right between the $T_H$ zone the and the $T_C$ zone poised to use this temperature difference as a heat engine.  The temperature gradient required in this case is $\frac{\pu{50 K}}{\pu{10^{-6} m}}$ for 50 million Kelvins per meter.

If the cell, like all known life, required water to work, this temperature gradient is not sustainable.  Water has a thermal diffusivity of $\alpha = \frac{\pu{0.14 mm^2}}{\pu{s}}$, which means it would take (very approximately) $\frac{L^2}{\alpha} \approx \frac{\pu{(10^{-6} m)^2}}{\pu{0.14\times 10^{-3} m^2//s}} \approx \pu{10 ns}$ for the temperature gradient to smooth itself out in the absence of very large external inputs.  Ten nanoseconds isn't very long for these hypothetical cells to be able to do any useful work!

A way out
Photosynthesis short-circuits this calculation because it is effectively using sunlight (effective temperature maybe ~ 5000 K) to do this work, because sunlight isn't in local thermal equilibrium with most solid materials on the Earth's surface.
Life based on sunlight-free chemical processes
Life can certainly live without sunlight.  It's not as obvious a statement as you might think though -- oxygen in our atmosphere comes from photosynthesis, so the lives of most familiar organisms (animals, fungi, etc) depends completely on sunlight, because they depend on oxygen for respiration.
However, certain types of chemoautotrophic bacteria can live without oxygen.  For example, 2.8 kilometers below the Earth's surface, scientists isolated a microbe called Desulforudis audaxviator, which probably obtains energy for its metabolism by using the hydrogen, sulfate, and formate formed from water and rocks by radiolysis from naturally occurring radioactive decay, primarily from uranium.
This isn't directly a type of metabolism powered by "heat", but since Earth's heat comes primarily from radioactive decay of heavy elements, and since this life is using some of the byproducts of that decay, maybe it's close enough?

The mechanism for photosynthesis has too many steps to be accomplished thermally from just CO2 and water, however, simple reactions involving CO2 and H2O can be promoted thermally.  Given enough time, energy, and varied conditions more complicated molecules will eventually be formed. It is thought that is what happened [is happening] over several billion years.

Without doing work on the system, water and carbon dioxide will not form sugars because of thermodynamic constraints (the equilibrium lies on the side of the smaller molecules).

That is what I said for a thermal activation process.  Photosynthesis is a series of steadystate processes and just by looking at the fields of corn and the green trees the final products are not the small molecules. There are no, or at least not many, equilibria involved in any of this because of the constant energy flux thru the system. Remember when equilibrium is reached nothing more happens so look forward to Fall.

I am acquiring down votes with no explanations.  I realize that my answer is brief but it is to the point as opposed to another answer that strings a series of pointless facts together that answer nothing together with blatant errors [Heat is Not infrared radiation]  and mixing of photosynthesis and respiration.

I did not down-vote, but I would like to down-vote the comment "The second law implies nothing" if I could.

I apologize, I was being too strict with connotations. My interpretation of the Second Law is that it is more specific. I agree that there is an entropy problem in the thermal activation of photosynthesis and photochemistry solves it by utilizing low entropy light changing the entropy change from chemical to energy changes. In the evolution of photosynthesis there probably was a transfer from thermal to photolytic or maybe not. the second Law does not rule it out; it does predict the better path.


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Can ice nucleation occur above 0 degrees Celsius?
Liquid water can be cooled below 0 degrees Celsius without immediately freezing because the formation of ice crystals requires nucleation. This made me think about the reverse situation: can ice nucleation occur above 0 degrees Celsius? It seems that there should be a statistical chance of water molecules aligning to create an ice crystal (even if that crystal is thermodynamically unstable and quickly returns to the liquid state). Is there a numerical way of describing the odds of nucleation sites forming at any given temperature?

Such temporary, molecular level scope changes occur in liquid water regularly for t < 3.98 °C.

Is it still theoretically possible above 3.98 °C, just increasingly uncommon due to thermodynamic and kinetic barriers?

Yes you'll get density fluctuations that may look ice-like, but since there is no thermodynamic driving force there will be no critical nucleus size and no crystallization above $T= 0^{\circ}\mathrm{C}$ at $P = 1\mathrm{bar}$.

So a few molecules might arrange themselves into an ice crystal structure, but it can never go beyond that?

Yes, there are several numerical (computer simulation) ways to quantify the odds of formation of nucleation sites. One of them is a modern molecular sampling technique called Metadynamics (For eg: See https://doi.org/10.1107/S2052252514027626). Here, the free energy of the nucleation process can be estimated along a strategically chosen crystallization pathway or "collective variable". The likelihood of crossing barriers along this free energy curve can be used to estimate the frequency of nucleation events. I believe the nucleation barrier would be much greater than thermal energy (kT) at 1 bar and T >273 K leading to very large barrier crossing time.. (that is never..)


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How can I calculate average interatomic distance between all identical elements in supercell?
Good afternoon! Maybe someone came across, I want to know the average interatomic distance between all identical elements in some supercell. In Vesta or a similar program, it shows only one piece at a time. Maybe someone knows in what program it is worth cranking? There probably is, but I couldn't find it on the forum. If it isn't possible in simple way, I want get all lengths.

Also I tried with Avogadro, but it only gave a list of interatomic distances without a possibility to calculate the average value.

I'm not sure I understand the question, but probably you might want to look at Olex2 and its envi command in combination with sel $<atom_label>. Note that Chemistry.SE and Stack Exchange sites in general are not forums.

I also suspect in Avogadro 1, that the distances don't consider unit cell boundaries. If you submit it as a feature request to the Avogadro forum it should be fairly easy to add, although that might not help in the immediate moment.


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Alternate formulation of h(P,T)?
I'm using the following handbook "Engineering and Chemical Thermodynamics 2nd edition" by Koretsky. In the book they have an alternate formulation for h(P,T) though it was never explained how they got it (source, equation 5.43).
I'm trying to prove:
$$dh=(c_p-{\beta}vP)dT + vdP$$
with $\beta$ the volumetric thermal expansion coefficient. I keep ending up with:
$$dh=Tds+vdP$$
$$dh=T(\frac{c_p}{T}dT-{\beta}vdP)+vdP$$
$$dh=c_pdT + (v-{\beta}vT)dP$$
How do you get the $-{\beta}vPdT$ term in the first equation?

Is it because $({{\partial}h}/{{\partial}T})_p = ({{\partial}u}/{{\partial}T})_p $ and $({{\partial}u}/{{\partial}T})_p = c_p - {\beta}vP$?

Your derivation is correct, of course.  There is no way that the equation you are trying to prove is correct, because it doesn't  deliver dh=CpdT at constant pressure.

It uses unusually small letters for H, S and V.

@Poutnik: in the book the use of uncapitalized letters are for intensive units, so unit of H is [kJ] while for h it's [kJ/mol]

@ChetMiller yeah, I tried to write it for an ideal gas so
$${\beta}=(1/v)({\partial}v/{\partial}T)_P =(1/v)(R/P)=1/T$$
Then, the first equation for an ideal gas would become:
$$dh=(c_p-R)dT+vdP{\ce{->}}???{\ce{->}}dh=c_pdT$$

dh is not equal to CvdT

It would be nice if you could confirm that the equation comes from a reputable textbook. The result might be due to some constraint which you have not mentioned.

This is from the aforementioned book on page 287: https://imgur.com/a/hhvTgII. Specifically equation (5.43). Equation (5.44) is also slightly amiss, as I think the Helmholtz free energy is volume dependent for IG (but that's a different question altogether).

In the provided source, the metioned equations (5.7) etc. are just the fundamental relations for dh, da, dg - e.g. dh=TdS + vdP.

The complete derivation shown in the image confirms that they messed it up. Expression 5.43 is not correct. You are welcome to post an answer.

Ok, if you are interested in bringing your own answer back to life perhaps the question needs to be edited too, and then the correct derivation which you performed can be moved to the answer, and elaborated upon. What might be useful to other visitors is to see that someone spotted this error in the textbook, and has posted a correction. If this is too much of a hassle then you can leave it alone, but I or someone else might take the liberty of performing these edits.

This is my first post on StackExchange, so I'm not entirely sure how to do this properly. If you or one of the mods could do it/show an example I'll know what to do for next time. Sorry for the trouble!


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Bromine test fails in alkene
My book ( OP Tondon - an indian author) mentions that Bromine-Carbon tetrachloride test fails for alkenes of the type C2R4(R is an alkyl substituent and the double bond is between the carbons). Is this true ? If yes, then why ? Shouldn't all alkenes change the colour of bromine ?

Is this an error on part of the author ?

Steric hindrance might come in, but the R groups would likely have to be large. 2,3-dimethyl-but-2-ene should react.

@OscarLanzi  I agree. This paper records that 2,3-dimethylbut-2-ene can be dibrominated with NDBA so hindrance is not a bar to its formation  https://pubs.rsc.org/en/content/articlelanding/1989/p1/p19890001381/unauth

Hax anybody tried this test on $\ce{C2R4}$ with $\ce{R = C(CH3)3}$ (or $ tert$-butyl) ?

As Oscar Lanzi correctly pointed out the predominance of steric effects on the reactivity of tetrasubstituted alkenes $(\ce{R1R2C=CR3R4})$ is evident, which is confirmed by the careful work by Grosjean, et al. in 1976 (Ref.1). They concluded that it is likely that for the bromination of tetrasubstituted alkenes, the importance of steric effects indicates a clear predominance of bromine-substituent interactions in the rate-determining step, which is the formation of bromonium ion intermediate (Ref.1).
As expected, bromination of tetrasubstituted alkenes $(\ce{R1R2C=CR3R4})$ where R groups would be the smallest (2,3-dimethyl-but-2-ene, $\ce{Me2C=CMe2}$) is evident, considering the experimental rate constant of the reaction at $\pu {25 ^\circ C}$ in methanol is $\pu{9.16\times 10^5 M-1 s-1}$ (Ref.1). It is also have evidence that the structure of alkene and solvent play a significant role in this reaction (Ref.2). For example, when one of the $\ce{Me}$ group in $\ce{Me2C=CMe2}$ has been changed to a $\ce{Et}$ group, the bromination reaction rate of $\ce{Me2C=CMeEt}$ (2,3-dimethyl-pent-2-ene) have changed to $\pu{6.5\times 10^5 M-1 s-1}$ under similar condition (Ref.1). Further, it is evident that the bromination of alkenes under much extreme steric hindrence has still been possible. For instance, when all of the $\ce{Me}$ groups in $\ce{Me2=CMe2}$ has been changed to more bulkier groups such as $\ce{Et}$ and $\ce{n-Pr}$ groups, the bromination reaction has still been progressed. For example, the rate of $\ce{Et2C=CEt(n-Pr)}$ (3,4-diethyl-hept-3-ene) have changed to $\pu{1.43\times 10^4 M-1 s-1}$ under identical conditions (Ref.1). Thus, it is safe to say that under much more extreme steric effect, the bromination would still be processed. Note that the experimental rate for the bromination of $\ce{Me2C=CMe(neo-Pe)}$ (2,3,5,5-tetramethyl-hex-2-ene) has to be found as $\pu{1.78\times 10^3 M-1 s-1}$ under identical conditions (Ref.1). Therefore, it is safe to say that the statement of OP's book, bromine test fails for tetra substituted-alkenes, is errornous.

References:

Daniel Grosjean, Gerard Mouvier, and Jacques E. Dubois, "Bromination of ethylenic compounds. 39. Predominance of steric effects on the reactivity of tetrasubstituted alkenes," J. Org. Chem. 1976, 41(24), 3872–3876 (DOI: https://doi.org/10.1021/jo00886a022).
Giorgio Modena, Franco Rivetti, and Umberto Tonellato, "Solvent and structure effects on the rates of bromine addition to acetylene derivatives. Analogies and differences in electrophilic additions to double and triple bonds," J. Org. Chem. 1978, 43(8), 1521–1526 (DOI: https://doi.org/10.1021/jo00402a007).

Not convinced by Mathew's argument. Using his own data, if one methyl group from $\ce{Me2C=CMe2}$ is replaced by a neopentyl group, the rate constant decreases from $\ce{9.16·10^5 M^{-1} s^{-1}}$ to $\ce{1.78·10^3 M^{−1} s^{−1}}$ So it is about $500$ times less. And this happens when only one methyl group is substituted by a cumbersome group. What will happen if all four methyl groups are replaced by cumbersome radicals like $neo$-pentyl or $tert$-butyl ?

First of all, those data is not my own. They are fron the Refrence 1 I sighted correctly. Second, I'd likr to see an example of an alkene compond containing tetra neo-pentyl groups. ^Third, I didn't claim all of tetra-substituted alkenes undergoes bromination reaction, but  most of them gives positive test, even by some with highly steric hindrence such as specific alkene of interest.


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Is 2-pyrone an aromatic compound?
I was told by my professor that α-pyrone is not an aromatic compound, but I count two π-electrons as well as one of the lone pairs on oxygen giving six total which would be one of Hückel's $4n + 2$ numbers. Does the carbonyl contribute to aromaticity?

Not everything that follows the 4n+2 rule is aromatic ... not to mention this rule is not so hot when we have a pendant pi-bonding atom.

related https://chemistry.stackexchange.com/questions/15272/is-tropone-aromatic

https://chemistry.stackexchange.com/questions/101265/aromaticity-of-heterocyclic-compound

The second one particularity for this link https://pubs.acs.org/doi/abs/10.1021/ja00750a055

This answer discusses this situation in connection with the closely related 2-pyridone. The $4n+2$ rule does not apply at all when there are pendant pi-bonding atoms, but in the case of a carbonyl oxygen atom (with its high electronegativity) we can consider a contributing structure where the carbonyl pi bond is polarized towards oxygen: $\ce{\overset{+}{C} -\overset{-}{O}}$. If this polarization leaves a ring satisfying the $4n+2$ rule, this zwitterion becomes a significant contributing structure and the molecule will show aromatic characteristics.
In the case of $2$-pyrone the zwitterion structure leaves six pi electrons in the conjugated ring and thus an aromatic contribution is expected. The proton nmr spectrum[1] shows proton NMR shifts greater than that in ethylene and closer to that in benzene, indicating aromatic character. At the same time Wikipedia reports that 2-pyrone

may participate in a variety of cycloaddition reactions to form bicyclic lactones. For example, it readily undergoes Diels-Alder reactions with alkynes producing, upon loss of carbon dioxide, substituted benzenes.[2]

Thus the compound has mixed aromatic and aliphatic characteristics, a common occurrence with molecules having an aromatic-zwitterion contribution from a pendant carbonyl oxygen.
References

Pirkle,W.H. & Dines,M. (1069). _J.Heterocyl. Chem. 6, 1 (1969).

Woodard BT, Posner G H (1999). "Recent Advances in Diels-Alder Cycloadditions Using 2-Pyrones". Advances in Cycloaddition. 5: 47–83.


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Fundamental doubts about energy levels vs. shells vs. subshells vs. orbitals
First in school I learnt that when supplied enough energy, the electron in a hydrogen atom will jump energy level(s), denoted by $n=1$, $n=2$ and so on for integral values of $n$. Then I learnt about shells, subshells and orbitals and what they have to do with the quantum numbers $n$, $l$ and $m_l$.

I'm confused about the difference between energy level and shell, if there is any. The quantum number $n$ tells us the shell, but we also use $n$ to talk about energy levels in hydrogen atom as well.
When we say that the electron in hydrogen jumps energy levels, does it mean that it is jumping shells? Is it going from the orbital in $1s$ subshell to the orbital in $2s$ subshell? If we supply even more energy, can it go from the orbital in $1s$ subshell to one of the three orbitals in the $2p$ subshell? Do my questions make sense, or am I wrong about some basic fundamentals?
A comment to a different question said:

"s orbitals have lower energy than p orbitals which have lower energy than d orbitals" This statement is only true when there is more than one electron.

Why is it not true in multi-electron atoms? Is the person referring to the Aufbau Principle i.e. the orbitals in the $4s$ subshell have lower energy than the orbitals in $3d$ subshell?
If I have used any terminology incorrectly, please do let me know.

In a single-electron atom (such as H or He+), the $l$ and $m$ quantum numbers make no difference, so all $n$ levels are the same energy. In multi-electron atoms that is no longer the case since electron-electron effects come in to play, causing all sorts of headaches in the transition metals. Also, to perhaps add to your confusion, since a photon has spin, the transition from 1$s$ to 2$s$ will be less favored than 1$s$ to 2$p$.

Could you please explain why the $l$ and $m$ quantum numbers not matter?

in the single electron case, the energy levels for a given $n$ are the same regardless of $l$ and $m$.

so if n=2 (i.e. 2nd shell) the energy levels of 2s and 2p are the same? i thought 2p > 2s in terms of energy

@JonCuster Under certain experimental conditions (in a magnetic field, in a non-homogeneous magnetic field) you can make hydrogen-spectral lines "split", i.e. the states are no longer degenerate, see e.g. here.

@KarstenTheis - of course, then there are additional terms to deal with.

Shell, level and orbital can often be used interchangeably although they do have different meanings.
The level represents an ordinal or numerical representation of the state and corresponding relative energy of an electron in an atom. Electrons occupy particular quantum states described by fixed quantum numbers and with fixed associated energies. Therefore you can refer interchangeably to the state, energy or quantum number(s) of an electron, keeping in mind the possibility of degeneracy, which occurs when various possible states share the same energy.
Shell suggests an onion-like atomic structure, in which electrons are wrapped around the nucleus in shells, with shells of lower energy electrons closer to the nucleus. This gels with Bohr's idea of electrons occupying fixed orbits about the nucleus. Despite being incorrect, the fixed orbit idea is useful when trying to visualize how energy levels and electron distributions (such as represented with orbitals, see below) in a hydrogen atom (or a hydrogenic atom, one with a single electron) are related. A particular shell is associated with a value of the principal atomic quantum number and may contain subshells, each representing different possible values of the angular momentum quantum numbers.
An orbital describes the spatial distribution of an individual electron and is typically depicted as a boundary that encompasses a volume in which an electron with particular quantum numbers (and corresponding energy) is found with a certain high probability. Higher energy electrons will generally have orbitals with probability extending further away from the nucleus, which is consistent with the idea of a shell.
As an example, when we refer to an electron as being in the 1s subshell (or equivalently in a 1s orbital), we are specifying its energy and angular momentum (but not spin) quantum numbers ($\ce{n=1, l=0, m_l=0}$), and thereby its energy and angular momenta (but not spin angular momentum). The energy of an electron in the 1s shell can be computed exactly as -13.6 eV (relative to a stationary free electron).
Electrons may undergo transitions ("jumps") between shells corresponding to different energy levels and quantum numbers, subject to transition rules. In other words, not all imaginable transitions are allowed, because various properties (specifically angular momentum) must be conserved, just like energy.
When an electron in a hydrogen atom is excited into a higher energy state (or level), its main quantum number must change, since for hydrogen atoms states with the same main quantum number but differing in angular momentum are degenerate in energy (see above).


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How can evaporation from a wash bottle be prevented?
We have an issue with our wash bottles. In our electronic lab, our students use isopropanol and acetone to wash manufactured parts. Unfortunately, we noticed a great deal of evaporation over time. After about one month, the wash bottles are empty.
Is there any better kind of container for such use? Or is there something that we are missing?

Evaporation, or use? If used infrequently, only fill with the amount needed.

I guess it is more evaporation than use

Use less bottles or smaller; perhaps closing completely when not in use.

BTW a month is supposed to be fast use? I doubt even a week would be.

Are the acetone and isopropyl in separate bottles?

Thé above bottle is for acetone. The marking says so. We have the same for isopropanol

First, do not put volatile inflammable substances inside a normal refrigerator as there is a very good chance you will get an explosion. As an example (my recollection, going back 30 years) some rats which had been euthanized with ether in the university where I worked were inside a plastic bag in a normal fridge. In the early hours of the morning there a spark from the internal thermostat exploded the fridge and set fire to the lab, burning it out. Sufficient ether had diffused from the bag to form an explosive mixture in the fridge. Following that, all lab fridges were fitted with external thermostats and marked as "safe".
Personally, I use drmoishe-pippik's 2nd suggestion when I'm cleaning paint brushes by putting the brush, in its container of solvent, inside a tin with a press-in lid (I don't use a plastic container as many solvents will permeate through plastic). I use the same approach with small containers of glue and in both cases, there is greatly reduced evaporation. This is the approach I would recommend for your acetone and IPA bottles.

for noting that these liquids can evaporate trough plastic. Using food-grade containers sometimes helps. Glass or metal are, of course, better, but they have their own practical issues.

Added +1 for safety comment! Acetone is less volatile than ether, but still of concern, even at freezer temperature, so I've modified my answer accordingly.

Acetone is very volatile at room temperature, isopropanol a bit less. Not only is the solvent evaporating, but the ratio of acetone:isopropanol changes, making it less effective.
A few suggestions:

Use a wash bottle with a tight-fitting cap for the nozzle:

Store the bottles inside a larger, completely-sealed container between uses.
Use a pump-type bottle (it might be hard to find one not attacked by the solvents).
Put a screw lid on the bottle if storing for a few days.
*Possibly, store tightly capped bottles in a freezer between uses. As noted by Ken Mercer, evaporation in a refrigerator could cause fire or explosion, but the vapor pressure at freezer temperature, ~240 K, is 10 mm Hg (1,300 Pa), less likely to cause a flammable air/acetone mix. That said, this might still be a fire hazard!

N.B. Acetone is highly flammable, and, as the SDS states, "Can be ignited under almost all ambient temperature conditions." One hopes the technicians use it in a hood designed for fire suppression.

What this answer sort of but doesn't quite say is that wash bottles are not really intended for storage of solvents.

"Store the bottles inside a larger, completely-sealed container between uses." - isn't it safer to store them in a well-ventilated area? Otherwise, come the time to open the container, it may well be filled with a pressurised explosive mixture.


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Effect of beta branching and carbocation rearrangement on SN1 reaction rates
My book (NCERT class 12 Chemistry) asks

Predict the order of reactivity of the four isomeric bromobutanes in SN1 and SN2 reactions.

The answer given for SN1 is

1-bromobutane < 1-bromo-2-methylpropane < 2-bromobutane < 2-bromo-2-methylpropane

My confusion is between (I) 1-bromobutane and (II) 1-bromo-2-methylpropane. The book explains

the carbocation derivative derived from (II) is more stable than derived from (I) because of a greater electron donating inductive effect of (CH3)2CH- group.

However, I have been taught that hyperconjugation is (usually) a stronger effect than induction. Since the carbocation from (I) would have a greater number of alpha hydrogens, that should be more stable.
However, the carbocation in (II) can rearrange to form a tertiary carbocation, whereas (I) can only rearrange to secondary. How significant will rearrangement be when considering the reaction rates? As an additional example, how will reaction rates of 2-bromopentane and 1-bromo-3-methylbutane (where tertiary carbocation is formed after two rearrangements) compare?

The stability of a carbocation is governed with a mixture of contributions from these electronic effects and not just a single one of them. The carbocation formed in (I) has 2 alpha Hydrogens while the one in (II) has 1 alpha Hydrogen + an electron donating group through +Inductive effect. There is not much loss of Hyperconjugative stability while there is extra stability added from +I effect of the methyl group.

Also, since the book specifically mentions path of SN1 mechanism here, clearly carbocation formation has to be considered(usually we won't consider a primary carbocation forming and rather proceed with an SN2 path) and (if) a carbocation will be formed, it will rearrange into a more stable one. The (I) would rearrange to give a secondary and the (II) would rearrange to give a tertiary carbocation. We can look it in this way as well and still it points to (II) being more stable

@PrajwalTiwari Is going from 2 hyperconjugating structures to 1 not a significant loss of hyperconjugative ability? Besides, the methyl group has one of the weakest +I effect, and inductive effects drop significantly with distance.
When I say "carbocation in (II)" I mean the primary carbocation formed initially, since the book uses stability arguments on that without considering rearrangement. Whether (II) gives SN1 at a faster rate and whether the carbocation initially formed from (II) is more stable are two different questions here.


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Reaction of alkane with alkene in acidic medium

This question was asked in jee mains 2022 june 25 shift 1. I have never seen such a kind of reaction before and also could not find it anywhere on internet. But I guessed that the proton will add to primary carbon of the double bond and form tert-butyl carbocation while isobutane will release proton to form isobutyl carbanion. They will then add to each other to form the product D. This is why I marked it.
However the released answer key shows that option B is correct. One possible thought that came in my mind was that bulky bases like tertbutoxide ion make Hoffman product major because they can't attack middle carbon due to steric hindrance. Same case happen here also with nucleophile. So, instead primary carbocation would form and get attacked to form that product in option B.
However I am not sure of myself. My query is not regarding this specific question but asking in general wether such kind of reactions with alkane and alkene actually happen? What is their mechanism? And most importantly how to decide wether a given base or nucleophile is bulky or not?

Firstly, isobutane will never just form an anion by itself. Never. Loss of the proton  must be a concerted reaction with the nucleophilic attack. Then, the bulky tertiary carbocation cannot attack the central tertiary carbon atom of the isobutane. Sterically bad on both sides. So IF a reaction takes place, B must be the product, primary cation attacks tertiary C.

I doubt any reaction takes place. Is there any literature precedent?

Yeah, coaching institutes say that isobutane will form 1° carbanion and then add to 2methylpropene to form that in option B. But that carbanion forming thing is also concerning me

The question paper can can easily found on internet on various websites by searching jee mains 2022 june 25 shift 1 question paper or answer key

Correction: What I described would  be an electrophilic attack of course. Probably just as unlikely as the spontaneous deprotonation, but at least the acid would make sense. A carbanion CANNOT exist in the presence of acidic protons long enough for ANY reaction other than just grabbing one of those acidic protons.

I do not recognize this set of reaction conditions to be any typical in an undergraduate chemistry course (I TA-ed for an honors two semester sequence in college so I am quite familiar still). I would sooner believe there was a typo in this question than believe that any of these given molecules can result from mixing these reagents together.

@JeffGustafson Well, this is definitely not something many academics care about, a reaction between two gases. I actually had a prof who gave a two-semester industrial chemistry lecture, but I think that's pretty rare. There was no exam or anything, not part of the curriculum.

This reaction is actually called as
"alkylation of alkenes". If I had known this term, I could have directly searched it on internet but won't have asked it on SE ;)
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/10%3A_Alkenes_and_Alkynes_I_-_Ionic_and_Radical_Addition_Reactions/10.10%3A_Alkylation_of_Alkenes
This website shows that according to the scientist P.D. Bartlett, a carbocation can react with a hydrocarbon with tertiary carbon by transferring hydride (H-). This is reminiscent of the familiar cannizaro reaction where also this hydride transfer takes place. A chain reaction occurs leading to formation of tertbutyl cation adding with isobutene to form a carbocation which then extracts hydride from isobutane, and then the cycle continues.
I have found similar explanation by coaching channels on YouTube, like this one.(watch from 17:33)
https://youtu.be/T6TUROPO4eM
Though it is in hindi,but you can still see the structures
The catalyst required could be any acid like sulphuric or hydrofluoric acid etc but this website says that this particular reaction can also occur without catalyst at high temperatures (and is just lesser feasible economically). https://www.sciencedirect.com/topics/physics-and-astronomy/alkylation#:~:text=Conventional%20paraffin%20(alkane)%E2%80%93olefin,yielding%202%2C3%2Ddimethylbutane.
As expected there is no carbanion formation. The reaction mechanism is therefore quite similar to free radical halogenation in the sense that there is an initiation step and then propagation step although there is no radical involved.

Very analogous: the cationic polymerisation of propene. https://www.thieme-connect.de/products/ebooks/lookinside/10.1055/b-0035-114994. With methylpropene, the post-addition tertiary cation is so sterically hindered that it can do nothing but accept a hydrogen.

@Karl I am unable to acess this

I wonder if there is also a chain termination step?

One immediate thought came into mind that when the octyl carbocation was formed, is not it possible that it is added to one more molecule of isobutene electrophilically to give a kind of dodecyl cation which after abstracting hydride forms dodecane isomer? And why can't this process continue even longer, just like polymerization? Can there be some some reason (for example steric hindrance) to show that such a reaction will be either kinetically or thermodynamically unfavorable?

Look here: https://pubchem.ncbi.nlm.nih.gov/compound/2_2_4_4-Tetramethylpentane#section=3D-Conformer That's a terribly distorted molecule. Accepting a hydride is much better entropically. But yea, the polymerisation works at low temperatures. Oligomers at 0°C, high Mw's only at -100. https://de.wikipedia.org/wiki/Polyisobutylen

Yes I can see 3 preview pages but it is in German

Let us continue this discussion in chat.


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How does one derive a KD from an equilibrium titration experiment? I am definitely making a mistake somewhere
Any help would be greatly appreciated. Thank you in advance.
If I have an antibody A and a target B, and experimentally titrate the antibody against a single concentration of B, and then measure the % of B that is bound after the solutions reach equilibrium, I should be able to determine the KD of the interaction.
Source below (and others) shows that determining KD is as simple as determining the EC50 of such a curve. from:https://www.sciencedirect.com/topics/nursing-and-health-professions/maximum-binding-capacity 50% occupancy

How does that square with the following equation for KD?
KD = ([A]*[B]) / [AB] where A is unbound antibody, B is unbound antigen, and AB is the bound complex
Let's say I have 10 pM of target antigen, 50% of which is bound when incubated to equilibrium with 10 pM of antibody. That would mean the antibody and antigen are each 50% bound and 50% unbound.
KD = 5pM * 5pM / 5pM KD = 5 pM
So I have calculated a 5 pM KD, not the 10 pM that I would think would be similar to the graph below (assuming they had pM units on the X axis) I'm assuming these types of example graphs refer to the input concentration, right?
Even more confusing, what if the antibody is a really tight binder, such that 10pM of antibody is able to essentially bind 100%, such that 10 pM antibody is able to bind 50% of 20pM of antigen. (1:1 binding assumed)
KD = ~0.001pM * ~10pM / ~10pM
Now you have a KD of 0.001pM, but the graph would suggest that 10 pM of antibody is necessary to bind to 50% of the target antigen.
I'm sure I'm missing something. If anyone could help it would be greatly appreciated.

So I have calculated a 5 pM KD, not the 10 pM that I would think would be similar to the graph below (assuming they had pM units on the X axis) I'm assuming these types of example graphs refer to the input concentration, right?

Your calculation is correct. The concentration in the graph refers to the free concentration of ligand of the receptor. The receptor will be present at very low concentration, and the ligand is often cheap (if it is an approved drug), so it is fine to use it in large excess. In those cases, free ligand concentration is similar to total ligand concentration. However, the graph could be labelled more accurately.

Let's say I have 10 pM of target antigen, 50% of which is bound when incubated to equilibrium with 10 pM of antibody. That would mean the antibody and antigen are each 50% bound and 50% unbound.

In the case of antibody and antigen, both binding partners might be expensive, so using them at a 1:1 concentration ratio might make sense. In this case, you would calculate the free concentration like you did. So when you have equal concentrations of complex and free antibody, the concentration of free antigen is equal to the dissociation constant. Or when you have equal concentrations of complex and free antigen, the concentration of free antibody is equal to the dissociation constant. In your example, that would be 5 pM.
The graphs
The graph has a couple of issues. There should be a unit on the concentration axis (the y-axis is a percentage of full binding, so that is correct without units, but should have a percentage sign). The concentration axis should be labeled as "free ligand concentration". Often, as in the graph below, the free ligand concentration is simply written as $[\mathrm{L}]$. If you want to specify the initial (or total) concentration, the symbol you would use is $[\mathrm{L}]_\mathrm{total}$.

The shape of the graph looks different because the concentration axis is linear, not logarithmic as in the graph posted in the question. The source for the graph and explanation is  http://people.reed.edu/~glasfeld/Chem391/notes/ChLigand_2015.pdf.
If you google "graph occupancy ligand concentration", you will find that many graphs do not carefully distinguish between free ligand concentration and total ligand concentration. This is probably because these two concentration are quite similar when the ligand is in large excess over the binding protein. There is also the occasional confusion between dissociation and association constant, so you have to use your critical thinking skills when coming across figures in the wild (and elsewhere).

Thank you! Much appreciated. My problem was that I did not understand that the X axis was free ligand, I thought it was total ligand input. I suppose my other problem is that the antibody in question is so high affinity that my result is stoichiometrically limited (looks like 25 pM KD, but only because antigen is at 50 pM). Detection becomes impossible when using lower concentrations of antigen.

Yes, it is hard to quantify very tight interactions unless you have a very sensitive assay for the reason you mention.


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Can we make chemical bonds using light, instead of breaking them?
I am trying to bridge a conceptual understanding of how light affects both breaking and forming bonds. When bonds are broken, energy is absorbed (endothermic process). Conversely, when bonds are made, energy is released (exothermic process).
We can calculate the necessary wavelength that is required to break bonds using the equation $\lambda=\hbar c/E.$ For example, light with a wavelength of about $\pu{241 nm}$ is needed to break the bonds of a $\ce{O2}$ molecule (assuming you know the bond dissociation energy).
Is it possible to use a specific wavelength to get the atoms to bond again? If so, is there an equation to calculate this for the atoms involved? I’ve been reading that lasers are used to cool atoms which leads me to think that since atoms are absorbing and reemitting the energy from the lasers to lower its kinetic energy, bonds could be formed when the it's low enough.
All I have been finding is how to break bonds using light. Is there light of a specific frequency that could be used to form a bond between given atoms in proximity?

You might find this article interesting.

Why not? https://phys.org/news/2017-02-blue-carbon-nitrogen-bonds-energetically-unfavorable.html .. https://www.labroots.com/trending/cell-and-molecular-biology/18573/using-green-light-create-molecular-bonds .. https://pubs.acs.org/doi/10.1021/jacs.1c09958

Light could help to rearrange bonds by providing activation energy of reaction. OTOH, Bonding of free, unbound atom needs to release, not to absorb energy. E.g. 2 H atoms do not combine to H2, if they cannot pass energy away, like by collision.

@NavaMoore, I think you are asking about the processes commonly known photodissociation (also photolysis) and photoassociation.

Also, regarding the photodissociation of $O_2$, it is not enough that the energy
of the photon can break the bond, the photon must also be able to couple the ground  state of the molecule to an excited state. I mean, there are other conditions that must be fulfilled.

Surely, if light can break bonds, it can make bonds as well.
We find some simple examples in high school textbooks as well, such as the chlorination of olefin compounds under the presence of UV light. This is an example of a photo-initiation reaction, where the first step is dependent on light. You can find more about this in this article1. Here is a schematic from the same article:

Do note that halogenation often occurs as a substitution reaction $\alpha$ to the double-bond. Under special circumstances (e.g., absence of $\alpha$-$\ce{H}$), halogenation can be made to occur as an addition reaction at the double-bond as well.
We can also consider some more interesting examples. Following reaction is from this study2

Another schematic from the same article:

This concept is called photo-dimerization3:

The photodimerization is an example of a direct photoreaction where every step for polymer build-up is initiated by an absorbed photon, thus every single reaction step is dependent on the quantum yield of the photoreaction (generally very much smaller than one).

References

Schönberg, A. (1968). Photohalogenation. In: Preparative Organic Photochemistry. Springer, Berlin, Heidelberg. DOI: 10.1007/978-3-642-87918-0_37
"Light- and heat-triggered reversible luminescent
materials based on polysiloxanes with anthracene groups" by Han et.al. RSC Adv., 2017,7, 56489-56495, DOI: 10.1039/c7ra12201b
The Encyclopedia of Materials : Science and Technology. by K.H.J Buschow, 2001, Pages 6946-6951

From a synthetic chemist's point of view, I agree that light can make bonds ... eventually. However, the absorbance of light typically weakens or breaks bonds. This can be the first step in making a bond, as you show nicely for the radical reactions involving chlorine. Because you would not (could not) isolate the intermediate and it is happening in the same reactions mixture, it looks like the light "made the bond". Great examples in the answer!

Is it possible to use a specific wavelength to get the atoms to bond again?

Not directly. The crucial step in photochemistry is the absorption of a photon. Usually, this results in a molecule in an excited electronic state. Depending on the molecule, this will make a bond weaker, or break a bond.
An example of a simple (the simplest) molecule is the hydrogen molecular ion, $\ce{H2+}$. It has a single electron, making things so simple you can actually calculate the electronic states from scratch. If you excite the electron, the molecule will fall apart. You can read more about it on this page.

I am trying to bridge a conceptual understanding of how light affects both breaking and forming bonds. When bonds are broken, energy is absorbed (endothermic process). Conversely, when bonds are made, energy is released (exothermic process).

As Poutnik states in the comments to the question, making a bond is exothermic, so adding energy through photons is not productive for bond formation. However, for more complex photoreactions, absorbance of a photon is often the first step.
The most famous examples are cyclo-additions, or electrocyclic reactions. Some work without light (often indicated by "$\Delta$" for just heating up) while others require photoactivation (often indicated by "ℎ$ \nu$" for light with appropriate wavelength).

In the photochemical reaction sketched above, the absorbed photon results in a electronic excitation that does not break the molecule, but changes in electronic symmetry, allowing the reaction to happen. The product in the example above also is in an excited electronic state.

Is there light of a specific frequency that could be used to form a bond between given atoms in proximity?

Yes, but not in a single-step reaction where the bond "is made by the light", like welding two metal pieces together. Instead, using the welding analogy, you partially melt one work piece, and then attach it to the second work piece in a separate step that would not happen without the first step. Like all analogies, this does not capture the whole story, of course.

Is it possible to use a specific wavelength to get the atoms to bond again?

If the bond-breaking reaction is photochemical and there is an intermediate where the product is electronically excited before you get the ground state product, you can try to excite it to that state again to get the bond-forming reaction. This is non-equilibrium chemistry, so you can choose the wavelength (either exciting the reactant or the product) to manipulate the direction of the reaction away from the equilibrium concentrations, like the example in the answer by ananta.

Such a reaction would likely be indirect, with the bond formed by an intermediate that the electromagnetic radiation produces.
One of the simplest examples of such a reaction involves the noble gas helium. When helium gas is exposed to an electrical discharge, the electromagnetic radiation in this discharge ionizes some of the atoms, which then combine with still-neutral helium atoms to form the $\ce{He_2^+}$ cation (see Wikipedia). This is a true covalently bonded species, with bond order 1/2, because two of the valence electrons are in the bonding orbital and only one is in an antibonding orbital.

I do not say it is not equivalent. In fact, after seeing the update, I am going to delete the prior comment as obsolete. I have updated no answer. Saying one answer is good does not mean the other one is not.


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Is there a liquid with refractive index 1.5 to 1.6 that is substitute to bromoform?
So I am tasked with making my countries standards sustainable, wherever possible and trying to address any of the 17 SDGs. I came across a test in a certain material specification code that uses microscope and bromoform of AR or GR grade for measuring glass content in the material. USEPA classifies bromoform as possible human carcinogen and I was wondering if it is just used for its' refractive index properties, we could substitute it with safer materials. I found that correct proportions of glucose or sucrose can give up to 1.4 to 1.45 refractive indices. I was wondering if there was a liquid with 1.6 to 1.7 which could be a substitute to bromoform?
Rest physical conditions remain normal as in Temperature is around 27C and material should be transparent and operate in Visible light spectrum etc.
Thanks in advance for all the efforts and help!

Diiodomethane has a refractive index of 1.741

Here are some liquids with high RI

Thanks guys! I'll check out the options and make the choice!

If you can make do with a a refractive index of ~1.5, look at immersion oils used for microscopy. Since they are widely used in laboratories without restrictions such as a fume hood, they're likely nontoxic, though consult the chemical safety data sheets (SDS) for the particular ones of interest.
Various companies offer higher nD liquids. For example, Cargille has sets of liquids ranging from 1.3 to 1.8. The SDS for Cargille Refractive Index Liquid Series A, which includes 1.6 nD, states, "IARC (International Agency for Research on Cancer): None of the ingredients are listed." That said, if you have questions on toxicity, contact the manufacturer for more specific information.


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Confusing concentration calculation about vivianite solubility
I'm confused about the results of [PO4] and [Fe(II)] calculated in a paper. The example they posted is shown in the figure below:

I tried to calculate the phosphate and Fe(II) concentration. The calculation process is below:
$\ce{ Fe3(PO4)2 8H2O <=> 3Fe^2+ + 2PO4^3- + 8H2O}$
Let's set the concentration of $\ce{Fe^2+}$ as 3x mol/L, concentration of $\ce{PO4^3-}$ as 2x mol/L, then:
(3x)^3*(2x)^2 = 10^-36,   which makes x = 2.47$*$10^-8
So concentration of $\ce{Fe^2+}$ is 3x = 7.42$*$10^-8 mol/L, concentration of $\ce{PO4^3-}$ is 2x = 4.95$*$10^-8 mol/L.
However, the paper said that Fe(II) and P concentration is 1.53$*$10^-5 mol/L and 1.02$*$10^-5 mol/L, respectively.
Here are my two questions:
(1) Why my calculation is not the same (even in the magnitude) as the paper's results?
(2) Why do they mention "pH = 7", I couldn't find where is this "pH" take part in the solubiity calcculation.

At pH 7, most of phosphate is in form of HPO3^2- and H2PO4-, with traces of PO4^3-. Therefore real solubility is much higher, compared to the simple  calculation, not considering other phosphate forms.


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Is there an equivalent term for favourable and non-favourable entropy?
For Gibbs energy, we have 'exergonic' or 'endergonic'
For Enthalpy, we have 'exothermic' or 'endothermic'
It seems logical there should be an equivalent pair of terms for entropy, but I can't seem to find them. Any help would be appreciated; if I'm just blind and haven't seen the terms apologies!

Hmm, not sure if I have ever heard any. Only  informal ones come to mind, related to the order level.

Curious - I feel like such terms would be very useful in describing reactions.

Exothermic or endothermic does not refer to whether a process is "favourable", only the direction of heat transfer, which in a process at constant T and p corresponds to the enthalpy change. The Gibbs energy change actually describes whether the total entropy (system plus surroundings) increases or decreases for a  process at constant T and p.

but does not direction of heat transfer indicate how thermodynamically favourable it is? surely reactions tend towards products with lower thermodynamic energy (if entropy is roughly the same on both sides) - since they would be more stable?

Yes but "tend" is a key word. Perhaps I am nitpicking but it's important that you be clear. You are looking for a word that describes whether the entropy change is positive or negative. Whether that is favourable is irrelevant.

Yes - I suppose you're right. Thanks for adding the tag!

Perhaps a better phrasing of the question would be: is there a term for an increase or decrease of system entropy during a reaction?


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Calculation of canonical orthogonalized basis from overlap matrix?
From Equation [3.169] in David Cook's Book , I understand that we use eigenvalues and eigenvectors of the overlap matrix to calculate an orthonormal matrix using the formula $$X=Us^{-1/2}$$, where $U$ is the matrix of eigenvectors and $s$ is the matrix of eigenvalues for the overlap matrix. But, when I tried to calculate the dot product between two vectors from the new matrix $X$, I do not get $\phi_i\phi_j=1$ or $0$ depending on if $i=j$ or $i\neq j$. Does this mean that I have to normalize $X$?

What values do you get? If you re not getting zero for different vectors normalisation won't fix that. But to be honest without seeing your code it will be very difficult to say what is going wrong.

@IanBush I have posted the code in Github Repo. The values were found to be close to 1.45. I did not post the code at first due to the size of the code (It is close to 750 lines now)

OK, I don't speak python, but normalisation won't fix not getting zero for different vectors - think about it, scaling one of the vectors by a constant C scales its dot product with another vector also by C, so the only way you can get zero is if the dot product is already zero, or C=0, and as the latter would give you the zero vector that is not allowed. Hence you should be getting zero for different vectors even if they are not yet normalised.

@IanBush I am currently reworking the code to check that I am not any mistakes in the code. Thanks for your time.

This is a pretty specific question, but essentially it lacks context and explanation of your thought process. Please Add a proper citation for 'Cooks Book' and it wouldn't hurt to just quote the equation either. Please also [edit] all relevant information into your question, this may include any code. I believe the character maximum is 16,000; so there should be plenty of room.

@Martin, I had previously provided the equation in-text, and I realize that it might have been missed in the first glance. I have since moved the equation outside as a separate block. I have also provided the link to the book that I was referring to. Just to add, I had also provided the link to the entire codebase, and did not post an MWE because it was not a standalone section.

It was not a requirement that X had to be orthogonal. The only requirement was that $X^{\dagger}SX$ must be orthonormal.


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Isolation is a Consequence of Definition
Let's consider a ballon, such that it has an adiabatic wall, and there is no exchange of heat with the surroundings. Now, let the balloon wall be completely impermeable to the movement of matter across the boundary. If so, then the ballon is theoretically an isolated system. However, if we stretch the balloon, to increase its volume, then the temperature of the air inside the balloon will decrease. Will that mean work is done by the system on the surroundings, despite it being isolated?
Here I thus feel that the concept of isolation is actually a consequence and highly bound by definition. The balloon continues to remain isolated unless, some work is done on it, leading to it losing its property of isolation if some work is done on it.
Is my argument correct?

It could be said it is "heat isolated", not "work isolated". There are  implied rigid walls for isolated systems.

You're trying to split a hair here. If the system is thermally isolated, you should say "thermally isolated", not something else. Be specific!

For a nice definition of an isolated system in thermodynamics see the Wikipedia article on isolated systems. The headline is:

In physical science, an isolated system is either of the following:

a physical system so far removed from other systems that it does not interact with them.
a thermodynamic system enclosed by rigid immovable walls through which neither mass nor energy can pass.

Your balloon is not an isolated system if it can do work on its surrounding or have work done on it by its surroundings.


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Confusing statement about the Pauli exclusion principle
While reading "Francis A. Carey, Richard J. Sundberg - Advanced Organic Chemistry Part A. Structure and Mechanisms-Springer (2007)", I came across the following:

The paragraph says that there are two major reasons for electron pairs repelling each other.

Electrostatic repulsion
Pauli exclusion principle

Now, the statement says that, in accordance with the Pauli exclusion principle, only two electron can occupy the same point in space and that those two electrons must have opposite spins. However, as far as I know, that is not the statement of Pauli exclusion principle, or is it equivalent to the following statement?

No two electrons can have the same four quantum numbers.

This is a tricky question, since we really shouldn't be talking about electrons occupying a single point in space (given the uncertainty principle). Also, wouldn't electrostatic repulsion be enough to assert that no two electrons can occupy the same point in space regardless of the spin?

Comments are not for extended discussion; this conversation has been moved to chat.

Although this is not a common way to represent the Pauli exclusion principle, it is indeed one consequence of it. Specifically, we start with the statement that a system of identical electrons must be antisymmetric with respect to any exchange of the electrons, which is the actual postulate that Pauli proposed.[The derivation of this bit is quite complicated and comes from relativistic quantum field theory.] That is, $$\psi(q_1, q_2, . . . q_n)=-\psi(q_2, q_1,. . . . q_n)$$ where each "$q_i$" is an electron described by a given set of quantum numbers.
Although we don't consider electrons to be localized at specific points in space, we can consider the value of their wavefunctions as a function of three spatial coordinates and a spin coordinate, i.e. $x,y,z$ and $m_s$ or $\theta, \phi, r$ and $m_s$. If these are equal for $q_1$ and $q_2$ (ie two particles have the same spatial location and spin), we would have that $$\psi(q_1, q_1, . . . q_n)=-\psi(q_1, q_1,. . . . q_n)$$ which is only true if $\psi =0$, which means that the wave function associated with this set of electrons has zero probability of being in a state where two electrons of the same spin also have the same spatial coordinates.
There's a more complete description of this in Levine's Quantum chemistry book in chapter 10.
UPDATE based on comments: Since $\psi$ is a continous function, the fact that it has zero amplitude when two electrons have the same spatial and spin values also means that the amplitude approaches zero as two electrons get closer together, which is a way of saying that the probability (amplitude squared) of the system existing in a state where two electrons are close together (but not in the same position) is low and approaches zero as the electrons get closer together (because it necessarily equals zero for the state where they are at the same point). We describe this result as "repulsion" of the electrons from one another.

Nice answer. +1

Since the OP specifically talks about electrons occupying the same point in space, I feel it should also be mentioned that certain wavefunctions can indeed have an increased probability for finding both electrons at the same point; see, https://en.wikipedia.org/wiki/Fermi_heap_and_Fermi_hole and https://www.bu.edu/quantum/notes/GeneralChemistry/FermniHolesAndHeaps.html.

@Antimon - Since the OP asked about electrons of the same spin, it seems only the hole is relevant, which I've updated. Could add a mention of heaps as a tangent I guess.

How you describe the Pauli principle depends on the level of theory you are teaching in general. If that level is too low, the Pauli principle becomes almost unrecognizable, and it is questionable why you would even introduce it. For example, it is a bit too subtle to be matched with the Bohr model (which is a one-electron model). On the other hand, we are perfectly happy writing boxes with one arrow up and one arrow down to show two electrons in the same orbital. Synthetic chemists sometimes use quantum chemical results more like a metaphor, anyway, and then use rules of thumb in practice.

An addendum I'll add to the answer if I edit it further, but for now a comment: the inability for two electrons to be in one point is sometimes called "Pauli repulsion" as distinct from "Pauli exclusion" which is used to refer specifically to the inability of two electrons with the same spin to occupy the same orbital, ie have the same four quantum numbers


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Can glycerin crack a glass when heated?
Is it safe to put glycerin in double wall glass candle ( between the walls ) for decor purposes. I know they put it in vapes and double glass wall coffee tumblers. Is there a possibility of it to crack the either inner wall( the one that will be heated by the wax first ) or outside one if heated by candle burning? In condition if chamber with glycerin is closed .Thank you !

Not chemically, but from usual, thermal or mechanical reasons.
Direct flame contact or frequent or regular thermal shocks can crack glass. Heating resistant glass significantly decreases the risk, but does not eliminate it. The prolonged history of heating, together with tiny surface damages, increases the risk.
Heated liquids can crack glass, if kept in closed container. Either directly by thermal dilation, either indirectly by vapour pressure. Minor water content can cause significant pressure if heated enough.


text:
IUPAC name for dicarboxylic acid with aldehyde and amine groups

The IUPAC name of the structure is:

(A) 3-amino-2-formyl butane-1, 4-dioic. Acid
(B) 3-amino-2, 3-dicarboxy propanal
(C) 2-amino-3-formyl butane-1, 4-dioic acid.
(D) 1-amino-2-formyl succinic acid

Why is the correct answer option C and not option A? Can't we also name this compound by IUPAC nomenclature as 2-formyl-3-amino butane-1, 4 dioic acid? If we take the right carboxylic acid functional group COOH as 1, then number the carbon atoms, then CHO takes 2, so formyl is 2, right? Also, doesn't formyl have greater seniority over amino?

Related: https://chemistry.stackexchange.com/questions/133262/the-iupac-name-of-a-multifunctional-carboxylic-acid-compound

The preferred name of the parent structure is butanedioic acid. The name ‘succinic acid’ is retained for general nomenclature with functionalization but no substitution is allowed (see Rule P-65.1.1.2.2); thus, it cannot be used in this case.
The terminal locants ‘1,4’ of butanedioic acid are not cited.

P-14.3.4.1 Terminal locants are not cited in names for mono- and dicarboxylic acids derived from acyclic hydrocarbons and their corresponding acyl halides, amides, hydrazides, nitriles, aldehydes, amidines, amidrazones, hydrazidines, and amidoximes, when unsubstituted or substituted on carbon atoms.

Simple substituent groups that are named by means of prefixes (such as ‘amino’ and ‘formyl’) are arranged alphabetically.

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, the correct alphanumerical order for the compound given in the question is x-amino-y-formylbutanedioic acid (not ‘x-formyl-y-aminobutanedioic acid’).
The locants x and y are used to indicate positions of the parent structure at which modifications represented by suffixes occur. The relevant rules concerning the numbering of locants for substituent prefixes are:

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; (…)

and

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

In accordance with Rule (f), the compound given in the question could be named as ‘2-amino-3-formylbutanedioic acid’ acid as well as ‘3-amino-2-formylbutanedioic acid’ since the locant set ‘2,3’ is the same in both cases. However, according to Rule (g), this example is named as 2-amino-3-formylbutanedioic acid rather than ‘3-amino-2-formylbutanedioic acid’ since amino is cited first as a prefix in the name.