Why is 1,2,3,4,5,8-hexahydronaphthalene more stable than 1,4,4a,5,8,8a-hexahydronaphthalene? Why is 1,2,3,4,5,8-hexahydronaphthalene (1) more stable than 1,4,4a,5,8,8a-hexahydronaphthalene (2)?

A follow-up question is about stability being estimated by counting the α-hydrogens.
In the first case, do we consider α-hydrogens individually for each double bond, or are common hydrogens only taken once?  If it is the former, then the answer would be 12, or else 8. Which one is correct?
If it is the latter, both have an equal number of hyperconjugating structures, i.e. H = 8. Therefore the stability of both must be equal. Is this wrong?

Disclaimer:
This post might be hard to read for some people. Some may even consider this a rant. You have been warned.
This answer will question the validity of the used models, as well as the basis for such exercises. (It will also point to a fundamental flaw of this specific exercise.) If you are a student, you might learn concepts that will later turn out to be inaccurate or outdated. This answer should give you some general context, but it will probably not be able to help you in an exam.

Introduction
I have previously stated in the comments how much I hate these exercises, I even called them really, really big garbage. There are several reasons for that. I am going to try to convince you step by step why I dislike these exercises, this one in particular.
Stability is not a very meaningful concept
There is a reason why there is no easy definition for stability. It is too fuzzy a concept with too many different connotations that basically everybody could understand something different. This is an ultimate flaw, and for these exercises, some kind of understanding of 'stability' is implied, but probably never properly defined.
Let's start with something tangible, one possible (and accepted) definition of stable; one that is so trivial you might have not even thought about it. Stable means something exists. In more official terms, i.e. in the gold book, it is called structural stability (10.1351/goldbook.ST07109):

Within the Born–Oppenheimer (adiabatic) approximation, this is associated with an energy minimum on a potential energy surface. Structural stability implies that any change in the coordinates of the nuclei can only increase the total energy.

I'm not absolutely happy with the above definition, as it can be read that only global minima are stable. If you start from a local minima, some changes in the coordinates will lead to a decrease in total energy; that is how different conformations work and eventually also reactions. So the changes implied by this definition should be reasonably small.
Obviously, this definition does not at all help us with the exercise. Both compounds are stable according to this definition, but we cannot say which one is 'more stable'.
From this we could go ahead and rank molecules according to their total energy. As long as we would be looking at isomers, this could actually work, but another question is how meaningful this approach is. I won't and cannot go into any detail or discussion how to obtain the total energy; for the purpose of this post just assume every molecule does have an exact total energy.
In the end the total energy is a number associated with a molecule. It will not tell you anything about the chemistry that molecule does. Almost all tangible and practical information is not contained in this number. Sure it offers an exact and consistent way of sorting things, but –in my opinion– an utterly useless one.
It is also problematic in a semantic way: you are using one term to describe a phenomenon which already has a name.
This definition also will not help in answering the question, because estimating this value is hard. Quantum Chemistry has plenty of options to estimate or calculate the expectation value of these compounds. Some are simple, based on empirical data, e.g. semi-empirical methods, some are a bit more complex, e.g. Hartree-Fock, Density Functional Approximations, some are very complex, e.g. Configuration Interaction, Coupled Cluster. Almost all of them you cannot do on a piece of paper in a reasonable timeframe.
That leads us to the point where we need another definition. The gold book actually has another definition for us (10.1351/goldbook.S05900):

As applied to chemical species, the term expresses a thermodynamic property, which is quantitatively measured by relative molar standard Gibbs energies. A chemical species $\ce{A}$ is more stable than its isomer $\ce{B}$ if $\Delta_\mathrm{r}G^\circ > 0$ for the (real or hypothetical) reaction $$\ce{A -> B}$$ under standard conditions. If for the two reactions:
\begin{align}
\ce{P &-> X + Y} & (\Delta_\mathrm{r}G^\circ_1)\\
\ce{Q &-> X + Z} & (\Delta_\mathrm{r}G^\circ_2)
\end{align}
$\Delta_\mathrm{r}G^\circ_1 > \Delta_\mathrm{r}G^\circ_2$, $\ce{P}$ is more stable relative to the product $\ce{Y}$ than is $\ce{Q}$ relative to $\ce{Z}$. Both in qualitative and quantitative usage the term stable is therefore always used in reference to some explicitly stated or implicitly assumed standard. The term should not be used as a synonym for unreactive or 'less reactive' since this confuses thermodynamics and kinetics. A relatively more stable chemical species may be more reactive than some reference species towards a given reaction partner.

This brings us back to the original dilemma, the implied understanding of stability. While we certainly can estimate the hypothetical relative molar standard Gibbs energy of the reaction, it actually won't tell us much about the chemistry. At least this definition will give us some handling on the exercise, i.e. the problem becomes finding 'numbers'.
Sometimes, but this is very wrong, stability is used as masquerading a different concept altogether, a kinetic concept. Stability may then wrongly be defined as the lower likelihood to react, in other words a lower reactivity means more stability. Now this brings back the already discussed semantic problem, but this time it is actually worse. There is also no reason to not use the proper terminology. The gold book defines reactive (reactivity) as follows (10.1351/goldbook.R05180):

As applied to a chemical species, the term expresses a kinetic property. A species is said to be more reactive or to have a higher reactivity in some given context than some other (reference) species if it has a larger rate constant for a specified elementary reaction. The term has meaning only by reference to some explicitly stated or implicitly assumed set of conditions. It is not to be used for reactions or reaction patterns of compounds in general. The term is also more loosely used as a phenomenological description not restricted to elementary reactions. When applied in this sense the property under consideration may reflect not only rate, but also equilibrium, constants.

So this is pretty much linked to the activation barrier. This is surely nothing that is asked about by the exercise, especially since no reaction conditions are mentioned. However, it felt necessary to be included here, since quite a few people will actually mix up the terms or even use them synonymously.
You might come across terms like air-stable which means you can handle a substance in a normal atmosphere without having to worry that it would decompose or react. This is exactly the problem that I mentioned before.
Another point worth mentioning is that some people use 'stable' to indicate whether a species is long-lived; this is also a wrong usage according to the above definitions. It becomes especially problematic when it is used synonymously with unreactive or even inert; but these are discussions for a later time.
In conclusion: There are two correct ways to define stable/stability. (1) Stable either means something exists as a local minimum on the potential energy surface. (2) As a relative concept, A is more stable than B if it costs energy to transform A into B.
Hyperconjugation and the problem of counting interactions
The gold book again gives us a definition for hyperconjugation (10.1351/goldbook.H02924), but I will only quote a portion:

In the formalism that separates bonds into σ and π types, hyperconjugation is the interaction of σ-bonds (e.g. C–H, C–C, etc.) with a π network. This interaction is customarily illustrated by contributing structures, [...] sometimes said to be an example of 'heterovalent' or 'sacrificial hyperconjugation', so named because the contributing structures contains one two-electron bond less than the normal Lewis formula [...].

The important takeaway from this is, that this is only valid within a formalism that separates into σ and π bonds. This concept does not (in this form) exist in Molecular Orbital Theory.
As previously mentioned, there are various ways to describe electronic structure. Many of them require more than the back of an envelope. Those descriptions are basically modeling the electron density at given nuclear positions; these are just mathematical tools to reproduce observables.
As such, hyperconjugation interactions are also just mathematical models and you cannot measure their stabilising factor. All numbers you might find online or in textbooks, are pretty much partitioning schemes of experimental data.
As a very simple example: You can measure the hydrogenation enthalpy of ethene and propene. You can then assume that the π-bond will (in first approximation) behave equally. The resulting energy difference can then be interpreted as the stabilising effect of hyperconjugation. You can do that for thousands of molecules and find some average energy value for these interactions.
That certainly is a valid scientific approach; the problem comes then if you neglect all the conditions that lead to this value and take it as a given constant.
It should be painfully obvious, that different σ-bonds will interact differently with other bonds. There is no reasonable justification to assume that a hyperconjugation interaction will always have the same stabilising effect. It should therefore come as no surprise that a carbon-hydrogen σ-bond located between two carbon-carbon π-bonds will have a totally different stabilising effect than a terminal carbon-hydrogen σ-bond.
Unfortunately, this is taught as some kind of holy grail of organic chemistry. Counting hyperconjugation contributors is at best a zeroth order approximation, an educated guess, a ballpark figure. Surely, you can use it to do a rough estimation, but you should always be aware about how imprecise this is.
It is also important that for a complete description of the electronic structure, you always need all resonance contributors, n.b. hyperconjugation is a more specialised form of resonance. Since these are hypothetical structures, they cannot be assigned an energy and they cannot be weighted that way. It really does need more involved methods to accurately describe this.
Focusing on this criterion alone also only ensures that you are neglecting any and all other possible influences. Dispersion will certainly have a different effect in one molecule than it will have in the other. There will be other mesomeric, or inductive effects; the conformational space will also be of concern.
So while $\pi \approx 3$ is a good guess to roughly estimate the size of a round window, you probably shouldn't use it to built your house.
The problem is the question
Now that we have gotten all of the general problems out of the way, there was one problem pointed out previously in the comments by user55119:

There is a cis/trans issue with the second structure. You might explore the heats of hydrogenation/combustion of hexenes that bear the substitution of the double bond. Welcome to ChemSE!

As if we hadn't already enough problems, the question is already ambiguous by specifying the molecules. In my personal opinion, this shows that the instructor did not spend enough time actually posing the exercise, really thinking about it. It shows that this exercise is targeted exactly at this specific counting exercise. To be honest, I would not have been surprised if the approximation had yielded a completely different result.
To be on the safe side, I performed a couple of calculations. First I ran a conformational analysis with CREST on the xtb level of theory, then I optimised the lowest structures with RI-BP86/def2-SVP in Orca. These are the relative electronic energies. (I did not bother with thermal corrections, since they are not accurate enough for this level, and qualitatively will probably not change the general order.
(1)          1-2-3-4-5-8-hexahydronaphtalene      0.0 kJ/mol
(2)    cis-1-4-4a-5-8-8a-hexahydronaphtalene     48.8 kJ/mol
(3)  trans-1-4-4a-5-8-8a-hexahydronaphtalene     26.8 kJ/mol

You can see from these numbers, especially between 2 and 3 where the number of hyperconjugation interactions is the same, that there are quite large differences. So using hydrogenation or combustion values is probably a better approach to this kind of question.
Additionally, those numbers don't tell you much about the molecules themselves or the chemistry they most likely will perform. In general I'd expect 1,4-cyclohexadiene derivatives to readily undergo oxidation to an aromatic species, but that is besides the point.
My personal opinion is that these counting models oversimplify chemistry beyond the point of usefulness. They do not further any understanding of molecular and electronic structure. It turns useful and educational models into formulaic use cases, which are applied without further understanding.


Rusting without water? I was under the assumption that corrosion (rust specifically) required water + oxygen to occur. But the word equation below states that only oxygen is necessary. I know this is true but I do not know the explanation behind it. Could someone help me understand? Apparently water magically appears from somewhere.

A middle to high school level of explanation would be ideal, not too advanced.

metal + oxygen ->  metal oxide

metal + oxygen + water -> metal oxide

It seems the rust reaction was not well explained.  The following is the formula for iron rusting with water.
$$\ce{4Fe + 3O2 -> 4Fe^3+_(aq) + 6O^2- _(aq) -> 2Fe2O3}$$
Water will dissolve the iron because often carbon dioxide is available in the atmosphere to make the water slightly acidic. With the help of acidity electrons can be transported much easier from from dissolved iron to dissolved oxygen. Where oxygen, iron, and water meet is most often where rust pits start to form.

Without water the formula is the following:
$$\ce{4Fe + 3O2 -> 2Fe2O3}$$
This reaction is very slow.  More energy is needed to initiate the reaction as in the case with burning steel wool, or another means is needed to transport electrons.  Dirt, grime, residues, or any number of things can transport electrons resulting in rust.


How to make krypton gas glow So hypothetically, if one was to buy krypton and wanted to make it glow, would a little tesla coil be enough to make it glow? If not, what would be enough? And would one be able to make it glow multiple times or just once, like will making it glow wear down the electrons in the noble gas or somethin’?

Great question! I don't believe it'll wear out in the short term. Light is produced when the electron is excited to a higher orbital and drops back down; this is completely reversible. Neon tubes work on the same principle and last for years!
Eventually (probably after many years) a whole bunch of effects will probably kick in; if there's some impurities in the krypton, they might react with the plasma; the walls of the container will degrade and become opaque, etc.
Whether a certain voltage will cause the krypton to glow seems to be a pretty complicated problem! I believe that would depend on the frequency of the tesla coil (V = dB/dT), how it's coupled to the gas (is there a wire in the tube?), and how large the volume of gas that you want to light.


Why do PBr3 and PCl3 have larger bond angles than PH3? According to the Colby Chemistry Database, the bond angle of $\ce{PBr3}$ and $\ce{PCl3}$ are around $101$ degrees, but the bond angle of $\ce{PH3}$ is $92$ degrees. (Source: PCl3 PBr3 PH3)

There was a previous question on Stack Exchange about the bond angle difference in $\ce{PF3}$ and $\ce{PH3}$ here: PF3 and PH3, but the accepted answer mentions back bonding as a factor to increase the bond angle in $\ce{PF3}$. I was skeptical about this answer, because in $\ce{PCl3}$ and $\ce{PBr3}$, there should be less back bonding than in $\ce{PF3}$ because bromine and chlorine are both much larger than fluorine and there will be worse orbital overlap between the halogen. Less back bonding should make the bond angle of $\ce{PCl3}$ and $\ce{PBr3}$ smaller than that of $\ce{PF3}$, if back bonding were the major factor here? In reality, that is not the case. Why not?

The larger angles can simply be explained as a result of repulsion between the larger atoms of $\ce{Br}$ and $\ce{Cl}$. Hydrogen atoms in $\ce{PH3},$ as the are so small, experience less repulsion as compared to $\ce{Br}$ atoms in $\ce{PBr3}$ or $\ce{Cl}$ atoms in $\ce{PCl3},$ therefore the larger bond angles in $\ce{PBr3}$ and $\ce{PCl3}.$


Reduction of carbamate with LAH The products of the reduction of esters with $\ce {LiAlH4}$ and the products of the reduction of amides with $\ce {LiAlH4}$ are vastly different. The former reduction cleaves the ester and produces two alcohols while the latter reduction produces an amine with the carbonyl group of the original amide replaced with $\ce {CH2}$. A carbamate seems to display both chemical behaviour of esters and amides. I am curious to know what would be the mechanism by which reduction of carbamate with $\ce {LiAlH4}$ takes place and what would be the products of such a reduction.

Carbamates are usually reduced to N-methyl groups. There are numerous examples:

J. Am. Chem. Soc. 2012, 134 (16), 6936–6939

Org. Lett. 2012, 14 (18), 4834–4837

But it is not always a given. In this next example, the nitrogen is part of a three-membered ring (aziridine). These nitrogens are better leaving groups than usual, cf. Ketone/aldehyde synthesis from N-acylazetidines or aziridines where the same kind of reactivity is observed:

Angew. Chem. Int. Ed. 2002, 41 (24), 4683–4685


Sulfuric acid from elemental sulfur and hydrogen peroxide Will this reaction proceed under standard conditions? I don't really care if it is quite slow, as long as it takes less than a month to finish:

$$\ce{S8(s) + 24 H2O2(aq) -> 8 H2SO4(aq) + 16 H2O(l)}$$

My rationale for this happening is that the elemental sulfur can be oxidized by hydrogen peroxide to form sulfur dioxide, which is further oxidized by hydrogen peroxide to form sulfur trioxide. This then reacts with water to form sulfuric acid.

You will likedly find it faster to perform the first oxidation by burning the elemental sulphur in a little tinfoil boat to produce gaseous sulphur dioxide. then set up some aperatus sucking the gasses produced to bubble up through the hydrogen peroxide to produce sulphuric acid. The peroxide does not oxidize sulphur dioxide to sulphur trioxide because it forms sulphuric acid directly.

Does require some aperatus but will be much faster than waiting for the peroxide. Calculate the gibs free energy if you are not sure if a reaction will be spontaneous.


What is the difference between pH and pKa? Is there any difference between the two, when talking of an acid? I am a bit confused. Please help me out.

$\mathrm{p}K_\mathrm{a}$ is the intrinsic property of an acid. It does not depend on how much acid there is in a solution‡. For example, acetic acid has a $\mathrm{p}K_\mathrm{a}$ of $4.76$. This $\mathrm{p}x$ means $-\log x,$ we can calculate $K_\mathrm{a}$ to be $10^{-4.76}$ is $1.7\times 10^{-5}$, which means that at equilibrium, $\frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = 1.7\times 10^{-5}$. On the other hand, $\mathrm{pH}$ is related to the actual amount of $\ce{H+}$ ions, so if you had more acetic acid, the solution would be more acidic ($\mathrm{pH}$ is lower). This is why on Wikipedia, you see $\mathrm{p}K_\mathrm{a}$ given for acids, rather than a $\mathrm{pH}$, because $\mathrm{pH}$ is amount dependent. Similarly, in experiments, we are more interested in $\mathrm{pH}$ because that tells us the actual acidity of a solution.

Also know that though $K_\mathrm{a}$ may always be constant,‡ percent ionization isn't; as there is more weak acid, they dissociate less.

‡ Just like the ideal gas law, aqueous equilibrium constants based on concentrations are for ideal solutions, but they is very accurate for most work.


How do I create a spreadsheet containing the coordinates and predicted bond angles, molecular properties, torsion properties, etc of a molecule? On a previous answer here (https://chemistry.stackexchange.com/a/30841/876) I was informed in the comments that OpenBabel & Python terminal could be used to print the data contained in the property matrices of Avogadro (more info can be found here How do I print property matrices in Avogadro?). I should be clear, however, most of the files I want to be able to do this for are CIF files where the bond angles and lengths are not explicitly mentioned in the files, so these properties may need to be predicted by the software. So I felt I couldn't be the only person interested in this so I thought I'd ask this question on this site.

I would like any answer to be suitable for those operating on Windows machines, if possible. I have Python & OpenBabel installed on my 64 bit Windows 7 PC, so I believe this shouldn't be a major issue, hopefully.


How to read mutual solubility curves of salts with a common ion? Came across a few graphs that represent the mutual solubility of two compounds in a solvent, for e.g., solubility of $\ce{KCl}$, and $\ce{NaCl}$ in $\ce{H2O}$.
On these particular graphs, I can see $x$-axis as $\pu{g}/\pu{100 mL}$ of $\ce{KCl}$, on $y$-axis as $\pu{g}/\pu{100 mL}$ of $\ce{NaCl}$. And the graph has multiple lines each representing a curve at a specific temperature.
I just wanted to understand how to read these type of graphs:

To me the graphs represent a point wise mechanism. For example, if I pick a spot on the $x$-axis, with $\pu{50 g}$ $\ce{KCl}$ at $\pu{80 ^\circ C}$, does that mean if I dissolve $\pu{50g}$ $\ce{KCl}$ at $\pu{80 ^\circ C}$,then I will not be able to dissolve $\ce{NaCl}$ in it?
Or does it mean if I have a starting condition where I have $\pu{50g}$ $\ce{KCl}$ at $\pu{80 ^\circ C}$, and I start adding $\ce{NaCl}$, then $\ce{KCl}$ will precipitate out at a slow rate till I reach point A at a very rapid rate?
The graph doesn't appear complete, or is at least missing additional info to make sense of it. I want to know what that info is.

My interpretation is:
I assume both cross solubilities are related to the volume of water used for the initial solution, not of the initial solution volume, i.e. both in $\pu{g}/\pu{100 mL}$ of water.
If you start at $x$-axis with the $\ce{KCl}$ solution, you can dissolve $\ce{NaCl}$, climbing upwards until you reach the curve for the given temperature. If you cross it, some $\ce{NaCl}$ will not dissolve, and some $\ce{KCl}$ may precipitate, because of the common ion effect.
The same of the initial $\ce{NaCl}$ solution, starting on $y$-axis, and dissolving $\ce{KCl}$, going to the right, again until you reach or cross the curve.
$\ce{KCl}$ solubility is lower, if some extra $\ce{Cl-}$ ions from $\ce{NaCl}$ are present, because of the solubility product $K_{\mathrm{sp}, \ce{KCl}} = a_\ce{K+} \cdot a_\ce{Cl-}$.
Similarly for $\ce{NaCl}$,  $K_{\mathrm{sp},\ce{NaCl}} = a_\ce{Na+} \cdot a_\ce{Cl-}$.
Problem is difficulty with both measuring and predicting ion activities or activity coefficients for concentrated ionic solution. Replacing activities with concentrations could be done only with lot of grains of salt and both eyes closed tightly.
I do not have clear interpretation of the about horizontal edge in the upper chart part. It looks like a kind of "precipitation eutecticum", an analogue to a binary mixture phase chart.

Feedback to comments:
If the system is at the cross-saturation curve, and there is excess of one solid salt, then due dynamic nature of processes, this salt dissolves and the other precipitates. The system is moving slightly along the curve, until for each of salts the rates of dissolving and precipitating are equal. This process is slow.
If a cross-saturated solution cools down, both salts precipitate, according to the ratio of kinetic rates of both processes, until the curve for new temperature is reached.
The ratio can be only very roughly estimated from

the chart position
the direction of the normal vector of the curve
the different ion mobility.


What is sprayed on chromatography paper when developing? When developing the chromatography paper, what chemical will you use to spray on the chromatography paper?

It depends on what you will use the chromatography paper for. Different testable substances will require different mediums. However, for a plain-simple chromatography without staining/substance identification, aluminium oxide (or silica) coated on a plate is used, however paper can also be used.


Why is this reaction spontaneous in standard conditions? To study if this reaction is spontaneous I proceed in this way:
$$\ce{2H_{3}O^{+} +Fe ->H_{2} +Fe^{2+} + 2H_{2}O}$$
First I write the two semireaction:
$$reduction: \ \ \ce{2H_{3}O^{+} +2e^{-} -> H_{2}} $$
$$oxidation: \ \ \ce{Fe -> 2e^{-} +Fe^{+}} $$
Then I write the scheme of the cell:
$$\ce{Fe|Fe^{2+}||H_{3}O^{+}|H_{2}}$$
I know that:
$$E^{0}_{cell}=E^{0}_{right}-E^{0}_{left}$$
$$E^{0}_{cell}=E^{0}_{H_{3}O^{+}|H_{2}}-E^{0}_{Fe|Fe^{2+}}=-E^{0}_{Fe|Fe^{2+}}=-0.44$$
But also:
$$\Delta G^{0}=-nFE^{0}_{cell}$$
So $$\Delta G^{0}>0$$
And the reaction is not spontaneous, while my textbook says it is, so I must be wrong. Someone could find my error?

It has been a while since I took electrochemistry so I'm not sure about the formal procedure that you're using.

You wrote the correct overall reaction.

$$\ce{2H3O+ + Fe -> H2 + Fe^{2+} +2H2O}\tag{1}$$

and you almost wrote the correct half cells reactions (should be $\ce{Fe^{2+}}$ in equation 3):

$$reduction:  \ \ \ce{2H3O+ + 2e− ->H2}\tag{2}$$

$$oxidation: \ \ \ce{Fe -> 2e^{-} +Fe^{2+}}\tag{3} $$
Then you wrote the scheme of the cell: $$\ce{Fe|Fe^{2+}||H_{3}O^{+}|H_{2}}\tag{4}$$ And an equation for the overall cell potential.
$$E^{0}_{cell}=E^{0}_{right}-E^{0}_{left}\tag{5}$$

Your problem is between equations 4 and 5. To subtract $E^{0}_{left}$ the potential must be written as the reduction potential for the half cell. So either you're supposed to be:

Adding the values for the right and left reactions in equation 5

-- or--

Writing both half cells as reduction potentials in equation 4.

The gist here is that the tabular values are always written as reduction potentials. So for iron:

$$\ce{2e^{-} +Fe^{2+} -> Fe }\quad\quad E^{0}_{red} = -0.44\tag{6} $$

I was taught to reverse the sign for an oxidation reaction.

$$\ce{Fe -> 2e^{-} +Fe^{2+}}\quad\quad E^{0}_{ox} = 0.44\tag{7} $$

Then add the reduction and oxidation potentials to get the overall cell potential.

$$E^{0}_{cell}=E^{0}_{red} + E^{0}_{ox}\tag{8}$$


Number of hydrogen bonds of organic compounds? How to count the organic molecules ' hydrogen bonds in average?

What are the numbers of hydrogen bonds formed among alcohols ,alkanoic acids ,aldehyde , ketones , amine ,amide , and ether when they are dissolved in water?

Hint you could write down general formula for eg for alcohols it $C_nH_{2n}O$ and also  due to covalent bonds the hydrocarbons arent dissolved in water.


67 failures at extracting acetaminophen from Excedrin(R) I have adapted the lab experiment described in this J. Chem. Ed. article for the organic chemistry lab course I teach. In the past year, the first year of implementation, I have had 67 failures (100% failure rate) on the extraction part of this experiment.

The experiment involves the isolation/separation of the active components in Excedrin(R): acetylsalicylic acid, acetaminophen, and caffeine.

The experiment is in two parts. Part 1 involves analysis by TLC and separation by column chromatography. This part works beautifully. Part 2 involves separation by successive extractions with increasingly strong bases. Acetylsalicylic acid was extracted with 10% aqueous Na2CO3. Then, acetaminophen was extracted with 3 M NaOH. Finally, caffeine was recovered by evaporation of the remaining organic layer.

In Part 2 We recovered lots of acetylsalicylic acid and caffeine, but no-one recovered any acetaminophen. Below is a list of things I tried in order to improve the results. None of them worked. I know from TLC analysis of the initial organic solution that acetaminophen was present. What happened to it?

Things I have tried:

Changing the concentration of both bases (up and down).
Using a weaker base for the first extraction (like NaHCO3 or Na3PO4).
Increasing the HCl concentration used to work up the second extraction.
No acetaminophen was extracted with the weak base, nor was any left
behind after the second extraction.

Where has my acetaminophen gone?

Did you notice any change of colour (e.g. to brown) and/or the formation of gunk upon adding the 3M NaOH?

I'd assume one would have to boil it, but is the following scenario conceivable under your conditions:

alkaline hydrolysis of the acetaminophen (paracetamol) to p-aminophenol
aerial oxidation to the corresponding quinone imine
partial polymerisation of the latter

You have confirmed that acetaminophen was initially there. A a change in the constitution of Excedrine since the publication of the article can thus be excluded. Isn't a reaction (hydrolysis or whatever) under your extraction not the most likely explanation for the vanished acetaminophen?


Can flames be colored black? Most of us here would know that flames can be colored by adding an appropriate salt to the 'fuel' or by simply introducing it into a flame. Boron and Barium salts give the flame a green tinge, Strontium colors it red, Sodium does gold, copper a bluish-green, etc, etc.
Now from my (not-so-in-depth) high-schooler level of understanding, the energy released by whatever is being burnt to produce the flame in the first place (as is usually the case), excites electrons in metal ions in the salt to higher energy levels. And when these electrons 'de-excite' and drop back to the original energy levels, they emit that extra energy as photons (whose frequencies are determined by the Planck equation $E = h\nu$ ) and these photons are what we perceive as the colored flame.
If I'm not mistaken the book Harry Potter and the Philosopher's stone mentions a 'black flame' of sorts, being one of the obstacles Harry has to circumvent in order to reach the Stone. (Artist's depiction)

Now, as I understand it, for something to be 'black', it must absorb all ( or at least a fairly significant amount of) visible light, encompassing a wide range of frequencies.
Going back to colored flames, I mentioned that the ions must emit light energy, not absorb it, in order to color the flame.
Armed with all this information, would anyone know if it really is possible to create 'black colored flames'? Has it ever been achieved?
[I tried thinking it out myself.....just gave me a headache]
Edit 1: And yes, I've made note of what @Mithoron mentioned, but what I mentioned was just conventional logic, it isn't universal though.
...Newtonian mechanics was 'conventional logic', but we all saw how that failed in relativistic scenarios. So I didn't want to take the risk, hence I've asked If it actually is possible...
Edit 2: Just because someone mentioned otherwise in an answer... I emphasise that by black flames I don't mean invisible flames.

This video shows how a black flame is achieved. If you illuminate the fire with a monochromatic light source (sodium vapor lamp) and introduce a species in the fire that absorbs that wavelength (sodium ions) then the fire will in fact appear black under the illumination. A screen shot is shown below:

Click image for video.


Use of significant figures in addition and subtraction My book mentions:

  The proper use of significant figures in addition and subtraction involves a comparison of only the absolute uncertainties of numbers. This means that only as many digits are retained to the right of the decimal point in the answer as the number with the fewest digits to right of the decimal

But I don't understand that why must the answer be rounded off after addition? Is the answer not accurate enough to be retained as it is ?

No. The answer isn't accurate enough. Say, for example I wish to add two numbers: $1.23$ and $2.367$.

I proceed like this, but I've no idea what the third digit after the decimal in $1.23$  is and so I represent it as $?$.

$$\begin{align*} 1 &.23? \\ + 2 &.367 \\ \hline 3 &.59? \\ \hline  \end{align*}$$

This is why the end result is rounded off with the number of digits equal to the number having the fewest number of digits you initially set off with.


What happens to potassium if potassium permanganate is reduced in a redox reaction? How can I identify spectator ions? I'm trying to get a fool-proof way of finding the products of a redox reaction. I can do both the oxidation and reduction the half reactions. For example in the potassium permanganate with hydrogen peroxide I found the net ionic equation to be:

$$\ce{5H2O2 + 2MnO4- +6H+ -> 2Mn^2+ + 8H2O +5O2}$$

However, this troubles me. First, what happens to the $\ce{K+}$ that was attached to the permanganate ion? My guess is that it hangs around like a spectator ion to partner up with one of the products at the end, but it can't attach to the reduced manganese because that is also positively charged.  Would it do this with the anion that was bound to the $\ce{H+}$ (e.g., $\ce{SO4^2-}$)? Would it attach to the $\ce{Mn^2+}$. Then, what does the $\ce{Mn^2+}$ cation bind to?

I get that only including the reacting species in the net ionic equation is supposed to make things easier, but it always seems to confuse me more. How do I decide what is important and what's not?

Spectator ions do indeed just ‘hang around’. They are unchanged during the reaction and after the reaction they just pair up with whatever they can find. Remember that redox (half) reactions are always charge-balanced (if you set it up correctly). Thus, including the spectator ions which also have a constant charge, the overall charge must be zero. Taking your equation $\text{(Redox1)}$ and its half reactions $\text{(Red1)}$ and $\text{(Ox1)}$:

$$\begin{align}\ce{MnO4- + 5 e- + 8 H+ &-> Mn^2+ + 4 H2O}\tag{Red1}\\
\ce{H2O2 &-> O2 + 2e- + 2H+}\tag{Ox1}\\
\ce{5 H2O2 + 2 MnO4- + 6H+ &-> 2Mn^2+ + 8 H2O + 5O2}\tag{Redox1}\end{align}$$

We added the permanganate as potassium permanganate so there should be a potassium ion. On the reactant sides of the equations $\text{(Red1)}$ and $\text{(Redox1)}$, it can happily hang around with permanganate but it seems a little out of place on the product side; see equations $\text{(Red2)}$ and $\text{(Redox2)}$:

$$\begin{align}\ce{KMnO4 + 5 e- + 8 H+ &-> Mn^2+ + 4 H2O + K+}\tag{Red2}\\
\ce{H2O2 &-> O2 + 2e- + 2H+}\tag{Ox2}\\
\ce{5 H2O2 + 2 KMnO4 + 6H+ &-> 2Mn^2+ + 8 H2O \\ &+ 5O2 + K+}\tag{Redox2}\end{align}$$

And obviously, that cannot be the end of the story. We need something for the product side — but we also need something to offset the protons. This is where the acid comes in. We generally write it simply as $\ce{H+}$, but there will always be a counterion. For example, we might have used sulfuric acid. In that case, instead of writing $\ce{H+}$ we can also include the counterion as $\ce{H2SO4}$, see the third set of equations:

$$\begin{align}\ce{2KMnO4 + 10 e- + 8 H2SO4 &-> 2 MnSO4 + 16 H2O\\ & + K2SO4 + 5 SO4^2-}\tag{Red3}\\
\ce{H2O2 + SO4^2- &-> O2 + 2e- + H2SO4}\tag{Ox3}\end{align}$$

As is, this seems kind of counter-intuitive since there are still naked sulfate ions. What balances those? It’s the electrons that we are transferring. Since the electrons cancel out when adding up the half-reactions, so do the sulfates, see $\text{(Redox3)}$:

$$\begin{multline}\ce{5 H2O2 + 2 KMnO4 + 3 H2SO4 -> \\ 2 MnSO4 + 8 H2O + 5 O2 + K2SO4}\quad\quad\tag{Redox3}\end{multline}$$

As you can see, only part of the sulfates pairs up with potassium; the remaining sulfates pair up with manganese. All in all, there are enough sulfates around for every species to be netural — as required by the law of conservation of charge.

Identifying spectator ions is rather easy: Simply, anything that doesn’t partake in the actual redox process can be determined a spectator ion. So if you realise that only the sulfite anion of sodium sulfite reacts, sodium is automatically a spectator ion and can be left out. Similarly, if you add nitric acid but no nitrogen oxides $\ce{NO_x}$ can be detected, the nitrate ion is a spectator ion.

I’ll admit, it gets a lot easier identifying spectator ions once you ‘get a feeling’. You can always imagine what a potential spectator ion could do in a redox reaction. For most, it’s either improbable (e.g. potassium being reduced) or would have a notable outcome (e.g. nitrate being reduced).


Why does coordinate covalent bond form? Coordinate covalent bonds are bonds on which both electrons from one atom.

But why does this happen? Some may think it is because one of the bonding atoms have strong electronegativity. But experimental evidence suggests otherwise.

For example, carbon monoxide. The EN of carbon is 2.6, and oxygen 3.4. This may lead you to think that carbon will donate a pair of electron, but on Princeton.edu:

  Carbon monoxide (CO) can be viewed as containing one coordinate bond and two "normal" covalent bonds between the carbon atom and the oxygen atom. This highly unusual description illustrates the flexibility of this bonding description. Thus in CO, carbon is the electron acceptor and oxygen is the electron donor.

Nevertheless, in some cases like NO2, it does follows that the less electronegative atom domates electron. But why??

We can share the women, we can share the wine
We can share what we got of yours, cause we done shared all of mine

"Jack Straw" Grateful Dead
But seriously now, the most important thing in the Princeton.edu article (which explains at the bottom that its content comes from Wikipedia) is
"The distinction from ordinary covalent bonding is artificial"
Does $\ce{H2}$ come from a hydride and a proton or from two hydrogen atoms?  The reason a molecule is stable is independent of a particular path by which it is made.
For $\ce{CO}$, it is only in your mind that oxygen brought more wine (electrons) to the party, but what really matters is that there is enough wine (electrons) at the party for everyone to be satisfied.
$\ce{CO}$ is isoelectronic with $\ce{N2}$ and $\ce{CN-}$; the reason for bonding is the same for each.


Deprotection of carboxybenzyl (Cbz) in the presence of Methyl ester? Deprotection of carboxybenzyl and methyl ester can easily be achieved by using ammonium formate followed by addition of 10% Pd/C and LiOH/MeOH respectively.

But how we can achieve deprotection of carboxybenzyl group in the presence of methyl ester?

Just leave out the last step, i.e., LiOH/MeOH and you should be just fine, the Cbz group gets removed under reductive conditions which leave the methyl ester unaffected. If you lose the methyl ester, you could just add it again afterwards.


Calculating the Gibbs Free Energy of Mixing This is a Physical Chemistry Question and comes from the Atkins Physical Chemistry book.

  Suppose that $2.0~\mathrm{mol}$ $\ce{H2}$ at $2.0~\mathrm{atm}$ and $25~^\circ\mathrm{C}$ and $4.0~\mathrm{mol}$ $\ce{N2}$ at $3.0~\mathrm{atm}$ and $25~^\circ\mathrm{C}$ are mixed at constant volume. Calculate $\Delta_\text{mix}G$. What would be the value of $\Delta_\text{mix}G$ had the pressures been identical initially?

The answers are given but I don't understand how to get to $-9.7~\mathrm{kJ}$ for the first part of the question (when initial pressures are different. The gases are initially at different pressures but does that make a difference. Can I still use Raoult's Law?

Let’s start with the last question: Calculate $\Delta_\text{mix}G$ when the pressures are equal in both compartments before mixing. Most physical chemistry textbooks treat this case (including Atkins), and the expression of $\Delta_\text{mix}G$ is:
$$\Delta_\text{mix}G= nRT(x_1\ln x_1 +x_2\ln x_2)$$ In this exercise, $1$ denotes hydrogen, $2$ denotes nitrogen $n=6\ \text{mol}$, $x_1=\frac{2}{6}$ and $x_2=\frac{4}{6}$. So, by substituting these values in the above expression, we find: $$\Delta_\text{mix}G= 6\ \text{mol}\times 8.314\ \frac{\text{J}}{\text{mol}\cdot\text{K}}\times 298\ \text{K}\times(\frac{2}{6}\ln\frac{2}{6} +\frac{4}{6}\ln\frac{4}{6})= -9462\ \text{J}$$ As $\Delta_\text{mix}G<0$ at constant temperature and pressure, this means that the mixing of two ideal gases is a spontaneous process .

Let’s now treat the first question with two different pressure in the two compartments. The initial Gibbs energy is $$G_\text{i}=n_1(\mu_1^0 +RT\ln p_1)+n_2(\mu_2^0 +RT\ln p_2)$$ Where $n_1=2\ \text{mol}$, $n_2=4\ \text{mol}$, $p_1=2\ \text{atm}$, $p_2=3\ \text{atm}$. $$G_\text{i}=2\ \text{mol}\times(\mu_1^0 +RT\ln2)+4\ \text{mol}\times(\mu_2^0 +RT\ln3)$$
When the partition is removed, and each gas occupies the volume $V_1+V_2$.
The partial pressure of hydrogen is $$p^\prime_1=\frac{2\ \text{mol}\times RT}{V_1+V_2}$$
The partial pressure of nitrogen is $$p^\prime_2= \frac{4\ \text{mol}\times RT}{V_1+V_2} $$  On the other hand: $$V_1=\frac{2\ \text{mol}\times RT}{2\ \text{atm}}=1\frac{\text{mol}}{\text{atm}}\times RT$$ $$V_2=\frac{4\ \text{mol}\times RT}{3\ \text{atm}}=\frac{4}{3}\frac{\text{mol}}{\text{atm}}\times RT$$ i.e.  $$V_1+V_2=\frac{7}{3}\frac{\text{mol}}{\text{atm}}\times RT$$
So $$p^\prime_1=\frac{2\ \text{mol}\times RT}{(7/3\ \text{mol/atm})RT}=\frac{6}{7}\ \text{atm}$$ $$p^\prime_2= \frac{4\ \text{mol}\times RT}{(7/3\ \text{mol/atm})RT}=\frac{12}{7}\ \text{atm} $$ The final Gibbs energy is $$G_\text{f}=n_1(\mu_1^0 +RT\ln p^\prime_1)+n_2(\mu_2^0 +RT\ln p^\prime_2)$$  $$G_\text{f}=2\ \text{mol}\times(\mu_1^0 +RT\ln\frac{6}{7})+4\ \text{mol}\times(\mu_2^0 +RT\ln\frac{12}{7})$$ The Gibbs energy of mixing is the difference between $G_\text{f}$ and $G_\text{i}$
$$\Delta_\text{mix}G= 2\ \text{mol}\times RT\ln\frac{6/7}{2} +4\ \text{mol}\times RT\ln\frac{12/7}{3}$$  $$\Delta_\text{mix}G= 2\ \text{mol}\times RT\ln\frac{6}{14} +4\ \text{mol}\times RT\ln\frac{12}{21}$$  $$\Delta_\text{mix}G= 2\ \text{mol}\times 8.314\frac{\text{J}}{\text{mol}\cdot\text{K}}\times 298\ \text{K}\times(\ln\frac{3}{7} +2\times\ln\frac{4}{7})= -9744\ \text{J}\approx{-9.7\ \text{kJ}}$$
In this case, the value of $\Delta_\text{mix}G$ is the sum of two contributions: the mixing itself and the changes of pressure of the two gases to their final total pressure.

I hope it’s clear now.


Can you have a meta stereocenter? Imagine a compound such as 1,3-dibromo-1,2,3-trichloropropane. Since the first and third carbons are attached to four different groups, they are both chiral centers. Assuming that both of these stereocenters are mirror images of one another (as in (1$S$-3$R$)-1,3-dibromo-1,2,3-trichloropropane), would the second carbon also be considered a stereocenter? If so, how would you differentiate the two enantiomers around that carbon using IUPAC nomenclature? Also, would such a carbon be optically active?

Yes, carbon-2 is a stereocenter, and by (arbitrarily) giving the R conformation higher priority than the S, we can assign its stereochemistry using s/r notation:

$\hspace{2cm}$

Pseudochirality simply means that a molecule has two substituents of a stereocenter are mirror images of each other, and the other substituents are either achiral or also mirror images of each other. Despite having multiple stereocenters, pseudochiral molecules are ahciral (optically inactive) because the mirror images of both the s and r conformations are identical.


Contribution of vibrational degrees of freedom The ratio $\displaystyle\frac{C_p}{C_v}$ for linear triatomic molecules including vibration contribution is given by: ?

Attempt:

We are aware that $C_v = \frac{fR}{2}$ and $C_p = C_v+ R$

Hence, $\gamma = 1 + \frac 2f$

Now, for $f$ we have:

$2$ rotational degrees of freedom
$3$ translational degrees of freedom
$3N - 5 = 3\times 3-5 = 4$ vibrational degrees of freedom (formula available here)

Thus, $f = 2+3+4 = 9$

And $\gamma = 1+ \frac 29  = 1.22$

But answer given is $\gamma = 1.154$

Someone in chat room told me that each degree of freedom of vibrational actually has two times contribution so we would have $2\times 4  = 8$ vibrational degres of freedom and a total of $f = 13$, which gives $\gamma = 1.154$. However then I see a contradiction here. If vibrational indeed contributed $2$ per degree then degree of freedom of methane would be $3+3+2\times(3\times 5-6) = 24$ not $3+3+ (3\times 5 - 6) = 15$.

Can someone please clear the confusion here?

You have to be careful to make a distinction between the coordinates needed to describe a molecule, and the degrees of freedom which the molecule can actually access. For this purpose, one might choose to define the number of degrees of freedom as the minimum number of spatial coordinates needed to fully describe the positions of atoms relative to a fixed reference frame. This will be $3N$ total coordinates which corroborates the other answer you link. We then recognize that motion for non-linear (linear) systems, motion in six (five) of these coordinates correspond to translation and rotation. Therefore, the other $3N-6$ ($3N-5$) coordinates describe the vibrations of the system.

Now, if one is concerned with the total energy of a system, then we should keep track of the Hamiltonian. In a classical context, the Hamiltonian is a function of all coordinates which are capable of holding energy in one form or another, $H(\mathbf{q},\mathbf{p})$. For molecules this means we must keep track of $3N$ spatial coordinates, $\mathbf{q}$, and $3N$ momentum coordinates, $\mathbf{p}$.

Now, if we want to know about the average kinetic energy of atom 1 in the $x$ direction, we perform an ensemble average

$$
\langle T_{x,1}\rangle=\frac{1}{2m}\frac{\int p_{x,1}^2e^{-\beta H(\mathbf{q},\mathbf{p})}d\mathbf{q}d\mathbf{p}}{\int e^{-\beta H(\mathbf{q},\mathbf{p})}d\mathbf{q}d\mathbf{p}}
$$

It is a nice exercise to show that if all the degrees of freedom in the Hamiltonian are quadratic, then this separates into lots of integrals over gaussians and the top and bottom will cancel in all coordinates except for the $x$ direction motion of atom 1. This integral will give you the average energy as $1/2kT$.

But, since the integral over the quadratic momentum and the quadratic position as in the harmonic potential are exactly the same, we have just shown that every type of energy which depends quadratically on a coordinate has an average energy of $1/2kT$.

Therefore, if we have a particle in 1-D subject to a harmonic potential, the Hamiltonian is,
$$
H=\frac{p^2}{2m}+\frac{1}{2}kx^2
$$

which means the average energy is $kT$ since there are two coordinates which the Hamiltonian depends on quadratically.

So, hopefully now it is clear why we double to count of vibrations and arrive at the answer,
$$
\gamma=1+\frac{2}{13}=1.154
$$

I know this was very long-winded and overly detailed, but I hope this gives you a little better picture into where the equipartition theorem comes from and how this convenient counting of coordinates we do is really a coincidence of the fact a harmonic potential and kinetic energy are each quadratic in their respective coordinates.

As an aside, the distinction I have made between coordinates and degrees of freedom is completely arbitrary. Indeed, people often refer to both a harmonic potential and the kinetic energy as quadratic degrees of freedom. I prefer this language, but then one must be careful to clarify between degrees of freedom of position and degrees of freedom of momentum because this is more coordinates than are needed to describe the system completely, so it seems there are more degrees of freedom than there really are.


Can aromatic rings be cyanated using cyanide as a nucleophile? For example, can I convert chlorobenzene into benzonitrile by using potassium cyanide as the reagent?

Chlorobenzene is generally inert to nucleophilic substitution. It was observed by Rosenmund and Braun that using an ethanolic solution of potassium cyanide and cuprous cyanide as catalyst at 200°C, chlorobenzene converts to benzoic acid. But if the ethanol-cyanide solvent is removed and temperature is raised, then the reaction proceeds to form cyanobenzene. Yield was further improved when ionic liquids like DMF was used. This is Rosenmund-Braun reaction.

Note: Working with cyanide is highly dangerous. Do not handle cyanide until and unless you are a professional chemist.

Reference

Wikipedia: Rosenmund-von Braun reaction
Will cyanide substitute bromine in bromobenzene?
Potassium ferrocyanide, $\ce{K4[Fe(CN)6]}$ can be used as a non-toxic source for cyanation.
There are loads of other catalyst for cyanation like $\ce{ZnCl2}$, palladium or nickel salts.[@Waylander]


Formation of Hydrogen - from a proton and an electron If I place a proton and an electron close to each other, can they form an atom of hydrogen? Why or why not? Please explain the scenarios where this is not likely/unlikely.

Personally I doubt it, since the electron would have to spin around the proton to call it an atom.

Can it be well-explained by classical mechanics?

How does quantum mechanics explain this sort of a possible bonding? I mean,an electron and a proton can't collide with each other because they would have had infinite attraction between them (that is they would act like magnets), and we know that this does not happen.

Is there something in QM that defines a minimum distance of some sort that is relevant to the proton-electron forming hydrogen?

Both in classical and quantum mechanics you have to get rid of a (rather large) amount of energy, i.e. the difference in potential energy from zero (=infinite distance) to the bound state (= e.g. Bohr radius), also called the binding energy.

Nature has no way to do that without a third particle taking the excess energy with it. In quantum mechanics there is the additional problem with the spin, but also classically that doesn't work if you assume elastic interaction of proton and electron.

So the answer is no.

(Inelastic interaction requires a zoo of additional assumptions, which I let fall victim to Occams razor here. Inverse photodissociation is a possibility, if overall energy and momentum are conserved.)


How is the reaction rate constant affected by concentration? Consider the concentration-time dependence graph for two first-order reactions.

  Which reaction has a larger rate constant?

  Unable to determine
  The reaction represented by the upper curve
  The reaction represented by the lower curve correct

The above question is confusing me.
It gives two concentrations vs. time curves, and asks which reaction has the larger rate constant. The answer is (3) the reaction represented by the lower curve.

I know that the reaction is first-order, hence
$$\mathrm{rate} = k[A].$$
Thus, algebra reveals that
$$[A] = \frac{\mathrm{rate}}{k}.$$
Thus, it makes sense that increasing the value of $k$ means dividing rate by a larger number, yielding a smaller concentration at any given time. This makes sense why the lower curve is the correct answer.

However, I also understand that the value of $k$, the rate constant, does not depend/is not correlated with the concentration of the reactant(s). Thus, is my logic not applicable? How else can you reason your way to the correct answer?

...I also understand that the value of k, the rate constant, does not depend/is not correlated with the concentration of the reactant(s).

It would be easier to represent this in calculus, but I assume you have not gotten there yet.

Anyway, you are correct, $k$ is a constant.

$$\mathrm{rate} = k [A]^x \implies \mathrm{rate} \propto k$$

thus as $k$ increases, the rate will increase for a given concentration of $\ce A$. The question did provide that $x = 1$ but this is not needed to solve the problem. Think of rate as:
$$\mathrm{rate} = -\frac{\Delta [A]}{\Delta t}$$
(the reaction is removing $\ce A$) which is in the form of a slope. thus the larger the rate value, the lower the slope of the curve. Rate is dependend on concentration, but for the same concentration, a reaction with a higher rate constant will proceed faster, thus remove $\ce A$ fast and have a more negative slope.

Lets compare for the two $k$ values, $k_{upper}$ and $k_{lower}$

$$\mathrm{rate}_{upper}  = k_{upper}[\ce A]= -\frac{\Delta [A]}{\Delta t} \\
\text{and }\\
\mathrm{rate}_{lower}  = k_{lower}[\ce A]= -\frac{\Delta [A]}{\Delta t} $$

Lets say arbitrarily that two reaction rates are: $k_{1} = \pu{2 s-1}$ and $k_{2} = \pu{4 s-1}$ and the instantaneous concentration is $[\ce A] = \pu{0.6M}$ then the reaction rates would be as follows:

$$\mathrm{rate}_{1}  = \pu{2 s-1}\times \pu{0.6M} = \pu{1.2 M s-1} = -\frac{\Delta [A]}{\Delta t} \implies \frac{\Delta [A]}{\Delta t} = -\pu{1.2 M s-1}\\
\text{and }\\
\mathrm{rate}_{2}  = \pu{4 s-1}\times \pu{0.6M} = \pu{2.4 M s-1} = -\frac{\Delta [A]}{\Delta t} \implies \frac{\Delta [A]}{\Delta t} = -\pu{2.4 M s-1}$$

Since a reaction of $\mathrm{rate}_2$ has a more negative slope, it must correspond to a curve  that decreases faster than a reaction of $\mathrm{rate}_1$ and the concentration would be lower sooner, thus $\mathrm{rate}_2$ is the lower curve and $\mathrm{rate}_1$ is the upper curve.


Why is the endo product the major product in a Diels-Alder reaction? I understand that the endo product is, at least in all cases in my textbook, the major product in a Diels-Alder reaction. However, I don't understand why this is the case.

The explanation in my textbook says that the endo product allows for more interaction between the diene and dienophile during the transition state.

I also found the following post, which suggests that secondary orbital interactions are the cause. Stereochemistry of product in Diels-Alder reaction

Then I found the following paper that indicates, at least to my understanding, that steric strain during the transition state is the main reason behind the endo rule (or at least for the example reaction in the article). https://www.ncbi.nlm.nih.gov/pubmed/24449044

So, I'm left wondering what is the reason for the endo rule?

What is the endo-selectivity of Diels-Alder reactions?
If the diene used in the Diels-Alder reaction has asymmetric substituents at the end carbons, and if the dienophile is unsymmetric, then two different isomers of the final adduct can form. The isomer, where the functional group(s) (usually carbonyl) on the alkene end up on the same side as the newly formed double bond, is called the endo-isomer [1]. The other one is the exo isomer. The naming of the isomers can be a little ambiguous in cases where there are multiple functional groups of different types on the starting alkene. Usually the electron-withdrawing group is considered when identifying which isomer is endo and which is exo.
 [2]
The experiments tell us that, in general, the endo-product is the one which is favoured kinetically i.e. it forms faster. This is not a rule as such, because there is no way to predict to what extent the endo product will be favoured. (In many cases, the exo product is the thermodynamic product i.e. more stable, so allowing the reaction to equilibriate will result in higher yield of exo product)
Why does this happen?
It's difficult to answer this question definitively, because there is still a lot of active research going on in the area. I will try to summarise the main effects which have been proposed as possible causes of the endo-selectivity.

Secondary orbital interaction

The interaction between the frontier molecular orbitals at the points where the new $\sigma$-bonds form are the first-order interactions. However, there are other interactions possible between the diene and dienophile, there are called second-order interactions, or secondary orbital interactions (SOI). This explanation was proposed by Woodward and Hoffmann, in the book The conservation of orbital symmetry[3].
The following diagram will be useful:
 [4]
As you can see, in the transition state leading to the endo-product, the carbonyl part of the alkene can interact with the back of the diene, and the interaction is favourable, as the symmetry matches. In the transition state leading to the exo-product, this interaction would not be available, as the carbonyl group is away from the diene. For a long time, SOI was considered as the right answer, and was extensively invoked in explaining not just Diels-Alder reaction, but other types of cycloaddition reactions as well.
However, there are a lot of evidences, both experimental and computational, which calls into question the validity of SOI. There is a good paper from Salvatella et.al. summarising them: "Do  Secondary  Orbital Interactions  Really  Exist?" Acc. Chem. Res., 2000, 33, 658.
For example, the dimerization of butadiene was shown to have slight endo preference.
 [5]
This was frequently used as an example of the SOI effect. However, computational studies at RHF and CASSCF level have found that the reaction proceeds via a stepwise mechanism. [6] If the mechanism is stepwise, not concerted, then it does not go through the constrained transition state which is required for the secondary orbital interactions to come into existence. So, we cannot invoke SOI for this endo-preference.

Steric effects

In certain Diels-Alder reactions, especially those involving the cyclopentadiene, steric effects have been proposed.
 [5]
Experimental data shows that when $\ce{R}$ = $\ce{CH3}$ and $\ce{X}$ = $\ce{CHO}$ the endo product is formed in $17.0%$ yield, i.e. the exo product now dominates. [7] This indicates that the steric repulsion of cyclopentadiene ring with $\ce{CH3}$ and the  is greater than that with $\ce{CHO}$, which is consistent with $\ce{CH3}$ being bulkier.
These observations can also be explained by SOI, but the trends in the endo-selectivity for a range of $\ce{R}$ and $\ce{X}$ groups seem to fit better to the steric argument.[5]
In a more recent study by Fernandez and Bickelhaupt, the selectivity was found to arise from activation strain, i.e. steric strain as the reactants approach the transition state.[12] They used $\mathrm{M05{\text -}2X/def2{\text -}TZVPP}$ to model the reaction of cyclopentadiene and maleic anhydride.

The above image, taken from their paper, shows the full reaction coordinate, with ZPE-corrected energies in kcal/mol and distances in Å. As you can see, the exo-reactant complex (exo-RC) is higher in energy than the endo one, suggesting that the strain comes into existence even before the TS. The extra strain in the exo-pathway is thought to arise from the steric repulsion between the methylene of cyclopentadiene and the oxygen of the anhydride.

Hydrogen bonds

Hydrogen bond-like interactions between C-H in alkene and the double bond has also been proposed as a cause of endo-selectivity in particular cases, such as the reaction of cyclopropene and butadiene.[8] Here, both steric effects and H-bond like interactions are thought to be present.
 [5]
This is somewhat different than the SOI interaction, because in SOI, only orbital overlap (delocalization) is considered. H-bond interactions are a sum of orbital overlap and electrostatic attraction.

Electrostatic forces

Electrostatic forces can be used to explain the unusual exo-selectivity of the furan+cyclopropenone addition.

In the exo-TS, there is a favourable electrostatic interaction between the electronegative O of furan and the electropositive carbonyl carbon of cyclopropenone. This makes the exo-isomer the preferred one. The exo-TS is about 1.81 kcal/mol lower in energy than the endo-TS, according to computational studies. [9]
The usual endo-selectivity of other Diels-Alder reactions can also be explained by the electrostatic repulsion between slightly electropositive terminal C-H of the diene, and the electropositive carbon of the electron-withdrawing group on dienophile:
 [5]

Solvent effects

Sustmann and others carried out computational studies on Diels-Alder reactions in both gas phase and with dielectric-continuum solvation.[10] They found that in the gas phase, the exo-TS is lower in energy i.e. kinetically favoured. Using a solvation model resulted in the endo-TS being the favoured one.
They suggested that solvent effect might be the cause of endo-selectivity. The endo-TS is generally more polar than the exo-TS, so it is stabilised to a greater extent in solvent. It is also observed that the preference for endo-TS increases the more polar the solvent is, which lends support to this argument. The gas phase exo-preference would also disprove the role of SOI effects, because they should be present regardless of solvation.
However, computational studies like these have to be considered carefully, because they were mostly carried out using HF or B3LYP methods, both of which severely underestimate dispersion forces (which are necessary to describe SOI). So, the results might be artificially skewed in the direction of exo-TS. [11]
Conclusion
All of this means that we are not really sure if there is a single cause that can explain the endo-selectivity in Diels-Alder reactions.
In an extensive experimental-computational study , Paddon-Row et. al.(2020) studied reactions of isotope labelled butadienes with mono- or di-substituted alkenes.[11] Surprisingly, it was found that mono-substituted alkenes (such as methyl acrylate) did not have a strong endo-preference when the reaction was done in benzene solution. Methyl acrylate and butadiene, for example gave 50:50 exo-endo product yield. Di-substituted alkenes showed higher endo-selectivity. In their analysis they suggested that SOI effects are important in some cases, whereas steric effects are forcing the selectivity in some other cases.
So, it might just be that different effects are working in different reactions, and somehow the end result is that endo-product is favoured in almost all cases. Then it might not be useful to look for one particular blanket cause of the selectivity, rather, each reaction has to be explained uniquely.

References
(1) Organic Chemistry, Clayden, Greeves, Warren, 2nd ed., Oxford University Press, p.884
(2) Image source: https://employees.csbsju.edu/cschaller/Reactivity/pericyclic/peri%20endo.htm
(3) The conservation of orbital symmetry, R B Woodward, R Hoffmann,   Weinheim : Verlag Chemie GmbH, 1971
(4) Image source: https://www.cpp.edu/~psbeauchamp/pdf/316_MO_DA_rxn.pdf
(5) Image source: Acc. Chem. Res., 2000, 33, 658
(6) Y. Li and K. N. Houk, J. Am. Chem. Soc., 1996, 118(4), 880–885
(7) Y. Kobuke, T. Fueno, J. Furukawa, J. Am. Chem. Soc., 1970, 92, 6548-6553
(8) M. Sodupe,R. Rios,V. Branchadell, J. Am. Chem. Soc., 1997, 119, 4232-4238
(9) S. M. Bachrach, J. Org. Chem., 1995, 60, 4395-4398
(10) T.  Karcher,  W.  Sicking, J. Sauer and R. Sustmann, Tetrahedron Lett., 1992, 33, 8027–8030
(11) W. J. Lording, T. Fallon, M. S. Sherburn, M. N. Paddon-Row, Chem. Sci., 2020, 11, 11915–11926
(12) I. Fernández, F. M. Bickelhaupt, J. Comput. Chem., 2014, 35, 371– 376. DOI: 10.1002/jcc.23500


Can you heat water with additives? I have been curious about this question for a while. If you want to warm up a large amount of water, is it feasible to do this by adding a substance that has an exothermic reaction with the water? What is the relevant chemistry one has to know to work out how much the water would be heated up by?

From the comments:

  What do you want to do with the hot water? Swim in it. I was thinking
  of thousands of liters

That's an interesting idea, but unfortunately, I don't think adding chemicals to a pool in order to heat it is a good idea (especially yellowish chemicals). The water temperature would drop in a few hours, tops, and you'd have to be constantly adding more chemicals to it. At some point, you'd have a pool of chemicals instead of a pool of water. Not to mention that the chemicals would probably end up being more expensive than the electricity or the gas bill to heat up the pool to the same temperature the conventional way.

A nice way of heating up pools is using solar water heating systems, and coupling them with the conventional heating systems, if necessary.

So... Is it feasible? I'd say it isn't. But since you said this was more of a thought experiment... The only scenario I can imagine where the best solution to heat up a pool is to add chemicals to it is: You have a small pool, very little time to heat it up, you want to heat it up to sauna-like temperatures, and although you'll only use it for a couple of hours you are willing to spend a lot of money to do so. In this case, I'd add some water to the pool, and then add the following reagents to it, making the following reaction:

$$\ce{NaOH + HCl -> NaCl + H2O} \\ \Delta H = -56\,200\ \mathrm{\frac{J}{mol}}$$

You can do the math, but I assure you it heats up pretty quickly. We used to neutralize concentrated acid in ~10 L plastic buckets, and if we "accidentally" neutralized too much at a time the bucket would start to heat up to a point you couldn't even touch it.

NaCl is kitchen salt, so no issues with toxicity if you do the math correctly. Salt water also makes you float better, which I think is a nice quality in a pool. There would be salt on the bottom, but I think that's nice too, makes it feel like the sea.


Equations for internal energy and their relevance to the derivation of the equation for entropy for perfect gases My lecture handout proves derives an expression for entropy thus:

  For a perfect gas $U=U(T)$ (i.e. is not dependent on volume; only temperature). Therefore, $(\frac{\partial U}{\partial V})_T=0$

  Therefore:

  Equation 1) $$dU=(\frac{\partial U}{\partial V})_TdV+(\frac{\partial U}{\partial T})_VdT=C_VdT$$ Since $(\frac{\partial U}{\partial T})_V=C_V$

  Now, for a reversible change:

  Equation 2) $$dU=dq_{rev}+dw_{rev}=dq_{rev}-PdV$$

  Equation 3) $$dq_{rev}=PdV+dU=\frac{nRT}VdV+C_VdT$$
  Then divide by $T$ to make the equation equal to entropy rather than heat:

  $$dS=\frac{nR}{V}dV+C_VdT$$

I have a couple of issues with this. Mainly stemming from the fact that $U=U(T)$ and therefore that $(\frac{\partial U}{\partial V})_T=0$. I understand it on a physical level why this is the case (no intermolecular forces). I am not asking for a proof of this; I just have an issue with the implications of this. Let me explain:

An equation for dU is:

$$dU=TdS-PdV$$

This is understandable. $TdS$ accounts for heat transfer in/out of a system and $PdV$ accounts for work done by/on the system. However, surely this equation implies that $U=U(S,T)$ since both $dS$ and $dT$ appear in the expression.

The crux of it is: how can $dU=C_VdT$ and $dU=TdS-PdV$. Please help me understand what's going on here.

Your equation $dU = TdS - pdV$ is valid for all pure substances, not just ideal gases.  It is telling you that if you knew an equation that explicitly defined $U$ as a function of $S$ and $T$, then you could calculate all other thermodynamic properties of that substance, i.e. if you knew that $U=f(S, V)$, then you could easily calculate $T$ and $p$ because $T=\frac{\partial f(S, V)}{\partial S}$ and $p=\frac{\partial f(S, V)}{\partial V}$.  This is because $S$ and $V$ are canonical variables for $U$.

Now let's apply the ideal gas equation $dU=C_V dT$.  You can integrate this to yield $U=C_V T + K$.  Unfortunately, this does not express $U$ as a function of $S$, but rather of $T$.  So how do we proceed?  This physics.se question has a good general answer.

But in this specific case things are bit easier.  Let's find $dS$ from the equation you were given.

$dS =\frac{1}{T}(dU + p dV)=\frac{1}{T}(C_V dT + \frac{RT}{V}dV)=C_V \frac{dT}{T}  + R \frac{dV}{V}$

Since you now know $dS$ and from the ideal gas law you also know $p dV$, you have an expression for $dU$ from which you can derive all thermodynamic quantities.


What is the difference between strontium-86 and strontium-90? I know that strontium ($\ce{Sr^90})$ has a higher mass number than its counterpart, but strontium ($\ce{Sr^86}$) is non-radioactive and not deadly to humans.

What makes $\ce{Sr^90}$ so different that it is cancerous and lethal to humans? Is it just the chemical makeup of the element that makes the difference?

For all practical purposes, Sr90 is chemically identical to Sr86. However, the extra neutrons crammed into Sr90 make it radioactive.

As mentioned in the comments above, alkaline earth metals (e.g. Ca, Sr, Ba) behave chemically in a similar fashion, and tend to deposit in bone like calcium does. This keeps the radioactive Sr90 in contact with the rapidly-growing hematopoietic material in bone, increasing the risk of cancers such as leukemia.


UV-VIS spectra of n-decanes I was wondering if anyone knows where I can find UV-VIS spectra of the n-decanes.  Specifically tetradecane and dodecane.  I can't seem to find such information using my regular search methods, so I was wondering if there was a common database that contains such things that I am simply unaware of (likely).

All I really need to know is if these chemicals will absorb radiation with wavelengths between 355-375 nm.  A source to back this up will only help.

Thanks,
Sam

To the best of my knowledge, ordinary alkanes do not typically absorb anywhere in the UV/Vis region. The organic molecules that do typically contain electrons in $\pi$- or non-bonding orbitals, which have energy level transitions corresponding to UV/Vis wavelengths. That said, I did find one paper1 describing absorptions around the far UV ambit, where $\lambda < 200\mathrm{nm}$.

DOI: 10.1021/jp903435c


Procedure for measuring EGCG I am conducting a school-based research project to determine the concentration of EGCG.
One of the problems I am facing is that most of the procedures I find on the internet are of a university research lab level, which isn't feasible for me. Interestingly however, I found a suitable procedure for titration of red wine.
Is there a similar method I could use to determine the concentration of EGCG

I am conducting a school-based research project into the antioxidative properties of green tea. Part of the procedure is to determine the concentration of antioxidant in green tea. This antioxidant is EGCG (a type of catechin).

First of all, you cannot determine the concentration of just EGCG (epigallocatechin gallate) alone, because any extract of green tea contains polyphenols other than EGCG, which include catechin, and/or other flavonoids such as flavanols, flavandiols, and phenolic acids. Most of the green tea polyphenols are flavonols, commonly known as catechins, which are flavan-3-ols (Ref.1):

As shown in the diagram, There are four kinds of catechins mainly find in green tea: epicatechin (EC), epigallocatechin (EGC), epicatechin-3-gallate (ECG), and epigallocatechin-3-gallate (EGCG). The amount of catechins varies in the original tea leaves due to differences in variety, origin, harvesting time, and growing conditions.  Also, the preparation of fresh green tea cannot totally extract catechins from the leaves, therefore, the amount determine by you may not reflect the absolute values determined through the complete extraction of leaves. Thus, your findings may differs from the accurate benefit of green tea.
Nonetheless, there are methods to determine polyphenol content in particular extract as the amount equivalent to a reference. Easily, your reference could be gallic acid. One easy method I found in literature is the estimation of total phenolic content and other oxidation substrates using Folin–Ciocalteu reagent* (Ref.2; PDF is attached for convenience):

Determination of Total Polyphenols by Folin-Ciocalteu Method: One milliliter of the tea extract was diluted with water to $\pu{100 mL}$. The diluted extract solution $(\pu{1.0 mL})$ was then mixed with $\pu{5.0 mL}$ of 50% Folin-Ciocalteu reagent. After $\pu{5 min}$, $\pu{4.0 mL}$ of 7.5% $\ce{Na2CO3}$ solution was added and mixed. Then the mixture was incubated at room temperature for $\pu{60 min}$. The absorbance of the solution was measured at $\pu{765 nm}$ with a Shimadzu UV-2550 UV-Vis spectrophotometer against a reagent blank. The determination was three times. The total polyphenol content was expressed as gallic acid equivalents (GAE, units: $\pu{g}/\pu{100 g}$ sample).

Preparation of Tea Extract: Each tea sample was ground to a fine powder and $\pu{1.0 g}$ of the powder was extracted with $\pu{25 mL}$ of methanol/water (7 : 3, v/v) in a water bath at $\pu{70 ^\circ C}$ for $\pu{10 min}$. The extraction was repeated twice and the extracts were combined. After filtration through a filter paper and a $\pu{0.45 \mu m}$ membrane filter (Millipore), the extract volume was adjusted to $\pu{100 mL}$.

According to the procedure, you need just a spectrophotometer in visible range (do not need an UV-Vis spectrophotometer, which is an expensive one). If you do not have a membrane filtration setup, you may extract leaves without powdering them, but need extra extracting time (may be pre-prepare extract overnight at $\pu{70 ^\circ C}$ using the same solvent system). If you want to check the green tea as we prepared them to drink, use that way but you may have to adjust reagent concentrations accordingly to test.
If you have a 96-well plate-reader and some micro-pipettors, there is a reliable high-input method is listed in Ref.3 for the same method (PDF of the paper has been attached to compare).

*Note: Folin–Ciocalteu reagent is a specific redox reagent containing phosphotungstic-phosphomolybdenum. The reaction forms a blue chromophore constituted by a phosphotungstic-phosphomolybdenum complex with phenols that can be quantified by visible-light spectrophotometry (Ref.4,5).
Gallic acid could be used as calibration standard for this determination and the total phenols could be reported as gallic acid equivalent (GAE). Other alternative is pyrogallol. Choose the concentration range enough to give absorbance 0.4-1.0 (at $\pu{760 nm}$) so that the assay is linear (Ref.5). Following is the published calibration curve for gallic acid using Folin-Ciocalteu Assay (Ref.6):

The procedure used: In brief, aliquots of a standard solution of gallic acid $(20, 40, 60, 80, \pu{100 mg L-1})$ were mixed with $\pu{500 \mu L}$ of Folin-Ciocalteu reagent (previously diluted 10-fold with distilled water) and $\pu{400 \mu L}$ of $\ce{Na2CO3}$ (7%). After $\pu{40 min}$ of incubation at room temperature $(\pu{23 \pm 2 ^\circ C})$, the absorbance was measured at $\pu{760 nm}$ using a Spectrophotometer against a blank sample.
Note that the temperature of $\pu{23 ^\circ C}$, time of $\pu{40-60 min}$, and  $\lambda = \pu{760 nm}$ are optimized condition for Folin-Ciocalteu assay (Ref.5). Also make sure that you add $\ce{Na2CO3}$ solution at least after $\pu{10 min}$ of the addition of Folin-Ciocalteu reagent (not before). Folin-Ciocalteu reagent is unstable under alkaline conditions.

References:

Sarah C. Forester and Joshua D. Lambert, "Antioxidant effects of green tea," Mol. Nutr. Food Res. 2011, 55(6), 844–854 (doi:10.1002/mnfr.201000641)(PDF).
Chunjian Zhao, Chunying Li, Shuaihua Liu, and Lei Yang, "The Galloyl Catechins Contributing to Main Antioxidant Capacity of Tea Made from Camellia sinensis in China," The Scientific World Journal 2014, Article ID 863984 (11 pages) (doi: https://doi.org/10.1155/2014/863984)(PDF).
Elizabeth A Ainsworth1 and Kelly M Gillespie, "Estimation of total phenolic content and other oxidation substrates in plant tissues using Folin–Ciocalteu reagent," Nature Protocols 2007, 2(4), 875-877 (doi:10.1038/nprot.2007.102)(PDF).
P. Schofield, D. M. Mbugua, and A. N. Pell, "Analysis of condensed tannins: a review," Animal Feed Science and Technology 2001, 91(1–2), 21-40 (doi: https://doi.org/10.1016/S0377-8401(01)00228-0).
Andressa Blainski, Gisely Cristiny Lopes, and João Carlos Palazzo de Mello, "Application and Analysis of the Folin Ciocalteu Method for the Determination of the Total Phenolic Content from Limonium Brasiliense L.," Molecules 2013, 18, 6852-6865 ( doi:10.3390/molecules18066852).
A. Bouyahya1, N. Dakka, A. Talbaoui, N. El Moussaoui, J. Abrini, and Y. Bakri, "Phenolic contents and antiradical capacity of vegetable oil from Pistacia lentiscus (L)," Journal of Materials and Environmental Sciences 2018, 9(5), 1518-1524 (doi: https://doi.org/10.26872/jmes.2018.9.5.167).


Importance of higher order exchange terms? The MP2 ground state energy of a molecule or solid can be written as

$$E^{(2)} = \frac{1}{2}\sum^{\text{occ.}}_{ij}\sum^{\text{virt.}}_{ab} \frac{\langle ij | ab \rangle [\langle ab|ij \rangle - \langle ab|ji\rangle]}{\varepsilon_i + \varepsilon_j - \varepsilon_a - \varepsilon_b}$$

where the first part of the numerator is commonly called "direct MP2" and the second part "exchange MP2", since i and j are exchanged.

However, for energy differences, it is often sufficient to calculate only the direct MP2 energy, since the contribution of the exchange part is mostly negligible. For the same reason the random phase approximation, which consists of these "direct terms" up to infinite order, works so well for many materials.

But are there good examples (materials or molecules) where the exchange term is important? I guess magnetic systems or situations where Pauli repulsion is important... but does someone know particular examples?

Exchange is important and is present in any system. To be more specific, in MP2 you already have exact exchange for the ground state from the zeroth-term(HF) while in the second-term (MP2) you have a correction to correlation plus some more exchange from the excited state.

MP2 allows the electrons to be partly in the unoccupied MO however, it is limited to double excitations. It turns out that you end up having a small correction to exchange and a big correction to correlation which is completely missing in HF. So the exchange correction will be small in every system due to the type of excitation treated in MP2.


Migratory preference in a rearrangement involving carbocations I just read about the Pinacol rearrangement and the following reaction was presented just after it. The book (Clayden) shows path 1 being taken.

My question is why is the second path not taken? I know that migratory aptitude of $\ce{Ph>H}$ but all we want is stability, so why can't the $\ce{H}$ migrate?

why is the second path not taken?

$\ce{H}$ can migrate, it is just that phenyl migrates better.  The reason that phenyl migrates better is because it is much better able to stabilize the developing carbocation center.  The phenyl group participates by what is termed neighboring group participation (see the "NGP by an aromatic ring" section).  It forms a phenonium ion.  Note the geometry of this ion, the former aromatic ring is perpendicular to the plane of the screen while the former $\ce{C-C}$ bond is contained in the plane of the screen.

(image source)

We can draw several resonance structures to describe the phenonium ion, this leads to it being especially stable.

(image source)

Such extra stabilization is not possible with a migrating hydrogen atom. Therefore the pathway involving the phenonium ion is lower energy (faster) than the pathway involving hydrogen migration.


Thermodynamically Reversible Processes: How does an infinitesimal step make a process reversible and how is maximum work derived from the process? My book says 'Maximum work can be obtained only from thermodynamically reversible processes,' but why is it so? What is the cause?

Actually to me the definition of reversible process is confusing. It says that at each step during the process, equilibrium is maintained. But, let's say,we have to expand a gas in a cylinder headed by piston from pressure $P_i$ to $P_f$. Suppose,the process is done with innumerable infinitesimal steps. Then, say at each step, $dp$ is changed in the pressure.

Before the step, gas and the piston were exerting same pressure $P_i$ . Now, when the piston is raised, gas does expand to equalize the pressure,which is $P_i - dp$ . So, at the end of the step, the gas & the piston will exert same pressure $P_i - dp$ and hence will be in mechanical equilibrium.

During the step,the gas and the piston was not in equilibrium;it was at the end of the step  the equilibrium is achieved. So, during each step of the process the gas would be at disequilibrium . So,will it really be a reversible process? It is contradictory with the definition. But it is true that during each step, the system is at disequilibrium though infinitesimally small by $dp$ . So, why does the definition tell that at each step,there is equilibrium?

So, I have three questions:

Why is maximum work only obtained from reversible processes?
According to the definition, at each step of the reversible process,there must be equilibrium. But during each step, there is infinitesimal disequilibrium. So, why does the definition say so in-spite of the disequilibrium during each step?
How do infinitesimal steps make a process reversible? I will be very grateful if anyone answer these three questions.

Please help.

I'll give a mathematical answer, and an analogy also for ease of thought. These are my own intuitions, so they may work for you.

My mathematical reasoning:

A lot of thermodynamic processes are phrased very carefully in terms of reversibility. Now my intuition of this is that reversibility is a consequence of the maths we must use.  Thermodynamics likes to look at continuous processes such that we can use integrals instead of summations in the calculations: this stands to reason since classical thermodynamics treats a collection of particles as moving, say, a piston continuously from one place to another, not in discrete steps.

Thus, one way to think about it is to think of reversibility as being necessary for an integral formulation. If we allow ourselves to try and think in infinitesimal steps, moving an infinitesimal step forward is just an infinitesimal away from going backwards and hence this is described as being reversible. In short, if a process really happens continuously then we must impose the condition of reversibility on it, otherwise the mathematics fails.

My analogy:

Why this gives maximum work can be seen by analogy with a real life classical system: the tug-of-war. If you're in a tug of war with some of your friends who are of equal size to you and of equal strength, you're going to put in maximum work in pulling the rope. If one side is of unequal size and strength it becomes much easier for the stronger side.

To be more serious, if your particles move your piston irreversibly (i.e. they never move backwards along the way) then no work has been expended in going backwards—only forwards. Reversible systems belie the work they've had to do since the physical outcome we actually see—the final outcome—shows the piston in the same place as the irreversible case, however what we haven't seen is the system battling backwards and forwards.

To conclude: if we imagine a set amount of work needs to be done to move a piston irreversibly from a point A to a point B, then the reversible case must provide maximum work since some of the time the system has battled going backwards and has still ended up in the same place as the irreversible case.

EDIT: How to Achieve Equilibrium and still have Movement

I think the source of your confusion lies in the statement 'it was at the end of the step of the equilibrium'. I know what you mean by this, but I don't think it's precise. Equilibrium doesn't mean to describe position as such, but the oscillation of position. Even if you've physically pushed your piston a distance ds, it can still be in equilibrium by the very fact that it can still move forwards and backwards in its next spot.

So in fact the periodicity of position is maintained at each step, which is what we mean by the equilibrium is maintained.

If we imagine a spring oscillating backwards and forwards, the movement of oscillation is what we mean here by a dynamic equilibrium. As long as this oscillation is maintained this equilibrium exists, however if we imagine our particles as springs, then the spring must be extended to reach its next spot. Hence in this sense we have equilibrium and movement can still occur.


Why does superglue ignite cotton? Cotton is mostly cellulose, a plant saccharide. Superglue refers to a class of cyanoacrylates. What is it about the two that causes ignition?

I'd imagine that the cellulose in cotton is in its linear form and not in its ring form since the ring-linear form interconversion only occurs in solution. Is it that the cellulose attacks the cyanoacrylate?

  The three hydroxyl groups, one primary and two secondary, in each repeating cellobiose unit of cellulose are chemically reactive

So, do the hydroxyl groups attack the carbonyl group in superglue?

Though the monomers in cyanoacrylate glues contain an ester, their polymerization doesn't rely on that ester group directly.  The Wikipedia article for cyanoacrylates shows the polymerization more clearly than I can easily explain in words.  The many hydroxyl groups in cellulose do start polymerization effectively, and the large surface area of cotton wool provides a large number of sites for the glue to cure.

When I read your question I was curious; I had never heard of this before.  I searched the internet and was only able to find one or two examples of people actually trying this successfully, and both of them looked dubious.  I've spilled superglue on cotton shirts and pants any number of times and while it cures almost instantly, it has never caught my shirt on fire.

I decided to try it myself, and I soaked a cotton ball in superglue to see what would happen.  The glue cured extremely rapidly, producing very irritating (colorless) fumes, and the mass of glue and cotton got warm, but it was nowhere near hot enough to ignite - my tap runs hotter than the glue got.  There may be a particular type of glue that cures significantly more exothermically, but I'm inclined to think that this is an internet hoax, and Popular Science agrees.


What is the intuition behind 'mol' as a unit 'symbol' Why are moles written with 'mol' as opposed to something which is easier to write with one character, or even two?

The symbol mol is due to Ostwald who was a very influential and respected physical chemist more than a century ago. In German, "mole" is "Mol". It is a shortform of Molekül.
I believe the Internet Archive has an English translation of his historically famous book of 19th century. This is where I recall reading this. The original reference is W. Ostwald Hand- und Hilfsbuch zur Ausführung physikochemischer Messungen. This book is available on Google Books (pg 278). A footnote appears there:

Um nicht stets den schleppenden Ausdruck Gramm -Molekulargewicht brauchen zu müssen, werde ich diese Größe in der Folge kurzweg Mol nennen, und erlaube mir, diesen Vorschlag zur allgemeinen Annahme zu empfehlen. Man schreibt ohnedies diese Größe häufig Mol., und braucht daher nur den Punkt fortzulassen.

In order not to always have to use the sluggish expression gram - molecular weight, I will simply call this quantity Mol in the following, and allow me to recommend this proposal for general acceptance. Anyway, this quantity is often written as Mol., so you only have to omit the point.
Translated via (www.deepl.com) with some manual changes.


Why doesn't EDTA have a double bond? EDTA (Ethylenediaminetetraacetic acid) has a structure like this

But why, as the name 'Ethylene' suggests, doesn't it should have a double bond between carbons?

A lot of non-systematic chemical names don't describe the composition of the compound, they describe what the compound is synthesised from. That is the case here since EDTA is derived from ethylene by adding two units of the diamine.

It isn't that uncommon in chemical names. Polyethylene, for example, is (simplifying only slightly) a long chain of carbons connected by single bonds. But it is made by polymerising ethylene. Hence has a name that doesn't describe what its structure is but where it came from.


Heat Transfer Coefficient Problem The Problem

An unusual reference gives the formula below for the heat transfer coefficient of an emulsion.
$$H = 0.0061\left(X^{0.6} + 0.038T\right)\cdot V^{0.35}$$
In the formula $T$ is the temperature in $\pu{^\circ F}$, $V$ is the flow rate of the emulsion in $\pu{\frac{ft}{min}}$, $X$ is a characteristic dimension in inches and $H$ is the heat transfer coefficient in $\pu{BTU h^-1 ft^2 ^\circ F}$.
Calculate the heat transfer coefficient in $\pu{W m-2 K-1}$ when emulsion at $\pu{30 ^\circ C}$ flows at $\pu{1.25 m s-1}$ and the characteristic dimension is $\pu{0.8 mm}$.

My Attempt

I converted the characteristic dimension ($X$) from $\pu{mm}$ to $\pu{in}$.

$$\pu{0.8 mm} ⋅ \frac{\pu{39.37 in}}{\pu{1000 mm}} = \pu{0.03 in}$$

I converted the temperature ($T$) from $\pu{^\circ C}$ to $\pu{^\circ F}$.

$$\frac{(n \ \pu{^\circ F} - 32)}{1.8} = 30 \ \pu{^\circ C} $$
$$n \ \pu{^\circ F} = \pu{86 ^\circ F}$$

I converted the flow rate of the emulsion ($V$) from $\pu{m/s}$ to $\pu{ft/min}$.

$$\frac{\pu{1.25 m}}{\pu{s}} ⋅ \frac{\pu{3.28 ft}}{\pu{m}} ⋅ \frac{\pu{60 s}}{\pu{min}} = \frac{\pu{246 ft}}{\pu{min}}$$

I plugged the values of $X$, $T$, and $V$ into the equation to solve for $H$.

$$H = 0.0061\left(0.03^{0.6}\ \mathrm{in} + 0.038(86\ \mathrm{^\circ F})\right)\left(\frac{246\ \mathrm{ft}}{\mathrm{min}}\right)^{0.35}$$
Is my work correct so far? If yes, I'm a little confused as to what I should do next. How do I simplify the units in my last step in order to then convert to a value with the desired units $\pu{W m-2 K^-1}$?

Is my work correct so far?

Yes, your conversions to the units for the inputs to the equation (#1 to #3) look correct to me.

How do I simplify the units in my last step in order to then convert to a value with the desired units...?

You don't.
This sort of expression embodies a purely empirical relationship -- it has no particular basis of connection to the underlying physical relationships between these various quantities.  The reason the units of each of the input quantities are specified ($^\circ\mathrm{F}$, $\mathrm{\frac{ft}{min}}$, etc.) is because you're supposed to use just the numerical values of those quantities, after conversion to the indicated units, and substitute those into the equation. So, instead of the equation you have in #4, it would just be:
$$
H = 0.0061 \cdot \left(0.03^{0.6} + 0.038 \cdot 86\right) \cdot 246^{0.35}
$$
Then, you take the purely numerical result you obtain for $H$ from this equation and tack on the indicated units; in this case, $\mathrm{BTU \cdot {hr}^{-1} \cdot {ft}^{-2} \cdot \ {^\circ F}^{-1}}$. From there, you can carry out a straightforward conversion to the metric units asked for in the final answer.
These sorts of empirical correlations are extremely common in chemical engineering practice, where one doesn't especially care if an equation has solid theoretical underpinnings, as long as it performs reasonably well in predicting quantities of interest.
To note, while you can construct internally consistent units for the various constants in the formula, they usually don't end up having any particular useful physical interpretation. For example, since it's only possible to add two quantities together if they have the same units, the $0.6$ exponent on $X$ means that the $0.038$ coefficient on $T$ would have to have a dimensional value of $0.038\ \mathrm{{in}^{0.6} \over {^\circ F}}$, which has no particular physical significance that I can determine. I'm not even sure what sense there is to be made of units such as "$\mathrm{{in}^{0.6}}$", "$\left(\mathrm{ft\over min}\right)^{0.35}$", etc.


Why does the barium ion in barium titanate move in its tetragonal crystal structure? Barium titanate is known in the materials science world to be a great dielectric. This is known to be because the $\ce{Ba^2+}$ ion sits off center in the middle of the tetragonal crystal structure.

How does this $\ce{Ba^2+}$ ion, or even the crystal structure itself, move in order to accompany a larger amount of charge? How much freedom of movement does this ion have in a solid crystalline structure? Additionally, does the $\ce{Ba^2+}$ ion naturally sit off-center in this configuration, or does it only behave that way when introduced to a charge?

This is known to be because the $\ce{Ba^2+}$ ion sits off center in the middle of the tetragonal crystal structure

First of all this is not correct. It is the $\ce{Ti^4+}$ ion that sits in a tetragonal off center configuration (black atom in perovskite structure below). Also this hold for any material with perovskite structure with the $\ce{ABC3}$ general formula where $\ce B$ is your center site, $\ce A$ is the corner sites and $\ce C$ is your face site counter ions $(\ce{BaTiO3, LaAlO3, NaMgI3, ...})$.

  Additionally, does the $\require{enclose}\enclose{horizontalstrike}{\ce{Ba^2+}} \ \ce{Ti^4+}$ ion naturally sit off-center in this configuration, or does it only behave that way when introduced to a charge?

It does sit off center all of the time in any of 6 directions on a $(001)$ plane but the many centers are randomly oriented which produces no net dipole moment. When charge is introduced, the $\ce{Ti^4+}$ are attracted toward the negative plate and will shift toward that plate creating a net dipole moment.

  How much freedom of movement does this ion have in a solid crystalline structure?

Very little, the $\ce{Ti^4+}$ is located between the plane of 4 oxygens (red atoms) and an oxygen atom perpendicular to the 4-atom plane. It can move between the 6 possibilities of this shift $(\pm x, \pm y, \pm z)$ but not between cells.

  How does this $\require{enclose}\enclose{horizontalstrike}{\ce{Ba^2+}} \ \ce{Ti^4+}$ ion, or even the crystal structure itself, move in order to accompany a larger amount of charge?

When the electric field is applied, the titanium ion receives enough force to squeeze between oxygen atoms to a new lower energy position (e.g. $-x$ to $+y$).


Distinguishing atom types How does one recognize lipophilic points, hydrogen bond acceptors and hydrogen bond donors in organic molecules? For example, provided this SMILES instance

O=C1NC(=O)SC1Cc3ccc(OCCN(c2ncccc2)C)cc3

I know it contains 2 lipophilic points, 5 hydrogen bond acceptors and 1 hydrogen bond donor. How can I determine them?

To identify hydrogen bond donors: Look for a polarized hydrogen (typically bonded to O, N, S)
Hydrogen bonds form from the interaction of a partially positively-charged hydrogen and a (partially) negatively-charged atom. Thus, to "donate" a hydrogen for hydrogen bonding, you need a polarized hydrogen. Hydrogens are polarized by being bonded to electronegative elements: in organic molecules that will typically be O, N, or S.

To identify hydrogen bond acceptors: Look for atoms with electron lone pairs, or otherwise partially negatively-charged sites on the molecule. (Typically O, N, S, or halides).
To form an interaction with a polarized hydrogen, you need a source of (partial) negative charge, which typically can come from electron lone pairs. The most common species you'll encounter in an organic molecule include O, N, S, or halides (F, Br, Cl, I).


Do hard drive platters contain toxic chemicals? So, I'm considering an Artsy Fartsy project with hard drive platters. Being mildly aware of the world around me, I've come to the conclusion that I probably shouldn't grind up modern (or any) electronics and drink it for tea. To that end, is there anything in hard drive platters that is significantly toxic, or are they fine lying around the house? I don't intend in grinding them or anything, I'll be using them whole, so dust inhalation presumably isn't a big issue.

I never opened a newer 2.5" hard disk, but I've read that these may contain glass platters.

However, I've seen and touched stacks of platters from old 3.5" hard disks - a friend of mine makes clocks from hard disk cases.

These platters have an aluminium core, on which another metallic layer, that serves as a kind of primer, and the magnetic material are applied by sputtering. The surface is covered by a protective polymer layer.

Platters have been drilled or cut in a lot of electronics projects, e.g. to make hdd clocks based on the persistence of vision effect. Apart from the usual safety rules for metal works (protect your eyes, don't touch the hot drill, etc.) one only has to make sure that the rotating platter is properly mounted and doesn't start flying around!

When the platters aren't mechanically altered and only painted, there is even less danger. The platters are not toxic and do not even have sharp corners - it is safe to keep them around in the living room.

On a side note, the magnets from these hard disks are more dangerous. These magnets are pretty strong and when playing with them, you can easily jam your fingers between them ;)


The proteins "eluted" or "were eluted"? In describing proteins, I've written "the proteins detached from the [Ni-IDA] matrix and were eluted from the column"

Should I use "were" or can something simply "elute" ?

eˈlute, v.
trans. To wash out, cleanse. Now spec.  [after German eluieren, in same sense] to wash (adsorbed matter) away from the substance that has adsorbed it.

It's transitive, so you elute something, rather than something eluting. Your original "were eluted" is correct.